TITLE: Congruence Subgroups as Open Subgroups of the Modular Group Under the Right Topology
QUESTION [12 upvotes]: 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?

REPLY [4 votes]: 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.