finite (we'll prove this at the end). Consider the quotient map . Since acts by holomorphic functions, the quotient is locally of the form and the quotient is a smooth Riemann surface. The quotient map is a branched cover, and we will see below that the ramification points correspond to the orbits that have a non trivial stabiliser. Let be the genus of . By the Riemann-Hurwitz formula, where the sum is over the ramification points for the quotient map . The ramification index at is just the order of the stabiliser group, since where the number of pre-images of (the number of points in the orbit), and . By definition of ramification points, for all ramification indices. Now call the righthand side and since we must have . Rearranging the equation we find: If then , and If , then and so that , If , then and if then , so that if then , so that , if then write . We may assume . if then so that , if then if then so that , if then so that . In conclusion, . To show that is finite, note that acts on the cohomology preserving the Hodge decomposition and the lattice . In particular, its action on gives a homomorphism with discrete image . In addition, the image preserves the natural non degenerate Hermitian inner product on . In particular the image is contained in the unitary group which is compact. Thus the image is not just discrete, but finite. It remains to prove that has finite kernel. In fact, we will prove is injective. Assume acts as the identity on . If is finite, then by the Lefschetz fixed-point theorem, This is a contradiction, and so is infinite. Since is a closed complex sub variety of positive dimension and is a smooth connected curve (i.e. ), we must have . Thus is the identity, and we conclude that is injective and is finite. Q.E.D. Corollary of the proof: A Riemann surface of genus has automorphisms if and only if is a branched cover with three ramification points, of indices 2,3 and 7. The idea of another proof and construction of the Hurwitz surfaces By the uniformization theorem, any hyperbolic surface X – i.e., the Gaussian curvature of X is equal to negative one at every point – is covered by the hyperbolic plane. The