In honour of Stephen Gelbart for his 60 th birthday. Marko Tadić

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1 In honour of Stephen Gelbart for his 60 th birthday GL(n, C)ˆ AND GL(n, R)ˆ Marko Tadić Abstract. The set Ĝ of equivalence classes of all the irreducible unitary representations of a locally compact group G plays the role of basic harmonics from the classical harmonic analysis. A fundamental question of harmonic analysis on a group G is to describe the unitary dual Ĝ of G. In generalizing the classical harmonic analysis on Euclidean spaces, it is natural to consider groups which look locally like Euclidean spaces, i.e. Lie groups. The best known Lie groups are GL(n, R) and GL(n, C). These groups have remarkably simple unitary duals (in particular GL(n, C)). For example, all the representations of GL(n, C)ˆ are induced by one-dimensional representations (the induction is a natural generalization of the induction from the case of finite groups, considered already by Schur and Frobenius). Despite this simplicity, there were no examples of relatively simple classification of unitary duals of reductive Lie groups of higher ranks. In this paper we present a relatively simple classification of unitary duals GL(n, C)ˆ and GL(n, R)ˆ, which may be accessible not only to specialists in the field. This classification of GL(n, R)ˆ and GL(n, C)ˆ can be viewed as a particular case of a general classification of a unitary duals of general linear groups over non-discrete locally compact fields (the statement of the classification theorem is same for all such fields, which raises a question of finding a proof along the same strategy in all the cases). Introduction In the concept of abstract harmonic analysis on a locally compact group G, the set Ĝ of equivalence classes of all the irreducible unitary representations of G plays the role of basic harmonics from the classical harmonic analysis. The set Ĝ is called the unitary dual of G and a fundamental question of harmonic analysis on a group G is to describe the unitary dual of G. A natural generalization of the classical harmonic analysis was development of harmonic analysis on groups which look locally like Euclidean spaces, i.e., on Lie groups. The best known Lie groups GL(n, R) and GL(n, C) have remarkably simple unitary duals (in particular GL(n, C)). For example, all the irreducible unitary representations of GL(n, C) are induced by a simple one-dimensional representations. Here the representations of the Gelfand-Naimark series constructed in [GfN] already at 1940-es, together with the Stein s complementary series constructed in [St] in 1960-es, exhaust all of the unitary dual. The induction that we use here is a natural generalization of the induction in the case of finite groups, which was already considered by Schur and Frobenius. Despite this simplicity, there were no examples of relatively simple classification of unitary duals of reductive Lie groups of higher ranks. In this paper we present a relatively simple classification of unitary duals of groups GL(n, C) and GL(n, R), which may be accessible not only to specialists in the field. The 1 Typeset by AMS-TEX

2 MARKO TADIĆ results about non-unitary duals which we are using in this paper are relatively simple, wellknown general facts of the representation theory of reductive Lie groups. The description of this classification was used in L. Clozel s Bourbaki talk [C] on the classification of the unitary duals of general linear groups over archimedean fields. This classification of GL(n, R)ˆ and GL(n, C)ˆ can be viewed as a particular case of a general classification theorem of unitary duals of general linear groups over non-discrete locally compact fields. Since the statement of the classification theorem is same for all these fields, this raises a question of finding a proof along the same strategy in all the cases (fields other than C and R were settled in [T3]; the unitary duals are considerably more complicated there). We shall now recall briefly of the history of this classification (and this paper). In we solved the unitarizability problem for general linear groups over non-archimedean fields ([T], [T3]; see also [T4]). Since the result of the classification in the non-archimedean case, as well as the main strategy of the proof, was independent of the nature of the locally compact field, we extended in [T1] our approach to the classification of unitary duals of general linear groups over archimedean fields R and C (as we already mentioned, the result of the classification is exactly the same in the archimedean as well as in the non-archimedean case). Our proof of the classification in [T3] relied on a result of Kirillov ([Ki]), whose proof there is incomplete (nevertheless, that result was used in several occasions; for example I. J. Vakutinski used it in [Va] to get a classification of the unitary dual of SL(3, R)). At the same time, D. Vogan made a classification of unitary duals of general linear groups over archimedean fields along completely different lines ([Vo]). Recently, E.M. Baruch in [Ba] obtained a complete proof of the Kirillov s claim (he used the ideas of the proof of the Harish-Chandra regularity theorem for eigendistributions on reductive Lie groups to complete the Kirillov s original paper [Ki] 1 ). Therefore, the classification in [T3] is now complete. In this paper, we are giving a revised (and slightly simplified) version of the classification of unitary duals of GL(n, R) and GL(n, C) from [T1] (as we already mentioned, this is the description of the classification of unitary duals used in the Clozel s Bourbaki talk). The only difference between [T1] and this paper is that here at several places technical details are simplified. The relative simplicity of the approach of [T1] to the unitary duals of GL(n, R) and GL(n, C), which may be accessible not only to specialists in the representation theory of reductive Lie groups, and the importance of these groups (and their unitary duals) in the mathematics, makes this approach still interesting. Now we shall describe the classification in detail. By F we shall denote the field R or C. The standard absolute value on R will be denoted by R, while the square of the standard absolute value on C will be denoted by C (in both cases F is the modulus character of F ). By standard parabolic subgroup of GL(n, F ) we shall mean a parabolic subgroup which contains upper triangular matrices. We shall consider Levi factors of standard parabolic subgroups which contain diagonal matrices. Levi subgroup M is always isomorphic to a direct product of general linear groups, say GL(n i, F ), i = 1,..., k. If σ is an irreducible representation of the Levi subgroup M of a standard parabolic subgroup P of GL(n, F ), 1 We mention possibility of this strategy in [T4] (see page 47).

3 GL(n, C)ˆ AND GL(n, R)ˆ 3 then Ind GL(n,F ) P (σ) will denote the representation of GL(n, F ) parabolically induced by σ from P (the induction that we consider is normalized, i.e., it carries unitary representations to the unitary ones). If we have irreducible representations σ i, i = 1,..., k of some general linear groups GL(n i, F ), then there is a unique GL(n, F ) where M = GL(n 1, F ) GL(n k, F ) is a Levi factor (here n = n 1 + +n k ). We can view σ = σ 1 σ σ k as a representation of M. Further, there is a unique standard parabolic subgroup P such that M is its Levi factor. Therefore, we shall denote Ind GL(n,F ) P (σ) also by Ind GL(n,F ) (σ), or simply by Ind(σ) (having in mind that we consider here only the parabolic induction of the general linear groups from standard parabolic subgroups, and that σ determines n). For a unitary character δ of F let u(δ, n) : GL(n, F ) F, g δ(detg). For an irreducible representation δ of GL(, R) which is square integrable modulo center, and a positive integer n, the parabolically induced representation (1) Ind ( det (n 1)/ F δ det (n 1)/ 1 F δ det (n 1)/ F δ) has a unique irreducible quotient. This quotient will be denoted by For 0 < α < 1/ denote u(δ, n). () π(u(δ, n), α) = Ind ( det α F u(δ, n) det α F u(δ, n)). Denote by B the collection of all possible representations u(δ, n) and π(u(δ, n), α), when δ runs over the set of all unitary characters of F if F = C (resp. δ runs over the union of the set of all unitary characters of F and the set of all equivalence classes of irreducible square integrable modulo center representations of GL(, F ) if F = R ), while n runs over all positive integers and 0 < α < 1/. Now we can state the classification of unitary duals of groups GL(n, F ) We can define the set B in the following uniform way (also for non-archimedean fields). To each irreducible square integrable modulo center representation δ of GL(m, F ) and a positive integer n we define representation u(δ, n) as a unique irreducible quotient of (1) (if m = 1, then δ is a character and u(δ, n)(g) = δ(detg)). Define π(u(δ, n), α) by formula (). Denote by B the collection of all possible representations u(δ, n) and π(u(δ, n), α), when δ runs over all equivalence classes of irreducible square integrable modulo center representations of GL(m, F ) for all positive integers m, while n runs over all positive integers and 0 < α < 1/. Recall that in the case that δ is an irreducible square integrable modulo center representation of GL(m, C), then m can be only 1 and δ is a unitary character of C. If δ is an irreducible square integrable modulo center representation of GL(m, R), then m can be only 1 or (if it is 1, then δ is a unitary character of R, and if it is, then δ is an irreducible square integrable modulo center representation of GL(, R)).

