User wox - MathOverflow

Azimuthal and polar integration of a 3D Gaussian

2012-09-27T17:20:38Z

Numerical evaluation of the following integral of a 3D gaussian $G$ seems to result in a 1D Gaussian $g$: $$\int_{0}^{2\pi}\int_{0}^{\pi}G(R,\phi,\theta)\sin\theta\ \text{d}\theta \ \text{d}\phi= g(R)$$ where the 3D Gaussian in spherical coordinates with mean $\mu$ and covariance matrix $\Sigma$ $$G(R,\phi,\theta)=H \exp\left[-\frac{1}{2}(X-\mu)^{T}\cdot \Sigma^{-1}\cdot (X-\mu)\right]$$ $$X=\begin{bmatrix}R\cos\phi\sin\theta\\ R\sin\phi\sin\theta\\ R\cos\theta\end{bmatrix}$$ and the 1D Gaussian with mean $R_\mu$ and variance $\sigma^2$ $$g(R)=H'\exp\left[-\frac{(R-R_\mu)}{2\sigma^2}\right]$$ The question is: is this true? I realize this is a forum for more fundamental mathematics, but I've asked this question on applied math and physics forums without any luck (link). I'd be grateful if someone can give me a hit on how to approach this. My attempt to solve this problem involved transforming $G$ so that $\mu=(0\ 0\ R_\mu)$ (adapting $\Sigma$ in the process) and substitute $z=\cos\theta$. However I couldn't solve the integral over $z$ and $\phi$.

Edit The special case where $\Sigma=\sigma^2I$ proves that this is not a Gaussian but it is very close. For $R_\mu\gg\sigma$ we find (link) $$\int_{0}^{2\pi}\int_{0}^{\pi}G(R,\phi,\theta)\sin\theta\ \text{d}\theta \ \text{d}\phi= \frac{g(R)}{RR_\mu}(1-\exp(-\frac{2RR_{\mu}}{\sigma^2}))\approx\frac{g(R)}{RR_\mu}$$ If I could find a similar result for any $\Sigma$, it would solve my problem.

Fourier transform in n-dim Euclidean and Minkowski space

2012-04-04T14:35:39Z

As far as I understood, the Fourier decomposition of a function $\boldsymbol{F}\colon\mathbb{R}^{n}\to\mathbb{R}^{m}$ where $\mathbb{R}^{n}$ is endowed with the Euclidean inner product $\left<\cdot,\cdot\right>$ is given by

$\boldsymbol{F}(\bar{x})=\int_{\mathbb{R}^{n}}{\tilde{\boldsymbol{F}}(\bar{\nu})e^{2\pi i \left<\bar{\nu},\bar{x}\right>}}{d\bar{\nu}}$

where $\tilde{\boldsymbol{F}}(\bar{\nu})=\int_{\mathbb{R}^{n}}{\boldsymbol{F}(\bar{x})e^{-2\pi i \left<\bar{\nu},\bar{x}\right>}}{d\bar{x}}$

How does this come about and for which functions does it apply? I'm not even able to find the right framework to work in (Hilbert spaces?). Secondly, could I just replace the Euclidean inner product by the Minkowskian inner product when in Minkowski space?

Subgroups of the Euclidean group as semidirect products

2011-10-10T10:08:09Z

Consider the Euclidean group $E(n)$ as the semidirect product for Euclidean vector space $\mathbb{E}^n$ with its orthogonal group $O(\mathbb{E}^n)$:

$E(n)=\mathbb{E}^n\rtimes O(\mathbb{E}^n)$

Then the following short exact sequence splits

$1\rightarrow \mathbb{E}^n\rightarrow E(n)\rightarrow O(\mathbb{E}^n)\rightarrow 1$

Now consider a subgroup G of the Euclidean group which translational subgroup T (all isometries in G with trivial linear part) can be identified with a lattice $\mathcal{L}^{n}$ in Euclidean vector space, i.e. all $\mathbb Z$-linear combinations of a chosen basis. The translation subgroup T is normal in G and we can write the short exact sequence

$1\rightarrow T\rightarrow G\rightarrow Q\rightarrow 1$

where quotient group $Q=G/T$. This short exact sequence splits iff $G=T\rtimes Q$. This is the case iff Q is isomorphic to the automorphism of the lattice so that we can write for example the following split short exact sequence

$1\rightarrow \mathcal{L}^{n}\rightarrow G\rightarrow Aut(\mathcal{L}^{n}) \rightarrow 1$

The question is: ~~why~~ is $G=T\rtimes Q$ iff $Q$ is isomorphic to $Aut(\mathcal{L}^{n})$?

