# Famous Theorems of Mathematics/Applied Mathematics

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

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

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