4 4 MARKO TADIĆ Theorem. (1) For σ 1,..., σ k B the parabolically induced representation Ind(σ 1 σ σ k ) is an irreducible unitarizable representation of a general linear group over F. If p is a permutation of {1,..., k}, then Ind(σ 1 σ σ k ) = Ind(σ p(1) σ p() σ p(k) ). () Each irreducible unitarizable representation π of a general linear group over F is equivalent to a parabolically induced representation from (1). Moreover, π determines the sequence σ 1,..., σ k (in B) up to a permutation. With the set B defined in the footnote, the above classification theorem holds for unitary duals of general linear groups over any locally compact non-discrete field. Now we shall describe briefly the content of the paper. In the first section, we introduce the notation used in the paper. We recall also the basic general facts that we need from the representation theory of reductive groups, with special emphasis on the general linear groups. The notation which we use in the rest of this paper is adapted to the case of the general linear groups. We define an algebra formed from the Grothendieck groups of finite length representations of all general linear groups (analogous algebra was introduced earlier by J. Bernstein and A.V. Zelevinsky in the non-archimedean case). Multiplication in this algebra corresponds to the parabolic induction. This algebra is very important in our approach. The second section presents a reduction of the classification of unitary duals of general linear groups to five statements. This section is short, but it is the most important part of our paper. At the beginning of this section, we formulate five technical statements, and then we show that they imply the above theorem. In the third section we check these statements when F = C, while in the fourth and fifth section we chech them in the case F = R. At the end, it is a pleasure to thank Dragan Miličić for his help and useful suggestions during writing this revision of [T3], and to G. Muić for his very careful reading of the revised version of the paper and his very useful remarks for improvements of the manuscript. 1. Algebra of representations We shall denote by F either R or C. The set of non-negative integers in Z is denoted by Z +, and the subset of positive ones is denoted by N. Denote G n = GL(n, F ), n Z +. Let K n be a maximal compact subgroup of G n. We shall take K n to be U(n) if F = C and K n = O(n) if F = R. The groups G n are considered as real Lie groups. Let g n be the Lie algebra of G n. A (g n, K n )-module will be callede simply Harish-Chandra module of G n (the modules which show up in this paper are always isomorphic to the K n -finite vectors of some continuous representation of G n on a Hilbert space). The category of all Harish-Chandra modules of G n of finite length will be denoted by HC(G n ). The set of all equivalence classes of irreducible Harish-Chandra modules of G n is

5 GL(n, C)ˆ AND GL(n, R)ˆ 5 denoted by G n. We shall identify an irreducible Harish-Chandra module with its class. The set of all unitarizable classes in G n is denoted by Ĝn. Let R n be the Grothendieck group of HC(G n ). The set G n will be identified with a subset of R n in a natural way. In that case, R n is a free Z-module over G n. We have a natural map π π ss, HC(G n ) R n, which we shall call semi-simplification. For π in HC(G n ) and σ G n, denote by n(σ, π) the multiplicity of σ in π. Then π ss = σ G n n(σ, π) σ. We say that π contains σ 0 G n (resp. π contains σ 0 G n with multiplicity one) if n(σ 0, π) 1 (resp. n(σ, π) = 1). If π contains σ 0, we shall write σ 0 π. Let P = MN be the standard parabolic subgroup of G n1 +n given by P (n1,n ) = {(g ij ) G n1 +n ; g ij = 0 if i > n 1 and j n 1 }. Denote by M (n1,n ) the block upper triangular matrices of type (n 1, n ). The unipotent radical of P (n1,n ) is denoted by N (n1,n ). Let σ i be an object in HC(G ni ) for i = 1,. The tensor product σ 1 σ is (g n1 g n, K n1 K n )-module. Since M is naturally isomorphic to G n1 G n ; g n1 g n is considered as Lie algebra of M and K n1 K n is considered as a maximal compact subgroup in M. We shall denote by σ 1 σ the Harish-Chandra module parabolically induced by σ 1 σ. Proposition of [Vo1] implies that σ 1 σ is of the finite length. For Harish-Chandra modules σ 1, σ, σ 3 of finite length we have (1-1) σ 1 (σ σ 3 ) = (σ 1 σ ) σ 3. The induction in stages implies this relation (Proposition of [Vo1]). Let R = R n. n 0 Then R is a graded commutative group. We shall call this grading the standard one. We shall define the structure of a graded ring on R. For it, it is enough to define a Z-bilinear mapping : R n1 R n R n1 +n, n 1, n Z +. Take s R n1 and t R n. Then we can write s = σ G n1 a σ σ, a σ Z, t = τ G n b τ τ, b τ Z, where a σ 0 and b τ 0 only for finitely many σ and τ. The above expressions are unique. Now we define s t = s σ b τ (σ τ) ss. σ G n1, τ G n Recall (σ 1 σ ) ss R n1 +n. In this way R becomes (associative) graded ring with unit (associativity follows from (1-1)) 3. For irreducible Harish-Chandra modules σ 1 and σ we have the following equality in R: (1-) σ 1 σ = σ σ 1. This relation is a consequence of Proposition of [Vo1] about induction from associated parabolic subgroups. Note that if σ 1 σ Irr, then σ 1 σ = σ σ 1. The relation (1-) implies the following 3 Note that we could introduce R n as the group of virtual characters of G n. Then the multiplication in R corresponds to parabolic induction of characters.

