Harmonic Analysis on Real Reductive Groups III. The Maass-Selberg Relations and the Plancherel...

Annals of Mathematics

Harmonic Analysis on Real Reductive Groups III. The Maass-Selberg Relations and thePlancherel FormulaAuthor(s): Harish-ChandraSource: Annals of Mathematics, Second Series, Vol. 104, No. 1 (Jul., 1976), pp. 117-201Published by: Annals of MathematicsStable URL: http://www.jstor.org/stable/1971058 .

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Annals of Mathematics 104 (1976), 117-201

Harmonic analysis on real reductive groups III. The Maass-Selberg

relations and the Plancherel formula By HARISH-CHANDRA

TABLE OF CONTENTS

Section 1. Introduction ................................................... 118

Part I. The c-, j- and p-functions 2. Some elementary results on integrals .......................... 120 3. A lemma of Arthur ........................................... 125 4. Induced representations ........................................ 127 5. Intertwining operators ......................................... 129 6. The mapping T-+XT ..131...................................... 131 7. The relation between induced representations and Eisenstein

integrals .................................................... 132 8. Some simple properties of E(P: f: v) ............................ 134 9. Proof of Theorem 7.1 .......................................... 136

10. An application of Theorem 7.1 ................................. 138 11. Some properties of the j-functions ............................. 139 12. The p-function in a special case ........... .................... 141 13. The p-function in the general case and irreducibility of

representations ............................................... 142

Part II. Maass-Selberg relations and the functional equations 14. The Maass-Selberg relations .................................... 146 15. Some preparatory remarks ..................................... 147 16. Proof of Theorem 14.1 ......................................... 148 17. The functional equations for E(P: f: v) ......................... 152 18. Relations between the c- and j-functions ....................... 155 19. Rationality of 0cQIP ............................................ 155

Part III. Explicit determination of the Plancherel measure 20. Evaluation of (0a)(P) ........................................... 157 21. The characters )O,p ............................................ 160 22. Computation of (9O,,,, Osa) ....................................... 162 23. The characters of the discrete series ........................... 163 24. Determination of p(w: v) in a special case ....................... 165 25. Extension of p, j and c on the complex space ................... 168 26. Some applications of Theorem 25.1 ............................. 170 27. The Plancherel formula for K-finite functions .................. 171 28. First reduction in the proof of Theorem 25.1 ................... 175 29. Remarks on notation ........................................... 176 30. Some auxiliary lemmas ......................................... 177 31. Recapitulation of some earlier results .......................... 178 32. Statement of Theorem 32.1 ..................................... 180 33. First step in the proof ............ . ........................... 181



118 HARISH-CHANDRA

34. Second step .185 35. Third step .185 36. The formula for . .(.: .) 187

Part IV. Commuting algebras of induced representations 37. A lemma on surjectivity .191 38. The centralizer of rF(2F) .192 39. Closer study of a special case .195 40. A bound for the intertwining number .196 41. Irreducibility of the fundamental series .197 42. Appendix .199

1. Introduction

This paper is devoted to a deeper study of the Eisenstein integrals, their functional equations and the relationship of the Plancherel measure to their asymptotic behavior. We have seen in [2 (j), Thm. 18.1] that the constant term of an Eisenstein integral is a finite sum of plane waves whose ampli- tudes are given by the corresponding c-functions. It is an important principle that the absolute values of these several c-functions are always equal to each other and the Plancherel measure p can be written down very simply in terms of their common value (Lemma 13.4). The situation is therefore exactly similar to what was found some years ago in the case of the symmetric space G/K [2 (b), p. 611].

The c-functions are intimately related to the intertwining operators for induced representations which I denote by j. The j's have a natural product formula, which can be transferred to the c's and finally to a. This reduces the computation of p to the special case of maximal parabolic subgroups and thus enables us to obtain an explicit formula for p (? 36). It turns out that [e, c and j can all be extended to meromorphic functions on the complex space H, which parameterizes the Eisenstein integrals. Moreover P is holomorphic and nonnegative on the real subspace A. Let A' denote the set of all regular points of A. Then for v e A', p(v) > 0 and the corresponding induced representation wp of G is irreducible. The decomposition properties of wp, when v is singular, are intimately related to the behavior of ( at 2. For example if dim W = 1 and p(0) = 0, then w0 is irreducible.

This paper is divided into four parts. After introducing the intertwining operators, we show that the Eisenstein integral E(P: *: v) is essentially a matrix coefficient of the corresponding induced representation 7p, (Theorem 7.1). This establishes a close relationship between the c's and the j's and enables us to continue the j's analytically to the real subspace H (S 11). We introduce the function a,,. on g by means of the formula

tt-(V)iPIP()ji;p(V) = 1 (v e a')



HARMONIC ANALYSIS ON REDUCTIVE GROUPS III 119

and show that (-4. is holomorphic on g and p4(v) > 0 for v e A'. This is first done under the assumption prk P = 1 (? 12). The general case then follows from the product formula for jp (55). In particular we obtain the relation (? 13)

11W _= a e 1 ( 4) vA, a

The irreducibility of pe,, (v e A') is a consequence of the fact that Pf,,(V) > 0 (?13).

Part II is devoted to a proof of the Maass-Selberg relations (5 14) which assert, roughly speaking, that all the plane waves, appearing in the constant term of a suitable eigenfunction, have the same intensity. So if one of them is missing, we can actually conclude that the eigenfunction itself is zero. The functional equations of the Eisenstein integrals (5 17) are an immediate consequence of this fact. These equations involve certain linear transformations ?cQIp(s: v) which are rational functions of v e H (5 19).

In Part III we come to the Plancherel formula and the determination of the Plancherel measure. Our first task is to compute (s, ,) (see Theorem 22.1) where EH,, is the character of the induced representation ip (v e -) and ba the wave-packet given by

Oa= 5 pw(v')a(v*)E(P: A: v)dv (a e

Let F be the Cartan subgroup of G associated to P. Then this calculation leads to a simple formula for 'Foa (Lemma 24.1). In the special case when F is fundamental in G, one immediately obtains a formula for s,,, by applying [2 (i), Lemma 17.5] to O. This shows, in particular, that p4a,(v) is a polynomial function of v in this case (Theorem 24.1).

Theorem 25.1 asserts that pal, extends to a meromorphic function on and this function is holomorphic on a tube domain around g where it has at most polynomial growth. This theorem has the following two important consequences. Firstly it enables us to prove that the c- and j-functions also extend to meromorphic functions on g,. Secondly it implies that the wave- packets Oba corresponding to a e C(2) lie in the Schwartz space of G. The second result is crucial for the proof of the Plancherel formula (5 27).

On account of the product formula for p,, it is enough, for the proof of Theorem 25.1, to consider the case when prk P = 1. By the earlier result fiw is a polynomial if rank G > rank K. Hence the important case is when rank G = rank K. Theorem 25.1 is proved by an explicit determination of p, in this case (5 36). The method used here is basically the same as that of [2 (d), 5 24].



120 HARISH-CHANDRA

Theorem 38.1 contains the main result of Part IV. We use it to obtain an upper bound for the self intertwining number of an induced representation (corollary of Lemma 40.2) and, as an application, we show that every representation in the fundamental series is irreducible (Theorem 41.1).

The work presented in this paper has been carried out over a period of several years and a brief account of some parts of it has appeared in two short notes [2 (g), (h)]. Moreover I have given lectures on it in Princeton during the spring of 1971 and again during the academic year 1974-1975. For recent work by other authors on similar questions see [1], [4 (b)], [9].

As far as possible we shall keep to the notation of [2 (i), (j)] and therefore any undefined symbols should be given the same meaning as there.

Historical note. Intertwining operators have been studied intensively for twenty years by Bruhat [6], Kunze and Stein [5], Shiffman [8] and Knapp and Stein [4]. The product formula for p was first obtained by Gindikin and Karpelevic [7] in the case of GIK. The corresponding results of this paper may therefore be regarded as natural generalizations of this earlier work.

Part I. The c-, j- and tt-functions

2. Some elementary results on integrals

Fix a special vector subgroup A of G [2 (i), p. 114]. A root of (g, a) is called reduced if ra is not a root for 0 < r < 1 (r e R). If P e C(A), we denote by ?(P) the set of all reduced roots of (P, A). If P,, P2 e C(A), put I(P2 I P1) = z(P2) nl (P,) and' d(P,, P2) = [?(P2 I Pi)] where P2 = 0(P2) as usual.

Then d is a metric on the finite set C(A) and the following lemma is an easy consequence of the definition.

LEMMA 1. Fix P1, P2 and P in CP(A). Then the following conditions are equivalent.

( 1) d(Pi, P2) = d(Pi, P) + d(P, P2), ( 2 ) ?(P2 PI) Y ?(P I Pi), ( 3) (P2 P1) is the disjoint union of I(P2 I P) and ?(P I P1). We say that P lies between P, and P2 if (1) holds. Moreover P1, P2 are

said to be adjacent if d(P1, P2) = 1. Define

IP21P1(2) = 5l exp {-(y + p1)(H,(n))}dn (V e a*) .

Here Pi e 9P(A) (P, = MANS, i = 1, 2), N1 = 0(N,), and pi = pp1, H1(x) = Hpl(x) (x e G). Moreover dn = d(N2 fn N,) [2 (i), S 7]. It is obvious that

I [S] denotes the number of elements in a set S.




0 < IP21P1()) ?

and Ipilpl(v) = 1.

LEMMA 2. Let P1, P2 and P be three elements in 9P(A) such that P lies between P1 and P2. Then

Ip21P1(V) = IP21P(V)IP1P1(V)

for v e a*.

Here the method of proof is the same as in [7]. First we need some preparation. Given Ho e g such that 0(H0) =-H0,

we associate to it a psgp Q of G as follows. Let q be the subspace spanned by all Xe g such that [Ho, X] = xX for some X > 0 (X e R). Then q is a parabolic subalgebra of g and Q is the normalizer of q in G. We say that Q is the psgp of G determined by Ho.

Fix P e 9P(A) and a e ?(P). Let ua denote the hyperplane a = 0 in a and let Za be the centralizer of oa in G. Put Ma = 0Za and *Pa = Ma n P. For X e a*, let gA denote, as usual, the space of all Xe g such that [H, X] = X(H)X for all H e a.

LEMMA 3. Ma is a reductive group satisfying the conditions of [2(i), ? 3] and *Pa is a psgp of Ma. Moreover the Langlands decomposition of *Pa ,is given by

P. = MAaNa

where aa = RHa and

=a =Ek1i ka

Fix an element Ho e Ca such that every root of (g, a) which vanishes at Ho is a multiple of a. Let Q = MQAQNQ be the psgp of G determined by H0. Then it is clear that Za = MQAQ and Ma = MQ. Therefore our first assertion follows from [2(i), Lem. 4.9].

Now 3a is the centralizer of Ho in g. Hence 5a D m + a and Ja n n = na. This implies that

5a = 0(nta) + m + a + na

Since a = Ca + RHa and Ca lies in the center of 3a, we conclude that ma = 0(nta) + m + aa + na .

This shows that m + aa + na is a parabolic subalgebra of ma. Since it is obviously the Lie algebra of Ma n P, it follows that *Pa is a psgp of Ma and its Langlands decomposition is MAaNa.

Let P1, P2 be two distinct elements in CP(A) and a an element in I(P2 I P1).



122 HARISH-CHANDRA

Then a e ?2(P1) and - a e I(P2). Hence if *P = Ma ln P1, it follows from Lemma 3 that

Ma n P2= *P

Now suppose P1, P2 are adjacent. Then I(P2 l P1) = {a} where a is a simple root of P1. Put

(P't A') = (P1, A),a} in the notation of [2(i), ? 6]. Then P' = M'A'N' and it is obvious that Mt = Ma. Since dim A = dim A' + 1, it follows that *P is a maximal psgp of M'.

LEMMA 4. Let P1, P2 be two adjacent elements in <3(A) and a the unique root in I(P2 I P1). Put (P', A') = (P1, A){(,} (P' = M'A'N') and *P = M' n Pi (*P = *M. *A. *N). Then

1(2)pl~v) = I~p-l~p(*2) (V e a*)

where *2 is the restriction of v on *a = RHa

We observe that *M = M, *N = Na and

n2 n a(n^) = o(nl)a

Hence N2 A N1 = *N. Put *K = K n M'. Then M' = *K.*P. Put *H(m') = H~p(m') for m' e M'. Then it is obvious that

H1(m') = *H(m') e *a = RH,,

for m' e M'. Put *p = pap. Since n1 = n' + nt, it follows that p, = *p on *a. Therefore since *N c M', we conclude that

1P21 P1()) = exp {-(*L)

+ p1)(H1(n))}dn

- exp {-(2 + p1)(*H(n))}dn = L-l~p(*V) .

Now we come to the proof of Lemma 2. If d(P2, P1) = 0, our statement is true since both sides are 1. So we assume that d(P2, P1) > 1. First assume that d(P2, P) = 1. Put (P', A') = (P, A)(a} (P' = M'A'N') where a is the unique root in I(P2 I P). Put *P = M' n P (*P *M. *A. *N). Then

M, n P2=**P, *M=M, *N=Na1 and *a =RHa, .

Moreover

IP21 A = P

from Lemma 4, where *2 is the restriction of v on *a. Now P= MAN, N = N'* N, N2N= NoaN' and Nor U Nor c M'. Since




a e ?(P2 I P) c ?(P2 P1), it follows that

NnN=NiflN,N2nNi=Na(N'fnN)= *N(NPANA).

Let dp denote the invariant measure on N'/N' n N1 such that

dpe-d(N' n N1) = dN'

We observe that N' = (N' n N1)(N' n N1) and the projection of N' on N'/N' n N1 defines a homeomorphism of N' n N1 onto N'/N' A N1. Moreover d(N' n N1) corresponds to dp under this homeomorphism. Hence

IP2 1P) 5 exp {-(i + p1)(Hl(n))}dn

= | _XN,/N ,exp + p1)(H1(** -n'))}d*Jdp . *NxN'/N'nN,

Here d*ni = d *N. Fix *n e *N and put *K = K n M'. Since M' = *K. *P. we have *n = *k*p where *k e *K, *p e *P. Then

Hi(*in') = Hl(*pn') = Hl(*pn'. *p-i) + Hi(*p) = Hl(*pn'.*p-i) + *H(*p)

for n' e N'. Here *H(m') = H*p(m') (m' e M') as before and we have to use the fact that *N = Na, c N1. Moreover *P = M. *AN, c M' n P1. Therefore *p normalizes both N' and N1. Hence

|N'lNi exp {-(i + p1)(H1(*fn'))}dp

= exp {- ( + P1)(*H(*P))}3N,/NnN1(*P1) 5 exp {- (P + p1)(H1(n'))}dp

where

dN,/N'lnN1(*P) = det (Ad(*p1))A, nt, I - Let *p = *m .*a *n (*m e *M, *a e *A, *n e *N). Then *H(*p) = log *a =

*H(*nf) and

3N'/N'fnN1(*P1) = det (Ad (*a-l)),n-fn1

Define p+ and p- in a* as follows.

p+(H) = 2 tr (adH)t nt , p-(H) = 2tr(adH)antn (H e a).

Since n, n n nA + in n nu and n = n A ni + i n Aii, it follows that

P1 = p+ + p-, p= p+ - p-

where p = pp. Therefore since n' n ii, = n n fii, we get

3N,/N'lnN1(*P-) = det (Ad(*a-1)),na, = det (Ad(*a)),n.1 = exp {2p-(*H(* p))} .



124 HARISH-CHANDRA

Hence

exp{-(@ + p)(*H(*p))}j3N ,NnNj(*p-l) = exp{-(v + p)(*H(*ii))}

and this shows that

IP21Pl(V) = |exp -(P + p)(*H(*n))}d*i. I exp{-@( + pj)(H1(n'))}d1ie

But since do-i corresponds to d(N' A N1) and N' A N1 = N n N1, we have

|N'/N'nl exp {-(v + p1)(H1(n'))}dp = 1pjp1(v)

This proves that IP1P(v) = ILpl*p(*v)1pjp1(v)

= 1p21P(V)1plp1(V)

The above result enables us to prove the following lemma by induction on d(P2, P1).

LEMMA 5. For a G I(P2 I P1), define *Pa = MAaNa as in Lemma 3. Then

IP2 P1(V) = HaIa6 (P2IP1) I*PaI*Pa(a) (e G a*)

where a is the restriction of v on RHa.

It is clear that Lemma 2 is an immediate consequence of Lemma 5. For P e C1(A), put

y(P) = Iplp(pp) = exp {-2pp(Hp(if))}dif.

We know from [2(j), Lem. 20.3] that 7(P) < oo. If a is a reduced root of (g, a) define

ca =infp <pp, a>

where P runs over all elements in CP(A) such that a e ?(P).

COROLLARY 1. Let P1, P2e 9C(A) and v e a*. Then IP21pl(v) depends only on the values of <v, a> for a e I(P2 I P1). Moreover IP21 P(1) < co if

<v, a> > a for all a e2I(P2 IP1).

The first part is obvious from Lemma 5. Now fix a e I(P2 P1) and suppose there exists an element P e CP(A) such that a e ?(P) and

<k, a> > <pp, a> .

Put rI = pp. Then since

IAIp-0 = 7y(P) < 00

and a e ?(P) = 1(P I P), we conclude from Lemma 5 that




I*iai*pJ(fta) < ?

Hence

I*fial*Pa(2)a) < Li*ai*Pa(f1Aa) <

Our assertion now follows from Lemma 5. For any reduced root a of (g, a), define

Pa = 2 Lklmk k

where mk = dim 9ka in the notation of Lemma 3. Then Pa = P*Pa Put

7a = fY(*Pa) = 5 exp {-2Pa(Ha(if))}dff Na

where Ha(y) = H*pa(Y) (U e Ma)

COROLLARY 2. In order that Ip2,,1(v) < ao, it is sufficient that

<v, a> > <Pat a>

for all a e I(P2I P1).

This is obvious from Lemma 5 since 7a < ?? Put

7P21P1 = llael (P21P1) fa

for P1, P2 e C1(A). We note that if P e CP(A) (P = MAN), then

Hi (a(nf)) = - Hp((i) (f e N)

and therefore 7(P) = 7(P). In particular 'i-a = 7a and this shows that YP-Ip

is independent of P e CP(A). Moreover we have the following result.

LEMMA 6. 7(P) is independent of P e 9P(A).

Fix P e 9P(A) and let (P0, A,) be a minimal p-pair in G such that (P0, A,) - (P, A). Put *P = M n PO. Then *P is a minimal psgp of M and by [2(j), Lem. 20.3]

7(po) = V(PM7P) @

Since any two minimal psgps of G are conjugate under K, it is clear that PY(P0) is independent of P0 e 9(A). Similarly for PY(*P). The statement of the lemma is now obvious.

3. A lemma of Arthur

Let w be a representation of G on a complex, locally convex and complete Hausdorff space V. As usual let 6(K) be the set of all equivalence classes of irreducible (finite-dimensional) representations of K. For any b e 6(K), let eb denote the character of b and put ab = d(b) conj eb where



126 HARISH-CHANDRA

d(b) = $b(1) is the degree of b. For a finite subset F of 6(K), put

CF = FbeF b

and consider the operator

EF =1 caF(k)7c(k)dk

on V. (Here dk is the normalized Haar measure on K.) Put VF = EF V. We say that Xr is admissible if dim VF < 00 for every F.

Suppose Xr is admissible. Put ZF = End VF. We extend T e ZF to a continuous linear transformation on V by setting

Tv = TEFv (v G V) .

Then 'ZF CZF if F' c-F. Put End V = UF ZF

where F runs over all finite subsets of 6(K). Then Endo V is a subalgebra of the algebra of all continuous endomorphisms of V.

Put 2 = Cc (G) and

fF = acF*Kf *K aF (f G 2)

where the convolution is over K. Let SF be the image of 2 under the mapping f F-f-. 2 is an algebra under convolution (over G) and 2F is a subalgebra of 2. Define

w(f)= f(x)7c(x)dx (fe?) G

where dx is the standard Haar measure on G [2(i), ? 7]. Then it is obvious that

EFzr(f)EF = 2(fF) .

Therefore the mapping f - w(f) (f G SF) may be regarded as a representation of SF on VF. We denote this representation by ZF.

For T e End0 V, the operator Tz(x) (x e G) is of finite rank. Put

fT(x) = tr TC(x) (x e G) . The following lemma, which I owe to J. Arthur, is very simple but it will be crucial for us.

LEMMA 1. Let z be an admissible representation of G on V. Fix a finite subset F of 6(K) and for T e End VF put fT(x) = tr Tz(x) (x G G). Then the following three statements are equivalent.

