One fact that I didn’t mention when discussing the curl operator is that the curl of a gradient is zero: . In our terms, this is a simple consequence of the nilpotence of the exterior derivative. Indeed, when we work in terms of -forms instead of vector fields, the composition of the two operators is , and is always zero.
So why do we bring this up now? Because one of the important things to remember from multivariable calculus is that the divergence of a curl is also automatically zero, and this will help us figure out what a divergence is in terms of differential forms. See, if we take our vector field and consider it as a -form, the exterior derivative is already known to be (essentially) the curl. So what else can we do?
We use the Hodge star again to flip the -form back to a -form, so we can apply the exterior derivative to that. We can check that this will be automatically zero if we start with an image of the curl operator; our earlier calculations show that is always the identity mapping — at least on with this metric — so if we first apply the curl and then the steps we’ve just suggested, the result is like applying the operator .
There’s just one catch: as we’ve written it this gives us a -form, not a function like the divergence operator should! No matter; we can break out the Hodge star once more to flip it back to a -form — a function — just like we want. That is, the divergence operator on -forms is .
Let’s calculate this in our canonical basis. If we start with a -form then we first hit it with the Hodge star:
Next comes the exterior derivative:
and then the Hodge star again:
which is exactly the definition (in coordinates) of the usual divergence of a vector field on .