6 6 MARKO TADIĆ 1.1. Proposition. The induction functor induces on R a structure of a commutative graded Z-algebra. For further information about the ring R, we describe the Langlands classification of irreducible (g n, K n )-modules. of R n. Denote Irr = Gn, Irr u = Ĝ n. n=0 n=0 Clearly, Irr is basis of Z-module R. Representations which are square integrable modulo center, will be called simply square integrable representations. The set of all square integrable classes in Ĝn, n 1, is denoted by D u (G n ) (this set is non-empty only if n = 1 for F = C, and if n = 1, for F = R). Set D u = Ĉ if F = C, and D u = R D u (G ) if F = R (i.e. D u = D u (G n )). Let F be n=1 the normalized absolute value on F. In the case of F = R, this is the standard absolute value, while in the complex case this is the square of the standard one. Define ν : G n R, ν(g) = det g F. Let D(G n ) = {ν α π; α R, π D u (G n )} and denote D = C if F = C, and D = R D u (G ) if F = R (i.e. D = n=0 D(G n )). If δ D, then e(δ) R and δ u D u are uniquely determined by the relation δ = ν e(δ) δ u. Let X be a set. A function f : X Z + with the finite support is called a finite multiset in X. The set of all finite multisets in X denoted by M(X). The set M(X) is an additive semigroup in a natural way. Let f M(X). Suppose that {x 1,..., x n } is the support of f. Then we shall write f also in the following way f = (x 1,..., x }{{} 1, x,..., x,... x }{{} n,..., x n ). }{{} f(x 1 )-times f(x )-times f(x n )-times If f M(X), then we write card(f) = x X f(x). We call card(f) the cardinal number of the multiset f. The number f(x) will be called the multiplicity of x in f. Let a M(D). Choose δ i D(G ni ), i = 1,..., k such that a = (δ 1,..., δ k ). After a renumeration, we can assume that e(δ 1 ) e(δ ) e(δ k ). Now the Harish-Chandra module δ 1 δ δ nk has a unique irreducible quotient, which will be denoted by L(a). This quotient is independent (up to an equivalence) of a renumeration which satisfies the above condition. Denote λ(a) = δ 1 δ k R. Then λ(a) contains L(a). The mapping a L(a), M(D) Irr is a bijection. This is a version of Langlands classification of non-unitary duals of GL(n) s. If π is in HC(G n ), then π denotes the contragredient of π, and π the complex conjugate (module) of π. We denote π by π +, and call it a Hermitian contragredient π. If π is isomorphic to π +, then π is called a Hermitian module. For a = (δ 1,..., δ n ) M(D) and α R denote ã = ( δ 1,..., δ n ), ā = ( δ 1,..., δ n ), a + = (δ + 1,..., δ+ n ), ν α a = (ν α δ 1,..., ν α δ n ). If δ D, then δ = ν e(δ) δ u, δ = ν e(δ) (δ u ), δ = ν e(δ) (δ u ), δ + = ν e(δ) δ u, ν α δ = ν e(δ)+α δ u.

7 1.. Proposition. For a M(D) and α R we have GL(n, C)ˆ AND GL(n, R)ˆ 7 L(a) = L(ā), L(a) + = L(a + ), L(a) = L(ã), ν α L(a) = L(ν α a). Proof. The first relation is obvious and it implies that the second and the third relation are equivalent. The second relation is proved in the proof of Theorem 7 in [Kn-Zu]. The fourth relation can be proved directly by constructing intertwining operator between induced modules ν α λ(a) and λ(ν α a), which induces an equivalence between ν α L(a) and L(ν α a) Proposition. The ring R is a Z-polynomial ring over indeterminates D. This means that {λ(a); a M(D)} is a Z-basis of R. Proof. This is well known fact because λ(a), a M(D) correspond to the standard characters which form a basis of the group of all virtual characters (for a fixed reductive Lie group). In fact, the proposition can be proved easily directly using [J1], and properties of Langlands classification. Lemmas 3.3 and 4.5 of this paper also imply the proposition Corollary. (i) The ring R is factorial. (ii) If δ D, then δ is prime. (iii) Let π R be a homogeneous element of the graded ring R, different from 0. Suppose that π = σ 1 σ for some σ 1, σ R. Then σ 1 and σ are homogeneous elements. (iv) The group of invertible elements in R is {L( ), L( )}. Note that L( ) is identity in R. Proof. Proposition 1.3 implies (i) and (ii). Proposition 1.3 implies that R is an integral domain. This implies (iii). From (iii) we obtain (iv) directly Remark. The mappings π π, π π, π π + and π ν α π induce automorphisms of graded ring R (this follows from Proposition 1.), which we shall denote respectively,, +, ν α : R R. The first three automorphisms are involutions. Each of these four automorphisms can be described by a permutation of indeterminates D. We shall say that f R is Hermitian if f = f +.. Formal approach to unitary dual of general linear group In this section F is either R of C. For δ D and n N denote a(δ, n) = (ν n 1 δ, ν n 1 1 δ,..., ν n 1 δ), u(δ, n) = L(a(δ, n)). If n = 0, then we take a(δ, 0) = and u(δ, 0) = L( ). Proposition 1. implies ν α u(δ, n) = u(ν α δ, n) for α R. If δ D u, then u(δ, n) + = u(δ, n). For σ Irr and α R denote π(σ, α) = (ν α σ) (ν α σ + ) R. Clearly, π(σ, α) is a Hermitian element of R. Note that π(σ, α) = π(σ, α) if σ is Hermitian. Now we shall introduce the following statements: (U0) If σ, τ Irr u then σ τ Irr u.

8 8 MARKO TADIĆ (U1) If δ D u and n N, then u(δ, n) Irr u. (U) If δ D u, n N and 0 < α < 1/, then π(u(δ, n), α) Irr u. (U3) If δ D and n N, then u(δ, n) is a prime element of the factorial ring R. (U4) If a, b M(D), then L(a) L(b) contains L(a + b) as a subquotient. Assuming (U0) (U4) to hold, we describe the unitary duals of general linear groups. In the following sections of this paper, we shall prove these statements. The proof of the following proposition is in section 8. of [T4]. For the sake of completeness, we present also here a (slightly modified) proof of it..1. Proposition. Suppose that (U0) - (U4) holds. Then Irr u is a multiplicative semigroup and it is a free abelian semigroup with a basis In other words: B = {u(δ, n), π(u(δ, n), α); δ D u, n N and 0 < α < 1/}. (i) If π 1,..., π i B, then π 1 π π i Irr u (ii) If π Irr u, then there exist π 1,..., π i Irr u, unique up to a permutation, such that π = π 1 π i. Proof. By (U0), Irr u is a multiplicative semigroup. The statements (U1) and (U) imply B Irr u. Therefore (i) holds. If π 1 π i = σ 1 σ j for some π 1,, π i, σ 1,, σ j B, then (U3) implies that i = j and that the sequences π 1,..., π i and σ,..., σ j differ up to a permutation. It remains to prove the existence of presentation in (ii). Let π Irr u. Choose a M(D) such that π = L(a). Since π is unitarizable, π is Hermitian i.e. π = π +. By Proposition 1. we have a = a +. Recall that for δ = ν e(δ) δ u D we have δ + = ν e(δ) δ u. Therefore we can find γ 1,..., γ n, δ 1,..., δ m D u, and positive numbers α 1,..., α m, such that we have the following equality of multisets a = (γ 1,..., γ m ) + m i=1 (να i δ i, ν α i δ i ) (cases m = 0 or n = 0 are possible). After a change of enumeration, we can assume that α 1,..., α u (1/)Z and α u+1,..., α m / (1/)Z for some 0 u m. Now introduce σ 1,..., σ v D u and positive numbers β 1,..., β v such that a = (γ 1,..., γ m ) + u i=1 (να i δ i, ν α i δ i ) + v j=1 (νβ j σ j, ν β j σ j ). Recall that γ 1,..., γ n, δ 1,..., δ u, σ 1,..., σ v D u and α 1,..., α u, β 1,..., β v are positive numbers such that α 1,..., α u (1/)Z and β 1,..., β v / (1/)Z (the case of n = 0 or u = 0 or v = 0 is possible). Take r 1,..., r v R and m 1,..., m v (1/)Z such that Clearly, m 1,..., m v 0. One gets directly that β j = r j + m j and 0 < r j < 1/ for j = 1,..., v. (ν α i δ i, ν α i δ i ) + a(δ i, α i 1) = a(δ i, α i + 1), i = 1,..., u, (ν m j δ j ) + ν 1/ a(σ j, m j ) = a(σ j, m j + 1), j = 1,..., v.

