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## Congruence Subgroups as Open Subgroups of the Modular Group Under the Right Topology

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.

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

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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 2010 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 2010 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 2010 at 2:46
@Pete: Over a year later, having learned more, I think you're basically right. – David Corwin Mar 22 2012 at 9:02

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 probleme 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.

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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.

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 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 2012 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 $\prod\limits_p SL_2(\mathbb{Q}_{p})$.

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