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This is connected to my question here. Let $K$ be a normal compact subgroup in a locally compact group $G$, $\widehat{K}$ the dual object for $K$, and $\mu_K$ the normed Haar measure on $K$ ($\mu_K(K)=1$). For each (unitary irreducible) representation $\sigma\in\widehat{K}$, $\sigma:K\to B(X_\sigma)$, let us consider the measure $$ \chi_\sigma(f)=\dim X_\sigma\cdot \int_K f(s)\cdot \text{tr}\ \sigma (s^{-1})\cdot\mu_K(\text{d}\ s). $$ From the identities $$ \chi_\sigma* \delta_t=\delta_t*\chi_\sigma\qquad(t\in G) $$ $$ \chi_\pi*\chi_\sigma=\begin{cases}0,& \pi\ne\sigma\\ \chi_\sigma, & \pi=\sigma \end{cases} \qquad(\pi,\sigma\in\widehat{K}) $$ it follows that for any unitary representation $T:G\to B(X)$ the maps $$ T_\sigma:G\to B(X)\quad\Big|\quad T_\sigma(t)=\widetilde{T_\sigma}(\delta_t)=\widetilde{T}(\chi_\sigma*\delta_t),\quad t\in G. $$ are also unitary representations of $G$ (here $\widetilde{T}: {\mathcal C}^\star(G)\to B(X)$ means the homomorphism of algebras corresponding to the representation $T:G\to B(X)$, ${\mathcal C}^\star(G)$ the algebra of measures with compact support, i.e. the stereotype dual space to the algebra ${\mathcal C}(G)$ of continuous functions on $G$ with the topology of uniform convergence on compact sets in $G$).


Under which conditions and in which sense $$ \widetilde{T}=\sum_{\sigma\in\widehat{K}} \widetilde{T_\sigma} \qquad ? $$

This would be simple if $$ \delta_{1_K}=\sum_{\sigma\in\widehat{K}} \chi_\sigma, $$ but the method of summing this series is a separate problem (as we discussed here).

Yulia Kuznetsova in her paper on duality for Moore groups writes (if I understand her correctly) that if $G$ is a SIN-group, i.e. can be presented as an extension $$ {\mathbb R}^n\times K\to G\to D $$ where $D$ is a discrete group, then for each norm-continuous unitary representation $T$ the series $\sum_{\sigma\in\widehat{K}} T_\sigma$ has only finite number of non-zero summands (and converges to $T$).

But I must say, I don't understand her explanation. I would appreciate very much if somebody could clarify this (or give an advice).

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