Homological branching law for $(\mathrm{GL}_{n+1}(F), \mathrm{GL}_n(F))$: projectivity and indecomposability

TL;DR

This paper investigates the homological properties of irreducible smooth representations of general linear groups over non-Archimedean fields, establishing indecomposability of Bernstein components and classifying projective restrictions.

Contribution

It introduces a new framework using derivatives to analyze the restriction of representations, proving indecomposability and classifying projective cases in the homological branching law.

Findings

Bernstein components are indecomposable upon restriction.

All projective irreducible representations are classified.

The branching law is determined in both directions.

Abstract

Let $F$ be a non-Archimedean local field. This paper studies homological properties of irreducible smooth representations restricted from $\mathrm{GL}_{n+1}(F)$ to $\mathrm{GL}_n(F)$. A main result shows that each Bernstein component of an irreducible smooth representation of $\mathrm{GL}_{n+1}(F)$ restricted to $\mathrm{GL}_n(F)$ is indecomposable. We also classify all irreducible representations which are projective when restricting from $\mathrm{GL}_{n+1}(F)$ to $\mathrm{GL}_n(F)$. A main tool of our study is a notion of left and right derivatives, extending some previous work joint with Gordan Savin. As a by-product, we also determine the branching law in the opposite direction.

Equations

$$\mathrm{Ext}^{i}_{G_{n}}(\pi,\pi^{\prime})=0,\mbox{ for $i\geq 1$},$$

$${\mathrm{Hom}}_{G_n}({\pi }_2,{\pi }_1{|}_{G_n})\ne 0$$

$${}^{-}\Delta =[{\nu }^{a+1}\rho ,\dots ,{\nu }^b\rho ]\text{ and }{\Delta }^{-}=[{\nu }^a\rho ,\dots ,{\nu }^{b-1}\rho ].$$

$$\mathrm{St}(\Delta {)}^{\vee }\cong \mathrm{St}({\Delta }^{\vee })\text{mbox}and⟨\Delta {⟩}^{\vee }\cong ⟨{\Delta }^{\vee }⟩.$$

$$\mathfrak{m}\mapsto ⟨\mathfrak{m}⟩,\text{mbox}and\mathfrak{m}\mapsto \mathrm{St}(\mathfrak{m})$$

$$⟨\Delta ⟩\times ⟨{\Delta }^{\prime }⟩\cong ⟨{\Delta }^{\prime }⟩\times ⟨\Delta ⟩,$$

$$\mathrm{St}(\Delta )\times \mathrm{St}({\Delta }^{\prime })\cong \mathrm{St}({\Delta }^{\prime })\times \mathrm{St}(\Delta ).$$

$${\mathrm{csupp}}_{\mathbb{Z}}(\pi )=\left\{{\nu }^c\rho :\rho \in \mathrm{csupp}(\pi ),c\in \mathbb{Z}\right\},$$

$$R_{n-i}=\left\{\left(\begin{array}{cc}g& x\\ 0& u\end{array}\right)\in G_n:g\in {\mathrm{GL}}_{n-i}(F),u\in U_i,x\in {\mathrm{Mat}}_{n-i,i}(F)\right\}.$$

$$\pi \mapsto {\mathrm{Ind}}_{R_{n-i}}^{G_n}\pi ⊠{\psi }_i,\text{mbox}and\pi \mapsto {\mathrm{ind}}_{R_{n-i}}^{G_n}\pi ⊠{\psi }_i,$$

$${}^{(i)}\pi :={\theta }_{n-i}({\theta }_n(\pi {)}^{(i)}),$$

$${\pi }^{[i]}={\nu }^{1/2}\cdot {\pi }^{(i)},\text{mbox}and{}^{[i]}\pi ={\nu }^{-1/2}\cdot {}^{(i)}\pi .$$

$${\mathrm{ind}}_{R_{n-i}}^{G_n}\pi ⊠{\psi }_i\cong \pi \times {\mathrm{ind}}_{U_i}^{G_i}{\psi }_i=\pi \times {\Pi }_i.$$

$${\mathrm{ind}}_{R_{n-i}^{-}}^{G_n}\pi ⊠{\psi }_i\cong \left({\mathrm{ind}}_{U_i}^{G_i}{\psi }_i\right)\times \pi ={\Pi }_i\times \pi .$$

$${\mathrm{Ind}}_{R_{n-i}}^{G_n}\pi ⊠{\psi }_i\cong \pi \times {\mathrm{Ind}}_{U_i}^{G_i}{\psi }_i.$$

$${\pi }^{(i)}\cong ({\pi }_{N_i}{)}_{U_i,{\psi }_i},$$

$${\mathrm{Hom}}_{G_{n+1-i}}({\pi }^{[i]},{}^{(i-1)}{\pi }^{\prime })\ne 0;$$

$${\mathrm{Ext}}_{G_{n+1-j}}^k({\pi }^{[j]},{}^{(j-1)}{\pi }^{\prime }))=0$$

$$0\subset {\pi }_n\subset \dots \subset {\pi }_0=\pi$$

$${\mathrm{Ext}}_{G_n}^k({\mathrm{ind}}_{R_{n-j+1}}^{G_n}{\pi }^{[j]}⊠{\psi }_{j-1},{\pi }^{\prime })\cong {\mathrm{Ext}}_{G_{n-j+1}}^k({\pi }^{[j]},{}^{(j-1)}{\pi }^{\prime })=0$$

$$\dim {\mathrm{Hom}}_{G_n}(\pi ,{\pi }^{\prime })$$

$$=\dim {\mathrm{Hom}}_{G_{n+1-i}}({\pi }^{[i]},{}^{(i-1)}{\pi }^{\prime })\ne 0.$$

$$\lambda (\mathfrak{m})\cong \mathrm{St}(\Delta )\times \lambda (\mathfrak{m}\setminus \left\{\Delta \right\}).$$

$$\mathrm{St}(\Delta )\hookrightarrow b(\Delta )\times \mathrm{St}({\Delta }^{-}).$$

$$\lambda (\mathfrak{m})$$

$$\cong b(\Delta )\times \lambda (\mathfrak{m}\setminus \left\{\Delta \right\}+{\Delta }^{-})$$

$$\lambda (\mathfrak{m}\setminus \left\{\Delta \right\}+{\Delta }^{-})\cong \mathrm{St}({\Delta }^{-})\times \lambda (\mathfrak{m}\setminus \left\{\Delta \right\}).$$

$$b(\Delta )\cong b({\Delta }_1)\cong \dots \cong b({\Delta }_k)$$

$$\mathrm{St}({\Delta }^{-})\times \lambda (\mathfrak{m}\setminus \left\{\Delta \right\})\cong$$

$$\cong$$

$${}^{(i)}\pi \cong {\theta }_{n-i}({\theta }_n(\pi {)}^{(i)})\cong {\theta }_{n-i}(({\pi }^{\vee }{)}^{(i)})\cong {\theta }_{n-i}(({}^{(i)}\pi {)}^{\vee })$$

Full text

1. Introduction

1.1.

Let $F$ be a non-Archimedean local field. Let $G_{n}=\mathrm{GL}_{n}(F)$. Let $\mathrm{Alg}(G_{n})$ be the category of smooth representations of $G_{n}$, and all representations in this paper are in $\mathrm{Alg}(G_{n})$. This paper is a sequel of [CS18b] in studying homological properties of smooth representations of $G_{n+1}$ restricted to $G_{n}$, which is originally motivated from the study of D. Prasad in his ICM proceeding [Pr18]. In [CS18b], we show that for generic representations $\pi$ and $\pi^{\prime}$ of $G_{n+1}$ and $G_{n}$ respectively, the higher Ext-groups

[TABLE]

which was previously conjectured in [Pr18]. This result gives a hope that there is an explicit homological branching law, generalizing the multiplicity one theorem [AGRS10], [SZ12] and the local Gan-Gross-Prasad conjectures [GGP12, GGP20].

The main techniques in [CS18b] are utilizing Hecke algebra structure developed in [CS19, CS18] and simultaneously applying left and right Bernstein-Zelevinsky derivatives, based on the classical approach of using Bernstein-Zelevinsky filtration on representations of $G_{n+1}$ restricted to $G_{n}$ [Pr93, Pr18]. We shall extend these methods further, in combination of other things, to obtain new results in this paper.

The first part of the paper is to study projectivity under restriction. A supercuspidal representation of a reductive group $G$ is projective and, restricting to a closed subgroup $H$ of $G$, is still projective. It is natural to study the same property in the relative framework, that is to classify irreducible representations of $G$ restricted to $H$ is projective.

In [CS18b], we showed that an essentially square-integrable representation $\pi$ of $G_{n+1}$ is projective when restricted to $G_{n}$. However, those representations do not account for all irreducible representations whose restriction is projective. The first goal of the paper is to classify such representations:

Theorem 1.1**.**

(=Theorem 3.7) Let $\pi$ be an irreducible smooth representation of $G_{n+1}$. Then $\pi|_{G_{n}}$ is projective if and only if

1. (1)

$\pi$* is essentially square-integrable, or* 2. (2)

$n+1$* is even, and $\pi\cong\rho_{1}\times\rho_{2}$ for some cuspidal representations $\rho_{1},\rho_{2}$ of $G_{(n+1)/2}$.*

There are also recent studies of the projectivity under restriction in other settings [APS17], [La17] and [CS18c].

A main step in our classification is to the following projectivity criteria:

Theorem 1.2**.**

(=Theorem 3.3 ) Let $\pi$ be an irreducible smooth representation of $G_{n+1}$. Then $\pi|_{G_{n}}$ is projective if and only if the following two conditions hold:

1. (1)

$\pi$* is generic; and* 2. (2)

$\mathrm{Hom}_{G_{n}}(\pi|_{G_{n}},\omega)=0$* for any irreducible non-generic representation $\omega$ of $G_{n}$.*

Theorem 1.2 turns the projectivity problem into a Hom-branching problem. This is a consequence of two things: (1) the Euler-Poincaré pairing formula of D. Prasad [Pr18] and (2) the Hecke algebra argument used in [CS18b] by G. Savin and the author. Roughly speaking, (1) is used to show non-projectivity while (2) is used to show projectivity.

The second part of the paper studies indecomposability of a restricted representation. It is clear that an irreducible representation (except one-dimensional ones) restricted from $G_{n+1}$ to $G_{n}$ cannot be indecomposable as it has more than one non-zero Bernstein components. However, the Hecke algebra realization in [CS18b, CS19] of the projective representations in Theorem 1.1 immediately implies that each Bernstein component of those restricted representation is indecomposable. This is a motivation of our study in general case, and precisely we prove:

Theorem 1.3**.**

(Theorem 5.1) Let $\pi$ be an irreducible representation of $G_{n+1}$. Then for each Bernstein component $\tau$ of $\pi|_{G_{n}}$, any two non-zero $G_{n}$-submodules of $\tau$ have non-zero intersection.

As a consequence, we have:

Corollary 1.4**.**

Let $\pi$ be an irreducible representation of $G_{n+1}$. Then any Bernstein component of $\pi|_{G_{n}}$ is indecomposable.

In Section 5.2, we explain how to determine which Bernstein component of $\pi$ is non-zero in terms of Zelevinsky segments, and hence Theorem 1.3 essentially parametrizes the indecomposable components of $\pi|_{G_{n}}$.

For a mirabolic subgroup $M_{n}$ of $G_{n+1}$, it is known [Ze80] that $\pi|_{M_{n}}$ is indecomposable for an irreducible representation $\pi$ of $G_{n+1}$. The approach in [Ze80] uses the Bernstein-Zelevinsky filtration of $\pi$ to $M_{n}$ and that the bottom piece of the filtration is irreducible. We prove that the Bernstein component of a bottom piece is indecomposable as a $G_{n}$-module, and then make use of left and right derivatives, developed and used to prove main results in [CS18b]. The key fact is that left and right derivatives of an irreducible representation are asymmetric. We now make more precise the meaning of ’asymmetric’. We say that an integer $i$ is the level of an irreducible representation $\pi$ of $G_{n}$ if the left derivative $\pi^{(i)}$ (and hence the right derivative ${}^{(i)}\pi$) is the highest derivative of $\pi$.

Theorem 1.5**.**

(=Theorem 5.2) Let $\pi$ be an irreducible smooth representation of $G_{n}$. Let $\nu(g)=|\mathrm{det}(g)|_{F}$. Suppose $i$ is not the level of $\pi$. Then $\nu^{1/2}\cdot\pi^{(i)}$ and $\nu^{-1/2}\cdot{}^{(i)}\pi$ have no isomorphic irreducible quotients whenever $\nu^{1/2}\cdot\pi^{(i)}$ and $\nu^{-1/2}\cdot{}^{(i)}\pi$ are non-zero. The statement also holds if one replaces quotients by submodules.

We remark that $\nu^{1/2}\cdot\pi^{(i)}$ and $\nu^{-1/2}\cdot{}^{(i)}\pi$ are shifted derivatives in the sense of [Be84], which has been used in loc. cit. to study unitary representations of $G_{n}$.

As a by-product of Theorem 1.5, we give a complete answer to the branching law in another direction:

Theorem 1.6**.**

(=Corollary 6.8) Let $\pi_{1},\pi_{2}$ be irreducible representations of $G_{n+1}$ and $G_{n}$ respectively. Then

[TABLE]

if and only if both $\pi_{1}$ and $\pi_{2}$ are one-dimensional and $\pi_{2}=\pi_{1}|_{G_{n}}$.

Computing the structure of a derivative of an arbitrary representation is a difficult question in general. Our approach is to approximate the information of derivatives of irreducible ones by some parabolically induced modules, whose derivatives can be computed via geometric lemma. On the other hand, the Speh representations behave more symmetrically for left and right derivatives, which motivates our proof to involve Speh representations.

Another key ingredient in proving Theorem 1.3 is a study on the submodule structure between left and right Bernstein-Zelevinsky inductions. We explain in Section 4 how the submodule structure of an induced module can be partly reflected from the module which induced from (see Proposition 4.23 for a precise statement). The study relies on the Hecke algebra structure of the Gelfand-Graev representation.

1.2. Organization of the paper

Section 2 studies derivatives of generic representations, which simplifies some computations for Theorem 1.1. The results also give some guiding examples in the study of this paper and [CS18b].

Section 3 firstly proves a criteria for an irreducible representation to be projective under restriction, and then apply this criteria to give a classification of such class of modules.

Section 4 develops a theory of the submodule relation between left and right Bernstein-Zelevinsky inductions.

Section 5 proves a main result on the indecomposability of an irreducible representation under restriction, which uses results in Section 4 and the asymmetric property of derivatives proved in Section 6.

Section 7 proves that the Bernstein-Zelevinsky induction preserves indecomposability. This partly generalizes Section 4.

In the first appendix, we explain how an irreducible representation appears as the unique submodule of a product of some Speh representations. In the second appendix, we provide some preliminaries on module theory.

1.3. Acknowledgements

This article is a part of the project in studying Ext-branching laws, and the author would like to thank Gordan Savin for a number of helpful discussions. Part of the idea on indecomposability for restriction was developed during the participation in the program of ’On the Langlands Program: Endoscopy and Beyond’ at IMS of National University of Singapore in January 2019. The author would like to thank the organizers for their warm hospitality. The author would like to thank the anonymous referee for useful comments.

2. Bernstein-Zelevinsky derivatives of generic representations

2.1. Notations

Let $G_{n}=\mathrm{GL}_{n}(F)$. All representations in this paper are in $\mathrm{Alg}(G_{n})$ and over $\mathbb{C}$, and we usually omit the adjective ’smooth’. Let $\mathrm{Irr}(G_{n})$ be the set of all irreducible representations of $G_{n}$ and let $\mathrm{Irr}=\sqcup_{n}\mathrm{Irr}(G_{n})$. For an admissible representation $\omega\in\mathrm{Alg}(G_{n})$, denote by $\mathrm{JH}(\omega)$ the set of (isomorphism classes of) irreducible composition factors of $\omega$.

Let $\rho$ be a cuspidal representation of $G_{l}$. Let $a,b\in\mathbb{C}$ with $b-a\in\mathbb{Z}_{\geq 0}$. We have a Zelevinsky segment $\Delta=[\nu^{a}\rho,\nu^{b}\rho]$, which we may simply call a segment. Denote $a(\Delta)=\nu^{a}\rho$ and $b(\Delta)=\nu^{b}\rho$. The relative length of $\Delta$ is defined as $b-a+1$ and the absolute length of $\Delta$ is defined as $l(b-a+1)$. We can truncate $\Delta$ form each side to obtain two segments of absolute length $r(b-a)$:

[TABLE]

Moreover, if we perform the truncation $k$-times, the resulting segments will be denoted by ${}^{(kl)}\Delta$ and $\Delta^{(kl)}$. We remark that the convention here is different from the previous paper [CS18b] for convenience later. If $i$ is not an integer divisible by $l$, then we set $\Delta^{(i)}$ and ${}^{(i)}\Delta$ to be empty sets. We also denote $\Delta^{\vee}=[\nu^{-b}\rho^{\vee},\nu^{-a}\rho^{\vee}]$. For a singleton segment $[\rho,\rho]$, we abbreviate as $[\rho]$. For $\pi\in\mathrm{Alg}(G_{l})$, define $n(\rho)=l$. For $c\in\mathbb{C}$ and a segment $\Delta=[\nu^{a}\rho,\nu^{b}\rho]$, define $\nu^{c}\Delta=[\nu^{a+c}\rho,\nu^{b+c}\rho]$.

For a Zelevinsky segment $\Delta$, define $\langle\Delta\rangle$ and $\mathrm{St}(\Delta)$ to be the (unique) irreducible submodule and quotient of $\nu^{a}\rho\times\ldots\times\nu^{b}\rho$ respectively. We have

[TABLE]

For two cuspidal representations $\rho_{1},\rho_{2}$ of $G_{m}$, we say that $\rho_{1}$ precedes $\rho_{2}$, denoted by $\rho_{1}<\rho_{2}$, if $\nu^{c}\rho_{1}\cong\rho_{2}$ for some $c>0$. We say two segments $\Delta$ and $\Delta^{\prime}$ are linked if $\Delta\not\subset\Delta$, $\Delta\not\subset\Delta^{\prime}$ and $\Delta\cup\Delta^{\prime}$ is still a segment. We say that a segment $\Delta$ precedes $\Delta^{\prime}$, denoted by $\Delta<\Delta^{\prime}$, if $b(\Delta)$ precedes $b(\Delta^{\prime})$; and $\Delta$ and $\Delta^{\prime}$ are linked. If $\Delta$ does not precede $\Delta^{\prime}$, write $\Delta\not<\Delta^{\prime}$.

