A metric is said to have harmonic curvature if the exterior derivative of the Ricci tensor vanishes. It is known that there exists manifolds with dRic=0 that are not Einstein. My question is whether or not there are examples of manifolds that admit a metric with harmonic curvature but don't admit an Einstein metric.
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1$\begingroup$ I find this question confusing. What does "dRic" mean? Is it the full covariant derivative of Ricci? If so, doesn't dRic = 0 imply harmonic curvature? Or are you asking a topological question: Does there exist a manifold that has metrics with harmonic curvature but none that are Einstein? $\endgroup$– Deane YangOct 27, 2013 at 16:57
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1$\begingroup$ I think by $dRic = 0$ the poster means $D_{i}R_{jk} = D_{j}R_{ik}$, since (Bianchi identities) this is equivalent to $D^{p}R_{ijkp} = 0$, which is sometimes called "harmonic curvature". $\endgroup$– Dan FoxOct 27, 2013 at 18:01
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$\begingroup$ Deane: Dan has the right idea. dRic is $D_i R_{jk} = D_j R_{ik}$. $\endgroup$– Viktor BundleOct 28, 2013 at 16:25
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1 Answer
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There are many $3$-manifolds that admit a metric with harmonic curvature that don't admit an Einstein metric because in dimension $3$, 'Einstein' is equivalent to constant sectional curvature, while in dimension $3$, harmonic curvature is the same as conformally flat with constant scalar curvature.
Since one can take connected sums of conformally flat manifolds and get conformally flat manifolds and since, by Schoen's solution of the Yamabe problem, every compact $3$-manifold has a conformal change to one with constant scalar curvature, one sees that there are many more compact $3$-manifolds that admit a metric with harmonic curvature than there are compact $3$-manifolds that have an Einstein metric.
answered Oct 27, 2013 at 23:21