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http://mathoverflow.net/questions/55575?sort=oldest
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## Why is the dual of a torus the same as its fundamental group?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
The set of continuous homomorphisms from a torus ${\mathbb T}^n = ({\mathbb R}/{\mathbb Z})^n \to {\mathbb R}/{\mathbb Z}$ can be identified with ${\mathbb Z}^n$ if we assign to each $k = (k_1, \ldots k_n) \in {\mathbb Z}^n$ the character $x \mapsto k \cdot x$.
The fundamental group of homotopy classes of loops ${\mathbb R}/{\mathbb Z} \to {\mathbb T}$ can also be identified with ${\mathbb Z}^n$ because each equivalence class (with base point at the origin) can be represented by $x \mapsto (k_1 x, k_2 x, \ldots, k_n x)$ for some $k \in {\mathbb Z}^n$.
My question is basically how much of a coincidence this isomorphism is. For one thing, it can't be too natural because the fundamental group pushes forward under a map, whereas the character group pulls back. So the natural question should be whether there is a natural relationship at the level of the first cohomology group instead of the fundamental group.
Of course, totally disconnected groups have interesting dual groups even though their cohomology is uninteresting as far as I know. And ${\mathbb R}^n$, being isomorphic to its dual but contractible, does not seem to exhibit a similar relationship.
But for a Lie group, say, I would like to know if there's a natural relationship between its representation theory (e.g. irreducible representations) on the one hand and its topology (e.g. cohomology) on the other hand. It might be no deeper than "well, the cohomology groups are representations".
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1
I guess you may be referring to Pontryagin duality. My impression is that results on the representations of Lie groups by Harish-Chandra and others are an attempt to generalize Pontryagin duality to the non-commutative category. – Agol Feb 16 2011 at 21:22
1
Re: "cohomology is uninteresting" see en.wikipedia.org/wiki/Group_cohomology – Sam Nead Feb 16 2011 at 21:55
## 3 Answers
The two are naturally dual lattices. The fundamental group of a torus $T$ can be canonically identified with the group (known as the cocharacter lattice) of $\it homomorphisms$ from the circle group to $T$, or equivalently the kernel of the (universal cover=exponential map) homomorphism from the Lie algebra $t$ to $T$. The characters of the torus on the other hand can be identified naturally with a subsest of $t^*$, i.e. characters of the Lie algebra which are integral on the kernel of $t\to T$, i.e. the dual to the cocharacters.
As for the general question about Lie groups it seems way too general (note that a connected group acts trivially on its cohomology groups, so the relation is not that..) One direction to read about is the relation between the cohomology of the group, that of its classifying space, group cohomology and invariant polynomials (or "Casimirs") on the Lie algebra.
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Thanks! Very helpful. – Phil Isett Feb 16 2011 at 4:42
@David: Do you know a good place to read about those things and their relations? – ndkrempel Feb 18 2011 at 1:08
### 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 sure if this is something you're looking for:
There is a natural bijection $[X,K(G,n)] \rightarrow H^n(X;G)$ for any commutative group $G$ and any CW-complex $X$.
In particular, for $G = \mathbb{Z}$ and $X=\mathbb{T}^n$ we have the bijection $[\mathbb{T}^n,\mathbb{R}/\mathbb{Z}]\rightarrow H^1(\mathbb{T}^n)\cong H_1(\mathbb{T}^n)\cong\pi_1(\mathbb{T}^n$).
[] To establish the bijection with $Hom(\mathbb{T}^n,S^1)$, we can use Pontryagin duality to establish $Hom(\mathbb{T}^n,S^1)=Hom(\widehat{S}^1, \widehat{\mathbb{T}}^n)=Hom(\mathbb{Z},\mathbb{Z}^n)=Hom(\mathbb{Z}^n,\mathbb{Z})$. But $[\mathbb{T}^n,S^1]=H^1(\mathbb{T}^n)=Hom(\pi_1\mathbb{T}^n,\mathbb{Z})=Hom(\mathbb{Z}^n,\mathbb{Z})$, giving us the desired bijection.
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I don't see why the set of homotopy classes $[\mathbb{T}^n,\mathbb{R}/\mathbb{Z}]$ should be equal to the set of group homomorphisms $\mathbb{T}^n \to \mathbb{R}/\mathbb{Z}$. (Also, some argument is needed to show that $\pi_1$ is abelian, e.g. the Eckman-Hilton lemma.) – Dan Petersen Feb 16 2011 at 8:35
2
The fact that $\pi_1$ is abelian is well-known and trivial to show by the product-formula... But yes there should be more to say on why the set of homotopy classes is equal to the set of continuous group homomorphisms... for this, a rather twisted way to go about it I will place in an edit. – Chris Gerig Feb 16 2011 at 20:42
Some of your equalities should be isomorphisms. – S. Carnahan♦ Feb 17 2011 at 3:07
This is just a minor elaboration on David Ben-Zvi's answer. You can see the duality between the fundamental group of $T$ and the character lattice by composing based loops $\mathbb{R}/\mathbb{Z} \to T$ with characters $T \to \mathbb{R}/\mathbb{Z}$. You end up with based loops in $\mathbb{R}/\mathbb{Z}$, whose homotopy classes are completely determined by an integer invariant, namely the degree. You can view the degree algebraically as the induced homomorphism on $\pi_1(\mathbb{R}/\mathbb{Z},0) \cong H_1(\mathbb{R}/\mathbb{Z}, \mathbb{Z})$, or geometrically as the winding number if you choose an isomorphism with an oriented circle in the plane. This yields a perfect pairing between $\pi_1(T,0)$ and $X^*(T)$.
In the nonabelian situation, one can see that the fundamental group of a connected topological group doesn't determine that much about the representation theory. There are plenty of simply connected groups with fairly complicated representation theory. However, the K-theory of the classifying space can say a lot about representations.
Regarding the title of the question, one should be careful about calling two groups "the same", even if they are isomorphic.
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http://math.stackexchange.com/questions/98237/approaching-the-circumference-of-a-circle?answertab=votes
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# Approaching the circumference of a circle
We had a lengthy discussion yesterday, on how to prove that the circumference of a circle is $2\pi r$. By using google, the most commonly found proof starts in the following way.
"Consider the regular n-gon inside the circle touching the circle at its vertices, and the regular n-gon outside the circle touching the circle at its edges. Then the circumference of the inner n-gon is smaller than the circumference of the circle, which is smaller than the circumference of the outer n-gon."
While this apparently is true for n-gons, it is not true for arbitrary geometric shapes inside and outside the circle. So we conjectured it was the convexity of the n-gon that is of importance. Which leads to the question:
Let $C$ be a circle with radius $r$, and $A_n$ be a family of convex sets with $A_n \subset C$ for each $n$. If, for $n \rightarrow \infty$, the volume of these sets converges to $\pi r^2$ (the volume of $C$), will the circumference necessarily converge to $2\pi r$?
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1
If the sets $A_n$ need not be closed, the answer is no. Let the left hemisphere of the circle be filled in, but without boundary. Let the right half of the hemisphere be the right half of an $n$-gon. Then, when $n \to \infty$, the area converges to $\pi r^2$, but the area converges to $\pi r$. – JavaMan Jan 11 '12 at 18:27
Note also that the question seems ill-posed. For example, one could probably also concoct sets where the volume converges but the circumference neet not converge. A proper formulation of the question may then ask one to prove that for closed convex sets $A_n \subset C$, we have $$\lim_{n \to \infty} vol(A_n) = \pi r^2 \quad \implies \quad \limsup_{n \to \infty} \text{ }Circ(A_n) = 2 \pi r$$ where $vol(A_n)$ and $Circ(A_n)$ obviously stand for the volume and circumference of $A_n$, respectively. – JavaMan Jan 11 '12 at 18:30
I'm actually not sure what the precise definition of the circumference of an arbitrary set $A \subset \mathbb{R}^2$ is, but I would assume its some function of its boundary, which should always be closed. In this case, in your example above the circumference would converge to $2 \pi r$, unless I'm mistaken? – Benno Jan 11 '12 at 18:41
Probably the right term would be "perimeter" instead of "circumference". In this case this is the one-dimensional Hausdorff measure of the boundary of the set. – Dirk Jan 11 '12 at 19:24
@Benno: You are right. My counterexample yields a perimeter of $2 \pi r$ if by perimeter you mean the $1$-dimensional Lebesgue measure of the boundary of the set $A = \lim_{n \to \infty} A_n$. In my counterexample, I was instead thinking of the left hemisphere as not contributing at all to the perimeter. This made intuitive sense to me at the time, but it depends on how you define perimeter. – JavaMan Jan 11 '12 at 22:15
show 3 more comments
## 1 Answer
Yes.
If you don't mind, I'll do the problem in $\mathbb R^d$. Write $B_2^d$ for the unit Euclidean ball (centred at the origin, radius 1). Let $A_n$, as before, be a sequence of convex subsets of $B_2^d$ with $V(A_n)\to V(B_2^d)$, where $V(\cdot)$ denotes volume. Write $$V_1(K,B_2^d) = V(\underbrace{K,\dotsc,K}_{d-1},B_2^d)$$ for that mixed volume. I will use the following facts about mixed volume:
1. $V_1(K,K) = V(K)$
2. $V_1(K,L)$ is increasing in both arguments.
3. The surface area of a convex body $K$ is $$S(K) = \lim_{\epsilon\downarrow 0} \frac{V(K+\epsilon B_2^d) - V(K)}{\epsilon} = dV_1(K,B_2^d)$$ (where $K+\epsilon B_2^d$ is the Minkowski sum).
4. Minkowski's first inequality: $$V(K)^{(d-1)/d} V(L)^{1/d} \le V_1(K,L)$$
Using (4), then (2) (we assumed $A_n\subset B_2^d$), then (1) yields $$V(A_n)^{(d-1)/d} V(B_2^d)^{1/d} \le V_1(A_n,B_2^d) \le V_1(B_2^d,B_2^d) = V(B_2^d)$$ Sending $n\to\infty$ yields, by the squeeze law, $$\lim_{n\to\infty} V_1(A_n,B_2^d) = V(B_2^d) = V_1(B_2^d,B_2^d)$$ Multiplying by $d$ and using (3) twice yields $$\lim_{n\to\infty} S(A_n) = S(B_2^d)$$ as desired.
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http://mathhelpforum.com/calculus/152614-find-value-function-continuous.html
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Thread:
1. find the value for which the function is continuous
f(x)= ln x/(x-1), if 0<x not =1
c , if x=1
and what kind of discontinuity is present if c does not have this value?
can someone help me with this problem?
2. This is nearly impossible to read...
Is it
$f(x) = \begin{cases}\frac{\ln{x}}{x - 1}\textrm{ if }x \in (0, 1)\cup (1, \infty)\\c\textrm{ if }x=1\end{cases}$
or
$f(x) = \begin{cases}\ln{\left(\frac{x}{x - 1}\right)}\textrm{ if }x \in (0, 1)\cup (1, \infty)\\c\textrm{ if }x=1\end{cases}$?
3. $f(x) = \begin{cases}\frac{\ln{x}}{x - 1}\textrm{ if } 0<x not equal to 1 \\c\textrm{ if }x=1\end{cases}$
thats how it is i dont know how to do a not equal sign
4. Originally Posted by dat1611
$f(x) = \begin{cases}\frac{\ln{x}}{x - 1}\textrm{ if } x>0 \, ; \, x \ne 1 \\c\textrm{ if }x=1\end{cases}$
fify
5. Originally Posted by dat1611
$f(x) = \begin{cases}\frac{\ln{x}}{x - 1}\textrm{ if } 0<x not equal to 1 \\c\textrm{ if }x=1\end{cases}$
thats how it is i dont know how to do a not equal sign
It is done by \neq
[LaTeX ERROR: Convert failed]
For the question at hand, you need to determine a value c such that the function is continuous. One way could be to just plug in 1 in the above definition with the natural log, but you cannot do that since you'd get something of the form 0/0. Hence you need to find the limit of [LaTeX ERROR: Convert failed] as x approaches 1. Or more succinctly, find
[LaTeX ERROR: Convert failed]
Once you find that, you can just set c equal to it. From the definitions of continuous functions, f(x) is continuous at point a if
[LaTeX ERROR: Convert failed]
6. ok can someone help with the actual problem
7. Do you know L'Hopital's rule? If you do, just apply to the numerator and denominator and it ' should be pretty easy.
8. Hello, dat1611!
$f(x)\;=\; \left\{ \begin{array}{cc}\dfrac{\ln x}{x-1}& x > 0,\;x \ne 1 \\ \\[-3mm]<br /> c & x=1 \end{array} \right$
Find the value of $c$ for which the function is continuous.
To be continuous at $x = 1$, we want: . $\displaystyle{\lim_{x\to1}\frac{\ln x}{x-1} \;=\;c }$
The left side goes to $\frac{0}{0}$ so we can apply L'Hopital.
. $\displaystyle{\lim_{x\to1}\frac{\frac{1}{x}}{1} \;=\;c \quad\Rightarrow\quad \lim_{x\to1}\frac{1}{x} \:=\:c$
Therefore: . $c \:=\:1$
9. so
$$\displaystyle \lim\limit_{x \to 1}\frac{\ln{x}}{x - 1}$<br />$ = 1
so the answers 1?
10. Yes!Although, it seems like you don't quite understand why. Do you need help with that?
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http://math.stackexchange.com/questions/180507/how-to-resize-an-image?answertab=oldest
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# How to resize an image?
I am not sure about the title of this question, so if someone knows an appropriate one, please rename it. It's a programming related question (but doesn't involve any programming). I posted it on stack overflow but didn't get any responses so I am trying here.
I need to map a piece of rectangle (x, y, width, height) of an (original)image onto a canvas by resizing the original image. Here's a picture that explain it better.
Here's another one, bit different, so you get a better idea:
I can scale and move the image however I want. How do i figure out the scale at which to resize the original image.
I had an idea, which was to figure the bigger dimension of the given rectangle, and use that to figure out the scale, so:
Given: imageWidth, imageHeight, rectWidth, rectHeight, canvasWidth, canvasHeight (let's ignore the offsets for now)
But that doesn't work in some cases. So I wondering what's the best way to do this. ,
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It looks like $imageHeight=rectHeight$. So just consider the ratio $\lambda=canvasHeight/imageHeight$. If you scale the image by $\lambda$ in both directions, then the height of the scaled image will now be $canvasHeight$. Is this what you mean? – dls Aug 9 '12 at 4:17
Please check out the new image I uploaded. I am looking for a general solution, to figure out, exactly which dimension dictates the scale. – PragmaOnce Aug 9 '12 at 4:26
## 1 Answer
If I understand your question correctly then you need to do this:
````scaleVert:=(rectHeight/rectWidht > canvasHeight/canvasWidth);
if scaleVert then
scale:=canvasHeight/rectHeight
else
slace:=canvasWidth/rectWidth;
````
The value of scale is ratio by which you have to scale your image.
The Boolean variable scaleVert is used to test whether you have to fit your rectangle vertically (i.e., the height of rectangle after scaling will be the same as the height of the canvas) or horizontally.
Notice that if `rectHeight/rectWidht = canvasHeight/canvasWidth` then it does not matter whether you scale vertically or horizontally - you obtain the same result. (The two rectangles are similar.)
If the rectangle is "thinner" (i.e., the ratio `rectHeight/rectWidht` is higher) you have to scale vertically and otherwise you have to scale horizontally.
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Thanks for this, will try it out. Also, if got a moment, can you explain why scaleVert simply can't be 'rectHeight > rectWidth'? – PragmaOnce Aug 9 '12 at 5:01
1
@PragmaOnce I've added a few lines of explanation to my post. – Martin Sleziak Aug 9 '12 at 5:04
Thanks! Works perfectly. Appreciate your explanation. – PragmaOnce Aug 9 '12 at 5:59
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http://math.stackexchange.com/questions/260962/boundedness-of-a-family-of-functions-in-c10-1?answertab=oldest
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# Boundedness of a family of functions in $C^{1}[0,1]$
Suppose we have a set $S$ which contains all functions $v \in C^{1}[0,1]$ so that $v(0) = 0$ and
$$\int_{0}^{1} |v'(x)|^2 dx = 1.$$
How can I show that $S$ is bounded with the infinity norm. That is, how can I show that
$$sup_{v\in S}\lVert v\rVert_{\infty}$$
exists.
I know that for each $v \in S$ there exists a constant $M(v)$ so that $|v(x)| \le M(v)$ for all $x \in [0,1]$ since $v$ is continuous on a closed interval $[0,1]$. What I fail to see is how $\int_{0}^{1} |v'(x)|^2 dx = 1$ helps me.
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Assume you have $v_n=n\cdot x$. Then $v_n\in C^1[0,1]$ and $v_n(0)=0$ is fulfilled for all $n$. But $\sup_{v\in S}||v||_{\infty} \geq \sup_{n}||v_n||_{\infty}\rightarrow \infty$. – macydanim Dec 17 '12 at 20:24
@macydanim, so you are saying that $sup_{v\in S}\lVert v\rVert_{\infty} = \infty$ and thus $S$ is not bounded? – user53477 Dec 17 '12 at 20:27
I have no idea what @macydanim is trying to say. Because $v_n(x) = nx$ certainly doesn't satisfy $\int_0^1 |v_n'(x)|^2 \,dx = 1$. – kahen Dec 17 '12 at 20:28
@kahen , exactly. I just wanted to provide an example that shows that the condition $\int_0^1 ||v'|| dx$ is important and what can go wrong otherwise. Sorry if that did not come through. – macydanim Dec 17 '12 at 20:30
@macydanim, so is there something very obvious that I might be missing from that integral? – user53477 Dec 17 '12 at 20:35
## 2 Answers
Let $v\in S$ and $x\in[0,1]$. Then $$|v(x)|=\Bigl|\int_0^xv'(t)\,dt\Bigr|\le\int_0^x|v'(t)|\,dt\le\Bigl(\int_0^x|v'(t)|^2dt\Bigr)^{1/2}\Bigl(\int_0^xdt\Bigr)^{1/2}\le\sqrt x\le1.$$ Use has been made of the Cauchy-Schwarz inequality and the fact that $$\int_0^x|v'(t)|^2dt\le\int_0^1|v'(t)|^2dt\le1.$$
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Thank you. I'm still wondering why I couldn't see that connection. – user53477 Dec 17 '12 at 21:05
From your expression we have that $-\sqrt{x} \le v(x) \le \sqrt{x}$. So $\sqrt{x} \in S$. But now we have a problem since $\int_{0}^{1} |\frac{d}{dx} \sqrt{x}|^2 dx = \int_{0}^{1} \frac{1}{2 \sqrt{x}}^2 dx = \int_{0}^{1} \frac{1}{4x} dx$ which does not converge. Do we really have $v(x) \le \sqrt{x}$? or perhaps $v(x) \lt \sqrt{x}$? – user53477 Dec 17 '12 at 22:57
We certainly have $|v(x)|\le\sqrt x$. Can you spot something wrong in the proof? But this does nor mean that $\sqrt x\in S$. – Julián Aguirre Dec 18 '12 at 10:03
Another one by contradiction.
Suppose $\sup_{v\in S} || v||_{\infty} =\infty$ That means there exists a sequence $(x_n)\rightarrow \tilde{x}>0$ such that $v(x_n)\rightarrow \infty$. Now the mean value theorem states, that $\exists \xi$ such that $\xi \in [0,{x_n}]$ and $$v'(\xi) = \frac{v(x_n)-v(0)}{{x_n} -0}=\frac{v({x_n})}{{x_n}}$$ So we know
$$|v'(\xi)| =|\frac{v({x_n})}{{x_n}}| \rightarrow |\frac{v(\tilde{x})}{\tilde{x}}|= \infty\, \text{ as } n \rightarrow \infty$$ Now as $v\in C^1[0,1]$ we know that $v'\in C^0[0,1]$ so $\int_0^1 || v'(x) || dx \rightarrow \infty$. Which is a contradiction.
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http://math.stackexchange.com/questions/220331/parameter-optimization-in-probabilistic-models/234967
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# Parameter optimization in probabilistic models
Task: Suppose we model a variable $y = Wx + \mu$ as a linear transformation of $x$ plus some Gaussian noise $\mu\sim\mathcal N(0,\sigma I)$. Our aim is to minimize the estimation error of $x$ given $y$ in terms of $W$, i.e. we want to minimize the entropy $H(x|y,W)$ as a function of $W$. Suppose that, during learning, we know $x$ for every observed state $y$: what is the optimal supervised update of the model parameters $W$?
The problem is: I can't wrap my head around the question how I come - in a rigorous way - from the estimation problem that I want to solve to the parameter updates that depend on the variable I want to estimate. It's probably a standard procedure but the problem is hard to nail down to Google-friendly buzzwords.
Thanks in advance!! blue2script
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## 1 Answer
I dont have answer to your question but i am interested in knowing what exactly does this statement in your question mean :
"Our aim is to minimize the estimation error of x given y in terms of W, i.e. we want to minimize the entropy $H(x|y,W)$ as a function of W"
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http://www.physicsforums.com/showthread.php?t=558684
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Physics Forums
## A Unitary Matrix and Hermitian Matrix
Its true that one can say a unitary matrix takes the form
$U=e^{iH}$
where $H$ is a Hermitian operator. Thats great, and it makes sense, but how can you compute the matrix form of $H$ if you know the form of the unitary matrix $U$. For example, suppose you wanted to find $H$ given that the unitary matrix is one of the familiar rotation matrices (2 x 2) for simplicity. Let's say
$U=\left(\begin{array}{cc} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \end{array}\right)$
What would the procedure be in finding the matrix form of $H$? I suppose you could start by finding the eigensystem of the unitary matrix. Then, upon normalizing the eigenbasis of $U$, somehow you could find the matrix representation of $H$. Any pointers or suggestions would be great.
PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Can you do it for diagonal matrices?? To extend it to nondiagonal matrices, notice that if $D=e^{iH}$, then $$ADA^{-1}=e^{iAHA^{-1}}$$
So after some fiddling I find that the appropriate Hermitian matrix takes the form $H=\left(\begin{array}{cc} 0 & i\theta \\ -i\theta & 0 \end{array}\right)$ If this is indeed correct then I think I have what I need.
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http://mathoverflow.net/questions/41931?sort=oldest
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## About isogeny theorem for elliptic curves
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$K$ a number field, $G_K$ its Galois group, $E_1, E_2$ two elliptic curves defined over $K$. The isogeny theorem says that if for some prime number $\ell$, The Tate modules (tensored with $\mathbb{Q}$) $V_{\ell}(E_1)$ is isomorphic to $V_{\ell}(E_2)$ as Galois modules. Then these two elliptic curves are isogenous.
My question is, when are these two curves isomorphic? Namely, what more invariants are needed to fully characterize the elliptic curve (besides the Tate modules). Thanks!
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If you are interested in isomorphisms over the algebraic closure: $E_1$ and $E_2$ are isomorphic (over $\overline{K}$) iff their $j$-invariants are the same. Or are you interested in isomorphisms over $K$? – felix Oct 12 2010 at 19:12
Yes, over K. Thank you. – natura Oct 12 2010 at 19:28
I made a mistake in the original post, using $T_{\ell}$ of elliptic curves. It should use $V_{\ell}$. Thanks to Emerton for pointing out. – natura Oct 13 2010 at 1:11
## 1 Answer
If all Tate modules (i.e., for all $\ell$) are isomorphic then they differ by the twist by a locally free rank $1$ module over the endomorphism ring of one of them. This is true for all abelian varieties but for elliptic curves we only have two kinds of possibilities for the endomorphism ring; either $\mathbb Z$ or an order in an imaginary quadratic field. In the first case there is only one rank $1$ module so the curves are isomorphic. In the case of an order we get that the numbe of twists is a class number.
Addendum: Concretely, we have that $\mathrm{Hom}(E_1,E_2)$ is a rank $1$ projective module over $\mathrm{End}(E_1,E_1)$ (under the assumption that the Tate modules are isomorphic) and then $E_2$ is isomorphic to $\mathrm{Hom}(E_1,E_2)\bigotimes_{\mathrm{End}(E_1)}E_1$ (the tensor product is defined by presenting $\mathrm{Hom}(E_1,E_2)$ as the kernel of an idempotent $n\times n$-matrix with entries in $\mathrm{End}(E_1)$ and $E_2$ is the kernel of the same matrix acting on $E_1^n$. Hence, given $E_1$ $E_2$ is determined by $\mathrm{Hom}(E_1,E_2)$ and every rank $1$ projective module appears in this way.
Addendum 1: Note that I was talking here about the $\mathbb Z_\ell$ (and not $\mathbb Q_\ell$ Tate modules. You can divide up the classification of elliptic curves in two stages: First you see if the $V_\ell$ are isomorphic (and there it is enough to look at a single $\ell$). If they are, then the curves are isogenous. Then the second step is to look within an isogeny class and try to classify those curves.
The way I am talking about here goes directly to looking at the $T_\ell$ for all $\ell$. If they are non-isomorphic (for even a single $\ell$ then the curves are not isomorphic and if they are isomorphic for all $\ell$ they still may or may not be isomorphic, the difference between them is given by a rank $1$ locally free module over the endomorphism ring. In any case they are certainly isogenous. These can be seen a priori as if all $T_\ell$ are isomorphic so are all the $V_\ell$ but also a posteriori essentially because a rank $1$ locally free module becomes free of rank $1$ when tensored with $\mathbb Q$.
Of course the a posteriori argument is in some sense cheating because the way you show that the curves differ by a twist by a rank $1$ locally free module is to use the precise form of the Tate conjecture: ```$$
\mathrm{Hom}(E_1,E_2)\bigotimes \mathbb Z_\ell = \mathrm{Hom}_{\mathcal G}(T_\ell(E_1),T_\ell(E_2))
$$``` which for a single $\ell$ gives the isogeny.
Note also that the situation is similar (not by chance) to the case of CM-curves. If we look at CM-elliptic curves with a fixed endomorphism ring, then algebraically they can not be put into bijection with the elements of the class group of the endomorphism ring (though they can analytically), you have to fix one elliptic curve to get a bijection.
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If I understand it correctly, in simpler terms, the above means: The two are isomorphic if and only if the kernel of the isogeny is a principal ideal in $\text{End} (E1,E2)$ (which is independent of which isogeny we take). Is this correct? – Dror Speiser Oct 12 2010 at 21:27
Also, in the spirit of the question, do we need all Tate modules to be isomorphic? Or is it enough that for a positive density of $\ell$'s they are isomorphic? – Dror Speiser Oct 12 2010 at 21:29
2
The kernel of the isogeny is finite and hence can not be an ideal in $\mathrm{End}(E_1)$. If $E_1$ and $E_2$ are just isogeneous, then their Tate modules are isomorphic for all but a finite number of $\ell$'s so density $1$ is not sufficient. – Torsten Ekedahl Oct 12 2010 at 21:33
Oops. Of course. The latter hints that there is actually a finite list of primes that need to be checked. I'll throw a guess once more: primes that divide conductor are enough? – Dror Speiser Oct 12 2010 at 21:42
1
Dear Basic, Your assertion that "The Tate modules ... are isomorphic ... if [they are so] for just one prime" is not true. (I don't have Serre's book in front of me, but I presume he states this after tensoring with $\mathbb Q$.) E.g. if $E_1 \to E_2$ via a cyclic 5 isogeny (think about $X_1(11) \to X_0(11)$) then this will induce an isomorphism of $\ell$-adic Tate modules for every $\ell \neq 5$, but will not induce an isomorphism of $5$-adic Tate modules. (It will induce an embedding of one into the other, and the image will be of index 5. Furthermore, the $5$-adic Tate modules will ... – Emerton Oct 12 2010 at 23:46
show 6 more comments
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http://mathoverflow.net/questions/34591?sort=newest
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Probability of generating the symmetric group
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The statement is simple:
What is the probability that a set of n-1 transpositions generates the symmetric group, $S_n$?
The motivation is that I remembered reading that this was an open problem somewhere on the internet, and then I solved it. I'm curious to see other people's solutions, because I think it's a nice problem, and don't quite believe that it is hard enough to be open.
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2
The motivation, "I remember reading that this was an open problem somewhere on the internet" does not strike one as very compelling. Isn't it true that $n-1$ transpositions, viewed as edges of a graph on $n$ vertices, generate $S_n$ if and only if the graph is a tree? So it's a question of tree enumeration. – Victor Protsak Aug 5 2010 at 8:31
Good point. How does it look like translating the bijection it terms of Pruefer sequences? (I fear it doesn't add much to the picture though) – Pietro Majer Aug 5 2010 at 11:42
1
I also remember that this "open problem" was one of the questions of the university entrance examination of a friend of mine. – Alekk Aug 5 2010 at 12:06
So I thought as well. I wonder what the actual statement was, or whether I had dreamt it. – Ryan Thorngren Aug 5 2010 at 16:29
1 Answer
A solution (assuming that all transpositions are distinct and are choosen uniformly among all ${n\choose 2}$ possible transpositions) can be given as follows:
A set of $n-1$ transpositions $(a_1,b_1),\dots,(a_{n-1},b_{n-1})$ on the set $\lbrace 1,\dots,n\rbrace$ generates the whole symmetric group of ${1,\dots,n}$ if and only if the graph with vertices $\lbrace 1,\dots,n\rbrace$ and edges $\lbrace a_i,b_i\rbrace$ is a tree.
The probability to generate $S_n$ is thus the same as the probability to get a tree with $n$ vertices $V$ when choosing randomly $n-1$ edges with endpoints in $V$. By Cayley's theorem, there are $n^{n-2}$ different trees with vertices $\lbrace 1,\dots,n\rbrace$. Since there are ${{n\choose 2}\choose n-1}$ different graphs with $n-1$ edges and vertices ${1,\dots,n}$, the probability is given by $n^{n-2}/{{{n\choose 2}\choose n-1}}$.
If repetitions are allowed, one gets $n^{n-2}/{{n\choose 2}+n-2\choose n-1}$ (assuming uniform probability for all distinct multisets).
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Or very roughly $\left(\frac{2}{e}\right)^n$. – Peter Shor Aug 5 2010 at 21:53
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http://mathoverflow.net/questions/18359/motivation-of-surgery/76981
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## motivation of surgery
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
an $n$-surgery on m dim manifold M is to cut out $S^n\times D^{m-n}$and replace it by $D^{n+1}\times S^{m-n-1}$. I want to know how this is invented? I do know that the effect of passing a critical point of index $n$ in $m$-manifold is equivalent to attach an $n+1$-handle $D^{n+1}\times D^{m-n-1}$.Now the boundary of $D^n\times D^{m-n}$ is $S^n\times D^{m-n}\cup D^{n+1}\times S^{m-n-1}$,i think there must be some close relation between the special form of $n$-surgery and handle.can someone help make this clear?
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5
Some help may be gained by looking at Milnor's "Lectures on the h-cobordism theorem," in the Princeton Yellow series. Also look at Fenn Rourke and Sanderson's book --- sorry I forgot the name. Finally, try to compare what is going on using surfaces with boundary --- disks with handles attached --- and consider $3$-manifolds, their Morse functions, and their non-critical levels. Favorite examples include lens spaces. One has to be meticulous about the parametrizations. FRS introduce the belt sphere, the attaching sphere, the core disk and the co-core disk. All of these ideas help! – Scott Carter Mar 16 2010 at 13:22
3
Please improve punctuation and capitalization. – Theo Johnson-Freyd Mar 16 2010 at 16:04
3
@Theo, to whom is your comment directed? In my comment, "Yellow" should be lower case. The phrase "sorry I forgot the name" should be enclosed in parenthesis. The phrase "handles attached --- and consider [...]" should be "handles attached. Consider [...]" I think you have enough points to edit the original question:-) – Scott Carter Mar 16 2010 at 21:18
I just read another paper of Milnor and Kervaire, "groups of homotopy spheres." In it they talk about surgery, only they call it "spherical modification", so I think it is early in the discussion. It is also highly enlightening because it shows the best set of algebraic-topological arguments about surgery that allowed one to compute cobordism groups. I'm not sure what the question is though. OP, I second Theo's comment: could you please articulate what you are asking and what you have been thinking in separate sentences? – Elizabeth S. Q. Goodman Oct 2 2011 at 6:40
## 3 Answers
The Wikipedia article on surgery theory explains this! In addition, the Edinburgh Surgery Theory Study Group provides all kinds of surgery-related materials, including YouTube videos (not for the squeamish). There is all kinds of surgery bric-a-brac on Surgery Bits and Pieces and also here.
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1
That is quite an impressive article you've (mostly) written! – S. Carnahan♦ Apr 17 2010 at 23:38
### 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 best answer is to cite Tim Perutz answer to my question Surgery and homology: a reference request:
"To say that a smooth, closed manifold N is obtained by surgery along a (framed) sphere in M is to say that there is a cobordism P from M to N and a Morse function $f\colon P\to [0,1]$, with $f^{−1}(0)=M$, $f^{−1}(1)=N$, and exactly one critical point c."
Critical point corresponds to a handle, $S^n\times D^{m-n}$ lies in $M$, $D^{n+1}\times S^{m-n-1}$ lies in $N$ and spheres $S^n$ and $S^{m-n-1}$ are intersections of stable and unstable manifolds for $c$ with corresponding level sets.
In other words: to obtain $N$ from $M$ by a surgery one can consider $M\times [0,1]$, glue a handle using $S^n\times D^{m-n} \subset M=M\times 1$, smooth the resulted cobordism and take a component of a boundary.
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3
Or "surgery is what happens to the boundary during a handle attachment" – Ryan Budney Mar 16 2010 at 18:51
Well said, Ryan! – Petya Mar 16 2010 at 23:19
This question has already been answered, but there's a tiny piece of intuition which I'd like to make explicit:
If you're thinking about a manifold in the PL world, surgery might look a bit arbitrary- why cut out and glue in those pieces and not others? Surgery's natural setting is the smooth world, where you're equipping a manifold with a Morse function $f\colon\, M\to \mathbb{R}$, and using information about critical points of $f$ to encode $M$.
It's actually a bit more involved than you might think it might be, but when you pass a critical point of $f$ you add a handle to $M$, and the boundary changes by surgery. For details, see answers to this question.
So really, surgery isn't an a-priori construction which somebody pulled from a hat- it is rather an operation which stems naturally and inevitably from Morse theory.
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http://mathhelpforum.com/trigonometry/86956-factoring-trig-function-print.html
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# factoring a trig function
Printable View
• May 2nd 2009, 01:30 PM
brentwoodbc
factoring a trig function
I can do the basic ones, but when they have a coefficient in front of the first term I dont know what to do.
$2cos^2x-7cosx-3$
I thought maybe you do
(2cosx-1)(cosx-3)
Edit nevermind That is right.
BUt is there a better way to solve these other than putting the first term in the first bracket and the guessing at multiples of the last term?
because with one like 6sin^2x+sinx-1 I get (6sinx-1)(sinx+1)
and its supposed to be(2sinx-1)(3sinx-1)
Im confused with factoring these, could I have some help please.
thanks.
• May 2nd 2009, 01:49 PM
e^(i*pi)
Quote:
Originally Posted by brentwoodbc
I can do the basic ones, but when they have a coefficient in front of the first term I dont know what to do.
$2cos^2x-7cosx-3$
I thought maybe you do
(2cosx-1)(cosx-3)
but that doesnt work.
Im confused with factoring these, could I have some help please.
thanks.
You can factorise it like a normal quadratic: Let u = cos(x) :
$2u^2 - 7u - 3 = 0$
Since the discriminant does not equal a perfect square this cannot be factorised rationally. You may solve it using the quadratic formula or completing the square:
$cos(x) = u = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$
• May 2nd 2009, 01:54 PM
brentwoodbc
Quote:
Originally Posted by e^(i*pi)
You can factorise it like a normal quadratic: Let u = cos(x) :
$2u^2 - 7u - 3 = 0$
Since the discriminant does not equal a perfect square this cannot be factorised rationally. You may solve it using the quadratic formula or completing the square:
$cos(x) = u = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$
its 2u^2 - 7u + 3 = 0 btw sorry
• May 2nd 2009, 02:01 PM
e^(i*pi)
Quote:
Originally Posted by brentwoodbc
thanks so when I find x=... how do I give my answer in factor form, he answer given is(2cosx-1)(cosx-3)
(My last post is only valid if the equation is equal to 0)
Do you mean
$<br /> 2cos^2x-7cosx-3<br />$ or
$<br /> 2cos^2x-7cosx+3<br />$?
Since the second one factors I will assume you meant that one:
Let u = cos(x) : $2u^2-7u+3$
This would equal (2u )(u ). The other numbers will be 1 and 3 and there will be two minus signs (because the last term is positive). The sum of 2 times one of them plus the other will be equal to -7.
Spoiler:
$(2u-1)(u-3)$
• May 2nd 2009, 02:03 PM
brentwoodbc
Quote:
Originally Posted by e^(i*pi)
(My last post is only valid if the equation is equal to 0)
Do you mean
$<br /> 2cos^2x-7cosx-3<br />$ or
$<br /> 2cos^2x-7cosx+3<br />$?
Since the second one factors I will assume you meant that one:
Let u = cos(x) : $2u^2-7u+3$
This would equal (2u )(u ). The other numbers will be 1 and 3 and there will be two minus signs (because the last term is positive). The sum of 2 times one of them plus the other will be equal to -7.
Spoiler:
$(2u-1)(u-3)$
thanks ya I made a mistake it is +3
Thank you. Im a bit rusty a factoring.
• May 2nd 2009, 02:06 PM
e^(i*pi)
Quote:
Originally Posted by brentwoodbc
thanks ya I made a mistake it is +3
Thank you. Im a bit rusty a factoring.
I forgot to mention it in my last post but don't forget to change u back to cos(x) (Hi)
All times are GMT -8. The time now is 12:47 PM.
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http://mathhelpforum.com/differential-geometry/192687-show-p-not-complete.html
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# Thread:
1. ## Prove that D is connected
Let $D \subset R$, and let f: D-->R be continuous. Prove that D is connected if { $x,f(x): x \in D$}, the graph of f, is a connected subset of $R^2$
2. ## Re: Prove that D is connected
Originally Posted by wopashui
Let $D \subset R$, and let f: D-->R be continuous. Prove that D is connected if { $x,f(x): x \in D$}, the graph of f, is a connected subset of $R^2$
This is an if and only if. Let $\Gamma_f$ denote the graph, note then that $D=\pi_1(\Gamma_f)$ where $\pi_1:R^2\to R$ is the canonical projection onto the first coordinate--why does this tell us $D$'s connected. Conversely, if $D$ is connected then $\Gamma_f=h(D)$ where $h: D\to R^2: x \mapsto (x,f(x))$--why is $h$ continuous and why does this tell us that $\Gamma_f$ is connected?
3. ## Re: Prove that D is connected
Originally Posted by Drexel28
This is an if and only if. Let $\Gamma_f$ denote the graph, note then that $D=\pi_1(\Gamma_f)$ where $\pi_1:R^2\to R$ is the canonical projection onto the first coordinate--why does this tell us $D$'s connected. Conversely, if $D$ is connected then $\Gamma_f=h(D)$ where $h: D\to R^2: x \mapsto (x,f(x))$--why is $h$ continuous and why does this tell us that $\Gamma_f$ is connected?
sorry, what is canonical projection ?
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http://math.stackexchange.com/questions/194206/is-there-a-generalized-formula-to-solve-the-nth-exponent-of-an-m-dimensional
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# Is there a generalized formula to solve the $n$th exponent of an $m$ dimensional square matrix?
I was trying to do something like a general formula, and testing, I came with this:
$$\left( \begin{array}{cc} a & b \\ a' & b' \\ \end{array} \right)^2=\left( \begin{array}{ccc} a^2+a'b & b(a+b') \\ a'(a+b') & b'^2+a'b \\ \end{array} \right)$$
I've also tried with a $3$x$3$ matrix, but i don't get anything.
I was trying to do something that can solve matrices like this with only one formula:
$$\left( \begin{array}{ccccc} a_{11}&a_{12}&a_{13}&…&a_{1j}\\ a_{21}&a_{22}&a_{23}&…&a_{2j}\\ a_{31}&a_{32}&a_{33}&…&a_{3j}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ a_{i1}&a_{i2}&a_{i3}&\dots&a_{ij} \end{array} \right)^{n}$$
And that's what I want to know.
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## 1 Answer
Yes. You compute the Jordan form of the matrix which is related to $A$ by a similarity transformation $J =P^{-1}AP$ thus $A = PJP^{-1}$ and it's easy to show $A^k = PJ^kP^{-1}$. It turns out the $J^k$ is simple to calculate. For example, if $A$ is diagonalizable then $J = diag(\lambda_1,\dots , \lambda_n)$ hence $J^k = diag(\lambda_1^k,\dots , \lambda_n^k)$. So, look up the Jordan form of a matrix to answer this question in general.
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http://mathoverflow.net/questions/61336?sort=votes
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Existence of sequence of examples of braking ‘Cancellation law in homeomorphic products’
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I know there are manifolds (with or without boundary) $A$ and $B$ such that $A\times C$ is homeomorphic to $B\times C$ but $A$ is NOT homeomorphic to $B$.
My question is (in the Diffeomorphism Category)
Is there any infinitely many $A_i$, with same dimension of course, which are pairwise non-diffeomorphic, but $A_i\times C$ become all diffeomorphic to each other.
1) What's the answer to the question under the assumption on $A_i$, $C$: Smooth Close manifolds.
2) What if we change diffeomorphic to homeomorphic?
3) Is there any example when we require $C$ to be Torus?
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3 Answers
The answer to 2 is yes, there is such an example. In
McMillan, D. R., Jr., Some contractible open $3$-manifolds. Trans. Amer. Math. Soc. 102 (1962), 373–382.
there is a construction of uncountably many topologically distinct, contractible (open) $3$-manifolds $M_\alpha$ such that $M_\alpha \times \mathbb R$ is homeomorphic to $\mathbb R^4$.
Take a look at this recent MO question and the Wikipedia article on the Whitehead manifold for some closely related material.
Edit. The answer to 3 is also yes, assuming by "torus" you mean $S^1$. On page 221 of
Vogt, E., Foliations of codimension $2$ with all leaves compact on closed $3$-, $4$-, and $5$-manifolds. Math. Z. 157 (1977), no. 3, 201–223.
you can find a construction of infinitely many pairwise non-homeomorphic closed Seifert 3-manifolds whose product with $S^1$ gives the same Seifert 4-manifold.
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1. Take any closed $4$-manifold with infinitely many smooth structures, and multiply it by a torus. The product has only finitely many smooth structures, in fact any manifold $M$ of dimension $\ge 5$ has at most finitely many smooth structures if $H^3(M;\mathbb Z_2)$ is finite.
2. Take any closed manifold $X$ of dimension $\ge 5$ such that the Whitehead group of $\pi_1(X)$ is infinite (e.g. this is the case if $\pi_1(X)$ is finite cyclic of order $5$ or $\ge 7$). Then there are infinitely many closed pairwise non-diffeomorphic manifolds that are h-cobordant to $X$. On the other hand, all these manifolds become diffeomorphic after multiplying by $S^1$ (because this operation makes Whitehead torsion vanish).
3. By contrast, if $X$, $X'$ are closed simply-connected manifold of dimension $\ge 5$ that become diffeomorphic after taking product with $S^1$, then $X$, $X'$ are diffeomorphic (this followed from the h-cobordism theorem in the universal cover of the product).
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Wall classified certain smooth closed 6-manifolds up to diffeomophism in his 1966 Inventiones paper. It follows that if ${M_i}$ is a collection of smooth spin, simply connected 4-manifolds that are pairwise not-diffeomorphic but all homeomorphic, then $M_i\times S^2$ are all diffeomorphic. Many collections ${M_i}$ are known, distinguished eg by Seiberg-Witten invariants, eg by performing log transforms on $K3$. So 1 is true.
In general, this is the kind of problem that surgery theory is good for. So there are examples of 2 and I think 3 also, look in Wall's book.
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http://mathoverflow.net/questions/23060/set-theory-and-model-theory/23065
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## Set theory and Model Theory
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This question probably doesn't make any sense, but I don't see why, so I ask it here hoping someone will illuminate the matter:
There is this whole area of study in Set Theory about the consistency, independence of axioms, etc. In some of these you use model theory (e.g. forcing) to prove results about set theory.
My question is: What is the foundation of this model theory we are using? We are certainly using sets to talk about the models, what some may call sets in the "meta"-mathematics, that is to say, the "real" mathematics.
But then, all these arguments in the end in are all about the theory of sets as a theory, and not the theory of sets as a foundation of math, since we are using these sets in the meantime. So our set theory is not about the foundation of math.
Am I right?
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The word got out that our resident logicians are pretty, pretty good! :P – Mariano Suárez-Alvarez Apr 30 2010 at 2:18
## 5 Answers
Your worries arise from asymmetry between how you view ordinary mathematics and how you view logic and model theory.
If it is the business of logic and model theory to provide foundations for the rest of mathematics then, of course, logicians and model theorists will not be allowed to use mathematical methods until they have secured them. But how might they accomplish this? The more we think about it, the more it becomes obvious that "securing the foundations of mathematics", whatever that means, is a task for philosophers at best and a form of mysticism at worst.
It is far more fruitful to think of logic and model theory as just another branch of mathematics, namely the one that studies mathematical methods and mathematical activity with mathematical tools. They follow the usual pattern of "mathematizing" their object of interest:
• observe what happens in the real world (look at what mathematicians do)
• simplify and idealize the observed situation until it becomes manageable by mathematical tools (simplify natural language to formal logic, pretend that mathematicians only formulate and prove theorems and do nothing else, pretend that all proofs are always written out in full detail, etc.)
• apply standard mathematical techniques
As we all know well, the 20th century logicians were very successful. They gave us important knowledge about the nature of mathematical activity and its limitations. One of results was the realization that almost all mathematics can be done with first-order logic and set theory. The set-theoretic language was adopted as a universal means of communication among mathematicians.
The success of set theory has lead many to believe that it provides an unshakeable foundation for mathematics. It does not, at least not the mystical kind that some would like to have. It provides a unifying language and framework for mathematicians, which in itself is a small miracle. Always remember that practically all classical mathematics was invented before modern logic and set theory. How could it exist without a foundation so long? Was the mathematics of Euclid, Newton and Fourier really vacouous until set theory came along and "gave it a foundation"?
I hope this explains what model theorists do. They apply standard mathematical methodology to study mathematical theories and their meaning. They have discovered, for example, that however one axiomatizes a given body of mathematics in first-order logic (for example, the natural numbers), the resulting theory will have unintended and surprising interpretations (non-standard models of Peano arithmetic), and I am skimming over a few technical details here. There is absolutely nothing strange about applying model theory to the axioms known as ZFC.
Or to put it another way: if you ask "why are model theorists justified in using sets?" then I ask back "why are number theorists justified in using numbers?"
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+1: Very well put Andrej! (Not really the right place to mention this since few will know what I'm talking about, but I also really liked your question regarding your colleague's wife.) – François G. Dorais♦ Apr 30 2010 at 7:49
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The last line is great. – Uri Andrews Apr 30 2010 at 8:11
I can't help to give -1 to the answer as it understates the importance of the problem of foundations and the role of mathematical logic in it (and I'm not even a logician or a philosopher!). – Qfwfq Sep 19 2011 at 18:05
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I disagree, naturally. Where exactly am I understating the "problem of foundations and the role of mathematical logic"? I am stating that it is not the role of mathematical logic to provide the sort of mystical, unshakeable foundations that many mathematicians think it ought to provide. That is a job for philosophers, if it is a job worth doing at all. If you read again, you will see that I praise the foundational achievements of modern logic. – Andrej Bauer Sep 20 2011 at 14:56
-1 for the same reason as unknowngoogle. – Joël Sep 28 2011 at 14:04
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I totally agree with the answers already given but I still want to say something to your question, which emphasizes probably the formalist side. To cut a long story short the foundation of model theory, for which you were asking for, is ZFC (at least ZFC is one possibility), but this doesn't mean that one must not use model theory to investigate ZFC itself:
As you know, one can code the symbols of first order logic within set theory and, as a consequence, the whole model theory can be carried out in ZFC. Thus the Löwenheim-Skolem Theorem, the Compactness Theorem and so on are theorems of ZFC. (Note that when e.g. the Compactness Theorem talks about a “set of first order formulas”, it in fact talks about the set of the coded formulas, i.e. about a set of sets).
Now you can apply these results of model-theory to the coded axioms of ZFC (there is nothing wrong with that since ZFC is stated in first order logic and the set of the coded axioms is well defined); still everything is done in the frame ZFC.
The Theorem from logic, which states that if $T$ is a 'set' of formulas, $\psi$ another formula and $M$ a model such that $M \models T$ and $M \models \lnot \psi$, then T cannot prove $\psi$, is a theorem in ZFC (again $\psi$ and $T$ in this theorem are in fact coded, i.e. sets and moreover the metamathematical statement “there exits no proof from $T$ for $\psi$” is in fact a well defined statement about sets which mimics characteristics from proofs inside ZFC).
And this Theorem can now be used to state independence results about ZFC in ZFC. The only difficulty is that by Gödels celebrated result, one cannot prove inside ZFC the existence of a model of ZFC. Therefore one always assumes $Con(ZFC)$, i.e. the coded form of the assertion: “ZFC is consistent”, which is equivalent to “there exists a model of ZFC”. Then manipulate this given model to obtain a model for $ZFC+ { \varphi }$ where $\varphi$ is an arbitrary interesting statement.
We can now prove things like: If ZFC is consistent then so is ZFC+ “Continuum Hypothesis fails” which is the famous result of Cohen shown by models but within ZFC.
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The use of forcing in Set Theory is to investigate the Zermelo-Fraenkel axioms and their consequences. This is a perfectly valid use of Model Theory — the Completeness Theorem says that a statement φ is a consequence of ZFC if and only if φ is true in every model of ZFC. If one can produce a forcing poset that forces φ to be false, then we know that if ZFC is consistent then φ is not a consequence of ZFC since any suitable model can be extended to a model of ZFC in which φ is false. If another forcing forces φ to be true, then we know that φ is independent of ZFC.
I'm gathering from your question that you're a Platonist (I'm agnostic but I'll play along). This demonstration of independence via forcing says little about the truth of φ in the Platonic Universe. It only says that further information than the axioms of ZFC is needed to determine the truth of φ. However, these investigations over the past half-century have led to some insight on what statements are indeed true in the Platonic Universe. For example, see these two articles by Woodin (AMS Notices, 2001) where he discusses the status of the Continuum Hypothesis.
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The thing is, aren't you using sets when you prove the Completeness Theorem? In the completeness theorem you are referring to theories which are sets of first order sentences, which themselves are made up of symbols from a set. Don't you need sets in the first place to do model theory? I mean, what are the foundations of model theory? – Enrique Apr 30 2010 at 2:48
Sure, but that's not a problem. If you're indeed a Platonist then sets simply exist, the Completeness Theorem is true, and everyone is happy! (If my impression was wrong and you're not a Platonist, then explain and I'll adjust my answer accordingly.) – François G. Dorais♦ Apr 30 2010 at 2:52
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In other words, from the Platonic view, sets exist and ZFC is simply what we know about them so far; ZFC is not a definition of sets (as it would be to a formalist) it's a fact of life. – François G. Dorais♦ Apr 30 2010 at 3:07
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Here's a +1 from a dyed in the wool formalist. – Harry Gindi Apr 30 2010 at 3:58
ok François, I think I see your point. – Enrique Apr 30 2010 at 6:14
First of all, I would like to say that your question is not only justified, but actually most welcome: it shows the King naked. Who is the King? You can call him Semantics,or the mathematical theory of Truth (notice the capital letter) if you wish.
There is a widespread usage among logicians of sentences such as " the true universe of sets", the "set of natural numbers", and so forth, as if these entities were crystal clear and thus indisputable. They aren't, at least to me. But even if they are, even if there is such a thing as mathematical intuition which informs us on the world of platonic math (what my teacher aptly named PLATO's ATTIC), the fact is this: as soon as we SPEAK about them, they just become syntax, no more no less (ie part of a frame syntactic theory which grounds them).
Model theory is a great branch of mathematics, no doubts about it. But it is formalized mathematics, and grounded in ZFC. Does that mean that all its results are empty, especially in the model theory of ZFC? I say absolutely no. Instead, it bespeaks of the uncanny capability of a powerful theory such as ZFC to "reflect" upon itself and the rest of mathematics. The king is naked, but a king nevertheless. Long live the king!
PS I just discovered the recent work of Joel D. Hamkins and a few others on the Multiverse: finally some people from the set theory ranks are moving away from a dogmatic and monolitic notion of truth (the ghost-like "true universe of sets" and its equally ghostly eternal properties) towards a new dynamic and contextual one. I think this is a beginning of a new era, if I am not mistaken....
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I must say that I have never in my studies heard any of my teachers speak of this "One universe of sets!", if it was mentioned at all (depending on the course of course) then it was mentioned as "a universe of sets" and nothing more. – Asaf Karagila May 30 2011 at 18:53
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I don't understand the reasoning behind this: "even if there is such a thing as mathematical intuition which informs us on the world of platonic math, the fact is this: as soon as we SPEAK about them, they just become syntax, no more no less." Would my speaking about you make you "just become syntax, no more no less"? Platonists would regard syntax and whatever we say about these objects as merely a help to communicating about them, not as having any influence on their platonic existence or nature. – Andreas Blass May 30 2011 at 21:28
I will not become syntax, but what you write about me, regardless of whether I am really there, IS syntax and nothing but syntax. If you formalize me in a suitable frame theory,say ZFC, and then argue about my properties, you may be guided by your inner knowledge of what I am, but at the end of the day, the validity of your arguments is purely a syntactical matter. Moreover: you may be admitted to Plato's attic, and I may not, but if your reasoning is syntactically correct I will have to agree with your deductions assuming your premises. – Mirco Mannucci May 30 2011 at 23:37
Do I remember the following correctly? Using a coding of formal logic (and ZFC itself) in ZFC the following is correct: If ZFC can prove that coded-ZFC coded-proves a coded-theorem, then ZFC proves the theorem.
In other words, working in the world of ZFC-coded model theory, your theorem's are still correct in the 'real' world -- whatever you philosophical inclination for 'real'.
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No I think that this direction is the false one. It's the other way round: If ZFC proves a statement then ZFC proves that the 'coded ZFC proves the coded statement'. If your suggested direction would be true it would cause trouble with Gödels Incompleteness Theorem. – Stefan Hoffelner Apr 30 2010 at 8:48
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http://math.stackexchange.com/questions/40059/is-fracm-1x-an-unbiased-estimator-of-theta-for-given-pdf
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# Is $\frac{m-1}{x}$ an unbiased estimator of $\theta$ for given pdf?
Let $X$ be a continuous random variable with pdf, $$f(x;\theta)=\frac {\theta^m.x^{m-1}e^{-\theta x}} {(m-1)!} ; x\geq0, \theta>0$$ Is $\frac{m-1}{x}$ an unbiased estimator of $\theta$ for given pdf?
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If this is a homework question, please indicate what have you tried/how you got this result. – Dennis Gulko May 19 '11 at 11:12
no, i got this from past papers. – amul28 May 19 '11 at 11:43
## 2 Answers
For $m \geq 2$, $$\begin{align} \newcommand{\e}{\mathbb{E}}\newcommand{\rd}{\,\mathrm{d}} \e \frac{m-1}{X} = \int_0^\infty \frac{m-1}{x} \frac{\theta^m x^{m-1}}{(m-1)!} e^{-\theta x} \rd x = \theta \int_0^\infty \frac{\theta^{m-1} x^{m-2}}{(m-2)!} e^{-\theta x} \rd x = \theta \end{align}$$ since the integrand after the second equality is simply a gamma pdf with parameters $m-1$ and $\theta$ and so must integrate to one.
This is closely related to your previous question, by the way.
This "trick" pops up quite frequently in introductory mathematical statistics, so it's worth learning and being able to recognize when it can be applied.
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yes, even i understood that after long thinking. Thank you. – amul28 May 19 '11 at 13:08
Apologies but how is it different, if it is, from a copy of my answer that was posted 1 hour earlier? ;-) – Luboš Motl May 19 '11 at 14:18
@Luboš Motl, The point of my post was to demonstrate a very common trick for solving these problems. Mainly, factoring out certain terms to leave an integral of a probability density. This is not the only place that such a trick is useful. The OP was interested in an analytical means for solving the problem "by hand", as stated in his comment to you. This provides that, whereas the use of Mathematica does not. – cardinal May 19 '11 at 14:58
The downvote seems a little harsh, too. Particularly, as I was trying to demonstrate a (simple) complementary technique. – cardinal May 19 '11 at 14:59
@Luboš, if the answer has additional details, and in this case it certainly does, it is ok to post it. Your answer is a good one too, but it has a drawback, since it relies on having Mathematica, which is not cheap. – mpiktas May 23 '11 at 10:15
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it's an unbiased estimator if the expectation value of $(m-1)/x$ is equal to $\theta$. The expectation value is calculated as the integral over $x$ $$\int_0^\infty{\rm d}x \frac{m-1}{x} f(x;\theta) = \theta$$ so that it works, the answer is Yes. After a few trivial substitutions, the integral is just the Euler integral for the Gamma functions. In Mathematica, you may type
Integrate[(m - 1)/x theta^m x^(m - 1) Exp[-theta x] / Gamma[m], {x, 0, Infinity}]
and decode the result $\theta$ from the output, while accepting the sensible inequalities. Mathematica didn't exactly cancel the Gamma functions (or factorials) in an elegant way but they do cancel.
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thanks, but i want to know any other easy way of solving this by hand, because finding the integral by hand is very tedious! – amul28 May 19 '11 at 11:47
If it's not for a homework then what is the reason why can't you do it with a Mathematica, moreover an integral is quite trivial through the definition of Gamma function. – Ilya May 19 '11 at 12:35
because this is a previous exam paper so i want to know how to solve it. – amul28 May 19 '11 at 13:06
I didn't do it with Mathematica first, it's just an ordinary Euler integral! Cardinal didn't do any extra steps to calculate the Euler integral, either. He just wrote that it's an elementary integral just like I did. – Luboš Motl May 19 '11 at 14:19
To prove that the Euler integral $\int_0^\infty dt\exp(-t)t^n$ is equal to $n!=\Gamma(n+1)$ for integer $n$, just integrate it by parts several times, so that the exponent above $t$ is always reduced by one. In this way, you will collect factors of $n, n-1$, and so on, and finally you collect the whole $n!$ and the remaining integral will be $\int_0^\infty dt \exp(-t)$ which is easily integrated to one. – Luboš Motl May 19 '11 at 14:21
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http://mathoverflow.net/questions/98452?sort=oldest
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## Conditional expectation and algebraic expressions
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Let $\mathcal{A}$ and $\mathcal{B}$ be two sub-$\sigma$-algebras in a measure space. To each one, there is a conditional expectation associated, respectively $E^\mathcal{A}$ and $E^\mathcal{B}$. Given the two $\sigma$-algebras, we can form a third one, $\sigma(\mathcal{A},\mathcal{B})$, generated by both, and consequently, its conditional expectation $E^{\mathcal{A},\mathcal{B}}$.
My question is: knowing only the first two conditional expectations as projection operators in the measurable function space, can we obtain the third one as an algebraic expression of the first two, as the limit of a polynomial, for example?
A first try was to think about them as geometrical projections and try to find a complementar conditional expectation and calculate it in a similar fashion to $A\cup B=(A^C\cap B^C)^C$, and define the intersection as the limit of $(E^\mathcal{A}E^\mathcal{B})^n$. But the expected complementar $1-E^\mathcal{A}+E$ fails to be a conditional expectation.
Thank you!
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## 2 Answers
Clearly not. Let the measure space be the uniform measure on {$1,2,3,4$}. $A$ allows you to discern whether the number is greater or less than $2.5$ or not. $B$ allows you to discern whether the number is 0 or 1 mod $2$. Let $f$ be $x^2-5x+6$, then $E^Af=1$, $E^B f=1$, $E^{A,B}f=f$. One can't write $f$ as any polynomial or olgebraic expression in $1$ and $1$.
Edit: If you want $A \cap B$ instead of $A \cup B$ you can use $\lim \dots E^A E^B E^A E^B E^A E^B$, if the limit exists, since that is the projection onto the subspace fixed by both $E_A$ and $E_B$, which is the subspace of functions defined over both $A$ and $B$, which is the subspace you want to project onto.
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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.
By restricting to algebraic expressions you treat conditional expectations as merely projections, perhaps self-adjoint ones. That is, you miss too much structure.
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http://math.stackexchange.com/questions/189223/asymmetric-normal-probability-distribution?answertab=active
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# Asymmetric Normal Probability Distribution
I'm looking for a continuous probability distribution a little bit like the normal distribution but asymmetric. In my opinion this distribution applies to phenomenons related to response time in environments marked by resource contention.
Examples I have in mind are:
• In real life: the time it takes for my bus to go from my home to my office in the morning. In average it's around 15 minutes. However the way this duration varies each side of the mean value is asymmetric: it can hardly be 10 minutes less than the average but can easily take 10 minutes more.
• In computer capacity planning (which is the real domain where I want to use it ;-). One transaction needs to take place in 5 seconds (average) but my QoS constraint is that 90% of the time it takes less than 15 seconds.
Here is a diagram to illustrate my ideal distribution.
In this last example I could approximate the distribution to a Gaussian distribution and decide that 90% is roughly equivalent to a 1.5 standard deviation.
However I'm curious to know whether there is probability distribution more adapted to my problem.
The end goal is to deduce what percentage of my resources should be free (e.g. each CPU core should in average be at least 50% free, disk controllers bandwidth should be 50% free, etc) in order to satisfy the 90% threshold constraints.
## Edit
I'm adding more information here because I'm not convinced that the Log-Normal distribution fits the bill.
Going back to the example of my Bus journey, there is a minimum travel time which depends on propagation law limits (dictated by highway code or physics).
Similarly in a computer system, when my request runs unhampered by concurrent usage of the available physical resource, one can probably observe consistently close response times. I term this minimum latency and I ascribe the variations above this minimum latency time to other concurrent requests in real life.
The important thing here is that when contention increases, mean, median and mode values all increase when $\sigma$ increases.
Here is another diagram to illustrate what I mean.
So it looks like the Rayleigh distribution seems closer to what I need. However it also looks like it lacks some kind of "$\mu$" parameter since I have three sizing conditions to satisfy:
• average response time: 5s.
• In the CDF when the cumulated probability = 0.9 then the response time is 15s.
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Try a Gamma distribution, maybe? – Dilip Sarwate Aug 31 '12 at 11:50
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I think when you fit your distribution to the Rayleigh distribution buy some kind of fitting criterion like minimum mean squared error, there should be somehow one to one correspondance between your parameters and the parameters provided by the fit, namely by that specific Rayleigh distribution. – Seyhmus Güngören Aug 31 '12 at 16:58
## 4 Answers
I suggest Rayleigh distribution as it is quite similar to your figure, however it starts from zero. But one can shift it how he/she wants to. It is the distribution of the amplitude of the complex gaussian random variable.
http://en.wikipedia.org/wiki/Rayleigh_distribution
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It looks like this is the closest thing to what I need. I've edited my post accordingly. Thx. – Alain Pannetier Aug 31 '12 at 14:45
@AlainPannetier it is my pleasure. – Seyhmus Güngören Aug 31 '12 at 14:48
Response accepted. The reservations I had were due to the fact that I wrongly assumed that both conditions ("1. average response time: 5s" and "2/ In the CDF when the cumulated probability = 0.9 then the response time is 15s") were to be satisfied exactly. Instead these are two different ways of computing the standard deviation. To size the system I need to select the most severe value. Experimentations on the real platform will show whether this distribution needs to be calibrated. – Alain Pannetier Aug 31 '12 at 16:37
Try the log-normal distribution. It is probably the simpliest distribution that mimic the behavior you are searching for. It is easily implementable and has been successfully applied to many applied mathematics fields.
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If $X$ is normal, consider $Y=e^X$. It will be distributed as $$\frac{1}{x}e^{-\ln(x)^2} = \frac{1}{x^{1+\ln(x)}}$$
This is quite similar to your picture.
As Sasha said, it's the log-normal distribution.
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– Sasha Aug 31 '12 at 12:05
@Xoff, From what I understand this log-normal distribution does not work completely. What I need is a distribution in which the mode shifts rightwards when $\sigma$ increases. In Log-normal distributions mode, median and mean values shift leftwards when $\sigma$ increases. Thanks anyway. Already upvoted. – Alain Pannetier Aug 31 '12 at 14:50
If your asymmetric random variable is defined on $\mathbb{R}$, as opposed to $\mathbb{R}^+$, you may want to look into Azzallini's skew-normal distribution. It is implemented in R (package sn), and in Mathematica (ref-page).
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http://mathhelpforum.com/calculus/79785-spherical-coordinates-print.html
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# spherical coordinates!!!
Printable View
• March 21st 2009, 10:32 AM
althaemenes
spherical coordinates!!!
Hi,
For the following problem, how would I setup the triple integral? Do I have to convert it to spherical coordinates first:
many thanks....