A multisegment is a multiset of segments. Let $\mathrm{Mult}$ be the set of multisegments. Let $\mathfrak{m}=\left\{\Delta_{1},\ldots,\Delta_{r}\right\}\in\mathrm{Mult}$. We relabel the segments in $\mathfrak{m}$ such that for $i<j$, $\Delta_{i}$ does not precede $\Delta_{j}$. The modules defined below are independent of the labeling (up to isomorphisms) [Ze80]. Define $\zeta(\mathfrak{m})=\langle\Delta_{1}\rangle\times\ldots\times\langle\Delta_{r}\rangle$. Denote by $\langle\mathfrak{m}\rangle$ the unique irreducible submodule of $\zeta(\mathfrak{m})$. Similarly, define $\lambda(\mathfrak{m})=\mathrm{St}(\Delta_{1})\times\ldots\times\mathrm{St}(\Delta_{r})$. Denote by $\mathrm{St}(\mathfrak{m})$ the unique quotient of $\lambda(\mathfrak{m})$. Both notions $\langle\mathfrak{m}\rangle$ and $\mathrm{St}(\mathfrak{m})$ give a classification of irreducible smooth representations of $G_{n}$ ([Ze80, Proposition 6.1], also see [LM16]) i.e. both the maps

[TABLE]

determine bijections from $\mathrm{Mult}$ to $\mathrm{Irr}$. For example, when $\mathfrak{m}=\left\{[\nu^{-1/2}],[\nu^{1/2}]\right\}$, $\langle\mathfrak{m}\rangle$ is the Steinberg representation and $\mathrm{St}(\mathfrak{m})$ is the trivial representation. The two notions $\langle\mathfrak{m}\rangle$ and $\mathrm{St}(\mathfrak{m})$ are related by the so-called Aubert-Zelevinsky duality, and Mœglin-Waldspurger algorithm.

We shall use the following standard fact several times (see [Ze80, Theorems 1.9, 4.2 and 9.7]): if two segments $\Delta,\Delta^{\prime}$ are not linked, then

[TABLE]

[TABLE]

Let $\pi\in\mathrm{Irr}(G_{n})$. Then $\pi$ is a subquotient of $\rho_{1}\times\ldots\times\rho_{r}$ for some irreducible cuspidal representations $\rho_{i}$ of $G_{n_{i}}$. The multiset $(\rho_{1},\ldots,\rho_{r})$, denoted by $\mathrm{csupp}(\pi)$, is called the cuspidal support of $\pi$. We also set

[TABLE]

which is regarded as a set (rather than a multiset).

2.2. Derivatives and Bernstein-Zelevinsky inductions

Let $U_{n}$ (resp. $U_{n}^{-}$) be the group of unipotent upper (resp. lower) triangular matrices in $G_{n}$. For $i\leq n$, let $P_{i}$ be the parabolic subgroup of $G_{n}$ containing the block diagonal matrices $\mathrm{diag}(g_{1},g_{2})$ ($g_{1}\in G_{i}$, $g_{2}\in G_{n-i}$) and the upper triangular matrices. Let $P_{i}=M_{i}N_{i}$ with the Levi $M_{i}$ and the unipotent $N_{i}$. Let $N_{i}^{-}$ be the opposite unipotent subgroup of $N_{i}$. Let $\nu:G_{n}\rightarrow\mathbb{C}$ given by $\nu(g)=|\mathrm{det}(g)|_{F}$. Let

[TABLE]

Let $R_{n-i}^{-}$ be the transpose of $R_{n-i}$.

We shall use $\mathrm{Ind}$ for normalized induction and $\mathrm{ind}$ for normalized induction with compact support. Let $\psi_{i}$ be a character on $U_{i}$ given by $\psi_{i}(u)=\overline{\psi}(u_{1,2}+\ldots+u_{i-1,i})$, where $\overline{\psi}$ is a nondegenerate character on $F$ and $u_{k,k+1}$ is the value in the $(k,k+1)$-entry of $u$. For any irreducible representation $\tau$ of $G_{n-i}$, we extend trivially the $G_{n-i}\times U_{i}$-representation $\tau\boxtimes\psi_{i}$ to a $R_{n-i}$-representation. This defines functors from $\mathrm{Alg}(G_{n-i})$ to $\mathrm{Alg}(G_{n})$ given by

[TABLE]

both of which will be called (right) Bernstein-Zelevinsky inductions. Similarly, one has left Bernstein-Zelevinsky inductions by using $R_{n-i}^{-}$ instead of $R_{n-i}$.

Let $\pi$ be a smooth representation of $G_{n}$. Following [CS18b], define $\pi^{(i)}$ to be the left adjoint functor of $\mathrm{Ind}_{R_{n-i}}^{G_{n}}\pi\boxtimes\psi_{i}$. Let $\theta_{n}:G_{n}\rightarrow G_{n}$ given by $\theta_{n}(g)=g^{-T}$, the inverse transpose of $g$. Define the left derivative

[TABLE]

which is left adjoint to $\mathrm{Ind}_{R_{n-i}^{-}}^{G_{n}}\pi\boxtimes\psi_{i}^{\prime}$, where $\psi_{i}^{\prime}(u)=\psi_{i}(u^{T})$ for $u\in U_{i}^{-}$. The level of an admissible representation $\pi$ is the largest integer $i^{*}$ such that $\pi^{(i^{*})}\neq 0$ and $\pi^{(j)}=0$ for all $j>i^{*}$. It follows from (2.3) that if $i^{*}$ is the level of $\pi$, then ${}^{(i^{*})}\pi\neq 0$ and ${}^{(j)}\pi=0$ for all $j>i^{*}$. When $i^{*}$ is the level for $\pi$, we shall call $\pi^{(i^{*})}$ and ${}^{(i^{*})}\pi$ to be the highest left and right derivatives of $\pi$ respectively, where we usually drop the term of left and right if no confusion.

We now define the shifted derivatives as follow: for any $i$,

[TABLE]

For the details of Bernstein-Zelevinsky filtrations, see [CS18b].

2.3. On computing derivatives

Let $\Pi_{i}=\mathrm{ind}_{U_{i}}^{G_{i}}\psi_{i}$ be the Gelfand-Graev representation. Using inductions by stages, we have that

[TABLE]

Let $\dot{w}_{0}\in G_{n}$ whose antidiagonal entries are $1$ and other entries are [math]. By the left translation of $\dot{w}_{0}$ on $\mathrm{ind}_{R_{n-i}^{-}}^{G_{n}}\pi\boxtimes\psi_{i}$, we have that

[TABLE]

We similarly have that

[TABLE]

Since the right derivative is left adjoint to $\mathrm{Ind}_{R_{n-i}}^{G_{n}}$, we consequently have:

[TABLE]

where $U_{i}$ is regarded as the subgroup $G_{n-i}\times G_{i}$ via the embedding $g\mapsto\mathrm{diag}(1,g)$. We shall use the later expression when computing derivatives in Section 3.2. We also have an analogous expression for left derivatives.

We shall often use the following lemma:

Lemma 2.1**.**

Let $\pi\in\mathrm{Alg}(G_{n+1})$ and let $\pi^{\prime}$ be an admissible representation of $G_{n}$. Suppose there exists $i$ such that the following conditions hold:

[TABLE]

and

[TABLE]

for all $j=1,\ldots,i-1$ and all $k$. Then $\mathrm{Hom}_{G_{n}}(\pi,\pi^{\prime})\neq 0$.

Proof.

The Bernstein-Zelevinsky filtration of $\pi$ gives that there exists

[TABLE]

such that $\pi_{i-1}/\pi_{i}\cong\mathrm{ind}_{R_{n-i+1}}^{G_{n}}\pi^{[i]}\boxtimes\psi_{i}$. Now, by the second adjointness property of derivatives [CS18b, Lemma 2.2], we have

[TABLE]

for all $k$ and $j$. Now a long exact sequence argument gives that

[TABLE]

∎

2.4. Subrepresentation of a standard representation

Lemma 2.2**.**

Let $\mathfrak{m}\in\mathrm{Mult}$. Suppose all segments in $\mathfrak{m}$ are singletons i.e. of relative length $1$. Then $\lambda(\mathfrak{m})$ has unique irreducible submodule and quotient. Moreover, the unique irreducible submodule is generic.

Proof.

By definitions, $\lambda(\mathfrak{m})=\zeta(\mathfrak{m})$ and hence has unique submodule and quotient. Since all segments in $\mathfrak{m}$ are singletons, the submodule is generic [Ze80]. ∎

It is known (see [JS83]111The author would like to thank D. Prasad for mentioning this reference in a disucssion.) that $\lambda(\mathfrak{m})$ always has a generic representation as the unique submodule. We shall prove a slightly stronger statement, using Zelevinsky theory:

Proposition 2.3**.**

Let $\mathfrak{m}\in\mathrm{Mult}$. Then $\lambda(\mathfrak{m})$ can be embedded to $\lambda(\mathfrak{m}^{\prime})$ for some $\mathfrak{m}^{\prime}\in\mathrm{Mult}$ whose segments are singletons. In particular, $\lambda(\mathfrak{m})$ has a unique irreducible submodule and moreover, the submodule is generic.

Proof.

Let $\rho$ be a cuspidal representation in $\mathfrak{m}$ such that for any cuspidal representation $\rho^{\prime}$ in $\mathfrak{m}$, $\rho\not<\rho^{\prime}$. Let $\Delta$ be a segment in $\mathfrak{m}$ with the shortest relative length among all segments $\Delta^{\prime}$ in $\mathfrak{m}$ with $b(\Delta^{\prime})\cong\rho$.

By definition of $\lambda(\mathfrak{m})$, we have that

[TABLE]

On the other hand, we have that

[TABLE]

Thus we have that

[TABLE]

We now explain the last isomorphism (2.7), and it suffices to show

[TABLE]

To this end, we write $\mathfrak{m}=\left\{\Delta,\Delta_{1},\ldots,\Delta_{k},\Delta_{k+1},\ldots,\Delta_{s}\right\}$ such that

[TABLE]

and $b(\Delta_{j})\not\cong b(\Delta)$ for $j\geq k+1$. Then we have that

[TABLE]

where the second isomorphism follows from applying (2.2) few times. Here we need to use our choice of $\Delta$, which guarantees that $\Delta^{-}$ and $\Delta_{j}$ are unlinked for $j=1,\ldots,k$.

Now $\lambda(\mathfrak{m}\setminus\left\{\Delta\right\}+\Delta^{-})$ embeds to $\lambda(\mathfrak{m}^{\prime})$ by induction for some $\mathfrak{m}^{\prime}\in\mathrm{Mult}$ with all segments to be singletons. Thus this gives that $b(\Delta)\times\lambda(\mathfrak{m}\setminus\left\{\Delta\right\}+\Delta^{-})$ embeds to $b(\Delta)\times\lambda(\mathfrak{m}^{\prime})\cong\lambda(\mathfrak{m}^{\prime}+b(\Delta))$, and so does $\lambda(\mathfrak{m})$ by (2.6).

The second assertion follows from Lemma 2.2.

∎

2.5. Derivatives of generic representations

Recall that the socle (resp. cosocle) of an admissible representation $\pi$ of $G_{n}$ is the maximal semisimple submodule (resp. quotient) of $\pi$.

Lemma 2.4**.**

Let $\pi\in\mathrm{Irr}(G_{n})$. Then the cosocle of $\pi^{(i)}$ (resp. ${}^{(i)}\pi$) is isomorphic to the socle of $\pi^{(i)}$ (resp. ${}^{(i)}\pi$).

Proof.

This is almost the same as the proof of [CS18b, Lemma 2.2]. More precisely, it follows from that for an irreducible $G_{n}$-representation $\pi$,

[TABLE]

and the fact [BZ76] that $\theta_{n-i}(\tau)\cong\tau^{\vee}$ for any irreducible $G_{n-i}$-representation $\tau$. The last isomorphism follows from [CS18b, Lemma 2.2]. ∎

Proposition 2.5**.**

Let $\pi\in\mathrm{Irr}(G_{n+1})$. The socle and cosocle of $\pi^{(i)}$ (and ${}^{(i)}\pi$) are multiplicity-free.

Proof.

Let $\pi_{0}$ be an irreducible quotient of $\pi^{(i)}$. Let $\pi_{1}$ be a cuspidal representation of $G_{i-1}$ which is not in $\mathrm{cupp}_{\mathbb{Z}}(\pi_{0})$. Then, by comparing cuspidal supports,

[TABLE]

for all $j$ and $k<i$. With a long exact sequence argument using Bernstein-Zelevinsky filtration (similar to the proof of Lemma 2.1), we have that

[TABLE]

Now one applies the geometric lemma to obtain a filtration on ${}^{(i-1)}(\pi_{0}\times\pi_{1})$, and then by comparing cuspidal supports, the only possible layer that can contribute the above non-zero Hom is $\pi_{0}$. Hence,

[TABLE]

The first dimension is at most one by [AGRS10] and so is the second dimension. This implies the cosocle statement by Lemma 2.1 and the socle statement follows from Lemma 2.4. ∎

A representation $\pi$ of $G_{n}$ is called generic if $\pi^{(n)}\neq 0$. The Zelevinsky classification of irreducible generic representations is in [Ze80], that is, $\mathrm{St}(\mathfrak{m})$ is generic if and only if any two segments in $\mathfrak{m}$ are unlinked. With Proposition 2.5, the following result essentially gives a combinatorial description on the socle and cosocle of the derivatives of a generic representation.

Corollary 2.6**.**

Let $\pi\in\mathrm{Irr}(G_{n+1})$ be generic. Then any simple quotient and submodule of $\pi^{(i)}$ (resp. ${}^{(i)}\pi$) is generic.

Proof.

By Lemma 2.4, it suffices to prove the statement for quotient. Let $\mathfrak{m}=\left\{\Delta_{1},\ldots,\Delta_{r}\right\}$ be the Zelevinsky segment $\mathfrak{m}$ such that

[TABLE]

Since any two segments in $\mathfrak{m}$ are unlinked, we can label in any order and so we shall assume that for $i<j$, $b(\Delta_{j})$ does not precede $b(\Delta_{i})$. Then geometric lemma produces a filtration on $\pi^{(i)}$ whose successive subquotient is isomorphic to

[TABLE]

where $i_{1}+\ldots+i_{r}=i$. The last module is isomorphic to $\lambda(\mathfrak{m}^{\prime})^{\vee}$, where

[TABLE]

If $\pi^{\prime}$ is a simple quotient of $\pi^{(i)}$, then $\pi^{\prime}$ is a simple quotient of one successive subquotient in the filtration, or in other words, $\pi^{\prime}$ is a simple submodule of $\lambda(\mathfrak{m}^{\prime})$ for a multisegment $\mathfrak{m}^{\prime}$. Now the result follows from Proposition 2.3. ∎

Remark 2.7**.**

One can formulate the corresponding statement of Proposition 2.5 for affine Hecke algebra level by using the sign projective module in [CS19]. Then it might be interesting to ask for an analogue result for affine Hecke algebra over fields of positive characteristics.

Here we give a consequence to branching law (c.f. [GGP20], [Gu18], [Ch20]):

Corollary 2.8**.**

Let $\pi\in\mathrm{Irr}(G_{n+1})$ be generic. Let $\pi^{\prime}\in\mathrm{Irr}(G_{n})$ and let $\mathfrak{m}\in\mathrm{Mult}$ with $\pi^{\prime}\cong\langle\mathfrak{m}\rangle$. If $\mathrm{Hom}_{G_{n}}(\pi,\langle\mathfrak{m}\rangle)\neq 0$, then each segment in $\mathfrak{m}$ has relative length at most $2$.

Proof.

Write $\mathfrak{m}=\left\{\Delta_{1},\ldots,\Delta_{r}\right\}$ such that $\Delta_{i}$ does not precede $\Delta_{j}$ if $i<j$. Let $\pi_{0}=\langle\mathfrak{m}\rangle$. By using the Bernstein-Zelevinsky filtration, $\mathrm{Hom}_{G_{n}}(\pi,\pi_{0})\neq 0$ implies that

[TABLE]

for some $i\geq 1$ [CS18b] (c.f. Lemma 2.1). Hence Corollary 2.6 implies that ${}^{(i-1)}\pi_{0}$ is generic for some $i\geq 1$. Since

[TABLE]

we have ${}^{(i-1)}\pi_{0}\hookrightarrow{}^{(i-1)}\zeta(\mathfrak{m})$ and so ${}^{(i-1)}\zeta(\mathfrak{m})$ has a generic composition factor. On the other hand, geometric lemma gives that ${}^{(i-1)}\pi_{0}$ admits a filtration whose successive quotients are isomorphic to ${}^{(i_{1})}\langle\Delta_{1}\rangle\times\ldots\times{}^{(i_{r})}\langle\Delta_{r}\rangle$ where $i_{1},\ldots,i_{r}$ run for all sums equal to $i-1$. Then at least one such quotient is non-degenerate and so in that quotient, all ${}^{(i_{k})}\langle\Delta_{k}\rangle$ are cuspidal. Following from the derivatives on $\langle\Delta\rangle$, $\Delta$ can have at most of relative length $2$. ∎

Another consequence is on the indecomposability of derivatives of generic representations.

Corollary 2.9**.**

Let $\pi\in\mathrm{Irr}(G_{n+1})$ be generic. Then the projections of $\pi^{(i)}$ and ${}^{(i)}\pi$ to any cuspidal support component have unique simple quotient and submodule. In particular, the projections of $\pi^{(i)}$ and ${}^{(i)}\pi$ to any cuspidal support component are indecomposable.

Proof.

For a fixed cuspidal support, there is a unique (up to isomorphism) irreducible smooth generic representation. Now the result follows from Proposition 2.5 and Corollary 2.6. ∎

The analogous statement is not true in general if one replaces an irreducible generic representation by an arbitrary irreducible representation. We give an example below.

