Pullback of a differential form (${\varphi }^{\ast }$)

A smooth map $\varphi :M\to N$ pulls a differential form on $N$ back to one on $M$ by applying its differential $d\varphi$ to each tangent vector argument. In particular, a parametrization of a curve pulls a one-form back to an ordinary integrand on an interval.

For $\alpha =xdy$ and the parabola $\varphi (t)=(t,t^2)$, $0\le t\le 1$, substitute $x=t$ and $dy=2tdt$ to obtainThe same endpoints gave $0$ and $1$ along the two square-edge paths in the introduction, so this third answer is consistent with path dependence. By contrast, ${\varphi }^{\ast }d(xy)=d(t^3)=3t^{2}dt$ gives integral $1$, the same endpoint difference as before.

The pullback of a differential form commutes with the wedge product of differential forms and the exterior derivative: ${\varphi }^{\ast }(\alpha \wedge \beta )={\varphi }^{\ast }\alpha \wedge {\varphi }^{\ast }\beta$ and ${\varphi }^{\ast }(d\alpha )=d({\varphi }^{\ast }\alpha )$. These identities make changes of coordinates and parametrizations compatible with differentiation and integration.

It is the alternating-tensor case of pullback.