# Questions tagged [lie-algebroids]

In differential geometry, Lie algebroids generalize on one hand Lie algebras, on the other hand the tangent bundle of a manifold: they are vector bundles equipped with an anchor map, i.e. a vector bundle morphism to the tangent bundle, and a Lie algebra structure on the space of sections subject to certain Leibniz rules. The integrated version of a Lie algebroid is a Lie groupoid. A purely algebraic version is a Lie-Rinehart algebra.

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### Special cases of Lie II for groupoids using elementary techniques

I asked a similar question on math.stackexchange but did not get any responses, so I thought I'd kick it up to mathoverflow. In Crainic and Fernandes's "Integrability of Lie Brackets" (and ...
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### A-Paths as morphisms of Lie Algebroids $TI\longrightarrow A$?

In the paper Integrability of Lie Brackets Marius Crainic and Rui Fernandes describe obstructions to integrate a Lie algebroid to a Lie groupoid. The process of integration relies on the construction ...
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### Differentiation of Lie $\infty$-groupoids

I've been trying to understand how to differentiate Lie $\infty$-groupoids to get a Lie $\infty$-algebroid. First of all, I will state the definitions that I'm assuming. A Lie $\infty$-groupoid is a ...
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### Almost but not quite a Lie algebroid: what is it?

In some calculations, I have arrived at the following algebraic structure, reminiscent of a Lie algebroid, but not quite. I have a real line bundle $E \to M$, on whose smooth sections $\Gamma(E)$ I ...
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### AKSZ sigma models for higher spin

The AKSZ framework constructs 2D sigma models in the BV formalism. Is there a generalization of the AKSZ approach to higher spin?
In the paper by Weinstein and Xu: Extensions of symplectic groupoids and quantization, J. Reine Angew. Math. 417 (1991), there are two versions of Lie groupoid cohomology. The same differential $\... 1answer 462 views ### What's an example of a commutative algebra over$\mathbb Q$that fails to satisfy this version of the “PBW theorem” In a recent question, I recalled the notion of differential operator, polyderivation, and principal symbol for a commutative algebra$A$over some fixed commutative ring$k$. (I will not repeat those ... 1answer 593 views ### For which algebras does \{Differential Operators\} satisfy a PBW-like theorem? Let$k$be a commutative ring,$A$a commutative$k$-algebra, and for some other part of why I'm asking this question I only care about the case when$k \supseteq \mathbb Q$. Recall the following ... 2answers 2k views ### Differential forms on an almost complex manifold Hello! Let$M$be an almost complex manifold. Let$TM$denote its tangent bundle. Then we have the decomposition$TM\otimes\mathbb{C}=T^{1,0}M\oplus T^{0,1}M$corresponding to the eigenvalues of the ... 3answers 1k views ### Examples of Lie Algebroids The concept of a Lie Algebroid is given an important geometric meaning in the framework of Generalized Complex Geometry. For reference, the (barebones) definition of a Lie Algebroid is a vector bundle ... 3answers 833 views ### Is there any relation between deformation and extension of Lie algebras? In a paper of A. Weinstein on the geometry of Poisson manifolds, he relates the formal linearization around a zero, p, of the Poisson bivector to extensions of the Lie algebra induced by the bivector ... 3answers 844 views ### Geometry and Integrability in Other Bundles Background: Suppose$E=TM$is the tangent bundle to some differentiable manifold$M^n$. If we specify some subbundle$D\subset TM$(distribution of$k$-planes) then there are two natural situations ... 2answers 310 views ### Is the cohomology of the corresponding Lie algebroid an invariant under equivalence of source-simply-connected Lie groupoids? Recall the related notions of Lie groupoid, Lie algebroid, generalized morphism of Lie groupoids, and cohomology of Lie algebroid. Henceforth, I will drop the word "Lie" for all those things listed ... 1answer 288 views ### When does a VBLA induce an isomorphism on Lie algebroid cohomology? This question is geared towards the experts, so I will only briefly gloss the definitions. Everything I say is in the category of finite-dimensional smooth manifolds, and whenever I say "$\mathbb Z$-... 1answer 785 views ### Do Lie algebroids pull back (along submersions)? There are more general definitions, but for my purposes a Lie algebroid on a smooth manifold$X$is a vector bundle$A \to X$, a map$\rho: A \to {\rm T}X$of vector bundles over$X$, and a bracket$[,...
I've read in many places, including the n-Lab page, that a Lie algebroid (which I think of as in the first definition on the n-Lab page) is the same as a vector bundle $A \to X$ and a (properties?) ...