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http://mathhelpforum.com/number-theory/164816-unique-integers.html
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# Thread:
1. ## unique integers
what exactly does it mean for integers to be unique? if i am supposed to prove that there exists UNIQUE positive m and n integers under a condition, can m and n be equal at some conditions? for them to be unique, does it just mean there is only one m and only one n each time the condition is met? as in if xsqrd= m, there would not be a unique solution m?
thanks!!!
2. better you post the problem - it would make the context clear, and a lot of times answer to what you have asked follows from there
3. Originally Posted by mremwo
what exactly does it mean for integers to be unique? if i am supposed to prove that there exists UNIQUE positive m and n integers under a condition, can m and n be equal at some conditions? for them to be unique, does it just mean there is only one m and only one n each time the condition is met? as in if xsqrd= m, there would not be a unique solution m?
thanks!!!
"Unique m and n" means there is only one pair of numbers, (m, n), that satifies the condition. It is quite possible that m and n are the same.
If you mean a pair, (m, n), such that $m^2= n$ then, no, that would not be unique- but not just because, for example, both of (-2, 4) and (2, 4) is such a pair. It is also not unique because (2, 4) and (3, 9) are such pairs.
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http://electronics.stackexchange.com/questions/12865/is-a-current-limiting-resistor-required-for-leds-if-the-forward-voltage-and-supp/12866
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# Is a current-limiting resistor required for LEDs if the forward voltage and supply voltage are equal?
For blue LEDs with a forward voltage of 3.3 V and supply voltage of 3.3 V, is a series resistor still needed to limit current?
Ohm's Law in this case says 0 Ω, but is this correct in practice?
Perhaps just a small value like 1 or 10 Ω just to be safe?
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## 7 Answers
No, it's not correct, if only because neither the LED nor the power supply are 3.3V. The power supply may be 3.28V, and the LED voltage 3.32V, and then the simple calculation for the series resistor doesn't hold anymore.
The model of a LED is not just a constant voltage drop, but rather a constant voltage in series with a resistor, the internal resistance. Since I don't have the data for your LED let's look at this characteristic for another LED, the Kingbright KP-2012EC LED:
For currents higher than 10mA the curve is straight, and the slope is the inverse of the internal resistance. At 20mA the forward voltage is 2V, at 10mA this is 1.95V. Then the internal resistance is
$R_{INT} = \dfrac{V_1 - V_2}{I_1 - I_2} = \dfrac{2V - 1.95V}{20mA - 10mA} = 5\Omega$.
The intrinsic voltage is
$V_{INT} = V_1 - I_1 \times R_{INT} = 2V - 20mA \times 5\Omega = 1.9V.$
Suppose we have a power supply of 2V, then the problem looks a bit like the original, where we had 3.3V for both supply and LED. If we would connect the LED through a 0$\Omega$ resistor (both voltages are equal after all!) we get a LED current of 20mA. If the power supply voltage would change to 2.05V, just a 50mV rise, then the LED current would be
$I_{LED} = \dfrac{2.05V - 1.9V}{5\Omega} = 30mA.$
So a small change in voltage will result in a large change in current. This shows in the steepness of the graph, and the low internal resistance. That's why you need an external resistance which is much higher, so that we have the current better under control. Of course, a voltage drop of 10mV over, say, 100$\Omega$ gives only 100$\mu$A, which will be hardly visible. Therefore also a higher voltage difference is required.
You always need a sufficiently large voltage drop over the resistor to have a more or less constant LED current.
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"Always"? There isn't a more efficient way to drive them? – endolith Oct 26 '11 at 13:38
1
@endolith - there are ways of controlling the current with a smaller current sense resistor and a transistor (BJT or MOSFET), but there the transistor replaces the usual resistor, and also needs the voltage drop. – stevenvh Oct 26 '11 at 14:38
1
There are also ways of controlling the current with a switch-mode current regulator, which will be more efficient, but that's probably not necessary for readers of this question. – Kevin Vermeer Apr 17 '12 at 12:24
You always need a current limiting device. When using a voltage source, you should always have a resistor, think about what happens when the voltage changes by a small amount. With no resistor, the LED current would shoot up (until you hit a thermal based limit due to the LED materials). If you had a current source, then you would not need a series resistor because the LED would run at the current source level.
It is also unlikely that the forward voltage of the LED is always exactly the same as the supply. There will be a range mentioned in the datasheet. So even if your supply exactly matched the typical forward voltage, different LEDs would run at vastly different currents, and hence brightnesses.
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5
Also think about what happens if the LED fails and shorts. Limiting resistor is not just a good idea, its the law! (well ok maybe not) – freespace Apr 11 '11 at 13:55
You need a voltage drop over the current limiting resistor for it to work. And that voltage drop should be substantial to avoid high currents when your 3.3V is a bit off (maybe 3.45V for a while). If you would drive a LED with 1V voltage drop across a resistor and the supply is 1V higher, you would have approx. the double current.
A LED needs a constant current to shine. However, a constant current source probably needs more than 3.3V for a blue LED, unless you're using a buck-boost version.
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The I-V relationship in a diode is exponential, so applying a voltage difference of 3.3 V +/- 5% to an LED with a nominal 3.3V drop is not going to result in a 5% variation in intensity.
If the voltage is too low, the LED may be dim; if the voltage is too high, the LED may be damaged. As Hans says, a 3.3V supply is probably not enough for a 3.3V LED.
When driving an LED, it is better to set the current, not the voltage, since the current has a more linear correlation with the light intensity. Using a series resistor is a good approximation of setting the current through the LED.
If you can't use a supply with enough headroom to allow a current-setting resistor, you might be able to use a current mirror. That still requires some voltage drop, but possibly not as much as you'd need for a resistor.
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A current mirror also has a required voltage drop over the transistor. It may be able to follow the current quite well (so you set a 10mA reference and get very close to that on the other side), but it will need a little bit of voltage to work. – Hans Apr 11 '11 at 17:14
@Hans good point, I updated to make that clearer – Andy Apr 11 '11 at 17:25
If the forward voltage and supply voltage are nearly equal, using a resistor will yield results which are very sensitive to variations in supply voltage or LED characteristics. If the resistor is sized to avoid damaging the LED if it turns out the supply voltage is at maximum and the LED intrinsic voltage is minimum, the LED will only light with a fraction of its possible brightness if the supply voltage is at its minimum and the LED intrunsic voltage is at its maximum.
Using some type of current-regulating circuit will yield much better results, although most simple current-regulating circuits have a certain amount of compliance voltage. Probably the easiest thing to do in many cases is use an LED driver chip with a built-in booster circuit. Some of those can do a good job regulating LED brightness independent of supply voltage.
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If the power source was exactly 3.3V and the voltage drop across the LED was 3.3V then you would not need a current limiting resistor. However, the world is not perfect and there are imperfections in everything!
You only can calculate an appropriate safety resistor value after accounting for the power supply configuration and variation in the LED forward voltage. If you think $V_{SOURCE}-V_{LED}$ may have an error/variation up to $0.5 \mbox{ }V$, then you would calculate the resistor value for that. As an example the case of $\pm 0.5\mbox{ }V$, which isn't unreasonable for a 10% 5V source:
$$\frac{V}{I} = \frac{0.5\mbox{ }V}{20 \mbox{ mA}} = 25\mbox{ }\Omega$$
Just note that this is probably not a good idea in practice, but it is possible.
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If you don't use a current limiting resistor on a LED, it will typically splice itself in two (personal experience). – Hans Oct 25 '11 at 18:41
LEDS can handle much more PEAK current than steady state. Study the LED datasheet and then PWM the LED within its PEAK duty cycle limits and you then will not need a resistor
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– stevenvh Apr 17 '12 at 12:58
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http://mathhelpforum.com/calculus/1574-x-intercept-tangent-parabola.html
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Thread:
1. x-intercept of tangent to parabola
Show that the tangent to the parabola Y=Ax^2 (for A does not equal 0) at the point where x = c will intersect the x axis at the point (c/2 , 0 . Where does it intersect the y axis.
2. Originally Posted by frozenflames
Show that the tangent to the parabola Y=Ax^2 (for A does not equal 0) at the point where x = c will intersect the x axis at the point (c/2 , 0 . Where does it intersect the y axis.
To find the tangent find the derivative of $y=ax^2$ at $x=c$ which is $\lim_{h\rightarrow 0} \frac{a(x+h)^2-ax^2}{h}=2ac$
Thus, the slope of the tangent line is $m=2ac$. Now use the formula for finding equations of line by knowing thier slopes and the point that they pass through. That formula is $y-y_0=m(x-x_0)$this tangent line has slope $m=2ac$ and passes through the point of tangentcy $x=c$ thus $y=ac^2$ Thus, the equation of the tangent line is $y-ac^2=2ac(x-c)$ thus, $y=2acx-2ac^2+ac^2=2acx-ac^2$ now set $y=0$ because this is the x-intercept thus $2acx-ac^2=0$ solve for $x$ thus $x=c/2$
Q.E.D.
3. Where does it intersect the y Axis?
4. Originally Posted by frozenflames
Where does it intersect the y Axis?
Easy since the equation of the tangent line is $y=2acx-ac^2$ the y-intersept is when $x=0$ thus, $y=-ac^2$
Q.E.D.
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http://www.physicsforums.com/showthread.php?t=46093
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Physics Forums
## "classic" problem
Here is a problem I have been trying in Maple, but to no avail.
I am trying to calculate the theta that will result in maximum distance for a projectile fired from off the edge of plateau onto another, lower surface. The projectile can be considered to be fired from the exact corner of the ledge so there is no minumim distance; a velocity of zero will result in the projectile moving no distance but falling onto the lower surface.
...........v[0]
______/
.........|
.........|______
(ignore the periods. the slash is the vector)
The method I have tried is to try to solve for t using the y-coordinates of motion, and then plug this back into the x-equation, which is linear. However, in trying to solve for t with the y-equations, I get two very nasty solutions (due to the quadratic solution) that I really am not going to do by hand, and Maple has decided not to cooperate. I am using Maple 9.5 and I know my way around, and I don't think there is any more I can do this way. If anyone knows of a better way to solve this, please let me know. Thanks.
PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus
Recognitions: Gold Member Science Advisor Staff Emeritus The angle for best range is always 45 degrees above the horizontal. See this page: http://hyperphysics.phy-astr.gsu.edu...raj.html#tra12 The range of a projectile is $$R = \frac{v_0^2 \sin{2 \theta}}{g}$$ Assuming $v_0$ is a constant, you're just solving for the maximum of $\sin{2 \theta}$. The maximum value of the sine function is one, when its argument is $\pi / 2$, or when $\theta = \pi / 4 = 45^o$. - Warren
Recognitions: Gold Member Science Advisor Staff Emeritus Ah, actually, it serves me right for not reading carefully enough. You're trying to solve for the case with a launch platform higher than the landing site. (cyrus, you're right, it does indeed affect the best angle). You're trying to sovle the general equation shown here: http://hyperphysics.phy-astr.gsu.edu...mgmec/tra9.gif Sorry about that. You should be able to just take the derivative of the expression with respect to theta and solve for its zeroes: $$R = v\cos(\theta) \left[ \frac{2 v \sin(\theta) y}{g} \pm \sqrt{ \frac{(v \sin(\theta))^2}{g^2} - \frac{2 y}{g}} \right]$$ - Warren
## "classic" problem
I was really distraught when you posted that! Thanks for the correction.
I did take the derivative and solve for zero. That proved to be quite unwieldy and difficult to simplify. Thank you for the confirmation, however. Is there any other way to solve this?
Recognitions: Gold Member Science Advisor Staff Emeritus Machinus, I'll put it through Mathematica when I get home. I agree in that I'm not about to try to solve that one by hand. - Warren
Recognitions: Science Advisor Staff Emeritus OK, it takes some fooling around, but my ancient version of maple gets it. First we solve for the time as a function of the height h, assumed here to be positve when the projectile drops $${\frac {{\it vy}+\sqrt {{{\it vy}}^{2}+2\,gh}}{g}}$$ then xmax is given by $${\frac {v\cos \left( \theta \right) \left( v\sin \left( \theta \right) +\sqrt {{v}^{2} \left( \sin \left( \theta \right) \right) ^{ 2}+2\,gh} \right) }{g}}$$ then we make the substitutions sin(theta) = tan/sqrt(1+tan^2), cos(theta)=1/sqrt(1+tan^2) and tell maple to simplify, we get $${\frac {v \left( v \left( \tan \right) +\sqrt {{v}^{2}{\tan}^{2}+2\,gh +2\,gh{\tan}^{2}} \right) }{ \left( 1+{\tan}^{2} \right) g}}$$ solving for the derivative of the above to be zero wrt tan, we get tan = v/sqrt(v^2+2gh) which is the value of the tangent of the angle theta.
How did you get such a simple t? Also, aren't there two solutions?
hey warren, that website is incorrect. It should be Vy/g, not 2Vy/g I think they made a mathematical error. I got the same equation as pervect. Yes, two solutions, but you want the one that is positive and largest. This will be the max time. so the distance x=vx *t will have the largest value.
Recognitions:
Science Advisor
Staff Emeritus
Quote by Machinus How did you get such a simple t? Also, aren't there two solutions?
I put
minus !!!!
y := vy*t - .5*g*t^2;
solve(y=-h,t);
into Maple. I also did a foo:=sub(t=<expr>,y);simplify(foo);
But I''d encourage you to double check, just on general principles.
Recognitions: Gold Member Science Advisor Staff Emeritus pervect: Something about your tans without arguments is deeply unsettling to me. - Warren
hey warren, did u see the error?
Recognitions:
Science Advisor
Staff Emeritus
Quote by chroot pervect: Something about your tans without arguments is deeply unsettling to me. - Warren
OK, use 'u' or something that's more innocuous if 'tan' without arguments is too unsettling....
Quote by Machinus Also, aren't there two solutions?
I forgot to mention - choose the time solution that's positive.
Well, thanks to all for the help, but my professor from Fermi has solved it for me relatively easily. I haven't seen his work yet but he presented the solution to another student who knew that I was working on the problem. Without further ado: $$\theta_{max}=\arctan(\frac{v_{0}}{\sqrt{v_{0}^2+2gh}})$$ If you will note, an encouraging check to this problem is to substitute zero for the height, and the formula reduces to $$\theta_{max}=\arctan(\frac{v_{0}}{v_{0}})$$ which is of course 45 degrees, the correct value from a projectile on a flat surface. Thank you professor Davenport!
We also got that anwser machinus, did you read what pervect posted, he solved it with maple
solving for the derivative of the above to be zero wrt tan, we get tan = v/sqrt(v^2+2gh)
Ah, yes, now I see that they are the same answer. I thought it was kind of unclear, but you are right. Thank you!
Dont thank me thank pervect, and I will thank him too, thanks. I thought that was an interesting problem. What did the angle you were looking for turn out to be?
The angle is dependent on the height. Right now I am working on making a nice-looking graph to display the results.
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How do you go from a quadratic function in...
Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
3 posts • Page 1 of 1
How do you go from a quadratic function in...
by Absolutely on Mon Apr 04, 2011 9:07 pm
generic form $ax^2+bx+c$ to the standard parabola form $a(x-h)^2+k$ by completing the square?
I know that the vertex of a parabola (h, k) is $h=\frac{-b}{2a}$ and $k=f(h)$ but I can't figure out how the first form gets to the second. I just want to know how the standard form is derived form the generic form by completing the square in a general way (meaning without numbers, just the letters to represent the constants); step-by-step please!
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Re: How do you go from a quadratic function in...
by Martingale on Mon Apr 04, 2011 10:27 pm
Absolutely wrote:generic form $ax^2+bx+c$ to the standard parabola form $a(x-h)^2+k$ by completing the square? I know that the vertex of a parabola (h, k) is $h=\frac{-b}{2a}$ and $k=f(h)$ but I can't figure out how the first form gets to the second. I just want to know how the standard form is derived form the generic form by completing the square in a general way (meaning without numbers, just the letters to represent the constants); step-by-step please!
assume $a\neq0$
$ax^2+bx+c$
$=a(x^2+\frac{b}{a}x)+c$
$=a\left(x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2\right)+c$
$=a\left(x+\frac{b}{2a}\right)^2-a\left(\frac{b}{2a}\right)^2+c$
$=a\left(x-\frac{-b}{2a}\right)^2-a\left(\frac{b}{2a}\right)^2+c$
$=a\left(x-\frac{-b}{2a}\right)^2+c-\frac{b^2}{4a}$
$=a\left(x-h\right)^2+k$
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Re: How do you go from a quadratic function in...
by Absolutely on Mon Apr 04, 2011 11:46 pm
Thanks! Seeing the algebraic manipulations behind these derivations makes the formulas more sensible to me.
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http://math.stackexchange.com/questions/149139/explaining-why-sin-and-cos-are-not-at-right-angles?answertab=votes
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Explaining why sin and cos are *not* at right angles
Disclaimer: I'm an engineer, not a mathematician
I recently had a fierce discussion (lots of blood) on electronics.stackexchange about phase shifts.
The impedance of a resistor is real, that of a coil imaginary ($j\omega L$), so that adding 3V across a resistor and 4V across a coil in series results in a 5V overall.
Now John (let's call him John) explained the same by using a sine and a cosine function, claiming they're at 90°. I can see where he gets this, and I tried the following to explain that the sine of a number and the cosine of a number are scalars, not vectors, so they can't have a phase difference:
$$\sin(\omega t) = \dfrac{e^{j \omega t} - e^{- j \omega t}}{2j}$$
This is easy for me to visualize: $e^{\omega t}$ and $e^{- \omega t}$ are phasors rotating in opposite directions in the complex plane. Their difference is a vector on the imaginary axis. Dividing by $j$ rotates that vector by $\pi /2$ clockwise, so that it moves from the imaginary axis to the real axis. And then it's a sine, a scalar. Sitting nicely next to the cosine, no phase difference.
My problem is that half a second ago it was still a vector. How does it become a scalar? Or do "being a vector on the real axis" and "being a scalar" mean the same thing?
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They are orthogonal, in the sense of the $L^2$ inner product. (This is why we can decompose functions into Fourier series!) – Zhen Lin May 24 '12 at 10:26
A vector is (by definition) an element of a vector space. $R$ forms a vector space over itself, so the distinction that something is either scalar or vector is not exactly valid. – fishlips May 24 '12 at 10:29
3
The sine of a number is a number and can't have a phase difference with another number. The sine as a function over the real line is another matter. The sine function and the cosine function obviously have a $90^\circ$ phase difference, or I don't know what phase difference is. – Rahul Narain May 24 '12 at 10:31
2
I think the comments show that there is some ambiguity in the meaning of "at right angles" - but also that orthogonality is a rich mathematical concept with many applications. The naïve thought that - in a right-angled triangle with hypotenuse 1, sin and cos are literally at right angles is related, as also is the fact that the projections of points on the unit circle onto orthogonal axes track sin and cos at right angles. – Mark Bennet May 24 '12 at 11:14
2
@user32112, you have to distinguish between the sine of a particular number and the sine function itself. Your argument is kind of like saying, it's not true that the President of the United States is elected every four years, because Barack Obama is the President of the United States, and Barack Obama is not elected every four years. – Rahul Narain May 24 '12 at 11:20
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3 Answers
I'm a mathematician, not an engineer, so this may not be the "right" answer to your question, but hopefully will give some insight into the mathematical language (and expands on fishlips' comment).
Typically, the word "vector" means an element of a vector space, for example $\mathbb{R}^3$, or the set of functions $\mathbb{R}\to\mathbb{R}$, and the word "scalar" means an element of a field, such as $\mathbb{R}$ or $\mathbb{C}$. The problem with this terminology is that fields are themselves vector spaces, so depending on your point of view an element of them can either be a vector (if you're thinking about the vector space structure) or a scalar (if you're thinking about the field structure).
The field $\mathbb{C}$ as in your example can be particularly complicated in this respect, as it is a field, a one-dimensional complex vector space, and a two-dimensional real vector space. My guess is it's the last interpretation that is most useful to you, so you should interpret $\sin(\omega t)$ as a vector that happens to lie in the real axis, and not necessarily as a scalar. (But maybe that's not the interpretation you want, hence the disclaimer at the beginning).
Zhen Lin's comment is also worth expanding on - as I mentioned above, sets of functions can be vector spaces. As $\sin$ and $\cos$ are both functions $\mathbb{R}\to\mathbb{R}$, they are vectors in the vector space consisting of all such functions, and after choosing an appropriate inner product (the $L^2$ inner product), these vectors are orthogonal to each other. (If you don't have the time or inclination to read through the inner product article, the relevant feature of inner products is that they tell you what it means for two vectors to be orthogonal.)
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To comment on this:
````I can see where he gets this, and I tried the following
to explain that sine and cosine are scalars, not vectors,
so they can't have a phase difference:
````
This is wrong. First off, the notions of 'scalar' and 'vector' aren't really relevant here (directly, anyways). The basic definition of phase is that it's a property of certain real-valued functions of the real numbers. Specifically, for any function $f$ that can be written as
$$f(t) = B + A \sin(\omega t + \phi)$$
then we define $\mathop{\mathrm{Phase}}(f) = \phi$. Phase is an "angular position" meaning that it's only modulo $2 \pi$ radians (or modulo 360 degrees, if you prefer angles to be measured that way).
As a point of grammar, the type of "sin" is "function", and the type of "sin 3" is "number". The type of "sin t" should also be a "number" (the number one gets by plugging "t" into "sin") -- but by a common abuse of notation is also used to denote the function "sin".
Now that I have the link, I think the meat of John's argument is as follows.
The main point is the sum-of-sines identity which is, if I haven't made an error along the way,
$$A \sin(\omega t) + B\sin(\omega t + \phi) = C \sin(\omega t + \theta)$$
where
$$C^2 = (A + B \cos \phi)^2 + (B \sin \phi)^2$$ $$\tan(\theta) = (B \sin \phi) / (A + B \cos \phi)$$.
The main point you can see is that if the relative phase between the two sine waves is given by $\phi = \pi/2$, then the amplitude of the sum of the two sine waves is given by
$$C^2 = A^2 + B^2$$
My knowledge of electrical engineering is limited and rusty -- but I believe the point is that the current across the resistor and the coil are indeed sine waves 90 degrees out of phase, and that you seek to add them.
One interesting thing to note is that this gives a justification for representing a sine wave
$$A \sin(\omega t + \phi)$$
by the plane vector
$$\langle A \cos \phi, A \sin \phi \rangle$$
because the formula shows that adding sine waves corresponds to adding vectors. Additionally, the magnitude of the wave is the magnitude of the vector, and the phase of the wave is the polar angle of the vector.
I know I knew this once upon a time, but I had long forgotten. It's still a neat fact, I think. :)
(A more boring, but somewhat more obvious statement of this same fact is that $\{ \cos \omega t, \sin \omega t \}$ are a basis for the set of sine waves with period $2 \pi / \omega$)
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You're right, the sine and cosine in the quote seem to refer to functions. I'll fix it. – stevenvh May 24 '12 at 11:28
The "boring statement" is interesting. I just posted about the Fourier series on electronics.stackexchange, and used this to explain an alternative definition (sum of only sines instead of sin + cos)! :-) – stevenvh May 24 '12 at 12:14
If you decide to view $e^{\omega t}$ as a vector (which is a useful point of view, but not the only one possible), then you should also interpret $j$ as a vector (namely, $(0,1)$).
Now, your fraction is the proportion of two parallel vectors which is a scalar.
On the other hand, if you prefer to interpret $j$ as rotation acting on the space, then you should also interpret $e^{\omega t}$ as rotation on the space. In that case, your expression turns out to be the function that is the scalar multiplication by $\sin$.
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http://physics.stackexchange.com/questions/32360/what-entities-create-a-gravitational-field/32372
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What entities create a gravitational field?
It is well known that masses create a gravitational field. Photons are affected by gravitation, but do they generate a gravitational field as well? What about the other gauge bosons?
Do gravitons create a gravitational field?
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1
– Qmechanic♦ Jul 19 '12 at 10:57
3 Answers
According to the general equivalence principle: Anything creates gravitational field. This is because Anything (except Nothing) carries a non-zero energy-momentum tensor, which is the source of gravitational field.
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Do gravitons create a gravitational field?
There's an interesting section in MTWs "Gravitation" describing how the GR equations can be arrived at by considering massless spin-2 field ("gravitons") in flat spacetime and iterating corrections from considering that the non-zero stress-energy tensor for this field is a source of this field, i.e., gravitons gravitate.
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The source of the spacetime curvature is the stress-energy tensor. One component of this is energy density, where mass is counted as energy using Einstein's famous equation $E = mc^2$. Photons possess energy so they do contribute to the curvature.
However there are other things, as well as energy density, that contribute to the curvature and therefore gravity. These are:
• momentum flux
• pressure
• shear stress
The W and Z gauge bosons are massive at low energies and contribute to gravity courtesy of this mass. Above the electroweak transition the bosons become massless but still possess the same energy so still contribute to gravity.
Gravitons are somewhat speculative since they're a concept arising from a canonical quantisation that isn't renormalisable. However if you consider the classical equivalent of the graviton to be simply the gravitational field (i.e. the curvature) then yes it does contribute to the curvature. This is known as the gravitational self energy.
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http://terrytao.wordpress.com/2011/08/04/localisation-and-compactness-properties-of-the-navier-stokes-global-regularity-problem/
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What’s new
Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao
# Localisation and compactness properties of the Navier-Stokes global regularity problem
4 August, 2011 in math.AP, math.MP, paper | Tags: concentration compactness, energy estimates, global well-posedness, homogenisation, localisation, Navier-Stokes equations
I’ve just uploaded to the arXiv my paper “Localisation and compactness properties of the Navier-Stokes global regularity problem“, submitted to Analysis and PDE. This paper concerns the global regularity problem for the Navier-Stokes system of equations
$\displaystyle \partial_t u + (u \cdot \nabla) u = \Delta u - \nabla p + f \ \ \ \ \ (1)$
$\displaystyle \nabla \cdot u = 0 \ \ \ \ \ (2)$
$\displaystyle u(0,\cdot) = u_0 \ \ \ \ \ (3)$
in three dimensions. Thus, we specify initial data ${(u_0,f,T)}$, where ${0 < T < \infty}$ is a time, ${u_0: {\bf R}^3 \rightarrow {\bf R}^3}$ is the initial velocity field (which, in order to be compatible with (2), (3), is required to be divergence-free), ${f: [0,T] \times {\bf R}^3 \rightarrow {\bf R}^3}$ is the forcing term, and then seek to extend this initial data to a solution ${(u,p,u_0,f,T)}$ with this data, where the velocity field ${u: [0,T] \times {\bf R}^3 \rightarrow {\bf R}^3}$ and pressure term ${p: [0,T] \times {\bf R}^3 \rightarrow {\bf R}}$ are the unknown fields.
Roughly speaking, the global regularity problem asserts that given every smooth set of initial data ${(u_0,f,T)}$, there exists a smooth solution ${(u,p,u_0,f,T)}$ to the Navier-Stokes equation with this data. However, this is not a good formulation of the problem because it does not exclude the possibility that one or more of the fields ${u_0, f, u, p}$ grows too fast at spatial infinity. This problem is evident even for the much simpler heat equation
$\displaystyle \partial_t u = \Delta u$
$\displaystyle u(0,\cdot) = u_0.$
As long as one has some mild conditions at infinity on the smooth initial data ${u_0: {\bf R}^3 \rightarrow {\bf R}}$ (e.g. polynomial growth at spatial infinity), then one can solve this equation using the fundamental solution of the heat equation:
$\displaystyle u(t,x) = \frac{1}{(4\pi t)^{3/2}} \int_{{\bf R}^3} u_0(y) e^{-|x-y|^2/4t}\ dy.$
If furthermore ${u}$ is a tempered distribution, one can use Fourier-analytic methods to show that this is the unique solution to the heat equation with this data. But once one allows sufficiently rapid growth at spatial infinity, existence and uniqueness can break down. Consider for instance the backwards heat kernel
$\displaystyle u(t,x) = \frac{1}{(4\pi(T-t))^{3/2}} e^{|x|^2/(T-t)}$
for some ${T>0}$, which is smooth (albeit exponentially growing) at time zero, and is a smooth solution to the heat equation for ${0 \leq t < T}$, but develops a dramatic singularity at time ${t=T}$. A famous example of Tychonoff from 1935, based on a power series construction, also shows that uniqueness for the heat equation can also fail once growth conditions are removed. An explicit example of non-uniqueness for the heat equation is given by the contour integral
$\displaystyle u(t,x_1,x_2,x_3) = \int_\gamma \exp(e^{\pi i/4} x_1 z + e^{5\pi i/8} z^{3/2} - itz^2)\ dz$
where ${\gamma}$ is the ${L}$-shaped contour consisting of the positive real axis and the upper imaginary axis, with ${z^{3/2}}$ being interpreted with the standard branch (with cut on the negative axis). One can show by contour integration that this function solves the heat equation and is smooth (but rapidly growing at infinity), and vanishes for ${t<0}$, but is not identically zero for ${t>0}$.
Thus, in order to obtain a meaningful (and physically realistic) problem, one needs to impose some decay (or at least limited growth) hypotheses on the data ${u_0,f}$ and solution ${u,p}$ in addition to smoothness. For the data, one can impose a variety of such hypotheses, including the following:
• (Finite energy data) One has ${\|u_0\|_{L^2_x({\bf R}^3)} < \infty}$ and ${\| f \|_{L^\infty_t L^2_x([0,T] \times {\bf R}^3)} < \infty}$.
• (${H^1}$ data) One has ${\|u_0\|_{H^1_x({\bf R}^3)} < \infty}$ and ${\| f \|_{L^\infty_t H^1_x([0,T] \times {\bf R}^3)} < \infty}$.
• (Schwartz data) One has ${\sup_{x \in {\bf R}^3} ||x|^m \nabla_x^k u_0(x)| < \infty}$ and ${\sup_{(t,x) \in [0,T] \times {\bf R}^3} ||x|^m \nabla_x^k \partial_t^l f(t,x)| < \infty}$ for all ${m,k,l \geq 0}$.
• (Periodic data) There is some ${0 < L < \infty}$ such that ${u_0(x+Lk) = u_0(x)}$ and ${f(t,x+Lk) = f(t,x)}$ for all ${(t,x) \in [0,T] \times {\bf R}^3}$ and ${k \in {\bf Z}^3}$.
• (Homogeneous data) ${f=0}$.
Note that smoothness alone does not necessarily imply finite energy, ${H^1}$, or the Schwartz property. For instance, the (scalar) function ${u(x) = \exp( i |x|^{10} ) (1+|x|)^{-2}}$ is smooth and finite energy, but not in ${H^1}$ or Schwartz. Periodicity is of course incompatible with finite energy, ${H^1}$, or the Schwartz property, except in the trivial case when the data is identically zero.
Similarly, one can impose conditions at spatial infinity on the solution, such as the following:
• (Finite energy solution) One has ${\| u \|_{L^\infty_t L^2_x([0,T] \times {\bf R}^3)} < \infty}$.
• (${H^1}$ solution) One has ${\| u \|_{L^\infty_t H^1_x([0,T] \times {\bf R}^3)} < \infty}$ and ${\| u \|_{L^2_t H^2_x([0,T] \times {\bf R}^3)} < \infty}$.
• (Partially periodic solution) There is some ${0 < L < \infty}$ such that ${u(t,x+Lk) = u(t,x)}$ for all ${(t,x) \in [0,T] \times {\bf R}^3}$ and ${k \in {\bf Z}^3}$.
• (Fully periodic solution) There is some ${0 < L < \infty}$ such that ${u(t,x+Lk) = u(t,x)}$ and ${p(t,x+Lk) = p(t,x)}$ for all ${(t,x) \in [0,T] \times {\bf R}^3}$ and ${k \in {\bf Z}^3}$.
(The ${L^2_t H^2_x}$ component of the ${H^1}$ solution is for technical reasons, and should not be paid too much attention for this discussion.) Note that we do not consider the notion of a Schwartz solution; as we shall see shortly, this is too restrictive a concept of solution to the Navier-Stokes equation.
Finally, one can downgrade the regularity of the solution down from smoothness. There are many ways to do so; two such examples include
• (${H^1}$ mild solutions) The solution is not smooth, but is ${H^1}$ (in the preceding sense) and solves the equation (1) in the sense that the Duhamel formula
$\displaystyle u(t) = e^{t\Delta} u_0 + \int_0^t e^{(t-t')\Delta} (-(u\cdot\nabla) u-\nabla p+f)(t')\ dt'$
holds.
• (Leray-Hopf weak solution) The solution ${u}$ is not smooth, but lies in ${L^\infty_t L^2_x \cap L^2_t H^1_x}$, solves (1) in the sense of distributions (after rewriting the system in divergence form), and obeys an energy inequality.
Finally, one can ask for two types of global regularity results on the Navier-Stokes problem: a qualitative regularity result, in which one merely provides existence of a smooth solution without any explicit bounds on that solution, and a quantitative regularity result, which provides bounds on the solution in terms of the initial data, e.g. a bound of the form
$\displaystyle \| u \|_{L^\infty_t H^1_x([0,T] \times {\bf R}^3)} \leq F( \|u_0\|_{H^1_x({\bf R}^3)} + \|f\|_{L^\infty_t H^1_x([0,T] \times {\bf R}^3)}, T )$
for some function ${F: {\bf R}^+ \times {\bf R}^+ \rightarrow {\bf R}^+}$. One can make a further distinction between local quantitative results, in which ${F}$ is allowed to depend on ${T}$, and global quantitative results, in which there is no dependence on ${T}$ (the latter is only reasonable though in the homogeneous case, or if ${f}$ has some decay in time).
By combining these various hypotheses and conclusions, we see that one can write down quite a large number of slightly different variants of the global regularity problem. In the official formulation of the regularity problem for the Clay Millennium prize, a positive correct solution to either of the following two problems would be accepted for the prize:
• Conjecture 1.4 (Qualitative regularity for homogeneous periodic data) If ${(u_0,0,T)}$ is periodic, smooth, and homogeneous, then there exists a smooth partially periodic solution ${(u,p,u_0,0,T)}$ with this data.
• Conjecture 1.3 (Qualitative regularity for homogeneous Schwartz data) If ${(u_0,0,T)}$ is Schwartz and homogeneous, then there exists a smooth finite energy solution ${(u,p,u_0,0,T)}$ with this data.
(The numbering here corresponds to the numbering in the paper.)
Furthermore, a negative correct solution to either of the following two problems would also be accepted for the prize:
• Conjecture 1.6 (Qualitative regularity for periodic data) If ${(u_0,f,T)}$ is periodic and smooth, then there exists a smooth partially periodic solution ${(u,p,u_0,f,T)}$ with this data.
• Conjecture 1.5 (Qualitative regularity for Schwartz data) If ${(u_0,f,T)}$ is Schwartz, then there exists a smooth finite energy solution ${(u,p,u_0,f,T)}$ with this data.
I am not announcing any major progress on these conjectures here. What my paper does study, though, is the question of whether the answer to these conjectures is somehow sensitive to the choice of formulation. For instance:
1. Note in the periodic formulations of the Clay prize problem that the solution is only required to be partially periodic, rather than fully periodic; thus the pressure has no periodicity hypothesis. One can ask the extent to which the above problems change if one also requires pressure periodicity.
2. In another direction, one can ask the extent to which quantitative formulations of the Navier-Stokes problem are stronger than their qualitative counterparts; in particular, whether it is possible that each choice of initial data in a certain class leads to a smooth solution, but with no uniform bound on that solution in terms of various natural norms of the data.
3. Finally, one can ask the extent to which the conjecture depends on the category of data. For instance, could it be that global regularity is true for smooth periodic data but false for Schwartz data? True for Schwartz data but false for smooth ${H^1}$ data? And so forth.
One motivation for the final question (which was posed to me by my colleague, Andrea Bertozzi) is that the Schwartz property on the initial data ${u_0}$ tends to be instantly destroyed by the Navier-Stokes flow. This can be seen by introducing the vorticity ${\omega := \nabla \times u}$. If ${u(t)}$ is Schwartz, then from Stokes’ theorem we necessarily have vanishing of certain moments of the vorticity, for instance:
$\displaystyle \int_{{\bf R}^3} \omega_1 (x_2^2-x_3^2)\ dx = 0.$
On the other hand, some integration by parts using (1) reveals that such moments are usually not preserved by the flow; for instance, one has the law
$\displaystyle \partial_t \int_{{\bf R}^3} \omega_1(t,x) (x_2^2-x_3^2)\ dx = 4\int_{{\bf R}^3} u_2(t,x) u_3(t,x)\ dx,$
and one can easily concoct examples for which the right-hand side is non-zero at time zero. This suggests that the Schwartz class may be unnecessarily restrictive for Conjecture 1.3 or Conjecture 1.5.
My paper arose out of an attempt to address these three questions, and ended up obtaining partial results in all three directions. Roughly speaking, the results that address these three questions are as follows:
1. (Homogenisation) If one only assumes partial periodicity instead of full periodicity, then the forcing term ${f}$ becomes irrelevant. In particular, Conjecture 1.4 and Conjecture 1.6 are equivalent.
2. (Concentration compactness) In the ${H^1}$ category (both periodic and nonperiodic, homogeneous or nonhomogeneous), the qualitative and quantitative formulations of the Navier-Stokes global regularity problem are essentially equivalent.
3. (Localisation) The (inhomogeneous) Navier-Stokes problems in the Schwartz, smooth ${H^1}$, and finite energy categories are essentially equivalent to each other, and are also implied by the (fully) periodic version of these problems.
The first two of these families of results are relatively routine, drawing on existing methods in the literature; the localisation results though are somewhat more novel, and introduce some new local energy and local enstrophy estimates which may be of independent interest.
Broadly speaking, the moral to draw from these results is that the precise formulation of the Navier-Stokes equation global regularity problem is only of secondary importance; modulo a number of caveats and technicalities, the various formulations are close to being equivalent, and a breakthrough on any one of the formulations is likely to lead (either directly or indirectly) to a comparable breakthrough on any of the others.
This is only a caricature of the actual implications, though. Below is the diagram from the paper indicating the various formulations of the Navier-Stokes equations, and the known implications between them:
The above three streams of results are discussed in more detail below the fold.
— 1. Homogenisation —
We first discuss the homogenisation results. Let us first give the fully periodic version of the periodic regularity conjectures:
• Conjecture 1.13 (Qualitative fully periodic regularity for periodic data) If ${(u_0,f,T)}$ is periodic and smooth, then there exists a smooth fully periodic solution ${(u,p,u_0,f,T)}$ with this data.
• Conjecture 7.5 (Qualitative fully periodic regularity for homogeneous periodic data) If ${(u_0,0,T)}$ is periodic, homogeneous, and smooth, then there exists a smooth fully periodic solution ${(u,p,u_0,0,T)}$ with this data.
The implications are then that Conjecture 1.4, Conjecture 1.6, and Conjecture 7.5 are equivalent, with Conjecture 1.13 implying any of the previous three conjectures.
The equivalence of Conjecture 1.4 and Conjecture 7.5 is fairly well known (it is explicitly made for instance in my previous paper). The only remaining nontrivial implication is the deduction of Conjecture 1.6 from Conjecture 1.4, i.e. the deduction of global regularity for the inhomogeneous periodic problem from global regularity for the homogeneous periodic problem.
The deduction relies on the trick of using an asymptotic version of the Galilean symmetry ${(u,p,u_0,f,T) \mapsto (\tilde u,\tilde p,\tilde u_0, \tilde f, T)}$, where
$\displaystyle \tilde u(t,x) := u(t,x-\int_0^t v(s)\ ds) + v(t)$
$\displaystyle \tilde p(t,x) := p(t,x-\int_0^t v(s)\ ds) - x \cdot v'(t)$
$\displaystyle \tilde u_0(x) := u_0(x) + v(0)$
$\displaystyle \tilde f(t,x) := f(t,x-\int_0^t v(s)\ ds),$
and ${v: [0,T]\rightarrow {\bf R}^3}$ is an arbitrary smooth velocity function. One can verify that this symmetry preserves the Navier-Stokes system of equations. One can apply this symmetry to some rapidly growing velocity, say ${v = 2wt}$ for some large vector ${w \in {\bf R}^3}$. Then the transformation does not change the initial data ${u_0}$, but introduces a rapid time oscillation to the forcing term ${f}$, replacing it by ${\tilde f(t,x) := f(t,x-wt^2)}$. As one sends ${w}$ to infinity in a suitable “irrational” direction, ${\tilde f}$ converges weakly to a function that is constant in space (this is an application of the Riemann-Lebesgue lemma). It turns out that the Navier-Stokes system of equations has enough smoothing (or compactness) properties that we may then (for ${w}$ large and irrational enough) approximate ${f}$ effectively as if it was a constant. Applying a further Galilean-type symmetry
$\displaystyle \tilde u(t,x) := u(t,x-\int_0^t v(s)\ ds) + v(t)$
$\displaystyle \tilde p(t,x) := p(t,x-\int_0^t v(s)\ ds)$
$\displaystyle \tilde u_0(x) := u_0(x) + v(0)$
$\displaystyle \tilde f(t,x) := f(t,x-\int_0^t v(s)\ ds) + v'(t),$
for a suitable ${v}$ (not the same as the previous ${v}$), we can then set ${\tilde f}$ to be identically zero, without adjusting the initial data ${u_0}$. Thus, we have effectively transformed the inhomogeneous problem to a homogeneous one, which gives the deduction of Conjecture 1.6 from Conjecture 1.4.
Note that at one stage we subtracted a linear term ${x \cdot v'(t)}$ from the pressure. This would destroy any periodicity properties the pressure had, and so this trick does not work in the fully periodic setting.
One interpretation of this result is that the partially periodic setting is not the right setting to discuss the inhomogeneous problem. (For the homogeneous problem, it is not difficult to use Galilean transformations to show that the global regularity problem for the partially periodic and fully periodic problem are equivalent.) Indeed, the Galilean invariance reveals a breakdown of uniqueness for the Navier-Stokes flow in the partially periodic setting. Once one fixes the pressure to be periodic also, it is possible (by a variety of means, e.g. energy methods) to recover uniqueness for the flow (modulo the rather trivial caveat that the Navier-Stokes equations only determine the pressure ${p}$ up to a time-dependent constant, which does not have any net impact on (1) since ${p}$ only appears through its gradient).
— 2. Concentration compactness —
The compactness results in the paper relate qualitative and quantitative versions of the Navier-Stokes regularity problem. This portion of the argument is based on my previous paper in this topic, which dealt with the periodic homogeneous case; in the non-periodic setting it also uses the concentration-compactness method as developed by Bahouri-Gerard, by Gerard, and by Gallagher.
In the periodic homogeneous setting, the results in my previous paper show the equivalence of Conjecture 1.4 (and hence Conjectures 1.6 and 7.5) with the following conjectures:
• Conjecture 7.4 (Global existence of ${H^1}$ mild solutions for homogeneous periodic data) If ${(u_0,0,T)}$ is periodic, homogeneous, and ${H^1}$, then there exists a periodic ${H^1}$ mild solution ${(u,p,u_0,0,T)}$ with this data.
• Conjecture 7.2 (Local quantitative regularity for homogeneous periodic ${H^1}$ data) If ${(u,p,u_0,0,T)}$ is a periodic homogeneous smooth solution, then one can bound ${\|u\|_{L^\infty_t H^1_x}}$ by a function of ${\|u_0\|_{H^1_x}}$ and ${T}$.
• Conjecture 7.3 (Global quantitative regularity for homogeneous periodic ${H^1}$ data) If ${(u,p,u_0,0,T)}$ is a periodic homogeneous smooth solution, then one can bound ${\|u\|_{L^\infty_t H^1_x}}$ by a function of ${\|u_0\|_{H^1_x}}$.
The equivalences were largely based on a compactness property of the periodic Navier-Stokes flow (which was inherited by a corresponding compactness property of the periodic heat equation), in that for short times, the flow map mapped weakly convergent data in ${H^1}$ to strongly convergent solutions, and in particular maps closed balls in ${H^1}$ (which are weakly compact) to compact subsets; because continuous functions on compact sets are bounded, we can then convert qualitative estimates to quantitative ones. The global form of quantitative regularity was obtained by using the fact (from energy estimates) that the energy dissipation (which is essentially the ${H^1}$ norm of the solution ${u}$) must eventually become small.
It turns out that the same method partially applies in the inhomogeneous setting (after stating the conjectures properly, see Conjectures 1.14 and 1.15 in the text); we will not detail this here. Note that the inhomogeneous equation can add energy to the system, as well as dissipate it out, and so we do not have the equivalence between local and global quantitative estimates in this setting.
The method also works in the homogeneous non-periodic setting (with the right formulations of the conjectures, which are Conjectures 1.9, 1.17, 1.18, 1.19 in the text), but with the key additional difficulty that the Navier-Stokes flow is no longer compact, but merely concentration-compact, due to the non-compact translation symmetry that is available for the space ${H^1_x({\bf R}^3)}$. (Concentration compactness is discussed in these previous blog posts.) One then has to deal with sequences of data that are not strongly convergent, but are essentially the superposition of a number of profiles that are being translated off to infinity in different directions. However, the theory for dealing with this is well-developed; in particular, the paper of Gallagher already treats this sort of profile decomposition for Navier-Stokes in the critical regularity regime, which is more difficult than the subcritical regularity regime considered here. As such, we were able to use standard techniques to obtain the equivalences here. The main point here is that if one superimposes two profiles that are translated to be sufficiently far from each other, then the nonlinear interactions between the two profiles are weak enough that one can use perturbative techniques to obtain an approximate principle of superposition; the evolution of the superimposed profiles is essentially just the superposition of the individual profiles.
— 3. Localisation —
The most novel arguments in the paper concern the the localisation results, which connect the periodic, Schwartz, smooth finite energy, and smooth ${H^1}$ categories to each other. Before stating the results precisely, let us consider a simpler (but still unsolved) situation, namely that of the supercritical nonlinear wave equation
$\displaystyle - \partial_{tt} u + c^2 \Delta u = u^7, \ \ \ \ \ (4)$
for some scalar field ${u: [0,T] \times {\bf R}^3 \rightarrow {\bf R}}$ with a given smooth initial position ${u(0,x) = u_0(x)}$ and (for simplicity) zero initial velocity ${\partial_t u(0,x) = 0}$, and where ${c>0}$ is a fixed constant (the speed of light). One can phrase the global regularity problem in this setting for periodic smooth ${u_0}$, Schwartz ${u_0}$, smooth finite energy ${u_0}$ (where the energy is now the ${H^1_x({\bf R}^3)}$ norm rather than the ${L^2_x({\bf R}^3)}$ norm), and so forth. But for this equation, all formulations of the problem are logically equivalent, thanks to the finite speed of propagation property. Among other things, this property asserts that if one has two solutions ${u, \tilde u}$ to the equation (4) that initially agree on a ball ${B(x_0,R)}$, then for subsequent times ${t>0}$, the solutions will continue to agree on a slightly smaller ball ${B(x_0,R-ct)}$ (of course, this statement becomes vacuous once ${t}$ is large enough). This property allows us, for instance, to deduce global regularity for arbitrary smooth data (of unlimited growth) from, say, the Schwartz data case. Indeed, to obtain a solution up to time ${T}$ for smooth data ${u_0}$, one could cover the domain ${{\bf R}^3}$ by balls ${B(x_i,T)}$ of radius ${T}$, then for each such ball ${B(x_i,T)}$, smoothly truncate the data ${u_0}$ to the ball ${B(x_i,3T)}$ in such a way that it still agreed with ${u_0}$ on ${B(x_i,2T)}$. This truncated data is Schwartz, and so by hypothesis can be extended to a smooth solution up to time ${[0,T] \times B(x_i,T)}$; from finite speed of propagation we see that these partially defined solutions agree with each other on their common domain of definition, and can thus be glued together to form a global solution for the original data. The same sort of argument (combined with the trick of embedding a ball such as ${B(x_i,T)}$ into a sufficiently large torus) lets one deduce the non-periodic global regularity conjecture for (4) from the periodic one.
The finite speed of propagation property for (4) is proven by energy estimates; one basically computes the energy at time ${t}$ on the ball ${B(x_0,R-ct)}$ and shows via integration by parts that this local energy is non-increasing in time.
The Navier-Stokes equation (1) unfortunately does not enjoy finite speed of propagation. However, due to the transport term ${(u \cdot \nabla)u}$ in (1), it is reasonable to expect that the solution propagates at velocity ${u}$ (this is of course consistent with the physical interpretation of ${u}$ as the velocity field, though with the caveat that ${u}$ is actually the particle velocity rather than the group velocity). As such, if one had an a priori bound on the ${L^1_t L^\infty_x}$ norm of ${u}$, this would suggest that solutions to the Navier-Stokes equation only propagate themselves by a bounded distance. This would, heuristically at least, allow one to repeat the above types of arguments to equate the various forms of the Navier-Stokes conjecture.
It turns out, somewhat remarkably, that an a priori bound on the norm ${L^1_t L^\infty_x}$ of ${u}$ is indeed available. This does not resolve the main difficulty in the global regularity problem – the lack of a controlled coercive quantity that is either subcritical or critical, as discussed in this post – because this norm is supercritical. Actually, one can guess at the existence of such a bound by using the amplitude-frequency heuristics, discussed in this post. If the Navier-Stokes solution ${u}$ has an amplitude ${A}$ and a frequency ${N}$ for a period of time ${\tau}$ over a volume ${V}$ of space, then the energy dissipation estimate
$\displaystyle \int_0^T \int_{{\bf R}^3} |\nabla u(t,x)|^2\ dx \leq \int_{{\bf R}^3} \frac{1}{2} |u_0(t,x)|^2\ dx$
(setting ${f=0}$ for simplicity) suggests that
$\displaystyle \tau N^2 A^2 V \lesssim 1,$
while the uncertainty principle (discussed in this post) suggests that ${V \gtrsim N^{-3}}$. Finally, the linear term ${\Delta u}$ and nonlinear term ${(u \cdot \nabla) u}$ have heuristic magnitudes about ${N^2 A}$ and ${A^2 N}$ respectively, so nonlinear-dominant behaviour should only occur when ${A \gtrsim N}$. Putting all this together, one soon calculates heuristically that
$\displaystyle \tau A \lesssim 1,$
which then predicts the ${L^1_t L^\infty_x}$ bound. It turns out that one can make this argument rigorous by a routine application of Littlewood-Paley theory.
Even with the rigorous bound on ${\|u\|_{L^1_t L^\infty_x}}$, though, there is still work to do to make the localisation results rigorous. The main technical tool used in this paper is a localisation result for the enstrophy
$\displaystyle \frac{1}{2} \int_{{\bf R}^3} |\omega(t,x)|^2\ dx.$
The precise local enstrophy estimate we derive is technical, but roughly speaking, this result asserts that if the enstrophy is initially small on some ball ${B(x_0,R)}$, then it will remain small on the ball ${B(x_0,R-O(1))}$. It is proven somewhat similarly to finite speed of propagation for wave equations, in that one computes the local enstrophy adapted to a ball ${B(x_0,R(t))}$ of shrinking radius and tries to keep this enstrophy from blowing up. Whereas ${R(t)}$ was previously shrinking at a constant speed ${c}$, though, now one needs to shrink the radius at a speed proportional to ${\|u(t)\|_{L^\infty_x}}$.
A key difficulty though arises from non-local effects. If one takes the curl of (1) one obtains the vorticity equation
$\displaystyle \partial_t \omega + (u \cdot \nabla) \omega = (\omega \cdot \nabla) u + \nabla \times f.$
The problem is with the nonlinear term ${(\omega \cdot \nabla) u}$. Using the divergence-free nature of ${u}$, one can solve for ${u}$ in terms of ${\omega}$ via the Biot-Savart law
$\displaystyle u = \Delta^{-1} \nabla \times \omega.$
Unfortunately this law is non-local; the value of ${u}$ at a given point ${x}$ can be influenced by the vorticity at quite distant parts of space. In particular, the velocity field ${u}$ inside the ball ${B(x_0,R(t))}$ is influenced by the vorticity outside of ${B(x_0,R(t))}$, which will not be controlled by local enstrophy. One can still control this influence using other expressions, such as the total energy, but these are supercritical quantities and if one relies on them too heavily, then it will be impossible to keep the local enstrophy under control.
To resolve this, one has to perform some delicate harmonic analysis (in particular, a Whitney decomposition of the ball ${B(x_0,R(t))}$, and local elliptic regularity), and to carefully choose the right cutoffs to define local enstrophy properly; it turns out that one can (barely) reduce the non-local effects mentioned earlier to the point where they can actually be controlled satisfactorily by the quantities one has to play with (namely, local enstrophy and the total energy).
Using the local enstrophy inequality, one can deduce global regularity for Schwartz or smooth ${H^1}$ data from the periodic global well-posedness result in ${H^1}$. The idea is to locate a large ball ${B(0,R)}$ outside of which the data has small enstrophy; using the local enstrophy inequality, combined with standard partial regularity methods, this keeps the solution uniformly smooth (spatially, at least) outside of a slightly larger ball, say ${B(0,2R)}$. One can then truncate this solution to be compactly supported in, say, ${B(0,3R)}$, at the cost of introducing a forcing term to compensate for the error terms introduced by the truncation. One can then embed the resulting object into a sufficiently large torus to deduce the smooth ${H^1}$ regularity problem from the periodic theory. A related argument (using weak compactness, as in the construction of Leray-Hopf weak solutions) almost allows one to construct global smooth solutions from finite energy data once one has smooth solutions from finite enstrophy data.
In both of these results, though, there is an annoying technical quirk that prevents the results from being stated as cleanly as one might initially hope, which is that the partial regularity methods alluded to earlier give plenty of regularity in space in regions of small enstrophy, but very little regularity in time. This seems to be an inherent feature of the Navier-Stokes problem: if a singularity or near-singularity occurs at one point ${x_0}$ in time, then this instantaneously propagates (via the incompressibility of the fluid) to create discontinuities in time (but not in space) for the pressure at points ${x}$ very far from ${x_0}$. (Think of how, for instance, a heart beat can cause instantaneous change in blood pressure throughout the body, or how a collision in Newton’s cradle instantaneously creates an effect at the other end of the cradle.) This effect prevents one from completely localising the effect of a singularity to a bounded region of space, and causes some unpleasant technicalities to be introduced into the various implications one can prove by this method. For instance, the solutions ${u}$ we construct from smooth finite energy data are not completely smooth; they are smooth for all positive times (thanks to parabolic regularising effects) and are ${C^1_t C^\infty_x}$ in a neighbourhood of the initial time ${t=0}$, but I was unable to get more regularity in time beyond ${C^1}$.
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## 9 comments
missed some word in the first sentence.:) [Corrected, thanks - T.]
5 August, 2011 at 2:54 am
mfrasca
Hi Terry,
In the first sentence of point 3.localisation there is a double “the”.
Best,
Marco
[Corrected, thanks - T.]
We have seen attempts at creating a “pseudo” pressure grad term that is actually the sum of the pressure grad and a “hidden” external force. Perhaps it would have been better to have put limits on the external supply of energy rather than the force which has turned out to be awkward to bind. For example, any point cannot exceed a predefined external energy extraction rate per unit volume (power draw density). This would be coupled with a total external energy supply limit.
6 August, 2011 at 11:51 pm
vydonhu
Thank Terry!
[...] and compactness properties of the Navier-Stokes global regularity problem“, discussed in this previous blog post. As it turns out, I was somewhat impatient to finalise the paper and move on to other things, and [...]
Hi Terry,
I have very much enjoyed your posts on the N-S equations. I have been thinking about these ‘annoying technical quirks’ of the NS equations. As a fluid physicist, it is at times deeply unsettling that I tacitly allow for effects to propagate instantaneously in the formulations as written and analyzed. I wanted to see if you (or others) worry about the non-physicality of these mathematical representations of fluids, and whether non-regularity could simply be an inevitable result of this physical contradiction.
This problem seems to arise when using the kinematic approximation of the Biot-Savart (valid only for a steady flow) in the dynamic NS equations for an unsteady process. I see more complexity being needed in the definition of vorticity in these flows. e.g. replacing Biot with something like: http://en.wikipedia.org/wiki/Jefimenko%27s_equations
This would be akin to how E-M fields were proven to change each other in time and space (Maxwell’s original 1861 proof was in terms of ‘molecular vorticies’!) I believe accurately describing the mechanism of local flow organization (in space and time) via the vorticity could be key to unlocking this process.
Well, just about any useful mathematical model makes nonphysical assumptions – for instance, fluids are almost always modeled by a continuum, when in reality they are composed of a huge number of interacting particles. But if the model is robust enough, one can still expect it to give an accurate prediction of reality, even if at an ontological level it is quite distinct.
With regard to the instantaneous propagation of the pressure, one could use the compressible Navier-Stokes equation instead of the incompressible one, which would presumably be a more faithful model, and one without the instantaneous pressure issue. But in the incompressible limit I believe that it is expected that solutions of the former equation should converge to that of the latter (assuming, of course, that solutions to the latter are regular). There is quite a bit of literature on these topics; see for instance this paper of Desjardins, Grenier, Lions, and Masmoudi.
My feeling though is that this non-locality of pressure, while technically annoying, is not the main factor in determining the regularity, stability, or validity of the Navier-Stokes flow. The nonlocal aspects of the Biot-Savart law primarily affect the low frequency (i.e. coarse scale) components of the flow, whereas all the interesting nonlinear and turbulent behaviour is occuring at fine scales. Of course, in this highly nonlinear equation the coarse scales and fine scales are not completely decoupled, and so the non-locality of the Biot-Savart law does have some impact on the fine-scale dynamics, but it is indirect and I would be surprised if it ends up playing a decisive role in the regularity theory (except at the initial time t=0, where as noted in the post, slowly decaying fine-scale behaviour in the velocity can cause instantaneous loss of time smoothness for the pressure).
Think EthanEagle may have a point, though I feel a bit like an acolyte commenting to the masters.
While I am certainly nowhere near of your level in mathematics, could we not posit that the mathematical model need not be treated as a per se continuum or at the level of abstraction that it is. Based on the 2-d NS solution it would seem that the pressure terms are the problem.
Just as a thought model suppose we remove the concept of pressure and treat the fluid mass as a bounded variable step size n dimensional matrix populated with point sources (containing all the variables with which a molecule would be endowed), with n degrees of freedom and thermally defined mean free path step sizes.
Introduce a planer heat source and planer cold sink on either side of the fluid mass and one would simultaneously have an expansion wave propagating through the medium from the planar heat source, an increase in the mean free path of the area surrounding the heat source, and a decrease in the mean free path of the area surrounding the cold sink. This would introduce a density gradient from one side of the mass to the other. The random motion of the molecules within the mass would engender a momentum flux in the sparse direction. If this were occurring with gravity the expansion wave would propagate across the upper portion of the mass and the momentum flux from dense to sparse would propagate along the lower portion, i.e. a natural convective loop. The steeper the gradient the stronger the flow except in the region where the expansion and momentum force interact and counterbalance each other in the middle of the counter rotating flow which would also be the region of minimum bulk velocity and maximum turbulence provided the walls of the “container” did not have physical properties.
With this thought model it is a natural extension that turbulent regions are a function of the presence of introduced local density imbalances combined with energy conditions under which the directionally chaotic energy of the molecules expresses itself in structured turbulent motion simply because the energy of the flow near the source of turbulence is insufficient to apparently suppress the random motion (except at mach the random motion would never be suppressed) but instead “interacts” with it imparting a degree of local structure to the essentially chaotic motion. This thought model would satisfy the no-slip condition and the formation of turbulent boundary layers near physical objects.
Simplified the model would parse the flow into uni-directional and omni-directional momentum flux components based on local and global density and energy conditions. From the continuum perspective this would not be dissimilar to the pressure-velocity relationship, i.e. when pressure is at a maximum omni-directional flux is at a maximum and when velocity is at a maximum unidirectional flux is at a maximum.
I have been working on ambient flows (mainly wind) for the last 6 years and the pressure conceptual model is particularly problematic in accurately explaining or predicting the dynamics of these types of flows under certain conditions in the absence of a large body of empirical data.
It may be worth noting that the conceptual framework of incompressible vs. compressible flows comes from late 19th and early 20th century aerodynamics and forced flow environments such as wind tunnels, planes, or pipes, i.e. either driving a fluid through or at a physical object or driving a physical object through a fluid. It was convenient when analyzing those types of flows (which are most of the flows we need to analyze) to deal with subsonic flows as incompressible. It makes sense as any “compressible” effects are negligible compared to the energy contained in the forced flow except in the boundary layer where, notably, turbulence develops.
Perhaps the solution to NS may be to reformulate the state equations without the pressure term. That would of course mean stepping way back and having to reformulate all the associated equations sans pressure.
Such an approach would also inherently solve the coarse vs fine issue based on the resolution of the applied matrix. Only real problem would be introducing the right basis functions to generate the chaotic turbulent behavior in regions of high turbulence. It would never be fully accurate at describing the “reality” of the turbulence as the turbulence in this model is inherently chaotic but it should do a decent job of coming quite close at the fine scale and be very accurate at the coarse scale.
12 September, 2011 at 5:13 am
Tim Nguyen
Hi Terry, it seems that your blog posts are not updating on your buzz (at least for me).
Cancel
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http://mathoverflow.net/questions/76413/torsion-subgroups-in-families-of-twists-of-elliptic-curves/76416
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## Torsion subgroups in families of twists of elliptic curves
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Here is the short version:
Fix an elliptic curve $E/\mathbb{Q}$. How does the torsion structure $E_d(\mathbb{Q})_{tors}$ vary, as $E_d$ runs through the quadratic twists of $E$?
Here is the longer version:
I have been playing with SAGE this morning. I inserted the elliptic curve ('11a1') $$E : y^2 + y = x^3 - x^2 - 10x - 20$$ which has rational torsion subgroup isomorphic to $\mathbb{Z}/5\mathbb{Z}$. I then computed its quadratic twist $E_d$ for all squarefree $d$ up to 2000, and observed $E_d(\mathbb{Q})_{tors}$ was always trivial.
Can it be that, in this particular family of quadratic twists, all but one of the curves have trivial torsion? Is this a general phenomenon?
(I ran this experiment for several other curves $E$ and got the same impression; that all but one of the curves in a family of twists have the same torsion structure.)
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1
For non-$2$-torsion, you have $$E_d(\mathbb{Q})[n] \oplus E(\mathbb{Q})[n] = E(\mathbb{Q}(\sqrt{d})[n]$$ if $n$ is odd. So you can ask what new torsion points arise if you make a quadratic extension. For a fixed $n$, only at most three different $d$ can have that. In most cases (maybe all), it will be none or one. – Chris Wuthrich Sep 26 2011 at 12:57
## 5 Answers
Theorem (originally due to Setzer?): Fix $E/\mathbb{Q}$ with $j(E)$ not 0 or 1728. Then for all but finitely many inequivalent twists $E_d$, the torsion subgroup $E_d(\mathbb{Q})_{tors}$ is isomorphic to `$E[2](\mathbb{Q})$`, so in particular $E_d(\mathbb{Q})_{tors}$ has order 1, 2, or 4. (Probably he also proved it for number fields.)
There's a paper of mine$^1$ with a much more general theorem using the theory of heights. I don't recall Setzer's proof except that it doesn't use heights.
Theorem: Let $K$ be a number field and let $A/K$ be an abelian variety with $\mu_n\subset {\rm Aut}(A)$. (This means we can twist $A$ by $n$'th roots of $d$.) Then every point $P\in A_d(K)$ satisfies one of the following two conditions:
• $P$ is fixed by a non-trivial $\zeta\in\mu_n$.
• $\hat h(P) \ge C_1(A)h^{(n)}(d) - C_2(A)$.
Here $\hat h$ is the canonical height relative to an ample symmetric divisor, and $h^{(n)}(d)$ is a sort of "$n$'th power free height," say equal to the minimum of $h(du^n)$ for $u\in K^*$. The constants depend on $A$, but are independent of $d$.
It follows from the theorem that after discarding finitely many `$d \in K^*/{K^*}^n$`, a point in $A_d(K)$ is either $1-\zeta$ torsion (hence $nP=O$), or its canonical height is positive, and hence it is nontorsion.
Of course, to describe more precisely what happens for the finitely many exceptional $d$ can be a delicate matter, as some of the other answers have indicated. I think it's interesting to see how one can approach the problem via heights or via representation theory.
$^1$ J.H. Silverman, Lower bounds for height functions, Duke Math. J. 51 (1984), 395-403.
EDIT: Fixed statement of first theorem. I'd originally written that "for all but finitely many inequivalent twists $E_d$, the torsion subgroup $E_d(\mathbb{Q})_{tors}$ has at most two elements." This is clearly false, since if $E$ has the form $E:y^2=(x-a)(x-b)(x-c)$ with $a,b,c\in\mathbb{Q}$, then `$E_{d}[2](\mathbb{Q})$` has order 4 for every twist.
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1
Thank you very much for your answer Joe, it's nice to see a result that works for abelian varieties generally. Also, if I may say, I found your two books on Elliptic Curves to be a great inspiration. – Giuseppe Oct 8 2011 at 22:30
2
Thanks for saying so. Authors always like to hear that their books are being read and enjoyed. – Joe Silverman Oct 10 2011 at 13:16
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Let me expand my comment above. While we believe that we expect this very very frequently, it is not always the case.
As I commented, we have $$E_d(\mathbb{Q})[n]\oplus E(\mathbb{Q})[n] = E(\mathbb{Q}(\sqrt{d}))[n]$$ if $n$ is odd. So you can ask what new torsion points arise if you make a quadratic extension. For a fixed $n$, only at most three different d can have that. In most cases, this will be at most one $d$.
But for instance the curve 98a3 has no $3$-torsion defined over $\mathbb{Q}$ but its twist by $21$ and $-7$ both have. So that is an explicit counter-example to the intuition that only one $E_d$ will have $3$ torsion.
In terms of the Galois representation $E[p]$, it is the question if the image of $\bar\rho_p$ from the absolute Galois group of $\mathbb{Q}$ to $GL_2(\mathbb{F}_p)$ has a quotient isomorphic to two copies of $\mathbb{Z}/2\mathbb{Z}$. For $p=3$, as in the case above, this can happen if the image is for instance the group of diagonal matrices. But obviously that is very rarely the case.
In summary, as JSE said, a twist can have a $p$-torsion point only if $E[p]$ is reducible (i.e. it has at least one isogeny of degree $p$ from $E$ defined over $\mathbb{Q}$) and there is only one if there is a unique such isogeny.
Of course, the $2$-torsion is different. All twists will have a $2$-torsion point if one is present in $E(\mathbb{Q})$ as the Galois group of any quadratic extension will act trivially on it.
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7
There's also the curious case of the curves of conductor $50$ where one twist has $3$-torsion and another twist as $5$-torsion. – Noam D. Elkies Sep 26 2011 at 15:05
Thank you for your answer Chris. May I ask why "only at most three different d" can give us new torsion points? Is there a reference for this? – Giuseppe Sep 26 2011 at 22:30
Hmmm. I can't remember why I said three. The question is how many isogenies of odd prime degree are defined over Q. Maybe the maximum is two. It is certainly finite. If you need the exact number, I could start to think about it – Chris Wuthrich Sep 28 2011 at 9:03
"Can it be that, in this particular family of quadratic twists, all but one of the curves have trivial torsion? Is this a general phenomenon?"
Indeed, it is very likely. The only way E_d[p] can have a rational point is if E[p] has a cyclic subgroup on which Galois acts through a quadratic character. For a "typical" elliptic curve, there will be so such p, since all the mod p Galois representations will be irreducible; for every non-Cm elliptic curve p, there are only finitely many p such that E[p] is reducible.
For each such p, the quadratic action of Gal(Q) on E[p] is what it is, and it tells you the unique d such that E_d[p] has a rational point.
So yeah, the phenomenon you're observing is very much to be expected.
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See Chris's comment above for a correction to my assertion that there is always a unique d such that E_d[p] has a rational point. – JSE Sep 27 2011 at 3:45
The maximum number of quadratic twists with $n$-torsion (n odd) that an elliptic curve over a number field $K$ can have is 2, and here is an easy proof.
You can see this by asking how many twists with $n$-torsion can an elliptic curve which already has $n$-torsion have? If this number is $k$, then clearly an elliptic curve without $n$-torsion can have $k+1$ twists with $n$-torsion.
Suppose now that $E/K,\ E(K)\supset \mathbb Z/ n\mathbb Z$, has 2 twists $E_{d_1}$ and $E_{d_2}$ with $n$-torsion. It follows from what Chris wrote that $$E(K(\sqrt{d_1},\sqrt{d_2}))[n]\supset E(K)[n] \oplus E_{d_1}(K)[n] \oplus E_{d_2}(K)[n]\supset (\mathbb Z/ n\mathbb Z)^3,$$ which is clearly impossible.
Note that the answer depends a lot on the number field $K$ and the number $n$ you choose. The possibility of 2 twists exists only over number fields $K$ such that the complete $n$-tosion can be defined over a quadratic extension of $K$. For example, it follows that over $\mathbb Q$, an elliptic curve can have at most 2 twists with 3-torsion, 1 twist with 5 or 7-torison and 0 twists with $p$ torsion for all primes $p>7$.
So, Giuseppe, your curve $E$ with 5-torsion, as any other curve with 5-torsion, cannot have a twist with 5-torsion over $\mathbb Q$.
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For non-2-torsion, you have Ed(Q)[n]⊕E(Q)[n]=E(Q(d√)[n] if n is odd. can you (Chris Wuthrich) give a proof or some reference where i can have it ??
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I think you are not meant to ask questions as answers. The non-trivial Galois element $\sigma$ of $\mathbb{Q}(\sqrt{d})/\mathbb{Q}$ acts on the right hand side. The subgroup fixed by $\sigma$ is clearly the $n$-torsion over $\mathbb{Q}$. Now if $\sigma$ sends a point $P$ to $-P$, then its $y$-coordinate is of the form $z\cdot\sqrt{d}$ for some rational number $z$. Now writing out the equation between $z$ and the $x$-coordinate of $P$ gives you exactly the Weierstrass equation of the twist of $E$ by $d$. – Chris Wuthrich May 2 at 10:36
Thanks Chris and sorry for that. – Suman May 2 at 10:46
More generally, it's easy to define a map $$E_d(\mathbb{Q}) \oplus E(\mathbb{Q}) \to E(\mathbb{Q}(\sqrt{d}))$$ whose kernel and co-kernel are $2$-groups. The result on $n$-torsion for odd $n$ is then immediate, and in addition one obtains the well-known and useful formula $$\operatorname{rank} E_d(\mathbb{Q}) + \operatorname{rank} E(\mathbb{Q}) \ = \operatorname{rank} E(\mathbb{Q}(\sqrt{d}))$$ – Joe Silverman May 2 at 14:15
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http://math.stackexchange.com/questions/190829/whats-the-right-way-to-make-a-change-of-variables-under-another-integral
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# What's the right way to make a change of variables under another integral?
Here's something that has me a little perplexed. I'm missing some step in making a change of variables under a multiple integral, where the new variable depends on more than one of the original ones.
I can't think of a simplified example that demonstrates the problem, so I'll just list the actual integral I'm trying to do:
$$\int_0^1\mathrm{d}\xi'\frac{1 + \xi'^2}{(1 - \xi')_+}\Biggl(\iint_{\mathbb{R}^2}\frac{\mathrm{d}^2\vec{t}_\perp}{(2\pi)^2}\exp\biggl[-\frac{t_\perp^2Q_s^2}{4}\biggr]\frac{e^{i\xi'\vec{k}_\perp\cdot\vec{t}_\perp}}{t_\perp^2} - \frac{1}{(2\pi)^2}\iint_{\mathbb{R}^2}\mathrm{d}^2\vec{r}'_\perp\frac{e^{-i\vec{k}_\perp\cdot \vec{r}'_\perp}}{r'^2_\perp}\Biggr)$$
Each of the two integrals over $\mathbb{R}^2$ is individually divergent because of a singularity at 0. But when I take the difference, the divergences will cancel out (perhaps only in some limiting sense, but that's really what I'm after).
