ON REDUCIBILITY POINTS BEYOND ENDS OF COMPLEMENTARY

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1 ON REDUCIBILITY POINTS BEYOND ENDS OF COMPLEMENTARY SERIES OF p-adic GL(n) MARKO TADIĆ arxiv: v1 [math.rt] 17 Oct 013 Abstract. In this paper we consider reducibility points beyond the ends of complementary series of general linear groups over a p-adic field, which start with Speh representations. We describe explicitly the composition series of the representations at these reducibility points. They are multiplicities one representations, and they can be of arbitrary length. We give Langlands parameters of all the irreducible subquotients. 1. Introduction Problems of reducibility of parabolically induced representations are very important in the harmonic analysis on reductive groups over local fields (they are of particular importance for the problem of unitarizability). They are also very important for the theory of automorphic forms for number of questions (among others, see section 6 of [33] for a reducibility interpretation of the local packets constructed in [1]). Closely related (usually very non-trivial) problem is the determination of the composition series at the reducibility points. Knowledge of composition series is equivalent to the corresponding character identity. In this paper we study such type of problems for general linear groups over a local non-archimedean field F. Speh representations are key representations in the classification of unitary duals of general linear groups (see [4]; for the archimedean case see [31]). One directly gets all the complementary series for general linear groups from the complementary series starting with Speh representations. Composition series at the ends of these complementary series are crucial in determining the topology of the unitary duals ([5]). The composition series at the ends of these complementary series played also crucial role in obtaining explicit formula for characters of irreducible unitary representations in terms of standard characters ([8]). Date: October 18, Mathematics Subject Classification. Primary: E50. Key words and phrases. non-archimedean local fields, division algebras, general linear groups, Speh representations, parabolically induced representations, reducibility, unitarizability. The author was partly supported by Croatian Ministry of Science, Education and Sports grant #

2 MARKO TADIĆ These complementary series terminate at the first reducibility point and the representations there have length two (when one starts with a Speh representation). It is a natural question to ask what are the composition series at the further reducibility points. There can exist a significant number of reducibility points beyond the end of complementary series. In all these reducibility points we completely determine the composition series, and give Langlands parameters of the irreducible subquotients. These representations are always multiplicity one representations (when one starts from a Speh representation), and can be of arbitrary length. For example, if we want to get a representation of length 1000 supported by the minimal parabolic subgroup, we shall need to start with Speh representations of GL( ,F) 1, and consider the complementary series of GL( ,F). It is interesting that it is very easy to write down Langlands parameters of all the irreducible subquotients (they are given by the simple formulas (1.1); see Theorem 1.). Now we shall describe more precisely the principal results of the paper. Put ν = det F, where F denotes the normalized absolute value on a local non-archimedean field F. For u,v Rsuchthatv u Z 0, andforanirreduciblecuspidal representation ρofgl(p,f), the set [ν u ρ,ν v ρ] = {ν u ρ,ν u+1 ρ,...,ν v 1 ρ,ν v ρ} is called a segment in cuspidal representations (of general linear groups). The representation ν u ρ is called the beginning of the segment [ν u ρ,ν v ρ]. We say that such a segment 1 precedes another segment, and write 1, if 1 is a segment different from 1 and, and if the beginnings of 1 and 1 are the same. Let = [ν u ρ,ν v ρ] be a segment in cuspidal representations. For z R, we denote ν z = {ν z ρ ;ρ }. Consider the representation Ind GL((v u+1)p,f) (ν v ρ ν v 1 ρ... ν u ρ), parabolically induced from the appropriate parabolic subgroup containing regular upper triangular matrices (see the second section). Then the above representation has a unique irreducible subrepresentation. This sub representation is essentially square integrable. It is denoted by δ( ). 1 This is the lowest rank in which we get length 1000 at some reducibility point beyond the complementary series starting with a Speh representation.

3 REDUCIBILITY POINTS BEYOND THE ENDS OF COMPLEMENTARY SERIES 3 Let a = ( 1,..., k ) be a finite multiset of segments in cuspidal representations. Write i = [ν u i ρ i,ν v i ρ i ], where δ( i ) is a representation of GL(n i,f) and ρ i are unitarizable irreducible cuspidal representations. Take a permutation σ of {1,...,k} such that Then the representation u σ(1) +v σ(1) u σ +v σ. Ind GL(n 1+ +n k,f) (δ( σ(1) )... δ( σ )), parabolically induced from the appropriate parabolic subgroup containing regular upper triangular matrices, has a unique irreducible quotient. We denote it by L(a). Then attaching a L(a) is one possible description of the Langlands classification of the non-unitary duals of groups GL(n, F) s (by multisets of segments in cuspidal representations). We shall use this version of Langlands classification for general linear groups in this paper. We add finite multisets of segments in obvious way: where a i = ( (i) 1,..., (i) k (i) ), i = 1,. a 1 +a = ( (1) 1,..., (1) k (1), () 1,..., () k () ), Let = [ν u ρ,ν v ρ] be a segment in cuspidal representations such that ρ is unitarizable and u+v = 0. Fix n Z 1. Then the representation u(δ( ),n) := L(ν (n 1)/,ν (n 1)/ 1,...,ν (n 1)/ ) is called a Speh representation. It is unitarizable and each irreducible unitary representation of a general linear group is constructed from several such representations in a simple way ([4]). Let k Z and denote d = card( ). We shall consider representations R t (n,d) (ρ) = Ind(ν k/ u(δ( ),n) ν k/ u(δ( ),n)). For k = 1, this is the end of complementary series. Denote Then and thus i = ν k/ (ν (n 1)/+i 1 ), Γ i = ν k/ (ν (n 1)/+i 1 ), i = 1,...,n. ν k/ u(δ( ),n) = L( 1,..., n ), ν k/ u(δ( ),n) = L(Γ 1,...,Γ n ), R t (n,d) (ρ) = Ind(L( 1,..., n ) L(Γ 1,...,Γ n )). Definition 1.1. For j = 0 and for 1 j n for which n Γ j, denote j r j (n,d) (ρ) = n j ( i+n j Γ i, i+n j Γ i )+ ( i,γ i+j ). (1.1)

4 4 MARKO TADIĆ Inother words, r 0 (n,d) (ρ) k = ( 1,..., n,γ 1,...,Γ n ), andfor1 j nforwhich n Γ j we get r j (n,d) (ρ) k by replacing in r 0 (n,d) (ρ) k = ( 1,..., n,γ 1,...,Γ n ) the part n j+1,..., n,γ 1,...,Γ j with n j+1 Γ 1, n j+1 Γ 1,..., n Γ j, n Γ j. One gets easily that n Γ j if and only if Theorem 1.. Let k Z 0. Then: max(n k +1,1) j min(n k +d,n). (1) The representation R t (n,d) (ρ) is a multiplicity one representation. It has a unique irreducible subrepresentation and unique irreducible quotient. The irreducible subrepresentation is isomorphic to L(r 0 (n,d) (ρ) ), Further, Rt (n,d) (ρ) and Rt (n,d) (ρ) ( k) have the same composition series. () For n+d k, and for k = 0, R t (n,d) (ρ) is irreducible. Then Rt (n,d) (ρ) = L(r 0 (n,d) (ρ) ). (3) For 0 < k < n+d, the composition series of R t (n,d) (ρ) consists of L(r i (n,d) (ρ) ), max(n k +1,1) i min(n k +d,n), together with L(r 0 (n,d) (ρ) ). The irreducible quotient of Rt (n,d) (ρ) is isomorphic to L(r min(n k+d,n) (n,d) (ρ) ). A general method of applying Jacquet modules for studying reducibility of parabolically induced representations and their composition series in the case of reductive p-adic groups was introduced in [30]. This method has been significantly improved by C. Jantzen (see [9] and several of his other papers) and by G. Muić (see [14] and [15]). The structure on representations of classical groups constructed using parabolic induction and Jacquet modules in [9], is very useful for this method for classical groups. In the case of general linear groups we have simpler situation. Such structure is given by Hopf algebra ([35]). Jacquet modules are quite useful for reducibility questions (see for example [35] or [7], but also [5], [11], [1] etc.). In the case of split general linear groups we have one additional very powerful tool which we do not have in the case of other classical groups. We have derivatives (see [35]). The main tool in our handling of the composition series that we consider in this paper, is a very simple and nice formula for the derivatives of Speh representations (described in.10). The formula was obtained in [10] (and conjectured in [6]). Actually, the formula of

