There is a several-part question I'm trying to do, which I'm hoping to get some feedback on the parts I've completed and some help on the parts I can't complete.
Suppose $M$ is a compact connected $3$-manifold and $\omega$ is a nowhere zero $1$-form defined on $M$. Suppose that the distribution $\ker \omega$ is integrable, and $\ker \omega = T \mathcal{F}$ for a foliation $\mathcal{F}$.
(a) Show that $\omega \wedge d\omega = 0$
Parts of this solution was taken from: Why is $\ker\omega$ integrable iff $\omega\wedge d\omega=0$?
Since $\ker \omega$ is integrable, we have $I(\ker \omega)$ is closed under $d$ (where $I$ is the annihilator ideal) so that by assumption $d \omega \in I(\ker \omega)$. We now want to show $\omega \wedge d \omega(X, Y, Z) = 0$ for any three vector fields. However, we can assume $Y, Z \in \ker \omega$ (why?) so that $\omega \wedge d \omega(X, Y, Z) = \omega(X) \wedge dw(Y, Z) = 0$. We didn't need $\omega$ to be a $1$-form right? Also, where did we use the fact that the kernel is a foliation?
(b) Use a partition of unity to show that there is a $1$-form $\alpha$ such that $d\omega = \alpha \wedge \omega$.
Let $(U_\alpha, \phi_\alpha)$ be a chart on $M$ and let $\psi_\alpha$ be a partition of unity so that $Supp(\psi_\alpha) \subset U_\alpha$ and $\sum_\alpha \psi_\alpha = 1$ for any point in $M$. We can define $\alpha$ on each $U_\alpha$ as...not sure how to continue
(c) Show that $d \alpha \wedge \omega = 0$.
\begin{equation} 0 = d^2 \omega = d(d\omega) = d(\alpha \wedge \omega) = (d\alpha) \wedge \omega - \alpha \wedge (d \omega) = (d\alpha) \wedge \omega - \alpha \wedge (\alpha \wedge \omega) = (d\alpha) \wedge \omega \end{equation}
where the 4th equality follows from the fact that $\omega$ is a $1$-form. On the other hand, the 3rd and 5th inequality follows form part (b). Thus, the result follows.
(d) Suppose that $\alpha'$ is some other $1$-form satisfying $d \omega = \alpha' \wedge \omega$. show that $\alpha' = \alpha+ g \omega$ for some function $g$, and that $\alpha \wedge d \alpha = \alpha' \wedge d \alpha'$.
We have $\alpha' \wedge \omega = d\omega = \alpha \wedge \omega$ so that $(\alpha' - \alpha) \wedge \omega = 0$. Thus $\alpha - \alpha'$ is a multiple of $\omega$ so that $\alpha' = \alpha + g \omega$ for some function $g$. For the other part, we have \begin{align} \alpha' \wedge d\alpha' &= (\alpha + g \omega) \wedge d(\alpha + g \omega) \\ &= (\alpha + g \omega) \wedge (d\alpha + gd \omega)\\ &= \alpha \wedge d\alpha + g\alpha \wedge dw + g \omega \wedge d \alpha + g^2 \omega \wedge d\omega\\ &= \alpha \wedge d\alpha \end{align}
where the last equality follows from all the other quantities being $0$.
(e) Suppose that $\omega'$ is a nowhere zero $1$-form and $\ker \omega = \ker \omega'$. If $d\omega' = \gamma \wedge \omega$, so that $\alpha d \alpha - \gamma \wedge d \gamma$ is exact.
I just need a form $\beta$ so that $d\beta = \alpha d \alpha - \gamma \wedge d \gamma$ but after playing around with a couple I can't seem to find one.