Problem:
Integrate the function https://instruct.math.lsa.umich.edu/...b7704d4671.png over the solid above z=0 and contained between spheres centered on the origin with radii 1 and 4, excluding everything in the first octant.
integral =
(You should find the actual value when you evaluate the integral!)
• March 21st 2009, 11:34 AM
HallsofIvy
Quote:
Originally Posted by althaemenes
Hi,
For the following problem, how would I setup the triple integral? Do I have to convert it to spherical coordinates first:
many thanks....
Problem:
Integrate the function https://instruct.math.lsa.umich.edu/...b7704d4671.png over the solid above z=0 and contained between spheres centered on the origin with radii 1 and 4, excluding everything in the first octant.
integral =
(You should find the actual value when you evaluate the integral!)
Because of the spherical symmetry, the limits of integration in spherical coordinates are very simple. We want to go completely around so $\theta$ goes from 0 to $2\pi$. We want to go from the z axis down to the xy-plane so $\phi$ goes from 0 to $\pi/2$. Finally, of course, $\rho$ goes from 1 to 4.
I presume you know that, in polar coordinates, $x= \rho sin(\phi)cos(\theta)$ and $y= \rho sin(\phi)sin(\theta)$ so $-5x+ 5y= -5\rho sin(\phi)cos(\theta)+ 5 sin(\phi)sin(\theta)$ $= 5sin(\phi)(sin(\theta)- cos(\theta))$
Of course, the differential of volume in spherical coordinates is $\rho^2 sin(\phi)d\rho d\theta d\phi$.
• March 21st 2009, 11:46 AM
Showcase_22
$\int_1^4 \int_0^\frac{\pi}{2} \int_0^{2\pi} 5sin (\phi) (sin (\theta)-cos (\theta)) \rho^2 sin(\phi)d \rho d\phi d\theta$
I was having trouble deciding what the $\rho$ limits should be until I read HallsofIvy's post. Thanks!
• March 21st 2009, 01:55 PM
althaemenes
HallsofIvy and Showcase_22 (Rock) !!! Thanks guys!!
http://www.mathhelpforum.com/math-he...4dd0a69c-1.gif
=> (105/2)*Pi-(105/2)*Pi*sin(Pi)*cos(Pi)-(105/2)*Pi*cos(Pi)^2
=> 0
All times are GMT -8. The time now is 12:52 AM.
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http://homotopical.wordpress.com/tag/homotopy-theory/
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# Posts Tagged ‘Homotopy theory’
## London workshop on arithmetic geometry and homotopy theory
Posted by Andreas Holmstrom on May 8, 2012
At Imperial College London, from May 30th to June 1st. Many interesting talks, but the main theme appears to be the recent work of Harpaz and Schlank in which they use etale homotopy theory to study rational points and in particular obstructions to the local-global principle. Their big preprint is available here.
Related to this theme there is also a longer summer course in Lausanne in July/August with lectures by Harpaz and Schlank, as part of the of the EPFL program on rational points and algebraic cycles.
## Book draft by Jardine on local homotopy theory
Posted by Andreas Holmstrom on September 15, 2011
Rick Jardine recently posted a draft on his webpage for a book on local homotopy theory. He writes: “This a partial, rough manuscript for a monograph, which is tentatively to be published by Springer-Verlag. The book is meant to be a basic account of the homotopy theories of simplicial sheaves and presheaves, and the stable homotopy theory of presheaves of spectra. Selected applications are included.” The pdf file is available here.
Posted in Random things found on the web | Tagged: Homotopy theory, simplicial, stable homotopy theory | Leave a Comment »
## Motivic homotopy theory in Bonn, and much more
Posted by Andreas Holmstrom on March 11, 2010
The conference list is now updated with lots of exciting events for the summer. A small sample:
• Higher Structures in Topology and Geometry IV, Göttingen, June 2-4
• Young Topologists Meeting, Copenhagen, June 16-20
• Homotopy theory and derived algebraic geometry, Toronto, Aug 30 – Sep 3
• Geometric aspects of motivic homotopy theory, Bonn, Sep 6-10
I already mentioned the Paris conference on Cohomology of algebraic varieties, Hodge theory, algebraic cycles, and motives in last week of April, but there is still time to register.
Posted in events | Tagged: algebraic cycles, Cohomology, conference, derived algebraic geometry, events, Hodge theory, Homotopy theory, motives, Motivic homotopy theory, topology, workshop | 3 Comments »
## Homotopical categories and simplicial sheaves
Posted by Andreas Holmstrom on May 20, 2009
(This is an expanded version of the 2nd part of a talk I gave last month. For the first part, see this post.)
Homotopical categories
The topic for this post is “homotopical categories”, and their role in algebraic geometry. I want to emphasize that I am very much in the process of learning about all these things, so this post is based more on interest and enthusiasm than actual knowledge. I hope to convey some of the main ideas and why they could be interesting, and come back to the details in many future posts, after having learned more. I apologize for not defining everything carefully, and for brushing the “stable” aspects of the theory, i.e. spectra and sheaves of spectra, under the carpet.
There are many different ways to speak of “homotopical categories”, and I only use this expression because I don’t know of a better thing to call them. The most well-known approach is the language of model categories, invented by Quillen and developed by many others. There are many excellent online introductions, for example Dwyer-Spalinski, Goerss-Schemmerhorn, and appendix A2 of Jacob Lurie’s book on higher topos theory, available on his webpage. Other languages are given by the many different approaches to higher categories; see the nLab page and the survey of Bergner. Still other languages include Segal categories, A-infinity categories, infinity-stacks, and homotopical categories in the precise sense of Dwyer-Hirschhorn-Kan-Smith.
Although I don’t want to go into the details of all these different homotopical/higher-categorical subtleties, I will try to list some of the basic features that “homotopical” categories typically have.
• A homotopical category should behave like a nice category of topological spaces.
• In particular, there should be a class of morphisms called weak equivalences, and:
• To any homotopical category $M$, one should be able to associate a “homotopy category” $H$ and a functor $M \to H$ which is universal among functors sending weak equivalences to isomorphisms. Morally, $H$ is obtained from $M$ by “formally inverting the weak equivalences”.
• A homotopical category should admit all limits and colimits, and also homotopy limits and homotopy colimits.
• A homotopical category should be enriched over some kind of spaces, i.e. for any two objects $A,B$, the set $Hom(A,B)$ should be a “space” in some sense, for example a simplicial set, a topological space, or a chain complex of abelian groups.
Simplicial objects
Before talking about algebraic geometry, we need to recall some “simplicial language”. The category $\Delta$ is defined as follows. Objects are the finite ordered sets of the form $[n]:= \{ 0,1,2, \ldots , n \}$. Morphisms are order-preserving functions $[m] \to [n]$, i.e. functions such that $x \leq y \implies f(x) \leq f(y)$. If $C$ is any category, we define the category $sC$of simplicial $C$-objects to be the category in which the objects are the contravariant functors from $\Delta$ to $C$, and the morphisms are the natural transformations of functors. There is a functor from $C$ to $sC$ given by sending an object $X$ of $C$ to the corresponding constant functor, i.e. the functor sending all objects to $X$ and all morphisms to the identity morphisms of $X$.
Some examples:
• Take $C = Set$, the category of sets. The above construction gives us the category $sSet$ of simplicial sets. This category is “sort of the same as the category $Top$ of topological spaces”. The precise statement is that there is a pair of adjoint functors which make $Top$ and$sSet$ into Quillen equivalent model categories; in particular, their homotopy categories are equivalent (as categories). For the purposes of algebraic topology, we can work with any of these categories. For example, we can define homotopy groups and various generalized homology and cohomology groups of a simplicial set. The inclusion of $C$ into $sC$ corresponds to viewing a set as a discrete topological space. A weak equivalence between two simplicial sets is a morphism inducing isomorphisms on all homotopy groups.
• Take $C = Ab$, the category of abelian groups. There is a forgetful functor from $sAb$ to the category $sSet$, induced by the forgetful functor from$Ab$ to$Set$. The Dold-Kan correspondence tells us that there is an equivalence between $sAb$ and the category of (non-negatively graded) chain complexes of abelian groups. Under this equivalence, homotopy groups of a simplicial abelian group correspond to homology groups of a chain complex.
• Take$C = k-Alg$, the category of$k$-algebras for a commutative ring$k$. Then there is some kind of Dold-Kan correspondence between simplicial algebras and DG-algebras. See Schwede-Shipley for precise statements.
• Take $C = Shv$, the category of sheaves of sets on some topological space or site. Then $sShv$ is the category of simplicial sheaves. This category can also be viewed as the category of sheaves of simplicial sets on the site. Any category of simplicial sheaves is a “homotopical category” (I am not making this precise here). For example, one way of defining weak equivalences is to say that a morphism of simplicial sheaves is a weak equivalence iff it induces weak equivalences of simplicial sets on all stalks.
Homotopical categories in algebraic geometry
Now to algebraic geometry. Through a few examples I want to argue that homotopical categories (in particular categories of simplicial sheaves) provide a useful and natural setting for certain aspects of algebraic geometry.
Firstly, let’s consider the general problem of viewing a cohomology theory as a representable functor. In algebraic topology, the Brown representability theorem says that any generalized cohomology group is representable, when viewed as a functor on the homotopy category $Hot$ of topological spaces. In other words, there is a space $K$ such that the cohomology of a space $X$ is given by $Hom(X,K)$, where the $Hom$ is taken in the homotopy category. Examples include the Eilenberg-MacLane spaces $K(G, n)$, which represent the singular cohomology groups $H^n(X, G)$, and the space $BU \times \mathbf{Z}$, which represents K-theory. The existence of a long exact sequence relating the cohomology groups for various $n$ corresponds to the fact that the different Eilenberg-MacLane spaces fit together to form a so called spectrum. The Brown representability theorem is best expressed using the language of spectra, i.e. stable homotopy theory, but I want to postpone a discussion of this to a future post. An interesting aspect of Brown representability for singular cohomology is that by identifying the coefficient group $G$ with the corresponding Eilenberg-MacLane space, the two arguments of a singular cohomology group $H^n(X, G)$, namely the space $X$ and the coefficient group $G$, suddenly are on equal footing. By this I mean that they both live in the same category of topological spaces, rather than in the two separate worlds of topological spaces and abelian groups, respectively.
In classical algebraic geometry, there is no analogue of Brown representability. Most cohomology theories are of the form $H^n(X, F)$, where $X$ is some kind of variety, and $F$ is a sheaf of abelian groups. One may ask if there is a way to express such a cohomology group as a representable functor. In order to obtain a picture parallell to the topological picture above, a necessary requirement is to have a homotopical category in which the variety $X$ and the sheaf $F$ both live as objects, “on equal footing”. One possibility for such a category is some category of simplicial sheaves. In order to explain how this works, let us fix some category $Var$ of varieties, for example the category smooth varieties over some base field $k$. Let us also fix some Grothendieck topology on this category, for example the Zariski topology, the Nisnevich topology, the etale topology, or some flat topology. This defines a site, and we can speak of sheaves on this site, i.e. contravariant functors on $Var$, satisfying a “glueing” or “descent” condition with respect to the given topology.
Since Grothendieck, we are familiar with the idea of identifying a variety with the sheaf of sets that it represents, by the Yoneda embedding. We mentioned earlier that for any category $C$, there is a functor $C \to sC$. Taking $C$ to be the category of sheaves of sets, we get a functor from sheaves of sets to simplicial sheaves. In particular, any variety can be viewed as a simplicial sheaf, by composing the Yoneda embedding with the canonical functor from sheaves of sets to simplicial sheaves.
We also want to show that a sheaf of abelian groups can be viewed as a simplicial sheaf. We can regard any abelian group as a chain complex, by placing it in degree zero, and placing the zero group in all other degrees. This gives an embedding of the category of abelian groups into the category of chain complexes, and by composing with the Dold-Kan equivalence we get a functor from abelian groups to simplicial sets. This induces a functor from sheaves of abelian groups to simplicial sheaves. More generally, any complex of sheaves of abelian groups can be viewed as a simplicial sheaf.
Now one could hope for an analogue of Brown representability, namely that the sheaf cohomology group $H^n(X, F)$ could be expressed as $Hom(X,F)$, where the Hom is taken in the homotopy category of simplicial sheaves. It seems to be the case that something along these lines should be true. For example, this nLab page on cohomology seems to imply that all forms of cohomology should be of this form, at least sheaf cohomology groups of the type just described. Also, Hornbostel has proved a Brown representability theorem in the setting of motivic homotopy theory.
There are many other phenomena in algebraic geometry which also seem to indicate that categories of simplicial sheaves might be more natural to study than the smaller categories of schemes and varieties we typically consider. Some examples (longer explanations of these will have to wait until future posts):
• It seems to be the case that almost any geometric object generalizing the concept of a variety can be thought of as a simplicial sheaf. Examples: Simplicial varieties, stacks, algebraic spaces.
• Deligne’s groundbreaking work on Hodge theory in the 70s (see Hodge II and Hodge III) uses in a crucial way that the singular cohomology of a complex variety can be defined on the larger category of simplicial varieties. Simplicial varieties are special cases of simplicial sheaves, and I believe it should be true that functors on simplicial varieties can be extended to simplicial sheaves.
• Simplicial varieties/schemes also pop up naturally in other settings. For example, Huber and Kings need K-theory of simplicial schemes for their work on the motivic polylogarithm.
• As already indicated, simplicial sheaves appears to be the most natural domain of definition for many different kinds of cohomology theories.
• Morel and Voevodsky’s A1-homotopy theory (also known as motivic homotopy theory) is based on categories of simplicial sheaves for the Nisnevich topology.
• Brown showed that Quillen’s algebraic K-theory can be thought of as “generalized sheaf cohomology”, where the coefficients is no longer a sheaf of abelian groups, but a simplicial sheaf.
• The work of Thomason relating algebraic K-theory and etale cohomology uses the language of simplicial sheaves.
• Simplicial sheaves provide a natural language for “resolutions”. For example, it gives a unified picture of the two methods for computing sheaf cohomology: Cech cohomology and injective resolutions.
• Simplicial sheaves seems to be the most natural language for descent theory.
• Toen‘s work on higher stacks can be formulated in terms of simplicial sheaves.
• Homotopy categories of simplicial sheaves can be thought of a generalization of the more classical derived categories of sheaves. The homotopical point of view seems to clarify some unpleasant aspects of the classical theory of triangulated categories.
See also the nLab entry on motivation for sheaves, cohomology, and higher stacks.
Questions
I hope to come back to many of these examples in detail. For now, I just want to list a few questions which I find intriguing.
• To define a category of simplicial sheaves, we must choose a Grothendieck topology. How does this choice affect the properties of the category we obtain? Morel and Voevodsky work with the Nisnevich topology, Huber and Kings work with the Zariski topology, and Toen (at least sometimes) works with some flat topology. For some purposes, it seems to be the case that we don’t need a topology at all, instead we can just work with simplicial presheaves. What is the role of the Grothendieck topology?
• Most of the above examples are developed for varieties over a base field of characteristic zero. Based on the above, it seems reasonable to believe that simplicial sheaves are useful in this case, but what if the base scheme is field of characteristic p, a local ring, a Dedekind domain, or something even more general? Is it the case that simplicial sheaves is the most natural language for understanding cohomology theories for arithmetic schemes, such as schemes which are flat and of finite type over $Spec(\mathbb{Z})$? Are simplicial sheaves important in number theory/Arakelov theory/geometry over the field with one element? What are the obstacles to “doing homotopy theory over an arithmetic base”?
Obviously I hope that there will be interesting answers to these questions, but I am still completely in the dark as to what these answers might be.
Posted in Uncategorized | Tagged: arithmetic geometry, arithmetic schemes, Brown representability, Cohomology, Deligne, generalized sheaf cohomology, higher categories, homotopical categories, Homotopy theory, Lurie, Morel, Motivic homotopy theory, Quillen, simplicial schemes, simplicial sheaves, stacks, Toen, Voevodsky | 3 Comments »
## Resources on motivic homotopy theory
Posted by Andreas Holmstrom on December 14, 2008
I found a really nice page today, maintained by Aravind Asok, with notes and resources on motivic homotopy theory. Worth checking out.
Posted in Random things found on the web | Tagged: Homotopy theory, Motivic homotopy theory | 2 Comments »
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http://mathhelpforum.com/discrete-math/69156-countability-proof.html
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# Thread:
1. ## Countability Proof
Fix $n >= 1$. Show that if A1,A2, . . . ,An are countable, then A1×A2× . . . × An is countable.
Does this require knowledge of the fact that N x N is countably infinite?
Thanks in advance.
2. Hello,
Originally Posted by h2osprey
Fix $n >= 1$. Show that if A1,A2, . . . ,An are countable, then A1×A2× . . . × An is countable.
Does this require knowledge of the fact that N x N is countably infinite?
Thanks in advance.
You can use the fact that $\mathbb{N}^n$ is countable. Which can be done by induction, and you'll have to use the fact that $\mathbb{N} \times \mathbb{N}$ is countable.
1.
You know that there exists an injective mapping $\phi_2 ~:~ \mathbb{N}^2 \to \mathbb{N}$
Let $n \geqslant 2$
Basis : for n=2, it's verified (I assume it's a know fact for you)
Inductive hypothesis : assume that there exists an injective mapping $\phi_n ~:~ \mathbb{N}^n \to \mathbb{N}$
Now, you have to prove that there exists an injective mapping $\phi_{n+1} ~:~ \mathbb{N}^{n+1} \to \mathbb{N}$
For this, define $\phi_{n+1}$ this way :
$\phi_{n+1}(x_1,\dots,x_n,x_{n+1})=(\phi_n(x_1,\dot s,x_n),x_{n+1})$
It is easy to show that it's injective, knowing that $\phi_n$ is injective.
2.
Now prove that there exists an injection $\psi$ from $A_1 \times \dots \times A_n$ to $\mathbb{N}^n$
Since $A_1,\dots,A_n$ are countable, there exist injections $\psi_i ~:~ A_i \to \mathbb{N}$
Define $\psi (x_1,\dots,x_n)=(\psi_1(x_1),\dots,\psi_n(x_n))$
Once again, it's easy to show that it's injective.
3.
$A_1 \times \dots \times A_n \stackrel{\psi}{\longrightarrow} \mathbb{N}^n \stackrel{\phi_n}{\longrightarrow} \mathbb{N}$
We know that $\psi, \phi_n$ are injective.
Now you just have to prove that the composite of two injective functions is injective, in particular $\phi_n \circ \psi$.
And you're done.
3. Originally Posted by h2osprey
Fix $n >= 1$. Show that if A1,A2, . . . ,An are countable, then A1×A2× . . . × An is countable.
Does this require knowledge of the fact that N x N is countably infinite?
Thanks in advance.
If you know that $\mathbb{N}\times \mathbb{N}$ is countable then it is easy to prove this result.
If $|A| = |B|= |\mathbb{N}|$ then $|A\times B| = |\mathbb{N}\times \mathbb{N}|$ is countable.
Thus, $|A_1\times ... \times A_n| = |\mathbb{N} \times ... \times \mathbb{N}|$.
We established that $\mathbb{N}^2$ was countable.
If $\mathbb{N}^k$, $k\geq 2$ is countable then $\mathbb{N}^{k+1} = \mathbb{N}^k \times \mathbb{N}$ is countable.
The rest follows by induction.
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http://mathoverflow.net/questions/93148?sort=newest
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## Unitary representations of Quantum Groups
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $\mathfrak{g}$ be a finite-dimensional complex simple Lie algebra and let $U_q(\mathfrak{g})$ be some incarnation of the quantized universal enveloping algebra of $\mathfrak{g}$; here I am assuming that $q \in \mathbb{C}^\times$ is not a root of unity. I am interested in unitarizability of representations of $U_q(\mathfrak{g})$ with respect to various $*$-structures.
The $*$-structures on $U_q(\mathfrak{g})$ have been classified; this can be found, for example, in section 9.4 of A Guide to Quantum Groups by Chari and Pressley, for example. For the $*$-structure known as the compact real form of $U_q(\mathfrak{g})$, it is known that each finite-dimensional irreducible representation $V_\lambda$ admits an invariant inner product, i.e. a positive-definite Hermitian form such that $$\langle av, w \rangle = \langle v, a^* w \rangle$$ for all $v,w \in V_\lambda$ and all $a \in U_q(\mathfrak{g})$.
### Question: for an arbitrary $*$-structure on $U_q(\mathfrak{g})$, has anybody classified the set of dominant integral weights $\lambda$ for which $V_\lambda$ admits an invariant inner product?
Chari and Pressley mention in passing in their book that this is an open question, but that was in 1995 and I thought I'd see if anybody had resolved it in the meantime. I would expect that for an arbitrary $*$-structure, most irreps do not have such an invariant inner product, since in the classical situation the corresponding real form of the group is not compact, so you can't just average over Haar measure as you do in the compact case.
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## 1 Answer
As far as I know, there is no such classification or even a good attempt. There are some results about unitary representations of quantum groups with the suitable choice of involution. L.Vaksman and his collaborators studied unitary representations for quantum analogs of $SU(n,n)$ (Most of these papers are in arxiv). Also there are papers of Klimyk, Pakulyak etc. But the classification is done only for the trivial case of quantum $SU(1,1)$.
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http://mathoverflow.net/questions/4266/sobolev-norms-of-eigenfunctions
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## Sobolev norms of eigenfunctions
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Let D be a domain in R^n, and let f be an eigenfunction of the Laplacian with Dirichlet boundary condition with eigenvalue $\lambda$. Assume that f has L^2 norm 1. I want to know if I can say anything about the Sobolev s-norm of f (interms of s \lambda and D) ?
In particular, I want to know if it is true that |f|_s is like \lambda^{\frac{s}{2}}.
Same question for the Neumann and dbar-Neumann boundary conditions.
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When you say Sobolev s-norm, are you working in the Hilbert space $H^{s}_{0}$ (At least for the Dirichlet BC)? Any conditions on $s$? – MLevi Nov 5 2009 at 16:30
## 3 Answers
There is an estimate of the form
|f|_s < C(D, \lambda, s)|f|_0
where |f|_0 = L^2 norm of f and |f|_s = Sobolev s-norm. There are different ways to get this estimate, depending on how sharp an estimate you need and the regularity of the boundary of D.
In particular, you can apply any a priori estimate for the Sobolev norm of the solution f to
Laplacian(f) = h
in terms of h and bootstrap to get whatever you want.
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Yes. What I am trying to figure out is the precise nature of the constant C(D,\lambda,s). e.g. is it true that C is like \lambda^{\frac{s}{2}} – Debraj Chakrabarti Nov 6 2009 at 8:03
This sounds right to me, and, if it weren't for the boundary terms, easily proved by integration by parts. I am sure the boundary integrals can be handled, if the boundary of D is smooth, but someone else will need to explain in more detail how. – Deane Yang Nov 6 2009 at 18:15
### 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.
Here are some of my thoughts on the question. Fix `$s\in(0,\frac{1}{2})$`. Then `$C:=\sup_{r\geq 0}\frac{(1+r^{s})^{2}}{1+r}$`. Notice then that `$\int(1+|\xi|^{s})^{2}||\widehat{f}(\xi)|^{2}d\xi\leq C\int(1+|\xi|)^{2}|\widehat{f}(\xi)|^{2}d\xi\int |\widehat{f}(\xi)|^{2}d\xi$`. Now, as mentioned by Deane, there may be some issues with the boundary `$\partial D$`. Suppose that `$f\in H^{1}_{0}(D)\cap H^{2}(D)$` and `$-\triangle f = \lambda f$` (We maybe be able to drop the second order regularity of `$f$` if more regularity is assumed on the boundary for example). After integrating by parts and using perhaps using some sort of Poincaré inequality (need some sort of boundedness for the domain), one can see by integration by parts that `$||f||_{H^{1}_{0}(D)}\sim\lambda$`. I THINK that `$||f||_{H^{s}(D)} \sim\big(\int(1+|\xi|^{s})^{2}|\widehat{f}(\xi)|^{2}d\xi\big)^{\frac{1}{2}}$`, but I'm not sure. In fact this might be another question... I'm not very familiar with fractional Sobolev spaces - much less on subsets of `$\mathbb{R}^{n}$`. If it were true (it should be true for `$s$` an integer - See Evans page 282), then your result would be that `$||f||_{H^{s}(D)}\lesssim_{D}C_{n}\lambda$` (modulo $D$ because of the Poincaré inequality - which would require some sort of boundedness of one coordinate). This was my first idea. I'm sure there are better ideas/results. I hope this helps.
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Sorry for the $C_{n}$. I meant $C$ – MLevi Nov 7 2009 at 7:25
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Is it possible to just prove the desired inequality for s equal to a positive integer (avoiding Fourier transform) and then infer the result for all s by interpolation? It seems to me that the desired estimates can be inferred from results or techniques presented in, say, Gilbarg-Trudinger or one of Stein's books on singular integral operators. But I'm not sure and don't have the time to check. I confess to having never learned any of this properly. – Deane Yang Nov 7 2009 at 18:33
@Deane You might be right. One may need to use some C-Z theory. – MLevi Nov 7 2009 at 20:39
Let me make the question more precise -- f is normalized to have L^2 norm 1.
Then f is in $H^1_0(D)$ and the Sobolev 1-norm of f is \lambda.
We want the norm of f in H^s(D).
I'll be glad if anybody can help.
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@Debraj: you should edit your original question, instead of providing clarification in an answer. It looks like you didn't register your account, so if you're having trouble logging in to edit, please read mathoverflow.net/questions/24/… – Scott Morrison♦ Nov 5 2009 at 21:21
@Debraj: Morrison has a good point, but now that your question has been made more clear, what do you impose on $s>0$? Have you thought of using the Fourier transform and trying some interpolation? – MLevi Nov 5 2009 at 21:30
Debraj: I would be surprised if you could obtain an explicit formula for the norm. More likely I would expect you can obtain a bound in $H^s$ for $0 < s < 1$ by an interpolation inequality as MLevi has suggested. You will probably have something like: $||u||_{H^s} \leq C ||u||_{L^2}||u||_{H^1}$. – Dorian Oct 14 2010 at 21:08
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http://mathhelpforum.com/discrete-math/145277-simple-graph-proof-question.html
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# Thread:
1. ## Simple graph proof question
Hi,
The exercise is as follows:
A simple graph, also called a strict graph, is an unweighted, undirected graph containing no graph loops or multiple edges. A well-known theorem states that the sum of the degrees of the vertices of a simple graph equals twice the number of edges of the graph.
Prove the following:
There is no simple graph with 12 vertices and 28 edges so that
(i) all vertices have degree 3 or 4
(ii) all vertices have degree 3 or 6
Thank you very much
2. Notice that $2(28)=56$.
So (i) is impossible.
For (ii) let s be the number of degree six vertices and t be the number of degree three.
Then $6s+3t=56~\&~s+t=12$.
What is wrong with that?
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http://physics.stackexchange.com/questions/3568/temperature-of-a-system-of-molecules?answertab=oldest
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# Temperature of a System of molecules
Suppose I have a closed system with N molecules in it which are vibrating and all motion equations (rotation, translation and vibration) of the system are known along with any EM field equations in the region. Given all these information how do we calculate (not measure) the temperature of this system?
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The concept of temperature is actually slightly different for microcanonical ensembles compared to canonical ensembles. – QGR Jan 22 '11 at 18:21
## 5 Answers
AIB, As you have read the temperature is a measure of the change of energy of a system to the change of the number of microstates, or:
$$T = \dfrac{\partial{E}}{\partial{S}}$$
Where: $$S = \ln \Omega$$
and $\Omega$ is the number of microstates.
However, it is important to understand what a microstate is. In general a microstate describes all of the exact positions and momentums of all the particles in a system (or particles in a box). The microstate is often thought of as a point in a 6N dimension space, where N is the number of particles and the factor of 6 is related to there being 3 position components in a 3D space and 3 momentum components in 3D space (however, one could certainly consider other parameter).
If we could know all of the positions and momentums of all the particles than we would know the precise microstate of the system. This precise microstate would have an exact energy associated with it. However, we do not have an ability to know the precise microstate of a system, so we must appeal to a statistical concept.
So at any one instant, there are a large number of microstates a system could be in. Again, each microstate being a unique description of the exact position and momentum of all particles in the box. What this also suggests, is that for any microstate, there are a large number of other microstates that would have the same energy associated with them.
The question asked was: "Given all these information how do we calculate (not measure) the temperature of this system?"
The answer to this question would be determined first by the range of all the free parameters and the values of those parameters that that would keep the energy constant. The number of combinations possible within the energy constraint is the number of microstates of the systems, the log of which is the entropy.
The next step is tricky, because we would have to ask how the energy changes with a change in entropy. This is where the partition function comes into play, assuming energy of each microstate is equivalent, if we look at the partion function we see:
$${Z} = \sum_i e^{-\frac{\partial{S}}{\partial{E}}{E_i}}$$
$$Z = Ne^{-\frac{1}{T}E_i}$$
$$N = Ze^{\frac{1}{T}E_i}$$
So we can see that the temperature is a variable that controls the growth of states if we keep the value of Z constant. If I know an exact microstate from which to derive the energy, then I still won't necessarily know anything about the temperature until I can partition the relavant space in question. In this sense, the temperture is a sort of hyperbolic phase factor which changes the range of states.
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I got the answer from here.
Temperature is a purely statistical concept. It is defined as:
1/(k T) = d[Log(Omega(E))]/dE
where Omega(E) is the number energy eigenstates in a small interval around the energy E of the system.
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Actually temperature is usually defined as $T = (\frac{\partial S}{\partial E})^{-1}$, where $S$ is the entropy of the system. – Joe Fitzsimons Jan 22 '11 at 7:35
But Boltzmann also taught us that S = log$\Omega$ :) – wsc Jan 22 '11 at 22:51
Dear AIB, the temperature can clearly be anything in the system you described. The temperature is not a universal constant calculable for each system: the temperature is a macroscopic quantity describing the state of the system.
Roughly speaking, the absolute temperature $T$ in Kelvins will be related to the kinetic energy of the average molecule by $E=5kT/2$ - assuming that two rotational degrees of freedom exist from the absolute zero to the temperature $T$ which they usually don't. The kinetic energy depends on the speed of the molecules - it is not calculable out of nothing.
The heat capacity factor $5/2$ above should be replaced by a number that depends on the type of the molecule but for complicated enough stuff, the dependence won't be quite linear, anyway.
Cheers LM
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The equipartition theorem only applies for nearly ideal gases, and weakly coupled harmonic oscillators. – QGR Jan 22 '11 at 18:17
@QGR: Where did you get that from? Equipartition works for arbitrary degrees of freedom provided the energy level spacing is small compared to kT. – Johannes Jan 23 '11 at 5:12
@Johannes: The equipartition theorem is only exact in the limit of decoupled harmonic oscillators, or free particles. For a strongly coupled system like a liquid, it breaks down. – QGR Jan 23 '11 at 7:34
@QGR: Ok, in systems in which you can't define the independent degrees of freedom you can't apply equipartition. I wouldn't call that 'breaking down' of equipartition. – Johannes Jan 23 '11 at 7:54
1
– Marek Jan 24 '11 at 0:12
show 3 more comments
You ofcourse understand that it's simply impossible to obtain all possible initial conditions (position, momentum and other internal degrees of freedom) for a large system. Furthermore, we actually have a good knowledge of the equations of motion that govern the dynamics of such a system. E.g. in classical physics you have some set of differential equations that follow from Newton's law.
So suppose that we did know the inital conditions. Than there is still the other practical impossibility of actually determining the state of the system at some later time. We simply don't know how to solve these equations of motions analytically for large systems. Let alone the fact that we cannot keep track of each individual particle.
This is why one turns to a statistical description of the system. In a statistical description you are not interested in what each individual particle is doing, but rather what the system is doing as a whole. So instead of making a statement like: particle 1 has momentum P and position X and particle 2 has some other properties, you would talk about the probability of having a particle in some state and another in some other state.
A key ingredient in this approach is the concept of time averaging. The probabities mentioned before arise by averaging over a typical timescale. You assume that within that timescale the system explores all possible states it can end up in, as in, the system sits in all possible configurations it can during that timescale. It explores the phase space of the system. The statistical state of the system is obtained through this time-averaging (which, mind you, is not done by actually starting with a state and then see what it does over a typical timescale. Rather, one assumes that the system will explore the phase space in some way, and you're not really interested in the way it does that).
What does this have to do with temperature? For that we need to introduce the concept of entropy first. The entropy is nothing more but the logarithm of the total number of states in the phase space that the system can evolve into. It is a measure of the size of the phase space that is available to the system. Again, this makes no reference to the particular state the system starts out in -- not the microscopic details anyways. What it does require is the total energy of the initial state. The reason should be clear: microscopically, the system can only evolve into other states with the same total energy. So when we talk about this time averaging, where the system explores the phase space, than the total energy serves as a constraint on the possible "explorations". There are other constraints as well, such as the volume and number of particles.
Energy conservation places a severe constraint on the subspace of the phase space available to the system. At the same time the entropy is precisely determined by the size of this subspace. And this is where temperature comes in. By changing the total energy, the total entropy is changed -- the system now explores a different subspace of the phase space. Temperature is nothing more but the rate of change of entropy. It is the derivative of the entropy with respect to the total energy, 1/T = dS/dE. So it is in essence a measure on how the size of the phase space changes when you a little energy to the system (or extract some).
To give a little more detail on this. You can also think of changing the entropy and see how the total energy of the system changes. You can also look at the change of total energy due to a change in the particle number, while keeping the entropy fixed. This leads to the chemical potential. Pressure and volume obey similar relations.
So back to your question. How does one, in principle, calculate the temperature of the system? The only gedankenexperiment I can come up with is as follows: You let the system evolve. Then you perform a time averaging over this motion. You come up with some average state of the system, and, more importantly, the total entropy of the system. Now you add a little energy to the system and you the same thing. The change of entropy, divided by the change of energy is 1/T.
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I think time averaging is only one way of approaching statistical mechanics. It assumes ergodic theory. – QGR Jan 22 '11 at 18:24
Does it mean that we can not associate temperature to a very small number of Hydrogen Atoms even in principle , even if we know their KE (upto precision permissible by uncertainty)? A set of hydrogen atoms with a given KE can have different temperature depending on how the system evolve? – AIB Jan 25 '11 at 3:59
For a gas of particles the microstates of the system have energy $E_i$, and the distribution of these states in a gas with temperature $T$ is given by $$Z~=~\sum_i e^{-E_i\beta}$$ with $\beta~=~1/kT$. It is not hard to see that $$F~=~ln(Z)/\beta$$ which is the Helmholtz free energy. The internal energy is $$U~=~-\Big(\frac{\partial Z}{\partial\beta}\Big).$$ The free energy $F~=~U~-~TS$ and $S~=~k(ln~Z~+~\beta U)$ can be used to write the partition function according to macroscopic varianbles $Z~=~e^{-F\beta}$. $\beta$ may then be computed by taking a derivative with the Helmholtz free energy $F$ $$\beta~=~-\frac{\partial Z}{\partial\beta}.$$
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The Boltzmann factor is $e^{-\beta E_i}$, not $e^{-E_i/\beta}$. – QGR Jan 23 '11 at 7:35
Yep, my mistake. I first had kT and I did in the editor a global change from kT to \beta. – Lawrence B. Crowell Jan 23 '11 at 13:18
This is a nice lesson on statistical mechanics, but doesn't answer the question. He is asking how to calculate T from the microscopic state, but your answer already starts with an ensemble. – Bruce Connor Jan 23 '11 at 20:05
Temperature really does not make sense for a microstate. Temperature is a measure which emerges with coarse graining, entropy and the whole nine yards of cloth. – Lawrence B. Crowell Jan 23 '11 at 22:24
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http://terrytao.wordpress.com/tag/abc-conjecture/
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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
# Tag Archive
You are currently browsing the tag archive for the ‘abc conjecture’ tag.
## The probabilistic heuristic justification of the ABC conjecture
18 September, 2012 in math.NT, math.PR | Tags: abc conjecture, probabilistic heuristic | by Terence Tao | 34 comments
There has been a lot of recent interest in the abc conjecture, since the release a few weeks ago of the last of a series of papers by Shinichi Mochizuki which, as one of its major applications, claims to establish this conjecture. It’s still far too early to judge whether this proof is likely to be correct or not (the entire argument encompasses several hundred pages of argument, mostly in the area of anabelian geometry, which very few mathematicians are expert in, to the extent that we still do not even have a full outline of the proof strategy yet), and I don’t have anything substantial to add to the existing discussion around that conjecture. (But, for those that are interested, the Polymath wiki page on the ABC conjecture has collected most of the links to that discussion, and to various background materials.)
In the meantime, though, I thought I might give the standard probabilistic heuristic argument that explains why we expect the ABC conjecture to be true. The underlying heuristic is a common one, used throughout number theory, and it can be summarised as follows:
Heuristic 1 (Probabilistic heuristic) Even though number theory is a deterministic subject (one does not need to roll any dice to factorise a number, or figure out if a number is prime), one expects to get a good asymptotic prediction for the answers to many number-theoretic questions by pretending that various number-theoretic assertions ${E}$ (e.g. that a given number ${n}$ is prime) are probabilistic events (with a probability ${\mathop{\bf P}(E)}$ that can vary between ${0}$ and ${1}$) rather than deterministic events (that are either always true or always false). Furthermore:
• (Basic heuristic) If two or more of these heuristically probabilistic events have no obvious reason to be strongly correlated to each other, then we should expect them to behave as if they were (jointly) independent.
• (Advanced heuristic) If two or more of these heuristically probabilistic events have some obvious correlation between them, but no further correlations are suspected, then we should expect them to behave as if they were conditionally independent, relative to whatever data is causing the correlation.
This is, of course, an extremely vague and completely non-rigorous heuristic, requiring (among other things) a subjective and ad hoc determination of what an “obvious reason” is, but in practice it tends to give remarkably plausible predictions, some fraction of which can in fact be backed up by rigorous argument (although in many cases, the actual argument has almost nothing in common with the probabilistic heuristic). A famous special case of this heuristic is the Cramér random model for the primes, but this is not the only such instance for that heuristic.
To give the most precise predictions, one should use the advanced heuristic in Heuristic 1, but this can be somewhat complicated to execute, and so we shall focus instead on the predictions given by the basic heuristic (thus ignoring the presence of some number-theoretic correlations), which tends to give predictions that are quantitatively inaccurate but still reasonably good at the qualitative level.
Here is a basic “corollary” of Heuristic 1:
Heuristic 2 (Heuristic Borel-Cantelli) Suppose one has a sequence ${E_1, E_2, \ldots}$ of number-theoretic statements, which we heuristically interpet as probabilistic events with probabilities ${\mathop{\bf P}(E_1), \mathop{\bf P}(E_2), \ldots}$. Suppose also that we know of no obvious reason for these events to have much of a correlation with each other. Then:
• If ${\sum_{i=1}^\infty \mathop{\bf P}(E_i) < \infty}$, we expect only finitely many of the statements ${E_n}$ to be true. (And if ${\sum_{i=1}^\infty \mathop{\bf P}(E_i)}$ is much smaller than ${1}$, we in fact expect none of the ${E_n}$ to be true.)
• If ${\sum_{i=1}^\infty \mathop{\bf P}(E_i) = \infty}$, we expect infinitely many of the statements ${E_n}$ to be true.
This heuristic is motivated both by the Borel-Cantelli lemma, and by the standard probabilistic computation that if one is given jointly independent, and genuinely probabilistic, events ${E_1, E_2, \ldots}$ with ${\sum_{i=1}^\infty \mathop{\bf P}(E_i) = \infty}$, then one almost surely has an infinite number of the ${E_i}$ occuring.
Before we get to the ABC conjecture, let us give two simpler (and well known) demonstrations of these heuristics in action:
Example 1 (Twin prime conjecture) One can heuristically justify the twin prime conjecture as follows. Using the prime number theorem, one can heuristically assign a probability of ${1/\log n}$ to the event that any given large integer ${n}$ is prime. In particular, the probability that ${n+2}$ is prime will then be ${1/\log(n+2)}$. Making the assumption that there are no strong correlations between these events, we are led to the prediction that the probability that ${n}$ and ${n+2}$ are simultaneously prime is ${\frac{1}{(\log n)(\log n+2)}}$. Since ${\sum_{n=1}^\infty \frac{1}{(\log n) (\log n+2)} = \infty}$, the Borel-Cantelli heuristic then predicts that there should be infinitely many twin primes.
Note that the above argument is a bit too naive, because there are some non-trivial correlations between the primality of ${n}$ and the primality of ${n+2}$. Most obviously, if ${n}$ is prime, this greatly increases the probability that ${n}$ is odd, which implies that ${n+2}$ is odd, which then elevates the probability that ${n+2}$ is prime. A bit more subtly, if ${n}$ is prime, then ${n}$ is likely to avoid the residue class ${0 \hbox{ mod } 3}$, which means that ${n+2}$ avoids the residue class ${2 \hbox{ mod } 3}$, which ends up decreasing the probability that ${n+2}$ is prime. However, there is a standard way to correct for these local correlations; see for instance in this previous blog post. As it turns out, these local correlations ultimately alter the prediction for the asymptotic density of twin primes by a constant factor (the twin prime constant), but do not affect the qualitative prediction of there being infinitely many twin primes.
Example 2 (Fermat’s last theorem) Let us now heuristically count the number of solutions to ${x^n+y^n=z^n}$ for various ${n}$ and natural numbers ${x,y,z}$ (which we can reduce to be coprime if desired). We recast this (in the spirit of the ABC conjecture) as ${a+b=c}$, where ${a,b,c}$ are ${n^{th}}$ powers. The number of ${n^{th}}$ powers up to any given number ${N}$ is about ${N^{1/n}}$, so heuristically any given natural number ${a}$ has a probability about ${a^{1/n - 1}}$ of being an ${n^{th}}$ power. If we make the naive assumption that (in the coprime case at least) there is no strong correlation between the events that ${a}$ is an ${n^{th}}$ power, ${b}$ is an ${n^{th}}$ power, and ${a+b}$ being an ${n^{th}}$ power, then for typical ${a,b}$, the probability that ${a,b,a+b}$ are all simultaneously ${n^{th}}$ powers would then be ${a^{1/n-1} b^{1/n-1} (a+b)^{1/n-1}}$. For fixed ${n}$, the total number of solutions to the Fermat equation would then be predicted to be
$\displaystyle \sum_{a=1}^\infty \sum_{b=1}^\infty a^{1/n-1} b^{1/n-1} (a+b)^{1/n-1}.$
(Strictly speaking, we need to restrict to the coprime case, but given that a positive density of pairs of integers are coprime, it should not affect the qualitative conclusion significantly if we now omit this restriction.) It might not be immediately obvious as to whether this sum converges or diverges, but (as is often the case with these sorts of unsigned sums) one can clarify the situation by dyadic decomposition. Suppose for instance that we consider the portion of the sum where ${c=a+b}$ lies between ${2^k}$ and ${2^{k+1}}$. Then this portion of the sum can be controlled by
$\displaystyle \sum_{a \leq 2^{k+1}} \sum_{b \leq 2^{k+1}} a^{1/n-1} b^{1/n-1} O( ( 2^k )^{1/n - 1} )$
which simplifies to
$\displaystyle O( 2^{(3/n - 1)k} ).$
Summing in ${k}$, one thus expects infinitely many solutions for ${n=2}$, only finitely many solutions for ${n>3}$ (indeed, a refinement of this argument shows that one expects only finitely many solutions even if one considers all ${n>3}$ at once), and a borderline prediction of there being a barely infinite number of solutions when ${n=3}$. Here is of course a place where a naive application of the probabilistic heuristic breaks down; there is enough arithmetic structure in the equation ${x^3+y^3=z^3}$ that the naive probabilistic prediction ends up being an inaccurate model. Indeed, while this heuristic suggests that a typical homogeneous cubic should have a logarithmic number of integer solutions of a given height ${N}$, it turns out that some homogeneous cubics (namely, those associated to elliptic curves of positive rank) end up with the bulk of these solutions, while other homogeneous cubics (including those associated to elliptic curves of zero rank, including the Fermat curve ${x^3+y^3=z^3}$) only get finitely many solutions. The reasons for this are subtle, but certainly the high degree of arithmetic structure present in an elliptic curve (starting with the elliptic curve group law which allows one to generate new solutions from old ones, and which also can be used to exclude solutions to ${x^3+y^3=z^3}$ via the method of descent) is a major contributing factor.
Below the fold, we apply similar heuristics to suggest the truth of the ABC conjecture.
Read the rest of this entry »
## Distinguished Lecture Series III: Shou-wu Zhang, “Triple L-series and effective Mordell conjecture”
4 May, 2007 in DLS, math.NT | Tags: abc conjecture, Arakelov theory, Beilinson-Bloch conjecture, Bogolomov-Miyoaka-Yau inequality, BSD conjecture, Faltings theorem, Shou-Wu Zhang | by Terence Tao | 3 comments
On Thursday Shou-wu Zhang concluded his lecture series by talking about the higher genus case $g \geq 2$, and in particular focusing on some recent work of his which is related to the effective Mordell conjecture and the abc conjecture. The higher genus case is substantially more difficult than the genus 0 or genus 1 cases, and one often needs to use techniques from many different areas of mathematics (together with one or two unproven conjectures) to get somewhere.
This is perhaps the most technical of all the talks, but also the closest to recent developments, in particular the modern attacks on the abc conjecture. (Shou-wu made the point that one sometimes needs to move away from naive formulations of problems to obtain deeper formulations which are more difficult to understand, but can be easier to prove due to the availability of tools, structures, and intuition that were difficult to access in a naive setting, as well as the ability to precisely formulate and quantify what would otherwise be very fuzzy analogies.)
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http://en.wikipedia.org/wiki/Identification_condition
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Identifiability
(Redirected from Identification condition)
In statistics, identifiability is a property which a model must satisfy in order for inference to be possible. We say that the model is identifiable if it is theoretically possible to learn the true value of this model’s underlying parameter after obtaining an infinite number of observations from it. Mathematically, this is equivalent to saying that different values of the parameter must generate different probability distributions of the observable variables. Usually the model is identifiable only under certain technical restrictions, in which case the set of these requirements is called the identification conditions.
A model that fails to be identifiable is said to be non-identifiable or unidentifiable. In some cases, even though a model is non-identifiable, it is still possible to learn the true values of a certain subset of the model parameters. In this case we say that the model is partially identifiable. In other cases it may be possible to learn the location of the true parameter up to a certain finite region of the parameter space, in which case the model is set identifiable.
Definition
Let ℘ = {Pθ: θ∈Θ} be a statistical model where the parameter space Θ is either finite- or infinite-dimensional. We say that ℘ is identifiable if the mapping θ ↦ Pθ is one-to-one:[1]
$P_{\theta_1}=P_{\theta_2} \quad\Rightarrow\quad \theta_1=\theta_2 \quad\ \text{for all } \theta_1,\theta_2\in\Theta.$
This definition means that distinct values of θ should correspond to distinct probability distributions: if θ1≠θ2, then also Pθ1≠Pθ2.[2] If the distributions are defined in terms of the probability density functions, then two pdfs should be considered distinct only if they differ on a set of non-zero measure (for example two functions ƒ1(x)=10≤x<1 and ƒ2(x)=10≤x≤1 differ only at a single point x=1 — a set of measure zero — and thus cannot be considered as distinct pdfs).
Identifiability of the model in the sense of invertibility of the map θ ↦ Pθ is equivalent to being able to learn the model’s true parameter if the model can be observed indefinitely long. Indeed, if {Xt}⊆S is the sequence of observations from the model, then by the strong law of large numbers,
$\frac{1}{T} \sum_{t=1}^T \mathbf{1}_{\{X_t\in A\}} \ \xrightarrow{a.s.}\ \operatorname{Pr}[X_t\in A],$
for every measurable set A⊆S (here 1{…} is the indicator function). Thus with an infinite number of observations we will be able to find the true probability distribution P0 in the model, and since the identifiability condition above requires that the map θ ↦ Pθ be invertible, we will also be able to find the true value of the parameter which generated given distribution P0.
Examples
Example 1
Let ℘ be the normal location-scale family:
$\mathcal{P} = \Big\{\ f_\theta(x) = \tfrac{1}{\sqrt{2\pi}\sigma} e^{ -\frac{1}{2\sigma^2}(x-\mu)^2 }\ \Big|\ \theta=(\mu,\sigma): \mu\in\mathbb{R}, \,\sigma\!>0 \ \Big\}.$
Then
$\begin{align} f_{\theta_1}=f_{\theta_2}\ &\Leftrightarrow\ \tfrac{1}{\sqrt{2\pi}\sigma_1}e^{ -\frac{1}{2\sigma_1^2}(x-\mu_1)^2 } = \tfrac{1}{\sqrt{2\pi}\sigma_2}e^{ -\frac{1}{2\sigma_2^2}(x-\mu_2)^2 } \\ &\Leftrightarrow\ \tfrac{1}{\sigma_1^2}(x-\mu_1)^2 + \ln \sigma_1^2 = \tfrac{1}{\sigma_2^2}(x-\mu_2)^2 + \ln \sigma_2^2 \\ &\Leftrightarrow\ x^2\big(\tfrac{1}{\sigma_1^2}-\tfrac{1}{\sigma_2^2}\big) - 2x\big(\tfrac{\mu_1}{\sigma_1^2}-\tfrac{\mu_2}{\sigma_2^2}\big) + \big(\tfrac{\mu_1^2}{\sigma_1^2}-\tfrac{\mu_2^2}{\sigma_2^2}+\ln\sigma_1^2-\ln\sigma_2^2\big) = 0 \end{align}$
This expression is equal to zero for almost all x only when all its coefficients are equal to zero, which is only possible when |σ1| = |σ2| and μ1 = μ2. Since in the scale parameter σ is restricted to be greater than zero, we conclude that the model is identifiable: ƒθ1=ƒθ2 ⇔ θ1=θ2.
Example 2
Let ℘ be the standard linear regression model:
$y = \beta'x + \varepsilon, \quad \operatorname{E}[\,\varepsilon|x\,]=0$
(where ′ denotes matrix transpose). Then the parameter β is identifiable if and only if the matrix E[xx′] is invertible. Thus, this is the identification condition in the model.
Example 3
Suppose ℘ is the classical errors-in-variables linear model:
$\begin{cases} y = \beta x^* + \varepsilon, \\ x = x^* + \eta, \end{cases}$
where (ε,η,x*) are jointly normal independent random variables with zero expected value and unknown variances, and only the variables (x,y) are observed. Then this model is not identifiable,[3] only the product βσ²∗ is (where σ²∗ is the variance of the latent regressor x*). This is also an example of set identifiable model: although the exact value of β cannot be learned, we can guarantee that it must lie somewhere in the interval (βyx, 1÷βxy), where βyx is the coefficient in OLS regression of y on x, and βxy is the coefficient in OLS regression of x on y.[4]
If we abandon the normality assumption and require that x* were not normally distributed, retaining only the independence condition ε⊥η⊥x*, then the model becomes identifiable.[3]
Software
In the case of parameter estimation in partially observed dynamical systems, the profile likelihood can be also used for structural and practical identifiability analysis.[5] An implementation of the Profile Likelihood Approach is available in the MATLAB Toolbox PottersWheel.
References
• Casella, George; Roger L. Berger (2001), Statistical inference (2nd ed.), ISBN 0-534-24312-6, LCCN 2001 QA276.C37 2001
• Hsiao, Cheng (1983), Identification, Handbook of Econometrics, Vol. 1, Ch.4, North-Holland Publishing Company
• Lehmann, E. L.; G. Casella (1998), Theory of point estimation (2nd ed.), Springer, ISBN 0-387-98502-6
• Reiersøl, Olav (1950), "Identifiability of a linear relation between variables which are subject to error", Econometrica (The Econometric Society) 18 (4): 375–389, doi:10.2307/1907835, JSTOR 1907835
• van der Vaart, A.W. (1998), Asymptotic Statistics, Cambridge University Press, ISBN 978-0-521-49603-2
Notes
1. ^ a b
2. Raue, A; Kreutz, C; Maiwald, T; Bachmann, J; Schilling, M; Klingmüller, U; Timmer, J (2009), "Structural and practical identifiability analysis of partially observed dynamical models by exploiting the profile likelihood", Bioinformatics 25 (15): 1923–9, doi:10.1093/bioinformatics/btp358, PMID 19505944.
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http://scicomp.stackexchange.com/questions/3307/recovering-coordinates-by-eigendecomposition-without-double-centering
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# Recovering coordinates by eigendecomposition without double-centering
Suppose an Euclidean distance $D\in\mathbb{R}^{n\times n}$ matrix between a set of $n$ objects is given. To obtain inner-products (which will be further be used to recover coordinates), entries of $D$ are squared, and the matrix is double-centered and scaled, ie., $K=-\frac{1}{2}JD^{(2)}J$, where matrix $J=I-\frac{1}{n}11^T$ defines the origin wrt which the inner-products are formed. So, the aim is to reconstruct coordinates $X$ that give rise to inner products $K$, $$K=-\frac{1}{2}JD^{(2)}J=JXX^TJ.$$ This is done by eigendecomposition of $K$.
Given the above equation, I wonder if one could use a shortcut $-\frac{1}{2}D^{(2)}=XX^T$, ie, obtain coordinates $X$ without double-centering $-\frac{1}{2}D^{(2)}$ (matrix $J$ is removed from the left and from the right).
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## 1 Answer
Your proposal will give meaningless results. This can be seen already with simple examples consisting of two points only. Indeed, the matrix that you want to decompose is not even positive semidefinite as it is nonzero but has zero diagonal entries.
Note that you can't cancel $J$ since it is singular. In general, $Ju=Jv$ implies $u=v$ if and only if $J$ has a trivial kernel, i.e., iff the rank of $J$ equals the number of columns of $J$.
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Note that I'm mainly interested in a "technical reasoning". Wouldn't a solution $X$ to i) $-\frac{1}{2}D^{(2)}=XX^T$ be also a solution to ii) $-\frac{1}{2}JD^{(2)}J=XX^T$, up to origin position. Basically, why could not one solve ii) to obtain a solution to i) (just cancel $J$ from the sides of both sides?) Note that diagonal entries would remain zero. – usero Sep 18 '12 at 13:22
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No. $X$ will usually have complex entries, and is useless. - Multiplication by $J$ on both sides makes the diagonal positive when the distance matrix is Euclidean. Note that ypu can't cancel $J$ since it is singular. – Arnold Neumaier Sep 18 '12 at 14:56
Your last sentence answered the question and many related ones. So, in case some $J$ is invertible (general case), one could cancel it. However, suppose a system i) $C^TACx=C^Tb$ is given, for $A\in\mathbb{R}^{n\times n}$ and a rectangular $C\in\mathbb{R}^{n\times m}$, $m<n$. In what manner could one rationalize that the solution to ii) $ACx=b$ is not a solution to i) (now, $C^T$ is rectangular, hence the singularity does not apply) – usero Sep 18 '12 at 15:10
@usero: $C^Tu=C^Tv$ implies $u=v$ if and only if $C^T$ has a trivial kernel, i.e., iff the rank of $C$ equals the number of rows. – Arnold Neumaier Sep 18 '12 at 17:32
I guess there is a misunderstanding: if I understand correctly, solving ii) cannot be replaced by solving i), but solving i) can be replaced by solving ii)? – usero Sep 18 '12 at 20:23
show 4 more comments
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http://math.stackexchange.com/questions/265925/where-do-cantor-sets-naturally-occur?answertab=active
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# Where do Cantor sets naturally occur?
Cantor sets in general of course have many interesting properties on their own, and are also often used as examples of sets with these properties, but do they naturally occur in any application?
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Not clear what you mean by "naturally," since the Cantor set is an uncountably infinite set, which means it may or may not occur in "nature." It can be thought of as being a "fractal," and "fractal" structures can occur in nature, although rarely in a pure form. – Thomas Andrews Dec 27 '12 at 13:55
To add to TA's nice comment, you can find items such as the study of rainfall and earthquakes as well as other areas employing fractals. Regards – Amzoti Dec 27 '12 at 14:07
With natural I meant occurring in areas of mathematics in any other way than by construction it specifically. – malin Dec 28 '12 at 20:21
## 6 Answers
Cantor sets appear naturally in dynamical systems all the time.
Example 1: Consider the map $f\colon \mathbb{R}\to \mathbb{R}$ given by $f(x) = 5x(1-x)$. In dynamics, we are interested in the behavior of points under iteration of the map $f$. In other words, if we start off with a given point $x_0\in \mathbb{R}$, we are interested in the sequence $\{f^{\circ n}(x_0)\}_{n \geq 1}$. It is not hard to show that if $x_0\notin [0,1]$, then $f^{\circ n}(x_0)\to -\infty$ as $n\to \infty$. This gives us a dichotomy:
1. Either $x_0$ is such that $f^{\circ m}(x_0)\notin [0,1]$ for some $m\geq 1$, in which case $f^{\circ n}(x_0)\to -\infty$, or
2. $x_0$ is such that $f^{\circ n}(x_0)\in [0,1]$ for all $n\geq 1$, i.e., the orbit of $x_0$ is bounded.
The second case is the most interesting. The set $B$ of points $x_0$ with bounded orbits is exactly $B = \bigcap_{n\geq 1} f^{-n}([0,1])$, which is a Cantor set. Moreover, the dynamics of $f$ on $B$ is easily described (see the next example).
Example 2: Let $A$ be a finite set, and let $S = A^{\mathbb{N}}$ be the set of all infinite sequences of elements of $A$. An element $s\in S$ is then $s = (s_0,s_1,s_2,\ldots)$ where $s_i\in A$ for each $i$. Define a map $\sigma\colon S\to S$ obtained by shifting the sequence once place to the left: $$\sigma(s_0,s_1,s_2,\ldots) = (s_1,s_2,s_3,\ldots).$$ The dynamics of $\sigma$ on $S$ models many interesting dynamical systems that appear in practice, which is incredibly useful, since the dynamics of $\sigma$ is so easy to understand. Moreover, if we equip $A$ with the discrete topology and $S$ with the product topology, then $S$ is a Cantor set! As an example, the dynamics of $f$ on the set $B$ in example 1 is isomorphic in a suitable sense to the dynamics of the left-shift map $\sigma$ on the space $S = \{0,1\}^\mathbb{N}$ of binary sequences.
Example 3: Let $f\colon \mathbb{C}\to \mathbb{C}$ be the map $f(z) = z^2 + c$, where $c$ is a given complex number. If $c$ lies outside the Mandelbrot Set, then the Julia Set of $f$, i.e., the set where the dynamics of $f$ is the most chaotic and interesting, is a Cantor set.
These are just a few examples in dynamics, but there are many more. I'd be interested in seeing more examples outside dynamics!
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2
Most frogs confine themselves to a well, but a few of them roam the vast ocean of truth and knowledge. Great answer! – Haskell Curry Dec 27 '12 at 15:09
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Very nice answer. Two pages readers might find interesting: kneading theory on scholarpedia and Ornstein's amazing isomorphism theorem showing that the entropy is a complete invariant of the shifts in example 2. – Martin Dec 27 '12 at 15:27
The rings of Saturn have a Cantor set-like pattern.
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The Cantor set has been experimentally observed with X-ray diffraction in connection with the Quantum Hall effect,
http://www.eng.yale.edu/reedlab/publications/24.pdf
I don't know the physical reasons behind this, you will probably have to read the literature to find out.
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I recently came across this bit of history concerning Mandelbrot's observation that a certain noisy signal could be thought of in terms of the Cantor Set.
"... But Mandelbrot's work eventually showed that the noise was both consistent and erratic, some kind of inescapable natural feature of the system that did not disappear with increased signal strength. But more remarkably he also showed that every burst of noise also contained within it bursts of clear signal (a situation he conceived of in terms of the Cantor set). Stranger still, he found that the ratio of periods of noise to periods of clean transmission remained constant, regardless of the scale of time used to plot the phenomenon (i.e. months, days, seconds)."
From: Benoit Mandelbrot
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Note that the ring of $p$-adic integers $\mathbb Z_p$ is homeomorphic to a Cantor set.
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4
+1, and to add on that... any perfect, zero-dimensional, compact metric space is homeomorphic to the Cantor set. – Asaf Karagila Dec 28 '12 at 0:21
Cantor sets naturally appear in logic. For example, if you have countably many propositional variables $x_1, x_2, ...$, then the set of possible truth-assignments to them has a natural topology which can be identified with the product topology on $\{ 0, 1 \}^{\mathbb{N}}$, which is homeomorphic to the Cantor set. This topology is compact, which is roughly speaking the topological meaning behind the compactness theorem.
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1
I'm not sure this is what "naturally" means. Either way if you are going to bring up logic then one can talk about descriptive set theory and its applications in model theory and logic, in those areas I think that the Cantor set has more applications than just the compactness theorem. – Asaf Karagila Dec 27 '12 at 23:59
@Asaf: well, maybe you can; I don't know anything about descriptive set theory. – Qiaochu Yuan Dec 28 '12 at 0:08
1
Well, you are in Berkeley... Get the hint! :-) – Asaf Karagila Dec 28 '12 at 0:12
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http://en.wikipedia.org/wiki/Surface_area
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Surface area
Surface area is the total area of the faces and curved surface of a solid figure. Mathematical description of the surface area is considerably more involved than the definition of arc length or polyhedra (objects with flat polygonal faces) the surface area is the sum of the areas of its faces. Smooth surfaces, such as a sphere, are assigned surface area using their representation as parametric surfaces. This definition of the surface area is based on methods of infinitesimal calculus and involves partial derivatives and double integration.
General definition of surface area was sought by Henri Lebesgue and Hermann Minkowski at the turn of the twentieth century. Their work led to the development of geometric measure theory which studies various notions of surface area for irregular objects of any dimension. An important example is the Minkowski content of a surface.
Definition of surface area
While areas of many simple surfaces have been known since antiquity, a rigorous mathematical definition of the area requires a lot of care.
$S \mapsto A(S)$
of a positive real number to a certain class of surfaces that satisfies several natural requirements. The most fundamental property of the surface area is its additivity: the area of the whole is the sum of the areas of the parts. More rigorously, if a surface S is a union of finitely many pieces S1, …, Sr which do not overlap except at their boundaries then
$A(S) = A(S_1) + \cdots + A(S_r).$
Surface areas of flat polygonal shapes must agree with their geometrically defined area. Since surface area is a geometric notion, areas of congruent surfaces must be the same and area must depend only on the shape of the surface, but not on its position and orientation in space. This means that surface area is invariant under the group of Euclidean motions. These properties uniquely characterize surface area for a wide class of geometric surfaces called piecewise smooth. Such surfaces consist of finitely many pieces that can be represented in the parametric form
$S_D: \vec{r}=\vec{r}(u,v), \quad (u,v)\in D$
with continuously differentiable function $\vec{r}.$ The area of an individual piece is defined by the formula
$A(S_D) = \iint_D\left |\vec{r}_u\times\vec{r}_v\right | \, du \, dv.$
Thus the area of SD is obtained by integrating the length of the normal vector $\vec{r}_u\times\vec{r}_v$ to the surface over the appropriate region D in the parametric uv plane. The area of the whole surface is then obtained by adding together the areas of the pieces, using additivity of surface area. The main formula can be specialized to different classes of surfaces, giving, in particular, formulas for areas of graphs z = f(x,y) and surfaces of revolution.
One of the subtleties of surface area, as compared to arc length of curves, is that surface area cannot be defined simply as the limit of areas of polyhedral shapes approximating a given smooth surface. It was demonstrated by Hermann Schwarz that already for the cylinder, different choices of approximating flat surfaces can lead to different limiting values of the area (Known as Schwarz's paradox.) [1] .[2]
Various approaches to general definition of surface area were developed in the late nineteenth and the early twentieth century by Henri Lebesguè and Hermann Minkowski. While for piecewise smooth surfaces there is a unique natural notion of surface area, if a surface is very irregular, or rough, then it may not be possible to assign any area at all to it. A typical example is given by a surface with spikes spread throughout in a dense fashion. Many surfaces of this type occur in the theory of fractals. Extensions of the notion of area which partially fulfill its function and may be defined even for very badly irregular surfaces are studied in the geometric measure theory. A specific example of such an extension is the Minkowski content of the surface.
Common formulas
Surface areas of common solids
Shape Equation Variables
Cube $6s^2 \,$ s = side length
Rectangular prism $2(\ell w + \ell h + wh) \,$ ℓ = length, w = width, h = height
All Prisms $2B + Ph \,$ B = the area of one base, P = the perimeter of one base, h = height
Sphere $4\pi r^2 \,$ r = radius of sphere
Spherical lune $2r^2\theta \,$ r = radius of sphere, θ = dihedral angle
Closed cylinder $2\pi r(h+r) \,$ r = radius of the circular base, h = height of the cylinder
Lateral surface area of a cone $\pi r \left(\sqrt{r^2+h^2}\right) = \pi rs \,$ $s = \sqrt{r^2+h^2}$
s = slant height of the cone,
r = radius of the circular base,
h = height of the cone
Full surface area of a cone $\pi r \left(r + \sqrt{r^2+h^2}\right) = \pi r(r + s) \,$ s = slant height of the cone,
r = radius of the circular base,
h = height of the cone
Pyramid $B + \frac{PL}{2}$ B = area of base, P = perimeter of base, L = slant height
Ratio of surface areas of a sphere and cylinder of the same Radius and Volume
A cone, sphere and cylinder of radius r and height h.
