Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Recently, I have been learning about nef line bundles. I know that when $X$ is projective or Moishezon, a line bundle $L$ over $X$ is said to be nef iff $$L.C=\int_{C}c_{1}(L)\ge 0$$ for every curve $C$ in $X$.

Demailly gave a definition of nefness that works on an arbitrary compact complex manifold, i.e., a line bundle $L$ over $X$ is said to be nef if for every $\varepsilon >0$ there exists a smooth hermitian metric $h_{\varepsilon}$ on $L$ such that its curvature $\Theta_{h_{\varepsilon}}(L)\ge -\varepsilon\omega$. For projective manifolds, Demailly's definition coincides with the above one given by integration (this is an easy consequence of Seshadri's ampleness criterion).

Question: Is this equivalence also true for Moishezon manifolds?

I don't know of any counterexamples. If it is not true, could someone give me a counterexample?

share|improve this question
3  
I've cleaned up the layout of your question a bit. We usually put spaces after periods and commas, because this makes sentences easier to read. –  S. Carnahan May 24 '11 at 4:46
    
The proof that Demailly's definition implies the classical one carries over word-for-word for Moishezon manifolds (or any manifold that admits curves). The other direction is less clear, as we only have big bundles on a Moishezon manifold, and I don't know of a Nakai-Moishezon type criterion for bigness. I think one can work something out, probably by replacing the smooth hermitian metric $h_\epsilon$ by a singular hermitian metric and passing to curvature currents, but I'm not sure that will be equivalent to Demailly's definition with smooth metrics. –  Gunnar Magnusson May 30 '11 at 11:47

2 Answers 2

up vote 3 down vote accepted

Yes, the equivalence is true (the second notion used to be called "metric nef" by some). This was an open problem for quite some time until it was solved in

M. Paun "Sur l'effectivité numérique des images inverses de fibrés en droites" Math. Ann. 310 (1998), no. 3, 411–421, see the Corollaire on page 412.

share|improve this answer
1  
Thank you very much.And I think I should take some time to learn some French first... –  Unknown Jun 17 '11 at 2:19

what is $\omega$ here? if it is a Kahler form, then Moishezon + Kahler implies projective, and as you said they are equivalent.

share|improve this answer
2  
$\omega$ is the Kahler form of a hermitian metric - i.e. it is the associated real $(1,1)$-form $\omega = -i \Im h$, but it need not be closed. –  Gunnar Magnusson May 30 '11 at 11:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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