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I am currently reading some of Mackey's work on unitary representation.

Given a locally compact group $G$ and a unitary representation $\pi : G\rightarrow U(H)$. As far as I understood it, the representation $\pi$ is primary, if the von Neumann algebra generated by $\pi(g)$ for all $g \in G$ is a factor, see http://mathworld.wolfram.com/PrimaryRepresentation.html. The representation $\pi$ is irreducible, if there does not exist a nontrivial $G$ invariant closed subspace $H' \subset H$, i.e. $\pi(g) h \in H'$ for all $g \in G$ and $h \in H'$.

See the comments: The countable sum of the same irreducible representation is primary.

When is a primary representations quasi equivalent to an irreducible one? Are they the same if the group is of type 1? Does the decomposition of the group von Neumann algbera $L(G)$ into factors correspond to the decomposition (as a direct integral) in isotypic components? Are the some nice examples, which illustrate that this is to much to hope for.

Motivating example is the Peter Weyl theorem, which states that every irreducible is finite dimensional and $$L(G) = \bigoplus_{\pi \; \in irr(G)} M_{dim(\pi)}( \mathbb{C}),$$ where the components $M_{dim(\pi)}( \mathbb{C})$ are the factors. Hence here the factors are quasi equivalent to an irreducible one.

Aside to the original question: Do all unitary representation appear in $L(G)$?

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The centre of a von Neumann algebra is a von Neumann algebra itself, so if it's non-trivial, it contains a non-trivial projection; as this projection commutes with all elements of the algebra, the range will be an invariant subspace. So irreducible implies primary. I don't know any more... –  Matthew Daws Apr 27 '11 at 8:28
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Btw: the Mathworld article cites an article of Knapp from the Notices, which is freely available: ams.org/notices/199605/knapp-2.pdf Do a search for "primary" and you'll see that a finite or countable sum of irreducible representations is primary. So, they are not the same, and there are easy examples... –  Matthew Daws Apr 27 '11 at 8:35
    
I am really unexperienced with this stuff... Wouldn't there be a lot of projections to each component in the center, if we allow a sum of irreducibles? If so, which of definitions is wrong? –  Marc Palm Apr 27 '11 at 12:11
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@pm: let $\pi$ be an irreducible representation, i.e. $\pi(G)' = \mathbb{C}1$. Consider the direct sum $\rho = \oplus_{i=1}^n \pi$. Then $\rho(G)'$ is equivalent to $M_n(\mathbb{C})$, so it is reducible and primary,as $\rho(G)' \cap \rho(G)'' = \mathbb{C}1$. The point is that you cannot reduce $\rho$ using central projections in $\rho(G)''$. There is a notion of quasi-equivalence of representations (see for example section III.5.1 in Blackadar, Operator Algebras). $\rho$ is quasi-equivalent to $\pi$. –  pasquale zito Apr 27 '11 at 13:03
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@pm: Yes, sorry, I certainly meant a direct sum of the same irreducible rep (my comment was a bit misleading in that regard). I do urge you to go and read Knapp's article in the Notices, as it seems pretty readable to me. –  Matthew Daws Apr 27 '11 at 19:07

3 Answers 3

up vote 3 down vote accepted

I would recommend reading parts of Jacques Dixmier's book: "$C^\ast$-algebras" (North Holland, 1977 - translated from the french version of 1969), especially Chapters 5 (irreducible and factor representations of $C^\ast$-algebras) and Chapter 13 (the analogue for locally compact groups).

The regular representation of a free group is not type I (as the commutant is not type I), while every irreducible representation is type I (since the commutant is $\mathbb{C}$). Hence a factor representation is not necessarily quasi-equivalent to an irreducible. It is true that, for a type I group, every factor representation is quasi-equivalent to an irreducible one (Proposition 5.4.11 in Dixmier). For the decomposition of the regular representation and the corresponding decomposition of $L(G)$, see Proposition 18.7.7 in Dixmier.

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As mentioned in the comments, a direct sum of unitarily equivalent irreducible representations is primary. If G is a Type I group then every primary representation arises in this way, and this characterizes Type I groups. Folland's "A Course in Abstract Harmonic Analysis" (chapter 7) contains a discussion of these points, as well as many references.

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As I'm leaving too many comments... For a compact group $G$, you write $$ L(G) = \bigoplus_{\pi\in\hat G} \dim(\pi)\pi. $$ This isn't quite correct. $L(G)$ is the von Neumann algebra generated by the left regular representation $\lambda:G\rightarrow B(L^2(G))$. What you mean is that $$ \lambda = \bigoplus_{\pi\in\hat G} \dim(\pi)\pi. $$ Perhaps I shouldn't write "=" here; this is isomorphism of representations. This means that $$L(G) = \prod_{\pi\in\hat G} \mathbb M_{\dim(\pi)}. $$ That is, the algebra of each irreducible is a full matrix algebra (because $\pi$ is finite-dimensional, and irreducible). We no longer worry about the $\dim(\pi)$ part, because the von Neumann algebras generated by $\pi$ or $n\pi$ are canonically isomorphic.

To be explicit-- if $\pi$ is irreducible, then $\pi$ is primary. But $\pi\oplus\pi$ is also primary, and obviously not irreducible. The point is that the projection onto the first or second coordinate shows that $\pi\oplus\pi$ is not irreducible. But these projections are not in the von Neumann algebra $L(\pi\oplus\pi)$. Indeed, this is generated by the operators of the form $\pi(g)\oplus\pi(g)$, and so weak continuity (or a bicommutant argument) shows that any member of $L(\pi\oplus\pi)$ is of the form $x\oplus x$. Hence the map $L(\pi) \rightarrow L(\pi\oplus\pi)$ sending $x\mapsto x\oplus x$ is an isomorphism.

Knapp's Notices article alludes to the fact that the left regular representation $\lambda$ of $\mathbb F_2$ is primary but not irreducible. It's a fun, and not too hard, exercise to show that $L(\mathbb F_2)$ is a factor; I don't know a good explanation of why $\lambda$ is not irreducible or even quasi-equivalent to a direct sum of irreducibles.

Finally, you ask if all irreducibles appear in $L(G)$. Again, I think you mean: do all irreducibles appear in $\lambda$? Well, the trivial representation occurs in $\lambda$ if and only if $\lambda$ is amenable; if $G$ is amenable, then all irreducibles occur in $\lambda$. Again, I'm not sure of a good reference-- maybe the recent book by Valette et al. on Kazhdan's property (T)

(I'm not quite an expert on this stuff, so I might have glossed over some technicalities and/or used the wrong language...)

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Thanks, I fixed the statement about the Peter Weyl theorem ... I must have confused representation space and intertwiner:\ –  Marc Palm Apr 28 '11 at 13:30

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