(1 ) The mapping T - fT is injective on End VF. (2 ) lr(?F) = End VF.




(3 ) 7rF is irreducible.

Put Z = End VF. Then Z is a simple algebra and obviously (2) and (3) are equivalent by Burnside's theorem. Now the bilinear form

(St T) X o trST (St T eZ)

is nondegenerate on Z. Let Z: be the space of all T e Z such that

tr Tr(,8) = 0

for all 98 e SF. Then Z: = {0} if and only if Wc(2F) = Z. Hence in order to prove that (1) is equivalent to (2), it is sufficient to verify that Z: is exactly the set of all T e Z such that fT = 0.

Fix T e Z and f e 2. Then it is obvious that

tr T7r(/9F) = tr Tir(,8) = 5 /(x)fT(x)dx-

Since fT is a continuous function on G, it is clear that fT = 0 if and only if tr Tr(,8) = 0 for all ,9 e 2. This shows that fT = 0 if and only if T e Z.

4. Induced representations

Let A be a special vector subgroup of G. Fix P e ?P(A) (P = MAN) and let a be a unitary and admissible representation of M on a Hilbert space U. Put KM = K n M and let 1 denote the space of the representation

7Cr = IndKM (a I K1m).

Then 1 is the Hilbert space consisting of all measurable functions f: K -i U such that:'

( 1 ) f(mk) = o(m)f(k) for all m e K, and k e K.

(2) 11f 11' I f(k)12 dk < o. Then 7rK(ko)f (ko e K) is defined to be the function

k - o f(kko) (k e K).

Clearly 7CK is a unitary representation of K on A. Fix b e 6(K) and for a e &(K,), let [b: 8] and [a: 8] denote the multiplicities of a in b and a re-

spectively. Since a is admissible, [a: 8] < co and we conclude from the Frobenius reciprocity theorem that

dim sDb = d(b) EJ 166 (KRX) [b: 8][a: ?] < 00

This shows that wC is admissible. Let T e End l'F where F is a finite subset of 6(K). Then since 7CK is

unitary, its adjoint T* is also in End 'F. Therefore End0 ' is stable under 2 We do not distinguish between two such functions which coincide almost everywhere.



128 HARISH-CHANDRA

the mapping T - T*. The same holds for End? U, since a is unitary. Put i = a*. Then for v e 2, we define a representation w = 7r, of G on

Q as follows. Put

a,(man) = a(m) exp {(- 1)12"v(log a)} (m e M, a e A, n e N).

Then a. is a representation of P on U. Let ID, be the Hilbert space of all measurable functions f: G -- U such that:2

( 1 ) f(px) = 8p(p)"o2a,(p)f(x) for all p e P and x e G.

(2) 11f 112 = |.I f(k) 12dk < 0. Here 8p is the module of P[2(j), ? 19]. Since G = PK, it is clear that II f II = 0 implies f = 0. For f e Dp ,, let hf denote the restriction of f on K. Then f - hf is a unitary isomorphism of Dp > onto I and we shall identify these two spaces under this mapping whenever it is convenient to do so. If f e Dp,, = I and y e G, then 7r(y)f is the function x F--f(xy) on G. It is obvious that w I K = 7CK and therefore w is admissible.

LEMMA 1. Let a e CQ(G) and h e D. Then h' = rp ,(a)h is given by

h'(k2) = |Ka(k2: k,)h(klcu)dk, (k2 e K), K

where

Ka(k2: k,) = | a(kMXan k-)a(m) exp {((- 1)1/2k + p)(log a)}dmdadn MXAXN

for k, k2e K. (Here p = pp.)

We observe that ra(k2: k1) is a bounded operator on U and, for a fixed u e U, the mapping

(k2t ki) E>ra(k2: kj~u of K x K into U is continuous.

It follows from the definition of w that

h'(k2) = a(x)h(k2X)dx= a(k-'x)h(x)dx

= a(k1-'man kl)8p(a)"12a>(ma)h(kl)dkldmdadn

= |a(k2: kl)h(kj')dkl-

COROLLARY 1. Fix hl, h2 e Q and a e Ce(G). Then the function

we to (h2, uctp (ae)hl)

2 We do not distinguish between two such functions which coincide almost everywhere.




is holomorphic on!Air

Since a has compact support, this is obvious from the above lemma.

COROLLARY 2. Fix h e 1, x e G and define

h'(k) = exp { - ((-1)1/2k + p)(H(y))}or(,(y))-'h(yr()-'k) (k e K)

where y = kx-'k-' and the notation is as in [2(j), Thm. 19.1]. Then h' = 7cp,,(x)h.

Since kx = y'k, this follows from the definition of 7r. For a e CQ(G), put d(x) = conj a(x-1) (x e G). Then it is easy to verify

that

tc(k2: kJ)* = |A d(klmank2)a(M) exp {((-1)1/2j + p)(log a)ldmdadn,

where the bar denotes conjugation in He with respect to A. Therefore

(h2, wr,>(a)hl) = (wrp,;(d)h2, hi)

for hi, h2 e A. In particular this shows that 7p,, is unitary if v e A. More- over we have the following result.

COROLLARY 3. Suppose v e A. Then

Ka(k2: kJ)* = ra(kI : k-1)

for aeC (G) and k1, k2eK.

We write

7T = Ind" a312a,

since 7C, is obtained by inducing to G from the representation dp2a, of P on U.

LEMMA 2. Let a, e C(G). Then

ta(m2k2: k1ml) = O(m2)ira(k2: k1)a(mj) (in1, m2 G KM)

and

AKa*p(k2: kJ) = |a(k2: k)Arp(k-1: k2)dk

for k1, k2 e K.

The first statement is obvious and the second follows from the fact that w(a*l9) = w(a)7r(9).

5. Intertwining operators

Let P, e 9'(A) (P, = MANS, i = 1, 2) and v e H. Then we define, at first formally, a linear mapping



130 HARISH-CHANDRA

JP21 P1(V): P1,pi ) 'QP2'

as follows. If h e ,Ppl, then h' = Jp 2lpl(v)h is given by

h'(x) = NPf1 h(inx)dni N2 n N

= ')'P21 ~1 flN2 h(nx)dp (x e G). P1 NlnN2\Nf2

Here dnf = d(N2 nl N1) and the invariant measure d, on N1 n N2\N2 is so normalized that

dN2 = d(N1 nAN2)d.

The constant 7p 21 P has been defined in Section 2. Fix nf e N2 n N1. Then we can write

f-1 = komoaono (ko e K, mo e M, aO e A, no N1)

and since a is unitary, we have3

{ 5 h(nik) 12 dk} < I| h I| exp {(vI - p1)(log aj}

Therefore the following lemma is obvious.

LEMMA 1. Suppose v is an element in !i, such that

IP21 P1( VI) < K

Then the above integral defines a bounded operator JP21PI( '). In fact

1 Jlpl(i))h II -< Jp-1 p~~ll-I) hI II JP2~1(h I? ~P2I PlIP21 P1( VI,) IhI for h eGp1.

COROLLARY. JP21P1(V)4rP1, (X) = ZP2.>(X)JP21P1(V) (x e G).

This is obvious since G acts by right translations.

LEMMA 2. Suppose P lies between P1 and P2 (P e 9P(A)). Then

JP21P1(V) = JP21P(V)JPIP1(V)

provided Ip21p1(-VI) < 00?

It follows from Lemma 2.2 that IP21P(-,I) and IpIp(- I) are both finite. Hence JP21p(v) and Jplpl(v) are defined. Since P lies between P1, P2, we have,

N2 A N1 = (N n N1)(N2 n N) .

Let dn, dn', dn" denote the measures d(N2 n N1), d(Nn N1), d(N2 n N) respectively. Then dn = dn'dn" (n = W'n"). Since 7p21P1 = P our assertion follows immediately by Fubini's theorem.

8 As usual vR and v, denote the real and imaginary parts of v [2(j), ? 17].




COROLLARY 1. Let g,(P2 I P1) be the open set consisting of all v e jg, such that

<KI + p?, a> < 0

for all a e I(P2 I P1). Then for a fixed h e &, the mapping

V I ' JP21pl(v)h

from !?,(P2 I P1) to I is holomorphic.

This is an easy consequence of Corollary 1 of Lemma 2.5 and what we have seen above. (See also Lemma 3 below.)

Fix a finite subset F of &(K). Then since JP21P1(V) commutes with 7CK, it leaves ?jF invariant. Let jP21 pl(v) denote the restriction of JP21l1(v) on IjF. Then jP21P1 is a holomorphic function from !?(P2 1 P1) to End ?DF.

COROLLARY 2.

7rP21 '(a)iP21 Pl(V) = jP21 p1(v)7rsp1,(a)

for a eGSF and v e a-,(P2! P1).

This is obvious from the corollary of Lemma 1. We shall see that if a is an irreducible and square-integrable representa-

tion of M, then jP21 extends to a meromorphic function on W, Fix P e 9(A) (P = MAN) and use the notation of [2(j), Thm. 19.1].

LEMMA 3. Suppose v e -,q P2 e 9i(A), h e I and IP21P(- VI) < oo* Then hi = JP21 p(v)h is given by the formula

h'(k) = Y-p1P _ exp {-((-1)I/2 + p (k e K), N2nN

where dn = d(N2 n N).

Fix ko e K and ff e N2fnlN. Then n = kman where k = ,c(f), m = p(f), a = exp H(nf) and n e N. Hence

h(ff -'k0) = h(a-'m-'k-k) = exp {-((-_1)12- + p)(log a)}a(m-')h(k-'kj) for h e p,, = . Therefore

| h(fiko)dii = - exp {-2(( -1) + p)(H(n))}of(,u))-'h(c(n)-'ko)dn N2fN N2ON

and this implies our assertion.

6. The mapping T --T

A function r from K x K to End' U will be called smooth if: (1) there exists a finite subset F of &(KM) such that r(K x K) c End UF, (2) r regarded as a function from K x K to End UF is K x K-finite.



132 HARISH-CHANDRA

Let X be the space of all smooth functions r from K x K to End? U such that

,c(m2k2: ktm1) = (m,2)jr(k2: k1)a(m1)

for k1, k2 e K and n1, m2 e KM. Put Z = End?&%.

LEMMA 1. There exists a linear bijection T F /CT of Z onto X such that

(Th)(k2)= 5 1(k2: k,)h(kl')dk1 (k2e K) K

for all he.D. Given ir e X, define a linear transformation T on I by

(Th)(k2) = j(k2: k,)h(k7')dk, (k2e K) K

for h e ,. It follows from the smoothness of j, that T e Z and T ? 0 unless r = 0. Conversely given T e Z, choose a finite subset F of &(K) such that T e End ?jF. Let hi (1 < i < r) be an orthonormal base for ?,F. For k1, k2 e K, consider the linear transformation r(k2: k1) in U given by

r(k2: k,)u = l hi(k2) ((T*hi)(k:1), u) (U e U) .

Since h, and T*hi are in '@F, jc(k2: k1) e End? Uand r is smooth. Then if h e ,

5K j(k2: k,)h(k:')dk, = l1g,<r h(k2)(hi, Th) = (Th)(k2)

It is easy to verify that r e KX and the lemma follows. We write wc(k) instead of 7rK(k) (k e K).

COROLLARY. (1) 1 CT*(k2: k) = ('r(1k': k1 ))*.

(2 ) 'CTjT2(k2: k1) = CT1(k2: k)'rT2(kAV: k,)dk. K

(3) trT= 5trKT(k:k-I)dk.

(4) Suppose T' = 7c(k1)T7c(k1) (k', k1 e K). Then

AT (k2: k) = ArT(k2kl: k11c).

Here T, T1, T2 e Z and k1, k2 e K. This is a simple exercise.

7. The relation between induced representations and Eisenstein integrals

In [2(i), ? 261, we have defined a double representation z of K on V = C-(K x K). It is unitary with respect to the norm




I V 12 = | I v(k,: k2) 12 dkldk2 (ve V). KxK

If F is a finite subset of 6(K), we denote by CF(K x K) = VF the subspace of all v e V such that

v= 5 aF(k)z-(k)vdk= 5 aF(k)v-r(k)dk. K K (See ? 3 for the definition of aF.) Then VF is stable under - and its dimen- sion is finite. Let ZF denote the restriction of z on VF.

Now suppose a is an irreducible and square-integrable representation of M on U. Then a is admissible. Let w denote the class of a in 62(M). If r is a unitary double representation of K on a finite-dimensional Hilbert space V, we put L((w) = Ce(M, -rM) as usual (see [2(j), ? 181). Then dim L((w) < oo.

Fix a finite subset F of 6(K) and write (V, z) for the pair (CF(Kx K), ZF)

introduced above. For. T e End S@F define fTe L((w) as follows. If m e M, kT(m) is the element v of V = CF(K x K) given by

v(k1: k2) = fT(kl: m: k2) = tr {I'T(k2: k,)a(m)} (k1, k2 e K) .

LEMMA 1. The mapping T frT from End 'DF to L(Co) is bijective.

Let T be an element in End 'DF such that fT = 0. Fix k1, k2 e K. Then tr (KT(k2: k1)a(m)) = 0

for all m e M. Since a is irreducible, we conclude from Lemma 3.1 that cT(k2: k1) = 0. This proves that 'IT = 0 and therefore T = 0 (Lem. 6.1).

Conversely let * be a given element in L((o). Then + may be regarded as a function on K x M x K. Let Fo be the set of all 3 e &(K,) such that [b: 3] > 0 for some b e F. Then Fo is a finite set. It follows from the definition of L(co) that, for fixed k1, k2 e K, the function

m i ) *(k,: m: k2)

on M is a finite linear combination of KM-finite matrix coefficients of a. Since a is irreducible, we conclude from Lemma 3.1 that there exists a unique element Ai(k2: k1) e Endo U such that

f(k1: m: k2) = tr (,c(k2: k1)a(m)) for all m e M. Since * e L(co), we conclude easily that tc(k2: k1) e End UFO and r lies in X. Therefore by Lemma 6.1, there exists a unique T e Endo0 such that r = AT. Applying part (4) of the corollary of Lemma 6.1, we conclude that T e End O., and this proves that * = *T.

Let z-' be the double representation of K on V = CF(K x K) defined as follows. If v e V and k1, k1 e K, then v' = z'(k1)vz-(k1) is given by



134 HARISH-CHANDRA

v'(kQ: k2) = v(kck,: k2k1) (ki, k2 e K).

LEMMA 2. Let k1, k2 e K and T e End 9F. Then

*fr(kI)Tir(k2)(7n) = Z- (ki)*IT(m)z-(k2)

for m e M.

This is an immediate consequence of part (4) of the corollary Lemma 6.1. Fix P e 9P(A) and v e H. Then for f e L(@o), we consider the Eisenstein

integral E(P: fr: v) [2(i), Hi 19, 26].

THEOREM 1. Let T e End 9F. Then

E(P: *1 Tv kl: x: k2) = tr Tar,, (klxk2)

for k1, k2 e K and x e G.

We shall give a proof of this theorem in Section 9. First we need some preparation which will also be useful later on.

8. Some simple properties of E(P: A: ')

Let r be a unitary double representation of K on a finite-dimensional Hilbert space. We make the following assumptions on V and r.

( 1 ) V is an associative algebra with an anti-involution u i- u* (u e V) over R so that'

(cu)* = j , (u * v)* = v* * u* (ce C, u, v e V) .

(2) r is compatible with multiplication in V [2(i), p. 154]. (3) There exists a linear mapping tr: V -a C such that

tr zl(k)u = tr ur(k) , tr (uv) = tr (vu) (u, v) = tr(u*v) (u, ve V, keK) .

The above assumptions imply the following relation.

(4) (z(k1)uz(k2))* = z-(k2-)u*z-(kl1) (ki, k2 e K, u e V) .

For if v e V, we have

((z(ki)uz((k2))*, v) = tr (z(k1)uz(k2) v) = tr (u. z-((k2)vz((k1)) = (u*, z-(k2)vz-(k1)) = (z(k-1)u*z(k1), v) .

We regard C(G, r) as an algebra under convolution. For a e C(G, r), define d(x) = a(x-1)* (x e G). Then fe- f is an anti-involution (over R) in C(G, r) and a similar statement holds for C(M, zd). If P e P(A) (P = MAN) and v e A, put

4 Here the bar denotes the complex conjugate.




fp)(m) = f(man) exp {-((-1)1/2 - p)(log a)}dadn (m e M) AXN

for f e C(G, 4). (Here p = pp.)

LEMMA 1. f-- f>'P) is a homomorphism of C(G, z-) into C(M, z,,) and

(f)"I = (f"P)-)

This is verified by a simple computation. Put L = 0C(M, z). Then dim L < oo [2(i), Thm. 27.3] and L is a two-

sided ideal in C(M, z,) which is stable under the anti-involution . More- over it has the structure of a Hilbert space under the norm

Il Ib * I 12 I *(M) 12 dm (+eL).

We note that

(0, u) = tr(5* +)(1) for 0, ' 1 L.

LEMMA 2. Let i e L, P e 9(A), v e g- and f e C(G, z-). Then: ( 1 ) f* E(P: fr: 2) = E(P: f,'P)*r: 2). (2) E(P: ur:2))*f= E(P:+ **f,'P': 2). ( 3) E(P: : )=E(P: *r: v)-. Fix g e C(G, z). Then if h = f * g, we have (see [2(j), ? 18])

(f * E(P: ": o), g) = (E(P: ": 2), h) = (r, P)) = (f, (f.(P))- * g(P)) = (f.P) * f, g P)) = (E(P: f.(P) * f: 2), g)

from Lemma 1. This holds in particular for all g e C,-(G. z) and so (1) follows. The proof of (2) is similar. So we come to (3). Let g e C-(G. z-). Then

(E(P: -: t)~, g) = conj (E(P: ": 2), g) = conj ("I, (P)'P) .

But (fr, (g)'P)) = (fr, (g(P))-) = conj (#, g9Pp) .

Hence

(E(P: t: 2)~- g) = (i, g'P)) = (E(P: v 2), g) .

This being true for all g e CG(G, zr), our assertion follows.

LEMMA 3. Let f e at(G, z) and let P be any psgp of G. Then

(fp)r = (f)p Let P = MAN. Then it follows from the definition of fp [2(i), t 21] that

li p I {d -(ma)f(m-'a-') - fp(m-'a-')} = 0 P



136 HARISH-CHANDRA

for m e MA. Therefore lima.+ {d-(ma)f(ma) - (fp)-(ma)} = 0.

P This proves that (f)p = (fp)l.

COROLLARY. Let P e 9P(A), v e !' and * e L. Then

c-IA~: V),+ = (cPjP(l: V)*)- in the notation of Theorm 18.1 of [2(j)].

This follows by applying the above lemma to f = E(P: v: 'r) and making use of part (3) of Lemma 2.

9. Proof of Theorem 7.1

Now take V = CF(K x K) and z = zF as before in Section 7. Define

trv = k v(k: k-)dk,

v*(kQ: k2) = coni v(k-1': ki') (k1, k2 e K) for v e V. Moreover define the product v = v1 v2 of v1, v2 e V by

v(kl: k2) = 5 vl(kl: k)v2(k-l: k2)dk.

Then all the conditions of Section 8 are fulfilled. It is obvious that L(w) is a two-sided ideal in L which is stable under .

Let d((o) denote the formal degree of wo. LEMMA 1. Let St T e End S'F. Then ( 1 ) frS * frT = () TS

(2) (IT)- = *T*I

(3) tr4r(1) =tr T, (4) (*s, *fT) = d(o)-1trS*T. Let B, C e End' U. Then it follows from the Schur orthogonality rela-

tions for co that

M tra(m-1)B tra(mm0)Cdm = d((o)-1tr (Boa(m0)C) for m eM. Hence if 0 =*S*T

0(k1: in0: k2) = | Is(k1: m: k)Tr(k-1: m-1m0: k2)dkdm SKxM

= S| dk tr a(m)rs(k: k1) tra(m1mO)ICT((k2: k1')dm R M

= d((o)-' S tr {rs(k: kl)a(mO)rT(k2: k-')}dk

= d())-' tr'ATS(k2: klc)a(m0) = d(w")-1fTS(kj: min: k2.)




from the corollary of Lemma 6.1. This proves (1). (2) and (3) also follow from the same corollary. (4) is a consequence of the fact that

(0, ir) = tr(s * +)(1) ( e, i eL) .

Define 2 and 2F as in Section 3 and put a(x) = a(x-1) (x e G) for a e ?.

For a e (2,F) define a e C,-(G. z) by [2(i), ? 26]:

a(k,: x: k2) = ac(kxk2) (k1, k2 e K, x e G) .