5 REDUCIBILITY POINTS BEYOND THE ENDS OF COMPLEMENTARY SERIES 5 E. Lapid and A. Mínguez is much more general - it is for ladder representations (defined in [10] ). Another tool that we use in this paper is the Mœglin-Waldspurger algorithm for the Zelevinsky involution ([13]). The content of the paper is as follows. The second section recalls the notation that we use in the paper. The third section contains preparatory technical results, while in the fourth section we define multisegments r i (n,d) (ρ). We determine composition series in the cases of disjoint beginnings in the fifth section. The sixth section gives composition series in the case of non-disjoint beginnings.. Notation and preliminaries We recall very briefly some notation for general linear groups in the non-archimedean case (one can find more details in [35] and [16])..1. Finite multisets. Let X be a set. The set of all finite multisets in X is denoted by M(X) (we can view each multiset as a functions X Z 0 with finite support; here finite subsets correspond to all functions S(C) {0, 1} with finite support). Elements of M(X) are denoted by (x 1,...,x n ) (repetitions of elements can occur; multiset does not change if we permute x i s). The number n is called the cardinality of (x 1,...,x n ), and it is denoted by card({x 1,...,x n }). On M(X) we have a natural structure of a commutative associative semi group with zero: (x 1,...,x n )+(y 1,...,y m ) = (x 1,...,x n,y 1,...,y m ). For x,y M(X) we write x y if there exists z M(X) such that x+z = y. Let π = L(a), where a = ( 1,..., k ) M(S(C)). Suppose that i j whenever i j. If π is supported by one cuspidal Z-line, then π is called a ladder representation.

6 6 MARKO TADIĆ.. Segments in C. Let F be a non-archimedean locally compact non-discrete field and F its modulus character. Denote G n = GL(n,F),n 0 (we take G 0 to be the trivial group; we consider it formally as the group of 0 0 matrices). The set of all equivalence classes of irreducible representations of all groups G n,n 0, is denoted by Irr. The subset of all cuspidal classes of representations G n,n 1, is denoted by Unitarizable classes in C are denoted by C u. Put C. ν = det F. Fix u,v R such that v u Z 0, and ρ C. Then the set [ν u ρ,ν v ρ] = {ν u ρ,ν u+1 ρ,...,ν v 1 ρ,ν v ρ} is called a segment in C. The set of all segments in cuspidal representations of general linear groups is denoted by S(C). Let = [ν u ρ,ν v ρ] S(C). The representation ν u ρ is called the beginning of the segment, and ν v ρ is called the end of the segment. We denote the beginning and the end by respectively. b( ) and e( ) For z R, denote We define and by ν z = {ν z ρ ;ρ }. if u < v. Otherwise we take =. = [ν u ρ,ν v 1 ρ] and = [ν u+1 ρ,ν v ρ] Segments 1, S(C) are called linked if 1 S(C) and 1 { 1, }. If the segments 1 and are linked and if 1 and 1 have the same beginnings, we say that 1 precedes. In this case we write 1.

7 REDUCIBILITY POINTS BEYOND THE ENDS OF COMPLEMENTARY SERIES 7.3. Multisegments. Let a = ( 1,..., k ) M(S(C)). Suppose that i and j are linked for some 1 i < j k. Let c be the multiset that we get by replacing segments i and j by segments i j and i j in a (we omit i j if i j = ). In this case we write c a. Using, we generate in a natural way an ordering on M(S(C)). For a = ( 1,..., k ) M(S(C)), denote a = ( 1,..., k ), a = ( 1,..., k ) M(S(C)) (again, we omit i and i if they are empty sets). Further, for a = ( 1,..., k ) M(S(C)) define k supp(a) = i M(C), where we consider in the above formula i s as elements of M(C). i=0 The multiset of all beginnings (reps. ends) of segments from a M(S(C)) is denoted by B(a) (resp. E(a)). Clearly, B(a),E(a) M(C). Take positive integers n and d and let ρ C. Denote where a(n,d) (ρ) = (ν n 1,ν n 1 +1,...,ν n 1 ) M(S(C)), (.) = [ν (d 1)/ ρ,ν (d 1)/ ρ]..4. Algebra of representations. The category of all smooth representations of G n is denotedbyalg(g n ). Thesetofallequivalence classesofirreduciblesmoothrepresentations of G n is denoted by G n. The subset of unitarizable classes in G n is denoted by Ĝ n. The Grothendieck group of the category of all smooth representations of G n of finite length is denoted by R n. It is a free Z-module with basis G n. We have the canonical mapping s.s. : Alg(G n ) R n. The set of all finite sums in R n of elements of the basis G n is denoted by (R n ) +. Set R = n Z 0 R n, R + = n Z 0 (R n ) +.

8 8 MARKO TADIĆ The ordering on R is defined by r 1 r r r 1 R +. An additive mapping ϕ : R R is called positive if r 1 r = ϕ(r 1 ) ϕ(r ). For two finite length representations π 1 and π of G n we shall write s.s.(π 1 ) s.s.(π ) shorter π 1 π..5. Parabolic induction. Let M (n1,n ) := {[ ] } g1 ;g 0 g i G i G n1 +n, and let σ 1 and σ be smooth representations of G n1 and G n, respectively. We consider σ 1 σ as a the representation [g1 ] σ 0 g 1 (g 1 ) σ (g ) of M (n1,n ). By σ 1 σ is denoted the representation of G n1 +n parabolically induced by σ 1 σ from M (n1,n ) (the induction that we consider is smooth and normalized). For three representations, we have (σ 1 σ ) σ 3 = σ1 (σ σ 3 ). (.3) The induction functor is exact and we can lift it in a natural way to a Z-bilinear mapping : R n1 R n R n1 +n, and further to : R R R. In this way, R becomes graded commutative ring. The commutativity implies that if π 1 π is irreducible for π i G ni, then π 1 π = π π Classifications of non-unitary duals. Let = [ν u ρ,ν v ρ} S(C). The representation ν v ρ ν v 1 ρ... ν u ρ. has a unique irreducible subrepresentation, which is denoted by δ( ), and a unique irreducible quotient, which is denoted by z( ). The irreducible subrepresentation is essentially square integrable, i.e. it becomes square integrable (modulo center) after twisting with a suitable character of the group. Let a = ( 1,..., n ) M(S(C)). Choose enumeration of i s such that for all i,j {1,,...,n} the following holds:

