Existence of special transversal on foliation

Question

This is a somewhat technical question about a line in Sharpe's book Differential Geometry: Cartan's Generalization of Klein's Erlangen Program in the proof of the structure theorem, Theorem 8.3 in chapter 2. We have assumed that the leaf space of a foliation $M/\sim$ is Hausdorff. Because of this, each leaf $\mathcal{L}$ is closed and is in particular an embedded submanifold. Using the hypotheses that a foliation has trivial holonomy (w.r.t. transversals) and each leaf has finitely generated fundamental group, he concludes that a given leaf has a transversal that meets each leaf at most once. How can I justify this?

Because each leaf is embedded, we can certainly choose a transveral intersecting a given leaf exactly once. But the tricky part is doing this uniformly for all the leaves meeting the transversal. If we replaced the assumption that leaves $\mathcal{L}$ have finitely generated fundamental group $\pi_1(\mathcal{L})$ with the stronger assumption that each leaf is compact, then Theorem 7.8 of Sharpe guarantees the existence of a tubular neighborhood $U$ of $\mathcal{L}$ with a leaf-preserving diffeomorphism $U\cong \mathcal{L}\times T$ for some transversal $T$, which would give a transversal meeting each leaf at most once.

The main application of finitely generated fundamental group and trivial holonomy I've seen is that, given a transveral of a leaf, we can pick a single open subset of the transversal where the entire fundamental group acts as the identity.

Thanks!

Answer 1

I think the result, as stated, is incorrect. Precisely, because the step in the proof you are talking about is not true. To see this, consider the following counterexample:

Consider the Moebius band $M := \mathbb{R}^2/\mathbb{Z}$ realized as quotient of the action $n \cdot (a, b) = (a + n, (-1)^nb)$. $M$ carries a regular foliation $\mathcal{F}$ with leaves given by $L_c = \{[a, b] : |b| = c\}$, where $c \geq 0$. I.e. this is the standard foliation of the Moebius band by a circle, where a single circle loops around $M$ once and all other circles loop around $M$ twice. Let us now write $M^* = M\setminus \{[0, 0]\}$ for the Moebius band with one point removed from the middle circle that loops around once. Then, we can pullback $\mathcal{F}$ along the inclusion $i : M^* \to M$. This is, again, a regular foliation.

Remark that the leaves of $i^*\mathcal{F}$ are either diffeomorphic to $\mathbb{R}$ or $\mathbb{S}^1$. So, their fundamental groups are finitely generated. Furthermore, all leaves can be separated by opens, so $M/\mathcal{F}$ is Hausdorff. Finally, the leaves $L_c$ for $c > 0$ have trivial holonomy as the foliation structure is isomorphic to the trivial foliation $(a, b) \times \mathcal{S}^1$. The leaf $L_0$ is diffeomorphic to $\mathbb{R}$; thus, it has trivial holonomy. So, according to the stated th, $i^*\mathcal{F}$ should be simple.

However, this is not the case. Notably the quotient map $q: M \to [0, \infty)$ is given by $[a, b] \mapsto |b|$. Note that for $a \neq 0$, we can indeed make this into a surjective submersion. This fixes the smooth structure on $(0, \infty) \subseteq [0, \infty)$. In turn, this shows that $[0, \infty)$ has to carry the standard smooth structure. However, then the map $q$ is not smooth. So, we see that the foliation $\mathcal{F}$ is not simple.

An alternative characterization of simple foliations is given in homework 8.8 in these notes by Rui Loja Fernandes. Here, we see that locally, being able to choose a transversal that intersects each leaf once precisely is what characterizes simple foliations.