Let $\Gamma$ be a discrete (locally compact) subgroup of a locally compact Lie group. Let $H = L^2(\mathbb R^n, \mathbb C)$. Assume we have a complex unitary representation $$\Phi : \Gamma \to \mathbb U (H). $$ Can we say anything about how $\Phi$ splits into a direct sum (integral?) of (irreducible) representations of $\Gamma$ of finite or infinite dimensions? Can someone point towards references for infinite dimensional unitary representations of locally compact groups?
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$\begingroup$ I don't have enough reputation to comment but this is in Dixmier's book C* algebras (chapter integration and disintegration I think) $\endgroup$– Omar MohsenCommented Nov 14 at 4:57
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5$\begingroup$ It’s not possible to decompose into irreducible representations in general. For example, for a discrete icc group $\Gamma$, its left regular representation generates a factor, so it cannot be decomposed as a direct integral, but the representation is not already irreducible. In general, what you can get is a direct integral decomposition into factorial representations (which, in the finite-dimensional case, does reduce to a direct sum of irreducible representations). This can be found in many books on $C^\ast$-algebras and von Neumann algebras. $\endgroup$– David GaoCommented Nov 14 at 5:02
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1$\begingroup$ Dixmier's book only achieves this for Type I Cstar algebras (at least if one is looking for a disintegration into a direct integral of irreducible representations). The only discrete groups that are Type I are those which are virtually abelian; this is a theorem of Thoma. I think you need to narrow down what you are hoping to achieve $\endgroup$– Yemon ChoiCommented Nov 14 at 22:24
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$\begingroup$ @YemonChoi Agreed. I was looking at Krillov's "Elements of the Theory of Representations" and the analogous result there is only stated locally compact groups with a countable basis. Hence, it only applies to countable discrete groups. Do you know if this assumption can be relaxed? I will continue digging into the literature. I understand I'll have to narrow down the type of discrete group I'm considering. $\endgroup$– user82261Commented Nov 14 at 22:36
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2$\begingroup$ @DavidGao Belatedly coming back to this: as soon as you allow G to be non-discrete, Type I examples are plentiful! all semisimple Lie groups, all nilpotent Lie groups, the Euclidean motion groups, various p-adic version ... Indeed, this is what allows one to do noncommutative harmonic analysis. $\endgroup$– Yemon ChoiCommented Nov 20 at 1:43
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I don't have enough reputation to comment but I wanted to add the following :
Like it was said in the comments in general this isn't possible but if you only care about the operator norm then this is always possible. In the sense that if $\pi$ is a unitary representation, there there exists a set of irreducible unitary representations $S$, such that if $a\in C^0_c(\Gamma)$ (or generally the $C^*$-algebra), then $\|\pi(a)\|=\sup_{\pi' \in S}\|\pi'(a)\|$.
Edit: Here is the proof :
1- You see the irreducible unitary representation $\pi$ as a representation $\pi:C^*\Gamma\to B(H)$.
2-The kernel $\ker(\pi)\subseteq C^*\Gamma$ is a closed $*$-ideal, and $A:=C^*\Gamma/\ker(\pi)$ is a $C^*$-algebra.
3-Since any injective $*$-morphism between $C^*$-algebras is an isometry, $\|a\|_A=\|\pi(a)\|_{B(H)}$ for any $a\in A$.
4-General theory of $C^*$-algebras tells that you that if $a\in A$, then $\|a\|_A$ is equal to $\sup \|\pi'(a)\|$ where the sup is over all irreducible unitary representations of $A$ (Dixmier Chapter 2)
5-By Dixmier (Chapter 3), set of irreducible unitary representations of $A$ is just a closed set of irreducible unitary representations of $\Gamma$. It is exactly those which factor through $\ker(\pi)$.
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