In order to extend the results from last week into the realm of classical vector analysis, we must introduce an additional operator called the exterior derivative and which is denoted by $$d$$. The exterior derivative can be thought of as the classical nabla $$\vec \nabla$$ on steroids since it is a bit more versatile and works in spaces of arbitrary dimension.

The defining property of the exterior derivative is that when it operates on a differential form $$\phi(x,y,z)$$ it produces a new differential form $$d\phi$$ of one higher order. So if $$\phi$$ is an $$n$$-form then $$d\phi$$ is an $$n+1$$-form. To understand how this operation is carried out, we start with its action on a function (i.e. a $$0$$-form). In this case

\[ df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz \; ,\]

which is just what is expected from basic calculus. The remaining piece of the definition is to see what the action of $$d$$ is on a $$1$$-form. Since it results in a $$2$$-form, we can suspect that may involve the wedge product and we would be right. Specifically, suppose that the one form is given by

\[ \phi = A_x dx \]

then

\[ d \phi = d (A_x dx) = \frac{\partial A_x}{\partial x} dx \wedge dx + \frac{\partial A_y}{\partial y} dy \wedge dx + \frac{\partial A_z}{\partial z} dz \wedge dx \; .\]

From the anti-symmetry property of the wedge product, this result simplifies to

\[ d \phi = \frac{\partial A_y}{\partial y} dy \wedge dx + \frac{\partial A_z}{\partial z} dz \wedge dx \; .\]

The generalization to more complex $$1$$-forms is obvious.

As a result of these properties, the exterior derivative satisfies $$d^2 = dd = 0$$ regardless of what it operates on.

We are now in the position to adapt the language of differential forms to some familiar results of vector calculus.

The core adaptations are for the divergence and curl. The remaining vector calculus identities more or less flow from these.

The expression for the divergence start with the expression for the corresponding one form

\[ \phi_A = A_x dx + A_y dy + A_z dz \; . \]

Apply the Hodge star operator first

\[ *\phi_A = A_x dy \wedge dz + A_y dz \wedge dx + A_z dx \wedge dy \; .\]

The application of the exterior derivative only produces derivatives for the excluded variable in the wedge product

\[ d*\phi_A = A_{x,x} dx \wedge dy \wedge dz + A_{y,y} dy \wedge dz \wedge dx + A_{z,z} dz \wedge dx \wedge dy \; , \]

where the term

\[ A_{x,x} = \frac{\partial A_x}{\partial x} \]

and so on for the other terms.

All of the wedge products are cyclic permutations of $$dx \wedge dy \wedge dz$$ so rearrangement involves no changes of sign. A further application of the Hodge star operator give the zero form

\[ *d*\phi_A = A_{x,x} + A_{y,y} + A_{z,z} \Leftrightarrow \vec \nabla \cdot \vec A \; .\]

The curl, which is handled in a similar way, is expressed as

\[ \vec \nabla \times \vec A \Leftrightarrow *d\phi_A \; .\]

The proof is reasonably easy. Start again with the expression for the corresponding form

\[ \phi_A = A_x dx + A_y dy + A_z dz \; \]

and then operate on it with the exterior derivative to get

\[ d \phi_A = A_{x,y} dy \wedge dx + A_{x,z} dz \wedge dx + A_{y,x} dx \wedge dy \\ + A_{y,z} dz \wedge dy + A_{z,x} dx \wedge dz + A_{z,y} dy \wedge dz \; \]

Organizing the terms in lexicographical order (i.e. $$dx \wedge dy$$, $$dx \wedge dz$$, and $$dy \wedge dz$$), accounting for changes in sign as terms are swapped, gives

\[ d\phi_A = (A_{y,x}-A_{x,y}) dx \wedge dy \\ + (A_{z,y}-A_{y,z}) dy \wedge dz \\ + (A_{x,z}-A_{z,x}) dz \wedge dx \; \]

The final step is to apply the Hodge star operator to arrive at

\[ *d\phi_A = (A_{y,x}-A_{x,y}) dz + (A_{z,y}-A_{y,z}) dx + (A_{x,z}-A_{z,x}) dy \; , \]

which proves the correspondence

\[ *d\phi_A \Leftrightarrow \vec \nabla \times \vec A \; .\]

Next column, I’ll expand on the algebra of differential forms just a bit and then will apply the results introduced here to prove a host of vector calculus identities.