IRREDUCIBLE CHARACTERS OF SEMISIMPLE LIE GROUPS IV ...Byinspection of (1.10), this gives 2 102-1190...

Vol. 49, No. 4 DUKE MATHEMATICAL JOURNAL (C) December 1982

IRREDUCIBLE CHARACTERS OF SEMISIMPLELIE GROUPS IV.

CHARACTER-MULTIPLICITY DUALITY

DAVID A. VOGAN, JR.

CONTENTS

1. Introduction 9432. Notation and assumptions on G 9493. Root systems with involutions 9554. The cross action on regular characters 9685. Cayley transforms: formal theory 9766. Formal Cayley transforms of 6 groups 9907. Cayley transforms of regular characters 9978. A standard form for blocks 10079. Cayley transforms of 6 groups 1015

10. Characterization of "(3’) 102011. Existence of ( 103412. The Kazhdan-Lusztig conjecture 103813. Duality for Hecke modules, and proof of the main theorem 104514. Duality and primitive ideals 105115. An L-group formulation 105516. Examples 1059

1. Introduction. Let G be a real linear reductive Lie group. (It is important toallow G to be disconnected; precise hypotheses on G are formulated in section2.) To each irreducible admissible representation of G, Langlands in [17] hasassociated a natural induced representation, of a kind we will call standard.Roughly speaking, the standard representations are non-unitarily induced fromdiscrete series representations. This association sets up a bijection between theirreducible representations and the standard ones. Write for the irreduciblerepresentation corresponding to the standard representation r. The standardrepresentations are fairly well understood---much better, at least, than theirreducible representations. One way to describe irreducible representations is towrite them as (finite) integer combinations of standard representations, in anappropriate Grothendieck group. That is, we look for an expression

E M(p, )p (M(p,) e Z).standard

(1.1)

Received June 14, 1982. Supported in part by an NSF grant MCS-8202127.

943

944 DAVID A. VOGAN

The expression on the right is called the formal character of . As was firstobserved by Zuckerman, the formal character of any irreducible admissiblerepresentation exists and is unique--see [25], Proposition 6.6.7. (The difficultproblem of computing the integers M(O,) explicitly will not directly concern ushere. For the moment it suffices to know that there is a conjecture on how tocompute them ([23]), which has been proved by Lusztig, Beilinson, and Bernsteinin some cases--see [24], and section 12 below.)The inverse problem is more elementary, and is just as important in

representation theory. It is the problem of decomposing a standard representa-tion into its irreducible constituents. That is, we seek an expression

] m(i,r) (m(, r) N (0, 1,2 )). (1.2)irreducible

Here m(, r) is the multiplicity of as a composition factor of r. Since r is not adirect sum of irreducible representations, (1.2) must also be understood in aGrothendieck group. Combining (1.1) and (1.2) gives

r m(, r)

m(, r)M(/x, b)#

By the uniqueness of formal characters,

M(/x,)m(,rr) 8.,, (1.3)

(the Kronecker ); that is, the matrices m and M are inverses of each other. Sothe two problems just describedmcomputing formal characters and computingmultiplicities--are equivalent. In an appropriate ordering of the standardrepresentations, M and m are upper triangular with one’s on the diagonal ([25],Proposition 6.6.7). This equivalence is therefore even quite accessible computa-tionally. Our goal in this paper is to show that sometimes the problems are notonly similar, but actually "the same". To understand what this should mean, letus consider first the simpler case of Verma modules for a complex semisimple Liealgebra g. Fix a Borel subalgebra b 9 + rt of g, with 9 a Cartan subalgebra andthe nil radical. Write O for half the sum of the roots of 9 in n, and W for theWeyl group. If w W, define

Z(w) (R) Cw0-o

to be the Verma module of highest weight wp- p, and L(w) to be its unique

IRREDUCIBLE CHARACTERS OF SEMISIMPLE LIE GROUPS IV 945

irreducible quotient. Then there are expressions

L(w)

_M(y,w)Z(y) (M(y,w) Z) (1.4)

yw

Z(w) , m(y,w)L(y) (m(y,w) N). (1.5)yw

Let w0 be the longest element of 14/’. Then Z(wo) is irreducible; that is,L(w0)= Z(wo). The module L(1) is the trivial representation, and the Weylcharacter formula (see [10]) may be interpreted to mean

L(1) (-1)’Y)Z(y), (1.6)yw

or

M(y, 1) (- 1)Y); (1.6’)

here l is the length function on W. For (2), these facts tell us everything.W is the two element group 1, w0), and we compute

L(w0) Z(w0) Z(w0) L(w0)(1.7)

L(1)- Z(1)- Z(wo) Z(1) L(1) + L(wo).

In terms of the matrices M and m this is

-1 (1.7’)

These formulas suggest the guess that, for general q,

M(y, w) (- 1)t(Y)-’(W)m(y, w).Applied to (1.6’), this would give

m(y, 1) 1, ally;

that is, it would suggest that Z(1) should contain each L(y) exactly once.Unfortunately, this is known to be false--for example, in i(4) (see [4]). Theeasiest result saying that something occurs with multiplicity one is Verma’stheorem that Z(wo) (which is equal to L(wo)) occurs exactly once in every Z(y);that is,

m(wo ),) 1, ally. (1.8)

In conjunction with (1.6’) and (1.7’), this suggests

M(y, w) (- 1)t(y)- l(W)m(WoW, Wo.);). (1.9)

946 DAVID A. VOGAN

Formula (1.9) was conjectured by Jantzen on the basis of extensive calculations.Kazhdan and Lusztig deduced it from their conjectured formulas for M(y, w) in[15]. As these formulas have now been proved ([3],[6]), (1.9) is true. It makesprecise the idea that the problems of computing m and M are exactly the same.Theorem 1.15 below extends (1.9) to real groups. To see what is involved in

that, we begin with G--SL(2, Iq). There are exactly three irreducible representa-tions of G on which the center Z(G) of G and the center () of U() both act asin the trivial representation. They are the trivial representation N0, and twodiscrete series representations 1 and 2. Write ,/7" for the corresponding standardrepresentations; % is a spherical principal series representation, and rrl ,rr2-- 2. The decomposition of % into irreducibles is well known; we find

2 2 7/"2

_(1.10)

--ff0 ’T/’0- ’T/’l ’T/’2 e/T0 N0 + -ffl "" 2"Here again there is an obvious conjecture like the one formulated after (1.7); buthere again that conjecture is false for other groups (for example SU(2, 1)). To getan analogue of (1.9), we must find three irreducible representations 0, 1,satisfying

M(qrj,i) f.ijm(i,Pj) (/j _+ 1). (1.11)

This says that the multiplicity with which i occurs in pj should equal thecoefficient of rrj in the formal character of i, up to sign. By inspection of (1.10),this gives

2 102- 1190 P2 2 q- 1190

Po Po Po Po

(1.12)

Now SL(2,1q) has only one standard representation having exactly twocomposition factors (the reducible unitary principal series); so these formulascannot hold for any three representations of SL(2, Iq). What one has to do is lookat a different group. The group needed here is sometimes called SL- (2, Iq):

SL (2, Iq) g GL(2, Iq) det g _+

0 )SL(2, Iq). (1 13)SL(2, FI) tO (0(The group PGL(2, Iq) could also be used, but SL +- (2, Iq) is closer to the one wewill actually construct in the proof of Theorem 1.15.) The Weyl group of acompact Cartan subgroup of SL +- (2,1q) has two elements; so by Harish-Chandra’s general theory (or trivial calculation), there is exactly one discreteseries representation 0 in which (g) acts as in a one dimensional representation.


On the other hand, the group has two components, so there are two onedimensional representations , 2. Writing Pi for the corresponding standardrepresentations, one easily verifies (1.12) (see [25], Chapter 1).

There is another subtlety here, however. The group SU(2,1) has sixrepresentations 6 having the same infinitesimal and central charactersas the trivial representation. It is an entertaining exercise to match them with sixrepresentations 1... 6 of SL(3, R) so that (1.11) holds. (The reader moreinterested in enlightenment than in entertainment may consult (16.3a) below.)However, SL(3, ) has a seventh irreducible representation 0 having the sameinfinitesimal and central character as the other i; o is an irreducible principalseries representation. So the matching does not seem to work perfectly. The pointis that 0 must be matched with the trivial representation of SU(3): the secondgroup involved depends not only on the first, but also on which representationsof the first are considered. To keep track of this in general, one needs the notionof blocks.

Definition 1.14. Block equivale,nce of irreducible (or standard) representationsof G is the equivalence relation B generated by

lB2 if m(,"/7"2) O.

The equivalence classes are called blocks.

There is a more natural definition in terms of Ext groups, but this is equivalent(see [25], Chapter 9).

THEOREM 1.15. (Theorem 13.13 below) Let G be a complex connectedreductive Lie group, and G a realform of G (in the weak sense of section 2). Fix ablock (,..., r) of irreducible representations of G, having the same

infinitesimal character as some finite dimensional representation of G. Write cforthe complex simply connected semisimple group whose root system is dual to that ofG Then there is a real form of and a block (,..., r) of irreduciblerepresentations of G, such that

M(qrj i) eijm (i Pj)

M(Oj i) eijm( rj)

for all i, j; here %. +_ 1. In the correspondence )i (’-’) i’ discrete series correspondsto Langlands quotients of principal series for split groups; and finite dimensionalrepresentations correspond to representations whose annihilator is a minimalprimitive ideal.

In this correspondence, ( will be specified essentially by the block ;), andthe block {i) will be specified by the real form G. The signs %. will be specifiedprecisely. The proof is entirely constructive, in the sense that ( and {i) areeasily computable in examples. Despite the symmetry of the conclusion in G and

948 DAVID A. VOGAN

(, the proof does not necessarily produce G and (i) when applied to ( and(i). The hypothesis that i have the same infinitesimal character as some finitedimensional representation is needed only because a proof of the conjecture of[23] has been published only in that case. J. Bernstein has informed me that theproof can be carried out in general, however, (see [24]). We will state a

generalization (Theorem 13.13) depending on the conjecture. In it, ( has theroot system dual to the integral root systems defined by i; otherwise theformulation is unchanged.The idea of the proof is to associate to the pair (G, i) (for each i) a collection

of additional structures on the root system R of G (Proposition 4.11; amongother things, an involutive automorphism /9 of R). These structures pass in a

natural way to the same kind of structures on the dual root system /--forexample,/9 is replaced by -0 on/. From the structures, one can reconstruct a

group G and a representation i (Theorem 11.1). All of the difficulties stem fromthe fact that and i are not uniquely defined. For example, if G SL-(2, R)and i is the trivial representation, then the construction says that ( should beSL(2, Iq), and i should be a discrete series; but it does not say which discreteseries. Even if we fix the infinitesimal character, the two candidates differ by anautomorphism of SL(2, Iq), so they cannot possibly be distinguished in anyintrinsic way. Most of our effort, therefore, is devoted to showing that all choicescan be made in a compatible way as i varies. We will choose two special (butnot quite canonical) elements and 2 of the block, and use them to constructd, 1’ and 2. All the other i will be obtained from 1 and 2 by variouscanonical operations (mostly related to Cayley transformssee section 7). Thesecan be duplicated in G to give the corresponding i in terms of 1 and 02(Theorem 10.1). All of the indeterminacy in the construction is thereforeconcentrated in the choice of ,’/rl, ’7/’2, )l’ and 02. Once the i are specified, one

only needs to verify some very simple formal properties of the correspondence;then Theorem 1.15 follows from [24] just as (1.9) follows from the Kazhdan-Lusztig conjectures (Proposition 13.12).

Section 16 is devoted to a fairly detailed description of several examples ofvarious aspects of the definitions and results of the main body of the paper. Thepaper might almost be read by beginning with these and referring back to thegeneral results only as they are needed to understand the examples. (It wascertainly written in very much that way, with proofs being modified until theirconclusions agreed with the examples.) Experienced readers may prefer toconsider their own favorite examples, but in any case one should avoid trying tomake sense of the results in some abstract sense.The form of Theorem 1.15 suggests that we should try to invoke the theory of

dual groups in some form; in particular, that we should use L(GC) instead of (c(see [5]). This is possible, at least if we require G to be precisely the set of realpoints of G. A reformulation of Theorem 1.15 along these lines is sketched insection 15. Lusztig has suggested that the reformulation ought to be much easierto prove, but I have not been able to see that. There are many technical


simplifications, however. (In this context it is also interesting that Langlands’L-packets play an important part in the paper (Definition 8.1).)

It is a pleasure to thank George Lusztig for many very helpful discussions, andparticularly for explaining how to deduce Theorem 1.15 from a formal dualitytheory.

2. Notation and assumptions on G. Unexplained notation will in generalfollow [25]; we recall here some of the main points. Fix once and for all aconnected reductive algebraic group G, and a connected real form Go c_ G c.Write g0 Lie(G0), and g Lie(G) for the complexification of g0. Choose aCartan involution 0 of G0, and write

go fo + P0

for the Cartan decomposition. Fix a non-degenerate symmetric invariant bilinearform } on go, preserved by 0, which is negative on 0 and positive on P0. Fixonce and for all a real Lie group G, with identity component Go, satisfying

G/Go is finite (2.1 a)

Ad(G) c_ Ad(G c) as automorphism groups of g (2.1b)

the Cartan subgroups of G are abelian. (2.1c)

(By a Cartan subgroup, we mean the centralizer in G of a Cartan subalgebra ofg0.) We summarize these conditions by saying that G is a real form of G c; notehowever that G need not be isomorphic to a subgroup of G c. The uniquemaximal compact subgroup of G containing K0 is called K. Put

Z(G) center of G

U(g) universal enveloping algebra of g

)(g) center of U(g).

Fix a Cartan subgroup H of G. Put

A(g, b) (roots of b in g);

we may regard these roots either as elements of b*, or as elements of thecharacter group H. Put

W(g, b) W(A(g, b)) complex Weyl group

W(G, H ) Weyl group of H in G (the real Weyl group)

(normalizer of H in G )/H

950 DAVID A. VOGAN

W(g, [9) acts on f9 or f*.. W(G,H) may be regarded as a subgroup of W(g, 3), butit also acts on H and H. If a A(g, f), the coroot 8 is

Define s W(g, 9) to be the reflection through a" if h 9", then

The roots are classified as real, complex, or imaginary, according to their valueson [90. If a is imaginary, then the root vectors X___ generate a subalgebra of gwhose intersection with g0 is isomorphic to l(2,1q) or to tt(2); we call a

noncompact (singular in Harish-Chandra’s terminology) or compact accordingly.We will say that H is split if all roots are real.

Suppose H is a Cartan subgroup, and a A(g, b) is either real or noncompactimaginary. Then the root vectors for a generate a subalgebra whose intersectionwith g0 is isomorphic to (2, iq); so we get a map

q" I(2, FI) - go. (2.2)

if a is real, we choose so that

’ 0 -1

and if a is imaginary, so that

(00 01) {artspace);

a(-10 01)"If H is 0-stable, we can and do also arrange

oo(x).

These conditions do not specify uniquely, but this is not a problem. Since G islinear, q exponentiates to

SL(2, lq) G. (2.3)

These maps are fundamental to the theory of Cayley transforms; they make itpossible to reduce many questions to SL(2, Iq). Put

%= (I),( -10 01)(2.4)

(-1 o) u.2 O 0 -1ma %

If H is 0-stable, % and m both belong to K. Clearly they have order 4 and 2 (or


2 and 1, if has a kernel) respectively. The element m does not depend on thechoice of , but changing may replace % by %m. If a is real and H is0-stable, then % normalizes H, and represents s W(G,H).

If b C_ is any Cartan subalgebra, we have a Harish-Chandra isomorphism

If 9", write

xxfor the composition of this isomorphism with evaluation at 2,. This identifies themaximal ideals in g(g) with Weyl group orbits in b*. We call and Xxnonsingular if (a,)v 0 for all a A(g, 9). If a positive root system A+ (g, ) isspecified, we call dominant if

a/

If (t,X Z, then X is said to be integral with respect to a. Put

w(x) W(R (X))(2.5a)

the system of integral roots and the integral We),l group for X. If X is nonsingular,put

(simple roots of R + (2.5b)

the positive integral roots, the simple integral roots, and the simple integralreflections.

It is convenient to be able to identify roots and Weyl groups on differentCartan subalgebras. There are natural ways to do this, but for our purposes it isenough to fix, once and for all, a Cartan subalgebra Oa c_ , and a positive rootsystem

(ma)+ ( m({,a) ma.

We call these the abstract Caftan subalgebra and the abstract positive root system.Fix once and for all a (A+ )a-dominant nonsingular weight

X e (a),, (2.6a)

952 DAVID A. VOGAN

and define

xRa- R(,a)_A

(R ) + R + (k ) C (Aa) +

1-Ia= I-I(X a)

Wa--- W(Xa)_W(,a)

sa’-- s(xa)_ W

(the abstract integral roots)

(the abstract positive integral roots)

(the abstract simple integral roots)(2.6b)

(the abstract integral Weyl group)

(the abstract integral simple reflections)

(notation as in (2.5)). Suppose now that b c_ g is any Cartan subalgebra, 9",and that Xx is equal to the fixed X defined above. Then there is a uniqueisomorphism

x :((a)*,ka) ---) (*, ) (2.7a)

which is inner for G. This isomorphism satisfies

x R --’) g(x)

x W W(X),(2.7b)

and so forth. If c A(, a) and w W(tg, ga), write

ctx ix(a A(g, ))(2.7c)

wx ix(w) W(6, ))

these are called the root corresponding to the abstract root a, etc. It is oftenimportant to understand how x changes when X is modified by W(, b). This istrivial, but the result is worth recording; if w W(, b), then

iwx(X w[ ix(x) (X ()")*)

(. e

iwx(O wix(o)w- (o W(, a))

Ow) wo)w-1 (o W({, a)).

(2.8)

We are going to consider only representations having the infinitesimalcharacter X defined in (2.6). Thus we are excluding representations with singularinfinitesimal character. In the context of results like Theorem 1.15, this is anatural thing to do. A representation of G with singular infinitesimal character


should (formally) correspond to one of t whose infinitesimal character is "atinfinity" in some sense. Such objects may very well exist, but there is at presentno theory of them.

Let H TA be a 0-stable Cartan subgroup of G; such notation will always beinterpreted to mean

T=Hf)K

A exp(t)o fq o).(2.9)

Then T is compact abelian (but may be disconnected), and A is a vector group.Recall from [21] or [25], Definition 6.6.1, the notion of a regular character of H;this is an ordered pair

r /, 7 * (2.10)

satisfying the following conditions. Write G a MA for the Langlands decompo-sition of the centralizer of A in G; thus T c_ M is a compact Cartan subgroup.The first assumption is

lt itS’ is nonsingular for A(m, t). (2.10a)

Write

and p (respectively p) for half the sum of the roots in A+ 0n, t) (respectivelyA / (rrt N f, t)). The second assumption is

dF + Pro-- 2Pmn" (2.lOb)

If H is split, these conditions mean that F may be arbitrary, and is required tobe its differential. If H is compact, then F must be a highest weight of the lowestK-type of a discrete series representation; and , is required to be theHarish-Chandra parameter of that discrete series.

Attached to 3’ there is a standard representation r(,) (see [21] or [25]); it maybe defined by parabolic induction from a discrete series representation on acertain cuspidal parabolic subgroup of the form P MAN, with MA as above.The group N is chosen so that the Langlands subquotients appear assubrepresentions of r(7); and we set

(7) maximal completely reducible subrepresentation of r(7), (2.11)

the Langlands subrepresentation of r(’/). When is nonsingular (the only case we

954 DAVID A. VOGAN

will consider), if(),) is irreducible. Define

’ set of regular characters of H

H { (r, ) /’ x x)(2.12)

{ v (r, ) ((o),, o) is Ge-conjugate to (9", ))

( V /’1 r(V) has infinitesimal character X)

(notation (2.6)). Occasionally we may write (,)o and fro(7) to emphasize thedependence on the group G.

THO, 2.13. (Langlandssee [25]). Suppose r has infinitesimalcharacter X. Then there is a O-stable Caftan subgroup H C_ G, and 3,1 ( l)xsuch that

/f (H2,3,2) is another pair of the same kind, then (1)____ (3,2) if and only if(H , 3,1) is conjugate to (H 2, 3, 2) under K.

This theorem says that performing formal operations on representations is thesame as performing formal operations on conjugacy classes of regular characters;and in fact this paper is full of Cartan subgroups and roots, and empty of Hilbertspaces and operators. All the work of relating the formalism to the behavior ofrepresentations is already done, and summarized in [24].

Suppose 3, Hx. Write

cl(H, 3,) ((H,3,)I(HI,3, ’) is conjugate to (H, 3,) by K), (2.14)

the equivalence class of (H,3,). Often we will write simply cl(’/). If 3,i (/_i),1,2, then we may identify ([91)* and (2)* by the isomorphism

(2.15)

In this way we can speak of (say) the root for 2 "corresponding" to some rootfor 3. We also define

m(3,1, 3,2) m(ff(,l),,z.(y2)), (2.16)

and similarly for M (see (1.1), (1.2)). Similarly, write

3,1 3,2 (/1) 7 (3,2). (2.17)


(Definition 1.14) This is called block equivalence of regular characters.

3. Root systems with involutions. One of the ingredients of the proof ofTheorem 1.15 is a detailed knowledge of the Weyl group of an arbitrary Cartansubgroup of G. Much of what we need is known, the most sophisticated resultsbeing those of [16]. Published proofs of these results are rarely in quite the formwe need, however, and can often be simplified substantially. Therefore, we willprove what we need more or less from scratch, assuming mainly some familiaritywith the case of compact or complex groups. In this section we deal with formalpreliminaries; the result we are aiming at is Proposition 4.14.

Definition 3.1. A (reduced) root system is a triple (V, ( ),R), satisfying thefollowing conditions.

(a) V is a finite dimensional real or complex vector space, (,) is anondegenerate symmetric bilinear form on V, and R C_ V- {0) is a finite subset.

(b) ( ) is positive definite on the real subspace spanned by R.(c) If a, fl R, then 2(a, fl/(a,a Z.(d) If a R, write s" Vo V,

x-

for the simple reflection through a. Then s preserves R.(e) If a R, then 2a R.

A possibly non-reduced root system is a triple (V, ( ),R) satisfying (a)-(d).The unqualified term "root system" will always mean a reduced root system. Weallow V to be complex (which is not the usual convention) so that (b*,( ), A(g, b)) will be a root system if b c_ g is a Cartan subalgebra; of course thisdoes not change anything serious. When V and ( ) are understood or notparticularly important, we will call R itself a root system.

Definition 3.2.(V, ( ),R), with

Suppose (V, ( ), R) is a root system. The dual root system is

we call the coroot corresponding to a. Put

W(R group generated by (s a R

the Weyl group of R (or/).

956 DAVID A. VOGAN

That/ is a root system and W(R)= W(/) follows from the easily verifiedformulas

sa s

with a, fle R. The root system of type B and C are dual to each other; allother simple roots systems are equivalent to their dual systems (but notnecessarily by the map a ). The map R --)/ is at the heart of Theorem 1.15;as we add more structure to R, we will always have to transfer it to/ in someway.

Definition 3.3. Suppose R is a root system. Put

L L(R ) Z-module generated by R c_ V,

the root lattice in V;

P=P(R)= {)t Vl(t,X)Z, allaR},

the integral weights in V,

Pl Pl(R) e f’l OL,

the small weight lattice in V. Clearly L C_ P c_ P. Write/, /;, etc. for the sameobjects defined using R, and define

Zo=Zo(R)={) V[(a,X)=O, allaR}

Z=Z(R)=P/z,

It is easy to see that

and therefore that

/; Pl + Z0, (3.4a)

Z Z + Zo (3.4b)

Suppose for definiteness that V is complex. Let G(R) be a complex simplyconnected reductive group (not algebraic) with a Cartan subalgebra b(R) in itsLie algebra (R), so that

(V,R) --(f(R )*, A((R ), f(R ))). (3.5a)


Then

Z(G(R )) Z(R ); (3.5b)

this group has identity component Zo(R), and component group ZI(R). We Willnot use these facts, but they motivate the notation. The group Z(R) is finite. It isthe product of the corresponding groups for the simple factors of R; and these inturn are well known (see for example [10], p. 68).

Definition 3.6. Suppose R is a root system. We will call a positive root system+ c_ R an ordering of R, or say that R is ordered if R + is specified. In this caseis also ordered, by

k+= (lcR +

Define

O o(R + ),R

A positive root a is simple for R + if it is not the sum of two positive roots; or,equivalently if is simple for/ /. Set

II II(R + ) set of simple roots of R +.

LEMMA 3.7. In the setting of Definition 3.6, pC PI(R), and [ge /51(R )el(). We have for a e R

a e R + (a,p) > O=,(a,[J) > 0

a simple <=> (, p) <=>(a, tS)= 1.

The statements about p are well known, and those for are the same onesapplied to/. It is not true in general that (a,) (,p), for example in B2.

One final general fact about root systems will be used often.

PROPOSITION 3.8 (Chevalley). Suppose R is a root system in V, and X c_ V.

DefineRX a e R (g,) O, all e X

Wx={wE W(R)IwX=X, allXeX

R(X) {a e R <,X> eZ, allX e X}

W(X)- {w W(R)IwA-XL(R),allAX}.

958 DAVID A. VOGAN

(Definition 3.3) Then R x and R(X) are root systems, and

Wx W(R x )

W(X) W(R(X)).

For a proof, see [7]. When X (h}, we write R x R , etc.

COaOLARY 3.9. In the setting of Definition 3.6, suppose w W(R). Thenwp O if and only if w 1.

For R P is empty by Lemma 3.7.

Definition 3.10. A root system with involution is a pair (R, 0), with R a rootsystem on V, and 0: V---) V an orthogonal linear transformation of order two,such that

OR R.

The dual root system with involution is the pair (/,- 0). Write W(R) for thesubgroup of W(R) commuting with 0; then W(R) W()-o). Define

R" R"(O) a e R lOa a)

Wn Wn(O) W(R n),the real roots and real Weft group for (R, 0); and

R in_.= R in(0)= (a C R IOa a)

WiFI wiFt(O ) W(R iFI),the imaginary roots and imaginary Weyl group for (R,O). The roots which areneither real nor imaginary are called complex. We call (R, O) quasisplit if R in O,fundamental if R O, and complex if it is both fundamental and quasisplit. Thecompact part of V is

V V(O ) x c V Ox x

the split part is

e vlox -x}.

The example of a graded root system of interest to us is the action of theCartan involution on the roots of a O-stable Cartan subalgebra. This accounts forsome of the terminology. Suppose (R,O) is a root system with involution. Wewant to describe W(R)a. We begin with a special case.

