Are there any cases of finite subgroups of $O_n(\mathbb{Q})$ not contained in not isomorphic to any subgroup of $O_n(\mathbb{Z})$?
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- 2$\begingroup$ As literally written, this is very easily seen to be false - taking a finite subgroup of $O_n(\mathbb Z)$ and conjugating it by a random element of $O_n(\mathbb Q)$ will very likely not be contained in $O_n(\mathbb Z)$. The question is more sensible if you ask about subgroups up to conjugation, or even isomorphism. But Noam's answer shows not even that is true. $\endgroup$Wojowu– Wojowu2021-10-09 19:25:23 +00:00Commented Oct 9, 2021 at 19:25
- $\begingroup$ @markvs That every finite subgroup of $O_n(\mathbb Q)$ is conjugate to/isomorphic to a subgroup of $O_n(\mathbb Z)$. $\endgroup$Wojowu– Wojowu2021-10-09 20:53:01 +00:00Commented Oct 9, 2021 at 20:53
- $\begingroup$ If $H$ is any non-central subgroup of $\mathrm{O}_n(\mathbf{Z})$, then it has some conjugate by some element of $\mathrm{O}_n(\mathbf{Q})$ that is not contained in $\mathrm{O}_n(\mathbf{Z})$. This is just because the normal closure of $\mathrm{O}_n(\mathbf{Z})$ in $\mathrm{O}_n(\mathbf{Q})$ is $\mathrm{O}_n(\mathbf{Q})$. $\endgroup$YCor– YCor2021-10-09 21:04:42 +00:00Commented Oct 9, 2021 at 21:04
- 1$\begingroup$ An immediate remark is that $\mathrm{O}_n(\mathbf{Z})$ is the signed permutation group, of order $2^n.n!$. But $\mathrm{O}_n(\mathbf{Q})$ also contains the symmetric group $\mathrm{Sym}(n+1)$ (acting on the $\sum=0$ hyperplane). And the latter doesn't embed into $\mathrm{Sym}(n)\ltimes C_2^n$ as soon as $n\ge 4$ (immediate using simplicity of a subgroup of index $2$ in $\mathrm{Sym}(n+1)$). $\endgroup$YCor– YCor2021-10-09 21:16:11 +00:00Commented Oct 9, 2021 at 21:16
- 1$\begingroup$ @YCor the quadratic form $\sum_{i=0}^n x_i^2$ on the hyperplane $\sum_{i=0}^n x_i = 0$ is not in general equivalent to any scaling of $\sum_{j=1}^n x_j^2$. (For example, when $n=2$ the hyperplane contains equilateral triangles but ${\bf Q}^n$ does not.) So the image of ${\rm Sym}(n+1)$ need not be contained in $O_n({\bf Q})$. $\endgroup$Noam D. Elkies– Noam D. Elkies2021-10-10 02:51:42 +00:00Commented Oct 10, 2021 at 2:51
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3 Answers
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3 Let $n=m^2$ be a square. Then the vector $(1,\dots,1)\in\mathbf{Q}^n$ has norm $m$, as does $(0,\dots 0,m)$. Hence by Witt's theorem there exists an element $u$ of $\mathrm{O}_n(\mathbf{Q})$ mapping $(0,\dots,0,m)$ to $(1,\dots,1)$. Hence $u$ maps orthogonals to orthogonals: it maps the hyperplane $\mathbf{Q}^{n-1}\times\{0\}$ of equation $x_n=0$ onto the hyperplane of equation $\sum x_i=0$. Let $S_n$ be the subgroup of $\mathrm{O}_n(\mathbf{Q})$ permutating coordinates. Then $S_n$ fixes $(1,\dots,1)$ and acts faithfully on its orthogonal. Therefore $u^{-1}S_nu$ acts faithfully on the orthogonal $\mathbf{Q}^{n-1}$.
Hence, for every square $n$, the group $\mathrm{O}_{n-1}(\mathbf{Q})$ contains a copy of the symmetric group $S_n$. But for every $n\ge 5$ there is no nontrivial homomorphism $S_n$ to the Weyl group $S_{n-1}\ltimes C_2^{n-1}$ of type $B_{n-1}$, which is isomorphic to $\mathrm{O}_{n-1}(\mathbf{Z})$.
This proves that for every square $n\ge 9$, there is a finite subgroup of $\mathrm{O}_{n-1}(\mathbf{Q})$, isomorphic to $S_n$, not embedding into $\mathrm{O}_{n-1}(\mathbf{Z})$.
- 1$\begingroup$ I think the argument does work for squares and that you have just dropped a square root somewhere: If $n=m^2$, then $|(1,1,\ldots,1)| = m = |(m,0,0,\ldots,0)|$. $\endgroup$David E Speyer– David E Speyer2021-10-10 14:15:44 +00:00Commented Oct 10, 2021 at 14:15
- $\begingroup$ @DavidESpeyer ah! and this was the way I initially posted :) I thought this double square was weird. I'll edit again, thanks. $\endgroup$YCor– YCor2021-10-10 14:19:52 +00:00Commented Oct 10, 2021 at 14:19
- $\begingroup$ Thanks for accepting this answer, but Noam Elkies's answer was posted earlier (and was already posted when I started writing this one). $\endgroup$YCor– YCor2021-10-11 14:14:02 +00:00Commented Oct 11, 2021 at 14:14
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0 If it's the same $n$ then yes this can happen. For example, the lattice $D_4$ (consisting of all integer vectors in ${\bf Z}^4$ with even sum) has more isometries than ${\bf Z}^4$. If we allow different $n$, then no, because every finite group $G$ is contained in the group of permutations of $G$, which in turn is contained in $O_{|G|}({\bf Z})$.
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1 A theorem which you might be looking for is that, if $G$ is any finite subgroup of $O_n(\mathbb{Q})$, then there is a lattice $\Lambda$ in $\mathbb{Q}^n$ which is preserved by $G$. However, that lattice does not have to be $\mathbb{Z}^n$.
Proof: Let $L$ be any lattice in $\mathbb{Q}^n$. Take $\Lambda = \sum_{g \in G} g L$.
- 2$\begingroup$ In the conjugacy result, the point is that not only this lattice does not have to be $\mathbf{Z}^n$, but it doesn't have to be the image of $\mathbf{Z}^n$ by any rational similarity (= rational matrix $A$ such that $AA^t$ is a nonzero scalar matrix). $\endgroup$YCor– YCor2021-10-10 13:31:30 +00:00Commented Oct 10, 2021 at 13:31