Proofs involving the Laplace–Beltrami operator

−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.

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:

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

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

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

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

where vol_n; 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.

Properties

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

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

Proof

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

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

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

$= f\, \Delta h$

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

${} + h \, \Delta f$

$= f \, \Delta h + (\partial_i f \, \partial^i h + \partial_i h \, \partial^i f){*\mathrm{vol}_n} + h \, \Delta f$

$= f \, \Delta h + 2 \partial_i f \, \partial^i h + h \, \Delta f$

where f and h are scalar functions.