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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 ?

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    $\begingroup$ I am not an expert, but I think this is precisely the kind of question the Bushnell-Kutzko theory of types was developed to settle. For example, in Fiona Murnaghan's lecture notes (available on her website; I don't think links can be included in comments) it's stated that if $\sigma$ is a $K$-type of some $G$-representation $\pi$, then all subquotients of $\pi$ will contain $\sigma$. It's my vague understanding that the theory of types says that this happens when $K$ is a $G$-cover of compact open subgroup $K_M$ of a Levi $M$. $\endgroup$ Apr 6, 2021 at 2:04
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    $\begingroup$ Many apologies, there is an uneditable typo in my first comment. If $\sigma$ is a $K$-type of $\pi$, then NOT all subquotients of $\pi$ will contain $\sigma$! $\endgroup$ Apr 6, 2021 at 15:07
  • $\begingroup$ @StefanDawydiak I see, thank you for the comments. My question arised precisely after reading on type theory via the papers by Moy-Prasad or Morris. I should give a look to Murnaghan's lecture notes, as they always are of very good quality ! $\endgroup$
    – Suzet
    Apr 8, 2021 at 9:44
  • $\begingroup$ @Suzet Your questions on StackExchange are often of research level. You should post them on Mathoverflow. You would be more likely to get answers $\endgroup$ Jul 1, 2021 at 11:06
  • $\begingroup$ @PaulBroussous I have been considering posting on mathoverflow lately for the reasons you point out. I'll take your advice and switch to it for questions of similar level, thanks ! $\endgroup$
    – Suzet
    Jul 1, 2021 at 12:46

1 Answer 1

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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 ${\rm c-Ind}_K^G 1_K$.

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  • $\begingroup$ Thanks a lot for this answer, this clears up a few confusions that I had, namely to what extent "being a type in $G$" really is a strong condition or results from clever use of Frobenius reciprocity formalism. So, to make sure that I get it right, with the example that you exhibit we see that the pair $(K,1_K)$ fails to be a type. However, the argument in Morris' paper would justify that descending to a Iwahori subgroup $I\subset K$ gives a level-$0$ $G$-type $(I,1_I)$, which is indeed a type under Bushnell and Kutzko's definition. [...] $\endgroup$
    – Suzet
    Jul 1, 2021 at 12:41
  • $\begingroup$ Then, $\mathrm{c-Ind}_K^G 1_K$ is a subspace of $\mathrm{c-Ind}_I^G 1_I$ by transitivity and exacteness of compact induction. So, every subquotient of the former is a subquotient of the latter. Since $(I,1_I)$ is a type, I may say that any such irreducible subquotient contains $1_I$ ; but it may not contain $1_K$. Is that alright ? $\endgroup$
    – Suzet
    Jul 1, 2021 at 12:41

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