# Is every singular foliation induced by a Lie algebroid?

Let $M$ be a smooth manifold.

A smooth distribution $D$ on $M$ is the union of a family $\{D_p \leq T_p M : p\in M\}$ of vector spaces such that there is a family $\mathcal C$ of smooth vector fields on $M$ satisfying $D_p = \text{span}\{X_p : X\in \mathcal C \}$ for every $p \in M$.

Remark that we do not ask the dimension of the fiber $D_p$ to be constant: we call a distribution regular if the dimension is constant, and singular if it is not.

We call a distribution $D$ integrable if for every point $p \in M$ there is a submanifold $S\subseteq M$ which is tangent to $D$ and satisfies $T_q S = D_q$ for every $q \in S$. In this case, it can be proved that the maximal connected integral manifolds of the distribution form a partition of $M$ into weakly embedded submanifolds of $M$, which we call the foliation associated to $D$.

Typical examples of integrable distributions are given by Lie algebroids: if $A\to M$ is a Lie algebroid over $M$ with anchor map $\rho : A\to TM$, then the image of $\rho$ is an integrable distribution on $M$. For example, every Poisson manifold has an integrable, possibly singular distribution given by the image of the Poisson bivector field $\Pi : T^*M\to TM$, and the induced foliation is precisely the symplectic foliation of the Poisson manifold.

The integrability problem for regular distributions is solved by the Frobenious theorem: a regular distribution $D$ is integrable if and only if it's involutive. A singular version of the Frobenious theorem can be stated in the following way: a (possibly singular) distribution $D$ is integrable if and only if there is a family of vector fields $\mathcal C$ which span $D$ pointwise, such that the flow of every element of $\mathcal C$ preserves $D$ (see Theorem 3.5.10 of this book for a more precise statement and a proof).

A sufficient condition for the integrability of a singular distribution $D$ is the following: there exists a module $\mathcal C$ of compactly supported vector fields spanning $D$ which is locally finitely generated and involutive. Some people calls such an object a Stefan-Sussman foliation.

I have two related questions:

1) Is it true that every integrable distribution is spanned by a module $\mathcal C$ of compactly supported vector fields which is locally finitely generated and involutive?

2) Is it true that every integrable distribution is the image of the anchor map of some Lie algebroid?

Clearly, (2) implies (1). There is people which believe that (2) is true, and I would like to know if this question is still open.

Thank you!

• A smooth (possibly singular) integrable distribution is the same as a "singular foliation" in the same sense as for holomorphic foliations, i.e. a sub sheaf-of-Lie-algebras of the tangent sheaf, right? Mar 27, 2017 at 15:21
• I am not sure, but I would say no. Take the distribution $D$ on $\mathbb R ^2$ which has rank 2 on points with $x >0$ and is generated by $\partial / {\partial x}$ for $x\leq 0$. This is a smooth distribution and is not integrable (the "leaves" have boundary). However, the sheaf $\Sigma$ of smooth vector fields tangent to $D$ is a sheaf of Lie algebras, in the sense that for every open $U\subseteq M$ the module $\Sigma (U)$ is closed under Lie brackets. Mar 27, 2017 at 15:36
• From what you say, I would say yes (as the sheaf of $\mathcal{C}^\infty_M$-modules $\Sigma$ is closed under Lie bracket). Mar 27, 2017 at 17:00
• But I don't understand your example: which are the vector fields generating $D$ (or $\Sigma$, which is the same up to a standard -at least in algebraic geometry- abuse of notation/terminology) on points of the form $(0,y)\in\mathbb{R}^2$? Mar 27, 2017 at 17:03
• I'm sorry, I gave the wrong definition of integrable distribution. $D$ is integrable iff for every $p\in M$ there exists a submanifold (without boundary) $S$ containing $p$ such that for all $q\in S$ we have $T_q S = D_q$. With this definition,you can immediately see that my counterexample is not integrable. However, a set of generators of $D$ is $\partial / {\partial x},f \partial/{\partial y}$ where $f:\mathbb R ^2 \to \mathbb R$ is a smooth function which is zero for $x\leq 0$ and nonzero for $x>0$. Mar 27, 2017 at 17:29

## 2 Answers

It is possible to define the Lie groupoid of a singular foliation and associates to it its Lie algebroid when it is smooth. This Lie algebroid satisfies the property 2.

https://projecteuclid.org/download/pdf_1/euclid.jdg/1090348356

http://users.uoa.gr/~iandroul/AS-holgpd-final.pdf

https://en.wikipedia.org/wiki/Lie_algebroid#Lie_algebroid_associated_to_a_Lie_groupoid

It is a conjecture of Androulidakis and M. Zambon (see ( Lavau's thesis p. 65) that not every singular foliation arises from a Lie algebroid. In the second reference, there is a condition which implies the smoothness of the holonomy groupoid.

https://tel.archives-ouvertes.fr/tel-01447963/document

https://www3.ubu.es/ifwgp2012/transparencias%20web/Zambon.pdf

• I have two comments. 1) In the article you cited, by singular foliation they mean what I call a Stefan-Sussman singular foliation. My definition seems to be more general, and in fact my first question asks precisely if the two definitions are equivalent. 2) I didn't read the whole article, but it seems to me that they associate to every Stefan-Sussman singular foliation a topological groupoid, which is not always Lie. For a topological groupoid, we don't have a well defined notion of Lie algebroid associated to the groupoid. Mar 27, 2017 at 9:26
• You may prefer the first reference for the definition of the associated Lie groupoid Mar 27, 2017 at 9:42
• If the foliation is almost regular, then there is a well defined notion of Holonomy groupoid of it. For example, the construction fails for the foliation on $\mathbb R ^3$ given by the spheres centered at the origin. (however, this foliation is indeed given by a Lie algebroid because there is a Poisson structure on $\mathbb R ^3$ which induces that foliation) Mar 27, 2017 at 9:57
• Can you give me a reference of Skandalis' conjecture? Mar 27, 2017 at 12:22
• In fact it is a conjecture of Androulidakis and M. Zambon see Lavau thesis p.65 Mar 27, 2017 at 14:26

For Stefan-Sussmann singular foliations, the answer is negative: See Prop. 1.3 in the following paper, for the construction of an explicit counterexample:

http://users.uoa.gr/~iandroul/AZsmooth_08DEc2011.pdf

Second, regarding the holonomy groupoid of a singular foliation: In fact, it turns out that it is a diffeological space, so one can do differential geometry with it. In particular, one can differentiate it in the sense of diffeological spaces (but this can be done explicitely) and obtain a Lie algebroid in a generalized sense. This Lie algebroid is nothing else than the original module $\cF$ of vector fields which defined the foliation in hand.

Of course this module $\cF$ will not be projective, unless the foliation is "almost regular" (this is the case studied by Debord). So the Serre-Swan theorem does not apply in general, which means that $\cF$ cannot be realized as the module of sections of some honest vector bundle. However, there does exist a "singular" bundle around -- its fiber at a point $x$ is the quotient of $\cF$ by the maximal ideal at x...