Timeline for Using topology to characterize embedded Lie subgroups of Lie groups.
Current License: CC BY-SA 2.5
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 4, 2011 at 13:09 | comment | added | Alain Valette | Even if it is not the same question, the problem seems related to mathoverflow.net/questions/63868/… | |
| Apr 22, 2011 at 8:43 | comment | added | Selene Routley | To avoid a "messy" immersion is I guess why some authors require a matrix group to be a closed subgroup of gl(n, K); I can see why this might be done in a first course, but Rossman's treatment (see comment after second answer below) does not make this requirement, IMHO better for thus being less cluttered and the situation really isn't more more complicated: you just have to be careful about which topology you are talking about. | |
| Apr 22, 2011 at 8:41 | comment | added | Selene Routley | I believe I have seen the terminology "Virtual Lie Subgroup" for the sitation where we have an immersion like the irrational slope subgroup. See "maths.dept.shef.ac.uk/magic/course_files/43/…;. I personally don't like the word, but it could be a useful mnemonic to say "beware: the group topology for the subgroup is not generally the same as the relative topology inherited from the surrounding group". | |
| Apr 11, 2011 at 22:37 | answer | added | Enrique Macias | timeline score: 0 | |
| Mar 1, 2010 at 21:32 | history | edited | Khalid Bou-Rabee | CC BY-SA 2.5 |
clarified title
|
| Mar 1, 2010 at 6:06 | history | edited | Khalid Bou-Rabee | CC BY-SA 2.5 |
fixed typo
|
| Mar 1, 2010 at 3:02 | comment | added | Theo Johnson-Freyd | I think in general the word "Lie subgroup" should not mean "embedded Lie subgroup". Rather, "embedded Lie subgroup" should continue to carry that extra adjective. The reason for preferring this terminology is because there is a bijection between Lie subalgebras of the Lie algebra of a give Lie group G and (immersed, but not embedded) connected Lie subgroups of G. But as the irrational line in the torus shows, not every Lie subalgebra integrates to an embedded Lie subgroup. But anyway, the question of when subgroups are embedded Lie is a good one. | |
| Feb 28, 2010 at 19:12 | history | edited | Khalid Bou-Rabee | CC BY-SA 2.5 |
clarified question
|
| Feb 28, 2010 at 2:22 | comment | added | José Figueroa-O'Farrill | Thanks for the clarification. I believe that this condition is known. (See below for a possible answer.) | |
| Feb 28, 2010 at 2:22 | answer | added | José Figueroa-O'Farrill | timeline score: 3 | |
| Feb 28, 2010 at 2:04 | history | edited | Khalid Bou-Rabee | CC BY-SA 2.5 |
Capitilzed "Lie" in title
|
| Feb 28, 2010 at 0:21 | history | edited | Khalid Bou-Rabee | CC BY-SA 2.5 |
Added a clarification in response to a post.
|
| Feb 27, 2010 at 22:29 | answer | added | Akhil Mathew | timeline score: 2 | |
| Feb 27, 2010 at 20:39 | comment | added | Khalid Bou-Rabee | Thanks for the question. My explanation is that Lie subgroups locally look like a single plane. So one might expect that as long as we rule out "local disconnectivity" and "small loops" we might have a subgroup that locally resembles a plane enough to be a Lie subgroup. | |
| Feb 27, 2010 at 19:15 | comment | added | José Figueroa-O'Farrill | At the risk of asking a silly question: why would you expect "semi-locally simply connected" to be able to replace "closed"? The only ever time I've come across that property is in the theorem about the existence of universal coverings, but probably I'm missing something... | |
| Feb 27, 2010 at 18:27 | history | asked | Khalid Bou-Rabee | CC BY-SA 2.5 |