ON UNITARY REPRESENTATIONS WITH REGULAR INFINITESIMAL CHARACTER by SUSANA A. SALAMANCA RIBA Sc.B. Universidad Autonoma Metropolitana Mexico D.F. Mexico (1980) Submitted of in Partial Fulfillment the Requirements the for Degree of Doctor of Philosophy the at MASSACHUSETTS INSTITUTE OF TECHNOLOGY MAY, 1986 @ Susana A. Salamanca Riba The author hereby grants to M.I.T. permission to reproduce and to distribute copies of this thesis document in whole or in part. Signature of Author .. ,. ..... Department Certified by .................... of Mathematics May 9, 1986 ................. gDavid A. Vogan Supervisor Accepted by ............ MASSACHUSETTS INSTITUTE OFTECHNOLOGY SEP 16 1986 L!BRAP S ARCHIVES Chairman, - - ----............... ............. Nesmith C. Ankeny Departmental Graduate Committee 2 Dedication Dedico esta tesis a mis padres y hermanos por el amor, apoyo y fe que me han prodigado durante especialmente a mi toda mi vida, hermana Lourdes con quien he compartido la misma experiencia de estudiantes en M.I.T. y que durante estos afios fue mi ejemplo y mi apoyo y que generosamente me dio su tiempo, comprension y numerosos consejos para el desarrollo de mi investigacion. 3 Acknowledgements It is a pleasure for me to thank my advisor David Vogan for introducing me to representation theory, for teaching me how to do research, for being always willing to share his knowledge, and for being a source of support and enthusiasm. I also want to thank all my friends constant encouragement and support, for providing and for teaching me how to do research, especially Jeff Adams, Jeremy Orloff, Roger Zierau, my sister Lourdes Salamanca Young, Devra Garfinkle, and Ranee Gupta. The latter three, together with Michele Vergne have been outstanding female models out the time I was a student at M.I.T. Barbasch for helpful for me I also through- thank Dan discussions and suggestions, and generous advice and interest in my work. Thank you Luis Casian for patiently discussing with me numerous parts of my thesis as well as problems other interesting in mathematics, for continuous advice and under- standing, and constant interest in my progress. Thank you again Roger Zierau and David Vogan for making research in mathematics so much fun. Finally I want to thank Phyllis Ruby for her constant interest and help during my stay at M.I.T., for her excellent work in typing this and Viola Wiley thesis. 4 ON UNITARY REPRESENTATIONS WITH REGULAR INFINITESIMAL CHARACTER by SUSANA A. SALAMANCA RIBA Submitted to the Department of Mathematics on May 2, 1986, in partial fulfillment of the requirements for the degree of Doctor of Philosophy. ABSTRACT G = SL(nR), For SU(p,q), and SP(n,R) we prove that every irreducible unitary representation of the same infinitesimal character as that involve case-by-case arguments features of these groups. So with of a finite dimen- sional, arises as cohomological parabolic one-dimensional unitary character. G induction from a The techniques used that do not use any it seems reasonable that these arguments could be extended in order special to hope to solve the problem for other groups. G For of G, 90 = k Lie(G), G WK as above let + P%, (7il) on which K be a maximal compact subgroup the Cartan decomposition of an irreducible Hermitian representation of Z(g) acts as on a finite dimensional module, the Harish-Chandra module of K-type of WK. We prove a Zuckerman module A that (N), q = t + u C g. subalgebra character X restricted to o= of V Thesis Supervisor: Title: (p@V ) either for some and IK (ji,V ) is a lowest isomorphic to 8-stable parabolic and one-dimensional unitary L = NG(q), G 7r or else the Hermitian form is indefinite. Dr. David A. Vogan, Jr. Professor of Mathematics 5 Contents Abstract 4 1. Introduction 6 2. Notation and Main Results 12 2.1 Structure Theory 12 2.2 Harish-Chandra Modules 14 2.3 Parabolic Subalgebras 16 2.4 Derived Functor Modules 17 2.5 The Modules A (X) 36 2.6 Reduction Step for 2.7 3. the Proof of Theorem 1.3 45 Methods 59 to Detect Nonunitarity G = SL(n,R) 66 3.1 Preliminary Notation 66 3.2 Construction of LY(X) for a Harish- Chandra Module X 4. 5. 3.3 Lowest K-types 3.4 Proof of 69 the Modules A (X) of Theorem 2.6.7 for G = SL(n,R) 72 75 G = SU(p,q) 82 4.1 Preliminary Notation 82 4.2 Computation of Lv(X) and XV(X) for a Harish-Chandra Module X 85 4.3 The Lowest K-types of 95 4.4 Parameters XV (AQ (X)) and LV(A ()) 100 4.5 Proof of Theorem 2.6.7 for G = SU(pq) 104 the Modules A q(X) G = SP(n,R) 130 5.1 Preliminary Notation 130 5.2 Consruction of 5.3 Lowest K-types of 5.4 Proof of Theorem 2.6.7 for G = SP(n,R) Bibliography LV(X) for any Module X the Modules A (N) 133 138 142 154 6 Chapter Introduction 1. those irreducible unitary The problem of classifying representations of a Lie group whose ter is the same as al, has infinitesimal charac- that of an irreducible interested many people for many years. representations are important because esting applications like the Here is an example: is a semisimple Lie group, subgroup such that F\G finite dimensional G-module, on L2 (F\G) F an irreducible m(7r,F), * = m(r,F)7r, u Matsushima's formula (see Borel-Wallach [1980], p. 223) is H (F,F) m (r,F)H*(g,K;WKOF = rEG Here g G irreducible, unitary representations the set of of G. (irW) a discrete then the right action of rEG u F is compact, and L2 (\G) G ch. VII.5-VII.6). gives a Hilbert space decomposition, with finite multiplicities with in inter- theory of automorphic forms for example, Borel-Wallach [1980], G These they appear (see If finite dimension- is Lie(G) ® C, u and WK is the Harish-Chandra 7 module of The relative Lie algebra cohomology (irt). the groups on the right are non-zero only when r mal character of is that of the same as XK the modules F. algebraic construction In Vogan-Zuckerman [1984], an of infinitesi- with non-vanishing cohomology groups in terms of cohomological parabolic induction is given. this way are conjec- derived functor modules constructed tured to exhaust a larger The family of unitary representations described below. to chapter 2 for precise definitions of We refer the terms used below. Let pact G be a reductive Lie group, subgroup, 90 = k 0 D bra (cf. L + u C g 2.3), OL = L, L = and that be a go. e-stable parabolic subalge- is OU = U, L, with denoting complex conjugation = L. q, n o Let a maximal com- the corresponding Cartan involution, and the Cartan decomposition of q = Let 0 K L dimensional be the normalizer of unitary character of q L. in A G, X (g,K)-module is a Harish-Chandra module constructed as Chapter 6 by cohomological parabolic and a oneA (N) in Vogan [1981] induction from the 8 one-dimensional unitary character X. (See Definitions 2.4.14, 2.5.2). The conditions on the infinitesimal character men- tioned above can be weakened to include a larger family of representations. If X is a Harish-Chandra module with infinitesimal x, character and g h C a Cartan subalgebra, then up Weyl-group orbit considerations -y E I' . h corresponds to a weight Choose a positive root system A+(g,h) such that is dominant. Conjecture 1.1. Suppose X is an irreducible unitary Harish-Chandra module such that A +(g,h). q x to Then, there are a r - p is dominant for e-stable parabolic subalgebra and a unitary one-dimensional character X of L such that X A (M). Some progress has been made when we assume that regular and T. integral. J. Enright [1979] then there exists a X. Also, Namely, for proved that (g,K) in Speh [1981], G if module is regular (N) following result the same result for is proved in this integral isomorphic to G = SL(n,IR) is proved. The is a complex group, i A -r thesis. 9 Theorem 1.2. and -r If G is SL(n,R), regular and integral, SU(p,q), then, for or some q SP(n,R) and X X = A (N). The proof from Speh's for is new and quite different original one. The proof involves SL(n,R) is by induction on the dimension of choosing an appropriate proper and embedding 9K as subgroup that unitarity of G and our The set up It L C G the Langlands submodule of a derived functor module induced from a representation of such a way G. L, in the information about unitarity or non- the representation of representation L can be carried up to IK' thesis is organized as follows. In Chapter 2 we the notation and results needed to restate and prove the result in the following form: Theorem 1.3. Suppose X is an irreducible Harish-Chandra module with regular integral infinitesimal character, equipped with a non-zero Hermitian form < >. Then, K-type V , either a) X b) There are a V ® p, = A (X), for some q, lowest-K-type such that A as above, or V6 and a 2 10 i = 1,2, HomK(V .,X) A 0 1 < and the restriction of , > to the sum V5 G V is indefinite. Sections 2.1 the results through 2.4 are devoted to notation and that will be needed for main issues are functor modules the definition of 9 (Y) the proof. the Zuckerman derived and Vogan's embedding of any irreducible Harish-Chandra module derived functor module. ties of these modules. into some Zuckerman We also give some useful proper- In Section 2.5 we define the modules some nice The two A (X) and prove features that we will use in later chapters. Sections 2.6 and 2.7, and Chapters 3 and 5 are the actual proof of Theorem 1.3. The main results are Theorems 2.6.7 and 2.6.8, which say that we can exhibit submodule of a derived functor module way that we can reduce the problem XL of the group L. Chapters I X (XL) to the as a in such a representation 3, 4 and 5 are the proof of Theorem 2.6.7. We argue by contradiction: embedding result we find another With the help of Vogan's 0-stable parabolic algebra and another Zuckerman module containing have to check several conditions that will ensure X. the subWe 11 reduction, but mainly 2.6.7 b) and c). the representation isomorphic to a module XL is not Then assuming that OL A O(N ) we prove non-unitarity on XL. For this we use q the properties of the A (N) modules discussed in 2.5 and some techniques discussed in 2.7, primarily Lemma 2.7.1. 12 Chapter 2 In this chapter we set up notation, state the basic results we will need and our main result, and provide a scheme for the proof. For undefined terms in this section see, Vogan [1981] Chapter 0. 