This question is something of a follow-up to Transformation formulae for classical theta functions .

How does one recognise whether a subgroup of the modular group $\Gamma=\mathrm{SL}_2(\mathbb{Z})$ is a congruence subgroup?

Now that's too broad a question for me to expect a simple answer so here's a more specific question. The subgroup $\Gamma_1(4)$ of the modular group is free of rank $2$ and freely generated by $A=\left( \begin{array}{cc} 1&1\\\ 0&1 \end{array}\right)$ and $B=\left( \begin{array}{cc} 1&0\\\ 4&1 \end{array}\right)$. If $\zeta$ and $\eta$ are roots of unity there is a homomorphism $\phi$ from $\Gamma_1(4)$ to the unit circle group sending $A$ and $B$ to $\zeta$ and $\eta$ resepectively. Then the kernel $K$ of $\phi$ has finite index in $\Gamma_1(4)$. How do we determine whether $K$ is a congruence subgroup, and if so what its level is?

In this example, the answer is yes when $\zeta^4=\eta^4=1$. There are also examples involving cube roots of unity, and involving eighth roots of unity where the answer is yes. I am interested in this example since one can construct a "modular function" $f$, homolomorphic on the upper half-plane and meromorphic at cusps such that $f(Az)=\phi(A)f(z)$ for all $A\in\Gamma_1(4)$. One can take $f=\theta_2^a\theta_3^b\theta_4^c$ for appropriate rationals $a$, $b$ and $c$.

Finally, a vaguer general question. Given a subgroup $H$ of $\Gamma$ specified as the kernel of a homomorphism from $\Gamma$ or $\Gamma_1(4)$ (or something similar) to a reasonably tractable target group, how does one determine whether $H$ is a congruence subgroup?


There is one answer in the following paper, along with a nice bibliography of other techniques:

MR1343700 (96k:20100) Hsu, Tim(1-PRIN) Identifying congruence subgroups of the modular group. (English summary) Proc. Amer. Math. Soc. 124 (1996), no. 5, 1351--1359.

  • $\begingroup$ Thanks Andy, that's exactly the sort of thing I was looking for. $\endgroup$ Apr 10 '10 at 16:32

Although Andy has the correct answer, I thought I would point out that there will only be finitely many groups of the type you consider that are congruence subgroups. Since these are abelian (cyclic) covers of your surface, they will contain the universal abelian cover (corresponding to the commutator subgroup of $\Gamma_1(4)$). By "strong approximation", this group will map onto all but finitely many congruence quotients of $SL_2(\mathbb{Z})$, and therefore so will any group containing it. Thus, only finitely many of your groups will be congruence.

Another way to see this is to notice that abelian covers have Cheeger constant approaching zero, and therefore first eigenvalue of the Laplacian $\lambda_1$ approaching zero by Buser's inequality. By Selberg's estimate, $\lambda_1\geq 3/16$ for a congruence subgroup. In principle, one could probably get explicit estimates on the index of a congruence abelian cover this way.

  • 2
    $\begingroup$ Thanks Ian, that's very interesting; it's a shame I can only tick one response. $\endgroup$ Apr 11 '10 at 8:16
  • 1
    $\begingroup$ I don't know a good reference for strong approximation, but in this context there's a very elementary argument one may make - see: ams.org/mathscinet-getitem?mr=1459136 Their argument shows that the group maps onto all but finitely many $SL_2(Z/p)$. This then implies that it maps onto all but finitely many $SL_2(Z/p^k)$, and therefore onto a finite-index subgroup of the pro-congruence completion (I don't know if this is standard terminology). $\endgroup$
    – Ian Agol
    Apr 11 '10 at 16:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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