Let $(M^n,g)$ be a smooth Riemannian manifold. It is well known that if $sec(M)\geq \kappa$ then $Ric(M)\geq (n-1)\kappa$.


If I understand correctly in dimensions $n\geq 4$ a lower bound on $Ric(M)$ does not imply a lower bound on $sec(M)$. However in $n=2$ this implication is true for trivial reason.


**Is it true that in $n=3$ a lower bound on $Ric(M)$ implies a lower bound on $sec(M)$?**



Relation between Ricci curvature and sectional curvature for 3-manifolds
This is definitely false. In dimension 3 if $\lambda\_1,\lambda\_2,\lambda\_3$ are eigenvalues of the curvature operator then Ricci curvatures of eigenvectors are $\lambda\_1+\lambda\_2, \lambda\_1+\lambda\_3, \lambda\_2+\lambda\_3$. If one of $\lambda\_i$'s is very negative but the other two are very positive then all these sums can be bounded below. Every algebraic curvature tensor can be realized at a point so locally this can definitely happen. Complete examples exist too. I can't seem to find a reference to an explicit example but it's well known that there are examples of metrics on $\mathbb R^3$ with $Ric\ge 0$ which have some negative sectional curvature. If you rescale such metric by a small number Ricci curvature remains nonnegative but sectional curvature can become arbitrary negative in some 2-planes.