2.1. Structure Theory for example, We will denote Lie groups by upper case roman letters such as G, H, L letters such as and complex Lie algebras by script g, h, L. We will make the distinction between the real Lie algebra of a Lie group and its complexification as follows: 90 = Lie Let 9 = 0 C, etc. U(g) = universal enveloping algebra of = center of Z(g) (G) and U(g.). Although we will eventually study connected simple g real linear Lie groups, we will consider connected reductive linear Lie groups. These are Lie groups real satisfy- ing: a) G b) go c) G is connected is a real has a reductive Lie algebra faithful finite dimensional representation 13 Let 00 8 be a Cartan involution of the Cartan decomposition of eigenspaces of go and +1 into the go go = k0 and -1 8. Fix once and for all a nondegenerate, invariant symmetric bilinear form on go. We will denote this and its various complexifications, restrictions and ( dualizations by , ). We may choose Cartan decomposition of go it so that the is orthogonal and > 0 ( , ) 1k0 ( , Let H be a Cart an subgroup of the roots of ht In general subalgebra of 9 then o in if g9 A(V,o) multiplicities). < 0. G. A = A(g,h) Denote by g. o, is an abelian reductive Lie and V is an ad(a)-stable subspace of is the set of weights of For any B C A(V,o3) o, let in V (with p(B) = a. aEB When there no confusion we will use A(V) H is a 8-stable Cartan subgroup, H = TA; with T = H If and is A (g,k) is 8-stable. n K, A = H n for A(V,o). then (exp po) = exp(t 0fp 0 ) 14 W = W(g,h) Let be the Weyl group of W(G,H) = NG(H)/H Let A+ & = h p = + n, (resp. c the corresponding Borel g and n K. be a set of positive roots of H t f subalgebra and Hc) c is C k 0 -0 be a Cartan subalgebra. to be the centralizer in in p = of Tc = Hc n K G. the other extreme, 0 G) K. abelian subalgebra and ft (resp. is called the fundamental or maximally compact Cartan subgroup of On with TcAc a Cartan subgroup of Hc g fc Define 8-stable, so we can write Hc then in p(n). Let t0. A +(g,) = NK(H)/H f h= if 0 p as ts + a 0 0 is a maximal is maximal abelian is also a Cartan subalgebra of centralizer Hs in G go. Its is a Cartan subgroup of G, the maximally split one. 2.2. Harish-Chandra Modules Let (r,l) be a continuous complex Hilbert space representation and *K the subset of W of K-finite g, 15 vectors. If (irl) is admissible, K-isotypic components the are is, if all the finite dimensional, then limit = 7r(x)v exists for all is XK a representation complex). in tx)v-v) , v E AK and defines a *K' module since we can complexify the (g,K) Also (7r(exp and go IK 1 lim t-+O x0 E representation of WK WK of that to a representation of AK is a g representation of the Harish-Chandra module of (r,l) (W K. being We call (cf. Harish- Chandra [1953]. Any irreducible unitary representation of G is admissible. Denote by by d(g,,K) A(g,K) the category of the category of admissible An irreducible (g,K) module (g,K) (g,K) modules and modules. is automatically admissible. A g scalars on module X. X is quasisimple if Z(g) Then we have a homomorphism x : Z(g) -- 4 C acts by 16 x(z)x = V(z)x called the infinitesimal character of Any irreducible If (r,X), are equivalent element of (g,K) module is quasisimple. (r'.X') E A(g,K) L Write G (g,K) modules. we say (7r',*') 2.3 X and X' the defined by = {L : X -- I X is complex linear and i'L = Lr}. set of equivalence classes of If (r,g), (r',*') irreducible are representations of they are infinitesimally equivalent if (rr K)' are equivalent. Parabolic subalgebras Let tc 0 k - ' Fix x E i(tc We say that (9,K)-module maps Hom9,K('r') = Homrn,K(X,X') G we if there is an invertible map which is an the set of for X. define a Let * e-stable parabolic subalgebra q as follows. 17 A(ttC) = {a E A(g,tc) <a,x> = 0} A(u) = A(.,tc) = (a E A(g,tc) <a,x> > 0} A(L) = t = D CX + t, a aEA(L) then q = L + u 2.4. Derived Functor Modules is CX @-stable. In this section we consider cation of Harish-Chandra modules a certain set of parameters Chandra module. a aEA(u) that part of the classifi- that consists of attaching to an irreducible Harish- We are going to exhibit each irreducible module as a submodule of a derived functor module. (gK) set We will first consider a particular modules when (g,K) G is quasisplit. of irreducible To define these groups we need some notation. Let a be a maximal abelian subalgebra and C p0 the corresponding connected subgroup of Let M = K Define in As = centralizer of s,a)= A(g/(i+as As G. in K. the nonzero roots g. Fix a positive system A+ C As As and let of as 18 = 0 0 and N + aEA~ the corresponding connected subgroup of Definition 2.4.1. S ^s A For a fixed representation , define the Hilbert {f : G = A6,V m E M; The action of a G -- V E A; on n V I E N f G. Definition 2.4.2. G abelian. M measurable; f(g); f K E L2 (K)}. and I(6@v) = f(g = Ind 1g 0 ) G (60v), the + a is p representation of of given by (T 6,V(g))f(gO) representation (6,V) space = a -(v+p) 6(m) f(gman) defines a G. Ps = MAsN. Define and RX (D is quasisplit if a induced 19 Hence, if subalgebra of of G go is quasisplit = i0 + a0 0 0 k HS = MAs = TsAs and is a Cartan a Cartan subgroup G. Therefore ^S H =homomorphisms Definition 2.4.3. a) d6 a0 n : H -- + C is trivial on b) Consider A a E A is ^s ^ M xA representation ^ M x 6 E M s (a) is fine if [g. the set A = {a E AS A root x real if ! a f AS a = s for some 0 E As real. Let Ma n K. | a(X) = 0} aa = {X E a and Choose an injection a a 0 : such that a 0 -1 Oa 0G = E a0 00o GA = MaAa; Ka = 20 0 1 Oa .0 E a-root [0 Put c) A a For included in ii) 1 ka a -1 representatioi i) space. 0] E K real a E 0 0 is fine G) (for has eigenvalues p( iZa) {0,+1} For each complex root a E i is 1(k a n K 6 E M If d) |I is is f ine, a Suppose G set A(b) = {i E p Im See Vogan [1981] 4.4.8. p to M is a p E K sum of is fine. Then fine representa- each occurring with multiplicity one. p E A(6) Say that irreducible occurs in is quasisplit and the restriction of M, 6 trivial. 909 ) fine representation and Proposition 2.4.5. tions of if (g.,K) Then there for some 6 module containing is a character fine. the V E As Homg,K(XI(6 @v)) A 0. Let K-type such that X be an ji. 21 For an arbitrary linear reductive Lie group we will define the notion of minimal (or lowest) K-type of a representation and attach to it certain parameters. next section we will then construct a (g,K) In the module with these parameters using a reduction to a quasisplit group. The irreducible representation with a subquotient of this module. that lowest For proofs of K-type is these results see Vogan [1981] Chapters 5 and 6. Fix a Cartan subalgebra t of kc , a positive root system A+(k) = A+(k,tc) A+(k)-dominant and a for its differential. weight hc positive root dominant. as in 2.1. system write k)) E )* (t E (Cc)* Then there exists a A+ (9,c) which makes See for example, Vogan [1981], p. Fix a noncompact imaginary root X i Define 2pc = 2p(A+ Let Cc;T for a root vector for the root x- =-U e-stable p + 2p c 239. E A(g9,hc). f. Put Write 22 Z P= XP+ = X_ centralizer of | (hc)= {x E h @ S= (t ) 0t3 Then 9 (ftc)' is reductive, Lie algebra 9P E is real Cartan subalgebra of g 0g Z in go (x) = O} (2.4.6) Z. $ <Z13 >. the subgroup Gp reductive linear and and of go. of G with h is a See Vogan [1981], p. 235. Proposition 2.4.7 (Vogan [1981], 5.3.3). A+(k,tc)-dominant weight ment Xv(p) ties: G - fix a Then XV(p) ./.. a) ,3r} .it E (tc M there having the following proper- dominant; and write is dominant for A+(g,hc), If we put -2<p c> , p+2p c> A +(g,h ) p = p(A + ( ,.c)) and of imaginary roots, A(g,hc) each is a unique ele- e-stable positive root system p + 2p making i C Tc, For there is a set satisfying 23 c V = p , then o < c. < 1, and XV(p) = XV(p) = p + 2pc a E A (g,hC) If b) <a ,P1> 4 0 then c) simple, The root or there P1 such tha t = Either al 1 e) Write A + (9.hc satisfy " .1 t n+9 a of is A + (g,hc) a + Oa. d) in equations and either it i s a complex simple root = 0, C g~ as i. is noncompact; Pi (aXv(p)> = 0, imaginary, and is for some + 2+v v. -p 1 ft =~ (2.4.6). 1,h 1 ) or Then and these same condi t ions c1 A 0. = hP the positive system its for subset g1 and {P2''13d the weight 24 Definition 2.4. 8. let For a A +(k)-dominant G qV(p) be the parabolic defined by cribed in 2.3. If we also define G v is a V(7r) - (a) is quasisplit L t V( i-2p(unp)) = (b) (c) If r E K G 'v(11)= q as desp qV qV(p) L0 G XV(p) ^c p E K-type of highest weight Proposition 2.4.9 (Vogan [1981] A+(k)-dominant and weight §5.3). = L + a. has highest weight Suppose p E T then is fine is Then p, r . We will now define a preordering on K. Def inition 2.4.10. a) as If 7r E K has highest weight in proposition 2.4.6. the If X is a nonzero lambda= E K | X(r) X 0 N = C( X,>. (gK)-module define r-isotypic component of {r put Define 1111lambda b) p E T and X. X(ir) Then the set vI"l lambda is minimal} to be 25 We define it is nonempty. types. c) LKT We will refer to XV(X) Define of X to be such a XV(p) = of lowest K- as an LKT. 7 p for G qV(X) and let the set a highest weight of a the parabolic associated be to 11. will When there is no confusion we parameters as Let tL qV be normalizer of T Li tion of VL c qV , so in these XV. the Levi factor o qV, f and LV the G. let be 7r the irreducible representa- generated by the fl K L and refer to p-weight space, inside flK. Let Lv V = V0 [A(R (Lfl4* E (LVfK) (2.4.11) R = dim u Notice p-2p(uflp) that L and choose = TA H 6 is a highest weight of is fine for a maximally split E T, r V By and c) proposition 2.4.9 b) Fix n o. Lv. Cartan subgroup of fine occurring in L L 26 This discrete (qV,HV' triple VL) is, by definition a 0-stable data attached to We have attached all representation X X. these parameters to a with LKT 7r. Now with these parameters we will exhibit a module that contains X set of as a subquotient. (9,K) We need more definitions. Definitior 2.4.12. quadruple (q,H,b,v) a) q = L + a b) L is a c) 6 E T d) If is (G,a> i) G ii ) that <X H = TA C L L and is a v E A. L + p(A(a,tc)) > 0 , > = 0 E t 6 and C h for all a E A(a,h) for all 13 E A(L,h). 0-stable data attached to together with any character of 0-stable L. the differential of the discrete ) i s a the conditions: Lv (o,,VHV'b G 0-stable parabolic subalgebra of is fine for XL E t X E A(g,K) 0-stable data for split Cartan subgroup of XG = Notice of with is quasisplit, and maximally then A set 0-stable data for G. v E A some are a set 27 In order to give a generalization of Proposition 2.4.5 for non-quasisplit groups we need to define the objects in which we are going to realize the Harish-Chandra modules of G. Definition 2.4.13 Zuckerman Functors. q = Let L + a C g be a 0-stable parabolic subalgebra as defined in 2.3 and subgroup corresponding to t0' is connected, then so are G Since the reductive Lie L C G K, T, L and L n K. Let Z be any (q,LK) module. Define pro (Z) = pro9,LnK (Z) = Hom (U(g),Z)Kfinite t Q U9)ZLnfl q LfK Od This is a (g,LK) Now, if W module. is any (g,LfK) I FW = {v E W FW is a is a left exact derived functors of (g,LfK) }. F : A(9,LfK) -- A(9,K) functor. The Zuckerman functors a dim U(k)-v < module and (g,K) module define module F then {(jflK (written W r ). admits an 1i That are the right is, if W is injective resolution 28 0 W 1 ---- 10 then we have a cochain complex 0 W Define FW=ker m i Im Definition 2.4.14. Then L n L K 00 FW - _ . Let We will define induction functor (oq)i . 0 q = t + u L C G and i-th as in 2.3. (connected) Lie group and compact subgroup of the 1 F W = FW. So is a reductive linear is a maximal - F"l 1I 0 G. cohomological parabolic : A(L,LfK) -- A(g,K), as follows. Let U act V E A(L,LfK). We make V E A(q,LfK) trivially. Let prog.,LK q, LK VO Adim C U) then ()(V) Q(V) = = i( F i(W). = 9 (V) by letting 29 We are now in a position to state the generalization of Proposition 2.4.5. moment it is convenient at to mention some properties of this these derived functor that we will need later. modules Suppose such However, = q + u are two parabolic 2 - 1 L1 t2 n1 L subalgebras that t 2' t set U= then = 1 + U C C2 K C L2 K; is a parabolic subalgebra of Proposition 2.4.15 (Zuckerman [1977]; With notation as above if that for n is an W L 2. Vogan [1981] 6.3.10). (L ,L fK) module such q X q [2 ]qW =0 then g [ q21 [ g 2 ]qo0 pqq 1pq() ) q Proposition 2.4.16 (Zuckerman [1977]; Vogan [1981] 6.3.11). Let L, q = L + u be a 9-stable parabolic a Cartan subalgebra. Let Y be an subalgebra; (L,LK) h C module 30 with infinitesimal character 1 (W) has infinitesimal Definition 2.4.17. Let be a set (definition 2.4.12) . P = TAN and for all a Then X + p(u) character Standard Representations. (q,H,6,v) such that X E h. of 0-stable data for H = TA C L Let and choose is a minimal parabolic G N C L subgroup of L in the corresponding positive system A(n,a) (Re v,a> < 0. IL(60v) Let Ind L p (60vol). 