Take the 2-minute tour ×

It occurred to me that a subgroup of the modular group $\Gamma$ is a congruence subgroup iff it contains a subgroup of the form $\Gamma(N)$, while a subgroup of a general topological group is open iff it contains an open subgroup. This suggests making a topology on the modular group $\Gamma$ with the subgroups $\Gamma(N)$ as a basis of open neighborhoods of the origin so that $\Gamma$ becomes a topological group. It would then follow that a subgroup of $\Gamma$ is a congruence subgroup iff it is open.

Furthermore, for any $\gamma \in \Gamma$ not equal to the identity, there exists $N$ such that $\gamma \notin \Gamma(N)$, so this topology is Hausdorff, even totally disconnected.

I was inspired in part by this thread and looked at this paper but could not find anything about this idea.

Has anyone considered this topology? Does it provide insight into the problem of determining whether a group is a congruence subgroup?

share|improve this question
My first reaction is that this is related to the adelic topology on $GL_2$, although exactly how would require some further thought on my part. Probably someone else here will be able to follow up on this before I get back to it... –  Pete L. Clark Aug 26 '10 at 16:48
The kernel of the map from the profinite completion of $\Gamma$ to the completion with respect to the congruence topology is called the $\textit{congruence kernel}$ and has been studied in connection with the congruence subgroup problem. –  Victor Protsak Aug 26 '10 at 18:37
@Pete: So does your comment hold up that it is somehow related to the adelic topology on $GL_2$? –  David Corwin Aug 29 '10 at 2:46
@Pete: Over a year later, having learned more, I think you're basically right. –  David Corwin Mar 22 '12 at 9:02

3 Answers 3

up vote 6 down vote accepted

To expand Henry Wilton's concise answer, the Congruence Subgroup Problem has a distinguished history including important work by Serre and a number of others (exploiting effectively the congruence topology). See for example: MR0272790 (42 #7671) 14.50, Serre, Jean-Pierre, Le probl`eme des groupes de congruence pour SL2. (French) Ann. of Math. (2) 92 1970 489–527.

This sort of topology on a group originates earlier, but the application here is highly original.

ADDED: Like many other journal articles, the one mentioned here by Serre is available in PDF format but only through JSTOR (or other library resource). There is a lot of literature, including my 1980 Springer Lecture Notes 789 Arithmetic Groups which cover some of the background as well as an expository account of Matsumoto's thesis.

share|improve this answer

It's called the congruence topology, and is (obviously) always at least as coarse as the profinite topology. If your group has the Congruence Subgroup Property (the modular group doesn't, but $SL_n(\mathbb{Z})$ does for $n>2$) then it's the same as the profinite topology.

A google search found, for instance, Section 7.3 of Algebraic theory of the Bianchi groups by Benjamin Fine.

share|improve this answer
Modulus Margulis rigidity, this is the same answer as mine. For $SL(2)$ you have to take the profinite completion with respect to a special sequence of subgroups. –  Marc Palm Mar 21 '12 at 15:50

You can take the profinite completion $\widehat{\mathbb{Z}} = \prod_p \mathbb{Z}_p$ of $\mathbb{Z}$, then open subgroups of $G( \widehat{\mathbb{Z}})$ correspond to congruence subgroups in $G(\mathbb{Z})$.

The identification is easy:

$$ \Gamma \; congruence \leftrightarrow K \; open$$

"$\rightarrow$": Assume you have a congruence subgroup $\Gamma \subset \Gamma(N)$, then we can consider $$ \Gamma / \Gamma(N) \subset SL_2(\mathbb{Z} / N) = \prod\limits_{p^k || N} SL_2(\mathbb{Z} / p^k ).$$ Define $K = K(\Gamma)$ as the pullback of $\Gamma / \Gamma(N)$ along the surjection $$p: \prod\limits_{p} SL_2(\mathbb{Z}_p) \rightarrow \prod\limits_{p^k || N} SL_2(\mathbb{Z} / p^k )$$

"$\leftarrow$": Pick $\Gamma = K \cap SL_2(\mathbb{Q})$, where $SL_2( \mathbb{Q})$ is diagonal subgroup $\prod\limits_p SL_2(\mathbb{Q}_{p})$.

share|improve this answer

Your Answer


