User wox - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T10:56:19Z http://mathoverflow.net/feeds/user/17551 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108267/azimuthal-and-polar-integration-of-a-3d-gaussian Azimuthal and polar integration of a 3D Gaussian Wox 2012-09-27T17:20:38Z 2012-09-28T12:58:34Z <p>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 $\sigma^2$ $$g(R)=H'\exp\left[-\frac{(R-R_\mu)}{2\sigma^2}\right]$$ <strong>The question is: is this true?</strong> 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 (<a href="http://math.stackexchange.com/questions/195210/azimuthal-and-polar-integration-of-a-3d-gaussian" rel="nofollow">link</a>). 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 $\phi$.</p> <p><strong>Edit</strong> 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 (<a href="http://math.stackexchange.com/questions/195210/azimuthal-and-polar-integration-of-a-3d-gaussian" rel="nofollow">link</a>) $$\int_{0}^{2\pi}\int_{0}^{\pi}G(R,\phi,\theta)\sin\theta\ \text{d}\theta \ \text{d}\phi= \frac{g(R)}{RR_\mu}(1-\exp(-\frac{2RR_{\mu}}{\sigma^2}))\approx\frac{g(R)}{RR_\mu}$$ If I could find a similar result for any $\Sigma$, it would solve my problem.</p> http://mathoverflow.net/questions/93127/fourier-transform-in-n-dim-euclidean-and-minkowski-space Fourier transform in n-dim Euclidean and Minkowski space Wox 2012-04-04T14:35:39Z 2012-04-04T14:35:39Z <p>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 $\left&lt;\cdot,\cdot\right>$ is given by</p> <p>$\boldsymbol{F}(\bar{x})=\int_{\mathbb{R}^{n}}{\tilde{\boldsymbol{F}}(\bar{\nu})e^{2\pi i \left&lt;\bar{\nu},\bar{x}\right>}}{d\bar{\nu}}$</p> <p>where $\tilde{\boldsymbol{F}}(\bar{\nu})=\int_{\mathbb{R}^{n}}{\boldsymbol{F}(\bar{x})e^{-2\pi i \left&lt;\bar{\nu},\bar{x}\right>}}{d\bar{x}}$</p> <p>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?</p> http://mathoverflow.net/questions/77682/subgroups-of-the-euclidean-group-as-semidirect-products Subgroups of the Euclidean group as semidirect products Wox 2011-10-10T10:08:09Z 2011-10-13T13:35:05Z <p>Consider the Euclidean group $E(n)$ as the semidirect product for Euclidean vector space $\mathbb{E}^n$ with its orthogonal group $O(\mathbb{E}^n)$:</p> <p>$E(n)=\mathbb{E}^n\rtimes O(\mathbb{E}^n)$</p> <p>Then the following short exact sequence splits</p> <p>$1\rightarrow \mathbb{E}^n\rightarrow E(n)\rightarrow O(\mathbb{E}^n)\rightarrow 1$</p> <p>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</p> <p>$1\rightarrow T\rightarrow G\rightarrow Q\rightarrow 1$</p> <p>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</p> <p>$1\rightarrow \mathcal{L}^{n}\rightarrow G\rightarrow Aut(\mathcal{L}^{n}) \rightarrow 1$</p> <blockquote> <p>The question is: <strike>why</strike> is $G=T\rtimes Q$ iff $Q$ is isomorphic to $Aut(\mathcal{L}^{n})$?</p> </blockquote> <p>FYI: in crystallography, G are called space groups in three dimensions. Space groups which are semidirect products, are called symmorphic.</p> <p>Some thoughts that might lead to the solution:</p> <p><strong>1.</strong> 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)</p> <p>$\quad\quad T=(t_q,q).T.(t_q,q)^{-1}$</p> <p>$\Leftrightarrow T= (t_q,q).\lbrace (t,id)\rbrace.(-q^{-1}.t_q,q^{-1})$</p> <p>$\Leftrightarrow T=\lbrace (q.t,id)\rbrace=q.T$</p> <p>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 <strong>$Q(\mathcal{L}^{n})$ is a subgroup of $Aut(\mathcal{L}^{n})$</strong>.</p> <p><strong>2.</strong> Consider a coset $(t_q,q).T\$ of T, then we can write</p> <p>$\quad\quad (t_q,q).T=\lbrace (t_q,q).(t,id): t\in \mathcal{L}^{n}\rbrace$</p> <p>$\Leftrightarrow (t_q,q).T=\lbrace (t_q+q.t,q): t\in \mathcal{L}^{n}\rbrace$</p> <p>$\Leftrightarrow (t_q,q).T=\lbrace (t_q+t',q): t'\in \mathcal{L}^{n}\rbrace$</p> <p>which means that each q belongs to exactly one coset of T so that <strong>$Q(\mathcal{L}^{n})$ is isomorphic to $Q$</strong>.