Timeline for Permutable (Lie) subgroups
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 20, 2015 at 16:07 | vote | accept | Duchamp Gérard H. E. | ||
| Jul 20, 2015 at 5:27 | comment | added | Duchamp Gérard H. E. | @grghxy OK, ${\rm{Lie}}(A)\cap {\rm{Lie}}(B)={\rm{Lie}}(A\cap B)$ is not hard to prove using one-parameter subgroups. I have a look to your answer later. | |
| Jul 20, 2015 at 3:17 | comment | added | grghxy | The property ${\rm{Lie}}(A)\cap{\rm{Lie}}(B)=0$ when $A\cap B=1$ is valid without hypotheses on $AB$ inside $G$. More generally, ${\rm{Lie}}(A)\cap {\rm{Lie}}(B)={\rm{Lie}}(A\cap B)$ for any two closed subgroups $A$ and $B$ of a Lie group $G$. (I haven't seen this in a textbook in that generality, but perhaps it is in a standard reference? It is not hard to prove, but requires some care since (i) $A \cap B$ might not be connected and (ii) when ${\rm{Lie}}(A)\cap{\rm{Lie}}(B)$ is exponentiated to a connected Lie subgroup $H$ of $G$ we don't know a-priori that $H$ has the subspace topology.) | |
| Jul 20, 2015 at 0:13 | answer | added | grghxy | timeline score: 3 | |
| Jul 19, 2015 at 22:52 | comment | added | José Figueroa-O'Farrill | You're right, of course. I will think further and delete my answer in the meantime. | |
| Jul 19, 2015 at 22:04 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
Added some hints of how to solve the question and where I am stuck
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| Jul 19, 2015 at 21:58 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
Added some hints of how to solve the question and where I am stuck
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| Jul 19, 2015 at 17:33 | answer | added | José Figueroa-O'Farrill | timeline score: 2 | |
| Jul 19, 2015 at 15:37 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
Formatting the first block
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| Jul 19, 2015 at 14:35 | history | asked | Duchamp Gérard H. E. | CC BY-SA 3.0 |