Accordingly, the way I tried to do this is to change variables in the second (inner) integral from $\vec{r}'_\perp$ to $-\xi'\vec{t}_\perp$ so that I could then subtract the integrands. I compute the Jacobian as
$$\mathrm{d}\xi'\mathrm{d}^2\vec{r}' = \begin{vmatrix}\frac{\partial\xi'}{\partial\xi'} & \frac{\partial\xi'}{\partial t_x} & \frac{\partial\xi'}{\partial t_y} \\ \frac{\partial r'_x}{\partial\xi'} & \frac{\partial r'_x}{\partial t_x} & \frac{\partial r'_x}{\partial t_y} \\ \frac{\partial r'_y}{\partial\xi'} & \frac{\partial r'_y}{\partial t_x} & \frac{\partial r'_y}{\partial t_y}\end{vmatrix}\mathrm{d}\xi'\mathrm{d}^2\vec{t} = \begin{vmatrix}1 & 0 & 0 \\ -t_x & -\xi' & 0 \\ -t_y & 0 & -\xi'\end{vmatrix}\mathrm{d}\xi'\mathrm{d}^2\vec{t} = \xi'^2\mathrm{d}\xi'\mathrm{d}^2\vec{t}$$
I notice that by making this change of variables, I change the second term from something independent of $\xi'$ to something dependent on $\xi'$. But leaving that possible issue aside for now... I get
$$\begin{gather}\int_0^1\mathrm{d}\xi'\frac{1 + \xi'^2}{(1 - \xi')_+}\Biggl(\iint_{\mathbb{R}^2}\frac{\mathrm{d}^2\vec{t}_\perp}{(2\pi)^2}\exp\biggl[-\frac{t_\perp^2Q_s^2}{4}\biggr]\frac{e^{i\xi'\vec{k}_\perp\cdot\vec{t}_\perp}}{t_\perp^2} - \frac{1}{(2\pi)^2}\iint_{\mathbb{R}^2}\mathrm{d}^2\vec{t}'_\perp\frac{e^{i\xi'\vec{k}_\perp\cdot \vec{t}'_\perp}}{t'^2_\perp}\Biggr)\\ \int_0^1\mathrm{d}\xi'\frac{1 + \xi'^2}{(1 - \xi')_+}\iint_{\mathbb{R}^2}\frac{\mathrm{d}^2\vec{t}_\perp}{(2\pi)^2}\biggl(\exp\biggl[-\frac{t_\perp^2Q_s^2}{4}\biggr] - 1\biggr)\frac{e^{i\xi'\vec{k}_\perp\cdot\vec{t}_\perp}}{t_\perp^2}\end{gather}$$
I can express the inner integral in polar coordinates as
$$\begin{gather}\int_0^1\mathrm{d}\xi'\frac{1 + \xi'^2}{(1 - \xi')_+}\int_0^{2\pi}\mathrm{d}\theta_t\int_0^\infty\frac{\mathrm{d}t_\perp\,t_\perp}{(2\pi)^2}\biggl(\exp\biggl[-\frac{t_\perp^2Q_s^2}{4}\biggr] - 1\biggr)\frac{e^{i\xi' k_\perp t_\perp\cos\theta_t}}{t_\perp^2}\\ \int_0^1\mathrm{d}\xi'\frac{1 + \xi'^2}{(1 - \xi')_+}\int_0^\infty\frac{\mathrm{d}t_\perp}{2\pi t_\perp}\biggl(\exp\biggl[-\frac{t_\perp^2Q_s^2}{4}\biggr] - 1\biggr)J_0(\xi' k_\perp t_\perp)\\ -\int_0^1\mathrm{d}\xi'\frac{1 + \xi'^2}{(1 - \xi')_+}\frac{1}{4\pi}\Gamma\biggl(0, \frac{k^2 \xi'^2}{Q_s^2}\biggr) \end{gather}$$
(the last equality comes from Mathematica, I haven't gotten to show it myself). But the true result is supposed to be
$$-\int_0^1\mathrm{d}\xi'\frac{1 + \xi'^2}{(1 - \xi')_+}\frac{1}{4\pi}\biggl[\Gamma\biggl(0, \frac{k^2 \xi'^2}{Q_s^2}\biggr) {\color{red}{+ \ln\xi'^2}}\biggr]$$
I can't figure out where in this procedure that extra logarithm (in red) is supposed to come in. I guess it's supposed to be something in the change of variables that does it, but given that I've computed the Jacobian determinant and also the new region of integration (still $\mathbb{R}^2$), I don't know what else there would be. And I would very much like to know how I can avoid getting caught by this issue in the future.
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## 1 Answer
Subtracting one divergent integral from another is not, without additional information, going to yield a meaningful result (even if the singularities are in some sense "the same"). To take a silly example, suppose you're asked to calculate $$\int_{0}^{1}\frac{dx}{x}-\int_{0}^{1}\frac{dt}{t}.$$ Obviously this should be zero. Right? But by making the change of variable $t=ax$ in the second integral, it is seen to also equal $$\int_{0}^{1}\frac{dx}{x}-\int_{0}^{1/a}\frac{dx}{x}=\int_{1/a}^{1}\frac{dx}{x}=\ln x\Big\vert_{1/a}^{1}=\ln a,$$ where $a$ is completely arbitrary. Hopefully the relevance to your problem is clear: changes of variables can change the overall result if the individual integrals are improper. In order to nail down a particular result, you need to specify the regularization procedure. Here, you might make the individual integrals meaningful by calculating $$\int_{\varepsilon}^{1}\frac{dx}{x}-\int_{\varepsilon}^{1}\frac{dt}{t}$$ and then letting $\varepsilon\rightarrow 0$. (The result would then be zero no matter what changes of variable were carried out midstream.) If you adopt the same procedure in your problem (i.e., restrict the first integral to $|t_\perp|>\varepsilon$ and the second to $|r'_{\perp}|>\varepsilon$, then make your change of variable and subtract the results, and finally let $\varepsilon$ go to $0$), you will get an additional contribution from the annular region between radii $\varepsilon$ and $\xi' \varepsilon$. Specifically, the new term is $$\int_{\varepsilon}^{\xi'\varepsilon}\frac{dt_\perp}{2\pi t_\perp} J_0\left(\xi'k_\perp t_\perp\right).$$ In the limit as $\varepsilon\rightarrow 0$, the Bessel function is irrelevant (it just goes to $1$), and the rest integrates to $$\frac{1}{2\pi}\ln \xi' = \frac{1}{4\pi}\ln\xi'^2,$$ which is the term you were missing.
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1
Great, that helps! I had noticed that regularization procedure gives the right answer, but I wasn't sure whether it was a valid way to proceed. Although... it seems like this transfers the arbitrariness to the regularization procedure, i.e. why would it be invalid to cut off one integral at $\epsilon$ and the other at $2\epsilon$, or $\xi'\epsilon$, etc.? (Maybe that should be a whole separate question?) – David Zaslavsky Sep 7 '12 at 17:38
1
Yes, the fact that the regularization procedure affects the answer shows that the initial expression isn't well-defined as it stands. But there can be physical reasons (assuming the motivating problem is physics-based?) to choose one regularization procedure over another. Often it's possible to embed the regularization into the original problem somehow... e.g., through an unknown but large energy scale... so that it will carry through to all of the divergent integrals that appear in the middle of the calculation. And if you're lucky (as here), you can still let it go to $\infty$ in the end. – mjqxxxx Sep 7 '12 at 20:25
Yep, this is a physics-based problem. Unfortunately I don't think there is a physically relevant scale $\varepsilon$ to use for regularizing these integrals, but I will have to look through the paper I'm getting this from in some more detail. Thanks! – David Zaslavsky Sep 10 '12 at 20:00
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http://www.reference.com/browse/wiki/Representation_theory_of_SL2(R)
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Definitions
# Representation theory of SL2(R)
In mathematics, the main results concerning irreducible of the Lie group SL2(R) are due to Gelfand and Naimark (1946), V. Bargmann (1947), and Harish-Chandra (1952).
## Structure of the complexified Lie algebra
We choose a basis H, X, Y for the complexification of the Lie algebra of SL2(R) so that iH generates the Lie algebra of a compact Cartan subgroup K (so in particular unitary representations split as a sum of eigenspaces of H), and {H,X,Y} is an sl2-triple, which means that they satisfy the relations
$\left[H,X\right]=2X, quad \left[H,Y\right]=-2Y, quad \left[X,Y\right]=H.$
One way of doing this is as follows:
$H=begin\left\{pmatrix\right\}0 & -i i & 0end\left\{pmatrix\right\}$ corresponding to the subgroup K of matrices $begin\left\{pmatrix\right\}cos\left(theta\right) & -sin\left(theta\right) sin\left(theta\right)& cos\left(theta\right)end\left\{pmatrix\right\}$
$X=\left\{1over 2\right\}begin\left\{pmatrix\right\}1 & i i & -1end\left\{pmatrix\right\}$
$Y=\left\{1over 2\right\}begin\left\{pmatrix\right\}1 & -i -i & -1end\left\{pmatrix\right\}$
The Casimir operator Ω is defined to be
$Omega= H^2+1+2XY+2YX.$
It generates the center of the universal enveloping algebra of the complexified Lie algebra of SL2(R). The Casimir element acts on any irreducible representation as multiplication by some complex scalar μ2. Thus in the case of the Lie algebra sl2, the infinitesimal character of an irreducible representation is specified by one complex number.
The center Z of the group SL2(R) is a cyclic group {I,-I} of order 2, consisting of the identity matrix and its negative. On any irreducible representation, the center either acts trivially, or by the non trivial character of Z, which represents the matrix -I by multiplication by -1 in the representation space. Correspondingly, one speaks of the trivial or nontrivial central character. The central character and the infinitesimal character of an irreducible representation of any reductive Lie group are important invariants of the representation. In the case of irreducible admissible representations of SL2(R), it turns out that, generically, there is exactly one representation, up to an isomorphism, with the specified central and infinitesimal characters. In the exceptional cases there are two or three representations with the prescribed parameters, all of which have been determined.
## Finite dimensional representations
For each nonnegative integer n, the group SL2(R) has an irreducible representation of dimension n+1, which is unique up to an isomorphism. This representation can be constructed in the space of homogeneous polynomials of degree n in two variables. The case n = 0 corresponds to the trivial representation. An irreducible finite dimensional representation of a noncompact simple Lie group of dimension greater than 1 is never unitary. Thus this construction produces only one unitary representation of SL2(R), the trivial representation.
The finite-dimensional representation theory of the noncompact group SL2(R) is equivalent to the representation theory of SU(2), its compact form, essentially because their Lie algebras have the same complexification and they are "algebraically simply connected". (More precisely the group SU(2) is simply connected and SL2(R) is not, but has no non-trivial algebraic central extensions.) However, in the general infinite-dimensional case, there is no close correspondence between representations of a group and the representations of its Lie algebra. In fact, it follows from the Peter-Weyl theorem that all irreducible representations of the compact Lie group SU(2) are finite-dimensional and unitary. The situation with SL2(R) is completely different: it possesses infinite-dimensional irreducible representations, some of which are unitary, and some are not.
## Principal series representations
A major technique of constructing representations of a reductive Lie group is the method of parabolic induction. In the case of the group SL2(R), there is up to conjugacy only one proper parabolic subgroup, the Borel subgroup of the upper-triangular matrices of determinant 1. The inducing parameter of an induced is a (possibly non-unitrary) character of the multiplicative group of real numbers, which is specified by choosing ε = ± 1 and a complex number μ. The corresponding principal series representation is denoted Iε,μ. It turns out that ε is the central character of the induced representation and the complex number μ may be identified with the infinitesimal character via the Harish-Chandra homomorphism.
The principal series representation Iε,μ (or more precisely its Harish-Chandra module of K-finite elements) admits a basis consisting of elements wj, where the index j runs through the even integers if ε=1 and the odd integers if ε=−1. The action of X, Y, and H is given by the formulas
$H\left(w_j\right) = jw_j$
$X\left(w_j\right) = \left\{mu+j+1over 2\right\}w_\left\{j+2\right\}$
$Y\left(w_j\right) = \left\{mu-j+1over 2\right\}w_\left\{j-2\right\}$
## Admissible representations
Using the fact that it is an eigenvector of the Casimir operator and has an eigenvector for H, it follows easily that any irreducible admissible representation is a subrepresentation of a parabolically induced representation. (This also is true for more general reductive Lie groups and is known as .) Thus the irreducible admissible representations of SL2(R) can be found by decomposing the principal series representations Iε,μ into irreducible components and determining the isomorphisms. We summarize the decompositions as follows:
• Iε,μ is reducible if and only if μ is an integer and ε=−(−1)μ. If Iε,μ is irreducible then it is isomorphic to Iε,−μ.
• I−1, 0 splits as the direct sum Iε,0 = D+0 + D−0 of two irreducible representations, called limit of discrete series representations. D+0 has a basis wj for j≥1, and D-0 has a basis wj for j≤−1,
• If Iε,μ is reducible with μ>0 (so ε=−(−1)μ) then it has a unique irreducible quotient which has finite dimension μ, and the kernel is the sum of two discrete series representations D+μ + D−μ. The representation Dμ has a basis wμ+j for j≥1, and D-μ has a basis w−μ−j for j≤−1.
• If Iε,μ is reducible with μ<0 (so ε=−(−1)μ) then it has a unique irreducible subrepresentation, which has finite dimension μ, and the quotient is the sum of two discrete series representations D+μ + D−μ.
This gives the following list of irreducible admissible representations:
• A finite dimensional representation of dimension μ for each positive integer μ, with central character −(−1)μ.
• Two limit of discrete series representations D+0, D−0, with μ=0 and non-trivial central character.
• Discrete series representations Dμ for μ a non-zero integer, with central character −(−1)μ
• Two families of irreducible principle series representations Iε,μ for ε≠−(−1)μ (where Iε,μ is isomorphic to Iε,−μ).
### Relation with the Langlands classification
According to the Langlands classification, the irreducible admissible representations are parametrized by certain tempered representations of Levi subgroups M of parabolic subgroups P=MAN. This works as follows:
• The discrete series, limit of discrete series, and unitary principle series representations Iε,μ with μ imaginary are already tempered, so in these cases the parabolic subgroup P is SL2 itself.
• The finite dimensional representations and the representations Iε,μ for ℜμ>0, μ not an integer or ε≠−(−1)μ are the irreducible quotients of the principal series representations Iε,μ for ℜμ>0, which are induced from tempered representations of the parabolic subgroup P=MAN of upper triangular matrices, with A the positive diagonal matrices and M the center of order 2. For μ a positive integer and ε=−(−1)μ the principal series representation has a finite dimensional representation as its irreducible quotient, and otherwise it is already irreducible.
## Unitary representations
The irreducible unitary representations can be found by checking which of the irreducible admissible representations admit an invariant positively-definite Hermitian form. This results in the following list of unitary representations of SL2(R):
• The trivial representation
• The two limit of discrete series representations D+0, D−0.
• The discrete series representations Dk, indexed by non-zero integers k. They are all distinct.
• The two families of irreducible principal series representation, consisting of the spherical principal series I+,iμ indexed by the real numbers μ, and the non-spherical unitary principal series I-,iμ indexed by the non-zero real numbers μ. The representation with parameter μ is isomorphic to the one with parameter −μ, and there are no further isomorphisms between them.
• The complementary series representations I+,μ for 0<|μ|<1. The representation with parameter μ is isomorphic to the one with parameter −μ, and there are no further isomorphisms between them.
Of these, the two limit of discrete series representations, the discrete series representations, and the two families of principal series representations are tempered, while the finite dimensional and complementary series representations are not tempered.
## References
• V. Bargmann, Irreducible Unitary Representations of the Lorentz Group, The Annals of Mathematics, 2nd Ser., Vol. 48, No. 3 (Jul., 1947), pp. 568-640
• Gelfand, I.; Neumark, M. Unitary representations of the Lorentz group. Acad. Sci. USSR. J. Phys. 10, (1946), pp. 93--94
• Harish-Chandra, Plancherel formula for the 2×2 real unimodular group. Proc. Nat. Acad. Sci. U.S.A. 38 (1952), pp. 337--342
• Roger Howe, Eng-Chye Tan, Nonabelian harmonic analysis. Applications of SL(2,R). Universitext. Springer-Verlag, New York, 1992. ISBN 0-387-97768-6
• Knapp, Anthony W. Representation theory of semisimple groups. An overview based on examples. Reprint of the 1986 original. Princeton Landmarks in Mathematics. Princeton University Press, Princeton, NJ, 2001. xx+773 pp. ISBN 0-691-09089-0
• Kunze, R. A.; Stein, E. M. Uniformly bounded representations and harmonic analysis of the 2×2 real unimodular group. Amer. J. Math. 82 (1960), pp. 1--62
• D. Vogan, Representations of real reductive Lie groups, ISBN 3-7643-3037-6
• N. R. Wallach, Real reductive groups I. ISBN 0-12-732960-9
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### Pion Decay and Fractional electric Charge disappeared, why?
Since the charged pions decay into two particles, a muon and a muon neutrino Fractional electric Charge disappeared, why? The decay proceeds by the weak interaction $W^{+}$ and can be visualized in ...
2answers
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### Could the fractional model of Quarks electric charge turn out to be false? [closed]
The delta baryons (also called delta resonances) are a family of subatomic hadron particles which have the symbols $\Delta^{++}$, $\Delta^{+}$, $\Delta^{0}$, and $\Delta^{−}$ and electric charges +2, ...
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### Is there an explanation for the 3:2:1 ratio between the electron, up and down quark electric charges?
I understand that the NNG formula relates $Q$, $I_3$, and $Y$ and can be derived in QCD; does this unambiguously predict the electric charge ratios without making assumptions about the definitions of ...
1answer
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### Similar masses and lifetimes of the $\Delta$ baryons
Why do the four spin 3/2 $\Delta$ baryons have nearly identical masses and lifetimes despite their very different $u$ and $d$ quark compositions?
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http://physics.stackexchange.com/questions/26888/on-shell-symmetry-from-a-path-integral-point-of-view/53070
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# On-shell symmetry from a path integral point of view
Normally supersymmetric quantum field theories have Lagrangians which are supersymmetric only on-shell, i.e. with the field equations imposed. In many cases this can be solved by introducing auxilary fields (field which don't carry dynamical degrees of freedom, i.e. which on-shell become a function of the other fields). However, there are cases where no such formulation is known, e.g. N=4 super-Yang-Mills in 4D.
Since the path integral is an integral over all field configurations, most of them off-shell, naively there is no reason for it to preserve the on-shell symmetry. Nevertheless the symmetry is preserved in the quantum theory.
Of course it is possible to avoid the problem by resorting to a "Hamiltonian" approach. That is, the space of on-shell field configurations is the phase space of the theory and it is (at least formally) possible to quantize it. However, one would like to have an understanding of the symmetries survival in a path integral approach. So:
How can we understand the presence of on-shell symmetry after quantization from a path integral point of view?
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Dear @Squark, surely you may write $N=4$ in the $N=1$ superspace, making the $N=1$ subalgebra manifest even off-shell and even in the path integral, can't you? The path integral for the $N=1$ language is trivialy equivalent to the $N=0$ "in components" formulation – the only slightly nontrivial statement behind this assertion is that the measure flip including the aux. fields doesn't spoil SUSY. So in this sense, I think that SUSY is manifest even in the non-SUSY $N=0$ "in components" formalism of the path integral, off-shell. If you see some problems with this conclusion, tell me details. – Luboš Motl Nov 5 '11 at 19:16
The equivalence between N=1 and N=0 is by integrating over the auxiliary fields, as far as I see. Hence it is not quite manifest in the N=0 language. For N=4 you can sure use the N=1 superspace, moreover the GIKOS approach apparently allows making an N=3 sub supergroup manifest. However this doesn't prove the whole symmetry is preserved. – Squark Nov 5 '11 at 20:19
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OK, I think I see the answer. Once you can prove the equivalence between N=0 and N=1 you can get N=4 by choosing different N=1 subsupergroups – Squark Nov 5 '11 at 20:27
Well, I need to chew a bit more on the N=4 SYM case, but, in the meanwhile, consider N=1 SYM in 10D. Already we have no off shell formulation and the N is minimal. – Squark Nov 5 '11 at 20:42
I was just wondering about the same question. On physicsforums somebody posted a citation from Sohnius: "This situation could in principle be disastrous for quantum corrections: there the fields must be taken off-shell, away from their classical paths through configuration space. What this does for on-shell supersymmetric theories is quite unclear Obviously, if there exists - even unknown to us - some unique off-shell version, there should be no problem. But what happens if the theory were intrinsically only on-shell supersymmetric (as in N=4 and N=8 supergravity) is unclear to date." Is this – user9744 Jun 8 '12 at 17:47
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## 3 Answers
How can we understand the presence of on-shell symmetry after quantization from a path integral point of view?
One can derive a Schwinger-Dyson equation associated with the current conservation, also known as a Ward identity; see e.g. Peskin and Schroeder, An Introduction to Quantum Field Theory, Section 9.6; or Srednicki, Quantum Field Theory, Chapter 22.
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I don't have access to the book. How do you derive it using the path integral if the symmetry only exists on shell? – Squark Nov 5 '11 at 20:43
– Simon Nov 5 '11 at 22:01
OK, and where is the answer to my question in Srednicki's book? In fact it seems to me he doesn't go beyond N=1 4D supersymmetry – Squark Nov 5 '11 at 23:20
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I interpreted the question(v1) as asking about on-shell symmetry on general grounds, cf. the title(v1). The references do not have any specific mentioning of $N=4$ 4D supersymmetry. – Qmechanic♦ Nov 6 '11 at 0:26
OK. So Peskin and Schroeder have a treatment of on-shell symmetry on general grounds? Which example they consider if not supersymmetry? Also, I still have no access to the book. – Squark Nov 11 '11 at 20:46
The point of view you give of quantizing phase space is not so good. If you have a classical phase space, there will be more than one quantum analog, differing by commutator terms that vanish classically. I never found the idea that you start with a classical system particularly lucid--- the quantum system is defined by the path integral, and the answer must always be found in the path integral itself.
In 4d, as you said in the comments, the problem is sidestepped by using N=1 superspace plus the appropriate R symmetry, which rotate the different SUSY's into each other, ensuring that the full SUSY is there. In order for the theory to be SUSY, you just have to check that the theory is N=1 and that the R-symmetry is good. This is usually manifest, since the R symmmetry is simple global rotation of fields into each other. But this doesn't work in 10 or 11 dimensions.
A part of the answer suggested by Qmechanic, the Schwinger Dyson equation (Heisenberg equation of motion) is obeyed by the field operators, so the variation really does vanish as an operator, in a certain sense described precisely below. But this is not a complete answer because it isn't true that everything that vanishes classically when the equations of motion are satisfied also vanishes quantum mechanically, in the path integral. The reason is that the quantities in the path integral can be multiplied at the same point in space and time, and there are singularities of the coinciding operator products which make the naive equations of motion fail in exactly the type of expressions that occur in the action.
For a trivial example, consider the product of two X operators
$$\langle x(t) x(t')\rangle = 0$$
In the trivial Euclidean free-particle (Brownian motion) 1-d (quantum mechanics, not field theory) path integral $S= \int \dot{x}^2$. The operator X obeys the equation of motion $\ddot{X}=0$, so the second derivative is zero, but of course it isn't at the coinciding point (or else propagator would vanish at all times with the reasonable boundary conditions). The correct second derivative of the correlation function is
$$\langle \ddot{x}(t) x(t')\rangle = \delta(t-t')$$
The equation of motion holds except at coincident points. The same is true in field theories for the same reason--- the propagator is sourced at the point where the two operators coincide by a delta-function with unit amplitude (if the fields are canonically normalized).
Further, since $\ddot{X}=0$, you might think that the following transformation is a symmetry
$$\dot x + \epsilon f(t)\ddot{X}$$
Since the variation of S to first order in $\epsilon$ is $\int \dot{x}\ddot{x}f(t)$. So naively, using the equation of motion $\ddot{x}=0$, you would naively find that this is a symmetry. But this is obvious nonsense--- if f is nonconstant (if f is constant, the transformation is a time translation) this is obviously not a symmetry of the action. since it is a time translation by a different amount at different times, and there is a unit of time defined by the diffusion.
The reason is that the $\ddot{x}$ is multiplied by a coinciding $\dot{x}$, and when there are coinciding quantities, the equations of motion are not necessarily satisfied. The actual variation in S is
$$\int f(t) {d\over dt} {\dot{x}^2\over 2}$$
and you see that it is a perfect derivative if f is constant, assuming Stratonovich convention for time derivatives--- centered differences--- and this is the symmetry of time translation in this case), but it is not a symmetry if f is not constant.
## When are equations of motion satisfied in a path integral?
When you have a path integral:
$$\int e^{iS} D\phi$$
The integral is invariant under a shift of integration variables $\phi(x)+\delta\phi(x)$, where $\delta\phi$ is an arbitrary function of x, since each integral is separately translation invariant. There is no determinant for this transformation--- it's just a shift in the integration variable by a constant at each time.
The change in the integrand when you do the shift is, given by expanding the thing to first order.
$$\int e^{iS(\phi+\delta\phi)} D\phi = \int e^{iS} D\phi + \int (\int {\delta S\over \delta \phi(x)}\delta\phi(x) d^dx) e^{iS} D\phi$$
and from this, you learn that the equations of motion are satisfied. This is the correct demonstration of the equations of motion.
$${\delta S\over \delta\phi} = 0$$
Now suppose you have some other insertions:
$$\int \phi_1(x_1) \phi_2(x_2) ... \phi_n(x_n) e^{iS} D\phi$$
Now doing the same shift, to first order, you find
$$\int (\phi_1 +\delta\phi(x_1)) ... (\phi_2+\delta\phi(x_2)) e^{iS} (1+\delta S) D\phi = \int e^{iS} D\phi$$
So you find that the equation of motion is satisfied in an insertion except at coinciding points:
$$\langle {\delta S\over \delta \phi}(x) \phi_1 ... \phi_n\rangle = \delta(x-x_1) + ... + \delta(x-x_n)$$
repeating this Ward-identity type argument for insertions of any type with lots of fields running around you find the following rules for using the equation of motion:
1. The equations of motion derived from varying $\phi$ are satisfied away from insertions.
2. the equation of motion derived from varying $\phi$ are satisfied even at cross-insertions of other fundamental fields (independent path-integrations) other than $\phi$.
3. the equations of motion fail only at those insertions which vary under a shift of the field whose equation of motion it is, and the failure is of the ward-identity type.
So it isn't true that anything that vanishes classically has no effect. But when you have a product of the equation of motion multiplying functions of fields which are different from the one you vary to get the equation of motion, these are still identically zero.
In the SUSY case, your variation involves products of fields with their SUSY partners, which are independent integration variables. The equations of motion are satisfied in the products that you reduce, so using the equations of motion in the SUSY closure is justified, but you check it theory by theory by hand, by looking to see which equation of motion you use, and what it is multiplying.
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First of all there is an error in the Schwinger-Dyson equation that was derived in the previous post. There will be n-1 fields on the right-hand side. In any event, the SD equations don't save you because the supercurrent has fields sitting at the same point, so these contact terms come in and spoil the equations of motion. I.e., you get the equation of motion when you differentiate the supercurrent, but it is multiplied by fields at the same points, so all these delta functions come in. Try it at home, you'll see what I mean.
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Perhaps you could run through a simple calculation to show people what you mean? – Muphrid Feb 4 at 21:03
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http://thecodeabode.com/
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Probabilistic Compositions
Last time, I wrote a post about the odds of reaching a certain space in Monopoly. An interesting result ended up coming out of this, and it is a consequence of the relationship between a sequence of die rolls and a partition. First, some definitions: a partition is any way of writing a positive number as a sum of positive numbers. For instance,
$4 \hspace{2 cm} 3+1 \hspace{2 cm} 2+2 \hspace{2 cm} 1+1+1+1$
are all distinct partitions of 4. Order doesn’t matter, so (1 + 3) is considered the same partition as (3 + 1). The simple question, “How many distinct partitions are there of a given number?” is notoriously difficult and has intrigued mathematicians for centuries. Landmark theorems have related the number of partitions to functions involving pentagonal numbers, infinite sums, π, e, and most recently, fractal behavior.
After wrapping up last week, I realized that a sequence of die rolls that gets you to a certain spot n spaces away defines a partition of n. For instance, if I have one die and roll a 4, then 3, then 4, to get to a spot 11 spaces away, that defines the partition 4 + 3 + 4 = 11. However, the rolls don’t uniquely define partitions, because a roll of 3, then 4, then 4 would give 3 + 4 + 4 = 11, which is the same partition. Thus in order to make use of this, we need to insist that order matter in the partitions. When order matters, a partition is called a composition. So we would say that in the above example, the two roll sequences give the same partition, but different compositions, of 11.
Last time, I found that the probability of landing on a space sufficiently far away is given by the reciprocal of the expected value of a single roll. For one die with r sides, that means the probability of making a sequence that adds up to n is $\frac{2}{r+1}$ .
Another way of viewing this probability is as a sum over the compositions of n, with each assigned a certain weight based on how “likely” they are to show up. The total probability of getting to n is equal to the sum of the probabilities of getting there any one way. And any way of getting there is a unique composition of n.
So what’s the probability of a certain composition showing up? Using a die with r sides, the odds of each specific roll occupying a place in the sequence is 1/r , so the probability of seeing a complete composition rolled is 1/r, multiplied by itself a number of times equal to the number of terms in the composition. Let’s call the number of terms the length of the composition and denote it $| c |$ for a composition c . That means that the probability of a certain composition c being the beginning of a sequence of die rolls is given by
$\displaystyle \frac{1}{r^{|c|}}$
Let $C_n$ denote the set of all compositions of n, and $C_{n, r}$ be the set of all the compositions of n involving only numbers less than or equal to r . These are the only compositions allowable since we’re rolling a die with r sides. Now we can finally equate our sum over compositions with the previous result to obtain:
$\displaystyle \lim_{n \to \infty}\sum_{c \in C_{n, r}} \frac{1}{r^{|c|}} = \frac{2}{r+1}$
This is pretty interesting … as long as n is pretty large, this sum over its compositions will come out to approximately $\frac{2}{r+1}$ . It’s well known that the number of compositions of any n is exactly $2^{n-1}$ , but I find this equation interesting because it has so much to do with the distribution of lengths of compositions.
Naturally, I wanted to check this result for some r. r = 1 is no fun at all, but let’s say r = 2. In our original problem, this means the probability of landing on any given space given a two-sided die (flipping a coin perhaps?) is 2/3. In our new equation, we should be able to find that
$\displaystyle \lim_{n \to \infty}\sum_{c \in C_{n, 2}} \frac{1}{2^{|c|}} = \frac{2}{3}$
Well, it’s never fun to deal with sums over elements of a set, so I wanted some way to describe just how many compositions of each length would show up in $C_{n, r}$ . Eventually I found a strategy: imagine any composition of n using only 1s and 2s. Let’s call a composition like this a 2-composition. The 2-composition ends in either a 1 or a 2. That means you could describe it as either a 2-composition of n – 1 with a 1 tacked on the end, or a 2-composition of n – 2 with a 2 at the end. And every 2-composition of n – 1 or n – 2 can be made into a 2-composition of n by this process. Therefore,
$\displaystyle \left|{C_{n, 2}}\right| = \left|{C_{n-1, 2}}\right| + \left|{C_{n-2, 2}}\right|$
Hmmm, the sum of the previous two terms? This is the Fibonacci sequence! Not bad. But we don’t just need to know the number of 2-compositions, we need to know the number of 2-compositions of each length. Well, with this same process of adding 1s and 2s to lesser compositions, we can uncover a pattern. In the following diagram, I’ll show a breakdown of the lengths of 2-compositions of the first few numbers, and use $\displaystyle n \bullet m$ to represent n compositions of length m .
$\displaystyle 1 \to 1 \bullet 1$
$\displaystyle 2 \to 1 \bullet 2 \hspace{2 cm} 1 \bullet 1$
$\displaystyle 3 \to 1 \bullet 3 \hspace{2 cm} 2 \bullet 2$
$\displaystyle 4 \to 1 \bullet 4 \hspace{2 cm} 3 \bullet 3 \hspace{2 cm} 1 \bullet 2$
$\displaystyle 5 \to 1 \bullet 5 \hspace{2 cm} 4 \bullet 4 \hspace{2 cm} 3 \bullet 3$
$\displaystyle 6 \to 1 \bullet 6 \hspace{2 cm} 5 \bullet 5 \hspace{2 cm} 6 \bullet 4 \hspace{2 cm} 1 \bullet 3$
$\displaystyle 7 \to 1 \bullet 7 \hspace{2 cm} 6 \bullet 6 \hspace{2 cm} 10 \bullet 5 \hspace{2 cm} 4 \bullet 4$
There’s something predictable in the distribution of lengths of 2-compositions. For instance, we can guess that there will be exactly 1 composition of 8 with length 8, and exactly 7 compositions of 8 with length 7. But how many compositions of length 6 will there be?
The concept of adding two previous terms to create a new term is not unique to the Fibonacci sequence. Pascal’s triangle is built using the same principle. Starting with a 1, each row is constructed so that each number is equal to the sum of the two numbers right above it. Here are the first few rows:
$1$
$1 \hspace{.5 cm}1$
$1 \hspace{.5 cm} 2 \hspace{.5 cm} 1$
$1 \hspace{.5 cm} 3 \hspace{.5 cm} 3 \hspace{.5 cm} 1$
$1 \hspace{.5 cm} 4 \hspace{.5 cm} 6 \hspace{.5 cm} 4 \hspace{.5 cm} 1$
$1 \hspace{.5 cm} 5 \hspace{.5 cm} 10 \hspace{.5 cm} 10 \hspace{.5 cm} 5 \hspace{.5 cm} 1$
Each row is created by adding terms from the previous rows. This is similar to our formulation for $\displaystyle \left|{C_{n, 2}} \right|$ , though not exactly the same. In fact, the numbers that represent the number of compositions using the numbers 1 and 2 form rows that bend diagonally in Pascal’s triangle. Compare the numbers found in the distribution from earlier to the colored rows in the triangle, where each row moves left to right, but curves upwards.
$1$
$1$ $. \hspace{.5 cm} 1$
$1$ $. \hspace{.5 cm} 2$ $. \hspace{.5 cm}1$
$1$ $. \hspace{.5 cm} 3$ $. \hspace{.5 cm} 3$ $. \hspace{.5 cm} 1$
$1$ $. \hspace{.5 cm} 4$ $. \hspace{.5 cm} 6$ $. \hspace{.5 cm} 4$ $. \hspace{.5 cm} 1$
$1$ $. \hspace{.5 cm} 5$ $. \hspace{.5 cm} 10$ $. \hspace{.5 cm} 10$ $. \hspace{.5 cm} 5$ $. \hspace{.5 cm} 1$
The distribution of compositions of certain lengths can be read straight from these diagonals of Pascal’s triangle. So how would one find, say, the number of 2-compositions of 6 with length 4? The sequence for the 2-compositions of 6 will start on the 6th row of Pascal’s triangle, and will start with the number of 2-compositions of 6 with length 6, followed by the number with length 5, and then length 4. So we go to the 6th row, then over 2 and up 2, and find 6. This fits our data: there are 6 2-compositions of 6 with length 4. In general, the number of 2-compositions of n with length k is given by $\displaystyle \binom{k}{n-k}$ .
Let’s go back to the equation we’re working on:
$\displaystyle \lim_{n \to \infty}\sum_{c \in C_{n, 2}} \frac{1}{2^{|c|}} = \frac{2}{3}$
Now that we know exactly how many times the 2-compositions of certain lengths show up, we can sum over those lengths instead of summing over the set of 2-compositions. All possible 2-composition lengths k are going be be between $\lceil \frac{n}{2} \rceil$ and $n$ , so we find that
$\displaystyle \lim_{n \to \infty} \sum_{k = \lceil \frac{n}{2} \rceil}^n \binom{k}{n-k} \frac{1}{2^k} = \frac{2}{3}$
This is a form that’s easy for a program to check. A small C program computes this very quickly and shows that even for small n , the sum is very close to two-thirds.
So there you have it: an interesting equation that links die rolls to compositions, by way of Pascal’s triangle!
Monopoly Odds
I’m going to use this post to thoroughly explain a good, real-world problem in a way that does not require any advanced math or programming knowledge. I’ve been wanting to write a post for my non-math-inclined friends that will hopefully show why I find solving these problems so rewarding. There’s some probability, a good bit of intro programming, and a lot of intuition. I hope it’s fun, and more mathy stuff will return next time (though this style may return as well if it is successful!).
I thought of this problem while playing Monopoly with a friend. Monopoly is a game where understanding odds is key to any good strategy, and information such as the most likely dice roll (a 7), the most landed-on properties (the oranges), and the chance of pulling an “Advance to Nearest Railroad” chance card (1/8) is crucial. I recently found myself in a situation where my opponent was 10 spaces away from landing on my most expensive property. I was disappointed that he wasn’t 7 spaces away, since a 7 is the most likely roll, but I realized there may be more to it than that. Maybe it was better for him to be 10 spaces away, since if he rolled short he would be more likely to hit my property on the roll after his next roll. I decided to compute the probability that he would land on my space, perhaps not immediately, but eventually.
As I mentioned before, 7 is the most likely roll of two six-sided dice. This is because for every number from 1 – 6, there is another number 1 – 6 for which the sum of the two is 7. The following diagram shows pairs of numbers from two dice which add to 7:
$\displaystyle 1 \to 6 \hspace{2 cm} 2 \to 5 \hspace{2 cm} 3 \to 4$
$\displaystyle 4 \to 3 \hspace{2 cm} 5 \to 2 \hspace{2 cm} 6 \to 1$
However, a number like 10 is less likely, because some numbers on the first die have no partner on the second die to add to 10. The missing partner is shown by an X.
$\displaystyle 1 \to \text{X} \hspace{2 cm} 2 \to \text{X} \hspace{2 cm} 3 \to \text{X}$
$\displaystyle 4 \to 6 \hspace{2 cm} 5 \to 5 \hspace{2 cm} 6 \to 4$
The probability of a number being rolled is equal to the number of dice pairings that sum to that number over the total number of pairings (for six-sided dice, this total possible number of pairings is 6 x 6 = 36). So the probability of rolling a 7 is 6/36, while the probability of rolling a 10 is 3/36.
In fact, the probabilities of rolling numbers from 2 to 12 look like a perfect staircase, up from 2 to 7 and down back again to 12.
To answer my question about whether my friend would land on my property 10 spaces away, we need to make a distinction. On one hand, there’s the chance of moving a certain number of spaces on one roll, and on the other hand there is the chance of moving the same distance in any number of rolls. I’m interested in the odds that he will travel 10 spaces in any number of rolls, since I only care that he lands on my property eventually.
So what are the possible ways he could move 10 spaces? Well, there is the chance that he rolls a 10 right away, that’s 3/36. Then there is the chance he rolls an 8, then a 2, which would be (5/36)*(1/36). More likely is the possibility that he rolls a 7 and then a 3 (6/36)*(2/36), or a 6 and then a 4 …. but wait, he could roll a 6 and then 4 either of two ways, either 6 then 4 or 6 then 2 then 2. What we really want to know is the odds of him rolling some number with one initial roll, then making up the difference with any number of rolls.
$\displaystyle 10$
$\displaystyle 8 \to 2$
$\displaystyle 7 \to 3$
$\displaystyle 6 \to 4 \hspace{1 cm} \text{or} \hspace{1 cm} 6 \to 2 \to 2$
$\displaystyle 5 \to 5 \hspace{1 cm} \text{or} \hspace{1 cm} 5 \to 3 \to 2 \hspace{1 cm} \text{or} \hspace{1 cm} 5 \to 2 \to 3$
$\displaystyle \text{etc.}$
Again, we need to add up the probabilities of rolling some number with one roll, then making up the difference with any number of rolls.
This is a perfect candidate for what’s called a recursive function. Here’s how it will work: say we are writing the function p_anyRolls(n), with which we want to find the probability that the player will land on a space that is n spaces away. The basic tactic will be to add up all the probabilities of all the ways that a sequence of rolls could add up to n, as shown above. Here is an outline of the algorithm:
1. Let p be a running tally of the probability of getting to n. Initially, set p = 0.
2. Add (chance of rolling a 2)*(chance of rolling (n-2) in any number of rolls) to p.
3. Add (chance of rolling a 3)*(chance of rolling (n-3) in any number of rolls) to p.
4. Add (chance of rolling a 4)*(chance of rolling (n-4) in any number of rolls) to p.
5. Repeat for 5 – 12.
6. p now holds the probability of landing on the n-th space.
Adding some function names, we now have p_anyRolls(n), and we can let p_oneRoll(n) be the probability of rolling a certain number n with one roll. This simplifies the algorithm:
1. Set p = 0 initially.
2. Add p_oneRoll(2)*p_anyRolls(n-2) to p.
3. Repeat for 3 – 12.
4. p now holds the probability of landing on the n-th space.
Wait, so p_anyRolls is a function that calls itself while running? This is the recursive part. Notice that each time p_anyRolls is called with an argument n, it only calls itself recursively with arguments less than n. So in evaluating, it will keep calling itself with smaller and smaller arguments until it reaches some base case. The recursion will only work if we add in some provision that if n is sufficiently small, p_anyRolls does not need to run itself recursively. Once the tower of function calls hits this number, the function can finally evaluate and return a value to the instance that asked for it, thus sending the answer rippling up through the ranks. So we’re going to add the provision that if n is equal to zero, the result is 1. (If you’re already there, the probability you get there is 1). It may seem a strange case to account for, but it makes the recursion work.
1. If n equals 0, stop the program and return the answer 1.
2. Otherwise, set p = 0 initially.
3. For each number “i” between 2 and 12, add p_oneRoll(i)*p_anyRolls(n-i) to p.
4. Return the answer p.
Notice that we also condensed the “Repeat” statements into a “For each” statement. This is a common way to do something many times over a certain range of values.
There’s only one more thing we need to consider: how hard is this going to be for the computer to do? Notice that if we compute p_anyRolls(10), the computer will need to compute p_anyRolls(8), p_anyRolls(7), p_anyRolls(6), etc.. Then, when computing p_anyRolls(11), the computer will need to compute each of these values again, and they’ll take longer and longer to compute as n gets bigger. A good idea would be to save the values we find so that we only need to compute them once. Our finished algorithm looks like this:
1. If n equals 0, stop the program and return the answer 1.
2. If n is a value we’ve already computed, stop and return the value we already know.
3. Set p = 0.
4. For each number “i” between 2 and 12, add p_oneRoll(i)*p_anyRolls(n-i) to p.
5. Save the fact that we’ve now computed p_anyRolls(n) to equal p.
6. Return the value of p.
If you’re curious, my code looks like this, in the language C:
````
double p_anyRolls(int n)
{
if(probToHit[n] >= 0)
return probToHit[n];
if(n == 0)
return 1.0;
double p = 0.0;
int cap = intMin(12, n);
int i;
for(i = 2; i <= cap; i++)
p += p_oneRoll(i) * p_anyRolls(n - i);
probToHit[n] = p;
return p;
}
````
Using this code, I printed out the likelihoods of landing on spots from 2 to 50 spaces ahead:
A few notes: 7 away is still the best place to be, with almost a 1 in 5 chance of landing there. The odds then dip before settling near this 0.1428 number. In fact, the odds are approaching 1/7, about 0.14286.
This makes sense intuitively. For spots very close to the player, the different roll probabilities mean a lot. But as the player moves toward distant spaces at an average of 7 spaces per roll, she will probably land on about 1/7 of the spaces … which means that each space has a 1/7 chance of being landed on.
There’s something interesting going on if you allow the possible rolls and probabilities to vary (ie, not just using two six-sided dice). The property still holds that the probability of landing on almost any given space is given by 1 over the average roll. So with one eleven-sided die, a space will have a 1/6 chance of being landed on; for a 99-sided die, a space will have a 1/50 chance. Even more, each sequence of rolls that sum to a number n determines a unique composition of n (an idea I hope to develop more in a later post).
Average Exponents and the Divisor Function
Here’s something I’ve been thinking about lately: patterns in the prime factorizations of natural numbers. Every number can be written in exactly one way as a product of powers of primes:
$\displaystyle n = p_1^{a_1}p_2^{a_2}\dotsm p_k^{a_k}$
This factorization is also linked to the number of proper divisors of n. If we want to know how many divisors n has, we are really asking how many numbers can be made from the same primes as n. If some d can be written as:
$\displaystyle d = p_1^{b_1}p_2^{b_2}\dotsm p_k^{b_k}, \hspace{10 mm} 0 \le b_i \le a_i$
then d clearly divides n, and every divisor of n can be written this way. So how many such d can there be? The only things to change are the exponents $\displaystyle b_i$ , and these can each be any number from $\displaystyle \{0, 1, 2, \ldots , a_i\}$ . This gives $\displaystyle a_i + 1$ choices for each exponent, so the total number of divisors of n is
$\displaystyle \prod_{i=1}^k \left(a_i + 1 \right)$
Before I go on, I should warn you that the following are just some observations on the behavior of factorizations and numbers of divisors. Perhaps someone reading this will have some insight, or has already investigated this. I have not managed to say anything meaningful about the divisor function, but I do find it linked to an expression that I think is interesting.
Because there are patterns in the values that these exponents tend to take. For instance, if I pick a number at random and ask you to guess the exponent of some prime in its factorization, what should you guess? It turns out that there is an elegant form for the expected value of these exponents.
Start with 2, and let’s say we’re picking from all natural numbers. Let $\displaystyle \mu(p)$ be the expected value of the exponent of p. Even numbers are divisible by 2, so the exponent of 2 for half of the numbers is at least 1. That gives $\displaystyle \mu(2) \ge 1/2$.
One fourth of numbers are divisible by 4, so they each contribute a 2 to the average. Think of them as contributing 1 along with the general even numbers, then an extra 1. Then one eighth of all numbers contribute again, and one sixteenth contribute a fourth time, etc. In the end we have
$\displaystyle \mu(2) = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dotsb = 1$
This can be done just the same with the other primes
$\displaystyle \mu(p) = \sum_{k=1}^\infty \frac{1}{p^k}$
This can be simplified in a cute way if you’ll trust me that it’s convergent. Multiply the sum by p to get
$\displaystyle p\mu(p) = 1 + \frac{1}{p} + \frac{1}{p^2} + \dotsb = 1 + \mu(p)$
$\displaystyle (p - 1)\mu(p) = 1$
$\displaystyle \mu(p) = \frac{1}{p-1}$
So the average exponent for each prime p is $\displaystyle \frac{1}{p-1}$ . I would end that sentence with an exclamation point, but I have gotten in trouble before with the whole excitement vs. factorial issue.
Well what can you do with these exponents? Back to the divisor function. Since we’re picking over all numbers, we can treat the exponents of prime terms like independent variables (if we were picking under say 100, an exponent of 2 like 6 would certainly affect the possibilities for 3). That means that expected values are preserved over addition and multiplication. So we can plug our expected exponents into the divisor formula to get an expected value for the number of divisors:
$\displaystyle \prod_{p \in \mathbb{P}} \left(1 + \frac{1}{p-1} \right)$
Wouldn’t it be a kick if it were convergent? Well, it’s divergent. In fact, the expanded sum contains every term of the form $\displaystyle \frac{1}{p-1}$ , which is certainly greater than the sum of all terms of the form $\displaystyle \frac{1}{p}$ , which by no proof of my own is divergent. So this is most definitely divergent.
So what does that get us? The average value of the divisor function grows with n , very slowly but without limit. And it is related to a very pretty product over primes. If you have some insight on the topic, leave a reply!
Differences Between Random Numbers
Have you ever taken a list of numbers, and produced a new list by taking the difference between each adjacent pair? For instance:
1 4 9 16 25 => 3 5 7 9
It’s easy to see that with the second list, along with the first number of the original list, we can reconstruct the entire original list. In fact, if we notice a pattern with the second list, we can add a number there and extend the original list
3 5 7 9 [11] => 1 4 9 16 25 [36]
I used to do this in church, where there were numbers hanging on the walls (Bible verses, I assume). By building these “difference lists,” I decided I could predict the next number in any sequence. Does it work? Well, the results are often interesting but not always very useful.
One thing I noticed while making difference lists is that the differences are generally substantially smaller than the original numbers. Take this list of 10 random numbers between 1 and 100, inclusive
(65 79 77 29 9 2 75 55 89 63) average value => 54.3
The difference list, using absolute values, is
(14 2 48 20 7 73 20 34 26) average value => 27.1
This example seems to point to a 2:1 ratio between the average value of the original list and the average value of the difference list, so I ran a couple of thousand simulations, and the resulting data were quite different. The average value of a difference list constructed from 10000 whole numbers chosen from 1 to 100 was 33.1. When the numbers were chosen from 1 to 300, the average difference was 99.5. Now I put forth the question: given two numbers chosen randomly from 1 to N, what is the expected value of their difference?
From the simulations, I knew I was looking for something in the range of N/3. Here’s how I worked it out.
My method was to build up the expected difference from probabilities. Think about picking the first number, then working out the likelihoods of the second number being a certain distance from the first. Let N = 100 for now, and say the first number picked is 30. If we’re working between 1 and 100, and the first number picked is 30, the distance can’t very well be over 70, can it? Every other difference between 0 and 70 is still possible, but the differences 1 through 30 are especially likely. If k is a possible difference somewhere between 1 and 30, then 30 – k and 30 + k are both candidates for a second number which would produce a difference of k. So any number less than the first number picked is twice as likely as other numbers to be the difference. In fact, this needs a little refining: if 70 is the first number picked, not all numbers under 70 are twice as likely, but we’ll see that our formula accounts for this nicely.
So here’s the strategy: there’s a 1/N chance of the first draw being a certain number k. Then we add up probabilities, with each possible difference having a 1/N baseline chance, and numbers that could be a difference on either side of k having a extra 1/N chance of being the difference. The formula looks like this:
$\displaystyle\frac{1}{N}\sum_{k=1}^{N}\frac{1}{N}\left(\sum_{j=1}^{N-k}j + \sum_{j=1}^{k-1}j\right)$
A bit of explanation: if k is less than or equal to N/2, then all the numbers from 0 to N - k are possible differences, and all numbers from 1 to k - 1 get the extra, bonus chance of being the difference. If k is greater than N/2, then all the numbers from 0 to k - 1 are possible differences, while all the numbers from 1 to N – k get the bonus chance. Since terms where j = 0 are worth nothing, we can let both sums start at 1 and take care of both cases! Their roles simply switch depending on the value of k.
I arrived at this formula several different ways, so I’m pretty sure it’s accurate. From here, it’s some fun algebra and use of the formulas for sums:
$\displaystyle\frac{1}{N}\sum_{k=1}^{N}\frac{1}{N}\left(\sum_{j=1}^{N-k}j + \sum_{j=1}^{k-1}j\right)$
$\displaystyle\frac{1}{N^2}\sum_{k=1}^{N}\left(\sum_{j=1}^{N-k}j + \sum_{j=1}^{k-1}j\right)$
$\displaystyle\frac{1}{N^2}\sum_{k=1}^{N}\left(\frac{(N-k)(N-k+1)}{2} + \frac{(k-1)k}{2}\right)$
$\displaystyle\frac{1}{2N^2}\sum_{k=1}^N\left(2k^2 - 2k - 2Nk + N^2 + N\right)$
$\displaystyle\frac{1}{N^2}\left(\sum_{k=1}^{N}k^2 - (N+1)\sum_{k=1}^{N}k + \sum_{k=1}^N\frac{N^2 + N}{2}\right)$
$\displaystyle\frac{1}{N^2}\left(\frac{N(N+1)(2N+1)}{6} - \frac{N(N+1)(N+1)}{2} + \frac{N^3 + N^2}{2} \right)$
$\displaystyle\frac{1}{N}\left(\frac{2N^2 - 2}{6}\right)$
$\displaystyle\frac{N^2 - 1}{3N}$
I rather like this. It might as well be N/3 for large values of N, but there’s a big difference when N is small. What if you have to know the average difference between two standard six-sided dice when thrown? Sure, 2 is a fine guess, but I’d have to say that 1.9444… works even better.
Can you think of a better proof of this, or a mistake in mine? Leave me a comment!
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Group (mathematics)
# Group (mathematics)
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Encyclopedia
In mathematics
Mathematics
Mathematics is the study of quantity, space, structure, and change. Mathematicians seek out patterns and formulate new conjectures. Mathematicians resolve the truth or falsity of conjectures by mathematical proofs, which are arguments sufficient to convince other mathematicians of their validity...
, a group is an algebraic structure
Algebraic structure
In abstract algebra, an algebraic structure consists of one or more sets, called underlying sets or carriers or sorts, closed under one or more operations, satisfying some axioms. Abstract algebra is primarily the study of algebraic structures and their properties...
consisting of a set together with an operation
Binary operation
In mathematics, a binary operation is a calculation involving two operands, in other words, an operation whose arity is two. Examples include the familiar arithmetic operations of addition, subtraction, multiplication and division....
that combines any two of its elements to form a third element. To qualify as a group, the set and the operation must satisfy a few conditions called group axiom
Axiom
In traditional logic, an axiom or postulate is a proposition that is not proven or demonstrated but considered either to be self-evident or to define and delimit the realm of analysis. In other words, an axiom is a logical statement that is assumed to be true...
s, namely closure
Closure (mathematics)
In mathematics, a set is said to be closed under some operation if performance of that operation on members of the set always produces a unique member of the same set. For example, the real numbers are closed under subtraction, but the natural numbers are not: 3 and 8 are both natural numbers, but...
, associativity
Associativity
In mathematics, associativity is a property of some binary operations. It means that, within an expression containing two or more occurrences in a row of the same associative operator, the order in which the operations are performed does not matter as long as the sequence of the operands is not...
, identity
Identity element
In mathematics, an identity element is a special type of element of a set with respect to a binary operation on that set. It leaves other elements unchanged when combined with them...
and invertibility
Inverse element
In abstract algebra, the idea of an inverse element generalises the concept of a negation, in relation to addition, and a reciprocal, in relation to multiplication. The intuition is of an element that can 'undo' the effect of combination with another given element...
. Many familiar mathematical structure
Mathematical structure
In mathematics, a structure on a set, or more generally a type, consists of additional mathematical objects that in some manner attach to the set, making it easier to visualize or work with, or endowing the collection with meaning or significance....
s such as number systems obey these axioms: for example, the integer
Integer
The integers are formed by the natural numbers together with the negatives of the non-zero natural numbers .They are known as Positive and Negative Integers respectively...
s endowed with the addition
Addition
Addition is a mathematical operation that represents combining collections of objects together into a larger collection. It is signified by the plus sign . For example, in the picture on the right, there are 3 + 2 apples—meaning three apples and two other apples—which is the same as five apples....
operation form a group. However, the abstract formalization of the group axioms, detached as it is from the concrete nature of any particular group and its operation, allows entities with highly diverse mathematical origins in abstract algebra
Abstract algebra
Abstract algebra is the subject area of mathematics that studies algebraic structures, such as groups, rings, fields, modules, vector spaces, and algebras...
and beyond to be handled in a flexible way, while retaining their essential structural aspects. The ubiquity of groups in numerous areas within and outside mathematics makes them a central organizing principle of contemporary mathematics.
Groups share a fundamental kinship with the notion of symmetry
Symmetry
Symmetry generally conveys two primary meanings. The first is an imprecise sense of harmonious or aesthetically pleasing proportionality and balance; such that it reflects beauty or perfection...
. A symmetry group
Symmetry group
The symmetry group of an object is the group of all isometries under which it is invariant with composition as the operation...
encodes symmetry features of a geometrical
Geometry
Geometry arose as the field of knowledge dealing with spatial relationships. Geometry was one of the two fields of pre-modern mathematics, the other being the study of numbers ....
object: it consists of the set of transformations that leave the object unchanged, and the operation of combining two such transformations by performing one after the other. Such symmetry groups, particularly the continuous Lie group
Lie group
In mathematics, a Lie group is a group which is also a differentiable manifold, with the property that the group operations are compatible with the smooth structure...
s, play an important role in many academic disciplines. Matrix group
Matrix group
In mathematics, a matrix group is a group G consisting of invertible matrices over some field K, usually fixed in advance, with operations of matrix multiplication and inversion. More generally, one can consider n × n matrices over a commutative ring R...
s, for example, can be used to understand fundamental physical
Physics
Physics is a natural science that involves the study of matter and its motion through spacetime, along with related concepts such as energy and force. More broadly, it is the general analysis of nature, conducted in order to understand how the universe behaves.Physics is one of the oldest academic...
laws underlying special relativity
Special relativity
Special relativity is the physical theory of measurement in an inertial frame of reference proposed in 1905 by Albert Einstein in the paper "On the Electrodynamics of Moving Bodies".It generalizes Galileo's...
and symmetry phenomena in molecular chemistry
Chemistry
Chemistry is the science of matter, especially its chemical reactions, but also its composition, structure and properties. Chemistry is concerned with atoms and their interactions with other atoms, and particularly with the properties of chemical bonds....
.
The concept of a group arose from the study of polynomial equations, starting with Évariste Galois
Évariste Galois
Évariste Galois was a French mathematician born in Bourg-la-Reine. While still in his teens, he was able to determine a necessary and sufficient condition for a polynomial to be solvable by radicals, thereby solving a long-standing problem...
in the 1830s. After contributions from other fields such as number theory
Number theory
Number theory is a branch of pure mathematics devoted primarily to the study of the integers. Number theorists study prime numbers as well...
and geometry, the group notion was generalized and firmly established around 1870. Modern group theory
Group theory
In mathematics and abstract algebra, group theory studies the algebraic structures known as groups.The concept of a group is central to abstract algebra: other well-known algebraic structures, such as rings, fields, and vector spaces can all be seen as groups endowed with additional operations and...
—a very active mathematical discipline—studies groups in their own right. To explore groups, mathematicians have devised various notions
Glossary of group theory
A group is a set G closed under a binary operation • satisfying the following 3 axioms:* Associativity: For all a, b and c in G, • c = a • ....
to break groups into smaller, better-understandable pieces, such as subgroup
Subgroup
In group theory, given a group G under a binary operation *, a subset H of G is called a subgroup of G if H also forms a group under the operation *. More precisely, H is a subgroup of G if the restriction of * to H x H is a group operation on H...
s, quotient group
Quotient group
In mathematics, specifically group theory, a quotient group is a group obtained by identifying together elements of a larger group using an equivalence relation...
s and simple group
Simple group
In mathematics, a simple group is a nontrivial group whose only normal subgroups are the trivial group and the group itself. A group that is not simple can be broken into two smaller groups, a normal subgroup and the quotient group, and the process can be repeated...
s. In addition to their abstract properties, group theorists also study the different ways in which a group can be expressed concretely (its group representation
Group representation
In the mathematical field of representation theory, group representations describe abstract groups in terms of linear transformations of vector spaces; in particular, they can be used to represent group elements as matrices so that the group operation can be represented by matrix multiplication...
s), both from a theoretical
Representation theory
Representation theory is a branch of mathematics that studies abstract algebraic structures by representing their elements as linear transformations of vector spaces, and studiesmodules over these abstract algebraic structures...
and a computational point of view
Computational group theory
In mathematics, computational group theory is the study ofgroups by means of computers. It is concernedwith designing and analysing algorithms anddata structures to compute information about groups...
. A particularly rich theory has been developed for finite group
Finite group
In mathematics and abstract algebra, a finite group is a group whose underlying set G has finitely many elements. During the twentieth century, mathematicians investigated certain aspects of the theory of finite groups in great depth, especially the local theory of finite groups, and the theory of...
s, which culminated with the monumental classification of finite simple groups
Classification of finite simple groups
In mathematics, the classification of the finite simple groups is a theorem stating that every finite simple group belongs to one of four categories described below. These groups can be seen as the basic building blocks of all finite groups, in much the same way as the prime numbers are the basic...
announced in 1983. Since the mid-1980s, geometric group theory
Geometric group theory
Geometric group theory is an area in mathematics devoted to the study of finitely generated groups via exploring the connections between algebraic properties of such groups and topological and geometric properties of spaces on which these groups act .Another important...
, which studies finitely generated groups as geometric objects, has become a particularly active area in group theory.
### First example: the integers
One of the most familiar groups is the set of integers Z which consists of the numbers
..., −4, −3, −2, −1, 0, 1, 2, 3, 4, ...
The following properties of integer addition
Addition
Addition is a mathematical operation that represents combining collections of objects together into a larger collection. It is signified by the plus sign . For example, in the picture on the right, there are 3 + 2 apples—meaning three apples and two other apples—which is the same as five apples....
serve as a model for the abstract group axioms given in the definition below.
1. For any two integers a and b, the sum
Summation
Summation is the operation of adding a sequence of numbers; the result is their sum or total. If numbers are added sequentially from left to right, any intermediate result is a partial sum, prefix sum, or running total of the summation. The numbers to be summed may be integers, rational numbers,...
a + b is also an integer. Thus, adding two integers never yields some other type of number, such as a fraction
Fraction (mathematics)
A fraction represents a part of a whole or, more generally, any number of equal parts. When spoken in everyday English, we specify how many parts of a certain size there are, for example, one-half, five-eighths and three-quarters.A common or "vulgar" fraction, such as 1/2, 5/8, 3/4, etc., consists...
. This property is known as closure
Closure (mathematics)
In mathematics, a set is said to be closed under some operation if performance of that operation on members of the set always produces a unique member of the same set. For example, the real numbers are closed under subtraction, but the natural numbers are not: 3 and 8 are both natural numbers, but...
under addition.
2. For all integers a, b and c, (a + b) + c = a + (b + c). Expressed in words, adding a to b first, and then adding the result to c gives the same final result as adding a to the sum of b and c, a property known as associativity
Associativity
In mathematics, associativity is a property of some binary operations. It means that, within an expression containing two or more occurrences in a row of the same associative operator, the order in which the operations are performed does not matter as long as the sequence of the operands is not...
.
3. If a is any integer, then 0 + a = a + 0 = a. Zero is called the identity element
Identity element
In mathematics, an identity element is a special type of element of a set with respect to a binary operation on that set. It leaves other elements unchanged when combined with them...
of addition because adding it to any integer returns the same integer.
4. For every integer a, there is an integer b such that a + b = b + a = 0. The integer b is called the inverse element
Inverse element
In abstract algebra, the idea of an inverse element generalises the concept of a negation, in relation to addition, and a reciprocal, in relation to multiplication. The intuition is of an element that can 'undo' the effect of combination with another given element...
of the integer a and is denoted −a.
The integers, together with the operation +, form a mathematical object belonging to a broad class sharing similar structural aspects. To appropriately understand these structures as a collective, the following abstract definition
Definition
A definition is a passage that explains the meaning of a term , or a type of thing. The term to be defined is the definiendum. A term may have many different senses or meanings...
is developed.
### Definition
A group is a set, G, together with an operation
Binary operation
In mathematics, a binary operation is a calculation involving two operands, in other words, an operation whose arity is two. Examples include the familiar arithmetic operations of addition, subtraction, multiplication and division....
• (called the group law of G) that combines any two elements a and b to form another element, denoted or ab. To qualify as a group, the set and operation, , must satisfy four requirements known as the group axioms:
Closure: For all a, b in G, the result of the operation, a • b, is also in G.
Associativity: For all a, b and c in G, (a • b) • c = a • (b • c).
Identity element: There exists an element e in G, such that for every element a in G, the equation e • a = a • e = a holds. The identity element of a group G is often written as 1 or 1G, a notation inherited from the multiplicative identity.
Inverse element: For each a in G, there exists an element b in G such that a • b = b • a = 1G.
The order in which the group operation is carried out can be significant. In other words, the result of combining element a with element b need not yield the same result as combining element b with element a; the equation
a • b = b • a
may not always be true. This equation does always hold in the group of integers under addition, because a + b = b + a for any two integers (commutativity
Commutativity
In mathematics an operation is commutative if changing the order of the operands does not change the end result. It is a fundamental property of many binary operations, and many mathematical proofs depend on it...
of addition). However, it does not always hold in the symmetry group below. Groups for which the equation a • b = b • a always holds are called abelian
Abelian group
In abstract algebra, an abelian group, also called a commutative group, is a group in which the result of applying the group operation to two group elements does not depend on their order . Abelian groups generalize the arithmetic of addition of integers...
(in honor of Niels Abel). Thus, the integer addition group is abelian, but the symmetry group in the following section is not.
The set G is called the underlying set of the group . Often the group's underlying set G is used as a short name for the group . Along the same lines, shorthand expressions such as "a subset of the group G" or "an element of group G" are used when what is actually meant is "a subset of the underlying set G of the group " or "an element of the underlying set G of the group ". Usually, it is clear from the context whether a symbol like G refers to a group or to an underlying set.
### Second example: a symmetry group
Two figures in the plane are congruent
Congruence (geometry)
In geometry, two figures are congruent if they have the same shape and size. This means that either object can be repositioned so as to coincide precisely with the other object...
if one can be changed into the other using a combination of rotation
Rotation (mathematics)
In geometry and linear algebra, a rotation is a transformation in a plane or in space that describes the motion of a rigid body around a fixed point. A rotation is different from a translation, which has no fixed points, and from a reflection, which "flips" the bodies it is transforming...
s, reflection
Reflection (mathematics)
In mathematics, a reflection is a mapping from a Euclidean space to itself that is an isometry with a hyperplane as set of fixed points; this set is called the axis or plane of reflection. The image of a figure by a reflection is its mirror image in the axis or plane of reflection...
s, and translation
Translation (geometry)
In Euclidean geometry, a translation moves every point a constant distance in a specified direction. A translation can be described as a rigid motion, other rigid motions include rotations and reflections. A translation can also be interpreted as the addition of a constant vector to every point, or...
s. Any figure is congruent to itself. However, some figures are congruent to themselves in more than one way, and these extra congruences are called symmetries
Symmetry
Symmetry generally conveys two primary meanings. The first is an imprecise sense of harmonious or aesthetically pleasing proportionality and balance; such that it reflects beauty or perfection...
. A square has eight symmetries. These are:
| | | | |
|--------------------------------------------------------------------------------------------------------------------------------|----------------------------|-----------------------------|-----------------------------|
| id (keeping it as is) | r1 (rotation by 90° right) | r2 (rotation by 180° right) | r3 (rotation by 270° right) |
| fv (vertical flip) | fh (horizontal flip) | fd (diagonal flip) | fc (counter-diagonal flip) |
| The elements of the symmetry group of the square (D4). The vertices are colored and numbered only to visualize the operations. | | | |
• the identity operation leaving everything unchanged, denoted id;
• rotations of the square around its center by 90° right, 180° right, and 270° right, denoted by r1, r2 and r3, respectively;
• reflections about the vertical and horizontal middle line (fh and fv), or through the two diagonal
Diagonal
A diagonal is a line joining two nonconsecutive vertices of a polygon or polyhedron. Informally, any sloping line is called diagonal. The word "diagonal" derives from the Greek διαγώνιος , from dia- and gonia ; it was used by both Strabo and Euclid to refer to a line connecting two vertices of a...
s (fd and fc).
These symmetries are represented by functions. Each of these functions sends a point in the square to the corresponding point under the symmetry. For example, r1 sends a point to its rotation 90° right around the square's center, and fh sends a point to its reflection across the square's vertical middle line. Composing two of these symmetry functions gives another symmetry function. These symmetries determine a group called the dihedral group
Dihedral group
In mathematics, a dihedral group is the group of symmetries of a regular polygon, including both rotations and reflections. Dihedral groups are among the simplest examples of finite groups, and they play an important role in group theory, geometry, and chemistry.See also: Dihedral symmetry in three...
of degree 4 and denoted D4. The underlying set of the group is the above set of symmetry functions, and the group operation is function composition
Function composition
In mathematics, function composition is the application of one function to the results of another. For instance, the functions and can be composed by computing the output of g when it has an argument of f instead of x...
. Two symmetries are combined by composing them as functions, that is, applying the first one to the square, and the second one to the result of the first application. The result of performing first a and then b is written symbolically from right to left as
b • a ("apply the symmetry b after performing the symmetry a").
The right-to-left notation is the same notation that is used for composition of functions.
The group table on the right lists the results of all such compositions possible. For example, rotating by 270° right (r3) and then flipping horizontally (fh) is the same as performing a reflection along the diagonal (fd). Using the above symbols, highlighted in blue in the group table:
fh • r3 = fd.
Group table
Cayley table
A Cayley table, after the 19th century British mathematician Arthur Cayley, describes the structure of a finite group by arranging all the possible products of all the group's elements in a square table reminiscent of an addition or multiplication table...
of D4
• id r1 r2 r3 fv fh fd fc
id id r1 r2 r3 fv fh fd fc
r1 r1 r2 r3 id fc fd fv fh
r2 r2 r3 id r1 fh fv fc fd
r3 r3 id r1 r2 fd fc fh fv
fv fv fd fh fc id r2 r1 r3
fh fh fc fv fd r2 id r3 r1
fd fd fh fc fv r3 r1 id r2
fc fc fv fd fh r1 r3 r2 id
The elements id, r1, r2, and r3 form a subgroup
Subgroup
In group theory, given a group G under a binary operation *, a subset H of G is called a subgroup of G if H also forms a group under the operation *. More precisely, H is a subgroup of G if the restriction of * to H x H is a group operation on H...
, highlighted in red (upper left region). A left and right coset
Coset
In mathematics, if G is a group, and H is a subgroup of G, and g is an element of G, thenA coset is a left or right coset of some subgroup in G...
of this subgroup is highlighted in green (in the last row) and yellow (last column), respectively.
Given this set of symmetries and the described operation, the group axioms can be understood as follows:
1. The closure axiom demands that the composition b • a of any two symmetries a and b is also a symmetry. Another example for the group operation is
r3 • fh = fc,
i.e. rotating 270° right after flipping horizontally equals flipping along the counter-diagonal (fc). Indeed every other combination of two symmetries still gives a symmetry, as can be checked using the group table.
2. The associativity constraint deals with composing more than two symmetries: Starting with three elements a, b and c of D4, there are two possible ways of using these three symmetries in this order to determine a symmetry of the square. One of these ways is to first compose a and b into a single symmetry, then to compose that symmetry with c. The other way is to first compose b and c, then to compose the resulting symmetry with a. The associativity condition
(a • b) • c = a • (b • c)
means that these two ways are the same, i.e., a product of many group elements can be simplified in any order.
For example, (fd • fv) • r2 = fd • (fv • r2) can be checked using the group table at the right
| | | | | |
|----------------|----|---------|----|------------------|
| (fd • fv) • r2 | = | r3 • r2 | = | r1, which equals |
| fd • (fv • r2) | = | fd • fh | = | r1. |
While associativity is true for the symmetries of the square and addition of numbers, it is not true for all operations. For instance, subtraction of numbers is not associative: (7 − 3) − 2 = 2 is not the same as 7 − (3 − 2) = 6.
3. The identity element is the symmetry id leaving everything unchanged: for any symmetry a, performing id after a (or a after id) equals a, in symbolic form,
id • a = a,
a • id = a.
4. An inverse element undoes the transformation of some other element. Every symmetry can be undone: each of the following transformations—identity id, the flips fh, fv, fd, fc and the 180° rotation r2—is its own inverse, because performing it twice brings the square back to its original orientation. The rotations r3 and r1 are each other's inverses, because rotating 90° and then rotation 270° (or vice versa) yields a rotation over 360° which leaves the square unchanged. In symbols,
fh • fh = id,
r3 • r1 = r1 • r3 = id.
In contrast to the group of integers above, where the order of the operation is irrelevant, it does matter in D4: fh • r1 = fc but r1 • fh = fd. In other words, D4 is not abelian, which makes the group structure more difficult than the integers introduced first.
## History
The modern concept of an abstract group developed out of several fields of mathematics. The original motivation for group theory was the quest for solutions of polynomial equations of degree higher than 4. The 19th-century French mathematician Évariste Galois
Évariste Galois
Évariste Galois was a French mathematician born in Bourg-la-Reine. While still in his teens, he was able to determine a necessary and sufficient condition for a polynomial to be solvable by radicals, thereby solving a long-standing problem...