LEMMA 2. Let a e (2F)f, Pe 9P(A), v e 9 and T e End 9F. Then aP * * = *fT7pp,,(5) t

*T * aP = *p,,(a)T

We write 7C = 7Cp, and observe that 2r(a) e End 9PF. Then we conclude from Lemma 4.1 that

Ar(a)(k2: k) = a(P)(kl: m: k2)a(m'1)dm

Therefore (cor. of Lem. 6.1)

tCTir()(k2: k1) = |T(k2: k ')Pr,(&)(k: kl)dk

= 5 aP((k1: m: k)CT(k2: k-l')a(m-')dkdm.

Hence if m0 e M,

*Tr(a)(kl: m(0: k2) = tr {/)T7Z()(k2: kl)o?(m2O)}

= aP((k1: m: k)tr{'TT(k2: k-l)a(m'-m0)}dlkdm

= a(P((k1: m: k)1T(k-l: m-1m0: k2)dkdm.

This shows that

kTit(a') = aP * - T -

The proof for the second statement is similar.

Now we come to the proof of Theorem 7.1. Put E(T) = E(P: PT: ) and r = spp. Then

E(T: 1) = 5 z(k)*T(l)z(k-')dk

and therefore

E(P: *T: : 1: 1: 1) = E(T: 1: 1: 1) = tr*T(l) = tr T from Lemma 1.

Now a - a is a bijective mapping of (2F)f onto C,-(G, z-) [2(i), ? 26]. Fix



138 HARISH-CHANDRA

a e (,F)f. Then

a*E(T) = E(P: aP)*T )= E(Tr(aQ))

from Lemmas 8.2 and 2. Put 0 = a*E(T). Then it follows from the above result that

0(1: 1: 1) = tr T7r(a') .

But

O(1) = a(x-') E(T: x)dx

and therefore

0(1: 1: 1) = a(x-lk-')E(T: k: x: 1)dkdx

= a(x-lk-')E(T: 1: kx: 1)dkdx

= 5 a(x-)E(T: 1: x: 1)dx .

Since

tr T7r(a) = 5a(x-1) tr T7r(x) . dx

and a is arbitrary in (2,F), we conclude that

E(T: 1: x: 1) = tr Tr(x)

and from this the assertion of Theorem 7.1 is obvious.

10. An application of Theorem 7.1

We keep to the notation of Section 9.

LEMMA 1. Fix Pe CP(A) and suppose v is an element in I' such that

cPIP(1: v) is infective on L(@o). Then 1Cp,,(2F) = End IF.

We note that, in view of Lemma 19.3 of [2(j)], v can actually be chosen so as to satisfy our condition.

Put r = 7p, ,and fT(x) = tr Tr(x) (x e G) for T e End IF. By Lemma 3.1, it is enough to verify that the mapping T F fT is infective on End IF. So fix T such that fT = 0 and put

0 = E(P: fT: V)).

Then 0 = 0 by Theorem 7.1 and therefore Up = 0. This implies that Cpp(1: V)WT = 0 and therefore AT = 0. But then we conclude from Lemma 7.1 that T= 0.

We shall use this lemma to prove later that Up,, is irreducible for v e




(see Lemma 13.3).

11. Some properties of the j-functions

For P, Q e 9J(A) and v e !C(Q I P), put jQP(2) = EFJQIP(v)EF as in Section 5. We note that gC(P I P) c !c(P) in the notation of Theorem 19.1 of [2(j)]. Put

c(P) = 'P-1P'(P)-

It follows from Lemma 2.6 that c(P) is independent of Pe 9J(A). Hence we may denote it by c(A) or c(G/A).

LEMMA 1. Fix P e 9i(A) and v e (P I P). Then

CPI (l: V)WT= c(A)=jp 64 T

for T e End 9F

Put j = jj-Ip(v). Then it follows from Lemma 5.3 that

IjT(k2: k1) = 5 exp {-((-1)"'12 + p)(H(n))}rT(kl: ma('f)-': jr(n)-'k2)dn .

Our assertion now follows from Theorem 19.1 of [2(j)]. The above result shows that j-lp(v) extends to a meromorphic function

on Hc(P) U 9-#). Moreover it is holomorphic at every point V e g'. There- fore we conclude from Corollary 1 of Lemma 4.1 and Corollary 2 of Lemma 5.2 that

;P( v)rp,, (a) =P,.()!jjpP( ) (a e 2F)

for v e g'. Let g" be the set of all v e g' where jjp-(v) is bijective. Since det cplp(l: V)

is not identically zero [2(j), Lem. 19.3], we conclude from Lemma 1 that " is a dense subset of g.

LEMMA 2. Fix v e s". Then there exists a scalar c > 0 such that

(jiIP(V))*jiP-IP() = c.

Put j = j -1(p'), 7p = 7rp, and 7r- = 7rz-. Then

7rF(a)j = j7rp(a) (a e 2F)

and therefore, taking adjoints, we get

j*7r-() = 7rP(a)j*

Here d(x) = conj a(x-') (x e G) and we make use of the fact that 7rp and 7r - are unitary. This shows that j*j commutes with 7rp(a) (a e 2F). But 7rP(2F) =

End gF by Lemma 10.1. Therefore j*j is a scalar. It is obvious that c > 0.



140 HARISH-CHANDRA

LEMMA 3. Let v e g'. Then

jPP(y) = (hP-P(V)) and

c pi (l: 2)T = c(A)*Tjp 1 (

for TeEnd .F.

Put i = ijip(v). Then we conclude from the corollary of Lemma 8.3, Lemma 9.1 and Lemma 1 that

Cal p(l: L)WT = (CPIX(1: V)T*)- = c(A)(IjT*)- =c(A)*Tj*

Hence it remains only to verify that jipl( ) = jke Replacing P by P in the above result, we get

CpIA(l: 2)T =c(A)*T(i p p W

On the other hand we have the following simple result.

LEMMA 4. Put S = E.. Then

E (PA: E) = (P A:

for all e;.

Assuming this for the moment, we continue with the proof of Lemma 3. Fix v e A' as before. Then

CpIp(l: V)Ws = cPji(l: 4)*r

from Lemma 4. But we have seen above that

cPIX(l: 4)*s = c(A)*j-lp(, Cplp(l: 4)*s = c(A)*(fpli-(&,))*.

In view of Lemma 7.1 this proves that jip-(v) = (jpli(v))*.

Now we come to the proof of Lemma 4 and use the notation of the proof of Lemma 2. It would be enough to consider the case when v e A". Then i = (Pj-lp() is bijective and

7r-(a)j = jicp(a) (a e SF) -

Therefore

trir-(a) = tr7rp(aT) (a e 2F)

or

tr EF71c;,,(x) = tr EF7P, P(x) (x e G).

The required result now follows from Theorem 7.1.

LEMMA 5. Fix P e @(A) and v e A;'. Then jiP(z(v), j1j(;(v) commute and




their product is a scalar operator.

This is obvious from Lemmas 2 and 3.

12. The fl-function in a special case

We assume in this section that dim A = prk G +1. Then 9P(A) = {P, P}.

LEMMA 1. Suppose ,F ? {O} and fix v e A'. Then there exists a number

p,(v) > 0 such that

Pi(V)jplpj(V)jp-lp(v) = P.(V)jp-lp(V)jplp(v) = 1

on IF. Moreover pet,(v) is independent of F.

Put j = ij-I(2). We claim that j ? 0. For otherwise suppose j = 0. Fix T e End IF and put

= E(P: AT V)).

Since 9J(A) = {P, P} and j = j*= 0, it follows from Lemmas 1 and 3 that

OP = - = 0.

On the other hand by [2(j), Lem. 18.3], p,- 0 for every psgp P' of G which is not associated to P. Therefore 0 = 0 [2(i), Lem. 25.2] and we conclude from Theorem 7.1 that

tr TOrp,(x) = 0 for all x e G. Substituting x = 1 and T = EF, we get

0 = trEF = dim 'F

which contradicts our hypothesis. Since j ? 0, we conclude from Lemmas 11.3 and 11.5 that

j*j = c

where c is a positive scalar. We take [e(,(v) = c-. Now we write JF and CF instead of j and c to emphasize their dependence

on F. Then

jFJF = CF.

Suppose F1, F2 are two finite subsets of 6(K) such that Fi? {0} (i = 1, 2). Put F = F, U F2. Then it is obvious that

jFj =EFijF

and so we conclude immediately that CF = CF, (i = 1, 2). This shows that CFj = CF2 and therefore pe,(v) is independent of F.

COROLLARY.

11 C ,,(1: 4)* 112 = II C-1p(j: V)) 112 = 1()'c(A)2 * 112



142 HARISH-CHANDRA

for v e!$' and e L(@o). This follows immediately from Lemmas 11.1, 11.3 and 1. It is obvious from what we have seen in Section 11 that A, extends to a

meromorphic function on $?(3) which is holomorphic on

LEMMA 2. p. is holomorphic on $.

Let a be the unique simple root of (P, A). Then Lemma 5.3 shows that jfipp(v) depends only on <K, a> and its value remains unchanged if we replace (G, P) by (0G, 'G n P). Hence we may assume that prk G = 0. Then dim A = 1 and therefore $ is the complex plane. We have therefore only to verify that A, does not have a pole at v = 0.

So let us assume that , = A, has a pole at zero. Fix T e End 9OF and put

= E(P: T: 0)

We know [2(j), ? 10] that

ip = limi0 Ep(P: AT: V) )

0i = lime0 EA(P: AT: V) (V e $') pointwise on MA. Moreover

I|I CP1 A : 2))*T I|| - I|| CP-J A : I)T |112 ?

from the corollary of Lemma 1. This shows that ip = 0i = 0 and we con-

clude, as in the proof of Lemma 1, that 0 = 0. But then

tr Tir, 0(x) = 0 (x e G)

from Theorem 7.1. Substituting x = 1 and T = EF, we get a contradiction.

13. The [e-function in the general case and irreducibility of representations

We return to the general case of Section 11. Let a be a reduced root of (g, a). Define Ma,, a,, as in Lemma 2.3. For v e A, let *v denote the restriction of v on a.. If we replace (G, A) by (Mar Aa) then the conditions of Section 12 are fulfilled. Put iaco: v) = [e.(*v) where the function A,, is defined as in Section 12 for the pair (Ma, M) in place of (G, M). Clearly [ea(w: a) =

[te..(w 4). Suppose P1, P2 are two adjacent elements in iP(A) and a is the unique

root in Y(P2 1 P1). We use the notation of Lemma 2.3 and put *P = P n fMa.

LEMMA 1. jP21 pl(v) and iPj1P2(V) commute and

Pfa(w: V)VP1XP2(VV)jP2XP1(V) =

Moreover ip2lp1(v) = (P11P2(v))* for 2 e -'.




*P = M. *A. *N where *A = Aa, *N = Na. Put *K =Mal nK and let

* be the representation space of

Ind` (a I KM).

For h e EF and i e K, define *hke *e by

*hk(*k) = h(*kl*k) (*k e *K)

We can choose a finite subset *F of 5(*K) such that *hk e * AF for all h e SpF and k e K. Since P2 f Ma = x it follows (see Lemma 5.3 and the proof of Lemma 2.4) that

(UP21 pl(v)h)k = *P1 *P(*v)(*hk)

for v e g0(P2 I P1). But then by analytic continuation this holds also for v e I'. Similarly

*(jPlP2(v)h) k =

j*PI*P-()(k) () G 9')

On the other hand

[ja(co: V)j*Pj*P(*2)jfij*P(*2) =

on *g*F from Lemma 12.1. Therefore

[a(o: V)*(jiP21P1(V)jiP1P2()h)k =hk (V e G )

for all h e 9PF and k e K. This proves that

X[((w): 2)jP21 P1(2)iP1 1P20)

on OF. We can obviously interchange P1 and P2 in this relation. Now fix v e I'. We know from Lemma 11.3 that

i*P-l*P(*V) = (i*PI*P( )

Since

(hi, h2) = (*hlk, *h2k)dk (hi, h2 e SF) K

it follows from what we have seen above that

(jP21P2 (v)hl h2) = (hi, jP11P2())h2)

and this completes the proof of the lemma. Given 3 > 0 and P1, P2 e 9P(A), let g0(P2 l P1, 3) denote the open set con-

sisting of all v e H- such that

<KI a> < 3

for all a e I(P2 1 P1). Note that g0(P2 1 P1, 3) is a convex set and it contains

g as well as the set g0(P2 I P1) of Corollary 1 of Lemma 5.2. Moreover

go(P21] P1, 3) = Hc if P, = P2.



144 HARISH-CHANDRA

THEOREM 1. Fix P1, P2e 9P(A). Then if 3 is sufficiently small, ji21 1 extends to a meromorphic function on !a,(P2 I P1, 3).

We have seen in Section 11 that this is true if P2 = P1. So in particular our assertion holds if dim A = 1. Now first assume that d(P2, P1) = 1 and keep to the notation of the proof of Lemma 1. Then if hl, h2 e DF and

91;(P21] PJ1)

(h2, jP21P1(v)hl) = K (*h2k, ji*P*P((*2)*hlk)dk

Since dim* A = 1, our theorem holds for j*-!*P. Since k I * hik (i=1, 2)

are continuous mappings of K into *D*F, it follows that our statement is true also for jP21P1

So now let us use induction on d(P2, P1) and suppose that d(P2, P1) ? 2. Then we can choose P e 9P(A) such that d(P2, P) = 1 and d(P, P1) = d(P2, P1) -1. Then by induction hypothesis the theorem holds for jP21P and jipip and we know from Lemma 5.2 that

jP21IP1(V) = jP21P(V))P 1P1(2)

for v e gF0(P2 I Pj). From this it is obvious that our theorem is true also for jP2IP1.

Fix 3 > 0 such that jP21 p extends to a meromorphic function on ;,(P2 I P1, 3) for all Pi, P2 e 9P(A).

COROLLARY. Let P1, P2, P be elements in 9P(A) such that P lies between P1 and P2. Then

jP21P1(V) = jP21P())jP1P1(V)

for v e g,(P2 I P1, 3).

This is obvious from what we have said above. Put

ePP21P1((w: )) = Ia eV(P2IP1) [ea((w: 2)

for P1, P2 e 9P(A). Clearly PP21pl = 11P11P2

LEMMA 2. Let P1, P2 e 9P(A) and v e a'. Then

PL11P2(): 1))jP1iP2(1))jP21P1()) = 1 iPP21pl() = (jPlIP2(V))*

This follows immediately from Lemma 1 and the corollary of Theorem 1. Note that p -1p((w: 2) is independent of P e 9P(A). Hence we denote it by

p(wo: 2). It is clear from Lemma 12.1 that p(wo: 2) > 0 for v e A'. We recall that the representation wp,1, (P e 9P(A), v e A?) is unitary (? 4).




LEMMA 3. 7r, p is irreducible for P e 9(A) and v e A'. Moreover its class is independent of P.

Fix a finite subset F of 6(K). In view of Lemma 3.1 and Theorem 7.1, it would be enough to verify that the mapping

T X ) E(P: AT: 2)

is injective on End DF* So suppose E(P: A T: ) = 0. Then

ijIp(2))T = 0

from Lemma 11.1 and therefore

T = #( )jwpjj(v)jp-jp(v)T = 0

from Lemma 2. This proves that rp, is irreducible. Now let P1, P2 e 9P(A). Then

iP2 I P1( ) P1 p,(a) = p 2,P(a)jP2I Pl(V)

for a e SF. (This follows immediately from Corollary 2 of Lemma 5.2 by analytic continuation.) Since iP21P (P) is bijective from Lemma 2, this shows that the two representations of SF on DF (defined by wf = wrpI,, i = 1, 2) are equivalent. Since w1, wZ2 are irreducible and unitary and we can choose F so that DF # {0}, this implies that w1 = r2

COROLLARY. Fix T e End F*. Then

E(P1: &TjP21P1(v): E(P2: *'P2Pi()T:)

for P1, P2 e CP(A) and ve A'.

For tr {TjP21 Pl(*~rPll ~) tr I PrP21 P(a)jP21 P1(V) I

= tr{jP21P1(v)Trp2,1,(a)}

for a e 2F. The required result now follows from Theorem 7.1. For P e 9P(A) and v e A', let C(P, 0, 2) denote the class of the irreducible

unitary representation Up @ = Indp P2ap

of G. Define t = w(G/A) as usual (see [2(j), Thm. 18.1]).

THEOREM 2. Let P1, P2 e 9P(A), w e 62(M), v e A' and s e S. Then

C(P1, w, 2) = C(P2, S0, SO)).

It is obvious that

C(P1, w, 2) = C(P1, s(, sv) .

Hence it is enough to consider the case s = 1 which is covered by Lemma 3.



146 HARISH-CHANDRA

LEMMA 4. Under the conditions of Theorem 2, cp21p1(s: v) defines a bijection of L(@) onto L(sw). Moreover

P(wO: V) I c 21lP1(s: 4)+ 112 = c(A)2 11 11 l2

for P e 9P(A) and * e L(@).

It is obvious that s defines a unitary mapping of L(@) onto L(sw). Hence, in view of [2(j), Lem. 18.1], it would be sufficient to consider the case s = 1.

Fix T e End QF* Then by the corollary of Lemma 3,

CP21lP(1: 4)fTl = CP21P2(1: 2))fT2

where T, = TjP21p,(v) and T2 = jP21lp(i)T. Put T3 = ji2IP2(y)T2. Then

CP2IP2(1: V))T2 = c(A)lrT3

from Lemma 11.1. Put

I SI2 = trSS* = trS*S

for S e End QF. Then

l T, 112 = PP21P1((0w )-1 I I T1 12

from Lemma 2. Similarly since T73 = j 2pl2(2)jp21pl(v)T, we have

I T3112 = trT T3 = P21P1(w: 2))'1e(w: 2))1 I I T112

= i(w: v)1 II T, 112

Since jp21pP(V) is bijective, this proves that

II CP21 P(2 V)+ 112 = L(W: 2)-lc(A)2 11 A 112

for * e L(a)).

Part II. Maass-Selberg relations and the functional equations

14. The Maass-Selberg relations

Let r be a unitary double representation of K on a finite-dimensional Hilbert space V. Fix a special vector subgroup A of G and define M, A, t as above. Let (G(G, r) and 0C(M, z-,) have their usual meanings [2(i)]. Then 0C(M, z-,) has a natural norm [2(j), ? 18].

THEOREM 1. Fix v e A' and let f be an element in (G(G, r) such that: ( 1 ) fp,, 0 for every psgp P' = M'A'N' of G such that A' is not con-

jugate to A under K. ( 2 ) For every P e 9P(A), we can choose elements ps e (M, z-,) (s e w)

such that




fp(ma) = s p6p ,,(m) exp {(-1)"2sv(log a)} (m e M, a e A).

Then

11 oPll 1 I= I IP2,s2 11

for Pi e rP(A) and s, e W (i = 1, 2).

By the Maass-Selberg relations we mean the equalities

IIH op 81l= IIH P282 11

of the above theorem.

COROLLARY. Suppose , = 0 for some pair (P, s). Then f = 0.

For then it follows from the Maass-Selberg relations that fp = 0 for every P e EP(A). Hence we conclude from Lemmas 21.1 and 25.2 of [2 (i)] that f = O.

The above corollary shows that the functions op, are uniquely determined by f. Hence we may denote them by fp,,.

15. Some preparatory remarks

We need some preparation before taking up the proof of Theorem 14.1. A unitary double K-module (or, more briefly, a module) is a pair (V, Z-)

consisting of a finite-dimensional complex Hilbert space V and a unitary double representation r of K on V. Two such modules (V, r) and (V', z-') are said to be isomorphic if there exists a K-isomorphism v F v' of V onto V' and a number c > 0 such that I v'I -c = v I for all v e V. We shall usually denote the module by r instead of (V, r). The meaning of the direct sum r eD r' of two modules z, r' is clear.

Let P be a statement about modules. We say P is additive if the following two conditions hold.

(1) If r and z' are isomorphic, then P(r) implies P(z'). (2) Suppose r = r' eD r". Then P(z) holds if and only if both P(z-') and

P(i/") hold. LEMMA 1. Suppose P is additive. Then in order that P be true for all

Z, it is sufficient that P(z) be true for all irreducible r.

This is obvious. For any finite subset F of d(K), define the module (VF, ZF) as in Section

7 so that VF = CF(K x K).

LEMMA 2. Let r be an irreducible module. Then r is isomorphic to a submodule of ZF for some F.

