Is every singular foliation induced by a Lie algebroid?

Question

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 Frobenius theorem: a regular distribution $D$ is integrable if and only if it's involutive. A singular version of the Frobenius 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 Rudolph and Schmidt - Differential Geometry and Mathematical Physics: Part I 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 call 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 are people which believe that (2) is true, and I would like to know if this question is still open.

Answer 1

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.

Debord - Holonomy Groupoids of Singular Foliations

Androulidakis and Skandalis - The holonomy groupoid of a singular foliation

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.

Lavau - Lie $\infty$-algébroïdes et feuilletages singuliers

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

Answer 2

$\newcommand\cF{\mathcal F}$For Stefan–Sussmann singular foliations, the answer is negative: See Prop. 1.3 in the following paper, for the construction of an explicit counterexample: Androulidakis and Zambon - Smoothness of holonomy covers for singular foliations and essential isotropy.

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 explicitly) 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 $x$….

Answer 3

Even though we may not be able to associate a Lie algebroid with a singular foliation, we can associate a Lie $\infty$-algebroid with a singular foliation (satisfying certain not so strange conditions). This result is due to Camille Laurent-Gengoux, Sylvian Lavau and Thomas Storbl published as "The universal Lie $\infty$-algebroid of a singular foliation", whose arXiv version is available at https://arxiv.org/abs/1806.00475

I will try to explain the basic idea that I understood.

A singular foliation $\mathcal{F}$ is a locally generated $C^\infty(M)$-submodule of $\mathfrak{X}(M)$. Just like any other $R$-module, this $C^\infty(M)$-module $\mathcal{F}$ will also have resolution, $$\cdots\rightarrow P_{-3}\rightarrow P_{-2}\rightarrow P_{-1}\rightarrow \mathcal{F}\rightarrow 0,$$ where $P_{-i}$ are projective $C^\infty(M)$-modules for every $i\geq 1$.

As the singular foliation $\mathcal{F}$ is coming from a geometric structure on the manifold $M$, it is natural to focus on resolutions that come from some geometric structures on $M$. This is where the Serra-Swan theorem comes for help. Whenever we have a finitely generated projective $C^\infty(M)$-module $P$, then, it has to be of the form $\Gamma(M,E)$ for some vector bundles $E\rightarrow M$.

Even though $\mathcal{F}$ is a locally finitely generated $C^\infty(M)$-module, we may not assure that there would be a projective resolution $$\cdots\rightarrow P_{-3}\rightarrow P_{-2}\rightarrow P_{-1}\rightarrow \mathcal{F}\rightarrow 0$$ where all these $P_{-i}$ are locally finitely generated. So, the question of $P_{-i}$ being finitely generated may be too much to hope for. They were able to get rid of this locally finitely generated and finitely generated issue by asking that $M$ is a compact manifold.

Such a projective resolution $$\cdots\rightarrow P_{-3}\rightarrow P_{-2}\rightarrow P_{-1}\rightarrow \mathcal{F}\rightarrow 0$$ gives vector bundles $E_{-i}\rightarrow M$ with $\Gamma(M,E_{-i})=P_{-i}$ and a morphism of vector bundles $E_{-i}\rightarrow E_{-i+1}$ coming from map of sections $P_{-i}\rightarrow P_{-i+1}$ (because of $C^\infty(M)$-linearity).

Thus, we have an exact sequence of vector bundles $$\cdots \rightarrow E_{-3}\rightarrow E_{-2}\rightarrow E_{-1}\rightarrow TM$$ such that the corresponding map of sections is the resolution of $\mathcal{F}$ that we mentioned above. This complex of vector bundles is what they have called as geometric resolution.

Under some mild conditions on the singular foliations, they were able to prove the following theorem

Theorem $2.4$ : A locally real analytic singular foliation admits a geometric resolution of length at most $\dim(M)+1$ over any relatively compact open subset of $M$.

It does not end here. I am not very sure but, may be the above result can be proved for any random locally finitely generated $C^\infty(M)$-submodule of $\mathfrak{X}(M)$, without having the condition of being closed under Lie bracket (in other words $\mathcal{F}$ having structure of a Lie bracket). For obvious reasons, this extra structure on $\mathcal{F}$ should reflect somewhere in the sequence of vector bundles $\cdots \rightarrow E_{-3}\rightarrow E_{-2}\rightarrow E_{-1}\rightarrow TM$. This should remind the notion of Lie $\infty$-algebroid and this is what they say in the next result.

Theorem $2.7$ : Let $\mathcal{F}$ be a singular foliation on $M$ which permits a geometric resolution. Then, there is a (universal) Lie algebroid structure on the resolution.

It also explains in what sense it is "universal".

In this sense, any singular foliation (with mild assumptions) comes from a Lie $\infty$-algebroid, which they call as "the Lie $\infty$-algebroid of a singular foliation".

I can add some more details if anyone wants to see them.