, extending prior work of Paolo Ruffini
Paolo Ruffini
Paolo Ruffini was an Italian mathematician and philosopher.By 1788 he had earned university degrees in philosophy, medicine/surgery, and mathematics...
and Joseph-Louis Lagrange, gave a criterion for the solvability of a particular polynomial equation in terms of the symmetry group
Symmetry group
The symmetry group of an object is the group of all isometries under which it is invariant with composition as the operation...
of its roots (solutions). The elements of such a Galois group
Galois group
In mathematics, more specifically in the area of modern algebra known as Galois theory, the Galois group of a certain type of field extension is a specific group associated with the field extension...
correspond to certain permutation
Permutation
In mathematics, the notion of permutation is used with several slightly different meanings, all related to the act of permuting objects or values. Informally, a permutation of a set of objects is an arrangement of those objects into a particular order...
s of the roots. At first, Galois' ideas were rejected by his contemporaries, and published only posthumously. More general permutation group
Permutation group
In mathematics, a permutation group is a group G whose elements are permutations of a given set M, and whose group operation is the composition of permutations in G ; the relationship is often written as...
s were investigated in particular by Augustin Louis Cauchy
Augustin Louis Cauchy
Baron Augustin-Louis Cauchy was a French mathematician who was an early pioneer of analysis. He started the project of formulating and proving the theorems of infinitesimal calculus in a rigorous manner, rejecting the heuristic principle of the generality of algebra exploited by earlier authors...
. Arthur Cayley
Arthur Cayley
Arthur Cayley F.R.S. was a British mathematician. He helped found the modern British school of pure mathematics....
's On the theory of groups, as depending on the symbolic equation θn = 1 (1854) gives the first abstract definition of a finite group
Finite group
In mathematics and abstract algebra, a finite group is a group whose underlying set G has finitely many elements. During the twentieth century, mathematicians investigated certain aspects of the theory of finite groups in great depth, especially the local theory of finite groups, and the theory of...
.
Geometry was a second field in which groups were used systematically, especially symmetry groups as part of Felix Klein
Felix Klein
Christian Felix Klein was a German mathematician, known for his work in group theory, function theory, non-Euclidean geometry, and on the connections between geometry and group theory...
's 1872 Erlangen program
Erlangen program
An influential research program and manifesto was published in 1872 by Felix Klein, under the title Vergleichende Betrachtungen über neuere geometrische Forschungen...
. After novel geometries such as hyperbolic
Hyperbolic geometry
In mathematics, hyperbolic geometry is a non-Euclidean geometry, meaning that the parallel postulate of Euclidean geometry is replaced...
and projective geometry
Projective geometry
In mathematics, projective geometry is the study of geometric properties that are invariant under projective transformations. This means that, compared to elementary geometry, projective geometry has a different setting, projective space, and a selective set of basic geometric concepts...
had emerged, Klein used group theory to organize them in a more coherent way. Further advancing these ideas, Sophus Lie
Sophus Lie
Marius Sophus Lie was a Norwegian mathematician. He largely created the theory of continuous symmetry, and applied it to the study of geometry and differential equations.- Biography :...
founded the study of Lie group
Lie group
In mathematics, a Lie group is a group which is also a differentiable manifold, with the property that the group operations are compatible with the smooth structure...
s in 1884.
The third field contributing to group theory was number theory
Number theory
Number theory is a branch of pure mathematics devoted primarily to the study of the integers. Number theorists study prime numbers as well...
. Certain abelian group
Abelian group
In abstract algebra, an abelian group, also called a commutative group, is a group in which the result of applying the group operation to two group elements does not depend on their order . Abelian groups generalize the arithmetic of addition of integers...
structures had been used implicitly in Carl Friedrich Gauss
Carl Friedrich Gauss
Johann Carl Friedrich Gauss was a German mathematician and scientist who contributed significantly to many fields, including number theory, statistics, analysis, differential geometry, geodesy, geophysics, electrostatics, astronomy and optics.Sometimes referred to as the Princeps mathematicorum...
' number-theoretical work Disquisitiones Arithmeticae
Disquisitiones Arithmeticae
The Disquisitiones Arithmeticae is a textbook of number theory written in Latin by Carl Friedrich Gauss in 1798 when Gauss was 21 and first published in 1801 when he was 24...
(1798), and more explicitly by Leopold Kronecker
Leopold Kronecker
Leopold Kronecker was a German mathematician who worked on number theory and algebra.He criticized Cantor's work on set theory, and was quoted by as having said, "God made integers; all else is the work of man"...
. In 1847, Ernst Kummer
Ernst Kummer
Ernst Eduard Kummer was a German mathematician. Skilled in applied mathematics, Kummer trained German army officers in ballistics; afterwards, he taught for 10 years in a gymnasium, the German equivalent of high school, where he inspired the mathematical career of Leopold Kronecker.-Life:Kummer...
led early attempts to prove Fermat's Last Theorem
Fermat's Last Theorem
In number theory, Fermat's Last Theorem states that no three positive integers a, b, and c can satisfy the equation an + bn = cn for any integer value of n greater than two....
to a climax by developing groups describing factorization into prime number
Prime number
A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. A natural number greater than 1 that is not a prime number is called a composite number. For example 5 is prime, as only 1 and 5 divide it, whereas 6 is composite, since it has the divisors 2...
s.
The convergence of these various sources into a uniform theory of groups started with Camille Jordan
Camille Jordan
Marie Ennemond Camille Jordan was a French mathematician, known both for his foundational work in group theory and for his influential Cours d'analyse. He was born in Lyon and educated at the École polytechnique...
's Traité des substitutions et des équations algébriques (1870). Walther von Dyck
Walther von Dyck
Walther Franz Anton von Dyck , born Dyck and later ennobled, was a German mathematician...
(1882) gave the first statement of the modern definition of an abstract group. As of the 20th century, groups gained wide recognition by the pioneering work of Ferdinand Georg Frobenius
Ferdinand Georg Frobenius
Ferdinand Georg Frobenius was a German mathematician, best known for his contributions to the theory of differential equations and to group theory...
and William Burnside
William Burnside
William Burnside was an English mathematician. He is known mostly as an early contributor to the theory of finite groups....
, who worked on representation theory
Representation theory
Representation theory is a branch of mathematics that studies abstract algebraic structures by representing their elements as linear transformations of vector spaces, and studiesmodules over these abstract algebraic structures...
of finite groups, Richard Brauer
Richard Brauer
Richard Dagobert Brauer was a leading German and American mathematician. He worked mainly in abstract algebra, but made important contributions to number theory...
's modular representation theory
Modular representation theory
Modular representation theory is a branch of mathematics, and that part of representation theory that studies linear representations of finite group G over a field K of positive characteristic...
and Issai Schur
Issai Schur
Issai Schur was a mathematician who worked in Germany for most of his life. He studied at Berlin...
's papers. The theory of Lie groups, and more generally locally compact group
Locally compact group
In mathematics, a locally compact group is a topological group G which is locally compact as a topological space. Locally compact groups are important because they have a natural measure called the Haar measure. This allows one to define integrals of functions on G.Many of the results of finite...
s was pushed by Hermann Weyl
Hermann Weyl
Hermann Klaus Hugo Weyl was a German mathematician and theoretical physicist. Although much of his working life was spent in Zürich, Switzerland and then Princeton, he is associated with the University of Göttingen tradition of mathematics, represented by David Hilbert and Hermann Minkowski.His...
, Élie Cartan
Élie Cartan
Élie Joseph Cartan was an influential French mathematician, who did fundamental work in the theory of Lie groups and their geometric applications...
and many others. Its algebraic counterpart, the theory of algebraic group
Algebraic group
In algebraic geometry, an algebraic group is a group that is an algebraic variety, such that the multiplication and inverse are given by regular functions on the variety...
s, was first shaped by Claude Chevalley
Claude Chevalley
Claude Chevalley was a French mathematician who made important contributions to number theory, algebraic geometry, class field theory, finite group theory, and the theory of algebraic groups...
(from the late 1930s) and later by pivotal work of Armand Borel
Armand Borel
Armand Borel was a Swiss mathematician, born in La Chaux-de-Fonds, and was a permanent professor at the Institute for Advanced Study in Princeton, New Jersey, United States from 1957 to 1993...
and Jacques Tits
Jacques Tits
Jacques Tits is a Belgian and French mathematician who works on group theory and geometry and who introduced Tits buildings, the Tits alternative, and the Tits group.- Career :Tits received his doctorate in mathematics at the age of 20...
.
The University of Chicago
University of Chicago
The University of Chicago is a private research university in Chicago, Illinois, USA. It was founded by the American Baptist Education Society with a donation from oil magnate and philanthropist John D. Rockefeller and incorporated in 1890...
's 1960–61 Group Theory Year brought together group theorists such as Daniel Gorenstein
Daniel Gorenstein
Daniel E. Gorenstein was an American mathematician. He earned his undergraduate and graduate degrees at Harvard University, where he earned his Ph.D. in 1950 under Oscar Zariski, introducing in his dissertation Gorenstein rings...
, John G. Thompson
John G. Thompson
John Griggs Thompson is a mathematician at the University of Florida noted for his work in the field of finite groups. He was awarded the Fields Medal in 1970, the Wolf Prize in 1992 and the 2008 Abel Prize....
and Walter Feit
Walter Feit
Walter Feit was a Jewish Austrian-American mathematician who worked in finite group theory and representation theory....
, laying the foundation of a collaboration that, with input from numerous other mathematicians, classified all finite simple group
Classification of finite simple groups
In mathematics, the classification of the finite simple groups is a theorem stating that every finite simple group belongs to one of four categories described below. These groups can be seen as the basic building blocks of all finite groups, in much the same way as the prime numbers are the basic...
s in 1982. This project exceeded previous mathematical endeavours by its sheer size, in both length of proof and number of researchers. Research is ongoing to simplify the proof of this classification. These days, group theory is still a highly active mathematical branch crucially impacting many other fields.
## Elementary consequences of the group axioms
Basic facts about all groups that can be obtained directly from the group axioms are commonly subsumed under elementary group theory. For example, repeated
Mathematical induction
Mathematical induction is a method of mathematical proof typically used to establish that a given statement is true of all natural numbers...
applications of the associativity axiom show that the unambiguity of
a • b • c = (a • b) • c = a • (b • c)
generalizes to more than three factors. Because this implies that parentheses can be inserted anywhere within such a series of terms, parentheses are usually omitted.
The axioms may be weakened to assert only the existence of a left identity and left inverses. Both can be shown to be actually two-sided, so the resulting definition is equivalent to the one given above.
### Uniqueness of identity element and inverses
Two important consequences of the group axioms are the uniqueness of the identity element and the uniqueness of inverse elements. There can be only one identity element in a group, and each element in a group has exactly one inverse element. Thus, it is customary to speak of the identity, and the inverse of an element.
To prove the uniqueness of an inverse element of a, suppose that a has two inverses, denoted l and r, in a group (G, •). Then
| | | | |
|----|----|-------------|-------------------------------------------------------------|
| l | = | l • 1G | as 1G is the identity element |
| | = | l • (a • r) | because r is an inverse of a, so 1G = a • r |
| | = | (l • a) • r | by associativity, which allows to rearrange the parentheses |
| | = | 1G • r | since l is an inverse of a, i.e. l • a = 1G |
| | = | r | for 1G is the identity element |
The two extremal terms l and r are equal, since they are connected by a chain of equalities. In other words there is only one inverse element of a. Similarly, to prove that the identity element of a group is unique, assume G is a group with two identity elements 1G and e. Then 1G = 1G • e = e, hence 1G and e are equal.
### Division
In groups, it is possible to perform division
Division (mathematics)
right|thumb|200px|20 \div 4=5In mathematics, especially in elementary arithmetic, division is an arithmetic operation.Specifically, if c times b equals a, written:c \times b = a\,...
: given elements a and b of the group G, there is exactly one solution x in G to the equation
Equation
An equation is a mathematical statement that asserts the equality of two expressions. In modern notation, this is written by placing the expressions on either side of an equals sign , for examplex + 3 = 5\,asserts that x+3 is equal to 5...
x • a = b. In fact, right multiplication of the equation by a−1 gives the solution x = x • a • a−1 = b • a−1. Similarly there is exactly one solution y in G to the equation a • y = b, namely y = a−1 • b. In general, x and y need not agree.
A consequence of this is that multiplying by a group element g is a bijection
Bijection
A bijection is a function giving an exact pairing of the elements of two sets. A bijection from the set X to the set Y has an inverse function from Y to X. If X and Y are finite sets, then the existence of a bijection means they have the same number of elements...
. Specifically, if g is an element of the group G, there is a bijection from G to itself called left translation by g sending h ∈ G to g • h. Similarly, right translation by g is a bijection from G to itself sending h to h • g. If G is abelian, left and right translation by a group element are the same.
## Basic concepts
To understand groups beyond the level of mere symbolic manipulations as above, more structural concepts have to be employed. There is a conceptual principle underlying all of the following notions: to take advantage of the structure offered by groups (which sets, being "structureless", do not have), constructions related to groups have to be compatible with the group operation. This compatibility manifests itself in the following notions in various ways. For example, groups can be related to each other via functions called group homomorphisms. By the mentioned principle, they are required to respect the group structures in a precise sense. The structure of groups can also be understood by breaking them into pieces called subgroups and quotient groups. The principle of "preserving structures"—a recurring topic in mathematics throughout—is an instance of working in a category
Category (mathematics)
In mathematics, a category is an algebraic structure that comprises "objects" that are linked by "arrows". A category has two basic properties: the ability to compose the arrows associatively and the existence of an identity arrow for each object. A simple example is the category of sets, whose...
, in this case the category of groups
Category of groups
In mathematics, the category Grp has the class of all groups for objects and group homomorphisms for morphisms. As such, it is a concrete category...
.
### Group homomorphisms
Group homomorphisms are functions that preserve group structure. A function a: G → H between two groups (G,•) and (H,*) is called a homomorphism if the equation
a(g • k) = a(g) * a(k)
holds for all elements g, k in G. In other words, the result is the same when performing the group operation after or before applying the map a. This requirement ensures that a(1G) = 1H, and also a(g)−1 = a(g−1) for all g in G. Thus a group homomorphism respects all the structure of G provided by the group axioms.
Two groups G and H are called isomorphic
Isomorphism
In abstract algebra, an isomorphism is a mapping between objects that shows a relationship between two properties or operations. If there exists an isomorphism between two structures, the two structures are said to be isomorphic. In a certain sense, isomorphic structures are...
if there exist group homomorphisms a: G → H and b: H → G, such that applying the two functions one after another
Function composition
In mathematics, function composition is the application of one function to the results of another. For instance, the functions and can be composed by computing the output of g when it has an argument of f instead of x...
in each of the two possible orders gives the identity function
Identity function
In mathematics, an identity function, also called identity map or identity transformation, is a function that always returns the same value that was used as its argument...
s of G and H. That is, a(b(h)) = h and b(a(g)) = g for any g in G and h in H. From an abstract point of view, isomorphic groups carry the same information. For example, proving that g • g = 1G for some element g of G is equivalent
Logical equivalence
In logic, statements p and q are logically equivalent if they have the same logical content.Syntactically, p and q are equivalent if each can be proved from the other...
to proving that a(g) * a(g) = 1H, because applying a to the first equality yields the second, and applying b to the second gives back the first.
### Subgroups
Informally, a subgroup is a group H contained within a bigger one, G. Concretely, the identity element of G is contained in H, and whenever h1 and h2 are in H, then so are and h1−1, so the elements of H, equipped with the group operation on G restricted to H, indeed form a group.
In the example above, the identity and the rotations constitute a subgroup R = {id, r1, r2, r3}, highlighted in red in the group table above: any two rotations composed are still a rotation, and a rotation can be undone by (i.e. is inverse to) the complementary rotations 270° for 90°, 180° for 180°, and 90° for 270° (note that rotation in the opposite direction is not defined). The subgroup test
Subgroup test
In Abstract Algebra, the one-step subgroup test is a theorem that states that for any group, a nonempty subset of that group is itself a group if the inverse of any element in the subset multiplied with any other element in the subset is also in the subset...
is a necessary and sufficient condition
Necessary and sufficient conditions
In logic, the words necessity and sufficiency refer to the implicational relationships between statements. The assertion that one statement is a necessary and sufficient condition of another means that the former statement is true if and only if the latter is true.-Definitions:A necessary condition...
for a subset H of a group G to be a subgroup: it is sufficient to check that for all elements g, h ∈ H. Knowing the subgroups
Lattice of subgroups
In mathematics, the lattice of subgroups of a group G is the lattice whose elements are the subgroups of G, with the partial order relation being set inclusion....
is important in understanding the group as a whole.
Given any subset S of a group G, the subgroup generated by S consists of products of elements of S and their inverses. It is the smallest subgroup of G containing S. In the introductory example above, the subgroup generated by r2 and fv consists of these two elements, the identity element id and fh = fv • r2. Again, this is a subgroup, because combining any two of these four elements or their inverses (which are, in this particular case, these same elements) yields an element of this subgroup.
### Cosets
In many situations it is desirable to consider two group elements the same if they differ by an element of a given subgroup. For example, in D4 above, once a flip is performed, the square never gets back to the r2 configuration by just applying the rotation operations (and no further flips), i.e. the rotation operations are irrelevant to the question whether a flip has been performed. Cosets are used to formalize this insight: a subgroup H defines left and right cosets, which can be thought of as translations of H by arbitrary group elements g. In symbolic terms, the left and right cosets of H containing g are
gH = {g • h, h ∈ H} and Hg = {h • g, h ∈ H}, respectively.
The cosets of any subgroup H form a partition
Partition of a set
In mathematics, a partition of a set X is a division of X into non-overlapping and non-empty "parts" or "blocks" or "cells" that cover all of X...
of G; that is, the union
Union (set theory)
In set theory, the union of a collection of sets is the set of all distinct elements in the collection. The union of a collection of sets S_1, S_2, S_3, \dots , S_n\,\! gives a set S_1 \cup S_2 \cup S_3 \cup \dots \cup S_n.- Definition :...
of all left cosets is equal to G and two left cosets are either equal or have an empty
Empty set
In mathematics, and more specifically set theory, the empty set is the unique set having no elements; its size or cardinality is zero. Some axiomatic set theories assure that the empty set exists by including an axiom of empty set; in other theories, its existence can be deduced...
intersection
Intersection (set theory)
In mathematics, the intersection of two sets A and B is the set that contains all elements of A that also belong to B , but no other elements....
. The first case g1H = g2H happens precisely when
If and only if
In logic and related fields such as mathematics and philosophy, if and only if is a biconditional logical connective between statements....
, i.e. if the two elements differ by an element of H. Similar considerations apply to the right cosets of H. The left and right cosets of H may or may not be equal. If they are, i.e. for all g in G, gH = Hg, then H is said to be a normal subgroup
Normal subgroup
In abstract algebra, a normal subgroup is a subgroup which is invariant under conjugation by members of the group. Normal subgroups can be used to construct quotient groups from a given group....
.
In D4, the introductory symmetry group, the left cosets gR of the subgroup R consisting of the rotations are either equal to R, if g is an element of R itself, or otherwise equal to U = fcR = {fc, fv, fd, fh} (highlighted in green). The subgroup R is also normal, because fcR = U = Rfc and similarly for any element other than fc.
### Quotient groups
In some situations the set of cosets of a subgroup can be endowed with a group law, giving a quotient group or factor group. For this to be possible, the subgroup has to be normal
Normal subgroup
In abstract algebra, a normal subgroup is a subgroup which is invariant under conjugation by members of the group. Normal subgroups can be used to construct quotient groups from a given group....
. Given any normal subgroup N, the quotient group is defined by
G / N = {gN, g ∈ G}, "G modulo N".
This set inherits a group operation (sometimes called coset multiplication, or coset addition) from the original group G: (gN) • (hN) = (gh)N for all g and h in G. This definition is motivated by the idea (itself an instance of general structural considerations outlined above) that the map that associates to any element g its coset gN be a group homomorphism, or by general abstract considerations called universal properties
Universal property
In various branches of mathematics, a useful construction is often viewed as the “most efficient solution” to a certain problem. The definition of a universal property uses the language of category theory to make this notion precise and to study it abstractly.This article gives a general treatment...
. The coset eN = N serves as the identity in this group, and the inverse of gN in the quotient group is (gN)−1 = (g−1)N.
• R U
R R U
U U R
Group table of the quotient group .
The elements of the quotient group are R itself, which represents the identity, and U = fvR. The group operation on the quotient is shown at the right. For example, U • U = fvR • fvR = (fv • fv)R = R. Both the subgroup R = {id, r1, r2, r3}, as well as the corresponding quotient are abelian, whereas D4 is not abelian. Building bigger groups by smaller ones, such as D4 from its subgroup R and the quotient is abstracted by a notion called semidirect product
Semidirect product
In mathematics, specifically in the area of abstract algebra known as group theory, a semidirect product is a particular way in which a group can be put together from two subgroups, one of which is a normal subgroup. A semidirect product is a generalization of a direct product...
.
Quotient groups and subgroups together form a way of describing every group by its presentation
Presentation of a group
In mathematics, one method of defining a group is by a presentation. One specifies a set S of generators so that every element of the group can be written as a product of powers of some of these generators, and a set R of relations among those generators...
: any group is the quotient of the free group
In mathematics, a group G is called free if there is a subset S of G such that any element of G can be written in one and only one way as a product of finitely many elements of S and their inverses...
over the generators
Generating set of a group
In abstract algebra, a generating set of a group is a subset that is not contained in any proper subgroup of the group. Equivalently, a generating set of a group is a subset such that every element of the group can be expressed as the combination of finitely many elements of the subset and their...
of the group, quotiented by the subgroup of relations. The dihedral group D4, for example, can be generated by two elements r and f (for example, r = r1, the right rotation and f = fv the vertical (or any other) flip), which means that every symmetry of the square is a finite composition of these two symmetries or their inverses. Together with the relations
r 4 = f 2 = (r • f)2 = 1,
the group is completely described. A presentation of a group can also be used to construct the Cayley graph
Cayley graph
In mathematics, a Cayley graph, also known as a Cayley colour graph, Cayley diagram, group diagram, or colour group is a graph that encodes the abstract structure of a group. Its definition is suggested by Cayley's theorem and uses a specified, usually finite, set of generators for the group...
, a device used to graphically capture discrete group
Discrete group
In mathematics, a discrete group is a group G equipped with the discrete topology. With this topology G becomes a topological group. A discrete subgroup of a topological group G is a subgroup H whose relative topology is the discrete one...
s.
Sub- and quotient groups are related in the following way: a subset H of G can be seen as an injective map , i.e. any element of the target has at most one element that maps to it. The counterpart to injective maps are surjective maps (every element of the target is mapped onto), such as the canonical map . Interpreting subgroup and quotients in light of these homomorphisms emphasizes the structural concept inherent to these definitions alluded to in the introduction. In general, homomorphisms are neither injective nor surjective. Kernel
Kernel (algebra)
In the various branches of mathematics that fall under the heading of abstract algebra, the kernel of a homomorphism measures the degree to which the homomorphism fails to be injective. An important special case is the kernel of a matrix, also called the null space.The definition of kernel takes...
and image
Image (mathematics)
In mathematics, an image is the subset of a function's codomain which is the output of the function on a subset of its domain. Precisely, evaluating the function at each element of a subset X of the domain produces a set called the image of X under or through the function...
of group homomorphisms and the first isomorphism theorem address this phenomenon.
## Examples and applications
Examples and applications of groups abound. A starting point is the group Z of integers with addition as group operation, introduced above. If instead of addition multiplication
Multiplication
Multiplication is the mathematical operation of scaling one number by another. It is one of the four basic operations in elementary arithmetic ....
is considered, one obtains multiplicative group
Multiplicative group
In mathematics and group theory the term multiplicative group refers to one of the following concepts, depending on the context*any group \scriptstyle\mathfrak \,\! whose binary operation is written in multiplicative notation ,*the underlying group under multiplication of the invertible elements of...
s. These groups are predecessors of important constructions in abstract algebra
Abstract algebra
Abstract algebra is the subject area of mathematics that studies algebraic structures, such as groups, rings, fields, modules, vector spaces, and algebras...
.
Groups are also applied in many other mathematical areas. Mathematical objects are often examined by associating
Functor
In category theory, a branch of mathematics, a functor is a special type of mapping between categories. Functors can be thought of as homomorphisms between categories, or morphisms when in the category of small categories....
groups to them and studying the properties of the corresponding groups. For example, Henri Poincaré
Henri Poincaré
Jules Henri Poincaré was a French mathematician, theoretical physicist, engineer, and a philosopher of science...
founded what is now called algebraic topology
Algebraic topology
Algebraic topology is a branch of mathematics which uses tools from abstract algebra to study topological spaces. The basic goal is to find algebraic invariants that classify topological spaces up to homeomorphism, though usually most classify up to homotopy equivalence.Although algebraic topology...
by introducing the fundamental group
Fundamental group
In mathematics, more specifically algebraic topology, the fundamental group is a group associated to any given pointed topological space that provides a way of determining when two paths, starting and ending at a fixed base point, can be continuously deformed into each other...
. By means of this connection, topological properties such as proximity
Neighbourhood (mathematics)
In topology and related areas of mathematics, a neighbourhood is one of the basic concepts in a topological space. Intuitively speaking, a neighbourhood of a point is a set containing the point where you can move that point some amount without leaving the set.This concept is closely related to the...
and continuity
Continuous function
In mathematics, a continuous function is a function for which, intuitively, "small" changes in the input result in "small" changes in the output. Otherwise, a function is said to be "discontinuous". A continuous function with a continuous inverse function is called "bicontinuous".Continuity of...
translate into properties of groups. For example, elements of the fundamental group are represented by loops. The second image at the right shows some loops in a plane minus a point. The blue loop is considered null-homotopic (and thus irrelevant), because it can be continuously shrunk
Homotopy
In topology, two continuous functions from one topological space to another are called homotopic if one can be "continuously deformed" into the other, such a deformation being called a homotopy between the two functions...
to a point. The presence of the hole prevents the orange loop from being shrunk to a point. The fundamental group of the plane with a point deleted turns out to be infinite cyclic, generated by the orange loop (or any other loop winding once
Winding number
In mathematics, the winding number of a closed curve in the plane around a given point is an integer representing the total number of times that curve travels counterclockwise around the point...
around the hole). This way, the fundamental group detects the hole.
In more recent applications, the influence has also been reversed to motivate geometric constructions by a group-theoretical background. In a similar vein, geometric group theory
Geometric group theory
Geometric group theory is an area in mathematics devoted to the study of finitely generated groups via exploring the connections between algebraic properties of such groups and topological and geometric properties of spaces on which these groups act .Another important...
employs geometric concepts, for example in the study of hyperbolic group
Hyperbolic group
In group theory, a hyperbolic group, also known as a word hyperbolic group, Gromov hyperbolic group, negatively curved group is a finitely generated group equipped with a word metric satisfying certain properties characteristic of hyperbolic geometry. The notion of a hyperbolic group was introduced...
s. Further branches crucially applying groups include algebraic geometry
Algebraic geometry
Algebraic geometry is a branch of mathematics which combines techniques of abstract algebra, especially commutative algebra, with the language and the problems of geometry. It occupies a central place in modern mathematics and has multiple conceptual connections with such diverse fields as complex...
and number theory
Number theory
Number theory is a branch of pure mathematics devoted primarily to the study of the integers. Number theorists study prime numbers as well...
.
In addition to the above theoretical applications, many practical applications of groups exist. Cryptography
Cryptography
Cryptography is the practice and study of techniques for secure communication in the presence of third parties...
relies on the combination of the abstract group theory approach together with algorithm
Algorithm
In mathematics and computer science, an algorithm is an effective method expressed as a finite list of well-defined instructions for calculating a function. Algorithms are used for calculation, data processing, and automated reasoning...
ical knowledge obtained in computational group theory
Computational group theory
In mathematics, computational group theory is the study ofgroups by means of computers. It is concernedwith designing and analysing algorithms anddata structures to compute information about groups...
, in particular when implemented for finite groups. Applications of group theory are not restricted to mathematics; sciences such as physics
Physics
Physics is a natural science that involves the study of matter and its motion through spacetime, along with related concepts such as energy and force. More broadly, it is the general analysis of nature, conducted in order to understand how the universe behaves.Physics is one of the oldest academic...
, chemistry
Chemistry
Chemistry is the science of matter, especially its chemical reactions, but also its composition, structure and properties. Chemistry is concerned with atoms and their interactions with other atoms, and particularly with the properties of chemical bonds....
and computer science
Computer science
Computer science or computing science is the study of the theoretical foundations of information and computation and of practical techniques for their implementation and application in computer systems...
benefit from the concept.
### Numbers
Many number systems, such as the integers and the rationals enjoy a naturally given group structure. In some cases, such as with the rationals, both addition and multiplication operations give rise to group structures. Such number systems are predecessors to more general algebraic structures known as rings
Ring (mathematics)
In mathematics, a ring is an algebraic structure consisting of a set together with two binary operations usually called addition and multiplication, where the set is an abelian group under addition and a semigroup under multiplication such that multiplication distributes over addition...
and fields
Field (mathematics)
In abstract algebra, a field is a commutative ring whose nonzero elements form a group under multiplication. As such it is an algebraic structure with notions of addition, subtraction, multiplication, and division, satisfying certain axioms...
. Further abstract algebra
Abstract algebra
Abstract algebra is the subject area of mathematics that studies algebraic structures, such as groups, rings, fields, modules, vector spaces, and algebras...
ic concepts such as module
Module (mathematics)
In abstract algebra, the concept of a module over a ring is a generalization of the notion of vector space, wherein the corresponding scalars are allowed to lie in an arbitrary ring...
s, vector space
Vector space
A vector space is a mathematical structure formed by a collection of vectors: objects that may be added together and multiplied by numbers, called scalars in this context. Scalars are often taken to be real numbers, but one may also consider vector spaces with scalar multiplication by complex...
s and algebras
Algebra over a field
In mathematics, an algebra over a field is a vector space equipped with a bilinear vector product. That is to say, it isan algebraic structure consisting of a vector space together with an operation, usually called multiplication, that combines any two vectors to form a third vector; to qualify as...
also form groups.
#### Integers
The group of integers Z under addition, denoted (Z, +), has been described above. The integers, with the operation of multiplication
Multiplication
Multiplication is the mathematical operation of scaling one number by another. It is one of the four basic operations in elementary arithmetic ....
instead of addition, (Z, ·) do not form a group. The closure, associativity and identity axioms are satisfied, but inverses do not exist: for example, a = 2 is an integer, but the only solution to the equation a · b = 1 in this case is b = 1/2, which is a rational number, but not an integer. Hence not every element of Z has a (multiplicative) inverse.
#### Rationals
The desire for the existence of multiplicative inverses suggests considering fractions
Fraction (mathematics)
A fraction represents a part of a whole or, more generally, any number of equal parts. When spoken in everyday English, we specify how many parts of a certain size there are, for example, one-half, five-eighths and three-quarters.A common or "vulgar" fraction, such as 1/2, 5/8, 3/4, etc., consists...
$\frac{a}{b}.$
Fractions of integers (with b nonzero) are known as rational number
Rational number
In mathematics, a rational number is any number that can be expressed as the quotient or fraction a/b of two integers, with the denominator b not equal to zero. Since b may be equal to 1, every integer is a rational number...
s. The set of all such fractions is commonly denoted Q. There is still a minor obstacle for the rationals with multiplication, being a group: because the rational number 0
0 (number)
0 is both a numberand the numerical digit used to represent that number in numerals.It fulfills a central role in mathematics as the additive identity of the integers, real numbers, and many other algebraic structures. As a digit, 0 is used as a placeholder in place value systems...
does not have a multiplicative inverse (i.e., there is no x such that x · 0 = 1), (Q, ·) is still not a group.
However, the set of all nonzero rational numbers Q \ {0} = {q ∈ Q, q ≠ 0} does form an abelian group under multiplication, denoted . Associativity and identity element axioms follow from the properties of integers. The closure requirement still holds true after removing zero, because the product of two nonzero rationals is never zero. Finally, the inverse of a/b is b/a, therefore the axiom of the inverse element is satisfied.
The rational numbers (including 0) also form a group under addition. Intertwining addition and multiplication operations yields more complicated structures called rings and—if division is possible, such as in Q—fields, which occupy a central position in abstract algebra
Abstract algebra
Abstract algebra is the subject area of mathematics that studies algebraic structures, such as groups, rings, fields, modules, vector spaces, and algebras...
. Group theoretic arguments therefore underlie parts of the theory of those entities.
#### Nonzero integers modulo a prime
For any prime number
Prime number
A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. A natural number greater than 1 that is not a prime number is called a composite number. For example 5 is prime, as only 1 and 5 divide it, whereas 6 is composite, since it has the divisors 2...
p, modular arithmetic
Modular arithmetic
In mathematics, modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" after they reach a certain value—the modulus....
furnishes the multiplicative group of integers modulo p
Multiplicative group of integers modulo n
In modular arithmetic the set of congruence classes relatively prime to the modulus n form a group under multiplication called the multiplicative group of integers modulo n. It is also called the group of primitive residue classes modulo n. In the theory of rings, a branch of abstract algebra, it...
. Its elements are integers not divisible by p, considered modulo
Modular arithmetic
In mathematics, modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" after they reach a certain value—the modulus....
p, i.e. two numbers are considered equivalent if their difference is divisible by p. For example, if p = 5, there are exactly four group elements 1, 2, 3, 4: multiple
Multiple (mathematics)
In mathematics, a multiple is the product of any quantity and an integer. In other words, for the quantities a and b, we say that b is a multiple of a if b = na for some integer n , which is called the multiplier or coefficient. If a is not zero, this is equivalent to saying that b/a is an integer...
s of 5 are excluded and 6 and −4 are both equivalent to 1 etc. The group operation is given by multiplication. Therefore, 4 · 4 = 1, because the usual product 16 is equivalent to 1, for 5 divides 16 − 1 = 15, denoted
16 ≡ 1 (mod 5).
The primality of p ensures that the product of two integers neither of which is divisible by p is not divisible by p either, hence the indicated set of classes is closed under multiplication. The identity element is 1, as usual for a multiplicative group, and the associativity follows from the corresponding property of integers. Finally, the inverse element axiom requires that given an integer a not divisible by p, there exists an integer b such that
a · b ≡ 1 (mod p), i.e. p divides the difference .
The inverse b can be found by using Bézout's identity
Bézout's identity
In number theory, Bézout's identity for two integers a, b is an expressionwhere x and y are integers , such that d is a common divisor of a and b. Bézout's lemma states that such coefficients exist for every pair of nonzero integers...
and the fact that the greatest common divisor
Greatest common divisor
In mathematics, the greatest common divisor , also known as the greatest common factor , or highest common factor , of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder.For example, the GCD of 8 and 12 is 4.This notion can be extended to...
equals 1. In the case p = 5 above, the inverse of 4 is 4, and the inverse of 3 is 2, as 3 · 2 = 6 ≡ 1 (mod 5). Hence all group axioms are fulfilled. Actually, this example is similar to (Q\{0}, ·) above, because it turns out to be the multiplicative group of nonzero elements in the finite field Fp, denoted Fp×. These groups are crucial to public-key cryptography
Public-key cryptography
Public-key cryptography refers to a cryptographic system requiring two separate keys, one to lock or encrypt the plaintext, and one to unlock or decrypt the cyphertext. Neither key will do both functions. One of these keys is published or public and the other is kept private...
.
### Cyclic groups
A cyclic group is a group all of whose elements are powers (when the group operation is written additively, the term 'multiple' can be used) of a particular element a. In multiplicative notation, the elements of the group are:
..., a−3, a−2, a−1, a0 = e, a, a2, a3, ...,
where a2 means a • a, and a−3 stands for a−1 • a−1 • a−1=(a • a • a)−1 etc. Such an element a is called a generator or a primitive element
Primitive root modulo n
In modular arithmetic, a branch of number theory, a primitive root modulo n is any number g with the property that any number coprime to n is congruent to a power of g modulo n. In other words, g is a generator of the multiplicative group of integers modulo n...
of the group.
A typical example for this class of groups is the group of n-th complex roots of unity
Root of unity
In mathematics, a root of unity, or de Moivre number, is any complex number that equals 1 when raised to some integer power n. Roots of unity are used in many branches of mathematics, and are especially important in number theory, the theory of group characters, field theory, and the discrete...
, given by complex number
Complex number
A complex number is a number consisting of a real part and an imaginary part. Complex numbers extend the idea of the one-dimensional number line to the two-dimensional complex plane by using the number line for the real part and adding a vertical axis to plot the imaginary part...
s z satisfying zn = 1 (and whose operation is multiplication). Any cyclic group with n elements is isomorphic to this group. Using some field theory
Field theory (mathematics)
Field theory is a branch of mathematics which studies the properties of fields. A field is a mathematical entity for which addition, subtraction, multiplication and division are well-defined....
, the group Fp× can be shown to be cyclic: for example, if p = 5, 3 is a generator since 31 = 3, 32 = 9 ≡ 4, 33 ≡ 2, and 34 ≡ 1.
Some cyclic groups have an infinite number of elements. In these groups, for every non-zero element a, all the powers of a are distinct; despite the name "cyclic group", the powers of the elements do not cycle. An infinite cyclic group is isomorphic to (Z, +), the group of integers under addition introduced above. As these two prototypes are both abelian, so is any cyclic group.
The study of abelian groups is quite mature, including the fundamental theorem of finitely generated abelian groups; and reflecting this state of affairs, many group-related notions, such as center
Center (group theory)
In abstract algebra, the center of a group G, denoted Z,The notation Z is from German Zentrum, meaning "center". is the set of elements that commute with every element of G. In set-builder notation,...
and commutator
Commutator
In mathematics, the commutator gives an indication of the extent to which a certain binary operation fails to be commutative. There are different definitions used in group theory and ring theory.-Group theory:...
, describe the extent to which a given group is not abelian.
### Symmetry groups
Symmetry groups are groups consisting of symmetries
Symmetry
Symmetry generally conveys two primary meanings. The first is an imprecise sense of harmonious or aesthetically pleasing proportionality and balance; such that it reflects beauty or perfection...
of given mathematical objects—be they of geometric nature, such as the introductory symmetry group of the square, or of algebraic nature, such as polynomial equations and their solutions. Conceptually, group theory can be thought of as the study of symmetry. Symmetries in mathematics
Symmetry in mathematics
Symmetry occurs not only in geometry, but also in other branches of mathematics. It is actually the same as invariance: the property that something does not change under a set of transformations....
greatly simplify the study of geometrical
Geometry
Geometry arose as the field of knowledge dealing with spatial relationships. Geometry was one of the two fields of pre-modern mathematics, the other being the study of numbers ....
or analytical objects
Analysis
Analysis is the process of breaking a complex topic or substance into smaller parts to gain a better understanding of it. The technique has been applied in the study of mathematics and logic since before Aristotle , though analysis as a formal concept is a relatively recent development.The word is...
. A group is said to act
Group action
In algebra and geometry, a group action is a way of describing symmetries of objects using groups. The essential elements of the object are described by a set, and the symmetries of the object are described by the symmetry group of this set, which consists of bijective transformations of the set...
on another mathematical object X if every group element performs some operation on X compatibly to the group law. In the rightmost example below, an element of order 7 of the (2,3,7) triangle group
(2,3,7) triangle group
In the theory of Riemann surfaces and hyperbolic geometry, the triangle group is particularly important. This importance stems from its connection to Hurwitz surfaces, namely Riemann surfaces of genus g with the largest possible order, 84, of its automorphism group.A note on terminology – the "...
acts on the tiling by permuting the highlighted warped triangles (and the other ones, too). By a group action, the group pattern is connected to the structure of the object being acted on.
In chemical fields, such as crystallography
Crystallography
Crystallography is the experimental science of the arrangement of atoms in solids. The word "crystallography" derives from the Greek words crystallon = cold drop / frozen drop, with its meaning extending to all solids with some degree of transparency, and grapho = write.Before the development of...
, space group
Space group
In mathematics and geometry, a space group is a symmetry group, usually for three dimensions, that divides space into discrete repeatable domains.In three dimensions, there are 219 unique types, or counted as 230 if chiral copies are considered distinct...
s and point group
Point group
In geometry, a point group is a group of geometric symmetries that keep at least one point fixed. Point groups can exist in a Euclidean space with any dimension, and every point group in dimension d is a subgroup of the orthogonal group O...
s describe molecular symmetries
Molecular symmetry
Molecular symmetry in chemistry describes the symmetry present in molecules and the classification of molecules according to their symmetry. Molecular symmetry is a fundamental concept in chemistry, as it can predict or explain many of a molecule's chemical properties, such as its dipole moment...
and crystal symmetries. These symmetries underlie the chemical and physical behavior of these systems, and group theory enables simplification of quantum mechanical
Quantum mechanics
Quantum mechanics, also known as quantum physics or quantum theory, is a branch of physics providing a mathematical description of much of the dual particle-like and wave-like behavior and interactions of energy and matter. It departs from classical mechanics primarily at the atomic and subatomic...
analysis of these properties. For example, group theory is used to show that optical transitions between certain quantum levels cannot occur simply because of the symmetry of the states involved.
Not only are groups useful to assess the implications of symmetries in molecules, but surprisingly they also predict that molecules sometimes can change symmetry. The Jahn-Teller effect
Jahn-Teller effect
The Jahn–Teller effect, sometimes also known as Jahn–Teller distortion, or the Jahn–Teller theorem, describes the geometrical distortion of non-linear molecules under certain situations. This electronic effect is named after Hermann Arthur Jahn and Edward Teller, who proved, using group theory,...
is a distortion of a molecule of high symmetry when it adopts a particular ground state of lower symmetry from a set of possible ground states that are related to each other by the symmetry operations of the molecule.
Likewise, group theory helps predict the changes in physical properties that occur when a material undergoes a phase transition
Phase transition
A phase transition is the transformation of a thermodynamic system from one phase or state of matter to another.A phase of a thermodynamic system and the states of matter have uniform physical properties....
, for example, from a cubic to a tetrahedral crystalline form. An example is ferroelectric materials, where the change from a paraelectric to a ferroelectric state occurs at the Curie temperature and is related to a change from the high-symmetry paraelectric state to the lower symmetry ferroelectic state, accompanied by a so-called soft phonon
Phonon
In physics, a phonon is a collective excitation in a periodic, elastic arrangement of atoms or molecules in condensed matter, such as solids and some liquids...
mode, a vibrational lattice mode that goes to zero frequency at the transition.
Such spontaneous symmetry breaking
Spontaneous symmetry breaking
Spontaneous symmetry breaking is the process by which a system described in a theoretically symmetrical way ends up in an apparently asymmetric state....
has found further application in elementary particle physics, where its occurrence is related to the appearance of Goldstone boson
Goldstone boson
In particle and condensed matter physics, Goldstone bosons or Nambu–Goldstone bosons are bosons that appear necessarily in models exhibiting spontaneous breakdown of continuous symmetries...
s.
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| BuckminsterfullereneBuckminsterfullereneBuckminsterfullerene is a spherical fullerene molecule with the formula . It was first intentionally prepared in 1985 by Harold Kroto, James Heath, Sean O'Brien, Robert Curl and Richard Smalley at Rice University... displaysicosahedral symmetryIcosahedral symmetryA regular icosahedron has 60 rotational symmetries, and a symmetry order of 120 including transformations that combine a reflection and a rotation.... | AmmoniaAmmoniaAmmonia is a compound of nitrogen and hydrogen with the formula . It is a colourless gas with a characteristic pungent odour. Ammonia contributes significantly to the nutritional needs of terrestrial organisms by serving as a precursor to food and fertilizers. Ammonia, either directly or..., NNitrogenNitrogen is a chemical element that has the symbol N, atomic number of 7 and atomic mass 14.00674 u. Elemental nitrogen is a colorless, odorless, tasteless, and mostly inert diatomic gas at standard conditions, constituting 78.08% by volume of Earth's atmosphere...H3HydrogenHydrogen is the chemical element with atomic number 1. It is represented by the symbol H. With an average atomic weight of , hydrogen is the lightest and most abundant chemical element, constituting roughly 75% of the Universe's chemical elemental mass. Stars in the main sequence are mainly.... Its symmetry group is of order 6, generated by a 120° rotation and a reflection. | CubaneCubaneCubane is a synthetic hydrocarbon molecule that consists of eight carbon atoms arranged at the corners of a cube, with one hydrogen atom attached to each carbon atom. A solid crystalline substance, cubane is one of the Platonic hydrocarbons. It was first synthesized in 1964 by Philip Eaton, a... C8CarbonCarbon is the chemical element with symbol C and atomic number 6. As a member of group 14 on the periodic table, it is nonmetallic and tetravalent—making four electrons available to form covalent chemical bonds...H8HydrogenHydrogen is the chemical element with atomic number 1. It is represented by the symbol H. With an average atomic weight of , hydrogen is the lightest and most abundant chemical element, constituting roughly 75% of the Universe's chemical elemental mass. Stars in the main sequence are mainly... features octahedral symmetryOctahedral symmetry150px|thumb|right|The [[cube]] is the most common shape with octahedral symmetryA regular octahedron has 24 rotational symmetries, and a symmetry order of 48 including transformations that combine a reflection and a rotation.... | Hexaaquacopper(II) complex ion, [CuCopperCopper is a chemical element with the symbol Cu and atomic number 29. It is a ductile metal with very high thermal and electrical conductivity. Pure copper is soft and malleable; an exposed surface has a reddish-orange tarnish...(OOxygenOxygen is the element with atomic number 8 and represented by the symbol O. Its name derives from the Greek roots ὀξύς and -γενής , because at the time of naming, it was mistakenly thought that all acids required oxygen in their composition...H2)6]2+. Compared to a perfectly symmetrical shape, the molecule is vertically dilated by about 22% (Jahn-Teller effect). | The (2,3,7) triangle group, a hyperbolic group, acts on this tilingTessellationA tessellation or tiling of the plane is a pattern of plane figures that fills the plane with no overlaps and no gaps. One may also speak of tessellations of parts of the plane or of other surfaces. Generalizations to higher dimensions are also possible. Tessellations frequently appeared in the art... of the hyperbolicHyperbolic geometryIn mathematics, hyperbolic geometry is a non-Euclidean geometry, meaning that the parallel postulate of Euclidean geometry is replaced... plane. |
Finite symmetry groups such as the Mathieu group
Mathieu group
In the mathematical field of group theory, the Mathieu groups, named after the French mathematician Émile Léonard Mathieu, are five finite simple groups he discovered and reported in papers in 1861 and 1873; these were the first sporadic simple groups discovered...
s are used in coding theory
Coding theory
Coding theory is the study of the properties of codes and their fitness for a specific application. Codes are used for data compression, cryptography, error-correction and more recently also for network coding...
, which is in turn applied in error correction
Forward error correction
In telecommunication, information theory, and coding theory, forward error correction or channel coding is a technique used for controlling errors in data transmission over unreliable or noisy communication channels....
of transmitted data, and in CD players. Another application is differential Galois theory
Differential Galois theory
In mathematics, differential Galois theory studies the Galois groups of differential equations.Whereas algebraic Galois theory studies extensions of algebraic fields, differential Galois theory studies extensions of differential fields, i.e. fields that are equipped with a derivation, D. Much of...
, which characterizes functions having antiderivative
Antiderivative
In calculus, an "anti-derivative", antiderivative, primitive integral or indefinite integralof a function f is a function F whose derivative is equal to f, i.e., F ′ = f...
s of a prescribed form, giving group-theoretic criteria for when solutions of certain differential equation
Differential equation
A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders...
s are well-behaved. Geometric properties that remain stable under group actions are investigated in (geometric)
Geometric invariant theory
In mathematics Geometric invariant theory is a method for constructing quotients by group actions in algebraic geometry, used to construct moduli spaces...
invariant theory
Invariant theory
Invariant theory is a branch of abstract algebra dealing with actions of groups on algebraic varieties from the point of view of their effect on functions...
.
### General linear group and representation theory
Matrix group
Matrix group
In mathematics, a matrix group is a group G consisting of invertible matrices over some field K, usually fixed in advance, with operations of matrix multiplication and inversion. More generally, one can consider n × n matrices over a commutative ring R...
s consist of matrices
Matrix (mathematics)
In mathematics, a matrix is a rectangular array of numbers, symbols, or expressions. The individual items in a matrix are called its elements or entries. An example of a matrix with six elements isMatrices of the same size can be added or subtracted element by element...
together with matrix multiplication
Matrix multiplication
In mathematics, matrix multiplication is a binary operation that takes a pair of matrices, and produces another matrix. If A is an n-by-m matrix and B is an m-by-p matrix, the result AB of their multiplication is an n-by-p matrix defined only if the number of columns m of the left matrix A is the...
. The general linear group GL(n, R) consists of all invertible n-by-n matrices with real
Real number
In mathematics, a real number is a value that represents a quantity along a continuum, such as -5 , 4/3 , 8.6 , √2 and π...
entries. Its subgroups are referred to as matrix groups or linear groups. The dihedral group example mentioned above can be viewed as a (very small) matrix group. Another important matrix group is the special orthogonal group SO(n). It describes all possible rotations in n dimensions. Via Euler angles
Euler angles
The Euler angles are three angles introduced by Leonhard Euler to describe the orientation of a rigid body. To describe such an orientation in 3-dimensional Euclidean space three parameters are required...
, rotation matrices are used in computer graphics
Computer graphics
Computer graphics are graphics created using computers and, more generally, the representation and manipulation of image data by a computer with help from specialized software and hardware....
.
Representation theory is both an application of the group concept and important for a deeper understanding of groups. It studies the group by its group action
Group action
In algebra and geometry, a group action is a way of describing symmetries of objects using groups. The essential elements of the object are described by a set, and the symmetries of the object are described by the symmetry group of this set, which consists of bijective transformations of the set...
s on other spaces. A broad class of group representation
Group representation
In the mathematical field of representation theory, group representations describe abstract groups in terms of linear transformations of vector spaces; in particular, they can be used to represent group elements as matrices so that the group operation can be represented by matrix multiplication...
s are linear representations, i.e. the group is acting on a vector space
Vector space
A vector space is a mathematical structure formed by a collection of vectors: objects that may be added together and multiplied by numbers, called scalars in this context. Scalars are often taken to be real numbers, but one may also consider vector spaces with scalar multiplication by complex...
, such as the three-dimensional Euclidean space
Euclidean space
In mathematics, Euclidean space is the Euclidean plane and three-dimensional space of Euclidean geometry, as well as the generalizations of these notions to higher dimensions...
R3. A representation of G on an n-dimension
Dimension
In physics and mathematics, the dimension of a space or object is informally defined as the minimum number of coordinates needed to specify any point within it. Thus a line has a dimension of one because only one coordinate is needed to specify a point on it...
al real vector space is simply a group homomorphism
ρ: G → GL(n, R)
from the group to the general linear group. This way, the group operation, which may be abstractly given, translates to the multiplication of matrices making it accessible to explicit computations.
Given a group action, this gives further means to study the object being acted on. On the other hand, it also yields information about the group. Group representations are an organizing principle in the theory of finite groups, Lie groups, algebraic group
Algebraic group
In algebraic geometry, an algebraic group is a group that is an algebraic variety, such that the multiplication and inverse are given by regular functions on the variety...
s and topological group
Topological group
In mathematics, a topological group is a group G together with a topology on G such that the group's binary operation and the group's inverse function are continuous functions with respect to the topology. A topological group is a mathematical object with both an algebraic structure and a...
s, especially (locally) compact group
Compact group
In mathematics, a compact group is a topological group whose topology is compact. Compact groups are a natural generalisation of finite groups with the discrete topology and have properties that carry over in significant fashion...
s.
### Galois groups
Galois groups have been developed to help solve polynomial equations by capturing their symmetry features. For example, the solutions of the quadratic equation
Quadratic equation
In mathematics, a quadratic equation is a univariate polynomial equation of the second degree. A general quadratic equation can be written in the formax^2+bx+c=0,\,...
ax2 + bx + c = 0 are given by
$x = \frac{-b \pm \sqrt {b^2-4ac}}{2a}.$
Exchanging "+" and "−" in the expression, i.e. permuting the two solutions of the equation can be viewed as a (very simple) group operation. Similar formulae are known for cubic and quartic equations, but do not exist in general for degree 5
Quintic equation
In mathematics, a quintic function is a function of the formg=ax^5+bx^4+cx^3+dx^2+ex+f,\,where a, b, c, d, e and f are members of a field, typically the rational numbers, the real numbers or the complex numbers, and a is nonzero...
and higher. Abstract properties of Galois groups associated with polynomials (in particular their solvability
Solvable group
In mathematics, more specifically in the field of group theory, a solvable group is a group that can be constructed from abelian groups using extensions...
) give a criterion for polynomials that have all their solutions expressible by radicals, i.e. solutions expressible using solely addition, multiplication, and roots
Nth root
In mathematics, the nth root of a number x is a number r which, when raised to the power of n, equals xr^n = x,where n is the degree of the root...
similar to the formula above.
The problem can be dealt with by shifting to field theory
Field theory (mathematics)
Field theory is a branch of mathematics which studies the properties of fields. A field is a mathematical entity for which addition, subtraction, multiplication and division are well-defined....
and considering the splitting field
Splitting field
In abstract algebra, a splitting field of a polynomial with coefficients in a field is a smallest field extension of that field over which the polynomial factors into linear factors.-Definition:...
of a polynomial. Modern Galois theory
Galois theory
In mathematics, more specifically in abstract algebra, Galois theory, named after Évariste Galois, provides a connection between field theory and group theory...
generalizes the above type of Galois groups to field extension
Field extension
In abstract algebra, field extensions are the main object of study in field theory. The general idea is to start with a base field and construct in some manner a larger field which contains the base field and satisfies additional properties...
s and establishes—via the fundamental theorem of Galois theory
Fundamental theorem of Galois theory
In mathematics, the fundamental theorem of Galois theory is a result that describes the structure of certain types of field extensions.In its most basic form, the theorem asserts that given a field extension E /F which is finite and Galois, there is a one-to-one correspondence between its...
—a precise relationship between fields and groups, underlining once again the ubiquity of groups in mathematics.
## Finite groups
A group is called finite if it has a finite number of elements. The number of elements is called the order
Order (group theory)
In group theory, a branch of mathematics, the term order is used in two closely related senses:* The order of a group is its cardinality, i.e., the number of its elements....
of the group. An important class is the symmetric group
Symmetric group
In mathematics, the symmetric group Sn on a finite set of n symbols is the group whose elements are all the permutations of the n symbols, and whose group operation is the composition of such permutations, which are treated as bijective functions from the set of symbols to itself...
s
SN, the groups of permutation
Permutation
In mathematics, the notion of permutation is used with several slightly different meanings, all related to the act of permuting objects or values. Informally, a permutation of a set of objects is an arrangement of those objects into a particular order...
s of N letters. For example, the symmetric group on 3 letters S3
Dihedral group of order 6
The smallest non-abelian group has 6 elements. It is a dihedral group with notation D3 and the symmetric group of degree 3, with notation S3....
is the group consisting of all possible orderings of the three letters ABC, i.e. contains the elements ABC, ACB, ..., up to CBA, in total 6 (or 3 factorial
Factorial
In mathematics, the factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n...
) elements. This class is fundamental insofar as any finite group can be expressed as a subgroup of a symmetric group SN for a suitable integer N (Cayley's theorem
Cayley's theorem
In group theory, Cayley's theorem, named in honor of Arthur Cayley, states that every group G is isomorphic to a subgroup of the symmetric group acting on G...
). Parallel to the group of symmetries of the square above, S3 can also be interpreted as the group of symmetries of an equilateral triangle.
The order of an element a in a group G is the least positive integer n such that a n = e, where a n represents
i.e. application of the operation • to n copies of a. (If • represents multiplication, then an corresponds to the nth power of a.) In infinite groups, such an n may not exist, in which case the order of a is said to be infinity. The order of an element equals the order of the cyclic subgroup generated by this element.
More sophisticated counting techniques, for example counting cosets, yield more precise statements about finite groups: Lagrange's Theorem
Lagrange's theorem (group theory)
Lagrange's theorem, in the mathematics of group theory, states that for any finite group G, the order of every subgroup H of G divides the order of G. The theorem is named after Joseph Lagrange....
states that for a finite group G the order of any finite subgroup H divides
Divisor
In mathematics, a divisor of an integer n, also called a factor of n, is an integer which divides n without leaving a remainder.-Explanation:...
the order of G. The Sylow theorems give a partial converse.
The dihedral group
Dihedral group
In mathematics, a dihedral group is the group of symmetries of a regular polygon, including both rotations and reflections. Dihedral groups are among the simplest examples of finite groups, and they play an important role in group theory, geometry, and chemistry.See also: Dihedral symmetry in three...
(discussed above) is a finite group of order 8. The order of r1 is 4, as is the order of the subgroup R it generates (see above). The order of the reflection elements fv etc. is 2. Both orders divide 8, as predicted by Lagrange's Theorem. The groups Fp× above have order .
### Classification of finite simple groups
Mathematicians often strive for a complete classification (or list) of a mathematical notion. In the context of finite groups, this aim quickly leads to difficult and profound mathematics. According to Lagrange's theorem, finite groups of order p, a prime number, are necessarily cyclic (abelian) groups Zp. Groups of order p2 can also be shown to be abelian, a statement which does not generalize to order p3, as the non-abelian group D4 of order 8 = 23 above shows. Computer algebra system
Computer algebra system
A computer algebra system is a software program that facilitates symbolic mathematics. The core functionality of a CAS is manipulation of mathematical expressions in symbolic form.-Symbolic manipulations:...
s can be used to list small groups, but there is no classification of all finite groups. An intermediate step is the classification of finite simple groups. A nontrivial group is called simple
Simple group
In mathematics, a simple group is a nontrivial group whose only normal subgroups are the trivial group and the group itself. A group that is not simple can be broken into two smaller groups, a normal subgroup and the quotient group, and the process can be repeated...
if its only normal subgroups are the trivial group
Trivial group
In mathematics, a trivial group is a group consisting of a single element. All such groups are isomorphic so one often speaks of the trivial group. The single element of the trivial group is the identity element so it usually denoted as such, 0, 1 or e depending on the context...
and the group itself. The Jordan–Hölder theorem exhibits finite simple groups as the building blocks for all finite groups. Listing all finite simple groups was a major achievement in contemporary group theory. 1998 Fields Medal
Fields Medal
The Fields Medal, officially known as International Medal for Outstanding Discoveries in Mathematics, is a prize awarded to two, three, or four mathematicians not over 40 years of age at each International Congress of the International Mathematical Union , a meeting that takes place every four...
winner Richard Borcherds
Richard Borcherds
Richard Ewen Borcherds is a British mathematician specializing in lattices, number theory, group theory, and infinite-dimensional algebras. He was awarded the Fields Medal in 1998.- Personal life :...
succeeded to prove the monstrous moonshine
Monstrous moonshine
In mathematics, monstrous moonshine, or moonshine theory, is a term devised by John Horton Conway and Simon P. Norton in 1979, used to describe the connection between the monster group M and modular functions .- History :Specifically, Conway and Norton, following an initial observationby John...
conjectures, a surprising and deep relation of the largest finite simple sporadic group
Sporadic group
In the mathematical field of group theory, a sporadic group is one of the 26 exceptional groups in the classification of finite simple groups. A simple group is a group G that does not have any normal subgroups except for the subgroup consisting only of the identity element, and G itself...
—the "monster group
Monster group
In the mathematical field of group theory, the Monster group M or F1 is a group of finite order:...
"—with certain modular functions, a piece of classical complex analysis
Complex analysis
Complex analysis, traditionally known as the theory of functions of a complex variable, is the branch of mathematical analysis that investigates functions of complex numbers. It is useful in many branches of mathematics, including number theory and applied mathematics; as well as in physics,...
, and string theory
String theory
String theory is an active research framework in particle physics that attempts to reconcile quantum mechanics and general relativity. It is a contender for a theory of everything , a manner of describing the known fundamental forces and matter in a mathematically complete system...
, a theory supposed to unify the description of many physical phenomena.
## Groups with additional structure
Many groups are simultaneously groups and examples of other mathematical structures. In the language of category theory
Category theory
Category theory is an area of study in mathematics that examines in an abstract way the properties of particular mathematical concepts, by formalising them as collections of objects and arrows , where these collections satisfy certain basic conditions...
, they are group object
Group object
In category theory, a branch of mathematics, group objects are certain generalizations of groups which are built on more complicated structures than sets...
s in a category
Category (mathematics)
In mathematics, a category is an algebraic structure that comprises "objects" that are linked by "arrows". A category has two basic properties: the ability to compose the arrows associatively and the existence of an identity arrow for each object. A simple example is the category of sets, whose...
, meaning that they are objects (that is, examples of another mathematical structure) which come with transformations (called morphism
Morphism
In mathematics, a morphism is an abstraction derived from structure-preserving mappings between two mathematical structures. The notion of morphism recurs in much of contemporary mathematics...
s) that mimic the group axioms. For example, every group (as defined above) is also a set, so a group is a group object in the category of sets
Category of sets
In the mathematical field of category theory, the category of sets, denoted as Set, is the category whose objects are sets. The arrows or morphisms between sets A and B are all functions from A to B...
.
### Topological groups
Some topological space
Topological space
Topological spaces are mathematical structures that allow the formal definition of concepts such as convergence, connectedness, and continuity. They appear in virtually every branch of modern mathematics and are a central unifying notion...
s may be endowed with a group law. In order for the group law and the topology to interweave well, the group operations must be continuous function
Continuous function
In mathematics, a continuous function is a function for which, intuitively, "small" changes in the input result in "small" changes in the output. Otherwise, a function is said to be "discontinuous". A continuous function with a continuous inverse function is called "bicontinuous".Continuity of...
s, that is, and g−1 must not vary wildly if g and h vary only little. Such groups are called topological groups, and they are the group objects in the category of topological spaces
Category of topological spaces
In mathematics, the category of topological spaces, often denoted Top, is the category whose objects are topological spaces and whose morphisms are continuous maps. This is a category because the composition of two continuous maps is again continuous...
. The most basic examples are the reals
Real number
In mathematics, a real number is a value that represents a quantity along a continuum, such as -5 , 4/3 , 8.6 , √2 and π...
R under addition, , and similarly with any other topological field such as the complex number
Complex number
A complex number is a number consisting of a real part and an imaginary part. Complex numbers extend the idea of the one-dimensional number line to the two-dimensional complex plane by using the number line for the real part and adding a vertical axis to plot the imaginary part...
s or p-adic numbers
P-adic number
In mathematics, and chiefly number theory, the p-adic number system for any prime number p extends the ordinary arithmetic of the rational numbers in a way different from the extension of the rational number system to the real and complex number systems...
. All of these groups are locally compact, so they have Haar measure
Haar measure
In mathematical analysis, the Haar measure is a way to assign an "invariant volume" to subsets of locally compact topological groups and subsequently define an integral for functions on those groups....
s and can be studied via harmonic analysis
Harmonic analysis
Harmonic analysis is the branch of mathematics that studies the representation of functions or signals as the superposition of basic waves. It investigates and generalizes the notions of Fourier series and Fourier transforms...
. The former offer an abstract formalism of invariant integral
Integral
Integration is an important concept in mathematics and, together with its inverse, differentiation, is one of the two main operations in calculus...
s. Invariance
Invariant (mathematics)
In mathematics, an invariant is a property of a class of mathematical objects that remains unchanged when transformations of a certain type are applied to the objects. The particular class of objects and type of transformations are usually indicated by the context in which the term is used...
means, in the case of real numbers for example:
for any constant c. Matrix groups over these fields fall under this regime, as do adele ring
Adele ring
In algebraic number theory and topological algebra, the adele ring is a topological ring which is built on the field of rational numbers . It involves all the completions of the field....
s and adelic algebraic group
Adelic algebraic group
In abstract algebra, an adelic algebraic group is a topological group defined by an algebraic group G over a number field K, and the adele ring A = A of K. It consists of the points of G having values in A; the definition of the appropriate topology is straightforward only in case G is a linear...
s, which are basic to number theory. Galois groups of infinite field extensions such as the absolute Galois group
Absolute Galois group
In mathematics, the absolute Galois group GK of a field K is the Galois group of Ksep over K, where Ksep is a separable closure of K. Alternatively it is the group of all automorphisms of the algebraic closure of K that fix K. The absolute Galois group is unique up to isomorphism...
can also be equipped with a topology, the so-called Krull topology, which in turn is central to generalize the above sketched connection of fields and groups to infinite field extensions. An advanced generalization of this idea, adapted to the needs of algebraic geometry
Algebraic geometry
Algebraic geometry is a branch of mathematics which combines techniques of abstract algebra, especially commutative algebra, with the language and the problems of geometry. It occupies a central place in modern mathematics and has multiple conceptual connections with such diverse fields as complex...
, is the étale fundamental group
Étale fundamental group
The étale fundamental group is an analogue in algebraic geometry, for schemes, of the usual fundamental group of topological spaces.-Topological analogue:In algebraic topology, the fundamental group\pi_1...
.
### Lie groups
Lie groups (in honor of Sophus Lie
Sophus Lie
Marius Sophus Lie was a Norwegian mathematician. He largely created the theory of continuous symmetry, and applied it to the study of geometry and differential equations.- Biography :...
) are groups which also have a manifold
Manifold
In mathematics , a manifold is a topological space that on a small enough scale resembles the Euclidean space of a specific dimension, called the dimension of the manifold....
structure, i.e. they are spaces looking locally like
Diffeomorphism
In mathematics, a diffeomorphism is an isomorphism in the category of smooth manifolds. It is an invertible function that maps one differentiable manifold to another, such that both the function and its inverse are smooth.- Definition :...
some Euclidean space
Euclidean space
In mathematics, Euclidean space is the Euclidean plane and three-dimensional space of Euclidean geometry, as well as the generalizations of these notions to higher dimensions...
of the appropriate dimension
Dimension
In physics and mathematics, the dimension of a space or object is informally defined as the minimum number of coordinates needed to specify any point within it. Thus a line has a dimension of one because only one coordinate is needed to specify a point on it...
. Again, the additional structure, here the manifold structure, has to be compatible, i.e. the maps corresponding to multiplication and the inverse have to be smooth.
A standard example is the general linear group introduced above: it is an open subset of the space of all n-by-n matrices, because it is given by the inequality
det (A) ≠ 0,
where A denotes an n-by-n matrix.
Lie groups are of fundamental importance in physics: Noether's theorem
Noether's theorem
Noether's theorem states that any differentiable symmetry of the action of a physical system has a corresponding conservation law. The theorem was proved by German mathematician Emmy Noether in 1915 and published in 1918...
links continuous symmetries to conserved quantities. Rotation
Rotation
A rotation is a circular movement of an object around a center of rotation. A three-dimensional object rotates always around an imaginary line called a rotation axis. If the axis is within the body, and passes through its center of mass the body is said to rotate upon itself, or spin. A rotation...
, as well as translations
Translation (geometry)
In Euclidean geometry, a translation moves every point a constant distance in a specified direction. A translation can be described as a rigid motion, other rigid motions include rotations and reflections. A translation can also be interpreted as the addition of a constant vector to every point, or...
in space
Space
Space is the boundless, three-dimensional extent in which objects and events occur and have relative position and direction. Physical space is often conceived in three linear dimensions, although modern physicists usually consider it, with time, to be part of a boundless four-dimensional continuum...
and time
Time
Time is a part of the measuring system used to sequence events, to compare the durations of events and the intervals between them, and to quantify rates of change such as the motions of objects....
are basic symmetries of the laws of mechanics
Mechanics
Mechanics is the branch of physics concerned with the behavior of physical bodies when subjected to forces or displacements, and the subsequent effects of the bodies on their environment....
. They can, for instance, be used to construct simple models—imposing, say, axial symmetry on a situation will typically lead to significant simplification in the equations one needs to solve to provide a physical description. Another example are the Lorentz transformation
Lorentz transformation
In physics, the Lorentz transformation or Lorentz-Fitzgerald transformation describes how, according to the theory of special relativity, two observers' varying measurements of space and time can be converted into each other's frames of reference. It is named after the Dutch physicist Hendrik...
s, which relate measurements of time and velocity of two observers in motion relative to each other. They can be deduced in a purely group-theoretical way, by expressing the transformations as a rotational symmetry of Minkowski space
Minkowski space
In physics and mathematics, Minkowski space or Minkowski spacetime is the mathematical setting in which Einstein's theory of special relativity is most conveniently formulated...
. The latter serves—in the absence of significant gravitation
Gravitation
Gravitation, or gravity, is a natural phenomenon by which physical bodies attract with a force proportional to their mass. Gravitation is most familiar as the agent that gives weight to objects with mass and causes them to fall to the ground when dropped...
—as a model of space time in special relativity
Special relativity
Special relativity is the physical theory of measurement in an inertial frame of reference proposed in 1905 by Albert Einstein in the paper "On the Electrodynamics of Moving Bodies".It generalizes Galileo's...
. The full symmetry group of Minkowski space, i.e. including translations, is known as the Poincaré group
Poincaré group
In physics and mathematics, the Poincaré group, named after Henri Poincaré, is the group of isometries of Minkowski spacetime.-Simple explanation:...
. By the above, it plays a pivotal role in special relativity and, by implication, for quantum field theories. Symmetries that vary with location
Local symmetry
In physics, a local symmetry is symmetry of some physical quantity, which smoothly depends on the point of the base manifold. Such quantities can be for example an observable, a tensor or the Lagrangian of a theory....
are central to the modern description of physical interactions with the help of gauge theory
Gauge theory
In physics, gauge invariance is the property of a field theory in which different configurations of the underlying fundamental but unobservable fields result in identical observable quantities. A theory with such a property is called a gauge theory...
.
## Generalizations
In abstract algebra
Abstract algebra
Abstract algebra is the subject area of mathematics that studies algebraic structures, such as groups, rings, fields, modules, vector spaces, and algebras...
, more general structures are defined by relaxing some of the axioms defining a group. For example, if the requirement that every element has an inverse is eliminated, the resulting algebraic structure is called a monoid
Monoid
In abstract algebra, a branch of mathematics, a monoid is an algebraic structure with a single associative binary operation and an identity element. Monoids are studied in semigroup theory as they are naturally semigroups with identity. Monoids occur in several branches of mathematics; for...
. The natural number
Natural number
In mathematics, the natural numbers are the ordinary whole numbers used for counting and ordering . These purposes are related to the linguistic notions of cardinal and ordinal numbers, respectively...
s N (including 0) under addition form a monoid, as do the nonzero integers under multiplication , see above. There is a general method to formally add inverses to elements to any (abelian) monoid, much the same way as is derived from , known as the Grothendieck group
Grothendieck group
In mathematics, the Grothendieck group construction in abstract algebra constructs an abelian group from a commutative monoid in the best possible way...
.
Groupoid
Groupoid
In mathematics, especially in category theory and homotopy theory, a groupoid generalises the notion of group in several equivalent ways. A groupoid can be seen as a:...
s are similar to groups except that the composition a • b need not be defined for all a and b. They arise in the study of more complicated forms of symmetry, often in topological
Topology
Topology is a major area of mathematics concerned with properties that are preserved under continuous deformations of objects, such as deformations that involve stretching, but no tearing or gluing...
and analytical
Mathematical analysis
Mathematical analysis, which mathematicians refer to simply as analysis, has its beginnings in the rigorous formulation of infinitesimal calculus. It is a branch of pure mathematics that includes the theories of differentiation, integration and measure, limits, infinite series, and analytic functions...
structures, such as the fundamental groupoid or stacks
Stack (descent theory)
In mathematics a stack is a concept used to formalise some of the main constructions of descent theory.Descent theory is concerned with generalisations of situations where geometrical objects can be "glued together" when they are isomorphic when restricted to intersections of the sets in an open...
