Is a simply connected Ricci-flat Kaehler manifold a Calabi-Yau manifold? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T15:06:02Z http://mathoverflow.net/feeds/question/109987 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109987/is-a-simply-connected-ricci-flat-kaehler-manifold-a-calabi-yau-manifold Is a simply connected Ricci-flat Kaehler manifold a Calabi-Yau manifold? Miguel 2012-10-18T08:06:08Z 2012-10-20T12:43:32Z <p>Hi,</p> <p>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.</p> <p>Miguel B. </p> http://mathoverflow.net/questions/109987/is-a-simply-connected-ricci-flat-kaehler-manifold-a-calabi-yau-manifold/110008#110008 Answer by Spiro Karigiannis for Is a simply connected Ricci-flat Kaehler manifold a Calabi-Yau manifold? Spiro Karigiannis 2012-10-18T12:23:41Z 2012-10-20T12:43:32Z <p>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.]</p> <p>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 <em>topologically</em> trivial. So there exists a nowhere vanishing <em>smooth</em> $(n,0)$-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</p> <p>$$\frac{\omega^n}{n!} = (-1)^{\frac{n(n-1)}{2}} \Omega \wedge \overline \Omega.$$</p> <p>Since $\Omega$ is type $(n,0)$ and the complex structure is integrable, then $\Omega$ will be holomorphic (and thus the canonical bundle is <em>holomorphically</em> 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 $SU(n)$.</p> <p>Compactness is needed to go the other way: Yau's theorem says that <em>if</em> $M$ 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.</p> <p>A good elementary reference is Chapter 6 of <A HREF="http://books.google.ca/books?id=c3P-YUD8GZQC&amp;printsec=frontcover#v=onepage&amp;q&amp;f=false" rel="nofollow">Compact Manifolds with Special Holonomy</A> by Dominic Joyce.</p>