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-algebras
Search options not deleted
user 118688
Lie algebras are algebraic structures which were introduced to study the concept of infinitesimal transformations. The term "Lie algebra" (after Sophus Lie) was introduced by Hermann Weyl in the 1930s. In older texts, the name "infinitesimal group" is used. Related mathematical concepts include Lie groups and differentiable manifolds.
4
votes
1
answer
319
views
Morphism of Lie algebras giving an action of Lie group on manifolds
Let $G$ be a Lie group and $\mathfrak{g}$ be its Lie algebra.
Let $ M$ be a manifold and $\mathfrak{X}(M)$ be its Lie algebra of vector fields on $M$.
Let $G\times M\rightarrow M$ be an action of t …
- 6,023
3
votes
2
answers
296
views
Lie's third theorem via graded geometry
Lie's third theorem : Given any finite dimensional Lie algebra $\mathfrak{g}$, there exists a Lie group $G$ whose Lie algebra is equal to $\mathfrak{g}$.
In one of the talks, speaker mentions that thi …
- 6,023
3
votes
3
answers
1k
views
Lie algebra bundle associated to a Lie group bundle
I was reading the paper Non abelian Differentiable gerbes (page 24) and came across notion of Lie algebra bundles associated to a Lie group bundle.
I am not comfortable with these notions and google …
- 6,023