Reading through Lee's introduction to smooth manifold, I bumped into this result:

$\textbf{Lemma 14.12.}$ Let $\cal F$ be a foliation of a smooth manifold $M$. The collection of tangent spaces to the leaves of $\cal F$ forms an involutive distribution on $M$.

I've tried to prove it, but have gotten stuck. A foliation is basically slicing $M$ into $k$-dimensional submanifolds (immersed submanifolds), called leaves, so that each $p\in M$ has a neighborhood $(U,\phi)$ which intersects each leaf either in the empty set or in a coutable union of "slices", i.e. connected pieces where the final $\dim M-k$ coordinates of $\phi$ are constant. This type of chart is called flat chart. I have to prove the tangent spaces to the foliation give a $k$-dimensional distribution on $M$. A distribution is assigning to each point a linear subspace $\Delta_p$ of dimension the distribution's dimension, in this case $k$, of the tangent space $T_pM$, in such a way that each $p\in M$ has a neighborhood $U$ with $k$ linearly independent fields $X_1,\dotsc,X_k$ on $U$ which are smooth and span $\Delta_q$ for all $q\in U$, i.e. $\Delta_q=\mathrm{span}(X_1(q),\dotsc,X_k(q))$ for all $q\in U$. That we have a subspace of the tangent space per point of $M$ is easy: take the tangent spaces to the leaves. The $k$ fields are the first problem. How do I construct them? I'm sure I must use the flat charts somehow. Perhaps the coordinate fields of the flat chart? Those may not span the subspaces though. So how do I find these fields? Involutiveness means that those $k$ fields have Lie brackets which stay in the distribution. If I can't show the fields exist, I certainly cannot show they fulfill this condition, so I am stuck here. How do I proceed?


There's no need to actually construct a local frame. To show a distribution is involutive you just need to show that for any two local sections of this distribution $X$ and $Y$, $[X,Y]$ is also a local section.

It might be easier to think in terms of flows instead of vector fields. Let $X$ and $Y$ be local sections of this distribution; that is, let $X$ and $Y$ (locally) be vector fields tangent to the foliation. Let $\phi_t$ be the flow of $X$. Then you know $$\frac{d}{dt}\bigg|_{t=0} \phi_t^* Y = \mathcal L_X Y = [X,Y].$$

Now, by definition of the flow $\phi_t^*$, it preserves the integral submanifolds (since $X$ is tangent to them). Hence it restricts to a diffeomorphism of (each of) the integral submanifolds, and takes tangent vectors in the integral submanifolds to tangent vectors in the integral submanifolds. Hence $\phi_t^*Y$ is tangent to the foliation for all $Y$, and thus $[X,Y]$ is tangent to the foliation as desired.

  • $\begingroup$ One little detail: AFAIK «it preserves the integral submanifolds» at least for small $t$. That is of course sufficient here because we are taking a limit (the derivative), but better note that AFAIK it might not preserve the submanifolds for any $t$. Also, this shows the distribution is involutive, but I think I need to show it is smooth, and that should require a local frame, right? $\endgroup$ – MickG Aug 10 '15 at 9:09
  • $\begingroup$ According to Lee's definition of distribution, it has to be a smooth vector bundle over the manifold, which is equivalent (Lemma 14.1) to having a smooth local frame. $\endgroup$ – MickG Aug 10 '15 at 9:12
  • $\begingroup$ @MickG If $\phi_t$ exists, then it preserves integral manifolds by definition (you're just following the integral curves). I didn't realize you wanted to show it was a distribution, period, too; use that there is a smooth chart around any point in which the foliation becomes the canonical foliation of $\Bbb R^n$ by $\Bbb R^k$. $\endgroup$ – user98602 Aug 10 '15 at 11:53
  • $\begingroup$ Oh right, it's not just a submanifold tangent to the field. Whoops. THe flat chart (smooth chart making the foliation into the canonical one) takes the tangent space to the foliation into $\mathbb{R}^k$, which is generated by coordinate vectors from 1 to $k$, which give the desired local frame when moved back to the manifold. Right? $\endgroup$ – MickG Aug 10 '15 at 13:29
  • 1
    $\begingroup$ @MickG: Precisely. $\endgroup$ – user98602 Aug 10 '15 at 14:14

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