# 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

## 3 Comments »

1. […] interesting to look at what happens when we apply the Hodge star twice. We just used the fact that in our special case of we always get back exactly what we started with. That is, in […]

Pingback by The Hodge Star, Squared « The Unapologetic Mathematician | October 18, 2011 | Reply

2. […] a Hodge star to work with, this tells us that we always have some functions on which are not the divergence of any vector field on . LD_AddCustomAttr("AdOpt", "1"); LD_AddCustomAttr("Origin", "other"); […]

Pingback by Compact Oriented Manifolds without Boundary have Nontrivial Homology « The Unapologetic Mathematician | November 24, 2011 | Reply

3. […] again with Maxwell’s equations, we see all these divergences and curls which, though familiar to many, are really heavy-duty equipment. In particular, they rely […]

Pingback by Maxwell’s Equations in Differential Forms « The Unapologetic Mathematician | February 22, 2012 | Reply