Every line bundle $L$ on a complex algebraic curve $X$ is of the form $\mathcal{O}(D)$, where $D$ is some divisor on $X$. This means $L$ has at least one nonzero meromorphic global section, i.e. $$H^0(L \otimes_{\mathcal{O}_X} \mathcal{M}_X) \neq 0$$ where $\mathcal{M}_X$ is the sheaf of meromorphic functions on $X$.

Is there any way to see this by directly calculating global sections of the sheaf $L \otimes_{\mathcal{O}_X} \mathcal{M}_X$ and without assuming $L \cong \mathcal{O}(D)$ for some divisor $D$? I've seen other arguments so I'm mostly just interested in a calculation of the above cohomology group.

More generally, how does one compute $H^0(V \otimes_{\mathcal{O}_X} \mathcal{M}_X)$, where $V$ is an arbitrary vector bundle on $X$?

  • $\begingroup$ The answer in this post answers your question: math.stackexchange.com/questions/1994463/… $\endgroup$ – user347489 Feb 22 at 9:13
  • $\begingroup$ @user347489 I don't think it does. I'd like to prove directly that $L \otimes_{\mathcal{O}_X} \mathcal{M}_X$ has a global section without assuming $L \cong \mathcal{O}(D)$. I'll edit to clarify. $\endgroup$ – leibnewtz Feb 22 at 9:18
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    $\begingroup$ Let $U$ an open where $X$ is trivialised, then $1 : U \to \mathcal O_{U}$ is a meromorphic section. $\endgroup$ – Nicolas Hemelsoet Feb 22 at 12:19
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    $\begingroup$ You should either speak of algebraic curves and rational functions or of Riemann surfaces and meromorphic functions, rather than mix up the algebraic and analytic categories. $\endgroup$ – Georges Elencwajg Feb 22 at 13:32

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