Proofs involving the Laplace–Beltrami operator

Short description: Proofs

Main article: Laplace–Beltrami operator

This article relates to proofs involving the Laplace–Beltrami operator.

## Contents

* 1 −div is adjoint to d
* 2 Laplace–de Rham operator
* 3 Properties
* 3.1 Proof
* 4 References

## −div is adjoint to d

The claim is made that −div is adjoint to d:

∫Mdf(X)ω=−∫Mfdiv⁡Xω

Proof of the above statement:

∫M(fdiv(X)+X(f))ω=∫M(fℒX+ℒX(f))ω

=∫MℒXfω=∫MdιXfω=∫∂MιXfω

If f has compact support, then the last integral vanishes, and we have the desired result.

## Laplace-de Rham operator

One may prove that the Laplace-de Rham operator is equivalent to the definition of the Laplace-Beltrami operator, when acting on a scalar function f. This proof reads as:

Δf=dδf+δdf=δdf=δ∂ifdxi

=−⋆d⋆∂ifdxi=−⋆d(εiJ|g|∂ifdxJ)

=−⋆εiJ∂j(|g|∂if)dxj∧dxJ=−⋆1|g|∂i(|g|∂if)voln

=−1|g|∂i(|g|∂if),

where voln; is the volume form and ε is the completely antisymmetric Levi-Civita symbol. Note that in the above, the italic lower-case index i is a single index, whereas the upper-case Roman J stands for all of the remaining n − 1 indices. Notice that the Laplace-de Rham operator is actually the negative Laplace-Beltrami operator; this minus sign follows from the conventional definition of the properties of the codifferential. Unfortunately, Δ is used to denote both; reader beware.

## Properties

Given scalar functions f and h, and a real number a, the Laplacian has the property:

Δ(fh)=fΔh+2∂if∂ih+hΔf.

### Proof

Δ(fh)=δdfh=δ(fdh+hdf)=⋆d(f⋆dh)+⋆d(h*df)

=⋆(fd⋆dh+df∧⋆dh+dh∧⋆df+hd⋆df)

=f⋆d⋆dh+⋆(df∧⋆dh+dh∧⋆df)+h⋆d⋆df

=fΔh

+⋆(∂ifdxi∧εjJ|g|∂jhdxJ+∂ihdxi∧εjJ|g|∂jfdxJ)

+hΔf

=fΔh+(∂if∂ih+∂ih∂if)⋆voln+hΔf

=fΔh+2∂if∂ih+hΔf

where f and h are scalar functions.

## References

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