The above formulas can be used to show that the surface area of a sphere and cylinder of the same radius and height are in the ratio 2 : 3, as follows.
Let the radius be r and the height be h (which is 2r for the sphere).
$\begin{array}{rlll} \text{Sphere surface area} & = 4 \pi r^2 & & = (2 \pi r^2) \times 2 \\ \text{Cylinder surface area} & = 2 \pi r (h + r) & = 2 \pi r (2r + r) & = (2 \pi r^2) \times 3 \end{array}$
The discovery of this ratio is credited to Archimedes.[3]
In chemistry
See also: Accessible surface area
Surface area is important in chemical kinetics. Increasing the surface area of a substance generally increases the rate of a chemical reaction. For example, iron in a fine powder will combust, while in solid blocks it is stable enough to use in structures. For different applications a minimal or maximal surface area may be desired.
In biology
See also: Surface-area-to-volume ratio
The inner membrane of the mitochondrion has a large surface area due to infoldings, allowing higher rates of cellular respiration (electron micrograph).
The surface area of an organism is important in several considerations, such as regulation of body temperature and digestion. Animals use their teeth to grind food down into smaller particles, increasing the surface area available for digestion. The epithelial tissue lining the digestive tract contains microvilli, greatly increasing the area available for absorption. Elephants have large ears, allowing them to regulate their own body temperature. In other instances, animals will need to minimize surface area; for example, people will fold their arms over their chest when cold to minimize heat loss.
The surface area to volume ratio (SA:V) of a cell imposes upper limits on size, as the volume increases much faster than does the surface area, thus limiting the rate at which substances diffuse from the interior across the cell membrane to interstitial spaces or to other cells. Indeed, representing a cell as an idealized sphere of radius r, the volume and surface area are, respectively, V = 4/3 π r3; SA = 4 π r2. The resulting surface area to volume ratio is therefore 3/r. Thus, if a cell has a radius of 1 μm, the SA:V ratio is 3; whereas if the radius of the cell is instead 10 μm, then the SA:V ratio becomes 0.3. With a cell radius of 100, SA:V ratio is 0.03. Thus, the surface area falls off steeply with increasing volume.
References
1. Rorres, Chris. "Tomb of Archimedes: Sources". Courant Institute of Mathematical Sciences. Retrieved 2007-01-02.
• Yu.D. Burago, V.A. Zalgaller, L.D. Kudryavtsev (2001), "Area", in Hazewinkel, Michiel, , Springer, ISBN 978-1-55608-010-4
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http://math.stackexchange.com/questions/5286/what-concepts-were-most-difficult-for-you-to-understand-in-calculus/233392
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# What concepts were most difficult for you to understand in Calculus?
I'm developing some instructional material for a Calculus 1 class and I wanted to know from experience for yourself, tutoring others, and/or helping people on this site where is the most difficulty in Calculus?
If you had any good methods of helping people that would be very helpful.
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7
Community wiki I suppose. – J. M. Sep 23 '10 at 9:02
Added "soft-question" tag. – Arturo Magidin Sep 23 '10 at 14:38
Who says I ever understood it? You never understand calculus,you just get used to it........LOL I'm kidding,but it speaks of a deep truth:Most of us spend the next 10 years after our first calculus class really trying to understand the subject. In fact,I'd go so far as to say that no one really understands calculus until they teach a real analysis course. – Mathemagician1234 Mar 17 '12 at 7:57
I always found optimization VERY difficult. There's a lot of nit-picky stuff that you have to keep in mind. – user39302 Sep 3 '12 at 3:19
## 11 Answers
The hardest thing for me was to understand what is meant when someone writes $\mathrm d x$.
I still don't know...
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+1. Yes, this bothered me so much at the time. I mean, you have dx appearing in integration, and then dy/dx as derivatives, and some texts also mention "differentials" df = f'(x) dx (so df is a function of x and dx, and dx is just a real number). – Jesse Madnick Sep 23 '10 at 9:38
Every year without fail, this is the hardest idea for my students to understand, especially in the context of anti-differentiation (as opposed to integration). – Alex Basson Sep 23 '10 at 11:07
There is a really superb explanation of this in A. Ya. Khinchin's Eight lectures on mathematical analysis, in which he thoroughly explodes the notion that it has no well-defined meaning. – MJD Jun 2 '12 at 2:33
$dx$ is just a piece of notation, and one never needs to use this notation. For the integral you can just write $\int_a^b f$, as is done in Spivak's Calculus on Manifolds, a book that is certainly very well respected. For the derivative you can just use $f'(x)$. $dx$ is not some kind of "infinitely small number", whatever that is supposed to mean. (More advanced math subjects may give a precise meaning to $dx$, but that is not something one needs to worry about when learning calculus.) – littleO Nov 9 '12 at 6:33
I really struggled with the $\epsilon-\delta$ definition of limits, especially for non-linear functions. This was also my first exposure to proof, as in: Prove that $$\lim_{x \to 2} (x^2 + 3) = 7$$ and I had a hard time with it at first. To be clear, computing these limits was no problem, but using the definition to prove they were correct really confused me.
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It is probably more difficult in higher ages; my first contact with epsilon/delta limits and integration was in middle school, at the age of 14... it was slightly confusing then, but it got easier when I took calculus in high school. It was a breeze finally at the university. – Paxinum Sep 23 '10 at 13:34
3
The problem with ε/δ proofs is that initially, almost nobody teaches why one should care --- that is to say, what it means. "Formalism" is not a good enough motivation, much as I like it now. This is unfortunate, as perhaps the same one-minute challenge-response skit, performed maybe three times in the course of a semester, would solidify the concept in people's heads. – Niel de Beaudrap Sep 23 '10 at 13:37
Somewhat like Niel, until I read Bressoud's treatment of $\epsilon-\delta$ as a "game", I was scratching my head for weeks at what the fuss was all about. – J. M. Sep 23 '10 at 14:40
3
@Paxinum Where did you go to school that you were learning epsilon and delta style limits in MIDDLE SCHOOL?!? Unless you were a remarkable prodigy,I can't imagine any middle school in the US teaching you that! – Mathemagician1234 Mar 17 '12 at 7:54
1
@Mathemagician1234 I was very lucky having access to a teacher willing to teach me, so I had 1 hr/week with real mathematics. Just a teacher being happy to having a student willing to learn some more advanced stuff. – Paxinum Mar 17 '12 at 23:13
show 3 more comments
At the school I was taught to look at the derivative as the instantaneous rate of change and that fit well with applications in physics. But later, when I was learning Economics in college, I had to learn to look at the derivative as the best linear (affine) approximation, and a differentiable function as a function which had 'good' linear approximations. That is also the intuition that generalizes to many variables. I wish it had been discussed in my early calculus classes.
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1
+1, for me, I gained a better geometric view of derivatives (and Taylor expansions in general) in the context of "polynomials that greatly resemble the function in the vicinity of the expansion point". – J. M. Sep 23 '10 at 15:04
I think the entire concept behind integration is hard to grasp for students who are not familiar with analysis. They tend to think of it only as the "inverse operation of derivatives", which is quite restrisctive, in my opinion.
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The inverse relationship between differentiation and integration, and understanding it from the graph.
And I still have not understood that part of calculus at all! :(
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1
There's a very nice picture you can draw to show why $\frac{d}{dx} \int_a^x f(s) \, ds = f(x)$. Also, keep in mind that $f'(t)$ is the instantaneous rate of change of $f$ at time $t$, and $f'(t) dt$ is a tiny change in $f$ during a very short period of time, and by adding up all the tiny changes you get the total change: $\int_a^b f'(t) \, dt = f(b) - f(a)$. – littleO Nov 9 '12 at 6:42
1
– littleO Nov 11 '12 at 23:32
By far the hardest thing for me was notation for partial derivatives. Having never been shown just how to actually interpret the symbols, I had great difficulty parsing what was actually meant by various expressions.
Eventually, I abandoned the more common notations entirely, and fell back on other notation we had been shown, like $f_1(x,y,x)$, which means "the derivative of $f$ in the first argument, evaluated at $(x,y,x)$".
These days, I have a deeper understanding of just what my problem was. Roughly speaking, $dx$ makes sense on its own, but $\partial/\partial x$ does not; the latter depends not on $x$, but on a curve $x$ is being viewed as a parameter for. (e.g. it suffices to view $x$ as a component in a coordinate system)
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I've always been fond of the notation Spivak uses for partial derivatives in Calculus on Manifolds: $D_i f(x)$ rather than $\frac{\partial f(x)}{\partial x_i}$. – littleO Nov 9 '12 at 6:38
Calculus was hard for me until I learned how to visualize things. Learning calculus only by writing symbols and solving problems with many $\varepsilon$'s will not make anyone understand it. If you learn to visualize all the basic concepts as limit, derivative, integration, etc. then the symbolic part is a lot easier.
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The area formula, namely, how could
$$\sum_{i=1}^{\infty} y_i \delta x_i=\int y dx$$
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I took differential calculus twice in two different schools. Not until years later did I realize that I had not known what a function is and that differential calculus is the study of one particular index of a point property of a function that produces another function of the same independent variable, and what the property is, and what index is used, and why, and that the derivative of a function is the result of an operation on a function called differentiation, and that the role of the limit is simply to carry out the operation and has nothing whatever to do with the basic idea. Defining the derivative as a limit completely obfuscated what it was really about. The insistence that mathematics is abstract and axiomatic buries all of the simple intriguing ideas. No wonder John von Neumann said "In mathematics you don't understand things. You just get used to them."
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Being more algebraically minded, I found it incredibly hard to follow all the techniques of integration, because rather soon those tend to become either very hand-wavy or very technical. I think I finished my degree without ever "calculating" a concrete non-trivial integral, such as doing integration by partial fractions or
I was just too scared about the "intuitive" notion with which various "dx", which at that point are merely a meaningless symbol, suddenly get replaced by some dy dx/dy - at that time, I never was able to assure myself that that I could make that rigorous.
If this is not clear, I found a quick example on wikipedia of something that I find scary even today; I quote:
"Integrating by this substitution: $cos(x) dx = d sin(x)$".
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For me,the hardest part of elementary calculus was infinite series and the idea of convergence. I learned it in an accelerated summer course taught by Elliott Mendelson and I remember going insane trying to absorb all the basic tests in one feverish night on vacation with my family in the Catskills.
The main reason infinite series was so difficult was because you can't really understand how they work-indeed, the very concepts involved-without a rigorous formulation of both real numbers and limits. Queens College was-and still sadly is,from what I hear-determined to use a pencil-pushing course with Stewart as the text.Of COURSE infinite series and sequences are going to be a garbled mess if you try to use hand waving to explain it!
In retrospect,I feel bad for Elliott-he became visibly frustrated at times trying to teach it to us via "handwaving" and endless sample calculations.I didn't understand at the time why he was frustrated.Of course,I realize now how difficult what he was trying to do was-especially in a full semester course that was crammed into 5 weeks in the dead of summer!
This is why unless I can teach it with some rigor, I may just skip it entirely when I teach calculus the first time.
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Uh-why was this answer removed?The question itself is rather subjective and personal,so I gave a personal and subjective answer. I didn't think it was any more so then any of the other responses.This site-like MO-can be very capricious sometimes. – Mathemagician1234 Mar 25 '12 at 0:15
I see now it hasn't been removed-just downvoted to death. Why is still a complete mystery to me-unless my fan club's been working overtime to spite me like children again. – Mathemagician1234 Mar 25 '12 at 3:10
Infinite series is part of the second semester of calculus at all of the (five) universities where I have studied. Arguably, Calc II is the hardest math class that a student will have had by that point. – The Chaz 2.0 Apr 9 '12 at 16:40
(And this question is about Calc I, if I read it correctly...) – The Chaz 2.0 Apr 9 '12 at 16:40
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http://math.stackexchange.com/questions/282271/scaled-lp-norm-and-geometric-mean
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# “Scaled $L^p$ norm” and geometric mean
The $L^p$ norm in $\mathbb{R}^n$ is \begin{align} \|x\|_p = \left(\sum_{j=1}^{n} |x_j|^p\right)^{1/p}. \end{align} Playing around with WolframAlpha, I noticed that, if we define the "scaled" $L^p$ norm in $\mathbb{R}^n$ to be
\begin{align} \| x \|_p = \left(\frac{1}{n}\sum_{j=1}^{n} |x_j|^p\right)^{1/p} \end{align} then \begin{align} \lim_{p \to 0} \|x\|_p &= \left( \prod_{j=1}^{n} |x_j| \right)^{1/n}, \end{align} which is the geometric mean of the coordinates' absolute values. This is interesting maybe because the $L^p$ norm doesn't have a nice limit at zero.
My questions:
1. How do I prove this?
2. Is this definition of "scaled $L^p$ norm" interesting, or known by another name, or used anywhere?
3. Is there any interesting reason to define the $L^0$ norm as the geometric mean, as above?
4. Further reading?
Thanks!
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## 2 Answers
I'll prove the general case in which $\mu$ is a positive measure on a space $X$ and $\mu(X) = 1$. Your particular case follows by setting $X = \{1, \ldots, n\}$ and $\mu(i) = 1/n$.
By definition: $$\|f\|_p = \left\{\int_X |f|^p \,d\mu\right\}^{1/p}$$
Lemma 1: If $0 < r < s < 1$, then $\|f\|_r \le \|f\|_s$.
Proof: $\varphi(x) = x^{s/r}$ is a convex function. Apply Jensen's inequality to $\int_X |f|^r \,d\mu$ to get: $$\left\{\int_X |f|^r \,d\mu\right\}^{s/r} \le \int_X |f|^s \,d\mu$$
Hence $\|f\|_r \le \|f\|_s$.
Lemma 2: If $0 < p < 1$, then $\int_X \log|f| \,d\mu \le \log \|f\|_p$.
Proof: $\log$ is a concave function. Apply Jensen's inequality to $\int_X |f|^p \,d\mu$ to get the desired inequality.
From lemmas 1 and 2, it follows that $\log\|f\|_{1/n}$ is decreasing and bounded from below. Therefore, it converges as $n \to \infty$.
To find the limit, apply the inequality $\log a \le a - 1$ with $a = \int_X |f|^{1/n} \,d\mu$ to get: $$\log \|f\|_{1/n} \le \int_X \frac{|f|^{1/n} - 1}{1/n} \,d\mu \tag{1}$$
Use L'Hôpital's rule to obtain $\lim_{x \to 0} \dfrac{a^x - 1}{x} = \log a$. Take the limit of (1) as $n \to \infty$ and apply the dominated convergence theorem to get: $$\lim_{n \to \infty} \log \|f\|_{1/n} \le \int_X \log|f| \,d\mu$$
Apply the squeeze theorem with lemma 2 to obtain: $$\lim_{n \to \infty} \log \|f\|_{1/n} = \int_X \log|f| \,d\mu$$
Since $\log$ is continuous, we conclude: $$\lim_{n \to \infty} \|f\|_{1/n} = \exp\left(\int_X \log|f| \,d\mu\right)$$
To answer your other questions, the "scaled norm" follows from the general case as I explained at the beginning of my answer. I've never seen the geometric mean called $L^0$. As for further readings, check out Rudin's Real and Complex Analysis or Folland's Real Analysis. The above is an exercise in one of them (I think the former).
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By the way, if you're only familiar with $L^p$ norms and Jensen's inequality in finite spaces, you can follow the same proof by replacing integrals with sums. Try it! – Ayman Hourieh Jan 19 at 22:40
Sorry for the delay, but thanks for your answer, your proof was very clear and easy to follow! – usul Jan 25 at 15:14
Scaled $L^p$ norm (or rather $\ell^p$, since you work with vectors) is known as Generalized mean. A bunch of interesting inequalities involving the means are found in the book Inequalities by Hardy, Littlewood, and Pólya.
The integral geometric mean $\exp\left(\frac{1}{2\pi}\int_0^{2\pi} \log f(\theta)\,d\theta \right)$ comes up in complex analysis, especially as it relates to operator theory and involves the name of Gabor Szegő. See the terse Wikipedia article on Szegő limit theorems and the not-at-all-terse book by Barry Simon Szego's Theorem and Its Descendants
In a visit back to his native Budapest, Pólya mentioned this conjecture to Szegő, then an undergraduate, and he proved the theorem below, published in 1915... At the time, Szegő was nineteen, and when the paper was published, he was serving in the Austrian Army in World War I
The book Banach spaces of analytic functions by Kenneth Hoffman presents this topic from the viewpoint of complex analysis without much operator theory. The quantity $\exp\left(\frac{1}{2\pi}\int_0^{2\pi} \log f(\theta)\,d\theta \right)$ turns out to be equal to $\inf_{p}\int|1-p|^2 f(\theta)\,d\theta$ where $p$ runs over all polynomials vanishing at $0$. In particular, this gives a criterion for the density of polynomials in weighted $L^2$ spaces.
In a rather different direction, the integral geometric mean comes up in number theory. If $p$ a complex polynomial, the quantity $\exp\left(\frac{1}{2\pi}\int_0^{2\pi} \log |p(\theta)|\,d\theta \right)$ is called the Mahler measure of $p$, denoted $M(p)$. Lehmer's conjecture asserts that there is a gap $(1,\mu)$ in the possible values of $M(p)$: that is, either $M(p)=1$ or $M(p)\ge \mu>1$. Conjecturally, $\mu$ is attained by the polynomial $$p(z)= z^{10}-z^9+z^7-z^6+z^5-z^4+z^3-z+1$$ But even the existence of such $\mu$ remains unknown, let alone its precise value. The Wikipedia article has a good list of references.
That said, please do not call the integral geometric mean "the $L^0$ norm". This term is ambiguous and misleading enough as it is.
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Thanks very much, this was very interesting! – usul Jan 25 at 15:14
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http://diracseashore.wordpress.com/2009/01/19/everyday-physics-boats-and-ice-float/
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# Shores of the Dirac Sea
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## Everyday physics: boats and ice float.
January 19, 2009 by dberenstein
As you might have noticed, some objects float and some others don’t. Here below I have a rendition of a boat and a cube of ice floating.
Various floating objects
Today, I will go on a bit about flotation. As a matter of fact, some of you might remember a puzzle with an egg I wrote down a while ago. Of course, most of you have probably heard of Archimedes Principle as describing flotation, so I will explain some aspects of how that principle comes about.
So the first thing you have is a fluid, and secondly you place an object on top of the fluid and you can ask if it will float or not. But this can depend on the shape of the object that is floating. If someone asks wether iron floats on water or not, the answer is that iron sinks to the bottom. Except that we have iron boats (nowadays they are mostly made out of steel). Indeed, Metal boats were a XIX century wonder and many thought they could not possibly float. Nowadays we are so used to it that we don’t stop to see the wonder of it.
In the end, the important concept is pressure. In particular, the liquid exert pressure on every object inserted in it. The pressure is defined as a force per unit area. It is exerted uniformly in all directions. This is a property that liquids have. So if you have an area that encloses some volume, you can expect that there is some pressure exerting force on the corresponding area.
Pressure is not uniform in a liquid because of gravity.
However, the force is perpendicular to the area in question, so if you have some complicated object, force will be exerted in all kinds of directions, because the direction of the surface of the object is changing. So let us stick to a rectangular -shaped like a box- object as it is more simple. We can also make it as small as we like.
On each area of the box there will be some force pushing it. If we look at the vertical direction, there is a force from the bottom pushing it up and there is also a force from the top pushing it down. In the vertical direction these are equal to
$F_z= -P(z+h)A + P(z)A\sim -\partial_z P(z) A h$
Where P is the local pressure at various points, A is the area of the base and h is the height. There will be similar expressions for the horizontal directions. So if we divide the formula by the volume, we find that there is a force density per unit volume equal to the gradient of the pressure. This is
$\vec f = -\nabla P$
In particular, consider a liquid in equilibrium in the presence of gravity. That means that the liquid is not moving and it is not experiencing acceleration. We have two forces acting on the liquid: gravity and the pressure gradients. There should be a balance of forces between these two. We this find that
$-\rho \vec g -\nabla P=0$
where $\rho$ is the mass density of the fluid.
Thus, we can solve for the pressure at various depths. We find that
$P(z) = P(0) -\rho g z$
So if we submerge a box, the top of the cube feels less pressure than the bottom. The total force is
$\vec F= (P(z+h)-P(z))A= \rho g Ah= W$
where we see that the right hand side is the weight of a box of the liquid (the displaced liquid as in Archimedes Principle).
This explains why a metal boat can float. So long as the displaced volume of water weighs more than the boat, the boat will float. This requires one to have a shape that guarantees that the boat is in the end lighter than the water. We do that by adding a lot of empty space filled with air.
Incidentally, since ice floats, the volume of liquid water that it displaces weighs more than the ice itself. This means that ice is less dense than water. So when water cools off below freezing: it expands. This expansion can cause excess pressure on water pipes. After all, the ice will be trying it’s best to expand to full size and may cause the water pipes to break.
There is one more thing to consider. Most objects get a bit smaller when compressed, but water is pretty incompressible. Thus, if you try to keep an object at a fixed depth and it goes a bit further down you will see that it will most probably start sinking, whereas if it goes a bit up it will most probably start to rise. This actually happens to fish and many have developed a swim bladder to help with this. The fish ‘empty’ the swim bladder if they go too high and fill it up if they are sinking too much. Thus, any equilibrium in depth they have is dynamical. Sharks on the other hand rely on hydrodynamic lift instead to accomplish this equilibrium.
### Like this:
Posted in hydraulics | 13 Comments
### 13 Responses
1. Weight is a manifestation of curvature, nothing else or more. Flat is heavy (steel plates sink), curved is light (steel bowls float). You will be voluntarily buying Curvature Credits, by law, to save the little children.
http://www.mazepath.com/uncleal/flat.htm
2. on January 20, 2009 at 2:18 am carlbrannen
I clicked down hoping to find an explanation for the dip in the water level near the iceberg but no.
3. Waves in the sea……
4. David,
If this is to much myth in perpetuity being supplied then by all means delete it.
I have an ole factory experiment that you might look at, too dispel some mythical superstition I have about weight.
This experiment was done in my youth as well, and again after some years in this factory by men. As stupid as it sounds it has been the motivator of my pursuance for understanding gravity better.
Hulse and Taylor is gravity wave generations?:) Soliton wave generation? Hmmm…..
The story:
The scale in the kitchen is supplanted by a large scale that will hold five men and one chair. One man is sitting and four are standing with their fingers to balance under the chair to lift. Before this lift ensues, each places his hand over top of the man in a hand over hand fashion, and then proceeds to lift.
IN our perception of the lift previously we were astounded to think that a man who weighted so much could be lifted by all, and we not feel this weight appropriately. So we sought to see if this “weight changes when we lifted him.”
AS we watch the scale, it did not move, yet our perception of his weight remained. I know perception is a strange thing( not reliable as a measure) yet, I had seen it go further in my youth. Was part of the incentive to get the men to try it.
Also….
IN thinking about your egg experiment some things came to mind about the “spaces in between the fluid,” versus ice etc…. and I wanted to know about how much water will rise depending on how much volume per tea spoon sugar can be added to the fluid to raise it’s level? The displacement value. So…..even in a liquid form, there is space?
The aluminum gravity bars seemed related in this context when thinking about resonance? Webber.
Best,
5. Just to go a little bit further here and I’ll leave you alone.
Hmmmm…….IN Gems, a refractive index, in water, an illusion of a bent pencil. Ligo operations, interferometry on a large scale, measuring light (speed of light constant in a vacuum)changes of light pinpointed to mirror? Kip Thorne valuation?
Grace satellite changing our understanding of the landscape of earth from our first pristine view of “stepping outside” the boundaries perceived of earth, now, in it’s most spherical shape “quite lumpy” in it’s measured valuation.
Not sure where I am going with this. I think I am basing on it on the assumption you are heading in the direction of explaining gravity waves? Gas in a box under gravity? If not, sorry.
Best,
6. on January 20, 2009 at 11:07 pm carlbrannen
Actually, the old aluminum gravity wave detectors were called Weber bars.
Joe Weber was quite a guy, a bit infamous for repeatedly observing gravity waves in error. I remember his recollections of the sinking of the carrier Lexington, on which he was stationed, in the battle of the Coral Sea, 1942. He said she was incandescent as she sank beneath the surface, but that the closest brush with death he experienced was on the island afterwards, when a monkey nearly nailed him with a coconut.
I took a couple classes from him, general relativity and an advanced statistical mechanics sort of class. I guess I should plug my general relativity simulator, which does test masses around a black hole in Schwarzschild and Painleve coordinates. Which reminds me, I need to update the java code because there’s an error in the Runge-Kutta algorithm and so it’s not as accurate as it could be (guilt).
7. on January 21, 2009 at 12:29 am Plato
Sorry for going off topic David.
Carl:Joe Weber was quite a guy, a bit infamous for repeatedly observing gravity waves in error.
Wow, that’s cool Carl. I mean knowing him and his history. What was his initial insight into using the Aluminum bars. Do you remember off hand?
I assume it was acoustical in nature and relevant thinking based on “sound values” were thought at one time to be held relevant in terms of Acoustic Hawking Radiation.
Renaud Parentani wrote something in this regard, phonon’s versus photons. Unruh showed something in this respect that phonon’s were not sensitive enough.
Do think Joe Webber thought this way in terms of gravitational wave detection? To this day I think people do think this way in terms of gravitational wave signatures, as being sound by nature? Wayne Chu for example.
Similarily, the laws of gravity and light seem totally dissimilar. They obey different physical assumptions and different mathematics. Attempts to splice these two forces have always failed. However, if we add one more dimension, a fifth dimension, to the previous four dimensions of space and time, then equations governing light and gravity appear to merge together like two pieces of a jigsaw puzzle. Light, in fact, can be explained in the fifth dimension. In this way, we see the laws of light and gravity become simpler in five dimensions.Kaku’s preface of Hyperspace, page ix, para 3.
Thinking of Susskind’s Gedanken experiment, and about the elephants, it would seem from the fifth dimensional perspective, to have some value in the photon description, in terms of valuating that Hawking Radiation from the blackhole?
Best,
8. [...] is very similar to the notion of pressure, as described in the previous post on flotation. Indeed, the units of stress and the units of pressure are the same. The main difference, however, [...]
9. on January 22, 2009 at 5:48 am carlbrannen
Plato,
It’s been a very long time, but I seem to recall that Weber’s was trained as an EE. Just a second. Let me check wikipedia for this… Yes, I recall correctly. He got into the US Naval Academy after “his senator visited me to make sure I wasn’t black”, which would be exactly in keeping with the kind of stories I remember him telling, self deprecating. It’s hard for me to imagine a better naval officer.
For an electrical engineer, it is very natural to think in terms of resonances. And so, he examined the resonant conditions of a cylinder. His detectors were only good for gravity waves of a very particular frequency, and their ability to detect also depended on their orientation. More recent resonance based detectors are spheres.
The only book on Weber I know is “Gravity’s Shadow” by Harry Collins, which is really about the sociology of gravity wave detectors. Of course I bought the book and read it because I knew him, but it turns out that it gives fascinating insights on how experimental and theoretical physics is done, from a sociological point of view. These transfer to other branches of physics.
One of the classes I took from him was a 2nd year statistical mechanics class (my favorite subjects were probability and group theory so I took a lot of these sorts of classes). It was mostly about the “fluctuation dissipation theorem”.
He assigned us a homework problem that I couldn’t solve, which was quite rare for me in grad school. I should write it up as a blog post. It was years before I forgave him for this problem.
Take a 75 ohm resistor. Use it to short out a length of 75-ohm coaxial cable. Attach the other end of the cable to a short dipole antenna. Place the antenna in a black body cavity at temperature T. Show that the resistor reaches an equilibrium temperature T.
When it came time to turn in the homework, only the two or three students who had asked him for hints were able to solve it. We were bending up our copies of JD Jackson calculating approximations of absorption cross sections and integrating over solid angles and frequencies, multiplied by the black body radiation curve, etc..
Instead, he was using a fact about antennas well known to electrical engineers but pretty much unknown to physics grad students, certainly nothing that had ever been discussed in class etc. One might look here for a clue.
25 years on, I’m still pissed over that homework assignment. So to make it even, I’m going to pass on the advice he gave us: “never start working on a calculation unless you already know the answer you will get”, which is a good short hand description of how he got into trouble with gravity waves.
10. Thanks Carl.
11. Wegener proposed that the continents floated somewhat like icebergs in water. Wegener also noted that the continents move up and down to maintain equilibrium in a process called isostasy.
Alfred Wegener
Just thought I would add this for consideration. Grace satellite does a wonderful job of imaging these features into a global perspective?
In the simplest example, isostasy is the principle of buoyancy observed by Archimedes in his bath, where he saw that when an object was immersed, an amount of water equal in volume to that of the object was displaced. On a geological scale, isostasy can be observed where the Earth’s strong lithosphere exerts stress on the weaker asthenosphere which, over geological time flows laterally such that the load of the lithosphere is accommodated by height adjustments.
Isostasy
12. on February 3, 2009 at 8:53 am rufus
hai..a have a project on preliminary design of amphibius vehicle..and i think it is an mazing project..the problem is the reliable calculation on how can i make sure that the car will be float..anyone can help?..all your opinion will be useful for me..tq
13. Dear Rufus:
Make a scaled toy model and submerge it in a container of water to the desired level. If you check how much does the water level rise, that tells you an estimate of the volume. After that, calculate the weight (with typical load included). If the weight divided by the displaced volume is too high, it will sink. Otherwise, there is a chance that it will work.
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http://math.stackexchange.com/questions/254953/algebraic-solution-of-x-3x-4
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# Algebraic solution of $x + 3^x < 4$
I solved graphically and found that $x + 3^x < 4$ is true for $x < 1$ but I can't find a way to prove it algebraiclly, any hints will be greatly appreciated!
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## 3 Answers
The left hand side is an increasing function of $x$. With $x=1$, its value is $4$.
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I suppose we just take that $3^x$ is strictly increasing from the fact that $3^x = e^{x ln(3)}$ and the properties of $e$, right? – Edgar Sánchez Dec 9 '12 at 23:58
Yes, or with smaller teaspoons, that $3>1$ so that $\ln 3>0$, and the exponential function (i.e., base $e$) is strictly increasing. – Harald Hanche-Olsen Dec 10 '12 at 8:48
The solution is $x<1$
Since $x+3^x$ is strictly increasing. Therefore for all $x<1$ we have $x+3^x<1+3^1=4$.
If $1\leq x$ we have $1+3^1\leq x+3^x$ because $x+3^x$ is increasing.
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Thank you, that was fast! – Edgar Sánchez Dec 9 '12 at 23:58
Having found the solution $1$ to $x+3^x=4$ toucan use the fact that the derivative is positive to show it is unique.
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It's a pre-calculus problem so we are not allowed to use derivatives yet, but thank you anyway! – Edgar Sánchez Dec 9 '12 at 23:59
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http://unapologetic.wordpress.com/2011/10/06/the-hodge-star-on-differential-forms/?like=1&source=post_flair&_wpnonce=671450efa2
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# The Unapologetic Mathematician
## The Hodge Star on Differential Forms
Let’s say that $M$ is an orientable Riemannian manifold. We know that this lets us define a (non-degenerate) inner product on differential forms, and of course we have a wedge product of differential forms. We have almost everything we need to define an analogue of the Hodge star on differential forms; we just need a particular top — or “volume” — form at each point.
To this end, pick one or the other orientation, and let $(U,x)$ be a coordinate patch such that the form $dx^1\wedge\dots\wedge dx^n$ is compatible with the chosen orientation. We’d like to use this form as our top form, but it’s heavily dependent on our choice of coordinates, so it’s very much not a geometric object — our ideal choice of a volume form will be independent of particular coordinates.
So let’s see how this form changes; if $(V,y)$ is another coordinate patch, we can assume that $U=V$ by restricting each patch to their common intersection. We’ve already determined that the forms differ by a factor of the Jacobian determinant:
$\displaystyle dx^1\wedge\dots\wedge dx^n=\det\left(\frac{\partial x^i}{\partial x^j}\right)dy^1\wedge\dots\wedge dy^n$
What we want to do is multiply our form by some function that transforms the other way, so that when we put them together the product will be invariant.
Now, we already have something else floating around in our discussion: the metric tensor $g$. When we pick coordinates $x^i$ we get a matrix-valued function:
$\displaystyle g^x_{ij}=g\left(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right)$
and similarly with respect to the alternative coordinates $y^i$:
$\displaystyle g^y_{ij}=g\left(\frac{\partial}{\partial y^i},\frac{\partial}{\partial y^j}\right)$
So, what’s the difference between these two matrix-valued functions? We can calculate two ways:
$\displaystyle\begin{aligned}g^x_{ij}&=g\left(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right)\\&=g\left(\sum\limits_{k=1}^n\frac{\partial y^k}{\partial x^i}\frac{\partial}{\partial y^k},\sum\limits_{l=1}^n\frac{\partial y^l}{\partial x^j}\frac{\partial}{\partial y^l}\right)\\&=\sum\limits_{k,l=1}^n\frac{\partial y^k}{\partial x^i}\frac{\partial y^l}{\partial x^j}g\left(\frac{\partial}{\partial y^k},\frac{\partial}{\partial y^l}\right)\\&=\sum\limits_{k,l=1}^n\frac{\partial y^k}{\partial x^i}\frac{\partial y^l}{\partial x^j}g^y_{kl}\end{aligned}$
That is, we transform the metric tensor with two copies of the inverse Jacobian matrix. Indeed, we could have come up with this on general principles, since $g$ has type $(0,2)$ — a tensor of type $(m,n)$ transforms with $m$ copies of the Jacobian and $n$ copies of the inverse Jacobian.
Anyway, now we can take the determinant of each side:
$\displaystyle\lvert g^x_{ij}\rvert=\left\lvert\frac{\partial y^i}{x^j}\right\rvert^2\lvert g^y_{ij}\rvert$
and taking square roots we find:
$\displaystyle\sqrt{\lvert g^x_{ij}\rvert}=\left\lvert\frac{\partial y^i}{x^j}\right\rvert\sqrt{\lvert g^y_{ij}\rvert}$
Thus the square root of the metric determinant is a function that transforms from one coordinate patch to the other by the inverse Jacobian determinant. And so we can define:
$\displaystyle\omega_U=\sqrt{\lvert g^x_{ij}\rvert}dx^1\wedge\dots\wedge dx^n\in\Omega^n_M(U)$
which does depend on the coordinate system to write down, but which is actually invariant under a change of coordinates! That is, $\omega_U=\omega_V$ on the intersection $U\cap V$. Since the algebras of differential forms form a sheaf $\Omega^n_M$, we know that we can patch these $\omega_U$ together into a unique $\omega\in\Omega^n_M(M)$, and this is our volume form.
And now we can form the Hodge star, point by point. Given any $k$-form $\eta$ we define the dual form $*\eta$ to be the unique $n-k$-form such that
$\displaystyle\zeta\wedge*\eta=\langle\zeta,\eta\rangle\omega$
for all $k$-forms $\zeta\in\Omega^k(M)$. Since at every point $p\in M$ we have an inner product and a wedge $\omega(p)\in A^n(\mathcal{T}^*_pM)$, we can find a $*\eta(p)\in A^{n-k}(\mathcal{T}^*_pM)$. Some general handwaving will suffice to show that $*\eta$ varies smoothly from point to point.
### Like this:
Posted by John Armstrong | Differential Geometry, Geometry
## 11 Comments »
1. [...] will be useful to be able to write down the Hodge star in a local coordinate system. So let’s say that we’re in an oriented coordinate patch [...]
Pingback by | October 8, 2011 | Reply
2. [...] (Pseudo)-Riemannian Metrics, Isometries, Inner Products on 1-Forms, The Hodge Star in Coordinates, The Hodge Star on Differential Forms, Inner Products on Differential [...]
Pingback by | October 8, 2011 | Reply
3. [...] want to start getting into a nice, simple, concrete example of the Hodge star. We need an oriented, Riemannian manifold to work with, and for this example we take , which we [...]
Pingback by | October 11, 2011 | Reply
4. [...] is a -form, while we want the curl of a vector field to be another vector field. But we do have a Hodge star, which we can use to flip a -form back into a -form, which is “really” a vector field [...]
Pingback by | October 12, 2011 | Reply
5. [...] use the Hodge star again to flip the -form back to a -form, so we can apply the exterior derivative to that. We can [...]
Pingback by | October 13, 2011 | Reply
6. [...] interesting to look at what happens when we apply the Hodge star twice. We just used the fact that in our special case of we always get back exactly what we [...]
Pingback by | October 18, 2011 | Reply
7. [...] we want another way of viewing this orientation. Given a metric on we can use the inverse of the Hodge star from on the orientation -form of , which gives us a covector defined at each point of . Roughly [...]
Pingback by | October 27, 2011 | Reply
8. [...] do not arise from taking the exterior derivatives of -forms. If is pseudo-Riemannian, so we have a Hodge star to work with, this tells us that we always have some functions on which are not the divergence of [...]
Pingback by | November 24, 2011 | Reply
9. [...] remember that we’re working in our standard with the standard metric, which lets us use the Hodge star to flip a -form into a -form, and a -form into a vector field! The result is exactly the field [...]
Pingback by | January 11, 2012 | Reply
10. [...] Now I want to juggle around some of these Hodge stars: [...]
Pingback by | February 22, 2012 | Reply
11. [...] we push ahead with the Faraday field in hand, we need to properly define the Hodge star in our four-dimensional space, and we need a pseudo-Riemannian metric to do this. Before we were [...]
Pingback by | March 7, 2012 | Reply
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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://physics.stackexchange.com/questions/328/angular-momentum-and-force/329
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# Angular Momentum and Force
I'm stuck on number 5. The answers to the first 4 are correct, but I dont know how to set up number 5. Any idea that I would have would require me having some kind of time information, but thats not given. Any suggestions?
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1
Our goal here is an answer conceptual questions as opposed to homework questions, so I would prefer omitting the picture and asking something like, "Suppose I have a small bit of mass stuck on the outside of a spinning hoop. I know by intuition that the hoop is essentially 'trying to throw the mass off', but I don't know how to calculate this effect. How would I find the force between the hoop and the mass?" – Mark Eichenlaub Nov 7 '10 at 23:51
## 2 Answers
I'm not sure I totally understand the question, but I can make an educated guess. Firstly, I envisage a person sitting on the merry-go round with their body pointed tangentially to the circle, in which cases no force in that direction is required.
However, if the person is free to move radially (can slip inwards/outwards), then a force is required to counteract the centripetal force. In other words,
$$F_{holding} = F_{centripetal}$$
Consider the appropriate equation for centripetal force in terms of your known variables, and you can then plug in the values to get the answer.
(P.S. Don't look at my edit unless you want the answer, best you try yourself first!)
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Question in English: what is the name of a radial force acting toward the center ? What is the name of the pseudo-force introduced in the rotating frame of reference ? Because you said "to counteract the centripetal force" ... I would say that the force we are talking about is the centripetal force... (in French we make a distinction "centripeDe" and "centripeTe"). – Cedric H. Nov 7 '10 at 22:33
Yes this was it, thanks! – maq Nov 7 '10 at 22:39
@fprime: Glad to help. This is a pretty good example of a homework question; I see too many bad ones. :) – Noldorin Nov 7 '10 at 22:41
@Cedric: Yes, this is a slight problem of language. I'm considering the whole problem from the inertial frame of the ground. In this case, there is only a centripetal force. A force needs to be exerted by the person to counteract this, otherwise the person would fly off. Of course, you can also consider the problem from the rotating reference frame and talk about the centrifugal pseudo-force. Just complicates things though, in my view. (Oh, and in English centripedal is not a word - maybe a common mispelling of centripetal. Confusing, if you speak French, I admit!) – Noldorin Nov 7 '10 at 22:45
OK thanks for this clarification ! So centripetal and centrifugal. – Cedric H. Nov 7 '10 at 22:46
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http://math.stackexchange.com/questions/201796/describe-the-locus-of-points-of-this-fraction?answertab=oldest
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# Describe the locus of points of this fraction
What's the locus of: $$\mathrm{Im} \bigg(\frac{z-z_{1}}{z-z_{2}}\bigg)=0$$
I tried to figure it out and I get that it's a line. But it looks to me it will be wrong.
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## 3 Answers
Recall that $\frac{z-z_1}{z-z_2}$ is a complex number whose argument is the angle $z_1 z z_2$. Saying this number is real, is saying the angle is either $0$ or $\pi$, which means $z$ is on the line that contains $z_1$ and $z_2$.
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Put $z=x+iy,z_1=a+ib,z_2=c+id$
$$\frac{z-z_1}{z-z_2}=\frac{x-a+i(y-b)}{x-c+i(y-d)}=\frac{(x-a+i(y-b))(x-c-i(y-d))}{(x- c)^2+(y-d)^2}$$
$$Imz(\frac{z-z_1}{z-z_2})=\frac{(y-b)(x-c)-(y-d)(x-a)}{(x- c)^2+(y-d)^2}$$
$\implies (y-b)(x-c)=(y-d)(x-a)\implies x(d-b)+y(a-c)=ad-bc$
linear equation of $x,y\implies$ 2-D line.
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So it would be a line... – Mykolas Sep 24 '12 at 19:10
If it would ask for real part being equal to 0, would it be a circle? @lab bhattacharjee – Mykolas Sep 24 '12 at 19:23
@Mykolas, $x^2-(a+c)x+ac+y^2-(b+d)y-bd=0\implies$ 2-D circle. – lab bhattacharjee Sep 24 '12 at 19:26
Thankyou very much @lab bhattacharjee – Mykolas Sep 24 '12 at 19:26
Since $\Im\left(\frac{z-z_1}{z-z_2}\right)=0\Leftrightarrow \left(\frac{z-z_1}{z-z_2}\right)=c\in\mathbb{R}-\{0\}$, so for $z_1=a+b\cdot\mathbb{i}$ and $z_2=a'+b'\cdot\mathbb{i}$,
$z-z_1=c\cdot (z-z_2)\Leftrightarrow \left(x-a,y-b\right)=c\cdot\left(x-a',y-b'\right)\Leftrightarrow\begin{cases}x-a=c(x-a')\\y-b=c(y-b')\end{cases}$
and $\frac{x-a}{y-b}=\frac{x-a'}{y-b'}\Leftrightarrow y=-\frac{x-a}{a-a'}b'+b\frac{x-a'}{a-a'}$ which is a line.
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http://math.stackexchange.com/questions/333487/suppose-a-subset-0-1-b-subset-1-2-we-do-not-assume-they-are-lebesgue
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# Suppose $A \subset [0,1]$, $B \subset [1,2]$. We do not assume they are Lebesgue measurable?
Suppose $A \subset [0,1]$, $B \subset [1,2]$ (we do not assume they are Lebesgue measurable). Show that $m^{*}(A \cup B) = m^{*}(A) + m^{*}(B)$.
Just learned this. Not sure about it. I used $m^{*}(A \cup B)-m^{*}(B) \le m^{*}(A)$. Please help with proof.
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This problem is actually very easy if you understand the definition of $m^{*}$ (I think). Do you? If not, what don't you understand? – Quinn Culver Mar 18 at 4:54
I agree with user67236 who says that we should help user67236. – Quinn Culver Mar 18 at 5:13
Here is what I know so far because we just learned some of it not much. If A is measurable and B is any set with A intersect B = empty set. Letting A intersect B = empty set (A intersect B)intersect A is equal to A+(A union B)-A = B. That is the only thing I learned so far in this topic. – 9959 Mar 18 at 5:38
You said $A$ and $B$ are not assumed measurable yet used some fact that begins "If $A$ is measurable". This seems like a problem, doesn't it? And I'm sure you've learned more in this topic; e.g. the definition of $m^*$. Do you have the definition of $m^{*}$? – Quinn Culver Mar 18 at 5:57
That is the problem. It was given as trick question that was not learned yet. That is why I used that statement. Thanks for seeing that. I think if they are non measurable,we could have A intersect B equal the empty set. m*(A union B) does not = m*(A)+m*(B). – 9959 Mar 18 at 6:04
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## 2 Answers
I assume $m^*$ denotes the outer Lebesgue Measure in $\mathbb{R}$. In this case, we already know from the definition that $$m^*(A \cup B) \leq m^*(A) + m^*(B).$$ Hence it suffices to show the inequality "$\geq$". By definition $$m^*(A) = \inf\{ \sum_{k=1}^\infty l(I_k): I_k \text{ open Intervalls and } A\subset\bigcup_{k=1}^\infty I_k\},$$ where for an interval $I = (a,b)$ we denote by $l(I) := b-a$ the length of the interval $I$. Now let $I_k$ be arbitrary open Intervalls such that $A\cup B\subset\bigcup_{k=1}^\infty I_k$. Without loss of generality $\sum_{k=1}^\infty l(I_k) < \infty$. Then we have $A\subset\bigcup_{k=1}^\infty (I_k\cap[0,1])$ and $B\subset\bigcup_{k=1}^\infty (I_k\cap[1,2])$ respectively. Furthermore $$\sum_{k=1}^\infty l(I_k) \geq \sum_{k=1}^\infty l(I_k \cap [0,1]) + \sum_{k=1}^\infty l(I_k \cap [1,2]).$$ It follows from here that $$\begin{align} m^*(A\cup B) &= \inf\{ \sum_{k=1}^\infty l(I_k): I_k \text{ open Intervalls and } A \cup B \subset\bigcup_{k=1}^\infty I_k\} \\ &\geq \inf\{ \sum_{k=1}^\infty l(I_k \cap [0,1]): I_k \text{ open Intervalls and } A\cup B\subset\bigcup_{k=1}^\infty I_k\} \\ &\quad+ \inf\{ \sum_{k=1}^\infty l(I_k \cap [1,2]): I_k \text{ open Intervalls and } A\cup B\subset\bigcup_{k=1}^\infty I_k\}\\ &\geq m^*(A) + m^*(B) \end{align}$$
Please note, that the proof heavily relies on the fact, that $I_k \cap [0,1]$ is again an interval. The separation of $A$ and $B$ into disjoint intervals is the crucial assumption. Thus, a necessary condition for the outer Lebesgue measure to have strict subadditivity for two disjoint sets $A$ and $B$ is, that for any two intervals $I$ and $J$ the intersection $I \cap J$ is not a null set (in this case empty set or a point) whenever $A\subset I$ and $B\subset J$.
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The set $E=[0,1]$ is Lebesuge measurable. According to the definition of measurability, $$m^*(A \cup B)=m^*((A \cup B) \cap E)+m^*((A \cup B) \cap E^c)$$
Simplifying a little:
$(A \cup B) \cap E=(A \cap E) \cup (B \cap E)=A \cup(B \cap E)$
$B \subset [1,2]$ implies $B \cap E$ has at most one point, and therefore $m^*((A \cup B) \cap E)=m^*(A)$
Similarly, you can show that $m^*((A \cup B) \cap E^c)=m^*(B)$ which proves what's required.
I've used this known result: If $F$ is any set with $m^*(F)=0$, then $m^*(E \cup F)=m^*(E)$ for any $E \subseteq \mathbb R$
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http://physics.stackexchange.com/questions/35118/can-the-montevideo-interpretation-of-quantum-mechanics-do-what-it-claims
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# Can the Montevideo interpretation of quantum mechanics do what it claims?
Partly inspired by the great responses to a my previous physics.SE question about "reversing gravitational decoherence, today I was rereading the intriguing papers by Gambini, Pullin, Porto, et al., about what they call the "Montevideo interpretation" of quantum mechanics. They've written lots of papers on this subject with partly-overlapping content; see here for a list.
The overall goal here is to try to identify, within the (more-or-less) known laws of physics, a source of decoherence that would be irreversible for fundamental physics reasons, rather than just staggeringly hard to reverse technologically. One can argue philosophically about whether anyone should care about that, whether such a decoherence source is either necessary or sufficient for "solving the measurement problem", etc. Here, though, I'm exclusively interested in the narrower issue of whether or not such a decoherence source exists.
Gambini et al.'s basic idea is easy to explain: quantum-gravity considerations (e.g., the Bekenstein bound) very plausibly put fundamental limits on the accuracy of clocks. So when performing a quantum interference experiment, we can't know exactly when to make the measurement---and of course, the energy eigenstates are constantly rotating around! So, for that reason alone (if no other!), we can think about any pure state we measure as "smeared out a little bit" into a mixed state, the off-diagonal entries in the density matrix a little bit less than maximal.
More precisely, Gambini et al. claim the following rough upper bound for the magnitudes of the off-diagonal elements. Here, T is the elapsed time between the beginning of the experiment and the measurement, Tplanck is the Planck time, and EA-EB is the difference in energy between the two things being kept in superposition (so that $\frac{E_A - E_B}{\hbar}$ is the Bohr frequency).
(1) $\left| \rho_{offdiagonal} \right| \lt \exp \left( -\frac{2}{3} T_{planck}^{4/3}T^{2/3} \left( \frac{E_A - E_B}{\hbar} \right)^{2} \right).$
If EA-EB were equal to (say) the mass-energy of a few million protons, then (1) could certainly lead to observable effects over reasonable timescales (like a second).
Now, it might be that there's an error in Gambini et al.'s analysis or in my understanding of it, or that the analysis relies on such speculative assumptions that one can't really say one way or the other. If so, please let me know!
If none of the above holds, though, then my question is the following:
Can the bound (1) really do anything like the work that Gambini et al. claim for it---that is, of preventing "macroscopic interference" from ever being observed? More concretely, is it really true that anything we'd intuitively regard as a "macroscopic superposition" must have a large EA-EB value, and therefore a relative phase between the two components that rotates at unbelievable speed? In principle, why couldn't we prepare (say) a Schrödinger cat in an energy eigenstate, with the alive and dead components having the same energy (so that EA-EB=0)? Would such a state not constitute a counterexample to what Gambini et al. are trying to do?
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## 4 Answers
Jorge Pullin sent me an email responding to my question and to Ron Maimon's criticisms. In case people are interested, I'm posting it here with Jorge's kind permission. I particularly appreciated his clarification that the Montevideo interpretation is not really an "interpretation" at all, but new physics that makes new and different testable predictions. In particular, in Jorge and Rodolfo's view, the argument they give involving clocks is "motivational" in character; they're not claiming it as a derivation from accepted principles of quantum gravity. --Scott
Let us start with your own question. We believe it is answered in this paper of Paz and Zurek.
Quantum limit of decoherence: Environment induced superselection of energy eigenstates. Juan Pablo Paz (Buenos Aires U.), Wojciech Hubert Zurek (Los Alamos). Nov 1998. 4 pp. Published in Phys.Rev.Lett. 82 (1999) 5181-5185 DOI: 10.1103/PhysRevLett.82.5181 e-Print: quant-ph/9811026 [quant-ph] PDF
In fact our approach may be considered as a completion of the environmental decoherence approach. Paz and Zurek have shown that the pointer basis is defined by the eigenvectors of the interaction Hamiltonian and the self-Hamiltonian. It is not clear if pointer states can be defined among the eigenstates belonging to the same eigenlevel of the dominant Hamiltonian. If that were the case our axioms would not lead to the production of events for a Schroedinger cat that would involve states of that type.
As for the points raised by Maimon, there are several, some explicit, some implicit, so we will try to treat them separately:
a) His first point is that there is no relative phase in two interacting systems because underlying the theory is ordinary quantum mechanics. We think this is a misunderstanding of what we do. Or perhaps we did not explain it clearly. Although one talks of an "interpretation of quantum mechanics" the Montevideo interpretation has new physics, described by the modified Schroedinger equation we have. This is clear, for instance, when we formulated it axiomatically. Incidentally, a convenient place to find our papers is at http://www.montevideointerpretation.com
As a result systems lose relative coherence. One can motivate the new physics in the impossibility to observe precisely ordinary quantum mechanics, and we did so in some of the papers, but at some point one has to admit it is new physics. The new physics emerges naturally from the relational descriptions that are suitable for generally covariant theories like general relativity. We showed in the paper by Torterolo et al. that in such a context the only thing one can compute are conditional probabilities between observables that evolve in the unobservable time. One could extend that calculation including a second clock. But even if one prepared both clocks in the same initial state, after some time the clocks will disagree on the times assigned to each measurement, contrary to what Maimon claims.
b) As for his point of the photons, if one could put the screen very very far away, we contend that indeed interference would disappear. But obviously in any feasible experiment the effect is too small to observe. We wish it would be easier, then it would make the whole paradigm experimentally testable. Unfortunately the only way to see the effect is to have Schroedinger cat type states with huge energy differences compared to what is available at the atomic level. Those are not easy to find in the lab.
c) As for the kool aid part, we agree with you as well. I doubt very many people, apart from some string zealots, will claim that string theory solves the problem of time in generally covariant systems. Let alone the fact that the real universe is not AdS!
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I did not read these papers as "quantum-gravity considerations" but about limits imposed by (quasi)classical gravitation on quantum experiments. They discuss mainly clocks (*) but I think for this discussion it is easier to consider distance measurements:
In an interferometer the distances between mirrors etc. have to be controlled with great precision to see an interference pattern. In order to control the distances between mirrors one needs to make them heavier and the limitation from (quasi)classical gravitation is that the mirrors cannot form a black hole.
So I think it misses the point to state that "you get a pattern that doesn't give a hoot about what time it is".
If one would perform an interference experiment with macroscopic bodies the effective wavelengths would be at or below the Planck length and one could not place the mirrors with sufficient precision to see anything. I think this is the kind of issue they are talking about and not a new quantum gravity theory. The fact that the S-matrix can be calculated in string theory is great, but does not really have anything to do with this imho.
(*) I guess the reason they consider clocks is because they want to get universal quantitative bounds on decoherence times and not just qualitative statements about a particular interference experiment.
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The issue I have with this interpretation (which is close to what they suggest, replacing time with space) is that it assumes that there is a decoherence caused by the uncertainty in the gravitational metric background. If this background is coherent vacuum, there should be no uncertainty at all. I understand the experimental setup makes it hard to measure interference for macroscopic objects, but if you have a zero-temperature mesoscopic grain, you should be able to observe interference when it is at rest, and then boost the whole system to a velocity where the wavelength is trans-Planckian – Ron Maimon Aug 29 '12 at 13:10
If they suggested the cosmological horizon has thermal gravitons that you can't shield, I would be on board--- I do think that the deSitter horizon makes it difficult to define quantum coherence for a deSitter universe, and I believe the proper description of dS space is by a thermal statistical ensemble, and no pure state. But the bounds from gravity alone in flat space are not consistent with an interpretation that gravity leads to fundamental decoherence. – Ron Maimon Aug 29 '12 at 13:12
In my example the issue is the uncertainty in the position of the mirror(s) (which has to withstand the recoil of the photon or Schroedinger's cat) and not uncertainty about the metric. – wolfgang Aug 29 '12 at 13:44
Also, if you move the particle (or Schroedinger's cat) very slowly through the interferometer to increase the wavelength, then the payoff is that the experiment takes much longer and you need to fight decoherence that much longer. If G & P did their calculation correctly (which I did not check) then there is a limit to what you can achieve by slowing down the particle or cat. – wolfgang Aug 29 '12 at 13:47
No it cannot. This is not an interpretation, nor a new theory, it is a misunderstanding. The papers are vacuous, and not in an interesting way.
### What time is it?
The basic idea is that we don't know what time it is, really. So we make a probability distribution for what time it is, and this changes the phase of the wavefunction by an amount proportional to however much we don't know the time. Then the authors claim that this uncertainty introduces a decoherence phenomenon into the Schrodinger equation, because the phases of energy eigenstates are shifted by an uncertain amount.
This is just plain wrong. The reason is that although we don't know what time it is, we know from the author's assumption that there is a consistent Schrodinger time (this is one of their axioms) that whatever time it is, it's the same time for any two things in the theory. So while there is an uncertainty in the phase of an isolated system from not knowing what time it is, there is no uncertainty introduced in the relative phase of two interacting systems, and there is no decoherence caused by this uncertainty, except through mistakes in analysis.
### Mistakes
These mistakes are subtly introduced by making a separation between "observer" and "system", and introducing the probability distribution for the clock reading only in those cases where the observer is interacting with the system, and when observers are interacting with each other, not when systems are interacting with systems.
For example, suppose you have a photon split by a beam splitter, with one part going through some glass, then through a double slit to measure interference. This will work, the beams stay coherent, because the photon that gets through doesn't excite any irreversible quantum in the glass. Notice that the glass is large and macroscopic, though.
Anyway, once the photons interfere, you get a pattern that doesn't give a hoot about what time it is, just on the relative phase-difference between the two photons, the path-difference in the optical system. So the photon didn't care that we don't know what time it is for the glass, because it just goes through the glass and not, and whatever time it is, it interferes with the other photon, which doesn't care what time it is either, because it's the same time as the other photon.
Note that the photon interacted with this enormous glass, and all the atomic absorption and emission events had to coherently come together even though for the glass, we don't know what time it is. The relative coherence is maintained throughout.
So when does the problem of time show up in these papers? It shows up when the observer entangles with the system, and at this point, the authors declare that the uncertainty in what time it is shows up as an uncertainty in the phase of the system that the observer measures.
If there is a second observer measuring something else, they introduce an uncertainty in the second observer's time. But then when the observers come to talk, the authors pretend that the two observers phase-shifts are uncorrelated, when in fact the uncertainty in what time it is is exactly the same for the two observers, because it is an uncertainty in the same global t variable that they both don't know.
So whatever the t-uncertainty for each observer, the coherence effects between the two observers are not washed out, unless you make the assumption that the actual global t is different for the two observers, an assumption that is at odds with the postulates of the theory, that there is a global time ticking down there underneath it all.
### There is no problem of time in S-matrix theory
Constrary to the authors' claims, string theory solves the problem of time definitively and for good, that's the whole point. The solution was the motivation for Heisenberg to introduce S-matrix theory in the first place, it allows you to make a theory in cases where space and time are unreliable.
An S-matrix theory doesn't give a detailed history of the events in the interior of spacetime, it only relates things on the boundary to other things on the boundary. It doesn't have a real local non-asymptotic time variable at all, so it can't describe time-dependent phenomenon, like the formation and evaporation of a quantum black hole in detail. This is why we are in the embarassing position of having essentially exact quantum description of forming and evaporating black holes while at the same time not being able to answer some of the simplest questions about this process.
So if you make a string scattering calculation, or an AdS/CFT calculation using boundary states, you don't have a problem of time on the interior, you can't, because time on the interior just doesn't appear in the description. It is at best reconstructed approximately from the quantum state on the boundary.
You might say "but then what about the problem of time on the boundary!", but the boundary theory is non-gravitational, and it doesn't have a time problem either. This is the miracle of string theory, and this is what makes it the only plausible candidate for quantum gravity--- the philosophical problems completely evaporate in S-matrix, it is as if they never existed.
You might object that there is a t-variable on the perturbative string world-sheet, but this is an artifact of the perturbative theory, of describing the string scattering process in detail using intermediate states, which you then interpret as localized in time. This interpretation is not completely good, you can't associate local operators to the string. If you do string field theory, you need to do it in light cone, and then the string story becomes more or less local along the light-front, but the light-front time variable is going diagonally in space time, and the string field is only telling a local story in the transverse coordinates to the light-cone pair. It stays nonlocal in the light-cone pair (time and one other coordinate).
If one were given the correct exact string S-matrix in our vacuum, there would be no t-variable in the S-matrix, only the S-matrix in and out state which does not reference any clocks at all. You might object that the S-matrix gives you phase shifts in outgoing waves, and to measure these phase shifts you might think you need a clock, but this is not so, since the relative phase between two states can be determined in principle by perfoming a second much-later scattering which can be approximated as two separate scatterings, and allowing the scattered products in different direction to interfere with each other to make fringes. The phase shifts of the original scattering now show up in the k-directions of the bragg diffraction of the two waves, and you can reconstruct the phase shift information from complicated scattering matrix data in principle without needing a clock on the interior, just by considering a more complicated in state.
This is exactly what you do for a photon--- you scatter it off a double slit to turn the difference in phase shift into a spatial diffraction pattern. This is not obscure at all, although it is hopeless to describe the appropriate s-matrix elements in detail for any realistic experiment.
This means that the problem of time cannot even be stated in S-matrix theory, and time is not treated differently from space, because neither is treated at all. This is the greatest virtues of string theory, and it is the reason that the S-matrix program was able to make such surprising progress in quantum gravity, which was not its original goal.
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Thanks, Ron. Offhand remark: to those who haven't fully drunk the Kool-Aid yet, not being able to state the problem of time, since you don't treat space or time at all, might not sound like such a wonderful "virtue"! I.e., if you want to say various things we care about are just "illusions," then the question shifts immediately to explaining those illusions and figuring out their basic properties, which you admit hasn't been done yet in the case of AdS/CFT. – Scott Aaronson Aug 29 '12 at 11:47
Moving on to your criticism of Gambini et al.: I imagine their response would be, sure, in the case of the photon travelling through glass, you'll see the interference for exactly the reasons you say. But when trying to superpose two macroscopic bodies, the situation is substantially different because of the huge energy differences involved (cf wolfgang's answer). Is there any merit to that claim? (For this question, let's forget about whether everything is "really" unitary in some boundary description to which we don't have access; I'm only interested in limits on actual experiments.) – Scott Aaronson Aug 29 '12 at 12:00
@ScottAaronson: The problem is that their hypothesis of fundamental decoherence is not justified by the mechanism they give--- not knowing what time it is. For observing coherence, you only need to know that there is a global time in which all phenomena are coherent. This is a postulate in their method, and if they omitted this postulate, and said that there was a real fundamental decoherence due to gravity, it wouldn't be necessarily inconsistent, but it would probably conflict with experiment (but it might be possible, it's just not what they do). – Ron Maimon Aug 29 '12 at 12:34
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... regarding string theory kool aid, which I drink with no reservation, reconstructing time and space from boundary data is indeed the main sticking point in accepting S-matrix theory, and it is why it took two decades from 1940-1960 before the theory could really take off. The 1960s literature is entirely devoted to the problem of reconstructing local physics from S-matrix data, and it is possible, although extremely difficult and philosophically involved. Although the "illusion of space" is not completely worked out in AdS/CFT, it is worked out enough to know for sure it works. – Ron Maimon Aug 29 '12 at 12:36
OK, thanks. I understood them to be saying: when you superpose two bodies with a huge difference in energy, the Bohr frequency condition implies that the relative phase between them really is rotating at an enormous speed, which then means that you really do need to know what time it is to great precision if you want to measure anything about the phase. This argument might still be wrong, but I don't see how one can refute it by pointing to their assumption of a global time ticking underneath everything. (And incidentally, I didn't see where they ever discussed two observers.) – Scott Aaronson Aug 29 '12 at 12:44
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Can the Montevideo interpretation of quantum mechanics do what it claims? Yes.
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Please back it up! – Qmechanic♦ Oct 26 '12 at 20:41
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http://math.stackexchange.com/questions/227691/show-the-usual-schwartz-semi-norm-is-a-norm-on-the-schwartz-space/227702
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Show the usual Schwartz semi-norm is a norm on the Schwartz space
Let $f \in C^\infty(\mathbb R)$. Define the semi-norm $$\|f\|_{a,b}=\sup_{x \in \mathbb R} |x^af^{(b)}(x)|$$ where $a,b \in \mathbb Z_+$, and $f^{(b)}$ is the $b$-th derivative of $f$.
Show $\|\cdot\|_{a,b}$ is a norm on the Schwartz space $S(\mathbb R)$.
I don't see how to proove this direction of the nonnegativity requirement for the norm $$\|f\|_{a,b} =0 \text{ implies } f=0.$$
If one of the derivative is zero, how can I infer that the original function is also zero?
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1 Answer
Let $f$ such that $\lVert f\rVert_{a,b}=0$. Then for each $x\neq 0$, we have $f^{(b)}(x)=0$, and as $f^{(b)}$ is continuous, $f^{(b)}(x)=0$ for all $x\in\Bbb R$. This implies that $f$ is a polynomial. As $f$ is assumed to be in the Schwartz space, $f$ vanishes at infinity, hence $f\equiv 0$.
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Very good proof. Thank you. – Nicolas Essis-Breton Nov 2 '12 at 18:04
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http://math.stackexchange.com/questions/109395/raising-a-power-series-by-a-power-series
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# Raising a power series by a power series?
I know addition and multiplication are well defined operations on formal power series. Now say you have two formal power series $F(x),G(x)\in R[[x]]$, with $R\supset\mathbb{Q}$ is the coefficient ring.
Is there a way to define $F(x)^{G(x)}$? Is there a standard well defined definition for this operation that hopefully satisfies the usual exponent laws? Thanks.
