Let $G\subset \mathrm{SL}\_2(\mathbb R)$ be a subgroup such that $\mathrm{SL}\_2(\mathbb Z)\subset G$.


What are the possible groups such that $\mathrm{SL}\_2(\mathbb Z)\subset G$ is of finite index? Is $G=\mathrm{SL}\_2(\mathbb Z)$ the only possibility?


What if we replace $\mathrm{SL}\_2$ by $\mathrm{Sp}\_{2n}$? 


**Question.** Suppose that $\mathrm{Sp}\_{2n}(\mathbb Z)\subset G$ is of finite index and $G\subset \mathrm{Sp}\_{2n}(\mathbb{R})$. Is $G= \mathrm{Sp}\_{2n}(\mathbb{Z})$?



Subgroups of $SL_2(\mathbb R)$ which contain $SL_2(\mathbb Z)$ as a finite index subgroup
Even for $Sp\_{2g}$ the only possibility is $Sp\_{2g}(\mathbb{Z})$. To see this, suppose $\Gamma \subset Sp\_{2g}(\mathbb{R})$ is a subgroup containing $Sp\_{2g}(\mathbb{Z})$ as a finite index subgroup. Suppose we prove that $\Gamma $ consists of rational symplectic matrices. The fact that $Sp\_{2g}(\mathbb{Z}\_p)$ is a $maximal$ compact subgroup of $Sp\_{2g}(\mathbb{Q}\_p)$ for all primes $p$ (and strong approximation) implies that $Sp\_{2g}(\mathbb{Z})$ is a maximal arithmetic subgroup of $Sp\_{2g}(\mathbb{Q})$ (edited: if $\Gamma \subset Sp\_{2g}(\mathbb{Q})$ contains $Sp\_{2g}(\mathbb{Z})$ as a finite index subgroup, then $\Gamma $ is the integral symplectic group ) and hence this implies that $\Gamma =Sp\_{2g}(\mathbb{Z})$. 


Thus it remains to prove that $\Gamma $ consists of rational matrices. This is essentially a Galois cohomology argument. Since $\Gamma$ contains $Sp\_{2g}(\mathbb{Z})$ as a finite index subgroup, in particular, there is a finite index subgroup $N\subset Sp\_{2g}(\mathbb{Z})$ $normalised$ by $\Gamma$. $N$ acts irreducibly on $\mathbb{C}^{2g}$ (for example, snce $N$ is Zariski dense in the complex symplectic group). Thus, by Burnside's theorem, there is a basis $e\_1,\cdots,e\_N$ of $2g\times 2g$ rational matrices consisting of elements of $N$. 


Suppose $x\in \Gamma \subset Sp\_{2g}(\mathbb{R})\subset Sp\_{2g}(\mathbb{C})$. Under conjugation by $x$, the basis $e\_i$ goes into elements of $N \subset Sp\_{2g}(\mathbb{Z})$ and hence $xe\_ix^{-1}$ is a rational linear combination of the matrices $e\_i$. If $\sigma \in Aut(\mathbb{C})$ then $\sigma (x)e\_i\sigma (x)^{-1}=xe\_ix^{-1}$. The irreducibility of $N$ now implies that the cocycle $x^{-1}\sigma (x)$ is a scalar matrix $\lambda$. Since $x$ preserves the symplectic form, this implies that $\lambda ^2=1$, and hence $\lambda =\pm 1$. If $\lambda =1$ for all $\sigma$ then $x$ is rational, and that is what we set out to prove. 


If the Galois cocycle $\lambda =\lambda \_{\sigma}= -1$, then we may write $\lambda =i^{-1}\sigma (i)$ (where $i^2=-1$) and hence $h=i^{-1}x$ is a rational matrix: $h^{-1}\sigma (h)=1$ for all $\sigma$. But, $\Gamma $ consists of real matrices, and hence $x$ is a real matrix, and hence, the equation $x=ih$ for some rational matrix $h$ is impossible. Thus, $\lambda =-1$ is ruled out.


So essentially, the fact that over the $p$-adic field $\mathbb{Q}\_p$, the group $Sp\_{2g}(\mathbb{Z}\_p)$ is maximal compact implies that $Sp\_{2g}(\mathbb{Z})$ is a subgroup of $Sp\_{2g}(\mathbb{R}$ such that every intermediate subgroup contains $Sp\_{2g}(\mathbb{Z})$ as an infinite index subgroup. 


