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 is a differentiable real-valued function on and is a tangent vector with point of application and vector part , then the differential of is the 1-form such that

(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 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

(where the are the natural frame field vectors at the point of application ). Therefore we have shown that

I emphasised a specific tangent vector argument 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

The exterior derivative of is the 2-form

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

Therefore we have shown that

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

which has a correspondence with the vector field of the type (1) in the problem, that is,

The exterior derivative of is the 3-form

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