# Reference on generators of subgroups of symplectic groups

We should start with the definition of the symplectic group for an arbitrary ring $R$. The symplectic group $Sp(g,R)$ is the subgroup of $SL(2g,R)$ such that all elements satisfy $M=J_g^t M J_g$ with $J_g$ being the canonical almost complex structure - or involution whatever you prefer to call it.

A generic element of $Sp(g,R)$ is denoted by $M=(A B \\ C D)$ where $A,B,C,D$ are $g\times g$ matrices.

Note bene, if $R$ is Euclidean then $Sp(g,R)$ is generated by the involution $J_g$ and the translations $\begin{pmatrix}I_g\ S \\ 0_g \ I_g \end{pmatrix}$ sending $Z$ to $Z+S$ where $S$ is a symmetric $g \times g$ matrix.

## Why am I interested in these generators ?

Well, first of all I am interested in modular forms. They are holomorphic functions $f:\mathbb{H}_g \to V$ transforming under a subgroup $\Gamma$ of a symplectic group as follows $$f(M \cdot Z)=j(M,Z)\cdot f(Z)\quad \quad \forall\ M \in \Gamma ,$$ where $j$ is a factor of automorphy. This means that $j: \Gamma \times \mathbb{H}_g \to GL(V)$ is holomorphic in the second variable and satisfies the cocycle relation $$j(MN,Z)=j(M,N \cdot Z) \cdot j(N,Z).$$

Hence, it suffices to check the first equation only for the generators of $\Gamma$.

Sometimes we have modular forms to a proper subgroup and even know how they transform under the full symplectic group. But they do not transform with a factor of automorphy. Examples would be the theta series $$f_a(Z):=\sum_{\nu \in \mathbb{Z}^g}{exp \left(2\pi i \left(\nu+\frac{a}{2}\right)^t Z \left(\nu+\frac{a}{2}\right) \right)}, \quad \quad a \in \mathbb{F}_2^g$$ They are modular forms to certain proper subgroups. But the action of $Sp(g,\mathbb{Z})$'s generators can be given quite easily, roughly speaking : the translations scale the thetas and involution returns a linear combination of all thetas. That way, it is possible to determine whether a polynomial in the different theta series is a modular form to the full modular group.

## The actual question

I would be very pleased if someone could give a reference for the generators of subgroups of $Sp(g,\mathbb{Z})=\Gamma_g$ like $$\Gamma_g[q]:= ker\left(Sp(g,\mathbb{Z})\to Sp(g,\mathbb{Z}/q\mathbb{Z})\right)$$ $$\Gamma_{g,0}[N]:=\left\{ M \in \Gamma_g : C \equiv 0 \mod N \right\}$$ $$\Gamma_{g}^{0}[N]:=\left\{ M \in \Gamma_g : B \equiv 0 \mod N \right\}$$ and others if you know them, too. In particular, I am interested in $g=2$. I guess this way it is faster and I cannot make any mistakes. As hinted above it would be also nice if these generators could be given in terms of the generators of $Sp(g,\mathbb{Z})$.

Thanks Tom

p.s. it would be nice if someone could help me fixing the brackets in the above definition of $\Gamma_{g,0}[N]$.

edit1 : added notation $M=(A B \\ C D)$

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Have you looked at Igusa's book "Theta functions" or Mumford's "Tata Lectures on theta I"? Plus I would also say that $Sp_{2g}R$ is generated by transvections (as you've described) and GL_g(R) via diagonals (A 0 \\ 0 {}^tA^{-1}). Just $J$ and transvections are not enough. – J. Martel Mar 5 '13 at 16:53
Maybe you could also clarify that you're considering M=(A B \\ C D). – J. Martel Mar 5 '13 at 16:56
Dear J, as you proposed I just clarified the notation $M=(A B \\ C D)$. With the few generators I was tkinking of E. Freitag 'Siegelsche Modulformen' (in Springer's Comprehensive Studies 254 ) appendix V pages 322-328. The proof relies on the fact that he finds for an EUCLIDEAN ring 'smaller' or 'easier to handle' sets of generators for $SL(g,R)$ and $GL(g,R)$. Now I'm having a look in the 2 books you mentioned. – Tom Mar 5 '13 at 17:22
Indeed, Mumford gives generators for $\Gamma_g$,$\Gamma_g[2]$ and $\Gamma_g[1,2]$ on pages 202-210. But to be honest I was hoping for more. – Tom Mar 5 '13 at 18:05