LEMMA 3.1. Let (R,0) be a root system with involution; and suppose it is

complex (Definition 3.10). Then we can write

R=RIUR2


a disjoint orthogonal union, in such a way that

0 "R R3_ (i 1,2)is an isomorphism. By this isomorphism,

W(R ) ((w, Ow) W W(R1))

W(R,).

In particular, W(R) is generated by the elements S,So, with a R.

Proof. Since there are no real roots, we can find x V such that (a, x) is anon-zero real number for all a R. Define

Since Ox x, OR + R +. Write

R R t2 t2 R ,an orthogonal union of simple subsystems. Of course 0 permutes these. Thelemma will follow if we show that no R is fixed by 0. Suppose to the contrarythat OR i= R i. Let fl (R i) + be the highest root of R with respect to theordering R + ([10, p. 54). By the uniqueness of the highest root, Off ft. Thiscontradicts the hypothesis that there are no imaginary roots, and proves theclaim. Q.E.D.

PROPOSITION 3.12. Suppose (R, 0) is a root with involution; use the notation ofDefinition 3.10. Fix positive systems (Rn) +, (R in) + for the real and imaginaryroots. Set

0il::l p((/ilR)

+)

as in Definition 3.6; and define

Rf-- ( e

RC= R q CI Rf

Wq= W(Rq), Wf-- W(Rf), We= W(R c)(a) (R q, 0), (RY, 0), and (R c, O) are root systems with involution; they are

quasisplit, fundamental, and complex, respectively.

960 DAVID A. VOGAN

(b) Wn and Will are normal subgroups of W(R).(c) w(,e )0= ( wC)O( ( w x w’)

ill

(w) wAll of these are semidirect products, with the second factor normal.

(We use imaginary roots and real coroots in the definition of R q and Rf onlyto preserve symmetry in the duality (R, O)---)(/,-O). This result remains true of

t5 II is replaced by pll.)

Proof. We have

so/9 preserves R q, Rf, and R c. By Lemma 3.7, every imaginary root a satisfies(a,0ill =/= 0; so R q does not meet R ill, and R q is quasisplit. Similarly (or byconsidering (/,-0)) Rf is fundamental, and (a) follows. For (b), supposea R II, and w W(R). Then 0a -a, and

O(w.) w(O.) w(- .) w..

so wa R ll. Since ws,w-= Sw,, it follows that Wll is normal in W(R). Theclaim for Will is identical. For the first formula of (c), suppose w W(R). Thenw-|(Rll)+ and w-l(Rill)+ are positive systems for the real and imaginary rootsrespectively; so we can find o Wll, z Will such that

W- ’((R ll)+

(Rill)+

) (o- l(R ll)+

,’/" l(R ill)+ ),or

((R ll) +, (Rill)+

) wo-l’i’-l((R ll)+, (Rill)+

).In particular, w wo-’- satisfies

Wl(ll, tO ill) --(ll, iDill).In the notation of Proposition 3.8,

R c R (R’’")

so that proposition implies that w W(R c)= Wc. Since, w, , and commutewith , w does as well; so

W WIOT ( Wc)OWNWiN.


The first formula will follow when we show that

( WC) N wRwiR { ).

But every element of (WC) fixes (, pm); so this follows from Corollary 3.9.Applying the first formula to R q gives

(wq)o (WC)o[X WR,

so the second follows; and the third is identical. Q.E.D.

The next structures to be introduced are Z/2Z gradings. The prototype here isthe separation of imaginary roots into compact and noneompaet ones (discussedin section 2).

Let R be a root system on a vector space V. A Z/2Z gradingDefinition 3.13.of R is a map

such that

R->Z/2Z- {0,1)

((, +/)= (.) + (/)

((.) (- .)

whenever a, fl, and ct + fl are in R. Put

R0 Ro() l(0),the even roots, and

R, R,(e) e-’(1),the odd roots. The pair (R, e)--or, equivalently, the triple (R, Ro, R)--is called aZ/2Z-graded root system. Define

Wo= Wo()= W(o)

w w() { w w( ) Wo oE(R ) set all Z/2Z gradings of R.

Since we will discuss only Z/2Z gradings, we may call them simply gradingswithout danger of confusion.

LEMMA 3.14. Suppose R is a root system; use the notation of Definitions 3.3 and3.13.

(a) There are natural bij’ections

E(R ) Hom(L(R ), Z/2Z)

,()/2,(),

962 DAVID A. VOGAN

defined as follo,ws. If E(R), extend by Z-linearity to a map from L(R) to

Z/2Z. If A el(R), define (A) E(R ) by

(et) (a,X) (mod 2Z).

This depends only on A mod 2/bI, and defines the last isomorphism.(b) Suppose R + is a positive system for R, with simple roots II. Then the first

bijection of (a) defines by restriction to II a bijection

() (z/2z)",the functions from II to Z/2Z.

This is very easy, and we leave it to the reader. Most of the subtleties we mustface center on the difference between WE and W0 in Definition 3.13. Fortunately,this is very well understood, mostly by the work of Knapp.

Suppose (R, ) is a Z/2Z graded root system. The universalDefinition 3.15.q-group for is

,) w(,)/Wo(,)

(notation as in Definition 3.13). An qL-group for is a subgroup

c_ w()/Wo();

or, equivalently, a subgroup W satisfying

Wo() c_ w, c_ w(o.

The phrasing of this definition assumes that Wo() is normal in WE(C). This isvery easy to check directly; but it is also a consequence of the following result.

PROPOSITION 3.16. Suppose (R, ) is a Z/2Z-graded root system. Then there isa natural homomorphism

t(.) w:()--) z,(.e ),

(Definitions 3.3 and 3.13) with kernel exactly Wo().

Proof. By Lemma 3.14, there is a A /I(R) such that

(a) (a,)) (mod 2Z).

Suppose w W(R), and a R. Then

(W-IIX) (W--l/x, X} (mod 2)

(a, w,} (mod 2).

IRREDUCIBLE CHARACTERS OF SEMISIMPLE LIE GROUPS IV

So w preserves R0 if and only if

(a, wX- k) --= 0 (moO 2)

for all a R; that is, if and only if

1/2(wX ,) /6,(R ).

So we get a map . w()- (R)(w) (wX- X)

Suppose now that/ /(R). By Proposition 3.8 applied to/,

w/ -/x L(/ ) L(h )

for all w W. In particular, for w W2(0

/(w) 1/2(wX- x) + (w )

=_w (mod/R).Since defines X mod 2/1, we get a well defined map

() W2(e)o b,(R )/L(R ) ZI(R )

(Q(w) x(w) (mod L(R )).To see that () is a group homomorphism, suppose o, z 12(0. Then

1/2(’rX- k) /61(R );

so by (3.17),

We now compute

(x- x) + (ox- x)

(mod (R)).

(mod/(R))(mod/ (R))

(mod/ (R))(,)() + (,)(o).

963

(3.17)

964 DAVID A. VOGAN

So (e) is a group homomorphism. Its kernel is

{w W(R)I (wX- X)

By Proposition 3.8 (applied to /), this is exactly the Weyl group of the rootsystem

(. (mod 2)}

0"Its Weyl group is by definition W0(e). Q.E.D.

COROLLARY 3.17. In the setting of Definition 3.15, W0(e) is a normal subgroupof W_(e). The quotient () is canonically isomorphic to a subgroup of Z(R ), andis therefore abelian.

This is in some ways the nicest result about (), and it will be importanttechnically. However, we sometimes need the more concrete informationprovided by Knapp’s cross section for W2/W0; it will allow us to reduce someproblems to SL(2, [q) and closely related groups.

Definition 3.18. A set {al,..., at) of roots in a root system R is calledstrongly orthogonal if o "+" Oj is never a root or zero for :/:j. It is calledsuperorthogonal if the only roots in the linear span of (ai} are (___ ai}, and theseare all distinct.

Suppose a, fl R, a :/:

__fl, and (a, fl > :/: 0. Then either a- fl is a root, or

a + fl is a root (or both). It follows that a strongly orthogonal set of roots isorthogonal. A superorthogonal set is obviously strongly orthogonal. As one seesin B2, however, neither of the converse implications is true in general.

LV.MMA 3.19. Suppose R is a root system. A set {al,..., at} C_ R issuperorthogonal if and only if there is a positive system R / in which the ol arenon-adjacent simple roots. Suppose {ai} is superorthogonal. If Ro c_ R is asubsystem not meeting a }, then

W({ -+- ITi}) n W(Ro) {1 },

Proof. Suppose that (Oi} is superorthogonal. Write V for the underlyingvector space of R, and V +/- for the orthogonal complement of (ai). Choosev V +/- so that if/3 R, then (v, fl)= 0 if and only if fl is the span of {ai).Thus

R # Rl<v, #>=0} U


Put

R +=(fl Rl(v, fl>>O} U

(If V is complex, we need to choose v OR, say, for this to make sense.)Obviously s preserves R + -(al} for each j, since s v v. So each aj is simple

Y.in R +. Conversely, a set of nonadjacent simple roots IS clearly superorthogonal,as the simple roots are a basis of the root lattice.

If w W(( _+ ai) ), then w fixes the element v defined above. By Proposition3.8 (applied to v and R0), if w W(Ro), then w is a product of reflections inR0 f3 ai}. As this set is empty, w 1. Q.E.D.

PROPOSITION 3.20 (Knapp). Suppose (R, ) is a Z/2Z-graded root system. Inthe notation of Definition 3.13 and 3.15, there is a superorthogonal set

(a, at} C_ R1() such that, if we set

Q W(e) n w({ _+ 11,""" 0/}),then

W2() Q W0(),a semidirect product with the second factor normal.

Proof. Fix a positive system Ro+() for the even roots, and write t9o for halfthe sum of the roots in it. Set

Q= (wE WlwRdThe proof of Proposition 3.12(a) shows that

W). Q cc Wo(Q;

it remains only to describe Q. Obviously

O (w W21 w00

Choose a positive system R + making 00 dominant, and put

R= (a Rl(c,p0) 0).

By Lemma 3.7 (applied to R0), R oo C_ R. Since P0 is dominant, R P0 is spanned bythe R +-simple roots (al,..., at) contained in it. Suppose a is adjacent to aj.Then a -I- aj RP_ R1. But a and aj also belong to R, so a @ Oj R0, acontradiction. So the a are all non-adjacent. By Lemma 3.19, they aresuperorthogonal; so

ROo ( -t- Oi)

By Proposition 3.8,

W(R) {w e wl WOo oo).

966

So

Definition 3.21.

DAVID A. VOGAN

Q=W2NW(R). Q.E.D.

Suppose R is a root system. A cograding of R is a map

g" R- Z/2Z

such that the map.-z/2z,

is a grading of/. Define

(Definition 3.13), and

() 8(,) (R)

(i 0,1)

(i 0,2).

g )= w,_(g )/Wo(g ).

(Definition 3.15). An q-group for is a subgroup

c_),or, equivalently, a group W1 satisfying

Wo( ) c w c w( ).

All the definitions and results about gradings will be applied to cogradings(with appropriate minor modifications) without comment.

Definition 3.22. Suppose R is a root system on V. A weak bigrading of R is atriple g (, ,/) such that

(a) O is an involution of R (Definition 3.10; use that notation).(b) E(Rin), ,(R R) (Definitions 3.13 and 3.21).

We write =/9(g), etc. The set of all bigradings of R is written (R). Thenotation of Definitions 3.10, 3.13, 3.15, and 3.21 will be extended to bigradings infairly obvious ways; for example, Wra(g) denotes the subgroup W2() ofW(Rin). A strong bjgrading of R is a five-tuple (/9, , Wn, g, W), such that ifwe write g (, , ), then g (R); and

Wo(g) c_ Wl_

W2(g).

The set of strong bigradings of R is written (R). Again we may write W(),O(g), etc.


To every representation of G with infinitesimal character X (cf. (2.6)), we aregoing to attach a strong bigrading of the abstract ordered integral root system(Ra,(Ra)+ ) (Definition 4.12). The duality of Theorem 1.15 will correspond tothe next definition.

Definition 3.23. Suppose R is a root system, and

The dual strong bigrading is, (-o,, w,, w) ( ).

Here

for a R R(O)__ [/ iR(_ O)]v; and similarly for .The main point is still that of Definition 3.10--replacing 0 on R by -O on/.

The other structure (which will not appear at all in complex groups, for example)can to some extent be regarded as a technicality.As was mentioned in the introduction, we are going to make use of several

ways of producing new representations from old. The first of these involves onlythe Weyl group, and we can describe the formal part of it now.

Definition 3.24. Suppose R is a root system. The cross action of the Weylgroup W(R) on strong bigradings is defined as follows. Suppose

g (O, , W,, W) (R ),

and w W(R ). Put

w g= (wOw-,w. ,,wW-’,w. LwW-).More explicitly, w. is defined as a function on the wow-l-imaginary roots by

(w. ,)()= ,(w- ).For this to make sense, we need to know that ct is wow- 1-imaginary if and only ifw-la is 0-imaginary; that is,

(wOw-)() .O(w-) w-.This is clear, and the rest of the verification that w , (R) is similar.Obviously the cross action is defined on (R) as well. We can regard E(R) and

968 DAVID A. VOGAN

/(R) as subsets of (R), by

eo (id, e,)(3.25)

--) (- id, , ),

where e E(R), (R), and is the unique function from the emtpy set toZ/2Z. In this identification, E(R) and/(R) are stable under the cross action; sowe have cross actions on E(R) and/(R).We will give a representation--theoretic interpretation of the cross action in

the next section. The cross action in E(R) may be slightly more familiar.Suppose R is isomorphic to the root system of a compact Cartan subalgebra o ofour real Lie algebra %. Then A(g, t) is graded, with the even roots the compact

ones (defined before (2.2)). So the isomorphism R A(g,t) defines a gradingE(R). But qr is not canonical; and if we modify it by an element w W(R)

(replacing by q w-), then e is replaced by w e. It is not very hard to checkthat this construction sets up a bijection

(real forms of with a compact Cartan - (3.26){orbits of the cross action of W(g, ha) on E(Aa))

here h h(g, )a) is the (abstract) root system defined before (2.6). This is relatedto Kac’s proof of the Cartan classification of all real forms (see [9], Chapter 10).(There is an important technicality to bear in mind in interpreting (3.26); see theremarks after Table 16.4(h).)

4. The cross action on regular characters.

Definition 4.1 ([25], Definition 8.3.1). Suppose 7 (F,) Hx (notation(2.12)), and w W() (notation 2.5(a)). Write

w=- ] na,x(,)

for some n Z; this is possible by Proposition 3.8. Define p,., P.nr as in (2.10),using w instead of ,. Write

p- 2pnt =pm- 2pmc

for some m Z. Define w ,[, w cross ,, by

w (w r,w) G Hx

w x r r- ]E (n. + m.)a ;here the roots a are interpreted as characters of H.


The most serious checking required in this defintion is of the fact that any sumof roots )t b* is the differential of a unique sum of roots in H. This is notobvious, since the particular expression na is certainly not unique. Adetailed discussion of the definition may be found in [25]. It arises naturally inthe study of tensor products of finite dimensional and standard representations;but for us, only formal properties matter.

/raDefinition 4.2. Suppose , Hx, and w (the abstract integral Weylgroup; see (2.6)). Recall the element wv W(7) of (2.7)(c); and define w ,, wcross by

w

the action on the right being that of Definition 4.1. (This is an action by (2.8),which also explains the inverse appearing in the definition.)

Because we used such a naive definition of the "abstract" Cartan Da, there istechnically a danger of ambiguity in this notation. It could happen that Dacoincides with [9, and so w might itself belong to W(). In this case thedefinitions of w ,/given by (4.1) and (4.2) differ. We could avoid this by callingthe second ,,a,,, for example; but since it will always be clear from contextwhich action is meant, this is unnecessary.

PROPOSITION 4.3 ([25], Theorem 9.2.11). If Hx and w Wa, then ()and (w ) belong to the same block (Definition 1.14).

Of course it suffices to prove this when w is a simple reflection, and that is thecase treated in [25].We turn now to the construction of a bigrading of the integral roots attached

to the regular character 7--(F,) Hx. The most delicate part is defining acograding of the real integral roots. Recall from (2.4) the element m Hattached to the real root a. Since m has order two,

r(m.) _+ 1. (4.4a)

If a,/3, and , are real roots, and +/ , then

mm mr (4.4b)

([25], Corollary 4.3.20). Thus (-1)) F(m) is a cograding of the real roots.However, the Jantzen-Zuckerman translation principle (see [25]) suggests thatany construction which has representationmtheore,tic meaning ought to beinvariant when ), is changed by a weight A H of a finite dimensionalrepresentation of G. Now for such A,

A(m) (-1)(&dA), (4.4c)

970 DAVID A. VOGAN

by calculation in SL(2, Iq). This indicates that we should consider the cograding

(-1)4(")= F(m.)(-1)(’),

now defined only on the integral real roots. The objection to this is more subtle"it is not well behaved under Langlands functoriality--that is, when G is changedin certain natural ways. (This problem manifests itself in more obvious ways--forexample, in the formulation of the result in [21] on reducibility of the standardrepresentation r(7). There what enters is not the sign of F(m,)(- 1)<’’>, but themore complicated invariant of Definition 4.5. However, the explanation of thisfact, as can be seen by inspecting the proof, is what happens when G is changed.)

Definition 4.5 ([25], Definition 8.3.11). Suppose H TA is a 0-stable Caftansubgroup of G, and , (F,) ’ is a regular character. For each real root a,define = +_1 by [25], Definition 8.3..11. if a is integral for , defineg(T)(a) 8(a) E Z/2Z by

(-)()= r(m.)(-)<’>this is the cograding of the real integral roots defined by T.

It is not obvious from the definition that g(3’) is a cograding. As is shown in[25], we can find a subgroup L c_ G (the centralizer of To), containing H, and aregular character for L

r0 (r0, 0)

so that(a) If a is real, (, ?- ?, 2Z(b) (.r(m.)= %r.(m.) (4.6)(c) :--1.

(The first two conditions hold for some larger class of subgroups L _D H, and arcrelated to the functoriality discussed earlier. The third depends on theassumption L G r.) By (4.6)(a)-(b), the maps g(3’) (defined for G) and g(3’o)(efined for L) coincide. By (4.6)(c), and (4.4)(a)-(b), g(3’o) is a cograding; so8(3t) is as well.

Definition 4.7. Suppose , ’. Define

(i

the real integral roots not satisfying the parity condition for 3’, and

R(-/) g (y)-’(1),

the real integral roots satisfying the parity condition for .Thus the "parity condition" on a real root relates the parity of (ti, > and the

sign of I’(m).


LEMMA 4.8 ([25], Lemma 8.3.17). Suppose ,/ H, and a is a real integral rootnot satisfying the parity condition. Then

Here the action on the left is the cross action of Definition 4.1; and that on the rightis the obvious action of the real Weyl group W(a, It) by conjugation.

LM 4.9 ([25], Lemma 8.3.3). Suppose H TA is a O-stable Caftansubgroup of G, and I’. Write G for the centralizer of A in G. Ifw W(G, H), then

wXy=w.y.

LEMMA 4.10. Suppose H, and w W(V) (notation 2.5(a)). Then

Ril(y) Ri(w )< ) (i 0, 1)

(Definition 4.7); that is, the cross action does not change the set of roots satisfyingthe parity condition.

Proof. This is immediate from Definitions 4.1 and 4.5, and (4.4c). (We haveonly to notice that if fl is real and a is imaginary, then (a,/3) 0. Therefore theterm ea(m,)ma in the definition of w F does not affect (w F)(m).)Q.E.D.

PROI’OSITIOS 4.11. Suppose H TA is a O-stable Cartan subgroup of G, andHx (notation (2.12)). Then defines in )* an ordered root system(R(y),R + ()) with strong bigrading g(),) (0(7),e(y), W(3),((),), Wl(’/))(Definition 3.22) as follows.

R + (y) (c R(V)I <a, V> > 0}O(V) O Cartan involution on )*

Rn(7) compact imaginary roots

Rn(V) noncompact imaginary roots

e(V) E(R in) corresponding grading (Definition 3.13)

gon(V), g(V), (Y) as in Definition 4.7

W(’y) W(R i) CI W( G, H )

Wl (,/) { w c W(R ) w x " w "r }

972 DAVID A. VOGAN

Proof. That e is a grading of the imaginary roots is an elementary fact,discussed before (3.26). Since the compact simple reflections lie in W(G,H) andgenerate (by definition) WR(7), we have WR(7)_ W(7). Since W(G,H)preserves the set of compact roots, W(7) c_ Wn(7). That g is a cograding wasestablished after Definition 4.5. The containments W0(7)c_ W(7) Wff(7)follow from Lemmas 4.8 and 4.10, respectively. Q.E.D.

Definition 4.12. Suppose 7 Hx. Use the isomohism v of (2.7) to define

ga() (a(),a(), W()a, ga(), W()a) e (R ),

the strong bigrading of the abstract root system defined by . Clearly this dependsonly on the conjugacy class of 7; so if is an irreducible admissiblerepresentation of G with infinitesimal character X, then we can define g() aswell.

When we construct the dual objects (, ) of Theorem 1.15, we will have

Ra() =IRa(G)]g(): [g()]

(the duality of Definition 3.23). Part of the definition of will be the requirement

(x )’= x .In order for this to be well-defined on conjugacy classes, we need to know thatw x V is conjugate to V if and only if w x is conjugate to . This suggests thefollowing definition.

Definition 4.13. Suppose y Hx. The cross stabilizer of y is the group

w,() { e w(,g) w()l x . }.Using i, we can define the corresponding group

w,() w".

PROPOSITION 4.14. In the setting of Definition 4.13 and Proposition 4.11, defineRC_ R as in Proposition 3.12 (using R +f3 R iq, R +f3 R for the positiveimaginary and real roots). Then

Wl(’) W(RC)] ( W[FI(’) X WlFl(’)),a semidirect product with the second factor normal.

Proof. By Proposition 3.12

w()= W(R)( (w’() x w()).


Now every element of W(G, H) has a representative in K, and so commutes with0. So by the definition of

w(r) w(o,H) w(r)

C_ W(R (will)(f)

We will show that Wl(’y) W(RC). Assuming this, the conclusion of theproposition is immediate from Lemma 4.9, and the definitions of Wll(7) andW(,). To prove the claim, recall the elements Sll and 5 ill of * which appear inthe definition of R c (Lemma 3.4). Choose elements ll a0, rill/to dual tothese; and set

(c centralizer of ll and r ill in G.

(This need not be a complex group, because of the presence of nonintegral realroots.) We have

A(Sc. t) {. A(. t) <’.5> (t,pia 0}Therefore

A(c, fg) R() R c.

We define a (C-regular character ,c of H by

o (r,)Fc F + (sum of positive compact imaginary roots)

Since there are no imaginary roots of H in (c, condition 2.10(b) shows that 7c isreally a (C-regular character. Also

R(7C) {a e h(C,b) 1(6,7c) e Z)

A(C, 9) CI R(),)= R c

since oiR C__ kills the roots of c. Finally, if w W(R c)0, then

w x .,,,c (w x ,)cw. ,c= (w.

by inspection of Definition 4.1. So we are reduced to proving the claim forthat is, we may assume that H is maximally split in the quasisplit group G, and

974 DAVID A. VOGAN

that there are no real integral roots. Fix a root a R c, and put

o (x 0 {.(x) 0() 0)G" centralizer of " in G.

The element w, s,so, of W(R c) has a representative in G"; and since theseelements generate W(RC) (Lemma 3.11), it is enough to show that w, .y

w, ,. So we may assume G G". In that case, 5(g,9) is spanned byorthogonal roots a and 0a of the same length. Since G is quasisplit, no rationalmultiple of a + 0a is a root; so A(g, ) is not of type B2. So it is of type A A l,

and G is complex. It follows that

H= z(). no;for every Ad(g)--inner automorphism of a complex reductive Lie algebra g0comes from Ad(G0). (Of course we only need this for SL(2, C).) Obviously w, .yand w, , agree on H0 and on Z(G), so they coincide. Q.E.D.

LEMMA 4.15. Suppose / H, w Wa, and y W(7). Then

?,(w x )= w x ()

g,(y x ,)= (,)

(notation (3.24), (4.1), (4.11), and (4.12)).

Proof. The second assertion follows from Lemma 4.10 and easy checking.The first is a consequence of (2.8) and the second. Q.E.D.

Although we will make no use of it, we may as well record here how Knapp’sresult on W(G,H) follows from Proposition 4.14.

PROPOSITION 4.16 (Knapp [16]). Let H TA be a O-stable Cartan subgroup ofG. Write R A(g, ), and use the notation of Definition 3.10 and Proposition 3.12.Then

(a) W(G,H)-(WC)Ot (WR W(GA,H))

-(W*)(W(O,H))Here GA denotes the centralizer ofA in H. The set

(b) Rq= (alla e Rq Ca*

is a (possibly non-reduced) root system in a*, and

(c) w( G, H )1o ( W* )1o w(* ).


Furthermore, in the notation of Definition 3.13,

(d) Wgn(e) C_ W(G’,H ) c_ Wn(e).

Here is the grading by compact and noncompact roots of the imaginary roots.

The group W(G’,H) depends on GIGo, and so cannot be specified in termsof Lie algebra data alone. In Proposition 10.20(a), we will describe exactly whichpossibilities can occur. A method for computing W(GA,H) is also given in [20].

Part (c) is Knapp’s result that W(G,A) is the Weyl group of a root system. Theroot system R q is in general smaller than Knapp’s "useful roots" ([16]), but thetwo can easily be shown to have the same Weyl group.

Proof of Proposition 4.16. Choose a regular character 3’ ’ with regularintegral infinitesimal character. Then R(3’)= R, so Proposition 4.14 impliesthat (WC) c_ W(G,H). (It would be more reasonable to say, "By the proofof Proposition 4.14, (wc)c_ W(G,H).") By the remarks after (2.4), WRC_W(G,H). Now (a) follows from Proposition 3.12. For (b), suppose__, fl Rq;write , fl R q for their restrictions to a. We must show that (2,/3) Z, and

sfl R q. If a is real, then a, and these follow from the fact that R q is a rootsystem. So suppose a =/= Oa. Since R q contains no imaginary roots, a + 0a is nota root. If a- 0a 3’ is a root, then it is real; and

Therefore

So we may as well suppose a +_ Oa are not roots. Then they generate an SL(2,12).Put

Wa SaSo W(Rq )0.By computation in SL(2, C),

By the first of these,

Sni wa[ R q.

976 DAVID A. VOGAN

By the second,

(,/) 2(R,

4(R, fl )/(a, a)2(a, fl >/(a, ct) 2(0ct, fl >/(Oct,<, #> <o, #> E z.

For (c), the argument just given also showed that

(wq)1o_w(,q )

_W(R)Olo,

By Proposition 3.12(c), (Wq) is generated by W(RC) and W(RR); so (c)follows. Part (d) is obvious (proof of Proposition 4.11). Q.E.D.

