# A problem involving the use of exterior derivatives of differential forms to re-express the classical gradient, curl and divergence operations

Modern differential geometry makes extensive use of differential forms and the concept of exterior derivatives of differential forms developed by the French mathematician Élie Cartan (1869-1951). A Wikipedia article about exterior derivatives of differential forms can be found here. As alluded to in this article, exterior derivatives of differential forms encompass a lot of results usually expressed in terms of vector fields in classical vector calculus. In particular, there is a duality between 1-forms, 2-forms and vector fields which allows the classical gradient, curl and divergence operations of vector calculus to be fully subsumed within the realm of exterior derivatives. In the present note I want to briefly explore how these three differentiation operations of vector calculus can be replaced with Cartan’s exterior derivative. The necessary notation and motivation for this are nicely encapsulated in the following problem which appears in Barrett O’Neill’s Elementary Differential Geometry book (Revised Second Edition, p.33):

This problem was also the subject of an interesting Mathematics Stack Exchange discussion which can be found here. The reader should attempt to solve this problem by himself/herself before reading my solution below.

To solve part (a), we use the fact that if $f$ is a differentiable real-valued function on $\mathbb{R}^3$ and $\bold{v}_p$ is a tangent vector with point of application $\bold{p}$ and vector part $\bold{v}$, then the differential $df$ of $f$ is the 1-form such that

$df(\bold{v}_p) = \sum v_i \frac{\partial f}{\partial x_i}(\bold{p}) = \sum \frac{\partial f}{\partial x_i}(\bold{p}) dx_i(\bold{v}_p)$

(where the last equality uses the fact that the differentials of the natural coordinate functions evaluated at a tangent vector are equal to the the coordinates $v_i$ of the vector part of the tangent vector). But using the correspondence (1) between 1-forms and vector fields in the problem we can then write

$df(\bold{v}_p) = \sum \frac{\partial f}{\partial x_i}(\bold{p}) dx_i(\bold{v}_p) \stackrel{\mathrm{(1)}}{\longleftrightarrow} \sum \frac{\partial f}{\partial x_i}(\bold{p}) U_i(\bold{p}) = \text{grad } f(\bold{p})$

(where the $U_i(\bold{p})$ are the natural frame field vectors at the point of application $\bold{p}$). Therefore we have shown that

$df \stackrel{\mathrm{(1)}}{\longleftrightarrow} \text{grad } f$

I emphasised a specific tangent vector argument $\bold{v}_p$ in the above solution but I will not do this in the solutions for (b) and (c) as the notation becomes too cumbersome. To solve part (b), we consider the 1-form

$\phi = f_1 dx_1 + f_2 dx_2 + f_3 dx_3$

The exterior derivative of $\phi$ is the 2-form

$d \phi = df_1 \wedge dx_1 + df_2 \wedge dx_2 + df_3 \wedge dx_3$

$=$

$\big(\frac{\partial f_1}{\partial x_1} dx_1 + \frac{\partial f_1}{\partial x_2} dx_2 + \frac{\partial f_1}{\partial x_3} dx_3 \big) \wedge dx_1$

$+ \big(\frac{\partial f_2}{\partial x_1} dx_1 + \frac{\partial f_2}{\partial x_2} dx_2 + \frac{\partial f_2}{\partial x_3} dx_3 \big) \wedge dx_2$

$+ \big(\frac{\partial f_3}{\partial x_1} dx_1 + \frac{\partial f_3}{\partial x_2} dx_2 + \frac{\partial f_3}{\partial x_3} dx_3 \big) \wedge dx_3$

$=$

$-\frac{\partial f_1}{\partial x_2} dx_1 dx_2 - \frac{\partial f_1}{\partial x_3} dx_1 dx_3$

$+ \frac{\partial f_2}{\partial x_1} dx_1 dx_2 - \frac{\partial f_2}{\partial x_3} dx_2 dx_3$

$+ \frac{\partial f_3}{\partial x_1} dx_1 dx_3 + \frac{\partial f_3}{\partial x_2} dx_2 dx_3$

$= \big( \frac{\partial f_2}{\partial x_1} - \frac{\partial f_1}{\partial x_2} \big) dx_1 dx_2 + \big( \frac{\partial f_3}{\partial x_1} - \frac{\partial f_1}{\partial x_3} \big) dx_1 dx_3 + \big( \frac{\partial f_3}{\partial x_2} - \frac{\partial f_2}{\partial x_3} \big) dx_2 dx_3$

But using the correspondence (2) between 2-forms and vector fields in the problem we can then write

$d \phi = \big( \frac{\partial f_2}{\partial x_1} - \frac{\partial f_1}{\partial x_2} \big) dx_1 dx_2 + \big( \frac{\partial f_3}{\partial x_1} - \frac{\partial f_1}{\partial x_3} \big) dx_1 dx_3 + \big( \frac{\partial f_3}{\partial x_2} - \frac{\partial f_2}{\partial x_3} \big) dx_2 dx_3$

$\stackrel{\mathrm{(2)}}{\longleftrightarrow}$

$= \big( \frac{\partial f_3}{\partial x_2} - \frac{\partial f_2}{\partial x_3} \big) U_1 + \big( \frac{\partial f_1}{\partial x_3} - \frac{\partial f_3}{\partial x_1} \big) U_2 + \big( \frac{\partial f_2}{\partial x_1} - \frac{\partial f_1}{\partial x_2} \big) U_3$

$= \text{curl } V$

Therefore we have shown that

$d \phi \stackrel{\mathrm{(2)}}{\longleftrightarrow} \text{curl } V$

Finally, to solve part (c) we can consider the 2-form

$\eta = f_1 dydz + f_2 dx dz + f_3 dx dy$

which has a correspondence with the vector field $V = \sum f_i U_i$ of the type (1) in the problem, that is,

$\eta \stackrel{\mathrm{(1)}}{\longleftrightarrow} V$

The exterior derivative of $\eta$ is the 3-form

$d \eta = df_1 \wedge dy dz + df_2 \wedge dx dz + df_3 \wedge dx dy$

Since products of differentials containing the same differential twice are eliminated, we see immediately that this reduces to

$d \eta = \big( \frac{\partial f_1}{\partial x} dx \big) dy dz + \big( \frac{\partial f_2}{\partial y} dy \big) dx dz + \big( \frac{\partial f_3}{\partial z} dz \big) dx dy$

$= \big(\frac{\partial f_1}{\partial x} + \frac{\partial f_2}{\partial y} + \frac{\partial f_3}{\partial z} \big) dx dy dz$

$= (\text{div } V) dx dy dz$