**Problem**. Let be a self map of a hyperbolic surface such that has two fixed points. Show that is a finite order automorphism, i.e. there exists such that .

**Solution.** Let be a covering map from the unit disk to . First of all, notice that due to Schwarz-Pick theorem such an is either

*A global isometry, i.e. a conformal isomorphisms of *
*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 cannot have two fixed points. Therefore let’s assume that is a covering map. Since the covering is universal, one can lift the map to the map .

Also note that . Assume that are the fixed points of and , perhaps composing with an automorphism of we can also assume that . Remember that is not strictly decreasing Poincare metric, the same applies to its lift , hence must be a rotation around the origin (it is not difficult to classify all automorpisms of the unit disk), for some real .

Note that , hence is in the fiber of and therefore the set is a discrete set which implies that 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:

,

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

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