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 ...
Tsemo Aristide's user avatar
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 ...
Iakovos's user avatar
  • 61
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 ...
Will Sawin's user avatar
  • 135k
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 ...
Dmitri Pavlov's user avatar
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: ...
unknownymous's user avatar
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
Niels's user avatar
  • 3,938
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 ...
Dmitri Pavlov's user avatar
2 votes
Accepted

Poisson structure on the dual Lie algebroid

Depending on you sign convention, this goes as follows. First you denote the bundle projection by $pr\colon E^* \longrightarrow X$. For a section $s \in \Gamma^\infty(E)$ you have a linear function $J(...
Stefan Waldmann's user avatar
2 votes
Accepted

Property of Lie algebroid morphism: $\#_B\circ \Phi=d\phi\circ \#_A$?

A vector is determined by its action on functions as a directional derivative. This is just the Lie derivative in degree $0$ mentioned at the end of $(iii)$. Together with the naturality $(ii)$ of $d$ ...
Sebastian Goette's user avatar
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)...
Praphulla Koushik's user avatar
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.)...
Neil Strickland's user avatar
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 $g_1=g_2$, ...
Dmitri Pavlov's user avatar
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 ...
unknownymous's user avatar
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 ...
Joel Villatoro's user avatar
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 &...
JeremyA's user avatar
  • 49
1 vote

A-Paths as morphisms of Lie Algebroids $TI\longrightarrow A$?

You may find it useful to look at Lectures on the Integrability of Lie Brackets by the same authors. I believe it takes a slower approach to the subject you seem to be reading about. As for the ...
Joel Villatoro's user avatar

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