# Killing exterior forms

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I define Killing exterior forms:

$$\nabla_X \alpha = X^* \wedge \beta$$

$$\nabla_X \beta =i(X) \alpha$$

Are Killing exterior forms, proper vectors of the Laplacian?

asked Sep 4, 2021

## 1 Answer

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it is the operator of Laplace – de Rham :


More generally, one can define a Laplacian differential operator on sections of the bundle of differential forms on a pseudo-Riemannian manifold. On a Riemannian manifold it is an elliptic operator, while on a Lorentzian manifold it is hyperbolic. The Laplace–de Rham operator is defined by

{\displaystyle \Delta =\mathrm {d} \delta +\delta \mathrm {d} =(\mathrm {d} +\delta )^{2},\;}

where d is the exterior derivative or differential and δ is the codifferential, acting as (−1)kn+n+1∗d∗ on k-forms, where ∗ is the Hodge star.

https://en.wikipedia.org/wiki/Laplace%E2%80%93Beltrami_operator
answered Sep 5, 2021 by anonymous

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