TITLE: Can we obtain Ricci curvature tensor from sectional curvature?
QUESTION [5 upvotes]: Let $(M,g)$ be a $3$-dimensional Riemannian manifold and $\{e_1,e_2,e_3\}$ an orthonormal frame. About sectional curvature of this manifold we know that
$$K(e_1,e_2)=a,\quad K(e_1,e_3)=b,\quad K(e_3,e_2)=c,$$
My question is : Can we determine Ricci Curvature from above assumption?
Thanks.

REPLY [5 votes]: You know only the components $R_{1212}, R_{2323}, R_{3131}$ (and those determined from these by the usual symmetries); so you cannot determine e.g. $R_{1231}$. Since knowing the Ricci tensor is the same thing as knowing the Riemann tensor in dimension $3$, this implies  that you cannot determine the full Ricci tensor.
We're tacitly relying on the fact that the usual (non-differential) symmetries of the Riemann tensor are the only constraints on a $4$-tensor in order for it to be the curvature (at a single point) of some metric  - to prove this, you just need the formula for the metric in normal coordinates.