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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16$\begingroup$ 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 PetruninCommented Jun 8, 2011 at 6:54
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2$\begingroup$ See this answer math.stackexchange.com/questions/47323/scalar-flat-metrics $\endgroup$– user21574Commented Jun 8, 2017 at 14:53
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2$\begingroup$ 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 $\endgroup$– user21574Commented Jun 8, 2017 at 15:30
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1$\begingroup$ 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 $\endgroup$– user21574Commented Jul 20, 2017 at 20:46
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2 Answers
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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. $\endgroup$ Commented Jun 20, 2011 at 2:28
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On a compact connected 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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1$\begingroup$ 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. $\endgroup$ Commented Nov 15, 2018 at 14:01
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$\begingroup$ @Bilateral See mathoverflow.net/a/294346/394 $\endgroup$ Commented Apr 11, 2020 at 19:03
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1$\begingroup$ If you are familiar with Ricci flow, the easiest argument to prove "On a compact connected manifold that does not carry a metric of positive scalar curvature, every scalar flat metric is Ricci-flat." is as follows: Assume that $g_t$, $t\in [0,\epsilon)$ is a solution of the Ricci-flow with $g_0$ scalar flat. Then it is a standard results within the theory of Ricci flow, thar either $g_0$ is Ricci-flat (and thus $g_0=g_t$ for all $t$) or $g_t$ has positive scalar curvature for all $t>0$. A proof of his should be contained in every advanced textbook on Ricci flow. $\endgroup$ Commented Oct 13, 2020 at 12:46