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## Return to Question
8 added 99 characters in body
EDIT: Based on comments, I've decided to essentially rewrite this question. My apologies to those who commented below, whose comments seem a bit off topic when you try to match them with the current question. I've also posted a better-thought out follow-up here.
Lots of people like to study torus actions on varieties and manifolds, and in particular, like to study their fixed points. Most people I know just think about fixed points as a set, but they have a canonical scheme structure which carries more information.
As Bcnrd points out below, this subscheme of a scheme $X$ over a field $k$ is defined by looking at the functor on $k$-algebras defined abstractly by the $(\mathbb{G}_m)_A$-invariant points of $X(A)$. As Dave Anderson points out, if $X=\mathrm{Spec}(R)$, then this is simply the subscheme defined by the ideal generated by all elements of non-zero weight $I=R^{>0}R+R^{<0}R$.
For example, consider $R=\mathbb{C}[x,y,z]/(xy=z^n)$ where $x$ has weight 1, $y$ has weight -1 and $z$ has weight 0. In this case $I=(x,y)$, so $X^{\mathbb{G}_m}=\mathrm{Spec} \:\:\:\mathbb{C}[z]/(z^n)$. So, as you can see, the fixed point scheme doesn't have to be reduced, though if $X^{\mathbb{G}_m}$ is 0-dimensional, this can only happen if $X$ is not regular at the corresponding fixed point.
Now, imagine $X$ has a $\mathbb{G}_m$-equivariant resolution of singularities $\tilde X$, with a compatible $\mathbb{G}_m$ action, and $\tilde X^{\mathbb{G}_m}$ also 0 dimensional.
Can I conclude anything about the length of $X^{\mathbb{G}_m}$ from knowing the length of $\tilde X^{\mathbb{G}_m}$ (which is just the number of fixed $k$-points by smoothness of $\tilde X$)? In a number of examples I'm looking at, these turn out to be equal, and I'm wondering how general a phenomenon this is.
For example, the example $R=\mathbb{C}[x,y,z]/(xy=z^n)$ has a resolution with $n$ fixed points, and indeed, that's the length of $X^{\mathbb{G}_m}$.
Now, the examples I'm looking at have special features which may or may not be revelant, but I mention them in case they strike a chord.
1. I'm looking at examples where $\tilde X$ is symplectic, and the $\mathbb{G}_m$-action is Hamiltonian.
2. Also in my examples, $\tilde X$ has a smooth $\mathbb{G}_m$-equivariant deformation $\tilde Y$, where the generic fiber is affine, and the fixed point scheme $\tilde Y^{\mathbb{G}_m}$ is flat and finite over the base.
7 added 1 characters in body
EDIT: Based on comments, I've decided to essentially rewrite this question. My apologies to those who commented below, whose comments seem a bit off topic when you try to match them with the current question.
Lots of people like to study torus actions on varieties and manifolds, and in particular, like to study their fixed points. Most people I know just think about fixed points as a set, but they have a canonical scheme structure which carries more information.
As Bcnrd points out below, this subscheme of a scheme $X$ over a field $k$ is defined by looking at the functor on $k$-algebras defined abstractly by the $(\mathbb{G}_m)_A$-invariant points of $X(A)$. As Dave Anderson points out, if $X=\mathrm{Spec}(R)$, then this is simply the subscheme defined by the ideal generated by all elements of non-zero weight $I=R^{>0}R+R^{<0}R$.
For example, consider $R=\mathbb{C}[x,y,z]/(xy=z^n)$ where $x$ has weight 1, $y$ has weight -1 and $z$ has weight 0. In this case $I=(x,y)$, so $X^{\mathbb{G}_m}=\mathrm{Spec} \:\:\:\mathbb{C}[z]/(z^n)$. So, as you can see, the fixed point scheme doesn't have to be reduced, though if $X^{\mathbb{G}_m}$ is 0-dimensional, this can only happen if $X$ is not regular at the corresponding fixed point.
Now, imagine $X$ has a $\mathbb{G}_m$-equivariant resolution of singularities $\tilde X$, with a compatible $\mathbb{G}_m$ action, and $\tilde X^{\mathbb{G}_m}$ also 0 dimensional.
Can I conclude anything about the length of $X^{\mathbb{G}_m}$ from knowing the length of $\tilde X^{\mathbb{G}_m}$ (which is just the number of fixed $k$-points by smoothness of $\tilde X$)? In a number of examples I'm looking at, these turn out to be equal, and I'm wondering how general a phenomenon this is.
For example, the example $R=\mathbb{C}[x,y,z]/(xy=z^n)$ has a resolution with $n$ fixed points, and indeed, that's the length of $X^{\mathbb{G}_m}$.
Now, the examples I'm looking at have special features which may or may not be revelant, but I mention them in case they strike a cordchord.
1. I'm looking at examples where $\tilde X$ is symplectic, and the $\mathbb{G}_m$-action is Hamiltonian.
2. Also in my examples, $\tilde X$ has a smooth $\mathbb{G}_m$-equivariant deformation $\tilde Y$, where the generic fiber is affine, and the fixed point scheme $\tilde Y^{\mathbb{G}_m}$ is flat and finite over the base.
6 Fixed up the description of the fixed-point scheme.
EDIT: Based on comments, I've decided to essentially rewrite this question. My apologies to those who commented below, whose comments seem a bit off topic when you try to match them with the current question.
Lots of people like to study torus actions on varieties and manifolds, and in particular, like to study their fixed points. Most people I know just think about fixed points as a set, but they have a canonical scheme structure which carries more information.
As Bcnrd points out below, this subscheme of a scheme $X$ over an algebraically closed a field $k$ is defined by looking at the sheaf functor on $k$-algebras defined abstractly by the $\mathbb{G}_m(A)$ invariant (\mathbb{G}_m)_A$-invariant points of$X(A)$. As Dave Anderson points out, if$X=\mathrm{Spec}(R)$, then this is simply the subscheme defined by the ideal generated by all elements of non-zero weight$I=R^{>0}R+R^{<0}R\$.
For example, consider $R=\mathbb{C}[x,y,z]/(xy=z^n)$ where $x$ has weight 1, $y$ has weight -1 and $z$ has weight 0. In this case $I=(x,y)$, so $X^{\mathbb{G}_m}=\mathrm{Spec} \:\:\:\mathbb{C}[z]/(z^n)$. So, as you can see, the fixed point scheme doesn't have to be reduced, though if $X^{\mathbb{G}_m}$ is 0-dimensional, this can only happen if $X$ is not regular at the corresponding fixed point.
Now, imagine $X$ has a $\mathbb{G}_m$-equivariant resolution of singularities $\tilde X$, with a compatible $\mathbb{G}_m$ action, and $\tilde X^{\mathbb{G}_m}$ also 0 dimensional.
Can I conclude anything about the length of $X^{\mathbb{G}_m}$ from knowing the length of $\tilde X^{\mathbb{G}_m}$ (which is just the number of fixed $k$-points by smoothness of $\tilde X$)? In a number of examples I'm looking at, these turn out to be equal, and I'm wondering how general a phenomenon this is.
For example, the example $R=\mathbb{C}[x,y,z]/(xy=z^n)$ has a resolution with $n$ fixed points, and indeed, that's the length of $X^{\mathbb{G}_m}$.
Now, the examples I'm looking at have special features which may or may not be revelant, but I mention them in case they strike a cord.
1. I'm looking at examples where $\tilde X$ is symplectic, and the $\mathbb{G}_m$-action is Hamiltonian.
2. Also in my examples, $\tilde X$ has a smooth $\mathbb{G}_m$-equivariant deformation $\tilde Y$, where the generic fiber is affine, and the fixed point scheme $\tilde Y^{\mathbb{G}_m}$ is flat and finite over the base.
5 added 53 characters in body
EDIT: Based on comments, I've decided to essentially rewrite this question. My apologies to those who commented below, whose comments seem a bit off topic when you try to match them with the current question.
Lots of people like to study torus actions on varieties and manifolds, and in particular, like to study their fixed points. Most people I know just think about fixed points as a set, but they have a canonical scheme structure which carries more information.
As Bcnrd points out below, this subscheme of a scheme $X$ over an algebraically closed field $k$ is defined by looking at the sheaf on $k$-algebras defined abstractly by the $\mathbb{G}_m(A)$ invariant points of $X(A)$. As Dave Anderson points out, if $X=\mathrm{Spec}(R)$, then this is simply the subscheme defined by the ideal generated by all elements of non-zero weight $I=R^{>0}R+R^{<0}R$.
For example, consider $R=\mathbb{C}[x,y,z]/(xy=z^n)$ where $x$ has weight 1, $y$ has weight -1 and $z$ has weight 0. In this case $I=(x,y)$, so $X^{\mathbb{G}_m}=\mathrm{Spec} \:\:\:\mathbb{C}[z]/(z^n)$. So, as you can see, the fixed point scheme doesn't have to be reduced, though if $X^{\mathbb{G}_m}$ is 0-dimensional, this can only happen if $X$ is not regular at the corresponding fixed point.
Now, imagine $X$ has a $\mathbb{G}_m$-equivariant resolution of singularities $\tilde X$, with a compatible $\mathbb{G}_m$ action, and $\tilde X^{\mathbb{G}_m}$ also 0 dimensional.
Can I conclude anything about the length of $X^{\mathbb{G}_m}$ from knowing the length of $\tilde X^{\mathbb{G}_m}$ (which is just the number of fixed $k$-points by smoothness of $\tilde X$)? In a number of examples I'm looking at, these turn out to be equal, and I'm wondering how general a phenomenon this is.
For example, the example $R=\mathbb{C}[x,y,z]/(xy=z^n)$ has a resolution with $n$ fixed points, and indeed, that's the length of $X^{\mathbb{G}_m}$.
Now, the examples I'm looking at have special features which may or may not be revelant, but I mention them in case they strike a cord.
1. I'm looking at examples where $\tilde X$ is symplectic, and the $\mathbb{G}_m$-action is Hamiltonian.
2. Also in my examples, $\tilde X$ has a smooth $\mathbb{G}_m$-equivariant deformation $\tilde Y$, where the generic fiber is affine, and the fixed point scheme $\tilde Y^{\mathbb{G}_m}$ is flat and finite over the base.
4 significant rewrite; added 1 characters in body
EDIT: Based on comments, I've decided to essentially rewrite this question. My apologies to those who commented below, whose comments seem a bit off topic.
Lots of people like to study torus actions on varieties and manifolds, and in particular, like to study their fixed points. Most people I recently in my research came upon know just think about fixed points as a variation on this that I've been stuck on some details ofset, but they have a canonical scheme structure which carries more information.
Given an action
As Bcnrd points out below, this subscheme of $\mathbb{G}_m$ on an affine a scheme $X=\mathrm{Spec}(R)$ (lets' say X$over an algebraically closed field), we can consider the set of closed points which are field$\mathbb{G}_m$-invariant, or we can study the scheme k$ is defined by looking at the ring $B=R^0/(R^0\cap R^{>0}R^{<0})$ where sheaf on $R^\star$ denotes k$-algebras defined abstractly by the set of elements$\mathbb{G}_m(A)$invariant points of$R$whose weights are in that set. Geometrically, this should correspond to the fixed X(A)$. As Dave Anderson points , but maybe with some funny scheme structure. Indeedout, the closed points of this scheme are just if $X^{\mathbb{G}_m}$, but not necessarily with X=\mathrm{Spec}(R)$, then this is simply the reduced scheme structure. Somehow, subscheme defined by the unreducedness ideal generated by all elements of this measures how singular the variety is at fixed points (for example, if you're smooth at a fixed point, it will be reduced).non-zero weight$I=R^{>0}R+R^{<0}R\$.
For example, consider $R=\mathbb{C}[x,y,z]/(xy=z^n)$ where $x$ has weight 1, $y$ has weight -1 and $z$ has weight 0. In this case $B=\mathbb{C}[z]/(z^n)$.I=(x,y)$, so$X^{\mathbb{G}_m}=\mathrm{Spec} \:\:\:\mathbb{C}[z]/(z^n)$. So, as you can see, the fixed point scheme doesn't have to be reduced, though if$X^{\mathbb{G}_m}$is 0-dimensional, this can only happen if$X\$ is not regular at the corresponding fixed point.
Now, imagine $X$ has a $\mathbb{G}_m$-equivariant resolution of singularities $\tilde X$, with isolated a compatible $\mathbb{G}_m$ fixed pointsaction, and $\tilde X^{\mathbb{G}_m}$ also 0 dimensional.
Can I conclude anything about the dimension length of $B=R^0/(R^0\cap R^{>0}R^{<0})$ X^{\mathbb{G}_m}$from knowing the number length of fixed points on the resolution$\tilde X\$? Especially, can I find upper bounds?
What I would love would be if X^{\mathbb{G}_m}$(which is just the number of fixed points in the resolution were the same as the dimension$k$-points by smoothness of$B$. \tilde X$)? In a number of examples I'm looking at, these turn out to be equal, and I'm wondering how general a phenomenon this is.
For example, the example $R=\mathbb{C}[x,y,z]/(xy=z^n)$ has a resolution with $n$ fixed points, and indeed, that's the dimension length of $B$.X^{\mathbb{G}_m}\$.
Now, the examples I'm willing to assume looking at have special features which may or may not be revelant, but I mention them in case they strike a large number of things about this situationcord.For example:
• I'm willing to assume that looking at examples where $\tilde X$ is symplectic, and the $\mathbb{G}_m$-action is Hamiltonian.
• I'm also willing to assume that
• Also in my examples, $\tilde X$ has a smooth $\mathbb{G}_m$-equivariant deformation $\tilde Y$, where the generic fiber is affine, and the fixed point scheme $\tilde Y^{\mathbb{G}_m}$ is flat and finite over the base.
• 3 deleted 78 characters in body
Lots of people like to study torus actions on varieties and manifolds, and in particular, like to study their fixed points. I recently in my research came upon a variation on this that I've been stuck on some details of.
Given an action of $\mathbb{G}_m$ on an affine scheme $X=\mathrm{Spec}(R)$ (lets' say over an algebraically closed field), we can consider the set of closed points which are $\mathbb{G}_m$-invariant, or we can study the subscheme scheme defined by the ideal ring $B=R^0/(R^0\cap R^{>0}R^{<0})$ where $R^\star$ denotes the set of elements of $R$ whose weights are in that set. Geometrically, this can be thought of as passing should correspond to the subscheme where positive weights vanish, and then passing tofixed fixed points, but maybe with some funny scheme structure. Indeed, the closed points of this subscheme scheme are just $X^{\mathbb{G}_m}$, but not necessarily with the reduced scheme structure. Somehow, the unreducedness of this measures how singular the variety is at fixed points (for example, if you're smooth at a fixed point, it will be reduced).
For example, consider $R=\mathbb{C}[x,y,z]/(xy=z^n)$ where $x$ has weight 1, $y$ has weight -1 and $z$ has weight 0. In this case $B=\mathbb{C}[z]/(z^n)$.
Now, imagine $X$ has a $\mathbb{G}_m$-equivariant resolution of singularities $\tilde X$, with isolated $\mathbb{G}_m$ fixed points.
Can I conclude anything about the dimension of $B=R^0/(R^0\cap R^{>0}R^{<0})$ from knowing the number of fixed points on the resolution $\tilde X$? Especially, can I find upper bounds?
What I would love would be if the number of fixed points in the resolution were the same as the dimension of $B$. For example, the example $R=\mathbb{C}[x,y,z]/(xy=z^n)$ has a resolution with $n$ fixed points, and indeed, that's the dimension of $B$.
1. I'm willing to assume that $\tilde X$ is symplectic, and the $\mathbb{G}_m$-action is Hamiltonian.
2. I'm also willing to assume that $\tilde X$ has a smooth $\mathbb{G}_m$-equivariant deformation, where the generic fiber is affine.
2 added 439 characters in body; edited title
# Understanding the unreducedness of "scheme-yasubschemesupportedon fixed points"
Lots of people like to study torus actions on varieties and manifolds, and in particular, like to study their fixed points. I recently in my research came upon a variation on this that I've been stuck on some details of.
Given an action of $\mathbb{G}_m$ on an affine scheme $X=\mathrm{Spec}(R)$ (lets' say over an algebraically closed field), we can consider the set of closed points which are $\mathbb{G}_m$-invariant, or we can study the subscheme defined by the ideal $R^0/(R^0\cap B=R^0/(R^0\cap R^{>0}R^{<0})$ where $R^\star$ denotes the set of elements of $R$ whose weights are in that set. Geometrically, this can be thought of as passing to the subscheme where positive weights vanish, and then passing tofixed points, but maybe with some funny scheme structure. Indeed, the closed points of this subscheme are just $X^{\mathbb{G}_m}$, but not necessarily with the reduced scheme structure. Somehow, the unreducedness of this measures how singular the variety is at fixed points (for example, if you're smooth at a fixed point, it will be reduced).
For example, consider $R=\mathbb{C}[x,y,z]/(xy=z^n)$ where $x$ has weight 1, $y$ has weight -1 and $z$ has weight 0. In this case $B=\mathbb{C}[z]/(z^n)$.
Now, imagine $X$ has a $\mathbb{G}_m$-equivariant resolution of singularities $\tilde X$, with isolated $\mathbb{G}_m$ fixed points.
Can I conclude anything about the dimension of $R^0/(R^0\cap B=R^0/(R^0\cap R^{>0}R^{<0})$ from knowing the number of fixed points on the resolution $\tilde X$? Especially, can I find upper bounds?
What I would love would be if the number of fixed points in the resolution were the same as the dimension of $B$. For example, the example $R=\mathbb{C}[x,y,z]/(xy=z^n)$ has a resolution with $n$ fixed points, and indeed, that's the dimension of $B$.
1. I'm willing to assume that $\tilde X$ is symplectic, and the $\mathbb{G}_m$-action is Hamiltonian.
2. I'm also willing to assume that $\tilde X$ has a smooth $\mathbb{G}_m$-equivariant deformation, where the generic fiber is affine.
1
# Understanding the unreducedness of "scheme-y fixed points"
Lots of people like to study torus actions on varieties and manifolds, and in particular, like to study their fixed points. I recently in my research came upon a variation on this that I've been stuck on some details of.
Given an action of $\mathbb{G}_m$ on an affine scheme $X=\mathrm{Spec}(R)$ (lets' say over an algebraically closed field), we can consider the set of closed points which are $\mathbb{G}_m$-invariant, or we can study the subscheme defined by the ideal $R^0/(R^0\cap R^{>0}R^{<0})$ where $R^\star$ denotes the set of elements of $R$ whose weights are in that set. Geometrically, this can be thought of as passing to the subscheme where positive weights vanish, and then passing tofixed points, but maybe with some funny scheme structure. Indeed, the closed points of this subscheme are just $X^{\mathbb{G}_m}$, but not necessarily with the reduced scheme structure. Somehow, the unreducedness of this measures how singular the variety is at fixed points (for example, if you're smooth at a fixed point, it will be reduced).
Now, imagine $X$ has a $\mathbb{G}_m$-equivariant resolution of singularities, with isolated $\mathbb{G}_m$ fixed points.
Can I conclude anything about the dimension of $R^0/(R^0\cap R^{>0}R^{<0})$ from knowing the number of fixed points on the resolution? Especially, can I find upper bounds?
1. I'm willing to assume that $\tilde X$ is symplectic, and the $\mathbb{G}_m$-action is Hamiltonian.
2. I'm also willing to assume that $\tilde X$ has a smooth $\mathbb{G}_m$-equivariant deformation, where the generic fiber is affine.
|
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|
http://en.wikipedia.org/wiki/Generalized_context-free_grammar
|
# Generalized context-free grammar
Generalized Context-free Grammar (GCFG) is a grammar formalism that expands on context-free grammars by adding potentially non-context free composition functions to rewrite rules.[1] Head grammar (and its weak equivalents) is an instance of such a GCFG which is known to be especially adept at handling a wide variety of non-CF properties of natural language.
## Description
A GCFG consists of two components: a set of composition functions that combine string tuples, and a set of rewrite rules. The composition functions all have the form $f(\langle x_1, ..., x_m \rangle, \langle y_1, ..., y_n \rangle, ...) = \gamma$, where $\gamma$ is either a single string tuple, or some use of a (potentially different) composition function which reduces to a string tuple. Rewrite rules look like $X \to f(Y, Z, ...)$, where $Y$, $Z$, ... are string tuples or non-terminal symbols.
The rewrite semantics of GCFGs is fairly straight forward. An occurrence of a non-terminal symbol is rewritten using rewrite rules as in a context-free grammar, eventually yielding just compositions (composition functions applied to string tuples or other compositions). The composition functions are then applied, reducing successively reducing the tuples to a single tuple.
## Example
A simple translation of a context-free grammar into a GCFG can be performed in the following fashion. Given the grammar in (1), which generates the palindrome language $\{ ww^R : w \in \{a, b\}^{*} \}$, where $w^R$ is the string reverse of $w$, we can define the composition function conc as in (2a) and the rewrite rules as in (2b).
1. $S \to \epsilon ~|~ aSa ~|~ bSb$
1. $conc(\langle x \rangle, \langle y \rangle, \langle z \rangle) = \langle xyz \rangle$
2. $S \to conc(\langle \epsilon \rangle, \langle \epsilon \rangle, \langle \epsilon \rangle) ~|~ conc(\langle a \rangle, S, \langle a \rangle) ~|~ conc(\langle b \rangle, S, \langle b \rangle)$
The CF production of abbbba is
S
aSa
abSba
abbSbba
abbbba
and the corresponding GCFG production is
$S \to conc(\langle a \rangle, S, \langle a \rangle)$
$conc(\langle a \rangle, conc(\langle b \rangle, S, \langle b \rangle), \langle a \rangle)$
$conc(\langle a \rangle, conc(\langle b \rangle, conc(\langle b \rangle, S, \langle b \rangle), \langle b \rangle), \langle a \rangle)$
$conc(\langle a \rangle, conc(\langle b \rangle, conc(\langle b \rangle, conc(\langle \epsilon \rangle, \langle \epsilon \rangle, \langle \epsilon \rangle), \langle b \rangle), \langle b \rangle), \langle a \rangle)$
$conc(\langle a \rangle, conc(\langle b \rangle, conc(\langle b \rangle, \langle \epsilon \rangle, \langle b \rangle), \langle b \rangle), \langle a \rangle)$
$conc(\langle a \rangle, conc(\langle b \rangle, \langle bb \rangle, \langle b \rangle), \langle a \rangle)$
$conc(\langle a \rangle, \langle bbbb \rangle, \langle a \rangle)$
$\langle abbbba \rangle$
## Linear Context-free Rewriting Systems (LCFRSs)
Weir (1988)[1] describes two properties of composition functions, linearity and regularity. A function defined as $f(x_1, ..., x_n) = ...$ is linear if and only if each variable appears at most once on either side of the =, making $f(x) = g(x, y)$ linear but not $f(x) = g(x, x)$. A function defined as $f(x_1, ..., x_n) = ...$ is regular if the left hand side and right hand side have exactly the same variables, making $f(x, y) = g(y, x)$ regular but not $f(x) = g(x, y)$ or $f(x, y) = g(x)$.
A grammar in which all composition functions are both linear and regular is called a Linear Context-free Rewriting System (LCFRS), a subset of the GCFGs with strictly less computational power than the GCFGs as a whole, which is weakly equivalent to multicomponent Tree adjoining grammars. Head grammar is an example of an LCFRS that is strictly less powerful than the class of LCFRSs as a whole.
## References
1. ^ a b Weir, David H. 1988. Characterizing mildly context-sensitive grammar formalisms. Dissertation, U Penn.
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|
http://physics.stackexchange.com/questions/30496/water-to-ice-expansion-in-1-textrmmm3-pit-pressure-on-the-pit-walls?answertab=oldest
|
Water to ice expansion in $1\textrm{mm}^3$ pit - pressure on the pit walls?
[EDITED] by mistake, the subject was regarding 1m^3 instead of 1mm^3. There should be a significant difference between the two...
A 1x1x1mm pit filled with water is frozen at a slow rate (1K/minute).
When the water becomes ice and expands 9%, I assume that it will be pressed out of the pit since it has nowhere else to go. But how much pressure does it apply to the pit walls?
We can assume that the water is impure and will freeze at 0 degrees C. Whether the pit material or the air is first to reach below 0 degrees C cannot be assumed.
As you can imagine, I need to know whether the pit may be damaged.
-
This is a hard problem as it depends on the local temperatures in and around the ice forming. Guessing from the geometry of the problem the expanding ice can push out some water at the top and the pressure should be negligible for a pit made out of a metal, crystal or similar material. – Alexander Jun 21 '12 at 11:07
1
This is impossible to solve without saying whether the pit will get cold first, or the air. If the water freezes from the top down, it will break the pit, if it freezes from the bottom up, nothing will happen. – Ron Maimon Jul 5 '12 at 6:54
Also note that your cooling rate is not quite right: the freezing process can (and probably should?) be considered to occur at $0^\circ\textrm{ C}$. The cooling rate that matters is the rate of removal of latent heat, i.e. something with dimensions of power. – Emilio Pisanty Jul 5 '12 at 13:48
Your diagram that says "Ice Expanded 9%": is that a given expansion along the vertical direction or are you referring to total volumetric expansion? This problem has too many unknowns to solve outright unless the height of the ice protrusion is known. (Also, technically the problem is statically non-determinate without knowing or assuming the hardness of the walls; perhaps assume they are perfectly rigid?) – nicholas Jul 5 '12 at 20:50
@nicholas. I recognize the point that with perfectly rigid walls, the water would just be pushed upwards. The walls in this case are reinforced plastic (top 0.5mm), rubber o-ring (bottom 0.5mm) and silicon at the bottom. I tried to define the problem as generic as possible, but I can now see that it is impossible to define the problem without including the hardness of the walls. – Christian Madsen Jul 11 '12 at 18:46
4 Answers
1 atm.
The atmosphere is pushing down on the water, holding it the box. When the freezing water expands, it pushes against the atmosphere and wins. Once the pressures are balanced, the volume comes to rest. There is no pressure on the side walls. If there was, the ice would expand if you took it out of the mold.
There is, of course, mg/(1 mm)^2 pressure down simply from the weight of water. That doesn't change when you freeze it.
-
"If there was, the ice would expand if you took it out of the mold." Yes, definitely. But why are we sure that that wouldn't happen? – Emilio Pisanty Jul 5 '12 at 14:07
Since the water expands in all directions, I don't understand how we get 1 atm on the side walls. – Christian Madsen Jul 11 '12 at 18:26
The water will push in all directions until it comes to equilibrium. If the water is pressing at more than 1 atm, it will expand upwards until it is pressing 1 atm at the walls. I'm assuming a slow, steady, non-interesting freeze. If the top freezes first then all bets are off. – 16BitTons Jul 13 '12 at 1:41
An exact answer of your problem will only come from an experiment. The exact conditions and temperature profiles can make all the difference.
For a slow and homogeneous cooling process the pressure should not exceed atmospheric pressure significantly, as the ratio between volume and surface area is not critical.
The highest measured pressure in frost weathering is 207 MPa, so it can be enormous if the water is enclosed and has only narrow paths to escape. There is a fairly large review article available from 1991 about frost heave in which several factors are discussed, such as the influence of different temperature gradients and of different the crack sizes. A more recent study is unfortunately behind a paywall but you can always ask the authors directly for a copy.
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Cool problem! A little phase transition thermodynamics, though more information will be necessary for a complete answer, e.g. ambient air temperature, container surface temperature, and the temperature range change for each, if applicable. You may have to bust out the old pchem book for this one.
Water will freeze first at the boundary where it's losing energy. Let's look at the scenarios:
1) water freezes from the top first because that's where it's losing energy the fastest. - the thicker the water layer gets, the greater chance you have of doing damage to the container, because the frozen layer is in place while the liquid water beneath is freezing and expanding. It will expand into the path of least resistance; it's likely that as the ice wall becomes thicker, the path of least resistance will be the material of your container, unless it's a very thick and strong material. What's it made out of? How thick is it?
2) water freezes against the container walls first (if the walls are really cold. Is this outside or in a lab?) - if each wall surface is the same temperature, then the water will freeze uniformly and expand against the liquid towards the air barrier. In this case, you're safe.
3) water freezes against the container and air boundaries with a liquid core. This result will be similar to case 1.
For the floor of your container, 1m^3 of Vienna standard mean ocean water has a density of 1000kg/m^3, and this will exert a pressure on it of P = F/A = mg/A = (1000 kg)(9.80665 m s^-2)/(1 m^2) = 9806 N m^-2 = 1.422 psi, from the weight alone. Liquids will exert a pressure on the side walls, but if you've got case 2, ice will not once it's solid.
That's the best I can do with the given information; temperature measurements from the walls and air, as well as information about the container will let us answer your question more concisely.
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in scenario 1, the ice surface would probably not get thicker in the "down" direction, but rather as pressure builds in the remaining liquid water push the ice cap upwards, maintaining atmospheric pressure. in scenario 2 a similar pushing would happen (possibly even allowing water to spill). For 1 and 2, given a sufficient cooling rate, no pressure would build. For scenario 3, in which we treat the cube as an expanding solid, stress could build against the container. – nicholas Jul 5 '12 at 20:55
The water follows the path of least resistance, which in this case is upwards. I think the result will be that the ice would finish up with a domed of slightly pointed appearance
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http://mathhelpforum.com/advanced-algebra/202520-how-find-if-neutral-idnty-element-exists-16-different-binary-operations.html
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1Thanks
• 1 Post By HallsofIvy
# Thread:
1. ## How to find if neutral/idnty element exists for 16 different binary operations of *?
Suppose I'm given an operation $*$ on a set $A\,\, \text{where}\,\, a,b \in A$ such that the operation between two elements of set $A$ is also an element of set $A$
Now there are $2 \times 2 \times 2 \times 2 = 16$ possible binary operations of $*$ on set $A$.
If I'm asked to find if there exists a neutral/identity element of a certain table where $a * a, a * b, b * a \text{ and } b * b$
are defined, how do I figure that out?
What I mean is suppose 1 of 16 table of operations looks like this:
$\begin{array}{|l||r|} \hline (a * a) & a \\ \hline (a * b) & b \\ \hline (b * a) & a \\ \hline (b * b) & b \\ \hline \end{array}$
How do I figure out if an identity element exist for this table? Also is it possible to find the identity/neutral element from this table?
2. ## Re: How to find if neutral/idnty element exists for 16 different binary operations of
That binary operation does not have an identity. The definition of identity is "e is an identity for operation "*" if and only if e*x= x*e= x for every x in in the set". Here, a*b= b but $b*a= a\ne b$ so a is NOT an identity. Similarly, b*a= a but $a*b= b\ne a$ so b is not an identity.
3. ## Re: How to find if neutral/idnty element exists for 16 different binary operations of
Again thanks.
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http://mathoverflow.net/questions/106543/p-adic-representations-of-groups/106548
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## p-adic representations of groups
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hi, is there a possibility to classify irreducible representations of (finite) groups over $\mathbb{Z_p}$ with $p\neq 2$ a prime? Can one get the number of the irred. representations here? I tried to solve this for "easy" groups like the cyclic or symmetric groups, but did not really get a result. Maybe there is a connection to representation over $\mathbb{Z}/p\mathbb{Z}$?
Best regards
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## 2 Answers
Usually, "irreducible" means having no subrepresentations. In the integral context, there is no such thing, since for any $\mathbb{Z}_p[G]$-module $M$, $pM$ is a proper submodule. The right question is to try and classify the indecomposable representations, i.e. those that cannot be written as direct sums of subrepresentations.
To find all indecomposable $\mathbb{Z}_p$-representations of a given finite group is a hard problem, in general, and a lot of work has been done on special cases. If $G$ is a cyclic group of order $n$, then it is known that there is a finite number of indecomposable $\mathbb{Z}_p[G]$-modules if and only if $p$ divides $n$ at most to the power 2. If $p^3|n$, then there are infinitely many indecomposable modules over $\mathbb{Z}_p$. See Heller, Representations of Cyclic Groups in Rings of Integers, II in The Annals of Mathematics, Vol. 77, No. 2. Many other special cases have been treated in the literature, e.g. dihedral groups of order $2p$ (M. P. Lee, Integral representations of dihedral groups of order 2$p$, Trans. American Math. Soc. 110 no. 2 (1964), 213-231).
There is indeed a strong connection to modular representations. For example if $G$ is a finite group and $M$ is a $\mathbb{Z}_p[G]$-module, then $M$ is decomposable if and only if $M/p^kM$ is a decomposable $\mathbb{Z}/p^k\mathbb{Z}[G]$-module for some $k$. Also, two modules $M$ and $N$ are isomorphic if and only if $M/p^kM$ is isomorphic to $N/p^kN$ for some (explicitly computable) $k$. For this and more results, see e.g. Curtis and Reiner, Representation theory of finite groups and associative algebra, §76.
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Thank you. In which articles could I find this for the dihedral group? – Matt E. Sep 6 at 21:24
I have added a reference. – Alex Bartel Sep 6 at 21:28
### 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.
What Alex says is correct, but there is another viewpoint, which is a bit easier in some ways. You can consider finitely generated `$\mathbb{Z}_{p}$`-free `$\mathbb{Z}_{p}G$`-modules which have no proper non-zero pure submodules. These are the modules which are irreducible when the ground ring is extended to $\mathbb{Q}_{p}.$ There are finitely many isomorphism types of such modules. It may be possible to deduce the structure of faithful such modules for the dihedral group using Clifford theory.
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http://mathoverflow.net/questions/57651/tame-abelian-tensor-categories
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## Tame abelian tensor categories
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
In the article "Tannaka duality for geometric stacks" (arxiv, see nlab for a summary) Jacob Lurie introduced the notion of a tame abelian tensor category. An abelian tensor category is called tame if every short exact sequence $0 \to M' \to M \to M'' \to 0$ with $M''$ flat remains exact after tensoring with an arbitrary object. Basically this means that flat objects behave in the usual way. For example it is easy to check that in a tame abelian tensor category every extension of flat objects is flat (Lemma 5.4).
Question 1: Do you know any further literature about tame abelian tensor categories?
Question 2: Can you give an explicit example of an abelian tensor category which is not tame? Jacob Lurie remarks that every abelian tensor category with enough flat objects is tame, because then we can define $Tor_1$ and use symmetry. So in particular, a counterexample does not have enough flat objects. [Tom Goodwillie has given an example below]
Question 3: Can you give an example of a cocontinuous tensor functor between tame abelian tensor categories which is not tame?
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What's a flat object? – Qfwfq Mar 7 2011 at 11:22
Flat module: en.wikipedia.org/wiki/Flat_module – David Roberts Mar 7 2011 at 12:07
1
Well it's an object $M$ such that the functor $M \otimes -$ is exact. See also Lurie's article for more definitions and background. – Martin Brandenburg Mar 7 2011 at 14:00
## 1 Answer
I don't know the definition: is the category of torsion abelian groups an example of an abelian tensor category? With $\otimes$ having its usual meaning? If so, then it seems that $\mathbb Q/\mathbb Z$ is flat (tensor product with anything is zero!) but the sequence
$0\to \mathbb Z/p \to \mathbb Q/\mathbb Z\to \mathbb Q/\mathbb Z\to 0$
does not remain exact when you tensor it with $\mathbb Z/p$.
EDIT: For a true example I need the $\otimes$ operation to have a unit object. You can artificially introduce a unit, but it won't help with this example, so first let me alter the example. Choose a domain $R$ that is a $\mathbb Q$-algebra and not a field. Let $K$ be the field of fractions. In the category of torsion $R$-modules $M\otimes (K/R)$ is always zero, and the sequence
$0\to R/fR \to K/fR\to K/R\to 0$
does not remain exact when tensored with $R/fR$. Now, to introduce a unit, consider the product of the category of $\mathbb Z$-modules and the category of torsion $R$-modules. Define
$(A,M)\otimes (A',M')=(A\otimes_{\mathbb Z}A',A\otimes_{\mathbb Z}M'\oplus A'\otimes_{\mathbb Z}M\oplus M\otimes_RM')$.
The object $(0,K/R)$ is flat (this uses that $K/R$ is a flat $\mathbb Z$-module), and the sequence
$0\to (0,R/fR) \to (0,K/fR)\to (0,K/R)\to 0$
does not remain exact when tensored with $(0,R/fR)$.
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Thanks! This answers 2). [The definition of abelian tensor categories is given in the nlab article which I have linked, for example] – Martin Brandenburg Jun 23 2011 at 14:40
Hmm ... My example doesn't qualify because there is no unit object for $\otimes$. – Tom Goodwillie Jun 23 2011 at 15:34
I didn't see your edit. This example works :). – Martin Brandenburg Jul 8 2011 at 9:36
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http://unapologetic.wordpress.com/2009/07/06/dirac-notation-iii/?like=1&source=post_flair&_wpnonce=d91e0e6d9a
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# The Unapologetic Mathematician
## Dirac Notation III
So we’ve got Dirac notation and it’s nice for inner product spaces, but remember we’re not just interested in vectors and vector spaces. We’re even more interested in transformations between vector spaces. So what happens when we throw a linear transformation into the mix?
To keep things simple, let’s just think of one vector space $V$ and a linear transformation $T:V\rightarrow V$. If we have a vector $v$, we can write it as the ket $\lvert v\rangle$. Then we can hit it with our transformation to get $Tv$. We can put this new vector into a ket as well: $\lvert Tv\rangle$. But we can also just write the transformation outside the ket, giving $T\lvert v\rangle$. Both of these notations make pretty good sense.
But what about bras? We can take the ket $\lvert Tv\rangle$ and flip it over into the bra $\langle Tv\rvert$. Now there’s no reason to believe that $T$ acts in exactly the same way on bra vectors as it does on ket vectors. So what does it do? Well, we can probe the bra $\langle Tv\rvert$ by seeing what happens when we feed it various kets. We pick a ket $\lvert w\rangle$ and pair up to form the bra-ket $\langle Tv\vert w\rangle$.
Now this is an inner product, and so we have the adjoint property, allowing us to rewrite it as $\langle v\vert T^*w\rangle$. This is the pairing of the bra $\langle v\rvert$ and the ket $\lvert T^*w\rangle$. But this latter ket we can also write as $T^*\lvert w\rangle$. So we could reasonably write out the pairing as $\langle v\rvert T^*\lvert w\rangle$. But we could then reasonably read this as the pairing of the ket $\lvert w\rangle$ with the bra $\langle v\rvert T^*$.
That is, when we have a transformation $T$ acting on a bra vector $\langle v\rvert$ to give the bra $\langle Tv\rvert$, we can see this as the adjoint of $T$ acting from the right in our notation. Then expressions like $\langle v\rvert T\lvert w\rangle$ can be interpreted either as $T$ acting on $w$ and then paired with $v$, or as $T^*$ acting on $v$, which then pairs with $w$. And this is actually no ambiguity, since the way adjoints and inner products interact guarantees that these both give the same answer in the end.
Incidentally, what is an expression like $\langle v\rvert T\lvert w\rangle$ anyway? It’s a matrix element of the transformation $T$. In fact, if we have a basis $\left\{\lvert i\rangle\right\}_{i=1}^n$ then the matrix components can be written in Dirac notation as $t_j^i=\langle i\rvert T\lvert j\rangle$. The Dirac notation makes clear the way that bra and ket vectors probe two different aspects of linear transformations.
In quantum physics, we see exactly this sort of probing going on all the time. Except that there, the physicsts talk about picking an “input state” $\lvert w\rangle$ and an “output state” $\langle v\rvert$, and thinking of a physical process as a linear transformation $T$ from a space of input states to a space of output states. The “transition amplitude” — connected with the probability of ending up at the given output when starting at the given input — is then given by $\langle v\rvert T\lvert w\rangle$. It’s just a matrix element of the linear transformation representing out physical process.
### Like this:
Posted by John Armstrong | Algebra, Linear Algebra
## 3 Comments »
1. I’m really enjoying this section Dirac notation. If I explain this to students, though, I might have to avoid using the phrase, “…we can probe the bra…”
Many thanks for the fine explanations.
Comment by | July 6, 2009 | Reply
2. [...] and Bilinear Forms on Inner Product Spaces in Dirac Notation Now, armed with Dirac notation, we can come back and reconsider matrices and forms. For our background, we’ve got an inner [...]
Pingback by | July 8, 2009 | Reply
3. [...] First, we hit each of these vectors with . The ket vector clearly becomes , while the bra vector becomes . Now using the form from before we [...]
Pingback by | July 24, 2009 | 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/tagged/black-holes+entropy
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# Tagged Questions
0answers
51 views
### Entanglement and Black holes
If you have two entangled quantum states, One state falls into a black hole and you measure the other state, What can you say about the state that has fallen into the black hole? If you have billions ...
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### Why isn't the Bekenstein-Hawking Entropy considered the quantum gravitational unification?
Based on the Bekenstein-Hawking Equation for Entropy, hasn't the relationship between quantum mechanics and gravity already been established.
1answer
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### Where and how is the entropy of a black hole stored?
Where and how is the entropy of a black hole stored? Is it around the horizon? Most of the entanglement entropy across the event horizon lies within Planck distances of it and are short lived. Is ...
2answers
160 views
### Information Loss in annihilation
The concept of information loss is usually discussed with respect to a black hole. My understanding is that whatever matter you put into the black hole, it has only 3 "hairs" and so one doesn't know, ...
1answer
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### Black hole entropy
Bekenstein and Hawking derived the expression for black hole entropy as, $$S_{BH}={c^3 A\over 4 G \hbar}.$$ We know from the hindsight that entropy has statistical interpretation. It is a measure ...
0answers
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### Relation between maximum entropy thermodynamics and entropy bounds
Here is a question which I've been thinking about, I'm sorry if it's little too vague, maybe you could point me to some source to read, if there's no clear direct answer. The whole concept of ...
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### Wasn't the Hawking Paradox solved by Einstein?
I just watched a BBC Horizon episode where they talked about the Hawking Paradox. They mentioned a controversy about information being lost but I couldn't get my head around this. Black hole ...
2answers
538 views
### About Susskind's claim “information is indestructible”
I really can't understand what Leonard Susskind means when he says that information is indestructible. Is that information really recoverable? He himself said that entropy is hidden information. ...
1answer
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### Area law for Entropy in Loop Quantum Gravity
In connection with the long saga of the (claimed) microscopic calculations of the Hawking-Bekenstein entropy in (3+1) Loop Quantum Gravity (LQG) and related approaches I have the following question: ...
0answers
145 views
### How is the logarithmic correction to the entropy of a non extremal black hole derived?
I`ve just read, that for non extremal black holes, there exists a logarithmic (and other) correction(s) to the well known term proportional to the area of the horizon such that \$S = \frac{A}{4G} + K ...
2answers
224 views
### black hole no-hair theorems vs. entropy and surface area
I was revisiting some old popular science books a while ago and two statements struck me as incompatible. No-hair theorems: a black hole is fully-described by just a few numbers (mass, spin etc) ...
0answers
88 views
### Geometric entropy vs entanglement entropy (dependent on curvature coupling parameter)
I have a quick question. In hep-th/9506066, Larsen and Wilczek calculated the geometric entropy (which I believe is just another name for entanglement entropy) for a non-minimally coupled scalar field ...
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http://physics.stackexchange.com/questions/54575/what-is-generalized-free-field?answertab=active
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What is generalized free field?
I came across the term generalized free field in a paper recently but I don't know its definition. Google leads to other papers which take it for granted and use it without defining it. It appears that O.W. Greenberg introduced the term in the paper Generalized free fields and models of local field theory, Ann. Physics 16. Unfortunately I can't access this paper. Can someone please explain it to me or give me a link to that paper? Thank you.
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I wonder where did the other answer with the link to Greenberg's paper go. Was it removed due to copyright violation? – Sudip Paul Feb 22 at 21:26
1 Answer
A generalized free field is one for which (modulo field redefinitions) the connected $n$-point functions $G_n(x_1,...,x_n)$ vanish whenever $n > 2$. This means, basically, that a generalized free field is one for which the Euclidean functional measure is Gaussian. The field is completely specified by its 2-point functions $G_2(x,y)$.
Generalized free fields are usually discussed using the parametrization given by the Kallen-Lehmann decomposition of the 2-point function $G_2(x,y)$, which says that (for scalar fields)
$G_2(x,y) = \int_0^\infty d\rho(m) \Delta_m(x-y)$,
where $\Delta_m(x-y)$ is the real-space propagator for a free real field of mass $m$ and $\rho$ is a positive measure.
This parametrization makes it easy to write down examples of generalized free scalar fields. Just pick a positive measure $\rho$ on the mass line $[0,\infty)$. The simplest example is the free field of mass $M$, which corresponds to $\rho(m) = \delta_M(m)$, the Dirac delta function supported at $M$. You get other examples by picking other measures. For example, you can take a purely continuous measure, like $d\rho(m) = \Theta_M(m)dm$ or $d\rho(m) = m^2dm$. (Here, $\Theta_M(m) = 0$ if $m <M$ and $1$ otherwise.)
These examples where the measure has continuos support are somewhat difficult to think of in the usual Lagrangian formalism; it's as if you have a continuum of fields of different mass which are all constrained to move together. But giving the correlation functions is enough to define a field theory; you can use Wightman's reconstruction theorem to recover the Hilbert space and the field operators.
If you're working in axiomatic field theory, then you usually impose some sort of growth conditions on your correlation functions. In the case of generalized free fields, these growth conditions translate into conditions on $G_2$, or equivalently on $rho$. For example, if the free field's Euclidean measure obeys the Osterwalder-Schrader axioms, then $\rho$ must be a tempered measure of polynomial growth at $\infty$ (and not too unreasonable behavior near $0$).
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Thanks. Can you give me an example of generalized free fields which are not free fields? – Sudip Paul Feb 21 at 4:46
2
In conformal field theory, people use these fields with some scaling dimension $\Delta$ as toy models for interacting theories. Through the AdS/CFT correspondence, they correspond (if I'm not mistaken) to massive free fields in AdS-space, so many properties can be calculated explicitly. – Vibert Feb 21 at 8:47
@SudipPaul: I've altered my answer; hopefully it is now clearer what non-trivial examples look like. – user1504 Feb 21 at 15:24
@Vibert Thanks, that's how I came upon this term in the first place. – Sudip Paul Feb 22 at 21:23
@user1504 Thanks. – Sudip Paul Feb 22 at 21:25
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http://en.wikipedia.org/wiki/Tagged_union
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# Tagged union
In computer science, a tagged union, also called a variant, variant record, discriminated union, disjoint union, or sum type, is a data structure used to hold a value that could take on several different, but fixed types. Only one of the types can be in use at any one time, and a tag field explicitly indicates which one is in use. It can be thought of as a type that has several "cases," each of which should be handled correctly when that type is manipulated. Like ordinary unions, tagged unions can save storage by overlapping storage areas for each type, since only one is in use at a time.
Tagged unions are most important in functional languages such as ML and Haskell, where they are called datatypes (see algebraic data type) and the compiler is able to verify that all cases of a tagged union are always handled, avoiding many types of errors. They can, however, be constructed in nearly any language, and are much safer than untagged unions, often simply called unions, which are similar but do not explicitly keep track of which member of the union is currently in use.
Tagged unions are often accompanied by the concept of a constructor, which is similar but not the same as a constructor for a class. Constructors produce a tagged union value, given the initial tag value and a value of the corresponding type.
Mathematically, tagged unions correspond to disjoint or discriminated unions, usually written using +. Given an element of a disjoint union A + B, it is possible to determine whether it came from A or B. If an element lies in both, there will be two effectively distinct copies of the value in A + B, one from A and one from B.
In type theory, a tagged union is called a sum type. Notations vary, but usually the sum type $A+B$ comes with two introduction forms $\text{inj}_1 : A \to A+B$ and $\text{inj}_2 : B \to A+B$. The elimination form is case analysis, known as pattern matching in ML-style programming languages: if $e$ has type $A+B$ and $e_1$ and $e_2$ have type $\tau$ under the assumptions $x:A$ and $y:B$ respectively, then the term $\texttt{case}\ e\ \texttt{of}\ x \Rightarrow e_1 | y \Rightarrow e_2$ has type $\tau$. The sum type corresponds to intuitionistic logical disjunction under the Curry–Howard correspondence.
An enumerated type can be seen as a degenerate case: a tagged union of unit types. It corresponds to a set of nullary constructors and may be implemented as a simple tag variable, since it holds no additional data besides the value of the tag.
## Advantages and disadvantages
The primary advantage of a tagged union over an untagged union is that all accesses are safe, and the compiler can even check that all cases are handled. Untagged unions depend on program logic to correctly identify the currently active field, which may result in strange behavior and hard-to-find bugs if that logic fails.
The primary advantage of a tagged union over a simple record containing a field for each type is that it saves storage by overlapping storage for all the types. Some implementations reserve enough storage for the largest type, while others dynamically adjust the size of a tagged union value as needed. When the value is immutable, it is simple to allocate just as much storage as is needed.
The main disadvantage of tagged unions is that the tag occupies space. Since there are usually a small number of alternatives, the tag can often be squeezed into 2 or 3 bits wherever space can be found, but sometimes even these bits are not available. In this case, a helpful alternative may be folded, computed or encoded tags, where the tag value is dynamically computed from the contents of the union field. Common examples of this are the use of reserved values, where, for example, a function returning a positive number may return -1 to indicate failure, and sentinel values, most often used in tagged pointers.
Sometimes, untagged unions are used to perform bit-level conversions between types, called reinterpret casts in C++. Tagged unions are not intended for this purpose; typically a new value is assigned whenever the tag is changed.
Many languages support, to some extent, a universal data type, which is a type that includes every value of every other type, and often a way is provided to test the actual type of a value of the universal type. These are sometimes referred to as variants. While universal data types are comparable to tagged unions in their formal definition, typical tagged unions include a relatively small number of cases, and these cases form different ways of expressing a single coherent concept, such as a data structure node or instruction. Also, there is an expectation that every possible case of a tagged union will be dealt with when it is used. The values of a universal data type are not related and there is no feasible way to deal with them all.
Like option types and exception handling, tagged unions are sometimes used to handle the occurrence of exceptional results. Often these tags are folded into the type as "reserved values", and their occurrence is not consistently checked: this is a fairly common source of programming errors. This use of tagged unions can be formalized as a monad with the following functions:
$\text{return}\colon A \to \left( A + E \right) = a \mapsto \text{value} \, a$
$\text{bind}\colon \left( A + E \right) \to \left(A \to \left(B + E \right) \right) \to \left( B + E \right) = a \mapsto f \mapsto \begin{cases} \text{err} \, e & \text{if} \ a = \text{err} \, e\\ f \, a' & \text{if} \ a = \text{value} \, a' \end{cases}$
where "value" and "err" are the constructors of the union type, A and B are valid result types and E is the type of error conditions. Alternately, the same monad may be described by return and two additional functions, fmap and join:
$\text{fmap} \colon (A \to B) \to \left( \left( A + E \right) \to \left( B + E \right) \right) = f \mapsto a \mapsto \begin{cases} \text{err} \, e & \text{if} \ a = \text{err} \, e \\ \text{value} \, f \, a' & \text{if} \ a = \text{value} \, a' \end{cases}$
$\text{join} \colon ((A + E ) + E) \to (A + E) = a \mapsto \begin{cases} \text{err} \, e & \mbox{if} \ a = \text{err} \, e\\ \text{err} \, e & \text{if} \ a = \text{value} \, \text{err} \, e \\ \text{value} \, a' & \text{if} \ a = \text{value} \, \text{value} \, a' \end{cases}$
## Examples
Say we wanted to build a binary tree of integers. In ML, we would do this by creating a datatype like this:
```datatype tree = Leaf
| Node of (int * tree * tree)
```
This is a tagged union with two cases: one, the leaf, is used to terminate a path of the tree, and functions much like a null value would in imperative languages. The other branch holds a node, which contains an integer and a left and right subtree. Leaf and Node are the constructors, which enable us to actually produce a particular tree, such as:
```Node(5, Node(1,Leaf,Leaf), Node(3, Leaf, Node(4, Leaf, Leaf)))
```
which corresponds to this tree:
Now we can easily write a typesafe function that, say, counts the number of nodes in the tree:
```fun countNodes(Leaf) = 0
| countNodes(Node(int,left,right)) =
1 + countNodes(left) + countNodes(right)
```
## Timeline of language support
### 1960s
In ALGOL 68, tagged unions are called united modes, the tag is implicit, and the `case` construct is used to determine which field is tagged:
`mode node = union (real, int, compl, string);`
Usage example for `union` `case` of `node`:
``````node n := "1234";
case n in
(real r): print(("real:", r)),
(int i): print(("int:", i)),
(compl c): print(("compl:", c)),
(string s): print(("string:", s))
out print(("?:", n))
esac```
```
### 1970s & 1980s
Although primarily only functional languages such as ML and Haskell (from 1990s) give a central role to tagged unions and have the power to check that all cases are handled, other languages have support for tagged unions as well. However, in practice they can be less efficient in non-functional languages due to optimizations enabled by functional language compilers that can eliminate explicit tag checks and avoid explicit storage of tags.
Pascal, Ada, and Modula-2 call them variant records (formally discriminated type in Ada), and require the tag field to be manually created and the tag values specified, as in this Pascal example:
```type shapeKind = (square, rectangle, circle);
shape = record
centerx : integer;
centery : integer;
case kind : shapeKind of
square : (side : integer);
rectangle : (length, height : integer);
circle : (radius : integer);
end;
```
and this Ada equivalent:
```type Shape_Kind is (Square, Rectangle, Circle);
type Shape (Kind : Shape_Kind) is record
Center_X : Integer;
Center_Y : Integer;
case Kind is
when Square =>
Side : Integer;
when Rectangle =>
Length, Height : Integer;
when Circle =>
Radius : Integer;
end case;
end record;
-- Any attempt to access a member whose existence depends
-- on a particular value of the discriminant, while the
-- discriminant is not the expected one, raises an error.
```
In C and C++, a tagged union can be created from untagged unions using a strict access discipline where the tag is always checked:
```enum ShapeKind { Square, Rectangle, Circle };
struct Shape {
int centerx;
int centery;
enum ShapeKind kind;
union {
struct { int side; } squareData;
struct { int length, height; } rectangleData;
struct { int radius; } circleData;
} shapeKindData;
};
int getSquareSide(struct Shape* s) {
assert(s->kind == Square);
return s->shapeKindData.squareData.side;
}
void setSquareSide(struct Shape* s, int side) {
s->kind = Square;
s->shapeKindData.squareData.side = side;
}
/* and so on */
```
As long as the union fields are only accessed through the functions, the accesses will be safe and correct. The same approach can be used for encoded tags; we simply decode the tag and then check it on each access. If the inefficiency of these tag checks is a concern, they may be automatically removed in the final version.
C and C++ also have language support for one particular tagged union: the possibly-null pointer. This may be compared to the `option` type in ML or the `Maybe` type in Haskell, and can be seen as a tagged pointer: a tagged union (with an encoded tag) of two types:
• Valid pointers,
• A type with only one value, `null`, indicating an exceptional condition.
Unfortunately, C compilers do not verify that the null case is always handled, and this is a particularly prevalent source of errors in C code, since there is a tendency to ignore exceptional cases.
### 2000s
One advanced dialect of C called Cyclone has extensive built-in support for tagged unions. See the tagged union section of the on-line manual for more information.
The variant library from Boost has demonstrated it was possible to implement a safe tagged union as a library in C++, visitable using functors.
```struct display : boost::static_visitor<void>
{
void operator()(int i)
{
std::cout << "It's an int, with value " << i << std::endl;
}
void operator()(const std::string& s)
{
std::cout << "It's a string, with value " << s << std::endl;
}
};
boost::variant<int, std::string> v = 42;
boost::apply_visitor(display(), v);
boost::variant<int, std::string> v = "hello world";
boost::apply_visitor(display(), v);
```
Scala has case classes:
```sealed abstract class Tree
case object Leaf extends Tree
case class Node(value: Int, left: Tree, right: Tree) extends Tree
val tree = Node(5, Node(1,Leaf,Leaf), Node(3, Leaf, Node(4, Leaf, Leaf)))
```
Because the class hierarchy is sealed, the compiler can check that all cases are handled in a pattern match:
```tree match {
case Node(x, _, _) => println("top level node value: " + x)
case Leaf => println("top level node is a leaf")
}
```
Scala's case classes also permit reuse through subtyping:
```sealed abstract class Shape(centerX: Int, centerY: Int)
case class Square(side: Int, centerX: Int, centerY: Int) extends Shape(centerX, centerY)
case class Rectangle(length: Int, height: Int, centerX: Int, centerY: Int) extends Shape(centerX, centerY)
case class Circle(radius: Int, centerX: Int, centerY: Int) extends Shape(centerX, centerY)
```
Opa has support for polymorphic tagged unions:
```type material = { wood } or { plastic } or { metal }
function burns(m) {
match (m) {
case {wood}
case {plastic}: true;
case {metal}: false;
}
}
```
## Class hierarchies as tagged unions
In a typical class hierarchy in object-oriented programming, each subclass can encapsulate data unique to that class. The metadata used to perform virtual method lookup (for example, the object's vtable pointer in most C++ implementations) identifies the subclass and so effectively acts as a tag identifying the particular data stored by the instance (see RTTI). An object's constructor sets this tag, and it remains constant throughout the object's lifetime.
Nevertheless, a class hierarchy involves true subtype polymorphism; it can be extended by creating further subclasses of the same base type, which could not be handled correctly under a tag/dispatch model. Hence, it is usually not possible to do case analysis or dispatch on a subobject's 'tag' as one would for tagged unions. Some languages such as Scala allow base classes to be "sealed", and unify tagged unions with sealed base classes.
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http://physics.stackexchange.com/questions/tagged/string+homework
|
# Tagged Questions
1answer
53 views
### The second resonance of string?
What is the relationship between "the second resonance " and string and the wavelength. Like in this question: if the length of the string is 2cm with second resonance, then what is wavelength?
0answers
241 views
### Neglecting friction on a pulley?
So, this is how the problem looks: http://www.aplusphysics.com/courses/honors/dynamics/images/Atwood%20Problem.png Plus, the pulley is suspended on a cord at its center and hanging from the ceiling. ...
0answers
168 views
### Shape of a string/chain/cable/rope?
The height of a string in a gravitational field in 2-dimensions is bounded by $h(x_0)=h(x_l)=0$ (nails in the wall) and also $\int_0^l ds= l$. ($h(0)=h(l)=0$, if you take $h$ as a function of arc ...
1answer
76 views
### Finding the acceleration of Block attached using tricky string setup
Below shown is a setup, and block B starts from rest and moves towards right with a constant acceleration. Does the acceleration differ for the blocks ? I am a bit confused because of the tricky ...
0answers
80 views
### How to figure out an elastic constant? [closed]
Im doing this study and i have this question which I'm not 100% sure on, got me pretty stumped. anyone think they can help me?! When a bowstring is pulled back in preparation in preparation for ...
1answer
262 views
### Lagrangian density for a Piano String
So I'm trying to do this problem where I'm given the Lagrangian density for a piano string which can vibrate both transversely and longitudinally. $\eta(x,t)$ is the transverse displacement and ...
1answer
263 views
### How do we find the frequency of wave propagated along the x-axis?
I don't know how to solve question like this: A transverse wave is propagated in a string stretched along the x-axis. The equation of the wave, in SI units, is given by:y = 0.006 cos π(46t - 12x). ...
1answer
198 views
### resonance frequency [closed]
A string has a mass per unit length of 9 10–3 kg/m. What must be the tension in the string if its second harmonic has the same frequency as the second resonance mode of a 2m long pipe open at one end? ...
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http://mathhelpforum.com/geometry/122013-stucked-part-ii-after-comleting-part-i.html
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Thread:
1. Stucked @ part ii after comleting part i
I don't know how to do the part ii.
The equation of a curve is $y = 4x^2 - 2kx + k$
i) Find the range of values of k if the curve does not meet the x-axis.
ii) Show that the line $y = x + 1$ intersects the curve for all real values of k.
Solution
i) Since curve does not meet x-axis, $b^2-4ac < 0$
$(-2k)^2-4(4)(k) < 0$
$4k^2-16k < 0$
$k(4k-16) < 0$
k < 0 or 4k < 16
k < 4
Range of Values of k is k < 4
ii) Stucked!
2. Does "Real Values" in part ii means that the numbers must be more than or equal to 0?
3. Observe:
We like to find out for wich k the following equation has solutions:
$4x^2-2kx+k = x+1$
$4x^2+(-1-2k)x+(k-1) = 0$
This equation has real solutions if: $D= (-1-2k)^2-16(k-1) = 4k^2-12k+17 \geq 0$
It's not hard to see that D > 0 for all real k.
Does "Real Values" in part ii means that the numbers must be more than or equal to 0?
No it means the numbers are not complex, that is of the form a+bi
where $i = \sqrt{-1}$
4. Originally Posted by Dinkydoe
Observe:
We like to find out for wich k the following equation has solutions:
$4x^2-2kx+k = x+1$
$4x^2+(-1-2k)x+(k-1) = 0$
This equation has real solutions if: $D= (-1-2k)^2-16(k-1) = 4k^2-12k+17 \geq 0$
It's not hard to see that D > 0 for all real k.
No it means the numbers are not complex, that is of the form a+bi
where $i = \sqrt{-1}$
What you have written is not sufficient to say that the Discriminant is always nonnegative.
You have $\Delta = 4k^2 - 12k + 17$
$= 4\left(k^2 - 3k + \frac{17}{4}\right)$
$= 4\left[k^2 - 3k + \left(-\frac{3k}{2}\right)^2 - \left(-\frac{3k}{2}\right)^2 + \frac{17}{4}\right]$
$= 4\left[\left(k - \frac{3k}{2}\right)^2 + \frac{17 + 9k^2}{4}\right]$
$= 4\left(k - \frac{3k}{2}\right)^2 + 17 + 9k^2$.
This is sufficient to say that the discriminant is nonnegative for all $k$.
5. What you have written is not sufficient to say that the Discriminant is always nonnegative
Ofcourse it's not fully sufficient, but a simple argument is:
$\Delta(k)= 4k^2-12k+17 = 0$ has no real solutions
since $\Delta(\Delta(k)) = 144-16\cdot 17 = -128 < 0$
Therefore by continuity of $\Delta(k)$ we have $\Delta(k)> 0$ or $\Delta(k) < 0$ for all k.
Since for all negative k we have $\Delta(k) > 0$ it must be the first case.
6. LOL, I don't understand the question neither do I understand the solutions you guys have provided.. But great job.
From my observation, is the way of completing part ii by subsituting values of k that are < 4?
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http://math.stackexchange.com/questions/193185/problem-with-the-definition-of-lebesgue-measure-and-borel-sets
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# Problem with the definition of Lebesgue Measure and borel sets!
The definition of lebesgue measure (in my textbook):
The set-function $\lambda^{n}$ on ($\mathbb{R}^{n}, \mathcal{B}(\mathbb{R}^{n})$) that assigns every half-open $[[a,b)) = [a_{1},b_{1}) \times \dots \times [a_{n},b_{n})\in\mathcal{J}$ the value: $\lambda^{n}([[a,b))):=\prod_{j=1}^{n}(b_{j}-a_{j})$ is called n-dimensional Lebesgue measure.
On the next page the book mentions that the Lebesque Measure is a measure on the Borel sets $\mathcal{B}(\mathbb{R})$
My question/problem:
According to the definition of a measure in the book. The measure must be a map between $\mathcal{B}(\mathbb{R})$ and $[0,\infty]$. But how can the defined $\lambda^{n}$ maps all the elements in $\mathcal{B}(\mathbb{R})$? When it only maps half-open rectangles? Can $\lambda^{n}$ map other kinds of set in $\mathcal{B}(\mathbb{R})$?
-
4
This is not only a definition, but also a theorem : there exists a unique measure on Borel sets which gives this value to each half-open rectangles. This is not really obvious, and is a consequence of Caratheodory's extension theorem. You should find details in textobooks on measure theory. – Ahriman Sep 9 '12 at 14:43
Usually, the Lebesgue measure is defined not only on Borel sets but also on all Lebesgue-measurabe sets. Note that the measure must be $\sigma$-additive. If you define it only on sets $[[a,b))$, its value on other sets will be defined uniquely. – Yury Sep 9 '12 at 14:52
– t.b. Sep 10 '12 at 10:10
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http://math.stackexchange.com/questions/54910/a-detail-about-mct-application?answertab=votes
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# A detail about MCT application
I have a indirect question about Monotone Class Theorem (MCT), in its functional form. Here is a version which should be sufficiently general for my purpose.
Functional Monotone Class Theorem :
Let H a collection of real valued and bounded functions such that:
- H is a vector space
- H contains constants
- If $f_n$ is an increasing sequence of real and bounded functions of H, admitting a limit $f$ that is a bounded real valued function, then $f$ belongs to H.
Now let K be another collection of real bounded functions which is stable by multiplication, and such that $K \subset H$.
Conclusion :
Then H contains all real bounded functions that belongs to $\sigma(K)$ (the sigma algebra generated by K).
The statement here is clear but I have a question about applications of it.
Usually the collection K is built by considering a collection P over real valued functions satisfying some propetry and intersecting it with the collections of bounded functions. Then using MCT with a nice space H, you have your conclusion for $\sigma(K)$.
My question is the follwing is there a general way or sufficient conditions to see if $\sigma(K)=\sigma(P)$. There is a trivial inclusion so the question is better stated as under what conditions does $\sigma(P) \subset \sigma(K)$ holds ?
The motivation behind this question comes from the fact that when I see MCT used in proofs of theorems, it is often the case that the theorem's conclusion is stated for the unbounded real valued functions collection P rather than K (P+the bounded property).
Best Regards
-
I think that Proposition 8.15 and Theorem 8.16 in these lecture notes answer your question to some extent. See also Nate Eldredge's answer here where I learned about the existence of these notes. – t.b. Aug 1 '11 at 11:35
@Theo Buehler: Hi thank's for the link to those rich and interesting Notes. But regarding the theorems that you quote I don't think they answer the question unfortunately (or I don't see how). – – TheBridge Aug 1 '11 at 12:53
I apologize for promising too much in haste, I didn't think very deeply about your question but it seemed close enough. – t.b. Aug 1 '11 at 13:14
@Theo Buehler: Indeed rather close and please do not apologize. As a matter of fact, the question isn't really directly about MCT, but rather under what conditions 2 sigma fields can be the same. My best guess would be that if for any $f$ in P (bounded or not) there's a sequence of functions in K that ponctually converge to $f$ then we are done. Best regards – TheBridge Aug 1 '11 at 13:40
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http://en.wikipedia.org/wiki/Pivot_element
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# Pivot element
The pivot or pivot element is the element of a matrix, or an array, which is selected first by an algorithm (e.g. Gaussian elimination, simplex algorithm, etc.), to do certain calculations. In the case of matrix algorithms, a pivot entry is usually required to be at least distinct from zero, and often distant from it; in this case finding this element is called pivoting. Pivoting may be followed by an interchange of rows or columns to bring the pivot to a fixed position and allow the algorithm to proceed successfully, and possibly to reduce round-off error.
Pivoting might be thought of as swapping or sorting rows or columns in a matrix, and thus it can be represented as multiplication by permutation matrices. However, algorithms rarely move the matrix elements because this would cost too much time; instead, they just keep track of the permutations.
Overall, pivoting adds more operations to the computational cost of an algorithm. These additional operations are sometimes necessary for the algorithm to work at all. Other times these additional operations are worthwhile because they add numerical stability to the final result.
## Examples of systems that require pivoting
In the case of Gaussian elimination, the algorithm requires that pivot elements not be zero. Interchanging rows or columns in the case of a zero pivot element is necessary. The system below requires the interchange of rows 2 and 3 to perform elimination.
$\left[ \begin{array}{ccc|c} 1 & -1 & 2 & 8 \\ 0 & 0 & -1 & -11 \\ 0 & 2 & -1 & -3 \end{array} \right]$
The system that results from pivoting is as follows and will allow the elimination algorithm and backwards substitution to output the solution to the system.
$\left[ \begin{array}{ccc|c} 1 & -1 & 2 & 8 \\ 0 & 2 & -1 & -3 \\ 0 & 0 & -1 & -11 \end{array} \right]$
Furthermore, in Gaussian elimination it is generally desirable to choose a pivot element with large absolute value. This improves the numerical stability. The following system is dramatically affected by round-off error when Gaussian elimination and backwards substitution are performed.
$\left[ \begin{array}{cc|c} 0.00300 & 59.14 & 59.17 \\ 5.291 & -6.130 & 46.78 \\ \end{array} \right]$
This system has the exact solution of x1 = 10.00 and x2 = 1.000, but when the elimination algorithm and backwards substitution are performed using four-digit arithmetic, the small value of a11 causes small round-off errors to be propagated. The algorithm without pivoting yields the approximation of x1 ≈ 9873.3 and x2 ≈ 4. In this case it is desirable that we interchange the two rows so that a21 is in the pivot position
$\left[ \begin{array}{cc|c} 5.291 & -6.130 & 46.78 \\ 0.00300 & 59.14 & 59.17 \\ \end{array} \right].$
Considering this system, the elimination algorithm and backwards substitution using four-digit arithmetic yield the correct values x1 = 10.00 and x2 = 1.000.
## Partial and complete pivoting
In partial pivoting, the algorithm selects the entry with largest absolute value from the column of the matrix that is currently being considered as the pivot element. Partial pivoting is generally sufficient to adequately reduce round-off error. However for certain systems and algorithms, complete pivoting (or maximal pivoting) may be required for acceptable accuracy. Complete pivoting considers all entries in the whole matrix, interchanging rows and columns to achieve the highest accuracy. Complete pivoting is usually not necessary to ensure numerical stability and, due to the additional computations it introduces, it may not always be the most appropriate pivoting strategy.
## Scaled pivoting
A variation of the partial pivoting strategy is scaled partial pivoting. In this approach, the algorithm selects as the pivot element the entry that is largest relative to the entries in its row. This strategy is desirable when entries' large differences in magnitude lead to the propagation of round-off error. Scaled pivoting should be used in a system like the one below where a row's entries vary greatly in magnitude. In the example below, it would be desirable to interchange the two rows because the current pivot element 30 is larger than 5.291 but it is relatively small compared with the other entries in its row. Without row interchange in this case, rounding errors will be propagated as in the previous example.
$\left[ \begin{array}{cc|c} 30 & 591400 & 591700 \\ 5.291 & -6.130 & 46.78 \\ \end{array} \right]$
## Pivot position
A pivot position in a matrix, A , is a position in the matrix that corresponds to a row–leading 1 in the reduced row echelon form of A. Since the reduced row echelon form of A is unique, the pivot positions are uniquely determined and do not depend on whether or not row interchanges are performed in the reduction process.
## References
• R. L. Burden, J. D. Faires, Numerical Analysis, 8th edition, Thomson Brooks/Cole, 2005. ISBN 0-534-39200-8
• G. H. Golub, C. F. Loan, Matrix Computations, 3rd edition, Johns Hopkins, 1996. ISBN 0-8018-5414-8.
• Fukuda, Komei; Terlaky, Tamás (1997). "Criss-cross methods: A fresh view on pivot algorithms". In Thomas M. Liebling and Dominique de Werra. Mathematical Programming: Series B (Amsterdam: North-Holland Publishing Co.) (Papers from the 16th International Symposium on Mathematical Programming held in Lausanne, 1997): 369–395. doi:10.1016/S0025-5610(97)00062-2. MR 1464775. Postscript preprint. Unknown parameter `|vomume=` ignored (help)
• Terlaky, Tamás; Zhang, Shu Zhong (1993). "Pivot rules for linear programming: A Survey on recent theoretical developments". Annals of Operations Research (Springer Netherlands). 46–47 (1): 203–233. doi:10.1007/BF02096264. ISSN 0254-5330. MR 1260019. CiteSeerX: .
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http://physics.stackexchange.com/questions/44967/mass-points-of-a-mass-spring-model
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# Mass points of a Mass-spring model
Let's say I have a mass spring model like the one in the picture below:
So, there are 3 parts of the spring joined together in an equilateral triangular manner. Each of the joints has a mass of $m$. The resting length of each of the springs is $l$. The top joint point of the spring is fixed to the ceiling.