9 REDUCIBILITY POINTS BEYOND THE ENDS OF COMPLEMENTARY SERIES 9 Then the representations if i j, then j < i. ζ(a) := z( 1 ) z( )... z( n ), λ(a) := δ( 1 ) δ( )... δ( n ) are determined by a up to an isomorphism (i.e., their isomorphism classes do not depend on the enumerations which satisfies the above condition). The representation ζ(a) has a unique irreducible subrepresentation, which is denoted by Z(a), while the representation λ(a) has a unique irreducible quotient, which is denoted by L(a). In this way we obtain two classifications of Irr by M(S(C)). Here, Z is called Zelevinsky classification of Irr, while L is called Langlands classification of Irr. It is well known (see [35]) that for a,b M(S(C)) holds Z(b) ζ(a) b a. Thecontragredient representation ofπ isdenotedby π. For S(C), set := { ρ;ρ }. If a = ( 1,..., k ) M(S(C)), then we put ã = ( 1,..., k ). Then L(a) = L(ã) and Z(a) = Z(ã). Analogous relations hold for Hermitian contragredients. The Hermitian contragredient of a representation π is denoted by π Classification of the unitary dual. Denote by B rigid = {Z(a(n,d) ρ) );n,d Z 1,ρ C u }. and B = B rigid {ν α σ ν α σ;σ B rigid,0 < α < 1/}. Then the unitary dual is described by the following: Theorem.1. ( [4]) (1) Let τ 1,...,τ n B. Then the representation π := τ 1... τ n is irreducible and unitary. () Suppose that a representation π is obtained from τ 1,...,τ n B in the same manner as π was obtained from τ 1,...,τ n in (1). Then π = π if and only if n = n and if the sequences (τ 1,...,τ n ) and (τ 1,...,τ n ) coincide after a renumeration.

10 10 MARKO TADIĆ (3) Each irreducible unitary representation of G m, for any m, can be obtained as in (1)..8. Duality - Zelevinsky involution. Define a mapping t : Irr Irr by Z(a) t = L(a),a M(S(C)). Extend t additively to R. Clearly, t is a positive mapping, i.e., satisfies: r 1 r = r t 1 rt. Obviously, z( ) t = δ( ), S(C). A non-trivial fact is that t is also multiplicative, i.e., a ring homomorphism (see [], [16] and [17]). Further, t is an involution. Define a t M(S(C)) for a M(S(C)) by the requirement (L(a)) t = L(a t ). We could also use the Zelevinsky classification to define t : M(S(C)) M(S(C)), and we would get the same involutive mapping. From this follows that Z((a t 1 +at )t ) Z(a 1 ) Z(a ). (.4) One can find more information about the involution in [16]. We shall point out one useful property of this involution, which is a consequence of the multiplicativety of t. Let P Z[X 1,...,X n ] 3 and a 1,...,a n M(S(C)). Then in R holds P(Z(a 1 ),...,Z(a n )) = 0 P(L(a 1 ),...,L(a n )) = 0. (.5).9. Algorithm of C. Mœglin and J.-L. Waldspurger. Let a M(S(C)) be nonempty. Fix ρ C and denote by X ρ (a) the set of all x R such that there exists a segment in a satisfying e( ) = ν x ρ. Now fix ρ such that X ρ (a). Let x = max(x ρ (a)), and consider segments in a such that e( ) = ν x ρ. Among these segments, choose one of minimal cardinality. Denote it by 1. This will be called the first stage of the algorithm. Consider now segments in a such that e( ) = ν x 1 ρ, and which are linked with 1. Among them, if this set is non-empty, choose one with minimal cardinality. Denote it by. One continues this procedure with ends x, x 3, etc., as long as it is possible. The segments considered in this procedure are 1,..., k (k 1). Let Γ 1 = [e( k ),e( 1 )] = [ν x k+1 ρ,ν x ρ] S(C). 3 In applications in the paper, P will be very simple polynomial of the form X 1 X X 3 X n.

11 REDUCIBILITY POINTS BEYOND THE ENDS OF COMPLEMENTARY SERIES 11 Let a be the multiset of M(C) which we get from a by replacing each i by i, i = 1,...,k (we omit those i for which i = ). If a is non-empty, we now repeat the above procedure with a as long as possible. In this way we get a segment Γ and (a ) M(S(C)). Continuing this procedure as long as possible, until we reach the empty set, we get Γ 1,...,Γ m S(C). Then by [13] we have This algorithm will be denoted by a t = (Γ 1,...,Γ m ). MWA. There is also a dual (or left ) version of this algorithm, denoted by MWA (see [34]). It is easy to show that for n,d Z 1 and ρ C. Z(a(n,d) (ρ) ) = L(a(d,n) (ρ) ) (.6).10. Derivatives on the level of R. The algebra R is a Z-polynomial algebra over {z( ); S(C)} (see [35]). Therefore, there exists a unique ring homomorphism D : R satisfying D(z( )) = z( )+z( ), S(C). Then D is a positive homomorphism ([35]). Let π be anirreducible representation of G n. Consider it asan element of R. Let π = Z(a), where a M(S(C)), and let Z(a ) be a representation of G l. Write D(π) = i 0 D i(π). Then D l (Z(a)) = Z(a ) and D j (Z(a)) = 0 for j < l (see [35]). The representation Z(a ) is called the highest derivative of Z(a). Observe that one easily reconstructs Z(a) from its highest derivative and its cuspidal support of a. One defines in a naturel way highest derivative of not-necessarily irreducible representation, like Z(a 1 )... Z(a k ). Then the highest derivative of Z(a 1 )... Z(a k ) is Z(a 1 )... Z(a k ). Observe that from the composition series of the above highest derivative we can reconstruct all the irreducible subquotients Z(a) of Z(a 1 )... Z(a k ) which satisfy card(a) = card(a 1 + +a k ). Moreover, the corresponding multiplicities also coincide. E. Lapid and A. Mínguez have obtained in [10] the formula for the derivative of the ladder representations (ladder representations are defined in [10]). Representations Z(a(n,d) ρ) )

12 1 MARKO TADIĆ are very special case of ladder representations. We shall now explain this formula in the case of representations Z(a(n,d) ρ) ) (this formula will play crucial role in our paper). Write a(n,d) (ρ) = ( 1,..., n ) in a way that 1... n. Then D(Z( 1,..., n )) = Z( 1,..., n )+Z( 1,,..., n )+ +Z( 1,..., n 1, n )+Z( 1,..., n). Remark.. Sometimes is useful to consider the positive ring homomorphism D : R R,π (D( π)). This homomorphism has analogous properties as D. It is positive. It sends z( ) z( )+z( ). Here the highest derivative of Z(a) is Z( a). Further one has D (Z(a(n,d) (ρ) )) = Z( 1,..., n )+Z( 1,..., n 1, n )+ +Z( 1,,..., n )+Z( 1,..., n ). 3. Some general technical lemmas 3.1. Representations. We shall consider the representations R(n,d) (ρ) (l) := ν l/ Z(a(n,d) (ρ) ) ν l/ Z(a(n,d) (ρ) ) = Z(a(n,d) (ν l/ ρ) ) Z(a(n,d) ( νl/ ρ) ), where n,d Z 1, l Z and ρ C. Observe that in R we have (R(n,d) (ρ) (l) )t = R(d,n) (ρ) (l) (3.7) and s.s.(r(n,d) (ρ) (l) ) = s.s.(r(n,d)(ρ) ( l) ). The formula for the highest derivative is h.d.(r(n,d) (ρ) ) = R(n,d 1)(ν 1/ ρ) (we take formally that R(n,0) (ρ) to be Z( )). In what follows, we shall always assume When n,d,k and ρ are fixed, we denote k Z 0. a = a(n,d) (ν k/ ρ), a + = a(n,d) ( νk/ ρ). Write segments 1,..., n of a in a way that 1... n,

