Given a 2-covariant tensor field $T$ on a pseudo-Riemannian manifold $(M, g)$, is it possible to solve the equation $$ \nabla \alpha = T $$ where $\alpha$ is a 1-form and $\nabla$ is the Levi-Civita connection of $g$? This seems like a fundamental question but I can't find anything that touches on this at all. It would also be interesting to replace the setting with a more general affine connection $\nabla$.
For instance, it is a classical problem in differential topology (answered by de Rham cohomology) whether or not a given 1-form $\omega$ on a manifold $M$ can be written as $\omega = df$ for some function $f$. At one rank higher, we have the question of if a 2-form $\omega$ on $M$ can be written as $d\alpha$ for a 1-form $\alpha$. My question concerns the analogous scenario in the space of symmetric tensor fields, rather than antisymmetric tensor fields (differential forms). In this case, we no longer have the exterior derivative $d$, so one natural differential operator to take its place is a connection $\nabla$.
I think the answer is no.
Even locally there is a problem for the solvability, let alone globally. I want to draw analogy to the exact 1-form question, that is, one rank lower to the proposed question. For a 1-form $\omega$ to have an $f$ such that $$\nabla f = df = \omega,$$ we need a local condition $d\omega=0$, but also globally $H^1(M, R)=0$.
Here is a concrete counter example, mainly addressing the global issue. Say $M=S^2$ with the standard metric. Let $T=dv$ be the volume form. Then for a 1-form $\alpha$, $$\nabla \alpha = T \Longrightarrow d\alpha = T.$$ The reason is that the skew-symmetrizaiton of $\nabla \alpha$, as a covariant 2-tensor, is the exterior differentiation, but $T$ is skew-symmetric already.
It is well-known such an $\alpha$ does not exist, by cohomological reasons.