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A Riemannian manifold with constant sectional curvature is Einstein. Why?

It's true the inverse?

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closed as off-topic by quid, TravisJ, Alex Provost, user99914, Leucippus Jul 16 '16 at 0:23

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By definition, a Riemannian manifold has constant sectional curvature if the sectional curvature $K$ is a constant that is independent of the point and $2$-plane chosen. If $R$ denotes the covariant curvature tensor and $g$ is the metric then, as a consequence of the definition of $K$, the components satisfy the relation $$R_{ljhk} = K(g_{lh}g_{jk}-g_{lk}g_{jh}).$$

Multiplication with the contravariant metric tensor $g^{jk}$ yields $$R_{lh} = K(ng_{lh} - g_{lh}) = K(n-1)g_{lh},$$

from which we conclude that our manifold is Einstein.

For a counterexample of the converse, note that $\mathbb{C}P^n$ is Einstein but its sectional curvature is not constant (except for the sphere $n=1$).

However I believe that the converse is true for manifolds of dimension $\leq 3$.

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    $\begingroup$ More precisely, $CP^n$ with its Fubini-Study metric. For the case of dimension $\le 3$ see A. Besse "Einstein manifolds", Prop. 1.120. $\endgroup$ – Moishe Kohan Jul 15 '16 at 17:43

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