Laplace-Beltrami operator/Proofs

-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::$int\_M\; (fmathrm\{div\}(X)\; +\; X(f))\; omega\; =\; int\_M\; (fmathcal\{L\}\_X\; +\; mathcal\{L\}\_X(f))\; omega$

::$=\; int\_M\; mathcal\{L\}\_X\; fomega\; =\; int\_M\; mathrm\{d\}\; iota\_X\; fomega\; =\; int\_\{partial\; M\}\; iota\_X\; fomega$

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:

:$Delta\; f\; =\; mathrm\{d\}delta\; f\; +\; delta,mathrm\{d\}f\; =\; delta,\; mathrm\{d\}f\; =\; delta\; ,\; partial\_i\; f\; ,\; mathrm\{d\}x^i$

::$=\; -\; *mathrm\{d\}\{*partial\_i\; f\; ,\; mathrm\{d\}x^i\}\; =\; -\; *mathrm\{d\}(varepsilon\_\{i\; J\}\; sqrt,partial^i\; f)\; mathrm\{vol\}\_n$

::$=\; -frac\{1\}\{sqrt\; partial^j\; h\; ,\; mathrm\{d\}x^J\; +\; partial\_i\; h\; ,\; mathrm\{d\}x^i\; wedge\; varepsilon\_\{jJ\}\; sqrt\; partial^j\; f\; ,\; mathrm\{d\}x^J)$

::::$+\; h\; ,\; Delta\; f$

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

:::$=\; f\; ,\; Delta\; h\; +\; 2\; partial\_i\; f\; ,\; partial^i\; h\; +\; h\; ,\; Delta\; f$

where "f" and "h" are scalar functions.