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Consider the symplectic group $\text{Sp}_{2g}(\mathbb{Z})$ over the integers. It has a classical root system $C_g$ and associated root subgroups $U_\varphi$ for $\varphi\in C_g$. These subgroups are defined by the Bourbaki tables.

Denote by $2U_\varphi$ the set of matrices $A^2$ where $A\in U_\varphi$. The symplectic group also has a maximal torus $T$, consisting of the diagonal matrices in $\text{Sp}_{2g}(\mathbb{Z})$.

The kernel $\Gamma(2)$ of the canoncical reduction map $\text{Sp}_{2g}(\mathbb{Z})\to\text{Sp}_{2g}(\mathbb{Z}/2\mathbb{Z})$ is the principal congruence subgroup of level $2$.

The question

I want to show that $\Gamma(2)$ is the group generated by $T$ and the $2U_\varphi$. The inclusion $\langle T,2U_\varphi:\varphi\in C_g\rangle\subset \Gamma(2)$ is trivial, by definition of the root subgroups and because the only nonzero entries of any matrix in $T$ are $1$ and $-1$. How do I prove the other inclusion?

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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), no. 9, Ai, A693–A695)
that the group $\Delta$ generated by $T$ and $2U_{\phi}$ (your notation) has finite index in $Sp_{2g}(\mathbb Z)$. You can now use the fact that $Sp_{2g}(\mathbb Z)$ has the congruence subgroup property, to conclude that $\Delta$ is a congruence subgroup; it is easily seen that its closure in the congruence completion $Sp_{2g}(\widehat{\mathbb Z})$ of $Sp_{2g}(\mathbb Z)$ contains $\prod Sp_{2g}(\mathbb Z_p)$, where the product is over all odd primes. At the prime 2, the group $\Delta$ and $\Gamma (2)$ both have the same closure; we then conclude that $\Gamma (2)=\Delta$ (if two congruence subgroups have the same congruence closure, then they are the same).

When $g=1$, this is a well known result (and is actually proved in Ahlfors' book on complex analysis; the group $\Gamma (2)$ modulo centre is interpreted there as the fundamental group of ${\mathbb P}^1\setminus \{0,1,\infty \}$).

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  • $\begingroup$ Thank you so much! It seems like it is exactly what I need. However, I don't completely your following argument. If we denote by $\overline{\Delta}$ the closure of $\Delta$ in the congruence completion, you claim that $\prod \text{Sp}_{2g}(\mathbb{Z}_p)\subset \overline{\Delta}$ where the product is over the odd primes. What do then mean with "At the prime $2$, the groups $\Delta$ and $\Gamma(2)$ have the same closure"? $\endgroup$
    – user341790
    Oct 29, 2021 at 7:55
  • $\begingroup$ Thanks again. I think I need to understand the congruence completion and congruence closure better. Do you have any useful references on this, by any chance? $\endgroup$
    – user341790
    Oct 29, 2021 at 11:02
  • $\begingroup$ I have found some references of my own, but more are always welcome of course. Is the following correct? To show that the congruence closures of $\Delta$ and $\Gamma(2)$ are equal, it is enough to show that the closures of their images under the maps $\text{Sp}_{2g}(\mathbb{Z})\to\text{Sp}_{2g}(\mathbb{Z}_p)$ for every prime $p$ are equal. $\endgroup$
    – user341790
    Oct 29, 2021 at 12:42
  • $\begingroup$ I see. Do you have a good reference to prove all this? I would really appreciate it! $\endgroup$
    – user341790
    Oct 29, 2021 at 13:22
  • $\begingroup$ Indeed. But don't I also need to know what the closure is of $T$? And how do I prove that $\Delta$ and $\Gamma(2)$ have the same closure in $\text{Sp}_{2g}(\mathbb{Z}_2)$? $\endgroup$
    – user341790
    Oct 29, 2021 at 14:34

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