I have the following problem: Let $X$ be a $n-$dim Kähler manifold with Ricci-flat Kähler form $\omega$. There is a known fact that then there exists a holomorphic (n,0)-form $\Omega$ such that $\frac{\omega^{n}}{n!} = (-1)^{\frac{n(n-1)}{2}}(i/2)^{n} \Omega \wedge \bar{\Omega}$. And that $\Omega$ is also parallel with respect to the Levi-civita connection. Does anyone know where I can find a proof of this assertion? I would be very thankful for answers.

greetings bruno


You need some more hypotheses for the existence of the $(n,0)$-form, in general it will exist only up to finite torsion. For example, an Enriques surface does not admit a nowhere zero holomorphic $(2,0)$-form. You can also construct explicit examples with products of two or three elliptic curves quotiented by a group generated by well chosen involutions.

When such an $n$-form exists, the usual way to prove that it is parallel is to use a Bochner technique. This is sketched in the first pages of Beauville's "Variétés Kahleriennes dont la premiere classe de Chern est nulle" (the proof might work for the general case, i.e., for $(n,0)$-forms that have some zeros by using that the norm in the Bochner technique will not be zero at all points). If I recall correctly, a more complete version of the same argument is given in "Calabi-Yau manifolds and their friends" and in the opening chapters of "Mirror symmetry".

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    $\begingroup$ does one maybe need the assumption that $X$ is compact ? $\endgroup$ – bruno Sep 13 '12 at 10:22
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    $\begingroup$ Yes: if $\tau$ is a holomorphic $p$-form then the Bochner formula gives $\Delta \| \tau \|^2 = \| D \tau \|^2$, where $\Delta$ is the Laplacian and $D$ the connection, so the function $x \mapsto \|\tau(x) \|^2$ is subharmonic on the manifold. From there we need compactness to deduce it is constant and thus that $\tau$ is parallel. $\endgroup$ – Gunnar Þór Magnússon Sep 13 '12 at 11:43
  • $\begingroup$ and what about for the first assumption. is it enough to assume that $X$ is compact? $\endgroup$ – bruno Sep 13 '12 at 12:53
  • $\begingroup$ The tricky part in all this is getting a holomorphic form $\Omega$. If you have that, I think it is automatically parallel: a volume form $\omega^n/n!$ defines a smooth hermitian metric $h$ on the canonical bundle by $$h(\alpha,\beta) \omega^n/n! = i^{n^2} \alpha \wedge \overline \beta.$$ If $i^{n^2} \Omega \wedge \overline \Omega = \omega^n/n!$ then $\|\Omega\|_h^2 = 1$, so $\Omega$ is parallel by the Bochner formula. So if you have the form, it is parallel, but the manifold may not be compact (example: $\mathbb C$, $\Omega = dz$). $\endgroup$ – Gunnar Þór Magnússon Sep 13 '12 at 14:37

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