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.


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…)