13 REDUCIBILITY POINTS BEYOND THE ENDS OF COMPLEMENTARY SERIES 13 and segments Γ 1...,Γ n of a + also in a way that Γ 1 Γ... Γ n. Now we introduce the following numbers (in 1 Z): Observe that A = d 1 n 1 k, B = d 1 n 1 k, C = d 1 + n 1 k, D = d 1 + n 1 k, A + = d 1 n 1 + k, B + = d 1 n 1 + k, C + = d 1 + n 1 + k, D + = d 1 + n 1 + k. 1 = [ν A ρ,ν B ρ], = [ν A +1 ρ,ν B +1 ρ],..., n = [ν C ρ,ν D ρ], Γ 1 = [ν A + ρ,ν B + ρ], Γ = [ν A ++1 ρ,ν B ++1 ρ],..., Γ n = [ν C + ρ,ν D + ρ]. Obviously, A B,C D, A C,B D and B A = D C. Analogous relations hold for A +,B +,C + and D +. It is well known that R(n,d) (ρ) (0) is irreducible (see [3], and also [4]). Observe that for D + A +, R(n,d) (ρ) is irreducible by [35]. In other words, for n+d k R(n,d) (ρ) is irreducible. Therefore, we can have reducibility (for k 0) only if 1 k n+d 1. (3.8) We shall assume in the rest of this section that (3.8) holds. Consider exponents that show up in the cuspidal supports of both a and a +. The cardinality of this set is D A + +1, which is n+d 1 k. (3.9)

14 14 MARKO TADIĆ 3.. Unique irreducible subrepresentation and quotient. Proposition 3.1. Suppose 1 k n+d 1. The representation R(n,d) (ρ) ( k) has a unique irreducible subrepresentation, and it is isomorphic to Z(a(n,d) (ν k/ ρ) +a(n,d) (νk/ ρ) ). (3.10) Further, it has a unique irreducible quotient, and it is isomorphic to Z((a(d,n) (ν k/ ρ) +a(d,n) (νk/ ρ) ) t ). (3.11) Both irreducible representations have multiplicity one in the whole representation. The representation R(n,d) (ρ) also have the unique irreducible subrepresentation and the unique irreducible quotient. Their position is opposite to the position in R(n,d) (ρ) ( k). Proof. Let t be any integer satisfying 0 t n. We have Z(a + ) ζ(a + ) and Z(Γ n t+1,...,γ n ) Z(Γ 1,...,Γ n t ) ζ(a + ). Since ζ(a + ) has a unique irreducible subrepresentation, and it is Z(a + ), we get that there exists an embedding Analogously, we get Z(a + ) Z(Γ n t+1,...,γ n ) Z(Γ 1,...,Γ n t ). Z(a ) Z( t+1,..., n ) Z( 1,..., t ). Now we shall specify t. If C +D A + +B + (i.e. n 1 k), we take t = n. In the opposite case A + +B + < C +D (i.e. k < n 1), t = k. Observe that this implies that Γ 1 = t+1,...,γ n t = n. Now we have Z(a + ) Z(a ) Z(Γ n t+1,...,γ n ) Z(Γ 1,...,Γ n t ) Z( t+1,..., n ) Z( 1,..., t ) = Z(Γ n t+1,...,γ n ) Z(Γ 1,...,Γ n t, t+1,..., n ) Z( 1,..., t ) since Z(Γ 1,...,Γ n t ) Z( t+1,..., n ) is irreducible (we are in the essentially unitary situation, from which we get irreducibility of the induced representation). Now Z(Γ n t+1,...,γ n ) Z(Γ 1,...,Γ n t, t+1,..., n ) Z( 1,..., t ) ζ(γ n t+1,...,γ n ) ζ(γ 1,...,Γ n t, t+1,..., n ) ζ( 1,..., t ) = ζ(a + +a ). Therefore, Z(a + ) Z(a ) has a unique irreducible subrepresentation, and it is Z(a + +a ).

15 REDUCIBILITY POINTS BEYOND THE ENDS OF COMPLEMENTARY SERIES 15 For the quotient setting, observe that R(n,d) (ρ) ( k) = L(a(d,n) (νk/ ρ) ) L(a(d,n) (ν k/ ρ) ). Now applying similar arguments as above (dealing with quotients instead of subrepresentations), we get that R(n,d) (ρ) ( k) has a unique irreducible quotient, and that this quotient is L(a(d,n) (νk/ρ) +a(d,n) (ν k/ρ) ) = Z((a(d,n) (νk/ρ) +a(d,n) (ν k/ρ) ) t ). To get the description of the irreducible quotient and the irreducible subrepresentation of, apply the contragredient to R(n,d)( ρ) R(n,d) (ρ) ( k). The proof of the proposition is now complete Highest derivatives of irreducible subquotients. Lemma 3.. Let Z(b) Z(a ) Z(a + ). Denote by c(b) the cardinality of b. Then n c(b) n, and we get the cuspidal support of the highest derivative of b from the cuspidal support of a + a + removing ends of all segments in a + and removing ends of the first c(b) n segments in a. Therefore, the multiset of ends of b consists of all ends of segments Γ i, and the ends of the first c(b) n ends of segments i. Proof. Recall that ) D(Z(a ) Z(a + )) = (Z( 1,..., n )+Z( 1,,..., n )+ +Z( 1,..., n ) ) (Z(Γ 1,...,Γ n )+Z(Γ 1,Γ,...,Γ n )+ +Z(Γ 1,...,Γ n ). Observe thattheendofthelast segment ina + isnotinthesupport ofthehighest derivative of b. Therefore, the highest derivative of Z(b) must be subquotient of the product, in which the second factor is Z(Γ 1,...,Γ n). Now the claim of the lemma follows directly Symmetry. Lemma 3.3. Let Z(b) Z(a ) Z(a + ). Suppose that ρ is unitarizable. Then τ τ + defines a bijection from the multiset of all ends of segments in b onto the multiset of all beginnings of segments in b, i.e. E(b) + = B(b).

16 16 MARKO TADIĆ Proof. Observe that Z(b + ) Z(a ) Z(a + ) since ρ is unitarizable (the unitarizability implies that the representation on the right hand side is a Hermitian element of the Grothendieck group since Z(a ) + = Z(a + )). Denote by c(b) the cardinality of b (we know n c(b) n from the previous proposition). Then the last proposition tells us that the multiset of all ends of b depends only on c(b). Obviously, b and b + have the same cardinality. Therefore, the multisets of the ends of b + and of b are the same, i.e. E(b) = E(b + ). Observe that one gets the multiset of the beginnings of b from the multiset of the ends of b + applying τ τ +, i.e. B(b) = E(b + ) +. Therefore E(b) + = E(b + ) + = B(b), i.e. τ τ + is a bijection from the multiset of all ends of segments in b onto the multiset of all beginnings of segments in b (it preserves the multisets) Key lemma. The following lemma will be crucial for exhaustion of composition series. Lemma 3.4. Let Z(b) Z(a ) Z(a + ). Suppose that ρ is unitarizable. (1) There exists a (multi)set b left = ( 1,..., n) such that b( i ) = b( i), i = 1,...,n, e( 1 ) e( 1) < < e( n) and that each segment of b left is contained in b, i.e. b left b. () There exists a (multi)set b right = (Γ 1,...,Γ n) such that e(γ i) = e(γ i ), i = 1,...,n, b(γ 1 ) < < b(γ n ) b(γ n) and that each segment of b right is contained in b, i.e. b right b. (3) Suppose that B(a ) and B(a + ) are disjoint and that the cardinality c(b) of b is strictly smaller then n. Denote l = n c(b) (i.e. c(b) = n l). Then (a) i = i, i = 1,...,n l. (b) Γ i = Γ i, i = l+1,...,n. (c) For i = 1,...,l, n l+i and Γ i are disjoint, n l+i Γ i is a segment, and n l+i = n l+i Γ i = Γ i. (d) b = ( 1,..., n l, n l+1,... n,γ l+1,...,γ n ). (e) d k (i.e. B < A + ) and k +1 n+d. (f) The multiset b as above is unique. Further, l is the maximal index such that n Γ l S(C). (g) In this situation we have (a t +a t +) t = b.