Let V be the representation space of r. Fix a linear function X # 0 on



148 HARISH-CHANDRA

V. For any v e V, consider the function s, on K x K given by

Kr,(k,: k2) = <X, z(k,)vz(k2)> (ki, k2 e K) Then if F is suitably chosen, sv e VF = CF(K x K) and

r: v I ;rV

is a K-homomorphism of V into VF. Hence ker r is a submodule of V. Since r is irreducible and X # 0, we conclude that r is infective.

Put

(u, v)o = (us, Kv) (u, v e V).

This is a positive-definite K-invariant scalar product in V. Since r is irreducible, this must be a multiple of the old scalar product. Hence the lemma.

COROLLARY. Let P be an additive statement on modules. Then in order to prove it for all modules, it is enough to verify P(ZF) for F = F1, F2, ** where Fj is an increasing sequence of finite subsets of 6(K) such that

6(K)== U1iFj. This is obvious from Lemma 2.

16. Proof of Theorem 14.1

Fix v e I' and let d(v) denote the set of all f e (G(G, r) which satisfy the hypothesis of Theorem 14.1.

LEMMA 1. Put L = 0C(M, zM) and fix P e 9P(A). Then

* 1 E(P::) is a bijection of L onto (X(v).

Note that the statements of this lemma and Theorem 14.1 are both additive in T. Hence it is enough to verify them for Z = F where F is a finite subset of 6(K) such that' F = F *.

So fix F and put Z = Z-F. We claim that Lemma 1 is equivalent to Theorem 14.1 in this case. We know [2(j), ? 18] that dim L < and

L = we'62(M) L(w)

where the sum is orthogonal. Fix * e L. Then

* = L, *U where *,,& is the component of * in L(w) and it follows from Lemma 13.4 that

5 b* is the class contragredient to be &(K) and F* is the image of F under the mapping b 1- b*.




C1 QIP(S: V) II2 = I CQIP(S V)* @11 2 =c(A)2 E p(w: ')-1 1 "k 112 for P, Q e 9P(A) and s e W. This shows that II cQIp(s: v)* II is independent of P, Q and s. Moreover cQJp(s: 2) is bijective on L.

It is obvious that E(P: f: 2) e di(v). Moreover since cplp(l: v) is bijective on L, it is clear that the mapping of Lemma 1 is infective. So the main assertion of our lemma is that it is surjective.

Now suppose Theorem 14.1 holds. Fix f e d(v) and put

* = cplp(l: 4)-'fp,,, g = f - E(P: A: v) .

Then g e (3(v) and g, l = 0. Hence by the corollary of Theorem 14.1, g = 0 and this proves the surjectivity in Lemma 1.

Conversely suppose Lemma 1 holds. Then f = E(P: A: 2) for some * e L. Since v e A', it is clear that

=Q,= CQJP(S. 2))

for Q e 9P(A) and s e W. Therefore

I1I OQ,8 = CQIp(S: 2))lf

is independent of (Q, s). This proves the equivalence of Lemma 1 and Theorem 14.1.

LEMMA 2. dim (3(p) < oo.

Fix f e (3(v) and P e (G(A). Since v e A', s2 (s e W) are all distinct and therefore the functions op,, are uniquely determined by f. We may therefore denote them by fp,8. Since fp = s(fp), it is clear that

fP,8= s(fQ,,)

where Q = P -'. This means that if fp, l = 0 for all Pe 9P(A), then f = 0. Let p = [9P(A)J. Then we conclude that

dim (i(v) < (dim L)P < oo

For a e 2F define a e Cc (G, z-) as in Section 9. (Recall that F = F*.) If f e G(v), it follows from [2(i), Thm. 21.2] that a* fe (3(v) and

(a*f)p,8 = aPI' *fp,8

in the notation of Section 9. Since a F-+ a is a homomorphism of SF into C,-(G, z), we may regard (3(v) as a left SF-module.

LEMMA 3. Fix P e 9P(A) and suppose f is an element in (X(v) such that fp,l = 0. Then f = 0.

Let B be the set of all f e (3(v) such that fp 1 = 0. We have to show that = {0}. Clearly B is an 2F-submodule of (3(v). If L # {0}, we can, in view



150 HARISH-CHANDRA

of Lemma 2, choose a simple submodule ?B # {0} of B. Fix Jf # 0 in @, and let Eon denote the orthogonal projection of L on L(@) for GO e 62(M). Then (see the proof of Lemma 2) we can choose Q e CP(A) and w e 62(M) such that E,(fo)Qwl # 0. Consider the mapping

f-> E~fQl

of 9o into L(@). If a e 2FI then

Ej(a*f)Ql = EW(a'Q) *fQl) = a'Q) *E fQ,,

for f e U0. Hence we conclude that the kernel of this mapping is an ?F-

submodule of U0. But ?0 is simple and therefore this mapping must be injective.

Note that an element f e (3(v) is a function on K x G x K. Since

f(kl: x: k2) = f(1: klxk2: 1) (kl, k2 e K, x e G),

we may also regard it as a function on G by setting f(x) = f(1: x: 1). Under this identification a *of corresponds to a * f (a e SF).

Given f e 90, there exists a unique element Sf e End QF such that

(IT, E~fQ,1) = d(w)-'tr T*Sf

for all T e End -F. (Here the notation is the same as in ? 9.) Then if a e ?F,

tr(T*Sa-f) = d(Wo)(*T, a(Q) * E fQ,J) = d(o)((&)'Q) * *fTi Eo fQl)

= d(Wo)(*T1rT(a,) E@fQl)

= tr(w(a-)T*.Sf) = tr(T*Sf@(a))

from Lemma 9.2. Here d = conj a and w = WQw,,,. This shows that

Sa*f = Sfw(@)

Put ax T = Tr(&) (a e 2F, T e End OF).

This makes End OF into a left 2F-module and then the mapping f l - Sf

defines an SF-homomorphism of 90 into End DF* Since 90 is simple and Sf0 # 0, this mapping is infective.

Let fi, 1 **, f4 be a base for U0. Put Si = Sfi and choose Tj e End DF such that trSiTj = 8ij (1 < i, j < r). Put

sOij(x) = tr(Sis(x)Tj) (x e G). Then if ae GF,




Sa*f = S,(a') = Fj tr (S2r(a")Tj)Sj

= Jo Sa(x-')b0j(x)dx -S,.

The mapping f 1 Sf being injective, this implies that

a*f= a(x-=)Oij(x)dx fj .

Therefore

(a* *f)(1)= (a* ij)(1)

for all a e SF. Put 2 = Cc (G) and define the mapping 38 H ,SF of 2 into SF

as in Section 3. It is obvious that

f = aF *Kf *KaF oij = aF*KRSij *KaF

Hence we conclude that

(I8 *f{)(1) = Sj (Ie * Ij1) l.f,(1) for ,Q e 2. This implies that

fA(x) = 0, ij(x)fj(1).

Now put

*ij = kTjSS 'ki = jj(l) (1 ij < r) . Then God e L(@) and

E(Q: *11: v) = Os from Theorem 7.1. So the above relation becomes

As = E(Q: *,: v) :(1< i < r) .

This shows that every f e Go0 is of the form

f = E(Q: *: v) with * e L(@). But then

fPl = CplQ(l: V)* = 0 .

This implies that + = 0 and hence f = 0. This proves that B0 = {0} and so we get a contradiction. This concludes the proof of Lemma 3.

Now we shall prove Lemma 1 for z = Z-F. We have only to verify the surjectivity. Fix f e d(v) and put

* = cplp(l: ')-1fp1, g = f - E(P: 1: v) .

It is clear that g e (3(v) and gpl = 0. Hence g = 0 from Lemma 3 and therefore



152 HARISH-CHANDRA

This completes the proof of Lemma 1 and hence also of Theorem 14.1.

17. The functional equations for E(P: i: v)

We return to the notation of Section 14. Put L = OC(M, zM). Then L is the orthogonal sum of L(w) for wo e &2(M).

LEMMA 1. Fix P1, P2 e 9P(A), s e ro, v e a' and wo e &2(M). Then cP21Pl(s: V)

defines a bijection of L(w) onto L(sw). Moreover

P(w: V)cP2IP1(s: ))*Cp2lpl(S: 2) =c(A)2

on L(w) for P e 9P(A).

It is easy to check that this statement is additive in r. Hence it follows from Lemma 13.4.

COROLLARY. [(sW): sv) = tt(wo: i) for s e ro.

Given w e G2(M), we can choose z so that L(w) 0 {0}. Fix P e 9(A) and put Q = P8. Then if v e ' and * e L(w), we have

p(Sw: S) I CQQ(1: S))Sr 11j2 = c(A)2 1j sf 112 = c(A)2 11 * 112

from the above lemma. On the other hand [2(j), Lem. 18.1]

S-lCQIQ(l: Si)s = cPjP(1: 2) .

Hence

C I Qc(1.: S2)Sb 12 = Ic P(1: V)* j12 = p(wo: 2)-)c(A)2 11 * 112

and the desired result follows. Put

0CQIP(S: 2) = CQjQ(l: S2))'CQIP(S: 2)

for P, Q e 9P(A) and v e B3)

LEMMA 2. E(Q: ?CQlp(s: v)): s8) = E(P: *: 2) for * e L and v e W?(3) It is enough to verify this for ' e L(w) and v e -'. Put

f = E(P: f: v) - E(Q: ocQIp(s: v)): sv) .

Then it is obvious that f e (2)) and

fQS = CQIP(S: V)) - CQIQ(l: S2))CQP(S: V)* =0 .

Hence we conclude from Theorem 14.1 that f = 0.

COROLLARY. Let P, P1, P2 e 9P(A) and s, t e t,. Then

0cP21lP(st: 2) = 0cP21P(s: t2)0cPP11(t: 2)

for v e ,(8).




Fix * e L and v e!2'. Then

E(P1: 1: v) = E(P: 0cplp(t: 4)1: tv)

from Lemma 2. Taking the constant terms of both sides along P2, we conclude that

cP2l1P(st: 4)+ = cP21P(s: tV)0cPP11(t: V)1

and this implies our assertion.

LEMMA 3. ?CQjp(s: v) maps L(@) into L(sw). Moreover if 2 e A', ?CQjp(s: 2)

is unitary.

The first statement is obvious from Lemma 1. Now fix v e I'. If 01, 02 are two distinct elements in 62(M), then L(o1) is orthogonal to L(w2). Hence it is sufficient to verify that

11 jCQIp(S: V)+ 112 = 11 j 112

for * e L(@). Put

0q = CQIP(S: 4)+ .

Then 0 e L(sw) and

CQIQ(l: s2)) = CQJPp(s: V)) .

But

CI CQQ(l: S2))q j j2 = c(A)2je(sw: sV)-l jj 0 112 I CQ I P(S: 2)+ jj2 = c(A)2e(w: 2)-1 jj I1 112

from Lemma 1. Since [(sw: sv) = p(w: 2) (cor. of Lemma 1), we conclude that 11011 = 11 * 11

LEMMA 4. Let P1, P2 e 9P(A) and suppose P' = M'A'N' is a psgp of G such that P'DP1U P2. Put *Pi = MnfPi(i =1,2)and *A=M'fnA. Let *to be the subgroup of all elements in tn which leave a'pointwise fixed. Then if Ve ',

0cP21P1(s 2) = 0c'P21*P1(s: *2)

for s e *to. Here *2 is the restriction of v on *a.

The proof depends on an auxiliary result on Eisenstein integrals. Fix P e 9P(A) and let P' = M'A'N' be a psgp of G containing P. Define *P Pn M'. Then *P = M. *A. *N where *A = An M'.

LEMMA 5.

E(P': E(*P: V: *v): 2') = E(P: b: 2) for * e L and v e H

Here *2 and 2' are the restrictions of v on *a and a' respectively.



154 HARISH-CHANDRA

It is enough to verify this for v e I. Fix f e Cc (G, z). Then

(E(P': E(*P: A: *v): 2'), f) = (E(*P: 1: *2), flP')) = (*, (f(PP'))(*P)) .

This follows immediately from the definition of the Eisenstein integral [2(i), ?191. But

(f(P'))(*P) fP)

and

(i, f(P)) = (E(P: i: 2), f)

and so our assertion is obvious. Now we come to the proof of Lemma 4. Since s leaves a' fixed, it may

be regarded as an element of to(M'/*A). Then so *v = *(sv) and we conclude from Lemma 2 that

E(*P1: A: *2) = E(*P2: 0C*P21*P1(s: *V))y *(s2)) .

Hence

E(P1: A: v) = E(P': E(*P1: v *,): V) = E(P': E(*P2: 'c*P21*Pl(s: *V)+: *(sV)): V) = E(P2: 'C*P21*Pl(S: *V)): S)

from Lemma 5. On the other hand

E(P1: *: 2) = E(P2: 'CP21P1(S: V)): S)

and so the desired result follows from Lemma 16.1.

LEMMA 6. Fix P, Q e 9P(A) and s e to. Then 'cQlp(s: v) is everywhere holomorphic and unitary on A.

In view of [2(j), Lem. 18.11, it is enough to consider the case s = 1. Then by the corollary of Lemma 2 we have the product formula

CQjP(l.: 2) = CQjP1(l: V)0cP11P(1: 2) (P1 e @(A)) .

Hence it is enough to consider the case d(Q, P) = 1. (Note that 0cplp(l: 2) = 1.) Let a be the unique root in 2(Q I P). Put *P = Ma, n P in the notation of Lemma 2.3. Then

CQjp(l. V) = Oc*p-~p(l: *2)

by Lemma 4 where *2 is the restriction of v on a,,, = RH,,. (We take P' = P.,, in Lemma 4.) Now 0c*p-~p(l: *v) is a meromorphic function of the complex variable p)a = <a, 2> defined for all sufficiently small values of I Im V) 1. Since it is unitary for pa real and #0, it is obvious that it cannot have a pole at pa = 0. This proves the lemma.




18. Relations between the c- and j-functions

We return to the notation of Section 13.

LEMMA 1. Fix T e End 'F, P, Q e CP(A) and v e !'. Then

CQIP(l. V)*T = IfT'

where

T= jpQ(V)-)TjPIQ(V)).

This follows from the corollary of Lemma 13.3 if we take into account Lemmas 13.2, 16.1 and 17.2.

COROLLARY.

CQIP(l. 4)T = kT1

where = c(A)j-lp(v)TjPQ(v).

Since cQIp(l: v) = cQQ(l: v)0cQIp(l: v), we conclude from Lemma 11.1 that

CQIp(l. V)T = Tj

where = c(A)j-JQ(v)jPQ(v)-'TjPQ(v) .

But P lies between Q and Q. Hence (corollary of Theorem 13.1)

jQIQ(V) = jQIP(V)jpIQ(V)

Therefore

T = c(A)jQjp(v)TjpjQ(v).

19. Rationality of 0CQ1p

Let r and L be as in Section 17.

LEMMA 1. Fix P e CP(A) and 2, V in A;'. Then the four linear transformations

cPIP~l ) p(l: l: ), cp-jP(l: '), cplp-(l: v')

commute with each other. Moreover

cPIP(l: 2)* = cp-1 (1: 2) , c -jP(1: 2')* = clp,(1: 2')*

It is easy to see that the above statement is additive in r. Moreover by Lemma 17.1 all these linear transformations map L(@) into itself. There- fore our assertion follows from Lemma 13.2 and the corollary of Lemma 18.1.

The following result is of some interest although we shall not make use



156 HARISH-CHANDRA

of it in this paper.

LEMMA 2. Fix P, Q e CP(A) and s e Ir. Then

V I ) 0CQIp(S: V)

defines a rational mapping of He into End L.

Again, in view of [2(j), Lem. 18.1], it is enough to consider the case s = 1. Moreover, as we have seen above, it would be sufficient to verify the following lemma in the notation of Section 18.

LEMMA 3. Fix T e End DF. Then 2 I - jpJQ(2))lTjPJQ(V)

defines a rational mapping of He into End OF.

Define 0 and S as usual (see [2(i), ? 11]) and let ( = K? & denote the space of all functions f: K -- 0 such that f(k) = 0 for all kc e K except a finite number. Then ( is a complex vector space. For k e K and g e 0, let k ? g denote the function f e @ given by f(k) = g, f(k') = 0 (k' # k). We turn @

into an associative algebra by introducing a multiplication as follows:

(kI c) g1)(12 0 g2) = 12 ? (Dg& *g2).

Put

=LUFO G1DFO

when FO runs over all finite subsets of 6(K). Then if ir = p we associate to it a representation ir0 of ( on .? as follows. If h e 0?,

0(k (lc g)h = ic(k)l(g)h (k ce K, g e ) .

(Recall that there is a natural action of ( on .0 under ir.) We shall usually write ir instead of ir .

Put A = K (0 c (. Then A is a subalgebra of (. Let EF denote the set of all x e A such that lr(i)h = 0 for every h e QF* Then 9F is a two-sided ideal in A (which is independent of P and P). Let IF be the set of all f e d such that

RF f C_ @ SF. Then OF is a subalgebra of ( and 'F is stable under 1r(OF). Let lrF = ZFP , denote the corresponding representation of 3F on 'F. If r is irreducible as a representation of G on I, then it is easy to see that lrF is an irreducible representation of OF on IFS

Now fix o e g'. Then by Lemma 13.3, WrF,P,,o is irreducible. Let T, (1 ? i < r) be a base for End OF. Then we can choose gi e OF such that




7rF,P,"0(gj) = T, (1 < i < r) . On the other hand, it is easy to deduce from Corollary 2 of Lemma 4.1 that for fixed g e OF and Q e CP(A),

) I- 7 FQ4Xg)

is a polynomial mapping from g- to End OF. Hence

rFP,(gi) = p pi(P)T, where pij are polynomial functions on H* Put

p(p) = det (pij(2)))ij~r

Then p is also a polynomial function on H. Moreover pii(po) = 3ij and hence

p(o0) = 1. This shows that p # 0. We can choose polynomial functions pi' on H- such that if

gi(v)= , pii()g6, then

7CF, P ,(gj(v)) = p(P)Tj (1 < i < r, v e .

Now let v e I'. Then

7cF,PV(g9(V))jPjQ(V)) = jPIQ())rF,Q,1(g9(V)) )

This follows immediately from the corollary of Lemma 5.1 by analytic continuation. Hence

jpjQ()-) TjjpIQ(v)) = P(V)) 1rFQv(9i(V))

provided p(v) # 0. Since the right hand side is rational in v, the lemma is proved.

Part III. Explicit determination of the Plancherel measure

20. Evaluation of (Oa) P)

Let d2) denote the Euclidean measure on = a* which is dual to the Haar measure da = dA on A. Define i(G/A) = v(P) (Lemma 2.6) and c(G/A) = c(P) = c(A) (? 11) for P e CP(A). Fix P1, P2 e CP(A) and * e L(@).

THEOREM 1. Put

Oa= 5 [(a): v))c(v)E(P1: A: 2))d2

for a e Cc (I). Then fo e C(G, z) and ( )(P2) - Ye2 S a(s'2))00 (s:

for 2) e A. Here 'Y = -(G/A) and c = c(G/A). Let Q be an open and relatively compact subset of A. Fix is e Cc(I)



158 HARISH-CHANDRA

such that 8 = 1 on Q and put

O(P: x) = f8(,Wep(: P)E(P1: A: v: x) (v e A, x e G).

Then if P e FP(A), v e A' and s e n,, we have

[ I P(O: )c )lp(s V)* 112 = C21P(: 2) I I I 112

from Lemma 17.1. Since t< is holomorphic on a (Lem. 12.2), we conclude from [2(j), Thms. 11.1 and 13.11 that OA e C(G, z) for a e Cc(Q). The same argument shows that a - OA is a continuous mapping of Cc(g) into C(G, 4).

Now first assume that a e Cc(I'). Then

OIP2)(ma) = exp {(-1)1"2V(log a)}

X v8_V -C2 2 P 8 a~s C(v),p(G): Sl)pp- (1: V)C - Ip (s: s-'v)*dv

from [2(j), Thm. 19.21 for m e M, a e A. But

P21P2(l: V)CP2IP1(s: S'2)) = ep21p2(l: I ) P 22p2(l P2) p (s s'l)

= p-2lp-2(l: V)C)2p2(1: V))C0-2p1(S: Is ')

from Lemma 19.1. Moreover

0p2lp-2(l: V)0C-21P1(s: s'2p) = CP2lP2(l: ))CP2lP-2(1: ))CP-21 P1(s: s 'p)

= CP2lP2(1 2))CP2lPl(S: S'2))

from the corollary of Lemma 17.2. Since 0p2I1pl(s: s-'l)+ e L(sw) (Lem. 17.3), we conclude from Lemmas 17.1 and 19.1 that

e 2p2(l: 2)C)2jp1(s: s 'p)+ = C2[e(s): 2))'0CP2IP1(s: S'V)) .

