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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.

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    $\begingroup$ Since $(M^{2n},J,\omega)$ is Kähler, its holonomy is contained in $U(n) \subset SO(2n)$. If in addition it is Ricci-flat, then the holonomy is contained in $SU(n)$ since the Ricci-form is essentially the determinant. Now $SU(n)$ leaves invariant a nonzero $(n,0)$-form and hence, by the holonomy principle, there is a parallel nonzero $(n,0)$-form $\Omega$. In particular, $\Omega$ is holomorphic. If properly normalised, you get the equation in the question, since both $\omega^n$ and $\Omega \wedge \bar\Omega$ are nonzero $(n,n)$-forms. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Oct 18, 2012 at 9:05
  • $\begingroup$ where does one use here simply conectedness? and does one use any compacness of $M$ ? $\endgroup$
    – Miguel
    Commented Oct 18, 2012 at 9:09
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    $\begingroup$ Dear José, don't we need simple connectedness to know that the reduced holonomy group is the entire holonomy group for this argument to work? I think a finite quotient of a simply connected C-Y manifold may not have any nonzero holomorphic $(n,0)$-form, while still admitting Ricci-flat metrics. $\endgroup$
    – Gunnar Þór Magnússon
    Commented Oct 18, 2012 at 9:15
  • $\begingroup$ You can relax the simply-connected requirement in odd dimensions. The universal covering space of a Ricci-flat Kähler manifold is Calabi-Yau (just apply the argument from Spiro's answer). A simple application of the Atiyah-Bott fixed point formula then gives that in odd dimensions, every Ricci-flat Kähler manifold is Calabi-Yau, and in even dimensions, it is either Calabi-Yau, or it has fundamental group $\mathbb{Z}_2$. $\endgroup$
    – Rhys Davies
    Commented Oct 19, 2012 at 10:22
  • $\begingroup$ @Rhys: what you say can't be correct, think about a bielliptic surface (finite quotient of a complex $2$-torus). It does not have trivial canonical bundle (so it is not "Calabi-Yau" in the OP's sense), but the fundamental group is not $\mathbb{Z}_2$. Also, I hope that by "odd dimension" you mean "odd complex dimension", since the manifolds are Kahler. $\endgroup$
    – YangMills
    Commented Oct 19, 2012 at 13:52

1 Answer 1

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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.

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  • $\begingroup$ Spiro, could you clarify how you conclude that $c_1$ is trivial? I can only see its triviality in de Rham cohomology, which doesn't see the torsion. If I understood correctly, you're not presupposing simple connectivity. $\endgroup$
    – Tim Perutz
    Commented Oct 18, 2012 at 13:40
  • $\begingroup$ @Tim: You are correct, we need simple connectivity to ensure that $\Lambda^{(n,0)}(M)$ is topologically trivial, since that ensures that $H^2(M, \mathbb Z)$ has no torsion. I will edit my post again. Thanks for catching that. $\endgroup$
    – Spiro Karigiannis
    Commented Oct 18, 2012 at 15:38
  • $\begingroup$ what do you mean by topologically trivial ? if the first chern class vanishes doesnt then follow that the cannonical bundle is trivial, hence we get a holomorphic $(n,0)-$form that trivializes it? $\endgroup$
    – Miguel
    Commented Oct 19, 2012 at 14:13
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    $\begingroup$ @Spiro: your sentence "which would give you holonomy $SU(n)$" is not literally correct, all you get is that the holonomy is contained in $SU(n)$ but it could be strictly smaller, such as $SU(p)\times SU(q)$ for $p+q=n$ (reducible case), $Sp(n/2)$ (hyperkahler case), etc. $\endgroup$
    – YangMills
    Commented Oct 19, 2012 at 15:42
  • $\begingroup$ @YangMills: Yes, of course, I should have been more clear. I will edit that. $\endgroup$
    – Spiro Karigiannis
    Commented Oct 20, 2012 at 12:42

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