$\DeclareMathOperator\SL{SL}$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.

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    $\begingroup$ Is it clear that there are no subgroups of Hilbert modular groups with this property? $\endgroup$
    – R W
    Jul 3 at 23:02
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    $\begingroup$ 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? $\endgroup$ Jul 4 at 9:06
  • $\begingroup$ @IliaSmilga: Google suggests you're currently at Oxford. I bet Emmanuel Breuillard can answer this. $\endgroup$
    – HJRW
    Jul 18 at 9:51
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    $\begingroup$ @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 ). $\endgroup$ Oct 11 at 13:43


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