I've seen that here https://www.math.uchicago.edu/~ngo/Shimura.pdf there's a theorem called Mumford's Vanishing Theorem (Theorem 2.2.2) which says:
Let $\mathcal{L}$ be a line bundle on $X$ (abelian variety) 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 $H^i(X,\mathcal{L})\neq (0)$. Moreover, $\mathcal{L}$ is ample if and only if $i(\mathcal{L})=0$.
This theorem has no proof here, and I'm intersted in understanding the last claim because, in the book "Abelian Varieties" of Mumford, the Vanishing Theorem doesn't say this result but it actually uses it implicity in the beginning of proof of the theorem at page 163 (old edition).
So my question is: why is it true that $\mathcal{L}$ is ample if and only if $i(\mathcal{L})=0$? In particular I'm intersted in the implication
$\mathcal{L}$ ample implies $i(\mathcal{L})=0$.
Thank you!
If $\mathcal{L}$ is ample, then it is nondegenerate (cf. page 84 of Mumford's book) and $h^0(\mathcal{L}^n)>0$ for some $n>0,$ where $h^q(\mathcal{L})=dim_k H^q(A, \mathcal{L}),$ so $i(\mathcal{L}^n)=0$ and hence also $i(\mathcal{L})=0$ by the Corollary of Mumford in page 159.