0-stable data be the principal We define the standard (q,H,6,v) in 2. 4.14 where We A +(k) module with = !q(IL(60v)) s = dim a n k. will now state some properties of these standard modules. (0= h0 (g,,K) by G(qv) as series representation Fix n k0 tC C0 kf a Cartan subalgebra containing and a positive root system = A + (tk,tc) u A(unk). A+ (Lfk,tc). Set 31 Proposition 2.4.18 (Vogan [1981] 6.5.9). be a set of 0-stable data for G, Let (q,H,6,v) d6 + p(u) XG = and then a) b) 1i(IL(6®v)) If 7r highest has infinitesimal is a K-type occurring in weight C 7rL (N G ,v). D 6 with q (IL( 0v)) then there exists an L n K-type rL 77 highest weight 6 chara cter 7L of such that IT and p = pL + 2p(unp) + n aa. aEA (A) n a EIN c) Moreover if term is is 7r zero and 7L a is LKT, then the fine for Suppose L (oV,H,6V) and to (b)). X. s (I X Then there is a character (6 Lv)) parameters is the standard (qV,H,6 ,VV), (g) is an irreducible a set of discrete and the proof of (9,K) module 0-stable data attached V E A (g,K) then summation L. Proposition 2.4.19 (Vogan [1981] 6.5.9 6.5.12 last such that if module with Homn,K(X, V( V))) is one dimensional. Lemma 2.4.20. subalgebra and Let Y q = C + a C g an (L,LK) be a module. 0-stable parabolic Write 32 s n = dim a k, (Y) XL = <N +p(u),a> > 0; X = 9 and a E A(u). Choose (Y). A +(k) Assume = A+(Lfk) U A(unk). L Suppose L n K) of Y A+(tnk,tc) with respect to and is dominant. is the highest weight of a LKT b) p + 2pc compatible with A +(k) 2pc = U + for A+(g), where a E A+ = If <a,-r> 0; pLfk A+(g) + = A+(L) U A(U). 2p(ufnp) + 2p(unk) + 2 pLfk is simple, then ) <4+2p c,a> = <p +2ptnk'a> + <2p(u),a> ii) 2 A+(k). and a E A +(g) If jL + 2p nk + 2p(u). = i) so that is dominant for is dominant A+(g) Suppose A+(L) Then p = p L + 2p(fnp) ji + the positive system that we choose a) Proof. (for <pL+2p t, a> a E A(u) hence <a,P> then, 0 + 0 for any simple root for 3 E A+(L). r, is 33 {fp,} C A(Lfp) Choose as in Proposition 2.4.7, such that xV xL L + 22p~fk tn1 -p c 43 1 + Osc (1 then p + 2p (2.4.21) = 1 + c 3 + 2p(a) then V <a , p + 2p c> = V V .x L + p(u.)> + (a,p (a, - p > 0 + 1. This proves b) of For simple, (a), it is enough lemma. to prove that then ( i) the If V ,4 -Y E A+(tfk) + 2pc > 2. if r E A+(k) is 34 < (,p + 2pc > = <( , L V <= + 2p(Lfk)> L ,1 Vr,2 + u) + 2p(Lfk)> + 0 > 2, since L is dominant for 1, -Y E A(ulk) If ii) A+(Lflk). < then, as for b): and it is an integer s ince W + 2pc ,p + 2pc exponentiates. q.e.d. Lemma 2.4.22. A+(g), dominant for Proof. In the By 2.4.21, setting of Lemma 2.4.20, then Xv(pG) = x P(U) - XV(p) the condi tions for in Proposition 2.4.7: Condition (a) Since A(u), satisfies XL + p(U) is c io . C We claim that p + p(U). p =XL + pi + 2pc - if A(u) holds by the definition of = -A(u) L + p(a). (V XL+ <aX + p(u)> > 0 and 2.4.7 (b) holds because V L (a,X + P(u)) = 0 for a E implies a E A(L). But then, <a,p(u)> = 0; Since simple roots A(g.,hc), for 2.4.7 (c) (d) (e) hence A(,h c) V <aP3.> A 0 are simple for some for hold. q.e.d. 35 K-type occurring in X with highest weight T1 lambda-norm of 77. We want to estima te the L qV= Let L 06V(Y)= = Hv L VV) , and Y to V If b) for in Definition 2.4. 12. V L is a set ,VV) is a K-type 8-stable data for of in X, then G. (<X(7 7 ),xV(n)> G ( X>. in b) c) If equality holds 77 a highest weight of a a LKT of then 77 = TL + 2p(ufnp) L(LfK)T of Y and V V is is X. d) Conversely if LKT of X Proof. For all L With no tation as in Lemma 2.4.20 L a) 8-stable data for as V+ Lemma 2. 4.23. a C Q + (Definition 2.4.12). xG = d 6 L + p(V) Let G + U1 a set of L H V, 6 attached L1 be a L. 1 C a1 + QV (7r,Z) let In the setting of Lemma 2.4.20, roots -q = 77L + 2p(unp), then a and equality holds. in (a) we only need to see that A(u). By hypothesis on L L L (qryH'V, 6vy), xL = dL + p( ) V <XG ,a> > 0 for 36 L V , a> > 0 ( H = TA for is maximally split a E A(ul), Cartan of 6 and L is fine. Now G X = If dL v + P(aV) = d6 L + += a E A(ul) 9 A(C) + p(U) L V a E A(u), L + p(U). then (XV + p(U),a> = If - L V Xv, a> > 0. by hypothesis in Lemma 2.4.20, G V (X ,a> > 0. The proof for b) and c) is exactly the proof of Lemma 6.5.6 in Vogan [1981]. q.e.d. 2.5. The Modules Let group, L q = be a connected real C + u C g the normalizer of Let of G A (N). L. X : Assume -- 4 that a q C reductive linear Lie 0-stable parabolic subalgebra and in G. Then L0 = Lie(L). be a one-dimensional representation 37 a) X (2.5.1) L We say b) is the differential character of L (XIc a> > 0 for all that X Definition 2.5.2. (call of a unitary it X also). a E A(utc is an admissible representation of With notation as above, we define Harish-Chandra module A (X) with the by = O () s = dim u n A (N) L. (Definition 2.4.14) k. Fix positive root systems A+(Lfk) A +() Then and are positive = A+(L,L), A +(k) A +() t-root = and compatible with A+(Lfk). = A+(Lfk) U A(unk) A+(q) = A+(L) U A(U) systems for k g, and Choose a fundamental Cartan subalgebra hc respectively. c + ac and a 38 positive root system A+(g,hc) A+(gh i c so = A +(g-).- 1 P = Then that aEA+ (g,hc) Proposition 2.5.3 I3EA'(g9) (Vogan-Zuckerman [1984]. Speh-Vogan [1980] and Vogan [1981]). weight (tc)*. in The See also Regard c X as a Let p = N ic a) P1. (g,K) + 2p(unp) E module A (N) (tc )* is the unique irreducible module satisfying: i) As a K-representation, K-type with highest weight ii) +p: is Z(g) Z(g) where contains X A (N) (z) b y the character = ( X+p)(f(z)) and the Harish-Chandra homomorphism. iii) Any K-type occurring in A (X) highest weight of the form T7 = X c + the p. acts on -- + C; A (N) 2p(up) + n 0 P. PEA(up) n EIN has a f 39 b) Proof. is the unique LKT of The infinitesimal -- S:L i Moreover C character of X + pt. is has infinitesimal A (M). the representation Then by Proposition 2.4.16 X + P character + p(c) A (X) So ii) p. holds. If (lowest) L = Nl L Choose n K-type L is the highest weight of the C . of 4. A (L) L xV then tc = making L (4L V(. ) pL + 2 = i c + 2ptk - Then dominant. p tLk Pt + c #i |c with Q But a sum of roots iL in = 0 so a E A(u) is IA(L) L. A(L) can be chosen so that Q is dominant. Then if simple, (L V + p(),a) > 0. In fact (L + p(u),a> = ( <N c + p(a),a> > <p(u),a> > 0. By Lemmas 2.4.20, 2.4.22, and 2.4.23, i) and b) hold. 40 The irreducibility and uniqueness of A work, and since we won't be using these facts we Speh-Vogan [1980]. take more (A) refer See also Vogan [1981]. 1.3 in Vogan [1984], w e have By 2.5.4 and Theorem following. In the above Proposition 2.5.5. A (A) 1,2, A(q.) Fix be A (k). Q A+(k) A. E t and + U . C g; Li admissible one-dimensional (Definition 2.5.1). A1(A ) A = We need a Lemma 2.5.7. = L. 1 1 Let 6-stable parabolic subalgebras such that representations of Proof. setting, the modules are unitarizable. Proposition 2.5.6. i = to X2 and A(2 U1 2 Then, ) nP = u2 n A. few lemmas: Suppose Q = L + U, q = t + U C g, 6-stable parabolic subalgebras, and admissible representations such that A : - C are the 41 (2.5.8) Then Proof. 1) q 2 q, 2) X . 3) a nafl . = ~to A (y). By induction by S but a D a. and A(L). n A() L t D that is, stages (Proposition 2.4.15), Lf(unk) (y) (dim ) q f L = L + a n I and, u n L c k by 3), so dim alt . (Cx) qnfl Hence Ms(ex) s(C) = A'~'(X) this proves the lemma. q q.e.d. By this lemma, we may assume proposition are maximal Lemma 2.5.9. that both with respect i in 's to conditions 1) In the above setting A(ct ink) = (a E A(g) I <a,AX + 2p(u op) > = 01}. the - 3). 42 Proof. Suppose a E A+ (k,tc) a) a f A(L.fk). b) <a,pi.> = 0, p 1 is a simple root so that 1 n A(z) = Span(A(L.),a) Let A(u) .fln). 1 = X. + 2p(u = A(g.) A(u.)\A(T) 11 We want to contr adict Breaking up A(u nilb) 1* the maximality of in maximal { 0; 0+a; (i.e. -y0 a qi strings +ra}, . .. - a, -0 + (r+i)a f A(ulp)) and using representation theory of oL(2) we can conclude that <a,2p(u . Ap> > 0 and we have equality if and onl y if under the the a three dimensional sub ialgebra a-root vector X is 1 a g. a But, by definition of X., 1 <a, A.>1 > 0. invariant that contains 43 So, a imply that (a) and (b) a. f P is invariant under and <a,X.> = 0 = <a,2p(u .lop)>. 1 w that Now we want to prove (2.5.10) 1 E A+(g,hc) If P n = A. n p. 1 P3 and = a then tc s If is not )3 in Hence A(u. flp) so it for A(k,t c), a i A(Lfk) and a = Y + 6 7 and again. -r - In fact, A+(g,hc). and a = root of -it a contradicts invariance under is an imaginary root of a simple for say, c' is complex, then the non-compact also E A+(g.,,c) [ A+(g,he). since a we can assume is that a is simple if -, then E A(u inP), -e 6 A(u .lo); 1 contradicting invariance 5 44 Consider a simple factor L. L0 Then is not orthogonal be a set of simple roots a = Pi Say n.P = for n P A 0. with some So 6 the string Hence satisfying t0 a. not contained in Let ni 0+1 containing 6 a. is adjacent to > 0 a. and such that n.<a,p .> < 0. is a non-compact root, and through {p1,02'...'13 Then there is a non-compact root <a,P> = a + P = to L0 1+1 P and t0 Suppose 1 t0 g T, 6 E A(u i po). is not complete. is compact and (C q) q is not maximal (2.5.8). This proves Lemma 2.5.9. q.e.d. We are now able to prove Proposition 2.5.6. By Lemma 2.5.9, ti 1 hence X\ + 2p(a 1) <2p(u.),P> = 0 (X. ,a> > 0, 1 - X2 for all a E A(u 1.). n k - 62 n k n k = 2 n k 2 ). + 2p(u P E A(L.) So But and (X.P> = <2p(u.),a> > 0, 45 A(Li) = {1 A(ui) = cX. + 2p(u.),P> = O} < E A(g,,tc) {a E A(g,tc) (X n fl Hence and X + 2p(u.),a> > 0}. = 2 - X2' This proves Proposition 2.5.6. q.e.d. 2.6. Reduction step for the proof of Theorem 1.3. We are now in a position to prove the main result stated in Chapter 1. We will argue by contradiction and reduction to a proper subgroup Suppose X E d(g.,K) Hermitian form < , realized as an A (N) >. L C G. is irreducible and has a We will assume module, but will X cannot be exhibit X as a Langlands submodule of some derived functor module induced from an (L,LK) module XL, making sure information can be carried over We need to keep track of to G and that this X. the existence of Hermitian forms at different steps of signatures on some finite sets of induction as well as of K-types. their 46 Recall from Voga n [1984] Hermitian dual of a Yh = module (9,K) : Y -- ), C If (Definition 2.10) I is a (g,K) on a (g.,K) An invariant, module Y < such is a pairing , > Y x Y- <x,ay+bz> = a(x,y> + b<x,z) <ax+bw, y> a, b) b E C, X E C x E Y}. symmetric Hermitian form that a) for Ef(x), < 00 ; module. 