JACQUET MODULES AND IRRREDUCIBILITY OF INDUCED REPRESENTATIONS FOR CLASSICAL p-adic GROUPS

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1 JACQUET MODULES AND IRRREDUCIBILITY OF INDUCED REPRESENTATIONS FOR CLASSICAL p-adic GROUPS CHRIS JANTZEN 1. Introduction Let G be a classical p-adic group of the type consided in [Mœ-T]. If T is an irreducible tempered representation of such a group and ρ an irreducible unitary supercuspidal representation of a general linear group, we can form the parabolically induced representation Ind G P (νy ρ T ), where ν = det as in [B-Z]. The main result in this paper is the determination for which y R the induced representation is reducible. The key technical result in establishing this is the determination of a certain Jacquet module subquotient µ {ν x ρ} (discussed later in this introduction and dealt with at length in the paper). The results are somewhat reminiscent of those on representations induced from discrete series in [Mu], though the approach is a bit different. Part of the purpose of this paper is, in fact, to demonstrate a different approach. The results could also serve as a starting point to a more general analysis of induced representations, with a long-term goal of understanding when a standard module reduces. However, our immediate purpose is in analyzing the duality operator of [Au], [S-S] for the groups under consideration. In fact, both the µ {ν x ρ} (T ) results and the reducibility results obtained from them are central to [J5]. To start, we recall a key definition. Let G n (F ) be from one of the families of classical groups under consideration (symplectic, odd special orthogonal, even orthogonal, unitary see section ). For 1 m n, we have a (maximal proper) standard parabolic subgroup with Levi factor M (m) = GL(m, F ) G n m (F ). For π a representation of G, set n µ (π) = r M(i),G(π), i=0 with the sum in R R[S] see section.1. This was originally defined in [T1], and has many useful properties, discussed in more detail in the next section. However, the definition is sufficient for the purposes at hand. In particular, it enables us to define the following (Definition [J4]): Definition 1.1. For X a set of (not necessarily unitary) supercuspidal representations of general linear groups, let f = f π (X) be the largest value such that a (minimal nonzero) Jacquet module of π has a term of the form ν x 1 ρ 1 ν x f ρ f... with ν x 1 ρ 1,..., ν x f ρ f X (where ν = det as in [Ze]). We let µ X(π) = i λ i θ i, where the sum is over all irreducible λ i θ i µ (π) for which the (minimal nonzero) Jacquet module of λ i contains a term of the form ν x 1 ρ 1 ν x f ρ f, with ν x 1 ρ 1,..., ν x f ρ f X Research supported in part by NSA grant H

2 CHRIS JANTZEN Suppose X has the property that ν x ρ X ν x ˇρ X. If π is irreducible, we have the following key properties (section 3.1 [J4]): (1.1) µ X (π) consists of a single representation, denoted λ π(x) θ π (X). π i G,M (λ π (X) θ π (X)) as unique irreducible subrepresentation. If µ X (π 1) = µ X (π ), then π 1 = π (follows from above). Note that we have X = 1 for most applications. To illustrate how these results can be used to analyze reducibility, we look at the example of i G,M (ν 1 ρ T ), T tempered (in the more interesting case, when the cuspidal reducibility is in 1 + N; see section.1). First, observe that we have π = L quot(ν 1 ρ; T ) the Langlands quotient of i G,M (ν 1 ρ T ) as the unique irreducible quotient. In the reducible case, we show there is also a tempered subrepresentation T. We do so by providing the data for T, then using Theorem 3.1 which gives µ {ν 1 (T ) and (1.1) to show that T i G,M (ν 1 ρ T ). In the irreducible case, Proposition 4.1 tells us that if i G,M (ν 1 ρ T ) has another irreducible subquotient, it must be tempered. Were there such a T, it is not difficult to show that for X = {ν 1, one would have θ T (X) = θ T (X). However, one can also use Theorem 3.1 and the observation that f T (X) = f T (X) + 1 (with f(x) the f in Definition 1.1) to show that exactly one of θ T (X) and θ T (X) is tempered, a contradiction. The arguments in this paper are ultimately built from the Mœglin-Tadić classification of discrete series ([Mœ-T]) and the machinery needed for that classification. We therefore make the assumptions needed for [Mœ-T]; the results then apply in the generality of [Mœ-T], that is, to the symplectic, odd special orthogonal, even orthogonal, and unitary groups considered there. We close by briefly describing the contents of this paper. In the next section, we introduce notation and give some background results. In section 3, we give the result for µ {ν x ρ} (T ), where T is tempered (Theorem 3.1). Section 4 contains the main reducibility results (Theorem 4.7; also Note 4.8). We close with an appendix, which offers a characterization of admissibility (in the sense of [Mœ-T]) designed to make claims of admissibility more transparent. We also take the opportunity to thank the referee for suggestions and corrections which helped improve the paper.. Notation and preliminaries.1. Notation and preliminaries. We first discuss some structure theory from [Ze] and [T1],[Ba]. First, let S(n, F ) denote the rank n member of one of the families of classical groups under consideration and set R = R(GL(n, F )) and R[S] = R(S(n, F )), n 0 n 0 where R(G) denotes the Grothendieck group of the category of smooth finite-length representations of G. We define multiplication on R as follows: suppose ρ 1, ρ are representations of GL(n 1, F ), GL(n, F ), resp. We have M = GL(n 1, F ) GL(n, F ) the Levi factor of a standard parabolic subgroup of G = GL(n, F ), where n = n 1 +n, and set τ 1 τ = i G,M (τ 1 τ ) (normalized parabolic induction see [B-Z]). This extends (after semisimplification) to give the multiplication : R R R. To describe the comultiplication on R, let M (i) denote the standard Levi factor for G = GL(n, F ) having M (i) = GL(i, F ) GL(n i, F ). For a representation τ of GL(n, F ), we define n m (τ) = r M(i),G(τ), i=0

3 IRREDUCIBILITY OF INDUCED REPRESENTATIONS 3 the sum of semisimplified Jacquet modules (lying in R R). This extends to a map m : R R R. We note that with this multiplication and comultiplication (and antipode map given by the Zelevinsky involution, a special case of the general duality operator of [Au],[S-S]), R is a Hopf algebra. There are two analogues for general linear groups of the µ X discussed in the introduction: m X and Xm. For an irreducible representation π, we let f = f π (X) (resp., g = g π (X)) be the largest value such that a minimal nonzero Jacquet module of π has a term of the form ν x 1 ρ 1 ν x f ρ f... (resp., of the form ν xg ρ g ν x 1 ρ 1 ) with all ν x i ρ i X, 1 i f (resp., 1 i g). The analogue of (1.1) holds without restriction on X (Lemma.1. [J]); we define m X and Xm accordingly. Recall that for a, b with a b and b a Z, δ([ν a ρ, ν b ρ]) denotes the generalized Steinberg representation associated to the segment [ν a ρ, ν b ρ], i.e., the unique irreducible subrepresentation of ν b ρ ν b 1 ρ ν a ρ ([Ze]). The unique irreducible subrepresentation of ν a ρ ν a+1 ρ ν b ρ is denoted ζ([ν a ρ, ν b ρ]). Note that ζ([ν a ρ, ν b ρ]) is dual to δ([ν a ρ, ν b ρ]). Next, suppose τ is a representation of GL(n 1, F ) and θ a representation of S(n, F ). We have M = GL(n 1, F ) S(n, F ) the Levi factor of a standard parabolic subgroup of G = S(n, F ), with n = n 1 +n, and set τ θ = i G,M (τ θ). If one extends to a map : R R[S] R[S], we have R[S] as a module over R. To describe its comodule structure, let M (i) = GL(i, F ) S(n i, F ), a standard Levi factor for G = S(n, F ). For a representation π of S(n, F ), we define n µ (π) = r M(i),G(π), i=0 the sum of (normalized) semisimplified Jacquet modules (lying in R R[S]). This extends to a map µ : R[S] R R[S]. In addition to µ X introduced earlier, there is another variant of this which is needed occasionally in what follows. For an irreducible representation λ of a general linear group and a representation π of one of the classical groups under consideration, we let µ λ (π) be the sum of everything in µ (π) having first factor isomorphic to λ. More precisely, if µ (π) = i λ i ξ i, we let µ λ (π) = i I λ λ i ξ i, where I λ = {i λ i = λ}. For unitary groups, let ξ denote the nontrivial element of the Galois group of the underlying quadratic extension. For a representation π of S(n, F ), we then define { π ξ for unitary groups, ˇπ = π otherwise, where denotes contragredient. Using this, we may give R[S] the structure of an M -module over R ([T1],[Ba],[Mœ-T]): Theorem.1. Define M : R R R by M = (m 1) (ˇ m ) s m, where m denotes the multiplication : R R R and s : R R R R the extension of the map defined on representations by s : τ 1 τ τ τ 1. Then µ (τ π) = M (τ) µ (π), where on the right hand side is determined by (τ 1 τ ) (τ θ) = (τ 1 τ) (τ θ). We now take a moment to review cuspidal reducibility values. Suppose ρ is an irreducible unitary supercuspidal representation of a general linear group and σ an irreducible supercuspidal representation of a classical group. If ρ = ˇρ, then ν x ρ σ is irreducible for all x R; if ρ = ˇρ, then there is a unique nonnegative x R such that ν x ρ σ reduces ([Si] and Corollary 4.4 [B-J1]), which we denote by red(ρ; σ). The values for red(ρ; σ) for Sp(n, F ) and SO(n + 1, F ) have been

