Let $(M, g)$ be a Riemannian manifold, $E$ a Hermitian vector bundle and $A$ a unitary connection over $E$ (i.e. the covariant derivative $d_A$ respects the inner product). The action of $d_A$ is extended to vector valued forms with values in $E$ by the Leibniz rule. If we introduce the covariant Hodge Laplacians acting on forms $C^\infty(M; E \otimes \Omega^k)$ for $k = 0,1, \dotso, \dim M$ $$\Delta_{A, k} := d_A^*d_A + d_Ad_A^*$$ Here $d_A^*$ is the formal adjoint of $d_A$ w.r.t. natural inner products. One can then compute that: $$\Delta_{A, k+1} d_A u = d_A \Delta_{A, k} u + [d_A^*, F(A)\wedge] u$$ Here, $F(A) = d_A^2 = dA + A\wedge A$ denotes the curvature two form and $[\cdot, \cdot]$ is the commutator.
My question is: when does the covariant Hodge Laplacian commute with the covariant derivative? In other words, when is the commutator $[d_A^*, F(A)\wedge]$ above equal to zero?
For example, this is the case when $A$ is flat. Is there a different condition on the connection, such that this is true? It is difficult for me to see at the moment, since the codifferential $d_A^*$ doesn't distribute nicely over a wedge product. However, the operator $[d_A^*, F(A)]$ transforms tensorially, so imposing the zero condition on it feels like a natural condition.
OK, so I think I got it know. After trying small examples, which showed the commutation does not hold unless $A$ is flat, we could try to compute the principal symbol of $P = [d_A^*, F(A)\wedge]$. We claim it's equal to: $$\sigma(P)(x, \xi) u = -i \iota_\xi F(A) \wedge u$$ Here $u$ is a vector valued differential form and $\iota_\xi$ is contraction. This follows easily from the fact that $\sigma(d_A^*) = \sigma(d^*) \otimes Id_E$ and $\sigma(d^*) w = -i\iota_\xi w$, which is well-known. Thus, evaluating the symbol $\sigma(P)$, we have \begin{align} \sigma(P) &= [\sigma(d^*) \otimes Id, F_{kj}(A) dx^k \wedge dx^j] = [\sigma(d^*), e_{kj}] \otimes F_{kj}(A)\\ &= -i\iota_{\xi}e_{kj}F_{kj}(A) = -\iota_\xi F(A) \end{align} Here $e_{ij} = dx^i \wedge dx^j$ and $F(A) = e_{ij} F_{ij}(A)$. So the operator $P$ vanishes (restricted to the space of $k$-forms for any $k$) iff $F(A) = 0$.