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If $A$ is an abelian locally compact group, the Plancherel measure on $\hat A$ is a Haar measure, so, up to scaling it is the unique invariant Radon measure.

Now for a nonabelian locally compact group $G$. Can the Plancherel measure on the unitary dual $\hat G$ be characterized by any (invariance) property? In a way that it is the unique (up to scaling) measure with property XYZ, where XYZ should not be the property that defines the Plancherel measure.

Just to make it clear: I do know the definition of the Plancherel measure. I am looking for a different characterization.

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If $G$ is a unimodular second countable Type I group, then the Plancherel measure is the unique measure $\mu$ such that

$$\|f\|_2^2 = \int_{\widehat{G}} \|\pi(f)\|_{\mathrm{HS}}^2 \mathrm{d}\mu(\pi).$$

for every $f \in \mathrm{L}^1(G) \cap \mathrm{L}^2(G)$. This appears as Theorem 18.8.2 in Dixmier's book on $C^*$-algebras.

When $G$ is not unimodular, the question becomes more complicated, because the Plancherel measure needs to be twisted by a section of a line bundle; see the paper of Duflo-Moore on the subject for the gory details. When $G$ is not second countable, I do not know of a published result; the technical details of direct integral theory are more difficult in this case and not standard. When $G$ is not Type I, the decomposition of the left regular representation into irreducibles is no longer unique, and some of the operators on the right-hand side of the formula will fail to have finite Hilbert-Schmidt norm.

Edit: The question has been clarified to indicate that it is not asking for this characterization of the Plancherel measure, but rather a characterization that is closer to the description in the abelian case of the Plancherel measure as a Haar measure on $\widehat{G}$. Since there is usually no algebraic structure on $\widehat{G}$ for which to consider invariance, a direct generalization is impossible.

The closest analogue is the characterization of the Plancherel measure as a unique co-invariant trace (or weight) on the von Neumann algebra $\mathcal{M}$ generated by the left-regular representation of $G$. Suppose $G$ satisfies the same hypotheses as above and $\Delta : \mathcal{M} \to \mathcal{M} \overline{\otimes} \mathcal{M}$ is the comultiplication on $\mathcal{M}$ given by $\lambda(s) \mapsto \lambda(s) \otimes \lambda(s)$. Then the Plancherel trace is the unique normal semifinite trace $\tau$ on $\mathcal{M}$ such that

$$\tau((\varphi \otimes \mathrm{id}) (\Delta(a))) = \tau(a)$$

for all $a \in \mathcal{M}_\tau^+$ and $\varphi \in \mathcal{M}_*$. A similar characterization holds for the Plancherel weight of an arbitrary locally compact group, or for the Haar weight of a locally compact quantum group. For proofs, see volume 2 of Takesaki or any of the literature on von Neumann algebraic quantum groups.

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  • $\begingroup$ Sorry, but that's exactly the kind of answer I did not want. $\endgroup$
    – user1688
    Commented Oct 30, 2017 at 9:04
  • $\begingroup$ @Corbennick I believe the disappointment is not a reason to vote down, an answer tries to help anyway. Maybe is it also a reason for you to clarify what you really want as an answer. Cameron gave you the unique characterization of the Plancherel measure (and it is a kind of invariance property since it is an averaging operator on the whole dual group). An by the way, the Plancherel measure is not a Haar measure (only Radon positive). $\endgroup$ Commented Oct 30, 2017 at 9:18
  • $\begingroup$ I wasn't meaning to say "unique characterization", but "characterization as the unique measure satisfying a certain invariance property". $\endgroup$ Commented Oct 30, 2017 at 9:29
  • $\begingroup$ @Desiderius Severus: The Plancherel measure is of course a Haar measure in the case of abelian group. This is the content of the Plancherel Theorem. $\endgroup$
    – user1688
    Commented Oct 30, 2017 at 12:14
  • $\begingroup$ @Corbennick Yet your question seems to be exactly in the nonabelian case, isn't it? $\endgroup$ Commented Oct 30, 2017 at 13:15

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