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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.

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    $\begingroup$ The standard exercise is that in the case of a $3$-dimensional manifold, the Ricci curvature totally determines the entire curvature tensor. But, either way, for that or for your exercise, the point is to count dimensions. Both are determined by a symmetric $3\times 3$ matrix. $\endgroup$ – Ted Shifrin Aug 14 '17 at 20:25
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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.

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  • $\begingroup$ Suppose $Ric(e_1,e_1)=f_1, Ric(e_2,e_2)=f_2, Ric(e_3,e_3)=f_3$. Then what is the value of $Ric(e_1,e_2)$? $\endgroup$ – C.F.G Aug 14 '17 at 15:55
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    $\begingroup$ Ok, that took me by surprise, I am pretty sure you can retrieve the Ricci tensor IF you got the value of $K$ for every 2-plane in the tangent space (that's nearly obvious since $K$ determines the curvature tensor entirely). I was pretty sure one could find those values having only the values for an orthonormal base, but maybe I'm wrong. I'll edit as soon as I can $\endgroup$ – Alessio Di Lorenzo Aug 14 '17 at 16:20
  • $\begingroup$ @AlessioDiLorenzo: $\rho$ is a self-adjoint operator on $\wedge^2 TM$. $\endgroup$ – Ted Shifrin Aug 15 '17 at 1:45
  • $\begingroup$ Of course, thanks for the correction! $\endgroup$ – Alessio Di Lorenzo Aug 15 '17 at 1:48
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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.

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  • $\begingroup$ If we knew that $R(X,Y)e_1=A(X,Y)+Ric(X,e_1)Y-Ric(Y,e_1)X$ for some $(1,2)$-tensor field $A$ then what we can say about Ricci? $\endgroup$ – C.F.G Aug 15 '17 at 5:04
  • $\begingroup$ what we can say about $Ric(e_1,X)$ for $X\perp e_1$? $\endgroup$ – C.F.G Aug 15 '17 at 6:43
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    $\begingroup$ Not sure exactly what you're getting at with your first question - do you mean to ask whether we can express $\mathrm Ric$ in terms of $a,b,c$ and $A$? For your second question, these are exactly the parts of $\mathrm Ric$ we don't know. $\endgroup$ – Anthony Carapetis Aug 15 '17 at 7:52

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