I know some examples of compact complex manifolds whose first Chern class does not have a definite sign (is neither negative, nor positive nor zero on all complex curves). I am looking for a necessary and sufficient condition that the first Chern class has a definite sign.
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4$\begingroup$ What do you mean by "has sign"? Do you mean "is negative"? $\endgroup$– MishaCommented Jun 15, 2012 at 23:31
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$\begingroup$ I revised my question $\endgroup$– user21574Commented Jun 15, 2012 at 23:34
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1$\begingroup$ Positive in the sense of Kahler geometry? You can look up Kodaira's embedding theorem. But I'm not sure if that's what you're after. $\endgroup$– Donu ArapuraCommented Jun 15, 2012 at 23:44
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$\begingroup$ Dear Donu, I mean on Kahler manifolds with complex dimension n $\endgroup$– user21574Commented Jun 16, 2012 at 0:51
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1 Answer
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The first Chern class of a Kaehler manifold is the cohomology class of the Ricci form. A sufficient condition that the first Chern class is positive (negative) is that the Ricci curvature is positive (negative). See Aubin's book Nonlinear Analysis on Manifolds: Monge-Ampere Equations. A sufficient condition for the first Chern class being nonnegative is that you find a holomorphic section of the anticanonical bundle which is nonzero on a dense open set. A necessary condition for the first Chern class to be positive (negative) is that the integral of the Ricci form is positive (negative) over all compact complex curves in the manifold.
answered Jun 16, 2012 at 6:22