Timeline for A question on lie groups( Lie algebras)
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jan 19, 2014 at 17:32 | comment | added | Ali Taghavi | @AlainValette Prof.Valette, thank you very much for$SU(3)$ example. I have a related question; What is a pure (Lie) algebraic interpretation for the following property of a finite dim. lie algebra $L$: For every lie group $G$ with $Li(G)\sim L$ and for every two non zero $X,Y \in L$, two foliations $F_{X}$ and $F_{Y}$ are topological equivalent, as two foliation on $G$. Are such lie algebras, classified? | |
| Jan 19, 2014 at 17:05 | review | Close votes | |||
| Jan 22, 2014 at 17:58 | |||||
| Jan 19, 2014 at 16:55 | comment | added | Sasha Anan'in | @Alain @ Ali Yes. Many compact groups which are not products contain a compact one-parameter subgroup as well as non-compact one. The cosets with respect to those provide desired foliations. | |
| Jan 19, 2014 at 16:45 | comment | added | Alain Valette | Take Sasha's example, and embed it into the simple Lie group $SU(3)$ (where $\mathbb{T}^2$ embeds as a maximal torus). | |
| Jan 19, 2014 at 16:37 | comment | added | Ali Taghavi | @SashaAnan'in Are you talking about quotient of 2- torous by its finite subgroup? in this case what is the resulting lie group(after quotient). Note that every compact commutative lie group is a torous. Could you please explain more precisely? | |
| Jan 19, 2014 at 16:30 | comment | added | Sasha Anan'in | Ok. So, you may try to take a product and quotient it by a finite normal subgroup so that the result is no more a product, but those compact and noncompact leafs survive. (Of course, the torus does not work anymore.) | |
| Jan 19, 2014 at 16:27 | comment | added | Ali Taghavi | @Sasha torous does not satisfies the condition in my question. it is product of $S^{1}$ by $S^{1}$ | |
| Jan 19, 2014 at 16:24 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
added 11 characters in body
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| Jan 19, 2014 at 16:24 | comment | added | Sasha Anan'in | Yes. The simplest example: take a torus $T:={\mathbb R}^2/{\mathbb Z}^2$ and a couple of tangent vectors at $0$. One can provide the foliation with compact leafs, the other may have all leafs dense. | |
| Jan 19, 2014 at 16:21 | comment | added | Peter Crooks | Such a Lie group $G$ is always isomorphic to $G\times\{e\}$, its product with the trivial group. Perhaps there are some refinements to be made to the question. | |
| Jan 19, 2014 at 16:08 | history | asked | Ali Taghavi | CC BY-SA 3.0 |