9 GL(n, C)ˆ AND GL(n, R)ˆ 9 The second relation implies (ν β j σ j, ν β j σ j ) + ν r j 1/ a(σ j, m j ) + ν 1/ r j a(σ j, m j ) = (ν r j+m j σ j, ν r j m j σ j ) + ν r j 1/ a(σ j, m j ) + ν 1/ r j a(σ j, m j ) = ν r j a(σ j, m j + 1) + ν r j a(σ j, m j + 1) for j = 1,..., v. In the rest of the proof we shall use the following property. Let a 1, a, M(D). Suppose that L(a 1 ), L(a ) are unitarizable. Then (U0) and (U4) implies L(a 1 ) L(a ) = L(a 1 +a ). By induction we obtain that L(a 1 ) L(a ) L(a k ) = L(a 1 + a + + a k ) if a 1,..., a k M(D) satisfy that L(a 1 ),..., L(a k ) Irr u. Now we shall finish the proof. We compute, using (U), (U3) and the above property π u i=1 u(δ,α i 1) v j=1 π(u(σ j, m j ), r j 1/) ( = L (γ 1,..., γ m ) + u i=1 (να i δ i, ν α i δ i ) + ) v j=1 (νβ j σ j, ν β j σ j ) u i=1 u(δ,α i 1) v j=1 π(u(σ j, m j ), r j 1/) ( = L (γ 1,..., γ n ) + u i=1 [(να i δ i, ν α i δ i ) + a(δ i, α i 1)] + v j=1 [(ν β j σ j, ν β j σ j ) + ν rj 1/ a(σ j, m j ) + ν 1/ r j a(σ j, m j )] ( u v ( = L (γ 1,..., γ n ) + a(δ i, α i + 1) + ν r j a(σ j, m j + 1) + ν r j a(σ j, m j + 1) )) i=1 j=1 = L(γ 1 ) L(γ n ) u i=1 L(a(δ i, α i + 1)) v j=1 π(l(a(σ j, m j + 1)), r j ) = u(γ 1, 1) u(γ n, 1) u i=1 u(δ i, α i + 1) v j=1 π(u(σ j, m j + 1), r j ). Thus, π divides u(γ 1, 1) u(γ n, 1) u i=1 u(δ i, α i + 1) v j=1 π(u(σ j, m j + 1), r j ). Now (U3) implies that π is a product of a subfamily of the modules u(γ i, 1), u(δ j, α j + 1), ν r k u(σ k, m k + 1), ν r k u(σ k, m k + 1), i = 1,..., n, j = 1,..., u, k = 1,..., v. The fact that π is Hermitian implies that π is a product of a subfamily of the modules u(γ i, 1), u(δ j, α j + 1), π(u(σ k, m k + 1), r k ). Thus, we have proved the existence of an expansion of π into a product of elements of B. This concludes the proof... Corollary. If (U0) (U4) hold, then the mapping Θ : (π 1,..., π n ) π 1 π n, M(B) Irr u, is an isomorphism of semigroups. In the rest of this paper we shall concentrate our attention to the proof of (U0) (U4), or give a reference where one can find proofs. Theorem 0.3 of [Bu] implies (Kirillov s) Conjecture 0.1 of the same paper and Theorem.1 of [Sa] implies (U0). Now we prove the remaining claims (U1) - (U4). We shall separately consider the complex and the real case. )

10 10 MARKO TADIĆ 3. Complex general linear group In the preceding section we have shown that (U0) (U4) imply a classification of the unitary dual of GL(n, F ). In this section we shall assume F to be C and we shall see that (U1) (U4) hold in this case (we have noticed above that (U0) holds). First we shall recall a number of basic facts from the representation theory of GL(n, C). We shall start with GL(1, C) and GL(, C). It is well known that D = G 1 = (C ), D u = Ĝ 1 = (C )ˆ (as we already have noticed). Let δ D u. This means that δ is a unitary character of C. Then u(δ, n) is just a unitary character g δ(det g), G n C, i.e. an one-dimensional unitarizable module of G n. Thus, (U1) holds. Further let 0 < α < 1/. The module π(u(δ, n), α) restricted to SL(n, C) is irreducible and unitarizable by [St] 4. This implies first π(u(δ, n), α) Irr. The module π(u(δ, n), α) is Hermitian, so its central character is unitary. Therefore π(u(δ, n), α) is unitarizable (which means that (U) holds). Now we shall introduce two parameterizations of D. If δ D, then there exist a unique n Z and β C such that δ(z) = z β (z/ z ) n = z β C (z/ z )n, z C = G 1. In this case we shall write δ = δ(β, n). Here denotes the usual absolute value on C, and we have = C. Note that δ(β, n) is a unitary character if and only if the real part of β is zero. The mapping C Z (C ) is an isomorphism, which gives a parameterization of D. Further, δ(β, u) + = δ( β, n) and e(δ(β, n)) = Re β. For given β C and n Z, there exist unique x, y C such that x + y = β and x y = n. Then we shall write δ(β, n) = γ(x, y). Therefore γ(x, y)(z) = z x+y (z/ z ) x y. In this way we obtain another parameterization of D by the set {(x, y) C ; x y Z}. Note that γ(x 1, y 1 )γ(x, y ) = γ(x 1 + x, y 1 + y ), γ(x, y) + = γ( ȳ, x) and e(γ(x, y)) = (1/) Re(x + y). We shall say that δ 1, δ D are linked if and only if δ 1 δ is reducible. The representation theory of GL(, C) implies that δ 1 and δ are linked if and only if there exist p, q Z such that pq > 0 and (δ 1 δ 1 )(z) = zp z q = γ(p, q)(z) for all z C (see [Go]). Then we have in R the equality δ 1 δ = L((δ 1, δ ))+ν 1 ν, where ν 1, ν D are defined by ν 1 (z) = ( z) q δ 1 (z), ν (z) = ( z) q δ (z) ([Go]). Further, then ν 1 ν is irreducible. If δ 1, δ, ν 1 and ν are as above, we shall write (ν 1, ν ) (δ 1, δ ). Now we shall interpret the above results in terms of the other parameterization of the characters of C. Let γ(x i, y i ) D, i = 1,. Then γ(x 1, y 1 ) and γ(x, y ) are linked if and only if If γ(x 1, y 1 ) and γ(x, y ) are linked, then x 1 x Z, and (x 1 x )(y 1 y ) > 0. (γ(x 1, y ), γ(x, y 1 )) (γ(x 1, y 1 ), γ(x, y )). Let (δ 1,..., δ n ) M(D). Suppose that δ i and δ j are linked for some 1 i < j n. Choose ν i, ν j D such that (ν i, ν j ) (δ i, δ j ). Then we shall write (δ 1,..., δ i 1, ν i, δ i+1,..., δ j 1, ν j, δ j+1,..., δ n ) (δ 1, δ, δ 3,..., δ n ). 4 More precise, the Harish-Chandra module that we use to see the unitarizability of π(u(δ, n), α) restricted to SL(n, C), is the module of K-finite vectors in the complementary series constructed in [St].

