8
votes
Is every singular foliation induced by a Lie algebroid?
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 ...
- 3,652
6
votes
Is every singular foliation induced by a Lie algebroid?
$\newcommand\cF{\mathcal F}$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 ...
- 61
4
votes
First examples of Lie-Rinehart algebras that are not coming from Lie algebroids
Yes, there are many such Lie-Rinehart algebras. To understand such examples we need to recall some classical geometric results.
Note that a $C^\infty(M)$- module is the space of global sections of a ...
4
votes
Connection between Grothendieck's homotopy hypothesis and Lie's second and third theorems?
There are analogues of Lie's theorems in homotopy theory, primarily for rational and $p$-adic homotopy types, as well as Lie ∞-groupoids, which can be seen as smooth homotopy types.
In the rational ...
- 37.8k
4
votes
Accepted
Is $\mathbb{P}^1$ the only smooth projective curve with a locally split tangent lie algebroid?
If $f: X \to \mathbb A^n$ is an etale map, then we can pull back vector fields on $\mathbb A^n$ to vector fields on $X$. This pullback operation is a lie algebroid homomorphism. Hence if we pull back ...
- 148k
3
votes
Accepted
Use of theory of Lie algebroids in (better) understanding of generalised complex structures
The compatibility conditions that you mention in the definition of a generalized complex structure are equivalent to the statement that the $+i$-eigenbundle $L$ of $J$ is a complex Dirac structure:
...
- 491
3
votes
references to learn the general theory Lie $\infty$-groupoids and Lie $\infty$-algebroids
There is no introductory book on Lie ∞-groupoids and ∞-algebroids analogous to Mackenzie's book.
The only book-length treatment that covers these subjects is Urs Schreiber's Differential cohomology in ...
- 37.8k
3
votes
Lie algebroid in algebraic geometry
I suggest having a look at
Beĭlinson, A.; Bernstein, J. A proof of Jantzen conjectures.
MR: Matches for: MR=1237825
§1.2 .
https://people.math.harvard.edu/~gaitsgde/grad_2009/BB%20-%20Jantzen.pdf
- 3,998
2
votes
Accepted
Regarding first order differential operator and derivative endomorphism
Substituting $f=f_1f_2$ in the definition of a derivative endomorphism immediately implies that $D_M$ is a derivation, using the fact that $g_1ψ=g_2ψ$ for all vector fields $ψ$ implies $g_1=g_2$, ...
- 37.8k
2
votes
Accepted
Lie algebroid associated to a vector bundle
Question 1. The two procedures indeed give the same Lie algebroid. One possible way of seeing this is by considering the flows of vector fields: a section of $T(GL(E))/GL(n)$ is a vector field on the ...
- 491
2
votes
Is every singular foliation induced by a Lie algebroid?
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)...
- 6,023
2
votes
Example of tensor category with non-simple unit $J\to \mathbb{1} \to Q$ and suitably extension $Q\to M\to J$
In general, if $N$ is a finitely generated $S$-module, and $J$ is an ideal with $JN=N$, it is a standard fact that there exists $u\in J$ with $(1-u)N=0$. (This is one incarnation of Nakayama's Lemma.)...
- 56.9k
1
vote
Special cases of Lie II for groupoids using elementary techniques
I suspect you've probably figured it out by now but here is an answer anyway:
Here is a sketch of the most low-tech way of accomplishing the task.
First, there is a bijection between groupoid ...
1
vote
Lie Groupoid of a Transitive Lie algebroid
If the base manifold M is connected, and A is such a Lie algebroid then the only orbit of A is M itself. It turns out that one can always find a Lie groupoid (called the Weinstein groupoid by Crainic &...
- 49
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