I'm playing with the electromagnetic field tensor. Heard about it? Yes, it's the very tensor that, well, pretty much makes Maxwell's equations redundant.

We start with an arbitrary smooth vector field, $A^\mu$, in four dimensions, one for time, three for space. In the language of differential forms, this vector field is also called a 1-form. (Well, strictly speaking, its the dual vector field, defined as $A_\mu=g_{\mu\nu}A^\nu$, that can serve as a 1-form, but that in no way detracts from the argument presented here.) To make a 2-form from the 1-form, we can apply the derivative operator:

\[d{\bf\mathrm{A}}=\partial_\mu A_\nu-\partial_\nu A_\mu=F_{\mu\nu}= \begin{pmatrix} 0&\partial_1A_0-\partial_0A_1&\partial_2A_0-\partial_0A_2&\partial_3A_0-\partial_0A_3\\ \partial_0A_1-\partial_1A_0&0&\partial_2A_1-\partial_1A_2&\partial_3A_1-\partial_1A_3\\ \partial_0A_2-\partial_2A_0&\partial_1A_2-\partial_2A_1&0&\partial_3A_2-\partial_2A_3\\ \partial_0A_3-\partial_3A_0&\partial_1A_3-\partial_3A_1&\partial_2A_3-\partial_3A_2&0 \end{pmatrix}=g_{\mu\nu}g_{\kappa\lambda}F^{\kappa\lambda}.\]

This tensor is totally antisymmetric, with 6 independent components. Let's label these components:

\[\begin{pmatrix} 0&E^1&E^2&E^3\\ -E^1&0&-B^3&B^2\\ -E^2&B^3&0&-B^1\\ -E^3&-B^2&B^1&0 \end{pmatrix}\]

Looks familiar? Of course through these labels, we defined the components of the electric and magnetic field.

What can we do with this tensor now? Why, we can apply the differential operator to it one more time. There are two ways to do so: we can compute the interior and the exterior derivative. Let's start with the interior derivative:

\[\partial_\nu F^{\mu\nu}=\begin{pmatrix} \partial_1E^2+\partial_2E^2+\partial_3E^3\\ -\partial_0E^1+\partial_3E^2-\partial_2E^3\\ -\partial_0E^2+\partial_1E^3-\partial_3E^1\\ -\partial_0E^3+\partial_2E^1-\partial_1E^2 \end{pmatrix}=\begin{pmatrix}\nabla\cdot{\bf\mathrm{E}}\\~\\\nabla\times{\bf\mathrm{B}}-\partial{\bf\mathrm{E}}/\partial t\\~\end{pmatrix}=\begin{pmatrix}\rho\\~\\{\bf\mathrm{j}}\\~\end{pmatrix}.\]

Yes, this is just the charge density and current. Computing the inner product with the derivative operator one more time tells us why these quantities are special: it is easy to check that the continuity equation applies, i.e., $\partial_\mu\partial_\nu F^{\mu\nu}=\partial\rho/\partial t+\nabla\cdot\vec{j}=0$, meaning that charge is conserved.

With the exterior derivative things get just as interesting, because we know from the calculus of differential forms that repeated application of the exterior derivative produces a null result, i.e., ${\rm d}^2A=0$. Can this be used to extract new and useful identities? Let's see. ${\rm d}^2A={\rm d}F$, which is a 4×4×4 totally antisymmetric matrix. Let's call it $M$. It has only four independent components:

\begin{align}M^{012}&=\partial_0F^{12}-\partial_0F^{21}+\partial_1F^{20}-\partial_1F^{02}+\partial_2F^{01}-\partial_2F^{10}=2(-\partial_0B^3-\partial_1E^2+\partial_2E^1),\\ M^{013}&=\partial_0F^{13}-\partial_0F^{31}+\partial_1F^{30}-\partial_1F^{03}+\partial_3F^{01}-\partial_3F^{10}=2(\partial_0B^2-\partial_1E^3+\partial_3E^1),\\ M^{023}&=\partial_0F^{23}-\partial_0F^{32}+\partial_2F^{30}-\partial_2F^{03}+\partial_3F^{02}-\partial_3F^{20}=2(-\partial_0B^1-\partial_2E^3+\partial_2E^2),\\ M^{123}&=\partial_1F^{23}-\partial_1F^{32}+\partial_2F^{31}-\partial_2F^{13}+\partial_3F^{12}-\partial_3F^{21}=2(-\partial_1B^1-\partial_2E^2-\partial_3E^3), \end{align}

All these should be identically zero of course. Now is the time to notice that $M^{012}=0$, $-M^{013}=0$, $M^{023}=0$ together express the following equation:

\[-\frac{\partial{\bf\mathrm{B}}}{\partial t}+\nabla\times{\bf\mathrm{E}}=0,\]

while the fourth equation translates into this:

\[\nabla\cdot{\bf\mathrm{B}}=0.\]

So what have we got? Basically we discovered that the first two of Maxwell's equations are merely the defining equations of the charge density and current, whereas the second pair are identities that hold for all smooth vector fields $A^\mu$.

In sum, the theory of electromagnetism is really just the geometric theory of an arbitrary smooth vector field.

As an added bonus, we can discover something new. Neither the charge density or current, nor the electric or magnetic fields would change if we were to modify $A$ by adding to it the gradient of a scalar field: $A\rightarrow A+\nabla f$. That is because a scalar field is nothing but a 0-form; its gradient is a 1-form; and applying the derivative operator for the second time to form $F^{\mu\nu}$ the contribution of $f$ will be identically zero (${\rm d}(A + {\rm d}f)={\rm d}A+{\rm d}^2f={\rm d}A$), so the expression for the electromagnetic field tensor will not change.

Ah, and one more thing. The way the exterior derivative is defined, were you to compute it in curved space, all Christoffel-symbols would drop out automatically. In other words, the values you get are independent of the differential operator you use: in particular, you are allowed to use the ordinary differential operator. The physical significance of this is that in the case of the electromagnetic field, both the equations used to define $\vec{E}$ and $\vec{B}$ and the field equations will automatically be satisfied in curved spacetime, under the conditions of general relativity.

Indeed, I believe this, namely that their definition is not dependent on the geometry of the underlying manifold, is one of the major strengths of differential forms.

One other observation that is evident from the above is worth mentioning. It concerns the somewhat arbitrary relabeling of the components of $F^{\mu\nu}$. Though I chose to call $\vec{E}$ and $\vec{B}$ vectors, it should be evident that they decidedly do NOT transform as 4-vectors under a change of coordinates. Indeed, it is possible at any point in space to choose a coordinate system in which either $\vec{E}$ or $\vec{B}$ (but usually not both) vanishes altogether. I have often seen $\vec{E}$ and $\vec{B}$ described as "real" physical quantities, as opposed to $A$, which is not "real" because it is undetermined to the extent that you can add $\nabla f$ to it. But what we see here suggests to me that $\vec{E}$ and $\vec{B}$ are far less "real" than $A$! (Arguably, though, the electromagnetic field tensor $F$ is real in the sense that it is a "proper" tensorial quantity.)