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).


1 Answer 1


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.