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Let $C$ be a curve of genus $g \geq 1$ and let $J^d$ be its degree $d$ Jacobian.

Inside of $J^{g-1}$ there is the Theta divisor $\Theta$, which can be defined in various ways; the quickest definition is probably: it's the image of the Abel-Jacobi map $C^{(g-1)} \to J^{g-1}$ sending an effective degree $g-1$ divisor to the corresponding line bundle. Picking an isomorphism $J^{g-1} \cong J^d$, we also write $\Theta$ for the corresponding divisor in $J^d$.

How to compute $H^\ast(J;\Theta)$, or $h^\ast(J;\Theta)$? Or alternatively, what is known about these groups?

I suspect this is something embarrassingly standard and/or obvious and/or well-known and/or classical, but I haven't been able to figure anything out. The only thing along these lines that I was able to figure out was how to compute the Euler characteristic $\chi(J;\Theta^k)$ where $k$ is an integer: By Hirzebruch-Riemann-Roch and the Poincare formula it's $$\int_J \operatorname{ch}(\Theta^k) = \int_J e^{k\theta} = \int_J k^g \theta^g / g! = k^g.$$

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    $\begingroup$ All cohomology groups but the zeroth vanish so the R-R formula gives what you want (see for instance David Mumford: Abelian varieties, Ch 16). $\endgroup$
    – Torsten Ekedahl
    Commented Aug 25, 2011 at 16:50
  • $\begingroup$ Why is $\theta^g = g!$? $\endgroup$
    – fish_monster
    Commented Aug 4, 2022 at 12:00

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this seems to be the kodaira vanishing theorem. i.e. any line bundle of form K+A where is ample, has no higher cohomology. for an abelian variety K is trivial, and Theta is ample. qed.

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    $\begingroup$ In characteristic zero yes, it turns out to be true irregardless (even though the Kodaira vanishing theorem is false in general). $\endgroup$
    – Torsten Ekedahl
    Commented Aug 26, 2011 at 3:58
  • $\begingroup$ Yes, and even in characteristic zero, Mumford's vanishing is a little stronger than the traditional statement of KVT. I.e. he shows one gets vanishing except in the degree equal to the number n of negative eigenvalues of a non singular curvature operator. The characteristic zero proof is almost the same as for KVT. I.e. the eigenvalues of the kth degree curvature operator are sums of k of the usual eigenvalues, which can be scaled on an abelian variety so that any positive one is larger than the sum of the negative ones. That does it in degree > n, and K=0 --> duality does the rest. QED. $\endgroup$
    – roy smith
    Commented Aug 27, 2011 at 17:25
  • $\begingroup$ that was a little sketchy. if you have read the proof by Bochner's inequality, as in Kodaira and Morrow, you know you get vanishing in degree k whenever the degree k curvature operator is positive, so this is the goal of the previous argument. I.e. if every positive eigenvalue is larger (in absolute value) than the sum of all negative ones, then any sum of k eigenvalues will be positive, as long as k > the number of negative eigenvalues. Thus you get vanishing in all degrees k and above. I learned this argument from notes of Deligne. $\endgroup$
    – roy smith
    Commented Sep 5, 2011 at 22:35

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