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