17 REDUCIBILITY POINTS BEYOND THE ENDS OF COMPLEMENTARY SERIES 17 (4) Suppose that B(a ) and B(a + ) are not disjoint (this is equivalent to A + C ). Further, assume that n d (this is equivalent to C B ). Then the cardinality c(b) of b is equal n. Proof. Observe that Z(b t ) Z(a t ) Z(a t +) = Z(a(d,n) (ν k/ ρ) ) Z(a(d,n) (νk/ ρ) ), which implies b t a t +at + = a(d,n)(ν k/ ρ) +a(d,n) (νk/ ρ). This implies that in b t there is unique segment such that b( ) = b( 1 ). This segment must have obviously at least n elements. Now MWA implies (1). Analogously follows () using the dual version of MWA (see [34]). Now we shall prove (3). We assume that conditions of (3) on a and a + hold (in particular, disjointness of beginning of segments is equivalent to C < A + ). Recall that by Lemma 3. the multiset of ends of b consists of all ends of segments Γ i, and the ends of the first n l ends of segments i. Now Lemma 3.3 implies that the multiset of beginnings of b consists of all beginnings of segments i, and the beginnings of the last n l ends of segments Γ i. Let ρ be an element of some i or Γ j. Then we can write ρ = ν x ρ for unique x 1 Z. In this case we write exp ρ (ρ ) = x, and we shall say that x is the exponent of ρ with respect to ρ. Denote by X the multiset of all exponents with respect to ρ of beginnings of segments in b. Write elements of X as b 1 < < b n l. Then by Lemma 3.3, X is the multiset of all exponents with respect to ρ of ends of segments in b. Write elements of X as e 1 < < e n l. Therefore, there exists a permutation σ of {1,...,n l} satisfying such that Now by (1) and () we must have This implies b i e σ(i), i = 1,...,n l, b = ([ν b 1 ρ,ν e σ(1) ρ],...,[ν b n l ρ,ν e σ(n l) ρ]) = ([ν b σ 1 (1) ρ,ν e 1 ρ],...,[ν b σ 1 (n l) ρ,ν e n l ρ]). σ(1) < < σ(n) and σ 1 (n l +1) < < σ 1 (n l). i σ(i),,i = 1,...,n and σ 1 (j) j, j = n l +1,...,n l.

18 18 MARKO TADIĆ To shorten the formulas below, for { 1,..., n,γ 1,...,Γ n } we use the following notation: b ρ ( ) := exp ρ (b( )) and e ρ ( ) := exp ρ (e( )). Since the cuspidal supports of a + a + and b must be the same, their cardinalities must be the same. The first cardinality is nd = (e ρ ( i ) b ρ ( i )+1)+ (e ρ (Γ i ) b ρ (Γ i )+1) Fromtheother side, using thefact that b i e σ(i), i = 1,...,n l, thesecond cardinality is n l n l n l n l n l (e σ(i) b i +1) = (e σ(i) +1) b i = (e i +1) n l = (e ρ ( i ) b ρ ( i )+1)+ (e ρ (Γ i ) b ρ (Γ i )+1)+ i=l+1 i=n l+1 Since the above two cardinalities must be the same, we have l (e ρ ( i ) b ρ ( i )+1)+ (e ρ (Γ i ) b ρ (Γ i )+1) = i=n l+1 This further implies e ρ ( i )+ Thus i.e. i=n l+1 l (e ρ (Γ i ) b ρ (Γ i )+1) = l e ρ ( n l+i )+ i=n l+1 e ρ ( i )+ i=n l+1 i=n l+1 l ( b ρ (Γ i )+1) = 0, l ( b ρ (Γ i )+1) = b i (e ρ (Γ i (n l) ) b ρ ( i )+1). (e ρ (Γ i (n l) ) b ρ ( i )+1). e ρ (Γ i (n l) ) = l e ρ (Γ i ). l (e ρ ( n l+i ) b ρ (Γ i )+1) = 0. Since all the numbers e ρ ( n l+i ) b ρ (Γ i )+1, i = 1,...,l are the same, we get that e ρ ( n l+i )+1 = b ρ (Γ i ), i = 1,...,l. (3.1) For i = l we get e ρ ( n )+1 = b ρ (Γ l ), i.e. D +1 = A + +l 1. Going to n,d,k-notation, we get d 1 + n 1 k +1 = d 1 n 1 + k +l 1, which gives d+n = k+l. This implies l = n+d k. Recall 1 l n. Therefore k +1 n+d, d k.

19 REDUCIBILITY POINTS BEYOND THE ENDS OF COMPLEMENTARY SERIES 19 The last inequality is equivalent to B < A +. We illustrate the situation by the following example: (3.13) Our next aim will be to prove that the permutation σ is the identity permutation. In the following considerations, we shall use the fact that the cuspidal supports of b and a + a + must coincide (in previous considerations we have used only the fact that their cardinalities must be the same). Consider the case B + 1 < A +. Suppose σ(1) 1. Then 1 < σ(1). Recall i σ(i), 1 i n. This implies that the multiplicity of ν B +1 ρ in the cuspidal support of b is strictly bigger then the multiplicity in the cuspidal support of a + a +. Thus, σ(1) = 1. In the same way we get σ() = if B + < A +. Continuing in this way, we get that σ(i) = i, if B +i < A +. We shall now find the maximal i such that B +i < A +. In n,d,k-notation, this condition becomes d 1 n 1 k + i < d 1 n 1 + k, i.e. d 1+i < k, which is equivalent to d+i k. Therefore since we know l = n+d k. σ(i) = i for all 1 i k d = n l, (3.14) This implies that σ carries {n l+1,...,n l} into itself. Since σ 1 is monotone on this set, we get that σ is the identity permutation. This completes the proof of (a) - (d) in (3). It remains to prove (g). Because of (.6), it is enough to prove (a(d,n) (ν k/ ρ) +a(d,n) (νk/ ρ) ) t = b. The proof of this relation is simple straight forward application of MWA, and we shall not present detail here. Rather we shall illustrate proof on example (3.13). We illustrate a t +at + by the drawing (3.15)