. Finally, it is possible to generalize any of these concepts by replacing the binary operation with an arbitrary n-ary
Arity
In logic, mathematics, and computer science, the arity of a function or operation is the number of arguments or operands that the function takes. The arity of a relation is the dimension of the domain in the corresponding Cartesian product...
one (i.e. an operation taking n arguments). With the proper generalization of the group axioms this gives rise to an n-ary group
N-ary group
In mathematics, an n-ary group is a generalization of a group to a set G with a n-ary operation instead of a binary operation. The axioms for an n-ary group are defined in such a way as to reduce to those of a group in the case .-Associativity:The easiest axiom to generalize is the associative law...
. The table gives a list of several structures generalizing groups.
## See also
• Abelian group
Abelian group
In abstract algebra, an abelian group, also called a commutative group, is a group in which the result of applying the group operation to two group elements does not depend on their order . Abelian groups generalize the arithmetic of addition of integers...
• Group ring
Group ring
In algebra, a group ring is a free module and at the same time a ring, constructed in a natural way from any given ring and any given group. As a free module, its ring of scalars is the given ring and its basis is one-to-one with the given group. As a ring, its addition law is that of the free...
• Group algebra
Group algebra
In mathematics, the group algebra is any of various constructions to assign to a locally compact group an operator algebra , such that representations of the algebra are related to representations of the group...
• Euclidean group
Euclidean group
In mathematics, the Euclidean group E, sometimes called ISO or similar, is the symmetry group of n-dimensional Euclidean space...
In mathematics, a group G is called free if there is a subset S of G such that any element of G can be written in one and only one way as a product of finitely many elements of S and their inverses...
• Finitely presented group
• Fundamental group
Fundamental group
In mathematics, more specifically algebraic topology, the fundamental group is a group associated to any given pointed topological space that provides a way of determining when two paths, starting and ending at a fixed base point, can be continuously deformed into each other...
• Non-abelian group
• Grothendieck group
Grothendieck group
In mathematics, the Grothendieck group construction in abstract algebra constructs an abelian group from a commutative monoid in the best possible way...
• Symmetry in physics
Symmetry in physics
In physics, symmetry includes all features of a physical system that exhibit the property of symmetry—that is, under certain transformations, aspects of these systems are "unchanged", according to a particular observation...
### General references
, Chapter 2 contains an undergraduate-level exposition of the notions covered in this article., Chapter 5 provides a layman-accessible explanation of groups.., an elementary introduction.......
### Special references
........ ... ........................
### Historical references
... . (Galois work was first published by Joseph Liouville
Joseph Liouville
- Life and work :Liouville graduated from the École Polytechnique in 1827. After some years as an assistant at various institutions including the Ecole Centrale Paris, he was appointed as professor at the École Polytechnique in 1838...
in 1843). .. ...
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http://mathhelpforum.com/number-theory/171644-multiplication-natural-numbers-proof.html
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# Thread:
1. ## Multiplication of Natural Numbers proof
Let m be an element of the Natural Numbers and n be an element of the Integers. If m*n is an element of the Natural Numbers, then n is an element of the Natural Numbers.
Any help with the above proof would be appreciated!
2. Originally Posted by jstarks44444
Let m be an element of the Natural Numbers and n be an element of the Integers. If m*n is an element of the Natural Numbers, then n is an element of the Natural Numbers.
One again, if you would post a list of axioms, definitions, and theorems you would receive better help.
Not having a such here a guess based upon the definition of $\mathbb{N}$ you posted before.
Suppose that $n\notin\mathbb{N}$
Based on the given we know that $n\in \mathbb{Z}$ so from the axiom $n=0\text{ or }-n\in \mathbb{N}$.
You have proven that $j>0$ for all $j\in\mathbb{N}$.
What would that say about $mn~?$
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http://mathoverflow.net/questions/117684?sort=newest
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## Are spectra really the same as cohomology theories?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $E \to F$ be a morphism of cohomology theories defined on finite CW complexes. Then by Brown representability, $E, F$ are represented by spectra, and the map $E \to F$ comes from a map of spectra. However, it is possible that the map on cohomology theories is zero while the map of spectra is not nullhomotopic. In other words, the homotopy category of spectra does not imbed faithfully into the category of cohomology theories on finite CW complexes. This is due to the existence of phantom maps:
Let $f: X \to Y$ be a map of spectra. It is possible that $f$ is not nullhomotopic even if for every finite spectrum $F$ and map $F \to X$, the composite $F \to X \stackrel{f}{\to} Y$ is nullhomotopic. Such maps are called phantom maps. For an explicit example, let $S^0_{\mathbb{Q}} = H\mathbb{Q}$ be the rational sphere. This is obtained as a filtered (homotopy) colimit of copies of $S^0$ and multiplication by $m$ maps. The universal coefficient theorem shows that there are nontrivial maps $S^0_{\mathbb{Q}} \to H \mathbb{Z}[1]$; in fact they are parametrized by $\mathrm{Ext}^1(\mathbb{Q}, \mathbb{Z}) \neq 0$. However, these restrict to zero on any of the terms in the filtered colimit (each of which is a copy of $S^0$).
In other words, the distinction between flat and projective modules is in some sense an algebraic analog of the existence of phantom maps. Given a flat non-projective module $M$ over some ring $R$, then there is a nontrivial map in the derived category $M \to N[1]$ for some module $N$. Now $M$ is a filtered colimit of finitely generated projectives -- Lazard's theorem -- and the map $M \to N[1]$ is "phantom" in that it restricts to zero on each of these finitely generated projectives (or more generally for any compact object mapping to $M$). So it should not be too surprising that phantom maps of spectra exist and are interesting.
Now spectra are analogous to the derived category of $R$-modules, but spectra also come with another adjunction: $$\Sigma^\infty, Omega^\infty: \mathcal{S}_* \leftrightarrows \mathcal{Sp}$$ between pointed spaces and spectra. They thus come with another distinguished class of objects, the suspension spectra. (Random question: what is the analog of a suspension spectrum in algebra?)
Definition: A map of spectra $X \to Y$ is hyperphantom if for any suspension spectrum $T$ (let's interpret that loosely to include desuspensions of suspension spectra), $T \to X \to Y$ is nullhomotopic.
In other words, a map of spectra is hyperphantom if the induced natural transformation on cohomology theories of spaces (not necessarily finite CW ones!) is zero.
Is it true that a hyperphantom map is nullhomotopic? Rudyak lists this as an open problem in "On Thom spectra, orientability, and cobordism." What is the state of this problem?
-
Nice question! Nothing substantive to say, but regarding your random side-question: wouldn't the natural analog of the suspension spectrum of a space in algebra be the free abelian group (or free R-module) on a set? – David Ben-Zvi Dec 31 at 3:20
@David: I'd say it would be a bounded below chain complex. Anyway, any answer to this question is valid I guess ;-) – Fernando Muro Dec 31 at 3:31
Perhaps, although it's not really clear to me why sets (in relation to the derived category of $R$-modules) should take the place of "spaces." The $\infty$-category of spectra can be described as the stabilization of the category of spaces, and the derived category of $R$-modules (if $R$ is commutative) can be described as the stabilization of $E_\infty$-algebras over and under $A$. That's a candidate, although $E_\infty$-algebras over and under the sphere don't seem to have much to do with spaces. – Akhil Mathew Dec 31 at 3:59
@Akhil - Sets take the place of spaces in relation to the abelian category of R-modules. For the derived category spaces (or simplicial sets) are still the analog, and your adjunction and Fernando's (the derived version of mine) are literally the same --- both special cases of an adjunction (suspension spectra for the sphere, Dold-Kan for a discrete ring) between spaces and R-module spectra for any $E_\infty$ ring spectrum. – David Ben-Zvi Dec 31 at 16:40
@David: OK, agreed. – Akhil Mathew Dec 31 at 21:28
## 2 Answers
Consider the periodic complex $K$-theory spectrum $KU$. The integral homology group $H_i(KU)$, the direct limit of
$$\dots \to H_{2n+i}(BU)\to H_{2n+2+i}(BU)\to\dots,$$ is a one-dimensional rational vector space if $i$ is even and trivial if $i$ is odd. It follows that $H^1(KU)$ is nontrivial. (It's $Ext(\mathbb Q,\mathbb Z)$.) But this can't be detected in the cohomology of suspension spectra, because $H^{2n+1}(BU)$ is trivial.
So that's an example of a "hyperphantom" map from $KU$ to the Eilenberg-MacLane spectrum $\Sigma H\mathbb Z$.
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1
This is very neat! This says that in some sense there are characteristic classes of stable vector bundles that vanish on any space but not on spectra. – Eric Wofsey Jan 1 at 8:55
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Nice, I should have remembered that. Notice how classical it is. – Peter May Jan 1 at 13:33
Thanks! This is very nice. – Akhil Mathew Jan 1 at 13:59
3
The integral homology of $KU$ is a favorite example, but I hadn't realized it was example of this phenomenon until I thought a while. – Tom Goodwillie Jan 1 at 15:54
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The answer to this question is in LMS (I.6.9 of http://www.math.uchicago.edu/~may/BOOKS/equi.pdf) and in McClure's contribution to BMMS (VII\S1 of http://www.math.uchicago.edu/~may/BOOKS/h_infty.pdf), which gives full details. Let $T = {T_i}$ be a prespectrum. There is a cylinder construction $ZT$ that gives a weakly equivalent $\Omega$-spectrum (I first defined it in 1968 http://www.math.uchicago.edu/~may/PAPERS/7.pdf).
It is the telescope of the desuspensions $\Sigma^{-i} \Sigma^{\infty} T_i$. There is no loss of generality in taking $X=ZT$ in your question. There results a $lim^{1}$ exact sequence of the form
$$0 \to lim^{1}[\Sigma^{1-i} \Sigma^{\infty} T_i,Y] \to [X,Y] \to \lim[\Sigma^{-i} \Sigma^{\infty} T_i,Y]\to 0.$$
This has nothing to do with finite CW spectra, a priori, and it can be viewed via the usual adjunctions as giving a precise measure of the difference between the stable homotopy category of spectra and the homotopy category of based spaces. McClure (VII\S4 op cit) gives a clear criterion for when the $lim^{1}$ term vanishes and examples where the criterion holds. It is obviously not to be expected that the $lim^{1}$ term vanishes in general. One way to construct counter-examples is to relate this $lim^{1}$ exact sequence with the one given by approximating $X$ by a CW-spectrum, but I'll leave that to the interested reader. Of course, the elements of this $lim^{1}$ term are your hyperphantom maps.
Added as an edit: In answer to Tom Goodwillie's comment, the adjunctions I referred to give that if $Y$ is an $\Omega$-spectrum with $i$th space $Y_i$, then $$[\Sigma^{-i} \Sigma^{\infty} T_i,Y] \cong [T_i,Y_i].$$ The brackets refer to spectra on left and based spaces on the right. Therefore the original $lim^{1}$ exact sequence can be rewritten as $$0 \to lim^{1}[T_i,Y_{i-1}] \to [X,Y] \to \lim[T_i,Y_i]\to 0.$$ The $lim$ and $lim^1$ terms are computed in terms of homotopy classes of maps of based spaces. That is what I had in mind with my sloppy statement about comparing homotopy categories. This is really a comparison between the stable homotopy category and the category of cohomology theories on based spaces, answering the original question.
Everyone go have fun: it's New Year's Eve (with a whole new meaning to the countdown to midnight).
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(Presumably "homotopy category of based spaces" should be "stable homotopy category of based spaces".) – Tom Goodwillie Dec 31 at 17:44
Tom, I edited in an elaboration in response to your comment. – Peter May Dec 31 at 18:36
Interesting. To elaborate for those who were initially confused (such as myself), to show that a map $X \to Y$ is hyperphantom, it suffices to show that $\Sigma^\infty \Omega^\infty X \to X \to Y$ is nullhomotopic (and the same with $X \to Y$ replaced by any (de)suspension) because the universal map from a suspension spectrum into $X$ is $\Sigma^\infty \Omega^\infty X \to X$. – Akhil Mathew Dec 31 at 21:17
1
However, while I may be misunderstanding something, it is still not clear to me how to construct an explicit example of a map $X \to Y$ satisfying this condition (that is, the above description of $X$ as a filtered colimit of suspension spectra is sufficiently special that one might imagine the $\lim^1$ terms would tend to vanish). – Akhil Mathew Dec 31 at 21:27
Akhil, that goes back before my time. One of the motivations for constructing the stable homotopy category back in early 1960's was the understanding that the category of cohomology theories cannot be triangulated. I can't remember counterexamples or even if I ever knew them. I'm pretty sure there was some discussion in Boardman's (unpublished) work. In any case, the non-triangulability has to be explained by non-vanishing lim^1 terms. – Peter May Jan 1 at 0:45
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http://mathoverflow.net/revisions/86476/list
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Return to Question
2 Added a missing _ in a formula
The short version:
Given non-zero real numbers $\alpha$ and $\beta$, can one prove the following estimate in a simple manner? Or does it follow from a well-known result on exponential sums? $$\sum_{n=1}^N \frac{1}{\sqrt{n}}e(\alpha n + \beta n \log n) = O_{\alpha,\beta}(\log N)$$ (Here $e(x)=e^{2 \pi i x}$) And, if so, can one replace big-O with little-o for certain $\alpha$ and $\beta$?
The background:
This peculiar question came from my recent study of the van der Corput transform (also called Process B or the method of van der Corput). The transform says that given sufficiently "nice" functions $f$ and $g$, with $f$ strictly increasing, $$\sum_{a\le n \le b} g(n) e(f(n)) \approx \sum_{f'(a)\le \nu \le f'(b)} \frac{g(x_\nu)}{\sqrt{f''(x_\nu)}} e(f(x_\nu)-\nu x\nu x_\nu +\tfrac{1}{8})$$ where $f'(x_\nu)=\nu$.
The error term in this transformation has been the subject of much study, but the simplest and most traditional is written as $$O( \lambda^{-1/2} + \log(f'(b)-f'(a)+1))$$ where $f''(x) \asymp \lambda$ on $[a,b]$ (for example, Theorem 8.16 in Iwaniec and Kowalski).
Out of curiosity, I considered the case when $g(n) = 1$, $f(n) = k(\tfrac{3}{2})^n$, $a=1$, and $b=x$. If such a sum could be shown to be $o(x)$ for all non-zero integer $k$, then this would imply the equidistribution of the fractional parts of $(\tfrac{3}{2})^n$. The van der Corput transform of such a sum looks like a constant multiple of the sum at the beginning of this problem, with $$\alpha = \frac{1-\log(k\log 3/2)}{\log(3/2)}$$ $$\beta = \frac{1}{\log(3/2)}$$ $$N=k\left(\frac{3}{2}\right)^x \log(3/2)$$
The traditional error terms in this case are, in fact, on the order of $x$, but I believe I have a method to reduce them, which would give the big-O bound at the top of this question as a simple corollary of an exceedingly complicated theorem. I ask this question of mathoverflow because I do not know whether a simpler proof exists, or whether there exists a little-o estimate, which might imply the equidistribution of the fractional parts of $(\tfrac{3}{2})^n$.
I would suspect that any little-o estimates would require very strong conditions on $\alpha$ and $\beta$, given that the sequence $2^n$ does not equidistribute modulo 1, but the van der Corput transforms of the associated exponential sums have very similar $\alpha$ and $\beta$ to the ones mentioned above.
1
Bounds on an exponential sum related to an equidistribution question
The short version:
Given non-zero real numbers $\alpha$ and $\beta$, can one prove the following estimate in a simple manner? Or does it follow from a well-known result on exponential sums? $$\sum_{n=1}^N \frac{1}{\sqrt{n}}e(\alpha n + \beta n \log n) = O_{\alpha,\beta}(\log N)$$ (Here $e(x)=e^{2 \pi i x}$) And, if so, can one replace big-O with little-o for certain $\alpha$ and $\beta$?
The background:
This peculiar question came from my recent study of the van der Corput transform (also called Process B or the method of van der Corput). The transform says that given sufficiently "nice" functions $f$ and $g$, with $f$ strictly increasing, $$\sum_{a\le n \le b} g(n) e(f(n)) \approx \sum_{f'(a)\le \nu \le f'(b)} \frac{g(x_\nu)}{\sqrt{f''(x_\nu)}} e(f(x_\nu)-\nu x\nu +\tfrac{1}{8})$$ where $f'(x_\nu)=\nu$.
The error term in this transformation has been the subject of much study, but the simplest and most traditional is written as $$O( \lambda^{-1/2} + \log(f'(b)-f'(a)+1))$$ where $f''(x) \asymp \lambda$ on $[a,b]$ (for example, Theorem 8.16 in Iwaniec and Kowalski).
Out of curiosity, I considered the case when $g(n) = 1$, $f(n) = k(\tfrac{3}{2})^n$, $a=1$, and $b=x$. If such a sum could be shown to be $o(x)$ for all non-zero integer $k$, then this would imply the equidistribution of the fractional parts of $(\tfrac{3}{2})^n$. The van der Corput transform of such a sum looks like a constant multiple of the sum at the beginning of this problem, with $$\alpha = \frac{1-\log(k\log 3/2)}{\log(3/2)}$$ $$\beta = \frac{1}{\log(3/2)}$$ $$N=k\left(\frac{3}{2}\right)^x \log(3/2)$$
The traditional error terms in this case are, in fact, on the order of $x$, but I believe I have a method to reduce them, which would give the big-O bound at the top of this question as a simple corollary of an exceedingly complicated theorem. I ask this question of mathoverflow because I do not know whether a simpler proof exists, or whether there exists a little-o estimate, which might imply the equidistribution of the fractional parts of $(\tfrac{3}{2})^n$.
I would suspect that any little-o estimates would require very strong conditions on $\alpha$ and $\beta$, given that the sequence $2^n$ does not equidistribute modulo 1, but the van der Corput transforms of the associated exponential sums have very similar $\alpha$ and $\beta$ to the ones mentioned above.
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http://mathhelpforum.com/differential-geometry/118341-eigenspace-lambda-1-bounded-linear-operator-its-adjoint.html
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# Thread:
1. ## Eigenspace of \lambda = 1 for a bounded linear operator and its adjoint
$H$ is a Hilbert space and $T \in B(H)$ is such that $||T||_{H} \leq 1$. Show that $Tx = x \iff T^{*}x = x$.
I have tried looking at this problem in several ways and I just cannot seem to see the significance of $||A|| \leq 1$ nor how to go about proving it. I understand how the adjoint is defined for a Hilbert space as a consequence of Reisz representation. I have tried to attack it using fixed point theorems. Any nudge in the right direction would be appreciated.
2. If $\Vert T \Vert <1$ then $I-T$ is invertible and so the only possible eigenvector of $T$ would be $0$. For the case $\Vert T\Vert =1$ I don't know how to prove this yet.
3. Originally Posted by jprobst
$H$ is a Hilbert space and $T \in B(H)$ is such that $||T||_{H} \leq 1$. Show that $Tx = x \iff T^{*}x = x$.
The important thing to know is that $\|T^*\|=\|T\|$. Therefore $\|T^*x\|\leqslant\|x\|$.
If $Tx=x$ then $\|T^*x-x\|^2 = \langle T^*x-x,T^*x-x\rangle = ...$ (multiply out the inner product and use the fact that $\langle T^*x,x\rangle = \langle x,Tx\rangle$) $... = \|T^*x\|^2 - \|x\|^2\leqslant0$. Hence $T^*x-x=0$.
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http://math.stackexchange.com/questions/tagged/standard-deviation
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http://mathhelpforum.com/math-challenge-problems/107413-number-puzzle-u.html
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# Thread:
1. ## A number puzzle for u
There are two numbers with the difference of 3 between them and the difference of their squares is 51.
Can you find the numbers???
2. Spoiler:
Suppose the two numbers are $a$ and $b$, with $a<b$.
Then the problem says that $b-a=3$ and $b^2-a^2=51$. Note that $b^2-a^2=(b-a)(b+a)$. Can you finish now?
3. Originally Posted by Bruno J.
Suppose the two numbers are $a$ and $b$, with $a<b$.
Then the problem says that $b-a=3$ and $b^2-a^2=51$. Note that $b^2-a^2=(b-a)(b+a)$. Can you finish now?
I suspect the OP was meant to be a challenge question (which is why the thread was moved to here and why I put spoiler tags around your reply).
4. @ Bruno: yes you are right,,,
But No answers yet??
5. Originally Posted by mr fantastic
I suspect the OP was meant to be a challenge question (which is why the thread was moved to here and why I put spoiler tags around your reply).
Then why post his solution immediately below?
6. X = 10
y = 7
7. Originally Posted by HallsofIvy
Then why post his solution immediately below?
I didn't want to delete the reply on the one hand but I didn't want the reply to be in plain sight on the other hand. So my third hand decided a good compromise would be to put spoiler tags around it.
8. Originally Posted by mr fantastic
So my third hand decided a good compromise would be to put spoiler tags around it.
Is 3 hands what makes you "fantastic"?
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http://math.stackexchange.com/questions/67417/product-of-integrals?answertab=votes
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# Product of Integrals
I'm given $\Psi(x,t)$ as a proposal for a wave function. $\Psi(x,t)=\int_{1}^{1+\Delta k} e^{i(kx-wt)} k^2 dk$
Now I try to compute $\Psi^*(x,t)\Psi(x,t)$ wich is the product
$(\int_{1}^{1+\Delta k} e^{-i(kx-wt)} k^2 dk) (\int_{1}^{1+\Delta k} e^{i(kx-wt)}k^2 dk)$
In wich way should I transform this to a double integral? Taking into account that $w=w(|k|)$
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There is no comment about $\Delta k$ so I assumed is a real number because $k$ is. – Nivalth Sep 25 '11 at 12:48
1
This is a product of Fourier integrals, so you can obtain its Fourier transform as the convolution of the two Fourier transforms. – joriki Sep 25 '11 at 13:01
In Fourier Transforms, as far as I know, the range of integration is $\mathbb{R}$ not just $(1,1+\Delta k)$ – Nivalth Sep 25 '11 at 13:04
## 1 Answer
It is easier to solve the one dimensional integral, and then to perform the multiplication:
$\int_1^{1+\Delta} e^{ikx-i\omega t} k^2 dk = - e^{-i\omega t} \frac{\partial}{\partial x^2}( \int_1^{1+\Delta} e^{ikx} dk) = - \Delta e^{-i\omega t} \frac{\partial}{\partial x^2}\left ( e^{[i(1+\Delta/2)x]} \frac{sin(\frac{\Delta x}{2})}{\frac{\Delta x}{2}}\right)$
All is left is to perform the differentiationwith respect to x.
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Thank you very much, nice idea! – Nivalth Sep 25 '11 at 15:13
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http://mathoverflow.net/questions/78147/walks-that-cannot-hit-the-boundary/78151
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## Walks that cannot hit the boundary
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I've casually proved, as application of some ideas that I am developing, a result that might be of interest in itself. I am completely new in this field and then I would like to ask your help to understand: 1) might it be of interest? 2) Is it trivial, in the sense that it can be proved directly? 3) is it well-known?
Let me fix a planar and regular setting, even if I could state the result in more general settings: let $n\geq1$ be a fixed integer and $P=[-n,n]^2\subseteq\mathbb Z^2$. Let me fix the following notation: given $(x,y)\in P$, I will denote by $A(x,y)$ the set formed by the following at most five points: $(x-1,y),(x,y),(x+1,y),(x,y-1),(x,y+1)$, where at most means that if one of those points does not belong to $P$, then I will not consider it.
Now, the situation is the following: for any point $p\in P$, let $\gamma_p$ a walk starting on $p$ and ending on $p^+$. I suppose that: if a walk hits the boundary, then it ends. In particular, if $p\in\partial P$, then $p^+=p$.
Definition: A flow is a family of walks $\gamma_p$, one for each $p\in P$, such that: for all $p\in P$, whenever $q\in A(p)$, then $q^+\in A(p^+)$.
My result would be: Given a flow of walks, there is at least one walk which does not hit the boundary.
I was thinking that it might be useful to prove that some walks are bounded, but I repeat that I am really new in this field.
Every comment is welcome and also references are appreciated.
Thanks in advance,
Valerio
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## 1 Answer
In fact, you can prove that $p^+=p$ for all $p$.
Actually, consider the shortest paths from $(-n,-n)$ to $(n,n)$ and from $(-n,n)$ to $(n,-n)$ passing through $p$. Taking `pluses' of them, you should also obtain the paths of the same lengths connecting the same points. Now from the first path you obtain that the sum of coordinates of $p^+$ is the same as for $p$, and from the second path you see that the difference of coordinates is the same for $p$ and $p^+$. Hence $p=p^+$.
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Jesus! I'm doing a mess. It was enough to draw a stupid picture to realize it! – Valerio Capraro Oct 14 2011 at 17:23
I was thinking that actually it would be enough for me that the points on the boundary are mapped bijectively in points on the boundary (veryfying the property in the definition). But probably also in this case the result remains trivial: it seems to me that in this case every square of the shape $[-k,k]^2$ is mapped in itself – Valerio Capraro Oct 14 2011 at 17:32
Well, if your map is a bijection on the boundary, then some two points should come to opposite vertices of the square. Then they should be the vertices themselves, otherwise there would exist a path between them which is shorter than $4n$. Hence it is also a bijection on the vertices, and by the same reasons of the shotrest path the whole map is some symmetry of the square. – Ilya Bogdanov Oct 14 2011 at 17:40
Yes, on the squares everything get very intuitive. Actually my proof would work for more general subsets of $\mathbb Z^2$; intuitively those subsets without holes whose boundary has a hole. Well, maybe I will put it just as a remark. Many thanks for the clarifications. – Valerio Capraro Oct 14 2011 at 21:43
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http://mathoverflow.net/questions/25875/generators-of-a-maximal-ideal-of-kx1-xn
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## Generators of a maximal ideal of k[X1,…,Xn]
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hi. Given $k$ an algebraically closed field, I know that that a maximal ideal $\mathfrak{m}$ of $A = k[X_1,\cdots,X_n]$ is just a $\langle X_1-a_1,\cdots,X_n-a_n \rangle$ (Nullstellensatz). Knowing that, it seems intuitive that $\mathfrak{m}$ can not be generated by less than $n$ elements. Is that true ? In that case, how can I show that ?
(Actually, I need that, just after using Nakayama's lemma, to show that $dim_{A/\mathfrak{m}=k} \mathfrak{m}/\mathfrak{m}^2$ is greater than $n$.)
(I've seen things that might be relevant like "Krull height theorem" but I think such things take place in a more general context and I have some difficulties, first to understand them, and second to adapt them...)
Thank you.
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## 4 Answers
Since you mentioned Krull's height theorem (= the generalized principal ideal theorem) and having difficulty applying it, I thought you or someone else might appreciate seeing how this works: it is quite straightforward.
The generalized principal ideal theorem is as follows: let $R$ be a Noetherian ring and $I$ a proper ideal of $R$ which can be generated by $n$ elements. Let $\mathfrak{p}$ be a prime ideal which is minimal among all primes containing $I$. Then $\mathfrak{p}$ has height at most $n$, that is, there do not exist prime ideals $\mathfrak{p}_0,\ldots,\mathfrak{p}_n$ such that
$\mathfrak{p}_0 \subsetneq \mathfrak{p_1} \subsetneq \ldots \subsetneq \mathfrak{p_n} \subsetneq \mathfrak{p}$.
[For a deduction of this from Krull's Principal Ideal Theorem, see e.g. Theorem 96 on p. 70 of http://math.uga.edu/~pete/integral.pdf.]
Let us apply this with $R = k[x_1,\ldots,x_n]$ and the ideal $I = \langle x_1 - a_1,\ldots,x_n - a_n \rangle$. $I$ is itself a maximal -- hence prime -- ideal, since $R/I \cong k$. Thus the generalized principal ideal theorem simply says that $I$ cannot be generated by fewer elements than its height. But its height is certainly at least $n$. No geometry is needed here: just define $\mathfrak{p}_0 = 0$ and for $1 \leq i \leq n$, $\mathfrak{p}_i = \langle x_1 - a_1,\ldots,x_i - a_i \rangle$.
Finally, a comment: I did not use that $k$ was algebraically closed per se but only worked with maximal ideals of this particular form. On the other hand, it is still true over an arbitrary field $k$ that every maximal ideal of $k[x_1,\ldots,x_n]$ has height $n$ and can be generated by $n$ elements (and no fewer, by Krull's theorem): see Corollary 130 on p. 83 of the document linked to above.
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Thanks a lot, that exactly answers my question ! – Laurent May 25 2010 at 18:22
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The classes of the $X_i - a_i$ are easily seen to be a basis for $\mathfrak{m}/\mathfrak{m}^2$.
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The dimension is exactly n: you are counting linear polynomials modulo terms of higher order.
The geometrical interpretation is that intersecting n - 1 hypersurfaces with a common point must give an algebraic set with a component of dimension at least one. If you insist on a commutative algebra approach, you are going to have to study those theorems ...
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Trying to write down something along the lines you were looking for, I would argue that over an algebraically closed field the set of common zeros of $k < n$ polynomials has codimension at most $k$ and so cannot be a single point $(a_1,\ldots,a_n)$. You can use Krull theorem indeed (e.g. check out Ravi Vakil's notes: http://math.stanford.edu/~vakil/725/class14.pdf), or argue more along the 19th century and prove by some sort of elimination that the set of common zeros depend on at least $n-k$ parameters...
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http://math.stackexchange.com/questions/tagged/intuition+integral
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http://unapologetic.wordpress.com/2007/04/30/tensor-products-of-modules/?like=1&source=post_flair&_wpnonce=1ba466d084
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# The Unapologetic Mathematician
## Tensor products of modules
The notion of a tensor product also extends to modules, but the generalization is not quite as straightforward as it was for direct sums.
We start with a ring $R$ and consider a right $R$-module $A_R$ and a left $R$-modules ${}_RB$. We consider functions $f:A\times B\rightarrow X$ which take an element of each module and return an element of some abelian group $X$. Such a function is called “middle linear” if
• $f(a_1+a_2,b)=f(a_1,b)+f(a_2,b)$
• $f(a,b_1+b_2)=f(a,b_1)+f(a,b_2)$
• $f(a\cdot r,b)=f(a,r\cdot b)$
That is, if it is bilinear as a function of abelian groups, and if we can pull the action of $R$ from the first argument to the second and back without changing the value of the function. This condition may seem a little artificial, but I’ll motivate it a bit more later.
The tensor product $A\otimes_RB$ is the abelian group for the universal middle linear function, just as the tensor product for abelian groups was the abelian group for the universal bilinear function. We show that such an object exists by a similar construction. Take the free abelian group generated by all elements of $A\times B$ and write a generator as $a\otimes b$. Then impose the relations $(a_1+a_2)\otimes b$, $a\otimes(b_1+b_2)$, and $(a\cdot r)\otimes b=a\otimes(r\cdot B)$. Then given any middle linear map $f:A\times B\rightarrow X$ there is a unique homomorphism of abelian groups $\bar{f}:A\otimes B\rightarrow X$ so that $f(a,b)=\bar{f}(a\otimes b)$.
In general, the tensor product over $R$ is just an abelian group — it eats $R$-module structures like $\hom$ does. That said, like $\hom$ plays well with extra module structures, so does tensor product. If ${}_SA_R$ is a right $R$-module and a left $S$-module, then $A\otimes_RB$ carries the structure of a left $S$-module. Indeed we can define $s\cdot(a\otimes b)=(s\cdot a)\otimes b$ and check that this action respects the relations we imposed. Similarly, if $A$ has an additional right $S$-module structure commuting with the action of $R$ then $A\otimes_RB$ is a right $S$-module, and the same goes for extra modules structures on $B$. Unlike in the case of $\hom$, no structures get “flipped over” in this process.
If $R$ is a commutative ring, then every module is both a left and a right $R$-module. Thus, the same is true of $A\otimes_RB$ — the tensor product eats the left module structure on $A$ and the right module structure on $B$, but leaves the other two structures.
Now for an example that should motivate the idea of a middle-linear map. Let $A_R$, ${}_SB_R$, and $C_R$ be right $R$-modules with an extra left $S$-module structure on $B$. Then $\hom(A,B)$ is a left $S$-module and $\hom(B,C)$ is a right $S$-module. We consider $f\in\hom(A,B)$ and $g\in\hom(B,C)$ and calculate the two composites
$\left[g\circ (s\cdot f)\right](a)=g(\left[s\cdot f\right](a))=g(s\cdot(f(a)))$
$\left[(g\cdot s)\circ f\right](a)=\left[g\cdot s\right](f(a))=g(s\cdot(f(a)))$
so we can use the $S$ action on either factor. This means that composition of these homomorphisms is a middle-linear function, and so defines a linear function $\hom(B,C)\otimes_S\hom(A,B)\rightarrow\hom(A,C)$.
### Like this:
Posted by John Armstrong | Ring theory
## 6 Comments »
1. [...] and in particular a zero object — the trivial -dimensional vector space . It also has a tensor product, which makes this a monoidal category, using the one-dimensional vector space itself as monoidal [...]
Pingback by | May 19, 2008 | Reply
2. [...] Since we’re looking at vector spaces, which are special kinds of modules, we know that has a tensor product structure. Let’s see what this means when we pick [...]
Pingback by | May 23, 2008 | Reply
3. [...] Remember that the structure is just a bilinear multiplication, which is just a linear map from the tensor square to . And we know that the basis for a tensor product consists of pairs of basis elements. So we [...]
Pingback by | July 28, 2008 | Reply
4. [...] concepts tied to linear algebra, and before we get into that we need to revisit an old topic: tensor powers and the subspaces of symmetric and antisymmetric tensors. Specifically, how do all of these [...]
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5. [...] finite groups and , and we have (left) representations of each one: and . It turns out that the tensor product naturally carries a representation of the product group . Equivalently, it carries a [...]
Pingback by | November 4, 2010 | Reply
6. [...] the tensor product of vector spaces is old hat by now, as is using the dual space . We’ll put them together by [...]
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://math.stackexchange.com/questions/254305/why-is-this-inequality-important
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# Why is this inequality important?
I'm reading Courant's What is Mathematics?
In the beginning, he's showing some proofs, there's a proof about An Important Equality and this important equality is:
$$(1+p)^n\geq 1+np$$
The book's author states the proof but he doesn't say why the inequality is important, I had a feeling that this inequality importance is by stabilishing the $\geq$ relation in $\mathbb{N}$, is this it's importance? If not, what is it?
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## 4 Answers
It is usually called Bernoulli's inequality. It frequently appears in mathematical contests. This inequality is important, because many other inequalities can be proved using it. Here is some person's Bachelor's thesis where it is mentioned among others; in particular, reference [60; Zhu H.] seems to offer some explanation with applications, but I could not find an on-line version.
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One place it is important in approximation. When $np \ll 1, (1+p)^n$ is greater than, but close to, $1+np$ as subsequent terms are of order $(np)^2$ and higher. Another is proving inequalities. $1+np$ is much simpler and easier to work with than $(1+p)^n$. If it is not so much smaller that your proposed inequality fails, it often gets you somewhere useful.
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It comes up in all kinds of surprising places. It comes from the binomial theorem: $$(1+p)^n=\sum_{k=0}^{n}\binom{n}{k}p^k$$From which you can easily infer that for any choice of $m\leq n$, $$(1+p)^n\geq \sum_{k=0}^m \binom{n}{k}p^k$$
Where in this case we are taking $m=1$.
Just to give you an example, here's a quick proof that $\lim_{n\rightarrow \infty} \sqrt[n]{p}=1$ for $p>1$. Let $x_n=\sqrt[n]{p}-1$ so that $\sqrt[n]{p}=x_n+1$. Then $p=(x_n+1)^n$ and we can use the inequality to obtain $$p=(x_n+1)^n\geq nx_n+1 \\ \frac{p -1}{n}\geq x_n>0$$
And so $x_n\rightarrow 0$ and $\sqrt[n]{p}=1-x_n\rightarrow1$.
It just comes up in surprising places.
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It can be used (as shown in "What is Mathematics") to prove that $n^{1/n} \to 1$ as $n \to \infty$ for integer $n$ by starting with $(1+1/\sqrt{n})^n$. This has been discussed here (by me, naturally), so I'll leave the details for the reader.
It can also be used to prove the arithmetic-geometric mean inequality for the case when $n-1$ of the $n$ items are the same. This, in turn, can be used to show that $\lim_{n \to \infty} (1+1/n)^n$ exists (for integer $n$).
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http://mathhelpforum.com/calculus/118683-total-length-astroid.html
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# Thread:
1. ## Total length of the astroid
This is the last section of my class and I just don't get it. I can't even get my calculator so graph it for me (probably my stupid though). Can anyone please explain how to answer this type of question? I have 5 questions on the length of a parametric curve to do. I just can't understand what the book is saying (again probably my stupid). Any help is greatly appreciated.
If f(θ) is given by: f(θ)=18cos3θ and g(θ) is given by: g(θ)=18sin3θ
Find the total length of the astroid described by f(θ) and g(θ).
(The astroid is the curve swept out by (f(θ),g(θ)) as θ ranges from 0 to 2π. )
2. $f(t)=18cos(3t), \;\ g(t)=18sin(3t)$
$f'(t)=-54sin(3t), \;\ g'(t)=54cos(3t)$
parametric arc length is given by $\int_{0}^{2\pi}\sqrt{(-54sin(3t))^{2}+(54cos(3t))^{2}}dt=54\int_{0}^{2\pi }dt$
I suppose this is what they are getting at. To graph this in parametric, your calculator must be in parametric mode. Parametrically, this is an ellipse, not an astroid. Unless I am missing something. The convention normally used is t instead if theta. Theta is commonly used when dealing with polar coordinates. Even in polar coordinates, this is a rose and not an astroid.
Please do not tell me you mean $f(t)=18cos^{3}(t), \;\ g(t)=18sin^{3}(t)$ and did not use ^ to mean power.
3. That is what the equation reads. All greek to me . Thanks for your help.
4. For an astroid, the equations should be $x(t)=18cos^{3}(t) \;\ and \;\ y(t)=18sin^{3}(t)$.
$x'(t)=-54sin(t)cos^{2}(t)$
$y'(t)=54sin^{2}(t)cos(t)$
Repeating, Parametric Arc length is give by $\int_{a}^{b}\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{d t})^{2}}dt$
Yours whittles down to
$54\int_{0}^{2\pi}|sin(t)cos(t)|dt$
Let's do it this way though. Since we have symmetry, we can integrate from 0 to Pi/2 and multiply by 4:
$216\int_{0}^{\frac{\pi}{2}}sin(t)cos(t)dt$
Can you integrate that?. See the graph?. That is an astroid. Astroid means star-shaped.
Attached Thumbnails
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http://en.m.wikibooks.org/wiki/Famous_Theorems_of_Mathematics/Logic
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# Famous Theorems of Mathematics/Logic
## Gödel's completeness theorem
In the following, we state two equivalent forms of the theorem, and show their equivalence.
Later, we prove the theorem. This is done in the following steps:
1. Reducing the theorem to sentences (formulas with no free variables) in prenex form, i.e. with all quantifiers ($\forall$ and $\exists$) at the beginning. Furthermore, we reduce it to formulas whose first quantifier is $\forall$. This is possible because for every sentence, there is an equivalent one in prenex form whose first quantifier is $\forall$.
2. Reducing the theorem to sentences of the form $\forall x_1 \forall x_2 ... \forall x_k \exists y_1 \exists y_2... \exists y_m \phi(x_1...x_k, y_1...y_m)$. While we cannot do this by simply rearranging the quantifiers, we show that it is yet enough to prove the theorem for sentences of that form.
3. Finally we prove the theorem for sentences of that form.
• This is done by first noting that a sentence such as $B = \exists x_1 \exists x_2 ... \exists x_k \exists y_1 \exists y_2... \exists y_m \phi(x_1...x_k, y_1...y_m)$ is either refutable or has some model in which it holds; this model is simply assigning truth values to the subpropositions from which B is built of. The reason for that is the completeness of propositional logic, with the existential quantifiers playing no role.
• We extend this result to more and more complex and lengthy sentences, Dn (n=1,2...), built out from B, so that either any of them is refutable and therefore so is φ, or all of them are not refutable and therefore each holds in some model.
• We finally use the models in which the Dn hold (in case all are not refutable) in order to build a model in which φ holds.
Theorem 1. Every formula valid in all structures is provable.
This is the most basic form of the completeness theorem. We immediately restate it in a form more convenient for our purposes:
Theorem 2. Every formula φ is either refutable or satisfiable in some structure.
"φ is refutable" means by definition "¬φ is provable".
### Equivalence of both theorems
To see the equivalence, note first that if Theorem 1 holds, and φ is not satisfiable in any structure, then ¬φ is valid in all structures and therefore provable, thus φ is refutable and Theorem 2 holds. If on the other hand Theorem 2 holds and φ is valid in all structures, then ¬φ is not satisfiable in any structure and therefore refutable; then ¬¬φ is provable and then so is φ, thus Theorem 1 holds.
### Proof of theorem 2: first step
We approach the proof of Theorem 2 by successively restricting the class of all formulas φ for which we need to prove "φ is either refutable or satisfiable". At the beginning we need to prove this for all possible formulas φ in our language. However, suppose that for every formula φ there is some formula ψ taken from a more restricted class of formulas C, such that "ψ is either refutable or satisfiable" → "φ is either refutable or satisfiable". Then, once this claim (expressed in the previous sentence) is proved, it will suffice to prove "φ is either refutable or satisfiable" only for φ's belonging to the class C. Note also that if φ is provably equivalent to ψ (i.e., (φ≡ψ) is provable), then it is indeed the case that "ψ is either refutable or satisfiable" → "φ is either refutable or satisfiable" (the soundness theorem is needed to show this).
There are standard techniques for rewriting an arbitrary formula into one which does not use function or constant symbols, at the cost of introducing additional quantifiers; we will therefore assume that all formulas are free of such symbols. In Gödel's paper, he uses a version of first-order predicate calculus which has no function or constant symbols to begin with.
Next we consider a generic formula φ (which no longer uses function or constant symbols) and apply the prenex form theorem to find a formula ψ in normal form such that φ≡ψ (ψ being in normal form means that all the quantifiers in ψ, if there are any, are found at the very beginning of ψ). It follows now that we need only prove Theorem 2 for formulas φ in normal form.
Next, we eliminate all free variables from φ by quantifying them existentially: if, say, x1...xn are free in φ, we form $\psi=\exists x_1 ... \exists x_n \phi$. If ψ is satisfiable in a structure M, then certainly so is φ and if ψ is refutable, then $\neg \psi = \forall x_1 ... \forall x_n \neg \phi$ is provable, and then so is ¬φ, thus φ is refutable. We see that we can restrict φ to be a sentence, that is, a formula with no free variables.
Finally, we would like, for reasons of technical convenience, that the prefix of φ (that is, the string of quantifiers at the beginning of φ, which is in normal form) begin with a universal quantifier and end with an existential quantifier. To achieve this for a generic φ (subject to restrictions we have already proved), we take some one-place relation symbol F unused in φ, and two new variables y and z.. If φ = (P)Φ, where (P) stands for the prefix of φ and Φ for the matrix (the remaining, quantifier-free part of φ) we form $\psi = \forall y (P) \exists z ( \Phi \wedge [ F(y) \vee \neg F(z) ] )$. Since $\forall y \exists z ( F(y) \vee \neg F(z) )$ is clearly provable, it is easy to see that $\phi=\psi$ is provable.
### Reducing the theorem to formulas of degree 1
Our generic formula φ now is a sentence, in normal form, and its prefix starts with a universal quantifier and ends with an existential quantifier. Let us call the class of all such formulas R. We are faced with proving that every formula in R is either refutable or satisfiable. Given our formula φ, we group strings of quantifiers of one kind together in blocks:
$\phi = (\forall x_1 ... \forall x_{k_1})(\exists x_{k_1+1} ... \exists x_{k_2}).......(\forall x_{k_{n-2}+1} ... \forall x_{k_{n-1}})(\exists x_{k_{n-1}+1} ... \exists x_{k_n}) (\Phi)$
We define the degree of $\phi$ to be the number of universal quantifier blocks, separated by existential quantifier blocks as shown above, in the matrix of $\phi$. The following lemma, which Gödel adapted from Skolem's proof of the Löwenheim-Skolem theorem, lets us sharply reduce the complexity of the generic formula $\phi$ for which we need to prove the theorem:
Lemma. Let k>=1. If every formula in R of degree k is either refutable or satisfiable, then so is every formula in R of degree k+1.
Comment: Take a formula φ of degree k+1 of the form $\phi = (\forall x)(\exists y)(\forall u)(\exist v) (P) \psi$, where $(P)\psi$ is the remainder of $\phi$ (it is thus of degree k-1). φ states that for every x there is a y such that... (something). It would have been nice to have a predicate Q' so that for every x, Q'(x,y) would be true if and only if y is the required one to make (something) true. Then we could have written a formula of degree k which is equivalent to φ, namely $(\forall x')(\forall x)(\forall y)(\forall u)(\exist v)(\exist y') (P) Q'(x',y') \wedge (Q'(x,y) \rightarrow \psi)$. This formula is indeed equivalent to φ because it states that for every x, if there is a y which satisfies Q'(x,y), then (something) holds, and furthermore, we know that there is such a y, because for every x', there is a y' which satisfies Q'(x',y'). Therefore φ follows from this formula. It is also easy to show that if the formula is false, then so is φ. Unfortunately, in general there is no such predicate Q'. However, this idea can be understood as a basis for the following proof of the Lemma.
Proof. Let φ be a formula of degree k+1; then we can write it as
$\phi = (\forall x)(\exists y)(\forall u)(\exist v) (P) \psi$
where (P) is the remainder of the prefix of $\phi$ (it is thus of degree k-1) and $\psi$ is the quantifier-free matrix of $\phi$. x, y, u and v denote here tuples of variables rather than single variables; e.g. $(\forall x)$ really stands for $\forall x_1 \forall x_2 ... \forall x_n$ where $x_1 ... x_n$ are some distinct variables.
Let now x' and y' be tuples of previously unused variables of the same length as x and y respectively, and let Q be a previously unused relation symbol which takes as many arguments as the sum of lengths of x and y; we consider the formula
$\Phi = (\forall x')(\exists y') Q(x',y') \wedge (\forall x)(\forall y)( Q(x,y) \rightarrow (\forall u)(\exist v)(P)\psi )$
Clearly, $\Phi \rightarrow \phi$ is provable.
Now since the string of quantifiers $(\forall u)(\exists v)(P)$ does not contain variables from x or y, the following equivalence is easily provable with the help of whatever formalism we're using:
$( Q(x,y) \rightarrow (\forall u )(\exists v)(P) \psi) \equiv (\forall u)(\exists v)(P) ( Q(x,y) \rightarrow \psi )$
And since these two formulas are equivalent, if we replace the first with the second inside Φ, we obtain the formula Φ' such that Φ≡Φ':
$\Phi' = (\forall x')(\exist y') Q(x',y') \wedge (\forall x)(\forall y) (\forall u)(\exists v)(P) ( Q(x,y) \rightarrow \psi )$
Now Φ' has the form $(S)\rho \wedge (S')\rho'$, where (S) and (S') are some quantifier strings, ρ and ρ' are quantifier-free, and, furthermore, no variable of (S) occurs in ρ' and no variable of (S') occurs in ρ. Under such conditions every formula of the form $(T)(\rho \wedge \rho')$, where (T) is a string of quantifiers containing all quantifiers in (S) and (S') interleaved among themselves in any fashion, but maintaining the relative order inside (S) and (S'), will be equivalent to the original formula Φ'(this is yet another basic result in first-order predicate calculus that we rely on). To wit, we form Ψ as follows:
$\Psi = (\forall x')(\forall x)(\forall y) (\forall u)(\exists y')(\exists v)(P)Q(x',y') \wedge (Q(x,y) \rightarrow \psi )$
and we have $\Phi' \equiv \Psi$.
Now $\Psi$ is a formula of degree k and therefore by assumption either refutable or satisfiable. If $\Psi$ is satisfiable in a structure M, then, considering $\Psi \equiv \Phi' \equiv \Phi \wedge \Phi \rightarrow \phi$, we see that $\phi$ is satisfiable as well. If $\Psi$ is refutable, then so is $\Phi$ which is equivalent to it; thus $\neg \Phi$ is provable. Now we can replace all occurrences of Q inside the provable formula $\neg \Phi$ by some other formula dependent on the same variables, and we will still get a provable formula. (This is yet another basic result of first-order predicate calculus. Depending on the particular formalism adopted for the calculus, it may be seen as a simple application of a "functional substitution" rule of inference, as in Gödel's paper, or it may be proved by considering the formal proof of $\neg \Phi$, replacing in it all occurrences of Q by some other formula with the same free variables, and noting that all logical axioms in the formal proof remain logical axioms after the substitution, and all rules of inference still apply in the same way.)
In this particular case, we replace Q(x',y') in $\neg \Phi$ with the formula $(\forall u)(\exists v)(P)\psi(x,y|x',y')$. Here (x,y|x',y') means that instead of ψ we are writing a different formula, in which x and y are replaced with x' and y'. Note that Q(x,y) is simply replaced by $(\forall u)(\exists v)(P)\psi$.
$\neg \Phi$ then becomes
$\neg ( (\forall x')(\exists y') (\forall u)(\exists v)(P)\psi(x,y|x',y') \wedge (\forall x)(\forall y) ( (\forall u)(\exists v)(P)\psi \rightarrow (\forall u)(\exists v)(P) \psi ) )$
and this formula is provable; since the part under negation and after the $\wedge$ sign is obviously provable, and the part under negation and before the $\wedge$ sign is obviously φ, just with x and y replaced by x' and y', we see that $\neg \phi$ is provable, and φ is refutable. We have proved that φ is either satisfiable or refutable, and this concludes the proof of the Lemma.
Notice that we could not have used $(\forall u)(\exists v)(P)\psi(x,y|x',y')$ instead of Q(x',y') from the beginning, because $\Psi$ would not have been a well-formed formula in that case. This is why we cannot naively use the argument appearing at the comment which precedes the proof.
### Proving the theorem for formulas of degree 1
As shown by the Lemma above, we only need to prove our theorem for formulas φ in R of degree 1. φ cannot be of degree 0, since formulas in R have no free variables and don't use constant symbols. So the formula φ has the general form:
$(\forall x_1...x_k)(\exists y_1...y_m) \phi(x_1...x_k, y_1...y_m)$
Now we define an ordering of the k-tuples of natural numbers as follows: $(x_1...x_k) < (y_1...y_k)$ should hold if either $\Sigma_k (x_1...x_k) < \Sigma_k (y_1...y_k)$, or $\Sigma_k (x_1...x_k) = \Sigma_k (y_1...y_k)$, and $(x_1...x_k)$ precedes $(y_1...y_k)$ in lexicographic order. [Here $\Sigma_k (x_1...x_k)$ denotes the sum of the terms of the tuple.] Denote the nth tuple in this order by $(a^n_1...a^n_k)$.
Set the formula $B_n$ as $\phi(z_{a^n_1}...z_{a^n_k}, z_{(n-1)m+2}, z_{(n-1)m+3}...z_{nm+1})$. Then put $D_n$ as
$(\exists z_1...z_{nm+1}) (B_1 \wedge B_2 ... \wedge B_n)$
Lemma: For every n, $\phi \rightarrow D_n$.
Proof: By induction on n; we have $D_k \equiv D_{k-1} \wedge (\exists z_1...z_{nm+1}) B_n \equiv D_{k-1} \wedge (\exists z_{a^n_1}...z_{a^n_k})(\exists y_1...y_m) \phi(z_{a^n_1}...z_{a^n_k}, y_1...y_m)$, where the latter equivalence holds by variable substitution, since the ordering of the tuples is such that $(\forall k)({a^n_1}...{a^n_k}) < (n-1)m + 2$. Each conjunct here obviously follows from φ.
For the base case, $D_1 \equiv (\exists z_1...z_{m+1}) \phi(z_{a^1_1}...z_{a^1_k}, z_2, z_3...z_{m+1}) \equiv (\exists z_1...z_{m+1}) \phi(z_1...z_1, z_2, z_3...z_{m+1})$ is obviously a corollary of φ as well. So the Lemma is proven.
Now if $D_n$ is refutable for some n, it follows that φ is refutable. On the other hand, suppose that $D_n$ is not refutable for any n. Then for each n there is some way of assigning truth values to the distinct subpropositions $E_h$ (ordered by their first appearance in $D_n$; "distinct" here means either distinct predicates, or distinct bound variables) in $B_k$, such that $B_k$ will be true when each proposition is evaluated in this fashion. This follows from the completeness of the underlying propositional logic.
We will now show that there is such an assignment of truth values to $E_h$, so that all $D_n$ will be true: The $E_h$ appear in the same order in every $D_n$; we will inductively define a general assignment to them by a sort of "majority vote": Since there are infinitely many assignments affecting $E_1$, either infinitely many make $E_1$ true, or infinitely many make it false and only finitely many make it true. In the former case, we choose $E_1$ to be true in general; in the latter we take it to be false in general. Then from the infinitely many n for which $E_1 through E_{h-1}$ are assigned the same truth value as in the general assignment, we pick a general assignment to $E_h$ in the same fashion.
This general assignment must lead to every one of the $B_k$ and $D_k$ being true, since if one of the $B_k$ were false under the general assignment, $D_n$ would also be false for every n > k. But this contradicts the fact that for the finite collection of general $E_h$ assignments appearing in $D_k$, there are infinitely many n where the assignment making $D_n$ true matches the general assignment.
Now from this general assignment which makes all of the $D_k$ true, we construct an interpretation of the language's predicates which makes φ true. The universe of the model will be the natural numbers. Each i-ary predicate $\Psi$ should be true of the naturals $(u_1...u_i)$ precisely when the proposition $\Psi(z_{u_1}...z_{u_i})$ is either true in the general assignment, or not assigned by it (because it never appears in any of the $D_k$).
In this model, each of the formulas $(\exists y_1...y_m) \phi(a^n_1...a^n_k, y_1...y_m)$, is true by construction. But this implies that φ itself is true in the model, since the $a^n$ range over all possible k-tuples of natural numbers. So φ is satisfiable, and we are done.
### Intuitive explanation
We may write each Bi as Φ(x1...xk,y1...ym) for some x-s which we may call "first arguments" and y-s which we may call "last arguments".
Take B1 for example. Its "last arguments" are z2,z3...zm+1, and for every possible combination of k of these variables there is some j so that they appear as "first arguments" in Bj. Thus for large enough n1, Dn1 has the property that the "last arguments" of B1 appear, in every possible combinations of k of them, as "first arguments" in other Bj-s within Dn. For every Bi there is a Dni with the corresponding property.
Therefore in a model which satisfies all the Dn-s, there are objects corresponding to z1, z2... and each combination of k of these appear as "first arguments" in some Bj, meaning that for every k of these objects zp1...zpk there are zq1...zqm which makes Φ(zp1...zpk,zq1...zqm) satisfied. By taking a submodel which contains only these z1, z2... objects, we have a model satisfying φ.
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## Gödel's incompleteness theorems
First Theorem: For any consistent formal, computably enumerable theory that proves basic arithmetical truths, an arithmetical statement that is true, but not provable in the theory, can be constructed. That is, any effectively generated theory capable of expressing elementary arithmetic cannot be both consistent and complete.
Second Theorem: For any formal recursively enumerable (i.e. effectively generated) theory T including basic arithmetical truths and also certain truths about formal provability, T includes a statement of its own consistency if and only if T is inconsistent.
### Modern Proof
Assuming that a computer exists, and that formal logic can be represented as a computer program, it is easy to prove Gödel's incompleteness theorem from basic lemmas in computer science.
1. The Quine lemma-- any computer program can include a subroutine that prints out the entire program's code. This allows any program to write itself into a string variable, or a large enough bigint.
proof: Let the printing subroutine be a standard quine, with additional data that contains all the rest of the code.
2. The Halting lemma: There does not exist a computer program PREDICT(P) which takes the code to program P and predicts whether P eventually halts.
proof: Write program SPITE, which prints itself into variable R, then calculates PREDICT(R). If the answer is R halts, Spite goes into an infinite loop. If the answer is R does not halt, SPITE halts. Since R is really SPITE in disguise, no matter what PREDICT says, the answer is wrong.
The incompleteness theorem: Suppose that an axiom system describes integers (or any other discrete structure), and that it has enough operations (addition and multiplication are sufficient) to describe the working of a computer. Then this system is either inconsistent or incomplete.
proof: Write DEDUCE to deduce all consequences of the axiom system. Let DEDUCE print its own code into the variable R, and search for the theorem R never halts, in the embedding of a computer inside the system. Only if DEDUCE finds this theorem does it halt.
If the axiom system proves that DEDUCE doesn't halt, it is inconsistent. If the system is consistent, DEDUCE doesn't halt and the axioms cannot prove it.
This argument doesn't explicitly demonstrate the incompleteness, because a consistent axiom system could still prove the $\omega$-inconsistent theorem that DEDUCE halts (even though it doesn't) without contradiction.
So write ROSSER: ROSSER prints its code into R, and searches deductions for either 1. R prints something out or 2. R never prints anything out. If it finds 1, it halts without printing anything. If it finds 2, It prints "Hello World!" to the screen and halts.
If the axiom system is consistent, it cannot prove either ROSSER eventually prints something nor the negation ROSSER does not print anything. So whatever its conclusions about DEDUCE, the axiom system is incomplete.
To prove the second incompleteness theorem, note that the consistency of the axioms proves that DEDUCE does not halt, so the axiom system cannot prove its own consistency.
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http://mathoverflow.net/questions/84749/transitivity-of-a-flow-and-its-time-1-map
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## Transitivity of a flow and its time-1 map
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $M$ be a closed manifold, $f:M\to M$ be a homeomorphism, and $\phi_t:M\to M$ be a flow. .
The map $f$ is said to be (point)-transitive if some orbit $\lbrace f^nx:n\in\mathbb{Z}\rbrace$ is dense in $M$.
The flow $\phi_t$ is said to be (flow)-transitive if some flow line $\lbrace\phi_tx:t\in\mathbb{R}\rbrace$ is dense in $M$.
My first question is:
• is there any sufficient condition under which the time-1 map of a transitive flow is still transitive?
In the following we assume $f$ is (point)-transitive and consider a special class of flows: suspensions of $f$.
Let $r\le R$ be positive numbers and $c:M\to[r,R]$ be a continuous suspension function. Let
$M_c=\lbrace(x,r):0\le r\le c(x),x\in M\rbrace/\sim$ where $(x,c(x))\sim(fx,0)$,
and $f_t:M_c\to M_c$ represented by the time-translation $(x,r)\mapsto(x,r+t)$.
According to our definition, the suspension flow $f_t:M_c\to M_c$ is (flow)-transitive.
If $c\equiv1$, it is easy to see that the time-1 map $f_1$ is not transitive (as a homeomorphism on $M_1$). So my second question is:
• is the time-1 map $f_1:M_c\to M_c$ (point)-transitive whenever $c$ is not constant?
Any proof or reference are good. Thank you!
As suggested by Zarathustra, it is worth to point out that the time-1 map of constant suspension flow with $c=\sqrt{2}$ is also (point)-transitive.
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You can take $c=\sqrt 2$ and your time one map will be transitive. – Zarathustra Jan 2 2012 at 18:48
Oh! Thank you! I will edit the question. – Pengfei Jan 3 2012 at 4:37
2
I don't think, in general, the time-$\sqrt{2}$ map of a suspension flow of a transitive map is transitive itself. In fact, consider the following example: let $f\colon\mathbb{T}^2\to\mathbb{T}^2$ be given by $f(x,y):=(x+\sqrt{3},y+\sqrt{2}) \mathrm{mod}\mathbb{Z}^2$. Then, $f$ is a minimal diffeomorphism. The suspension flow of $f$ is $\Phi_t\colon\mathbb{T}^3\to\mathbb{T}^3$ given by $\Phi_t(x,y,z)=(x+t\sqrt{3},y+t\sqrt{2},z+t)$. In this case $\Phi_{\sqrt{2}$ is not transitive. For me it is not clear at all that any transitive suspension flow admits a time-t transitive map. – Alejandro Feb 3 2012 at 16:12
@Alejandro Although the time-1 map of $\Phi_t$ is isomorphic to $f$, but I don't think $\Phi_t$ is the suspension of $f$. It should be $\phi_t:(x,y,z)\to(x,y,z+t)\text{mod}\sim$ where $(x,y,c(x,y))\sim(x+\sqrt{3},y\sqrt{2},0)$. If $c\equiv\sqrt{2}$, then $\phi_1$ is transitive. – Pengfei Feb 4 2012 at 15:23
1
@Pegnfei: I'm sorry, I changed your notation and that produced some confusion. However, I think my example can be rewritten following your notation: consider the map $f : \mathbb{T}^2\ni (x,y) \mapsto (x+\sqrt{2},y+\qrt{3})\in\mathbb{T}^2$. The suspension flow of $f$ with $c\equiv \sqrt{2}$ is isomorphic to the flow $\Phi_t(x,y,z)=(x+t,y+t\sqrt{3/2},z+\frac{t}{\sqrt{2}})$ and hence, we get $\Phi_1(x,y,z)=(x+1,y+\sqrt{3/2},z+\frac{1}{\sqrt{2}})$, which is not transitive. – Alejandro Feb 5 2012 at 7:13
show 6 more comments
## 1 Answer
I think the answer for the second question is no. We can construct a counterexample as follows:
Let $f\colon M\to M$ be an arbitrary transitive homeomorphism and $u\colon M\to\mathbb (0,1/4)$ be an arbitrary non-constant continuous function. Then, let's define $$c(x):=u(x)-u(f(x))+1,\quad\forall x\in M,$$ and consider the flow $f_t\colon M_c\to M_c$ as above.
We claim $f_1$ is not transitive. In fact, given any $x\in M$ and any $t\in (0,1/4)$ it holds: $$f_1(x,u(x)+t)=(x,u(x)+t+1)=(x,c(x)+u(f(x))+t)=(f(x),u(f(x))+t).$$ That means that every compact set $U_t:=\lbrace (x,u(x)+t) : x\in M\rbrace \subset M_c$ is $f_1$-invariant and hence it cannot be transitive. Notice the function $c$ is not constant because $f$ is transitive $u$ is not constant itself.
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Oh Alejandro! You saved my life! This example is so great! Thank you! – Pengfei Feb 18 2012 at 12:45
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http://physics.stackexchange.com/questions/29785/is-the-escape-velocity-at-geosynchronous-earth-orbit-0km-hr?answertab=active
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# Is the escape velocity at Geosynchronous Earth Orbit 0km/hr?
Follow-up question to How long must escape velocity be maintained?
Is the escape velocity at GSO 0?
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1
What do you mean by GEO? – Bernhard Jun 8 '12 at 22:16
Geosynchronous Earth Orbit is what I had in mind (+: But we can do without the mention of Earth there – Everyone Jun 8 '12 at 22:30
## 2 Answers
No. Any circular orbital velocity is about 70% ($1/\sqrt{2}$) of the required escape velocity.
To find circular orbital velocity, equate the centripetal force to the force of gravity: $$\frac{m v^2}{r} = G \frac{ M m}{r^2} \rightarrow \boxed{ v_\textrm{circ} = \sqrt{\frac{GM}{r}}}$$
To find escape velocity, equate the magnitude of the potential energy to that of the kinetic energy (i.e. to find zero total energy): $$\frac{1}{2}m v^2 = G \frac{ M m}{r} \rightarrow \boxed{ v_\textrm{esc} = \sqrt{2\frac{GM}{r}}}$$
Edit: where $G$ is Newton's Constant, $m$ is the mass of the object, $M$ is the mass of the gravitating body, and $r$ is the separation between the centers of mass.
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1
Where G = acceleration due to gravity, M is the mass of the body in orbit, and r is the altitude at which velocity is to be determined? – Everyone Jun 9 '12 at 6:59
Not quite---I added a clarification. Generally big-'G' is Newton's gravitational constant, while little-'g' is the acceleration due to gravity. And 'r' is the distance between the center of earth (the center of mass of the earth) and the altitude at which the velocity is desired. – zhermes Jun 9 '12 at 8:34
No, it isn't. If it were, then Geosynchronous Earth Orbit wouldn't be an orbit.
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http://en.wikipedia.org/wiki/Cube_root
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# Cube root
This article may be expanded with text translated from the corresponding article in the Catalan Wikipedia. (June 2012) Click [show] on the right to read important instructions before translating. Google's machine translation is a useful starting point for translations, but translators must revise errors as necessary and confirm that the translation is accurate, rather than simply copy-pasting machine-translated text into the English Wikipedia. Do not translate text that appears unreliable or low-quality. If possible, verify the text with references provided in the foreign-language article. After translating, `{{Translated|ca|Arrel cúbica}}` must be added to the talk page to ensure copyright compliance.
Plot of y = $\sqrt[3]{x}$ for $x \ge 0$. Complete plot is symmetric with respect to origin, as it is an odd function. At x = 0 this graph has a vertical tangent.
In mathematics, a cube root of a number, denoted $\sqrt[3]{x}$ or x1/3, is a number a such that a3 = x. All real numbers (except zero) have exactly one real cube root and a pair of complex conjugate roots, and all nonzero complex numbers have three distinct complex cube roots. For example, the real cube root of 8 is 2, because 23 = 8. All the cube roots of −27i are
$\sqrt[3]{-27i} = \begin{cases} \ \ \ \ \ \ \ \ \ \ \ 3i \\ \ \ \frac{3\sqrt{3}}{2}-\frac{3}{2}i \\ -\frac{3\sqrt{3}}{2}-\frac{3}{2}i. \end{cases}$
The cube root operation is not associative or distributive with addition or subtraction.
The cube root operation is associative with exponentiation and distributive with multiplication and division if considering only real numbers, but not always if considering complex numbers, for example:
$(\sqrt[3]{8})^3 = 8$
but
$\sqrt[3]{8^3} = \begin{cases} \ \ 8 \\ -4+4i\sqrt{3} \\ -4-4i\sqrt{3}. \end{cases}$
## Formal definition
The cube roots of a number x are the numbers y which satisfy the equation
$y^3 = x.\$
### Real numbers
The three cube roots of 1
If x and y are real, then there is a unique solution and so the cube root of a real number is sometimes defined by this equation. If this definition is used, the cube root of a negative number is a negative number.
If x and y are allowed to be complex, then there are three solutions (if x is non-zero) and so x has three cube roots. A real number has one real cube root and two further cube roots which form a complex conjugate pair. This can lead to some interesting results.
For instance, the cube roots of the number one are:
$\sqrt[3]{1} = \begin{cases} \ \ 1 \\ -\frac{1}{2}+\frac{\sqrt{3}}{2}i \\ -\frac{1}{2}-\frac{\sqrt{3}}{2}i. \end{cases}$
The last two of these roots lead to a relationship between all roots of any real or complex number. If a number is one cube root of any real or complex number, the other two cube roots can be found by multiplying that number by one or the other of the two complex cube roots of one.
### Complex numbers
Plot of the complex cube root together with its two additional leaves. The first picture shows the main branch which is described in the text
Riemann surface of the cube root. One can see how all three leaves fit together
For complex numbers, the principal cube root is usually defined by
$x^{1/3} = \exp ( \tfrac13 \ln{x} )$
where ln(x) is the principal branch of the natural logarithm. If we write x as
$x = r \exp(i \theta)\,$
where r is a non-negative real number and θ lies in the range
$-\pi < \theta \le \pi$,
then the principal complex cube root is
$\sqrt[3]{x} = \sqrt[3]{r}\exp ( \tfrac13 i\theta ).$
This means that in polar coordinates, we are taking the cube root of the radius and dividing the polar angle by three in order to define a cube root. With this definition, the principal cube root of a negative number is a complex number, and for instance $\sqrt[3]{-8}$ will not be $-2$, but rather $1 + i\sqrt{3}.$
This limitation can easily be avoided if we write the original complex number x in three equivalent forms, namely
$x = \begin{cases} r \exp \bigl(i (\theta) \bigr), \\ r \exp \bigl(i (\theta + 2\pi) \bigr), \\ r \exp \bigl( i (\theta - 2\pi) \bigr). \end{cases}$
The principal complex cube roots of these three forms are then respectively
$\sqrt[3]{x} = \begin{cases} \sqrt[3]{r}\exp \bigl( i ( \tfrac13 \theta) \bigr), \\ \sqrt[3]{r}\exp \bigl( i ( \tfrac13 \theta + \tfrac23 \pi ) \bigr), \\ \sqrt[3]{r}\exp \bigl( i ( \tfrac13 \theta - \tfrac23 \pi ) \bigr). \end{cases}$
In general, these three complex numbers are distinct, even though the three representations of x were the same. For example, ∛-8 may then be calculated to be -2, 1 + i√3, or 1 - i√3.
In programs that are aware of the imaginary plane, the graph of the cube root of x on the real plane will not display any output for negative values of x. To also include negative roots, these programs must be explicitly instructed to only use real numbers.
## Cube root on standard calculator
From the identity
$\frac{1}{3} = \frac{1}{2^2} \left(1 + \frac{1}{2^2}\right) \left(1 + \frac{1}{2^4}\right) \left(1 + \frac{1}{2^8}\right) \left(1 + \frac{1}{2^{16}}\right) \dots$
there is a simple method to compute cube roots using a non-scientific calculator, using only the multiplication and square root buttons, after the number is on the display. No memory is required.
• Press the square root button once. (Note that the last step will take care of this strange start.)
• Press the multiplication button.
• Press the square root button twice.
• Press the multiplication button.
• Press the square root button four times.
• Press the multiplication button.
• Press the square root button eight times.
• Press the multiplication button...
This process continues until the number does not change after pressing the multiplication button because the repeated square root gives 1 (this means that the solution has been figured to as many significant digits as the calculator can handle). Then:
• Press the square root button one last time.
At this point an approximation of the cube root of the original number will be shown in the display.
If the first multiplication is replaced by division, instead of the cube root, the fifth root will be shown on the display.
### Why this method works
After raising x to the power in both sides of the above identity, one obtains:
$x^{\frac{1}{3}} = x^{\frac{1}{2^2} \left(1 + \frac{1}{2^2}\right) \left(1 + \frac{1}{2^4}\right) \left(1 + \frac{1}{2^8}\right) \left(1 + \frac{1}{2^{16}}\right) \cdots}.$ (*)
The left hand side is the cube root of x.
The steps shown in the method give:
After 2nd step:
$x^{\frac{1}{2}}$
After 4th step:
$x^{\frac{1}{2} (1 + \frac{1}{2^2})}$
After 6th step:
$x^{\frac{1}{2} (1 + \frac{1}{2^2}) (1 + \frac{1}{2^4})}$
After 8th step:
$x^{\frac{1}{2} (1 + \frac{1}{2^2}) (1 + \frac{1}{2^4}) (1 + \frac{1}{2^8})}$
etc.
After computing the necessary terms according to the calculator precision, the last square root finds the right hand of (*).
### Alternative method
The above method requires the calculator to have a square root button. Having a simple method of calculating the square root the following function converges fast to the result:
$x_{i+1} = \tfrac{4}{3}\sqrt[4]{a x_{i}} - \tfrac{1}{3}x_i.$
Where with each iteration the result comes closer to the cube root of a.
The method requires fewer iterations than Halley's method but needs more calculations, hidden in determining the square roots. Because of the fast convergence, an initial approximation of 1 suffices.
## Numerical methods
Newton's method is an Iterative method that can be used to calculate the cube root. For real floating point numbers this method reduces to the following iterative algorithm to produce successively better approximations of the cube root of $a$:
$x_{i+1} = \frac{1}{3} \left(\frac{a}{x_i^2} + 2x_i\right).$
The method is simply averaging three factors chosen such that $x_i \times x_i \times \frac{a}{x_i^2}=a$ at each iteration.
Halley's method improves upon this with an algorithm that converges more quickly with each step, albeit consuming more multiplication operations:
$x_{i+1} = x_i \left(\frac{x_i^3 + 2a}{2x_i^3 + a}\right).$
With either method a poor initial approximation of $x_0$ can give very poor algorithm performance, and coming up with a good initial approximation is somewhat of a black art. Some implementations manipulate the exponent bits of the floating point number; i.e. they arrive at an initial approximation by dividing the exponent by 3. This has the disadvantage of requiring knowledge of the internal representation of the floating point number, and therefore a single implementation is not guaranteed to work across all computing platforms.