4 4 CHRIS JANTZEN determined (assuming certain conjectures) in [Mœ1] and [Zh]; in the generic case, it is known that they must lie in {0, 1, 1} ([Sh1],[Sh]). Further, in the quasi-split, characteristic zero case, the reducibility values are now known to be half-integral ([Art], [Mœ]). We next review the Casselman criterion for S(n, F ) (see [Ca],[Wa], which extends easily to the non-connected group O(n, F )). Suppose π is an irreducible representation of S(n, F ). Suppose ν x 1 ρ 1 ν x kρ k σ r M,G (π) has ρ i an irreducible unitary supercuspidal representation of GL(m i, F ) for i = 1,..., k, σ an irreducible supercuspidal representation of S(m, F ), and x 1,..., x k R. The Casselman criterion tells us that if π is tempered, the following hold: m 1 x 1 0 m 1 x 1 + m x 0. m 1 x 1 + m x + + m k x k 0. Conversely, if these inequalities hold for any such ν x 1 ρ 1 ν x kρ k σ (i.e., ρ i an irreducible unitary supercuspidal representation of GL(m i, F ) and σ an irreducible supercuspidal representation of S(m, F )) appearing in a Jacquet module of π, then π is tempered. The criterion for squareintegrability is the same except that the inequalities are strict. We also take a moment to review the Langlands classification ([B-W],[Si1],[Kon]; also the appendix of [B-J1] for the non-connected group O(n, F )). We work in the subrepresentation setting of the Langlands classification as it is the most convenient for applying Jacquet module methods. Suppose τ 1,..., τ k are irreducible tempered representations of general linear groups and x 1 < < x k. Then the induced representation ν x 1 τ 1 ν x kτ k has a unique irreducible subrepresentation which we denote L(ν x 1 τ 1,..., ν x kτ k ). Every irreducible admissible representation of a general linear group may be written in this way, and the data ν x 1 τ 1 ν x kτ k are unique. Turning to classical groups, if τ 1,... τ k are irreducible tempered representations of general linear groups, τ an irreducible tempered representation of S(n, F ), and x 1 < < x k < 0, the representation ν x 1 τ 1 ν x kτ k τ has a unique irreducible subrepresentation which we denote L(ν x 1 τ 1,..., ν x kτ k ; τ). Further, any irreducible admissible representation of a classical group may be written in this way, and the data ν x 1 τ 1 ν x kτ k τ are again unique. The next lemma is Lemma 5.5 of [J1]. Lemma.. Suppose π is an irreducible representation of G, λ an irreducible representation of M and π i G,M (λ). If L > M, then there is an irreducible representation ρ of L such that (1) π i G,L (ρ) () ρ is a subquotient of i L,M (λ). We pause to note that for X = {ρ} with ρ = ˇρ, the hypotheses for (1.1) do not hold. In this case, the first property in (1.1) holds, but only up to multiplicity. The second property also holds, but the third does not. For the first two, the proofs are essentially the same; for the third, an easy counterexample occurs when ρ σ is reducible the two components are inequvalent but have µ {ρ} the same. Lemma.3. Suppose f π (ν x ρ) = f. If π (ν x ρ) f λ 1 λ k T with λ 1 λ k T satisfying the conditions for Langlands data (subrepresentation setting), then (up to multiplicity if x = 0). µ {ν x ρ} (π) = (νx ρ) f L(λ 1,..., λ k ; T )

5 IRREDUCIBILITY OF INDUCED REPRESENTATIONS 5 Proof. By Lemma., π (ν x ρ) f θ for some irreducible θ λ 1 λ k T ; necessarily θ = θ π (ν x ρ). Also, as µ {ν x ρ} (π) = (νx ρ) f θ (up to multiplicity if x = 0), and r M,G (π) (ν x ρ) f λ 1 λ k T (by Frobenius reciprocity), it follows from Lemma and Remark [J4] that r M,G (θ) λ 1 λ k T (where θ is a representation of G and M < G is the appropriate standard Levi). However, the only irreducible subquotient of λ 1 λ k T containing λ 1 λ k in its Jacquet module is L(λ 1,..., λ k ; T ) ([B-J] or [J1]), finishing the lemma... The extended Mœglin-Tadić classification. In this section, we review the extension of the construction of [Mœ-T] to tempered representations. The extension used here is from [J4]; we also note the somewhat different extension available in [T4]. Recall that the Mœglin-Tadić classification is a bijective correspondence between (equivalence classes of) discrete series for a family of classical groups and (equivalence classes of) admissible triples. An admissible triple is a triple of the form (Jord, σ, ε). Here Jord consists of pairs (ρ, a), with ρ an irreducible unitary supercuspidal representation of a general linear group and a N subject to a parity condition from ρ, σ the partial cuspidal support (the supercuspidal representation of a classical group which appears in any minimal nozero Jacquet module term), and ε a function defined on a subset of Jord (Jord Jord) taking values in {±1} which essentially distinguishes between discrete series having the same supercuspidal support (or equivalently, by Lemma.1.1 [J4], the same Jord). Information about induced representations into which the discrete series embeds is also encoded in the data. This classification and its properties have been summarized in [Mœ-T], many of the references for this paper (e.g., [T],[Mu1],[Mu],[T3],[J3],[T4],[J4]), as well as many other places. We forgo doing so again and simply refer the reader to these sources, as well as to the characterization of admissibility in the appendix of this paper. To extend the Mœglin-Tadić classification to tempered representations, we first consider the elliptic case. Suppose (.1) T ell δ([ν c 1 +1 ρ 1, ν c 1 1 ρ 1 ]) δ([ν c l +1 ρ l, ν c l 1 ρ l ]) δ, with δ a discrete series for a classical group. Let (Jord(δ), σ, ε δ ) be the Mœglin-Tadić data for δ, with S δ Jord(δ) (Jord(δ) Jord(δ)) the domain for ε δ. Intuitively, we construct Jord(T ) from Jord(δ) by adding two copies each of (ρ 1, c 1 ),..., (ρ l, c l ) (one for each end of δ([ν c i +1 ρ i, ν c i 1 ρ i ]), even if c i = 1). More precisely, we introduce a fourth datum, m T the multiplicity so have T associated to (Jord(T ), σ, ε T, m T ). Then, Jord(T ) = Jord(δ) {(ρ 1, c 1 ),..., (ρ l, c l )} and m T (ρ, a) = { 1 if (ρ, a) Jord(δ), if (ρ, a) = (ρ i, c i ) for some i. Again, we have ε T : S T {±1}, with the domain S T Jord(T ) (Jord(T ) Jord(T )). We have S T S δ, and ε T Sδ = ε δ. The additional values of ε T effectively distinguish the l components of δ([ν c 1 +1 ρ 1, ν c 1 1 ρ 1 ]) δ([ν c l +1 ρ l, ν c l 1 ρ l ]) δ. In particular, we have the following extension of the basic embedding property of [Mœ-T]: for (ρ, a) Jord(T ), let a be the largest value of b < a satisfying (ρ, b) Jord(δ) if it exists. Then (Proposition.3. [J4])