11 GL(n, C)ˆ AND GL(n, R)ˆ 11 Let a, b M(D). Then we write a < b if there exist c 1,..., c k M(D), k, such that a = c 1 c c c k 1 c k = b (for a, b M(D) we write a b if a = b or a < b). We shall see later that is a partial ordering on M(D) (and we shall examine some properties of ). Let a M(D). We say that a = (δ 1,..., δ n ) is written in a standard order if e(δ 1 ) e(δ ) e(δ n ) (we shall usually write elements of M(D) in a standard order). If this is the case we define e(a) = (e(δ 1 ), e(δ ),..., e(δ n )) R n. Let x = (x 1,..., x n ) R n and y = (y 1,..., y m ) R m. We shall write x y if and only if n = m and k i=1 x i k i=1 y i holds for each k = 1,..., n. It is obvious that is a partial ordering on n 0 R n. For x = (x 1,..., x n ) R n denote Tr(x) = x x n. For a M(D), we define Tr(a) to be Tr(e(a)). We have now a simple technical lemma regarding the notation that we have just introduced. 3.1 Lemma. (i) Let a, b M(D) and a b. Then the gradings of L(a) and L(b) in R are the same (recall that R = n=0r n is in a natural way a graded ring, and we can view Irr as a subset of R). (ii) Fix a M(D). The set of all b M(D) such that a b or b a is finite. (iii) Suppose that a b for a, b M(D). Then e(a) e(b) and Tr a = Tr b. We have a < b if and only if e(a) < e(b). (iv) The relation a b is a partial ordering on M(D). (v) Let a i, b i M(D), i = 1,. Suppose that a i b i, i = 1,. Then a 1 + a b 1 + b. We have a 1 + a = b 1 + b if and only if a i = b i for i = 1,. Proof. The definition of on M(D) implies (i). Let a = (γ(x 1, y 1 ),..., γ(x n, y n )). Suppose that b = (γ(x 1, y1),..., (x n, yn)) M(D) satisfies a b or b a. Then x i {x 1,..., x n } and yi {y 1,..., y n }. This implies (ii). Let a, b M(D) satisfy a b. Then a simple verification gives e(a) < e(b) and Tr(a) = Tr(b). This implies (iii). The claim (iii) implies (iv). Let a i, b i M(D), a i b i, i = 1,. Then the definition of on M(D) implies a 1 + a b 1 + b. If a i = b i, i = 1,, then clearly a 1 + a = b 1 + b. Suppose now a 1 + a = b 1 + b. Let a 1 < b 1. Then there exists c M(D) such that a 1 c b 1, and thus a 1 + a c + a b 1 + a b 1 + b. Therefore a 1 + a < b 1 + b, which contradicts to a 1 + a = b 1 + b since is a partial ordering on M(D). This completes the proof of (v). 3. Lemma. Let a, b M(D). Then λ(a) contains L(b) if and only if b a. Proof. Suppose b a. We shall prove that λ(a) contains L(b) by the induction with respect to the partial ordering on M(D) (this is possible by (ii) of Lemma 3.1). Let c be an element in M(D). Then by definition of L(c), λ(c) contains L(c). Suppose that a is an minimal element of M(D). Then b = a and by the above remark, λ(a) contains L(b) = L(a). Let a M(D) be arbitrary. We suppose that the claim λ(a) contains L(b) holds for all a M(D) such that a < a. Let b M(D), b a. If b = a, then λ(a ) contains L(b) = L(a ). Thus, we need to consider only the case of b < a. By the definition

12 1 MARKO TADIĆ of <, there exists c M(D) such that b c a. One sees directly, using commutativity and associativity of R, that each π Irr which is contained in λ(c), is also contained in λ(a ). Now the inductive assumption implies that λ(a ) contains L(b). Suppose now that λ(a) contains L(b). Applying Example 3.16 of [SpVo] (or Corollary 3.15 of of the same paper), we know that either L(a) = L(b) i.e. a = b, or L(b) is in the kernel of some factor of the long intertwining operator. In our situation it means that either L(a) = L(b) i.e. a = b or there exist c a such that λ(c) contains L(b), because the kernels of the factors of the long intertwining operator attached to λ(a) have form λ(c) for c a (see Lemma 3.8 of [SpVo]). Now we get b a using induction on a, with respect to the ordering of M(D) Lemma. Fix a M(D). (i) There exist m a b N for b a, such that λ(a) = b a ma b L(b) holds in R. Further, m a a = 1. (ii) There exist m (a,b) Z for b a, such that L(a) = b a m (a,b)λ(b). We have m (a,a) = 1. (iii) Suppose that c M(D) satisfies c < a. Let c be adjacent to a, i.e. there does not exist b M(D) such that c < b < a. Then m (a,c) 0. (iv) Let c M(D) satisfies c < a. Suppose that for d M(D) such that d a we have e(c) e(d). Then a is adjacent to c (and m (a,c) 0). Proof. Lemma 3. and the fact that L(a) has multiplicity one in λ(a), imply (i). We shall show (ii) and (iii) simultaneously by the induction with respect to a M(D). If a M(D) is minimal, then L(a) = λ(a) by (i). Therefore (ii) holds. Note that there does not exist c such that c < a. Therefore (iii) also holds. Let a M(D) be arbitrary. From (i) and the inductive assumption we have (3-1) L(a) = λ(a) b<a ma b L(b) = λ(a) b<a ma b ( λ(b) + d<b m (b,d)λ(d) ). After a gathering of the terms in the above expansion of L(a), we obtain (ii). Suppose that m (a,b) = 0. Since m a b > 0 for all b < a, (3-1) implies that there exist b M(D) such that b < b < a. This proves (iii). Suppose c M(D) satisfies (iv). Assume that c is not adjacent to a. Then there exist b, d M(D) such that c < b d a. Then e(c) < e(d) by (iii) of Lemma 3.1. This contradiction proves (vi). The following proposition is just (U4) Proposition. If a, b M(D), then L(a) L(b) contains L(a + b) as a subquotient. The multiplicity is one.

