Riemannian manifold, $\alpha \in \Omega^p(M)$ parallel implies $\alpha$ is closed?

Question

Let $M$ be a Riemannian manifold, and let $\alpha \in \Omega^p(M)$ be parallel; i.e. suppose $\nabla \alpha = 0$ where $\nabla$ is the Levi-Civita connection. Does it necessarily follow that $\alpha$ is closed?

Answer 1

Yes I believe so. We have the formula that $$(p+1)\nabla_{[a}\alpha_{b\dots c]}=(d\alpha)_{ab\dots c}\in \Omega^{p+1}(M)$$This is in abstract index notation. So $\nabla\alpha=0$ implies the antisymmetric part is also $0$, so $d\alpha=0$