## Self maps of hyperbolic surfaces with two fixed points

Problem. Let $f$ be a self map of a hyperbolic surface $X$ such that $f$ has two fixed points. Show that $f$ is a finite order automorphism, i.e. there exists $n$ such that $f^{\circ n} = id_X$.

Solution. Let $\pi$ be a covering map from the unit disk $D$ to $X$. First of all, notice that due to Schwarz-Pick theorem such an $f$ is either

• A global isometry, i.e. a conformal isomorphisms of $X$
• A covering map, but is not one-to-one, i.e. a local isometry
• Strictly decreases the Poincare metric

It is easy to see that in the last case $f$ cannot have two fixed points. Therefore let’s assume that $f$ is a covering map. Since the covering is universal, one can lift the map $f$ to the map $\tilde{f}: D \to D$.

Also note that $\pi \circ \tilde{f} = f \circ \pi$. Assume that $p,q$ are the fixed points of $f$ and $\pi(0)=p$, perhaps composing $\tilde{f}$ with an automorphism of $D$ we can also assume that $\tilde{f}(0)=0$.  Remember that $f$ is not strictly decreasing Poincare metric, the same applies to its lift $\tilde{f}$, hence $\tilde{f}$ must be a rotation around the origin (it is not difficult to classify all automorpisms of the unit disk), $\tilde{f}(z)=e^{2\pi i \alpha}$ for some real $\alpha$.

Note that $\pi \circ \tilde{f}^{\circ n} = f^{\circ n} \circ \pi$, hence $\tilde{f}^{\circ n}(\tilde{q})$ is in the fiber of $q$ and therefore the set $\{ \tilde{f}^{\circ n}(\tilde{q}) \in N \}$ is a discrete set which implies that $\alpha$ is rational and we are done.

P.S. Actually, if we were talking about self-coverings of compact surfaces, the situation would be easier: for compact Riemann surfaces we do have Riemann-Hurwitz formula:

$2g(x)-2=n(2g(Y)-2)+\sum_{j=1}^n (e_j-1)$,

where $f:X \to Y$ is a ramified covering of degree $n$. If $latex X=Y$ then we would obtain that either $g=1$ or $n=1$. Inother words, for compact Riemann surfaces of genus $>1$ the only self-covering is a biholomorphism.