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Apr 11, 2012 at 21:49 vote accept Sergio A. Yuhjtman
Apr 11, 2012 at 18:55 comment added Miek Messerschmidt You are perfectly correct Yemon, thanks a lot actually for pointing it out to me! I should have read more carefully! Wallach treats things way more specifically than I remembered from reading him a few years ago. Quoting: "Also, rather than treating the general 'tame' case (ie, Type I) we study CCR algebras. Since real reductive groups satisfy the CCR condition, for the purposes of this book this is no ..." :P My mistake. [blushes morosely, skulks away, mumbling "RTFM" to himself].
Apr 11, 2012 at 16:32 comment added Yemon Choi Miek, thanks for the update, but separability is not the issue. I don't know Wallach's book, but if he is focusing on reductive groups then one would expect better results than in the general case. Roughly speaking, the problem for e.g. the free group on 2 generators is that you can disintegrate the left reg rep (which s a factor rep) into irreducibles in two completely different ways (a failure of unique prime factorization, in some sense). So while you may be right to say you can decompose as an integral of irreps, this doesn't have the properties you expect unless group is Type I
Apr 11, 2012 at 13:27 history edited Miek Messerschmidt CC BY-SA 3.0
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Apr 11, 2012 at 11:40 comment added Miek Messerschmidt @Yemon: Thank you for your comments. Now that you mention it, Wallach (from whence my most of my familiarity with the subject stems) does everything assuming separability of the group, C*-algebra and Hilbert space! Your point is well made: That things can go wrong if we're not careful! Thank you for the reference. @Matt: I thought I once saw the weak/strong continuity equivalence somewhere!
Apr 11, 2012 at 10:38 comment added Yemon Choi Regarding my first comment, I think one can find some details in Alain Robert's book in the LMS Lecture Notes series books.google.ca/…
Apr 11, 2012 at 10:34 comment added Yemon Choi @Matt: on 2nd thought it's surely not the earliest reference in the Hilbert space case, which must have been known to Godement, Segal, Gelfand et al. But it may be one of the earliest to write it down explicitly for the Banach case
Apr 11, 2012 at 10:24 comment added Matthew Daws @Yemon: Do you think that is the first (earliest) reference? It's the first one I know... @Everyone: I should of course have said "bounded linear maps of $E$" above...
Apr 11, 2012 at 10:20 comment added Yemon Choi Regarding integration of G-reps and the inverse operation of getting G-reps from reps of the Cstar algebra, see also Chapter 2 of B. E. Johnson, Mem AMS 127 (1972) -- that does it for $L^1$ but by its construction the full group Cstar algebra has as its unitary reps precisely the unitary reps of $L^1(G)$
Apr 11, 2012 at 10:12 comment added Yemon Choi Also, even in the non-compact abelian case, decomposition into irreducibles is delicate, or rather, reconstructing the rep from irreducible constituents needs integrals.
Apr 11, 2012 at 10:10 comment added Matthew Daws Weakly and strongly continuous group representations are the same-- formally, if $E$ is a Banach space, and $\theta$ is a group homomorphism from a locally compact group into the group of invertible linear maps of $E$, then $\theta$ is strongly continuous if and only if it is weakly continuous. The reference which is on my desk is Section 1, Chapter X of Takesaki, Volume 2.
Apr 11, 2012 at 10:10 comment added Yemon Choi You direct integral claim is, IIRC, going to run into problems with groups that are not Type I. You can decomposes as a direct integral of factor reps but that's not the same thing...
Apr 11, 2012 at 9:27 history answered Miek Messerschmidt CC BY-SA 3.0