Example 2.10**.**

Let $\mathfrak{m}=\left\{[1,\nu],[\nu^{-1},1],[1],[\nu^{-1},\nu]\right\}$. We note that

[TABLE]

We compute the components of $\langle\mathfrak{m}\rangle^{(2)}$, at the cuspidal support $\left\{\nu^{-1},\nu^{-1},1,1,1,\nu\right\}$. In this case, there are four composition factors whose Zelevinsky multisegments are:

[TABLE]

[TABLE]

Note that the socle and cosocle coincide by Lemma 2.4. Note $\langle\mathfrak{m}_{1}\rangle$ and $\langle\mathfrak{m}_{3}\rangle$ are in socle (and so cosocle). $\langle\mathfrak{m}_{2}\rangle$ cannot be a submodule or quotient. Thus the only possible structure is two indecomposable modules. One of them has composition factors of $\langle\mathfrak{m}_{3}\rangle$ (with multiplicity 2) and $\langle\mathfrak{m}_{2}\rangle$. Another one is a simple module isomorphic to $\langle\mathfrak{m}_{1}\rangle$.

3. Projectivity

3.1. Projectivity Criteria

We need the following formula of D. Prasad:

Theorem 3.1**.**

[Pr18]** Let $\pi_{1}$ and $\pi_{2}$ be admissible representations of $\mathrm{GL}_{n+1}(F)$ and $\mathrm{GL}_{n}(F)$ respectively. Then

[TABLE]

where $\mathrm{Wh}(\pi_{1})=\pi_{1}^{(n+1)}$ and $\mathrm{Wh}(\pi_{2})=\pi_{2}^{(n)}$.

Lemma 3.2**.**

Let $\pi$ be an irreducible $G_{n+1}$-representation. If $\pi^{(i)}$ has a non-generic irreducible submodule or quotient for some $i$, then there exists a non-generic representation $\pi^{\prime}$ of $G_{n}$ such that $\mathrm{Hom}_{G_{n}}(\pi,\pi^{\prime})\neq 0$. The statement still holds if we replace $\pi^{(i)}$ by ${}^{(i)}\pi$.

Proof.

By Lemma 2.4, it suffices just to consider that $\pi^{(i)}$ has a non-generic irreducible quotient, say $\lambda$. Now let

[TABLE]

where $\tau$ is a cuspidal representation such that $\tau$ is not an unramified twist of a cuspidal representation appearing in any segment in $\mathfrak{m}$. Here $\mathfrak{m}$ is a multisegment with $\pi\cong\langle\mathfrak{m}\rangle$. Now

[TABLE]

for $j<n(\tau)$ and any $k$ since $\tau$ is in $\mathrm{cupp}({}^{(j-1)}\pi^{\prime})$ whenever it is nonzero while $\tau\not\in\mathrm{cupp}(\pi^{[j]})$. Moreover, ${}^{(n(\tau)-1)}\pi^{\prime}$ has a simple quotient isomorphic to $\nu^{1/2}\lambda$. This checks the Hom and Ext conditions in Lemma 2.1 and hence proves the lemma. The proof for ${}^{(i)}\pi$ is almost identical with switching left and right derivatives in suitable places. ∎

Theorem 3.3**.**

Let $\pi\in\mathrm{Irr}(G_{n+1})$. Then the following conditions are equivalent:

1. (1)

$\pi|_{G_{n}}$* is projective* 2. (2)

$\pi$* is generic and any irreducible quotient of $\pi|_{G_{n}}$ is generic.*

Proof.

For (2) implying (1), it is proved in [CS18b]. We now consider $\pi$ is projective. All higher Exts vanish and so $\mathrm{EP}(\pi,\pi^{\prime})=\mathrm{dim}~{}\mathrm{Hom}_{G_{n}}(\pi,\pi^{\prime})$ for any irreducible $\pi^{\prime}$ of $G_{n}$. If $\pi^{\prime}$ is an irreducible quotient of $\pi$, then $\mathrm{EP}(\pi,\pi^{\prime})\neq 0$ and hence $\pi$ is generic by Theorem 3.1. But Theorem 3.1 also implies $\pi^{\prime}$ is generic. This proves (1) implying (2).

∎

3.2. Classification

Definition 3.4**.**

We say that an irreducible representation $\pi$ of $G_{n+1}$ is relatively projective if either one of the following conditions holds:

1. (i)

$\pi$ is essentially square-integrable; 2. (ii)

$\pi$ is isomorphic to $\pi_{1}\times\pi_{2}$ for some cuspidal representations $\pi_{1},\pi_{2}$ of $G_{(n+1)/2}$ with $\pi_{1}\not\cong\nu^{\pm 1}\pi_{2}$.

In particular, a relatively-projective representation is generic. The condition $\pi_{1}\not\cong\nu^{\pm 1}\pi_{2}$ is in fact automatic from $\pi$ being irreducible.

We can formulate the conditions (i) and (ii) combinatorially as follows. Let $\pi\cong\mathrm{St}(\mathfrak{m})$ for a multisegment $\mathfrak{m}=\left\{\Delta_{1},\ldots,\Delta_{r}\right\}$. Then (i) is equivalent to $r=1$; and (ii) is equivalent to that $r=2$, and $\Delta_{1}$ and $\Delta_{2}$ are not linked, and the relative lengths of $\Delta_{1}$ and $\Delta_{2}$ are both $1$.

Lemma 3.5**.**

Let $\pi\in\mathrm{Irr}(G_{n+1})$ be not relatively projective. Then there exists an irreducible non-generic representation $\pi^{\prime}$ of $G_{n}$ such that $\mathrm{Hom}_{G_{n}}(\pi,\pi^{\prime})\neq 0$.

Proof.

It suffices to construct an irreducible non-generic representation $\pi^{\prime}$ satisfying the Hom and Ext properties in Lemma 2.1.

Let $\mathfrak{m}=\left\{\Delta_{1},\ldots,\Delta_{r}\right\}$ be a multisegment such that $\pi\cong\mathrm{St}(\mathfrak{m})$. We divide into few cases.

Case 1: $r\geq 3$; or when $r=2$, each segment has relative length at least $2$; or when $r=2$, $\Delta_{1}\cap\Delta_{2}=\emptyset$. We choose a segment $\Delta^{\prime}$ in $\mathfrak{m}$ with the shortest absolute length. Now we choose a maximal segment $\Delta$ in $\mathfrak{m}$ with the property that $\Delta^{\prime}\subset\Delta$. By genericity, $\nu^{-1}a(\Delta)\notin\Delta_{k}$ for any $\Delta_{k}\in\mathfrak{m}$. Let

[TABLE]

where $\tau$ is a cuspidal representation so that $\mathrm{St}(\mathfrak{m}^{\prime})$ is a representation of $G_{n}$ and $\tau$ is not an unramified twist of any cuspidal representation appearing in a segment of $\mathfrak{m}$. To make sense of the construction, it needs the choices and the assumptions on this case. Let $k=n(a(\Delta)),l=n(\tau)$. Let

[TABLE]

Now as for $i<k+l+1$, either $\nu^{-1/2}a(\Delta)$ or $\tau$ appears in the cuspidal support of ${}^{(i-1)}\pi$, but not in that of $\nu^{1/2}\pi^{(i)}$,

[TABLE]

and all $j$. By Corollary 2.6, $\nu^{1/2}\cdot\pi^{(k+l+1)}$ has a simple generic quotient isomorphic to $\nu^{1/2}\mathrm{St}({}^{-}\Delta)$. On the other hand, a submodule structure of ${}^{(k+l)}\pi^{\prime}$ can be computed as follows:

[TABLE]

Here the non-zeroness comes from the fact that $\mathrm{St}(\mathfrak{m}^{\prime})$ is the unique quotient of $\lambda(\mathfrak{m}^{\prime})$, and the isomorphism follows from Frobenius reciprocity. Since taking the derivative is an exact functor, we have that $\mathrm{St}(\nu^{1/2}\cdot{}^{-}\Delta)$ is a subrepresentation of ${}^{(k+l)}\mathrm{St}(\mathfrak{m}^{\prime})$ (see Section 2.3). Thus we have

[TABLE]

Case 2: $r=2$ with $\Delta_{1}\cap\Delta_{2}\neq\emptyset$ and one segment having relative length $1$ (and not both having relative length $1$ by the definition of relatively-projective type). By switching the labeling on segments if necessary, we assume that $\Delta_{1}\subset\Delta_{2}$. Let $p$ and let $l$ be the absolute and relative length of $\Delta_{2}$ respectively. Let

[TABLE]

where $\tau$ is a cuspidal representation of $G_{k}$ (here $k$ is possibly zero) so that $\mathrm{St}(\mathfrak{m}^{\prime})$ is a $G_{n}$-representation. Note that $\pi^{\prime}$ is non-generic. By Corollary 2.6 and geometric lemma, a simple quotient $\nu^{1/2}\cdot\pi^{(p)}$ is isomorphic to $\nu^{1/2}a(\Delta_{1})$. Similar computation as in (3.8) gives that a simple module of ${}^{(p-1)}\pi^{\prime}$ is isomorphic to $\nu^{1/2}a(\Delta_{1})$. This implies the non-vanishing $\mathrm{Hom}$ between those two $G_{n+1-p}$-representations.

We now prove the vanishing $\mathrm{Ext}$-groups in order to apply Lemma 2.1. Now applying the Bernstein-Zelevinsky derivatives $(j=1,\ldots,p-1)$, unless $a(\Delta_{1})\cong b(\Delta_{2})$, we have that $\nu^{3/2-l}a(\Delta_{1})$ is a cuspidal support for ${}^{(j-1)}\pi^{\prime}$ whenever ${}^{(j-1)}\pi^{\prime}$ is nonzero and is not a cuspidal support for $\nu^{1/2}\pi^{(j)}$. It remains to consider $a(\Delta_{1})\cong b(\Delta_{2})$. We can similarly consider the cuspidal support for $\nu^{-1/2}a(\Delta_{1})$ and $\nu^{1/2}a(\Delta_{1})$ to make conclusion. ∎

Lemma 3.6**.**

Let $\pi\in\mathrm{Irr}(G_{n+1})$. If $\pi$ is (generic) relatively-projective, then $\pi|_{G_{n}}$ is projective.

Proof.

When $\pi$ is essentially square-integrable, it is proved in [CS18b]. We now assume that $\pi$ is in the case (2) of Definition 3.4. It is equivalent to prove the condition (2) in Theorem 3.3. Let $\pi^{\prime}\in\mathrm{Irr}(G_{n})$ with $\mathrm{Hom}_{G_{n}}(\pi,\pi^{\prime})\neq 0$. We have to show that $\pi^{\prime}$ is generic. Note that the only non-zero derivative of $\pi^{(i)}$ can occur when $i=n+1$ and $\frac{n+1}{2}$.

Case 1: $\mathrm{Hom}_{G_{(n+1)/2}}(\pi^{[(n+1)/2]},{}^{((n-1)/2)}\pi^{\prime})\neq 0$. For (1), let

[TABLE]

for some irreducible cuspidal representations $\rho_{1},\rho_{2}$ of $G_{(n+1)/2}$ with $\rho_{1}\not\cong\nu^{\pm 1}\rho_{2}$, and

[TABLE]

By a simple count on dimensions, we must have $\Delta_{k}^{\prime}\cong\nu^{1/2}\rho_{1}$ or $\cong\nu^{1/2}\rho_{2}$ for some $k$. Using dimensions again, we have for $l\neq k$, $[\rho_{1}]$ and $[\rho_{2}]$ are unlinked to $\Delta_{l}^{\prime}$ and so

[TABLE]

for $r=1$ or $2$. Then

[TABLE]

which implies

[TABLE]

Thus $\mathrm{St}(\mathfrak{m}^{\prime}\setminus\left\{\Delta_{k}^{\prime}\right\})$ is generic and so is $\pi^{\prime}$.

Case 2: $\mathrm{Hom}_{G_{(n+1)/2}}(\pi^{[(n+1)/2]},{}^{(n-1)/2}\pi^{\prime})=0$. We must have

[TABLE]

and so ${}^{(n)}\pi^{\prime}\neq 0$. Hence $\pi^{\prime}$ is generic. ∎

We now achieve the classification of irreducible representations which are projective when restricted from $G_{n+1}$ to $G_{n}$:

Theorem 3.7**.**

Let $\pi\in\mathrm{Irr}(G_{n+1})$. Then $\pi|_{G_{n}}$ is projective if and only if $\pi$ is relatively projective in Definition 3.4.

Proof.

The if direction is proved in Lemma 3.6. The only if direction follows from Lemma 3.5 and Theorem 3.1. ∎

One advantage for such classification is that those restricted representations admit a more explicit realization as shown in [CS18b]:

Theorem 3.8**.**

Let $\pi,\pi^{\prime}\in\mathrm{Irr}(G_{n+1})$. If $\pi$ and $\pi^{\prime}$ are relatively projective, then $\pi|_{G_{n}}\cong\pi^{\prime}|_{G_{n}}$. In particular, $\pi|_{G_{n}}$ is isomorphic to the Gelfand-Graev representation $\mathrm{ind}_{U_{n}}^{G_{n}}\psi_{n}$.

Proof.

We have shown that $\pi$ and $\pi^{\prime}$ have to be generic. Then we apply [CS18b, Corollary 5.5 and Theorem 5.6]. ∎

4. Submodule structure of Bernstein-Zelevinsky filtrations

Several insights come from the affine Hecke algebra realization of Gelfand-Graev representations. We shall first recall those results.

4.1. Affine Hecke algebras

Definition 4.1**.**

The affine Hecke algebra $\mathcal{H}_{l}(q)$ of type $A$ is an associative algebra over $\mathbb{C}$ generated by $\theta_{1},\ldots,\theta_{l}$ and $T_{w}$ ($w\in S_{l}$) satisfying the relations:

1. (1)

$\theta_{i}\theta_{j}=\theta_{j}\theta_{i}$; 2. (2)

$T_{s_{k}}\theta_{k}-\theta_{k+1}T_{s_{k}}=(q-1)\theta_{k}$, where $q$ is a certain prime power and $s_{k}$ is the transposition between the numbers $k$ and $k+1$; 3. (3)

$T_{s_{k}}\theta_{i}=\theta_{i}T_{s_{k}}$, where $i\neq k,k+1$ 4. (4)

$(T_{s_{k}}-q)(T_{s_{k}}+1)=0$; 5. (5)

$T_{s_{k}}T_{s_{k+1}}T_{s_{k}}=T_{s_{k+1}}T_{s_{k}}T_{s_{k+1}}$.

Let $\mathcal{A}_{l}(q)$ be the (commutative) subalgebra generated by $\theta_{1},\ldots,\theta_{l}$. Let $\mathcal{H}_{W,l}(q)$ be the subalgebra generated by $T_{s_{1}},\ldots,T_{s_{l-1}}$. Let $\mathrm{sgn}$ be the $1$-dimensional $\mathcal{H}_{W,l}(q)$-module characterized by $T_{s_{k}}$ acting by $-1$.

It is known from [Lu89, Proposition 3.11] that the center $\mathcal{Z}_{l}$ of $\mathcal{H}_{l}(q)$ has a basis $\left\{z_{M}=\sum_{w\in S_{l}}\theta_{1}^{i_{w(1)}}\ldots\theta_{l}^{i_{w(l)}}\right\}_{M}$, where $M=(i_{1},\ldots,i_{l})$ runs for all $l$-tuples in $\mathbb{Z}^{l}/S_{l}$.

Bernstein decomposition asserts that

[TABLE]

where $\mathfrak{R}(G_{n})$ is the category of smooth $G_{n}$-representations, $\mathfrak{B}(G_{n})$ is the set of inertial equivalence classes of $G_{n}$ and $\mathfrak{R}_{\mathfrak{s}}(G_{n})$ is the full subcategory of $\mathfrak{R}(G_{n})$ associated to $\mathfrak{s}$ (see [BK00]). For a smooth representation $\pi$ of $G_{n}$, define $\pi_{\mathfrak{s}}$ to be the projection of $\pi$ to the component $\mathfrak{R}_{\mathfrak{s}}(G_{n})$.

For each $\mathfrak{s}\in\mathfrak{B}(G_{n})$, [BK93] and [BK00] associate with a compact group $K_{\mathfrak{s}}$ and a finite-dimensional representation $\tau$ of $K_{\mathfrak{s}}$, such that the convolution algebra

[TABLE]

is isomorphic to the product $\mathcal{H}_{n_{1}}(q_{1})\otimes\ldots\otimes\mathcal{H}_{n_{r}}(q_{r})$ of affine Hecke algebra of type $A$, denoted by $\mathcal{H}_{\mathfrak{s}}$. For a smooth representation $\pi$ of $G_{n}$, the algebra $\mathcal{H}(K_{\mathfrak{s}},\tau)$ acts naturally on the space $\mathrm{Hom}_{K_{\mathfrak{s}}}(\tau,\pi)\cong(\tau^{\vee}\otimes\pi)^{K_{\mathfrak{s}}}$. This defines an equivalence of categories:

[TABLE]

By abuse notation, we shall identify $\mathrm{Hom}_{K_{\mathfrak{s}}}(\tau,\pi)$ with $\pi_{\mathfrak{s}}$ under (4.10) and consider $\pi_{\mathfrak{s}}$ as $\mathcal{H}_{\mathfrak{s}}$-module.

Let $\mathcal{A}_{\mathfrak{s}}=\mathcal{A}_{n_{1}}(q_{1})\otimes\ldots\otimes\mathcal{A}_{n_{r}}(q_{r})$. Let $\mathcal{H}_{W,\mathfrak{s}}=\mathcal{H}_{W,n_{1}}(q_{1})\otimes\ldots\otimes\mathcal{H}_{W,n_{r}}(q_{r})$. Let

[TABLE]

as an $\mathcal{H}_{W,\mathfrak{s}}$-module. Note that in [CS19], we proved when $\mathfrak{s}$ is a simply type, but the generalization to all types follows from [BK00] and a simple generalization of [CS19, Theorem 2.1]. The center of $\mathcal{H}_{\mathfrak{s}}$ is equal to $\mathcal{Z}_{1}\otimes\ldots\otimes\mathcal{Z}_{r}$, where each $\mathcal{Z}_{k}$ is the center of $\mathcal{H}_{n_{k}}(q_{k})$.

Recall that $\Pi_{n}=\mathrm{ind}_{U_{n}}^{G_{n}}\psi_{n}$. We may simply write $\Pi$ for $\Pi_{n}$ if there is no confusion.

