Irreducible subquotients of compact induction

Question

Let $G$ be a locally profinite group and let $K$ be a compact open subgroup. All the representations are assumed to be smooth and complex. Let $\sigma$ be an irreducible representation of $K$.

Is it true that the irreducible subquotients of the compactly induced representation $$\mathrm{c-Ind}_{K}^{G} \sigma$$ are precisely those irreducible representations $\pi$ of $G$ which contain $\sigma$ upon restriction to $K$ ?

Sure enough, such an irreducible representation $\pi$ of $G$ is a subquotient of $\mathrm{c-Ind}_{K}^{G} \sigma$ ; it is in fact even a quotient. Indeed, by Frobenius reciprocity we have $$\mathrm{Hom}_G(\mathrm{c-Ind}_{K}^{G} \sigma,\pi) \simeq \mathrm{Hom}_K(\sigma,\pi_{|K})$$ and the right-hand side is non-zero by hypothesis. Thus, there is a non-zero $G$-map $\mathrm{c-Ind}_{K}^{G} \sigma\to \pi$, which must be surjective as $\pi$ is irreducible. So, the representation $\pi$ is isomorphic to the quotient of $\mathrm{c-Ind}_{K}^{G} \sigma$ by the kernel of the map.

The same argument also shows that any irreducible quotient of $\mathrm{c-Ind}_{K}^{G} \sigma$ contains $\sigma$ when restricted to $K$.

Now, I noticed that any irreducible subspace of $\mathrm{c-Ind}_{K}^{G} \sigma$ also contains $\sigma$ when restricted to $K$, if I'm not mistaken. Indeed, let $\pi$ be such an irreducible subspace. In particular, it is also a subspace of $\mathrm{Ind}_{K}^{G} \sigma$. By Frobenius reciprocity, we have $$\mathrm{Hom}_G(\pi,\mathrm{Ind}_{K}^{G} \sigma) \simeq \mathrm{Hom}_K(\pi_{|K},\sigma)$$ The left-hand side is non-zero as we have the inclusion morphism ; thus the right-hand side neither is zero. But now, $K$ is an open compact subgroup so $\pi_{|K}$ is $K$-semisimple. It implies that we have an isomorphism $$\mathrm{Hom}_K(\pi_{|K},\sigma) \simeq \mathrm{Hom}_K(\sigma,\pi_{|K})$$ So that $\pi$ does contain $\sigma$ when restricted to $K$.

Now, what about a more general irreducible subquotient $\pi$ ? It means that we have two $G$-subspaces $W\subset V \subset \mathrm{c-Ind}_{K}^{G} \sigma$ such that $\pi \simeq V/W$ as $G$-representations. By the same argument as above, I see that both $V$ and $W$ contain $\sigma$ when restricted to $K$, but it looks like I can't have information on the multiplicity. The $\sigma$-isotypic compotents could cancel each other when forming the quotient $V/W$.
Am I missing something or is my statement false ?

Answer 1

Here I assume that $G$ is the group of $F$-rational points of a connected reductive group defined over $F$.

The answer to your question is negative. To see this use Proposition 3.3 of Bushnell and Kutzko's "smooth representations of reductive $p$-adic groups: structure theory via types"(1997). I give some explanations.

If $e$ is an idempotent of the Hecke algebra $H(G)$ of $G$, define $R_e (G)$ to be the full subcategory of the category of all smooth representations of $G$ whose objects are those $V$ that are generated by $e\star V$ as $G$-modules. A particular element of $R_e (G)$ is $H(G)\star e$, where $G$ acts by left translation on functions. Proposition 3.3 says in particular that the following assertions are equivalent:

(1) $R_e (G)$ is closed under subquotients,

(2) Every subquotient of $H(G)\star e$ lies in $R_e (G)$.

Now take $G={\rm GL}(2,F)$, and $e$ to be $\frac{1}{\mu (K)}\, 1_K$, the idempotent attached to $K={\rm GL}(2,{\mathfrak o}_F)$ ($\mu$ is the Haar measure defining the convolution product on $H(G)$). Then $R_e (G)$ is not closed under subquotients (read the last paragraph of section 2 of loc. cit.). On the other hand $H(G)\star e$ is nothing other than the compactly induced representation ${\rm c-Ind}_K^G 1_K$.