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Given a locally compact group $G$, does there exist a measure $\nu$ on the conjugacy classes $conj(G)$ such that for $f \in C_c(G)$ $$ \int_G f(g) d \mu_G(g) = \int_{conj(H)} \int_{G / G_\gamma} f(g\gamma g^{-1}) d \mu_\gamma(g) d \nu(\gamma),$$ where $G_\gamma$ is the centralizer of $\gamma$ in $G$ and $\mu_\gamma$ is a measure on $G/ G_\gamma$? Btw., what would be the image of the operator $P : C_c(G) \rightarrow ?$, where $$ Pf(\gamma) = \int_{G / G_\gamma} f(g\gamma g^{-1}) d \mu_\gamma(g).$$ Do the conjugacy classes come with a natural topology? What is the relation to irreducible representations?

Paul Garrett has adressed the case of a reductive group, but I am interested in more general groups.

A more general question: Given an action of $G$ on a measurable space $X$, when does there exist a measure $\mu$ such that its translates $\mu_g$, where $\mu_g(A) =\mu(g^{-1}A)$, are equivalent measures?

Mark Schwarzmann has adressed this question in the comments stating that for continuous action of amenable groups on compact metrizable spaces, we actually obtain invariant measure, meaning that $\mu(gA)= \mu(A)$. I am asking for less in more general context. Note: That exy=istence (not uniqueness) of quaiinvariant measure is true for transcendental actions (=one orbit) and also when each orbits are dense + "extra condition".

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One general fact: If $G$ is amenable then every (continuous) action of $G$ on a compact metrizable space has a $G$-invariant Borel probability measure (and in particular quasi-invariant). Of course, in general the quotient space here won't be compact or metrizable, and I suppose that you're interested in non-amenable groups as well. –  Mark Jul 28 '11 at 15:45

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As in Weyl's treatment of compact Lie groups, Gelfand-Naimark for $GL_n(\mathbb C)$, Harish-Chandra's treatment of characters of reductive Lie groups $G$: the regular semi-simple elements $g$ form a set of full measure in the group, and the centralizer $Z(g)$ of regular semi-simple $g$ includes a maximal torus. For compact $G$, all these centralizers are conjugate, so $G/Z(g)$ is isomorphic as $G$-subspace (with conjugation) of full measure in $G$, and Weyl's character formula and dimension formula fall out. For complex reductive (following Gelfand-Naimark and Harish-Chandra) there's again a single conjugacy class. For real reductive, Hirai and Harish-Chandra had to worry about "patching conditions" for characters at the boundaries/interstices between the finitely-many conjugacy classes.

(A similar structural thing happens in p-adic reductive groups...)

Edit: in some further detail... with $A$ the diagonal subgroup in $G=U(n)$, for example, and $G'$ the full-measure subset of $G$ consisting of regular semi-simple elements, for a _conjugation_invariant_ function $f$, $\int_G f = \int_{G'} f = \int_{G/A} \int_A f(gag^{-1}) = \int_A f(a) (\int_{G/A}1)$. A similar computation works, for example, in $GL_n(\mathbb C)$, or whenever there is a single conjugacy class, with the orbital integrals of general $f$ appearing. In Harish-Chandra's formulations of Plancherel in terms of characters, this integration formula is it, indeed. (I don't think it's really about Cartan decomposition or other of the standard decompositions...)

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To rephrase your answer in order to see if I can follow: Yes for reductive groups with values in local fields, where it is sufficient to give a measure on the regular semisimple conjugacy classes here, which can be given the structure of a maximal torus. Is there a connection with the Cartan decomposition of the Haarmeasure and hence with the Plancherel measure here? Away from the Lie groups, every locally compact group has a Plancherel theorem, so it would be really nice to have an correspondance between conj. classes and irred. repr. here, wouldn't it? –  Marc Palm Jul 28 '11 at 15:03
    
Yes, if possible! :) I have trouble visualizing "the general case" of topological groups and their repn theory, having already enough trouble with reductive Lie groups and p-adic groups! :) But, perhaps we're just missing a concept...? –  paul garrett Jul 29 '11 at 16:33

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