2.6.1. Def inition Y dim U(k)-f f(Nx) = the x, y, z, = a(x,y> + b<w,y> w E Y. <(U+iV)x,y> = -<x,(U-iV)y> U, V E go c) <k-x,y> = <x,k d) <x,y> = <y,x>. -y> y, x E Y. k E K. 47 < The radical of > Rad(, It on Y is clear are given by f = fh . Yh -- + Y. Proposition 2.6.2. Then X {x E Y = that is (x,Y> = 0}. < invariant symmetric Hermitian forms (gK) maps : f Y Yh - such that Moreover we have Suppose X E sd(g,K) is irreducible. admits a non-zero invariant Hermitian form if and only if h In this any case the Hermitian form is non-degenerate and two such forms differ by multiplication by a real constant. Proposition 2.6.3. (4V,HV,5,vV) a Let set of X E d(g,K) be irreducible and @-stable data attached to X, so Xh if that dim Hom ,K(X, (see Proposition 2.4.19). and only if there (I V( Let is an element 6 @V)) Hv = TA. = 1 Then X 48 w E W(L,A) = In such that WD = -v. and 65 this case we get a Hermitian form on X from a form on L g (I V(6V(v )). vV). QV This result is essentially due to Knapp and Zuckerman [1976]. A formulation close to this one is in Vogan [1984], Corollary 2.15. Corollary 2.6.4. Let X E 4(g,K), with a non-zero Hermitian form Let q = t + u that L 3 be a LV, data attached and X. X , >. Write 0-stable parabolic U C U to < irreducible, endowed qV =V(X). subalgebra such (qV,HV,6V,vV), a 0-stable Write L = 9 t iq(IL - (Vvv)). Then xh L Proof. This has a Hermitian form is a formal < , > consequence of Proposition 2.6.3. q.e.d. 49 Proposition 2.6.5. parabolic subalgebra. with a q = Fix L + u C g, Suppose (possibly degenerate) Y E a A(L,LnK) 8-stable is equipped invariant Hermitian form L Then G < > is a natural invariant Hermitian form s h h [As(Y)] on Proof. 6.1.5 there Recall from Vogan [1981] Chapter 6, Definition the functors : A(L,LfK) ind -- +* A(g,LfK). Q ind9Y = U(g) 0_ Y. 1% Q Write V : eiY = VY = p where yj functors ind9(Y@A t op a) A (g ,LfK) -- + A (g,K) (cf. A(,K) A(L,LfK) -- are the Zuckerman's Definition 2.4.14). By hypothesis, Set we have a map L 0:Y-*y h Y = Y @ Atop . 50 This induces a map ind(Y) - pro h and s G h By Theorem 5.3 (Enright-Wallach) in Vogan [1984] L2s-e h h Let < , > : VY + (VsY) h be the natural pairing given by Definition 2.10 in Vogan [1984]. Define (u,v>G = <u,O v>. This gives an invariant Hermitian form on (cfr. fs(M the proof of Corollary 5.5, Vogan [1984]). q.e.d. 51 Definition 2.6.6. If Z E A(g,K) and 6 E K, write Z(6) = Hom K(V 6 ,Z). Then (2.6.7) 0 Z Z(6) 0 V . 6EK If we fix a positive definite form on inherits a Hermitian form. non-zero Hermitian form z(6)), Z(6) for where Write < Suppose , >. Write the multiplicity of ( > , sgn(< , >IZ(6)) V6 is positive the signature of < Z V6 , Z(6) is equipped with a p(b) (resp. q(6), in the subspace of (resp. negative or zero). , > on Z(6) as = (p(b),q(6),z(6)). Then write, formally sgn(< , >) = (p(6),q(6),z(6)). 6EK We will prove in the next chapters the following result. Theorem 2.6.7. and X E A(g,K) Let G = SL(n,R), SU(p,q) or SP(n,R) irreducible, endowed with a non-zero 52 < invariant Hermitian form > , and regular integral infinitesimal character. If are a X = A ,(X'), for any L a) X and 9 X is '. an Then there (L,LfK)-module such that i = 1,2 6 (LfK)-types and q = t + u, 0-stable parabolic and XL q' the unique irreducible submodule of A (XL)' occurs only once as a composition factor of (XL). h b) is endowed with a Hermitian form XL Write (pL'qLzL) L >L 0. Then signature. x0 q and 2 0. A+(k) = A+(Lfk) U A(unk). Choose , L L L) c) for its < L has highest weight = p L1, P L+ + 2p(n) 6. Then, if A (k) is dominant. Chapters 3, result. Assume this for Using X. on this the moment. We need to check that < , > the proof of this result, we want to prove non-unitarity of induced on X; that for the L nK the corresponding the signature of same as that of the Hermitian form < < , >G by Proposition 2.6.5 is a multiple of q(XL)h Theorem 2.6.7, that 4, 5 will be devoted to , the form on L > on the types satisfying c) of K types occur in X K-types is the these L 6. 1* and 53 Theorem 2.6.8. Suppose X E d(9.K) < . a non-zero Hermitian form stable and unique XL an (L,LK) (LfK)-type of LL L + 2p(ulp) XL in >L X(6) A 0 >IX(6)] , t if 6 A(unk) X and Q L L with highest weight p, Sign ( q(XL), Ots(X) is dominant for highest weight = u + module such that < non-zero Hermitian form q Let >. irreducible submodule of once as composition factor is irreducible and has X is 0the occurs only X L E (LfK) AL be has a is an such that then if 6 E K p has and = Sgn< , >L L Proof. Applying the appropriate definitions and results K q and n k we have maps 91A : A(Lk,LfK) vi : A(Lk,LK) Qflk A(k,K) A A(k,K). - qfk If Y E A(L,LfK) there are natural maps - pro Y indk qflk Y pro qnk ind9Y. Q to 54 These induce (k,K)-maps i r i qYq 4 kY Se y .jY Then, the following diagram is commutative h 2s-i [i Y]h r It h i Yh h 912s-i _ ____ qnk The isomorphisms across are Theorem 5.3 in Vogan [1984] for and (G,q,) Arguing as (K,qk), respectively. in the Proof of 2.6.5 (for K) maps qnk * . k Y pro k h Y qflk :~kink Yn_ qflk tk .2s -nu Y - gs qnk Yh and we have the following commutative diagram we have 55 G s es Y h sY) h Ih (2.6.10) Ir K ps K s s Yh (Qnk~h nk And we have a Hermitian form on Is h ;fnk (Y) flk <xy>K 0K = ro Got, Since [1984], L _ K and by Proposition 6.10 in Vogan is a unitary map, <x,y>K = (Lxy>G (2.6.11) (2.6.12) Write sign(< , > ) = (PK'qK"zK) sign(< , >G =(G'q G"zG) PK' q, -[ ZK -4IN 'G G', qG, N and again sign(< , > ) = (PL q LzL); By 2.6.11, zL (LfK) > IN 56 ZG(6 ) > zK( 6 ) The main ingredient is in the proof of the following result due to T. Proposition 2.6.8 Enright. Proposition 2.6.13 (Enright [1984]). Let q = L + a 0- stable parabolic. Let pL + 6L C (LnK) L with highest weight Set p = 2p(unp). a) If 4 is not A(u 1k)-dominant, 2 b) If w is representation of 0.flk A(unk) K with Y(6) = 0. dominant, write highest weight L qK then = L( 6 zK06) = ZL( 6 ) L 6 E K p. Then for the 57 For a proof of Lemma 2.6.14. S is result see Vogan this Suppose V is a module of [1984] 6.5-6.8. finite length and irreducible. Assume a) of S C V factor V. b) Any non-zero c) S , > W C V contains to scalars, Vh has a unique Hermitian form and S = Vh/rad(< The proof of this Theorem 2.6.8. (2.6.15) and S. is equipped with a Hermitian form. Then, up < occurs exactly once as a composition , >1). lemma is standard. We can now prove By Proposition 2.6.13 and 2.6.11 pG( L( 6 ) qG(6) L(L ) zG(6) zL56 L Apply Lemma 2.6.14 to V = 9 (XL) and S = X. 58 We know that a) part of < , - c) hold in this Lemma since they are our assumptions on >G X 0 X. We also know that by 2.6.15. Hence, we have the following result: Proposition 2.6.16. In < X So ( , >GIX the setting of Theorem 2.6.8 , >GX = c< ]h /rad(< =- [p(X > , , >G is nondegenerate and has signature sgn(< , >) = (pG'qG ) q.e.d. It is now straightforward to prove Theorem 1.3. Theorem 2.6.7, proved in chapters 3-5 for our groups question, we have that the hypotheses true and by 2.6.15 p(G 6 and qG(6 2 ) > 0 > 0 Using in in Theorem 2.6.8 are 59 and < the form , > on X is indefinite too. q.e.d. 2.7. Methods to detect non-unitarity. To prove Theorem 2.6.7, we will need a few that we will discuss here. techniques Fix a positive root system A+(k). Lemma 2.7.1 (Parthasarathy's Dirac operator See Borel-Wallach [1980] 11.6.1.1.) unitary representation of G and Let 9K inequality. (r,A) be a its Harish-Chandra module. Fix a positive A+(k) and a weight p E t-root system k-type t'c. 6 occurring in Pn = p(A+(A)) g acting dominant on for with highest E (C )* PC= p(A(k)) E C WK compatible with Write p = p(A+ (9)) Let A+(g) be IK' (tc)* = P - PC the eigenvalue of and A +(k). w E W(k,t) Then (c)*. the Casimir operator of making w(P-pn 60 <W(A-pn) Lemma 2.7.2. K-type < 1) 6, There (00Ln) + Pc> Suppose the Dirac >. some choice of for is a inequality fails A+ (p). Then V6 occurring in ri k-type <P,P>. > C0 + with a non-zero, invariant X E A(g,K) Let Hermitian form on a + Pc' ® p such that S,>V6 is indefinite. 2) Suppose is Hermitian symmetric with a one- G/K compact center, dimensional with the property that decomposition of g z, so that we can choose g = k @ P+ into the eigenspaces ik 0 the is 0, +1, -1 of respectively. Set p on 6 fails V6 ® P n = p(A(p±)). for p-, Then, if there is a the Dirac k-type such that >I(V is - z E indefinite. 6 V F) rq inequality occurring in 61 Proof. Recall definition of from Borel-Wallach [1980], V defined over < positive definite inner product the unitary structure on also, (Y,H) = v E V x, the , >. R, finite with a Write ( , > for such that - <x,((V)y> y E S(V). the definition of the Dirac operator D(ves) H 0 S H 0 S : (go 0 ko)-module and a unitary (2.7.3) S(V) (T(v)x,y> D for §6, the space of spinors of a (7,S(V)), dimensional vector space Recall II = S = S(PO). r(Xa)v ® i(X-a)s. aEA(k) Since m-V A+ (where then m = 2 dim ac/2]) is o(p-pn) occurring in V +(k) (cfr. Borel-Wallach [1984] II §6) the highest weight of a @ V n C H ® S. k-representation 62 f Let = v ® s Write also product on be a weight vector for , < H @ S; D for the G(p-p tensor product ). inner then the proof of Lemma 2.7.1 shows that 0 > <DE,Df>DD (<W(p-pn) So Df Df 0 + Pc 'W(WPn)+Pc> - c 0 PD' - and = (Xa)v ® r(X-a)s E p.V 6 S C H ® S. aEA(p) This gives a non-zero map p ® V. So Hom k(l V6, H) X 0. pov 6. a Let positive definite this means E = Im a. that ( Since > is < >5 is indefinite on V 6 @ E. This proves a) For b) p(p+) and of the simply observe + is a lemma. that p - p~n = p + pn; representation of A(k) <p ,P> = 0. k. pn Hence if 1 E 63 So V + is one-dimensional. Since p + a is not a pn weight of S, for a E A(p+)' V+ is killed by r(X ) pn and (2.7.3) becomes, for [ E V6 0 V _ Pn Df = r(Xa)v ® (X-a )s aEA(p ) so (p Df E ).V @ S C H @ S. Similarly for p~. q.e.d. Lemma 2.7.4. group. Let G be a connected, reductive linear Lie Assume that rank G = rank K. Then, any representation with real has a Hermitian Proof. G of a principal character form. By Proposition 2.6.3 it lemma for infinitesimal is enough quasisplit and a Langlands series I(6@v) Hs =T the subrepresentation 6 ® v with a maximally split Cartan subgroup to prove ss A a character of 64 Since G is equal {ay, ... ,ak} of that , Hs G, since then B Hence strongly orthogonal ref 1ections simple real roots such is the maximally split Cartan subgroup of spans s w , = Lie(A s a w = sa if B = rank there is a subset ... sa is acts by the product of simple -1 on As and by the LL. 1 T . identity on Recall nA(2,R) -- from Definition 2.4.3 the maps + a ino. Consider Sa : 0a the exponentiated map SL(2,R) > Ma set 1 m a = 0 0 -1 (cfr. Vogan [1981] page 172). connected, Ts w E M'/M = W, But m Ra =m then there is Recall -a Then, is generated by O-m =m. a = M. E a a TM U 0 E M'I aa U -1 since G is (ma a real}. such that w-a a the elements 6 E M and v E A. Then Let 65 (w6)(ma) = 6(w-ma) = 6(ma W-6 and Hence w5 = 6. Since I(60v) infinitesimal character, Also since This is IT 0 o A = -1 V = 6. is assumed to have real is real. then -Ov = -V = -V. the condition of Proposition 2.6.3 for the existence of a Hermitian form. q.e.d. 66 G = SL(n,IR) Chapter 3. Preliminary Notation. 3.1. G = SL(2n,R); To fix notation consider is the odd case similar. G = {g E GL(2n,R) The maximal compact I subgroup det g = K = SO(2n,R) = {g E G The ko = 0 is I G of is gtg corresponding Lie algebras are go = If K 1}. {X E gL(2n,R) I trace X = 0} X + tX = 0} = oa(2n,IR). X E the Cartan involution defined by then p0 = {X E 90 The Cartan subgroup of K I X _ tX}. is O(X) = -tX, 67 rr(e1) Tc 10 E IR; r(n) sin cosO. -sin ." cos 0. 0. 1 l and if ri ry A = r. E [R det g = 1 1 < r n rn then Hc =TcAC is a maximally compact Cartan of G. 0 -06 0 o 02 C 0 0. E Rl> 2 0 -0 n en 0 68 a1 a1 a2 C a0 a2 2 a. = 0 1 x = a. a E llR n a n h 0 = t0 + aC a0 Complexifying: g = ot(2n,C) and k = oo(2n,C). Define e. E i(t j 0 -0 j 0 )* = 1,2,. .. ,n, by 0 0 0 e. 0 2g. = 0 0 -0 Then the roots of tc in k, A n and n 0 g are, ly: A (k, t') = f±( i ±ek) 1 jo.. 1 < j < k < n} respective- 69 A(t,tc) = {±2 ee; (e A(,,tc) = {±2 ee; can be + e A +(k,tc) = (a 1 a 2 . .. . an) Computation of Ii = Let of a LKT of (a,a X. n as a root 1 < j in j< k ( 1 1 < k < n}. is 2. Then n}. K a1 a2 >...