# The symmetry group of $\mathbb Z^d$

Let $d \ge 1$, and consider the integer lattice $\mathbb Z^d$. This is a homogeneous space, in the manner of the Erlangan Programm.

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.

Let $V = \{\pm \mathrm e_1, \cdots, \pm \mathrm e_d\}$ 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 $V$.

The group $G$ should then consist of translations and rotations. Does that mean that it is a semidirect product of $\mathbb Z^d$ and $H$?

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Why is it not $SO(n, \mathbb{Z})\backslash SL(n, \mathbb{Z})?$ – Igor Rivin Oct 13 '12 at 18:31
@Igor: $\text{SO}(n, \mathbb{Z})$ doesn't act transitively (e.g. it fixes $0$). – Qiaochu Yuan Oct 13 '12 at 18:38
@Tom: Yes, I agree with you. Now I'm not sure what you're puzzled about, but what I was puzzled about is why $H$ seems smaller than I expected. I think the reason is that you've picked an inner product on $\mathbb Z^n$ (the usual one) and it doesn't have that many symmetries, since any element has only $2n$ nearest neighbours. You would get different answers by choosing a different inner product. – Hugh Thomas Oct 13 '12 at 19:28
@Qiaochu: True, I guess I did not really understand the question: $G$ is supposed to be the FULL symmetry group of a lattice, which $SL(n, \mathbb{Z})$ is not... – Igor Rivin Oct 13 '12 at 19:30
@Igor: ah, I misinterpreted your notation; I meant to say $\text{SL}_n(\mathbb{Z})$. @Tom: the question strikes me as ambiguous. You ask for a symmetry group but don't specify what structure you want to preserve. – Qiaochu Yuan Oct 13 '12 at 19:55

If $G=\mathbb Z^d \rtimes H$ for some group $H$ that acts on $\mathbb Z^d$, there is indeed a natural bijection $G/H \cong \mathbb Z^d$.

As Qiaochu says, what $H$ would be depends on the structure you want to preserve.

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.

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.

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.

All the structures are chosen to be translation-invariant so that their symmetry group can act transitively on $\mathbb Z^d$.

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Thank you, Will! – Tom LaGatta Oct 14 '12 at 17:31

1. You require your symmetries to preserve distances. Then $H=O(d,Z)$, and $G$ is the Galilean group over the integers.

2. You require your symmetries to preserve affine structure ($H=GL(d,Z)$)

3. Minor variations involving determinants.

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Thank you, Scott! – Tom LaGatta Oct 14 '12 at 17:31

As far as I read (See page 138 of R.W.Sharpe), 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.

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.

But you didn't say which structure on $\mathbb{Z}^d$ you want the symmetries to preserve...

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]

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Thank you, Qfwfq! – Tom LaGatta Oct 14 '12 at 17:31