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Can you put a Ricci flat metric on the $n$-sphere, $n>4$?

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I am have not been following the subject recently, but I think, the existence of a Ricci flat $n$-sphere is an open problem for $>3$. I think any such metric will necessarily have generic holonomy, and no examples of compact simply-connected Ricci-flat manifolds generic holonomy are known. See references in mathoverflow.net/questions/16818/… and especially Berger's "Panaramic view of Riemannian geometry". –  Igor Belegradek Jun 18 '11 at 3:11
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It seems that such metric does not admit an isometric $S^1$-action. I.e. the metric if exists has almost no symmetry. Therefore it is unlikely that one can construct it. –  Anton Petrunin Jun 18 '11 at 16:07
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Compact $G_2$ manifolds also have finite symmetry groups, by the same argument as for the other holonomy groups: a vector field preserving the Riemannian metric would have to be dual to a harmonic 1-form. By Bochner, a harmonic 1-form must be parallel, and so the holonomy reduces, by deRham splitting theorem, to a product of holonomy groups, and the manifold is, up to finite covering, a product with a product metric. –  Ben McKay Jun 20 '11 at 8:19
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It is worthwhile to note that there can be no Killing vector fields for such a metric, because that would imply the existence of a non-trivial harmonic one-form. –  Viktor Bundle Jun 20 '11 at 16:09
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@Igor: In fact, the holonomy of any metric on the $n$-sphere is $SO(n)$, not just the Ricci-flat ones; i.e., the $n$-sphere does not admit any metric of reduced holonomy. –  Robert Bryant Jun 21 '11 at 19:57

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