# Simple discrete subgroups of Lie groups

Upon Ian Agol's suggestion, I separated this question from the one on non-residual finiteness in

Non-residually finite matrix groups

Question. Are there infinitely generated simple discrete subgroups of $SL(n, {\mathbb R})$, $n\ge 3$?

Note: Infinitely generated means not admitting a finite generating set.

Background: By Malcev's theorem, every finitely generated matrix group is residually finite. In particular, it is either finite or non-simple. On the other hand, there are (say, countable) simple matrix groups like $PSL(n, {\mathbb Q})$. The question is what happens in the case of discrete infinitely generated groups. On one hand, arguments similar to Malcev's appear to fail since there are discrete non-residually finite subgroups of $SO(3,1)$, see Agol's examples in Non-residually finite matrix groups

Moreover, as Ian observes, one can construct groups like this which have no nontrivial finite quotients. However, an easy ping-pong argument shows the following:

Lemma. There are no simple discrete infinite subgroups $\Gamma$ in rank 1 Lie groups.

Proof. The statement is clear if $\Gamma$ is free, so suppose that it is not. Take a high power of a hyperbolic element in $\Gamma$, then its normal closure in $\Gamma$ will be free and, hence, a proper normal subgroup of $\Gamma$. If $\Gamma$ has no hyperbolic elements, then it is virtually nilpotent and, hence, not simple.

This argument, however, will fail miserably for discrete subgroups in Lie groups of rank $\ge 2$ since the latter contain irreducible lattices $\Gamma$. Such lattices (by Margulis' normal subgroups theorem) contain no infinite normal subgroups of infinite index. Of course, Margulis' argument will not work for infinitely generated groups, but the point is that any proof of non-simplicity of discrete subgroups $\Gamma$ of higher rank Lie groups would have to use the fact that $\Gamma$ is not a lattice. (The only way to exploit this fact that I can see at this point is that an infinitely generated group $\Gamma$ does not have Property (T), but it is unclear to me how to use this even to get started on proving non-simplicity.)

So, maybe there are simple infinite discrete subgroups of higher rank Lie groups $G$. The problem is that there are not that many ways to construct discrete infinitely generated subgroups in such $G$'s. I know only of two reasonably general constructions:

(1). Some form of ping-pong (used, for instance, by Margulis and Soifer to construct maximal subgroups of matrix groups). However, the resulting groups will be free products, so never simple.

Addendum: I am not claiming here that maximal subgroups $H$ (in a f.g. group $G$) constructed by Margulis and Soifer are not free products, only that the key step in the construction of $H$ is a free product $F$ (actually, a free group in the main paper of Margulis and Soifer). Then the subgroup $H\subset G$ is a maximal proper subgroup of $G$ containing $F$: This last step of the proof is nothing by Zorn's lemma and does not contain any constructive information. (All what we know about $H$ is that it is maximal and has infinite index in $G$.) In particular, $H$ itself is useless as far as constructing simple subgroups of matrix groups. Incidentally, it is an open problem, going back to Prasad and Tits: Does, say, $SL(n, {\mathbb Z}), n\ge 3$, contain a maximal subgroup which is not (virtually) a nontrivial free product. Margulis and Soifer prove that, for $n\ge 4$, one can get a maximal subgroup $H$ containing a free product of ${\mathbb Z}^2$ and of a free group.

(2). Vinberg's theorem on discreteness of groups generated by linear reflections in faces of convex "polyhedral" cones (such cones would have to have infinitely many faces in our situation, so polyhedrality has to be taken with the grain of salt). Of course, Coxeter groups one obtains as the result are not simple, but the only obvious normal subgroup is of index 2, so maybe one can obtain a virtually simple group (i.e., a group containing a simple subgroup of finite index) in this fashion. Could Coxeter groups with finite-dimensional Davis complex be virtually simple? Hard to imagine, but, at this point, this seems to be the best hope for an example of a discrete infinite simple matrix group.

Update: Actually, one can show that (infinite) discrete Coxeter subgroups of Lie groups are never virtually simple. Even more, any discrete subgroup $\Gamma$ of $SL(n, {\mathbb R})$, which contains an abstract rank 1 diagonalizable element, is never virtually simple (same proof as in the case of rank 1 Lie groups works). Here abstract rank 1 element of an (abstract) group $\Gamma$ is an infinite order element $\gamma$ so that for any $k\ne 0$ the centralizer of $\gamma^k$ in $\Gamma$ is virtually cyclic with uniform bound in "virtually." So, maybe one can show that all discrete Zariski dense subgroups of $SL(n, {\mathbb R})$ without abstract rank 1 elements, are lattices by running some Margulis-type argument...

• Why does a discrete subgroup contain a lattice? Mar 2, 2012 at 17:24
• Maybe it would be reasonable to start with the following a priori easier question: find a discrete subgroup $G$ of $SL(n,\mathbb{R})$ and an infinite subgroup $H$ of $G$, so that $H$ is contained in the normal closure in $G$ of any of its nontrivial elements. [Clearly, no residually finite $G$ has such a subgroup].
– YCor
Mar 2, 2012 at 18:19
• Igor, a discrete subgroup need not contain a lattice. However, any proof of non-simplicity would have to deal with Margulis-type phenomenon. For instance, one cannot give a "proof" of the following type: Take a very proximal (i.e., both $\gamma^{\pm 1}$ have maximal eigenvalue of multiplicity 1) element $\gamma$ of a discrete group $\Gamma$. Then take the normal closure $N$ in $\Gamma$ of $\gamma^m$ for large $m$. Then $N$ is a proper normal subgroup of $\Gamma$. This argument works nicely in rank 1 case but appears to fail in higher rank because of lattices and non-RF subgroups. Mar 2, 2012 at 18:31

This is not an answer; but Margulis and Soifer construct maximal subgroups of lattices $\Gamma$ . Contrary to what Misha says, they are not free products (nor necessarily free groups), but they are maximal subgroups which CONTAIN free subgroups which map onto every finite quotient of the lattice $\Gamma$. It is entirely possible that maximal subgroups are simple (I do not know a proof, one way or another). Aakumadula
• @Aakumadula: If you read what I said in my question, I am not (and was not) claiming that maximal subgroups $H$ constructed by Margulis and Soifer, are free. Incidentally, as far as I know, pretty much nothing is known about nature of the groups $H$, even in the rank 1 case. (Except, as Yves observes, they are RF and are not simple.) For instance, if you start with $G$, a torsion-free uniform lattice in $O(n,1)$, $n\ge 3$, it is unknown if one can find a non-free maximal subgroup $H\subset G$. For non-uniform $G$, one can find nonfree $H$ as Margulis-Soifer do in their "Uspekhi" paper. May 27, 2012 at 14:28
• Oops: In the case of $O(n,1)$ we now also know that there are maximal infinite index subgroups $H$ in an arbitrary uniform lattice, so that $H$ contains a closed surface subgroup (by combining Kahn-Markovic results with Margulis-Soifer arguments). So, the open question is: Find maximal infinite index subgroups which are not free products of surface groups and free groups. May 27, 2012 at 14:55
• I don't understand @Misha's comment: are you talking just about $O(n, 1)$? For higher rank, Margulis-Soifer construct maximal subgroups which contain higher rank abelian subgroups, so are not free products of surface and free groups. Jun 22, 2013 at 16:45