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I am thinking of a possible generalization of Einstein metrics (or a possible characterization of Einstein metrics) on four-dimensional manifolds, \begin{equation*} \mathrm{Ric}\circ\mathrm{Ric}=\lambda^2 g\circ g, \end{equation*} for some $\lambda\in\mathbb{R}$, where $\circ$ is the Kulkarni-Normizu product.

If $g$ is Einstein, then obviously it satisfies this condition. I can show that if $\lambda\not=0$, or if $\lambda=0$ and $R=0$, then $g$ is Einstein. If $\lambda=0$ and $R\not\equiv0$, then the Ricci curvature has eigenvalues $0, 0, 0, R$, I guess this is impossible, or are there any examples?

I was wondering is there any previous work on this condition? Thank you very much.

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I don't really know about any previous literature on the topic. But here is an example of a metric that satisfies the condition above but is not Einstein (in sloppy notation, but I hope it's clear anyway):

Let $M=\mathbb{R}_{+}\times S^2\times S^1$ and $g=dr^2+r^2(g_{S^2}+g_{S^1})$ where $\text{Ric}(g_{S^2})=2g_{S^2}$. Then \begin{equation*} \text{Ric}(g)=\text{Ric}(g_{S^2})-2(g_{S^2}+g_{S^1})=-2g_{S^1}. \end{equation*} One can check that $\text{Ric}\circ\text{Ric}=0$.

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    $\begingroup$ But this metric is incomplete, is it possible to find a complete one? $\endgroup$
    – user38600
    May 5, 2014 at 18:10

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