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?

  • $\begingroup$ I'm also interested in the case of lattices in $\mathrm{SL}_2(\mathbb{R})$. $\endgroup$ Dec 17, 2013 at 3:13
  • $\begingroup$ 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. $\endgroup$ Dec 17, 2013 at 4:01
  • $\begingroup$ (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). $\endgroup$ Dec 17, 2013 at 4:02
  • $\begingroup$ 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. $\endgroup$
    – YCor
    Dec 17, 2013 at 11:22

1 Answer 1


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.

  • 3
    $\begingroup$ 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. $\endgroup$ Dec 17, 2013 at 8:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.