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For a Lie group $G$ with compact Lie subgroup $K$, we say that $(G,K)$ is a pair of Gelfand type if the representation $L^2(G/K)$ of $G$ is multiplicity free, that is, if it is a direct integral of distinct irreducible representations.

Can there exist a pair of dual representations in $L^2(G/K)$ for a Gelfand pair $(G,K)$?

Or do there exist non-self dual irreducible representations in $L^2(G/K)$?

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    $\begingroup$ What about $(G,K) = (\mathbf R/\mathbf Z,0)$? $\endgroup$ – Mikael de la Salle Mar 23 '18 at 21:21
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I gave in the comments the example of the Gelfand pair $(G,K) = (\mathbf R/\mathbf Z,0)$ for which every non-trivial irreducible representation arises together with its (distinct) dual representation. But actually this is completely general, at least when $G$ is compact: the representation $L^2(G/K)$ is self-dual. So by uniqueness of the decomposition into irreducible representations (here I use compactness), for every irreducible representation appearing in this decomposition, its dual representation also appears. Of course, the self-dual irreducible representations in $L^2(G/K)$ appear only once. This is for example the case for $(G,K)=(SO(3),SO(2))$, where all the irreducible representations of $SO(3)$ are self-dual.

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  • $\begingroup$ If in addition we assume that $K$ is a maximal compact subgroup, then do we still get non-self dual irreps in $L^2(G/K)$? $\endgroup$ – Alesandro Levi Mar 26 '18 at 11:19
  • $\begingroup$ @AlesandroLevi Yes, essentially same example with $(G,K) = (\mathbf R,0)$. Almost every character of $\mathbf R$ is not self dual. $\endgroup$ – Mikael de la Salle Mar 26 '18 at 13:10

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