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Let $(M,g)$ be a Riemannian manifold and $Ric$ its Ricci curvature. Let $\{e_1,\cdots, e_n\}$ be specific orthonormal frame and in this frame we have: $$Ric(e_i,e_i)>0\quad \forall i=1,...,n$$

My question is

Does this condition imply positive Ricci curvature tensor?i.e., $Ric(X,X)>0$ for all vector field $X$?

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  • $\begingroup$ It is true for Einstein manifolds $\endgroup$ – Semsem Sep 8 '17 at 19:02
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This is really a linear algebra question - you know that the diagonal of a symmetric matrix is positive, and you want to know whether that matrix is positive-definite. The answer is no: consider for example $$ \mathrm{Ric}(e_i,e_j) = \left[\begin{matrix} 1 & -2 \\ -2 & 1 \end{matrix}\right]_{ij},$$ which has $$\mathrm{Ric}(e_1+e_2, e_1+e_2) = -2 .$$

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