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](https://mathoverflow.net/questions/78984/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. 



Hurwitz's automorphisms theorem for infinite genus Riemann surfaces
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. 


