Timeline for When is a finite dimensional real or complex Lie Group not a matrix group
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 8, 2011 at 23:50 | vote | accept | Selene Routley | ||
| May 7, 2011 at 23:15 | comment | added | José Figueroa-O'Farrill | @Igor: Ignore my comment on the Heisenberg group. I was thinking about the infinite-dimensionality of unitary representations, but of course as a unipotent group it has a three-dimensional faithful representation. | |
| May 7, 2011 at 18:42 | answer | added | Tom Goodwillie | timeline score: 3 | |
| May 7, 2011 at 18:38 | comment | added | Marcel Bischoff | with the universal cover of for example $\mathrm{SO}_e(4,2)$ it is the same as for $\mathrm{SL}(2,\mathbb R)$, I guess | |
| May 7, 2011 at 18:04 | answer | added | Jeffrey Adams | timeline score: 12 | |
| May 7, 2011 at 16:45 | answer | added | Alain Valette | timeline score: 25 | |
| May 7, 2011 at 16:33 | comment | added | Igor Belegradek | @José: you can delete a comment, and repost the edited version. Also what Heisenberh group are you talking about? The usual one (see en.wikipedia.org/wiki/Heisenberg_group) is by definition is a group of matrices. | |
| May 7, 2011 at 15:41 | comment | added | José Figueroa-O'Farrill | In my comment, when I said $SL(2,\mathbb{R})$ I meant, of course, its universal cover! (How I wish I could edit comments!) | |
| May 7, 2011 at 15:03 | comment | added | José Figueroa-O'Farrill | I'm not sure about 3) but the Heisenberg group and also $SL(2,\mathbb{R})$ were known early on not to possess faithful finite-dimensional representations. The answers to 4) are Yes, Yes (if connected) and No, respectively. The exponential map is surjective for a compact connected Lie group and $SL(2,\mathbb{C})$ is noncompact (also complex), simply connected and yet it's a matrix group. | |
| May 7, 2011 at 14:59 | comment | added | José Figueroa-O'Farrill | In 4) I suppose you mean "compact group" and not "complex group"? | |
| May 7, 2011 at 14:16 | comment | added | Benjamin Hayes | Can't you get the statement that compact Lie groups are linear more or less from Peter-Weyl? This is at least writes your compact Lie group as an inverse limit of Lie groups $G_{n}.$ It seems that once $\dim G_{n}=\dim G_{m}$ for $m\geq n,$ the maps $G_{m+1}--->G_{m}--->G_{m-1}--->...--->G_{n},$ will have to be covers and this process can't be nontrivial for very long. | |
| May 7, 2011 at 14:02 | history | asked | Selene Routley | CC BY-SA 3.0 |