5. Cayley transforms: formal theory. In this section we begin the study ofoperations on graded (and, eventually, bigraded)root systems, which correspondto changing Cartan subgroups in G. On the level of root systems with involution,we will simply be modifying the involution 0 by an involution in W whichcommutes with 0. The effect of this on bigradings is complicated, and willoccupy the next two sections.

LEMMA 5.1 (D. Garfinkle). Let R be a root system, and a R. Put

R= fl RI<, #>--0}

Define g: R --> Z/2Z by

fl ) 0 if a +_ fl are not roots

) if + ae oots.

Then g E(R) (Definition 3.13).

Proof. Suppose fl, 3’ and fl + 3’ are all roots orthogonal to a. First, assumethat g(fl) g(y) 0; we must show that g( fl + 3’) 0. Suppose not; that is, thata + fl + V is a root. Since <fl + 3’, fl + V> is positive, either <fl, fl + 3’ > 0 or

<3’, fl / 3’> > 0. Suppose for definiteness that the first holds. Since fl R ,<#, + # + r> <#, # + r> > o;

SO

a+y=(a+fl+3’)-fl

contradicting g(3’) 0.Next, suppose that g(fl)= 0, g(3’)= 1. We want to show that g(fl + 3’)= 1.

Suppose not; that is, that g( fl + 3’) 0. Applying the first case with fl, 3’, fl + 3’)replaced by (-fl, fl + 3’, 3’), we deduce that g(3’)= 0; a contradiction.


Finally, suppose that g(fl)= g(y)= 1, but that g(fl + 7)= 1. Then a _+ fl,a +_ 7, and a _+ (fl + 7) are all roots. It follows that fl, ),, and/3 + , all have thesame length as a, and that a + fl has twice that length. So/3 and 7 generate anA2, and (/,) 1. Since 7 R, we compute

+ + +

This contradicts the axioms for a root system. Q.E.D.

Definition 5.2 (Schmid [20]). Suppose (R, e) is a graded root system, a R,and e(a)= 1. Set R= {/3 R l(a, fl)= 0}. The Cayley transform of (R,e)through a is the pair c(R,e)= (R,c(e)), with

(,) g;

here g is the grading of Lemma 5.1. More explicitly,

Rg fl Rol a +_ fl are not roots}

tO {/9 R ll (a, fl ) 0, and a _+ fl are roots}

R[ fl R, a _+ fl are not roots

tO fl R01 (a, fl) 0, and a +_ fl are roots}.

An "explanation" of this definition may be found in the proof of Lemma 10.9.

Definition 5.3. Suppose (R,) is a graded root system. An ordered setS a at} c_ R is called admissible if the iterated Cayley transform

cS-- Ca C a

is defined. That is, we assume that e(al)= 1, and that {a at} is anadmissible sequence in R ’.

In particular, an admissible sequence must be orthogonal. A stronglyorthogonal sequence (Definition 3.18) contained in R e-l(1) is automaticallyadmissible; this is clear from Definition 5.2. These are the most importantexamples, as we will show that any iterated Cayley transform is equivalent to onegiven by such a sequence (Corollary 5.7).

LEMMA 5.4. Suppose a, fl and 7 are orthogonal roots in R, and that a +_ fl anda +_ are roots. Then fl +_ ,[ are roots as well..

Proof. Since a, fl, and -/are orthogonal,

978 DAVID A. VOGAN

So

(- + ) (- - v) + v

is a root. Q.E.D.

LEMMA 5.5. Suppose S , 1) is an orthogonal sequence of roots in a

graded root svstem (R, ). Set

g s= (flgl(fl, ai=O, alli

n I(j j +- are roots

(i= 1,...,/); and

n, l{ j a -/ are roots }[ fl R s ).

Then S is admissible (Definition 5.3) if and only if e(ai)= n "" (mod 2) forl. In that case,

c s(e)(fl) (fl) + n (mod 2).

This is clear from Definition 5.2.

LEMMA 5.6. Suppose (R, ) is a graded root system, and S (a... at) is anadmissible sequence which is not strongly orthogonal. Choose < j as small as

possible so that a a + aj and aj a aj are both roots. Then

s,= {, ;, ;, ,)is an admissible sequence; and

Proof. Since S and S’ have the same span, R S= R s’. Write nk, nt as inLemma 5.5, and n,, n for S’. What must be shown is that

nk+(ak)--n:+e(ct) (mod2) (k= 1,...,1)

n/+(fl)---n+(fl) (mod2), all flR s.We consider only the first of these; the second is formally identical. If k < i, theassertion is obvious. Since ai’ and aj are long, they are strongly orthogonal to theother ak; so n nj 0. By the minimality of (i, j) and Lemma 5.5,

n O, .(oli)

nj= l, e(ttj) O.


So (ai’)=(aj)= 1, and the claim follows for k=i,j. If i<k<j, theminimality of (i, j) guarantees n n, 0. Suppose k > j. If a +__ aj and a

___a

are not roots, then nk n’k, and the claim holds. Otherwise, Lemma 5.4guarantees that ak is strongly othogonal to neither a nor aj. Thereforen n 2, and the claim still holds. Q.E.D.

COROLLARY 5.7. Suppose (R,) is a graded root system, and S c_ R is anadmissible sequence. Then there is a strongly orthogonal subset S’ of R -1(1),having the same span as S, such that c s() c s’().

Proof. Suppose S is not strongly orthogonal. Lemma 5.6 provides a newadmissible sequence with the same span and same Cayley transform. The newsequence consists of longer roots than the first. If it is not strongly orthogonal werepeat the process. Since the roots keep getting longer, this ends in a finitenumber (at most [l/2]) of steps. Q.E.D.

We want to show that R s depends only on the span of the admissible sequenceS. The following proposition is the key to that fact, and we will have many otheruses for it as well.

PROPOSITION 5.8. Let (R,) be a graded root system, and suppose(al,... am) and (1 n) are maximal admissible sequences (Definition5.3) in R. Then m n, and there is an element w W(Ro) such that

W<al am>-- < [1 n>"(We write (X for the span of a set X.) If both (ai) and flj) are stronglyorthogonal, then we can arrange

an equality of unordered sets.

Proof. Write S,S’ for the two sequences. To say that they are maximalamounts to saying that R s and R s’ have the trivial grading; that, cS(e) andcS’(e) are identically zero. By Corollary 5.7, we can replace (ai) and (flj) bystrongly orthogonal sequences having the same span and the same Cayleytransform. By the description of maximality just given, the new sequences willalso be maximal. So we may as well assume that (ai) and (fl.) are stronglyorthogonal, and try to prove the last claim; for everything else will follow from it.Clearly we may assume that R is simple and spans V. Write Ri---e-l(/),Wo W(Ro) as usual. We proceed by induction on the dimension of V. Ifdim V---2 and R v 0, then it is easily checked that (R, Ro) is one of the pairs(A 2,A 1), (B2,A 1) (with A consisting of short roots), (B2,A x A 1) (with A )< Athe long roots), or (G2,A A ). For G2 the proposition is easily checked; so weassume R is not of type G2 for simplicity. The difficulties are caused mostly byroots a having the following special properties (corresponding to rank two

980 DAVID A. VOGAN

subsystems of the second type mentioned above)"

((a) 1, a is short, and a 1/2( fl + f12), with fliorthogonal long roots satisfying e( fig 1.

Step I. Suppose a, fl R, e(a)= 1, and (a, fl) va O. Then either fl (Woa)(the span of Woa), or a satisfies (,). If also (a,a) (fl, fl) and e(fl)= 1, then+_ fl Woa. Obviously this can be reduced to the case dim V-- 2, and checkedcase by case.

Step II.fails. Set

If e(a)= 1, then either (Woa) V, or a satisfies (,). Suppose (,)

(W0a) n R

+/- (/3 e RI(B,

By Step I, R U +/-, an orthogonal union of root systems. Since R is simple,+/- is empty.

Step III. fie(a)= e(fl)= l, ad (a,a (fl, fl, the fl Woa. Supposefirst that (,) fails. By Step II, there is a E W0 such that (fl,a) 0. Nowapply Step I.

Next, suppose (,) holds; choose fl accordingly. By Step II applied toinstead of a, we can find a E W0 so that (fl, flwe may assume that (fl, fl) > 0. As the fl are lon and fl (which has the samelength as a) is short, this ives

(#2, fl) 1,0, or -1.

If(fl2, fl) is orO, then

so fl W0a by Step I. So suppose (#2, fl)= -1. Then

By the Cauchy-Schwarz inequality, fl =(fll- f12); so fl + f12. Sincee(a) e(fl) (f12) 1, this is impossible.


Step IV. If there are two root lengths in R l, then one of the a is long. Supposenot. Fix a long root a R. By Step II, we may assume (after replacing a by+_wa, for some w W0) that (a, al)> O. Since (a,a)> (al, al) a- a is ashort root. Since e(a)= e(a)= 1, e(a- a)= 0. Suppose a is also adjacent tosome other ai; say (possibly after replacing a by -ai) (a, ai) > O. Then a a isa root. Since a is orthogonal to ai,

(a a, a a) (a, a) > 0;

SO

(a ,) (a ai)= a a R.

This contradicts the assumption that a at) is strongly orthogonal. So{a at, a-al) is orthogonal, and a-a R0. We claim that thissequence is admissible. This will contradict the maximality of S, and prove theclaim. Since a

_(a- al) are roots, it suffices (by Lemma 5.5) to show that

a- a is strongly orthogonal to all the other ai. Since a al) is stronglyorthogonal, and (a,a- al) is not, this follows from Lemma 5.4. (The alertreader will have noticed that this last argument is essentially the standard one onequine pigmentation.)We now complete the proof of the proposition. By Steps III and IV, we may

assume (after renumbering and acting by W0 on S) that a +__ ill. Defineusing a as in Lemma 5.1, and set

-’(0) C_ R ’.

Then

ca(e)[. el,i,.

The sequences (a2 am), (f12 n) therefore satisfy the hypotheses ofthe proposition for (t’, e[,). By induction, there is a w W0(t’) c_ W0 such that

W(__a2-+ am) (’+’2, -+" & )

up to order. Since w

_W(t’), Wa a -+- ill. Q.E.D.

COROLLARY 5.9. Suppose S,S’ are admissible sequences in the graded root

system (R, e), and that S and S’ have the same span. Then c s(e) c S’(e).

Proof. By Corollary 5.7, we may assume that S= (a, at} andS’= fl,..., fit} are both strongly orthogonal. Write 8 c_ V for their mutualspan, and R N 8, a graded root system on 8. By Proposition 5.8 applied to ,there is a w W((I)0) c_ W(Ro) such that w _+ ai) flj). Suppose fl R s

3’ R 1(3’, 8) 0}. Then wfl =/3, since w is a product of reflections fixing/3.

982 DAVID A. VOGAN

In the notation of Lemma 5.5, therefore,

n I( fl - Oi are roots 1I(il w(/3

_ai) are roots)l

l( l/3 _+ WOl are roots)l

I(jl/3_

fl. are roots }1

n.By Lemma 5.5, cS()= cS’(). Q.E.D.

Definition 5.10. Suppose (R, ) is a graded roots system on V. A subspace, c_ V is called e-admissible if it is the span of some admissible sequenceS (al,..., at) (Definition 5.3); or, equivalently, if it is the span of a stronglyorthogonal subset (ill,..., fit) of R(). (Lemma 5.6). The Cayley transform of(R, ) through is the graded root system ca(R, ) (R , c()) defined by

c(R ) R 7 e R (V,) 0),

c

(Definition 5.3) with S c_ any admissible sequence. (This is well defined byCorollary 5.9.) Cayley transforms of cogradings are defined similarly, and writ-ten ca.

We turn now to some technical results about the case when the span of all theroots is an admissible subspace.

PV.O’OSmON 5.11. Suppose (R, e) is a graded root system such that the spanof all roots is an admissible subspace (Definition 5.10). Fix an admissible sequenceS (a a whose span is 8. Call a positive root system R + S-compatible iffor each i, one of the roots +_ a is dominant for R + fq (span of al,..., oti).

(a) There exists an S-compatible positive root system R +; an), other S-compatible positive system is of the form wR +, with w a product of reflections in S.

(b) There exists a positive root system, all of whose simple roots fl satisfy(fl)= 1.

(c) If + is as in (b), then there is a w Wo(e) such that w + is S-compatible.

The proof uses the following lemma. A much quicker proof of the lemma canbe given with the machinery of group representations, by comparing Gelfand-Kirillov dimensions in the Hecht-Schmid character identity. The proof givenhere is elementary, but not very illuminating.

LEM] 5.12. Suppose (R, ) is a graded root system, R + is a positive system, ais R +-simple, and e(a)= 1. Set

(R)+= R + (’IR ,


and assume that every simple root fl of (R)+ satisfies c(e)(fl)= 1. Then eitherR + or sR + has this same property for .

Proof. Every simple root fl of R / which is orthogonal to a is simple in(R)+, and so satisfies c()(fl)= 1. Since/3- a is not a root (as a and fl areR +-simple), (fl)= 1. So we need only consider simple roots of R / adjacent toa. Passing to an appropriate subsystem of R, we may assume that every simpleroot of R / is adjacent to a. This gives rise to about ten cases, which are alltreated in much the same way; so we will discuss only two of the morecomplicated ones. Suppose first that R is of type D4, and that 3’1, 3’2, and ")’3 arethe other R/-simple roots. The (R)+-simple roots are a + 3’i + 3’j, with( i, j) c_ 1,2, 3 ). By hypothesis,

(O "1-3’i + 3’j)"--1, all (i,j) c_ (1,2,3).

Since e(t)-- 1, it follows that

e(Vi)--e(3’j), all (i,j) c_ (1,2,3).

So all (3’i) are 1, or all are 0. In the first case we are done. In the second, replaceR + by s,,R +. Then 3’i is replaced by s3’i 3’i " O; and

(’(Sot3’i) ’(3’i) ""=0+1=1,

as desired.Next, suppose R is of type C3, with a the middle root, 3’1 the short end, and 3’2

the long end. Then a + 3’2 is simple in (R)+, and not strongly orthogonal toso (a + 3’2)= 0. It follows that (3’2)= 1. If (3’)= 1, we are done. Otherwise,replace R / by s(R / ). This does not affect the first part of the argument, soe(s3"2) 1. But 3’1 is replaced by s(3’)= 3’1 + a; so

=0+1=1,

as desired. Q.E.D.

LEMMA 5.13. Let R be a root system, and ( a, at) S an orthogonal set

of roots spanning R. Set

RJ= (aRl(a,a,>=O,i=j+ ,l)

JR--- (t Rl(a,(i)=O,i= 1,... ,j)

(aj+ 1, at) (q R

984 DAVID A. VOGAN

(a) Let R + be an S-compatible positive root system (defined as in Proposition5.11). Define a sequence {kl, kt) of +_ i’s by kjaj R +. Then

(1) qjaj is dominant for (RJ)+(2) R + ct R lthere is a j with (a, ai) 0 for > j, and (, hjaj) > O)(3) kjaj is simple in (j- l)R +

(b) Suppose k {k, kt} is a sequence of +_ l’s. Define R + by (a2) above.Then R + is S-compatible.

Proof. Part (a l) is the definition of an S-compatible positive root system. For(a2), suppose a R. Clearly there is a unique j so that a Rj Rj- 1. Thismeans that

( Ol, Oli) --0, > j

By (al), a R + if and only if (a,.,l,jaj) is positive. For (a3), suppose

a. B + ’, with B and , both in -R . Write

fl E mi(dPili), [ E ni(dPili),i>j i>j

with mi, n l. Since qjaj fl + , m n for > j. Since fl and , are both inR /, (a2) now guarantees that m n 0 for > j; that is, that fl and , are bothmultiples of jaj. Since R is a reduced root system, this is a contradiction.Assertion (b) is obvious. Q.E.D.

Proof of Proposition 5.11. Part (a) is immediate from Lemma 5.13(a2), (b). Toprove (b), we are going to find a sequence (tl, (l) of

__l’s, with the

following property. Define R +, JR as in Lemma 5.13, and give JR the grading

The property we want is:

Every simple root of (J-1)R + belongs to (J-1)Rl(J-le).We will choose . and verify (,) by downward induction on j. Put t 1; then(,) holds for j since (a at) is an e-admissible sequence. For theinductive step, suppose (,) holds for some j + 1. We have

Je CaJ (j- le), (** )

By Lemma 5.13(a3), Ojaj is simple in J-1R +, with either choice of hj. By Lemma5.12, (**), and the inductive hypothesis, we can make this choice so as to satisfy(,). This completes the induction; and forj 1, (,) is Proposition 5.1 l(b).

For (c), we proceed by induction on l. Choose an S.compatible positive systemR + as in (b). Define w2 W(R) by

w2R + R +.


Since all simple roots in R + and/ + are in R(e), w2 must preserve e; that is,w2 W:(e) (Definition 3.13). Define

fli= wlai (i= 1,...,l)

Since w2 W., S’ is an admissible sequence. What we will actually do isproduce an element w W(Ro) such that wfli- + ot for all i; clearly thissuffices. Since S’ is admissible, fll R1; and a W(R)fl1. By Step III in theproof of Proposition 5.8, we can find a wo W(Ro) so that

O -I- W0I"After replacing/ / by w0/ + (which replaces w2 by ww 1, and fli by Wofli) wemay therefore assume that a

___ill. Therefore R’ R ’. Since w).fll

___a 1,

and w W, w2 preserves R’ and its grading. This is also true for s,w. One ofthese elements fixes a l, and so lies in W(R’); calling it w;, we havew W(R’), and a w for > 2. By inductive hypothesis, we can findw’o W(R’) such that w’o Bi +- ai for > 2. We claim that w (acting on R ’) isthe restriction to R ’ of an element w of W(Ro) satisfying wa +_a 1. Thiselement w will then satisfy the requirements of the lemma.The claim will be proved for any element of W(R’); so it is enough to

consider a reflection st, with y R’. If 3’ and a are strongly orthogonal, then3’ R0, and sr W(Ro), as desired. Otherwise 3’ R l, and 3’ +__ a are roots; asa R1, 3’ a t Ro. Set wv s/olSt_l W(Ro). Clearly

WyO O

as we wished to show. Q.E.D.COROLLARY 5.14. Suppose (R,e) is a graded root system, and S=

{ al,... am} is a maximal admissible sequence in R (Definition 5.3). Then everyelement of W(e)/ Wo(e) (Definition 3.13) has a representative which is a product ofreflections in S.

Proof. Choose a superorthogonal set ill,..., ill} c_ Rl(e) as in Proposition3.20, and extend it to a maximal admissible sequence {ill,..., fin}. ByProposition 5.8, the spans of {oti} and .} are conjugate by W0(e); so we may aswell suppose that they coincide. By Proposition 3.20, every element ofW2(e)/W0(e has a representative which is a product of certain sg.. Therefore wemay as well replace R by the common span of ai} and flj}. By Proposition5.11, we can choose an S-admissible positive system R + so that all simple rootsbelong to R 1. Fix w W(e), and set

R + =w-lR +

then / / has the same property, by the definition of W(R). Proposition 5.12

986 DAVID A. VOGAN

now guarantees that w is of the form WsWo, with w0 W(R0), and ws a productof reflections in S, as required. Q.E.D.

Next, we want to be able to recover a graded root system from a Cayleytransform of it. This is not quite possible, but we can determine precisely theextent to which it is.

Definition 5.15. Suppose R is a root system on V, and c_ V. Set

c(R ) g a g (a,) 0).

Fix a grading e E(R 8), and define

da(e) (e E(R )[ g is e-admissible, and e ca(e)}

the inverse Cayley transform of e.PROPOSITION 5.16. In the setting of Definition 5.15, set

a

Then d(a) either is empty, or consists of exactly .one orbit of W() in its crossaction on E(R) (Definition 3.24). More explicitly, suppose =(R, Ro, R) and--(R, Ro, R) are two gradings of R such that is admissible for both gradings,and

ca() ca() .Then there is an element w W() such that

wRo Ro, wR R.

The last assertion really is a more explicit version of the first; for if such a wexists then is -admissible if and only if it is ’-admissible, and in that case theCayley transforms coincide. This is obvious.

Proof. We proceed by induction on dim. Since the span of the roots isadmissible in either grading of , Proposition 5.11 shows that we can findpositive systems +,+ for which every simple root belongs to (I) or to lrespectively. Choose w W() taking / to /; then clearly

w, x

So, replacing by w x , we may assume that


Now choose a root a (I)

_R f) R, and let " be the orthogonal complement

of a in 8. Then

by Definition 5.10. By inductive hypothesis, we can find wsuch that

w" x o(()= -().

Since w" fixes a, it is clear that

w x ()= (w" x );

so, replacing by w" x , we may assume that

c"(() c"(’). (5.17)

Now choose a positive root system R +

for some choice of w (1, s), we havemaking a dominant. We will show that

(w X e)( fl ’( fl ) (5.18)

for every simple root fl for R +. This will complete the proof. If fl is orthogonalto a, then (5.18) follows from (5.17), with either choice of w. Therefore we onlyneed to consider the simple roots fl adjacent to a; so we may as well assume R issimple. If R is of type A 1, then a fl, and e(a)= ’(a)= by the choice ofOtherwise there are exactly one or two simple roots adjacent to (but distinctfrom) a; we consider these cases separately.

First suppose there are two such roots, fl and fiE. Then R is of type A, (n > 2),and

,, #, + #= + Y v,,

with the ’)t simple roots orthogonal to a. We know that

and we have already seen that

It follows that

(#l) -I- (#2) ’(#l) -I- ’( #2)"

If (#)= ’(#1), this implies that (#2)"--’(#2), and (5.18) holds with w 1.

988 DAVID A. VOGAN

Otherwise, choose w s. Then

(.,,, x ,0( ,i , (.,o,,)(, )

e(#i)

so (5.18) holds.Next, suppose a is adjacent to a unique simple root ft. If e(fl) ff(fl), we are

done; so suppose e(fl)= ff(fl). If (&, fl) is odd, then

and (5.18) holds with w =s as in the first case. So we may assume that(&, fl)= 2. Write

# +,with the sum over simple roots yj orthogonal to a. We know that e(y)= if(y); so

() () m((#) (#)) (mod )

m

since we are assuming (fl)= ff(fl). But we also have e(a)= if(a)= 1; so mis necessarily even. On the other hand,

since a is orthogonal to all the yj, and (&, fl)= 2. So m 1, a contradiction.Q.E.D.

Definition 5.19. Suppose R is a root system, and g (O, e, ) e (R) is a weakbigrading of R (Definition 3.22). We use the notation of Definitions 3.10, 3.13,3.15, and 3.21. Suppose V,(O) is -admissible (Definition 5.10). Since isspanned by an orthogonal set of roots, the orthogonal involution w with -1eigenspace belongs to Wa. Define

c(O) O 0 w.Since V,(O), O and w commute; so 0 is an involutive automohism of R.Furthermore,

v’,(o) (x e r,(o) <x,> o).

IRREDUCIBLE CHARACTERS OF SEMISIMPLE LIE GROUPS IV

Define

989

co(g"(o))

( e g"(0) (,) 0)c(8 )= 8 e (R"(O)).

(Definition 5.10). Recall from Definition 5.15 the inverse Cayley transform

Define the Ca]le] transform of tke bigrading g tkrougk ,c( g) C 6(R ),

to be the set of weak bigradings of R defined by

(g) (0, d(),

Similarly, if u V is e-admissible, we can define c"(g).PROPOSITION 5.20. Suppose R is a root system, g (0,e,) (R), andV(0) is -admissible. Write

W() W().

(a) The Cafley transform c(g) 6(R) (Definition 5.19) either is empty, orconsists of a single orbit under the cross action of W(g) (Definition 3.24).

(b) If w W(R) is arbitral, then

x (g)=( x g).

(c) Suppose is the dual weak bigrading of (Defintion 3.23). Then

Part (a) is immediate from Proposition 5.16. The other assertions are clear, andare included mainly to set the mood for increasingly complicated versions ofthem which will appear later. From a formal point of view, the main shortcomingof this formulation is that there is no discussion of groups. They presentenough difficulties to warrant a section of their own.

990 DAVID A. VOGAN

6. Formal Cayley transforms of X groups.

L,MMA 6.1. Suppose (R,) is a graded root system, and c_ V is an

e-admissible subspace (Definition 5.10). Set

w(,) (w w()l w )

Wo(,) w:(,) w0

c(W:()) (w s W(R)lwl= 1,wl W(,)l)

(Wo()) (w W(R)lwl= ,wl Wo(,)l).

Then

w:(())_c(W:())_ c(W0(())_ Wo(c())

Proof. Suppose w W2(e, 8). Choose a strongly orthogonal sequence(al,..., at} spanning 8. Then wS is another admissible sequence spanning

so by Corollary 5.9,

()-- ()

(w-’ )

(w-’ x )

w- c(().

It follows that the automorphism of R defined by w preserves ca(e). Therefore

The second inclusion is obvious. For the last, we simply elaborate on the end ofthe proof of Proposition 5.11. Suppose a R; it is enough to show thats, ca(Wo(e)). If a R0, then s, W0(e), and s, on 8; so the claim isobvious. If a R l, then Lemma 5.5 shows that there is a root fl 8 N R suchthat a is orthogonal to fl, and a +_. fl are roots. Since a and fl are in R l,

a +__ fl R0. The element w, s,+,s,_, is the orthogonal involution witheigenspace spanned by a and fl; so w W0(e,8), and wla sl, It followsthat s ca(Wo(e)), as required. Q.E.D.

Definition 6.2. Suppose (R,e, Wl) is a graded root system with o group(Definition 3.15), and 8 is an e-admissible subspace (Definition 5.10). DefineW(e,) and ca(Wl) in analogy with the notation for W0 and W2 in Lemma 6.1.


The Cayley transform of (R,e, W,) through is (,c(e),caW,)). Similarly wedefine Cayley transformations of cogradings with groups. By Lemma 6.1,(R , ca(e), ca(Wl)) is a graded root system with o group.

The main subtlety we must face in extending this to bigradings is in thedefinition of some sort of inverse Cayley transform. We have a natural Cayieytransform defined on o groups, but it is not injective" if W1 and W; are distinct

groups for (R, e), then it can happen that ca(W1)= ca(W). Of course we hadthe same situation for gradings; but the various gradings in the inverse image ofone could all correspond to the same group G. Changing o groups correspondsroughly to changing the component group G/Go; so a multi-valued inverseimage is not natural. We begin with the machinery needed to remedy this.

Definition 6.3. A graded root system (R,e) is called principal if there is a

positive root system R / such that for every simple root a, e(a)= 1. It is calledadmissible if the span of the roots is an admissible subspace (Definition 5.11).

Obviously principal gradings of any root system exist, and constitute a singleorbit of W(R) in the cross action of Definition 3.24. By Definition 5.11, a rootsystem can have an admissible grading only if it is spanned by an orthogonal setof roots; and not every root system has this property. (An example is the systemof type A2.) With these comments in mind, we can state the basic result aboutprincipal gradings.

