Picture below is from Hamilton's Three manifolds with positive Ricci curvature. I know why $R(u,v,u,v)>0$. Since the secional curvature of sphere is positive, I have $$ \frac{R(u,v,u,v)}{|u\wedge v|^2} = K(u,v)>0 $$ therefore, I agree $R(u,v,u,v)>0$. But, why $R(u,u)>0$ on sphere ?
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1First you cannot deduce $R(u,v,u,v)>0$ just from $K>0$ in general. Second Ricci curvature is sum of sectional curvature. see this post – C.F.G Sep 13 '21 at 04:26
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Since it is just the sphere, why don't you just calculate and check? – Arctic Char Sep 13 '21 at 08:11
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@C.F.G Sorry, why I can't deduce $R(u,v,u,v)>0$ from $K>0$ ? I fail to find my mistake. – Enhao Lan Sep 13 '21 at 11:52
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@lanse7pty: you can ask this as a separate post. – C.F.G Sep 13 '21 at 13:07
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Let $u_1\in T_pM$ with $|u_1|=1$ and extend it to an onb $u_1,\dots,u_n$. Then
$$R(u_1,u_1)= \sum_{i=1}^nR(u_1,u_i,u_1,u_i)=\sum_{i=2}^nR(u_1,u_i,u_1,u_i)=\sum_{i=2}^nK(u_1,u_i)>0$$
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