Now, if I were to pull both the lower points of the spring model such that each of the springs extend proportionally by $\Delta l$ change of length, and release. Now, I want to find an equation for Point A (indicated in the picture above) under gravity and sprint forces when springing back to position. The assumption is all the angles remain at $60$ degrees in every iteration when it springs back.
What I did is:
Let $k$ be the elasticity of the spring.
Then, for the X-axis component of the equation,
$k \cdot \Delta l \cdot cos(60) + k \cdot \Delta l = m\cdot a_x$
The acceleration of the spring going back to original x position would then be dividing both sides by the mass $m$.
For the Y-axis component of the equation with consideration of gravity as $g$,
$k \cdot \Delta l \cdot sin(60) + k \cdot \Delta l - mg = m\cdot a_y$
Similar to the X-axis, I thought I would consider the total of the extended length plus the projection from the x-axis onto the y-axis, and then minus the gravity resistance.
However, it turns out that I am wrong for the Y-axis component. The given answer is:
X: $k \cdot \Delta l \cdot cos(60) + k \cdot \Delta l = m\cdot a_x$
Y: $mg - k \cdot \Delta l \cdot sin(60) = m\cdot a_y$
I don't understand why is it so for the Y-axis, especially when the the gravity turns out to minus the projection and the extended $\Delta l$ is not added as part of the force.
-
## 1 Answer
you have taken the right approach by summing the forces and setting that equal to mass times acceleration.
For the Y-component you have included a $k$ $⋅$ $\Delta{l}$ which should not be there. In the X-direction, this term represents the force contribution from the bottom spring (the horizontal one), however this spring does not contribute at all to forces in the Y-direction (because it lies completely in the X-direction) thus has no effect on the Y-axis equation.
Also, in the diagram you provided, the Y-axis points down. You wrote your equations as though the Y-axis points up, and this has caused a sign error. Gravity points down along the positive Y-direction, thus the $mg$ term should be positive. An extension of the spring by $\Delta$$l$ leads to a restoring force that will pull point A in the negative Y-direction, thus the $k$ $⋅$ $\Delta{l}$ $⋅$ $sin(60)$ term must have a minus sign.
-
Thanks! For the part in which $k \cdot \Delta l$ should not be in the Y-component, it is because it is assumed that the spring is being pulled in the x-direction along the x-axis and therefore has no effect on the Y-axis equation? In other words, if I were to drag diagonally, $k \cdot \Delta l$ will have to be included in the Y-component of the equation? – xenon Nov 24 '12 at 22:37
We aren't assuming that the springs get pulled in the X-direction. You mentioned in your introduction that we "...pull both the lower points of the spring model such that each of the springs extend proportionally by $Δl$ change of length, and release." You also assumed "all the angles remain at 60 degrees". This second assumption ensures that the lower spring will always remain horizontal and will never contribute to forces in the Y-direction. To visualize this, just imagine the equilateral triangle oscillating in size while maintaining its shape. It will always be equilateral. – cb3 Nov 24 '12 at 22:46
Ohh...in other words, if it was pulled in such a way that the lower spring no longer remains horizontal, ie, slanted, the Y-component will have to include the $k \cdot \Delta l$, am I right? – xenon Nov 24 '12 at 22:56
Almost...you would have to include the sine of the angle of the slant. – cb3 Nov 24 '12 at 23:01
Thank you so much for the explanation. I still have one last doubt. Sorry. Earlier, you said $k \cdot \Delta l$ is a term representing the force contribution from the bottom spring. Wouldn't the $k \cdot \Delta l \cdot sin(60)$ term more like the contribution from the bottom spring since it is a projection from the bottom spring? As you said, if there is a slant, then the sine of the angle would give the length to extend in the Y-direction. Did I just confuse myself? – xenon Nov 24 '12 at 23:22
show 2 more comments
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http://mathoverflow.net/questions/83337/on-product-of-some-spaces/83340
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## on product of some spaces
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I know that $X$ is a topological space, $R$ is the real line, $S^n$ is n-sphere and $X\times R$ is diffeomorphic to $S^n\times R$. Is it true that $X$ is homeomorphic to $S^n$?
-
2
You said "diffeomorphic". Does this mean you know $X$ is a manifold? – Todd Trimble Dec 13 2011 at 14:40
## 2 Answers
If $X$ is a smooth manifold (and this is the only case when you can speak of a diffeomorphism between $X\times \mathbb R$ and $\mathbb S^n\times\mathbb R$) then this is true by Poincare. If $X$ is not assumed to be a manifold then this is false. For example, there is a theorem of Edwards that if $Y$ is a closed $(n-1)$-dimensional manifold and a homology sphere then $X$ equal to suspension of $Y$ satisfies that $X\times \mathbb R$ is homeomorphic to $\mathbb S^n\times\mathbb R$. There are many examples of homology spheres already in dimension $3$ which are not spheres so any of them will work.
-
Great thanks for answer. Of course $X\times R$ is homeomorphic (not diffeomorphic) to $S^n\times R$. – Olga Dec 14 2011 at 7:01
### 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.
This question (or a close relative) is discussed and answered in
http://mathoverflow.net/questions/26385/when-factors-may-be-cancelled-in-homeomorphic-products
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http://mathhelpforum.com/discrete-math/143707-set-builder-notation.html
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# Thread:
1. ## Set builder notation
In the beginning part of my book. There was a solution which says:
Given a set of negative integers, $S$
, where $S \subset \mathbb{Z}^-$, we can build a set $T$ of positive numbers by a set builder notation such as
$T=\{x \in Z: -x \in S\}$.
Question:
Since
$x$ is only a dummy variable, I am wondering whether it's possible at all to built the set $T$ with $S \subset \mathbb{N}$?
2. Originally Posted by novice
In the beginning part of my book. There was a solution which says:
Given a set of negative integers, $S$
, where $S \subset \mathbb{Z}^-$, we can build a set $T$ of positive numbers by a set builder notation such as
$T=\{x \in Z: -x \in S\}$.
Question:
Since
$x$ is only a dummy variable, I am wondering whether it's possible at all to built the set $T$ with $S \subset \mathbb{N}$?
You can, but then T would be a subset of the negative integers.
3. Originally Posted by gmatt
You can, but then T would be a subset of the negative integers.
Did you notice that $S$ is a subset of negative integers?
That means what about $T$?
4. Originally Posted by Plato
Did you notice that $S$ is a subset of negative integers?
That means what about $T$?
I found that set notations can be very tricky. I have many dumb questions. Apparently, they also teach me the most. It made me think.
I am thinking: Since elements of $S$ are negative, where $S \subset \mathbb{Z}^-$, and that $-x\in S$, then $x$ ought to be positive, i.e. $x>0$, so $T=\{x\in Z: x>0\}$.
I contemplated about the question you asked--and of course, the poster immediately before you too helped me think. So, I wrote on a piece of paper
If $S\not= \emptyset$ and $S \subset \mathbb{N}$, then the elements of $S$ cannot be negative, despite $T$ being defined as
$T=\{x\in \mathbb{Z}: -x\in S\}$. Since
$-x >0$, it has to be nothing else but $x<0$,
so $T=\{x\in \mathbb{Z}: x<0\}$ as gmatt said.
I am a little too slow, but I will have another 70 years (I meant 70 more years) to learn.
Thanks to both of you.
5. Originally Posted by Plato
Did you notice that $S$ is a subset of negative integers?
That means what about $T$?
I'm not exactly sure what you mean, if $S \subset \mathbb{N}$ then it can't be a subset of the negative integers. I assumed what the OP meant was that he was switching the meaning of $S$, not adding another condition onto it ( since trivially if $S \subset \mathbb{N}$ and $S \subset \mathbb{Z}^-$ then $S = \emptyset$)
6. Originally Posted by novice
I found that set notations can be very tricky. I have many dumb questions. Apparently, they also teach me the most. It made me think.
I am thinking: Since elements of $S$ are negative, where $S \subset \mathbb{Z}^-$, and that $-x\in S$, then $x$ ought to be positive, i.e. $x>0$, so $T=\{x\in Z: x>0\}$.
I contemplated about the question you asked--and of course, the poster immediately before you too helped me think. So, I wrote on a piece of paper
If $S\not= \emptyset$ and $S \subset \mathbb{N}$, then the elements of $S$ cannot be negative, despite $T$ being defined as
$T=\{x\in \mathbb{Z}: -x\in S\}$. Since
$-x >0$, it has to be nothing else but $x<0$,
so $T=\{x\in \mathbb{Z}: x<0\}$ as gmatt said.
I am a little too slow, but I will have another 70 years to learn.
Thanks to both of you.
Well to be clear, if $S \subset \mathbb{N}$ then $T$ are all the elements of $S$ negated...
i.e.
if
$S = \{ s_1, s_2, ... \} \subset \mathbb{N}$
then
$T = \{ -s_1, -s_2, ... \} \subset \mathbb{Z}^-$
7. Originally Posted by gmatt
I'm not exactly sure what you mean, if $S \subset \mathbb{N}$ then it can't be a subset of the negative integers.
Originally Posted by novice
Given a set of negative integers, $S$, where $S \subset \mathbb{Z}^-$, we can build a set $T$ of positive numbers by a set builder notation such as
$T=\{x \in Z: -x \in S\}$
Did you bother to read the OP?
It clearly states that $S$ is a set of negative integers.
8. Originally Posted by gmatt
I'm not exactly sure what you mean, if $S \subset \mathbb{N}$ then it can't be a subset of the negative integers. I assumed what the OP meant was that he was switching the meaning of $S$, not adding another condition onto it ( since trivially if $S \subset \mathbb{N}$ and $S \subset \mathbb{Z}^-$ then $S = \emptyset$)
Sorry, I didn't mean to confuse anyone. I had one thing and I asked myself a question by switching $S\subset \mathbb{Z}^-$ to $S\subset \mathbb{Z}^+$. Experimenting with things by taking off the head and tail, and putting tail where the head was and head where the tail was and see what shape I would get. I am groping like a blind man.
9. Originally Posted by Plato
Did you bother to read the OP?
It clearly states that $S$ is a set of negative integers.
Sorry, sir, I wasn't ignoring your question. I realized that I have confused you. I should have made it more clear that I was switching things around for experiment.
While I was switching things around, I confused myself and then asked question, but I have learned quite a bit though.
If there is any consolation, you do help me learn.
10. Originally Posted by novice
Sorry, sir, I wasn't ignoring your question. I realized that I have confused you. I should have made it more clear that I was switching things around for experiment.
While I was switching things around, I confused myself and then asked question, but I have learned quite a bit though.
If there is any consolation, you do help me learn.
Yes, that is the way I interpreted things, I think Plato might not have realized that the first part of your post was a statement and the later part of it was the question.
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http://mathematica.stackexchange.com/questions/3963/animating-a-voronoi-diagram?answertab=votes
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# Animating a Voronoi Diagram
edit: Excellent answers have been provided and I made an animation which is suitable for my use, however, all the examples rely on bitmap/rasterized data; is there a vector based approach?
I would like to animate the formation of a voronoi network from a set of semi-random points.
````points = Table[{i, j} + RandomReal[0.4, 2], {i, 10}, {j, 10}];
points = Flatten[points, 1];
````
The final `VoronoiDiagram` can be easily plotted with `DiagramPlot` in the `ComputationalGeometry` package.
````Needs["ComputationalGeometry`"]
voronoi = DiagramPlot[points, TrimPoints -> 50, LabelPoints -> False];
````
I want to animate a series of circles growing outwards uniformly from each of the points until they intersect to form the voronoi network.
````ExpandingCircles[r_, points_] :=
Graphics@{Point /@ points, Circle[#, r] & /@ points}
plots = ExpandingCircles[#, points] & /@ {0.1, 0.2, 0.3, 0.4, 0.5};
GraphicsGrid@Partition[Join[plots, {voronoi}], 3]
````
Similar to that progression but in mine the circles overlap. I want them to stop growing as they hit the adjacent circle to form the voronoi network but I can't figure out how to do this.
Based on @R.M.s pointing out @belisarius answer I've tried this:
````GraphicsGrid@
Partition[
ColorNegate@
EdgeDetect@
Dilation[ColorNegate@Binarize@Rasterize@Graphics@Point@points,
DiskMatrix[#]] & /@ Range[1, 24, 3], 4]
````
However, I can't get them to merge into the voronoi structure.
Somewhat like this video (http://www.youtube.com/watch?v=FlkrBSh4514) except all of mine start growing at the same point in time.
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5
– rm -rf♦ Apr 5 '12 at 19:52
@R.M That answer of mine was for a more difficult requirement, because the OP there wanted a final result looking like a "natural" tissue. I think the current answers are ok for this question – belisarius Apr 6 '12 at 1:06
Re the edit: There is a natural vector-based approach. It relies on the fact that each Voronoi cell "belongs" to a specific point. (This is not the case for higher-order Voronoi diagrams.) Therefore, intersecting the disk of radius $r$ around each point $i$ with that point's Voronoi cell produces the desired collection of (non-overlapping) polygons at stage $r$ in the animation. Mathematica is not well suited to computing and displaying intersections of disks and polygons. – whuber Apr 6 '12 at 16:05
## 2 Answers
The first step is to rasterize the points, so let's just start there as an example:
````n = 512;
g = Image[Map[Boole[# > 0.001] &, RandomReal[{0, 1}, {n, n}], {2}]]
````
The trick is to exploit the distance image. Almost all the work is done here (and it's fast):
````i = DistanceTransform[g] // ImageAdjust // ImageData;
````
We need a little more precomputation of the final boundaries. Rasterizing a vector-based Voronoi tessellation would be faster, but here's a quick and dirty solution:
````mask = Image[WatershedComponents[Image[i]]]
````
Now the animation is instantaneous: it's done simply by thresholding the distances. (Colorize it if you like.) Have fun!
````Manipulate[
ImageMultiply[
Image[MorphologicalComponents[Image[Map[1 - Min[c, #] &, i, {2}]], 1 - c]], mask],
{c, 0, 1}
]
````
-
Wonderful approach! And how would you add colors to that? I was thinking of using your result as a mask on a colorized Voronoi diagram, would that work? – F'x Apr 5 '12 at 21:56
I haven't tested it--and am out of time today--but I imagine another application of `MorphologicalComponents` followed by `Colorize` would do it. – whuber Apr 5 '12 at 21:58
Ah--it's not quite that simple, because the colors determined by `Colorize` change from frame to frame. Instead, consider colorizing the components of the "mask" itself, and then masking that with the thresholded distances. This will keep the colors constant. Not only that, it will be even faster because each frame requires only a threshold and multiply, nothing more. – whuber Apr 5 '12 at 22:05
I know think of at least one way of doing it slowly and in a bitmap approach:
````img[p_, r_] := Module[{f, closest, color, colors, n, t},
n = 250;
colors =
List @@@ {Red, Green, Blue, Yellow, Orange, Pink,
RGBColor[0, 0, 0], Cyan, Magenta, Brown, Purple};
color[i_] := Module[{c},
c = colors[[1 + Mod[i, Length@colors]]];
If[i == 0, {1, 1, 1}, c]
];
closest[pt_] := First@Ordering[Norm[pt - #] & /@ p, 1];
Image[Table[Module[{res, pt},
pt = {(i + 0.5)/n, (j + 0.5)/n};
res = closest[pt];
color[If[Norm[pt - p[[res]]] < r, res, 0]]
], {i, 1, n}, {j, 1, n}]]
]
````
which is then used as:
````points = RandomReal[1, {25, 2}];
t = Table[img[points, r], {r, 0.03, 0.4, 0.03}];
Export["growth.gif", t];
````
What it does is for each point of the graph, determine both what's the closest point in the points set, and whether it's close enough to actually be inside it growth radius. Animate that on a fixed set of points with an increasing radius, and I believe you're there.
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lang-mma
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http://mathhelpforum.com/discrete-math/171740-choice-functions-collections.html
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# Thread:
1. ## Choice functions for collections...
Find if possible a choice function for each of the following collections, without using the choice axiom:
a. The collection A of nonempty subsets of Z+ (positive integers)
b. The collection B of nonempty subsets of Z (integers)
c. The collection C of nonempty subsets of the rational numbers Q
d. The collection D of nonempty subsets of X^(omega), where X = {0, 1}
2. Originally Posted by iamthemanyes
Find if possible a choice function for each of the following collections, without using the choice axiom:
a. The collection A of nonempty subsets of Z+ (positive integers)
b. The collection B of nonempty subsets of Z (integers)
c. The collection C of nonempty subsets of the rational numbers Q
d. The collection D of nonempty subsets of X^(omega), where X = {0, 1}
For (a) , for example , $\forall\,X\in A\,,\,\,f(X):=$ the minimal element of $X$ wrt the usual order
on the natural numbers.
You try now the other ones by yourself.
Tonio
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http://physics.stackexchange.com/questions/21232/is-the-form-of-the-lagrangian-relevant-before-the-renormalization-procedure
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# Is the form of the Lagrangian relevant before the renormalization procedure?
In the renormalization procedure, is writing things like
$$\varphi=\sqrt{Z_{\varphi}}\ \varphi_R\ ,\ \ m_0^2=Z_m\ m_R^2\ ,\ \ g_0=Z_g \mu^{\epsilon}\ g_R$$ and $$Z_i=1+\sum_{\nu=1}^\infty C_i^{(\nu)}(m_R,\mu,\Lambda\text{ or }\epsilon)·g_R^\nu\ , \ \ \ \ \ i=\varphi, m, g$$ really more than just an arbitrary ansatz?
I have no idea what principle people follow, when people have a Lagrangian, say for QED and then write down Lagrangians in the to-be-renormalized-stage. There seems to be a motivation to make them look similar to the old Lagrangian before introducing that coupling constrant expansion - and why in $g$, not other variables like $m$? Hence they write things like $m_{old}=c·m_{new}$, which seems faily conservative, because it doesn't introduce new terms, beyond maybe counter terms that look structurally list the old ones. But as far as I can see, the theory really just starts with the Lagrangian, which contains the to be found $Z$-expressions. You don't use the Lagrangian before that, do you? At least not beyond tree graphs. Therefore I think you could just begin with a buch of terms, with object that have to be fitted by renormalization. The theory effectively seems just to start with the non-bare object.
From all the possible 'unphysical numbers' in the expansion for the (finite number of) $Z$-terms, why does only the 'scale' $\mu$ survive? Do all scheme leave one number open, and if yes, why? I don't get the what this object '$\mu$' is, at all.
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– Vladimir Kalitvianski Jan 19 at 19:27
## 2 Answers
Yeah, it's just an ansatz.
For most practical purposes, it doesn't matter precisely which Lagrangian you work with, because most of the physical values you're computing only depend on the large size asympotics of the correlation functions, i.e., on the universality class of the Lagrangian. You can add any reasonably small non-renormalizable perturbation to your short distance action, and you won't noticeably change the asymptotics you're computing.
It's a convenient ansatz though, because it means that you have a lot less stuff to keep track of. You're saying that the short distance physics has pretty much the same character as the long distance physics, up to rescalings. If you can satisfy the ansatz, you get nicer computations. But of course, you can't always satisfy the ansatz. If you're seeing non-renormalizable interactions in the long distance physics, you should expect to see new physics arising before you get to much smaller distance scales.
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In 4 spacetime dimensions, an unrenormalized Lagrangian is meaningless, but serves as a template for the construction of a family of renormalizable theories. The reason is that the interactions are not relatively compact compared to the Hamiltonian of the free theory, so one can only add an infinitesimal amount of them.
Thus one never has, as in classical mechnaics, ''a Lagrangian'' but a family of basic operators whose linear combinations give the bare Lagrangians.
Because the coupling constant(s) $g$ must be infinitesimal, the expansion must be in terms of $g$. To proceed, one makes all coefficients arbitrary functions of $g$, regularizes to get finite approximations, and then adjusts the functions in such a way that a limit in which the regularization disappears, can be performed.
This determines the coefficient functions to a large extent, leaving a small vector space of coefficent functions.
In renormalizable theories, this space of is finite-dimensional, and can be parameterized by the renormalized masses and coupling constants. Thus one gets a few-parameter family of physical theories, and the parameters can then be adjusted to match experimental data and determine the ''correct'' theory.
But the space of parameters has always one more dimension than the space of theories, because of the particular way regularization and renormalization cooperate. Thus one gets 1-parameter families of equivalent theories, which are identified via an equivalence relation defined by the renormalization group equations.
In nonrenormalizable theories, this space is infinite-dimensional, but at any fixed energy scale, only finitely many coefficients must be taken into account, and again, the parameters can be adjusted to match experimental data.
Thus there is nothing arbitrary in the procedure, except for the particular way the coefficients are represented in specific calculations, and the particular way the solution set is parameterized. How to do these steps is more or less accidental, and chosen partly for historical reasons, partly because it simplifies subsequent calculations.
See also my tutorial paper Renormalization without infinities - a tutorial, which discusses renormalization on a much simpler level than quantum field theory, and Chapter B5: Divergences and renormalization of my theoretical physics FAQ.
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http://mathhelpforum.com/math-topics/15252-rational-irrational-numbers.html
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# Thread:
1. ## Rational and Irrational numbers
Now, don't think this is just an ordinary question! Here goes:
a and b are irrational numbers
a + b is rational
a x b is rational
Work out possible values for a and b where they are positive irrational numbers.
I am sure that I have gotten down all the details, any help will be much appreciated, thank you!
2. Originally Posted by Geometor
Now, don't think this is just an ordinary question! Here goes:
a and b are irrational numbers
a + b is rational
a x b is rational
Work out possible values for a and b where they are positive irrational numbers.
I am sure that I have gotten down all the details, any help will be much appreciated, thank you!
a = (1 + sqrt(2))
b = -sqrt(2)
then a and b are irrational numbers and...
a + b = (1 + sqrt(2)) + (-sqrt(2)) = 1 which is rational
a = sqrt(2)
b = 1/sqrt(2)
then a and b are both irrational and...
a*b = sqrt(2)*1/sqrt(2) = 1 which is rational
i should be using this oppurtunity to try out Latex, i don't know how to use it
EDIT: o sorry, i forgot a and b must be positive, so the first example is incorrect, i'll come up with another one
a = 10 - sqrt(5) + sqrt(2)
b = sqrt(5) - sqrt(2)
then a + b = 10 which is rational and both a and b are positive irrational numbers
3. Originally Posted by Jhevon
a = (1 + sqrt(2))
b = -sqrt(2)
then a and b are irrational numbers and...
a + b = (1 + sqrt(2)) + (-sqrt(2)) = 1 which is rational
a = sqrt(2)
b = 1/sqrt(2)
then a and b are both irrational and...
a*b = sqrt(2)*1/sqrt(2) = 1 which is rational
i should be using this oppurtunity to try out Latex, i don't know how to use it
EDIT: o sorry, i forgot a and b must be positive, so the first example is incorrect, i'll come up with another one
a = 10 - sqrt(5) + sqrt(2)
b = sqrt(5) - sqrt(2)
then a + b = 10 which is rational and both a and b are positive irrational numbers
Many thanks but I don't think that is correct since
10 - sqrt(5) + sqrt(2) times sqrt(5) - sqrt(2) is irrational :S
thanks for trying, lol this is a really complex problem ay?
is this a trick question and it might be impossible?
4. Originally Posted by Geometor
Now, don't think this is just an ordinary question! Here goes:
a and b are irrational numbers
a + b is rational
a x b is rational
Work out possible values for a and b where they are positive irrational numbers.
$a=1-\sqrt{2} \mbox{ and }b=1+\sqrt{2}$
5. Originally Posted by ThePerfectHacker
$a=1-\sqrt{2} \mbox{ and }b=1+\sqrt{2}$
erm... the question requires "positive" numbers and I believe that 1- sq. root 2 is negative?
thanks for helping anyway
6. Originally Posted by Geometor
Many thanks but I don't think that is correct since
10 - sqrt(5) + sqrt(2) times sqrt(5) - sqrt(2) is irrational :S
thanks for trying, lol this is a really complex problem ay?
is this a trick question and it might be impossible?
o, we have to come up with a pair of numbers that satisfy both condtions at the same time? i thought we could use different examples for each.
EDIT: TPH came up with an example
7. Originally Posted by Geometor
erm... the question requires "positive" numbers and I believe that 1- sq. root 2 is negative?
thanks for helping anyway
So then change it a little bit,
$a = 2 - \sqrt{2} \mbox{ and } b = 2 + \sqrt{2}$
What is so hard?
8. Originally Posted by Geometor
Now, don't think this is just an ordinary question! Here goes:
a and b are irrational numbers
a + b is rational
a x b is rational
Work out possible values for a and b where they are positive irrational numbers.
I am sure that I have gotten down all the details, any help will be much appreciated, thank you!
Because their sum is rational $a$ and $b$ are of the form:
$a = A + x$,
$b = B - x$
where $A$ and $B$ are rational and x is irrational.
Then:
$<br /> a\ b = AB + (-A+B)x - x^2 = C<br />$
where C is rational.
Now we have demanded that A, B and C be rational, but we may as well
demand that they be integers (the derivation of rational solutions from
integer solutions is fairly elementary and left to the reader).
Anyway we have:
$<br /> x^2 + (A-B)x + (C- AB) = 0<br />$
and $x$ is irrational. However:
$<br /> x=\frac{-(A+B) \pm \sqrt{(A-B)^2 - 4(C-AB)}}{2}=\frac{-(A+B) \pm \sqrt{(A+B)^2 - 4C}}{2}<br />$
which is irrational only if $q= (A+B)^2 - 4C$ is not a perfect
square.
So here is our algorithm for finding solutions:
Choose an integer $K$, and an integer $C$ such that $K^2 - 4C$ is not a perfect square,
then choose $A$ and $B$ so that $A+B=K$, then:
$<br /> x=\frac{-(A+B) \pm \sqrt{(A+B)^2 - 4C}}{2}<br />$
is an irrational number such that if:
$a=A+x$
$b=B-x$
then $a$ and $b$ are irrational and $a+b$ is an integer as is $a \times b$
Of course this does not guarantee that both $a$ and $b$ are positive.
RonL
9. thanks for the help! sorry, can only thank one person a day I believe?
I find this very confusing though :S
10. Originally Posted by Geometor
thanks for the help! sorry, can only thank one person a day I believe?
I find this very confusing though :S
No you can thank as many as you want.
RonL
11. Originally Posted by Geometor
thanks for the help! sorry, can only thank one person a day I believe?
how come?
12. Originally Posted by Jhevon
how come?
oh woops, nevermind it's just that I can't thank on the same post so I got mixed up
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http://math.stackexchange.com/questions/242606/how-to-compute-the-hilbert-symbol?answertab=oldest
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# How to compute the Hilbert symbol?
Let $a,b,c \in \mathbb Q_2^*$, where $\mathbb Q_2$ denotes the 2-adic field. How to compute the Hilbert symbol $(b^2-4ac,2a)$?
By definition, $(b^2-4ac,2a)=1$, if the equation $z^2-(b^2-4ac)x^2-2ay^2=0$ has a non-trivial solution and $(b^2-4ac,2a)=-1$ otherwise.
I know, that for $\alpha=2^mu$ and $\beta=2^n v$ (with $u,v\in U$) we have $(\alpha,\beta)=(-1)^{\varepsilon (u)\varepsilon (v)+m\omega (v)+n\omega (u)}$.
But I don't think, that it is helpful, because I need to write $b^2-4ac$ as $2^k w$ with $w\in U$.
Are there other ways to compute it?
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## 1 Answer
On $\Bbb Q_p$ the Hilbert symbol $(a,b)$ depends only on the classes of $a$ and $b$ modulo $(\Bbb Q_p^\times)^2$. There are eight such classes when $p=2$. So, if nothing better, you can try to obtain the classes of $b^2-4ac$ and $2a$ modulo $(\Bbb Q_2^\times)^2$ depending on $a$, $b$ and $c$.
When $a$ and $b$ are in $\Bbb Q$, a way to compute $(a,b)_2$ is to compute $(a,b)_\infty$ and $(a,b)_p$ for odd $p$ (which is somewhat simpler) and use the fact that $$\prod_{p\leq\infty}(a,b)_p=1.$$
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http://physics.stackexchange.com/questions/49867/cyclic-coordinates-in-hamiltonian-mechanics/49873
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# Cyclic Coordinates in Hamiltonian Mechanics
I was reading up on Hamiltonian Mechanics and came across the following:
If a generalized coordinate $q_j$ doesn't explicitly occur in the Hamiltonian, then $p_j$ is a constant of motion (meaning, a constant, independent of time for a true dynamical motion). $q_j$ then becomes a linear function of time. Such a coordinate $q_j$ is called a cyclic coordinate.
The above quote is taken from p. 4 in Ref. 1.
What I don't understand is why $q_j$ is a linear function of time if $p_j$ is constant in time. In other words, why does $p_j$ constant in time imply partial $\frac{\partial H}{\partial p_j}$ is a constant? (In particular, $\frac{\partial H}{\partial p_j}$ could depend on any of the other coordinates or momenta.)
Reference:
1. Patrick Van Esch, Hamilton-Jacobi Theory in Classical Mechanics, Lecture notes. The pdf file is available from the author's homepage here.
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## 1 Answer
OP is right. The text has an error. A cyclic coordinate $q_j$ does not have to be an linear function of $t$.
Example: Consider two canonical pairs $(q,p)$ and $(Q,P)$ with Hamiltonian $H= p Q +P$.
Then $q$ is cyclic, and therefore $p$ is a constant of motion.
$\dot{Q} =\frac{\partial H}{\partial P}=1$, so $Q$ is a linear function of time.
$\dot{q}= \frac{\partial H}{\partial p} = Q$, and hence $q$ is a quadratic function of time.
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http://mathhelpforum.com/calculus/59073-ugh-optimazation.html
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# Thread:
1. ## ugh, optimazation...
i just have issues with the algebra, i can take the derivative, but then im stuck. thanks for any help in advance!
An electric current, , in amps, is given by
where is a constant. What are the maximum and minimum values of I?
2. Originally Posted by purplegirl1818
i just have issues with the algebra, i can take the derivative, but then im stuck. thanks for any help in advance!
An electric current, , in amps, is given by
where is a constant. What are the maximum and minimum values of I?
Solve dI/dt = 0.
Show all the work you've done and where you get stuck.
By the way, there's also a non-calculus way of doing this that involves re-writing the expression for current as $I = A \cos (\omega t + \phi)$.
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http://physics.stackexchange.com/questions/937/how-does-gravity-escape-a-black-hole
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# How does gravity escape a black hole?
My understanding is that light can not escape from within a black hole (within the event horizon). I've also heard that information cannot propagate faster than the speed of light. It would seem to me that the gravitational attraction caused by a black hole carries information about the amount of mass within the black hole. So, how does this information escape? Looking at it from a particle point of view: do the gravitons (should they exist) travel faster than the photons?
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1
This question doesn't really make sense as far as I'm concerned. For a start, it's trying to consider gravity from a quantum perspective (which we don't understand) rather than GR. – Noldorin Nov 16 '10 at 17:45
37
It seems like a reasonable question for someone who has only heard stories about physics to ask. It shows good analytical thinking and deserves an answer. (I would do it, but I only have a vague notion that it has to do with the self-energy of the curvature of spacetime around the black hole, rather than the mass inside it, causing the black hole's gravitational field.) – Mark Eichenlaub Nov 16 '10 at 18:00
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@Noldorin: I, on the other hand, think it is a brilliant question. Also, your last statement is correct only if by we you mean I :-) – Marek Nov 16 '10 at 19:01
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@Marek: You are either being very arrogant or joking here. No-one understands quantum gravity properly; I say that with confidence! – Noldorin Nov 16 '10 at 19:09
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@Noldorin: I was only half-joking. It's true that it's not understood properly (and certainly not by me) but they (as in best theoretical physicist) know something about it. And working under assumption that string theory is a correct (which seems reasonable) they know actually a great deal about it. – Marek Nov 16 '10 at 19:29
show 10 more comments
## 13 Answers
Well, the information doesn't have to escape from inside the horizon, because it is not inside. The information is on the horizon.
One way to see that is from the fact that from the perspective of an observer outside the horizon of a black hole, nothing ever crosses the horizon. It asymptotically gets to the horizon in infinite time (as it is measured from the perspective of an observer at infinity).
An other way to see that is the fact that from the boundary conditions on the horizon you can get all the information you need to describe the space-time outside but that is something more technical.
Finally, since classical GR is a geometrical theory and not a quantum field theory*, gravitons is not the appropriate way to describe it.
*To clarify this point, GR can admit a description in the framework of gauge theories like the theory of electromagnetism. But even though electromagnetism can admit a second quantization (and be described as a QFT), GR can't.
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3
Good answer, this addresses many of the issues. – Noldorin Nov 16 '10 at 18:05
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Last paragraph is just wrong, please remove it: GR is very similar to other (classical) gauge theories and there are reformulations where it is completely equivalent. Also gravitons are precisely what describes gravity in any reasonable quantum theory of gravity. And there is also no problem in quantizing e.g. gravitational waves (the quanta of which are gravitons) on curved background in the very same manner we quantize EM waves to get photons. – Marek Nov 16 '10 at 18:59
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I didn't say it is not a gauge theory. It is not a quantum field theory. Electromagnetism is also a gauge theory, one that admits a second quantization. I am not aware of a similar process for GR. – Vagelford Nov 17 '10 at 10:50
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– Daniel Grumiller Feb 10 '11 at 1:35
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Yes. GR is an effective quantum field theory. It is non-renormalizable, like many other effective quantum field theories. However, I think for the current context quantum considerations are actually irrelevant. The static 1/r potential comes from classical considerations. – Daniel Grumiller Feb 10 '11 at 13:00
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There are some good answers here already but I hope this is a nice short summary:
Electromagnetic radiation cannot escape a black hole, because it travels at the speed of light. Similarly, gravitational radiation cannot escape a black hole either, because it too travels at the speed of light. If gravitational radiation could escape, you could theoretically use it to send a signal from the inside of the black hole to the outside, which is forbidden.
A black hole, however, can have an electric charge, which means there is an electric field around it. This is not a paradox because a static electric field is different from electromagnetic radiation. Similarly, a black hole has a mass, so it has a gravitational field around it. This is not a paradox either because a gravitational field is different from gravitational radiation.
You say the gravitational field carries information about the amount of mass (actually energy) inside, but that does not give a way for someone inside to send a signal to the outside, because to do so they would have to create or destroy energy, which is impossible. Thus there is no paradox.
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Note that there is NO NEED to indroduce any quantum mechanics AT ALL into this discussion. That's why I specifically said "electromagnetic radiation" and "gravitational radiation", not "photons" or "gravitons". – Keenan Pepper Nov 17 '10 at 2:47
Hello Keenan, what about static magnetic field? – Georg Jan 18 '11 at 11:29
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+1 for being the only answer a non-physics major can understand. – Anonymous Type Mar 10 '11 at 0:02
Let's get something out of the way: let's agree not to bring gravitons into this answer. The rationale is simple: when you talk about gravitons you imply a whole lot of things about quantum phenomena, none of which is really necessary to answer your main question. In any case, gravitons propagate with the very same speed as photons: the speed of light, $c$. This way we can focus simply in Classical GR, ie, the Differential Geometry of Spacetime: this is more than enough to address your question.
In this setting, GR is a theory that says how much curvature a space "suffers" given a certain amount of mass (or energy, cf Stress-Energy Tensor).
A Black Hole is a region of spacetime that has such an intense curvature that it "pinches out" a certain region of spacetime.
In this sense, it's not too bad to understand what's going on: if you can measure the curvature of spacetime, you can definitely tell whether or not you're moving towards a region of increasing curvature (ie, towards a block hole).
This is exactly what's done: one measures the curvature of spacetime and that's enough: at some point, the curvature is so intense that the light-cones are "flipped". At that exact point, you define the Event Horizon, ie, that region of spacetime where causality is affected by the curvature of spacetime.
This is how you make a map of spacetime and can chart black holes. Given that curvature is proportional to gravitational attraction, this sequence of ideas completely addresses your doubt: you don't have anything coming out of the black hole, nor anything like that. All you need is to chart the curvature of spacetime, measuring what happens to your light-cone structure. Then, you find your Event Horizon and, thus, your black hole. This way you got all the information you need, without having anything coming out of the black hole.
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Suppose, very hypothetically of course, that some extra mass were suddenly created inside the black hole. Would the spacetime curvature outside the black hole change? I realize this is an unphysical process, but if the hand of God reached down and created a large lump of stuff just inside the event horizon, what do the equations of GR tell us about whether we would we be able to tell about the event from outside the event horizon? – Mark Eichenlaub Nov 16 '10 at 19:40
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The thing to note is that curvature is not something that lives only inside the Black Hole: this is a property of spacetime as a whole, and that's what counts. Global, topological, properties are very non-intuitive things. ;-) – Daniel Nov 16 '10 at 19:50
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But then doesn't information go from inside the black hole to outside it? I could send morse code by turning my mass on and off, right? – Mark Eichenlaub Nov 16 '10 at 20:05
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@MarkE: only if you were God. Look, the bottom-line is that we're dealing with classical GR, and not Quantum Gravity nor its effects. And, within the framework of classical GR, it's simply not possible for you to change any of the properties (charge, mass, angular momentum) of a black hole from the inside of it. A black hole is simply a "sink" of gravitational fields. – Daniel Nov 16 '10 at 20:25
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@Nogwater: As far as we know, the mass itself is at the singularity, for some definition of "is" (namely that the proper time between crossing the event horizon and reaching the singularity is finite). But, speaking in vague terms, the information of how much mass there is gets "imprinted" on the horizon, it does not "fall" down to the singularity along with the mass. In more precise form, this is called the holographic principle. – David Zaslavsky♦ Nov 17 '10 at 18:30
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The problem here is a misunderstanding of what a particle is in QFT.
A particle is an excitation of a field, not the field itself. In QED, if you set up a static central charge, and leave it there a very long time, it sets up a field $E=k{q \over r^2}$. No photons. When another charge enters that region, it feels that force. Now, that second charge will scatter and accelerate, and there, you will have a $e^{-}->e^{-}+\gamma$ reaction due to that acceleration, (classically, the waves created by having a disturbance in the EM field) but you will not have a photon exchange with the central charge, at least until it feels the field set up by our first charge, which will happen at some later time.
Now, consider the black hole. It is a static solution of Einstein's equations, sitting there happily. When it is intruded upon by a test mass, it already has set up its field. So, when something scatters off of it, it moves along the field set up by the black hole. Now, it will accelerate, and perhaps, "radiate a graviton", but the black hole will only feel that after the test particle's radiation field enters the black hole horizon, which it may do freely. But nowhere in this process, does a particle leave the black hole horizon.
Another example of why the naïve notion of all forces coming from a Feynman diagram with two pairs of legs is the Higgs boson—the entire universe is immersed in a nonzero Higgs field. But we only talk about the 'creation' of Higgs 'particles' when we disturb the Higgs field enough to create ripples in the Higgs field—Higgs waves. Those are the Higgs particles we're looking for in the LHC. You don't need ripples in the gravitational field to explain why a planet orbits a black hole. You just need the field to have a certain distribution.
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I think the best explanation that can be given is this: you have to discern between statical and dynamical properties of the space-time. What do I mean by that?
Well, there are certain space-times that are static. This is for example the case of the prototypical black-hole solution of GTR. Now, this space-time exists a priori (by definition of static: it always was there and always will be), so the gravity doesn't really need to propagate. As GTR tells us gravity is only an illusion left on us by the curved space-time. So there is no paradox here: black holes appear to be gravitating (as in producing some force and being dynamical) but in fact they are completely static and no propagation of information is needed. In reality we know that black-holes are not completely static but this is a correct first approximation to that picture.
Now, to address the dynamical part, two different things can be meant by this:
• Actual global change of space-time as can be seen e.g. in the expansion of the universe. This expansion need not obey the speed of light but this is in no contradiction with any known law. In particular you cannot send any superluminal signals. In fact, opposite is true: by too quick an expansion parts of universe might go too far away for even their light to ever reach us. They will get causally disconnected from our sector of space-time and to us it will appear as if it never existed. So it shouldn't be surprising that no information can be communicated.
• Gravitational waves, which is a just fancy name for the disturbances in the underlying space-time. They obey the speed of light and the corresponding quantum particles are called gravitons. Now these waves/particles indeed wouldn't be able escape from underneath the horizon (in the precisely the same way as any other particle, except for Hawking radiation, but this is a special quantum effect).
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I think it's helpful to think about the related question of how the electric field gets out of a charged black hole. That question came up in the (now-defunct) Q&A section of the American Journal of Physics back in the 1990s. Matt McIrvin and I wrote up an answer that was published in the journal. You can see it at https://facultystaff.richmond.edu/~ebunn/ajpans/ajpans.html .
As others have pointed out, it's easier to think about the question in purely classical terms (avoiding any mention of photons or gravitons), although in the case of the electric field of a charged black hole the question is perfectly well-posed even in quantum terms: we don't have a theory of quantum gravity at the moment, but we do think we understand quantum electrodynamics in curved spacetime.
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While in many ways the question was already answered, I think it should be empasized that on the classical level, the question is in some sense backwards. The prior discussion of static and dynamic properties especially comes very close.
Let's first examine a toy model of a spherically-symmetric thin shell of dust particles collapsing into a Schwarzschild black hole. The spacetime outside of the the shell will then also be Schwarzschild, but with a larger mass parameter than the original black hole (if the shell starts at rest at infinity, then just the sum of the two). Intuitively, the situation is analogous to Newton's shell theorem, which a more limited analogue in GTR. At some point, it crosses the horizon and eventually gets crushed out of existence at the singularity, the black hole now gaining mass.
So we have the following picture: as the shell collapses, the external graviational field takes on some value, and as it crossed the horizon, the information about what it's doing can't get out the horizon. Therefore, he gravitational field can't change in response to shell's further behavior, for this would send a signal across the horizon, e.g., a person riding along with the shell would be able to communicate across it by manipulating the shell.
Therefore, rather than gravity having a special property that enables it to cross the horizon, in a certain sense gravity can't cross the horizon, and it is that very property that forces gravity outside of it to remain the same.
Although the above answer assumed a black hole already, that doesn't matter at all, as for a spherically collapsing star the event horizon begins at the center and stretches out during the collapse (for the prior situation, it also expands to meet the shell). It also assumes that the situation has spherical symmetry, but this also turns out to not be conceptually important, although for far more complicated and unobvious reasons. Most notably, the theorems of Penrose and Hawking, as it was initially thought by some (or perhaps I should say hoped) that any perturbation from spherical symmetry would prevent black hole formation.
You may also be wondering about a related question: if the Schwarzschild solution of GTR is a vacuum, does it make sense for a vacuum to bend spacetime? The situation is somewhat analogous to a simpler one from classical electromagnetism. Maxwell's equations dictate how the electric and magnetic fields change in response to the presence and motion of electric charges, but the charges alone do not determine the field, as you can always have a wave come in from infinity without any contradictions (or something more exotic, like an everywhere-constant magnetic field), and in practice these things are dictated by boundary conditions. The situation is similar in GTR, where the Einstein field equation that dictates how geometry are connected only fixes half of the twenty degrees of freedom of spacetime curvature.
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In my opinion this is an excellent question, which manages to puzzle also some accomplished physicists. So I do not hesitate to provide another, a bit more detailed, answer, even though several good answers exist already.
I think that at least part of this question is based upon an incomplete understanding on what it means to mediate a static force from a particle physics point of view. As others have mentioned in their answers already, you encounter a similar issue in the Coulomb problem in electrodynamics.
Let me answer your question from a field theory point of view, since I believe this concurs best with your intuition about particles being exchanged (as apparent from the way you phrased the question).
First, no gravitational waves can escape from inside the black hole, as you hinted already in your question.
Second, no gravitational waves have to escape from inside the black hole (or from the horizon) in order to mediate a static gravitational force.
Gravity waves do not mediate the static gravitational force, but only quadrupole or higher moments.
If you want to think about forces in terms of particles being exchanged you can view the static gravitational force (the monopole moment, if you wish) as being mediated by "Coulomb-gravitons" (see below for the analogy with electrodynamics). Coulomb-gravitons are gauge degrees of freedom (so one may hesitate to call them "particles"), and thus no information is mediated by their "escape" from the black hole.
This is quite analog to what happens in electrodynamics: photon exchange is responsible for the electromagnetic force, but photon waves are not responsible for the Coulomb force.
Photon waves do not mediate the static electromagnetic force, but only dipole or higher moments.
You can view the static electromagnetic force (the monopole moment, if you wish) as being mediated by Coulomb-photons. Coulomb-photons are gauge degrees of freedom (so one may hesitate to call them "particles"), and thus no information is mediated by their "instantaneous" transmission.
Actually, this is precisely how you deal with the Coulomb force in the QFT context. In so-called Bethe-Salpeter perturbation theory you sum all ladder graphs with Coulomb-photon exchanges and obtain in this way the 1/r potential to leading order and various quantum corrections (Lamb shift etc.) to sub-leading order in the electromagnetic fine structure constant.
In summary, it is possible to think about the Schwarzschild and Coulomb force in terms of some (virtual) particles (Coulomb-gravitons or -photons) being exchanged, but as these "particles" are actually gauge degrees of freedom no conflict arises with their "escape" from the black hole or their instantaneous transmission in electrodynamics.
An elegant (but perhaps less intuitive) way to arrive at the same answer is to observe that (given some conditions) the ADM mass - for stationary black hole space-times this is what you would call the "black hole mass" - is conserved. Thus, this information is provided by boundary conditions "from the very beginning", i.e., even before a black hole is formed. Therefore, this information never has to "escape" from the black hole.
On a side-note, in one of his lectures Roberto Emparan posed your question (phrased a bit differently) as an exercise for his students, and we discussed it for at least an hour before everyone was satisfied with the answer - or gave up ;-)
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Could you provide a link to the lecture? – Ehryk Aug 19 '12 at 2:22
@Ehryk: no, sorry - there were no lecture notes, just a series of blackborad lectures – Daniel Grumiller Aug 22 '12 at 17:24
Gravitation doesn't work the way light does (which is why quantum gravity is hard).
A massive body "dents" space and time, so that, figuratively speaking, light has a hard time running uphill. But the hill itself (i.e. the curved spacetime) has to be there in the first place.
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A bit vague, but on its way to being a good answer. – Noldorin Nov 16 '10 at 18:06
But, you could also ask "How does electric force escape a charged black hole", which would be an equally valid question. – Jerry Schirmer Feb 10 '11 at 15:10
The various theories - QED, GTR, classical electromagnetism, quantum loop gravity, etc - are all different ways to describe nature. Nature is what it is; theories all have defects. So far as saying whether gravity resembles electromagnetism in some way or not, is just blowing warm air about how humans think and not saying anything substantial about physical reality.
So what if we don't have a full grasp of quantum gravity? Gravitons are a sensible concept, and a key part in some unified (or semi-unified) field theories. It might get tricky because unlike other quantum particles, gravitons are a part of the curvature of spacetime and the relations of nearby lightcones, as they fly through said spacetime. We can sort of ignorer that for now. The question is good, and can be answered in terms of quantum theory and gravitons. We just don't know, given the existing state of physics knowledge, how far we can push the idea.
When charged particle attract or repel, the force is due to virtual photons. Photons like to travel at the universal speed c, but they don't have to. Heisenberg says so! You can break the laws of conservation of energy and momentum as much as you like, but the more you deviate, the shorter the time span and smaller the bit of space in which you violate these laws. For the virtual photons connecting two charged particles, they've got the room between the two particles, and a time span matching that at lightspeed. These not running waves with a well-defined wavelength, period or phase velocity. This ill-defined velocity can be faster than c or less equally well. In QED, the photon propagator - the wavefunction giving the probability amplitude of a virtual photon connecting (x1, t1) to (x2, t2) is nonzero everywhere - inside and outside the past and future light cones, though becoming unlimited in magnitude on the light cones.
So gravitons, if they are that much like photons, can exist just fine outside the horizon and inside. They are, in a rough sense, as big as the space between the black hole and whatever is orbiting or falling into it. Don't picture them as little energy pellets flying from the black hole center (singularity or whatever) - even with Heisenberg's indulgence, it's just not a matter of small particles trying to get through the horizon the wrong way. A graviton is probably already on both sides!
For a more satisfying answer, I suspect it takes knowing the math,Fourier transforms, Riemann tensors and all that.
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The holographic principle gives a clue, as pointed out by David Zaslavsky. The Schwarzschild metric element $g_{tt}~=~1 – r_0/r$, for $r_0~=~2GM/c^2$ gives a proper distance called the delay coordinate $$r^*~=~r~+~r_0 ln[(r-r_00)/r_0]$$ which diverges $r^*~\rightarrow~-\infty$ as you approach the horizon. What this means is that all the stuff which makes up the black hole is never seen to cross the horizon from the perspective of a distant outside observer. The clock on anything falling into a black hole is observed to slow to a near stop and never cross the horizon. This means nothing goes in or out of the black hole, at least classically. So there really is not problem of gravity escaping from a black hole, for as observed from the exterior nothing actually ever went in.
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1
"Proper distance" means taking the length of a curve in a spatial hyperslice, but $r^*$ is not produced in the surface of constant Schwarzschild time, so it's unclear what you're referring to. For radial light rays, $\Delta t = \pm\Delta r^*$, which is relevant to "not seeing", but this comes from $g_{rr}$, not $g_{tt}$. "Nothing goes in or out of the black hole" is just wrong, although it was the view before the mid-1960s, with falling objects slowing and stopping at the infinite redshift surface. (EG: acceleration in Minkowski; things obviously cross the horizon without been seen to do so.) – Stan Liou Jan 18 '11 at 5:00
I should have said proper interval. The toroise coordinate indicates that to see something from the horizon it is seen from the "infinite past." Nothing can be directly observed to actually reach the event horizon. – Lawrence B. Crowell Jan 18 '11 at 19:19
There are observational evidences of matter crossing the horizon of black holes and simply increasing the BH mass, in binary systems. Because the typical spectral fingerprint of the shock wave heating in similar systems but with a white dwarf as accreting object instead of a BH, is absent. Whatever happens to proper time of the accreting matter, it crosses the horizon. – Eduardo Guerras Valera Dec 5 '12 at 21:40
The black hole does "leak" information, but it is not due to "gravitions, but in the form of the Hawking radiation. It has its basis in quantum mechanics, and is a thermal sort of radiation with extremely low rate. This also means that the black hole is slowly evaporates, but on a time scale that is comparable to the age of the universe.
The origin of this radiation can be described in a little bit hand-waving way as such: due to quantum fluctuations, there's particle-antiparticle pair creation going on in the vacuum. If such a pair-creating happens on the horizon, one of the pair can fall into the black hole while the other can escape. To preserve the total energy (since the vacuum fluctuations are around 0) with a particle now flying away, its fallen pair has to have a negative energy from the black hole's point of view, thus it is effectively losing mass. The outside observer perceives this whole process as "evaporation".
This radiation has a distribution as described by a "temperature", which is inversely proportional to the black hole's mass.
Might want to check out http://en.wikipedia.org/wiki/Hawking_radiation and other sources for more details...
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According to Lynden-Bell and J. Katz, at the Racah Institute of Physics, Hebrew University of Jerusalem, there is no mass inside the black hole. The mass resides completely in the gravitational field around it.
They say that all the mass is in the self-energy of the curvature of spacetime around the hole just as alluded to by @Mark Eichenlaub. Information doesn't have to escape from mass within the black hole. There isn't any mass there.
Here are a couple of quotes and a link to the paper:
"... the field energy outside a Schwarzschild black hole totals Mc^2. In this sense all the energy remains outside the hole."
" ... all these formulae lead to all the black hole's mass being accounted for by field energy outside the hole."
http://adsabs.harvard.edu/full/1985MNRAS.213P..21L
I'm not qualified to evaluate the paper but it seemed relevent to the discussion.
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## protected by David Zaslavsky♦Jan 13 '11 at 22:20
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http://mathhelpforum.com/advanced-algebra/136513-zero-ring.html
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# Thread:
1. ## Zero ring
This feels like a dumb question, but I'll ask it anyway. Is it possible for a ring to have a principle ideal $I$ generated by a nonzero element such that $I = \{0\}$?
Edit: Nevermind, I figured this out right after I posted it. This would be the case in a zero ring (not the trivial ring).
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http://mathhelpforum.com/trigonometry/176130-trig-word-problem-angle-depression.html
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# Thread:
1. ## Trig word problem - angle of depression.
Here is the word problem.
A bird on top of a vertical pole looked South at an angle of depression of 35 Degrees, saw Icing, Maya's bunny on the ground. The bird turned and looked East, and at an angle of depression of 40 degrees saw, Notowada, Maya's turtle on the ground. The distance between Icing and Ntwoada is 60m. Find the height of the pole given this information.
I have diagrammed this out and see it perfectly I just don't know how to calculate this given the information.
I can't see how i can use Sin Rule, Cos rule or anything...Can someone show me the process of doing this. My diagram looks like a right triangular pyramid.
Please clearly explain your work so I can understand why you are performing certain operations.
Thank you so much in advance!
Sincerely,
Raymond
2. Originally Posted by raymac62
Here is the word problem.
A bird on top of a vertical pole looked South at an angle of depression of 35 Degrees, saw Icing, Maya's bunny on the ground. The bird turned and looked East, and at an angle of depression of 40 degrees saw, Notowada, Maya's turtle on the ground. The distance between Icing and Ntwoada is 60m. Find the height of the pole given this information.
I have diagrammed this out and see it perfectly I just don't know how to calculate this given the information.
I can't see how i can use Sin Rule, Cos rule or anything...Can someone show me the process of doing this. My diagram looks like a right triangular pyramid.
Please clearly explain your work so I can understand why you are performing certain operations.
Thank you so much in advance!
Sincerely,
Raymond
let h = pole height , x = distance from base of the pole to bunny , y = distance from base of the pole to turtle
$\tan(35) = \dfrac{h}{x}<br />$
$\tan(40) = \dfrac{h}{y}<br />$
$x^2 + y^2 = 60^2$
solve the system of equations for $h$
3. Originally Posted by skeeter
let h = pole height , x = distance from base of the pole to bunny , y = distance from base of the pole to turtle
$\tan(35) = \dfrac{h}{x}<br />$
$\tan(40) = \dfrac{h}{y}<br />$
$x^2 + y^2 = 60^2$
solve the system of equations for $h$
Hi Skeeter,
Thanks for adressing my question. I have never solved systems of equations with 3 variables, let alone in such a complex form. I know we are dealing with a 3 dimension problem here. Could you please demonstrate and explain how to solve the system of equations and solve for H. I have been obsessing over this question all day and I am so perplexed.
Thanks in advance,
Ray
4. $\tan(35) = \dfrac{h}{x} \implies x = \dfrac{h}{\tan(35)}$
$\tan(40) = \dfrac{h}{y} \implies y = \dfrac{h}{\tan(40)}$
$x^2 + y^2 = 60^2$
$\left[\dfrac{h}{\tan(35)}\right]^2 + \left[\dfrac{h}{\tan(40)}\right]^2 = 60^2$
$\dfrac{h^2}{\tan^2(35)} + \dfrac{h^2}{\tan^2(40)} = 60^2$
$h^2 \left[\dfrac{1}{\tan^2(35)} + \dfrac{1}{\tan^2(40)}\right] = 60^2$
$h^2[\cot^2(35) + \cot^2(40)] = 60^2$
$h^2 = \dfrac{60^2}{\cot^2(35)+\cot^2(40)}$
$h = \dfrac{60}{\sqrt{\cot^2(35) + \cot^2(40)}}$
5. ## adjacent/opposite ?? h/x?
Originally Posted by skeeter
$\tan(35) = \dfrac{h}{x} \implies x = \dfrac{h}{\tan(35)}$
$\tan(40) = \dfrac{h}{y} \implies y = \dfrac{h}{\tan(40)}$
$x^2 + y^2 = 60^2$
$\left[\dfrac{h}{\tan(35)}\right]^2 + \left[\dfrac{h}{\tan(40)}\right]^2 = 60^2$
$\dfrac{h^2}{\tan^2(35)} + \dfrac{h^2}{\tan^2(40)} = 60^2$
$h^2 \left[\dfrac{1}{\tan^2(35)} + \dfrac{1}{\tan^2(40)}\right] = 60^2$
$h^2[\cot^2(35) + \cot^2(40)] = 60^2$
$h^2 = \dfrac{60^2}{\cot^2(35)+\cot^2(40)}$
$h = \dfrac{60}{\sqrt{\cot^2(35) + \cot^2(40)}}$
Why: $\tan(35) = \dfrac{h}{x}$ ?
That's adjacent/opposite. I thought tan A was opposite/adjacent.
Did you mean to say $\tan(55) = \dfrac{h}{x}$ and $\tan(50) = \dfrac{h}{y}$ ??
6. Originally Posted by raymac62
Why: $\tan(35) = \dfrac{h}{x}$ ?
That's adjacent/opposite. I thought tan A was opposite/adjacent.
Did you mean to say $\tan(55) = \dfrac{h}{x}$ and $\tan(50) = \dfrac{h}{y}$ ??
no, I did not mean that. remember alternate interior angles? the angle of depression = the angle of elevation. look at the sketch ...
Attached Thumbnails
7. I see! Thank you very much.
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http://mathoverflow.net/questions/102624/multiplicity-one-conjecture/107264
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## Multiplicity one conjecture
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I recently became interested in Maass cusp forms and heared people mentioning a "multiplicity one conjecture". As far as I understood it, it says that the dimension of the space of Maass cusp form for fixed eigenvalue should be at most one.
Since Maass cusp forms always are defined for a Fuchsian lattice, I wonder 1) for which lattices this conjecture had been conjectured? 2) what is the motivation for this conjecture? 3) to whom this conjecture is due? 4) is it published somewhere? 5) is it proven in some cases?
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## 3 Answers
I rethought your question and have discovered a partial answer for 1) and 2). I add this as a disjoint answer, since my other answer adresses a totally different (negative) issue.
In Sarnak's article http://web.math.princeton.edu/sarnak/baltimore.pdf, he recalls one famous conjecture (Conjecture I, due to himself) that $\Gamma \backslash \mathbb{H}$ should have very few Maass forms for "most" Fuchsian lattices $\Gamma$.
Btw, he attributes the simplicity conjecture (Conjecture 3) for $SL_2(\mathbb{Z})$ to Cartier.
Wolpert (Theorem I) has shown that Sarnak's conjecture would follow if the simplicity conjecture holds for the congruence subgroup $\Gamma(2)$.
Also GH last conjecture that the multiplicity is uniformly bounded in $N$ would suffice for Sarnak's conjecture, but current knowledge is that the multiplicity of an eigenvalue of magnitude $T$ is at most $\ll_N \sqrt{T}/ \log T$ and not $\ll_N 1$.
In fact, Sarnak conjectures that there exist $\Gamma$ with only finitely many Maass wave forms, but this does not follow from the simplicity conjecture for $\Gamma(2)$.
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This conjecture is usually stated for $\mathrm{SL}_2(\mathbb{Z})$, and it is widely open. I think it is folklore, and is stated in several papers, e.g. in Luo: Nonvanishing of $L$-values and the Weyl law (before (3)).
The motivation, I think, is similar as with the conjecture for the multiplicity of the Riemann zeta zeros. The belief is that there is no "accidental" algebraic independence among the eigenvalues of the Laplacian or the zeros of an automorphic $L$-function. For example, the Laplacian eigenvalue $1/4$ is expected to "come from" an even Galois representation, while the zero $1/2$ is expected to "come from" rational points of infinite order on an abelian variety. For $\mathrm{SL}_2(\mathbb{Z})$ or $\zeta(s)$ we don't know of any object that would "impose" any algebraic independence on the data, hence we believe that in those cases the data is entirely transcendental.
For congruence subgroups $\Gamma_0(N)$ the "multiplicity one conjecture" is false, because the eigenvalue $1/4$ is known to occur with multiplicity for some $N$'s. The known examples come from even Galois representations. I think it is safe to believe that the multiplicities are bounded for any $N$.
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Is it also know for newforms of $\Gamma_0(N)$? – Marc Palm Jul 19 at 14:29
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I think the eigenvalue 1/4 occurs with multiplicity even among newforms of some level. All we need is two even Galois representations with the same Artin conductor. – GH Jul 19 at 19:16
Multiplicity one refers to something else, related but much weaker.
For the analogue question for lattices, there are trivial counter examples: Induction by steps for example suggests on the level of Lie groups $$Ind_{\Gamma(N)} ^{PSL_2(\mathbb{R})} 1 \cong Ind_{PSL_2(\mathbb{Z})} ^{PSL_2(\mathbb{R})} Ind_{\Gamma(N)}^{PSL_2(\mathbb{Z})} 1$$ and e.g. by the Peter-Weyl theorem, we know that $$Ind_{\Gamma(N)}^{PSL_2(\mathbb{Z})} 1$$ the multiplicity of an irreducible representation equals its dimension.
Note that GH's example is less trivial, since $$Ind_{\Gamma_0(N)}^{PSL_2(\mathbb{Z})} 1$$ decomposes with multiplicity one.
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Could you elaborate to what it does refer, please? – Ruedi Meier Aug 15 at 17:49
For every prime, you get also an eigenvalue. Multiplicity one theorem states that their exists only one eigenfunction with all the same eigenvalues. More rigorously put, you find this here: en.wikipedia.org/wiki/Multiplicity-one_theorem – Marc Palm Sep 4 at 9:33
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http://mathoverflow.net/questions/8957/homotopy-groups-of-lie-groups/8960
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Homotopy groups of Lie groups
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Several times I've heard the claim that any Lie group $G$ has trivial second fundamental group $\pi_2(G)$, but I have never actually come across a proof of this fact. Is there a nice argument, perhaps like a more clever version of the proof that $\pi_1(G)$ must be abelian?
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Isn't for semi-simple Lie Groups isn't it true that $\pi_3(G)=\mathbb{Z}$? I would be happy to see explanations of this may be as a by-product of this discussion. I suppose this is important to see why the "level" of Chern-Simons theory is quantized. – Anirbit Jun 30 2010 at 14:47
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@Anirbit: that cannot be true, for the product of semi-simple Lie groups is semisimple,and the $\pi_3$ of a product is the product of the $\pi_3$s. It is true for simple groups, though. – Mariano Suárez-Alvarez Feb 18 at 5:19
7 Answers
I don't know of anything as bare hands as the proof that $\pi_1(G)$ must be abelian, but here's a sketch proof I know (which can be found in Milnor's Morse Theory book. Plus, as an added bonus, one learns that $\pi_3(G)$ has no torsion!):
First, (big theorem): Every (connected) Lie group deformation retracts onto it's maximal compact subgroup (which is, I believe, unique up to conjugacy). Hence, we may as well focus on compact Lie groups.
Let $PG = \{ f:[0,1]\rightarrow G | f(0) = e\}$ (I'm assuming everything is continuous.). Note that $PG$ is contracitble (the picture is that of sucking spaghetti into one's mouth). The projection map $\pi:PG\rightarrow G$ given by $\pi(f) = f(1)$ has homotopy inverse $\Omega G =$Loop space of G = $\{f\in PG | f(1) = e \}$.
Thus, one gets a fibration $\Omega G\rightarrow PG\rightarrow G$ with $PG$ contractible. From the long exact sequence of homotopy groups associated to a fibration, it follows that $\pi_k(G) = \pi_{k-1}\Omega G$
Hence, we need only show that $\pi_{1}(\Omega G)$ is trivial. This is where the Morse theory comes in. Equip $G$ with a biinvariant metric (which exists since $G$ is compact). Then, following Milnor, we can approximate the space $\Omega G$ by a nice (open) subset $S$ of $G\times ... \times G$ by approximating paths by broken geodesics. Short enough geodesics are uniquely defined by their end points, so the ends points of the broken geodesics correspond to the points in $S$. It is a fact that computing low (all?...I forget)* $\pi_k(\Omega G)$ is the same as computing those of $S$.
Now, consider the energy functional $E$ on $S$ defined by integrating $|\gamma|^2$ along the entire curve $\gamma$. This is a Morse function and the critical points are precisely the geodesics**. The index of E at a geodesic $\gamma$ is, by the Morse Index Lemma, the same as the index of $\gamma$ as a geodesic in $G$. Now, the kicker is that geodesics on a Lie group are very easy to work with - it's pretty straight forward to show that the conjugate points of any geodesic have even index.
But this implies that the index at all critical points is even. And now THIS implies that $S$ has the homotopy type of a CW complex with only even cells involved. It follows immediately that $\pi_1(S) = 0$ and that $H_2(S)$ is free ($H_2(S) = \mathbb{Z}^t$ for some $t$).
Quoting the Hurewicz theorem, this implies $\pi_2(S)$ is $\mathbb{Z}^t$.
By the above comments, this gives us both $\pi_1(\Omega G) = 0$ and $\pi_2(\Omega G) = \mathbb{Z}^t$, from which it follows that $\pi_2(G) = 0$ and $\pi_3(G) = \mathbb{Z}^t$.
Incidentally, the number $t$ can be computed as follows. The universal cover $\tilde{G}$ of $G$ is a Lie group in a natural way. It is isomorphic to a product $H\times \mathbb{R}^n$ where $H$ is a compact simply connected group.
H splits isomorphically as a product into pieces (all of which have been classified). The number of such pieces is $t$.
(edits)
*- it's only the low ones, not "all", but one can take better and better approximations to get as many "low" k as one wishes.
**- I mean CLOSED geodesics here
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I love your mouth-watering picture of sucking spaghetti ! Sei di origine italiana ? – Georges Elencwajg Dec 15 2009 at 8:38
This is a beautiful proof. – Saul Glasman Dec 15 2009 at 9:54
While I have a great grandfather from Italy, I'm from the US. All the credit for the proof SHOULD go to Milnor - I just copied his ideas ;-) – Jason DeVito Dec 15 2009 at 13:46
The Omega's aren't displaying correctly. – Noah Snyder Dec 15 2009 at 16:13
Noah: They work for me. – Kevin Lin Dec 15 2009 at 18:46
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The elementary proof that $\pi_1$ is abelian applies more generally to H-spaces (spaces $X$ with a continuous multiplication map $X \times X \to X$ having a 2-sided identity element) without any assumption of finite dimensionality, but infinite-dimensional H-spaces can have nontrivial $\pi_2$, for example $CP^\infty$ (which can be replaced by a homotopy equivalent topological group if one wants, as Milnor showed). Thus finite-dimensionality is essential, so any proof would have to be significantly less elementary than for the $\pi_1$ statement. It is a rather deep theorem of W.Browder (in the 1961 Annals) that $\pi_2$ of a finite-dimensional H-space is trivial.
Hopf's theorem that a finite-dimensional H-space (with finitely-generated homology groups) has the rational homology of a product of odd-dimensional spheres implies that $\pi_2$ is finite, but the argument doesn't work for mod p homology so one can't rule out torsion in $\pi_2$ so easily. It's not true that a simply-connected Lie group is homotopy equivalent to a product of odd-dimensional spheres. For example the mod 2 cohomology ring of Spin(n) is not an exterior algebra when n is sufficiently large. For SU(n) the cohomology ring isn't enough to distinguish it from a product of spheres, but if SU(n) were homotopy equivalent to a product of odd-dimensional spheres this would imply that all odd-dimensional spheres were H-spaces (since a retract of an H-space is an H-space) but this is not true by the Hopf invariant one theorem. There are probably more elementary arguments for this.
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Thanks for correcting my statement! I have ammended my answer below. This explains why I did not remember the proof! I had misremembered the theorem :( – José Figueroa-O'Farrill Dec 16 2009 at 1:34
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This is very fascinating. Is there a known example of a manifold $M$ such that $\pi_2(Diff(M))$ is nontrivial? – Mircea Mar 17 2012 at 7:07
There's a proof that $\pi_2(G)$ is trivial for compact semi-simple Lie groups in section 8.6 of Pressley and Segal's Loop Groups. They say "This proof is in essence the same as Bott's Morse theory proof" but it has some differences in treatment even if the substance is the same. The major difference is that they don't approximate $\Omega G$ by a finite dimensional manifold but by an infinite dimensional one.
In slightly more detail, the idea of the proof is to find a Grassmannian model for $\Omega G$. This is done by considering the action of $G$ on $L^2(S^1;\mathfrak{g}_\mathbb{C})$ and then taking the restricted Grassmannian of this space. Within that, one can identify a sub-Grassmannian that is diffeomorphic to $\Omega G$. They then find a cell decomposition of this Grassmannian and analyse that. One of the important pieces is to consider the subgroup of polynomial loops in $\Omega G$. This corresponds to a certain sub-Grassmannian and it's easy to see that for this Grassmannian then all the cells are of even dimension, whence $\pi_1(\Omega_{\operatorname{pol}} G)$ is trivial. The final step is thus to show that the two Grassmannia (corresponding to $\Omega G$ and $\Omega_{\operatorname{pol}} G$) are homotopy equivalent. Then $\Omega G$ and $\Omega_{\operatorname{pol}} G$ are homotopy equivalent and so $\pi_1(\Omega G)$ is trivial.
Thus $\pi_2(G)$ is trivial since $\pi_2(G) = \pi_1(\Omega G)$ (incidentally, one doesn't need the long exact sequence for fibrations to see that $\pi_k(X) = \pi_{k-1}(\Omega X)$; that's either by definition or by using the adjunction $[\Sigma X, Y] \cong [X, \Omega Y]$).
As I said, Pressley and Segal say that this is in essence the same as Bott's proof; meaning that it proceeds by a cell decomposition based on "energy". However, it treats the infinite dimensional spaces as infinite dimensional spaces so I like it! Also, Grassmannia are more obviously structured so the cell decomposition may be simpler to see and understand in the Grassmannian model for the loop group.
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Here's another proof based on the structure of the flag variety G/T of G. A compact Lie group G has a maximal torus T, and G is a principal T-bundle over the quotient G/T. Borel showed that G/T is a complex manifold, and gave a CW decomposition of it with no odd-dimensional cells. (This is not deep but still astonishing, and the start of a long story; I like the context given by Hirzebruch's eulogy for Borel, available on page 9 here.)
Since pi_2(T) = 0, we have an exact sequence
0 --> pi_2(G) --> pi_2(G/T) --> pi_1(T)
We can conclude immediately that pi_2(G) is torsion-free, since pi_2(G/T) = H_2(G/T) is a free group on the 2-cells in G/T. After Allen's answer (Hopf's theorem) this shows pi_2(G) = 0.
With a little more Lie theory one can show directly that the connecting map pi_2(G/T) --> pi_1(T) is injective. pi_1(T) has a linearly independent subset of simple coroots, and the 2-cells in G/T are indexed by simple roots. The connecting homomorphism matches these up in the natural way, which one can see by considering rank 1 subgroups (subgroups of the form SU(2) or PSU(2)) of G. As a consequence you get a formula for pi_1(G) in terms of roots and coroots.
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This is proved in the book Representations of compact Lie groups by Bröcker and tom Dieck and reviewed here. It is Proposition 7.5 in Chapter V. The proof is for compact, connected Lie groups, but any connected Lie group has the homotopy type of its maximal compact subgroup. (Everything here is for finite-dimensional Lie groups, of course.)
Edit: Maybe I should add, given that two of the other answers mention similar proofs, that the one in this book does not use Morse theory. It uses only basic covering space techniques once it is shown that $\pi_2(G)$ is isomorphic to $\pi_2(G_r)$, where $G_r$ are the regular elements, itself not a difficult lemma.
Edit: The following is wrong! It is only true rationally, which is why I didn't remember having seen a proof of the general case :)
Also if you believe that simply-connected compact Lie groups have the homotopy type of a product of odd spheres, then this follows. It is fairly easy to see that this is the case rationally, but do not remember whether the general statement is hard to prove.
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For the classical Lie groups, I think that an easy way to obtain the result is through the fibrations:
SO(n-1)-->SO(n)--> S^n,
SU(n-1)-->SU(n)--> S^(2n-1)
SP(n-1)-->SP(n)--> S^(4n-1)
and the homotopy long exact sequence and pi_m(S^n) = 0 for m less than n, and pi_2(SO(2)) = pi_2(SU(2)=0 and the isomorphism of SP(2) and SO(5).