6 6 CHRIS JANTZEN (.) (1) if m(ρ, a) = 1, ε ((ρ, a), (ρ, a )) = 1 T δ([ν a +1 ρ]) θ for some irreducible θ. () if m(ρ, a) =, ε ((ρ, a), (ρ, a )) = 1 T δ([ν a +1 ρ]) δ([ν a +1 ρ]) θ for some irreducible θ. Other basic properties of the Mœglin-Tadić classification also have counterparts in the extension to the tempered case; we forgo including them here but include citations when used. For more general tempered representations i.e., no longer assuming elliptic we construct Jord(T ), m T in a similar manner. Write (.3) T = δ([ν d 1 +1 ρ 1, ν d 1 1 ρ 1]) δ([ν dm+1 ρ m, ν dm 1 ρ m]) T ell (irreducibly induced) with T ell elliptic tempered. We construct Jord(T ), m T from Jord(T ell ), m Tell by adding one copy each of (ρ 1, d 1), ( ˇρ 1, d 1 ),..., (ρ m, d m ), ( ˇρ m, d m ) (so two copies of (ρ i, d i) are added if ˇρ i = ρ i ); S T and ε T match S Tell and ε Tell (noting that the corresponding induced representation is irreducible so we do not have components to distinguish). We then have an extension of the above embedding: for (ρ, a) Jord(T ell ), let a < a be the largest value such that (ρ, a ) Jord(T ell ) if it exists. Then ε ((ρ, a), (ρ, a )) = 1 T δ([ν a +1 a +1 ρ]) δ([ν ρ]) θ }{{} m(ρ,a) This is not explicitly stated in [J4], but is a straightforward consequence of (.) and the irreducibility of (.3). We add one notational convention: as defined in [Mœ-T], Jord ρ = {a (ρ, a) Jord }. By abuse of notation, we also allow this to be interpreted as {(ρ, a) (ρ, a) Jord } as this interpretation is useful when dealing with the domain of ε. Before closing, we note one immediate consequence of the definitions in section.3 [J4] which is needed later: Lemma.4. Suppose T is an elliptic tempered representation with (ρ, a) Jord(T ) and m(ρ, a) =. Let T be the elliptic tempered representation with T δ([ν a+1 ρ]) T. Then, T satisfies Jord(T ) = Jord(T ) \ {(ρ, a)} with m T and ε T given by restriction (and partial cuspidal supports the same). 3. Jacquet modules of tempered representations In this section, we prove the main Jacquet module result of this paper, Theorem 3.1. The elliptic case is covered by Proposition 3.4 (for a > ) and Lemma 3.7 (for a = 1, ). The general tempered case follows from (.3) and the elliptic tempered case. Note that the discrete series case is already known (Theorem 3.. [J4] or Theorem 8.5 [Ma-T]). We begin by stating the result, then proceed to prove it in a sequence of lemmas. Theorem 3.1. Suppose T is irreducible tempered with (ρ, a) Jord(T ). If a >, we have the following: (1) If ρ = ˇρ, then µ a 1 (T ) = (ν ρ) m T (ρ,a) L(δ([ν a+1 ρ]) m T (ρ,a) ; T 1 ),

7 IRREDUCIBILITY OF INDUCED REPRESENTATIONS 7 where m T1 (ρ, a) = m T1 (ˇρ, a) = 0, m T1 (ρ, b) = m T (ρ, b) for all other (ρ, b), and ε T1 = ε T (noting that (ρ, a) S T ). () If ρ = ˇρ but a 1 red(ρ; σ) mod 1, µ a 1 (T ) = (ν ρ) m T (ρ,a) T where m T (ρ, a) = 0, m T (ρ, a ) = m T (ρ, a), m T (ρ, b) = m T (ρ, b) for all other (ρ, b), and ε T = ε T (noting that (ρ, a) S T ). (3) If ρ = ˇρ with a 1 red(ρ; σ) mod 1 and either (i) (ρ, a ) Jord, or (ii) ε T (ρ, a)ε T (ρ, a ) 1 = 1, then µ a 1 (T ) = (ν ρ) m T (ρ,a) T 3, where m T3 (ρ, a) = 0, m T3 (ρ, a ) = m T (ρ, a)+m T (ρ, a ), and m T (ρ, b) = m T (ρ, b) for all other (ρ, b). If m T (ρ, a ) > 0, ε T3 is just given by restriction; if m T (ρ, a ) = 0, it is given by substituting (ρ, a ) for (ρ, a). More precisely, if m(ρ, a ) = 0, ε T3 is determined by the following changes: ε T3 (ρ, a ) = ε T (ρ, a) if defined, and ε T3 (ρ, a )ε T3 (ρ, b) 1 = ε T (ρ, a)ε T (ρ, b) 1 for all other b. (4) If ρ = ˇρ with a 1 red(ρ; σ) mod 1, a = a, and ε T (ρ, a)ε T (ρ, a ) 1 = 1, then µ (T ) = { (ν a 1 ρ) m T (ρ,a) 1 T 4 if m T (ρ, a) odd (ν a 1 ρ) m T (ρ,a) 1 L(δ([ν a+1 ρ]); T 5 ) if m T (ρ, a) even. Here, m T4 (ρ, a) = 1, m T4 (ρ, a ) = m T (ρ, a )+m T (ρ, a) 1, m T4 (ρ, b) = m T (ρ, b) for all other (ρ, b), and ε T4 = ε T ; for T 5, m T5 (ρ, a) = 0, m T5 (ρ, a ) = m T (ρ, a )+m T (ρ, a), m T5 (ρ, b) = m T (ρ, b) for all other (ρ, b), and ε T5 the restriction of ε T. If a =, we have the following: (5) If ρ = ˇρ, then µ {ν 1 (T ) = (ν 1 ρ) m T (ρ,) L((ν 1 ρ) m T (ρ,) ; T 1 ), where m T1 (ρ, a) = m T1 (ˇρ, a) = 0, m T1 (ρ, b) = m T (ρ, b) for all other (ρ, b), and ε T1 (noting that (ρ, ) S T ). (6) If ρ = ˇρ but red(ρ; σ) 1 mod 1, = ε T µ {ν 1 (T ) = (ν 1 ρ) m T (ρ,) T where m T (ρ, ) = 0, m T (ρ, b) = m T (ρ, b) for all other (ρ, b), and ε T (ρ, a) S T ). (7) If ρ = ˇρ with red(ρ; σ) 1 mod 1 and ε T (ρ, ) = 1, then = ε T (noting that µ {ν 1 (T ) = (ν 1 ρ) m T (ρ,) T 3, where m T3 (ρ, ) = 0, m T (ρ, b) = m T (ρ, b) for all other (ρ, b), and ε T 3 just the restriction of ε T. (8) If ρ = ˇρ with red(ρ; σ) 1 mod 1 and ε T (ρ, ) = 1, then { µ (T ) = (ν 1 ρ) m T (ρ,) 1 T 4 if m T (ρ, ) odd, {ν 1 ρ} (ν 1 ρ) m T (ρ,) 1 L(ν 1 ρ; T 5 ) if m T (ρ, ) even. Here, m T4 (ρ, ) = 1, m T4 (ρ, b) = m T (ρ, b) for all other (ρ, b), and ε T4 = ε T ; for T 5, m T5 (ρ, ) = 0, m T5 (ρ, b) = m T (ρ, b) for all other (ρ, b), and ε T5 just the restriction of ε T.

8 8 CHRIS JANTZEN If a = 1, we have the following: (9) If ρ = ˇρ, then µ {ρ} (T ) = (ρ)m T (ρ,1) T 1, where m T1 (ρ, 1) = m T1 (ˇρ, 1) = 0, m T1 (ρ, b) = m T (ρ, b) for all other (ρ, b), and ε T1 (noting that (ρ, 1) S T ). (10) If ρ = ˇρ but red(ρ; σ) 0 mod 1, = ε T µ {ρ} (T ) = m T (ρ,1) (ρ) m T (ρ,1) T where m T (ρ, 1) = 0, m T (ρ, b) = m T (ρ, b) for all other (ρ, b), and ε T (ρ, 1) S T ). (11) If ρ = ˇρ with red(ρ; σ) 0 mod 1, then { µ {ρ} (T ) = m T (ρ,1) (ρ) m T (ρ,1) T 3 if m T (ρ, 1) even, m T (ρ,1) 1 (ρ) m T (ρ,1) 1 T 3 if m T (ρ, 1) odd, = ε T (noting that { 0 if mt (ρ, 1) even where m T3 (ρ, 1) =, m 1 if m T (ρ, 1) odd T3 (ρ, b) = m T (ρ, b) for all other (ρ, b), and ε T3 = ε T if m T (ρ, 1) odd, ε T3 the restriction of ε T if m T (ρ, 1) even. The following lemma is covered by the results of [Mu], but corrects an error. We remark that the results could also be obtained using the approach from section 4 for the tempered case, noting that the Jacquet module results obtained in this section for tempered representations (and used in the arguments in section 4) are known in the discrete series case (Theorem 3.. [J4]). Lemma 3.. Let δ be a discrete series representation and a 1. Then ν a 1 ρ δ is reducible if and only if one of the following occurs: (1) a >, (ρ, a ) Jord(δ), and either (i) (ρ, a) Jord(δ) or (ii) ε δ (ρ, a)ε δ (ρ, a ) = 1, () a = and either (i) (ρ, ) Jord(δ) with red(ρ; σ) 1 mod 1, or (ii) ε δ(ρ, ) = 1, (3) a = 1 and (ρ, a) Jord(δ) with red(ρ; σ) 0 mod 1. Lemma 3.3. Let T be an elliptic tempered representation. If (ρ, a) Jord(T ), we have f T (ν a 1 ρ) = 0. If (ρ, a) Jord(T ), we have the following: (1) If a exists, f T (ν a 1 ρ) = { mt (ρ, a) if either a > a + or ε T (ρ, a)ε T (ρ, a ) 1 = 1, m T (ρ, a) 1 if a = a and ε T (ρ, a)ε T (ρ, a ) 1 = 1. () If a does not exist, f T (ν a 1 ρ) = { mt (ρ, a) if either a >, or a = and ε T (ρ, a) = 1, m T (ρ, a) 1 if either a = 1, or a = and ε T (ρ, a) = 1. Proof. That f T (ν a 1 ρ) = 0 when (ρ, a) Jord(T ) follows immediately from the embeddings of T into induced representations (equation (.1)) and the corresponding property for discrete series (e.g., Remark 1.3. [J4]). More generally, we have f T (ν a 1 ρ) m T (ρ, a) for a > 1. For (1), if a > a +, we use (3.1) T δ([ν c 1 +1 ρ 1, ν c 1 1 ρ 1 ]) δ([ν c h +1 ρ h, ν c h 1 ρ h ]) δ,

