# Specific line bundle over complex manifold implies Kähler?

Let $L$ be a holomorphic line bundle on complex manifold $X$, such that it admits a hermitian structure whose Chern connection has positive curvature. Is $X$ then Kähler?

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Yes, by definition of Kahler manifold. Moreover by Kodaira embedding theorem such complex manifold is projective – Dmitri Jul 18 '11 at 21:26
My definition of Kahler does not refer to these line bundles. Sorry if I am overseeing something. – Matt Fahrad Jul 18 '11 at 22:44
Matt: Kahler usually means that you have a metric $h_{ij}$ such that the associated form $\omega = const. \sum h_{ij}dz_i\wedge d\bar z_j$ is closed. You can work backwards to see that any closed positive $(1,1)$-form $\omega$ is a Kahler form. In your case, let $\omega$ be the Chern form of your positive line bundle $L$. Does that make it clearer? – Donu Arapura Jul 19 '11 at 0:32

As Dima said, it is much more: in fact it is projective. But let me give you some more insights on this kind of questions.

I shall give you the definition of four different classes of compact complex manifolds.

1. Projective manifold: closed submanifold of some complex projective space.
2. Moishezon manifold: compact complex manifold such that the field of meromorphic functions on it has transcendence degree equal to its complex dimension.
3. (Compact) Kähler manifold: compact complex manifold carrying a Kähler form, that is a closed positive smooth (1,1)-form.
4. Manifold in the Fujiki class ($\mathcal C$): compact complex manifold bimeromorphic to a compact Kähler manifold.

A Moishezon manifold can be shown to be bimeromorphic to a projective manifold, so that -in some sense- Moishezon manifolds are with respect to projective manifolds as manifolds in the Fujiki class ($\mathcal C$) are with respect to Kähler manifolds.

It turns out, that one can characterize these four classes in terms of cohomological properties (these characterizations reflect again this relation between projective-Moishezon and Kähler-Fujiki). Here is the characterization for you:

1. A compact complex manifold is projective if and only if it carries a (1,1) rational cohomology class which can be represented by a positive (1,1)-form (or equivalently if it carries a positive hermitian line bundle). This is the content of Kodaira's embedding theorem.
2. A compact complex manifold is Kähler if and only if it carries a (1,1) real cohomology class which can be represented by a positive (1,1)-form. This is almost the definition.
3. A compact complex manifold is Moishezon if and only if it carries a (1,1) rational cohomology class which can be represented by a (1,1) Kähler current, that is a (1,1)-closed positive current which is bounded from below by a (non necessarily closed) smooth positive (1,1)-form (or equivalently if it carries a big line bundle).
4. A compact complex manifold is in the Fujiki class ($\mathcal C$) if and only if it carries a (1,1) real cohomology class which can be represented by a (1,1) Kähler current. This is the content of a theorem by Demailly-Paun.
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I find myself objecting more and more to the definition of a Kahler manifold as one carrying a hermitian metric with closed Kahler form. It's a very nice characterization for calculations, but it tells nothing about where this property comes from or what it means. I much prefer saying a Kahler metric is one whose Chern connection is torsion free, i.e. the Chern connection is the Levi-Civita connection of the real part of the hermitian metric, because the question of when that happens at least arises fairly naturally. – Gunnar Þór Magnússon Dec 16 '11 at 10:01
... Now given a hermitian metric $h$ with Kahler form $\omega$, we write $\tau$ for its torsion tensor. Then we have $h \circ \tau = \partial \omega$, where composition with $h$ denotes raising of indices. This makes clear the equivalence between "Chern connection torsion free" and $d \omega = 0$. – Gunnar Þór Magnússon Dec 16 '11 at 10:04
The question did not ask about compact manifolds. You need compactness as well as a positive line bundle to ensure projectivity. – Ben McKay Feb 20 '12 at 22:13
Ben, incidentally I know that a projective manifold has to be compact (closed in a compact is compact, right?)... This question did not state either that the manifold had to be non-compact. I just treated one aspect of the question: the compact case. – diverietti Feb 21 '12 at 7:49
@Gunnar: Do you have a reference which takes the approach you mention (in particular, shows $h\circ\tau = \partial\omega$)? – Michael Albanese Jun 27 '12 at 8:37