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.

The principal congruence subgroup of level q is defined as $$ \Gamma_g[q]:= ker\left(Sp(g,\mathbb{Z})\to Sp(g,\mathbb{Z}/q\mathbb{Z})\right).$$

The actual question

I am searching for a reference that the only element of finite order in $\Gamma_2[q]$ for $q \geq 3$ is the identity matrix. Or expressed in a formula : $$\forall\ q \geq 3 \quad\forall\ M \in \Gamma_2[q] : \quad M^n=I \Longrightarrow M=I .$$

The application

The above result implies that elements of finite order in $\Gamma_2[2]$ are of order 2. A reference on this would also be appreciated. This allows me to apply a theorem on involutions on matrices of finite order in $\Gamma_2[2]$.


This is a well known argument (for any even $g \geq 2$, and for other integral matrix groups) . If $x$ is an element in the level $q$ congruence subgroup, then $\frac{x-I}{q}$ is an integral matrix, so has all its eigenvalues algebraic integers. If $\alpha$ is any eigenvalue of $x,$ then $\alpha$ is a root of unity as $x$ has finite order. On the other hand, $\frac{\alpha - 1}{q}$ is also an algebraic integer, and note that its absolute value is less than $1$ as $q \geq 3$. The same applies to any of its algebraic conjugates. As was known to W. Burnside, the only algebraic integer with all algebraic conjugates of absolute value less than $1$ is $0$ ( for otherwise, the geometric mean of their absolute values would be at least $1$, hence the arithmetic mean of those eigenvalues would greater than $1$, which is not the case). Hence all eigenvalues of $x$ are $1$, and $x =I,$ as $x$ has finite order.

  • 2
    $\begingroup$ Very nice proof ! Could you do me a small favor and publish it in Inventiones such that I have a descent reference ? $\endgroup$
    – Tom
    Jun 13 '13 at 8:28
  • $\begingroup$ I am not sure I know the best reference for this well-known fact. Maybe it it is in one of Serre's books. $\endgroup$ Jun 13 '13 at 17:32
  • $\begingroup$ In the end, I used some parts of your proof. As you put quite some effort in answering my question(s) I shall accept your answer. $\endgroup$
    – Tom
    Jun 28 '13 at 12:47

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