FYI: in crystallography, G are called space groups in three dimensions. Space groups which are semidirect products, are called symmorphic.

Some thoughts that might lead to the solution:

1. Since T is normal in G we can write for every isometry $(t_q,q)\in G\quad$ ($t_q$: translational component, q linear component)

$\quad\quad T=(t_q,q).T.(t_q,q)^{-1}$

$\Leftrightarrow T= (t_q,q).\lbrace (t,id)\rbrace.(-q^{-1}.t_q,q^{-1})$

$\Leftrightarrow T=\lbrace (q.t,id)\rbrace=q.T$

So q is a permutation of T. Since T is isomorphic (as a free $\mathbb Z$-module) to $\mathcal{L}^{n}$ and since q is orthogonal ($q\in O(\mathbb{E}^n)$), we find that $q\in Aut(\mathcal{L}^{n})$. This means that the set of all linear parts of G, which we'll call $Q(\mathcal{L}^{n})$ is a subgroup of $Aut(\mathcal{L}^{n})$.

2. Consider a coset $(t_q,q).T\ $ of T, then we can write

$\quad\quad (t_q,q).T=\lbrace (t_q,q).(t,id): t\in \mathcal{L}^{n}\rbrace$

$\Leftrightarrow (t_q,q).T=\lbrace (t_q+q.t,q): t\in \mathcal{L}^{n}\rbrace$

$\Leftrightarrow (t_q,q).T=\lbrace (t_q+t',q): t'\in \mathcal{L}^{n}\rbrace$

which means that each q belongs to exactly one coset of T so that $Q(\mathcal{L}^{n})$ is isomorphic to $Q$.

3. From 1. and 2. we find that in any case (i.e. also when G is not a semidirect product) $Q$ is isomorphic to a finite subgroup of $Aut(\mathcal{L}^{n})$.

4. If we have $G=T\rtimes Q$, there exists a homomorphism from $Q$ to $Aut(T)$ and since T is isomorphic to $\mathcal{L}^{n}$ (as a free $\mathbb Z$-module)

$Hom:Q\rightarrow Aut(\mathcal{L}^{n})$

Conjugacy in $GL(n,\mathbb Z)$

2011-10-04T14:08:12Z

How can I determine whether $A_1,A_2\in GL(n,\mathbb Z)$ conjugate in $GL(n,\mathbb Z)$ and if they are, how can I find a $P\in GL(n,\mathbb Z)$ for which $A_2 = P^{-1}.A_1.P$ ?

In $GL(n,\mathbb Q)$ one could achieve this by checking if the Frobenius normal forms (FNF) are equal and if they are

$\quad\quad FNF_2 = FNF_1$

$\Leftrightarrow P_2^{-1}.A_2.P_2=P_1^{-1}.A_1.P_1$

$\Leftrightarrow A_2=M^{-1}.A_1.M\quad\quad\quad M=P_1.P_2^{-1}$

I found an algorithm which gives the FNF of a matrix with P a matrix of integers. Is there an way of performing subsequent elementary similarity transformations on $P_i$ (and hence also on $P_i^{-1}$) until $P_i\in GL(n,\mathbb Z)$ while also checking whether it is even possible to arrive at such a $P_i$?

Finite subgroup of $Gl(n,\mathbb Z)$ and congruences

2011-09-02T15:55:14Z

Suppose we have an invertible matrix q in a finite subgroup $Q$ of $Gl(n,\mathbb Z)$, the group of all invertible integer matrices. Now I want to find all $x\; mod\; \mathbb Z^n$ for which

$(q+q^2+q^3+...+q^m).x = 0\quad mod\; \mathbb Z^n$

where $m$ is the order of $q$ in the finite subgroup $Q$ of $Gl(n,\mathbb Z)$ so that $q^m=1$. I tried using the Smith normal form so that

$(q+q^2+q^3+...+q^m) = U.D.V$

where $U,V$ in $Gl(n,\mathbb Z)$ and $D$ the Smith normal form, so we have to solve

$D.V.x=0\quad mod\; \mathbb Z^n$

Since $D.V$ is diagonal, $x$ must have rational components unless the diagonal element is zero. Now my question is, what is the maximal denominator of the components in $x$ ? So what is the maximal absolute value in $D.V$ ?I think this must be $m$, but I can't figure out why.

Edit: Let me clarify why I expect x to be rational with an upper bound on the denominator. Suppose G is a subgroup of the Euclidean Group with isometries (t,q) as elements (t: translational part, q: linear part). The subgroup T which contains all isometries in G with trivial linear part is a normal subgroup of G. Suppose now that T can be identified with a $\mathbb Z$-lattice in $\mathbb R^n$, then G/T is isomorph with a finite subgroup Q of $GL(n,\mathbb Z)$. Crystallographers call G a space group and Q a point group.

