Let $M$ be a compact hyperbolic Riemann surface. Is there a simple way to show that the automorphism group $Aut(M)$ of conformal self-mappings of $M$ is a finite group?

Recall that a hyperbolic Riemann surface is one whose universal cover is $\mathbb D$.

I was hoping there is some way to use basic covering space theory (especially the various Galois-like relations between groups and covers) and the fact the group of automorphisms of the unit disk is $PSL(2,\mathbb R)$, since we know $\mathbb D/ G\cong M$, where $G$ is the covering group of $M$. But it has been a while since I've studied covering space theory, and I'm not sure if this idea can be made to work. In particular, I'm not sure how to make use of the hypothesis that $M$ is compact.

  • $\begingroup$ en.wikipedia.org/wiki/Hurwitz's_automorphisms_theorem $\endgroup$ Commented Aug 3, 2014 at 19:29
  • $\begingroup$ @QiaochuYuan I should probably clarify. I know about Hurwitz's bounded. I was wondering if there is a simpler way to obtain the weaker statement that the automorphism group is finite. $\endgroup$
    – Potato
    Commented Aug 3, 2014 at 19:31

2 Answers 2


You can do this using the theory of covering spaces. Though I don't know of any 'direct' approach.

By a careful but rather elementary treatment of subgroups of $PSL(2, \mathbb{R})$, you can show that, up to conjugation, the only discrete subgroups which act freely on the unit disc and have positive dimensional normalizer are the groups generated by $\begin{pmatrix} a & 0 \\ 0 & a^{-1} \end{pmatrix}$ with $a\in \mathbb{R}- \{0\}$ or the group generated by $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$.

The quotient in the first case gives you the annuli. In the second case you get the punctured unit disc. (Thus these are the only hyperbolic riemann surfaces whose automorphism group is of positive dimension)

Hence any compact hyperbolic surface X has discrete automorphism group, which then has to be finite because of compactness of X. (Any orbit in X is discrete, hence finite since X is compact)

  • $\begingroup$ Great, this is exactly what I was looking for. Do you know of a reference for the facts you mention, or a place where this proof is given in detail? $\endgroup$
    – Potato
    Commented Aug 4, 2014 at 0:22
  • $\begingroup$ Not really, sorry. The only book where I know for certain that something in this direction is done, is "Riemannsche Flächen" by Klaus Lamotke (which is only available in German I think). Otherwise you should probably look for books/lecture notes that treat Fuchsian groups (discrete subgroups of PSL(2,R) ). $\endgroup$
    – Rieux
    Commented Aug 4, 2014 at 14:13

The result follows also from Bochner's theorem, because for surfaces negative Gaussian curvature is equivalent to negative Ricci curvature, and because of $g\ge 2$, that $M$ has a hyperbolic metric $g$ with $Aut(M)=Isom^+(M,g)$.

Theorem (Bochner). Let $(M^n,g)$ be a compact Riemannian manifold of dimension $n$, with negative definite Ricci curvature everywhere, then the isometry group of $M$ is finite.

Here we use that $M$ is compact to conclude that also $Isom(M)$ is compact. Then we can proceed to show that the connected component $of Isom(M)$ which contains the identity, $Isom^0(M)$, is just the identity.


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