> a n-1 > IaI|}. LV(X) for a Harish-Chandra module X. 2 ... an) E i be Fix a positive root a1 as ek 1 < j < k < n} e < n; I1 ± ek {e < n; identified with the set {p = 3.2. 1 < e | ±(e ±ek) The multiplicity of Choose ±ek) a2 2 the highest weight system A +(k) so that |a ni ... in 3.1. To obtain LY(X) = LV(p) 2pc = Let p + 2pc phism of tion of A +(qc) dominant. K in 2.4.8 we need: (2n-2,2n-4,...,2,0). be a 8-stable positive system making After conjugating by an outer automor- we may assume that A+ (,hc) as to tc is an 0. Then the restric- 70 A+ (g t ) Write t. {e.; k' 2e±| e j < k}. 1 < j,k,O < n; for the set of simple roots restricted O(g.,tc) to Then *(gtc) = {e1 - e 2 ; e2 - e 3 ;...; en-1 - en; 2e I. Let p + 2pc = (x 1 x 2 . n.). We can form an array with the coordinates of p + 2pc by grouping them into maximal blocks of elements decreasing by 2. is, That if (3.2.1) (a ,.. ,a.1 ,a 2 ,...,a 2 r 1 times where a1 times ... ,a rt times 0 ... R 0) times > a2 > ...> at > 0. Then since the array would (3.2.2) r2 a in1 m 1 -2 the coordinates of 2pc decrease by two, look like ... m 1 -2p1 +1 m2 mi2 -2 ... m 2 -2 p2 + 2R-2,...,2,01. . 71 Proposition 3.2.3. K of a representation of weight p + 2pc a) Suppose that = x has E and it 0 is the highest that the picture for Then is as in 3.2.2. Xv(p) p the form ' 2 r1 2 .. x t ...xt ,0... 0) ' rt r2 R where X. = a. . .1 i) ii) So b) t = if > x 2 >..>xt > 0. x 0 if [yp()l d oA (r xt\ > 0 at= xt Proof. 1. is isomorphic ,1 ) (D oA (r 2,C) to, ( (D...@ either A (r tC) (D oA (2R.R), or A0.... (r 1C) if 1 a t(rt , ) ( ot(2(R+rt ' ' S0. Notice that 2pc - p = (-1.-,...,-1). Define 72 { }I = {2e . I <p {1n-}+=1{2en-j+l XV(p) = p + 2p Then (3.2.4), orthogonal {e 1 -e 2 Moreover the to Xv(p) ; 2 e3 'e' - 2 c P. 2 i J - p + - and the conditions a) are obvious. > = -c. + 2p -p,2e has < 0}. the form e) of Proposition 2.4.7 subset of simple roots is r -1~r 1 } U {er er1+1'..er +r2-1 r +r2 U...U {...;e n-1 en;e2d. This spans (Ar r 1 since A the root system ) r1 D (A r r2 the roots - e A e 2- ) (.. A (Ar -l @A r ) CD A 2 k' rs 1 s1 2k involved are complex roots and therefore occur restrictions of twice in A(g,Lc). Now the proposition is clear. q.e.d. 3.3. Lowest K-types of the modules We will give some criteria representation of modules A (X). K is A (M). to determine when a the LKT of one of the (g,K) 73 q = t + a parabolic subalgebra 6-stable to construct a Recall from 2.3 that x E it. we need a weight Suppose x = (x 1 . . . (3.3.1) x x . r 1 times where x1 q = q(x) = Write x as x, ... x times r2 t L(x) + a( x) for 0) R times times rt > x2 0,. x t...,i >0 the parabolic defined by in 2.3. Clearly ( 2p(ank) = (s 1s ,0.. sy .. r2 r s. = 2(n-r 1-.. with J R rt - .-r j.) - .0) r. - 1 (3.3.2) 2p(ulp) and = (u .u ... r I u r = 2(n- r u ... u ,0 ...0) R r 2z -..- t r+)1 r + 74 Proposition 3.3.3. ji Let be as (3.2.1) and suppose it in is the highest weight of a representation of is the LKT of a (g,K)-module a) a - b) at > 2R + rt ai+ 1 r (X) A V Then K. if and only if + r i+1 and X + 2p(unp) orthogonal t C-roots in the tc the roots of to u. in , X = (X, . . . . a at a. - a. t SJ+l1 a + = X 2(n-r1 x -.. is is positive t rt .- rt-) r r .' R > 0. t .. 2 + 2(n-r -. .- j X .'.'I' ,0 ,...,0) 2' ' '2' x1 Then = That is, r2 and and it t in Hence and b). satisfying 2.5.1 a) L character of p Then (X). is the weight of a one-dimensional X and A is the LKT of an V Suppose Proof. + 1. -rt r. + 1 ) - + 1 > 2R + rt l)- + 1 = x j - x j+1 - r j + 1 + 2r ,j + r. j - 1 > - r. + r j+ . 3 J+ 75 4 Conversely, suppose b) is a weight and satisfying a) then we can define and q = q(p) A. = a. J 31 p Then 2(n-r -r. will be the LKT of ) + r. - 1. A (A). q.e.d. 3.4. G = SL(nR). Proof of Theorem 1.6.7 for Suppose X E d4(g;K) infinitesimal character is as Y E of a LKT of highest weight in Theorem 2.6.7 with (/c)* X. Write IL Considering what the weights in 1. 2R + r 2. at + 1 ® t the in 3.3.1. look like, we > a.t > 2R + rt + 1. By the conditions given in 3.3, A then (A), Therefore, in case Case V as (itc)* study 2 cases: will an p E and 1 X p if is the LKT of is in case 2. the first thing we must do is not unitary: 1. We will use the V following result. is verify that 76 Lemma 3.4.1. + r 1. Le t inequality fails p= be as that Suppose Proof. 1 a. 1 for s= The hypotheses on a i+ - Then Dirac operator 1. (n,n-l,...,.1). p imply that (x+t-l,..x+t-l, x+t-2,...,x+t-2. that 1 < x < 2R + R + 1 (4) ..x...,x,0.. .0) r2 rN Note = 11 - x rt implies R < 2R + at in 3.3.1 and suppose and R that R + r t- - x > -R. Now, Pn = (n,n-l,...,R+rtR+rt-1,...,R+1,R,R-1....,.1), so pn (n-x-t+ln-x-t....,R+rt-x R+rt x-1,...,R+1-x,R,R-1,...,1). By (*), the sequence of R + rt - x, integers R + rt - x - 1,....R + 1 - x 77 overlaps the sequence R,R-1,R-2,...,-R+1,-R. Clearly, the first n-R coordinates of pn - decrease by steps of at most one. So if W E WK is such that then the coordinates of w(P-p o(p-pn) in the is dominant, will be a sequence of integers decreasing by at most one, and ) ending in latter case, there must be 0 or repetitions +1; in the sequence. Since P it = (n-l,n-2,...,R+1,R,R-1,...,2,1,0) follows that W(4-P n.P ' c WA-P )(-n Hence <w((-pn) + pc < n' Pc> n><< n)+Pc n' Pn < '. Pn+p C'n+pC> = Pp>. q.e.d. Now to prove nonunitarity the minimal 1 for all integer i > i0 ' in for case 1, {l,2,...,t} take such that i0 a. 1 to be - a i+ + = 78 K = r 0+1 + r 0+2 +...+ Let and S= ot(r 1+r 2 +...+ri L C L Then rt + R ) @ ot(2K,R) = L@ and by Proposition 2.4.15, Langlands quotient of 1 E (I6 OvVV)) QVL (6 if L2* X is the and if we set XL = Snf[ILv(6V vY)), then X is the Langlands quotient of L gg(X and a) V)) of Theorem 2.6.7 holds. Also, by Corollary 2.6.4, h XL has a Hermitian form < L Write pL = p - 2p(unp). Then jL is a LKT of By 3.3.2, 2p(unp) = (2n-(r 1 +.. .+ri )+l, ... ,2n-(r +.. r +r2+...+r So SL(2K,R) .+r )+1,0,...,0). K 0 = ISL(2K,R) 2 XL 79 and by Lemma 3.4.1, the Dirac inequality Lemma 2.7.2, there is a that makes the Hermitian form The roots A(L 2 lp) in V 2 K-type < V 2 in >L , fails on By . 2 indefinite. are {(0.... 1,0... 0 +1 0.. .0),(0... 0,+2,0...0)}. K It is clear that A(L 2 lp) if is also dominant for j there = a + P is dominant ai +1 - A(nlk). for f3 E some p + K-type the 2 Hence note that if 1...,.t, Theorem 2.6.7 follows a, j+1 > r. +j+1 an for - the LKT of an A (X) and to prove. So, assume that a. 0 - ai + 1 a. then we hav e is nothing Set 2) = 1. For case 2, all C2 r then, since for case K there is < r + r0 +1' K = r + r2 +...+ rt 0 t such that and t = at(K,C) @ oL(2R,R). Note that o L(2R,R), at t ; > 2. LV Hence and XL= = OL(r 1 ,C) and we can find Theorem 2.6.7 holds. UL L t In fact, IL ( 6 V®VV), as if XL G... (D s.t. oO = t n in Definition ot(rtC) (a) V = in v + t n 2.4.17, we can 80 choose to be XL XL since = L then, by Proposition 2.4.15 (XL where A(a) = A(v)\A(t) and L + U. q = By Corollary 2.6.4, form ( h L admits a non-zero Hermitian >L Write XL2 X where XL as the exterior L is an XL. tensor product (Li,L fnK) 0 XL L = XL module, 1 L1 = oc~(K,O) By Theorem 6.1 p - inequality fails on 2p(unp) L2 = SL(2R,R). in Enright [1979], and especially its proof (pp. 518-523), if Dirac and XL the is not an lowest and 1 = M= 4 L 1 L1 A ,(X') K-type. then Write L = 81 By Lemma 2.7.2 there V (L 1 fK)-type is an with 1 r1 = Ti 1 + P for (3.4.1) Then p + P If P E A(L 1 lPL). a. - for all a i+1 + i X ri+1 0 2. is dominant. Otherwise take K' r 2 = with iEB B = {i = 1,...,t Then apply Enright's result to | 3.4.1 holds}. the rest. q.e.d. 82 G = SU(p,q) Chapter 4. Preliminary Notation 4.1. n = p + q. Let m for the conjugate transpose of q,p I0 K = {g E G X E od.t(n,C) = A 0] B; 0 BJ [A I X C I . is G A E U(p), B E U(q)}. 0 p 0 E = 0 + 0 B = q notation: the usual with g0 |I g = of K Then the maximal compact subgroup 0 =[I 0 G = {g E SL(n, C) Also, the We define A. matrix A and GL(a,C), for the identity matrix I Write ot(nC) I -I A E u(p). q D E L(q) D skewhermitian; -B *= C A k0 = E go X 0 I A E u(p), D E u(q)}. in 83 If 8 is the Cartan involution defined by 8(X) = -X and {X E 0 go 0(X) = -X then L I B X 0= B .B arbitrary G 1o Tc = matrix . 0. The compact Cartan Subgroup of H c= p x q is n e ,e n) g = diag(e 0. = 0 I J j=1 Then, t = diag(i( 1 ..... n)) E , 8 |I j=1 Complexifying everything we have: = 0 84 9 = o L(n,C) 0" k = o(gL(p) gL(q)) = {X = A E gL(p), 0 Di D E g-L(q) nn t = {diag(z ,...,zn zi = 0}. j=1 K can be identified with the space {1 = (al , . . . ,ap I ap+1 , . . a p+) correspond to in > an j ej E IR If we denote by the dual basis ... Rn . then ,an) I ai = 0 a 1, .r.. nf = aP; a. - aE Z. the elements of the roots of t in g the set A(9) = A(qt) = {e, - ej A(k) = A(k,t) = I t 9 J; -, j 1 n}. Also {eL - e i 1 L, J p} U {ek - em I p < k, m n} 85 the compact imaginary roots of = A(,t) A(l) = (+ (e -e p+j) t and I 1< e p; . 1 t in g-. the noncompact imaginary roots of LY(X) Computation of 4.2. i weight of a A+(k) so = (a 1 ,a2 ,. . . aplb1 ,b 2 , ... q} for a Harish- ,b ) be the highest Fix the positive system X. K-type of lowest that a 1 > ... > a ; b > ... > b . to obtain an explicit expression of We want parameters XY(X), and j X. Chandra module Let g. and XV L attached to pi the by 2.4.7 and 2.4.8. 2pc = (p-l,p-3,....-p+lq-lq-2,...,-q+l). Let p + 2 pC = (xlx 2 '. . xp I3 1 'Y2''''Yq) We can form an array of two rows with the coordinates of 86 p + 2pc left to the y. so that right as follows: the are in the second; right in the array. j S > xi x are in the first rows; and terms decrease from j+1 xi+3 >i+2 > k j+2 k+1 j+3.. ''' Xk Xk+l look like: xi+l i+2 ~~-* y.y i+3 yj+2 +1 This array gives a choice of positive roots A+(g,t), left to For example, if we have >i+l the array would ... they are aligned in decreasing order from compatible with A+(k). roots are given by the arrows. That is, the j+3 A+ = simple In the preceding example, they would be ... ei - e ;+j; ep+j - i+1; e i+2 ... ek ~ ep+j+ 2 ;p+j+2 e 1+3 e i+l "p+j+1 e ''. e k+l ;ek+1 -e i+2; 87 Because the terms in each row decrease by at entire array is a union of blocks of least 2, the the following five types. 1. 2. r r-2 r r-2 r r-2k ... r-2 ... r-1 3. r-2k r-3 r r-2k r-2k-l ... r-2 r-l 4. r-2 r-3 r-2k+l r-2 r r Zr+1 r-2k ... r-l ... r-2k+1 I 5. r r+l ... r-1 From now on we will drop since r-2k r-2 the ordering of r-2k+l the arrows the roots is clear r -2k- in the pictures, from the 88 arrangement of agree on choosing also refer the order prescribed to them as |\ Example. p + 2pc, the coordinates of p = 7, Let _ q = 8, provided we in block 1. We will etc. and = (2,2,2,-2,-2,-2,-311,1,0,0,0,-2,-3,-3) then, 2pc = (6,4,2,0,-2,-4,-617,5,3,1,-1,-3,-5,-7) p + 2p c = (8,6,4,-2,-4,-6,-98,6,3,1,-1,-5,-8,-10). We obtain 8 6 8 6 the following picture 4 p (g .g 1 r 1 times where r. number of is as -9 -10 8 p + 2p c the picture of coordinates of = -5 3 Using -6/ 4 - we can split the follows ... ... rt g times the number of f f f ' t s1times st times p-coordinates and q-coordinates making up the i-th f)t t .. s. block the of the 89 array of p + 2p c, and gt r. > 0, s. 1 1 9t 9 f 2> f1 i = 1. 0, . .. 