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?
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On
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.
This is just an attempt to find something useful to solve the problem.
We know that $$ Ric_m(x,y)=\sum^n_{i=1}R_m(x,e_i,y,e_i)\qquad x,y\in T_{M,m}. $$ Now, the sectional curvature is $$ K(x,y) = R(x,y,x,y), $$ if $x,y$ are orthonormal. Hence, we can see that $$ Ric_m(x,x)=\sum^n_{i=1}R_m(x,e_i,x,e_i)=\sum^n_{i=1}K_m(x,e_i). $$
Now, we know that our curvature tensor $R$ can be seen as a self-adjoint operator $\rho$ on $\wedge^2T_M$, and $$ K_m(x,y) = \frac{\rho_m(x\wedge y,x\wedge y)}{g_m(x\wedge y,x\wedge y)}. $$ Although $\{e_1\wedge e_2,e_1\wedge e_3,e_2\wedge e_3\}$ is a basis for $\wedge^2T_{M,m}$ in our case, we are in the same position as before: we have got only the values for the basis and that does not enable us to find the entire operator. Maybe the fact that our manifold is three dimensional lets us have some other information, but at this point I would say that the assumptions are not enough, but I have yet to find a counterexample.