Let $X$ be a non-compact Riemann surface with universal covering $\mathbb H$ and suppose that the fundamental group of $X$ is an arithmetic subgroup of $\mathrm{Aut}(\mathbb H) = \mathrm{PSL}_2(\mathbb R)$.

Up to a finite etale cover, $X$ is an etale cover of $Y(2)$. This is the same as saying that the fundamental group of $X$ is a subgroup of $P\Gamma(2)$ up to restricting to some finite index subgroup.

My question is whether it is really necessary to pass to an etale cover of $X$.

How do I construct an example of an $X$ as above for which $\pi_1(X)$ is not contained in $\Gamma(2)$?

  • 2
    $\begingroup$ What is your definition of arithmetic? $\endgroup$
    – Igor Rivin
    Oct 24 '14 at 15:55
  • 1
    $\begingroup$ I guess $\Gamma(2)$ is the kernel of mod 2 reduction ... then what would you think of $H/\Gamma(3)$ ? $\endgroup$
    – few_reps
    Oct 24 '14 at 16:08
  • $\begingroup$ @few_reps Is $P\Gamma(3)$ torsion free? If not, then it is not the fundamental group of its quotient. In any case, probably $\Gamma(p)$ is torsion free for large (odd) primes $p$, so this does answer the question. $\endgroup$
    – Stephan29
    Oct 24 '14 at 16:15
  • $\begingroup$ $\Gamma(p)$ is torsion free for any prime $p$, except $p=2$. $\endgroup$
    – few_reps
    Oct 24 '14 at 16:18
  • $\begingroup$ (and the projection $\Gamma(p)\to P\Gamma(p)$ is an isomorphism for $p>2$) $\endgroup$
    – few_reps
    Oct 24 '14 at 16:31

For $n>2$, the kernel $\Gamma(n)$ of the mod. $n$ reduction $\mathrm{SL}_2(\mathbf Z)\to \mathrm{SL}_2(\mathbf Z/n\mathbf Z)$ is torsion free (and is in fact a free group). This result is often attributed to Minkovski.

In particular, it acts freely on Poincaré half plane $\mathbb H$, so that the groups $\pi_1(\mathbb H/\Gamma(n))$ and $\Gamma(n)$ are isomorphic.

Now for $n$ odd, $\Gamma(n)$ is not a subgroup of $\Gamma(2)$. The intersection of these two subgroups is $\Gamma(2n)$, which gives the coverings $$\mathbb H/\Gamma(n)\leftarrow \mathbb H/\Gamma(2n)\rightarrow \mathbb H/\Gamma(2)$$ alluded to in your question.


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