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Let G be a locally compact Hausdorff group, and C*(G) the full group C* algebra.

I found in some books that "representation theory of both is the same". Can this be expressed as "the categories are equal"?

(Here "representation of G" means a unitary weakly continuous representation on a Hilbert space, and representations of C*(G) are nondegenerate. Morphisms in those categories are the interwiner bounded operators).

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Well "equal" is not a wellbehaved notion. For categories the question if they are equivalent is the more useful one. –  Johannes Hahn Apr 10 '12 at 17:04
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In this case they would be isomorphic. –  Sergio A. Yuhjtman Apr 10 '12 at 17:17
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Looking at the equivalence of categories, you see that in this case, it is actually an isomorphism of categories. What Johannes points out is that this observation is less useful than one could expect at first. (In analogy, you can find spaces which are homeomorphic instead of just being homotopy equivalent. This can be useful, but most of the time -if you study homotopy theory- it is not.) Talking about "isomorphic" objects as being "the same" or "equal" is also a bit problematic in general, but in this case I see no confusion that could arise. –  Andreas Thom Apr 10 '12 at 18:18
    
@Sergio: Do you also ask for the construction of the isomorphism between these categories? –  Martin Brandenburg Apr 10 '12 at 18:30
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Yes, intertwiners are preserved. I'm pretty sure Dana Williams does this in "Crossed products of C*-algebras". –  Miek Messerschmidt Apr 10 '12 at 19:00
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up vote 4 down vote accepted

Following Marc's suggestion, I'll make my comments into a more extended answer.

The answer is yes. Every unitrary representation of $G$ can be "integrated" to get a representation of $C^{\ast}(G)$ and every representation of $C^{\ast}(G)$ can be "separated" to get a representation of $G$ (the terms 'integrating' and 'separating' come from the more general crossed product theory, and are the functors between the said categories). These constructions are mutual inverses and preserve a whole bunch of properties, one of which is intertwining operators, therefore the categories are isomorphic. You could take a look at Proposition 2.40 in "Crossed prducts of $C^*$-algebras" by Dana P. Williams (who does it for only equivalences) or Proposition 5.5 in this paper (http://arxiv.org/abs/1104.5151) by some colleagues of mine (who do it a lot more generally, but they do specifically include the case for intertwiners which is what Sergio was asking about in the comments).

One small difference is that in both these references strong instead of weak continuity of the group representation is assumed. I don't think this is too much of a problem, as long as we can pull bounded operators through integrals (however we might choose define those). But do correct me if I'm wrong!


PS: As Andreas mentions, this isomorphism might not be particularly useful. For abelian and compact groups this is the case, but it does become useful in the general case!

We often want to answer the following question:

Can we write every unitary group representation as 'built up' from irreducible representations? (representations with no non-trivial invariant closed subspaces)

For the representations of abelian and compact groups we can go through the motions and show that this is indeed the case without ever needing to resort to this isomorphism of categories (take a look at the Peter-Weyl theorem for compact groups on wikipedia). However, for a unitary representation on a seperable Hilbert space of a separable non-abelian, non-compact, locally compact groups things become more difficult but still works. To get a result, the rough idea is as follows: We transfer everything to the category of representations of $C^\ast (G)$, and invoke von Neumann algebra theory to show that every such representation can be written as a "direct integral" of irreducibles and then transfer back to the category of group representations to conclude that every such unitary group representation can be reduced to a direct integral of irreducables. Details are in Chapter 14 of "Real reductive groups. II" by Nolan R. Wallach

UPDATE: As Yemon pointed out things are not as simple as first thought, and what I remembered reading in Wallach is not what I actually read.

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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... –  Yemon Choi Apr 11 '12 at 10:10
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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. –  Matthew Daws Apr 11 '12 at 10:10
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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 $L^1(G)$ –  Yemon Choi Apr 11 '12 at 10:20
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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/… –  Yemon Choi Apr 11 '12 at 10:38
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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 –  Yemon Choi Apr 11 '12 at 16:32
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