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

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...

• "From here and here, I know that there is a nowhere zero 1-form..." This is only true if the $3$-manifold is orientable. – PVAL-inactive Feb 26 '16 at 22:40
• Actually, it's possible to have a $2$-dimensional subbundle of $TM$ that admits no such form $\alpha$ (called a defining form) even if $M$ is orientable. As this answer pointed out, the necessary and sufficient condition is that the normal bundle of $\xi$ is trivial. For example, there's a nontrivial $1$-dimensional subbundle of the tangent bundle to $\mathbb R^3$ minus the $z$-axis, and the $2$-dimensional bundle orthogonal to that has no global defining form. – Jack Lee Feb 26 '16 at 23:23

3 Answers

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.

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.

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.