# The Unapologetic Mathematician

## 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