There are only finite many conjugacy classes of finite subgroups in $GL(n,\mathbb Z)$, so there are only finite many point groups up to conjugacy in $GL(n,\mathbb Z)$. Now I want to understand why from this finite number of point groups, a finite number of (non-equivalent) space groups can be deduced. If we write G as the union of cosets of T

$G=\bigcup_{i=1}^{|Q|}(t_{qi},q_{i})T$

we see that (composition of two isometries and q belongs to exactly one coset)

$t_{q_1.q_2}=t_{q_1}+q_1.t_{q_2} \quad mod\ \mathbb Z^n$

So we know that $t_{q}$ is a real vector $0\leq t_{q}<1$. Using the previous property we also find that (m order of q)

$(t_{q},q)^{m}=(q^{1}\cdot t_{q}+\cdots+q^{m}\cdot t_{q},q^m)\in (0,id)T$

$\Leftrightarrow (q^{1}+\cdots+q^{m})\cdot t_{q}=0\quad mod\ \mathbb{Z}^{n}$

If an appropriate origin is chosen in Euclidean space, $t_{q}$ should be rational with maximal denominator $m$. Maybe investigating $(t_{q},q)^{m}$ is not the best way to find bounds on $t_{q}$?

Answer by Wox for Finite subgroup of $Gl(n,\mathbb Z)$ and congruences

2011-09-07T15:56:07Z

Edit: This is a secondary question on how Ralph's solution can be simplified by choosing an appropriate origin in Euclidean space.

Ralph's solution to my original question, in the context of space groups, states that an isometry $(x,q)$ in a space group $G$ with linear part $q\in Q< GL(Z^n)$, must have a translational part x for which

$X_q=\lbrace x\in\mathbb R^n: (q^{1}+\cdots+q^{m})\cdot x\in \mathbb Z^n\rbrace =Col(q-1)+\frac{1}{m}(Null(q-1)\cap \mathbb Z^n)$

Note first that from the composition of isometries we find that

$(t_1,q_1)(t_2,q_2)=(t_1+q_1\cdot t_2,q_1\cdot q_2)$

$\Leftrightarrow X_{q_{1}\cdot q_{2}}=X_{q_{1}}+(q_{1}-1)\cdot X_{q_{2}}$

This means that we must only consider the $X_q$ for the generators of the finite group $Q< GL(Z^n)$ (i.e. the point group).

After a shift of origin in Euclidean space, i.e. an affine transformation $(v,1)$ with $v\in \mathbb R^n$, we can write that

$X_q'=Col(q-1)+\frac{1}{m}(Null(q-1)\cap \mathbb Z^n)+(q-1)\cdot v$

Since $(q-1)\cdot v\in Col(q-1)$, we can find for every $u\in Col(q-1)$ a vector $v\in \mathbb R^n$ for which $(q-1)\cdot v=-u$. Thus for a proper choice of origin we can write for a generator q

$X_q=\frac{1}{m}(Null(q-1)\cap \mathbb Z^n)$

so that $t_q = X_q\ mod\ \mathbb Z^n$ is a rational number with maximal denominator $|Q|$ (which is the maximal possible m). The question is now, can we find one $v\in \mathbb R^n$ so that this simplification can be done for all $X_q$? For this, the column spaces $Col(q-1)$ for generators q of Q should be linear independent. If we call $S$ the generating set of Q, then this can be expressed as

$\forall q,p\in S: Col(q-1)\cap Col(p-1)=\lbrace 0 \rbrace$

Is this true?

Comment by Wox

2012-09-27T20:24:33Z

I tried that already without result.

Comment by Wox

2012-04-04T14:50:15Z

Too basic, I get it :-). But could you maybe give me an idea in which direction I should be looking, that would save me a lot of time.

Comment by Wox

2011-10-12T09:56:31Z

A thought on $Aut_O(L)$ vs $Aut(L)$. If a lattice is defined as a finitely generated free Z-module with positive definite symmetric bilinear form, then I can just use $Aut(L)$ which covers what you note as $Aut_O(L)$, right?