</p> <p><strong>3.</strong> From <strong>1.</strong> and <strong>2.</strong> we find that in any case (i.e. also when G is not a semidirect product) <strong>$Q$ is isomorphic to a finite subgroup of $Aut(\mathcal{L}^{n})$</strong>.</p> <p><strong>4.</strong> If we have $G=T\rtimes Q$, there exists a <strong>homomorphism from $Q$ to $Aut(T)$</strong> and since T is isomorphic to $\mathcal{L}^{n}$ (as a free $\mathbb Z$-module)</p> <p>$Hom:Q\rightarrow Aut(\mathcal{L}^{n})$</p> http://mathoverflow.net/questions/77133/conjugacy-in-gln-mathbb-z Conjugacy in $GL(n,\mathbb Z)$ Wox 2011-10-04T14:08:12Z 2011-10-05T12:46:33Z <p>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 $A_2 = P^{-1}.A_1.P$ ?</p> <p>In $GL(n,\mathbb Q)$ one could achieve this by checking if the Frobenius normal forms (FNF) are equal and if they are</p> <p>$\quad\quad FNF_2 = FNF_1$</p> <p>$\Leftrightarrow P_2^{-1}.A_2.P_2=P_1^{-1}.A_1.P_1$ </p> <p>$\Leftrightarrow A_2=M^{-1}.A_1.M\quad\quad\quad M=P_1.P_2^{-1}$ </p> <p>I found an <a href="http://hal.archives-ouvertes.fr/docs/00/32/37/05/PDF/Ozello.Patrick_1987_these.pdf" rel="nofollow">algorithm</a> 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 $P_i$?</p> http://mathoverflow.net/questions/74370/finite-subgroup-of-gln-mathbb-z-and-congruences Finite subgroup of $Gl(n,\mathbb Z)$ and congruences Wox 2011-09-02T15:55:14Z 2011-09-10T01:04:26Z <p>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</p> <p>$(q+q^2+q^3+...+q^m).x = 0\quad mod\; \mathbb Z^n$</p> <p>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</p> <p>$(q+q^2+q^3+...+q^m) = U.D.V$</p> <p>where $U,V$ in $Gl(n,\mathbb Z)$ and $D$ the Smith normal form, so we have to solve</p> <p>$D.V.x=0\quad mod\; \mathbb Z^n$</p> <p>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.</p> <p><strong>Edit:</strong> 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.</p> <p>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</p> <p>$G=\bigcup_{i=1}^{|Q|}(t_{qi},q_{i})T$</p> <p>we see that (composition of two isometries and q belongs to exactly one coset)</p> <p>$t_{q_1.q_2}=t_{q_1}+q_1.t_{q_2} \quad mod\ \mathbb Z^n$</p> <p>So we know that $t_{q}$ is a real vector $0\leq t_{q}&lt;1$. Using the previous property we also find that (m order of q)</p> <p>$(t_{q},q)^{m}=(q^{1}\cdot t_{q}+\cdots+q^{m}\cdot t_{q},q^m)\in (0,id)T$</p> <p>$\Leftrightarrow (q^{1}+\cdots+q^{m})\cdot t_{q}=0\quad mod\ \mathbb{Z}^{n}$</p> <p>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 $t_{q}$?</p> http://mathoverflow.net/questions/74370/finite-subgroup-of-gln-mathbb-z-and-congruences/74765#74765 Answer by Wox for Finite subgroup of $Gl(n,\mathbb Z)$ and congruences Wox 2011-09-07T15:56:07Z 2011-09-08T17:21:38Z <p><strong>Edit: This is a secondary question on how Ralph's solution can be simplified by choosing an appropriate origin in Euclidean space.</strong></p> <p>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 $q\in Q&lt; GL(Z^n)$, must have a translational part x for which </p> <p>$X_q=\lbrace x\in\mathbb R^n: (q^{1}+\cdots+q^{m})\cdot x\in \mathbb Z^n\rbrace =Col(q-1)+\frac{1}{m}(Null(q-1)\cap \mathbb Z^n)$</p> <p>Note first that from the composition of isometries we find that</p> <p>$(t_1,q_1)(t_2,q_2)=(t_1+q_1\cdot t_2,q_1\cdot q_2)$</p> <p>$\Leftrightarrow X_{q_{1}\cdot q_{2}}=X_{q_{1}}+(q_{1}-1)\cdot X_{q_{2}}$</p> <p>This means that we must only consider the $X_q$ for the generators of the finite group $Q&lt; GL(Z^n)$ (i.e. the point group).