# Are there free and discrete subgroups of $\mathrm{SL}_2(\mathbb{R}) \times \mathrm{SL}_2(\mathbb{R})$ that are not Schottky on any factor?

$$\DeclareMathOperator\SL{SL}$$Does there exist a free and discrete subgroup $$\Gamma < \SL_2(\mathbb{R}) \times \SL_2(\mathbb{R})$$ such that neither $$\pi_1(\Gamma)$$ nor $$\pi_2(\Gamma)$$ is free and discrete (where, as you may guess, $$\pi_{1,2}: \SL_2(\mathbb{R}) \times \SL_2(\mathbb{R}) \to \SL_2(\mathbb{R})$$ denote the canonical projections onto each of the factors)?

First, let me say a word about where this is coming from. There is the following long-standing problem, which has been haunting me for years: for which values of $$\lambda$$ is the group $$\Gamma_\lambda$$ generated by $$(\begin{smallmatrix} 1 & \lambda \\ 0 & 1 \end{smallmatrix})$$ and $$(\begin{smallmatrix} 1 & 0 \\ 1 & 1 \end{smallmatrix})$$ free? It has been originally asked for rational values of $$\lambda$$.

A natural approach to try to answer it is to then use the fact that such a group $$\Gamma_\lambda$$ has a discrete embedding into the product of $$\SL_2(\mathbb{R})$$ with all of the $$\SL_2(\mathbb{Q}_p)$$'s for which $$p$$ divides the denominator of $$\lambda$$.

This product seems however quite hard to study. It might be easier to first study the case where $$\lambda$$ is an algebraic integer (say quadratic), in which case $$\Gamma_\lambda$$ "merely" embeds into $$\SL_2(\mathbb{R}) \times \SL_2(\mathbb{R})$$. (If nothing else, real numbers are, at least, more familiar than $$p$$-adics...) Of course if one of its projections is Schottky (which happens iff either $$\lambda$$ or its conjugate has absolute value $$\geq 4$$), then $$\Gamma_\lambda$$ is automatically Schottky. But now suppose that neither of the projections is Schottky. Then can we conclude anything at all?

Also, a remark: if you drop the "free" assumption, then an example is very easy to produce. Just take $$\SL_2(\mathbb{Z}[\sqrt{2}])$$, with its embedding into $$\SL_2(\mathbb{R}) \times \SL_2(\mathbb{R})$$ given by the two embeddings of $$\mathbb{Z}[\sqrt{2}]$$ into $$\mathbb{R}$$. It is discrete (but not free), but both of its projections are dense.

• Is it clear that there are no subgroups of Hilbert modular groups with this property?
– R W
Jul 3 at 23:02
• Well, no - in fact nothing is clear to me here so far :-) Is there any recipe for constructing free subgroups of such groups, other than having them play ping-pong on one of the factors? Jul 4 at 9:06
• @IliaSmilga: Google suggests you're currently at Oxford. I bet Emmanuel Breuillard can answer this.
– HJRW
Jul 18 at 9:51
• @HJRW: I finally got around to asking him. He does not know the answer, but pointed me to a paper that is very much relevant: arxiv.org/pdf/2202.04027.pdf . (There is also a follow-up now: arxiv.org/pdf/2308.07785.pdf ). Oct 11 at 13:43