Also useful is this generalized continued fraction, based on the nth root method:
If x is a good first approximation to the cube root of z and y = z − x3, then:
$\sqrt[3]{z} = \sqrt[3]{x^3+y} = x+\cfrac{y} {3x^2+\cfrac{2y} {2x+\cfrac{4y} {9x^2+\cfrac{5y} {2x+\cfrac{7y} {15x^2+\cfrac{8y} {2x+\ddots}}}}}}$
$= x+\cfrac{2x \cdot y} {3(2z-y)-y-\cfrac{2\cdot 4y^2} {9(2z-y)-\cfrac{5\cdot 7y^2} {15(2z-y)-\cfrac{8\cdot 10y^2} {21(2z-y)-\ddots}}}}.$
The second equation combines each pair of fractions from the first into a single fraction, thus doubling the speed of convergence. The advantage is that x and y are only computed once.
## History
See also: Doubling the cube#History
In 499 CE Aryabhata, a great mathematician-astronomer from the classical age of Indian mathematics and Indian astronomy, gave a method for finding the cube root of numbers having many digits in the Aryabhatiya (section 2.5).[1]
## References
1. Aryabhatiya Marathi: आर्यभटीय, Mohan Apte, Pune, India, Rajhans Publications, 2009, p.62, ISBN 978-81-7434-480-9
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http://en.wikipedia.org/wiki/Non-circular_gear
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# Non-circular gear
Non-circular gear example
Another non-circular gear
A non-circular gear (NCG) is a special gear design with special characteristics and purpose. While a regular gear is optimized to transmit torque to another engaged member with minimum noise and wear and with maximum efficiency, a non-circular gear's main objective might be ratio variations, axle displacement oscillations and more. Common applications include textile machines,[1] potentiometers, CVTs (continuously variable transmissions),[2] window shade panel drives, mechanical presses and high torque hydraulic engines.[1]
A regular gear pair can be represented as two circles rolling together without slip. In the case of non-circular gears, those circles are replaced with anything different from a circle. For this reason NCGs in most cases are not round, but round NCGs looking like regular gears are also possible (small ratio variations result from meshing area modifications).
Generally NCG should meet all the requirements of regular gearing, but in some cases, for example variable axle distance, could prove impossible to support and such gears require very tight manufacturing tolerances and assembling problems arise. Because of complicated geometry, NCGs are most likely spur gears and molding or electrical discharge machining technology is used instead of generation.
## Mathematical description
Ignoring the gear teeth for the moment (i.e. assuming the gear teeth are very small), let $r_1(\theta_1)$ be the radius of the first gear wheel as a function of angle from the axis of rotation $\theta_1$, and let $r_2(\theta_2)$ be the radius of the second gear wheel as a function of angle from its axis of rotation $\theta_2$. If the axles remain fixed, the distance between the axles is also fixed:[3]
$r_1(\theta_1)+r_2(\theta_2)=a\,$
Assuming that the point of contact lies on the line connecting the axles, in order for the gears to touch without slipping, the velocity of each wheel must be equal at the point of contact and perpendicular to the line connecting the axles, which implies that:[3]
$r_1\,d\theta_1=r_2\,d\theta_2$
Of course, each wheel must be cyclic in its angular coordinates. If the shape of the first wheel is known, the shape of the second can often be found using the above equations. If the relationship between the angles is specified, the shapes of both wheels can often be determined analytically as well.[3]
It is more convenient to use the circular variable $z=e^{i\theta}$ when analyzing this problem. Assuming the radius of the first gear wheel is known as a function of z, and using the relationship $dz=iz\,d\theta$, the above two equations can be combined to yield the differential equation:
$\frac{dz_2}{z_2}=\frac{r_1(z_1)}{a-r_1(z_1)}\,\frac{dz_1}{z_1}$
where $z_1$ and $z_2$ describe the rotation of the first and second gears respectively. This equation can be formally solved as:
$\ln(z_2)=\ln(K)+\int\frac{r_1(z_1)}{a-r_1(z_1)}\,\frac{dz_1}{z_1}$
where $\ln(K)$ is a constant of integration.
1. ^ a b
2. ^ a b c
## Further reading
• Noncircular Gears: Design and Generation by Faydor L. Litvin, Alfonso Fuentes-Aznar, Ignacio Gonzalez-Perez, and Kenichi Hayasaka
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http://math.stackexchange.com/questions/106019/calculate-all-the-equivalent-martingale-measure/106244
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# calculate all the equivalent martingale measure
Under the assumption of no arbitrage without vanish risk, in an incomplete market $(\Omega,{\cal F}, P)$, the set of equivalent martingale measure is NOT empty, i.e. ${\cal P} = \{Q: Q \sim P\}\neq \emptyset.$
My question is: in the following simplified market with one stock which is driving by two independent Brownian Motions and one bond, i.e.
$$dS_t = S_t(\mu dt + \sigma_1 dW_1(t) + \sigma_2 dW_2(t))$$ $$dB_t = rB_tdt, \mbox{ } B_0 = 1$$
How to calculate all the equivalent martingale ${\cal P}.$ We suppose that $\mu,\sigma_1,\sigma_2, r$ are constants.
One approach in my mind is using another stock to complete the market, i.e. we suppose there is another stock $\tilde{S}$ with parameters, $\tilde{\mu}, \tilde{\sigma_1},\tilde{ \sigma_2}$ such that $$d\tilde{S_t} = \tilde{S_t}(\tilde{\mu} dt + \tilde{\sigma_1} dW_1(t) + \tilde{\sigma_2} dW_2(t)).$$
Then, following the classic method, we could get the equivalent martingale measures described by parameters, $\mu,\sigma_1,\sigma_2, r, \tilde{\mu}, \tilde{\sigma_1},\tilde{ \sigma_2}.$
But, how could I know the equivalent martingale measure obtained by above approach are the set of all the equivalent martingale measures in this financial market?
Any suggestion, reference books, or papers are welcome. Thanks.
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– learningmath Feb 5 '12 at 22:16
## 2 Answers
I'm afraid your market is complete here at least if all assets can only involve $S$ and $r$, so here only 1 Equivalent Martingale measure exists.
You can see this by seeing that you can express the dynamic of $S$ with only one Brownian motion as you have : $dS_t/S_t=\mu dt + \sqrt{\sigma_1^2+\sigma_2^2}dB_t$
where $B_t=\frac{\sigma_1 W^1_t+\sigma_2 W^2_t}{\sqrt{\sigma_1^2+\sigma_2^2}}$ is Brownian motion.
So you are back in a Black Scholes framework which is a complete market framework.
The situation would have be different if you had supposed that there exist assets that would involve only $W_1$ and/or $W_2$ separatly in there dynamics.
Best Regards
Ps : I agree with learningmath that this question would have been suitable for quantstackexchange forum.
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It is not complete since the martingale measure is NOT unique. – Jun Feb 7 '12 at 1:35
Jun : Yes it is, as long as you don't have assets driven by W^1 or W2 serparately as I have shown in my post. – TheBridge Feb 7 '12 at 7:54
The question is ambiguously worded. If the "martingale" in "equivalent martingale measure" is understood relative to the filtration generated by the stock price process $S$, then there is but one equivalent martingale measure (EMM), as noted by @TheBridge. If the filtration $({\mathcal F}_t)_{t\ge 0}$ is taken to be that generated by $W_1$ and $W_2$, then there are many EMMs. Suppose (for simplicity) that $\mu=0$ and $\sigma_1=\sigma_2=1$. Then for each real $\alpha$ the measure $Q_\alpha$ defined on ${\mathcal F}_t$ by $$dQ_\alpha/dP := \exp(\alpha W_1(t)-\alpha W_2(t)- \alpha^2t)$$ is an equivalent martingale measure. And there are more: If $H$ is a bounded $({\mathcal F}_t)$-predictable process then, on ${\mathcal F}_t$, $$dQ/dP :=\exp\left(\int_0^t H_s dW_1(s) - \int_0^t H_s dW_2(s) -\int_0^t H_s^2 ds\right)$$ defines an EMM.
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Yes. The filtration is generated by $W_1, W_2.$ Thanks. – Jun Feb 9 '12 at 3:22
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http://mathhelpforum.com/pre-calculus/10064-graph.html
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# Thread:
1. ## Graph
Graph y = sqrt{x^2}, y = x, y |x|, and y = (sqrt{x})^2, NOTING which graphs are the SAME.
NOTE:
What is the mirror-like image of the graph y = x?
2. Originally Posted by symmetry
Graph y = sqrt{x^2}, y = x, y |x|, and y = (sqrt{x})^2, NOTING which graphs are the SAME.
NOTE:
What is the mirror-like image of the graph y = x?
y = (sqrt(x))^2 is the same as --> y = x; y = sqrt(x^2) --> y = abs(x)
The mirror image is y = -x
3. Originally Posted by symmetry
Graph y = sqrt{x^2}, y = x, y |x|, and y = (sqrt{x})^2, NOTING which graphs are the SAME.
NOTE:
What is the mirror-like image of the graph y = x?
I'm not going to graph these, you should be able to do that yourself. However I will note that only two of these graphs are the same. The trick is in the last graph: $\sqrt{x^2}$ and $(\sqrt{x})^2$ are "apparently" the same since each reduces to x, but note that in the first expression the domain is $(-\infty, \infty )$ and in the last second expression is $[0, \infty )$.
$y = \sqrt{x^2}$ and $y = x$ aren't the same either since $\sqrt{.}$ only returns a positive value. It is for this very reason that $y = \sqrt{x^2}$ and $y = |x|$ have the same graph.
Originally Posted by symmetry
What is the mirror-like image of the graph y = x?
Depends on which "mirror" you are talking about. The typical mirror "planes" are the x and y axes, and the lines y = x and y = -x.
y = x reflected over the x-axis is y = -x.
y = x reflected over the y-axis is y = -x.
y = x reflected over the line y = x is y = x.
y = x reflected over the line y = -x is y = x.
-Dan
4. Originally Posted by AfterShock
y = (sqrt(x))^2 is the same as --> y = x; y = sqrt(x^2) --> y = abs(x)
The mirror image is y = -x
I've posted a graph of both functions. $y = \sqrt{x^2}$ is in solid red and $y = x$ is the dotted blue. It's hard to see the overlap in the first quadrant, but you know it's there.
-Dan
Attached Thumbnails
5. Originally Posted by topsquark
I've posted a graph of both functions. $y = \sqrt{x^2}$ is in solid red and $y = x$ is the dotted blue. It's hard to see the overlap in the first quadrant, but you know it's there.
-Dan
Indeed; as far as I could tell, he was asking for the mirror image of just y = x, not the original reflected over y = x.
Wasn't very clear.
6. Originally Posted by AfterShock
Indeed; as far as I could tell, he was asking for the mirror image of just y = x, not the original reflected over y = x.
Wasn't very clear.
Actually I wasn't very clear, sorry. I was responding to your comment that $y = (\sqrt{x})^2$ and $y = x$ have the same graph. But I graphed the wrong function. These aren't the same because the first only has the domain $[0, \infty )$, whereas the second has a domain of all real numbers.
-Dan
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http://physics.stackexchange.com/questions/48587/how-to-calculate-the-darcy-weissbach-friction-factor-for-shear-thinning-laminar
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# How to calculate the Darcy-Weissbach friction factor for shear thinning laminar flow in a pipe?
The Darcy-Weissbach friction factor for laminar flow would be $\frac{64}{Re}$
Now, having a shear thinning (non-newtonian) fluid where the viscosity is not constant how do I arrive at $Re$?
To know an apparaent viscosity, I'd need to know the shear rate, but that is not constant over the diameter of the pipes. Obviously I need to make allowances anyway (like assuming that my fluid obeys a power law over the relevant shear rates), so the method doesn't neet to be uber-exact. Bu I will want to know where I'm off.
Googling this, I only ound numerical/CFD solutions to far more complex problems an I couldn'T draw my answer from there.
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http://en.wikipedia.org/wiki/Raoult's_law
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# Raoult's law
Established by François-Marie Raoult in 1882,[1] Raoult's law states:
The vapour pressure of an ideal solution is directly dependent on the vapour pressure of each chemical component and the mole fraction of the component present in the solution.[2]
Once the components in the solution have reached equilibrium, the total vapour pressure p of the solution is:
$p = p^{\star}_{\rm A} x_{\rm A} + p^{\star}_{\rm B} x_{\rm B} + \cdots$
and the individual vapour pressure for each component is
$p_i = p^{\star}_i x_i$
where
pi is the partial pressure of the component i in the mixture (in the solution)
p*i is the vapor pressure of the pure component i
xi is the mole fraction of the component i in the mixture (in the solution).
If a pure solute which has zero vapor pressure (it will not evaporate) is dissolved in a solvent, the vapor pressure of the final solution will be lower than that of the pure solvent.
This law is strictly valid only under the assumption that intermolecular forces between unlike molecules are equal to those between similar molecules: the conditions of an ideal solution. Comparing measured vapor pressures to predicted values from Raoult's law provides information about the true relative strength of intermolecular forces. If the vapor pressure is less than predicted (a negative deviation), fewer molecules of each component than expected have left the solution in the presence of the other component, indicating that the forces between unlike molecules are stronger. The converse is true for positive deviations.
Raoult’s law assumes ideal behavior based on a simple picture just as the ideal gas law does. The ideal gas law is very useful as a limiting law. As the interactive forces between molecules and the volume of the molecules approach zero, so the behavior of gases approach ideality. Raoult’s law similarly assumes that the physical properties of the components are identical. The more similar the components are, the more their behavior approaches that described by Raoult’s law. For example, if the two components differ only in isotopic content, then the vapor pressure of each component (i) will be equal to the vapor pressure of the pure substance $p^{\star}_i$ times the mole fraction in the solution. This is Raoult’s law.
Vapor pressure of a binary solution which obeys Raoult's law. The black line shows the total vapor pressure as a function of the mole fraction of component B, and the two green lines are the partial pressures of the two components.
Using the example of a solution of two liquids, A and B, if no other gases are present, then the total vapor pressure p above the solution is equal to the weighted sum of the "pure" vapor pressures of the two components, pA and pB. Thus the total pressure above solution of A and B would be
$p = p^{\star}_{\rm A} x_{\rm A} + p^{\star}_{\rm B} x_{\rm B}.$
Since the sum of the mole fractions is equal to one,
$p = p^{\star}_{\rm A} (1-x_{\rm B}) + p^{\star}_{\rm B} x_{\rm B} = p^{\star}_{\rm A} + (p^{\star}_{\rm B}-p^{\star}_{\rm A}) x_{\rm B}$
which is a linear function of the mole fraction xB as shown in the graph.
## Relationship to thermodynamics
Raoult’s law was originally discovered as an approximate experimental law. Using Raoult’s law as the definition of an ideal solution, it is possible to deduce that the chemical potential of each component is given by
$\mu _i = \mu_i^{\star} + RT\ln x_i\,$,
where $\mu_i^{\star}$ is the chemical potential of component i in the pure state. This equation for the chemical potential may then be used to derive other thermodynamic properties of an ideal solution. (see Ideal solution)
However a more theoretical thermodynamic definition of an ideal solution is one in which the chemical potential of each component is given by the above formula. It is then possible to re-derive Raoult’s law as follows:
If the system is at equilibrium, then the chemical potential of the component i must be the same in the liquid solution and in the vapor above it. That is,
$\mu _{i,{\rm liq}} = \mu _{i,{\rm vap}}.\,$
Assuming the liquid is an ideal solution, and using the formula for the chemical potential of a gas, gives:
$\mu _{i,{\rm liq}}^{\star} + RT\ln x_i = \mu_{i,{\rm vap}}^\ominus + RT\ln \frac{{f_i }} {{p^\ominus }}$
where ƒi is the fugacity of the vapor of $i$ and $^\ominus$ indicates reference state.
The corresponding equation for pure $i$ in equilibrium with its (pure) vapor is:
$\mu _{i,{\rm liq}}^{\star} = \mu _{i,{\rm vap}}^\ominus + RT\ln \frac{{f_i^{\star}}} {{p^\ominus }}$
where * indicates the pure component.
Subtracting both equations gives us
$RT\ln x_i = RT\ln \frac{{f_i }}{{f_i^{\star} }}$
which re-arranges to
$f_i = x_i f_i^{\star}.$
The fugacities can be replaced by simple pressures if the vapor of the solution behaves ideally i.e.
$p_i \approx x_i p_i^{\star}$
which is Raoult’s Law.
## Ideal mixing
An ideal solution can be said to follow Raoult's Law but it must be kept in mind that in the strict sense ideal solutions do not exist. The fact that the vapor is taken to be ideal is the least of our worries. Interactions between gas molecules are typically quite small especially if the vapor pressures are low. The interactions in a liquid however are very strong. For a solution to be ideal we must assume that it does not matter whether a molecule A has another A as neighbor or a B molecule. This is only approximately true if the two species are almost identical chemically. We can see that from considering the Gibbs free energy change of mixing:
$\Delta_{\rm mix} G = nRT(x_1\ln x_1 + x_2\ln x_2).\,$
This is always negative, so mixing is spontaneous. However the expression is, apart from a factor –T, equal to the entropy of mixing. This leaves no room at all for an enthalpy effect and implies that ΔmixH must be equal to zero and this can only be if the interactions U between the molecules are indifferent.
It can be shown using the Gibbs–Duhem equation that if Raoult's law holds over the entire concentration range x = 0–1 in a binary solution then, for the second component, the same must also hold.
If the deviations from ideality are not too strong, Raoult's law will still be valid in a narrow concentration range when approaching x = 1 for the majority phase (the solvent). The solute will also show a linear limiting law but with a different coefficient. This law is known as Henry's law.
The presence of these limited linear regimes has been experimentally verified in a great number of cases. In a perfectly ideal system, where ideal liquid and ideal vapor are assumed, a very useful equation emerges if Raoult's law is combined with Dalton's Law.
$x_i = \frac{y_i p_{\rm total}}{p_{i,{\rm eqm}}}\,$,
where $x_i$ is the mole fraction of component $i$ in the solution and $y_i$ is its mole fraction in the gas phase. This equation shows that, for an ideal solution where each pure component has a different vapor pressure, the gas phase will be enriched in the component with the higher pure vapor pressure and the solution will be enriched in the component with the lower pure vapor pressure. This phenomenon is the basis for distillation.
## Non-ideal mixing
Raoult's Law may be adapted to non-ideal solutions by incorporating two factors that will account for the interactions between molecules of different substances. The first factor is a correction for gas non-ideality, or deviations from the ideal-gas law. It is called the fugacity coefficient ($\phi$). The second, the activity coefficient ($\gamma$), is a correction for interactions in the liquid phase between the different molecules.
This modified or extended Raoult's law is then written:
$p_{i} \phi_i = p_i^{\star} \gamma_i x_i.\,$
## Real solutions
Many pairs of liquids are present in which there is no uniformity of attractive forces i.e. the adhesive and cohesive forces of attraction are not uniform between the two liquids, so that they show deviation from the Raoult's law which is applied only to ideal solutions.
### Negative deviation
Positive and negative deviations from Raoult's law. Maxima and minima in the curves (if present) correspond to azeotropes or constant boiling mixtures.
If the vapor pressure of a mixture is lower than expected from Raoult's law, there is said to be a negative deviation. This is evidence that the adhesive forces between different components are stronger than the average cohesive forces between like components. In consequence each component is retained in the liquid phase by attractive forces which are stronger than in the pure liquid so that its partial vapor pressure is lower.
For example, the system of chloroform (CHCl3) and acetone (CH3COCH3) has a negative deviation[3] from Raoult's law, indicating an attractive interaction between the two components which has been described as a hydrogen bond.[4] The system hydrochloric acid - water has a large enough negative deviation to form a minimum in the vapor pressure curve known as a (negative) azeotrope, corresponding to a mixture which evaporates without change of composition.[5]
### Positive deviation
When the cohesive forces between like molecules are greater than the adhesive forces, the dissimilarities of polarity or internal pressure will lead both components to escape solution more easily. Therefore, the vapor pressure will be greater than expected from the Raoult's law, showing positive deviation. If the deviation is large, then the vapor pressure curve will show a maximum at a particular composition and form a positive azeotrope. Some mixtures in which this happens are (1) benzene and methyl alcohol, (2) carbon disulfide and acetone, and (3) chloroform and ethanol.
## Electrolytes solutions
The expression of the law for this case includes the van't Hoff factor.
## References
Chapter 24, D A McQuarrie, J D Simon Physical Chemistry: A Molecular Approach. University Science Books. (1997)
E. B. Smith Basic Chemical Thermodynamics. Clarendon Press. Oxford (1993)
1. Barnett, Martin K. (2001), A Dictionary of the History of Science, Informa Health Care, p. 287 . Extract.
2. A to Z of Thermodynamics by Pierre Perrot. ISBN 0-19-856556-9
3. P. Atkins and J. de Paula, Physical Chemistry (8th ed., W.H. Freeman 2006) p.146
4. K. Kwak, D.E. Rosenfeld, J.K. Chung and M.D. Fayer, J. Phys. Chem. B (2008), 112, 13906–13915
5. Atkins and de Paula, p.184
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http://physics.stackexchange.com/questions/tagged/unitarity?sort=active&pagesize=15
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# Tagged Questions
The unitarity tag has no wiki summary.
2answers
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### Is there a symmetry associated to the conservation of information?
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### why nontrivially space-like connected event horizons do not respect unitarity?
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### Krauss operators for random unitary
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### Constructing the exponential form of a unitary operator
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### Unitarity of S-matrix in QFT
I am a beginner in QFT, and my question is probably very basic. As far as I understand, usually in QFT, in particular in QED, one postulates existence of IN and OUT states. Unitarity of the S-matrix ...
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### Information loss
First time poster! I just burnt a piece of paper containing a 5 digit number I made up randomly and as far as I am concerned no one else will ever be able to retrieve the information contained on ...
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### Holstein-Primakoff and Dyson-Maleev representation
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### Information Loss in annihilation
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### Validity of Cutkosky cutting rules for fermions
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### Does unitarity apply in between measurements?
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### When can a classical field theory be quantized?
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http://mathoverflow.net/questions/50544/some-dirichlet-series-questions/50688
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## Some Dirichlet series questions.
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I asked this question on m.SE in an attempt to find out the right words to say for these questions I am about to ask.
In his great answer, Matthew Emerton explained that (cuspidal) automorphic L-functions correspond to the Dirichlet series with "nice" properties like having a reflection equation, a meromorphic continuation to the entire complex plane and a suitable analogue of the Riemann hypothesis. This leads me to my first question:
1) Are there Dirichlet series that cannot be classified as (cuspidal) automorphic L-functions, yet still possess a critical line of nontrivial zeroes?
Now, I come to the question I had been meaning to ask here. It is known that functions like Riemann $\zeta$ and the Ramanujan Dirichlet series admit a "Riemann-Siegel" decomposition; that is, letting $\sigma$ denote the position of the "critical line" of the Dirichlet series $g(s)$, they can be expressed as
$$g(\sigma+it)=z(t)\exp(-i\vartheta(t))$$
where $z(t)$ and $\vartheta(t)$ are "Riemann-Siegel" functions corresponding to the Dirichlet series $g(s)$. The value of $z(t)$ is that it eases the task of finding nontrivial zeroes of the corresponding Dirichlet series (essentially helping to verify its corresponding "hypothesis"). My question now is
2) Do all (cuspidal) automorphic L-functions have a "Riemann-Siegel" decomposition? If not, what restrictions are there for them to possess such a decomposition?
My motivation is more of curiosity than anything else. Hopefully this is not too elementary a question!
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## 2 Answers
In answer to #2, the Riemann-Siegel formula was developed by Siegel based on unpublished posthumous notes of Riemann, the Nachlass. It writes the Hardy function $Z(t)$ as a sum of two finite series of length about $\sqrt{t}$, up to a small error. The two finite series show explicitly the functional equation, i.e one is in $1/2+it$, the other $1/2-it$. Riemann-Siegel replaced earlier $O(t)$ methods of computation such as Euler-Maclaurin summation. Edwards book on Riemann's Zeta Function, reprinted by Dover, is a good source.
Extending this to other Dirichlet series is complicated by the fact that the coefficients are so unruly (the zeta function is the Dirichlet generating function of the constant sequence.) Davies, in "An approximate functional equation for Dirichlet L-functions." Proc. Roy. Soc. Ser. A 284 1965 pp. 224–236. considered the case of Dirichlet $L$-functions and did computations of zeros. Motohashi did $\zeta(s)^2$, which is the Dirichlet generating function of the divisor function. There are some other results, but if constant in the big Oh error in the truncation is not made explicit, they are not useful for computation.
A more modern approach which applies generally to automorphic $L$-functions uses something called an 'Approximate Functional Equation', see section 5.2 of the book Analytic Number Theory by Iwaniec and Kowalski. Iwaniec elsewhere has described this as "a Dirichlet series representation ... tempered by a test function which makes the series rapidly convergent. Formulas of this type are known in the literature as 'approximate functional equations'. [T]his is a somewhat misleading name, because we need exact expressions ... We rather think of these as a kind of Poisson's summation formulas." The test function replaces the sharp cutoff of the a finite series in Riemann-Siegel, which necessitates a corresponding error estimate, with a rapidly decaying test function of the user's choice. I believe this form is due to Lavrik, see MR0188170 (32 #5609). This was first used for computations by PJ Weinberger in "On small zeros of Dirichlet $L$-functions." Math. Comp. 29 (1975), pp. 319–328.
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So, what was so special with, for instance, the Ramanujan Dirichlet series that enabled it to have a(n admittedly rather complicated) "Riemann-Siegel" decomposition? I'd think $\tau(n)$ is an even more unruly function than a Dirichlet character... – J. M. Dec 29 2010 at 2:52
3
Yes, and that result was extremely technically difficult. Also, Gritsenko did the case of the product of two Dirichlet $L$-functions, which in the case of two real characters, are the $L$-functions of genus characters on the class group. Riemann-Siegel can be extended a lot but as I said above, without explicit constants in the big Oh, it's not useful for computation. The 'Approximate Functional Equation' method generalizes Riemann-Siegel, and is much easier. – Stopple Dec 29 2010 at 23:23
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
In answer to #1, I'm not quite sure what you're asking but I'll point out the Epstein zeta function $\zeta_Q(s)=\sum^\prime_{(m,n)}Q(m,n)^{-s}$ attached to, say, a positive definite binary quadratic form $Q(x,y)=ax^2+bxy+cy^2$ has a functional equation under $s\to 1-s$, but except in extraordinary cases (class number 1) they do not possess an Euler product over primes. Setting $z=\frac{-b+i\sqrt{D}}{2a}$ gives a point in the complex upper half plane (these are more or less what Heegener points are) and makes the Epstein zeta function an Eisenstien series. Thus these are automorphic, but not cuspidal.
This has been a area of research on the failure of the Riemann Hypothesis, they are know to have real zeros in $(1/2,1)$ if $\sqrt{d}/a$ is big enough. Up to large height (depending on $d$) the complex zeros are on the critical line.
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Hmm, okay. Is there a Dirichlet series that isn't automorphic but still has "nice" properties, or is that condition absolutely necessary? – J. M. Dec 30 2010 at 3:59
'Nice' is a little vague. From a Dirichlet series $\sum_n a_n n^{-s}$ one can formally construct a function $f(z)=\sum_n a_n\exp(2\pi i nz)$. This $f$ is invariant under $z\to z+1$, i.e automorphic under the action of the matrix $T=\left(\begin{smallmatrix} 1&1\\0&1\end{smallmatrix}\right)$. This is one of the two canonical generators of the modular group $SL(2,\mathbb Z)$. Automorphy under the action of the other, $S=\left(\begin{smallmatrix} 0&1\\-1&0\end{smallmatrix}\right)$, will correspond to the Dirichlet series having a functional equation. – Stopple Dec 31 2010 at 21:45
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http://math.stackexchange.com/questions/32100/is-there-a-third-dimension-of-numbers/32103
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# Is there a third dimension of numbers?
Is there a third dimension of numbers like real numbers, imaginary numbers, [blank] numbers?
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12
– Aryabhata Apr 10 '11 at 16:30
Wow, that is annoying. Are three-dimensional numbers even possible/necessary? – awesomeguy Apr 10 '11 at 16:42
1
Quaternions are four-dimensional, but they are used in computer graphics to deal with 3-D situations. – André Nicolas Apr 10 '11 at 16:54
6
A good hint that there is no three dimensional numbers is the fact that $(a_1^3 + b_1^3 + c_1^3)(a_1^3 + b_2^3 + c_2^3)$ doesn't always equal the sum of three cube numbers (the axiom that product of two three cubes is another three cube). The quaternions barely satisfy the condition for four dimensional except for commutativity. – Mark Apr 10 '11 at 20:27
4
EDIT for above comment: I accidentally wrote cubes for some reason. I meant the product of three squares $(a_1^2+b_1^2+c_1^2)(a_2^2+b_2^2+c_2^2)$ doesn't always equal the sum of three square numbers. It's well known in any elementary number theory book that Euler proved its true for product of four square hinting that four dimensional numbers exist. – Mark Apr 10 '11 at 22:36
## 7 Answers
Alas, there are no algebraically coherent "triplexes". The next step in the construction as has been said already are "quaternions" with 4 dimensions.
Many young aspiring mathematicians have tried to find them since Hamilton in the 19th century. This impossibility links geometric dimensionality, fundamental properties of polynomial equations, algebraic systems and many other aspects of mathematics. It is really worth studying.
A quite recent book by modern mathematicians which details all this for advanced college undergraduates is Numbers by Ebbinghaus, Hermes, Hirzebruch, Koecher, Mainzer, Neukirch, Prestel, Remmert, and Ewing.
However, the set of quaternions with zero real part is an interesting system of dimension 3 with very interesting properties, linked to the composition of rotations in space.
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You may also find of interest some more general results besides the mentioned Frobenius Theorem. Weierstrass (1884) and Dedekind (1885) showed that every finite dimensional commutative extension ring of $\mathbb R$ without nilpotents ($\rm\:x^n = 0 \ \Rightarrow\ x = 0\:$) is isomorphic as a ring to a direct sum of copies of $\rm\:\mathbb R\:$ and $\rm\:\mathbb C\:.\:$ Wedderburn and Artin proved a generalization that every finite-dimensional associative algebra without nilpotent elements over a field $\rm\:F\:$ is a finite direct sum of fields. Such structure theoretic results greatly simplify classifying such rings when they arise in the wild. For example, I applied a special case of these results last week to prove that a finite ring is a field if its units $\cup\ \{0\}$ comprise a field of characteristic $\ne 2\:.\:$ For another example, a sci.math reader once proposed an extension of the real numbers with multiple "signs". This turns out to be a very simple case of the above results. Below is my 2009.6.16 sci.math post on this topic.
The results in Eitzen's paper Understanding PolySign Numbers the Standard Way, characterizing Tim Golden's so-called PolySign numbers as ring direct sums of $\mathbb R$ and $\mathbb C$, have been known for over a century and a half. Namely that $\rm\:P_n =\: \mathbb R[x]/(1+x+x^2+\:\cdots\: + x^{n-1})\$ is isomorphic to a certain ring direct sum of $\:\mathbb R$ and $\:\mathbb C\:,\:$ is just a special case of more general results due to Weierstrass and Dedekind in the 1860s. These classic results are so well-known that you will find them mentioned even in many elementary textbooks on number systems and their generalizations. For example, in Numbers by Ebbinghaus et.al. p.120:
Weierstrass (1884) and Dedekind (1885) showed that every finite dimensional commutative ring extension of R with unit element but without nilpotent elements, is isomorphic to a ring direct sum of copies of R and C.
Ditto for historical expositions, e.g. Bourbaki's Elements of the History of Mathematics, p. 119:
By 1861, Weierstrass, making precise a remark of Gauss, had, in his lectures, characterized commutative algebras without nilpotent elements over R or C as direct products of fields (isomorphic to R or C); Dedekind had on his side reached the same conclusions around 1870, in connection with his "hypercomplex" conception of the theory of commutative fields, their proofs were published in 1884-85 [1,2]. [...] These methods rely above all on the consideration of the characteristic polynomial of an element of the algebra relative to its regular representation (a polynomial already met in the work of Weierstrass and Dedekind quoted earlier) and on the decomposition of the polynomial into irreducible factors.
Nowadays these fundamental results are merely special cases of more general structure theories for algebras that are part of any first course on algebras (but not always met in a first course on abstract algebra). A web search turns up more on the subsequent history, e.g. excerpted from
Y. M. Ryabukhin, Algebras without nilpotent elements, I,
Algebra i Logika, Vol. 8, No. 2, pp. 181-214, March-April, 1969
Algebras without nilpotent elements have been studied long ago. So, Weierstrass characterized in his lectures in 1861 finite-dimensional associative-commutative algebras without nilpotent elements over the field of real or complex numbers as finite direct sums of fields. To be exact, some nonessential restrictions have there been imposed. In 1870 Dedekind removed those nonessential restrictions. The following theorem of Weierstrass-Dedekind is now considered as a classical one: every finite-dimensional associative-commutative algebra without nilpotent elements over a field F is a finite direct sum of fields. The results of Weierstrass and Dedekind (for the case when F is the field of complex or real numbers) have been published in [1,2]. The results of works of Molien, Cartan, Wedderburn and Artin [3-6] imply that Dedekind's theorem holds for any field F. Moreover, the following theorem of Wedderburn-Artin holds: every finite-dimensional associative algebra without nilpotent elements over a field F is a finite direct sum of fields." [...]
1. K. Weierstrass, "Zur Theorie der aus n Haupteinheiten gebildeten complexen Grossen," Gott. Nachr. (1884).
2. R. Dedekind, "Zur Theorie der aus n Haupteinheiten gebildeten complexen Grossen," Gott. Nachr. (1885).
3. F. Molien, "Ueber Systeme hoherer complexer Zahlen," Math. Ann., XLI, 83-156 (1893).
4. E. Cartan, "Les groupes bilineaires et les systemes de nombres complexes," Ann. Fac. Sci., Toulouse (1898).
5. J. Wedderburn, "On hypercomplex numbers," Proc. London Math. Soc. (2), VI, 349-352 (1908).
6. E. Artin, "Zur Theorie der hyperkomplexen Zahlen," Abh. Math. Sere. Univ. Hamburg, 5, 251-260 (1927).
and excerpted from its sequel
Y.M. Ryabukhin, Algebras without nilpotent elements, II,
Algebra i Logika, Vol. 8, No. 2, pp. 215-240, March-April, 1969
In [1] we proved structural theorems on the decomposition of algebras without nilpotent elements into direct sums of division algebras; certain chain conditions were imposed on these algebras.
Yet it is possible to prove structural theorems also without imposing any chain conditions. In this case the direct sums are replaced by subdirect sums and instead of division algebras we shall consider algebras without zero divisors.
The first structural theorem of this kind is apparently the classical theorem of Krull [2]:
Any associative-commutative ring without nilpotent elements can be represented by a subdirect sum of rings without zero divisors. Krull's theorem was subsequently extended to the case of any associative ring. This was done by various authors and in various directions. In [3], Thierrin came very close to a final generalization of Krull's theorem to the associative, but not commutative case. The final result was obtained in [4]:
Any associative ring without nilpotent elements can be represented by a subdirect sum of rings without zero divisors. At the Ninth All-Union Conference on General Algebra (held at Gomel'), I. V. L'vov reported an even stronger result:
Any alternative ring without nilpotent elements can be represented by a subdirect sum of rings without zero divisors.
It could be assumed that the theorem on decomposition into a subdirect sum of algebras without zero divisors holds for any ring. Yet this assumption is erroneous (see [1]), since there exists a finite-dimensional simple, special Jordan algebra without nilpotent elements that has zero divisors and cannot therefore be decomposed into a subdirect sum of algebras (or rings) without zero divisors.
There naturally arises the following question: what conditions must a ring without nilpotent elements satisfy to permit its representation by a subdirect sum of rings without zero divisors?
In this paper we answer this question:
An algebra R over an associative-commutative ring F with unity can be represented by a subdirect sum of rings without zero divisors, iff it is a conditionally associative algebra without nilpotent elements.
Let us recall that an algebra R is said to be conditionally associative, iff we have in R the conditional identity x(yz) = 0 iff (xy)z = 0.
We say a (not necessarily associative) algebra R does not have nilpotent elements, iff in R we have the conditional identity x^2 = 0 iff x = 0.
From this theorem we easily obtain the above-mentioned results of [2-4], as well as the result of L'vov (it suffices to take as the ring F the ring Z of integers). [...]
1. Yu. M. Ryabukhin, "Algebras without nilpotent elements,I," this issue, pp. 215-240.
2. W. Krull, "Subdirect representations of sums of integral domains," Math. Z., 52, 810-823 (1950).
3. J. Thierrin, "Completely simple ideals of a ring," Acad. Belg. Bull. C1. Sci., 5 N 43, 124-132 (1957}.
4. V. A. Andrunakievich and Yu. M. Ryabukhin, "Rings without nilpotent elements in completely simple ideals," DAN SSSR, 180, No. 1, 9 (1968).
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Bill, something seems to be wrong with the title of Dedekind's paper. Two blanks are missing. When you add these, the title becomes identical to Weierstrass' which also seems weird to me. – Rasmus Apr 10 '11 at 20:15
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@Ras No, it's correct. The papers do in fact have the same titles. – Gone Apr 10 '11 at 21:04
Interesting, was it sort of a collaboration between them? – Rasmus Apr 11 '11 at 7:15
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Wow. This must have taken hours! Sadly, I am limited to (+1)... – The Chaz 2.0 Jul 13 '11 at 2:51
Every finite-dimensional division algebra over $\mathbb{R}$ is one of $\mathbb{R}$, $\mathbb{C}$ or $\mathbb{H}$. This is what is called the Frobenius Theorem. You may refer to here for details.
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Every finite dimensional associative division algebra over $\mathbb R$ is one of the above. The octonions $\mathbb O$ is also a division algebra, but it is not associative. – Calle Apr 10 '11 at 17:29
@sis440: You mean Frobenius. More broadly, you might also like to lookup theorems by Dickson, Artin and Wederburn which gives a broader picture of the classification of algebra. – ogerard Apr 10 '11 at 17:35
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Oh, I forgot that division algebra need not be associaitive... You are right, $\mathbb{O}$ has to be added if associativity is omitted. – sos440 Apr 10 '11 at 17:35
You might look up quaternions.
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This answer is useful. – The Chaz 2.0 Jul 13 '11 at 2:52
At the time, I did not know how to insert links. – André Nicolas Jul 13 '11 at 3:26
I was hovering over the +1 and it said "This answer is useful". I agree(d), and clicked! ..and quoted the popup :) – The Chaz 2.0 Jul 13 '11 at 3:39
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Will try hovering over one of your answers, and do what the gnome in the machine asks me to do. – André Nicolas Jul 13 '11 at 4:31
In addition to complex numbers and quaternions, you might want to look up Clifford Algebras which encapsulate both and extend to arbitrary dimension. Complex and quaternios are sub-algebras of the Clifford Algebras over $\mathbb{R}^2$ and $\mathbb{R}^3$ respectively.
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There are of course, examples of R3, in the form X, Y, Z, which behave as three separate real numbers, with an invarient when X -> Y -> Z -> X applies.
Such numbers turn up in the study of the heptagon and enneagon. The integer systems are discrete points forming a lattice in this space, the invariants cycle through the solutions to the heptagonal ($x^3-x^2+2x-1=0$) and enneagonal equations, which transforms eg {7} to {7/3} to {7/2}.
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This answer makes no sense. – Zev Chonoles♦ 2 days ago
I'm pretty sure that the process is similar to the process of replacing x+iY by x-iy, but through the solutions of 3d. If you replace i² = -1 by j² = +1, you get a hyper-complex space, which transforms a+b sqrt(5) into a-b sqrt(5). It can be done over any number of dimensions, if the values solve an integer polynomial of the same order. – wendy.krieger 2 days ago
If you think of the dimensions for numbers as going real numbers (1st dimension), fuzzy numbers (2nd dimension), then the 3rd dimension ends up fuzzy numbers of dimension two. For more details see A. Kaufmann and M. M. Gupta's Introduction to Fuzzy Arithmetic or George and Maria Bojadziev's Fuzzy Sets, Fuzzy Logic, Applications.
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http://mathoverflow.net/questions/59717/non-linearity-of-the-consistency-strength-ordering-in-zf/59800
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## (Non?)-linearity of the consistency strength ordering in ZF
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Much of the research taking place in set theory, is related to the classification of sentences according to their consistency strength relative to ZF. In order to be more specific, we say that for all sentences $\sigma,\tau\in Sent$, $\sigma$ has less consistency strength than $\tau$ (or $\sigma\leq_{cons}\tau$) iff $cons(ZF+\tau)\rightarrow cons(ZF+\sigma)$. My main question (which I believe should have a negative answer) is the following: $$\mbox{ Is } \leq_{cons}\mbox{ linear? If not is there a natural counterexample?}$$ In fact I have several similar questions concerning $\leq_{cons}$ and I would be glad if someone could provide me with some references. Thanks!
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I'm not even close to know enough for a proper answer, but I know that some large cardinal axioms might not be linearly ordered. You can find a graph in Kanamori's The Higher Infinite, and see there for yourself some of the possibly-non-linear sentences. – Asaf Karagila Mar 27 2011 at 13:43
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Google gave me this: websupport1.citytech.cuny.edu/faculty/vgitman/… I believe it is the same one as in Kanamori. – Tanmay Inamdar Mar 27 2011 at 16:06
## 6 Answers
Marios, this is indeed a fascinating topic.
The consistency strength hierarchy is not linearly ordered. One can produce counterexamples by variants of Gödel sentences or of Rosser sentences. It is actually an interesting exercise to produce explicit examples of a pair of Rosser-like sentences $\phi,\psi$ both independent of ZFC, such that ZFC$+\phi$ and ZFC$+\psi$ are incomparable, by letting $\phi$ refer to the length of proofs of consistency of ZFC$+\psi$, and letting $\psi$ refer to the length of proofs of the consistency of ZFC$+\phi$. Explicit examples can be found, I believe, in Per Lindström's "Aspects of Incompleteness" (Lecture Notes in Logic 10).
Harvey Friedman has worked on this question, and expanded on the idea above. I believe it is his result that one can even embed a universal countable partial order in the consistency strength degrees or in the degrees of interpretability. I'm pretty sure he has commented on this in the FOM list, but I don't seem to be able to find his relevant postings right now. (I once gave a talk on this years ago, but cannot find my slides either. Old age...)
On the other hand, everybody agrees that these examples are not natural (see for example, the beginning of the section "The Gödel hierarchy" in this paper by Simpson, or page 54 of "Fixing Frege" by John P. Burgess). It is a remarkable empirical phenomenon that we indeed have comparability for natural theories. We expect this to always be the case, and a significant amount of work in inner model theory is guided by this belief. But although we have many striking partial results in this direction, it is not quite a theorem, in part because natural is not a formal concept.
The standard way by which we establish comparability of statements is by using forcing and the technique of inner models to measure a given statement against the large cardinal hierarchy, identifying this way the large cardinal companion of the statement. Different statement can then be compared by comparing their companions.
Hugh Woodin, in his book on Pmax suggests a way of formalizing this observation, by explaining how to associate Universally Baire sets to theories, the point being that under appropriate determinacy assumptions, the Wadge degrees of any two such sets are comparable, and the resulting ordering coincides precisely with the interpretability or consistency strength orderings. (The universally Baire sets ultimately trace back to the complexity of certain iteration strategies.) This, of course, requires that we work under additional large cardinal assumptions beyond ZFC, so it only applies to theories of high consistency strength (past $\omega$ Woodin cardinals, for example). This remarkable approach has thus the drawback that it does not explain why we find the phenomenon in the low levels of the hierarchy, the ones that are precisely amenable to current inner model theoretic techniques.
At the moment there are a few notable examples of theories that we do not quite know how to fit into the linear ordering of consistency strength. This is most likely due to our present technical limitations. For example, strongly compact cardinals are expected to either coincide or sit just below supercompact cardinals consistency-wise. However, their behavior is rather atypical and they are not really as well understood as other (strong) large cardinals.
For a nice introduction to these ideas, examples, and a far reaching program that touchesof Woodin's recent work on "ultimate $L$", see these slides of a talk by John Steel at the Workshop on Set Theory and the Philosophy of Mathematics, Oct. 2010.
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Minor typo: Wadge, not Wedge. In the fifth paragraph. (Sorry for the nitpicking!) – Tanmay Inamdar Mar 28 2011 at 5:49
Hehe. Thanks, fixed. – Andres Caicedo Mar 28 2011 at 6:44
Thank you very much Andres! This is truely a very fascinating part of mathematics I wasn't aware of. It provides motivation and a nice background for what is taking place in set theory. – Marios Koulakis Apr 6 2011 at 12:15
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There are some results of Woodin that indicate why large cardinal axioms of a certain kind might indeed be linearly ordered by consistency strength. In Kanamori's chart there are some possibilities for nonlinearity, but I think this is just lack of knowledge.
I am not aware of any result that actually says that two natural statements are incomparable in consistency strength over ZF. Note however that this might be a problem of our current proof technology.
We have good technology to show that the consistency of a statement $\sigma$ implies the consistency of $\tau$ (all over ZF). Namely, start with a model of ZF+$\sigma$ and use forcing and/or inner models to construct a model of ZF+$\tau$. The second incompleteness theorem tells us how we can get a stronger theory from a consistent theory $T$, namely by adding the axiom "$T$ is consistent".
But how could we prove that two given statements are incomparable in consistency strength?
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The order $\le_{cons}$ isn't linear. There are Godel Lob provability logic $GL$. It's a modal logic with one modality. $GL$ is complete with respect to arithmetic semantic. It was shown by Solovay. Formula $\square(\Diamond p \to \Diamond q) \lor \square(\Diamond q \to \Diamond p)$ is not theorem of $GL$. So by completeness theorem (it can be proved for $ZF$) there are propositions $A$ and $B$ such that $ZF \not \vdash (Con({\ulcorner} A{\urcorner}) \to Con({\ulcorner} B{\urcorner}))$ and $ZF\not \vdash (Con({\ulcorner} B{\urcorner}) \to Con({\ulcorner} A{\urcorner}))$ But such $A$ and $B$ provided by Solovay proof are complex. I don't know any "natural" example.
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For theories one can define their proof theoretic ordinal, which is a countable ordinal that is (roughly) the smallest ordinal that the theory cannot prove is well ordered. If one orders theories by the size of the proof theoretic ordinal, this gives a total order on theories because ordinals are totally ordered. In practice the size of the proof-theoretic ordinal is a good measure of the consistency strength of a theory (roughly speaking, proving that the proof theoretic ordinal is well ordered tends to be about as strong as proving that the theory is consistent) so consistency strength usually gives a total order on theories.
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My understanding is that the large cardinals are indeed linearly ordered, which is a remarkable fact. Of course, it is not a theorem (nor could it ever be?), but merely an empirical fact; still, there is no known explanation for this completely surprising coincidence.
On the other hand, there are sentences $\sigma$ such that $T+\sigma>T$ and $T+\neg\sigma>T$. These are called "Double jump sentences" and though they are rare, a number of them exist. Thus, $\le_{cons}$ is not linear, though the counterexamples are quite contrived (none of the known ones would be called "natural" by a mathematician).
You can read more about Double jump sentences (and find references) here:
http://plato.stanford.edu/entries/independence-large-cardinals/#IntHie
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Terminological note: "Orey sentence" isn't another name for a double jump sentence. It refers instead to when there is no jump: $T + \sigma \equiv T + \neg\sigma \equiv T$. – Ed Dean Mar 28 2011 at 1:00
You are entirely right; I stand corrected. – Alex Lupsasca Mar 30 2011 at 19:39
Here is the relevant posting on the FOM list by Harvy Friedman can be found here: http://www.cs.nyu.edu/pipermail/fom/1998-August/001929.html The result that the consistency strength ordering has an infinite set of incomparable elements is found in corollary 4. It seems to me by the way that his proof of corollary 4 only holds for models of PA+Con(PA), not models of PA+notCon(PA)
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http://mathoverflow.net/revisions/49474/list
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## Return to Question
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Solutions to this equation, however, are not unique. If we fill any unit real interval with some function, then the equation entails the function at the whole real line (provided $f$ is defined there.). So there are as many solutions as there are functions on the real numbers -- an uncountable $\beth_2$ possible solutions (which is even bigger than the continuum itself at $\beth_1$). In general, given any solution $F$, we can express any other $G$ as $G(x) = F(x) + \theta(x)$, where $\theta(x)$ is a 1-periodic "wobble" function.
And thus our problem is: is there a "good" or "natural" some solution to of this equation for a given function $f$? f$, which is more "good" or "natural" than others? To attempt to ease the problem, it seems best to impose some restrictions on the possible candidate pool of input function functions$f$, f$ to consider, as in general, the tamer the function, the easier the analysis of the equation and the more techniques are available. But we don't want too much up-front restriction, or the method could be methods available may then become relatively useless (e.g. a method that only summed linear functions and nothing else would have little to no use.).
There was a method proposed by Markus Mueller in an article called "Fractional sums and Euler-like identities" that attempts to do something like this, but it suffers from some drawbacksits restrictions seem a little too stiff. For example, it does not appear the method is of any use in summing, e.g.
So we want
What is wanted here is to come up with some kind of "grand unified theory" of these kinds of sums: a solution for the functional equation that would be able to recover most if not all of these kind of formulae, and also enable the summation of many other functions.
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where $a$ and $b$ are non-integer fractional, real, or even complex numbers, at least for a sufficiently well-behaved target function $f$ defined on at least one of those domains.
What does that mean, exactly? Well, I consider this the problem of constructing a "summation operator" $\Sigma$ which is an "inverse" of the unit forward difference operator $\Delta$, but with both acting on functions on the real numbersor complex continuum instead of just on the integers. This relationship is the same as how the integral is the inverse of the derivative. That is, but applying $\Sigma$ to a function $f$ is equivalent to solving the functional equation $F(z + 1) - F(z) = f(z)$ for $F$. With such an $F$ in hand, we can then say $\sum_{n=a}^b f(n) = F(b+1) - F(a)$. And thus our problem is: is there a "good" or "natural" solution to this method I'm considering that equation for a given function $f$? To attempt to ease the problem, it is defined seems best to impose some restrictions on the complex numbers possible input function $f$, as wellin general, the tamer the function, the easier the analysis of the equation and the more techniques are available. But we don't want too much restriction, or the method could be useless.
There was a method proposed by Markus Mueller in an article called "Fractional sums and Euler-like identities" that attempts to do something like this, but it suffers from some drawbacks. For example, it does not appear the method is of any use in summing, e.g.
The idea is to try
So we want to come up with some kind of "grand unified theory" of these kinds of sums, : a solution for the functional equation that would enable sense to be made out able to recover most if not all of the fractal sum for a wide variety these kind of functions formulae, and from which results like also enable the one above could be derived or proven to be consistent withsummation of many other functions.
Anyway, after experimenting with some methods
For my approach, I decided it'd suppose that $f$ be easiest to limit attention to analytic functions, trying to find at least a "sum operator" that yields analytic sums, on the holomorphic function of a complex numbers. The basic idea is to find such an operatorvariable, call it $\Sigma$, such and that $\Delta \Sigma f = f$F$is to be as well. This is still pretty broad and does not uniquely determine a solution, where$\Delta$but is tight enough to maximize the unit forward difference operatoravailable tools for analysis. If At this point, though, I'll suppose even further that$F = \Sigma f$, then and$\sum_{n=a}^{b} f(n) = F(b+1) - F(a)$F$ be entire. Then we can see if the techniques generalize to non-entire cases.
Add: The justification for considering this approach as "natural" is based on two approaches. One is Faulhaber's formula, which gives a sum of powers by the Bernoulli polynomials, and this sum has a simple uniqueness criterion: it sends polynomials to polynomials. One can then apply this to Taylor series. The trouble is, that such a method looks only to work on an extremely limited set of analytic functions: entire functions of exponential type less than $2\pi$. This limit seems a bit too onerous. This is one of the "some methods" I mentioned as having experimented with for defining the continuum sum. It is somewhat long to give the whole derivation, but for $e^{uz}$, summed from $0$ to $z-1$, and $|u| < 2\pi$, it yields $\frac{e^{uz} - 1}{e^u - 1}$. Another justification is much simpler. We know that $\Delta e^{uz} = \left(e^u - 1\right) e^{uz}$. Thus, it'd seem sensible to presume $\Sigma \left(e^u - 1\right) e^{uz} = e^{uz}$. This leads to (assuming $\Sigma$ is linear), $\Sigma e^{uz} = \frac{e^{uz}}{e^u - 1}$, and then the sum from $0$ to $z-1$ is $\frac{e^{uz} - 1}{e^u - 1}$. I suppose one could get a third justification is in that both of these methods give the same result. Finally, because $\Sigma$ is linear, it is no big step to obtain the result for periodic functions.
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Hi.
I have this idea about developing what I call a "continuum sum", that is, a method to "add up a non-integer number of terms", i.e. to see if there is a "natural" way to assign a meaning to the expression
$\sum_{n=a}^b f(n)$
where a $a$ and b $b$ are non-integer fractional, real, or even complex numbers, at least for a sufficiently well-behaved target function f$f$ defined on at least the real numbers, but for this method I'm considering that it is defined on the complex numbers as well.
There was a method proposed by Markus Mueller in an article called "Fractional sums and Euler-like identities" that attempts to do this, but it suffers from some drawbacks. For example, it does not appear the method is of any use in summing, e.g.
$\sum_{n=a}^b n! = \sum_{n=a}^b \Gamma(n+1)$
though a solution does exist, namely
$\sum_{k=1}^{n} k! = \frac{-e + \mathrm{Ei}(1) + \pi i + E_{n+2}(-1) \Gamma(n+2)}{e}$
using the exponential integral and En-function.
The idea is to try to come up with some kind of "grand unified theory" of these kinds of sums, that would enable sense to be made out of the fractal sum for a wide variety of functions and from which results like the one above could be derived or proven to be consistent with.
Anyway, after experimenting with some methods, I decided it'd be easiest to limit attention to analytic functions, trying to find a "sum operator" that yields analytic sums, on the complex numbers. The basic idea is to find such an operator, call it $\Sigma$, such that $\Delta \Sigma f = f$, where $\Delta$ is the unit forward difference operator. If $F = \Sigma f$, then $\sum_{n=a}^{b} f(n) = F(b+1) - F(a)$.
The approach I settled on (how I got to this is omitted here for the sake of brevity), is using Fourier series. If $f(z)$ is a periodic function with period $P$, then
$f(z) = \sum_{n=-\infty}^{\infty} a_n e^{\frac{2\pi i}{P} nz}$.
Now we have a simple formula for the sum of an exponential: $\sum_{n=0}^{z-1} e^{un} = \frac{e^{un} - 1}{e^u - 1}$. We can apply this to the above to get
$\sum_{n=0}^{z-1} f(n) = \sum_{n=-\infty}^{\infty} \frac{a_n}{e^{\frac{2\pi i}{P} n} - 1} \left(e^{\frac{2\pi i}{P} nz} - 1\right)$
with the term at $n = 0$ on the right (for which the given expression fails directly with a division by 0) interpreted as $a_0 z$. This series can converge even when the given function fails some of the mentioned requirements, however it does not work for functions with harmonics of period 1.
My idea, then, was to consider a sequence of periodic entire functions $f_i$ that converge to a given function $f$ that's entire and aperiodic, but not necessarily of exponential type less than $2\pi$. Then take their continuum sums by the above formula and take the limit. If this limit exists, call it the continuum sum of $f$ itself. The questions I have, then, are, what conditions are needed on $f$ for this to work, and also, more importantly, is the limit independent of the chosen sequence of functions, and if so, what is the proof, and if not, what is a counterexample? This is why I mean by it being "viable" or not. If the limit doesn't work, this isn't of much use. I'm not necessarily interested in a complete proof but more on advice about how one would go about approaching a proof of this, useful reference material, etc. as I'd like to do some of it myself. However, if the hypothesis is false, I'd like a full counterexample.
Add: The justification for considering this as "natural" is based on two approaches. One is Faulhaber's formula, which gives a sum of powers by the Bernoulli polynomials, and this sum has a simple uniqueness criterion: it sends polynomials to polynomials. One can then apply this to Taylor series. The trouble is, that such a method looks only to work on an extremely limited set of analytic functions: entire functions of exponential type less than $2\pi$. This limit seems a bit too onerous. This is one of the "some methods" I mentioned as having experimented with for defining the continuum sum. It is somewhat long to give the whole derivation, but for $e^{uz}$, summed from $0$ to $z-1$, and $|u| < 2\pi$, it yields $\frac{e^{uz} - 1}{e^u - 1}$. Another justification is much simpler. We know that $\Delta e^{uz} = \left(e^u - 1\right) e^{uz}$. Thus, it'd seem sensible to presume $\Sigma \left(e^u - 1\right) e^{uz} = e^{uz}$. This leads to (assuming $\Sigma$ is linear), $\Sigma e^{uz} = \frac{e^{uz}}{e^u - 1}$, and then the sum from $0$ to $z-1$ is $\frac{e^{uz} - 1}{e^u - 1}$. Yet I suppose one could get a third justification is that both of these methods give the same result. Finally, because $\Sigma$ is linear, it is no big step to obtain the result for periodic functions.
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Add: The justification for considering this as "natural" is based on two approaches. One is Faulhaber's formula, which gives a sum of powers by the Bernoulli polynomials, and this sum has a simple uniqueness criterion: it sends polynomials to polynomials. One can then apply this to Taylor series. The trouble is, that such a method looks only to work on an extremely limited set of analytic functions: entire functions of exponential type less than $2\pi$. This limit seems a bit too onerous. This is one of the "some methods" I mentioned as having experimented with for defining the continuum sum. It is somewhat long to give the whole derivation, but for $e^{uz}$, summed from $0$ to $z-1$, and $|u| < 2\pi$, it yields $\frac{e^{uz} - 1}{e^u - 1}$. Another justification is much simpler. We know that $\Delta e^{uz} = \left(e^u - 1\right) e^{uz}$. Thus, it'd seem sensible to presume $\Sigma \left(e^u - 1\right) e^{uz} = e^{uz}$. This leads to (assuming $\Sigma$ is linear), $\Sigma e^{uz} = \frac{e^{uz}}{e^u - 1}$, and then the sum from $0$ to $z-1$ is $\frac{e^{uz} - 1}{e^u - 1}$. Yet a third justification is that both of these methods give the same result.
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Hi.
I have this idea about developing what I call a "continuum sum", that is, a method to "add up a non-integer number of terms", i.e. to see if there is a "natural" way to assign a meaning to the expression
$\sum_{n=a}^b f(n)$
where a and b are non-integer fractional, real, or even complex numbers, at least for a sufficiently well-behaved target function f.
There was a method proposed by Markus Mueller in this an article called "Fractional sums and Euler-like identities" that attempts to do this, but it suffers from some drawbacks. For example, it does not appear the method is of any use in summing, e.g.
$\sum_{n=a}^b n! = \sum_{n=a}^b \Gamma(n+1)$
though a solution does exist, namely
$\sum_{k=1}^{n} k! = \frac{-e + \mathrm{Ei}(1) + \pi i + E_{n+2}(-1) \Gamma(n+2)}{e}$
using the exponential integral and En-function.
The idea is to try to come up with some kind of "grand unified theory" of these kinds of sums, that would enable sense to be made out of the fractal sum for a wide variety of functions and from which results like the one above could be derived or proven to be consistent with.
Anyway, after experimenting with some methods, I decided it'd be easiest to limit attention to analytic functions, trying to find a "sum operator" that yields analytic sums, on the complex numbers. The basic idea is to find such an operator, call it $\Sigma$, such that $\Delta \Sigma f = f$, where $\Delta$ is the unit forward difference operator. If $F = \Sigma f$, then $\sum_{n=a}^{b} f(n) = F(b+1) - F(a)$.
The approach I settled on (how I got to this is omitted here for the sake of brevity), is using Fourier series. If $f(z)$ is a periodic function with period $P$, then
$f(z) = \sum_{n=-\infty}^{\infty} a_n e^{\frac{2\pi i}{P} nz}$.
Now we have a simple formula for the sum of an exponential: $\sum_{n=0}^{z-1} e^{un} = \frac{e^{un} - 1}{e^u - 1}$. We can apply this to the above to get
$\sum_{n=0}^{z-1} f(n) = \sum_{n=-\infty}^{\infty} \frac{a_n}{e^{\frac{2\pi i}{P} n} - 1} \left(e^{\frac{2\pi i}{P} nz} - 1\right)$
with the term at $n = 0$ on the right (for which the given expression fails directly with a division by 0) interpreted as $a_0 z$. This series can converge even when the given function fails some of the mentioned requirements, however it does not work for functions with harmonics of period 1.
My idea, then, was to consider a sequence of periodic entire functions $f_i$ that converge to a given function $f$ that's entire and aperiodic, but not necessarily of exponential type less than $2\pi$. Then take their continuum sums by the above formula and take the limit. If this limit exists, call it the continuum sum of $f$ itself. The questions I have, then, are, what conditions are needed on $f$ for this to work, and also, more importantly, is the limit independent of the chosen sequence of functions, and if so, what is the proof, and if not, what is a counterexample? I'm not necessarily interested in a complete proof but more on advice about how one would go about approaching a proof of this, useful reference material, etc. as I'd like to do some of it myself. However, if the hypothesis is false, I'd like a full counterexample.
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# Is this method of "fractional sums" using a Fourier series viable?
Hi.
I have this idea about developing what I call a "continuum sum", that is, a method to "add up a non-integer number of terms", i.e. to see if there is a "natural" way to assign a meaning to the expression
$\sum_{n=a}^b f(n)$
where a and b are non-integer fractional, real, or even complex numbers, at least for a sufficiently well-behaved target function f.
There was a method proposed by Markus Mueller in this article called "Fractional sums and Euler-like identities" that attempts to do this, but it suffers from some drawbacks. For example, it does not appear the method is of any use in summing, e.g.
$\sum_{n=a}^b n! = \sum_{n=a}^b \Gamma(n+1)$
though a solution does exist, namely
$\sum_{k=1}^{n} k! = \frac{-e + \mathrm{Ei}(1) + \pi i + E_{n+2}(-1) \Gamma(n+2)}{e}$
using the exponential integral and En-function.
The idea is to try to come up with some kind of "grand unified theory" of these kinds of sums, that would enable sense to be made out of the fractal sum for a wide variety of functions and from which results like the one above could be derived or proven to be consistent with.
Anyway, after experimenting with some methods, I decided it'd be easiest to limit attention to analytic functions, trying to find a "sum operator" that yields analytic sums, on the complex numbers. The basic idea is to find such an operator, call it $\Sigma$, such that $\Delta \Sigma f = f$, where $\Delta$ is the unit forward difference operator. If $F = \Sigma f$, then $\sum_{n=a}^{b} f(n) = F(b+1) - F(a)$.
The approach I settled on (how I got to this is omitted here for the sake of brevity), is using Fourier series. If $f(z)$ is a periodic function with period $P$, then
$f(z) = \sum_{n=-\infty}^{\infty} a_n e^{\frac{2\pi i}{P} nz}$.
Now we have a simple formula for the sum of an exponential: $\sum_{n=0}^{z-1} e^{un} = \frac{e^{un} - 1}{e^u - 1}$. We can apply this to the above to get
$\sum_{n=0}^{z-1} f(n) = \sum_{n=-\infty}^{\infty} \frac{a_n}{e^{\frac{2\pi i}{P} n} - 1} \left(e^{\frac{2\pi i}{P} nz} - 1\right)$
with the term at $n = 0$ on the right (for which the given expression fails directly with a division by 0) interpreted as $a_0 z$. This series can converge even when the given function fails some of the mentioned requirements, however it does not work for functions with harmonics of period 1.
My idea, then, was to consider a sequence of periodic entire functions $f_i$ that converge to a given function $f$ that's entire and aperiodic, but not necessarily of exponential type less than $2\pi$. Then take their continuum sums by the above formula and take the limit. If this limit exists, call it the continuum sum of $f$ itself. The questions I have, then, are, what conditions are needed on $f$ for this to work, and also, more importantly, is the limit independent of the chosen sequence of functions, and if so, what is the proof, and if not, what is a counterexample? I'm not necessarily interested in a complete proof but more on advice about how one would go about approaching a proof of this, useful reference material, etc. as I'd like to do some of it myself. However, if the hypothesis is false, I'd like a full counterexample.
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http://physics.stackexchange.com/questions/8951/what-are-the-ideal-centers-of-mass-and-rotation-for-a-steadicam
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# What are the ideal centers of mass and rotation for a steadicam?
For those who don't know what a steadicam is, I'll explain it in two points to give a basic idea. If you know what a steadicam is, skip to point 3.
1) A steadicam is a stabilizing mount for a motion picture camera, which mechanically isolates the operator's movement from the camera, allowing a very smooth shot even when the operator is moving quickly over an uneven surface.
2) How does it work (image courtesy of wsclater):
Red dots = control masses, Blue dot = center of rotation
The idea behind a steadicam is that the center of the mass is taken out of the camera itself. Therefore, when moving / panning the whole system (steadicam + camera), the motion of the camera itself is more fluid. This can be seen in action in this amazing youtube example.
3) The question:
How do we determine the ideal distance of the center of rotation from the camera?
How do we determine the ideal distance of the center of mass from the center of rotation?
What's the ideal length of the handle (the point where we hold the system), i.e. how far should we hold the system from its center of rotation?
Disclaimer: This is neither a matter of profit, nor a matter of homework. I'm just trying to build some of those for my own personal use, especially to find out how it really works, as I'm kind of amazed from this.
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As a point of nomenclature the red dots in your image are almost certainly not "centers of mass" for any useful system as physicists use that term. They appear to be masses added to the system in order to control where the (single!) center of mass of the whole system is located. You might call these "control masses" or something similar. – dmckee♦ Apr 21 '11 at 0:32
sry, fixed. feel free to edit my post if there are any more inconsistencies – Richard Rodriguez Apr 21 '11 at 0:36
## 2 Answers
You need to know two things to design the system:
1. Where you want the systems's center of mass to be.
2. How to calculate such a thing.
Having read the wikipedia and watch some video I'm guess the answer to (1) is "a short distance directly below the gimbal" (for both high and low mode, though that means slightly different places).
How to calculate the CoM for a set of point masses.
• Establish a coordinate system. Say $z$ increases up and starts at the camera platform, and $x$ increase to the front and starts at the location of the gimbal. This makes $y$ increase to the left if we want the usual convention, and we'll start it at the gimbal as well.
• Find the (three dimensional!) location and mass of each part of the system (remember so far I'm only talking about little dense masses) and call these $(x,y,z)_i$ and $m_i$ where $i$ is just an index that runs over all you bits. Write
$$x_{com} = \frac{\sum_i x_i m_i}{\sum_i m_i}$$
and an identical equation for $y$ and $z$. And now we know the location of the the Center of Mass.
For continuous objects, we could turn those sums into integrals, but you probably don't want to do that. Especially as you don't know the mass distribution of the camera.
Instead I will suggest finding the center of mass for each object (and this is why the wsclater link calls the red dots center of mass, they are for each object), by the simple expedient of balancing or hanging them.
• If you hang an object from any point, it's CoM will lie below the point you hang it from.
• If you balance an object on a thin edge, the objects CoM will lie above the edge.
Combine several such measurements and you can find the three diminsional position of the CoM.
Now here's the trick. Once you know the CoM for an extended object, you can treat the whole thing as a point mass at that location (for the purposes of calculating the CoM of a system that includes the object).
OK. Making progress, but were, still not done.
You know where you want the CoM, and need to find where to put the masses. Which is backwards from what we did above.
You need to control three values (the $x$, $y$, and $z$ positions of the CoM), and you have three equations. Which means that you can have only three adjustable parameters in your system.
Side note: The value of $y$ is easy, if you position the camera so that it's CoM is centered. Then, because everything else is symmetric, it happens automatically. So I'm going to only worry about $x$ and $z$
What you do is fix all the position relationships in the system except for two. Maybe the height of the lower control mass and the horizontal position of the middle one. Those two you make variables. Now you write down the CoM equations above *with those two variables in there and the desired values for the location of the CoM. The resulting system is solvable. Do the algebra and you're done.
But leave some ability to fine tune things (say by putting the masses on threaded rods). This lets you only worry about getting close with the calculation. The pictured one appears to use a flexible support to provide the adjustability. That will work, but may require constant tinkering.
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To find the centre of mass of an object, you find the centre of gravity which is at the same point within acceptable error for your camera.
Hang any apex of your steadycam and a plumb-line from a point in space. Trace the path the plumb line takes from one apex to another point. Repeat this for another two apexes. Where all three traces intersect is the centre of mass of your steadycam.
Google "centre of gravity plumbline" for more details.
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http://physics.stackexchange.com/questions/16653/why-is-beta-negative-decay-more-common-than-beta-positive/16654
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# Why is beta negative decay more common than beta positive?
In simple terms, why is beta negative decay more common than beta positive?
I know it's something to do with occuring inside/outside the nucleus - but I can't find a simple, easy to understand explanation!
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## 2 Answers
The beta-decay may be "locally" reduced to a decay of a proton or a neutron inside a nucleus. The beta-minus decay contains the microscopic process $$n\to p + e^- + \bar \nu_e + O(1{\rm \,MeV})$$ where the last term indicates the rough increase of the kinetic energy of the decay products. On the other hand, the beta-plus decay contains the process $$p + O(1{\rm \,MeV})\to n + e^+ + \nu_e$$ which means that the proton has to acquire some extra energy if it wants to decay to a neutron and a positron. In realistic beta-plus decays, it takes it from the surrounding nucleons in the nucleus.
Obviously, decays to lighter products where the energy conservation allows to give the final products some extra kinetic energy are more frequent than decays which only occur if an extra energy is found at the beginning. For example, among the bare processes above (for free nucleons), only the decay of the neutron may occur. The proton is stable (ignoring very infrequent processes linked to grand unification).
At the end, the inequality between the two types of the decay boils down primarily to the fact that the neutron is heavier than the proton which subsequently boils down to the fact that the down-quark is heavier than the up-quark (the rest masses).
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2
And there is K-capture, a kind of beta positive. – Georg Nov 7 '11 at 11:07
I haven't done anything with the kinetic energy of the decay products before - and yet this is still a question we have. Is there a simpler way of explaining it without discussing kinetic energy? – Parachuting Panda Nov 7 '11 at 11:10
Dear @Georg, you should be careful about the terminology. K-capture is sometimes called "inverse beta decay" but "beta negative decay" – despite the similar origin of "negative" and "inverse" – has to include positron emission. Comparisons of beta decays emitting electrons or positrons with K-capture is a subtle question because the two classes are qualitatively different. – Luboš Motl Nov 7 '11 at 11:11
Dear @ParachutingPanda, could you please be more specific about which concepts we are allowed to use in the answer if the concept of energy is already too difficult? – Luboš Motl Nov 7 '11 at 11:12
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Nope, @ParachutingPanda, both proper beta-decays, positive and negative, occur inside the nucleus. I don't know how to simplify this thing more than the trivial assertion it is now. Can you please define "simple"? AdamRedwine: all beta-decays are ultimately mediated by W-bosons. (There are also similar interactions mediated by Z-bosons, but they only occur - with a few irrelevant exceptions - when there is an electromagnetic interaction of the same kind as well, and the electromagnetic one is much more important in such cases.) W-bosons only become important at high energies, 100 GeV or so... – Luboš Motl Nov 7 '11 at 12:55
show 3 more comments
Beta-minus decay occurs in nuclei with an excess of neutrons, while beta-plus decay takes place in neutron-deficit nuclei. A lot of natural background radiation on Earth is due to fission or alpha-decay of heavy radioactive elements. The remains of fission or alpha-decay are neutron-rich nuclei, so beta-minus decay is more common on Earth.
Whereas on stars beta-plus decay is typical, because neutron-deficit nuclei are produced in nuclear fusion.