PROPOSITION 6.4. Let R be a root system. Suppose ,’c_ V are subspacesspanned by orthogonal sets of roots, and is maximal with respect to this property.

(a) ’ is conjugate by W(R) to a subspace of 8.(b) is equal to the span of R if and only if ZI(R ) (Definition 3.3) is a product

of copies of Z/27’.(c) A grading of R is admissible if and only if it is principal, and is the span

of R.(d) If e is a principal grading, then the image of the injection W2(e)/Wo(e)

-ZI(R ) (Corollary 3.17) is exactly the subgroup of elements of order 2. Inparticular, if e is admissible, the map is an isomorphism.

(e) A grading e of R is principal if and only if every sequence of orthogonal rootsis W(R)-conjugate to an e-admissible sequence.

Before proving this, we will show how to use it to define Cayley transforms ofbigradings with cA groups.

LEMMA 6.5. Let R be a root system. In the notation of Definition 3.3,Zl(R)--[Zl()] ^.

Here [ZI(J)] denotes the character group of the finite abelian group Zl(/ ).This is an elementary fact about lattices, which is well known.

992 DAVID A. VOGAN

COROLLARY 6.6. Suppose (, R) is an admissible graded root system, and is anadmissible cograding of R. There are natural isomorphisms

W)(Q/ Wo(,) Z,(R ) 2[ Z,( )] *--[ W2( )/Wo( )] ’.

Therefore, to ever subgroup c_ W2()/Wo(), there is naturally associated a

subgroup

Ann g c_ W2(g )/W0(g

defined by

Anne (X W2( )/Wo(g )Ix(r)= for all r W2(e)/Wo());here x(r)C is defined by the isomorphisms above. The correspondenceo% ___)Ann is bijective and order-reversing on the set of subgroups of the groups inquestion; and

Ann(Ann

This is immediate from Proposition 6.4(d) and Lemma 6.5.We can now define Cayley transforms of bigradings with e groups. Probably

this could be done in some generality; but we will confine ourselves to a simplecase adequate for our purposes (the assumption that e is trivial belov).

Definition 6.7. Suppose g (0,e, wR,(, W)is a strong bigrading of theroots sytem R (Definition 3.22), c_ V(O) is ( admissible, and the grading e istrivial. Define the Cayley transform of the bigrading g through ,

c_

to be the set of strong bigradings

g (0, o, W"(g),, c(

such that, if g (fl, (, () (R), then(a) ga (0a,,Sa) ca(g) (Definition 5.19)(b) ca(W) is given by Definition 6.2.(c) Put (I)= R (q, and W()= W((I)); thus ((I),(lo) is an admissible

cograding, and ((I), O]o) is an admissible grading. Put

)= ( wp w2c

and define

Ann c_ W2((rlo)/ W0(olo)


by Corollary 6.6. Define W(ol) to be the preimage of Anno in W:(ol,). Thenour last condition on a is that

w(go) w() w(ol).

If u c gc(O ) is e-admissible and g is trivial then cU(g) is defined analogously.

LEMMA 6.8. In the setting of Definition 6.7, fix ga (Oa, o, a) ca(g). Then

(a) Won r’l W() Wo( ) rq W() Wo(iWo(o) w()= Wo(ol)

(b) There is a natural inclusion

w()/ Wo() - w(l)/ Wo(l).

(c) The condition (c) of Definition 6.7 is satisfied by at most one qL groupWR(ga) for o.

(d) The conclusions of Proposition 5.20 hold also for Cayley transforms of strongbigradings.

Proof. For (a), write

x vI (,) 0).

Because of Proposition 3.8, we have

w(a ) ( w w(&) wl }for any root system A c_ R. Thus

W(a n ) W(a) n W().

Taking A= Ron() and A Rgn(o), (a) follows. For (b), there is a naturalinclusion

w:(o) n w()/ Wo(o) n w()--) w:(o)/ Wo(o). (,)

By definition, ca(o)= is trivial; so is maximal admissible for o. By Corollary5.14, (,) is an isomorphism. By (a), there is a natural inclusion

w(o) n w()/ Wo(o) n w()-) w(ol)/ w0(ol).

Now (,) gives (b). Part (e) follows. Part (d) is a consequence of (c) andProposition 5.20. Q.E.D.

Since the inclusion of Lemma 6.8(b) is not onto, it can easily happen that ca(g)is empty even when ca(g is not.We turn now to the proof of Proposition 6.4.

994 DAVID A. VOGAN

LEMMA 6.9. Suppose (R,c) is a principal graded root system (Definition 6.3),and R + is a positive system for which each simple root a satisfies ()= 1. Let flbe the highest root of a simple factor of R, not of type A 2,+ 1.

(a) ((fl)=(b) c(e) is principal. More precisely, every simple root a of (R)+ R R +

(notation 5.1) is simple (in R + ), and strongly orthogonal to fl so

Proof. Part (a) says that if we write

nact,simple

then n, is odd. This is easily checked (see [10], p. 66); n, + is the Coxeternumber, which is well known to be even except in type A2, + . (In the setting ofMcKay’s beautiful ideas in [19], this can be deduced from the fact the only finitesubgroups of SU(2) of odd order are cyclic.) In (b) the second formulation isobvious, and gives the first statement. Q.E.D.

We need an analogous fact for A2n+ 1"

LEMMA 6.10. Suppose (R,() is a principal graded root system, and R / is a

positive system for which each simple root a satisfies ((a) 1. Let fl be an extremalsimple root in a simple factor of R of type A,.

(a) c(fl)= 1.(b) c() is principal. More precisely, every simple root of (R/) + is simple in

R +, and strongly orthogonal to fl; so

Proof. Part (a) is trivial, and is included only to preserve the analogy withLemma 6.9. For (b), we may assume R is simple. Then one knows (or can check)that R # is of type A,__. Since there are clearly n 2 roots a as described (thosenot adjacent to fl), they must be all the simple roots of (R )/. Q.E.D.

LEMMA 6.11. Suppose (R,) is a graded root system, a R, and e(a)= 1. If() is principal, then so is .Proof. Choose (R") + making c"(e)(fl)= for each (R")+-simple root ft.

Regard (R")/ as defined by an ordering of

v (x vI (-,,) o}.

Every root but a has a non-zero restriction to V"; so this ordering defines aunique positive root system R / containing a. Clearly a is simple for R /. ByLemma 5.12, is principal. Q.E.D.


PROPOSITION 6.12. Suppose (R, e) is a principal graded root system, a R, ande(a) 1. Then ca(e) is principal.

Proof. We may assume R is simple. By Step III in the proof of Proposition5.8, we can replace a by any root fl of the same length, satisfying e(fl)= 1. If ais long, Lemmas 6.9 and 6.10 provide a fl which works; so assume there are tworoot lengths, and a is short. Let fl0 be any long root satisfying e(fl0)= l; thisexists by the definition of principal, since there must be a long simple root ineach positive system. Necessarily fl0 is adjacent to some short root a0 (that is,(ao, flo) > 0); so if we put O 0- 0’ 1 CtO- 1’ then ao and a are short,and flo ao + al. So (perhaps after interchanging ao and a)

e(ao) O, e(a,)

e(fl0) e(fl,) 1.

The sequences (a,a0) and (ill, fl0) are admissible, with the same span; so

The right side is principal by the first part of the proof; so the left side is as well.By Lemma 6.11, c’,(e) is principal, as we wished to show. Q.E.D.

Proof of Proposition 6.4. Part (a) is well known, and can be proved by asimplified version of the proof of Proposition 5.8. We leave this to the reader. For(b), suppose first that <a,..., cq> (with the c orthogonal) spans R. Then, inthe notation of Definition 3.3,

(R ) 5,= _, qiai qi O,<x,fl>zforallflRi=l

c alli )

So

{2t _qiil2qi_

Z, all i}

Z,(R ) =/I(R )/L(R ) c_ 1/2 ,(R )/[_,(R ),

which is the "only if" part of (b). The "if" part (which we will not need in anycase) can be checked case by case: the simple root systems not spanned byorthogonal sets of roots are:

g A,, (n > 1); Z(R) Z/(n + 1)ZR D2n+l (n > 1); ZI(I) Z/4Z

996 DAVID A. VOGAN

We prove (e) next, by induction on ]R]. Suppose e is principal, and (a, at)is an orthogonal sequence of roots. Since every root is W(R)-conjugate to asimple one, we may assume e(a) 1. By Proposition 6.12, c’(e) is principal. Byinduction, we can replace (a2,..., at) by a W(R’) conjugate which isc’(e)-admissible. By definition, this means that (a at) is e-admissible. Forthe converse, suppose (R, e) has the property in question. If R is non-empty, thene cannot be trivial; so fix a with e(a)= 1. We claim that (R’,c’(e)) has thedesired property. Let (a2,..., at) be an orthogonal sequence in R . Byhypothesis, there is a wE W(R) with (wa, wa2 war) admissible. Inparticular, e(wa)= 1. By Step III in the proof of Proposition 5.8, we can find

1421 Wo(e) so that

W1WOl "+’t9l.

Acting by w does not change the property of being admissible, so (_+ a,wwa2,..., WlWat) is admissible. Now wwa is orthogonal to a, so s, fixes

wwai. Taking y ww or s,ww as needed, we get

y(a, a2 Ol) (owiwo2... w1woQ).

The first property shows that y W(R), and the second that (a, ya2,..., yat)is e-admissible. By the definition of admissible, it follows that (ya2 yat) isc(e)-admissible. Thus ca(e) has the desired property, and is principal byinduction, proving (e).The only part of (c) which is not immediate from (e) is that an admissible

grading is principal. We proceed by induction on R I. Choose an e-admissiblesequence (a ,al) spanning R. Then c’(e) is obviously admissi-ble--(a2 at) is c’(e)-admissible and spans R--so c’(e) is principal byinduction. By Lemma 6.11, e is principal.

Consider now (d). Choose a positive system R + for which every simple root asatisfies e(a)= 1. Fix z Z(R) of order 2, and a representative " /6(R).Thus 2" L(R); so

qt, 2q Z. (6.13a)simple

Choose /b(R) as in the proof of Proposition 3.16. By our assumptions on eand R /,

(X,a) --= (mod 2) (a simple) (6.13b)

We seek an element w W(R) such that

1/2(wh- h) " (mod(R )); (6.14)

by Proposition 3.16, such an element will automatically lie in W2(e), and will


satisfy

as desired. Define

(,)(w)

F= (a simple lq Z).We claim that F is orthogonal. To see this, order the simple roots as (al, al)in such way that each o is adjacent to at most one of its successors, and thelengths are increasing. Suppose F is not orthogonal. Choose ci, cj F, with < jso that (ai, aj) :/: O, and is minimal with these properties. Then, writing qm q

(m,i) (mod Z)adjacent to iF

am adjacent to

F

mi,j

The first term here is 1, and the second 1/2 (since (%, %) (ai,ai)). Suppose mis in the third summation. If m < i, then the pair (am,ai) contradicts theminimality assumption on (i, j). Since j > i, a can have no other adjacentsuccessors; so m i. So the last sum is empty, and

ai) (rood Z),

contradicting the assumption that f (R). So F is orthogonal. Put

then (6.14) is immediate from (6.13) and the choice of F. Q.E.D.

7. Cayley transforms of regular characters. In [21] or [25], Cayley transformsare defined which take regular characters on one Cartan subgroup H to thoseon a second Cartan H2, such that H IN H2 has codimension one in both Hi.Here we want something more general (and therefore easier to define). Althoughwe will make little explicit use of the special case, a familiarity with it (seeSection 8.3 of [25]) would be helpful.

Definition 7.1. Let H l= T IA be a 0-stable Cartan subgroup of G, andc_ a a subspace spanned by an orthogonal set of (real) roots. (To make sense of

998 DAVID A. VOGAN

this, we identify b and b* using (,).) Define a2 to be the orthogonalcomplement of 8 in a 1. Write

M2‘42 G

for the Langlands decomposition of the centralizer of .42 in G; then TS is aCartan subgroup of M 2. Because of the assumption on 8, M2 also contains a0-stable compact Cartan subgroup T2 which is unique up to conjugacy byM2N K; we can and do choose T- so that

T2 T(.Thus

H2= T2‘42 c(H l)is a Cartan subgroup of G, the Cayley transform of H through 8. Put

L L(8) centralizer of T in M2.

Then L contains T IS as a split Cartan subgroup, and T2 as a compact Cartansubgroup; and H2 is defined up to conjugation by L tq K.

We want a map c from ( l), to (2),. Its characteristic property should be(very roughly speaking) that (c(,)) is a composition factor of r(,) whenever, (HI)’. Always one should think of the case when H is split and H2 iscompact. Then r(7) is a principal series representation, and r(c(’/)) is a discreteseries. Now most principal series contain no discrete series at all; and those thatdo generally contain several. So c can only be defined on a subset of ( 1),, andthere it will be multivalued. To describe the domain, we need to recall thenotation of Definition 4.7.

LEMMA 7.2. In the setting of Definition 7.1, suppose H 2 is compact, and H is

split (that is, that G L). Fix 3/ (H 1). Then there is a h (H2) belongin to

the same block as y (Definition 1.14) if and only if the real integral roots R "(y)span (al)*, and the cograding (l) is admissible (Definition 6.3); that is, there is asubset a 1, ar) C_ R() such that (i) is strongly orthogonal and spans (a 1)..

If such a q, exists, then q’ (I2) belongs to the same block as y if and only if(a) lz

or, equivalently, if and only if

(b) ’ W(g, 2) .(Definition 4.1).

Proof. The conditions for the existence of follow from the theory of thecompactness of a block ([25], Proposition 9.2.12). So suppose exists. Fix


,, (2). By Proposition 4.3,

(b) q,’ and - in same block (a).

So assume (a); it is enough to prove (b). Since and q/ define the sameinfinitesimal character, we can find w W(g,b2) with ’= w; here we write

q--(, q) for the regular character as usual. Define q w q. Then q q/, so

’1,0 1 ,0.By hypothesis,

Since H: H02. Z(G) (because H: is compact), these give , ,’. Q.E.D.

To extend this result to the general case, we need a rather messy definitionfrom [23] or [25]. We will not give all of the definition here, but a summaryfollows.

Definition 7.3. Let q + tt be a 0-stable Levi decomposition of a 0-stableparabolic subalgebra of g, and assume that ; here bar denotes the complexstructure defined by the real form g0. (From now on we may say "let q + ube a 0-stable parabolic subalgebra of g," with the remaining assumptionsunderstood.) Let L be the normalizer of in G (a reductive group withcomplexified Lie algebra I). Suppose H TA C_ L is a 0-stable Cartan subgroup.Write

20(u) sum of the roots of in u,

an element of *. Define m as in (2.10), and fix a positive system A+ (re, t). Thenthere is a naturally defined character

(o, A+ (m, t), ) ;this is the character of T on the vector space V* defined by Lemma 8.1.1 of [25].(Here stands for "twist". The definition of V*, which is complicated, is what weare omitting.) Now suppose T=(F,)(/’) is a regular character, andA+ (m, t) is the positive system defined by ]t (see (2.10)). Put

7. - p(u)

r.I r rl r

v0 (r.,L).

1000 DAVID A. VOGAN

[,EMMA 7.4. In the setting of Definition 7.3, "to (/,)L; that is, the map ,[-

defines a correspondence between regular characters of H with respect to G, andthose with respect to the smaller group L. Suppose now that , h (H’)x

(a) /()= 6(y) (Definition 4.5); that is V and V define the same cograding ofthe real integral roots (all of which lie in A(I, )).

(b) If o and cho lie in the same block for L, then and h lie in the same block

for G.

Proof. That , is a regular character for L is Lemma 8.1.2 of [25]. Part (a) isLemma 8.3.12 of [25]. Part (b) is not explicitly stated in [25], but is established inthe proof of Proposition 9.3.1 there. Q.E.D.

The converse of (b) is false: -/, and need not even have the sameinfinitesimal character when , and do, since W(, 9) is smaller than W(, 9).The correspondence ,--> 7, on the level of group representations, is a simple(but only partly understood) case of Langlands’ functoriality ideas. Since theparabolic c is not defined over Iq, the correspondence is not realized by ordinaryinduction. See [25] and [21] for some details about it.

Definition 7.5. In the setting of Definition 7.1, choose a 0-stable parabolic-- / tt of I122; here we want to be the centralizer of in m2, as in Definition7.1. Suppose 3’ (1). We say that 3’ (R)(ca) (the domain of ca) if isadmissible (definition 5.11) with respect to the cograding g(/) of Definition 4.5.(To make sense of this, we identify with a subspace of b* using ( ).) Suppose-/ 6(c). The Cayley transform of through 8, ca(y), is the set of all in (2),satisfying the following conditions. Write (notation after (2.12))

Then we want(a) @IA2

lie in the same block for L.(b) @ and .LEMMA 7.6. In the setting of Definition 7.5, suppose , (F,?) (R)(ca), and(.) (2). Then c() if and only if

(a) and are conjugate by Ad(I); and(b) [ z<) differs from rl by a sum of roots of Z(L) in .Proof. The necessity of (a) and (b) is straightforward, and we leave it to the

reader. So assume (a) and (b). The twists z of Definition 7.3 are sums of roots ofT in . It follows that conditions (a) and (b) hold with and replaced by3’, ’o. By Lemma 7.2, it follows that there is a regular character (xI,,ff)

[(2),]z, lying in the same block for L as ,,, and satisfying


Since regular characters in the same block agree on the center of the group, Fand H agree on Z(L). By (b),

lz) H + mc] (**)a A(o,2) Z(L)

By (*), the differential of ( ma)lZ(L) is zero. Choose a positive system + (, )2)compatible with . Then we can write

m ot n#fl.fl simple

The simple roots in are zero on the Lie algebra of Z(L), and the rest of thesimple roots are linearly independent there. We therefore conclude from thedifferentiated version of (**) that ma is a sum of roots in L Since these aretrivial on Z(L), (, ,) gives

Since T2 is a compact Cartan in L, this condition and (,) give q ft. By thechoice of q, hypothesis (b) of Definition 7.5 follows. Hypothesis 7.5(a) isimmediate from condition (a) of the lemma, since Ad(D centralizes a2. Q.E.D.

COROLLARY 7.7. In the setting of Definition 7.5, suppose y (R)(ca) andw Wa. Then

x w x

(notation 4.2)).

This follows from Lemma 7.6 and the defintion of w y.

PROPOSITION 7.8. In the setting of Definition 7.5, suppose (R)(ca). Thenca(y) C_ (H2) consists of exactly one orbit of W( + ,) under the cross action.

Every element of ca(y) belongs to the same block as y.

Proof. The first claim follows from Lemma 7.2(b) and the definitions. For thesecond, suppose q, ca(7); define q, 3, as in Definition 7.5. By that definition,

belong to the same block forL. By Lemma 7.4(b), q and belong toq and ythe same block for M2. Since also qblA2 ylA2, the proposition is a consequenceof the following simple result.

LEMMA 7.9. Suppose P MAN is a parabolic subgroup of G. Let Hi-- TiA(i 1,2) be O-stab& Cartan subgroups of G, satisfying A

_A (so that H C MA).

Fix y ( )’, (2),. Write

Assume that(a) and are nonsingular.

1002 DAVID A. VOGAN

(b) ’[A [A, and(c) 71 and 61 lie in the same block for M.Then and lie in the same block for G.

Proof. By Lemma 9.2.7 of [25], we may as well assume that t(6 l) is asubquotient of %t(7 l) (notation (2.11)). By induction by stages, Indea[%t(7 l) (R)

VIA (R) 1] has the same composition series as r(6); and (6) is a subquotient ofinde[et(6 1)()61,, () 1]. So e(6)is a subquotient of ,r(,/). By Definition 1.14, 7and are in the same block. Q.E.D.

Next, we want to give another perspective on the definition of. ca(,). Beforedoing this, we should recall exactly how canonical ca(7) is. The only choiceinvolved is in the construction of H2; and this is unique up to conjugationbyL N K. In some sense, the Cayley transform ought to be defined on conjugacyclasses of regular characters. The next definition accomplishes this.

Definition 7.10. Suppose (/l), and a a (the abstract Cartansubalgebra--see (2.6)). Write iv(a) (notation (2.7)). We say that (R)(ca.) if

is a(7)-admissible (Definition 4.12); or, equivalently, if (R)(ca). In this casewe define

the Cayley transform of through . In the notation of (2.14), it is clear that

depends only on c/(); we may therefore call it ca.(cl()).LEMMA 7.11. Suppose 7 I’, a C_ a, and w W (notation 2.6). Then

w (R)(Cw,o).. (R)(C,o);and if these hold, then

Cw,o(w x w x

This is immediate from Corollary 7.7.There are two other basic properties of Cayley transforms which we need.

Although they appear more or less unrelated, it is most convenient to prove themsimultaneously. The first relates our present definitions to the formal ones givenearlier, and the second concerns iterated Cayley transforms.

LEMMA 7.12. In the setting of Definition 7.10, suppose 7 (R)(cao), ande# cao(7). Then, in the notation of Definition 4.12

oo(,) (Definition 5.19)

(R")n(q) { fl (Ra)m(7)l<fl,>-- O} (Ra)Fl(]t)a"

(Definition 5.1O)


LEMMA 7.13. Suppose a____ l.Ia ])Da_ a (orthogonal direct sum), y (I)’x,and ,/ (R)(c.) N (R)(cuo). Then

(a) c,o() @(CDa)(b) c,(ca()) c().We will need an auxiliary formal result.

LEMMa 7.14. Suppose (R, e) is a graded root system on V, and u 3 t c_ V(orthogonal direct sum). Assume that and u are e-admissible. Then t is

cu(e)-admissible, and

Proof. Choose an e-admissible sequence U= (al,..., a,) spanning u. ByProposition 5.8 applied to R t3 ,, U can be extended to an e-admissible sequenceS (a ar,ar+l,’’’, a) spanning . Then V= (ar+l,..., a) spans theorthogonal complement t of tt in . The lemma is now obvious from Definition5.10. Q.E.D.

We prove Lemmas 7.12 and 7.13 at the same time, by induction on dim . Ifthis is zero, there is nothing to prove; so suppose dim, > 0, and the results areknown for lower dimensions.

Proof ofLemma 7.13. If u or Da is zero, there is nothing to prove; so supposeboth are non-zero. Then dim tt < dima; so if, c.a(y), we have

by the inductive hypothesis for Lemma 7.12. Now (a) follows from Lemma 7.14(or rather its analogue for cogradings, which is equivalent). For (b), we define, u, t c_ bt as in Definition 7.10. Choose the Cartan subgroups H2 c(H 1) andn3 cu(H l) in such a way that

Then H2 c(H 3). Since

we find

.41---_ UA3_-SA2= UVA 2

AIA3A2.

L()A 2 centralizer of TA 2

L(u)A 3 centralizer of TAL()A 2 centralizer of TA 2,

1004 DAVID A. VOGAN

As these groups share the Cartan H3, we deduce that

Using Lemma 7.6, one sees that (b) is a formal consequence of (a), (,), and (**).Q.E.D.

Proof of Lemma 7.12. The first two assertions are immediate. For the third,the case dima= is (9.2.15) of [25]. In general, the first conclusion ofProposition 7.8 shows that we may replace by any other element of co().Choose an admissible sequence S (al, a spanning a, and choose

The iterated Cayley transform is defined at each step by the case dima 1, and

by the case dim= and Definition 5.10. By Lemma 7.13 (which is nowavailable for this dimension of a), belongs to c.(7). Q.E.D.

We turn now to the "co" theol. In the formal version of Section 5, resultsabout cogradings were equivalent to results about gradings, because of theduality of Definition 3.23. Now, however, we are trying to create such a dualityfor regular characters. All the results must therefore be proved again, often bysomewhat different methods. This is not particularly hard--in fact most of it willbe left to the readerbut it is not trivial.

Definition 7.15. Let H= TZA be a 0-stable Cartan subgroup of G, andu t a subspace spanned by a strongly orthogonal set of noncompact(imagina) roots. That is, we assume that u is -admissible, with the grading ofProposition 4.11. Put

M centralizer of f in G,

so that H is a Cartan subgroup of M . Let M2A be the Langlandsdecomposition of the centralizer of A in G, and put

L L(u)=M M.By [20], for example, there is a 0-stable Cartan subgroup

c(H)= H= TA ;here is defined above, so H is split in M . H is unique up to conjugacy byL & K; we call it the Cayley transform of H through u. Write

S A q M2, Lie(S ).


Then

CI , + CI2,

an orthogonal direct sum, and H2- c(H1).Notice that now we need to worry about the grading even to define cU(H); for

ca, the grading entered only in the definition c(,). This is complemented by thefact that c will be defined on all of (2).

Definition 7.16. In the setting of Definition 7.15, suppose (/2). Usingthe notation of Definition 7.5, we define the Cayley transform of through u,c"(q), to consist of those () such that

(a) yla la2and(b) yo lie in the same block for L.

In the definition, () is a discrete series representation of L, sinceH L T is compact. By Harish-Chandra’s subquotient theorem for L,() is a constituent of some ordina principal series rpresentation (y) of

belongs to [(TS)’], and lies in theL; the corresponding regular character Y0is justified" by Corolla 8.13 ofsame block as by definition. The notation yo

[25], Vo is really constructed for some V [(TiS)’], which can then beextended to y (D) by the requirement (7.16)(a). Then y c(), so theCayley transform is nonempty. Suppose V cU(). The theo of compactness ofblocks ([25], Proposition 9.2.12) guarantees that must be (y)-admissible, andtherefore (v)-admissible by Lemma 7.4. In light of the formal resemblancesbetween Definitions 7.16 and 7.5, Lemma 7.6 implies the following result.

LE 7.17. In the setting of Definition 7.15, suppose 0 (D2). Then

{ 63(c,)1 and are conjugate by

Ad(i), and rl differs from ql Z(L)

by a sum of roots of Z(L) in g).

COROLLARY 7.18.W Wa. Then

In th setting of Definition 7.5, suppose q (i?t) and

c"(w x w x

This is proved in the same way as Corollary 7.7.

PROPOSITION 7.19. In the setting of Definition 7.5, suppose 0 (/2). Then

c(,O) consists of the W(LA,H l) conjugates of a single orbit of W( + a2, bl) on(H )’ in the cross action. Every element of cU(qO belongs to the same block as q.

1006 DAVID A. VOGAN

Proof. Just as in Proposition 7.8, one reduces to the case G L. Then thesecond claim is part of the definition of cU(q0, and the first follows from Lemma9.3.12 of [25]. Q.E.D.

COROLLARY 7.20. In the setting of Definition 7.5, suppose q (/_2), and(R) (c). Then

(a) cc()= W(I + a2, 92)(b) c"c,(/) W(LA 2, n ). W(I + ct2, 92) ].

