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In chapter 5 of the book "Einstein Manifolds", Arthur Besse discusses the possibility to find the metric $g$ when knowing the Ricci curvature tensor $Ric(g)$ ($=R_{ij}$).

But what do we know about the problem of finding the metric by knowing the mixed Ricci tensor $R^i{}_j$?

The two problems seem to me different, because in the absence of the metric, the two tensors are not equivalent. A first step may be to find all possible symmetric tensors $g_{ij}$ which make $g_{ij}R^j{}_k$ symmetric, but this is far from solving the problem, because then $Ric(g)$ is not necessarily equal to $g_{ij}R^j{}_k$. So the next step may be to take each of these possible tensors $g_{ij}R^j{}_k$ and see if they are equal to $Ric(g)$, by using the known results for the $R_{ij}$ problem. But this seems much more complicated than the $R_{ij}$ problem. Perhaps it should be a more direct way.

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    $\begingroup$ This system is overdetermined (it's $n^2$ equations for $\tfrac12n(n{+}1)$ unknowns) and generally not solvable.The $0$-th order equations on $g$ that you have identified as coming from the symmetry conditions are only the first step (and one must deal with them). The algebraic nature of the Ricci tensor regarded as an endomorphism of $TM$ plays a key role in the behavior of the solutions: For example, if $R^i_j = f\delta^i_j$ for some $f$, and $n>2$, then $f$ must be constant, or there is no solution. (When $f$ is constant, all Einstein metrics (with the right constant) satisfy the equation.) $\endgroup$ Nov 27, 2015 at 11:36

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