Product Rule for differential forms

Question

For a $p$-form, $\alpha$, and a $q$-form, $\beta$, why is the exterior derivative product rule given by $$d(\alpha\wedge\beta)=d\alpha\wedge\beta+(-1)^p\alpha\wedge d\beta,$$ with $(-1)^p$? Wikipedia says that it comes from the graded product (Leibniz) rule, but those Wikipedia articles give this relation as a definition that applies to differential forms.

Both resources I am using to learn about differential forms give this relation without proof. How can I mathematically prove it and/or physically/geometrically see it?

Answer 1

To get the rough idea of why the factor is needed, you should realize that the differential should be antisymmetric in all the objects it is constructed from—not just the $\alpha$ and $\beta$, but also the $d$. That's, in a sense, what ensures that we always have $d^{2}(\cdots)=0$. Another suggestive way to think about it is to think of the index notation, in which the inclusion of an additional derivative $d$ becomes a $\partial_{\mu}$, with the whole expression having to be antisymmetrized over the existing tensorial indices of the $\alpha$ and $\beta$ and the new index $\mu$.

In this light, think of what happens when we take a differential of a wedge product $\alpha\wedge\beta=(dx_{1}\wedge dx_{2}\wedge\cdots\wedge dx_{p})\wedge\beta$. To advance the $d$ to act on the $\beta$, it has to pass through all $p$ of the basis $dx_{j}$ differential forms, effectively picking up a factor of $(-1)$ each time.

Answer 2

In principle very easy, but sesquipedalian in writing, especially at blackboards.

Use associativity, aka bracket flatness of $\bigwedge$ as a multinear function on forms $$a\wedge (b\wedge (c\wedge d))=\bigwedge(a,b,c,d)$$ product rule $$d\bigwedge(a,b,c,d)= \sum_{i=1}^4 \ d_i\ \bigwedge(a,b,c,d)$$ where $d_i$ denotes a partial derivative to be applied at position $i$ and "all arguments except the $i^{th}$ kept constant",

complete antisymmetry, especially $$...\wedge \lambda \ a \wedge .... \wedge \mu \ a ... = 0$$ for any $a$ form and $\lambda, \mu$ scalars.

Now apply the $d$ per factor $a_i$ by a jump of $a_i \ dx^i$ to the left over $i-1$ factors $$d_i\ \bigwedge(a_1,a_2\dots,a_n)=(-1)^{i-1} d_1 \bigwedge(a_i, a_1,\dots , a_{i-1}, a_{i+1} , a_n)$$ and let the 2-form $$d (a_i dx^i ) = \sum_{k\neq i} \partial_k a_i dx^k \wedge dx^i $$ jump back to its original place.

All subforms forms of even degree can be shifted freely without sign change. So put is back to its original place.

In practice, one collects the product of all inner coefficients to the right as a single coefficient of a pure basis n-form $\bigwedge(dx^a\dots \ dx^z)$. The sum of these has to be sorted

 Wedge[x_, y__] :> Signature[{x, y}] Sort[Wedge[x, y]]

It was Grassmans great ability of abstraction and Fubinis detailed analysis of multidimensional integrals, that established the principle of integrating $$\int\dots\int f(x) \bigwedge_{i}(dx_{i_k})$$ has always be done over the leftmost factor, differentiation creates a new left factor, and a permutation of the two leftmost factors changes signs by inverting the integration direction over the edges of a rectangle.

Of course, fundamentally critical against concepts difficult to teach, learn and apply correctly, the community of applied math users, especially physicists, has a more simple rule: use ordinary products of differentials of coordinates, just at hand, and apply the signs at the end of the calculation, by hand rules, left handed or right handed.

What is missing in the elementary definiton of Leibniz 1-d integral is Stokes theorem: What looks like a 1-d integral is an area integral over a 2-form

$$\int_a^b\ f(x) \ dx = \int_{a<x<b, 0 < y < f(x)} dx \wedge dy = \int_{-\infty}^\infty y f^{-1}(dy)$$ where $f^{-1}(\text{set})$ denotes the measure of all intervals along the total of the x-axis mapped by f onto the interval $dy$