Since we are interested only in conjugacy classes of regular characters, theextra conjugation action on the outside in Corollary 7.20(b) does not seriouslydetract from the symmetry of the result.

Definition 7.21. Suppose q (H)x, and a C a. Write u i7(a) (notation2.7). We say that q (R)(c *a) if a is ea(q)-ad-missible (Definition 4.12); or,equivalently, if u c_ b actually lies in the compact part 2, and satisfies thehypotheses of Definition 7.15. In this case we define

c,(,) c.(,t,).

Just as in Definition 7.10, we write

c(cZ(,))

LEMMA 7.22. ^2 a Wa.Suppose H )x, C_ ba, and w Then

w 4, (cW)., (R)(c);and if these hold,

wO(w x )= w x c().

This is a consequence of Corollary 7.18.

PROPOSITION 7.23. Suppose ,/ (H 1),, (/. 2), and a a. Then thefollowing conditions are equivalent.

(a) 7 (R)(c0o) and cl(qO co(cl(T))(b) q (R)(ca) and cl(t) c(cl(q)).

Write W(a) W(R n ) as usual. If (a) and (b) hold, then(c) c"co(cl(l))= W(a) x cl()(d) CaC"(cl())= W()a) X cl().

This follows from Lemma 7.17 and Corollary 7.20.


LEMMA 7.24. In the setting of Definition 7.21, suppose (R)(ca) andy c(). Then, in the notation of Definition 4.12,

0 (y) C3a (0)(t) Definition 5.19)

(R )i(y) ( _. (R )iFl(,) (,a) O) (R )iFl(*)a

(Definition 5.20)

Just as before, it is most convenient to prove this with

LEMMA 7.25. Suppose a Ha Da a (orthogonal direct sum), 0 (I)’x,and k (R)(ca) fq (R)(c’) Then

(a) cDa() O(CLia)(b) c(c’()) c_ c().The two lemmas are proved simultaneously, as before. In this setting, the case

of Lemma 7.24 when dim lis due to Schmid [20]; his proof appears in [25],Proposition 5.3.1.

PROPOSITION 7.26. Suppose a a, and y I2I’(a) If 7 (R)(ca), then

ga(ca(y)) Co(ga(7)) (Definitions 4.12, 5.19);

that is, the set of weak bigradings of R defined by the elements of c,(y) is

precisely the (formal) Cayley transform of the weak bigrading defined by ,through a.

(b) If y (R) (c"), then

ga(c(y))=ca(ga(y)).

Proof. By Lemmas 7.12 and 7.24, and (a)*(b) in Proposition 7.23, we find

ga(ca(y)) C_ C.( g(y)),

and similarly for (b). Now equality follows from Proposition 7.23(c) and (d), andProposition 5.20. Q.E.D.

The problem of extending this result to strong bigradings is most convenientlydealt with later.

8. A standard form for blocks. As was mentioned in the introduction, weintend to describe each block in terms of two special elements of it. Themachinery to do this is now in place. Because the proofs are tedious, we willbegin this section by formulating all the major results before proving any ofthem.

1008 DAVID A. VOGAN

Definition 8.1. Suppose q Hx. The c-packet of q, cp(q,), consists of allconjugacy classes of regular characters meeting the orbit of q, under the crossaction of wiR(q), the Weyl group of imaginary roots (notation 4.11). (Here cstands for compact.) That is

1)() ( wiB() cl(@) ).If C ha, and @ E @(c) (Definition 7.11), we set

W() cp(c()) c();

the last containment is Proposition 7.8. Similarly, we define

rp() wFl() ) cl()

Analogous definitions are made with q, replaced by a bigrading of an abstractroot system.

If r() is a discrete series, then (r(7) 7 cp()) consists of all other discreteseries with the same infinitesimal and central character as r(q,); andrp() cl(), the conjugacy class of . If r(q,) is a principal series representationfor a split group, then (r(7)[ 7 rp()) consists of all other principal seriessharing a composition factor with r(q,). In general, cp() coincides with theLanglands L-packet of q, (on the level of group representations). We have notused his terminology in order to keep clear the duality between c-packets andr-packets.A number of facts about packets are obvious but worth stating.

LEMMA 8.2. The c-packets partition each block with infinitesimal character X;that is, haying a common c-packet is an equivalence relation. The abstractcograding a(f) is constant on c-packets; so if q @(c), then cp(q) c_ (R)(c).Analogous results hold for r-packets.

We will leave the verification of this to the reader. For our purposes, the mostimportant property of packets is

PROPOSITION 8.3. An r-packet and a c-packet can intersect in at most oneconjugacy class.

We will parametrize each block by parametrizing the r-packets and c-packetsseparately, and describing which have a non-empty intersection.

A!Definition 8.4. Suppose y Hx. In the notation of Proposition 4.11, we saythat 3’ is almost minimal if the cograding (7) is trivial, and almost maximal if thegrading e(7) is trivial.

Thus y is almost maximal exactly when H is maximally split in G; this meansthat r(7) is an ordinary principal series representation. If r(7) is a discrete series,


or even a fundamental series representation, then is almost minimal; but theconverse is not true. This is dual to the fact that a maximally split Cartan neednot be split, or even quasisplit.

THEOREM 8.5. Fix a block B of regular characters with infinitesimal characterX (cf. (2.17)). Then we can find regular characters 3’i B (i 1,2), such that

(a) 3’1 is almost maximal, and 3"2 is almost minimal.These elements are uniquely determined up to conjugacy and the cross action of Wa.Fix any other element q B. Then we can find i c_ ba, w W (i 1,2), suchthat

(b) w 3’! (C.l), w2 X y2 { o--]O(c..2)(c) q, E c,(cl(w 3"1)) rq c2(cl(w2 3’)) (Definitions 7.10 and 7.21).

In particular, we can find #, w so that(d) cp()= cp,(w 3") rp()= rp2(w2

3, 2) (Definition 8.1).

Now Proposition 8.3 shows that B is parametrized by quadruples2), via Theorem 8.5(d). To make this more precise (so that it can be used in ourproposed duality theory) we need a formal description of which quadruplesoccur, and which define the same

PROPOSITION 8.6.and

Define

Suppose 3’ Hx is almost maximal, 8, u C_ al, w, y Wa,

w x (c),

a(,) {x 9a 6(Oa(w X 3")).X, --X}.(Definitions 5.19 and 4.12)

Wct(g)T waIT[(g) }.

Then the following conditions are equivalent(a) cp(w X y)= cp,(y X y)(b) There is a W() such that ru , and w-zy W(y) (Definition

4.13).Analogous results hold for c when is almost minimal.

The point of this result is not that it is particularly understandable, but that itis phrased entirely in terms of the strong bigrading a(y) (in light of thedescription of W(y) in Proposition 4.14).

PROPOSITION 8.7. Fix a block B, and y (i l, 2) as in Theorem 8.5. Supposei a, W Wa, and

WI X vle @(C,), w2X 2e @(c2).Then the following conditions are equivalent.

(a) Cp,(w X V 1) n rp2(w2 X V2) 0(b) Cpl(W X ga(Tl)) n rp2(w2 X ga(T2)) O.

1010 DAVID A. VOGAN

Here in (b) we use the notion of packets for (weak) bigradings mentioned in

Definition 8.1.

We now begin the proofs.

Proof of Proposition 8.3. Suppose i Cp(()) ("] rp(y), i= 1,2. Since and 2are both in cp(q),

t(’) t(w ’),with some w wiR( ). Since they are both in rp(,),

ct() ct(y ’),with some y WR(j ). Thus

so y

cZ(w x t ’)= cz(y x t ’),--Iw f WI( 1) (Definition 4.13). By Proposition 4.14, w Wn( ), so

cZ() d(w x ’)= cZ( ’),

as we wished to show. Q.E.D.

For Theorem 8.5, we need some a priori control on the size of B. This isprovided by

THEOREM 8.8 ([25], Theorem 9.2.11). The blocks of regular characters with

infinitesimal character X are the smallest sets closed under conjugacy, the crossaction of Wa, and the Cayley transformations of Definitions 7.10 and 7.21.

aLEMMA 8.9. Suppose q H.. Choose a maximal (q,)-admissible subsvace2C_ , and a maximal (q)-admissible subspace c_ (Definitions 4.12 and25.10). Then q (R)(c f (R)(c2); choose .[ c (q,), y c2(q) (Definitions 7.10

and 7.21).(a) 3, is almost maximal, 2 is almost minimal, (R)(c,), 2 (R)(c) and

cl(,) co,( d(’)) n c( c(,))(b) The W" cross orbits W" X cl(y ) and W X cl(T2) depend only onW cl(d?) (and not on an), of the other choices in the definition of yi).

Proof. That belongs to the domains of the Cayley transforms in questionfollows from the choice of . That -1, is almost maximal and 3,

2 almost minimalis a consquence of Proposition 7.26 (compare the remark at the beginning ofProposition 5.8). Now (a) follows from the first part of Proposition 7.23. For (b),suppose tt is maximal admissible for .a(w X (), and c"’(w x q,). We want toshow that ( W x c/(3,). By Lemma 7.22, w-tt is maximal admissible for


IHIea(), and w x c (p). By Proposition 5.8, there is a y wlq(t)a; suchthat w-u =y. By Proposition 4.14, y W(9)a; so cl(q)=cl(y x 9), andProposition 7.23 gives

c (cZ(y x

=y x

Therefore (wy) X c(cl(e)); and by definition, ,,/1 (l(di))). By Proposition7.19, l W X c/(/l), as we wished to show. The proof for is identical.

Q.E.D.

LEMMA 8.10. The regular characters y of Lemma 8.9 depend only on the block

of , up to conjugacy and the cross action of Wa.

Proof. By Lemma 8.9(b), W X cl(’ i) depends only on W X C/(). ByTheorem 8.8, it therefore suffices to prove the following claim" suppose there isanother regular character q, a subspace ttc_Oa, that @ (R)(c), and thatq c(q). Then and define the same elements 3, i, up to conjugacy and Wa.(Once this is proved, Lemma 7.23 will give the analogous result for cu, byinterchanging and q.) Consider first /l; say we define it by choosing gmaximal admissible for ea(q), and choose

We claim that this same ,,1 works for . Since tt is a()-admissible, u iscontained in the -1 eigenspace of oa(q). Similarly is contained in the +eigenspace, so and tt are orthogonal. We know that

By Lemma 7.12, a(q)= ,(ga(tl)); and tt is ga(+)-admissible by assumption.The inductive form of the definition of admissible now guarantees that @ u isga(yl)-admissible. By Lemma 7.13, c,u(-l). By Lemma 7.23, 81 tt ise2()-admissible, and ,l c’.(). So can be obtained from by theconstruction of Lemma 8.7, as claimed. The case of /2 is identical. Q.E.D.

Theorem 8.5 is a consequence of Lemmas 8.9 and 8.10.

Proof of Proposition 8.6. If cp(w t), then

a() eigenspace of 0 ()

WO() W(,/,)".(8.11)

1012 DAVID A. VOGAN

Now assume (b). Since w-zy Wl(,)a,

() ((w-’y) )

Therefore

ct(w )= ct((y) ).

e,(w )= ce( (y ))"r x cp(y x )

(by Lemma 7.11). Therefore,

--’ x p(w x .y)= p.(y x .y).

By (8.11), the hypothesis - Wa(8) gives (a).Next assume (a). This means that there is an element $ 6 c(cl(w y)), and a

o 6 W<), such that

o-’,/, ,,(,(y x -y)).

Using Lemma 7.11, we can rewrite this as

Since y is almost maximal, w x y and oy x y are as well; so and ott must bemaximal ()-admissible subspaces of ha. By Proposition 5.8, we can findoo W)R(,I,) with

By Proposition 4.14, cl(o x )=cl(); so by Lemma 7.11,

8(C[(f)) 00 X C’( C[(O1 X

Oo x c"(_cl(oooy x ,);

the last containment is (,) and Proposition 7.23. On the other hand,,0 c(cl(w y)); so the left side above is

W() X cl(w X ")/)

by Proposition 7.8. So we can find o W() so that

Z(w x ) cz((O,OoO)y x ,).


Now o was chosen in W(); and o0 and o belong to W() by (8.11). So if weset z olo0o, then - W()

cZ(w x cz( y x

Finally, 01-- ; SO (**) gives

These three conditions are (8.6)(b).

Proof of Proposition 8.7. The implication (a)(b) is obvious; so assume (b).Replacing ,; by wi .y i, we may assume wi= 1. By Proposition 7.26, we canchoose

so that

cp,(y’) (8.12a)

ga() U_ rp2( ga(y2)). (8.12b)

By (8.12a), we can write

4= o q, o

By (8.12b), we can find 02 so that

wiFl()a, (/) C,(I). (8.12C)

Therefore

ga(02 X l)) C2( ga(y2)), o2 WFI(*)a. (8.12d)

ga(o dp) C2( ga(y2)), o O"102 (8.12e)

This last inclusion means that 2 is ga(o q))-admissible; so by Proposition 5.20,

c2( ga(o h))= W(,-q’2) ga(,y2). (8.12f)

We may therefore choose q so that

ga()= ga(/2).

(8.13a)

(8.13b)

If q were conjugate to 72, we could easily get the conclusion of 8.7(a), by readingback through the construction of q. It need not be, however. Two things will saveus. First, we made several choices in getting q, so we can modify it a littlewithout changing its properties. Second, Lemma 8.10 guarantees that there is an

1014 DAVID A. VOOAN

element w W such that

cl(y-) cl(w +). (8.13c)We will use (8.13b) to deduce that w has a very particular form. Then w x q willbe one of the allowable modifications of q mentioned above. By 8.13(b)-(c),

w x ga()= ga(g,). (8.14a)

In particular, w commutes with 0a(ff). By Proposition 3.12,

W---q’lT2, 1 Wc()a, T’ wi()a, T2 Wn() (8.14b)

(in the notation of Proposition 3.12). Since + is almost minimal, a() is trivial;SO Proposition 4.14 gives

WI()a, 1 W()a. (8.14c)

Therefore we may replace w by zl, and get (from (8.13c))

(v:) (w +), w w"()". (8.15)

By (8.12c) and Lemma 7.1 l,

o co,(o 7); (8.16a)

so o is e(o )-admissible. By (8.13a) and the inductive nature of thedefinition of admissible,

2 and o are e()-admissible. (8.16b)

Since 7 is almost maximal,o2 is a maximal ()-admissible subspace. ByCorolla 5.14 and (8.15), we can write

w ,%0, ’ w(o’), w(), w w6"()". (8.6c)

By Proposition 4.14, w W(); so (8.15) becomes

ct(v) (w ), w ’:, ’ w(o’), : w(). (8.7)

By Proposition 7.8 and (8.13a),

s c:(o ). (8.18a)

Now os and 2 are orthogonal, so s fixes 2. By Lemma 7.11,

sis2 x c:(so x )= c(o(o-lso) x ). (8.18b)

Now s W(o), so

’= o-’’o w(’). (8.8)


By (8.12c) and Proposition 7.8,

By (8.17), (8.18b), and (8.18d), we get

,z, :((o ’)),

By Proposition 7.23, this can be written

But o o= (see (8.12)); so

l). (8.18d)

’ e,(3,’). (8.19a)

(o’ x ’)e (ob-’x :((*)), e’e 0,(’),(8.19b)

o’ w;(o), ,,e w().But (8.18c)-(8.18d) shows that and ’ define the same abstract real andimaginary roots; so (8.19b) amounts to

Thus

p,(v’) n p:(vb o. O.E.B.

9. Cayley transforms of @ groups. Just as in the formal theory, there are aneasy case and a hard case.

LEMMA 9.1. Suppose Hx, C_ a, , (R)(c), and q c(y). Then

w,"()= (w())(Definitions 4.13 and 6.2). That is,

w()= { w w(6). w(v) wl0 }.The analogous result holds for c.The proof that Wl(q)a contains the indicated elements is along the lines of the

proof of Proposition 8.3; and the proof that it contains only these is like theproof of Proposition 8.6. We therefore leave this to the reader. (The analogue forc, which can be proved in the same way, is due to Schmid [20].)

PROPOSITION 9.2. In the setting of Lemma 9.1, suppose also that is almostmaximal. Then

ga((,,[)) C3(

1016 DAVID A. VOGAN

(Definition 6.7). In particular, if q c(y), then

s W2F(.y)a [,-] W() / W(’) ( W()]and

are isomorphic by the natural map of Corollary 6.6. Analogous results hold for cwhen y is almost minimal.

By Lemma 9.1 and Definition 6.7, we only have to prove the naturalisomorphism s . Recall the group L of Definition 7.1. By the definition ofc(7) and the groups, we can immediately reduce to the case G L. That is,we make

Assumption 9.3. G contains both a compact Cartan H= T2, and a splitCartan H T’A ’. We have y ( 1), (2). The infinitesimal character Xis integral. Write span of the abstract roots. The () and g() are anadmissible grading and cograding (Definition 6.3); and c(7), T c(). Wewill define @ and @c as in Proposition 9.2. Define

G ) Ad(G )/Ad(G0).

We will prove Proposition 9.2 by proving that and c are both naturallyisomohic to . A little notation is helpful.

Definition 9.4. In the general setting (2.1), put

G" x GlAd(x) Ad(Go)

Ad- l(Ad(Go))

G 2 ( x GClAd(x) normales go g}

G’= (x GClAd(x) Ad(G)}o ( x GClAd(x) Ad(Go) }

go) G/Go, the universal group of go

G) GI/G GIG # Ad(G)/Ad(Go) C go).

We say that G is small if @(G) }, and large if @(G) %($0).

PROPOSITION 9.5. Suppose G contains a compact Cartan subgroup T2, and(,T )x" There are natural isomorphisms

(a) W()/ Wg() (o)(b) W(O)/ W() (G) (,ottio, 4. ).


Proof. Every component of G 2 meets the normalizer of t2, since the compactCartan is unique up to conjugacy by G0. Write 72 for the centralizer of T2 in G.It is well known (and easy to check) that Ad(F) Ad(2F0)--that is, that72c G--and that

w(6,) w(, t) w"(v).Therefore

alaO w(a, )lW(aO,) w(a, f)lw"().It is easy to verify that every Weyl group element preserving the notion ofcompact root has a representative in G c normalizing g0; so

w(e:, f:) w,(o).This proves (a); and (b) is similar. Q.E.D.

PROPOSITION 9.6. Suppose G contains a split Cartan subgroup H T1A 1. Put

(HI)#= G # M H

I centralizer of )l in G 2 (notation 9.4).

Then there are natural isomorphisms(a) fill[ill n G](b) (/l rl GI)/(/I n G) HI/(H1)# o(G)(c) "tL(io) PI(R)/(L(R) + 2ill(R)) ZI(R)/2ZI(R)

(notation (3.3), using R 5(6a, oa)). Now suppose y () is integral. Recall

from Proposition 3.16 the map. w:( ()) Zl( ) [ Zl()]"

defined by the cograding g(y). Use (c) to identi [(o)]" as a subgroup of[ZI(R)]" (namely the elements of order 2); thus we have

(d) f: W2((v)) [(o)] .Then

(e) Wl(g(7)) -l(Ann (G))and so we have a natural suqec{ions

() (o)[w:((v))/ Wo((v))]"(g) (6) w2((v))/w((v))]

If (y) is a principal grading (Definition 6.3), these maps are isomorphisms.

Proof. Since the split Cartan is unique up to conjugation by Go, evecomponent of G 2 has a representative normalizing . As all roots of are real,

W(Go,H 10 Go)-- W({,l);

1018 DAVID A. VOGAN

so in fact every component of G - meets . Now (a) and (b) are clear. For (c),list the simple roots of fg in g0 (for some positive system) as (a cq). Tospecify an automorphism of go centralizing fgo, we must specify its eigenvalue oneach root vector Xi; and these may be any real number. The positive realeigenvalues are realized by Ad(H01), So

Ad(#’)/Ad(#J) [’,(R )/::,(R ),

the correspondence being

if (-1)<"’x>= a(h)/la(h)l.

Now ln GO is generated by d, Z(G), and the elements m of (2.4). Since

a(m) (- 1)<"’/i>

(by (4.4c)), (c) follows.For (e), suppose w Wz((7)). We want to show that w x 3’ w-), if and

only if (w) is trivial on (G). So fix r (G), and a representative x of r in T(see (b)). We will show that

w X ),(x)= (w .T)(x)co(w)(r)= 1.

Clearly w can be replaced by any wwo, with w0 W0(g(3’)). By Proposition 3.20,we may therefore assume that there is a superorthogonal set fl fls, suchthat

(’Y)( fli 1, all/

w=II To say that x is a representative of r means that, if X ibm(R) is a representativeof r in the isomorphism of (c), then

a(x) (- 1)<"’x>, all a e R.

By the definition of j (proof of Proposition 3.16),

(w)(r) ( I’[ sl,)(r

II (- 1)

II/ i(x).


On the other hand, by [25], proof of Lemma 8.3.17, we have

proving (,). Parts (f) and (g) follow; and the last claim is Proposition 6.4(d).Q.E.D.

PROPOSITION 9.7. Suppose we are in the setting (9.3). Consider the diagram (ofisomorphisms)

Here the map on the left if Proposition 9.6(f), that on the right is Proposition 9.5(a),and that on the bottom is Corollar 6.6. This diagram commutes, and identifies thesubgroups @s on the left, on the right, and @(G) on top.

Proof. We need only worry about commutativity; then the last assertion is aconsequence of Propositions 9.5(b) and 9.6(e). For the commutativity, all mapsare really defined inside G c, without direct reference to the group G. So we canreduce to the case when g is simple. By Corollary 6.6, all the groups areisomorphic to Z(R), which is of course the fundamental group of Ad(G). Sinceit must also be a product of copies of Z/2Z, we see by inspection of cases that(g0) has order two except when g is of type D2,. Since the arrows are allisomorphisms, the diagram commutes except perhaps in this case. So suppose g isof type DEn thus g0 0(2n, 2n). We may take

Gc= SO(4n, O) _D SOe(2n,2n) G.

As representatives of generators of /1,(%)= G2/G we can take the diagonalmatrix r with entries

[(- 1, 1,..., 1),(- 1,..., 1)]and the matrix

(0Now the maps in the proposition can be computed explicitly on these elements,and one finds that the diagram commutes. The simple but tedious details are leftto the reader. Q.E.D.

One can also prove this proposition by observing that the maps are in anappropriate (very restricted) sense functorial in G: one considers the category of

1020 DAVID A. VOGAN

subgroups containing both H and H2, and satisfying (9.3) for appropriate twistsof and ,. One can show easily that there is such a subgroup locally isomorphicto a product of SL(2, Fl)’s; so the proposition is reduced to the trivial case ofSL(2, iq). This is a slightly more satisfactory argument, but the necessaryverifications seem to me to be a little worse than for the proof given first.

In light of the discussion after its statement, Proposition 9.2 follows fromProposition 9.7.

10. Characterization of ga(). We begin by using what we have proved so farabout the structure of blocks to formalize the problem of constructing G.

THEOREM 10.1. Suppose r is another real reductive group, with abstract Cartana. Fix a regular infinitesimal character ( for , and a corresponding weight

(a), as in (2.6). Suppose that the integral root system R(k a) is isomorphic to, by an isomorphism taking (1 a)+ to R + ( a). Fix such an isomorphism, and use

it to identify W and W(R(a)), and the span of R in (b’)* with the span of R(a)in (ba),. Write a instead of R(a), using the isomorphism.Now fix blocks B, for G and G, with infinitesimal characters X and . Suppose

there are elements B, [l , such that(a) [ra(’/l)] "= ra(’ 1) (Definitions 3.23 and 4.12)(b) is almost maximal. (Definition 8.2).

Then there is a bijection ("duality")

d" conjugacy classes in B---> conjugacy classes in ,with the following properties. Suppose B, w Wa, and c_ a is in the span ofR a; write also for the corresponding subspace of a.

(C) ga(Cl())] ga(d(cl()))(d) If d? (R)(c), then

d(c( cl(,))) c*(d( cl(q))),and similarly for c.

(e) d(cl(w ,))= w d(cl()).

Theorem 1.15 will be deduced fairly easily from the formal properties (c)-(e)of the bijection, using [24]. In (b), we could just as well have taken to bealmost minimal. Probably the result is true without hypothesis (b), but provingthis would require a little more work.

Proof. By (a), .1 is almost minimal. Choose a maximal (l)-admissiblesubspace 0 of ha; then the corresponding subspace 0 of is maximal()-admissible by (a). Fix t2c,o(’)tl). By Proposition 9.2, there is a

.2 C,(.l), such that

ga(,y2)1"-- ga(,2).


By (10.1)(a) and Proposition 4.14, we get

(a)

(b)

(c)

(d)

a(3"i) ],= ga(i)

Wl(3"i)a-" Wl(i)

3’1 and 2 are almost maximal

l and 3’2 are almost minimal.

(10.2)

We define the duality map d by

d[ .p,(w1 X 3’1)n rp2(w2 x 3’2)] rp’(wlX 1) n ..2(w2 X ,,2)

whenever both sides are defined and non-empty. Using (10.2), Theorem 8.5, andPropositions 8.6 and 8.7, we see that this is a well defined bijection. ByProposition 9.2, it satisfies (10.1)(c). By Lemmas 7.11 and 7.22, it satisfies(10.1)(e). Part (d) is a little tricky, and requires two lemmas.

LEMMA 10.4. Suppose U D G a (orthogonal direct sum). Then the domain

of the composition c .cp, is

6(c cp.) @(c) n 6(c.) C_ @(c)

Suppose that 7 is almost maximal, and lies in this domain. Then

c,(p.(,s)) { e cp(,) n (’) ’(ga()) n CjOtt(la(3’) :=/= 0}.

Analogous results hoM for c rp".

Proof. By Lemma 7.13, c,(cp.(y)) C_ c1(3’). So by Proposition 7.23,

,(.(y))_

{#. ,(,) n (’) c’(ga(’l:)) n clpt(ga(3’)) ::/= O}.

So suppose ,/, belongs to the right hand side. The proof that ,/, belongs to the leftside is a straightforward imitation of the proof of Proposition 8.7, except at onepoint. In that proposition, we knew that 3,

2 (the analogue of ) was almostminimal; but need not be. This was used in one place. We constructed anelement q, and wanted to know that

for some w wiR(3,2)a. This was done in two steps (see (8.13c) and (8.14)), bothof which used the almost maximality of k. In the present case, the assumption

1022 DAVID A. VOGAN

cp(7) (which is a single orbit of wiR()a) will give

cl(w )= cl(q), some w Win()for the analogously constructed . Details are left to the reader. Q.E.D.

LMMA 10.5. Suppose q and e are regular characters, and q (R)(c). Then thefollowing conditions are equivalent.

(a) c(cl(q))(b) (R)(ca), c(cp(q)), and q c(rp(e/)).

