1
$\begingroup$

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 $\Gamma_2[2]$.

$\endgroup$

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
3
  • 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, 2013 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, 2013 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, 2013 at 12:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.