# free subgroups of $SL_2(\mathbb{Z[i]})$

The group $SL_2(\mathbb{Z})$ contains many free subgroups, for example all of the principal congruence subgroups for $n\geq 3$ and the subgroup $\left\langle \left(\begin{array}{cc} 1 & 2\\ 0 & 1 \end{array}\right),\left(\begin{array}{cc} 1 & 0\\ 2 & 1 \end{array}\right)\right\rangle$ which is almost the principal congruence subgroup for n=2, and more over it has a finite index in $SL_2(\mathbb{Z})$. In addition, when projecting $SL_2(\mathbb{Z})$ onto $SL_2(\mathbb{Z}/p)$, the restriction to these subgroups is also surjective as long as $p \nmid n$.

Is a similar phenomena true for $SL_2(\mathbb{Z}[i])$, or more generally when we substitute $\mathbb{Z}$ with the ring of integers $\mathcal{O}_k$ for some number field $k$?

EDIT: I look for free subgroups which are hopefully finite index (for example the congruence subgroups). Since finite index means they are a lattice, this enables me to count the number of lattice points in a ball of radius R asymptotically as R goes to infinity. If this is not possible, I would like to find free subgroups which are large in the sense that they have large growth.

• You should be more specific, the answer depends on the exact meaning of "phenomena". Are you asking for existence of free subgroups? Subgroups generated by powers of two given unipotent elements not generating a solvable subgroup? Then yes. Are you asking for free subgroups of finite index? Then no (ever). – Misha Sep 12 '16 at 10:14
• A possible interpretation of the question is whether every lattice contains a profinitly dense free group. Then the answer should be yes, but I am not sure what is the right refernce (Margulis-Soifer?). – Uri Bader Sep 12 '16 at 10:19
• @Misha I was hoping that the principal congruence subgroups in $SL_2(\mathbb{Z}[i])$ would be free, or at least virtually free. Why aren't there any finite index free subgroups? – Ofir Sep 12 '16 at 10:39
• @Prometheus: One way to see this is to observe that $SL(2, Z[i])$ contains free abelian subgroups of rank 2. – Misha Sep 12 '16 at 11:40
• As Misha and Uri Bader have observed, you cannot get free subgroups of finite index. If you view $SL_2(\mathbb{Z}[i])$ as a subgroup of $SL_2(\mathbb{C})$ the latter sen as a REAL group, then free subgroups of infinite index which are Zariski dense in the real group do exist (again, by Margulis Soifer, as referenced by Uri Bader, if you like). Then a result of Nori and Weisfeiler says that these free subgroups surject onto $SL_2(O_K/p)$ for any non-zero prime ideal in $O_K$ where $O_K$ is the ring of integers in the imaginary quadratic $\mathbb{Q}(i)$. – Venkataramana Sep 12 '16 at 12:50