Mod-$p$ maximal compact inductions do not have irreducible admissible subrepresentations

TL;DR

This paper proves that mod-$p$ maximal compact inductions in p-adic split reductive groups lack irreducible admissible subrepresentations, clarifying the structure of these representations in the mod-$p$ setting.

Contribution

It establishes a new non-existence result for irreducible admissible subrepresentations within mod-$p$ maximal compact inductions.

Findings

Maximal compact inductions in mod-$p$ setting have no irreducible admissible subrepresentations.

Provides insight into the structure of mod-$p$ representations of p-adic groups.

Clarifies limitations of irreducibility in the context of mod-$p$ representation theory.

Abstract

Let $p$ be a prime number. We show in this short note that mod-$p$ maximal compact inductions of a $p$-adic split reductive group do not have irreducible admissible subrepresentations.

Full text

1 Introduction

Let $F$ be a non-archimedean local field of residue characteristic $p$, and $G$ be a $p$-adic split reductive group defined over $F$. Let $K$ be a hyperspecial maximal compact subgroup of $G$. Let $\sigma$ be an irreducible smooth $\overline{\mathbf{F}}_{p}$-representation of $K$. We show in this short note:

Theorem 1.1**.**

The compactly induced representation $\textnormal{ind}^{G}_{K}\sigma$ does not have irreducible admissible subrepresentations.

2 Proof of Theorem 1.1

Let $\sigma$ be an irreducible smooth $\overline{\mathbf{F}}_{p}$-representation of $K$.

Proof of Theorem 1.1.

Assume $\pi$ is an irreducible admissible $\overline{\mathbf{F}}_{p}$-representation of $G$ contained in $\textnormal{ind}^{G}_{K}\sigma$, i.e., we are given a $G$-embedding

$\iota:\pi\hookrightarrow\textnormal{ind}^{G}_{K}\sigma$.

Take an irreducible smooth $\overline{\mathbf{F}}_{p}$-representation $\sigma_{1}$ of $K$ contained in $\pi$. By Frobenius reciprocity, we get a non-zero $G$-map $\theta$ in the space $\text{Hom}_{G}(\textnormal{ind}^{G}_{K}\sigma_{1},\pi)$, which will be denoted by $H(\sigma_{1},\pi)$. Since $\pi$ is admissible, the space $H(\sigma_{1},\pi)$ is finite dimensional. By composition, the space $H(\sigma_{1},\pi)$ is a right module over the spherical Hecke algebra $\mathcal{H}(K,\sigma_{1}):=\text{End}_{G}(\textnormal{ind}^{G}_{K}\sigma_{1})$. As $G$ is split, the algebra $\mathcal{H}(K,\sigma)$ is commutative ([Her11b, Corollary 1.3]). Therefore we may replace $\theta$ by an eigenvector. That is to say, there is a character $\chi:\mathcal{H}(K,\sigma_{1})\rightarrow\overline{\mathbf{F}}_{p}$ so that $\theta$ is annihilated by the kernel of $\chi$.

Now under our assumption, the composition $\iota\circ\theta$ is a non-zero map in $\text{Hom}_{G}(\textnormal{ind}^{G}_{K}\sigma_{1},\textnormal{ind}^{G}_{K}\sigma)$. We take a non-zero $T$ in the kernel of $\chi$. We get

$(\iota\circ\theta)\circ T=\iota\circ(\theta\circ T)=0$.

As $T$ and $\iota\circ\theta$ are both non-zero, we get a contradiction by the argument of [Her11a, Corollary 6.5]. ∎

Remark 2.1**.**

Note that we have assumed $\pi$ is admissible in the theorem. However, in certain cases such an assumption is not necessary, say $G=GL_{2}/U(2,1)$, as in both cases a weaker substitute, i.e., the existence of Hecke eigenvalues ([BL94], [Xu18]), is available. Note also that Daniel Le proved recently that there are non-admissible irreducible mod-$p$ smooth representations of $GL_{2}(\mathbf{Q}_{p^{3}})$ ([Le18]).

Remark 2.2**.**

One interest of Theorem 1.1 is to compare it with the complex case: when $G$ is a classical group, a complex maximal compactly induced representation of $G$ might often happen to be irreducible.