Are 3 dimensional Einstein manifolds isotropic?

In A. L. Besse, "Einstein Manifolds", it is shown that

A 2 or 3 dimensional pseudo-Riemannian manifold is Einstein if and only if it has constant sectional curvature.

My question is if this does imply that 2, trivially, and 3 dimensional Einstein manifolds are isotropic? I think this follows by definition, but I have seen apparent contradicting assertions regarding what is an isotropic manifold and I would like to ask for clarification.

Obs: Indeed there seems to be some ambiguity with the definition of Isotropic Manifold. The one I was taught is:

A (semi) Riemannian manifold $$\left(\mathcal{M},g\right)$$, where $$g$$ the metric tensor, is isotropic at a point $$p\in\mathcal{M}$$ if its sectional curvature is a constant K for every section $$\Pi\subset T_{p}\mathcal{M}$$. If $$\left(\mathcal{M},g\right)$$ is isotropic at all points then, it is called isotropic.

Is this definition not equivalent to the one in Wikipedia's article? Using this definition we do find that a flat torus is isotropic, as seems reasonable to expect.

• There is no such implication. While it is true that simply connected manifolds of constant sectional curvature (i.e. space forms) are isotropic, not every Riemannian manifold of constant sectional curvature is a space form. The flat torus has zero sectional curvature and is not isotropic. For more details, see this Wikipedia page Jan 7, 2023 at 2:49
• I don't know if it's relevant, but there are Riemannian manifolds which have constant curvature but are not isotropic, e.g., lens spaces. Do these also work in the pseudo-Riemannian setting? Jan 7, 2023 at 2:49
• We need a definition of isotropic. The wikipedia page gives a different definition than I am used to. The flat torus is absolutely isotropic, as far as I am concerned: At each point, all sectional curvatures are equal. Jan 7, 2023 at 3:31
• @TedShifrin I expect the same, a flat torus should definitely be isotropic, yet Wiki disagrees. I have included the definition that I've been taught back in the day, yet I can not find it in any book...
– PML
Jan 7, 2023 at 11:09

I checked a fair number of books on Riemannian (and semi-Riemannian) geometry. Not all of these define isotropic (semi)-Riemannian manifolds, but all that do agree on the following definition (same as the Wikipedia article):

1. A Riemannian manifold $$M$$ is called isotropic if for every $$x\in M$$ and every pair of unit vectors $$u, v\in T_xM$$ there exists an isometry $$f: M\to M$$ fixing $$x$$ and sending $$u$$ to $$v$$.

2. A semi-Riemannian manifold $$M$$ is called isotropic if for every $$x\in M$$ and every pair of non-null vectors $$u, v\in T_xM$$, satisfying $$=$$, there exists an isometry $$f: M\to M$$ fixing $$x$$ and sending $$u$$ to $$v$$.

Thus, a manifold of constant curvature need not be isotropic. However, if you assume that a constant curvature Riemannian manifold is complete and simply-connected, then it is isotropic. (I am unsure about semi-Riemannian manifolds of constant curvature.)

References:

Berger, Marcel, A panoramic view of Riemannian geometry, Berlin: Springer (ISBN 3-540-65317-1/hbk). xxiii, 824 p. (2003). ZBL1038.53002.

Gallot, S.; Hulin, D.; Lafontaine, J., Riemannian geometry, Universitext. Berlin etc.: Springer-Verlag. XI, 248 p.; DM 48.00 (1987). ZBL0636.53001.

Lee, John M., Riemannian manifolds: an introduction to curvature, Graduate Texts in Mathematics. 176. New York, NY: Springer. xv, 224 p. (1997). ZBL0905.53001.

O’Neill, Barrett, Semi-Riemannian geometry. With applications to relativity, Pure and Applied Mathematics, 103. New York-London etc.: Academic Press. xiii, 468 p. {\$} 45.00 (1983). ZBL0531.53051.

Wolf, J. A., Spaces of constant curvature, New York-St. Louis-San Francisco-Toronto-London-Sydney: McGraw-Hill Book Comp. XV, 408 p. (1967). ZBL0162.53304.

Edit. I can only say that the definition given to you (in lecture notes?) is nonstandard. The first part ("isotropic at a point") is consistent with the standard definitions, especially the one in Volume 3 of Spivak's book, where he uses the terminology an isotropic point rather than isotropic at a point.

However, the second part of the definition (of an isotropic manifold) given to you contradicts the standard one. What's worse, is that your definition is not equivalent to the standard one even under the assumption that the manifold is complete and simply-connected:

(a) If a connected semi-Riemannian manifold (of dimension $$\ge 3$$) is isotropic at each point (in the sense of your definition), then it has constant sectional curvature (Schur's theorem). But this is, of course, false in dimension 2. Applying this, together with the Poincare-Hopf theorem, one obtains that if a complete, simply-connected Riemannian manifold has dimension at least 3 and is isotropic at each point, then the manifold is isometric to one of the standard spaces of constant curvature (the Euclidean space or sphere of some radius or hyperbolic space of some constant negative curvature). From this, it follows that the manifold is isotropic in the standard sense.

(b) However, there are connected Riemannian manifolds isotropic in the standard sense which do not have constant curvature: These are various rank 1 symmetric spaces. The simplest example is the complex-projective space $$CP^n$$ ($$n\ge 2$$) with the Fubini-Study metric.

I can only say that it is regretful that your professor gave you a definition inconsistent with the standard one in the literature.

Spivak, Michael, A comprehensive introduction to differential geometry. Vol. 3, Boston, Mass.: Publish or Perish, Inc. V, 474 p. (1975). ZBL0306.53001.

• Interesting. Spivak was my source (and it was from him that I learned this material). Kobayashi-Nomizu do not define the term and just prove Schur's Theorem without a term for the hypothesis. Since I gave away almost all my library when I retired, I cannot look through a pile of books. Funnily, Chern, Chen, Lam call this a point "wandering" if the sectional curvatures at the point are all equal. Jan 7, 2023 at 5:45
• Thank you very much for your time and effort, for my purposes, imposing that the manifold is simply-connected and complete is enough, Nonetheless, I've edited the question to include a definition that I was taught. Using that, It does not seem to need the manifold to be complete, though. I'm I wrong and everything is equivalent?
– PML
Jan 7, 2023 at 11:12
• PS: In Barrett O’Neill's book, the author does not seem to require that the manifold is complete and simply connected. For instance, the autor only requires the isotropic manifold to be path-connected so that it is also homogeneous.
– PML
Jan 7, 2023 at 11:23
• @PML: For your last comment, I did not say that an isotropic manifold is required to be complete or simply connected. There are examples of non-simply connected isotropic manifolds (such as real projective spaces). But there is a theorem that an isotropic Riemannian manifold is necessarily complete. If you look closely, I said that a complete simply connected manifold of constant curvature is isotropic (which is just a consequence of the Poincare-Hopf theorem). Jan 7, 2023 at 13:39
• @TedShifrin: Somehow I forgot to check Spivak's books! He does not define isotropic manifolds, but does define isotropic points, as those with the same curvature along all planes (just as you said). Thus, he is careful not to contradict the standard terminology. Jan 7, 2023 at 14:00