Does there exist an Einstein manifold which is not conformally flat, which is to say one which has non-vanishing Weyl tensor. If so, what is a good example.
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7$\begingroup$ Since I don't do Riemannian geometry anymore, I might have this wrong. It seems to me that if the traceless Ricci and Weyl tensors vanish, then the Riemann curvature has to be of the form $R_{ijkl} = k(g_{ik}g_{jl}-g_{il}g_{jk})$, where, by the Bianchi identities, $k$ is constant. So conformally flat Einstein metrics have constant sectional curvature. If this is correct, then the simplest non-conformally-flat Einstein metric that comes to mind is complex projective space with complex dimension greater than $1$. $\endgroup$– Deane YangCommented May 29, 2014 at 13:34
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6$\begingroup$ @DeaneYang: Indeed, you are correct; conformally flat and Einstein implies constant sectional curvature in dimensions greater than $2$. The Fubini-Study metric on $\mathbb{CP}^2$ is Einstein but not conformally flat. $\endgroup$– Robert BryantCommented May 29, 2014 at 14:38
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2$\begingroup$ Another example is $S^2\times S^2$ with the standard product metric. $\endgroup$– user38600Commented May 29, 2014 at 23:24
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4 Answers
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All conformally flat homogeneous riemannian manifolds are symmetric spaces, by a result of Takagi. All homogeneous riemannian manifolds of dimension $\leq 11$ admit Einstein metrics, by results of Wang and Ziller. Most homogensous riemannian manifolds are not symmetric spaces.
References:
Takagi, H.: Conformally flat Riemannian manifolds admitting a transitive group of isometries I, II. Tohoku Math. J. 27, 103–110(I), 445–451(II) (1975)
McKenzie Y. Wang, Wolfgang Ziller: Existence and non-existence of homogeneous Einstein metrics, Inventiones mathematicae, 1986, Volume 84, Issue 1, pp 177-194
answered May 29, 2014 at 13:15
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K3 surface is Einstein, by Calabi-Yau theorem, but it is not conformally flat; this can be seen e.g. from topology, or from the reference that José Figueroa-O'Farrill has given above.
answered May 29, 2014 at 15:41
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Does your question refer to both Riemannian and semi-Riemannian spaces? If so, then the Schwarzschild metric is an example. It's a vacuum solution to the Einstein field equations (with zero cosmological constant), so it's an Einstein manifold. It has a nonvanishing Weyl tensor.
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By the standard curvature decomposition, for an Einstein metric, if the Weyl curvature vanishes, then the sectional curvature is constant, so other than space forms, all Einstein manifolds are not conformally flat.
Sorry I didn't see Deane Yang's answer, he already got this answer.
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1$\begingroup$ Oops. I forgot that Einstein already implies constant scalar curvature. $\endgroup$ Commented May 30, 2014 at 0:27