But pe(sw: 2) = e(o): s-'v) (cor. of Lem. 17.1). Therefore

O'P2)(ma) = YC2 exp {( - 1)" 2V(log a)} a 6la(s-'2)0c 21p1(s: s-l2))d2

and so the statement of the theorem is true in this case. Now let a e Cc-(). Put f = Oa. Then we have seen above that

f e C(G, z). Moreover if P' is a psgp of G with prk P' > dim A, then f (P' = 0 (cor. of Thm. 13.2 of [2(j)]). Fix 2 ea and put g= f(P2)* We claim that g e0C(M, TM) = L. Let *P # M be a psgp of M (*P = *M. *A. *N) and P' the corresponding psgp of G contained in P2. Then P' = M'A'N' where

M'= *M, A' = *A.A, N' = *N.N2

Hence if *2 e (*a)* and 2' = v + *2, it is clear that g(*P) = P)= 0

This shows that g(*P) = 0 and therefore g e L. Fix v e a and consider the mapping




T,: a - (0,)(P2)

of Cc(O) into L. We have seen above that a F 0 is a continuous mapping of Cc(Z) into C(G, zr). Hence [2(i), Lem. 16.1] T, may be regarded as a distribution on a with values in L. (Recall that dim L < Ao.)

Fix vo e g-, (0o e 62(M) and put

T~~(a) = Eo(0,)IPo2)

where E.( is the orthogonal projection of L on L(w0o). Write T= T=OO. Then T is a distribution on Z with values in L(0o).

Extend a to a 0-stable Cartan subalgebra t of g such that tR = a and tc Cm (see [2(i), ? 81 for notation). Let X = X. and x0 = 'X, be the elements in (-1)1/2t, which correspond to the infinitesimal characters of w and a0 respectively. For z e 3, define the polynomial functions p(z) and po(z) on as follows:

p(z): V -+Y(Z: X + (_1)1/2V), po(z): - I 7(z: X0 + (-_1)"12) (Ve )

where " = ", (see [2(i), S 11] for notation).

LEMMA 1. T(p(z)a) = po(z: o)T(a) for a e CX (9;) and z e 3.

Note that

zE(P: A: v) = p(z: v)E(P: A: ,)

from [2(i), Lem. 19.1]. Hence

Z2a = Op(z)a

and therefore

T(p(z)a) = Eo)(Z0J(P2)

But

(Za) (P2) = P(Z)(-a) (P2)

where , = ,PP2 in the notation of [2(i), Lem. 16.11. Hence

T(p(z)a) = po(z: vo)T(a) .

COROLLARY 1. If v) e Supp T, then X + (-1)l/2V = -5(X0 + (_1)1/2 vo) for some s e W(g/l).

This is obvious from the above result.

COROLLARY 2. Suppose vo e H'. Then Supp T c {Js e en

Put W = W(g/l) and suppose v e Supp T. Then by Corollary 1, v = s'Vo for some s' e W. Put Ho = H,0, H = H,. Then Ho and H are two points in a which are conjugate under W and therefore also under K [2(i), Lem. 5.11.



160 HARISH-CHANDRA

Let 30 and 3 respectively be the centralizers of Ho and H in g. Then dim 3 = dim 30 and 3 D m + a = 30 since P0 e A'. This shows that = = = m + a. Fix k e K such that H = Hok. Then k normalizes 3. Since a is the intersection of the center of 3 with 4, k also normalizes a. Hence k defines an element s e tt and sHO = H. This proves that v = spo.

Now fix v e A' and consider the distribution

T,: a H - (,0)(P2)

on a with values in L. By Corollary 2 above, Supp T, c {s2)}e.X. Fix ,8 e Cc (a') such that 8 = 1 on some neighborhood of the set {sP}Je0. Then T,(a) = T,(,a) and therefore

I P2) = YC2 a a(s'V)0CP21 P(s: s'1p)+ (a eC

from the result already proved above. But for a fixed a, both sides are continuous functions of v (with values in L). For the left side this follows from [2(i), Lem. 16.11 and for the right side from Lemma 17.6. Therefore since the equality holds for v e A', it must hold for all v e A. This proves Theorem 1.

We state the following simple lemma for later use.

LEMMA 2. Let A' be a special vector subgroup of G such that dim A' > dim A. Then if P' e 9P(A') and ac e C,-(g-),

O P' = 0

unless A' is conjugate to A under K.

This follows immediately from the corollary of [2(j), Thm. 13.21.

21. The characters 9w,,

Fix P e 5>(A), W e 62(M), v e 9ic and put Xr = 1rp <, . We intend to deter- mine the character 9,, of ir.

LEMMA 1. There exists an integer N > 1 such that

dim fib < Nd(b)2

for all b e &(K).

We have seen in Section 4 that

dim .lb = d(b) [bG6(K) [b 3[w: c31

Since a) is the class of an irreducible unitary representation of M, we can choose an integer N > 1 such that

[a): 31] < Nd(3)

for all 3 e &(KM). Then




dim fb < Nd(b) FMS(KM) d(3)[b: 8J < Nd(b) 2

It follows from Lemma 1 (see [2(a)]) that for fe CQ(G), the operator ic(f) is of the trace class and the mapping

0": f trs(f) (feC,-(G)), is a central distribution on G.

LEMMA 2. For f e CQo(G), put

f (x) = 5 f(kxk-')dk (xeG).

Then

91(f) = 0Xf )-P)

where 80, denotes the character of the class w.

We recall that

a( )(m) = XI oa(man) exp {-((-1)1/2V - p)(log a)}dadn (m e M) AXN

for a eC,-(G) and veH For a finite subset F of 6(K), define fF as in Section 3. Then it is clear

that

f = limF fF

in CQ(G), where F runs through an increasing sequence of finite subsets F, of 6(K) such that

Us F, = (K).

Now fix F and put a = fF. Then T = ir(a) lies in End 'F and

,r(r) = tr T = 5 trtT(1: k-1)dk

from the corollary of Lemma 6.1. But

TT(k2: k) = Ka(k2: k1)

= a(k N('man lk-')a(m) exp {((- 1)1/2 + p)(log a)}dmdadn

from Lemma 4.1. Hence

Eor(a) = 0,((aO)-P)

and the required result follows by taking limits with respect to F. Define t and x, as in the proof of Lemma 20.1 and put " = -o, COROLLARY. ze1, = Y(Z: X. + (-1)1/2V)9, for z e 3.



162 HARISH-CHANDRA

This is an immediate consequence of the above result and [2(i), Lem. 16.1].

We can now conclude from [2(i), Thm. 11.1] that 0, is a function which we denote by 0P,,p.

LEMMA 3. Op,,,, is independent of P e CP(A).

Fix f e C-(G). Then it is obvious from Lemma 2 that

V 1 Ojp(f)

is a holomorphic function on We Now fix P1, P2 e 5>(A) and put

F(p)) = Op vwAf) -EPV,@Af) -

Then F is a holomorphic function on Zc which is zero on A' by Theorem 13.2. Hence F = 0.

In view of the above lemma, we can write O,, instead of Opp. LEMMA 4. Let s e tv and v e Gil Then

e0@,8' = o,

This is deduced in the same way from Theorem 13.2. Now suppose v e A. Then it follows from Lemma 2 and [2(i), Lem. 16.11

that 0,, is a tempered distribution on G. Define

(cup, f) = 5 conj O.,(x) f(x)dx (f e (G, V))

In view of [2(i), Cor. 2, p. 1391, this is justified. Similarly put

(OW, g) = conj 0,(m) * g(m)dm (g e e(M, V))

LEMMA 5. Let f e C(G, z), p e v and P e 9P(A). Then

(90>, f) = Fo(OoP f,'P')

Here F0 is the operator on V given by

Fov = |(k)vr(k-')dk (ye V).

This is obvious from Lemma 2.

22. Computation of (90,, 0) Fix W0 e 62(M), * e L(wo), P e 9'(A) and put

O = 5 [c(wo: P))a(P))E(P: fr: P))dv)

for ac e Cc(@). Let tv(wo) be the subgroup consisting of all s e t such that sWo = ()0o




THEOREM 1. Fix a e Cc(I) and put

f (wa: V) = (O., 0)

for w e 62(M) and v e A. Then (1) f(sw0: sP) = f(w): V) (s eG ). (2) f(wo: P) = 0 unless w = so, for some s e o. (3) f (w0,: P) = Y2d(w0)F)-1F0(l) A, e(w0) a(sp).

Here d(0)o) denotes the formal degree of o)o and -i and c have the same meaning as in Theorem 20.1.

Put f = -. Then

f(w: 2) = (0,,, f) = F#(O, f(P)) = IC2 s a(s-'v)F0(O., Ocplp(s: s2)*)

from Theorem 20.1. Since 0cplp(s: s')/ e L(swo), the second statement is obvious and the first follows from Lemma 21.4. Moreover it is clear that

f(w)0: ) = YC2 Ehso a(s-'2)F0(8,,,0, cplp(s: s'2)*) where >0 = w(o)

On the other hand, by [2(f), Lem. 81],

d(w0)(O0,, At) = 0 (1) for ' e L(0o). Hence

Fo(Owoy 0CPIP(s: s'2))) = d(o)0)-F00cpjp(s: s-lv))(l) for s e n0. Fix s e to and put

*= I0CPIP(s: s8'2))'&

Then it follows from the definition of the Eisenstein integral that

E(P: r': 2): 1) = Fo*'(1) . On the other hand we conclude from its functional equation (Lem. 17.2) that

E(P: ': 2): 1) = E(P: 'r: s't): 1) = Fo*(1). Therefore

f (w)0: 2) = C2 d(wo)F Fo/(1) E a(sp) and this proves the theorem.

23. The characters of the discrete series We assume in this section that rank G = rank K and recall some facts

about the discrete series of G. As far as possible we shall keep to the notation of [2(i), ? 271.



164 HARISH-CHANDRA

Let B be a Cartan subgroup of G contained in K and B * the set of all irreducible characters of B. Fix an order in (-1)/2b* and for b* e B* define ?X(b*) and ty(b*) as in [2 (i), ? 27]. Let B*' be the subset of those b* e B* where tY(b*) # 0. The group W(G/B) operates on B*'.

THEOREM 1. There is a unique bijective mapping from 62(G) to W(GIB)\B*' with the following property. Suppose w) e 62(G) and b* e B*' correspond under this mapping. Then

'A(b)0,(b) = (-l)q sign ty(b*)* We W(G/B) s(s)<sb*, b>

for b e B'. Here q = 2 dim GIK and 90 is the character of w.

This follows immediately from [2(i), Thm. 27.1 and Cor. 1, p. 1811. Put

'@D(b) = 'A(b)0,(b) (b e B')

Then '(D. extends to an analytic function on B.

LEMMA 1. Let f e O(G). Then

'Ff (b) = Ge'2(G) (e0, f)"D,(b) (b e B')

This is a restatement of the corollary of [2(i), Lem. 27.41. Put ti' = e--Pt o eP and define the constant CG as in [2(i), Lem. 27.5].

Then

'Ff(1; tux) = (- )qCGf(1)

for f e C(G). As usual, let d(w) (W e &2(G)) denote the formal degree of w).

COROLLARY. Suppose W e &2(G) corresponds to b* e B*'. Then

d(w)) = c' I t(b*) I [W(GIB)ld(b*)

Moreover f(l) = Ewer;2(G) d(w)(93@, f) for f e 0C(G).

This is obvious from Corollary 1 of [2(i), Lem. 27.51. Let V and z be as in Section 14. Put L = 0C(G, z). Then dim L < o and

L is the orthogonal sum of L(w) (w) e &2(G)). For f e C(G, V) and x e G let r(x)f denote the function y l * f(yx) (y e G).

LEMMA 2. Given f e C(G, z) and ) e &2(G), let EJf denote the function

x ! ) d(wx)(O., r(x)f ) (x eG).

Then E. is a continuous projection of C(G, z) onto L(w) and

(fi, Ew f2) = (Ewfi. f2)




for f, f2 e C(G, ). Fix ir e c) and let D be the representation space of ir. We can choose a

finite subset F of &(K) such that aF *f = f for all f e C(G, 4). Put

OWF =aF* 0.

Then

OWF(x) = trEFr(x) (x e G),

in the notation of Section 3. Therefore )W,F e 0C(G) and it is clear that

(O., r(x)f) = f(y) conj E,,F(YX-1) *dy

Taking into account the Schur orthogonality relations for w, we conclude that E., is a continuous projection of C(G, z) onto L(@). The last statement of the lemma follows from the above formula and Fubini's theorem.

24. Determination of [(o): v) in a special case

Now we return to the general case. Fix a 0-stable Cartan subgroup A of G and use the notation of [2(i), ? 171. Then A = AIAR. Fix P e P(AR)

(P = MARN), w0 e 62(M), ' e L(wo) and put 5 = aR.

LEMMA 1. Put

0a = [((oo: v)a(v)E(P: A: 2))dv

for a e C(-). Then

'Foa(aia2) = -12d(o)-'F0'*(1) E,,,. "(P8,,(a1) | a(v) exp {(- 1)"'2s(log a2)}dv

for a, e A' and a2 e AR.

Here Y, c and F0 have the same meaning as in Theorem 22.1, '"Pw0 is defined as in Section 23 and A' = A'(I) n K.

Fix a2 e AR and put f = 0 and

g(m) -f fP(ma) (m e M) .

Then by [2(i), Lem. 8.2],

'Ff (a~a2) = Fo a'F,(ai) (a, e A')

where

'Fg(ai) - 'FM/AI(a,) = 'A.(a1) g(malm-')dm . M On the other hand, it follows from Lemma 20.2 that g e C(M, zMX) =L. Hence

'Fg(ai) = EWee2(X) (Ow, g)'PO(al)



166 HARISH-CHANDRA

by Lemma 23.1.

LEMMA 2. Let cO e 62(M). Then

(K g)= (O, fgP) exp {(-l)"'2.(log a2)}dv

It is clear that

g(m) = \f'P)(m) exp {(-_1)2,V(log a2)}dv'.

Since Supp a is compact, we conclude from Theorem 20.1 that

adv 5 6O,(m) I I f'p' (m) I dm < oo

Hence our assertion follows from Fubini's theorem. This shows that

'Ff(ala2) = je62(M) (a) f (o: ) exp {(-1)o2v(log a2)}dv

=~~~ ~~~ 8>> )@() (sajo: O) exp {(-)1/2.V(1g a2)ldv

in the notation of Theorem 22.1. (Here WO = w(ajo).) But

f(soO: s) f(oO: s-'V)

and therefore

'Ff(ala2) = [wj' aes 1D8wo(a1) \ f(coo: ) exp {(-1)/2S(log a2)}dv

( 2d~o0)'FO/(1) ,8, 'D8W0(a1) \ a(v) exp {(-1)1/2sV(log a2)}dv

from Theorem 22.1. This proves Lemma 1.

LEMMA 3. Let r be a d-stable Cartan subgroup of G such that dim rR > dim AR. Then

'Fly = 0

for a e C-(O) unless r is conjugate to A in G.

Suppose r is not conjugate to A. Then FR and AR are not conjugate under K. Hence if Q e P(rFR), we conclude from Lemma 20.2 that 0(Q) = 0. This implies our assertion (see [2(i), Lem. 8.2]).

Let Q be the set of all positive roots of (g, a). Then

Q = QI U QR U QC

where Q1, QR, Qc respectively are the subsets of imaginary, real and complex roots in Q. Put




tI = laeeQ, Ha, aR

= leQR Ha +

= llaeQc Ha

Then t Ia = V = VIVR+

LEMMA 4. [Q,] is even and

IV8 = V+

for any element s e W(g/a) which maps a into itself.

Without loss of generality we may assume that a root a of (g, a) is positive if its restriction on aR is a root of (P, AR). Then if a e Q?, it is clear that a # -Oa and -Oa e Q,. This shows that [Q,] is even. Note that if A e (-1)l/2a*, then

<-Oa, A> =-<a, OA> =-conj <a, A>

and therefore

(-1)P-v+(A) > 0

where 2p= [Q,] Now fix s as above. Then s permutes the complex roots among themselves and therefore -,=r'x+ where '= ?1. Choose A e (-1)1/2a* such that f+(A) # 0. Then since s-'A is also in (-1)1/2a*, it is clear from the above result that )7 = 1.

Let a* be an element in A`' which corresponds to CO, (under Thm. 23.1 applied to M). Put X = X(a*) e (-1)12a, .

THEOREM 1. Suppose a is fundamental in g. Then

e(o0: p) = IC2[>]~ XL j ii+(X + ( _1)1/22)) |

for i e

Here CG and cM are positive constants for G and M respectively corresponding to (-l)qe of [2 (i), Thm. 37.1]. Also we recall that a is fundamental if a, is a Cartan subalgebra of f.

In proving this theorem, we may obviously assume that the set Q of positive roots of (g, a) is chosen as above. Put

- = e-Pr' o ePI

where 2p, is the sum of all roots in Q,. Then

'Ff (1; C) = (-l)gGcf (1)

for f e C(G). Here

qG= dim GIK - rank G + rank K} .

Put qX = a dim MIK,. Since G = PK, it is clear that



168 HARISH-CHANDRA

dim G/K = dimP/P nK = dim MIKM + dim AR + dim N. Moreover since a is fundamental in g,

rank G - rank K = dim a - dim a, = dim aR.

Also QR is empty and dim N = [Q,] = 2p. Therefore

qG = qM + p. Now ti = W(G/A)/W(M/A,). Put

a) = ,'(D8,0(aj) exp {(- 1)"'2sv(log a2)} (al e AI, a2 e AR) for v e A. Then we conclude from the results of Section 23 that

qua (1; Br') - (-i)gMc~doO0) FseW(G/A)/W(M/AI) t+(SX + (_1)1/2sV)

- (-l)qGCM[1Jd(coo) I ,-V+(X + (-1)1/22)) l

Put f = se in the notation of Lemma 1. Then

(-l)qG Cf(1) = 'Ff(1; tr') = yc2d(a0o)-'Fo*(1) x(z)$v(1; ui')dv

= (-l)qGM[t7c2 Fo*(1) e a(v) I ii+(x + (-1)1/2V) I dj

from Lemma 1. On the other hand

f (1) = C5(l) |(o0: 2)a(v)F0*(1)dv'.

Since a e CC(Z) is arbitrary, we conclude that

4a(oo0: )F0*(1) = CMCG1[jljyC2 I -V+(X + (-1)1/2p) I F0(1) for v e A. Theorem 1 will be proved if we can choose z and f in such a way that Fo*(1) ? 0.

Take z = zF (see ? 7) where F is a finite subset of 6(K) such that # {O}. (Here Q is defined as in Section 4 corresponding to a e C)o.) Choose

T e End O(F such that tr T # 0 (for example T = EF) and put * = FT. Then A e L(Coo). Let v = Fo*(1). Then

V(1: 1) = |*(k: 1: k-')dk = trlcT(k-1: k)dk = tr T # 0

from the definition of *T (? 7) and the corollary of Lemma 6.1. This shows that Fo*(1) # 0.

25. Extension of A, j and c on the complex space

Let us now use the notation of Section 13 and consider the function P,(V) = P (o: p) (p' e 0).

THEOREM 16. p, extends to a meromorphic function on Air Moreover 8 Cf. the first conjecture of [2(b), S 16].




we can choose 3 > 0 such that the following two conditions hold. (1) ,e, is holomorphic on (2) There exist numbers c, r > 0 such that

[P((O2a) I < c(l + | pR J)"

for all v e W3)

Fix P e 9P(A). Then (see ? 13)

P((o: V) = llcre P) Pei:() .)

Hence it is clear that it would be sufficient to verify the above theorem under the assumption that prk G = 0 and dim A = 1. This will be done a little later in Sections 28 and 36. For the moment we assume Theorem 1 and proceed to derive its consequences.

LEMMA 1. Fix P, Q e @P(A) and s e S. Then jQjp(2v) and CQlp(s: ,) extend to meromorphic functions on Air

Here the notation is the same as in Section 18. In view of the results of Section 18, it is enough to prove this for jQjp.

Moreover by the corollary of Theorem 13.1, we are reduced to the case when d(Q, P) = 1. This in turn reduces to the case when dim A = 1 and Q = P (see the proof of Lemma 13.1). Let a be the unique element in I(P) in this case. Then a- may be identified with the complex plane under the mapping ,v - <k', a>. We know that

X(o: >)jP1)jPl(V) = 1

and ji;p is holomorphic if Im <k', a> < 0, while jpjl is holomorphic if Im<k', a> > 0 (see Lemmas 11.1 and 12.1). Our assertion now follows immediately from Theorem 1.

