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Let M a riemannian manifold. How can I show that the hodge-laplace-operator of a function $f$ is the negative of the laplace-operator?

edited Jun 23 2010 at 15:02

Charles Matthews
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asked Jun 23 2010 at 12:11

Differentialgeometer
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Andrey Rekalo · Accepted Answer · 2010-06-23 13:14:09Z UTC

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A rather short proof can be found here.

I assume you are interested in the case when $f$ is a scalar function. Otherwise the Hodge Laplacian differs from the Laplace–Beltrami operator not only by a sign due to the Ricci curvature. See the Weitzenböck identity.

edited Jun 23 2010 at 13:14

answered Jun 23 2010 at 12:34

Andrey Rekalo
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Your link to the Weitzenböck formula is broken, btw. It should be en.wikipedia.org/wiki/Weitzenböck_identity – Willie Wong Jun 23 2010 at 12:44

Thank you, Willie. I stand corrected. – Andrey Rekalo Jun 23 2010 at 12:48

Connection between the Hodge laplacian and the Laplace operator

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