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Hurwitz's automorphisms theorem states that for a compact Riemann surface $X$ the cardinality of $Aut(X)$, the group of holomorphic automorphisms, is bounded above by $84(g(X)-1)$ and is therefore finite. From an earlier post on MO (Riemann surfaces that are not of finite type) one cannot expect $Aut(X)$ to be finite when $X$ is of infinite genus. When can one expect $Aut(X)$ to be discrete (we give $Aut(X)$ the compact-open topology)? My conjecture is that when $X$ is hyperbolic and non-prolongable, $Aut(X)$ is discrete. Here non-prolongable means that we cannot imbed $X$ conformally into another Riemann surface $\tilde{X}$, such that $\tilde{X} - X$ has a non-empty interior.

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up vote 8 down vote accepted

The action is discrete if $X$ is hyperbolic and is not a disk or annulus. By uniformization, $X=\mathbb{H}^2/\Gamma$ for some discrete subgroup $\Gamma< PSL_2(\mathbb{R})$. Let $\Lambda < PSL_2(\mathbb{R})$ be the normalizer of $\Gamma$, then $Aut(X)\cong \Lambda/\Gamma$ since any conformal automorphism of $X$ must be an isometry of the hyperbolic metric, and therefore has some lift to $PSL_2(\mathbb{R})$. Since $\Gamma$ is discrete, $\Lambda$ is a closed subgroup of $PSL_2(\mathbb{R})$. If $\Lambda$ is discrete, then $Aut(X)=\Lambda/\Gamma$ is discrete. Otherwise, if $\Lambda$ is not discrete, then $\Lambda$ is a Lie subgroup of $PSL_2(\mathbb{R})$. The only (non-discrete) Lie subgroups of $PSL_2(\mathbb{R})$ are elementary or $PSL_2(\mathbb{R})$. In the first case, $\Gamma$ is elementary, so $\Gamma$ is abelian, and $X$ is a disk or annulus. In the second case, $\Gamma$ must be trivial since $PSL_2(\mathbb{R})$ is simple.

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It seems that this question is addressed in:

On the action of the mapping class group for Riemann surfaces of infinite type

Ege FUJIKAWA, Hiroshige SHIGA, and Masahiko TANIGUCHI Source: J. Math. Soc. Japan Volume 56, Number 4 (2004), 1069-1086.

(open access)

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