Is there a concrete example of a $4$ tensor $R\_{ijkl}$ with the same symmetries as the Riemannian curvature tensor, i.e.
\begin{gather\*}
R\_{ijkl} = - R\_{ijlk},\quad R\_{ijkl} = R\_{jikl},\quad R\_{ijkl} = R\_{klij}, \\
R\_{ijkl} + R\_{iklj} + R\_{iljk} = 0.
\end{gather\*}
for which there is no metric for which it is the Riemannian curvature tensor?


The existence of such a curvature was already shown by [Robert Bryant](https://mathoverflow.net/questions/202211/equations-satisfied-by-the-riemann-curvature-tensor), however, I'm looking for a concrete example.



Example of a curvature with no associated metric
A simple example (which just uses Deane Yang/Robert Bryant's idea) is to consider any space of dimension at least three and consider the tensor field
$$ R\_{ijkl} = f(x)(\delta\_{ik}\delta\_{jl}-\delta\_{il}\delta\_{jk})$$
where $f(x)$ is your favorite function which changes sign and whose derivative is non-vanishing when $f=0$ (thanks to Robert Bryant for the correction). When $f=0$, this curvature tensor has constant sectional curvature $0$, no matter what metric we choose.


However, we can now apply the proof of [Schur's lemma](https://en.wikipedia.org/wiki/Schur%27s_lemma_(Riemannian_geometry)) (i.e., trace the second Bianchi identity twice) to see that when $f$ vanishes, we have that
$$dR=\frac{n}{2}dR$$
where $R$ is the scalar curvature.
As such, the differential of the scalar curvature must vanish when $f=0$. However, no matter which metric we pick, there is no way to make this happen if the differential of $f$ is nonzero as it goes from being positive to negative.


Edit: My original answer had a mistake which was pointed out in the comments. Here is a revised version which uses the same idea which should (hopefully) work.


