Proofs involving the Laplace–Beltrami operator

Short description: Proofs

Main page: Laplace–Beltrami operatorThis 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:

[math]\displaystyle{ \int_M df(X) \;\omega = - \int_M f \, \operatorname{div} X \;\omega }[/math]

Proof of the above statement:

[math]\displaystyle{ \int_M (f\mathrm{div}(X) + X(f)) \omega = \int_M (f\mathcal{L}_X + \mathcal{L}_X(f)) \omega }[/math]

[math]\displaystyle{ = \int_M \mathcal{L}_X f\omega = \int_M \mathrm{d} \iota_X f\omega = \int_{\partial M} \iota_X f\omega }[/math]

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:

[math]\displaystyle{ \Delta f = \mathrm{d}\delta f + \delta\,\mathrm{d}f = \delta\, \mathrm{d}f = \delta \, \partial_i f \, \mathrm{d}x^i }[/math]

[math]\displaystyle{ = \- {\star}\mathrm{d}{{\star}\partial_i f \, \mathrm{d}x^i} = \- {\star}\mathrm{d}(\varepsilon_{i J} \sqrt{|g|}\partial^i f \, \mathrm{d}x^J) }[/math]

[math]\displaystyle{ = \- {\star}\varepsilon_{i J} \, \partial_j (\sqrt{|g|}\partial^i f)\, \mathrm{d} x^j \wedge \mathrm{d}x^J = \- {\star} \frac{1}{\sqrt{|g|}} \, \partial_i (\sqrt{|g|}\,\partial^i f) \mathrm{vol}_n }[/math]

[math]\displaystyle{ = -\frac{1}{\sqrt{|g|}}\, \partial_i (\sqrt{|g|}\,\partial^i f), }[/math]

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:

[math]\displaystyle{ \Delta(fh) = f \, \Delta h + 2 \partial_i f \, \partial^i h + h \, \Delta f. }[/math]

### Proof

[math]\displaystyle{ \Delta(fh) = \delta\,\mathrm{d}fh = \delta(f\,\mathrm{d}h + h\,\mathrm{d}f) = {\star}\mathrm{d}(f{{\star}\mathrm{d}h}) + {\star}\mathrm{d}(h{*\mathrm{d}f}) }[/math]

[math]\displaystyle{ = {\star}(f\,\mathrm{d}{\star}\mathrm{d}h + \mathrm{d}f \wedge {\star}\mathrm{d}h + \mathrm{d}h \wedge {\star}\mathrm{d}f + h\,\mathrm{d}{\star}\mathrm{d}f) }[/math]

[math]\displaystyle{ = f{\star}\mathrm{d}{\star}\mathrm{d}h + {\star}(\mathrm{d}f \wedge {\star}\mathrm{d}h + \mathrm{d}h \wedge {\star}\mathrm{d}f) + h{\star}\mathrm{d}{\star}\mathrm{d}f }[/math]

[math]\displaystyle{ = f\, \Delta h }[/math]

[math]\displaystyle{ {} + {\star}(\partial_i f \, \mathrm{d}x^i \wedge \varepsilon_{jJ} \sqrt{|g|} \partial^j h \, \mathrm{d}x^J + \partial_i h \, \mathrm{d}x^i \wedge \varepsilon_{jJ} \sqrt{|g|} \partial^j f \, \mathrm{d}x^J) }[/math]

[math]\displaystyle{ {} + h \, \Delta f }[/math]

[math]\displaystyle{ = f \, \Delta h + (\partial_i f \, \partial^i h + \partial_i h \, \partial^i f){{\star}\mathrm{vol}_n} + h \, \Delta f }[/math]

[math]\displaystyle{ = f \, \Delta h + 2 \partial_i f \, \partial^i h + h \, \Delta f }[/math]

where f and h are scalar functions.

## References

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