Let $\Sigma$ be a compact Riemann surface and $L \to \Sigma$ be a given(!) line bundle, with $c_1(L) \geq N(P) > 0 $ where $N(P)$, depending on $P$ as soon to be explained, is an integer as large as you desire, but also fixed.
I want to know wether $L$ admits a meromorphic section with exactly $P$ poles, counting multiplicity. If you can prescribe them, it would be even better, but it is no must.

I am aware of the fact that you can come up with a meromorphic section having at least $1$ pole (not necesarrily simple) at a prescribed point (see Forster, lectures on Riemann surfaces theorem 29.16). That being said, we can pick $P$ distinct points, say $a_1,...,a_P \in \Sigma$, which would then give sections $s_i: \Sigma \to L$ having a pole precisely at $a_i$ (order unnknown, might be double or triple) and define $$ s=\sum_{i=1}^P s_i $$
which gives me a meromorphic section with at least $P$ poles. In this case, the information on the chern class would just yield the number of zeroes $N$ via the formula $N-P=\langle c_1(L), \Sigma \rangle$, a completely worthless information for my purpose. I am aware of many closeley related theorems (e.g. Riemann Roch, which only gives me meromorphic functions/Holomorphic section) and the "pole problem" is solveable in the complex plane. I feel like this should be a well-studied problem, since thanks to my chern class/degree of my line bundle being high enough (say higher than genus of $\Sigma$ plus some correction term) I should be able to deduce it from a known theorem.

I am grateful for any reference/hint anyone could give me.


1 Answer 1


Yes, this follows from Riemann-Roch and your assumption that the degree is sufficiently large. We'll slightly swap language to talk about things in terms of divisors.

First, recall that Riemann-Roch states that $l(D)-l(K-D)=\deg D + 1 -g$. Combining this with the fact that a divisor of negative degree has no global sections, we observe that once $\deg D > \deg K$ we get $l(K-D)=0$, and so $l(D+P)=l(D)+1$ for any point $P$.

We can use this to solve the problem as follows. For a fixed divisor $D$ with $\deg D > \deg K+1$, we let $$S=\{D'\in\operatorname{Div} \Sigma \mid D' \text{ effective and } \exists P\in\Sigma \text{ with } D'+P=D\}.$$ Then for each $D'\in S$, we see that $l(D')=l(D)-1$, and as $S$ is finite, the union of the vector spaces $\mathcal{L}(D')$ inside $\mathcal{L}(D)$ cannot be the whole space. Therefore there are meromorphic functions in $\mathcal{L}(D)$ which have poles exactly on $D$.

  • $\begingroup$ To be honest, I didnt know Riemann-Roch extended to meromorphic sections like that. I was stuck on the holomorphic section version for ages and even deduced the existence of a meromorphic section with a pole of order at most(!) the prescribed one, but failed to conclude like you did. Thanks for your answer and your time! $\endgroup$
    – F. Conrad
    Nov 23, 2022 at 23:18
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    $\begingroup$ @F.Conrad I think the key idea to keep in mind for this kind of problem is that meromorphic sections of $\mathcal L$ with poles along a divisor $D$ correspond to holomorphic sections of the twist $\mathcal L(D)$. $\endgroup$ Nov 24, 2022 at 1:43
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    $\begingroup$ Tabes is exactly correct - see for instance this post potentially explaining things a bit more. $\endgroup$
    – KReiser
    Nov 24, 2022 at 3:12
  • $\begingroup$ Thanks for the input, I will check it out. $\endgroup$
    – F. Conrad
    Nov 24, 2022 at 8:44

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