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One can also extend this idea to $G_2$ and $F_4$. $G_2$ sits in the sequences $SU(3)\rightarrow G_2\rightarrow S^6$ and $F_4$ sits in the sequence $Spin(9)\rightarrow F_4\rightarrow CaP^2$, where $CaP^2$ is the Cayley plane. I don't know of similar fibrations for $E_6, E_7,$ or $E_8$ though. – Jason DeVito Dec 16 2009 at 2:46
The Cayley plane $\mathbb{C}aP^2$ is given also as a homogeneous space of $E_6$ mod a parabolic subgroup $P$ of $E_6$. This argument can be used to prove that $E_6$ is also connected with the above method. To be more presice, we have $\mathbb{C}aP^2=E_6/U(1)\times Spin(10)$. – math3.14159 Feb 1 2012 at 2:45
In fact, one can reduce the problem to these cases with some big theorems (inc. classification of simple Lie groups). see math.binghamton.edu/somnath/Notes/GSS5.pdf – Ma Ming Apr 27 2012 at 9:27
There is a sketch of the proof using the maximal torus as an exercise in Greub-Halperin-Vanstone's book.
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http://www.physicsforums.com/showthread.php?t=382374
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Physics Forums
## Inversion of this Vandermonde matrix
I was trying to expand a three and more parameter functions similarly to the two-parameter case f(x,y)=(f(x,y)+f(y,x))/2+(f(x,y)-f(y,x))/2.
Anyway, to do the same for more parameters I need to solve
[tex]
\begin{pmatrix}
1 & 1 & 1 & \dotsb & 1\\
1 & \omega & \omega^2 & \dotsb & \omega^{n-1}\\
1 & \omega^2 & \omega^4 & \dotsb & \omega^{2(n-1)}\\
1 & \omega^3 & \omega^6 & \dotsb & \omega^{3(n-1)}\\
\vdots & &&& \vdots \\
1 & \omega^{n-1} & \omega^{2(n-1)} & \dotsb & \omega^{(n-1)(n-1)}
\end{pmatrix}\mathbf{x}=
\begin{pmatrix}
1 \\ 0 \\ 0 \\ 0 \\ \vdots \\ 0
\end{pmatrix}
[/tex]
with $\omega=\exp(2\pi\mathrm{i}/n)$
Is there a closed form expression for x?
EDIT: Oh, silly me. I realized it's a discrete Fourier transform. So is this the correct way to expand then? In the 3 parameter case the solution would be
[tex]
f(x,y,z)=\frac13(f(x,y,z)+f(y,z,x)+f(z,x,y))+\frac13\left(f(x,y,z)+\ome ga f(y,z,x)+\omega^*f(z,x,y)\right)+\frac13\left(f(x,y,z)+\omega^*f(y,z,x) +\omega f(z,x,y)\right)[/tex]
with $\omega=\exp(2\pi\mathrm{i}/3)$?
Now I'm just wondering why I get linear dependent terms when I consider the real part only?
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Recognitions: Science Advisor You're on the right track. It is convenient to write this equation in matrix notation as $$Wx=b$$ where b is the vector you wrote on the RHS. Since W is full-rank and non-singular, it is invertible and the solution to your problem is $$x=W^{-1}b$$ . You need the inverse of W, but this is easy since we know from the properties of the discrete Fourier transform (DFT) that the individual vector columns of W are independent. Each is an orthonormal basis vector of the DFT. In words, this follows because the component of signal at one frequency (cosine or sine at w_m) is independent to that at every other frequency w_n. In fact, $$W^{-1} = W^{\dagger}$$ that is, W is Hermitian (the dagger is the conjugate transpose operation) and unitary (you get the identity matrix in the next equation) $$WW^{\dagger}=I$$ . You can perform this multiplication explicitly it to see that this is so. Accordingly $$x=W^{\dagger}b$$ and you can write out the components of x explicitly.
I just wonder if that way to decompose a multiparameter function makes sense. It's nicely symmetrical and generalizes to higher dimensions. However it introduces complex number where the initial function might actually be real only. And also I haven't included odd parity permutations of the function arguments...
Recognitions:
Science Advisor
## Inversion of this Vandermonde matrix
Sorry, I'm not following your comments. I provided the closed solution to Wx=b, where W is complex, which was the question asked in your first post. Are you looking for something else?
I did that exercise to find a way to extend the rule f(x,y)=(f(x,y)+f(y,x))/2+(f(x,y)-f(y,x))/2 to higher dimensions (I didn't explain the connection; just mentioned it in the intro). I assumed some cyclic symmetry for the final form of f(x,y,z)=... and with the help of the discrete Fourier transform (which I didnt recognise at first), I can find some "decomposition". Now I wasn't sure if the decomposition is useful this way.
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http://mathoverflow.net/questions/102586?sort=oldest
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## How else can we describe the volume of a lagrangian submanifold in a Kahler manifold?
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Suppose $(V^{2g}, g, \omega, J)$ is an almost Kahler manifold. ie. $(V,\omega)$ is a symplectic manifold with $\omega$-compatible almost complex structure $J$ ($J$ is a symplectomorphism) and such that $\omega(\cdot, J\cdot)$ coincides with a riemannian metric $g$ on $V$.
Suppose further that $L$ is a lagrangian submanifold of $V$. Then $L$ has a volume, as defined with respect to the riemannian metric $g$.
Nonetheless I want to believe that there still is an alternative expression of the metric quantity $vol(L,g)$ which exploits both the (almost) Kahler structure and the fact that the submanifold $L$ is lagrangian (ie. totally isotropic) and that we have a canonical decompostion $TM|L=TL \oplus T JL$, ie. the normal bundle of $L$ is precisely $JL$.
Can anybody testify or provide some evidence or enlightenment on this matter?
Alternatively, an even simpler question: if we suppose $V$ is closed, then can we give an expression to the volume of $V$ (wrt g) in terms of $vol(L,g)$ which depends essentially on the circumstances of $L$ being lagrangian and $g=\omega(\cdot, J\cdot)$?
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Your definition of a Kahler manifold should also include the assumption that $J$ is integrable. What you defined is usually called an "almost Kahler" structure. – YangMills Jul 18 at 21:49
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Also, $\omega^g$ is a $2g$-form so it cannot yield a volume form on any $g$-dimensional submanifold, Lagrangian or not. – YangMills Jul 18 at 21:51
sure, forgot to add integrable. And of course $\omega^g$ is not a volume form on any $g$-dimensional submanifold. I was hasty and should have given $g/2$ (when $g$ even) to emphasize just that $\omega$ vanishes on $L$, so it ''appears'' to give to no data on $J$. I'll correct this typos. – J. Martel Jul 18 at 23:03
## 2 Answers
This is probably closer in spirit to what you're looking for than what you've received in the comments. If $(V^{2m}, J, \omega, g)$ is Calabi-Yau (which for me means that $J$ is integrable, and the first Chern class $c_1(V) = 0$), then one can say much more. In this case there exists a holomorphic nowhere vanishing $(m,0)$ form $\Omega$, called the "holomorphic volume form." The form $\Omega$ is unique up to multiplication by a nowhere vanishing holomorphic function. We don't need to assume that $\Omega$ is parallel, but Yau's theorem does tell us (if $V$ is compact) that we can change the metric, keeping the Kaehler class $[\omega]$ unchanged, to make $\Omega$ parallel (and consequently also the new metric will be Ricci-flat.) But I have digressed.
In such a situation, if $L$ is Lagrangian, then it is well known that the restriction of $\Omega$ to $L$ is equal to $e^{i \theta} \mathrm{Vol_L}$, where $\mathrm{Vol}_L$ is the volume form of $L$ (with the induced metric) and $e^{i \theta}$ is the "phase" of the Lagrangian, where $\theta : L \to \mathbb R/ (2 \pi \mathbb Z)$ is a smooth, multivalued function on $L$. In addition, the mean curvature $H$ of $L$ in $V$ is given by $H = J \nabla \theta$. So the minimal Lagrangian submanifolds (vanishing mean curvature) correspond to those with constant phase function $\theta = \theta_0$. In this case, if $L$ is compact, then the volume of $L$ is given by
$\mathrm{Vol} (L) = \int_L \mathrm{Vol}_L = \int_L e^{-i \theta_0} \Omega = e^{- i \theta_0} [\Omega] \cdot [L],$
which is topological. (It looks complex, but it's actually real, because $[\Omega]$ is a class in $H^g(V, \mathbb C)$.) If you prefer, you can just replace $\Omega$ by $e^{- i \theta_0} \Omega$ to get rid of the phase factor.
Such "minimal Lagrangian" submanifolds, whose volume is purely topological, are also called special Lagrangian submanifolds, and are widely studied in calibrated geometry and differential geometric approaches to mirror symmetry. The best place to start looking is the text "Riemannian Holonomy Groups and Calibrated Geometry" by Dominic Joyce and its multiple references.
I'm not sure how much of this will extend to the case of $J$ non-integrable and $c_1(V) \neq 0$. I'd have to think about it.
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I appreciate your answer very much Spiro. This formula for the volume is impressive, although i do not really know what it means. In my own case, I am looking at the volumes of lagrangians which are rational inside a given symplectic lattice. More precisely, let's fix the standard Kahler structure $(\mathbb{R}^{2g}, \omega, J)$. We say $\Lambda$ is a \emph{symplectic lattice} if $\Lambda=A\mathbb{Z}^{2g}$ for $A \in Sp{2g}\mathbb{R}$. Notice that $\omega$ will be integral and unimodular on $\Lambda$. Let's say a lagrangian $\ell$ in $\mathbb{R}^{2g}$ is $\Lambda$-\emph{rational} if – J. Martel Aug 23 at 4:20
$\ell \cap \Lambda$ is a cocompact lattice in $\ell$. In this sense we have a volume $vol(\ell, \Lambda):=vol(\ell / \ell \cap \Lambda)$. Hence the volume of $\ell$ in $\Lambda$ is the volume of the compact embedded lagrangian $\ell / \ell \cap \Lambda$ in the torus $\mathbb{R}^{2g} / \Lambda$ equipped with its flat Kahler structure. Now we have a problem here, right? Since my assumptions are exactly that this torus is actually a principally polarized abelian variety. Now whatever... – J. Martel Aug 23 at 4:27
that means', it at least says that the first Chern class of the torus is nonzero (since $\omega$ is nontrivial on the torus). But we do have of course $c_1 \mathbb{R}^{2g}=0$, and i am computing volumes of lagrangian paralellipipeds. Now at this point i should like to know precisely what' $\Omega$ is, what are the phases $\theta$ of these lagrangian parallelipipeds, and what is the computation that yields $[\Omega] \cdot [L]$? What confuses me is why the formula does not yield $Vol(L)=0$ for all lagrangian submanifolds in $\mathbb{R}^{2g}$? I'm admitting my own ignorance – J. Martel Aug 23 at 4:32
because i truly do not understand why $[L] \neq 0$. ANyway, i need to think more on this. I'll look at your reference. Thank you again for your answer, it is tremendously motivating. – J. Martel Aug 23 at 4:33
I'll try to clear up some of your confusions, but I am also confused by some of your comments. First of all, $c_1$ of a torus is zero. (I'm talking about the first Chern class (with complex or real coefficients) of its tangent bundle. I don't know anything about principally polarized abelian varieties. You might be saying that $c_1$ has torsion? I am pretty sure that doesn't matter here. On $\mathbb R^{2g} \cong \mathbb C^g$, the form $\Omega$ is just $dz^1 \wedge \cdots \wedge dz^g$. This descends to the quotient $\mathbb C^g / \Lambda$. – Spiro Karigiannis Aug 23 at 13:27
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Take an open rectangle in $\mathbb R^2$ with side lengths $a$ and $b$. This is a Riemannian manifold, thus symplectic, and has an almost complex structure. The center line is a Lagrangian submanifold and has volume $a$.
Two versions of this whole ensemble are symplectomorphic if $a_1b_1=a_2b_2$. So the volume of the center line is not invariant under symplectomorphisms.
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Ok, so what? This is no answer to my question. If we fix a Kahler structure on $(V,g)$, then every lagrangian submanifold has a well-defined volume (with respect to the Kahler structure). Your rectangle has varying Kahler structure, so of course lagrangians have different length. – J. Martel Jul 18 at 23:17
Rereading my question, I see that the point of your example is that one cannot expect volume to be a measure of only $\omega$, ie. at some point we have to refer to the metric $\omega(\cdot, J\cdot)$. – J. Martel Jul 18 at 23:21
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Do you, after rereading, feel that there is a question still to be answered? (If it's the second question, then could you please clarify what you are looking for? You seem to request, if $vol(L,g)$ is not invariant under symplectomorphisms, an alternate measure which is not invariant under symplectomorphisms.) – Will Sawin Jul 18 at 23:26
I just want to know $vol(L,g)$ in another way. There may be possible another expression of $vol(L,g)$ which depends essentially on the circumstances of $L$ lagrangian, and $g=\omega(\cdot, J\cdot)$. OR does a Kahler structure on a symplectic manifold yield a different functional on lagrangians? – J. Martel Jul 18 at 23:32
2
If I'm given, say, a region in $\mathbb R^n$ I want to compute the volume of, I can give you a Lagrangian submanfiold of $\mathbb R^{2n}$ with the same volume. So if you have some easy way to compute the volumes of Lagrangian submanifolds, then I have an easy way to compute the volume of everything else. – Will Sawin Jul 19 at 13:48
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http://unapologetic.wordpress.com/2009/02/02/
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# The Unapologetic Mathematician
## Upper-Triangular Matrices
Until further notice, I’ll be assuming that the base field $\mathbb{F}$ is algebraically closed, like the complex numbers $\mathbb{C}$.
What does this assumption buy us? It says that the characteristic polynomial of a linear transformation $T$ is — like any polynomial over an algebraically closed field — guaranteed to have a root. Thus any linear transformation $T$ has an eigenvalue $\lambda_1$, as well as a corresponding eigenvector $e_1$ satisfying
$T(e_1)=\lambda_1e_1$
So let’s pick an eigenvector $e_1$ and take the subspace $\mathbb{F}e_1\subseteq V$ it spans. We can take the quotient space $V/\mathbb{F}e_1$ and restrict $T$ to act on it. Why? Because if we take two representatives $v,w\in V$ of the same vector in the quotient space, then $w=v+ce_1$. Then we find
$T(w)=T(v+ce_1)=T(v)+cT(e_1)=T(v)+c\lambda_1e_1$
which represents the same vector as $T(v)$.
Now the restriction of $T$ to $V/\mathbb{F}e_1$ is another linear endomorphism over an algebraically closed field, so its characteristic polynomial must have a root, and it must have an eigenvalue $\lambda_2$ with associated eigenvector $e_2$. But let’s be careful. Does this mean that $e_2$ is an eigenvector of $T$? Not quite. All we know is that
$T(e_2)=\lambda_2e_2+c_{1,2}e_1$
since vectors in the quotient space are only defined up to multiples of $e_1$.
We can proceed like this, pulling off one vector $e_i$ after another. Each time we find
$T(e_i)=\lambda_ie_i+c_{i-1,i}e_{i-1}+c_{i-2,i}e_{i-2}+...+c_{1,i}e_1$
The image of $e_i$ in the $i$th quotient space is a constant times $e_i$ itself, plus a linear combination of the earlier vectors. Further, each vector is linearly independent of the ones that came before, since if it weren’t, then it would be the zero vector in its quotient space. This procedure only grinds to a halt when the number of vectors equals the dimension of $V$, for then the quotient space is trivial, and the linearly independent collection $\{e_i\}$ spans $V$. That is, we’ve come up with a basis.
So, what does $T$ look like in this basis? Look at the expansion above. We can set $t_i^j=c_{i,j}$ for all $i<j$. When $i=j$ we set $t_i^i=\lambda_i$. And in the remaining cases, where $i^gt;j$, we set $t_i^j=0$. That is, the matrix looks like
$\displaystyle\begin{pmatrix}\lambda_1&&*\\&\ddots&\\{0}&&\lambda_d\end{pmatrix}$
Where the star above the diagonal indicates unknown matrix entries, and the zero below the diagonal indicates that all the entries in that region are zero. We call such a matrix “upper-triangular”, since the only nonzero entries in the matrix are on or above the diagonal. What we’ve shown here is that over an algebraically-closed field, any linear transformation has a basis with respect to which the matrix of the transformation is upper-triangular. This is an important first step towards classifying these transformations.
Posted by John Armstrong | Algebra, Linear Algebra | 19 Comments
## 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://mathhelpforum.com/advanced-algebra/142738-quotient-groups.html
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# Thread:
1. ## Quotient Groups
Let H be a normal subgroup of finite group G. If the order of the quotient groups $G/H$ is m, prove that $g^m$ is in H for all $g \in G$.
So since H is normal, there are no distinctions between left and right cosets in G.
G is a finite group. Let o(G) = n. The o(G/H) = m.
Don't know where to go after this...
2. Hint: Given any group $G$ of order $n$ and any $g \in G$, $g^n = e$.
In the case of $G/H$, what does this tell you about the coset $g^mH$?
3. Originally Posted by spoon737
Hint: Given any group $G$ of order $n$ and any $g \in G$, $g^n = e$.
In the case of $G/H$, what does this tell you about the coset $g^mH$?
Since $G/H$ is of finite order m, does that imply that $g^mH = eH = H$?
But if $g^mH = H$ then $g^m$ must $\in H$ by theorem (If H is any subgroup of a group G, then hH = H = Hh, $h \in H$).
4. Originally Posted by MissMousey
Since $G/H$ is of finite order m, does that imply that $g^mH = eH = H$?
But if $g^mH = H$ then $g^m$ must $\in H$ by theorem (If H is any subgroup of a group G, then hH = H = Hh, $h \in H$).
Yes, that is correct. $\left|G/H\right|=[G:H]$ and so $\left(gH\right)^{[G:H]}=g^{[G:H]}H=\underbrace{H}_{\text{id elem.}}$ and what you said finishes it.
5. Originally Posted by Drexel28
Yes, that is correct. $\left|G/H\right|=[G:H]$ and so $\left(gH\right)^{[G:H]}=g^{[G:H]}H=\underbrace{H}_{\text{id elem.}}$ and what you said finishes it.
*dances* Thank you!
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http://mathhelpforum.com/differential-geometry/125480-solved-complex-polynomial-division.html
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# Thread:
1. ## [SOLVED] Complex Polynomial Division
However, this is wrong. It should be $z^2+zi-1$. I'm not sure how one would arrive here, however.
2. Wait, nevermind. I see it now. Stupid mistake. Sorry.
3. Originally Posted by davismj
However, this is wrong. It should be $z^2+zi-1$. I'm not sure how one would arrive here, however.
Try using the Difference of Two Cubes Rule:
$z^3 + i = z^3 - (-i)$
$= z^3 - i^3$
$= (z - i)(z^2 + iz + i^2)$
$= (z - i)(z^2 + iz - 1)$.
So $\frac{z^3 + i}{z - i} = \frac{(z - i)(z^2 + iz - 1)}{z - i}$
$= z^2 + iz - 1$.
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http://mathhelpforum.com/discrete-math/134356-logically-equivalent-print.html
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# logically equivalent
Printable View
• March 17th 2010, 06:15 PM
questionboy
logically equivalent
can anyone explain to me how to prove that p → q and not q → not p are logically equivalent without truth tables
• March 17th 2010, 08:19 PM
Soroban
Hello, questionboy!
Quote:
Prove that $p\to q$ and $\sim\!q \to \;\sim\!p$ are logically equivalent without truth tables
$p \to q \;\equiv\;\sim\!p \vee q$ . . I call it "Alternate Definition of Implication" (ADI).
. . $\begin{array}{ccccc}<br /> \sim\!q \to \:\sim\!p & \equiv & q \:\vee \sim\!p & &\text{ADI} \\ \\<br /> & \equiv & \sim\!p \vee q && \text{Comm.} \\ \\<br /> & \equiv & p \to q && \text{ADI}<br /> \end{array}$
• March 18th 2010, 12:32 AM
Seppel
Hello!
Quote:
Originally Posted by Soroban
Hello, questionboy!
$p \to q \;\equiv\;\sim\!p \vee q$ . . I call it "Alternate Definition of Implication" (ADI).
. . $\begin{array}{ccccc}<br /> \sim\!q \to \:\sim\!p & \equiv & q \:\vee \sim\!p & &\text{ADI} \\ \\<br /> & \equiv & \sim\!p \vee q && \text{Comm.} \\ \\<br /> & \equiv & p \to q && \text{ADI}<br /> \end{array}$
You should also mention the use of Double Negation.
Best wishes,
Seppel
• March 18th 2010, 05:55 AM
emakarov
This is fine, but expressing $p\to q$ as $\neg p\lor q$ misses an important subtlety. One can define $\neg p$ as $p\to\bot$ where $\bot$ denotes a contradiction (e.g., 0 = 1). Indeed, $p$ and $\neg p$ imply $\bot$ by Modus Ponens, and deriving $\bot$ from $p$ proves $\neg p$ by Deduction theorem (or implication introduction).
Then $\neg q\to\neg p$ becomes $(q\to\bot)\to(p\to\bot)$. Now, it is surprising that this formula is implied by $p\to q$ without using the fact that $\bot$ is a contradiction. I.e.,
$(p\to q)\to\Big((q\to r)\to(p\to r)\Big)$
is the transitivity of $\to$, and it is derivable for any $r$ using only basic rules about implication.
This suggests that $p\to q$ implies $\neg q\to\neg p$ in a much stronger sense than just in Boolean logic.
All times are GMT -8. The time now is 01:23 AM.
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http://mathhelpforum.com/calculus/26308-minimums.html
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# Thread:
1. ## minimums
HELP !!! ><><
A printed page of a book is to have side margins of 1cm, a top margin of 2 cm and a bottom margin of 3 cm. IT is to contain 200cm^2 of printed matter, Find the dimensions of the page if the area of the paper used is to be a minimum
2. Originally Posted by chibiusagi
HELP !!! ><><
A printed page of a book is to have side margins of 1cm, a top margin of 2 cm and a bottom margin of 3 cm. IT is to contain 200cm^2 of printed matter, Find the dimensions of the page if the area of the paper used is to be a minimum
let the width of the page be $x$, let the height of the page be $y$.
then, the width of the printed area is $(x - 2)$ ....why is that?
and the length of the printed area is $(y - 5)$ .....why is that?
since the area of the printed material is 200 $\mbox{cm}^3$
we have $(x - 2)(y - 5) = 200$ .........this is our constraint equation. we can use this to solve for one variable in terms of the other. for instance, solving for x we get: $x = \frac {200}{y - 5} + 2$
now, the area of the page is given by:
$A = xy$
$\Rightarrow A = \left( \frac {200}{y - 5} + 2 \right)y$ ...........this is what we want to minimize
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http://math.stackexchange.com/questions/tagged/approximation+functions
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# Tagged Questions
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http://physics.stackexchange.com/questions/35605/dynamical-operator-and-sun1
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# dynamical operator and $SU(n+1)$
I want to know precisely by example
1. what is dynamical operator?
2. what is the relationship between dynamical operators and the $SU(n+1)$
3. How to show all the eigen states of a dynamical operator form a complete set of basis vectors of a Hilbert Space?(Any quantum state of a quantum system can be regarded as a vector in the Hilbert space)
-
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http://physics.stackexchange.com/questions/tagged/relativity
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http://mathoverflow.net/questions/8684/homotopy-pullbacks-and-homotopy-pushouts
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## Homotopy pullbacks and homotopy pushouts
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I have a good grasp of ordinary pullbacks and pushouts; in particular, there are many categorical constructions that can be seen as special cases: e.g., equalizers/coequalizers, kernerls/cokernels, binary products/coproducts, preimages,...
I know the (a?) definition of homotopy pullbacks/pushouts, but I am lacking two things: examples and intuition. So here are my questions:
1. What are the canonical examples of homotopy pullbacks/pushouts? E.g., in the category of pointed topological spaces the loop space $\Omega X$ is a homotopy pullback of the map $\ast \to X$ along itself.
2. How should I think about homotopy pullbacks/pushouts? What is the intuition behind the concept?
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Pullbacks and pushouts should just be the homotopy limits and colimits of an appropriately categorified span (cospan?) – Harry Gindi Dec 12 2009 at 16:23
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I don't know whether this is helpful or confusing, but every ordinary (co)limit is also a homotopy (co)limit! (It is a homotopy (co)limit in the topological category formed by taking the ordinary category and giving its Hom sets the discrete topology. Note that this is not how we normally make Top into a topological category.) – Reid Barton Dec 12 2009 at 17:58
## 4 Answers
You can think of the pushout of two maps f : A → B, g : A → C in Set as computing the disjoint union of B and C with an identification f(a) = g(a) for each element a of A. We could imagine forming this as either the quotient by an equivalence relation, or by gluing in a segment joining f(a) to g(a) for each a, and taking π0 of the resulting space. If two elements a, a' of A satisfy f(a) = f(a') and g(a) = g(a'), the pushout is unaffected by removing a' from A. The homotopy pushout is formed by gluing in a segment joining f(a) to g(a) for each a and not forgetting the number of ways in which two elements of B ∐ C are identified; instead we take the entire space as the result. It is the "derived" version of the pushout.
In general you can think of the homotopy pushout of A → B, A → C as the "free" thing generated by B and C with "relations" coming from A. But it's important that the "relations" are imposed exactly once, since in the homotopical/derived setting we keep track of such things (and have "relations between relations" etc.)
Another, possibly more familiar example: In a derived category, the mapping cone of a morphism f : A → B is the homotopy pushout of f and the zero map A → 0. This certainly depends on A, even when B is the zero object: it is the suspension of A.
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Very nice! It reminds me of the introduction to Lurie's thesis, where he talks about Bezout's theorem. Do you have a similar picture for the pullback? – Alberto García-Raboso Dec 12 2009 at 17:25
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Well, I more or less repeated it from the beginning of one of his talks :) As for homotopy pullbacks, of course they are formally dual, but maybe more helpful is this: The ordinary fiber product of X and Y over Z is the set of points of X x Y which have the same image in Z. In homotopy theory, we cannot talk about equality of points in a space--or rather, we have an entire space of "ways in which points may be considered equal", namely, the space of paths in Z between the two points. If you write out the common definition for homotopy pullbacks in spaces you'll see this is what it computes. – Reid Barton Dec 12 2009 at 17:36
So I see why the homotopy pushout is a derived version of the ordinary pushout: the latter is π_0 of the former (at least in your example). I fail to see the corresponding statement for the homotopy pullback: it seems to me that the ordinary pullback is a subspace of the homotopy pullback. Is this correct? – Alberto García-Raboso Dec 12 2009 at 18:01
5
That is true with the usual formula for the homotopy pullback. The properly analogous statement would be for pullbacks of sets, but in that case the two notions of pullback agree. A more typical example is comparing the pullback of vector spaces A -> C <- B to the homotopy pullback in unbounded chain complexes. The homotopy pullback will be the ordinary pullback in degree 0, and in (homological) degree -1 it should be something like the cokernel of the map A+B -> C. Generally homotopy limits have "extra stuff in negative degrees", which we can't see in topological spaces. – Reid Barton Dec 12 2009 at 18:22
I see a DG-scheme-type of thing going on... very interesting. Thanks for all your wonderful (and swift) explanations! – Alberto García-Raboso Dec 12 2009 at 18:42
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Here's a simple geometric example for a homotopy pushout. This is stolen from the Dwyer-Spalinski paper on model categories.
We first look at the following diagram: pt <-- S^1 --> pt. The pushout of this diamgram is just a point.
Now look at D^2 <-- S^1 --> D^2 where the maps are the inclusion at the boundary. The pushout of this diagram is S^2.
What one notices is that the point and D^2 have the same homotopy type, but the pushout of the two diagrams have different homotopy types! So ordinary pushouts don't go well with homotopy theory.
The right thing to do is to always replace the maps by cofibrations before computing pushouts. This is precisely what the homotopy pushout does.
Replacing maps by cofibrations before computing pushouts is not just philosophy but actually a theorem. If you want to do things correctly you should put a model category strutcure on the diagram category such that the constant functor preserves fibrations and trivial fibrations and then take the derived colimit functor. Since you have a model category structure on the diagram category you replace diagrams cofibrantly! If you work out all the details you get replacing a diagram means exactly to replace the maps by cofibrations before taking the pushout.
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For (2): they are the needed modification of the non-homotpy ones when you want the result to change only by an homotopy equivalence ff you change the input to the construction by homotopy equivalences.
Later: My intuition comes from the following image: say you have a space $X$ on which $\mathbb Z$ acts, and you want to take the quotient. I look at the homotopy version as the result of tacking the orbits: you attach a thread (a copy of $\mathbb R$) to each orbit... If later you want to take the "real" quotient, you just need to pull the threads, and the orbits contract to a point (I know it makes a funny noise when you do this!). When you take homotopy quotients by another group $G$, you need to get yourself a "$G$-shaped thread", which is what we write $EG$.
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Yet that does not quite give me an intuition on what they are, only on why they are necessary in some contexts. – Alberto García-Raboso Dec 12 2009 at 16:22
OK, I see the picture, but what homotopy pullback/pushout is that? – Alberto García-Raboso Dec 12 2009 at 16:44
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Usually, it isn't one: it's the "homotopy orbits" which is the homotopy colimit over a diagram with one object with automorphism group G. When G = Z, it happens to be the homotopy pushout of the diagram X <- X -> X where the left map is the identity and the right map is the action of a generator of Z. – Reid Barton Dec 12 2009 at 16:59
Oops, that last sentence above is certainly false. Exercise: fix it! – Reid Barton Dec 13 2009 at 0:20
A list of definitions, general formulas and types of examples is at nLab: homotopy limit.
In particular there is a whole subsection on homotopy pullbacks.
Closely related material (further examples, if you wish) is at fibration sequence.
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Thanks for the links. I'll be sure to check them out. – Alberto García-Raboso Dec 15 2009 at 19:35
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http://www.physicsforums.com/showthread.php?p=3856432
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Physics Forums
## number sequence 15, 101, 8, 86, 9699 ...
I've been struggling with this number sequence for some time now, and i can't find the pattern, can anyone help?
the sequence is: 0, 15, 101, 8, 86, 9699, 6008, ... what comes after?
any thoughts?
I used linear regression to produce$$x_n = \frac{ -2009823672425 x_{n-1}^2 + 1317576645654994 x_{n-2}^2 + 19713318671536597 x_{n-1} - 131718609410997380 x_{n-2} - 290871998770928574} {43322484909156}$$though, even if it is a perfect fit, it's probably not what the puzzle designer had in mind, as the next value is$$\frac {10226107345860276146148} {3610207075763}$$and the rest plunge down the negative.
There are an infinite number of possible numbers that could be the next term in that sequence. Problems like this are always silly because you have to guess what the person writing the problem intended.
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http://physics.stackexchange.com/questions/30613/what-is-spontaneous-symmetry-breaking-in-quantum-gauge-systems/32575
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# What is spontaneous symmetry breaking in QUANTUM GAUGE systems?
Wen's question What is spontaneous symmetry breaking in QUANTUM systems? is cute, but here's an even cuter question. What is spontaneous symmetry breaking in QUANTUM GAUGE systems?
There are some gauge models where the Higgs phase and the confinement phase are continuously connected, you know. Why are the Higgs phase and the confinement phase identical in Yang-Mills-Higgs systems? You mention cluster decomposition? Sorry, because bilinear operators have to be connected by a Wilson line.
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## 4 Answers
Spontaneous breaking of a gauge symmetry is breaking the global symmetry. All gauge symmetries are not symmetries in the local part, if you do a local gauge transformation, you have the same exact state, But the global part of the gauge symmetry is physical, and leads to a Noether current. when you break the gauge symmetry, you are breaking the global part.
For example, in quantum electrodynamics, if you perform a phase rotation of the electron field by a constant factor $e^{i\theta}$, you do nothing to the vector potential. This transformation produces a nontrivial Noether current, it is the electric current. You can break this global symmetry with no problem, just using a charged condensate. This is the Higgs mechanism for U(1), or superconductivity.
A completely local version would be to do a local phase rotation by $\theta(x)$ where $\theta$ is a function which is nonzero only in a compact region. This transformation adds a gradient to A, and now if you do Noether's prescription, you find a conserved charge which is identically zero.
For gauge theories, the Noether currents derived from gauge tranformations are of the kind that they reduce to boundary integrals. In any theory, Noether currents are such that the charge can be evaluated at the initial time surface and the final surface, and you get the same answer. In a gauge theory, these charges can be evaluated on a large sphere on the initial surface and the final surface, using only the gauge field. In electrodynamics, this is Gauss's law. The Noether charge becomes a boundary object in a gauge theory, which is determined by the asymptotic values of the fields.
This is discussed to some extent on the Wikipedia page on infraparticle.
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Since gauge symmetry is not a symmetry in quantum theory, there is no spontaneous symmetry breaking of QUANTUM GAUGE symmetry.
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You can still break the global part. The gauge transformations on the boundary are physical, although if you only do a gauge transformation on the interior, it's unphysical, just a redundancy. – Ron Maimon Jun 23 '12 at 8:17
@Ron: The gauge symmetry is not a symmetry in quantum theory since both the local and the global gauge transformations are do-nothing transformations when act on quantum states. The quantum states simply do not change under those transformations, regardless if the system has boundary or not. So even the global part on the boundary is a do-nothing transformation and cannot break. – Xiao-Gang Wen Jun 23 '12 at 12:48
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@Xiao-Gan-Wen: This is true of the local symmetry, but not the global symmetry. You can see this because the Noether prescription for a global phase rotation in a U(1) symmetric theory still gives a nontrivial conserved current. This is true of energy momentum in gravity, of supercurrents in supergravity, of every gauge current. Further, the global phase rotations are physically inequivalent--- you can translate a system relative to the boundaries, even if translations are gauged. – Ron Maimon Jun 23 '12 at 20:00
I absolutely agree with @RonMaimon. The true symmetry group is the global group that is equivalent to the gauge group $G$ modulo $G_*$, where $G_*$ is the redundancy group, that is, the gauge group whose elements go to the identity in the boundaries. – drake Jul 26 '12 at 23:58
@drake: In my language for the situation that you described, $G_*$ is the gauge symmetry which cannot break. $G/G_*$ is the physical symmetry which can break. – Xiao-Gang Wen May 7 at 11:53
It's all about RELATIVE symmetry breaking in QUANTUM GAUGE theories. RELATIVE to the Higgs field direction, symmetry is broken. This is a novel concept of RELATIVE symmetry breaking. A spin-0 helium atom quantum state might not break rotational symmetry, but RELATIVE to the direction of one of the electrons relative to the nucleus, the rotational symmetry for the other electron is RELATIVELY broken.
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It's only RELATIVE to the boundary value of the Higgs field. You can still arbitrarily phase rotate the Higgs direction in the interior. – Ron Maimon Jul 23 '12 at 7:50
Gauge symmetries are redundancies of the description, not a part of physics; gauge fields have surplus structures (e.g., non-physical polarizations) one brings in to describe the system more conveniently (for instance in a local and manifestly Lorentz covariant form). You can describe a gauge system in a language that does not have gauge symmetry at all. A famous example is AdS/CFT: you can describe an SU(N) gauge theory at large N by a theory of gravity; there is no SU(N) on the gravity side. Note that the global part of a gauge symmetry is as non-physical as its local part. It is however the case that ordinary gauge field theories (like QCD; not like GR) do have, aside from the non-physical gauge symmetry, a physical global symmetry which yields charge conservation. This physical global symmetry looks exactly like the global part of the (non-physical) gauge transformation, but one should differentiate between the two: gauge transformations are just like changing coordinates, but the physical global transformation involves changing the dynamical degrees of freedom. (In a similar situation in GR one should differentiate between the non-physical diff invariance and physical symmetries due to the existence of Killing vectors. The global part of the diff group is not only poorly defined, it is as non-physical as its local part. It is the isometries of the spacetime that are of physical significance. For example the existence of a conserved energy is due to the existence of a time-like Killing vector, not the meaningless "global part of the diff group".) (The case of non-abelian gauge theories with topological sectors has one further subtlety: your physical degrees of freedom are the equivalence classes of gauge field configurations that are continuously deformable to each other within each class.)
Gauge symmetry being non-physical, gauge symmetry breaking is also not a matter of physics, but a matter of changing the description. In the abelian Higgs model (which resembles superconductivity) for instance, if one changes their description from A) 2 components of scalar Higgs + 2 polarizations of the photon, to B) 1 component of Higgs + (1 component of Higgs + 2 components of the photon), one is breaking (or sacrificing) the symmetry in the initial description, to more clearly see the mass spectrum of the theory.
However, the "Higgs mechanism" is not non-physical as the above paragraph might initially suggest. It is physical in the following sense. Abelian gauge theories have three well-known phases: Landau phase, Coulomb phase, and the Higgs phase. Massless electrodynamics is an example that is always in the Landau phase (with logarithmic charge fall-off). Massive electrodynamics has a Landau regime for distances less than $m_{e}^{-1}$ and a Coulomb regime (with $1/r$ potential) for large distances. The abelian Higgs model has a Coulomb phase (with $m_{v}=0$) if the Higgs potential is minimum at the origin, and a Higgs phase (with $m_{v}\neq 0$) if the Higgs potential is minimum away from the origin. It is due to this dynamical phase transition that the change of description comes useful to understand the mass spectrum. (Look at John Preskill's note http://www.theory.caltech.edu/~preskill/ph230/notes2000/230Lectures27-29-Page347-402.pdf)
The important point to remember is: In the phenomena commonly referred to as spontaneous breaking of gauge symmetries, no physical (real) symmetry breaks. Physically, no global or local symmetry gets broken, only a phase transition happens that one can track by the mass of the vector boson as the order parameter. In the previous sentence, by local symmetry I mean some physical local symmetry (as the word "Physically" at the beginning of the sentence shows), like conformal symmetry in 2D; I'm not even thinking about non-physical symmetries like gauge symmetry, because they do not have anything to do with the nature.
I have limited the discussion to the abelian Higgs model, but the essence of the argument is the same for non-abelian gauge fields as well. If this answer has not been illuminating, I recommend reading the aforementioned notes by John Preskill.
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http://www.physicsforums.com/showthread.php?p=4170568
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Physics Forums
## scalar product square matrix hermitian adjoint proof
1. The problem statement, all variables and given/known data
If M is a square matrix, prove:
(A, MB) = (adj(M)A, B)
where (A, MB) denotes the scalar product of the matrices and adj() is the adjoint (hermitian adjoint, transpose of complex conjugate, M-dagger, whatever you want to call it!)
2. Relevant equations
adj(M)=M(transpose of the complex conjugate)
adj( ) = adjoint
scalar product : A dot B = (A, B) = adj(A)B
I'm supposed to first write (A, MB) = (MB,A)^(complex conjugate) and apply the definitions of the scalar product above and adj(M)...and am warned to take complex conjugates carefully
3. The attempt at a solution
My attempted work is this:...i didnt follow the hint above because i just applied the inner product definition as in the picture and got both sides to be equivalent
[IMG][/IMG]
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Recognitions: Gold Member Science Advisor Staff Emeritus Are you given that these matrices are all 3 by 3? Because all you prove this for.
It doesnt say, I listed everything about the problem. I'm still not sure what to do
## scalar product square matrix hermitian adjoint proof
Have you learned that $(a,Mb)=(Mb)^\top a$ and that $(Mb)^\top = b^\top M^\top$?
Of course the formula is the same for conjugate transpose as well (ie you can replace the T by *).
No I haven't learned that yet. What I have to do has something to do with the hint in OP but I can't make sense of
this is as far as i got using the hint... [IMG] free photo hosting[/IMG] M is supposed to be a square matrix. The hint says after I applied the first step and the definitions of the scalar product and M(adjoint), I need to take complex conjugates? Where did I go wrong, or what do I have to do next in this proof? Many thanks
Wait a second, here. I think you actually do know those formulas I gave you before, but the notation you are using doesn't seem to be what the standard notation is. Usually T means transpose, * means Hermitian adjoint (or conjugate transpose, both mean the same) and a bar means complex conjugate. Also, I believe that your formula is incorrect for $(A,B)$. It should be $B^*A$ (in "standard notation") or $B^\top A$ (in your notation.) Whose notation is this, btw? Is it yours? Your prof's? Your book's?
its my books notation as seen here. [IMG] upload pics[/IMG] The difference in notation arises from the fact my book uses column matrix representation for vectors instead of row matrix representation , you think?
I made a mistake in my work picture above...quick correction here. Still do not know what to do next [IMG] photo sharing websites[/IMG]
So, using your notation, $M^\dagger A,B) = (M^\dagger A)^\dagger B = A^\dagger MB$. Now do you see what to do? Of course, this didn't really use the hint, but it looks like it works. Again, just out of curiosity, what is the title of this book? Is it a Mathematical Physics book? I am interested because I have never seen the scalar product defined that way in a math book (that's not to say it hasn't happened, I just haven't seen it.)
Yea it's a "tutorial" math methods in physics book written by a professor for us, it's broken into 2 parts (semesters) and I also have Mary boas math methods book but couldn't find any help in there for this problem. Many thanks for the help !!! So what you just posted is basically the proof I did in the very first picture, right ? Would that very first picture I posted consist of the whole proof, like what you just hinted at, which seems correct ?
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http://unapologetic.wordpress.com/2010/07/06/the-radon-nikodym-theorem-statement/?like=1&source=post_flair&_wpnonce=5092e672ce
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# The Unapologetic Mathematician
## The Radon-Nikodym Theorem (Statement)
Before the main business, a preliminary lemma: if $\mu$ and $\nu$ are totally finite measures so that $\nu$ is absolutely continuous with respect to $\mu$, and $\nu$ is not identically zero, then there is a positive number $\epsilon$ and a measurable set $A$ so that $\mu(A)>0$ and $A$ is a positive set for the signed measure $\nu-\epsilon\mu$. That is, we can subtract off a little bit (but not zero!) of $\mu$ from $\nu$ and still find a non-$\mu$-negligible set on which what remains is completely positive.
To show this, let $X=A_n\uplus B_n$ be a Hahn decomposition with respect to the signed measure $\nu-\frac{1}{n}\mu$ for each positive integer $n$. Let $A_0$ be the union of all the $A_n$ and let $B_0$ be the intersection of all the $B_n$. Then since $B_0\subseteq B_n$ and $B_n$ is negative for $\nu-\frac{1}{n}\mu$ we find
$\displaystyle0\leq\nu(B_0)\leq\frac{1}{n}\mu(B_0)$
for every positive integer $n$. This shows that we must have $\nu(B_0)=0$. And then, since $\nu$ is not identically zero we must have $\nu(A_0)=\nu(X\setminus B_0)>0$. By absolute continuity we conclude that $\mu(A_0)>0$, which means that we must have $\mu(A_n)>0$ for at least one value of $n$. So we pick just such a value, set $A=A_n$, and $\epsilon=\frac{1}{n}$, and everything we asserted is true.
Now for the Radon-Nikodym Theorem: we let $(X,\mathcal{S},\mu)$ be a totally $\sigma$-finite measure space and let $\nu$ be a $\sigma$-finite signed measure on $\mathcal{S}$ which is absolutely continuous with respect to $\mu$. Then $\nu$ is an indefinite integral. That is, there is a finite-valued measurable function $f:X\to\mathcal{R}$ so that
$\displaystyle\nu(E)=\int\limits_Ef\,d\mu$
for every measurable set $E$. The function $f$ is unique in the sense that if any other function $g$ has $\nu$ as its indefinite integral, then $f=g$ $\mu$-almost everywhere. It should be noted that we don’t assert that $f$ is integrable, which will only be true of $\nu$ is actually finite. However, either its positive or its negative integral must converge or we wouldn’t use the integral sign for a divergent integral.
Let’s take a moment and consider what this means. We know that if we take an integrable function $f$, or a function whose integral diverges definitely, on a $\sigma$-finite measure space and define its indefinite integral $\nu$, then $\nu$ is a $\sigma$-finite signed measure that is absolutely continuous with respect to the measure against which we integrate. What the Radon-Nikodym theorem tells us that any such signed measure arises as the indefinite integral of some such function $f$. Further, it tells us that such a function is essentially unique, as much as any function is in measure theory land. In particular, we can tell that if we start with a function $f$ and get its indefinite integral, then any other function with the same indefinite integral must be a.e. equal to $f$.
### Like this:
Posted by John Armstrong | Analysis, Measure Theory
## 12 Comments »
1. [...] Radon-Nikodym Theorem (Proof) Today we set about the proof of the Radon-Nikodym theorem. We assumed that is a -finite measure space, and that is a -finite signed measure. Thus we can [...]
Pingback by | July 7, 2010 | Reply
2. [...] Radon-Nikodym Theorem for Signed Measures Now that we’ve proven the Radon-Nikodym theorem, we can extend it to the case where is a -finite signed [...]
Pingback by | July 8, 2010 | Reply
3. [...] Radon-Nikodym Derivative Okay, so the Radon-Nikodym theorem and its analogue for signed measures tell us that if we have two -finite signed measures and with [...]
Pingback by | July 9, 2010 | Reply
4. [...] the proof itself, the core is based on our very first result about absolute continuity: . Thus the Radon-Nikodym theorem tells us that there exists a function so [...]
Pingback by | July 14, 2010 | Reply
5. [...] we can combine this with the Radon-Nikodym theorem. If is a measurable function from a measure space to a totally -finite measure space so that the [...]
Pingback by | August 2, 2010 | Reply
6. [...] that , and thus that is a (signed) measure. It should also be clear that implies , and so . The Radon-Nikodym theorem now tells us that there exists an integrable function so [...]
Pingback by | September 3, 2010 | Reply
7. In the argument, when you said 0<=nu(Bo)<=1/n*mu(Bo). It is not clear to me why, 0<=nu(Bo). I tried to search through many books, but not found any where.
Comment by | December 29, 2012 | Reply
8. The measures $\mu$ and $\nu$ are not signed measures here.
Comment by | December 29, 2012 | Reply
9. I’m also confused. Following suneel’s comment, though I understand that mu and nu are not signed measures, but why would that specifically mean that 0<=nu(Bo)?
Comment by Alice | January 6, 2013 | Reply
10. Go back to the definition of a measure: “an extended real-valued, non-negative, countably additive set function $\mu$ defined on an algebra $\mathcal{A}$, and satisfying $\mu(\emptyset)=0$.” By definition, measures are nonnegative.
Comment by | January 6, 2013 | Reply
11. Now my question is different, Radon-Nikodym theorem is for signed measure and we are using this lemma(which is only for positive measure) in the proof of Radon-Nikodym, I am loosing some connection here? Thanks for clarification.
Comment by | January 7, 2013 | Reply
12. There’s a separate version of Radon-Nikodym for signed measures. Look up at comment #2, which is the pingback from a later post called “The Radon-Nikodym Theorem for Signed Measures”.
Comment by | January 7, 2013 | 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://math.stackexchange.com/questions/177181/how-to-measure-the-volume-of-rock?answertab=oldest
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How to measure the volume of rock?
I have a object which is similar to the shape of irregular rock like this
I would like to find the volume of this. How to do it?
If I have to find the volume, what are the things I would need. eg., If it is cylindrical, I would measure length and diameter. But, it is irregularly shaped. Like the above rock.
Where should I start? Couple of google search says something related to integration and contours. Somebody pls give me some handle :) I would say i'm very beginner level in math.
Many Thanks :)
Edit: 60 to 70% accuracy would be helpful.
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14
You could always use Archimedes' solution: submerge it in water, and measure the volume of water displaced. – Nate Eldredge Jul 31 '12 at 14:14
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For something as irregular as your example rock, I think Nate's solution is the only sensible one. – John Wordsworth Jul 31 '12 at 14:15
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The object is tumor inside the stomach. @Qiaochu Yuan a 3D angiogram – Prince Ashitaka Jul 31 '12 at 14:26
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– draks ... Jul 31 '12 at 14:28
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What kind of measuring tools do you have then? I assume you can see it in fair detail if you're comparing it to the rock. How do you know the shape/rough size of it in the first place? – Robert Mastragostino Jul 31 '12 at 14:29
show 5 more comments
3 Answers
As your comment indicates, you're not interested in rocks so much as tumors.
One possible approach is to use a tomographic technique. Many medical imaging tools image the body using tomography: that is, examining the body one "slice" at a time.
If you have access to such tools, or can derive such an example, then what you want to do is cut the rock/tumor into many slices along some axis, and then compute the area of the tumor at that slice. This is a bit easier to do that doing it in three dimensions.
Then, you move forward by some $\Delta z$ along the $z$-axis (or whatever axis), and repeat.
Multiply each surface area by $\Delta z$, sum them, and you will get a good estimate. The estimate is better the smaller your $\Delta z$.
To compute the area of each "slice", you can do many things: fit a simpler shape to the data, perform Monte Carlo integration, or decompose the shape into a series of piecewise linear segments.
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2
– J. M. Jul 31 '12 at 16:34
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I think typically the slices from tomographies are digital images consisting of pixels. In that case determining the size is easy: Just count all the pixels which belong to the tumor in all the slices, and multiply this with the area each pixel covers in each slice, and with the distance between slices. – celtschk Jul 31 '12 at 18:15
1
@celtschk Good point. If you have a digital image, you can use a simple threshold method to get the "area". A simple MATLAB script could be something like length(find(IMG > thresh))/prod(size(a)). – Arkamis Jul 31 '12 at 18:22
If you don't want to use Archimedes' solution, and you know the material, you can look up the density and weigh the rock.
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The easiest way (if you have a 3d model) is simple Monte Carlo integration. One builds a big bounding box around the "rock", and then randomly chooses a large points inside the box. The ratio of points inside the "rock" to the the number of total points, is the ratio of volumes. Since you can easily calculate the box's volume, you now also know the volume of the rock.
The task of finding whether or not a point is inside the rock is in general complex, but is simplified if you can assume that the rock is convex. In that case, you test each triangle that belongs to your 3D mesh: if the point is in the right side for all triangles, the point is inside.
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http://physics.stackexchange.com/users/14105/mark-mitchison?tab=activity
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# Mark Mitchison
reputation
112
bio
website www3.imperial.ac.uk/…
location United Kingdom
age
member for 7 months
seen 2 hours ago
profile views 188
Currently doing a PhD in Controlled Quantum Dynamics at Imperial College and the University of Oxford.
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| | 1,691 reputation | website | www3.imperial.ac.uk/… | member for | 7 months |
| 112 badges | location | United Kingdom | seen | 2 hours ago | |
# 343 Actions
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| 2h | comment | Is every quantum measurement reducible to measurements of position and time?Also see the comments of Dan Stahlke and Jess Riedel for a slightly better explanation of what I was trying to say :) |
| 2h | comment | Is every quantum measurement reducible to measurements of position and time?But on matters of theoretical principle, it clearly should be possible to phrase the outcome of any measurement process in terms of final positions of particles, just by pushing the quantum/classical boundary back a few steps further. Often it only matters that one can do something in principle for a particular theoretical manipulation to be valid. |
| 2h | comment | Is every quantum measurement reducible to measurements of position and time?You make a good point about electromagnetic field intensities, but it is easy to shift the emphasis again to positions. After all, "reading off measurements with our eyes" involves conformation changes of certain proteins within retinal cells. These in turn stimulate an action potential as different ionic species are pumped across the membranes of neurones, and you can keep going like this all the way until you hit the question of what is consciousness anyway. All of these involves changes in positional degrees of freedom. Of course, in practical terms this view hugely complicates matters. |
| 2h | comment | Is every quantum measurement reducible to measurements of position and time?I'll admit this is a matter of opinion, but I think you have missed Feynman's point. The question is where do you draw the boundary between quantum and classical, or more precisely, how you fix the boundary conditions of the quantum path integral. Depending on this choice you can say that the "outcome" was a set of positions, a set of field strengths, a set of spin orientations, whatever. As long as all of these results are consistent with the same physics between "observations", the choice is irrelevant - QM does not give a prescription for where exactly the observation takes place. |
| 3h | revised | Can a diver swim a short distance in great depths without being physically crushed by the pressure?added 41 characters in body |
| 3h | comment | Can a diver swim a short distance in great depths without being physically crushed by the pressure?Hence my "pretty much no risk" as opposed to no risk whatsoever ;) But yes, this is absolutely correct, thanks for the clarification. Not only would the time have to be short, but you only have a single lungful of air. Even if you could remain at depth long enough for the gas in your lungs to chemically equilibrate with your slowest tissues, I doubt that you would get bent on ascent. But I haven't done any calculations to prove it :) |
| 4h | revised | Can a diver swim a short distance in great depths without being physically crushed by the pressure?added 12 characters in body |
| 4h | answered | Can a diver swim a short distance in great depths without being physically crushed by the pressure? |
| 6h | revised | Is every quantum measurement reducible to measurements of position and time?added 79 characters in body |
| 6h | answered | Is every quantum measurement reducible to measurements of position and time? |
| May8 | revised | Slinky reverb: the origin of the iconic Star Wars blaster soundadded 2 characters in body |
| May8 | asked | Slinky reverb: the origin of the iconic Star Wars blaster sound |
| May5 | comment | Do we always ignore zero energy solutions to the (one dimensional) Schrödinger equation? |
| May4 | comment | Do we always ignore zero energy solutions to the (one dimensional) Schrödinger equation?@BenCrowell But all the basis states $|x + 2n \pi\rangle$ with $n$ integer are physically equivalent. I would argue that the correct variance of the position is finite ($2\pi / \sqrt{3}$ if my algebra's right). Presumably the apparent conflict with the canonical position-momentum uncertainty relation is resolved by properly deriving the analogous expression on a compact space with periodic BCs. I'm going to bed but will try to formalise this tomorrow. Or leave it as a challenge... |
| May3 | comment | Coulomb interaction and conservation lawsYou're welcome, hopefully Emilio's detailed response has answered your question. |
| May3 | comment | Coulomb interaction and conservation laws@DaPhil I disagree, in condensed matter physics the natural basis to choose is plane (Bloch) waves because the free Hamiltonian is invariant under (discrete) translations. In quantum chemistry the "free" problem is obviously not translation invariant and momentum is not conserved. Choosing a basis of momentum eigenstates to make the translation invariance of the Coulomb potential manifest would be a bizarre convention. Far more important is the nuclear Coulomb potential, which is not translation invariant (in the Born-Oppenheimer approximation), so molecular orbitals are a natural basis set. |
| Apr27 | comment | On the Aharonov-Bohm effect, and the reality of the classical fields |
| Apr23 | answered | Ratio of Size of Atom to Size of Nucleus |
| Apr23 | revised | Free 1d proton in magnetic fieldadded 194 characters in body |
| Apr23 | revised | Free 1d proton in magnetic fieldadded 427 characters in body |
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http://mathhelpforum.com/math-topics/201452-assignment-problem.html
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# Thread:
1. ## Assignment problem
If I write a mathematical algorithm. ( something that might become part of a computer program) and I need to assign a value to a variable. I normally just write something like this "set x to something" or "set x=something"
But in math you have notations for everything. Is there a notation or this?
2. ## Re: Assignment problem
Originally Posted by mariusg
If I write a mathematical algorithm. ( something that might become part of a computer program) and I need to assign a value to a variable. I normally just write something like this "set x to something" or "set x=something"
But in math you have notations for everything. Is there a notation or this?
As far as I know there is no universal agreement on tbis.
However, some computer algebra systems use $a:= x$ to mean $a$ is assigned the value $x$.
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http://math.stackexchange.com/questions/71435/optimizing-with-absolute-value-objective-function
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# Optimizing with Absolute Value Objective Function
max : $w = |q^T y|$
subject to
$A y \leq b$
$y \geq 0$
Please describe how one could solve the non-linear programming prob. above by using linear programming methods.
I tried changing $y$ to $y' - y''$ in the constraints and $y' + y''$ for the objective function. However, my Excel solver says that "the cells do not converge". How should I solve this?
Thanks a bunch!
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the maximum value of $w$ may be infinite. do you have any information about $A,b,q$? – Ilya Oct 10 '11 at 13:42
no, but there's a hint saying: "Try breaking it into 2 linear programming problems. Then, could you think of combining them into just 1 problem?" – John Oct 10 '11 at 13:46
## 2 Answers
To follow the advise given to you, consider two problems: $$\begin{cases} w^+ &= q^Ty^+\to\max, \\ Qy^+&\leq0, \\ Ay^+&\leq b, \\ y^+&\geq 0. \end{cases}$$ and
$$\begin{cases} w^- &= q^Ty^-\to\max, \\ Qy^-&\geq0, \\ Ay^-&\leq b, \\ y^-&\geq 0. \end{cases}$$
Then $w = \max\{w^+,w^-\}$. Here matrix $Q = (q\quad0\quad\dots\quad0)^T$. With such decomposition you just consider two possible cases for the absolute value.
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Thanks for your answer! However, is there any way to solve this using just 1 linear programming problem? – John Oct 10 '11 at 17:05
@John: I'm afraid, I don't know such a way. – Ilya Oct 10 '11 at 17:29
Hint: let $y = y_1 + y_2$ where $q^Ty_1 \le 0$ and $q^Ty_2 \ge 0$.
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http://mathoverflow.net/questions/105415?sort=oldest
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## What is the theoretical interest of finding closed-form sols. of infinite series?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hi,
I was reading this when I came across Gourevitch's conjecture.
My understanding is that solutions to these series are of practical interest. If one encounters such a series, being able to solve it exactly is more practical than having to solve it numerically.
But, not being a mathematician, I simply can't imagine what the theoretical implications of proving such conjectures are.
What are they?
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3
The theoretical implication is that this is really cool, and understanding why it is true is likely to require understanding something deep. – Igor Rivin Aug 24 at 20:05
2
Isn't this like porn? – Mariano Suárez-Alvarez Aug 24 at 20:44
4
Voted to close. As you seem to be an undergraduate in chemistry or related, it is unlikely we can convince you of the value of this, and I don't think anyone should try. If you find further questions, try math.stackexchange.com/questions – Will Jagy Aug 24 at 21:49
8
@WillJagy How absurd, pejorative and close minded. Please tell me you're not serious. – CHM Aug 25 at 1:09
3
CHM - Please don't be offended by the votes to close. This is not personal! Mathoverflow is specifically for questions that arise in mathematical research, as the FAQ makes clear. Your question is a great one, but as you admit is not a question that arose out of mathematical research. You might find that math.stackexchange.com would be a better venue. – HW Aug 25 at 7:47
show 9 more comments
## 4 Answers
Let $k \ge 2$ be an ineteger. Consider the series `$$ \sum_{n=1}^{\infty} \frac{1}{n^k} $$` this is typically denoted $\zeta(k)$ as it is the value of the Riemann zeta function at $k$.
Now, Euler showed that for even $k$ this is equl to `$$q_k \pi^k $$` where `$q_k$` is some rational number (that one can also describe explicty).
This is on the one hand an interesting fact and would also allow to calculate approximations of the powers of $\pi$ but what I actually want to get at is that from this it follows that if one knows that $\pi$ is transcendental then one gets directly that $\zeta(k)$ is transcendental and in particular not a rational number.
So this is for even $k$. What about odd $k$? Say $k=3$. Is this rational or irrational? This question was open for a long time until it was proved at the end of the 1970s by Apéry.
How does this prove go (very roughly!):
He first showed that `$$ \zeta(3) = \frac{5}{2} \sum_{i=1}^{\infty} \frac{ (-1)^n}{n^3 C (2n , n )} $$` where $C(2n,n)$ is just the obvious binomial coefficient, which I fail to be able to type properly.
So one could say he evaluated the series on the right in 'a closed form'; showing that its values is something already known/defined.
Then based on this he derived some sequences of rational numbers that converge to $\zeta(3)$ so fast that it is impossible for $\zeta(3)$ to be rational itself thus proving the irrationality of $\zeta(3)$.
So, in order to show that $\zeta(3)$ is irrational he first needed to show that it is equal to the limit of this (other) series, or put differently to evaluate this series; not only the first simpler one.
It would be interesting to be able to do something like this for other odd numbers, but so far noone knows how to do so and the irrationality of $\zeta$ at any other odd positive integers is unknown (although there are results that assert that among certain collections of them there are at least some irrational ones).
Thus finding an evaluation of a series can be used to infer something theoretical on its value.
This is not always the motivation, but sometimes it is the case that the point is not so much to know the value of the series in order to replace it in some compuation say, but rather to use the series as a form of describing its value by simpler buildingblocks and thereby allowing to learn something (new) on the value.
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You could use(^{2n}_{n}) to get $(^{2n}_{n})$ – Lunasaurus Rex Aug 25 at 3:19
3
To get $\binom{2n}{n}$ I've used \binom{2n}{n}. – Américo Tavares Aug 25 at 10:09
Thanks for the hints. (I will refrain from editing at least for the moent; but good to know for the future). – quid Aug 26 at 1:19
### 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.
Having the close form for a series of functions is also important:
1. Closed forms can sometimes be used to find exact solutions to differential equations.
2. Closed forms allow for accurate estimates on errors in approximations of special functions. Some of these are used in your calculator.
3. Closed forms can sometimes produce proofs of non-series results. For example, the equation $$e^{ix} = \cos(x) + i\sin(x)$$ is proven by using three Taylor series.
4. This is a bit incestuous, but having the closed form for one series allows for manipulation to produce the closed form of other series.
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the identity in 3 is much more interestingly proven in other ways! :-) – Mariano Suárez-Alvarez Aug 25 at 4:56
I can recommend reading "Closed forms: what they are and why we care" by Jon Borwein and Richard Crandall, the article is to appear in Notices Amer. Math. Soc. 60 (2013).
Edit (Dec 2012). The paper has just appeared in the Notices: pdf. Together with the sad news about Richard Crandall: he passed away.
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1
Wow. Anyone who's ever asked for a "closed form solution" should read this paper. Very eye-opening on a topic that I think most of us take for granted. – Aeryk Aug 26 at 2:08
HERE is an example in MO. For the first question, I could evaluate the series in closed form, so I could compute with their known properties, and thus answer the question. For the second question, the series were just ${}_2F_1$ series, and I did not know their properties, so I could not answer the question. Despite the numerical evidence in the graphs.
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http://mathhelpforum.com/differential-geometry/144652-surjective-function.html
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# Thread:
1. ## Surjective Function
Let $X=\mathbb{R}$ then define an equivalence relation $\sim$ on $X$ s.t.
$x\sim y$ if and only if $x-y\in\mathbb{Z}$
Show that $X/\sim\cong S^1$
So denoting the elements of $X/\sim$ as $[t]$
The function
$f([t])=\exp^{2\pi ti}$ defines a homemorphism.
$f([t])$ is continuous since $\exp^{2\pi ti}=cos(2\pi t)+isin(2\pi t)$ which is the sum of continous functions.
Letting $f([x])=f([y])\Rightarrow\exp^{2\pi xi}=\exp^{2\pi yi}\Rightarrow2\pi xi=2\pi yi\Rightarrow x=y$ so injective.
Now $f([t])=\exp^{2\pi ti}=z$ for $z\in\mathbb{C}$ s.t. $|z|=1$
Then $t=\frac{-i}{2\pi}log(z)=\frac{-i}{2\pi}(log|z|+iArg(z))$
Since $|z|=1$ we have $log|z|=0$ so $t=\frac{1}{2\pi}Arg(z)$ which is in $X/\sim$ so surjective.
Therefore a bijection.
Not sure how to show $f^{-1}$ continuous?
Is this correct? Any input would be great. Thanks
2. $f([t])=\exp^{2\pi ti}$ defines a homemorphism.
True. But the mapping is only well defined (independent of representative from the equivalence class) because $e^{2\pi i t}$ is periodic over the integers.
$f([t])$ is continuous (trivial)
Letting $f([x])=f([y])\Rightarrow\exp^{2\pi xi}=\exp^{2\pi yi}\Rightarrow2\pi xi=2\pi yi\Rightarrow x=y$ which implies injective.
Saying it is slightly more subtle, the above is true ONLY because of the construction of $X/\sim$. Say more.
And yes it is surjective. But, why do it this way? Really you aren't done, you haven't proven the inverse function is continuous.
Have you proven that $X/\sim$ is Hausdorff? Then, reverse the direction of your mapping and you only have to show that your function is continuous (you already proved it was bijective) and thus you will have a continuous bijection from a compact space into a Hausdorff space and thus automatically a homeomorphism.
That said, you've already done everything else. Might as well finish it.
3. Originally Posted by ejgmath
$f([t])$ is continuous since $\exp^{2\pi ti}=cos(2\pi t)+isin(2\pi t)$ which is the sum of continous functions.
Just a remark in taste. Although what you said is true I think it is personally more enlightening to realize that $\mathbb{C}\approx\mathbb{R}^2$ and so you're mapping will be continuous if and only if $\alpha:X/\sim\to\mathbb{R}^2:[t]\mapsto \left(\cos(2\pi [t]),\sin(2\pi[t])\right)$ from where continuity is much more clear since the coordinate functions $\pi\alpha:X/\sim\to\mathbb{R}:[t]\mapsto\cos(2\pi[t])$ are continuous. This isn't altogether clear (at least to me) but remember the defining characteristic of quotient maps. Namely, $\pi\alpha$ is continuous if and only if so is $\pi\circ\alpha\circ\xi:\mathbb{R}\to\mathbb{R}:t\m apsto \cos(2\pi[t])=\cos(2\pi t)$. There, now it is OBVIOUS that it's continuous
EDIT: $\xi:\mathbb{R}\to \mathbb{R}/\mathbb{Z}:t\mapsto [t]$ was meant to be the quotient map.
P.S. Interestingly enough we can go further. If we consider $\mathbb{S}^1$ to be both a topological space with the usual topology and a group under complex multiplication, we give $\mathbb{R}/\mathbb{Z}$ the quotient topology and the group structure as a quotient group, and finally give $\text{SO}(2)]$ the normal topology and matrix multiplication it is true that $\mathbb{R}/\mathbb{Z}\overset{\text{T.G.}}{\cong}\mathbb{S}^1 \overset{\text{T.G.}}{\cong}\text{SO}(2)$ where the T.G. means topological group isomorphism.
4. Thanks for the responses, after writing my post I realised I had missed out the inverse function continuity.
Just a quick question, if I let $f^{-1}(z)=\frac{1}{2\pi}Arg(z)$ that would clearly create the equivalence class for any $t$ that generates $z$.
But what would you say about it's continuity? Is $Arg(z)$ continuous by definition?
5. Originally Posted by ejgmath
Thanks for the responses, after writing my post I realised I had missed out the inverse function continuity.
Just a quick question, if I let $f^{-1}(z)=\frac{1}{2\pi}Arg(z)$ that would clearly create the equivalence class for any $t$ that generates $z$.
But what would you say about it's continuity? Is $Arg(z)$ continuous by definition?
Honestly, I'm not too sure if something gets screwed up. I'll have to check. Have you thought about proving that $X/\sim$ is compact (it's not too hard) and then you're homeomorphism is immediate.
6. Yeh, I did consider that but the (revision) question is specifically looking for the whole continuous bijection business.
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http://math.stackexchange.com/questions/tagged/banach-algebras+analysis
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# Tagged Questions
0answers
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### Open map in Banach algebra
I'm having trouble showing a certian function is open and can be extended. Let $\Omega$ be a completely regular topological space and $A=C_b(\Omega)$ the space of all complex-values bounded ...
2answers
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### Banach-algebra homeomorphism.
Let $A$ be a commutative unital Banach algebra that is generated by a set $Y \subseteq A$. I want to show that $\Phi(A)$ is homeomorphic to a closed subset of the Cartesian product \$ ...
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### Is the space of bounded functions with the Supremum norm a Banach Algebra
X is an arbitrary , non empty set, B(X) the set of bounded functions $f:X\rightarrow \mathbb{R}$ and $||f||_\infty = \sup_{x\in X }|f(x)|$. Is $(B(X),||.||_\infty )$ a Banach Algebra? My attempt ...
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### Prove that if $A$ is algebra generated by $\sin(x)$ and $\cos(x)$ then $A = \{ f\in C_b(\mathbb R ) : f(t) = f(t + 2\pi )$ for all $t \in \mathbb R\}$ [duplicate]
Possible Duplicate: Finding a closed subalgebra generated by functions. Let $A$ be the uniformly closed subalgebra of $C_b(\mathbb{R} )$ generated by $\sin(x)$ and $\cos(x)$. Prove: \$A = ...
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http://www.wall.org/~aron/blog/fields/
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comments on physics and theology
Fields
by Posted on November 10, 2012
What is the world made out of? In the most usual formulations of our current best theories of physics, the answer is fields. What are those?
Well, if you know what a function is, you're already most of the way there. A function, you will recall, is a gadget where, for any number you input, you can get a number out as an output. We can write $f(x)$ where $x$ is the number you input, and $f(x)$ is the number you output. The function $f$ itself is the rule for going from one to the other, e.g. For example $f(x) = \sqrt{\sin x^2 + 1}$.
Now, nothing stops you from having a function that depends on multiple numbers as input; for example the function $f(x,\,y) = xy^2 + x^3y$ depends on two input variables, $x$, and $y$. If there are $D$ input numbers, then the $D$-dimensional space of possible combinations of input numbers is called the domain of the function.
Also nothing stops you from having the output be a set of several numbers. In this case we would need some sort of subscript $i$ to refer to the different possible output numbers. For example, if we had a function with one input number $x$ and three output numbers $y$, then we could write $f_i(x)$, where $i$ takes the values 1, 2, or 3. Then $f_i(x)$ would really be just a package of three different functions: $f_1(x)$, $f_2(x)$, and $f_3(x)$. So if you specify the input $x$, you get three output numbers $(f_1, f_2, f_3)$. If there are $T$ different output numbers, the $T$ dimensional space of possible outputs is called the target space.
Now a field is just a function whose domain is the points of spacetime. For example, the air temperature in a room may vary from place to place, and it may also change with time. So if you imagine checking all possible points of space in the room at all possible times, you could describe this with a temperature field $T(t, x, y, z)$. However, the temperature field isn't a fundamental entity that exists on its own. It subsists in a medium (air) and describes its motion. When the air molecules are moving around quickly in a random way, we say it's hot, and when they start to move around slower, we say it's getting chilly. An example of a field which actually is fundamental (as far as we know) would be the electromagnetic field. This has 6 output numbers, since the electric field can point in any of the 3 spatial directions, and the magnetic field also has 3 numbers.
For a while in the 19th century, scientists were confused about this. They thought that electromagnetic waves had to be some sort of excitation of some sort of stuff, which they called the aether. That's because they were assuming (based on physical intuitions filtered through Newtonian mechanics) that matter is something solid and massy, which interacts by striking or making contact with other things. The 20th century scientific advances partly came from realizing that its okay to describe things with abstract math. Any kind of mathematical object you write down satisfying logically consistent equations is OK, as long as it matches experiment. So electromagnetic waves don't have to be made out of anything. They just are, and other things are (partly) made out of them.
In our current best theory of particle physics, the Standard Model, there are a few dozen different kinds fields, and all matter is explained as configurations of these fields. I can't tell you exactly how many fields there are, because it depends on how you count them. Not counting the gravitational field, there are 52 different output numbers corresponding to bosons, and 192 different output numbers corresponding to fermions (Don't worry about what these terms mean yet). So you could say that there are 244 different fields in Nature, each with one output number.
That sounds awfully complicated. But there's also a lot of symmetries in the Standard Model which relate these output numbers to each other. This includes not only the Poincaré group of spacetime symmetries, but also various internal symmetries related to the dynamics of the strong, weak, and electromagnetic forces. They are called internal because they don't move the points of spacetime around. Instead they just mutate the different kinds of output numbers into each other.
So normally, particle physicists just package the output numbers into sets, such that the numbers in each set are related by the various kinds of symmetry. (For example, the 6 different numbers of the electromagnetic field are related by rotations and Lorentz boosts.) Each of these sets is called a field. In future posts I'll give more details about the different kinds of fields. As always, questions are welcome.
UPDATE: I forgot to include the 4 vector components of the spin-1 gauge bosons, so the numbers of degrees of freedom of the bosons were wrong before. Note to Experts: These are the "off-shell" degrees of freedom before taking into consideration constraints or gauge symmetry. Note to Non-Experts: the numbers in this post are just for flavor, in order to give you the sense that there are a LOT of different fields in Nature. You won't need to understand how I got these numbers in order to enjoy future posts!
