Does every ‘curvature’ tensor induce a metric? So we know that given a Riemannian manifold $(M,g)$ we can calculate its Riemannian curvature $R$. This covariant $4$-tensor field then satisfies some important symmetries
\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*}
The main question of the global existence of a metric for a prescribed curvature was already explained by Robert Bryant (thank you to @Deane Yang for the summary). However, this answer only describes why such a curvature exists. Hence, I'd like to know whether there is a concrete example of a curvature that is not generated by a metric.
 A: Here's a quick summary. The answers provided in the link cited by @RBega2 have more details.

*

*Given a curvature-like tensor $R$ at a point, there always exists a metric whose curvature tensor at that point is equal to $R$.

*In general, this is system of nonlinear second order PDEs, where the unknown functions are the components of the metric tensor and there is an equation for each component of the curvature tensor.

*In dimensions 4 and higher, there are more equations than unknown functions and therefore there are no solutions unless further conditions are imposed on the curvature-like tensor. These conditions are poorly understood.

*In dimension 3, the number of equations is equal to the number of unknown functions. Again, the necessary and sufficient conditions are not known. However, as Robert Bryant explains in his answer, he proved that if the curvature-like tensor satisfies a nondegeneracy condition and is real analytic, then there is a local real analytic solution. Later, DeTurck and I extended this to the smooth category.

*As far as I know, nothing is known about the global problem. This system of PDEs is quite nasty. A fundamentally new advance in how to solve system of PDEs like this is needed.