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I would just add that "neutron heavy" or "neutron deficit" is relative to the island of stability, not N=Z. I remember that there is one process for one direction and 2 processes for the other direction, so in that context I think you're missing one. Internal capture for N-1 and Z+1? Anyway, that's not fully in the scope of the question. – AlanSE Nov 7 '11 at 13:43
– voix Nov 7 '11 at 14:21
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http://mathoverflow.net/questions/71970/quartic-diagonal-as-a-sum-of-squares-of-quadratic-forms/72058
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## quartic diagonal as a sum of squares of quadratic forms
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I would appreciate if someone can point out to the literature related to characterizing the set of all different ways to write real quartic diagonal $\sum \limits_{k=1}^n x_k^4, x \in \mathbb{R^n}$ as a sum of squares of real quadratic forms. Murray Marshall in his book "Positive polynomials and sums of squares" show that quartic diagonal is in the interior of the cone of sum of squares. Does anyone knows details about such representation.
In particular, suppose $\sum \limits_{k=1}^n x_k^4= \sum_p (x^T A_p x)^2$ ($x$ is a column vector, and $x^T$ its transpose, $A_p \in \mathbb{R^n \times R^n}$) then what $Q\in \mathbb{R^n \times R^n}$ can be represented by the expression $\sum_p (x^T A_p x - {x^*}^T A_p x^*)^2= \sum \limits_{k=1}^n x_k^4 - 2x^T Q x+ const$, $x^*$ is a point in $\mathbb{R^n}$. More specifically, whether $Q$ is dense around identity matrix (in a small neighborhood).
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## 1 Answer
I passed your question on to a friend who knows about these things, and he replied,
The theorem that $\sum_{k=1}^m x_k^{2r}$ is interior to the sum of squares of appropriate degree, can be found, with proof, in a paper of R. M. Robinson: Some definite polynomials which are not sums of squares of real polynomials, Izdat. "Nauka" Sibirsk. Otdel. Novosibirsk (1973), 264-282, Selected questions of algebra and logic (a collection dedicated to the memory of A. I. Mal'cev), Abstract in Notices Amer. Math. Soc. 16 (1969), p. 554.
My friend also recommended the Memoir by Bruce Reznick, which can be found, scanned, on Bruce's webpage at UIUC.
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Thanks a lot. Nevertheless this answer does not provide any specific information about quartic polynomials of interest. As far as I know no one was studying them in details. – mkatkov Oct 17 2011 at 7:52
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http://math.stackexchange.com/questions/183100/computing-the-differential-of-the-map-phi-m-2-mathbbc-times-m-2-mathbbc?answertab=active
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# Computing the differential of the map $\phi: M_2(\mathbb{C})\times M_2(\mathbb{C})\rightarrow M_2(\mathbb{C})$
Let $\phi: M_2(\mathbb{C})\times M_2(\mathbb{C})\rightarrow M_2(\mathbb{C})$ be the map $$(B_1,B_2)\mapsto [B_1,B_2]$$ which takes two $2\times 2$ matrices to its Lie bracket.
Then why does $d\phi_{(B_1,B_2)}:M_2(\mathbb{C})\times M_2(\mathbb{C})\rightarrow M_2(\mathbb{C})$ send $$(D_1,D_2)\mapsto [B_1,D_2]+[D_1,B_2]?$$
Taking $B_1=(g_{ij})$ and $B_2=(h_{ij})$, I do not think taking the partials of the Lie bracket $[B_1,B_2]=$ $$\left[ \begin{array}{cc} g_{12} h_{21} - g_{21} h_{12} & -g_{12} h_{11} + g_{11} h_{12} - g_{22} h_{12} + g_{12} h_{22} \\ g_{21} h_{11} - g_{11} h_{21} + g_{22} h_{21} - g_{21} h_{22} & g_{21} h_{12} - g_{12} h_{21} \\ \end{array}\right]$$ is a clever way to figure out the map.
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## 2 Answers
Taking differentials is all about looking at what happens to your map upon a very small perturbation. So compute the bracket $$[B_1 + \epsilon D_1, B_2 + \epsilon D_2]$$
and look at the coefficient of $\epsilon$.
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Thanks Qiaochu! I will give that a try. Definitely better than working with coordinates... – math-visitor Aug 16 '12 at 6:02
I know that we "ignore" higher order perturbations $O(\epsilon^2)$, but why are we ignoring $[B_1,B_2]$ that appears in the expansion as well? – math-visitor Aug 16 '12 at 6:09
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@math-visitor: because it's a constant. We only care about how the expression changes as $\epsilon$ changes. – Qiaochu Yuan Aug 16 '12 at 6:11
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Another way to say that is we ignore $[B_1,B_2]$ because that's the constant part of our estimate, which we'd subtract out before computing the limit of a difference quotient in a more elementary computation. The differential's just the linear portion-if we wanted the quadratic part of the estimate, we'd take $[D_1,D_2]$ and forget above both the constant $\textit{and}$ the linear part. – Kevin Carlson Aug 16 '12 at 9:29
Congratulations on your graduation Qiaochu! Just one more question: how can one see that the dimension of the cokernel of $d\phi_{(B_1, B_2)}$ equal to 2? It is equal to 2 right? – math-visitor Aug 18 '12 at 2:25
This has actually nothing to do with Lie brackets nor Lie algebras!
Given finite-dimensional complex (or real) vector spaces $E,F,G$ and a bilinear map $f:E\times F\to G$, the differential of $f$ at $(a,b)\in E\times F$ is given by the formula (whose proof follows directly from the definition)
$$df_{(a,b)} (x,y)=f(a,y)+f(x,b)$$
The only thing you have to check in your case is that the bracket is bilinear, which is obvious in the concrete case of matrices and in the abstract theory of Lie algebras is an axiom .
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Thank you for the generalization Georges! This is a very good remark and an answer especially if others want to do similar computations for other bilinear maps. One question though: for the above example, shouldn't the dimension of the image of $d\phi_{(B_1, B_2)}$ equal 2? How can one see that? – math-visitor Aug 18 '12 at 18:34
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Dear math-visitor, no, the dimension of the image is not always 2: what makes you believe that? For example for $B_1=B_2=0$ the differential is zero and so its image has dimension zero. – Georges Elencwajg Aug 18 '12 at 22:10
Very good point. Thank you Georges. – math-visitor Aug 18 '12 at 23:16
You are welcome, math-visitor. – Georges Elencwajg Aug 18 '12 at 23:22
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http://www.physicsforums.com/showthread.php?p=1623716
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Physics Forums
Cosmology Question: Expansion of the Universe
1. The problem statement, all variables and given/known data
So the other day in my cosmology class we talked about whether smaller objects like a solar system, a star or a galaxy expand with the rest of the universe. While the answer is obviously no, since local gravitational forces can overcome that expansion, our professor thought it would be kinda amusing the calculate the effect of that expansion on an object like a human being if we were to expand along with the universe.
So we're asked to find out how long it would it take us to expand by 1mm, given the current value of the Hubble parameter of Ho = 72km/s/Mpc, and assuming for the sake of this problem that the Hubble parameter is constant even if the time we find is very large.
We're also have to find how tall we have to be for that expansion to take one year.
I know it looks like a silly problem but I am just having a hard time finding a place to start thinking about it.
Thanks for taking the time.
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Not takers?
I'm not a cosmologist but I'll take a complete shot in the dark at it. It seems like the Hubble parameter value means that something which is one megaparsec in size will grow to become one megaparsec and 72km in size after one second. Is that right? If that's true it seems like you'd have to rig up an algebraic equality with ratios, the $$\frac{a}{b} = \frac{c}{d}$$ kinda thing, or an expression with calculus to account for compound growth Hope I'm not misunderstanding and ridiculously oversimplifying the problem.⚛
Cosmology Question: Expansion of the Universe
Quote by Moneer81 1. The problem statement, all variables and given/known data So the other day in my cosmology class we talked about whether smaller objects like a solar system, a star or a galaxy expand with the rest of the universe. While the answer is obviously no, since local gravitational forces can overcome that expansion, our professor thought it would be kinda amusing the calculate the effect of that expansion on an object like a human being if we were to expand along with the universe. So we're asked to find out how long it would it take us to expand by 1mm, given the current value of the Hubble parameter of Ho = 72km/s/Mpc, and assuming for the sake of this problem that the Hubble parameter is constant even if the time we find is very large. We're also have to find how tall we have to be for that expansion to take one year. I know it looks like a silly problem but I am just having a hard time finding a place to start thinking about it. Thanks for taking the time.
Hubble's law is $$v= \frac{\Delta x}{\Delta t} = H_0 d$$
So you simply need to isolate delta x. This gives by how much the distance between two objects which were initially a distance "d" apart will have increased after a time delta t. (This assumes of course that delta t is not so large that H_0 will have changed value appreciably during that time interval)
Quote by CaptainQuasar I'm not a cosmologist but I'll take a complete shot in the dark at it. It seems like the Hubble parameter value means that something which is one megaparsec in size will grow to become one megaparsec and 72km in size after one second. Is that right? If that's true it seems like you'd have to rig up an algebraic equality with ratios, the $$\frac{a}{b} = \frac{c}{d}$$ kinda thing, or an expression with calculus to account for compound growth Hope I'm not misunderstanding and ridiculously oversimplifying the problem.⚛
The logic is correct. And it gives the same equation that I provided in my post.
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http://www.physicsforums.com/showthread.php?t=149298
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Physics Forums
## Total energy of a damped oscillator
Is it possible to express the total energy of a damped linear oscillator as a function of time? I'm confused here. I'd like to find E(t). As the oscillation is damped, dE/dt should everywhere be negative (energy being dissipated as radiation or heat). By setting E(t) equal to zero, shouldn't I be able to solve for the time at which the energy of the oscillating system is zero, and thus the time at which the system stops oscillating? And shouldn't this time be finite?
Is there another way to find the time at which the damped oscillator will stop oscillating?
Thanks!
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Recognitions: Science Advisor Yes. The peak amplitude of the oscillation, that is, the envelope, decays exponentially. Since average or rms energy is related to the peak amplitude, it also decays exponentially. In theory it never exactly reaches zero so you can't say when the oscillator "stops." In practice you can say it stops when the amplitude is comparable to thermal noise or some other criterion. It is more common to specify the time constant, which is the time for the envelope to decay to 1/e of its initial amplitude.
Well... you can simlpy solve the differential equation for a damped oscillator, then use $$E=\frac{1}{2}kx^2+\frac{1}{2}m\dot{x}^2$$
## Total energy of a damped oscillator
An ideal damped oscillator won't stop oscillating until infinite time has elapsed. However, practically the easiest way to find the time when the damped oscillator will top oscillating would be to determine the time constant, sqrt(m/k), and then multiply it by five because after five time constants the motion will be reduced to 1% (or something close to that) of its initial amplitude.
Recognitions:
Science Advisor
Quote by Signifier As the oscillation is damped, dE/dt should everywhere be negative (energy being dissipated as radiation or heat).
The above posts are correct in saying the oscillation continues for ever, but dE/dt is not always negative.
Taking a mechanical damped oscillator for example, with equation of motion M x'' + C x' + K x = 0, the energy is dissipated by the the damper. The work is (force times velocity) = C x'^2. That is zero twice every cycle, when the velocity becomes zero.
If you evaluate E from tim_lou's equation you will get an exponential decay multiplied by a something looking like (A + B sin pt), which is a curve that "wobbles" around the "average" exponential decay in the energy.
BTW, when you consider the different solutions for under-damped, over-damped or critically damped oscillation, you get different solution. I think there won't be any wobbling in the over-damped or critically damped cases.
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http://mathhelpforum.com/pre-calculus/45363-solving-system.html
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# Thread:
1. ## Solving system
I tried to simplify it down and go to this point:
I don't know if I did it right. From there, I have no idea what to do T_T The problem asks for a exact answer so everything must stay in radicals. I also don't know how to solve it using matrices...
2. from the first equation ...
$\sqrt{3x} = 2y-1$
$3x = (2y-1)^2$
$x = \frac{(2y-1)^2}{3}$
sub for x in the second equation ...
$\frac{2(2y-1)^2}{3} + 5y^2 = 12$
$2(2y-1)^2 + 15y^2 = 36$
expand, collect like terms, and solve the quadratic for y ... then go back and evaluate for x. note that x > 0.
3. I solved for y and got
I plugged it into the x equation to solve for x and it looks like this:
Im not sure how to break that down =O Do I multiply the 2 by both the 4 and root 3192? What about the plus and minus?
4. why bother? leave those ugly solutions as they are.
use your calculator to check the two sets of solutions (x,y) in the original system.
5. I’m pretty sure my y is correct. However when I plugged x and y into one of the equations in my calculator, it didn’t work O.o I plugged it in using parentheses in the right places too.
Thanks for the help!
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http://math.stackexchange.com/questions/229497/f-mathbfr-rightarrow-mathbfr-monotone-increasing-rightarrow-f-is/229502
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$f: \mathbf{R} \rightarrow \mathbf{R}$ monotone increasing $\Rightarrow$ $f$ is measurable
Problem. Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a monotone increasing function. Show that $f$ is measurable.
Solution. We know that the set of discontinuites of any monotone increasing function $f$ is measure zero (since it is at most countable). We define a continuous function $g$ such that $g(x)=f(x)$ except the discontinuous points of $g$. Then $g(x)=f(x)$ almost everywhere. Note that any continuous function $g: \mathbf{R}\rightarrow \mathbf{R}$ is measurable. Also note that if $g$ is measurable and $f=g$ almost everywhere, then $f$ is measurable. Hence we conclude that $f$ is measurable.
Is my solution correct? Thanks.
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4
Unless $f$ is continuous, such a $g$ can't exist. – JSchlather Nov 5 '12 at 6:41
2 Answers
It is true that $f$ is continuous almost everywhere, but it is not true that there exists a continuous function $g:\mathbb R\to\mathbb R$ such that $f=g$ almost everywhere, unless $f$ is already continuous, as Jacob said in a comment. Note that the left-hand and right-hand limits of $f$ exist everywhere, and if they are not equal for $f$, then they cannot be equal for any function equal to $f$ almost everywhere. E.g., the characteristic function of $[0,\infty)$ is monotone and not equal almost everywhere to a continuous function.
(The characteristic function of the rationals is equal almost everywhere to a continuous function, but is continuous nowhere. This shows from the other direction why "continuous almost everywhere" and "equal almost everywhere to a continuous function" are very different.)
Modifying your work to something correct takes more effort than the simpler and clearer solution given by Cass. Let $D$ be the set of discontinuities of $f$. Then $D$ is countable, hence of measure $0$. The restriction $f|_D$ is measurable on $D$ because every subset of $D$ is measurable, and the restriction $f|_{\mathbf R\setminus D}$ is measurable on $\mathbf R\setminus D$ because it is continuous. You can show this implies that $f$ is measurable.
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If $f$ is increasing, the set {$x:f(x)>a$} is an interval for all $a$, hence measurable. By definition (Royden's), the function $f$ is measurable.
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What about my solution? – Erno Nemecsek Nov 5 '12 at 7:37
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http://mathoverflow.net/questions/122532/semiring-naturally-associated-to-any-monoid
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## Semiring naturally associated to any monoid?
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For any monoid $M$, we can naturally construct a semiring $S$ as follows:
1. Let the additive monoid of $S$ be the free commutative monoid on $M$
2. Let the multiplicative monoid of $S$ be $M$
Then, if you make multiplication distribute over addition, you get a semiring.
This has an extremely simple interpretation: the underlying additive monoid can be interpreted as the set of finite multisets over the elements of $M$, with addition being just union of multisets. Then, semiring multiplication between multisets $A$ and $B$ is simply the multiset you get if you apply the original binary operation of $M$ pairwise to all elements in $A \times B$.
This construction is rather natural. Does it have a name, or is it well-known? I've found it interesting because it's arisen organically in music theory, where the semifield associated with a certain free abelian group representing musical intervals has the beautifully clear interpretation of being a semifield of musical chords.
There are a few nice variations on this idea:
1. You can instead force the free commutative monoid to be idempotent, so that it now has a natural interpretation as the set of finite subsets of $M$, rather than the set of finite multisets over it.
2. Given any semiring $S'$, you can use this construction on each of its monoids, giving you an algebraic structure with three operators that are totally ordered with respect to distributivity.
Do any of these things have names, and/or are they well-known?
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But for the mistakes pointed out by Noah below, all of this (and much more) is found, e.g., in the 3rd chapter of Golan's Semirings and their Applications. – Salvo Tringali Feb 21 at 15:32
1
There is a forgetful functor from semirings to monoids given by taking the multiplicative monoid, and this construction is its left adjoint. Similar constructions give the group ring, group algebra, etc. – Qiaochu Yuan Feb 21 at 21:51
## 2 Answers
It is at least sometimes called a "monoid semiring" by analogy with "group ring". As such it would be notated $S = \mathbb{N_0}[M]$ (or $\mathbb{N}[M]$ depending how you define things).
By the way, the ring $\mathbb{Z}[M]$ you define in #1 is a commutative ring, but not a field because the element $2$ (the identity of M plus itself in $S$) has no inverse. Even if you use $\mathbb{Q}[M]$ in place of $\mathbb{Z}[M]$, you still do not necessarily get a field. For example, let $x$ denote a generator of $\mathbb{Z}$. Then $1+x\in \mathbb{Q}[\mathbb{Z}]$ has no multiplicative inverse.
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I think it is more or less always called by this name except for by people who say "rig" instead of semiring. – Benjamin Steinberg Feb 21 at 16:36
Thanks for correcting the error, Noah, I hadn't realized that. Also, aha, never thought this could be viewed as a simple monoid generalization of the group ring! Thanks for that. – Mike Battaglia Feb 21 at 19:30
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
You are describing free constructions between finitary varieties.
A finitary variety is an equationally defined class of algebras for (i) an arbitrary set $\Sigma$ of operation symbols each $\sigma \in \Sigma$ having finite arity, (ii) an arbitrary set of equations $E$ consisting of pairs $(\phi_1,\phi_2)$ where each $\phi_i$ is a term built from the operation symbols and some fixed countable set of variables.
Then the induced finitary variety $\mathcal{V}$ is a category. Its objects are sets equipped with the operations from $\Sigma$ that satisfy the equations $E$. Its morphisms are those functions between the carriers that preserve each operation in $\Sigma$. Composition is the usual composition of functions.
Now, suppose $(\Sigma_1,E_1)$ specify the variety $\mathcal{V}_1$ and $(\Sigma_2,E_2)$ specify the variety $\mathcal{V}_2$.
In the case where $\Sigma_1 \subseteq \Sigma_2$ and $E_1 \subseteq E_2$, then there is a free construction $F : \mathcal{V}_1 \to \mathcal{V}_2$ which is the left adjoint of (i.e. uniquely determined by) the forgetful functor $U : \mathcal{V}_2 \to \mathcal{V}_1$ which merely forgets the additional operations.
Examples:
1. Let $(\Sigma_1,E_1)$ be the usual axiomatisation of monoids, so that $\mathcal{V}_1$ is the variety of monoids. Let $\Sigma_2$ also contain the additional semiring operations (you are adding +,0) and $E_2$ also contain the relevant equations (e.g. that + is a commutative monoid). Then the induced free functor $F : \mathcal{V}_1 \to \mathcal{V}_2$ is what you describe: it constructs a free commutative monoid and forces the original monoid to distribute over it in the 'simplest' way.
2. One can do the analogous thing but now with abelian groups and rings. As the above comment states, the latter are not the same thing as fields (which do not form a variety in the sense above). Or one could view the original abelian group as being a commutative multiplication to get a free construction to commutative rings.
3. Similarly one can go from monoids to idempotent semirings.
4. Your third example again follows: you are extending the signature and operations.
Two more things.
Firstly the free algebra construction $F : \mathsf{Set} \to \mathcal{V}$ is a special case: the category of sets is the finitary variety with no operations and no equations. So for example, the free monoid construction is covered.
Secondly the conditions $\Sigma_1 \subseteq \Sigma_2$ and $E_1 \subseteq E_2$ are a special case of a more general construction: a translation between algebraic theories, which again induces a unique free construction $F$ as above.
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http://mathhelpforum.com/algebra/207093-induction-proof-de-moivres-formula.html
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# Thread:
1. ## induction proof of de moivres formula
Is this correct and enough proof
$(cosx+isinx)^n=cos(nx)+isinx$ is true so i put 1 and its true now ima prove for
$S(k+1) = (Cosx+isinx)^k+1= (Cosx+isinx)^k(cosx+isinx)$ now we can rewrite that ^k with a formula i cant remember as
$(coskx+isinkx)(cosx+isinx)$
and now i foctour out cos x and sin x
$Cos x ( k+1) + iSinx(k+1)$
is this enough
2. ## Re: induction proof of de moivres formula
You "can't remember"? You just wrote it two lines above!
For n a positive integer, yes, that will be a valid proof once you use trig identities to show how you got from $(cos(kx)+ isin(kx))(cos(x)+ i sin(x))$ to $cos((k+1)x)+ i sin((k+1)x)$.
DeMoivre's formula is true for n any number, of course, not just positive integers. It would think the simplest way to prove it is to use the fact that $cos(x)+ i sin(x)= e^{ix}$.
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http://en.wikipedia.org/wiki/Wormhole
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# Wormhole
From Wikipedia, the free encyclopedia
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For other uses, see Wormhole (disambiguation).
"Einstein-Rosen Bridge" redirects here. For the EP by Venetian Snares, see Einstein-Rosen Bridge (EP).
General relativity
$G_{\mu \nu} + \Lambda g_{\mu \nu}= {8\pi G\over c^4} T_{\mu \nu}$
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A wormhole, also known as an Einstein-Rosen Bridge is a hypothetical topological feature of spacetime that would be, fundamentally, a "shortcut" through spacetime. For a simple visual explanation of a wormhole, consider spacetime visualized as a two-dimensional (2D) surface. If this surface is folded along a third dimension, it allows one to picture a wormhole "bridge". (Please note, though, that this is merely a visualization displayed to convey an essentially unvisualisable structure existing in 4 or more dimensions. The parts of the wormhole could be higher-dimensional analogues for the parts of the curved 2D surface; for example, instead of mouths which are circular holes in a 2D plane, a real wormhole's mouths could be spheres in 3D space.) A wormhole is, in theory, much like a tunnel with two ends each in separate points in spacetime.
There is no observational evidence for wormholes, but on a theoretical level there are valid solutions to the equations of the theory of general relativity which contain wormholes. Because of its robust theoretical strength, a wormhole is also known as one of the great physics metaphors for teaching general relativity. The first type of wormhole solution discovered was the Schwarzschild wormhole which would be present in the Schwarzschild metric describing an eternal black hole, but it was found that this type of wormhole would collapse too quickly for anything to cross from one end to the other. Wormholes which could actually be crossed in both directions, known as traversable wormholes, would only be possible if exotic matter with negative energy density could be used to stabilize them. (Many physicists such as Stephen Hawking,[1] Kip Thorne,[2] and others[3][4][5] believe that the Casimir effect is evidence that negative energy densities are possible in nature.) Physicists have not found any natural process which would be predicted to form a wormhole naturally in the context of general relativity, although the quantum foam hypothesis is sometimes used to suggest that tiny wormholes might appear and disappear spontaneously at the Planck scale,[6][7] and stable versions of such wormholes have been suggested as dark matter candidates.[8][9] It has also been proposed that if a tiny wormhole held open by a negative-mass cosmic string had appeared around the time of the Big Bang, it could have been inflated to macroscopic size by cosmic inflation.[10]
The American theoretical physicist John Archibald Wheeler coined the term wormhole in 1957; however, in 1921, the German mathematician Hermann Weyl already had proposed the wormhole theory, in connection with mass analysis of electromagnetic field energy.[11]
This analysis forces one to consider situations...where there is a net flux of lines of force, through what topologists would call "a handle" of the multiply-connected space, and what physicists might perhaps be excused for more vividly terming a "wormhole".
—John Wheeler in Annals of Physics
"Embedding diagram" of a Schwarzschild wormhole (see also below)
## Definition
The basic notion of an intra-universe wormhole is that it is a compact region of spacetime whose boundary is topologically trivial but whose interior is not simply connected. Formalizing this idea leads to definitions such as the following, taken from Matt Visser's Lorentzian Wormholes.
If a Minkowski spacetime contains a compact region Ω, and if the topology of Ω is of the form Ω ~ R x Σ, where Σ is a three-manifold of the nontrivial topology, whose boundary has topology of the form ∂Σ ~ S2, and if, furthermore, the hypersurfaces Σ are all spacelike, then the region Ω contains a quasipermanent intra-universe wormhole.
Characterizing inter-universe wormholes is more difficult. For example, one can imagine a 'baby' universe connected to its 'parent' by a narrow 'umbilicus'. One might like to regard the umbilicus as the throat of a wormhole, but the spacetime is simply connected. For this reason wormholes have been defined geometrically, as opposed to topologically, as regions of spacetime that constrain the incremental deformation of closed surfaces. For example, in Enrico Rodrigo’s The Physics of Stargates a wormhole is defined informally as
a region of spacetime containing a "world tube" (the time evolution of a closed surface) that cannot be continuously deformed (shrunk) to a world line [(the time evolution of a point)].
## Schwarzschild wormholes
An artist's impression of a wormhole from an observer's perspective, crossing the event horizon of a Schwarzschild wormhole which bridges two different universes. The observer originates from the right, and another universe becomes visible in the center of the wormhole’s shadow once the horizon is crossed, the observer seeing light that has fallen into the black hole interior region from the other universe; however, this other universe is unreachable in the case of a Schwarzschild wormhole, as the bridge always collapses before the observer has time to cross it, and everything that has fallen through the event horizon of either universe is inevitably crushed in the singularity.
Lorentzian wormholes known as Schwarzschild wormholes or Einstein–Rosen bridges are connections between areas of space that can be modeled as vacuum solutions to the Einstein field equations, and which are now understood to be intrinsic parts of the maximally extended version of the Schwarzschild metric describing an eternal black hole with no charge and no rotation. Here, "maximally extended" refers to the idea that the spacetime should not have any "edges": for any possible trajectory of a free-falling particle (following a geodesic) in the spacetime, it should be possible to continue this path arbitrarily far into the particle's future or past, unless the trajectory hits a gravitational singularity like the one at the center of the black hole's interior. In order to satisfy this requirement, it turns out that in addition to the black hole interior region which particles enter when they fall through the event horizon from the outside, there must be a separate white hole interior region which allows us to extrapolate the trajectories of particles which an outside observer sees rising up away from the event horizon. And just as there are two separate interior regions of the maximally extended spacetime, there are also two separate exterior regions, sometimes called two different "universes", with the second universe allowing us to extrapolate some possible particle trajectories in the two interior regions. This means that the interior black hole region can contain a mix of particles that fell in from either universe (and thus an observer who fell in from one universe might be able to see light that fell in from the other one), and likewise particles from the interior white hole region can escape into either universe. All four regions can be seen in a spacetime diagram which uses Kruskal–Szekeres coordinates.
In this spacetime, it is possible to come up with coordinate systems such that if you pick a hypersurface of constant time (a set of points that all have the same time coordinate, such that every point on the surface has a space-like separation, giving what is called a 'space-like surface') and draw an "embedding diagram" depicting the curvature of space at that time, the embedding diagram will look like a tube connecting the two exterior regions, known as an "Einstein–Rosen bridge". Note that the Schwarzschild metric describes an idealized black hole that exists eternally from the perspective of external observers; a more realistic black hole that forms at some particular time from a collapsing star would require a different metric. When the infalling stellar matter is added to a diagram of a black hole's history, it removes the part of the diagram corresponding to the white hole interior region, along with the part of the diagram corresponding to the other universe.[12]
The Einstein–Rosen bridge was discovered by Albert Einstein and his colleague Nathan Rosen, who first published the result in 1935. However, in 1962 John A. Wheeler and Robert W. Fuller published a paper showing that this type of wormhole is unstable if it connects two parts of the same universe, and that it will pinch off too quickly for light (or any particle moving slower than light) that falls in from one exterior region to make it to the other exterior region.
The motion through a Schwarzschild wormhole connecting two universes is possible in only one direction. The analysis of the radial geodesic motion of a massive particle into an Einstein–Rosen bridge shows that the proper time of the particle extends to infinity. Timelike and null geodesics in the gravitational field of a Schwarzschild wormhole are complete because the expansion scalar in the Raychaudhuri equation has a discontinuity at the event horizon, and because an Einstein–Rosen bridge is represented by the Kruskal diagram in which the two antipodal future event horizons are identified. Schwarzschild wormholes and Schwarzschild black holes are different, mathematical solutions of general relativity and Einstein–Cartan–Sciama–Kibble theory of gravity. Yet for distant observers, both solutions with the same mass are indistinguishable. These results suggest that all observed astrophysical black holes may be Einstein–Rosen bridges, each with a new universe inside that formed simultaneously with the black hole. Accordingly, our own Universe may be the interior of a black hole existing inside another universe.[13]
According to general relativity, the gravitational collapse of a sufficiently compact mass forms a singular Schwarzschild black hole. In the Einstein–Cartan–Sciama–Kibble theory of gravity, however, it forms a regular Einstein–Rosen bridge. This theory extends general relativity by removing a constraint of the symmetry of the affine connection and regarding its antisymmetric part, the torsion tensor, as a dynamical variable. Torsion naturally accounts for the quantum-mechanical, intrinsic angular momentum (spin) of matter. The minimal coupling between torsion and Dirac spinors generates a repulsive spin–spin interaction which is significant in fermionic matter at extremely high densities. Such an interaction prevents the formation of a gravitational singularity. Instead, the collapsing matter reaches an enormous but finite density and rebounds, forming the other side of the bridge.[14]
Before the stability problems of Schwarzschild wormholes were apparent, it was proposed that quasars were white holes forming the ends of wormholes of this type.[citation needed]
While Schwarzschild wormholes are not traversable in both directions, their existence inspired Kip Thorne to imagine traversable wormholes created by holding the 'throat' of a Schwarzschild wormhole open with exotic matter (material that has negative mass/energy).
## Traversable wormholes
Image of a traversable wormhole that connects the place in front of the physical institutes of Tübingen University with the sand dunes near Boulogne sur Mer in the north of France. The image is calculated with 4D raytracing in a Morris–Thorne wormhole metric, but the gravitational effects on the wavelength of light have not been simulated.[15]
Lorentzian traversable wormholes would allow travel in both directions from one part of the universe to another part of that same universe very quickly or would allow travel from one universe to another. The possibility of traversable wormholes in general relativity was first demonstrated by Kip Thorne and his graduate student Mike Morris in a 1988 paper. For this reason, the type of traversable wormhole they proposed, held open by a spherical shell of exotic matter, is referred to as a Morris–Thorne wormhole. Later, other types of traversable wormholes were discovered as allowable solutions to the equations of general relativity, including a variety analyzed in a 1989 paper by Matt Visser, in which a path through the wormhole can be made where the traversing path does not pass through a region of exotic matter. However, in the pure Gauss–Bonnet gravity (a modification to general relativity involving extra spatial dimensions which is sometimes studied in the context of brane cosmology) exotic matter is not needed in order for wormholes to exist—they can exist even with no matter.[16] A type held open by negative mass cosmic strings was put forth by Visser in collaboration with Cramer et al.,[10] in which it was proposed that such wormholes could have been naturally created in the early universe.
Wormholes connect two points in spacetime, which means that they would in principle allow travel in time, as well as in space. In 1988, Morris, Thorne and Yurtsever worked out explicitly how to convert a wormhole traversing space into one traversing time.[2] However, according to general relativity it would not be possible to use a wormhole to travel back to a time earlier than when the wormhole was first converted into a time machine by accelerating one of its two mouths.[17]
### Raychaudhuri's theorem and exotic matter
To see why exotic matter is required, consider an incoming light front traveling along geodesics, which then crosses the wormhole and re-expands on the other side. The expansion goes from negative to positive. As the wormhole neck is of finite size, we would not expect caustics to develop, at least within the vicinity of the neck. According to the optical Raychaudhuri's theorem, this requires a violation of the averaged null energy condition. Quantum effects such as the Casimir effect cannot violate the averaged null energy condition in any neighborhood of space with zero curvature,[18] but calculations in semiclassical gravity suggest that quantum effects may be able to violate this condition in curved spacetime.[19] Although it was hoped recently that quantum effects could not violate an achronal version of the averaged null energy condition,[20] violations have nevertheless been found,[21] so it remains an open possibility that quantum effects might be used to support a wormhole.
### Faster-than-light travel
The impossibility of faster-than-light relative speed only applies locally. Wormholes allow superluminal (faster-than-light) travel by ensuring that the speed of light is not exceeded locally at any time. While traveling through a wormhole, subluminal (slower-than-light) speeds are used. If two points are connected by a wormhole, the time taken to traverse it would be less than the time it would take a light beam to make the journey if it took a path through the space outside the wormhole. However, a light beam traveling through the wormhole would always beat the traveler. As an analogy, running around to the opposite side of a mountain at maximum speed may take longer than walking through a tunnel crossing it.
### Time travel
Main article: Time travel
The theory of general relativity predicts that if traversable wormholes exist, they could allow time travel.[2] This would be accomplished by accelerating one end of the wormhole to a high velocity relative to the other, and then sometime later bringing it back; relativistic time dilation would result in the accelerated wormhole mouth aging less than the stationary one as seen by an external observer, similar to what is seen in the twin paradox. However, time connects differently through the wormhole than outside it, so that synchronized clocks at each mouth will remain synchronized to someone traveling through the wormhole itself, no matter how the mouths move around.[22] This means that anything which entered the accelerated wormhole mouth would exit the stationary one at a point in time prior to its entry.
For example, consider two clocks at both mouths both showing the date as 2000. After being taken on a trip at relativistic velocities, the accelerated mouth is brought back to the same region as the stationary mouth with the accelerated mouth's clock reading 2004 while the stationary mouth's clock read 2012. A traveler who entered the accelerated mouth at this moment would exit the stationary mouth when its clock also read 2004, in the same region but now eight years in the past. Such a configuration of wormholes would allow for a particle's world line to form a closed loop in spacetime, known as a closed timelike curve.
It is thought that it may not be possible to convert a wormhole into a time machine in this manner; the predictions are made in the context of general relativity, but general relativity does not include quantum effects. Analyses using the semiclassical approach to incorporating quantum effects into general relativity have sometimes indicated that a feedback loop of virtual particles would circulate through the wormhole with ever-increasing intensity, destroying it before any information could be passed through it, in keeping with the chronology protection conjecture. This has been called into question by the suggestion that radiation would disperse after traveling through the wormhole, therefore preventing infinite accumulation. The debate on this matter is described by Kip S. Thorne in the book Black Holes and Time Warps, and a more technical discussion can be found in The quantum physics of chronology protection by Matt Visser.[23] There is also the Roman ring, which is a configuration of more than one wormhole. This ring seems to allow a closed time loop with stable wormholes when analyzed using semiclassical gravity, although without a full theory of quantum gravity it is uncertain whether the semiclassical approach is reliable in this case.
### Inter-universe travel
A possible resolution to the paradoxes resulting from wormhole-enabled time travel rests on the many-worlds interpretation of quantum mechanics. In 1991 David Deutsch showed that quantum theory is fully consistent (in the sense that the so-called density matrix can be made free of discontinuities) in spacetimes with closed timelike curves.[24] However, later it was shown that such model of closed timelike curve can have internal inconsistencies as it will lead to strange phenomena like distinguishing non orthogonal quantum states and distinguishing proper and improper mixture.[25][26] Accordingly, the destructive positive feedback loop of virtual particles circulating through a wormhole time machine, a result indicated by semi-classical calculations, is averted. A particle returning from the future does not return to its universe of origination but to a parallel universe. This suggests that a wormhole time machine with an exceedingly short time jump is a theoretical bridge between contemporaneous parallel universes.[27] Because a wormhole time-machine introduces a type of nonlinearity into quantum theory, this sort of communication between parallel universes is consistent with Joseph Polchinski’s discovery of an “Everett phone” in Steven Weinberg’s formulation of nonlinear quantum mechanics.[28]
## Metrics
Theories of wormhole metrics describe the spacetime geometry of a wormhole and serve as theoretical models for time travel. An example of a (traversable) wormhole metric is the following:
$ds^2= - c^2 dt^2 + dl^2 + (k^2 + l^2)(d \theta^2 + \sin^2 \theta \, d\phi^2).$
One type of non-traversable wormhole metric is the Schwarzschild solution (see the first diagram):
$ds^2= - c^2 \left(1 - \frac{2GM}{rc^2}\right)dt^2 + \frac{dr^2}{1 - \frac{2GM}{rc^2}} + r^2(d \theta^2 + \sin^2 \theta \, d\phi^2).$
## In fiction
Main article: Wormholes in fiction
Wormholes are a common element in science fiction as they allow interstellar, intergalactic, and sometimes interuniversal travel within human timescales. They have also served as a method for time travel.
## Notes
1. "Space and Time Warps". Hawking.org.uk. Retrieved 2010-11-11.
2. ^ a b c Morris, Michael; Thorne, Kip; Yurtsever, Ulvi (1988). "Wormholes, Time Machines, and the Weak Energy Condition". Physical Review Letters 61 (13): 1446–1449. Bibcode:1988PhRvL..61.1446M. doi:10.1103/PhysRevLett.61.1446. PMID 10038800.
3. Sopova; Ford (2002). "The Energy Density in the Casimir Effect". 66 (4): 045026. arXiv:quant-ph/0204125. Bibcode:2002PhRvD..66d5026S. doi:10.1103/PhysRevD.66.045026.
4. Ford; Roman (1995). "Averaged Energy Conditions and Quantum Inequalities". 51 (8): 4277–4286. arXiv:gr-qc/9410043. Bibcode:1995PhRvD..51.4277F. doi:10.1103/PhysRevD.51.4277.
5. Olum (1998). "Superluminal travel requires negative energies". 81 (17): 3567–3570. arXiv:gr-qc/9805003. Bibcode:1998PhRvL..81.3567O. doi:10.1103/PhysRevLett.81.3567.
6. Thorne, Kip S. (1994). . W. W. Norton. pp. 494–496. ISBN 0-393-31276-3.
7. Ian H., Redmount; Wai-Mo Suen (1994). "Quantum Dynamics of Lorentzian Spacetime Foam". 49 (10): 5199. arXiv:gr-qc/9309017. Bibcode:1994PhRvD..49.5199R. doi:10.1103/PhysRevD.49.5199.
8. Kirillov, A.A.; E.P. Savelova (21 February 2008). "Dark Matter from a gas of wormholes". 660 (3): 93. arXiv:0707.1081. Bibcode:2008PhLB..660...93K. doi:10.1016/j.physletb.2007.12.034.
9. Rodrigo, Enrico (30 November 2009). "Denouement of a Wormhole-Brane Encounter". 18 (12): 1809. arXiv:0908.2651. Bibcode:2009IJMPD..18.1809R. doi:10.1142/S0218271809015333.
10. ^ a b John G. Cramer, Robert L. Forward, Michael S. Morris, Matt Visser, Gregory Benford, and Geoffrey A. Landis (1995). "Natural Wormholes as Gravitational Lenses". 51 (6): 3117–3120. arXiv:astro-ph/9409051. Bibcode:1995PhRvD..51.3117C. doi:10.1103/PhysRevD.51.3117.
11. Coleman, Korte, Hermann Weyl's Raum - Zeit - Materie and a General Introduction to His Scientific Work, p. 199
12. "Collapse to a Black Hole". Casa.colorado.edu. 2010-10-03. Retrieved 2010-11-11. This is a tertiary source that clearly includes information from other sources but does not name them.
13. Poplawski, Nikodem J. (2010). "Radial motion into an Einstein–Rosen bridge". 687 (2–3): 110–113. arXiv:0902.1994. Bibcode:2010PhLB..687..110P. doi:10.1016/j.physletb.2010.03.029.
14. Poplawski, Nikodem J. (2010). "Cosmology with torsion: An alternative to cosmic inflation". Phys. Lett. B 694 (3): 181–185. arXiv:1007.0587. Bibcode:2010PhLB..694..181P. doi:10.1016/j.physletb.2010.09.056.
15. Other computer-rendered images and animations of traversable wormholes can be seen on this page by the creator of the image in the article, and this page has additional renderings.
16. Elias Gravanis; Steven Willison (2007). "`Mass without mass' from thin shells in Gauss-Bonnet gravity". Phys.Rev.D75:084025,2007 75 (8). arXiv:gr-qc/0701152. Bibcode:2007PhRvD..75h4025G. doi:10.1103/PhysRevD.75.084025.
17. Thorne, Kip S. (1994). . W. W. Norton. p. 504. ISBN 0-393-31276-3.
18. Fewster, Christopher J.; Ken D. Olum, Michael J. Pfenning (10 January 2007). "Averaged null energy condition in spacetimes with boundaries". 75 (2): 025007. arXiv:gr-qc/0609007. Bibcode:2007PhRvD..75b5007F. doi:10.1103/PhysRevD.75.025007.
19. Visser, Matt (15 October 1996). "Gravitational vacuum polarization. II. Energy conditions in the Boulware vacuum". 54 (8): 5116. arXiv:gr-qc/9604008. Bibcode:1996PhRvD..54.5116V. doi:10.1103/PhysRevD.54.5116.
20. Graham, Noah; Ken D. Olum (4 September 2007). "Achronal averaged null energy condition". 76 (6): 064001. arXiv:0705.3193. Bibcode:2007PhRvD..76f4001G. doi:10.1103/PhysRevD.76.064001.
21. Urban, Douglas; Ken D. Olum (1 June 2010). "Spacetime averaged null energy condition". 81 (6): 124004. arXiv:1002.4689. Bibcode:2010PhRvD..81l4004U. doi:10.1103/PhysRevD.81.124004.
22. Thorne, Kip S. (1994). . W. W. Norton. p. 502. ISBN 0-393-31276-3.
23. Deutsch, David (1991). "Quantum Mechanics Near Closed Timelike Lines". 44 (10): 3197. Bibcode:1991PhRvD..44.3197D. doi:10.1103/PhysRevD.44.3197.
24. Brun et.al (2009). "Localized Closed Timelike Curves Can Perfectly Distinguish Quantum States". 102 (21): 210402. Bibcode:2009PhRvL..102.210402. doi:10.1103/PhysRevLett.102.210402.
25. Pati, Chakrabarty, Agrawal (2011). "Purification of mixed states with closed timelike curve is not possible". 84 (6): 062325. arXiv:1003.4221. Bibcode:2011PhRvA..84f2325P. doi:10.1103/PhysRevA.84.062325. Unknown parameter `|bibicode=` ignored (help)
26. Rodrigo, Enrico (2010). The Physics of Stargates. Eridanus Press. p. 281. ISBN 0-9841500-0-5.
27. Polchinski, Joseph (1991). "Weinberg’s Nonlinear quantum Mechanics and the Einstein-Podolsky-Rosen Paradox". 66 (4): 397. Bibcode:1991PhRvL..66..397P. doi:10.1103/PhysRevLett.66.397.
## References
• DeBenedictis, Andrew and Das, A. (2001). "On a General Class of Wormhole Geometries". 18 (7): 1187–1204. arXiv:gr-qc/0009072. Bibcode:2001CQGra..18.1187D. doi:10.1088/0264-9381/18/7/304.
• Dzhunushaliev, Vladimir (2002). "Strings in the Einstein's paradigm of matter". 19 (19): 4817–4824. arXiv:gr-qc/0205055. Bibcode:2002CQGra..19.4817D. doi:10.1088/0264-9381/19/19/302.
• Einstein, Albert and Rosen, Nathan (1935). "The Particle Problem in the General Theory of Relativity". 48: 73. Bibcode:1935PhRv...48...73E. doi:10.1103/PhysRev.48.73.
• Fuller, Robert W. and Wheeler, John A. (1962). "Causality and Multiply-Connected Space-Time". 128: 919. Bibcode:1962PhRv..128..919F. doi:10.1103/PhysRev.128.919.
• Garattini, Remo (2004). "How Spacetime Foam modifies the brick wall". 19 (36): 2673–2682. arXiv:gr-qc/0409015. Bibcode:2004gr.qc.....9015G. doi:10.1142/S0217732304015658.
• González-Díaz, Pedro F. (1998). "Quantum time machine". 58 (12): 124011. arXiv:gr-qc/9712033. Bibcode:1998PhRvD..58l4011G. doi:10.1103/PhysRevD.58.124011.
• González-Díaz, Pedro F. (1996). "Ringholes and closed timelike curves". 54 (10): 6122–6131. arXiv:gr-qc/9608059. Bibcode:1996PhRvD..54.6122G. doi:10.1103/PhysRevD.54.6122.
• Khatsymosky, Vladimir M. (1997). "Towards possibility of self-maintained vacuum traversable wormhole". 399 (3–4): 215–222. arXiv:gr-qc/9612013. Bibcode:1997PhLB..399..215K. doi:10.1016/S0370-2693(97)00290-6.
• Krasnikov, Serguei (2006). "Counter example to a quantum inequality". 46: 195. arXiv:gr-qc/0409007. Bibcode:2006GrCo...12..195K.
• Krasnikov, Serguei (2003). "The quantum inequalities do not forbid spacetime shortcuts". 67 (10): 104013. arXiv:gr-qc/0207057. Bibcode:2003PhRvD..67j4013K. doi:10.1103/PhysRevD.67.104013.
• Li, Li-Xin (2001). "Two Open Universes Connected by a Wormhole: Exact Solutions". 40 (2): 154–160. arXiv:hep-th/0102143. Bibcode:2001JGP....40..154L. doi:10.1016/S0393-0440(01)00028-6.
• Morris, Michael S., Thorne, Kip S., and Yurtsever, Ulvi (1988). "Wormholes, Time Machines, and the Weak Energy Condition". 61 (13): 1446. Bibcode:1988PhRvL..61.1446M. doi:10.1103/PhysRevLett.61.1446. PMID 10038800.
• Morris, Michael S. and Thorne, Kip S. (1988). "Wormholes in spacetime and their use for interstellar travel: A tool for teaching general relativity". 56 (5): 395–412. Bibcode:1988AmJPh..56..395M. doi:10.1119/1.15620.
• Nandi, Kamal K. and Zhang, Yuan-Zhong (2006). "A Quantum Constraint for the Physical Viability of Classical Traversable Lorentzian Wormholes". 9: 61–67. arXiv:gr-qc/0409053. Bibcode:2004gr.qc.....9053N.
• Ori, Amos (2005). "A new time-machine model with compact vacuum core". 95 (2). arXiv:gr-qc/0503077. Bibcode:2005PhRvL..95b1101O. doi:10.1103/PhysRevLett.95.021101.
• Roman, Thomas, A. (2004). "Some Thoughts on Energy Conditions and Wormholes". arXiv:gr-qc/0409090 [gr-qc].
• Teo, Edward (1998). "Rotating traversable wormholes". 58 (2). arXiv:gr-qc/9803098. Bibcode:1998PhRvD..58b4014T. doi:10.1103/PhysRevD.58.024014.
• Visser, Matt (2002). "The quantum physics of chronology protection by Matt Visser". arXiv:gr-qc/0204022 [gr-qc]. An excellent and more concise review.
• Visser, Matt (1989). "Traversable wormholes: Some simple examples". 39 (10): 3182–3184. arXiv:0809.0907. Bibcode:1989PhRvD..39.3182V. doi:10.1103/PhysRevD.39.3182.
Retrieved from "http://en.wikipedia.org/w/index.php?title=Wormhole&oldid=555105027"
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Physics Forums
## Vertical Spring
Let's say we have a vertical spring. As it oscillates, the energy bounces between gravitational potential energy, spring potential energy, and kinetic energy, right? Why does my lab say that spring potential energy = kinetic energy at any moment in time? Is that a mistake on their part?
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Gravitational potential energy is probably negligible. More accurately, the difference in gravitational potential energy between different positions is negligible. You know the mass on your spring, calculate its gravitational potential energy at different positions and see how it compares to the spring's energy (presuming you know the spring constant).
I question the assertion based on the idea that when the mass bounces across the spring's neutral point, it will have zero potential energy while the kinetic energy can be non-zero.
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## Vertical Spring
What they should be saying is that the total energy, PE + KE, is conserved and constant in time.
Quote by ModusPwnd Gravitational potential energy is probably negligible. More accurately, the difference in gravitational potential energy between different positions is negligible. You know the mass on your spring, calculate its gravitational potential energy at different positions and see how it compares to the spring's energy (presuming you know the spring constant).
Can someone just confirm that this is correct? I'm about to submit a lab tomorrow morning and I'm too tired at the moment to do what he suggested.
Edit: Never mind. I understand why they ignored gravitational potential energy now... If you measure the distance from the hanging equilibrium position (without the mass) opposed to the equilibrium position with mass then gravitational potential energy is incorporated into spring potential energy.
Edit2: Never mind. I think my lab is wrong. I'm so confused.
Quote by tahayassen Can someone just confirm that this is correct? I'm about to submit a lab tomorrow morning and I'm too tired at the moment to do what he suggested. Edit: Never mind. I understand why they ignored gravitational potential energy now... If you measure the distance from the hanging equilibrium position (without the mass) opposed to the equilibrium position with mass then gravitational potential energy is incorporated into spring potential energy. Edit2: Never mind. I think my lab is wrong. I'm so confused.
If your lab script say PE = KE at all times then it is wrong. Otherwise there would be no transfer of energy and hence no bouncing of the spring!
(With a contrived definition of where PE = 0 you could set up a situation where |PE| = KE, but I'm sure that wasn't their intention).
You're also right about the GPE. If your spring is nice and linear, then you can simplify things by just considering PE (a combination of GPE and spring PE).
Quote by mickybob If your spring is nice and linear, then you can simplify things by just considering PE (a combination of GPE and spring PE).
What would the point of that be? :| They would still be two separate terms, right?
Quote by tahayassen What would the point of that be? :| They would still be two separate terms, right?
Yes, but only one will be time dependent.
If you consider first a horizontal spring with a mass on the end (i.e. ignore gravity). The total energy, at any time, is:
$E_{Total}= E_{K} + E_{SP}$
where $E_{K}$ is kinetic energy and $E_{SP}$ is spring potential energy. More explicitly we have:
$E_{Total} = 1/2 mv^{2} + 1/2 kx^{2}$
where $v$ is the velocity of the mass, and $x$ is the extension of the spring. Assuming we're not losing energy through friction etc., then the total energy is constant from the Conservation of Energy Principle. We can confirm that by substituting in expressions for $x$ and $v$ if we want.
So we have quite a simple situation. In terms of problem solving, we know that:
$1/2 mv^{2} + 1/2 kx^{2} =$ constant
at all times, which is a pretty useful identity.
What I want to show is that we can recover this simple situation even when gravity is involved. I'll do this by working relative to the initial, equilibrium extension of the spring due to gravity.
When the spring is in equilibrium with gravity, forces are balanced - the downward force of gravity matches the upwards force of the spring. So:
$kl = mg$
where $l$ is the equilibrium extension of the spring and $g$ is the acceleration due to gravity. So the initial extension is:
$l = mg/k$
which we'll need shortly.
Now, we set the spring bouncing. At any time we have the total energy:
$E_{Total} = E_{K} + E_{GP} + E_{SP}$
where $E_{GP}$ is gravitational potential energy.
To simplify things, I'm going to choose $E_{GP} = 0$ at the equilibrium position - I'm free to choose this anywhere I like.
Now we can say:
$E_{Total} = 1/2 mv^{2} - mgx + 1/2 k(x + l)^{2}$
where $x$ is the extension beyond the equilibrium extension at any give time. So the total extension is $x + l$, which is why that expression appears in the term for the spring potential energy.
Now, expand out the final term:
$E_{Total} = 1/2 mv^{2} - mgx + 1/2 kx^{2} + kxl + 1/2 kl^{2}$
substitute in the expression we calculated for the equilibrium extension, $l$ earlier:
$E_{Total} = 1/2 mv^{2} - mgx + 1/2 kx^{2} + mgx + 1/2 kl^{2}$
cancel the $mgx$ terms, and we have:
$E_{Total} = 1/2 mv^{2} + 1/2 kx^{2} + 1/2 kl^{2}$
So now, as you say, we still have three terms instead of two. The first describes the kinetic energy as a function of velocity, and hence of time. The second describes the potential energy as a function of extension from equilibrium, and so also as a function of time. And the final term describes the energy as a function of the equilibrium extension which is a constant in time.
So, in terms of the time evolution of the system, we have got back to a nice, simple, identity:
$1/2 mv^{2} + 1/2 kx^{2}=$ constant
where $x$ is now the extension from equilibrium.
This makes some problems easier to solve.
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http://mathhelpforum.com/advanced-algebra/37377-give-example-non-normal-subgroup-group.html
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# Thread:
1. ## give an example of non-normal subgroup of a group
Give an example of a subgroup of a group that is not normal
2. Originally Posted by szpengchao
Give an example of a subgroup of a group that is not normal
Any subgroup of $S_5$ which is not $\{ e\}, A_5, S_5$ would not be normal. This provides a lot of examples.
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http://mathoverflow.net/questions/88317/coproduct-on-coordinate-ring-of-finite-algebraic-group/88320
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## Coproduct on coordinate ring of finite algebraic group
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I'm reading Mukai's book "An introduction to invariants and moduli", and I am having trouble understanding one of his examples. It is example 3.49 on page 101.
The setup is as follows. Let $G$ be a finite group, considered as an algebraic group over a field $k$. The coordinate ring of $G$ is then just the set of functions $G \rightarrow k$ with the usual pointwise addition and multiplication. This can be identified with the group ring $k[G]$ in the obvious way (an element $[g] \in k[G]$ corresponds to the function $G \rightarrow k$ that takes $g$ to $1$ and $h$ to $0$ for $h \neq g$). Under this identification, it seems to me that the coproduct is the function
$$\phi : k[G] \rightarrow k[G] \otimes k[G]$$
$$\phi([g]) = \sum_{h \in G} [h] \otimes [h^{-1} g]$$
However, Mukai asserts that if $G$ is the finite cyclic group of order $n$, then the coordinate ring of $G$ is $k[t]/(t^n-1)$ with the coproduct $t \mapsto t \otimes t$. These do not seem like the same thing to me -- what am I doing wrong?
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2
What Mukai has written down here is the co-ordinate ring of the finite group scheme $\mu_n$ that parametrizes $n^\text{th}$-roots of unity. As people have noted below, this is the dual to the group scheme $\mathbb{Z}/n\mathbb{Z}$, whose co-ordinate ring will have the co-multiplication you describe. – Keerthi Madapusi Pera Feb 13 2012 at 2:03
## 4 Answers
I think the author accidentally described the dual of the Hopf algebra you're thinking of. Finite group rings are usually endowed with multiplication $(g,h)\mapsto gh$ and comultiplication $g \mapsto g\otimes g$ (see here).
The coordinate ring $k[G]$ is obtained by dualizing. Then $g \mapsto g\otimes g$ becomes $e_g^2 = e_g$, where $e_g$ is the function on $G$ that maps $g$ to $1$ and all other group elements to $0$. Comultiplication will look exactly the way you described it (i.e. $e_g \mapsto \sum_h e_{gh^{-1}}\otimes e_h$).
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The Hopf algebra structure here involves a coproduct taking the function `$f$` on `$G$` to `$\sum_i g_i \otimes h_i$`, where `$f(xy) = \sum_i f_i(x) g_i(y)$` when `$x,y \in G$`. Whatever Mukai is doing for a cyclic group should be consistent with this formulation of the coproduct, but I'm unfamiliar with his book.
More generally, this kind of formalism occurs when you consider a finite group scheme as in Jantzen Representations of Algebraic Groups (AMS, 2003), I.2.3.
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The book presumably assumes the field $k$ contains $n$ distinct roots of unity (in particular, that the characteristic of $k$ is coprime to $n$). Then you get a $k$-algebra isomorphism between $k[x]/(x^n-1)$ (isomorphic to the group ring $k[G]$ by sending $x$ to a generator) and the coordinate ring $\bigoplus_{g \in G} k$ by a finite Fourier transform. This reflects the fact that finite abelian groups are Pontryagin self-dual.
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It might be that the author accidentally described the dual of the Hopf algebra as Florian Eisele suggested, however in this case the Hopf algebra is self dual so k[G] is actualy isomorphic to $k[t]/(t^n−1)$. The isomorphism is not compliantly canonical, it becomes canonical if $G$ is the group of n-s roots of $1$. Then it is given by $[g] \mapsto \sum_1^n g^i t^i$.
So may be this is what the author meant.
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Are you sure $k[\mathbb Z/n\mathbb Z]$ is self-dual? If $char(k)=p$ and $n=p$, then $k[t](t^p-1)$ is not semisimple as an algebra, but its dual (the coordinate ring $k\oplus \ldots\oplus k$) is. – Florian Eisele Feb 13 2012 at 2:08
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http://mathoverflow.net/questions/11219/is-there-a-the-arithmetic-of-elliptic-curves-for-shimura-curves
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## What is a good roadmap for learning Shimura curves?
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I am interested in learning about Shimura curves. Unlike most of the people who post reference requests however (see this question for example), my problem is not sorting through an abundance of books but rather dealing with what appears to be an extreme paucity of sources.
Anyway, I'm a graduate student and have spent the last year or so thinking about the arithmetic of orders in quaternion algebras (and more generally in central simple algebras). The study of orders in quaternion algebras seems to play an important role in Shimura curves, and I'd like to study these connections more carefully.
Unfortunately, it has been very difficult for me to find a good place to start. I only really know of two books that explicitly deal with Shimura curves:
• Shimura's Introduction to the arithmetic theory of automorphic functions
• Alsina and Bayer's Quaternion Orders, Quadratic Forms, and Shimura Curves
Neither book has been particularly helpful however; the first only mentions them briefly in the final section, and the second has much more of a computational focus then I'd like.
Question 1: Is there a book along the lines of Silverman's The Arithmetic of Elliptic Curves for Shimura curves?
I kind of doubt that such a book exists. Thus I've tried to read the introductory sections of a few papers & theses, but have run into a problem. There seem to be various ways of thinking about a Shimura curve, and it has been the case that every time I look at an article I'm confronted with a different one. For example, this talk by Voight and this paper by Milne. By analogy, it seems to be a lot like trying to learn class field theory by switching between articles with ideal-theoretic statements and articles taking an adelic slant without having a definitive source which tells you that both are describing the same theorems.
My second question is therefore:
Question 2: Can anyone suggest a 'roadmap' to Shimura curves? Which theses or papers have especially good expository accounts of the basic properties that one needs in order to understand the literature.
Clearly I need to say something about my background. As I mentioned above, I'm an algebraic number theorist with a particular interest in quaternion algebras. I don't have the best algebraic geometry background in the world, but have read Mumford's Red Book, the first few chapters of Hartshorne and Qing Liu's Algebraic Geometry and Arithmetic Curves. I've also read Silverman's book The Arithmetic of Elliptic Curves and Diamond and Shurman's A First Course in Modular Forms.
Thanks.
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11
Alas, the answer to Question 1 is definitely no, and I don't know a good answer for Question 2. I just mention that the article Carayol, Henri. Sur la mauvaise réduction des courbes de Shimura. (French) [Bad reduction of Shimura curves] Compositio Math. 59 (1986), no. 2, 151--230. MR0860139 is something of a classic, and a basic reference, but it is also quite difficult. – JS Milne Jan 9 2010 at 6:53
11
+1 Milne. I learnt a lot from Carayol, which is really the only place for Shimura curves over totally real fields. Over Q there are other places to look. Wait until Pete Clark wakes up and he'll probably tell you good places to start over Q. – Kevin Buzzard Jan 9 2010 at 8:34
3
@Buzzard. I plus-ed your comment for the remark on Pete Clark waking up. – Anweshi Jan 9 2010 at 14:51
@Ben, I retitled your question to reflect more general second question; it includes the first one as well. Feel free to revert! – Ilya Nikokoshev Feb 4 2010 at 19:27
## 2 Answers
First of all, Kevin is being quite modest in his comment above: his paper
Buzzard, Kevin. Integral models of certain Shimura curves. Duke Math. J. 87 (1997), no. 3, 591--612.
contains many basic results on integral models of Shimura curves over totally real fields, and is widely cited by workers in the field: 22 citations on MathSciNet. The most recent is a paper of mine:
Clark, Pete L. On the Hasse principle for Shimura curves. Israel J. Math. 171 (2009), 349--365.
http://math.uga.edu/~pete/plclarkarxiv7.pdf
Section 3 of this paper spends 2-3 pages summarizing results on the structure of the canonical integral model of a Shimura curve over `$\mathbb{Q}$` (with applications to the existence of local points). From the introduction to this paper:
"This result [something about local points] follows readily enough from a description of their [certain Shimura curves over Q] integral canonical models. Unfortunately I know of no unique, complete reference for this material. I have myself written first (my 2003 Harvard thesis) and second (notes from a 2005 ISM course in Montreal) approximations of such a work, and in so doing I have come to respect the difficulty of this expository problem."
I wrote that about three years ago, and I still feel that way today. Here are the documents:
1) http://math.uga.edu/~pete/thesis.pdf
is my thesis. "Chapter 0" is an exposition on Shimura curves: it is about 50 pages long.
2) For my (incomplete) lecture notes from 2005, go to
http://math.uga.edu/~pete/expositions.html
and scroll down to "Shimura Curves". There are 12 files there, totalling 106 pages [perhaps I should also compile them into a single file]. On the other hand, the title of the course was Shimura Varieties, and although I don't so much as attempt to give the definition of a general Shimura variety, some of the discussion includes other PEL-type Shimura varieties like Hilbert and Siegel moduli space. These notes do not entirely supercede my thesis: each contains some material that the other omits.
When I applied for an NSF grant 3 years ago, I mentioned that if I got the grant, as part of my larger impact I would write a book on Shimura curves. Three years later I have written up some new material (as yet unreleased) but am wishing that I had not said that so directly: I would need at least a full semester off to make real progress (partly, of course, to better understand much of the material).
Let me explain the scope of the problem as follows: there does not even exist a single, reasonably comprehensive reference on the arithmetic geometry of the classical modular curves (i.e., $X_0(N)$ and such). This would-be bible of modular curves ought to contain most of the material from Shimura's book (260 pages) and the book of Katz and Mazur Arithmetic Moduli of Elliptic Curves (514 pages). These two books don't mess around and have little overlap, so you get a lower bound of, say, 700 pages that way.
Conversely, I claim that there is some reasonable topology on the arithmetic geometry of modular curves whose compactification is the theory of Shimura curves. The reason is that in many cases there are several ways to establish a result about modular curves, and "the right one" generalizes to Shimura curves with little trouble. (For example, to define the rational canonical model for classical modular curves, one could use the theory of Fourier expansions at the cusps -- which won't generalize -- or the theory of moduli spaces -- which generalizes immediately. Better yet is to use Shimura's theory of special points, which nowadays you need to know anyway to study Heegner point constructions.) Most of the remainder concerns quaternion arithmetic, which, while technical, is nowadays well understood and worked out.
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I can't add much to this fairly comprehensive answer, so I'm going to leave this as a comment. Miyake's book "Modular forms" also covers some ground related to automorphic forms on Shimura curves. It is mostly upper-half-plane stuff in a manner similar to more elementary references than you're looking for with respect to modular curves. However, it at least has the benefit of covering the Eichler-Selberg trace formula (and, correspondingly, some about CM points) for Shimura curves in some detail. – Tyler Lawson Jan 9 2010 at 12:50
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@TL: That's a useful remark. The trace formula is certainly something that should be in "The Bible of Modular Curves" and is not found in either Shimura or Katz-Mazur. – Pete L. Clark Jan 9 2010 at 13:00
@Pete: This response is really fantastic. Thanks so much! – Ben Linowitz Jan 9 2010 at 13:43
1
@Pete Clark. Perhaps for the situation of moduli of elliptic curves the expose of Deligne-Rapoport ought to be more accessible than Katz-Mazur. – Anweshi Jan 9 2010 at 15:02
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
[I decided my previous answer was long enough, so I'm adding this one separately and making it Community Wiki. Feel free to add to it!]
Other people's PhD theses that have nice expositions on Shimura curves include:
David Helm (Berkeley 2003)
Bruce Jordan (Harvard 1981)
David Roberts (Harvard 1989)
Victor Rotger Cerda (Universitat de Barcelona 2002)
John Voight (Berkeley 2005)
[Edit (Emerton):] Ken Ribet's Inventiones 100 article describes certain instances of Shimura curves over $\mathbb Q$, including some relations with orders in quaternion algebras, and some information about $p$-adic uniformization and their bad reduction at primes describing the discriminant. Like all of Ribet's papers, it is a masterpiece of exposition.
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http://mathhelpforum.com/discrete-math/97743-serious-trouble-finding-big-o-print.html
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serious trouble finding big O
Printable View
• August 11th 2009, 06:27 PM
Roclemir
serious trouble finding big O
Original statement:
$f(n)=(n^3+4log_2 n)/(n+4)$
I'm sure the n+4 in both the denominator and numerator have something to do with it, but if i make $n$in the denominator a $n^3$, wouldn't that then make it smaller the the LHS?
I'm suppose to prove
$f(n)=0(n^2)$
which means i need the inequality to make the RHS bigger than the LHS
• August 11th 2009, 09:08 PM
CaptainBlack
Quote:
Originally Posted by Roclemir
Original statement:
$f(n)=(n^3+4log_2 n)/(n+4)$
I'm sure the n+4 in both the denominator and numerator have something to do with it, but if i make $n$in the denominator a $n^3$, wouldn't that then make it smaller the the LHS?
I'm suppose to prove
$f(n)=0(n^2)$
which means i need the inequality to make the RHS bigger than the LHS
For $n>1$ (which makes everything of interest here positive):
$f(n)=\frac{n^3+4\log_2 n}{n+4}<\frac{n^3+4\log_2 n}{n}$ $=n^2+\frac{4 \log_2 (n)}{n}$
Then as $\frac{4 \log_2 (n)}{n} \to 0$ as $n \to \infty$ for large enough $n$ we have:
$\frac{4 \log_2 (n)}{n}<n^2$
So for large enough $n$ :
$f(n)<2n^2$
CB
• August 11th 2009, 11:17 PM
Roclemir
thanks mate, just on one line...
Thanks. All makes sense except that one line starting with "Then as... we have:", hate to be a noob, but can you please explain what you mean with the arrows and stuff thanks.
• August 12th 2009, 12:01 AM
Gamma
Beautiful CB, just filling in some details for him
That notation means as n approaches infinity, $\frac{4log_2(n)}{n}$ becomes arbitrarily close to 0. It is read the limit of $\frac{4log_2(n)}{n}$ as n approaches infinity is 0.
So given any $\epsilon >0$, there exists an integer N for which, $|\frac{4log_2(n)}{n}|<\epsilon$ for all n>N.
In particular, if $\epsilon=1$, there exists an N, such that for all n>N, $\frac{4log_2(n)}{n}<1$, this means we clearly have the identity $\frac{4log_2(n)}{n}<1<n^2\Rightarrow n^2+\frac{4log_2(n)}{n}<n^2+n^2=2n^2$.
Thus $|\frac{x^2+\frac{4log_2(n)}{n}}{n^2}|\leq 2$ for all n>N. Which is the definition of $f(n)=O(n^2)$
All times are GMT -8. The time now is 02:42 AM.
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http://math.stackexchange.com/questions/271035/calculating-a-multivariate-integral-via-level-sets
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# calculating a multivariate integral via level sets
I'm considering the possibility of calculating an integral of the form $\int_{S_n} f(x_1,\dots,x_n) dx_1\dots dx_n$ via level sets, where $S_n$ is the domain of integration. In my problem everything is real valued, and $f:R^n\to R$. Also, consider $f$ smooth, with no holes, etc.
Now, consider $g$ a real number that solves the equation $g=f(x_1,\dots,x_n)$, i.e., a level set. Then, I think one can write the relation \begin{equation} \int_{S_n}f(x_1,\dots,x_n) dx_1\dots dx_n=\int g\rho(g)dg \end{equation} where $\rho(g)dg$ is the density distribution of $g$, or roughly speaking the number of solutions of the level set equation for $g$. A simple example: consider calculating the volume of a half sphere (radius $r$) centered at $(a,a,0)$. Then, the integral on the left hand of the identity is $\int dx \int dy \sqrt{r^2-(x-a)^2-(y-a)^2}$, and the right hand is $\int dz 2\pi\sqrt{r^2-z^2}z$ where $g=z$ and $\rho(g=z)=2\pi\sqrt{r^2-z^2}$.
First, is this statement roughly correct? I suspect it's related to the coarea formula, but am not entirely sure as this is not my area of expertise. Does this result have a particular name?
Also, it seems to me that $\rho(g)$ would relate to the Jacobian (modulus) in some way, although I don't think it's just equal. Any ideas?
Finally, even if all of this is correct, in a concrete case it seems to me that unless $f$ is a very simple example, it is going to be difficult to make concrete calculations in most cases. However, I should still whether there are some techniques out there to deal with this analytically. Or should one consider numerical techniques?
Thanks
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It's roughly correct, but you are right that it's usually very difficult to compute the new integral. Sometimes, when the domain of integration is at least partially bounded by level sets of $f$, the method can help. – mrf Jan 5 at 17:32
@mrf By the way, is there a name for this result? I'm not entirely sure it's this coarea formula, although it seems clearly related. Also, $\rho(g)$ probably relates to a Jacobian or Hausdorff measure, I imagine. I've amended the question a bit to add these questions. – Ed Wolf Jan 5 at 20:27
Small example added – Ed Wolf Jan 7 at 11:50
## 1 Answer
Your example is the same as the "cylindrical shell method," of which your method is a generalization. In general, $\rho(g)$ will be equal to the $(n-1)$-dimensional content of the intersection of $S_n$ and the level set $f({\bf x}) = g$, that is, the integral of $1$ over the intersection.
As with most changes of variables, this technique would be helpful if it suited $f$ and $S_n$ particularly well -- for instance, if $\rho(g)$ is easy to find and $g\,\rho(g)$ easy to integrate. See also the comment to the original post by @mrf.
If a numerical answer is all your after, then numerical techniques are probably sufficient and faster. But if you're interested in theoretical purposes or to get formulas for particular integrals, then numerical approximation won't help much, except perhaps to shed some light in a dark corner, after Hamming's dictum, "the purpose of computing is insight...."
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@ Michael E2 The problem I have is relatively complicated and the solutions $f(\boldsymbol{x})=g$ are difficult to obtain, but the multivariate integral is worse. Numerical analysis may shed insight as suggested by Hamming, but I may need divine inspiration to finish... Thank you – Ed Wolf Jan 8 at 9:51
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http://mathematica.stackexchange.com/questions/tagged/group-theory+graphs-and-networks
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# Tagged Questions
3answers
497 views
### Find cycles of graphs with both directed and undirected edges
I need to enumerate all the simple cycles of a graph which has both directed and undirected edges, where we can treat the undirected edges as doubly directed. (Specifically, I am looking at the Cayley ...
1answer
413 views
### Visualizing Rubik's Graph
After the August 2010 discovery that the diameter of the Rubik graph is 20, I wanted to make a way to visualize Rubik's graph. Since there are about $4.3 \times 10^{19}$ vertices in this graph, it is ...
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http://en.wikipedia.org/wiki/Lucas_number
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# Lucas number
Not to be confused with Lucas sequences, which are a generic class of sequences to which the Lucas numbers belong.
The Lucas numbers or Lucas series are an integer sequence named after the mathematician François Édouard Anatole Lucas (1842–1891), who studied both that sequence and the closely related Fibonacci numbers. Lucas numbers and Fibonacci numbers form complementary instances of Lucas sequences.
## Definition
Similarly to the Fibonacci numbers, each Lucas number is defined to be the sum of its two immediate previous terms, i.e. it is a Fibonacci integer sequence. However, the first two Lucas numbers are L0 = 2 and L1 = 1 instead of 0 and 1, and the properties of Lucas numbers are therefore somewhat different from those of Fibonacci numbers.
The Lucas numbers may thus be defined as follows:
$L_n := \begin{cases} 2 & \text{if } n = 0; \\ 1 & \text{if } n = 1; \\ L_{n-1}+L_{n-2} & \text{if } n > 1. \\ \end{cases}$
The sequence of Lucas numbers begins:
$2,\;1,\;3,\;4,\;7,\;11,\;18,\;29,\;47,\;76,\;123,\; \ldots\;$(sequence in OEIS).
