Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with lie-algebroids
Search options answers only
not deleted
user 78
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.
4
votes
Examples of Lie Algebroids
This is not particularly an answer to your question. Beyond Lie algebroids arising from Poisson structures (which Dima Shlyakhtenko has already mentioned in the comments), I think the Wikipedia list …
- 54.6k
2
votes
Is there any relation between deformation and extension of Lie algebras?
There are big pictures that I'll let others describe. Here's a little picture which cogeneralizes the Weinstein remark. (To "cogeneralize" is to make more specific, rather than less.)
Recall that a …
- 54.6k
0
votes
Is the cohomology of the corresponding Lie algebroid an invariant under equivalence of sourc...
Here is a counterexample to the second question. Let $\mathfrak g$ be the nonabelian two-dimensional Lie algebra, thought of as an algebroid with one object, and $G$ its simply-connected group, thoug …
- 54.6k