The symmetry group of $\mathbb Z^d$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T12:01:46Z http://mathoverflow.net/feeds/question/109549 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109549/the-symmetry-group-of-mathbb-zd The symmetry group of $\mathbb Z^d$ Tom LaGatta 2012-10-13T18:23:43Z 2012-10-13T21:32:18Z <p>Let $d \ge 1$, and consider the integer lattice $\mathbb Z^d$. This is a homogeneous space, in the manner of the Erlangan Programm. </p> <p>I would like to write $\mathbb Z^d = G / H$, where $G$ is the symmetry group of the lattice and $H$ is the stabilizer of a point, but I do not see readily how to do so. This should be an easy question, so maybe one of you can answer it quickly.</p> <p>Let <code>$V = \{\pm \mathrm e_1, \cdots, \pm \mathrm e_d\}$</code> denote the $2d$ standard basis vectors in $\mathbb Z^d$, and let $H$ be the group consisting of lattice orthogonal matrices from $V$. i.e., an element $\beta \in H$ describes an orthogonal basis of $\mathbb Z^d$, in the sense that it is a matrix whose columns are an independent set from $V$.</p> <p>The group $G$ should then consist of translations and rotations. Does that mean that it is a semidirect product of $\mathbb Z^d$ and $H$?</p> http://mathoverflow.net/questions/109549/the-symmetry-group-of-mathbb-zd/109557#109557 Answer by S. Carnahan for The symmetry group of $\mathbb Z^d$ S. Carnahan 2012-10-13T21:14:11Z 2012-10-13T21:14:11Z <p>The answer to your last question is "yes" if:</p> <ol> <li><p>You require your symmetries to preserve distances. Then $H=O(d,Z)$, and $G$ is the Galilean group over the integers.</p></li> <li><p>You require your symmetries to preserve affine structure ($H=GL(d,Z)$)</p></li> <li><p>Minor variations involving determinants.</p></li> </ol> http://mathoverflow.net/questions/109549/the-symmetry-group-of-mathbb-zd/109558#109558 Answer by Will Sawin for The symmetry group of $\mathbb Z^d$ Will Sawin 2012-10-13T21:16:11Z 2012-10-13T21:16:11Z <p>If $G=\mathbb Z^d \rtimes H$ for some group $H$ that acts on $\mathbb Z^d$, there is indeed a natural bijection $G/H \cong \mathbb Z^d$.</p> <p>As Qiaochu says, what $H$ would be depends on the structure you want to preserve.</p> <p>The obvious $H$ to choose is $GL_d(\mathbb Z)$. $\mathbb Z^d \rtimes GL_d(\mathbb Z)$ is the symmetry group of $\mathbb Z^d$ preserving the affine structure.</p> <p>Another choice would be $O_d(\mathbb Z)$ for the standard quadratic form on $\mathbb Z^d$, which would give the group preserving the standard metric structure.</p> <p>If you want the group to consist of translations and rotations, you need to get rid of the reflections, which you do by setting $H=SO_d(\mathbb Z)$. This preserves a metric and an orientation.</p> <p>All the structures are chosen to be translation-invariant so that their symmetry group can act transitively on $\mathbb Z^d$.</p> http://mathoverflow.net/questions/109549/the-symmetry-group-of-mathbb-zd/109559#109559 Answer by Qfwfq for The symmetry group of $\mathbb Z^d$ Qfwfq 2012-10-13T21:24:34Z 2012-10-13T21:32:18Z <p>As far as I read (See page 138 of <a href="http://books.google.it/books?hl=it&amp;lr=&amp;id=Ytqs4xU5QKAC&amp;oi=fnd&amp;pg=PR5&amp;dq=sharpe+differential+geometry&amp;ots=xqfoRPI5pn&amp;sig=tqZhS-cZxisB5MxLJDbgVbBawaI" rel="nofollow">R.W.Sharpe</a>), the Erlangen program, strictly speaking, describes connected homogeneus manifolds $X$ as $G/H$ where $G$ is a Lie group considered as the "automorphism group" of a geometry on $X$ and $H$ is the stabilizer of a point. </p> <p>First of all, the space $\mathbb{Z}^d$ is a non-connected zero dimensional manifold, so I don't know how much we can say it fits the Erlangen program. Anyways, the "full symmetry group" as a manifold (even as a Riemannian manifold, being zero dimensional) is then simply the full symmetric group (i.e. set theoretic permutations) $G=\mathfrak{S}(\mathbb{Z}^d)$, and $H$ the stabilizer of any point.</p> <p>But you didn't say which structure on $\mathbb{Z}^d$ you want the symmetries to preserve...</p> <p>If you want to preserve the distance induced by the Euclidean norm on $\mathbb{R}^d$, then you can take $G=O(d,\mathbb{Z})\ltimes\mathbb{Z}^d$ and $H=O(d,\mathbb{Z})$. [I see that S.Carnahan has posted the same suggestion right before me. Edit: and also Will Sawin]</p>