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Let $G$ be a noncompact simple Lie group, and $G'$ a noncompact reductive subgroup of $G$ such that $(G,G')$ is a symmetric pair. If $\pi$ is an infinitely dimensional unitary representation of $G$, does the restriction $\pi|_{G'}$ contains a trivial representation of $G'$ as a subrepresentation? In other words, whether $\mathrm{Hom}_{G'}(1,\pi|_{G'})=\{0\}$ always holds or not?

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    $\begingroup$ The Howe Moore ergodicity says that if $\pi $ is a non-trivial unitary irrep of $G$, the matrix coefficients $g \mapsto (\pi (g) (v),w)$ tend to zero as $g$ tends to infinity. Hence if a vector is fixed under a non-compact subgroup, it will be fixed by the whole group and is therefore zero. $\endgroup$
    – Venkataramana
    Commented Sep 13, 2018 at 5:30
  • $\begingroup$ @Venkataramana That is great! Thank you so much, professor Venkataramana! $\endgroup$
    – Hebe
    Commented Sep 13, 2018 at 5:52
  • $\begingroup$ no problem! Glad to be of help $\endgroup$
    – Venkataramana
    Commented Sep 13, 2018 at 6:03

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