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If $f:M \to N$ is a submersion with connected fibers, then the fibers of $f$ foliate $M$. This is called a simple foliation of $M$ and the leaf space can be identified with $N$. Suppose that a foliation $\mathfrak{F}$ on $M$ has that the fundamental group of each leaf is finitely generated. Then one can show that $\mathfrak{F}$ is simple if and only the leaf space is Hausdorff and the holonomy group of each leaf is trivial.

Question 1: Is there a characterization of when a foliation $\mathfrak{F}$ is simple which works without demanding the above finitely generated condition on the fundamental groups of the leaves?

Note:

If $f:M \to N$ is any submersion (perhaps with disconnected fibers), then the connected componts of the fibers of $f$ foliate $M$. In this case, the leaf space can be identified with the etale-space of a sheaf over $N$, so has a (possibly non-Hausdorff) manifold structure.

Question 2:

Is there a characterization of when the leaf space of a foliation $\mathfrak{F}$ has the structure of a (possibly non-Hausdorff) manifold?

Remark: Simply removing the Hausdorff condition from the other characterization does not work, since e.g., the Kronecker foliation of the torus has no holonomy and each leaf is simply connected (diffeomorphic to $\mathbb{R}$ even), but the leaf space has the indiscrete topology.

Please Note:

I'm aware that the leaf space should be a manifold if and only if the holonomy groupoid is Morita equivalent to a manifold, but I would like a translation of this into more classical data.

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  • $\begingroup$ Is there something about foliations that demands that leaves are connected? I've noticed this in other papers. Why not just foliate by disconnected manifolds (perhaps not of pure dimension)? $\endgroup$
    – David Roberts ♦
    Commented Nov 11, 2014 at 6:18
  • $\begingroup$ @DavidRoberts: By definition, the leaves of a foliation are connected. If you have a "non-connected" foliation, by passing to connected components, you have a foliation with connected leaves. You may argue that this is merely convention, but I think this only seems like this from considering foliations inudced by a submersion. If you start with the definition of a foliation as an involutive subbundle of the tangent bundle, I don't know how to make sense of leaves that are not connected. $\endgroup$
    – David Carchedi
    Commented Nov 11, 2014 at 6:35
  • $\begingroup$ But why is the definition like that? What theorem fails? If every foliation that arises from a subbundle has connected leaves, that's no reason to demand foliations always have connected leaves, since there are natural foliations that don't cf the case of a submersion. Also, if you have a proper Lie groupoid, its space of objects is foliated by orbits, its quotient space is foliated by conjugation classes of stabilisers etc. You can specify you want the connected components of orbits to be the leaves, but why? $\endgroup$
    – David Roberts ♦
    Commented Nov 11, 2014 at 8:12
  • $\begingroup$ One of the many (equivalent) ways of defining a foliation is as an involutive subbundle of the tangent bundle. If you want, take it as a theorem that this definition is equivalent to the foliation chart definition. This theorem would fail if you removed the connected hypothesis. You make a compelling point about (regular) Lie groupoids, but you could rephrase what yo are saying as a foliation together with a compatible groupoid action- the groupoid action descends to the space of leaves (when the leaves are defined via connected components). $\endgroup$
    – David Carchedi
    Commented Nov 11, 2014 at 8:23
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    $\begingroup$ Right exactly, but, now how do you, from the data of such a chart, construct leaves which are not connected? If you have an idea, I'm interested. Btw, I have a stacky explanation about why one should consider connected leaves, but, as this is going well off topic perhaps it's better to go to email? $\endgroup$
    – David Carchedi
    Commented Nov 11, 2014 at 23:21

1 Answer 1

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Firstly, for reference, the result

"Suppose that a foliation F on M has that the fundamental group of each leaf is finitely generated. Then one can show that F is simple if and only the leaf space is Hausdorff and the holonomy group of each leaf is trivial."

appears to come from Sharpe's Differential Geometry: Cartan's Generalization of Klein's Erlangen Program, theorem 2.8.3. However, this result is incorrect. See this answer on MSE. To see where the proof goes wrong, note that we have the following claim in the proof:

Let $L$ be a leaf of the foliation $\mathfrak{F}$. (Here, $L$ is potentially closed or even stronger, we may assume that $M/\mathfrak{F}$ is Hausdorff.) Let $p \in L$ and let $T$ be a transversal through $p$. Then, Sharpe states: "If $\pi_1(L)$ is finitely generated and the holonomy is trivial, we may assume that $T$ is so small that it meets each leaf at most once." However, as the counterexample in the answer on MSE shows, this is actually not true.

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 is precisely what characterizes simple foliations. Furthermore, this characterisation also works for non-Hausdorff quotients.

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