Let $G$ be a non-compact simple Lie group, and $G'$ a reductive subgroup of $G$. Suppose that $\pi$ is a non-trivial (hence, infinite dimensional) irreducible unitary representation of $G$ on a Hilbert space, then the restriction $\pi|_{G'}$ of $\pi$ to $G'$ is (analytic) discretely decomposable if $\pi|_{G'}$ is decomposed as a Hilbert direct sum of irreducible representations of $G'$; namely, it contains no continuous spectrums.

Denote by $\mathfrak{g}$ and $\mathfrak{g}'$ the complexified Lie algebras of $G$ and $G'$ respectively. Take a maximal compact subgroup $K$ of $G$ such that $K':=K\cap G'$ is a maximal compact subgroup of $G'$.

Denote by $\pi_K$ the underlying (unitarizable simple) $(\mathfrak{g},K)$-module of $\pi$. Then the restriction $\pi_K|_{\mathfrak{g}'}$ of $\pi_K$ to $\mathfrak{g}'$ is (algebraic) discretely decomposable if $\pi_K|_{\mathfrak{g}'}$ is decomposed as a direct sum of simple $(\mathfrak{g}',K')$-modules.


It is well-known that if $\pi_K|_{\mathfrak{g}'}$ is algebraic discretely decomposable, then $\pi|_{G'}$ is analytic discretely decomposable. However, if $\pi|_{G'}$ is analytic discretely decomposable, is $\pi_K|_{\mathfrak{g}'}$ algebraic discretely decomposable?


Actually, it is known that $\pi_K|_{\mathfrak{g}'}$ is algebraic discretely decomposable if and only if there exists a simple simple $(\mathfrak{g}',K')$-module $\tau'_{K'}$ such that $\mathrm{Hom}_{\mathfrak{g}',K'}(\tau'_{K'},\pi_K|_{\mathfrak{g}'})\neq\{0\}$. Now $\pi|_{G'}$ is analytic discretely decomposable, and take a direct summand $\tau'$ which is a subrepresentation. Can $\tau'_{K'}$ be embeded into $\pi_K|_{\mathfrak{g}'}$? I think it not correct because the element in $\tau'_{K'}$ is not necessarily $K$-finite, right?

I did not find explicit answers for this point in the papers focusing on branching law. I shall be grateful if experts here may share some ideas.


Check the home page of Toshiyuki Kobayashi (link), download the earlier papers of his, over there you fill an answer.

Edit: in a 2017 note of Duflo-Galina-Vargas (link behind paywall), you will find a proof that the answer to your question is yes for discrete series and any pair $(G,G')$. For $A_q(\lambda)$ and $(G,G')$, you can find the answer in the work of Kobayashi.

  • 1
    $\begingroup$ Thank you for your comments, professor Vargas. I tried to go through the papers such as "The restriction of Aq(λ) to reductive subgroups (I) (II)"(1993,1995), "Discretely decomposable restrictions of unitary representations of reductive Lie groups—examples and conjectures - Analysis on homogeneous spaces and representation theory of Lie groups"(1997), and so on, written by professor Kobayashi, but did not find the explicit answer or some counter-examples. I shall be grateful if you may give more hints. Thanks! $\endgroup$
    – Hebe
    Feb 20 '18 at 6:12

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