Is a simply connected Ricci-flat Kaehler manifold a Calabi-Yau manifold?

2012-10-18T08:06:08Z

Hi,

I have the following question: Let $(M,\omega, J)$ be a simply connected Kaehler manifold with Ricci-flat Kaehler metric. How can one show that $M$ is a Calabi-Yau manifold. By Calabi-Yau manifold I mean that there exists a holomorphic $(n,0)-$form $\Omega$ such that the following equation is satisfied: $\frac{\omega^{n}}{n!} = (-1)^{\frac{n(n-1)}{2}}(\frac{i}{2})^{n} \Omega \wedge \bar{\Omega}$. Should one put the assumption on $M$ to be compact? But what kind of compactness? With or without boundary? Is this necessary? Does this also work without any compactness assumption? Where can I find a proof of this? Is there any reference? Or is this question too trivial? I hope that someone has the answer and also hope for a lot of replys. Thanks in advance.

Miguel B.

Answer by Spiro Karigiannis for Is a simply connected Ricci-flat Kaehler manifold a Calabi-Yau manifold?

2012-10-18T12:23:41Z

José is correct, with the caveat that Gunnar mentioned - you need simple-connectedness to know that reduced holonomy = holonomy. Below I expand a bit more on the details. [Thanks to Tim Perutz for catching errors in the initial version of this answer.]

Notice that the OP did not ask for $\Omega$ to be parallel or even closed. The following is true: If $(M, J, g, \omega)$ is Ricci-flat Kaehler, then the image of the first Chern class $c_1 (M)$in $H^2 (M, \mathbb R)$ vanishes, so that if $\pi_1(M) = 0$, then $H^2(M, \mathbb Z)$ has no torsion, and thus the canonical bundle $\Lambda^{n, 0} (M)$ is topologically trivial. So there exists a nowhere vanishing smooth $(n,0)$-form $\Omega$ that trivializes the canonical bundle. By consideration of type, $\Omega \wedge \overline \Omega$ is a nonvanishing $(n,n)$-form, so by rescaling $\Omega$ by a nowhere vanishing complex valued function, one gets for "free" the identity that

$$ \frac{\omega^n}{n!} = (-1)^{\frac{n(n-1)}{2}} \Omega \wedge \overline \Omega.$$

Since $\Omega$ is type $(n,0)$ and the complex structure is integrable, then $\Omega$ will be holomorphic (and thus the canonical bundle is holomorphically trivial) if and only if it is closed. Since $M$ is Ricci-flat, the Bochner theorem tells you that an $(n,0)$ form is closed if and only if it is parallel, which would give you holonomy contained in $SU(n)$.

Compactness is needed to go the other way: Yau's theorem says that if $M$ is compact Kaehler and $c_1 (M) = 0$, then there exists a unique Ricci flat Kaehler metric in each Kaehler class. There are noncompact examples where uniqueness fails. I don't know as much as I should about the literature on existence in the noncompact case, but the papers of Tian-Yau should have the answer.

A good elementary reference is Chapter 6 of Compact Manifolds with Special Holonomy by Dominic Joyce.

Is a simply connected Ricci-flat Kaehler manifold a Calabi-Yau manifold? - MathOverflow

Is a simply connected Ricci-flat Kaehler manifold a Calabi-Yau manifold?

Answer by Spiro Karigiannis for Is a simply connected Ricci-flat Kaehler manifold a Calabi-Yau manifold?