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It is stated in Joyce's 'Compact Manifolds with Special Holonomy' (P.124) that if $M$ is a compact Riemannian symmetric space, then $M$ is Ricci flat implies $M$ is flat. I am having trouble seeing why this is true.

In fact, could I not choose a Ricci flat Lie group, such as $G_2$ (which has holonomy $G_2$, see Holonomy of Lie groups, hence flat) to get a counterexample?

It would be great if anyone can explain Joyce's statement and see what is wrong with my proposed counterexample. Thanks!

Chi Cheuk Tsang
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1 Answers1

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I think you also need simply connected and irreducible to conclude that. simply connected + irreducible implies Einstein see[*]section2, Cor3. However in this case the sectional curvature is positive if the symmetric space is compact, see[*] Theorem 6.

[*] "Lecture Notes on Symmetric Spaces J.-H. Eschenburg"

Alan Zhang
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  • Symmetric spaces of compact type have non-negative sectional curvature. They have strictly positive curvature exactly when they are rank $1$, so many examples attain zero sectional curvature. – Max Dec 29 '20 at 23:48