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

Question

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 1

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 problème 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 2

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 3

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

Answer 4

It's worth mentioning that there is a general definition of the "congruence subgroup topology" on the automorphism group of a profinite group $\widehat{A}$: namely, the open subgroups of $\text{Aut}(\widehat{A})$ under this topology are generated (as a topology) by the subgroups $$\Gamma[K]: = \ker(\text{Aut}(\widehat{A})\rightarrow\text{Aut}(\widehat{A}/K))$$ as $K$ varies over finite index characteristic subgroups of $\widehat{A}$. You can find this for example in Ribes/Zalesskii's book Profinite Groups $(\S4.4)$. In the same way we also obtain a "congruence subgroup topology" on $\text{Out}(\widehat{A})$.

For a general abstract discrete group $G$, for every abstract group $A$ and a homomorphism $\phi : G\rightarrow\text{Aut}(A)$ (or $\text{Out}(A)$), one obtains the notion of "congruence subgroups" on $G$ (relative to $\phi$), which are by definition those subgroups which contain a preimage of some $\Gamma[K]$ under the map

$$G\stackrel{\phi}{\longrightarrow}\text{Aut}(A)\subset\text{Aut}(\widehat{A}),$$ where the hat denotes profinite completion.

For the modular group $G := \text{SL}_2(\mathbb{Z})$, one can naturally take $A = \mathbb{Z}^2$, and consider the natural homomorphism $$\phi : \text{SL}_2(\mathbb{Z})\rightarrow\text{Aut}(\mathbb{Z}^2)$$ Then, $\text{Aut}(\widehat{\mathbb{Z}^2}) = \text{GL}_2(\widehat{\mathbb{Z}})$, the finite index characteristic subgroups of $\widehat{\mathbb{Z}}^2$ are $K_n := n\widehat{\mathbb{Z}}\times n\widehat{\mathbb{Z}}$, and the preimage of $\Gamma(K_n)$ in $\text{SL}_2(\mathbb{Z})$ is just the classical principal congruence subgroup $\Gamma(n)$.

However, for $G = \text{SL}_2(\mathbb{Z})$, there is another natural choice for $\phi$, given by noting that if $F_2$ denotes the free group of rank 2, then $\text{Out}(F_2)\cong\text{GL}_2(\mathbb{Z})$. From this, one obtains a natural map $$\text{SL}_2(\mathbb{Z})\stackrel{\phi'}{\longrightarrow}\text{Out}(F_2)\subset\text{Out}(\widehat{F_2})$$ The preimages of $\Gamma[K]$ as $K$ varies over finite index characteristic subgroups of $\widehat{F_2}$ then generate the congruence subgroups (relative to $\phi'$) in the sense described above.

The cool thing is that while $\text{SL}_2(\mathbb{Z})$ doesn't have the congruence subgroup property relative to $\phi$, it does relative to $\phi'$. In particular, there are finite index subgroups of $\text{SL}_2(\mathbb{Z})$ which are congruence relative to $\phi'$, but not congruence in the usual sense (ie relative to $\phi$). There is a paper of Bux-Ershov-Rapinchuk which has a nice description of this phenomenon, though the result they prove is originally due to Asada.

In fact, the notion of congruence relative to $\phi$ gives rise to the usual moduli interpretations of "congruence" modular curves, whereas the more general notion of congruence relative to $\phi'$ gives rise to the moduli interpretations for "noncongruence" modular curves.