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stankewicz
stankewicz
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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) \hookrightarrow \mathrm{Aut}(\Gamma \setminus \mathcal{H}^*)$$N(\Gamma) \to \mathrm{Aut}(\Gamma \setminus \mathcal{H}^*)$ whose kernel is $\Gamma$. If you have no cusps, then you can say something else.

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 $N(\Gamma) \hookrightarrow \mathrm{Aut}(\Gamma \setminus \mathcal{H}^*)$ whose kernel is $\Gamma$. If you have no cusps, then you can say something else.

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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stankewicz
stankewicz
  • 3.6k
  • 26
  • 39

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 $N(\Gamma) \hookrightarrow \mathrm{Aut}(\Gamma \setminus \mathcal{H}^*)$ whose kernel is $\Gamma$. If you have no cusps, then you can say something else.