# Existence of a global section for principal polarization $\mathcal L$ on abelian variety $X/k$.

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).

• Show vanishing of higher cohomlogy by Kodaira vanishing. – Samir Canning Jul 4 at 15:23
• @SamirCanning Thanks, I was just writing an answer using Kodaira vanishing. – Ruben Jul 4 at 15:24

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 $$\chi(\mathcal L) = \text{dim}_k\text{H}^0(X,\mathcal L)$$.
• 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 4 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 4 at 16:06