Theorem 4.2**.**

[CS19]** For any $\mathfrak{s}\in\mathfrak{B}(G_{n})$, the Bernstein component of the Gelfand-Graev representation $\Pi_{\mathfrak{s}}$ is isomorphic to $\mathcal{H}_{\mathfrak{s}}\otimes_{\mathcal{H}_{W,\mathfrak{s}}}\mathrm{sgn}_{\mathfrak{s}}$.

Lemma 4.3**.**

Let $\pi\in\mathrm{Alg}(G_{n})$ be admissible. Then

[TABLE]

Proof.

It suffices to prove for $\pi\in\mathrm{Irr}(G_{n})$. Let $\mathfrak{s}\in\mathfrak{B}(G_{n})$ such that $\pi\in\mathfrak{R}_{\mathfrak{s}}(G_{n})$. Now $\pi_{\mathfrak{s}}$ is an irreducible finite-dimensional $\mathcal{H}_{\mathfrak{s}}$-module, but there is no finite dimensional submodule for

[TABLE]

since there is no finite-dimensional submodule of $\mathcal{A}_{\mathfrak{s}}$ as $\mathcal{A}_{\mathfrak{s}}$-module. Hence the Hom space is zero. ∎

Remark 4.4**.**

One can generalize Lemma 4.3 to a connected quasisplit reductive group $G$ with a non-compact center.222This is also pointed out by the referee. The author would like to thank the referee for that. For an irreducible $\pi\in\mathfrak{R}_{\mathfrak{s}}(G)$,

[TABLE]

where $D$ is the Aubert-Schneider-Stuhler-Zelevinsky dual, and $d(\mathfrak{s})$ is the cohomological dimension of the Bernstein block $\mathfrak{R}_{\mathfrak{s}}(G)$. The last isomorphism follows from [NP20] (also see [SS97, Ch16]). Since $G$ has non-compact center, $d(\mathfrak{s})\geq 1$. Hence, last Ext is zero by the projectivity of $\Pi$ [CS18]. The admissible case follows from the irreducible case.

4.2. Inertial equivalence classes

For the following discussions, see, for example, [Be92] and [BK00]. An inertial equivalence class $\mathfrak{s}$ of $G_{n}$ can be represented by a pair $[G_{m_{1}}\times\ldots\times G_{m_{r}},\rho_{1}\boxtimes\ldots\boxtimes\rho_{r}]$, where $m_{1}+\ldots+m_{r}=n$ and each $\rho_{k}$ is a cuspidal $G_{m_{k}}$-representation. Two pairs

[TABLE]

represent the same inertial equivalence class if and only if $r=s$ and there exists a permutation $\sigma\in S_{r}$ such that

[TABLE]

and

[TABLE]

for some unramified character $\chi_{k}$ of $G_{m_{k}}$ ($k=1,\ldots,r$).

Let

[TABLE]

[TABLE]

Then $\pi_{1}\times\pi_{2}$ lies in $\mathfrak{R}_{\mathfrak{s}}(G_{n_{1}+n_{2}})$, where

[TABLE]

From this, one deduces the following two lemmas:

Lemma 4.5**.**

Let $\pi_{1}\in\mathfrak{R}_{\mathfrak{s}}(G_{n_{1}})$ for some $\mathfrak{s}\in\mathfrak{B}(G_{n_{1}})$ and let $\pi_{2}\in\mathrm{Alg}(G_{n_{2}})$. Fix $\mathfrak{t}\in\mathfrak{B}(G_{n_{1}+n_{2}})$ with $(\pi_{1}\times\pi_{2})_{\mathfrak{t}}\neq 0$. There exists a unique $\mathfrak{s}^{\prime}\in\mathfrak{B}(G_{n_{1}+n_{2}})$ such that $\pi_{1}\times(\pi_{2})_{\mathfrak{s}^{\prime}}\cong(\pi_{1}\times\pi_{2})_{\mathfrak{t}}$.

Lemma 4.6**.**

Let $\pi_{1}\in\mathrm{Alg}(G_{n_{1}})$ and let $\pi_{2}\in\mathrm{Alg}(G_{n_{2}})$ with $n_{1},n_{2}\neq 0$. Then any irreducible subquotient of $\pi_{1}\times\pi_{2}$ is not cuspidal.

4.3. Indecomposability of Gelfand-Graev representations

Proposition 4.7**.**

For any $\mathfrak{s}\in\mathfrak{B}(G_{n})$, any two non-zero submodules $\pi_{1},\pi_{2}$ in $(\mathrm{ind}_{U_{n}}^{G_{n}}\psi_{n})_{\mathfrak{s}}$ has non-zero intersection. In particular, $(\mathrm{ind}_{U_{n}}^{G_{n}}\psi_{n})_{\mathfrak{s}}$ is indecomposable.

Proof.

By Theorem 4.2,

[TABLE]

Since $\mathcal{A}_{\mathfrak{s}}$ is commutative, any two non-zero $\mathcal{A}_{\mathfrak{s}}$-submodules of $\mathcal{A}_{\mathfrak{s}}$ have non-zero intersection. Hence any two non-zero submodules of $\Pi_{\mathfrak{s}}$ also have non-zero intersection. ∎

Remark 4.8**.**

We give another proof for the indecomposability of $\Pi_{\mathfrak{s}}$ as below, whose argument can be applied to other connected quasisplit reductive groups $G$. For any $\mathfrak{s}\in\mathfrak{R}(G)$, [BH03] showed that $\mathrm{End}_{G}(\Pi_{\mathfrak{s}})$ is isomorphic to the Bernstein center $\mathfrak{Z}_{\mathfrak{s}}$ of $\mathfrak{R}_{\mathfrak{s}}(G)$. Since $\mathfrak{R}_{\mathfrak{s}}(G)$ is an indecomposable category, we also have $\mathfrak{Z}_{\mathfrak{s}}$ is indecomposable as a $\mathfrak{Z}_{\mathfrak{s}}$-module. This implies that $\mathrm{End}_{G}(\Pi_{\mathfrak{s}})$ is indecomposable as a $\mathfrak{Z}_{\mathfrak{s}}$-module and hence $\Pi_{\mathfrak{s}}$ is indecomposable in $\mathfrak{R}_{\mathfrak{s}}(G)$.

4.4. Jacquet functors on Gelfand-Graev representations

Lemma 4.9**.**

Let $P=LN$ be the parabolic subgroup containing upper triangular matrices and block-diagonal matrices $\mathrm{diag}(g_{1},\ldots,g_{r})$ with $g_{k}\in G_{i_{k}}$, where $i_{1}+\ldots+i_{r}=n$. Then $(\Pi_{n})_{N}\cong\Pi_{i_{1}}\boxtimes\ldots\boxtimes\Pi_{i_{r}}$.

Proof.

Let $w$ be a permutation matrix in $G_{n}$. Then $w(N)\cap U_{n}$ contains a unipotent subgroup $\left\{I_{n}+tu_{k,k+1}:t\in F\right\}$ for some $k$ if and only if $w(N)\not\subset U_{n}^{-}$. Here $u_{k,k+1}$ is a matrix with $(k,k+1)$-entry $1$ and other entries [math]. For any such $w$, it gives that $PwB$ is the same unique open orbit in $G_{n}$. Now the geometric lemma in [BZ77, Theorem 5.2] gives the lemma. ∎

4.5. Some lemmas

For $\pi\in\mathrm{Irr}$, let $\mathrm{csupp}^{\prime}(\pi)$ be the cuspidal support of $\pi$ but without multiplicities, and in particular, $\mathrm{csupp}^{\prime}(\pi)$ is a set. For an admissible representation $\pi$ of $G_{n}$, let

[TABLE]

Lemma 4.10**.**

Let $\omega\in\mathrm{Alg}(G_{n_{1}})$ be admissible with $n_{1}\neq 0$. Then, for any irreducible subquotient $\lambda$ of $\Pi_{n_{2}}\times\omega$, $\mathrm{csupp}^{\prime}(\lambda)\cap\mathrm{csupp}^{\prime}(\omega)\neq\emptyset$. Similarly, for any irreducible subquotient $\lambda$ of $\omega\times\Pi_{n_{2}}$, $\mathrm{csupp}^{\prime}(\lambda)\cap\mathrm{csupp}^{\prime}(\omega)\neq\emptyset$.

Proof.

We only sketch the proof for the first statement. Suppose $\lambda$ is an irreducible subquotient of $\Pi_{n_{2}}\times\omega$. We consider that

[TABLE]

for some cuspidal representations $\rho_{1},\ldots,\rho_{r}$. The adjointness gives a Jacquet functor associated to a unipotent subgroup $N$:

[TABLE]

Using the geometric lemma on $(\Pi_{n_{2}}\times\omega)_{N}$ to obtain a filtration, we then use $\rho_{1}\boxtimes\ldots\boxtimes\rho_{r}$ appears in a subquotient of $(\Pi_{n_{2}}\times\omega)_{N}$ (also see Lemma 9.1) to obtain that $\rho_{i}\in\mathrm{csupp}^{\prime}(\omega)$ for some some $i$. ∎

Corollary 4.11**.**

We use the same setting as in the previous lemma. Let $\pi$ be a submodule of $\Pi_{n_{2}}\times\omega$, and let $\pi^{\prime}$ be a submodule of $\omega\times\Pi_{n_{2}}$. Then

1. (1)

for any irreducible subquotient $\tau$ of $\pi$, $\mathrm{csupp}^{\prime}(\tau)\cap\mathrm{csupp}^{\prime}(\omega)\neq\emptyset$; and 2. (2)

for any irreducible subquotient $\tau^{\prime}$ of $\pi^{\prime}$, $\mathrm{csupp}^{\prime}(\tau^{\prime})\cap\mathrm{csupp}^{\prime}(\omega)\neq\emptyset$.

4.6. Bernstein center

The Bernstein center of a category $\mathfrak{R}$ is defined as the endomorphism ring of the identity functor in $\mathfrak{R}$. Denote by $\mathfrak{Z}_{n}$ the Bernstein center of $\mathfrak{R}(G_{n})$. For $\mathfrak{s}\in\mathfrak{B}(G_{n})$, denote by $\mathfrak{Z}_{\mathfrak{s}}$ the Bernstein center of $\mathfrak{R}_{\mathfrak{s}}(G_{n})$, which is a finitely-generated commutative algebra [BD84, Theorem 2.13]. We have

[TABLE]

It follows from [BD84, Corollaire 3.4] that for any maximal ideal $\mathcal{J}$ of $\mathfrak{Z}_{n}$ and any finitely-generated $\pi$ of $G_{n}$, $\pi/\mathcal{J}(\pi)$ is admissible. For more properties of the Bernstein center, see, e.g. [BD84], [BH03], [Ha14].

Lemma 4.12**.**

Let $\omega\in\mathrm{Alg}(G_{n_{1}})$ be admissible. Fix $\tau\in\mathrm{Irr}(G_{n_{1}+n_{2}})$ such that $\mathrm{csupp}^{\prime}(\omega)\cap\mathrm{csupp}^{\prime}(\tau)=\emptyset$. Let $\mathcal{J}$ be the maximal ideal in $\mathfrak{Z}_{n_{1}+n_{2}}$ which annihilates $\tau$.

1. (1)

Let $\pi$ be a submodule of $\Pi_{n_{2}}\times\omega$. Then $\pi/\mathcal{J}(\pi)=0$. 2. (2)

Let $\pi^{\prime}$ be a submodule of $\omega\times\Pi_{n_{2}}$. Then $\pi^{\prime}/\mathcal{J}(\pi^{\prime})=0$.

Proof.

We only prove (1) and a proof for (2) is similar. Each Bernstein component of $\Pi$ is finitely generated [BH03] and so each Bernstein component of $\pi$ is also finitely-generated [BD84]. Thus $\pi/\mathcal{J}(\pi)$ is admissible [BD84] and is annihilated by $\mathcal{J}$. If $\pi/\mathcal{J}(\pi)$ is non-zero, then $\pi$ has a subquotient with irreducible composition factors isomorphic to $\tau^{\prime}$ with $\mathrm{csupp}(\tau^{\prime})=\mathrm{csupp}(\tau)$ [BD84, Proposition 2.11]. However, this contradicts Corollary 4.11(1). ∎

4.7. Geometric lemma on $\omega\times\Pi_{i}$

Let $\pi_{1}\in\mathrm{Alg}(G_{n_{1}})$ and let $\pi_{2}\in\mathrm{Alg}(G_{n_{2}})$. Consider $N_{j}\subset G_{n_{1}+n_{2}}$ for some $j$. The geometric lemma [BZ77] gives that $(\pi_{1}\times\pi_{2})_{N_{j}}$ admits a filtration:

[TABLE]

such that, for each $r-1\geq k\geq 0$,

[TABLE]

where $P$ is the parabolic subgroup in $G_{n-j}\times G_{j}$ containing matrices $\mathrm{diag}(g_{1},g_{2},g_{3},g_{4})$ with $g_{1}\in G_{n_{1}-k}$, $g_{2}\in G_{n_{2}-j+k}$, $g_{3}\in G_{k}$ and $g_{4}\in G_{j-k}$, and upper triangular matrices, and $((\pi_{1})_{N_{k}}\boxtimes(\pi_{2})_{N_{j-k}})^{w}$ is a $P$-representation with underlying space $(\pi_{1})_{N_{k}}\boxtimes(\pi_{2})_{N_{j-k}}$ such that the action of the unipotent part is trivial and the action of the Levi part $G_{n_{1}-k}\times G_{n_{2}-j+k}\times G_{k}\times G_{j-k}$ is given by:

[TABLE]

where $v_{1}\in(\pi_{1})_{N_{k}}$ and $v_{2}\in(\pi_{2})_{N_{j-k}}$. Here we consider the space to be zero if $k>n_{1}$ or $j-k>n_{2}$.

Lemma 4.13**.**

Let $\omega\in\mathrm{Alg}(G_{n-i})$ be admissible. Let $\tau\in\mathrm{Irr}(G_{i})$ such that $\mathrm{csupp}^{\prime}(\omega)\cap\mathrm{csupp}^{\prime}(\tau)=\emptyset$. Let $\mathcal{J}^{\prime}$ be the maximal ideal in $\mathfrak{Z}_{i}$ which annihilates $\tau$, and set $\mathcal{J}=\mathcal{J}^{\prime}\otimes\mathfrak{Z}_{n-i}\subset\mathfrak{Z}_{i}\otimes\mathfrak{Z}_{n-i}$. Set $N=N_{n-i}$. Then $(\omega\times\Pi_{i})_{N}$ admits, as $G_{i}\times G_{n-i}$-representations, a short exact sequence

[TABLE]

such that

1. (1)

$\kappa_{2}/\mathcal{J}(\kappa_{2})=0$, and 2. (2)

$\kappa_{1}/\mathcal{J}(\kappa_{1})$* is admissible and is isomorphic to $\tau\boxtimes\omega$ for some $\tau\in\mathrm{Alg}(G_{i})$.*

Proof.

We explicate the filtration (4.11) in this situation. The bottom layer takes the form

[TABLE]

(Note that both $N_{n-i}$ on $\omega$ and $N_{0}$ on $\Pi_{i}$ are trivial groups.) By applying $\mathcal{J}$ on $\kappa_{1}$, we obtain (2).

Now set $\kappa_{2}=\pi/\kappa_{1}$. In order to prove $\kappa_{2}/\mathcal{J}(\kappa_{2})=0$, we continue to consider the filtration (4.11). Now $\kappa_{2}$ admits a filtration whose successive quotients take the form:

[TABLE]

for $k<n-i$. Since $\omega$ is admissible, $(\omega)_{N_{k}}$ admits a (finite) filtration of irreducible $G_{n-i-k}\times G_{k}$-representations. Thus we obtain a finite filtration of $\kappa_{2}$ whose successive quotients take the form

[TABLE]

where $\tau^{\prime}\in\mathrm{Irr}(G_{n-k-i})$ and $\tau^{\prime\prime}\in\mathrm{Irr}(G_{k})$ satisfying

[TABLE]

Note that $G_{n-k-i}$ is not the trivial group (i.e. $n-k-i\neq 0$) and $\mathrm{csupp}^{\prime}(\tau^{\prime})\cap\mathrm{csupp}^{\prime}(\tau)=\emptyset$, and so by Lemma 4.12, $(\tau^{\prime}\times\Pi_{2i+k-n})/\mathcal{J}^{\prime}(\tau^{\prime}\times\Pi_{2i+k-n})=0$. Thus, successive quotients in (4.12) give

[TABLE]

Thus we have proved that $\kappa_{2}$ admits a filtration

[TABLE]

such that $Q_{k}/\mathcal{J}(Q_{k})=0$, where $Q_{k}=F_{k}/F_{k-1}$ for $k=1,\ldots,l$. Now the short exact sequence

[TABLE]

gives a right exact sequence

[TABLE]

Thus inductively, we have $F_{k}/\mathcal{J}(F_{k})=0$ for all $k$, including $k=l$. This proves (1).

∎

Lemma 4.14**.**

Keep the same setting as the previous lemma. Let $\pi$ be a non-zero submodule of $\omega\times\Pi_{i}$. Then $\pi_{N}$, as a $G_{i}\times G_{n-i}$-representation, admits a short exact sequence:

[TABLE]

such that $\kappa_{2}^{\prime}/\mathcal{J}(\kappa_{2}^{\prime})=0$ and moreover, $\kappa_{1}^{\prime}$ is a submodule of $\Pi_{i}\boxtimes\omega$.

Proof.

We use the notations in the proof of Lemma 4.13. Thus we have a short exact sequence for $(\omega\times\Pi_{i})_{N}$. Hence, intersecting with $\pi$, we have a short exact sequence for $\pi_{N}$:

[TABLE]

We can then further obtain a fine filtration on $\kappa_{2}^{\prime}$ with each successive quotient $Q^{\prime}$ satisfying:

[TABLE]

(see (4.12)). Since $\mathrm{csupp}^{\prime}(\tau)\cap\mathrm{csupp}^{\prime}(\tau^{\prime})=\emptyset$, one can follow the argument in Lemmas 4.10 and 4.12 to conclude that $Q^{\prime}/\mathcal{J}(Q^{\prime})=0$.

∎

We also derive another consequences, and later on we will remove an assumption of the following lemma in Corollary 4.22.

