Timeline for Is there a "geometric" language that describes the equivalence groupoid of a foliated manifold?
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| Oct 26, 2021 at 9:20 | comment | added | Iakovos Androulidakis | There are plenty of singular foliations beyond the "constant rank" cases (regular and almost regular). (Actually, they are the majority of foliations around!) But a soon as you drop the constant rank assumption, the holonomy groupoid is no longer smooth. It's just a topological groupoid - its topology is very bad, in some sense it reflects the nature of the singularity. | |
| Oct 26, 2021 at 9:16 | comment | added | Iakovos Androulidakis | For these "almost regular" foliations, the partition to leaves is no longer equivalent to the module of vector fields. The holonomy groupoid depends on the choice of vector fields, not on the partition to leaves. This is the price you have to pay for allowing singularities... | |
| Oct 26, 2021 at 9:14 | comment | added | Iakovos Androulidakis | What I say, is that there is another "constant rank" class of foliations, the "almost regular ones". As partitions to leaves, the leaf dimension drops. Precisely, this dimension is constant only on a dense open subset $U$. But if you define such a foliation using differential equations (i.e. vector fields), then you have a $C^{\infty}(M)$-module of vector fields which is projective. Thanks to the Serre-Swan theorem, it is the module of sections of a vector bundle. The fibre at $p$ of this bundle is a vector subspace of $T_p M$ only when $p \in U$. It's a Lie algebroid (integrable). | |
| Oct 26, 2021 at 9:07 | comment | added | Iakovos Androulidakis | Well, explicitly you say: "Take a foliated manifold, which for my question might as well have constant rank." The wikipedia article in the link is about regular foliations. I guess you think of a foliated manifold as such, as a partition to (immersed) submanifolds whose dimension is constant. But even if you think of the collection of tangent bundles of these sub manifolds, it gives an involutive sub bundle of TM. The Frobenius theorem says that the two viewpoints are equivalent, this is what "constant rank" means really. | |
| Oct 24, 2021 at 14:21 | comment | added | Theo Johnson-Freyd | Ok, I clearly cannot read my own writing. Yes, I agree, I had asked about the constant rank case, it says so right there! | |
| Oct 24, 2021 at 11:47 | comment | added | Iakovos Androulidakis | Sorry, in the last sentence I wanted to write "...tangent to them", not "target to them". So, if you take the partition of $\R^2$ to the origin and its complement (2 leaves), and you consider all the vector fields tangent to these leaves, you obtain the infinitesimal generators of the $GL(2)$ action I mentioned before. But there are other modules as well, for instance the inf. generators of the action of $\C^*$, which is a projective module. Another one is the inf generators of the $SL(2,\R)$ action. It's not a projective module, but "less" singular than the $GL(2)$ action (smaller isotropy). | |
| Oct 24, 2021 at 11:29 | comment | added | Iakovos Androulidakis | What all this says is: If you want to have a groupoid which: a) is a Lie groupoid b) orbits are the leaves c) is "minimal" (meaning smallest isotropy) - then the only case is for an almost regular foliation. If your foliation is "too singular", then you are forced to pass to topological groupoids. The point here is that "too singular" has nothing to do with the dimension of the leaves, as the previous examples show. This is because, in the truly singular case, there is no longer a bijective correspondence between the partition to leaves and the module of vector fields target to them. | |
| Oct 24, 2021 at 11:27 | comment | added | Iakovos Androulidakis | A "truly" singular foliation is a non-projective $F$. (e.g action of $GL(2,\R)$ on $\R^2$.) The holonomy groupoid in this case is NEVER a Lie groupoid, just a topological groupoid. There is also the graph, a quotient of this groupoid. I suppose the standard reference is my paper "The holonomy groupoid of a singular foliation". | |
| Oct 24, 2021 at 11:24 | comment | added | Iakovos Androulidakis | Theo In your question I read "constant rank" so I thought it was about regular foliations. Maybe by "constant rank" you mean the "almost regular" case. The easiest way to see this, is to start from the vector fields.That's a $C^{\infty}(M)$-submodule $F$ of which is projective. So $F$ is the sections of a Lie algebroid (just think of the Serre-Swan theorem), with anchor map injective on an open dense subset. (e.g. action of $\C^{\ast}$ on $\R^2$). For such foliations the holonomy Lie groupoid integrates F. Its smooth structure is in Debord's work. The graph may not be a Lie groupoid. | |
| Oct 11, 2021 at 11:27 | comment | added | Theo Johnson-Freyd | Hi Lakovos, Thanks. What you write is of course completely correct. It's hard for me to remember what I knew and didn't know 10 years ago, but I think this was things past-me knew. What I was trying to ask about in my question was exactly the non-regular case. | |
| Oct 10, 2021 at 6:42 | review | Late answers | |||
| Oct 10, 2021 at 9:42 | |||||
| S Oct 10, 2021 at 6:22 | review | First answers | |||
| Oct 10, 2021 at 10:43 | |||||
| S Oct 10, 2021 at 6:22 | history | answered | Iakovos Androulidakis | CC BY-SA 4.0 |