I am looking for some examples of computing the dimension of the first sheaf cohomology for smooth projective surfaces. To be more precisely, let $X$ be a smooth, projective surface. Let $D$ be an invertible coherent sheaf. I hope to know examples, beside of different kinds of vanishing theorem, in which people could really compute the dimension of the first cohomology $H^1(X, \mathcal{O}_X(D))$. Here by the word "compute", it could either mean to find a concrete integer, or it could mean to relate this dimension with some invariants of $X$ (or $D$ as well). However, I am not looking for an expression such as the Riemann–Roch formula, unless the $0$th and the $2$nd sheaf cohomology in the example can be computed concretely. Any comments, suggestions, or references are welcome.
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$\begingroup$ By "beside of different kinds of vanishing theorem", do you mean just that you are looking for examples where the cohomology is non-trivial? $\endgroup$– LSpiceCommented Jul 17, 2023 at 16:28
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2$\begingroup$ Hi @LSpice, exactly! I am looking for nontrivial examples. Thanks for helping me to clarify my question. $\endgroup$– Leo DCommented Jul 17, 2023 at 16:54
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$\begingroup$ There are so many examples that it is hard to choose. Here is one: suppose your surface admits an elliptic fibration $X\rightarrow \mathbb{P}^1$, and let $F$ be a fiber. Then $\mathscr{O}(F)=p^*\mathscr{O}_{\mathbb{P}^1}(1)$, from which one gets $h^0(nF)=n+1$. We have $h^0(K-nF)=0$ for $n$ large, hence $h^1(nF)= n+1-\chi (\mathscr{O}_X)$ by Riemann-Roch. $\endgroup$– abxCommented Jul 17, 2023 at 18:22
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$\begingroup$ @abx Thanks for this nice example. So can I ask if every example you know comes from a similar method? That is, we compute both $h^0(F)$ and $h^2(F)$, and then use Riemann-Roch. $\endgroup$– Leo DCommented Jul 18, 2023 at 5:27
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$\begingroup$ There are other ways to compute $h^1$. For instance, suppose $X$ is regular ($h^1(\mathscr{O}_X)=0$), and $D$ is a disjoint union of $r$ integral curves. Then the exact sequence $$0\rightarrow \mathscr{O}_X(-D)\rightarrow \mathscr{O}_X\rightarrow \mathscr{O}_D\rightarrow 0$$ gives $h^1(-D)=r-1$. $\endgroup$– abxCommented Jul 18, 2023 at 6:34
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