Let $X$ be an abelian variety.

In "Mumford, Abelian Varieties" the Riemann-Roch Theorem has the following form:

For all line bundles $\mathcal{L}$ on $X$, if $\mathcal{L}\cong\mathcal{O}_X(D)$, we have $\chi(\mathcal{L})=\frac{(D^g)}{g!}$, $\chi(\mathcal{L})^2=\deg\phi_\mathcal{L}$, where $(D^g)$ is the $g$-fold self-intersection number of $D$.

They have told me (here and here) that I should use this form of the Riemann-Roch Theorem and the Kodaira Vanishing Theorem to prove the following equivalent results:

  • $\mathcal{L}$ ample implies $\dim H^0(X,\mathcal{L})>0$;
  • $\mathcal{L}$ ample implies $i(\mathcal{L})=0$.

Where $i(\mathcal{L})$ comes from the Vanishing Theorem in Mumford:

Let $\mathcal{L}$ be a line bundle on $X$ such that $K(\mathcal{L})$ is finite. There exists a unique integer $i=i(\mathcal{L})$, $0\leq i(\mathcal{L})\leq g=\dim X$, such that $H^p(X,\mathcal{L})=(0)$ for $p\neq i$ and $H^i(X,\mathcal{L})\neq (0)$.

Can someone give me more details of how to do that? Thanks!

  • $\begingroup$ If you use Kodaira vanishing, both your bulletted statements are immediate. I am not sure what exactly is your confusion. $\endgroup$ – Mohan May 30 '16 at 13:23

Let $L$ be an ample line bundle on an abelian variety $X$ of dimension $g$. Then, by Kodaira vanishing we see that $H^i(X,L)=H^i(X,\omega_X\otimes L)=0$ for all $i>0$. In particular $\chi(X,L)=\sum_{i=0}^g(-1)^i h^i(X,L)=h^0(X,L)$. Now, since $L$ is ample we also know that $(L^g)\ne 0$ and then Riemann-Roch implies that $h^0(X,L)\ne 0$.


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