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No, even $x^x$ doesn't make sense in the power series ring ! – Georges Elencwajg Feb 14 '12 at 21:55
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At least, $F(x)$ probably needs to have a nonzero constant term, since otherwise you expect something like a logarithmic singularity (think of the expression as $\exp(G(x)\ln F(x))$). But then, I would suspect (but don't know) that this standard formula will provide a positive answer. – Harald Hanche-Olsen Feb 14 '12 at 22:00
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It interesting to note that $${\left( {1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \cdots } \right)^x} = 1 + x^2 + \frac{{{{\left( {{x^2}} \right)}^2}}}{{2!}} + \frac{{{{\left( {{x^2}} \right)}^3}}}{{3!}} + \cdots$$ i.e $(e^x)^x = e^{x^2}$ – Peter Tamaroff Feb 18 '12 at 23:30
## 2 Answers
Composition of formal power series can be done via the power-series version of Faà di Bruno's formula. In that way one can find the series for $\log_e F(x)$. Then one can say $$F(x)^{G(x)} = e^{G(x)\log_e F(x)}.$$ Multiplication of formal power series is a well-known operation. Exponentiation can be done via the exponential formula.
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Provided $F(x)$ has a nonzero constant term. – Did Feb 14 '12 at 23:58
There is no general , universally accepted answer to this problem .If the exponent is a function of some variable , you cannot be restricted to the Real Number World . You have to get into the Complex Number World. f(x)=f(x)*1 =f(x)*exp(i*2*n*pi) f(x)^g(x) = [f(x)^g(x)] *[exp(i*2*n*pi*g(x))]
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What does this have to do with power series? – Martin Argerami Nov 28 '12 at 4:51
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http://www.physicsforums.com/showthread.php?t=437669
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Physics Forums
Page 1 of 2 1 2 >
Associates and their Norms
Hello PhysicsForums:
I have two things I'd like to learn more about:
1. I was noticing that the norms of +-1+-i (which are associates) are the same. The same goes for +-(1-2i). Is there a general reason explaining why any two integers which are associates have the same norm?
2. Assume that 'a' is a set of quadratic integers in Q[Sqrt(-3)] so that $$\bar{a}$$ and 'a' are associates. Can someone help describe what this set would be like when this occurs?
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Recognitions: Gold Member 1. It isn't true in general that associates have the same norm. For example, consider Q($\sqrt{2}$). Then u = 1 + $\sqrt{2}$ is a unit. Let a = 2 + $\sqrt{2}$. Then ua is an associate of a and ua = 4 + $3\sqrt{2}$. However, N(a) = (2 + $\sqrt{2}$)(2 - $\sqrt{2}$) = 2, but N(ua) = (4 + $3\sqrt{2}$)(4 - $3\sqrt{2}$) = -2 This fails because norms in Q($\sqrt{2}$) can have negative values. However, if d < 0 and squarefree, then norms of elements in Q($\sqrt{d}$) are $\geq$ 0. Then the norm of any unit = 1 and the product formula for norms implies that associates have the same norm. 2. I don't understand what you mean by the phrase so that $\bar{a}$ and 'a' are associates. Could you please elaborate? Thanks.
Quote by Petek 1. It isn't true in general that associates have the same norm. For example, consider Q($\sqrt{2}$). Then u = 1 + $\sqrt{2}$ is a unit. Let a = 2 + $\sqrt{2}$. Then ua is an associate of a and ua = 4 + $3\sqrt{2}$. However, N(a) = (2 + $\sqrt{2}$)(2 - $\sqrt{2}$) = 2, but N(ua) = (4 + $3\sqrt{2}$)(4 - $3\sqrt{2}$) = -2 This fails because norms in Q($\sqrt{2}$) can have negative values. However, if d < 0 and squarefree, then norms of elements in Q($\sqrt{d}$) are $\geq$ 0. Then the norm of any unit = 1 and the product formula for norms implies that associates have the same norm. 2. I don't understand what you mean by the phrase so that $\bar{a}$ and 'a' are associates. Could you please elaborate? Thanks.
1. Is saying that any two integers which are associates the same thing as saying two associates? I was referring to the Norm of these aforementioned integers being equal, not the norm of the associates being equal. (I hope that makes sense)
2. I meant to say that $\bar{a}$ and $a$ are associates. If I would re-write it:
Assume that $a$ is a set of quadratic integers in Q[Sqrt(-3)] so that $\bar{a}$ and $a$ are associates. Can someone help describe what this set would be like when this occurs?
Recognitions:
Gold Member
Associates and their Norms
1. You originally observed that (1 + i) and -(1 + i) were associates and had the same norm (that's not exactly what you wrote, but I think is what you meant). You also observed that (1 -2i) and -(1 -2i) were associates and had the same norm. You wondered whether there was an explanation why any two integers that were associates had the same norm. I gave an example that showed this wasn't always true. However, I pointed out that your statement was true if the norm of any integer was non-negative. Here's how to prove that:
First, it's easy to see that, in general, the norm of a unit must be either +1 or -1. Since we're assuming that all norms are non-negative, then any unit must have norm = 1. Now suppose that a and b are associates. Then there exists a unit u such that a = ub. Therefore,
N(a) = N(ub) = N(u) N(b) [Product rule for norms] = N(b)
since N(u) = 1. It remains to show that if d < 0, then the norm of any integer in Q($\sqrt{d}$) is non-negative. That's a simple calculation.
I hope this makes my previous post more clear.
2. $a$ is a set, not a number, so you need to define what you mean by $\bar{a}$ and what you mean by two sets being associates. I could try to guess, but prefer that you give your definition. I'm not trying to be obstinate, but your usage isn't standard, AFAIK.
Quote by Petek 1. You originally observed that (1 + i) and -(1 + i) were associates and had the same norm (that's not exactly what you wrote, but I think is what you meant). You also observed that (1 -2i) and -(1 -2i) were associates and had the same norm. You wondered whether there was an explanation why any two integers that were associates had the same norm. I gave an example that showed this wasn't always true. However, I pointed out that your statement was true if the norm of any integer was non-negative. Here's how to prove that: First, it's easy to see that, in general, the norm of a unit must be either +1 or -1. Since we're assuming that all norms are non-negative, then any unit must have norm = 1. Now suppose that a and b are associates. Then there exists a unit u such that a = ub. Therefore, N(a) = N(ub) = N(u) N(b) [Product rule for norms] = N(b) since N(u) = 1. It remains to show that if d < 0, then the norm of any integer in Q($\sqrt{d}$) is non-negative. That's a simple calculation. I hope this makes my previous post more clear. 2. $a$ is a set, not a number, so you need to define what you mean by $\bar{a}$ and what you mean by two sets being associates. I could try to guess, but prefer that you give your definition. I'm not trying to be obstinate, but your usage isn't standard, AFAIK.
First off, thank you for explaining that 1st point to me, I really appreciate it!
2. $$\bar{a}$$ is simply the conjugate of $a$. It is to my understanding that since $a$ is a set of quadratic integers of the form $a+b*\sqrt{-3}$, that $\bar{a}$ would be a set of integers of the form $a-b*\sqrt{-3}$.
Associate: If $\alpha$ and $\beta$ are non-zero integers in $Q[\sqrt{d}]$ such that $\alpha = \beta*\epsilon$, where $\epsilon$ is a unit, then $\alpha$ is said to be an associate of $\beta$. In other words, $\alpha$ is only an associate of $\beta$ if and only if $\alpha / \beta$ is a unit.
Recognitions: Gold Member Is this your question? Let a be an integer in Q($\sqrt{-3}$). Under what conditions is $\bar{a}$ an associate of a? If so, start by describing both the integers and units of Q($\sqrt{-3}$). Hint: There are more of them than you might think.
Quote by Petek Is this your question? Let a be an integer in Q($\sqrt{-3}$). Under what conditions is $\bar{a}$ an associate of a? If so, start by describing both the integers and units of Q($\sqrt{-3}$). Hint: There are more of them than you might think.
No, I already know what an associate is. I'm simply saying this:
If $a$ is a set of quadratic integers in Q($\sqrt{-3}$) and $\bar{a}$ is the associate set of quadratic integers with respect to $a$, what would the set of quadratic integers in $a$ have to look like? I essentially would like to know a description of $a$.
Recognitions: Gold Member Let's fix some notation. Perhaps that will help me to understand your question. Let K = Q($\sqrt{-3}$). Let $O_{K}$ denote the integers of K. Also, let's use uppercase letters to denote sets and lowercase to denote elements of sets. Let A $\subseteq\\O_{K}$ (A is a subset of the integers of K.) Let $$\overline{A}$$ = {$$\bar{a}$$: a $$\in$$ A} (the set of conjugates of elements in A) Let A~ = {ua: a$$\in$$A and u is a unit in $$O_K$$} (the set of all associates of elements in A) As I understand it, you want A to be a subset of integers in K that satisfies certain relationships with its set of conjugates and set of associates. Is that correct? If so, what is the relationship?
Quote by Petek Let's fix some notation. Perhaps that will help me to understand your question. Let K = Q($\sqrt{-3}$). Let $O_{K}$ denote the integers of K. Also, let's use uppercase letters to denote sets and lowercase to denote elements of sets. Let A $\subseteq\\O_{K}$ (A is a subset of the integers of K.) Let $$\overline{A}$$ = {$$\bar{a}$$: a $$\in$$ A} (the set of conjugates of elements in A) Let A~ = {ua: a$$\in$$A and u is a unit in $$O_K$$} (the set of all associates of elements in A) As I understand it, you want A to be a subset of integers in K that satisfies certain relationships with its set of conjugates and set of associates. Is that correct? If so, what is the relationship?
You are correct. I don't know what you mean by asking me what is the relationship. I'm trying to determine what the set of $A$ will be like though so that the conditions are met.
Recognitions: Gold Member Let me rephrase your original question using the above notation: Assume that A $$\subseteq \\ O_K$$ is such that $$\overline{A}$$ and A are associates. Can someone help describe what this set would be like when this occurs? So the condition on the set A is that "$$\overline{A}$$ and A are associates." In a prior post, I asked for your definition of what it means for two sets to be associates. You replied with the definition of what it means for two integers in $O_K$ to be associates. So, suppose that A and B are subsets of $O_K$. Then A and B are said to be associates if ... . Please supply the rest of the definition*. I know what it means for two numbers to be associates, but not what you mean by saying that two sets are associates. Thanks. * For example, a possible definition would be that A and B are said to be associates if for every a $\in$ A, there exists a b $\in$ B such that a and b are associates.
Quote by Petek * For example, a possible definition would be that A and B are said to be associates if for every a $\in$ A, there exists a b $\in$ B such that a and b are associates.
That is exactly what I'm saying! :-)
Recognitions:
Gold Member
Quote by Brimley That is exactly what I'm saying! :-)
Good! So let's define A = {a $\in O_K$: a is an associate of some element in $$\overline{A}$$}
Now, try to prove the following:
1. Any rational integer (that is, any element of Z) belongs to A.
2. Any unit in $O_K$ belongs to A.
3. 1 + $\sqrt{-3} \in$ A.
Can you find any other elements of A?
Quote by Petek Good! So let's define A = {a $\in O_K$: a is an associate of some element in $$\overline{A}$$} Now, try to prove the following: 1. Any rational integer (that is, any element of Z) belongs to A. 2. Any unit in $O_K$ belongs to A. 3. 1 + $\sqrt{-3} \in$ A. Can you find any other elements of A?
+-1 (Units)
$+-1+*\sqrt{-3}$
+-2
$+-\sqrt{-3}$
I believe those all work, and I know that each shown above and its negative are associates, however $1 +*\sqrt{-3}$ and $1 - \sqrt{-3}$ are not associates.
Recognitions: Gold Member You need to work on your notation: That extra i isn't necessary. And, 1 + $\sqrt{-3}$ and 1 - $\sqrt{-3}$ are associates. There are 6 units in $O_K$. Two are +1 and -1. Can you find the other 4?
Quote by Petek You need to work on your notation: That extra i isn't necessary. And, 1 + $\sqrt{-3}$ and 1 - $\sqrt{-3}$ are associates. There are 6 units in $O_K$. Two are +1 and -1. Can you find the other 4?
My bad! I totally goofed that up, but I'm glad you knew what I meant.
Perhaps 2 of them are 0+-i ? I'm not sure what would the others be.
Recognitions: Gold Member No problem. I'll be mostly offline for the next few days. Take a look at the Wikipedia articles on Quadratic Integers and Eisenstein Integers. Use the article on Quadratic Integers to determine the elements of $O_K$. Figure out why the Eisenstein integers are relevant to your situation. I'll check back next week to see how you did. Perhaps someone else will add to the discussion.
Quote by Petek No problem. I'll be mostly offline for the next few days. Take a look at the Wikipedia articles on Quadratic Integers and Eisenstein Integers. Use the article on Quadratic Integers to determine the elements of $O_K$. Figure out why the Eisenstein integers are relevant to your situation. I'll check back next week to see how you did. Perhaps someone else will add to the discussion.
If someone could help me with this, I'd really appreciate it. I looked at those Wiki's and I was totally lost.
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http://physics.stackexchange.com/questions/33750/is-temperature-an-extensive-property-like-density
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# Is temperature an extensive property, like density?
I was thinking about it some time ago, and now that I've discovered this site I would like to ask it here because I couldn't work it out then.
I know that the higher temperature the air in my room has, the more energy the molecules have. But temperature isn't energy because otherwise we'd be measuring temperature in joules, and we don't. And then temperature would depend on the number of molecules in the room, and that doesn't make any sense. So what I thought temperature had to be was the total energy that the molecules in the room have divided by something, for example the number of molecules or the volume of the room. If it was the latter, then temperature would be exactly like density, only with energy instead of mass. But in any case, I went to Wikipedia and tried to see if I could understand what they said about temperature. I didn't understand too much, but I saw that they used something called entropy to define temperature. I couldn't understand the article on entropy at all, but I think it means my thinking must have been incorrect because otherwise they would mention something simple like this in the article. Could you please explain it to me?
EDIT: Here's why I thought it should be the total energy divided by the volume rather than by the number of particles: because if we divided energy by a number, it would still be energy, and we measure energy in joules, not kelvins.
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Actually, you do hear physicists talk about "a temperature of 0.01eV" or "an energy of 5000K". The conversion factor is the Boltzmann constant: $k_B T$ has units of energy. This is similar to $E = m c^2$ (and again, phrases like "a mass of 0.511eV" are very common). The difference between these two situation is that $k_b T$ is simply a characteristic energy scale of your system - most of the molecules in a gas have an energy of about $k_b T$ - while $E = m c^2$ is an exact equivalence. – poorsod Aug 11 '12 at 21:12
## 3 Answers
Temperature is related to average energy per degree of freedom via the equipartition theorem. For example, as the kinetic energy is quadratic in the velocity and corresponds to three degrees of freedom (the three spatial directions), on average each molecule will have a kinetic energy of $\frac32k_BT$ where $k_B$ is the Boltzmann constant.
This means that temperature and energy are indeed intimately related (temperature can be considered a measure of average energy) and thus there are so-called natural systems of units which set $k_B=1$. This means that temperature will have the same unit as energy, eg electron volt in case of particle physics.
While the equipartition theorem probably provides the most intuitive visualization of temperature, it also has its problems: As Arnold Neumaier points out, it only holds under specific assumptions and in particular breaks down in case of non-ergodic systems or in cases where continuity is no longer a good approximation for quantized energy levels.
An example of such a system would be a non-atomic gas, which adds quantized internal degrees of freedom to the mix. The heat capacity of diatomic gases provides a good illustration for this as it can be derived classically via the equipartition theorem. A reasonably good explanation of the measurements is that 'quantum mechanical' degrees of freedom do not contribute at low temperatures and start approching the classical contribution as temperature increases. Rotational degrees of freedom contribute almost fully at room temperature, whereas the vibrational degrees of freedom only contribute for heavier molecules as the spacing of the vibrational energy levels depends on the reduced mass of the system.
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Everyone knows that room temperature is about 1/40 eV, right? – dmckee♦ Aug 8 '12 at 20:33
Note that this is valid for an ideal gas only! – Arnold Neumaier Aug 11 '12 at 12:24
@Arnold: that would be news to me - note that I'm only talking about kinetic energy, so this should be valid for any non-relativistic system – Christoph Aug 11 '12 at 13:08
@Christoph: Indeed, Wikipedia says that it holds for all classical ergodic systems, and gives a derivation. But there are many classical systems that are not ergodic. – Arnold Neumaier Aug 11 '12 at 13:12
@Christoph: I read the whole Wikipedia article. The successful examples are all about ideal cases: ideal gas, ideal solution, ideal crystal, in which the relevant degrees of freedom are independent. At the end of the ''solids'' section they say that the equipartition theorem often breaks down. – Arnold Neumaier Aug 11 '12 at 13:30
show 15 more comments
In equilibrium, temperature and energy are rigorously related by the first law of thermodynamics.
For one mole of a simple system (chemically homogeneous matter), the first law relates for arbitrary reversible processes the infinitesimal changes $dU$, $dS$, and $dV$ of the internal energy $U$, the entropy $S$ (a measure of microscopic complexity), and the volume $V$, respectively, by the differential equation $dU=TdS-PdV$, where the temperature $T$ and the pressure $P$ appear as factors.
The term $TdS$ is the infinitesimal amount of heat generated or absorbed by the process, which gives some intuitive meaning for entropy as a suitably normalized measure of heat. The term $PdV$ is the infinitesimal amount of mechanical work done by or on the system.
Thus temperature is an ''intensive'' quantity analogous to the pressure. Equilibrium is characterized by constancy of the intensive quantities. Just as pressure differences cause mechanical motion, so temperature differences cause heat flow.
In statistical mechanics, one learns how to express $U$ as a function $U=U(S,V)$ of $S$ and $V$ that depends on the microscopic structure of the material. Keeping $V$ constant (no mechanical work done), the first law implies $T=\partial U(S,V)/\partial S$; keeping $S$ constant (no heat exchange), the first law implies $P=\partial U(S,V)/\partial V$.
For an ideal gas, exact formulas can be derived, leading to the ideal gas law, which say that $PV/T$ is a universal constant $R$ (independent of the composition of the gas).
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Consider a huge number of particles, each one is characterized by its energy and velocity. The number is so great that one can not access the speed or energy of a test particle. Physicists have devised a mean to gain some information about this set of particles. They do average position, mass, momentum, energy, ... But to effectively compute this average one needs to know how many particles do have a specific energy. It is the same situation when you compute the average of your marks at school. You need to know how many tests you have got A, B, c, etc.. There is a function, called distribution function which precisely tells us how many particles do have a specific velocity. When the system is in equilibruium, this function depends on a parameter that can be identified with the temperature T. For equilibrium, the macroscopic temperature coincides with the average kinetic energy of a particle. Out of equilibrium temperature may not be unique or even defined. o you get it right for the most common case. For the unit, temperature at the microscopic scale always appears combined with kb, the Boltzmann constant, and is often given in eV (1 ev=1.6 10^-19 J)
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http://physics.stackexchange.com/questions/19459/time-reversal-symmetry-and-t2-1
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# Time reversal symmetry and T^2 = -1
I'm a mathematician interested in abstract QFT. I'm trying to undersand why, under certain (all?) circumstances, we must have $T^2 = -1$ rather than $T^2 = +1$, where $T$ is the time reversal operator. I understand from the Wikipedia article that requiring that energy stay positive forces $T$ to be represented by an anti-unitary operator. But I don't see how this forces $T^2=-1$. (Or maybe it doesn't force it, it merely allows it?)
Here's another version of my question. There are two distinct double covers of the Lie group $O(n)$ which restrict to the familiar $Spin(n)\to SO(n)$ cover on $SO(n)$; they are called $Pin_+(n)$ and $Pin_-(n)$. If $R\in O(n)$ is a reflection and $\tilde{R}\in Pin_\pm(n)$ covers $R$, then $\tilde{R}^2 = \pm 1$. So saying that $T^2=-1$ means we are in $Pin_-$ rather than $Pin_+$. (I'm assuming Euclidean signature here.) My question (version 2): Under what circumstances are we forced to use $Pin_-$ rather than $Pin_+$ here?
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– Qmechanic♦ Jan 21 '12 at 13:31
## 1 Answer
There are two possible answers to why $T^2=-1$:
a) Why not. The total phase of a quantum state is unphysical. So a symmetry may be realized as a projective representation. Here T may be viewed as a projective representation of time reversal $T_{phy}$ which satisfy $T^2_{phy}=1$.
b) If we define the time reversal symmetry to be realized as a regular representation in a many-body systems with $T^2=1$, the symmetry operations that act on fractionalized quasiparticles may be realized projectively, with $T^2_{quasi}=-1$.
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http://mathhelpforum.com/advanced-algebra/119939-nomenclature.html
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# Thread:
1. ## Nomenclature
How exactly does this read:
(prove that) R[x]/<x^2+2> (is isomorphic to C)
And what exactly does the "/" imply algebraicly?
Thanks.
2. ...
3. Here, you are creating the complex numbers! This is very elegant. The $\mathbb{R}$ shows that you are in the reals, so $x^2$ can have any coefficient in the reals. BUT, $x^2=-2$ There is nothing when squared in the reals equal to 2, so you have to go imaginary. So, in the multiplication table, whenever you see an $x^2$ replace it with a -2. The $/$ means you are "modding out by"
4. sfspitfire23 give us such a elegant explanation.
So, follow his thread, and you will find the isomorphism.
Good luck!
5. Originally Posted by elninio
How exactly does this read:
(prove that) R[x]/<x^2+2> (is isomorphic to C)
It reads:
(prove that) the quotient ring of $\mathbb R[x]$ [the ring of polynomials with real coefficients] by the ideal generated by $x^2+2$ (is isomorphic to $\mathbb C)$
To prove this, given any $f(x)\in\mathbb R[x],$ use the division algorithm to write $f(x)=q(x)(x^2+2)+ax+b$ where $q(x)\in\mathbb R[x]$ and $a,b\in\mathbb R$ are uniquely determined by $f(x).$ Define a map $\varphi:\mathbb R[x]\to\mathbb C$ by $\varphi\left(f(x)\right)=b+i\sqrt2a.$ All that remains is to show that $\varphi$ is a ring epimomophism (surjective homomorphism) with kernel $\left\langle x^2+2\right\rangle.$
Originally Posted by elninio
And what exactly does the "/" imply algebraicly?
In general if $R$ is a ring and $I$ is an ideal of $R,$ $R/I$ denotes the quotient ring of $R$ by $I$:
$R/I\ =\ \{r+I:r\in R\}$
where given a fixed $r\in R,$ $r+I$ denotes the additive coset $\{r+j:j\in I\}.$
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http://physics.stackexchange.com/questions/33063/does-the-expression-of-the-orbital-magnetic-dipole-moment-have-c?answertab=active
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# Does the expression of the orbital magnetic dipole moment have $c$?
The orbital magnetic dipole moment of a particle with mass $m$ and charge $q$ can be shown to be related to the orbital angular momentum through the equation
$$\displaystyle \boldsymbol\mu_L=\frac{q}{2m}\bf{L}.$$
One of the quantum mechanics textbooks has instead the same equation but with the speed of light in the denominator,
$$\displaystyle \boldsymbol\mu_L=\frac{q}{2mc}\bf{L}.$$
The same book also defined the spin magnetic dipole moment for the electron with a $c$,
$$\displaystyle \boldsymbol\mu_S=-g\frac{e}{2m_ec}\bf{S},$$
where $g$ is the usual Lande factor (~2).
Is the speed of light appearing in those equations just a typo? (but it is not in the book errata)
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There is a typo in first sentence. – user10001 Jul 28 '12 at 21:54
@dushya OOPS! Corrected, thanks a lot. – Revo Jul 28 '12 at 22:21
## 1 Answer
This is difference in the unit system. The former uses SI units, where the latter uses cgs/Gauss system.
In Gaussian units, unlike SI units, the electric field E and the magnetic field B have the same dimension. This amounts to a factor of c difference between how B is defined in the two unit systems, on top of the other differences.
(http://en.wikipedia.org/wiki/Gaussian_units)
This extra factor of $c$ extends to many magnetic quantities in Gaussian system.
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Bohr magneton has different expressions in different unit systems, one with c and one without, but not the magnetic dipole moment. The magnetic dipole moment by definition equals the current times the area enclosed by the current loop. If one starts from this definition and try to write it in terms of angular momentum, one will never get a factor of c! – Revo Jul 29 '12 at 23:05
– C.R. Jul 30 '12 at 0:47
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http://math.stackexchange.com/questions/30850/significance-of-matrix-vector-multiplication/30855
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# Significance of Matrix-Vector multiplication
Can someone give me an example illustrating physical significance of the matrix-vector multiplication?
1. Does multiplying a vector by matrix transforms it in some way?
2. Do left & right multiplication signify two different things?
3. Is matrix a scalar thing? (EDIT)
Thank you.
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## 3 Answers
The physical significance depends on the matrix. The primary point is that multiplication by a matrix (in the usual sense, matrix on the left) represents the action of a linear transformation. We'll work with one basis throughout which we'll use to represent our matrices and our vectors. Just note that because of linearity, if we have a vector $x = c_1e_1 + \ldots + c_ne_n$, and a linear transformation $L$, then $$\begin{eqnarray*} L(x) &=& L(c_1e_1 + \ldots + c_ne_n) \\ &=& L(c_1e_1) + \ldots + L(c_ne_n) \\ &=& c_1L(e_1) + \ldots + c_nL(e_n) \end{eqnarray*}.$$
This means that any linear transformation is uniquely determined by its effect on a basis. So to define one, we only need to define its effect on a basis. This is the matrix
$$\left(L(e_1) \ldots L(e_n)\right) = \left( \begin{array}{c} a_{11} & \ldots & a_{1n} \\ \ldots & \ldots & \ldots \\ a_{n1} & \ldots & a_{nn} \end{array} \right)$$
where $a_{ij}$ is the $i$'th compenent of $L(e_j)$.
Let's call this matrix $M_L$. We want to define multiplication of $M_L$ and some vector $x$ so that $M_L \cdot x = L(x)$. But there's only one way to do this. Because the $j$'th column of $M_L$ is just $L(e_j)$ and in light of our decomposition of the action of $L$ in terms of the $L(e_j)$, we can see that
$$M_L \cdot x = \left( \begin{array}{c} a_{11} & \ldots & a_{1n} \\ \ldots & \ldots & \ldots \\ a_{n1} & \ldots & a_{nn} \end{array} \right) \cdot \left( \begin{array}{c} c_1 \\ \ldots \\ c_n \end{array} \right)$$
must equal
$$c_1\left( \begin{array}{c} a_{11} \\ \ldots \\ a_{n1} \end{array} \right) + \ldots + c_n\left( \begin{array}{c} a_{1n} \\ \ldots \\ a_{nn} \end{array} \right) = \left( \begin{array}{c} c_1a_{11} + \ldots + c_na_{1n} \\ \ldots \\ c_1a_{n1} + \ldots + c_na_{nn} \end{array} \right)$$
which is the standard definition for a vector left-multiplied by a matrix.
EDIT: In response to the question "Is a matrix a scalar thing". Kind of but no.
If you consider the most basic linear equation in one variable, $y = mx$, where everything in sight is a scalar, then a matrix generalizes the role played by $m$ to higher dimensions and a vector generalizes the role played by $y$ and $x$ to higher dimensions. But matrices don't commute multiplicatively. So that's one big thing that's different. But they're strikingly similar in a lot of ways. We can define the function of matrices $f(A) = A^2$ and we can differentiate it with respect to $A$. When we do this in one variable with the map $f(x) = x^2$, we get the linear map $f_x'(h) = 2xh$ but when we do it with matrices, we get the linear map $f_A'(H) = AH + HA$. If matrices commuted, then that would just be $2AH$!
EDIT2:
My "add comment" button isn't working for some reason. The $e_j$'s are a basis, $e_1, \ldots, e_n$. I think the best thing to do would be to wait for your teacher to get around to it. I sometimes forget that people introduce matrices before vector spaces and linear transformations. It will all make much more sense then. The main point of a basis though is that it's a set of vectors so that every vector in the given space can be written as a unique linear combination of them.
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what do you mean by e's in the definition of vector of x? (Excuse me, I've an idea of vectors which're matrices of (nx1) or (1xn) dimension, So, I'm finding it difficult to get this representation of vectors) AND what do you mean by 'basis' (really I'm poor on this, so). – Amit L Apr 4 '11 at 9:11
hey, unfortunately, I've no teacher for this & I'm trying to do this on my own (the linear algebra, probability & statistics for machine learning, which too I'll try to self-learn). So would you suggest some material that will satisfy my need, provided I'm in the situation mentioned? – Amit L Apr 4 '11 at 9:38
– knucklebumpler Apr 4 '11 at 9:45
– knucklebumpler Apr 4 '11 at 9:50
thanks. How about that Gilbert Strang Book? MIT prof., I guess...? – Amit L Apr 4 '11 at 9:59
I believe that parts 2 and 3 of your question have been answered well. I'd like to take a stab at part 1, though the other answers to this part are probably better.
There's an interesting way of thinking of the application of a matrix to a vector using the Singular Value Decomposition (SVD) of a matrix. Let A be an $m \times n$ rectangular matrix. Then, the SVD of A is given by $A = U \Sigma V^T$, where $U$ is an $m \times m$ unitary matrix, $\Sigma$ is an $m \times n$ diagonal matrix of so-called singular values and $V$ is an $n \times n$ unitary matrix. For more on the SVD, check out the Wikipedia article: http://en.wikipedia.org/wiki/Singular_value_decomposition
That same article contains proof that every matrix has an SVD. Given that fact, we can now think of matrix vector multiplication in terms of the SVD. Let $\bf x$ be a vector of length $n$. We can write the matrix-vector multiplication as ${\bf b} = A {\bf x}$. But, $A{\bf x} = U\Sigma V^T {\bf x}$.
Since $V$ is unitary, $V {\bf x}$ does not change the magnitude of ${\bf x}$. Unitary matrices applied to a vector only change the direction of the vector (rotate it by some angle). The product $V^T {\bf x}$ rotates ${\bf x}$.
$\Sigma$ is a diagonal matrix. Its entries directly multiply the corresponding entries of the vector ${\bf x}$, thus scaling the vector (increasing its length) along the axis around which $V$ rotated ${\bf x}$. However, remember that the rotated vector $V^T{\bf x}$ is a vector of length $n$ while $\Sigma$ has dimensions $m \times n$. This means that $\Sigma$ is also embedding the vector $V^T{\bf x}$ in an $m$-dimensional space, i.e., changing the dimensions of the vector. If $m=3$ and $n=2$, for example, $\Sigma$ scales the 2D vector $V^T{\bf x}$ in 2 dimensions and then "places" it in a 3D space.
Finally, we have the product of the unitary matrix $U$ with the $m$-dimensional vector $\Sigma V^T{\bf x}$. $U$ rotates that vector in the $m$-dimensional space.
Every matrix thus potentially rotates, scales and embeds and then again rotates a vector when applied to a vector. When $m=n$, of course, a matrix-vector product doesn't involve any embedding- simply a rotation, scaling and another rotation. Like a little assembly line.
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I will interpret physical significance geometrically.
1. Does multiplying a vector by matrix transforms it in some way?
Yes. It performs a sequence of translations, rotations, and scalings. If you keep studying linear algebra you will learn how to write down the formulas for each of these operations given a matrix.
1. Do left & right multiplication signify two different things?
Yes: only one of these is defined. If you have an $m\times n$ matrix $M$ and a vector $V\in\mathbb{R}^n$ then only the multiplication $M V$ is defined.
1. Is matrix a scalar thing? (EDIT)
A matrix is most definitely not a scalar.
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then is matrix a vector of vectors, possibly? Like 3x2 matrix, a row vector of two column vectors (each of length 3) or a column vector of three row vectors (each of length 2) ? Help! – Amit L Apr 4 '11 at 9:03
Yes, you can think about it like this. There are a lot of equivalent ways to think about it :). – Glen Wheeler Apr 4 '11 at 12:47
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http://mathoverflow.net/questions/114824/how-to-solve-a-specific-multivariate-recurrence-relation-or-general-ones/114833
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## How to solve a specific multivariate recurrence relation (or general ones)
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How do you solve this recurrence (or multivariate recurrences in general)? Note that $p\in[0,1]$ and $n\in\mathbb{N}$ are given constants, where $np\leq 1$.
$$f:(\mathbb{N}\cup\{0\})\times(\mathbb{N}\cup\{0\})\rightarrow[0,1]$$ Base case: $f(0,b)=(1-np)^b$ $\forall$ $b\geq 0$.
$f(a,b)=f(a-1,b-1)[n-(a-1)]p+f(a,b-1)[1-(n-a)p]$, if $a\leq \min{(n,b)}$.
$f(a,b)=0$, otherwise.
Note: The recursive definition could just be applied for $a\leq b$. The function values for $n< a\leq b$ (if $n < b$) can be set to $0$ afterwards.
I have attempted using a multivariate generating function, but it doesn't yield a good simplification. Also looked into somehow manipulating the formula for multivariate taylor series, but to no avail.
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The constraint $a \leq \min(n,b)$ can be replaced by simply $a \leq b$ (values $f(a,b)$ for $a>n$ can be set to zero later). The reason is that if the first argument of $f(,)$ in the l.h.s. of the recurrent formula is $\leq n$, then so are the first arguments of $f(,)$ in the r.h.s. – Max Alekseyev Nov 29 at 5:05
What is $f(0,b)$ for $b>0$ ? – Max Alekseyev Nov 29 at 5:17
Oops, $f(0,b)$ for $b>0$ is now specified in the question. I suppose I will also modify the constraint to make working with it simpler. You're right, those values can be fixed to $0$ later. – unknown (google) Nov 29 at 6:36
## 2 Answers
I get $$f(a,b) = \frac{ (1-np)^b n!}{(n-a)!} \sum_{j=0}^b \binom{b}{j} \left\{ j \atop a \right\} \left(\frac{p}{1-np}\right)^j,$$ where $\left\{ j \atop a \right\}$ is a Stirling number of the second kind. I tested the formula in Mathematica against your recurrence for different values of $n$ and $p$, and they agree. Unfortunately I'm having trouble pasting the Mathematica input and output that verifies that agreement on this site without it turning into gobbledygook.
To get the formula I used Theorem 6 (which is actually due to Neuwirth) in my paper "On Solutions to a General Combinatorial Recurrence," Journal of Integer Sequences 14 (9): Article 11.9.7, 2011. The paper is about solution techniques for solving certain multivariate recurrences. Your recurrence happens to be in one of the forms for which the techniques work.
There might be a way to simplify the summation involving binomial coefficients and Stirling numbers, but I don't see it right now.
Added: The formula I used is the following.
Theorem. Suppose $R(n,k)$ satisfies the recurrence $$(\alpha(n-1) + \beta k + \gamma)R(n-1),k) + (\beta' + \gamma')R(n-1,k-1) + [n=k=0].$$ Then $$R(n,k) = \left(\prod_{i=1}^k (\beta' i + \gamma') \right) \sum_{i=0}^n \sum_{j=0}^n \left[ n \atop i \right] \binom{i}{j} \left\{ j \atop k \right\} \alpha^{n-i} \beta^{j-k} \gamma^{i-j}.$$ (Here, $0^0$ is taken to be $1$.)
For the OP's recurrence, we have $\alpha = 0, \beta = p, \gamma = 1-np, \beta' = -p,$ and $\gamma' = (n+1)p$.
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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.
Read Bender and Orszag (Advanced Mathematical Methods for Scientists and Engineers), and you will be enlightened.
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http://mathoverflow.net/questions/39485/are-there-motives-which-do-not-or-should-not-show-up-in-the-cohomology-of-any-s/39592
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Are there motives which do not, or should not, show up in the cohomology of any Shimura variety?
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Let $F$ be a real quadratic field and let $E/F$ be an elliptic curve with conductor 1 (i.e. with good reduction everywhere; these things can and do exist) (perhaps also I should assume E has no CM, even over F-bar, just to avoid some counterexamples to things I'll say later on). Let me assume that $E$ is modular. Then there will be some level 1 Hilbert modular form over $F$ corresponding to $E$. But my understanding is that the cohomology of $E$ will not show up in any of the "usual suspect" Shimura varieties associated to this situation (the level 1 Hilbert modular surface, or any Shimura curve [the reason it can't show up here is that a quaternion algebra ramified at precisely one infinite place must also ramify at one finite place]).
If you want a more concrete assertion, I am saying that the Tate module of $E$, or any twist of this, shouldn't show up as a subquotient of the etale cohomology of the Shimura varieties attached to $GL(2)$ or any of its inner forms over $F$ (my knowledge of the cohomology of Hilbert modular surfaces is poor though; I hope I have this right).
But here's the question. I have it in my head that someone once told me that $E$ (or perhaps more precisely the motive attached to $E$) should not show up in the cohomology of any Shimura variety. This is kind of interesting, because here is a programme for meromorphically continuing the L-function of an arbitrary smooth projective variety over a number field to the complex plane:
1) Observe that automorphic forms for GL_n have very well-behaved L-functions; prove that they extend to the whole complex plane. [standard stuff].
2) Prove the same for automorphic forms on any connected reductive algebraic group over a number field [i.e. prove Langlands functoriality]
3) Prove that the L-functions attached to the cohomology of Shimura varieties can be interpreted in terms of automorphic forms [i.e. prove conjectures of Langlands, known in many cases]
4) Prove that the cohomology of any algebraic variety at all (over a number field) shows up in the cohomology of a Shimura variety. [huge generalisation of Taniyama-Shimura-Weil modularity conjecture]
My understanding is that this programme, nice though it looks, is expected to fail because (4) is expected not to be true. And I believe I was once assured by an expert that the kind of variety for which problems might occur is the elliptic curve over $F$ mentioned above. At the time I did not understand the reasons given to me for why this should be the case, so of course now I can't reproduce them.
Have I got this right or have I got my wires crossed?
EDIT (more precisely, "addition"): Milne's comment below seems to indicate that I did misremember, and that in fact I was probably only told what Milne mentions below. So in fact I probably need to modify the question: the question I'd like to ask now is "is (4) a reasonable statement?".
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The strategy of Blasius-Rogawski to construct a motive for Hilbert modular forms (base change to unitary then transfer to U(3) then to an inner form) is indeed not known to succeed in this case (because of the shape of the L-packets). I don't know if this strategy and its close cousin are known to fail (way back, Blasius and Rogawski were actually "cautiously optimistic" that it should succeed by transfer to U(4)). But you surely knew this, as you also surely know that the meromorphic continuation of the L-function of E as in your introduction is known anyway. – Olivier Sep 21 2010 at 12:31
I was pretty sure that one couldn't attach a motive to the level 1 eigenform. I know nothing about L-packets or U(4). I know the meromorphic continuation is known for general E/F---this is because E is potentially modular. But even proving E is potentially modular doesn't realise it in the cohomology of a Shimura variety. In fact in the situation above I assumed E was modular, so analytic continuation will be known in this setting. I wanted to emphasize that it wasn't the modularity that was the problem I was interested in, it was the Shimura variety issues. – Kevin Buzzard Sep 21 2010 at 12:46
[clarification: of course in the setting above one can attach a motive because I started with $E$; I mean that in general given a level 1 eigenform I agree that it might be hard to attach a geometric object, when $F$ has even degree] – Kevin Buzzard Sep 21 2010 at 12:56
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@Kevin Buzzard: Thanks for this nice summary 1–4 of the Langlands programme! – norondion Sep 21 2010 at 12:59
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Blasius has pointed out that the naive generalization of the modularity conjecture fails --- there exist elliptic curves over number fields that are not quotients of the albanese of any Shimura variety --- but I don't know of any reason why the more general version (4) can't be true. (Blasius 2004 MR2058605). – JS Milne Sep 21 2010 at 13:09
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3 Answers
Here is an example of an elliptic curve $E$ over $Q(\sqrt{997})$ of conductor 1: $[ 0, w, 1, -24w - 289, -144w - 2334 ]$, where $w=(1+\sqrt{997})/2$ (thanks to Lassina Dembelle; this curve even has rank 2!). Shimura's paper "Construction of class fields and zeta functions of algebraic curves" suggests (according to MathSciNet) how to construct a Shimura variety of dimension 2 that isn't a curve but is associated to the relevant quaternion algebra. Shimura lets the quaternion algebra act on the product of two copies of the upper half plane instead of 1, and is able to show the relevant variety is defined over Q by using Siegel modular forms. Perhaps the cohomology of $E$ shows up there? I don't know.
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Hi William. This is a nice example to have here, but unfortunately I and probably many people don't know what the bracket notation means. I guess they're the coefficients of a Weierstrass equation, but in which order? – James Borger Sep 21 2010 at 23:23
@James Borger: A Weierstrass model $y^2 + a_1 xy + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6$ for an elliptic curve is written $[a_1, a_2, a_3, a_4, a_6]$. – Jamie Weigandt Sep 21 2010 at 23:59
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P.S.: I would upvote William, but he's currently at 389 reputation, and I have the feeling he wants to stay there. – Jamie Weigandt Sep 22 2010 at 0:00
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Hey William. In fact there are lots of conductor 1 examples in the literature. The earliest one I know about is $y^2+xy+e^2y=x^3$ over $\mathbf{Q}(\sqrt{29})$ discovered by Tate, with $e=(5+\sqrt{29})/2$; this is mentioned in Serre's 1972 Inventiones article on the image of Galois in the Tate module. Richard Pinch's thesis has a bunch in, IIRC, and one of these was proved to be modular by an explicit computation, by Socrates and Whitehouse, in 2004. – Kevin Buzzard Sep 22 2010 at 9:13
As for the Shimura variety you mention, from what you write I think you are talking about the Hilbert modular surface attached to the data: these are built over the complexes precisely by acting on two copies of the upper half plane rather than one and have canonical models over $\mathbf{Q}$. I believe that one can check that the cohomology of $E$ does not show up in the cohomology of these surfaces. The interesting cohomology of the surface will be in the middle dimension, and the weights in the middle dimension exclude the possibility that the curve can show up there. – Kevin Buzzard Sep 22 2010 at 9:15
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The question may be precised depending on what you call "show up". More precisely:
1. Concerning the cohomology of Shimura varieties there are two points of view: intersection cohomology or ordinary cohomology (this is of course the same for compact Shimura varieties).
2. Concerning the fact that the cohomology shows up in something we can add an option: it may show up potentially.
3. Then we can add another option: to proove it shows up in the Tannakian sub category of motives generated by the motives of Shimura varieties (even weaker (?): the class in the K_0 of motives of your variety is a virtual combination of the classes of motives showing up in Shimura varieties).
Let's first say we look at intersection cohomology. As stated your hope 4) is false for trivial reasons: the only Artin motives showing up in the intersection cohomology of Shimura varieties are the abelian one...In fact by purity they show up only in the H^0 that has been computed by Deligne and is abelian.
Of course if you put option (2) in my list this counterexample disappears.
Now you may say: yes but we can twist an Artin motive by a CM character and ask the same question. This is where I come to the following point: you're saying that because the twisting operation that is a particular case of Langlands functoriality is a known Langlands functoriality. Where I want to come is that in fact if you suppose Langlands functoriality known then the fact that your variety shows up in the Tannakian category generated by motives of Shimura varieties implies its L function is automorphic (tensor product functoriality).
If you suppose Langlands functoriality and your variety shows up potentially in the motive of a Shimura variety then its L-function is automorphic (existence of automorphic induction which implies for example Artin conjecture).
About the intersection cohomology of Shimura varieties: it is now pretty well understood and I think there is no reason why any variety would show up potentially in it. More precisely the Langlands parameters of automorphic representations showing up in the intersection cohomology of Shimura varieties factor through some representation $r_\mu:\;^L G_E\rightarrow GL_n$ where $G$ is the group attached to the Shimura variety (well to be more serious I woud have to invoke cohomological Arthur's parameter but it would take 5 hours to write this in details). Thus I clearly think the class of varieties that show up potentially in the cohomology of Shimura varieties has some serious restrictions...
Now there is another thing I did not speak about: the cohomology of non-compact Shimura varieties that may not be pure. For this little is known and it may be possible some interesting Galois representations that do not show up in the intersection cohomology of Shimura varieties show up in the cohomology...I know some people are looking at this (I won't give any name, even if I'm tortured) but as I said up to now little is known.
Well, I will stop here since this is an endless story and you can speak about this during hours...
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I agree with Laurent that the question should be precised before attempting any answer. One precision is, as he said, are we talking of the intersection cohomology (which is the same as the L^2 cohomology, and which will see the motives attached to the discrete automorphic representations of the group G defining the Shimura varieties), or of the ordinary cohomology (which sees all the cuspidal automorphic representations, some of the discrete non-cuspidal, and some others, no one knows exactly which)? – Joël Sep 22 2010 at 4:04
That said, I'm pretty confused. What motives are supposed to appear (directly or potentially) in the cohomology of Shimura varieties? If we assume the motive regular, that is with distinct Hodge numbers, shouldn't this has a simple answer? Take F=Q, for example. Shouln't any regular motive that it is a twist of its dual appear in a Shimura variety (orthogonal, or symlectic, or unitary after restriction to quadratic imaginary field)? What about the converse? Now, another queston (that might be stupid): isn't any motive over Q (and with coefficients in Q) dual to a Tate twist of itself? – Joël Sep 22 2010 at 4:22
As I mentioned to Kevin, you two should also ask Clozel about my comment, and let us know here if my memory (or my understanding at the time) is (or was) completely stupid. I'd quite like to know myself. – Minhyong Kim Sep 22 2010 at 5:32
@Laurent: I agree I should make it precise. The reason I didn't make it precise initially was that I was "fishing" for a precise statement that I couldn't quite remember, so it was to my advantage to be as vague as possible! I already found the answer to that in Milne's comment, so then I had to change the question a bit and I just figured I would leave it to see if anyone could formulate a precise negative result (e.g. "the etale cohomology of a Shimura variety always has this property, hence this Galois representation can never be a subquotient"). – Kevin Buzzard Sep 22 2010 at 9:24
The reason I didn't mention the "potential" issue was because of the following construction: if $L/K$ is a finite Galois extension then I can consider something like $G=Res_{L/K}(GL(1))$ (or even $GL(0)$ if Deligne's axioms allow it; I forget) and get Shimura varieties whose $H^0$ is abelian over $L$ but which still give the Artin Galois representation over $K$ that I want as a subquotient. In general you're abelian over the reflex field but you can control the reflex field! – Kevin Buzzard Sep 22 2010 at 9:27
show 4 more comments
Let me expand and hopefully clarify my first comment about the more specific question of whether the cohomology of a modular elliptic curve with everywhere good reduction shows up in the cohomology of a Shimura variety.
My (very limited) understanding is that the first thing to check is whether or not it shows up in the cohomology of a Picard modular surface. This is not known to happen, but I don't know if this is known to not happen. Then, one could try the following strategy: base-change to a quadratic field, base-change to $U(2)$, extension to $U(2)\times U(n)$, endoscopic transfer to $U(2+n)$ (now available, I think) and then switch to an inner form. Or in other words, one could try to look for a motive on a Picard modular variety.
As far as I understand, and this is not far at all, this strategy works when one is able to 1) construct motives on Picard varieties attached to some suitable Picard automorphic representations 2) Show that the series of operations described in the previous paragraph can yield a suitable automorphic representation. When $n=1$, we know that one can obtain a suitable representation when starting with a Hilbert modular form provided it has weight greater than 2 or a finite place at which it is Steinberg, so our case of interest is excluded. The main obstruction for 2) to work is that the Galois representations arising in the cohomology of Picard varieties will have dimensions determined by (roughly) the degree of freedom available for automorphic types at infinite places. We want this dimension to be 2, so there is a restriction there. Going to higher $n$ allows more flexibility to fiddle with these degree of freedom, so it might help.
All of this is shameful stealing from the articles of Blasius-Rogawski, which are highly recommended reading.
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## What do these new symbols mean...?I can't start this without knowing
Oh if f = 1, then P = 1.
Can I jsut state that or should I write up more on nCr?
Mentor That can't be right. P(x) doesn't depend on f(x) at all. Try again.
In the summation I get $$\sum_{k=0}^{n} \binom{n}{k} x^k (1 - x)^{n-k}$$
Oh wait, there are n - k + 1 terms...shoot I forgot how to do this. I am guessing that means the sum sums to 1.
Mentor
Quote by flyingpig So I must find $$P = \binom{n}{k} x^k (1 - x)^{n-k} = 1$$
Why? Where did you get this from?
Quote by vela Why? Where did you get this from?
Yeah scratch that I aws still thinking of multiplying and stuff. Forgot to get rid of it.
In $$\sum_{k=0}^{n} \binom{n}{k} x^k (1 - x)^{n-k}$$ It can only be 1 when k = 0, and n = 0... Is that what I Have to show? This is due tomorrow and I m still scratching my head
In the picture, f(x) was shown to be $$f(\frac{k}{n})$$ does this imply that k = n since f(x) = 1? If so then $$\frac{n!}{n!(0!)} x^n (1-x)^0 = x^n$$ Darn something is still off.
Mentor Do you understand sigma notation?
Quote by vela Do you understand sigma notation?
It means sum
EDIT: oh wait, are you referring when I said that n = k? Oh okay, so the sum is $$\sum_{n=0}^{n} \{whatever is here} = 1$$?
EDITING..not actually too sure of the property above. Looking through my old calc text
Mentor I'm asking you what is $$\sum_{k=0}^{n} \binom{n}{k} x^k (1 - x)^{n-k}$$ shorthand for? In other words, if you write the sum out, what do you get? I'm asking because you're making a bunch of guesses that make no sense if you understand what the notation means.
$$\binom{n}{k} x^k (1 - x)^{n-k} = P$$ $$\sum_{k=0}^{n} P$$ I am sorry for being so slow. wEDIT:writing out the sum...
I wrote the first three terms $$\sum_{k=0}^{n} \binom{n}{k} x^k (1 - x)^{n-k} = (1-x)^n + \frac{n!}{(n-1)!}x(1-x)^{n-1} + \frac{n!}{2!(n-2)!}x^2 (1-x)^{n-2}...$$
Mentor Ok, good. So you see how k isn't a variable you can really mess with, right? It takes on the values 0 to n just to generate the terms in the sum. In fact, when you expand the sum out, there is no k appearing anymore because it was just a dummy variable. So you can't do stuff like assume restrict k to just one value to try get the result you want. Similarly, you're asked to evaluate the sum for any value of n, so you can't set n to any one value. Now, are you familiar with the binomial theorem? That is, what is the expansion of (a+b)n?
$$(a + b)^n = \sum_{k=0}^{n}\binom{n}{k} a^{n-k} b^k$$ By observations I took x = b 1 - x = a So that a + b = 1 1^n = 1 forever, so that completes the problem!!! YES THANK YOU VELA, Kurt, and micro and all who helped *blows kisses*
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## Something I don't understand about String Theory
Hello, I have read a few books on the subject of String theory, including The Elegant Universe and Fabric of the Cosmos by Brian Greene (that guy is AWESOME!). Anyway, Mr. Greene may have answered my questions somewhere in his books, but the other day, I heard someone say that strings are tiny vibrating strands of energy. Correct me if I'm wrong, but isn't everything in the universe, including energy supposed to be made up of strings, according to this theory? So, if I'm right in saying that, there seems to be a paradox here. How can strings be pure energy if they make up energy?
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
I think that the statement is Everything is made of energy, including strings.'' Or, perhaps, Strings are made of energy, electrons are made of strings. Therefore electrons are made of energy.'' A==B, B==C, therefore A==C. Sorry if this wasn't the profound answer you were looking for :)
Hmmm...that makes sense. Although if it's true, doesn't it mean that our whole definition of energy is rendered obsolete since a string is composed of energy and nobody knows what makes up strings since they're fundamental??????????
## Something I don't understand about String Theory
I don't think so. Before you read Elegant Universe'' and knew anything about string theory, were you ok with the notion of electrons being made out of energy? If so, then nothing has changed :)
Hmmmmm.........yes that is a good point. But still, I always thought that strings made up fermions (matter particles) and bosons (force particles) and bosons were considered energy and a string was simply a fundamental particle that made up both of them depending on their vibration configurations. What is really going on here?
Well the different configurations'' of strings can make up bosons or fermions. And if bosons are energy, what about heat? Or sound. You can have heat energy or sound energy right?
Blog Entries: 19 Recognitions: Science Advisor According to string theory, strings are fundamental entities. By being fundamental, they are not made of something else (such as energy). Instead, they possess energy as one of its properties. More precisely, energy of a string is determined by its shape and velocity, which, of course, are easily conceivable properties of a string. In more conventional theories which assume that particles or fields (rather than strings) are fundamental entities, the same can be said for particles or fields. (Of course, particles do not have a shape.)
Blog Entries: 6 Recognitions: Gold Member It'ld be interesting to track how the concept of "made of energy" appears. Probably it comes from some interpretation of Einstein work. Before E=mc2, the concept of Energy was still a secondary object, near of its original definition as "ability to produce Work". In a comment to my post at Dorigo's blog, Kea has recalled some old say of Heisenberg (It should be nice to have the exact reference), about the concept of fundamental being meaningless in quantum field theory, as particles can disintegrate into others. On the other hand, a string can not disintegrate except into another modes of the same string.
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Quote by arivero It'ld be interesting to track how the concept of "made of energy" appears. Probably it comes from some interpretation of Einstein work. Before E=mc2, the concept of Energy was still a secondary object, near of its original definition as "ability to produce Work".
I would say that the Einstein work has not changed the concept of energy, but of mass. Now mass is also a kind of "ability to produce work".
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Quote by arivero On the other hand, a string can not disintegrate except into another modes of the same string.
That is not strictly true. Namely, a string can split into two or more strings. Even a classical string can do that:
http://xxx.lanl.gov/abs/hep-th/9502049
But, if you define a single string as ANY function of the form
$$X^{\mu}(\sigma,\tau)$$
including discontinuous functions as well, then a splitted string can also be thought of as a single string.
Like Demystifier also wrote: If you consider the general question of any distinguishable "structure" or "object", wether we call it string, point or anything else, energy is best throught of as a property of the object, and since usually any object is described in terms of it's properties, the object and it's properties are inseparable. To give a description of what an object is, independent of it's properties really makes no sense. The set of distinguishable properties IS the best description of the object we have. I personally think of energy as a measure of an objects relative significance or potential influence in a particular sense. An object with zero energy is insignificant, while a high energy object has the potential to have a much higher impact on your state of information. I think it may be possible to give a information theoretic meaning of energy as a measure of relational capacity that therefore bounds entropies. I have been fascinated by the speculative entropy bounds and it's relations to mass and area, and I think there is some deeper stuff in there yet to be uncovered. Any observer in my thinking must have en information equivalent of "energy" or storage capacity, which limits it's impact on the environment, and also provides intertia to resist impacts. An interesting question is the unification of memory capacity of an observer vs maximum information capacity of a physical object. There is seemingly interesting relations between storage capacity, information capacity limits of uncertainty as well as physical mass and energy. I don't think we understand all this quite yet. /Fredrik
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Quote by Demystifier That is not strictly true. Namely, a string can split into two or more strings. Even a classical string can do that: http://xxx.lanl.gov/abs/hep-th/9502049
Hey, thank very much! I am interested on results on string decay rates... the idea being to investigate if it could be possible to explain the equality of reduced decay widths of composite bosons and elementary ones (ej pion, J/Psi and Z0) by reinterpretation as open and closed versions of the same string.
Quote by Demystifier But, if you define a single string as ANY function of the form $$X^{\mu}(\sigma,\tau)$$ including discontinuous functions as well, then a splitted string can also be thought of as a single string.
Yes, that is a point. But most importantly, it implies that all the sheet pieces of the worldsheet are solution of the same dynamical equations, so also in this sense it can be say that the splitted strings are of the same kind that the original one.
Quote by Demystifier According to string theory, strings are fundamental entities. By being fundamental, they are not made of something else (such as energy). Instead, they possess energy as one of its properties. More precisely, energy of a string is determined by its shape and velocity, which, of course, are easily conceivable properties of a string. In more conventional theories which assume that particles or fields (rather than strings) are fundamental entities, the same can be said for particles or fields. (Of course, particles do not have a shape.)
I guess the proper statement is that, for example, a photon is an excitation of an electromagnetic field, and that excitation costs some energy to make. Liger is referring to the tiny vibrating bands of pure energy'' description that Brian Greene uses, I think.
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Quote by BenTheMan I guess the proper statement is that, for example, an electron is an excitation of an electromagnetic field, and that excitation costs some energy to make.
You mean photon (not electron) I guess.
Quote by Demystifier That is not strictly true. Namely, a string can split into two or more strings. Even a classical string can do that: http://xxx.lanl.gov/abs/hep-th/9502049 But, if you define a single string as ANY function of the form $$X^{\mu}(\sigma,\tau)$$ including discontinuous functions as well, then a splitted string can also be thought of as a single string.
Well, string theory (in contrast to string field theory) does not a priori include string splitting or other interactions; just as in relativistic quantum mechanics (sometimes called the "first quantized" framework), you must include the processes of splitting and other interactions with other strings "by hand". The string interaction diagrams are then smooth 2-dimensional surfaces (described by the functions mentioned above), one for each genus ("number of holes" in the surface). You wouldn't say there is one string in a decay just because it's described by a single surface, e.g., just as you wouldn't in a particle decay that is diagrammed as a single connected graph (though, there is only one type of fundamental string). Anyway, as has already been covered here, in string theory the idea is that strings (or the string field) are to be fundamental objects as elementary particles (or quantum fields) were considered.
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Quote by Demystifier According to string theory, strings are fundamental entities. By being fundamental, they are not made of something else (such as energy). Instead, they possess energy as one of its properties. More precisely, energy of a string is determined by its shape and velocity, which, of course, are easily conceivable properties of a string.
Demy, thanks for clearing that up. I can't imagine how Brian Greene or anyone else would say strings are "tiny vibrating strands of energy" except drunk at a cocktail party. But that leaves open the question
What is space made of?
Strings are fundamental, not made of anything, but they are in space. They can't vibrate unless space has some metric (geometric) properties. So what is geometry made of? Is space made of strings too?
What is mass, and in particular, what do we mean when we talk about the mass and spin of an elementary particle? Well, mass and spin simply label the irreducible representations of the Lorentz group under which the mathematical objects describing the particles transform. What is energy? Well, energy is defined to be the property that is conserved in a system when it is invariant under translations in time. What is momentum? It is the property that is conserved when a system is invariant under translations in space. The point is that on the most fundamental level, we can only think about these properties relative to a specific mathematical description of our universe. This is simply the best we can do. Of course we can talk about such properties in more qualitative or descriptive terms as people are trying to in this thread, but this is just imagery, which of course can be very useful, but it is still just imagery.
Quote by marcus Strings are fundamental, not made of anything, but they are in space. They can't vibrate unless space has some metric (geometric) properties. So what is geometry made of? Is space made of strings too?
Einstein said that the geometry of spacetime is just the large-structural property of the gravitational field. At very high energies, these structural properties as they’re described by the gravitational field equations of General Relativity break down and we need to understand gravity in a way that goes beyond classical physics. In lqg we have the idea of a spin-network and in string theory the quantum of the gravitational field is a string, as are all the quanta of the four known fundamental interactions according to string theory.
We also need to understand how high energy gravitational fields congeal at lower energies to once again yield solutions of the Einstein equations, i.e., a geometrical description of the gravitational field. String theory is a clear success in this regard, but the situation with lqg etc is not so clear, which is a bit odd since lqg provides a description of the gravitational field which even at high energies has a simple geometrical interpretation. On the other hand, whether or not nature requires a background-independent description of gravity is an open question. It ‘s difficult to see how such a formulation wouldn’t be required, but who knows? I don’t think there are many people who believe such a description will ever prove to be unnecessary.
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# Thread:
1. ## Simultaneous Equation [C1]
I've been going through a textbook and I've came across Simultaneous equations by this point. I can do many, such as:
c = 7d - 2 [1]
3c - 4d = 11 [2]
To do this I subbed c in [2] and did this method:
3(7d-2) - 4d = 11
21d - 4d - 6 = 11
17d = 17
d = 1
And then went back to 3c - 4d = 11
I found that 3c = 15
and c = 5.
However the last question is:
6x - 5y = 27
3y = 1 - 5x
Now I'm pretty sure you don't use the same method as last time so what can I do here
2. For example: from the second equation: $y=\frac{1-5x}3$ and then substitute this into the first one: $6x-\frac 53(1-5x)=27\leftrightarrow 18x-5(1-5x)=81\leftrightarrow 18x-5+25x=81\leftrightarrow 43x=86\leftrightarrow x=2$ substitute this into $y=\frac{1-5x}3$: $y=\frac {1-5\cdot 2}3=-3$.
3. rearrange so that the x and y on same side
6x - 5y = 27
3y + 5x = 1
then multiply top equation by 3(for y) and bottom equation by 5 so you get
18x - 15y = 81
15y + 15x = 5
now i think you will be able to solve it
4. i made a slight mistake, the bottom equation should say 15y + 25x = 5
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# Polar Equations
### From Math Images
Revision as of 15:30, 12 July 2011 by Chanj (Talk | contribs)
A polar rose (Rhodonea Curve)
This polar rose is created with the polar equation: $r = cos(\pi\theta)$.
A polar rose (Rhodonea Curve)
Fields: Algebra and Calculus
Created By: chanj
# Basic Description
Polar equations are used to create interesting curves, and in most cases they are periodic like sine waves. Other types of curves can also be created using polar equations besides roses, such as Archimedean spirals and limaçons. See the Polar Coordinates page for some background information.
# A More Mathematical Explanation
Note: understanding of this explanation requires: *calculus, trigonometry
[Click to view A More Mathematical Explanation]
## Rose
The general polar equations form to create a rose is UNIQ5c9d636c480d0bf4-math-00000001-Q [...]
[Click to hide A More Mathematical Explanation]
## Rose
The general polar equations form to create a rose is $r = a \sin(n \theta)$ or $r = a \cos(n \theta)$. Note that the difference between sine and cosine is $\sin(\theta) = \cos(\theta-\frac{\pi}{2})$, so choosing between sine and cosine affects where the curve starts and ends. $a$ represents the maxium value $r$ can be, i.e. the maximum radius of the rose. $n$ affects the number of petals on the graph:
• If $n$ is an odd integer, then there would be $n$ petals, and the curve repeats itself every $\pi$.
• If $n$ is an even integer, then there would be $2n$ petals, and the curve repeats itself every $2 \pi$.
• If $n$ is a rational fraction ($p/q$ where $p$ and $q$ are integers), then the curve repeats at the $\theta = \pi q k$, where $k = 1$ if $pq$ is odd, and $k = 2$ if $pq$ is even.
• If $n$ is irrational, then there are an infinite number of petals.
Below is an applet to graph polar roses:
If you can see this message, you do not have the Java software required to view the applet.
## Other Polar Curves
Archimedean Spirals
Archimedes' Spiral $r = a\theta$ The spiral can be used to square a circle and trisect an angle. Fermat's Spiral $r = \pm a\sqrt\theta$ This spiral's pattern can be seen in disc phyllotaxis.
Hyperbolic spiral$r = \frac{a}{\theta}$ It begins at an infinite distance from the pole, and winds faster as it approaches closer to the pole. Lituus $r^2 \theta = a^2$It is asymptotic at the $x$ axis as the distance increases from the pole.
Limaçon[1]
The word "limaçon" derives from the Latin word "limax," meaning snail. The general equation for a limaçon is $r = b + a\cos(\theta)$.
• If $b \geq 2a$, then the curve is convex.
• If $2a > b > a$, then it is dimpled.
• If $b = a$, then it becomes a cardioid.
• If $b = a/2$, then it is a trisectrix.
$r = \cos(\theta)$ $r = 0.5 + \cos(\theta)$ | Cardioid $r = 1 + \cos(\theta)$ $r = 2 + \cos(\theta)$
## Finding Derivatives
Consider the polar curve $r = f(\theta)$.
If we turn it into parametric equations, we would get:
• $x = r \cos(\theta) = f(\theta) \cos(\theta)$
• $y = r \sin(\theta) = f(\theta) \sin(\theta)$
Using the method of finding the derivative of parametric equations and the product rule, we would get:
$\frac{dy}{dx} = \frac{\frac{dr}{d\theta} \sin(\theta) + r \cos(\theta)}{\frac{dr}{d\theta} \cos(\theta) - r \sin(\theta)}$
## Finding Areas and Arc Lengths
To find the area of a sector of a circle, where $r$ is the radius, you would use $A = \frac{1}{2} r^2 \theta$.
Therefore, for $r = f(\theta)$, the formula for the area of a polar region is:
$A = \int_a^b\! \frac{1}{2} r^2 d\theta$
The formula to find the arc length for $r = f(\theta)$ and assuming $r$ is continuous is:
$L = \int_a^b\! \sqrt{r^2 + {\bigg(\frac{dr}{d\theta}\bigg)} ^2}$ $d\theta$
# Why It's Interesting
Polar coordinates are often used in navigation, such as aircrafts. They are also used to plot gravitational fields and point sources. Furthermore, polar patterns are seen in the directionality of microphones, which is the direction at which the microphone picks up sound. A well-known pattern is a cardioid.
## Possible Future work
• More details can be written about the different curves, maybe they can get their own pages.
• Applets can be made to draw these different curves, like the one on the page for roses.
# Teaching Materials
There are currently no teaching materials for this page. Add teaching materials.
