$\ker \alpha$ is involutive if and only if $\alpha \wedge d\alpha = 0$ Let $M$ be a manifold and $\alpha \in \Omega^1(M)$ is a nowhere-vanishing one-form. I have to show that $\ker \alpha$ is involutive if and only if $\alpha \wedge d\alpha = 0$ but I'm having some trouble to prove this. I found this question - Why is $\ker\omega$ integrable iff $\omega\wedge d\omega=0$? - but I'm having a hard time proving this in the case when $M$ is a manifold of an arbitrary dimension.
 A: Let's think for a moment about what it means to say $\omega\wedge d\omega = 0$ when $\omega\in\Omega^1(M)$. If you consider extending $\omega$ to a basis $\{\omega=\omega_1,\omega_2,\dots,\omega_n\}$ for the $1$-forms locally, you can easily convince yourself that $d\omega = \omega\wedge\eta$ for some $1$-form $\eta$, again locally. Then (locally) it is clear that $d\omega(X,Y) = 0$ whenever $\omega(X)=\omega(Y)=0$. Thus, $\omega\wedge d\omega = 0$ implies involutivity. 
On the other hand, if the distribution is involutive, then you already know that $d\omega(X,Y) = 0$ whenever $\omega(X)=\omega(Y)=0$. But then this tells us that $d\omega = \omega\wedge\eta$ for some $1$-form $\eta$; for, if not, we would have $$d\omega=\sum\limits_{i=2\\i<j}^n f_{ij}\omega_i\wedge\omega_j$$ for some functions $f_{ij}$, with some coefficient $f_{i'j'}$ nonzero (at a particular point). Choosing $X,Y$ so that $\omega_{i'}(X)=1$ and $\omega_k(X)=0$ for all $k\ne i'$ (including $k=1$) and $\omega_{j'}(Y)=1$ and $\omega_k(Y)=0$ for all $k\ne j'$ (likewise) shows that $d\omega(X,Y)\ne 0$.
