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How do we prove that a compact Kahler manifold whose 1st Chern class vanishes admits a globally defined nowhere vanishing volume form? Thanks.

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    $\begingroup$ Why are "we" interested in this question? $\endgroup$
    – Igor Rivin
    Apr 24, 2014 at 21:07

1 Answer 1

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The vanishing of the $c_1$ of a compact Kahler manifold does not imply that the canonical bundle is trivial. The converse is true.

If $S$ is a hyperelliptic surfaces, that is a finite quotient of a complex $2$-dimensional torus (elliptic fibration over an elliptic curve), then $c_1(S) = 0$ but $K_S$ is not trivial. On the other hand $12K_S$ is trivial.

If $X$ is a Kahler manifold and $c_1(X) = 0$ then $nK_X$ is trivial for some integer $n$. You can find this in Theorem 3 of this paper:

http://iopscience.iop.org/0025-5726/8/1/A02

On the other hand if you require that the structure group of $X$ can be reduced from $U(n)$ to $SU(n)$ you can proceed as follws. By Calabi-Yau theorem there exists a Ricci-flat Kahler metric. This yields a flat connection on the canonical bundle. Now, you consider the Albanese map. In this case it is a locally trivial fibration with Calabi-Yau fibers having trivial first Betti number. By Bochner's theorem the holomorphic $1$-forms are parallel. If the canonical bundle is trivial on the base and on the fiber, then it is trivial. For the base this follows from the fact that it is a torus. The fiber is Calabi-Yau and its first cohomology group vanishes. Now, one concludes by Bogomolov's decomposition theorem.

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