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One can average a class function $f:G\to\Bbb C$ for a finite group $G$ by interpreting $f$ as a complex-valued function on the space ${\rm cl}(G)$ of conjugacy classes and computing the expectation $\Bbb E(f(C))$, where $C$ is a class-valued random variable whose distribution is pulled from the uniform one on $G$.

In light of $\sum(\dim\rho)^2=|G|$ (where the sum is over complex irreps $\rho$), we could equally attach the weights $(\dim\rho)^2/|G|$ to irreps and turn $\widehat{G}$ into a probability space. This is how we get the Plancherel measure on a finite group. But what question is the Plancherel measure an answer to? Historically, we know the Lebesgue and more generally the Haar measure answer the question of translation-invariant measures, and do so nicely (existence and uniqueness). Is there some a priori system of nice properties we might have liked a measure on irreps to have, which we could have later discovered the Plancherel measure uniquely satisfies?

For $G$ locally compact abelian, the complex irreps comprising $\widehat{G}$ are $1$-dimensional and form a group under tensoring - this is just the dual group, which has its own topology (the compact-open topology), and the Plancherel measure is just the Haar measure. However if $G$ is nonabelian, the irreps can be higher-dimensional and so aren't closed under tensor product. (Pontryagin duality generalizes to Tannaka-Krein duality, where we look at symmetry of the tensor operation on the full category of reps to recover $G$, at least as I understand it.) I may be muddying the waters with infinite nonabelian $G$ though, as I don't know how to define the Plancherel measure in that context.

We know conjugacy classes are dual to irreducible representations (see Qiaochu's answer here; the tl;dr version is that they are identified with the maxspecs of two algebras in a dual pairing). They are not generally in bijection, but one situation where they are in bijection is with finite symmetric groups. There might be some process that is "dual" to averaging class functions (which is where the boring measure on the set of conjugacy classes appears) in which the Plancherel measure is useful. This could also potentially explain why there are applications of the Plancherel measure on symmetric groups to combinatorial/probabilistic questions but I haven't seen any applications with other groups.

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    $\begingroup$ I am wondering if this question would be a better fit for MathOverflow? $\endgroup$
    – Thomas
    Commented Jul 4, 2014 at 15:16

2 Answers 2

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What you ultimately want to study is (upon fixing a Haar measure in the noncompact case) the left regular representation $\lambda\colon G\to U(L^2(G,\mathrm{Haar}))$. Now, general theory tells us that while it's not always possible to decompose $L^2(G)$ as a direct sum of irreducible reprenentations (this already fails for $G=\mathbb Z$), it is always possible to decompose it as a direct integral of irreducible representations (which are parametrised by the unitary dual $\widehat G$ of $G$). Now, if $G$ is unimodular and type I, the direct integral decomposition (with respect to both left and right actions of $G$) is as follows: $$ L^2(G) \cong \int_{\widehat G} H_\pi\,d\mu(\pi), $$ where $H_\pi = \pi\otimes \pi^*$, and its understanding requires, in particular, to determine the measure $\mu$ on $\widehat G$ such that the above becomes an isometric isomorphism. The unique measure with this property is called the Plancherel measure of $G$ (associated to a given Haar measure). Equivalently, it's the unique measure such that $$ \|f\|_2^2 = \int_{\widehat{G}} \|\pi(f)\|_{\mathrm{HS}}^2 \mathrm{d}\mu(\pi),\quad f\in L^1(G,\mathrm{Haar})\cap L^2(G,\mathrm{Haar}). $$

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In the case, $G = S^1$ we have the class functions $f \in L^2(G) = L^2(S^1)$ can be written as the sum over irreudicible representations $\theta \mapsto e^{2\pi i n \theta}$ giving Fourier series:

$$ f(x) = \sum_{n \in \mathbb{Z}} \hat{f}(n) e^{2\pi i n x} = \mathbb{E}[\hat{f}(n)e^{2\pi i n x}]$$

In a sense we reconstructing our function by the "average" over all representations.


We could do without any probabilistic interpretation at all, but has proven beneficial in certain contexts.

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    $\begingroup$ For LCA $G$, the Plancherel measure is the Haar measure on the dual group $\widehat{G}$ (as I remarked), which indeed has a nice interpretation via a "sum-of-waves" formula $$f(g)=\int_{\widehat{G}}\widehat{f}(\chi)\chi(g)d\mu(\chi). \tag{$\circ$}$$ But this fails to be correct if $G$ is finite nonabelian and $\mu$ the Plancherel measure. Indeed, the correct version of $(\circ)$ in that context would be when $\mu$ is the counting measure and $f$ is a class function, both automatically true if $G$ is abelian. Something more general must be going on than just what happens for abelian groups. $\endgroup$
    – anon
    Commented Jul 5, 2014 at 13:03
  • $\begingroup$ @blue Do you want class functions only or all of $L^2(G)$? Using the Peter Weyl theorem we have that $$L^2(G) = \bigoplus_{V \in \mathrm{Irr}(G)} V \otimes \overline{V}$$ so any function in $L^2(G)$ can be written as the sum of "matrix elements" of the representations of $G$. Taking the trace of both sides we get $$|G| = \sum (\dim V)^2 $$ which returns Plancherel measure under normalization. $\endgroup$
    – cactus314
    Commented Jul 5, 2014 at 13:28
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    $\begingroup$ Indeed, PW gives us the sum-of-square-dimensions formula for finite groups which in turn tells us the Plancherel measure is in fact a measure. This is all cart-before-horse, though: we know there are weights attached to each irrep which add up to an invariant of $G$ (its order), and hence we have a measure. But if this measure is to be "naturally occurring," there needs to be a way to specify an unknown measure in terms of desirable properties for it to have, only to discover later we're actually thinking of the Plancherel measure. This is the content of my question. $\endgroup$
    – anon
    Commented Jul 5, 2014 at 13:37
  • $\begingroup$ In light of the Fourier expansion formula $(\circ)$ ... which is valid both (i) when $G$ is abelian and $\widehat{G}$ is given the Haar measure, and (ii) when $\widehat{G}$ is countable (e.g. if $G$ is compact and second-countable) with the counting measure ... we might seek an even more general measure for which $(\circ)$ holds. This is the content of this excellent question. This is the reverse direction of mine, though, and our questions don't answer each other, as I noted in my 1st comment. $\endgroup$
    – anon
    Commented Jul 5, 2014 at 13:52
  • $\begingroup$ @blue In a sense, there are really only two groups. Permutation groups $S_n$ & unitary groups $SU(n)$ which are related by Schur-Weyl duality. All other groups can be gotten by embedding into a permutation group or matrix group. This an exaggeration. Plancherel measure is natural since its dual to uniform measure on $G$ is natural. To make this more interesting, consider group actions on sets $X$ other than $G$ itself, & induce maps $g: L^2(X) \to L^2(X)$. $\endgroup$
    – cactus314
    Commented Jul 5, 2014 at 14:01

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