Timeline for The symmetry group of $\mathbb Z^d$
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 14, 2012 at 17:32 | comment | added | Tom LaGatta | Thank you, everybody. You're right that as stated, my question is ambiguous. However, the answers clarify this ambiguity so I'm satisfied. | |
| Oct 14, 2012 at 17:30 | vote | accept | Tom LaGatta | ||
| Oct 13, 2012 at 22:10 | comment | added | Noam D. Elkies | The group $H$ of lattice symmetries fixing the origin is the finite group of symmetries of the hyperoctahedron (cross-polytope) spanned by $\pm e_i$, which is the semidirect product (here also a wresth product) $({\bf Z}/2{\bf Z})^d \rtimes S_d$. The group $G$ is then generated by $H$ and the normal subgroup of $G$ consisting of translations by integer vectors; thus $G$ is the semidirect product ${\bf Z}^d \rtimes H$. | |
| Oct 13, 2012 at 21:24 | answer | added | Qfwfq | timeline score: 1 | |
| Oct 13, 2012 at 21:16 | answer | added | Will Sawin | timeline score: 2 | |
| Oct 13, 2012 at 21:14 | answer | added | S. Carnahan♦ | timeline score: 1 | |
| Oct 13, 2012 at 19:55 | comment | added | Qiaochu Yuan | @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. | |
| Oct 13, 2012 at 19:53 | comment | added | Hugh Thomas | @Tom: If you didn't want to choose any inner product at all, though, then $H$ is $GL(n,\mathbb Z)$. | |
| Oct 13, 2012 at 19:30 | comment | added | Igor Rivin | @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... | |
| Oct 13, 2012 at 19:28 | comment | added | Hugh Thomas | @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. | |
| Oct 13, 2012 at 18:38 | comment | added | Qiaochu Yuan | @Igor: $\text{SO}(n, \mathbb{Z})$ doesn't act transitively (e.g. it fixes $0$). | |
| Oct 13, 2012 at 18:31 | comment | added | Igor Rivin | Why is it not $SO(n, \mathbb{Z})\backslash SL(n, \mathbb{Z})?$ | |
| Oct 13, 2012 at 18:23 | history | asked | Tom LaGatta | CC BY-SA 3.0 |