Let $M$ be a (closed if necessary) Riemannian manifold with Levi-Civita connection
Let $\nabla^*$ be the formal adjoint of $\nabla$ with respect to the $L^2$ inner product. Let $\Delta=\nabla^*\nabla$ denote the Laplacian.
Question 1: In general, do we have $\nabla\circ\Delta$=$\Delta\circ\nabla$, as operators $C^\infty(M)\to C^\infty(M)\otimes C^\infty(T^*M)$?
Question 2: If not, is this relation true if the metric is flat?
Comment: I feel that the answer to Q2 at least must be yes, but I am not good with these computations. So I would appreciate it if someone could work through the computation (or perhaps share a reference) for the commutator $[\nabla,\Delta]$ involving the curvature terms.
Best Answer
As it has been mentioned in comment, by Bochner's formula for function $f$ the answer is clear:
$$\Delta\nabla_if = \nabla_i\Delta f + \mbox{R}_{ij}\nabla_jf.$$
Obviously in flat metrics we have $\mbox{R}_{ij}=0$ and ..