We now use the notation of Section 22. Fix f e L(@), P e @P(A) and put

0(y: x) = e(oj: ,)E(P: A: ,: x) (v 'e A, x e G) .

Extend a to a 8-stable Cartan subalgebra 1 of g such that tR = a and tI c m. Let X be an element of (- )1/2w which corresponds to the infinitesimal character of c.

LEMMA 2. 0 is a function on a x G of type II'(X) in the terminology of [2(j), ? 9].

It is easy to deduce from Theorem 1 and [2(j), Lem. 17.1] that 0 is of type II(X) (see [2(j), ? 8]). Hence it remains to verify that 0 satisfies the condition of [2(j), Thm. 11.1].

Fix Q e 9P(A), v e A' and s e w. Then



170 HARISH-CHANDRA

fQ"(,s) = P(o: 2))CQIP(S: A)+

in the notation of [2(j), Thm. 11.1]. Hence

IIfQ,8(V) 112 = C2/(o: V) I I * 112

from Lemma 17.1. Since p,,, is holomorphic on I, it is clear that IIfQ,,(v) remains locally bounded on a and this proves Lemma 2.

26. Some applications of Theorem 25.1

We keep to the notation of Section 22. Fix P e 9(A), o eG 2(M) and ire L(@).

THEOREM 1. Put

Oa= X (o: 4)ax()E(P: A :')dv

for a e C(g). Then a - Oa is a continuous mapping of C(s) into C(G, z).

This is an immediate consequence of Lemma 25.2 and the corollary of [2(j), Thm. 13.1].

COROLLARY 1. Fix Q e 9(A) and v e Z-. Then

(0a)(Q) - Ye2 E (s ')0CQlP(s: S ')+

for all a e C(s). Put

T(a) = (5a)(Q)- se2 gEw6 a(s'2)0CQlp(s: S'V)+

for a e C(s). Then T is a tempered distribution on g (with values in L). Since Cc(@) is dense in C(s), we conclude from Theorem 20.1 that T = 0.

COROLLARY 2. The statements of Theorem 22.1 and Lemma 24.1 remain true for a e C(s).

This is proved in the same way.

Let P' = M'A'N' be a psgp of G and f an element in C(G, 4). Recall that f(P') 0 means that

XM (q(m'), f ')P(m'a'))dm' = 0

for all q e 0(M', zM,) and a' e A' (see [2(i), ? 20]). Let CA(G, z) denote the space of all f e e(G, z) such that f (P') - 0 unless A' is conjugate to A under K. It follows from [2(i), Lem. 20.2] that CA(G, z) is a closed subspace of (G, z).

LEMMA 1. The function Oa of Theorem 1 lies in CA(G, z).




Put f = Oa and assume that A' is not conjugate to A under K. Then by [2(i), Lem. 20.2], it is enough to verify that

(E(P': i': V), f) = 0

for *' e L' = 0C(M', TM') and V e (a')*. Since

(E(P': 'i': V), f) = (*'y fp(jP)

and the right hand side is a continuous function of V, it is sufficient to consider the case when <V, a'> + 0 for every root a' of (P', A'). Moreover dim L' < and

L' = Ie22(M') L'(co')

Hence without loss of generality, we may assume that *' e L'(o'). Extend a and a' to 8-stable Cartan subalgebras tj and t' respectively of g such that

p =a, tc m and R = a', c m'. Choose Xe(-1)1"2I and X'e (1)1/2I* corresponding to the infinitesimal characters of o and )' respectively. Put

7,= and r = 7 Then

zE(P': ag': V) = Y'(z: X' + (-1)1/2XP')E(P': i': v') (z e ;3) from [2(i), Lem. 19.1]. Similarly if we denote by p(z) the polynomial function

'-7(Z: X + (-1)12 )

on A, we have

zf = Op(z) a

Put

T(a) = (E(P': *': V), Oa) (a e

Then T is a tempered distribution on a and

T(p(z)a) = 7'(z: X' + (-1)1/2V/) T(a) (z e &3)

Hence if v e Supp T, it follows that X + (- 1)1/2V is conjugate to X' + (- 1)1/2 V

under G,. But since this implies that a and a' are conjugate under K [2(i), Lem. 29.2], we conclude that T = 0. This proves the lemma.

27. The Plancherel formula for K-finite functions

Fix V and z as in Section 14 and put L = (M, m). For fe (G, z-), define a function f from 62(M) x a x G to V by

f (co: V: x) = (e),0 r(x)f) = (eOp, I(x-')f) Here (O e 62(M), V e A, x e G and r(x)f, I(x)f respectively are the functions y - f(yx), y - f(x-ly) (y e G). It is easy to verify that



172 HARISH-CHANDRA

f (o: v: klxk2) = z-(k,)f (o: v: x)z-(k2) for k,, k2 e K. Hence it follows from the corollary of Lemma 21.1 that f (o: V) is a K-finite eigendistribution of 23 and therefore it lies in C-(G, z).

THEOREM 1. Fix P e 9I(A) (P = MAN), v e g- and, for f e C(G, z), define

gf (P: m) = fAP (m) =X f(man) exp {-((-)1/2 - p)(log a)}dadn

(m e M) and

If((O. v: m) = d(wo)(Oa, r(m)gf(v)) e 6,(M)) Then *f((o. v) e L(o) and

d(o)f (o0: V) = E(P: *f(o: ,): o).j

Moreover let *f((o) denote the function

2) [- 1ff((O: 2))

on Z-. Then f ~- f (%o) is a continuous mapping of C(G, z) into C(s) ? L((o). We observe that gf e (M, z,,) from [2(i), Lem. 16.1] and therefore

*f((o. 2) e L(o) by Lemma 23.2. Fix * e L(a) and put

hf(2) = (i, *f ((O. 2)) = 5 (*(m), f. 'P')(m))dm

= exp {-(-1)"2)(log a)}da H (ik(m), f'P'(ma))dm

Then it is clear that hf e C(s) and f ~ hf is a continuous mapping of C(G, Z-) into C(s). Since dim L(o) < dim L < o, we conclude that f *f(co) is a continuous mapping of C(G, z) into C(s) ? L((o).

Now fix v e g and x e G. Then

f(c: 2: x) = I(t-x)f)

=KMANconj O<,(m) .f(xkmank-') Kx~xAxNV x exp {I- ((-1)1/2 -p)(log a)}dkdmdadn

from Lemma 21.2. Fix k e K and let xk = k1mnaon( (ko e K, mO e M, aO e A, no e N). Then

| f(xkmank-') exp {-((-1)1/2 -p)(log a)}dadn AxlV

= Z(ko){ f(momaoan) exp {-((1)/2 _-p)(log a)}dadn}z(k-') AxlV

= exp {((_ 1)1/2 -_p)(log ao)}z(ko)gf((v: mom)z(k-') . Hence




d(w) conj O,(m) f(xkmank-') exp {-((-1)"/v - p)(log a)}dmdadn MXAXNV

= *f(v: xk)zc(k-) exp {((-1)1/2V- p)(H(xk))} and therefore

d(o)f(c: v: x) = E(P: f%(co: V): V: x) .

This proves the theorem. The function f%(co) (on g x M) depends on P e 9P(A). Hence we denote

it by *f (P: o).

LEMMA 1. Fix P1, P2 e 9P(A), s e W and w) e G2(M). Then

*f f(P2: Sw: S>) = 0cP2,1l(s: ))*f (Pl: O :)

for f eC(G, z) and v c . It would be enough to verify this for v e g'. It follows from Lemma

21.4 that f (sco: sv) = f (co: 4). Therefore

E(P2: /f(P2: sco: sv): sv) = E(P1: /f (Pl: i: >): s) = E(P2: 0CP2l1P(s: 'V)/f (Pl: C: 4): S)

from Lemma 17.2. Our assertion follows from Lemma 16.1. Put

f = c-2r'~to-ld((o) | ((o: i)f(w: ,v)dv

for fe C(G, z-) and Gw e e2(M). (Here Y and c have the same meaning as in Theorem 20.1.) Then it follows from Theorem 1 and the results of Section 26 that f ~ f, is a continuous linear mapping of C(G, z-) into eA(G, zr).

Let Q be the set of all Gw e C2(M) such that L(co) ? {0}. Since dim L < oo, it is clear that Q is a finite set which is stable under w. Moreover it follows from Theorem 1 that f, = 0 for f e C(G, z-) unless a) e Q. Put

fA = 'WE&2(MWf' (fee( G, z)). Then f * fA is also a continuous linear mapping of C(G, z-) into CA(G, zr).

LEMMA 2. Let f e C(G, r), cw e 62(M) and s e a. Then

fAW = f. - This is an immediate consequence of the relations

f(sc: sV) = f(c: V), e(Sco: sV) = e(60: V) (> e ) (see Lem. 21.4 and the corollary of Lem. 17.1).

LEMMA 3. Fix f e C(G, z), Pc 9P(A), cw, cw' e &2(M), e L(co') and v e : Then



174 HARISH-CHANDRA

(E(P: ) f ) =[:D(a)F-l(E(P: *:,v, )f) if o' = so for some s e a, 0 otherwise.

Here iv(co) denotes, as before, the stabilizer of co in to. We know from Theorem 1 that

Ifs = C-27-1[ ep(o: V)E(P: *f(P: o: k): v)dv .

Therefore

(E(P: A:,v), fV)=(, f,"

= Iw]' E86e (*i cOp|p(s: S'2))f (P: o: s-'2))

from Corollary 1 of Theorem 26.1. Hence applying Lemma 1, we get

(E(P: A,: ;), fo) = [n,-' t86W (*, *ff(P: swa: v))

If the right hand side is not zero, we can choose so e tv such that cl' = so(0.

Since f, = f80,, there is no loss of generality in assuming that so = 1. Then

(E(P: A: ,), fo) = [w: *o)J'(+, 'f(P: oj:,v)) .

On the other hand

(E(P: *: k), f) = (f, f'P) = (',f (P: o: ))

from Lemma 23.2 and so this proves Lemma 3.

COROLLARY. (E(P: *: k), fA) = (E(P: A: k), f) for all * e L.

It is enough to verify this for * e L(a') for a given Oc' e &2(M). Then

(E(P: A: v), fA) = XE6,2(M) (E(P: : ), fo) = SJ w1>(St ((P: A: f), 8w

= [iv: bt(a')](E(P: A: 2), fm) = (E(P: A 2), f)

from Lemmas 2 and 3. Let rl, * , r, be a complete set of 8-stable Cartan subgroups of G, no

two of which are conjugate. Put Ai = (ri)R. Then if i ? j, Ai is not conjugate to Aj under K. Let C denote the set (rl, * r,) and S the set (Alp .. Ar).

THEOREM 2. f = EA 1S fA for all fe (G, z).

Put

9 =

EA16 fA

Then g e C(G, r). In order to prove that g = 0, it is enough to verify that g(PI - 0 for every psgp P of G [2(i), Lem. 20.1]. Fix a psgp P = MAN of G.




We may obviously assume that 'C(M, zM) = {o}. Then we conclude from [2(i), Thm. 18.1] that A is the split component of a 0-stable Cartan subgroup of G. Hence we can choose k e K such that Ak e S. Since our problem is not substantially changed if we replace P by Pk, we may assume that A e S. Now

g (P) = f(P) - A' eS (fA ')(

andfA' e CA'(G, z). Therefore (fA,)(P) 0 if A' ? A. Hence it would be enough to verify that

(f -fA) P 0 .

But this follows from [2(i), Lem. 20.2] and the corollary of Lemma 3. We recall that

fA(x) = c(G/A)-2`(G/A)-'to(G/A)]-' Ewet 2(H) d(a) 5 e(o: ) r(x)f)dv

for f e C(G, z). Here A e S and P = MAN is a psgp in 9'(A). Observe that

c(G/A) = Y(G/A) = [w(G/A)] = ,ci(w: 0) = 1

if A = {1}. Evaluating both sides in Theorem 2 at 1 we get the following result.

THEOREM 3. Let f e C(G, z). Then

f(l) = EAeS c(G/A) 27(G/A)V[t(G/A)L]' E"6S2(M) d(co)

X ,cl(c0: )(0W,, yf)dv .

This is just the Plancherel formula for K-finite functions in e(G). In fact fix a finite subset F of d(K) and let CF(G) be the space of all f e e(G) such that f= aF*f* aF. Then CF(G) may be identified with C(G, z) for Z = ZF* [2(i), 26] and therefore the above theorem holds for all f Ge CF(G). Moreover 0, , being the character of a unitary representation, is a distribution of positive type. Since

UF CF(G)

is dense in L2(G), it is easy to derive from Theorem 3 the usual Plancherel formula for functions in L2(G).

We shall see in another paper that the statement of Theorem 2 can be extended to the case when dim z = oo (see [2(g), Thm. 12]).

28. First reduction in the proof of Theorem 25.1

We now come to the proof of Theorem 25.1. As we saw in Section 25, it is enough to consider the case when prk G = 0 and dim A = 1. We shall



176 HARISH-CHANDRA

compute /i,. in this case explicitly. Extend a to a 0-stable Cartan subalgebra t) of g such that a = TR and

fIcm. Then

rank G = dim t) = 1 + dim VI .

If rank G > rank K, we conclude that 4I is a Cartan subalgebra of f and therefore 1 is fundamental in g. Then by Theorem 24.1, pces is actually a polynomial function on He and so the statement of Theorem 25.1 is true in this case.

So it remains to consider the case when rank G = rank K.

29. Remarks on notation

In this section we fix some notation. Let A be a 0-stable Cartan subgroup of G. Then A = A, a AR. Fix compatible orders in a* and a* + (-1)112a*

and let Q be the set of all positive roots of (g, a). Define QR, Q, and Q, as in Section 24. Put

'A(a) = IIaeQ (1-e(a-1)) (a e A)

and define 'A, and A+ as in [2(i), ? 17]. If a e A we write a = ala2 (a, e AI, a2e AR). Put

es(a) = sign HaeQR(a.) a(log a2) (a e A' = A n G') .

Here G' is the set of all regular elements of G and QR(al) is the set of all a e QR such that $,(a1) = 1. Let

P 2 aQ a. PI, a 2jQI *

LEMMA 1. 'AI(a)A+(a) = sR(a)'A(a) exp {p(log a2)} for a e A'.

This is proved in the same way as [2(d), Lemma 12]. For f e C(G), define

'Ff(a) = 'A1(a)A+(a) f(xax-l)dx* (a e A'),

where dx* is the invariant measure on GIAR normalized as usual [2 (i), p. 116]. Put

= aCeQ Ha 9 ' = e PIT o eP'

Now suppose G is connected and acceptable [2(c), ? 18]. Then we define

A(a) = dp(a)'A(a) (a e A)

and

Ff(a) = p(al)'Ff(a) = es(a)A(a) / f(xax-l)dx* (a e A')




for f e C(G). Then

Ff(a; t) = $p(aj)'Ff(a; ti,')

LEMMA 2. conj A(a) = (- l)+[QR(al)]A(a) (a e A) where r = 2(dim g - rank g) and QR(al) is the set of all roots a e QR such that a(aj) = 1.

This is a simple consequence of Lemma 1 (cf. [2(e), p. 309]).

30. Some auxiliary lemmas

Let us now assume that rank G = rank K. Fix a Cartan subgroup B of K and let A be a 8-stable Cartan subgroup of G with dim AR = 1. Let 3 be the centralizer of a,. in g. Then 3 = a, + I where I = [3, 3] is a three-dimensional simple algebra of noncompact type. Clearly n = a, + I t f is a Cartan subalgebra of g. We say that a is related to b (or A is related to B) if 3 n t = b. It is clear that we can always choose k e K' such that Ak is related to B.

A real root of (g, a) may be regarded as a root of (I, aR). Hence (g, a) has exactly one positive real root which we denote by a. Put H' = 2 a a l-2 * Ho. This can be completed to a base (H', X', Y') for I over R such that

[H', X'] = 2X', [H', Y'] = -2Y', [X', Y'] = H' .

Moreover we can assume that 0 maps (H', X', Y') into (-H', - Y', -X'). Put y = exp (- 1)1"2(2/4) ad (X' + Y') e GC and suppose that a is related

to b. Then y-1aC = b and fi = ay-' is a root of (g, b). Put HA' = 2 1 ,81-2.Hp. Then

H= (-1)"/2(X_-Y'), b=aI + R(X'-Y') .

Put b, = a, and b2 = (-1)12RHA. Then fi may be regarded as a root of (I, b2) and therefore

IC = CHp + CX, + CX_

is the usual notation. Moreover we may assume, without loss of generality that a. = R(Xp + X_,).

LEMMA 1. Let Bp denote the kernel of $p in B. Then A, n B = Be.

Let a e A, n B. Then a centralizes X, + X-p e aR. Hence ,p(a) = 1. Con- versely if a e Bp, it is clear that a centralizes a. This shows that AI n B = Be.

COROLLARY. A, = Bp if G is connected.

Let E be the centralizer of b1 in K. Since K is connected and b1 c I, it follows that E is connected. But t n 3 = b. Hence E = B. Since A, = A nK Kc E = B, we conclude that A, = Be.



178 HARISH-CHANDRA

Put B, = exp b, (i = 1, 2). Then B1, B2 are closed subgroups of B [2(d), p. 99] and B1B2 = B0. Since b, nb,0 = {O}, A1 A B2 is a finite group.

LEMMA 2. Put 11 = exp (2r(X' - Y')). Then A, n B2 is a cyclic group with ̂ 1 as generator. Moreover 7 centralizes I.

Since B, is connected, we may assume for our proof that G is connected. Then by the above corollary, A, = Be. Put

b, = exp (-(-1)1/2tHp) (t e R)

Then b, e Bp if and only if t e Z7r. Since 7 = bar this proves our first assertion. The second follows from the fact that 7 e Be.

Put

p = 2<p, a> G Z

where p is half the sum of all positive roots of (g, a). Let P = MARN be a psgp in 9P(AR).

LEMMA 3. 7 centralizes m + a1,. Moreover if G is connected and acceptable,

e = (-l)Pa

We know [2(i), p. 119] that

Ad (^) = exp (Z(_1)1/2 adH')

and therefore 7 centralizes m + a,,. Now suppose G is connected and acceptable. Let j: G -. G, be an accept-

able complexification of G. Then, by the same argument,

j(7) = exp {r(-1)'2H11

and therefore

= exp {w(- 1)1"2p(H')} = (_ 1)Pa.

31. Recapitulation of some earlier results

In this section we recapitulate some results of [2(d)]. So suppose that rank G = rank K and G is connected and acceptable. Fix B as in Section 30.

Let A be a 0-stable Cartan subgroup of G. We denote by Z(A) the subgroup of all a e A, with the following property. If j: G --+ G. is any complexification of G [2(c), ? 18], then j(a) e exp (-1)"2QR. Then Z(A) is a finite subgroup of A, [2(d), p. 99]. Put AR = Z(A) AR.

We now use the notation of Section 29. Put

VA = V 9 tO = Ia eQ0 Ha 9 AA=A




where Q0 is the complement of QR in Q. For f e C(G) define

Fft (a) = SR(a)A&A(a) | f (xax-l)dx*

GIAR

as before and put

SA(a) = e(a)(-1)[QR(a1)J (a e A')

It is known that ?oFfA extends (uniquely) to a continuous function on A. Let OA denote the restriction of rvFfA on AR. Then f H OA is a continuous mapping of C(G) into C(AR) [2(f), ? 20].

Put

GA = UxeG xA'x'

Then GA is an open subset of G and

f(x)dx = [W(G/A)]-l I I&A(a) 2 da f(xax-l)dx* GA A G/AR

for f e C(G) [2(f), p. 110]. Define L, Ei and TP (\ e L) as in [2(d), ? 8]. Then (see [2(d), ? 15])

v(X)%(f) = p.v. D-lPVGfdx

= F{A1 [ W(G/A)]-' Pr(a)SA(a)FfA(a; 11A)da

for X e L and f e C-(G). Here {A} denotes a complete set of 0-stable Cartan subgroups A of G no two of which are conjugate. We assume that B is contained in {A}.

LEMMA 1. Fix A and fe C-o(G). Then

ie lA da2 | Pr(a)6A(a)FfA(a; VA)dal < oa

Here da1 is the normalized Haar measure on the compact group A, and da2 = dAR so that da = dalda2. This lemma is an immediate consequence of [2(d), Lem. 15].

Put

SA(f) = RfeL [ W(G/A)]' |A P(a)1SA(a)FfA(a; tZA)da

for fe CQ(G). LEMMA 2. There exists a tempered distribution TA on AR such that

SA(f) = TA(4A)

for all f e CQ(G).



180 HARISH-CHANDRA

This follows from [2 (d), ? 14] (see also [2 (f), ? 20]).

