0
$\begingroup$

Let $\Gamma \subseteq PSL_2(\mathbb{R})$ be a Fuchsian group, possibly containing elliptic elements. Is it true that $N(\Gamma) / \Gamma$, where $N(\Gamma)$ the normalizer of $\Gamma$ in $PSL_2(\mathbb{R})$, is isomorphic to the automorphism group of $\Gamma \backslash \mathcal{H}^*$?

Here, $\mathcal{H}^*$ is the union of the upper-half plane and the set of cusps of $\Gamma$.

If so, can you point me to a reference, please?

$\endgroup$
3
$\begingroup$

The compact Riemann surface $\Gamma\backslash\mathcal H^*$ may have authomorphisms that do not preserve the set of points corresponding to the cusps of $\Gamma$, so I believe the answer is no.

$\endgroup$
  • 3
    $\begingroup$ Say, the compactified quotient could be the 2-sphere, whose group of automorphisms is $PSL(2,C)$, while the quotient $N(\Gamma)/\Gamma$ is always finite. $\endgroup$ – Misha Feb 25 '13 at 20:38
  • $\begingroup$ @Misha: pls post this as an answer. $\endgroup$ – Marc Palm Feb 25 '13 at 21:06
  • $\begingroup$ @Marc: stankewicz already wrote this as an answer, so I hope it will suffice. $\endgroup$ – Misha Feb 25 '13 at 23:51
1
$\begingroup$

How about $\Gamma = PSL_2(\mathbf{Z})$, where $N(\Gamma)/\Gamma)$ is the trivial group but the automorphism group of $\Gamma \setminus \mathcal{H}^*$ is infinite?

In general the best you can do is that you have a map $N(\Gamma) \to \mathrm{Aut}(\Gamma \setminus \mathcal{H}^*)$ whose kernel is $\Gamma$. If you have no cusps, then you can say something else.

$\endgroup$
  • $\begingroup$ D'oh! Sorry Misha, completely didn't see your answer. I will gladly make this answer CW if requested. $\endgroup$ – stankewicz Feb 25 '13 at 22:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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