riemann surface and discontinuous group action

If G is a group that acts properly discontinuously on a Riemann surface X , than we can give to the quotient X/G a structure of Riemann surface such that the projection p:X→X/G is holomorphic. How can I prove that?

• $G$ should act holomorphically right? and $p$ will be a covering map I guess. – Marso Jun 19 '12 at 13:56
• Yes! The theorem says: "Let $G$ be a group acting properly discontinuously on a Riemann surface $X$. Then the quotient space $X/G$ is a Riemann surface and the canonical projection map $\pi:X\rightarrow X/G$ is holomorphic branched covering map of degree $|G|$. The fixed points of $G$ are the branch points of $\pi$." I can't understand how to prove this... – angy Jun 19 '12 at 17:33
• could you tell me the book or source of the problem? – Marso Jun 19 '12 at 20:47

A (slightly terse) proof of a very similar result occurs as Proposition 13 in these notes. The key point is that every fixed point of $G$ admits a local holomorphic coordinate in which the map is simply multiplication by a root of unity. This is true because every holomorphic map $f$ in one variable is conformally equivalent near a fixed point $z_0$ to the linear map $z \mapsto f'(z_0)(z-z_0)$.
Note that the result does not hold for holomorphic maps in dimension greater than one -- not even for finite order linear maps of $\mathbb{C}^2$! For instance consider the map $(z,w) \mapsto (-z,-w)$. The ring of invariant functions is $\mathbb{C}[z^2,zw,w^2] \cong \mathbb{C}[a,b,c]/(b^2-ac)$, so the origin (the only fixed point of the map) is a singular point of the quotient.
I found two completely different proofs. The shorter is here: http://tinyurl.com/cg4plxx at page 9. I think, the main problem for me is to understand why the fixed points of $G$ are the branch points of $\pi$.