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Let $D = \text{diag}(1,d)\in M_{2}(\mathbb{Z})$ be a $2\times 2$ matrix, where $d$ is an odd integer. We define the subgroup $\Gamma_D\subset M_{4}(\mathbb{Z})$ as: $$\Gamma_D := \left\lbrace R\in M_{4}(\mathbb{Z}) \: | \: R \left(\begin{matrix} 0 & D \\ -D & 0 \end{matrix}\right) R^t = \left(\begin{matrix} 0 & D \\ -D & 0 \end{matrix}\right) \right\rbrace,$$ and the subgroup $\Gamma_D(D)\subset\Gamma_D$ as: $$\Gamma_2(1,d):= \left\lbrace \left(\begin{matrix} a & b \\ c & d \end{matrix}\right)\in \Gamma_D \: | \: a-Id_2 \equiv b \equiv c \equiv d-Id_2 \equiv 0 \: mod(D) \right\rbrace,$$ where $A\equiv 0\: mod(D)$ means that $A = D\cdot B$ for some $B\in M_2(\mathbb{Z})$.

Does there exists an exact sequence

$$1 \to \Gamma_2(2,2d) \to \Gamma_2(1,d) \to Sp(4,\mathbb{Z}/2\mathbb{Z}) \to 1,$$

where the last map is the reduction $mod\ 2$ of the entries of the matrices? In order to prove this I wanted to check the index of the level subgroups inside $\Gamma_2$. I know that there exists a formula by Igusa for the level $n$ subgoups, and I wonder if also these indexes are known, or at least easily computable.

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  • $\begingroup$ What is the definition of $\Gamma_g(1,d)$? $\endgroup$ Commented Jun 17, 2014 at 21:00
  • $\begingroup$ I added the precise definition of $\Gamma_2(1,d)$ in the question. $\endgroup$
    – Puzzled
    Commented Jun 17, 2014 at 21:57
  • $\begingroup$ The strong approximation theorem shows that $Sp_{2n}(\mathbf Z)$ projects onto all quotients $Sp_{2n}(\mathbf Z/m)$. This implies your mod 2 reduction is onto (if I understand the question). $\endgroup$
    – few_reps
    Commented Jun 17, 2014 at 22:01
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    $\begingroup$ (Now that I've seen the definition, I'm less sure ... but at least it should be a good point to start) $\endgroup$
    – few_reps
    Commented Jun 17, 2014 at 22:07
  • $\begingroup$ What does mod (D) mean? $\endgroup$ Commented Jun 18, 2014 at 2:48

1 Answer 1

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Yes, there is a short exact sequence

$$ 1 \to \Gamma_2(2,2d) \to \Gamma_2(1,d) \to Sp(4,\mathbb{Z}/2\mathbb{Z}) \to 1. $$

One can obtain this sequence as follows. Consider the group scheme $G := Sp(f)$ defined over $\mathbb{Z}$ where $f$ is the symplectic form defined above, i.e. $f(u,v) = u^t\begin{pmatrix} 0 & D \\ -D & 0 \end{pmatrix}v$.

The group of $\mathbb{Z}$-points is $G(\mathbb{Z}) = \Gamma_D$, this is, the subgroup of matrices in $GL_4(\mathbb{Z})$ which preserve the symplectic form. As a group scheme over $\mathbb{Q}$ the group $G$ is just the symplectic group. However, over $\mathbb{Z}$ they are not isomorphic. Fortunately, since $d$ is odd, we have an isomorphism $G \times \mathbb{Z}_2 \cong Sp(4)\times \mathbb{Z}_2$ over the ring of 2-adic integers. In particular, $G(\mathbb{Z}_2)$ surjects onto $Sp(4,\mathbb{Z}/2\mathbb{Z})$.

So it suffices to verify that $\Gamma_2(1,d)$ is dense in $G(\mathbb{Z}_2)$, which follows from strong approximation. (For instance, the principal congruence subgroup $\Gamma_D(d) \subset \Gamma_2(1,d)$ is dense in $G(\mathbb{Z}_2)$ -- since $d$ is odd we obtain this group by applying a congruence condition away from $2$).

The same argument applies to all primes away from $d$, so at these primes you may also compute the index of congruence subgroups just as in the symplectic group.

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  • $\begingroup$ Very nice proof. As far as I understand the same proof would not hold for even $d$, right? $\endgroup$
    – IMeasy
    Commented Jun 30, 2014 at 20:04
  • $\begingroup$ @IMeasy : The argument works at every prime where the symplectic form $f$ is regular (since in this case it is isomorphic to the standard symplectic form). So if $d$ is even, the argument fails for the prime $2$. $\endgroup$ Commented Jul 1, 2014 at 8:58

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