# Why is the index $i(\mathcal{L})$ of an ample line bundle on an abelian variety equal to $0$?

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!

• The answer is the same as the answer to your question about nonvanishing of $H^0$ for an ample line bundle on an abelian variety: Kodaira vanishing and Riemann--Roch. – Nefertiti May 29 '16 at 13:35
• Thank you @Nefertiti but I still don't get how. Could you please give me some more details? – Poecilia May 29 '16 at 14:01