One of the basic properties of the Laplacian is that given a compact Riemannian manifold-with-boundary (to which all this ${\mathrm{div}, \mathrm{grad}, \Delta}$ business applies equally), then for ${u}$ vanishing on the boundary, the ${L^2}$ inner product ${(\Delta u, u)}$ is fairly large relative to ${u}$. As an immediate corollary, if ${u}$ satisfies the Laplace equation ${\Delta u= 0}$ and vanishes on the boundary, then ${u}$ is identically zero.

It turns out that the proof of this will require the divergence theorem. This is a familiar fact from multivariable calculus, but it generalizes to ${n}$-dimensions nicely as a corollary of Stokes theorem and some of the other machinery thus developed.

So, let’s choose an oriented Riemannian manifold ${M}$ of dimension ${n}$ with boundary ${\partial M}$. There is a volume form ${dV}$ because of the choice of orientation globally defined. On ${\partial M}$, there is an induced Riemannian metric and an induced orientation, with a corresponding volume form ${dS}$ on ${\partial M}$. If ${X}$ is a compactly supported vector field, the divergence theorem states that

$\displaystyle \boxed{ \int_M \mathrm{div} X dV = \int_{\partial M} dS ,}$ (more…)

Just for fun, let’s see what all this actually means in local coordiantes.  It’s a good idea to get our hands dirty too.  To me at least, this makes the operators seem somewhat more friendly.

Choose local coordinates ${x^1, \dots, x^n}$. Let

$\displaystyle g_{ij}(x) = < \partial_i, \partial_j >.$

Let ${g = \det( g_{ij})}$. Note that ${g>0}$.

Henceforth, we will use the Einstein summation convention: all repeated indices are to be summed over, unless otherwise stated. For instance, the inner product is the 2-tensor

$\displaystyle \frac{1}{2} g_{ij} dx^i \otimes dx^j.$

Div

Now let ${X = X^j \partial_j}$ be a vector field. We want to compute ${\mathrm{div} X}$ in terms of the quantities ${X^i}$ and ${g}$. First, it will be necessary to compute the form ${dV}$. I claim that, if we take ${x^i}$ to be oriented coordinates, then

$\displaystyle dV = g^{1/2} dx^1 \wedge \dots \wedge dx^n.$ (more…)

I’ve set a tentative goal of heading towards the solution of the Dirichlet problem on compact manifoldw-with-boundary and Hodge theory; these will require various preliminaries, since of course it will be more fun to do it this way than to restrict to open sets in $\mathbb{R}^n$.  Today I will go through some of the basics of how the well-known operators from multivariable calculus work more generally on a Riemannian manifold, which will be necessary in the sequel.   (I shamelessly took the title of the post from a book I haven’t read.)

Recall the well-known operator on functions of ${n}$-variables, the Laplacian

$\displaystyle \Delta f := \sum_i \frac{ \partial^2f}{\partial x_i^{2}}.$

The problem is, ${\Delta}$ doesn’t transform nicely with respect to changes in coordinates, and since we want to define the Laplacian on manifolds, this causes a problem. However, it can be done on a Riemannian manifold. The idea is to use the formula

$\displaystyle \Delta = \mathrm{div} (\mathrm{grad} ),$

which is immediate from the definitions on ${\mathbb{R}^n}$. The key point is that ${\mathrm{div} }$ and ${\mathrm{grad}}$ make sense on any Riemannian manifold—from, respectively, vector fields to functions and from functions to vector fields. (more…)