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.


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