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I have the following questiom: let $X$ and $Y$ be two different points (represented by Riemann surfaces) in the Teichmuller space $T_g$ of genus $g \geq 2$ Riemann surfaces. Then of course $X$ and $Y$ are homeomorphic and not bi-holomorphically equivalent. My question is, whether there exists a holomorphic covering from $X$ to $Y.$ Namely, is there a topological covering $p: X \to Y$ which is holomorphic with respec to the complex structures of $X$ and $Y$? Why or why not?

Thanks in advance!

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@silktomath: If you like one of the answers, you should click the "accept" button. If you like several of the answers, you should click the earliest one that you like. – Lee Mosher Jan 19 '13 at 15:32
Lee, there is a very good chance I first heard about Teichmuller Space from you. – Adam Epstein Jan 19 '13 at 16:52
Could be.. Could be. :-) – Lee Mosher Jan 20 '13 at 0:52
up vote 7 down vote accepted

As $g\geq 2$, it follows by Riemann-Hurwitz that any topological covering $X\rightarrow Y$ is a homeomorphism, and any holomorphic homeomorphism is biholomorphic.

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Thanks a lot for this! – silktomath Jan 19 '13 at 12:02
To express this a little differently, this holds for $X,Y$ if and only if they are in the same orbit of the mapping class group on $T_g$. – Lee Mosher Jan 19 '13 at 13:30

As Adam said in his answer, by Riemann Hurwitz, if the genus is greater than 1, every holomorphic covering must be biholomorphic. But biholomorphic maps certainly exist. (Contrary to what is stated in the question, different points of the Teichmuller space can be bihilomorphically equivalent).

They form a group acting on the Teichmuller space called the Modular group, which acts on the Teichmuller space (and in general has fixed points). The factor over this group is called the moduli space, and the group itself has been very well studied.

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That is just what I said in my comment to Adam Epstein's answer. – Lee Mosher Jan 19 '13 at 15:27

Moreover, by the same reason there is no ramified covering $f\colon X\to Y$ either.

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