Lemma 4.15**.**

Keep the same setting as the previous two lemmas. Again let $\pi$ be a non-zero submodule of $\omega\times\Pi_{i}$. Suppose $\pi_{N}/\mathcal{J}(\pi_{N})\neq 0$. Then there exists a non-zero submodule $\tau$ of $\pi_{N}$ such that

1. (1)

$\tau\hookrightarrow\Pi_{i}\boxtimes\omega$, and 2. (2)

regard $\tau$ as $G_{n-i}$-representation via the embedding $g\mapsto\mathrm{diag}(I_{i},g)$,

[TABLE]

for some submodule $\omega^{\prime}$ of $\omega$.

Proof.

We keep using the notations in the proof of the previous lemma. Note that (2) follows from (1). For (1), it suffices to show that $\kappa_{1}^{\prime}$ in Lemma 4.14 is non-zero. Indeed, we have the following right exact sequence:

[TABLE]

The last term is zero by Lemma 4.14. Now $\pi_{N}/\mathcal{J}(\pi_{N})\neq 0$ implies $(\kappa_{1}^{\prime})/\mathcal{J}(\kappa_{1}^{\prime})\neq 0$. Hence $\kappa_{1}^{\prime}\neq 0$. ∎

4.8. Geometric lemma on $\Pi_{i}\times\omega$

We can obtain a parallel version of Lemma 4.15 for $(\Pi_{i}\times\omega)_{N_{n-i}}$. Since we do not really need that, we only state without a proof.

Lemma 4.16**.**

Let $\omega\in\mathrm{Alg}(G_{n-i})$ be admissible. Let $\tau\in\mathrm{Irr}(G_{i})$ such that $\mathrm{csupp}^{\prime}(\omega)\cap\mathrm{csupp}^{\prime}(\tau)=\emptyset$. Let $\mathcal{J}^{\prime}$ be the maximal ideal in $\mathfrak{Z}_{i}$ which annihilates $\tau$, and set $\mathcal{J}=\mathcal{J}^{\prime}\otimes\mathfrak{Z}_{n-i}\in\mathfrak{Z}_{i}\otimes\mathfrak{Z}_{n-i}$. Set $N=N_{n-i}$. Then $(\Pi_{i}\times\omega)_{N}$ admits a short exact sequence:

[TABLE]

such that $\eta_{1}/\mathcal{J}(\eta_{1})=0$. Moreover, suppose $\pi$ is a submodule of $\Pi_{i}\times\omega$. If $\pi_{N}/\mathcal{J}(\pi_{N})\neq 0$, then $\pi\cap\eta_{1}\neq\pi$.

We now prove a lemma that is needed later:

Lemma 4.17**.**

Let $\omega\in\mathrm{Alg}(G_{n-i})$ be admissible. Let $\pi$ be a submodule of $\Pi_{i}\times\omega$. Set $N=N_{n-i}$. Then there exists a short exact sequence on $\pi_{N}$:

[TABLE]

such that, as $G_{i}\times G_{n-i}$-representations,

1. (1)

$\eta_{2}$* embeds to $\Pi_{i}\boxtimes\omega$;* 2. (2)

regard as $G_{n-i}$-representations via the embedding $g\mapsto\mathrm{diag}(I_{i},g)$, $\eta_{1}$ does not have any admissible $G_{n-i}$-submodule;

In particular, regard $\pi_{N}$ as $G_{n-i}$-representation, if $\tau\in\mathrm{Irr}(G_{n-i})$ embeds $\pi_{N}$, then $\tau$ is an irreducible submodule of $\omega$.

Proof.

Again we consider the filtration from the geometric lemma (4.11). We refine the filtration by incorporating the the filtrations on Jacquet functors of $\omega$ by simple composition factors. We also use Lemma 4.9. Then we obtain a filtration on $(\Pi_{i}\times\omega)_{N}$:

[TABLE]

such that $\eta_{0}^{\prime}/\eta_{1}^{\prime}\cong\Pi_{i}\boxtimes\omega$ and for $k>1$,

[TABLE]

for some $l\geq 1$ and $\omega^{\prime}\in\mathrm{Irr}(G_{l})$ and $\omega^{\prime\prime}\in\mathrm{Irr}(G_{n-i-l})$.

We set $\eta_{1}=\eta_{1}^{\prime}\cap\pi_{N}$ and $\eta_{2}=\pi_{N}/(\eta_{1}^{\prime}\cap\pi_{N})$. It remains to check that the short exact sequence

[TABLE]

satisfies the required properties. (1) is immediate from the definition of $\eta_{2}$.

Now we consider (2). Suppose $\tau\in\mathrm{Irr}(G_{n-i})$ such that

[TABLE]

and we shall arrive a contradiction. Then

[TABLE]

for some $k\geq 2$. Since

[TABLE]

we have that, by the expression (4.13),

[TABLE]

for some $\omega^{\prime\prime}\in\mathrm{Alg}(G_{n-i-l})$. However, the last Hom is zero as $l>0$ (see Lemma 4.3 as well as the proof of Lemma 4.18 below). This gives a contradiction and thus we have (2). ∎

4.9. Intersection properties of Bernstein-Zelevinsky layers

The main tool of this section is the exactness of Jacquet functor and the existence of certain irreducible subquotient in Corollary 4.20, which reduces the problem of studying submodules of a representation to certain submodules of the Jacquet module. The key property needed for Gelfand-Graev representation is Lemma 4.19, in which such construction is most conveniently done in the setting of affine Hecke algebra.

The following result can be viewed as a generalization of the fact that the Gelfand-Graev representation does not admit an admissible submodule.

Lemma 4.18**.**

For $i\neq j$, let $\omega,\tau$ be admissible $G_{n-i}$-representation and $G_{n-j}$-representation respectively. There is no non-zero isomorphic submodule of $\mathrm{ind}^{G_{n}}_{R_{n-i}}\omega\boxtimes\psi_{i}$ and $\mathrm{ind}_{R_{n-j}^{-}}^{G_{n}}\tau\boxtimes\psi_{j}$.

Proof.

We only prove for $i>j$ and the case for $j<i$ is similar. Then we have that $(\mathrm{ind}^{G_{n}}_{R_{n-i}}\omega\otimes\psi_{i})_{N_{i}}$ admits a filtration that successive quotients take the form

[TABLE]

Similarly, $(\mathrm{ind}^{G_{n}}_{R_{n-i}^{-}}\tau\boxtimes\psi_{j})_{N_{i}}$ admits a filtration that successive quotients take the form

[TABLE]

where $\tau^{\prime}$ is an admissible $G_{n-i-l}$-representation and $\tau^{\prime\prime}$ is an admissible $G_{i-j+l}$ representation. Since Jacquet functor is exact, it reduces, by Lemma 9.3, to see that there is no subrepresentation for

[TABLE]

for all $k,l$.

In the case that $k=0$ and $l=j$; or $k=n-i$ and $l=0$, a certain element in the Bernstein center annihilates $\omega^{\prime\prime}$ or $\omega^{\prime}$, but not on $\Pi_{i}$ or $\Pi_{n-i}$ respectively (by [BH03]). Now to prove other cases, we notice that either $k>l$ or $i-k>j-l$ (which happens when $l\geq k$). In the first case, we apply $N_{k}\subset G_{n-i}$ on the first factor, and in the second case, apply $N_{i-k}$ on the second factor. Then repeat the process and the process terminates as each process the value on $i$ decreases.

∎

Lemma 4.19**.**

Let $\mathfrak{s}\in\mathfrak{B}(G_{n})$. Let $\tau=\langle\mathfrak{m}\rangle=\zeta(\mathfrak{m})$ be an irreducible principle series in $\mathfrak{R}_{\mathfrak{s}}(G_{n})$, where $\mathfrak{m}\in\mathrm{Mult}$ with all segments to be singletons. Then there exists a $G_{n}$-filtration on $\Pi_{\mathfrak{s}}$:

[TABLE]

such that

1. (1)

each successive quotient $\pi_{r-1}/\pi_{r}$ is admissible and

[TABLE]

and; 2. (2)

[TABLE]

Proof.

We shall prove this by the Hecke algebra model of the Gelfand-Graev representation in Section 4.1, and we shall use those notations. Let $\mathcal{Z}$ be the center of $\mathcal{H}_{\mathfrak{s}}$ and let $\mathcal{J}$ be the maximal ideal in $\mathcal{Z}$ annihilating $\tau_{\mathfrak{s}}$. By Theorem 4.2,

[TABLE]

We consider the filtration given by $\pi_{k}=\mathcal{J}^{k}\Pi\cong\mathcal{J}^{k}\mathcal{H}_{\mathfrak{s}}\otimes_{\mathcal{H}_{W,\mathfrak{s}}}\mathrm{sgn}$. Since $\mathcal{H}_{\mathfrak{s}}$ is finitely-generated as a $\mathcal{Z}$-module, each successive quotient of the filtration from $\pi_{k}$ is finite-dimensional. With the fact that $\tau_{\mathfrak{s}}$ is the only irreducible $\mathcal{H}_{\mathfrak{s}}$-module annihilated by $\mathcal{J}$, we have that any irreducible composition factor of $\pi_{k}/\pi_{k-1}$ is isomorphic to $\tau_{\mathfrak{s}}$. This gives the corresponding property (1) in the group side. For (2), we have that $\pi_{k}|_{\mathcal{A}_{\mathfrak{s}}}\cong\mathcal{J}^{k}\mathcal{A}_{\mathfrak{s}}$. Now $\bigcap_{k}\mathcal{J}^{k}=0$ and $\mathcal{A}_{\mathfrak{s}}=\mathcal{Z}a_{1}+\ldots+\mathcal{Z}a_{k}$ for finitely many $a_{1},\ldots a_{k}\in\mathcal{A}_{\mathfrak{s}}$, and this gives that $\bigcap_{r}\pi_{r}=0$ as desired.

∎

Lemma 4.20**.**

Let $\mathfrak{s}\in\mathfrak{B}(G_{n-i})$ and let $\mathfrak{t}\in\mathfrak{B}(G_{i})$. Let $\omega$ be an irreducible representation in $\mathfrak{R}_{\mathfrak{s}}(G_{n-i})$. Let $\pi$ be a non-zero submodule of $\omega\times(\Pi_{i})_{\mathfrak{t}}$. Let $\tau\in\mathfrak{R}_{\mathfrak{t}}(G_{i})$ be an irreducible principle series as in Lemma 4.19 and further assume that $\mathrm{csupp}_{\mathbb{Z}}(\tau)\cap\mathrm{csupp}_{\mathbb{Z}}(\omega)=\emptyset$. Then there exists a submodule $\pi^{\prime}$ of $\omega\times(\Pi_{i})_{\mathfrak{t}}$ such that

1. (1)

the projection from $\pi$ to $(\omega\times(\Pi_{i})_{\mathfrak{t}})/\pi^{\prime}$ is non-zero; and 2. (2)

the image $\pi/(\pi^{\prime}\cap\pi)$ of the projection is admissible and any irreducible subquotient of the image is isomorphic to $\omega\times\tau$.

Remark 4.21**.**

1. (1)

The condition that $\mathrm{csupp}_{\mathbb{Z}}(\tau)\cap\mathrm{csupp}_{\mathbb{Z}}(\omega)=\emptyset$ implies that $\tau\times\omega$ is irreducible. 2. (2)

One can replace an irreducible representation $\omega$ with an admissible representation $\omega$ in $\mathfrak{R}_{\mathfrak{s}}(G_{n-i})$. Then one replaces $\mathrm{csupp}_{\mathbb{Z}}(\omega)$ by $\cup\mathrm{csupp}_{\mathbb{Z}}(\omega^{\prime})$, where the union runs for all $\omega^{\prime}\in\mathrm{JH}(\omega)$; and replace the last sentence ”irreducible subquotient isomorphic to $\omega\times\tau$” with ”irreducible subquotient isomorphic to $\omega^{\prime}\times\tau$ for some $\omega^{\prime}\in\mathrm{JH}(\omega)$”. In order to prove such statement, one considers a filtration on the admissible representation $\omega$:

[TABLE]

with irreducible successive quotients. This induces a filtration on $\omega\times(\Pi_{i})_{\mathfrak{t}}$. Intersecting $\pi$ with $\tau_{k}\times(\Pi_{i})_{\mathfrak{t}}$, we obtain a filtration for $\pi$. Now we consider the largest $k$ such that $\pi\cap(\tau_{k}\times(\Pi_{i})_{\mathfrak{t}})\neq\pi\cap(\tau_{k-1}\times(\Pi_{i})_{\mathfrak{t}})$, and obtain

[TABLE]

This embedding reduces to the original Lemma 4.20 and one can obtain the required submodule. 3. (3)

There are only countably many elements in $\mathrm{csupp}_{\mathbb{Z}}(\omega)$ and so such $\tau$ always exists.

Proof.

Let

[TABLE]

be a filtration of $(\Pi_{i})_{\mathfrak{t}}$ with the properties in Lemma 4.19. Then we have a filtration

[TABLE]

whose successive subquotients, by exactness of parabolic induction, are admissible with composition factors isomorphic to $\omega\times\tau$. Since

[TABLE]

we must have that the projection of $\pi$ to $(\omega\times(\Pi_{i})_{\mathfrak{s}})/(\omega\times\lambda_{k}^{\prime})$ is non-zero for sufficiently large $k$. Take such $\pi^{\prime}=\omega\times\lambda_{k}^{\prime}$. Hence, we obtain property (1). Now (2) follows from (1) and the description of the filtration above since $\pi/(\pi^{\prime}\cap\pi)$ embeds to $(\omega\times(\Pi_{i})_{\mathfrak{t}})/\pi^{\prime}$. ∎

We now can show the existence of certain irreducible $(1\times G_{n-i})$-submodule in $(\omega\times\Pi_{i})_{N_{n-i}}$.

Corollary 4.22**.**

Let $0\neq\omega\in\mathrm{Alg}(G_{n-i})$ be admissible. Set $N=N_{n-i}$. Let $\pi$ be a non-zero submodule of $\omega\times\Pi_{i}$. Regard $\pi_{N}$ as a $G_{n-i}$-representation via the embedding $g\mapsto\mathrm{diag}(I_{i},g)$,

[TABLE]

for some irreducible submodule $\omega^{\prime}$ of $\omega$.

Proof.

In view of Lemma 4.15, we have to find an ideal $\mathcal{J}$ in the Bernstein center such that $\pi_{N}/\mathcal{J}(\pi_{N})\neq 0$ and some additional properties. To this end, we write

[TABLE]

such that each $\omega_{k}\in\mathfrak{R}_{\mathfrak{s}_{k}}(G_{n-i})$ for some $\mathfrak{s}_{k}$ and $\mathfrak{s}_{k}\neq\mathfrak{s}_{l}$ if $k\neq l$. Then

[TABLE]

Thus, by using lemma 9.2, there exists a non-zero submodule $\widetilde{\pi}$ of $\pi$ such that

[TABLE]

for some $k^{*}$. By projecting $\widetilde{\pi}$ to a Bernstein component if necessary, we also assume that $\widetilde{\pi}$ lies in $\mathfrak{R}_{\mathfrak{u}}(G_{n})$ for some $\mathfrak{u}\in\mathfrak{B}(G_{n})$. Now $\mathfrak{u}$ and $\mathfrak{s}_{k^{*}}$ also determine a $\mathfrak{t}\in\mathfrak{B}(G_{i})$ such that

[TABLE]

(see Lemma 4.5). Now we can choose $\tau$ satisfying assumptions in Lemma 4.19 (see Remark 4.21 (2) and (3)). Thus we also have the condition that

[TABLE]

Now we set $\mathcal{J}^{\prime}\subset\mathfrak{Z}_{i}$ the maximal ideal annihilating $\tau$, and set $\mathcal{J}=\mathcal{J}^{\prime}\boxtimes\mathfrak{Z}_{n-i}$, an element in $\mathfrak{Z}_{i}\otimes\mathfrak{Z}_{n-i}$. Now, by Lemma 4.20, there is a surjection:

[TABLE]

and so a surjection:

[TABLE]

Since $(\tau\times\omega)_{N}/\mathcal{J}((\tau\times\omega)_{N})\neq 0$, we obtain $\widetilde{\pi}_{N}/\mathcal{J}(\widetilde{\pi}_{N})$. Now, by Lemma 4.15, we obtain

[TABLE]

for some irreducible submodule $\omega^{\prime}$ of $\omega_{k^{*}}$. This proves the corollary. ∎

Proposition 4.23**.**

Let $\omega_{1},\omega_{2}\in\mathrm{Alg}(G_{n-i})$ be admissible and non-zero. If $\mathrm{ind}_{R_{n-i}}^{G_{n}}\omega_{1}\boxtimes\psi_{i}$ and $\mathrm{ind}_{R_{n-i}^{-}}^{G_{n}}\omega_{2}\boxtimes\psi_{i}$ have isomorphic non-zero submodules, then $\omega_{1}$ and $\omega_{2}$ have isomorphic non-zero submodules.

Proof.

Let $0\neq\pi\in\mathrm{Alg}(G_{n})$ such that

[TABLE]

By Corollary 4.22, there exists an irreducible submodule $\omega^{\prime}$ of $\omega_{1}$ such that

[TABLE]

By Lemma 4.17, $\omega^{\prime}$ is an irreducible submodule of $\omega_{2}$. This proves the proposition.

∎

4.10. Strong indecomposability

A $G_{n}$-representation $\pi$ is said to be strongly indecomposable if any two non-zero submodules of $\pi$ have non-zero intersection.

We now prove a main property of a Bernstein-Zelevinsky layer. In Section 7, we shall prove a variation, which says that the Bernstein-Zelevinsky induction preserves indecomposability.

Lemma 4.24**.**

Let $\pi_{1}\in\mathrm{Alg}(G_{n_{1}})$ and $\pi_{2}\in\mathrm{Alg}(G_{n_{2}})$. Suppose $\pi_{1}$ is strongly indecomposable and $\pi_{2}$ is irreducible. Then $\pi_{1}\boxtimes\pi_{2}$ is also strongly indecomposable, as a $G_{n_{1}}\times G_{n_{2}}$-representation.

Proof.

This follows from that any submodule of $\pi_{1}\boxtimes\pi_{2}$ is the subspace $\tau\boxtimes\pi_{2}\subset\pi_{1}\boxtimes\pi_{2}$ for some submodule $\tau$ of $\pi_{1}$. ∎

Lemma 4.25**.**

Let $\pi_{1},\pi_{2},\pi$ be in $\mathrm{Alg}(G_{n})$ such that $\pi_{1}\hookrightarrow\pi$ and $\pi_{2}\hookrightarrow\pi$. Then $(\pi_{1}\cap\pi_{2})_{N}\cong(\pi_{1})_{N}\cap(\pi_{2})_{N}$. Here the later intersection is taken in $\pi_{N}$.