# Related Links
### Additional Resources
Polar Coordinates
Cardioid
# References
Wolfram MathWorld: Rose, Limacon, Archimedean Spiral
Wikipedia: Polar Coordinate System, Archimedean Spiral, Fermat's Spiral
1. ↑ Weisstein, Eric W. (2011). http://mathworld.wolfram.com/Limacon.html. Wolfram:MathWorld.
2. ↑ 2.0 2.1 Stewert, James. (2009). Calculus Early Transcendentals. Ohio:Cengage Learning.
Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.
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http://mathhelpforum.com/discrete-math/193225-no-ways-seat-around-table-when-seats-numbered.html
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# Thread:
1. ## No. of ways to seat around a table when seats are numbered
Three single men, two single women and two families take their places at a round table. Each of the two families consists of two parents and one child. Find the number of possible seating arrangements if the seats are numbered and each child sits between their parents.
My answer: 7!2!2!=20160 but answer is 31680
2. ## Re: No. of ways to seat around a table when seats are numbered
Originally Posted by Punch
Three single men, two single women and two families take their places at a round table. Each of the two families consists of two parents and one child. Find the number of possible seating arrangements if the seats are numbered and each child sits between their parents.
My answer: 7!2!2!=20160 but answer is 31680
There are eleven seats. Place the youngest child at the table and his/her parents on either side in $\boxed{11\cdot 2}$ ways.
There are six places to seat the other child because its parents must be able to sit on each side: $\boxed{6\cdot 2}$ ways.
Then there are $\boxed{5!}$ to seat the others.
That product gives the correct answer.
3. ## Re: No. of ways to seat around a table when seats are numbered
@Plato: is it preferred that someone post a separate thread for each problem when all are related?
I feel like we might be helping you more if we find a way to explain how to attack these problems rather than illustrating each answer. Is this for a class? Are there any problems you have done where you did get the right answer by yourself?
4. ## Re: No. of ways to seat around a table when seats are numbered
Originally Posted by Plato
There are eleven seats. Place the youngest child at the table and his/her parents on either side in $\boxed{11\cdot 2}$ ways.
There are six places to seat the other child because its parents must be able to sit on each side: $\boxed{6\cdot 2}$ ways.
Then there are $\boxed{5!}$ to seat the others.
That product gives the correct answer.
Why is it not possible to treat a family as a cluster? Such that we treat a family as one so there are now 7 people. So 7! ways of arranging them. And the parents can rotate, since there are 2 families, 2!2!.
Therefore 7!2!2!. Although i understood your workings, I dont see how this is wrong
5. ## Re: No. of ways to seat around a table when seats are numbered
Originally Posted by Punch
Why is it not possible to treat a family as a cluster? Such that we treat a family as one so there are now 7 people. So 7! ways of arranging them. And the parents can rotate, since there are 2 families, 2!2!. Therefore 7!2!2!. Although i understood your workings, I dont see how this is wrong
There are several questions there.
1) I did treat the family as a cluster. But in this case the child must sit between the two parents that can be done in only two ways.
2) If this were a row of eleven seats then would be only nine places to seat the first child. Then there are a number cases for the second depending on the placement of the first.
3) But this is a circular arrangement with eleven seats. So it has eleven choices for the first child and six for second. Being circular makes the question easier that if it were linear.
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http://mathoverflow.net/questions/73246/why-is-mathbbr-infty-defined-the-way-it-is/73262
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## Why is $\mathbb{R}^{\infty}$ defined the way it is?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I've been thinking about Grassmannians recently. Think of $\mathbb{R}^k$ as a $k$-dimensional vector space. Let $\text{Gr}_n(\mathbb{R}^k)$ denote the Grassmannian of all $n$-dimensional vector subspaces of $\mathbb{R}^k.$ (This is a compact, Hausdorf topological manifold of dimension $n(k-n)$.) Let
$\Gamma^n(\mathbb{R}^k) := \{ (X,v) : X \in \text{Gr}_n(\mathbb{R}^k) \text{ and } v \in X \} .$
There's a standard idea of a vector bundle $\pi : \Gamma^n(\mathbb{R}^k) \twoheadrightarrow \text{Gr}_n(\mathbb{R}^k)$ given by $\pi(X,v) := X.$ This bundle has the nice property that lots of other bundles can be realised as sub-bundles of it. There is a more general definition, where we use $\mathbb{R}^{\infty}$ in place of $\mathbb{R}^k$. My question is about why we define $\mathbb{R}^{\infty}$ the way we do.
We define $\mathbb{R}^{\infty}$ as the set of infinite sequences $(x_1,x_2,x_3,\ldots)$ where each $x_i \in \mathbb{R}$ and only finitely many of the $x_i$ are non-zero. We identify $\mathbb{R}^k$ with the sequences of the form $(x_1,\ldots,x_k,0,0,\ldots),$ and then topologize $\mathbb{R}^{\infty}$ as the direct limit of the sequence $\mathbb{R}^1 \subset \mathbb{R}^2 \subset \mathbb{R}^3 \subset \ldots$ Then we get the universal bundle $\pi : \Gamma^n(\mathbb{R}^{\infty}) \twoheadrightarrow \text{Gr}_n(\mathbb{R}^{\infty}).$
My question is why do we insist that only finitely many of the $x_i$ are non-zero for each $(x_1,x_2,x_3,\ldots) \in \mathbb{R}^{\infty}$? I understand that it gives a countably infinite dimensional vector space, but that's a result of the definition; it doesn't explain why we define it the way we do. I suspect that it's related to the topology, but I don't really know.
Edit: The context is the OP is reading Milnor and Stasheff.
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Also it may be good to keep in mind, that this setup gives $$Gr_n(\mathbb{R}^\infty)= \bigcup Gr_n(\mathbb{R}^k)$$ as a directed union. So it is well approximated by finite dim objects. – Donu Arapura Aug 19 2011 at 22:25
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@Ryan I think the question is interesting for the beginning graduate student. It raises a lot interesting questions about topology and limits. Even though most of us have thought through these, I see no harm in leaving it open, unless the OP wants to accept the answers in the comments. – Scott Carter Aug 19 2011 at 22:33
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@Scott: I agree some beginning grad students might be interested in this question, but that's far from making it an appropriate MO question. – Ryan Budney Aug 19 2011 at 22:55
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"Why not spend more time trying to understand what people are asking?" I could turn this round and say "why not spend more time crafting a more localized question that is less open-ended?" Note that I did not vote to close your question. The fact that the answer you wanted is "It's a CW complex" is only obvious with hindsight. People work with spaces that are not CW complexes, so how is one meant to have known that this was the desired answer? I reiterate that while your original question may have been clear to you, it was not clear what kind of answer you were expecting or desiring. – Yemon Choi Aug 21 2011 at 4:33
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Belatedly: meta thread meta.mathoverflow.net/discussion/1119/… – Yemon Choi Aug 21 2011 at 4:50
show 19 more comments
## 5 Answers
This way it's a CW-complex, so studiable by standard tools of algebraic topology (pretty close to Donu's comment).
For a striking example of how these things can depend on the definition, look at the infinite unitary groups $U(\infty)$ vs. $U({\mathbb H})$, where the first is defined as $\bigcup U(n)$ and the second as the group of all unitary operators on Hilbert space. Then $U(\infty)$ has very interesting homotopy ($\pi_n(U(\infty)) = 0$ if $2|n$, $= {\mathbb Z}$ if not), whereas $U({\mathbb H})$ is contractible.
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I must admit that I'm not happy with the example you chose! $U(\infty)$ is not the unitary group of $\sum_{\infty} \mathbb{R}$. You can take a Hilbert space $\mathbb{H}$ and form a very nice closed subgroup of $U(\mathbb{H})$ which has the homotopy type of $U(\infty)$. What's more, this one is a smooth (infinite dimensional) manifold, unlike $U(\infty)$. There are lots of models for $BU(\infty)$, each with their own advantage, but I think that several of them are CW complexes. The advantage of this model that springs to mind is that it is fairly easy to show that (ctd) – Andrew Stacey Aug 20 2011 at 7:39
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Thank you very much Allen, that makes sense. And thanks to Donu too; I wasn't saying he was wrong, I was just expressing a confusion. I appreciate both of your efforts. I'm glad I managed to get an answer despite other people's efforts to shut me out. – Fly by Night Aug 20 2011 at 18:29
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Dear Allen: Is there a simple reason why $U(\mathbb{H})$ is contractible? – Akhil Mathew Aug 22 2011 at 12:00
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@Akhil: That is the content of Kuiper's theorem. – Ulrich Pennig Aug 22 2011 at 16:20
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Akhil: Kuiper's paper is well worth a read on that. One key step is that you can write a Hilbert space as a countable sum of copies of itself. Then if you consider a map in to the unitary group from a sphere, you can shift it so that it takes values in just one factor, then do the Mazur-swindle (is that it's name? I forget) which cancels what happens in that factor by something in the next, then that gets cancelled by something in the third, and so on. It really is a neat proof. – Andrew Stacey Aug 22 2011 at 18:03
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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.
There's been quite a bit of discussion of this thread in the meta thread. I'd like to take a stab at answering my interpretation of what the question is getting at. In a sense there's at least two things going on, and that's part of why I think there's been so much discussion.
To take a step back from Grassmannians, I'd like to mention why one might be interested in $\mathbb R^\infty$. The Whitney embedding theorem states that every continuous function $f : N \to \mathbb R^k$ can be approximated uniformly by a $C^\infty$-smooth embedding provided $k \geq 2n+1$ where the dimension of $N$ is $n$. Moreover, any two embeddings $N \to \mathbb R^k$ are isotopic provided $k \geq 2n + 2$. So if you wanted to, you could replace the class "$n$-manifolds up to diffeomorphism" with "isotopy classes of $n$-dimensional submanifolds of $\mathbb R^k$ provided $k \geq 2n+2$".
A key nice result about the weak topology on $\mathbb R^\infty$ is that any continuous function from a compact space to $\mathbb R^\infty$ has an image in $\mathbb R^k$ for some $k$. So from the perspective of the Whitney embedding theorem above, "$n$-dimensional manifolds up to diffeomorphism" is precisely "$n$-dimensional submanifolds of $\mathbb R^\infty$ up to isotopy". The key thing here is the ambient space is now independent of the dimension of the manifold you're talking about. This is pretty much exactly what's going on with the Grassmannians.
Given a vector bundle $p : E \to B$ over a finite-dimensional space $B$ (say a manifold or a CW-complex), there exists a classifying map for the bundle, meaning $p$ is isomorphic to the pull-back of the tautological bundle over $G_{n,k} \equiv G_k(\mathbb R^n)$. $n$ is just some sufficiently large integer. Although people don't state it this way, this theorem is basically the Whitney embedding theorem but for vector bundles. Because it's saying that up to isomorphism, $E$ is a collection of pairs $(b,v)$ where $b \in B$ and $v \in \chi^{-1}(h(b))$, where $\chi : E_{n,k} \to G_{n,k}$ is the tautological bundle over $G_{n,k}$. $h : B \to G_{n,k}$ the classifying map. In a sense we've "embedded" the vector bundle in Euclidean space, well, we've made the fibers as subspaces of Euclidean space.
But again, you have the "for some $n$ sufficiently large" thing. And like with manifolds $n$ has an upper bound in terms of the dimension of $B$ (if $B$ is finite dimensional). Since it's sometimes awkward to carry-around these "for sufficiently large $n$" statements, you take the limit space $G_{\infty,k}$ and now your statement is far more clean, because any $k$-dimensional vector bundle over any space $B$ is the pull back of some map $B \to G_{\infty,k}$. The point is that $G_{\infty,k}$ is a universal space -- independent of $B$ or the vector bundle over $B$. The weak topology is exactly what allows us to ensure this happens.
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Perhaps the following is so obvious that no one saw fit to mention it in a comment:
Let $X$ be a paracompact topological space. With the definition the way it is, homotopy classes of maps from $X$ to the $n$-Grassmannian are in bijection with isoclasses of rank $n$ vector bundles on $X$. (If you haven't got to them yet, this is thms 5.6 and 5.7). This is what really gets used in application to Stiefel-Whitney classes.
I don't know what is classified by the other definition of the $n$-Grassmannian you suggest ($n$ dimensional subspaces of the product of infinitely many copies of the reals). But certainly for your purposes (working through Milnor+Stasheff) this is the point.
The fact that the usual Grassmannian is a direct limit is used in several technical lemmas in chapter 5, but I have never tried pushing the construction you suggest through.
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That result only depends on the homotopy type of the Grassmannian, and all the different possibilities have the same homotopy type, so it doesn't address the actual question as to why Milnor and Stasheff use this particular model. – Andrew Stacey Aug 22 2011 at 18:04
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Andrew, I thought the question was pretty clear (and don't understand all the fuss): why define the Grassmannian of n-planes using the direct sum instead of direct product? The cleanest possible answer is: because using direct product doesn't classify vector bundles on paracompact spaces. Unfortunately, it's not clear to me that this is true, but it might be worth thinking about for a little while for someone who is interested. – SPG Aug 24 2011 at 12:55
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SPG I didn't (and still don't) see anything that implies that we are comparing the direct sum with the direct product. It seems more "Why the direct sum as opposed to ... err ... anything else?". There certainly are other things that work just as well. (I don't know, either, whether the direct product is one of them.) But that's one of the issues I have with this question: it is possible to read it in too many different ways and each has a subtly different answer. – Andrew Stacey Aug 26 2011 at 13:32
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From the question: "My question is why do we insist that only finitely many of the xi are non-zero for each (x1,x2,x3,…)∈R∞?" If we do not insist this, we get the direct product, do we not? This seems unambiguous to me. – SPG Aug 29 2011 at 14:40
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In fact, that's the only sentence in the OP's question with a question mark after it, so I'm even more puzzled by your assertion that you don't see anything that implies we are comparing the direct sum with the direct product. – SPG Aug 29 2011 at 14:43
show 4 more comments
Obviously tastes/opinions vary, but I think some ambiguous, or insufficiently localized, or not publishable-research-y enough, but nevertheless valuable to (a significant demographic of) research mathematicians... In fact, sometimes these questions are exactly the "dumb, non-research" questions that "everyone" (anyway, many people) have asked themselves... and not received a cogent answer.
One cliched-but-important (in my opinion) point is the "naive category-theory" explanation of why $\mathbb R^\infty$ is "defined" to be what it is. This does raise the entirely legitimate meta-meta-question of why we "have definitions", and "who is authorized to make them"... to which the easy answer (in my opinion, with some hindsight) is that, not merely must definitions capture the phenomena of important examples (or else the definitions are dumb), but, actually, as it seems to happen very often, re-ordering the "definition...theorem" sequence to "(mapping-)characterization..." rewrites the narrative so that the required/desired property is written in mildly category-theoretic terms, and the technical bit is perhaps proof-of-existence.
That is, a (typically, set-theoretic) "definition" is actually just _one_specific_construction_ of an object whose important features (if it exists at all) are completely determined by its interactions with other objects. That is, its characterization is "category theoretic" rather than "set theoretic".
(Yes, this is an advertisement for a certain little bit of category theory, though it is not a paid advertisement, insofar as I do not at all advocate formal category theory, nor would I advocate allocating one's personal resources to fretting over reconciliation of set theory and category theory... e.g., Grothendieck's "universes" and large cardinals? Fun, but likely not refering directly to one's original issue...)
So, rewriting the question about "why is the definition of $\mathbb R^\infty$ what it is?", we are required to ask what function this thing should have. Well, it is almost immediate that it should be the _ascending_union_ of the $\mathbb R^n$'s. That is, (upon reflection!) it is a (filtered) colimit (a.k.a. "inductive limit"). That is, it should/must/does have certain diagrammatic/mapping properties... as opposed to goofy set-theoretic constructional details.
Issues about infinite-dimensional Grassmannians... infinite-dimensional simplicial complexes... do share that basic feature, namely, that there is a mapping property (if only ascending-union sorts of (filtered) colimit properties) that are relevant.
Truly, the above viewpoint seems to me to be extraordinarily efficient/effective as explanatory device...
(And, one more time, questions that fail to be "documentable research" sometimes are far more interesting and useful to "us" than more focused ones... Of course, this is not a general rule...)
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If we don't insist that all but finitely many entries are non-zero, then we get the (set-theoretic) product of countably many copies of ${\mathbb R}$. If we look at the degree-n Grassmanian of this object, then this is not the object defined in Milnor Stasheff. Hence there exists at least one theorem that is true for the object they define, which is not true for the object we have just defined.
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I very rarely downvote. However, this answer is a role model for an unuseful answer. (Which may make it of some use, after all.) – Gil Kalai Aug 30 2011 at 9:00
Yemon, your answer may be intended rhetorically but is it evidently correct? Could it be that the Grassmannians of $n$-dimensional subspaces of the sum and the product, with their tautological vector bundle, are isomorphic in the category of spaces with vector bundles? – Tim Perutz Aug 30 2011 at 16:30
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http://math.stackexchange.com/questions/35264/congruences-for-fermat-quotients
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# Congruences for Fermat Quotients
If $p$ is a prime number and $a$ is relatively prime to $p$, then by Fermat's Little Theorem, the Fermat quotient $q_p(a) = (a^{p-1}-1)/p$ is an integer. A well-known collection of theorems beginning with the work of Wieferich shows that if the first case of Fermat's Last Theorem holds for $p$, then $q_p(a)$ is divisible by $p$ for every prime number $a \leq 89$.
I have stumbled across some congruences between Fermat quotients, and haven't turned up similar ones in a Google search. I hope someone here is an expert on these and can tell me 1) if these and/or similar congruences are known, and 2)where I can find other proofs than my own, either simple enough to post here or written up somewhere.
Here are some examples: If $p = 2^a-1$ is a Mersenne prime, so $a$ is prime, then $q_p(2) \equiv 2q_a(2) \pmod{p}$. In particular, since $q_a(2) < p$, a Mersenne prime is not a Wieferich prime (i.e., $q_p(2)$ is not divisible by $p$), a well-known result but not by this proof, I think.
If $p = 2^n-3$ is prime for some $n$, then $3nq_p(2) + 1 \equiv 3q_p(3) \pmod{p}$, so in particular by the result mentioned above, the first case of FLT must hold for all such $p$.
If $p = 3^n-4$ is prime for some $n$, then $4nq_p(3) + 1 \equiv 8 q_p(2) \pmod{p}$, so again, the first case of FLT must hold for all such $p$.
I have proved a few others of this nature and seem to have a way to generate more if I wish.
Thanks in advance for any information that you can provide.
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## 1 Answer
I absolutely am not an expert, but I did informally study Fermat quotients a few years ago, and I don't recall seeing this sort of thing in the literature, so I will make a mild guess that these are not known. Unfortunately I don't expect this approach will help much with FLT because e.g. primes of the form $2^n-3$ are probably extremely sparse (possibly there are only finitely many; I'm not sure how/whether the heuristics for Mersenne primes would carry over).
If you are interested, there is another class of quotients whose indivisibility by $p$ implies the first case of FLT for $p$; I don't recall if they have an established name, but one might call them Lucas quotients. For $m\equiv 1\bmod 4$ and a particular Lucas sequence $L$ depending on $m$, one has the congruence $$L_{p-\left(\frac{m}{p}\right)}\equiv 0\bmod p$$ (note that $L_0=0$; there is an obvious analogy with $a^{p-1}\equiv a^0\bmod p$), and the first case of FLT holds for $p$ if $p$ does not divide the quantity $$\frac{L_{p-\left(\frac{m}{p}\right)}}{p}$$ The only reference I'm aware of for this fact is this paper by Granville (p.13, equation 6.9).
In the particular instance of $m=5$, the Lucas sequence is just the Fibonacci numbers, and the above criterion is called the Wall-Sun-Sun criterion. There has been more in-depth investigation of this case; no Wall-Sun-Sun primes (i.e. $p^2\mid F_{p-\left(\frac{p}{5}\right)}$) are currently known.
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Thank you Zev. I'll wait to see if anyone else chimes in. If not, then I'll pick your answer and I guess repost at MO. I wasn't aware of the Wall-Sun-Sun criterion and the paper by Granville, so I'll have some fun reading if I find the time. Thanks again! – Barry Smith Apr 27 '11 at 19:56
@Barry: No problem, I just wanted to chime in since I happened to have some (small) amount of interaction with them before. Hope you have better luck on MO! – Zev Chonoles♦ Apr 28 '11 at 1:16
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http://math.stackexchange.com/questions/193939/error-bound-of-a-integration
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# error bound of a integration
I have a function, which is $f(x) = e^{1/x}$. I want to calculate the error bound for the `Trapezoid Rule`, which formula is:
$$|E|\leq K\frac{(a-b)^3}{12\cdot n^2}$$
where $|f''(x)|\leq K$. What's the value of $K$ for the above function? If I calculated $f''(x)$ correctly, it should be: $$f"(x) = \frac{e^{1/x}}{x^3}\left(\frac 1x + 2\right).$$
Please correct me if I'm wrong. The boundaries are $[1, 2]$.
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Can you write down $|f''(x)|$ explicitly first? – Alex Sep 11 '12 at 3:37
@Sam Done! I've updated my question – philippe Sep 11 '12 at 3:39
Alright, now you have it explicitly. You want an upper bound on it's absolute value. What happens to the second derivative when $x\rightarrow 0$ and $x\rightarrow\pm\infty$? – Alex Sep 11 '12 at 3:40
it goes to infinite when `x -> 0`, and goes to 0 when `x-> infinite` – philippe Sep 11 '12 at 3:42
This is the time to consider what interval you are integrating the function over. Does it include 0? Notice btw that $e^{1/x}$ has no limit as $x\rightarrow 0$ since from the positive side, it's infinity but from the negative side it's 0 – Alex Sep 11 '12 at 3:43
show 3 more comments
## 1 Answer
If you look at $f''$ you should be able to convince yourself that it is decreasing over $[1,2]$. In that case, the maximum value is $f''(1)$, which is then $K$
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http://mathoverflow.net/questions/85090/coloring-mathbbzk-and-a-fixed-point-theorem
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## Coloring $\mathbb{Z}^k$ and a fixed point theorem
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This is potentially another approach to this question. I put it as an update there, but perhaps it would be better to post it separately. If we color $\mathbb{Z}^k$ with the $\ell_\infty$ metric in fewer than $k+1$ colors, then there are always arbitrary long monochromatic paths. This follows from the classical results about the HEX game. It is known that this result is "equivalent" (both results can be easily deduced from each other) to the Brouwer fixed point theorem (see also a discussion here ). Now consider the $\mathbb{Z}^k$ with the $\ell_1$-metric and a similar question as before: can we color it in fewer than $k+1$ colors, so that there are no arbitrary long monochromatic $3$-paths (i.e. we are allowed to jump by 1 or by 2 or by 3 in the $\ell_1$-metric).
Question. Is this statement equivalent (in the above sense) to a fixed point theorem.
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http://mathhelpforum.com/number-theory/186551-prove-equation-solvable-print.html
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# Prove that this equation is solvable
Printable View
• August 22nd 2011, 01:23 PM
alexmahone
Prove that this equation is solvable
Given any integers a, b, c and any prime p not a divisor of ab, prove that $ax^2+by^2\equiv c\ (mod\ p)$ is solvable.
• August 24th 2011, 01:04 AM
Opalg
Re: Prove that this equation is solvable
Quote:
Originally Posted by alexmahone
Given any integers a, b, c and any prime p not a divisor of ab, prove that $ax^2+by^2\equiv c\ (mod\ p)$ is solvable.
I can only prove this by quoting a theorem from a recent research paper. However, the theorem is not that hard. Essentially the same result seems to have been proved a long time ago, in this 1893 article by J C Fields (after whom the Fields Medal is named).
If c is a multiple of p then we can take x = y = 0. So we can assume that $c\in\mathbb{Z}_p^*$ (the multiplicative group of nonzero residues mod p). Also, the result is easy to prove in the case p=2. So assume that p is an odd prime.
Let R be the set of quadratic residues mod p, and let N be the set of quadratic non-residues. According to Theorem 2.1 in this paper (pdf file), every element of $\mathbb{Z}_p^*$ belongs to each of the three sets R+R, R+N and N+N. But R is a subgroup of $\mathbb{Z}_p^*$. Its cosets are R and N. The sets $\{ax^2:x\in\mathbb{Z}_p^*\}$ and $\{by^2:y\in\mathbb{Z}_p^*\}$ are also cosets of R. Thus the set $\{ax^2 +by^2:x,y\in\mathbb{Z}_p^*\}$ is equal to one of the sets R+R, R+N or N+N, and therefore contains c.
All times are GMT -8. The time now is 10:45 AM.
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http://math.stackexchange.com/questions/148332/non-isomorphic-atomless-boolean-algebras
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Non-isomorphic atomless Boolean algebras
All countable atomless algebras are isomorphic. Can one give an example of a pair of mutually non-isomorphic atomless Boolean algebras of cardinaliy continuum?
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I'm glad that you were happy with my answer but I suggest that you wait a little bit before accepting it (maybe a day or two) because others might have a lot more to say than I do and accepting an answer generally discourages others to look at questions and/or write up answers. – t.b. May 22 '12 at 19:21
Well, at least I can verify ccc condition in the algebras you mentioned, so this is fairly good example! – MarkNeuer May 22 '12 at 19:56
4 Answers
Take $\mathfrak{A}$ to be the Lebesgue measure algebra and $\mathfrak{B} = P\mathbb{N}/[\mathbb{N}]^{\lt \omega}$, the quotient algebra of $P\mathbb{N}$ modulo the ideal of finite sets. Then both are atomless and have cardinality continuum but they are not isomorphic because $\mathfrak{A}$ is ccc while $\mathfrak{B}$ isn't.
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Did you mean the Borel measure algebra? The Lebesgue one is strictly larger than $\mathcal P(\mathbb N)/\mathrm{Fin}$. – Asaf Karagila Dec 29 '12 at 23:12
For me the measure algebra of a measure space is the algebra of measurable sets modulo the ideal $\mathcal{N}$ of null sets. Thus, $\mathfrak{A} = \mathfrak{L}/\mathcal{N} \cong \mathfrak{B} / (\mathfrak{B} \cap \mathcal{N})$ has cardinality $\mathfrak{c}$. Otherwise ccc wouldn't tell $\mathfrak{A}$ and $\mathfrak{B}$ apart, would it? :) – t.b. Feb 3 at 9:19
Hmm, that's the second time someone mentions this to me recently. – Asaf Karagila Feb 3 at 9:58
I suppose it's a matter of background and culture. I've seen reduced measure algebra used for the same thing (I think Givant/Halmos use that). However, it doesn't make much sense to me to use the term measure algebra as a synonym of $\sigma$-algebra and from this point of view $\Sigma/\mathcal{N}$ is pretty much the only possible interpretation. It seems that for analysts $\Sigma/\mathcal{N}$ is the interesting thing to consider, not $\Sigma$ itself. – t.b. Feb 3 at 10:12
Let $\langle X,\tau\rangle$ be a topological space. A set $U\in\tau$ is a regular open set iff $\operatorname{int}_X\,\operatorname{cl}_XU=U$. Let $\mathrm{RO}(X)$ be the family of regular open subsets of $X$. For $U,V\in\mathrm{RO}(X)$ define
$$\begin{align*} &U\land V=U\cap V,\\ &U\lor V=\operatorname{int}_X\,\operatorname{cl}_X(U\cup V),\\ &\lnot U=X\setminus\operatorname{cl}_XU,\text{ and}\\ &U\le V\text{ iff }U\subseteq V\;; \end{align*}$$
it’s well-known that this makes $\mathrm{RO}(X)$ a complete Boolean algebra. Clearly this algebra is atomless if $X$ has no isolated points. In particular, $\mathrm{RO}(\Bbb R)$ is atomless. Since $\Bbb R$ is second countable, it’s clear that $|\mathrm{RO}(\Bbb R)|\le 2^\omega$. On the other hand, for any $A\subseteq\Bbb Z$ the set $$\bigcup_{n\in A}\left(n-\frac14,n+\frac14\right)$$ is regular open, and there are clearly $2^\omega$ such sets, so $|\mathrm{RO}(\Bbb R)|=2^\omega$. In short, $\mathrm{RO}(\Bbb R)$ is a complete, atomless Boolean algebra of power $2^\omega$.
Now let $\mathscr{B}=\wp(\omega)/[\omega]^{<\omega}$, the quotient of the power set algebra of $\omega$ by the ideal of finite subsets of $\omega$. Since $|\wp(\omega)|=2^\omega$, and $\left|[\omega]^{<\omega}\right|=\omega$, it’s clear that $|\mathscr{B}|=2^\omega$. It’s also clear that $\mathscr{B}$ is atomless. However, $\mathscr{B}$ is not complete, so it most be distinct from $\mathrm{RO}(\Bbb R)$.
To see that $\mathscr{B}$ is not complete, let $\{A_n:n\in\omega\}$ be a partition of $\omega$ into infinite subsets. For each $A\subseteq\omega$ denote by $\widehat A$ its equivalence class in $\mathscr{B}$. If $\widehat S$ is any upper bound in $\mathscr{B}$ for $\{\widehat{A_n}:n\in\omega\}$, $|A_n\setminus S|<\omega$ for each $n\in\omega$, so for each $n\in\omega$ we may choose $s_n\in S\cap A_n$. Let $T=\{s_n:n\in\omega\}$, and let $S\,'=S\setminus T$; clearly $|A_n\setminus S\,'|<\omega$ for each $n\in\omega$, so $\widehat{S\,'}$ is an upper bound for $\{\widehat{A_n}:n\in\omega\}$, but it’s also clear that $\widehat{S\,'}<_{\mathscr{B}}\widehat{S}$.
(One could also note that $\mathrm{RO}(\Bbb R)$ is ccc, while $\mathscr{B}$ isn’t, but t.b. already used that idea, so I thought that I’d do something different.)
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Let $B_0$ be the free Boolean algebra generated by $2^{\aleph_0}$ generators. Observe that it has $2^{\aleph_0}$ elements and by construction descending chains of length $\omega$. On the other hand take the language of boolean algebras and add $2^{\aleph_0}$ many constants, $(c_\alpha)_{\alpha<2^{\aleph_0}}$. Take the theory that contains the theory of atomless Boolean algebras plus the sentences $c_\alpha<c_\beta$ for all $\alpha>\beta$. This is certainly finitely satisfiable by the countable atomless Boolean algebra. Hence by compactness it is satisfiable and it has a model $B_1$. In fact by Lowenheim-Skolem theorem you can make $B_1$ of size $2^{\aleph_0}$.
Now $B_0$ and $B_1$ are not isomorphic since one has descending sequences of size $\omega$ only while the other contains a descending sequence of size $2^{\aleph_0}$.
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I'll try:
If I'm not mistaken, the cardinality of the completion of the countably infinite atomless Boolean algebra is that of the continuum.
For the second example, take the Boolean algebra freely generated by the set of all reals.
But I'm not altogether sure those are not isomorphic, so this answer is not fully complete.
Later edit: Apostolos has now posted my second example plus another, which he shows is not isomorphic to it.
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http://mathoverflow.net/questions/11306/character-table-does-not-determine-group-vs-tannaka-duality/122367
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## Character table does not determine group Vs Tannaka duality
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From the example $D_4$, $Q$, we see that the character table of a group doesn't determine the group up to isomorphism. On the other hand, Tannaka duality says that a group $G$ is determined by its representation ring $R(G)$.
What is the additional information contained in $R(G)$ as opposed to the character table?
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You might be interested in this related question: mathoverflow.net/questions/500/… – Steven Sam Jan 10 2010 at 10:31
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The representation ring is determined by the character table, since direct sum of representations is pointwise sum of characters and tensor product of representations is pointwise multiplication of characters. – Qiaochu Yuan Jan 10 2010 at 11:21
## 7 Answers
If I'm not mistaken, the extra information is not contained in the representation ring, you have to look at the category of representations. In particular, you want to look at the representation category equipped with its forgetful functor to vector spaces. Then the group can be recovered as the automorphisms of this functor. Here's a blogpost I wrote which may be helpful: http://concretenonsense.wordpress.com/2009/05/16/tannaka%E2%80%93krein-duality/
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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.
Let me try and give as low-level an explanation as I can, in case you're scared of all this "monoidal category" stuff.
A complex representation of a finite group $G$ is just a module for the group ring $R=\mathbf{C}[G]$. The two groups you mention with the same character table have isomorphic group rings as well (by some abstract structure theorem for simple $R$-modules), so, if you fix an isomorphism between the group rings you have a way of moving from a representation of one to a representation of the other! Let's call this way of moving between representations of the groups "the trick". The representation theory of both groups is very close because we can use the trick.
However the way to spot the difference in the representation theory of the two groups is to use the tensor product. If $M$ and $N$ are representations of $G$, then one can give $M\otimes_\mathbf{C} N$ the structure of a representation of $G$, but this construction must use "more than the abstract $R$-module structure of $M$ and $N$". What I mean by that is this: if $R$ is a random non-commutative $\mathbf{C}$-algebra and $M$ and $N$ are two $R$-modules, there's no natural way (that I know of) to give $M\otimes_\mathbf{C}N$ an $R$-module structure. The way this works with $R=\mathbf{C}[G]$ is that you let $\mathbf{C}$ act in the obvious way (i.e. on one of the factors) and let $G$ act diagonally (i.e. on both of the factors at once). This is a big asymmetry and we're using the fact that $R=\mathbf{C}[G]$ (i.e. that $G$ gives a basis of $R$) rather than just its abstract ring structure.
Because the groups you mention have the same character tables, it's true that if $M$ and $N$ are modules for one, then using the trick they're modules for the other. But the shocker is this: the trick commutes with tensor products! (at least up to isomorphism). Because the character of the tensor product is the product of the characters! So applying the trick to $M\otimes N$ gives a module that happens to be isomorphic to the tensor product of the tricks applied to $M$ and $N$. So now the representation theory of the groups is looking really close.
But the trick is not canonical, it involves a random choice of isomorphism between the two representation rings, and so when you actually start drawing bunches of natural isomorphisms that Tannakian categories are supposed to satisfy, they don't commute with the trick, because the "(at least up to isomorphism)" remark comes back to haunt you. This is my understanding of the explicit point where the subtlety is buried. I should say however that I figured all this out myself once and I might be wrong. If I'm wrong I will no doubt shortly be put right!
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Thanks to both of you! – norondion Jan 10 2010 at 11:26
It might be worthwhile to supplement Kevin Buzzard's answer by describing exactly what goes wrong in the case of $Q$ and $D_4$. Let $V$ and $W$ be the two dimensional irreducible representations of $Q$ and $D_4$ respectively.
The representation $V \otimes V$ is isomorphic to `$1 \oplus \chi_1 \oplus \chi_2 \oplus \chi_3$`, where the $\chi_i$ are the nontrivial (one-dimensional) characters of $Q$. Similarly, $W \otimes W$ is isomorphic to `$1 \oplus \eta_1 \oplus \eta_2 \oplus \eta_3$` where the $\eta_i$ are the nontrivial (one-dimensional) characters of $D_4$. The reason these formulas look the same is that the representation rings of $Q$ and $D_4$ are isomorphic.
However, we cannot build this isomorphism out of maps of vector spaces which commute with tensor product. Specifically, we cannot find an isomorphism $\alpha : V \to W$ of vector spaces so that the induced map $\alpha \otimes \alpha : V \otimes V \to W \otimes W$ carries $1$ to $1$ and $\chi_i$ to $\eta_i$. (Of course, I haven't told you how to label the $\chi_i$ and $\eta_i$. My statement is that there is no choice of labeling for which you can do this.)
This is very easy to see. The map $\alpha \otimes \alpha$ will carry the anti-symmetric elements of $V \otimes V$ to the anti-symmetric elements of $W \otimes W$. But $\wedge^2 V$ is the trivial representation, and $\wedge^2 W$ is not! So isomorphisms of the form $\alpha \otimes \alpha$ can't carry $1$ to $1$.
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Interesting. Out of curiosity, is the structure of R(G) as a lambda-ring enough to distinguish finite groups? – Qiaochu Yuan Jan 10 2010 at 16:38
I was wondering that myself! I don't know. – David Speyer Jan 10 2010 at 16:54
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Thanks David. I knew from the Tannakian formalism that "something like this had to be true", but seeing the explicit example is really great. I feel like I should steal from your answer and Theo's answer to make my answer "even better" with edits, but on the other hand I could just save myself a job and do nothing because your answers won't go away. I think this is great: it seems to me that we now have completely concrete illustrations of how the categories are similar and how they differ in this case. – Kevin Buzzard Jan 10 2010 at 20:58
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Qiaochu's question about Lambda rings is answered below by Mariano; the answer is no. – David Speyer Jan 10 2010 at 21:46
The Lambda ring structure is enough to show that R(D_4) is different from R(Q) via Adams operations built from the Lambda ring structure. – Paul Pearson Feb 20 at 2:02
Regarding a question which arose in comments to another answer: The lambda ring structure is on $RG$ is not enough to reconstruct the group. Dade has given examples (MathSciNet review here; paper here) of pairs of groups which have the same character table with the same power maps, and from this it follows that the whole lambda ring structure is the same.
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Nice! I don't think I would have found that. I've added the link for you. – David Speyer Jan 10 2010 at 21:44
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Here's a link to the paper: dx.doi.org/10.1016/0021-8693(64)90002-X – Steven Sam Jan 10 2010 at 22:10
Thanks Steven! I've edited that in the main answer. – David Speyer Jan 11 2010 at 16:46
Edit: Somehow I totally misread the question. I talked about the group algebra $\mathbb C[G]$, which is not at all the same as the character ring $R(G)$. Over $\mathbb C$ (or any other field of characteristic 0), $R(G)$ is naturally a subalgebra of $(\mathbb C[G])^*$, which is the algebra of functions on $G$ with pointwise multiplication, and now the comultiplication encodes the group structure. On the other hand, it is not a subbialgebra: the coproduct of a class function is not a class function.
Anyway, original post below, with the obviously wrong things struck out. So it's really an answer to Kevin, rather than anything else.
Well, it depends on what you mean by "$R(G)$". I won't address TK duality, and most of what I'll say is essentially a follow-up to Kevin's answer, rather than an answer in its own right. Also, I'm only going to address finite groups and their finite-dimensional representations. Also, for me the word "ring" means (associative, unital, noncommutative) "$\mathbb C$-algebra".
Recall that a complex representation of $G$ is the same as an algebra representation of $\mathbb C[G]$. Let $R$ be a ring. As Kevin says, it's in general impossible to define an $R$-module structure on $M\otimes N$ when $M,N$ are $R$-modules. (When $R$ is abelian, which is not the case here, one can define a tensor product $M \otimes_R N$, but that's not the tensor product of representations anyway.) What would a tensor product of modules require? It would require a rule that assigns to each $r\in R$ and each pair $M,N$ of $R$-modules an endomorphism of $M\otimes N$, of course, and we should impose all sorts of axioms that force the tensor product to be well-behaved. Among other things, it's much easier if the endomorphism is an element of the tensor product $\text{End}(M) \otimes \text{End}(N) \subseteq \text{End}(M\otimes N)$. And we already have some distinguished elements of $\text{End}(M)$ and $\text{End}(N)$, namely the action of $R$.
So one way to try to construct a well-behaved tensor product on the category of $R$-modules is to find a nice map $\Delta: R \to R\otimes R$. Then the axioms for this map that assure that the tensor product is good are that $\Delta$ be an algebra homomorphism, and that it be "coassociative": $(\text{id}\otimes \Delta)\circ \Delta = (\Delta \otimes \text{id})\circ \Delta)$. Let's suppose that there's also a distinguished "trivial" representation $\epsilon: R \to \text{End}(\mathbb C) = \mathbb C$; if this is to be the monoidal unit, then we'd need $(\text{id}\otimes \epsilon) \circ \Delta = \text{id} = (\epsilon \otimes \text{id})\circ \Delta$. The maps $\Delta, \epsilon$ satisfying these axioms define on $R$ the structure of a bialgebra.
By the way, the map is called "$\Delta$" because if $G$ is a group (or monoid) and $R = \mathbb C[G]$, then the map $R \to R\otimes R$ given on the basis $G$ by the diagonal map $\Delta: g \mapsto g\otimes g$ is such a structure.
Then here's a cool fact. Define an element $r\in R$ to be grouplike if $\Delta(r) = r\otimes r$. Then the grouplike elements are a multiplicative submonoid of $R$. And when $R = C[G]$, the grouplike elements are precisely $G$.
So my answer to your question is that "the additional information contained in $R(G)$ as opposed to the character table" is its bialgebra structure.
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Thanks Theo, that's a great comment. You've elucidated a point which I only had a much vaguer appreciation of when I wrote my answer. – Kevin Buzzard Jan 10 2010 at 20:54
The representation ring $R(G)$ from the question is not the group algebra, as far as I can see, and it is the latter which is a bialgebra. – Mariano Suárez-Alvarez Jan 11 2010 at 6:44
oops! I wasn't thinking. – Theo Johnson-Freyd Jan 11 2010 at 16:23
An important point to make here is that the tensor product structure on R(G) comes from the comultiplication on C[G], which means you then have to ask yourself: where did the multiplication go? – Qiaochu Yuan Jan 11 2010 at 16:42
It might be worth explaining why you shouldn't expect $R(G)$ to tell you everything about a group. $R(G)$ is naturally isomorphic to the ring of class functions $G \to \mathbb{C}$ (the functions constant on conjugacy classes) under pointwise addition and multiplication, and as such the information it contains is precisely the multiset of sizes of each the number of conjugacy classes of $G$. That's it! No other information. (Note that $D_4$ and $Q$ both have conjugacy classes of sizes $1, 1, 2, 2, 2$ five conjugacy classes.)
In other words, the abstract structure of the representation ring actually gives you less information than the character table; the character table at least hands you a distinguished basis of $R(G)$. Without this basis, $R(G)$ can't even tell you what tensor products of representations look like.
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How do you get the class sizes from the abstract ring $R(G)\subseteq\mathbb C^G$? It is a semisimple commmutative $\mathbb C$-algebra, so its only invariant is the dimension. – Mariano Suárez-Alvarez Jan 11 2010 at 16:59
Whoops. I seem to have been secretly assuming that R(G) came with the dual basis on Hom(G, C), which is obviously wrong. Thanks! – Qiaochu Yuan Jan 11 2010 at 17:19
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That's what R(G) \otimes C can tell you. If you don't tensor with C, you can get a little more. For example, R(Z/2 x Z/2) is not isomorphic to R(Z/4) because, after tensoring with a field k of characteristic 2, the former becomes k[u,v]/<u^2, v^2> and the latter becomes k[u]/u^4. But, morally, I agree with this answer. – David Speyer Jan 11 2010 at 17:37
An RWTH Aachen thesis by Helena Skrzipczyk [1992] is referred to by Eick and J. Mu"ller J. ALg 304 (2006) On Brauer pairs. She gives several examples of non-isomorphic groups with bijections between the character tables and power maps. They are the smallest order Brauer pairs assuming such to be p-groups.
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http://math.stackexchange.com/questions/207955/prove-that-operatornamespec-sqrt2-contains-infinitely-many-powers-of-2
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# Prove that $\operatorname{Spec}\sqrt2$ contains infinitely many powers of $2$.
$\newcommand{\spec}{\operatorname{Spec}}\spec\sqrt2=\{\lfloor k\sqrt2\rfloor: k \ge 0\}$.
I have no idea of how I can prove the statement in the question.
Prove that $\spec\sqrt2$ contains infinitely many powers of $2$.
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The most obvious place to start would be to enumerate the first few values of $\lfloor k \cdot \sqrt{2}\rfloor$ to get a feel for how it behaves. The next most obvious is to devise an algorithm for solving $\lfloor k \cdot \sqrt{2} \rfloor = 2^n$ for $k$. – Hurkyl Oct 5 '12 at 21:10
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What a strange notation. Why $\mathrm{Spec} \sqrt{2}$? – Alexander Shamov Oct 5 '12 at 21:12
@Alexander: Graham, Knuth, & Patashnik, Concrete Mathematics, refer to this sequence as the spectrum of $\sqrt2$. – Brian M. Scott Oct 5 '12 at 21:12
As stated, this question is obviously true. Do you mean infinitely many integer powers of 2? – Graphth Oct 5 '12 at 21:14
@Graphth: Of course. – Brian M. Scott Oct 5 '12 at 21:15
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## 1 Answer
Let $k=\lceil 2^n\sqrt 2\rceil$. Then $2^n\sqrt 2<k<2^n\sqrt 2+1$. In fact we have either $2^n\sqrt 2<k<2^n\sqrt 2+\frac12$ or $2^n\sqrt 2+\frac12<k<2^n\sqrt 2+1$, depending on the ($n+1)$st binary digit of $\sqrt 2$ (which becomes the first digit of $2^n\sqrt 2$). Since $\sqrt 2$ is irrational, there are infinitely many $k$ such that $$2^n\sqrt 2<k<2^n\sqrt 2+\frac12$$ holds. Together with $k\sqrt 2 -1<\lfloor k\sqrt 2\rfloor <k\sqrt 2$ we find $$2^n\cdot 2-1 <\lfloor k\sqrt 2\rfloor <2^n\cdot 2+\frac{\sqrt 2}2,$$ hence $\lfloor k\sqrt 2\rfloor=2^{n+1}$.
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Thanks for the answer, but I see a lot of unexplained things, which I don't seem to be able to comprehend :( . (n+1)st binary digit is the one counted from the left end, right? And, how does this bit influence the range of k like you posted? Infinitely many k? But k is a natural number, right? How would k take infinitely many values in between 2 real numbers differing by 1/2? – learner Oct 5 '12 at 21:55
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http://math.stackexchange.com/questions/tagged/fibration+category-theory
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# Tagged Questions
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### Beck Chevalley condition and maps of adjunctions
Suppose we have a split fibration $p : \mathbb{E}\to\mathbb{B}$ with (split) simple products. To fix notation, this means that for every projection $\pi_{I,J} : I\times J \to I$ in the base category, ...
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### Monomorphisms and fibrations are preserved by pullback
I just came across a strange property of morphisms that are preserved under pullbacks, and it made me wonder. Consider a model category $\mathcal{M}$. Because the fibrations are exactly the maps that ...
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http://www.bioscience.ws/encyclopedia/index.php?title=Logical_validity
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Logical validity
Validity - Wikipedia, the free encyclopedia
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# Validity
(Redirected from Logical validity)
For other uses, see Validity (disambiguation).
In logic, an argument is valid if and only if its conclusion is logically entailed by its premises and each step in the argument is logical. A formula is valid if and only if it is true under every interpretation, and an argument form (or schema) is valid if and only if every argument of that logical form is valid.
## Validity of arguments
An argument is valid if and only if the truth of its premises entails the truth of its conclusion and each step, sub-argument, or logical operation in the argument is valid. Under such conditions it would be self-contradictory to affirm the premises and deny the conclusion. The corresponding conditional of a valid argument is a logical truth and the negation of its corresponding conditional is a contradiction. The conclusion is a logical consequence of its premises.
An argument that is not valid is said to be "invalid".
An example of a valid argument is given by the following well-known syllogism (also known as modus ponens):
All men are mortal.
Socrates is a man.
Therefore, Socrates is mortal.
What makes this a valid argument is not that it has true premises and a true conclusion, but the logical necessity of the conclusion, given the two premises. The argument would be just as valid were the premises and conclusion false. The following argument is of the same logical form but with false premises and a false conclusion, and it is equally valid:
All cups are green.
Socrates is a cup.
Therefore, Socrates is green.
No matter how the universe might be constructed, it could never be the case that these arguments should turn out to have simultaneously true premises but a false conclusion. The above arguments may be contrasted with the following invalid one:
All men are mortal.
Socrates is mortal.
Therefore, Socrates is a man.
In this case, the conclusion does not follow inescapably from the premises. All men are mortal, but not all mortals are men. Every living creature is mortal; therefore, even though both premises are true and the conclusion happens to be true in this instance, the argument is invalid because it depends on an incorrect operation of implication. Such fallacious arguments have much in common with what are known as howlers in mathematics.
A standard view is that whether an argument is valid is a matter of the argument's logical form. Many techniques are employed by logicians to represent an argument's logical form. A simple example, applied to two of the above illustrations, is the following: Let the letters 'P', 'Q', and 'S' stand, respectively, for the set of men, the set of mortals, and Socrates. Using these symbols, the first argument may be abbreviated as:
All P are Q.
S is a P.
Therefore, S is a Q.
Similarly, the third argument becomes:
All P are Q.
S is a Q.
Therefore, S is a P.
An argument is formally valid if its form is one such that for each interpretation under which the premises are all true, the conclusion is also true. As already seen, the interpretation given above (for the third argument) does cause the second argument form to have true premises and false conclusion (if P is a not human creature), hence demonstrating its invalidity.
## Valid formula
Main article: Well-formed formula
A formula of a formal language is a valid formula if and only if it is true under every possible interpretation of the language. In propositional logic, they are tautologies.
## Validity of statements
A statement can be called valid, i.e. logical truth, if it is true in all interpretations.
## Validity and soundness
Validity of deduction is not affected by the truth of the premise or the truth of the conclusion. The following deduction is perfectly valid:
All animals live on Mars.
All humans are animals.
Therefore, all humans live on Mars.
The problem with the argument is that it is not sound. In order for a deductive argument to be sound, the deduction must be valid and all the premises true.
## Satisfiability and validity
Main article: Satisfiability and validity
Model theory analyzes formulae with respect to particular classes of interpretation in suitable mathematical structures. On this reading, formula is valid if all such interpretations make it true. An inference is valid if all interpretations that validate the premises validate the conclusion. This is known as semantic validity.1
## Preservation
In truth-preserving validity, the interpretation under which all variables are assigned a truth value of 'true' produces a truth value of 'true'.
In a false-preserving validity, the interpretation under which all variables are assigned a truth value of 'false' produces a truth value of 'false'.2
## n-Validity
A formula A of a first order language $\mathcal{Q}$ is n-valid iff it is true for every interpretation of $\mathcal{Q}$ that has a domain of exactly n members.
### ω-Validity
A formula of a first order language is ω-valid iff it is true for every interpretation of the language and it has a domain with an infinite number of members.
## References
1. L. T. F. Gamut, Logic, Language, and Meaning: Introduction to logic, p. 115
• Barwise, Jon; Etchemendy, John. Language, Proof and Logic (1999): 42.
• Beer, Francis A. "Validities: A Political Science Perspective", Social Epistemology 7, 1 (1993): 85-105.
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http://unapologetic.wordpress.com/2010/05/12/topological-vector-spaces-normed-vector-spaces-and-banach-spaces/?like=1&source=post_flair&_wpnonce=671450efa2
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# The Unapologetic Mathematician
## Topological Vector Spaces, Normed Vector Spaces, and Banach Spaces
Before we move on, we want to define some structures that blend algebraic and topological notions. These are all based on vector spaces. And, particularly, we care about infinite-dimensional vector spaces. Finite-dimensional vector spaces are actually pretty simple, topologically. For pretty much all purposes you have a topology on your base field $\mathbb{F}$, and the vector space (which is isomorphic to $\mathbb{F}^n$ for some $n$) will get the product topology.
But for infinite-dimensional spaces the product topology is often not going to be particularly useful. For example, the space of functions $f:X\to\mathbb{R}$ is a product; we write $f\in\mathbb{R}^X$ to mean the product of one copy of $\mathbb{R}$ for each point in $X$. Limits in this topology are “pointwise” limits of functions, but this isn’t always the most useful way to think about limits of functions. The sequence
$\displaystyle f_n=n\chi_{\left[0,\frac{1}{n}\right]}$
converges pointwise to a function $f(x)=0$ for $n\neq0$ and $f(0)=\infty$. But we will find it useful to be able to ignore this behavior at the one isolated point and say that $f_n\to0$. It’s this connection with spaces of functions that brings such infinite-dimensional topological vector spaces into the realm of “functional analysis”.
Okay, so to get a topological vector space, we take a vector space and put a (surprise!) topology on it. But not just any topology will do: Remember that every point in a vector space looks pretty much like every other one. The transformation $u\mapsto u+v$ has an inverse $u\mapsto u-v$, and it only makes sense that these be homeomorphisms. And to capture this, we put a uniform structure on our space. That is, we specify what the neighborhoods are of $0$, and just translate them around to all the other points.
Now, a common way to come up with such a uniform structure is to define a norm on our vector space. That is, to define a function $v\mapsto\lVert v\rVert$ satisfying the three axioms
• For all vectors $v$ and scalars $c$, we have $\lVert cv\rVert=\lvert c\rvert\lVert v\rVert$.
• For all vectors $v$ and $w$, we have $\lVert v+w\rVert\leq\lVert v\rVert+\lVert w\rVert$.
• The norm $\lVert v\rVert$ is zero if and only if the vector $v$ is the zero vector.
Notice that we need to be working over a field in which we have a notion of absolute value, so we can measure the size of scalars. We might also want to do away with the last condition and use a “seminorm”. In any event, it’s important to note that though our earlier examples of norms all came from inner products we do not need an inner product to have a norm. In fact, there exist norms that come from no inner product at all.
So if we define a norm we get a “normed vector space”. This is a metric space, with a metric function defined by $d(u,v)=\lVert u-v\rVert$. This is nice because metric spaces are first-countable, and thus sequential. That is, we can define the topology of a (semi-)normed vector space by defining exactly what it means for a sequence of vectors to converge, and in particular what it means for them to converge to zero.
Finally, if we’ve got a normed vector space, it’s a natural question to ask whether or not this vector space is complete or not. That is, we have all the pieces in place to define Cauchy sequences in our vector space, and we would like for all of these sequences to converge under our uniform structure. If this happens — if we have a complete normed vector space — we call our structure a “Banach space”. Most of the spaces we’re concerned with in functional analysis are Banach spaces.
Again, for finite-dimensional vector spaces (at least over $\mathbb{R}$ or $\mathbb{C}$) this is all pretty easy; we can always define an inner product, and this gives us a norm. If our underlying topological field is complete, then the vector space will be as well. Even without considering a norm, convergence of sequences is just given component-by-component. But infinite-dimensional vector spaces get hairier. Since our algebraic operations only give us finite sums, we have to take some sorts of limits to even talk about most vectors in the space in the first place, and taking limits of such vectors could just complicate things further. Studying these interesting topologies and seeing how linear algebra — the study of vector spaces and linear transformations — behaves in the infinite-dimensional context is the taproot of functional analysis.
## 9 Comments »
1. [...] We can also turn around and define what it means for a sequence to be uniformly Cauchy almost everywhere — for any there is some so that for all we have . Unpacking again, there is some measurable set so that for all . It’s straightforward to check that a sequence that converges uniformly a.e. is uniformly Cauchy a.e., and vice versa. That is, the topology defined by the essential supremum norm is complete, and the algebra of essentially bounded measurable functions on a measure space is a Banach space. [...]
Pingback by | May 14, 2010 | Reply
2. I’d known for a long time that a proper linear subspace (i.e. a subspace in the usual linear algebra sense) of an infinite dimensional normed space can be fairly exotic (e.g. they can be dense), but I had no idea how exotic linear subspaces could be until a few years ago. For example, in a finite dimensional normed space, each proper linear subspace is closed and nowhere dense. On the other hand, in EACH infinite dimensional separable Banach space there exist proper linear subspaces that are dense and **, where ** can be any of the following:
F_sigma_delta,
but not F_sigma or G_delta
G_delta_sigma,
but not F_sigma or G_delta
F_sigma_delta AND G_delta_sigma,
but not F_sigma or G_delta
F_sigma_delta_sigma,
but not F_sigma_delta or G_delta_sigma
G_delta_sigma_delta,
but not F_sigma_delta or G_delta_sigma
F_sigma_delta_sigma AND G_delta_sigma_delta,
but not F_sigma_delta or G_delta_sigma
Moreover, the 2nd order examples, represented by the first 3, and the 3rd order examples, represented by the last 3, are just the tip of the iceberg. There are analogous 4th order examples, 5th order examples, …, n’th order examples, … In fact, such examples can be found at each of the w_1 (first uncountable ordinal) Borel levels. Also, there exist proper linear subspaces that are not Borel, and worse.
I did a little literature research on this topic about 6 years ago and finally got around to writing an essay on it about 4 years ago:
“Exotic normed linear subspaces” (18 August 2006)
http://tinyurl.com/2fd7gl6
Comment by Dave L. Renfro | May 14, 2010 | Reply
3. [...] it turns out that there is no metric which gives this sense of convergence, but we still refer to a sequence as being “Cauchy in [...]
Pingback by | May 19, 2010 | Reply
4. [...] We can now introduce a norm to our space of integrable simple functions, making it into a normed vector space. We [...]
Pingback by | May 28, 2010 | Reply
5. [...] This shows that any mean Cauchy sequence of integrable functions is also mean convergent to some function, and thus the space of integrable functions equipped with the norm is a Banach space. [...]
Pingback by | June 8, 2010 | Reply
6. [...] it so happens that we can define a norm on this space. Indeed, taking the Jordan decomposition, we must have both and , and thus . We [...]
Pingback by | June 28, 2010 | Reply
7. [...] functions on a measure space , which we called or . We’ve seen that this gives us a complete normed vector space — a Banach space. This is what we’d like to [...]
Pingback by | August 26, 2010 | Reply
8. [...] of an essentially bounded function. We’ll now write this as , suggesting that it’s a norm. And it’s clear that , and that if and only if almost everywhere. Verifying the triangle [...]
Pingback by | August 30, 2010 | Reply
9. [...] Linear Transformations In the context of normed vector spaces we have a topology on our spaces and so it makes sense to ask that maps between them be [...]
Pingback by | September 2, 2010 | Reply
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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://openwetware.org/index.php?title=Physics307L:People/sosa/formallabreport&oldid=271345
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# Physics307L:People/sosa/formallabreport
### From OpenWetWare
Revision as of 23:01, 15 December 2008 by David E. Sosa (Talk | contribs)
# Determining the Elementary Charge Via Millikan's Oil Drop Experiment
Author: David Sosa
Experimentalists: Manuel Franco Jr. and David Sosa
Location: Junior Lab, Department of Physics & Astronomy, University of New Mexico
Albuquerque, NM 87131
dsosa55@unm.edu
## Abstract
The measurement of the fundamental electric charge is of extreme importance for physics because electrons and its characteristics play an essential role in many physical phenomena such as electricity, magnetism, quantum mechanics, thermal conductivity, etc [1].In this experiment we tried to demonstrate the existence of an elementary charge and at the same time attempted to measure its magnitude. In order to achieve these two objectives, we replicated Robert Millikan's oil drop experiment [5]. In this experiment we measure the force on tiny oil droplets suspended against gravity between two metal electrodes. Knowing the electric field, the charge on the droplet could be determined. Although it was well known that electric charge was quantized before Millikan's experiment in 1913 [8], the Millikan experiment provided the most accurate measurement of the electron charge to that day. In the replication of the experiment we used the least squares method to obtain the value $(1.63\pm 0.02)\times 10^{-19}\;C$ with a difference of $2.14%\;$ with the accepted value of $1.602\times 10^{-19}\;C$ [7] This result suggests that we were able to replicate Millikan's Experiment.
## Introduction
It is difficult to say when the idea of an elementary charge exactly came to be, but Robert Millikan himself gives the credit to Benjamin Franklin for writing about it around 1750 [2]. It was also theorized several times during the 19th century, by a number of physicists including the Irish physicist George Johnstone Stoney, who would name this elementary charge, electron [9]. But it was not until 1887 that Sir Joseph John Thomson obtained the definite experimental proof of the existence of an elementary subatomic particle with negative charge, he would call corpuscle [8]. This experiment would eventually award him with the Nobel Prize in 1906 [10]. After this discovery, Thomson along one of his co-workers J.S.E. Townsend proceeded to measure the charge of the "corpuscles". The result they obtained was $1.000\times 10^{-19}\;C$ [3]. Millikan began to work in his oil drop in 1906 and continued for 7 years until he published his final paper in 1913.[3] The oil drop method was superior to the method devised by Thomson and Townsend mainly for 3 reasons:
• Oil evaporates much slower than water, allowing a more consistent measurements.[3]
• Drops can be studied one at the time unlike the water cloud.
• In following the oil drop over many ascents and descents, he could measure the drop as it lost or gained electrons.
With the Oil Drop experiment, Millikan obtained a value of $1.592(17)\times 10^{-19}\;C$ for the charge of the electron, which is less than one percent from the actual accepted value of $1.602(49)\times 10^{-19}\;C$ [7]. For this report we attempted to recreate Millikan's experiment.
### Other Methods to Measure the Elementary Charge
#### From the Josephson and von Klitzing constants
This method is the most accurate method to measure the elementary charge. This method consists in two effects in quantum mechanics: The Josephson effect, voltage oscillations that arise in certain superconducting structures; and the quantum Hall effect, a quantum effect of electrons at low temperatures, strong magnetic fields, and confinement into two dimensions.[11] [12]
The Josephson constant is:
$K_J = \frac{2e}{h}$ (where h is Planck's constant).
The von Klitzing constant is
$R_K = \frac{h}{e^2}$ It can be measured directly using the quantum Hall effect.
From these two constants, the elementary charge can be deduced:
$e = \frac{2}{R_K K_J}$
#### In terms of the Avogadro constant and Faraday constant
The elementary charge can also be determined using the following formula:
$e=\frac{F}{N_A}\;$
where
$F\;$ is Faraday's Constant and it is defined as the amount of charge that must pass through a solution to electrolytically deposit a mole of a singly charged, or monovalent element contained in the solution.
$N_A\;$ is Avogadro's Constant and it is defined as the number of atoms or molecules contained in a mole, which is defined as a mass in grams equal to the atomic or molecular weight of a substance.
Faraday's constant can be determined using Faraday's 1st law of electrolysis and Avogadro's constant is the number of atoms or molecules needed to make up a mass equal to the substance's atomic or molecular mass, in grams.
#### Shot noise
Shot noise is when the passing of electron is not a continual random flow, but rather a discrete flow in which electrons pass just one at the time. By carefully measuring the noise of the current, the elementary charge can be measured. [12]
#### Automated Millikan Oil Drop
This is a modern version of Millikan's Oil Drop experiment. This automated device was used in this paper to search for the fractional charge particles required for unification in all the superstring models. There are clear differences in this new setup and the old one The apparatus uses a digital charged coupled device (CCD) camera interfaced to a computer. It also uses silicone instead of oil, because silicone has a low vapor pressure and the right viscosity to generate stable drops. Also the falling of the drops is controlled by an electric pulse that regulates the falling rate. It is valuable to see how Millikan experiment for determining the charge of the electron is being used to explore the current forefront of physics. [4] [5]
## Materials and Methods
Figure 1. Overall view of the Millikan Apparatus (taken from Physics 307L)
Figure 2. Close up of the apparatus (taken from Physics 307L)
Figure 3. Close up diagram of the apparatus (PASCO)
Figure 4. High voltage set up connection (PASCO)
Figure 5. Droplet viewing chamber. (PASCO)
Figure 6. Ionization level settings.(PASCO)
• Millikan oil drop apparatus(PASCO scientific Model AP-8210).
• Tel-Atomic 50V & 500V supply.
• Fluke 111 True RMS Multimeter
• Mineral oil(Squib #5597)
• Atomizer(Included with the apparatus)
• Smiec 0-25mm 0.01mm Micrometer
• Stopwatch
### Set Up
First we looked for a comfortable position to set the apparatus, so we could take measurements during 3 straight hours. We used some books to achieve the best position.Although the final position of the apparatus was not entirely comfortable, it did not stop us from continuing. Then, as the manual indicated, we made sure the Millikan apparatus was correctly leveled, by using the leveling gauge and the adjusting feet. Then we used the micrometer to measure the spacer, which is equal to the plate separation.We did not get a good result the first time, as our analysis seemed to indicate, so we went a second time to measure it and with the help of Dr. Koch we obtained the correct measurement. We then opened the Droplet viewing chamber to clean it from left oil from previous experiments.This is important because as Dr. Koch and I would find out, if the chamber is full of oil, the droplets won't be seen. We then reassembled the Droplet viewing chamber according to figure 5
With the lights off we proceeded to calibrate the microscope. First we adjusted the viewing scope and the the vertical filament adjustment knob by unscrewing the focusing wire from its storage place on the platform and inserting it into the hole in the center of the top capacitor plate. According to the manual, the best view is achieved we the light is brightest on the wire in the area of the reticle and when the wire is focused. Manuel and I then connected the power supply on the $500 V\;$ side to the plate voltage connectors. We measure to voltage to get exactly $500 V\;$. Then we measured the resistance by connecting the multimeter to the thermistor connectors. We obtained a value of $2.09M\Omega\;$. We would used this value later to obtained the temperature of the room. With these preparations and measurements of constants done, we proceeded to observe the droplets.
### Experiment
We prepared the atomizer by spraying oil into a tissue. As Aram instructed us, it is possible to hear when the atomizer is spraying oil or not. One has to pay close attention.We moved the ionization source lever to the Spray Droplet Position to allow air to escape from the chamber. We pointed the atomizer 90° into the chamber, sprayed and slowly released. We should have selected our drops based on the approximate criteria given by the manual;
Excess Electrons Fall time (s) Rise time (s)
1 15 15
2 15 7
3 15 2
but unfortunately we didn't proceed that way. Instead we selected a droplet without any regard for its excess electrons and started taking several measurements of it fall and rise times.
