Non-residually finite matrix groups

Question

By Malcev's theorem, every finitely generated linear group is residually finite (RF). On the other hand, say, the group of rational numbers is linear, but is not residually finite. Thus, one has to impose some restrictions on linear groups to avoid this example. Discreteness sounds like a reasonable hypothesis. It is a nice and easy exercise to show that every discrete subgroup of $PSL(2, {\mathbb R})$ is residually finite.

Question 1. (This question came from Fanny Kassel) Are there non-RF discrete subgroups of $PSL(2, {\mathbb C})$?

On the other hand, almost surely, there are no simple discrete infinite subgroups $\Gamma$ of rank 1 Lie groups. (Take a high power of a hyperbolic element in $\Gamma$, then its normal closure in $\Gamma$ should have infinite index in $\Gamma$.) This argument fails however in the higher rank case because of the Margulis' normal subgroups theorem for higher rank lattices.

Question 2. Are there infinite simple discrete subgroups of $SL(n, {\mathbb R})$?

It is hard to imagine that such things could exist, but I see no way to rule them out...

Answer 1

The answer to Question 1 is yes.

For each prime $p>2$, take the von Dyck group $D(p,p,\infty)$, generated by two rotations of order $p$ whose product is a parabolic $q_p$. These groups lie in $PSL(2,\mathbb{R})< PSL(2,\mathbb{C})$. We may normalize the parabolic element in each of these to be $q_p: z\mapsto z+1$. For each $p$, we may conjugate $D(p,p,\infty)$ by a parabolic element $\alpha_p: z\mapsto z+ir(p)$, where $r(p)>0$ is some sequence chosen to grow fast enough so that the resulting group $\Gamma=\langle \alpha_p^{-1} D(p,p,\infty) \alpha_p, p\ prime \rangle$ is isomorphic to the abstract infinite amalgamated product $D(3,3,\infty) \ast_{q_3=q_5} D(5,5,\infty) \ast_{q_5=q_7} D(7,7,\infty) \ast \cdots$. This is proved using the Klein combination theorem.

To see that $\Gamma$ is not residually finite, we see that $q_3$ is contained in the kernel for any homomorphism $\varphi:\Gamma \to K$, where $K$ is finite. Choose $p> |K|$, then $\varphi_{|D(p,p,\infty)}=Id$, since any element of order $p$ must be sent to the identity. So $q_p$ must lie in the kernel, but $q_p=q_3$ from the amalgamated product structure, so $\varphi(q_3)=1$.

With a bit more care (such as choosing $D(p_i,q_i,\infty)$ where $p_i,q_i$ are coprime primes $\to \infty$ as $i\to \infty$), one may guarantee that the group has no finite quotients.

Answer 2

The answer to question 2 is no [edit :] when the group is finitely generated. It follows from Malceev's result. In the case of finitely presented, here is a simple proof.

Proof : Let $G$ be a group defined by a finite number $n$ of generators and a finite number of relations.

Let $k$ be a field. Pick an element $g$ in $G$.

The set of morphisms $\varphi:G\to GL_m(k)$ such that $\varphi(g)\ne 1$ can be seen as an algebraic subvariety of $M_m(k)^n$. In fact this variety is defined over $\mathbf Z$.

So if this variety admits a point over a field $k$ of characteristic $0$, then it has to have points over infinitely many finite fields.

Thus if the group is simple and infinite, the variety is empty.