Instead of the principal congruence subgroup $\Gamma _g(q)$, suppose we consider the subgroup $\Delta _g(q)$ consisting of matrices $u= \pmatrix {I_g & S \\ 0 & I_g}$ where $S$ is congruent to the zero matrix modulo $q$, and the conjugates of $u$ by the involution $J_g$. Then this subgroup has finite index in $\Gamma _g(q)$ if $g\geq 2$. This is a well known theorem of Bass, Milnor and Serre (this is in an IHES paper where they also prove the congruence subgroup property for $Sp_g({\mathbb Z})$). If you want some invariance properties for your function, and you can check that it is invariant under these generators, then you know your function lives on (i.e. is invariant under) some congruence subgroup, because of the Bass-Milnor-Serre Theorem. This approach is used by Borel-Wallach in their book "Continuous cohomology, discrete subgroups, ..." (see the chapter on the Weil representation), precisely to prove that certain theta series are invariant under a congruence subgroup of the integral symplectic group.

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I think you might be interested in the article "Simple Graded Rings of Siegel Modular Forms, Differential Operators and Borcherds Products" by H. Aoki and T. Ibukiyama published in International Journal of Mathematics, Vol. 16, No. 3 (2005) 249–279. In Lemma 6.2 on pg. 265 they prove: "For any natural number $N$, the group $\Gamma_0(N)$ is generated by the above four kinds of matrices":

$T = \begin{pmatrix} 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{pmatrix}$

$u(x)=\begin{bmatrix} 1 & 0 & 0 & 0\\ x & 1 & 0 & 0\\ 0 & 0 & 1 & -x\\ 0 & 0 & 0 & 1 \end{bmatrix}, x\in \mathbb{Z}$

$u(S) = \begin{bmatrix} 1_2 & S\\ 0 & 1_2 \end{bmatrix}, S={}^tS\in M_2(\mathbb{Z})$

$C(a,b,c,d)=\begin{bmatrix} a & 0 & b & 0\\ 0 & 1 & 0 & 0\\ cN & 0 & d & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$ where $\begin{bmatrix} a & b \\ cN & d\end{bmatrix}\in\Gamma_0^{(1)}(N)$

(Sorry for the laTexing; I couldn't get matrices to work).

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you can put backticks $...$  so that you get matrices. I had the same problem.See the instructions at the faq on "how to write math". – Venkataramana Mar 6 '13 at 0:28
Dear Aakumadula, thanks for helping with the syntax! Sorry, I haven't had time to look up your reference in detail, yet ! Dear Nathan, that's a really nice one ! But doesn't $u(1)$ generate all $u(x)$s ? And can't we pick only finitely many $C(a,b,c,d)$s as $\Gamma_0^{(1)}(N)$ is finitely generated ? – Tom Mar 11 '13 at 18:03

I just wanted to share a tiny part of the solution with you. The group $\Gamma_{2,0}[2]$ is generated by the matrices $\begin{pmatrix}I_g & S \\ 0_g & I_g \end{pmatrix}$ where $S=S^t$,

$\begin{pmatrix}I_g & 0_g \\ 2 \cdot S & I_g \end{pmatrix}$ where $S=S^t$

and $\begin{pmatrix}U^t & 0_g \\ 0_g & U^{-1} \end{pmatrix}$ where $U \in GL(2,\mathbb{Z})$. The reference is http://arxiv.org/abs/1001.0324 page 6.

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