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*}

**Now suppose that we start with some covariant $4$-tensor field $R$ that satisfies these equations, would we always be able to find a metric globally such that its Riemannian curvature tensor is $R$?**

If this were possible, would there be a formula to obtain the metric from the curvature tensor?

If this were not possible, what would be an example of a ‘curvature’ tensor that does not have an associated metric?