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

13$\begingroup$ 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 simplyconnected Ricciflat manifolds generic holonomy are known. See references in mathoverflow.net/questions/16818/… and especially Berger's "Panaramic view of Riemannian geometry". $\endgroup$ – Igor Belegradek Jun 18 '11 at 3:11

5$\begingroup$ 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. $\endgroup$ – Anton Petrunin Jun 18 '11 at 16:07

3$\begingroup$ @Igor, you right, there is no logic, but no $S^1$action either :) $\endgroup$ – Anton Petrunin Jun 19 '11 at 13:25

4$\begingroup$ 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 1form. By Bochner, a harmonic 1form 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. $\endgroup$ – Ben McKay Jun 20 '11 at 8:19

14$\begingroup$ @Igor: In fact, the holonomy of any metric on the $n$sphere is $SO(n)$, not just the Ricciflat ones; i.e., the $n$sphere does not admit any metric of reduced holonomy. $\endgroup$ – Robert Bryant Jun 21 '11 at 19:57