That is, is it true that there does not exist a lattice in $G = {\rm SO}(n,1)$ which contains the group of integer points of $G$ as a proper subgroup (obviously then of finite index)? if such a lattice exists, what is the maximal possible index? I know that ${\rm SL}(n,\mathbb{Z})$ is a maximal arithmetic subgroup of ${\rm SL}(n,\mathbb{R})$ so the answer is yes if $n=2$, what is known for larger $n$? this is probably related to hyperbolic manifolds of minimal volume but I can't locate an appropriate reference... any suggestions will be appreciated :)
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1$\begingroup$ Take a look at the Section 11.4 of the book by Maclachlan and Reid "Arithmetic of hyperbolic 3-manifolds". They discuss maximality in the SO(3,1) case, but there is a decent chance that it will work in general as well. (You may have to deal with general affine buildings though, not just trees.) $\endgroup$– MishaCommented Apr 12, 2013 at 23:16
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$\begingroup$ Here's a link to a paper discussing minimal volume cusped orbifolds: ams.org/mathscinet-getitem?mr=2326144 $\endgroup$– Ian AgolCommented Apr 16, 2013 at 0:58
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$\begingroup$ You can find a detailed answer to this question in ams.org/mathscinet-getitem?mr=2124587, ams.org/mathscinet-getitem?mr=2352518, and ams.org/mathscinet-getitem?mr=2974199. You will only need small parts of the papers. As Misha said, for even n it is always true but for certain odd n there is an index 2 extension. $\endgroup$– Misha BelolipetskyCommented Jun 6, 2013 at 15:54
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Here is a proof for even $n$, I am not quite sure about odd $n$ since $G=SO(n,1)$ is not an adjoint group and weak approximation might fail in this case (I do not know enough about this, you would need to ask experts, like Andrei Rapinchuk). First, note that the commensurator of $G({\mathbb Z})$ in $G({\mathbb R})$ is $G({\mathbb Q})$ (as it has to preserve the set of rational lines in the light cone: They are the fixed points of unipotent elements of $G({\mathbb Z})$).
Claim. Let $G=SO(n,1)$, $n$ is even. Then $G({\mathbb Z})$ is a maximal subgroup of $G({\mathbb Q})$ and, hence, in $G({\mathbb R})$.
Proof. It is known that $$ G({\mathbb Z}_p)= G({\mathbb Q}_p)\cap SL_{n+1}({\mathbb Z}_p)$$ `
is a maximal compact subgroup of $G({\mathbb Q}_p)$ for all $p$. By weak approximation (since $G$ is an adjoint group), $G({\mathbb Q})$ is dense in $G({\mathbb Q}_p)$, which implies that $G({\mathbb Z})$ is dense in $G({\mathbb Z}_p)$. If $G({\mathbb Z})$ is not a maximal subgroup of $G({\mathbb Q})$ and $G({\mathbb Z})<\Gamma< G({\mathbb Q})$ is a larger discrete subgroup, then $\Gamma$ is not contained in $G({\mathbb Z}_p)$ for some prime $p$. The closure of $\Gamma$ in $G({\mathbb Q}_p)$ is compact (since $\Gamma$ is a finite extension of $G({\mathbb Z})$) and strictly contains $G({\mathbb Z}_p)$, which contradicts maximality of $G({\mathbb Z}_p)$. qed
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$\begingroup$ Given your observation that a maximal lattice containing $G(Z)$ is in $G(Q)$, one then need only show that if $A\in SL_n(Q)$ and $A^k\in SL_n(Z)$ for some non-zero $k$, then $A\in SL_n(Z)$. $\endgroup$– Ian AgolCommented Apr 16, 2013 at 22:16
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$\begingroup$ Ian: Yes, maybe it is an elementary fact, I just do not know how to prove this without invoking p-adic completions. Algebraic groups over global fields is not really my cup of tea... $\endgroup$– MishaCommented Apr 17, 2013 at 2:36
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$\begingroup$ Misha, Ian – many thanks! and what is a reference for the maximality of $G({\mathbb Z}_p)$? $\endgroup$ Commented Apr 17, 2013 at 16:40
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2$\begingroup$ Ian: by the way the claim in your comment is clearly not true, for example take $A = \begin{pmatrix}1 & 1/2\\ 0 & 1\end{pmatrix}$ :( $\endgroup$ Commented Apr 17, 2013 at 16:54
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$\begingroup$ Dima: I used Bruhat's paper "p-adic groups" in Proceedings "Algebraic Groups and Discontinuous Subgroups" (1966); he worked out the example of p-adic groups preserving quadratic forms. There are probably other references too. $\endgroup$– MishaCommented Apr 17, 2013 at 17:03