Proof. By Proposition 7.23, (a)(b). (Notice that c(cp(q)) is defined, byLemma 8.2). So assume (b). Then we can write

N=+-,an orthogonal direct sum, with

9 + + eigenspace of 0 (()+ (3 + eigenspace of/9 (t))- eigenspace of 0 (cO)

(10.6)

b- eigenspace of 0(k).

Since e(ep(q)), there is a w + such that

w + W(+ ), w + c(cl(qO). (10.7a)

Since c(rp(e/)), Proposition 7.23 provides an element

v c(,) n p();

so there is a w- with

w- W(9- ), (w-)-’ q e(cl(q0). (10.7b)

By Proposition 7.8, (10.7a)- (10.7b) imply that there is a w0 with

wo W(), (W-Wow+ ) cl+ cl. (10.7c)

By (10.6) and (10.7a)-(10.7c),

W- ( wR()a, WOW+ wiR(I)a. (10.7d)

By Proposition 4.14 and (10.7c)-(10.7d), we deduce that

(W-)-Ix clt cl6,


By (10.7b), therefore,

q c(c;()). Q.E.D.

We return now to the proof of (10.1d). The idea of the proof is this. Duality dis defined to respect Cayley transforms based on the almost maximal or almostminimal elements; so we must express c in terms of these. Lemmas 10.4 and 10.5accomplish this. So suppose ,# (R)(c0), and q is another regular character. Wemust show that

q co(cl(/))<=d(/) c(d(cl(+))). (*)

By (lO.lc), we may as well assume that (R)(c), and d(cl(/)) (R)(co). Write

cl(q0 cp,(w’ X ’) n rp(w? x 1.Thus

cp( ) cp ,(w’ x

By Lemma 10.4, the definition of d, and (10.1 c), we conclude that

d[ c cp() ] c rp(d( cl())).Using also the analogous formula for c0, we see that (,) follows from Lemma10.5. Q.E.D.

Now we can direct all our efforts to proving the existence of a 1 satisfying(10.1a). To do this, we need to describe abstractly which bigradings can arisefrom regular characters.

PROPOSITION 10.8. Suppose H TA is a O-stable Cartan subgroup of G, and

H. Write

g g() (0,e(),/ (q)) (0,,,) e 6(R())

for the bigrading of Proposition 4.11. Write bn,p for the half sums of positiveintegral imaginary coroots (respectively, real roots); and identify with * using(,).

(a) There is an element

x /,(R(#0) t (/to)

(Definition 3.3)such that

+ =_ (mod 2), a

_R i(,).

1024 DAVID A. VOGAN

(b) There is an element

such that

+ =- (mod 2), a e R ().

The proof involves the following fairly well-known fact.

LEMMA 10.9. Suppose is a complex reductive Lie algebra, b c_ is a Cartansubalgebra, 0 is an involution of A(, b), and (AIR) + is a set of positive imaginaryroots (Definition 3.10). Then there is an involution o of such that

(a) b is a-stable, and ol 0(b) If a is a (AiR)+-simple root, then oX -X.

Proof. .We proceed by induction on IAiRI. Suppose first that this set is empty.Then we can find a positive system A+(g,) stable under -0. By thefunctoriality of the correspondence between Dynkin diagrams and semisimpleLie algebras, we get an involution o0 of such that o09 b, and o01 -0. Wemay arrange for o0 to permute the simple root vectors X, for a Chevalley basis;then o0 will commute with the principal automorphism o of g (see [10],Proposition 14.3) which is on b. Then o o0o has the desired properties.

Next, suppose AiR is non-empty. Choose a simple imaginary root fl, and sett sO. Then the/-imaginary roots are the 0-imaginary roots orthogonal to fl,By induction, we can find an involution 6 of g satisfying the conditions of thelemma for 0 and . Write

CZ SL(2, C) -> Ad()

for the root SL(2) (compare (2.3)-(2.4)) with

x 0 ) Ad()q0 x -1

Since fffl -fl, 5 preserves the image of /, and pulls back to an involution ofSL(2, G) which is h--> h- on diagonal matrices. Therefore there is a non-zeroa G such that

(’[(I)t(g)J=(I)t-a 0 -’ (10.10)

a 0

(Here we write for the action of tL)Let x be a square root of -a2, and define

--X-1 0


a representative of s/ in W(g,b). We will choose o to be one of the twoautomorphisms

Obviously o preserves , and oi[ s(=/9. To see that o is an involutiveautomorphism, we compute (say for ol)

By (10.10),

-a -1 0 -x -1 0 a -1 0

b/( 0 a2x -1 )-a-2x 0

Since x2 a2, x a2x- 1. SO

x 0 ="Therefore,

as required. Define gradings

i E(AiR(0 )), ’ E(AiR( ff )).(Definition 3.13) by

(- 1)’;(’) eigenvalue of o on

and similarly, for . Since

it is clear that

-1Ol-- OflOotOfl

We claim that also

’--" Cfl(.i) (10.1 la)

1026 DAVID A. VOGAN

(Definition 5.2). By Lemma 5.12 and the choice of t, it will follow that one of theE satisfies

+Ei(a)= 1, all c(AiR) simple.

Then o o satisfies the conclusions of the lemma. Now by Definition 5.2 andthe definition of oi, (10. la) is equivalent to

if a and fl are strongly orthogonalif a and fl are orthogonal but not

strongly orthogonal.(10.1 lb)

Now a and fl are strongly orthogonal exactly when (SL(2, G)) fixes X, so thefirst case is obvious. In the second case, X, is the zero weight vector in a threedimensional representation of /(SL(2,G)). Since / acts by -1 on such aweight vector (as is obvious in the adjoint representation, for example), thesecond case follows. Q.E.D.

The dual of Lemma 10.9 is

LEMMA 10.12. Suppose H TA is a O-stable Cartan subgroup of G, andA+ (, )) is a fixed positive system. Suppose o )* is a A+ (, ))-dominantweightof a finite dimensional representation of G. Then there is a "/0 (F0, o) H’ suchthat

(a) 3’0 20 + o(A+ (, [9))(b) If a is a (AR)+-simple root, then g(v0)(a)= (Definition 4.5).

Proof. Suppose first that there are no real roots. Let A0 be a weight of afinite dimensional representation of G, with dA0 2o. Define

r01 aol + 2o(a+ (o, t))

r01 aol0 Xo + (a/ (, t)).

Obviously ’0 (F0,?o) ’, and satisfies the lemma. We now proceed byinduction on IAI, using the Cayley transform c ’ of Definition 7.16 in place ofthe construction used in the previous lemma. Details (which are easier in thiscase) are left to the reader. Q.E.D.

Lemma 10.12 is easier than Lemma 10.9 because it is obvious that every realreductive group has a fundamental series of representations. Lemma 10.9 wouldhave been just as easy if we had invoked the dual fact that every real reductivegroup has a quasisplit inner form. We essentially proved that fact at thebeginning of the proof of Lemma 10.9.


A more direct proof can be given for Lemma 10.12. It is shown in [24] that theregular characters satisfying (10.12a) are parametrized naturally by (H/Ho) ^. Inthis parametrization, one can take 70 corresponding to the trivial character of/4//40.

Proof of Proposition 10.8.such that o1 0[, and

For (a), use Lemma 10.9 to find an involution o of

,(o)() , a R ilq(t/)) simple.

(Here we define e(o) like e(q), using o in place of 0. Thus if/3 is any imaginaryroot, oXt (- 1)’()X .) Then

e(o)(a) =--- (a,[9 iR) (mod 2). (10.13)

The automorphism 0/9 of g is trivial on 9, so it is of the form Ad(exp(x0)) forsome x0 belonging to the span of the roots in t; that is

0 o Ad(exp x0) (10.14a)

Now

0 2 o Ad(exp Xo)]2o[ o. Ad(exp Xo)]Ad(exp Xo)

Ad(exp(Oxo + Xo)).

Thus

Ox + x I2rriZ, all a A(g, ). (10.14b)

Put

Oxo + xox2ri /I(A({’ )); (10.14c)

the last inclusion is (10.14b). Obviously x satisfies the first condition of (10.8)(a).For the second, suppose a is an imaginary root, and X a root vector. ThenO O, SO

(,, Xo) (,, Oxo + Xo)

i,rr(a,x>. (10.14d)

1028 DAVID A. VOGAN

Therefore

(- )’)x oxo[Ad(exp(xo))X

o[exp<ot, xo>Xo,

o[(-1)<’>X] by (lO.lad)

(-l)<’x>( I)’()()X

(-l)’(>x (-l)<’,+’>x by (I0.I).

is proves the second condition of (10.8)(a).The proof of (b) is analogous, but requires a technical obseation (formula

(10.16) below) to get started. Choose an orthogonal set { fl), fl}such that

m is maximal for orthogonal sets of roots, and (I0.15a)

{ , } is ,tongly othogonal in R (). (I0. Sb)

If i " (g, ), then

< z,

so i Ra(0) This is not the case, by (I0.15b); so {i} is stronglyorthogonal in A(q, b).By Froposition 6.4(e), there is a principal cograding )of As(S, ) so that

g,(,) , all i. (10.1Sc)

By the definition of principal cograding, there is a system (A+ )a(q, ) of positivereal roots, such that if we write 9 for half its sum, then

g,() <a,> (mod 2). (10.1Sd)

By (I0.15a)-(I0.15c), there is a maximal orthogonal set in R a(0) which isgi(o)-admissible. By the progf of Froposition 6.4(e), it follows that l]R()principal. The cograding g0 E(R a(O)) defined by

g0() <a,> (mod 2) (0.S)

is obviously principal; so it differs from )]() by an element w of W(Ra(O)).ter acting on (A + )(, ) by w, we conclude that

<,a> <,a> (mod ), all e R(O). (10.16)


Now extend (A + )R(g, b) to any positive system A+ (g, b), and fix any ’o as inLemma 10.12. By (10.12b)

By (10.16),

Now (10.18) gives

By Definition 4.5,

(-1)i(*)(") (-1)i(’o)(")ro(m,)p(m,)-l( 1)(,Vo-)(10.18a)

(-1)g(>(") Fo(m,)O(m,)-( l)<a,ro-*+o%for a R R(g)); the second equality uses (10.17). Since G is linear and T iscompact, there is a weight A H of a finite dimensional representation of G,such that

Write , dA. By (4.4c),

Al --[ro- o]l . (lO.18b)

A(m,) (- I)<’x>. (lO.18c)

(- 1)(’)(’) (- l)<’Lx-(v-’)+o% (10.19)

Let y be the projection of - (?0- )) on the span of the roots. By (10.18b),y a’. As the differential of a weight of a finite dimensional representation, h isintegral; and ?0 is as well by Lemma 10.12(a). So the integral roots for y areexactly R((h), proving the first condition of Proposition 10.8(b). The secondcondition is (10.19). Q.E.D.

Finally, we need to describe the 0% groups which can be attached to regularcharacters.

PROPOSITION 10.20. Suppose H TA is a O-stable Cartan subgroup of G, andHx. Use the notation of Proposition 4.11. Choose a set (al,..., a) of

superorthogonal imaginary roo for e() as in Proposition 3.14, and (ill,superorthogonal real roots for (/) similarly. Define

(I) { fl h(g, {)) fl It span ((/i))

( -z)xI, ( fl h(g, b) fl la e span ( flj

(To)(a) --= <,p> (a e R ()). (10.17)

1030 DAVID A. OGAN

For fl (, put

and for fl , put

Put

H();’c w’(v)J

group generated by o/[ fl C (b )

group generated by r/ fl C )

W(’,l) Q + Wg(

w+ ()= (2 + wo(v).(a) For all fl tb, o[3 WR(7); that is

w() c w() c w{"() c

If G is small (Definition 9.4), then

() w,() w(.)

(d) W{R(’) W(,).

If G is large (Definition 9.4), then

(e) W{n(7) wiR(Y)

(f) W(3t) W+ (,),

In order to reduce this result to subgroups of G, we need an elementary andwell-known structural fact.

LMMA 10.21. Suppose to C_ to, % C_ o are abelian subalgebras of go. Write G t,Ga for their respective centralizers.

(a) If G is small, so is G.(b) If G is large, so is G.Proof. For (a), recall that G is small exactly when Ad(G) is connected. Now

Ad(Gt) is obviously a homomorphic image of Ad(G); so it is enough to prove


that Gt is connected whenever G is. But this is immediate from the Cartandecomposition G K. exp(o) and the corresponding fact for compact groups.

For (b), we may as well assume G Ad(G)c_ Ad(g)= G c. Extend a0 to amaximal abelian subalgebra a of :Po. Define

F= {xexp(ia’)lx2=1} c_Gc.

Suppose x F. Since x exp(ia’), 2 x-1; here bar is the complex structure.Since x2= 1, it follows that x 2. So x normalizes go, and (since G is large),F C_ G; and therefore also F C_ G a.Now suppose g Gc normalizes G"0; we will show that

Ad( g)l,o e Ad(GgF) C_ Ad(Ga), (,)

proving that Ga is large. Extend % to a Cartan subalgebra ) t + ao c_ go.Obviously it is maximally split, and therefore unique up to conjugation in G; sowe may assume (after changing g inside gG) that g normalizes b’. ByProposition 4.15

W( G, H ) W(GO H i GO)

SO we may even assume that g centralizes ’. Therefore g belongs to (Hm)c.Write g exp(X), X m.

Since Ad(g) normalizes m0 ggm, the (necessarily imaginary) roots of m in rrt

must take imaginary values on X. So we may modify X by an element of t(which doesn’t change gGg) and assume that X centralizes m. Choose a system ofpositive restricted roots (roots of ct in g) so that A(g, am) is spanned by simpleroots 21 ,2r. List the remaining simple restricted roots as 2r+l,..., 2+s.Write

moA(,am)

A(",am)

Now the group M0 c_ G has an open orbit on each g; so since g centralizes m,Ad(g) is a non-zero real number 2(g) on each g, for 2 A(a, am). (For theroots not in , g may not preserve the real root space , so we cannot define2(g) in general.) After modifying g by A m, we may assume that all the 2(g) are__. 1. Now choose Y iag’ so that

<2i,Y>=-Z-i-r, i=l,...,r and 2/(g)=-I

(2 Y> 0 otherwise

Then g’= exp(Y) has the following properties. First, Ad(g’) agrees with Ad(g)

1032 DAVID A. VOGAN

on all the gg’ (i 1,. , r) and so on all of gg0. Second, g’ obviously belongs toF. This proves (.). Q.E.D.

The following proof assumes some familiarity with the structure of 6X groupsin products of SL(2, Iq)’ssee [25], Section 1.5.

Proof of Proposition 10.20. Clearly it suffices to prove (c)-(f). For (c), let L bethe centralizer of TO in G. By Lemma 10.21(a), L is also small. We can replaceby a regular character y, for L, defined in Definition 7.3. (Of course, we have tochoose act.) This does not change W2 (Lemma 7.4(a)) and it is easy to checkthat it does not change W either. So we may as well assume that H ismaximally split on G. Since (G)= ), (c) follows from Proposition 9.6(g). For(e), we consider the centralizer of A0 in G, and use Proposition 9.5 instead.Now consider (d). Let t be the orthogonal complement of the span of the ai,

and let G be the centralizer of in G. By Proposition 3.14,

W() W()(W() W(l, )).In this way, and again using Definition 7.3, we can reduce to the case G G ;for this does not change the root system (I). Since a are superorthogonal, we nowhave

+_ .,)and all of these are noncompact. Write u for the span of the a. Define a newCartan subgroup

H

(Definition 7.15); this i is really the same as the one defined above. Since Td iscentral in G, H is split; that is, all roots are real. Write (5) for the real rootscorresponding to the a. Define maps

" TI---(+/-I) r, I(t) (Si(t))

’" Wn(/)--> +__ 1)’, ’(w)---- ((--1)m’) if w 1-Is.,Both these maps are related only to the group GA, which is locally a product of rcopies of SL(2, R). By [25], (1.5.21), and " have the same image. Since G issmall, T is generated by T, Z(G), and the various rag, for A(g, b). Clearly

is trivial on T and Z(G); and

((-1)

with fl A(g, ) the root corresponding to ft. This proves (d).


For (f), set CI orthogonal complement of the flj. Just as in (d), we may passto the centralizer of al. As this centralizer has .A as a direct factor, we may evenassume al (0). Thus

gR(y) AR(g, b)= ( --flj ),and

(y)( fig. ) 1, allj.

Again it is helpful to write g for the span of the flj, and to introduce

c(H) T

as in Definition 7.1. Write/ for the (noncompact) roots of in g correspondingto the fl.. Define

" T- _+ 1)’, (t) (fli(t))

using the fli as before. The study of Wl(7) can be reduced to the centralizer ofTO in G, which is a product of copies of SL(2, q). If we identify WR with the dualgroup of +_ 1) in the obvious way, then

W(y) Ann(T)

([25], Lemma 1.5.22). We can also define

’(W(G, TI) W({ ---/ ))) --> ( -+-

as above; and " and have the same image. So we have to show that theannihilator of the image of " is generated exactly by the various z. So supposew W(( _+/g)). Since G is large, w W(G, T 1) if and only if w preserves thenotion of compact. Write

Then

w;=jA

Since all the fl are noncompact, we see that

w W(G, T l) E ( J’’ ) 27, all/.jA

On the other hand, in the pairing of Wn with

___)r mentioned earlier,

jA

1034 DAVID A. VOGAN

SO

w w(e,

as we wished to show. Q.E.D.

all fl,

11. Existence of (.

THEOREM 11.1. Suppose y Hx. Let r denote a complex simply connectedsemisimple Lie group with root system (see (2.6)). More precisely, we fix in theLie algebra ( of c a Cartan subalgebra a; and we fix once and for all an

identification of ( with the span of R in a, taking A(c, a) to a. In particular,we identify W with w(C,a).

(a) There is a real form o of , with Cartan involution and a -stable Cartansubalgebra o, such that we have an isomorphism

span of R (y) in

satisfying(1) (A(1), )) =/(y)(2) i(DIA) Olspa, of <>

If we write for the grading of the A(, )-imaginary roots given by , then, inducesa cograding i() on R() by (1). We can arrange

(3) i()= g(V)(b) Define

(2 ( x c ClAd(x) normalizes o)centralizer of in 2

Then we can find a regular character 2 of I 2, with regular integral infinitesimalcharacter satisfying

[g(7)]’=g"(#2)

(Definition 3.23).(c) There is a group c_ r 2 offinite index having the following property. Write

IYI IYI2 f) r, /= /21h. Then .In particular, if 7 is almost minimal, we get

ga() ga(,).


Proof. Let be any Cartan subalgebra of . Using appropriate innerautomorphisms to identify with a and 9 with 9a, we get

---> span of R(V) in

A(t], )--> R (y)

Since R(y) is 0-stable, we can define/ on to satisfy (a2) of the theorem. Nowuse Lemma 10.9 to construct an involution a of g, satisfying

oX X, c [Ain(, )] +simple.

Here we use the identification

i" AIR({], ) = R(,/)

to define (A/R) +. If (o) is the grading of AiR defined by o, then the correspondingcograding 0 of R R(,) obviously satisfies

g0(a) (t, pR (mod 2).

By Proposition 10.8(b), there is an elementy a’, satisfying

(, y) Z, all a R()

g (a) g0(a) + (a, y) (mod 2).(11.2)

Let)3 denote the element of corresponding toy under the map i. Then

( fl, )5) Z, all fl h(, )(11.3)

Define

0 o Ad(exp irfi).

By the first part of (11.3), o and Ad(exp iry) commute; so by the second, 0 is aninvolution. Because of the second formula in (11.2), it is easy to check (a3) of thetheorem. Any involution of a complex semisimple Lie algebra is the complexifiedCartan involution for a unique real form, so the rest of (a) follows. The proof of(b) is entirely similar, using Lemma 10.12 and Proposition 10.8(a). At some pointone has to lift an integral weight of to a weight of a finite dimensionalrepresentation of (c; there one uses the assumption that G c is simplyconnected. Details may be left to the reader.

For (c), a generalization of Proposition 9.6(e) is needed.

1036 DAVID A. VOGAN

LEMMA 11.4. Suppose H2= TA is a O-stab& Cartan subgroup of a realreductive group G 9_, and 2 (H:),x. Then we can find a subgroup Gc_ G 2, offinite index, with the following properties. Write

HI H2 fq G Yl

(b) There is a natural surjective homomorphism

(c) Suppose G c_ G is a subgroup offinite index, and H H lf’l G,Then

w,() { w w=(v’)I (g)](w) 1, all g G }(d) Suppose X is any subgroup of W(V) containing WI({). Then there is a

subgroup G as in (c), such that W(V)= X.

Proof. There are natural maps r" G2-> G2/Go, i" H2/(H2FI Go)-> G2/Gowhich are surjective and injective respectively. Put

G’ r-’(i(H2/H2 fl Go)).Then H H2, and for every G c_ G as in (c), there is a natural surjection

G-> H/H fq GO (11.5)

Put L centralizer of TO in G 1. Choose a 0-stable parabolic cl + u c_ g, andfor L (Definition 7.3). It is clear fromuse it to construct a regular character 7,

the definition that if w WR(3,1) c_ W(L,H), then

(w r’)o= w (r2),By Lemma 7.4,

(i 1,2).

Fix h H 1. Then for w WR(),),

[(w x r)(h)] [(w. Fo)(h)]-’=[(w x r’)(h)] [(w. r’)(h)]-’ (11.6)

by Definition 7.3. Denote their common value by [g’(h)](w). Then defines agroup homomorphism


(It is not obvious that g(h) is a character of W2(T); but by (11.6), it is enough tocheck this for L. There it follows from the definition of the homomorphism ofProposition 9.6(e).) If G is as in (c), then clearly

W(,) Ann((H))

(w W(’,l) l[g(h)](w)= 1, all h H). (11.7)

Taking G GO (the identity component of G ) and applying Proposition10.20(c), we conclude in particular that is trivial on H N Go; so it is defined onH1/H 0 Go. Taking G G 1, we see that Ann((H))= W(y); so

a surjection. Composing g with the surjection of (11.5), we get the map s of theProposition. Parts (a)-(c) are clear from (11.7), and (d) is a consequence of (c):take

G s-l(AnnX). Q.E.D.

To prove Theorem 11.1(c), recall from Proposition 10.20(a) that

w/+ ()_w()

_w().

The group WR(,) depends only on the grading e(,) (Definition .13); socondition (b) of Theorem 11.1 says that

w() w(:).The group W/+R () is defined using all of A(g, 9) (and not just the integral roots).If we had used only R(), we would obviously have gotten a smaller group

x c_ w ().

By the symmetry of the definitions in Proposition 10.20, and Theorem 11.1(b),we see that (if we transfer everything to a)

X W ([)a.This shows that

WIR+ (’2)a W(Iq()_

W21q(’2)

By its definition, (: is large (Definition 9.4). By Proposition 10.20(f),

(11.8)

w,() w+ (.:).

1038 DAVID A. VOGAN

By Lemma 11.4(d) and (11.8), we can find G c_ G 2 of finite index, so that

W()a= WR(,)a. Q.E.D.

THEOREM 11.9. Suppose G is a real reductive group (cf. (2.1)) andB ((,),..., "(Vr)} is a block of irreducible admissible representat!ons of Ghaving the fixed nonsingular infinitesimal character X of (2.6b). Write a for thespan of the abstract integral roots R a. Fix a complex simply connected semisimpleLie group r c, with Lie algebra , having abstract Cartan subalgebra a androot system

m(, a) ja.

(We use this identification constantly below.) Then there is a real form o of , anda subgroup G of c, with Cartan involution l; and a block J

((’),..., (r)} of irreducible admissible representations of , having thefollowing properties. Write d" B ---> [1 for the bijection taking Yi to [i.

(a) (’i) has nonsingular integral infinitesimal character--that is, the same

infinitesimal character as some finite dimensional representation of .(b) The strong bigradings of R and h defined by and i (Definition 4.12) are

dual in the sense of Definition 3.23:

a(’[i) ]"= a(i) all i.

(c) If C_ - , then 7i (R)(ca) if and only if i 0(c) (Definitions 7.10 and7.16); and in that case

d(c( cl([i))) ca(cl(i))

That is,

C( cl([i)) { cl[j clj e Ca(cl(i)) ).Analogous results hold for ca.

(d) If w Wa, then

d( t(w x v,))= ct(w

This is immediate from Theorems 10.1 and 11.1" given B, we fix an almostminimal 3’i0 B, and construct (, io using Theorem 11.1, to satisfy (a) and (b)for i---i0. Then the existence of the bijection d with the stated properties isTheorem 10.1.

12. The Kazhdan-Lusztig conjecture

Definition 12.1.. Fix a block B of regular characters for G with infinitesimalcharacter X, and a constant c0(l/2)Z. If H= TA is a 0-stable Cartan


subgroup of G, and 7 Hx, define the integral length of ‘/by

ll(‘/)--’-1/2]{a R + () Oa g + ())1 + 1/2dimA co

We assume co is such that lI(‘/) is always an integer. In [25], Definition 8.1.4,lI(‘/) was defined with co equal to half the dimension of the split part of afundamental Caftan in G. It was asserted incorrectly there that this makes lI(‘/)an integer; that is the case if ‘/is integral, but not in general. That there is such achoice of co follows from Lemma 8.6.13 and Theorem 9.2.11 of [25]; these showthat for any co

I(‘/) =-- l I(‘/’) (mod Z), all ‘/,‘/’ B.

It is inconvenient to normalize I in any uniform way: to make some formulasnicer, we want to use different kinds of normalizations for G and 0.

Definition 12.2. Suppose ‘/ Hx. Write l-I(,/) for the simple roots of R +

The ,-invariant of is the subset ,(‘/) c_ l-I(‘/) defined by

(v) ( e rI(v) ,, is imaginary and e(‘/)(a) 0; or

a is complex and -Oa R / (‘/); or a is real

and g ()(a) 1}.

The abstract ’-invariant is the corresponding subset .a(‘/)c_ 1-I (cf. (2.6b)). Wemay identify either of these sets with the corresponding simple reflections inS(‘/) c_ W(‘/) or S c_ Wa, respectively--cf. (2.5b), (2.6b).

Definition 12.3 ([24], Definition 6.4). In the setting of Definition 12.1,consider the ring Z[u,u -l] of finite Laurent series. The Hecke module of B,(B), (or simply ) is the free Z[u,u -t] module with basis (cl(‘/) ‘/ B ). Anelement of 9re(B) is thus a formal expression

av‘/,av 7[u,u-l], ‘/ B,

with conjugate ‘/’s identified. For each s Sa, we define a Z[u,u-l]-linear map

as follows. It is enough to define each Ts‘/ (‘/ B). Write a I-I(‘/) for thesimple root corresponding to ‘/. There are several cases.

(a) a imaginary, (a)= 0. Ts‘/(a*) a real, (a)= 0. Ts(b) a complex, Oa R + (‘/). T‘/ s(b*) a complex, Oa q R + (7). Ty u(s(c) a imaginary, (a) 1, s W(n(y). (In the terminology of [25], a is type I).

