Are the categories of representations of G and C*(G) isomorphic? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T01:23:00Z http://mathoverflow.net/feeds/question/93681 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93681/are-the-categories-of-representations-of-g-and-cg-isomorphic Are the categories of representations of G and C*(G) isomorphic? Sergio A. Yuhjtman 2012-04-10T16:39:39Z 2012-04-11T13:27:32Z <p>Let G be a locally compact Hausdorff group, and C*(G) the full group C* algebra.</p> <p>I found in some books that "representation theory of both is the same". Can this be expressed as "the categories are equal"?</p> <p>(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).</p> http://mathoverflow.net/questions/93681/are-the-categories-of-representations-of-g-and-cg-isomorphic/93746#93746 Answer by Miek Messerschmidt for Are the categories of representations of G and C*(G) isomorphic? Miek Messerschmidt 2012-04-11T09:27:43Z 2012-04-11T13:27:32Z <p>Following Marc's suggestion, I'll make my comments into a more extended answer. </p> <p><strong>The answer is yes.</strong> 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 <em>a lot</em> more generally, but they do specifically include the case for intertwiners which is what Sergio was asking about in the comments). </p> <p>One small difference is that in both these references <em>strong</em> 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! </p> <hr> <p>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! </p> <p>We often want to answer the following question:</p> <blockquote> <p>Can we write every unitary group representation as 'built up' from irreducible representations? (representations with no non-trivial invariant closed subspaces)</p> </blockquote> <p>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 <strong>seperable</strong> Hilbert space of a <strong>separable</strong> 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</p> <p>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. </p>