## Data and Analysis
The complete data taken during this experiment and also the analysis can be seen in this Excel Spreadsheet
We analyzed out data first by obtaining the fall and rise time for all the measurements we made. We used the straightforward formula $V=\frac{d}{t}$. We were able to use this formula because the droplets reach terminal velocity quickly and therefore travel at a constant rate after that.
Figure 7.The chart from the manual used to find the temperature of the air in the capacitor from the resistance in the thermistor.
Figure 8. The chart from the manual used to find the viscosity of the air in the capacitor from the temperature of the air.
To find the elementary charge we first found the radius of the droplet by using the formula:
$a=\sqrt{\left(\frac{b}{2 p}\right)^2+\left(\frac{9\eta v_f}{2g\rho}\right)}-\frac{b}{2p}$
Knowing the radius and with the density of oil given, it was straightforward to get the mass of each drop:
$m=\frac{4}{3}\pi a^3 \rho$
With the mass of the droplet, we calculated the charge carried by each droplet with:
$q=\frac{mg(v_f+v_r)}{Ev_f}$
Plugging everything we obtain:
$q = {\frac{4}{3}}\pi \rho g[\sqrt{(\frac{b}{2p})^2 +\frac{9 \eta v_f}{2g\rho}}-\frac{b}{2p}]^3\frac{v_f + v_r}{Ev_f}$
The symbols presented in the previous equations are:
$q\;$ The charge of the electron
$p\;$ Barometric pressure = $8.33\times 10^{4}Pa\;$
$d\;$ Distance between capacitor plates $7.59mm\;$
$g\;$ Acceleration due to gravity- $9.8 \frac{m}{s^2}$
$b\;$ Constant $8.44\times 10^{-3} Pa\;$
$a\;$ Radius in drop measured in meters
$\rho = 886 \;\frac{kg}{m^3}$ Mineral Oil density
$\eta\;$ Viscosity of air
$V\;$ Potential difference across the plates in Volts
$v_r\;$ Rise velocity
$v_f\;$ Falling velocity
$E\;$ Electric field (found with $\frac{V}{d})$
When we finally obtain a charge for the for each drop by relating all the constants and previously obtained values, we can get an average charge for each electron. Looking at this charge, we can assign a multiple of $-e\;$ to each one. I assigned a multiple of to each of these measured charges. I also obtained an average when several measurements were made for the same drop. I also had to change the calculations in order to account for the change in temperature that we measure two different days. Because we made measurements two different days, we have to different temperatures for each day. Since the temperature was not exactly as given in the Thermistor Resistance Table it was necessary to plot that data and the insert our values for the resistance to get the temperature for the two day. Once I had the temperature, I used this online Gas Viscosity Calculator to get the viscosity for each day. It was not necessary to use Figure 8. Also the barometric pressure for each day is going to be different. I looked for the different barometric pressures for each day in Albuquerque here. I used Excel to organaize and process all the data taken. I used the Excel command Linest to obtain the value of the elementary charge. I also used the Polyfit command to obtain the values of the temperature according to the resistance each day Figure 9 .
Figure 9: Temperature Vs Resistance
## Results and Discussion
Figure 1: Charge vs. Multiples of e
The result obtained using Linear Regression was $(1.63\pm 0.02)\times 10^{-19}\;C$ with a difference of $2.14%\;$ with the accepted value of $1.602\times 10^{-19}\;C$. This result suggests that we were able to replicate Millikan's experiment succesfully.
## Acknowledgments
First and foremost I would like to thank Dr. Koch for teaching this course and helping us learn more about experimental physics. I also would like to thank Lab Assistant Aram Gragossian for his helpful inputs in the experiments. I also thank my lab parter Manuel Franco JR who was crucial in the achievement of a good result for this experiments. In addition, I thank Linh Le from last year for his contribution in this experiment. I would also like to thank all my partners from this year, because I used many of their ideas and feedback to complete my formal report.
## References
[1] How atoms work http://science.howstuffworks.com/atom.htm
[2] The electron and the light-quant from the experimental point of view ,Nobel Lecture, May 23, 1924 http://nobelprize.org/nobel_prizes/physics/laureates/1923/millikan-lecture.pdf
[3] Determination of the Charge on an Electron http://dbhs.wvusd.k12.ca.us/webdocs/AtomicStructure/Determine-electron-charge.html
[4] D. Loomba et. al, Search for Free Fractional Electric Charge Elementary Particles Using an Automated Millikan Oil Drop Technique, 1999. http://prola.aps.org/pdf/PRL/v84/i12/p2576_1
[5] Millikan R.A, On the elementary electrical charge and the Avogadro constant, 1913 http://authors.library.caltech.edu/6438/1/MILpr13b.pdf
[6] Quarter Electrons May Enable Exotic Quantum Computer, Scientific American,2008 http://www.sciam.com/article.cfm?id=quarter-electrons-may-enable-quantum-computer
[7] Official value of the elementary charge. http://physics.nist.gov/cgi-bin/cuu/Value?e
[8] Thomson J.J ,Cathode Rays, Philosophical Magazine, 44, 293, 1897. http://web.lemoyne.edu/~GIUNTA/thomson1897.html
[9] Of the "Electron" or Atom of Electricity, G. Johnstone Stoney, 1894. http://dbhs.wvusd.k12.ca.us/webdocs/Chem-History/Stoney-1894.html
[10] J.J. Thomson ,The Nobel Prize in Physics 1906, Nobel Lecture http://nobelprize.org/nobel_prizes/physics/laureates/1906/thomson-lecture.html
[11] B. D. Josephson. The discovery of tunnelling supercurrents. Rev. Mod. Phys. 1974; 46(2): 251-254. http://prola.aps.org/abstract/RMP/v46/i2/p251_1
[12] Ando, Tsuneya; Matsumoto, Yukio; Uemura, Yasutada (1975). "Theory of Hall Effect in a Two-Dimensional Electron System". J. Phys. Soc. Jpn. 39: 279–288. doi:10.1143/JPSJ.39.279.
[13] Horowitz, Paul and Winfield Hill, The Art of Electronics, 2nd edition. Cambridge (UK): Cambridge University Press, 1989, pp. 431-2.
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http://mathhelpforum.com/calculus/110202-function-limits-f-g.html
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# Thread:
1. ## Function limits-f and g
Let $h(x)=2f(x) - g(x)$, where $f$and $g$ are functions satisfying:
$\lim_{x\to3}f(x)=-2$,
$\lim_{x\to3}g(x)=4$,
$\lim_{x\to7}f(x)=6$,
$\lim_{x\to7}g(x)=-5$.
Find the following:
a. $\lim_{x\to3}h(x)$
b. $\lim_{x\to7}(g(x)/h(x))$
c. $\lim_{x\to3}f(x)*g(x+4)$
Here are the answers I have..but I'm not sure if I'm doing the problems correctly or if my answers are correct. An explanation would be great!
a. -8
b. -5/17
c. -16
2. Originally Posted by live_laugh_luv27
Let $h(x)=2f(x) - g(x)$, where $f$and $g$ are functions satisfying:
$\lim_{x\to3}f(x)=-2$,
$\lim_{x\to3}g(x)=4$,
$\lim_{x\to7}f(x)=6$,
$\lim_{x\to7}g(x)=-5$.
Find the following:
a. $\lim_{x\to3}h(x)$
b. $\lim_{x\to7}(g(x)/h(x))$
c. $\lim_{x\to3}f(x)*g(x+4)$
Here are the answers I have..but I'm not sure if I'm doing the problems correctly or if my answers are correct. An explanation would be great!
a. -8
b. -5/17
c. -16
a) Correct.
b) Correct.
c) Incorrect.
Note that when $x \to 3, g(x + 4) \to g(7) = \lim_{x \to 7}g(x)$.
So $\lim_{x \to 3}f(x)\cdot g(x + 4) = \lim_{x \to 3}f(x)\cdot\lim_{x \to 3}g(x + 4)$
$= \lim_{x \to 3}f(x)\cdot\lim_{x \to 7}g(x)$
$= -2\cdot(-5)$
$= 10$.
3. Also, what additional information would you need to know about g in order to prove that there is a value between 3 and 7, such that $g(x)=0$?
4. Originally Posted by live_laugh_luv27
Also, what additional information would you need to know about g in order to prove that there is a value between 3 and 7, such that $g(x)=0$?
Isn't it obvious that if $\lim_{x \to 3}g(x) = 4$ and $\lim_{x \to 7}g(x) = -5$, then the function must cross the x-axis somewhere in between in order to allow for the change in sign?
5. yeah it is, now that I look at it. So by using IVT, you need to know the information that g(x) crosses the x axis at some point, which would make g(x) = 0?
6. Originally Posted by live_laugh_luv27
yeah it is, now that I look at it. So by using IVT, you need to know the information that g(x) crosses the x axis at some point, which would make g(x) = 0?
Correct.
Edit: This is assuming, of course, that the function $g(x)$ is continuous on the interval $x\in (3, 7)$.
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http://physics.stackexchange.com/questions/27141/some-questions-about-the-spectral-function?answertab=active
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# Some questions about the spectral function
If you think that this question is likely to get closed then please do not answer and only say that in the comments since this system doesn't let me delete the question once it has answers.
Everytime I am faced with an analysis of the spectral function it looks like a "new" unintuitive set of jugglery with expectation values and I am unable to see a general picture of what this construction means. I am not sure that I can frame a coherent single question and hence I shall try to put down a set of questions that I have about the idea of spectral functions in QFT.
• I guess in a Dirac field theory one defines the spectral function as follows,
$\rho_{ab}(x-y) = \frac{1}{2} <0|\{\psi_a(x),\bar{\psi}_b(y)\}|0>$
Also I see this other definition in the momentum space as,
$S_{Fab}(p) = \int _0 ^\infty d\mu^2 \frac{\rho_{ab}(\mu^2)}{p^2 - \mu ^2 + i\epsilon}$
Are these two the same things conceptually? I tried but couldn't prove an equivalence.
(I define the Feynman propagator as $S_{Fab} = <0|T\big(\psi_a(x)\bar{\psi}_b(y)\big)|0>$)
• Much of the algebraic complication I see is in being able to handle the quirky "_" sign in the time-ordering of the fermionic fields which is not there in the definition of the Feynman propagator of the Klein-Gordon field (..which apparently is seen by all theories!..) and to see how the expectation values that one gets like $<0|\psi_a(0)|n><n|\bar{\psi}_b(0)|0>$ and $<0|\bar{\psi}_b(0)|n><n|\psi_a(0)|0>$ and how these are in anyway related to the Dirac operator $(i\gamma^\mu p_\mu +m)_{ab}$ that will come-up in the far more easily doable calculation of the spectral function for the free Dirac theory.
• Is it true that for any QFT given its Feynman propagator $S_F(p)$ there will have to exist a positive definite function $\rho(p^2)$ such that a relation is satisfied like,
$S_F(p) = \int _0^\infty d\mu ^2 \frac{\rho(\mu^2)} {p^2 - \mu^2 +i\epsilon}$
So no matter how complicatedly interacting a theory for whatever spin it is, its Feynman propagator will always "see" the Feynman propagator for the Klein-Gordon field at some level? (..all the interaction and spin intricacy being seen by the spectral function weighting it?..)
• One seems to say that it is always possible to split the above integral into two parts heuristically as,
$$\begin{eqnarray}S_F(p) &=& \sum (\text{free propagators for the bound states})\\ &&+ \int_\text{states} \big( (\text{Feynman propagator of the Klein-Gordon field of a certain mass})\\ && ~~~~~~~~~~~~~~\times(\text{a spectral function at that mass})\big)\end{eqnarray}$$
Is this splitting guaranteed irrespective of whether one makes the usual assumption of "adiabatic continuity" as in the LSZ formalism or in scattering theory that there is a bijection between the asymptotic states and the states of the interacting theory - which naively would have seemed to ruled out all bound states?
To put it another way - does the spectral function see the bound states irrespective of or despite the assumption of adiabatic continuity?
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## 1 Answer
You would like to state the Källen-Lehman representation through the Fermonic field commutator, that is $\{\bar\psi(x),\psi(y)\}$ but, as far as the proof goes, you can only use this in a time-ordered way. I mean, you should use $\theta(x_0-y_0)\psi(x)\bar\psi(y)-\theta(y_0-x_0)\bar\psi(y)\psi(x)$. This will grant the due appearance of the Klein-Gordon propagator in the final formula. When you will do that, the standard view, seen through states and bound states, holds true.
About adiabatic continuity, you will always get a weighted sum of free propagators with all the spectrum of the theory, free and bound states, that is in agreement with such a hypothesis. The effect of the interaction will be coded in the weights and the spectrum itself.
Finally, positivity of the spectral function can only be granted, and a proof holds, when the states behave in a proper way. This is not exactly the case for a gauge theory and some of the difficulties arising in proving the existence of a mass gap can be tracked back to a problem like this. E.g. see this book by Franco Strocchi.
Further clarification: When you insert the operator generating a translation in the bosonic field, the same is somewhat different for the spinorial case. You will get
$$U^\dagger\psi U=S\psi$$
with $S$ the one I think you studied in the proof of Lorentz invariance of the Dirac equation. Now, you are almost done. This will give for you matrix element
$$\langle 0|\psi(0)|\alpha\rangle=\sqrt{Z}u(\alpha)$$
being $\alpha$ running both on momenta and spin. You are practically done as, using the known relations $\sum_s u\bar u= \gamma\cdot p+m$ and $\sum_s v\bar v= \gamma\cdot p-m$, you will get back Källen-Lehman representation.
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The "problem" is with the minus sign that time-ordering of the fermionic fields introduces. After one introduces a complete set one is left with two terms which look like, $<0|\psi_a(x)|n><n|\bar{\psi_b(y)}|0>$ and $<0|\bar{\psi_b(y)}|n><n|\psi_a(x)|0>$. In the corresponding terms of the calculation with scalar fields these terms are equal but for the Dirac field I don't know whether these are related or not..things would be simple if one could argue that these two terms differ by exactly a negative sign -- but i don't know either way. – user6818 Nov 18 '11 at 0:38
I do not know if you mean this. When you use translational invariance of the vacuum, that is you consider $\psi(x)=e^{ipx}\psi(0)e^{-ipx}$ and $e^{ipx}|0\rangle=|0\rangle$, you should also consider in this case the spin contribution with respect to the scalar field. This contribution, managed through standard formula like $\sum u\bar u=\gamma p+m$ will permit you to recover the standard contribution $iZ/\gamma p-m$ and in the end you will get the K-L representation for Fermions. – Jon Nov 20 '11 at 10:20
I thought of this but am not clear as to how or where the spin contribution is going to come from. If you can show as to how the $\sum u\bar{u}$ - the free Dirac field term is going to emerge from the matrix elements I typed above. – user6818 Nov 20 '11 at 20:30
I added a clarification about in the answer. – Jon Nov 20 '11 at 22:05
Thanks for your efforts. I have tried doing everything you are saying. When I insert that general $U$ in that matrix element then I will get those factors of $S$ but that doesn't help me understand as to why one should get the free-field $u_\alpha$ out of it? The main query of the question! And also why do you say that just translations should affect the spinor fields any non-trivially? I would think that only when one rotates does the non-triviality of the spin come into play - translations are always a phase! – user6818 Nov 21 '11 at 0:59
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http://math.stackexchange.com/questions/42149/compressing-the-mandelbrot-set
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Compressing the Mandelbrot set
This question may not have a definitive answer. However, if someone is able to illuminate the topic for me, I would be very grateful.
The Mandelbrot set is the set obtained from the quadratic recurrence equation{1}:
$$\begin{equation} z_{n+1}=z_n^2 + c \end{equation}$$
I'm sure most of you know what the graphical representation of the Mandelbrot set looks like, so I won't post a picture of it here.
Question
Have there been any attempts to derive the Mandelbrot set equation purely from it's graphical representation?
I would imagine that this would involve some sort of machine learning process which searches through program space trying to find a correct program with the smallest Kolmogorov complexity{2}.
What branch of mathematics works on solving this type of problem?
Thank you.
{1}: http://mathworld.wolfram.com/MandelbrotSet.html
{2}: http://en.wikipedia.org/wiki/Kolmogorov_complexity
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As far as I am aware, there is no computationally precise way of encoding the Mandlebrot set except for the definition of the Mandlebrot set (although there are slightly different theoretical descriptions which are easily equivalent). You can't compress a collection of data if you don't at least have some finite yet inefficient way of representing it. The "graphical representation" of the Mandlebrot set is not actually such a representation, it is just a series of approximations. – Aaron May 30 '11 at 16:39
– lhf May 30 '11 at 16:44
2 Answers
In a certain sense the answer is yes -- look at Hubbard and Douady's work concerning "external angles" and "Hubbard trees". Modulo a conjecture about local path-connectedness I believe they have a very explicit topological model of the Mandelbrot set which in some sense is derived from a "picture" of it.
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What graphical representation? The Mandelbrot set looks different at different resolutions. For a fixed resolution (and a fixed iteration threshold), you could try fractal image compression using iterated function systems, but I doubt the compression will be better than the definition. See this for an attempt.
One could say that the wonder of the Mandelbrot set is that so much information is compressed in such a simple definition. In that sense, I don't think you can compress the Mandelbrot set further.
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Regarding your second paragraph, that is indeed my point. The Mandelbrot set equation is the simplest compression of the Mandelbrot set. My question is, are there techniques for deriving the Mandelbrot set equation from the Mandelbrot set of points. – Ncarlson May 30 '11 at 16:51
@NCarlson You can't derive anything from the Mandelbrot set of points unless you have some way of deciding what it means to be in the Mandlebrot set of points. If we throw the definition away, what would be the starting point? For a slightly different example, suppose that you wanted to characterize the prime numbers, but you wanted to ignore that they were the prime numbers. You could ask "What is something that generates this set?", but if you throw away the definition of prime, how do you describe the set in the first place? It is infinite. – Aaron May 30 '11 at 16:59
@Aaron, the goal is to find the Mandelbrot set equation from the Mandelbrot set of points. In the case of primes, given a finite the finite set {2, 3, 5, 7, 11, 13}, there are numerous programs which will produce this set. A polynomial approximation would be one program. However, if our finite set of primes has 1,000,000 elements, then a polynomial approximation would be much more complex than a program which stated "A set of natural numbers where each element is only divisible by itself and 1". – Ncarlson May 30 '11 at 17:20
@Aaron, " if you throw away the definition of prime, how do you describe the set in the first place?". We assume that a (finite) set is given to us as some sort of binary input. – Ncarlson May 30 '11 at 17:22
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http://mathhelpforum.com/advanced-applied-math/140892-matrix-estimation-problem.html
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# Thread:
1. ## Matrix estimation problem
Imagine the following problem: we have a set of n matrix equations in the form of :
[b1] = [A] * [b0]
[b2] = [A] * [b1]
etc.
vertical vectors [b0], [b1], ... are GIVEN. We try to estimate matrix A. As there are many equations (more than cells in matrix A) the system has no solutions.
Is there any method that would allow to concisely write target function to estimate A by, for instance, minimizing sum of squares or applying other optimization technique?
([b1] - [A] * [b0])^2 + ([b2] - [A] * [b1])^2 + ... -> minimize
I have to input explicitly and symbolically the derivative of target function to the software.
Kind regards and thanks for help
J
2. Have you explored singular value decomposition?
3. ## Matrix estimation problem
Yes I did, but I couldn't get to the desired closed matrix formula. I believe that it is possible to explore the conditions and eventually use singular value decomposition, but at this moment I wasn't able to do it.
Thanks,
4. ## Diagonalization?
This may or may not be helpful, but whenever I see matrix exponentiation, I start to think spectrum and eigenvalues. Now, you've got $b_{n}=A^{n}b_{0}$. If you could diagonalize $A$, then you could find an orthogonal matrix $P$, and a diagonal matrix $B$ such that $A=P^{-1}BP$. Then, by a telescoping series, you'd have $A^{n}=P^{-1}B^{n}P$, and $B^{n}$ is easy to calculate (just exponentiate the numbers on the main diagonal). Like I said, this idea may or may not be able to help you.
I should also point out that just because your system has more equations than unknowns does not imply that the system is not solvable. You might have loads of redundant equations. You might even, if you eliminate enough redundant equations, end up with an under-determined system. Certainly, if the problem was generated by a particular matrix $A$ and starting vector $b_{0}$, then you've got yourself a consistent system that might well be solvable.
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http://physics.stackexchange.com/questions/35409/why-is-a-nucleus-isotropic
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# Why is a nucleus isotropic?
I believe in Neutron Scattering the neutrons after hitting a nucleus can bounce in any of 360*3 dimensions -> 1080 degrees?
Why is this so? Shouldn't it only bounce "off" the neutron in approximately the same "direction" that it came in such as when a particle bounces off a mirror -> because of the cross-section ...
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2
To describe the angular distribution in 3-D space, you should use solid angle, with a maximum of $4\pi$. $360^\circ\times 3$ does not make sense. – C.R. Sep 2 '12 at 2:00
@KarsusRen: can you elaborate? I'm thinking essentially setting the center of the neutron as the origin of a spherical coordinate system. – Eiyrioü von Kauyf Sep 2 '12 at 21:52
Which is exactly what he meant. $360 \equiv 0\ \text{mod}\ 360$. So in your example $1080 \equiv 360\ \text{mod}\ 360$ – CHM Sep 3 '12 at 5:45
Leaving aside the odd misconceptions in the question, not all nuclei are isotropic: polarized observables are big business at transition energies these days. – dmckee♦ Sep 4 '12 at 17:27
## 3 Answers
Physics is not a matter of beliefs, but of measurements with their errors and the analysis of those data according to theoretical( mathematically expressed) models.
In two body scattering, two in two out, the scattering takes place in a plane, because of momentum conservation whether classically or in the quantum mechanical mircrocosm. Thus the angle of scatter is one and goes from 0 to 360 degrees.
If one looks at the data, there is no isotropy. The angular distributions are analyzed (page 6 in the link) in a series of Legendre polynomials, which are a function of the angle theta.
edit after rereading the question:
Each individual scatter of a neutron on a nucleus will have a specific plane in the center of mass, and thus the angle theta will show the functional form of the interaction, and not be isotropic after aligning the scattering planes . The distribution in the angle phi which will define the rotation to align the scattering planes will be isotropic. Because of momentum conservation as mentioned above any phi is equally probable.
There cannot be total isotropy because there are nuclear resonances which are probed when a neutron scatters off a nucleus and the spins involved create in the crossection necessarily a function of theta.
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can you elaborate on which phi and theta - different conventions.... – Eiyrioü von Kauyf Sep 2 '12 at 21:55
Did you look at the link I gave? It is the normal spherical coordinates in which Legendre polynomials are expressed. In the experiment, which is really the trump card, they have a number of theta angles all around phi, where they measure, find isotropy, and extrapolate from that that they can use the Legendre polynomials for the functional form of the scattering angular dependence. – anna v Sep 3 '12 at 3:37
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Differential cross sections are introduced precisely to quantify the percentage of particles that scatter in a given direction. If all the directions are possibles, they do not have the same differential cross section value. And indeed, the higest value is for scattering directions close to the incoming beam.
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This is misleading--- you are talking about a case (forward scattering) which swamps all off beam scattering in the limit that the neutron wavefunction is spread out. The non-delta-function part of forward scattering is not larger by orders of magnitude than other directions for a localized potential, it is just the limit of off-forward scattering, and this is how the optical theorem works. – Ron Maimon Sep 2 '12 at 22:07
The neutron at low momentum is quantum mechancal during the scattering, and the bouncing off is by wave mechanics, not by bouncing off. The neutron gets reflected into all directions by the small region with a potential, and the amount of scattering is given by the Born approximation to lowest order in the momentum of the neutron:
$$A(k-k') = -i\tilde V(k-k')$$
For scattering from incoming direction k to outgoing direction k'. You decorate this with phase-space factors to take into account the size of the neutron wave. If the neutron wave is large, nearly all the scattering is in the forward direction, and this is expressed by writing the S-matrix as;
$$S = 1 + i A$$
Where the "1" gives a contribution to the scattering which is $\delta(k-k')$ for the case of single-particle scattering. If you smear this with the incoming wavepacket, you get the outgoing wavepacket, which is mostly the incoming wave, plus a spherical outgoing wave. The delta-function guarantees that as you approach a plane-wave limit, there will be no scattering, because the neutron wavefunction area will be much larger than the nuclear area.
This scattering, when the neutron wavefunction is larger wavelength than the nucleus (almost always in real life) leads to a spherical scattering which is roughly isotropic, which you add to the incoming wave to get the full outgoing wave. The imaginary part of the scattering in the forward direction only subtracts some weight from the wavefunction that keeps going forward, and unitarity guarantees that this imaginary part is equal to the scattering probability in all directions added together.
This is covered in scattering theory in most quantum mechanics books, but the treatment in Gribov's Regge theory book "The Theory of Complex Angular Momentum" is most instructive to my mind.
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But Ron, all this is fine and good as an introduction to scattering in the S matrix format, but how about experimental evidence (which I linked in my answer) that shows non isotropy in theta due to nuclear resonances? It would have been isotropic if there were no further interactions open to the scatter which the experiment demonstrates. It is not isotropic in the real world! – anna v Sep 3 '12 at 4:35
@annav: I am talking about energies lower than the ones required for exciting resonances. I get the sense tha OP is asking about slow or thermal neutrons, not KeV/MeV neutrons. – Ron Maimon Sep 3 '12 at 5:50
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http://physics.stackexchange.com/questions/2516/mass-of-particle-near-light-speed-in-a-medium
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# Mass of particle near light speed in a medium
I am trying to get a common understanding from these two previous questions:
Does the increase of mass occur only if the particle approaches c (speed of light in a vacuum) or if it simply approaches the speed of light in its current medium? For example, does the mass of charged particles increase during Cherenkov radiation?
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## 2 Answers
Yes, the increase of mass occurs only when a particle approaches $c$ (speed of light in vacuum).
$c$ is fundamental in Special Relativity, not because it is the speed of photons, but because it is the constant speed in the universe (the only speed invariant to boosts). Just because macroscopic light is transmitted at a lower speed inside a particular medium, that doesn't mean that the fundamental speed of Special Relativity is any different. Even inside mediums where light travels more slowly, all relativistic effects happen when a particle approaches $c$.
Since Cherenkov radiation (CR) is just an effect related to the speed of light in a medium (and not to $c$), it doesn't have anything to do with mass increase either. Though CR and mass increase can happen simultaneously to a particle, they are independent (the first does not imply the second, and vice-versa).
Second, about the increase of mass.
It has been a historical habit to say that a particle's mass increases as $m=\gamma m_0$ when its velocity approaches $c$. That is not very appropriate. While it may seem convenient to define this relativistic mass, it's not a good habit.
First, because it's confusing to some people. There are physically intuitive ways to explain to a student why time intervals must stretch and why space intervals have to contract, but there's no way at all to explain why a particle's mass should increase.
Second, it's also not accurate. The defined relativistic mass parameter does not sustain the properties you would expect from a mass under close analysis. (I have a reference for this, but the pdf file somehow got corrupted in the last 8 years. I'm looking for a copy.)
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1
This doesn't mention Cherenkov radiation at all? – Noldorin Jan 5 '11 at 14:41
@Noldorin: I said that the speed of light in a medium has nothing to do with relativistic effects, thus Cherenkov radiation has nothing to do with them either. I'll add that to the answer. – Bruce Connor Jan 5 '11 at 14:56
Special relativity is based on the concept of a single absolute "speed limit" for the universe, from all (inertial) reference frames. That "speed limit" is the constant $c$, which is equivalent to the speed of light in a vacuum. The speed of light in other materials is not fundamental in any way.
So the simple answer is, the two points you bring up are quite orthogonal. Sure, a particle undergoing Cherenkov radiation is likely to be going a high percentage of the speed of the light, and would thus have increased mass. (Note that the Cherenkov radiation would decrease its energy and thus relativistic mass somewhat, due to the EM interaction with the surrounding molecules of the material). At the end of the day, consider 1. as a relativistic effect and 2. as a electromagnetic one. They're both related of course, but there's no direct connection that I think you're looking for.
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Good answer but the part about Cherenkov radiation is not quite correct (depending on the point of view). The particle loses energy via EM work done on the molecules of the medium (essentially creating dipole moments by distorting electron orbitars). This work is then soon converted to coherent Cherenkov radiation as those dipoles return to their equilibrium position. – Marek Jan 5 '11 at 12:58
@Marek: Ok, thanks for the clarification - I wasn't sure on that point. I suppose the energy comes from both the travelling particle and the electrons in the molecules of the medium? Will edit... – Noldorin Jan 5 '11 at 14:41
well, I'd say all the energy ultimately comes from the particle (and there is indeed measurable energy loss due to this effect; but it is few orders smaller than the usual ionization losses so it's not terribly important). Anyway, I am no expert on these matters; hopefully someone else can clarify this. – Marek Jan 5 '11 at 15:00
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http://physics.stackexchange.com/questions/28741/one-plaquette-action-and-su2s-irreducible-representations
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# One-Plaquette Action and SU(2)'s Irreducible Representations
I have a typical single-plaquette partition function for a gauge-field $$Z=\int [d U_{\text{link}}] \exp[-\sum_{p} S_{p}(U,a)]$$ with $U$ as the product of the the $U$'s assigned to each link around a plaquette. Now the $U$'s are irreducible representations of my group elements, which in my case is SU(2), and lets take the 1/2 representation as an example, then define the character as $\Xi_{r}\equiv \text{Tr}[U]$ which for our case is $\Xi_{1/2}=\text{Tr}[U]$. Now I have to take the product of these representations, however (and here's my question), how do I know which group elements/that-element's-representation to assign to each link?
I'm not sure how to compute the Trace without knowing first how to do the product of the link's representations, but I don't even know how to assign the elements to the links.
Thanks,
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Okay. Thanks for that ;) – kηives May 22 '12 at 3:09
There is no representation on a link, there is a 2 by 2 matrix, an element of SU(2), on each link. The trace of the product is the matrix trace of the matrix product. Do you want to know how to simulate this? You need a metropolis step, and there are tricks for making good updates over big regions. – Ron Maimon May 22 '12 at 6:24
## 1 Answer
I'm not completely sure what OP is asking(v1). However here is my interpretation.
How do I know which group elements [...] to assign to each link?
The group element $U_{\ell}\in SU(2)$ affiliated with a link $\ell\in L$ is not fixed. One is supposed to integrate over all possible group values of $U_{\ell}\in SU(2)$. Phrased differently, the link variables $(U_{\ell})_{\ell\in L}$ are the dynamical variables of the model.
The integration measure in the integral for $Z$ reads
$$[dU]~=~ \prod_{\ell\in L} dU_{\ell}$$
Here $dU_{\ell}$ typically denotes the Haar measure for $SU(2)$.
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Oh, duh, just like stat-mech. Thanks. – kηives May 22 '12 at 0:39
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http://math.stackexchange.com/questions/2584/combine-n-normal-distribution-probability-sets-in-a-limited-float-range
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# Combine n Normal distribution Probability Sets in a limited float range
I've got a fixed solution space defined by a minimum and maximum float which has no divisible slice.
You then have 0..N Normal distribution probability sets which I need to combine.
In the end I neeed
Method to define probability set with a numeric range (not sliceable) and 0..N Gaussian functions A function which can generate a random number in the range as defined by the calculated probabilities. Also I know it is possible that some combinations will generate a zero solution space.
Now I'm thinking the way to do it is take the normalised probability density functions and multiply them to get a new space then normalising the result. I just can't seem to break it down into algorithmic form.
Any ideas?
Extra Clarification
Let's examine the following height rules.
STATUS Height [ 1.2 : 2.4 ] MODIFIER Gender[ Male ] {Height [ 1.9 ~ 0.4 ] } MODIFIER Nation[ Red ] { Height [ 1.7 ~ 0.2 ] }
Now assuming that we have a man from the Red nation we need to resolve these. We know that the original status declaration defines the entire possibility space. We cannot leave that space.
Now basically I need to find a way of combining these to get a new probability. What I meant by slice was because its a floating point number I can't break it into a set of elements of calculate the new probability for each element. I can't slice it up into pieces.
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Kimau, could you please clarify your question? In particular, what do you mean with slice? – Rasmus Aug 16 '10 at 16:52
If you have trouble coming up with a way to clarify, a simple example (with n at 2 or 3) might help. – Jonathan Fischoff Aug 16 '10 at 19:20
Added clarification. I would add a bounty but I don't have enough points yet on this stack :) – Kimau Aug 17 '10 at 12:55
## 2 Answers
First let's make sure I understood your question correctly:
• You have a probability function that is expressed as a sum/product of N parametrized one-dimensional gaussians, each with different mean and standard deviation.
• You want to generate stochastic variables according to this distribution.
Is this correct?
If this is the case, I reccommend you use a variation of rejection sampling. The recipe is quite straightforward, but you might have to iterate a bit before you get an answer out of it. This is the basic outline.
1. You generate a uniformly distributed random number in your desired interval, x.
2. You calculate the value of your probability distribution, p(x)
3. You generate another uniformly distributed random number between 0 and 1, q
4. If q < p(x), return x
5. If not, start from step 1.
No matter how large the temptation, do not re-use q for different iterations.
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– Kimau Aug 21 '10 at 8:17
I'm sure it's just not so easy to explain. If you manage to give us a problem statement that's more or less unambiguous, I'm sure we can send you home with at least an approximation to a solution. I still have no idea what the problem is with dividing or "slicing" as you call it the interval. I appreciate the image, but I do know what a few Gaussians look like. What I do not understand is what you want to know about them. – drxzcl Aug 21 '10 at 15:09
I think it's card on the tables time if you want to solve this problem. Tell us exactly what you are doing and what you want to know. Example: There are three crops of string beans, Dutch, Japanese and Argentinian. They have different lengths, distributed along known Gaussians f1,f2,f3. I have a field with 30% Dutch, 50% Japanese and 20% Argentinian. What proportion of string beans will be below 7cm in length? – drxzcl Aug 21 '10 at 15:12
It's a rules based procedural generation system for social interactions. The original floating point range is defined then a number of user defined rules can be applied based on situation. So as in the question example the valid height range of a character is [1.2 : 2.4]. Each nation has an average height and variance, same for gender, job and a few other factors. Its for a scripting system which needs to handle it in a generic fashion as the number of rules is unknown and the significant digits is not known. – Kimau Aug 25 '10 at 12:33
And what do you want to know about it? – drxzcl Aug 25 '10 at 14:35
First up, you should accept Ranieri's answer, which is obviously correct. The question of "slicing" seems like a temporary misunderstanding that shouldn't stand in your way. Rejection sampling works with continuous distributions.
Second, it seems you are attempting to randomly generate individuals from a population with some number of subgroups, each of which has an expected distribution of some continuous traits. Since the traits are distributed normally, $\forall{x}: P(x) > 0$, so it is notionally possible to have outlier individuals with any trait combination. However, it is very likely that arbitrary distributional constraints will not be satisfiable at the population level.
For example, imagine that you have a population divided into two genders and two colours, and specified to be 50/50 male/female and 50/50 red/blue. Then require that some trait, say foolishness, is distributed $N(1,0.1)$ for each of the subgroups men, red and blue, but $N(0.5, 0.1)$ for women. Is this possible? I'm pretty sure not. And that's without even trying to clip values into a fixed range.
So, I suspect that you might need a somewhat better model for your probability distribution. My first inclination would be to weight the different criteria in some fashion to prioritise the distribution of some traits over others. However, I'd have to think that through properly, which I don't have time to do just now. Maybe later...
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Thanks, I think I just wandered down a rabbit hole of logic and was so deep down it I couldn't remember the original reason. Which is a rules based system for helping procedural content generation. I think I may just need to re-examine my requirements because as you say normal distribution may not be best. – Kimau Aug 26 '10 at 11:22
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http://mathoverflow.net/questions/86760/on-the-generalisation-of-the-laplace-transform
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## On the generalisation of the Laplace transform
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I consider a measure transform $A$ given by $$A\mu(x) = \int\limits_{\mathbb{R}^n_{+}} e^{-g(x,y)} \mu(dy)$$ where $g(x,y)$ is some positive smooth function, $\mu$ is a Borel measure. Is it a well-known transform? Where can I read about it? I'm interested in conditions of injectivity and in description of it's image.
If it is very general we can consider only homogeneous of order $1$ and concave functions $g(x,y)$ such that $g(x,y) = g(x_1 y_1,x_2 y_2, \ldots, x_n y_n)$.
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2
There is a such a large literature on Laplace and related transforms on cones (e.g. the positive orthant) that it is hard to know where to start. The book of Stein and Weiss might be one place. There is a TAMS paper of O. Rothaus entitled "Some properties of Laplace transforms of measures" which directly addresses some of your questions, although surely there are other sources. – Dan Fox Jan 27 2012 at 8:20
I think this is too general to have any useful answer, without extra conditions on $g$; although I'd be glad to be wrong, since any nontrivial general results on this would be very interesting. If you know $g$ satisfies a differential equation, for example, you could start to do stuff. – Zen Harper Jun 4 at 9:12
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http://mathoverflow.net/questions/99411?sort=newest
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## Sequence of permutations without a fixed point
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I need to find $m$ permutations $A_1,..,A_m$ where each $A_i$ is a permutation on $n$ objects such that any of the compositions $A_jA_{j+1}..A_{j+k}$; $1 \leq j \leq n-1$ and $j+k \leq m$ does not have a fixed point.
Does such a sequence of permutation always exist? If yes, is there a good way to generate such permutations? Do we have a name for such a sequence of permutations? Thanks for your time.
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## 1 Answer
Of course such permutations exist if $n > m$ (just take each $A_i$ to be the same $n$-cycle). On the other hand if $m \le n$, a pigeonhole argument should give some pair of $A_1, A_1 A_2, A_1A_2A_3, \ldots$ mapping element $1$ to the same element, inducing a fixed point in the quotient of this pair. Are you sure this question is well-posed?
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Yes, you are right. This was easy after all... Thanks. – Hardik Udeshi Jun 13 at 8:46
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http://cms.math.ca/Reunions/ete12/abs/oa
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Réunion d'été SMC 2012
Hôtels Regina Inn et Ramada (Regina~Saskatchewan), 2 -4 juin 2012 www.smc.math.ca//Reunions/ete12
Algèbres des opérateurs
Org: Martin Argerami, Juliana Erlijman et Remus Floricel (Regina)
[PDF]
BERNDT BRENKEN, University of Calgary
Universal C*-algebras of *-semigroups and the C*-algebra of a partial isometry [PDF]
A structure of the C*-algebra of a partial isometry is described in terms of a Cuntz-Pimsner C*-algebra associated with a C*-correspondence; this can be viewed as a form of crossed product C*-algebra for an action by a completely positive map on a non-unital C*-algebra. The C*-algebras involved occur as universal C*-algebras associated with contractive *-representations, and complete order *-representations, of certain *-semigroups.
MAN-DUEN CHOI, University of Toronto
The Taming of the Shrew with Positive Linear Maps [PDF]
I look into the full structure of positive linear maps between matrix algebras. In particular, I wish to tame the quantum entanglements, from the pure mathematical point of view. Note that the research work along these lines, has been proven to be useful to the foundation of abstract quantum information in the light of (the reality of) quantum computers.
This is an expository talk. No knowledge of quantum information or operator algebras is assumed.
ANDREW DEAN, Lakehead University
Classification of C*-dynamical systems [PDF]
We shall discuss the problem of classifying C*-dynamical systems up to outer conjugacy.
HEATH EMERSON, University of Victoria
Fredholm modules and boundaries of hyperbolic groups [PDF]
We describe the construction of finitely summable Fredholm modules over the crossed-product C*-algebras of Gromov hyperbolic groups acting on their boundaries. These Fredholm modules are homologically nontrivial (yield nonzero maps on K-theory) and encode in an analytic way the canonical invariant Holder geometry that exists on the boundary of any such group. The degree of summability is computed, and shown to agree with the Hausdorff dimension of the boundary, and we will describe how to compute the induced maps on K-theory using several methods, e.g. via Connes' Chern character in cyclic cohomology, and, in the case of classical hyperbolic groups, via characteristic classes. This is joint work with Bogdan Nica.
DOUG FARENICK, University of Regina
Ando's numerical radius theorem revisited [PDF]
A classic theorem in operator theory is Ando's result on the structure of Hilbert space operators with numerical radius no larger than one. One way to view Ando's theorem is that it provides a solution to a certain matrix completion problem, where the matrix entries are Hilbert space operators. It is natural to ask whether Ando's theorem has a purely C*-algebraic formulation. Furthermore, by taking the view that Ando's theorem is a result concerning matrix completions, one can formulate a multivariable version of Ando's theorem and pose the question: what are the C*-algebras for which the multivariable form of Ando's theorem holds? The answer: precisely those C*-algebras that have the weak expectation property. This is joint work with Ali Kavruk and Vern Paulsen.
GILAD GOUR, University of Calgary
Closed formula for the relative entropy of entanglement [PDF]
A quantum state (positive semi-definite matrix) acting on a tensor product of two Hilbert spaces, is called a product state if it can be written as a tensor product of two quantum states. A separable state is a convex combination of product states, that describes a composite physical system with no quantum entanglement. Entanglement of a non-separable (i.e. entangled) quantum state is measured by the relative entropy "distance" of the state to the convex set of separable states. This distance is therefore called the relative entropy of entanglement (REE). Since it is NP hard to determine whether a quantum state is separable or not, the convex optimization problem posed by the REE can not be solved analytically. However, in this talk, I will show that a closed formula exists for the inverse problem. That is, for a quantum state on the boundary of the set of separable states, there is a closed formula for all the entangled state for which this state is the closest separable state (CSS). In addition I will show that if an entangled state is full rank, then its CSS is unique. My talk is based on a joint work with Shmuel Friedland.
SIRI-MALÉN HØYNES, NTNU
Toeplitz dynamical systems and their K-theory [PDF]
We will show that the family of Toeplitz systems can be associated to simple dimension groups with non-trivial rational subdimension groups. Furthermore, we will present a class of examples which has a particularly nice Bratteli diagram presentation.
CRISTIAN IVANESCU, MacEwan University
A Krein-Milman type theorem for C*-algebras [PDF]
Positive unital maps between C*-algebras form a convex set whose extreme points are unital *-homomorphisms. K. Thomsen showed for a large class of homogeneous C*-algebras that the closed convex hull of *-homomorphisms is the set of unital positive maps. Later L. Li showed a major improvement of Thomsen's result which proved to be essential to the classification program for C*-algebras. In our work, we plan to extend these results to certain subhomogeneous C*-algebras.
LAURENT MARCOUX, University of Waterloo
Almost invariant subspaces [PDF]
A closed subspace of a Banach space $\mathcal{X}$ is almost-invariant for a collection $\mathcal{S}$ of bounded linear operators on $\mathcal{X}$ if for each $T \in \mathcal{S}$ there exists a finite-dimensional subspace $\mathcal{F}_T$ of $\mathcal{X}$ such that $T \mathcal{Y} \subseteq \mathcal{Y} + \mathcal{F}_T$. In this paper, we study the existence of almost-invariant subspaces of infinite dimension and codimension for various classes and sets of Banach and Hilbert space operators.
This is joint work with A. Popov and H. Radjavi.
SHAWN MCCANN, Univ. of Calgary
$C^*$-algebras Associated with Topological Group Quivers [PDF]
We shall provide a quick survey of properties concerning $C^*$-algebras associated with topological quivers with edge set $\Gamma$, a locally compact group, and vertex set $\Omega_{\alpha,\beta}(\Gamma)=\{(x,y)\in\Gamma\times\Gamma\,\vert\, \alpha(y)=\beta(x)\}$ where $\alpha$ and $\beta$ are continuous endomorphisms on $\Gamma.$ In particular, we are interested in the case when $\Gamma$ is the $d$-torus and both $\alpha$ and $\beta$ are matrices with integer coefficients.
YASUHIKO SATO, University of Oregon, Kyoto University
Strict comparison and Z-absorption of nuclear C*-algebras [PDF]
X. Jiang and H. Su constructed a unital separable simple infinite-dimensional nuclear C*-algebra, called the Jiang-Su algebra, whose K-theoretic invariant is isomorphic to that of the complex numbers. The Jiang-Su algebra has recently become to play a central role in Elliott’s classification program for nuclear C*-algebras. In our research of its structure, we proved that tensorial absorption of the Jiang-Su algebra, strict comparison, and property (SI) are equivalent for any unital separable simple infinite-dimensional nuclear C*-algebra with finitely many extremal traces. This result provides a partial answer to Toms-Winter's conjecture.
EUGENIU SPINU, University of Alberta
Domination problem in the non-commutative setting [PDF]
We will consider the classical Domination Problem for Banach lattices in the non-commutative setting. Let X and Y be ordered Banach spaces and $0<T<S$ are operators from X to Y. Assume that S belongs to a certain class of operators (ideal of compact, weakly compact and Dunford-Petis operator). Does T belong to the same class? We will consider the case when either X or Y is a C*-algebra or a non-commutative function space.
AMI VISELTER, University of Alberta
Locally compact quantum groups and amenability [PDF]
In this talk we will introduce the Kustermans-Vaes definition of locally compact quantum groups (LCQGs), and discuss a few specific types of LCQGs. Afterwards, we will review the definition of amenability for locally compact groups, present its generalization(s) to LCQGs, and relate several problems of current research connected with these notions.
DILIAN YANG, University of Windsor
Type III von Neumann algebras associated with rank 2 graphs [PDF]
Let $\mathbb{F}_\theta^+$ be a rank 2 graph, where $\theta$ is a permeation encoding the factorization property in the rank 2 graph, and $\omega$ be a distinguished faithful state associated with its graph C*-algebra. In this talk, we will discuss when the von Neumann algebra induced from the GNS representation of $\omega$ is a factor and its type.
## Commandites
Nous remercions chaleureusement ces commanditaires de leur soutien.
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http://mathoverflow.net/questions/120365/first-eigenvalue-of-delta-on-kaehler-manifold-with-ricci-ge-k
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## First eigenvalue of $\Delta$ on Kaehler manifold with $Ricci\ge k$.
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Let $M$ be a Kaehler manifold of complex dimension $n$. Let $\Delta$ be the real Laplacian of the underline Riemannian manifold. Let's assume the Ricci curvature of $M$ satisfies $\text {Ric}\ge k>0$. Denoted the first eigenvalue of $\Delta$ by $\lambda_1$. Then we have $$\lambda_1\ge 2k.$$ Notice that this estimate is better than the non-Kaehler case namely Lichnerowicz's estimates for Riemannian manifolds would yield the lower bound: $$\lambda_1\ge \frac{2n}{2n-1}k.$$
Some authors attributes this results to Lichnerowicz: G´eom´etrie des groupes de transformations, Travaux et Recherches Math´ematiques, III. Dunod, Paris, 1958.
Some author attributes it to Udagawa: Compact Kaehler manifolds and the eigenvalues of the Laplacian Colloq. Math. 56 (1988), no. 2, 341–349.
Also Urakaw also give a proof of this lower bound in "Stability of harmonic maps and eigenvalues of the Laplacian" Trans of AMS 1987. However the proof given in this paper is not direct and depends on the more sophisticated Harmonic map theory, which is not the way I prefer.
However I can't find the first two references. So my question is: who is the first to prove this theorem and is there any book or reference that contains the proof.(Chavel's book on eigenvalues does not contain Kaehler case.)
I suppose the proof is not hard based on the length of paper of Udagawa.
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## 1 Answer
I am not sure about the history of this result. You can find in Thierry Aubin's book "Some nonlinear problems in Riemannian geometry", Theorem 4.20. He attributes it to his own 1978 paper.
The proof is really simple, though. If you write $\Delta f=g^{i\overline{j}}f_{i\overline{j}}$ for the complex Laplacian, and assume that $\Delta f=-\lambda f$ with $\lambda$ the first eigenvalue of $\Delta$ (which is half of the first eigenvalue of the real Laplacian), then commuting covariant derivatives you can easily see that $$\int_M (\Delta f)^2=\int_M g^{i\overline{j}}g^{p\overline{q}}f_{i\overline{j}} f_{\overline{q}p}=-\int_M g^{i\overline{j}}g^{p\overline{q}}f_{i} f_{\overline{q}p\overline{j}}=-\int_M g^{i\overline{j}}g^{p\overline{q}}f_{i} f_{\overline{q}\overline{j}p} +\int_M R^{i\overline{j}} f_i f_{\overline{j}}$$ $$=\int_M |f_{ij}|^2+\int_M R^{i\overline{j}} f_i f_{\overline{j}}.$$ If you assume that $\mathrm{Ric}\geq k>0$, then $R_{i\overline{j}}\geq kg_{i\overline{j}}$ and so $$\lambda^2\int_M f^2 =\int_M(\Delta f)^2\geq k\int_M |\partial f|^2=k\lambda\int_M f^2,$$ and so $\lambda\geq k$.
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@YangMills, I was looking at your answer to a question here: mathoverflow.net/questions/120286/…. It seems there is factor $2$ in front of the Real part. But why there is no $2$ in the first equality before $\int_M (\Delta f)^2$. I think this term comes from integration by part of the Real part of your answer. Did I miss something? – Ralph Feb 1 at 20:16
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http://mathoverflow.net/questions/88692?sort=oldest
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## Why should I care about Heegaard-Floer theory?
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I would like to collect a list of applications of Heegaard-Floer theory. By applications, I don't mean things like "it can detect the unknot" or "it can detect knot genus". Algorithms for these kinds of things have been known since the '60's and '70's. Instead, I mean two kinds of things.
1. Questions that make no reference to Heegaard-Floer theory that can be answered using Heegaard-Floer theory (and, preferably, cannot be answered in other ways).
2. Things about knots or 3-manifolds that can be computed with Heegaard-Floer theory for which algorithms did not previously exist.
I'm asking this question here in response to a large number of talks about Heegaard-Floer theory I've attended over the years. It seems like a hot subject and lots of talented young people are working on it, but most of the talks I've attended about it addressed what seemed to me to be technical questions internal to the subject. And when I've asked the speakers this question, I never seem to get a good answer. But since it is such a hot subject, I assume there must be some killer applications.
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## 7 Answers
I don't think you can say that detecting the unknot and detecting knot genus have been long understood just because there were algorithms for them. You want effective algorithms, or ways of computing infinite families of examples.
Aside from that, Ozsvath and Szabo had early applications to questions about which three manifolds can be obtained from which other manifolds by what kinds of surgeries. The Ozsvath-Szabo contact invariant has a number of applications, for example to understanding symplectic fillings. There is much more, too much for me to completely follow, so that anything I say would be very incomplete.
In addition, most of the applications of Seiberg-Witten theory (e.g. the Thom Conjecture, distinguishing various non-diffeomorphic but homeomorphic smooth manifolds) can be re-proved using the Ozsvath-Szabo theory. Since the two theories are known to be equivalent, it is a matter of taste whether you want to regard these as applications of Seiberg-Witten or Ozsvath-Szabo.
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Indeed! In a similar vein, algorithms for computing the homotopy groups of spheres have been known for ages. – Mariano Suárez-Alvarez Feb 17 2012 at 8:01
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Yes, but computing Heegaard Floer invariants is still not very efficient. I'm not sure about knot genus calculations, but tables of isotopy classes of knots were computed up to 16 crossings using classical methods. My understanding is that the algorithms for Heegaard Floer computations are not computationally practical for that number of crossings. Thus the fact that they detect the unknot (much weaker than separating isotopy classes) adds nothing computational to the picture. – Mitya Feb 18 2012 at 4:37
Frequently new methods of computation gives new insight. This is certainly the case for homotopy groups of spheres. – Sean Tilson Feb 19 2012 at 1:56
### 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.
Taken from the introduction to A Tour of Bordered Floer Theory (by Lipshitz, Ozsvath, D. Thurston):
http://arxiv.org/pdf/1107.5621v1.pdf
Heegaard Floer homology, introduced in a series of papers of Zoltán Szabó and the second author, has become a useful tool in 3- and 4-dimensional topology. The Heegaard Floer invariants contain subtle topological information, allowing one to detect the genera of knots and homology classes; detect fiberedness for knots and 3-manifolds; bound the slice genus and unknotting number; prove tightness and obstruct Stein fillability of contact structures; and more. It has been useful for resolving a number of conjectures, particularly related to questions about Dehn surgery. It is either known or conjectured to be equivalent to several other gauge-theoretic or holomorphic curve invariants in low-dimensional topology, including monopole Floer homology, embedded contact homology, and the Lagrangian matching invariants of 3- and 4-manifolds. Heegaard Floer homology is known to relate to Khovanov homology, and more relations with Khovanov-Rozansky type homologies are conjectured.
(references provided in the paper for each statement above)
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The $d$ invariant coming from the rational grading can be seen as an enhancement of the torsion linking form on a rational homology $3$-sphere. Just like the regular torsion linking form of 3-manifolds, this gives an obstruction to embedding rational homology $3$-spheres in the $4$-sphere (or any homology $4$-sphere). Similarly, this gives obstruction to $spin^c$-cobordism. Sometimes these obstructions are completely new, so it's a genuinely novel application where there were no comparable tools beforehand.
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May I put in a word for Kim Froyshov? He used Seiberg-Witten theory, in the days before either SW Floer or Heegaard Floer theory, to define an invariant of homology 3-spheres, and gave applications to 4-manifold topology. [The Seiberg-Witten equations and four-manifolds with boundary, Math. Res. Lett. 3 (1996), no. 3, 373–390.] Ozsvath-Szabo's d-invariant was inspired by the Froyshov invariant, and is conjecturally equal to it. Of course, the big Heegaard Floer package helps one use this invariant to full advantage. – Tim Perutz Feb 17 2012 at 15:18
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Actually, I think the recent work of Kutluhan-Lee-Taubes on SW vs. Heegaard Floer implies that the d-invariant really is the Froyshov invariant. – Tim Perutz Feb 17 2012 at 15:21
It provides lots of computable invariants in contact geometry, in particular the contact invariant defined by Ozsvåth and Szabó via open books and the Giroux correspondence. For example, on Seifert fibered spaces the question of whether tight contact structures exist was completely solved by Lisca and Stipsicz, and the classification was completed in many of these cases in several other papers; and Ghiggini used it to exhibit contact structures which are strongly fillable but not Stein fillable.
Similarly, the LOSS invariant (named after Lisca-Ozsváth-Stipsicz-Szabó) and the related, easily computable Ozsváth-Szabó-Thurston grid diagram invariants of Legendrian and transverse knots were used by Ng-Ozsváth-Thurston to successfully distinguish many pairs of knots in the standard contact $S^3$, and been used to prove other properties such as the fact due to Etnyre and Vela-Vick that the complement of the binding of any open book of any contact structure has no Giroux torsion. (According to recent work of Baldwin--Vela-Vick--Vértesi, these invariants are the same for knots in the standard $S^3$.)
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Ghiggini has shown that the only knot giving rise to the Poincare homology sphere by Dehn filling is the trefoil knot, as an application of Heegaard Floer homology.
Another important point is that now it is known that certain types of Heegaard Floer homology are computable, starting with Sarkar-Wang. Since it is known to be equivalent to other Floer homologies (Seiberg-Witten, Embedded Contact Homology), this allows one to compute these other invariants. The hope is to be able to get combinatorial formulae for invariants of 4-manifolds from this, which are related to Seiberg-Witten invariants. Unfortunately, sometimes the combinatorial formulae are more intricate and less well-motivated than the geometric definitions, or have the geometry suppressed.
As you noted, many talks on Heegaard Floer homology are getting quite technical, especially the bordered Floer theory. I believe that people think the added complexity is worth the investment, since it should yield new insights (and already is). But this also means that the motivation for some technical results might not be currently apparent.
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Since you talk about surgeries, Agol, let me mention Josh Greene's work on the Berge conjecture (following earlier work of Ozsváth and Szabó, and Rasmussen, among others) and the cabling conjecture. Also, more recently he worked on Conway mutations. – Marco Golla Feb 17 2012 at 22:18
Kronheimer, Mrowka, Ozsváth, and Szabó obtained a new proof of Gordon and Luecke's Knot Complement Theorem using monopole Floer homology. That's pretty good, I think. They also proved that $\mathbb{RP}^3$ can't be obtained by surgery on a nontrivial knot (I don't know if there are any other proofs now).
Kronheimer, Mrowka, Ozsváth, Szabó, Monopoles and lens space surgeries. Ann. of Math. (2) 165 (2007), no. 2, 457–546.
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Yi Ni used Heegaard Floer Homology to prove, among many other things, that a knot admitting a lens-space surgery is fibred. I believe that no 'conventional' proof of this is known.
Ni, Yi, Knot Floer homology detects fibred knots. Invent. Math. 170 (2007), no. 3, 577–608.
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http://math.stackexchange.com/questions/300575/integration-involving-the-absolute-function
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# Integration Involving the Absolute Function
How do I integrate the double integral of the form $|x^2-y|$ with the boundaries $-1\leq x\leq 1$ and $-1\leq y\leq 1$?
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Separate the integral into two regions, one with $x^2>y$ and one with $x^2 \leq y$. – copper.hat Feb 11 at 21:28
## 1 Answer
You can break it into two integrals, one inside the parabola where $y \gt x^2$ and one outside where $y \lt x^2$ So $$\int_{-1}^1\int_{-1}^1 |x^2-y|\; dy \; dx=\\ \int_{-1}^1\int_{-1}^{x^2}x^2-y \;dy \; dx+\int_{-1}^1\int_{x^2}^{1}y-x^2 \;dy \; dx$$
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Sweet! Thanks for that! – user61881 Feb 11 at 22:25
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http://mathoverflow.net/questions/87923?sort=newest
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## Sequences without long arithmetic progressions
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First, a bit of notation. If we have an arithmetic progression $a, a+k, a+2k, \ldots, a+(n-1)k$ we will call $k$ the distance, and $n$ the length.
While trying to find an example for a paper I'm writing in ring theory, I was led to ask the question: Is there a sequence of 0's and 1's for which if there is an arithmetic progression in the sequence which is constantly 0 (or 1), is there a bound for the length in terms of the distance?
I found that the answer is yes, and the Thue-Morse sequence works. Modifying the ideas of Corollary 2 in "Thue-Morse at multiples of an integer" (available here), we see that the length of a constant arithmetic progression on the Thue-Morse sequence of distance $k$ is bounded by $32k^3$.
So, here are my questions for you experts.
(1) Is there an easier sequence where one can prove this is true (possibly with citation in the literature)?
(2) If not, is there a straightforward citation for this fact for the Thue-Morse sequence? (The reference I gave above works for arithmetic progressions which start near the front of the sequence. But to get an arbitrary arithmetic progression, you need to increase the bound given in the paper a little, and also give a supplementary argument.)
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## 1 Answer
How about this? Fix your favourite irrational number $\phi$. I like the golden mean. Let $x_s=[ s\phi ] - [(s-1)\phi]$ ($[t]$ means the integer part of $t$). These sequences are called Sturmian sequences.
Of course $x_s$ is 1 if and only if $s\phi \bmod 1$ lies in $[0,\phi)$.
Now for any $a$ and $k$, you're asking whether $(a+jk)\phi\bmod 1$ lies in $[0,\phi)$ for all $0\le j < n$ or lies in $[\phi,1)$ for all $0\le j < n$.
Provided $S_{k,n}:=\lbrace jk\phi\bmod 1\colon 0\le j < n\rbrace$ is $\delta$-dense in the circle, where $\delta=\min(\phi,1-\phi)$ this cannot happen.
This means you can "compute" the maximum length as a function of $k$, namely $L_{max}(k)=\max\lbrace n\colon S_{k,n}$ is not $\delta$-dense$\rbrace$. In the case of golden mean, $L_{max}(k)$ grows linearly in $k$, but I can't write down the proof of that here (the margin is too small).
Some more stuff as requested by OP
Claim: If $d(k\phi,\mathbb Z/6)=\epsilon$ then $L_{max}(k)<1/\epsilon$.
Proof: Let $k\phi=a/6+\epsilon$, where $|\epsilon|<1/12$. Notice that $\delta<1/3$. If $a=0$, then $1/\epsilon$ steps produces an $\epsilon$-dense subset of the circle. If $a=1$ or 5, then 12 steps produces a $\delta$-dense subset of the circle. If $a=3$, then $1/\epsilon$ steps produces a $2\epsilon$-dense subset of the circle [because $2k\phi$ is $2\epsilon$-close to $\mathbb Z$] and if $a=2$ or 4, then $1/\epsilon$ steps produces a $3\epsilon$-dense subset of the circle.
Next notice $d(k\phi,\mathbb Z/6)\le d(6k\phi,\mathbb Z)$. From Hardy and Wright (5th ed, Theorem 194) plus a simple argument you get $d(k\phi,\mathbb Z)\le 1/(Ak)$ for a suitable $A$.
Combining these ingredients gives the linear growth of $L_{max}$.
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If the margin is too small, how about giving me a reference? :-) – Pace Nielsen Feb 8 2012 at 19:56
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http://blog.eqnets.com/2010/02/
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# Equilibrium Networks
Science, networks, and security
## Martingales from finite Markov processes, part 1
15 February 2010
In an earlier series of posts the emerging inhomogeneous Poissonian nature of network traffic was detailed. One implication of this trend is that not only network flows but also individual packets will be increasingly well described by Markov processes of various sorts. At EQ, we use some ideas from the edifice of information theory and the renormalization group to provide a mathematical infrastructure for viewing network traffic as (e.g.) realizations of inhomogeneous finite Markov processes (or countable Markov processes with something akin to a finite universal cover). An essentially equation-free (but idea-heavy) overview of this is given in our whitepaper “Scalable visual traffic analysis”, and more details and examples will be presented over time.
The question for now is, once you’ve got a finite Markov process, what do you do with it? There are some obvious things. For example, you could apply a Chebyshev-type inequality to detect when the traffic parameters change or the underlying assumptions break down (which, if the model is halfway decent, by definition indicates something interesting is going on–even if it’s not malicious). This idea has been around in network security at least since Denning’s 1986-7 intrusion detection article, though, so it’s not likely to bear any more fruit (assuming it ever did). A better idea is to construct and exploit martingales. One way to do this to advantage starting with an inhomogeneous Poisson process (or in principle, at least, more general one-dimensional point processes) was outlined here and here.
Probably the most well-known general technique for constructing martingales from Markov processes is the Dynkin formula. Although we don’t use this formula at present (after having done a lot of tinkering and evaluation), a more general result similar to it will help us introduce the Girsanov theorem for finite Markov processes and thereby one of the tools we’ve developed for detecting changes in network traffic patterns.
The sketch below of a fairly general version of this formula for finite processes is adapted from a preprint of Ford (see Rogers and Williams IV.20 for a more sophisticated treatment).
Consider a time-inhomogeneous Markov process $X_t$ on a finite state space. Let $Q(t)$ denote the generator, and let $P(s,t)$ denote the corresponding transition kernel, i.e. $P(s,t) = U^{-1}(s)U(t),$ where the Markov propagator is
$U(t) := \mathcal{TO}^* \exp \int_0^t Q(s) \ ds$
and $\mathcal{TO}^*$ indicates the formal adjoint or reverse time-ordering operator. Thus, e.g., an initial distribution $p(0)$ is propagated as $p(t) = p(0)U(t).$ (NB. Kleinrock‘s queueing theory book omits the time-ordering, which is a no-no.)
Let $f_t(X_t)$ be bounded and such that the map $t \mapsto f_t$ is $C^1.$ Write $t_0 \equiv 0$ and $t_m = t.$ Now
$f_t(X_t)-f_0(X_0) \equiv f_{t_m}(X_{t_m})-f_{t_0}(X_{t_0})$
$= \sum_{j=0}^{m-1} \left[f_{t_{j+1}}(X_{t_{j+1}}) - f_{t_j}(X_{t_j})\right],$
and the Markov property gives that
$\mathbb{E} \left(f_{t_{j+1}}(X_{t_{j+1}}) - f_{t_j}(X_{t_j}) \ \big| \ \mathcal{F}_{t_j}\right)$
$= \sum_{X_{t_{j+1}}} \left[f_{t_{j+1}}(X_{t_{j+1}}) - f_{t_j}(X_{t_j})\right] \cdot P_{X_{t_j},X_{t_{j+1}}}(t_j,t_{j+1}).$
The notation $\mathcal{F}_t$ just indicates the history of the process (i.e., its natural filtration) at time $t.$ The transition kernel satisfies a generalization of the time-homogeneous formula $P(t) = e^{tQ}:$
$P_{X_{t_j},X_{t_{j+1}}}(t_j,t_{j+1})$
$= \delta_{X_{t_j},X_{t_{j+1}}} + (t_{j+1} - t_j) \cdot Q_{X_{t_j},X_{t_{j+1}}}(t_j) + o(t_{j+1} - t_j)$
so the RHS of the previous equation is $t_{j+1} - t_j$ times
$\frac{f_{t_{j+1}}(X_{t_j}) - f_{t_j}(X_{t_j})}{t_{j+1} - t_j} + \sum_{X_{t_{j+1}}} f_{t_{j+1}}(X_{t_{j+1}}) \cdot Q_{X_{t_j},X_{t_{j+1}}}(t_j)$
plus a term that vanishes in the limit of vanishing mesh. The fact that the row sums of a generator are identically zero has been used to simplify the result.