2 - t. Write X -2= V1) in Proposition 2.4.7 as Proposition 4.2.1. 1. = XV(X) in chapter 2. The expression for has XV(p) the form V I times r 1 times rt times t s times where X1 2. Let > X2 t* t sttimes t 90 A(Ly) = A(Lv, tc) = {a E A I XV, a (@EA(L xa) = 0}. Then tv(X) V) = Example. r1 Note r3 Let @ u(r t,sd) ... In our current example we have , (2,2, 2 = o(u(r 1 ,s1 ) ( c r2 -2,-2, -2, -3 r4 r5 , | 1,1, 0 , r6 s1 s2 s3 0 , s -2, s5 -3,-3) s6 0. = X\ = p + 2pc = - p (1,1,1,-l,-2,-2,-3|1 2,2,1,0,-i, -2.,-3, -3). Then 1 V2 r and tv (X) r2 -1, r4 -2,-2, -3 r5 r6 ~Y2'2' |I ,0 s1 1 s2 , 0 , -1, s3 s -2, -3,-3 s5 s6 91 Proof. Take a block of the form 1. Then coordinatewise we have I... p + 2pc = p = (... ,s,s-2,.. .,s-2k,... I...,s-l,s-3,...,s-2k-1,...) (...,r,r-2,..., r-2k,... x V = p + 2pc = - ,r,r-2,...,r-2k,...) p (. ..,r-s,r-s,...,r-s... | ... ,r-s+l,r-s+1,...,r-s+1,...) Let e m-e p+;ep+em+1 be the set of .. ;"m+k e p++k} simple roots making up this block. I i = 1......k+1}. -- e . {03i = e m m+1-1p+e+i-1 Then -c. 1 = -> V 1 = -1 . Choose 92 Also <,6 > = 6 P j and c. V Let {(1..... 13 noncompact of be 0. i the subset of imaginary r positive roots chosen this way from all blocks the form 1. subset of } k.'...'Pk 1 i 2 For blocks 2 we can choose the simple roots. same kind of In this case V c. = -IV ,. = 0 since coordinatewise, we have p + 2p = (...,r,r-2,...,r-2k,...| ... ,r-l,r-3,...,r-2k+1,.. p = (. . .,s,s-2....,.s-2k,.. .I ... Vy = A .. similar ,r-s,r-s,...,.r-s, ... ,.s-1,s-3,... |... situation occurs in cases s-2k+1 ,...) ,r-s,r-s, 3-5. Choose all the 93 simple roots For of the form cases 4 and 5 choose all For all clear is is a set involved in blocks 3. those of the form ep+b of that = -<X (,9i the union > IT of strongly orthogonal V these 5 kinds of subsets simple imaginary noncompact + 2 V c.fp.. 1 1 Pi Ell - d) of Proposition 2.4.7 are clear. Let A+ = a E A+ ja,p > = 0} A+ \{eiem;em-eji;e A i-e ;e +,-e.} im m i p+e p~ep+e 1 = 0 then t ea. = 0. Def ine roots. a) - these subsets c It ep+b - ea 1 {(x can be . . . 0 identified with +x x p Ix p+1' the set n ) E t 0 IXm =Xp+L p == 94 So if i 1 = = A Ig it a +b = (a1., To verify e) . . ., . 2 a m+b 2 ,apb..,..., , ,..,b). of proposition 2.4.7 in chapter 2 we use the following lemmas. Lemma 4.2.2 (Schmid [1975]) 1 + PC = p(A (k 1 (see Vogan [1981] p. 1 , t )), A +A (g 1 ,#L 1 ) = A (g) n 13 and + 11 p 1 = p(A+(g. , h )) then 1 C So 1 = (2p ) ' 247). If 95 1 ~"1 1 1 x= 2pc + 1 - p V it1 = (Vogan [1981], Lemma 4.2.3 e) So 2.4.7, Now it 4.3. 1 -c 2 11 + p p. 249). 1 + 2p 1 is dominant c ). A + (q , for V is is clear. straightforward to verify Proposition 4.2.1. The Lowest K-types of the Modules A (N). In this section we obtain necessary and sufficient conditions of the g-modules If parabolic replacing it for a representation of is of x = (x to be the LKT of one A (N). , . . . ,oxn) E i(t subalgebra x K q(x) = )* L(x) + u(x) by a conjugate under the form we obtain a as W(KT), 0-stabl e in 2.3. A f ter we may ass ume 96 x = (x 1 ,...,x ,x2''' p1 x 2' Pt P2 x , x .. , x .. t'' qq Et such that x1 > x2 >' Pi ' q j t '; > and Pi = p q and -0 = q. Then L(x) = o(u(pj,qj) 2q ( u(p 2 ) .. E D t't Clearly 2p(aflt) = (rl,...,rl,r 2 2 S. s ,...,s q1 and ,1s ,r ..... ,rt ... ,r ... 1t ,...,'s q2 , ,...s , ...,s q t ) ' t 97 2p(ank) = (sl,...,s ,s2.... s 2,. . ,st,...,st p2 rr,...,r ,r2 , . q where - +q2+.. = S = -(pl+p q n Wri te pi, and their 2 p E L(t 0c) represen tation of .t,...,.rt). +.. .+p be qt .+q_) _) + q+1 + + pj+ + 1 S+ qt + Pt. the highest w eight of a K. k = (qfk)(W). = p + 2p(ufk). p.' r2,. q2 r Let Pt co mpact parts 2p(unk) as above, (Note that for Q(W) A 0'(W') but coincide.) Wri te ' = (z,...,Z ,Z 2 ,...,z k 2 , . . . . . Za,... k2 za ka Z , .. Z aa .. ,. ,...,. ) ............................ Proposition 4.3.1. e.. n n. 1 Then +1 + n.. i+1~ p In the above is the LKT of an setting, A (N) iff let z. n. 1 = k. + 1 i+1- 98 t c-roots the is positive on t the roots of to is orthogonal X b). satisfying 2.5.1 a), L, dimensional character of Then in L in and it a. By Proposition 2.5.6 and its proof we may assume p q' n determines and if a(p') k and p' determines satisfies 2.5.1, X X = (X 1 , , 2' '' k .., ,X 1' C'. So then x 2''''' a''''' k2 1 a ka .. a xa .2' aX, a 2 with x p that Suppose z. = N = + X + 1x2 >..>xt'* 2p(aup), p' then (-n -n2-,..-n ) X = + ni+1 + + 2p(u); ... + nt and z. I -z i+1 L + a the weight of a unitary admissible one- (tc *, X E and a = some for w = X + 2p(ulp) Suppose that Proof. = i i+1 n. 1 + n. i+1 n i + n i+1* that ' = 99 On the other hand i f then let o' X.= z p + n1 + satisfies z. - z ... - 1+1 I + n,_, + n > n. - I and 1+. .. +nt) (n+ = q(p'), i.e. '= {a E A(g.,tc) I <p',a> (a E A(, = A(U') Then I tc) > O} U <p',a> = 0} U A(L'). if , 9x tI .t9. k 1 1 t > 0 <X,A(L )> = 0 and p = p' - 2p(ua'k) = X + 2p(u'np). q. e.d. 1+1' 100 4.4. XV(A (X)) The parameters Aq (X) = g By definition since By definition 2.5.1, highest weight ji = X + By Lemma 2.4.23, first X (M)) s = dim a where is perpendicular to The LKT of A n k. the has (X) 2p(unfk). GL X v(A ()) LV(CX) + p(a). = Assume that N = (0.. .010.. .0) (p-1,. ..- p+l|q-l,....-q+1) X + 2p c picture (say p which gives a q) p-3. p-i L = g and q p then tv (A (C.), X E [center(L)]J. L, roots of and q-1 q-3 ... -q+1 q-1 q-3 ... -q+1 p+ (if (4.4.1) p = q+2k) or else: p-l / p-3 ... q-4 q-2 q q-1 q-3 ... -q -p+ ... -q+1 (if p = q+2k+l) (4.4.1) 101 Then p-q-1 p-q-3 2 2 1 0 2 -p+q+1 1 2'''.'.2 0 0 .q q (p XV (C ) = q (mod 2)) = p- 0.. - 0 0 ...O ; 2 q+1 q (p So X = if give (a,...,ala,...,a) and t E q+1 = g, (mod 2)) will X + 2pc the same picture and Xv(ex) = [a + 2 ,a + + a + 2 a.. .a, q+6 1 ...a S .. p 6 q+e = 1,0. q Now suppose a1 a2 2 .. a a taI t = a) . .(G ... ( t'q t)) 102 and X = (a 1. ... ,a1 ,a 2 ,. . . .,a 2 ,. . . p1 . at,...,ati p2 Pt al,... ,a ,a2,. . a2 ... ,at,...,ad q2 1 - qt Clearly 2 k = (pl-1,....-p. +,p2 'p t-1 , ' ,-p t+1| q -1....,-ql+l,... ,q t-1 So on the (p ,q )-coordinates we have a similar picture and xv c) I. .a.+ .. - p.q.i-1 1+6.11 1+6. 1 1 Il ,...a.+ ,a....a,a... 2 2 .. .. a . , 1 - 1 a -1 2 2 a q - (p.q.i-1) 1 1 2 +61 1 P. ... |I... a .... a .i... I qi if say, p. picture for = q. + 2k + e., 1 X + 2 pLflk 6. 1 = 0,1, k > 0. Also the will have pictures like (4.4.1) or, 103 if q1 p . . . z-1 . y z- 2p 1 + 1 z-2 z-2p- .. - 2 p- ... 3 .. (4.4.2) y y-2 ... z z-2 ... z-2p +1 z z-2 ... z-2p +1 A... Now p(a) = (u ,...,u ,...,Ut,...,ut|u .. ,.. Pt ,u 1 .ut,....u t qt .11 So Sd. (u(1) LY(Aq(X)) = (N)1 d. 1 ,s) x u(r X ((1)) ] t C '. 1=1 ((U(Pitq i)) where r ( F = 1 if p= q = min(p ,q + 1 (mod 2) + 1) and . p > qq 1; = 0 104 otherwise) s. = min(p.,q.) + (6. = 1 if 1 p. q. + 1 1 2d 4.5. i + 1 and r. I + 1 Proof of Theorem 2.6.7. Suppose X E d(g,K) infinitesimal character = 1 coordinates of = i p. ' 2 Y1 > 2> p E (Cc) X. ' ' ' ... t' .. xt Pt y ,y 2 q1 > x2 and let than that of Section 4.2: (x 1. . . . . x 1 , x 2 '' x1 1 slightly different splitting of the y9 ... that q.. G = SU(p,q). ), p2 so + 1 otherwise) 1 in Theorem 2.6.7, with r E (c) the highest weight of a LKT of Let's consider a 1 for is as 6. = 0 > p ; q. s. 6. ' .y 2 . . . . . . .. q2 > t s 'ys' ys) qs be 105 p , q but here necessarily It of p is compatibl'e with convenient splitting is not this that is, the blocks given by to draw a picture of with the same blocks obtained from In the example 11 But now, of section 4.2 (2 2 2 -2 -2 -2 -3 Pi p2 p3 p + 2p C means p is | 1 1 0 0 0 q1 q2 -2 -3 -3) q3 4 first W. We may assume that either x + p-1 > y1 + q-1 or x + p-1 = y 1 + q-1 otherwise we can interchange -3 -3 to study what happens around the coordinates of coordinates the -2/ the splitting of We are going 2pC' -2 - 0 0 this p + -3 A 2 [ p = > 0, and p and pl q. qi, p1 106 If for < p p1 we can have the following configurations p p1 1. x x 1**x x Y1 ...Y y q -1 2 x2 '.' ' ' Y2 p1 or x x1 x2 Y 11 1 2. xl -- - y_1 where y.i-i 3. ... xl *... with x 1 x ... y1 i-i ... > y.1 > z. x1 x1 _ .. y i-i y.1i z. y.i-i > y.1 > - ... x1 x2 ... y.1 z '' ' 1 ... Y x x2 y. z 107 4. xl x yi y1 x 5. > 1 x x. ... x. z y ... y y2 x ... > z j . x .. x. z ... y1 y > x x x 1 .. y2 y > z. The blocks in these pictures are simple factors of LvX).Note in case that 1. ji if is the LKT of an So we have to check reductive subgroup L C G, tion of A A (X), Dirac the representation of the indefiniteness of that o must be X as the Langlands in the case of an the He rmitian form is preserved the derived functor, and in the case when we don't have an V L we coming from a representa- but in such a way that, the signature of (N), under L; (X), we can find a that and embed submodule of a Zuckerman module A (Lfl) 7L inequali ty fails on L, and the form the on with highest weigh SIL + 2p(unp) is dominant p L , the LKT of (LfK)-types involved in L, and occurring = AL + P in will be such 108 On the other hand, non-unitarity. in In each case a group 1, making sure All for cases 2. that a) this will - c) - 5. we need L will be to prove found as of Theorem 2.6.7 hold. reduce the problem to the case p1 = p. In this case we have two configurations 6. x x 1 1 yI Yi ... x ...- y *** x1 '' k yk' x Y Case 6. x z ... > yk > Z. x 7. .*** x 1... ... can be y x ... ys x . x Ys included in either 2. or 3. and case 7. will be dealt with similarly. Note that as soon as we have shown that a) 2.6.7 holds, then by Lemma 2.7.4 of Theorem the representation of in question, as well as its Hermitian dual, L have a Hermitian form. For is 1, either Then =V - Lv + t = o(u(pj,q a) let q, or L D L L n av' a' a)), here qa 0. and by Proposition 2.4.15, if = L n 109 DE[ q Assume that XL Od0 =A9, (X (X )) QV Ly ~ ®(XL ) ) LV A1 (N). Define Q2 by a2 a + au 1 %2 ~ 41 + a a1 + t = a. + Then (A O 1d ck2 Proposition 2.4.15 again. To prove see that that 9 2(C.) (X,a> > 0 is a module a E A(a for all A2(ex) 2 ). we need to But, by Lemma 2.4.22, xV - X + p(a) + P afL = X + + p(a) + Hence By A (XL) Proposition 2.4.16, is N + P, and it the p - res res infinitesimal character of is regular. 110 Hence, is dominant X\ res + pLre,a> > 0 KX+p,a> = (X particular, a E A(u if <XV+P Lres , a> 2 = X + p 4-* for all is dominant a positive. In ) 0. <presa> a> + (X Now 2p(nlp) = (q-a**q-q a'-q............,l-,......P-P1 'P l ' 1 Qap''1 '''9a p-p 1 Pi P-Pi qa 1 9 9Pl) l -a Set p=L i-2p(unp) = (U ,...,u ,...ut,...,utivi''''' Pi with u1 > ui+1' By c) LKT of j Hence, by p1 < p q1 s,. , s) qs >vj+ of Lemma 2.4.23 the module Since Pt 1''''' pL is the highest weight of a XL. then L A G induction, Theorem 1.3 and dim L implies < dim G. that there exists 111 an L n K-type in V L V L 11 77 o V L (Lfp) such that, on 1~1 77 the Hermitian form is indefinite. L p are p- U {+(0... 0 0.. .0 1 if 0.. .0 0.. .0) qa 1 0.. .0 q qa qa = 4a 10.. .0 0.. .0 -1 0...0) P-Pi Pi the roots 0.. .010.. .0 -1 ±(0.. .0 1 0...0 A(Llp) = n L the weights in Now, a 1 or A(Lflp) = (0.. .0 0.. .0 P-Pi p1 if 1 0.. .010.. .0 -1 0.. .0) q = 0. q A highest weight of an then of the form candidates pL + for L n K-type in some C V L t A(tnp.). for highest weights are the weights o (fl) So the is 112 L .,u 77~ ( 3 +1,u.,... u.... . .V p.- 1 .1k p x p1 k q V -k V k- ... ) q - qk X qa or L TL = (u 1 +1,u ... p1i u .Iva ...v a 1',.. v -1...). a qa -1 p1 -1 So T7 = (x 1 . .. x 1 ,... x .+1 x x.. yk .. k Yk 1 '.. p .-1 or Ti = (x +1 x. ... p- ,x1,x 2 .'