Are there Ricci-flat riemannian manifolds with generic holonomy? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T22:31:59Z http://mathoverflow.net/feeds/question/16818 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16818/are-there-ricci-flat-riemannian-manifolds-with-generic-holonomy Are there Ricci-flat riemannian manifolds with generic holonomy? José Figueroa-O'Farrill 2010-03-02T02:50:13Z 2010-03-02T21:55:17Z <p>This may well be an open problem, I'm not sure.</p> <p>In Berger's classification (refined by Simons, Alekseevsky, Bryant,...) of the holonomy representations of irreducible non-symmetric complete simply-connected riemannian manifolds, there are some cases which imply Ricci-flatness: namely, $\mathrm{SU}(n)$ (Calabi-Yau) in dimension $2n$, $\mathrm{Sp}(n)$ (hyperkähler) in dimension $4n$, $G_2$ in dimension $7$ and $\mathrm{Spin}(7)$ in dimension $8$.</p> <p>A natural question is the converse: whether Ricci-flatness implies a reduction of the holonomy. The other holonomy representations are known not to be Ricci-flat: $\mathrm{Sp}(n)\cdot \mathrm{Sp}(1)$ (quaternionic kähler) is known to be Einstein with nonzero scalar curvature, and in the case of $\mathrm{U}(n)$ in dimension $2n$ (Kähler) it is known that if a Kähler manifold is Ricci-flat then the holonomy is contained in $\mathrm{SU}(n)$, so that it is Calabi-Yau. So the remaining question is whether there exists any Ricci-flat riemannian manifolds with generic holonomy $\mathrm{SO}(n)$ in dimension $n$.</p> <p>I would like to know the present status of this question and if it's still open what the experts think: do people expect examples of Ricci-flat riemannian manifolds with generic holonomy?</p> <p><strong>Bonus question</strong>: How about if the manifold is pseudoriemannian?</p> <hr> <p><strong>Added</strong></p> <p>Thanks to Igor's answer below, here are some further remarks.</p> <p>The question needs to be refined. The riemannian analogue of the Schwarzschild metric on $\mathbb{R}^2 \times S^2$ is an example of a complete, simply-connected noncompact Ricci-flat metric with generic holonomy. So the question is about <em>compact</em> examples.</p> <p>In fact, in Berger's 2003 book <em>A panoramic view of Riemannian geometry</em> (page 645) one reads at the bottom of the page:</p> <blockquote> <p>It remains a great mystery that no Ricci flat compact manifolds are known which do not have one of these special holonomy groups.</p> </blockquote> http://mathoverflow.net/questions/16818/are-there-ricci-flat-riemannian-manifolds-with-generic-holonomy/16824#16824 Answer by Chris Schommer-Pries for Are there Ricci-flat riemannian manifolds with generic holonomy? Chris Schommer-Pries 2010-03-02T03:54:12Z 2010-03-02T21:55:17Z <p>The Ricci curvature is a local quantity, so I am only going to focus on the case that the local holonomy group is SO(n). Philosophically, the local holonomy group and the curvature of a connection attempt to measure the same thing. On the one hand, the curvature is the infinitesimal comparison of parallel transporting in two directions in differing orders. The heuristic picture is just an infinitesimal parallelogram. The local holonomy, on the other hand, measures the actual change along null-homotopic paths. Thus it is not too surprising that these two invariants are closely related. </p> <p>One reflection of this is captured by the Ambrose-Singer holonomy theorem. Roughly this theorem says that the lie algebra of the (local) holonomy group must be large enough to accomodate all the parallel translates of the curvature tensor evaluated on all 2-planes at a given point. In fact the lie algebra of the holonomy group consists of precisely these translates (for each loop and each 2-plane we get an endomorphism of $T_xM$). </p> <p>On the other hand, the curvature of a torsion free connection (such as the Levi-Civita connection) must satisfy the first and second Bianchi identities. This can also be related to the holonomy group and the space of tensors which satisfy these conditions becomes smaller and smaller as the Lie group becomes larger. </p> <p>Berger's classification theorem works by playing these two opposing forces off each other. The holonomy group must at the same time be both small and large and only certain groups satisfy both requirements. <strike>I'm fairly certain that local SO(n) holonomy is incompatible with being Ricci flat</strike>. <strong>Edit</strong>: This guess was wrong. See Igor's answer for a counterexample. </p> http://mathoverflow.net/questions/16818/are-there-ricci-flat-riemannian-manifolds-with-generic-holonomy/16864#16864 Answer by Igor Belegradek for Are there Ricci-flat riemannian manifolds with generic holonomy? Igor Belegradek 2010-03-02T15:22:15Z 2010-03-02T15:22:15Z <p>I am not an expert but the question: <p> "Does there exist a simply-connected closed Riemannian Ricci flat $n$-manifold with $SO(n)$-holonomy?" </p>is a well-known open problem. Note that Schwarzschild metric is a complete Ricci flat metric on $S^2\times\mathbb R^2$ with holonomy $SO(4)$, so the issue is to produce compact examples; I personally think there should be many. The difficulty is that it is hard to solve Einstein equation on compact manifolds. If memory serves me, Berger's book "Panorama of Riemannian geometry" discusses this matter extensively.</p>