About Aron Wall
I am a postdoctoral researcher studying quantum gravity and black hole thermodynamics at UC Santa Barbara. Before that, I studied the Great Books program at St. John's college Santa Fe, and got my Ph.D. in physics from U Maryland.
This entry was posted in Physics. Bookmark the permalink.
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http://en.wikipedia.org/wiki/Sierpinski_triangle
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# Sierpinski triangle
Sierpinski triangle
Generated using a random algorithm
Sierpinski triangle in logic: The first 16 conjunctions of lexicographically ordered arguments
The columns interpreted as binary numbers give 1, 3, 5, 15, 17, 51... (sequence in OEIS)
The Sierpinski triangle (also with the original orthography Sierpiński), also called the Sierpinski gasket or the Sierpinski Sieve, is a fractal and attractive fixed set named after the Polish mathematician Wacław Sierpiński who described it in 1915. However, similar patterns appear already in the 13th-century Cosmati mosaics in the cathedral of Anagni, Italy,[1] and other places, such as in the nave of the Roman Basilica of Santa Maria in Cosmedin.[2]
Originally constructed as a curve, this is one of the basic examples of self-similar sets, i.e. it is a mathematically generated pattern that can be reproducible at any magnification or reduction.
Comparing the Sierpinski triangle or the Sierpinski carpet to equivalent repetitive tiling arrangements, it is evident that similar structures can be built into any rep-tile arrangements.
## Construction
An algorithm for obtaining arbitrarily close approximations to the Sierpinski triangle is as follows:
Note: each removed triangle (a trema) is topologically an open set.[3]
1. Start with any triangle in a plane (any closed, bounded region in the plane will actually work). The canonical Sierpinski triangle uses an equilateral triangle with a base parallel to the horizontal axis (first image).
2. Shrink the triangle to ½ height and ½ width, make three copies, and position the three shrunken triangles so that each triangle touches the two other triangles at a corner (image 2). Note the emergence of the central hole - because the three shrunken triangles can between them cover only 3/4 of the area of the original. (Holes are an important feature of Sierpinski's triangle.)
3. Repeat step 2 with each of the smaller triangles (image 3 and so on).
Note that this infinite process is not dependent upon the starting shape being a triangle—it is just clearer that way. The first few steps starting, for example, from a square also tend towards a Sierpinski triangle. Michael Barnsley used an image of a fish to illustrate this in his paper "V-variable fractals and superfractals."[4]
The actual fractal is what would be obtained after an infinite number of iterations. More formally, one describes it in terms of functions on closed sets of points. If we let $d_a$ note the dilation by a factor of ½ about a point a, then the Sierpinski triangle with corners a, b, and c is the fixed set of the transformation $d_a$ U $d_b$ U $d_c$.
This is an attractive fixed set, so that when the operation is applied to any other set repeatedly, the images converge on the Sierpinski triangle. This is what is happening with the triangle above, but any other set would suffice.
If one takes a point and applies each of the transformations $d_a$, $d_b$, and $d_c$ to it randomly, the resulting points will be dense in the Sierpinski triangle, so the following algorithm will again generate arbitrarily close approximations to it:
Start by labeling p1, p2 and p3 as the corners of the Sierpinski triangle, and a random point v1. Set vn+1 = ½ ( vn + prn ), where rn is a random number 1, 2 or 3. Draw the points v1 to v∞. If the first point v1 was a point on the Sierpiński triangle, then all the points vn lie on the Sierpinski triangle. If the first point v1 to lie within the perimeter of the triangle is not a point on the Sierpinski triangle, none of the points vn will lie on the Sierpinski triangle, however they will converge on the triangle. If v1 is outside the triangle, the only way vn will land on the actual triangle, is if vn is on what would be part of the triangle, if the triangle was infinitely large.
Animated creation of a Sierpinski triangle using the chaos game
Animated construction of a Sierpinski triangle
Or more simply:
1. Take 3 points in a plane to form a triangle, you need not draw it.
2. Randomly select any point inside the triangle and consider that your current position.
3. Randomly select any one of the 3 vertex points.
4. Move half the distance from your current position to the selected vertex.
5. Plot the current position.
6. Repeat from step 3.
Note: This method is also called the Chaos game. You can start from any point outside or inside the triangle, and it would eventually form the Sierpinski Gasket with a few leftover points. It is interesting to do this with pencil and paper. A brief outline is formed after placing approximately one hundred points, and detail begins to appear after a few hundred.
Sierpinski triangle using IFS
Or using an Iterated function system
An alternative way of computing the Sierpinski triangle uses an Iterated function system and starts by a point at the origin (x0 = 0, y0 = 0). The new points are iteratively computed by randomly applying (with equal probability) one of the following three coordinate transformations (using the so-called chaos game):
xn+1 = 0.5 xn
yn+1 = 0.5 yn; a half-size copy
This coordinate transformation is drawn in yellow in the figure.
xn+1 = 0.5 xn + 0.25
yn+1 = 0.5 yn + 0.5 $\sqrt{3}\over 2$; a half-size copy shifted right and up
This coordinate transformation is drawn using red color in the figure.
xn+1 = 0.5 xn + 0.5
yn+1 = 0.5 yn; a half-size copy doubled shifted to the right
When this coordinate transformation is used, the triangle is drawn in blue.
Or using an L-system — The Sierpinski triangle drawn using an L-system.
bitwise AND - The 2D AND function, z=AND(x,y) can also produce a white on black right angled Sierpinski triangle if all pixels of which z=0 are white, and all other values of z are black.
bitwise XOR - The values of the discrete, 2D XOR function, z=XOR(x,y) also exhibit structures related to the Sierpinski triangle. For example, one could generate the Sierpinski triangle by setting up a 2 dimensional matrix, [rows][columns] placing the uppermost point on [1][n/2], then cycling through the remaining cells row by row the value of the cell being XOR([i-1][j-1],[i-1][j+1])
Other means — The Sierpinski triangle also appears in certain cellular automata (such as Rule 90), including those relating to Conway's Game of Life. The automaton "12/1" when applied to a single cell will generate four approximations of the Sierpinski triangle.
If one takes Pascal's triangle with 2n rows and colors the even numbers white, and the odd numbers black, the result is an approximation to the Sierpinski triangle. More precisely, the limit as n approaches infinity of this parity-colored 2n-row Pascal triangle is the Sierpinski triangle.
## Properties
The Sierpinski triangle has Hausdorff dimension log(3)/log(2) ≈ 1.585, which follows from the fact that it is a union of three copies of itself, each scaled by a factor of 1/2.[5]
The area of a Sierpinski triangle is zero (in Lebesgue measure). The area remaining after each iteration is clearly 3/4 of the area from the previous iteration, and an infinite number of iterations results in zero. Intuitively one can see this applies to any geometrical construction with an infinite number of iterations, each of which decreases the size by an amount proportional to a previous iteration.[citation needed]
## Analogues in higher dimensions
A Sierpinski square-based pyramid and its 'inverse'
A Sierpiński triangle-based pyramid as seen from above (4 main sections highlighted). Note the self-similarity in this 2-dimensional projected view, so that the resulting triangle could be a 2D fractal in itself.
The tetrix is the three-dimensional analogue of the Sierpinski triangle, formed by repeatedly shrinking a regular tetrahedron to one half its original height, putting together four copies of this tetrahedron with corners touching, and then repeating the process. This can also be done with a square pyramid and five copies instead. A tetrix constructed from an initial tetrahedron of side-length L has the property that the total surface area remains constant with each iteration.
The initial surface area of the (iteration-0) tetrahedron of side-length L is $L^2 \sqrt{3}$. At the next iteration, the side-length is halved
$L \rightarrow { L \over 2 }$
and there are 4 such smaller tetrahedra. Therefore, the total surface area after the first iteration is:
$4 \left( \left( {L \over 2} \right)^2 \sqrt{3} \right) = 4 { {L^2} \over 4 } \sqrt{3} = L^2 \sqrt{3}.$
This remains the case after each iteration. Though the surface area of each subsequent tetrahedron is 1/4 that of the tetrahedron in the previous iteration, there are 4 times as many—thus maintaining a constant total surface area.
The total enclosed volume, however, is geometrically decreasing (factor of 0.5) with each iteration and asymptotically approaches 0 as the number of iterations increases. In fact, it can be shown that, while having fixed area, it has no 3-dimensional character. The Hausdorff dimension of such a construction is $\textstyle\frac{\ln 4}{\ln 2}=2$ which agrees with the finite area of the figure. (A Hausdorff dimension strictly between 2 and 3 would indicate 0 volume and infinite area.)
## Trivia
In an interview with KCRW's Bookworm, David Foster Wallace stated that his novel Infinite Jest structurally resembles the Sierpinski triangle.[6]
## References
1.
2. Michael Barnsley, et al. PDF (2.22 MB)
3. Falconer, Kenneth (1990). Fractal geometry: mathematical foundations and applications. Chichester: John Wiley. p. 120. ISBN 0-471-92287-0. Zbl 0689.28003.
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http://physics.stackexchange.com/questions/3527/newtons-cradle/37658
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# Newton's cradle
Why, when one releases 2 balls in Newton's cradle, two balls on the opposite side bounce out at approximately the same speed as the 1st pair, rather than one ball at higher speed, or 3 balls at lower speed?
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## 6 Answers
Contrary to what is stated in many textbooks, energy-momentum conservation alone cannot explain the behavior of Newton’s cradle. For N balls we have two equations and N final velocities to calculate. Hence, conservations laws can do the job only for N=2. This means that if we want to give an explanation of the cradle behavior based on conservation laws, we have to split the N-ball collision into a sequence of two-ball collisions, as done in some of the answers given to the question.
The problem with this approach is that it assumes that initially the balls are not touching one another, which is not the case in most cradles. One might argue that it doesn’t really matter that the balls are initially in contact, only that the collision time (the time during which momentum is transferred from one ball to the other) is much smaller than the time it takes for the mechanical perturbation to cross one ball. However, the typical collision time is of the order of 0.1 ms, much bigger than the propagation time of about 0.01 ms. Under such conditions one should expect wave propagation along the ball chain to play a significant role in the cradle’s dynamics.
There is a nice paper,
F. Herrmann and P. Schmälzle. Simple explanation of a well‐known collision experiment. Am. J. Phys. 49, issue 8, pp. 761 (1981). doi: 10.1119/1.12407. Open access version at F. Herrmann's Karlsruher Physikkurs website.
showing that wave propagation along the ball line must be dispersion-free if we are to find equal numbers of incoming and outgoing balls. The authors also provide an interesting picture of how the system decides how many balls it is going to send away. Basically, the first impact produces two wave pulses moving in opposite directions along the ball line. The pulses are reflected at the ends of the line and meet again at the point where the ball chain breaks up. It is easy to see that this break-up point is symmetrical (with respect to the middle of the line) to the impact point, which explains why the number of outgoing balls is equal to that of the incoming ones.
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Thanks for the nice literature reference! It would be amusing to create a (Non-)Newtonian cradle from some dispersive material. – nibot Feb 1 '11 at 16:37
2
– Fabian Jul 7 '11 at 19:50
Open access link now dead :( – Qmechanic♦ Sep 18 '12 at 10:25
It conserves both energy and momentum in the collision at the same time.
By design, when the balls collide the strings that hold them up are vertical (assuming balls are only swung from one side). This means there are no horizontal forces from the string on the balls so linear momentum in the direction of swing must be conserved in the collision. Energy is also nearly conserved provided not too much noise and heat are produced.
Consider the two-ball case where each ball has mass $m$ and velocity $v$ at the time of collision. The kinetic energy will be $E = mv^2$ and the momentum $p = 2mv$. So suppose $n$ balls were to fly off the other side with velocity $u$. The energy would be $\frac{n}{2} mu^2$ and the momentum $nmu$. So by conservation we must have $mv^2 = \frac{n}{2} mu^2$ and $2mv = nmu$. It is not difficult to see that the only solution to these equation together is $n=2$ and $u=v$.
A more complete analysis would rule out solutions with different balls leaving at different speeds, but I think this is enough to demonstrate the principle.
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Thank you too Philip. This also is very insightful. – jack mosevich Jan 21 '11 at 22:50
It's not the only way to conserve momentum and kinetic energy. If you pull back two balls and drop them against three, for example, you could conserve momentum and energy by having the sets of balls act as solid blocks, so that the two balls bounced back the way they came from with some diminished speed and the three balls pushed ahead at another speed. – Mark Eichenlaub Jan 30 '11 at 20:02
@Mark is right, of course, but this analysis serves to show that the observed behavior is plausible, when many people find it counter intuitive. – dmckee♦ Mar 10 '11 at 23:28
I agree with Mark and dmckee – Philip Gibbs Mar 11 '11 at 12:08
Let's start with an observation from billiards. Say the red ball is stationary and you hit it dead-on with the cue ball at speed $v$. The cue ball will stop and the red ball will continue on at speed $v$. Lubos gave a nice, simple description of this in his answer. You can see it happening at the beginning of this video.
Now imagine a row of balls (red-orange-yellow-green-blue-violet, say) lined up perfectly.
You hit the red one dead-on with the cue, and the cue stops and the red ball goes ahead. After just a moment, the red ball hits the orange ball. The red one stops and the orange on goes on ahead. Then the orange one hits the yellow one and stops, etc. Ultimately the violet ball comes out of the line, and all the other balls in the line have just moved down the table a little way. One ball in, one ball out. You can see something pretty much like this happen 50 seconds into the same video (it's not quite perfect, though).
Now we do it again. Line up all the balls, but this time roll the cue ball towards the red ball, and roll the 8-ball right after it.
The cue ball strikes the red ball and stops; the red ball goes forward, as before. The red ball hits the orange ball. This time, that's not the only thing happening. Simultaneously with that collision, the 8-ball runs into the stopped cue ball. After these collisions there are two balls in motion - the orange ball in front, and the cue ball, which has now had two collisions, one in front and one from behind.
This process repeats, going down the line, always with two balls in motion with one stationary ball in between. Near the end, the two moving balls are the violet (last) ball and green (third-from-last) ball. The violet ball is free, but the green ball impacts the blue one, and that's the last collision. The result is that the blue and violet balls come away from the line up. Everything else is stopped. Two balls went in, and two come out.
From here it is not hard to see that we could send in three, four, or even 100 balls (in theory) and get the same number out again.
The difficulty with this explanation is that in Newton's cradle, the balls physically touch one another, and so the stop-start argument does not apply in the same way. This simple analysis should make Newton's cradle plausible, but a full argument relies on studying the continuum mechanics associated with the collisions between solid bodies. This paper, mentioned by Georg in the comments, provides such an analysis.
Note: I didn't watch the YouTube video all the way through and don't necessarily vouch for everything it says about physics. I just wanted examples of those shots. Also, the balls in billiards are rolling, which can sometimes make a difference compared to Newton's cradle, but here we will assume it does not.
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Hello Mark, there is one weak point in Your otherwise brilliant explanation. I think it is not obvious, that having smaller and smaller distance between balls in the end leads to the real cradle. Being in contact is something different than having the tiniest distance. I think one has to go to the real mechanics within the spheres (compression, sound wave propagation) to get the "innermost truth" :=) I assume that when one analyses the elastic reaction within the sphere, the main result will be, that it takes enough time, to make the spheres act as if there was some distance. – Georg Jan 30 '11 at 11:53
@Georg I don't think there's a big problem with thinking about balls in contact as being essentially the same as balls separated by a small distance in this particular scenario. Even if we imagine the balls had infinite bulk modulus, the result would be the same. We could discuss things like the speed of sound in the balls and wave motion, and that would be interesting, but it would also be beyond the scope of the original question. Those considerations would be more important if the question were about how long after the impact on the left hand side the balls took to come out on the right. – Mark Eichenlaub Jan 30 '11 at 12:07
Hello Mark to me it is not so obvious. I'd like to see a "movie" of the strain inside the sphere, for me the efficency of momentum and energy transfer between steel or glas spheres still is a wonder. – Georg Jan 30 '11 at 13:39
@Georg It sounds like you're asking about why the collisions are elastic. I agree that may not be obvious, but it's a different issue. I wanted as simple an answer as possible here. It's helpful to note that the speed of sound in the balls is on the order of thousands of meters per second, while in these experiments the speed of the balls themselves is only meters per second. If we moved to larger, softer, less-elastic balls moving at higher speeds, we'd see a higher fraction of energy dissipated. – Mark Eichenlaub Jan 30 '11 at 18:51
@Georg After thinking about it some more, I realize there's more to the point you made than I originally acknowledged. If the balls are touching, one might wonder why, if you pull out one ball from the left and let it drop, you don't get all four of the other balls to bounce up together to the right and the single ball to bounce elastically back to the left, as would happen if the balls were welded together. – Mark Eichenlaub Jan 30 '11 at 19:00
show 4 more comments
First, I have a free iPhone app for Newton's cradle - called Kinetic Balls - which has about 500 skins. ;-)
The events in the cradle are composed of many steps although they occur quickly after each other. In particular,
if you have five balls, the first ball starts by colliding with the second ball, the second ball hits the third ball, the third ball hits the fourth ball, the fourth ball hits the fifth ball. OK?
So let's study what happens when the first ball from the left hits the second ball from the left. The speed of the first ball is $v$ before it collides with the second one. Now, it simplifies things to look at the collision from the center-of-mass system of the first two balls.
Needless to say, the center of mass is moving by the speed which is the average of the speeds of the first two (equally heavy) balls. Because the first one is moving by $v$ to the right and the second one has the speed equal to $0$, the average is $v/2$, OK?
Before the collision, the first ball is moving by the speed $v-v/2=v/2$ relatively to the center-of-mass frame, and the second ball is moving by the speed $0-v/2=-v/2$ relatively to the center-of-mass frame.
Now, what happens after the collision? The balls can't penetrate through each other so their relative speed has to change the sign: the speeds get simply reversed. They change the sign. The situation is completely symmetric with respect to the exchange of the two balls combined with the reflection of "left" and "right" in the space. Consequently, the two final velocities (in the center-of-mass frame) must still be equal up to the opposite sign. The overall magnitude is guaranteed to be the same as before the collision, to conserve the kinetic energy $2\times mv^2/2$. The signs have to be opposite to one another - but also opposite to the initial ones because the balls can't penetrate one another.
So in the center-of-mass system, after the collision, the left ball will be moving by the speed $-v/2$ and the right ball will be moving by $+v/2$. Transforming back from the center-of-mass frame to the lab frame, we have to add $v/2$ again. So the first ball final speed will be zero, while the second ball's final speed will be $v/2+v/2=v$. OK?
Now, the second ball is approaching the third ball at speed $v$ while the third ball is at rest. Take these two balls' rest frame.
Now, repeat the fairy-tale above four times - I could do it, it would be very exciting, but I want to save the server's hard disks - and you will end up with the situation in which the first four balls are at rest and the fifth one is moving by the speed $v$ to the right.
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Thanks Lubos. Very insightful. – jack mosevich Jan 21 '11 at 22:50
Always pleasure, Jack. – Luboš Motl Jan 22 '11 at 8:51
I forgot to say that even if the balls are touching each other, the answer "1,2,3,4" at rest and "5" taking the original speed of "1" is the only right answer. Any distribution of the momentum among several balls would lower the total energy and violate its conservation law. I guess that it's what Phil is explaining in more detail. Well, the momentum $p$ is distributed to $k_ip$ where $k_i$ sum up to one. Then the energy is the sum of $k_i^2$ times the initial energy, but unless all of the $k_i$'s are zero but one of them which is one, the sum of squares is smaller than one - energy is lost. – Luboš Motl Jan 22 '11 at 8:53
You've made it pretty clear why, if you pull back one ball, one ball comes out. Since it only discussed that case, though, your answer doesn't address the question, which is why the same number of balls come out as go in. It should be easy to adapt the discussion to cover that point. – Mark Eichenlaub Jan 30 '11 at 9:52
You are modelling a system with one ball released, whereas the OP asked for one with two, so you are not answering the question. – Sklivvz♦ Jan 30 '11 at 9:55
Given that the initial balls do not bounce back, the original question can be answered directly as the top response shows, but he needed to specify that he was assuming the other balls do not bounce back or otherwise move. I can explain it without math: more or fewer balls can't come off at a slower or faster speed because energy and momentum must be conserved, and they are each a different function of velocity. If you change the velocity, and if no other balls move, then you would be violating one of the principles of conservation. For example, if only 1 ball came off with twice the velocity, momentum would be the same (conserved) as the incoming 2 balls, but there would be twice as much energy in the outgoing ball as there was in the 2 incoming balls. Conversely, if 4 balls came off with 1/2 the velocity, momentum would be conserved, but the energy would be half of what was put it in.
However, it is much more difficult to answer the question if other balls are allowed to bounce back in the other direction.
The answers based on separation are not correct and one poster hinted at how the device can really be explained. It is only a happen accident that the series of independent strikes calculation is correct for multiple balls of equal weight. Imagine 3 balls with the one in the middle that is much heavier. The first ball will bounce off, which is not what happens when they are touching. So the calculation is wrong for a fundamental reason. When they are touching, the middle heavy ball is in a compressed state, not yet pushing back on the first ball to make it bounce back before it is releasing energy to the last ball. This can't be modeled with one striking pair followed by another. To see everything more clearly, model the balls as big, weak springs.
From wiki: "At the point of collision, two shock waves propagate, one forward and one backward. They are carrying all the kinetic energy of the formerly-moving balls as potential energy in the elastic compression of the metal. The wave going backward is reflected forward when it reaches the beginning of the chain of balls, coming behind the wave that was moving forward first. The first wave reaches the end and is reflected back and it collides with the delayed wave at a point symmetrical to the initial point of collision, forcing the balls apart, releasing the potential energy as kinetic energy.[8] This explanation is more complicated if balls of equal weight are given different lengths or if the balls have different weights, but the solution for the final velocities can be found by examining the compression and expansion of the metals as a result of the shock waves. The conservation of energy and momentum are adequate to explain the system if the potential energy and transition time is included in the conservation of energy calculations."
Now to answer the original question: As the other poster said, the shock wave meets at a point symmetrical to the original collision. However, the answer is more complicated if the balls at the end are of equal weight but different in length. The shock waves will not meet to begin with, but the expanding system of metal will eventually cause the equal-weight balls to separate so that the initial balls do not bounce back.
Important Update/Correction:
In looking at this in much more detail, I've found that no one understands how the cradle works. The reflected shock wave theory comes from: http://www.physikdidaktik.uni-karlsruhe.de/publication/ajp/Ball-chain_part1.pdf
but these researchers followed this paper up with this: http://www.physikdidaktik.uni-karlsruhe.de/publication/ajp/Ball-chain_part2.pdf
and then others: http://www-astro.physics.ox.ac.uk/~ghassan/newton_cradle_2.pdf
in which they show the collision time is 10 times slower than sound (shock) wave propagation which indicates "hertzian compression" (not simply shock waves) describes the cradle by including the springiness of the interfaces between the balls. There are 4 interfaces in 5 balls, so energy and momentum give the solution for the final velocity of 2 balls and then the ratio of the 3 subsequent interfaces to the first interface gives the 3 variables needed to solve for the other 3 balls. They apply F=ma to each ball where it is in contact with its 1 or 2 neighbors, subject to a psuedo-spring force (hertzian) in the surface contact of F=kx^1.5 rather than F=kx. This gives a set interdependent differential equations which I'll state for the most general case of different masses and different surface spring constants (youngs modulus) at the end. This may agree with experiment for the case of 1 ball striking 4 that are in contact, but it gives a solution for the case of 2 striking 3 that is not correct. I solved this system of equations in excel and got the same result as this paper: http://www.damtp.cam.ac.uk/user/hinch/publications/PRSLA455_3201.pdf in which ball 4 and 5 velocities would theoretically be 0.80 and 1.14 times the velocity of the 2 incoming balls. The paper asserts that you can observe the difference in velocity, but they fail to point out that this is a 2 times difference in kinetic energy and therefore the height reached by the 5th ball is 2 times more than the 4th, which is not what happens at all. (If 5th ball comes out 30 degrees max, the 4th ball would theoretically reach only 21 degrees, which is not what happens: they both reach 30 degrees at the same time). Here are the F=ma equations that have to be solved with runge-kutta, if the Hertzian theory is correct. I did it in excel.
x is displacement from at-rest point
m1a1= - A*(x1-x2)^1.5
m2a2 = A*(x1-x2)^1.5 - B*(x2-x3)^1.5
m3a3 = B*(x2-x3)^1.5 - C*(x3-x4)^1.5
m4a4 = C*(x3-x4)^1.5 - D*(x4-x5)^1.5
m5a5 = D*(x4-x5)^1.5
where
A=(2/((1/k1)^(2/3)+(1/k2)^(2/3)))^1.5
B=(2/((1/k2)^(2/3)+(1/k3)^(2/3)))^1.5
C=(2/((1/k3)^(2/3)+(1/k4)^(2/3)))^1.5
D=(2/((1/k4)^(2/3)+(1/k5)^(2/3)))^1.5
For plugging into runge-kutta: a=dV(t)/dt and x=t*V(t)
A, B, C, D are the net effective spring constant between each pair of balls for the surface compression.
So for 2 balls or more balls striking the others, no one knows how it works. Even as late as 2004, these papers were incorrectly described as the complete solution: http://www.maths.tcd.ie/~garyd/Publications/Delaney_2004_AmJPhys_Rocking_Newtons_Cradle.pdf
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– steve Jul 7 '11 at 15:50
OK, after all that I think I found the answer: in a real cradle there must be at least 40 microns separating the balls in order to treat them as simple independent collisions that can be solved simply with conservation of energy and momentum. This is dispersion free energy propagation. I can combine this initial separation with the Hertzian differential equations and get the right answer, but the Hertzian complexity is redundant if they are indeed separated by at least 40 microns and the initial velocity is less than 1 m/s. Less than 40 microns requires the full hertzian solution – steve Jul 7 '11 at 20:53
People answering questions like this make the dreadful mistake of falling back to the simplistic maths found in physics books, forgetting that the simplistic maths does not represent either atomic mechanisms or the actual way in which momentum works in the real world.
Newton's cradle forces us to consider some very interesting consequences of physics in the real world. The fact that the cradle uses spheres means, for instance, that energy transfer occurs through a surface contact point that (in theory) tends to zero. The spheres are designed to approximate a perfectly elastic material.
Here is a question. What would happen if one sphere at the end was replaced with an identical sized sphere of twice the mass? Lifting this ball would surely be like lifting two ordinary ones, but would the collision lead to two spheres at the other end lifting off?
The solution lies in the theory of waves (shock waves travelling through metal, just as sound waves travel through air). Normally, an n-ball collision causes n-balls to lift at the other end, because the shock wave can be thought of as n-balls long, travelling down the row of balls once the initial collision occurs. It is important to note that the ball collision speed must always be much lower than the speed of propagation of the shock wave through the metal, in order for the cradle to behave as expected.
To go back to the question of the heavier ball, well in this case the greater mass in the same sized sphere implies a higher density, which will change the speed at which a shockwave propagates through the material. What happens when a sound-wave moves between materials of different density? The pitch and the wavelength change. The same process gives us refraction in a prism for the wave that is light.
At an atomic level, the shock-wave concept is going to be difficult to conceive, since the energy in this case enters each new sphere through a tiny point of contact. The 'width' of the energy travelling down the line of balls will determine how many balls depart at the other end, the energy being essentially kinetic in nature. Inertia prevents the kinetic impulse from significantly displacing the balls until only the end balls contain the kinetic energy, and thus can begin moving.
Every physicist should have one mantra they repeat over and over! "The model is not reality." The model is a mathematical tool we construct to predict behaviour under certain constraints (constraints we may not even as yet understand). We see people struggle with the Newton Cradle problem because they insist on believing that simplistic models must never be superseded by more sophisticated understanding. Just look at the nonsense that demands the balls have a minimum gap between one another so the simple model can even be applied in the first place. Look at the fools who think it is a n-ball problem (disallowing balls of differing sizes or density), as if the underlying physics has anything whatsoever to do with macro-collections of trillions of atoms in a particular Human pleasing shape.
Wave theory is fundamental to modern physics. People who care to contemplate the dynamics of moving bodies should always be prepared to switch to a deeper understanding of energy propagation through various materials when the problem clearly switches from very simple mathematical models, to models that must take into account underlying mechanisms caused by the nature of atoms (and their bonds). Newtonian physics at the macro level will only provide correct solutions if the key assumptions remain valid. Newton's cradle is an example of an experiment where Newtonian physics must be considered at an atomic level instead (the creation of the shock-wave). Macro level maths (a very simplified model of reality) will not give correct results. Hacking the wrong model (by requiring a x micron gap between the balls) to try to get the wrong model to kind of work, maybe, in some very limited set of conditions, is a scientific atrocity.
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Physics Forums
## Questions on Cross Product
First question:
a x b = -b x a
Why is this so?
As I understand, a major purpose of the cross product (if not, the purpose) is to find a third vector that is perpendicular to two other vectors simultaneously. Let's say a x b = c. Shouldn't the answer really be, a x b = +/- c? Since, of course, both c and -c are perpendicular to a and b simultaneously.
The situation of the sqrt(4) = +/- 2 is analogous to this.
Second question:
The cross product is said to be something that only works in 3 dimensional space. In 2D, it is said to be not applicable. Take two vectors in 2D that are just the negative of each other. Obviously a line can be drawn straight up relative to these two vectors that is perpendicular to both (visualize an upside down T). Since the cross product's main purpose is to find a third vector that is perpendicular to two other vectors, can it be said that the cross product fails in 2 dimensional space?
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Quote by doctorjuice a x b = -b x a Why is this so?
This follows easily from the definition. What definition have you been taught?
Quote by doctorjuice As I understand, a major purpose of the cross product (if not, the purpose) is to find a third vector that is perpendicular to two other vectors simultaneously. Let's say a x b = c. Shouldn't the answer really be, a x b = +/- c? Since, of course, both c and -c are perpendicular to a and b simultaneously.
You may have to define the word "should".
Torque is a good example of the use of the cross product in physics. When you use a screwdriver on a screw, you're applying torque that's equal to a cross product. The direction of that vector contains the information about which way you're turning the screw. Angular momentum is another good example. In this case, the direction of the vector corresponds to the direction of the rotation. So that direction isn't irrelevant.
Quote by doctorjuice The cross product is said to be something that only works in 3 dimensional space. In 2D, it is said to be not applicable. Take two vectors in 2D that are just the negative of each other. Obviously a line can be drawn straight up relative to these two vectors that is perpendicular to both (visualize an upside down T). Since the cross product's main purpose is to find a third vector that is perpendicular to two other vectors, can it be said that the cross product fails in 2 dimensional space?
You may also have to define the word "fail".
I can't say that the cross product fails in 2-dimensional space, since there is no cross product on ##\mathbb R^2##. But you are right that there are no vectors that are perpendicular to two linearly independent vectors.
Recognitions: Homework Help 0=(a+b)x(a+b)=a x a+a x b+b x a+b x b=a x b+b x a if a x b = c t c is perpendicular to both a and b for any t The cross product is chosen to preserve distance and to be right handed. There are generalizations of the cross product to other spaces, none of them have all the properties of the n=3 case, one reason is that several function are the same only for this case for example consider functions $$V \times V \rightarrow V \\ V \times V \rightarrow V^{n-2} \\ V \times V^{n-2} \rightarrow V \\$$ only when n=3 can we hope these functions are the same There is the Seven dimensional cross product which is the closest.
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## Questions on Cross Product
Quote by doctorjuice The situation of the sqrt(4) = +/- 2 is analogous to this.
The definition is that the square root is always nonnegative. Thus by definition, we have $\sqrt{4}=2$. We do not define $\sqrt{4}=-2$ or $\sqrt{4}=\pm 2$. We only give one value to the square root because we want it to be a function.
Blog Entries: 5 Recognitions: Homework Help Science Advisor Similarly, for the cross product, the definition is such that the cross product is "right-handed". That is, if you take a right handed coordinate system (turning the fingers of your right hand from the x to the y-axis your thumb points along the z-axis) then ex x ey = ez (or, in alternative notation, $\hat i \times \hat j = \hat k$) rather than -ez.
Quote by Fredrik This follows easily from the definition. What definition have you been taught?
I've been taught the same definition, I was just asking why it was defined this way, I didn't see the reason for it at first. There are actually a couple reasons, I believe (and I would like to hear confirmation or refutal of these reasons):
1) If a x b = +/-c (instead of +c), then the expression a x b would be ambiguous. By itself, this is not a major problem but once complicated calculations with cross products occur this problem becomes very large.
Analogous to sqrt(4), when it is part of complex calculations, if it were to ambiguously mean +/-2, then the whole mathematical expression of a complex operation involving multiple sqrt(4) terms would be very ambiguous and you could come up with many different answers for the same problem.
2) Even if the sign of c does not affect the end result in mathematics, in physics it does, and the information is very important in physics (as you said in other parts of your post).
You may also have to define the word "fail". I can't say that the cross product fails in 2-dimensional space, since there is no cross product on ##\mathbb R^2##. But you are right that there are no vectors that are perpendicular to two linearly independent vectors.
I think you misunderstood this part of my post. I was referring to two vectors in 2D, that go in opposite directions (so a and -a would be a pair of such vectors). If a vector is drawn straight up in 2D relative to these two vectors (think an upside-down T, with the left part being vector a, right part being vector b (or -a) and the up part being a vector perpendicular to both, vector c) then that vector is perpendicular to both a and b in 2D space simultaneously.
Since a major purpose of the cross-product is to find a vector that is perpendicular to two vectors simultaneously, I was making the assertion that the cross product fails at fulfilling its purpose in 2D space, for this specific situation.
I would love to hear yours and anybody else's thoughts on what I have said here.
Blog Entries: 5 Recognitions: Homework Help Science Advisor 1) is definitely true. It's just a matter of convention, and just like we chose sqrt(4) to give 2 and not -2 because that's often easier, we also choose (1,0,0) x (0,1,0) to be (0,0,1) and not (0,0,-1) because it's usually easier. You could view the definition of sqrt as just some function, which happens to have the property that it squares to its argument without claiming that there are no other such numbers, you could see the definition of the cross product in a similar way. I'm not sure about 2) - I think quite the opposite is true. It does not matter which way you define it, but if you get it the other way around you might also want to change other conventions so you don't end up writing minus signs all the time.
A formal cross product is defined for $\mathbb{R}^3$ and $\mathbb{R}^7$. There is a projective cross product for $\mathbb{R}^2$, which returns a directed scalar. Given $\mathbf{u}=\langle u_1,u_2\rangle$ and $\mathbf{v}=\langle v_1,v_2\rangle$, the "cross product" is $\mathbf{u}\times_2\mathbf{v}=(\langle u_1,u_2,0\rangle\times_3\langle v_1,v_2,0\rangle)\cdot\langle0,0,1\rangle$, where $\times_n$ is the cross product in $\mathbb{R}^n$.
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Quote by doctorjuice I think you misunderstood this part of my post.
Sorry about that. I just saw that you were asking about cross products in 2D and jumped to the wrong conclusion about what you wanted to ask. I guess I should have actually read the question.
Quote by doctorjuice I was referring to two vectors in 2D, that go in opposite directions (so a and -a would be a pair of such vectors). If a vector is drawn straight up in 2D relative to these two vectors (think an upside-down T, with the left part being vector a, right part being vector b (or -a) and the up part being a vector perpendicular to both, vector c) then that vector is perpendicular to both a and b in 2D space simultaneously. Since a major purpose of the cross-product is to find a vector that is perpendicular to two vectors simultaneously, I was making the assertion that the cross product fails at fulfilling its purpose in 2D space, for this specific situation. I would love to hear yours and anybody else's thoughts on what I have said here.
You could certainly define a "product" of two linearly dependent vectors in ℝ2 that has a vector perpendicular to both as the result. For example, you could define
$$(a_1,a_2)(b_1,b_2)=\operatorname{sgn}(a_1b_1)(a_2,-a_1)$$ where sgn is the sign function. But this seems rather pointless. This product also has very little in common with the cross product on ℝ3.
Recognitions: Homework Help 1&2)The convention does not matter, but using multiple conventions would lead to errors. This is the same in mathematics and physics. As for finding a vector that is perpendicular, as I mentioned above the cross product serves several purposes simultaneously in 3-space this cannot be done in other spaces. One generalization gives the (unique up to scalar multiplication) vector perpendicular to n-1 given vectors. In 2-space this means we take one vector and return a vector perpendicular to it. In 11-space this means we take ten vectors and return a vector perpendicular to it. Only in 3-space can we take two vectors and return one vector perpendicular to it.
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| | Introductory Physics Homework | 1 |
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http://mathhelpforum.com/pre-calculus/20426-composite-function.html
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Thread:
1. Composite function
If $f(x)=\frac{1}{x-1}$ for $x>2$ and $f(x)=x+1$ for $2 \geq x$, find $f(a-2)$ in terms of a.
Normally I would simply substitute a in, but I don't know if that is the right way to go about things?
I think I should have considered pre-calc a little more carefully before going into calc
2. Originally Posted by DivideBy0
If $f(x)=\frac{1}{x-1}$ for $x>2$ and $f(x)=x+1$ for $2 \geq x$, find $f(a-2)$ in terms of a.
Normally I would simply substitute a in, but I don't know if that is the right way to go about things?...
Hello,
of course your plan is OK - but you have to plug in the term (a-2) into the conditions too:
$f(x)=\left \{\begin{array}{lr}\frac{1}{x-1} & x>2 \\x+1 & x\leq 2\end{array}\right.$
Now you get:
$f(a-2)=\left \{\begin{array}{lr}\frac{1}{a-2-1} & a-2>2 \\a-2+1 & a-2\leq 2\end{array}\right.$ After simplifying a little bit you have:
$f(a-2)=\left \{\begin{array}{lr}\frac{1}{a-3} & a>4 \\a-1 & a\leq 4 \end{array}\right.$
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http://www.mathclique.com/2012/03/two-step-equation-word-problems.html
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## Linkbar
### Two-Step Equation Word Problems Worksheets
Two–step equation word problem is a simple word problem involving both multiplying a certain factor to the unknown variable or quantity and either addition or subtraction of a constant to this product. To solve two-step equation word problems, follow the following steps:
1. Assign a variable for the unknown quantity or number.
2. Set up the equation that best translate the word problem.
3. Solve the defining equation.
4. State the final answer.
Below are examples illustrating how to solve one-step equation word problem:
1. When 5 is added to twice a number, the result is 13. What is the number?
Solution:
Let $$a$$ be the number. The equation is $$2a+5=13$$. Solving for $$a$$ we get
$\begin{array}{c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c} &2a+5&=&13\\ &2a+5-5&=&13-5 \\&2a&=&8\\ &\therefore a&=&4 \end{array}$
Therefore, the number is 4.
2. If 3 is subtracted from half a number, the result is 24. Find the number.
Solution:
Let $$n$$ be the number. The equation is $$\displaystyle \frac{n}{2}-3=24$$. Solving for $$n$$ we get
$\begin{array}{c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c} &\displaystyle \frac{n}{2}-3&=&24\\ &\displaystyle \frac{n}{2}&=&21 \\ &\therefore n&=&42 \end{array}$
Therefore, the number is 42.
3. Mary has $$3x$$ pesos in her wallet. If her mother would give her additional 200 pesos, she will have 230 pesos in all. How much money does she originally have?
Solution:
Since the unknown amount is already as $$x$$, so the equation is $$3x+200=230$$. Solving for $$x$$ we get
$\begin{array}{c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c} &3x+200&=&230 \\ &3x&=&30&\textup{subtract 200 to both sides of the equation} \\ &\therefore x&=&10&\textup{divide both sides by 3} \end{array}$
Therefore, Mary originally has P10 in her wallet.
Practice Exercises
Practice Exercises
1. If an item is subject to 20% discount, what is the selling price if it was originally marked at 500 pesos?
2. If 10 is subtracted from three times number, the difference would be 5. What is the number?
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http://rip94550.wordpress.com/2009/10/26/color-from-spectrum-to-tristimulus/
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# Rip’s Applied Mathematics Blog
## Color: from spectrum to tristimulus
October 26, 2009 — rip
## Introduction
Edit 30 Oct 2009:
This is the first post about an example on p. 160 of Wyszecki & Stiles (see the bibliography).
and find another edit about the components wrt the dual basis.
End edit.
As always, if my chatter is confusing, just start following the examples.
In a sense, what I am about to do now is one of the most important color calculations I will show you.
How do we get from a spectrum to tristimulus values?
I am going to get from a spectrum to XYZ values, and then to xyz values. Recall that XYZ come from xyz bar tables, which in turn come from rbar, gbar, bbar tables by a change of basis. Then xyz values come from XYZ by making the sum = 1.
For more about the CIE, see the previous post.
The question before us is to get from a spectrum to XYZ (and then to xyz).
What we are talking about is the perception of color, rather than the reproduction. Sort of. We wil use “the standard observer” rather than my eyes or your eyes, but we are going to translate an arbitrary spectrum into 3 scalar values, which can then be transformed to some RGB values in a color space. But not in this post.
This example comes from w&s (Wyszecki & Stiles, in the bibliography), page 160. Believe it or not, I am going to work it four ways — but I will give you the full details for only two of them, pictures for all four of them.
The original problem uses data at 10 nm intervals, and it uses the 1964 CIE. Working this example as they did showed me that there were 3 typos in their solution.
First, however, I will work the problem using the data at 20 nm intervals, with the 1964 CIE. This will show you explicitly how the calculations proceed. At 20 nm intervals, I can display the data itself.
Second, I will work the original problem as they posed it. I will not get their answers, so I will also show you their typos.
Third and fourth, I will again work the problem at both 10 nm and 20 nm intervals, but using the 1931 CIE. As will be my ongoing custom, I will show you all the numbers only for the 20 nm case. Ultimately, what choice we make isn’t going to matter a lot. And that’s worth knowing.
We are given a reflectance spectrum for a plastic object illuminated by daylight. In fact, we will use “standard illuminant D65″ for “daylight”. The “reflectance spectrum” is the reflection from a perfectly uniform — perfectly equal energy — illuminant.
Let me say that another way. We are going to assume that the object is illuminated by one spectrum (“daylight” D65) and that the reflected spectrum which we perceive is the combination of the D65 spectrum and the so-called reflectance spectrum of the object.
The reflectance spectrum is a property of the object; by combining it with any arbitrary illumination spectrum, we get the reflected spectrum.
We are asked, in the given example, to find the XYZ tristimulus values, and the xy chromaticity coordinates, using the 1964 CIE color matching functions. That is, we will use the 1964 xyzbar tables.
(I didn’t plan to show this to you, but prudence dictated that I at least work it out. When it turned out that they had slightly wrong answers… well, I have to show it to you.)
Here’s what we will do. We will find the reflected spectrum, i.e. what is reflected from the object illuminated by D65 rather than by an equal energy illumninant. Then we will apply the xyzbar color matching functions to the reflected spectrum to get XYZ tristimulus values. Then we will get the normalized xyz values.
For more about the color matching functions, our matrices A for each problem, see the first Cohen post, and the second Cohen post.
## CIE 1964 at 20 nm
Here is the reflectance spectrum $\beta\$, from 380 to 780 at 20 nm intervals. As will be my custom, I am taking a subset of their values — I am not averaging the numbers. That is, their example provides data every 10 nm, but I am using half of that given data. For one thing, as I said, that gives us matrices of manageable size.
$\beta = \begin{array}{c} 0.102 \\ 0.348 \\ 0.46 \\ 0.475 \\ 0.462 \\ 0.454 \\ 0.478 \\ 0.564 \\ 0.663 \\ 0.691 \\ 0.693 \\ 0.79 \\ 0.845 \\ 0.853 \\ 0.853 \\ 0.853 \\ 0.852 \\ 0.849 \\ 0.79 \\ 0.712 \\ 0.625\end{array}$
Here’s a picture of that reflectance spectrum:
A word about scaling. It is appropriate that the reflectance spectrum be less than, or on the order of magnitude of, 1. At any given frequency, the amount reflected can range from significantly less to significantly more (since light absorbed at one frequency can be emitted at another) than the incident spectrum — but we dont want to be multiplying by a hundred. That would be bad.
Here is the D65 spectrum, from 380 to 780 nm at 20 nm intervals:
$D65 = \begin{array}{c} 50. \\ 82.8 \\ 93.4 \\ 104.9 \\ 117.8 \\ 115.9 \\ 109.4 \\ 104.8 \\ 104.4 \\ 100. \\ 95.8 \\ 90. \\ 87.7 \\ 83.7 \\ 82.2 \\ 78.3 \\ 71.6 \\ 61.6 \\ 75.1 \\ 46.4 \\ 63.4\end{array}$
Another word about scaling. The D65 spectrum ranges from 50 to more than 100. I have no idea what units it is measured in — but the reflected spectrum will be in the same units, whatever they are.
Here is a picture of that illumninant spectrum:
As I said, the scale is 100 times that of the reflectance spectrum. This doesn’t matter for the computations. If, however, I want to show both of these on the same axes, I will need to scale these values — but for the graph only, not for the computations.
Here are the 1964 xyzbar tables from 380 to 780 nm, at 20 nm intervals. This is the A matrix for this problem.
$A = \left(\begin{array}{ccc} 0.0002 & 0. & 0.0007 \\ 0.0191 & 0.002 & 0.086 \\ 0.2045 & 0.0214 & 0.9725 \\ 0.3837 & 0.0621 & 1.9673 \\ 0.3023 & 0.1282 & 1.7454 \\ 0.0805 & 0.2536 & 0.7721 \\ 0.0038 & 0.4608 & 0.2185 \\ 0.1177 & 0.7618 & 0.0607 \\ 0.3768 & 0.962 & 0.0137 \\ 0.7052 & 0.9973 & 0. \\ 1.0142 & 0.8689 & 0. \\ 1.124 & 0.6583 & 0. \\ 0.8563 & 0.3981 & 0. \\ 0.4316 & 0.1798 & 0. \\ 0.1526 & 0.0603 & 0. \\ 0.0409 & 0.0159 & 0. \\ 0.0096 & 0.0037 & 0. \\ 0.0022 & 0.0008 & 0. \\ 0.0005 & 0.0002 & 0. \\ 0.0001 & 0. & 0. \\ 0. & 0. & 0.\end{array}\right)$
And here is a picture of the color matching functions:
Those are our ingredients: the reflectance spectrum, the illiminant spectrum, and the color matching functions.
Now, let’s get the reflected spectrum. We multiply the values pointwise… i.e. at every wavelength, what is actually reflected is the product of the reflectance spectrum and the illuminant spectrum.
(Perhaps I should be calling this the perceived spectrum, since it is what our standard observer is seeing. I should not call it the incident spectrum, because that would be a sensible alternative to “the illimination spectrum”, in this case D65. Incident illumination on an object gives us a reflected spectrum, which our standard observer perceives.)
The result of multiplying the reflectance spectrum $\beta$ and the illuminant spectrum D65 is
$S = \begin{array}{c} 5.1 \\ 28.8144 \\ 42.964 \\ 49.8275 \\ 54.4236 \\ 52.6186 \\ 52.2932 \\ 59.1072 \\ 69.2172 \\ 69.1 \\ 66.3894 \\ 71.1 \\ 74.1065 \\ 71.3961 \\ 70.1166 \\ 66.7899 \\ 61.0032 \\ 52.2984 \\ 59.329 \\ 33.0368 \\ 39.625\end{array}$
and the 3 spectra look like this (after I divide the D65 and S by 100, otherwise the reflectance spectrum is effectively the x-axis):
Let’s be clear: the reflectance spectrum is in green, the illuminant spectrum is in blue, and the product, at each wavelength, is the reflected spectrum, in black.
Before we deal with the reflected spectrum, however, I want to deal with the illumination itself. If we apply the color functions to the D65 spectrum, that is we compute
$A^T\ D65\$,
we get
{554.645, 582.565, 631.147}.
Are those our tristimulus values? No way. These numbers depend on the scale of the spectrum, and the number of components of the vectors. Somehow we have to normalize them, and we will do that for the total energy received.
As an aside, which I will try to remember to repeat in subsequent posts…. We have spectra and color-matching functions which live in vector spaces. The result of applying color matching functions to spectra is a set of 3 scalars, i.e. 3 real numbers. But our vector spaces are not bounded spaces, so those 3 numbers are not, in principle, bounded. They are simply the components of a spectrum wrt a basis. Edit: oops. Careful! They are components of the fundamental of the spectrum. End edit. OTOH, we have color spaces, such as RGB, in which the RGB values are bounded (whether between 0 and 1, or 0 and 100, or 0 and 255, they are limited).
We need to move from the unbounded results of vector operators to the bounded values of color spaces.
But how do we do that?
Well. Recall that ybar, the second column of A, is the photopic response (the total response of the cones) of the standard observer. This is incredibly useful: the second column of AT tells us the total energy perceived. If we were doing calculus, we would compute the areas under all three curves. But we’re doing linear algebra, and the appropriate number is the middle one.
We get XYZ values for the illumnination perceived by the observer by dividing those 3 numbers by the middle number, 582.565:
XYZ = {0.952075, 1., 1.08339}.
Let’s keep going with this. How do we get xyz coordinates from XYZ?
Normalize the sum of the XYZ to 1. The sum is 3.03547 … so we get that x,y,z are
xyz = {0.31365, 0.329438, 0.356911}.
Since all three values are close to 1/3, this is a plausible number for illumination.
Now we can do the same thing with the reflected spectrum. That is, we apply the matrix $A^T$ to the reflected spectrum S:
{386.796, 381.417, 293.87}.
We are not going to divide by this middle number — this is the reflected spectrum, not the illimination spectrum. We need to divide by the same number as before, i.e. by the photopic response to the illumination, not the photopic response to the reflection. That number was 582.565, and now we divide our 3 values by that to get X,Y,Z…
XYZ = {0.663953, 0.654719, 0.504441}.
And, as before, we compute xyz values, too: we normalze the XYZ so that their sum is 1. Since the sum is 1.82311 … we get
xyz = {0.364186,0.359122,0.276692}.
This isn’t all that far from the illuminant.
Those dots are the outline of the 1964 chromaticity diagram. I got it the same way as the 1931, just with different data. Get the xyz tables — the ones where the rows add up to 1 — and just plot y versus x.
The red, green, and blue dots show the pure primaries for the 1964 standard. The white dot is (x,y) for D65, and the yellow dot is (x,y) for the reflected sprectrum. (Oh, I don’t think the reflection is yellow, but it’s rather closer to the light source than to much else.)
## CIE 1964 at 10 nm
Now let me show you the pictures, and the answers, for 10 nm intervals. In fact, let me juxtapose the new pictures (on the right) with the old (on the left):
Here are the old and new reflectance spectra:
Here are the D65 spectra:
And here are the color matching functions (xyzbar tables) from 380 to 780 nm The data on the right is the A matrix for this problem.
We have the same ingredients: the reflectance spectrum, the illiminant spectrum, and the color matching functions.
Now, let’s get the reflected spectrum. We multiply the values pointwise… i.e. at every wavelength, what is actually reflected is the product of the reflectance spectrum and the illuminant spectrum.
The result of multiplying $\beta$ and D65 together looks like this (after I divide the D65 and S by 100, otherwise the reflectance spectrum is effectively the x-axis):
As before, the reflectance spectrum is in green, the illuminant spectrum is in blue, and the product, at each wavelength, is the reflected spectrum, in black.
First, we apply the color functions to the D65 spectrum, that is we compute
$A^T D65\$,
and we get
{1102.11, 1162.02, 1247.78}.
Here we see quite clearly that these numbers depend on the scale of the spectrum and the number of values: they are about double what the previous values were. Oh, of course: our vectors are twice as long. (Even though the XYZ and xyz values will be comparable, these initial 3 values will not always be comparable. As I said, they depend on the dimension of our spectra and on the scaling of the illuminant spectrum.)
Once again, we need to scale those numbers, dividing each of them by the middle number, which is the total perceived illumination.
So we get XYZ values for the illumninant by dividing those 3 numbers by the middle number, 1162.02:
XYZ = {0.948445, 1., 1.0738}.
How do we get xy coordinates? Normalize the sum to 1. The sum is 3.02225 … so we get
xyz = {0.313821, 0.330879, 0.3553}.
As before, this is a plausible number for illumination.
Now we do the same thing with the reflected spectrum. That is, we apply the matrix $A^T$ to the reflected spectrum S:
{769.263, 761.189, 581.056}.
Once again, we divide by the same number as before, i.e. by the photopic response to the illumination, not to the reflection. (That number was 1162.02.)
Now we divide our 3 values by that to get X,Y,Z…
XYZ = {0.662005, 0.655057, 0.50004}.
And, as before, we compute xy values, too: we normalze the XYZ so that their sum is 1. Since the sum is 1.8171 … we get
xyz = {0.364319, 0.360495, 0.275185}.
You can’t tell much from these two pictures; the final picture in this post will plot (almost) all of the xy answers. The numbers using 20 nm intervals were not much different: for the XYZ values of the reflected spectrum, we had:
XYZ = {0.663953, 0.654719, 0.504441}.
The corresponding answers using 10 nm intervals were:
XYZ = {0.662005, 0.655057, 0.50004}.
Pretty darn close. 10 nm or 20 nm, both using CIE 1964, not a whole lot of difference.
Now, what were their answers? They got sums of
{758.6, 760.8, 580.8}
… while I had
{769.263, 761.189, 581.056}.
We do not agree. What’s going on? I found that there is a problem in their work at 640 nm.
For a wavelength of 640 nm, we agree on the values of all the ingredients. We agree that the reflectance spectrum value is 0.853 … the D65 illuminant spectrum value is 83.7 … the values of that row of A are {0.4316, 0.1798, 0.}.
The pointwise products of these numbers are…
{30.8146, 12.837, 0.}
but what they show is
30.2 12.6 0.
Two typos. I have no idea why. And they appear to be the only typos pertaining to individual wavelengths. Of course, it is possible that the error is in the reflectance spectrum instead, but I have no other data-source to compare that to. In any event, it is clear that their entries (30.2 and 12.6) are not in fact the product of the inputs they show.
There’s another problem. I pasted the column that leads to the X-column-sum into a spreadsheet and computed the sum. For the data they have shown, the first sum is really 768.6 rather than 758.6 . That’s the third typo.
(In addition, they round off at each wavelength, before adding the columns: this leads to some round-off error. That’s why we differ on the sum that leads to Z; no typos there, they just rounded before adding.)
But, where they show X,Y,Z as (change of scale, decimals versus percentages; ignore it) …
XYZ = {65.3, 65.5, 50.}
the correct answers are
XYZ = {66.2005, 65.5057, 50.004}
and even using 20 nm intervals instead of 10, I got
XYZ = {66.3953, 65.4719, 50.4441}.
For x,y, they show
xy = {0.3612, 0.3623}
but correct (at 10 nm) is
xy = {0.364319, 0.360495}.
Just working at half the resolution (20 nm), I got
xy = {0.364186, 0.359122}.
I daresay their mistakes are worse than my cutting the resolution in half.
## CIE 1931 10 nm
To switch from 1964 at 10 nm to 1931 at 10 nm requires only that I change the A matrix. (I have everything else in memory at 10 nm intervals. The reflectance, reflected, and D65 spectra do not change.)
Let me show 1931 on the right, for comparison with the 1964 color matching functions on the left. Note that the vertical scales are different.
Although they have not changed, I show again the reflectance spectrum and the illuminant spectrum for this case:
Go ahead, curse me out. After showing different cases side-by-side, I have just switched and shown side-by-side graphs for two different things for one and the same case.
Those are our ingredients: the reflectance spectrum, the illiminant spectrum, and the color matching functions. Only the color functions have changed, from 1964 to 1931.
Now, the pointwise product of the illuminant and reflectance spectra will be exactly the same reflected spectrum.
Let’s be clear: the reflectance spectrum is in green, the illumninant spectrum is in blue, and the product, at each wavelength, is the reflected spectrum — in black.
As before, we apply the color matching functions to the D65 spectrum, that is we compute
$A^T\ D65\$,
getting
{1004.46, 1056.89, 1150.02}.
These numbers are comparable to the previous case: we are using vectors of the same dimension.
Once again, we need to scale those numbers, dividing each of them by the middle number, which is the total perceived illumination.
So we get XYZ values or the illumninant by dividing those 3 numbers by the middle number:
XYZ = {0.950398, 1., 1.08812}.
To get xy coordinates, we normalize the sum to 1. The sum is 3.03852 … so we get
xyz = {0.312783, 0.329108, 0.358109}.
Now we do the same thing with the reflected spectrum. That is, we apply the matrix $A^T$ to the reflected spectrum S:
{707.724, 707.564, 537.101}.
Once again, we divide by the same number as before, i.e. by the photopic response to the illumination, not to the reflection. That number was 1056.89 .
So we divide our 3 values by that to get X,Y,Z…
XYZ = {0.669631, 0.669479, 0.508191}.
And, as before, we compute xy values, too: we normalze the XYZ so that their sum is 1. Since the sum is 1.8473 … we get xyz as
xyz = {0.362491, 0.362409, 0.275099}.
Again, this isn’t all that far from the illuminant. We’re getting very similar answers.
The CIE boundaries look pretty similar, but the triangles — the gamut of the 3 primaries — are different. And recall that the line x + y = 1 is tangent to the right-side boundary; furthermore, for the 1931 CIE, one side of the triangle is very, very close to the tangent line.
## CIE 1931 20 nm
Having shown you pictures with the 1931 CIE and 10 nm intervals, let me close by showing you the numerical data, as well as pictures, for the 1931 CIE 20 nm solution.
To switch to 20 nm intervals, I need to rebuild the color matching functions. Here’s the data:
$A = \left(\begin{array}{ccc} 0.0014 & 0. & 0.0065 \\ 0.0143 & 0.0004 & 0.0679 \\ 0.1344 & 0.004 & 0.6456 \\ 0.3483 & 0.023 & 1.7471 \\ 0.2908 & 0.06 & 1.6692 \\ 0.0956 & 0.139 & 0.813 \\ 0.0049 & 0.323 & 0.272 \\ 0.0633 & 0.71 & 0.0782 \\ 0.2904 & 0.954 & 0.0203 \\ 0.5945 & 0.995 & 0.0039 \\ 0.9163 & 0.87 & 0.0017 \\ 1.0622 & 0.631 & 0.0008 \\ 0.8544 & 0.381 & 0.0002 \\ 0.4479 & 0.175 & 0. \\ 0.1649 & 0.061 & 0. \\ 0.0468 & 0.017 & 0. \\ 0.0114 & 0.0041 & 0. \\ 0.0029 & 0.001 & 0. \\ 0.0007 & 0.0003 & 0. \\ 0.0002 & 0.0001 & 0. \\ 0. & 0. & 0.\end{array}\right)$
and here are the graphs (20 nm and 10 nm side by side):
I need to restore the reflectance spectrum $\beta$ and the illuminant spectrum. As you recall, they were
$\beta = \begin{array}{c} 0.102 \\ 0.348 \\ 0.46 \\ 0.475 \\ 0.462 \\ 0.454 \\ 0.478 \\ 0.564 \\ 0.663 \\ 0.691 \\ 0.693 \\ 0.79 \\ 0.845 \\ 0.853 \\ 0.853 \\ 0.853 \\ 0.852 \\ 0.849 \\ 0.79 \\ 0.712 \\ 0.625\end{array}$
and
$D65 = \begin{array}{c} 50. \\ 82.8 \\ 93.4 \\ 104.9 \\ 117.8 \\ 115.9 \\ 109.4 \\ 104.8 \\ 104.4 \\ 100. \\ 95.8 \\ 90. \\ 87.7 \\ 83.7 \\ 82.2 \\ 78.3 \\ 71.6 \\ 61.6 \\ 75.1 \\ 46.4 \\ 63.4\end{array}$
The reflected spectrum was the pointwise product of those, namely
$S = \begin{array}{c} 5.1 \\ 28.8144 \\ 42.964 \\ 49.8275 \\ 54.4236 \\ 52.6186 \\ 52.2932 \\ 59.1072 \\ 69.2172 \\ 69.1 \\ 66.3894 \\ 71.1 \\ 74.1065 \\ 71.3961 \\ 70.1166 \\ 66.7899 \\ 61.0032 \\ 52.2984 \\ 59.329 \\ 33.0368 \\ 39.625\end{array}$
Those are our ingredients: the reflectance spectrum, the illuminant spectrum, and the color matching functions. Only the color functions have changed, from 1964 to 1931.
The product of illumninant spectrum and reflectance spectrum is the same as shown before for 20 nm intervals:
As before, the reflectance spectrum is in green, the illumninant spectrum is in blue, and the product, at each wavelength, is the reflected spectrum — in black.
As before, we apply the color matching functions to the D65 spectrum, that is we compute
$A^T\ D65\$,
getting
{506.693, 529.764, 581.089}.
These numbers are smaller than the previous case: we are using vectors of half the dimension.
Once again, we need to scale those numbers, dividing each of them by their middle number, which is the total perceived illumination.
So we get XYZ values or the illumninant by dividing those 3 numbers by the middle number, getting:
XYZ = {0.956451, 1., 1.09688}.
We get xy coordinates by normalizing the sum to 1. The sum is 3.05333 … so we get
xyz = {0.313248, 0.327511, 0.359241}.
Now we do the same thing with the reflected spectrum. That is, we apply the matrix $A^T$ to the reflected spectrum S, getting:
{356.817,354.641,271.109}.
Once again, we divide by the same number as before, i.e. by the photopic response to the illumination, not to the reflection. That number was 529.764, so we get XYZ are
XYZ = {0.673539, 0.669432, 0.511754}.
And, as before, we normalze the XYZ so that their sum is 1. Since the sum is 1.85472 … we get
xyz = {0.363148 ,0.360933, 0.275919}.
OK, let’s compare.
Here are the unscaled sums (including their incorrect answers). Reading down, these are my 1964 at 20, then at 10, their answers, my 1931 at 10, then at 20.
$\begin{array}{ccc} X & Y & Z \\ 386.796 & 381.417 & 293.87 \\ 769.263 & 761.189 & 581.056 \\ 758.6 & 760.8 & 580.8 \\ 707.724 & 707.564 & 537.101 \\ 356.817 & 354.641 & 271.109\end{array}$
and the XYZ values scaled by the photopic response…
$\begin{array}{ccc} X & Y & Z \\ 0.663953 & 0.654719 & 0.504441 \\ 0.662005 & 0.655057 & 0.50004 \\ 0.653 & 0.655 & 0.5 \\ 0.669631 & 0.669479 & 0.508191 \\ 0.673539 & 0.669432 & 0.511754\end{array}$
and the xyz values
$\begin{array}{ccc} x & y & z \\ 0.364186 & 0.359122 & 0.276692 \\ 0.364319 & 0.360495 & 0.275185 \\ 0.3612 & 0.3623 & 0.2765 \\ 0.362491 & 0.362409 & 0.275099 \\ 0.363148 & 0.360933 & 0.275919\end{array}$
There’s not a whole lot of difference there, whether we use 1931 or 1964, or 10 nm or 20 nm, or make a few mistakes!
To look at it, here are my 4 computed xy, “white points”, for the illuminant, and the 5 xy points for the reflections.
(I didn’t grab their white point — the x,y coordinates — for D65, so only my 4 computed values are shown.)
The most important single computation in all this was to scale the values by the middle sum, i.e. by the total energy perceived by the observer. To be specific, we applied the ybar function to the illuminant spectrum.
I know, it looks just like the dot product of two vectors. But Cohen taught us that the color functions are really linear functionals. We want to visualize that computation as the matrix product of a row (the ybar) and a column (the D65 spectrum).
Of course, only such a trick as that division by the middle number could distract us from the marvelous computation of 3 numbers in the first place. We get from a spectrum to 3 scalars by applying 3 color matching functions (the transpose of our A matrix) to the reflected spectrum, which is also the spectrum incident on the eyes of the standard observer.
The first 3 numbers we get are, as we saw before and will see again, the (potentially unbounded) components of the incident spectrum wrt the dual basis spectra. The second set of 3 numbers are tristimulus values in XYZ space. The third set of numbers were xyz coordinates for the chromaticity chart.
The first 3 numbers depend on the dimensionality of our spectra, how many samples we have. In fact, we compute two such sets: one for the illuminant and one for the reflection. The middle number of the 3 for the illuminant sets the scale for white = (1,1,1).
The second set, XYZ, is found by dividing the first 3 numbers by that middle number. That is true whether we have XYZ for the illuminant or for the reflection: divide either by the total energy of the illuminant.
The third set, xyz, is found by dividing the XYZ numbers by their sum — instead of by their middle, or anybody’s middle.
I must emphasize that we have seen 4 different choices for the A matrix, and those were all “xyz bar” values! “The” A matrix is hardly unique.
More to come. You better believe it.
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Posted in color. Tags: color. 8 Comments »
### 8 Responses to “Color: from spectrum to tristimulus”
1. Says:
June 16, 2011 at 12:03 pm
dear rip, ive read this article of yours twice now, over a 2 day period. i am not having difficulty understanding the details of your calculations. rather i am struggling with trying to understand what the overall highest level problem is that you are setting out to solve.
my confusion involves the date you refer to as ‘reflectance spectrum’. can you please describe exactly what the ‘reflectance spectrum’ actually means in a real world example?
thanks,
greg
2. rip Says:
June 16, 2011 at 3:40 pm
Hi Greg,
Thanks for asking. I hope I understand the questions:
1- what is the highest level question I am answering? At this point, to find the XYZ tristimulus values of a spectrum seen by a standard observer.
2- what is the reflectance spectrum?
Let me quote Wyszecki & Stiles. “… an object-color stimulus is selected which involves an opaque plastic sample with given spectral reflectance factors $\beta(\lambda)\$… and… CIE standard illuminnant D65.
So we’re computing the XYZ values for a piece of plastic under standard illuminnant D65. What they call “spectral reflectance factors”, I call the reflectance spectrum. It is property of the piece of plastic. It specifies the ratio of the light reflected to the light incident, as a function of wavelength.
Want to solve the very same problem for a piece of wood instead of a piece of plastic? Use the reflectance spectrum for that piece of wood instead of the one for the piece of plastic.
3. Says:
June 17, 2011 at 11:30 am
so the first graph we see on this page would be a typical spectrophotometer plot of (in this case,) the reflectance of a piece of plastic.
the second graph we see on this page would be a typical spectrophotometer plot of (in this case,) the CIE ‘Standard Illuminant D65′.
the third graph represents the CIE 1931 2 degree ‘Standard Observer’.
the overall concept here stated in english (to make it easy for me to understand):
‘if this piece of plastic (exhibiting this reflectance characteristic) is illuminated by a light source (Standard Illuminant D65, in this case), we can estimate the human visual response one might have (when looking at this piece of plastic) – by performing the calculations to follow…’
we do this in 2 steps.
we first calculate the (relative) ‘reflected spectrum’, by multiplying the 1×21 vector (beta) times the 1×21 vector (D65). the answer to this multiplication is a 1×21 vector ‘reflected spectrum’, expressed in relative units.
question 1: Was there anything magical about the choice to divide this multiplied vector product by 100? in fact, if one looks at the 1×21 D65 vector value, we do see one value there thats ‘spot on’ = 100 is this coincidental – or is there some historical background that tells one what frequency value of a D65 spectral content represents 100%?
question 2: i believe for the purpose of this article you chose to plot all 3 curves in the same graph, and you simply made an editorial decision to rescale the multiplied vector product – merely so all 3 curves could be seen in the graph. in my mind, im still comfortable that the multiplied vector producct does not have to exist within any contstrained value domain. what causes me to be lost here in my thinking about this is your statement “and the 3 spectra look like this (after I divide the D65 and S by 100, otherwise the reflectance spectrum is effectively the x-axis)” im lost by the part of the statement that says “otherwise the reflectance spectrum is effectively the x-axis”. might you possibly try to explain this to me?
beyond my questions, it seems that to continue solving the problem, the second step is to convert the reflected spectrum into terms of the standard human visual response – we do this by multiplying the 1×21 vector ‘reflected spectrum’ times the 3×21 matrix CIE 1931 2deg xbar, ybar, zbar color matching functions.
thats enough for my brain to deal with at this time. i am hoping you can confirm my understanding of this (up to this point), and possibly try to help with the 2 questions im still wrestling with.
Greg
we just basically multiply the theoretical light source (as D65 light sources do not exist in reality) is that when one multiplies the ‘spectral reflectance factor of the object’ (beta), times the ‘Standard Illuminant D65′, one reads is ISO 13655 (page 12)times the third
4. Says:
June 17, 2011 at 11:32 am
ps – Yesterday, i just purchased the Wyszecki & Stiles book at Amazon, so once i receive this, i will review the pages you reference here… hopefully this will help as well…
5. rip Says:
June 18, 2011 at 10:26 am
Hi Greg,
“so the first graph we see on this page would be a typical spectrophotometer plot of (in this case,) the reflectance of a piece of plastic.”
Yes, but… the reflected spectrum of that piece of plastic illuminated by an equal-energy source, and you might choose to divide by 100, so that each number represents the fraction of incident light reflected at that wavelength.
“question 1: Was there anything magical about the choice to divide this multiplied vector product by 100? in fact, if one looks at the 1×21 D65 vector value, we do see one value there thats ‘spot on’ = 100 is this coincidental – or is there some historical background that tells one what frequency value of a D65 spectral content represents 100%?”
I recently saw… but cannot locate… a statement that the illuminant data is scaled to 100 at some specific frequency. If I find it again, I’ll post it.
As for the division by 100, it was just to have all three graphs visible on the same scale. See below.
“otherwise the reflectance spectrum is effectively the x-axis”. If you have two curves ranging from 0 to 100, and one curve ranging from 0 to 1, and you plot all three on a scale of 0 to 100, the 0-1 curve will be squished up against the x-axis.
“the second step is to convert the reflected spectrum into terms of the standard human visual response – we do this by multiplying the 1×21 vector ‘reflected spectrum’ times the 3×21 matrix CIE 1931 2deg xbar, ybar, zbar color matching functions.”
You got it.
Thanks for asking. You’re probably not the only reader a bit overwhelmed.
6. Says:
June 20, 2011 at 7:31 am
thanks for explaining your “effectively the x-axis” statement. now i understand what you were aiming for with that one!
thanks for explaining that ‘divide by 100′ was done purely to show all sets of data in the same graph!
as to the question ‘why in D65 standard illuminant did we find one value to be = 100?’, i read since my last post here that there is some notion of our human vision x_bar, y_bar, z_bar (short, medium, long cone response) – that the y_bar (medium cone response) also happens to be useable for overall luminosity when we view any given scene, and that if one were to graph the sensitivity of the y_bar sensitivity – the peak wavelength of our human visual sensitivity MIGHT possibly be the same corellated wavelength that would serve as the 100% mark within any ‘standard illuminant’ (this is my best guess). I believe this is around 560nm. (in fact, if you check your chart for D65 at 20nm steps, you’ll find the 100% mark is at 560nm).
there is only one topic i am now hazy about (in your last post to me). if ok, perhaps we can discuss this one newly brought up topic. your second paragraph…
“Yes, but… the reflected spectrum of that piece of plastic illuminated by an equal-energy source, and you might choose to divide by 100, so that each number represents the fraction of incident light reflected at that wavelength.”
so that i can understand this clearly, i’ll restate this in a more literal way, in the way my brain sees, and understands this. if i make mistakes along the way, please interject to clear up any of my misunderstanding or confusion.
if i look at your first graph (of this page), and i see (what appears in my mind to be) a typical spectral energy distribitution graph – as could be output from any spectrograph driver software being used in ‘relative count’ mode. for the purpose of your ‘downstream’ calculations presented in this page, it seems pretty irrelevent the actual scale used here for the y values (as in later calcs, you merely ‘normalize’ the scale of the multiplied terms). but as you eluded to in this paragraph, that if i wanted to get the y values, for smaller wavelength ‘steps’, i would need to divide the y values presently shown in the chart by ’100′. im struggling with this statement. maybe you made this statement for brevity without going into lots of details to keep it easy to understand. and maybe youll hate me for being so literal here – but i am truly wanting to understand all aspects of this ‘color stuff’ – so thats why i ask here abou tthis. but technically, isnt the idea here that if we graph such a reflectance spectral ‘graph’. that with detail, the idea is that if one were to calculate the ‘area under this graph’ – this area would represent the ‘total’ of the graph (in this case the wavelength steps are at 20 nm). if we wanted to know how much each frequency ‘band’ contributed to the overall ‘total’ response – we would need to (in the case of 380 to 400 nm, for example). calculate the area under the graph (between 380 and 400 nm) and divide this smaller area by the ‘total’ area – the result would be the % contribution of the first band 380 to 400 nm. we could in fact, step through the entire graph for each 20 nm band to do this. we would then ‘double check’ our answer by adding up each % contribution for each band, and the total should = 100%.
we could do this for ANY selection of lower frequency and upper frequency boundaries, using the exact same methodology.
technically, if what i present here is correct, i dont think thats the same as merely dividing the y scale by 100,as you stated – but then perhaps i am not seeing this the same way that you are…
again, thanks for your patience here with me.
by the way, if i have not stated yet. i used to be a photographer (in film days). i used to undersatnd how to control film response pretty well, having access to a testing situation and color densitometer, I used to characterize each of the films that i used, and was able to master the use fo film. now that photography is largely digital – i am wanting to try to understand ALL that is color, so that i can apply that knowledge to trying to better understand and control color digital photography and eventually color management.
