Timeline for Is a simply connected Ricci-flat Kaehler manifold a Calabi-Yau manifold?
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| Oct 19, 2012 at 15:45 | comment | added | YangMills | @Rhys: thanks for the precis. Unfortunately what you say is still not correct, because even if $M$ is compact, Ricci-flat Kahler with finite fundamental group, all that you can conclude about its universal cover is that it has holonomy contained (or equal) to $SU(n)$, but it might be strictly smaller, when $\tilde{M}$ is not irreducible, or it is hyperkahler, etc. What you say applies when $\tilde{M}$ has holonomy precisely equal to $SU(n)$. | |
| Oct 19, 2012 at 14:30 | comment | added | Rhys Davies | @YangMills: Thank-you for pointing that out. I was implicitly assuming a finite fundamental group (also compactness, to be certain!); in this case, what I said is true, and for surfaces, the only examples are K3 and Enriques surfaces. And yes, I am always referring to the complex dimension; I probably should have stated that explicitly. | |
| Oct 19, 2012 at 13:52 | comment | added | YangMills | @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. | |
| Oct 19, 2012 at 10:22 | comment | added | Rhys Davies | 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 $\mathbb{Z}_2$. | |
| Oct 18, 2012 at 12:23 | answer | added | Spiro Karigiannis | timeline score: 5 | |
| Oct 18, 2012 at 9:15 | comment | added | Gunnar Þór Magnússon | 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. | |
| Oct 18, 2012 at 9:09 | comment | added | Miguel | where does one use here simply conectedness? and does one use any compacness of $M$ ? | |
| Oct 18, 2012 at 9:05 | comment | added | José Figueroa-O'Farrill | 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. | |
| Oct 18, 2012 at 8:06 | history | asked | Miguel | CC BY-SA 3.0 |