9 IRREDUCIBILITY OF INDUCED REPRESENTATIONS 9 with δ discrete series. If m T (ρ, a) = 1, then (ρ, a) Jord(δ); the result then follows from Proposition 3.. [J4] and a commuting argument (resp., µ argument) for f T (ν a 1 ρ) = 1 (resp., f T (ν a 1 ρ) = 0). If m T (ρ, a) =, then without loss of generality, suppose (ρ, a) = (ρ h, c h ). Then, (3.) T δ([ν c 1 +1 ρ 1, ν c 1 1 ρ 1 ]) δ([ν c h 1 +1 ρ h 1, ν c h 1 1 ρ h 1 ]) T h, for some irreducible T h δ([ν c h +1 ρ h, ν c h 1 ρ h ]) δ. The result when f T (ν a 1 ρ) = m T (ρ, a) then follows from Lemma [J4] and a commuting argument. Now, suppose a = a + and ε T (ρ, a)ε T (ρ, a ) 1 = 1. It follows from Proposition.3. [J4] that f T (ρ, a) < m T (ρ, a); an easy commuting argument with (3.1) then tells us f T (ν a 1 ρ) 1 f T (ν a 1 ρ) = 1. We now look at (). For a = 1, the result follows from (3.1): if m T (ρ, 1) =, then (ρ i, c i ) = (ρ, 1) for some i, and the result is clear. If m T (ρ, 1) = 1, then (ρ, 1) Jord(δ). Therefore, µ {ρ}(δ) = 0 by the Casselman criterion, from which f T (ρ) = 0 follows. For a = and ε T (ρ, ) = 1, the result follows from Lemma.3.5 [J4]. If ε T (ρ, ) = 1, it follows from Lemma.3.5 [J4] that f T (ν 1 ρ) < m T (ρ, ). If ε T (ρ, ) = 1 and m T (ρ, ) = 1, we have (ρ, ) Jord(δ). That ε δ (ρ, ) = ε T (ρ, ) = 1 f δ (ν 1 ρ) = 0 follows directly from the Mœglin-Tadić classification; that f T (ν 1 ρ) = 0 then follows from (3.1). If m T (ρ, ) = and ε T (ρ, ) = 1, we must have (ρ i, c i ) = (ρ, ) in (3.). It then follows (easy commuting argument) that f T (ν 1 ρ) 1 f T (ν 1 ρ) = 1. If a > and m T (ρ, a) = 1, it follows from (3.1), Proposition.1. [J4] and and easy commuting argument that f T (ν a 1 ρ) = 1. If m T (ρ, a) =, the result follows from (3.), Lemma [J4], and an easy commuting argument. Proposition 3.4. Let T be an elliptic tempered representation, (ρ, a) Jord(T ). Suppose f T (ν a 1 ρ) 0 (see Lemma 3.3) and a >. (1) If (ρ, a ) Jord(T ), µ a 1 (T ) = (ν ρ) m T (ρ,a) T 1 where T 1 has Jord(T 1 ) = (Jord(T ) {(ρ, a )}) \ {(ρ, a)}, m T1 (ρ, a ) = m T (ρ, a), and remaining multiplicities matching those for T. We obtain ε T1 by substituting (ρ, a ) for (ρ, a): ε T1 is determined by ε T1 (ρ, a ) = ε T (ρ, a) if defined, ε T1 (ρ, b)ε T1 (ρ, a ) 1 = ε T (ρ, b)ε T (ρ, a) 1, and ε T1 ST1 S T = ε T ST1 S T. () If (ρ, a ) Jord(T ) with ε T (ρ, a)ε T (ρ, a ) 1 = 1, µ a 1 (T ) = (ν ρ) m T (ρ,a) T, where Jord(T ) = Jord(T ) \ {(ρ, a)}, m T (ρ, a ) = m T (ρ, a ) + m T (ρ, a), remaining multiplicities the same as for T, and ε T = ε T ST. (3) If (ρ, a ) Jord(T ) with ε T (ρ, a)ε T (ρ, a ) 1 = 1, in order to have f T (ν a 1 ρ) 0, we must have m T (ρ, a) = (and then f T (ν a 1 ρ) = 1). Further, µ a 1 a+1 (T ) = ν ρ L(δ([ν ρ]); T3 ), where T 3 has Jord(T 3 ) = Jord(T ) \ {(ρ, a)} and m T3, ε T3 given by restriction. Proof. The values of f T (ν a 1 ρ) are given in Lemma 3.3. We start with (3), which is the easiest of the remaining cases. Here, we have T δ([ν a+1 ρ]) T3 ν a 1 ρ δ([ν a+1 ρ]) T3

10 10 CHRIS JANTZEN with T 3 as in the statement of the proposition. Lemma.3 then gives the result for (3). We now turn to (1), in which case f T (ν a 1 ρ) = m T (ρ, a). Write T δ([ν a 1 +1 ρ 1, ν a 1 1 ρ 1 ]) δ([ν a k +1 ρ k, ν a k 1 ρ k ]) δ. In what follows, we essentially reduce to the case k = 1, which is then addressed in Lemma 3.5 below. Thus, suppose k > 1. First, for 1 j k, let Then, Λ j = δ([ν a 1 +1 ρ 1, ν a 1 1 ρ 1 ]) δ([ν a 1 +j 1 ρ j 1, ν a j 1 1 ρ j 1 ]) δ([ν a j+1 +1 ρ j+1, ν a j+1 1 ρ j+1 ]) δ([ν a k +1 ρ k, ν a k 1 ρ k ]). T Λ j δ([ν a j +1 ρ j, ν a j 1 ρ j ]) δ T Λ j T j for some irreducible T j δ([ν a j +1 ρ j, ν a j 1 ρ j ]) δ. By definition, ε T Sj = ε Tj, where S j denotes the domain for T j. Write µ (T ) = (ν a 1 ρ) m T, where m = m T (ρ, a). Set δ j = { δ if mt (ρ, a) =, δ if m T (ρ, a) = 1, where for m T (ρ, a) = 1, the δ is that of Theorem 3.. [J4] (noting that this δ is elliptic tempered, but need not be square-integrable). We define Λ j = Λ j if m T (ρ, a) = 1 or (ρ j, a j ) = (ρ, a). If not, we let i be such that (ρ i, a i ) = (ρ, a) and replace δ([ν a i +1 ρ i, ν a i 1 ρ i ]) with δ([ν a+3 ρ]) in the definition of Λ j (with the rest of the terms remaining the same). Finally, set a j = { aj if (ρ j, a j ) = (ρ, a) Then a j if not. T Λ j a j δ([ν +1 ρ j, ν a j 1 ρ j ]) δ j T Λ j T j for some irreducible T j a j δ([ν +1 ρ j, ν a j 1 ρ j ]) δ j. Again, we have ε T S j = ε T j. Fix j. First, suppose m T (ρ, a) = 1 or (ρ j, a j ) = (ρ, a). Note that in this case, Jord(T j ) = (Jord(T j ) \ {(ρ, a)}) {(ρ, a )}. If we knew that µ (T j ) = (ν a 1 ρ) m T (ρ,a) T j, the k = 1 case (Lemma 3.5 below) would imply that ε T S j is as claimed. As the values of ε T on j S j determine ε T (section.3 [J4]), the result follows for this case. Thus to finish this case, it remains to show µ (T j ) = (ν a 1 ρ) m T (ρ,a) T j. Observe that we have Λ j = Λ j and T Λ j T j Λ j (ν a 1 ρ) m T (ρ,a) θ Tj = (ν a 1 ρ) m T (ρ,a) Λ j θ Tj,

