-div is adjoint to d

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

${\displaystyle \int _{M}df(X)\;\omega =-\int _{M}f\,\operatorname {div} X\;\omega }$

Proof of the above statement:

${\displaystyle \int _{M}(f\mathrm {div} (X)+X(f))\omega =\int _{M}(f{\mathcal {L}}_{X}+{\mathcal {L}}_{X}(f))\omega }$

${\displaystyle =\int _{M}{\mathcal {L}}_{X}f\omega =\int _{M}\mathrm {d} \iota _{X}f\omega =\int _{\partial M}\iota _{X}f\omega }$

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

${\displaystyle \Delta f=\mathrm {d} \delta f+\delta \,\mathrm {d} f=\delta \,\mathrm {d} f=\delta \,\partial _{i}f\,\mathrm {d} x^{i}}$

${\displaystyle =-*\mathrm {d} {*\partial _{i}f\,\mathrm {d} x^{i}}=-*\mathrm {d} (\varepsilon _{iJ}{\sqrt {|g|}}\partial ^{i}f\,\mathrm {d} x^{J})}$

${\displaystyle =-*\varepsilon _{iJ}\,\partial _{j}({\sqrt {|g|}}\partial ^{i}f)\,\mathrm {d} x^{j}\,\mathrm {d} x^{J}=-*{\frac {1}{\sqrt {|g|}}}\,\partial _{i}({\sqrt {|g|}}\,\partial ^{i}f)\mathrm {vol} _{n}}$

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

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:

${\displaystyle \Delta (fh)=f\,\Delta h+2\partial _{i}f\,\partial ^{i}h+h\,\Delta f.}$