d-squared equals zero

Here is the most quietly powerful equation in the whole subject: apply the exterior derivative twice in a row and you always get zero. In symbols, d(d omega) = 0 for every form omega — usually written d squared = 0. It says that the boundary of a derivative is empty, that taking d once and then again destroys everything.

Why is it true? On a 0-form f you get d(df) = d(f_x dx + f_y dy + ...), and when you expand it the mixed second partials appear in pairs that differ by a sign because of the antisymmetry of the wedge: a term f_xy dy^dx meets a term f_yx dx^dy, and since the mixed partials are equal (Clairaut's theorem, for smooth f) and dy^dx = -dx^dy, they cancel exactly. The same cancellation, driven by equality of mixed partials against antisymmetry of forms, kills d squared on forms of every degree. So d squared = 0 is, at heart, the statement that partial derivatives commute, dressed in the language of forms.

This one identity is the source of the classical vector-calculus facts curl of grad equals zero and divergence of curl equals zero — both are just d squared = 0 read off at degree 0 and degree 1. It is also the linchpin of de Rham cohomology: because d squared = 0, every exact form (something that is d of a form) is automatically closed (killed by d), which sets up the comparison between closed and exact forms that detects holes in a space. Without d squared = 0 there would be no homology, no cohomology, and no clean Stokes' theorem.