Proof.

We have the natural projection $p:\pi_{1}\cap\pi_{2}\rightarrow\pi_{N}$ as linear spaces. Since the image of the projection lies in both $(\pi_{1})_{N}$ and $(\pi_{2})_{N}$, the projection factors through the embedding $(\pi_{1})_{N}\cap(\pi_{2})_{N}$ to $\pi_{N}$. We have the following commutative diagram:

[TABLE]

Now taking the Jacquet functor in the diagram, $p$ induces an isomorphism from $(\pi_{1}\cap\pi_{2})_{N}$ onto $(\pi_{1}\cap\pi_{2})_{N}\subset\pi_{N}$. Thus the map from $(\pi_{1})_{N}\cap(\pi_{2})_{N}$ to $(\pi_{1}\cap\pi_{2})_{N}$ is surjective. Hence we obtain an isomorphism between $(\pi_{1})_{N}\cap(\pi_{2})_{N}$ and $(\pi_{1}\cap\pi_{2})_{N}$. ∎

Theorem 4.26**.**

Let $\omega\in\mathrm{Irr}(G_{n-i})$. For any $\mathfrak{s}\in\mathfrak{B}(G_{n})$, $(\mathrm{ind}_{R_{n-i}}^{G_{n}}\omega\boxtimes\psi_{i})_{\mathfrak{s}}$ is strongly indecomposable whenever it is non-zero.

Proof.

Let $\pi_{1}$ and $\pi_{2}$ be non-zero submodules of $(\omega\times\Pi_{i})_{\mathfrak{s}}$. Thus we have injections:

[TABLE]

Following the same argument as in Corollary 4.22 (which uses Lemma 4.19), for $j=1,2$, we have a non-zero submodule $\tau_{j}$ of $(\pi_{j})_{N}$ such that

[TABLE]

The Bernstein component $\mathfrak{R}_{\mathfrak{s}}(G_{n})$ and the Bernstein component for $\omega$ determines that

[TABLE]

for some $\mathfrak{t}\in\mathfrak{B}(G_{i})$ (c.f. Lemma 4.5). Now, as $(\Pi_{i})_{\mathfrak{t}}\boxtimes\omega$ is strongly indecomposable (Lemma 4.24 and Proposition 4.7), we conclude

[TABLE]

To show $\pi_{1}\cap\pi_{2}\neq 0$, it suffices to show that $(\pi_{1}\cap\pi_{2})_{N}\neq 0$ as the Jacquet functor is exact. By Lemma 4.25, it is equivalent to show that $(\pi_{1})_{N}\cap(\pi_{2})_{N}\neq 0$, which indeed follows from (4.14):

[TABLE]

∎

5. Indecomposability of restricted representations

5.1. Indecomposability of restriction

We now prove our main result, and the proof is in a similar spirit to that of Theorem 4.26:

Theorem 5.1**.**

Let $\pi\in\mathrm{Irr}(G_{n+1})$. Then for each $\mathfrak{s}\in\mathfrak{R}(G_{n})$, $\pi_{\mathfrak{s}}$ is strongly indecomposable whenever it is nonzero i.e. for any two non-zero submodules $\tau,\tau^{\prime}$ of $\pi_{\mathfrak{s}}$, $\tau\cap\tau^{\prime}\neq 0$.

Proof.

There exists a Bernstein-Zelevinsky $G_{n}$-filtration on $\pi$ with

[TABLE]

such that

[TABLE]

We also have a $G_{n}$-filtration on $\pi$ with

[TABLE]

such that

[TABLE]

Let $i^{*}$ be the level of $\pi$. For $\mathfrak{s}\in\mathfrak{B}(G_{n})$ such that $\pi_{\mathfrak{s}}\neq 0$, we also have $(\pi_{i^{*}-1})_{\mathfrak{s}}\neq 0$ (see Section 5.2 for the detail). For notation simplicity, we set $\tau=(\pi_{i^{*}-1})_{\mathfrak{s}}$ and $\tau^{\prime}=({}_{i^{*}-1}\pi)_{\mathfrak{s}}$, both regarded as subspaces of $\pi$.

Let $\omega$ and $\gamma$ be two non-zero submodules of $\pi_{\mathfrak{s}}$.

Claim: $\omega\cap\tau\neq 0$.

Proof of the claim: Suppose not. Then, the natural projection gives an injection

[TABLE]

By Lemma 4.18, there is no isomorphic submodules between $\pi_{\mathfrak{s}}/\tau$ and $\tau^{\prime}$. This implies $\omega\cap\tau^{\prime}=0$. Hence, we also have an injection:

[TABLE]

By (5.15), (5.17), (5.19), (5.20) and Lemma 9.3, there exists a $G_{n}$-representation which is isomorphic to submodules of

[TABLE]

for some $j,k<i^{*}$. By Lemma 4.18 again, we must have $j=k$. However, Proposition 4.23 contradicts to the following Theorem 5.2 below (whose proof does not depend on this result). This proves the claim.

Since $\omega$ is an arbitrary submodule of $\pi$, we also have $\gamma\cap\tau\neq 0$. Since $\pi^{(i^{*})}$ is irreducible, $\tau$ is strongly indecomposable by Theorem 4.26. This implies $\omega\cap\gamma\neq 0$.

∎

Theorem 5.2**.**

Let $\pi\in\mathrm{Irr}(G_{n+1})$. If $i$ is not the level for $\pi$, then $\pi^{[i]}$ and ${}^{[i]}\pi$ do not have an isomorphic irreducible quotient, and also do not have an isomorphic irreducible submodule whenever the two derivatives are not zero.

The proof of Theorem 5.2 will be carried out in Section 6. Note that the converse of the above theorem is also true, which follows directly from the well-known highest derivative due to Zelevinsky [Ze80, Theorem 8.1].

5.2. Non-zero Bernstein components

Let $\pi\in\mathrm{Irr}(G_{n+1})$. In order to give an explicit parametrization of indecomposable components of $\pi|_{G_{n}}$, we also have to determine when $\pi_{\mathfrak{s}}\neq 0$ for $\mathfrak{s}\in\mathfrak{B}(G_{n})$. Indeed this can be done as follows. Write $\pi\cong\langle\mathfrak{m}\rangle$ for a multisegment $\mathfrak{m}=\left\{\Delta_{1},\ldots,\Delta_{k}\right\}$. Let $\pi^{\prime}$ be the (right) highest derivative of $\pi$. Then we obtain a multiset

[TABLE]

This multiset determines a cuspidal $G_{k_{1}}\times\ldots\times G_{k_{p}}$-representation $\rho_{1}\boxtimes\ldots\boxtimes\rho_{p}$. Now we pick positive integers $k_{p+1},\ldots,k_{q}$ such that $k_{1}+\ldots+k_{q}=n$, and pick cuspidal representations $\rho_{p+1},\ldots,\rho_{q}$ of $G_{k_{p+1}},\ldots,G_{k_{q}}$ respectively. Then for the inertial equivalence class

[TABLE]

we have that $\pi_{\mathfrak{s}}\neq 0$, which follows from that the bottom Bernstein-Zelevinsky layer $(\pi^{[i^{*}]}\times\Pi_{i^{*}-1})_{\mathfrak{s}}\subset\pi_{\mathfrak{s}}$ is non-zero. Here $i^{*}$ is the level of $\pi$.

Indeed for any $\mathfrak{s}\in\mathfrak{B}(G_{n})$ with $\pi_{\mathfrak{s}}\neq 0$, $\mathfrak{s}$ arises in the above way. To see this, we need the following lemma:

Lemma 5.3**.**

Let $\pi\in\mathrm{Irr}(G_{n+1})$. Let $i^{*}$ be the level of $\pi$. Then, for any $i\leq i^{*}$ with $\pi^{[i]}\neq 0$ and for any $\omega\in\mathrm{JH}(\pi^{[i]})$, $\mathrm{csupp}(\pi^{[i^{*}]})\subset\mathrm{csupp}(\omega)$ (counting multiplicities).

Proof.

Let $\mathfrak{m}\in\mathrm{Mult}$ such that $\pi\cong\langle\mathfrak{m}\rangle$. Now by definition, we have that

[TABLE]

and so $\langle\mathfrak{m}\rangle^{[i]}\hookrightarrow\zeta(\mathfrak{m})^{[i]}$.

For any segment $\Delta$, set $\Delta^{[0]}=\nu^{1/2}\Delta$ and set $\Delta^{[-]}=\nu^{1/2}\Delta^{-}$. We also set $\mathfrak{m}=\left\{\Delta_{1},\ldots,\Delta_{k}\right\}$ and set

[TABLE]

Now the geometric lemma gives a filtration on $\zeta(\mathfrak{m})^{[i]}$ whose successive quotients are isomorphic to $\zeta(\mathfrak{n})$ for some $\mathfrak{n}\in\mathcal{M}$ (see Lemma 6.4 below). Since $\langle\mathfrak{m}\rangle^{[i]}\hookrightarrow\zeta(\mathfrak{m})^{[i]}$ as discussed before, any $\omega\in\mathrm{JH}(\langle\mathfrak{m}\rangle^{[i]})$ is a composition factor of $\zeta(\mathfrak{n})$ for some $\mathfrak{n}\in\mathcal{M}$. Hence, $\mathrm{csupp}(\omega)=\cup_{\Delta\in\mathfrak{n}}\Delta$ (counting multiplicities).

On the other hand, $\pi^{[i^{*}]}=\langle\left\{\Delta_{1}^{[-]},\ldots,\Delta_{k}^{[-]}\right\}\rangle$. Hence $\mathrm{csupp}(\pi^{[i^{*}]})\subset\mathrm{csupp}(\omega)$ for any $\omega\in\mathrm{JH}(\pi^{[i]})$. ∎

Now we go back to consider that $\mathfrak{s}\in\mathfrak{B}(G_{n})$ with $\pi_{\mathfrak{s}}\neq 0$. By the Bernstein-Zelevinsky filtration,

[TABLE]

for some $i\leq i^{*}$. Now, as a similar manner to what we did for the bottom layer above, we could determine (abstractly) all possible $\mathfrak{s}^{\prime}\in\mathfrak{B}(G_{n})$ with $(\pi^{[i]}\times\Pi_{i-1})_{\mathfrak{s}^{\prime}}\neq 0$. Then by Lemma 5.3, one sees that,

[TABLE]

5.3. Indecomposability of Zelevinsky induced modules

Let $\mathfrak{m}\in\mathrm{Mult}$. In [BZ77], it is shown that the restriction of $\zeta(\mathfrak{m})$ to the mirabolic subgroup is strongly indecomposable. One may expect the following conjecture, which is a stronger statement of Theorem 5.1:

Conjecture 5.4**.**

Let $\mathfrak{m}$ be a multisegment for $n+1$. Then any Bernstein component of $\zeta(\mathfrak{m})|_{G_{n}}$ is strongly indecomposable.

We remark that $\zeta(\mathfrak{m})^{\vee}|_{G_{n}}$ is not strongly indecomposable in general. An example is $\mathfrak{m}=\left\{[\nu^{-1/2}],[\nu^{1/2}]\right\}$. In this case, the short exact sequence:

[TABLE]

where $\Delta=[\nu^{-1/2},\nu^{1/2}]$. Restriction to $G_{1}$ gives a split sequence since $\mathrm{St}(\Delta)|_{G_{1}}$ is projective (see Theorem 3.7). Hence the Iwahori component of $\zeta(\Delta)^{\vee}|_{G_{1}}$ is not indecomposable.

6. Asymmetric property of left and right derivatives

We are going to prove Theorem 5.2 in this section. The idea lies in two simple cases. The first one is a generic representation. Since an irreducible generic representation is isomorphic to $\lambda(\mathfrak{m})\cong\mathrm{St}(\mathfrak{m})$ for $\mathfrak{m}\in\mathrm{Mult}$ (with the property that any two segments in $\mathfrak{m}$ are unlinked), a simple counting on cuspidal supports of derivatives can show Theorem 5.2 for that case. The second one is an irreducible representation whose Zelevinsky multisegment has all segments with relative length strictly greater than $1$. In such case, one can narrow down the possibility of irreducible submodule of the derivatives via the embedding $\langle\mathfrak{m}\rangle^{(i)}\hookrightarrow\zeta(\mathfrak{m})^{(i)}$ and ${}^{(i)}\langle\mathfrak{m}\rangle\hookrightarrow{}^{(i)}\zeta(\mathfrak{m})$, and use geometric lemma to compute the possible submodules of derivatives of $\zeta(\mathfrak{m})^{(i)}$ and ${}^{(i)}\zeta(\mathfrak{m})$. The combination of these two cases seems to require some extra work. The strategy is to use Speh representations, which can be viewed as a generalization of generalized Steinberg representations, and then apply Lemma 8.4 to obtain information on submodules.

6.1. Union-intersection operation

Let $\mathfrak{m}=\left\{\Delta_{1},\ldots,\Delta_{r}\right\}$. For two segments $\Delta$ and $\Delta^{\prime}$ in $\mathfrak{m}$ which are linked, the process of replacing $\Delta$ and $\Delta^{\prime}$ by $\Delta\cap\Delta^{\prime}$ and $\Delta\cup\Delta^{\prime}$ is called the union-intersection process. It follows from [Ze80, Chapter 7] that the Zelevinsky multisegment of any irreducible composition factor in

[TABLE]

can be obtained by a chain of intersection-union process. For a positive integer $l$, define $N(\mathfrak{m},l)$ to be the number of segments in $\mathfrak{m}$ with relative length $l$.

Lemma 6.1**.**

Let $\mathfrak{m}=\left\{\Delta_{1},\ldots,\Delta_{r}\right\}\in\mathrm{Mult}$. Let $\mathfrak{m}^{\prime}$ be a Zelevinsky multisegment obtained from $\mathfrak{m}$ by a chain of union-intersection operations. Then there exists a positive integer $l$ such that

[TABLE]

and for any $l^{\prime}>l$,

[TABLE]

Proof.

We shall prove inductively on the number of union-intersection operations to obtain $\mathfrak{m}^{\prime}$ from $\mathfrak{m}$. We assume the targeted statement is true for $X\geq 2$ number of union-intersection operations. Let $\mathfrak{m}^{\prime\prime}$ be a new multisegment obtained from $\mathfrak{m}$ by $X+1$ number of union-intersection operations. Then we can find a multisegment $\mathfrak{m}^{\prime}$ obtained from $\mathfrak{m}$ by $X$ number of union-intersection operations and $\mathfrak{m}^{\prime\prime}$ is obtained from $\mathfrak{m}^{\prime}$ by one union-intersection operation.

By inductive hypothesis, we can find a positive number $l_{X}$ such that

[TABLE]

and for any $l^{\prime}>l_{X}$,

[TABLE]

Now let $\Delta_{i},\Delta_{j}$ be the multisegments in $\mathfrak{m}^{\prime}$ involved in the union-intersection operation to obtain $\mathfrak{m}^{\prime\prime}$. In particular, $\Delta_{i}$ and $\Delta_{j}$ are linked. Let $l_{0}$ be the relative length of $\Delta_{i}\cup\Delta_{j}$. If $l_{0}\geq l_{X}$, set $l=l_{0}$, and otherwise set $l=l_{X}$. Now it is straightforward to check that such $l$ satisfies the required properties.

It remains to prove for $X=1$. This can be proved by a similar argument as in the last paragraph. ∎

Indeed, the proof of Lemma 6.1 gives the following:

Lemma 6.2**.**

Let $k\geq 2$. Let $\mathfrak{m}_{1},\ldots\mathfrak{m}_{k}\in\mathrm{Mult}$ such that for $i=2,\ldots,k$, each $\mathfrak{m}_{i}$ is obtained from $\mathfrak{m}_{i-1}$ by one-step of union-intersection operation. Suppose $\Delta_{i},\Delta_{i}^{\prime}$ are two segments in $\mathfrak{m}_{i-1}$ involved in the union-intersection operation to obtain $\mathfrak{m}_{i}$. Let $l_{i}$ be the relative length of $\Delta_{i}\cup\Delta_{i}^{\prime}$. Then there exists $l$ such that $l\geq l_{i}$ for all $i$, and

[TABLE]

and, for any $l^{\prime}>l$,

[TABLE]

6.2. Speh multisegments

Definition 6.3**.**

Let $\Delta$ be a segment. Let

[TABLE]

We shall call $\mathfrak{m}(m,\Delta)$ to be a Speh multisegment. Define

[TABLE]

which will be called a Speh representation. In the literature, it is sometimes called essentially Speh representation. Denote by $L(\mathfrak{m}(m,\Delta))$ the relative length of $\Delta$.

We similarly define

[TABLE]

and

[TABLE]

Let $l=n(\rho)$. It follows from [Ta87, LM14] (also see [CS19]) that

[TABLE]

and $u(m,\Delta)^{(k)}$ is zero if $l$ does not divide $k$. Applying (2.3), we have that

[TABLE]

and ${}^{(k)}u(m,\Delta)=0$ if $l$ does not divide $k$.

6.3. A notation for multisegment

For a multisegment $\mathfrak{m}=\left\{\Delta_{1},\ldots,\Delta_{r}\right\}$, define

[TABLE]

[TABLE]

We shall need the following lemma:

Lemma 6.4**.**

Let $\mathfrak{m}\in\mathrm{Mult}$. Then $\zeta(\mathfrak{m})^{(i)}$ (resp. ${}^{(i)}\zeta(\mathfrak{m})$) admits a filtration whose successive quotients are isomorphic to $\zeta(\mathfrak{n})$ for some $\mathfrak{n}\in\mathfrak{m}^{(i)}$ (resp. $\mathfrak{n}\in{}^{(i)}\mathfrak{m}$).

Proof.

