# The Unapologetic Mathematician

## The Curl Operator

Let’s continue our example considering the special case of $\mathbb{R}^3$ as an oriented, Riemannian manifold, with the coordinate $1$-forms $\{dx, dy, dz\}$ forming an oriented, orthonormal basis at each point.

We’ve already see the gradient vector $\nabla f$, which has the same components as the differential $df$:

$\displaystyle\begin{aligned}df&=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz\\\nabla f&=\begin{pmatrix}\displaystyle\frac{\partial f}{\partial x}\\\displaystyle\frac{\partial f}{\partial y}\\\displaystyle\frac{\partial f}{\partial z}\end{pmatrix}\end{aligned}$

This is because we use the metric to convert from vector fields to $1$-forms, and with respect to our usual bases the matrix of the metric is the Kronecker delta.

We will proceed to define analogues of the other classical differential operators you may remember from multivariable calculus. We will actually be defining operators on differential forms, but we will use this same trick to identify vector fields and $1$-forms. We will thus not usually distinguish our operators from the classical ones, but in practice we will use the classical notations when acting on vector fields and our new notations when acting on $1$-forms.

Anyway, the next operator we come to is the curl of a vector field: $F\mapsto\nabla\times F$. Of course we’ll really start with a $1$-form instead of a vector field, and we already know a differential operator to use on forms. Given a $1$-form $\alpha$ we can send it to $d\alpha$.

The only hangup is that this is a $2$-form, while we want the curl of a vector field to be another vector field. But we do have a Hodge star, which we can use to flip a $2$-form back into a $1$-form, which is “really” a vector field again. That is, the curl operator corresponds to the differential operator $*d$ that takes $1$-forms back to $1$-forms.

Let’s calculate this in our canonical basis, to see that it really does look like the familiar curl. We start with a $1$-form $\alpha=Pdx+Qdy+Rdz$. The first step is to hit it with the exterior derivative, which gives

$\displaystyle\begin{aligned}d\alpha=&dP\wedge dx+dQ\wedge dy + dR\wedge dz\\=&\left(\frac{\partial P}{\partial x}dx+\frac{\partial P}{\partial y}dy+\frac{\partial P}{\partial z}dz\right)\wedge dx\\&+\left(\frac{\partial Q}{\partial x}dx+\frac{\partial Q}{\partial y}dy+\frac{\partial Q}{\partial z}dz\right)\wedge dy\\&+\left(\frac{\partial R}{\partial x}dx+\frac{\partial R}{\partial y}dy+\frac{\partial R}{\partial z}dz\right)\wedge dz\\=&\frac{\partial P}{\partial y}dy\wedge dx+\frac{\partial P}{\partial z}dz\wedge dx\\&+\frac{\partial Q}{\partial x}dx\wedge dy+\frac{\partial Q}{\partial z}dz\wedge dy\\&+\frac{\partial R}{\partial x}dx\wedge dz+\frac{\partial R}{\partial y}dy\wedge dz\\=&\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)dy\wedge dz\\&+\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)dz\wedge dx\\&+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\wedge dy\end{aligned}$

Next we hit this with the Hodge star. We’ve already calculated how the Hodge star affects the canonical basis of $2$-forms, so this is just a simple lookup to find:

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

which are indeed the usual components of the curl. That is, if $\alpha$ is the $1$-form corresponding to the vector field $F$, then $*d\alpha$ is the $1$-form corresponding to the vector field $\nabla\times F$.

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Posted by John Armstrong | Differential Geometry, Geometry

## 7 Comments	»

1. [...] 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 [...]

Pingback by | October 13, 2011 | Reply

2. [...] if we define as another -form then we know it corresponds to the curl . But on the other hand we know that in dimension we have , and so we find as well. Thus we [...]

Pingback by | November 23, 2011 | Reply

3. [...] The Curl Operator (unapologetic.wordpress.com) Advertisement LD_AddCustomAttr("AdOpt", "1"); LD_AddCustomAttr("Origin", "other"); LD_AddCustomAttr("LangId", "1"); LD_AddCustomAttr("Autotag", "technology"); LD_AddCustomAttr("Tag", "kronecker-product"); LD_AddCustomAttr("Tag", "leopold-kronecker"); LD_AddSlot("wpcom_below_post"); LD_GetBids(); Rate this: Share this:TwitterPrintFacebookLinkedInEmailDiggRedditStumbleUponLike this:LikeBe the first to like this post. This entry was posted in Uncategorized and tagged Kronecker product, Leopold Kronecker by Adrian McMenamin. Bookmark the permalink. [...]

Pingback by | January 14, 2012 | Reply

4. [...] we flip over to the language of differential forms, we know that the curl operator on a vector field corresponds to the operator on -forms, while the gradient operator corresponds [...]

Pingback by | February 18, 2012 | Reply

5. [...] 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 on the [...]

Pingback by | February 22, 2012 | Reply

6. How does the curl generalize to curved 3-space?

Comment by john archer | April 10, 2012 | Reply

7. It’s pretty much the same as in flat 3-space, except that the metric we use to translate back and forth to the language of differential forms — and to define the Hodge star — now varies.

Comment by | April 11, 2012 | Reply

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