Laplace operator/Proofs

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

∫_Mdf(X) ω =  − ∫_Mf div X ω

Proof of the above statement:

∫_M(fdiv(X)+X(f))ω = ∫_M(fℒ_X+ℒ_X(f))ω

 = ∫_Mℒ_Xfω = ∫_Mdι_Xfω = ∫_∂Mι_Xfω

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 Rahm operator is equivalent to the definition of the Laplace-Beltrami operator, when acting on a scalar function f. This proof reads as:

Δf = dδf + δ df = δ df = δ ∂_if dxⁱ

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

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

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

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.

Properties

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

Δ(fh) = f Δh + 2∂_if ∂ⁱh + h Δf.

Proof

Δ(fh) = δ dfh = δ(f dh+h df) =  * d(f*dh) +  * d(h*df) 

 =  * (f d*dh+df∧*dh+dh∧*df+h d*df)

 = f * d * dh +  * (df∧*dh+dh∧*df) + h * d * df

 = f Δh

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

 + h Δf

 = f Δh + (∂_if ∂ⁱh+∂_ih ∂ⁱf)*vol_n + h Δf

 = f Δh + 2∂_if ∂ⁱh + h Δf

where f and h are scalar functions.