11 IRREDUCIBILITY OF INDUCED REPRESENTATIONS 11 noting that θ Tj is tempered. Since µ (T ) = (ν a 1 ρ) m T (ρ,a) T, we have T Λ j θ Tj θ Tj = T j, as needed. Now, suppose (ρ j, a j ) (ρ, a) and m T (ρ, a) =. Note that in this case, we have Jord(T j ) = Jord(T j ) and must show T j = T j. We have T Λ j T j µ (T ) (ν a 1 ρ) Λ j T j T Λ j T j T j = T j, as needed. This finishes (1). For (), the argument is fairly simple if either (ρ, a) or (ρ, a ) has multiplicity two. If m T (ρ, a) =, we have T δ([ν a+1 ρ]) T, where the data for T are obtained by removing both copies of (ρ, a) and restricting ε T (Lemma.4). As µ (δ([ν a+1 ρ]) T ) = (ν a 1 ρ) (δ([ν a+3 ρ]) T ), and f T (ν a 1 ρ) =, it follows that (noting the irreducibility of δ([ν a+3 µ a 1 (T ) = (ν ρ) (δ([ν a+3 ρ]) T ). ρ]) T ) As δ([ν a+3 ρ]) T = T, the result follows. If m T (ρ, a) = 1 but m T (ρ, a ) =, the argument is similar: by Proposition [J4], we have µ T δ([ν a+3 ρ]) T, where the data for T are obtained by removing one copy each of (ρ, a) and (ρ, a ) and restricting ε T. Here, we have ( ) ( ) δ([ν a+3 ρ]) T = ν a 1 ρ δ([ν a+3 ρ]) T µ (T ) = ν a 1 ρ ( δ([ν a+3 ) ρ]) T. Again, the conclusion follows from the observation that T = δ([ν a+3 ρ]) T. If m T (ρ, a) = m T (ρ, a ) = 1, the argument is based on combining a pair of embeddings. First, by Proposition [J4], we have T δ([ν a+3 ρ]) T, with T having Jord(T ) = Jord(T ) \ {(ρ, a), (ρ, a )} and m T, ε T T δ([ν a+3 ρ]) T ν a 1 δ([ν a+3 ρ]) T. given by restriction. Then,

12 1 CHRIS JANTZEN By Lemma.3, letting θ T = θ T (ν a 1 ρ), we then have (3.3) θ T δ([ν a+3 ρ]) T. Note that it follows from Equation (3.3) that θ T (ν a 1 ρ) is tempered. In the embedding T δ([ν a 1 +1 ρ 1, ν a 1 1 ρ 1 ]) δ([ν a k +1 ρ k, ν a k 1 ρ k ]) δ T, we have (ρ, a), (ρ, a ) Jord(δ T ). By Theorem 3.. [J4], we have δ T ν a 1 ρ T, with T having Jord(T ) = Jord(δ T ) \ {(ρ, a)}, m T (ρ, a ) = with remaining multiplicities and ε T given by restriction. Then, by a commuting argument (noting the obvious irreducibilities involved) Again, by Lemma.3, T δ([ν a 1 +1 ρ 1, ν a 1 1 ρ 1 ]) δ([ν a k +1 ρ k, ν a k 1 ρ k ]) ν a 1 ρ T = ν a 1 ρ δ([ν a 1 +1 ρ 1, ν a 1 1 ρ 1 ]) δ([ν a k +1 ρ k, ν a k 1 ρ k ]) T. (3.4) θ T δ([ν a 1 +1 ρ 1, ν a 1 1 ρ 1 ]) δ([ν a k +1 ρ k, ν a k 1 ρ k ]) T. Recall ([J4]) that ε θt is defined as follows: we have θ T δ([ν a+3 ρ]) δ([ν a 1 +1 ρ 1, ν a 1 1 ρ 1 ]) δ([ν a k +1 ρ k, ν a k 1 ρ k ]) δ θt. For notational convenience, let (ρ 0, a 0 ) = (ρ, a ). T i δ([ν a i +1 ρ i, ν a i 1 ρ i ]) δ θt such that Then, for i = 0,..., k, there is a unique θ T δ([ν a 0 +1 ρ 0, ν a 0 1 ρ 0 ]) δ([ν a i 1 +1 ρ i 1, ν a i 1 1 ρ i 1 ]) δ([ν a i+1 +1 ρ i+1, ν a i+1 1 ρ i+1 ]) δ([ν a k +1 ρ k, ν a k 1 ρ k ]) T i. If S i denotes the domain for δ([ν a i +1 ρ i, ν a i 1 ρ i ]) δ θt, by definition (section.3 [J4]) we have ε θt Si = ε Ti. Observe that it then follows directly from (3.4) and the description of the data of T that ε θt S0 = ε T = ε T S0. Also, it follows directly from (3.3) and the description of the data of T that ε θt Si for i = 1,..., k. Combining these, we have = ε T Si = ε T Si ε θt Si = ε T Si for i = 0,..., k. As ε θt is determined by its values on S θt [Jord ρi (θ T ) Jord ρi (θ T ) Jord ρi (θ T )], i = 0, 1,..., k, it suffices to show that ε θt Si determines ε θt on S θt [Jord ρi (θ T ) Jord ρi (θ T ) Jord ρi (θ T )] for i = 0, 1,..., k. Fix i {0, 1,..., k}. If red(ρ i ; σ) = 0 or red(ρ i ; σ) 1 mod 1, then ε θt is defined on Jord ρi (θ T ); by Lemma.3.1 [J4], its values there determine ε θt on S θt [(Jord ρi (θ T ) (Jord ρi (θ T ) Jord ρi (θ T ))]. If red(ρ i ; σ) N, then ε θt is not defined on individual elements of Jord ρi (θ T ). However, it still follows from Lemma.3.1 [J4] that ε θt is determined on S θt [Jord ρi (θ T ) (Jord ρi (θ T ) Jord ρi (θ T ))] by its values on S i, i = 0,..., k (more precisely, those

13 IRREDUCIBILITY OF INDUCED REPRESENTATIONS 13 having ρ j = ρi ) as long as Jord ρi (δ θ ). (To see this, fix (ρ i, b) Jord(δ θt ) and observe that for ρ j = ρi, ε θt (ρ i, a i )ε θt (ρ i, a j ) 1 = ε θt (ρ i, a i )ε θt (ρ i, b) 1 ε θt (ρ i, b)ε θt (ρ i, a j ) 1.) However, that Jord ρi (δ θt ) follows for red(ρ i ; σ) 1 as in Example [Mœ-T], finishing this case and the proposition. Lemma 3.5. Proposition 3.4 holds in the case k = 1 and (ρ, a) satisfies either (i) a < a (or does not exist), or (ii) ε T (ρ, a)ε T (ρ, a ) 1 = 1. Proof. Write µ (T ) = (ν a 1 ρ) m T, where m = m T (ρ, a). Note that as in the proof of Proposition 3.4, it suffices to show that ε T is as claimed. Further, if we let S T (ρ) be that part of S T supported on Jord ρ (T ) (Jord ρ (T ) Jord ρ (T )), and similarly for S δ (ρ), S T (ρ), etc., it suffices to show that ε T ST (ρ) is as claimed. We note that the case where (ρ, a ) exists is covered by Lemma [J4]. Thus we assume (ρ, a ) does not exist. The proof is broken into three main cases: (1) Jord ρ (T ) = {(ρ, a)} with red(ρ; σ) half-integral, () Jord ρ (T ) = {(ρ, a)} with red(ρ; σ) integral, (3) (ρ, a + ) exists, where a + is the smallest value of b > a satisfying (ρ, b) Jord if it exists. Case 1: Jord ρ (T ) = {(ρ, a)} with red(ρ; σ) half-integral Observe that by supercuspidal support considerations (Lemma.4.1 [J4]), m T (ρ, a ) = m. Now, in one direction, ε T (ρ, a ) = 1 (Lemma.3.5 [J4]) T δ([ν 1 ρ]) m θ T (ν a 1 ρ) m δ([ν 1 ρ]) m θ. By Frobenius reciprocity, r M,G (T ) (ν a 1 ρ) m δ([ν 1 ρ]) m θ for the appropriate standard Levi factor M. As the only irreducible representation of a general linear group having (ν a 1 ρ) m δ([ν 1 ρ]) m in its Jacquet module is δ([ν 1 ρ]) m, we get µ (T ) δ([ν 1 It now follows from Note.4. and Lemma.3.5 [J4] that ε T (ρ, a) = 1. In the converse direction, ε T (ρ, a) = 1 (Lemma.3.5 [J4]) T δ([ν 1 ρ]) m θ (ν a 1 ρ) m δ([ν 1 ρ]) m θ r M,G (T ) (ν a 1 ρ) m δ([ν 1 ρ]) m θ ρ]) m θ. for the appropriate standard Levi factor M. As µ (T ) = (ν a 1 ρ) m T. It follows that µ (T ) δ([ν 1 ρ]) m θ. As above, this then implies ε T (ρ, a ) = 1, as needed. Case : Jord ρ (T ) = {(ρ, a)} with red(ρ; σ) integral First, observe that Jord ρ (T ) 1, so Jord ρ (δ) 1. This implies red(ρ; σ) {0, 1} (e.g., see section 14.4 [Mœ-T]). If red(ρ; σ) = 1, we have S T (ρ) Jord ρ (T ) Jord ρ (T ). Thus, ε = (see Example [Mœ-T], e.g.) and there is nothing to prove. We therefore assume red(ρ; σ) = 0 in what follows. Recall that in this case, one makes an arbitrary enumeration of components ρ σ = τ 1 (ρ; σ) τ 1 (ρ; σ) (see [T], [T3]) as part of the classification of discrete series.