13 GL(n, C)ˆ AND GL(n, R)ˆ 13 Proof. We compute in R L(a) L(b) = ( λ(a) + ) ( c<a m (a,c) λ(c) λ(b) + ) d<b m (b,d) λ(d) = λ(a) λ(b) + d<b m (b,d) λ(a) λ(d) + c<a m (a,c) λ(c) λ(b) = L(a + b) + u<a+b ma+b u L(u) + d<b (b,d)( m u a+d ma+d u L(u) + c<a, b<d m (a,c) m (b,d) λ(c) λ(d) ) + c<a m (a,c)( u c+b mc+b u ) L(u) ) + ( c<a, d<b m (a,c) m (b,d) u c+d mc+d u L(u). This and (v) of Lemma 3.1 implies that L(a) L(b) contains L(a + b) with multiplicity one. Now we return to u(δ, n), δ D. Let δ = γ(x, y). Then a(δ, n) = (ν n 1 δ, ν n 1 1 δ,..., ν n 1 δ) = (γ(x+ n 1, y + n 1 n 1 ), γ(x+ 1,, y + n 1 1),..., γ(x n 1, y n 1 )). In the rest of this section, we shall consider R as a polynomial ring in indeterminates D. 3.5 Lemma. Let δ, δ D. Then the degree of the polynomial u(δ, n) in the indeterminate δ is either zero or one. Proof. We know u(δ, n) = a a(δ,n) m (a(δ,n),a)λ(a). Let a 0 = (γ(x 1, y 1 ),..., γ(x n, y n )) a(δ, n). The formula for a b (in terms of γ-coordinates ) implies {x 1,..., x n } = {x + n 1, x + n 1 1,..., x n 1 }. Therefore, x 1,..., x n are all different, which implies that γ(x 1, y 1 ),..., γ(x n, y n ) are all different. This implies the lemma. Denote For 1 i < j n let Y i (i, j) = γ(x+ n 1 X i = γ(x + n i, y + n i), i = 1,..., n. n 1 +1 i, y + +1 j), Y j(i, j) = γ(x+ n 1 n 1 +1 j, y + +1 i). Note that (Y i (i, j), Y j (i, j)) (X i, X j ). Denote a 0 = a(δ, n) = (X 1, X,..., X n ) and a i,j = (X 1,..., X i 1, Y i (i, j), X i+1,..., X j 1, Y j (i, j), X j+1,..., X n ). Then a i,j a 0 for all 1 i < j n.

14 14 MARKO TADIĆ 3.6. Lemma. For 1 p n 1, a p,p+1 is adjacent to a 0. Proof. First note e(a 0 ) = ( n 1, n 1 1,, n 1 (3-) e(y i (i, j), Y j (i, j)) = ( n 1 i+j x+y ) + Re( ) (1, 1,..., 1), n 1 + 1, i+j x+y + 1) + Re ( ) (1, 1). Suppose that there exists some d M(D) such that a p,p+1 < d a 0. Then e(a p,p+1 ) < e(d) by (iii) of Lemma 3.1. There exists 1 i < j n such that d = a i,j. Thus e(a p,p+1 ) < e(a i,j ). The definition of the ordering on R n and (3-) implies p i. Using the fact that Tr a p,p+1 = Tr a i,j we obtain that j p + 1. Therefore (i, j) = {p, p + 1} i.e. i = p, j = p + 1, which implies a p,p+1 = a i,j and e(a p,p+1 ) = e(d). This contradicts our assumption. The proof of the lemma is now complete. For a moment, we shall interrupt our study of u(δ, n) s and consider a simple general lemma, which we shall use later in the proof of (U3). Suppose that R is a polynomial ring over the set of indeterminates D. Let ψ : D Z be any function. We can then define a grading gr ψ on R specifying gr ψ (d) = ψ(d) for d D. In this way we get a Z-grading on R, which we shall call ψ-grading. If we take ψ 1, then we get the usual grading by total degree of polynomials. If we take for ψ the characteristic function of a fixed d D, then we get the grading by the degree in d Lemma. Suppose that a non-constant T R has the property that each d D has the degree in T equal to 0 or 1. Let V be the set of d D which have degree one in T and let T = m S c mm be a shortest expansion of T into the sum of monomials d 1 d... d l, where l N and d i D. Suppose that T factors as a product of two non-constant polynomials P 1 and P from R. Then there exists a partition V = V 1 V into two non-empty sets which satisfies the following condition: if T is homogeneous for some ψ-grading, then the function m d V 1,d m ψ(d) on S is constant (here d m denotes: d divides m). Proof. Denote by V i the set of all d D which have degree at least one in P i (then the degree of d V i is exactly one, since R is an integral domain). The condition on degrees of d D in T implies that V 1 and V are disjoint (R is an integral domain). Clearly, V = V 1 V. Since P i are non-constant, V i are non-empty. Suppose that T is homogeneous for some ψ-grading. Since R is integral domain, P i must be then also homogeneous for ψ-grading. This easily implies that V 1 satisfies the condition of the lemma. Now we shall go back to our study of u(δ, n) R, where δ = γ(x, y). We shall consider two gradings bellow. First consider the ψ-grading for ψ 1. Now (ii) of Lemma 3.3 implies that elements of Irr are homogeneous for this grading. This is the same grading as the standard one. Define φ : γ(x, y ) x y, D Z. First one checks directly that a b implies gr φ (λ(a)) = gr φ (λ(b)). This implies gr φ (λ(a)) = gr φ (λ(b)) if a b. This and (ii) of Lemma 3.3 imply that elements of Irr are homogeneous also for φ-grading.

15 3.8. Proposition. Elements u(δ, n) are prime in R. GL(n, C)ˆ AND GL(n, R)ˆ 15 Proof. Suppose that some u(δ, n) is not prime. Then u(δ, n) = P 1 P for some P i R which are not invertible in R. Note that P i must be homogeneous for the standard grading of R = n=0r n. Write a shortest expansion u(δ, n) = m S c m m as in Lemma 3.7. First note that by (ii) of Lemma 3.3, at least one c m is 1. Therefore, P i are non-constant polynomials (homogeneous for the standard grading). We remind the reader that R is a Z-polynomial ring. By (ii) of Corollary 1.4 we know that n. The above discussion and Lemma 3.5 imply that T = u(δ, n) satisfies the conclusion of Lemma 3.7. Then there exist V 1 and V be as in Lemma 3.7. Further note that c λ(a0 ) 0 by (i) of Lemma 3.3, and c ap,p+1 0 for p = 1,..., n 1 by Lemma 3.6 and (iii) of Lemma 3.3. In other words, a 0 and a p,p+1 are in S. Denote v 1 (ψ) = card {i; 1 i n and X i V 1 } = 1. d V 1,d λ(a 0 ) If this would be 0, then V 1 would be empty (since this is independent of λ(a) S by Lemma 3.7). This and the fact that V is non-empty imply 1 v 1 (ψ) n 1. Therefore there exists p {1,..., n 1} such that {X p, X p+1 } V i for i = 1,. Then X p V 1 and X p+1 V, or X p V and X p+1 V 1. Without lost of generality, we can assume that the first possibility holds. We know by the property of V 1 from Lemma 3.7 that v 1 (ψ) = d V 1, d λ(a p,p+1 ) 1 = d V 1 \{X p,x p+1 }, d λ(a p,p+1 ) 1 = d V 1 \{X p,x p+1 }, d λ(a 0 ) 1 + j {p,p+1}, Y j (p,p+1) V 1 1. This implies ( ({X1,..., X n } V 1 ) \{X p } ) {Y j (p, p + 1)} = {d V 1 ; d λ(a p.p+1 )} for some j {p, p + 1}. Now applying Lemma 3.7 to φ-grading, we get d (({X 1,...,X n } V 1 )\{X p }) {Y j (p,p+1)} φ(d) = d {X 1,...,X n } V 1 φ(d), which implies φ(y j (p, p + 1)) = φ(x p ), for some j {p, p + 1}. Note that φ(x p ) = x y, φ(y p (p, p + 1)) = x y + 1 and φ(y p+1 (p, p + 1)) = x y 1. From this we see that we can not have φ(y j (p, p + 1)) = φ(x p ). completes the proof. We have seen that all (U0) (U4) hold in the complex case. 4. Real general linear group I This contradiction In the rest of this paper, we shall assume that F = R. First we shall parameterize D. The signum characters of R will be denoted by sgn (clearly sgn D u ). Now we shall recall of some simple facts from the representation theory of GL(, R) (see [Go]). Let δ 1, δ G 1. If δ 1 δ is irreducible, then δ 1 δ = L((δ 1, δ )). Further, δ 1 δ is reducible if and only if there exist p Z\{0} such that δ 1 (t)δ (t) 1 = t p sgn(t) for t R = G 1.

