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Suppose $X$ is a variety of dimension $n$ and $D$ is a divisor on $X$. Is there straightforward way to compute the dimension of $H^0(X,D)$? If $X$ is a curve, we can easily do this with Riemann-Roch. I know there is a version of Riemann-Roch for surfaces, but you are still left with an $H^1$ even after using Serre Duality (and this is still only for dimension $2$).

ponchan
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  • There is no straightforward way in general (I am assuming your divisor D is effective) – Cranium Clamp Dec 17 '23 at 04:27
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    Positivity and vanishing theorems are very useful in this kind of situation. See Lazarsfeld's "Positivity in Algebraic Geometry I" for a collection of these kinds of results. For example, see my answer here for the situation when $D$ is nef. – Daniel Dec 17 '23 at 17:09

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The higher-dimensional Riemann-Roch is telling you something about the Euler characteristic of the line bundle. Serre duality only implicates two of these cohomology groups. So as soon as more than three cohomology groups are involved, you're not going to get something clean about $H^0$ that's not also talking about other $H^i$ in some way. [deleting wrong remark about eliminating the $H^1$ term for surfaces]

hunter
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  • So is there another way to get info about $H^0$? Or $H^0(nD)$ for some $n$ in the general case? Can you elaborate on your last sentence? – ponchan Dec 16 '23 at 22:19
  • Sorry, just saw your request! For $H^0$, or $H^0(nD)$, in general this is very hard but @Daniel in the comments makes the excellent point that what you often want is a nice vanishing theorem to control the higher cohomology. – hunter Dec 22 '23 at 13:56
  • As to your second question, I cannot elaborate it because it is wrong :-) erasing it. – hunter Dec 22 '23 at 14:01