14 14 CHRIS JANTZEN In the notation of Definition.3.6 [J4], ε T (ρ, a ) = η T λ T (δ([νρ, ν a 3 ρ]) m ; τ η (ρ; σ)) T (ν a 1 ρ) m λ T (δ([νρ, ν a 3 ρ]) m ; τ η (ρ; σ)) = λ (ν a 1 ρ) m T (δ([νρ, ν a 3 ρ]) m ; τ η (ρ; σ)) r M,G (T ) λ (ν a 1 ρ) m (ν a 3 ρ) m (νρ) m τ η (ρ; σ), with the irreducibility of (ν a 1 ρ) m λ immediate from supercuspidal support considerations. Now, observe that by the Langlands classification, L((ν a+1 ρ) m, (ν a+3 ρ) m,..., (ν 1 ρ) m ; τ η (ρ; σ)) is the unique irreducible representation containing (ν a+1 ρ) m (ν a+3 ρ) m (ν 1 ρ) m τ η (ρ; σ) in its Jacquet module. It then follows from duality that T (δ([νρ, ν a 1 ρ]) m ; τ η (ρ; σ)) is the unique irreducible representation containing (ν a 1 ρ) m (ν a 3 ρ) m (νρ) m τ η (ρ; σ) in its Jacquet module. Thus, µ (T ) λ T (δ([νρ, ν a 1 ρ]) m ; τ η (ρ; σ)). By Lemma..8 [J4] for m = and a similar argument when m = 1, this then implies as needed. ε T (ρ, a) = η, Case 3: (ρ, a + ) exists We first consider the case m T (ρ, a + ) = 1. In one direction, ε T (ρ, a + )ε T (ρ, a ) = 1 T δ([ν a 1 ρ, ν a + 1 ρ]) λ T (ν a 1 ρ) m δ([ν a 1 ρ, ν a + 1 ρ]) λ = δ([ν a 1 ρ, ν a + 1 ρ]) (ν a 1 ρ) m λ δ([ν a 3 ρ, ν a + 1 ρ]) (ν a 1 ρ) m+1 λ ε T (ρ, a + )ε T (ρ, a) 1 = 1. In the other direction, by Proposition [J4] ε T (ρ, a + )ε T (ρ, a) 1 = 1 T δ([ν a+1 ρ, ν a + 1 ρ]) T δ([ν a 1 ρ, ν a + 1 ρ]) δ([ν a+1 ρ]) T (Lemma.) T δ([ν a 1 ρ, ν a + 1 ρ]) ξ for some irreducible ξ δ([ν a+1 ρ]) T. From this embedding and the irreducibility of δ([ν a 1 ρ, ν a + 1 ρ]) ν a 1 ρ, we see that f T (ν a 1 ρ) = f ξ (ν a 1 ρ), so f ξ (ν a 1 ρ) = m. It then follows

15 IRREDUCIBILITY OF INDUCED REPRESENTATIONS 15 that ξ (ν a 1 ρ) m ξ for some irreducible ξ. Then, T δ([ν a 1 ρ, ν a + 1 ρ]) (ν a 1 ρ) m ξ = (ν a 1 ρ) m δ([ν a 1 ρ, ν a + 1 ρ]) ξ r M,G (T ) (ν a 1 ρ) m δ([ν a 1 ρ, ν a + 1 ρ]) ξ for the appropriate standard Levi factor M. It then follows from (1.1) that µ (T ) δ([ν a 1 ρ, ν a + 1 ρ]) ξ (Note.4. [J4]) ε T (ρ, a + )ε T (ρ, a ) 1 = 1, as needed. It remains to address the case m T (ρ, a + ) =. Note that as k = 1, this has m T (ρ, a) = 1, so (ρ, a) Jord(δ). The implication ε T (ρ, a + )ε T (ρ, a ) 1 = 1 ε T (ρ, a + )ε T (ρ, a) 1 = 1 is similar to the argument when m T (ρ, a + ) = 1. In the other direction, noting that ε T (ρ, a + )ε T (ρ, a) 1 = 1 µ δ([ν a+1 a + 1 (T ) = δ([ν a+1 ρ, ν a + 1 ρ]) (δ([ν a+1 ρ]) δ), ρ,ν ρ]) µ δ([ν a+1 a + 1 ((δ([ν a++1 ρ, ν a + 1 ρ]) δ) = δ([ν a+1 a + 1 ρ, ν ρ]) (δ([ν a+1 ρ]) δ). ρ,ν ρ]) a 1 has the right-hand side irreducible (since (ρ, a) Jord(δ)). As f a+1 δ([ν ρ,ν a 1 (ν ρ) = 3, we ρ]) δ get T δ([ν a+1 ρ, ν a + 1 ρ]) (ν a 1 ρ) 3 θ (Lemma.) T δ([ν a+1 ρ, ν a + 1 ρ]) λ 1 (ν a 1 ρ) θ for some irreducible θ and some irreducible λ 1 δ([ν a+1 ρ, ν a + 1 ρ]) ν a 1 ρ. The two possibilities for λ 1 are δ([ν a 1 ρ, ν a + 1 ρ]) and L(ν a 1 ρ, δ([ν a+1 ρ, ν a + 1 ρ])). In either case, we have ν a 1 ρ λ 1 is irreducible ([Ze] and Lemma 3.6 below). We now have T δ([ν a+1 ρ, ν a + 1 ρ]) (ν a 1 ρ) λ 1 θ (Lemma.) T λ ν a 1 ρ λ 1 θ = ν a 1 ρ λ λ 1 θ for some irreducible λ δ([ν a+1 ρ, ν a + 1 ρ]) ν a 1 ρ. Now, if λ = L(ν a 1 ρ, δ([ν a+1 ρ, ν a + 1 ρ])), the embedding above would immediately imply f T (ν a 1 ρ), a contradiction. Thus, λ = δ([ν a 1 ρ, ν a + 1 ρ]). Since δ([ν a 1 ρ, ν a + 1 ρ]) λ 1 is irreducible for either possibility for λ 1 ([Ze], Lemma [J]), we have T ν a 1 ρ δ([ν a 1 ρ, ν a + 1 ρ]) λ 1 θ = ν a 1 ρ λ1 δ([ν a 1 ρ, ν a + 1 ρ]) θ.

16 16 CHRIS JANTZEN Applying the same argument again tells us λ 1 m T (ρ, a + ) = 1, we now get as needed. T ν a 1 ρ δ([ν a 1 ρ, ν a + 1 ρ]) θ r M,G (T ) ν a 1 ρ δ([ν a 1 ρ, ν a + 1 ρ]) θ µ (T ) δ([ν a 1 ρ, ν a + 1 ρ]) θ (Note.4. [J4]) ε T (ρ, a + )ε T (ρ, a ) 1 = 1, Lemma 3.6. The following representations are irreducible: (1) ν a 1 ρ L(ν a 1 ρ, δ([ν a+1 ρ, ν b 1 ρ])) for b a, () L(ν a+1 ρ, δ([ν a+3 ρ]), ν a 1 ρ) ν a 1 ρ for a, (3) L(ν a+1 ρ, δ([ν a+3 ρ])) ν a 1 ρ for a. = δ([ν a 1 ρ, ν a + 1 ρ]). Arguing as in the case Proof. (1) is a special case of Lemma 4.1 (or more generally, Proposition 4.3) of [Ma1]. For a =, () and (3) follow from [Ze]. For a >, (3) is immediate from Lemma [J]. This leaves () when a >. In this case, observe that by Lemma [J], the only possible irreducible subquotients of L(ν a+1 ρ, δ([ν a+3 ρ]), ν a 1 ρ) ν a 1 ρ are the following: (1) π 1 = L(ν a+1 ρ, δ([ν a+3 ρ]), ν a 1 ρ), () π = L(ν a+1 ρ, δ([ν a+3 ρ]), ν a 1 ρ) (3) π 3 = L(δ([ν a+1 ρ]), ν a 1 ρ) (4) π 4 = L(δ([ν a+1 ρ]), ν a 1 ρ) (i.e., π 4 = δ([ν a+1 ρ]) ν a 1 ρ). By the Langlands classification, π 1 appears with multiplicity one. We now show that the remaining possibilities cannot occur. Observe that by Proposition.1.4 [J], L(ν a+1 ρ, δ([ν a+3 ρ]), ν a 1 ρ) has f(ν a 1 ρ) = 0, so any irreducible subquotient of L(ν a+1 ρ, δ([ν a+3 ρ]), ν a 1 ρ) ν a 1 ρ has f(ν a 1 ρ) 1. However, Proposition.1.4 [J] tells us f π (ν a 1 ρ) = f π4 (ν a 1 ρ) =. Thus, they do not occur. Further, m (π 3 ) contains a term of the form δ([ν a+1 ( ) ρ])..., whereas m L(ν a+1 ρ, δ([ν a+3 ρ]), ν a 1 ρ) ν a 1 ρ does not (as the ν a+3 3 ρ appears after the ν a+1 ρ in any Jacquet module term). Irreducibility follows. Lemma 3.7. Suppose T is elliptic tempered. (1) If (ρ, ) Jord(T ), then µ {ν 1 (T ) = { (ν 1 ρ) m T (ρ,) T 1 if ε T (ρ, ) = 1 (ν 1 ρ) m T (ρ,) 1 L(ν 1 ρ; T 1 ) if ε T (ρ, ) = 1 where Jord(T 1 ) = Jord(T ) \ {(ρ, )} with remaining multiplicities matching those of T and ε T1 given by restriction. Note that if ε T (ρ, ) = 1, we must have m T (ρ, ) = to have f T (ν 1 ρ) 0. () If (ρ, 1) Jord(T ), then we must have m T (ρ, 1) = to have f T (ρ) = m T (ρ, 1) 1 nonzero. In this case µ {ρ} (T ) = ρ T,

