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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?

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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.

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    $\begingroup$ 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, 2013 at 20:38
  • $\begingroup$ @Misha: pls post this as an answer. $\endgroup$
    – Marc Palm
    Feb 25, 2013 at 21:06
  • $\begingroup$ @Marc: stankewicz already wrote this as an answer, so I hope it will suffice. $\endgroup$
    – Misha
    Feb 25, 2013 at 23:51
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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.

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  • $\begingroup$ D'oh! Sorry Misha, completely didn't see your answer. I will gladly make this answer CW if requested. $\endgroup$
    – stankewicz
    Feb 25, 2013 at 22:38

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