Closed $3$-manifold, $2$-dimensional subbundle of this manifold, is this form exact or not?

Question

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?

Answer 1

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:

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