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Let $\widehat{SL(2,\mathbb{Z})}$ be the profinite completion of $SL(2,\mathbb{Z})$. Let $\Gamma(N)$ denote the typical principal congruence subgroup of $SL(2,\mathbb{Z})$ (ie, all matrices congruent to the identity mod $N$). Let $\overline{\Gamma(N)}$ denote its closure in $\widehat{SL(2,\mathbb{Z})}$.

Can we describe generators for $\bigcap_{N\ge 1}\overline{\Gamma(N)}$? (At first I thought this intersection is trivial, but since $SL(2,\mathbb{Z})$ has noncongruence subgroups, the congruence subgroups do not form a fundamental system of neighborhoods of the identity in $\widehat{SL(2,\mathbb{Z})}$, so now I'm rather uncertain...)

What about $\bigcap_{N\ge 1}\overline{\Gamma_1(N)}$?

($\Gamma_1(N)$ is the subgroup consisting of matrices which mod $N$ are upper triangular unipotent).

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It is a result of Melnikov that the congruence kernel $ker\{ \widehat{SL_2(\mathbb{Z})}\to SL_2(\hat{\mathbb{Z}})\} \cong \hat{F}_\omega$, the free profinite group on a countable number of generators.

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  • $\begingroup$ Do you know if the map $\widehat{ SL_2(\mathbb{Z})}\rightarrow SL_2(\widehat{\mathbb{Z}})$ is surjective? $\endgroup$
    – Will Chen
    Dec 31, 2014 at 23:42
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    $\begingroup$ Yes, that is basically the Chinese remainder theorem, I believe, plus the fact that SL_2Z surjects SL_2 F_p $\endgroup$
    – Ian Agol
    Jan 1, 2015 at 4:09

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