Write $\mathfrak{m}=\left\{\Delta_{1},\ldots,\Delta_{r}\right\}$. We shall assume that for $i<j$, $b(\Delta_{i})\not\leq b(\Delta_{j})$ and if $b(\Delta_{i})\cong b(\Delta_{j})$, then $a(\Delta_{i})\not<a(\Delta_{j})$. By definition,

[TABLE]

By geometric lemma, we have that $\zeta(\mathfrak{m})^{(i)}$ admits a filtration whose successive quotients are:

[TABLE]

for $i_{1}+\ldots+i_{r}=i$. Now the lemma follows from that the product $\langle\Delta_{1}\rangle^{(i_{1})}\times\ldots\times\langle\Delta_{r}\rangle^{(i_{r})}$ is isomorphic to $\zeta(\mathfrak{n})$ for $\mathfrak{n}=\left\{\Delta_{1}^{(i_{1})},\ldots,\Delta_{r}^{(i_{r})}\right\}$. ∎

6.4. Proof of Theorem 5.2

By Lemma 2.6, it suffices to prove the statement for submodules of the derivatives.

Let $\mathfrak{m}$ be the Zelevinsky multisegment with $\pi\cong\langle\mathfrak{m}\rangle$. We shall assume that any cuspidal representation in each segment of $\mathfrak{m}$ is an unramified twist of a fixed cuspidal representation $\rho$, i.e.

[TABLE]

We shall prove that Theorem 5.2 for such $\pi$. The general case follows from this by writing an irreducible representation as a product of irreducible representations of such specific form.

Step 1: First approximation using Lemma 6.4. Let $\pi^{\prime}$ be a common isomorphic irreducible quotient of $\pi{}^{[i]}$ and ${}^{[i]}\pi$. (Here we assume that $\pi{}^{[i]}$ and ${}^{[i]}\pi$ are non-zero.) Recall that we have that

[TABLE]

Since taking derivatives is an exact functor, $\nu^{1/2}\cdot\pi^{(i)}$ embeds to $\nu^{1/2}\cdot\zeta(\mathfrak{m})^{(i)}$ and so does $\pi^{\prime}$.

By Lemma 6.4, there is a filtration on $\zeta(\mathfrak{m})^{(i)}$ given by $\zeta(\mathfrak{m}^{(i_{1},\ldots,i_{r})})$ for $i_{1}+\ldots+i_{r}=i$, where $i_{k}=0$ or $n(\rho)$. Then $\pi^{\prime}$ is isomorphic to the unique submodule of $\nu^{1/2}\zeta(\mathfrak{m}^{(i_{1}^{\prime},\ldots,i_{r}^{\prime})})$ for some $(i_{1}^{\prime},\ldots,i_{r}^{\prime})$ and so $\pi^{\prime}\cong\nu^{1/2}\langle\mathfrak{m}^{(i_{1}^{\prime},\ldots,i_{r}^{\prime})}\rangle$. Similarly, $\pi^{\prime}$ is isomorphic to $\nu^{-1/2}\langle{}^{(j_{1}^{\prime},\ldots,j_{r}^{\prime})}\mathfrak{m}\rangle$ for some $j_{1}^{\prime}+\ldots+j_{r}^{\prime}=i$.

Suppose $i$ is not the level of $\pi$. Then there exists at least one $i_{k}^{\prime}=0$ and at least one $j_{k}^{\prime}=0$. Among all those segment $\Delta_{k}$ with either $i_{k}^{\prime}=0$ or $j_{k}^{\prime}=0$, we shall choose $\Delta_{k^{*}}$ has the largest relative (and absolute) length. Denote the relative length of $\Delta_{k^{*}}$ by $L$.

**Step 2: Second approximation using Lemma 8.4 **

We write $\mathfrak{m}$ as the sum of Speh multisegments

[TABLE]

satisfying properties in Proposition 8.3.

Let

[TABLE]

be all the Speh multisegments appearing in the sum (6.23) such that $L(\mathfrak{m}_{k})=L$. For each $\mathfrak{m}_{k}$, write $\mathfrak{m}_{k}=\mathfrak{m}(m_{k},\Delta_{k})$ and define $b(\mathfrak{m}_{k})=\nu^{(m_{k}-1)/2}b(\Delta_{k})$. We shall label $\mathfrak{m}_{k}$ in the way that $b(\mathfrak{m}_{k})\not<b(\mathfrak{m}_{l})$ for $k<l$. Furthermore, the labelling satisfies the properties that

1. ($\Diamond$)

for any $\mathfrak{m}_{p}$ and $p<q$, $\mathfrak{m}_{p}+\Delta$ is not a Speh multisegment for any $\Delta\in\mathfrak{m}_{q}$. 2. ($\Diamond\Diamond$)

if $\mathfrak{m}_{p}\cap\mathfrak{m}_{q}\neq\emptyset$ and $p\leq q$, then $\mathfrak{m}_{q}\subset\mathfrak{m}_{p}$.

Let $\mathfrak{n}_{1}$ be the collection of all segments in $\mathfrak{m}\setminus(\mathfrak{m}_{1}+\ldots+\mathfrak{m}_{r})$ which satisfies either (1) $b(\mathfrak{m}_{1})<b(\Delta^{\prime})$ or (2) $b(\mathfrak{m}_{1})\cong b(\Delta^{\prime})$. Define inductively that $\mathfrak{n}_{k}$ is the collection of all segments $\Delta^{\prime}$ in $\mathfrak{m}\setminus(\mathfrak{m}_{1}+\ldots+\mathfrak{m}_{r}+\mathfrak{n}_{1}+\ldots\mathfrak{n}_{k-1})$ that satisfies the property that either (1) $b(\mathfrak{m}_{k})<b(\Delta^{\prime})$, or (2) $b(\mathfrak{m}_{k})\cong b(\Delta^{\prime})$. (It is possible that some $\mathfrak{n}_{k}$ is empty.)

By Lemma 8.4, we have a series of embeddings:

[TABLE]

For simplicity, define, for $k\geq 0$,

[TABLE]

Step 3: Approximation on right derivatives. Then for each $k$, we again have an embedding:

[TABLE]

As $\lambda_{k}$ is a product of representations, we again have a filtration on $\lambda_{k}^{(i)}$. This gives that $\pi^{\prime}$ embeds to a successive quotient of the filtration:

[TABLE]

with $p^{k}_{1}+\ldots+p^{k}_{k}+q_{1}^{k}+\ldots+q^{k}_{k}+s^{k}=i$,

[TABLE]

Lemma 6.5**.**

1. (1)

$i_{k}=0$* for some $\Delta_{k}$ in $\mathfrak{m}$ with relative length $L$ (see the choice of $L$ in Step 1).* 2. (2)

Following above notations, there exists a $k^{\prime}\geq 1$ such that at least one of $q^{k^{\prime}}_{l}$ is not equal to the level of $\langle\mathfrak{m}_{l}\rangle$.

Remark 6.6**.**

Similarly, we have $j_{k}=0$ for some $\Delta_{k}$ in $\mathfrak{m}$ with relative length $L$.

Proof.

We first assume (1) to prove (2). It suffices to show that when $k^{\prime}=r$, at least one of $q^{k^{\prime}}_{l}$ is not equal to the level of $\langle\mathfrak{m}_{l}\rangle$. Suppose not. Let $\langle\mathfrak{m}_{i}\rangle^{-}$ be the highest derivative of $\langle\mathfrak{m}_{i}\rangle$. Then we obtain an embedding:

[TABLE]

Now set $\mathfrak{m}_{i}^{-}$ to be the multisegment such that $\langle\mathfrak{m}_{i}^{-}\rangle\cong\langle\mathfrak{m}_{i}\rangle^{-}$. By definitions, $\mathfrak{m}_{i}^{-}$ is still a Speh multisegment. Now, by Lemma 6.4, we have that

[TABLE]

where $\widetilde{\mathfrak{n}}_{a}\in\mathfrak{n}_{a}^{(p_{i}^{k})}$. From the construction of $\mathfrak{n}_{a}$, we can check that those $\widetilde{\mathfrak{n}}_{i}$ satisfies the conditions in Lemma 8.4. Hence, by Lemma 8.4,

[TABLE]

Now we write $\mathfrak{n}_{a}=\left\{\Delta_{a,1},\ldots,\Delta_{a,r(a)}\right\}$ for each $a$. Then we have that

[TABLE]

where each $p_{a,k}=0$ or $n(\rho)$.

Claim: For any segment $\Delta_{a,k}$ with relative length at least $L+1$, $p_{a,k}=n(\rho)$.

Proof of claim: Recall that we also have $\pi^{\prime}\cong\nu^{1/2}\cdot\langle\mathfrak{m}^{(i_{1},\ldots,i_{r})}\rangle$ in Step 1. Hence,

[TABLE]

Let $L^{**}$ be the largest relative length among all the relative length of segments in $\mathfrak{m}$. If $L^{**}=L$, then there is nothing to prove (for the claim). We assume $L^{**}>L$. In such case, there is no segment in $\mathfrak{m}^{(i_{1},\ldots,i_{r})}$ has relative length $L^{**}$. This implies that all segments $\Delta_{a,k}$ with relative length $L^{**}$ have $p_{a,k}=n(\rho)$. With the use of (6.24) and

[TABLE]

we can proceed to the length $L^{**}-1$ in a similar fashion. Inductively, we obtain the claim.

Now we go back to the proof of the lemma. By using the claim (and the definitions of $\mathfrak{m}_{i}$),

[TABLE]

On the other hand,

[TABLE]

by using the hypothesis in the lemma. However, $\mathfrak{m}=\widetilde{\mathfrak{n}}_{1}+\mathfrak{m}_{1}^{-}+\ldots+\widetilde{\mathfrak{n}}_{k}+\mathfrak{m}_{k}^{-}+\widetilde{\mathfrak{n}}_{k+1}$ and this gives a contradiction. This proves (2) modulo (1).

It remains to prove (1). From our choice of $L$, either

[TABLE]

or

[TABLE]

however, since $\nu^{1/2}\cdot\mathfrak{m}^{(i_{1},\ldots,i_{r})}\cong\nu^{-1/2}\cdot{}^{(j_{1},\ldots,j_{r})}\mathfrak{m}$, we have that

[TABLE]

Thus, we must have that $N(\mathfrak{m}^{(i_{1},\ldots,i_{r})},L)\geq N(\mathfrak{m},L+1)+1$. But this forces (1).

∎

Now let $\bar{k}$ be the smallest number such that at least one of $q^{\bar{k}}_{l}$ is not equal to the level of $\langle\mathfrak{m}_{l}\rangle$. Now we shall denote such $l$ by $l^{*}$, and so $\bar{k}\geq l^{*}$.

We use similar strategy to further consider the filtrations on each $\mathfrak{n}_{a}$ by geometric lemma. For that we write, for each $a$,

[TABLE]

and

[TABLE]

Here $r(a)$ is an index counting the number of segments in $\mathfrak{n}_{a}$ depending on $a$.

Then we again have an embedding

[TABLE]

where, for $a=1,\ldots,\bar{k}$,

[TABLE]

with $p_{a,1}+\ldots+p_{a,r(a)}=p_{a}^{\bar{k}}$ and each $p_{a,b}=0$ or $n(\rho)$, and

[TABLE]

with $p_{\bar{k}+1,1}+\ldots+p_{\bar{k}+1,r(\bar{k}+1)}=s_{\bar{k}}$, with each $p_{\bar{k}+1,b}=0$ or $n(\rho)$.

Step 4: Computing some indexes $p_{a,b}$ and approximating the number of special segments by union-intersection operations. We claim (*) that if $\Delta_{a,b}$ has a relative length at least $L+1$, then $p_{a,b}=n(\rho)$. This is indeed similar to the proof of Lemma 6.5 and the main difference is that we do not have an analogous form of (6.24) (since we cannot apply Lemma 8.4). Instead, we can obtain this from Lemma 6.2 (and its proof of Lemma 6.1).

Now from our choice of $\bar{k}$, we have that $\langle\mathfrak{m}_{l^{*}}\rangle^{(q^{\bar{k}}_{l^{*}})}$ is not a Speh representation. We can write

[TABLE]

for a certain $\Delta^{*}$ with relative length $L$ and some $x$. By ($\Diamond$), $\nu\Delta^{*}\notin\mathfrak{m}_{l}$ for any $l\geq l^{*}$ from our labelling on $\mathfrak{m}_{l}$. Rephrasing the statement, we get the following statement:

(**) $\nu^{1/2}\Delta^{*}\notin\nu^{-1/2}\mathfrak{m}_{l}$* for any $l\geq l^{*}$ . *

Now with (*), we have that $\pi^{\prime}$ is a composition factor of $\zeta(\mathfrak{m}^{\prime\prime\prime})$, where $\mathfrak{m}^{\prime\prime\prime}$ is some multisegment containing all the segments $\Delta^{-}$ (counting multiplicities) with $\Delta$ in $\mathfrak{m}$ that has relative length at least $L+1$ and containing an additional segment $\nu^{1/2}\Delta^{*}$, and we shall call the former segments (i.e. the segment in the form of $\Delta^{-}$) to be special for convenience.

We can apply the intersection-union process to obtain the Zelevinsky multisegment for $\pi^{\prime}$ from $\mathfrak{m}^{\prime\prime\prime}$. However in each step of the process, any one of the two segments involved in the intersection-union cannot be special. Otherwise, by Lemma 6.2, there exists $l\geq L+1$ such that the number of segments in the resulting multisegment with relative length $l$ is more than the number of segments in $\mathfrak{m}^{(i_{1},\ldots,i_{r})}$. (Here we also used that $L(\mathfrak{m}^{(i_{1},\ldots,i_{r})},l)=L(\mathfrak{m}^{\prime\prime\prime},l)$ for any $l\geq L+1$.) Hence we obtain the following:

(***) *the number of segments $\Delta$ in $\nu^{1/2}\cdot\mathfrak{m}^{(i_{1},\ldots,i_{r})}$ such that $\nu^{1/2}\Delta^{*}\subset\Delta$ is at least equal to one plus the number special segments in $\mathfrak{m}^{\prime\prime\prime}$ satisfying the same property *

Step 5: Approximation on left derivatives. Now we come to the final part of the proof. We now consider $\nu^{-1/2}\cdot{}^{(i)}\pi$. Following the strategy for right derivatives, we have that for each $k$

[TABLE]

This gives that $\pi^{\prime}$

[TABLE]

with $u^{k}_{1}+\ldots+u^{k}_{k}+v^{k}_{1}+\ldots+v^{k}_{k}+w^{k}=i$. Let $\widetilde{k}$ be the smallest integer such that at least one of $v^{\widetilde{k}}_{l}$ is not equal to the level of $\langle\mathfrak{m}_{l}\rangle$. (For the existence of $\widetilde{k}$, see Lemma 6.5.) We shall denote such $l$ by $\widetilde{l}$.

Step 6: the case that $\widetilde{k}\geq\bar{k}$ and determining the number of segments.

We firstly consider the case that $\widetilde{k}\geq\bar{k}$. Similar to left derivatives, we have that

[TABLE]

where

[TABLE]

with $u_{a,1}+\ldots+u_{a,r(a)}=u_{a}$ and each $u_{a,b}=0$ or $n(\rho)$. Since we assume that $\widetilde{k}\geq\bar{k}$, we have that ${}^{(v_{l})}\langle\mathfrak{m}_{l}\rangle$ is a highest derivative for any $l\leq\bar{k}-1$ and so is a Speh representation, and we can apply Lemma 8.4(1). Hence the unique subrepresentation of

[TABLE]

is isomorphic to

[TABLE]

where $\langle\widehat{\mathfrak{m}}_{l}\rangle={}^{(v_{l})}\langle\mathfrak{m}_{l}\rangle$. Similar to (*) for right derivatives (but the proof could be easier here), we obtain the analogous statement for those $\widehat{\mathfrak{n}}_{a}$. Now if

[TABLE]

such that $\Delta=\nu^{1/2}\Delta^{*}$, then we must have that $\Delta=\nu^{-1/2}\cdot{}^{-}\Delta_{0}$ or $\nu^{-1/2}\cdot\Delta_{0}$ for some segment $\Delta_{0}$ in $\mathfrak{m}$. For the latter case to happen, we must have that $\Delta_{0}\in\mathfrak{m}_{l}\subset\mathfrak{o}_{\bar{k}}$ with $l\geq\bar{k}\geq l^{*}$, but, by (**), the possibility $\Delta=\nu^{-1/2}\Delta_{0}$ cannot happen. Thus we must have that $\Delta$ is a special in the same sense as the discussion in right derivatives. This concludes the following:

(****) *The number of segments $\Delta$ in $\nu^{-1/2}(\widehat{\mathfrak{n}}_{1}+\widehat{\mathfrak{m}}_{1}+\ldots+\widetilde{\mathfrak{n}}_{\bar{k}-1}+\widehat{\mathfrak{m}}_{\bar{k}-1}+\widehat{\mathfrak{o}}_{\bar{k}})$ with the property that $\nu^{1/2}\Delta^{*}\subset\Delta$ is equal to the number of special segments satisfying the same property. *

Now the above statement contradicts to (***) since both Zelevinsky multisegments give an irreducible representation isomorphic to $\pi^{\prime}$.

Remark 6.7**.**

The isomorphism (6.25) is a key to obtain the equality in (**), in comparison with the inequality in (*).

Step 7: the case that $\bar{k}>\widetilde{k}$. Now the way to get contradiction in the case $\bar{k}\geq\widetilde{k}$ is similar by interchanging the role of left and right derivatives. We remark that to prove the analogue of (), one uses ($\diamond$). And to obtain the similar isomorphism as (6.25), one needs to use Lemma 8.4(2). We can argue similarly to get an analogue of () and (***). Hence the only possibility that $\nu^{1/2}\cdot\pi^{(i)}$ and so $\nu^{-1/2}\cdot{}^{(i)}\pi$ have an isomorphic irreducible quotient only if $i$ is the level for $\pi$. This completes the proof of Theorem 6.4.

6.5. Branching law in opposite direction

Here is another consequence of the asymmetric property on the Hom-branching law in another direction:

Corollary 6.8**.**

Let $\pi^{\prime}\in\mathrm{Irr}(G_{n})$. Let $\pi\in\mathrm{Irr}(G_{n+1})$.

1. (1)

Suppose $\pi$ is not $1$-dimensional. Then

[TABLE] 2. (2)

Suppose $\pi$ is $1$-dimensional. Then

[TABLE]

if and only if $\pi^{\prime}$ is also $1$-dimensional and $\pi^{\prime}=\pi|_{G_{n}}$.

Proof.

