Let $M$ be a closed $3$-manifold, and let $\xi$ be a $2$-dimensional subbundle of $TM$. I know the following.


* There is a nowhere zero $1$-form $\alpha$ on $M$ with $\alpha(X) = 0$ for any vector field $X$ which is a section of $\xi$.
* Any two $1$-forms $\alpha$, $\alpha'$ with this property satisfy $\alpha = f\alpha'$ for some smooth nowhere zero function $f$.
* $\xi$ is integrable if and only if $\alpha \wedge d\alpha = 0$ for any $\alpha$ as above.
* For integrable $\xi$, we can write $d\alpha = \omega \wedge \alpha$ for some $1$-form $\omega$.
* Any two choices $\omega$, $\omega'$ satisfying this equation have $\omega' - \omega = g\alpha$ for some smooth function $g$.


Assuming $\xi$ is integrable, let $\alpha$, $\alpha'$ be two $1$-forms as above, and let $d\alpha = \omega \wedge \alpha$ and $d\alpha' = \omega' \wedge \alpha'$ for some $1$-forms $\omega$, $\omega'$. I have two questions.


1. Is $\omega \wedge d\omega - \omega' \wedge d\omega'$ an exact form or not?
2. What is the geometric meaning behind $\omega \wedge d\omega - \omega' \wedge d\omega'$ being an exact or nonexact form?


**Edit.** Now that Tsemo has mentioned that this form is indeed exact, could anybody supply a direct proof of its exactness?



Closed $3$-manifold, $2$-dimensional subbundle of this manifold, is this form exact or not?
In fact $\omega\wedge d\omega$ is closed its cohomology class is a well-known invariant of foliation named the Godbillon-Vey invariant. Thus $\omega\wedge d\omega-\omega'\wedge d\omega'$ is exact.


See for example this paper. <http://homepages.math.uic.edu/~hurder/papers/54manuscript.pdf>


See also p.3 of this paper


<http://perso.ens-lyon.fr/ghys/articles/invariantgodbillonvey.pdf>


The geometric meaning of the Godbillon-Vey invariant has been described by Thurston:


W. 
THURSTON, 
Non 
cobordant foliations 
of 
S3, 
Bulletin 
A.M.S. 
78 
(1972) 
511-514


To show that the class of $\omega\wedge d\omega$ does not depend of the choices, firstly, you keep $\alpha$ fixed and let $\omega$ varies. You have $d\alpha =\omega\wedge\alpha =\omega'\wedge \alpha$. This implies that $(\omega'-\omega)\wedge \alpha=0$. Thus $\omega'=\omega +f\alpha$ where $f$ is a function. We deduce that


$\omega'\wedge d\omega' =(\omega +f\alpha)\wedge d(\omega+f\alpha) = \omega\wedge d\omega+\omega\wedge d(f\alpha)+f\alpha\wedge d\omega+f\alpha\wedge d(f\alpha)$.


You have $f\alpha\wedge d(f\alpha) = f\alpha\wedge(df\wedge\alpha+f d\alpha)=0$ since $\alpha\wedge d\alpha =0$.


By definition of the differential, $\omega\wedge d(f\alpha)+f\alpha\wedge d\omega =\omega\wedge d(f\alpha) -d\omega\wedge f\alpha =-d(\omega \wedge f\alpha)$.


We deduce that $\omega'\wedge\omega' =\omega\wedge\omega -d(\omega\wedge d(f\alpha))$. 


Now let varies $\alpha$, you have $\alpha' = f\alpha$ where $f$ is a function which does not vanish. 


$d\alpha' =df\wedge\alpha+f\wedge d\alpha =df\wedge\alpha + f\omega\wedge\alpha$. Let $\omega' =d(Log\mid f\mid)+\omega)$. We have:


$\omega'\wedge\alpha' = (d(Log\mid f\mid+\omega)\wedge (f\alpha)=(df/f +\omega)\wedge (f\alpha) =df\wedge\alpha+f\omega\wedge\alpha=d\alpha'$


We have:


$\omega'\wedge d\omega' =(\omega+dLog\mid f\mid)\wedge d\omega =\omega\wedge d\omega +d(Log\mid f\mid d\omega)$. 


