The Unapologetic Mathematician

Mathematics for the interested outsider

The Divergence Operator

One fact that I didn’t mention when discussing the curl operator is that the curl of a gradient is zero: \nabla\times(\nabla f)=0. In our terms, this is a simple consequence of the nilpotence of the exterior derivative. Indeed, when we work in terms of 1-forms instead of vector fields, the composition of the two operators is *d(df), and d^2 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 1-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 1-form back to a 2-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 *^2 is always the identity mapping — at least on \mathbb{R}^3 with this metric — so if we first apply the curl *d and then the steps we’ve just suggested, the result is like applying the operator d**d=dd=0.

There’s just one catch: as we’ve written it this gives us a 3-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 0-form — a function — just like we want. That is, the divergence operator on 1-forms is *d*.

Let’s calculate this in our canonical basis. If we start with a 1-form \alpha=Pdx+Qdy+Rdz then we first hit it with the Hodge star:

\displaystyle*\alpha=Pdy\wedge dz+Qdz\wedge dx+Rdx\wedge dy

Next comes the exterior derivative:

\displaystyle d*\alpha=\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\right)dx\wedge dy\wedge dz

and then the Hodge star again:

\displaystyle*d*\alpha=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}

which is exactly the definition (in coordinates) of the usual divergence \nabla\cdot F of a vector field F on \mathbb{R}^3.

October 13, 2011 Posted by | Differential Geometry, Geometry | 3 Comments

   

Follow

Get every new post delivered to your Inbox.

Join 366 other followers