Free subgroups of the Hilbert modular group

Question

Let $F$ be a real quadratic field with ring of integers $\mathfrak{o}_F$. What is known about free subgroups of the Hilbert modular group $\mathrm{SL}_2(\mathfrak{o}_F)$? Does it contain free subgroups that are not cyclic and are not contained in $\mathrm{SL}_2(\mathbb{Z})$? For example, assuming $2$ is innert in $F$, is the principal congruence subgroup of level $2$ ever free?

Answer 1

I guess you'll modify the question again, but clearly it contains free subgroups not contained in $\mathrm{SL}_2(\mathbf{Z})$, namely conjugates of free subgroups of $\mathrm{SL}_2(\mathbf{Z})$.

No subgroup of finite index is free: indeed each $\Gamma$ as in the question is an irreducible lattice in $\mathrm{SL}_2(\mathbf{R})$ (use the two embeddings $F\to\mathbf{R}$) and such a group is known to be just-infinite by work of Margulis: every normal subgroup, other than $\{1\}$ and $\{\pm 1\}$ has finite index. An easier way to prove this is to use that $\Gamma$ has an infinite solvable subgroup of exponential growth (the subgroup of upper triangular matrices in $\Gamma$: use Dirichlet's unit theorem to prove this).

$\Gamma$ also contains finitely generated free subgroups that are Zariski-dense (in $\mathrm{SL}_2(\mathbf{R})\times\mathrm{SL}_2(\mathbf{R})$) and hence not virtually conjugate into $\mathrm{SL}_2(\mathbf{Z})$. I think this follows from results of Margulis-Soifer.