3
$\begingroup$

A uniform lattice in a locally compact group $G$ is a discrete subgroup $\Gamma\subset G$ such that $G/\Gamma$ is compact. My question is whether a uniform lattice exists in the group $$ G={\mathbb R}^2\rtimes SL_2({\mathbb R}). $$ This group is unimodular, so an obvious criterion is satisfied. It also admits a lattice (=discrete subgroup of finite covolume), for example the group of integer valued points ${\mathbb Z}^2\rtimes SL_2({\mathbb Z})$, but this is not cocompact.

$\endgroup$
  • $\begingroup$ Btw, it is unknown whether $G$ is quasi-isometric to any finitely generated group (probably it isn't). $\endgroup$ – YCor Feb 15 '16 at 13:24
  • $\begingroup$ It is a little intriguing. There do exist uniform lattices in $({\mathbb R }^2 \times {\mathbb R}^2 ) \rtimes SL_2({\mathbb R})^2$ but not in the example you have asked (as YCor's answer tells you) $\endgroup$ – Venkataramana Feb 15 '16 at 15:02
  • $\begingroup$ @Venkataramana how do you construct it? $\endgroup$ – YCor Feb 15 '16 at 15:41
  • 1
    $\begingroup$ Let $D$ be a quaternionic central division algebra over $K={\mathbb Q }[{\sqrt 2}]$ which splits over all the real places of $K$. Let $O$ be an order in $D$. Then the semi-direct product of $O$ wth $SL_1(O)$ is a co-compact lattice in the group I have written. $\endgroup$ – Venkataramana Feb 15 '16 at 23:29
  • 1
    $\begingroup$ (After talking with Yves Benoist): there are much simpler examples. $V\rtimes S$ has cocompact lattices as soon as $V$ is definable over some $\mathbb{Q}$-isotropic form of $S$. In particular, $\mathbb{R}^3\rtimes\mathrm{SL}_2(\mathbb{R})$ and $(\mathbb{R}^2\oplus\mathbb{R}^2)\rtimes\mathrm{SL}_2(\mathbb{R})$ both have cocompact lattices. $\endgroup$ – YCor Feb 18 '16 at 12:30
6
$\begingroup$

No, there's no uniform (=cocompact) lattice in $\mathbf{R}^n\rtimes\mathrm{SL}_n(\mathbf{R})$ for any $n\ge 2$. Up to the action by automorphisms of $\mathrm{GL}_n(\mathbf{R})$, all lattices are contained in $\mathbf{R}^n\rtimes\mathrm{SL}_n(\mathbf{Z})$, which is not cocompact.

Indeed, let $\Gamma$ be a lattice. In a connected Lie group, the intersection of a lattice with the amenable radical is a lattice in the amenable radical. So $\Gamma\cap\mathbf{R}^n$ is a lattice in $\mathbf{R}^n$; hence modulo a global automorphism induced by some element of $\mathrm{GL}_n(\mathbf{R})$, we can assume that $\Gamma\cap\mathbf{R}^n=\mathbf{Z}^n$. The subgroup $\Gamma$ is then contained in the normalizer of $\Gamma\cap\mathbf{R}^n=\mathbf{Z}^n$, and this normalizer is equal to $\mathbf{R}^n\rtimes\mathrm{SL}_n(\mathbf{Z})$.


(Edit; added to reflect the comments:) On the other hand, $\mathbf{R}^3\rtimes\mathrm{SL}_2(\mathbf{R})$ admits cocompact lattices. Indeed, let $q=X^2+Y^2-7Z^2$. Let $G(q)(K)=K^3\rtimes\mathrm{SO}(q)(K)$. Then $q$ is $\mathbf{Q}$-anisotropic; so $G(\mathbf{Z})$ is a cocompact lattice in $G(\mathbf{R})=\mathbf{R}^3\rtimes \mathrm{SO}(q)(\mathbf{R})$. Pulling back, we get a cocompact lattice in $\mathbf{R}^3\rtimes \mathrm{SL}_2(\mathbf{R})$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks Yves, can you give me a reference for this property of the amenable radical? $\endgroup$ – user1688 Feb 15 '16 at 10:39
  • $\begingroup$ In this case this follows from Corollary 8.28 in Raghunathan's book "Discrete subgroups of Lie groups" (Springer 1972). $\endgroup$ – YCor Feb 15 '16 at 13:12
  • $\begingroup$ Comment by Wolfgang Globke I'm a bit late to the party, but just want to remark that the reference given by YCor (Corollary 8.28 in Raghunathan) is actually known to be incorrect. Counterexamples are known, first due to Starkov (1984). Luckily, the statement is still correct if the Levi subgroup of G has no compact factors, which is the case here (see e.g. Starkov's book "Dynamical systems on homogeneous spaces", Section E). $\endgroup$ – YCor Jan 23 '17 at 16:36
  • $\begingroup$ Thanks for noticing. Actually I was aware of it, but since the proof is correct in the mentioned case, I was happy with it. The general correct statement is that in a connected Lie group, every lattice intersects the amenable radical in a lattice. The same statement with "solvable radical" instead (as given in Raghunathan) is trivially false: e.g. in $\mathbf{R}\times\mathrm{SO}(3)$ the subgroup generated by $(1,g)$ where $g$ has infinite order yields a counterexample. $\endgroup$ – YCor Jan 23 '17 at 16:39
1
$\begingroup$

I'm a bit late to the party, but just want to remark that the reference given by YCor (Corollary 8.28 in Raghunathan) is actually known to be incorrect. Counterexamples are known, first due to Starkov (1984).

Luckily, the statement is still correct if the Levi subgroup of G has no compact factors, which is the case here (see e.g. Starkov's book "Dynamical systems on homogeneous spaces", Section E).

(Apologies for not using the "comment" function, but I don't have permission to write comments yet.)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I copied your post as a comment. $\endgroup$ – YCor Jan 23 '17 at 16:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy