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Let $(M,g)$ be a Riemannian manifold and assume that for all orthonormal $v,z$ the sectional curvatures is bounded from below i.e. $K(v,z) \geq C$, where $C > 0$. Is it in this case possible for the Ricci curvature to vanish? Or is this condition, on the sectional curvature, very strong? Sorry if the question is too trivial :).

Gunam

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Wikipedia says that you also get a lower bound on the Ricci curvature. –  Sam Nov 10 '12 at 17:00
    
how exactly? do you know? –  Gunam Nov 10 '12 at 17:12

1 Answer 1

You basically just have to look at the definitions: Given a unit-length tangent vector $x\in T_pM$, we obtain the Ricci curvature of $x$ at $p$ by extending $x=z_n$ to an orthonormal basis $z_1, \ldots, z_n$. And then $$\mathrm{Ric}_p(x) = \frac{1}{n-1}\sum_{i=1}^{n-1} \langle R(x,z_i)x,z_i\rangle$$ where $R$ denotes the Riemannian curvature tensor. On the other hand, the sectional curvature $K(x, z_i)$ for $i<n$ is given by (remember the $z_i$ are orthonormal) $$K(x, z_i) = \frac{\langle R(x,z_i)x,z_i\rangle}{\Vert x\Vert\, \Vert z_i\Vert-|\langle x, z_i\rangle|^2}=\langle R(x,z_i)x,z_i\rangle$$ Since $K(x,z_i) \ge C$ by your assumption, we obtain $$\mathrm{Ric}_p(x) = \frac{1}{n-1}\sum_{i=1}^{n-1} \langle R(x,z_i)x,z_i\rangle \ge C > 0$$ So you get a lower bound on the Ricci curvature.

Attention: All sign conventions are as can be found in DoCarmo's "Riemannian Geometry".

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