It is well-known that a pseudo-Riemannian manifold $(M,g)$ admits a (unique) torsion-free linear connection $\stackrel{\circ}{\nabla}$ that is compatible with the given metric, i.e. $\stackrel{\circ}{\nabla} g = 0$.
Is it true that there also exists a curvature-free metric-compatible connection $\nabla$? Or under what circumstances?
Edit:
About my conjecture below:
In the refined case of a parallelizable pseudo-Riemannian manifold, does there always exist a curvature-free metric-compatible linear connection?
Take as an counterexample the maximally-extended Schwarzschild spacetime $(M,g)$. It is parallelizable as the underlying manifold is homeomorphic to $S^2 \times \mathbb R^2$. Suppose there exists a metric-compatible curvature-free connection $\nabla$. Since $M$ is simply-connected, the holonomy group $\operatorname{Hol}^{\nabla}(p)$ vanishes for any point $p \in M$. We thus can construct a global orthonormal frame using the parallel transport of $\nabla$. However, no such global orthonormal frame exists. A contradiction.