Let's work on $(\mathcal{M},\texttt{g},\nabla)$, an $n$-dimensional Lorentzian manifold $\mathcal{M}$ with metric tensor field $\texttt{g}$ and Levi-Civita connection $\nabla$. Let $\{e^{a}\}$ denote a $\texttt{g}$-orthonormal cobasis with dual $\texttt{g}$-orthonormal basis $\{X_{a}\}$ satisfying $e^{a}(X_{b})=\delta^{a}_{b}$.
Consider a $p$-form $\alpha$ on $\mathcal{M}$, then it is straightforward to show that if $\alpha$ is $\nabla$-parallel (i.e. $\nabla_{X}\alpha=0$ for all vector fields $X$) then $\alpha$ is both closed ($d\alpha=0$) and co-closed ($\delta\alpha=0$), where $d$ is the exterior derivative and $\delta$ is the co-derivative. This follows immediately from the relations between $d$, $\delta$ and $\nabla$: \begin{align*} d \,\equiv\,e^{a} \wedge \nabla_{X_{a}} \qquad\text{and}\qquad \delta \,\equiv\, (-1)^{n+1}i_{X^{a}}\nabla_{X_{a}} \end{align*} where $X^{a}=\texttt{g}^{ab}X_{b}$. However, it is not true that every closed and co-closed $p$-form is covariantly constant (it is straightforward to find examples of this). So my question is the following:
What conditions are required (if any such conditions exist) such that a closed and co-closed $p$-form $\alpha$ is $\nabla$-parallel on $\mathcal{M}$?
I am assuming something like "a closed and co-closed $p$-form is $\nabla$-parallel if $\mathcal{M}$ is flat", but have no idea of where to look or how to prove any sort of conjecture on this. I guess an answer on a Riemannian manifold would be just as interesting (and may give an idea at the result on a semi-Riemannian manifold). Thanks!