Timeline for When does a unitary Hilbert space rep of a reductive Lie group decompose into a direct sum of irreps with finite multiplicities?
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 12, 2011 at 15:47 | answer | added | paul garrett | timeline score: 4 | |
| Jun 9, 2010 at 4:42 | answer | added | Victor Protsak | timeline score: 8 | |
| Jun 8, 2010 at 23:24 | answer | added | Marty | timeline score: 12 | |
| Jun 8, 2010 at 19:11 | comment | added | Kevin Buzzard | Thanks BCnrd. I've never learnt the theory of integrating Hilbert-space valued functions! I'm very much a newcomer to this area. So all that's left is that I'm waiting for someone to come along and tell me that I can't possibly call $H$ a Hilbert-Schmidt representation, and/or giving me a general class of reps to which L^2(Gamma\G) belongs and each of which decompose into irreps with finite multiplicities... | |
| Jun 8, 2010 at 18:48 | comment | added | BCnrd | Kevin, measure theory works with integrands valued in Hilbert spaces, so the integral you wish to use ($\int_ G f(g) g.v {\rm{d}}g$) is perfectly meaningful since $g \mapsto f(g)g.v$ is a continuous $H$-valued function on $G$ with compact support. If one goes back to how Hilbert-valued integration is constructed (using step functions), say as in Lang's book "Real and Functional Analysis", that gives an answer of the sort you should like for your final "technical point". | |
| Jun 8, 2010 at 18:42 | history | asked | Kevin Buzzard | CC BY-SA 2.5 |