1
$\begingroup$

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.

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ 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. $\endgroup$
    – naf
    May 24, 2013 at 9:18
  • $\begingroup$ Yes, I need simplicity, thank you. And thank you for pointing me at Iyer. $\endgroup$
    – Charles Siegel
    May 27, 2013 at 2:23

1 Answer 1

Reset to default
3
$\begingroup$

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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ The base point free part also holds for 4-folds by the work of Kawamata. $\endgroup$
    – Fei YE
    May 26, 2013 at 5:24
  • $\begingroup$ 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 $n\geq 3$. $\endgroup$
    – Charles Siegel
    May 27, 2013 at 2:19
  • 1
    $\begingroup$ 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). $\endgroup$
    – Hacon
    May 27, 2013 at 6:04
  • $\begingroup$ @hacon: thanks for the reference, that's almost exactly the sort of thing I need $\endgroup$
    – Charles Siegel
    May 29, 2013 at 1:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.