20 0 MARKO TADIĆ Now MWA applied to a t +a t + can be illustrated by: General proof goes in the same way. (3.16) It remains to prove (4). The proof is a simple modification of the proof of (3). We assume that conditions of (3) on a and a + hold (in particular, A + C, which means that the beginnings of segments are not disjoint). Suppose c(b) < n. Write c(b) = n l. Then Write 1 l n. b = ( 1,..., n l ). Since the cuspidal supports of a + a + and b must be the same, their cardinalities must be the same. The first cardinality is nd = (e ρ ( i ) b ρ ( i )+1)+ (e ρ (Γ i ) b ρ (Γ i )+1) From the other side, using the fact that b ρ ( i ) e ρ( i ), i = 1,...,n l, the second cardinality is n l n l (e ρ ( i) b ρ ( i)+1) = (e ρ ( i)+1) n l b ρ ( i). (3.17) The multiset of ends of b consists of all ends of segments Γ i, and the ends of the first n l ends of segments i. Further, the multiset of beginnings of b consists of all beginnings of segments i, and the beginnings of the last n l ends of segments Γ i. Using this fact, and permuting elements in the sums, we easily get that we can write (3.17) as n l (e ρ ( i ) b ρ ( i )+1)+ (e ρ (Γ i ) b ρ (Γ i )+1)+ (e ρ (Γ i (n l) ) b ρ ( i )+1). i=l+1 i=n l+1 The above two cardinalities must be the same, which implies l (e ρ ( i ) b ρ ( i )+1)+ (e ρ (Γ i ) b ρ (Γ i )+1) = i=n l+1 This further implies e ρ ( i )+ i=n l+1 l (e ρ (Γ i ) b ρ (Γ i )+1) = i=n l+1 i=n l+1 (e ρ (Γ i (n l) ) b ρ ( i )+1). e ρ (Γ i (n l) ) = l e ρ (Γ i ).

21 Thus i.e. REDUCIBILITY POINTS BEYOND THE ENDS OF COMPLEMENTARY SERIES 1 l e ρ ( n l+i )+ i=n l+1 e( i )+ l ( b ρ (Γ i )+1) = 0, l ( b ρ (Γ i )+1) = l (e ρ ( n l+i ) b ρ (Γ i )+1) = 0. Since all the numbers e ρ ( n l+i ) b ρ (Γ i )+1, i = 1,...,l, are the same, we get that e ρ ( n l+i )+1 = b ρ (Γ i ), i = 1,...,l. (3.18) For i = l we get e ρ ( n )+1 = b ρ (Γ l ), i.e. D +1 = A + +l 1. Going to n,d,k-notation, we get d 1 + n 1 k +1 = d 1 n 1 + k +l 1, which gives d+n = k+l. This implies l = n+d k. Recall 1 l n. Therefore k +1 n+d, d k. The last inequality is equivalent to B < A +. From the other side, we suppose A + C and C B, which implies A + B. This contradicts to B < A +. The proof is now complete. 4. Definition of multisegments representing composition series Denote r 0 (n,d) (ρ) k = ( 1,..., n,γ 1,...,Γ n ). For 1 j n we define multisegments r j (n,d) (ρ) k whenever n Γ j. Observe that n Γ j if and only if n j+1 Γ 1. This is the case if and only if 1 b ρ (Γ 1 ) b ρ ( n j+1 ) d. The last condition becomes 1 ( d 1 n 1+ k ) ( d 1 n 1 k +n j) d. Therefore if j 0, the multisegments r j (n,d) (ρ) k are defined for indexes Then we get r j (n,d) (ρ) k by replacing in max(n k +1,1) j min(n k +d,n). r 0 (n,d) (ρ) k = ( 1,..., n,γ 1,...,Γ n )

22 MARKO TADIĆ the part n j+1,..., n,γ 1,...,Γ j with n j+1 Γ 1, n j+1 Γ 1,..., n Γ j, n Γ j (we omit if it shows up in the above formula and the formulas below). In other words we have n j j r j (n,d) (ρ) = ( i+n j Γ i, i+n j Γ i )+ ( i,γ n i+1 ) = n j j ( i+n j Γ i, i+n j Γ i )+ ( i,γ i+j ) for j = 0 and for 1 j n for which n Γ j. Suppose that and Γ are segments such that their intersection is non-empty. Then obviously (( Γ),( Γ) ) = ( Γ, Γ ) (as above, we omit if it shows up). This implies that if d and n Γ j such that n Γ j. h.d.(z(r j (n,d) (ρ) )) = ν 1/ Z(r j (n,d 1) (ρ) ) (4.19) 5. Composition series - disjoint beginnings of segments We continue with the notation of the previous section. In this section we shall assume that C < A +, i.e. {b( 1 ),...,b( n )} {b(γ 1 ),...,b(γ n )} =. We also assume A + D +1 (recall that for A + > D +1 the representation R(n,d) (ρ) is irreducible). In the n,d,k-notation these two conditions become d 1 + n 1 k < d 1 n 1 + k and d 1 n 1 + k d 1 + n 1 k +1, i.e. n k, (i.e. k < n+d). k n+d 1 Now we examine when n Γ j. This is the case if and only if A + +j 1 D +1.

23 REDUCIBILITY POINTS BEYOND THE ENDS OF COMPLEMENTARY SERIES 3 The last condition becomes d 1 n 1 + k d 1 +j 1 + n 1 k +1, i.e. k+j d+n. Therefore, the multisegments r j (n,d) (ρ) are defined for indexes in the case that we consider in this section. 0 j min(n+d k,n) Observe that in the case k d, r j (n,d) (ρ) are defined for all 0 j n, and then we have r 0 (n,d) (ρ) = ( i,γ n i+1 ). r 1 (n,d) (ρ) = ( n 1 n Γ 1, n Γ 1 )+ ( i,γ i+1 )... n j j r j (n,d) (ρ) = ( i+n j Γ i, i+n j Γ i )+ ( i,γ i+j ).... r n (n,d) (ρ) = ( i Γ i, i Γ i ). In the case d < k, let j is the greatest index for which n Γ j is still a segment (with the assumptions of the present section, this implies n Γ j ). Then only multisegments r 0 (n,d) (ρ),...,r j(n,d) (ρ) are defined (i.e. the first j +1 terms above are defined). Proposition 5.1. Let A + D +, i.e. k n+d (otherwise, R(n,d) (ρ) is irreducible4 ). Suppose C < A +, i.e. Then n k. (1) R(n,d) (ρ) is a multiplicity one representation. () Let B < A +. In that case Then R(n,d) (ρ) d k n+d. has length n+d+1 k, and its composition series consists of Z(r i (n,d) (ρ) ), 0 i n+d k. 4 Observe that also for k = n+d we have irreducibility.

24 4 MARKO TADIĆ (3) Suppose A + B, i.e. k < d. Then R(n,d) (ρ) has length n+1, and its composition series consists of Z(r i (n,d) (ρ) ), 0 i n. Proof. One can easily see that if any of the claims of the above proposition holds for unitarizable ρ, then it holds for any ν α ρ,α R (for that n,d and k). This follows from the factthat ingeneral holds ν α Z(a) = Z(ν a a), ν α (π 1 π ) = (ν α π 1 ) (ν α π ), ν α (R(n,d) (ρ) ) = R(n,d) (να ρ) and ν α (r i (n,d) (ρ) ) = r i (n,d) (να ρ). We shall first prove (). During proving (), we shall prove also that all the multiplicities are one. We fix n and k. The proof will go by induction with respect to d. Observe that k n d k. For d = k n (i.e. k = n+d), we know that R(n,d) (ρ) is irreducible, and isomorphic to Z(r 0 (n,d) (ρ) ). Therefore, we have the basis of the induction. Suppose d = 1. Then k n+1 and n k. If k = n+1, we have observed above that () holds. Let k = n. Then easily follows from [35] that (), together with multiplicity one, holds. Now in the rest of the proof, it is enough to consider the case of d > 1. Let k n < d k, and suppose that () holds for d 1, together with the multiplicity one claim (in (1)). We shall now show that () holds also for d (together with the multiplicity one claim). Recall for d. Also for h.d.(r(n,d) (ρ) )) = ν 1/ R(n,d 1) (ρ) ) h.d.(z(r j (n,d) (ρ) )) = ν 1/ Z(r j (n,d 1) (ρ) ) 0 j n+d k 1. From the first relation and the inductive assumption follows that R(n,d) (ρ) ) has precisely n+d k irreducible subquotients π = Z(a) such that the cardinality of a is n. It also implies that all these subquotients have multiplicity one. The inductive assumption tells us that the irreducible sub quotients of the highest derivative are ν 1/ Z(r j (n,d 1) (ρ) ), 0 j n+(d 1) k. Now the second relation above implies that these subquotients Z(a) are representations Z(r j (n,d) (ρ) ), 0 j n + (d 1) k. Now Proposition 3.1 and Lemma 3.4, (3), (g) imply also that Z(r n+d k (n,d) (ρ) ) is irreducible subquotient, and that the multiplicity of this subquotient is one. It remains to prove that these are all the irreducible subquotients. Let π = Z(a) be arbitrary irreducible subquotient of