COROLLARY 1. SA is a tempered distribution on G. Moreover if f e e(G) and Ff = 0 then SA(f) = 0.

This is obvious.

COROLLARY 2. e L t-k(X)0(f) = E{A} SA(f) (f e C(G)).

For f e C -(G) this is clear from the formula for tf()03,(f) given above. But since both sides are tempered distributions [2 (e), ? 29], this relation must hold for all f e e(G).

Fix f e Cc(G). Then by [2 (d), ? 15],

SB(f) = 2 .eL 5B TBFfBdidb = Ff(1; tCfB) = (-1)qGf (1)

where q and CG have the same meaning as in Section 23. Therefore if {A}+ denotes the complement of B in {A}, we get the following result.

LEMMA 3. ( 1) CGf(M) = 1:2eL T(X)E2(f) - S{A +SA(f) for al f e (G).

32. Statement of Theorem 32.1

We now drop all restrictions on G except that rank G = rank K. Fix a Cartan subgroup B of K and let A be a 0-stable Cartan subgroup of G with dim AR = 1. Let a be the unique positive root of (g, a). Define (H', X', Y') and put

Yr = exp c(X'-_ Ye) , 2<p, ap> exp 7r(X I P I1a 12

as in Section 30. Let C denote the cyclic group generated by -i and F the order of Y.

For f e (G), put

(f) F ;/r ;/ce C 5 $)r(t)Xr(c')'FfA(cht; zt'r)dtA

Here ht = exp tH' (t e R) and r runs over all integers such that

0<r<2F, r-paFmod2.

Xr is the character of C given by

XrO') = (- 1)Pa exp ((_ 1)1/2 r2L)

and

exp {(i - t}+ exp{ ( -

exp t - exp (-t)




It is easy to see that for a fixed c e C, t I 'FfA(cht: zTA)

is a Schwartz function on R which changes sign under the reflexion t -t. Therefore IA is a tempered distribution on G.

We recall that if H is a topological group, then HI denotes the connected component of 1 in H. Let P = MARN be a psgp of G in i(AR). We intend to prove the following theorem.

THEOREM 1. Let f be a function in C(G) satisfying the following two conditions:

(1) (E),, f) = o for allwe G42(G). (2) If F is a 0-stable Cartan subgroup of G with dim FR > 1, then

'Ff = 0 unless F is conjugate to A in G. Then

(-1)qCGf(1) = [G: Go *A] [ W(M/BA) I a IA(f2

The proof proceeds in three steps. First we show that

SA= [W(GIB)] 2 IA [W(M/AI)] I a I

provided G is connected and acceptable. In the second step we drop accept- ability and in the third step connectedness.

33. First step in the proof

So let us assume that G is connected and acceptable and keep to the notation of Sections 30, 31 and 32.

LEMMA 1. Define A and B as in Section 32. Then

SA-- W(G )] 2 IA.

[W(MI/A1)J aI The proof requires some preparation. Since A will be kept fixed, we

drop it from the notation whenever it is convenient to do so. Also we may assume that A is related to B.

LEMMA 2. Fix f e Cc (G) and X e L. Then

[W(GIA)]-l T2(a)eS(a)Ff(a; V)da

= - [ W(M/AI)]-J I I H' 5 Idt P 2(alht)Ff (alht; t,)da. 0 AI Since a is a real root of (g, a), the corresponding Weyl reflexion sa lies

in W(G/A). Hence



182 HARISH-CHANDRA

w(GIAR) = W(G/A)/ W(M/Aj)

is of order 2. Also a is the only positive real root and $a(t) = 1 by Lemma 30.2. Therefore

SA(alht) = -sign t .

Moreover dht = 1 H' II dt and the functions TP(a) and SA(a)Ff(a; V) are invariant under Sa. Our assertion is an immediate consequence of these facts.

Now

Pt(a) = E8eW(G/B) $82(aj) exp {- I (s)Y(log a2) I} where y has the same meaning as in Section 30. (This is proved in the same way as [2(d), Lem. 52].) Hence

T(ajht) = ,eW(GI/B) $2(aj) exp {-t I s,(HA) I}

for t > 0. Since L is stable under W(G/B), we conclude from [2(d), Lem. 15] that

2 e L 5 dt 5 'P(ajht)Ft(ajht; t)da,

- [W(G/B)] e: L 5 exp I X(Ha) I t}dt $2(a,)Ff(aht; ?f)da, .

On the other hand II H' = 2 a '1. Therefore we get the following result.

LEMMA 3. Let f e Cc0(G). Then

SA(f) = [W(G/B)] 2 JA(f)

where

JA(f) = Pate 5 exp {-I (Ha) I t}dt | 2(a,)Ff(alht; tr)da, .

We shall now transform the above expression for JA(f). Let L1 be the

lattice consisting of all X e L such that <x, f> = 0. Fix pu e L, t e R and put

g(al) = [C]-' ceC $P(caj)Ff(cajht; t) (a, e AI)

where C = A, n B2 is the cyclic group generated by -i. Then g is a continuous function on A, which may also be regarded as a function on AI/C. On the other hand L1 may be identified with the character group of B/B2 = AI/C.

Since the Fourier series of g converges absolutely [2(d), p. 100], we conclude that

g(1) = L A $2(aj)g(aj)daa

Thus we have obtained the following result.




LEMMA 4. Fix p e L and f e CQoo(G). Then

2e LI XA Ap+2(al)Ff(alht; lf)da, = [C]-' ,ceC $P(c)Ff(ch,; iV)

for t eR.

Let LO be the additive subgroup of L generated by L, and fi. Then

nfeLo ker $2 = B2 nfB = B2 n AI = C. Hence L/LO may be regarded as the character group of C. Therefore both C and L/LO are cyclic groups of order F.

Put

bt= exp {-(-1)2tH} (t e R)

We have seen in Section 30 that bt e Bp = A, if and only if t e Zw. Since b, = Yi, it follows that bt = 1 if and only if t e ZwF. Hence there exists a character X of B2 such that

X(bt) = exp {-(-1)"/22t/F} .

Since X extends to a character of B, we can choose p e L such that p(Hp) = 2/F. Clearly rfp (0 r < F) are incongruent mod LO. Since [L/LO] = F, every element in L can be written uniquely in the form

rp +X+kt3 (0<r<FkeZ,%eL1).

Hence we conclude from Lemma 4 and [2(d), Lem. 15] that

JA(f)= F' 1Or<F Ekez 5f exp (-2 jr + ki t)

X C erc-+kQC)Ff (Cht; -v)dt But

=k $p(Y) = exp {-(-1)"22rr/F} Ff(cht; z) = p(c) Ff(cht; v') (c e C)

and 1),(O) = (- 1Ya from Lemma 30.3. Moreover since yF =1

PaF 0- mod 2.

Define the character X2T of C by

X2T(r = (-1)Pa exp {(-1)1"22rr/F} (r e Z). Then we get

JA(f) = F' ;OTr<F EkeZ 5m exp (-2 |r + kJ t)

x :cGC X2r(C-)'Ff(cht; t')dt But



184 HARISH-CHANDRA

Sk6Zexp (-2 ? + k| t) = Ek20exp{-2(j ? + )t}

+ k2exP{-2(Ik -_ t

= )2r(t)

for t > 0 and 0 ? r < F. On the other hand for a fixed c e C,

t l ) Ffl(cht; -v') (t e R)

is an odd function of t lying in C-(R) [2(d), Cor. 3, p. 103]. Hence 00

5D2r(t) I Ff(cht; A') I dt < cc

and therefore we can interchange the summation over k with integration over t. This shows that

JA(f) = F-' 1:0r<F 5 $2r(t) IceC X2r(c')'Ff(Cht; t')dt

= IAMf -

Since both SA and IA are tempered distributions on G, Lemma 1 now follows from Lemma 3.

Now suppose Z is a finite subgroup lying in the center of G and Go = G/Z. Put C1 = C n Z and let Yio denote the image of -Y in Go. Similarly let C0 denote the image of C in Go and Fo the order of Ero. Then Fo is the least positive integer such that -Yo e C1 and it is clear that yFo is a generator of C1.

Now suppose f e C(G) is actually a function on G/Z. Then 'Ff(cht; Vr') depends only on c mod C1. Fix r (0 < r < F). Then

EceC1 X2r(C) = 0

unless X2r = 1 on C1, i.e., unless

ro PaFo mod 2

where ro = 2rF0/F. Note that 0 < ro < 2FO and

(-) = (- 1)Pa exp {(_-1)1'2r0/F0

if this condition holds. Hence

F-Or<F ( $2(t) E ceC X2r(c')'Ff (cht; v') =FO-' E, (D?(t) ECO eco X?(co-')Ff (coht; -v )

Here s runs over all integers such that

0 < s < Fo0, sPaFomod2

and XJ? is the character of C0 given by




XZO(C) = (- 1)-'a exp {( - 1)"'2si/Fo} -

Moreover $D? is obtained by replacing (r, F) by (s, Fo) in the definition of $, and 'Ff is to be regarded as a function on A/Z on the right hand side.

34. Second step

Let us now drop the condition that G be acceptable. Fix f e e(G). Then the discussion at the end of Section 33 shows that the value of IA(f ) remains unchanged if we go over to an acceptable, finite covering group G of G and regard f as a function on G.

Now for any 0-stable Cartan subgroup A of G with dim AR = 1, define

SAW= [W[W(G/B)] 2 IA(f) (f E eG)) E

in the notation of Section 32. Let {A}, denote a complete set of such Cartan subgroups of G, no two of them being conjugate.

THEOREM 1. Let f be an element in (G) such that: (1) e,(f) = o for all woee2(G). (2 ) 'Ff = O for every 0-stable Cartan subgroup F of G with dim FR > 1.

Then

(-i)qCGf(l) = SLAX1 SA(f)

Let G be an acceptable finite covering group of G. Then if we regard f as a function on G, it continues to satisfy all the conditions of the hypothesis. Moreover C5 = CG. Hence it is enough to consider the case when G is acceptable. But then our assertion follows immediately from Lemmas 31.3, 33.1 and [2(f), Thm. 16].

35. Third step

We now come to the proof of Theorem 32.1. Let a, = a, a2, ..., a, be a complete set of 8-stable Cartan subalgebras of g such that:

(1) they are all conjugate under G, (2) no two of them are conjugate under Go, (3) ai is related to b (see ? 30) for 1 < i < p.

Let P = MARN be a psgp of G in 9P(AR). Choose yi e K such that Ad (yi)a = ai (1 ? i < p, y1 = 1). Let Ai denote the Cartan subgroup of G corresponding to ai. Put

Mi= yMiyi-, Mo=Mtn Go, Aio=AinGO.

Then Ao is a Cartan subgroup of Go and

A- = yiAovy-, Mo = yM0y71



186 HARISH-CHANDRA

where Ao = A n G?, Mo = M n G?. Let ai be the unique positive real root of (g, aQ). Then

[W(Mo/AjoI)] = [W(Mo/AoI)] I ai I = I a I (1 < i < p) . For f e C(G), put

f (x) = f(kxk-l')dk (x e G) R

and let fo denote the restriction of f on Go. Then it is obvious that

IA) = IAi(f) = IAO(fO) (1 < i < p).

Now fix f as in Theorem 32.1. Then by [2(i), Lem. 27.6], fo satisfies the hypothesis of Theorem 34.1. Since fo(l) = f(l) and CGO = CO, we conclude that

(-1) CGf(M) = -Eji.P SAiO(fo)

But

SAto(fO) = - W (GM/B 0) 2 _IA o(fo)

- [W(G?/B0)] 2 IAW [W (MO/AOI) I a A

Hence

( 1) CGf(1) P [W(GM/IB)] 2 IA2

[W(Mo/A1) aA

Let A be the normalizer of a in G. Then it is obvious that

p = [G: AGO] = [G:G0

But

[A: A n G?] = [AI: AO] - [W(G/A)] [A Ao [A:AnGI -[A ni GO: A0] [W(GO/A0)J[: O

Hence

p = [G: GO-A] [W(G0/AO)] [W(G/A)J

Since

[w(G/A)] = [tv(G0/Ao)] = 2,

we conclude that

(-)qcGf (1) = [G: Go *A] [W(G0/B0)] 2 I2 A [TW(M/A c)I a IAff

This completes the proof of Theorem 32.1.




36. The formula for ,(wO: V)

We keep to the notation of Section 32. Fix fi e Cc(@) and put

f = p= |((o: v)fl(v)E(P: *: v)dv

in the notation of Lemma 24.1.

LEMMA 1. f satisfies the conditions of Theorem 32.1.

Fix w e 62(G) and put

T(fl) = (Ew), 0qi) (fi e CCo(M)).

Then T is a distribution on H (with values in V). It is easy to verify that Supp T is empty (see the proof of Lemma 26.1) and therefore T = 0. Our assertion now follows from Lemma 24.3.

Let Q be the set of all positive roots of (g, a). Then QR = {a} and r =

a?ILu1+ in the notation of Section 24. For a* e A* define the distribution O.. on M corresponding to [2(i), p. 177]. Since L = 0C(M, zM) has finite dimen- sion, it is clear that there exists a finite subset F* of A*' such that

(Oa*p * ) = O

for a* e A* and *o e L unless a* e F,. Put

f(a*: 2)) = Fo(6a-w f(P)) (a* e A*, v e ;)

where F0 has the same meaning as in Lemma 21.5. It follows from Lemma 20.2 that f'P) e L. Hence f(a*: v) = 0 unless a* e F,.

LEMMA 2.

'Ff(a) = Late Xi<a*, al> |f(a*: v) exp {(-1)"2v(log a2)}dv for a e A.

Here the argument is the same as in the proof of Lemma 24.1. Fix a2 e AR and put

g(m) = f(ma2)((m e M) . Then g e L and therefore [2(i), Lem. 18.2]

'Ff(aja2) = IF AI(a,) = Ea*eA I FO(6a*, g)<a*, a,>

But

(a g) = (6ar, f.(P)) exp {(- 1)",2V(log a2)}dv

and so our assertion follows. For a* e Aj*, define x(a*) e (-1)1"2a* as in [2(i), ? 27] and put

V(a*, 2) = V-(X + (-1)1/21), srI(a*) = VoI(X), V+(a*, 2) = Vr+(x + (-1)1/2')

(2 e )



188 HARISH-CHANDRA

with X = %(a*).

COROLLARY.

'Ff(a; ur') = En~eAi <a*, al> lr(a*, i)f^(a*: v) exp {(-1)"22)(loga2)}d2.

This is obvious. Let C * be the character group of C (see g 32). Fix a* e AI* and X G C*.

Since Aa is an irreducible representation of AI with character a*, we denote by [a*: X] the multiplicity of X in the restriction of Oa* to C. It is clear that

[CJ' CeC <a*, c>X(cl) = [a*: X]

We recall that F is the order of C. Let S denote the set of all integers r such that

O < r < 2F, r _ paF mod 2.

Then we have defined in Section 32 a bijective mapping r i X, of S onto C *. Put (DX, = B,.. Then

IA( f ) = X e C* [C ]- rc e c Dx(t)x(c-)'Ff(cht; t')dt

Put

= 2<2), a>= a 2

and recall that f(a*: 2) = 0 unless a* e F,. Moreover for a fixed c e C, 'Ff(cht; Vr') is an odd function of t. Therefore we conclude from the corollary of Lemma 2 that

IA(f) = Evaes; Exec. [a*: X] dt ( 1)l/ V(a*: 1)f(a*: v)$Dx(t) sin vj t d.

LEMMA 3. 00

\ dt (-1)l12t z(a*: v)f (a*: v))$(t) sin v t* dv

=( _)v+ 1 CX | t(a*) | o(X i) t1+(a*, v) If(a*: v)dv,

where

P = 1 [QC] ' a o- 2<2), a>

and

,po(X: 2) = any sinh wrvcosh any( - )D (X(7) + X01 2




Since

(-1)l/2t(a*, V) =-I a 2 I(a*)V+(a* V) 2

and

T+(a*, V) = (-1)P +(a*, v) this follows immediately from the corollary of Lemma 42.1.

Fix a real number t > -1 and put z sinh z

cosh z + t Then u is a meromorphic function on the complex plane which is holomorphic on the real axis. For a* e A*, define

fo(a*:v) = d(a*)-ltr Jw asinh w | cosh rv, - 0l) {oa*(Y) + Va*( I)}I

where ma is, as before, an irreducible representation of A, with character a*. Then ,u0(a*) is a meromorphic function on H, which is holomorphic on A.

COROLLARY.

IA(f) = (_-1)P+' fAae A* d(a*)V0(a*) p,(a*: 2) I V+(a*, v) I f(a*: v)dv . 4 3,

This follows immediately from what we have seen above if we observe that

d(a*)fa(a*: 2) = EXec* [a*: X]fo(X: v) )

Define q, and qM as before (see the proof of Theorem 24.1). LEMMA 3. p + 1 = qG -a.

For P = MARN and dim N = [QR] + [Q,] = 1 + 2p. Moreover

dim G/K = dim P/Kn P = dim M/KM + dim AR + dim N. Therefore

2qG = 2qm + 2 + 2p

and this implies our assertion. Fix a* e A*' and let w be the element in 62(M) which corresponds to a*

under Theorem 23.1. Put

(#o:) =) [W(MI/AI)]-l seW(M/AJ) u0(sa*: V) Tf+(&o: 2) = z+(a*, v) .

(We recall that, by Lemma 24.4, zr+ is invariant under W(M/AI).) Now



190 HARISH-CHANDRA

d(w) = jMI,(a*) I [ W(MIAI)]d(a*) from the corollary of Lemma 23.1. Hence

d(a*)V1(a*)f (a*: -) (-1) JCMV W(M/A1)]'d((o)f (co: v) in the notation of Theorem 22.1. Therefore we conclude from Lemma 3 and the corollary of Lemma 2 that

IA(f) = (-)GCM Id() 2)a ) I [ +(w: v) I f (a): )d2) 4 w 2 () d ao

Let sa be the Weyl reflexion corresponding to the root a of (g, a). Then sa = Ad((k) on a where

k = exp ( (X' - Y'))

in the notation of Section 30. Since -i = k2, it is easy to verify that

<saa*, c> = <a*, c> (a* e A*, c e C) and therefore

/eo(saa*: ;)) = fe0(a*: 2)) = [e0(a*:-I) .

This implies that

Pu(W: V) = Pu(sw: V) = P,0(w: sV) (a)e 62(M), s e ') Hence we conclude from Theorem 22.1 that

IA(f) = (-1)cG a4 [o: r(a(o0)]d((o0) 5 p[(w?: 2) I +(wo0: 2) I f (wo: v)dv 4a

= (- 1)wC I2 YI(G/AR)Fog(1) 5 u0(w90: v) I u+(w0: 2) I 8(v)dv .

(We observe that c(G/AR) = 1 since dim AR = 1.) On the other hand

f(1) = Fo/(1) | ((wo: ))/3(v))dv).

Therefore since 8 is an arbitrary element in Cc(@), we conclude from Theorem 32.1 (see also the end of ? 24) that

Xf(o00 o) - CM [G: G0 A][ W(G 0/B 0)] I a I jY(GIAR). [o(wo: 2) I u+(w0: 2)) I

CG W(M/A1)J

The assertion of Theorem 25.1 can now be verified immediately from the definition of ,eo((0) if we take into account Lemma 4 below.

For s > 0 let D, denote the region consisting of all z e C with I Im z I < s. Let U denote the set of all meromorphic functions u on the complex plane




satisfying the following conditions. There exist numbers s>0, s > 0 (depend- ing on u) such that:

(1) u is holomorphic on Do, (2) suPz.D I u(z) I (1 + I Z l) <

For a real number t > -1, put

u,(z) = z sinh z cosh z + t

LEMMA 4. U is a ring which contains all polynomials as well as the functions u, (t > -1).

This is obvious.

Part IV. Commuting algebras of induced representations

37. A lemma on surjectivity

We return to the notation of Section 14. Fix a psgp P = MAN in 9P(A), a) e 62(M) and v. e j = a*. Let ro = r(, v,) be the subgroup of all s e r such that sw = w and sv, = 'v. Put L = E'((M, zM) and define L(w) as usual [2(j), ? 18]. Then L(w) is stable under 0c,1,(s: v,) for s e o (Lemmas 17.3 and 17.6). Let X(s) denote the restriction of 0cpl,(s: v,) on L(w). Then we conclude from the corollary of Lemma 17.2 that X is a representation of ro on L(w).

