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Is the sheaf of meromorphic functions on a (connected) compact Riemann surface constant? I am refering here to meromorphic functions in the sense of complex analysis and not to those of algebraic geometry, where I know this to hold. I do not imagine it to be the case, but then how does one go about proving the correspondence between divisors and invertible sheaves? The algebro-geometric proof I know (the one in Hartshorne) relies crucially on the sheaf of meromorphic functions being constant ...

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No, the sheaf of meromorphic functions on a Riemann surface is definitely not constant.

For example if $X=\mathbb P^1(\mathbb C)=\mathbb C\cup \{ \infty \}$ the restriction map $res:\mathcal M (\mathbb P^1)\to \mathcal M (\mathbb C)$ is not surjective.
Indeed the exponential function $exp(z)=e^z$ satisfies $exp\in \mathcal M (\mathbb C)$ but is not in the image of $res$ since the exponential function has an esssential singularity at $\infty$ and is thus not meromorphic there.

Aside
Actually meromorphic functions on a compact Riemann surface are rational: $\mathcal M(X)=Rat (X)$.
This is a special case of the so called GAGA Principle and reinforces the above argument since it then follows that actually $\mathcal M( P^1(\mathbb C))=\mathbb C(z)$.
However I didn't want to invoke that more advanced result in my elementary proof at the level of an introduction to function theory of a complex variable.

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  • $\begingroup$ Dear Georges, you are absolutely right. I went over my old Riemann surface notes (taken from Miranda's book) to see why I remembered this fact incorrectly, and I see that Miranda defines two types of holomorphic/meromorphic sheaves: algebraic and analytic. Since in algebraic geometry one is used to the meromorphic function sheaf (that is, the rational function sheaf) being locally constant, I somehow fused the algebraic and analytic sheaves into one in my mind. Thank you very much for your clarification. $\endgroup$ – rfauffar Feb 20 '14 at 14:53
  • $\begingroup$ Dear Robert, I'm impressed by the honesty shown in your comment. We all make mistakes but not all of us admit it as gracefully as you. $\endgroup$ – Georges Elencwajg Feb 20 '14 at 19:37
  • $\begingroup$ Dear Georges, thank you for your kind words. $\endgroup$ – rfauffar Feb 20 '14 at 22:21

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