# Moving exterior differential/derivative inside a wedge product

Assumptions: Let $$M$$ be smooth $$m$$-manifold. (If needed: Let $$M$$ be orientable and then oriented. Let $$M$$ be compact. Let $$(M,g)$$ be a Riemannian manifold.)

Let $$\Omega^jM$$ be the set of smooth $$k$$-forms on $$M$$, for $$j=0, 1, ..., m$$. Let $$d_j: \Omega^jM \to \Omega^{j+1}M$$ be exterior differential / derivative on $$\Omega^jM$$ (based on $$d: \Omega(M) \to \Omega(M)$$, with $$\Omega(M)$$ $$:= \bigoplus_{j=0}^{m} \Omega^jM$$).

Let $$k \in \{0, 1, ..., m\}$$. Let $$(\alpha, \gamma) \in \Omega^kM \times \Omega^{m-(k+1)}M$$.

Observations:

1. $$d_k \alpha \wedge \gamma$$ is a smooth top form (aka smooth $$m$$-form)
2. $$(-1)^{1+k^2} \alpha \wedge d_{m-(k+1)}\gamma$$ is a smooth top form (aka smooth $$m$$-form)

Question 1: Assuming the above observations are correct, are they equal?

Question 2: In general, can we just move exterior differential/derivative through wedge products and just multiply $$(-1)^{\text{something}}$$?

Question 3: In anything above, are we assuming any additional things on $$M$$ like orientable/oriented/compact/Riemannian?

Question 4: If no to question 1, then do each of the 2 forms at least have equal integrals, i.e. the values we get when we plug each into $$\int_M$$ are equal? Here, we now suppose $$M$$ is orientable and then oriented and I guess compact (otherwise I guess we have to assume the forms have compact support or something).

Context: This comes from some definitions and propositions leading to Hodge decomposition theorem, including the definition of Hodge star operator, but I'm trying to see if I understand the non-Hodge parts correctly. ($$\gamma$$ is actually the image of some $$\beta \in \Omega^{k+1}M$$ under the Hodge-star operator.)

Here is an attempt of an answer.

Question 1 There is no need for an equality like that. What is true is that $$d\left(\alpha\wedge \gamma \right) = d\alpha \wedge \gamma + (-1)^{\deg\alpha}\alpha \wedge d\gamma$$

And assuming your equality to be true will lead to an assumption on $$d(\alpha\wedge\gamma)$$

Here is a concrete counter-example: \begin{align} \alpha &= dx^1 & \gamma = x^2dx^3\wedge\cdots\wedge dx^n \\ d\alpha \wedge \gamma &= 0 & \alpha \wedge d\gamma = dx^1\wedge\cdots\wedge dx^n \end{align}

Question 2 the answer is no. See above.

Question 3 above, the computations are local, so it does not depend on compactness or orientability: extend the counterexample by zero outside a chart.

Question 4 the answer is still no: in the counterexample above, $$d\alpha\wedge \gamma = 0$$, thus has zero integral, but $$\alpha\wedge d\gamma$$ is a volume form on an orientable manifold, it has non-zero integral.

Regarding @JanBohr's answer, (which leads to two self-refereing answers), I have to add that in case $$M$$ is oriented, then Stokes theorem states that $$\int_M d(\alpha\wedge \gamma) = \int_{\partial M} \alpha\wedge \beta$$ and thus, $$\int_M d\alpha \wedge \gamma = (-1)^{\deg \alpha+1}\int_{M}\alpha\wedge d\gamma + \int_{\partial M}\alpha\wedge \gamma$$ and thus there is (up to sign) an equality as soon as $$M$$ has no boundary or $$\alpha\wedge \gamma$$ is zero on $$\partial M$$.

• Thanks DIdier_!
– BCLC
Dec 15, 2020 at 12:50

One of the defining properties of the exterior differential is the Leibniz rule $$d(\alpha\wedge \gamma)=d\alpha\wedge \gamma+(-1)^{k} \alpha\wedge d\gamma,$$ where $$k$$ is the degree of $$\alpha$$, see e.g. on wikipedia. This holds true for arbitrary smooth manifolds, no need for a Riemannian metric or orientation. As $$k$$ and $$k^2$$ have the same parity, the right hand side in the previous display is exactly the difference between your two $$m$$-forms. In particular they are equal iff $$\alpha \wedge \gamma$$ is closed. The integral over both $$m$$-forms, say if $$M$$ is oriented and compact, is the same just because the integral of an exact form is zero by Stokes' theorem.

Regarding @DIdier_'s counterexample for question 4: This is a situation where the boundary integral in Stokes' theorem does not vanish (for any smooth domain in $$\mathbb{R}^n$$). Above I avoid this problem by assuming $$M$$ to be boundaryless. Another way out is to assume that $$\alpha$$ and $$\gamma$$ have compact support in the interior.

• Thanks Jan Bohr! It seems the only thing I'm missing is the $k$ vs $k^2$. Perhaps I possibly made a mistake in that I should've got $k$ instead of $k^2$ and then I can use antiderivation property of $d$ and Stokes' Theorem, but now you say something like parity...
– BCLC
Dec 15, 2020 at 12:48
• AAAAHHHH $k$ is even/odd if and only if (resp) $k^2$ is even/odd ?
– BCLC
Dec 15, 2020 at 12:49
• @BCLC if $k$ is even, then $k^2$ is even. Conversly, if $k=2n+1$ is odd , then $k^2 = 2(n^2 + 2n) +1$ is odd. Dec 15, 2020 at 12:59
• right thanks! @DIdier_
– BCLC
Dec 15, 2020 at 13:01