16 16 MARKO TADIĆ If δ 1 δ is reducible, then there exist γ(δ 1, δ ) D such that δ 1 δ = L((δ 1, δ )) + γ(δ 1, δ ). The mapping (δ 1, δ ) γ(δ 1, δ ) from the set of all pair in G 1 G 1 such that δ 1 δ reduces, into D\ G 1, is surjective. We have γ(δ 1, δ ) = γ(δ 1, δ ) if and only if {δ 1, δ } = {δ 1, δ } or {δ 1, δ } = {δ 1sgn, δ sgn}. The relation δ 1 δ = L((δ 1, δ )) + γ(δ 1, δ ) implies γ(δ 1, δ ) + = γ(δ 1 +, δ+ ). We have also e(γ(δ 1, δ )) = 1 (e(δ 1) + e(δ )). Thus γ(δ 1, δ ) D u if and only if e(δ 1 ) + e(δ ) = 0. Now we shall introduce another parameterization of elements of D. First we shall make a short preparation. Let γ(δ 1, δ ) D, where δ 1, δ are characters of R. By the definition of γ(δ 1, δ ) we know that (δ 1 δ 1 )(t) = tp sgn(t) for some p Z\{0} (here t R ). Write δ i (t) = t α i (sgn(t)) m i, where α i C, m i {0, 1} for i = 1,. Now (δ 1 δ 1 )(t) = tp sgn(t) implies t α 1 α sgn (t) m 1 m = t p sgn(t) = t p (sgn(t)) p+1. Thus α 1 α = p and (sgn) m 1 m = (sgn) p+1. The last relation implies m 1 m p+1(mod ). Therefore, if we denote by δ1(t) = t α 1, then γ(δ 1, δ ) = γ(δ1, δ) where δ = δ (sgn) m 1. Note that δ(t) = t α sgn(t) m sgn(t) m 1 = t α sgn(t) m 1 m = t α sgn(t) α 1 α +1. Fix x, y C satisfying x y Z\{0}. Define δ 1 (t) = t x, δ (t) = t y sgn(t) x y+1. Then δ 1 δ 1 (t) = tx y sgn(t), and therefore we can define γ(x, y) by γ(x, y) = γ(δ 1, δ ). From our above discussion follows γ(x, y) = γ(y, x) (using γ(δ 1, δ ) = γ(δ sgn, δ 1 sgn)). Further γ(x, y) = γ(x, y ) {x, y} = {x, y }. We have γ(x, y) + = γ( x, ȳ), e(γ(x, y)) = Re( x + y ), ν α γ(x, y) = γ(x + α, y + α), α R. For x C and ε Z/Z denote γ ε (x)(t) = t x (sgn(t)) ε, t G 1. Now (x, y) γ ε (x), C (Z/Z) G 1 is a bijection, which parameterizes G 1. Clearly e(γ ε (x)) = Re(x). Now the following lemma holds Lemma. The representation γ ε (x) γ ε (x ) / Irr x x Z\{0} and x x + 1 ε ε (mod ). If γ ε (x) γ ε (x ) reduces, then we have γ ε (x) γ ε (x ) = L((γ ε (x), γ ε (x ))) + γ(x, x ). Now we shall describe the infinitesimal characters of the modules L(a), a M(D). Let δ D. Then either δ = γ ε (x) for some x C, ε Z/Z, or δ = γ(x, y ) for some x, y C,

17 GL(n, C)ˆ AND GL(n, R)ˆ 17 x y Z\{0}. Now we define χ(δ) M(C) by χ(γ ε (x)) = (x), χ(γ(x, y )) = (x, y ). For a = (δ 1,..., δ n ) M(D) we define χ(a) by the formula χ(a) = χ(δ 1 ) + + χ(δ n ). We consider the standard grading on R. If π R n, then we shall write n = gr(π). For a M(D), define gr(a) = gr(l(a)). With this definition, we have gr(a) = cardχ(a). Let a n g n be the Lie algebra of the subgroup A n of all diagonal elements in G n. Let a C n and g C n be complexifications of these two algebras. The universal enveloping algebras of a C n and g C n are denoted by U(a C n) and U(g C n) respectively. The center of the algebra U(g C n) is denoted by Z(g C n). We consider the Harish-Chandra homomorphism ξ : Z(g C n) U(a C n). Let (a C n) be the space of all complex linear functionals on a C n. For λ (a C n), let ξ λ : Z(g C n) C be the composition of ξ with the evaluation at λ. Let A 0 n be the connected component of the group A n containing identity, and M the torsion subgroup of A n. The normalizer of A n in K n is denoted by M. Set W = M /M. Now W acts on g C n and a C n. As it is well known, every homomorphism of Z(g C n) into C is obtained as ξ λ for some λ (a C n). Also ξ λ = ξ µ if and only if W λ = W µ. We identify g C n in a natural way with the Lie algebra of all complex n n matrices. Then a C n is the subalgebra of all diagonal matrices in g C n. Now W acts on a C n by permutations of diagonal elements and W is isomorphic to the permutation group of order n. Let λ (a C n). Then there exist λ 1,..., λ n C such that λ : diag(x 1,..., x n ) λ 1 x λ n x n Note that λ (λ 1,..., λ n ) is an isomorphism of (a C n) onto C n. We shall identify these two vector spaces using this isomorphism. In this identification, W acts by the permutations of coordinates. Therefore (a C n) /W can be naturally identified with the set of all multisets (λ 1,..., λ n ) in C (of cardinal number n). For λ = (λ 1,..., λ n ) C n, let λ : A 0 n C be the character λ : diag(a 1,..., a n ) a λ aλ n n. The mapping λ λ is a group isomorphism of (a C n) onto the group (A 0 n) of all continuous homomorphisms of A 0 n into C. Fix a = (γ ε1 (x 1 ),..., γ εn (x n )) M( G 1 ). Let µ (a C n) corresponds to (x 1,..., x n ) C n under the above identification. Then by Lemma of [Vo1], λ(a) has infinitesimal character, and it is equal to ξ µ. For a M(D) such that gr(a) = n, denote χ(a) = (x 1,..., x n ). Then by Lemma 4.1 there exist ε 1,..., ε n Z/Z such that π Irr which is contained in λ(a), is contained in λ((γ ε1 (x 1 ),..., γ εn (x n ))). Therefore the infinitesimal character of λ(a) is ξ µ, where µ is as above. This implies: 4.. Lemma. Let a, b M(D) with gr(a) = gr(b). Then L(a) and L(b) have the same infinitesimal character if and only if χ(a) = χ(b). Now we shall describe a necessary conditions that λ(a) contains L(b) for a, b M(D). Let a = (δ 1,..., δ m ) M(D). We say that a = (δ 1,..., δ m ) is written in a standard order if e(δ 1 ) e(δ ) e(δ m ). Suppose that a = (δ 1,..., δ m ) is written in a standard order. Let gr(a) = n. Define e(a) = (e(δ 1 ),..., e(δ 1 ), e(δ ),..., e(δ ),..., e(δ m ),..., e(δ m )) R n. }{{}}{{}}{{} gr(δ 1 ) times gr(δ ) times gr(δ m ) times Clearly, e(a) is uniquely determined by a.

