Differential Forms

The wedge product, the exterior derivative, and d^2 = 0.

The idea

Differential forms are the expressions that appear under integral signs, such as $P\,dx + Q\,dy$, treated as algebraic objects with a product and a derivative of their own. On the plane, a smooth function is a $0$-form; an expression $\omega = P\,dx + Q\,dy$, with $P$ and $Q$ smooth functions of $x$ and $y$, is a $1$-form; and an expression $h\,dx \wedge dy$ is a $2$-form.

The product is the wedge product $\wedge$, and its defining rule is antisymmetry: $dx \wedge dy = -\,dy \wedge dx.$ Setting $dy = dx$ gives $dx \wedge dx = -\,dx \wedge dx$, so $dx \wedge dx = 0$. The sign records orientation: $dx \wedge dy$ stands for an oriented area element, sweeping the same patch in the opposite order reverses its sign, and an area spanned by one direction twice collapses to zero.

The derivative is the exterior derivative $d$, which raises a $k$-form to a $(k+1)$-form. On a $0$-form it gives $df = f_x\,dx + f_y\,dy$; on a $1$-form it gives $d\omega = (Q_x - P_y)\,dx \wedge dy.$

Theorem (The exterior derivative squares to zero).

For every smooth function $f$ on the plane, $d(df) = 0$: applying $d$ twice always yields zero, written $d^{2} = 0$.

Writing out $d(df)$ produces the coefficient $f_{xy} - f_{yx}$, which vanishes for smooth $f$ because the two orders of mixed partial differentiation agree. This identity is the source of the vector-calculus fact that a gradient field has no curl.

Ways to work on it

• Walkthrough. The wedge product, the exterior derivative, and d(df) = 0.
• Practice. Compute the coefficient of d for a polynomial 1-form at a point.
• Hardest. Evaluate d for a two-variable form, then identify closed versus exact.

Not sure where to start? Take the ten-question placement test.