Rip, what is your interest here in color? I’m thankful that you have this interest and are so willing to help others.
Greg
• rip Says:
June 21, 2011 at 3:25 pm
Hi Greg,
Let me quote you: “if i wanted to get the y values, for smaller wavelength ‘steps’, i would need to divide the y values presently shown in the chart by ’100′.”
No, that’s not what I meant. I wasn’t talking about changing the step size. Instead, If you measure a reflected spectrum under an equal intensity source, and if the numbers go up to about 100, you may want to divide them by 100 – so that what I call the reflectance spectrum goes up to about 1 — so that when you multiply a source spectrum by the reflectance spectrum you get numbers up to 100, instead of up to 10,000.
7. Says:
June 22, 2011 at 9:28 am
i understand what you meant. again its a form of scaling. thanks.
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http://mathoverflow.net/questions/84465/semiclassical-expansions-of-eigenvalues-of-schrodinger-operators/89237
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## Semiclassical expansions of eigenvalues of Schrödinger operators
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Considering Schrödinger operators $$H(\hbar) = \hbar \Delta + V$$ where $V$ is some potential, perturbation theory tells that the eigenvalues of $H(\hbar)$ are holomorphic on some region containing the positive real axis, but don't necessarily have an analytic continuation to the origin.
However, one tries to find asymptotic expansions in $\hbar$ in $0$, thus hopes that $$\lambda(\hbar) \sim \lambda_0 + \hbar \lambda_1 + \hbar^2 \lambda_2 + \dots.$$ Usually one finds those expansions by WKB-methods, belonging to eigenfunctions that concentrate at the minimums of $V$.
Now, there are a lot of theorems about the behavior of this and statements that tell you that the expansions you get by WKB-methods actually belong to a "true" eigenvalue, but to me it seems that nobody ever tries to answer the question, if every eigenvalue of $H(\hbar)$ (considered as holomorphic function in $\hbar$ on some region $U$ with $0 \in \partial U$) actually admits an asymptotic expansion at $0$. Instead, it seems, that this is always automatically assumed.
But this is per se is not clear to me at all; it means for example that $0$ is not a branch point of $\lambda(\hbar)$.
Do you know about this question? I can't even have a guess if it is true that every eigenvalue admits an asymptotic expansion in the general case or if that is true only if $V$ is "well-behaved", like for example having no nondegenerate minima (and always assuming that it is such that the spectrum is discrete). Neither do I have an idea how one would prove something like that so far.
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## 2 Answers
If your model is the anharmonic oscillator $$V=x^2+x^4$$ one can do a simple unitary transformation based on scaling $x\rightarrow \lambda x$, $\frac{d}{dx}\rightarrow \lambda^{-1} \frac{d}{dx}$ to reduce your problem to that of the analyticity of $\sqrt{\hbar}E_n(\sqrt{\hbar})$ where $E_n(\beta)$ are the eigenvalues of the Hamiltonian $$-\Delta+x^2+\beta x^4\ .$$ This has been studied for instance in B. Simon, "Coupling constant analyticity for the anharmonic oscillator", Ann. Phys. 58 (1970), 76-136. You might find the kind of methods you need in this article or in the ones referring to it, say on Google Scholar.
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There is no specific potential given. Just possibly some condition that it is morse, i.e. that all critical points are nondegenerate. – Kofi Feb 23 2012 at 22:32
@Kofi: sure. But whatever general theorem you are looking for, it should apply to this example. – Abdelmalek Abdesselam Feb 24 2012 at 15:38
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If you look at:
Microlocal WKB expansions by A. Martinez (available through cite seer), you will see that this is a fairly popular subject, where many results have been obtained by Sjostrand et al.
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I didn't now that specific paper, but the older works by Helffer and Sjöstrand. As far as I could see is, that Martinez "only" proves similar results for a wider class of operators. They always start with formal "local" eigenvectors and eigenvalues and then prove that this is the asymptotic expansion of an actual eigenvector. What I am wondering about is the question, if it is not possible to prove by abstract methods that all eigenvalus admit an asymptotic expansion at zero, just by looking at the operator itself. – Kofi Dec 28 2011 at 22:36
Ah, I see... Interesting question... – Igor Rivin Dec 28 2011 at 23:29
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http://mathoverflow.net/questions/84549/do-projective-hypersurfaces-contain-projective-toric-varieties/84567
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## Do projective hypersurfaces contain projective toric varieties?
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Is there an example of a smooth projective hypersurface in $\mathbb{P}^n_k$ ($k=\overline{k}$) that does not contain any projective toric varieties (edit: of positive dimension)? Or is it the case that every such hypersurface will contain a projective toric variety (edit: of positive dimension)?
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A point is an example of a toric variety. Maybe you would like to refine your question? – David Speyer Dec 29 2011 at 21:51
Yes, thanks, I edited. – Mahdi Majidi-Zolbanin Dec 29 2011 at 21:53
5
Let $n=2$, and let the hypersurface be an elliptic curve. Its only subvariety of positive dimension is itself, and it is not toric since all toric varieties are rational. – Alexander Woo Dec 29 2011 at 21:57
## 3 Answers
As Alexander Woo said in a comment, toric varieties are rational. Now, it turns out that projective hypersurfaces have strong hyperbolicity-type properties. This properties have been established by several authors in the last decades.
First, in 1986 Clemens showed that if $X$ is a generic hypersurface of degree $d \ge 2$ in $\mathbb P^{n+1}$, then $X$ does not admit an irreducible family $f\colon\mathcal C\to X$ of immersed curves of genus $g$ and fixed immersion degree $\deg f$ which cover a variety of codimension less than $D = ((2 -2g)/ \deg f) + d - (n + 2)$. As an immediate consequence, one gets, for example, that there are no rational curves on generic hypersurfaces $X$ of degree $d \ge2n + 1$ in $\mathbb P^{n+1}$.
Two years later, Ein studied the Hilbert scheme of $X \subset G$, a generic complete intersection of type $(m_1,\dots,m_k)$ in the Grassmann variety $G = G(r,n+2)$. As a remarkable corollary one gets that any smooth projective subvariety of $X$ is of general type if $m_1 + m_2 +\cdots+ m_k \ge\dim X + n + 2$. It is also proved that the Hilbert scheme of $X$ is smooth at points corresponding to smooth rational curves of "low" degree.
In 1996, Voisin had the idea of regarding the hypersurfaces in family and to use the positivity property of the tangent bundle of the family itself. Her main result is the following theorem which improves Ein's result in the case of hypersurfaces:
Let $X\subset\mathbb P^{n+1}$ be a hypersurface of degree $d$. If $d\ge 2n-\ell+ 1$, $1 \le\ell\le n - 2$, then any $\ell$-dimensional subvariety $Y$ of $X$ has a desingularization $\widetilde Y$ with an effective canonical bundle. Moreover, if the inequality is strict, then the sections of $K_{\widetilde Y}$ separate generic points of $\widetilde Y$.
The bound is now sharp and, in particular, the theorem implies that generic hypersurfaces in $\mathbb P^{n+1}$ of degree $d\ge 2n$, $n\ge 3$, contain no rational curves. The method also gives an improvement of a result of Xu as well as a simplied proof of Ein's original result.
Lastly, let me cite a result by Pacienza in 2004: this paper gives the sharp bound $d\ge 2n$ for a general projective hypersurface $X$ of degree $d$ in $\mathbb P^{n+1}$ containing only subvarieties of general type, for $n\ge 6$. This result improves the aforesaid results of Voisin and Ein.
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This is very helpful, thank you. – Mahdi Majidi-Zolbanin Dec 30 2011 at 4:42
Just an addendum, as long as I know, what I wrote is in the case $k=\mathbb C$. I don't know if it is true in an arbitrary algebraically close field... – diverietti Dec 30 2011 at 9:45
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It was shown by H. Clemens (1986) that a general hypersurface of degree $d$ in $\mathbb P^n$ does not contain any rational curves if $d$ is sufficiently large, specifically for $d \geq 2n-1$ for $n \geq 3$. Such a hypersurface can never contain a toric subvariety since any such subvariety must contain a rational curve.
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Thank you Parsa. – Mahdi Majidi-Zolbanin Dec 30 2011 at 4:41
If $X$ is an abelian variety then it contains no rational curves. Indeed, $\Omega_X$ is generated by global sections, hence if $f:P^1 \to X$ is a map then the image of the morphism $f^*\Omega_X \to \Omega_{P^1}$ is generated by global sections as well. But $\Omega_{P^1}$ has no global sections at all, so the image is $0$, which means that the map $f$ is constant.
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Even easier (and a little bit more general), if $X=\mathbb C^n/\Lambda$ is a complex torus, then $X$ does not admit any non-constant map from $\mathbb P^1$. In fact, any such map would lift to a proper non-constant map $\mathbb P1\to\mathbb C^n$. In particular, its image would be a positive dimensional compact analytic subset of $\mathbb C^n$, impossible. – diverietti Dec 31 2011 at 2:35
This may be a silly question, but what is an example of an abelian variety of dimension $n$ that can be embedded as a hypersurface in some $\mathbb{P}^{n+1}$ for $n≥4$? – Mahdi Majidi-Zolbanin Jan 2 2012 at 0:54
Sorry, I didn't notice the hypersurface condition. Of course you cannot get an abelian variety as a hypersutface in $P^{n+1}$ unless $n = 1$, this follows immediately from Lefschetz Theorem. – Sasha Jan 2 2012 at 11:24
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http://terrytao.wordpress.com/tag/ehud-hrushovski/
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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 ‘Ehud Hrushovski’ tag.
## Reading seminar 4: “Stable group theory and approximate subgroups”, by Ehud Hrushovski
5 November, 2009 in math.CO, math.LO, Logic reading seminar | Tags: Ehud Hrushovski, definability, henry towsner | by Terence Tao | 5 comments
This week, Henry Towsner concluded his portion of reading seminar of the Hrushovski paper, by discussing (a weaker, simplified version of) main model-theoretic theorem (Theorem 3.4 of Hrushovski), and described how this theorem implied the combinatorial application in Corollary 1.2 of Hrushovski. The presentation here differs slightly from that in Hrushovski’s paper, for instance by avoiding mention of the more general notions of S1 ideals and forking.
Here is a collection of resources so far on the Hrushovski paper:
• Henry Towsner’s notes (which most of Notes 2-4 have been based on);
• Alex Usvyatsov’s notes on the derivation of Corollary 1.2 (broadly parallel to the notes here);
• Lou van den Dries’ notes (covering most of what we have done so far, and also material on stable theories); and
• Anand Pillay’s sketch of a simplified proof of Theorem 1.1.
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## Reading seminar: “Stable group theory and approximate subgroups”, by Ehud Hrushovski
15 October, 2009 in math.CO, math.LO, Logic reading seminar | Tags: Freiman's theorem, Ehud Hrushovski, definability, type, Isaac Goldbring | by Terence Tao | 23 comments
One of my favorite open problems, which I have blogged about in the past, is that of establishing (or even correctly formulating) a non-commutative analogue of Freiman’s theorem. Roughly speaking, the question is this: given a finite set ${X}$ in a non-commutative group ${G}$ which is of small doubling in the sense that the product set ${X \cdot X := \{ xy: x, y \in X \}}$ is not much larger than ${X}$ (e.g. ${|X \cdot X| \leq K|X|}$ for some ${K = O(1)}$), what does this say about the structure of ${X}$? (For various technical reasons one may wish to replace small doubling by, say, small tripling (i.e. ${|X \cdot X \cdot X| = O( |X| )}$), and one may also wish to assume that ${X}$ contains the identity and is symmetric, ${X^{-1} = X}$, but these are relatively minor details.)
Sets of small doubling (or tripling), etc. can be thought of as “approximate groups”, since groups themselves have a doubling constant ${K := |X \cdot X|/|X|}$ equal to one. Another obvious example of an approximate group is that of an arithmetic progression in an additive group, and more generally of a ball (in the word metric) in a nilpotent group of bounded rank and step. It is tentatively conjectured that in fact all examples can somehow be “generated” out of these basic examples, although it is not fully clear at present what “generated” should mean.
A weaker conjecture along the same lines is that if ${X}$ is a set of small doubling, then there should be some sort of “pseudo-metric” ${\rho}$ on ${G}$ which is left-invariant, and for which ${X}$ is controlled (in some suitable sense) by the unit ball in this metric. (For instance, if ${X}$ was a subgroup of ${G}$, one would take the metric which identified all the left cosets of ${X}$ to a point, but was otherwise a discrete metric; if ${X}$ were a ball in a nilpotent group, one would use some rescaled version of the word metric, and so forth.) Actually for technical reasons one would like to work with a slightly weaker notion than a pseudo-metric, namely a Bourgain system, but let us again ignore this technicality here.
Recently, using some powerful tools from model theory combined with the theory of topological groups, Ehud Hrushovski has apparently achieved some breakthroughs on this problem, obtaining new structural control on sets of small doubling in arbitrary groups that was not previously accessible to the known combinatorial methods. The precise results are technical to state, but here are informal versions of two typical theorems. The first applies to sets of small tripling in an arbitrary group:
Theorem 1 (Rough version of Hrushovski Theorem 1.1) Let ${X}$ be a set of small tripling, then one can find a long sequence of nested symmetric sets ${X_1 \supset X_2 \supset X_3 \supset \ldots}$, all of size comparable to ${X}$ and contained in ${(X^{-1} X)^2}$, which are somewhat closed under multiplication in the sense that ${X_i \cdot X_i \subset X_{i-1}}$ for all ${i > 1}$, and which are fairly well closed under commutation in the sense that ${[X_i, X_j] \subset X_{i+j-1}}$. (There are also some additional statements to the effect that the ${X_n}$ efficiently cover each other, and also cover ${X}$, but I will omit those here.)
This nested sequence is somewhat analogous to a Bourgain system, though it is not quite the same notion.
If one assumes that ${X}$ is “perfect” in a certain sense, which roughly means that there is no non-trivial abelian quotient, then one can do significantly better:
Theorem 2 (Rough version of Hrushovski Corollary 1.2) Let ${X_0}$ be a set of small tripling, let ${X := X_0^{-1} X_0}$, and suppose that for almost all ${l}$-tuples ${a_1, \ldots, a_l \in X}$ (where ${l=O(1)}$), the conjugacy classes ${a_i^X := \{ x^{-1} ax: x \in X \}}$ generate most of ${X}$ in the sense that ${|a_1^X \cdot \ldots \cdot a_l^X| \gg |X|}$. Then a large part of ${X}$ is contained in a subgroup of size comparable to ${X}$.
Note that if one quotiented out by the commutator ${[X,X]}$, then all of the conjugacy classes ${a_i^X}$ would collapse to points. So the hypothesis here is basically a strong quantitative assertion to the effect that the commutator ${[X,X]}$ is extremely large, and rapidly fills out most of ${X}$ itself.
Here at UCLA, a group of logicians and I (consisting of Matthias Aschenbrenner, Isaac Goldbring, Greg Hjorth, Henry Towsner, Anush Tserunyan, and possibly others) have just started a weekly reading seminar to come to grips with the various combinatorial, logical, and group-theoretic notions in Hrushovski’s paper, of which we only have a partial understanding at present. The seminar is a physical one, rather than an online one, but I am going to try to put some notes on the seminar on this blog as it progresses, as I know that there are a couple of other mathematicians who are interested in these developments.
So far there have been two meetings of the seminar. In the first, I surveyed the state of knowledge of the noncommutative Freiman theorem, covering broadly the material in my previous blog post. In the second meeting, Isaac reviewed some key notions of model theory used in Hrushovski’s paper, in particular the notions of definability and type, which I will review below. It is not yet clear how these are going to be connected with the combinatorial side of things, but this is something which we will hopefully develop in future seminars. The near-term objective is to understand the statement of the main theorem on the model-theoretic side (Theorem 3.4 of Hrushovski), and then understand some of its easier combinatorial consequences, before going back and trying to understand the proof of that theorem.
[Update, Oct 19: Given the level of interest in this paper, readers are encouraged to discuss any aspect of that paper in the comments below, even if they are not currently being covered by the UCLA seminar.]
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http://mathoverflow.net/questions/110345/does-a-power-series-converging-everywhere-on-its-circle-of-convergence-define-a-c
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## Does a power series converging everywhere on its circle of convergence define a continuous function?
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Consider a complex power series $\sum a_n z^n \in \mathbb C[[z]]$ with radius of convergence $0\lt r\lt\infty$ and suppose that for every $w$ with $\mid w\mid =r$ the series $\sum a_n w^n$ converges .
We thus obtain a complex-valued function $f$ defined on the closed disk $\mid z\mid \leq r$ by the formula $f(z)=\sum a_n z^n$.
My question: is $f$ continuous ?
This is a naïve question which looks like it should be answered in any book on complex analysis.
But I checked quite a few books, among which the great treatises : Behnke-Sommer, Berenstein-Gay, Knopp, Krantz, Lang, Remmert, Rudin, Stein-Shakarchi, Titchmarsh, ... .
I couldn't find the answer, and yet I feel confident that it was known in the beginning of the twentieth century.
Edit
Many thanks to Julien who has answered my question: Sierpinski proved (in 1916) that there exists such a power series $\sum a_n z^n$ with radius of convergence $r=1$ and associated function $f(z)=\sum a_n z^n$ not bounded on the closed unit disk and thus certainly not continuous.
It is strange that not a single book on complex functions seems to ever have mentioned this example.
On the negative side, I must confess that I don't understand Sierpinski's article at all!
He airdrops a very complicated, weird-looking power series and proves that it has the required properties in a sequence of elementary but completely obscure computations.
I would be very grateful to anybody who would write a new answer with a few lines of explanation as to what Sierpinski is actually doing.
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You are right, Abel's theorem only give the convergence $\lim\limits_{t \rightarrow 1-}f( \alpha t) = f(\alpha)$ for $|\alpha|=1$, not continuity on the circle. Excuse my quick fire:) – Marc Palm Oct 22 at 16:51
Dear @Mrc, there is nothing for me to excuse: on the contrary, I want to thank you for your so quickly trying to help. In the books I browsed there are indeed many results in the style of Abel's theorem, but they seem not to be sufficient to answer my question. – Georges Elencwajg Oct 22 at 17:02
2
Did you try a trigonometric series $\sum c_n e^{inx}$ converging to a discontinuous function ? (The problem is that a priori the sum is over $n \in \mathbf{Z}$.) – François Brunault Oct 22 at 17:08
This is a very interesting question! I've also wondered about this myself while thinking about this question on M.SE : math.stackexchange.com/questions/209171/… I, too, did not find anything about this in any complex analysis textbook I know. I find it quite impressive that Sierpinski seemed to know the answers to a very large quantity of these simple-sounding, natural, yet quite difficult questions. – Malik Younsi Oct 23 at 13:54
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Here is a complex analysis book that mentions (yet just a note it seems) the result: An introduction to classical complex analysis, vol 1, by R. B. Burckel (Birkhäuser, 1979), see page 81. (Note: I did not know this either, I did not even know the book; it merely turned up in a search for the title of the mentioned paper.) – quid Oct 23 at 15:50
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## 3 Answers
I searched all over for an answer to this question back in my student days. I found the answer in a paper by Sierpinski, "Sur une série potentielle qui, étant convergente en tout point de son cercle de convergence,représente sur ce cercle une fonction discontinue ", which is featured in his collected works, see here, p282) and apparently was published in 1916.
It does confirm your expectation that this was known in the beginning of the twentieth century (I don't know whether it's the first proof or not, but from the paper it's clear that Sierpinski thought the result to be new).
EDIT: I just realized that not everybody speaks French ;-) so, to be clear: Sierpinski produces an example where the function converges everywhere on the unit circle but is discontinuous on the circle.
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@quid: thanks for fixing the link! I thought my url was OK but it was adding %20 instead of spaces - is that just a copy/paste issue? – Julien Melleray Oct 22 at 17:44
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here is a Zentralblatt/JFM review (of course, in German :)) of the paper in question: zentralblatt-math.org/zmath/en/search/… "Beispiel einer Potenzreihe, die den Einheitskreis als Konvergenzkreis hat, überall auf demselben konvergiert und dortselbst unstetig ist in der Art, dass sie in der Umgebung des Punktes 1 unbeschränkt ist. (Prof. Rademacher)" – Dima Pasechnik Oct 23 at 6:31
Thanks a lot for your perfect answer, Julien! By the way, what made you think of this problem in your student days? Are you still interested in these questions? – Georges Elencwajg Oct 23 at 11:53
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You're welcome! It was just idle curiosity that got me and a friend interested in this question (we were preparing for the "agregation" - meaning that for a few months we were going over undergraduate material and trying to think about what kind of questions one could ask about it). At this time I discovered Sierpinski's work and was fascinated -I still am, in some ways; he was an incredible mathematician) . As it turns out, my research interests are close to some of Sierpinski's (related to descriptive set theory) but nowhere near complex analysis (though I still find it a beautiful topic). – Julien Melleray Oct 23 at 12:42
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Cher Julien: je suis heureux que les jurys n'aient jamais posé cette question à mes agrégatifs. Et que ces derniers ne me l'aient jamis posée non plus... :-) – Georges Elencwajg Oct 23 at 12:56
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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.
This answer is in response to final sentence, "I would be very grateful to anybody who would write a new answer with a few lines of explanation as to what Sierpinski is actually doing". In fact, it is easy to construct power series converging on the circle of convergence, but are unbounded. For example, $$f(z)=\sum_{n=1}^\infty\frac1{n^5(1+in^{-3}-z)}$$ defines a function whose power series expansion has radius of convergence 1 and converges everywhere on the unit circle, but is unbounded in a neighbourhood of 1.
A method of constructing such functions is as an infinite sum $$f(z)=\sum_{n=1}^\infty f_n(z).$$ Here, $f_n(z)$ are chosen to have a power series expansion converging everywhere on the closed unit ball. Let $f^{(r)}_n(z)$ denote the sum of the first $r$ terms in the power series expansion of $f_n$. We need to arrange it so that $f^{(r)}(z)\equiv\sum_nf_n^{(r)}(z)$ converges on the closed unit ball, and that $f(z)=\lim_{r\to\infty}f^{(r)}(z)$ holds. That is, we need to be able to commute the limit $r\to\infty$ with the summation over $n$. A sufficient condition to be able to do this is that $\sum_n\sup_r\lvert f^{(r)}_n(z)\rvert < \infty$, for all $\lvert z\rvert\le1$. That this allows us to commute the summation with the limit is just a special case of dominated convergence.
Next, to ensure that $f(z)$ is unbounded on the unit ball, we want to choose $f_n$ such that there exists $q_n$ in the closed unit ball with $f_n(q_n)$ large, and such that it does not get cancelled out in the summation, so that $f(q_n)$ is large and diverges as $n\to\infty$.
For example, choose positive reals $\delta_n,\epsilon_n$ tending to zero, and setting $a_n=1+i\epsilon_n$, and $$f_n(z)=\frac{\delta_n}{a_n-z}=\sum_{m=0}^\infty \delta_na_n^{-m-1}z^m.$$ These are all well-defined as power series with radius of convergence greater than 1. Furthermore, the partial sums are $$f^{(r)}_n(z)=\delta_n\frac{1-(z/a_n)^r}{a_n-z},$$ which are bounded by $2\delta_n/\lvert a_n-z\rvert$. As $a_n\to1$, this is bounded by a multiple of $\delta_n$ for each fixed $z\not=1$, so the dominated convergence condition is satisfied when $\sum_n\delta_n$ is finite. On the other hand, if $z=1$, then $\lvert a_n-z\rvert=\epsilon_n$, so the dominated convergence condition is satisfied everywhere whenever $\sum_n\delta_n/\epsilon_n$ is finite. Next, $f_n(z)$ achieves its largest value on the unit ball at $q_n=a_n/\lvert a_n\rvert$, and its real part there is given by $$\Re f_n(q_n)=\frac{\delta_n}{\sqrt{1+\epsilon_n^2}(\sqrt{1+\epsilon_n^2}-1)}\ge\frac{2\delta_n}{\epsilon_n^2\sqrt{1+\epsilon_n^2}}.$$ As $f_m(z)$ has positive real part for all $m$, this bound also holds for $f(q_n)$, and we get that $f$ is unbounded whenever $\delta_n/\epsilon_n^2\to\infty$. These conditions are satisfied by taking $\epsilon_n=n^{-3}$ and $\delta_n=n^{-5}$.
Alternatively, for an example closer to Sierpinski's, consider choosing a sequence $a_n\to1$ on the unit circle and positive reals $K_n$, and set $$f_n(z)=K_n2^{-n}\sum_{k=0}^{2^n-1}a_n^{2^n-1-k}z^k=2^{-n}K_n\frac{a_n^{2^n}-z^{2^n}}{a_n-z}.$$ The partial sums of the power series expansion of $f_n(z)$ are bounded by $2^{1-n}K_n/\lvert a_n-z\rvert$, so the dominated convergence condition is satisfied for $z\not=1$ so long as $\sum_n2^{1-n}K_n$ is finite. Sierpinski chooses $a_n=(n^2-1+2ni)/(n^2+1)$ so that $a_n-1$ goes to zero at rate $1/n$. The dominated convergence condition is therefore satisfied whenever $\sum_n2^{-n}K_nn$ is finite.
Now, $f_n(z)$ is maximized at $z=a_n$ where $\lvert f_n(a_n)\rvert=K_n$. So, $$\lvert f(a_n)\rvert\ge K_n-\sum_{m\not=n}\frac{2^{1-m}K_m}{\lvert a_m-a_n\rvert}.$$ As $a_m-a_n$ is bounded below by a multiple of $1/m^2$, the summation on the right is bounded whenever $\sum_m2^{-m}K_mm^2$ is finite, and $f(a_n)$ is unbounded if we also take $K_n$ going to infinity. Sierpinski takes $K_n=n^2$ here. Finally, in Sierpinski's example, he multiplies $f_n$ by $z^{2^n}$. This changes nothing, except to separate out the non-zero terms of the power series of $f_n(z)$, so that the power series of $f(z)$ can be written easily term by term.
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Thanks, George! – Georges Elencwajg Dec 22 at 9:49
Just to complete the previous answer, Sierpiński's example is mentioned (without details, though) in at least one book, namely An introduction to classical complex analysis by R. B. Burckel (Vol. 1, Chap. 3, p. 81).
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2
A power series that converges everywhere on its circle of convergence must necessarily define a discontinuous function. For if D is the closed disk of radius r whose boundary is that circle, then D is compact and any continuous function defined on D must be bounded everywhere on D. But if the power series is bounded throughout the interior of D then r cannot be its radius of convergence. So the question should really be "can such a power series exist?" and Sierpinski has given an example of one. – Garabed Gulbenkian Oct 23 at 19:26
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@Garabed Gulbenkian: Perhaps I am confused, but wouldn't $\sum x^n/n^2$ we a powerseries converging everywhere on its circle of convergence? – quid Oct 23 at 20:24
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@Garabed Gulbenkian : I'm not sure I understand. It is easy to give an example of a power series that converges everywhere on its circle of convergence, just take $\sum_n z^n/n^2$. Why does the fact that the series is bounded in $D$ implies that its radius of convergence cannot be $r$? – Malik Younsi Oct 23 at 20:30
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@quid : Well if you are confused, then you are not the only one ;) – Malik Younsi Oct 23 at 20:32
2
Probably, Gulbenkian thinks that the only reason a series has radius of convergence $r$ is that there must be a pole at this distance. Alas, other kinds of singularity (such as branch points) exists... – Feldmann Denis Oct 23 at 22:47
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http://mathhelpforum.com/advanced-algebra/192136-null-space-matrix-similar-columns.html
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# Thread:
1. ## Null space for Matrix with similar columns
Hi,
I just read that the null space for the following matrix
[1 0 1;
5 4 9;
2 4 6]
is the line of all points x = c, y = c, z = -c.
I was wondering what the null space for the matrix with three similar columns would be e.g.
[1 1 1;
0 0 0;
0 0 0]
(c, -c, 0), (-c, 2c, -c), etc are all part of the null space for this matrix but they don't form a line nor a plane.
Any help would be much appreciated.
Regards
2. ## Re: Null space for Matrix with similar columns
the matrix:
$\begin{bmatrix}1&1&1\\0&0&0\\0&0&0\end{bmatrix}$
has rank 1, so it has null space of dimension 2. it's clear that (x,y,z) is in this nullspace iff x+y+z = 0, that is if (x,y,z) is of the form: (s,t,-s-t),
which is the plane: s(1,0,-1) + t(0,1,-1). a basis is {(1,0,-1),(0,1,-1)}.
both of your example points lie in this plane. simply take s = c, t = 0, for the first one, and s = -c, t = 2c for the second.
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http://mathoverflow.net/questions/109411/are-there-any-fake-3-tori
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## Are there any fake 3-tori?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hsiang, W.-c.; Shaneson, J. L. Fake tori, the annulus conjecture, and the conjectures of Kirby. Proc. Nat. Acad. Sci. U.S.A. 62 1969 687–691.
The paper above classified all fake tori for dimension $\ge 5$. How about low dimension?
To be precise: Let $M^n$ be a topological manifold of dimension $n=3, 4$, which has the same homotopy type of the standard torus $T^n$. My question is whether $M^n$ is homeomorphic to the standard torus?
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Could you remind us what definition of "fake" you're using? – Ryan Budney Oct 11 at 20:27
6
Presumably this answers your question: if a 3-manifold has fundamental group isomorphic to $\mathbb Z^3$ then it is homeo/diffeo/PL equivalent to $(S^1)^3$. The follows from geometrization. – Ryan Budney Oct 11 at 20:28
@Ryan Fake torus means the manifold which is homotopic to the torus. Could you please give more details on how to derive it from geometrization? Or does it follows from some easiler fact other than this big theorm? – J. GE Oct 11 at 20:36
This is the reference you want then: en.wikipedia.org/wiki/Moise%27s_theorem it's a theorem of Moise that topological, PL and smooth manifolds all have a unique compatible structure among these three categories. – Ryan Budney Oct 11 at 20:44
By "homotopic" do you mean homeomorphic or homotopy-equivalent? My previous comment answers the homeomorphic interpretation. If you really mean homotopy-type then you need geometrization to show the manifold is prime. Once you know the manifold is prime, it's an old theorem of Waldhausen's. – Ryan Budney Oct 11 at 20:46
show 5 more comments
## 1 Answer
@Ryan answers the question in three dimensions (and such a result cannot hold without the Poincare conjecture, since otherwise you could take a connected sum with a fake $S^3.)$ In general, this is a special case of the Borel Conjecture, which is known to hold in dimension four for groups of subexponential growth, such as $\mathbb{Z}^4$ (in the topological category). For more, see @Igor Belegradek's answer to this question, and references therein.
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Thanks Igor. Could you point to me the paper where the Borel Conjecture was proved for subexponential growth. Since in Belegradek's answer he mentioned it is true for "closed 4-manifolds homotopy equivalent to an infranil 4-manifold". – J. GE Oct 12 at 8:51
This is in Freedman and Quinn's book "Topology of 4-manifolds". The reference for the first paper it appears in should be there, as well. – Ryan Budney Oct 12 at 15:50
That's great, thanks Ryan. – J. GE Oct 13 at 6:21
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http://mathoverflow.net/revisions/10400/list
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## Return to Question
3 added 2 characters in body
Although the question is easy to pose, I think some background will help to motivate it, so I'll start with it.
Consider variables $X=(X_1, \ldots, X_n)$ over a field $K$ and the elementary symmetric functions $T=(T_1, \ldots, T_n)$ in $X$. In other words $X$ are the roots of the polynomial $Y^n + T_1 Y^{n-1} + \cdots + T_n$.
A polynomial $f$ in $X$ is symmetric is $f(s X) = f(X)$ for any permutation $s$. Here $s X := (X_{s(1)}, \ldots, X_{s(n)})$. Then a basic fact is that if $f(X)$ is symmetric, then $f(X) = g(T)$, for some polynomial $g$.
It is reasonable to define an alternating polynomial to be $f$ that satisfy $f(s X) = sign(s) f(X)$, where $sign(s) = \pm 1$ is the signature. The "elementary" alternating polynomial is the Vandermonde polynomial `$V(X) = \prod_{i<j} (X_j-X_i)$`, and any other alternating polynomial can be expressed as a polynomial in $T$ and $V$.
Note that $V$ is a square root of the discriminant $\Delta$ of $Y^n + T_1 Y^{n-1} + \cdots + T_n$ and the discriminant has an explicit formula in terms of $T$ using the Sylvester matrix.
That definition for alternating polynomials gives nothing interesting in characteristic $2$ (because then $1=-1$). The only definition that makes sense to me in characteristic $2$ is: $f$ is alternating if $f(s X) = f(X) + add.sign(s)$. Here $add.sign(s) = 0,1$ is the additive signature, i.e., equals $1$ if $s$ is odd and $0$ if $s$ is even.
I already figured out what is the "elementary" alternating polynomial $u$ u/V$and what is the Artin-Schreir equation it satisfies: `$u(X) = \sum_{s \ {\rm is\ even}} X^{n-1}_{s(1)} \cdots X^0_{s(n)}$` and it satisfies the Artin-Schreier equation$X^2 + X = \frac{u(X) u(s_0 X)}{\Delta}$, where$s_0$is any odd permutation (e.g., transposition), and$\Delta$is again the discriminant. (Note that$u(X) + u(s_0 X) = V\$.)
My question is: Does there exist a nice formula for $\frac{u(X) u(s_0 X)}{\Delta}$ in terms of $T$?
2 Fix an equation
# ExcplicitExplicit expression of an alternating polynomial in characteristic $2$?
Although the question is easy to pose, I think some background will help to motivate it, so I'll start with it.
Consider variables $X=(X_1, \ldots, X_n)$ over a field $K$ and the elementary symmetric functions $T=(T_1, \ldots, T_n)$ in $X$. In other words $X$ are the roots of the polynomial $Y^n + T_1 Y^{n-1} + \cdots + T_n$.
A polynomial $f$ in $X$ is symmetric is $f(s X) = f(X)$ for any permutation $s$. Here $s X := (X_{s(1)}, \ldots, X_{s(n)})$. Then a basic fact is that if $f(X)$ is symmetric, then $f(X) = g(T)$, for some polynomial $g$.
It is reasonable to define an alternating polynomial to be $f$ that satisfy $f(s X) = sign(s) f(X)$, where $sign(s) = \pm 1$ is the signature. The "elementary" alternating polynomial is the Vandermonde polynomial `$V(X) = \prod_{i<j} (X_j-X_i)$`, and any other alternating polynomial can be expressed as a polynomial in $T$ and $V$.
Note that $V$ is a square root of the discriminant $\Delta$ of $Y^n + T_1 Y^{n-1} + \cdots + T_n$ and the discriminant has an explicit formula in terms of $T$ using the Sylvester matrix.
That definition for alternating polynomials gives nothing interesting in characteristic $2$ (because then $1=-1$). The only definition that makes sense to me in characteristic $2$ is: $f$ is alternating if $f(s X) = f(X) + add.sign(s)$. Here $add.sign(s) = 0,1$ is the additive signature, i.e., equals $1$ if $s$ is odd and $0$ if $s$ is even.
I already figured out what is the "elementary" alternating polynomial $u$ and what is the Artin-Schreir equation it satisfies: `$u(X) = \sum_{s \ {\rm is\ even}} X^{n-1}_{s(1)} \cdots X^0_{s(n)}$` and it satisfies the Artin-Schreier equation $X^2 + X = \frac{u(X) u(s_0 X)}{\Delta}$, where $s_0$ is any odd permutation (e.g., transposition), and $\Delta$ is again the discriminant. (Note that $u(X) + u(s_0 X) = V$.)
My question is: Does there exist a nice formula for $\frac{u(X) u(s_0 X)}{\Delta}$ in terms of $T$?
1
# Excplicit expression of an alternating polynomial in characteristic $2$?
Although the question is easy to pose, I think some background will help to motivate it, so I'll start with it.
Consider variables $X=(X_1, \ldots, X_n)$ over a field $K$ and the elementary symmetric functions $T=(T_1, \ldots, T_n)$ in $X$. In other words $X$ are the roots of the polynomial $Y^n + T_1 Y^{n-1} + \cdots + T_n$.
A polynomial $f$ in $X$ is symmetric is $f(s X) = f(X)$ for any permutation $s$. Here $s X := (X_{s(1)}, \ldots, X_{s(n)})$. Then a basic fact is that if $f(X)$ is symmetric, then $f(X) = g(T)$, for some polynomial $g$.
It is reasonable to define an alternating polynomial to be $f$ that satisfy $f(s X) = sign(s) f(X)$, where $sign(s) = \pm 1$ is the signature. The "elementary" alternating polynomial is the Vandermonde polynomial \$V(X) = \prod_{i
Note that $V$ is a square root of the discriminant $\Delta$ of $Y^n + T_1 Y^{n-1} + \cdots + T_n$ and the discriminant has an explicit formula in terms of $T$ using the Sylvester matrix.
That definition for alternating polynomials gives nothing interesting in characteristic $2$ (because then $1=-1$). The only definition that makes sense to me in characteristic $2$ is: $f$ is alternating if $f(s X) = f(X) + add.sign(s)$. Here $add.sign(s) = 0,1$ is the additive signature, i.e., equals $1$ if $s$ is odd and $0$ if $s$ is even.
I already figured out what is the "elementary" alternating polynomial $u$ and what is the Artin-Schreir equation it satisfies: `$u(X) = \sum_{s \ {\rm is\ even}} X^{n-1}_{s(1)} \cdots X^0_{s(n)}$` and it satisfies the Artin-Schreier equation $X^2 + X = \frac{u(X) u(s_0 X)}{\Delta}$, where $s_0$ is any odd permutation (e.g., transposition), and $\Delta$ is again the discriminant. (Note that $u(X) + u(s_0 X) = V$.)
My question is: Does there exist a nice formula for $\frac{u(X) u(s_0 X)}{\Delta}$ in terms of $T$?
|
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|
http://mathhelpforum.com/trigonometry/94428-show-area-large-square-diagram-can-written-x-y.html
|
Thread:
1. Show that the area of the large square in the diagram can be written as (x+y)², and..
Show that the area of the large square in the diagram can be written as (x+y)², and also as z²+2xy?
HERE IS THE DIAGRAM http://img43.imageshack.us/img43/9428/trig.png
Do not use pythagoras theorem.
b) Show how these results can be used to prove pythagoras theorem
2. The side of the larger square is x+y
Hence area of the large square= $(x+y)^2$
3. prove pythagoras theorem
Thanks. How do I Show how these results can be used to prove pythagoras theorem
4. and also how do i show it can be written as z²+2xy?
5. Inside the larger square,
there is a square of side z, and four right triangles of side x,y.
total area=square of side z+4*area of triangles
= $z^2+4*\frac{1}{2}xy$
= $z^2+2xy$
pythagoras theorem: press spoiler
Spoiler:
$(x+y)^2=z^2+2xy$
$x^2 + y^2 =z^2$
|
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|
http://www.haskell.org/haskellwiki/index.php?title=User:Michiexile/MATH198/Lecture_10&diff=31701&oldid=31626
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User:Michiexile/MATH198/Lecture 10
From HaskellWiki
(Difference between revisions)
Line 1: Line 1:
IMPORTANT NOTE: THESE NOTES ARE STILL UNDER DEVELOPMENT. PLEASE WAIT UNTIL AFTER THE LECTURE WITH HANDING ANYTHING IN, OR TREATING THE NOTES AS READY TO READ. IMPORTANT NOTE: THESE NOTES ARE STILL UNDER DEVELOPMENT. PLEASE WAIT UNTIL AFTER THE LECTURE WITH HANDING ANYTHING IN, OR TREATING THE NOTES AS READY TO READ.
+ This lecture will be shallow, and leave many things undefined, hinted at, and is mostly meant as an appetizer, enticing the audience to go forth and seek out the literature on topos theory for further studies.
- ===Power objects=== + ===Subobject classifier===
- ===Classifying objects=== + One very useful property of the category <math>Set</math> is that the powerset of a given set is still a set; we have an internal concept of ''object of all subobjects''. Certainly, for any category (small enough) <math>C</math>, we have a contravariant functor <math>Sub(-): C\to Set</math> taking an object to the set of all equivalence classes of monomorphisms into that object; with the image <math>Sub(f)</math> given by the pullback diagram
+ :[[Image:SubobjectFunctorMorphismPullback.png]]
- ===Topoi=== + If the functor <math>Sub(-)</math> is ''representable'' - meaning that there is some object <math>X\in C_0</math> such that <math>Sub(-) = hom(-,X)</math> - then the theory surrounding representable functors, connected to the Yoneda lemma - give us a number of good properties.
- ===Internal logic=== + One of them is that every representable functor has a ''universal element''; a generalization of the kind of universal mapping properties we've seen in definitions over and over again during this course; all the definitions that posit the unique existence of some arrow in some diagram given all other arrows.
+
+ Thus, in a category with a representable subobject functor, we can pick a representing object <math>\Omega\in C_0</math>, such that <math>Sub(X) = hom(X,\Omega)</math>. Furthermore, picking a universal element corresponds to picking a subobject <math>\Omega_0\hookrightarrow\Omega</math> such that for any object <math>A</math> and subobject <math>A_0\hookrightarrow A</math>, there is a unique arrow <math>\chi: A\to\Omega</math> such that there is a pullback diagram
+ :[[Image:SubobjectClassifierPullback.png]]
+
+ One can prove that <math>\Omega_0</math> is terminal in <math>C</math>, and we shall call <math>\Omega<math> the ''subobject classifier'', and this arrow <math>\Omega_0=1\to\Omega</math> ''true''. The arrow <math>\chi</math> is called the characteristic arrow of the subobject.
+
+ In Set, all this takes on a familiar tone: the subobject classifier is a 2-element set, with a ''true'' element distinguished; and a characteristic function of a subset takes on the ''true'' value for every element in the subset, and the other (false) value for every element not in the subset.
+
+ ===Defining topoi===
+
+ '''Definition''' A ''topos'' is a cartesian closed category with all finite limits and with a subobject classifier.
+
+ It is worth noting that this is a far stronger condition than anything we can even hope to fulfill for the category of Haskell types and functions. The functional proogramming relevance will take a back seat in this lecture, in favour of usefulness in logic and set theory replacements.
+
+ ===Properties of topoi===
+
+ The meat is in the properties we can prove about topoi, and in the things that turn out to be topoi.
+
+ '''Theorem''' Let <math>E</math> be a topos.
+ * <math>E</math> has finite colimits.
+
+ ====Power object====
+
+ Since a topos is closed, we can take exponentials. Specifically, we can consider <math>[A\to\Omega]</math>. This is an object such that <math>hom(B,[A\to\Omega]) = hom(A\times B, \Omega) = Sub(A\times B)</math>. Hence, we get an internal version of the subobject functor. (pick <math>B</math> to be the terminal object to get a sense for how global elements of <math>[A\to\Omega]</math> correspond to subobjects of <math>A</math>)
+
+ ((universal property of power object?))
+
+ ====Internal logic====
+
+ We can use the properties of a topos to develop a logic theory - mimicking the development of logic by considering operations on subsets in a given universe:
+
+ Classically, in Set, we would say that a ''predicate'' corresponds to the set of elements on which the predicate holds true. Thus, say, for statements about integers, the set <math>\{n\in\mathbb Z: n>0\}</math> would correspond to the predicate <math>n>0</math>.
+
+ We could then start to define primitive logic connectives as set operations; the intersection of two sets is the set on which '''both''' the corresponding predicates hold true, so <math>\& = \cap</math>. Similarily, the union of two sets is the set on which either of the corresponding predicates holds true, so <math>| = \cup</math>. The complement of a set, in the universe, is the negation of the predicate, and all other propositional connectives (implication, equivalence, ...) can be built with conjunction (and), disjunction (or) and negation (not).
+
+ So we can mimic all these in a given topos:
+
+ We say that a ''universe'' <math>U</math> is just an object in a given topos.
+
+ A ''predicate'' is a subobject of the universe.
+
+ Given predicates <math>P, Q</math>, we define the ''conjunction'' <math>P\wedge Q</math> to be the pullback (pushout?)
+ :[[Image:ToposConjunction.png]]
+
+ This mimics, closely, the idea of the conjunction as an intersection.
+
+ We further define the ''disjunction'' <math>P\vee Q</math> to be the pushout (pullback?)
+ :[[Image:ToposDisjunction.png]]
+
+ And we define ''negation'' <math>\neg P</math> by (...)
+
+ We can expand this language further - and introduce predicative connectives.
+
+ Now, a statement on the form <math>\forall x. P(x)</math> is usually taken to mean that on all <math>x\in U</math>, the predicate <math>P</math> holds true. Thus, translating to the operations-on-subsets paradigm, <math>\forall x. P(x)</math> corresponds to the statement <math>P = U</math>.
+
+ ((double check in Awodey & Barr-Wells!!!!))
+
+ So we can define a topoidal <math>\forall x. P(x)</math> by ((diagrams))
+
+ And similarily, the statement <math>\exists x. P(x)</math> means that <math>P</math> is non-empty.
+
+ ====Examples====
+
+ ====Sheaves, topology and time sheaves====
===Exercises=== ===Exercises===
Revision as of 18:43, 22 November 2009
IMPORTANT NOTE: THESE NOTES ARE STILL UNDER DEVELOPMENT. PLEASE WAIT UNTIL AFTER THE LECTURE WITH HANDING ANYTHING IN, OR TREATING THE NOTES AS READY TO READ.
This lecture will be shallow, and leave many things undefined, hinted at, and is mostly meant as an appetizer, enticing the audience to go forth and seek out the literature on topos theory for further studies.
Contents
1 Subobject classifier
One very useful property of the category Set is that the powerset of a given set is still a set; we have an internal concept of object of all subobjects. Certainly, for any category (small enough) C, we have a contravariant functor $Sub(-): C\to Set$ taking an object to the set of all equivalence classes of monomorphisms into that object; with the image Sub(f) given by the pullback diagram
If the functor Sub( − ) is representable - meaning that there is some object $X\in C_0$ such that Sub( − ) = hom( − ,X) - then the theory surrounding representable functors, connected to the Yoneda lemma - give us a number of good properties.
One of them is that every representable functor has a universal element; a generalization of the kind of universal mapping properties we've seen in definitions over and over again during this course; all the definitions that posit the unique existence of some arrow in some diagram given all other arrows.
Thus, in a category with a representable subobject functor, we can pick a representing object $\Omega\in C_0$, such that Sub(X) = hom(X,Ω). Furthermore, picking a universal element corresponds to picking a subobject $\Omega_0\hookrightarrow\Omega$ such that for any object A and subobject $A_0\hookrightarrow A$, there is a unique arrow $\chi: A\to\Omega$ such that there is a pullback diagram
One can prove that Ω0 is terminal in C, and we shall call $\Omega<math> the ''subobject classifier'', and this arrow <math>\Omega_0=1\to\Omega$ true. The arrow χ is called the characteristic arrow of the subobject.
In Set, all this takes on a familiar tone: the subobject classifier is a 2-element set, with a true element distinguished; and a characteristic function of a subset takes on the true value for every element in the subset, and the other (false) value for every element not in the subset.
2 Defining topoi
Definition A topos is a cartesian closed category with all finite limits and with a subobject classifier.
It is worth noting that this is a far stronger condition than anything we can even hope to fulfill for the category of Haskell types and functions. The functional proogramming relevance will take a back seat in this lecture, in favour of usefulness in logic and set theory replacements.
3 Properties of topoi
The meat is in the properties we can prove about topoi, and in the things that turn out to be topoi.
Theorem Let E be a topos.
• E has finite colimits.
3.1 Power object
Since a topos is closed, we can take exponentials. Specifically, we can consider $[A\to\Omega]$. This is an object such that $hom(B,[A\to\Omega]) = hom(A\times B, \Omega) = Sub(A\times B)$. Hence, we get an internal version of the subobject functor. (pick B to be the terminal object to get a sense for how global elements of $[A\to\Omega]$ correspond to subobjects of A)
((universal property of power object?))
3.2 Internal logic
We can use the properties of a topos to develop a logic theory - mimicking the development of logic by considering operations on subsets in a given universe:
Classically, in Set, we would say that a predicate corresponds to the set of elements on which the predicate holds true. Thus, say, for statements about integers, the set $\{n\in\mathbb Z: n>0\}$ would correspond to the predicate n > 0.
We could then start to define primitive logic connectives as set operations; the intersection of two sets is the set on which both the corresponding predicates hold true, so Failed to parse (lexing error): \& = \cap . Similarily, the union of two sets is the set on which either of the corresponding predicates holds true, so $| = \cup$. The complement of a set, in the universe, is the negation of the predicate, and all other propositional connectives (implication, equivalence, ...) can be built with conjunction (and), disjunction (or) and negation (not).
So we can mimic all these in a given topos:
We say that a universe U is just an object in a given topos.
A predicate is a subobject of the universe.
Given predicates P,Q, we define the conjunction $P\wedge Q$ to be the pullback (pushout?)
Image:ToposConjunction.png
This mimics, closely, the idea of the conjunction as an intersection.
We further define the disjunction $P\vee Q$ to be the pushout (pullback?)
Image:ToposDisjunction.png
And we define negation $\neg P$ by (...)
We can expand this language further - and introduce predicative connectives.
Now, a statement on the form $\forall x. P(x)$ is usually taken to mean that on all $x\in U$, the predicate P holds true. Thus, translating to the operations-on-subsets paradigm, $\forall x. P(x)$ corresponds to the statement P = U.
((double check in Awodey & Barr-Wells!!!!))
So we can define a topoidal $\forall x. P(x)$ by ((diagrams))
And similarily, the statement $\exists x. P(x)$ means that P is non-empty.
4 Exercises
No homework at this point. However, if you want something to think about, a few questions and exercises:
1. blah
|
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|
http://mathoverflow.net/questions/32689?sort=votes
|
## How should a homotopy theorist think about sheaf cohomology?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
As a student of homotopy theory or algebraic topology, I have a certain outlook as to how one ought to think of a cohomology theory. There are axioms that help us with rudimentary computations, there are some spectral sequences and then there is Brown representability. This is far away from the starting point of looking at the singular (co)chain complex or a simplicial complex and trying to compute its homology, it seems a bit more refined. Even the ring structure is a bit clearer, it comes from the fact that we are mapping into a ring object.
There are more and more instances where i feel like i would benefit from understanding a bit more of sheaf cohomology than just "it's the derived functor of the global sections functor of a sheaf." This is a tidge helpful, but it does not really help too much with computations from my point of view. It feels like resolutions of sheaves are large hard objects mostly because sheaves contain so much data.
My question is essentially the following:
1. Are there homotopy theorists out there who have over come these feelings? what advice do you have? In fact any advice that someone might have that understands the uses of sheaf theory in homotopy theory would be helpful.
2. Are there things resembling the Eilenberg-Steenrod axioms for sheaf cohomology? not directly due to their classification theorem, but things that help you to compute the Sheaf cohomology like a MVS sequence or what have you. I mostly would like help in doing computations in the way that the E-S axioms do? so things like the Grothendieck-Riemann-Roch Theorem, which i am told can be used in such a way.
3. Is there a book that goes through explicit toy computations of sheaf cohomology? Are there toy examples you would suggest for getting to be more comfortable with these things? Are there examples that live on simpler spaces than schemes? these might help a bit more than others
Some addendums : I think i need to make a few comments. I am not well versed in algebraic geometry. I find this to be a fault of mine and this is an attempt to help bridge the gap. Suggestions for references are appreciated. I really appreciate all of the excellent answers so far! thanks for your time
EDIT MAIN QUESTION: I think what i am really asking about is the six functor formalism, but i don't really know since i don't know what that is. A friend started explaining it to me and it seemed like what i am looking for, but he said he did not feel confident in writing an answer explaining them. Hopefully someone will see this edit and give a fun answer.
-
2
Sheaf cohomology is in some sense orthogonal to ordinary homotopy theory. What this perspective does is allow you to study coefficient systems that vary in a nontrivial way over a base space (or topos). Homotopy theory works instead by sticking to plain spaces (where your "base space" is a point), and trying to study in depth the complexities that can occur in that situation. These both come together when studying something like sheaves of spaces or simplicial sets or spectra, where you might have families of cohomology theories varying nontrivially. – Tyler Lawson Jul 20 2010 at 20:30
1
2) and 3) Bredon's Sheaf Theory has some of the things you're looking for, and is dense but readable. As for toy examples, computing the cohomology of some simple sheaves via the Cech complex is illuminating, as is playing with sheaves on posets. I am by no means an expert, though. – Daniel Litt Jul 20 2010 at 20:37
12
A thing that you may want to keep in mind is that often sheaf cohomology tells you more about the sheaf than about the space. For example, a sheaf $\mathcal{F}$ on a subspace $X \subset Y$ can be extended to $0$ on $Y$ and cohomology will not change. – Andrea Ferretti Jul 20 2010 at 21:00
1
For (3), look at, for instance, the sections on Cech cohomology and cohomology of projective space in Chapter III of Hartshorne. Eisenbud's commutative algebra book probably has lots more good examples and exercises. – Kevin Lin Jul 20 2010 at 21:42
6
Andrea -- you probably want the subspace to be closed, or else the cohomology may well change. – algori Jul 20 2010 at 22:35
show 2 more comments
## 3 Answers
Hi Sean,
I think I great place to inhabit, be it for calculations or for conceptual understanding, is the derived category of sheaves (of abelian groups say) $D(X)$ on your topological space $X$. Here are the basic players:
1. For any $X$ and integer $i$, functors $H^i: D(X) \rightarrow$ sheaves of abelian groups on X (cohomology sheaves); and
2. For a map $f: X \rightarrow Y$, adjoint maps $f^* :D(Y) \rightarrow D(X)$ and $f_* :D(X) \rightarrow D(Y)$ (derived pullback and pushforward).
Then, for instance, the $i^{th}$ cohomology of a sheaf is just $H^i p_*$, where $p: X \rightarrow pt$.
Lots of your favorite computational tools carry over to this setting. For instance Meyer-Vietoris: if $X = U_1 \cup U_2$ with inclusions $j_1: U_1 \rightarrow X$ and $j_2: U_2 \rightarrow X$ and $j_{12}: U_1 \cap U_2 \rightarrow X$ and $F$ is a sheaf on $X$, then there is a distinguished triangle
$${j_{12}}_* j_{12}^* F \rightarrow {j_1}_* j_1^* F \oplus {j_2}_* j_2^* F \rightarrow F \rightarrow$$
(usual M-V follows by taking F constant and applying $p_*$).
What's a reference? I learned from BBD (Faisceaux Pervers, Asterisque 100), which is great, but maybe more scheme-y than you want.
P.S.: Since you're a homotopy theorist, maybe I'll mention what I think is a great perspective on what the derived category D(X) is. Basically it's just like the ordinary category of sheaves of abelian groups, except you do everything homotopically. So instead of abelian groups you take (excuse me) HZ-module spectra, and you make your sheaves satisfy homotopical descent (this homotopical descent is really the origin of things like M-V above). To make this perspective rigorous it's helpful to use infinity-category theory, as in Lurie's book Higher Topos Theory. Using this approach, one doesn't "derive" things in the sense of starting with an abelian situation and invoking derived categories and functors; rather one makes "derived" definitions, and then all of the natural operations are automatically "derived": for instance the derived f_* and f^* can be given, in the infinity-categorical setting, by the same formulas as the ordinary f_* and f^* in the classical (abelian) setting.
-
### 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 comments on the questions.
1. Sheaves may look like they are very big but they are not. There is a large class of sheaves, constructible sheaves, which are combinatorial objects. Here is how one can think about them. Imagine a polyhedron; associate an abelian group $A(\triangle)$ to each face $\triangle$ and a a morphism of groups $A(\triangle_1)\to A(\triangle_2)$ whenever $\triangle_2$ is a codimension 1 face of $\triangle_1$; assume that the two possible ways of getting from an $n$-face to an $n-2$ face coincide. Then one can introduce a cochain differential by the usual formula. Most sheaves that come up in real life topological problems arise in this way. For example, if all restriction mappings are isomorphisms, we get a local system.
2. The cup product in cohomology is very natural from the sheaf theory perspective as well. We have $H^i(X,A)=Hom_{D^+X}(\underline{A},\underline{A}[i])$ where $A$ is a commutative ring, $\underline{A}$ is a constant sheaf on $X$ with stalk $A$ and $\underline{A}[i]$ is $\underline{A}$ shifted $i$ steps to the left. Since shifting gives an isomorphism of the derived category, we have $Hom_{D^+X}(\underline{A},\underline{A}[j])=Hom_{D^+X}(\underline{A}[i],\underline{A}[i+j])$and so we get a bilinear map $$Hom_{D^+X}(\underline{A},\underline{A}[i])\times Hom_{D^+X}(\underline{A},\underline{A}[j])\to Hom_{D^+X}(\underline{A},\underline{A}[i+j])$$ which is essentially just the composition of morphisms in the derived category. This gives the cup product.
1. (Are there homotopy theorists...) Probably.
2. I am not aware of a version of Eilenberg-Steenrod axioms for sheaves, but this is a good question. One way to state it more precisely would be: give a criterion for a cohomological functor on $D^+X$ to be represented by the constant sheaf. In any case one can use Mayer-Vietoris (at least for an open cover or a closed cover of a reasonable space) to compute sheaf cohomology. Moreover, the long exact sequence for $i:X\subset Y$ a closed subset comes from the distinguished triangle $j_! j^{-1}\to id\to i_\ast i^{-1}\to\cdots$ where $j:Y\setminus X\subset Y$ is the open embeding of the complement.
3. For a general picture on sheaves and related stuff I would recommend two books by Gelfand and Manin. One is Methods of homological algebra and the other is called simply Homological algebra (a survey in Springer's Encyclopedia of math sciences). These books are excellent but a word of warning: there are some typos, missing conditions in some theorems etc.
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1
While constructible sheaves are combinatorial as you say, the constructible derived category, which is the interesting object from viewpoint of sheaf cohomology is not! It is not the derived category of constructible sheaves, but the full subcategory of the derived category of all sheaves consisting of objects with constructible cohomology. Therefore complicated sheaves are always present in the backstage. – Jan Weidner Mar 13 2012 at 21:25
This is mostly an answer to 3:
While I mostly use sheaves arising in topology (and since your question explicitly involves homotopy theory, that might also be your point of view), I think it's a useful learning tool to do some reading about how sheaves come up and are used in complex analysis and Hodge theory. The big benefit there is that most of the sheaves there are already nice (fine, soft,...), so one can do a lot without having to take resolutions by sheaves one has less control over. Wells's book Differential Analysis on Complex Manifolds might be a good starting point, or even Voisin's book on Hodge Theory (which is more advanced).
In the topological world, I second Bredon and Gelfand/Manin (the second edition of Methods of Homological Algebra is in better shape than the first regarding typos). Also Dimca has a book Sheaves in Topology, which is a nice introduction to the derived category point of view that includes some useful examples and exercises.
One last note: in doing actual computations, the exact sequences coming from adjunction triangles (see e.g. Section 2.4 of Dimca) are often your best friends :-)
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http://physics.stackexchange.com/questions/3081/why-cant-a-piece-of-paper-of-non-zero-thickness-be-folded-more-than-n-times
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# Why can't a piece of paper (of non-zero thickness) be folded more than “n” times?
Updated:
"In order to fold anything in half, it must be `π` times longer than its thickness, and that depending on how something is folded, the amount its length decreases with each fold differs."
– Britney Gallivan, the person who determined that the maximum number of times a paper or other finite thickness materials can be folded = 12.
Mathematics of paper folding explains the mathematical aspect of this.
I would like to know the physical explanation of this. Why is it not possible to fold a paper more than "n" (12) times?
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2
– iamsid Jan 16 '11 at 21:03
1
I have done this trick: use a large sheet of tissue paper. The small number of folds is an interesting demonstration of exponential growth, of course, but it isn't really a physics topic, so I have voted to close. – dmckee♦ Jan 16 '11 at 21:06
1
@iamsid What's your point? I don't think it's rude to link you there. It's a humorous way of telling you that you should have at least made a token effort towards research. – Mark Eichenlaub Jan 16 '11 at 21:44
2
Do the close-voters have a physical explanation for this 8-fold limit? I don't know about the mythbusters episode referenced in @Simon's answer, but I do know that this is a universal limit. I remember trying this several times when I was younger. Now it would be great to have a physical reason for this limit. – user346 Jan 16 '11 at 23:44
3
@dmckee IMO geometry is physics. So a geometrical explanation is perfectly satisfactory for me :) – user346 Jan 17 '11 at 22:19
show 8 more comments
## 2 Answers
I remember that the question in your title was busted in Mythbusters episode 72. A simple google search also gives many other examples.
As for single- vs alternate-direction folding, I'm guessing that the latter would allow for more folds. It is the thickness vs length along a fold that basically tells you if a fold is possible, since there is always going to be a curvature to the fold. Alternate-direction folding uses both flat directions of the paper, so you run out of length slightly slower. This would be a small effect since you have the linear decrease in length vs the exponential increase in thickness.
Thanks to gerry for the key word (given in a comment above). I can now make my above guess more concrete. The limit on the number of folds (for a given length) does follow from the necessary curvature on the fold. The type of image you see for this makes it clear what's going on
For a piece of paper with thickness $t$, the length $L$ needed to make $n$ folds is (OEIS) $$L/t = \frac{\pi}6 (2^n+4)(2^n-1) \,.$$ This formula was originally derived by (the then Junior high school student) Britney Gallivan in 2001. I find it amazing that it was not known before that time... (and full credit to Britney). For alternate folding of a square piece of paper, the corresponding formula is $$L/t = \pi 2^{3(n-1)/2} \,.$$
Both formulae give $L=t\,\pi$ as the minimum length required for a single fold. This is because, assuming the paper does not stretch and the inside of the fold is perfectly flat, a single fold uses up the length of a semicircle with outside diameter equal to the thickness of the paper. So if $L < t\,\pi$ then you don't have enough paper to go around the fold.
Let's ignore a lot of the subtleties of the linear folding problem and say that each time you fold the paper you halve its length and double its thickness: $L_i = \tfrac12 L_{i-1} = 2^{-i}L_0$ and $t_i = 2 t_{i-1} = 2^{i} t_0$, where $L=L_0$ and $t=t_0$ are the original length and thickness respectively. On the final fold (to make it n folds) you need $L_{n-1} \leq \pi t_{n-1}$ which implies $L \leq \frac14\pi\,2^{2n} t$. Qualitatively this reproduce the linear folding result given above. The difference comes from the fact you lose slightly over half of the length on each fold.
These formulae can be inverted and plotted to give the logarithmic graphs
where $L$ is measured in units of $t$. The linear folding is shown in red and the alternate direction folding is given in blue. The boxed area is shown in the inset graphic and details the point where alternate folding permanently gains an extra fold over linear folding.
You can see that there exist certain length ranges where you get more folds with alternate than linear folding. After $L/t = 64\pi \approx 201$ you always get one or more extra folds with alternate compared to linear. You can find similar numbers for two or more extra folds, etc...
Looking back on this answer, I really think that I should ditch my theorist tendencies and put some approximate numbers in here. Let's assume that the 8 alternating fold limit for a "normal" piece of paper is correct. Normal office paper is approximately 0.1mm thick. This means that a normal piece of paper must be $$L \approx \pi\,(0.1\text{mm}) 2^{3\times 7/2} \approx 0.3 \times 2^{10.5}\,\text{mm} \approx .3 \times 1000 \, \text{mm} = 300 \text{mm} \,.$$ Luckily this matches the normal size of office paper, e.g. A4 is 210mm * 297mm.
The last range where you get the same number of folds for linear and alternate folding is $L/t \in (50\pi,64\pi) \approx (157,201)$, where both methods yield 4 folds. For a square piece of paper 0.1mm thick, this corresponds to 15cm and 20cm squares respectively. With less giving only three folds for linear and more giving five folds for alternating. Some simple experiments show that this is approximately correct.
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Approximatation:
$a(n)$ - Area after $n$ folds.
$t(n)$ - Thickness after $n$ folds.
$d(n) = k\frac{t(n)}{d(n)}$ - Difficulty after $n$ folds.
$a(n+1) = \frac{1}{2} a(n) \rightarrow a(n) = c_1 2^{1-n}$
$t(n+1) = 2 t(n) \rightarrow t(n) = c_2 2^{n-1}$
$d(n) = k4^{n-1}$
Physics:
You can't fold an atom.
Area of atom $= a(N) = c_1 2^{1-N}$
Solve for $N$.
No paper can be folded more than N times. :)
So, its very very very very very difficult. After folding a few times. ;)
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http://math.stackexchange.com/questions/227515/the-product-of-digits-equal-to-the-sum-of-digits/227532
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# The product of digits equal to the sum of digits
How to find the number(or numbers ) that has $4$ digits, the product of these digits equal to the sum of these digits ?
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Not sure what does this have to do with the number itself? You are just looking for 4 digits that summed or multiplied give the same result? – gt6989b Nov 2 '12 at 13:21
@gt6989b yes but may be there is more than one such number . – htm Nov 2 '12 at 13:25
2
I'd think if there is one, you could get other ones by rotating them? I.e. if 1234 is ok, then 4321 must be ok as well, no? Or am I misunderstanding something? – gt6989b Nov 2 '12 at 13:26
@gt6989b yes you are right . – htm Nov 2 '12 at 13:32
I wonder why everybody is excluding zero as a digit up front, `0000` seems like an obvious first shot to me? – Frerich Raabe Nov 19 '12 at 9:22
## 4 Answers
First of all, let's observe that all of the digits of such a number cannot be the same. You can just manually check that numbers $1111$, $2222$ and so on don't suit us. It is also clear that all of the digits should be non-zero.
Now suppose that we have such a number. Let $a,\,b,\,c,\,d$ be its digits written in non-ascending order: $a \geqslant b \geqslant c \geqslant d$. Then we have $$abcd = a + b + c + d.$$
From this we have an inequality: $$a\cdot bcd < 4a.$$ This inequality is strict, because at least one of $b, c, d$ is strictly smaller than a. So we have: $$bcd < 4,$$ which is the same as saying $$bcd \leqslant 3.$$ This only leaves us with 3 possible combinations for $(b, c, d)$: $(1, 1, 1)$, $(2, 1, 1)$ and $(3, 1, 1)$.
If $b=c=d=1$, then $a\cdot 1 \cdot 1 \cdot 1 = a + 1 + 1 + 1$, which can't be true.
If $b=2$ and $c=d=1$, then $a \cdot 2 \cdot 1 \cdot 1 = a + 2 + 1 + 1$, which means that $a=4$. This gives us one possible solution: $a=4, b=2, c=d=1$.
If $b=3$ and $c=d=1$, then $a \cdot 3 \cdot 1 \cdot 1 = a + 3 + 1 + 1$, which is impossible.
So, the only solution is $a=4$, $b=2$, $c=d=1$. There are $12$ numbers with such digits.
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Why do you exclude zero up front, `0000` would work as well, no? – Frerich Raabe Nov 19 '12 at 9:24
@FrerichRaabe I think that technically $0000$ is not a 4-digit number. It is my understanding that 4-digit numbers all lie between $1000$ and $9999$. $0050$, for instance, is a 2-digit number, because its standard form is $50$. From this point of view, $0000$ is a 1-digit number, because the standard notation for it is $0$. – Dan Shved Nov 19 '12 at 11:22
You can narrow your search rapidly:
1. no digits $0$;
2. at least one digit $1$ (otherwise the product exceeds the sum easily);
3. at least two digits greater than $1$ (otherwise the sum now exceeds the product);
4. exactly two digits greater than $1$ (the product of three such digits would exceed their sum by at least $2$).
So we're looking for pairs of digits in $\{2,3,\ldots,9\}$ whose product exceeds their sum by exactly $2$ (the number of digits $1$ we need to throw in). If one of them is $2$, the other must be $4$. If the smallest of the pair is at least $3$, then their product exceeds their sum by at least $3$, so this cannot happen.
So all in all there is essentially one solution, but since you asked for numbers , the $12$ permutations of the digits of $1124$ give you all solutions.
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i observed a pattern in these numbers. 22 123 1124 11125 111126 1111127 and derived a formula for this.
if the last two digits are assumed to be a and b and for an n digit number there will be n-2 1's and a and b are to be found out using the below formula.
a=(b+n-2)/(b-1)
where a and b are from 2 to 9 which need to be evaluated manually for b=2 to 9
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Does this give a number as the question asks? – vonbrand Jan 29 at 17:32
Yes.. if the number of digits is 4 a=(b+2)/(b-1) and number of 1's in the number is 2.so the number would be 11ab.Now we have to solve for a and b using a=(b+2)/(b-1).which gives integer values for b=2 or b=4 and a=4 when b is 2 and a=2 when b=4.so the number would be 1124 and all permutations of it like 1124,1142,1214,1412 and so on. – Syam Kumar Feb 1 at 7:56
Must be various combinations of $1,1,2,4$. I don't think there are any other combinations, I looked at all small numbers...
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Can we find the number by using algebraic method ? – htm Nov 2 '12 at 13:41
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http://mathoverflow.net/questions/78788/dimension-of-moduli-space-in-lagrangian-floer-homology
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## Dimension of moduli space in Lagrangian Floer homology
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $(M,\omega)$ be symplectic manifold with $\omega=c_{1}=0$ on $\pi_{2}M$. Let $\Lambda\subseteq M$ be Lagrangian submanifold. Let $H:M\times S^{1}\rightarrow\mathbf{R}$ be Hamiltonian and $J$ be compatible almost complex structure.
Let $\xi(t)$ be non-degenerate contractible 1-periodic orbit of $H$. Let $\mathcal{M}$ be space of maps $u:]-\infty,0]\times S^{1}\rightarrow M$ which satisfy $u_{s}+Ju_{t}=JX_{H_{t}}(u)$ and $\lim_{s\rightarrow-\infty}u(s,t)=\xi(t)$ and $u(0,\cdot)\subseteq\Lambda$.
For generic $J$, $\mathcal{M}$ is manifold of dimension $-i(\xi)$, where $i(\xi)\in\mathbf{Z}$ is Conley-Zehnder index of $\xi$.
I have two question: (1) please explain why dimension of $\mathcal{M}$ is $-i(\xi)$. This seems strange, as it apparently does not depend on $\Lambda$? (2) is similar statement still true if instead of Lagrangian $\Lambda$ use any submanifold $P$ of $M$ - then maybe $\mathcal{M}$ has expected dimension $-i(\xi) + dim(P) - dim(M)/2$?
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## 2 Answers
(1) The virtual dimension is only equal to $-i(\xi)$ once you have chosen enough trivializations and so forth that it works out that way. This formula hides that this choice of trivialization is what sees $\Lambda$. Take a look at, for instance, the Bourgeois-Ekholm-Eliashberg paper on the surgery formula for contact homology for a discussion of choice of trivializations. I will give a very fast summary. (It is also discussed in Eliashberg-Givental-Hofer's introduction to SFT, and in a number of other papers on the subject of contact homology etc.)
