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Is that true that if a Riemannian manifold $(M,g)$ has parallel Ricci tensor i. e., $\nabla Ric=0$, then the manifold has constant scalar curvature?

I've seen this result with some aditional hypothesis and I'd like to know if it is true in this case as well, if it is a open problem or if there is some counter-example.

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    $\begingroup$ $scal = \mathrm{tr} Ric$ and $\nabla \mathrm{tr} = \mathrm{tr} \nabla$. $\endgroup$ – Alexander Thumm Sep 14 '13 at 15:24
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    $\begingroup$ Sure; scalar curvature is the trace of $Ric$, so it's also parallel and thus constant. $\endgroup$ – user8268 Sep 14 '13 at 15:25
  • $\begingroup$ Thank you Alexander and user8268, what I've seen is another notion of parallelism. $\endgroup$ – Myself Sep 14 '13 at 15:29
  • $\begingroup$ what does $\nabla\text{Ric}=0$ mean? $\endgroup$ – rmdmc89 Jun 6 at 17:59

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