$2$-dimensional subbundle of tangent bundle of closed $3$-manifold integrable if and only if $\alpha \wedge d\alpha = 0$?

Question

Let $M$ be a closed $3$-manifold, and let $\xi$ be a $2$-dimensional subbundle of $TM$. From here and here, I know that there is a nowhere zero $1$-form $\alpha$ on $M$ with $\alpha(X) = 0$ for any vector field $X$ which is a section of $\xi$, and that any two $1$-forms $\alpha$, $\alpha'$ with this property satisfy $\alpha = f\alpha'$ for some smooth nowhere zero function $f$.

My question now is, do we have that $\xi$ is integrable, i.e. tangent to the leaves of a foliation $\mathcal{F}$, if and only if $\alpha \wedge d\alpha = 0$ for any $\alpha$ as above?

Edit. I was wondering if anybody could supply a direct proof in this case? The full Frobenius theorem seems a bit overpowered...

Answer 1

Fix a nowhere zero $1$-form $\alpha$ satisfying $\alpha(X) = 0$ for any section $X$ of $\xi$. Select a nowhere zero section $s$ of $\nu$; we can do this because $\nu$ is trivial. Observe that $\alpha \wedge d\alpha = 0$ if and only if $(\alpha \wedge d\alpha)(s_1, s_2, s) = 0$ for all nonzero sections $s_1$, $s_2$ of $\xi$. We have$$(\alpha \wedge d\alpha)(s_1, s_2, s) = \alpha(s_1)s\alpha(s_2, s) - \alpha(s)d\alpha(s_1, s_2) = -\alpha(s)d\alpha(s_1, s_2),$$where we have utilized the fact that $\alpha(s_1) = 0$. Therefore, this is always zero if and only if $d\alpha(s_1, s_2) = 0$. We have$$d\alpha(s_1, s_2) = X(\alpha(s_2)) - Y(\alpha(s_1)) - \alpha([s_1, s_2]) = -\alpha([s_1, s_2]),$$which is zero if and only if $[s_1, s_2]$ is a section of $\xi$, i.e. if and only if sections of $\xi$ are closed under Lie bracket. It follows from the Frobenius theorem that this is precisely the criterion for integrability.

Now, assume $\alpha \wedge d\alpha = 0$.Because $\alpha$ is nowhere vanishing, we can write $d\alpha$ as a linear combination of forms of the type $\alpha \wedge \omega$, where $\omega \in \text{Hom}(\nu, \mathbb{R})$, and forms of the type $\omega_1 \wedge \omega_2$, where $\omega_i \in \text{Hom}(\nu, \mathbb{R})$ and $\omega_i$ are nonzero. We have to demonstrate that the $\omega_1 \wedge \omega_2$ part equals zero. Denote this part of the sum by $\Sigma$. Since $\alpha \wedge d\alpha = 0$, it follows that $\alpha \wedge \Sigma = 0$. However, $\alpha$ is outside the span of forms like $\omega_i$, so we have $\Sigma = 0$, as desired.

Answer 2

The to your question is yes as long as the 2-plane distribution is orientable. The more general theorem (the $n$-dimensional analogue) is called the Frobenius theorem. It is quite easy to find many proofs of this theorem online, and one is in Lee's Introduction to Smooth Manifolds.

Answer 3

It is, very much precisely, the Frobenius theorem. If you want to prove the Frobenius theorem in the special case of a codimension 1 distribution on a 3-manifold, feel free.

Suppose $\xi = \ker \alpha$. What, precisely, is $(\alpha \wedge d\alpha)(X,Y,Z)$?

Let's start with the forward direction. Assume $\xi$ is integrable. Then it's also involutive, and if $\alpha(X)$ and $\alpha(Y)$ are zero, then so is $\alpha([X,Y])$. Now if $X,Y,Z$ are a local frame with $\alpha(X)=\alpha(Y) = 0$ (and we may as well assume $\alpha(Z) = 1$), then \begin{align}(\alpha \wedge d\alpha)(X,Y,Z) &= \alpha(X)d\alpha(Y,Z) + \alpha(Y)d\alpha(Z,X)+\alpha(Z)d\alpha(X,Y)\\ &= d\alpha(X,Y) = 0.\end{align} Because $X,Y,Z$ was a local frame, we see that $\alpha \wedge d\alpha$ is globally zero. The converse is essentially the same; let $X,Y,Z$ be as above, and our computation holds until $d\alpha(X,Y) = 0$; instead we have $d\alpha(X,Y) = -\alpha([X,Y])$; but we assumed that $\alpha \wedge d\alpha = 0$, so $\alpha([X,Y])$ must be zero, and hence the distribution is involutive, and by Frobenius it's integrable.