Laplace-Beltrami operator/Proofs

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Main article: Laplace operator

1 -div is adjoint to d
2 Laplace-de Rham operator
3 Properties
- 3.1 Proof

[edit] -div is adjoint to d

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

$\int_M df(X) \;\omega = - \int_M f \, \operatorname{div} X \;\omega$

Proof of the above statement:

$\int_M (f\mathrm{div}(X) + X(f)) \omega = \int_M (f\mathcal{L}_X + \mathcal{L}_X(f)) \omega$

$= \int_M \mathcal{L}_X f\omega = \int_M \mathrm{d} \iota_X f\omega = \int_{\partial M} \iota_X f\omega$

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

[edit] 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:

$\Delta f = \mathrm{d}\delta f + \delta\,\mathrm{d}f = \delta\, \mathrm{d}f = \delta \, \partial_i f \, \mathrm{d}x^i$

$= - *\mathrm{d}{*\partial_i f \, \mathrm{d}x^i} = - *\mathrm{d}(\varepsilon_{i J} \sqrt{|g|}\partial^i f \, \mathrm{d}x^J)$

$= - *\varepsilon_{i J} \, \partial_j (\sqrt{|g|}\partial^i f)\, \mathrm{d} x^j \, \mathrm{d}x^J = - * \frac{1}{\sqrt{|g|}} \, \partial_i (\sqrt{|g|}\,\partial^i f) \mathrm{vol}_n$

$= -\frac{1}{\sqrt{|g|}}\, \partial_i (\sqrt{|g|}\,\partial^i f),$

where ω 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 minus the 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.

[edit] Properties

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

$\Delta(fh) = f \, \Delta h + 2 \partial_i f \, \partial^i h + h \, \Delta f.$

[edit] Proof

$\Delta(fh) = \delta\,\mathrm{d}fh = \delta(f\,\mathrm{d}h + h\,\mathrm{d}f) = *\mathrm{d}(f{*\mathrm{d}h}) + *\mathrm{d}(h{*\mathrm{d}f})\;$

$= *(f\,\mathrm{d}*\mathrm{d}h + \mathrm{d}f \wedge *\mathrm{d}h + \mathrm{d}h \wedge *\mathrm{d}f + h\,\mathrm{d}*\mathrm{d}f)$

$= f*\mathrm{d}*\mathrm{d}h + *(\mathrm{d}f \wedge *\mathrm{d}h + \mathrm{d}h \wedge *\mathrm{d}f) + h*\mathrm{d}*\mathrm{d}f$

$= f\, \Delta h$

$+ *(\partial_i f \, \mathrm{d}x^i \wedge \varepsilon_{jJ} \sqrt{|g|} \partial^j h \, \mathrm{d}x^J + \partial_i h \, \mathrm{d}x^i \wedge \varepsilon_{jJ} \sqrt{|g|} \partial^j f \, \mathrm{d}x^J)$