The Conley-Zehnder index is associated to a path of symplectic matrices starting at the identity and ending at a matrix without any eigenvalue equal to $1$. In order to obtain a path of matrices from a periodic orbit, you must trivialize $TM$ over $\xi$. In particular, the Conley-Zehnder index depends on trivialization.
The general index formula, however, has a Conley-Zehnder index term and a Maslov index term (and a relative Chern number term, if your boundary trivializations don't extend to a trivialization of $u^*TM$). The CZ index and the Maslov index each depend on trivialization. The difference/sum does not.
In this case, what you have implicitly done is trivialized $u^*TM$ so that the Lagrangian loop $u|_{{0}\times S^1}^*TL$ in $u|_{{0}\times S^1}^*TM$ has Maslov index $0$. Note that $u^*TM$ is trivial because $u$ is a surface with boundary. Your choice of trivialization on the boundary now extends to one of $u^*TM$. Finally, by the asymptotic convergence theorem, this induces a trivialization of $\xi^*TM$. You use this trivialization to compute the Conley-Zehnder index.
(2) Orbicular is correct. You can see the problem in a very simple toy problem. Consider finding holomorphic maps from the disk $D$ to $\mathbb{C}$. A totally real (actually Lagrangian) boundary condition is given by asking the boundary of the disk to map to the unit circle -- this has a 3 dimensional solution space. Let's take $P = \mathbb{C}$, say. Start with the standard map on the disk given by $u(z) = z$. Now take any simple closed curve in $\mathbb{C}$ that is close to the unit circle. Then, there exists a 3-parameter family of holomorphic disks with boundary on that curve. The space of such curves is infinite dimensional, showing that my linearized problem has an infinite dimensional kernel.
To see the problem for $P$ of lower dimension, we want to look in $\mathbb{C}^2$. Here, choose a generic circle for the boundary condition. For general choice of circle, you get a boundary value problem you cannot solve. You can work out by hand the infinite codimension condition you get on the circle that actually admits any solutions at all.
-
2
Let me add two more references - Appendix C of McDuff and Salamon's big book explains why the index of the operator on the disc obtained by capping off the Hamiltonian orbit end of the half cylinder is given by the Maslov index of the Lagrangian loop $u^*T\Lambda$ (wrt the chosen trivialization) - and Matthias Schwarz' thesis has a very detailed proof (see Section 3.3) as to why the index of the operator on the cap is given by the CZ index. Since the index is additive, as Sam says the index on the half cylinder is [index of capped cylinder] - [index of cap], which gives the required answer. – Will Merry Oct 22 2011 at 9:52
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
For a general submanifold P the problem you pose is not a Fredholm problem. In particular, you cannot expect your moduli space to be finite-dimensional. This has to do with admissible boundary conditions for the Cauchy-Riemann equation (buzzword: totally real boundary condition).
I am not so sure about the dimension of the moduli space. However I do know that the dimension can be read off from the Fredholm index of the linearized problem. I would expect some Maslov index to appear.
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http://cms.math.ca/Events/summer12/abs/car
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2012 CMS Summer Meeting
Regina Inn and Ramada Hotels (Regina~Saskatchewan), June 2 - 4, 2012 www.cms.math.ca//Events/summer12
Cluster Algebras and Related Topics
Org: Ralf Schiffler (Connecticut) and Hugh Thomas (UNB)
[PDF]
IBRAHIM ASSEM, Université de Sherbrooke
On the first Hochschild cohomology group of a cluster-tilted algebra [PDF]
This is a joint work with Ralf Schiffler and Maria Julia Redondo. Given a cluster-tilted $k$-algebra $B$, we study its first Hochschild cohomology group $HH^1(B)$ with coefficients in the $B$-$B$ bimodule $B$. If $C$ is a tilted algebra such that $B$ is the relation-extension of $C$, then we show that if $C$ is constrained, or else if $B$ is tame, then $HH^1(B)$ is isomorphic, as a $k$-vector space, to the direct sum of $HH^1(C)$ with $k^{n_{B,C}}$ where $n_{B,C}$ is an invariant linking the bound quivers of $B$ and $C$. In the representation-finite case, $HH^1(B)$ can be read just by looking at the quiver of $B$.
JUAN CARLOS BUSTAMANTE, Université de Sherbrooke
Hochschild cohomology and the derived class of $m$-cluster tilted algebras of type $\mathbb{A}$ [PDF]
Joint work with V. Gubitosi, from Sherbrooke, see http://arxiv.org/abs/1201.4182
For a given integer $m$ and a hereditary algebra $H$, the $m$-cluster category of $H$, $\mathcal{C}_m(H)$ is obtained from the derived category $\mathcal{D}(H)$ by identifying the Auslander-Reiten translation with the $m^{\rm th}$ power of the shift $[1]^m:=[m]$. In $\mathcal{C}_m(H)$ there are cluster tilting objects, whose endomorphisms algebras are called $m$-cluster tilted algebras.
The aim of this work is to classify the algebras that are derived equivalent to $m$-cluster tilted algebras of type $\mathbb{A}$. The first result states that a connected algebra $A=kQ/I$ is derived equivalent to an $m$-cluster tilted algebra of type $\mathbb{A}$ if and only of it is gentle, having exactly $|Q_1|-|Q_0|+1$ oriented cycles of length $m+2$ each of which has full relations. We then prove:
{\bf Theorem:} Let $A = kQ/I$ and $A' = k'Q'/I'$ be connected algebras derived equivalent to $m$-cluster tilted algebras of type $\mathbb{A}$. Then (among others) the following conditions are equivalent. \begin{enumerate} \item $A$ and $A'$ are derived equivalent, \item $A$ and $A'$ are tilting-cotilting equivalent, \item ${\rm HH}^\ast(A)\simeq {\rm HH}^\ast(A)$ and $K_0(A)\simeq K_0 (A')$ \item $\pi_1(Q, I) \simeq \pi_1 (Q' , I' )$ and $|Q_0| = |Q'_0|$. \end{enumerate} Our approach differs from previous works on related topics in the fact that we use the Hochschild cohomology ring as a derived invariant. We shall discuss about the proof and derive some consequences, among which we recover previous known results.
LUCAS DAVID-ROESLER, University of Connecticut
Representation theory using surfaces [PDF]
We introduce a class of finite dimensional gentle algebras, surface algebras, coming from certain partially triangulated surfaces. The construction of these algebras is motivated by the combinatorial relationship between iterated tilted algebras of Dynkin type and the cluster-tilted algebras of Dynkin type given admissible cuts of quivers. We will discuss some of the ways we can study the representation theory of these algebras using only data coming from the surface.
MICHAEL GEKHTMAN, University of Notre Dame
Cremmer-Gervais Cluster Algebras [PDF]
I will report on an ongoing joint project with M. Shapiro and A. Vainshtein devoted to a conjectural correspondence between classes in the Belavin-Drinfeld classification of of Poisson-Lie structures on a simple Lie groups and cluster algebra structures in the ring of regular functions on the group. I will concentrate on the case associated with the Cremmer-Gervais r-matrix, that is the farthest away from the standard Poisson-Lie structure in SL(n) and describe the corresponding cluster algebra.
MAX GLICK, University of Michigan
Singularity confinement for the pentagram map [PDF]
The pentagram map, introduced by R. Schwartz, is a birational map on the configuration space of polygons in the projective plane. We study the singularities of the iterates of the pentagram map. We show that a typical singularity disappears after a finite number of iterations, a confinement phenomenon first discovered by Schwartz. We provide a method to bypass such a singular patch by directly constructing the first subsequent iterate that is well defined on the singular locus under consideration. The key ingredient of this construction is the notion of a decorated (twisted) polygon, and the extension of the pentagram map to the corresponding decorated configuration space.
KIYOSHI IGUSA, Brandeis University
Distinguished triangles in continuous cluster categories. [PDF]
Cluster categories are triangulated categories which categorify cluster algebras. We give recognition principles for distinguished triangles in the continuous cluster category using continuous deformations of triangles and show that these also give recognition principles for distinguished triangles in the standard cluster categories of type $A_n$ and $A_\infty$. I will begin with a brief review of the continuous Frobenius category and end with explicit formulas for which mapping make $A\to B\to C\to A[1]$ into a distinguished triangle. This is joint work with Gordana Todorov from Northeastern University.
KYUNGYONG LEE, Wayne State University
Positivity in rank 2 cluster algebras [PDF]
We present a canonical basis for the cluster algebra associated to any skew-symmetrizable $2\times 2$ integer matrix. This is joint work with Li Li, Paul Sherman and Andrei Zelevinsky.
GREGG MUSIKER, University of Minnesota
Graph theoretical formulas for certain periodic quivers [PDF]
We present work in progress on combinatorial formulas for cluster variables arising from periodic cluster algebras. Here, we mean periodic in the sense of Fordy and Marsh. This includes formulas for cluster variables with principal coefficients associated to the Gale-Robinson sequence and the Aztec Diamond quiver.
DYLAN RUPEL, University of Oregon
Quantum Cluster Characters via Hall-Ringel Algebras [PDF]
It was recognized in the original papers of Caldero-Chapoton and Caldero-Keller that the multiplication formulas for cluster characters resemble the multiplication in the dual Hall-Ringel algebra. In this talk we will realize the quantum cluster character via a natural algebra homomorphism from the Hall-Ringel algebra to quasi-commuting Laurent polynomials. This is joint work with Arkady Berenstein.
VASILISA SHRAMCHENKO, University of Sherbrooke
Cluster automorphisms [PDF]
We introduce the notion of cluster automorphism of a given cluster algebra as a $\mathbb Z$-automorphism of the cluster algebra that sends a cluster to another and commutes with mutations. We study the group of cluster automorphisms in detail for acyclic cluster algebras and cluster algebras from surfaces, and we compute this group explicitly for the Dynkin and Euclidean types.
GORDANA TODOROV, Northeastern University
Morphisms detemined by objects. [PDF]
Abstract: The notion of Morphism Determined by Object was introduced by Maurice Auslander in order to construct and classify morphisms, mostly in module categories. A recent renewed interest in this notion is coming from C. M. Ringel for module categories and also from H. Krause, who extended and studied this notion in triangulated categories. In this talk, I will discuss functorial interpretation of the morphisms determined by objects in terms of the socle of the cokernel functor induced by the given morphism. I will give applications in cluster categories and other related categories.
ADAM-CHRISTIAN VAN ROOSMALEN, University of Regina
Cluster categories associated to new hereditary categories [PDF]
Given a finite quiver, one can associate a cluster category by considering orbits of the bounded derived category of finite dimensional representations. In this talk, we want to replace the original quiver by a suitable small category such that the orbit construction still makes sense, thus obtaining new examples of 2-Calabi-Yau categories with cluster tilting subcategories. We will consider some examples where one can use combinatorics to describe the cluster tilting subcategories, as is done by Holm and Jørgensen in the case of the infinite Dynkin quiver $A_\infty$ using triangulations of the $\infty$-gon.
## Event Sponsors
Support from these sponsors is gratefully acknowledged.
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http://mathhelpforum.com/differential-geometry/107647-proving-ratio-test-fails.html
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# Thread:
1. ## Proving the ratio test fails.
Can anyone prove that the ratio test is inconclusive when
$\lim_{n \rightarrow \infty} |\frac{a_{n+1}}{a_{n}}| = 1,$ or
$\lim_{n \rightarrow \infty} |\frac{a_{n+1}}{a_{n}}| DNE.$
2. For the first one take $a_n= \frac{1}{n}$ and $b_n= \frac{1}{n^2}$ both limits are 1, but one converges and the other diverges. For the second I'm not sure...
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http://mathhelpforum.com/discrete-math/143330-bijection.html
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# Thread:
1. ## Bijection
Prove that there is a bijection between (0,1) and (b,infinity) for any real number b.
A function is needed. I've though about the function f(x)=b/x but it won't work if b<=0. Any idea?
2. Originally Posted by TitaniumX
Prove that there is a bijection between (0,1) and (b,infinity) for any real number b.
A function is needed. I've though about the function f(x)=b/x but it won't work if b<=0. Any idea?
How about f(x) = b - 1 + 1/x?
3. Originally Posted by TitaniumX
Prove that there is a bijection between (0,1) and (b,infinity) for any real number b.
A function is needed. I've though about the function f(x)=b/x but it won't work if b<=0. Any idea?
$f (x)= tan(\pi x - \pi/2)$ is (I believe) standard for expanding $(0, 1)$ to $\mathbb{R}$, so I would use this.
This can be relatively easily manipulated to give you the function you want.
(Half it to give you a bijection between $(0, 1)$ and $(0, \infty)$, then scale).
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http://physics.stackexchange.com/questions/45027/time-contraction
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# Time Contraction
This is my first time posting on this site. I am a computer programmer that stumbled across a physics text book and have a question on special relativity. So firstly, I understand that there is no preferential inertial frame of reference. Secondly, as a A body travels faster relative to another body B at rest, body A experiences time dilation, effectively resulting in a forward jump in time. Thirdly, the speed at which a body in relative motion can travel is limited by the speed of light (c).
Due to the the first premise (that there is no preferential frame), could we not also argue that the body at rest is actually travelling at a negative velocity relative to the other body? Thus, if we consider the situation in from this reversed perspective, a time contraction is then experienced to the body moving at relative negative velocity. For instance, if we consider a person on Earth as being at rest (0 km/h), then could we think of a person standing on Pluto (which has an orbital velocity of 0.159 relative to Earch) as having a negative velocity relative to that of the person stationary on the surface of Earth, thus experiencing time contraction relative to the time experienced on Earth?
This can be confirmed by substituting a negative value in the standard time dilation equation for u^2. This results in a value greater than 1 occurring as the divisor, causing the numerator to 'shrink' in value, thus resulting in a 'time contraction' rather than time dilation.
With some searching, I have come across a number of examples of time dilation. However, all of them seem to assume that a body on rest is traveling at 0 km/h. Rather, I argue that such a body would be 'relatively' at rest, as from a greater perspective, this body is subject to the turning of the Earth about its axis, the movement of the Earth along its orbit around the Sun, the Sun around the Milky Way, and finally the Milky Way away from other galaxies due to the expansion of the universe.
If my above reasoning is correct, then it seems that if there was a body moving at 0 km/h relative to the expansion of the universe, then the Earth would be travelling faster than that body, which means that the Earth would experience time dilation relative to this body, and that body would experience time contraction relative to the Earth. If time dilation equates to a forward jump in time, then it stands to reason that time contraction would likewise equate to a backwards jump in time. Although, it occurs to me that perhaps it is more useful to not think of the body moving backwards in time, but rather to think of it as being stationary relative to the expansion of space-time, thus time is moving forward past the body experiencing time contraction.
I know I am not a physicist, but I was wondering if anyone could tell me if my above speculations are correct, or that I'm an idiot, or if all of this is already well documented somewhere (which I have yet to discover).
Any clarification would be greatly appreciated.
-
You need to use -(v^2) instead of (-v)^2. The reason for this is that practically, you can't have time dilation occurring from both inertial frame perspectives, otherwise you would never have the twin paradox. – ralfe Nov 24 '12 at 22:37
## 1 Answer
Any clarification would be greatly appreciated.
So firstly, I understand that, firstly, there is no preferential inertial frame of reference. Secondly, as a body travels faster relative to another body at rest, the body in relative motion experiences time dilation, effectively resulting in a forward jump in time. Thirdly, the speed at which a body in relative motion can travel is at the speed of light (c).
It's true that there is no preferred inertial frame of reference (IRF). However, your second point is, at best, confused.
Since there is no preferred IRF, there isn't a body in relative motion. To have relative motion, you need at least two bodies.
Relative motion means that the distance between the two bodies changes with time. While body A can claim that it is body B that is moving, body B has an equally valid claim that it is body A that is moving.
Thus, when it comes to time dilation, you must understand that one body doesn't "experience" it while the other doesn't.
This is important: According to A, B's clocks run slow; according to B, A's clock run slow. Neither A or B "experience" time dilation. They simply observe that the other's clocks run at slower rate then their own. (They also note that the other's clocks are not synchronized but, while ultimately important, it's not needed now).
Likewise, according to A, B's rulers are contracted; according to B, A's rulers are contracted. Neither A or B "experience" length contraction. They simply observe that the other's rulers are shorter than theirs.
So, your speculations reveal a fundamental misunderstanding of Special Relativity. Time dilation and length contraction are perfectly symmetrical.
(As more a less an side and to your third point, no material object can be observed travelling at the speed of $c$. The speed can be arbitrarily close but it must be less than $c$.)
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Thank you for the clarification. As I said, I'm no physicist, I just came across this stuff the other day and had some thoughts. The text book I was looking at said that in the twin paradox the younger twin jumped forward in time. Would this not be relative to the older twin? If so, could you not state that the older twin jumped backwards in time relative to the younger twin? – ralfe Nov 25 '12 at 1:03
Properly understanding the twin "paradox" involves only recognizing that the stay at home twin never changes reference frames while the space travelling twin changes from an outgoing to an ingoing reference frame halfway through his journey. The world lines of the two twins are then not congruent and thus, the lapse of proper time along the world lines are different. See: en.wikipedia.org/wiki/… – Alfred Centauri Nov 25 '12 at 1:16
Thank you once again. That does make a lot of sense. My next question is then what if the space travelling twin were to change reference frames to an inertial frame slower than than that of the stay at home twin? – ralfe Nov 25 '12 at 1:38
A slower frame of reference according to... what? – Alfred Centauri Nov 25 '12 at 1:50
So, if the surface of Earth where the stay-at-home twin lives is moving at a certain speed, then I am asking how would the time dilation equation be interpreted if the space travelling twin were to remain stationary in the space ship relative to the Sun, thus in a 'slower' frame of reference relative to the stay-at-home twin? The Earth would continue rotating and travelling around the Sun, thus travelling at a velocity greater than that of the spaceship. – ralfe Nov 25 '12 at 1:59
show 4 more comments
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http://nrich.maths.org/public/leg.php?code=149&cl=3&cldcmpid=6853
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# Search by Topic
#### Resources tagged with Area similar to History of Trigonometry - Part 2:
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##### Other tags that relate to History of Trigonometry - Part 2
Area. Mathematical reasoning & proof. Mixed trig ratios. Countability. Famous mathematicians. History of measurement. History of computing. History of mathematics. History of number systems. Infinity.
### There are 86 results
Broad Topics > Measures and Mensuration > Area
### Six Discs
##### Stage: 4 Challenge Level:
Six circular discs are packed in different-shaped boxes so that the discs touch their neighbours and the sides of the box. Can you put the boxes in order according to the areas of their bases?
### From All Corners
##### Stage: 4 Challenge Level:
Straight lines are drawn from each corner of a square to the mid points of the opposite sides. Express the area of the octagon that is formed at the centre as a fraction of the area of the square.
### Square Pizza
##### Stage: 4 Challenge Level:
Can you show that you can share a square pizza equally between two people by cutting it four times using vertical, horizontal and diagonal cuts through any point inside the square?
### Max Box
##### Stage: 4 Challenge Level:
Three rods of different lengths form three sides of an enclosure with right angles between them. What arrangement maximises the area
### Salinon
##### Stage: 4 Challenge Level:
This shape comprises four semi-circles. What is the relationship between the area of the shaded region and the area of the circle on AB as diameter?
### Towers
##### Stage: 3 Challenge Level:
A tower of squares is built inside a right angled isosceles triangle. The largest square stands on the hypotenuse. What fraction of the area of the triangle is covered by the series of squares?
### Percentage Unchanged
##### Stage: 4 Challenge Level:
If the base of a rectangle is increased by 10% and the area is unchanged, by what percentage (exactly) is the width decreased by ?
### Appearing Square
##### Stage: 3 Challenge Level:
Make an eight by eight square, the layout is the same as a chessboard. You can print out and use the square below. What is the area of the square? Divide the square in the way shown by the red dashed. . . .
### Framed
##### Stage: 3 Challenge Level:
Seven small rectangular pictures have one inch wide frames. The frames are removed and the pictures are fitted together like a jigsaw to make a rectangle of length 12 inches. Find the dimensions of. . . .
### Take Ten
##### Stage: 3 Challenge Level:
Is it possible to remove ten unit cubes from a 3 by 3 by 3 cube made from 27 unit cubes so that the surface area of the remaining solid is the same as the surface area of the original 3 by 3 by 3. . . .
### Semi-square
##### Stage: 4 Challenge Level:
What is the ratio of the area of a square inscribed in a semicircle to the area of the square inscribed in the entire circle?
### Tri-split
##### Stage: 4 Challenge Level:
A point P is selected anywhere inside an equilateral triangle. What can you say about the sum of the perpendicular distances from P to the sides of the triangle? Can you prove your conjecture?
### Disappearing Square
##### Stage: 3 Challenge Level:
Do you know how to find the area of a triangle? You can count the squares. What happens if we turn the triangle on end? Press the button and see. Try counting the number of units in the triangle now. . . .
### Gutter
##### Stage: 4 Challenge Level:
Manufacturers need to minimise the amount of material used to make their product. What is the best cross-section for a gutter?
### Equilateral Areas
##### Stage: 4 Challenge Level:
ABC and DEF are equilateral triangles of side 3 and 4 respectively. Construct an equilateral triangle whose area is the sum of the area of ABC and DEF.
### An Unusual Shape
##### Stage: 3 Challenge Level:
Can you maximise the area available to a grazing goat?
### Circle-in
##### Stage: 4 Challenge Level:
A circle is inscribed in a triangle which has side lengths of 8, 15 and 17 cm. What is the radius of the circle?
### Isosceles Triangles
##### Stage: 3 Challenge Level:
Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
### Efficient Packing
##### Stage: 4 Challenge Level:
How efficiently can you pack together disks?
### Curvy Areas
##### Stage: 4 Challenge Level:
Have a go at creating these images based on circles. What do you notice about the areas of the different sections?
### Partly Circles
##### Stage: 4 Challenge Level:
What is the same and what is different about these circle questions? What connections can you make?
### Inscribed in a Circle
##### Stage: 3 Challenge Level:
The area of a square inscribed in a circle with a unit radius is, satisfyingly, 2. What is the area of a regular hexagon inscribed in a circle with a unit radius?
### Rhombus in Rectangle
##### Stage: 4 Challenge Level:
Take any rectangle ABCD such that AB > BC. The point P is on AB and Q is on CD. Show that there is exactly one position of P and Q such that APCQ is a rhombus.
### Hallway Borders
##### Stage: 3 Challenge Level:
A hallway floor is tiled and each tile is one foot square. Given that the number of tiles around the perimeter is EXACTLY half the total number of tiles, find the possible dimensions of the hallway.
### Isosceles
##### Stage: 3 Challenge Level:
Prove that a triangle with sides of length 5, 5 and 6 has the same area as a triangle with sides of length 5, 5 and 8. Find other pairs of non-congruent isosceles triangles which have equal areas.
### Semi-detached
##### Stage: 4 Challenge Level:
A square of area 40 square cms is inscribed in a semicircle. Find the area of the square that could be inscribed in a circle of the same radius.
### Pick's Theorem
##### Stage: 3 Challenge Level:
Polygons drawn on square dotty paper have dots on their perimeter (p) and often internal (i) ones as well. Find a relationship between p, i and the area of the polygons.
### Areas of Parallelograms
##### Stage: 3 Challenge Level:
Can you find the area of a parallelogram defined by two vectors?
### Muggles Magic
##### Stage: 3 Challenge Level:
You can move the 4 pieces of the jigsaw and fit them into both outlines. Explain what has happened to the missing one unit of area.
### Bound to Be
##### Stage: 4 Challenge Level:
Four quadrants are drawn centred at the vertices of a square . Find the area of the central region bounded by the four arcs.
### Two Circles
##### Stage: 4 Challenge Level:
Draw two circles, each of radius 1 unit, so that each circle goes through the centre of the other one. What is the area of the overlap?
### Dissect
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http://physics.stackexchange.com/questions/3611/why-does-a-ballerina-speed-up-when-she-pulls-in-her-arms?answertab=active
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# Why does a ballerina speed up when she pulls in her arms?
My friend thinks it's because she has less air resistance but I'm not sure.
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## 4 Answers
This is a fairly long answer, but I thought it would be fun to analyze the skater thinking solely about forces. Angular momentum comes in at the end, when it pops up inevitably. I'll give a qualitative answer describing the forces in the system and how they cause spinning, then a quantitative answer to calculate the spin rate.
Qualitative Answer
As an ice skater pulls in her arms a legs, her arms and legs exert a torque on her body, causing her to spin faster.
Set a plate in front of you. Try pushing on at in various places and in various directions. You can imagine an arrow pointing in the direction of the force, starting from the point where you're pushing. If this arrow points towards the center of the plate, the plate will not rotate. Otherwise, it will.
In this picture, the red arrows indicate forces on the gray plate. The force labeled "Spin" will induce some spinning in the plate (in addition to accelerating it as a whole) because the dotted line "j" of that force does not go through the center of the plate. The force labeled "NoSpin" will only accelerate the plate and not cause any spinning because it lies on a line that passes through the center of the plate.
There is nothing special about the circular plate in this example. Any other shape would work as well, but you would need to define the center point by the center of mass.
To see that the skater spins up as she pulls in her arms, we'll need to find the forces exerted on her body while she pulls her arms in. Then we can see whether these forces point straight towards the center of her body or not.
We'll model an ice skater as a circle of mass $M$ with a massless stick pointing through it, and two more circles of mass $m$ on either end of the stick. The skater can move the small circles inward or outward as she spins. Here's a picture of the setup, along with the path her "arms" (the small circles) trace out if she does not pull them in at all while spinning.
If you imagine holding something heavy in your arms while spinning, your arms would feel as if they're being pulled out of their sockets. In fact, they are. You arms exert a force on the weights straight inwards towards your body, and the weights exert equal and opposite force straight outward.
Each force is color-coded according to which body it acts on. $F_1$, for example, is the force exerted on the skater by the red arm. (This force is really exerted on the stick, which is rigidly attached to the skater.)
Both the blue forces lie on a line passing through the center of the skater, so the skater's spin rate doesn't change in this scenario. As long as she leaves her arms out, she'll remain spinning the same speed (neglecting friction or other losses of energy).
Now we imagine the skater pulling her arms in. If you watch the path the "arms" (small circles) trace out, you see a spiral shape.
Here, we show the two arms with both their past and future trajectories traced out. The arms are spinning around the skater while simultaneously being pulled in.
It will be more difficult to find the forces involved here. The arms are no longer moving in simple circles. However, at any given instant, there is a particular circle along which a given arm appears to move. This is the osculating circle. As before, there's a force on the arm pointing in towards the center of the osculating circle.
This is not the whole story, because we can no longer assume that the speed of the arms is constant. Hence there may also be a force on the arms in the direction of their motion. In the next picture, we'll draw only the forces on one arm, just to keep things from getting too cluttered.
This picture is zoomed in on one arm. The blue circle is still the skater. The green circle is the osculating circle. $F_r$ is the centripetal force towards the center of the osculating circle that curves the arm's path, and $F_t$ is a (small) tangential force in the direction of motion that speeds the arm up.
Without knowing just how fast the arms are moving, how much they're accelerating, and the equation describing their trajectory, it's hard to tell precisely what the sizes of these forces will be. However, we can see that the forces in general have no obligation to point towards the skater's center any more. She can change her spin rate because the forces may not necessarily lie on lines that pass through her center.
In order to see just what these forces are, we need to do an quantitative analysis.
Quantitative Answer
The plan of this answer is to find the forces on the arms as they spiral in, then use Newton's third law to find the forces the arms exert on the stick. Next, we'll relate these forces to the rate of change of the energy of the skater. The energy of the skater can also be calculated from her motion directly, so we'll do that, take a time derivative, and compare to our previous expression. This will reveal a conserved quantity, the angular momentum, which allows us to find the skater's rotation rate as a function of the initial conditions and the final distance of the arms from her center - the details of the shape of the spiral and how fast the arms are pulled in do not matter. Finally, we'll see that the skater spins faster and faster as she pulls her arms in.
We'll use polar coordinates to describe the positions of the arms. The radial coordinate $r$ of arm 1 is some function of its angular coordinate $\theta$.
$$r = f(\theta)$$
With the definition $\omega = \dot{\theta}$, we have
$$\dot{r} = f'\omega$$ $$\ddot{r} = f''\omega^2 + f'\dot{\omega}$$
The acceleration in polar coordinates is
$$\vec{a} = \hat{r}(\ddot{r} - r \dot{\theta}^2) + \hat{\theta}(2\dot{r}\dot{\theta} ^2 + r\dot{\omega})$$
The force on this arm is given by Newton's second law, $\vec{F} = m\vec{a}$. The force the arm exerts on the stick is the negative of this, by Newton's third law. This force of the arm on the stick is what we're interested in.
As the stick spins, the force of the arm on the stick does work on the stick, which goes into the kinetic energy of the skater (the stick itself has no mass). The velocity of the point on the stick where the force is applied is $\vec{v} = f\omega\hat{\theta}$. The power delivered by this force, doubled to include the work done by the other arm, is
$$P = \vec{F} \cdot \vec{v} = -2m\omega f(2f'\omega^2 + f\dot{\omega})$$
This is the rate of change of the kinetic energy of the skater. That kinetic energy is
$$T = \frac{1}{2}M (R \omega)^2$$
with $R$ the radius of the skater. If we equate the power to the time derivative of the kinetic energy, we get
$$MR^2\dot{\omega} = -2mf(2f'\omega^2 + f\dot{\omega})$$
If you stare at this and scratch your head a moment, it's mathematically equivalent to
$$\frac{\textrm{d}}{\textrm{d}t}\left(\omega(MR^2 + 2mf^2)\right) = 0$$
This means we've discovered something that doesn't change in time - a conserved quantity. It's called the angular momentum, and the part $MR^2 + 2mf^2$ is called the moment of inertia. We denote the angular momentum by $L$ and (switching back from $f$ to $r$) write
$$L = \omega(MR^2 + 2mr^2)$$
Because $L$ is a constant, we can find $\omega$.
$$\omega = \frac{L}{MR^2 + 2mr^2}$$
We've got the angular frequency as a function of $r$ only - the precise function form of $f$ didn't matter, and neither did how quickly we traversed the path. As long as we know $L$ from the initial conditions, we've solved the problem.
From
$$\frac{\textrm{d}\omega}{\textrm{d}r} = \frac{-4m L r}{(MR^2+2mr^2)^2}$$
we see that, assuming $L$ is positive, as $r$ goes down, $\omega$ goes up, so the skater goes faster and faster as she pulls in her arms.
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Yes, I was considering mentioning this to my students as an explanation for why it is that W=Fd works to increase her KE when she pulls in those arms. – Carl Brannen Jan 24 '11 at 5:33
@Carl What age range are your students? – Mark Eichenlaub Jan 24 '11 at 5:40
Maybe 19 to 65. They're typically working on 2-year degrees in drafting or electronics technician. Teaching them physics is a humbling experience. – Carl Brannen Jan 24 '11 at 5:52
A very simple explanation is the following: the arms of the ballerina are pulled outwards by the centripetal force she experiences by spinning. When she pulls her arms in, she is doing work by more than counteracting this force and this is what makes her spin faster.
This is due to the fact that the spin velocity is related to the effort that the ballerina makes in pulling in her arms. The closer the arms, the more force she has to use to pull them in future or keep them in position, and the faster she spins.
When I was in high school, we did a more hardcore version of this experiment by sitting on a chair that could spin, holding two heavy weights outwards. Then someone would spin the poor test subject and ask him to pull in the weights... The results were quite scary... :-)
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It has nothing to do with how hard she pulls in her arms (although it is related to the work done). If she pulls harder they come in faster, but her total change in angular velocity is the same. – Joe Fitzsimons Jan 22 '11 at 21:10
@Joe, what's your problem dude? This is a qualitative answer. The energy she uses to pull in the arms is exactly the energy she gains by spinning faster. – Sklivvz♦ Jan 22 '11 at 21:16
@Sklivvz: This isn't a personal attack, it simply that the line "This is due to the fact that the spin velocity is a function of how hard you pull your arms in." is false, and I was pointing that out. It's an unfortunate coincidence that I have problems with your other question in parallel. I've been trying to raise the quality of the answers by flagging errors. – Joe Fitzsimons Jan 22 '11 at 21:19
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It's a beautiful qualitative answer, an important one, and one that I give to my students regularly (of course I also talk about conservation of angular momentum). And thanks to the other answer providers, Steven is one of my students. We put the question onto Stack Exchange at the start of class at 9AM PST and the system gave an answer before the lab was over. Nice demonstration of a useful tool for students. – Carl Brannen Jan 22 '11 at 21:30
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The problem here is that second paragraph, which is phrased in a way that seems to reverse the causality. I can see how you might arrive at that wording from the correct understanding of the physics, but it's a really awkward formulation that makes it sound like the strength of the pull determines the speed, where later answers suggest the real intent was something closer to "the force required to pull her arms in is greater when she's spinning faster." – Chad Orzel Jan 22 '11 at 21:50
show 12 more comments
Joe's answer is of course right and I gave it +1. However, let me say some slightly complementary things.
Whenever the laws of physics don't depend on the orientation in space, a number known as the angular momentum is conserved. For a rotating body - including the body of a lady - the angular momentum $J$ may be written as the product of the moment of inertia $I$ and the angular frequency $\omega$ (the number of revolutions per second, multiplied by $2\pi=6.28$): $$J = I \omega$$ The moment of inertia $I$ is approximately equal to $$I = MR^2$$ where $M$ is the mass and $R$ is equal to the weighted average distance of the atoms (weighted by the mass) from the axis. (More precisely, I need to compute the average $R^2$.)
It's up to you whether she is spinning clockwise or counter-clockwise.
So if the ballerina pulls in her arms, she becomes closer to the axis, and $R$ decreases. Her mass $M$ doesn't change but the moment of inertia $I$ decreases, too. Because $J=I\omega$ has to be conserved and $I$ decreased, $\omega$ inevitably increases.
You may also explain the increased angular frequency of the rotation in terms of forces and torques. If the arms move closer to the axis, they exert a torque on the ballerina that speeds her up. I would need some cross products here but I am afraid that wouldn't be fully appreciated.
These issues were also discussed here:
Why do galaxies and water going down a plug hole spin?
Why do galaxies and bathtub whirlpools spin?
Cheers LM
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You're definitely better at expository answers +1. – Joe Fitzsimons Jan 22 '11 at 17:36
As a side note regarding the new ballerina picture, it seems there is a right and a wrong answer, since the picture appears to use perspective rather than a parallel projection. – Joe Fitzsimons Jan 22 '11 at 18:27
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Hah, I've seen this image so many times... every time I'm sure she's spinning clockwise. Is it even possible to argue she's spinning the other way? – Noldorin Jan 22 '11 at 18:31
Just one correction: \omega is actually the angular velocity in this context. – Nick Pascucci Jan 22 '11 at 18:34
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@Noldorin: only thing by which you could distinguish the rotation besides colors and texture (which are both absent here) is depth. But the relative changes in distance are too small here and you don't see the biggest changes anyway because they are hidden by the body. As for the nose/eyes that's again a matter of interpretation. For ccw. rotation you see them appear from the left side. – Marek Jan 24 '11 at 8:43
show 9 more comments
No, it's caused by conservation of angular momentum. Reducing air resistance won't cause her (or anything else) to speed up without an external force.
Like linear momentum ($m v$), angular momentum ($r \times mv$) is a conserved quantity, where $r$ is the vector from the center of rotation. For a skater holding a static pose, for each particle making up her body, the contribution in magnitude to the total angular momentum is given by $m_i r_i v_i$. Thus bringing in her arms reduces $r_i$ for those particles. In order to conserve angular momentum, there is then an increase in the angular velocity.
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http://math.stackexchange.com/questions/79283/dimension-of-a-span-of-matrix-powers
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# Dimension of a span of matrix powers
Let $A \in M^{C}_{n\times n}$ be a square matrix with a minimal polynomial of degree $k$. What would be $\dim(\mathrm{Span}\{I,A,A^{2},A^{3},...\})$? I think it's $k$ but I'm not sure exactly how to prove it.
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Wouldn't it be nice if the span you are interested in were isomorphic to $\mathbb C[X]/(minimal \; polynomial)$ ? Hmmm, let me see, how would we prove that... – Georges Elencwajg Nov 5 '11 at 18:27
## 2 Answers
Suppose that $c_0I_n+c_1A+c_2A^2+\cdots+c_{k-1} A^{k-1}=0$. Then if $f(x)=c_0+c_1x+c_2x^2+\cdots+c_{k-1} x^{k-1}$, we have $f(A)=0$. But the minimal polynomial has degree $k>k-1$. Therefore, $f(x)=0$ and so $c_0=\cdots=c_{k-1}=0$. This shows that the set $\{I_n, A, A^2, \cdots, A^{k-1} \}$ is linearly independent.
Now suppose $A$ has minimal polynomial $g(x)=a_0+a_1x+\cdots+a_kx^k$. This means $A^k = a_k^{-1}(a_0I_n+\cdots+a_{k-1}A^{k-1})$ and in general $A^m = a_k^{-1}(a_0I_n+\cdots+a_{k-1}A^{k-1})A^{m-k}$ for all $m \geq k$. Thus powers of $A$ bigger than $A^{k-1}$ can always be replaced with lower powers. Thus $\{I_n, A, A^2, \cdots, A^{k-1} \}$ spans as well.
Therefore, the dimension of the span of powers of $A$ is exactly the degree of its minimal polynomial.
Edit: To elaborate on the spanning arugment.
Suppose that $A^n = c_0I_n+c_1A+\cdots+c_\ell A^\ell$ with $c_\ell\not=0$ and $\ell$ as small as possible. Then if $\ell \geq k$, we can write $A^\ell=a_k^{-1}(a_0I_n+\cdots+a_{k-1}A^{k-1})A^{\ell-k}$ and so $A^n = c_0I_n+c_1A+\cdots+c_{\ell-1} A^{\ell-1} + c_\ell a_k^{-1}(a_0I_n+\cdots+a_{k-1}A^{k-1})A^{\ell-k}$. We have expressed $A^n$ using $I_n,A,\dots,A^{\ell-1}$ so $\ell$ was not minimal (contradiction). Thus $\ell<k$.
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It cannot be more than $k$, because the minimal polynomial allows us to write every $A^n$ with $n\ge k$ as a linear combination of lower powers. On the other hand, it cannot be less than $k$ because then there would be a non-trivial linear relation among $I, A, A^2, \ldots, A^{k-1}$, which would be a polynomial with $p(A)=0$ of degree $<k$, contradicting the minimality of the minimal polynomial.
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http://mathoverflow.net/questions/123055/orthogonal-base-in-unimodular-lattice/123065
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## orthogonal base in unimodular lattice
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $\Lambda$ be an unimodular lattice with a quadratic form $(-,-)$ of signature $(m,n)$ , $m,n>0$.
I know that, fixed a base $e_1,\cdots,e_{m+n}$ for $\Lambda$, the matrix which has entries $a_{i,j}=(e_i,e_j)$ has determinant ugual to $\pm 1$.
I wonder if in this case it is always possibile to find an orthogonal base, i.e. a base $f_1,\cdots,f_{m+n}$ with $(f_i,f_i)=\pm 1$ and $(f_i,f_j)=0$.
I think yes, but i can't really prove it.
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3
The hyperbolic lattice cannot be diagonalized with an integral basis. – Atsushi Kanazawa Feb 27 at 1:23
## 1 Answer
Given that $\Lambda$ is unimodular and indefinite, this can be done if and only if $\Lambda$ is odd (i.e. iff the diagonal entries $a_{i,i}$ are not all even). This follows from Milnor's classification. A couple of references where this is worked out are Serre's "A course in arithmetic" and Milnor and Husemoller's "Symmetric bilinear forms".
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I'm interested in the case $\Lambda=3H \oplus -2E_8$, where $H$ is the hyperbolic plane, so i think this is the case you are talking about, right? – rick Feb 27 at 23:56
no, that one's an even lattice. so it doesn't have an orthogonal basis. – Abhinav Kumar Feb 28 at 0:19
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http://math.stackexchange.com/questions/250222/indecomposable-l-module
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# Indecomposable L-module
I have the following exercice which I have be trying to solve:
Let L be a Lie algebra and $r:L\rightarrow gl_3(F)$ a representation of L such that $im(r)=t_3(F)$ (the upper triangular matrices). Show that the L-module associated is indecomposable.
Thank you very much for your help!
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What is $t_3(F)$? – Brad Dec 3 '12 at 20:34
the upper triangular matrices – Nre Dec 3 '12 at 20:41
## 1 Answer
To be explicit, the associated $L$-module is $F^3$, and the action of $L$ is the usual action of $t_3(F)$.
Hint: Suppose that this $L$-module is decomposable, so that $F^3 = V\oplus W$ for nonzero submodules $V$ and $W$. Then one of these submodules, say $V$, is one-dimensional as an $F$-vector space. Being an $L$-submodule of $F^3$, it must be invariant under the action of $t_3(F)$, as is $W$. Use this information to derive a contradiction.
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http://mathhelpforum.com/pre-calculus/36609-finding-distance-between-parallel-lines.html
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# Thread:
1. ## Finding The Distance Between Parallel Lines...?
How do you find the distance between the two parallel lines:
y = 4x+6
y = 4x-9
A lot of people think it's 15, which sounds plausible, but the answer REALLY is 3.64. My teacher says so. My textbook says so. But I don't know how to arrive at that answer in a step by step process...
Is anyone smart enough to do this question?
2. Originally Posted by AlphaRock
How do you find the distance between the two parallel lines:
y = 4x+6
y = 4x-9
A lot of people think it's 15, which sounds plausible, but the answer REALLY is 3.64. My teacher says so. My textbook says so. But I don't know how to arrive at that answer in a step by step process...
Is anyone smart enough to do this question?
$0=4x+6\Rightarrow{x=\frac{-3}{2}}$
$0=4x-9\Rightarrow{x=\frac{9}{2}}$
$\frac{3}{2}+\frac{9}{2}=3.75$
3. Originally Posted by Mathstud28
$0=4x+6\Rightarrow{x=\frac{-3}{2}}$
$0=4x-9\Rightarrow{x=\frac{9}{2}}$
$\frac{3}{2}+\frac{9}{2}=3.75$
I also thought it'd be 3.75, but suprisingly, it's 3.64.
My friend explained this to me hour ago, but I accidentally exited the conversation.
4. Originally Posted by AlphaRock
I also thought it'd be 3.75, but suprisingly, it's 3.64.
My friend explained this to me hour ago, but I accidentally exited the conversation.
It is 3.75...I think your book may be wrong
5. Originally Posted by Mathstud28
It is 3.75...I think your book may be wrong
Probably, but if it helps... it included tangent that opp is 4 times bigger than four or something. And that the tangent = slope.
and x^2 + #x^2 = 225
17x^2 = 225
4.12x = 15
x = 3.64
These are all that I can remember. I can't remember the steps or understanding in between. Perhaps you can.
6. I REALLY don't know where the X^2 or 17 came from. If somebody could explain everything clearly... I would appreciate it.
7. What Mathstud found was the horizontal distance between two points of the two lines which isn't necessarily the shortest distance.
Recall the distance formula from a point to the line (presumably the shortest):
$d = \frac{|Ax_{0} + By_{0} + C|}{\sqrt{A^{2} + B^{2}}}$
where (x_{0},y_{0}) is your point and Ax + By + C = 0 is your line.
Now, let's pick a point from y = 4x + 6 ... let's say (0,6). This will be our point.
From our other line y = 4x - 9, we must but it in standard form (Ax + By + C = 0):
$y = 4x - 9 \: \Rightarrow \: 0 = 4x - y - 9$
Now apply the formula:
$d = \frac{|4(0) + (-1)(6) + (-9)|}{\sqrt{(4)^{2} + (-1)^{2}}} \approx 3.638$
8. Also, note that the shortest distance between two lines is a perpendicular line that intersects both.
Start with y = 4x + 6. Let's use the point (0,6). We want to find the equation of a line perpendicular to y = 4x + 6 that goes through (0,6). Perpendicular lines have negative reciprocal slopes, so our line will have a slope of -1/4 - note that it has the same y-intercept.
So our equation is y = (-1/4)x + 6. To see where that line intersects our OTHER line (in other words, to find the point on the other line closest to (0,6), set the two equal to one another.
(-1/4)x + 6 = 4x - 9. Solve for x. It isn't pretty, but you get x = 60/17. Plug that in to either our perpendicular line's equation or the second equation to get y, which is 87/17. So now we have these two points, (0,6) and (60/17, 87/17), and we know, since they're on a perpendicular line that intersects our two original lines, that they're as close as possible. But how close?
The distance formula:
${\sqrt{\frac{60}{17}^2 + \frac{15}{17}^2}\approx 3.638}$
Admittedly, this is a longer method than the one o_O proposed, but I never remember the equation for the distance between a point and a line - this particular method relies on a little more "basic" geometrical concepts and formulas. And, having multiple methods to solve a problem is never a bad thing.
9. The triangles in my diagram are similar.
Using this fact you can label one triangle with sides 15, 4x and x and then use Pythagoras' theorem.
$x^2+(4x)^2=15^2$
$17x^2=15^2$
$x=\frac{15}{\sqrt{17}}$
$x=3.638..$
10. Does this apply to any dimension.if we have a1*x1+a2*x2+...an*xn=b1 and a1*x1+a2*x2+...an*xn=b2.
11. ## Yes
Yes, the formula works for higher dimensions as well
If we have a hyperplane wx=b, then the distance from point x1 (which is a vector) to the line is given by:
|wx-b|/||w||^2
Remember w is also a vector of weights so ||w||= w1^2+w2^2+.....
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http://en.wikipedia.org/wiki/Credible_interval
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# Credible interval
Bayesian statistics
Theory
Techniques
Uses
In Bayesian statistics, a credible interval (or Bayesian confidence interval) is an interval in the domain of a posterior probability distribution used for interval estimation.[1] The generalisation to multivariate problems is the credible region. Credible intervals are analogous to confidence intervals in frequentist statistics.[2]
For example, in an experiment that determines the uncertainty distribution of parameter $t$, if the probability that $t$ lies between 35 and 45 is 0.95, then $35 \le t \le 45$ is a 95% credible interval.
## Choosing a credible interval
Credible intervals are not unique on a posterior distribution. Methods for defining a suitable credible interval include:
• Choosing the narrowest interval, which for a unimodal distribution will involve choosing those values of highest probability density including the mode.
• Choosing the interval where the probability of being below the interval is as likely as being above it. This interval will include the median.
• Assuming the mean exists, choosing the interval for which the mean is the central point.
It is possible to frame the choice of a credible interval within decision theory and, in that context, an optimal interval will always be a highest probability density set.[3]
## Contrasts with confidence interval
A frequentist 95% confidence interval of 35–45 means that with a large number of repeated samples, 95% of the calculated confidence intervals would include the true value of the parameter. The probability that the parameter is inside the given interval (say, 35–45) is either 0 or 1 (the non-random unknown parameter is either there or not). In frequentist terms, the parameter is fixed (cannot be considered to have a distribution of possible values) and the confidence interval is random (as it depends on the random sample). Antelman (1997, p. 375) summarizes a confidence interval as "... one interval generated by a procedure that will give correct intervals 95 % of the time".[4]
In general, Bayesian credible intervals do not coincide with frequentist confidence intervals for two reasons:
• credible intervals incorporate problem-specific contextual information from the prior distribution whereas confidence intervals are based only on the data;
• credible intervals and confidence intervals treat nuisance parameters in radically different ways.
For the case of a single parameter and data that can be summarised in a single sufficient statistic, it can be shown that the credible interval and the confidence interval will coincide if the unknown parameter is a location parameter (i.e. the forward probability function has the form $\mathrm{Pr}(x|\mu) = f(x - \mu)$ ), with a prior that is a uniform flat distribution;[5] and also if the unknown parameter is a scale parameter (i.e. the forward probability function has the form $\mathrm{Pr}(x|s) = f(x/s)$ ), with a Jeffreys' prior $\scriptstyle{\mathrm{Pr}(s|I) \;\propto\; 1/s}$ [5] — the latter following because taking the logarithm of such a scale parameter turns it into a location parameter with a uniform distribution. But these are distinctly special (albeit important) cases; in general no such equivalence can be made.
## References
1. Edwards, W., Lindman, H., Savage, L.J. (1963) "Bayesian statistical inference in statistical research". Psychological Review, 70, 193-242
2. Lee, P.M. (1997) Bayesian Statistics: An Introduction, Arnold. ISBN 0-340-67785-6
3. O'Hagan, A. (1994) Kendall's Advance Theory of Statistics, Vol 2B, Bayesian Inference, Section 2.51. Arnold, ISBN 0-340-52922-9
4. Antelman, G. (1997) Elementary Bayesian Statistics (Madansky, A. & McCulloch, R. eds.). Cheltenham, UK: Edward Elgar ISBN 978-1-85898-504-6
5. ^ a b Jaynes, E. T. (1976). "Confidence Intervals vs Bayesian Intervals", in Foundations of Probability Theory, Statistical Inference, and Statistical Theories of Science, (W. L. Harper and C. A. Hooker, eds.), Dordrecht: D. Reidel, pp. 175 et seq
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http://stats.stackexchange.com/questions/17400/estimator-of-mu-for-gaussian-variable-with-penalty
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# Estimator of mu for gaussian variable with penalty
I have a question about the following problem: estimate $\mu=\mu_1,...\mu_n$ when $Y_i \sim N(\mu_i,\sigma^2)$ using a ridge like penalty
$$\min_\mu \sum_i(Y_i-\mu_i)^2 + \lambda\sum_i\mu_i^2$$
I have calculated the closed form decomposition of the generalized cross validation error and I know the optimal $\lambda$ depends on $\mu$ and $\sigma$.
My question is how calculate the optimal $\lambda$ and the cross validation error with a cross validation procedure ? I don't see how to do that.
Thanks for your help
-
I'm not sure I quite understand your question yet. Is $\mu$ a vector, as in $\mu = (\mu_1,\ldots,\mu_n)$? And, is the notation $Y_i \to N(\mu_i,\sigma^2)$ meant to indicate that $Y_i$ has a normal distribution with mean $\mu_i$ and $\sigma^2$. (I ask because this is usually denoted by $\sim$, i.e., `\sim` and so I want to make sure your question is being interpreted properly.) The description after the display equation could also be improved with a little more elaboration, I think. – cardinal Oct 22 '11 at 13:18
hi, thanks for the answer. Your right , i had to use \sim... sorry. In fact i ask the question for $\mu$ constant and more specifically if $\mu$ is not constant like this kind of model: $$Y_t = \mu_t + \epsilon_1$$ $$\mu_t = \mu_{ t-1} + \epsilon_2$$ where $\epsilon_1$ and $\epsilon_2$ are two independents gaussian noise. like times series as example. I know Kalman filter can estimate the vector $\mu$, i ask me if there is an other way thanks to cross validation – Trevor Oct 24 '11 at 22:22
What is the relation between $\mu$ and $\mu_i$? Without it the expression $\min_{\mu}f(\mu_1,...,\mu_n,Y_1,...,Y_n)$ does not make sense. – mpiktas Oct 25 '11 at 11:07
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http://www.territorioscuola.com/wikipedia/en.wikipedia.php?title=Symmetry
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More results on: Download PDF files on: Download Word files on: Images on: Video/Audio on: Download PowerPoint on: More results from.edu web: Map (if applicable) of:
Symmetry - Wikipedia, the free encyclopedia
# Symmetry
For other uses, see Symmetry (disambiguation).
Sphere symmetrical group o.
Leonardo da Vinci's Vitruvian Man (ca. 1487) is often used as a representation of symmetry in the human body and, by extension, the natural universe.
Symmetric arcades of a portico in the Great Mosque of Kairouan also called the Mosque of Uqba, in Tunisia.
Symmetry (from Greek συμμετρεῖν symmetreín "to measure together") has two meanings. The first is a vague sense of harmonious and beautiful proportion and balance.12 The second is an exact mathematical "patterned self-similarity" that can be demonstrated with the rules of a formal system, such as geometry or physics.
Although these two meanings of "symmetry" can sometimes be told apart, they are related, so they are here discussed together.23
Mathematical symmetry may be observed
• with respect to the passage of time;
• as a spatial relationship;
• through geometric transformations such as scaling, reflection, and rotation;
• through other kinds of functional transformations;4 and
• as an aspect of abstract objects, theoretic models, language, music and even knowledge itself.56
This article describes these notions of symmetry from four perspectives. The first is symmetry in geometry, which is the most familiar type of symmetry for many people. The second is the more general meaning of symmetry in mathematics as a whole. The third describes symmetry as it relates to science and technology. In this context, symmetries underlie some of the most profound results found in modern physics, including aspects of space and time. The fourth discusses symmetry in the humanities, covering its rich and varied use in history, architecture, art, and religion.
The opposite of symmetry is asymmetry.
## In geometry
The most familiar type of symmetry for many people is geometrical symmetry. Formally, this means symmetry under a sub-group of the Euclidean group of isometries in two or three dimensional Euclidean space. These isometries consist of reflections, rotations, translations and combinations of these basic operations.7
### Reflection symmetry
Main article: reflection symmetry
A butterfly with bilateral symmetry
Reflection symmetry, mirror symmetry, mirror-image symmetry, or bilateral symmetry is symmetry with respect to reflection.
In 1D, there is a point of symmetry; in 2D there is an axis of symmetry, and in 3D a plane of symmetry. An object or figure which is indistinguishable from its transformed image is called mirror symmetric (see mirror image).
The axis of symmetry of a two-dimensional figure is a line such that, if a perpendicular is constructed, any two points lying on the perpendicular at equal distances from the axis of symmetry are identical. Another way to think about it is that if the shape were to be folded in half over the axis, the two halves would be identical: the two halves are each other's mirror image. Thus a square has four axes of symmetry, because there are four different ways to fold it and have the edges all match. A circle has infinitely many axes of symmetry, for the same reason.
If the letter T is reflected along a vertical axis, it appears the same. This is sometimes called vertical symmetry. One can better use an unambiguous formulation; e.g., "T has a vertical symmetry axis" or "T has left-right symmetry".
The triangles with this symmetry are isosceles, the quadrilaterals with this symmetry are the kites and the isosceles trapezoids.
For each line or plane of reflection, the symmetry group is isomorphic with Cs (see point groups in three dimensions), one of the three types of order two (involutions), hence algebraically C2. The fundamental domain is a half-plane or half-space.
Bilateria (bilateral animals, including humans) are more or less symmetric with respect to the sagittal plane.
### Point reflection and other involutive isometries
Reflection symmetry can be generalized to other isometries of m-dimensional space which are involutions, such as
(x1, … xm) ↦ (−x1, … −xk, xk+1, … xm)
in certain system of Cartesian coordinates. This reflects the space along a m−k-dimensional affine subspace. If k = m, then such transformation is known as point reflection, which on the plane (m = 2) is the same as the half-turn (180°) rotation; see below.
Such "reflection" keeps orientation if and only if k is even. This implies that for m = 3 (as well for other odd m) a point reflection changes orientation of the space, like mirror-image symmetry. That's why in physics the term P-symmetry is used for both point reflection and mirror symmetry (P stands for parity).
### Rotational symmetry
Main article: rotational symmetry
Rotational symmetry is symmetry with respect to some or all rotations in m-dimensional Euclidean space. Rotations are direct isometries; i.e., isometries preserving orientation. Therefore a symmetry group of rotational symmetry is a subgroup of E+(m).
Symmetry with respect to all rotations about all points implies translational symmetry with respect to all translations, and the symmetry group is the whole E+(m). This does not apply for objects because it makes space homogeneous, but it may apply for physical laws.
For symmetry with respect to rotations about a point we can take that point as origin. These rotations form the special orthogonal group SO(m), the group of m × m orthogonal matrices with determinant 1. For m = 3 this is the rotation group SO(3).
In another meaning of the word, the rotation group of an object is the symmetry group within E+(m), the group of direct isometries; in other words, the intersection of the full symmetry group and the group of direct isometries. For chiral objects it is the same as the full symmetry group.
Laws of physics are SO(3)-invariant if they do not distinguish different directions in space. Because of Noether's theorem, rotational symmetry of a physical system is equivalent to the angular momentum conservation law. See also rotational invariance.
### Translational symmetry
Main article: Translational symmetry
Translational symmetry leaves an object invariant under a discrete or continuous group of translations $\scriptstyle T_a(p) \;=\; p \,+\, a$.
### Glide reflection symmetry
A glide reflection symmetry (in 3D: a glide plane symmetry) means that a reflection in a line or plane combined with a translation along the line / in the plane, results in the same object. It implies translational symmetry with twice the translation vector. The symmetry group is isomorphic with Z.
### Rotoreflection symmetry
In 3D, rotoreflection or improper rotation in the strict sense is rotation about an axis, combined with reflection in a plane perpendicular to that axis. As symmetry groups with regard to a roto-reflection we can distinguish:
• the angle has no common divisor with 360°, the symmetry group is not discrete
• 2n-fold rotoreflection (angle of 180°/n) with symmetry group S2n of order 2n (not to be confused with symmetric groups, for which the same notation is used; abstract group C2n); a special case is n = 1, inversion, because it does not depend on the axis and the plane, it is characterized by just the point of inversion.
• Cnh (angle of 360°/n); for odd n this is generated by a single symmetry, and the abstract group is C2n, for even n this is not a basic symmetry but a combination. See also point groups in three dimensions.
### Helical symmetry
A drill bit with helical symmetry.
See also: Screw axis
Helical symmetry is the kind of symmetry seen in such everyday objects as springs, Slinky toys, drill bits, and augers. It can be thought of as rotational symmetry along with translation along the axis of rotation, the screw axis. The concept of helical symmetry can be visualized as the tracing in three-dimensional space that results from rotating an object at an even angular speed while simultaneously moving at another even speed along its axis of rotation (translation). At any one point in time, these two motions combine to give a coiling angle that helps define the properties of the tracing. When the tracing object rotates quickly and translates slowly, the coiling angle will be close to 0°. Conversely, if the rotation is slow and the translation is speedy, the coiling angle will approach 90°.
Three main classes of helical symmetry can be distinguished based on the interplay of the angle of coiling and translation symmetries along the axis:
Infinite helical symmetry
If there are no distinguishing features along the length of a helix or helix-like object, the object will have infinite symmetry much like that of a circle, but with the additional requirement of translation along the long axis of the object to return it to its original appearance. A helix-like object is one that has at every point the regular angle of coiling of a helix, but which can also have a cross section of indefinitely high complexity, provided only that precisely the same cross section exists (usually after a rotation) at every point along the length of the object. Simple examples include evenly coiled springs, slinkies, drill bits, and augers. Stated more precisely, an object has infinite helical symmetries if for any small rotation of the object around its central axis there exists a point nearby (the translation distance) on that axis at which the object will appear exactly as it did before. It is this infinite helical symmetry that gives rise to the curious illusion of movement along the length of an auger or screw bit that is being rotated. It also provides the mechanically useful ability of such devices to move materials along their length, provided that they are combined with a force such as gravity or friction that allows the materials to resist simply rotating along with the drill or auger.
n-fold helical symmetry
If the requirement that every cross section of the helical object be identical is relaxed, additional lesser helical symmetries become possible. For example, the cross section of the helical object may change, but still repeats itself in a regular fashion along the axis of the helical object. Consequently, objects of this type will exhibit a symmetry after a rotation by some fixed angle θ and a translation by some fixed distance, but will not in general be invariant for any rotation angle. If the angle (rotation) at which the symmetry occurs divides evenly into a full circle (360°), the result is the helical equivalent of a regular polygon. This case is called n-fold helical symmetry, where n = 360°; e.g., double helix. This concept can be further generalized to include cases where $\scriptstyle m\theta$ is a multiple of 360° – that is, the cycle does eventually repeat, but only after more than one full rotation of the helical object.
Non-repeating helical symmetry
This is the case in which the angle of rotation θ required to observe the symmetry is irrational. The angle of rotation never repeats exactly no matter how many times the helix is rotated. Such symmetries are created by using a non-repeating point group in two dimensions. DNA is an example of this type of non-repeating helical symmetry.citation needed
### Non-isometric symmetries
A wider definition of geometric symmetry allows operations from a larger group than the Euclidean group of isometries. Examples of larger geometric symmetry groups are:
• The group of similarity transformations; i.e., affine transformations represented by a matrix A that is a scalar times an orthogonal matrix. Thus homothety is added, self-similarity is considered a symmetry.
• The group of affine transformations represented by a matrix A with determinant 1 or −1; i.e., the transformations which preserve area.
This adds, e.g., oblique reflection symmetry.
• The group of all bijective affine transformations.
• The group of Möbius transformations which preserve cross-ratios.
This adds, e.g., inversive reflections such as circle reflection on the plane.
In Felix Klein's Erlangen program, each possible group of symmetries defines a geometry in which objects that are related by a member of the symmetry group are considered to be equivalent. For example, the Euclidean group defines Euclidean geometry, whereas the group of Möbius transformations defines projective geometry.
### Scale symmetry and fractals
Scale symmetry refers to the idea that if an object is expanded or reduced in size, the new object has the same properties as the original. Scale symmetry is notable for the fact that it does not exist for most physical systems, a point that was first discerned by Galileo. Simple examples of the lack of scale symmetry in the physical world include the difference in the strength and size of the legs of elephants versus mice, and the observation that if a candle made of soft wax was enlarged to the size of a tall tree, it would immediately collapse under its own weight.
A more subtle form of scale symmetry is demonstrated by fractals. As conceived by Benoît Mandelbrot, fractals are a mathematical concept in which the structure of a complex form looks similar or even exactly the same no matter what degree of magnification is used to examine it. A coast is an example of a naturally occurring fractal, since it retains roughly comparable and similar-appearing complexity at every level from the view of a satellite to a microscopic examination of how the water laps up against individual grains of sand. The branching of trees, which enables children to use small twigs as stand-ins for full trees in dioramas, is another example.
This similarity to naturally occurring phenomena provides fractals with an everyday familiarity not typically seen with mathematically generated functions. As a consequence, they can produce strikingly beautiful results such as the Mandelbrot set. Intriguingly, fractals have also found a place in CG, or computer-generated movie effects, where their ability to create very complex curves with fractal symmetries results in more realistic virtual worlds.
## In mathematics
Main article: Symmetry in mathematics
In formal terms, we say that a mathematical object is symmetric with respect to a given mathematical operation, if, when applied to the object, this operation preserves some property of the object. The set of operations that preserve a given property of the object form a group. Two objects are symmetric to each other with respect to a given group of operations if one is obtained from the other by some of the operations (and vice versa).
### Mathematical model for symmetry
The set of all symmetry operations considered, on all objects in a set X, can be modeled as a group action g : G × X → X, where the image of g in G and x in X is written as g·x. If, for some g, g·x = y then x and y are said to be symmetrical to each other. For each object x, operations g for which g·x = x form a group, the symmetry group of the object, a subgroup of G. If the symmetry group of x is the trivial group then x is said to be asymmetric, otherwise symmetric.
A general example is that G is a group of bijections g: V → V acting on the set of functions x: V → W by (gx)(v) = xg−1(v)] (or a restricted set of such functions that is closed under the group action). Thus a group of bijections of space induces a group action on "objects" in it. The symmetry group of x consists of all g for which x(v) = xg(v)] for all v. G is the symmetry group of the space itself, and of any object that is uniform throughout space. Some subgroups of G may not be the symmetry group of any object. For example, if the group contains for every v and w in V a g such that g(v) = w, then only the symmetry groups of constant functions x contain that group. However, the symmetry group of constant functions is G itself.
In a modified version for vector fields, we have (gx)(v) = h(g, xg−1(v)]) where h rotates any vectors and pseudovectors in x, and inverts any vectors (but not pseudovectors) according to rotation and inversion in g, see symmetry in physics. The symmetry group of x consists of all g for which x(v) = h(g, xg(v)]) for all v. In this case the symmetry group of a constant function may be a proper subgroup of G: a constant vector has only rotational symmetry with respect to rotation about an axis if that axis is in the direction of the vector, and only inversion symmetry if it is zero.
For a common notion of symmetry in Euclidean space, G is the Euclidean group E(n), the group of isometries, and V is the Euclidean space. The rotation group of an object is the symmetry group if G is restricted to E+(n), the group of direct isometries. (For generalizations, see the next subsection.) Objects can be modeled as functions x, of which a value may represent a selection of properties such as color, density, chemical composition, etc. Depending on the selection we consider just symmetries of sets of points (x is just a Boolean function of position v), or, at the other extreme; e.g., symmetry of right and left hand with all their structure.
For a given symmetry group, the properties of part of the object, fully define the whole object. Considering points equivalent which, due to the symmetry, have the same properties, the equivalence classes are the orbits of the group action on the space itself. We need the value of x at one point in every orbit to define the full object. A set of such representatives forms a fundamental domain. The smallest fundamental domain does not have a symmetry; in this sense, one can say that symmetry relies upon asymmetry.
An object with a desired symmetry can be produced by choosing for every orbit a single function value. Starting from a given object x we can, e.g.:
• Take the values in a fundamental domain (i.e., add copies of the object).
• Take for each orbit some kind of average or sum of the values of x at the points of the orbit (ditto, where the copies may overlap).
If it is desired to have no more symmetry than that in the symmetry group, then the object to be copied should be asymmetric.
As pointed out above, some groups of isometries are not the symmetry group of any object, except in the modified model for vector fields. For example, this applies in 1D for the group of all translations. The fundamental domain is only one point, so we can not make it asymmetric, so any "pattern" invariant under translation is also invariant under reflection (these are the uniform "patterns").
In the vector field version continuous translational symmetry does not imply reflectional symmetry: the function value is constant, but if it contains nonzero vectors, there is no reflectional symmetry. If there is also reflectional symmetry, the constant function value contains no nonzero vectors, but it may contain nonzero pseudovectors. A corresponding 3D example is an infinite cylinder with a current perpendicular to the axis; the magnetic field (a pseudovector) is, in the direction of the cylinder, constant, but nonzero. For vectors (in particular the current density) we have symmetry in every plane perpendicular to the cylinder, as well as cylindrical symmetry. This cylindrical symmetry without mirror planes through the axis is also only possible in the vector field version of the symmetry concept. A similar example is a cylinder rotating about its axis, where magnetic field and current density are replaced by angular momentum and velocity, respectively.
A symmetry group is said to act transitively on a repeated feature of an object if, for every pair of occurrences of the feature there is a symmetry operation mapping the first to the second. For example, in 1D, the symmetry group of {…, 1, 2, 5, 6, 9, 10, 13, 14, …} acts transitively on all these points, while {…, 1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15, …} does not act transitively on all points. Equivalently, the first set is only one conjugacy class with respect to isometries, while the second has two classes.
### Symmetric functions
Main article: symmetric function
A symmetric function is a function which is unchanged by any permutation of its variables. For example, x + y + z and xy + yz + xz are symmetric functions, whereas x2 – yz is not.
A function may be unchanged by a sub-group of all the permutations of its variables. For example, ac + 3ab + bc is unchanged if a and b are exchanged; its symmetry group is isomorphic to C2.
### In logic
A dyadic relation R is symmetric if and only if, whenever it's true that Rab, it's true that Rba. Thus, "is the same age as" is symmetrical, for if Paul is the same age as Mary, then Mary is the same age as Paul.
Symmetric binary logical connectives are and (∧, or &), or (∨, or |), biconditional (if and only if) (↔), nand (not-and, or ⊼), xor (not-biconditional, or ⊻), and nor (not-or, or ⊽).
## In science and nature
Further information: Patterns in nature
### In physics
Main article: Symmetry in physics
Symmetry in physics has been generalized to mean invariance—that is, lack of change—under any kind of transformation, for example arbitrary coordinate transformations. This concept has become one of the most powerful tools of theoretical physics, as it has become evident that practically all laws of nature originate in symmetries. In fact, this role inspired the Nobel laureate PW Anderson to write in his widely read 1972 article More is Different that "it is only slightly overstating the case to say that physics is the study of symmetry." See Noether's theorem (which, in greatly simplified form, states that for every continuous mathematical symmetry, there is a corresponding conserved quantity; a conserved current, in Noether's original language); and also, Wigner's classification, which says that the symmetries of the laws of physics determine the properties of the particles found in nature.
### In physical objects
#### Classical objects
Although an everyday object may appear exactly the same after a symmetry operation such as a rotation or an exchange of two identical parts has been performed on it, it is readily apparent that such a symmetry is true only as an approximation for any ordinary physical object.
For example, if one rotates a precisely machined aluminum equilateral triangle 120 degrees around its center, a casual observer brought in before and after the rotation will be unable to decide whether or not such a rotation took place. However, the reality is that each corner of a triangle will always appear unique when examined with sufficient precision. An observer armed with sufficiently detailed measuring equipment such as optical or electron microscopes will not be fooled; he will immediately recognize that the object has been rotated by looking for details such as crystals or minor deformities.
Such simple thought experiments show that assertions of symmetry in everyday physical objects are always a matter of approximate similarity rather than of precise mathematical sameness. The most important consequence of this approximate nature of symmetries in everyday physical objects is that such symmetries have minimal or no impacts on the physics of such objects. Consequently, only the deeper symmetries of space and time play a major role in classical physics; that is, the physics of large, everyday objects.
#### Quantum objects
Remarkably, there exists a realm of physics for which mathematical assertions of simple symmetries in real objects cease to be approximations. That is the domain of quantum physics, which for the most part is the physics of very small, very simple objects such as electrons, protons, light, and atoms.
Unlike everyday objects, objects such as electrons have very limited numbers of configurations, called states, in which they can exist. This means that when symmetry operations such as exchanging the positions of components are applied to them, the resulting new configurations often cannot be distinguished from the originals no matter how diligent an observer is. Consequently, for sufficiently small and simple objects the generic mathematical symmetry assertion F(x) = x ceases to be approximate, and instead becomes an experimentally precise and accurate description of the situation in the real world.
#### Consequences of quantum symmetry
While it makes sense that symmetries could become exact when applied to very simple objects, the immediate intuition is that such a detail should not affect the physics of such objects in any significant way. This is in part because it is very difficult to view the concept of exact similarity as physically meaningful. Our mental picture of such situations is invariably the same one we use for large objects: We picture objects or configurations that are very, very similar, but for which if we could "look closer" we would still be able to tell the difference.
However, the assumption that exact symmetries in very small objects should not make any difference in their physics was discovered in the early 1900s to be spectacularly incorrect. The situation was succinctly summarized by Richard Feynman in the direct transcripts of his Feynman Lectures on Physics, Volume III, Section 3.4, Identical particles. (Unfortunately, the quote was edited out of the printed version of the same lecture.)
… if there is a physical situation in which it is impossible to tell which way it happened, it always interferes; it never fails.
The word "interferes" in this context is a quick way of saying that such objects fall under the rules of quantum mechanics, in which they behave more like waves that interfere than like everyday large objects.
In short, when an object becomes so simple that a symmetry assertion of the form F(x) = x becomes an exact statement of experimentally verifiable sameness, x ceases to follow the rules of classical physics and must instead be modeled using the more complex, and often far less intuitive, rules of quantum physics.
This transition also provides an important insight into why the mathematics of symmetry are so deeply intertwined with those of quantum mechanics. When physical systems make the transition from symmetries that are approximate to ones that are exact, the mathematical expressions of those symmetries cease to be approximations and are transformed into precise definitions of the underlying nature of the objects. From that point on, the correlation of such objects to their mathematical descriptions becomes so close that it is difficult to separate the two.
### Generalizations of symmetry
If we have a given set of objects with some structure, then it is possible for a symmetry to merely convert only one object into another, instead of acting upon all possible objects simultaneously. This requires a generalization from the concept of symmetry group to that of a groupoid. Indeed, A. Connes in his book "Non-commutative geometry" writes that Heisenberg discovered quantum mechanics by considering the groupoid of transitions of the hydrogen spectrum.
The notion of groupoid also leads to notions of multiple groupoids, namely sets with many compatible groupoid structures, a structure which trivialises to abelian groups if one restricts to groups. This leads to prospects of higher order symmetry which have been a little explored, as follows.
