Exterior multiplication with parallel form commutes with Laplacian Let $(M,g)$ be a Riemannian manifold, let $\nabla$ be the Levi-Civita connection, and let $\Delta=dd^*+d^*d$ be the Laplacian. Suppose $\omega \in \Omega^*(M)$ is a parallel differential form (i.e. $\nabla \omega=0$), is it true that $\Delta (\alpha \wedge \omega)=\Delta(\alpha) \wedge \omega$?
I have seen this fact being used a few times in literature, but I cannot find or work out a full proof myself. The sources I have consulted simply say 'use Weitzenbock's formula', but I cannot see how the formula can prove this. Any help is appreciated!
 A: Let's denote the regular Laplacian on $\mathbb{R}^n$ acting on functions by $\tilde{\Delta}$. The product rule for $\tilde{\Delta}$ says that
$$ \tilde{\Delta}(\alpha \omega) = \tilde{\Delta}(\alpha) \omega + 2 \nabla \alpha \cdot \nabla \omega + \alpha \tilde{\Delta}(\omega). $$
This implies that if $\tilde{\Delta}(\omega)(0) = 0$ and $(\nabla \omega)(0) = 0$ then
$$ \tilde{\Delta}(\alpha \omega)(0) = \tilde{\Delta}(\alpha)(0) \cdot \omega(0). $$
The general cases reduces to the observation above by using normal coordinates. Note first that if $\nabla \omega = 0$ implies that $\Delta \omega = 0$. Let $p \in M$ and choose normal coordinates $\varphi \colon U \rightarrow V$ around $p \in U \subseteq M$ where $V \subseteq \mathbb{R}^n$, $\varphi(p) = 0$ and $\varphi = (x^1,\dots,x^n)$. Write $\omega = \sum_{I} \omega_I dx^I$ and $\alpha = \sum_{J} \alpha_J dx^J$ and set $\tilde{\omega_I} = \omega_I \circ \varphi^{-1}, \tilde{\alpha_J} = \alpha_J \circ \varphi^{-1}$. Then
$$ 0 = \nabla_{\partial_j}(\omega)(p) = \sum_{I} \frac{\partial \tilde{\omega_I}}{\partial x^j}(0) dx^I|_{p}$$
which implies that
$$ \frac{\partial \tilde{\omega_I}}{\partial x^j}(0) = 0 $$
for all $1 \leq j \leq n$ and all $I$. Similarly,
$$ 0 = \Delta(\omega)(p) = \sum_{I} -\tilde{\Delta}(\tilde{\omega_I})(0) dx^I|_{p} $$
which implies that $\tilde{\Delta}(\tilde{\omega_I})(0) = 0$ for all $I$. Finally,
$$ \Delta(\alpha \wedge \omega)(p) = \sum_{I,J} \Delta 
 \left( \alpha_J \omega_I dx^J \wedge dx^I \right)(p) = \sum_{I,J} -\tilde{\Delta} \left( \tilde{\alpha}_J \tilde{\omega}_I \right)(0) dx^J \wedge dx^I|_{p}  = \\\sum_{I,J} -\tilde{\Delta}(\tilde{\alpha}_J)(0) \tilde{\omega_I}(0) dx^J \wedge dx^I|_{p} = \left( \sum_{I} \tilde{\Delta}(\tilde{\alpha}_J)(0) dx^J \right) \wedge \left( \sum_{J} \tilde{\omega_I}(0) dx^I|_{p} \right) = \Delta(\alpha)(p) \wedge \omega(p).$$
A: A rather simple proof based on graded Lie superalgebra and by computing the commutator of  linear operators $L_{\omega}: \eta\mapsto \eta\wedge \omega$ and $\Delta$ (i.e. $[L_\omega,\Delta]$) then using Jacobi identity of graded Lie superalgebra, can be found in
Verbitsky, Misha, Manifolds with parallel differential forms and Kähler identities for $G_{2}$-manifolds, J. Geom. Phys. 61, No. 6, 1001-1016 (2011). ZBL1214.58002.
corollary 2.9 page 6. According to this paper, the wanted result is due to S. S. Chern.