All Fibonacci-like integer sequences appear in shifted form as a row of the Wythoff array; the Fibonacci sequence itself is the first row and the Lucas sequence is the second row. Also like all Fibonacci-like integer sequences, the ratio between two consecutive Lucas numbers converges to the golden ratio.
## Extension to negative integers
Using Ln−2 = Ln − Ln−1, one can extend the Lucas numbers to negative integers to obtain a doubly infinite sequence:
..., −11, 7, −4, 3, −1, 2, 1, 3, 4, 7, 11, ... (terms $L_n$ for $-5\leq{}n\leq5$ are shown).
The formula for terms with negative indices in this sequence is
$L_{-n}=(-1)^nL_n.\!$
## Relationship to Fibonacci numbers
The Lucas numbers are related to the Fibonacci numbers by the identities
• $\,L_n = F_{n-1}+F_{n+1}=F_n+2F_{n-1}$
• $\,L_{m+n} = L_{m+1}F_{n}+L_mF_{n-1}$
• $\,L_n^2 = 5 F_n^2 + 4 (-1)^n$, and thus as $n\,$ approaches +∞, the ratio $\frac{L_n}{F_n}$ approaches $\sqrt{5}.$
• $\,F_{2n} = L_n F_n$
• $\,F_{n+k} + (-1)^k F_{n-k} = L_k F_n$
• $\,F_n = {L_{n-1}+L_{n+1} \over 5}$
Their closed formula is given as:
$L_n = \varphi^n + (1-\varphi)^{n} = \varphi^n + (- \varphi)^{- n}=\left({ 1+ \sqrt{5} \over 2}\right)^n + \left({ 1- \sqrt{5} \over 2}\right)^n\, ,$
where $\varphi$ is the Golden ratio. Alternatively, as for $n>1$ the magnitude of the term $(-\varphi)^{-n}$ is less than 1/2, $L_n$ is the closest integer to $\varphi^n$ or, equivalently, the integer part of $\varphi^n+1/2$, also written as $\lfloor \varphi^n+1/2 \rfloor$.
Conversely, $\varphi^n = {{L_n + F_n \sqrt{5}} \over 2}$.
## Congruence relations
If Fn ≥ 5 is a Fibonacci number then no Lucas number is divisble by Fn.
Ln is congruent to 1 mod n if n is prime, but some composite values of n also have this property.
## Lucas primes
A Lucas prime is a Lucas number that is prime. The first few Lucas primes are
2, 3, 7, 11, 29, 47, 199, 521, 2207, 3571, 9349, ... (sequence in OEIS).
If Ln is prime then n is either 0, prime, or a power of 2.[1] L2m is prime for m = 1, 2, 3, and 4 and no other known values of m.
## Lucas polynomials
In the same way as Fibonacci polynomials are derived from the Fibonacci numbers, the Lucas polynomials Ln(x) are a polynomial sequence derived from the Lucas numbers
## References
1. Chris Caldwell, "The Prime Glossary: Lucas prime" from The Prime Pages.
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http://unapologetic.wordpress.com/2011/06/15/the-lie-derivative/?like=1&source=post_flair&_wpnonce=c8d80d4293
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# The Unapologetic Mathematician
## The Lie Derivative
Let’s go back to the way a vector field on a manifold $M$ gives us a “derivative” of smooth functions $f\in\mathcal{O}M$. If $X\in\mathfrak{X}M$ is a smooth vector field it has a maximal flow $\Phi$ which gives a one-parameter family $\Phi_t$ of diffeomorphisms, which we can think of as “moving forward along $X$ by $t$.
Now given a smooth function $f$ we use this as if we were taking a derivative from all the way back in single-variable calculus: measure $f$ at $p\in M$, flow forward by $\Delta t$ and measure $f$ at $\Phi_{\Delta t}(p)$, take the difference, divide by $\Delta t$, and take the limit as $\Delta t$ approaches zero:
$\displaystyle\begin{aligned}\lim\limits_{\Delta t\to0}\frac{f\left(\Phi_{\Delta t}(p)\right)-f(p)}{\Delta t}&=\lim\limits_{\Delta t\to0}\frac{f\left(\Phi(\Delta t,p)\right)-f\left(\Phi(0,p)\right)}{\Delta t}\\&=\frac{\partial}{\partial t}f\left(\Phi(\Delta t,p)\right)\Big\vert_{t=0}\\&=\left[df(p)\right]\left(\Phi_*\left(\frac{\partial}{\partial t}(t,p)\Big\vert_{t=0}\right)\right)\\&=\left[df(p)\right]\left(X(\Phi(0,p))\right)\\&=\left[df(p)\right]\left(X(p)\right)\\&=Xf(p)\end{aligned}$
Note that even if $X$ is not complete we do always have some interval around $t=0$ on which $\Phi_t$ is defined and this difference quotient makes sense.
So far this is just a complicated (but descriptive!) way of restating something we already knew about. But now we can take this same approach and apply it to other vector fields. So if $Y\in\mathfrak{X}M$ is another smooth vector field, we define the “Lie derivative” of $Y$ by $X$ as:
$\displaystyle(L_XY)_p=\lim\limits_{\Delta t\to0}\frac{\Phi_{-\Delta t*}\left(Y_{\Phi_{\Delta t}(p)}\right)-Y_p}{\Delta t}$
Again we evaluate $Y$ at both $p$ and $\Phi_{\Delta t}(p)$, but here’s where a trick comes in: we can’t compare these two vectors directly, since they live at different points on $M$, and thus in different tangent spaces. So in order to compensate we use the flow itself to move backwards from $\Phi_{\Delta t}(p)$ back to $p$, and use the derivative $\Phi_{-\Delta t*}$ to carry along the vector $Y_{\Phi_{\Delta t}(p)}$.
We can come up with an alternate version of this formula by using similar techniques to those above:
$\displaystyle\begin{aligned}(L_XY)_p&=\lim\limits_{\Delta t\to0}\frac{\Phi_{-\Delta t*}\left(Y_{\Phi_{\Delta t}(p)}\right)-Y_p}{\Delta t}\\&=\lim\limits_{\Delta t\to0}\frac{\Phi_{-\Delta t*}\left(Y_{\Phi_{\Delta t}(p)}\right)-\Phi_{0*}\left(Y_{\Phi_0(p)}\right)}{\Delta t}\\&=\frac{\partial}{\partial t}\Phi_{-t*}\left(Y_{\Phi_t(p)}\right)\Big\vert_{t=0}\end{aligned}$
That is, if we define the curve $c(t)=\Phi_{-t*}\left(Y_{\Phi_t(p)}\right)$ in the tangent space $\mathcal{T}_pM$ then $(L_XY)_p=c'(0)$. This would seem to make it live in the tangent space to $c(0)$ — that is, in $\mathcal{T}_{Y_p}\mathcal{T}_pM$ — but remember that since $\mathcal{T}_pM$ is a vector space we identify it with all of its tangent spaces. Thus $(L_XY)_p\in\mathcal{T}_pM$, just like $Y_p$ is.
### Like this:
Posted by John Armstrong | Differential Topology, Topology
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1. [...] all well and good to define the Lie derivative, but it’s not exactly straightforward to calculate it from the definition. For one thing, it [...]
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2. [...] last time, the fact that for all means that is -invariant. That is, . But this implies that the Lie derivative vanishes, and we know that [...]
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3. [...] we mentioned last time, Math Fail showed how to spot what your colleagues did over the summer, The Unapologetic Mathematician gave a nice introduction to the Lie derivative, and on Computational Complexity there was a list of computer scientists that recently found a new [...]
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4. [...] Armstrong: The Lie Derivative, Brackets and [...]
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5. [...] defined the Lie derivative of one vector field by another, . This worked by using the flow of to compare nearby points, and [...]
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://johncarlosbaez.wordpress.com/2012/05/27/symmetry-and-the-fourth-dimension-part-2/
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Azimuth
Symmetry and the Fourth Dimension (Part 2)
The Coxeter group of the cube
Coxeter groups are a huge amount of fun. Normally their delights are reserved for people who have already studied group theory. But I don’t think that’s fair. You don’t need to know music theory to enjoy a piece by Bach. And you don’t need to know group theory to enjoy Coxeter groups.
In fact, it’s probably better to learn theories after falling in love with some examples. Imagine a world where you had to learn music theory before listening to music. A world where everyone studied music theory in elementary school, high school and college, but only people who majored in music were allowed to listen to the stuff. In that world, people might say they hate music… just as in this world they say they hate math.
So, here goes:
Last time I showed you that any Platonic solid has a bunch of symmetries where we reflect it across planes. These planes, called mirrors, all intersect at the center of the solid. If we take a sphere and slice it with these mirrors, it gets chopped up into triangles, and we get a pattern called a Coxeter complex.
If we start with the cube, here’s the Coxeter complex we get:
For artistic reasons, half the triangles are colored blue and half are colored black. But it’s not just pretty: there’s also math here. If we take any black triangle and reflect it across any mirror, we get a blue triangle… and vice versa.
Instead of taking a sphere and slicing it with mirrors, we can start with the cube itself. Here’s what we get:
It’s not quite as pretty (especially because I drew it), but it makes certain games easier to play. These games involve picking one triangle and calling it our favorite. It doesn’t matter which. But we have to pick one… so how about this:
Each different symmetry of the cube sends this triangle to a different triangle. This instantly lets us count the symmetries: there are 48, since each of the cube’s 6 faces has been chopped into 8 triangles.
But even better, we get a vivid picture of the symmetries of the cube! Let’s see how this works.
Any triangle in the Coxeter complex has three corners:
• one corner is a vertex of the cube,
• one corner is the center of an edge of the cube,
• one corner is the center of a face of the cube.
Here’s how it works for our favorite triangle:
Now the real fun starts. We can move from any triangle to a neighboring one in three ways:
1) We can change which vertex of the cube our triangle contains. Starting from our favorite triangle, we get this blue triangle:
Note: the blue triangle touches the same edge of the cube as the black one. It also lies on the same face. Only the vertex has changed!
What have we actually done here? We’ve reflected our triangle across a mirror in a way that changes which vertex of the cube it contains. Let’s call this way of flipping a triangle V.
2) We can change which edge of the cube our triangle touches. Starting from our favorite triangle, we get this yellow triangle:
Note: the yellow triangle contains the same vertex of the cube. It also lies on the same face. Only the edge has changed!
What have we actually done here? We’ve reflected our triangle across a mirror in a way that changes which edge of the cube it touches. Let’s call this way of flipping a triangle E.
3) We can change which face of the cube our triangle lies on. Starting from our favorite triangle, we get this green triangle:
Note: the green triangle contains the same vertex of the cube. It also touches the same edge. Only the face has changed.
What have we actually done here? If you can’t guess, you must be asleep: we’ve reflected our triangle across a mirror in a way that changes which face of the cube it lies on! Let’s call this way of flipping a triangle F.
By repeating these three operations—changing the vertex, edge or face—we can get to any triangle starting from our favorite one. We can even use this trick to label all the triangles. Let’s call our favorite triangle 1:
Let’s call its neighbors F, E and V, since we use those three reflections to get to these new triangles:
Starting with these, we can get more triangles by changing the vertex, edge or face:
See what I’m doing? We get the triangle VE by starting with the triangle V and then changing which edge of the cube it contains. We get EF by starting with E and then changing which face it lies on. And so on.
However, there’s a ‘problem’. See the triangle VF? We got there from the triangle V by changing which face it lies on. But we could also get there another way! We could start at F and then change which vertex this triangle contains. So we could equally well call this triangle FV.
Luckily, in math nothing is really a problem once you understand it. This is why math is more fun than real life: merely understanding a problem makes it go away. We’ll just say that
VF = FV
So, we can use either label for this triangle: it doesn’t matter.
More deeply, if you start with any triangle, change the vertex it contains and then change the face it lies on, you get the same triangle as if you first change the face and then the vertex. That’s what the equation VF = FV really means. It’s a fact of geometry: a general fact about Platonic solids.
Let’s go a bit further:
I’m using the same rules; check to make sure I did everything right! There’s another little ‘problem’, though: see the triangle labelled FEF? We got there from FE by changing which face of the cube our triangle lies on. But we could also get there starting from EF by changing the edge. So really we have
FEF = EFE
But this is not a general fact about Platonic solids: it shows up because in the cube, three faces and three edges meet at each vertex. That’s why the equation has three F’s and three E’s.
We can go on even further, but you can already see where the next problem will show up. See that unlabelled triangle in the front face of the cube? At the next stage we’ll want to label it VEVE, but we’ll also want to label it EVEV. So:
VEVE = EVEV
Again, this is not a general fact about Platonic solids! It shows up because the cube has square faces, so four vertices and four edges touch each face. That’s why the equation has four V’s and four E’s.
We almost have enough equations to avoid all future problems. But there are a few more that are so obvious you may have overlooked them. Suppose we change which vertex of the cube our triangle contains, and then do this again. We get back where we started! For example, first we go from the black one to the blue one:
and then we go back to black.
So, we have
VV = 1
This says switching vertices twice gets us back where we started. Similarly, we have
EE = 1
and
FF = 1
And now, although I haven’t proved it to you, we have a complete set of equations to give each triangle an unambiguous name… or more precisely, an unambiguous element of the ‘Coxeter group’ of the cube. Two different expressions, like EFE and FEF, give the same element of the Coxeter group if we can get from one to the other using our equations. For example, in the Coxeter group we have
FEFVVEFE = FEFEFE = FEFFEF = FEEF = FF = 1
Coxeter groups of Platonic solids
We can do this stuff for other Platonic solids, too. The Coxeter group of the octahedron is secretly the same as that of the cube, since they’re dual. The only difference is that the names F and V get switched, because faces of the cube correspond to vertices of the octahedron, and vice versa! Similarly for the icosahedron and dodecahedron. So, I mainly have two puzzles for you:
Puzzle 1: Find equations defining the Coxeter group of the tetrahedron.
Puzzle 2: Find equations defining the Coxeter group of the dodecahedron.
If these seem hard, let’s reflect a bit more on what we did for the cube. For the cube we have
VF = FV
because of this picture:
Similarly, we have
EFE = FEF
because of this picture:
And finally, we have
VEVE = EVEV
because of this picture:
This should make it easy to solve the puzzles. We can also phrase the solutions in a different way:
Puzzle 3: Show that that Coxeter groups of the tetrahedron, cube and dodecahedron can be completely described by the equations
V2 = E2 = F2 = 1
and
(VE)a = (VF)b = (EF)c = 1
for some integers a, b, and c.
The story doesn’t stop here—far from it! Later we’ll meet Coxeter groups for the higher-dimensional analogues of Platonic solids, which are called regular polytopes. And we’ll use them to classify so-called uniform polytopes obtained by chopping vertices, edges, faces and so on off the regular ones. For example, the cuboctahedron:
can be gotten either by chopping the corners off a cube, or chopping the corners of an octahedron! We can classify such shapes using Coxeter diagrams, which are based on Coxeter groups. So, there’s no shortage of fun stuff to do… in fact, there’s way too much!
Actually, that’s the main problem with mathematics, once you start actually doing it. There’s just too much fun stuff.
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17 Responses to Symmetry and the Fourth Dimension (Part 2)
1. qinglong1397 says:
Reblogged this on 公丘子 and commented:
I like it. Thanks to John Baez
2. Greg Egan says:
I’m a bit confused about the notational conventions here. On the diagrams, it looks as if you’re multiplying reflections on the left: for example, the triangle you label EF is the triangle reached by taking the triangle labelled 1, multiplying by (the reflection that takes 1 to) F, then multiplying by (the reflection that takes 1 to) E. The triangle you label FE is reached from 1 by multiplying first by (the reflection that takes 1 to) E, then multiplying by (the reflection that takes 1 to) F.
But in the text, you seem to describe the labels as if you’re multiplying on the right:
[S]ee the triangle labelled FEF? We got there from FE by changing which face of the cube our triangle lies on.
• Greg Egan says:
After re-reading this I think I understand what you’re getting at in the text: you’re describing these labels without referring to the multiplication of reflections at all, purely in terms of the incidence relations between the triangles and features of the cube.
So the triangle FEF in this scheme is defined as the triangle that intersects the same vertex and edge of the cube as FE, while intersecting a different face. That this also corresponds to the triangle reached by applying group elements F then E then F to the triangle labelled 1 is a separate fact. So I should probably try to prove that both ways of interpreting the labels are equivalent.
• Greg Egan says:
If S is any string of Es, Fs and Vs interpreted as reflections multiplied together and then applied to the triangle 1, the triangles SE, SV and SF will each bear the same relationship to the triangle S as the triangles E, V and F, respectively, bear to the triangle 1, in terms of which features of the cube they have in common. Any symmetry of the cube will preserve incidence relations, so applying S (as a symmetry) to, say, triangle 1 and triangle E will yield triangles S and SE that intersect different edges of the cube but share the same vertex and face.
So the two ways of interpreting the labels are compatible. Sorry to be thick, this is pretty obvious in retrospect.
• John Baez says:
Your first comment got me scared for a minute, because I tend to mix up left and right, and I also have a habit of tinkering with standard conventions—an explosive combination. Since I was trying to make everything easy for nonmathematicians, I wanted a convention where flipping our favorite triangle to one touching a new Edge and then to one lying on a new Face is called EF, and so on.
After re-reading this I think I understand what you’re getting at in the text: you’re describing these labels without referring to the multiplication of reflections at all, purely in terms of the incidence relations between the triangles and features of the cube.
Right.
So the triangle FEF in this scheme is defined as the triangle that intersects the same vertex and edge of the cube as FE, while intersecting a different face.
Right. Do you think there’s any way for me to make this clearer without bringing in phrases that will be mysterious to mathematicians, like ‘multiplying reflections’? I’ll probably try to turn all these posts into an article someday… Even if I’m aiming my discussion at nonmathematicians, I don’t want mathematicians to be confused by it!
That this also corresponds to the triangle reached by applying group elements F then E then F to the triangle labelled 1 is a separate fact. So I should probably try to prove that both ways of interpreting the labels are equivalent.
Indeed, this is a somewhat confusing issue, the sort I have to work out each time—just like remembering whether to do things earlier or later when ‘daylight savings time’ starts in the US… especially when phoning my mom from Singapore!
All these questions are related to the ‘active versus passive coordinate transformation’ issue that physicists love to talk about—and mathematicians who don’t take physics courses seem never to have heard of.
We can think of a chosen triangle on the cube as putting coordinates on the cube. Then we can think of doing V, E or F to the triangle as changing these coordinates while leaving the cube fixed. This is what physicists would call a ‘passive’ coordinate transformation.
There are annoying left-right issues involved in switching from the ‘active’ versus ‘passive’ conventions, since an active transformation on some space
$x \mapsto g(x)$
has the following effect on functions on that space, e.g. coordinate functions:
$f \mapsto f \circ g$
So, you can pile this confusion on top of any caused by my nonstandard conventions when pondering this issue.
• John Baez says:
I think I see one way I can reduce confusion: don’t say “let’s call this reflection E…”, or at least say it some better way.
• Greg Egan says:
I must confess that I didn’t read the text carefully the first time, because the diagrams alone convinced me that you were multiplying on the left with reflections called E, V and F. People coming at this post without any preconceptions wouldn’t assume that.
But since inevitably some people reading your posts will have had some prior exposure to group theory, maybe it would help them if you give a hint near the start that you’re playing a different game, even if it turns out to be isomorphic to the one they’re expecting!
What’s nice about your way of generating the labels is that it’s incredibly easy to see what they should be, whereas applying a given reflection to a triangle that doesn’t even touch the reflection plane takes a lot more effort.
• John Baez says:
Thanks—yes, I think this ‘triangle-flipping’ approach makes the whole subject easier. It’s not really new to me: it’s implicit in the theory of Coxeter complexes, buried under layers of jargon.
In this approach, the triangulated cube is the symmetry group of the cube! In other words: once we’ve chosen our favorite triangle, each triangle is named by the group element that carries our favorite triangle to that one.
This made me think about how the 2-sphere can’t be made into a Lie group. The triangulated cube seems like a ‘good attempt at solving an impossible problem’, since the surface of a cube is topologically a 2-sphere, and we’ve made the set of triangles into a group. The group operation is even ‘continuous’ in some funny sense: triangles touching the identity (our favorite triangle) act as operations that move a triangle to one that touches it! Alas, I don’t know what concept of ‘continuity’ this is an example of.
Actually, the triangulated dodecahedron is an even better attempt, in the sense that we get more triangles.
• Mike Stay says:
The rotation subgroup is also continuous, since a pair of reflections leaves one vertex in common. It’s also a subgroup of SO(3).
• John Baez says:
I don’t really get what you mean, Mike. Of course the rotational symmetries of anything 3-dimensional form a subgroup of SO(3), and yes, a rotational symmetry of a cube leaves an antipodal pair of these triangle corners fixed. But I don’t know what you mean by “the rotation group is also continuous”.
• Mike Stay says:
I just meant that it’s continuous in your “funny sense”, and rotations seem more amenable to becoming a Lie group than reflections.
• John Baez says:
I don’t see how to subdivide the cube’s surface into polygons, one for each of its 24 rotational symmetries, such that the symmetries corresponding to the polygons touching the polygon for the identity act to map any other polygon to one that touches it. You seem to be hinting that you can do it.
I _can_ do it for rotational/reflectional symmetries of the cube.
Some relevant bits of topology:
The rotation group of Euclidean 3-space is SO(3), which is topologically the projective 3-space $\mathbb{R}P^3.$ The rotation/reflection group of Euclidean 3-space is O(3), which is topologically two disjoint copies of the projective 3-space $\mathbb{R}P^3.$
Jim Stasheff proved some very general theorems saying stuff like: there’s no way to make any space that’s homotopy equivalent to $S^2$ into a topological group.
3. John Baez says:
Over on G+, Shanthanu Bhardwaj has solved the puzzles. Click here and read his comment if you want your fun to be spoiled by seeing the solutions.
4. [...] John Baez wants you to fall in love with the Coxeter group of the cube [...]
5. Daniel says:
Dear Sir,
I was really inspired and overwhelmed by the website you and your colleagues run (http://math.ucr.edu/home/baez/README.html). I think that it is really noble of you to just let the general public access the.pdf files of your books for free and to offer your valuable advice at no cost, when you could have charged them for it. I wish you and your colleagues all the best, and once again, thank you so much for taking your valuable time to maintain and update this blog and the above website.
• John Baez says:
Thanks! I didn’t go into mathematics to make money, and in fact I’m a lot more well-off than I ever expected to be. Given all that, it seems ridiculous to charge money for my books… especially since these books never made much money in the first place, and illegal Russian websites offer most scientific books for free anyway.
But anyway, I’m glad you’re enjoying all the material on that website.
6. [...] in Part 2, we looked at operations that flip these triangles around. I showed you how it works for the cube. [...]
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http://physics.stackexchange.com/questions/6377/top-spun-up-with-string-under-tension-problem
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# Top spun up with string under tension problem [duplicate]
Possible Duplicate:
I have a top with an unknown mass. It has a moment of inertia of 4.00 * 10^-7 kgm^2 a string is wrapped around the top and pulls it so that its tension is kept at 5.57 N for a distance of .8 m.
Could somebody help me derive some equations to help with this? Or to get me in the right direction? I have been trying to derive some sort of equations from KEr = 1/2 * I * w^2 but I cant get anywhere without ending up at radius = radius or mass = mass.
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please give a better title – Mark Eichenlaub Mar 5 '11 at 0:21
– Georg Mar 6 '11 at 13:06
## marked as duplicate by mbq♦Mar 6 '11 at 20:08
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
## 1 Answer
Hint: The work done on the top is force times distance. So given the MOI you can find $\omega$.
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http://math.stackexchange.com/questions/166792/need-a-formula-for-a-quadratic-spline
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# Need a formula for a quadratic spline
I'm trying to reproduce some results from a paper and I need an explicit formula for a specific quadratic spline to do so.
The problem is, I've only got a plot of it. The quadratic spline is from figure 2 (a) of Stephane Mallats paper "Singularity Detection and Processing with Wavelets" in IEEE Transactions on Information Theory, Vol. 38, No. 2. March 1992. Link to journal. Link to PDF.
The spline has a positive lobe lobe from $[-1,0]$ and a negative lobe from $[0,1].$ It appears to go to zero from $[-2,-1]$ and from $[1,2].$ It also acheives a peak values of about $0.66$ or $-0.66$ in either lobe as measured in imageJ. This spline is bound to be a wavelet so it's necessarily $0$ on the average and is compactly supported from $[-2,2]$ (possibly a smaller interval if it goes to zero where I think it does.)
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That doesn't look very quadratic to me. Does "quadratic spline" have some technical meaning in your field that is different from the usual "curve parameterized on the general form $t\mapsto(at^2+bt+c, dt^2+et+f)$"? – Henning Makholm Jul 5 '12 at 2:32
This is a spline, and is piecewise quadratic. It should be thought of as a piecewise quadratic polynomial smoothly joined. There is an upward parabola prior to t =0 and a downward parabola following t= 0. – ncRubert Jul 5 '12 at 4:55
– J. M. Jul 5 '12 at 5:34
A function of the form $-ax \exp(-bx^2)$ with $a,b >0$ will yield a very similar curve. – Peter Tamaroff Jul 5 '12 at 19:23
## 2 Answers
It looks something like this. $$\cases{0 & $t < -1$\cr 1.481481482+ 2.962962963\,t+ 1.481481482\,{t}^{2}&$-1 \le t<- 0.55$\cr 1.285416667+ 2.250000000\,t+ 0.833333333\,{t}^{2}&$-0.55 \le t<- 0.35$\cr 0.3666666622- 3.000000024\,t- 6.666666700\,{t}^{2}&$-0.35 \le t<- 0.25$\cr - 5.933333324\,t- 12.53333330\,{t}^{2}&$-0.25 \le t<0$\cr - 5.933333324\,t+ 12.53333329\,{t}^{2}&$0 \le t< 0.25$\cr - 0.3666666564- 3.000000069\,t+ 6.666666780\,{t}^{2}&$0.25 \le t< 0.35$\cr - 1.285416675+ 2.250000040\,t- 0.8333333750\,{t}^{2}&$0.35 \le t< 0.55$\cr - 1.481481475+ 2.962962942\,t- 1.481481467\,{t}^{2}&$0.55 \le t < 1$\cr 0 & $t \ge 1$}$$
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So, the spline comes from the reference in the comment above? Should I be able to work out the formula in the time domain by examining the appendix of that paper and doing some Fourier transforms? – ncRubert Jul 5 '12 at 16:09
In Maple 16: > CurveFitting[Spline]([[-1,0],[-0.55,0.3],[-0.35,0.6],[-0.25,0.7],[0,0],[0.25,-0.7],[0.35,-0.6],[0.55,-0.3],[1,0]],t,degree=2,knots=data); – Robert Israel Jul 5 '12 at 16:19
From appendix A of the Mallat paper linked to in the comment by J.M., the smoothing spline figure is the Fourier transform of the following function:
$\hat{\psi}(\omega) = i \omega (\mathrm{sinc}(\omega/4) )^4$
where
$\mathrm{sinc}(\omega) = \frac{\sin(\omega)}{\omega}$ and $\omega = 2 \pi f$
The FT of a sinc function is simply a rectangle of width 1
$\mathcal{F}^{-1}\{sinc(\omega)\} = Rect(\omega)$
Multiplication becomes convolution:
$\mathcal{F}^{-1}\{sinc(\omega)^4\} = Rect(t) \star Rect(t) \star Rect(t) \star Rect(t)$
Working out these four convolutions gives a curve defined piecewise from $t \epsilon [-2,2]$:
$\mathcal{F}^{-1}\{sinc(\omega)^4\} =$
$t^3/6 + t^2 + 2t + 4/3,\text{ } -2 \le t \le -1$
$-t^3/2 - t^2 + 2/3,\text{ } -1 <t \le 0$
$t^3/2 - t^2 + 2/3,\text{ } 0<t \le 1$
$-t^3/6 + t^2 - 2t + 4/3,\text{ } 1 <t \le 2$
The FT of a scaled sinc is a scaled Rect having width 1/4 and height 4:
$\mathcal{F}^{-1}\{sinc(\omega/4)\} = 4 Rect(4t)$
The extra factor of $i\omega$ is equivalent to a derivative in the time domain according to:
$\frac{d}{dt}(f(t) ) = i \omega \mathcal{F}^{-1}\{\hat{f}(\omega)\}$
Scaling the convolved sincs gives a curve defined piecewise from $t \epsilon [-1/2,1/2]$, which we then take a derivative of to get the final result:
$\mathcal{F}^{-1}\{i \omega sinc(\omega)^4\} =$
$4*((4t)^2/2 + 2*(4t) + 2), \text{ } -1/2 \le t \le -1/4$
$4*(-3/2*(4t)^2 - 2(4t) ) ,\text{ } -1/4 <t \le 0$
$4*(3/2*(4t)^2 - 2*(4t)) ,\text{ } 0<t \le 1/4$
$4*(-(4t)^2/2 + 2*(4t) - 2),\text{ } 1/4 <t \le 1/2$
The plot looks like this:
The problem with this spline is that it is half the width of the plot shown in figure 2 of Mallat's paper. I believe the difference between the two is just a scaling factor though.
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http://physics.stackexchange.com/questions/27230/from-vertex-function-to-anomalous-dimension
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# From vertex function to anomalous dimension
• In a $d$ dimensional space-time, how does one argue that the mass dimension of the $n-$point vertex function is $D = d + n(1-\frac{d}{2})$?
• Why is the following equality assumed or does one prove that a function $f$ will exist such that, $\Gamma^n(tp,m,g,\mu) = f(t)\Gamma^n(p,m(t),g(t),\mu)$?
..in writing the above I guess one is assuming that all the inflowing momentum are equal to $p$ but if that is so then what is the meaning of again specifying the "renormalization scale" of $\mu$?..Does this somehow help fix the values of the functions, $m(t)$ and $g(t)$ at some value of $t$?..If yes, how?..
If the above equality is assumed as a property of the vertex function then it naturally follows that it satisfies the differential equation,
$[-t\frac{\partial}{\partial t} + m[\gamma_m(g)-1]\frac{\partial}{\partial m} + \beta(g)\frac{\partial}{\partial g} + [D-n\gamma(g)]]\Gamma^n(tp,m,g,\mu)=0$
where one defines,
$\beta(g) = t\frac{\partial g(t)}{\partial t}$
$m[\gamma_m(g)-1] = t\frac{\partial m(t)}{\partial t}$
$D-n\gamma(g) = \frac{t}{f}\frac{\partial f}{\partial t}$
and the last equality integrates to give, $f(t) = t^De^{-n\int_1^t \frac{\gamma(g(t))}{t}dt}$ and $-n\int_1^t \frac{\gamma(g(t))}{t}dt$ is defined as the "anomalous dimension".
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http://mathoverflow.net/questions/87382?sort=oldest
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## Detecting equivalences of (infinity) categories by nerves
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I have two questions:
1. Is there a way to tell if a functor $F:C \to D$ between two small categories is an equivalence in terms of the map $$N(F):N(C) \to N(D)$$ between simplicial sets? More generally, can we test separately when a functor $F$ is: essentially surjective, full, faithful etc., by an easy to verify property of $N(F)$? I am not interested in any answer involving applying the left-adjoint to $N;$ I am really looking for a description using simplicial sets.
2. If $\varphi:X \to Y$ are quasi-categories, is there way of saying when $\varphi$ is a weak equivalence in the Joyal model structure (categorical equivalence in the language of Lurie) akin to "full and faithful and essentially surjective?". I am most interested in definitions which are easily checkable, so not definitions like "the induced map on mapping complexes..." since these are hard to compute in practice. The definition I have seen just says that $\varphi$ becomes a weak equivalence of simplicial categories after applying the left-adjoint to the homotopy coherent nerve, which is somehow not a very satisfying definition. Is there a more explicit description, not involving simplicial categories?
I hope this question is not too vague. Any comments to improve the wording etc. are welcome. Thanks!
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Are we allowed to use a factorisation system on sSet? Or how about translating the canonical model structure on Cat (where equivalences are the weak equivalences) to sSet? Fully faithful should be easy - consider maps right orthogonal to the nerve of the inclusion {1,2} --> {1->2} – David Roberts Feb 3 2012 at 0:06
Thanks for the comment about fully faithful. Do you know if this holds for quasi-categories too? As far as translating the canonical model structure, do you mean transferring along the nerve-adjunction? If so, I'd rather avoid applying the left adjoint to $N$... – David Carchedi Feb 3 2012 at 0:09
We wouldn't need the full model structure, just the result that an equivalence of catgories factors as an acyclic cofibration followed by an acyclic fibration. An acyclic fibration should be easy to characterise, it is an isofibration which is fully faithful. An isofibration is, at the level of simplicial sets, a map with the right lifting property with respect to the nerve of {1} --> {1\simeq 2}. An acyclic cofibration is an equivalence which is injective on objects. So we just need to characterise injective-on-0-cells maps which are ess. surj. (cont..) – David Roberts Feb 3 2012 at 6:34
(essentially surjective isofibrations are surjective on objects) – David Roberts Feb 3 2012 at 6:34
## 1 Answer
Just a few ideas/observations...
For 1:
a. A functor is an isomorphism if and only if the induced map is an isomorphism of simplicial sets.
b. So a functor is an equivalence if and only if the induced map on the nerves of skeleta is an isomorphism. (This isn't very helpful though...)
c. Any map of simplicial sets $NC \rightarrow ND$ gives us a functor $C \rightarrow D$ (just look at the map on 0-simplices, 1-simplices, and 2-simplices to see what to do). So a map $NC \times \Delta^1 \rightarrow ND$ corresponds to functor $F: C \times [1] \rightarrow D$ (since $N[1] = \Delta^1$), but this is precisely a natural transformation of functors $C \rightarrow D$. Thus, if we have $NC \rightarrow ND$ and $ND \rightarrow NC$ such that the composites are "homotopic" to the identity (where I mean, use $\Delta^1$), then the original functor is an equivalence.
d. None of what we have said so far is any easier than just proving your original map is an equivalence. One thing you could do, that allows for homotopy theory, is associate to $C$ a stronger invariant. For example, take the bisimplicial set call it $\mathfrak{N}C$ given by: $\mathfrak{N}C_k = N(\text{iso }C^{[k]})$ (where "iso C" means "maximal groupoid" or "just take isomorphisms as your morphisms." This is a special case of a construction of Rezk's). It turns out that a functor is an equivalence if and only if the induced map of these special nerves is a weak equivalence.
e. I don't think that checking it's a weak equivalence works for the usual nerve, since (for example) any category with a final object has a nerve that's weakly equivalent to a point. However it is probably the case that if the induced map on nerves is a categorical equivalence then the original categories are equivalent. This brings us to point 2...
For 2
a. I'm not sure if there is any way of showing that a functor between $\infty$-categories is an equivalence (in general) that does not amount to showing the induced map on mapping spaces is an equivalence. However, you can do all of this without mentioning simplicial categories. Given an $\infty$-category, $\mathcal{C}$, we can define something (or several things) equivalent to the mapping space between two objects in $\mathfrak{C}[\mathcal{C}]$; let Hom_C^R(x,y) be the simplicial set defined by requiring that a map $\Delta^n \rightarrow \text{Hom}_{\mathcal{C}}^R(x,y)$ be a map $z: \Delta^{n+1} \rightarrow \mathcal{C}$ such that $d_0z$ is the constant diagram on $y$ and $z$ evaluated on the initial vertex is $x$. (Actually this might be $\text{Hom}^L$, I can't remember off the top of my head). Anyway, this turns out to be a Kan complex (i.e. a space), and Lurie/Joyal show that it's the same as what you get after doing $\mathfrak{C}$. See Lurie for more.
b. Given the description above, we can say a functor is an equivalence if it induces a weak equivalence of Kan complexes for all mapping spaces as we defined above (i.e. fully faithful), and essentially surjective (i.e. every object is equivalent to something in the image). This isn't easy to do but...
c. In general, an equivalence of $\infty$-categories is kind of a big deal. In particular, it should be at least as hard as writing down various Quillen equivalences of model categories, right? Maybe slightly easier, but still quite hard.
Anyway, I realize this probably isn't what you want, but it's the best I can do at the moment! I'll let you know if I think of anything more helpful (but probably someone brighter will end up posting something more helpful soon...)
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I should point at that in 1d, only $\mathfrak{N}C_0$ and $\mathfrak{N}C_1$ really matter, so that might cut down on computation. – Dylan Wilson Feb 3 2012 at 5:25
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(A proof of my suspicion in (e) can be found, along with many relevant fun facts, in this paper of by Emily Riehl: math.uchicago.edu/~eriehl/topic.pdf) – Dylan Wilson Feb 3 2012 at 6:26
Thanks Dylan. This does go in the right direction. And besides, my question was vague :p. – David Carchedi Feb 3 2012 at 17:49
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http://bayesianthink.blogspot.com/2012/08/understanding-bayesian-inference.html
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# Probability Puzzles
## Tuesday, August 21, 2012
### Understanding Bayesian Inference
Understanding Bayesian approaches to estimating probabilities are important. Often people don't get the full import of it and/or fail to see the consequences of not estimating it the right way. Most books that discuss it have confusing terminology to explain something that is fairly simple. Estimating Bayesian probabilities for events are relatively easier to understand, while those involving hypotheses aren't so even though the math involved is the exact same! In this blog, I'll try to explain the concept, but more importantly show the formula that any student can use so that they don't have "think through" the Bayesian logic all the time. However for some problems, it might be better to enumerate out the cases. To develop an intuition for this, you need to keep an eye out for new information that could be coming in and altering our belief in an existing hypothesis.
Here is the structure of the problem you will almost always run into. For simplicity let us assume you have a two hypotheses H0 and H1. There are probabilities associated with them P(H0) and P(H1). There comes along a piece of evidence E and you want to update your probability measures. The formula you could use is as simple as
..and thats it! This works, always. You just need to be able to map your problem to this framework.
To demonstrate this here is an example worked out.
Q. There are two boxes. Box A has 2 white coins and 1 black coin. Box B has 1 white coin and 2 black coins. If a user picks a box at random, what is the probability that box A was chosen? If the user reveals the coin to be a white one, what is the probability that box A was chosen?
A. The first half is simple. No extra information is revealed. The probability is simply 50%. The second half gets a little more interesting. As there are only two boxes to be chosen, knowing the probability of one implies you know the other.
So let's cast it in the framework indicated above. The hypothesis H0 is that box A was chosen. P(H0) is the "prior", that is the probability that box A was chosen prior to any new knowledge or evidence. This we know is 50% which is the same as P(H1).
The next piece is to understand P(E | H0). This is the probability that you would see the "evidence" given that the hypothesis H0 is true. In this case its relatively easy to estimate this as 2/3. This also means that P(E | H1) is 1/3. Plugging all these into the equation above gives,
The intuition behind this is also easy to follow. As the person revealed a white ball it is more likely it came from box A than B. To further reinforce that intuition, think what would you have concluded if box A had a 100 white coins.
The above is a relatively simple exercise in Bayesian inference. While most real world problems can be mapped to this frame work the difficulty comes in
• Realizing there is a Bayesian "trap" hidden somewhere
• If there is, casting it to the above framework
Here is another example of a scenario where Bayesian thinking comes into play often, that of tests for diseases.
Q. Assume there is a disease D, that has a test T. Overall 2% of the population get the disease. If a person actually has the disease, the test is right 90% of the time. If the person does not have the disease, the test could still show as positive 20% of the time. If the test shows positive for a person, what is the probability that the person has the disease.
A. Here, the hypotheses are "No Disease" and "Disease", named H0 and H1 respectively. With no prior evidence we know that P(H0) is 98% and P(H1) = 100% - 98% = 2%. Now, there is new evidence (E) that the test is showing up as a positive. So let us see how each of the parts would fit in.
We want to estimate P(H1|E), we know P(H1) & P(H0). Additionally we know P(E|H1) = 90% and thus P(E | H0) = 10%. Simply plug them all in again.
Notice, that even though the test is right 90% of the time the person actually has just a 15.51% chance of having it given the test proved positive. The intuition here is that it is a rare disease and it would take a very accurate test to confirm it.
All is fine in such scenarios where the numbers are nicely given to us. The Bayesian angle is becomes elusive when it is not put forth cleanly. The next example demonstrates that.
Q. A man has two children. One of them is a boy, what is the probability that the other is a girl?
A. You might be tempted to say 50%. You be wrong! Here is why. Your evidence (E) here is "one of the children is a boy". The hypothesis you want the probability for is H0 = "Other child is a girl". This makes H1 = "Other child is a boy". The values for P(H0) = P(H1) = 1/2. We want to estimate P(H0 | E). To estimate this, notice P(E | H1) = 1/4, as there is exactly one way this is possible. Now we are all set, simply plug it into the formula (again!)
$$P(H_{0}|E) = \frac{\frac{1}{2}\times \frac{1}{2}}{\frac{1}{2}\times \frac{1}{2} + \frac{1}{4}\times \frac{1}{2}} = \frac{2}{3}$$
This is one example where it is likely easier to visualize the problem. In the diagram below, the left hand side shows the situation without any information, and the right hand side shows the information provided and how it ends up encapsulating the relevant cases. It is easier to see why the probability is 66% from this figure.
No discussion on Bayesian inference is complete without a mention of the Monty Hall problem. The problem statement is quite simple, you are shown 3 doors behind one of which there is a treasure. You are allowed to pick one door, but not open it. Once you pick a door, a door which does not contain the treasure is opened. You are allowed to stick to your choice or switch. What should you do? The simplest explanation is as follows: If you switch the probability that you will win is 1/2, else it is 1/3, so you must always switch.
Clearly not all fit the "formula" framework. Some of these problems can solved more easily by using the conventional counting method. Perhaps the most startling of Bayesian puzzles to hit the web is the Tuesday Birthday problem. It is a very subtle variant of the boy/girl problem mentioned above, but with a startling result. The problem is "A man has two children. One of them is a boy born on a Tuesday. What is the probability that the other child is a boy?". The link I mention above (and other sources on the web) describe the solution and I'll try to describe it in my own words here.
If the first child is a boy born on a Tuesday, then the second child can be either Boy/Girl and could be born on any of the 7 days. This yields 14 cases (7 x 2). If the second child is a boy born on a Tuesday, then, just as the previous argument, the first child can be a Boy/Girl born on any of the 7 days yielding 7 x 2 = 14 cases. However, both sets have a case of a Boy-Boy. The total 14 + 14 = 28 double counts this case. So in reality we have 28 - 1 = 27 cases. Next of these 27 cases, we need to know how many have two boys in them. We can apply the same logic. If the first child is a boy born on a Tuesday, the second boy child can be born on any of the 7 days giving 7 cases. Same logic applies if the second child is a boy born on a Tuesday, but like before we need to subtract one because the case of boy-boy is counted twice. This gives a total of 13 cases where there are two boys. So the required probability is 13/27.
If you are creative you can extend this to make your own tricky problems. What happens if you change day of week to month of year? If you follow the train of thought above, you will arrive at 23/47, which is slightly greater than 13/27.
If you are looking to buy some books in probability here are some of the best books to learn the art of Probability here are some great books to own
This book is a great compilation that covers quite a bit of puzzles. What I like about these puzzles are that they are all tractable and don't require too much advanced mathematics to solve.
This is a book on algorithms, some of them are probabilistic. But the book is a must have for students, job candidates even full time engineers & data scientists
Good read. Overall Poker/Blackjack type card games are a good way to get introduced to probability theory
Easily the most expensive book out there. So if the item above piques your interest and you want to go pro, go for it.
Well written and easy to read mathematics. For the Poker beginner.
An excellent resource (students/engineers/entrepreneurs) if you are looking for some code that you can take and implement directly on the job.
This is a good book if you are new to statistics & probability while simultaneously getting started with a programming language. The book supports R and is written in a casual humorous way making it an easy read. Great for beginners. Some of the data on the companion website could be missing.
Posted by RumpelStiltSkin at 7:30 PM
#### 3 comments:
1. On second child problem - (1/4) / (1/4 + 1/8) = 2/3, not 1/3.
Same is obvious from the picture - 'second child is girl' happens on two boxes out of three.
2. Typo, corrected. Thanks for pointing out
3. Another typo... In the disease/test question, P(E | H0) should be 20% instead of 10% (of course, you can change the false positive precentage from 20% to 10% in the question definition).
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http://mathhelpforum.com/differential-equations/58771-power-series-differential-equations.html
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Thread:
1. Power Series/Differential Equations
Find the first four nonzero terms in each of two linearly independent power series solutions about the origin.
What do you expect the radius of convergence to be for each solution?
$(cosx)y^{''} +xy^{'} -2y=0$
Set the coefficients near zero and first terms to 1
I am not exactly sure what the bolded sentence means. Also, how should I go about starting the problem? Is it best to start writing out the series for each term? like
Cosx = $1-\frac{x^2}{2!} + \frac{x^4}{4!} + ...$
and so on.
Thanks for any help!
2. Originally Posted by Oblivionwarrior
Find the first four nonzero terms in each of two linearly independent power series solutions about the origin.
What do you expect the radius of convergence to be for each solution?
$(cosx)y^{''} +xy^{'} -2y=0$
Set the coefficients near zero and first terms to 1
I am not exactly sure what the bolded sentence means. Also, how should I go about starting the problem? Is it best to start writing out the series for each term? like
Cosx = $1-\frac{x^2}{2!} + \frac{x^4}{4!} + ...$
and so on.
Thanks for any help!
Start by assuming that the solution to this problem is $y=\sum_{n=1}^{\infty}a_nx^n$. Then substitute, differentiate, adjust indicies so they are all the same, and then solve the ensuing recurrence relation.
3. Ok, ill try that out. What about that bolded sentence? Is that just talking about where I begin once I got the recurrence formula?
4. Originally Posted by Oblivionwarrior
Ok, ill try that out. What about that bolded sentence? Is that just talking about where I begin once I got the recurrence formula?
I'm not entirely sure.
5. I think the equation $\cos(x)y''+xy'-2y=0$ would be quite a challenge to solve by substituting $\sum_{n=0}^{\infty}a_nx^n$ because of the $\cos(x)$ term: that's going to involve a Cauchy product of sums for the first term which can be done but is messy. I think in this particular case, just need to differentiate the differential equation to find the first few coefficients: Let $y(0)=a$ and $y'(0)=b$ and write it as:
$y''=-\frac{x}{\cos(x)}y'+\frac{2}{\cos(x)}y$
Now we seek a solution about the origin in which everything is well-behaved until we reach the first singular point at $x=\pi/2$. Therefore we can expect a solution of the form:
$y(x)=\sum_{n=0}^{\infty}y^{(n)}(0)\frac{x^n}{n!};\ quad R=\pi/2$
We already know what the first two terms are. Now differentiate to get the third term:
$\frac{d}{dx}y''\Bigg|_{x=0}=\frac{d}{dx}\left(-\frac{x}{\cos(x)}y'+\frac{2}{\cos(x)}y\right)_{x=0 }$
and substitute the values for $y(0),y'(0),y''(0)$.
Do this again to get the fourth term. However that's only one series but I think the two linearly independent solutions are made up of the separate sums of the even and odd powers of x which would mean you'd need to calculate about 8 derivatives. Might get messy and maybe there is another way.
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http://math.stackexchange.com/questions/280070/simplex-volume-in-terms-of-the-radius-of-the-inscribed-sphere
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# Simplex volume in terms of the radius of the inscribed sphere?
It is well known that for the area of a triangle $A$ we have $$A=r\cdot s,$$ where $s$ is the semiperimeter, and $r$ is the radius of the inscribed circle.
Is there an analogue for the higher-dimensional case. In other words, can I express the volume of a $d$-simplex in terms of the radius of its inscribed sphere and the volume of its boundary? If such a formula exists, what are the references?
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## 1 Answer
The same holds, that in $\mathbb{R}^n$, $A = r \frac {V_b}{n}$, where $A$ is the volume of your simplex, and $V_b$ is the volume of your boundary.
To see why this works, simply show that a simplex with base of volume $B$ and height $r$ has volume $r \frac {B}{n}$, and then add up over all faces.
Hint: The constant $\frac {1}{n}$ comes from $\int x^{n-1}\, dx = \frac {x^n}{n}$.
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I see, one just decomposes the simplex into smaller simplices, spanned by the insphere center and the faces. However, the $d$-simplex volume should be $r\frac{B}{d!}$. – A.Schulz Jan 16 at 16:47
@A.Schulz I disagree with the $d!$. As a reminder, the tetrahedron has volume $\frac {1}{3} A \cdot h$. You might be confusing this with the volume of the standard $n-$simplex, which is$\frac {1}{n!}$. This comes about because it has a base in $n-1$ dimensions with volume $\frac {1}{(n-1)!}$, and a height of 1, so according to the formula it has volume $\frac {1}{n!}$. – Calvin Lin Jan 16 at 16:59
You are right. I was confused by the standard formula for the simplex volume. – A.Schulz Jan 16 at 17:38
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http://mathoverflow.net/revisions/54929/list
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## Return to Question
2 added 1 characters in body; edited tags
Hi everyone. My recent work has me developing software to compute in $H^\ast(G/P)$, where $G$ is a complex connected semisimple algebraic group and $P$ is a standard parabolic subgroup (usually, $B$ or a maximal $P$). While my programs are built on sound theory, one can never be to too sure. It's always good practice to check your work.
I'm looking for references of multiplication tables of these cohomology rings.
In particular, I'm interested in cases where $G$ is NOT simply-laced (that is, Lie type $B_n$, $C_n$, $F_4$, $G_2$), though simply-laced tables would be nice too. Any tables would depend on a choice of additive basis for $H^\ast(G/P)$. I typically use cohomology classes either Hom-dual or Poincare dual to the usual Schubert varieties living in $G/P$, and I like to parameterize my Schubert varieties with $W^P$, the minimal length coset representatives of $W/W_P$ where $W$ is the Weyl group and $W_P$ is the Weyl group of the Levi associated to $P$. Tables using this convention would be great. Of course, tables in any basis would be fine. :D
Thanks so much.
1
# Multiplication tables for H*(G/P)?
Hi everyone. My recent work has me developing software to compute in $H^\ast(G/P)$, where $G$ is a complex connected semisimple algebraic group and $P$ is a standard parabolic subgroup (usually, $B$ or a maximal $P$). While my programs are built on sound theory, one can never be to sure. It's always good practice to check your work.
I'm looking for references of multiplication tables of these cohomology rings.
In particular, I'm interested in cases where $G$ is NOT simply-laced (that is, Lie type $B_n$, $C_n$, $F_4$, $G_2$), though simply-laced tables would be nice too. Any tables would depend on a choice of additive basis for $H^\ast(G/P)$. I typically use cohomology classes either Hom-dual or Poincare dual to the usual Schubert varieties living in $G/P$, and I like to parameterize my Schubert varieties with $W^P$, the minimal length coset representatives of $W/W_P$ where $W$ is the Weyl group and $W_P$ is the Weyl group of the Levi associated to $P$. Tables using this convention would be great. Of course, tables in any basis would be fine. :D
Thanks so much.
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http://mathoverflow.net/questions/91789/non-linear-lie-group
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## Non-linear Lie group [closed]
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Possible Duplicate:
Complex Lie group without faithful real representations?
We know that for a matrix (linear) Lie group $G$, we define it to be a closed subgroup of $GL(n,\mathbb{C})$. But Lie groups are defined as manifolds in $\mathbb{R}^n$ for some $n$, in general. The question is that, do we know any Lie group which is not a matrix Lie group? Thank you very much.
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I googled it, and find at the introduction of the paper (Denis Luminet, Alain Valette, Faithful Uniformly Continuous Representations of Lie Groups,J. London Math. Soc. (1994) 49 (1): 100-108.), said the following: Although any connected real lie group G is locally isomorphic to some linear group, No nontrivial covering group of $SL_2(R)$ is linear. – Xiaolei Wu Mar 21 2012 at 4:20
4
For reference this was asked (and answered in the same traditional form!) at math.stackexchange.com/questions/122612/… – Mariano Suárez-Alvarez Mar 21 2012 at 4:41
This is a special case of a question that was already asked. – S. Carnahan♦ Mar 21 2012 at 11:42
## 2 Answers
The traditional example is the universal cover of $SL(2,\mathbb{R})$. You can look e.g. at the wikipedia article on $SL(2,\mathbb{R})$.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Copied from http://planetmath.org/encyclopedia/ExamplesOfNonMatrixLieGroup.html
While most well-known Lie groups are matrix groups, there do in fact exist Lie groups which are not matrix groups. That is, they have no faithful finite dimensional representations.
For example, let $H$ be the real Heisenberg group
$$H=\{\begin{pmatrix} 1 & a & b\newline 0&1&c\newline 0 &0 &1\end{pmatrix}\mid a,b,c\in\mathbb{R} \},$$
and $\Gamma$ the discrete subgroup
$$\Gamma=\{\begin{pmatrix} 1 & 0 & n\newline0&1&0\newline 0 &0 &1\end{pmatrix}\mid n\in\mathbb{Z}\}.$$
The subgroup $\Gamma$ is central, and thus normal. The Lie group $H/\Gamma$ has no faithful finite dimensional representations over $\mathbb{R}$ or $\mathbb{C}$.
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http://mathoverflow.net/questions/93055/hypercohomology-of-a-complex-of-sheaves-that-might-be-acyclic-or-might-not/93087
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## Hypercohomology of a complex of sheaves that might be acyclic (or might not)
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Back again, check this out, let $X$ be a topological space and let $F^{\bullet}$ be a cochain complex of sheaves, I'm trying to compute the cohomology of the complex of global sections of the sheaves
$F^0(X) \rightarrow F^1(X) \rightarrow F^2(X) \rightarrow \cdots$.
I'm trying to prove specifically that the cohomology of this complex is 0. Now there's a possibility that the sheaves $F^q$ MIGHT be ACYCLIC but I have yet to prove that, when I brought up this possibility some of you guys pointed me to Hypercohomology (which I thank you for). I was reading up on Hypercohomology and I'm kind of lost and don't know how to even get started on how to compute the Hypercohomology groups. Here's basically what I want to know:
1 - Let's say the sheaves $F^q$ are in fact acyclic, what I got from what I was reading in this case was
$\mathbb{H}^n(X,F^{\bullet}) \cong H^n(H^0(X,F^{\bullet}))$,
but $H^0(X,F^q) \cong \Gamma(X,F^q)$, so
$\mathbb{H}^n(X,F^{\bullet}) \cong H^n(\Gamma(X,F^\bullet))$,
which is what I'm looking for, so all I have to do is obtain $\mathbb{H}^n(X,F^{\bullet})$? And this is the cohomology of the complex $CF^\bullet(X) = tot(C^\bullet(F^\bullet)(X))$, where $CF^n(X)= \oplus_{p+q=n} C^p(F^q)(X)$ right?
And $F^\bullet(X) \rightarrow C^1(F^\bullet)(X) \rightarrow C^2(F^\bullet)(X) \rightarrow C^3(F^\bullet)(X) \rightarrow \cdots$ is an acyclic resolution. This is all part of the definition of the Cartan-Eilenberg resolution right? So how can I go from
$F^\bullet(X) \rightarrow C^1(F^\bullet)(X) \rightarrow C^2(F^\bullet)(X) \rightarrow C^3(F^\bullet)(X) \rightarrow \cdots$ is an acyclic resolution to
$CF^\bullet(X)$ is an acyclic resolution?
2 - How do I even go about this if I don't know if the sheaves $F^q$ are acyclic?
Any ideas?
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## 2 Answers
I'm not sure I completely understand your question, but:
There is a double complex with $\Gamma(C^pF^q)$ in the $(p,q)$ place.
By taking cohomology first vertically and then horizontally, or vice versa, you get two spectral sequences.
One spectral sequence has $E_1^{p,q}=H^q(X,F^p)$. The other has $E_2^{p,q}=H^p(X,H^q(F^\bullet))$. Both converge to the hypercohomology of $F^\bullet$.
Without more information about $F$, that's pretty much all you can say. Of course if you do have some information about the terms in one of these spectral sequences, you might be able to use it, together with the fact that both sequences have the same abutment, to deduce something about some of the terms in the other.
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Thank you, yeah, I feel like I don't have much to go on to solve this, does the topology of $X$ affect at all the computation of these groups, $X$ is a topological vector spaces, or the fact that the sheaves $F^q$ are sheaves of vector spaces? – Samuel Mf Apr 5 2012 at 2:57
If "sheaves of vector spaces" means "sheaves of sections of vector bundles", and (certainly at least if everything is finite dimensional), you are done, since over a TVS every vector bundle is trivial so all of your $F^p$ are surely acyclic. – Steven Landsburg Apr 5 2012 at 3:52
You mean if the sections of the sheaves $F^q$ on a set $U$ are continuous sections $g:U \rightarrow E$ where $E$ is a vector bundle? the TVS might be infinite-dimensional, thank you – Samuel Mf Apr 5 2012 at 17:51
All vector bundles on any contractible space are trivial. – Steven Landsburg Apr 5 2012 at 17:55
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I am not entirely sure what you are asking, but I think it is essentially about how one can compute $\mathbb{H}^n(X,F^{\bullet})$.
As you point out, if the $F^i$ are acyclic, then indeed
$\mathbb{H}^n(X,F^{\bullet}) \cong H^n(\Gamma(X,F^\bullet))$,
This is of course not true if they are not acyclic. Now this is the point where you have to do what you have to do when you want to compute any derived functor: You have to replace your object by an injective (or acyclic) resolution.
So perhaps your question is: What's a resolution of a complex? ? (nested question marks!) But it seems that you know about the total complex of "mashing" together the resolutions of the individual sheaves.
So perhaps you just need a simple hint: For each $F^q$ take an arbitrary acyclic resolution $A^{q,\bullet}$ and then put them together into one (big) complex, all of whose entries are acyclic. Then use this to compute $\mathbb H$ as you did when you had an acyclic complex.
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Thank you, as I was saying, I wonder if the fact that $X$ is a topological vector space or the type of sheaf i.e. the $F^q$'s are sheaves of vector spaces help at all in the computation, I'm taking a plunge into this – Samuel Mf Apr 5 2012 at 3:02
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http://mathoverflow.net/revisions/70594/list
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## Return to Answer
2 edited body
Hi,
Your Question:When are the dg-Lie algebra structures on Hochschild cochains: HCH∗(C∗(Ω(M),Q),C∗(Ω(M),Q))[1]≅HCH∗(C∗(M,Q),C∗(M,Q))[1] quasi-isomorphic ?
this is always true.
Step 1: From my paper with Felix and Thomas, looking at the proof, you can see that dg-Lie algebra structures on Hochschild cochains: $HCH∗(\Omega C_*(M),\Omega C_*(M))[1]≅HCH∗(C∗(M,Q),C∗(M,Q))[1]$ are quasi-isomorphic Here $\Omega C_*(M)$ is the Adams Cobar construction on the coalgebra C_*(M).
Step 2: There is an quasi-isomorphism of chains algebras called Adams cobar equivalence $\Theta:\Omega C_*(M)\rightarrow C∗(Ω(M)$. In our paper, we prove (very short proof) that this quasi-isomorphism $\Theta$ induces an isomorphism of Gerstenhaber algebras between $HH∗(C∗(Ω(M),Q),C∗(Ω(M),Q))$ and $HH∗(\Omega C_*(M),\Omega C_*(M))$. In particular, we have an isomorphism of graded Lie algebras. You want a dg-Lie algebra isomorphism on Hochschild cochains: HCH∗(C∗(Ω(M),Q),C∗(Ω(M),Q))[1] and HCH∗(\Omega C_(M),\Omega C_(M)). This is true. One of my coauthor had a proof. But it is not in our paper, since I thought he it was not interesting and too complicated. But if I remember well, Hamilton and Lazarev proved it in a paper following our paper. I think that Keller proved also in the paper you quote "Derived Invariance of Higher Structures of the Hochschild complex".
ps: There is two versions of my paper with Felix and Thomas, the published squezeed version valid only over a field, and the arxiv longer version with more details.
1
Hi,
Your Question:When are the dg-Lie algebra structures on Hochschild cochains: HCH∗(C∗(Ω(M),Q),C∗(Ω(M),Q))[1]≅HCH∗(C∗(M,Q),C∗(M,Q))[1] quasi-isomorphic ?
this is always true.
Step 1: From my paper with Felix and Thomas, looking at the proof, you can see that dg-Lie algebra structures on Hochschild cochains: $HCH∗(\Omega C_*(M),\Omega C_*(M))[1]≅HCH∗(C∗(M,Q),C∗(M,Q))[1]$ are quasi-isomorphic Here $\Omega C_*(M)$ is the Adams Cobar construction on the coalgebra C_*(M).
Step 2: There is an quasi-isomorphism of chains algebras called Adams cobar equivalence $\Theta:\Omega C_*(M)\rightarrow C∗(Ω(M)$. In our paper, we prove (very short proof) that this quasi-isomorphism $\Theta$ induces an isomorphism of Gerstenhaber algebras between $HH∗(C∗(Ω(M),Q),C∗(Ω(M),Q))$ and $HH∗(\Omega C_*(M),\Omega C_*(M))$. In particular, we have an isomorphism of graded Lie algebras. You want a dg-Lie algebra isomorphism on Hochschild cochains: HCH∗(C∗(Ω(M),Q),C∗(Ω(M),Q))[1] and HCH∗(\Omega C_(M),\Omega C_(M)). This is true. One of my coauthor had a proof. But it is not in our paper, since I thought he was not interesting and too complicated. But if I remember well, Hamilton and Lazarev proved it in a paper following our paper. I think that Keller proved also in the paper you quote "Derived Invariance of Higher Structures of the Hochschild complex".
ps: There is two versions of my paper with Felix and Thomas, the published squezeed version valid only over a field, and the arxiv longer version with more details.
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http://mathhelpforum.com/pre-calculus/76247-absolute-values-question.html
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# Thread:
1. ## Absolute Values Question?
ok so i have never been given a quesiton like this and im trying to figure it out for the first time. here it is
Solve |2x-3/4x+1| = > 2
2. Originally Posted by jamman790
ok so i have never been given a quesiton like this and im trying to figure it out for the first time. here it is
Solve |2x-3/4x+1| = > 2
one way to go, is to square both sides. you get
$\frac {(2x - 3)^2}{(4x + 1)^2} \ge 2$
now solve for $x$. be sure to check if there are any extraneous solutions.
another approach is to account for the inside of the absolute values to be positive and negative. that is, solve the following two inequalities separately:
$\frac {2x - 3}{4x + 1} \ge 2$ and $- \frac {2x - 3}{4x + 1} \ge 2$
3. It says...show all solutions?...but what does that mean?
4. Originally Posted by Jhevon
one way to go, is to square both sides. you get
$\frac {(2x - 3)^2}{(4x + 1)^2} \ge 2$
now solve for $x$. be sure to check if there are any extraneous solutions.
another approach is to account for the inside of the absolute values to be positive and negative. that is, solve the following two inequalities separately:
$\frac {2x - 3}{4x + 1} \ge 2$ and $- \frac {2x - 3}{4x + 1} \ge 2$
I believe you'll find if you square both sides then the inequation becomes
$\frac{(2x - 3)^2}{(4x + 1)^2}\geq 4$.
5. Originally Posted by Prove It
I believe you'll find if you square both sides then the inequation becomes
$\frac{(2x - 3)^2}{(4x + 1)^2}\geq 4$.
but of course. silly mistake on my part
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http://physics.stackexchange.com/questions/tagged/reference-request
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# Tagged Questions
Use this tag for questions seeking a single specific paper or a short, non-open-ended list of references, like "What paper first discovered X?", "Where can I find the original derivation of X?", or "What is the canonical source for X?" etc.
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### Research Paper for relativistic mechanics [closed]
O. Klein, Z. Phys. , 37, 895906, 1926. W. Gordon, Z. Phys. , 40, 117133, 19261927. A. P. French, Special Relativity (The M.I.T. Introductory Physics Series), 1968. I need to download these paper. ...
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### Nicholas Kollerstrom article on the history of Calculus
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### Reference request for the Aharonov-Bohm effect
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### Dynamical supersymmetry breaking and Witten index
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http://mathhelpforum.com/calculus/187022-confusion-about-total-derivative.html
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Thread:
1. Confusion about Total Derivative
Hello everyone, I was a little confused about something and hope that someone can help me.
Let $U \subseteq \mathbb{R}$ be open, and if a function $f : U \rightarrow \mathbb{R}$ is differentiable on $U,$ then for each $u\in U$ there must exist a linear map $Df(u):\mathbb{R} \rightarrow \mathbb{R}$ such that
$\displaystyle\lim_{h\rightarrow 0} \displaystyle\frac{| f(u+h)-f(u) - Df(u)h|}{|h|} = 0$
Then this is true if and only if $Df(u) = f'(u) = \lim_{h \rightarrow 0} \displaystyle\frac{f(u+h)-f(u)}{h}$. So doesn't this mean that if we can differentiate f, it's then differentiable as a linear map? I know that sounds stupid, but I'm trying to think of the result that if the partial derivatives of f are continuous, then f is differentiable in this special case. So is continuity of the derivative not important in the real case?
What about parametric equations, for example $f: \mathbb{R} \rightarrow \mathbb{R}^2$ where $f(x) = (y_1(x), y_2(x))$. On wikipedia it just says that the derivative of f is $f'(x) = (y_1'(x), y_2'(x)).$ But do the derivatives of $y_1, y_2$ only have to exist for f to be differentiable? Does it matter if they're not continuous?
Thanks for any help
2. Re: Confusion about Total Derivative
Sorry, I've discovered that this is a known result:
If $\gamma:\mathbb{R} \rightarrow \mathbb{R}^m$ is differentiable if and only if its component functions are differentiable in the sense of single-variable calculus.
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http://mathpages.blogspot.com/2008/02/how-to-square-circle.html
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# Math Pages Blog
God used beautiful mathematics in creating the world.
## Friday, February 15, 2008
### How to square a circle
A few days ago, I wrote a post in which I introduced a simple and elegant method to "square" a circle using only compass and straightedge. The solution in that post gave a value for $%5Cpi$ that was accurate to 7 decimal places.
One of the readers left a comment asking for a way to get a better approximation for $%5Cpi$, using only compass and straightedge.
While it is impossible to square a circle, because $%5Cpi$ is transcendental, there exists a simple algorithm to get a construction for any desired approximation. In this post I will not describe a specific construction, but a general algorithm for finding a construction, given an error you want to get.
Lets look on the following circle:
This is a unit circle, with the point $sqrt%7B%5Cpi%7D$ marked in red. Our goal is to get a construction that will give us a point B on the radius so that the distance between B and the red dot is as small as we want.
Point C is totally random - it can be any point on the circle or even outside of the circle. I put it only for convenience.
Step one : Connect C with O, the center of the circle.
Step two: The red dot is to the right of O, so mark a dot $I_%7B1%7D$ at the center of OD. Connect this new dot with C.
Step three: The red dot is still to the right of $I_%7B1%7D$, so mark a dot $I_%7B2%7D$ at the center of $I_%7B1%7D$D. Connect it with C.
Step four: Now the red dot is between $I_%7B1%7D$ and $I_%7B2%7D$. Again mark the dot $I_%7B3%7D$ at the center of this segment. Connect it with C.
At this point the length of the line A$I_%7B3%7D$ is close to $%5Csqrt%7B%5Cpi%7D$. If this is not good enough continue in the some way. With each step you will be getting closer and closer to the red dot.
You should get something like this:
It is very easy to see on this picture that each step brings you closer, and it happens relatively fast.
After you will find the sequence of steps that bring you close enough to the red dot, simply repeat it without marking the dot first.
Lastly, an exercise to the reader: prove that is impossible to square the circle this way.
Posted by Anatoly
Labels: history, math
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http://mathoverflow.net/questions/80871/how-many-idempotent-elements-are-in-z-m-closed
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## How many idempotent elements are in Z_m [closed]
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
How many idempotent elements are in Z_m or How many roots have this polynomial in Z_m
f(x)=x^2 +x procedure of proof is important for me.
-
That looks too much like homework: -1. – Julien Puydt Nov 14 2011 at 6:43
maybe you meant x^2-x? Also, by Z_m do you mean Z/mZ or p-adics?(or something else entirely? – B. Bischof Nov 14 2011 at 6:43
## 1 Answer
Idempotent elements are roots of $g(x)=x^2-x$; my answer will apply equally well to $f(x)=x^2+x$.
The important step is the Chinese Remainder Theorem: one way of stating it is that if $m=p_1^{r_1}\times\cdots\times p_k^{r_k}$ is factored into powers of distinct primes, then the ring $Z_m$ is equal to the direct product of rings $Z_{p_1^{r_1}} \times \cdots \times Z_{p_k^{r_k}}$. So it suffices to count the number of roots in each ring $Z_{p_i^{r_i}}$ and then multiply those numbers together to obtain the number of roots in $Z_m$.
I think it will be easy to convince yourself that the polynomial $g(x)=x^2-x$ has exactly two roots in any ring of the form $Z_{p^r}$.
(By the way, the isomorphism between $Z_m$ and $Z_{p_1^{r_1}} \times \cdots \times Z_{p_k^{r_k}}$ is completely explicit, so this even gives a way to construct the idempotent elements, not just count them.)
-
why polynomial g(x)=x 2 −x has exactly has two roots in any ring of the form Z p r ? – david Nov 14 2011 at 7:05
@george martin: please don't answer homework here. – Martin Brandenburg Nov 14 2011 at 10:00
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http://math.stackexchange.com/questions/163326/finding-a-subgroup-of-psl-211
|
# Finding a subgroup of $PSL_2(11)$
Here is my problem:
Let $X=\begin{pmatrix} 10 & 8 \\ 8 & 1 \end{pmatrix}$ and $Y=\begin{pmatrix} 5 & 7 \\ 5 & 5\end{pmatrix}$ be two elements of $SL_2(11)$. Find a subgroup of $PSL_2(11)$ isomorphic to $A_5$.
I know that $A_5$ has a presentation as $$A_5=\langle x,y|x^2=y^3=(xy)^5=1\rangle$$ and $PSL_2(11)=\displaystyle\frac{SL_2(11)}{\{\lambda I|\lambda^2=1, \lambda\in GF^*(11) \}}$. How can I use $X$ and $Y$? Any help will be appreciated. Thanks
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1
Have you looked at the powers of $X$ and $Y$? For instance, what is $X^2$ in $SL_2(11)$? What is it in $PSL_2(11)$? – Steven Stadnicki Jun 26 '12 at 15:20
1
I would just verify that the images of X and Y in PSL(2,11) satisfy the relations of A5 (and that the image of X is not the identity, so that you don't have a proper quotient of A5). It is slightly nicer numbers if you use [-1,-3;-3,1] and [5,-4;5,-6] for X and Y. – Jack Schmidt Jun 26 '12 at 15:31
@StevenStadnicki: I wanted to take $<X,Y>$ as possible subgroup here, but I hesitated. Because I don't know "if $X$ and $Y$ staisfy in relations of $A_5$ so we have an image of it" or not. – Babak S. Jun 26 '12 at 15:51
1
If $X$ and $Y$ satisfy the defining relations of $A_5$ then you can define a homomorphism $\phi : A_5 \ \to\ <X,Y>$, simply by setting $\phi(x)=X$ and $\phi(y)=Y$. Clearly $\phi$ is surjective, and since the kernel of a homomorphism is always a normal subgroup you just need one final observation... – newguy Jun 26 '12 at 15:52
@newguy: Ok thanks.Let me think. – Babak S. Jun 26 '12 at 16:18
show 1 more comment
## 1 Answer
Note that $X^2 = -I$, and $Y^3 = I$, so in $PSL_2(11)$, the coset containing $X$ has order $2$, and the coset containing $Y$ has order $3$ (since neither $X$ nor $Y$ is in $SZ_2(11) = \{I,-I\}$).