17 IRREDUCIBILITY OF INDUCED REPRESENTATIONS 17 where Jord(T ) = Jord(T ) \ {(ρ, 1)} with remaining multiplicities matching those of T and ε T given by restriction. Proof. The claim for () follows directly from the definition of Jord(T ). For (1), the proof in the case where ε T (ρ, ) = 1 is similar to that for Proposition 3.4 (3); the proof in the case where ε T (ρ, ) = 1 and m T (ρ, ) = is similar to that of Proposition 3.4 () in the case where m(ρ, a) =. The proof of (1) in the case where ε T (ρ, ) = 1 and m T (ρ, ) = 1 reduces to k = 1 in the same manner as in the proof of Proposition 3.4 () when m(ρ, a) = 1. Thus we are reduced to showing (1) in the case where ε T (ρ, ) = 1, m T (ρ, ) = 1, and k = 1. In this case, we have T δ([ν c+1 ρ, ν c 1 ρ]) δ for some discrete series δ. Then, writing µ {ν 1 (δ) = ν 1 ρ δ as in Theorem 3.. [J4], we have T δ([ν c+1 ρ, ν c 1 ρ]) ν 1 ρ δ = ν 1 ρ δ([ν c+1 ρ, ν c 1 ρ]) δ µ {ν 1 (T ) = ν 1 ρ T for some irreducible T δ([ν c+1 ρ, ν c 1 ρ]) δ. T clearly has the same partial cuspidal support as T ; that Jord(T ) and m T are as described follows from supercuspidal support considerations (Lemma.4.1 [J4]). It remains to show that ε T is as described. For this, it suffices to show that ε T (ρ, b) = ε T (ρ, b) for (ρ, b) Jord(T ) (so b > ). We do this in two cases. Case 1: (ρ, b ) does not exist in Jord(T ). Noting that m T (ρ, b) = m T (ρ, b), Lemma.3.5 [J4] tells us T ε T (ρ, b) = 1 T δ([ν 1 ρ, ν b 1 ρ]) m T (ρ,b) θ ν 1 ρ δ([ν 1 ρ, ν b 1 ρ]) m T (ρ,b) θ = δ([ν 1 ρ, ν b 1 ρ]) m T (ρ,b) ν 1 ρ θ δ([ν 3 ρ, ν b 1 ρ]) m T (ρ,b) (ν 1 ρ) m T (ρ,b)+1 θ (Lemma.) T δ([ν 3 ρ, ν b 1 ρ]) m T (ρ,b) θ ε T (ρ, b)ε T (ρ, ) 1 = 1 (Lemma.3.1 [J4]) ε T (ρ, b) = 1 as we have ε T (ρ, ) = 1. Conversely, ε T (ρ, b) = 1 ε T (ρ, b)ε T (ρ, ) 1 = 1 T δ([ν 3 ρ, ν b 1 ρ]) m T (ρ,b) θ.

18 18 CHRIS JANTZEN Further, by iterating Proposition 3.4, we see that θ is tempered with m θ (ρ, ) = m T (ρ, b) + 1 and ε T (ρ, ) = 1. In particular, f θ (ν 1 ρ) = 1 + m T (ρ, b). Thus, noting T ν 1 ρ T by (1.1), T δ([ν 3 ρ, ν b 1 ρ]) m T (ρ) (ν 1 ρ) m T (ρ,b)+1 θ µ (T ) δ([ν 1 ρ, ν b 1 ρ]) m T (ρ) ν 1 ρ λ for some irreducible λ µ (ν 1 ρ T ) δ([ν 1 ρ, ν b 1 ρ]) m T (ρ,b) ν 1 ρ λ µ (T ) δ([ν 1 ρ, ν b 1 ρ]) m T (ρ,b) λ by a standard µ analysis. As m T (ρ, b) = m T (ρ, b), this implies ε T (ρ, b) = 1 by the same argument as in Corollary..7 [J4], as needed. Case : (ρ, b ) exists in Jord(T ). By Lemma.3.1 [J4] and the preceding case, to show ε T (ρ, b) = ε T (ρ, b), it suffices to show ε T (ρ, b)ε T (ρ, b ) 1 = ε T (ρ, b)ε T (ρ, b ) 1. The proof that ε T (ρ, b)ε T (ρ, b ) 1 = 1 ε T (ρ, b)ε T (ρ, b ) 1 = 1 is very similar to the proof that ε T (ρ, b) = ε T (ρ, b) in the previous case. The converse direction is easier than in the previous case: ε T (ρ, b)ε T (ρ, b ) 1 = 1 µ (T ) δ([ν b +1 ρ, ν b 1 ρ]) m(ρ,b) λ for some irreducible λ µ (ν 1 ρ T ) δ([ν b +1 ρ, ν b 1 ρ]) m(ρ,b) λ µ (T ) δ([ν b +1 ρ, ν b 1 ρ]) m(ρ,b) λ for some irreducible λ, (as 1 { b +1, b +3,..., b 1 }). As m T (ρ, b) = m T (ρ, b), it then follows that ε T (ρ, b)ε T (ρ, b ) 1 = 1, as needed. 4. Reducibility results In this section, we determine when ν a 1 ρ T is reducible. We focus on the case where (ρ, a) Jord with a red(ρ; σ) + 1 mod, with the final result given in Theorem 4.7; Note 4.8 discusses the (much simpler) case when this fails. We start with a general lemma, then turn to the task of analyzing ν a 1 ρ T. Proposition 4.1. Suppose π (ν a+1 ρ) k δ([ν a+1 ρ]) l T is irreducible. Then, π has the form π = L((ν a+1 ρ) k, δ([ν a+1 ρ]) l ; T ) with k + l k + l. Proof. Write π = L( ; T ). If =, we are done. Thus, suppose δ([ν d+1 ρ, ν c 1 ρ]), c < d, is a generalized Steinberg representation appearing in. Then, if we write µ (T ) = h λ h θ h, it follows from Theorem.1 that δ([ν d+1 ρ, ν c 1 ρ]) (ν a+1 ρ) x 1 (ν a 1 ρ) x ( k s=1 δ([ν is+1 ρ, ν a 3 ρ]) δ([ν j s ρ, ν a 1 ρ]) ) λ h,

