3
$\begingroup$

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 $\Gamma(2)$?

$\endgroup$
6
  • 2
    $\begingroup$ What is your definition of arithmetic? $\endgroup$
    – Igor Rivin
    Oct 24, 2014 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, 2014 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, 2014 at 16:15
  • $\begingroup$ $\Gamma(p)$ is torsion free for any prime $p$, except $p=2$. $\endgroup$
    – few_reps
    Oct 24, 2014 at 16:18
  • $\begingroup$ (and the projection $\Gamma(p)\to P\Gamma(p)$ is an isomorphism for $p>2$) $\endgroup$
    – few_reps
    Oct 24, 2014 at 16:31

1 Answer 1

9
$\begingroup$

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.

$\endgroup$

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.