Partial Integration of the Wedge Product

Question

Is it in general true that $$ A \wedge dB = -dA \wedge B $$

Answer 1

Remember that the usual integration by parts is just the integrated product rule. We can figure out if the above formula holds in the exact same way.

If $A$ and $B$ are real valued differential forms then then you have the Leibniz rule: $$d(A\wedge B)=dA\wedge B + (-1)^{|A|}A\wedge dB,$$ where $|A|$ denotes the degree of $A$. So in general the equation you have does not hold, but it will hold (up to a sign depending on the degree of the form) if $d(A\wedge B)=0$. In particular, this holds if $A$ is a smooth function $0$-form, and if $B$ is the volume form; this is the case we integrate to derive the usual integration by parts.

For a concrete counterexample, consider the manifold $\mathbb{R}^4$ and the differential forms $A=z~dx$ and $B=w~dy$. Then $A\wedge B = zw~dx\wedge dy$, and so $d(A\wedge B)=w ~dz\wedge dx\wedge dy + z~dw\wedge dx\wedge dy$, and so we should not expect $A\wedge dB = -dA\wedge B$. Indeed, we have \begin{align} dA & = dz\wedge dx,\\ dB & = dw\wedge dy,\\ A\wedge dB & = z ~ dx\wedge dw\wedge dy,\\ dA\wedge B & = w ~ dz\wedge dx\wedge dy. \end{align}