Congruence Subgroups as Open Subgroups of the Modular Group Under the Right Topology - MathOverflow

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

2010-08-26T16:36:52Z

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?

Answer by HW for Congruence Subgroups as Open Subgroups of the Modular Group Under the Right Topology

2010-08-26T16:53:35Z

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.

Answer by Jim Humphreys for Congruence Subgroups as Open Subgroups of the Modular Group Under the Right Topology

2010-08-26T17:06:27Z

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.

Answer by Marc Palm for Congruence Subgroups as Open Subgroups of the Modular Group Under the Right Topology

2012-03-21T15:47:57Z

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