Moving exterior differential/derivative inside a wedge product

derivativesdifferential-formsdifferential-geometryexterior-algebrariemannian-geometry

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.)

Best Answer

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 $\partial M$.

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