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 ...


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.

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.