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?

7$\begingroup$ Yes, by definition of Kahler manifold. Moreover by Kodaira embedding theorem such complex manifold is projective $\endgroup$– Dmitri PanovCommented Jul 18, 2011 at 21:26

$\begingroup$ My definition of Kahler does not refer to these line bundles. Sorry if I am overseeing something. $\endgroup$– Matt FahradCommented Jul 18, 2011 at 22:44

$\begingroup$ 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? $\endgroup$– Donu ArapuraCommented Jul 19, 2011 at 0:32
1 Answer
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.
 Projective manifold: closed submanifold of some complex projective space.
 Moishezon manifold: compact complex manifold such that the field of meromorphic functions on it has transcendence degree equal to its complex dimension.
 (Compact) Kähler manifold: compact complex manifold carrying a Kähler form, that is a closed positive smooth (1,1)form.
 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 projectiveMoishezon and KählerFujiki). Here is the characterization for you:
 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.
 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.
 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).
 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 DemaillyPaun.

1$\begingroup$ 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 LeviCivita connection of the real part of the hermitian metric, because the question of when that happens at least arises fairly naturally. $\endgroup$ Commented Dec 16, 2011 at 10:01

1$\begingroup$ ... 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 $d \omega = 0$. $\endgroup$ Commented Dec 16, 2011 at 10:04

$\begingroup$ The question did not ask about compact manifolds. You need compactness as well as a positive line bundle to ensure projectivity. $\endgroup$ Commented Feb 20, 2012 at 22:13

$\begingroup$ 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 noncompact. I just treated one aspect of the question: the compact case. $\endgroup$ Commented Feb 21, 2012 at 7:49

$\begingroup$ @Gunnar: Do you have a reference which takes the approach you mention (in particular, shows $h\circ\tau = \partial\omega$)? $\endgroup$ Commented Jun 27, 2012 at 8:37