Now $XY = \begin{bmatrix}2&0\\1&6 \end{bmatrix}$ and some short calculations show that:
$(XY)^2 = \begin{bmatrix}4&0\\8&3 \end{bmatrix}$, $(XY)^3 = \begin{bmatrix}8&0\\8&7 \end{bmatrix}$, $(XY)^4 = \begin{bmatrix}5&0\\1&9 \end{bmatrix}$, $(XY)^5 = \begin{bmatrix}10&0\\0&10 \end{bmatrix} = -I$
so the coset containing $XY$ has order $5$.
Letting $X',Y'$ be the images of $X,Y$ (respectively) in $PSL_2(11)$, we see that $\langle X',Y' \rangle$ is isomorphic to $A_5$.
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http://math.stackexchange.com/questions/257712/a-problem-with-regard-to-wiener-process?answertab=votes
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# A problem with regard to Wiener process
Let $W$ be a Wiener process and $U_x$ is the amount of time spent below $x$ during time interval $(0,1)$. Hence $U_x=\int\limits_0^1I_{\{W(t)<x\}}dt$. My question is: what is the probability density function of $U_x$? Thank you.
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1
– TheBridge Dec 13 '12 at 12:12
@TheBridge I'm not the owner of this question. I know that this one is a consequence of Lévy's arcsine law. – saz Dec 13 '12 at 12:18
@ saz : my mistake, but I can't edit my comment. best regards. – TheBridge Dec 13 '12 at 14:49
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|
http://mathematica.stackexchange.com/questions/8782/multidimensional-array-reduction-through-summation-over-one-of-its-dimensions?answertab=votes
|
# Multidimensional array reduction through summation over one of its dimensions
## 1. Introduction
I am using an array of dimension 3 (might become more) to store some values. I would like to implement a function that takes as argument the array and a couple of numbers smaller than the array dimension and returns a standard matrix (dim=2) which is obtained by reduction or "projection" (I don't really know how to call it) of the array along all the "other" dimensions. First a quick example with an array of dimension/depth 3
````data=Table[10 (i - 1) + 3 (j - 1) + k, {i, 2}, {j, 3}, {k, 4}]
--> data = {{{1, 2, 3, 4}, {4, 5, 6, 7}, {7, 8, 9, 10}},
{{11, 12, 13, 14}, {14, 15, 16, 17}, {17, 18, 19, 20}}}
````
My desired function would do, for example, the following
````Myfunction[data,{1,2}] = {{10,22,34},{50,62,74}}
````
After this brief introduction, let me explain where I've arrived. First let's generalize a little bit, as it is my goal, eventually.
## 2. Definitions
Let $M \in \mathbb{R}^{\Pi d_i}$ be an array of dimension $N \in \mathbb{N}$.
Let $d_i \in \mathbb{N}, \quad i=1,\dots,N\quad$ be the sublengths of each of the dimensions.
What I will call in the following the projection of the array along the dimension $q$ is the following surjective function : $$\textrm{For } q\leq N \in \mathbb{N}_0,\quad P(M,q) : \mathbb{R}^{\Pi d_i} \rightarrow\mathbb{R}^{\frac{\Pi d_i}{d_q}} : M(n_1,\dots,n_N) \mapsto \sum_{j=1}^{n_q}M(n_1,\dots,n_{q-1},j,n_{q+1},\dots,n_N)$$
## 3. Goal
The goal is to write a function that takes any multidimensional array as well as a list of all the dimensions that will be "kept" in the final array. For the sake of any representation, the obtained array should be of dimension $\leq$ 2 but I would like to stay as general as I (we) could.
The dimensions ordering should simply be the one used to implement the array, and the code should be consistent in order to keep the dimensions in the same order at any time.
## 4. Implementation
````ProjectedColumns[len_,cols_]:= Complement[Range[len],cols]
````
ProjectedColumns is a function that returns a list of the dimensions to project, where `cols` is the list of dimensions to be kept and len is the dimension of the array.
````CompoundProjection[data_, cols_] :=
Module[{len, vect},
len = Depth[data] - 1;
vect = ProjectedColumns[len, cols];
Do[
SingleProjection[data, vect[[i]]],
{i, Length[vect]}
]
]
````
CompoundProjection is a function that will project all the dimensions sequentially in order to arrive to the final result (The projection is commutative).
````SingleProjection[data_, dimnumber_] := ???
````
Now what I need and I don't manage to get is the function that will actually perfom a one-dimensional projection. In my mind, I would need to have a number of `Do` loops which equals the dimension of the array.
In that case, I simply parse through the array and do the summation over the one I'm interested in. Is it possible to set up such a structure with all the commands that Mathematica offers and that I probably don't know ?
## 5. In summary
1. Has anyone followed this nonsense ?
2. Is there a Mathematica command that makes what I want directly ?
(If Yes then No to the previous questions, then I'm happy)
1. Is there a way to improve the correctness or elegance of what I did ?
2. Is there a way to intricate a dynamic number of Do loops ?
3. Or, in general, is there another way to achieve the goal described in 3.
-
1
Maybe you can start from this : `Map[Total , data, {2}]`, which will do what you seek with `Myfunction` – b.gatessucks Jul 27 '12 at 18:07
1
Seems that @b.gatessucks is right. But I would like to see a few more examples for the `{1,2}` parameter in your first code snippet – belisarius Jul 27 '12 at 18:10
## 3 Answers
Did you know about the second argument of `Total`, which lets you sum up element at a certain level, which in practice means along a certain dimensions?
For example, if you want to keep levels 1 and 2, and sum up along level 3, you can use
````Total[data, {3}]
(* ==> {{10, 22, 34}, {50, 62, 74}} *)
````
Or sum up along dimension 1:
````Total[data, {1}]
(* ==> {{12, 14, 16, 18}, {18, 20, 22, 24}, {24, 26, 28, 30}} *)
````
This is the same as `removeDimensions[data, {3, 1}]`.
-
Reading this I though initially that you could actually get all the desired functionality out of just Total and complement, however you run into trouble with it if you want to remove for example dimension 1 and 4 in a 6 dimensional structure, and then you need to reorder the arguments, or call Total continually while recalculating the dimensions on each pass. – jVincent Jul 27 '12 at 18:27
@jVincent Yes, for that it's necessary to call `Total` two times. – Szabolcs Jul 27 '12 at 19:03
1
`Total` seems to be a good tool indeed. I've come up with a way of using it recursively. You don't need to relaculate anything in the process. You just apply the reduction beginning by the deepest dimensions and then it works fine. – Pschoofs Jul 30 '12 at 9:08
I may be misunderstanding what you need, however consider this. If you have an structure of nested lists such as your data, then summing across the deepest list is easily accomplished using `Map[Total,data,{-2}]`. So, as long as you want to remove dimensions from "the back" you are good to go. And if we need to remove for example the second dimension, then you can just transpose it to the back, and then remove it:
```` Map[Total,Transpose[data,{1,3,2}],{-2}]
````
Summing out more then one dimension would then just be pushing them all to the back, and applying Total at a higher level and summing all the way down. Which can be done as:
```` transposeOrder[data_, dimensions_] :=
Join[dimensions, Complement[Range[Depth[data] - 1], dimensions]]
removeDimensions[data_, dimensions_] /; (Length[dimensions] < Depth[data] - 1) :=
Map[Total[#, Infinity] &,
Transpose[data, transposeOrder[data, dimensions]],
{-1 + Length[dimensions] - (Depth[data] - 1)}]
````
And for the case where you aren't removing any dimensions, I suppose it should just fall back to reordering them:
```` sumOutDimensions[data_, def_] /; Length[def] == Depth[data] - 1 := Transpose[data, def]
````
Now these functions output the results with dimensions in the order given by the input, if you want to retain the order of the data structure, you could just sort the arguments to it.
```` removeDimensions[data, {1, 2}]
````
````{{10, 22, 34}, {50, 62, 74}}
````
-
Here's a version using `Flatten` to push all the "unwanted" dimensions down to the innermost level before applying the desired function.
````keepDimensions[data_, dims_List, func_:Total]:=
Map[func,Flatten[data,Thread@{dims}~Join~{Complement[Range@ArrayDepth@data,dims]}],{-2}]
````
Examples :
````data=Table[10 (i-1)+3 (j-1)+k,{i,2},{j,3},{k,4}]
````
````{{{1,2,3,4},{4,5,6,7},{7,8,9,10}},{{11,12,13,14},{14,15,16,17},{17,18,19,20}}}
````
````keepDimensions[data, {1,2}]
````
````{{10,22,34},{50,62,74}}
````
Any combination of dimensions can be kept
````keepDimensions[data, {1,3}]
````
````{{12,15,18,21},{42,45,48,51}}
````
Kept dimensions can be re-ordered
````keepDimensions[data, {3,1}]
````
````{{12,42},{15,45},{18,48},{21,51}}
````
An optional third argument can be supplied to do something other than summation:
````keepDimensions[data, {1,3}, Max]
````
````{{7,8,9,10},{17,18,19,20}}
````
-
lang-mma
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|
http://mathhelpforum.com/advanced-statistics/88689-p-m-f.html
|
# Thread:
1. ## p.m.f.
Let x and y have the joint p.m.f. given by:
$<br /> \begin{pmatrix}&&&x&&\\y&1&2&3&4&5\\3&0.02&0.03&0. 06&0.15&0.07\\2&0.02&0.05&0.08&0.12&0.06\\1&0.05&0 .05&0.06&0.13&0.05\end{pmatrix}<br />$
a) Find the marginal p.m.f.s and the means and variances.
b) Are x and y independent? Why?
c) Find the covariance and the correlation coefficient of x and y.
I appreciate any help. I don't know how a problem like this is done so can't do the rest of them. Thank you so much
2. SUM across to get the values, that is where the name marginal comes from.
The $P(Y=3)=.02 + .03 + .06 + .15 + .07$
Do that across and down (for x's distribution).
IN order for these two rvs to be independent... for EVERY position the sum across and the sum down's product must equal the probability of that position.
The shortcut formula for covariance is $Cov(X,Y)=E(XY)-E(X)E(Y)$ and if the rvs are indep, this is zero.
The definition of correlation is ${Cov(X,Y)\over \sigma_x\sigma_y}$ and if the rvs are indep, this is zero.
3. Thank you!
I am also supposed to:
a) sketch the support of X and Y
b) Find the best-fitting line and draw it on the figure
I have no examples so I don't know what is being asked for. Do you have any ideas of what they're looking for?
4. the support are just the points that have positive probability...
(1,1),..., (5,1)
(1,2),..., (5,2)
(1,3),..., (5,3).
I'm not sure what is meant by the best fitting line.
I can plot a least squares line through these 15 points.
Are we supposed to weight these points via the probabilities?
5. I'm sorry, this question is still killing me! So I just sum across to get 3 marginal p.m.f.s for y and sum down to get 5 p.m.f.s for x? Then how would I calculate the means and variances for this type of problem?
"IN order for these two rvs to be independent... for EVERY position the sum across and the sum down's product must equal the probability of that position."
My probability is just out of 15? For example 0.05 occurs 4 times so P(0.05)=4/15?
I'm totally lost on the covariance and correlation coefficient.
For the graph, I simply graphed the 15 points.
Thank you for everything. I'm sorry I am having such a hard time grasping this concept.
|
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|
http://mathhelpforum.com/pre-calculus/118985-finding-inverse-function.html
|
# Thread:
1. ## Finding inverse function
In the theory of relativity, the mass of a particle with velocity v is
$m=f(v)=\frac{m_{0}}{\sqrt{1-\frac{v^2}{c^2}}}$
where $m_{0}$ is the rest mass of the particle and c is the speed of light in vacuum. Find the inverse function of f.
I've tried to make v the subject and I got this,
$\displaystyle v^{2}=c^{2}\left(1-\frac{m^{2}_{0}}{m^{2}}\right)$
When I square root both sides to get v, should I choose the +ve or the -ve square root?
The v here is just the speed and no direction is involved here, am I right?
2. Originally Posted by acc100jt
In the theory of relativity, the mass of a particle with velocity v is
$m=f(v)=\frac{m_{0}}{\sqrt{1-\frac{v^2}{c^2}}}$
where $m_{0}$ is the rest mass of the particle and c is the speed of light in vacuum. Find the inverse function of f.
I've tried to make v the subject and I got this,
$\displaystyle v^{2}=c^{2}\left(1-\frac{m^{2}_{0}}{m^{2}}\right)$
When I square root both sides to get v, should I choose the +ve or the -ve square root?
The v here is just the speed and no direction is involved here, am I right?
This is the formula for mass increase:
$m=f(v)=\frac{m_{0}}{\sqrt{1-(\frac{v}{c}})^2}$
m= mass when obeject is moving at speed v
$m_0$= mass when object is not moving
So, now you're solving for velocity. The velocity CAN be negative since it is a vector. Given that the direction can be 180 degrees away from the reference, it can be considered negative. It really depends on your problem and the value you are interested in.
3. I'm not really good at physics or anything, but I'm making a guess.
In physics, the values are highly directional. To really know the eventual signs of the values, one must first understand the equation.
As we also know, the rest mass of a particle is, as far as I have learnt, positive; consequently, the mass of a particle with velocity v is always positive.
Let the rest mass of the particle be M.
And, (m) = (+)/(+) = (+)
More importantly, you have to know the inequality shared between M & m.
Firstly,
let g(v) = [1 - (M^2)/(m^2)]^(1/2), and the range: (0,1)
The range is in consideration that no particle can attain the speed of light.
Hence, let [1 - (M^2)/(m^2)]^(1/2) be k.
=> m = M/k
when k tends to 1, m ≈ M > M
when k tends to 0, m ≈ ∞ > M
Hence, the inequality is this: m > M
Hence, as we both know, physics is very particular about the nature and direction of all matters, or whatever in the universe, we choose signs; in other words, this is not a matter of mathematics where we show a full range of values, instead, in physics, we show the full range of possible, and desired range.
Following your method, by square rooting the expression, I will arrive at this:
v = [(c)^(1/2)][1-(M^2)/(m^2)]^(1/2)
Using what we know about the values of M & m as positive values, and the fact that c is merely speed, and thus 'positive'.
Since M < m
(M^2)/(m^2) is always < 1, hence,
1-(M^2)/(m^2) is always positive.
(v) = (+)(+) = (+), hence v is always positive.
There you have it, you should most probably choose the +ve square root, that is if you always define it in the direction it goes, or it moves in a linear fashion you defined it positive.
4. Originally Posted by acc100jt
In the theory of relativity, the mass of a particle with velocity v is
$m=f(v)=\frac{m_{0}}{\sqrt{1-\frac{v^2}{c^2}}}$
where $m_{0}$ is the rest mass of the particle and c is the speed of light in vacuum. Find the inverse function of f.
I've tried to make v the subject and I got this,
$\displaystyle v^{2}=c^{2}\left(1-\frac{m^{2}_{0}}{m^{2}}\right)$
When I square root both sides to get v, should I choose the +ve or the -ve square root?
The v here is just the speed and no direction is involved here, am I right?
Since you want an inverse function, f has to be restricted so that it's 1-to-1. The restriction used will determine whether you want the +ve or -ve square root in your inverse function. The restriction v > 0 => +ve root.
5. I see, so it is my choice to choose the restriction. Thanks
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|
http://math.stackexchange.com/questions/107885/parity-condition
|
# Parity Condition
In the proof for the NP-completeness of Edge-colouring (paper), there is an intermediate result used called the parity condition, which is formulated in a lemma.
Quoting from the paper:
Let $G$ be a cubic, 3-edge colored graph and $V' \subseteq V(G)$ a set of vertices in $G$. Let $E' \subseteq E(G)$ be the set of edges of $G$ which connect $V'$ to the remainder of the graph. If the number of edges of color $i$ in $E'$ is $k_i$ ($i = 1, 2, 3$), then
$$k_1 \equiv k_2 \equiv k_3 \pmod 2$$
Proof. If $E_{12}$ is the set of edges of $G$ which are colored with color 1 or 2, the $E_{12}$ consists of a collection of cycles. Thus $E_{12}$ meets $E'$ in an even number of edges, and so $k_1 + k_2 \equiv 0 \pmod 2$. Similarly $k_2 \equiv k_3 \pmod 2$. $\Box$
I'm having trouble getting the intuition for the proof. Why does $E_{12}$ necessarily consist of cycles? And why does that fact lead us that $E_{12}$ necessarily meets $E'$ in an even number of edges?
-
## 1 Answer
Since $G$ is cubic (i.e., 3-regular) and 3-edge-colored, every vertex is adjacent to exactly one edge of each color. Therefore $E_{12}$ is 2-regular, and every edge in a finite 2-regular graph must be part of a cycle that is disjoint from the rest of the graph (start from the chosen edge, and continue following the only path you can without doubling back until you reach your starting point).
Assign, arbitrarily, a direction to each of the cycles in $E_{12}$. The cycle passes into $V'$ exactly as many times as it passes out of $V'$, so it contributes an even number of edges to $E_{12}\cap E'$.
-
Thank you! I was way off with my original thinking. – Sami Lehtinen Feb 10 '12 at 19:53
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http://mathhelpforum.com/advanced-statistics/167114-liminf-product-print.html
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# liminf of a product
Printable View
• December 30th 2010, 12:57 AM
amheissan
liminf of a product
I am trying to prove the following:
$\underbrace{(\lim \inf f_n)}_f\underbrace{(\lim\inf g_n)}_g\leq \underbrace{\lim\inf (f_ng_n)}_k$.
Here's what I have so far:
FTSOC assume $fg>k$. We know that $f_n<f-\epsilon$ and $g_n<g-\epsilon$ for only finitely many $n$. Then $f_ng_n<fg-\epsilon(f+g-\epsilon)$ for only finitely many $n$. This implies $f_ng_n\geq fg-\epsilon(f+g-\epsilon)$ almost always. So, $f_ng_n\geq k-\epsilon(f+g-\epsilon)$ almost always.
But this doesn't get me anywhere. Where am I going wrong? If I let $\epsilon=\min\{f,g\}$, I just get $f_ng_n\geq 0$. So, this tells me nothing.
Any help would be appreciated. (Even though it doesn't state it, I am assuming that all $f_n, g_n$ are nonnegative.)
• December 30th 2010, 07:50 AM
snowtea
I don't think that is true.
Consider $f_n$ the sequence $-1,0,-1,0,...$
and $g_n$ the sequence $2,0,2,0,...$
so $f_ng_n$ is $-2, 0,-2, 0,...$
Now $(\lim \inf f_n)(\lim\inf g_n) = -1 \times 0 = 0$
$\lim\inf (f_ng_n) = -2$
• December 30th 2010, 08:59 AM
amheissan
Quote:
Originally Posted by amheissan
...I am assuming that all $f_n, g_n$ are nonnegative.
You are right that it wouldn't hold if were allowed to have a function as you have described, but we are limited to non-negative functions.
Any more thoughts on proving this property? Thanks!
• December 30th 2010, 09:18 AM
Drexel28
Quote:
Originally Posted by amheissan
You are right that it wouldn't hold if were allowed to have a function as you have described, but we are limited to non-negative functions.
Any more thoughts on proving this property? Thanks!
Note that by definition $\displaystyle \liminf_{n\to\infty}f_n(x)=\inf \left\{\lim_{n\to\infty}f_{n_k}(x):f_{n_k}(x)\text { is a subsequence of }f_n(x)\right\}$ and similarly for $\displaystyle \liminf_{n\to\infty}g_n(x)$. Use the fact then that $\inf\left(A\right)\inf\left(B\right)\leqslant \inf\left(AB\right)$.
• December 30th 2010, 09:33 AM
amheissan
Okay, but how would you prove $\inf(A)\inf(B)\leq \inf (AB)$? I'm really just not seeing this property.
• December 30th 2010, 12:17 PM
snowtea
For only nonnegative sequences, Drexel28 is giving you the hint to only look at the inf of the set of values at any subsequence (ignore the order of the sequence).
Let $AB = \{ab | a\in A, b\in B\}$.
Consider any element $a\in A$ and $b\in B$, $ab \geq \inf(A)\inf(B)$ (by definition since everything is nonnegative).
Also, remember that the inf of a subset of AB must be larger than the inf of AB
All times are GMT -8. The time now is 03:42 AM.
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http://physics.stackexchange.com/questions/tagged/symmetry+hamiltonian-formalism
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http://math.stackexchange.com/questions/222915/what-taking-gcd-of-two-even-numbers-gives
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# What taking gcd of two even numbers gives
What does taking the gcd of two even numbers $y$ and $z$ give? Does it give another indefinite even number $x$?
-
without further knowledge about $y$ and %z$, all we can say, is that$\cgd(y,z)\$ is even. But that does not make it indefinite. – Hagen von Eitzen Oct 28 '12 at 18:50
what if both numbers are odd? – Gladstone Asder Oct 28 '12 at 18:57
If the numbers are odd then the GCD is odd, since the GCD divides both of the numbers $y$ and $z$. – Tom Oldfield Oct 28 '12 at 19:00
## 3 Answers
Any two positive integers will have a greatest common divisor which can be computed using the Euclidean algorithm, so it is a definite number. If the numbers are even, the gcd is even as well.
-
Generally $\rm\ \ gcd(2j,2k)\, =\, 2\,gcd(j,k)\ \$ and $$\rm\:gcd(2j\!+\!1,2k\!+\!1)\, =\, gcd(2(j\!-\!k),2k\!+\!1)\, =\, gcd(j\!-\!k,2k\!+\!1).$$
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Imagine you have $y$ and $z$ and you make two lists: one contains all divisors of $y$ and the other contains all divisors of $z$. The greatest common divisor is just the largest number that is common to both lists.
In particular, both lists will contain the number $1$, since $1$ divides any number. Thus, $\gcd(y,z)$ is defined and is at least $1$ for any integers $y$ and $z$.
If $y$ and $z$ are both odd, it is possible that the only number appearing on both lists is $1$ (for example, $\gcd(3,5) = 1$). It is also possible for the greatest common divisor to be much larger (for example, $\gcd(10,15) = 5$).
If both $y$ and $z$ are even, the we know $2$ is a divisor of both $y$ and $z$ (this is what it means to be even), so the greatest common divisor is at least $2$. As for the odd case, however, this can be exact (e.g. $\gcd(6,8) = 2$) or very far from the truth (e.g. $gcd(32,48) = 16$).
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http://www.physicsforums.com/showthread.php?p=4003030
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Physics Forums
## Developing Inner Product in Polar Coordinates via metric
Hey all,
I've never taken a formal class on tensor analysis, but I've been trying to learn a few things about it. I was looking at the metric tensor in curvilinear coordinates. This Wikipedia article claims that you can formulate a dot product in curvilinear coordinates through the following:
$\textbf{u} \cdot \textbf{v} = g_{ij} u^i v^j$ where $g_{ij}$ is the covariant metric tensor for the coordinate system.
In particular, I'm interested in 2-D polar coordinates, therefore, $g_{ij} = \left( \begin{array}{cc} 1 & 0 \\ 0 & r^2 \end{array} \right)$ which makes sense: the polar coordinates are orthogonal so $g_{ij}$ is diagonal, and an infinitesimal change $ds$ occurs with a proportional change in $dr$ and varies with $r$ in $d \theta$. Now I expand the expression for the dot product and using Einstein summation:
$\textbf{u} \cdot \textbf{v} = u^1 v^1 + (r)^2 u^2 v^2$
Now this really doesn't make much sense to me. To me, this implies that the length of a vector in polar coordinates depends on $\theta$, which isn't true, the length is contained entirely within the $r$ component. If we transform a vector in polar to Cartesian and then write our dot product, we find $\textbf{u} \cdot \textbf{v} = u^1 v^1 \cos(u^2 - v^2)$, but I don't see this happening any time soon with this metric and the given dot product formulation. Is there any way to get the same dot product using the metric of the coordinate space? If not, why isn't this working? Thanks!
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Hey gordon831 and welcome to the forums. Lets consider going from euclidean to polar. x = rcos(theta), y = rsin(theta). r = SQRT(x^2 + y^2), theta = arctan(y/x). dr/dx = x/SQRT(x^2+y^2), dr/dy = y/SQRT(x^2+y^2), dtheta/dx = - y / (x^2 + y^2), dtheta/dy = x / (x^2 + y^2) So from this our Jacobian for this system has the values [dr/dx, dr/dy; dtheta/dx, dtheta/dy] where g = J^2. Is this what you got for your tensor where the transformation is from (x,y) to (r,theta)?
So here's how I developed the metric. One way to develop a covariant metric of a coordinate space is to dot the covariant bases: $g_{ij} = \textbf{e}_i \cdot \textbf{e}_j$ (Source) see (9) To do the dot product, I use my bases defined in Cartesian coordinates so the dot product is simply $a_i b_i$. My bases in Cartesian coordinates are: $\vec{e}_r = \frac{\partial x}{\partial r} \hat{e}_x + \frac{\partial y}{\partial r} \hat{e}_y = \cos \theta \hat{e}_x + \sin \theta \hat{e}_y$ $\vec{e}_\theta = \frac{\partial x}{\partial \theta} \hat{e}_x + \frac{\partial y}{\partial \theta} \hat{e}_y = -r \sin \theta \hat{e}_x + r \cos \theta \hat{e}_y$ With $\vec{e}_\theta$ remaining un-normalized. This gives the covariant metric tensor for polar coordinates: $g_{ij} = \left( \begin{array}{cc} 1 & 0 \\ 0 & r^2 \end{array} \right)$ I went through using the Jacobian you described, and I got the inverse tensor: $g_{ij}^{-1} = \left( \begin{array}{cc} 1 & 0 \\ 0 & \frac{1}{r^2} \end{array} \right)$ Which makes sense. Let Cartesian components be denoted with $x$ and polar be $q$, your Jacobian is expressed as $\frac{\partial q^i}{\partial x^j}$, but the dot product described above would yield $\frac{\partial x^i}{\partial q^j}$ - the inverse.
## Developing Inner Product in Polar Coordinates via metric
Quote by gordon831 $\textbf{u} \cdot \textbf{v} = u^1 v^1 + (r)^2 u^2 v^2$ Now this really doesn't make much sense to me. To me, this implies that the length of a vector in polar coordinates depends on $\theta$, which isn't true, the length is contained entirely within the $r$ component. If we transform a vector in polar to Cartesian and then write our dot product, we find $\textbf{u} \cdot \textbf{v} = u^1 v^1 \cos(u^2 - v^2)$, but I don't see this happening any time soon with this metric and the given dot product formulation. Is there any way to get the same dot product using the metric of the coordinate space? If not, why isn't this working? Thanks!
What you need to understand is that position vectors transform according to the full nonlinear transformation--notions of length for them are not considered or handled by the metric.
The metric allows you to take dot products of vectors that themselves depend on position, however--i.e. that are fields. That is, at some $(r,\theta)$, the vectors $u$ and $v$ point in some directions.
Simple case: let $u = 2e_r + 3e_\theta$, and let's find $u \cdot u$. The metric (and common logic) would tell you that the answer is $(u)^2 = 4 + 9(r)^2$. You see, $r$ refers to where this vector is located on our coordiante system, so quite the contrary, it's not $\theta$ that enters into anything but $r$. The reason this happens is because $e_\theta$ is not a unit vector. Its magnitude everywhere is $r$. Saying $(u)^2 = 4+9(r)^2$ is a perfectly sensible result.
Okay, so when I first read your response I was really confused about what it meant for a vector to be "located" at $(r,\theta)$ and I think I'm still a little confused. By specifying $(r, \theta)$, we are defining our basis vectors. If we normalize $e_r$ and $e_\theta$, we see that $\theta$ is the Cartesian equivalent of rotating the vector about the positive z-axis by $\theta$. But what does it mean to specify an $r$ where a vector is located?
Yes, the unit vectors depend on $(r,\theta)$. This is precisely what I meant. Hence, you cannot just say you have a vector with polar components. One must specify at which location $(r,\theta)$ that the unit vectors should be evaluated at. To me, that's just simply saying what the vector's position on the coordinate space is. Unlike cartesian coordinates, in order to make sense of a vector in a polar coordinate basis, you must know where this vector lies. Let me give a concrete example. Let's say you have $u = u^1 e_r + u^2 e_\theta$ and similarly for $v$. The metric tells us that $$u \cdot v = u^1 v^1 + (r)^2 u^2 v^2$$ We can verify this by writing out the expressions for $e_r$ and $e_\theta$ in a cartesian basis. $$e_r = e_x \cos \theta + e_y \sin \theta \\ e_\theta = r(-e_x \sin \theta + e_y \cos \theta)$$ Do you follow this so far? I hope so. In particular, pay attention to $e_\theta$. It increases in magnitude as you get further away from the origin. This leads to some strange results. For example, if you had the vector $e_y$ and wanted to move it smoothly along the x-axis, its expression in polar would be $e_y = (1/r) e_\theta$. Its apparent component of $e_\theta$ would decrease as the vector is transported along the axis, even though the full magnitude of the vector is not changing. At any rate, you can compute the dot products of these vectors. $$e_r \cdot e_r = \cos^2 \theta + \sin^2 \theta = 1\\ e_r \cdot e_\theta = r(-\cos \theta \sin \theta + \cos \theta \sin \theta) = 0 \\ e_\theta \cdot e_\theta = r^2 (\sin^2 \theta + \cos^2 \theta) = r^2$$ Exactly like you'd expect. Anyway, I'm not exactly sure what's confusing you. I think you still don't grasp that this can't be used for position vectors. You need to picture, say, a vector with its tail fixed to a specified point (like $r=1, \theta=\pi/4$, say) and pointing in some arbitrary direction (not necessarily the radial). That vector's components are evaluated in terms of the basis vectors at that point where the tail is fixed. Hopefully that helps.
Okay, so let me see if I've got this. Let's say I have two vectors, $u$ which has a length of 1 and parallel to the Cartesian x-axis and $v$ which has a length of 1 and is rotated $\frac{\pi}{4}$ rads counter-clockwise from the x-axis. These are position vectors, so the metric doesn't really apply here. Now, I'm going to write these vectors in terms of the basis. For convenience, I'll write both vectors as if they were located at $(1,0)$ on the polar coordinate system. I will find: $$u = e_r$$ $$v = \frac{1}{\sqrt{2}}e_r + \frac{1}{\sqrt{2}}e_\theta$$ Which leads to: $$u \cdot v = \frac{1}{\sqrt{2}}$$ Which is correct. But isn't it arbitrary what I choose for $(r,\theta)$? I mean, depending on the choice the vector components change because the basis changes. At a more abstract level, I can see that $\theta$ will never be significant in the length of a vector, because it's not a factor in the dot product. Sorry to be so bothersome, I've never considered that the tail of a vector might be important - I thought vectors were considered "free" or that their initial points held no significance.
Yes, it's arbitrary what position you choose for this example, but this becomes meaningful when you're talking about vector fields that naturally have a notion of what position they're at.
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curvilinear, inner product, metric, polar, tensor
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http://mathoverflow.net/questions/44860?sort=oldest
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## orbits in locally compact group
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As everyone knows if $x\in S^1$, then the set ${ x^n }$ is either finite or dense. Under which condition is true for any other locally compact group, i.e if $G$ is a locally compact group, and $x\in G$ is a non-trivial element, then the set $x^n$ is either finite or dense.
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@Mohammad: in the future, keep in mind that it's generally considered impolite to edit your question in such a way as to make existing answers unintelligible. – Qiaochu Yuan Nov 4 2010 at 19:24
My apology, you are completely right, Sorry about that. – Mohammad Nov 4 2010 at 19:28
I added the tag "topological-groups" – rpotrie Nov 4 2010 at 23:41
Nice question. You might want to read about Burnside's problem: en.wikipedia.org/wiki/Burnside%27s_problem . It doesn't give an answer to your question, but investigates situation more closely in the case of discrete groups. – Łukasz Grabowski Nov 5 2010 at 17:14
## 4 Answers
I don't think ever for a Lie group $G$. If $G$ is not compact then it will have a closed subgroup isomorphic to $\mathbb{R}$, which rules it out. If it is compact, then it will have a proper subgroup isomorphic to $S^1$ which again rules it out.
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For Lie groups, the only one with this property is $S^1$.
To see this, consider a one parameter subgroup given by exponential of a vector in the lie algebra, and the closure of this subgroup should be abelian.
Since the only abelian lie group with the property you are asking is $S^1$, this concludes.
For other groups, I believe it should also be true, but I don't know. The argument above, shows that if the group is not abelian, then it does not hold (since the closure of an orbit is an abelian closed subgroup), but I don't know if abelian topological groups are all known (the ones I know, the only one verifying your property is $S^1$).
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This is not true even for compact abelian groups. If you take $G = S^1 \times S^1$ and $(x,y) \in G$ is an element such that the orbit of $x$ in $S^1$ is dense and the orbit of $y$ in $S^1$ is finite, then the orbit of $(x,y)$ in $S^1 \times S^1$ will be neither.
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It is easy to produce groups $G$ in which the set ${x^n}$ is finite for every $x\in G$. E.g, let $F$ be any finite group; then $G=F^{\mathbb{N}}$ is a compact, totally disconnected group with this property. So in the OP we may assume that $G$ contains some element generating an infinite cyclic dense subgroup.
I claim that, in this case, $G$ is indeed isomorphic to $S^1$. First, $G$ is locally compact abelian. Here is a result I found in the book by W. Rudin, "Fourier analysis on groups" (Wiley, 1962). Say that a locally compact abelian group is {\it monothetic} if it contains a dense cyclic subgroup. Theorem 2.3.2: every monothetic group is either compact or isomorphic to $\mathbb{Z}$ (as a topological group). Coming back to the OP, our group $G$ is monothetic. By the theorem just quoted, $G$ must be compact (as $\mathbb{Z}$ clearly does not satisfy the assumptions of the OP).
Let $\hat{G}$ be the dual of $G$, a discrete, infinite abelian group. By Pontrygin duality, it is enough to prove that $\hat{G}$ is infinite cyclic. Dualizing the assumptions in the OP, we see that every homomorphism $\hat{G}\rightarrow S^1$ either has finite image, or is injective. In particular, $\hat{G}$ is just infinite (infinite, but every proper quotient is finite). It is not very difficult to see that a just infinite, abelian group is infinite cyclic (see McCarthy, Donald, Infinite groups whose proper quotient groups are finite. I. Comm. Pure Appl. Math. 21 1968 545–562).
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http://mathoverflow.net/questions/14440/how-can-i-conclude-that-i-live-in-a-solar-system/14447
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How can I conclude that I live in a solar system?
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Well, this is an awkward question and I don't know if it is mathematical enough for MO (I'm sorry if not) but I'll try it: What observations in the coordinate system centered in my fixed position on earth are necessary to conclude that the earth (and the planets) move (approximately) in ellipses around the sun and that earth is rotating around itself?
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Thumbs down to the votes to close! Boooo. – Kevin Lin Jul 10 2010 at 21:26
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Uh,looking up in the daytime is a rather unambigous clue......LOL Just kidding. It's actually an excellent question since all of us grew up at a time in history when it's more or less taken as obvious.Considering it from first principles,it really isn't. Also,there's so much mythology and apocrophya surrounding the gradual development of the heliocentric model that culminated in classical mechanics,that most of us nonexperts simply have erroneous ideas of how it was actually proven by the Ancient Greeks and thier sucessors. – Andrew L Jul 11 2010 at 1:01
9 Answers
You can use a Foucault pendulum to determine that you are on a rotating planet. If you set it up on the North or South pole it will complete one rotation in one day.
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Very good point. – Ilya Grigoriev Feb 7 2010 at 1:06
Indeed! I obviously missed school. – S1 Feb 7 2010 at 1:08
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Ernst Mach would have claimed that you get the same result if the earth stood still and the stars were rotating. Without assuming certain basic laws of physics (Newton, Einstein), it seems that you cannot deduce anything at all. – Franz Lemmermeyer Feb 7 2010 at 7:10
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Take a look at Terence Tao's pdf slides (4.3 Mb), http://terrytao.files.wordpress.com/2009/09/cosmic-distance-ladder2.pdf. Kepler makes an appearance and there's much more besides.
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Very nice lecture -- well worth the time. – Sam Nead Mar 17 2010 at 13:40
I think that you can write down the motion of planets in any coordinate system you like. Since prehistoric times, astronomers have recorded the motion of planets, and before Kepler everybody used the coordinate system where the earth is fixed. I don't know if they ever wrote down equations of the planets' motion, but they could certainly predict it.
Of course, the equations of motion in this coordinate system would be very messy and inconvenient. The reason that we say that "planets move around the sun" is, IMHO, the fact that in the coordinate system centered at the sun, the equations become so simple and easy to understand.
The one "specific observation" that, to my mind, shows how confusing and inconvenient to calculate the motion of planets is from an Earth-centered system, is the retrograde motion of planets. When viewed from the earth, planets usually move in one direction in the sky (when viewed at the same time but on different days). Sometimes, however, planets like Mars move backwards in the sky. I think this has to do with the angular velocity of the Earth moving around the sun being greater than that of Mars (anybody has a better explanation?).
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I believe that is the correct reason, Ilya. – Harry Gindi Feb 6 2010 at 22:56
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@Harry: Here's what confuses me: why doesn't Mars then move backwards always? @Arminius: Next, you try to write down equations to predict the exact motion, and try to find the coordinate system in which it's easiest. I'm not sure if there are any qualitative phenomena that will guide you, although if you invent some other "wrong" coordinate system, there will probably be something qualitatively weird with it. To get it right, you might need Kepler-level intelligence. – Ilya Grigoriev Feb 6 2010 at 23:06
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@Arminius: Of course, there might be a better answer to your question, especially if we instead ask: how did Kepler justify his theory? What exact observations did he use? But I don't know, and I suspect that even if you study this in detail, you might not find an answer that will satisfy you exactly. It's easy to explain why a theory is good, but very hard to explain how to make up a good theory. – Ilya Grigoriev Feb 6 2010 at 23:09
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@Ilya " What exact observations did he use?" --- he used observations of Mars orbit by Tycho Brage. @Arminius I suggest you to read the Keplers book "Astronomia nova". Answer should be there. – Petya Feb 6 2010 at 23:36
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@Ilya: Sorry, I did not see your comment until after I had posted my comment. To answer your question: Yes, it is certain that Copernicus was aware of Aristarchus's priority because the original draft of his 1543 book has survived and it included a passage that refers to Aristarchus, which Copernicus later crossed out so as not to compromise the originality of his theory. – Marko Amnell Mar 17 2010 at 18:02
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Evidence that the Earth is spinning about its axis: the Coriolis effect. The coordinate system fixed to a specific spot on the Earth's surface is not inertial, that is, Newton's first law does not hold.
Evidence that the Earth and other planets are moving about the Sun in elliptical orbits: agreement of the projected elliptical motions with observations in the night sky.
There is no fundamental reason to choose the Sun as the center of the solar system coordinates. Any point will do, including the Earth or any other planet. However, as noted by such luminaries as Copernicus, Galileo, Kepler and other MO respondents, choosing the Sun as the center simplifies things quite a bit.
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I think Copernicus' role in all this is a bit more complicated than is often assumed -- I attended several history of mathematics lectures some years ago, which argued in semi-jocular but also semi-serious fashion that his heliocentrism was mystical and not scientific. But good answer, nonetheless. – Yemon Choi Feb 7 2010 at 0:02
I find the parallax effect parallax effect especially convincing evidence. Parallax is the shifting of lines of sight due to translation, eg by waiting half an earth year at which point theory tells we have moved about 16 light minutes around the sun from where we were.
Regarding retrograde motions: as Ilya said, Kepler's 2nd law closer planets move faster. Now draw two circles centered at 'Sun' with a point 'Earth' moving on the inner circle and another point 'Mars' moving more slowly but in the same sense, say counterclockwise on the outer circle. Drow a line between the two moving points. That line indicates how Mars looks, viewed from earth, relative to the distant stars.
How does the line move? Put the Sun at the origin. If the order is Sun-Earth-Mars, with Earth and Mars on the positive x axis, then the slope of the line is decreasing. But put the order Earth-Sun-Mars with Earth on the negative x axis, Mars on the positive x -axis. The slope of said line is now increasing. One is `prograde' the other 'retrograde'.
Finally, the explanation of elliptical versus circular motion is more of an Occam's razor business. Originally we had
Ptolemy's ''epicycles''-- in essence Fourier series. Ptolemy had earth at the solar system center and each planet moving on a system of nested circles, as in $z(t) = r_1 e ^{i \omega_1 t} + r_2 e^{i \omega_2 t} + r_3 e^{i \omega_3 t} + ...$, $r_1 > r_2 > \ldots$. Ptolemy needed 20 to 30 circles to account for observations. Kepler realized that but putting the sun at the 'center' and having the planets move in slightly eccentric ellipses with focus, a bit off from the sun, sweeping out ''equal areas in equal times'', he could account for all of Ptolemy's data plus Brahe's much more detailed data.
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An interesting (and very old) argument in favour of heliocentrism is based on estimates of the relative sizes of the Earth and Sun.
Actually, Aristarchus of Samos estimated that the Sun is six to seven times wider than the Earth (and therefore over 200 times more voluminous). These calculations arguably led him to conclude that it made more sense for the Earth to be moving than for the huge Sun to be moving around it.
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That's how I learned it was suggested-although it took nearly another 2000 years for it to become the dominant worldview once Aristotle's ideas began to take hold in the Western World,which favored geocentrism. The influence of the Christian Church further reinforced this. Something else we can thank the Church for........ – Andrew L Jul 11 2010 at 1:06
@ Andrew: there are plenty of other "churches", including the lack of any, to be thanked for many uglier things. Comments on this subject are not math related and I think should not belong here. – Yaakov Baruch Jul 11 2010 at 11:08
But I thank the Church too. – timur May 13 2011 at 3:31
All the answers given are essentially correct - motion in inertial and non-inertial coordinate systems is governed by quite different laws, which are pretty easy to observe. Recall, that Newton's laws (at least the first two) hold only in inertial coordinate systems.
I just wanted to point out that this is a fundamental reason why we choose the Sun as a center - this coordinate system is more inertial. Not just because "equations look simpler". "Occam's razor business" also has nothing to do with this.
Question itself is a bit anachronic - people knew this many centures ago.
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Stellar aberration, the change in the apparent positions of more or less most of the night sky with the seasons, directly due to the velocity of the earth in its orbit. Note - aberration, which is related to the idea that raindrops appear to fall slanted in a moving vehicle, is not the same as parallax, e.g. has no relation to the distance of the stellar source, etc. :-)
See http://en.wikipedia.org/wiki/Aberration_of_light . Predicted by Bradley in 1725. First measured > 0 by Bessel in 1838, with unpublished successful observations by Henderson 5 years earlier. See http://thonyc.wordpress.com/2010/01/09/we-know-it-moves-but-can-you-prove-it/ .
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Isn't the whole idea why we accept the 'fact' that the Earth revolves around the sum based on how science works?
You have a theory about how something works, do some experiments, and if you don't get a contradiction, you do not reject the theory. I think that if you investigated any of the theories that placed the Earth at the center of the Universe, you would eventually find a contradiction. When you place the Sun at the center of the Solar System, no contradictions develop ( except for the thing about the behavior of Mercury - ask Eddington and Einstein).
Actually, when you listen to phyicists about why they accept quantum mechanics, they will say that it is a theory that has never encountered a contradiction over the past eighty years. However, they are still looking to a theory to unit it with general relativity, but that is another story.
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The question is asking "what theory? what experiments?" – Qiaochu Yuan Jul 10 2010 at 16:57
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http://physics.stackexchange.com/questions/tagged/metric-tensor+units
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Tagged Questions
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Einstein tensor in Friedmann equations : where is the missing $c^2$?
I would like to demonstrate the several forms of the Friedmann equations WITH the $c^2$ factors. Everything is fine ... apart that I have a missing $c^2$ factor somewhere. In all the following $\rho$ ...
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http://physics.stackexchange.com/questions/16281/what-is-the-probability-distribution-for-the-angle-of-an-approximate-laserbeam
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# what is the probability distribution for the angle of an approximate laserbeam
I'm trying to simulate the light distribution characteristics from a Gaussian laser beam, but having difficulty with the angular distribution.
I need to generate a large number of points on an x/y plane, along with pointing vectors down the z-axis, such that their aggregate approximates the power distribution of a laser beam. The distribution on the x/y plane is a Gaussian, while the pointing vector down the z-axis approximates the laser divergence.
For example, a beam with a 1 mm beam waist with a 1.5 mrad divergence has the following beam irradiance, E(r):
$E(r) = exp(-r^2/b^2)/(\pi b^2)$
where $r$ is the radius from the beam center, $b$ is the beam waist (1/e radius). Therefore, I can sample from this distribution by the following equation:
$r = b\sqrt(-ln(1-\mathbb{R}))$
where $\mathbb{R}$ is a random number uniformly distribution on [0,1], to get the radius from the origin on the beams starting point on the x/y plane.
OK, so the question I'm having trouble with, is how do I randomly choose the polar angle $\theta$ of divergence such that it approximates the laser divergence of 1.5 mradians? Do I just choose the polar angle to be uniform on $[0,1.5e^{-3}]$ radians?
I guess I'm getting confused by whether the distribution should be uniform over polar angle, or uniform irradiance over the solid angle, and how to sample from that. Below is an illustration to help sort things out. I'm trying to determine how to distribute the polar angle in order to approximate the laser divergence.
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## 1 Answer
This process is treated in this paper: http://www.springerlink.com/content/2ww1gtp5cvbrerhm/fulltext.pdf
The relevant equations are:
$x_s=\frac{w(-d)}{\sqrt{2}}\text{Erf}^{-1}(2r_1-1)$
$y_s=\frac{w(-d)}{\sqrt{2}}\text{Erf}^{-1}(2r_2-1)$
$x_f=\frac{w_0}{\sqrt{2}}\text{Erf}^{-1}(2r_3-1)$
$y_f=\frac{w_0}{\sqrt{2}}\text{Erf}^{-1}(2r_4-1)$
These give the x and y positions of rays at the beam waist ($x_f$) and at a point far away ($x_s$), with d much greater than the Rayleigh range. $w(z)$ is the standard gaussian beam waist formula, and the $r_i$ are uniform random variates on (0,1).
EDIT: Realized I didn't really address your divergence issue. The above is valid for a collimated diffraction limited beam, so to add excess divergence you could apply the ABCD matrix for a thin lens to the rays in order to create a diverging beam.
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Thanks for the added clarification. I just read through the paper and it doesn't seem obvious how to get an angular divergence from the equations for position. In other words, I want to create a group of x,y points with angles, not a group of x/y points on the receiver plane. – gallamine Oct 31 '11 at 13:59
@gallamine - just get the x/y points at the focal plane, then use the x/y points for those rays at the farfield plane to calculate the angles. – user2963 Oct 31 '11 at 14:05
– user2963 Oct 31 '11 at 14:12
thanks. The only difficulty is that those equations only apply to distances >> Rayleigh range, and I'm working at shorter distances. I'm thinking now I'll use a Gaussian distribution on the start plane, with perfect colimation then use the Thin Lens ray transfer matrix to diverge them. Make sense? I'll try and post an answer. – gallamine Oct 31 '11 at 15:21
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– user2963 Oct 31 '11 at 16:06
show 1 more comment
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http://mathoverflow.net/questions/77175?sort=newest
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## Taking “Zooming in on a point of a graph” seriously.
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
In calculus classes it is sometimes said that the tangent line to a curve at a point is the line that we get by "zooming in" on that point with an infinitely powerful microscope. This explanation never really translates into a formal definition - we instead approximate the tangent line by secant lines.
I seem to have found a way to obtain tangent lines (and more) by taking "zooming in" seriously.
Example 1
Take the curve $y = x(x-1)(x+1)$.
I want to find an equation for the tangent line to this curve at the origin. So I zoom in on the origin with a microscope of magnification power $c$ (i.e. I stretch both vertically and horizontally by a factor of $c$) to obtain
$\frac{y}{c} = \frac{x}{c}(\frac{x}{c} - 1)(\frac{x}{c}+1)$.
Multiplying through by $c$ I have
$y = x(\frac{x}{c} - 1)(\frac{x}{c}+1)$
Now letting my magnification power go to infinity I have
$y = -x$
Which is the correct answer.
Example 2
Take the curve $y = x^2$.
I want to find an equation for the tangent line to this curve at the point (3,9). I first rewrite the equation as
$(y-9) + 9= ((x-3) + 3)^2$
so that I am focusing on the appropriate point. To zoom on this point with magnification $c$ I have
$\frac{y-9}{c} + 9 = (\frac{x-3}{c} + 3)^2$.
$\frac{y-9}{c} + 9 = \frac{(x-3)^2}{c^2} + 6\frac{x-3}{c} + 9$ Multiplying through by $c$ I have
$y - 9 = \frac{(x-3)^2}{c} + 6(x-3)$
Now letting my magnification power $c$ go to infinity I have
$y - 9 = 6(x-3)$
Which is the correct answer.
Example 3
Here is the example which actually motivated me to consider this at all:
Take the curve $y^2 = x^2(1 - x)$.
This is a cubic curve with a singularity at the origin, and so it doesn't really have a well defined tangent line. It sort of looks like it should have two tangent lines (y = x, and y = -x), but it is a little bit tricky to formalize this. Let's see what "zooming in" does:
$\frac{y^2}{c^2} = \frac{x^2}{c^2}(1 - \frac{x}{c})$
$y^2 = x^2(1 - \frac{x}{c})$
Letting $c$ go to infinity I have
$y^2 = x^2$, or $(y-x)(y+x) = 0$, which is the pair of lines I desired.
My Questions
1. Do any books take this approach when developing the derivative?
2. I would imagine that algebraic geometers do this kind of thing formally. Is there a more rigorous analogue of the prestidigitation I engage in above? Where would I look to read up on such things?
p.s. It would be nice to illustrate each of these examples with a little movie of the "zooming in" process, but I am not sure how to put such things on MO. Any hints?
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Off the top of my head: on the one hand, this looks to me like "just" being the local linear approximation definition of the derivative (cf. multivariable defns), but I've not seen it presented/motivated this way before. In partial answer to your Q2, I guess you could work in the ring $R[\epsilon]$ with $\epsilon^2=0$. – Yemon Choi Oct 4 2011 at 21:02
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In a sense, the $\epsilon$-$\delta$ definition of differentiability is doing exactly this zooming-in process. But instead of talking about the zoomed-in version of the function, you're saying $f(x)$ is inside the "cone" $f(a)+f'(a)(x-a) \pm \epsilon (x-a)$ where epsilon can be made arbitrarily small -- and once you make that choice you suitably restrict $|x-a|$. – Ryan Budney Oct 4 2011 at 21:05
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This sort of homothetic scaling is the standard way to think about tangent objects in geometric measure theory. You are just looking at the tangent cone to the graph of your function. – Rbega Oct 4 2011 at 21:09
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This is explored seriously in Keisler's nonstandard calculus text Elementary Calculus; his treatment of the microscope methodology is based on work of Keith Stroyan. – Robert Haraway Oct 4 2011 at 21:11
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I find your approach to derivatives really cool! Have you tried to interpret what you are doing in terms of projective geometry? (the substitution x---> x/c; y ---> y/c really reminds me of the homogenization process) – Tommaso Centeleghe Oct 4 2011 at 21:23
show 5 more comments
## 8 Answers
In algebraic geometry, this construction is known as the tangent cone to the graph. More generally, suppose we have the zero set of any polynomial $f(x,y) = 0$, and assume $f(0,0)=0$. Then we can write
$f(x,y) = a_m (x,y) + a_{m+1}(x,y) +a_{m+2}(x,y) +\cdots$
where $a_i(x,y)$ is a homogeneous polynomial of degree $i$ and $a_m$ is nonzero. The zero set of $a_m$ is called the tangent cone to the curve at the origin. It is a product of $m$ linear forms (over $\mathbb{C}$), and $m=1$ exactly when the zero set is smooth at the origin. In this case, the tangent cone coincides with the tangent space.
From your point of view, when we substitute $x\mapsto x/c$ and $y\mapsto y/c$ it is clear that the term left in the limit is $a_m$.
We can of course find tangent cones at other points of the zero set by changing coordinates.
In general, for a smooth function $f$ you should be able to take a multivariate Taylor expansion and read off the tangent cone from the lowest degree part. This is where the difficulty comes in for actually defining the tangent line in terms of the tangent cone in a calculus class, as computing the Taylor expansion demands we already have a notion of derivative. This difficulty is obviously not seen in the case of polynomials, although recentering the Taylor expansion of a polynomial at a different point is perhaps easiest done with the aid of derivatives.
Higher dimensional analogues are also available without any real work, although in the singular case the tangent cone is much more interesting than just a union of hyperplanes: it will be a cone over some variety. The homogeneous polynomial $a_m(x_1,\ldots,x_n)$ typically doesn't factor into a product of linear forms when $n>2$.
Tangent cones are treated in any reasonable introduction to algebraic geometry, such as Harris' "First course" book or Shafarevich.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Hugh Thurston explored this point of view (or, rather, the tangent-cone reformulation described by Jack Huizenga) in a number of articles in Amer. Math. Monthly and Math. Mag. in the mid-'60's: http://www.ams.org/mathscinet/search/publications.html?pg1=AUCN&s1=Thurston%2C+H%2A&co1=AND&pg2=ALLF&s2=tangent.
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@L Spice: I tried to edit your post by fixing the second link but the MO editor did not like the "H*" for the first name in the author search. I changed it to "Hugh" but that drops two out of the five found items: the two papers in the Monthly. – Abdelmalek Abdesselam Feb 5 at 17:45
@Abdelmalek Abdesselam, I hadn't noticed that the link was problematic; thanks! I think I've fixed it now by percent-encoding the *. – L Spice Feb 5 at 18:25
For freshman calculus students, I try to explain we're just putting a "microscope" to the different curves and we usually get one of four pictures.
But picturing the kinds of pathologies that can occur is much harder as you move beyond calculus. The Cantor function. We can picture the Cantor function as flat if you zoom in outside the Cantor set, but the "devil's staircase" if you zoom in at a Cantor set point. Brownian motion will never calm down" no matter how much you zoom in -- it does not have bounded variation.
It would be nice to have a picture book of these pathologies.
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I would say you have not applied the "microscope" to the first of your examples: it would become a horizontal line... Also the last example seems not to be quite right (as a result of a zooming in process). – Qfwfq Oct 5 2011 at 19:11
Also, the four pictures don't seem to be exhaustive of all the possible behaviours of "elementary functions". – Qfwfq Oct 5 2011 at 19:18
no they are not... but freshman students are not even aware of these yet. – John Mangual Oct 5 2011 at 19:40
As AmbroseH commented, Keisler's book takes this pedagogical approach. The catch is that his book is old, out-of-print, and uses the infintesimal approach. He talks of "infinite microscopes" and "infinite telescopes" (the latter for horizontal and vertical asymptotes of curves) and has illustrations (sadly without color or animation). You can download the book from his website (Creative Commons License).
You can find more recent notes (not full textbooks yet) that sketch how to teach calculus using various flavors of infinitesimals (without sacrificing a rigorous mathematical foundation):
1. Keisler uses Robinson-style nonstandard analysis. A more recent variation on this is relativized nonstandard analysis. Instead of two "levels," standard and nonstandard, it's turtles all the way down. See these slides for an introduction and citations.
2. A very different flavor is smooth infinitesimal analysis, which uses nilsquare infinitesimals (and may use intuitionistic logic). Here is one exposition.
3. The idea of nilsquare infinitesimals generalizes to nilpotent infinitesimals. This is discussed here, for example.
Since you tagged your question algebraic geometry, my guess is that you'd prefer option 2 or 3.
Finally, you might find this relevant; there's an even older (but not out of print) calculus book that has wonderful pedagogy but implicitly throws around nilsquares without any pretense of rigor.
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An animation of your first example, $y=x(x−1)(x+1)$. The animation is flawed in that it jumps jarringly when the tick labels reach the boundary. I removed the ticks and labels, which gains smoothness at the expense of magnitude feedback. (The previous version is here.) And I limited the number of frames so that the file would not be too huge (it is ~1MB now). Frame rate is browser and processor dependent. At best this gives an idea of what a more professional animation might look like.
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Awesome. – Steven Gubkin Oct 5 2011 at 16:55
The tangent space $T_pM$ of a Riemannian manifold $M$ with inner metric $d$ can be realized as the pointed Gromov-Hausdorff limit as $\lambda\rightarrow\infty$ of $(\lambda M, p)$ where $\lambda M$ is $M$ with metric $\lambda d$.
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I haven't seen any Calculus books take this approach. Possibly, this is because it is not clear how to proceed in the case of functions that are not polynomials. Consider even the really nice function $y = e^x$ and say you are interested in the tangent line to the graph at (0,1).
Using your scheme and the fact that $e^{a+b} = e^a\cdot e^b$, one writes $\frac{y-1}{c} = e^\frac{x}{c}$. This leaves us with
$$y = 1 + c \cdot e^\frac{x}{c}$$
While taking the limit as $c \to \infty$ certainly gives the correct answer $y = 1+x$, I am not sure how to actually compute this limit without using either Taylor series or L'Hospital's rule, both of which already require you to know the derivative of $e^x$.
Of course you will run into similar difficulties if you try to compute the derivative of this function from the limit-based definition directly. If I remember correctly, standard textbooks on Calculus get around this annoyance either by declaring the answer without proof or by using implicit differentiation and $\ln(y) = x$. It is not too difficult to obtain the derivative of the natural log function from first principles.
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Well, if you say what $e$ is without attempting to motivate it, then it's easy enough to give a proof. That's the approach of the calculus book I'm using (Stewart). – Frank Thorne Oct 4 2011 at 23:46
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Small mistake: you should have `$\frac{y-1}{c} + 1 = e^{\frac{x}{c}}$` so that `$y = 1 + c(e^{\frac{x}{c}} - 1) \to 1+x$`. – Faisal Oct 5 2011 at 1:23
Nonstandard analysis handles things like $\lim_{c\rightarrow\infty} c\cdot e^{\frac{x}{c}}$ by first defining real powers as standard parts of finite hyperrational powers. My approach is to prove that the standard part of $\frac{a^\delta-1}{\delta}$, where $\delta$ is an arbitrary nonzero rational infinitesimal and $a$ is a standard postive real, equals $\int_1^a\frac{dx}{x}$. Then I define $e$ via $\int_1^e\frac{dx}{x}=1$. The pay-off is that $d(e^x)/dx=e^x$ becomes true by definition. – David Milovich Oct 5 2011 at 17:50
As Rbega says in the comments, if you are really keen to see this rescaling idea put to use in a more rigorous or advanced way, then you can look at some Geometric Measure Theory. While it will look very technical compared to this (because it is designed for potentially badly-behaved or very weakly-defined geometric objects), this sort of homothetic blowing up is standard for defining tangent objects to things. You get a weak kind of convergence of the rescalings of your original object to the tangent object, which, depending on the circumstances, may well (or perhaps will hopefully) display some sort of rigidity, e.g. it may be have to be a cone. It is rigorous and yes you can indeed end up with things like the union of two lines as your tangent object.
In the special case of the graph of a differentiable function, the tangent object at a point will indeed be the graph of the affine function associated with the derivative at the point.
I don't know of any books which take this approach pedagogically, in the development of calculus though.
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I felt I should add that, one of the things that happens with working with such a weak notion of tangent is that tangent objects need not be unique. In other words instead of scaling by a continuous variable $\epsilon\to 0$ you scale by a sequence of numbers $\epsilon_i\to 0$ then a different sequences of $\epsilon_i'\to 0$ might give you a different tangent object. Showing this can't happen in "good" situations is one of the hardest problems in the field... – Rbega Oct 4 2011 at 23:47
Cool! I will have to check out some books on Geometric Measure Theory. – Steven Gubkin Oct 5 2011 at 16:53
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http://math.stackexchange.com/questions/215059/two-non-isomorphic-root-field-extension-of-the-field
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# two Non isomorphic root field-extension of the field.
Does there exist two non isomorphic minimal field extension ( root field) of $f= \frac {x^{64}-x}{x(x-1)} \in F_2[x]$ .
I may be using wrong word here saying minimal field extension but in german its said as "minimal würzelkörper" .
Any help would be appreciated .
Attempt : I tried factorizing using maple into irreducible factors . And using the fact that if we have a field of degree $p^n$ then the subfield must have degree $p^m$ where $m$ divides $n$ , and if this subfield contains the root of the irreducible polynomial then its the minimal root field ( field extension ) . Where am i going wrong ?
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## 1 Answer
I think you may be referring to "splitting field" = (the/a) minimal extension field containing all the roots of a given polynomial over some certain field, which you haven't provided. Anyway:
$$\frac{x^{64}-x}{x(x-1)}=\frac{x(x-1)(x^{62}+x^{61}+...+x+1)}{x(x-1)}=x^{62}+x^{61}+...+x+1$$
Now, $\,62=2^6\Longrightarrow\,$ the splitting field of $\, x^{64}-x\,$ over the prime field $\,\Bbb F_2:=\Bbb Z/2\Bbb Z\,$ is the (unique, up to isomorphism of fields) field $\,\Bbb F_{2^6=64}\,$ of degree $\,6\,$ over $\,\Bbb F_2\,$.
Since in the above simplification we've cancelled the linear factors for $\,x=0,1\,$ and we must have these two elements in any field, we can see that the minimal extension field of $\,\Bbb F_2\,$ that contains ll the roots of the original rational function is the forementioned field $\,\Bbb F_{64}\,$ .
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@DanAntonio : Sir, I am just talking about the extension field not the splitting field. How can i find two extensions $F(\alpha)$ and $F(\beta)$ which are not isomorphic ?? – Theorem Oct 16 '12 at 22:58
Extension of what field? What does the expression $\,\frac{x^{64}-x}{x(x-1)} \,$ have to do with all this? In general, for a field $\,\Bbb F\,$ and two elements $\,\alpha, \beta\,$ in some algebraic extension, the fields $\,\Bbb F(\alpha)\cong\Bbb F(\beta)\,$ iff $\,\alpha,\beta\,$ are roots of the same irreducible polynomial in $\,\Bbb F[x]\,$ . If one of $\,\alpha,\beta\,$ is not algebraic over $\,\Bbb F\,$ then things can be even harder... – DonAntonio Oct 17 '12 at 2:58
Sir sorry for less information , here the field is $F_2$ , $f\in F_2[x]$ – Theorem Oct 17 '12 at 5:32
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http://mathoverflow.net/questions/25848?sort=oldest
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## Shortest Paths on fractals
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
How can one find shortest paths between 2 specified points on fractals, or (since I'm pretty sure this is quite complicated) make useful generalizations about them?
Since the above question is broad, how about this one: What is the general formulation (in a direct equation, recursive formulation, or other form) for distance between 2 points on the sierpinski carpet?
Obviously for some fractals all points are infinite. Identifying these is often easy, but are there any edge cases where it's hard to decide whether all paths are infinite length? And if so, how does one decide?
Edit: This question was inspired, by the way, by this thread on a different website (where it became clear that it was beyond the average math knowledge there). http://echochamber.me/viewtopic.php?f=3&t=40348#p1618494 That particular post shows paths(whose presence is recursive) in the carpet.
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You might be interested in [this](portal.acm.org/citation.cfm?id=1596550.1596555) paper which discusses an application of shortest paths on the Sierpiński graphs, ie. the finite graphs approximating the Sierpiński gadget. – Dan Piponi May 25 2010 at 16:49
## 2 Answers
As to the border case. An example that you might like to consider is given by the blanc-mange curve, $f_{\lambda}:\mathbb{R}\to\mathbb{R}$, that for any value of $0\leq\lambda\leq 1,$ is defined as the unique bounded solution of the fixed point functional equation
$f(x)=\mathrm{dist }(x,\mathbb{Z})+\lambda f(2x)$
(by the contraction principle there is a unique such function; it is continuous and 1-periodic, with an immediate series expansion coming from the iteration).
Consider $f_{\lambda}$ on the unit interval. If you take $\lambda=1/4$ you find a parabola; with $\lambda < 1/2$ it's Lipschitz (hence the graph has a finite length) with constant $(1-2\lambda)^{-1}$; if $\lambda > 1/2$ it's Hoelder continuous with an exponent depending from $\lambda$. The parameter 1/2 is critical: you find a curve that is not Lipschitz, but it's Hoelder of all exponents $\alpha>0$, precisely, it has a modulus of continuity ct|log(t)|, and looking at it a bit more closely, it is not of bounded variation on any nontrivial interval (so the graph is not rectifiable even locally), nor is BV for any $\lambda \geq 1/2.$
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As to the geodesic distance in the Sierpiński carpet S. A nice problem there is! Though it seems difficult. The geodesic distance function d(x,y) should satisfy a functional equation due to the self-similarity of S. But I don't see enough information to identify it. Note that two opposite corners (0,0) and (1,1) may be connected easily because the straight segments from (0,0) to (0,1/2) and from (0,1/2) to (1,1) are completely included in S. This make a length 2sqrt(5)/3 that in this case is clearly minimal. I imagine that S contain a dense set of line-segments, indeed. – Pietro Majer May 25 2010 at 14:51
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
For the Sierpinski carpet, using the metric induced by its embedding in $\mathbb{R}^2$ is quite difficult computationally. There are alternative metrics: for fractals arising from hyperbolic iterated function systems (of which the Sierpinski carpet is one), one can identify points via its tops code space, which involves determining which subcopies of the Sierpinski carpet a point lies in. It's not difficult to place a metric on the tops code space of such a fractal, and this in some sense tells us how "close" two points are, in terms of how many iterations of the function system have to occur before the two points are sent to different subcopies of the fractal. It's a bit difficult to make this more precise without going into a fair bit of detail; I'm pretty sure this is all covered in Michael Barnsley's textbook Fractals Everywhere though. Note also of course that this only works for fractals arising from hyperbolic iterated function systems, which includes other fractals like the Koch curve and the Menger sponge.
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http://mathematica.stackexchange.com/questions/5519/solving-an-equation-and-finding-a-couple-of-results-given-that-they-are-integer/5520
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# Solving an equation and finding a couple of results given that they are integer
My problem is to find all solutions to an equation with 2 variables between two values that define the range of my search. e.g., $(x+y-y²)/(1-y-x²)=0$ with solutions ${x,y} \in \mathbb{Z}$.
-
## 2 Answers
The following might be of use:
````Reduce[x + y == y^2 && x^2 + y != 1, {x, y}, Integers]
````
To ease things, we consider the numerator and denominator separately. Here, we are asking Mathematica for conditions on integer `x` and `y` such that the numerator is zero, and the denominator isn't.
````Solve[x + y == y^2 && x^2 + y != 1, {x, y}, Integers]
````
is useful as well.
If you need examples, you can use
````FindInstance[x + y == y^2 && x^2 + y != 1, {x, y}, Integers, 10]
````
Change `10` to the number of examples you wish to generate.
-
thanks the tip of separating numerator and denominator is great ;) – sol May 14 '12 at 10:36
and for the FindInstance 10 that's what I needed! – sol May 14 '12 at 10:47
You can obtain all integer solutions with `Reduce`:
````Reduce[(x + y - y^2)/(1 - y - x^2) == 0, {x, y}, Integers]
````
which gives
If you also want to restrict to solutions satisfying $x<15$ and $y<42$ then this works
````Reduce[(x + y - y^2)/(1 - y - x^2) == 0 && x < 15 && x < 42, {x,y}, Integers]
````
-
ok thanks and if I want to condition it so that x and y are between [O,1] (how to condition the solving process?) – sol May 14 '12 at 10:29
you can do that in a way analogous to the second example I gave – acl May 14 '12 at 10:34
yep thanks Acl ^^ – sol May 14 '12 at 10:37
lang-mma
|
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|
http://windowsontheory.org/2012/11/27/755/
|
## A Research Blog
by
Computational learning is full of problems that are deceptively simple to state but fiendishly hard to solve. Perhaps none more so than the problem of learning Juntas, posed by Avrim Blum and Pat Langley. Its the kind of problem which seems well suited for a polymath endeavor, Dick Lipton likes to say that you could think about it while running for the bus. Alas, I haven’t ridden many buses here in the Bay Area (and coincidentally haven’t had any good ideas about this problem).
But what I’d like to talk about in this post is a variant of the problem which I think is kind of cute. The question is from a paper by Mossel, O’Donnell and Servedio, this particular formulation came up in discussions with Ryan O’Donnell, Amir Shpilka and Yi Wu (sorry guys, I know you’d rather not be associated with this one).
Let us begin with the problem of learning Juntas. The setup is that of learning an unknown Boolean function under the uniform distribution: we are given random examples from $\{0,1\}^n$ labelled by some function $f:\{0,1\}^n \rightarrow \{0,1\}$. Our goal is to learn $f$. This is only possible if we assume that $f$ has some nice structure. In the Junta learning problem, we are told that $f$ only depends on k out of the n input variables, where we think of $k \leq \log n$ (or just anything $\omega(1)$). In fact, just find one relevant variable and you’re done.
It is trivial to solve the problem in time $O(n^k)$. The first non-trivial algorithm for this problem was given by Mossel, O’Donnell and Servedio, it runs in $n^{\omega k/(\omega +1)}$, where $\omega$ is the matrix
multiplication exponent. So it will never do better than $n^{2k/3}$. But (as one of the authors was kind enough to point out), it gives you something interesting even for $\omega =3$. Or 4. Or more. An algorithm witha better running time than MOS (and better than $n^{2k/3}$) was recently found by (Greg) Valiant. Better bounds are known for symmetric functions due to Lipton, Markakis, Mehta and Vishnoi and Shpilka and Tal.
Here is a bare-bones sketch of the MOS algorithm. A natural approach to finding a relevant variable is to compute correlations between the n input variables and $f$, if we find a correlation we have a relevant variable. If that fails, we could try pairs of variables, triples of variables and so on, essentially running the low degree algorithm of Linial, Mansour and Nisan. A cheeky way to defeat this algorithm (at least for a while) is to take some function, say Majority on $2k/3$ bits and XOR it with Parity on $k/3$ bits. This has the effect of making the resulting function uncorrelated with every subset of size less than $k/3$. However, note that the resulting function is not very complex when viewed as a polynomial over GF[2]. Even if Majority had full degree over GF[2] (it does not), the resulting function would only have degree $2k/3$, since the Parity part does not increase degree at all.
The MOS algorithm relies on the insight that these are (essentially) the only functions that evade the low-degree algorithm: any such function must have low degree as a polynomial over GF[2], and is easy to learn using polynomial interpolation over GF[2]. The precise result they use is due to Seigenthaler from 1984.
Much as I love polynomials over GF[2], one can’t help wondering how robust this approach is. Suppose we were trying to learn a Boolean function but over an ternary alphabet. Would the right analogue be polynomials over GF[3]? Over an alphabet of size 4 (A,T,G,C), should the polynomials be over GF[4] or $Z_4$? And I’d rather not think about alphabets of size 6.
The larger alphabet is mentioned as an open problem in MOS, and as far as I know, no one has managed to do anything with it (it is quite likely that no one has tried too hard). But I think this might turn out to be a worthwhile digression that could shed some light on the original problem. Here is a particular formulation of the alphabet 4 question that I like:
Learning $k \times 2$ juntas: There are n pairs of variables, $(X_1,Y_1),\ldots,(X_n,Y_n)$. We
are given uniform examples of an unknown function $f$ which depends only on k pairs out of n. Our goal is to learn $f$.
It is trivial to do it in time $O(n^k)$. One could view it as a $2k$-junta, but would need an algorithm for the $k$- junta problem that runs faster than $n^{k/2}$. Is it possible to do better? More ambitiously, can one match the running time for the k-junta problem, perhaps by a better reduction to learning $k$-juntas?
Finally, I am told there is money on offer (not from me) for solving the following seemingly innocuous problem: suppose you know that the function is Majority on $2k/3$ bits XORed with Parity on $k/3$ bits. All you need to do is find out which the relevant bits are. It is possible to do this in time $O(n^{k/3})$, in fact there are a couple of ways. Can you do any better?
|
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|
http://mathoverflow.net/questions/109707/lyapunov-function-on-chains
|
## Lyapunov function on chains
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
How can I prove that if $x R y$ (for each $e$ there is a $e$-chain from x to y) then for each lyapunov function $L$ we have $L(x)>= L(y)$.
-
Homework? Read the FAQ – Anthony Quas Oct 15 at 11:06
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|
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