Timeline for Compact subgroups of the unitary group of operators in a hilbert space
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 26, 2013 at 17:06 | answer | added | Martin | timeline score: 0 | |
| Feb 14, 2013 at 8:09 | answer | added | Narutaka OZAWA | timeline score: 15 | |
| Feb 13, 2013 at 23:42 | comment | added | Jesse Peterson | @Andras Batkai: Actually, the unitaries with the weak topology do not form a compact group. In fact, the weak and strong topologies give the same relative topology on the space of unitaries. | |
| Feb 13, 2013 at 22:44 | comment | added | Amin | According to Wikipedia, yes : "Gleason, Montgomery and Zippin characterized Lie groups amongst locally compact groups, as those having no small subgroups." | |
| Feb 13, 2013 at 22:26 | comment | added | Alain Valette | The unitary group $U(H)$ of a Hilbert space $H$, with the norm topology, has no small subgroup (its intersection with the ball of radius 1 centered at the identity operator, contains only the trivial subgroup). Therefore the same holds for a compact subgroup $G$ of $U(H)$. Isn't there a result related to Hilbert's 5th problem (by von Neumann maybe?) that implies that $G$ is a Lie group? (a not necessarily connected Lie group, of course). I've no time to look it up today... | |
| Feb 13, 2013 at 22:10 | answer | added | Rami | timeline score: 1 | |
| Feb 13, 2013 at 21:41 | history | edited | Rami |
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| Feb 13, 2013 at 21:21 | comment | added | Alain Valette | For the strong topology, there is no characterization: any compact group $G$ embeds into the unitary group of $L^2(G)$ via the left regular representation. | |
| Feb 13, 2013 at 20:59 | history | asked | Nicolas Börger | CC BY-SA 3.0 |