# Irreducible subquotients of compact induction

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 ?

• 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 $M$. Apr 6, 2021 at 2:04
• 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 $\sigma$! Apr 6, 2021 at 15:07
• @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 ! Apr 8, 2021 at 9:44
• @Suzet Your questions on StackExchange are often of research level. You should post them on Mathoverflow. You would be more likely to get answers Jul 1, 2021 at 11:06
• @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 ! Jul 1, 2021 at 12:46

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

• 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. [...] Jul 1, 2021 at 12:41
• 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 ? Jul 1, 2021 at 12:41