18 18 MARKO TADIĆ We define a partial ordering on R n as before (recall (x 1,..., x n ) (y 1,..., y n ) k i=1 x i k i=1 y i for all 1 k n). For x = (x 1,..., x n ) R n define Tr(x) = x x n. If a M(D), then we define Tr(a) = Tr(e(a)) Lemma. Let a, b M(D). Suppose that L(b) is contained in λ(a). Then gr(a) = gr(b), χ(a) = χ(b), Tr(a) = Tr(b) and e(b) e(a). Further, a b if and only if e(b) < e(a). Proof. The claim gr(a) = gr(b) is obvious. Since L(b) and λ(a) have the same infinitesimal character, χ(a) = χ(b) is a consequence of Lemma 4.. Let c = (δ 1,..., δ m ) M(D). After a renumeration we can suppose that δ i = γ(x i, y i ) for 1 i < k, and δ i = γ εi (z i ) for k i m, for some 1 k m + 1, x i, y i, z i C, ε i Z/Z. Then χ(c) = (x 1, y 1, x, y,..., x k 1, y k 1, z k,..., z m ). Now we have Tr(c) = Re( k 1 i=1 (x i + y i ) + m i=k z i). Therefore, Tr(c) depends only on χ(c). Now χ(a) = χ(b) (which we have observed) implies Tr(a) = Tr(b). Let gr(a) = n. Denote β i = (1, 1,..., 1, 0,..., 0) i }{{} n (1, 1,..., 1, 1,..., 1) Rn, i = 1,..., n 1. i times We consider on R n the standard scalar product < (x 1,..., x n ), (y 1,..., y n ) >= n i=1 x iy i. Now by Proposition 4.13 of [BlW] we have < β i, e(b) > < β i, e(a) > for i = 1,..., n 1. Also, if all n 1 above inequalities are actually equalities, then a = b by the same proposition. Note that < (x 1,..., x n ), β i >= x x i i n Tr((x 1,..., x n )). Since Tr(a) = Tr(b), we obtain that e(b) e(a), and further that a b if and only if e(b) < e(a) Lemma. (i) Let a M(D). The set of all b M(D) such that χ(a) = χ(b), is finite. (ii) Let a i, b i M(D), i = 1,. Suppose that e(b i ) e(a i ) for i = 1,. If e(a 1 + a ) = e(b 1 + b ), then e(a i ) = e(b i ) for 1 = 1,. Proof. The claim (i) is a direct consequence of the definition of χ(b), b M(D). Suppose that a i, b i M(D) satisfy e(b i ) e(a i ) for i = 1,. Let e(a 1 ) e(b 1 ) or e(a ) e(b ). Without lost of generality, we can assume that e(a 1 ) e(b 1 ). Now e(b 1 ) < e(a 1 ). Write e(a 1 ) = (e 1,..., e n ), e(a ) = (e n+1,..., e m+n ), e(b 1 ) = (f 1,..., f n ), e(b ) = (f n+1,..., f m+n ). Choose a permutation σ of {1,..., n + m} such that f σ 1 (1) f σ 1 () f σ 1 (n+m). Now e(b 1 + b ) = (f σ 1 (1),..., f σ 1 (n+m)). We can choose σ as above, which satisfies 1 σ 1 (i) < σ 1 (j) n i < j, and m + 1 σ 1 (i) < σ 1 (j) n + m i < j. Now e(b 1 ) < e(a 1 ) implies that e(b 1 + b ) = (f σ 1 (1),..., f σ 1 (n+m)) < (e σ 1 (1),..., e σ 1 (n+m)). Since (e σ 1 (1),..., e σ 1 (n+m)) e(a 1 + a ) (which one can easily check), we have e(b 1 + b ) < e(a 1 + a ). This proves (ii) of the lemma. Let a = (δ 1,..., δ k ) M(D). For b M(D) we shall say b a if there exist 1 i < j k and c M(D) such that b = (δ 1, δ i 1, δ i+1,..., δ j 1, δ j+1,..., δ k ) + c, where L(c) is a subquotient of δ i δ j satisfying c (δ i, δ j ). Using Lemma 4.3, it is a simple combinatorial exercise to show that e(b) < e(a). This implies that if we generate by an ordering < on M(D), that it is really an ordering.

19 GL(n, C)ˆ AND GL(n, R)ˆ 19 Suppose that λ(a) contains L(b) and a b (a, b M(D)). The end of 3 of [SpVo] (more precisely, Corollary 3.15, Lemma 3.8 and Theorem 3.7 of [SpVo]) implies that L(b) is contained in λ(a 1 ), for some a 1 a. Continuing this analysis as in the second paragraph of the proof of Lemma 3., we would get b < a Lemma. Fix a M(D). (i) There exist m a b Z + for b M(D), b a, such that λ(a) = b a ma b L(b). Further, m a a = 1. (ii) There exist m (a,b) Z for b M(D), b a, such that L(a) = b a m (a,b)λ(b). We have m (a,a) = 1. Proof. The fact that L(a) has multiplicity one in λ(a) ([BlW]), Lemma 4.3 and the discussion preceding the proposition imply (i). We prove (ii) in the same way as (ii) of Lemma 3.3 (by induction in a finite set). Now we can see that (U4) holds Proposition. If a, b M(D), then L(a) L(b) contains L(a+b), and the multiplicity is one. Proof. The proof of the above proposition is very similar to the proof of Proposition 3.4 (use (ii) of Lemma 4.4 instead of (v) of Lemma 3.1). We shall now get additional information regarding expansions considered in Lemma Lemma. Let a, b, c M(D). If b a, then m a b 1. Let c a and suppose that c satisfies the following condition (4-1) for each d M(D), such that d a (and d c), we have e(c) e(d). Then m (a,c) 0. Proof. The fact that m a b 1 if b a follows directly from the last proposition (or the factorization of the long intertwining operator). For the other claim of the lemma we shall use relation (3-1) (which obviously holds also here). Fix some c M(D) for which c a and m (a,c) = 0. Then m a c 1, and (3-1) implies that there must exist d M(D) satisfying c < d < a and m a d 1. Suppose additionally that c satisfies (4-1). Take d such that d d a. Then d c since c < d d. The last inequality implies e(c) < e(d ). This contradicts (4-1). Thus, m (a,c) 0. Now we shall observe that (U1) holds for GL(n, R) (B. Speh proved it in [Sp]) Proposition. Representations u(δ, n) are unitarizable if δ D u and n N. Proof. If gr(δ) = 1, then u(δ, n) is a unitary character of G n obtained by composing δ with the determinant homomorphism. Thus, u(δ, n) is unitarizable. It remains to consider the case gr(δ) =. Then δ = γ(x, y) for some x, y C such that x y Z\{0} and Re(x+y) = 0. If x + y = 0, then Theorem of [Sp] implies unitarizability of u(γ(x, y), n). From γ(x, y) = ν x+y γ( x y, y x x+y ) we obtain u(γ(x, y), n) = ν u(γ( x y, y x x+y ), n). Since ν is a unitary character of G n, u(γ(x, y), n) is unitarizable.