The automorphisms of a set, or a set with some structure, form a group, which models a homotopy 1-type. The automorphisms of a group G naturally form a crossed module $\scriptstyle G \;\to\; \mathrm{Aut}(G)$, and crossed modules give an algebraic model of homotopy 2-types. At the next stage, automorphisms of a crossed module fit into a structure known as a crossed square, and this structure is known to give an algebraic model of homotopy 3-types. It is not known how this procedure of generalising symmetry may be continued, although crossed n-cubes have been defined and used in algebraic topology, and these structures are only slowly being brought into theoretical physics.78
Physicists have come up with other directions of generalization, such as supersymmetry and quantum groups, yet the different options are indistinguishable during various circumstances.
### In biology
Further information: symmetry (biology) and facial symmetry
### In chemistry
Main article: molecular symmetry
Symmetry is important to chemistry because it explains observations in spectroscopy, quantum chemistry and crystallography. It draws heavily on group theory.
## In history, religion, and culture
In any human endeavor for which an impressive visual result is part of the desired objective, symmetries play a profound role. The innate appeal of symmetry can be found in our reactions to happening across highly symmetrical natural objects, such as precisely formed crystals or beautifully spiraled seashells. Our first reaction in finding such an object often is to wonder whether we have found an object created by a fellow human, followed quickly by surprise that the symmetries that caught our attention are derived from nature itself. In both reactions we give away our inclination to view symmetries both as beautiful and, in some fashion, informative of the world around us.citation needed
### In social interactions
People observe the symmetrical nature, often including asymmetrical balance, of social interactions in a variety of contexts. These include assessments of reciprocity, empathy, apology, dialog, respect, justice, and revenge. Symmetrical interactions send the message "we are all the same" while asymmetrical interactions send the message "I am special; better than you." Peer relationships are based on symmetry, power relationships are based on asymmetry.9
### In architecture
The ceiling of Lotfollah mosque, Isfahan, Iran has rotational symmetry of order eight and eight lines of reflection.
Leaning Tower of Pisa
The Taj Mahal has bilateral symmetry.
Another human endeavor in which the visual result plays a vital part in the overall result is architecture. Both in ancient and modern times, the ability of a large structure to impress or even intimidate its viewers has often been a major part of its purpose, and the use of symmetry is an inescapable aspect of how to accomplish such goals.
Just a few examples of ancient architectures that made powerful use of symmetry to impress those around them included the Egyptian Pyramids, the Greek Parthenon, the first and second Temple of Jerusalem, China's Forbidden City, Cambodia's Angkor Wat complex, and the many temples and pyramids of ancient Pre-Columbian civilizations. More recent historical examples of architectures emphasizing symmetries include Gothic architecture cathedrals, and American President Thomas Jefferson's Monticello home. The Taj Mahal is also an example of symmetry.10
An interesting example of a broken symmetry in architecture is the Leaning Tower of Pisa, whose notoriety stems in no small part not for the intended symmetry of its design, but for the violation of that symmetry from the lean that developed while it was still under construction. Modern examples of architectures that make impressive or complex use of various symmetries include Australia's Sydney Opera House and Houston, Texas's simpler Astrodome.
Symmetry finds its ways into architecture at every scale, from the overall external views, through the layout of the individual floor plans, and down to the design of individual building elements such as intricately carved doors, stained glass windows, tile mosaics, friezes, stairwells, stair rails, and balustrades. For sheer complexity and sophistication in the exploitation of symmetry as an architectural element, Islamic buildings such as the Taj Mahal often eclipse those of other cultures and ages, due in part to the general prohibition of Islam against using images of people or animals.1112
### In pottery and metal vessels
Persian vessel (4th millennium BC)
Since the earliest uses of pottery wheels to help shape clay vessels, pottery has had a strong relationship to symmetry. As a minimum, pottery created using a wheel necessarily begins with full rotational symmetry in its cross-section, while allowing substantial freedom of shape in the vertical direction. Upon this inherently symmetrical starting point cultures from ancient times have tended to add further patterns that tend to exploit or in many cases reduce the original full rotational symmetry to a point where some specific visual objective is achieved. For example, Persian pottery dating from the fourth millennium BC and earlier used symmetric zigzags, squares, cross-hatchings, and repetitions of figures to produce more complex and visually striking overall designs.
Cast metal vessels lacked the inherent rotational symmetry of wheel-made pottery, but otherwise provided a similar opportunity to decorate their surfaces with patterns pleasing to those who used them. The ancient Chinese, for example, used symmetrical patterns in their bronze castings as early as the 17th century BC. Bronze vessels exhibited both a bilateral main motif and a repetitive translated border design.131415
### In quilts
Kitchen Kaleidoscope Block
As quilts are made from square blocks (usually 9, 16, or 25 pieces to a block) with each smaller piece usually consisting of fabric triangles, the craft lends itself readily to the application of symmetry.16
### In carpets and rugs
Persian rug.
A long tradition of the use of symmetry in carpet and rug patterns spans a variety of cultures. American Navajo Indians used bold diagonals and rectangular motifs. Many Oriental rugs have intricate reflected centers and borders that translate a pattern. Not surprisingly, rectangular rugs typically use quadrilateral symmetry—that is, motifs that are reflected across both the horizontal and vertical axes.1718
### In music
Major and minor triads on the white piano keys are symmetrical to the D. (compare article) (file)
Symmetry is not restricted to the visual arts. Its role in the history of music touches many aspects of the creation and perception of music.
#### Musical form
Symmetry has been used as a formal constraint by many composers, such as the arch (swell) form (ABCBA) used by Steve Reich, Béla Bartók, and James Tenney. In classical music, Bach used the symmetry concepts of permutation and invariance.19
#### Pitch structures
Symmetry is also an important consideration in the formation of scales and chords, traditional or tonal music being made up of non-symmetrical groups of pitches, such as the diatonic scale or the major chord. Symmetrical scales or chords, such as the whole tone scale, augmented chord, or diminished seventh chord (diminished-diminished seventh), are said to lack direction or a sense of forward motion, are ambiguous as to the key or tonal center, and have a less specific diatonic functionality. However, composers such as Alban Berg, Béla Bartók, and George Perle have used axes of symmetry and/or interval cycles in an analogous way to keys or non-tonal tonal centers.
Perle (1992) explains "C–E, D–F♯, [and] Eb–G, are different instances of the same interval … the other kind of identity. … has to do with axes of symmetry. C–E belongs to a family of symmetrically related dyads as follows:"
| | | | | | | |
|----|----|----|----|----|----|----|
| D | D♯ | E | F | F♯ | G | G♯ |
| D | C♯ | C | B | A♯ | A | G♯ |
Thus in addition to being part of the interval-4 family, C–E is also a part of the sum-4 family (with C equal to 0).
| | | | | | | | | | | | | | |
|----|----|----|----|----|----|----|----|----|----|----|----|----|----|
| + | 2 | | 3 | | 4 | | 5 | | 6 | | 7 | | 8 |
| 2 | | 1 | | 0 | | 11 | | 10 | | 9 | | 8 | |
| 4 | | 4 | | 4 | | 4 | | 4 | | 4 | | 4 | |
Interval cycles are symmetrical and thus non-diatonic. However, a seven pitch segment of C5 (the cycle of fifths, which are enharmonic with the cycle of fourths) will produce the diatonic major scale. Cyclic tonal progressions in the works of Romantic composers such as Gustav Mahler and Richard Wagner form a link with the cyclic pitch successions in the atonal music of Modernists such as Bartók, Alexander Scriabin, Edgard Varèse, and the Vienna school. At the same time, these progressions signal the end of tonality.
The first extended composition consistently based on symmetrical pitch relations was probably Alban Berg's Quartet, Op. 3 (1910). (Perle, 1990)
#### Equivalency
Tone rows or pitch class sets which are invariant under retrograde are horizontally symmetrical, under inversion vertically. See also Asymmetric rhythm.
### In other arts and crafts
The concept of symmetry is applied to the design of objects of all shapes and sizes. Other examples include beadwork, furniture, sand paintings, knotwork, masks, musical instruments, and many other endeavors.
### In aesthetics
Main article: Symmetry (physical attractiveness)
The relationship of symmetry to aesthetics is complex. Certain simple symmetries, and in particular bilateral symmetry, seem to be deeply ingrained in the inherent perception by humans of the likely health or fitness of other living creatures, as can be seen by the simple experiment of distorting one side of the image of an attractive face and asking viewers to rate the attractiveness of the resulting image. Consequently, such symmetries that mimic biology tend to have an innate appeal that in turn drives a powerful tendency to create artifacts with similar symmetry. One only needs to imagine the difficulty in trying to market a highly asymmetrical car or truck to general automotive buyers to understand the power of biologically inspired symmetries such as bilateral symmetry.
Another more subtle appeal of symmetry is that of simplicity, which in turn has an implication of safety, security, and familiarity.citation needed A highly symmetrical room, for example, is unavoidably also a room in which anything out of place or potentially threatening can be identified easily and quickly.citation needed For example, people who have grown up in houses full of exact right angles and precisely identical artifacts can find their first experience in staying in a room with no exact right angles and no exactly identical artifacts to be highly disquieting.citation needed Symmetry thus can be a source of comfort not only as an indicator of biological health, but also of a safe and well-understood living environment.
Opposed to this is the tendency for excessive symmetry to be perceived as boring or uninteresting. Humans in particular have a powerful desire to exploit new opportunities or explore new possibilities, and an excessive degree of symmetry can convey a lack of such opportunities.citation needed Most people display a preference for figures that have a certain degree of simplicity and symmetry, but enough complexity to make them interesting.20
Yet another possibility is that when symmetries become too complex or too challenging, the human mind has a tendency to "tune them out" and perceive them in yet another fashion: as noise that conveys no useful information.citation needed
Finally, perceptions and appreciation of symmetries are also dependent on cultural background. The far greater use of complex geometric symmetries in many Islamic cultures, for example, makes it more likely that people from such cultures will appreciate such art forms (or, conversely, to rebel against them).citation needed
As in many human endeavors, the result of the confluence of many such factors is that effective use of symmetry in art and architecture is complex, intuitive, and highly dependent on the skills of the individuals who must weave and combine such factors within their own creative work. Along with texture, color, proportion, and other factors, symmetry is a powerful ingredient in any such synthesis; one only need to examine the Taj Mahal to powerful role that symmetry plays in determining the aesthetic appeal of an object.
Modernist architecture rejects symmetry, stating only a bad architect relies on symmetry;citation needed instead of symmetrical layout of blocks, masses and structures, Modernist architecture relies on wings and balance of masses. This notion of getting rid of symmetry was first encountered in International style. Some people find asymmetrical layouts of buildings and structures revolutionizing; other find them restless, boring and unnatural.
A few examples of the more explicit use of symmetries in art can be found in the remarkable art of M.C. Escher, the creative design of the mathematical concept of a wallpaper group, and the many applications (both mathematical and real world) of tiling.
## See also
Symmetry in statistics
• Skewness, asymmetry of a statistical distribution
Symmetry in games and puzzles
Symmetry in literature
Moral symmetry
Other
## References
1. Penrose, Roger (2007). Fearful Symmetry. City: Princeton. ISBN 978-0-691-13482-6.
2. ^ a b
3. Mainzer, Klaus (2005). Symmetry And Complexity: The Spirit and Beauty of Nonlinear Science. World Scientific. ISBN 981-256-192-7.
4. Symmetric objects can be material, such as a person, crystal, quilt, floor tiles, or molecule, or it can be an abstract structure such as a mathematical equation or a series of tones (music).
5. ^ a b
6. n-category cafe – discussion of n-groups
7. Derry, Gregory N. (2002). What Science Is and How It Works. Princeton University Press. pp. 269–. ISBN 978-1-4008-2311-6.
8. see ("Fugue No. 21," pdf or Shockwave)
9. Arnheim, Rudolf (1969). Visual Thinking. University of California Press.
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http://unapologetic.wordpress.com/2009/10/21/
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# The Unapologetic Mathematician
## Bad Language
I know I’ve been really lazy about my blogroll, and I should get around to that sometime. But there’s a new one I have to point out right away: The Language of Bad Physics (as noted in the comments, now on WordPress at this address). Like Frank Costanza on Festivus, she’s got a lot of problems with you people (physics writers) and how mathematical terms get mangled and confounded in the physics literature. It’s one big Airing of Grievances, and you’d do well to listen up.
My only complaint is that it’s one of those horrid, ugly Blogspot pages and not a nice $\LaTeX$-enabled WordPress page. But maybe it’s early enough to get her to switch
Posted by John Armstrong | Uncategorized | 2 Comments
## Smoothness
With Clairaut’s theorem comes the first common example of a smoothness assumption. It’s a good time to say just what I mean by this.
Let’s look at an open region $X\subseteq\mathbb{R}^n$. We can now define a tower of algebras of functions on this set. We start by setting out the real-valued functions which are continuous at each point of $X$, and write this as $C^0(X)$. It’s an algebra under pointwise addition and multiplication of functions.
Next we consider those functions which have all partial derivatives at every point of $X$, and these partial derivatives are themselves continuous throughout $X$. We’ve seen that this will imply that such a function has a differential at each point of $X$. This gives us a subalgebra of $C^0(X)$ which we write as $C^1(X)$. That is, these functions have “one continuous derivative”, or are “once (continuously) differentiable”.
Continuing on, we consider those functions which have all second partial derivatives, and that these second partials are themselves continuous at each point of $X$. Clairaut’s theorem tells us that the mixed second partials are equal, since they’re both continuous, and we can define the second differential. These functions form a subalgebra of $C^1(X)$ (and thus a further subalgebra of $C^0(X)$) which we write as $C^2(X)$. These functions have “two continuous derivatives”, or are “twice (continuously) differentiable”.
From here it’s clear how to proceed, defining functions with higher and higher differentials. We get algebras $C^3(X)$, $C^4(X)$, and so on. We can even define the infinitely differentiable functions $C^\infty(X)$ to be the limit (in the categorical sense) of this process. It consists of all the functions that are in $C^k(X)$ for all natural numbers $k$. Taking any directional derivative (with a constant direction) of a function in $C^k(X)$ lands us in $C^{k-1}(X)$, although such differentiation sends $C^\infty(X)$ back into itself.
Is this the end? Not quite. Just like in one variable we have analytic functions. Once we’re in $C^\infty(X)$ and we have all higher derivatives we can use Taylor’s theorem to write out the Taylor series of our function. But this may or may not converge back to the function itself. If it does for every point in $X$ we say that the function is analytic on $X$. The collection of all analytic functions forms a subalgebra of $C^\infty(X)$ which we write as $C^\omega(X)$.
It’s interesting to observe that at each step, “most” functions fail to fall into the finer subalgebra, just like “most” points on the $x$-$y$ plane are not on the $x$-axis. An arbitrary function selected from $C^4(X)$ will actually lie within $C^5(X)$ with probability zero. An arbitrary infinitely differentiable function is analytic with probability zero. Pretty much every example we show students in calculus classes is infinitely differentiable, if not analytic, and yet such functions make up a vanishingly small portion of even once differentiable functions.
So, what does it mean to say that a function is “smooth”? It’s often said to mean that a function is in $C^\infty(X)$ — that it has derivatives of all orders. But in practice, this seems to actually be just a convenience. What “smooth” actually means is a subtler point.
Let’s say I’m working in some situation where I’m going to be taking first and second partial derivatives of a function in a region, and I’m going to want the mixed partials to commute by Clairaut’s theorem. If I say that $f$ is infinitely differentiable on $X$, this will certainly do the trick. But I’ve excluded a huge number of functions. All I really need is for $f$ to fall into $C^2(X)$, of which $C^\infty(X)$ is an incredibly tiny subalgebra.
In practice, then, “smooth” effectively means “has enough derivatives to do what I want with it”. It’s a way of saying that we understand that it’s possible to come up with pathological cases which break the theorem we’re stating, but as long as we have sufficiently many derivatives (where “sufficiently many” is some fixed natural number we don’t care to work out in detail) the pathological cases can be excluded. Saying that “smooth” means “infinitely differentiable” accomplishes this goal, and it’s usually easier than trying to stomach the idea that “smoothness” is a highly context-dependent term of art rather than a nice, well-defined mathematical concept.
Posted by John Armstrong | Analysis, Calculus | 6 Comments
## 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://mathhelpforum.com/geometry/153725-finding-internal-angles-polygon.html
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# Thread:
1. ## Finding the internal angles of a polygon
Hi there,
This might seem a simple question, but I suck at maths. I need to cut some ply to fill a gap behind my boat kitchen and I have measured the lengths of the sides. I'd like to have the angles to help me draw a decent template to cut from, but I'm not sure how.
A = 1055mm
B = 399mm
C = 860mm
D = 73mm
And the angle between D and A is a right angle.
Any help welcome.
Thanks.
2. You'll need to draw a line from the vertex between B and C, to the vertex between A and D.
You will then need to use a combination of the sine and cosine rules to find the missing sides and angles...
3. It would help if you are working with triangles.
First draw a line from angle AB to angle CD. This will divide the shape into 2 triangles.
Solve this unknown length (call this E) by using the Pythagorus Theorum (Because it has a right angle):
$A^2 + D^2 = E^2$ (I substituted your letters into it)
$1055^2 + 73^2 = E^2$
$E \approx 1057.52\mbox{mm}$
Now we find the angle of BC by using the cosine rule:
$\mbox{angle BC} = cos^{-1}\left(\dfrac{B^2+C^2-E^2}{2BC}\right)$
$\mbox{angle BC} = cos^{-1}\left(\dfrac{399^2+860^2-1057.52^2}{2 * 399 * 860}\right)$
$\mbox{angle BC} \approx 108.657^{\circ}$
To solve the angle AB, we do the same thing, but must divide it into 2 parts.
$\mbox{angle AE} = cos^{-1}\left(\dfrac{A^2+E^2-D^2}{2AE}\right)$
$\mbox{angle AE} = cos^{-1}\left(\dfrac{1055^2+1057.52^2-73^2}{2 * 1055 * 1057.52}\right)$
$\mbox{angle AE} \approx 3.958^{\circ}$
$\mbox{angle EB} = cos^{-1}\left(\dfrac{E^2+B^2-C^2}{2EB}\right)$
$\mbox{angle EB} = cos^{-1}\left(\dfrac{1057.52^2+399^2-860^2}{2 * 1057.52 * 399}\right)$
$\mbox{angle EB} \approx 50.3976^{\circ}$
$\mbox{angle AB} = \mbox{angle AE} + \mbox{angle EB}$
$\mbox{angle AB} = 3.958^{\circ} + 50.3976^{\circ}$
$\mbox{angle AB} \approx 54.3556^{\circ}$
To find angle CD, we must subtract the other 3 angles by 360.
$\mbox{angle CD} = 360^{\circ} - 54.3556^{\circ} - 108.657^{\circ} - 90^{\circ}$
$\mbox{angle CD} \approx 106.9874^{\circ}$
4. That's fantastic. Thank you very much.
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http://physics.stackexchange.com/questions/51875/current-without-voltage-and-voltage-without-current?answertab=oldest
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# Current without Voltage and Voltage without Current?
At school I've always learned that you can view Current and Voltage like this:
The current is the flow of charge per second and the Voltage is how badly the current 'wants' to flow.
But I'm having some trouble with this view. How can we have a Voltage without a current? There is nothing to 'flow', so how can it be there? Or is it 'latent' voltage, I mean is the voltage just always there and if a current is introduced it flows?
Also, I believe you can't have current without voltage. This to me seems logical from the very definition of current. But if you have a 'charge' without a voltage, doesn't it just stay in 1 place? Can you view it like that? If you introduce a charge in a circuit without a voltage it just doesn't move?
-
## 1 Answer
What flows is not the voltage but the charge, and that flow is called current. Voltage can be without a current, if you have a single charge, that charge induces a voltage in all space, even if it's empty. Voltage, in the most physical way, is a scalar field that determines the potential energy per unit charge at every point in space.
Now, you can't have currents without voltages because if there's a current there's a charge moving, and every charge produces a voltage, but you can have currents without voltage differences in space. For example, if you have a charged sphere, and you make it rotate, the charge will be on the surface and by rotating the sphere you will have a current on the surface, but the voltage is the same in all surface. Also magnetization of materials can induce currents by the same way.
If you introduce a charge in a circuit without a voltage it just doesn't move?
That's true, it won't move, unless you have some changing magnetic field that may introduce "voltage differences" between the same point, making $\nabla\times E\not =0$, although that wouldn't be electrostatic voltage the way you're seeing it.
-
The only exception is that superconductors can carry a current without a voltage. – Jerry Schirmer Jan 22 at 15:33
@JerrySchirmer Really, isn't it that they just have super low resistence? I mean, I know that a very very very low voltage could produce a huge current, but would a superconductor in total absence of voltage difference produce a current? – MyUserIsThis Jan 22 at 16:25
2
superconductors can maintain eddy currents flowing in rings with no externally supplied voltage. – Jerry Schirmer Jan 22 at 17:51
And i should say that they don't match up with ordinary classical expectations about conductors, because superconductivity is a fundamentally quantum mechanical phenomenon. – Jerry Schirmer Jan 22 at 17:58
|
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|
http://noncommutativegeometry.blogspot.com/2012/07/operatorscalar-fusion-in-finite.html
|
# Noncommutative geometry
## Wednesday, July 11, 2012
### Operator/scalar fusion in finite characteristic and remarkable formulae
Let $E$ be a curve of genus $1$ over the rational field $\mathbf Q$. One of the glories of mathematics is the discovery that (upon choosing a fixed rational point "$\mathbf O$") $E$ comes equipped with an addition which makes its points over any number field (or $\mathbf R$ or $\mathbf C$) a very natural abelian group. (In the vernacular of algebraic geometry, one calls $E$ an "abelian variety" of dimension $1$ or an "abelian curve".)
Built into this setup is a natural tension between the two different avatars of the integers $\mathbf Z$ which now arise. On the one hand, an integer $n$ is an element of the scalars $\mathbf Q$ over which our curve $E$ lies; on the other hand, $n$ is also an operator on the group formed by the elliptic curve (and, in fact, it is well known that this operator is actually a morphism on the elliptic curve).
One would, somehow, like to form a ring that encompasses both of these avatars. An obvious way to do this would be to form ${\mathbf Z}\otimes {\mathbf Z}$ but, alas, this fails as this tensor product is simply $\mathbf Z$. I have always thought, perhaps naively, that one of the motivations in studying ${\mathbf F}_1$ was the hope that progress could be made here....
In any case, in finite characteristic we are blessed with more flexibility. Let $q$ be a power of a prime $p$ and let ${\mathbf F}_q$ by the field with $q$-elements with $A:={\mathbf F}_q[\theta]$ the polynomial ring in the indeterminate $\theta$. In the 1970's, soon after he defined elliptic modules (a.k.a., Drinfeld modules) Drinfeld was influenced by the work of Krichever to define an associated vector bundle called a "shtuka". In order to do so, Drinfeld worked with the $2$-dimensional algebra $A\otimes_{\mathbf F_q} A$ which precisely combined the roles of operator and scalar. Soon after that, Greg Anderson used this algebra to develop his higher dimensional analog of Drinfeld modules (called "$t$-modules"); in particular, Anderson's theory allowed one to create a good category of "motives" out of Drinfeld modules which is, itself, equipped with a good notion of a tensor product.
One can associate to Drinfeld modules analogs of classical special functions such as $L$-series, gamma functions; etc. Classical theory leads to the expectation that these gamma functions should somehow be related to the $L$-series much as gamma functions are "Euler-factors at infinity" in classical algebraic number theory. But so far that has not been the case and the connection, if one exists, remains unknown.
The basic Drinfeld module is the rank $1$ module $C$ discovered by L. Carlitz in the 1930's (in a triumph of old school algebra!); it is a function field analog of the algebraic group ${\mathbf G}_m$ and its exponential is a function field analog of the classical exponential function. Let $\tau(z):=z^{q}$ be the $q$-th power mapping with $\tau^i$ defined by composition; the Carlitz module is then the $\mathbf F_q$-algebra map defined by $C_\theta:=\theta \tau^0+\tau$. Using Anderson's notion of a tensor product, Greg and Dinesh Thakur rapidly defined, and studied, the $n$-tensor power $C^{\otimes n}$ of the Carlitz module in "Tensor powers of the Carlitz module and zeta values," Ann. of Math. 132 (1990), 159–191. In particular, they defined the following marvelous function
$$\omega (t):=\theta_1 \prod_{i=0}^\infty \left(1-\frac{t}{\theta^{q^i}}\right)^{-1}\,,$$
where $\theta_1$ is a fixed $(q-1)$-st root of $-\theta$. Notice that $\omega(t)$ is obviously the reciprocal of an entire function and, in that, it reminds one of Euler's gamma function.
However, much more profound is the result of Anderson/Thakur (loc. cit.) that $\lim_{t\mapsto\theta}(t-\theta)\omega(t)$ is the period $\tilde{\xi}$ of the Carlitz module. Here one can't help but be reminded of the famous equality $\Gamma(1/2)=\sqrt \pi$; so one is led to view $\omega(t)$ as yet another function field manifestation of the notion of a gamma function. Indeed, in a tour de force, "Determination of the algebraic relations among special $\Gamma$-values in positive characteristic," (Ann. of Math. (2) (2004), 237-313), Anderson, Dale Brownawell, and Matt Papanikolas used $\omega(t)$ to establish virtually all the transcendence results one would want of the geometric gamma function.
So it was apparent, to me anyway, that this magical $\omega(t)$ should also make itself known in the theory of characteristic $p$ $L$-series. However, I simply did not see how this could happen. This impasse was recently broken by some fantastic results of Federico Pellarin ("Values of Certain $L$-series in positive characteristic," Ann. of Math. to appear, http://arxiv.org/abs/1107.4511) and these results precisely provide the operator/scalar fusion mentioned in the title of this blog!
So I would like to finish by describing some of Federico's results, and also those of my student Rudy Perkins in this regard. They both are obtaining all sorts of beautiful formulae of the sort one might find in the famous book by Whittaker and Watson which is very exciting and certainly bodes very well for the future of the subject. But before doing so, we do need one more result of Anderson/Thakur.
As in my previous blog put $K:={\mathbf F}_q((1/\theta))$ with the canonical absolute value. Put ${\mathbf T}:=\{\sum_{i=0}^\infty a_it^i\}$ where $\{a_i\}\subset K$ and $a_i\to 0$ as $i \to \infty$; so $\mathbf T$ is simply the Tate algebra of functions with coefficients in $K$ converging on the closed unit disc.
The algebra $\mathbf T$ comes equipped with two natural operators: First of all, the usual hyperdifferential operators act on $\mathbf T$ via differentation with respect to $t$ in the standard fashion. Now let $f(t)=\sum a_it^i\in \mathbf T$; we then set $\tau (f):=\sum a_i^qt^i$ and call it the
"partial Frobenius operator" (in an obvious sense). Note that, in this setting, $\tau$ is actually $\mathbf F_q[t]$-linear. Note also, because we are in characteristic $p$ these operators commute.
Anderson and Thakur look at the following partial Frobenius equation on $\mathbf T$: $\tau \phi=(t-\theta)\phi$ (N.B.: $t-\theta$ is the "shtuka function" associated to the Carlitz module). The solutions to this equation clearly form an $\mathbf F_q[t]$-module and the remarkable result of A/T is that this module is free of rank $1$ and generated by $\omega(t)$.
One can rewrite the fundamental equation $\tau \omega=(t-\theta)\omega$ as
$$(\theta \tau^0+\tau)\omega=t\cdot \omega\,;$$
in other words, if we use the partial Frobenius operators to extend the Carlitz module to $\mathbf T$ then $\omega$ trivializes this action. So if $f(\theta)\in A$ one sees immediately that $C_f\omega=f(t)\omega$.
Abstracting a bit, if $t$ is a scalar, then one defines the "quasi-character" $\chi_t(f):=f(t)$ simply by evaluation. It is Federico's crucial insight that this quasi-character is exactly the necessary device to fuse both the scalars and operators in the theory of characteristic $p$ $L$-series by defining the associated $L$-series $L(\chi_t,s)$ (in the standard fashion). These functions have all the right analytic properties in the $s$-variable and also have excellent analytic properties in the $t$-variable!
(The reader might have imagined, as I did at first, that the poles $\{\theta^{q^i}\}$ of $\omega(t)$ are too specialized to be associated to something canonical. However, we now see that these poles correspond to the quasi-characters $f(\theta)\mapsto f(\theta^{q^i})=f(\theta)^{q^i}$ and so are completely canonical...)
The introduction of the variable $t$ is, actually, a realization of the notion of "families" of $L$-series. Indeed, if $t$ belongs to the algebraic closure of $\mathbf F_q$, then $\chi_t$ is a character modulo $p(\theta)$, where $p(\theta)$ is the minimal polynomial of $t$.
Theorem: (Pellarin) We have $(t-\theta)\omega(t)L(\chi_t,1)= -\tilde{\xi}\,.$
And so $\omega(t)$ makes its appearance in $L$-series! (One is also reminded a bit of Euler's famous formula $e^{\pi i}=-1$.) Now let $n$ be a positive integer $\equiv 1$ mod $(q-1)$.
Theorem: (Pellarin) There exists a rational function $\lambda_n\in {\mathbf F}_q(t,\theta)$ such that
$$(t-\theta)\omega(t)L(\chi,n)=\lambda_n \tilde{\xi}^n\,.$$
In "Explicit formulae for $L$-values in finite characteristic" (just uploaded to the arXiv as http://arxiv.org/abs/1207.1753), my student Rudy Perkins gives a simple closed form expression for these $\lambda_n$ as well as all sorts of connections with other interesting objects (such as the Wagner expansion of $\mathbf F_q$-linear functions, recursive formulae for Bernoulli-Carlitz elements, etc.).
So the introduction of $\chi_t$ has opened the door to all sorts of remarkable results. Still, the algebraic closure of $K$ is such a vast thing (with infinitely many extensions of bounded degree etc.), that there may be other surprises we do not yet know. Moreover, we do know that the algebras of measures can be interpreted as hyperdifferential operators on $\mathbf T$. Where are they in the game Federico started?
Posted by at 1:27 AM
#### 3 comments:
Robert Smart said...
The mathematical formulas didn't display correctly in safari, but are ok in firefox.
Hamid said...
Just at once, after realizing that the power of a_i in tau(f) is not (q_t)^i, everything suddenly fit together and I actually then went after the original papers of Laurent Lafforgue generalizing shtoukas of Drinfeld for Gl_n, reassured that I can do it, that I can find out what it says. Only if I could get my hands on these papers and Kamionkowski's paper on dark matter and wimps.
willson said...
I think A right fusion frame is a sequence of orthogonal projection operators that sum to a scalar multiple of identity operator. I appreciate your given information, useful to all.
kite in geometry
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|
http://www.reference.com/browse/light-quantum
|
Definitions
# photon
[foh-ton] /ˈfoʊtɒn/
or light quantum
Minute energy packet of electromagnetic radiation. In 1900 Max Planck found that heat radiation is emitted and absorbed in distinct units, which he called quanta. In 1905 Albert Einstein explained the photoelectric effect, proposing the existence of discrete energy packets in light. The term photon came into use for these packets in 1926. The energies of photons range from high-energy gamma rays and X rays to low-energy infrared and radio waves, though all travel at the same speed, the speed of light. Photons have no electric charge or rest mass and are the carriers of the electromagnetic field.
Encyclopedia Britannica, 2008. Encyclopedia Britannica Online.
In physics, a quantum (plural: quanta) is an indivisible entity of a quantity that has the same units as the Planck constant and is related to both energy and momentum of elementary particles of matter (called fermions) and of photons and other bosons. The word comes from the Latin "quantus," for "how much." Behind this, one finds the fundamental notion that a physical property may be "quantized", referred to as "quantization". This means that the magnitude can take on only certain discrete numerical values, rather than any value, at least within a range. There is a related term of quantum number.
A photon is often referred to as a "light quantum." The energy of an electron bound to an atom (at rest) is said to be quantized, which results in the stability of atoms, and of matter in general. But these terms can be a little misleading, because what is quantized is this Planck's constant quantity whose units can be viewed as either energy multiplied by time or momentum multiplied by distance.
Usually referred to as quantum "mechanics," it is regarded by virtually every professional physicist as the most fundamental framework we have for understanding and describing nature at the infinitesimal level, for the very practical reason that it works. It is "in the nature of things", not a more or less arbitrary human preference.
## Development of quantum theory
Quantum theory, the branch of physics which is based on quantization, began in 1900 when Max Planck published his theory explaining the emission spectrum of black bodies. In that paper Planck used the system of units he invented the previous year. The consequences of the differences between classical and quantum mechanics quickly became obvious. But it was not until 1926, by the work of Werner Heisenberg, Erwin Schrödinger, and others, that quantum mechanics became correctly formulated and understood mathematically. Despite tremendous experimental success, the philosophical interpretations of quantum theory are still widely debated.
Planck was reluctant to accept the new idea of quantization, as were many others. But, with no acceptable alternative, he continued to work with the idea, and found his efforts were well received. Eighteen years later, when he accepted the Nobel Prize in Physics for his contributions, he called it "a few weeks of the most strenuous work" of his life. During those few weeks, he even had to discard much of his own theoretical work from the preceding years. Quantization turned out to be the only way to describe the new and detailed experiments which were just then being performed. He did this practically overnight, openly reporting his change of mind to his scientific colleagues, in the October, November, and December meetings of the German Physical Society, in Berlin, where the black body work was being intensely discussed. In this way, careful experimentalists (including Friedrich Paschen, O.R. Lummer, Ernst Pringsheim, Heinrich Rubens, and F. Kurlbaum), and a reluctant theorist, ushered in a momentous scientific revolution.
### The quantum black-body radiation formula
When a body is heated, it emits radiant heat, a form of electromagnetic radiation in the infrared region of the EM spectrum. All of this was well understood at the time, and of considerable practical importance. When the body becomes red-hot, the red wavelength parts start to become visible. This had been studied over the previous years, as the instruments were being developed. However, most of the heat radiation remains infrared, until the body becomes as hot as the surface of the Sun (about 6000 K, where most of the light is green in color). This was not achievable in the laboratory at that time. What is more, measuring specific infrared wavelengths was only then becoming feasible, due to newly developed experimental techniques. Until then, most of the electromagnetic spectrum was not measurable, and therefore blackbody emission had not been mapped out in detail.
The quantum black-body radiation formula, being the very first piece of quantum mechanics, appeared Sunday evening October 7, 1900, in a so-called back-of-the-envelope calculation by Planck. It was based on a report by Rubens (visiting with his wife) of the very latest experimental findings in the infrared. Later that evening, Planck sent the formula on a postcard, which Rubens received the following morning. A couple of days later, he informed Planck that it worked perfectly. At first, it was just a fit to the data; only later did it turn out to enforce quantization.
This second step was only possible due to a certain amount of luck (or skill, even though Planck himself called it "a fortuitous guess at an interpolation formula"). It was during the course of polishing the mathematics of his formula that Planck stumbled upon the beginnings of Quantum Theory. Briefly stated, he had two mathematical expressions:
• (i) from the previous work on the red parts of the spectrum, he had x;
• (ii) now, from the new infrared data, he got x².
Combining these as x(a+x), he still has x, approximately, when x is much smaller than a (the red end of the spectrum); but now also x² (again approximately) when x is much larger than a (in the infrared). The formula for the energy E, in a single mode of radiation at frequency λ, and temperature T, can be written
$E = frac\left\{h lambda\right\}\left\{e^\left\{frac\left\{h lambda\right\}\left\{k T\right\}\right\} - 1\right\}$
This is (essentially) what is being compared with the experimental measurements. There are two parameters to determine from the data, written in the present form by the symbols used today: h is the new Planck's constant, and k is Boltzmann's constant. Both have now become fundamental in physics, but that was by no means the case at the time. The "elementary quantum of energy" is hλ. But such a unit does not normally exist, and is not required for quantization.
## Beyond electromagnetic radiation
While quantization was first discovered in electromagnetic radiation, it describes a fundamental aspect of energy not just restricted to photons.
### The birthday of quantum mechanics
From the experiments, Planck deduced the numerical values of h and k. Thus he could report, in the German Physical Society meeting on December 14, 1900, where quantization (of energy) was revealed for the first time, values of the Avogadro-Loschmidt number, the number of real molecules in a mole, and the unit of electrical charge, which were more accurate than those known until then. This event has been referred to as "the birth of quantum mechanics".
## References
• J. Mehra and H. Rechenberg, The Historical Development of Quantum Theory, Vol.1, Part 1, Springer-Verlag New York Inc., New York 1982.
• Lucretius, "On the Nature of the Universe", transl. from the Latin by R.E. Latham, Penguin Books Ltd., Harmondsworth 1951. There are, of course, many translations, and the translation's title varies. Some put emphasis on how things work, others on what things are found in nature.
• M. Planck, A Survey of Physical Theory, transl. by R. Jones and D.H. Williams, Methuen & Co., Ltd., London 1925 (Dover editions 1960 and 1993) including the Nobel lecture.
|
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|
http://www.sagemath.org/doc/reference/plane_curves/sage/schemes/elliptic_curves/ell_number_field.html
|
# Elliptic curves over number fields¶
An elliptic curve $$E$$ over a number field $$K$$ can be given by a Weierstrass equation whose coefficients lie in $$K$$ or by using base_extend on an elliptic curve defined over a subfield.
One major difference to elliptic curves over $$\QQ$$ is that there might not exist a global minimal equation over $$K$$, when $$K$$ does not have class number one. Another difference is the lack of understanding of modularity for general elliptic curves over general number fields.
Currently Sage can obtain local information about $$E/K_v$$ for finite places $$v$$, it has an interface to Denis Simon’s script for 2-descent, it can compute the torsion subgroup of the Mordell-Weil group $$E(K)$$, and it can work with isogenies defined over $$K$$.
EXAMPLE:
```sage: K.<i> = NumberField(x^2+1)
sage: E = EllipticCurve([0,4+i])
sage: E.discriminant()
-3456*i - 6480
sage: P= E([i,2])
sage: P+P
(-2*i + 9/16 : -9/4*i - 101/64 : 1)
```
```sage: E.has_good_reduction(2+i)
True
sage: E.local_data(4+i)
Local data at Fractional ideal (i + 4):
Reduction type: bad additive
Local minimal model: Elliptic Curve defined by y^2 = x^3 + (i+4) over Number Field in i with defining polynomial x^2 + 1
Minimal discriminant valuation: 2
Conductor exponent: 2
Kodaira Symbol: II
Tamagawa Number: 1
sage: E.tamagawa_product_bsd()
1
```
```sage: E.simon_two_descent()
(1, 1, [(i : 2 : 1)])
```
```sage: E.torsion_order()
1
```
```sage: E.isogenies_prime_degree(3)
[Isogeny of degree 3 from Elliptic Curve defined by y^2 = x^3 + (i+4) over Number Field in i with defining polynomial x^2 + 1 to Elliptic Curve defined by y^2 = x^3 + (-27*i-108) over Number Field in i with defining polynomial x^2 + 1]
```
AUTHORS:
• Robert Bradshaw 2007
• John Cremona
• Chris Wuthrich
REFERENCE:
• [Sil] Silverman, Joseph H. The arithmetic of elliptic curves. Second edition. Graduate Texts in Mathematics, 106. Springer, 2009.
• [Sil2] Silverman, Joseph H. Advanced topics in the arithmetic of elliptic curves. Graduate Texts in Mathematics, 151. Springer, 1994.
class sage.schemes.elliptic_curves.ell_number_field.EllipticCurve_number_field(x, y=None)¶
Bases: sage.schemes.elliptic_curves.ell_field.EllipticCurve_field
Elliptic curve over a number field.
EXAMPLES:
```sage: K.<i>=NumberField(x^2+1)
sage: EllipticCurve([i, i - 1, i + 1, 24*i + 15, 14*i + 35])
Elliptic Curve defined by y^2 + i*x*y + (i+1)*y = x^3 + (i-1)*x^2 + (24*i+15)*x + (14*i+35) over Number Field in i with defining polynomial x^2 + 1
```
conductor()¶
Returns the conductor of this elliptic curve as a fractional ideal of the base field.
OUTPUT:
(fractional ideal) The conductor of the curve.
EXAMPLES:
```sage: K.<i>=NumberField(x^2+1)
sage: EllipticCurve([i, i - 1, i + 1, 24*i + 15, 14*i + 35]).conductor()
Fractional ideal (21*i - 3)
sage: K.<a>=NumberField(x^2-x+3)
sage: EllipticCurve([1 + a , -1 + a , 1 + a , -11 + a , 5 -9*a ]).conductor()
Fractional ideal (6*a)
```
A not so well known curve with everywhere good reduction:
```sage: K.<a>=NumberField(x^2-38)
sage: E=EllipticCurve([0,0,0, 21796814856932765568243810*a - 134364590724198567128296995, 121774567239345229314269094644186997594*a - 750668847495706904791115375024037711300])
sage: E.conductor()
Fractional ideal (1)
```
An example which used to fail (see trac #5307):
```sage: K.<w>=NumberField(x^2+x+6)
sage: E=EllipticCurve([w,-1,0,-w-6,0])
sage: E.conductor()
Fractional ideal (86304, w + 5898)
```
An example raised in #11346:
```sage: K.<g> = NumberField(x^2 - x - 1)
sage: E1 = EllipticCurve(K,[0,0,0,-1/48,-161/864])
sage: [(p.smallest_integer(),e) for p,e in E1.conductor().factor()]
[(2, 4), (3, 1), (5, 1)]
```
gens(verbose=0, lim1=5, lim3=50, limtriv=10, maxprob=20, limbigprime=30)¶
Returns some generators of this elliptic curve. Check rank() or rank_bounds() to verify the number of generators.
Note
The optional parameters control the Simon two descent algorithm; see the documentation of simon_two_descent() for more details.
INPUT:
• verbose – 0, 1, 2, or 3 (default: 0), the verbosity level
• lim1 – (default: 5) limit on trivial points on quartics
• lim3 – (default: 50) limit on points on ELS quartics
• limtriv – (default: 10) limit on trivial points on elliptic curve
• maxprob – (default: 20)
• limbigprime – (default: 30) to distinguish between small and large prime numbers. Use probabilistic tests for large primes. If 0, don’t use probabilistic tests.
OUTPUT:
The linearly independent elements given by the Simon two-descent.
Note
For non-quadratic number fields, this code does return, but it takes a long time.
EXAMPLES:
```sage: K.<a> = NumberField(x^2 + 23, 'a')
sage: E = EllipticCurve(K, '37')
sage: E == loads(dumps(E))
True
sage: E.gens()
[(-1 : 0 : 1), (1/2*a - 5/2 : -1/2*a - 13/2 : 1)]
```
Here is a curve with two-torsion, so here the algorithm does not uniquely determine the rank:
```sage: Qrt5.<rt5>=NumberField(x^2-5)
sage: E=EllipticCurve([0,5-rt5,0,rt5,0])
sage: E.gens()
[(3/2*rt5 + 5/2 : -9/2*rt5 - 15/2 : 1), (-1/2*rt5 + 3/2 : 3/2*rt5 - 9/2 : 1), (0 : 0 : 1)]
```
IMPLEMENTATION:
Uses Denis Simon’s PARI/GP scripts from http://www.math.unicaen.fr/~simon/.
global_integral_model()¶
Return a model of self which is integral at all primes.
EXAMPLES:
```sage: K.<i> = NumberField(x^2+1)
sage: E = EllipticCurve([i/5,i/5,i/5,i/5,i/5])
sage: P1,P2 = K.primes_above(5)
sage: E.global_integral_model()
Elliptic Curve defined by y^2 + (-i)*x*y + (-25*i)*y = x^3 + 5*i*x^2 + 125*i*x + 3125*i over Number Field in i with defining polynomial x^2 + 1
```
trac #7935:
```sage: K.<a> = NumberField(x^2-38)
sage: E = EllipticCurve([a,1/2])
sage: E.global_integral_model()
Elliptic Curve defined by y^2 = x^3 + 1444*a*x + 27436 over Number Field in a with defining polynomial x^2 - 38
```
trac #9266:
```sage: K.<s> = NumberField(x^2-5)
sage: w = (1+s)/2
sage: E = EllipticCurve(K,[2,w])
sage: E.global_integral_model()
Elliptic Curve defined by y^2 = x^3 + 2*x + (1/2*s+1/2) over Number Field in s with defining polynomial x^2 - 5
```
trac #12151:
```sage: K.<v> = NumberField(x^2 + 161*x - 150)
sage: E = EllipticCurve([25105/216*v - 3839/36, 634768555/7776*v - 98002625/1296, 634768555/7776*v - 98002625/1296, 0, 0])
sage: E.global_integral_model()
Elliptic Curve defined by y^2 + (33872485050625*v-31078224284250)*x*y + (2020602604156076340058146664245468750000*v-1871778534673615560803175189398437500000)*y = x^3 + (6933305282258321342920781250*v-6422644400723486559914062500)*x^2 over Number Field in v with defining polynomial x^2 + 161*x - 150
```
global_minimal_model(proof=None)¶
Returns a model of self that is integral, minimal at all primes.
Note
This is only implemented for class number 1. In general, such a model may or may not exist.
INPUT:
• proof – whether to only use provably correct methods (default controlled by global proof module). Note that the proof module is number_field, not elliptic_curves, since the functions that actually need the flag are in number fields.
OUTPUT:
A global integral and minimal model.
EXAMPLES:
```sage: K.<a> = NumberField(x^2-38)
sage: E = EllipticCurve([0,0,0, 21796814856932765568243810*a - 134364590724198567128296995, 121774567239345229314269094644186997594*a - 750668847495706904791115375024037711300])
sage: E2 = E.global_minimal_model()
sage: E2 # random (the global minimal model is not unique)
Elliptic Curve defined by y^2 + a*x*y + (a+1)*y = x^3 + (a+1)*x^2 + (368258520200522046806318444*a-2270097978636731786720859345)*x + (8456608930173478039472018047583706316424*a-52130038506793883217874390501829588391299) over Number Field in a with defining polynomial x^2 - 38
sage: E2.local_data()
[]
```
See trac #11347:
```sage: K.<g> = NumberField(x^2 - x - 1)
sage: E = EllipticCurve(K,[0,0,0,-1/48,161/864]).integral_model().global_minimal_model(); E
Elliptic Curve defined by y^2 + x*y + y = x^3 + x^2 over Number Field in g with defining polynomial x^2 - x - 1
sage: [(p.norm(), e) for p, e in E.conductor().factor()]
[(9, 1), (5, 1)]
sage: [(p.norm(), e) for p, e in E.discriminant().factor()]
[(9, 1), (-5, 2)]
```
has_additive_reduction(P)¶
Return True if this elliptic curve has (bad) additive reduction at the prime $$P$$.
INPUT:
• P – a prime ideal of the base field of self, or a field element generating such an ideal.
OUTPUT:
(bool) True if the curve has additive reduction at $$P$$, else False.
EXAMPLES:
```sage: E=EllipticCurve('27a1')
sage: [(p,E.has_additive_reduction(p)) for p in prime_range(15)]
[(2, False), (3, True), (5, False), (7, False), (11, False), (13, False)]
sage: K.<a>=NumberField(x^3-2)
sage: P17a, P17b = [P for P,e in K.factor(17)]
sage: E = EllipticCurve([0,0,0,0,2*a+1])
sage: [(p,E.has_additive_reduction(p)) for p in [P17a,P17b]]
[(Fractional ideal (4*a^2 - 2*a + 1), False),
(Fractional ideal (2*a + 1), True)]
```
has_bad_reduction(P)¶
Return True if this elliptic curve has bad reduction at the prime $$P$$.
INPUT:
• P – a prime ideal of the base field of self, or a field element generating such an ideal.
OUTPUT:
(bool) True if the curve has bad reduction at $$P$$, else False.
Note
This requires determining a local integral minimal model; we do not just check that the discriminant of the current model has valuation zero.
EXAMPLES:
```sage: E=EllipticCurve('14a1')
sage: [(p,E.has_bad_reduction(p)) for p in prime_range(15)]
[(2, True), (3, False), (5, False), (7, True), (11, False), (13, False)]
sage: K.<a>=NumberField(x^3-2)
sage: P17a, P17b = [P for P,e in K.factor(17)]
sage: E = EllipticCurve([0,0,0,0,2*a+1])
sage: [(p,E.has_bad_reduction(p)) for p in [P17a,P17b]]
[(Fractional ideal (4*a^2 - 2*a + 1), False),
(Fractional ideal (2*a + 1), True)]
```
has_good_reduction(P)¶
Return True if this elliptic curve has good reduction at the prime $$P$$.
INPUT:
• P – a prime ideal of the base field of self, or a field element generating such an ideal.
OUTPUT:
(bool) – True if the curve has good reduction at $$P$$, else False.
Note
This requires determining a local integral minimal model; we do not just check that the discriminant of the current model has valuation zero.
EXAMPLES:
```sage: E=EllipticCurve('14a1')
sage: [(p,E.has_good_reduction(p)) for p in prime_range(15)]
[(2, False), (3, True), (5, True), (7, False), (11, True), (13, True)]
sage: K.<a>=NumberField(x^3-2)
sage: P17a, P17b = [P for P,e in K.factor(17)]
sage: E = EllipticCurve([0,0,0,0,2*a+1])
sage: [(p,E.has_good_reduction(p)) for p in [P17a,P17b]]
[(Fractional ideal (4*a^2 - 2*a + 1), True),
(Fractional ideal (2*a + 1), False)]
```
has_multiplicative_reduction(P)¶
Return True if this elliptic curve has (bad) multiplicative reduction at the prime $$P$$.
Note
See also has_split_multiplicative_reduction() and has_nonsplit_multiplicative_reduction().
INPUT:
• P – a prime ideal of the base field of self, or a field
element generating such an ideal.
OUTPUT:
(bool) True if the curve has multiplicative reduction at $$P$$, else False.
EXAMPLES:
```sage: E=EllipticCurve('14a1')
sage: [(p,E.has_multiplicative_reduction(p)) for p in prime_range(15)]
[(2, True), (3, False), (5, False), (7, True), (11, False), (13, False)]
sage: K.<a>=NumberField(x^3-2)
sage: P17a, P17b = [P for P,e in K.factor(17)]
sage: E = EllipticCurve([0,0,0,0,2*a+1])
sage: [(p,E.has_multiplicative_reduction(p)) for p in [P17a,P17b]]
[(Fractional ideal (4*a^2 - 2*a + 1), False), (Fractional ideal (2*a + 1), False)]
```
has_nonsplit_multiplicative_reduction(P)¶
Return True if this elliptic curve has (bad) non-split multiplicative reduction at the prime $$P$$.
INPUT:
• P – a prime ideal of the base field of self, or a field element generating such an ideal.
OUTPUT:
(bool) True if the curve has non-split multiplicative reduction at $$P$$, else False.
EXAMPLES:
```sage: E=EllipticCurve('14a1')
sage: [(p,E.has_nonsplit_multiplicative_reduction(p)) for p in prime_range(15)]
[(2, True), (3, False), (5, False), (7, False), (11, False), (13, False)]
sage: K.<a>=NumberField(x^3-2)
sage: P17a, P17b = [P for P,e in K.factor(17)]
sage: E = EllipticCurve([0,0,0,0,2*a+1])
sage: [(p,E.has_nonsplit_multiplicative_reduction(p)) for p in [P17a,P17b]]
[(Fractional ideal (4*a^2 - 2*a + 1), False), (Fractional ideal (2*a + 1), False)]
```
has_split_multiplicative_reduction(P)¶
Return True if this elliptic curve has (bad) split multiplicative reduction at the prime $$P$$.
INPUT:
• P – a prime ideal of the base field of self, or a field element generating such an ideal.
OUTPUT:
(bool) True if the curve has split multiplicative reduction at $$P$$, else False.
EXAMPLES:
```sage: E=EllipticCurve('14a1')
sage: [(p,E.has_split_multiplicative_reduction(p)) for p in prime_range(15)]
[(2, False), (3, False), (5, False), (7, True), (11, False), (13, False)]
sage: K.<a>=NumberField(x^3-2)
sage: P17a, P17b = [P for P,e in K.factor(17)]
sage: E = EllipticCurve([0,0,0,0,2*a+1])
sage: [(p,E.has_split_multiplicative_reduction(p)) for p in [P17a,P17b]]
[(Fractional ideal (4*a^2 - 2*a + 1), False), (Fractional ideal (2*a + 1), False)]
```
height_pairing_matrix(points=None, precision=None)¶
Returns the height pairing matrix of the given points.
INPUT:
• points - either a list of points, which must be on this curve, or (default) None, in which case self.gens() will be used.
• precision - number of bits of precision of result (default: None, for default RealField precision)
EXAMPLES:
```sage: E = EllipticCurve([0, 0, 1, -1, 0])
sage: E.height_pairing_matrix()
[0.0511114082399688]
```
For rank 0 curves, the result is a valid 0x0 matrix:
```sage: EllipticCurve('11a').height_pairing_matrix()
[]
sage: E=EllipticCurve('5077a1')
sage: E.height_pairing_matrix([E.lift_x(x) for x in [-2,-7/4,1]], precision=100)
[ 1.3685725053539301120518194471 -1.3095767070865761992624519454 -0.63486715783715592064475542573]
[ -1.3095767070865761992624519454 2.7173593928122930896610589220 1.0998184305667292139777571432]
[-0.63486715783715592064475542573 1.0998184305667292139777571432 0.66820516565192793503314205089]
sage: E = EllipticCurve('389a1')
sage: E = EllipticCurve('389a1')
sage: P,Q = E.point([-1,1,1]),E.point([0,-1,1])
sage: E.height_pairing_matrix([P,Q])
[0.686667083305587 0.268478098806726]
[0.268478098806726 0.327000773651605]
```
Over a number field:
```sage: x = polygen(QQ)
sage: K.<t> = NumberField(x^2+47)
sage: EK = E.base_extend(K)
sage: EK.height_pairing_matrix([EK(P),EK(Q)])
[0.686667083305586 0.268478098806726]
[0.268478098806726 0.327000773651605]
```
```sage: K.<i> = QuadraticField(-1)
sage: E = EllipticCurve([0,0,0,i,i])
sage: P = E(-9+4*i,-18-25*i)
sage: Q = E(i,-i)
sage: E.height_pairing_matrix([P,Q])
[ 2.16941934493768 -0.870059380421505]
[-0.870059380421505 0.424585837470709]
sage: E.regulator_of_points([P,Q])
0.164101403936070
```
integral_model()¶
Return a model of self which is integral at all primes.
EXAMPLES:
```sage: K.<i> = NumberField(x^2+1)
sage: E = EllipticCurve([i/5,i/5,i/5,i/5,i/5])
sage: P1,P2 = K.primes_above(5)
sage: E.global_integral_model()
Elliptic Curve defined by y^2 + (-i)*x*y + (-25*i)*y = x^3 + 5*i*x^2 + 125*i*x + 3125*i over Number Field in i with defining polynomial x^2 + 1
```
trac #7935:
```sage: K.<a> = NumberField(x^2-38)
sage: E = EllipticCurve([a,1/2])
sage: E.global_integral_model()
Elliptic Curve defined by y^2 = x^3 + 1444*a*x + 27436 over Number Field in a with defining polynomial x^2 - 38
```
trac #9266:
```sage: K.<s> = NumberField(x^2-5)
sage: w = (1+s)/2
sage: E = EllipticCurve(K,[2,w])
sage: E.global_integral_model()
Elliptic Curve defined by y^2 = x^3 + 2*x + (1/2*s+1/2) over Number Field in s with defining polynomial x^2 - 5
```
trac #12151:
```sage: K.<v> = NumberField(x^2 + 161*x - 150)
sage: E = EllipticCurve([25105/216*v - 3839/36, 634768555/7776*v - 98002625/1296, 634768555/7776*v - 98002625/1296, 0, 0])
sage: E.global_integral_model()
Elliptic Curve defined by y^2 + (33872485050625*v-31078224284250)*x*y + (2020602604156076340058146664245468750000*v-1871778534673615560803175189398437500000)*y = x^3 + (6933305282258321342920781250*v-6422644400723486559914062500)*x^2 over Number Field in v with defining polynomial x^2 + 161*x - 150
```
is_global_integral_model()¶
Return true iff self is integral at all primes.
EXAMPLES:
```sage: K.<i> = NumberField(x^2+1)
sage: E = EllipticCurve([i/5,i/5,i/5,i/5,i/5])
sage: P1,P2 = K.primes_above(5)
sage: Emin = E.global_integral_model()
sage: Emin.is_global_integral_model()
True
```
is_isogenous(other, proof=True, maxnorm=100)¶
Returns whether or not self is isogenous to other.
INPUT:
• other – another elliptic curve.
• proof (default True) – If False, the function will return True whenever the two curves have the same conductor and are isogenous modulo $$p$$ for all primes $$p$$ of norm up to maxp. If True, the function returns False when the previous condition does not hold, and if it does hold we attempt to see if the curves are indeed isogenous. However, this has not been fully implemented (see examples below), so we may not be able to determine whether or not the curves are isogenous..
• maxnorm (integer, default 100) – The maximum norm of primes $$p$$ for which isogeny modulo $$p$$ will be checked.
OUTPUT:
(bool) True if there is an isogeny from curve self to curve other.
EXAMPLES:
```sage: x = polygen(QQ, 'x')
sage: F = NumberField(x^2 -2, 's'); F
Number Field in s with defining polynomial x^2 - 2
sage: E1 = EllipticCurve(F, [7,8])
sage: E2 = EllipticCurve(F, [0,5,0,1,0])
sage: E3 = EllipticCurve(F, [0,-10,0,21,0])
sage: E1.is_isogenous(E2)
False
sage: E1.is_isogenous(E1)
True
sage: E2.is_isogenous(E2)
True
sage: E2.is_isogenous(E1)
False
sage: E2.is_isogenous(E3)
True
```
```sage: x = polygen(QQ, 'x')
sage: F = NumberField(x^2 -2, 's'); F
Number Field in s with defining polynomial x^2 - 2
sage: E = EllipticCurve('14a1')
sage: EE = EllipticCurve('14a2')
sage: E1 = E.change_ring(F)
sage: E2 = EE.change_ring(F)
sage: E1.is_isogenous(E2)
True
```
```sage: x = polygen(QQ, 'x')
sage: F = NumberField(x^2 -2, 's'); F
Number Field in s with defining polynomial x^2 - 2
sage: k.<a> = NumberField(x^3+7)
sage: E = EllipticCurve(F, [7,8])
sage: EE = EllipticCurve(k, [2, 2])
sage: E.is_isogenous(EE)
Traceback (most recent call last):
...
ValueError: Second argument must be defined over the same number field.
```
Some examples from Cremona’s 1981 tables:
```sage: K.<i> = QuadraticField(-1)
sage: E1 = EllipticCurve([i + 1, 0, 1, -240*i - 400, -2869*i - 2627])
sage: E1.conductor()
Fractional ideal (4*i + 7)
sage: E2 = EllipticCurve([1+i,0,1,0,0])
sage: E2.conductor()
Fractional ideal (4*i + 7)
sage: E1.is_isogenous(E2)
Traceback (most recent call last):
...
NotImplementedError: Curves appear to be isogenous (same conductor, isogenous modulo all primes of norm up to 1000), but no isogeny has been constructed.
sage: E1.is_isogenous(E2, proof=False)
True
```
In this case E1 and E2 are in fact 9-isogenous, as may be deduced from the following:
```sage: E3 = EllipticCurve([i + 1, 0, 1, -5*i - 5, -2*i - 5])
sage: E3.is_isogenous(E1)
True
sage: E3.is_isogenous(E2)
True
sage: E1.isogeny_degree(E2)
9
```
is_local_integral_model(*P)¶
Tests if self is integral at the prime ideal $$P$$, or at all the primes if $$P$$ is a list or tuple.
INPUT:
• *P – a prime ideal, or a list or tuple of primes.
EXAMPLES:
```sage: K.<i> = NumberField(x^2+1)
sage: P1,P2 = K.primes_above(5)
sage: E = EllipticCurve([i/5,i/5,i/5,i/5,i/5])
sage: E.is_local_integral_model(P1,P2)
False
sage: Emin = E.local_integral_model(P1,P2)
sage: Emin.is_local_integral_model(P1,P2)
True
```
isogeny_degree(other)¶
Returns the minimal degree of an isogeny between self and other, or 0 if no isogeny exists.
INPUT:
• other – another elliptic curve.
OUTPUT:
(int) The degree of an isogeny from self to other, or 0.
Warning
Not all isogenies over number fields are yet implemented. Currently the code only works if there is a chain of isogenies from self to other of degrees 2, 3, 5, 7 and 13.
EXAMPLES:
```sage: x = QQ['x'].0
sage: F = NumberField(x^2 -2, 's'); F
Number Field in s with defining polynomial x^2 - 2
sage: E = EllipticCurve('14a1')
sage: EE = EllipticCurve('14a2')
sage: E1 = E.change_ring(F)
sage: E2 = EE.change_ring(F)
sage: E1.isogeny_degree(E2)
2
sage: E2.isogeny_degree(E2)
1
sage: E5 = EllipticCurve('14a5').change_ring(F)
sage: E1.isogeny_degree(E5)
6
```
kodaira_symbol(P, proof=None)¶
Returns the Kodaira Symbol of this elliptic curve at the prime $$P$$.
INPUT:
• P – either None or a prime ideal of the base field of self.
• proof – whether to only use provably correct methods (default controlled by global proof module). Note that the proof module is number_field, not elliptic_curves, since the functions that actually need the flag are in number fields.
OUTPUT:
The Kodaira Symbol of the curve at P, represented as a string.
EXAMPLES:
```sage: K.<a>=NumberField(x^2-5)
sage: E=EllipticCurve([20, 225, 750, 625*a + 6875, 31250*a + 46875])
sage: bad_primes = E.discriminant().support(); bad_primes
[Fractional ideal (-a), Fractional ideal (7/2*a - 81/2), Fractional ideal (-a - 52), Fractional ideal (2)]
sage: [E.kodaira_symbol(P) for P in bad_primes]
[I0, I1, I1, II]
sage: K.<a> = QuadraticField(-11)
sage: E = EllipticCurve('11a1').change_ring(K)
sage: [E.kodaira_symbol(P) for P in K(11).support()]
[I10]
```
local_data(P=None, proof=None, algorithm='pari')¶
Local data for this elliptic curve at the prime $$P$$.
INPUT:
• P – either None or a prime ideal of the base field of self.
• proof – whether to only use provably correct methods (default controlled by global proof module). Note that the proof module is number_field, not elliptic_curves, since the functions that actually need the flag are in number fields.
• algorithm (string, default: “pari”) – Ignored unless the base field is $$\QQ$$. If “pari”, use the PARI C-library ellglobalred implementation of Tate’s algorithm over $$\QQ$$. If “generic”, use the general number field implementation.
OUTPUT:
If $$P$$ is specified, returns the EllipticCurveLocalData object associated to the prime $$P$$ for this curve. Otherwise, returns a list of such objects, one for each prime $$P$$ in the support of the discriminant of this model.
Note
The model is not required to be integral on input.
For principal $$P$$, a generator is used as a uniformizer, and integrality or minimality at other primes is not affected. For non-principal $$P$$, the minimal model returned will preserve integrality at other primes, but not minimality.
EXAMPLES:
```sage: K.<i> = NumberField(x^2+1)
sage: E = EllipticCurve([1 + i, 0, 1, 0, 0])
sage: E.local_data()
[Local data at Fractional ideal (i - 2):
Reduction type: bad non-split multiplicative
Local minimal model: Elliptic Curve defined by y^2 + (i+1)*x*y + y = x^3 over Number Field in i with defining polynomial x^2 + 1
Minimal discriminant valuation: 1
Conductor exponent: 1
Kodaira Symbol: I1
Tamagawa Number: 1,
Local data at Fractional ideal (-3*i - 2):
Reduction type: bad split multiplicative
Local minimal model: Elliptic Curve defined by y^2 + (i+1)*x*y + y = x^3 over Number Field in i with defining polynomial x^2 + 1
Minimal discriminant valuation: 2
Conductor exponent: 1
Kodaira Symbol: I2
Tamagawa Number: 2]
sage: E.local_data(K.ideal(3))
Local data at Fractional ideal (3):
Reduction type: good
Local minimal model: Elliptic Curve defined by y^2 + (i+1)*x*y + y = x^3 over Number Field in i with defining polynomial x^2 + 1
Minimal discriminant valuation: 0
Conductor exponent: 0
Kodaira Symbol: I0
Tamagawa Number: 1
```
An example raised in #3897:
```sage: E = EllipticCurve([1,1])
sage: E.local_data(3)
Local data at Principal ideal (3) of Integer Ring:
Reduction type: good
Local minimal model: Elliptic Curve defined by y^2 = x^3 + x + 1 over Rational Field
Minimal discriminant valuation: 0
Conductor exponent: 0
Kodaira Symbol: I0
Tamagawa Number: 1
```
local_information(P=None, proof=None)¶
code{local_information} has been renamed code{local_data} and is being deprecated.
local_integral_model(*P)¶
Return a model of self which is integral at the prime ideal $$P$$.
Note
The integrality at other primes is not affected, even if $$P$$ is non-principal.
INPUT:
• *P – a prime ideal, or a list or tuple of primes.
EXAMPLES:
```sage: K.<i> = NumberField(x^2+1)
sage: P1,P2 = K.primes_above(5)
sage: E = EllipticCurve([i/5,i/5,i/5,i/5,i/5])
sage: E.local_integral_model((P1,P2))
Elliptic Curve defined by y^2 + (-i)*x*y + (-25*i)*y = x^3 + 5*i*x^2 + 125*i*x + 3125*i over Number Field in i with defining polynomial x^2 + 1
```
local_minimal_model(P, proof=None, algorithm='pari')¶
Returns a model which is integral at all primes and minimal at $$P$$.
INPUT:
• P – either None or a prime ideal of the base field of self.
• proof – whether to only use provably correct methods (default controlled by global proof module). Note that the proof module is number_field, not elliptic_curves, since the functions that actually need the flag are in number fields.
• algorithm (string, default: “pari”) – Ignored unless the base field is $$\QQ$$. If “pari”, use the PARI C-library ellglobalred implementation of Tate’s algorithm over $$\QQ$$. If “generic”, use the general number field implementation.
OUTPUT:
A model of the curve which is minimal (and integral) at $$P$$.
Note
The model is not required to be integral on input.
For principal $$P$$, a generator is used as a uniformizer, and integrality or minimality at other primes is not affected. For non-principal $$P$$, the minimal model returned will preserve integrality at other primes, but not minimality.
EXAMPLES:
```sage: K.<a>=NumberField(x^2-5)
sage: E=EllipticCurve([20, 225, 750, 625*a + 6875, 31250*a + 46875])
sage: P=K.ideal(a)
sage: E.local_minimal_model(P).ainvs()
(0, 1, 0, a - 33, -2*a + 64)
```
period_lattice(embedding)¶
Returns the period lattice of the elliptic curve for the given embedding of its base field with respect to the differential $$dx/(2y + a_1x + a_3)$$.
INPUT:
• embedding - an embedding of the base number field into $$\RR$$ or $$\CC$$.
Note
The precision of the embedding is ignored: we only use the given embedding to determine which embedding into QQbar to use. Once the lattice has been initialized, periods can be computed to arbitrary precision.
EXAMPLES:
First define a field with two real embeddings:
```sage: K.<a> = NumberField(x^3-2)
sage: E=EllipticCurve([0,0,0,a,2])
sage: embs=K.embeddings(CC); len(embs)
3
```
For each embedding we have a different period lattice:
```sage: E.period_lattice(embs[0])
Period lattice associated to Elliptic Curve defined by y^2 = x^3 + a*x + 2 over Number Field in a with defining polynomial x^3 - 2 with respect to the embedding Ring morphism:
From: Number Field in a with defining polynomial x^3 - 2
To: Algebraic Field
Defn: a |--> -0.6299605249474365? - 1.091123635971722?*I
sage: E.period_lattice(embs[1])
Period lattice associated to Elliptic Curve defined by y^2 = x^3 + a*x + 2 over Number Field in a with defining polynomial x^3 - 2 with respect to the embedding Ring morphism:
From: Number Field in a with defining polynomial x^3 - 2
To: Algebraic Field
Defn: a |--> -0.6299605249474365? + 1.091123635971722?*I
sage: E.period_lattice(embs[2])
Period lattice associated to Elliptic Curve defined by y^2 = x^3 + a*x + 2 over Number Field in a with defining polynomial x^3 - 2 with respect to the embedding Ring morphism:
From: Number Field in a with defining polynomial x^3 - 2
To: Algebraic Field
Defn: a |--> 1.259921049894873?
```
Although the original embeddings have only the default precision, we can obtain the basis with higher precision later:
```sage: L=E.period_lattice(embs[0])
sage: L.basis()
(1.86405007647981 - 0.903761485143226*I, -0.149344633143919 - 2.06619546272945*I)
sage: L.basis(prec=100)
(1.8640500764798108425920506200 - 0.90376148514322594749786960975*I, -0.14934463314391922099120107422 - 2.0661954627294548995621225062*I)
```
rank(verbose=0, lim1=5, lim3=50, limtriv=10, maxprob=20, limbigprime=30)¶
Return the rank of this elliptic curve, if it can be determined.
Note
The optional parameters control the Simon two descent algorithm; see the documentation of simon_two_descent() for more details.
INPUT:
• verbose – 0, 1, 2, or 3 (default: 0), the verbosity level
• lim1 – (default: 5) limit on trivial points on quartics
• lim3 – (default: 50) limit on points on ELS quartics
• limtriv – (default: 10) limit on trivial points on elliptic curve
• maxprob – (default: 20)
• limbigprime – (default: 30) to distinguish between small and large prime numbers. Use probabilistic tests for large primes. If 0, don’t use probabilistic tests.
OUTPUT:
If the upper and lower bounds given by Simon two-descent are the same, then the rank has been uniquely identified and we return this. Otherwise, we raise a ValueError with an error message specifying the upper and lower bounds.
Note
For non-quadratic number fields, this code does return, but it takes a long time.
EXAMPLES:
```sage: K.<a> = NumberField(x^2 + 23, 'a')
sage: E = EllipticCurve(K, '37')
sage: E == loads(dumps(E))
True
sage: E.rank()
2
```
Here is a curve with two-torsion, so here the bounds given by the algorithm do not uniquely determine the rank:
```sage: Qrt5.<rt5>=NumberField(x^2-5)
sage: E=EllipticCurve([0,5-rt5,0,rt5,0])
sage: E.rank()
Traceback (most recent call last):
...
ValueError: There is insufficient data to determine the rank -
2-descent gave lower bound 1 and upper bound 2
```
IMPLEMENTATION:
Uses Denis Simon’s PARI/GP scripts from http://www.math.unicaen.fr/~simon/.
rank_bounds(verbose=0, lim1=5, lim3=50, limtriv=10, maxprob=20, limbigprime=30)¶
Returns the lower and upper bounds using simon_two_descent(). The results of simon_two_descent() are cached.
Note
The optional parameters control the Simon two descent algorithm; see the documentation of simon_two_descent() for more details.
INPUT:
• verbose – 0, 1, 2, or 3 (default: 0), the verbosity level
• lim1 – (default: 5) limit on trivial points on quartics
• lim3 – (default: 50) limit on points on ELS quartics
• limtriv – (default: 10) limit on trivial points on elliptic curve
• maxprob – (default: 20)
• limbigprime – (default: 30) to distinguish between small and large prime numbers. Use probabilistic tests for large primes. If 0, don’t use probabilistic tests.
OUTPUT:
lower and upper bounds
Note
For non-quadratic number fields, this code does return, but it takes a long time.
EXAMPLES:
```sage: K.<a> = NumberField(x^2 + 23, 'a')
sage: E = EllipticCurve(K, '37')
sage: E == loads(dumps(E))
True
sage: E.rank_bounds()
(2, 2)
```
Here is a curve with two-torsion, so here the algorithm gives bounds on the rank:
```sage: Qrt5.<rt5>=NumberField(x^2-5)
sage: E=EllipticCurve([0,5-rt5,0,rt5,0])
sage: E.rank_bounds()
(1, 2)
```
IMPLEMENTATION:
Uses Denis Simon’s PARI/GP scripts from http://www.math.unicaen.fr/~simon/.
reduction(place)¶
Return the reduction of the elliptic curve at a place of good reduction.
