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3
votes
Accepted
The principal congruence subgroup of the symplectic group over the integers
If we assume that $g \geq 2$, then it is known by a Theorem of Tits
( Tits, Jacques :
Systèmes générateurs de groupes de congruence. (French. English summary)
C. R. Acad. Sci. Paris Sér. A-B 283 (1976 …
12
votes
Subgroups of Sp(2g,Z) that map onto all Sp(2g,Z/m)
The answer to your question as to whether $G=Sp_{2g}({\mathbb Z})$ is NO. $Sp_{2g}({\mathbb Z})$ contains a pro-finitely dense FREE subgroup (hence has infinite index) which is also finitely generat …
18
votes
Accepted
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. S …