25 REDUCIBILITY POINTS BEYOND THE ENDS OF COMPLEMENTARY SERIES 5 R(n,d) (ρ) ). If cardinality of a is n, then we have seen that it must be one of Z(r j(n,d) (ρ) ), 0 j n+(d 1) k. Supposethatthecardinalityofaisstrictly smaller thenn. Nowwe shall findindexlsuch thate ρ ( n )+1 = b ρ (Γ l ), i.e. n 1 +d 1 k +1 = n 1 d 1 +k+l 1, which implies l = n+d k. Then (3) of Lemma 3.4 implies that π = Z(r n+d k (n,d) (ρ) ). This completes the proof of (). Now we shall prove (3). In the same time we shall prove also that all the multiplicities are one. This and the first part of the proof, will then give (1). We prove (3) by induction with respect to d. Observe that we need to prove for k < d. Actually, we shall prove the claim of (3) (together with multiplicity one property) by induction for k d. For d = k, claim (3) holds by () (observe that in this case n+d k is n, and we have n + 1 irreducible subquotients). Also the multiplicity one holds in this case. Let k < d, and suppose that our claim holds for d 1. Then in the same way as in the first part of the proof, looking at the highest derivative of R(n,d) (ρ) ), the inductive assumption implies that Z(r j (n,d) (ρ) ), 0 j n are subquotients, and all have multiplicity one. Further, (4) of Lemma 3.4 and the condition A + B (which we assume in (3)) imply that these are all the irreducible subquotients. The proof of the proposition is now complete. Summing up, we get the following: Theorem 5.. Let k Z 0. (1) Representations R(n,d) (ρ) () For (3) For R(n,d) (ρ) R(n,d) (ρ) and R(n,d)(ρ) ( k) n+d k, is irreducible, and R(n,d)(ρ) = Z(r 0 (n,d) (ρ) ). n k n+d, have the same composition series. is a multiplicity one representations. Its composition series consists of Z(r i (n,d) (ρ) ), 0 i min(n,n+d k).

26 6 MARKO TADIĆ (4) The representation R(n,d) (ρ) has a unique irreducible subrepresentation as well as a unique irreducible quotient. The irreducible quotient is isomorphic to Z(r 0 (n,d) (ρ) ), and the irreducible sub representation is isomorphic to Z(r min(n,n+d k) (n,d) (ρ) For R(n,d) (ρ) ( k), we haveopposite situation regardingirreducible subrepresentation and quotient. Proof. It remains to prove only the claim regarding the irreducible sub representation in (4). For d k d+n 1, this follows from (g) in (3) of Lemma 3.4. Therefore, we shall suppose k < d. For this, one needs to prove (r 0 (d,n) (ρ) )t = r n (n,d) (ρ). One can get this directly from MWA. We shall give here a different argument. Observe that Now we have r 0 (d,n) (ρ) = a(d+k,n)(ρ) +a(d k,n) (ρ). Z(a(d+k,n) (ρ) +a(d k,n) (ρ) ) t = Z(a(d+k,n) (ρ) ) t Z(a(d k,n) (ρ) ) t = Z(a(n,d+k) (ρ) ) Z(a(n,d k) (ρ) ) = Z(a(n,d+k) (ρ) +a(n,d k) (ρ) ), since the unitary parabolic induction is irreducibility for general linear groups. This implies that (r 0 (d,n) (ρ) )t = a(n,d+k) (ρ) +a(n,d k) (ρ). One gets directly that r n (n,d) (ρ) = a(n,d+k)(ρ) +a(n,d k) (ρ). This completes the proof that (r 0 (d,n) (ρ) )t = r n (n,d) (ρ). Considering thehermitian contragredients, we get that R(n,d) (ρ) ( k) hasopposite irreducible quotient and subrepresentation from their position in R(n,d) (ρ). The proof is now complete. Remark 5.3. For a set X, denote by X its partitive set (i.e. the set of all subsets of X). Denote by L the set of all non-zero subrepresentations of R(n,d) (ρ). Since R(n,d)(ρ) is a multiplicity one representation, the mapping J : V {i;z(r i (n,d) (ρ) ) V}, L {0,...,min(n,n+d k)} is injective. The knowledge of the image J(L) of J would imply the complete understanding of the lattice of subrepresentations of R(n,d) (ρ). Regarding this lattice, up to now we have proved the following: ).

27 REDUCIBILITY POINTS BEYOND THE ENDS OF COMPLEMENTARY SERIES 7 (1) each element in J(L) contains 0; () {0,...,min(n,n+d k)} is the only element in J(L) which contains min(n,n+d k). Denote l = min(n,n+d k) and X l = {{0},{0,1},...{0,1,...,l}}. The above two facts obviously imply J(L) = X l, for l. Considering the highest derivatives, one can inductively prove that this holds also for l 3. Similarly, we can describe the lattice of sub representations for remaining R(n,d) (ρ) (i.e. in the case when the beginnings of segments are not disjoint). 6. Composition series - non-disjoint beginnings of segments We continue with the notation of the previous sections. In this section we shall assume that A + C. Passing to the n,d,k-notation, this becomes d 1 n 1 + k d 1 + n 1 k, i.e. k < n. Recall that for j 0, we defined r j (n,d) (ρ) k whenever n Γ j. Then we get r j (n,d) (ρ) k by replacing in r 0 (n,d) (ρ) k = ( 1,..., n,γ 1,...,Γ n ) the part n j+1,..., n,γ 1,...,Γ j with n j+1 Γ 1, n j+1 Γ 1,..., n Γ j, n Γ j. Recall that n Γ j if and only Proposition 6.1. Let A + C, i.e. Then (1) R(n,d) (ρ) max(n k +1,1) j min(n,n+d k). k < n. is a multiplicity one representation.

