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-groups
Search options not deleted
user 32022
Lie Groups are Groups that are additionally smooth manifolds such that the multiplication and the inverse maps are smooth.
20
votes
2
answers
2k
views
Is every topological (resp. Lie-) group the isometrygroup of a metric space (resp. Riemannia...
The isometry group of a metric space is a topological group (with the compact open topology). The isometry group of a Riemann Manifold is a Liegroup. (Thm. of Steenrod-Myers)
So, is every topological …
- 2,974
8
votes
cohomology of BG, G compact Lie group
Just for completeness, here's another argument without spectral sequences via rational homotopy theory.
Recall a theorem of Hopf, which states that the rational cohomology of a path-connected H-space …
- 2,974
5
votes
$\pi_{2n-1}(\operatorname{SO}(2n))$ element represents the tangent bundle $TS^{2n}$, not tor...
For $n=1$, the answer to your question is negative, as explained by Gregory Arone in the comments.
In the cases $n\neq 1,2,4$, there is the following easy argument:
The long exact sequence of the fib …
- 2,974
4
votes
connected compact semisimple lie group finite fundamental group
Every connected Liegroup, which has a semisimple Liealgebra with a definite Killing form is compact.
The Liealgebra of a compact Liegroup is always the direct sum of an semisimple and abelian Liealgeb …
- 2,974
2
votes
Decomposition of solvable Lie group
As Yves Cornulier already said: Your presumed statement is wrong.
Any connected, linear, solvable Lie group over the reals is the semi-direct product of a compact abelian subgroup and a simply connec …
- 2,974