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It is a theorem of Selberg that a lattice $\Gamma$ in a linear group has a torsion-free subgroup of finite index. Page 64 in 'Introduction to Arithmetic Groups' by Dave Morris asserts these can be realized as follows. Since $\Gamma$ is finitely generated, $\Gamma \subseteq \mathrm{SL}_k(R)$ where $R = \mathbb{Z}[a_1, \ldots,a_n]$ for some $a_1, \ldots, a_n \in \mathbb{C}$. If $\mathfrak{m}$ is a maximal ideal in $R$ then the kernel of reduction modulo $\mathfrak{m}$ is torsion-free and has finite index in $\Gamma$ since $\mathrm{SL}_k(R/\mathfrak{m})$ is finite. For the purposes of this question let $k=2$.

In the case $\Gamma = \mathrm{SL}_2(\mathbb{Z})$ doing this gives the principal congruence subgroups, which are actually free for levels $\geq 3$. So my question is, when do lattices in $\mathrm{SL}_2(\mathbb{C})$ have a free subgroup of finite index? Relatedly, for which commutative rings $R$ does $\mathrm{SL}_2(R)$ have a free subgroup of finite index?

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I'm also interested in the case of lattices in $\mathrm{SL}_2(\mathbb{R})$. – burtonpeterj Dec 17 '13 at 3:13
OK, if you're interested in $\text{SL}_2(\mathbb{R})$, then I'll repost my deleted comment. A lattice $\Gamma$ in $\text{SL}_2(\mathbb{R})$ has a free subgroup of finite index if and only if $\Gamma$ is nonuniform (that is, when $\mathbb{H}^2/\Gamma$ is not compact). Indeed, by passing to a finite-index subgroup of $\Gamma$, we can assume that $\Gamma$ is torsion-free, which implies that $\pi_1(\mathbb{H}^2/\Gamma) \cong \Gamma$. If $\mathbb{H}^2 / \Gamma$ is noncompact, then $\mathbb{H}^2/\Gamma$ is a noncompact surface and thus has a free fundamental group. – Andy Putman Dec 17 '13 at 4:01
(continued) However, if $\mathbb{H}^2 / \Gamma$ is compact, then $\Gamma$ is the fundamental group of a closed surface. Finite covers of closed surfaces are closed surfaces, so in this case all finite-index subgroups of $\Gamma$ are themselves surface group (so not free). – Andy Putman Dec 17 '13 at 4:02
For your last question: let $R$ be a commutative ring. If $SL_2(R)$ is virtually free then so is its its upper unipotent subgroup, which is the abelian group $(R,+)$. Hence either $R$ is finite, or is, as an additive group, isomorphic to $\mathbf{F}\oplus$(finite). In the latter case, $R$ has a unique maximal finite ideal $M$ such that $R/M=\mathbf{Z}$ as a ring. Conversely for such a ring, the projection to $SL_2(R/M)=SL_2(\mathbf{Z})$ is onto with finite kernel, which implies that $SL_2(R)$ is virtually free. – YCor Dec 17 '13 at 11:22
up vote 3 down vote accepted

A co-compact lattice in $SL_2({\mathbb C})$ cannot contain a finite index free subgroup because of cohomological reasons. The cohomological dimension of the finite index torsion free subgroup is 3. \vskip 5mm

A non-cocompact lattice $\Gamma $ in $SL_2({\mathbb C})$, after a conjugation, intersects the upper triangular unipotent matrices in a lattice; in particular, it contains a free abelian subgroup of rank two; therefore, $\Gamma $ cannot contain a finite index free subgroup.

For the same reason, a finite index subgroup of $SL_2(R)$ , for a commutative ring $R\subset {\mathbb C}$, contains the upper triangular unipotent matrices of the form $$\begin{pmatrix} 1 & Nx \cr 0 & 1\end{pmatrix}$$ where $N$ is a non-zero integer and $x\in R$. So, unless $R={\mathbb Z}$, the group $SL_2(R)$ cannot contain a finite index free subgroup.

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The virtual cohomological dimension of a non-cocompact lattice in $\mathrm{SL}_2(\mathbb{C})$ is 2, hence the argument for uniform lattices also works for nonuniform ones. Alternalively, one could unse the fact that the Euler characteristic of any finite-volume hyperbolic three--manifold is zero, while that of a nonabelian free group is negative. – Jean Raimbault Dec 17 '13 at 8:44

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