Example of a curvature with no associated metric 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, however, I'm looking for a concrete example.
 A: @AntonPetrunin's comment points to, I think, another way to describe the counterexample given by Robert Bryant in his answer.
Consider a curvature-like tensor
$$
R_{ijkl}(dy^i\wedge dy^j)(dy^k\wedge dy^l)
$$
where $(y^1, \dots, y^n)$ are coordinates in a neighborhood of $0$. If $R$ is the curvature tensor of a Riemanian metric, then there exists a change of coordinates $y=\phi(x)$, such that $\phi(0) = 0$, $\partial_i\phi^j(0) = \delta_i^j$, and, with respect to the coordinates $x= (x^1, \dots, x^n)$, the Christoffel symbols vanish at $0$. It follows by the second Bianchi identity that at the point $0$,
$$
\partial_mR_{ijkl}-\partial_lR_{ijkm} = 0.
$$
Now consider a curvature-like tensor $R$ in a neighborhood of $0$ such that $R(0) = 0$ but for some choice of $i,j,k,l,m$,
$$
\partial_mR_{ijkl} - \partial_lR_{ijkm} \ne 0.
$$
You can now verify that this inequality will still hold at $0$ for any change of coordinates. Therefore, this tensor cannot be the curvature tensor of a Riemannian metric.
A: 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 (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.