The second statement is trivial. We consider the first one. Since $\pi$ is not one-dimensional, the level of $\pi$ is not $1$ by Zelevinsky classification. By Theorem 5.2 and Proposition 2.5, $\pi^{[1]}$ and ${}^{[1]}\pi$ have no common irreducible submodule if $\pi^{(1)}\neq 0$ and ${}^{(1)}\pi\neq 0$. Then at least one of

[TABLE]

On the other hand, we have that for all $i\geq 2$, by Frobenius reciprocity,

[TABLE]

and

[TABLE]

However, in the last two equalities, the Hom in the right hand side is always zero (see Lemma 4.3). Now applying a Bernstein-Zelevinsky filtration implies the corollary.

∎

7. Preserving indecomposability of Bernstein-Zelevinsky induction

We shall use the following criteria of indecomposable representations:

Lemma 7.1**.**

Let $G$ be a connected reductive $p$-adic group. Let $\pi$ be a smooth representation of $G$. The only idempotents in $\mathrm{End}_{G}(\pi)$ are [math] and the identity (up to automorphism) if and only if $\pi$ is indecomposable.

Proof.

If $\pi$ is not indecomposable, then any projection to a direct summand gives a non-identity idempotent. On the other hand, if $\sigma\in\mathrm{End}_{G}(\pi)$ is a non-identity idempotent (i.e. $\sigma(\pi)\neq\pi$), then $\pi\cong\mathrm{im}(\sigma)\oplus\mathrm{im}(1-\sigma)$. ∎

Lemma 7.2**.**

Set $\Pi=\Pi_{k}$. Let $\mathfrak{s}\in\mathfrak{B}(G_{k})$. Then ${}^{(k)}(\Pi_{\mathfrak{s}})\cong\mathfrak{Z}_{\mathfrak{s}}$, where $\mathfrak{Z}_{\mathfrak{s}}$ is the Bernstein center of $\mathfrak{R}_{\mathfrak{s}}(G_{k})$.

Proof.

We have

[TABLE]

where the second isomorphism follows from the second adjointness theorem of [CS18b], and the third isomorphism follows from the Bernstein decomposition. Now the lemma follows from that the final endomorphism is isomorphic to $\mathfrak{Z}_{\mathfrak{s}}$ by [BH03]. ∎

Theorem 7.3**.**

Let $\pi$ be an admissible indecomposable smooth representation of $G_{n-i}$. For each $\mathfrak{s}\in\mathfrak{B}(G_{n})$, the Bernstein component $(\mathrm{ind}_{R_{n-i}}^{G_{n}}\pi\boxtimes\psi_{i})_{\mathfrak{s}}$ is indecomposable.

Proof.

We first prove the following:

Lemma 7.4**.**

Let $\pi\in\mathrm{Alg}(G_{n-i})$ be admissible. Then, as endomorphism algebras,

[TABLE]

where $\mathfrak{Z}_{i}=\prod_{\mathfrak{s}\in\mathfrak{B}(G_{i})}\mathfrak{Z}_{\mathfrak{s}}$ is the Bernstein center.

Proof.

Let $\mathfrak{s}_{1},\mathfrak{s}_{2}\in\mathfrak{B}(G_{n-i})$. Let $\pi_{1},\pi_{2}$ be two admissible $G_{n-i}$ representations in $\mathfrak{R}_{\mathfrak{s}_{1}}(G_{n-i})$ and $\mathfrak{R}_{\mathfrak{s}_{2}}(G_{n-i})$ respectively. Then

[TABLE]

We remark that in order to apply the second adjointness theorem [CS18b, Lemma 2.2] , we need to get the first isomorphism. For the third isomorphisms, see discussions in Section 4.2.

Now we apply the geometric lemma on ${}^{(i)}(\pi_{2}\times(\Pi_{i})_{\mathfrak{u}})$ and we obtain a filtration on ${}^{(i)}(\pi_{2}\times(\Pi_{i})_{\mathfrak{u}})$ whose successive quotients are of the form

[TABLE]

where $j+k=i$. Using Lemma 4.3,

[TABLE]

unless $j=0$ and $k=i$. Combining the above isomorphisms, we now have

[TABLE]

We remark that the third and last isomorphisms require the admissibility of $\pi_{1}$ and $\pi_{2}$. The second isomorphism follows from Lemma 7.2. One further traces the above isomorphisms to see that they give an isomorphism preserving the algebra structure and we omit the details. Now we specialize to $\pi_{1}=\pi_{2}=\pi$ and we obtain the lemma when each of $\pi_{1},\pi_{2}$ lying in one Bernstein component.

The general case follows by writing $\pi=\oplus_{\mathfrak{s}}\pi_{\mathfrak{s}}$. ∎

Now the theorem follows from Lemmas 7.1 and 7.3 since the right hand side has only 0 and 1 as idempotents if and only if the left hand side does. ∎

Let $M_{n}$ be the mirabolic subgroup in $G_{n+1}$, that is the subgroup containing all matrices with the last row of the form $(0,\ldots,0,1)$. We have the following consequence restricted from $M_{n}$ to $G_{n}$. Here $G_{n}$ is viewed as the subgroup of $M_{n}$ via the embedding $g\mapsto\mathrm{diag}(g,1)$.

Corollary 7.5**.**

Let $\pi$ be an irreducible smooth representation of $M_{n}$. Then for any $\mathfrak{s}\in\mathfrak{B}(G_{n})$, $\pi_{\mathfrak{s}}$ is indecomposable.

Proof.

This follows from [BZ77, Corollary 3.5] and Theorem 7.3. ∎

8. Appendix: Speh representation approximation

8.1. Two lemmas

Some studies on parabolic induction can be found in [Ta15, LM16]. We give a proof of the following specific cases, using the theory of derivatives. We use notations in Section 6.2.

Lemma 8.1**.**

Let $m\in\mathbb{Z}$. Let $\Delta=[\nu^{a}\rho,\nu^{b}\rho]$. If $\Delta^{\prime}=[\nu^{k}\rho,\nu^{l}\rho]$ such that

[TABLE]

then

[TABLE]

and is irreducible.

Proof.

When $-(m-1)/2+a-k\notin\mathbb{Z}$, this case is easier and we omit the details. We now consider $-(m-1)/2+a-k\in\mathbb{Z}$.

The statement is not difficult when $\Delta$ is singleton $[\rho]$ because

[TABLE]

and $\mathrm{St}(\Delta^{\prime})\times\langle\Delta^{\prime\prime}\rangle\cong\langle\Delta^{\prime\prime}\rangle\times\mathrm{St}(\Delta^{\prime})$ whenever $\Delta^{\prime\prime}\subset\Delta^{\prime}$. (Indeed, one can also prove the latter fact by similar arguments as below by noting that the Zelevinsky multisegment of any simple composition factor in $\mathrm{St}(\Delta^{\prime})\times\langle\Delta^{\prime\prime}\rangle$ contains a segment $\widetilde{\Delta}$ with $b(\widetilde{\Delta})\cong b(\Delta^{\prime})$ and at least one segment $\widehat{\Delta}$ with $b(\widehat{\Delta})\cong b(\Delta^{\prime\prime})$.)

We now assume $\Delta$ is not a singleton. We consider two cases:

1. (1)

Case 1: $l=b+\frac{m-1}{2}$. Suppose $\tau$ is a composition factor of $a(m,\Delta)\times\langle\Delta^{\prime}\rangle$ with the associated Zelevinsky multisegment $\mathfrak{m}$. Then we know that at least two segments $\Delta_{1},\Delta_{2}$ in $\mathfrak{m}$ takes the form $b(\Delta_{1})\cong b(\Delta_{2})\cong\nu^{\frac{m-1}{2}+b}\rho$. If $\tau^{(i^{*})}$ is the highest derivative of $\tau$, then we know that the cuspidal support of $\tau^{(i^{*})}$ does not contain $\nu^{\frac{m-1}{2}}\rho$. We also have that $\tau^{(i^{*})}$ is a composition factor of $(u(m,\Delta)\times\langle\Delta\rangle)^{(i^{*})}$. The only possibility is that $i^{*}=m+1$ i.e $(u(m,\Delta)\times\langle\Delta^{\prime}\rangle)^{(i^{*})}=u(m,\Delta^{-})\times\langle(\Delta^{\prime})^{-}\rangle$, which the latter one is irreducible by induction. This proves the lemma. Since taking derivative is an exact functor, we have that $u(m,\Delta)\times\langle\Delta^{\prime}\rangle$ is irreducible. Using the Gelfand-Kazhdan involution [BZ76, Section 7], we have that $\langle\Delta^{\prime}\rangle\times u(m,\Delta)\cong u(m,\Delta)\times\langle\Delta^{\prime}\rangle$. 2. (2)

Case 2: $l<b+\frac{m-1}{2}$. The argument is similar. Again suppose $\tau$ is a composition factor of $u(m,\Delta)\times\langle\Delta^{\prime}\rangle$. Using an argument similar to above, we have that the level of $\tau$ is either $m+1$ or $m$. However, if the level of $\tau$ is $m$, then $\tau^{(m)}$ would be $u(m,\Delta^{-})\times\langle\Delta^{\prime}\rangle$, which is irreducible by induction. Then it would imply that the number of segments in Zelevinsky multisegment of $\tau^{(m)}$ is $m$ and contradicts that the number of segments for the Zelevinsky multisegment of the highest derivative of an irreducible representation $\pi$ must be at most that for $\pi$. Hence, the level of $\tau$ must be $m+1$. Now repeating a similar argument as in (1) and using the induction, we obtain the statements.

∎

Lemma 8.2**.**

Let $\Delta$ be a segment. Then

[TABLE]

is irreducible.

Proof.

The statement is clear if $\Delta$ is a singleton. For the general case, we note by simple counting that the Zelevinsky multisegment of any composition factor must contain a segment $\Delta_{1}$ with $b(\Delta_{1})\cong\nu^{(m-1)/2}b(\Delta)$ and at least one segment $\Delta_{2}$ with $b(\Delta_{2})\cong\nu^{(m-1)/2}b(\Delta^{-})$. Then one proves the statement by a similar argument using highest derivative as in the previous lemma.

∎

8.2. Speh representation approximation

For a Speh multisegment

[TABLE]

define $b(\mathfrak{m})=b(\Delta)$.

Proposition 8.3**.**

Let $\mathfrak{m}=\left\{\Delta_{1},\ldots,\Delta_{k}\right\}$. Then there exists Speh multisegments $\mathfrak{m}_{1},\ldots\mathfrak{m}_{r}$ satisfying the following properties:

1. (1)

$\mathfrak{m}=\mathfrak{m}_{1}+\ldots+\mathfrak{m}_{r}$; 2. (2)

For each Speh multisegment $\mathfrak{m}_{i}$ and any $j>i$, there is no segment $\Delta$ in $\mathfrak{m}_{i}$ such that $\mathfrak{m}_{j}+\left\{\Delta\right\}$ is a Speh multisegment; 3. (3)

$\langle\mathfrak{m}\rangle$* is the unique submodule of $\langle\mathfrak{m}_{1}\rangle\times\ldots\times\langle\mathfrak{m}_{r}\rangle$;* 4. (4)

$b(\mathfrak{m}_{i})\not\leq b(\mathfrak{m}_{j})$* if $i<j$* 5. (5)

if $\mathfrak{m}_{i}\cap\mathfrak{m}_{j}\neq\emptyset$ and $i\leq j$, then $\mathfrak{m}_{j}\subset\mathfrak{m}_{i}$.

Proof.

We shall label the segments $\mathfrak{m}$ in the way that for $i<j$, either (i) $b(\Delta_{i})\not<b(\Delta_{j})$ and (ii) if $b(\Delta_{i})=b(\Delta_{j})$, then $a(\Delta_{i})\not<a(\Delta_{j})$. Let $\Delta=\Delta_{1}$. Let $k$ be the largest integer ($k\geq 0$) such that $\Delta,\nu^{-1}\Delta,\ldots,\nu^{-k}\Delta$ are segments in $\mathfrak{m}$. We claim that

[TABLE]

and moreover $\langle\mathfrak{m}\rangle$ is the unique submodule of $\langle\mathfrak{m}^{\prime}\rangle\times\langle\mathfrak{m}\setminus\mathfrak{m}^{\prime}\rangle$. By induction, the claim proves (1) and (3), and the remaining ones follow from the inductive construction.

We now prove the claim. We shall prove by induction that for $i=0,\ldots,k$,

[TABLE]

where $\mathfrak{m}_{i}=\left\{\Delta,\nu^{-1}\Delta,\ldots,\nu^{-i}\Delta\right\}$. When $i=0$ the statement is clear from the definition. Now suppose that we have the inductive statement for $i$. To prove the statement for $i+1$. We now set $\mathfrak{n}$ to be the collection of all segments $\Delta^{\prime}$ in $\mathfrak{m}$ such that $b(\Delta^{\prime})=b(\Delta),\nu^{-1}b(\Delta),\ldots,\nu^{-i}b(\Delta)$ and $\nu^{-i}a(\Delta)$ proceeds $a(\Delta^{\prime})$. We also set $\bar{\mathfrak{n}}$ to be the collection of all segments $\Delta^{\prime}$ in $\mathfrak{m}$ such that $b(\Delta^{\prime})=b(\Delta),\nu^{-1}b(\Delta),\ldots,\nu^{-i}b(\Delta)$ and $a(\Delta^{\prime})$ precedes $\nu^{-i-1}a(\Delta)$. By using (2.1 ) several times, we have that $\zeta(\mathfrak{n})\times\zeta(\bar{\mathfrak{n}})\cong\zeta(\bar{\mathfrak{n}}+\mathfrak{n})$ and

[TABLE]

From our construction (and $i\neq k$), we have that $\nu^{-i-1}\Delta$ is a segment in $\mathfrak{m}\setminus(\mathfrak{n}+\bar{\mathfrak{n}}+\mathfrak{m}_{i})$ and

[TABLE]

On the other hand

[TABLE]

The first and last isomorphism is by Lemma 8.3. The injectivity in forth line comes from the uniqueness of submodule in $\zeta(\mathfrak{m}_{i+1})$. The second isomorphism follows from again by (2.1). Combining (8.27), the above series of isomorphisms and the inductive case, we have:

[TABLE]

This gives the desired injectivity by using the uniqueness of submodule of $\zeta(\mathfrak{m})$ and proves the claim.

∎

We shall need a variation which is more flexible in our application.

Lemma 8.4**.**

Let $\mathfrak{m}=\mathfrak{m}(m,\Delta)$ be a Speh multisegment. Let $\mathfrak{n}_{1}$ be a Zelevinsky multisegment such that for any segment $\Delta^{\prime}$ in $\mathfrak{n}_{1}$ satisfying $b(\Delta)<b(\Delta^{\prime})$ or $b(\Delta)\cong b(\Delta^{\prime})$. Let $\mathfrak{n}_{2}$ be a Zelevinsky multisegment such that for any segment $\Delta^{\prime}$ in $\mathfrak{n}_{2}$ satisfying either one of the following properties:

1. (1)

$b(\Delta)\not<b(\Delta^{\prime})$; or 2. (2)

$b(\Delta)\not<b(\Delta^{\prime})$, or if $b(\Delta)<b(\Delta^{\prime})$, then $(\Delta^{\prime})^{-}=\Delta$.

Then

[TABLE]

In particular, $\zeta(\mathfrak{n}_{1})\times\langle\mathfrak{m}\rangle\times\zeta(\mathfrak{n}_{2})$ has unique submodule isomorphic to $\langle\mathfrak{n}_{1}+\mathfrak{m}+\mathfrak{n}_{2}\rangle$.

Remark 8.5**.**

Case (2) covers case (1). But for the purpose of clarity of an argument used in Section 6.4, we divide into two cases. We also recall that the case $b(\Delta)\not\leq b(\Delta^{\prime})$ includes the possibility $b(\Delta)\cong b(\Delta^{\prime})$.

Proof.

For all cases, we have that

[TABLE]

Using (8.26) for (1) we obtain the lemma. For (2), let $\mathfrak{n}^{\prime}$ be all the segments in $\mathfrak{n}_{2}$ with the property that $(\Delta^{\prime})^{-}\cong\Delta$. Then we have that

[TABLE]

By Lemma 8.2, we have that

[TABLE]

which proves the lemma. ∎

9. Appendix

Let $G$ be a connected reductive group over a non-Archimedean local field.

Lemma 9.1**.**

Let $\pi$ be a smooth representation of $G$. Let $\pi$ admits a filtration

[TABLE]

Suppose $\pi$ admits an irreducible subquotient $\tau$. Let $\pi_{k}=\pi_{k}^{\prime}/\pi_{k-1}^{\prime}$ for $k=1,\ldots,r$. Then there exists $s$ such that $\pi_{s}$ contains an irreducible subquotient isomorphic to $\tau$.

Proof.

Let $\lambda_{1},\lambda_{2}$ be submodules of $\pi$ such that $\lambda_{2}/\lambda_{1}=\tau$. Now we have a filtration on $\lambda_{p}$ ($p=1,2$):

[TABLE]

Now $\lambda_{2}/\lambda_{1}$ admits a filtration

[TABLE]

where $\gamma_{k}=(\pi_{k}^{\prime}\cap\lambda_{2})/(\pi_{k}^{\prime}\cap\lambda_{1})$ and so

[TABLE]

This implies $\pi_{k+1}^{\prime}\cap\lambda_{2}$ and so $\pi_{k+1}^{\prime}$ has irreducible subquotient isomorphic to $\tau$.

∎

Lemma 9.2**.**

Let $\pi$ be a non-zero smooth representation of $G$. Let $\tau$ be a non-zero $G$-submodule of $\pi$. Suppose $\pi$ admits a filtration on

[TABLE]

Then there exists a non-zero $G$-submodule $\tau^{\prime}$ of $\tau$ such that

[TABLE]

for some $k$.

Proof.

Let $k$ be the smallest positive integer such that

[TABLE]

Define $\tau^{\prime}=\tau\cap\pi_{k}$. The non-zero $G$-submodule $\tau^{\prime}$ embeds to $\pi_{k}/\pi_{k-1}$ as desired. ∎

Lemma 9.3**.**

Let $0\neq\pi\in\mathrm{Alg}(G)$. Suppose $\pi$ admits two $G_{n}$-filtrations:

[TABLE]

and

[TABLE]

Then there exists a non-zero $\tau\in\mathrm{Alg}(G)$ such that for some $i,j$,

[TABLE]

Proof.

This follows by considering the smallest integer $k$ such that $\pi_{1}\cap\pi^{\prime}_{k}\neq 0$. ∎