INPUT:
• place – a prime ideal in the base field of the curve
OUTPUT:
An elliptic curve over a finite field, the residue field of the place.
EXAMPLES:
```sage: K.<i> = QuadraticField(-1)
sage: EK = EllipticCurve([0,0,0,i,i+3])
sage: v = K.fractional_ideal(2*i+3)
sage: EK.reduction(v)
Elliptic Curve defined by y^2 = x^3 + 5*x + 8 over Residue field of Fractional ideal (2*i + 3)
sage: EK.reduction(K.ideal(1+i))
Traceback (most recent call last):
...
ValueError: The curve must have good reduction at the place.
sage: EK.reduction(K.ideal(2))
Traceback (most recent call last):
...
ValueError: The ideal must be prime.
sage: K=QQ.extension(x^2+x+1,"a")
sage: E=EllipticCurve([1024*K.0,1024*K.0])
sage: E.reduction(2*K)
Elliptic Curve defined by y^2 + (abar+1)*y = x^3 over Residue field in abar of Fractional ideal (2)
```
regulator_of_points(points=[], precision=None)¶
Returns the regulator of the given points on this curve.
INPUT:
• points -(default: empty list) a list of points on this curve
• precision - int or None (default: None): the precision in bits of the result (default real precision if None)
EXAMPLES:
```sage: E = EllipticCurve('37a1')
sage: P = E(0,0)
sage: Q = E(1,0)
sage: E.regulator_of_points([P,Q])
0.000000000000000
sage: 2*P==Q
True
```
```sage: E = EllipticCurve('5077a1')
sage: points = [E.lift_x(x) for x in [-2,-7/4,1]]
sage: E.regulator_of_points(points)
0.417143558758384
sage: E.regulator_of_points(points,precision=100)
0.41714355875838396981711954462
```
```sage: E = EllipticCurve('389a')
sage: E.regulator_of_points()
1.00000000000000
sage: points = [P,Q] = [E(-1,1),E(0,-1)]
sage: E.regulator_of_points(points)
0.152460177943144
sage: E.regulator_of_points(points, precision=100)
0.15246017794314375162432475705
sage: E.regulator_of_points(points, precision=200)
0.15246017794314375162432475704945582324372707748663081784028
sage: E.regulator_of_points(points, precision=300)
0.152460177943143751624324757049455823243727077486630817840280980046053225683562463604114816
```
Examples over number fields:
```sage: K.<a> = QuadraticField(97)
sage: E = EllipticCurve(K,[1,1])
sage: P = E(0,1)
sage: P.height()
0.476223106404866
sage: E.regulator_of_points([P])
0.476223106404866
```
```sage: E = EllipticCurve('11a1')
sage: x = polygen(QQ)
sage: K.<t> = NumberField(x^2+47)
sage: EK = E.base_extend(K)
sage: T = EK(5,5)
sage: T.order()
5
sage: P = EK(-2, -1/2*t - 1/2)
sage: P.order()
+Infinity
sage: EK.regulator_of_points([P,T]) # random very small output
-1.23259516440783e-32
sage: EK.regulator_of_points([P,T]).abs() < 1e-30
True
```
```sage: E = EllipticCurve('389a1')
sage: P,Q = E.gens()
sage: E.regulator_of_points([P,Q])
0.152460177943144
sage: K.<t> = NumberField(x^2+47)
sage: EK = E.base_extend(K)
sage: EK.regulator_of_points([EK(P),EK(Q)])
0.152460177943144
```
```sage: K.<i> = QuadraticField(-1)
sage: E = EllipticCurve([0,0,0,i,i])
sage: P = E(-9+4*i,-18-25*i)
sage: Q = E(i,-i)
sage: E.height_pairing_matrix([P,Q])
[ 2.16941934493768 -0.870059380421505]
[-0.870059380421505 0.424585837470709]
sage: E.regulator_of_points([P,Q])
0.164101403936070
```
simon_two_descent(verbose=0, lim1=5, lim3=50, limtriv=10, maxprob=20, limbigprime=30)¶
Computes lower and upper bounds on the rank of the Mordell-Weil group, and a list of independent points. Used internally by the rank(), rank_bounds() and gens() methods.
INPUT:
• verbose – 0, 1, 2, or 3 (default: 0), the verbosity level
• lim1 – (default: 5) limit on trivial points on quartics
• lim3 – (default: 50) limit on points on ELS quartics
• limtriv – (default: 10) limit on trivial points on elliptic curve
• maxprob – (default: 20)
• limbigprime – (default: 30) to distinguish between small and large prime numbers. Use probabilistic tests for large primes. If 0, don’t use probabilistic tests.
OUTPUT:
(lower, upper, list) where lower is a lower bound on the rank, upper is an upper bound (the 2-Selmer rank) and list is a list of independent points on the Weierstrass model. The length of list is equal to either lower, or lower-1, since when lower is less than upper and of different parity, the value of lower is increased by 1.
Note
For non-quadratic number fields, this code does return, but it takes a long time.
ALGORITHM:
Uses Denis Simon’s PARI/GP scripts from http://www.math.unicaen.fr/~simon/.
EXAMPLES:
```sage: K.<a> = NumberField(x^2 + 23, 'a')
sage: E = EllipticCurve(K, '37')
sage: E == loads(dumps(E))
True
sage: E.simon_two_descent()
(2, 2, [(-1 : 0 : 1), (1/2*a - 5/2 : -1/2*a - 13/2 : 1)])
sage: E.simon_two_descent(lim1=3, lim3=20, limtriv=5, maxprob=7, limbigprime=10)
(2, 2, [(-1 : 0 : 1), (-1/8*a + 5/8 : -3/16*a - 9/16 : 1)])
```
```sage: K.<a> = NumberField(x^2 + 7, 'a')
sage: E = EllipticCurve(K, [0,0,0,1,a]); E
Elliptic Curve defined by y^2 = x^3 + x + a over Number Field in a with defining polynomial x^2 + 7
sage: v = E.simon_two_descent(verbose=1); v
courbe elliptique : Y^2 = x^3 + Mod(1, y^2 + 7)*x + Mod(y, y^2 + 7)
points triviaux sur la courbe = [[1, 1, 0], [Mod(1/2*y + 3/2, y^2 + 7), Mod(-y - 2, y^2 + 7), 1]]
#S(E/K)[2] = 2
#E(K)/2E(K) = 2
#III(E/K)[2] = 1
rang(E/K) = 1
listpointsmwr = [[Mod(1/2*y + 3/2, y^2 + 7), Mod(-y - 2, y^2 + 7), 1]]
(1, 1, [(1/2*a + 3/2 : -a - 2 : 1)])
sage: v = E.simon_two_descent(verbose=2) # random output
K = bnfinit(y^2 + 7);
a = Mod(y,K.pol);
bnfellrank(K, [0,0,0,1,a]);
courbe elliptique : Y^2 = x^3 + Mod(1, y^2 + 7)*x + Mod(y, y^2 + 7)
A = 0
B = Mod(1, y^2 + 7)
C = Mod(y, y^2 + 7)
LS2gen = [Mod(Mod(-5, y^2 + 7)*x^2 + Mod(-3*y, y^2 + 7)*x + Mod(8, y^2 + 7), x^3 + Mod(1, y^2 + 7)*x + Mod(y, y^2 + 7)), Mod(Mod(1, y^2 + 7)*x^2 + Mod(1/2*y - 1/2, y^2 + 7)*x - 1, x^3 + Mod(1, y^2 + 7)*x + Mod(y, y^2 + 7))]
#LS2gen = 2
Recherche de points triviaux sur la courbe
points triviaux sur la courbe = [[1, 1, 0], [Mod(1/2*y + 3/2, y^2 + 7), Mod(-y - 2, y^2 + 7), 1]]
zc = Mod(Mod(-5, y^2 + 7)*x^2 + Mod(-3*y, y^2 + 7)*x + Mod(8, y^2 + 7), x^3 + Mod(1, y^2 + 7)*x + Mod(y, y^2 + 7))
symbole de Hilbert (Mod(2, y^2 + 7),Mod(-5, y^2 + 7)) = -1
zc = Mod(Mod(1, y^2 + 7)*x^2 + Mod(1/2*y - 1/2, y^2 + 7)*x + Mod(-1, y^2 + 7), x^3 + Mod(1, y^2 + 7)*x + Mod(y, y^2 + 7))
symbole de Hilbert (Mod(-2*y + 2, y^2 + 7),Mod(1, y^2 + 7)) = 0
sol de Legendre = [1, 0, 1]~
zc*z1^2 = Mod(Mod(2*y - 2, y^2 + 7)*x + Mod(2*y + 10, y^2 + 7), x^3 + Mod(1, y^2 + 7)*x + Mod(y, y^2 + 7))
quartique : (-1/2*y + 1/2)*Y^2 = x^4 + (-3*y - 15)*x^2 + (-8*y - 16)*x + (-11/2*y - 15/2)
reduite: Y^2 = (-1/2*y + 1/2)*x^4 - 4*x^3 + (-3*y + 3)*x^2 + (2*y - 2)*x + (1/2*y + 3/2)
non ELS en [2, [0, 1]~, 1, 1, [1, 1]~]
zc = Mod(Mod(1, y^2 + 7)*x^2 + Mod(1/2*y + 1/2, y^2 + 7)*x + Mod(-1, y^2 + 7), x^3 + Mod(1, y^2 + 7)*x + Mod(y, y^2 + 7))
vient du point trivial [Mod(1/2*y + 3/2, y^2 + 7), Mod(-y - 2, y^2 + 7), 1]
m1 = 1
m2 = 1
#S(E/K)[2] = 2
#E(K)/2E(K) = 2
#III(E/K)[2] = 1
rang(E/K) = 1
listpointsmwr = [[Mod(1/2*y + 3/2, y^2 + 7), Mod(-y - 2, y^2 + 7), 1]]
v = [1, 1, [[Mod(1/2*y + 3/2, y^2 + 7), Mod(-y - 2, y^2 + 7)]]]
sage: v
(1, 1, [(1/2*a + 3/2 : -a - 2 : 1)])
```
A curve with 2-torsion:
```sage: K.<a> = NumberField(x^2 + 7, 'a')
sage: E = EllipticCurve(K, '15a')
sage: v = E.simon_two_descent(); v # long time (about 10 seconds), points can vary
(1, 3, [...])
```
A failure in the PARI/GP script ell.gp (VERSION 25/03/2009) is reported:
```sage: K = CyclotomicField(43).subfields(3)[0][0]
sage: E = EllipticCurve(K, '37')
sage: E.simon_two_descent()
Traceback (most recent call last):
...
RuntimeError:
*** at top-level: ans=bnfellrank(K,[0,0,1,
*** ^--------------------
*** in function bnfellrank: ...eqtheta,rnfeq,bbnf];rang=
*** bnfell2descent_gen(b
*** ^--------------------
*** in function bnfell2descent_gen: ...riv,r=nfsqrt(nf,norm(zc))
*** [1];if(DEBUGLEVEL_el
*** ^--------------------
*** array index (1) out of allowed range [none].
An error occurred while running Simon's 2-descent program```
tamagawa_exponent(P, proof=None)¶
Returns the Tamagawa index of this elliptic curve at the prime $$P$$.
INPUT:
• P – either None or a prime ideal of the base field of self.
• proof – whether to only use provably correct methods (default controlled by global proof module). Note that the proof module is number_field, not elliptic_curves, since the functions that actually need the flag are in number fields.
OUTPUT:
(positive integer) The Tamagawa index of the curve at P.
EXAMPLES:
```sage: K.<a>=NumberField(x^2-5)
sage: E=EllipticCurve([20, 225, 750, 625*a + 6875, 31250*a + 46875])
sage: [E.tamagawa_exponent(P) for P in E.discriminant().support()]
[1, 1, 1, 1]
sage: K.<a> = QuadraticField(-11)
sage: E = EllipticCurve('11a1').change_ring(K)
sage: [E.tamagawa_exponent(P) for P in K(11).support()]
[10]
```
tamagawa_number(P, proof=None)¶
Returns the Tamagawa number of this elliptic curve at the prime $$P$$.
INPUT:
• P – either None or a prime ideal of the base field of self.
• proof – whether to only use provably correct methods (default controlled by global proof module). Note that the proof module is number_field, not elliptic_curves, since the functions that actually need the flag are in number fields.
OUTPUT:
(positive integer) The Tamagawa number of the curve at $$P$$.
EXAMPLES:
```sage: K.<a>=NumberField(x^2-5)
sage: E=EllipticCurve([20, 225, 750, 625*a + 6875, 31250*a + 46875])
sage: [E.tamagawa_number(P) for P in E.discriminant().support()]
[1, 1, 1, 1]
sage: K.<a> = QuadraticField(-11)
sage: E = EllipticCurve('11a1').change_ring(K)
sage: [E.tamagawa_number(P) for P in K(11).support()]
[10]
```
tamagawa_numbers()¶
Return a list of all Tamagawa numbers for all prime divisors of the conductor (in order).
EXAMPLES:
```sage: e = EllipticCurve('30a1')
sage: e.tamagawa_numbers()
[2, 3, 1]
sage: vector(e.tamagawa_numbers())
(2, 3, 1)
sage: K.<a>=NumberField(x^2+3)
sage: eK = e.base_extend(K)
sage: eK.tamagawa_numbers()
[4, 6, 1]
```
tamagawa_product_bsd()¶
Given an elliptic curve $$E$$ over a number field $$K$$, this function returns the integer $$C(E/K)$$ that appears in the Birch and Swinnerton-Dyer conjecture accounting for the local information at finite places. If the model is a global minimal model then $$C(E/K)$$ is simply the product of the Tamagawa numbers $$c_v$$ where $$v$$ runs over all prime ideals of $$K$$. Otherwise, if the model has to be changed at a place $$v$$ a correction factor appears. The definition is such that $$C(E/K)$$ times the periods at the infinite places is invariant under change of the Weierstrass model. See [Ta2] and [Do] for details.
Note
This definition is slightly different from the definition of tamagawa_product for curves defined over $$\QQ$$. Over the rational number it is always defined to be the product of the Tamagawa numbers, so the two definitions only agree when the model is global minimal.
OUTPUT:
A rational number
EXAMPLES:
```sage: K.<i> = NumberField(x^2+1)
sage: E = EllipticCurve([0,2+i])
sage: E.tamagawa_product_bsd()
1
sage: E = EllipticCurve([(2*i+1)^2,i*(2*i+1)^7])
sage: E.tamagawa_product_bsd()
4
```
An example where the Neron model changes over K:
```sage: K.<t> = NumberField(x^5-10*x^3+5*x^2+10*x+1)
sage: E = EllipticCurve(K,'75a1')
sage: E.tamagawa_product_bsd()
5
sage: da = E.local_data()
sage: [dav.tamagawa_number() for dav in da]
[1, 1]
```
An example over $$\mathbb{Q}$$ (trac #9413):
```sage: E = EllipticCurve('30a')
sage: E.tamagawa_product_bsd()
6
```
REFERENCES:
• [Ta2] Tate, John, On the conjectures of Birch and Swinnerton-Dyer and a geometric analog. Seminaire Bourbaki, Vol. 9, Exp. No. 306.
• [Do] Dokchitser, Tim and Vladimir, On the Birch-Swinnerton-Dyer quotients modulo squares, Annals of Math., 2010.
torsion_order()¶
Returns the order of the torsion subgroup of this elliptic curve.
OUTPUT:
(integer) the order of the torsion subgroup of this elliptic curve.
EXAMPLES:
```sage: E = EllipticCurve('11a1')
sage: K.<t> = NumberField(x^4 + x^3 + 11*x^2 + 41*x + 101)
sage: EK = E.base_extend(K)
sage: EK.torsion_order()
25
```
```sage: E = EllipticCurve('15a1')
sage: K.<t> = NumberField(x^2 + 2*x + 10)
sage: EK = E.base_extend(K)
sage: EK.torsion_order()
16
```
```sage: E = EllipticCurve('19a1')
sage: K.<t> = NumberField(x^9-3*x^8-4*x^7+16*x^6-3*x^5-21*x^4+5*x^3+7*x^2-7*x+1)
sage: EK = E.base_extend(K)
sage: EK.torsion_order()
9
```
```sage: K.<i> = QuadraticField(-1)
sage: EK = EllipticCurve([0,0,0,i,i+3])
sage: EK.torsion_order()
1
```
torsion_points()¶
Returns a list of the torsion points of this elliptic curve.
OUTPUT:
(list) A sorted list of the torsion points.
EXAMPLES:
```sage: E = EllipticCurve('11a1')
sage: E.torsion_points()
[(0 : 1 : 0), (5 : -6 : 1), (5 : 5 : 1), (16 : -61 : 1), (16 : 60 : 1)]
sage: K.<t> = NumberField(x^4 + x^3 + 11*x^2 + 41*x + 101)
sage: EK = E.base_extend(K)
sage: EK.torsion_points()
[(16 : 60 : 1),
(5 : 5 : 1),
(5 : -6 : 1),
(16 : -61 : 1),
(t : 1/11*t^3 + 6/11*t^2 + 19/11*t + 48/11 : 1),
(-3/55*t^3 - 7/55*t^2 - 2/55*t - 133/55 : 6/55*t^3 + 3/55*t^2 + 25/11*t + 156/55 : 1),
(-9/121*t^3 - 21/121*t^2 - 127/121*t - 377/121 : -7/121*t^3 + 24/121*t^2 + 197/121*t + 16/121 : 1),
(5/121*t^3 - 14/121*t^2 - 158/121*t - 453/121 : -49/121*t^3 - 129/121*t^2 - 315/121*t - 207/121 : 1),
(10/121*t^3 + 49/121*t^2 + 168/121*t + 73/121 : 32/121*t^3 + 60/121*t^2 - 261/121*t - 807/121 : 1),
(1/11*t^3 - 5/11*t^2 + 19/11*t - 40/11 : -6/11*t^3 - 3/11*t^2 - 26/11*t - 321/11 : 1),
(14/121*t^3 - 15/121*t^2 + 90/121*t + 232/121 : 16/121*t^3 - 69/121*t^2 + 293/121*t - 46/121 : 1),
(3/55*t^3 + 7/55*t^2 + 2/55*t + 78/55 : 7/55*t^3 - 24/55*t^2 + 9/11*t + 17/55 : 1),
(-5/121*t^3 + 36/121*t^2 - 84/121*t + 24/121 : 34/121*t^3 - 27/121*t^2 + 305/121*t + 708/121 : 1),
(-26/121*t^3 + 20/121*t^2 - 219/121*t - 995/121 : 15/121*t^3 + 156/121*t^2 - 232/121*t + 2766/121 : 1),
(1/11*t^3 - 5/11*t^2 + 19/11*t - 40/11 : 6/11*t^3 + 3/11*t^2 + 26/11*t + 310/11 : 1),
(-26/121*t^3 + 20/121*t^2 - 219/121*t - 995/121 : -15/121*t^3 - 156/121*t^2 + 232/121*t - 2887/121 : 1),
(-5/121*t^3 + 36/121*t^2 - 84/121*t + 24/121 : -34/121*t^3 + 27/121*t^2 - 305/121*t - 829/121 : 1),
(3/55*t^3 + 7/55*t^2 + 2/55*t + 78/55 : -7/55*t^3 + 24/55*t^2 - 9/11*t - 72/55 : 1),
(14/121*t^3 - 15/121*t^2 + 90/121*t + 232/121 : -16/121*t^3 + 69/121*t^2 - 293/121*t - 75/121 : 1),
(t : -1/11*t^3 - 6/11*t^2 - 19/11*t - 59/11 : 1),
(10/121*t^3 + 49/121*t^2 + 168/121*t + 73/121 : -32/121*t^3 - 60/121*t^2 + 261/121*t + 686/121 : 1),
(5/121*t^3 - 14/121*t^2 - 158/121*t - 453/121 : 49/121*t^3 + 129/121*t^2 + 315/121*t + 86/121 : 1),
(-9/121*t^3 - 21/121*t^2 - 127/121*t - 377/121 : 7/121*t^3 - 24/121*t^2 - 197/121*t - 137/121 : 1),
(-3/55*t^3 - 7/55*t^2 - 2/55*t - 133/55 : -6/55*t^3 - 3/55*t^2 - 25/11*t - 211/55 : 1),
(0 : 1 : 0)]
```
```sage: E = EllipticCurve('15a1')
sage: K.<t> = NumberField(x^2 + 2*x + 10)
sage: EK = E.base_extend(K)
sage: EK.torsion_points()
[(8 : 18 : 1),
(3 : -2 : 1),
(8 : -27 : 1),
(t : t - 5 : 1),
(1/2 : 5/4*t + 1/2 : 1),
(-t - 2 : 2*t + 8 : 1),
(-7 : -5*t - 2 : 1),
(-1 : 0 : 1),
(-2 : 3 : 1),
(-13/4 : 9/8 : 1),
(-2 : -2 : 1),
(t : -2*t + 4 : 1),
(-7 : 5*t + 8 : 1),
(-t - 2 : -t - 7 : 1),
(1/2 : -5/4*t - 2 : 1),
(0 : 1 : 0)]
```
```sage: K.<i> = QuadraticField(-1)
sage: EK = EllipticCurve(K,[0,0,0,0,-1])
sage: EK.torsion_points ()
[(-2 : -3*i : 1), (0 : -i : 1), (1 : 0 : 1), (0 : i : 1), (-2 : 3*i : 1), (0 : 1 : 0)]
```
torsion_subgroup()¶
Returns the torsion subgroup of this elliptic curve.
OUTPUT:
(EllipticCurveTorsionSubgroup) The EllipticCurveTorsionSubgroup associated to this elliptic curve.
EXAMPLES:
```sage: E = EllipticCurve('11a1')
sage: K.<t>=NumberField(x^4 + x^3 + 11*x^2 + 41*x + 101)
sage: EK=E.base_extend(K)
sage: tor = EK.torsion_subgroup()
sage: tor
Torsion Subgroup isomorphic to Z/5 + Z/5 associated to the Elliptic Curve defined by y^2 + y = x^3 + (-1)*x^2 + (-10)*x + (-20) over Number Field in t with defining polynomial x^4 + x^3 + 11*x^2 + 41*x + 101
sage: tor.gens()
((16 : 60 : 1), (t : 1/11*t^3 + 6/11*t^2 + 19/11*t + 48/11 : 1))
```
```sage: E = EllipticCurve('15a1')
sage: K.<t>=NumberField(x^2 + 2*x + 10)
sage: EK=E.base_extend(K)
sage: EK.torsion_subgroup()
Torsion Subgroup isomorphic to Z/4 + Z/4 associated to the Elliptic Curve defined by y^2 + x*y + y = x^3 + x^2 + (-10)*x + (-10) over Number Field in t with defining polynomial x^2 + 2*x + 10
```
```sage: E = EllipticCurve('19a1')
sage: K.<t>=NumberField(x^9-3*x^8-4*x^7+16*x^6-3*x^5-21*x^4+5*x^3+7*x^2-7*x+1)
sage: EK=E.base_extend(K)
sage: EK.torsion_subgroup()
Torsion Subgroup isomorphic to Z/9 associated to the Elliptic Curve defined by y^2 + y = x^3 + x^2 + (-9)*x + (-15) over Number Field in t with defining polynomial x^9 - 3*x^8 - 4*x^7 + 16*x^6 - 3*x^5 - 21*x^4 + 5*x^3 + 7*x^2 - 7*x + 1
```
```sage: K.<i> = QuadraticField(-1)
sage: EK = EllipticCurve([0,0,0,i,i+3])
sage: EK.torsion_subgroup ()
Torsion Subgroup isomorphic to Trivial group associated to the Elliptic Curve defined by y^2 = x^3 + i*x + (i+3) over Number Field in i with defining polynomial x^2 + 1
```
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|
http://physics.stackexchange.com/questions/tagged/beam?sort=votes&pagesize=15
|
# Tagged Questions
The beam tag has no wiki summary.
0answers
151 views
### Mirrors and light beam divergence technology limits
There are many applications for orbital space mirrors in astronomy (better telescopes) and space propulsion (solar power for deep space probes), but this is limited by the minimum beam divergence ...
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### Beam splitters- Direction of use
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2answers
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### What is the mathematical formulation for buckling?
Argument: Buckling is an engineering concept that can only be applied to thin columns with compressive loading. (Is it possible to) Prove the above sentence right or wrong with mathematical ...
1answer
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### Euler's buckling formula applicable for impact calculations?
$$F = \frac{\pi^2 EI}{(KL)^2}$$ Is Euler's buckling formula applicable for impact calculations, considering speeds relevant for a car or aircraft crash? If there is a level where the formula ...
1answer
335 views
### Free boundary conditions
I am trying to simulate liquid film evaporation with free boundary conditions (in cartesian coordinates) and my boundary conditions are thus: $$\frac{\partial h}{\partial x} = 0, \qquad (1)$$ ...
1answer
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### Transparent boundary condition. Beam propagation method
I am interested in the finite-difference beam propagation method and its applications. I try to solve the Helmholtz equation. At first, i would like to solve numerically it for the easiest case, ...
0answers
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### Photon Statistics of a Coherent Gaussian Beam [closed]
Assume that a 100 pW He-Ne single-mode last emits light at 633 nm in a TEM00 Gaussian beam. (a)What is the mean number of photons crossing a circle of radius equal to the waist radius of the beam ...
0answers
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### Help With Difficult Deductive Proof [closed]
Suppose we have a Gaussian beam with a complex envelope expressed by the following equation 1: $$\tag{1} A_G(x,y,z) = \frac{A_1}{q(z)} e^{-ik \frac{x^2 + y^2}{2q(z)}}$$ where $$q(z) = z+iz_0$$ ...
1answer
110 views
### Width of Gaussian Beam and Refractive Index
I know that in free space, the width of a Gaussian beam can be written as $W=W_0\sqrt{1+(\frac{z}{z_0})^{2}}$. However, I was wondering if it was possible to express this width as a function of ...
0answers
41 views
### EigenMode expansion for beam propagation
I want to understand how to apply EigenMode expansion method (http://www.photond.com/files/docs/PW03_eme_paper.pdf) for beam propagation on a system of lenses. The interface between two mediums of ...
1answer
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### How to calculate beam spread of a non-point light source via an aspheric lens
I need to determine the angle, or rate of divergence of light from a single aspheric lens when I place a non-point light source (e.g. LED array) at a given distance from the lens which is less than ...
1answer
148 views
### muon neutrino momentum distribution
muon neutrino momentum distribution I have read the public data of T2K ,KEK to find this subject, I'm curiously that it's coincides with my prediction perfectly: The neutrino get its momemtum by its ...
|
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|
http://en.wikipedia.org/wiki/Surjective
|
# Surjective function
(Redirected from Surjective)
"Onto" redirects here. For other uses, see wikt:onto.
A surjective function from domain X to codomain Y. The function is surjective because every point in the codomain is the value of f(x) for at least one point x in the domain.
In mathematics, a function f from a set X to a set Y is surjective (or onto), or a surjection, if every element y in Y has a corresponding element x in X given by f(x) = y. Multiple elements of X might be turned into the same element of Y by applying f.
The term surjective and the related terms injective and bijective were introduced by Nicolas Bourbaki,[1] a group of mainly French 20th-century mathematicians who wrote a series of books presenting an exposition of modern advanced mathematics, beginning in 1935. The French prefix sur means over or above and relates to the fact that the image of the domain of a surjective function completely covers the function's codomain.
## Definition
A surjective function is a function whose image is equal to its codomain. Equivalently, a function f with domain X and codomain Y is surjective if for every y in Y there exists at least one x in X with $f(x)=y$. Surjections are sometimes denoted by a two-headed rightwards arrow ( ↠ rightwards two headed arrow),[2] as in f : X ↠ Y.
Symbolically,
If $f\colon X \rightarrow Y$, then $f$ is said to be surjective if
$\forall y \in Y, \, \exists x \in X, \;\; f(x)=y$.
## Examples
A non-surjective function from domain X to codomain Y. The smaller oval inside Y is the image (also called range) of f. This function is not surjective, because the image does not fill the whole codomain. In other words, Y is colored in a two-step process: First, for every x in X, the point f(x) is colored yellow; Second, all the rest of the points in Y, that are not yellow, are colored blue. The function f is surjective only if there are no blue points.
For any set X, the identity function idX on X is surjective.
The function f : Z → {0,1} defined by f(n) = n mod 2 (that is, even integers are mapped to 0 and odd integers to 1) is surjective.
The function f : R → R defined by f(x) = 2x + 1 is surjective (and even bijective), because for every real number y we have an x such that f(x) = y: an appropriate x is (y − 1)/2.
The function g : R → R defined by g(x) = x2 is not surjective, because there is no real number x such that x2 = −1. However, the function g : R → Rnn defined by g(x) = x2 (with restricted codomain) is surjective because for every y in the nonnegative real codomain Y there is at least one x in the real domain X such that x2 = y.
The natural logarithm function ln : (0,+∞) → R is a surjective and even bijective mapping from the set of positive real numbers to the set of all real numbers. Its inverse, the exponential function, is not surjective as its range is the set of positive real numbers and its domain is usually defined to be the set of all real numbers. The matrix exponential is not surjective when seen as a map from the space of all n×n matrices to itself. It is, however, usually defined as a map from the space of all n×n matrices to the general linear group of degree n, i.e. the group of all n×n invertible matrices. Under this definition the matrix exponential is surjective for complex matrices, although still not surjective for real matrices.
The projection from a cartesian product A × B to one of its factors is surjective.
In a 3D video game vectors are projected onto a 2D flat screen by means of a surjective function.
Interpretation for surjective functions in the Cartesian plane, defined by the mapping f : X → Y, where y = f(x), X = domain of function, Y = range of function. Every element in the range is mapped onto from an element in the domain, by the rule f. There may be a number of domain elements which map to the same range element. That is, every y in Y is mapped from an element x in X, more than one x can map to the same y. Left: Only one domain is shown which makes f surjective. Right: two possible domains X1 and X2 are shown.
Non-surjective functions in the Cartesian plane. Although some parts of the function are surjective, where elements y in Y do have a value x in X such that y = f(x), some parts are not. Left: There is y0 in Y, but there is no x0 in X such that y0 = f(x0). Right: There are y1, y2 and y3 in Y, but there are no x1, x2, and x3 in X such that y1 = f(x1), y2 = f(x2), and y3 = f(x3).
## Properties
A function is bijective if and only if it is both surjective and injective.
If (as is often done) a function is identified with its graph, then surjectivity is not a property of the function itself, but rather a relationship between the function and its codomain. Unlike injectivity, surjectivity cannot be read off of the graph of the function alone.
### Surjections as right invertible functions
The function g : Y → X is said to be a right inverse of the function f : X → Y if f(g(y)) = y for every y in Y (g can be undone by f). In other words, g is a right inverse of f if the composition f o g of g and f in that order is the identity function on the domain Y of g. The function g need not be a complete inverse of f because the composition in the other order, g o f, may not be the identity function on the domain X of f. In other words, f can undo or "reverse" g, but cannot necessarily be reversed by it.
Every function with a right inverse is necessarily a surjection. The proposition that every surjective function has a right inverse is equivalent to the axiom of choice.
If f : X → Y is surjective and B is a subset of Y, then f(f −1(B)) = B. Thus, B can be recovered from its preimage f −1(B).
For example, in the first illustration, there is some function g such that g(C) = 4. There is also some function f such that f(4) = C. It doesn't matter that g(C) can also equal 3; it only matters that f "reverses" g.
• Another surjective function. (This one happens to be a bijection)
• A non-surjective function. (This one happens to be an injection)
• Surjective composition: the first function need not be surjective.
### Surjections as epimorphisms
A function f : X → Y is surjective if and only if it is right-cancellative:[3] given any functions g,h : Y → Z, whenever g o f = h o f, then g = h. This property is formulated in terms of functions and their composition and can be generalized to the more general notion of the morphisms of a category and their composition. Right-cancellative morphisms are called epimorphisms. Specifically, surjective functions are precisely the epimorphisms in the category of sets. The prefix epi is derived from the Greek preposition ἐπί meaning over, above, on.
Any morphism with a right inverse is an epimorphism, but the converse is not true in general. A right inverse g of a morphism f is called a section of f. A morphism with a right inverse is called a split epimorphism.
### Surjections as binary relations
Any function with domain X and codomain Y can be seen as a left-total and right-unique binary relation between X and Y by identifying it with its function graph. A surjective function with domain X and codomain Y is then a binary relation between X and Y that is right-unique and both left-total and right-total.
### Cardinality of the domain of a surjection
The cardinality of the domain of a surjective function is greater than or equal to the cardinality of its codomain: If f : X → Y is a surjective function, then X has at least as many elements as Y, in the sense of cardinal numbers. (The proof appeals to the axiom of choice to show that a function g : Y → X satisfying f(g(y)) = y for all y in Y exists. g is easily seen to be injective, thus the formal definition of |Y| ≤ |X| is satisfied.)
Specifically, if both X and Y are finite with the same number of elements, then f : X → Y is surjective if and only if f is injective.
### Composition and decomposition
The composite of surjective functions is always surjective: If f and g are both surjective, and the codomain of g is equal to the domain of f, then f o g is surjective. Conversely, if f o g is surjective, then f is surjective (but g, the function applied first, need not be). These properties generalize from surjections in the category of sets to any epimorphisms in any category.
Any function can be decomposed into a surjection and an injection: For any function h : X → Z there exist a surjection f : X → Y and an injection g : Y → Z such that h = g o f. To see this, define Y to be the sets h −1(z) where z is in Z. These sets are disjoint and partition X. Then f carries each x to the element of Y which contains it, and g carries each element of Y to the point in Z to which h sends its points. Then f is surjective since it is a projection map, and g is injective by definition.
### Induced surjection and induced bijection
Any function induces a surjection by restricting its codomain to its range. Any surjective function induces a bijection defined on a quotient of its domain by collapsing all arguments mapping to a given fixed image. More precisely, every surjection f : A → B can be factored as a projection followed by a bijection as follows. Let A/~ be the equivalence classes of A under the following equivalence relation: x ~ y if and only if f(x) = f(y). Equivalently, A/~ is the set of all preimages under f. Let P(~) : A → A/~ be the projection map which sends each x in A to its equivalence class [x]~, and let fP : A/~ → B be the well-defined function given by fP([x]~) = f(x). Then f = fP o P(~).
## See also
Look up surjective, surjection, or onto in Wiktionary, the free dictionary.
## Notes
1. "Injection, Surjection and Bijection", Earliest Uses of Some of the Words of Mathematics, Tripod .
2. "Arrows – Unicode". Retrieved 2013-05-11.
3. Goldblatt, Robert (2006) [1984]. Topoi, the Categorial Analysis of Logic (Revised ed.). Dover Publications. ISBN 978-0-486-45026-1. Retrieved 2009-11-25.
## References
• Bourbaki, Nicolas (2004) [1968]. Theory of Sets. Springer. ISBN 978-3-540-22525-6.
|
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|
http://en.wikipedia.org/wiki/A_New_Kind_of_Science
|
# A New Kind of Science
A New Kind of Science
Author(s) Stephen Wolfram
Country US
Language English
Publisher Wolfram Media
Publication date 2002
Media type Print
Pages 1197
ISBN ISBN 1-57955-008-8
A New Kind of Science is a best-selling,[1] controversial book by Stephen Wolfram, published in 2002. It contains an empirical and systematic study of computational systems such as cellular automata. Wolfram calls these systems simple programs and argues that the scientific philosophy and methods appropriate for the study of simple programs are relevant to other fields of science.
## Contents
### Computation and its implications
The thesis of A New Kind of Science (NKS) is twofold: that the nature of computation must be explored experimentally, and that the results of these experiments have great relevance to understanding the natural world, which is assumed to be digital. Since its crystallization in the 1930s, computation has been primarily approached from two traditions: engineering, which seeks to build practical systems using computations; and mathematics, which seeks to prove theorems about computation. However, as recently as the 1970s, computing has been described as being at the crossroads of mathematical, engineering, and empirical traditions.[2][3]
Wolfram introduces a third tradition, which seeks to empirically investigate computation for its own sake, and asserts that an entirely new method is needed to do so. To Wolfram, traditional mathematics was failing to meaningfully describe the complexity seen in the systems he examined.
### Simple programs
The basic subject of Wolfram's "new kind of science" is the study of simple abstract rules—essentially, elementary computer programs. In almost any class of computational system, one very quickly finds instances of great complexity among its simplest cases. This seems to be true regardless of the components of the system and the details of its setup. Systems explored in the book include, amongst others, cellular automata in one, two, and three dimensions; mobile automata; Turing machines in 1 and 2 dimensions; several varieties of substitution and network systems; primitive recursive functions; nested recursive functions; combinators; tag systems; register machines; reversal-addition. For a program to qualify as simple, there are several requirements:
1. Its operation can be completely explained by a simple graphical illustration.
2. It can be completely explained in a few sentences of human language.
3. It can be implemented in a computer language using just a few lines of code.
4. The number of its possible variations is small enough so that all of them can be computed.
Generally, simple programs tend to have a very simple abstract framework. Simple cellular automata, Turing machines, and combinators are examples of such frameworks, while more complex cellular automata do not necessarily qualify as simple programs. It is also possible to invent new frameworks, particularly to capture the operation of natural systems. The remarkable feature of simple programs is that a significant percentage of them are capable of producing great complexity. Simply enumerating all possible variations of almost any class of programs quickly leads one to examples that do unexpected and interesting things. This leads to the question: if the program is so simple, where does the complexity come from? In a sense, there is not enough room in the program's definition to directly encode all the things the program can do. Therefore, simple programs can be seen as a minimal example of emergence. A logical deduction from this phenomenon is that if the details of the program's rules have little direct relationship to its behavior, then it is very difficult to directly engineer a simple program to perform a specific behavior. An alternative approach is to try to engineer a simple overall computational framework, and then do a brute-force search through all of the possible components for the best match.
Simple programs are capable of a remarkable range of behavior. Some have been proven to be universal computers. Others exhibit properties familiar from traditional science, such as thermodynamic behavior, continuum behavior, conserved quantities, percolation, sensitive dependence on initial conditions, and others. They have been used as models of traffic, material fracture, crystal growth, biological growth, and various sociological, geological, and ecological phenomena. Another feature of simple programs is that, according to the book, making them more complicated seems to have little effect on their overall complexity. A New Kind of Science argues that this is evidence that simple programs are enough to capture the essence of almost any complex system.
### Mapping and mining the computational universe
In order to study simple rules and their often complex behaviour, Wolfram believes it is necessary to systematically explore all of these computational systems and document what they do. He believes this study should become a new branch of science, like physics or chemistry. The basic goal of this field is to understand and characterize the computational universe using experimental methods.
The proposed new branch of scientific exploration admits many different forms of scientific production. For instance, qualitative classifications are often the results of initial forays into the computational jungle. On the other hand, explicit proofs that certain systems compute this or that function are also admissible. There are also some forms of production that are in some ways unique to this field of study. For example, the discovery of computational mechanisms that emerge in different systems but in bizarrely different forms.
Another kind of production involves the creation of programs for the analysis of computational systems. In the NKS framework, these themselves should be simple programs, and subject to the same goals and methodology. An extension of this idea is that the human mind is itself a computational system, and hence providing it with raw data in as effective a way as possible is crucial to research. Wolfram believes that programs and their analysis should be visualized as directly as possible, and exhaustively examined by the thousands or more. Since this new field concerns abstract rules, it can in principle address issues relevant to other fields of science. However, in general Wolfram's idea is that novel ideas and mechanisms can be discovered in the computational universe—where they can be witnessed in their clearest forms—and then other fields can choose among these discoveries for those they find relevant.
### Systematic abstract science
While Wolfram promotes simple programs as a scientific discipline, he also insists that its methodology will revolutionize essentially every field of science. The basis for his claim is that the study of simple programs is the minimal possible form of science, which is equally grounded in both abstraction and empirical experimentation. Every aspect of the methodology advocated in NKS is optimized to make experimentation as direct, easy, and meaningful as possible, while maximizing the chances that the experiment will do something unexpected. Just as this methodology allows computational mechanisms to be studied in their cleanest forms, Wolfram believes the process of doing so captures the essence of the process of doing science—and allows that process's strengths and shortcomings to be directly revealed.
Wolfram believes that the computational realities of the universe make science hard for fundamental reasons. But he also argues that by understanding the importance of these realities, we can learn to use them in our favor. For instance, instead of reverse engineering our theories from observation, we can enumerate systems and then try to match them to the behaviors we observe. A major theme of NKS is investigating the structure of the possibility space. Wolfram feels that science is far too ad hoc, in part because the models used are too complicated and/or unnecessarily organized around the limited primitives of traditional mathematics. Wolfram advocates using models whose variations are enumerable and whose consequences are straightforward to compute and analyze.
### Philosophical underpinnings
Wolfram believes that one of his achievements is not just exclaiming, "computation is important!", but in providing a coherent system of ideas that justifies computation as an organizing principle of science. For instance, he argues that the concept of computational irreducibility (that some complex computations are not amenable to short-cuts and cannot be "reduced"), is ultimately the reason why computational models of nature must be considered in addition to traditional mathematical models. Likewise, his idea of intrinsic randomness generation—that natural systems can generate their own randomness, rather than using chaos theory or stochastic perturbations—implies that computational models do not need to include explicit randomness.
Based on his experimental results, Wolfram has developed the Principle of Computational Equivalence, which asserts that almost all processes that are not obviously simple are of equivalent sophistication. From this vague principle Wolfram draws a broad array of concrete deductions that he takes to reinforce many aspects of his theory. Possibly the most important among these is an explanation as to why we experience randomness and complexity: often, the systems we analyze are just as sophisticated as we are. Thus, complexity is not a special quality of systems, like for instance the concept of "heat", but simply a label for all systems whose computations are sophisticated. Wolfram claims that understanding this makes the "normal science" of the NKS paradigm possible.
At the deepest level, Wolfram believes that like many of the most important scientific ideas, the Principle of Computational Equivalence allows science to be more general by pointing out new ways in which humans are not "special"; that is, it has been thought that the complexity of human intelligence makes us special, but the Principle asserts otherwise. In a sense, many of Wolfram's ideas are based on understanding the scientific process—including the human mind—as operating within the same universe it studies, rather than somehow being outside it.
#### Principle of computational equivalence
The principle states that systems found in the natural world can perform computations up to a maximal ("universal") level of computational power. Most systems can attain this level. Systems, in principle, compute the same things as a computer. Computation is therefore simply a question of translating input and outputs from one system to another. Consequently, most systems are computationally equivalent. Proposed examples of such systems are the workings of the human brain and the evolution of weather systems.
### Applications and results
There are a vast number of specific results and ideas in the NKS book, and they can be organized into several themes. One common theme of examples and applications is demonstrating how little complexity it takes to achieve interesting behavior, and how the proper methodology can discover this behavior.
First, there are several cases where the NKS book introduces what was, during the book's composition, the simplest known system in some class that has a particular characteristic. Some examples include the first primitive recursive function that results in complexity, the smallest universal Turing Machine, and the shortest axiom for propositional calculus. In a similar vein, Wolfram also demonstrates a large number of simple programs that exhibit phenomena like phase transitions, conserved quantities and continuum behavior and thermodynamics that are familiar from traditional science. Simple computational models of natural systems like shell growth, fluid turbulence, and phyllotaxis are a final category of applications that fall in this theme.
Another common theme is taking facts about the computational universe as a whole and using them to reason about fields in a holistic way. For instance, Wolfram discusses how facts about the computational universe inform evolutionary theory, SETI, free will, computational complexity theory, and philosophical fields like ontology, epistemology, and even postmodernism.
Wolfram suggests that the theory of computational irreducibility may provide a resolution to the existence of free will in a nominally deterministic universe. He posits that the computational process in the brain of the being with free will is actually complex enough so that it cannot be captured in a simpler computation, due to the principle of computational irreducibility. Thus while the process is indeed deterministic, there is no better way to determine the being's will than to essentially run the experiment and let the being exercise it.
The book also contains a vast number of individual results—both experimental and analytic—about what a particular automaton computes, or what its characteristics are, using some methods of analysis.
One specific new technical result in the book is a description of the Turing completeness of the Rule 110 cellular automaton. Rule 110 can be simulated by very small Turing machines, and such a 2-state 5-symbol universal Turing machine is given. Wolfram also conjectures that a particular 2-state 3-symbol Turing machine is universal. In 2007, as part of commemorating the fifth anniversary of the book, a \$25,000 prize was offered for a proof of the (2, 3) machine's universality.[4]
## NKS Summer School
Every year, Wolfram and his group of instructors[5] organizes a summer school.[6] The first four summer schools from 2003 to 2006 were held at Brown University. Later the summer school was hosted by the University of Vermont at Burlington with the exception of the year 2009 that was held at the Istituto di Scienza e Tecnologie dell’Informazione of the CNR in Pisa, Italy. After seven consecutive summer schools more than 200 people have participated, some of whom continued developing their 3-week research projects as their Master's or Ph.D theses.[7] Some of the research done in the summer school has resulted in publications.[8][9]
## Reception
A New Kind of Science received extensive media publicity for a scientific book, with articles being written in publications such as The New York Times,[10] Newsweek,[11] Wired,[12] and The Economist.[13] It was a best-seller, and reviewed in a range of scientific journals; several themes emerged in these reviews. Some scientists were critical of the book for the lack of modesty or originality [14][15] where others found valuable insights and refreshing ideas.[16][17] More recently, Wolfram answered criticisms in a series of blog posts [18] including an analysis of the number and type of citations the book has received after 10 years of its publication.[19]
### Scientific philosophy
A key tenet of NKS is that the simpler the system, the more likely a version of it will recur in a wide variety of more complicated contexts. Therefore, NKS argues that systematically exploring the space of simple programs will lead to a base of reusable knowledge. However, many scientists believe that of all possible parameters, only some actually occur in the universe. For instance, of all possible permutations of the symbols making up an equation, most will be essentially meaningless. NKS has also been criticized for asserting that the behavior of simple systems is somehow representative of all systems.
### Methodology
A common criticism[by whom?] of NKS is that it does not follow established scientific methodology[who?]. NKS does not establish rigorous mathematical definitions,[20] nor does it attempt to prove theorems; and most formulas and equations are written in Mathematica rather than standard notation.[21] Along these lines, NKS has also been criticized for being heavily visual, with much information conveyed by pictures that do not have formal meaning.[17] It has also been criticized for not using modern research in the field of complexity, particularly the works that have studied complexity from a rigorous mathematical perspective.[15] And it has been criticized for misrepresenting chaos theory: "Throughout the book, he equates chaos theory with the phenomenon of sensitive dependence on initial conditions (SDIC)."[22]
### Utility
NKS has been criticized for not providing specific results that would be immediately applicable to ongoing scientific research.[17] There has also been criticism, implicit and explicit, that the study of simple programs has little connection to the physical universe, and hence is of limited value. Steven Weinberg has pointed out that no real world system has been explained using Wolfram's methods in a satisfactory fashion.[23]
### Principle of computational equivalence
The PCE has been criticized for being vague, unmathematical, and for not making directly verifiable predictions.[21] It has also been criticized for being contrary to the spirit of research in mathematical logic and computational complexity theory, which seek to make fine-grained distinctions between levels of computational sophistication, and for wrongly conflating different kinds of universality property[21]. Moreover, critics such as Ray Kurzweil have argued that it ignores the distinction between hardware and software; while two computers may be equivalent in power, it does not follow that any two programs they might run are also equivalent.[14] Others suggest it is little more than a rechristening of the Church-Turing thesis.[22]
### The fundamental theory (NKS Chapter 9)
Wolfram's speculations of a direction towards a fundamental theory of physics have been criticized as vague and obsolete. Scott Aaronson, Assistant Professor of Electrical Engineering and Computer Science at MIT, also claims that Wolfram's methods cannot be compatible with both special relativity and Bell's theorem violations, and hence cannot explain the observed results of Bell test experiments.[24]
Edward Fredkin and Konrad Zuse pioneered the idea of a computable universe, the former by writing a line in his book on how the world might be like a cellular automaton, and later further developed by Fredkin using a toy model called Salt.[25] It has been claimed that NKS tries to take these ideas as its own. Jürgen Schmidhuber has also charged that his work on Turing machine-computable physics was stolen without attribution, namely his idea on enumerating possible Turing-computable universes. [26]
In a 2002 review of NKS, the Nobel laureate and elementary particle physicist Steven Weinberg wrote, "Wolfram himself is a lapsed elementary particle physicist, and I suppose he can't resist trying to apply his experience with digital computer programs to the laws of nature. This has led him to the view (also considered in a 1981 paper by Richard Feynman) that nature is discrete rather than continuous. He suggests that space consists of a set of isolated points, like cells in a cellular automaton, and that even time flows in discrete steps. Following an idea of Edward Fredkin, he concludes that the universe itself would then be an automaton, like a giant computer. It's possible, but I can't see any motivation for these speculations, except that this is the sort of system that Wolfram and others have become used to in their work on computers. So might a carpenter, looking at the moon, suppose that it is made of wood."[27]
According to Gerard 't Hooft, "Both the bosonic string theory and superstring theory can be reformulated in terms of a special basis of states, defined on a space-time lattice with lattice length $2\pi\sqrt{\alpha'.}$ The evolution equations on this lattice are classical. This allows for a cellular automaton interpretation of superstring theory."[28]
### Natural selection
Wolfram's claim that natural selection is not the fundamental cause of complexity in biology has led some to state that Wolfram does not understand the theory of evolution.[29]
### Originality and self-image
NKS has been heavily criticized as not being original or important enough to justify its title and claims.
The authoritative manner in which NKS presents a vast number of examples and arguments has been criticized as leading the reader to believe that each of these ideas was original to Wolfram;[22] in particular, one of the most substantial new technical results presented in the book, that the rule 110 cellular automaton is Turing complete, was not proven by Wolfram, but by his research assistant, Matthew Cook. However, the notes section at the end of his book acknowledges many of the discoveries made by these other scientists citing their names together with historical facts, although not in the form of a traditional bibliography section. This is generally considered insufficient in scientific literature.
Additionally, it has been pointed out that the idea that very simple rules often generate great complexity is already an established idea in science, particularly in chaos theory and complex systems.[15] Some have argued[who?] that the use of computer simulation is ubiquitous, and instead of starting a paradigm shift NKS just adds justification to a paradigm shift that has been undertaken. Wolfram's NKS might then seem as one of the books explicitly describing this shift.
## See also
• Scientific reductionism
• Calculating Space
• Fredkin Finite Nature Hypothesis
• Marcus Hutter's "Universal Artificial Intelligence" algorithm
## References
1.
2. Wegner, Peter (1976). "Research Paradigms in Computer Science". Proceedings of the 2nd International Conference on Software Engineering. San Francisco, CA, USA: IEEE Press. pp. 322–330.
3. Denning, Peter J.; et al. (1989). "Computing as a Discipline". Communications of the ACM 32 (1): 9–23. doi:10.1145/63238.63239.
4. "The Wolfram 2,3 Turing Machine Research Prize". Archived from the original on 15 May 2011. Retrieved 2011-03-31.
5. Rowland (2008). "A natural prime-generating recurrence". Journal of Integer Sequences 11 (08). arXiv:0710.3217.
6. Johnson, George (9 June 2002). "'A New Kind of Science': You Know That Space-Time Thing? Never Mind". The New York Times. Retrieved 28 May 2009.
7. Levy, Stephen (27 May 2002). "Great Minds, Great Ideas". Newsweek. Archived from the original on 16 April 2009. Retrieved 28 May 2009.
8. Levy, Stephen (June 2002). "The Man Who Cracked The Code to Everything ...". Wired magazine. Archived from the original on 27 May 2009. Retrieved 28 May 2009.
9. "The science of everything". The Economist. 30 May 2002. Retrieved 28 May 2009.
10. ^ a b Kurzweil, Ray. "Review:Reflections on Stephen Wolfram’s A New Kind of Science". Retrieved 13 May 2002.
11. ^ a b c
12. Rucker, Rudy (November 2003). "Review: A New Kind of Science". : 851–861. Retrieved 28 May 2009.
13. ^ a b c Berry et al., Michael (May 2002). "Review: A Revolution or self indulgent hype? How top scientists view Wolfram". . Retrieved 14 Aug 2012.
14.
15.
16. Bailey, David (September 2002). "A Reclusive Kind of Science". Computing in Science and Engineering: 79–81. Retrieved 28 May 2009.
17. ^ a b c Gray, Lawrence (2003). "A Mathematician Looks at Wolfram's New Kind of Science". Notices of the AMS 50 (2): 200–211.
18. ^ a b c
19. Weiss, Peter (2003). "In search of a scientific revolution: controversial genius Stephen Wolfram presses onward". Science News.
20. Aaronson, Scott (2002). "Book Review of A New Kind of Science". Quantum Information and Computation 2 (5): 410–423.
21.
22. Weinberg, S. (24 October 2002). "Is the Universe a Computer?". The New York Review of Books.
23. 't Hooft, G. (15 Sep 2012). "Discreteness and Determinism in Superstrings". arxiv.org.
24. Lavers, Chris (3 August 2002). "How the cheetah got his spots". London: The Guardian. Retrieved 28 May 2009.
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http://math.stackexchange.com/questions/11373/relationship-between-subrings-and-ideals
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# Relationship between subrings and ideals
I just want to make sure that I understand these two objects correctly.
Let $R$ be a ring. A subset $R'\subseteq R$ is a subring provided $R'$ is an additive subgroup that is closed with respect to multiplication.
A subset $I\subseteq R$ is an ideal provided $I$ is an additive subgroup such that $rI\subseteq I$ and $Ir\subseteq I$ for all $r\in R$ (where $rI$ and $Ir$ are defined in the "obvious" way).
Now, in my notes, for the definition of a subring, I didn't specify what I meant by "with respect to multiplication." Looking back now, I assume I meant left multiplication.
So then, if this is correct, an ideal is a subring that is closed with respect to both left and right multiplication.
Is this correct?
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Do your rings have identities? The usual definition of a subring requires that it also have a multiplicative identity, although this isn't universal. – Qiaochu Yuan Nov 22 '10 at 19:17
We have not specified that a ring or subring need a multiplicative identity – Bey Nov 22 '10 at 19:44
@Qiaochu: It really depends on whether you are talking about "subrings-with-1" or "subrings" (i.e., subrings in the category of rings with identity, and subrings in the category of rings). Think of the identity as a unary operation in the former. – Arturo Magidin Nov 22 '10 at 19:53
## 4 Answers
A left-ideal is an abelian subgroup (under addition) of the ring which is closed under left-multiplication by elements of the ring, and not just elements in the ideal. In the commutative case we identify left and right multiplication and then an ideal is a subring of the ring. However, that is not enough. Since the subring requires closure under multiplication of elements contained in itself while a left or right ideal requires closure under corresponding multiplication by elements of the ring.
For example, consider the polynomial ring $R=k[x]$ in one variable over a field $k$. Then $R$ is a commutative ring with identity. $k$ is a subring of $R$ being a field. However, any ideal that contains any element of $k$ is actually the whole ring (since if it contains an element of $k$, it contains $1$ by multiplying it by it's inverse and hence contain every element of $R$). So being an ideal is a stronger requirement.
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Since a ring R is also an Abelian group, is it necessary to specify that subrings or ideals are Abelian subgroups? Certainly we have to stipulate they're subgroups under +, but isn't automatic that they're Abelian? – Bey Nov 22 '10 at 19:43
@Bey: You're right, that's superfluous. You just need that they are subgroups under addition. – Timothy Wagner Nov 22 '10 at 19:47
Remember from groups and vector spaces the notions of "sub-group" and of "sub-space": you must have a subset which is also "closed" with respect to the operations. That is, if you take elements from the subset, and you subject them to the operations you have, you should end up with results that are also in your subset.
With groups, that means that if a subset $S$ of a group $G$ is going to be a group, then it must be the case that whenever $s,t\in S$, then $st\in S$ ("closed under multiplication"; if you take two things in $S$ and multiply them, you don't need to go "outside of $S$" to find the answer, it is all already inside $S$; no need to go outside); and if $s\in S$, then you need $s^{-1}\in S$ ("closed under inverses"); and you need $e\in S$ (closed under the operation that gives the identity element).
With vector spaces, if a subset $W$ of a vector space $V$ is going to be a subspace, you need that $\mathbf{0}\in W$ (closed under the operation that gives the identity element); that if $w_1,w_1\in W$, then $w_1+w_2\in W$ (closed under vector addition); and that if $w\in W$ and $\alpha\in F$, then $\alpha w\in W$ (closed under scalar multiplication).
The same is true for rings: if you have a ring $R$, then in order for a subset $S$ to be a subring you need it to be (i) closed under addition; (ii) closed under additive inverses; (iii) contain the identity element of the sum; and (iv) closed under multiplication (if $a,b\in S$, then $ab\in S$). Saying "is a subgroup" is the same as saying (i), (ii), and (iii).
Ideals, however, are a bit more than simply subrings. They play exactly the analogous role to rings as normal subgroups play to groups. You may recall that a normal subgroup $N$ of a group $G$ is a "subgroup-with-something-extra": not only is it "closed" under the three usual operations of the group (identity element, addition, and inverses), it must also be "closed" under a bunch of other operations (conjugation).
Similarly with ideals: not only must they be "closed" under the four usual operations of the ring (additive identity element, addition, additive inverses, and multiplication), it must also be "closed" under a bunch of other operations (that are like the scalar multiplications in the vector space case): left multiplication by any element of $R$, and right multiplication by any element of $R$. Just like "normal subgroup" is "subgroup with something extra", likewise ideals are "subrings with something extra."
The problem with saying "an ideal is a subring that is closed under left and right multiplication" is that saying "closed under left and right multiplication" only says that if $a,b\in I$, then $ab\in I$; you really want to add the coda of "...left and right multiplication by elements of the ring."
Examples abound. Take $R=\mathbb{R}[x]$, the polynomials with coefficients in the real numbers, and consider $\mathbb{Z}[x]$, the polynomials with integer coefficients. The latter is a subggroup of $R$, and if $p(x),q(x)\in\mathbb{Z}[x]$, then the product $p(x)q(x)$ is also in $\mathbb{Z}[x]$. So $\mathbb{Z}[x]$ is a subring of $R$. It is not, however, an ideal of $R$, because $x\in\mathbb{Z}[x]$, and $\pi\in R$, but their product $\pi x\notin\mathbb{Z}[x]$.
Or take $S=\mathbb{R}[x^2]$, the polynomials with real coefficients in which only even powers of $x$ occur. This is a subring of $R$ but not an ideal, because $x^2 \in S$, $x\in R$, but $x(x^2)\notin S$.
On the other hand, $S=\{p(x)\in R\mid p(0)=0\}$, the collection of all polynomials with constant term equal to $0$, is both a subring and an ideal: if $p(x)\in S$ and $q(x)\in R$, then $q(x)p(x)\in S$ since $q(0)p(0)=0$ (and $p(x)q(x)=q(x)p(x)$). That is, $q(x)S\subseteq S$ and $Sq(x)\subseteq S$.
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In the case of the multiplication operation in a (sub)ring the phrase 'with respect to multiplication' doesn't refer to any 'left multiplication': In a ring there is just one multiplication which may be or not be commutative. Being closed with respect to multiplication means that if $a$ and $b$ belong to $R'$, then also $ab$ (and $ba$ by exchanging the roles of $a$ and $b$) belong to $R'$.
In the non-commutative situation we often distinguish the products $ab$ and $ba$ of two elements $a$ and $b$ by saying that we 'multiply $b$ on left/right by $a$'.
Ideal is more than a subring in the sense that we require that products stay in the ideal even when multiplying by elements that don't belong to the ideal.
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"...we require that products stay in the ideal even when multiplying by elements that don't belong to the ideal." This is what I was forgetting that caused some confusion for me. Thanks =) – Bey Nov 22 '10 at 19:48
In noncommutative rings, the terms "left ideal" and "right ideal" are used indicate if you are multiplying from the left or right. The term "ideal" usualy means that it is a left and right ideal at the same time. In commutative ring theory, which is probably what you are studying, this notion of left and right is meaningless since ab and ba are always the same.
Simply put, an ideal is a very special type of subring, with the added property that if a is in the ideal and r is in the ring, r times a is in the ideal (even if r is not in the ideal). Subrings don't have this extra property in general.
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http://math.stackexchange.com/questions/128797/having-trouble-understanding-the-concept-of-mixing-in-dynamical-systems/128870
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# Having trouble understanding the concept of “mixing” in dynamical systems.
I'm trying to understand the concept of mixing in dynamical systems theory, especially when the system in question has a measure-preserving flow. Here's how the condition is expressed mathematically: If $\mu$ is the measure and $\phi$ is the flow, then for all subsets $A$ and $B$ of positive measure, $\lim_{t \rightarrow \infty}\mu(\phi^{t}(B) \cap A) = \mu(B) \times \mu(A)$.
Now suppose $B$ is an arbitrary set with measure greater than 0 and less than 1. If the flow is measure preserving, then for all $t$, $\mu(\phi^{t}(B)) = \mu(B)$. Pick $A = \lim_{t \rightarrow \infty}\phi^{t}(B)$. Then, $\mu(A) = \mu(B)$. It follows that $\lim_{t \rightarrow \infty}\mu(\phi^{t}(B) \cap A) = \mu(A) = \mu(B)$.
So if the dynamics is mixing, then we will have $\mu(B) = \mu(B) \times \mu(B)$. But this is only possible if $\mu(B)$ is 0 or 1, contradicting our initial assumption.
Isn't this a problem with the definition of mixing? Is the definition in my source wrong? Or am I doing something wrong?
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It just occurred to me that maybe the problem is that for a mixing system my choice of $A$ does not correspond to a measurable set. Is this the case? If so, how do we square that with the fact that the dynamics is measure-preserving? – Tarun Apr 6 '12 at 20:09
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I'm afraid your entire second paragraph doesn't make much sense. What does $\lim_{t \to \infty} \phi^t(B)$ mean? Why should $\mu(A) = \mu(B)$ imply that $\mu(A \cap B) = \mu(B \cap B)$? – ccc Apr 6 '12 at 20:37
$\lim_{t \rightarrow \infty} \phi^{t}(B)$ is the set of points to which $B$ evolves under the flow in the infinite time limit. – Tarun Apr 6 '12 at 22:13
As for the second part, I agree that I've put it in an unnecessarily confusing manner. The point is that in the infinite limit, $\phi^{t}(B)$ and $A$ are the same set of points, so their intersection is just going to be that set. So I should have said $\lim_{t \rightarrow \infty} \mu(\phi^{t}(B) \cap A) = \mu(A) = \mu(B)$. The rest of the argument follows. Does this work? – Tarun Apr 6 '12 at 22:19
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Your definition of $\lim_{t \to \infty} \phi^t(B)$ still doesn't make sense, and is most likely the source of your misunderstanding. Given a point $x$, what does it mean for $x$ to be in $\lim_{t \to \infty} \phi^t(B)$? How about a concrete example: $B$ is the set of points in the unit circle subtended by central angle $\pi/2$ radians, and $\phi$ is the rotation of the circle by $1$ radian. What is $\lim_{t \to \infty} \phi^t(B)$? – ccc Apr 6 '12 at 22:42
show 2 more comments
## 2 Answers
There seems to be suspicious reasoning as pointed out in the comments when you define $A$ and then conclude that since $\mu(A)=\mu(B)$ then $\lim_{t\rightarrow\infty}\mu(\phi^t(B)\cap A)=\mu(A)$.
Have you encountered ergodic theory? Your mixing condition is more formally referred to as "strong mixing" which implies ergodicity, in that $\phi_t$ must be ergodic. By definition a flow is ergodic when $A=\phi_{-t}(A)$ implies $\mu(A)=0,1$.
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You may want to think about the notion of independent events in probability theory: two events are independent if $Pr(A \cap B) = Pr(A)\cdot Pr(B).$
So the definition you give says that in the large time limit, the events of a point being in $\phi^t(B)$ and in $A$ are independent. So however $A$ and $B$ are positioned with respect to one another, after a long time $t$, the position of $\phi^t(B)$ is completely independent of the position of $A$.
Intuitvely, the points in $B$ are being completely mixed throughout the set, independently of where they were originally positioned. Hence the term mixing.
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http://mathhelpforum.com/calculus/142222-strange-question-improper-integrals.html
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Thread:
1. Strange question with improper integrals
First off, the limits of integration are both infinites, and I've only worked with one so far, with the other being a number. What changes?
And finally, the exponent within the exponent of e throws me off. There's a way to integrate that entire thing using substitution, but I'm not sure how. Can anyone tell me or at least link me to some tutorial online that explains it?
2. Write it as $\lim_{b \to \infty} \int^{b}_{-b} x^{5}e^{-x^{6}} \ dx$ and make the substitution $u = -x^{6}$
3. Originally Posted by Archduke01
First off, the limits of integration are both infinites, and I've only worked with one so far, with the other being a number. What changes?
And finally, the exponent within the exponent of e throws me off. There's a way to integrate that entire thing using substitution, but I'm not sure how. Can anyone tell me or at least link me to some tutorial online that explains it?
As $(-x^6)'=-6x^5$ , this is a very easy integral:
$\int\limits^\infty_{-\infty}x^5\,e^{-x^6}\,dx=$ $\lim_{a\to\infty\,,\,b\to -\infty}-\frac{1}{6}\int\limits^a_b-6x^5\,e^{-x^6}\,dx$ $=\lim_{a\to\infty\,,\,b\to -\infty}-\frac{1}{6}\,\left[e^{-x^6}\right]^a_b = 0$ , which shouldn't be too surprising since the function is odd...
Tonio
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http://mathhelpforum.com/advanced-math-topics/30897-can-we.html
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# Thread:
1. ## Can we?
Can we simplify the following fraction as :-
$(\frac{x^2 + 3xy + 2}{x^2 + y^2 +2}) = 0$
$(x^2 + y^2 +2) * (\frac{x^2 + 3xy + 2}{x^2 + y^2 +2}) = 0 * (x^2 + y^2 +2)$
$<br /> x^2 + 3xy + 2 = 0$
and then further simplification....
A friend of mine was contradicting this approach and was saying that if the denominator contains variables then you simply can't do it like this. His solution however was that :-
the solution would be that :-
Either, $x^2 + 3xy + 2 = 0$ or $x^2 + y^2 +2 = infinity$
What is the real solution ?
2. Hello,
This can be interesting... But i think the second method is strange. We look for the x and y that verify the equation. No x and y will make the denominator infinite (i mean no determined x nor y)...
3. I also told him that his method looks weird and that intuitively one would just multiply both sides and get the result.
4. The problem is mainly in the fact that 1/infinite is more a convention than an exact result...
Am not sure, but i see the thing like this
5. Originally Posted by Altair
Can we simplify the following fraction as :-
${\color{red}}(\frac{x^2 + 3xy + 2}{x^2 + y^2 +2}) = 0$
$x^2 + 3xy + 2 = 0$
The above is certainly correct.
A fraction is zero if and only if its numerator is zero.
One of infinitely many solution is (1,-1).
6. Another thing I told him was that according to his concept of either ${x^2 + 3xy + 2} =0$ or $\frac{1}{x^2 + y^2 +2} = 0$ is not feasible here because it can only be done in case of factors.
7. Originally Posted by Altair
Another thing I told him was that according to his concept of either ${x^2 + 3xy + 2} =0$ or $\frac{1}{x^2 + y^2 +2} = 0$ is not feasible here because it can only be done in case of factors.
Well of course.
In $\frac{1}{x^2 + y^2 +2} = 0$ the numerator is not zero, just as I said.
8. What if the denominator tends to infinity ? In this case if both x and y are infinity ?
9. Originally Posted by Altair
What if the denominator tends to infinity ? In this case if both x and y are infinity ?
That is a nonsense statement. Infinity is not a number!
Infinity is concept. Some wags even say it is a place where mathematicians hid their ignorance.
10. It's not nonsense, it's a perfectly reasonable question about limits. If you let either x or y approach either positive or negative infinity, the denominator will approach positive infinity. How that effects the overall fraction depends more specifically on how you choose x and y.
On the original question, your solution does leave a step unstated, I'm not sure whether you skipped it or not. In cancelling out the denominator, you make the assumption that it isn't zero. In the real numbers, it can't be, since it has to be at least 2, but on similar problems that could cause mistakes. If there was any chance it could be zero, it would be better to deal with that case explicitly.
11. Going back over the limit question more carefully, I see that there's no well-defined value of the fraction when both x and y tend to infinity. Following two different paths, you can get two different answers. Setting x=y and taking the limit as x->oo gives 2, but setting x=3y and taking the same limit gives 9/5.
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http://mathoverflow.net/questions/57166/computational-software-in-algebraic-topology
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## Computational software in Algebraic Topology?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I was wondering if there is any good software out there that allows you to do specific computations in algebraic topology. For example:
• Create a simplicial complex/set and ask questions about its homology, cohomology;
• Build manifolds using handle decompositions;
• Calculate homotopy limits, colimits.
Something quite flexible and robust in the vein of MAGMA
Thank you.
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Except for your simplicial question, the rest appears to be too vague. What do you mean by "build manifolds using handle decompositions"? What kind of data structure do you want the computer to store, and once it's "built" what do you want the computer to do with it? Similarly, what do you mean by "calculate homotopy limits and colimits"? of what kind of objects and what constitutes a calculation? For simplicial complexes there is a variety of software packages, see for example the bottom of: math.uiuc.edu/~nmd/computop – Ryan Budney Mar 2 2011 at 20:22
Related question: mathoverflow.net/questions/53595/… Also: chomp.rutgers.edu/advanced/programs.php and: orms.mfo.de/class_tree I'm sure I'm missing many other resources but Google brings up many. There's also a few other MO threads related to this topic... – Ryan Budney Mar 2 2011 at 20:28
Thank you for your reply. Yes, I had found some of those links already. I was actually looking for something more comprehensive. Something that allows you to various calculations within the same program... in the vein of MAGMA. I guess the program Kenzo comes closest. – Joris Weimar Mar 2 2011 at 20:58
There are people who have used MAGMA (and are starting to use SAGE) to do algebra computations that have a lot of topological relevance. (I have Adams spectral sequences in mind.) – Sean Tilson Mar 3 2011 at 22:41
## 3 Answers
There are several programs that answer to your first demand whilst the others, as Ryan says, are a bit more vague. There are books written on computational homology (and its applications) for instance, see http://chomp.rutgers.edu/ and the computational homology project. For simplicial complexes, the Plex routines written for Matlab are at http://comptop.stanford.edu/u/programs/plex/ and that leads to a lot of other interesting programs for which see http://comptop.stanford.edu/ and follow links. The main problems are always speed of computing with large simplicial complexes. (Work by Edelsbrunner and collaborators is good for some of this.)
For homotopy colimits, it seems likely that the only programs that might go some way are related to Kenzo project: see http://www-fourier.ujf-grenoble.fr/~sergerar/Kenzo/
but that is more difficult to use.
There are programs for detecting (small) handles used in computer graphics, but I cannot say anything about them.
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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.
Sage allows you to play with simplicial complexes and their (co)homology.
http://www.sagemath.org/doc/reference/sage/homology/simplicial_complex.html
http://www.sagemath.org/doc/reference/sage/homology/examples.html
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2
Sage also handles $\Delta$-complexes (which are essentially simplicial sets with no degeneracy maps) and cubical complexes. – John Palmieri Mar 3 2011 at 6:30
(1) The Computational Homology Project offers free software CHomP that will compute homology of simplicial complexes, at least with finite field coefficients.
(2) Jplex and Dionysus, from the computational topology group at Stanford, are good for quickly computing persistent homology of Rips and Cech complexes, etc. This might be especially useful, for example, if you had points sampled from a manifold.
(3) Afra Zomorodian has apparently recently written some code for computing homology of clique (i.e. flag) complexes very quickly and with small memory requirement by going through calculations involving simplicial sets, but I don't know if the code can compute homology of arbitrary simplicial sets, and (more importantly) I don't know if the code is publicly available.
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