This may well be an open problem, I'm not sure.

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 $8$.

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 $n$.

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?

Bonus question: How about if the manifold is pseudoriemannian?


Thanks to Igor's answer below, here are some further remarks.

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 compact examples.

In fact, in Berger's 2003 book A panoramic view of Riemannian geometry (page 645) one reads at the bottom of the page:

It remains a great mystery that no Ricci flat compact manifolds are known which do not have one of these special holonomy groups.


2 Answers 2


I am not an expert but the question:

"Does there exist a simply-connected closed Riemannian Ricci flat $n$-manifold with $SO(n)$-holonomy?"

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.

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    $\begingroup$ Berger's book is pretty old by now, but I think the situation remains as you describe. As far as I know, every construction of a closed Ricci-flat manifold proceeds by building a metric with a holonomy group that implies Ricci-flat. The holonomy assumption implies algebraic conditions on the curvature tensor that in turn play crucial roles in the elliptic estimates used to prove existence. In some sense, you're replacing a second order elliptic PDE by a stronger first order system that implies the second order PDE. $\endgroup$
    – Deane Yang
    Mar 2, 2010 at 15:28
  • $\begingroup$ Is the book by Berger that you mention the same as A panoramic view of Riemannian geometry? $\endgroup$ Mar 2, 2010 at 16:48
  • $\begingroup$ Another question: the Schwarzschild metric I know is lorentzian, so it has holonomy $\mathrm{SO}(3,1)$. Hence this answers the bonus question! This prompts another question: are there noncompact Ricci-flat riemannian manifolds with generic holonomy? $\endgroup$ Mar 2, 2010 at 16:55
  • $\begingroup$ Jose, yes this is the book. Another standard source is "Surveys in Differential Geometry: Essays on Einstein manifolds" by International Press which was also published a decade ago. I do not think there have been any major breakthroughs since then in regard to your question. $\endgroup$ Mar 2, 2010 at 16:56
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    $\begingroup$ Jose, the Schwarzschild metric I know is Riemannian. :) Look e.g. in Petersen's "Riemannian Geometry" text for detailed study of Riemannian Schwarzschild metric $\endgroup$ Mar 2, 2010 at 17:01

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.

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 $T_xM$).

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.

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. I'm fairly certain that local SO(n) holonomy is incompatible with being Ricci flat. Edit: This guess was wrong. See Igor's answer for a counterexample.

  • $\begingroup$ Right. So the question is then whether there is a proof of the statement that Ricci-flatness reduces the holonmy. $\endgroup$ Mar 2, 2010 at 12:09
  • $\begingroup$ The Berger theorem is purely a local theorem. Do you two believe that Ricci-flat in a neighborhood of a point implies reduced holonomy? I don't know the answer (I recall asking Robert Bryant this question but don't recall the answer), but it seems unlikely to me. The Ricci-flat condition is, morally speaking, a determined system of PDE's, so in the real analytic category, it can be solved using Cauchy-Kovalevski with arbitrary initial data along a hypersurface. This seems like too much flexibility for reduced holonomy. $\endgroup$
    – Deane Yang
    Mar 2, 2010 at 14:09
  • $\begingroup$ @Deane: Ricci-flatness in a neighbourhood does not imply reduced holonomy, now that I think about it. Imagine taking the round sphere in $\mathbb{R}^3$ and ironing out a corner in a way that the metric is still smooth (though perhaps not analytic). The sphere is flat in some neighbourhood where I applied the iron, but the holonomy group is not zero: consider a lasso based at a flat point, but where the loop is on the part of the sphere which is still curved. $\endgroup$ Mar 2, 2010 at 16:28
  • $\begingroup$ @José: But your example is not everywhere Ricci-flat. The local holonomy near a flat point is certainly still zero. $\endgroup$
    – Deane Yang
    Mar 2, 2010 at 16:37
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    $\begingroup$ @José: While it's true that, in general, holonomy is not local, when the connection is real-analytic and the underlying manifold is simply-connected, it is local. In other words, if you have a real-analytic connection on a simply-connected manifold, you can, in theory, compute the holonomy just by examining the germ of the connection at a single point. Since Ricci-flat metrics are real-analytic (by the work of DeTurck and Kazdan), it follows that, for a simply-connected, Ricci-flat manifold, you can compute the holonomy locally (at least in theory). $\endgroup$ Apr 23, 2011 at 14:21

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