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In Huybrecht's book on Fourier-Mukai transforms the following argument is used in a proof

Since $(X,\mathcal L)$ is a principally polarized abelian variety one has a unique global section $s : \mathcal O_X \to \mathcal L$.

The only things that seem to be used here is that a $\mathcal L$ is a principal polarization, so an ample line bundle such that $\chi(\mathcal L) = 1$.

This argument seems very brief to me. Can anyone explain to me this argument in more detail?

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This seems to follow from the fact (Mumford, Abelian Varieties) that $\chi(\mathcal L) = \text{dim}_k\text{H}^0(X,\mathcal L)$ (of which I have no proof).

Ruben
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1 Answers1

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Samir and Ruben have mentioned in the comments that this follows from Kodaira vanishing. I'll type out an answer (I myself found this instructive).

Kodaira vanishing in its algebraic form asserts that for $X$ a smooth projective $k$-scheme of dimension $d$ over a field of characteristic zero, and $\mathcal{L}$ ample, we have $$ H^q(X, \mathcal{L} \otimes \Omega^p_{X/k} ) = 0 \qquad \text{ for } \qquad p+q > d. $$ Since $X$ is an abelian variety, its cotangent sheaf is trivial, i.e. $\Omega_{X/k} \cong \mathcal{O}_X$. But then in particular we have $H^q(X, \mathcal{L} \otimes \Omega^d_{X/k} ) \cong H^q(X, \mathcal{L} ) = 0 $ for $d+q>d$, i.e. for $q>0$. It then follows by definition of the Euler characteristic of a sheaf, $$ \chi(\mathcal{L}) = (-1)^i \sum_i \dim_k H^i(X, \mathcal{L}) $$ that $\chi(\mathcal L) = \text{dim}_k\text{H}^0(X,\mathcal L)$.

mathphys
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  • Thank you for your trouble, I was already typing an answer using Hartshorne's version of Kodaira vanishing. One thing I find confusing with your version of Kodaira vanishing is this: What if I set $p = d + 1$? Then your statement says $\text{H}^i(X, \mathcal L) = 0$ for $d + 1 + q > d$, i.e., $q > -1$. In other words, that implies $\text{H}^0(X, \mathcal L) = 0$. – Ruben Jul 04 '19 at 15:57
  • Oh, I guess $\Omega_X^p = 0$ for $p > d$. The notation confused me, since $\Omega_{X/k}^d := \bigwedge^d \Omega_{X/k}$ and $\omega_X := \bigwedge^{\dim X} \Omega_{X/k}$, our proofs are really the same. I'll delete mine and accept yours. – Ruben Jul 04 '19 at 16:06
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    Oh sorry, I didn't realise you were the OP when you commented! No worries, yes the Hartshorne way is probably more to the point!. As for your question, yes I think the reasoning you've given above is correct. Thanks for accepting! :) – mathphys Jul 04 '19 at 16:07