How to understand $d^2=0$ in differential form?

Question

How to understand $d^2=0$ in differential form without a simple proof from the definition?

Answer 1

A differential $2$-form can be understood in terms of integral operators, specifically what happens when you integrate them over (directed) 2-surfaces. Stoke's theorem, the higher-dimensional analog of the fundamental theorem of calculus, tells us that

$$ \int_S d\omega = \int_{\partial S} \omega $$

where $\partial S$ is the boundary of $S$.

(if $\omega$ is a $0$-form -- i.e. a function -- and $S$ is a path, this is just the fundamental theorem of calculus: 'integrating' a function over a directed point is just plugging it in, and affixing the correct sign out front)

This formula, in turn, may be understood as viewing $d\omega$ as a gadget that tells us how $\omega$ varies in different directions, and viewing $S$ as being described by taking a curve that starts on half the boundary and sweeps across the surface until it reaches the other boundary. (it may also help to think of $S$ as being decomposed into infinitesimal surfaces, where this picture motivates the formula for the infinitesimal surfaces, and then adding them all up cancels out the internal boundaries, just leaving the boundary of $S$) (also, think of how you compute integrals over two-dimensional surfaces as double integrals in multivariable calculus)

Now, we can easily see that $d^2 = 0$:

$$ \int_S d^2 f = \int_{\partial S} df = \int_{\partial \partial S} f $$

and if you have the intuition that the boundary of a surface is made out of loops, then you'll see that the loops themselves don't have any endpoints, and so the rightmost integral is zero.

And $\omega = 0$ is the only $2$-form with the property that $\int_S \omega = 0$ for all $S$.

Answer 2

Henri Cartan, one of the greatest 20th century mathematicians wrote:
...utinam intelligere possim ratinationes pulcherrimae quae e propositione concisa DE QUADRATO NIHILO EXAEQUARI fluunt

In the unlikely event that you don't speak Latin, o amice, this means:
...if I could only understand the beautiful consequences following from the concise proposition $d ^2=0$

In good English: there is no easy answer.

Answer 3

To develop some intuition, you might find it helpful to go back to the physics that inspired the definitions of grad, div and curl: electromagnetism. There are nice physical reasons why we would expect $d^2$ to be zero on both $0$-forms/scalar fields, $\nabla \times \nabla \phi = \vec{0}$, and $1$-forms/vector fields, $\nabla\cdot\nabla\vec{A} = 0$.

(There's a good discussion of such ideas in A Student's Guide to Maxwell's Equations by Daniel Fleisch. Maybe others can point to an on-line discussion.)