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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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
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$\begingroup$ 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
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$\begingroup$ 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
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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 projective-Moishezon and Kähler-Fujiki). 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 Demailly-Paun.
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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 Levi-Civita 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
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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
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$\begingroup$ 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
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$\begingroup$ 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 non-compact. I just treated one aspect of the question: the compact case. $\endgroup$ Commented Feb 21, 2012 at 7:49
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$\begingroup$ @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