**[INTRODUCTION]**

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.

**[QUESTION]**

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?

**[THOUGHT]**

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.