. 1 lyl. . .Yi yl -1 ... ) 1 - qa are dominant. This completes the proof of Theorem 2.6.7 for this case. To prove the following lemmas. result for cases 2. 5. we need the 113 Lemma 4.5.1. G = U(m,m) If the Dirac operator inequality fails then pPo = (,i.t.e. Proof. Write And = pI ( 1 a as W 1 ,..., | .. pc + As 1 ' ,.. m W(p -p s n ) + PC C(p~O-pn) = is a with , ps E ) d = (x...xjx...x) = m m m (0,-i,...,.-m+ljm-1,m-2,...,.1,0) + p + c (OI-P+)+p >= n c <c' X for (Cfr. 2.7.1.) I, E (center 9)* Sc If p = (a+1,.. and c + (g,K)-module with s -n~c' n C' infinitesimal character then (-,-(> And ( s +)+p nP > <P'c4 c> + W P)+p> (p, p> < (p,p>. q. e.d. 114 Lemma 4.5.2. If w G = U(m+1,m), = (b+1,...,b+1lb...b). m+1 Then Dirac operator inequality m + fails -m-1 c as p = m Write 4-1 p s s=F-:M. 1' 2 and for m -m-1 m+1 Proof. m = m+1 b + 2m+1 s + 4 '' m+1 2m+ 1 b+ m+1 '''''2m+ 1 as m+1 2m+l'.'.' b+ -p +) ( same argument to show -P )+p > < (Pp> but pc ~ m m-2 2 2' -m M-1 m+1 2m+1 By the in the preceding Lemma we only need ( b+ M+1 2m+1'' in Lemma 4.5.1. m+1 2m+l -m+1 . p = (m,m- 1 ,...,-m 2 ) that 115 P P+ n + c 11-s .,-m+1 (1,0,-i, = -m+1 -(m+1) 2m+1 ' + .. 2m+1 lemma. the this proves m,m-1,..1) q.e.d. for each case 2 - Now, so 5 we will choose a subgroup cases discussed in the problem to the that we can reduce L Lemmas 4.5.1, 4.5.2. 2. the Remember that we have xi ,.x 1 following picture: x 1...x x2 y. i y.1-1 yy. Z -y s r where y _, t Choose L obviously = Let = UV - a' = u V n L UI L = Normalizer of Let = _ )@u(r,r)Ou(p-pl,s)) o(u(pl-r,ql+... C t. U L > z. > y Q = and q in G, t + U. then S(U(p 1 -r,ql+...+q_ )xU(r,r)xU(p-pl,s)). 116 By Proposition 2.4.15 there of such that L is X XL the Langlands subrepresentation of Adim the standard module is a representation Q unk (XL). Now, 2p(tLnp) = (a,....a b,...,b r p-r By Lemma 2.4.23 LKT of C,....,C |I p-p1 e,...,e f,...,f). d,...,d ql+...+q pLL = p-2p(anp) is _-1 r the highest weight of a XL. We claim now that pL U(r,r) = (x+l,...,x+1|x...,x). r In fact s r 117 PLIU(rr) - -ql-...-q. = (x 1 +q 1 +...+q _+sjp-p1 +r... _-s,..., x 1 +ql+...+q._-sly-p+2p,-r,...,y.-p+2p,-r). We know from the picture for + p - x 2p 1 A + q - = y that 2(q 1 +...+q y. + s - So x + q, +...+ ,_, - s - y + p - (q +.. .+q 2p, V L D v L indefinite. Now if L+ necessarily P = (0.. .0 -1 1 1 0.. .0). 3 j + 1. 1. p E >L such that the Hermitian form is 1 = _1+r) + r = Hence, by Lemma 4.5.1 and 2 of Lemma 2.7.2, A(u(r,r)fpl) + _1 +r) on is dominant, 118 Also A + P = is dominant (x, for . . . .,x 1 ,x 1 - 1,x 2 '... 1y1 '''yi- 1 yi+1,y....) A(unk). This proves Theorem 2.6 .7 for case 2. For 3, the picture that we had is p-p yi-1 x1 x1 -x yi yi'' .y d Let x = that d p-p 1 q in G, s if that q-d-r S(X) in 2.3, and L = --U(p1- r-1,d) (D u(r+1,r) (D u(p-pl,q-d-r). t / Lv. = dim u (q ,H Y V r c) E it* 0 then However, Proposition 8.2.15 of Vogan [1981] p. 545 gives us that where I a.. .a b.. .b C... 0-stable parabolic subalgebra as t Note q-d-r r+1 Normalizer of > yi > > z. i-1 z b.. .b c.. .c pl-r-1 define a x2 r (a.. .a 1 then fl k. X V V v7 ) Vf LS (XL)) = In fact, all we need is the to check is 0-stable data attached to V~ L 3 Hv s, [ Y = and that q contains some Borel V subalgebra of This qV. is clear by infinitesimal the picture character of Y is + 2p of -T = (X . Then the v) E hV. V V. Since 119 r + 07 = for (2 Xv,0), As before, a E A(u). 2p(unpl) pL = p - that if it is straightforward to verify (a+l,. .. a+1Ia.. .a). r+1 Lemma 4.5.2 and 2) of , < > ,a> > 0 then pL U(r+1,r) = Hermitian form <X enough to show that it is is r Proposition 2.7.2 indefinite on imply th at V L ( L +P the with P = (0,...,0 -111,0,...,0) E A(u(r+l,r)fpl). p + P Also, is again dominant for A(unk). Hence Theorem this case. 2.6.7 also holds for 4 and 5 are solved in exactly th e same way as 2 and 3, p using So + and reduced the problem to I have the case pi = p. 6 can be included in either 2 or 3. For 7 write the picture for W = W is (a a.. .alb1 .. .b ,b2..b2...bt...bt 120 r1 q 1 +6 a.. .a a.. .a b with e. = 0, J r2 qt+st a.. .a . .b rt+ 1 a.. t...a t. .a t 1 t+1 t r. + p= q. 6 + . 1 t qj- 1 Then = (a.. .ala+s ... a+s , a+s 1 q 2 .. .a+s . .. a+s ... a+s qt p sk = -(r 1 +.. .+rk) + (r k+1+. ..+r t+1) - In fact, ( 1+62+.. .+F k1) + from the above picture for p, (ek+1 +. .. +6 td. we know that: ) 121 a + p-2(r +...+rk) 1 2(q 1 - +61+ + 6 k-1)-1 = bk + q-2(q 1 +...+qk-1)-1+Ek k Sbk = a + p-q - k-i 2 r 1 = a - 6 2 j - (rl+...+rk) + (rk+I +...+r t+1l (61+...+6k-1) + Note that assume if q = 0 Fk 1 for all i > 1 (Fk+ 1 +...+Etd this . is case 1. So L to which I can q + t > 2. As before, I want to find a group apply some reduction argument. (*) Suppose Then let > r1 r t+1 L = U(s,q set s = r1 + 1) x U(p-s),q-q1 ). U(r1 ) x U(q 1 +el,q 1 ) x U(r x...x U(qt+Ft'qt 2 ) So verify L 3 LVs a) 61 Note ) + r that LV x U(rt+1)- and again, we can use Lemma 2.4.15 to of Theorem 2.6.7. 2p(alp) = (q-q1 ,..q-q1 ,-q 1 s .. . p-s -q Ip-s ... p-s,-s...-s qi q-q 1 122 Y + (N By Lemma 2.4.22 and 2.4.23 if infinitesimal character of is the infinitesimal character of In fact, by definition of (X Ia> > 0 (4.5.3) Write = U(s,q), L1 L L1 For = before; restriction of However, this r2 . = V XV(p) C A(uv). regular a+sl-p+s). it could be possible the minimal value of the the Cartan of Lr integral. is not possible for all involve all values of instead. v of then (x,x-l,...,x-r 1+1,w,w,...,w,x-r - 1,x-r -2.. .z r_1 to inequality as we have done to the split part of Therefore, we need to If x r 1, r2 by simply using V a E A(u) ,a-q+qlja+s1 -p+s ... of Dirac -TLIL that makes A(uv), for all some values of that is, XL' then (a-q+q , ... the failure prove (Nv-P(L),v) L then 9 (XL), is the ,V) q 1+e1 x-r 1....x-r q1 ... ) r ,1 123 s where x = a Hs If and = D + + (n-1) = = 2 p 1 + r1 1+6 - 1-1 2 r . Cartan subgroup of 1 1. v v 2* 1 (Xv, D) 1 0.. q regular 2 0 v 1 . .. 2 0.. .0 | 62 +r 2 1 -1) 1 3 2 1 1 , -q2 a+s +6 -1+r1 2 Ir => V 1 2 a + a-(s s 2 V1 1 si > max[2i +e1 -1) 2 = -r 1 + r 2 +...+ By (+*), ... integral we need: 6 s Lv then (0...0 0 + v -1 2 s = a 2 -v a + +6 + 2 61 +r To make a is a maximally split s vyEA (n-1) - = a - TsAs = 1 s1 w = a + z - + p r + rt+1 0. S0 62 +.. 1) .+ > si + Et* 611 + r 11. r -1 1 , s +72+ r 124 vl=~ 1 Let + 6L- V. Now, p (U) + 2 + r1 = (XV-P(U))jL + j J1 n-s-q 1 n-s-q 1 2 ' IL V 2' ' ' n-s-q 2 n-s-q '''''2 2 = s1 a+ 2 -1 s+q 2- Ir+ 1 -n 2 s1 - ,..a+T 2 1+ s+q2-n 2 ' r1 s s+q 1 -n a+2 2 s s+q 2 1 -n 2 q1 s + 6 -1 2 s+q -n 2 s1 ,aT s+q -n 2 s 1. . + s+q n 2 -2 1 +r2 We would like to prove that -n s ...... ,a+- s+q 1 -n s1 a +-2+ s+q the K-type with Highest weight 1 2 2 125 = (a+1,a,...,a p- a+s 1 a+s 1, a+s q, ,... -1,a+s 2 ... a+s q in ... a+s the representation X and that the Hermitian indefinite on form is V It t 24 q1 -l occurs a+s is enough to prove the 11 @ V. 77 failure of Dirac operator inequality on p1 L 1 and p (t ) = p(Lfk) - 2 '' s q So L IL P 1 q1 a+ql-q+2, Since q1 s s . . . , a+ql-q+-2 a+s +s-p-,...,a+sl+s-p- ) 126 s-1 = s-3 ''' L IL A ( 1 1 -s+ 1 ql-1 2 ' 2 L nk 2 q 1 +s-1 ,..., 2 a+ql-q+ and - (-da)) 2 s-q a+s-p Then if X = +1 2 ( v-PG )) IL1 , 6- 1S s+q1 -n 2 . 2+ y = a + +11 q 1 -s+1 s+q.-1 a+s 1 -p+ 2 Let 1 2 = ) + Pc(L) a+q 1 -q+ -q 2 - + 1 6 2-+r1 ,y+ -1 6-1 r 1 ,..y 2 r1 y Y+ y .... q1 If w E WK 6 6 -1 2 2 61 such that -1 65... y+ 2 2 r2 wY is dominant, then y . y) 127 -1 6 -1 - 2 ir 1 + +ql-1... y+s 1+ 2 1 +5, 6 + (0'r, = + y 1 6-16y+ 2 -1-1 11 'r1 ,. . .. 1,y+ 2+ 2 2 6 1,...,y+ -1 2 2 r2 r y-s - 2 r 1 -6,y-s1 - r -6-q 2 +1J Also 4 L ) + p+ flk i +qI,..,y+ y+ 2 2 r'' +1, 1 6 y+ -2 -1 6 r , + -1 6 -1 6 2 r r2 2 To prove what we want -1 ,+ - it is enough - to prove 2 that 2 128 (~EA4) < l' W r1 < IL i > p + ) L Pn ILi But this P k' + Pt ) 1 is equivalent to Ity+si j=1 + 2 1+ 1 +-+2 r 1+6+ji-l1 2 [ Using that Ic I > |d I -1 - 2 6- 6 + Lys 2 - 6 2 1'r 1+j (b+c)2 + (b-c) 2 - (b+d)2 - we conclude that (**) holds 6-1 s1 + 2 6 ir 1 +6+j-l -2 r 1-j (b-d)2 > 0 -1 > then j > 0. 6 6 1 > 0 + j - Now and s1 > 0 * s1 + 2 t 1 if . +j H ence -1 + 2 +ry 1 + j 6 + e 1 -1 2 > 0. if 6 > 0 Since j - > 0. ink > > 21 2 1 ' -qt) 1. if r. > r t+1, L = U(r t+q t+e t+rt+1'qt) and repeat the we choose x U(p-(r t same argument for this t+F t+rt+1 case. + 129 Note 6tI that, since s + r = r ... 2 + 0 or some So r. 1 < > 0; this reduces j, i - rt+1 = r t+1 r1 then (**) will also hold if 0t + + 62 +...+ and some 6. > t. to the case r ql+e 1 a.. .a a.. .a 1 q2 r a.. .a b b 2 ... b a.. .a q, q2 > 0 2 q1 But conclude But by symmetry, using that the case r 1 > r t+1' we can e 1= 0. then we have r a...a q r a.. .a a.. .a b.. .b q With which we have dealt before. the same way as case 1. for p1 This is solved in < p. This proves Theorem 2.6.7. for G = SU(p,q). q.e.d. 130 G = SP(n,IR) Chapter 5. Preliminary Notation 5.1. Im Let GL(a,C). in identity matrix be the We def ine G = SP(n,R) tg {g E SL(2n, IR) = .- I n The maximal compact K subgroup K = SP(n,[R) n of =. 0 nI I0J 0 11 n 1n} 0 is G U(2n) 2 U(n). Also I 9' that = oP(n,IR) X E oC(2n,R) It Kn .-- I n nj +0 = 0 n is g'0 and = if = X 0 is = A B A C tA A,B,C E gt(n,IR), B, C symmetric the Cartan involution defined by (x) = -tX, 0 131 = {X C o p(n,R) A A = -tA A0 X = . , go 1 8(X) A B" B t A. {X = n) tB B = U(n) A -B d(O tX = X}. B] {x = If | - = -X |I A, B symmetric}. then = the compact n n Cartan subgroup of cos 8 1. 0 H G is 0 n 8 1. 81 hc 0 = 0 n 8. E lRl = Tc 0 cos ++ [R d( . .. X 8 ) 0 -d( 81 ... C sin 8 its Lie algebra is 0 t 0 cos 81 -sin and 0 cos 6 = -sin sin 81 0n R = 0 nl The complexification of these Lie Algebras gives tc 0 132 Ii g = op(n,C) = {X E oL(2n,C) n tX I + 0 0] -I } 0n X n A x = = B = tB, C = tC C k 2 gt(n,C) d(z 0 h x = _ z. E C d(z1 . K , n 0 can be identified with the {p .. .z ) = (a 1 .. .an) I a 1 space a2 . . a n; a.1 E Z}. x1 0 c n If -N 1 0 n j Define, for of t in g A(g)= = = 1,2,...,n, e.(X) = ix.. 3 3 Then, the roots are A(,,tc) {±e e k; -2e I J,k,e = 1,2,...,n; j < k} 133 also A(k) = A(k,tC) the compact imaginary A(P) = A(p,tc) = t(e j-ek) roots of = {±(e +ek); j < k < n}, 1 tc in ± 2 e, | 1 < j < k ( n; 1 < e < n}, the non-compact 5.2. imaginary roots of Computation of As A+(k) for in g. for any module the preceding cases, X. fix a positive root so that if A = then is (a 1 1 a 2 Ii + 2 . . . . . an) a1 a2 pc = = an' (xx2' (n-1,n-3,...,-n+3,-n+1). .. ''n). Choosing a positive Weyl Chamber for corresponds ... A+(k)-dominant and 2p Let LY(X) tc to forming an array of A(g.,hc) two rows with the system 134 absolute value of in decreasing order as are aligned x ,...,x x2 x1 If r the coordinate s of are ... > > X they follows: > xr+1 in the first row so that + 2p >xn ... then _x n _x n,'*- x r1 en-1' t r+1 xn' in in left to right in the the second and they all decrease from array. For example, X...> -x. 1 -x.jt-1 if we have > xi+1 > x+ k+1 >i+2 k ~ > ... > -x r1= '= -x.- r+1 -rj-2 x r1> r- - x r= 0 r the array would look like . x. Xk ) Xi+2 xi+1 J-1 x As for the case of A+ j-2 Xr-1 r+1 the choice of arrows SU(p,q), gives a positive root system .. xk+1 = A+ ( c), compatible A +(k). with Again, the entire array is a union of blocks of the following types. 1. m m-2 ... 20 m m-2 ... 2 / 0 135 2. m m-2 ... 1 m m-2 ... 11 m+3 N*/ 3. m+2 ... 2 ... m m-2 m m-2... 2 m m+2 or M+3.m 4. m+3 ... m n+2 ... .. / 1 ... m 1 .. or m+2 m m ... ... 