Exterior derivative leibniz rule geometric

Question

I‘ve watched the video of Keenan Crane on the exterior derivative https://youtu.be/jeiDXhCiF44 where he explains that the product rule of differential forms just describes how little volumes change. So I‘ve drawn a picture which is almost the classic visualization of the product rule for functions. Here are $v$ and $w$ 1-forms which are visualized as vectors. The product rule now states that: $$\mathrm{d}(v\wedge w)= \mathrm{d}v\wedge w - v\wedge \mathrm{d} w $$ But than the geometric interpretation brakes down because when we view the wedge product of $v$ and $w$ as a (equivalence-class of) orientated parallelogram the second term should have a plus sign because of the same orientation. So where is the mistake?

Answer 1

Crane's heuristic is a good one (and a common one), but the precise analogy breaks down, indeed, here. This is not the correct product rule when you work with actual vectors (i.e., vector functions). There is no sign. The point is that differentiating a vector gives a mixed sort of object — a vector-valued $1$-form, and $d(u\wedge v)$ will be a $2$-vector valued $1$-form. For example, if $$u = f(t)e_1 \quad\text{and}\quad v = g(t)e_1 + h(t)e_2,$$ then $u\wedge v = f(t)h(t) e_1\wedge e_2$ and $$d(u\wedge v) = (f'(t)h(t)+f(t)h'(t))dt\, e_1\wedge e_2 = du\wedge v \color{red}{+} u\wedge dv,$$ because $dt$ commutes with vectors. (Most mathematicians will write $dt\otimes e_1$ and $dt\otimes e_1\wedge e_2$ when unlike objects — here, $1$-forms and vectors or $2$-vectors — are combined multiplicatively.)

On the other hand, when you work with differential forms, the basis $1$-forms interfere with one another and the sign comes in from interchanging them. For example, if $$\omega = f(x,y,z)dx \quad\text{and}\quad \eta = g(x,y,z)dy,$$ then $\omega\wedge\eta = f(x,y,z)g(x,y,z)dx\wedge dy$ and $$d(\omega\wedge\eta) = (\frac{\partial f}{\partial z}g+f\frac{\partial g}{\partial z})dz\wedge dx\wedge dy;$$ note the other partials do not appear because $dx\wedge dx=0$ and $dy\wedge dy=0$. Continuing the computation, $$ (\frac{\partial f}{\partial z}g+f\frac{\partial g}{\partial z})dz\wedge dx\wedge dy = (\frac{\partial f}{\partial z}dz\wedge dx)\wedge (g dy) \color{red}{-} (f\,dx)\wedge(\frac{\partial g}{\partial z}dz\wedge dy),$$ because $dx\wedge dz\wedge dy = -dz\wedge dx\wedge dy$. Thus, with actual $1$-forms, we have $$d(\omega\wedge\eta) = d\omega\wedge\eta \color{red}-\omega\wedge d\eta,$$ as promised.