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Michael Albanese
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Assume that $(M^n,g)$ is an $n$ dimensional ($n>=3$$n \geq 3$)closed closed Riemannian manifold with constant scalar curvature and $Ric_g$ nonnegative. Then is $g$ Einstein?

Assume that $(M^n,g)$ is an $n$ dimensional ($n>=3$)closed Riemannian manifold with constant scalar curvature and $Ric_g$ nonnegative. Then is $g$ Einstein?

Assume that $(M^n,g)$ is an $n$ dimensional ($n \geq 3$) closed Riemannian manifold with constant scalar curvature and $Ric_g$ nonnegative. Then is $g$ Einstein?

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Mathboy
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Assume that $(M^n,g)$ is an n-dimensional $n$ dimensional ($n>=3$)closed Riemannian manifold with constant scalar curvature and Ric_g$Ric_g$ nonnegative. Then is $g$ Einstein?

Assume that $(M^n,g)$ is an n-dimensional ($n>=3$)closed Riemannian manifold with constant scalar curvature and Ric_g nonnegative. Then is $g$ Einstein?

Assume that $(M^n,g)$ is an $n$ dimensional ($n>=3$)closed Riemannian manifold with constant scalar curvature and $Ric_g$ nonnegative. Then is $g$ Einstein?

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Mathboy
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when constant scalar curvature implies Einstein?

Assume that $(M^n,g)$ is an n-dimensional ($n>=3$)closed Riemannian manifold with constant scalar curvature and Ric_g nonnegative. Then is $g$ Einstein?