3 ... m-i 1, or ... ... 1 .. m+3m 5. 2 m ... m-1 2/ 2 4 ... 3 (5 is a particular case of 4.) 6. All blocks of 0 not containing Again, using by the blocks If that p + 2pgc the five or types discussed for 1. the picture , p + 2pc gives SU(p,q), split the coordinates determines as follows. of 136 p1 entries Pt entries q entries qt entries 5.2.1 with B a block of some = (a ... a 1 where m block B, Examp e: .at.. .at . is times pt times p1 c2 CI ' c 1 c 2 .Cm btibts b ...b1 ) ... m entries q t times a. cm. (2 2p c= 2 pc (8 2 2 1 1 1 6 4 2 0 -2 = (10 1 1 -1), -4 -6 -8), -3 -5 8 6 3 1 -1 6Z 3 3 gives U Then set if A + this 1-5, times the total number of coordinates composing the = since type B 1 1 1 -9) 137 = (2 2 1 1 2 1 1 1 -1) as in 2.4.7. m p1 the splitting that we want. is Write =Y11) - t xv (4) (NJ... gives figure 5.2.1 then = 1 times p1 yp p Ei i(c If Proposition 5.2.2. - x= t A2***' 2*' P2 times p2 -x t'' t- -x 1 . N1 ) m times (D (D...@ 2- a (p 1, q 1) (p) 0.. .0 A t'''. t (pt'qt) (D op(m, R) with x1 > x2 Proof. a) -c i = (p E A +( P + 2pc > 0. that Observe If t Se. - p. p tc) is such that + ek - 1 <p + 2pC - p + i c 2i i then and e. 0 P3 ek > < 0, 10 if 138 then eo - should be ek included in the set {f P} of Proposition 2.4.7. b) Suppose that + p - a root system generated by + q 1 > -b 1. - Then the the roots involving the (pi, ql) coordinates: n -e 2 ;e+e 2 n-1' {ei+en ne+;- is isomorphic Ap.+ql-1. to elliptic element then 1i Ap 1 +q is Except U(p 1 t C Since L. K Lowest x = Hence the real form of for these extra considerations, the proof of corresponding result for Let centralizes an ,ql). this proposition is analogous 5.3. L (x...x 1. p . the the one for SU(p,q). types of x E i(tc) to We may assume , x t' . . . . A the modules that (N). it is of the .xt 0.. .0 Pt m -x t ..'* - -x"1 qt x > xx2 > q1 ' t > 0. form 139 Write A(L) A (U) as in 2.3. {a C A(g.,hc) I <a,x> = = (a E A(,,hc) I (a,x> = > Then a) L b) 2p(ufnp) = (n-q1 +1. ' u(pl,ql) 19 u 2'2) 2 . (9...G 1 (D oMp(m,IR); b t'qt) .n-ql+1,n-2q1 -q 2 +1 1 ..n-2q,-q 2 +1... P2 (5.3.1) n- 2 (q n-2(qi+. . . N t-1 +...+qt )-qt+1,p-q,...,p-q m Pt -1-n+2(p 1+. .. +pt-1 t. ..- 1-n+2(p. +... .+pt-1t qt -1-n+p 1*, . q1 1-n+p ) . 140 2p(unk) = c) (n-p 1 .. .n-p 1 ,n-2p1 -p 2 . . .n-2p 1 -p 2 ... P1 n-2(pl+. . . pt_ )-pt p2 ,. . . ,n-2(p l+. ..+pt_ )-p ,q-p,. . ..q-p Pt -n+2(q 1 +. . m .+q t 1 )+qt.. .- n+2(ql+. ..+q t-)+qt...,-n+q -t Set n d) = 2p(u) p n+q t1 + q . = (2n-n +1... 2n-n +1,2n-2n -n 1 1 * 2 +1 ... 2n-2n -n +1 1 P1 2n-2(n1 +n 2. . 2 P2 .+nt-)-nt+,...,2n-2(n 1 +n 2 .. .+nt 1 )-nt+1, Pt 0 .. .0,-2n+2(n1 +.. .+n -)+n t-l.. .-2n+2(n +...+nt_,)+nt-1 m ... t -2n+n -1...-2n+n -1) q1 Now suppose a that representation of 4 E K. i(tc)* is the highest weight of 141 By determine a compact parabolic subalgebra f n p the proof of Proposition 2.5.6 we may use q n to t n k = + k k. 2p(ank) = 2p(A(unk)). Set Suppose that p + 2p(alk) = (a,... a. r pi is If hence - A - a > nt + 2m + 1. aii+1 p = X + ... n .-- setting, a,). si st the above a X1 = N x x1 In the LKT of some and Proof. m rt Proposition 5.3. 2. then 0 ... 0 -at...-at...-a .. at...at set n. = r + (N) + ni+1 2p(W1p), then - x ***x t0.. x > 0 the coordinates of i + 2p(ank) = X + 2p(u) give s. 142 X + 2n - 2(n 1 +...+ni _)-n+1-(X.+1+2n-2(n +...+n.)-n +1+1) 11+ Ni1 ~ + ni+1 i+1 + n > n + ni+1 and xt + 2n - 2(n + .+ nt-) - nt + Conversely, suppose we are given conditions of parabolic the theorem. S= a. - A + 2m + 1 satisfying the be the Set 2n + 2(n +...+n So > n q = t + a Let p + 2p(alk). defined by 1 ) + n <X,A(u)> > 0, <xA(t)> and = A + = 0, 2p(unp). q.e.d. 5.4. Proof of Theorem 2.6. 7 for Let X character LKT of X. as G = SP(nR). in Theorem 2.6.7, with infinitesimal -r E (h ) Suppose A E (tc X ), the highest weight of a is not a module A (A). Let 143 Yi) = LV = ( '(pl 9q 1 ) U (p 'q 2 p = 1 p (cfr. 5.2.2) and t1 ) $... t2 =O t = L1 @ Define by a C uV Then q D qV u = S(pt , q q = 2 q1 . = u(p,q), then 2 U + , Set (m,[R) 2LV (uvnt). and by Proposition 2.4.15, a) of Theorem 2.6.7 holds. Now let XL be an (L,LfK)-modul e occurs only as composition factor of XL as the exterior tensor product that 31 (XL). XL =XL < >L X can We 0 X L2 see with (t ,LifK)-module. an XL. such 1 That Xh L has a Hermitian form follows from Lemma 2.7.4. Lemma 5.4.1. 0 C. L - XL L1 ~A X0 for some q0 C L 1 ; X0 ) A D Proof. By Theorem 2.6.7, b) and c) (proved for and Theorem 2.6.8, if (LinK) , j = 1,2, XL/ A (N) K-types of XL1 SU(p,q)) then there are such that 6. E 144 < ,> i 4= IV L 4 1V 2is L. A(L lp) indefinite. L i = such that < and 2p(ulp). - ,>L = (x .. ,xp+ 1 ' .,x Then V 1 that if there is is indefinite on V 1 E If Moreover, we know P E the sum . xp+M+1'' .'. x p+m ' m) and 2p(unp) = (n-q+l,...,n-q+1,p-q....,p-q,-l-n+p...-1-n+p) then, since A(L 1 no) it is q m p clear = ±{(ei+e ) that + P then - p+m if is dominant + < p, p + m < i 1 is dominant for for A (k), unless n} j A(L fk), x p = x then that xp+ 1 ' xp L Note and henc e Suppose or p+m+l* Suppose So p+1 that p p2 C that 2p(u p)IL2 is fine and = 2p(alp)I s aL2 XL ((0.. .0);(l... 1,0. i2 is of the th en form is a principa l series. 2 145 2 1.. .1,0.. .0) = m-a a 2 Then for TI (0.. .0,-1...-1), = m-a XL2 , > 0. a with V 2 is also a LKT of a by Frobenius reciprocity. So Since if 77 = 17 A 1 + T2 + 2p(unp), is also dominant, it V 1? follows is a LKT o f xp+m - that X. 1 > p+m+1 Suppose iI that + P3 is dominant for 3 E some If A(L 1np) n U(pt'q t). 'B = (0.. .0 1 0... 00...0 1 0...0), P-pt since x p+m - 1> 13 = (0.. p+m+l1 0 ... m Pt p + since m = 17L pL is dominant. If 0 0 -1 0.. .0 0.. .0 -1 0.. p then, P q then qt +t Tj + 0 q~ is a dominant candidate. If argument and Ap2 =(0 ... 0,-1...-1),9 shows m-a a that x -1 p If 2 = (0... 0) xp+m ~ xp+m+l with a > 0, a similar xp+1* then both differences must be strictly positive. x p - x p+1 This is 146 clear from the pictures of p+ 2pC q.e.d. Lemma 5.4.2. A o(N 0 ) In the above setting, assume some for q 0 & t1 and X true if we assume : C q Then, Theorem 2.6.7 is that XL o. that > 2 p - x p+1i (5.4. 3) and Proof. if .+ P Suppose first x p+m + - 2 that m+1 m+1 2 ' ' ' ' ' 2 ] , 2 2 .. 0 -1 0.. V 2 p. 2 Then 1,0.. .0). + Pt 2nk> ( (p,P>. n is a .0 -1), into a (0,0.. .0 -2)} m m-a a making (1,1,..... + n( + pL 2lk' f3 E {(0. 2. an easy calculation shows in Lemma 2.7.2, there By 2) p+m+1 space on which ( is +j3 indefinite. Moreover Similarly if A + P is A +(k)-dominant, 2 by (5.4.3). then m-a a 147 ( E ((1,0...0 1 0.. .0);(2,0... m-a Now, if = j2 a (0... 0) then the Dirac operator inequalit y fails for any choice of unless r and, obvi ously, Now of if is XL p + f Yit2 is also dominant = representation. L2 and 6 = V Hence I(6, XL L2 V ) m+1 m+1 for P E A(t+ l 2). y = vIt (In fact, the is a principal series 1) trivial; = then, the Langlands subquotient , the trivial representation p(A+(t 2 t)), p if in particular, 2 2 pn = L2 . V the Lang lands submodule o f 1XL2 ) A(XL Q(XL2 (XL is X = d(A q By 0 (X 0 )) ® M (trivial induction by stages, X contradicting our assumptions on This proves representation). q the is an A (N), X. lemma.. q.e.d. 148 To finish the proof of Theorem 2.6.7, suppose now that x < - x 1. Lemma 5.4.4. = 1 x Under and x p+m- the hypothesis of Lemma 5.4.2, > Xp+m+ 2, if x then Theorem 2.6.7 is true. Proof. that The assumptions on the picture of the coordinates of ji + 2pc around Ii imply the coordinates involved is either ... m+3 ... m ... m-2 m-1 m+4 or M+ ... m+4 ... that m m-2 ... Im m-2 ... is, p + 2pc = (...m+5 m+3 m m-2 -m+1 ... | -m-4 ... or p + 2 pc = (...m+5 m+3 | m ... -m | -m-4 ... ) ) p - 149 Observe K-type that LV = that gives np) - 2p(a M-2 rn-i is p2 = p2 = f ine and that the fine the picture m is is (1,1,....,.1,0. .0) ... ... and the fine m m-2 ... m m-2 ... K-type that gives (0...0). Arguing as in the proof of Lemma 5.4.2. we can find, in both cases P E ((0.. .0 -1,0.. 0 -1);(0...0 -2)} as we want. q.e.d. Assume now that 0 x - xp+1 1I (5.4.5) p+m xx p+m+1 150 We want to contradict infinitesimal character Since we have an have the assumption that is regular and integral. -r on Recall L = U(p,q) x SP(m,R) = U(pq), L A (X)-module for some control that the we -Y. and L ; L = t TI (U(p.,q.))J x SP(m,IR). W e may assume qt. pt By i=1 the computation in 4.3, either V U(ptqt) S(X t+s,X t+s-1...X t+1 X t''* s X tX t-l...X t-s qt+1 It.'X t) s qt or XV U(pt'qt 1 t+s, ... Xt+ 1 t*' t t 2 qt t-st t qt and V U(pt ,qt ) Inside = (0...0 v 1. SP(n,R) Vt 0... 0 | -v this gives 1 ...- d 151 (Xt+s,.. . Xt-' t (0...0 v1 ... v If is -Y xt + regular t t - xt s .'x' t-s 0 I -Nt** x~t) v0 , .. *v 1* integral + s, > 0 > -Nt + -x t v v , '*t 11 + t -1 >s qt S> t v xt x > Claim. If Proof. The picture can be of 1. . . . .-- 2. (5.4.5) m-1 m+2 ... n+3 types. m-2 m m+2 i + 2pc for the following . . . . satisfies p m . .. rn-i M m .. . m-2 m-2 + q at s + qt then x t- - s < 1. around these coordinates 152 3. 4. ... m+4 m+ ... m+4 m+2 ... m+4 m+2 m-1 m-3 ... m+4 m+2 m-1 m-3 ... ... m+3 5. ... m+4 m-2 / or ... m . . ... m-2 1 r+3 m ... m+3 m ... ... \rn-i i+ So we either have ... (considering that 5 and 2, and 3 and 1 are symmetric) p + 2pc = p = and (...m+k+2,m+k -m-k...) -m-k+1...) m+k+2,m+k (... or A + 2pC = ( ... ... with In both m+2 m+1 | | -m-2...) cases we get xV This proves = (.. .1 1 the claim. 0.. .0 -1 -1...). 153 This 2pc reduces gives, ... quasisplit. = 0. But then, p + at worst, m+2 Because if q to the case when q. = 0, So, m+4 p. = we have This concludes M-1 ... m 1, x p+m since - U(p.,q.) xp+m+ > is 2! the proof of Theorem 2.6.7. q.e.d. 154 Bibliography A. Borel-N. Wallach [1980] "Continuous cohomology, discrete subgroups and representations of reductive groups". Annals of Mathematics Studies, 94 Princeton University Press. J. Dixmier [1977] "Enveloping algebras". North Holland Publishing Co., Amsterdam, N. Y., Oxford T. J. Enright [1979] "Relative Lie algebra cohomology and unitary representations of complex Lie groups". Duke Math. J. 46, 513-525. [1984] "Unitary representations for two real forms of a semisimple Lie algebra: a theory of comparison", in Proc. Special Year in Harmonic Analysis, University of Maryland 1024, 1-29. Lecture Notes in Math., Springer-Verlag, Berlin-Heidelberg-New York-Tokyo. A. Knapp-G. Zuckerman [1976] "Classification theorems for representations of semisimple Lie groups", in non-commutative Harmonic Analysis. Lecture Notes in Math. 587, 138-159, Springer-Verlag, Berlin-Heidelberg-New York. W. Schmid [1975] "On the characters of discrete series (the Hermitian symmetric case)". Inventiones Math. 30, 47-144. B. Speh [1981] "Unitary representations of SL(n,IR) and the cohomology of congruence subgroups". In noncommutative harmonic analysis and Lie groups 880, 483-505, J. Carmona and M. Vergne (eds.), Lecture Notes in Math., Springer-Verlag, Berlin-Heidelberg-New York. B. Speh-D. Vogan [1980] "Reducibility of generalized principal series representations", Acta Math. 145, 227-229. 155 D. Vogan [1981] "Representations of real reductive Lie groups", Birkhauser, Boston-Basel-Stuttgart. [1984] "Unitarizability of certain series of tions", Annals Math. 120, 141-187. D. representa- Vogan-G. Zuckerman [1984] "Unitary representations with non-zero cohomology", Compositio Mathematica 53, 51-90. G. Zuckerman [1977] "On construction of representations by derived functors." Handwitten Notes.