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!


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 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.