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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?

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$G$ should act holomorphically right? and $p$ will be a covering map I guess. – Un Chien Andalou 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? – Un Chien Andalou Jun 19 '12 at 20:47
up vote 3 down vote accepted

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 want to attribute the observation that the differing behavior in (complex) dimension one and dimension greater than one can be understood in terms of classical invariant theory to my colleague Robert Varley. Last semester we team taught a course on modular curves and he introduced this perspective.

Finally, note also that if the action is not only properly discontinuous but free then in all dimensions the quotient is a complex manifold, and this is a much easier result.

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I found two completely different proofs. The shorter is here: at page 9. I think, the main problem for me is to understand why the fixed points of $G$ are the branch points of $\pi$.

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