Summing over $j$ and taking the limit as the mesh of the the partition goes to zero shows that
$\boxed{\mathbb{E} \left(f_t(X_t)-f_0(X_0)\right) = \mathbb{E} \int_0^t \left(\partial_s + Q(s)\right)f_s \circ X_s \ ds.}$
That is,
$M_t^f := f_t(X_t)-f_0(X_0)- \int_0^t \left(\partial_s + Q(s)\right)f_s \circ X_s \ ds$
is a local martingale, or if $Q$ is well behaved, a martingale.
This can be generalized (see Rogers and Williams IV.21 and note that the extension to inhomogeneous processes is trivial): if $X$ is an inhomogeneous Markov process on a finite state space $\{1,\dots,n\}$ and $g : \mathbb{R}_+ \times \{1,\dots,n\} \times \{1,\dots,n\} \times \Omega \longrightarrow \mathbb{R}$ is such that $(t, \omega) \mapsto g(t,j,k,\omega)$ is locally bounded and previsible and $g(t,j,j,\omega) \equiv 0$ for all $j,k,$ then $M_t^g(\omega)$ given by
$\sum_{0 < s \le t} g(s,X_{s-},X_s,\omega) - \int_{(0,t]} \sum_k Q_{X_{s-},k}(s) \cdot g(s,X_{s-},k,\omega) \ ds$
is a local martingale. Conversely, any local martingale null at 0 can be represented in this form for some $g$ satisfying the conditions above (except possibly local boundedness).
To reiterate, this result will be used to help introduce the Girsanov theorem for finite Markov processes in a future post, and later on we’ll also show how Girsanov can be used to arrive at a genuinely simple, scalable likelihood ratio test for identifying changes in network traffic patterns.
1 Comment | Communications security, Equilibrium Networks, Mathematics, Networks, Nonrandom bits | Permalink
Posted by eqnets
## Random bits
12 February 2010
1 Comment | Communications security, Networks, Random bits | Permalink
Posted by eqnets
10 February 2010
Posted by eqnets
## Random bits
4 February 2010
Hacking for Fun and Profit in China’s Underworld
Google + NSA Information Assurance Directorate
“Every user in the world is convinced they need security features, not security procedures.”
Advanced Persistent Threat highlighted by DNI; Mandiant report gives details. Mandiant have coined the APT term, and it’s probably because they deal with that sort of thing constantly: they’re very good at what they do. We hired them for internal test and eval work as well as usability input as our software began taking shape, and I came away impressed. It’s not surprising to see them tackling high-profile events.
Quantum energy teleportation
1 Comment | Communications security, Equilibrium Networks, Random bits, Science | Permalink
Posted by eqnets
## Random bits
2 February 2010
“The Internet is a Hobbesian ‘state of nature’ where anything goes, where even government attacks maintain ‘plausible deniability,’ and where 80 percent of industrial control software is hooked into an IP network.”
Congressional Research Service overview of cybersecurity legislation, executive initiatives, and options (PDF)
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http://en.wikipedia.org/wiki/Saddle_points
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# Saddle point
(Redirected from Saddle points)
A saddle point on the graph of z=x2−y2 (in red)
Saddle point between two hills (the intersection of the figure-eight $z$-contour)
In mathematics, a saddle point is a point in the domain of a function that is a stationary point but not a local extremum. The name derives from the fact that the prototypical example in two dimensions is a surface that curves up in one direction, and curves down in a different direction, resembling a saddle or a mountain pass. In terms of contour lines, a saddle point in two dimensions gives rise to a contour that appears to intersect itself.
## Mathematical discussion
A simple criterion for checking if a given stationary point of a real-valued function F(x,y) of two real variables is a saddle point is to compute the function's Hessian matrix at that point: if the Hessian is indefinite, then that point is a saddle point. For example, the Hessian matrix of the function $z=x^2-y^2$ at the stationary point $(0, 0)$ is the matrix
$\begin{bmatrix} 2 & 0\\ 0 & -2 \\ \end{bmatrix}$
which is indefinite. Therefore, this point is a saddle point. This criterion gives only a sufficient condition. For example, the point $(0, 0)$ is a saddle point for the function $z=x^4-y^4,$ but the Hessian matrix of this function at the origin is the null matrix, which is not indefinite.
In the most general terms, a saddle point for a smooth function (whose graph is a curve, surface or hypersurface) is a stationary point such that the curve/surface/etc. in the neighborhood of that point is not entirely on any side of the tangent space at that point.
The plot of y = x3 with a saddle point at 0
In one dimension, a saddle point is a point which is both a stationary point and a point of inflection. Since it is a point of inflection, it is not a local extremum.
## Other uses
In dynamical systems, a saddle point is a periodic point whose stable and unstable manifolds have a dimension that is not zero. If the dynamic is given by a differentiable map f then a point is hyperbolic if and only if the differential of ƒ n (where n is the period of the point) has no eigenvalue on the (complex) unit circle when computed at the point.
In a two-player zero sum game defined on a continuous space, the equilibrium point is a saddle point.
A saddle point is an element of the matrix which is both the largest element in its column and the smallest element in its row.
For a second-order linear autonomous systems, a critical point is a saddle point if the characteristic equation has one positive and one negative real eigenvalue.[1]
## References
• Gray, Lawrence F.; Flanigan, Francis J.; Kazdan, Jerry L.; Frank, David H; Fristedt, Bert (1990), Calculus two: linear and nonlinear functions, Berlin: Springer-Verlag, pp. page 375, ISBN 0-387-97388-5
• Hilbert, David; Cohn-Vossen, Stephan (1952), Geometry and the Imagination (2nd ed.), New York: Chelsea, ISBN 978-0-8284-1087-8
• von Petersdorff, Tobias (2006), "Critical Points of Autonomous Systems", Differential Equations for Scientists and Engineers (Math 246 lecture notes) []
• Widder, D. V. (1989), Advanced calculus, New York: Dover Publications, pp. page 128, ISBN 0-486-66103-2
•
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http://www.gap-system.org/Manuals/doc/ref/chap44_mj.html
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[MathJax off]
### 44 Matrix Groups
Matrix groups are groups generated by invertible square matrices.
In the following example we temporarily increase the line length limit from its default value 80 to 83 in order to get a nicer output format.
```gap> m1 := [ [ Z(3)^0, Z(3)^0, Z(3) ],
> [ Z(3), 0*Z(3), Z(3) ],
> [ 0*Z(3), Z(3), 0*Z(3) ] ];;
gap> m2 := [ [ Z(3), Z(3), Z(3)^0 ],
> [ Z(3), 0*Z(3), Z(3) ],
> [ Z(3)^0, 0*Z(3), Z(3) ] ];;
gap> m := Group( m1, m2 );
Group(
[
[ [ Z(3)^0, Z(3)^0, Z(3) ], [ Z(3), 0*Z(3), Z(3) ],
[ 0*Z(3), Z(3), 0*Z(3) ] ],
[ [ Z(3), Z(3), Z(3)^0 ], [ Z(3), 0*Z(3), Z(3) ],
[ Z(3)^0, 0*Z(3), Z(3) ] ] ])
```
#### 44.1 IsMatrixGroup (Filter)
For most operations, GAP only provides methods for finite matrix groups. Many calculations in finite matrix groups are done via so-called "nice monomorphisms" (see Section 40.5) that represent a faithful action on vectors.
##### 44.1-1 IsMatrixGroup
`‣ IsMatrixGroup`( grp ) ( category )
The category of matrix groups.
#### 44.2 Attributes and Properties for Matrix Groups
##### 44.2-1 DimensionOfMatrixGroup
`‣ DimensionOfMatrixGroup`( mat-grp ) ( attribute )
The dimension of the matrix group.
##### 44.2-2 DefaultFieldOfMatrixGroup
`‣ DefaultFieldOfMatrixGroup`( mat-grp ) ( attribute )
Is a field containing all the matrix entries. It is not guaranteed to be the smallest field with this property.
##### 44.2-3 FieldOfMatrixGroup
`‣ FieldOfMatrixGroup`( matgrp ) ( attribute )
The smallest field containing all the matrix entries of all elements of the matrix group matgrp. As the calculation of this can be hard, this should only be used if one really needs the smallest field, use `DefaultFieldOfMatrixGroup` (44.2-2) to get (for example) the characteristic.
```gap> DimensionOfMatrixGroup(m);
3
gap> DefaultFieldOfMatrixGroup(m);
GF(3)
```
##### 44.2-4 TransposedMatrixGroup
`‣ TransposedMatrixGroup`( matgrp ) ( attribute )
returns the transpose of the matrix group matgrp. The transpose of the transpose of matgrp is identical to matgrp.
```gap> G := Group( [[0,-1],[1,0]] );
Group([ [ [ 0, -1 ], [ 1, 0 ] ] ])
gap> T := TransposedMatrixGroup( G );
Group([ [ [ 0, 1 ], [ -1, 0 ] ] ])
gap> IsIdenticalObj( G, TransposedMatrixGroup( T ) );
true
```
##### 44.2-5 IsFFEMatrixGroup
`‣ IsFFEMatrixGroup`( G ) ( category )
tests whether all matrices in G have finite field element entries.
#### 44.3 Actions of Matrix Groups
The basic operations for groups are described in Chapter 41, special actions for matrix groups mentioned there are `OnLines` (41.2-12), `OnRight` (41.2-2), and `OnSubspacesByCanonicalBasis` (41.2-15).
For subtleties concerning multiplication from the left or from the right, see 44.7.
##### 44.3-1 ProjectiveActionOnFullSpace
`‣ ProjectiveActionOnFullSpace`( G, F, n ) ( function )
Let G be a group of n by n matrices over a field contained in the finite field F. `ProjectiveActionOnFullSpace` returns the image of the projective action of G on the full row space $$\textit{F}^{\textit{n}}$$.
##### 44.3-2 ProjectiveActionHomomorphismMatrixGroup
`‣ ProjectiveActionHomomorphismMatrixGroup`( G ) ( function )
returns an action homomorphism for a faithful projective action of G on the underlying vector space. (Note: The action is not necessarily on the full space, if a smaller subset can be found on which the action is faithful.)
##### 44.3-3 BlowUpIsomorphism
`‣ BlowUpIsomorphism`( matgrp, B ) ( function )
For a matrix group matgrp and a basis B of a field extension $$L / K$$, say, such that the entries of all matrices in matgrp lie in $$L$$, `BlowUpIsomorphism` returns the isomorphism with source matgrp that is defined by mapping the matrix $$A$$ to `BlownUpMat`$$( A, \textit{B} )$$, see `BlownUpMat` (24.13-3).
```gap> g:= GL(2,4);;
gap> B:= CanonicalBasis( GF(4) );; BasisVectors( B );
[ Z(2)^0, Z(2^2) ]
gap> iso:= BlowUpIsomorphism( g, B );;
gap> Display( Image( iso, [ [ Z(4), Z(2) ], [ 0*Z(2), Z(4)^2 ] ] ) );
. 1 1 .
1 1 . 1
. . 1 1
. . 1 .
gap> img:= Image( iso, g );
<matrix group with 2 generators>
gap> Index( GL(4,2), img );
112
```
#### 44.4 GL and SL
(See also section 50.2.)
##### 44.4-1 IsGeneralLinearGroup
`‣ IsGeneralLinearGroup`( grp ) ( property )
`‣ IsGL`( grp ) ( property )
The General Linear group is the group of all invertible matrices over a ring. This property tests, whether a group is isomorphic to a General Linear group. (Note that currently only a few trivial methods are available for this operation. We hope to improve this in the future.)
##### 44.4-2 IsNaturalGL
`‣ IsNaturalGL`( matgrp ) ( property )
This property tests, whether a matrix group is the General Linear group in the right dimension over the (smallest) ring which contains all entries of its elements. (Currently, only a trivial test that computes the order of the group is available.)
##### 44.4-3 IsSpecialLinearGroup
`‣ IsSpecialLinearGroup`( grp ) ( property )
`‣ IsSL`( grp ) ( property )
The Special Linear group is the group of all invertible matrices over a ring, whose determinant is equal to 1. This property tests, whether a group is isomorphic to a Special Linear group. (Note that currently only a few trivial methods are available for this operation. We hope to improve this in the future.)
##### 44.4-4 IsNaturalSL
`‣ IsNaturalSL`( matgrp ) ( property )
This property tests, whether a matrix group is the Special Linear group in the right dimension over the (smallest) ring which contains all entries of its elements. (Currently, only a trivial test that computes the order of the group is available.)
```gap> IsNaturalGL(m);
false
```
##### 44.4-5 IsSubgroupSL
`‣ IsSubgroupSL`( matgrp ) ( property )
This property tests, whether a matrix group is a subgroup of the Special Linear group in the right dimension over the (smallest) ring which contains all entries of its elements.
#### 44.5 Invariant Forms
##### 44.5-1 InvariantBilinearForm
`‣ InvariantBilinearForm`( matgrp ) ( attribute )
This attribute describes a bilinear form that is invariant under the matrix group matgrp. The form is given by a record with the component `matrix` which is a matrix $$F$$ such that for every generator $$g$$ of matgrp the equation $$g \cdot F \cdot g^{tr} = F$$ holds.
##### 44.5-2 IsFullSubgroupGLorSLRespectingBilinearForm
`‣ IsFullSubgroupGLorSLRespectingBilinearForm`( matgrp ) ( property )
This property tests, whether a matrix group matgrp is the full subgroup of GL or SL (the property `IsSubgroupSL` (44.4-5) determines which it is) respecting the form stored as the value of `InvariantBilinearForm` (44.5-1) for matgrp.
##### 44.5-3 InvariantSesquilinearForm
`‣ InvariantSesquilinearForm`( matgrp ) ( attribute )
This attribute describes a sesquilinear form that is invariant under the matrix group matgrp over the field $$F$$ with $$q^2$$ elements, say. The form is given by a record with the component `matrix` which is is a matrix $$M$$ such that for every generator $$g$$ of matgrp the equation $$g \cdot M \cdot (g^{tr})^f = M$$ holds, where $$f$$ is the automorphism of $$F$$ that raises each element to its $$q$$-th power. ($$f$$ can be obtained as a power of the `FrobeniusAutomorphism` (59.4-1) value of $$F$$.)
##### 44.5-4 IsFullSubgroupGLorSLRespectingSesquilinearForm
`‣ IsFullSubgroupGLorSLRespectingSesquilinearForm`( matgrp ) ( property )
This property tests, whether a matrix group matgrp is the full subgroup of GL or SL (the property `IsSubgroupSL` (44.4-5) determines which it is) respecting the form stored as the value of `InvariantSesquilinearForm` (44.5-3) for matgrp.
##### 44.5-5 InvariantQuadraticForm
`‣ InvariantQuadraticForm`( matgrp ) ( attribute )
For a matrix group matgrp, `InvariantQuadraticForm` returns a record containing at least the component `matrix` whose value is a matrix $$Q$$. The quadratic form $$q$$ on the natural vector space $$V$$ on which matgrp acts is given by $$q(v) = v Q v^{tr}$$, and the invariance under matgrp is given by the equation $$q(v) = q(v M)$$ for all $$v \in V$$ and $$M$$ in matgrp. (Note that the invariance of $$q$$ does not imply that the matrix $$Q$$ is invariant under matgrp.)
$$q$$ is defined relative to an invariant symmetric bilinear form $$f$$ (see `InvariantBilinearForm` (44.5-1)), via the equation $$q(\lambda x + \mu y) = \lambda^2 q(x) + \lambda \mu f(x,y) + \mu^2 q(y)$$, see [CCNPW85, Chapter 3.4]. If $$f$$ is represented by the matrix $$F$$ then this implies $$F = Q + Q^{tr}$$. In characteristic different from $$2$$, we have $$q(x) = f(x,x)/2$$, so $$Q$$ can be chosen as the strictly upper triangular part of $$F$$ plus half of the diagonal part of $$F$$. In characteristic $$2$$, $$F$$ does not determine $$Q$$ but still $$Q$$ can be chosen as an upper (or lower) triangular matrix.
Whenever the `InvariantQuadraticForm` value is set in a matrix group then also the `InvariantBilinearForm` (44.5-1) value can be accessed, and the two values are compatible in the above sense.
##### 44.5-6 IsFullSubgroupGLorSLRespectingQuadraticForm
`‣ IsFullSubgroupGLorSLRespectingQuadraticForm`( matgrp ) ( property )
This property tests, whether the matrix group matgrp is the full subgroup of GL or SL (the property `IsSubgroupSL` (44.4-5) determines which it is) respecting the `InvariantQuadraticForm` (44.5-5) value of matgrp.
```gap> g:= Sp( 2, 3 );;
gap> m:= InvariantBilinearForm( g ).matrix;
[ [ 0*Z(3), Z(3)^0 ], [ Z(3), 0*Z(3) ] ]
gap> [ 0, 1 ] * m * [ 1, -1 ]; # evaluate the bilinear form
Z(3)
gap> IsFullSubgroupGLorSLRespectingBilinearForm( g );
true
gap> g:= SU( 2, 4 );;
gap> m:= InvariantSesquilinearForm( g ).matrix;
[ [ 0*Z(2), Z(2)^0 ], [ Z(2)^0, 0*Z(2) ] ]
gap> [ 0, 1 ] * m * [ 1, 1 ]; # evaluate the bilinear form
Z(2)^0
gap> IsFullSubgroupGLorSLRespectingSesquilinearForm( g );
true
gap> g:= GO( 1, 2, 3 );;
gap> m:= InvariantBilinearForm( g ).matrix;
[ [ 0*Z(3), Z(3)^0 ], [ Z(3)^0, 0*Z(3) ] ]
gap> [ 0, 1 ] * m * [ 1, 1 ]; # evaluate the bilinear form
Z(3)^0
gap> q:= InvariantQuadraticForm( g ).matrix;
[ [ 0*Z(3), Z(3)^0 ], [ 0*Z(3), 0*Z(3) ] ]
gap> [ 0, 1 ] * q * [ 0, 1 ]; # evaluate the quadratic form
0*Z(3)
gap> IsFullSubgroupGLorSLRespectingQuadraticForm( g );
true
```
#### 44.6 Matrix Groups in Characteristic 0
Most of the functions described in this and the following section have implementations which use functions from the GAP package Carat. If Carat is not installed or not compiled, no suitable methods are available.
##### 44.6-1 IsCyclotomicMatrixGroup
`‣ IsCyclotomicMatrixGroup`( G ) ( category )
tests whether all matrices in G have cyclotomic entries.
##### 44.6-2 IsRationalMatrixGroup
`‣ IsRationalMatrixGroup`( G ) ( property )
tests whether all matrices in G have rational entries.
##### 44.6-3 IsIntegerMatrixGroup
`‣ IsIntegerMatrixGroup`( G ) ( property )
tests whether all matrices in G have integer entries.
##### 44.6-4 IsNaturalGLnZ
`‣ IsNaturalGLnZ`( G ) ( property )
tests whether G is $$GL_n(ℤ)$$ in its natural representation by $$n \times n$$ integer matrices. (The dimension $$n$$ will be read off the generating matrices.)
```gap> IsNaturalGLnZ( GL( 2, Integers ) );
true
```
##### 44.6-5 IsNaturalSLnZ
`‣ IsNaturalSLnZ`( G ) ( property )
tests whether G is $$SL_n(ℤ)$$ in its natural representation by $$n \times n$$ integer matrices. (The dimension $$n$$ will be read off the generating matrices.)
```gap> IsNaturalSLnZ( SL( 2, Integers ) );
true
```
##### 44.6-6 InvariantLattice
`‣ InvariantLattice`( G ) ( attribute )
returns a matrix $$B$$, whose rows form a basis of a $$ℤ$$-lattice that is invariant under the rational matrix group G acting from the right. It returns `fail` if the group is not unimodular. The columns of the inverse of $$B$$ span a $$ℤ$$-lattice invariant under G acting from the left.
##### 44.6-7 NormalizerInGLnZ
`‣ NormalizerInGLnZ`( G ) ( attribute )
is an attribute used to store the normalizer of G in $$GL_n(ℤ)$$, where G is an integer matrix group of dimension n. This attribute is used by `Normalizer( GL( n, Integers ), G )`.
##### 44.6-8 CentralizerInGLnZ
`‣ CentralizerInGLnZ`( G ) ( attribute )
is an attribute used to store the centralizer of G in $$GL_n(ℤ)$$, where G is an integer matrix group of dimension n. This attribute is used by `Centralizer( GL( n, Integers ), G )`.
##### 44.6-9 ZClassRepsQClass
`‣ ZClassRepsQClass`( G ) ( attribute )
The conjugacy class in $$GL_n(ℚ)$$ of the finite integer matrix group G splits into finitely many conjugacy classes in $$GL_n(ℤ)$$. `ZClassRepsQClass( G )` returns representative groups for these.
##### 44.6-10 IsBravaisGroup
`‣ IsBravaisGroup`( G ) ( property )
test whether G coincides with its Bravais group (see `BravaisGroup` (44.6-11)).
##### 44.6-11 BravaisGroup
`‣ BravaisGroup`( G ) ( attribute )
returns the Bravais group of a finite integer matrix group G. If $$C$$ is the cone of positive definite quadratic forms $$Q$$ invariant under $$g \mapsto g Q g^{tr}$$ for all $$g \in \textit{G}$$, then the Bravais group of G is the maximal subgroup of $$GL_n(ℤ)$$ leaving the forms in that same cone invariant. Alternatively, the Bravais group of G can also be defined with respect to the action $$g \mapsto g^{tr} Q g$$ on positive definite quadratic forms $$Q$$. This latter definition is appropriate for groups G acting from the right on row vectors, whereas the former definition is appropriate for groups acting from the left on column vectors. Both definitions yield the same Bravais group.
##### 44.6-12 BravaisSubgroups
`‣ BravaisSubgroups`( G ) ( attribute )
returns the subgroups of the Bravais group of G, which are themselves Bravais groups.
##### 44.6-13 BravaisSupergroups
`‣ BravaisSupergroups`( G ) ( attribute )
returns the subgroups of $$GL_n(ℤ)$$ that contain the Bravais group of G and are Bravais groups themselves.
##### 44.6-14 NormalizerInGLnZBravaisGroup
`‣ NormalizerInGLnZBravaisGroup`( G ) ( attribute )
returns the normalizer of the Bravais group of G in the appropriate $$GL_n(ℤ)$$.
#### 44.7 Acting OnRight and OnLeft
In GAP, matrices by convention act on row vectors from the right, whereas in crystallography the convention is to act on column vectors from the left. The definition of certain algebraic objects important in crystallography implicitly depends on which action is assumed. This holds true in particular for quadratic forms invariant under a matrix group. In a similar way, the representation of affine crystallographic groups, as they are provided by the GAP package CrystGap, depends on which action is assumed. Crystallographers are used to the action from the left, whereas the action from the right is the natural one for GAP. For this reason, a number of functions which are important in crystallography, and whose result depends on which action is assumed, are provided in two versions, one for the usual action from the right, and one for the crystallographic action from the left.
For every such function, this fact is explicitly mentioned. The naming scheme is as follows: If `SomeThing` is such a function, there will be functions `SomeThingOnRight` and `SomeThingOnLeft`, assuming action from the right and from the left, respectively. In addition, there is a generic function `SomeThing`, which returns either the result of `SomeThingOnRight` or `SomeThingOnLeft`, depending on the global variable `CrystGroupDefaultAction` (44.7-1).
##### 44.7-1 CrystGroupDefaultAction
`‣ CrystGroupDefaultAction` ( global variable )
can have either of the two values `RightAction` and `LeftAction`. The initial value is `RightAction`. For functions which have variants OnRight and OnLeft, this variable determines which variant is returned by the generic form. The value of `CrystGroupDefaultAction` can be changed with with the function `SetCrystGroupDefaultAction` (44.7-2).
##### 44.7-2 SetCrystGroupDefaultAction
`‣ SetCrystGroupDefaultAction`( action ) ( function )
allows one to set the value of the global variable `CrystGroupDefaultAction` (44.7-1). Only the arguments `RightAction` and `LeftAction` are allowed. Initially, the value of `CrystGroupDefaultAction` (44.7-1) is `RightAction`.
generated by GAPDoc2HTML
|
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|
http://physics.stackexchange.com/questions/tagged/mass+equivalence-principle
|
# Tagged Questions
2answers
163 views
### Why equivalence principle is principle and not law?
We can prove that the inertial mass and the gravitational mass should be the same (equivalence principle) from the $f=mg=ma$ then $g=a$, so we have equivalence law! But why we said equivalence ...
5answers
959 views
### Why Gravity attracts all objects with the same speed?
Why Gravity attracts all objects with the same speed? Is this question was solved? What is the exact answer?
1answer
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### Are there any well-known theories successfully unifying the inertial and gravitational mass?
From what little I know of general relativity, the equality of inertial and gravitational mass is an axiom of the theory. I suspect that this precludes GR from unifying them in the same sense as ...
2answers
126 views
### Inertial Mass of a scalar field
Does it make sense to talk of the inertial mass of a scalar field? By the equivalence principle, it must be equal to its gravitational mass. We know that the scalar field contributes towards the ...
1answer
558 views
### How do Einstein’s equations support mass gain in particle accelerators? [duplicate]
Possible Duplicate: Why does the mass of an object increase when its speed approaches that of light? A charged particle that is accelerated through a particle accelerator like CERN ...
0answers
83 views
### Masses of all the particles in the Standard Theory [duplicate]
Possible Duplicate: If photons have no mass, how can they have momentum? I'm sure this question has been asked here before but I wasn't able to find it clearly answered in one q/a session. ...
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|
http://www.physicsforums.com/showthread.php?p=4247166
|
Physics Forums
## The vector nature of Angular Momentum
Here is an animation from Wikipedia : http://en.wikipedia.org/wiki/File:Torque_animation.gif
The angular momentum is given by the Cross product of r and p
We can see that the direction would be perpendicular to the direction of rotation of the particle (as shown in the animation)
I don't think this really makes sense, how is the vector nature of angular momentum justified ?
How can one get an intuitive sense about the direction of angular momentum ?
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
I think that direction of angular momentum is only a convention. Someone thought of that rule, it seemed practical since intensity of cross product of r and p really defines its amplitude. I am not sure, but think you could define another rule and get same physical results. Such thing is with phasor diagrams, you choose phase of one quantity to be zero, and according to that do everything else. This is only my opinion.
Having angular momentum as a vector comes in handy when you want to explain gyroscopic precession. Maybe someone with more knowledge than me can give better examples :)
## The vector nature of Angular Momentum
The vector comes into play when you're trying to solve for conservation of angular momentum. The vectors must all add to zero.
Quote by hms.tech We can see that the direction would be perpendicular to the direction of rotation of the particle (as shown in the animation)
It is parallel to the axis of rotation
Quote by hms.tech how is the vector nature of angular momentum justified ?
You need to express magnitude and axis of rotation.
Quote by hms.tech Here is an animation from Wikipedia : http://en.wikipedia.org/wiki/File:Torque_animation.gif The angular momentum is given by the Cross product of r and p We can see that the direction would be perpendicular to the direction of rotation of the particle (as shown in the animation) I don't think this really makes sense, how is the vector nature of angular momentum justified ? How can one get an intuitive sense about the direction of angular momentum ?
You use the right hand rule. For A x B, point your four fingers along A, then rotate your hand until they point along B. Your thumb sticking up tells you the direction of the cross product. The cross product of A and B is always perpendicular to both, just like your thumb sticking up is perpendicular to your four fingers.
If you want to think about it some more, you might ask, why the right hand rule, why not a left hand rule? You could use either, and everything would still make sense. That's because the angular momentum vector is not truly a vector, it is a "pseudovector", one that depends on which hand you use. Well, the laws of physics don't depend on which hand you use, and true vectors don't depend on which hand you use, so true vectors are, in a sense, more "real" than pseudovectors.
For calculation purposes, pseudovectors are nice, just three components that transform almost like a vector. But when you want to do theoretical work, you might not want to deal with the artificiality of pseudovectors. The bottom line is that pseudovectors are better represented by antisymmetric 3x3 matrices (antisymmetric tensors). Instead of a pseudovector [x,y,z] you use $$\left[\begin{matrix} 0 & z & -y \\-z & 0 & x \\ y & -x & 0 \end{matrix}\right]$$ This tensor transforms the same way no matter what, no worry about which hand you need to use, and its better for theoretical work, its the "real thing", unlike the more concise pseudotensor.
Quote by hms.tech Here is an animation from Wikipedia : http://en.wikipedia.org/wiki/File:Torque_animation.gif The angular momentum is given by the Cross product of r and p We can see that the direction would be perpendicular to the direction of rotation of the particle (as shown in the animation) I don't think this really makes sense, how is the vector nature of angular momentum justified ? How can one get an intuitive sense about the direction of angular momentum ?
It's a convenient way of describing torque that happens to work well with the rest of mathematics. Nothing much deeper than that I think.
I guess a good thing to think about is how else you would do it?
angular momentum and torque while having vectorial nature are pseudovectors i.e. under an inversion they don't get reflected. L=r×p replacing r by -r ,gives inversion.In p=m dr/dt,it becomes -.so overall L does not change direction. edit:oops,someone else also written it.
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Quote by hms.tech I don't think this really makes sense, how is the vector nature of angular momentum justified ?
The effect of a rotating object's angular momentum and the way it will interact with other objects and forces depends just as much on the direction of its axis of rotation as its MI and rate of rotation - so it can only be described fully using a vector. This is the same as for linear momentum (also a vector), where the mass, direction of travel and speed (velocity) are needed for full description.
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http://physics.stackexchange.com/questions/16018/does-a-photon-have-a-rest-frame
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# Does a photon have a rest frame?
Quite a few of the questions given on this site mention a photon having a rest frame such as it having a zero mass in its rest frame. I find this contradictory since photons must travel at the seed of light in all frames according to special relativity
Does a photon have a rest frame?
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Not in vacuum, but the question makes sense in a transparent medium. Experts, what does an observer see in a moving RF in a medium if its velocity is c/n? – Vladimir Kalitvianski Oct 21 '11 at 18:12
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I understand that in a medium you are dealing with a wave packet of photons, and not a single photon. In the lab frame, each photon in the packet moves at the speed c, but the packet's group speed is c/n. If you shot a pulse of light into a medium and then followed it at the speed c/n, you would just see a Lorentz-transformed packet of photons. Each photon will still move at the speed c, but the group speed of the packet would be zero. And no, a photon does not have a rest frame, if special relativity applies. – drlemon Oct 21 '11 at 19:40
A one-photon propagation mode is also possible in a medium, hence a photon is a wave train of a finite length, no? – Vladimir Kalitvianski Oct 21 '11 at 20:43
@drlemon: your comment is not right--- there is no packet--- it happens photon by photon. – Ron Maimon Oct 21 '11 at 20:52
@Ron and Vladimir: In E&M textbooks, the speed of light in a medium is derived for a monochromatic wave. The physical reasoning for slowing down the speed of light in a medium is that atoms polarize and produce their own E/M fields, interfering with the incident wave. The total E/M field is that of a monochromatic wave traveling at the speed c/n. There are no individual photons in this analysis. An individual photon will be scattered by atoms back and forth, and will, in general, have some stochastic transport. On average it will travel at c/n, but it will not just slow down to c/n. – drlemon Oct 21 '11 at 21:39
show 5 more comments
## 3 Answers
Short answer: no.
Explanation:
Many introductory text books talk about "rest mass" and "relativistic mass" and say that the "rest mass" is the mass measured in the particles rest frame.
That's not wrong, you can do physics in that point of view, but that is not how people talk about and define mass anymore.
In the modern view each particle has one and only one mass defined by the square of it's energy--momentum four vector (which being a Lorentz invariant you can calculate in any inertial frame): $$m \equiv p^2 = (E, \vec{p})^2 = E^2 - \vec{p}^2$$
For a photon this value is zero. In any frame, and that allows people to reasonably say that the photon has zero mass without needing to define a rest frame for it.
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I agree completely with @dmckee and would only add that for any particle the elapsed time experienced by that particle in it's rest frame is called the proper time and can be calculated (in units where $c=1$) by any observer as $$d\tau^2 = dt^2 - d\vec{x}^2$$ and for a photon in a vacuum the proper time is always identically $0$. So photons do not experience any passage of time so in that sense also, they do not have a rest frame. – FrankH Oct 21 '11 at 19:58
And in QM the photon energy is $\hbar\omega$ and $\omega$ in a medium is the same, so $m_{photon}=0$. – Vladimir Kalitvianski Oct 21 '11 at 20:41
Your answers are right,a solitary photon has no rest frame, nonetheless I find quite interesting to note that a system of massless particles(such as photons) can have a nonzero mass provided that all the momenta are not oriented in the same axis and that for such systems zero momentum frames CAN actually be defined.
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Not at all. Rest frame is a concept that does not exist in nature. Had it would exist, nature wouldn't be causal. A photon propagating through medium does not 'move' in a speed smaller than the speed of light in vacuum. It simply interacts electromagnetically with the medium and these interactions slows down its propagation through the medium.
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"Rest frame is a concept that does not exist in nature." That's a strange way of stating things. If (in SR) in some frame $L$ you observe a (massive) particle moving at a speed $v < c$, you can most definitely pass to some frame $L'$ in which the particular doesn't move. – Gerben Oct 22 '11 at 11:21
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http://math.stackexchange.com/questions/174145/what-is-the-relation-between-the-representation-of-a-hecke-algebra-and-the-repre
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What is the relation between the representation of a Hecke algebra and the representation of the Coxeter group?
Given a Coxeter group $W$, there is a corresponding Hecke algebra (Iwahori-Hecke algebra). There are many results on the representation of the Hecke algebra. But why is this motivated?
How is the representation of the Hecke algebra related to the representation of the Coxeter group? How is the representation of the Hecke algebra related to the represenation of an algebraic group when the Coxeter group $W$ is the Weyl group of this algebraic group? Or, why are we studying Hecke algebras?
Thanks a lot.
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There is no simple short answer to your question. Connection with rep. of Coxeter group: Let $\mathcal{H}_q(W)$ be the Hecke algebra corresponding to $W$, this is a deformation of the group algebra of $W$, i.e. when $q=1$, we get the group algebrw of $W$. Connection to algebraic group, this comes from connection via corresponding Lie algebra's universal envelop, $U_q(\mathfrak{g})$. In the type A case, this is connected through Schur-Weyl duality. I am not expert in this aspect perhaps someone else can give a more elaborate answer. – Aaron Jul 23 '12 at 17:40
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Kazhdan-Lusztig polynomial is the keyword to look for if you want to some connection between universal enveloping algebra/algebraic group. – Aaron Jul 23 '12 at 17:47
@Aaron: Thank you very much. – ShinyaSakai Aug 8 '12 at 11:33
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http://math.stackexchange.com/questions/44014/rsa-is-it-easy-to-find-the-public-key-from-the-secret-key/44017
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# RSA: is it easy to find the public key from the secret key?
I know that it is not easy to find the secret key from the public key . Is it relatively easy to find the public key from the secret key for scientists in the field?
If the answer is yes then is there any cryptography system such that only a particular person can encode and any one in public can decode? For an example scenario , a king writes his order and encodes it using his (secret) public key . Then his soldiers decode the code using the (open) secret key to check if it was truly written by the king.
Thank you very much.
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don't forget to accept an answer if you find that it adequately answers your question – AnkurVijay Jun 8 '11 at 6:54
## 5 Answers
For standard uses of RSA, the secret key is hard to find, and the public key is trivial. This is because most implementations use 65537 as the public key (the reason has to do with its primality and the efficiency of repeated squaring for $e=65537=2^{16}+1$).
Many systems will rechoose $p$ and $q$ if $\phi(pq)$ is not coprime to 65537, meaning that the public exponent is never anything else.
Several other answers note that $e$ and $d$ are interchangeable, but this ignores both practice and several important cryptanalytic results. In particular, even if we allow $e$ to vary, it can still be found given $d$ if it's sufficiently small. This is due to the fact that Coppersmith's algorithm can find small roots of polynomials modulo a composite of unknown factorization using LLL and some added cleverness. The bound keeps increasing, but (from memory) I believe if $e < n^{.33}$ it can be found from $n$ and $d$ (here $n=pq$). So saying $e$ and $d$ are equivalent is false (and dangerous). There is a conjecture that the "true" bound is $n^{.5}$, meaning that for security the secret exponent would have to be above $\sqrt{n}$.
Note: For those truly interested, Boneh has a nice survey of these results (and many others). And it's quite readable by non-cryptographers who have a moderately strong math background.
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"Many systems" will rechoose $p$ and $q$? I would hope all systems, otherwise nothing works at all! – TonyK Jun 8 '11 at 7:47
I'm uncertain about the exponents of $n$, do you mean $\sqrt[3]{n}$ and $\sqrt{n}$? If so, you can use `\sqrt[k]{n}` for the $k$-th root on $n$. – Asaf Karagila Jun 8 '11 at 12:57
@TonyK Some systems will try other $e$-values beside 65537 since this is faster than having to start all over with new $p$ and $q$ (whose generation is the slowest part of making RSA keys), but most systems will stick with $e=65537$ and just regenerate $p$ and $q$; these days even commodity computers are so fast you wouldn't notice the time-loss for even a 2048-bit modulus. – Fixee Jun 9 '11 at 2:07
@Asaf: In the survey I linked in my answer, Boneh shows that if a known exponent is $< n^{0.292}$, you can efficiently obtain the other exponent. A more recent result lifts this bound to $n^{.33}$ (which is close to, but not exactly, $\sqrt[3]{n}$ of course). Boneh notes that the likely correct bound is $n^{0.5}$ but that it was still open at the time of the survey. AFAIK, this remains open. – Fixee Jun 9 '11 at 2:15
Thank you very much for the answer. – seven_swodniw Jun 13 '11 at 18:24
No, finding the secret key from the public key is just as hard, in the abstract, as finding the secret key from the public key.
What you suggest is in fact the standard way in which RSA is used for "signing" messages. In order to authenticate a message (to make sure the message was sent by me), I encode it using my private key, which means that anyone can use the public key to decode it and find out that it was truly sent by me (because, presumably, nobody knows my private key except me).
But this has nothing to do with whether one can obtain the public key from the secret key; it has to do with the way RSA works (the roles of the public and private key are symmetric).
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Thank you for the answer. – seven_swodniw Jun 8 '11 at 5:34
No - they are equally difficult to find. In particular, the encryption key and the decryption key are modular inverses. The difficulty in finding the decryption key from the public key lies in the fact that they're inverses... mod $\varphi (n)$, and so one must factor or be incredibly witty to find the inverse.
If one chose backwards, and distributed the 'decryption' key instead and kept the 'encryption' key for oneself, the security would be the exact same. (I put them in quotes to show that they were the original decryption and encryption keys).
But to your second question, there is a method similar to RSA called digital signatures, where only the person who knows the secret key can distribute a message that everyone in public can decode. It's called a signature because it guarantees authenticity
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Thank you for the answer. – seven_swodniw Jun 8 '11 at 5:34
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@seven: no problem! (but we love upvotes and accepted answers, too - makes us feel special) – mixedmath♦ Jun 8 '11 at 5:35
It depends what you mean by "private key". If the private and public keys are $(n,d)$ and $(n,e)$, with $ed=1 \mod \varphi(n)$, then the other answers are correct - it's just as hard to get $d$ from $e$ as $e$ from $d$. But the key was generated as $n=pq$ for two primes $p$ and $q$, and in practice the numbers $p,q$ are saved along with the private key. (This allows much faster computations using the Chinese Remainder Theorem.) And if you know $p$ and $q$, then you can calculate $d$ from $e$ (or $e$ from $d$) easily -- it's just a modular inverse.
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In fact, in the standards that are used for computers (like PKCS7, 8 etc) the public key is (essentially) $(n,e)$ while the private key (for RSA) is, besides the also allowed $(n,d)$, (essentially) $(n,e,d,p,q, d \mod{p-1}, d \mod{q-1}, q^{-1} \mod {p})$, a very redundant set, but as said, convenient for computations. The private key info should be stored encrypted anyway, and a few bytes more... – Henno Brandsma Jun 8 '11 at 7:19
+1 Yes, exactly. Typically the primes p and q are stored in the secret private key file. The public key (n,e) is trivial to calculate from that: n = pq, and e is usually fixed at 65537. – David Cary Jun 12 '11 at 15:35
Even if you don't know p or q, if you have the private key (n,d), the public key (n, e) is usually (n, 65537). Even when e is not 65537, e is almost always a small number (compared to d), so it almost always takes a short amount of time to find e by brute force testing all possible values. – David Cary Jun 12 '11 at 15:47
Thank you very much for the answer. – seven_swodniw Jun 13 '11 at 18:24
No, the two are symmetrical (or interchangeable). So if it is hard to get the secret key from the public one, then it is equally hard to get the public from the private.
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Thank you for the answer. – seven_swodniw Jun 8 '11 at 5:34
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http://www.quantitativepoker.com/2011/01/utility-part-1-basics.html
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# Quantitative Poker
math beyond the odds
## Saturday, January 22, 2011
### Utility, Part 1: The Basics
Before we can dive into any models of various poker decisions, we first need to establish the building blocks of models for rational preferences under uncertainty.
In economics, the foundation of any approach to any decision made under uncertainty is expected utility theory, which quantifies risk aversion through establishing a correspondence with diminishing marginal utility of wealth.
Those familiar with the basics of utility can skip to the end of this post, but should stay tuned for the upcoming parts, where I will build upon these fundamentals in ways which are less common in abstract theoretical models, but which are specifically well-suited for practical poker applications.
Motivation
The common heuristic approach to decision-making in poker is to make decisions in order to maximize one's expected value, with volatility ("variance", as it is usually not-quite-accurately termed) an unquantified afterthought, managed through heuristic rules, if at all. People understand that less risk is preferable to more risk as long as expected value remains the same, but there is usually little consideration given to quantifying the value of risk relative to expected value.
How much expected value should a decision-maker be willing to give up in order to reduce the variance of a random payoff by a certain amount? More generally, how do rational decision-makers value the tradeoff between expectation and risk?
General Utility Functions
Preferences over different levels of wealth are quantified by assigning a utility function to each person or entity, a function which maps a level of wealth to a level of overall personal satisfaction derived from that wealth. The usual assumptions on a utility function are that it is:
• Increasing — Everyone prefers more money to less money.
• Continuous — There's no specific amount of wealth that is suddenly much more preferable to a slightly smaller amount of wealth.
• Concave — The slope of the function is decreasing. As one has more wealth, an additional dollar is less valuable, e.g. a poor person is much happier finding \$100 than a millionaire is. This is equivalent to the individual being risk-averse, rather than risk-neutral or risk-seeking.
Any function which satisfies these conditions is a potentially reasonable utility function. The precise form of the function will depend on the individual's specific preferences for different levels of wealth and, as we will see, his specific risk preferences.
Isoelastic Utility
One basic example is the isoelastic utility function, given by
Notice that, for ρ=0, this is simply the identity function, which represents no diminishing marginal utility of wealth and no aversion to risk. As ρ increases, the marginal utility of wealth becomes more diminishing, so ρ can be seen as a parameterization of risk aversion. A higher value of ρ means a higher aversion to risk.
For ρ=0.5, this function looks like this:
This function satisfies all of the desired properties. Though the scale of this plot does not indicate it well, the function is always less than that of the identity function, so this utility function can be thought of a means of "discounting" wealth in a way that accounts for diminishing marginal utility of wealth. Note, however, that it is not necessary that the scale of the function match that of the wealth; we shall see that the particular values taken by the utility function are irrelevant for decision-making, as they get mapped back into dollars after accounting for the different random payoffs of an opportunity.
The isoelastic utility function is said to represent constant relative risk aversion (CRRA), as the individual's aversion to risk is always proportional to his wealth. With higher wealth, he is less averse to risk. This is a desirable property and is generally fairly consistent with real-life decisions and the rules of thumb that most poker players use in managing bankroll requirements as they move up in stakes.
Exponential Utility
Another simple example is the exponential utility function, given by
For c=1/150000, this function looks like this:
The exponential utility function is said to represent constant absolute risk aversion (CARA), as the individual's aversion to risk is always constant regardless of his wealth. In practice, few people would exhibit constant absolute risk aversion, as we should expect that most rational individuals' risk aversion should decrease as wealth increases, though perhaps not according to the proportional scale of the CRRA utility function.
The exponential utility function is bounded from above, but that does not mean that an individual with this utility function has any upper bound to the amount of wealth he prefers. We will see, however, that this does make the individual less likely to take risks for large amounts of money.
Utility and Risk Aversion
Let's say an individual who has a net worth of \$500,000 and isoelastic utility with ρ=0.5 (defined on his net worth) is given the opportunity to bet all \$500,000 on the flip a fair coin, receiving a payoff of \$1,000,000 if it comes up heads and being broke if it comes up tails. What is the value to him of taking the bet? The expected value in the amount of wealth he will have after taking the bet is clearly \$500,000, but the expected utility of this random payoff is given by
Since the utility function is continuous and increasing, there is a unique dollar value, known as the certainty equivalent, that yields the same expected utility as any random payoff. It is the unique solution of the equation:
Here, the certainty equivalent is \$250,000. So while a completely risk-neutral individual should be indifferent between betting his \$500,000 net worth on this flip or not, the risk-averse individual with these particular preferences would rather have \$250,000 for certain than bet his \$500,000 on the flip. Since having \$500,000 for certain is even better than having \$250,000 for certain, the risk-averse individual of course passes on this opportunity. He would only be willing to spend his net worth to have a 50/50 chance at having either \$1,000,000 and \$0 if his net worth were less than \$250,000.
To get a feel for the practical implications of each of these two basic forms of utility functions, we can look at the certainty equivalents for similar situations of betting one's net worth on a coin flip, for varying values of net worth. The 1st column is the payoff for winning the coin flip (twice the net worth), the 2nd column is the certainty-equivalent value for the individual with isoelastic utility (with parameter ρ=0.5), and the 3rd column is the certainty-equivalent value for the individual with exponential utility (with parameter c=1/150000):
So, for example, an individual with exponential utility (with parameter c=1/150000) would only be willing to spend \$4,916.68 on a 50/50 chance of winning \$10,000.
A few simple observations:
• For isoelastic utility, the certainty equivalent is always a fixed percentage of the expected value of the coinflip. This is true regardless of what we set the parameter ρ equal to. So an individual with isoelastic utility is willing to bet his entire net worth on any weighted coinflip with fixed probability of winning (or on any 50/50 coinflip with a fixed percentage overlay, as in the example here), regardless of his wealth. This is unlikely to reflect any real person's preferences for such opportunities, but might be OK when we consider situations where the bet is for less than one's net worth.
• The certainty equivalents under exponential utility decrease significantly when more money is at risk. While the certainty equivalents in the table above for the smaller flips seem to be roughly in line with what most well-bankrolled poker players (with CARA utility, one's net worth relative to the bet size does not matter) would be willing to pay for these coinflips, most would likely be willing to pay more for the \$1M flip. This disparity can't be rectified by playing with the parameter c; if we reduce c enough that the player would be willing to pay something somewhat closer to \$500,000 for the \$1M flip, then the certainty equivalents for the smaller flips become extremely close to the pure expected values.
So these two simple utility functions may each be imperfect for capturing real-life risk preferences, at least for individuals risking their entire net worth in the case of isoelastic utility.
These two utility functions are the most commonly-used in mathematical models due to their desirable analytical properties, but for the purposes of making practical poker decisions, where the discrete-time nature of poker opportunities makes it unlikely that the methods of calculus would lead to nice analytical solutions in many models anyway, we should be fine with choosing any admissible utility function that can be evaluated numerically. If we look at more practical situations where only a portion of one's net worth is at risk, we might be able to find a good fit with the isoelastic utility function, or we might be better-served by building some sort of ugly-but-practical "hybrid" utility function.
Eventually, we'll use the methods of utility functions to look at the following questions:
• When players have practical and tax-conscious utility preferences, how much effective rake are we really paying for our chance at the glory of the WSOP Main Event title?
• What sort of approximate hand-by-hand utility functions should the Loose Cannon on the PokerStars Big Game have?
• How can a backer and a player formulate a split of a payoff in a way which is optimal for each of their personal risk preferences?
• Full Tilt takes \$1 out of the pot if you want to run it twice; under what conditions would we prefer to pay this fee to reduce volatility?
• Does whether or not we would take a certain risk ever depend on how many opportunities we will be given to play that game? In particular, is it a fallacy to manage the risk in a unique opportunity differently because we are unable to "reach the long run" with it?
But first, coming up next...
• Part 2: Finding or constructing a utility function that accurately represents practical risk preferences for poker players
• Part 3: Effects of taxation — and they're BIG ones
#### 9 comments:
1. The idea of bounded utility (as in your exponential example) and of decreasing utility are well-studied in economics. But gamblers seem to experience increasing marginal utility. Without that, why would anyone play a slot machine or buy a lottery ticket? Is it essential to your theory of poker that utility diminish?
2. The idea of having higher wealth leading to higher-ROI opportunities is something I had never thought about on as general a level as you describe. That you do better in pure-EV terms by taking a -EV gamble before investing in something paying compound interest is counterintuitive but very clear, and that has some interesting implications. The overall perspective of such a rational gamble being a simple way of reallocating wealth between people with similar investment goals is quite elegant. I wonder how many of such cases that occur in practice would be crowded out when risk aversion is factored in.
One example along these lines that I did think about once is if a skilled poker player is in a casino and is looking at a poker game that he thinks he can crush, but the buyin is \$500 and, despite having a sufficient bankroll, he only has \$250 on him at the time, he'll probably have a higher expected utility than just going home by instead trying to quickly double up the \$250 at roulette and playing in the poker game if he does (and going home otherwise).
As to the implications of this on marginal utility, I would say that the typical utility functions with decreasing marginal utility should be generally applicable to rational people in a vacuum (i.e. without specific investment goals, and without external investment opportunities where having more wealth would improve your ROI in a convex way). If one were a player with one of these specific investment goals, you could tweak your utility function to have a big jump at the target level of wealth, and all practical stuff you would do with it should still work nice.
I was planning on saying this: I think it's important to, at least in theory, demand that all rational players in a poker game have concave utility functions. In some sense, the rules of poker demand that decisions in the game are "supposed to" be made when they are +EV in the pure-EV sense. When players are risk averse, slightly -EV opportunities get passed up on, but that's not necessarily a big perturbation to the in-game decisions, especially because most players can be expected to be at least slightly risk-averse.
But it occurs to me that there's not really any fundamental difference between the perturbation of shutting out slightly-plus-EV moves and the perturbation of adding in slightly-minus-EV moves. So maybe it is fair to consider it.
The easy answer is that anyone who would have increasing marginal utility and would take a slightly-minus-EV move in poker is also someone who would gain utility by playing small-edge house games, so they could just play those instead of playing poker... but, if they're good at poker and are +EV in the game due to their skill, they would do better by playing the game and also taking the slightly-minus-EV opportunities. Poker would serve as a means of matching such rational risk-seeking people up with each other, just as the examples in your blog.
As to why anyone ever chooses to play a slot machine or lottery ticket, I think the other possible explanations are much more probable than legitimately rational decisions of individuals with increasing marginal utility. For one, such a legitimately rational person would do better by seeking out a similar person and gambling with them with no house edge. I think the vast majority of casino gambling is instead completely covered by factors such as willingness to pay for the entertainment of the process of gambling, heuristic probability biases, or just plain ignorance. My personal opinion is that there is more than enough evidence for all three of these explanations that economics need not necessarily search for other explanations, and for the first one, it can certainly be rational to pay for the entertainment value for those who happen to enjoy gambling.
I still feel like these are corner cases that don't need to be directly accounted for in poker theory, but you've given me a lot to think about. Your blog is very interesting and I shall be following it!
3. So, it *does* change your winning strategy if you have marginal increasing utility, in that you'd take -EV gambles. Is that a losing strategy against players who are more conservative? If it happens that some players take -EV gambles and some don't, there may be words for this in Poker lingo.
4. It would change your winning strategy and make your overall strategy have lower EV, but may keep you +EV overall. It will indeed be directly sacrificing money to all other opponents, and the conservative or non-risk-seeking opponents will all benefit. It is certainly a losing perturbation to one's strategy.
More importantly for the point I was making, even if adding the losing gambles makes the otherwise skilled player -EV in poker, he should still have a higher ROI than he would in any casino game; he only takes the -EV gambles in the poker game which favors his utility function (i.e. ones where he is getting the minimum level of odds that he'd demand for any gamble, such as a casino game), and he still gets plenty of actual +EV opportunities along the way, so on average, he should do better than the "minimum odds threshold" induced by his utility function.
Most commonly, the word for a player who takes -EV gambles in poker lingo is just "fish". In my personal experience and conjecture, there should be very few good players who would ever knowingly make a move that was -EV within the game. Maybe the risk-seeking players you describe tend to not become good poker players as often, or maybe good poker players are so conditioned to focusing entirely on EV that they neglect to take -EV gambles that may favor their own interests out of discipline.
5. You point out the difference between a -EV move in the game and on the other hand viewing the whole poker game from buy-in to cashing out as a gamble. I believe that we can take a -EV move on the whole game as part of a winning investment strategy. I wonder whether -EV moves within the game are a bad choice because other players will punish that player. I guess you think that they definitely will take advantage of that player. That makes sense.
I'd like to view the chip stack as wealth and moves in the game as investments made. A player with more chips has an advantage, I hear, and can make better investments. If so, then the -EV argument applies: if you make a -EV move and either lose or win and invest the money well conservatively, the average of these two strategies is better than simply staying behind in chips and making comparably worse bets than your chip-leading opponent.
If I try to manufacture a situation in which the opponents cannot punish me.. maybe if I am the last player to decide whether to call or fold, on the last round of a hand. Even in this case, I will show my cards, which will reveal my -EV move. So, I could very much believe that although a -EV move can be advantageous if winnings can be invested conservatively, in competitive games it will be punished.
How much of an advantage it is to have extra chips? Suppose you make two investments: you enter two players in two poker games. After some time, one of your players is winning and the other player is losing. The casino offers you the option to transfer some chips from one player to the other. Should you transfer chips to the player who is doing badly, or to the player who is doing well? This reflects the relative marginal ROI of the two situations.
6. Replace: the average of these two outcomes is better / the average of these two strategies is better
7. Actually, that a player with more chips has an advantage in *cash game* poker is a popular misconception. In fact, it can only be a disadvantage, though the effect is often negligible. If you have \$1,000 and the rest of the table has \$100 each, you have no in-game advantage, as you'll never get all-in for more than \$100. If, however, you have \$100 and the rest of the table has \$1,000 each, you are playing the exact same game as if the other players all had \$100 each, *except* that if you get all-in in a multiway pot, the other players might bet each other out after you are all-in, which can only increase your chances of winning the main pot.
In tournaments, I have always assumed a model of concave "chip stack utility". Conventional wisdom is that in almost all tournament situations, you need better than a 50% chance to commit all of your chips. This doesn't directly extend logically to a concave chip stack utility function, but I suspect that this is still the case except for possibly some special cases... I can't think of any right now.
There might be some advantages outside of the game where you might be willing to risk your stack with only a 49% chance to win, such as the example I discussed earlier where you want to move up to a soft higher-stakes game but don't have enough cash on hand. But, for a cash game, there's definitely no in-game advantage for a big stack that would cause in-game utility to be anything but concave (usually very close to linear, and only slightly concave for the effect of multiway all-ins that I discussed).
8. OK... if there is no move-by-move advantage to having a large stack, then I agree that no one should take a -EV move. And yet, you are right that I had expected the winnign player to get a move-by-move advantage. I had imagined that the player with the bigger stack might sit comfortably behind that large stack of chips and play only the best hands, while the player with fewer chips might feel, move by move, the pressure of having to pay the ante, forcing the low-stack player to accept less-desirable bets. But as you say, if one player has 1000 chips, and the next-richest player has only 100, then this hand is played as though all players had only 100 chips, with the difference that the player with 1000 cannot go "all-in" and the player with 1000 chips cannot lose in this hand.
Analogous to an "interest rate" accruing to a stack of chips is the ante draining smaller stacks faster than they drain big stacks. The amount to which the ante bothers you is -A/(PS) (total ante -- big blind plus small blind /(number of players times your chip stack)). If you always fold, you'll see PS/A hands before you run out of money, and surely we'd all prefer that number to be higher. That interest rate is concave; the "compound interest rate" accruing after n moves is -nA/PS which is also negative and concave in S. When interest is concave in wealth, -EV moves take you from your current value to a linear combination of the values you'd have if you won or lost and are always bad. So... never make a -EV move in Poker when S > 5A (where I picked 5 because I can't imagine worrying about being forced to gamble on my last hand, when 5P hands remain before that hand comes).
Consider your penultimate move -- on the next move you'll be forced to bet everything in order to pay a blind. At this moment, the interest curve might be locally convex: if you win, you get some breathing space and a lower per-hand fee, at least for the next few hands; if you lose, you still face a 100% fee on the next turn. So the "interest-rate versus wealth" curve is -100%, -100%, -20% (A/PS). So if to call right now has an EV below 0 but above the EV of a random hand (what you expect next turn), then you should call now with tomorrow's blinds-money. And, anticipating this, on the *previous* hand, if you see a -EV option which has EV better than you expect to get on your penultimate hand, then you should take it. And so on, arguing back; since we're cutting "between" the next-hand's expected value and 0 every turn, the -EV bets we expect to take are half as big each turn we go, so very soon we should ignore this and just say "Taking a 0EV move on your last 4 or 5 hands, to avoid taking a -EV move in your last 2 or 3 hands." These -EV moves work because if you do nothing you will be charged a high fee in ante; your only choices are to bet the money into a -EV move close to 0EV or wait until the blinds come, or choose to bet the money on a random hand when the blinds come.
9. I'm not totally sure that it's the case here, but the idea that the proportion of one's stack that a blind or ante represents is something that affects play is usually a misconception as well. The common forms it appears in are stuff like "I've only got 10 times the big blind left, I should just go with any decent hand I get" or "In a shorthanded game, hand selection changes because the blinds come around quicker". Neither of these comes to the wrong conclusion, but they are likely the wrong approach.
For a cash game, at least, the usual goal being to maximize one's expected value (utility) on every hand, one would not concern onesself with geometric growth rates of the percentage of stack that an ante represents. I think your example makes sense from an expected utility perspective if the player's entire net worth is at risk in the poker game, but if the player is playing a cash game for only a small portion of his net worth, he could just add on extra money if he wanted his small stack to be bigger.
For a tournament, your approach would make sense if your expected utility of prize payout (i.e. your position in the tournament) was concerned with the geometric growth of your stack, but I'm not sure how close that is to reality.
The hypothetical perfect map from tournament chip stack to expected utility of prize payout is certainly one that takes future hands into consideration. It would consider that, by folding your tiny stack first to act, you'll be putting it all in blind on the next hand. There could definitely be inflection points in this scenario for small stack sizes where your *chip stack* utility became convex at certain points, but I'm not convinced that they would exist at all. They probably don't matter very much practically, in that they might only exist for a narrow window of tiny stack sizes, but I'm not sure.
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http://johncarlosbaez.wordpress.com/2011/10/23/a-math-puzzle-coming-from-chemistry/
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# Azimuth
## A Math Puzzle Coming From Chemistry
I posed this puzzle a while back, and nobody solved it. That’s okay—now that I think about it, I’m not sure how to solve it either!
It seems to involve group theory. But instead of working on it, solving it and telling you the answer, I’d rather dump all the clues in your lap, so we can figure it out together.
Suppose we have an ethyl cation. We’ll pretend it looks like this:
As I explained before, it actually doesn’t—not in real life. But never mind! Realism should never stand in the way of a good puzzle.
Continuing on in this unrealistic vein, we’ll pretend that the two black carbon atoms are distinguishable, and so are the five white hydrogen atoms. As you can see, 2 of the hydrogens are bonded to one carbon, and 3 to the other. We don’t care how the hydrogens are arranged, apart from which carbon each hydrogen is attached to. Given this, there are
$2 \times \displaystyle{ \binom{5}{2} = 20 }$
ways to arrange the hydrogens. Let’s call these arrangements states.
Now draw a dot for each of these 20 states. Draw an edge connecting two dots whenever you can get from one state to another by having a hydrogen hop from the carbon with 2 hydrogens to the carbon with 3. You’ll get this picture, called the Desargues graph:
The red dots are states where the first carbon has 2 hydrogens attached to it; the blue ones are states where the second carbon has 2 hydrogens attached to it. So, each edge goes between a red and a blue dot. And there are 3 edges coming out of each dot, since there are 3 hydrogens that can make the jump!
Now, the puzzle is to show that you can also get the Desargues graph from a different kind of molecule. Any molecule shaped like this will do:
The 2 balls on top and bottom are called axial, while the 3 around the middle are called equatorial.
There are various molecules like this. For example, phosphorus pentachloride. Let’s use that.
Like the ethyl cation, phosphorus pentachloride also has 20 states… but only if count them a certain way! We have to treat all 5 chlorines as distinguishable, but think of two arrangements of them as the same if we can rotate one to get the other. Again, I’m not claiming this is physically realistic: it’s just for the sake of the puzzle.
Phosphorus pentachloride has 6 rotational symmetries, since you can turn it around its axis 3 ways, but also flip it over. So, it has
$\displaystyle{ \frac{5!}{6} = 20}$
states.
That’s good: exactly the number of dots in the Desargues graph! But how about the edges? We get these from certain transitions between states. These transitions are called pseudorotations, and they look like this:
Phosphorus pentachloride really does this! First the 2 axial guys move towards each other to become equatorial. Beware: now the equatorial ones are no longer in the horizontal plane: they’re in the plane facing us. Then 2 of the 3 equatorial guys swing out to become axial.
To get from one state to another this way, we have to pick 2 of the 3 equatorial guys to swing out and become axial. There are 3 choices here. So, we again get a graph with 20 vertices and 3 edges coming out of each vertex.
Puzzle. Is this graph the Desargues graph? If so, show it is.
I read in some chemistry papers that it is. But is it really? And if so, why? David Corfield suggested a promising strategy. He pointed out that we just need to get a 1-1 correspondence between
• states of the ethyl cation and states of phosphorus pentachloride,
together with a compatible 1-1 correspondence between
• transitions of the ethyl cation and transitions of phosphorus pentachloride.
And he suggested that to do this, we should think of the split of hydrogens into a bunch of 2 and a bunch of 3 as analogous to the split of chlorines into a bunch of 2 (the ‘axial’ ones) and a bunch of 3 (the ‘equatorial’ ones).
It’s a promising idea. There’s a problem, though! In the ethyl cation, a single hydrogen hops from the bunch of 3 to the bunch of 2. But in a pseudorotation, two chlorines go from the bunch of 2 to the bunch of 3… and meanwhile, two go back from the bunch of 3 to bunch of 2.
And if you think about it, there’s another problem too. In the ethyl cation, there are 2 distinguishable carbons. One of them has 3 hydrogens attached, and one doesn’t. But in phosphorus pentachloride it’s not like that. The 3 equatorial chlorines are just that: equatorial. They don’t have 2 choices about how to be that way. Or do they?
Well, there’s more to say, but this should already make it clear that getting ‘natural’ one-to-one correspondences is a bit tricky… if it’s even possible at all!
If you know some group theory, we could try solving the problem using the ideas behind Felix Klein’s ‘Erlangen program’. The group of permutations of 5 things, say $S_5,$ acts as symmetries of either molecule. For the ethyl cation the set of states will be $X = S_5/G$ for some subgroup $G.$ You can think of $X$ as a set of structures of some sort on a 5-element set. The group $S_5$ acts on $X,$ and the transitions will give an invariant binary relation on $X,$ For phosphorus pentachloride we’ll have some set of states $X' = S_5/G'$ for some other subgroup $G'$, and the transitions will give an invariant relation on $X'$.
We could start by trying to see if $G$ is the same as $G'$—or more precisely, conjugate. If they are, that’s a good sign. If not, it’s bad: it probably means there’s no ‘natural’ way to show the graph for phosphorus pentachloride is the Desargues graph.
I could say more, but I’ll stop here. In case you’re wondering, all this is just a trick to get more mathematicians interested in chemistry. A few may then go on to do useful things.
This entry was posted on Sunday, October 23rd, 2011 at 10:30 am and is filed under chemistry, mathematics, puzzles. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.
### 37 Responses to A Math Puzzle Coming From Chemistry
1. JM Allen says:
Does it not become very natural if you label the hydrogens on the ethyl cation in a different way?
Instead of labelling them as which carbon they’re on ignore the carbons altogether and label them according to whether they’re in a set of 3 or a set of 2 (maybe set is the wrong word, I just thought it would be less confusing than “group” – interpret in the non-maths-y sense) – this is exactly the same labelling as for phosphorus pentachloride.