1040 DAVID A. VOGAN

Then c"(T) consists of a single element (cf. [25], Definition 8.3.6), and we set

x +

(c*) a real, (c)= 1, s wR(T). (Then c is Type II in the sense of [25]). Inthis case c(T) has just one element ([25], Definition 8.3.16), and we set

T,T (u 1),/- s X T + (u 1)c(T).

(d) a imaginary, (a)= 1, s E WR(y). (In this case a is type II). Then c"(T)We defineconsists of two elements ,/+_, with T_+ s x T_v_.

Tsy= y+ T+

(d*) a real, {(a) 1, s e Wl(T). (Now a is type I.) Then c(T) consists of twoelements T, with -f- s x -f Put

(u + (u- 1)( 2 +

We will explain the pairing of cases ((a) with (a*), etc.) in due course.Although we have no need of it, the Hecke algebra is floating around in the

background somewhere; so we should recall its definition. For more about it inthis context see [15].

Definition 12.4. The Hecke algebra of Wa, .a, is the Z[u, u -l] algebra withgenerators Twlw wa), subject to the relations

Tw,w2 Tw,Tw2 (w Wa, l(WlW2) l(w,) -- /(w2))

(T "{" 1)(T U)--’0 ( SO).

Here denotes the length function on W.

It turns out that %a is a free Z[u,u -l] module with basis Twlw W). Thefirst relation in Definition 12.4 shows that the various Ts (s S a) generate

PROPOSITION 12.5. The operators T of Definition 12.3 come from an action ofthe.Hecke algebra %a on aYe(B).

For Aa integral, this is proved in [18]. The general case is a consequence of thefollowing result.

LEMMA 12.6. In the setting of Definition 12.3, we can find a second group G 2

and a block B 2 for G 2, such that(a) The representations in B 2 have integral infinitesimal character.(b) The abstract ordered root systems

(R a, (R")+) and (A(g2, ([2)a),A + (g2, (2)a))are isomorphic. Henceforth we identify them, and so also the abstract Weyl groups.


(c) There is a bijection B--) B 2 which commutes with Cayley transforms andthe cross action of Wa, and satisfies

ga(y) ga(i(y))

(Definition 4.12).(d) The induced isomorphism i: 3rC(B) Y(B) respects the operators

Proof. Apply Theorem 11.9 twice; this gives G 2 satisfying (a)-(c). Now (d)follows by inspection of the definition of T,. Q.E.D.

We will use this lemma from time to time to extend purely combinatorialresults proved for the integral case in [18] or [24] (sometimes using algebraicgeometry in a way which does not obviously extend) to the general case; theproof of Proposition 12.5 is typical. In the future we will simply refer to [18] or[24], even though the statements there cover only the integral case. We turn nowto the definition of the Bruhat order.

Definition 12.7. In the setting of Definition 12.1, suppose q,q/ B, ands B a. We say that q q/if and only if

(a) ll(q/) lI(q) l, and(b) q/appears in Tq (Definition 12.3).

Suppose q,q/. It is easy to check that if a R + (q,) is the correspondingsimple root, then either

(a) a is complex, Oa R l(q0, and q/ cl(s q0; or

(b) ct is real, g (q0(a) 1, and q’ c,(cl(qO)(12"8/

(Definition 7.5). In this case we also have the relations

s x q,-->s x ,l,’, s x ,#->,#’, q,-->s X ,l,’.

(The existence of case (b) is of course the source of all our woes.) Theseconditions can also be formulated in terms of the R / (q/)-simple root a’corresponding to s" we get

(a) a’ is complex, Oa’ R + (q/), and q cl(s x ’); or

(b) ct’ is imaginary, e(q)(a) 1, and q ca(cl(qd)).(12.8’)

Definition 12.9. In the setting of Definition 12.1, the Bruhat order on B is thesmallest ordering such that

m((q)),’(y)) :/: 0oq < y

(notation 1.2). More explicitly, q < y if and only if there is a sequenceq q0, q, qr Y, such that m((@i_ l),’n’(ti)) =/= 0 for 1,..., r.

1042 DAVID A. VOGAN

LEMMA 12.10. The Bruhat order is an order relation: that is,(a) < 7 < Y. We also have(b) ,< y/(,) < l(3’)(c) < 3’ and lI()= 11(3") 3’.

Proof. Obviously it suffices to prove (b) and (c); and for these, we mayassume m(e(),rr(3’)):#: 0. Then they are precisely Proposition 8.6.19 of [25].

LEMMA 12.11. The Bruhat order may be defined using the formal charactermultiplicities M(r(d:), (3’)) in place of m((), r(3")).

Proof. The definition of < may be reformulated as follows: , < 3’ if andonly if we cannot index B as

g (I’2’"""’ r )’

with 3’ preceding , and still have the matrix

rn (m,).) (m((;), r(j )))be upper triangular. (The hard part is only if; so suppose it is false that < 3’. Toconstruct the required indexing of B, list first all the elements less than(compatibly with < ), then 3’, then all the elements less than , but not less thanthen , then everything else. This has 3’ preceding ,, but makes m uppertriangular.) Set

M (M(r(i ),z(j ))).By (1.3), m M- ; so m is upper triangular if and only if M is. Q.E.D.

In the case of Verma modules, the Bruhat order is usually defined directly onthe Weyl group, and the formulation of Definition 12.9 is a theorem (see [8]).Essentially because Cayley transforms of regular characters are multi-valued, noeasy generalization of this approach is available to us. Even a posteriori, I knowof no simple combinatorial description of the Bruhat order on a block. To makecomputations, a larger relation was introduced in [24]; we use a slight variant ofit here.

Definition 12.12. Suppose 3’ B. Define r (3’) to be the smallest subset of Bcontaining 3’, and closed under the cross action of simple real reflectionss S(3’) (cf. (2.5), (2.6)). (This is smaller than the r-packet rp(3’) of Definition8.1, which was closed under the cross action of WR(3’). WR(3’) is not generatedby simple real reflections in general, because the simple roots of (R a)+ (3’) neednot be simple for all of R / (y).) The Bruhat r-order is the smallest relation on Bcontaining the Bruhat order, and constant on the various r (3’). We write it as. For duality, we will also have to consider the sets c (3’) (closed under the

ccross action of simple imaginary roots) and the c-order <.


These two orders do admit straightforward descriptions which we will recordhere.

LEMM 12.13. The Bruhat r-order is the smallest order relation on B which isconstant on the r (7) (Definition 12.12), and has the following property. Supposed, B are related as follows" there is an s Sa, ./, B, with - 7’; and at leastone of the following conditions is satisfied.

(a) q-y’, or

(b) there is a q’ B, with q- q, and q’ ’. Then q 7.

The Bruhat c-order has an analogous description with ( , --) replacing ( ,- ).

In order not to interrupt the development, we postpone the proof to the end ofthe section.

LEr, 12.14 ([24], Lemma 6.8, and [18]). In the setting of Definition 12.3,there is a unique Z-linear map D ]L(B)---)3L(B) with the following properties.Define Rq, Z[u, u 1] by

Then we require(a) D(um)= u-ID(m), all rn 631L(b) D((T + l)m) u-l(T + 1)O(m), all rn(c) R(d) R,v 4 0 only if q < (Definition 12.12).

This map has also the following properties.(e) Rq,_v is a polynomial in u, of degree at most 11(y)_ lI(q).(f) D"- is the identity.(g) The specialization of D to u is the identity.

LEMM 12.15 ([24], Corollary 6.12 and Theorem 7.1).12.14, fix B. Then there is a unique element

In the setting of Lemma

of 6"Y(B), with the following properties(a) DCv u- ’’ v)Cv(b) Pry(c) P,v =/= 0 only if q < V(d) If qv 7, then e,v is a polynomial in u, of degree at most 1/2(ll(y)

t()- ).

In [24], two methods for computing the polynomials P,v are discussed. Bothare quite cumbersome in practice.

1044 DAVID A. VOGAN

CONJECTURE 12.16 (The Kazhdan-Lusztig conjecture for real reductivegroups). Fix a block B of regular characters for G, with nonsingularinfinitesimal character X ((2.10)-(2.12)); and define polynomials .P,v(u) byLemma 12.15 above. Then

M(th, Y) (-1)%)-’(v)P,v(1 )

(notation (1.1), (12.1)); that is,

(Y) E (-

in the Grothendieck group of admissible finite length representations of G.

THeOreM 12.17 (Lusztig, Beilinson-Bernstein; see [24], Theorems 1.6 and1.12). Conjecture 12.16 is true whenever the infinitesimal character of X is

integral; or, more generally, whenever there are no non-integral roots in the rational

span of the integral roots. The polynomials Por have non-negative coefficients.To say that X is integral means R= A(q, b) (notation 2.6)). The hypothesis

for the "more generally" statement may be formulated as

OR n a(,) R.This case may be reduced to the first by the methods of [21]. The samearguments reduce Conjecture 12.16 to the case

OR h(, ).

This is the case which Bernstein informs me he can handle; so when the dustsettles, Conjecture 12.16 will presumably be proved. In any case, Theorem 12.17already covers all cases when Gc is of type A,, for example.We will now give the proof of Lemma 12.13. The non-formal part is contained

in the next result.

LEMMA 12.18. Suppose k and 7 are distinct elements of B, and m((,),r(7))4 O. Then we can find a 7’ B and s S with 7- 7’ (Definition 12.7). For anysuch s, at least one of the following conditions is satisfied.

(a) m((,), r(7’)) 4= 0

(b) There is a ’ B, -q’ and m((ck’), r(7’)) 4= O.

(c) The relation 7- 7’ is of the form (12.8b); and either (a) or (b) holds withs 7’ replacing 7’.

Conversely, suppose and 7 are distinct, 7- 7’, and at least one of (a)-(c) holds.Then either < 7, or s is a real reflection for 7, and < s 7.

This simply summarizes formally the algorithm for controlling the compositionseries of r(7) given in [25], proof of Proposition 8.6.19.


Proof of Lemma 12.13. Write < for the relation described by the lemma;here stands for lemma. We will prove that

r<7=>ff < 7 (12.19)

by induction on ll(y). So suppose , 7; we want to show that ’h < 7. Bydefinition of <, we can find g, B, 7 r (7), such that

(a) m((), r(7)) :/: 0, and

(b) either ,/, < q, or r(t) r().(12.20)

We claim that q < 7. Since < is defined to be constant on r (7), we may as wellassume 7 . By Lemma 12.18, q and 7 are in the relationship required byLemma 12.13; so q < 7. Now lZ(q)< li(7); so (12.20)(b) and the inductive

hypothesis guarantee that either <q, or r(,/,)= r(). Now the desired

relation < 7 follows from the transitivity of <.Conversely, suppose < 7. By the definition in Lemma 12.13, we can find a

k B, 37 r (7), such that

(a)

(b)

and 7 are in the relationship described by Lemma 12.13, and

either , < q, or (,/,) r(q).(12.21)

By the converse direction of Lemma 12.18, 7; and the argument is completedas above. The statements about are proved in the same way. Q.E.D.

13. Duality for Hecke modules and proof of the main theorem.

LEMMA 13.1. In the setting of Lemmas 12.14 and 12.15, we have the followingadditional information

(a) If P,v =/= O, then d? < 7

(b) If Rq,v =/= O, then < 7.

Proof. Since all of the objects in question depend only on the combinatorialstructure of the block (see Lemma 12.13), Lemma 12.6 allows us to assume thatRa= A(,ga), and therefore (by Theorem 12.17) that Pv has non-negativecoefficients, and _Now (a) is a consequence of Lemma 12.11. In fact we get the stronger statement

that < may be defined by Pov in place of M(O, 7). Write Qv for the inverse

1046 DAVID A. OGAN

matrix to P,v. By the proof of Lemma 12.11, Q,v can be used to define <; so

,{= Q,v(u)C,

By Lemma 12.15,

Dy u-t’(*)Qov(u-1)CqC

"-U

Comparing this with Lemma 12.14 gives

Rr (-1)’’(*)- ’’(v) ut’(v)-,’(OQ+v( u l) e,,(u)

Now assertion (b) of the lemma is immediate. Q.E.D.

Definition 13.3. In the setting of Definition 12.3, the dual Hecke module(B) (or simply @) is the free Z[u, u-1] module having the same basis as 631L(B)(that is, {cl(-)l B }), endowed with the following structures. When we wantto regard - as a basis element of ", we write it as . Put

(a) 11() 11(7)Define a Z[u, u-l]-linear pairing from Y(B) I(B) into Z[u, u- 1] by

(b) (’,)= { O, =/= y

1)/’(Y), t

Use a superscript to the left to denote the transpose with respect to this pairing"if A is a Z-linear operator on 63L(B), then tA is the unique Z-linear operator on(I(B) satisfying

For s S, define T" IL by(d) 7, -tT +(u- 1).

Recall the map D of Lemma 12.14. Write bar for the unique automorphism ofZ[u,u -] sending u to u -l. Define/. o- by the requirement

(e) (Dx, y) (x, by) (x U, y 6’(6)To see that (e) makes sense, fix y @, and consider the map

z[.,u-’], <Dx,

Clearly fy is Z[u,u-t]-linear; so it is given by

for a unique


LEMMA 13.4. The map )" -->L of Definition 13.3(e) has the followingproperties. Define elements R,i Z[u, u- l] by

B

Then(a) 6(um)= u-fi(m), all m .(b) D(T + 1)= u-’(L + 1)(c) k(d) R 0 only if > y (Definition 12.12).

Proof. Part (a) is equivalent to

(x,D(um)) (x,u-lD(m)), all x3lL, m.By Definition 12.10(e), this equation can be rewritten as

(uD(x),m) (D(u-t,x),m), all x3, m.In this form it is a consequence of Lemma 12.14(a). By a similar calculation, (b)amounts to

(- T + u)D(x)= D[(- T + u)(u-lx) ], all x

for t(s) Ts + (u 1). To prove this, write the right side as

(13.5)

-D[(T + 1)(u-Ix)] + D[(u + 1)(u-’x)]-u-’(r, + 1)D(u-’x) + (1 + u)Dx-(r, + 1)Dx + (1 + u)Dx(- rs + u)Dx.

(by (12.14)(b) and (a))(by (12.14)(a))

This proves (13.5), and so (b). For (c) and (d), we may as well compute thematrix of/5. We have

k. ut’(’)( 1) [coefficient of

u -t’(v) (Dq, ;/)

u-t’(v)(_ 1)t’(,t,)-t’(v) coefficient of 3’ in Dq(13.6)

k, u t’(*) -t’(V)Rv,(u-1).

1048 DAVID A. VOGAN

Part (c) follows from (13.6) and Lemma 12.14(c), and part (d) from(13.6) andLemma 13.1(b). Q.E.D.

Although we will not use the fact, Lemma 12.14(e) and (13.6) shows that R+ isa polynomial, of degree at most lI(-)- lI(,) here we are also using Definition13.3(a).

LEMMA 13.7. In the setting of Definition 13.3, there are unique elements

v17 B of such that (if we define C+ as in Lemma 12.15)

(Cq,,v) ((-1)/’(v), 0=7, (1)0 Ova7

More explicitly, define P+v as in Lemma 12.15, and write (Q+v)for the matrixinverse of (P+v). Then

v ] ( 1)l%)- "(v)Qv,t,. (2)

Write

(3)

so that

Then(a) Dtv u-t"+)(

v

C

(c) P+ : 0 only if + > V(d) If q:/: , then ++ is a polynomial in u, of degree at most 1/2(lI(z{)

lX()- 1).

Proof. The statements through (4) are obvious, since P is upper triangular forc> (Lemma 13. l(a)). For (a), we compute

c+, <DC+,

by Lemma 12.15(a), and Definition 13.3(a), (e). Now (a) follows from (1) of the


lemma. Since P is upper triangular for > with l’s on the diagonal, Q P- isas well; so (b) and (c) follow. Part (d) amounts to

deg Q,v < 1/2(lI(y)- l’()- 1), ve y; (13.8)

we will prove this by induction on li(7). Since Pry Q** l, the y entry of theequation PQ is

0=

0=Qr+Pr+ QPv(13.9)

By inductive hypothesis and Lemma 12.15(d),

deg(Qq,,P,r) < 1/2(1’(7) l’(/) + lX(/) zl(f) 2)

1/2(I’(T) l’() 1) 1/2

whenever q < q < 7. Now Lemma 12.15(d) shows that every term of (13.9)except perhaps Qv satisfies the degree estimate in question; so (13.8) follows.Q.E.D.

PROPOSITION 13.10. Fix a block B {cl(Y1),... cl(Yr) of regular charactersfor the real reductive group G, having nonsingular infinitesimal character. Fix asecond group and a block B (cl(/),..., c/()} satisfying the conclusions(a)-(d) of Theorem 11.9.

(a) The integral length function for (Definition 12.1) may be normalized sothat y) -lI(,) flof all B.

(b) If we identify J(B) with () (Definitions 12.3 and 13.3) bv identifyingtheir bases as indicated by the notation, then the operators T for 63L(B) are

identified with .(c) If , B, then d? if and only if .Proof. Part (a) is an easy consequence of Theorem 1.9(b) and Definition

12.1; in fact one only needs to know that oa(y)- --0(). Part (c) follows fromLemma 12.1 3; for (1 1.9)(c), (d), show that

s sT ,y’ "’ >’ (13.11)

(Definition 12.17). To prove (b), notice that (coefficient of in tT)=(-l)t’()-t’(v) (coefficient of in T), by Definition 13.3. This allows us tocompute case by case, using Definition 12.3. By Theorem 1.9(b), 7 is case (x)of that definition if and only if - is in case (x*). As an example, suppose 7 is incase (d); write c’(3,)= 3,.a+_. Then 7 appears with coefficient in T7, and with

1050 DAVID A. VOGAN

coefficient u in T,7+ (case (c*)). Since also ll(y+ lI(y) + 1, we compute

c()

By Definition 13.3(c), we get

/=(u-2)-+ (u- 1)4.

By Theorem 11.9(c), this can be written

#s= (u- 1)6+(u-2)’.

This is the formula for Ts" in Definition 12.3(d*). The other cases are similar.Q.E.D.

PROPOSITION 13.12. In the setting of Proposition 13.10, suppose the integrallength function on is normalized in accordance with Proposition 13.10(a), and6y(B is identified with q]L(a as in (13.10)(b).

(a) The operator ) of Definition 13.3(e)coincides with the operator O fordefined by Lemma 12.14.

(b) The elements v of Lemma 13.7 coincide with the elements C: for B ofLemma 12.15.

Proof. In light of Proposition 13.10, Lemmas 13.4 and 13.7 show that/) and

-’v satisfy the defining properties of D and C.. Q.E.D.

THEOREM 13.13. Let G be a real reductive group, and B (c/(Y1) cl(Yr))a block of regular characters of G having nonsingular infinitesimal character. Let Gbe a second reductive group, and B (c/(l) cl([r)) a block for ; andassume that the bijection Yi -- i satisfies conditions (a)-(d) of Theorem 11.9. Thenthe inverse of the Kazhdan-Lusztig matrix (P,v) (Lemma 12.15) is the matrix

(a) (P,v)- ((- 1)t’(*)-t’(v)P,.In partjcular, suppose that the Kazhdan-Lusztig conjecture (12.16) holds for ( G, B )and G, B ), so that

(b)

Then

M(rr(),(y)) (-1)t’(’/’)-t’(v)P,t,v(1 )

M(rr(6), ()) (-1)’%)-"()P$(1).

m((), r(y)) P6(1)

(-1)"(*)-"()M(r(),


This is immediate from Proposition 13.12, Lemma 13.7(3) and (1.3). WithTheorem 11.9 (and Theorem 12.17), it gives Theorem 1.15.

14. Duality and primitive ideals. In this section, we will lay the foundation fora study of certain Weyl group representations attached to blocks. This isconnected with the problems studied in [1] and [2] for complex groupsmFourierinversion of unipotent orbital integrals and related matters. They will be pursuedin a future paper with Barbasch.

Definition 14.1.Define

In the setting of Lemma 12.15, suppose q, 7 B and q < 7.

/(q, 7) coefficient of U l/2(ll()-l’(d)- 1) in Pv.

L.MMA 14.2 ([18], Section 5). In the setting of Lemma 12.15, suppose 7 B,sSa.

(a) If s ra(7), then (Definitions 12.2, 12.7)

(b) /f s ra(7), then

( rs u)C o

LEMMA 14.3. In the setting of Theorem 13.3, suppose q, 7 B, q < /, s S a,s ’a(q), and s q .a(y). Then

v)

Proof. By Proposition 13.12, Lemmas 14.2 and 13.7,

/(, 7) u 1/2(’’()-’’(*)- l) [coefficient of C, in (T + 1)Cv]

u-l/2(l,()-l,()-l)[<(rs .. 1)Cv, )](-1)u-’/z(t’()-t’()-’)[{Cv,(tL + 1)>](-1),’()

u-’/’t’()-t’()-’)[<g ,(u g )e,>](-1)--(-- 1)l’()u-l/2(l’()-l’() -1)

[coefficient of C in (T + 1)C](- 1)’)

1052 DAVID A. VOGAN

Here we are using several obvious facts. First, since :/: 7, we may replaceT u by Ts + in the next to last line. The result is trivial (/z(qs, 3/)=/z(-, ,)

0) unless lI(,) lI(7) is odd; so -(- 1)’() is equal to (- 1)’(v) this is usedat the end. Q.E.D.

Definition 14.4. Fix a block B of representationswith infinitesimal characterX. Write Z(B) for the free Z-module with basis B. We identify Z(B) with theGrothendieck group of the category of finite length admissible representationsgenerated by the r(7), by sending 7 to the class of r(7). Make W act on Z(B)by the coherent continuation representation ([25], Definition 7.2.28); write t(w)for the operator corresponding to w Wa. Explicitly ([25], Chaper 8), fix 7 B,s S a, and label cases as in Definition 12.3. Then

(a) t(s)7 -7

(a*)

(b)(b*)

(c) t( )r x r

(c*)

(d)

(d*) t(s)7 7

LEMMA 14.5. In the setting of Definitions 12.3 and 14.4,/et 91L(B)-#-> Z(B) bethe Z-linear map defined by

.(umT) (--1)l’(/)7.Then

(a) For rn 91L(B s S

(- Tm)= t(s)e(m).

(b) If the Kazhdan-Lusztig conjecture holds for B, then

e(Cv) (-1)"v)(7(el. (12.16)).

Proof. Part (a) is easily checked by comparing Definitions 12.3 and 14.4. Part(b) is a reformulation of Conjecture 12.16. Q.E.D.

La

Definition 14.6. In the setting of Definition 14.4, the LR preorder < on B isthe smallest preorder relation with the following property. Fix w W, 7 B,


and write

Then we require that

The cone over is

Define

LRa, 4:Oy <q,.

LR

LR/R(y) span of ((q) Y < q) -- Z(B ).

Thus fLR(,y) carries a representation of Wa, by restricting the coherentLR LR

continuation representation on B. We define LR equivalence by , if andonly if

LR LR

The double cell of " is

We must also define

LR

(v)

and ’V/R (V), vL+R (Y) in analogy with /R (7). Clearly ’vL+R (y) is t( wa)-invariant,and we define a representation of W of %/R ({) by the natural isomorphism

We call this a double cell representation. When G is complex, all of thesenotations can be refined into separate "left" and "right" pieces--see [2], for

example. The notation here is consistent with [2], although < and so on are notdefined in general.

LEMMA 14.7. Suppose the Kazhdan-Lusztig conjecture holds for B, y B, ands S a. Then, in the notation of Definition 14.1,

1054 DAVID A. VOGAN

if S . ’/’a(7); and

if s ’ra(7).

This is clear from Lemma 14.2 and 14.5.

COROLLARY 14.8. Suppose the Kazhdan-Lusztig conjecture holds for B. ThenLR< is the smallest preorder satisfying

LR7(a) if 7’- , then 7 <

LR(b) /f 7, B, s ’a(7), S .a(), and/(, 7) 0, then < .COROLLARY 14.9. In the setting of Theorem 13.13 and Definition 14.6, suppose

the Kazhdan-Lusztig conjectures hold for B and [1. If 7, d B, thenLR LR

(a) 7 < if and only if < z?. In particular

Define a pairing between Z(B ) and Z( by

((Y), ()) 6ov(- 1)

Then if m 7(B ), n Z( and w W we have(b) (t(w)m,n) (- 1)t(w) (m,t(w)n).

In particular, there is an isomorphism(c) cvtR (7) vLR (’)* (R) (sgn), with sgn the sign representation of Wa.

Proof. Part (a) follows from Corollary 14.8 and Lemma 14.2. It is enough toprove (b) when w s sa; and then it follows from Lemmas 14.7 and 14.2. Part(c) is immediate from (b). Q.E.D.

THEOREM 14.10 (King [26]).map (the character polynomial)

In the setting of Definition 14.6, there is a natural

ch LR(7) -->

here the right side is the symmetric algebra of ba. We have(a) ch((7)) is a homogeneous, Wa-harmonic polynomial.(b) Up to a non-zero constant, ch((7)) is Joseph’s Goldie rank polynomial

([12]) for the primitive ideal Ann((7))c U().(c) If w W,

ch(t(w)(7)) w" ch((7)).

COROLLARY 14.11. The double cell representation of W on cfLR (7)containsJoseph’s Goldie rank representation associated to Ann((7)).


One would like to show that every constituent of cLR (,y) lies in the samedouble cell in W ([2], Definition 2.12). There are certain minor technicalobstacles to proving this, but probably they can be overcome.

15. An L-group formulation. In this section, we will sketch without proofsanother approach to the definition of ( in Theorem 1.15. This method is adaptedonly to algebraic groups (and not to the minor modifications of them usuallyallowed, such as the class of groups defined in section 2). Notation will not beparticularly consistent with the rest of the paper, nor even with our mainreferences [5] and [22].

Let G be a reductive algebraic group over G. Fix once and for all Borel andCartan subgroups

B [D H (the abstract Borel subgroup, etc. )

m m({, )a) (ma) ._ m(, a).(15.1a)

These specify a root datum ([22])

(X*(H ),Aa, X,(H ), (Aa)’). (15.1b)

Here X*(Ha) is the lattice of rational characters of Ha, and X,(Ha) is the latticeof one parameter subgroups of Ha. We may regard X*(Ha) as contained in(tja) *, and X,(Ha) as contained in ba. Fix a dual group ([5])

LO0_L(Ba )0_ L(Ha )0 (15.2a)

with root datum

qF" (X,(Ha ), (Aa)",X*(Ha ), Aa). (15.2b)

This means that we are fixing isomorphisms

X,(H )_x,(L(Ha )o), (15.2c)

and so on.Roughly speaking, our objects of study will be pairs (G (R), r), with G c_ G

a reductive algebraic group of the same rank as G, G (R) a real form of G 1, andr an irreducible admissible representation of G (F{). (Actually we will consider aslightly smaller class of pairs; G will be among other things the identitycomponent of the centralizer of a semisimple element of G.) To each pair we willassociate a dual pair of the same sort in LG. However, technicalities abound.First, we will actually study only conjugacy classes of pairs. Second, therepresentations will be determined only up to "translation" in the sense ofJantzen and Zuckerman (see [25], Chapter 7). A more serious problem is evidentin this example: take G G SL(2, G), G (R) SL(2, Iq). Let ,ffl be a

1056 DAVID A. VOGAN

holomorphic discrete series representation of G l({), and r the correspondingantiholomorphic representation. Then the pairs (G (lq), r) and (G (Iq), r2) areconjugate under G (by the matrix ( _0/)); so we cannot hope to formulate a resultlike Theorem 1.15 in this setting. The cure proposed here is to consider alongwith G l(lq) and r certain additional "rigidifying" data.

