Let $(M,g)$ be a Riemannian manifold with vanishing Ricci curvature. Does this mean that every component of the Ricci curvature tensor vanshes, i.e., $R_{ij}=0$?
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1Yes indeed as we know that $Ric=R_{ij}dx^i\otimes dx^j$ [Using Einstein summation convention]. Further $dx^i\otimes dx^j$ is a basis element so $Ric=0 \iff R_{ij}=0$. – Sujit Bhattacharyya Oct 05 '19 at 06:38
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May I know what is your definition of "vanishing Ricci curvature"? – Arctic Char Oct 05 '19 at 20:40