Is there a rigorous integral definition of the exterior derivative analogous to the way the gradient, divergence & curl in vector analysis can be defined in integral form?

Furthermore can it be formulated before stating & proving Stokes theorem?

Finally, a beautiful classical argument for the integral definition of the divergence intuitively existing is given in Purcell (page 78), & it leads to the quickest (one line, coordinate independent!) proof of the Divergence theorem I've ever seen. How does one make the limiting process rigorous, & will making the limiting process rigorous justify that one-line proof?



1 Answer 1


Let $\omega$ be a differential $k$-form.

$$d\omega(v_1,v_2,...v_k,v_{k+1}) = \lim_{t_j \to 0}\dfrac{1}{t_1t_2 \dots t_{k+1}}\int_{bP_t} \omega$$

where $P_t$ is the parallelpiped formed by $t_1v_1,t_2v_2, \dots, t_{k+1}v_{k+1}$, and $bP_t$ is the oriented boundary of this parallelepiped.

Observe that this definition even applies to $0$-forms:

$$df(v)\big|_p = \lim_{t \to 0}\dfrac{1}{t} \left( f(p+tv) - f(p)\right)$$

since the boundary of a line segment is just two points oriented in opposite directions, and integration on a point is just evaluation.

This definition, which can be established as soon as you define integration of a form, is essentially an "infinitesmal" version of Stoke's theorem. Stoke's theorem seems a lot less mysterious when this is you definition of the exterior derivative! Although I cannot see your "one line proof" on google books, I am guessing it is basically this? It really is just summing up lots of little local oriented parallelepipeds, and observing that (due to orientations) the interior faces cancel out. That together with the fact that the result holds for incredibly small parallelepipeds by the definition above gives the result.

It also is very clear from this perspective why $d\circ d = 0$: if you already have Stoke's theorem, you can see that $dd\omega$ is defined by an integral which vanishes for all $t$ (applying stoke's theorem to the integral gives an integral over the boundary of the boundary of a parallelepiped, which is empty).

What is perhaps less clear from this definition is why $d\omega$ is a form. It is clearly alternating, by the definition of integration over an oriented boundary. It respects scalar multiplication in each slot just by variable substitution of the limiting variables. The hard part is seeing why it respects vector addition in each slot. I leave that as a fun exercise.

  • $\begingroup$ Fantastic stuff! You don't have a book reference for this by any chance do you? $\endgroup$
    – bolbteppa
    Dec 21, 2013 at 1:01
  • $\begingroup$ Unfortunately no. I am in the process of writing (essentially) a book for an online multivariable calculus class I am helping to create which uses forms. We will use this approach to motivative the exterior derivative. A MO answer also gives this perspective and cite's V. Arnol'd's book on classical mechanics. So you could have a look there. I never found the typical characterization very satisfying, especially how it treats functions and forms as different cases. This approach gives a unfied definition for all levels of forms. $\endgroup$ Dec 21, 2013 at 1:05
  • 1
    $\begingroup$ Another great thing is that it really makes connection to curl and divergence very apparent. I mean, exterior derivative of a one form is just integrating around a loop, so of course it is measuring a "curl" of some sort. $\endgroup$ Dec 21, 2013 at 1:07
  • 1
    $\begingroup$ @Bye_World Not yet. ximera.osu.edu/course/kisonecat/m2o2c2/course has just the differential calculus, and treats higher order derivatives as symmetric tensors. It does not reach differential forms and integration. Hopefully soon! $\endgroup$ May 27, 2015 at 22:58
  • 1
    $\begingroup$ $d\omega$ is defined this way in Hubbard's Vector Calculus book. $\endgroup$ Jun 4, 2015 at 13:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.