Let $L$ be a pseudoeffective line bundle on a complex manifold $X$ then is there a singular hermitian metric of $L$ which its curvature is not semipositive?

3$\begingroup$ Well, yes, even if $L$ is ample one can take some K\"ahler metric $\omega$ and subtract some large multiple of $dd^c\phi$ for some $\phi$. If you choose a decent $\phi$ this will no longer be semipositive. $\endgroup$ – Ruadhaí Dervan Jul 12 '16 at 10:33

$\begingroup$ @Jarek: When you say "not semipositive", do you mean "nowhere semipositive"? Then the trivial bundle on the projective line is pseudoeffective, but any connection on it has semipositive curvature somewhere: no, trivially. On the other hand, if you mean "not semipositive at some point", then Ruadhai's comment applies: yes. $\endgroup$ – Ben McKay Jul 12 '16 at 12:28
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There is a singular Hermitian metric on $L$ whose curvature is semipositive. See Demailly, On the cohomology of pseudoeffective line bundles, for a very clear survey.