# When is an ample line bundle on an abelian variety base point free?

So, any line bundle $L$ on an abelian variety $X$ determines a type $(d_1,\ldots,d_g)$ where $d_i|d_{i+1}$. It's well known that if $d_1\geq 3$ then $L$ defines an embedding, that if $L$ has no fixed components, $L^{\otimes 2}$ is very ample, and for $d_1=2$ we have base point freeness.

What about the case where $d_1=1$? Certainly it's not always base point free, for instance, in the case of a principal polarization. But what about nonprincipal polarizations? Are there any known criteria for determining if they are base point free or not? And if there is a base locus, for determining it (which presumably requires knowledge of $L$, not just the type of $L$)?

The only result I can find in this direction is that for an abelian surface with a $(1,d)$ polarization, we have no base points for $d\geq 3$ and for $d=2$ we have exactly 4 base points, but what about higher dimensional $X$?

EDIT: As pointed out below, I want to assume that $X$ is simple.

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I think what you say for abelian surfaces is not true without extra assumptions, like simplicity or genericity (consider the case of a product of elliptic curves). There are some results known in higher dimensions with similar assumptions; see, for example. papers of J. Iyer. – ulrich May 24 '13 at 9:18
Yes, I need simplicity, thank you. And thank you for pointing me at Iyer. – Charles Siegel May 27 '13 at 2:23

According to Fujita's conjecture, if $L$ is an ample divisor on a smooth complex variety $X$ of dimension $n$, then

1. $K_X+mL$ is basepoint-free for $m\geq n+1$, and

2. $K_X+mL$ is very ample for $m\geq n+2$.

Of course, if $X$ is an abelian variety, then $K_X=0$.

This conjecture is known for surfaces by the work of Reider and for threefolds by the work of Ein and Lazarsfeld.

I don't know if there are any more results of this kind specific to abelian varieties, but it is presumably an easier case and one may use more tools.

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The base point free part also holds for 4-folds by the work of Kawamata. – Fei YE May 26 '13 at 5:24
Unfortunately, the case I need information on is when the defined polarization has type with $d_1=1$ and this gives $d_1=m>n+1$, so doesn't help. Also, for abelian varieties we can do better: if $L$ is ample, then $L^2$ is bpf and $L^n$ is very ample when $n\geq 3$. – Charles Siegel May 27 '13 at 2:19
Just to mention that Fujita's conjecture also has a version for $K_X+L$ where you assume something like $L^d\cdot V>n$ for all irreducible subvarieties where $0<d\leq dim X$. If you replace $>n$ by a quadratic bound then this was proven by Angehrn-Siu. For general $(X,L)$ there is a relevant result in arXiv:alg-geom/9306004 (I am sure there are others). – Hacon May 27 '13 at 6:04
@hacon: thanks for the reference, that's almost exactly the sort of thing I need – Charles Siegel May 29 '13 at 1:46