28 8 MARKO TADIĆ () Suppose B < A +. In that case Then R(n,d) (ρ) together with Z(r 0 (n,d) (ρ) ). (3) Suppose A + B, i.e. Then R(n,d) (ρ) together with Z(r 0 (n,d) (ρ) ). d k < n. has length d+1, and its composition series consists of Z(r i (n,d) (ρ) ), n k +1 i n k +d, k < d. has length k +1, and its composition series consists of Z(r i (n,d) (ρ) ), n k +1 i n, Proof. We shall first prove (), proving in the same time also that all the multiplicities are one. This statement (with the multiplicity one claim) will be denoted by () +. We fix n and k and prove () + by induction with respect to d. Recall d < k. For d = 1, 1 k < n implies that R(n,1) (ρ) is a multiplicity one representation of length two. In its composition series is obviously Z(r 0 (n,1) (ρ) ). Further, a composition factor is also Z((a t +at + )t ), which one directly computes using MWA, and the result is r n k+1 (n,1) (ρ). This completes the prooffor d = 1. Also we have thebasis ofthe induction. Suppose 1 < d k, and suppose that () + holds for d 1. Recall for d. Also for h.d.(r(n,d) (ρ) )) = ν 1/ R(n,d 1) (ρ) ) h.d.(z(r j (n,d) (ρ) )) = ν 1/ Z(r j (n,d 1) (ρ) ) max(n k +1,1) j min(n,n+d k 1), and also for j = 0 holds the above relation for the highest derivative. Observe that k < n implies n k +1. Also d k implies n+d k n. From the first relation and inductive assumption follows that R(n,d) (ρ) ) has precisely d irreducible subquotients π = Z(a) such that the cardinality of a is n, and that all have multiplicity one. Now the second relation above and the inductive assumption imply that theses subquotients are representations Z(r j (n,d) (ρ) ), n k+1 j n k+d 1. Now

29 REDUCIBILITY POINTS BEYOND THE ENDS OF COMPLEMENTARY SERIES 9 Proposition 3.1 imply also that Z((r 0 (d,n) (ρ) )t ) is irreducible subquotient, and that the multiplicity of this subquotient is one. We need to show that (r 0 (d,n) (ρ) )t = r n k+d (n,d) (ρ). This follows in the same way as we have proved (e) in (3) of Lemma 3.4. We illustrate again by drawing, instead going into technical details: (6.0) (6.1) (6.) It remains to prove that these are all the irreducible subquotients. For this, it is enough to prove that the length of R(n,d) (ρ) ) is d+1 (counting also multiplicities). To prove this, it is enough to prove that the length of R(d,n) (ρ) is d + 1. Note that we can apply (3) of Proposition 5.1 to R(d,n) (ρ) ) (observe that the roles of n and d are switched). That proposition tells us that the length is d+1, which we needed. Now we shall prove (3). As above, in the same way we introduce (3) +. Suppose n = 1. Then k = 0, and the claim obviously holds. Suppose n > 1, and that the claim holds for n < n. We shall now prove (3) for this n by induction with respect to d. Recall that we need to prove for k < d. Actually, we shall prove the claim of (3) + by induction for k d. For d = k, claim (3) + holds by () +. Let k < d, and suppose that our claim holds for d 1. Then in the same way as in the first part of the proof, looking at the highest derivative of R(n,d) (ρ), the inductive assumption implies that Z(r j(n,d) (ρ) ), n k+1 j n, together

30 30 MARKO TADIĆ with Z(r 0 (n,d) (ρ) ) are subquotients, and that all these irreducible subquotients have multiplicity one. We also know that these are all the irreducible subquotients represented by multisegments of cardinality n. Supposen d (i.e. C B ). Inthiscase, since thecondition A + B holds(weassume it in (3)), we can apply (4) of Lemma 3.4, which says that these are all the irreducible subquotients. This completes the proof in this case. Suppose d < n (i.e. B < C ). We have seen that we have k+1 irreducible subquotients. For them, we want to prove that they exhaust the composition series of R(n,d) (ρ). Obviously, it is enough to prove that R(n,d) (ρ) ) has length k+1. For this, it is enough to show that (R(n,d) (ρ) )t = R(d,n) (ρ) ) has length k +1. Now consider R(d,n) (ρ) and denote n = d, d = n. In the same way as we have introduced numbers A ±,B ±,C ±,D ± for the triple n,d,k, we introduce A ±,B ±,C ±,D ± for the triple n,d,k. Observe that A ± = A ±,B ± = C ±,C ± = B ± and D ± = D ±. Then n < d (i.e. C < B ). Since A + C,A + B and B < C, we have A + B,A + C and C < B. Therefore, R(n,d ) (ρ) = R(d,n)(ρ) is covered by (3), and the inductive assumption implies that the length of this representation is k +1 (recall n = d < n). The proof of the proposition is now complete. Theorem 6.. Let k Z 0. (1) Representations R(n,d) (ρ) and R(n,d)(ρ) ( k) are multiplicity one representations. () For k < n, the composition series of R(n,d) (ρ) are given by have the same composition series. They Z(r i (n,d) (ρ) ), n k +1 i min(n k +d,n), together with Z(r 0 (n,d) (ρ) ). (3) The representation R(n,d) (ρ) has a unique irreducible subrepresentation and also a unique irreducible quotient. The irreducible subrepresentation is isomorphic to Z(r 0 (n,d) (ρ) ). The irreducible quotient is isomorphic to Z(r min(n k+d,n)(n,d) (ρ) ). For R(n,d) (ρ) ( k), we haveopposite situation regardingirreducible subrepresentation and quotient.

31 REDUCIBILITY POINTS BEYOND THE ENDS OF COMPLEMENTARY SERIES 31 Proof. It remains to prove only that the claim regarding the irreducible quotient in (3). For d k(< n), this is proved in the proof of previous proposition. Therefore, we shall suppose k < d. For this, one needs to prove (r 0 (d,n) (ρ) )t = r n (n,d) (ρ). One can get this directly from MWA. We can give here also a different argument like in the proof of Theorem 5.. Again r 0 (d,n) (ρ) = a(d+k,n)(ρ) +a(d k,n) (ρ). Further, as before we have This implies Direct checking gives Z(a(d+k,n) (ρ) +a(d k,n) (ρ) ) t = Z(a(n,d+k) (ρ) +a(n,d k) (ρ) ). The proof is now complete. (r 0 (d,n) (ρ) )t = a(n,d+k) (ρ) +a(n,d k) (ρ). r n (n,d) (ρ) = a(n,d+k)(ρ) +a(n,d k) (ρ). Remark 6.3. (1) The formula (.5) implies that the clams in Theorems 5. and 6. hold in the same form if one use Langlands classification instead of Zelevinsky classification. () Observe that the theory of types for general linear groups over finite dimensional central division algebras over F developed in [18] - [1], together with the theory of covers from [7], should relatively easy imply that Theorems 5. and 6. hold also for general linear groups over finite dimensional central division algebras over F, after (small) necessary modification in the notation (for these modifications of notation see [7]). We have not checked all details of this implication (one can see [] and [4] for such applications of [18] - [1] and [7]). References [1] Arthur, J., The Endoscopic Classification of Representations: Orthogonal and Symplectic Groups, preprint ( [] Aubert, A. M., Dualité dans le groupe de Grothendieck de la catégorie des représentations lisses de longueur finie d un groupe réductif p- adique, Trans. Amer. Math. Soc. 347 (1995), ; Erratum, Trans. Amer. Math. Soc 348 (1996), [3] Badulescu, A. I., On p-adic Speh representations, Bulletin de la SMF, to appear. [4] Badulescu, A. I., Henniart, G., Lemaire, B. and Sécherre, V., Sur le dual unitaire de GL r (D), Amer. J. Math. 13 no. 5 (010), [5] Badulescu, A. I., Lapid, E. and Míngues, A., Une condition suffisante pour l irreducibilite d une induite parabolique de GL(m, D), Ann. Inst. Fourier, to appear. [6] Bernstein, J., P-invariant distributions on GL(N) and the classification of unitary representations of GL(N) (non-archimedean case), Lie Group Representations II, Lecture Notes in Math. 1041, Springer- Verlag, Berlin, 1984,