</p> <p>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</p> <p>$X_q'=Col(q-1)+\frac{1}{m}(Null(q-1)\cap \mathbb Z^n)+(q-1)\cdot v$</p> <p>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</p> <p>$X_q=\frac{1}{m}(Null(q-1)\cap \mathbb Z^n)$</p> <p>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</p> <p>$\forall q,p\in S: Col(q-1)\cap Col(p-1)=\lbrace 0 \rbrace$</p> <p>Is this true?</p> http://mathoverflow.net/questions/108267/azimuthal-and-polar-integration-of-a-3d-gaussian Comment by Wox Wox 2012-09-27T20:24:33Z 2012-09-27T20:24:33Z I tried that already without result. http://mathoverflow.net/questions/93127/fourier-transform-in-n-dim-euclidean-and-minkowski-space Comment by Wox Wox 2012-04-04T14:50:15Z 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. http://mathoverflow.net/questions/77682/subgroups-of-the-euclidean-group-as-semidirect-products/77882#77882 Comment by Wox Wox 2011-10-12T09:56:31Z 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? http://mathoverflow.net/questions/77682/subgroups-of-the-euclidean-group-as-semidirect-products/77882#77882 Comment by Wox Wox 2011-10-12T09:48:32Z 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. http://mathoverflow.net/questions/77682/subgroups-of-the-euclidean-group-as-semidirect-products Comment by Wox Wox 2011-10-11T10:00:47Z 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 $Aut(\mathcal{L}^n)$. http://mathoverflow.net/questions/77682/subgroups-of-the-euclidean-group-as-semidirect-products Comment by Wox Wox 2011-10-11T08:57:17Z 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... http://mathoverflow.net/questions/77682/subgroups-of-the-euclidean-group-as-semidirect-products Comment by Wox Wox 2011-10-11T08:55:31Z 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: &quot;... 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.&quot; (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. http://mathoverflow.net/questions/77682/subgroups-of-the-euclidean-group-as-semidirect-products Comment by Wox Wox 2011-10-10T16:19:58Z 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 &quot;p2&quot; (<a href="http://en.wikipedia.org/wiki/Wallpaper_group#Group_p2" rel="nofollow">en.wikipedia.org/wiki/Wallpaper_group#Group_p2</a>). I know (<a href="http://it.iucr.org/Cb/ch1o4v0001/" rel="nofollow">it.iucr.org/Cb/ch1o4v0001</a>) 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 (<a href="http://www.crystallography.fr/mathcryst/pdf/havana/Souvignier_syllabus.pdf" rel="nofollow">crystallography.fr/mathcryst/pdf/havana/&hellip;</a> page 7,17 and 28). http://mathoverflow.net/questions/77682/subgroups-of-the-euclidean-group-as-semidirect-products Comment by Wox Wox 2011-10-10T14:18:53Z 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. http://mathoverflow.net/questions/77682/subgroups-of-the-euclidean-group-as-semidirect-products Comment by Wox Wox 2011-10-10T14:15:41Z 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 $GL(n,\mathbb Z)$. http://mathoverflow.net/questions/77682/subgroups-of-the-euclidean-group-as-semidirect-products Comment by Wox Wox 2011-10-10T14:10:15Z 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. http://mathoverflow.net/questions/77682/subgroups-of-the-euclidean-group-as-semidirect-products Comment by Wox Wox 2011-10-10T13:03:25Z 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). http://mathoverflow.net/questions/77682/subgroups-of-the-euclidean-group-as-semidirect-products Comment by Wox Wox 2011-10-10T12:56:32Z 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 $\lbrace id,\sigma\rbrace$. http://mathoverflow.net/questions/77133/conjugacy-in-gln-mathbb-z/77139#77139 Comment by Wox Wox 2011-10-05T16:03:13Z 2011-10-05T16:03:13Z @HW: I only need to consider finite subgroups of $GL(n,\mathbb Z)$. http://mathoverflow.net/questions/77133/conjugacy-in-gln-mathbb-z Comment by Wox Wox 2011-10-05T08:57:14Z 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)?