19 IRREDUCIBILITY OF INDUCED REPRESENTATIONS 19 where a+1 i s a+1. We first claim that d = a. If ν d+1 ρ came from and i s j s a+1 (ν a+1 ρ) x 1, we would have d = a, as claimed. We clearly cannot have ν d+1 ρ coming from (ν a 1 ρ) x. If d+1 = i s + 1, then we must have c 1 a 3 c a. Therefore, d a, so i s a+1 i s = a+1. This again corresponds to d = a. If j s = d+1, we must have c 1 a 1 c a. Therefore, d a +. This forces j s a 1, a contradiction. Thus we cannot have j s = d+1. Similarly, ν d+1 ρ cannot come from λ h as λ h must have nonnegative central exponent (by the Casselman criterion). Next, we claim that c 1 = a+1 or a 3 c 1. Again, consider the possible sources of ν ρ. If it came from (ν a+1 ρ) x 1 or δ([ν is+1 ρ, ν a 3 ρ]), we would have c 1 = a+1 or a 3, as claimed. The condition c < d = a immediately eliminates (ν a 1 ρ) x and δ([ν js ρ, ν a 1 ρ]) as possible sources. The remaining possibility is that ν c 1 ρ comes from λ h. As λ h must contribute to δ([ν a+1 ρ, ν c 1 must satisfy the Casselman criterion, we have λ h = δ([ν e+1 ρ, ν c 1 δ([ν a+1 ρ, ν e 1 ρ]) to be accounted for. We have e+1 c 1 ρ]) and ρ]) with e c < a. This leaves a 3 e 1, so < a 3. Therefore, ρ]). The ρ. Then e = a. Since a c < a, none of (ν a 1 ρ) x, δ([ν is+1 ρ, ν a 3 ρ]), δ([ν js ρ, ν a 1 ρ]) can contribute to δ([ν a+1 ρ, ν e 1 only remaining possibility is that δ([ν a+1 ρ, ν e 1 ρ]) = ν a+1 we must also have c = a, finishing the claim. As for the bounds on k, l, from the discussion above, each lower segment end of ν a+1 ρ appearing in (the GL part of) L((ν a+1 ρ) k, δ([ν a+1 ρ]) l ; T ) comes from a ν a+1 ρ in (ν a+1 ρ) k or δ([ν a+1 ρ]) l. It then follows that k + l k + l. We attack the reducibility conditions for ν a 1 ρ T, a >, through a sequence of lemmas. Lemma 4.. If m T (ρ, a ) >, then ν a 1 ρ T is reducible. Proof. Write T = δ([ν a+3, ν a 3 ρ]) T (irreducible as m T (ρ, a ) > ). Then ν a 1 ρ T = ν a 1 ρ δ([ν a+3, ν a 3 ρ]) T, whose reducibility follows from that of ν a 1 ρ δ([ν a+3, ν a 3 ρ]). Lemma 4.3. Suppose (ρ, a ) Jord. Then ν a 1 ρ T is irreducible. Proof. From Proposition 4.1, the only possible irreducible subquotients of ν a 1 ρ T are L(ν a 1 ρ; T ), L(δ([ν a+1 ρ]); T ), and T. The first occurs with multiplicity one by the Langlands classification. The second has a term of the form ν a 3 ρ... in its s (1), so cannot occur. Thus if the third does not occur, irreducibility follows. That the third cannot occur essentially reduces to supercuspidal support considerations. In the notation of Lemma.4.1 [J4], if a 3 red(ρ; σ), 0 = m T (ρ, a ) = n T ( ρ, a 3 ) n T ( ρ, a 1 ) ( n T ρ, a 3 ) ( = n T ρ, a 1 ), where n T (ρ, x) denotes the number of times ν ±x ρ appears in a term in the supercuspidal support (i.e., if ν x 1 ρ 1 ν x ρ ν xk 1 ρ k σ is in the minimal Jacquet module for T, then n T (ρ, x) = {i ρ i = ρ and xi = ±x} ). A tempered subquotient of T would have an extra ν ± a 1 ρ in its supercuspidal support, so ( m T (ρ, a ) = n T ρ, a 3 ) ( n T ρ, a 1 ) ( = n T ρ, a 3 ) ( (n T ρ, a 1 ) ) + 1 = 1

20 0 CHRIS JANTZEN a contradiction. The argument for a 3 < red(ρ; σ) is similar. The lemma now follows. Lemma 4.4. Suppose m T (ρ, a ) > 0 and either (1) m T (ρ, a) = 0, or () ε T (ρ, a)ε T (ρ, a ) 1 = 1. Then, ν a 1 ρ T is reducible. Proof. Let T have data given by m T (ρ, a ) 1 if (ρ, b) = (ρ, a ), m T (ρ, b) = m T (ρ, a) + 1 if (ρ, b) = (ρ, a), m T (ρ, b) otherwise, and ε T the restriction of ε T to S T (if m T (ρ, a) > 1, we have Jord(T ) = Jord(T ) and the restriction is an equality). We claim that such a T exists, i.e., that the data given is admissible. Recall that for (Jord(T ), σ, ε T, m T ) to be admissible, we need the underlying triple (Jord(T ) ds, σ, (ε T ) ds ) to be admissible in the [M-T] sense, where Jord(T ) ds = {(ρ, b) Jord(T ) b red(ρ; σ) + 1 mod and m T (ρ, b) odd} and (ε T ) ds is given by restriction of ε T, which in turn is the restriction of ε T. The admissibility may then be checked in cases based on the parity of m T (ρ, a) and m T (ρ, a ). In all four cases, admissibility is immediate from Proposition A.1 in the appendix. Now, observe that by Theorem 3.1, µ (T ) = (ν a 1 ρ ) mt (ρ,a)+1 T, where m T (ρ, a ) = m T (ρ, a )+m T (ρ, a), m T (ρ, a) = 0, and ε T given by restriction. Observe that Theorem 3.1 also tells us µ (T ) = (ν a 1 ρ) m T (ρ,a) T, so µ (ν a 1 ρ T ) = (ν a 1 ρ ) mt (ρ,a)+1 T. By Lemma., T ν a 1 ρ ((ν a 1 ρ) m T (ρ,a) T ) T ν a 1 ρ λ for some irreducible λ (ν a 1 ρ) m T (ρ,a) T. Since f T (ν a 1 ρ) 1+f λ (ν a 1 ρ), we have f λ (ν a 1 ρ) m T (ρ, a). Further, (λ) = since f T (ν a 1 ρ) m T (ρ, a) = 0, we have f λ (ν a 1 ρ) = m T (ρ, a). Therefore, µ (ν a 1 ρ) m T (ρ,a) T λ = T by (1.1). As we also have L(ν a+1 ρ; T ) ν a 1 follows. ρ T, reducibility Lemma 4.5. Suppose m T (ρ, a ) = and (ρ, a) Jord with ε T (ρ, a)ε T (ρ, a ) 1 = 1. Then, ν a 1 ρ T is reducible. Proof. Let π = L(ν a+1 ρ; T ). Observe that if ν a 1 ρ T were irreducible, we would have π = ν a 1 ρ T, hence f π (ν a 1 ρ) = 1 + f T (ν a 1 ρ). Thus, to show reducibility, it suffices to show f π (ν a 1 ρ) = f T (ν a 1 ρ). Note that f T (ν a 1 ρ) = m T (ρ, a) 1 by Theorem 3.1. As m T (ρ, a ) =, we have π ν a+1 ρ T ν a+1 ρ δ([ν a+3 ρ]) T, where the data for T is obtained by removing the two copies of (ρ, a ) and restricting ε T (Lemma.4). By Lemma., ) π L (ν a+1 ρ, δ([ν a+3 ρ]) T

21 IRREDUCIBILITY OF INDUCED REPRESENTATIONS 1 or π δ([ν a+1 ρ]) T. As the latter has L(δ([ν a+1 ρ]); T ) as its unique irreducible subrepresentation, the former must hold. Now, let m = m T (ρ, a) = m T (ρ, a). Since (ρ, a ) Jord(T ), we have T T as in Theorem 3.1. Thus, π L (ν a+1 ρ, δ([ν a+3 ρ]) ) (ν a 1 ρ ) m T. Applying Lemma. and Lemma 3.6, we have ) π (ν a 1 m 1 ( ) ρ L ν a+1 ρ, δ([ν a+3 a 1 ρ]), ν ρ T or π (ν a 1 ρ ) m 1 L ( ν a+1 ρ, δ([ν a+3 ρ]) ) T. If the latter held, we would have ( ) π ν a 1 m 1 a+1 ρ ν ρ δ([ν a+3 ρ]) T ( ) = ν a+1 ρ ν a 1 m 1 a+3 ρ δ([ν ρ]) T ( ) ν a+1 ρ ν a 1 m a+3 ρ δ([ν ρ]) T. ( ν a 1 ρ) m T, ( ) As µ (π) = ν a+1 {ν a+1 ρ T, this would imply r M,G (T ) ν a 1 m a+3 ρ δ([ν ρ]) T, contradicting f T (ν a 1 ) = m 1. Thus, the former holds. However, as (ρ, a) Jord(T ) and M (L ( ν a+1 ρ, δ([ν a+3 ρ]), ν a 1 tells us f π (ν a 1 ρ) = m 1, i.e., f π (ν a 1 ρ) = f T (ν a 1 ρ), as needed. ρ )) contains no terms of the form ν a 1 ρ..., the former Lemma 4.6. Suppose m T (ρ, a ) = 1 and (ρ, a) Jord with ε T (ρ, a)ε T (ρ, a ) 1 = 1. Then, ν a 1 ρ T is irreducible. Proof. From Proposition 4.1, there are three possible forms for an irreducible subquotient of ν a 1 ρ T : L(ν a+1 ρ; T ), L(δ([ν a+1 ρ]); T ), and T. The first appears with multiplicity one by the Langlands classification. Thus if we show the second and third do not appear, irreducibility follows. First, consider T. On one hand, T ν a 1 ρ T f T (ν a 1 ρ) 1 + f T (ν a 1 ρ) = m T (ρ, a). On the other hand, it follows from Lemma.4.1 [J4] that m T (ρ, a) = m T (ρ, a) + 1 and m T (ρ, a ) = m T (ρ, a ) 1 = 0. This implies f T (ν a 1 ρ) = m T (ρ, a) = m T (ρ, a) + 1, a contradiction. Thus, we cannot have T. We now consider L(δ([ν a+1 ρ]); T ). Were this to occur, we would have δ([ν a+1 ρ]) T M (ν a 1 ρ) µ (T ).