Definition 15.3. A real datum for G is five-tuple g (B, H, o, t, X), such that(a) H c_ B are Cartan and Borel subgroups of G(b) o is an involution of H preserving A(, b). Write t3, a b(-o) for the

+ and eigenspaces of o.

(c) Let [X*(H)] denote the invariants of o in X*(H). Put

X (H) z}

=[x,(n)]Then t/Xo(H).

(d) With notation analogous to (c), X o*/X-(H); here X-O(H) X*(H).A dual real datum for G is a five-tuple (CB,H, 6, [, ), such that (in thenatural isomorphism (0), determined by B, CB , and (15.2c)) 6 is thenegative transpose of , [ X, and t.

Definition 15.4. Suppose g (B, H, o, t, X) is a real datum for G. A realization

of g is a four-tuple (Gl(F),B,t,-r), with the following properties. Chooserepresentatives o t, X a* for and X.

(a) G is the identity component of the centralizer of exp(4rit)(b) G is endowed with a real structure so that H is defined over Iq; and there

is a Cartan involution 0 of G (for G l(Iq)) which fixes H, such that 01, o.

(c) 3’ is a regular character of H(Iq) for G(d) Write (tS1)iR for half the sum of the positive imaginary (that is, o-fixed)

coroots of b in l, and e for the grading of the imaginary roots defined by thenotion of compact root for G l(Iq). Then we require

(O[) (O, ( 1) iR + 2t) (mod 2Z).

(e) Write (p I)R for half the sum of the 2x-integral positive real roots of in 1.Choose a positive system (/R)+ for all the positive real roots, such that if tR ishalf its sum, then (c, p R fiR) 2Z, all a real and x-integral; this is possible by(10.16). Now choose a 0-stable parabolic Q= LU of G 1, with L

_H the

centralizer of t. Write p(u) for half the sum of the roots of 9 in tt, and t forp(u) + tR. Write H(Ft) TA as in (2.9), and write 2p(u fq ) for the determinantof the action of T on the -1 eigenspace of in u. Then we require that there


should be a X*(H) such that if 3’ (F, ).(cf. (2.10)),

2X +/z + t3

rl + 20(. n

(, c> > 0 for a A + (gl, ), 2X0 integral.

PROPOSITION 15.5. Every real datum for G has a realization, which is unique upto conjugation in G and Jantzen-Zuckerman translation on the regular characterinvolved; and the realization determines the real datum uniquely.

So far this is not particularly difficult. The next step is to define the crossaction (of the 2x-integral Weyl group of H in G 1), and the Cayley transforms,for real data and their realizations. (The operations on realizations do not changeG l(Iq).) This is in the spirit of section 7, and we leave it to the reader.

Definition 15.6. Blocks of real data are the smallest sets closed underconjugation in G, Cayley transforms, and the cross action.

This is a stronger requirement than the obvious one of asking only that thedata have realizations with the same G 1(R), and regular characters in the sameblock.

PROPOSITION 15.7. Suppose g and g2 are real data in the same block for G,with realizations on the same G (Iq); and suppose these realizations differ by Cayleytransforms and the cross action inside G (Iq). Then g and g2 are conjugate under G

if and only if the regular characters ’ and 2for G ’(lq) are conjugate under G

This is hard; it is proved in the same way as Theorem 10.1.

Definition 15.8. Suppose g and g2 are real data in the same block for G.Choose realizations on a common G (Iq), differing by Cayley transforms and thecross action inside G (Iq). Write 3, , 3,

2 for the corresponding regular characters;and set

m(g’, g2)=

M(g’, g2)=

(notation (1.1), (1.2)). If gl and g2 are in different blocks, we put m(g 1, g2)M(gl, g2)= 0.

This definition makes sense, even on conjugacy classes of real data, byProposition 15.7.

PROPOSITION 15.9. Suppose g and g2 are real data for G, and ,, ,2 are dualreal data for LG (Definition 15.3). Assume that the Kazhdan-Lusztig conjecture

1058 DAVID A. VOGAN

12.16 holds for all real groups which appear. Then

M( gl, g2) e( gl, g2)m ,2, l),with e(gl, g)= +_ 1; and we can determine the sign.

This proposition is just a formal combinatorial result about manipulating realdata, and is proved by the methods of sections 12 and 13. It is interesting onlybecause of Proposition 15.7, which relates it to multiplicities and characterformulas in real reductive (algebraic) groups.

Example 15.10. Suppose G SL(2). Put B upper triangular matrices,H diagonal matrices, o on H, X 0. We identify and tS* with (3, in sucha way that & corresponds to in iS, and a to 2 in kS*. Then

X,(H)=Z, X*(H)=Z

We consider real data

gt=(B,H,o,t,X),

with B, H, o, and X as above; thus

tc/z.

If 2t 1/2Z, then G= H, and things are dull. There are four other cases:t= 0, 1/4, 1/2,3/4. If t= 1/4 or 3/4, then we can find a realization

(SU(2), B, T, -),)

of gt; here r(y) is the trivial representation of SU(2). These two labellings of thetrivial representation of SU(2) belong to different blocks. The dual real datahave realizations

(PGL(2, R),LB , I, "_),

here r( +__ are irreducible principal series representations which agree on theidentity component of PGL(2, R). (The need to distinguish these two representa-tions gives rise to the need to label the trivial representation of SU(2) by 4 or

3/4.)If 0 or 1/2, then gt has a realization (SU(1, 1),B,t,’/+ ), with r(, + a

holomorphic discrete series representation for SU(1, 1). Now go and gl/2 are inthe same block for G, but these two realizations do not differ by Cayleytransforms and the cross action inside SU(1, 1). However, we can choose arealization (SU(1,1),B;,0,7 -) of a conjugate of go; here r(’-) is anantiholomorphic discrete series. This realization differs from the previous one of

gl/2 by the cross action inside SU(1, 1).


16. Examples.

Example 16.1. Weft groups of SL(n, I:1). We can choose representatives for theCartan subgroups of SL(n, lq) as follows. Choose r < [n/2]. Then there is aunique conjugacy class of Cartan subgroups with r-dimensional compact part. Asa representative, we choose Hr TrA r; here

Ar d(eX,,eX,,eX2, eX2 eXr, eXr, ey’ eY"-2r) 122Xi + _, Yj

T ( d(t(,),..., t(,r),e f’n-2r)Ii [:t, t(dp)(16.1a)

cos sin (/}ti _+_ 1, {[i-

sine, cos i=

(Here and throughout, d applied to a set of numbers (or square matrices) will bethe matrix with these numbers on the diagonal.) The standard identification ofthe complexified Cartan subalgebra (Dr)* with

gives

{t)’-- 2t)iei cnl 2t)i’-O}

n-2r }(cr) * xi(e2i-] + e:zi) + N y/e./ + 2ri--1 j=l

(16.1b)(tr) * {2 doi(e2i_ e2i)).

The Cartan involution 0 acts by sending v to -v, then interchanging the first rpairs of coordinates:

-[(v2ie2i_+VEi_le2i)+ vjejj. (16.1c)or(l)iei)i=1 j=n-2r+

In the notation of Definition 3.10, we have

R A(i,[?r) {e e.li=/:j}

(e ej[i,j > n 2r + 1}(16.1d)

R iFl_. { -I-(e2i 1-- e2i) < < r).

1060 DAVID A. VOGAN

If the root system is orderedProposition 3.12,

lexicographically, then in the notation of

t5R= (0,..., 0,(n 2r- 1)/2,(n 2r- 3)/2,..., (n 2r- 1)/2)

n -2r

2 [(n 2r- 2j + 1)/2]e2r+jj=l

10ilq (1/2,- 1/2, 1/2,- 1/2,..., 1/2,- 1/2,0... 0)

1/2(e2i-1i=1

e2i)(16.1e)

R c (e2k e2t), (e2-’R q Rcu R n

e2t_l) < k,l < r}

Rf= {e,- ejl i,j < 2r}

The subgroups of SL(n, gl) corresponding to R , R q, and Rf are (up to center)SL(r, ), SL(r, (3) SL(n 2r, FI), and SL(2r, Iq). Hence

(wC)- &, the permutation group on r letters; (16.1f)

it acts by permuting {el, e3,... e2r_l} and {e2, e2r simultaneously. Also,

WFI Sn_2r (16.1g)

acting by permuting the last n 2r coordinates, and

WiF! --(Z/27)r, (16.1h)

acting by changing signs on the q’i coordinates of (16. a). To compute Wnmtheelements of WiR having representatives in G--we use Proposition 10.20(a), (d).The roots (a ar) of that Proposition are just

Oti e2i-1-- ezi (i= 1,..., r).

These span if, so, in the notation of Proposition 10.20, R. We must computethe elements ot Wn(y). Suppose first that n > 2r. Then if fl e2j- e,,, wehave

1, i=j(a, fl> 0 j


So in this case Q + WiR, and (by Proposition 4.16)

wFI= WiFt

W(G,H S < (Sn_2r X (Z/2Z)r) n > 2r (16.1i)

W(BCr) x Sn_2r

Here W(BCr) is the Weyl group of Br or C; this last isomorphism is by thestandard embedding of that Weyl group in $2. Next, suppose n 2r. If e and 6are 0 or 1, j :/: k, and fl e2j_ e2k_, then.

_+1, i=j ork(al’ fl)=

0 j,k

Therefore

e+’-{Sv,...Sv2, lyjRiR)_ WiR

WR Q+ (Z/2Z)r-’

wiR/WIq Z/2Z.

So

W(G,H ) S V, (Z/2Z)r-1 l n 2r (16.1j)W(Dr)

(This is also clear from the fact that H is fundamental, so that W(G,H)W(K, Tr).) Of course replacing SL(n, FI) by GL((n,R) makes W equal to

Wm in this case by Proposition 10.20(e).

Example 16.2. The block of the trivial representation of SL(3, R). The Cartansubgroups H (split) and// (fundamental) are as in Example 16.1. Any regularcharacter corresponding to the same infinitesimal character as the trivialrepresentation must have differential half the sum of a set of positive roots. OnH, W(G,H)= W(g, ), so (up to conjugacy) we can consider only the regularcharacters ,f0 with differential

(1,0, 1) p; (16.2a)

we are introducing coordinates as in (16.1b). Since H has four connectedcomponents, there are four such regular characters. We write them as

,o 0y(1,,2,3 (H)

1062 DAVID A. VOGAN

with the convention that

",t,(d(tl, t2, t3))= I-Ieit$

Since d(Si) is in SL(3, R) only if H8 1, we have

We have

(Yl,l.,)) trivial representation; (16.2b)

for since H is split, rr(3,,) is just the principal series representation induced fromthe character 3,, of H. By Theorem 8.5, the other regular characters of H in thesame block as ,o are obtained by the cross action of the Weyl group; so we willnow compute this. Suppose a e e2. Then

so by Definition 4.1

as a character of H. Nowsap’-p--a;

,o x r r

so s 7o looks like 7_,,_=,) on diagonal matrices with entries _+ 1. However,we have also changed to sO. This we can reverse by conjugation by s; and wecompute

s x 3,, conjugate to 3’-,2,-,,,,, (a e,- e2). (16.2c)

Similarly,

s/ x 3,, conjugate to 3,,._,,,_,2 (fl e2 e3). (16.2d)

Using these, we compute

cross orbit of -t, yl, To_ 0

1,- 1,1’ Yl,- 1,- 1)" (16.2e)

On H l, W(G,H)= WiR has order 2 ((16.1i)); in the coordinates (16.1b), thenon-trivial element of W(G,H) interchanges the first two coordinates. Since His connected, there are just three regular characters (up to conjugacy) withdifferential conjugate to O. We choose

yl (0, 1, 1)

(1,0, 1) (16.2f)

3 (1,- 1,0)


FIGURE 16.2

as representatives. Since H is connected, the cross action of IV(g, b) is just theusual action on , so we will not say more about it. We would, however, like tospecify Cayley transforms explicitly. So let a be the abstract simple rootcorresponding to e e2 on b. If G(7,) is defined, and 3’ belongs to it, then theroot a on kS1 corresponding to a must be imaginary. By (16. ld), this forces, 2., that is,

c(’y) , all ,,o E D(c) (16.2g)

The domain of c is easily computed to be (-/, y0_ ,-1,1 We arrive finally at Fig.16.2 for the block of the trivial representation in SL(3, Iq). In it, the abstract Weylgroup S3 is acting on II2 as usual: a e e2, and the reflection s fixes the linethrough exp(5ri/6); fl e2 e3, and s fixes exp(irr/6). The cross action of theWeyl group is then the geometric one on the points representing regularcharacters; and Cayley transforms are shown by arrows.

1064 DAVID A. VOGAN

We can say a little more about the nature of the representations (3:). As wasmentioned already, (7,1,1) is the trivial representation. The only other unitaryrepresentation among the six is the irreducible standard representation

if(q, 3) r(:3), (16.2h)

which is a tempered fundamental series representation. Both (3,) and (72) areLanglands quotients of reducible (nonunitary) fundamental series representa-tions; we have

(16.2h)

in the Grothendieck group. Both (’y_l, 1,1) and r(3, 1,--1,--l) are Langlandsquotients of ordinary non-spherical principal series representations. Their formalcharacters are

(16.2h)

Finally, the formal character of the trivial representation is

(16.2h)

All of the formulas (16.2h) are well known, but we may as well outline theirderivation from the Kazhdan-Lusztig conjecture. We first tablulate the objectsof Definitions 12.2 and 12.3, normalizing to be zero on ./3.

TABLE 16.2(i)

T3 0 yl y22; (u- 1)2, + u),3 ,l + ,o,_,_ + 3,o./2 fl 2 + + (U- 1) + ul,-- l,l

2 a (u o + (u 1) 7l,-- l,l )-l,-l,l l,- l,l

,_ (u + (u 1)l)v,_ ,_7 2 a, fl (u 1)7 7 (u 1)7 o,_ ,_ + (u 1)7l,- l,l

The Bruhat r-order of Definition 2.12 turns out simply to be the partial order


defined by 1I(3,). The map D may be computed by the algorithm of [24]; one gets

TABLE 16.20)

Dy3 y3DV u-113,1 (u 1)3,3]D3, u-[3,- (u 1)3,3]

D3,0_l,_l,1 U-2[3,0_l,_l,1 (U 1)3, -I- (U U)3,3]D3,1o, 2[ (u 1)y-1,-1 U- 3,1,-l,-1 -" u2 U)3,3]

D3, u-213,1 (u 1)3, (u 1)3,1 + (u 2u + 1)3,3]

Of course it is elementary (but tedious) to verify that the map D defined byTable 16.2(j) satisfies the defining conditions for D in Lemma 12.14. Theformulas are organized so that the polynomials R,v (Lemma 12.14) may be read

0 U2off easily. It should be observed that if 3,2, 3, 3,_. ,-1,, then Rv u.

For complex groups of rank two (that is, the case originally studied by Kazhdanand Lusztig) one always had Rv=(u-l)k, with k= lI(3,) lI(d). Moregenerally, the constant term of Rv is always _+ in the case of complex groups(if Rv 4 0); and even that fails here.From Table 16.2(j), it is fairly easy to compute the elements Cv of Lemma

12.15. They are:

TABLE 16.2(k)

3,3 3,33,1 3,1 + 3,33,2 3,2 + 3,3

o__ o_,- ,1 ,- +3,10,_ 1,_ 3,10,_ 1,_ _{_

In particular, the non-zero P,v are all or 0, just as in the rank two complexcase. The important distinction from the complex case is that q < 3, does notimply P,v 4: 1", take for example q 3,3, 3’ 3,o_ ,_,. This says that the relation"occurs in the formal character of" is not transitive; 3,3 occurs in the formalcharacter of 3,2, and 3,2 in the formal character of 3,o_ but 3,3 does not occur1,- 1,1,

in the formal character of 3,o_ By applying Theorem 15, we deduce that1,- 1,1"

the relation "is a composition factor of" is not transitive in general. Thisphenomenon is well known, as it occurs in all of the rank one groups butSL(2, Iq) and SL(2, ).The formulas (16.2h) follow from Table 16.2(k) by Theorem 12.17.

1066 DAVID A. VOGAN

Example 16.3. The block of the trivial representation of SU(2, 1). This block isin duality with the one of Example 16.2 in the sense of Theorem 1.15. To provethat, one computes the strong bigradings attached to a non-holomorphic discreteseries, and to the trivial representation (Definition 4.12), and verifies that theyare dual to those attached to 3 and /2 for SL(3, lq) (Example 16.2). Then theduality in question follows from Theorem 10.1. Write for the SU(2, 1) regularcharacter dual to y for SL(3, Iq). Then

(2) trivial representation

() Langlands quotient of a principal series

(2) Langlands quotient of another principal series

(o_ ,,-,,,) holomorphic discrete series(16.3a)

(’ l,- 1,- 1) anti-holomorphic discrete series

r() non-holomorphic discrete series.

By Theorem 10.1, Figure 16.2 applies to these representations as well. We leaveto the reader the task of computing D and the various C.; by the results ofsection 13, one only has to perform some simple manipulations with Tables16.2(j)-(k). An interesting feature of this example is that all of the rr() areunitary representations of SU(2, 1); and we saw that only two of the (3’) forSL(3, R) were unitary. For this reason and others, it seems very unlikely that theduality relationship has anything to do with unitarity.

Example 16.4. The gradings of D4. We realize R D4 in R4 (with basis (ei))as usual, by

R e + [i vj). (16.4a)

Then

group of permutations and even numbers ofW(R ) W

sign changes on the basis e ); (16.4b)

this group has order 192. Put

R + (e +- eli <j)

1"I--- (e eg.,e2 e3,e3 e4,e3 + e4) {Ctl, cto,Ct2,ct3).(16.4c)

The numbering of the simple roots in II is explained by the Dynkin diagram,


which is

t

(16.4d)

(This is not the usual numbering of the simple roots, but it is a natural one.)By Lemma 3.14, a grading e E(R) is determined by its restriction to (ai),

which may be arbitrary. If dp--(d?o, dPl,2,3)(Z/2Z)4, then we define

% E(R) by

th(Oi) i" (16.4e)

We may represent a grading e by putting (ai) at the a vertex of the Dynkindiagram; thus

e(O,l,O,l (’--)1 00

To compute the orbit of the cross action on E(R) (Definition 3.24), suppose a

and/3 are simple roots, and e(a)= 1. Then

e( fl if (5, fl) even(16.4f)(s,e)(fl)=

1-e(fl) if(t,fl)odd.

Thus s e agrees with e except on the vertices adjacent to a. Using this, wecompute easily that the orbits of the cross action on E(r) are

0 11 10 }(gp= O10 1’

(I=( 1000’ 1100’ 0111’ 001)1ii (00 0 O1 0 11 10 }1’ 1’ 0’ 0

(16.4g)

01 0 10 0 )(III 0 00’ 0’

(o)Ct O00

(Herep stands for principal, and for trivial.) All of these arise as the grading bycompact and noncompact roots in some real form G of S0(8,13). The relevant

1068 DAVID A. VOGAN

real forms are tabulated below. In each case the set R0 of compact roots isspecified.

TABLE 16.4(h)

Orbit in E(R) G Ro compact roots

Ce SO(4, 4)I SO(2, 6)II SO*(8)(III SO*(8)t SO(8)

- el -+ e2, -- e3 -+ e4) D2 Dg_ (A l)4

(-e2-+- e3,_+e2+_ e4,_e3- e4) D3A(el ejl i, j < 3) U (e + e41i < 3)-----A(ei- ejl =/=j} A

Ro= R= D

We stated at (3.26) that the cross orbits on E(R) were in one-to-onecorrespondence with equal rank real forms of G c. This appears to contradictTable 16.4(h), since SO*(8) is attached to both (II and (iii; and in any caseSO*(8) S0(2, 6), which is attached to i. The point is that the isomorphismS0(2, 6) SO*(8) cannot be realized by conjugation in S0(8, G); and neithercan the isomorphism between the two versions of SO*(8) used for gii and Iii.By "real forms" at (3.26), we meant "real forms up to G C-conjugacy. To replacethis with "real forms up to isomorphism," we must expand W to includeautomorphisms of the Dynkin diagram. Here 0I, 0I, and (III differ by theautomorphisms which rotate the Dynkin diagram.

Set

0 q=lO0 =000 =00 et=O O0e, 01 O’ II III O’ 0

tg W eC_E(R) (eE(R)). (16.4i)

Since the various R0(e are tabulated in Table 16.4(h), we can immediately writedown W0(e)= W(Ro). The cardinality of W2(e) may be computed by theformula

wor the order of a finite homogeneous space. We would like also to specify theinclusion

Wz(Q/W0() 91,() Z(R

of Proposition 3.16. First we must describe Z(R). In the notation of Definition


3.3, we compute

/ lattice of coroots

((X1, X2 ,X ,X4) I Z4 2 Xi--- 0 (mod 2Z))/6 (2 R4 I)ki_ )tj Z, all i, j)

()kl,)k2,Yk3,)k41 all)k 7, or all)t/ 1/27)(16.4j)

Z(R)=/5//= (O,(1,O,O,O),(1/2,1/2,1/2,-1/2),(,,,)){O, zl,z:,z3);

here we write for x + . In the following table, the second column gives anelement of /5(R) defining e (cf. Lemma 3.14 and Propositon 3.16); (e) isexhibited as a subgroup of Z(R) (Proposition 3.16); and we give representativesof W2(e)/ Wo(e). For that purpose, we write s s,i.

TABLE 16.4(k) q groups for D4

GradingRepresentatives of generators of(e) in W2(e); (xl,x. ,x3, x4)--)--

ep (1, 1,0,0) (Z/2Z) Z(R)= (O, Zl, Z2,Z3} (x1, -x2,x3, -x4) (x3,x4,x1, x2)

(1, 0, 0, 0) S4 W(A3) (0, Z1} X --)’ X

t[ii (_,,, ) S4-- W(A3) {O, z2} x---)-x

t[Ii (,,,) S4-- W(A3) {, Z3} X---)--X

t (0, 0, 0, 0) W W(D4) {0}

The easiest way to calculate (e) is by first calculating [%(e)l as describedafter 16.4(i). The elements exhibited in the last column obviously belong to (e),and so generate it by counting. Now the formulas for (e) are immediate fromthe definition of the map in Proposition 3.16.

Example 16.5. Cayley transforms of the principal grading of C,. RealizeR C in [21n as usual, with

R +__ e +_ ]i=/=j} t_) +_2ei}

W(R ) W permutations and sign changes of ei)

S, , (Z/2Z) (the hyperoctahedral group) (16.5a)

R + (e +_ ejli <j} U (2ei}I-[ "-(e --ei+ 11 < i< n- 1} t_) {2e,}.

1070 DAVID A. VOGAN

We consider the grading E(R) defined in any of the following equivalentways. Put h (1, 1,..., 1). Then

Then

e(a)=--(a,X) (mod 2)

f.( +__ (e .-I- ej)) 1, f.(e ej) 0

Ro(f.) e An_

RI(,) e .q- ).

(16.5b)

Wo(e.) W(A,_) S,, c_ W

W2() group generated by Wo(() and -Id.(16.5c)

There are two kinds of one dimensional Cayley transforms. The simpler is(Definition 5.2, Lemma 5.1)

ot 2era, R .-t-- ’i .-t-- ej i, j vem), trivial(16.5d)

More interesting is the second kind:

a er + e

R’= {+_el+_ ejli, j=/= r,s} .3 -+-(e es)

=g trivial on R

(er- e)=(’)

c()l: lCa(.)(er- es)-"

(er-

It is not hard to convince oneself that every admissible sequence is (up topermutation and sign changes) conjugate under W0() to one of the followingform"

S Sp,q,r-’ {2ei,..., 2ep,ep+ + ep+2,ep+ ep+2,ee+2q-I + et,+2q,el+2q-- et,+2q,el:,+2q+l + e/,+2q+2, (16.5f)ep+2q+3 + ep+2q+4, ep+2q+2r_ + ep+ lq+2r}"


For this sequence S, we have

R s= (+_ei+. ejli, j>p+2q+2r (ee+aq+j_,-ep+q+2jll < j < r)

(16.5g)

The algorithm of Lemma 5.6 replaces Sp,q, by Sp+2,q_ 1,r; SO if we put

S Sp+2q,O,r (16.5h)

then S’ is strongly orthogonal, and

cS(Q=cS’(e). (16.5h)

Since the roots ep+ 2q+2j- ep+2q+2j lie in R0(e), the correspondingreflections sa belong to W0(e). In cs(e), however, flj is an odd root; so

s(e)). This observation leads, in the context of Lemma 6.1, tos Wo(c "

W0(Cg(,)) W(hn_(p+2q+2r)_l)

C(W0()) W(An-(p+2q+ 2r)- 1) X (Z/2Z)

(8 span of Sp,q,r (cf. (16.5f))).

(16.5i)

Using Lemma 9.1 (case of c) and Proposition 4.16, one can continue from hereto determine the Weyl groups of all the Cartan subgroups of Sp(n, FI). We leavethis to the reader.

Example 16.6. Connection with the "very strange formulas". Let G be thegroup SL(2, FI), and consider the regular character /0 for the split Cartansubgroup, such that r(7) is irreducible and has the same infinitesimal characteras the trivial representation of G. One checks easily that the cograding g(,{0) istrivial; that is, the roots do not satisfy the parity condition. Thus { ,0} is a blockby itself, and r(3,) is an irreducible (non-spherical) principal series. The dualgroup of ( of Theorem 1.15 is SU(2) and r() is the trivial representation. AsSU(2) is connected, we have

Now consider G c

(Definition 9.4. ) (16.6a)

--SL(2, (3), and define

1072 DAVID A. VOGAN

as usual. The Iwasawa decomposition of G c is now

GC= AN. (16.6c)

This verifies a conjecture of P. Sally that AN may be regarded as a complexsimple Lie group; I am grateful to him for supplying this example.

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IRREDUCIBLE CHARACTERS OF SEMISIMPLE LIE GROUPS IV ...Byinspection of (1.10), this gives 2 102-1190...

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