Holomorphic quasi-positive line bundles on a complex manifold $M$ are line bundles whose chern class can be represented by a closed $(1,1)$-form which is quasi-positive, that is, non-negative at all points on M and strictly positive at least at one point on $M$. Positive line bundles are defined similarly.

Now my question is: Are holomorphic quasi-line bundles on a compact complex manifold $M$ positive? The answer is negative, because there are non-projective Moishezon manifolds. A Moishezon manifold has a quasi-positive line bundle. So, quasi-positive line bundles on a non-projective Moishezon manifold cannot be positive.

So now I ask a related question: Are holomorphic quasi-line bundles on a compact Kähler manifold $M$ positive? The answer seems to be negative. But can one give an example, that is, a quasi-positive line bundle that is not positive? Or prove it, if it is true. Thank you.

  • $\begingroup$ Consider a projective manifold with line bundle $L$. Then $L$ is nef if and only if it admits a semipositive metric, and it is ample if and only if it admits a positive metric. Moreover, if $L$ is big then a semipositive metric will be positive at some point. But a big and nef divisor need not be ample. (Take the pullback of an ample divisor by a blow up.) $\endgroup$ Oct 14, 2015 at 19:48
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    $\begingroup$ @Vesselin Dimitrov: the statement that $L$ is nef if and only if it admits a semipositive metric is false, see example 1.7 here www-fourier.ujf-grenoble.fr/~demailly/manuscripts/dps1.pdf $\endgroup$
    – YangMills
    Oct 17, 2015 at 5:48
  • $\begingroup$ @YangMills: Yes, the $\Rightarrow$ implication does not hold - thank you for correcting me and pointing to to that counterexmpe! This is not however used in the example I gave, which is the same one as yours (a pullback). $\endgroup$ Oct 17, 2015 at 17:28
  • $\begingroup$ @Vesselin Dimitrov: sorry, for some reason I did not notice that you already gave that example in your comment, and I repeated it in my answer... $\endgroup$
    – YangMills
    Oct 18, 2015 at 4:21
  • $\begingroup$ Why is it true that for a $L$ with semi-positive metric, big implies (and is equivalent to?) being positive at some point? $\endgroup$
    – user125639
    Feb 22, 2022 at 15:11

1 Answer 1


Take $X$ any smooth projective variety of dimension at least $2$, let $H$ be an ample (i.e. positive) line bundle on $X$, let $\pi:Y\to X$ be the blow-up of a point in $X$ and let $L=\pi^*H$. Then $L$ is quasipositive (being the pullback of a positive line bundle by a holomorphic map which is an isomorphism on a Zariski open set) but not positive (its intersection number with the exceptional divisor of $\pi$ is zero), and $Y$ is a projective manifold (in particular Kähler).

  • $\begingroup$ YangMills, Vesselin: Nice answer. Just want to confirm whether the following is true: suppose x is a point of the blow-up arbitrarily close to the exceptional divisor (but not lying in it), that means there is no complex curve C (or complex submanifold in general) passing through x which is homologous to any curve (or 2k-dimensional cycle in general respectively) within the exceptional divisor, otherwise the intersection of L and C is strictly positive. $\endgroup$
    – Wai
    Oct 19, 2015 at 21:28
  • $\begingroup$ I mean the intersection number of L and C is strictly positive. $\endgroup$
    – Wai
    Oct 19, 2015 at 21:37
  • $\begingroup$ That's right, in the case that we described (the blowup of a point), if a curve is homologous to another curve which is contained in the exceptional divisor E then its intersection number with L must be zero, while if a curve passes through a point outside E then its intersection number with L is strictly positive (and the same holds for submanifolds instead of curves). $\endgroup$
    – YangMills
    Oct 21, 2015 at 3:23

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