I would like a help with understanding the formula for exterior derivative. Suppose that we have a differential 1-form on $\mathbb{R}^2$ given by $\omega(x)(v) = (a(x)dx + b(x)dy)(v)$. In other words $\omega$ is a mapping of the type $\mathbb{R}^2 \to \text{Hom}(\mathbb{R}^2, \mathbb{R})$, to every point we assign a linear mapping of the type $\mathbb{R}^2 \to \mathbb{R}$.

Now, I would like to derive the formula for $d\omega$. This means that I do not want to start with the axioms for the $d$ and I do not want to start with the formula and prove it. I want to naturally get to the fact that

$$d\omega(x)(u)(v) = \left (\frac{\partial b}{\partial x}(x) - \frac{\partial a}{\partial y}(x)\right )dx \wedge dy(v, u).$$

For example, for a 0-form $f\colon \mathbb{R}^2 \to \mathbb{R}$, we naturally want to approximate it every point by a linear mapping. So, we naturally get to the fact that the limit

$$\lim_{\|h\|\to 0}\frac{f(x+h)-f(x) - L(h)}{\|h\|}$$

must be $0$, for some linear mapping $L$ at a particular point $x$. The next step is to prove that the mapping $L$ has to be given by the $1\times 2$ matrix

$$\left (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right ).$$

Is it possible to do a similar analogy of linear approximation for the 1-form $\omega$, or in general for a $k$-form on $\mathbb{R}^n$?


In particular I would like to know why

$$d\omega(x)(v)(u) = (d\omega(x)(v))(u) - (d\omega(x)(u))(v).$$

This is in some textbooks a definition and it can be simplified to the formula for $d\omega(x)(v)(u)$ stated above.


If you want a limit-based definition for higher forms, I think basing it on the generalized Stoke's theorem is the way to go:

$$ \int_{B(P, r)} \mathrm{d}\omega = \oint_{S(P, r)} \omega $$

where $B(P,r)$ is the disk around $P$ with radius $r$, and $S$ is its boundary: the circle around $P$ with radius $r$.

If $\mathrm{d}\omega = f \mathrm{d}x \mathrm{d}y$, then you can obtain the values of $P$ by integrating over small disks:

$$ f(P) = \lim_{r \to 0} \frac{1}{\pi r^2} \oint_{S(P, r)} \omega $$

You could replace disks and circles with other suitable shapes, such as squares.

If I understand correctly, I believe the way you want to conceive the derivative doesn't actually relate to the exterior derivative — what you are conceiving, I think, is the covariant derivative, or maybe of connections.

  • $\begingroup$ Thank you. As far as I know, the exterior derivative is supposed to be something like a directional derivative. I would like to derive the formula for the exterior derivative without the axioms. Your approach already uses the exterior derivative, I would like to define it as a generalized directional derivative. $\endgroup$ – pizet Apr 25 '18 at 17:40
  • $\begingroup$ @pizet: Any motivation for a formula for the exterior derivative (whether definition or theorem or just a heuristic argument) is going to be based on what we think the exterior derivative should be, because that's what we're trying to describe. As I indicated in my addendum, your idea of "generalized directional derivative" makes me think the thing you have in mind is some other notion from differential geometry. $\endgroup$ – Hurkyl Apr 25 '18 at 17:45
  • $\begingroup$ Thank you, I have added an edit to my original post. So, if I understand it right, the real motivations for the definitions of differential forms, exterior algebra and exterior derivative is to define it in such a way that the Stokes theorem works? $\endgroup$ – pizet Apr 25 '18 at 17:55
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    $\begingroup$ @pizet: There are lots of angles into the theory; different people will find different starting points appealing. I happen to be fond both of the purely algebraic formulation and of the "n-forms are 'infinitesimal' n-dimensional surface integrals". Stokes' theorem happens to be a well-known particularly compact way to express how we think $n$-dimensional integrals should relate to derivatives, which is why I cite it. $\endgroup$ – Hurkyl Apr 25 '18 at 18:01
  • $\begingroup$ Thank you. Is it possible to prove that for the Stokes theorem to hold, we have to have exactly this definition of the exterior derivative? $\endgroup$ – pizet Apr 25 '18 at 18:04

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