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

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

We will prove a more general fact. Whenever ${f^i}$ is an oriented basis for the cotangent space (not necessarily orthonormal), and ${g_{ij} = }$, ${g}$ is defined as before:

$\displaystyle dV = g^{1/2} f^1 \wedge \dots \wedge f^n.$

Indeed, this is clear when ${f^i}$ is orthonormal. And it is clear that the above quantity is actually invariant under changes of frame—this is a fact of linear algebra.

So now, with this we can tackle the problem of computing ${\mathrm{div} X}$. We have:

$\displaystyle L_X (dV) = di_X (dV) + i_X d(dV).$

I’m not terribly thrilled with this notation. ${dV}$ looks like the exterior differential of some ${V}$, which it is not. However, this seems to be the norm. Anyhow, ${d(dV)}$ is zero, not because ${d^2=0}$, but rather because ${dV}$ is an ${n}$-form. As a result, we are reduced to computing ${d i_X(dV)}$, which turns out to be rather straightforward. First of all:

$\displaystyle i_X(dV) = \sum_j (-1)^{j-1} X^j g^{1/2} dx^1 \wedge \dots \widehat{dx^j} \dots \wedge dx^n,$

by definition of how the wedge product works and the interior product. (I guess this is not completely standard. If you use a different definition, you’d get a constant factor added.) Now if we apply ${d}$ to this we find:

$\displaystyle \sum_j \partial_j( X^j g^{1/2}) dx^1 \wedge \dots \wedge dx^n ,$

and as a result:

$\displaystyle \boxed{ \mathrm{div} X = g^{-1/2} \partial_j( X^j g^{1/2}).}$

From this the property ${\mathrm{div}(Xf) = f \mathrm{div} X + Xf}$ is clear, since ${Xf = X^i \partial_i f}$.

Incidentally, to say that ${\mathrm{div} X = 0}$ is to say that the flows of ${X}$ preserve the volume, so it makes sense to call ${X}$ incompressible.

Now we want to do the same for ${\mathrm{grad}}$. First, ${df = \partial_i f dx^i}$. The transformation of a vector ${v_i dx^i}$ to ${v^i \partial_i}$ works by ${v^i = g^{ij} v_j}$, where ${g^{ij}}$ is the inverse of the matrix ${g_{ij}}$, I claim. This can be seen as follows:

$\displaystyle < g^{ij} v_j \partial_i , w^k \partial_k> = g^{ji} w^k v_j g_{ik} = \delta_{jk} v_j w^k = v_j w^j .$

What this rather uninspiring statement means is that ${ g^{ij} v_j \partial_i}$ acts as functional on the tangent space via the inner product in the same way that the cotangent vector ${v_i dx^i}$ does. In particular, it proves the claim I just made. (Cf. also the Wikipedia article on raising and lowering indices.) So

$\displaystyle \mathrm{grad} f = g^{ij} \partial_j f \partial_i .$

The Laplacian

Finally, as a result, we can get the representation of ${\Delta f}$ in local coordinates:

$\displaystyle \mathrm{div} \ \mathrm{grad} f = \mathrm{div}( g^{ij} \partial_j f \partial_i ) = g^{-1/2} \partial_i \left( g^{1/2} g^{ij} \partial_j f \right).$

In particular, we find that ${\Delta}$ as an operator is ${g^{ij} \partial_i \partial_j f}$ plus something solely first order. Since ${g^{ij}}$ is a positive-definite matrix, this will imply (I will later define this) that the Laplacian is an elliptic operator.