Comment by Wox

2011-10-12T09:48:32Z

Great stuff Ralph! The proof for the third statement would be certainly useful. My effort: consider $T=\lbrace (t,0),\forall t\in L\rbrace$ and $Q=G/T=\lbrace (t_q+L,q),q\in Q(L),t_q\in \mathbb R^n\rbrace$ where $Q(L)$ the finite group of all linear parts of isometries in G, with $Q(L)\leq Aut_{O}(L)\leq GL(\mathbb{n,Z})$. For a symmorphic space group $G=L\rtimes Q(L)$ which has $\lbrace(0,q),q\in Q(L)\rbrace$ as subgroup so that $Q=G/T=\lbrace (L,q),q\in Q(L)\rbrace$. So for each $Q(L)$ (finite subgroup of $GL(\mathbb{n,Z})$) there is exactly one symmorphic space group.

Comment by Wox

2011-10-11T10:00:47Z

Since $Aut(\mathcal{L}^n)$ is isomorphic to a finite subgroup of $GL(n,\mathbb Z)$ and since there are only finite many conjugacy classes of finite subgroups in $GL(n,\mathbb Z)$, there are only finite many $Aut(\mathcal{L}^n)$ up to conjugacy. In three dimensions, there are 73 conjugacy classes, so 73 different $Aut(\mathcal{L}^n)$. However there are also 73 different symmorphic space groups in three dimensions. Therefore I thought there must be a 1-1 correspondence between $G=T\rtimes Q$ (the symmorphic space groups) and $Aut(\mathcal{L}^n)$.

Comment by Wox

2011-10-11T08:57:17Z

I'm also struggling with the example you gave. It doesn't seem to correspond to any known space group...

Comment by Wox

2011-10-11T08:55:31Z

I only have a reference for this statement: if $Q\cong Aut(\mathcal{L}^n)$ and $T\cong \mathcal{L}^n$ then $G=T\rtimes Q$. on page 28: "... But Aut(L) is a finite subgroup of $GL(n,\mathbb Z)$, hence it is a point group itself, namely of the symmorphic space group with point group Aut(L) and translation lattice L." (A point group is what I called $Q(\mathcal{L}^{n})$.) To prove already this statement would be great. Is there any stronger statement than this? In any case, not as strong as I proposed, as you showed with your counterexample.

Comment by Wox

2011-10-10T16:19:58Z

You're right. I must be missing something basic... In $\mathbb R^2$ the group we're talking about is "p2" (en.wikipedia.org/wiki/Wallpaper_group#Group_p2). I know (it.iucr.org/Cb/ch1o4v0001) that this should be a symmorphic space group, which means that $\lbrace id, \sigma\rbrace$ should be the full automorphism of the translational subgroup T of G. And since $T\cong \mathcal{L}^2$ also of the lattice. I also found some information on the subject (crystallography.fr/mathcryst/pdf/havana/… page 7,17 and 28).

Comment by Wox

2011-10-10T14:18:53Z

Note that these matrices need not to be orthogonal anymore. They are only orthogonal if the lattice basis is orthonormal.

Comment by Wox

2011-10-10T14:15:41Z

When choosing a lattice basis (basis so that all vectors have coordinates in $\mathbb Z^n$), $Aut(\mathcal{L}^{n})$ is given by a finite subgroup of $GL(n,\mathbb Z)$.

Comment by Wox

2011-10-10T14:10:15Z

The group of all linear maps of the lattice (module homomorphisms $Hom:\mathcal{L}^{n}\rightarrow\mathcal{L}^{n}$) that are invertible and that preserve the inner product inherited from $\mathbb R^n$. In short: all orthogonal transformations of the lattice.

Comment by Wox

2011-10-10T13:03:25Z

I think the confusion might originate from $\mathcal{L}^n\cong\mathbb{Z}^n$. The automorphism group of $\mathbb{Z}^n$ is $GL(n,\mathbb{Z})$ isn't it? However in this context, also the inner product of Euclidean vector space must be preserved. For example the automorphism group of the square lattice in $\mathbb R^2$ has order 8 (signed permutations).

Comment by Wox

2011-10-10T12:56:32Z

Why is this a counterexample? If I'm not mistaken the group you describe has cosets $(0,id)T$ and $(0,\sigma)T$ so $Q=G/T\cong\lbrace id,\sigma\rbrace$. The lattice isomorphic to T is the triclinic lattice. The automorphism group of the triclinic lattice is $\lbrace id,\sigma\rbrace$.

Comment by Wox

2011-10-05T16:03:13Z

@HW: I only need to consider finite subgroups of $GL(n,\mathbb Z)$.

Comment by Wox

2011-10-05T08:57:14Z

@Mark: I believe this is the Latimer-MacDuffee theorem? This can be applied when the two matrices have the same irreducible characteristic polynomial. Can this be extended to matrices in $GL(n,\mathbb Z)$? And if it can, how does it allow me to retrieve similarity transform M (see question)?