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What are the examples of Riemannian manifolds that have zero scalar curvature but non zero ricci curvature? Is there any sort of classification of such manifolds?

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    $\begingroup$ There are a LOT of examples, first thing comes to mind is a products of unit sphere and surface of constant curvature $-1$. This condition is too soft (opposite of rigid), you can not expect to have a classification. $\endgroup$ – Anton Petrunin Jun 8 '11 at 6:54
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    $\begingroup$ See this answer math.stackexchange.com/questions/47323/scalar-flat-metrics $\endgroup$ – user21574 Jun 8 '17 at 14:53
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    $\begingroup$ To generalize your question in Kähler $M$, If $ω$ a Kähler metric of constant scalar curvature with $\pi c_1(M)=λ[\omega]$,, then $\omega$ is Kähler-Einstein metric. See Proposition 2.12 in the book of Gang Tian springer.com/in/book/9783764361945 $\endgroup$ – user21574 Jun 8 '17 at 15:30
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    $\begingroup$ Let for symplectic manifold $(X,\omega)$ we have $ [ω]=λ⋅c_1(X)$ for some $λ∈R_{>0}$, such manifolds are called monotone symplectic manifold. Fukaya category of a monotone symplectic manifold are very important to verify HMS $\endgroup$ – user21574 Jul 20 '17 at 20:46
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To generalize Anton's comment a little, I should add that with the appropriate choice of $l$ and $k$, the product manifold $S^l \times N^k$ will have the property that you are looking for, where $N^k$ has hyperbolic $k$-dimensional half-space space as its cover. You can find the formulas for all of the geometric quantities related to these sorts of products in Chang, Han, Yang "On a class of locally conformally flat manifolds". This particular combination of manifolds can be used to construct many examples of manifolds with interesting curvature.

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  • $\begingroup$ Also, keep in mind that in three dimensions Einstein implies constant curvature, so the three sohere carries a scalar flat metric that is not Ricci flat. $\endgroup$ – Viktor Bundle Jun 20 '11 at 2:28
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On a compact manifold that does not carry a metric of positive scalar curvature, every scalar flat metric is Ricci-flat. Thus on such manifolds there are no such metrics.

If a manifold carries a metric of positive scalar curvature then it also carries a metric of zero scalar curvature. I assume that for dimension at least 3 one could extend this statement to saying that it even admits a scalar flat metric with non-zero Ricci curvature, but I have no proof at hand currently. What I know for sure is: there are many manifolds having an obstruction against Ricci-flat metrics and admitting a metric of positive scalar curvature. On Ricci-flat manifolds, the first Betti number is at most the dimension, and if it is the dimension then the manifold is flat. This yields manifolds without Ricci-flat metrics and many of them carry a metric of positive scalar curvature.

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    $\begingroup$ Could you please give a reference about the statement "On a compact manifold that does not carry a metric of positive scalar curvature, every scalar flat metric is Ricci-flat."? Thanks. $\endgroup$ – Bilateral Nov 15 '18 at 14:01
  • $\begingroup$ @Bilateral See mathoverflow.net/a/294346/394 $\endgroup$ – José Figueroa-O'Farrill Apr 11 at 19:03

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