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Let $(M,g)$ be a closed, smooth, Riemannian manifold of dimension $n>4$. Suppose both the scalar curvature and norm of the Ricci tensor are constant. In addition suppose that $g$ satisfies the following condition: $(\frac{1}{2} \Delta Ric_{ij} + Ric^{ml}W_{milj} - \frac{3n}{4(n-1)^2}Ric_{ij} -\frac{2}{n-2}Ric^l_j Ric_{il}) = \lambda g_{ij}.$

Here $r$ is the scalar curvature, $Ric$ is the Ricci tensor, $W$ is the Weyl tensor, $\Delta = g^{-1}\nabla\nabla$, and $\lambda$ is a real number. This condition is the Euler-Lagrange equation of a quadratic Riemannian functional. Notice that Einstein metrics satisfy it. Can one conclude that $g$ is Einstein?

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What about the Cartesian product of the standard sphere and a flat torus? – Deane Yang Jun 21 '11 at 14:13
Yang: That does the job. I'm going to add another hypothesis to it, to see if it forces Einstein-ness – Viktor Bundle Jun 21 '11 at 19:59
First, does this equation assume some kind of normalization of the metric? I do not understand why the terms do not all scale the same. Second, are you sure that my original counterexample, properly scaled, does not still work? The first two terms vanish, and you just get a quadratic polynomial. If you set $\lambda = 0$, then one of the roots is zero. The other is a number depending on $n$. If the latter is positive, use a sphere and otherwise hyperbolic space instead of the sphere. – Deane Yang Jun 22 '11 at 14:27
Deane: I like your argument. But why does the second term vanish? – Viktor Bundle Jun 24 '11 at 21:07

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