The pseudorotations and the hydrogen-swapping are then both just the action of turning the 3-set into a 2-set by moving one hydrogen. This is a little more difficult to see with the phosphorus pentachloride but the two equatorial hydrogens that you choose to physically “move” are the ones “staying in their set” and then one that stays physically stationary is the one that “swaps into” the other set.
No group theory needed, just thinking about them in a very slightly different way and it’s obvious.
• John Baez says:
Nice!!! That was fast!
I think there’s a wee bit more work to be done, though…
Labelling each hydrogen according to whether it’s in the set of 3 or the set of 2 does not completely specify a state of the ethyl cation. Since we’re treating the carbons are distinguishable, one extra bit of information is needed: which carbon has 3 hydrogens attached to it and which has 2?
Here ‘bit’ is used in the technical sense: binary digit. This extra bit is what brings the number of states up from
$\displaystyle{ \binom{5}{2} = 10 }$
to
$2 \times \displaystyle{ \binom{5}{2} = 20 }$
However, this bit of information always changes when we make a transition. Indeed, this bit says the color of the dots here:
and each edge goes between a red dot and a blue one.
Over on the phosphorus pentachloride side, there’s also an extra bit of information, in addition to the labelling saying which chlorines are axial and which are equatorial.
So, we’ve reduced the puzzle to this mini-puzzle: what is this extra bit, and does it always change when we do a pseudorotation? If it does, we’re done.
By the way, there’s another way of counting states in both molecules, where we ignore this extra bit. If we used that way, we’d have 10 states instead of 20… and we’d be done with this puzzle.
• Twan van Laarhoven says:
Does the extra bit come from the reflectional symmetry of the phosphorus pentachloride? So, if a and b are axial, with a above, then the equatorial (b c d) clockwise is not the same as (d c b) clockwise.
• John Baez says:
Yes, this extra bit does not change when we rotate the phosphorus pentachloride molecule—see my reply to Peter Morgan below—but it does change when we reflect it. The question is whether it changes when we pseudorotate it!
• Twan van Laarhoven says:
Just thinking out loud.
Let me write [a,b,c,d,e] for the first image in this post labeled in reading order. So a and e are the axial atoms, bcd are equatorial in counter clockwise order. We have by rotational symmetry that: [a,b,c,d,e] = [a,c,d,b,e] = [a,d,b,c,e] = [e,d,c,b,a] = [e,c,b,d,a] = [e,b,d,c,a]. I will always use the lexicographically smallest representation.
Then the pseudorotation in the image in this post is [a,b,c,d,e] |-> [b,a,e,c,d].
According to a simple program I wrote, there are 10 different states reachable from [1,2,3,4,5] in an even number of moves:
[[1,2,3,4,5],[1,2,4,5,3],[1,2,5,3,4],[1,3,5,4,2],[2,1,3,5,4],[2,1,4,3,5],[2,1,5,4,3],[3,1,2,4,5],[3,1,5,2,4],[4,1,3,2,5]]
and 10 in an odd number of moves:
[[1,2,3,5,4],[1,2,4,3,5],[1,2,5,4,3],[1,3,4,5,2],[2,1,3,4,5],[2,1,4,5,3],[2,1,5,3,4],[3,1,2,5,4],[3,1,4,2,5],[4,1,2,3,5]]
The bit could be the parity of the permutation (a,b,c,d,e). That is because pseudorotation has odd parity, while all symmetry permutations have even parity.
———————————-
{-# LANGUAGE NoMonomorphismRestriction #-}
import Data.List
import qualified Data.Set as Set
perm xs = sort \$ perm1 xs ++ perm1 (reverse xs)
perm1 [a,b,c,d,e] = [[a,b,c,d,e],[a,c,d,b,e],[a,d,b,c,e]]
cannon = head . perm
rot [a,b,c,d,e] = [b,a,e,c,d]
close :: Ord a => (a -> [a]) -> a -> Set.Set a
close f = go Set.empty
where
go s a
| a `Set.member` s = s
| otherwise = foldl go (Set.insert a s) (f a)
nextStates = map (cannon . rot) . perm
s0 = [1..5]
states = close nextStates s0
sEven = close (concatMap nextStates . nextStates) s0
sOdd = close (concatMap nextStates . nextStates) (nextStates s0!!0)
parity xs = (`mod`2) \$ sum[ if x>y then 1 else 0 | x:ys <- tails xs, y<-ys ]
• John Baez says:
Excellent! The puzzle is solved!
More later… now it’s my bedtime. I’d like to try to find a way to solve this puzzle using plain English.
• Peter Morgan says:
Good way of looking at it, Twan. Now, however, the pseudorotation looks to me to be a bit of a red herring as far as counting states is concerned. States are the same if they can be reached by rotations; rotations generate cycles containing 6 elements, [a,b,c,d,e] = [a,c,d,b,e] = [a,d,b,c,e] = [e,d,c,b,a] = [e,c,b,d,a] = [e,b,d,c,a], therefore there are 20 equivalence classes=states.
The pseudorotation plus *either* the rotation [a,b,c,d,e]->[a,c,d,b,e] *or* the rotation [a,b,c,d,e]->[e,b,d,c,a] generate cycles with 120 elements, with no grading. The pseudorotation plus the reflection [a,b,c,d,e]->[e,b,c,d,a] generates a cycle with 120 elements, with an odd-even grading. All of these facts have no bearing on the counting of states, but they show that the pseudorotation is sufficient to generate all transformations between the 20 equivalence classes under rotations.
There are also, a propos of nothing, 20 equivalence classes under pseudorotations, which generate cycles containing 6 elements, or 40 equivalence classes under even numbers of pseudorotations, which generate cycles containing 3 elements, [a,b,c,d,e] = [a,b,e,c,d] = [a,b,d,e,c].
I hope this rehearsal as I see it now looks OK. Thanks for the puzzle, John. Not sure yet how to present equivalence classes in plain English.
For the ethyl cation, a more-or-less comparable formalism has a list of two lists, one containing 2 elements, the other 3, [[1,2],[3,4,5]], which we take *not* to be equivalent to [[3,4,5],[1,2]]. On the other hand, the permutations of 3,4,5 and of 1,2 form an equivalence class of 12 elements. Hence, we have $2\times 5!/3!/2!$ states. Not sure how to introduce an equivalence to the phosphorus pentachloride case.
2. Peter Morgan says:
We get 20 in the ethyl cation case by “pretending” that the carbon atoms are distinguishable, so that we get the *2 in Binomial(5,2)*2 as the number of states, and the set of states is $S_5/S_3/S_2\times S_2$. Equally, we can “pretend” that the two axial positions in potassium pentachloride are distinguishable, so that the number of states is multinomial(5,1,1), and the set of states is $S_5/S_3$. Alternatively, we can “pretend” that even and odd permutations of the equatorial positions in potassium pentachloride are distinguishable, so that the number of states is binomial(5,2)*2 and the set of states is $S_5/((S_3/S_2)\times S_2)$.
In all three cases, other things being equal, all 20 states are equivalent to each other. You seem to be postulating a model dynamics of discrete moves between states at each discrete time step (without any of the probabilistic weighting that we’ve seen in earlier posts). Within a class of models of this type, we could reasonably define the ethyl cation model and the potassium pentachloride model as equivalent, as isolated discrete dynamical models, just if the single step transition graphs for the two discrete dynamics are, on some definition of graph equivalence, equivalent graphs (your geometrical description of the potassium pentachloride dynamics, however, introduces intermediate states that have a different symmetry structure, aiming at an equivalence of the ethyl cation dynamics with a subgraph of the potassium pentachloride dynamics, which has what I suppose could be called a $\mathbb{Z}_2$ grading).
I suppose that for any sort of detailed modeling we will have to find ways to modify these structures to add the effect of another molecule, of an electromagnetic field, or of some other perturbation to the single-molecule model, and that the 20 states will then not be equivalent, or that the number of states might change as the model symmetries change. In any case, I hope that at some level of detail an empirically accurate model will distinguish between an ethyl cation and a potassium pentachloride molecule.
I’m not sure you’ll think this solves the puzzle?
• John Baez says:
Peter wrote:
… we can “pretend” that the two axial positions in phosphorus pentachloride are distinguishable, so that the number of states is multinomial(5,1,1), and the set of states is $S_5/S_3$. Alternatively, we can “pretend” that even and odd permutations of the equatorial positions in potassium pentachloride are distinguishable,
Nice!!! But…
You just mentioned two very tempting ways to introduce an extra bit of information into the phosphorus pentachloride problem. But we don’t get to pick how to introduce that extra bit of information: the puzzle as stated tells us how we have to do it!
The puzzle says that two molecules of phosphorus pentachloride with chlorine atoms labelled 1,2,3,4,5 count as being in the same state if we can rotate one to the other, carrying the labelling of one to the labelling of the other.
So, the two axial positions are not distinguishable: we can always rotate the molecule so that the ‘top’ chlorine becomes the ‘bottom’ one, and vice versa.
Similarly, the two cyclic orderings of the equatorial chlorines are not distinguishable: we can always rotate the molecule so that the ‘clockwise’ cyclic ordering becomes the ‘counterclockwise’ one, and vice versa.
However, you’ll note that flipping the molecule over to make the top chlorine the bottom one also changes the clockwise cyclic ordering into the counterclockwise one.
So while neither of your ways of introducing an extra bit of information is the way prescribed by the problem, the sum of these two bits is. (Here I’m adding bits mod 2.)
So the mini-puzzle is whether this bit inevitably changes when we do a pseudorotation.
I was just about to work this out when I came back to my computer and saw your comment. It’s nice to see we’re thinking about the same stuff. This problem has been bugging me for weeks, but now it’s almost solved.
• Peter Morgan says:
So, there are 40 states in the state space for the potassium pentachloride case, which contains a 20 state subspace that is closed under pairs of reflections. Pseudorotations are not rotations, but they are one (of many) paths that implement reflections. I note that the pseudorotations are paths that preserve the lengths of the potassium-chlorine bonds, but not, except at the endpoints, the angles between them.
• John Baez says:
My state space for phosphorus pentachloride has 20 states, not 40. I defined it here and also in Part 14. I’m sure you’re doing something interesting, but it’s dinner-time so I’ll read it later!
By the way it’s phosphorus, not potassium. Potassium can’t hang on to 5 chlorines!
• Peter Morgan says:
Phosphorous-Potassium, Duh!
There’s a sense in which your state-space is infinite-dimensional, if you include pseudo-rotations as continuous paths between the 20.
I hope learning how to divide and multiply by 2 in different ways is doing me some good. Forget the 40.
• John Baez says:
Peter wrote:
You seem to be postulating a model dynamics of discrete moves between states at each discrete time step (without any of the probabilistic weighting that we’ve seen in earlier posts).
Well, my plan for ‘Part 15′ of the Network Theory series is to look at:
• a Markov chain model, where time is discrete, and the molecule has probabilities to hop from one state to another at each time step,
and also
• a Markov process model, where time is continuous, and the molecule has ‘probabilistic rates’ of hopping from one state to another.
The second one is a bit more realistic, but the two models are mathematically related in a nice way. I also plan to look at quantum analogues of both models, where instead of probabilities we have amplitudes.
None of this is especially tied to the particular molecules I’m talking about now. We could start with any graph and study models of this sort. However, one nice feature of these particular molecules is that they’re so symmetrical that the transition probabilities, or rates, must be the same for every edge of the graph! This simplifies these models a bit… and if I went far enough, which I probably won’t, it’d let us use the representation theory of the group $S_5 \times S_2$ to solve these models.
Within a class of models of this type, we could reasonably define the ethyl cation model and the potassium pentachloride model as equivalent, as isolated discrete dynamical models, just if the single step transition graphs for the two discrete dynamics are, on some definition of graph equivalence, equivalent graphs…
Yes, that’s the idea here: abstract away from the molecule and look at the graph.
In fact the real-world ethyl cation is nothing like what I’ve drawn here. As I explained in Part 14, it really looks like this:
However, transitions in the purely imaginary ethyl cation I discussed in this puzzle are easier to visualize than pseudorotations of phosphorus pentachloride! So, I’ll talk about ethyl cations that look like this:
even though they don’t exist. Translate the results to the phosphorus pentachloride picture, if you like…
• Eric says:
Just a quick note…
Who is to say Markov processes are more realistic than Markov chains? Markov processes could represent continuum approximations to something that is fundamentally finitary.
3. Hudson Luce says:
maybe you should substitute a chlorine atom for the positive charge in the ethyl cation – think about ethyl chloride instead…
4. Hudson Luce says:
the trouble is that the three hydrogen atoms on the one carbon of the ethyl cation are *not* distinguishable (C3 symmetry), and the two hydrogens on the other carbon are not distinguishable either (C2 symmetry); similarly, the two axial chlorines on the PCl5 molecule have C2 symmetry, and the three equatorial chlorines have C3 symmetry, but the connectivity is different – in ethyl cation you’ve got a two-carbon chain, while in PCl5 you’ve got single P atom with 5 chlorine atoms on it. Substituents (chlorines, hydrogens, etc) tend to situate themselves so as to minimize Coulombic repulsions. Your picture with the “cyclic” C-C-H structure is an average – it makes the two carbons indistinguishable – they have “half-ownership” of one of the hydrogens – not the highly strained full ownership that the picture depicts.
• John Baez says:
Thanks for all the info, Hudson! Yes, Bacharach claims the ethyl cation actually looks like this:
and I’m treating all sorts of things as distinguishable that actually aren’t. So, this was a math puzzle with chemistry serving as a rather flimsy excuse, not an actual chemistry puzzle. Still it may get some mathematicians interested in graph theory problems coming from chemistry!
• Danail Bonchev and D.H. Rouvray, eds., Chemical Graph Theory: Introduction and Fundamentals, Taylor and Francis, 1991.
• Nenad Trinajstic, Chemical Graph Theory, CRC Press, 1992.
• R. Bruce King, Applications of Graph Theory Topology in Inorganic Cluster Coordination Chemistry, CRC Press, 1993.
The second one is apparently the magisterial tome of the subject. The prices on these books are absurd: for example, Amazon sells the first for \$300, and the second for \$222. Luckily some universities will have them…
• Walter Blackstock says:
The experimental evidence favours the non-classical structure.
Andrei, H.-S.; Solcà, N.; Dopfer, O., “IR Spectrum of the Ethyl Cation: Evidence for the Nonclassical Structure, Angew. Chem. Int. Ed. 2008, 47, 395-397
5. Hudson Luce says:
I looked at the Bacharach reference and see that his calculations represent the ethyl cation as an ethylene molecule, H2C=CH2, with a proton associated halfway between the two carbons, and bonded to neither. This gives the ethyl cation the same symmetry on average as ethylene, which I think is D2h, C2 with a mirror plane of symmetry between the two carbons (C2(z) C2(y) C2(x) i σ(xy) σ(xz) σ(yz)), whereas phosphorus pentachloride is D3h (2C3 3C2 σh 2S3 3σv)
6. Johan Swanljung says:
If you decide on a way to uniquely identify states, you can match them up to your picture of the Desargues graph and each one will match either a blue or a red dot. So the extra bit is whether the state belongs to the blue dot set or the red dot set. This requires that you manually work out all of the possible transitions, but it does work. Unsatisfying perhaps.
I can also see a way to calculate a bit that will flip for each transition. (I teach high school physics and my group theory and abstract algebra is very rusty, so excuse me if I express things clumsily). Arrange the chlorine atoms in a cycle (1 2 3 4 5). The axial atoms are either adjacent in the cycle or not. Call that bit A.
Using the cycle we can assign any two atoms an order by requiring that the distance in the cycle between the earlier and later atoms is at most two. Rotate the atom so that the axis of symmetry is vertical and so that the top atom is the earlier of the two axial atoms. Then, looking down, the equatorial atoms are in clockwise order or not. Call that bit B.
The bit defined by A XOR B will flip in any transition.
I don’t have a good proof of this, but staring at some representative examples is convincing:
1
345
2
Here 1 and 2 are axial (and adjacent in the cycle) and 345 are equatorial in clockwise order looking down (to make the representation unique I’ll start with the smallest integer). There are exactly 3 possible transitions. If we replace 1 and 2 with two adjacent atoms (3 and 4 or 4 and 5), the equatorial order will reverse. If we replace them with the two non-adjacent atoms (3 and 5) then the equatorial order stays the same, but adjacency flips.
Another example, starting from non-adjacent axial atoms:
1
245
3
Again, flipping adjacency preserves equatorial order. If adjacency stays the same, order flips. I hope this makes sense to someone other than me.
• Johan Swanljung says:
In the second example, remember that 5 is earlier than 2 in the cycle.
7. Cristi says:
There is a way to represent each of the 20 states of the ethyl cation as an ordered pair made with the numbers {1, 2, 3, 4, 5}. Also, there is a way to represent each of the 20 states of the phosphorus pentachloride as such an ordered pair. And the transitions correspond via these representations.
The key is given by this wu xing diagram:
An ordered pair corresponds to an arrow from the wu xing diagram, together with a sign: a plus sign telling that the pair has the same order as the arrow, or a minus sign if the order is the inverse one.
Let’s label the five hydrogen atoms of the the ethyl cation by the numbers {1, 2, 3, 4, 5}, and the two carbon ones by the plus and minus signs {+, -}. We associate to each of the 20 states the pair formed by the two numbers representing the two hydrogen atoms linked to the carbon with two hydrogen atoms. To find the order, we look up in the wu xing diagram the arrow labeled with the same numbers as the pair of hydrogen atoms. If the carbon atom is labeled with the plus sign, we order them by the order given by the arrow. If the carbon atom is labeled with the minus sign, we take the inverse order. For example, if the hydrogen atoms labeled with {1, 2, 3} are connected to the carbon atom labeled with the plus sign, and the hydrogen atoms labeled with {4, 5} are connected to the carbon atom labeled with the minus sign, the ordered pair used to represent the state of the molecule is (5, 4).
Let’s now represent the 20 states of the phosphorus pentachloride. We need first to define the sign of a triple in the wu xing diagram. It is made by the product between two signs. The first sign is given by the orientation of the triangle made by the pair: it is plus iff the orientation is counterclockwise. The second sign is given by minus one to the number of arrows from the triangle which are on the contour of the wu xing diagram. There are two kinds of triangles: the obtuse ones, with two arrows on the contour, and the acute ones, with one arrow on the contour. We multiply the signs and obtain the sign of the triple.
Back to the 20 states of the phosphorus pentachloride. We label with the same numbers {1, 2, 3, 4, 5} the five atoms. We represent the state by the pair given by the two axial atoms. The order is obtained by the following rule. We consider in the 3D representation of the molecule, a vector connecting the two axial atoms, oriented as in the wu xing diagram. With the wu xing diagram, we calculate the sign of the triple made by the three numbers labeling the equatorial atoms. We then multiply with the orientation of the triangle made by the three equatorial atoms in the 3D picture, seen from the direction in which the vector is pointing. If the resulting sign is plus, the pair representing the state is given by the orientation of the arrow connecting the labels of the two axial atoms. If it is minus, we reverse the order indicated by the arrow.
For example, let’s say that the bottom axial atom is labeled with 1, the top one with 2, and the three equatorial atoms are labeled in counterclockwise order with 3, 4, 5. The wu xing diagram shows that the arrow goes from 1 to 2. The orientation of the triangle made with 3, 4, 5 is clockwise in the wu xing diagram, and it is obtuse, so its sign is minus. Unlike the 3D diagram in which the sign is plus, the triangle being CCW. Therefore, the state should be represented by the pair (2, 1).
The next step is to see that the transitions coincide. The rule of the transition, common to the two types of molecules, is the following: an ordered pair can go in another ordered pair, if they don’t have common elements. In addition, we invert the sign. In other words, if a pair is oriented the same as the arrow represented it, then the transition will give a pair which is oriented in the inverse direction than that given by the arrow representing it, and conversely.
• John Baez says:
Cool! Why do you call that a “wu xing diagram”? Was some diagram like that used in China for some reason? My wife studies Chinese philosophy, religion, and history, so I’m curious. It looks almost like the Petersen graph:
but not quite, since the Petersen graph has 10 vertices instead of just 5.
• John Baez says:
Oh, I guessed it and confirmed it: wu xing refers to ‘five phase theory’, roughly the Chinese version of the Western ‘four elements’, although the phases were more mutable than elements.
• Cristi says:
Thanks! And you guessed well.
There is a simpler way to make the correspondence between the 20 states of the ethyl cation and those of the phosphorus pentachloride. The idea is that, instead of labeling the carbon atoms of the ethyl cation, we can use the two possible orientations of the three hydrogen atoms connected to the same carbon atom.
We start with a phosphorus pentachloride state. We replace the central atom with a carbon, and the five terminal atoms with hydrogenes. We draw the vector connecting the axial atoms, oriented as in the wu xing diagram. Then, we replace the axial atom towards which the vector points, with a carbon atom to which we link the two axial atoms. This operation transforms a phosphorus pentachloride state in a ethyl cation state, with the triple of hydrogen atoms oriented the same as in the phosphorus pentachloride state. We now can check that the transition diagrams are isomorphic.
P.S. I have thought at 1) one proof using a regular 5-cell in the four dimensional space, and hyperplanes separating its vertices, and 2) another one using the 20 vertices of a labeled dodecahedron which I constructed to represent the multiplication table of the even permutations of 5 elements (http://www.unitaryflow.com/2009/06/polyhedra-and-groups.html). I can write them down if you are interested, but I see that you already have enough proofs!
• Stefan says:
Just wondering: Earth absorbs water, water extinguishes fire, fire melts metal, metal cuts wood, but how does wood overcome earth?
Makes for a nice rock-paper-scissors game.
• Stefan says:
…ok, I looked it up: wood parts earth, as in tree roots cracking stone.
• Cristi says:
“Makes for a nice rock-paper-scissors game.”
Well, maybe because the rock-paper-scissors game is Chinese too (http://en.wikipedia.org/wiki/Rock_paper_scissor#History). Indeed, for 4 or 5 players the rock-paper-scissors game does not offer enough options to avoid repetitions, but wu-xing offers :)
8. John Baez says:
It was sort of silly for me to post this puzzle both here and on the n-Category Café, but I was desperate for help. Here’s a really nice answer from Tracy Hall over at the n-Café. He answers the main puzzle and also my subsidiary puzzle, namely: what graph do we get if we discard the extra bit of information that says which carbon here has 3 hydrogens attached to it and which has 2?
The answer is the Petersen graph:
Tracy wrote:
As some comments have pointed out over on Azimuth, in both cases there are ten underlying states which simply pick out two of the five pendant atoms as special, together with an extra parity bit (which can take either value for any of the ten states), giving twenty states in total. The correspondence of the ten states is clear: an edge exists between state A and state B, in either case, if and only if the two special atoms of state A are disjoint from the two special atoms of state B. This is precisely one definition of the Petersen graph (a famous 3-valent graph on 10 vertices that shows up as a small counterexample to lots of naïve conjectures). Thus the graph in either case is a double cover of the Petersen graph—but that does not uniquely specify it, since, for example, both the Desargues graph and the dodecahedron graph are double covers of the Petersen graph.
For a labeled graph, each double cover corresponds uniquely to an element of the Z/2Z cohomology of the graph (for an unlabeled graph, some of the double covers defined in this way may turn out to be isomorphic). Cohomology over Z/2Z takes any cycle as input and returns either 0 or 1, in a consistent way (the output of a Z/2Z sum of cycles is the sum of the outputs on each cycle). The double cover has two copies of everything in the base (Petersen) graph, and as you follow all the way around a cycle in the base, the element of cohomology tells you whether you come back to the same copy (for 0) or the other copy (for 1) in the double cover, compared to where you started.
One well-defined double cover for any graph is the one which simply switches copies for every single edge (this corresponding to the element of cohomology which is 1 on all odd cycles and 0 on all even cycles). This always gives a double cover which is a bipartite graph, and which is connected if and only if the base graph is connected and not bipartite. So if we can show that in both cases (the fictitious ethyl cation and phosphorus pentachloride) the extra parity bit can be defined in such a way that it switches on every transition, that will show that we get the Desargues graph in both cases.
The fictitious ethyl cation is easy: the parity bit records which carbon is which, so we can define it as saying which carbon has three neighbors. This switches on every transition, so we are done. Phosphorus pentachloride is a bit trickier; the parity bit distinguishes a labeled molecule from its mirror image, or enantiomer. As has already been pointed out on both sites, we can use the parity of a permutation to distinguish this, since it happens that the orientation-preserving rotations of the molecule, generated by a three-fold rotation acting as a three-cycle and by a two-fold rotation acting as a pair of two-cycles, are all even permutations, while the mirror image that switches only the two special atoms is an odd permutation. The pseudorotation can be followed by a quarter turn to return the five chlorine atoms to the five places previously occupied by chlorine atoms, which makes it act as a four-cycle, an odd permutation. Since the parity bit in this case also can be defined in such a way that it switches on every transition, the particular double cover in each case is the Desargues graph—a graph I was surprised to come across here, since just this past week I have been working out some combinatorial matrix theory for the same graph!
The five chlorine atoms in phosphorus pentachloride lie in six triangles which give a triangulation of the 2-sphere, and another way of thinking of the pseudorotation is that it corresponds to a Pachner move or bistellar flip on this triangulation–in particular, any bistellar flip on this triangulation that preserves the number of triangles and the property that all vertices in the triangulation have degree at least three corresponds to a pseudorotation as described.
• TH says:
Is there a nice place for the uninitiated to learn about cohomology of graphs?
• John Baez says:
Do you know about cohomology of anything else, like topological spaces or simplicial complexes? A graph is a special case of either of those.
I’m not trying to intimidate you by boosting the level of generality—honest. It’s just that if you happen to know some other kinds of cohomology, learning about the cohomology of graphs may be a lot easier than you think! But cohomology of graphs is, in fact, about the easiest kind of all.
If you only want to learn about the cohomology of graphs, you could learn it from the end of week293, where I apply it to electrical circuits, or from this book:
• P. Bamberg and S. Sternberg, A Course of Mathematics for Students of Physics vol. 2, Chap. 12: The theory of electrical circuits, Cambridge University, Cambridge, 1982.
which takes more time but goes into more detail.
Sometime fairly soon my Network Theory notes on this blog will get into electrical circuits, and then I’ll have to explain the cohomology of graphs.
9. John Baez says:
Here’s a sketch of a solution that’s not too technical.
Puzzle. Show that the graph with states of a trigonal bipyramidal molecule as vertices and pseudorotations as edges is indeed the Desargues graph.
Answer. To be specific, let’s use iron pentacarbonyl as our example of a trigonal bipyramidal molecule:
It suffices to construct a 1-1 correspondence between the states of this molecule and those of the ethyl cation, such that two states of this molecule are connected by a transition if and only if the same holds for the corresponding states of the ethyl cation.
Here’s the key idea: the ethyl cation has 5 hydrogens, with 2 attached to one carbon and 3 attached to the other. Similarly, iron carbonyl has 5 carbonyl groups, with 2 axial and 3 equatorial. We’ll use this resemblance to set up our correspondence.
There are various ways to describe states of the ethyl cation, but here’s the best one for us. Number the hydrogens 1,2,3,4,5. Then a state of the ethyl cation consists of a partition of the set {1,2,3,4,5} into a 2-element set and a 3-element set, together with one extra bit of information, saying which carbon has 2 hydrogens attached to it. This extra bit is the color here:
What do transitions look like in this description? When a transition occurs, two hydrogens that belonged to the 3-element set now become part of the 2-element set. Meanwhile, both hydrogens that belonged to the 2-element set now become part of the 3-element set. (Ironically, the one hydrogen that hops is the one that stays in the 3-element set.) Moreover, the extra bit of information changes. That’s why every edge goes from a red dot to a blue one, or vice versa.
So, to solve the puzzle, we need to show that the same description also works for the states and transitions of iron pentacarbonyl!
In other words, we need to describe its states as ways of partitioning the set {1,2,3,4,5} into a 2-element set and a 3-element set, together with one extra bit of information. And we need its transitions to switch two elements of the 2-element set with two of the 3-element set, while changing that extra bit.
To do this, number the carbonyl groups {1,2,3,4,5}. The 2-element set consists of the axial ones, while the 3-element set consists of the equatorial ones. When a transition occurs, two of axial ones trade places with two of the equatorial ones, like this:
So, now we just need to figure out what that extra bit of information is, and why it always changes when a transition occurs!
Here’s how you calculate this extra bit. Hold the iron pentacarbonyl so that the axial guy with the lower number is pointing up, like this:
In this example the axial guys are numbered 2 and 4, so we hold the molecule so that the lower number, 2, is pointing up.
Then, looking down, see whether you can get the numbers of the three equatorial guys to increase as you read them going around clockwise, or counterclockwise. That’s your bit of information. In this example, you can read the numbers 1, 3, 5 as you go clockwise.
It’s easy to see that this bit of information doesn’t change when we rotate the iron carbonyl molecule, so we have a well-defined way of getting a bit from a state. On the other hand, this bit always changes when a transition occurs. For example:
At the end if you hold this molecule so the guy labelled 1 is pointing up, you can read the numbers 2, 4, 5 as you go around counterclockwise. So, the bit has changed.
This completes the proof except for checking that the bit always changes when a transition occurs. We leave this as a further small puzzle for the reader.
10. Johan Swanljung says:
That bit doesn’t always change. Try the one with 1 and 2 as axial and 3, 4, 5 clockwise looking down. Make the transition to 3 and 5 as axial and you’ll get 1, 2, 4 clockwise looking down, so no bit change. It’s more complicated than this. I started out thinking like this, but ended up with what I posted in my last comment.
• John Baez says:
1
345
2
and wind up with
3
234
5
Zounds!
• Johan Swanljung says:
It seems to me that there’s a necessary symmetry missing when you use the numerical order. Without it the relationship between up/down order and clockwise/counterclockwise order is inconsistent. Using a cycle and defining order in terms of shortest distance in the cycle (so 5 is after 3 but before 1 and 2) fixes this. But then there is also a difference if the axial atoms are consecutive in the cycle or 2 steps apart (those are the only two possibilities). If you think of that and the cw/ccw as bits, then one of those two bits flips in each transition. There’s probably a nicer way to formulate that but I think it’s correct.
11. John Baez says:
Okay, I think the following solution is correct. This time I used Twan van Laarhoven’s idea for computing that extra bit of information. It not only works better than my earlier wrong approach; it’s also easy to give a full proof that it works.
Puzzle. Show that the graph with states of a trigonal bipyramidal molecule as vertices and pseudorotations as edges is indeed the Desargues graph.
Answer. To be specific, let’s use iron pentacarbonyl as our example of a trigonal bipyramidal molecule:
It suffices to construct a 1-1 correspondence between the states of this molecule and those of the ethyl cation, such that two states of this molecule are connected by a transition if and only if the same holds for the corresponding states of the ethyl cation.
Here’s the key idea: the ethyl cation has 5 hydrogens, with 2 attached to one carbon and 3 attached to the other. Similarly, the trigonal bipyramidal molecule has 5 carbonyl grops, with 2 axial and 3 equatorial. We’ll use this resemblance to set up our correspondence.
There are various ways to describe states of the ethyl cation, but this is the best for us. Number the hydrogens 1,2,3,4,5. Then a state of the ethyl cation consists of a partition of the set {1,2,3,4,5} into a 2-element set and a 3-element set, together with one extra bit of information, saying which carbon has 2 hydrogens attached to it. This extra bit is the color here:
What do transitions look like in this description? When a transition occurs, two hydrogens that belonged to the 3-element set now become part of the 2-element set. Meanwhile, both hydrogens that belonged to the 2-element set now become part of the 3-element set. (Ironically, the one hydrogen that hops is the one that stays in the 3-element set.) Moreover, the extra bit of information changes. That’s why every edge goes from a red dot to a blue one, or vice versa.
So, to solve the puzzle, we need to show that the same description also works for the states and transitions of iron pentacarbonyl!
In other words, we need to describe its states as ways of partitioning the set {1,2,3,4,5} into a 2-element set and a 3-element set, together with one extra bit of information. And we need its transitions to switch two elements of the 2-element set with two of the 3-element set, while changing that extra bit.
To do this, number the carbonyl groups 1,2,3,4,5. The 2-element set consists of the axial ones, while the 3-element set consists of the equatorial ones. When a transition occurs, two of axial ones trade places with two of the equatorial ones, like this:
So, now we just need to figure out what that extra bit of information is, and why it always changes when a transition occurs.
Here’s how we calculate that extra bit. Hold the iron pentacarbonyl molecule vertically with one of the equatorial carbonyl groups pointing to your left. Remember, the carbonyl groups are numbered. So, write a list of these numbers, say (a,b,c,d,e), where a is the top axial one, b,c,d are the equatorial ones listed in counterclockwise order starting from the one pointing left, and e is the bottom axial one. This list is some permutation of the list (1,2,3,4,5). Take the sign of this permutation to be our bit!
Let’s do an example:
Here we get the list (2,5,3,1,4) since 2 is on top, 4 is on bottom, and 5,3,1 are the equatorial guys listed counterclockwise starting from the one at left. The list (2,5,3,1,4) is an odd permutation of (1,2,3,4,5), so our bit of information is odd.
Of course we must check that this bit is well-defined: namely, that it doesn’t change if we rotate the molecule. Rotating it a third of a turn gives an even permutation of the equatorial guys and leaves the axial ones alone, so this is an even permutation. Flipping it over gives an odd permutation of the equatorial guys, but it also gives an odd permutation of the axial ones, so it too is an even permutation. So, rotating the molecule doesn’t change the sign of the permutation we compute from it. The sign is thus a well-defined function of the state of the molecule.
Next we must to check that this sign changes whenever our molecule undergoes a transition. For this we need to check that any transition changes our list of numbers by an odd permutation. Since all transitions are conjugate in the permutation group, it suffices to consider one example:
Here we started with a state giving the list (2,5,3,1,4). The transition took us to a state that gives the list (3,5,4,2,1) if we hold the molecule so that 3 is pointing up and 5 to the left. The reader can check that going from one list to another requires an odd permutation. So we’re done.
12. Hauta says:
I made a system of labels for each molecule:
Label the hydrogens of the ethyl cation with letters a through e. The state is labeled with the hydrogen pair and an orientation – if it was ab – cde, this is ab+. If it was cde – ab, label as ab-.
Fill out the graph this way, and label transitions (edges) with the hydrogen that changes place.
For the pentachloride we’ll have a similar system. Label each of the outer atoms with letters a through e. Each state gets a label that is the axis pair, in alphabetical order. Use alphabetical order of the axial atoms to determine an orientation – up or down. If the ordering of the axis matches the orientation, it’s +. If it does not, it’s -.
Psuedorotations will be labeled by the axial atom that does not participate.
Using the diagram for the original system, plot the edges and their labels on a second graph. Put the ab+ state in the same position as the original graph. Now, just go around using the moves already written as psuedorotations. These will be valid psuedorotations and will not repeat. If you fill in the missing edges, they’ll match the original graph.
This does not distinguish the states of the pentachloride into two sets the way the original system did – the original graph alternates the +/- around the graph (blue/red states), while this one does not.
It’s possible some similar system would fix this, but I’m done with it for now.
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http://mathematica.stackexchange.com/questions/1489/how-to-determine-how-much-of-a-table-is-generated/1507
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# How to determine how much of a table is generated?
If I'm generating a table from a slow function, like this:
````foo = Table[SuperSlowExpression, {1000}]
````
is there a way to determine after submission (so it is too late to instrument the call) to find out how many values already have been generated (so that when it takes longer than expected I can find out whether to wait a bit longer or to abort the calculation)?
So is there still a way to get at the information? Note that there is no iteration variable in the call which I could read.
Optimally, I'd also want to get the already calculated values, of course.
Edit:
Since this seems to have been consistently overlooked:
This is about a calculation which is already running. It is no longer possible to make changes to the call of `Table` or to the expression inside.
-
2
I don't know about after, but if you already know that you have `SuperSlowExpression` to begin with, it seems worthwhile to set up a cache variable beforehand, and then modify `SuperSlowExpression` to store to the cache (with e.g. `Append[]`/`AppendTo[]`) with each iteration within `Table[]`. – J. M.♦ Feb 8 '12 at 10:41
## 5 Answers
Given your current situation, there is another option that might help if you have set `"Enable notebook history tracking"` in `Preferences > Advanced`:
Then, you go to `Cell > Notebook history` and navigate to your currently evaluating cell and look at the time stamp when you last edited it. Chances are that you edited it just prior to evaluating (note that if you open an old notebook and execute the cell right away, this might not work, because the last edit timestamp will not be what you want it to be).
Now find the current time from your system clock. Assuming you know how long it takes for one evaluation of `SuperSlowExpression`, you can now simply do:
$$\mathrm{approx\ progress=\frac{current\ time - last\ edited\ time}{time\ for\ one\ evaluation}}$$
This will give you a rough estimate of the progress — enough to make a decision whether to wait a bit longer or to give up.
Remember, this should be a last ditch option! There are many ifs and buts here, but seeing as all the other options require an iterator and you don't have one (and your `Table` is running), it's worth trying...
-
1
Thank you. This seems indeed to be the best option available. And the history tracking was up to now unknown to me (it turned out that it already was enabled, though). While your advice came too late for me (I stopped the calculation before I saw it; however a later, monitored run showed me that it was indeed the correct thing to do), it's really what I should have done (with estimating the time of one step by simply entering a subsession and timing a single evaluation of the expression). – celtschk Feb 9 '12 at 14:10
Yes, there is. You can use `Monitor`
````Monitor[
Table[Pause[1]; i, {i, 10}],
i
]
````
Some explanations
`Table` uses dynamic scoping to localize its variable (`i` in this case), just like `Block`. This means that while `Table` is evaluating, `i` has a "global" value that one can inspect even from outside the `Table` if we interrupt the evaluation.
On way of interrupting the evaluation is to request `i`'s value on the pre-emptive link (i.e. using `Dynamic` expressions) We could just put `Dynamic[i]` somewhere and it would count up just like `Manipulate` while the `Table` is evaluating.
Another way of interrupting the evaluation is through entering a `Dialog`: You can use the `Evaluation -> Interrupt evaluation...` menu item, choose `Enter subsession`. You will get a prompt where you can inspect the kernel state (including the value of `i`, `Stack[]`, etc.). Evaluate `Return[]` to resume he evaluation.
I have been using both of these techniques extensively to monitor very lengthy evaluations. The second one even seems to work for parallel evaluations, and least in those cases where I tried it.
Iteration variable and already calculated values
If there is no iteration variable in the `Table`, just insert one, and use `Table[..., {i, 1000}]`. Then we can use `Monitor` or `Dynamic[i]`.
It is not possible to get the already calculated values. If you want to do this (it would often be desirable to be able to!), you need to implement your own version of `Table`. I asked about how to implement an interruptible parallel `Table` or `Map` in this question. The solution I got does not allow inspecting the so far calculated results, but it lets me interrupt the calculation, and continue from where I stopped later.
A `table` function that allows monitoring partial results
As an example, I provide a custom `table` function that allows monitoring the partial results.
Try evaluating
````Dynamic[table`Results[]]
table[Pause[1]; i^2, {i, 10}]
````
``table`Results[]`could be used in`Monitor` too.
I am using a linked list to collect the results for better performance than what `AppendTo` would give. I cannot use `Sow`/`Reap` here because they don't allow inspecting the partial results (at least I am not aware of this being possible).
Note: If you use nested `table`s, only the inner one can be monitored, but all of them will work correctly.
The code:
````ClearAll[table, table`result, table`Results]
SetAttributes[table, HoldAll]
table`result = {};
table[expr_, iterator__] :=
Block[{table`result, table`elem},
table`result = {};
Do[
table`result = {table`result, table`elem[expr]},
iterator
];
table`Results[]
]
table`Results[] := Block[{table`res, table`elem},
table`res = Flatten[table`result];
table`elem = Identity;
table`res
]
````
-
"Note that there is no iteration variable in the call which I could read." - so, how does one use `Monitor[]` here? – J. M.♦ Feb 8 '12 at 10:47
1
As I said, it's a calculation which already started (had I had an idea that it could take that long, I would have instrumented it accordingly). That is, I cannot wrap it any more into a monitor or modify the expression. However a +1 for the idea to use a linked list instead of a built-in list for speed. – celtschk Feb 8 '12 at 11:39
@celtschk So you are running the calculation right now? You're out of luck then. I've been there. – Szabolcs Feb 8 '12 at 12:05
@celtschk For a few more custom Table functions (conditional Table and Abortable table), see my posts here: stackoverflow.com/questions/6367932/…, and here: stackoverflow.com/questions/6470625/mathematica-table-function/…. They are based on `Reap` and `Sow` and don't allow the "real-time" monitoring, but can be modified using @Szabolcs's suggestion, to allow that. They however accept a general syntax of `Table`, including the multi-dimensional case. – Leonid Shifrin Feb 8 '12 at 12:27
1
@celtschk In this context, the reason I gave you that link is not the memoization part but the part of backing things up (persistency), via SQL or disk, so that you don't lose the data in case of Aborts, crashes, etc. – Leonid Shifrin Feb 8 '12 at 14:12
show 8 more comments
Similar, with a progress bar:
````tEnd = 10;
ProgressIndicator[Dynamic[i], {0, tEnd}]
Table[Pause[1]; i, {i, tEnd}]
````
-
As I said, the calculation is already started and does not have an iteration variable. But thanks anyway. – celtschk Feb 8 '12 at 11:32
1
@celtschk, you can execute a FrontEndTokenExecute with "SubsessionEvaluateCells" and incrementing and i=0 via \$Post. ;-) But you know what: stop the execution and add a i in your table. – ruebenko Feb 8 '12 at 12:30
If with "stop" you mean "interrupt", then telling how to add an `i` after the fact would be exactly an answer to my question. If by "stop" you mean "abort" then that's exactly what I'd like to avoid. I'd hate to lose more than half a day of calculation while thinking that maybe it might have finished the next minute if I didn't kill it. Of course if the answer to my question should be "it's not possible" then of course that is what I'll have to do (actually with a new run I'll then make sure to get the actual data early). – celtschk Feb 8 '12 at 13:01
If you're using a function that doesn't use iterators such as `Scan` or `Fold` you could consider using `PrintTemporary`, for example
````f[x_] := (Pause[2]; Sin[x]);
res = {};
Module[{i = 0, temp},
Scan[(NotebookDelete[temp];
temp = PrintTemporary[++i];
AppendTo[res, f[#]]) &, Range[20]]]
````
-
1
If we're already setting a variable to a value, then we can `Monitor` again... – Szabolcs Feb 8 '12 at 12:08
@Szabolcs I tried that but couldn't get it to work. – Heike Feb 8 '12 at 12:24
@Szabolcs Ignore that last comment, I just realized I made a mistake when I tried to Monitor before – Heike Feb 8 '12 at 12:26
Suppose you're evaluating something slightly different:
````Table[Pause[0.1](*SuperSlowExpression*), {i, 1000}]
````
You can create a cell containing `i`, and use Evaluation > Evaluate in Subsession to see the current value of `i` without interrupting the main evaluation. In the case where you don't have an iterator variable to inspect, then the question is whether there is some state or property in your `SuperSlowExpression` that could be evaluated to get an idea of how much progress has been made.
-
1
I suggested this as well. But a question, since I don't really know anyone else who uses this actively: is this safe when doing parallel evaluations, especially in the case when TCPIP links are used between kernels? (Is there a chance that something will time out in a MathLink connection, messing up things irreparably?) – Szabolcs Feb 8 '12 at 16:55
lang-mma
|
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|
http://mathhelpforum.com/algebra/103099-discriminant-question.html
|
# Thread:
1. ## Discriminant question
Find the range of values of a for which the following equations have two real roots:
x^2 - 2ax + 3a = 0
I've got it down to a^2-3a>0, where do i go from here?
Thanks
2. The equation $a^2-3a=0$ has the roots $a_1=0, \ a_2=3$
Then $a^2-3a\geq 0$ if $a\in(-\infty,0]\cup[3,\infty)$
|
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http://mathoverflow.net/questions/79435?sort=votes
|
## Question about 0-dimensional Polish spaces
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hello everybody,
I'm stuck with proving (or disproving) the following statement.
Statement: For every $0$-dimensional Polish space $(X,\mathcal{T}\ )$, and a countable basis of clopen sets $\mathcal{B}$ for $\mathcal{T}$, every open set is the disjoint union of clopen sets in $\mathcal{B}$.
Every open set is the union $O=\cup_{n}B^{0}_{n}$ of the basic clopen sets contained in it (say ordered with a given numbering of $\mathcal{B}$). The idea is to make it a disjoint union by considering, iteratively,
$O= B^{0}_{1} \cup O^{1}$
where $O^{1} = O\setminus B^{0}_{1}$, which is open. Then again we have
$O^{1}=\cup_{n}B^{1}_{n}$.
So consider $O^{2}= O^{1}\setminus B^{1}_{1}$
etcetera. The resulting union
$\cup_{m} B^{m}_{1}$
is open. However, I'm stuck in proving that, in general, a point $x\in O$ ends up necessarily in some $B^{k}_{1}$, for $k\in \mathbb{N}$, i.e., I can't prove that
$O= \cup_{m}B^{m}_{1}$.
Googling around I found this interesting paper [1]. The author says that it is a known fact (unfortunately he doesn't give a reference) that for every $0$-dimensional Polish space, the Borel sets are generated from the clopens by closing under countable disjoint unions and complements. This does not solve my problem, but still, I would be interested in reading a proof. Could you point me to some relevant literature?
Thanks in advance,
[1] Abhijit Dasgupta. Constructing $\Delta^{0}_{3}$ using topologically restrictive countable disjoint unions.
-
## 2 Answers
Disjointify from the bottom up instead of from the top down, as you do in Real Analysis 1.
The question is at the level of homework, but I give a hint because you identify yourself and explain what you have tried, including looking at the literature, and I see why you are stuck on the problem.
EDIT 11.1.11: First an apology. Like Henno, I misread your question.
Let's examine what happens when $X$ is compact. Your question has an affirmative answer when $X$ is countable. Observe that it is enough to show that if $X$ is a countable compact metric space and $\mathcal{B}$ is a basis of clopen sets, then $X$ itself is the disjoint union of sets from $\mathcal{B}$. Indeed, from this special case you get that every clopen set is the disjoint union of basic clopen sets, and every open set is the disjoint union of clopen sets.
Now every countable compact metric space is homeomorphic to a ray $[1,\alpha]$ of ordinals, so you can use transfinite induction. Take a basic clopen set $A$ that contains $\alpha$ and apply the inductive hypothesis to `$X\sim A$`.
However, with some help from Gideon Schechtman I checked that your question has an negative answer when $X$ is the Cantor set `$\{-1,1\}^\Bbb{N}$`. Consider $X$ as a compact group with Haar measure and let $\mathcal{B}$ be the collection of clopen sets that have, for some $n$, measure `$2^{-n} + 2^{-n-1} = 3\cdot 2^{-n-1}$`. It is easy to check that this is a base for the topology. If $X$ were a disjoint union of sets from $\mathcal{B}$ then by compactness it would be just a finite disjoint union, which would imply that $1$ is a finite sum of numbers of the form `$3\cdot 2^{-n-1}$`.
-
Nice one; I was also looking for some base in the Cantor space as well for a counterexample. – Henno Brandsma Nov 1 2011 at 22:19
I kept getting lost in the complicated set theoretic combinatorics of bases but eventually realized that I could avoid the combinatorics entirely. Compactness is important in the argument. At first I thought the irrationals would be easier, but there also the combinatorics gave me problems. In fact, the simple bases for the irrationals I looked at did not give a counterexample. – Bill Johnson Nov 1 2011 at 23:32
### 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.
If you have such a union of $O = \cup_n B_n$, consider the sets $B'_n = B_n \setminus \cup_{i=0}^{n-1} B_i$, where $B'_0 = B_0$. Show these sets are disjoint, clopen, and have the same union as the original sets, as for every $x \in O$ there is a first index $n(x)$ such that $x \in B_{n(x)}$.
-
I agree: an answer should be an answer. Hints should be left as comments. – Todd Trimble Oct 29 2011 at 12:41
Hello Henno. Thanks for your ansewer. Your argument seems conclusive only if every clopen set of the form B'_{n} is itself expressible as a disjoint union of basic clopens. Note that in the statement of my question I didn't say that the basis is a boolean algebra. Of course I see it is possible to iterate the same procedure to "disjointify" B'_{n} (for every n>0), but again I don't see why a point should end up in a basic clopen at any finite stage. – Matteo Mio Oct 29 2011 at 18:07
@Matteo: do you really need exactly that statement? In other words, is there a problem with modifying the statement, taking into account that every basis of clopen sets can be refined to a basis of clopen sets that is a Boolean algebra, and working with a such a refinement? – Todd Trimble Oct 29 2011 at 20:00
@Todd: yes. Sometimes it is convenient to work with a basis which is not closed under, say, unions or intersections etcetera, even if the basis can always be refined to a boolean algebra. I'm not sure if the assertion, as it is formulated in the question, holds or not. – Matteo Mio Oct 29 2011 at 22:19
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http://www.physicsforums.com/showthread.php?s=d6ac94b40ef73978d413d8e6e579f7dd&p=4194368
|
Physics Forums
## Python and Rutherford Scattering
Hello everyone
I have been supplied with this eqn
$$n=\frac{Nat}{16r^2}(\frac{2Ze^2}{4\pi\epsilon_0E_K})^2cosec^4(\frac{\ph i}{2})$$
for Rutherford scattering.
$$N$$ is the number of alpha particles incident on a unit area of foil
$$t$$ is the thickness of the foil
$$a$$ is the number of atoms per unit volume within the foil
$$r$$ is the distance of the detector away from the collision
$$Z$$ is obviously the atomic number of the foil's material
and $$E_k$$ is the kinetic energy of the incident alpha particles
I need to use python to plot a graph of number of counts 'n' versus the scattering angle phi. To do this I need to create a GUI with entry fields for some of the above variables. This bit is not the problem.
The problem is so far I cannot get the eqn to give a useful plot.These figures are not based on an actual experiment which I have had to carry out. These are all theoretical values for the purpose of the computer program.
Could anyone suggest reasonable values for the above variables. Z is not necessary as I'm using 79 (gold). For example I've no idea how many atoms per unit volume there would be in a sheet of foil prepared for one of these experiments. Then if these reasonable values don't plot well either I will know that it is my code that needs looking at (which I'm sure is fine at the mo).
Any help would be greatly...greatly appreciated.
PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
Recognitions:
Science Advisor
Quote by maximus123 For example I've no idea how many atoms per unit volume there would be in a sheet of foil prepared for one of these experiments.
You can work that out easily enough from the density and the atomic mass. The atomic mass is 197 grams per mole, and the density is approx 19.3 grams per cm^3.
So there are 6.02E23/197 atoms per gram, and therefore 6.02E23/197*19.3 atoms per cm^3. This comes to about 5.9E22 by my quick calculation.
Recognitions:
Science Advisor
Quote by maximus123 The problem is so far I cannot get the eqn to give a useful plot.These figures are not based on an actual experiment which I have had to carry out. These are all theoretical values for the purpose of the computer program.
Well all of those constants essentially just boil down to a scaling factor for the cosec(theta/2) term anyway. So they are in effect irrelevant to the actual shape of the plot.
My problem with that equation is that the density of strikes goes to infinity as theta goes to zero, which doesn't seem right. Unless it is an approximation that is only valid over a limited range of theta perhaps. Do you have any more information on the origins of that formula?
## Python and Rutherford Scattering
like uart said, the basic shape of the plot is not going to change if all you are doing is plotting phi vs n.
here is what it looks like using Wolfram.
Thanks a lot uart. Not sure how my department has come up with the equation, tis probably some approximation for the purposes of the python exercise we have to do. You're right about the division by zero, I've had to make sure my range of angles does not include zero (which is stupid because most angles in this experiment would be zero) Any way I achieved my plot, an exponentially decaying curve. Thanks again.
Mentor
Quote by maximus123 Thanks a lot uart. Not sure how my department has come up with the equation, tis probably some approximation for the purposes of the python exercise we have to do.
It looks like a standard Rutherford scattering formula that you can find in textbooks and on Wikipedia. The result n is probably the number of scattered particles per m2 of detector area. It assumes point-like projectiles (alpha particles) and targets (nuclei).
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http://quant.stackexchange.com/tags/artificial-intelligence/hot?filter=all
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## Hot answers tagged artificial-intelligence
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### Multilayer Perceptron (Neural Network) for Time Series Prediction
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It's a bit unclear to me what you're trying to do, and maybe a better place to ask your question is Stats.SE but I would encourage you to go and have a look at this online class on machine learning which provides an implementation of the backpropagation algorithm. You can either register or hit preview and go to the NN:Learning chapter, but I would ...
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http://sbseminar.wordpress.com/2009/03/25/hall-algebras-and-donaldson-thomas-invariants-i/
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## Hall algebras and Donaldson-Thomas invariants I March 25, 2009
Posted by Joel Kamnitzer in Algebraic Geometry, conferences, homological algebra, quantum groups, things I don't understand.
trackback
I would like to tell you about recent work of Dominic Joyce and others (Bridgeland, Kontsevich-Soibelman, Behrend, Pandaripande-Thomas, etc) on Hall algebras and Donaldson-Thomas invariants. I don’t completely understand this work, but it seems very exciting to me. This post will largely be based on talks by Bridgeland and Joyce that I heard last month at MSRI.
In this post, I will concentrate on different versions of Hall algebras. Let us start with the most elementary one. Suppose I have an abelian category $\mathcal{A}$ which has the following strong finiteness properties: namely $Hom(A,B)$ and $Ext^1(A,B)$ are finite for any objects $A, B$. Then one can define an algebra, called the Hall algebra of $\mathcal{A}$, which has a basis given by isomorphism classes of objects of $\mathcal{A}$ and whose structure constants $c_{[M], [N]}^{[P]}$ are the number of subobjects of $P$ which are isomorphic to $N$ and whose quotient is isomorphic to $M$.
The main source of interest of Hall algebras for me is the Ringel-Green theorem which states that if you start with a quiver $Q$, then the Hall algebra of the category of representation of $Q$ over a finite field $\mathbb{F}_q$ is isomorphic to the upper half of the quantum group corresponding to $Q$ at the parameter $q^{1/2}$.
The obvious question concerning Hall algebras is to come up with a framework for understanding them when the Hom and Ext sets are not finite. This is what Joyce has done and he has applied it where $A$ is the category of coherent sheaves on a Calabi-Yau 3-fold.
Let us begin by reprasing the above definition. Let $M$ be the groupoid of objects of $\mathcal{A}$ (ie the same category but we throw away all morphisms which are not isomorphisms). Then we have a subgroupoid $M^{(2)}$ of $M \times M \times M$ which consists of short exact sequences. Then we can think of the Hall algebra as the convolution algebra you get from the various projections from $M^{(2)}$ to $M$. The “groupoid” nature of this becomes evident when we do the pushforward.
Now, let us generalize. We will need that our abelian category $\mathcal{A}$ comes equipped with some additional structure, namely there should be a moduli stack of objects $\mathcal{M}$. This moduli stack is usually easy to construct in examples. For quiver representations, it is simply the quotient stack corresponding to the quiver variety. For coherent sheaves on some variety $X$, it is the stack whose $S$ points are coherent sheaves on $X \times S$, flat over $S$.
Then we have a substack $\mathcal{M}^{(2)}$ of $\mathcal{M} \times \mathcal{M} \times \mathcal{M}$ for short exact sequences. Now we can do the same construction of convolution algebra as before except that we work with constructible functions on our stack. This means functions on the set of $k$ points of our stack (note this is just iso classes of objects) which are constructible in the sense of algebraic geometry (ie using constant functions on substacks). Thus we get a Hall algebra denote $CF(\mathcal{A})$.
Joyce states that when $\mathcal{A}$ is the category of representations of a quiver, the Hall algebra $CF(\mathcal{A})$ coincides with one defined by Lusztig and hence is isomorphic to the upper half of the universal envelopping algebra for the Lie algebra corresponding to the quiver.
However, for the application to DT invariants, a different (and more sophisticated) Hall algebra is required. Joyce considers what he calls stack functions. This is the vector space generated by isomorphism classes of stack functions over $\mathcal{M}$ modulo “motivic relations”. The relation with the constructible function is that you can think of the characteristic function of a substack as just that substack mapping to $\mathcal{M}$. Once Joyce proves that there is push-forward and pullback for these stack functions, he defines the Hall algebra of stack functions $SF(\mathcal{A})$ in a similar way.
In the case of quiver representations, he shows that $SF(\mathcal{A})$ is the upper half of the quantum group — here the parameter q is the “motivic variable”, corresponding to the affine line. This links up nicely with the finite field construction. Joyce also says that this fits with Lusztig’s perverse sheaf construction, although I don’t see how.
If I get around to writing part II of this post, I will explain how one constructs a map of algebras from $SF(\mathcal{A})$ to a twisted group algebra of $K(\mathcal{A})$ in the case where $\mathcal{A}$ is the category of coherent sheaves on a Calabi-Yau 3-fold and how one can use this to say something about DT invariants.
## Comments»
1. Joel Kamnitzer - March 26, 2009
I didn’t really say anything above about what it meant to quotient by “motivic relations”. As a first approximation, you quotient by the relation of disjoint union. There there is also some business about identifying the affine line with a variable q. Maybe, someone who understands motivic stuff better can explain this a bit better.
2. David Ben-Zvi - March 26, 2009
Great post! this stuff seems to me incredibly exciting, like the future of representation theory is being devised..
A couple of comments:
- the positive half of the quantum group always sits inside the Hall algebra of the corresponding quiver, but this is an isomorphism only for ADE quivers. Otherwise you get Lusztig’s “composition algebra”, a subalgebra generated by characteristic functions of components of
the moduli (or so I learned from Schiffmann’s great Lectures on Hall Algebras).
- Any abelian category has a moduli stack of objects, this is not an additional structure. Given a k-linear category C (here k can be Z), we can talk about R-families of objects in C for any k-algebra R, which are objects of C tensored up with R. We can also formulate flatness as an exactness property. This gives the functor of points of the moduli stack. In fact you don’t need the category to be abelian: in a derived category (construed correctly, i.e. with an enriched/dg structure) you get a moduli stack of objects without negative self-Exts
(Lieblich) and a moduli higher-stack of all objects (Toen-Vaquie)..
Maybe a nicer way to say this is that the functor (abelian categories) to (stacks) giving the moduli of objects can be characterized as the right adjoint of the functor (stacks) to (abelian categories) sending a stack to its category of coherent sheaves.. the idea is a functor from sheaves on X to your given category A gives an X-family of objects of A, namely the images of all the skyscraper sheaves on X.
- Finally, another name to mention in your list is Toledano-Laredo (for his joint work with Bridgeland). Both of them have given great expository “GRASP” talks on related topics at UT which are available online at http://www.math.utexas.edu/~benzvi/GRASP.html
3. Aaron Bergman - March 26, 2009
The talks at the recent miniconference at the KITP are also online at
http://online.kitp.ucsb.edu/online/duality09/
Definitely worth checking out.
4. Joel Kamnitzer - March 26, 2009
David -
1. Joyce also defines a composition subalgebra in the general case to be the subalgebra generated by the functions supported on the isoclasses of simple objects.
2. It is a nice, simple construction that you describe for producing a moduli stack of objects. I heard about that construction from Xinwen — I think that it will be used in his forthcoming paper with Frenkel about actions of algebraic groups on categories. Do you know of anywhere else where it is written up? Is the stack that you get this way always an algebraic (Artin) stack? (I guess I should look in those Lieblich and Toen-Vaquie papers you mention.)
In Joyce’s papers, he doesn’t use construction of a moduli stack of objects. Rather, he postulates it as extra data, which must satisfy certain axioms. I wonder if it amounts to the same thing in the end or if it differs in some examples.
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http://math.stackexchange.com/questions/202835/schwarz-lemma-extension/202876
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# Schwarz Lemma Extension
Let $f: \Omega \rightarrow \Omega$ be holomorphic where $\Omega \subset \mathbb{C}$ is a bounded region containing 0. If $f(0)=0$ and $f'(0)=1$, does $f(z)=z$?
This is true if $\Omega$ is a disk centered at 0 but does it hold if $D$ is only bounded?
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## 2 Answers
There is a substantial generalization of Schwarz' Lemma valid for arbitrary maps $f:\ \Omega\to\Omega$ of a Riemann surface into itself. It says that any such map which is not a conformal automorphism of $\Omega$ actually decreases the hyperbolic distance between points. This implies that if $|f'(z)|=1$ at some fixed point $z$ then $f$ is a conformal automorphism. This can be proven by lifting the map $f$ to a map $\tilde f:\ \tilde\Omega\to\tilde\Omega$ of the universal cover of $\Omega$. The latter is ("in most cases") conformally equivalent to the unit disk, where Pick's invariant version of Schwarz' Lemma can be applied. See here, p. 27:
Huber, Heinz: Analytische Abbildungen Riemannscher Flächen in sich. Comm. Math. Helv. 27 (1953), 1–73.
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Find a conformal map $h$ that maps $\Omega$ to the unit disc $D$, fixing $0$. Then $g:= h\circ f\circ h^{-1}:D\to D$ is holomorphic, with $g(0)=0$ and $g'(0)=1$.
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but $\Omega$ is just a bounded region containing zero. It might not be simply connected. If $\Omega$ was simply connected and open then I agree that such a conformal map exists by Riemann mapping theorem. I am not sure such a map exists. – Shankara Pailoor Sep 26 '12 at 14:29
you are right, I missed this condition.. – Berci Sep 27 '12 at 9:44
@Berci Well with this we can conclude that $f(h(z))=h(az)$ for some $a \in \mathbb{C}$ with $|a|=1$. How we can conclude that $a=1$ ? – Daniel Oct 8 '12 at 6:29
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http://math.stackexchange.com/questions/247425/is-there-a-vector-space-that-cannot-be-an-inner-product-space?answertab=oldest
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# Is there a vector space that cannot be an inner product space?
Quick question: Can I define some inner product on any arbitrary vector space such that it becomes an inner product space? If yes, how can I prove this? If no, what would be a counter example? Thanks a lot in advance.
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The easy answer is, no. But I think you want a non-trivial inner product. – vesszabo Nov 29 '12 at 17:49
@rschwieb I meant $\langle a,b\rangle =0$ for every $a,b$ is trivial. (My above comment is a joke only :-) ) So yes, what is interesting the question of the existence of positive definite inner product. As I see, the answer of Christian is complete. – vesszabo Nov 29 '12 at 21:47
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I'm sorry, in all lectures I have attended so far, the inner product is positive by definition, which is why I did not specify it. – Huy Nov 30 '12 at 8:10
@vesszabo Sorry, I should have kept my mutterings to myself! I didn't really have anything interesting to say either way :) – rschwieb Nov 30 '12 at 14:06
## 2 Answers
How about vector spaces over finite fields? Finite fields don't have an ordered subfield, and thus one cannot meaningfully define a positive-definite inner product on vector spaces over them.
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I'm not sure, but can one not just use the well-ordering theorem? – Huy Nov 29 '12 at 18:45
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You can't get an ordering compatible with the ring structure. – Zhen Lin Nov 29 '12 at 18:56
@ZhenLin How about the ring of integers? – chaohuang Nov 29 '12 at 23:04
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The ring of integers isn't a field. – Zhen Lin Nov 29 '12 at 23:22
I'm assuming the ground field is ${\mathbb R}$ or ${\mathbb C}$, because otherwise it's not clear what an "inner product space" is.
Now any vector space $X$ over ${\mathbb R}$ or ${\mathbb C}$ has a so-called Hamel basis. This is a family $(e_\iota)_{\iota\in I}$ of vectors $e_\iota\in X$ such that any $x\in X$ can be written uniquely in the form $x=\sum_{\iota\in I} \xi_\iota\ e_\iota$, where only finitely many $\xi_\iota$ are $\ne 0$. Unfortunately you need the axiom of choice to obtain such a basis, if $X$ is not finitely generated.
Defining $\langle x, y\rangle :=\sum_{\iota\in I} \xi_\iota\ \bar\eta_\iota$ gives a bilinear "scalar product" on $X$ such that $\langle x, x\rangle>0$ for any $x\ne0$. Note that in computing $\langle x,y\rangle$ no question of convergence arises.
It follows that $\langle\ ,\ \rangle$ is an inner product on $X$, and adopting the norm $\|x\|^2:=\langle x,x\rangle$ turns $X$ into a metric space in the usual way.
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Perhaps you could expand on what convergence w.r.t. (the metric generated by) your inner product means? – kahen Nov 29 '12 at 19:36
– kahen Nov 29 '12 at 19:41
@kahen: See my edit. – Christian Blatter Nov 30 '12 at 9:16
6
@vesszabo: As Christian explain, given the axiom of choice, any arbitrary vector space can admit an inner product. In some sense just having a vector space is too floppy. It is more interesting to ask your question for a more rigid class of spaces, the topological vector spaces. In which case the answer is no: you cannot always find a inner product compatible with the given topology. There exists TVSs which are not normable or metrizable. – Willie Wong♦ Nov 30 '12 at 9:30
@WillieWong It's a valuable remark. Indeed, a vector space structure alone is (usually) not "too much" to work with it. – vesszabo Dec 1 '12 at 11:04
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