Let Eo denote the projection of C(M, zM) on L(w) given by (Lemma 23.2),

EO~g: mJ | dOK), r(m>)g) (en e M) for g e e'(M, EM)

LEMMA 1. Let L(w, v) denote the space of all * e L(w) such that

0cPA1(s: V0)* = *

for all s e o(w, 4o). Then f v- EWof (P' is a surjective mapping of e(G, r) onto L(w, 4o)

Fix * e L(w) and f e e(G, z). Then if s e bo,

(*, 0cPIP(s: vo)EfJ(oP) = (Ocpip(s-1: vo)* f(P))

from the results of Section 17. But the right hand side

= (E(P: Ocplp(s-1: v0): o)p, f) = (E(P: *: v4o) f) from Lemma 17.2,

=(*, f.'oP') = (*, E~f.'oP') .

This being true for all g e L(w), we conclude that EQf'op e L(w, 4o). Conversely fix * e L(w, so) and choose an open neighborhood U of vo in

such that UfnsU= 0 for s e o(w) unless s e t0. Put



192 HARISH-CHANDRA

f = p(a): v)a((v)E(P: *: v)dv

where a e Cc ( U). Then f e C(G, r) and

E, fP) = 702 E )a(s-lv)0cplp(s: s'ulp) (v e )

from Theorem 20.1. Therefore

Eef.'oP- = 7c2[Ioj]a(vo)*

and if we choose a so that 7c2[ 0oja(v0) = 1, we get EQ f,'P) = A. This proves the lemma.

COROLLARY. The mapping

f A- - ) E, f.Pf) from C-(G, r) to L(w, v.) is surjective.

Since Cc (G, r) is dense in C(G, r) and dim L(w, o) < dim L < ao, this is an immediate consequence of the above lemma.

Fix a finite subset F of &(K) and take r = rF. Let us now use the notation of Lemma 9.2.

LEMMA 2. Let WrF denote the representation of SF on "IDF defined by w = XpspO Then under the bijection

T | - AT

of End ?!F onto L(@), 7CF(2F) corresponds to L(wo, O).

Let a e (2F)f and T e End 'F. Then

(E,,a(P)) *AT = aP * AT= #TrF(v)

from Lemma 9.2. Put T = I where I is the identity mapping of ?'F. Then we conclude from Lemma 9.1 that

Eta(P- = d(a4 /r. a)

Our assertion now follows from the corollary of Lemma 1.

38. The centralizer of 7F(2F)

We keep to the notation of Lemma 37.2. Fix s e >o(@) and choose a representative y in K for s. Recall that a e so and U is the representation space of a. Since so = w, we can choose a unitary transformation Ly in U such that

or(y-'my) = L;'or(m)Ly (me M) .

We define a unitary operator S on ! as follows. If h e ID, then

(Sh)(k) = Lyh(y-'k) (k e K) .




It is clear that S commutes with w(k) (k e K) and therefore maps 'QF into itself.

LEMMA 1. Let T e End 'F. Then

SI/T = ISTS-1

Put T' = STS-1. We shall compute its kernel 'CT' (Lemma 6.1). Let he,,. Then

(T'h)(k2) = Ly(TS-'h)(y-'k2)

= LYcT(Y-'k2: k,)(S-'h)(kl')dk,

= 5 LYIc(Y-lk2: k,)L-'h(ykT')dk,

= Lcr(Y-'k2: kly)L-'h(kT')dk,.

This implies that

tcT,(k2: k1) = LYlcT(Y-1k2: kly)L-' (ki, k2 e K)

and therefore

IT'(kl: m: k2) = tr {LrcT(Y-'k2: kly)Lj 'a(m)} = tr PrT(Y-1k2: kly)o(y-'my)} = rT(kly: y-1my: y'-k2) (m e M).

Hence

IT'(m) = Z(Y)IT(Y'1MY)Z(Y-1)

and this proves that IT' = S/T.

Recall that L(w) is an algebra under convolution (S 9).

COROLLARY. Fix s e ro(w), t e 9 and P1, P2e 9P(A). Then 0cp21 p(s: V)

defines an automorphism of the algebra L(w).

It follows from [2(j), Lem. 18.1] that

0CP21 PI(s: V) - s~cP,,jP,(1: V)

where P3 = P28-1. Hence it would be enough to consider the case s = 1. If v e la', our assertion follows from Lemmas 9.1 and 18.1. The rest is obvious from Lemma 17.6.

We return to the notation of Section 37.

LEMMA 2. Fix o e -. Then for every s e W(w, vo), we can choose t, e GL(F) such that:

(1) 0CPIP(S: VO)T = I/8Tr,- 1for all T e End Q,

(2 ) 8*7 -= 1.



194 HARISH-CHANDRA

Here "'* denotes the adjoint of i. Fix s e >o(a, v0) = >o. Then 0c,1,(s: v) defines an automorphism of L(@)

which, by Lemma 9.1, corresponds to an automorphism of the simple algebra End F Our assertion now follows immediately if we observe that 0c,1,(s: v0) is unitary.

COROLLARY. The mapping s 1-- Y8 defines a projective representation of Wb(w, 40

This is obvious. Let r be the subspace of End !F spanned by ". (s e to) Then r is a sub-

algebra of End QF and r* = r. THEOREM 1. r is exactly the centralizer of WF(2F) in End F Let E be the centralizer of r in End F Fix T e End !F. Then T e X if

and only if Y8T17' = T for all s e took But in view of Lemmas 2 and 37.2, this is equivalent to the condition that T e wF(2F). Hence Z = wF(2F). But since r = r*, we know that r is the centralizer of Z in End !F. This proves the theorem.

Let x(s) (s e to) denote the restriction of 'c,1,(s: v0) on L(@). We have seen in Section 37 that X is a representation of tv, Let t'0(F) denote the kernel of X. It is obvious that dim r ? [tv0: tv0(F)].

We recall that w = is a unitary representation of G on ,. Let Z

denote the algebra of all bounded operators on Q which commute with w(x) for all x e G. Z is called the commuting algebra of w and dimZ the self intertwining number of w.

LEMMA 3. dimZ _? maxF [tv,: Lo,(F)]. Moreover w is irreducible if and only if '0 so (F) for all F.

For any F, let ZF denote the centralizer of wrF(2F) in End SF If T e X,

then T commutes with wr(k) (k e K) and therefore S'!F is stable under T. Let TF denote the restriction of T on SDF. Then TF G ZF. Let Z(F) denote the subspace of all T G Z such that TF = 0. Then

dim Z/Z(F) < dim ZF < [ 0: a'0(F)J

from Theorem 1. Note that Z(F2) c Z(F1) if F1 c F2. Moreover UF S'!F is dense in ! and therefore

nFlF(F) = 0} . This implies that

dim Z = maxF dim Z/Z(F) < maXF [rol: to(F)] .

Now suppose w is irreducible. Then WF is irreducible and therefore, by




Theorem 1, F contains only scalar operators. Hence we conclude from Lemma 2 that Wo0(F) = W,

LEMMA 4. Fix s e ro such that sv'0 = v. Then

0cP21P1(s: VO) = OcP21P(l: VO)0cPIP(s: Vo)OCPlP1(l: V)

for P1. P2 e 9P(A).

This follows from the corollary of Lemma 17.2.

COROLLARY. wo(F) is independent of P e CP(A).

This is clear from Lemma 4 if we observe that

oCQjp(l: Vo)o0pQ(l: VO) = 1

for Q e C(A).

39. Closer study of a special case

We return to the notation of the beginning of Section 37.

LEMMA 1. Suppose dim A = 1 and 4w((: vo) = 0. Then o = 0, [u] = 2, s0) = w and 0cpip(s: 0) = 1 on L(@) for s e tv.

Since p(w: v) > O for v e!A', it is clear that prk G = 0 and vo = 0. Let a be the unique simple root of (P, A). Put H = 2 1 a I-2*Ha and identify g with R under the mapping v v ,(H). Define a, = exp tH (t e R) and regard t as a coordinate on A. Fix * 0 in L(@). Since 4(G): 0) = 0, we conclude from Lemma 17.1 that cp1p(l: v)+ has a pole at v = 0. Let k 2 1 be the order of this pole. Then we have the power-series expansion

)kCpIp(l: 2)) = '0 + V*1 + + ..

around v = 0. Here *o, 1, . * * are in L(@) and '0 # 0. Put f(v) = E(P: *: 4). If [ti] = 1, we conclude that

vkfp() = v'CpIp(l: v)+ exp {( - 1)1"2it} V 0).

Making v tend to zero, we deduce from [2(j), Lem. 10.7] that 'r = 0. This contradiction proves that [t] = 2. Let tD = (1, 8). Then we conclude again from Lemma 17.1 that

, CpIp(s: V)4 = Hoi +

')# +*

where , , *. are in L(sa). Hence

Vkfp(V) = VkepIp(l: P)+ exp {(- 1)1"2Vt} ?+ ,kcpp(s: V)+ exp {-(-1)1'22t} ('*)o + +) ? {(*1 ) + (-1)1/2(*o -)t}_ mod v)2

Since k ? 1, we conclude from [2(j), Lem. 10.71 that '0 ? 'r = 0 and this implies that 8a) = w. We claim that k = 1. For otherwise we would have



196 HARISH-CHANDRA

r + +/ ? (1)1/2(o _ - I)t = 0

This being true for all t, we could conclude that *o - = 0. This would imply that -0 = *' = 0, giving a contradiction. Hence k = 1 and

fp (v) _ 2( -1)1"2Vt'o + {(*r + +&) mod V2 .

Therefore

fp(O) = 2( -1)1"2t1o + i1 + -1 . Since '0 ? 0, this shows that fp(O) ? 0. This proves that the mapping

F-- E(P: A: 0) is injective on L(@). Let C denote the restriction of 0cplp(s: 0) on L((o). Since 82 = 1, it is clear

that C2 = 1. We claim that C = 1. For otherwise we could choose 0 ? 0 in L(ao) such that C* = - . Then

E(P: *: 0) = E(P: CA: 0) = -E(P: *: 0) from Lemma 17.2. Hence E(P: i: 0) = 0, contradicting the result obtained above. This completes the proof of the lemma.

COROLLARY. Under the above conditions p(co: LJ) has a zero of order 2 at v = 0.

This is an immediate consequence of Lemma 17.1 and the fact that k = 1.

40. A bound for the intertwining number

Let us now return to the general case of Section 37. Let I denote the set of all reduced roots of (g, a) and ?o. the subset of those a e I for which pa()(O: VO) = 0.

LEMMA 1. ?o. is stable under b0 = IP((, vo).

Let a e I and v e !. Then (see ? 13) pta(ao: L) depends only on <v, a> and per((o: v) > 0 unless <v, a> = 0. Now fix a e ?o. and s e wo. Then we have to verify that sa e ?o. Choose P e 9I(A) and let a1, a2, *. . , a, be all the distinct elements in ?(P). We may assume that a = a1. Let j be the unique index such that sa = ?ap. We have to show that aj e .

Fix v e ! such that

<a, v> = ?, <a,, v> # O (2 <-! i <fr) . Then

pa(o: i) = pa(O: 0) = pa(O: Vo) = 0 .

Hence

p(a: i1) = j [(ai((O V) = 0

But so = o and so we conclude from the corollary of Lemma 17.1 that




0 = p(c: i)= (so: si) = p(c: si). This implies that

fPa(a): SL1) = 0 for some i. But it is clear that

<sv, a,> 0

unless i = j. Hence

0 = 0aj(6Qo: s8) = Pa((O: 0) = paj(o: iP)

This proves that aj e Coo For any a e ?, let Sa denote the reflexion (in a) in the hyperplane a = 0

so that

saH = H-22a(H) Ia-2.Ha (Hea). LEMMA 2. Suppose a e 2o Then sa e o.

Define Ma, Aa as in Lemma 2.3 and apply Lemma 39.1 to (Ma, Aa) in place of (G, A). Since Pa(C): 1o) = 0, we conclude that <v4, a> = 0, sa e t, and SaG) = (0.

Let tmo. be the subgroup of W,, generated by sa for all a e 2o. Then it follows from Lemma 1 that Wtx. is a normal subgroup of W,. Define ti0(F) as in Section 38.

LEMMA 3. rOO c wDo(F).

Fix a e 1oo and choose P e CI(A) such that a is a simple root of (P, A). In view of the corollary of Lemma 38.4, it is sufficient to verify that 0cPip(sa: Lo) = 1 on L(o). Fix * e L(o) and put *P = Ma n P. Since <v4, a> = 0, we conclude from Lemmas 17.4 and 39.1 that

0cPIp(5a: 1)0+* = 0c-pI p(sa: 0)* =

This proves the lemma.

COROLLARY. Fix Pe Ci(A). Then the self intertwining number of

7Cp,',,O cannot exceed [I0: U0,].

This is obvious from Lemma 38.3. It follows from [3] and [4] that when G is linear this intertwining numbers is equal to [IDO IoD].

41. Irreducibility of the fundamental series As an application of the above theory, we shall prove a result which can

be regarded as a generalization of a theorem of Wallach [10, Thm. 4.1] and Zelobenko [11] on complex groups.

Let B be a 8-stable Cartan subgroup of G which is fundamental. Fix



198 HARISH-CHANDRA

P e g(BR). Then P = MBRN.

THEOREM 1. Fix 0) e 62(M) and v e =b*. Then O is irreducible.

Put A = BR. It follows from Lemma 38.3 and the corollary of Lemma 38.4 that the statement of the theorem is independent of P e 52(A). Hence we may assume that <v, a> > 0 for all a e 2(P). Let (D denote the set of simple roots a of (P, A) such that <K, a> = 0. Put (P', A') = (P, A).. Then P' =M'A'N'. Let

(*P,*A)=(M'nP,MPAA), *W = (M'/*A).

Then *w may be regarded as the subgroup consisting of those elements of t, which leave a' fixed. Let a(v) be the set of all s e n, such that sv = v. It is easy to verify that w(v) = *w (see the proof of [2(j), Lem. 22.3]). Hence

ID0 = W((@, V) C

and therefore

0C PAS: V-) = ?C*Pl*P(S: ?) (S e= SD)

from Lemma 17.4. So it would be sufficient to verify that 0c*p1p(s: 0) = 1 on L(ao) for s e w,~ Now M' n B is clearly a fundamental Cartan subgroup of M' and it remains to show that the representation w' of M' corresponding to (*Pt ao, 0) (obtained by inducing from *P to M') is irreducible. Thus we are reduced to the case v = 0.

Fix an element bV e B7' which corresponds to GO under Theorem 23.1. Put X x(b*) so that X e (-1)"2b/. Let 52 denote the centralizer of H2 in g. Since bV e B7', it is clear that (52, b) has no imaginary roots. On the other hand b is fundamental in g and therefore (g, b) has no real roots. This shows that every root of (8A, b) is complex.

Put W = W(g/b) and W(x) = W(52/b). Then W(x) is the stabilizer of x in W. Let WO be the subgroup of all elements in W which commute with the Cartan involution 8. Then

I = Wo/ W1

where W1 = W(m1/b) (ml = m + a). Note that W(x) n w1 = {1} since (62, b) has no imaginary root. Put Wo(x) = Wo n W(x) and let ID(X) be the image of WO(x) in to. Since Wo(x) n wl = {1}, the projection of WO on Wl defines an isomorphism of WO(x) on tv(x).

LEMMA 2. w((o) c I(x).

Fix s e wv(@) and choose u e WO such that s is the image of a in r = W0/ W. Since s8o = 0), it is clear that ax is conjugate to X under W1.




Hence we can choose t e W1 such that tu e Wo n w(x) = Wo(x). This proves that 8 e tG(%).

LEMMA 3. Let a be a reduced root of (j,, a). Then the reflexion sa lies in >x(X). Moreover >v(X) is generated by all such elements sa.

Since (52, b) has no imaginary roots, a = bR is maximal abelian in i2 n P. This implies the lemma.

Let a0 be a reduced root of (5d, a). Then there exists a root ,8 of (it, b) such that ,8 = a0 on a. Let a be the reduced root of (g, a) corresponding to ao. Put

(*P. *A) = (Ma n Pt Ma n A)

where Ma is defined as in Lemma 2.3. Since Be is a Cartan subgroup of K, it is clear that

*B=BnMa =*A.B

is a fundamental Cartan subgroup of Ma. Therefore by Theorem 24.1,

pa((O: 0) = Ca | II, x(H,) I where ca is a positive number and -i runs over all positive complex roots of (ma, *b). On the other hand there exists a number r>1 such that ,8 = a0=ra on a. Therefore ,8 is a root of (ma, *b). But since ,8 is a root of (5d, b), it is complex and x(H^) = 0. This shows that Pa((a: 0) = 0 and therefore ga e boo. Since sa = 8ao, we conclude from Lemma 3 that

ID(%) C byC ID(() .

Combining this with Lemma 2, we get

b()= bD0 = bte@))

and therefore 7r,,, o is irreducible from the corollary of Lemma 40.3.

42. Appendix LEMMA 1.

0 exp at + exp (-at) sin xtdt = w sinh rX -a exp t - exp (-t) cosh rx + cos wa

for a,xeR and a1I<1. Fix a. Then

exp at + exp (-at) ? 4exp{(Ia -1) ItI} exp t - exp (-t) -

for I t I > 1. Hence the integral converges and defines a continuous function of X. Hence it is enough to consider the case X > 0.



200 HARISH-CHANDRA

Fix X> 0 and put

f(z) - exp {Q( + ix)z} exp z - exp (-z)

where z is a complex variable. Then f has simple poles at z = nsii (n e Z) and

Res,=,,, f(z) = - exp {(a + iX)nwi} = exp {Q( - 1)nri -n7r}. 2 2

Now suppose z = x + iy (x, yeR) and IxI > 1, y > 0. Then

If(z) I < 2 exp {(ax - Xy) - I x I} < 2 exp {-(1 - Ia 1) xl - Xy} .

Let N be a large positive integer, T a large positive real number and s a small positive real number. Consider the rectangle in the complex plane with the four vertices - T, T, T + (N + 1)wi, - T + (N + 1)wi. Introduce a small semicircular dent of radius s around z = 0 so as to exclude the origin. Integrate f(z) along this contour and let s -+0, T -+ 00, N .-+ oo Then we get

p.v. exp {(a + iX)t} dt = ? + 7i , exp {(a - 1)nwri - nwX}{ exp t-exp (-t) 2

7ri sinh wx - i sin wca 2 cosh wx + cos wca

Therefore exp at sin7xtdt - sinhwx

exp t - exp (-t) 2 cosh 7rx + cos wra and this implies our assertion.

As usual let C(R) denote the Schwartz space of R.

COROLLARY. Let f e C(R). Then 0 expat + exp(- at)d \ xf(x)sinxtdx = 7 sinh 7 - exp t - exp (-t) - -co cosh 7X + cos wra

for -1 < a < 1. Put

g(t)= 5 xf(x) sin xtdx (teR).

Then g e Q(R) and g(O) = 0. Therefore

5 coth t I I g(t) I dt < .0

This shows that the left side in the corollary is a continuous function of a. On the other hand cos ra > -1 and




X sinh r = , coth ix X

cosh wx -1 2 Therefore

0? XsinhrX <Xcoth ?rx <c(1 + i) (xeR) cosh wx + cos ac- 2

where c is a suitable constant. Therefore the right hand side of the corollary is also a continuous function of a. Hence it is enough to verify the equality when IaI < 1.

But if I a I < 1, it is clear that the double integral on the left is absolutely convergent. Therefore the required result follows from Lemma 1 by Fubini's theorem.

THE INSTITUTE FOR ADVANCED STUDY, PRINCETON, N.J.

REFERENCES

[1 ] J. ARTHUR, Intertwining integrals for cuspidal parabolic subgroups, preprint. [21 HARISH-CHANDRA, (a) Representations of semisimple Lie groups, III, Trans. A.M.S. 76

(1954), 234-253. (b) Spherical functions on a semisimple Lie group, II, Amer. J. Math. 80 (1958), 553-613. (c) Invariant eigendistributions on a semisimple Lie group, Trans. A.M.S. 119 (1965),

457-508. (d) Two theorems on semisimple Lie groups, Ann. of Math. 83 (1966), 74-128. (e) Discrete series for semisimple Lie groups, I, Acta Math. 113 (1965), 241-318. (f) Discrete series for semisimple Lie groups, II, Acta Math. 116 (1966), 1-111. (g) Harmonic analysis on semisimple Lie groups, Bull. A.M.S. 78 (1970), 529-551. (h) On the theory of the Eisenstein integral, in Lecture Notes in Math., Vol. 266, pp.

123-149, Springer-Verlag, 1971. (i) Harmonic analysis on real reductive groups, I, J. Funct. Anal. 19 (1975), 104-204. (j) Harmonic analysis on real reductive groups II, to appear in Inv. Math.

[(3] A. KNAPP, Commutativity of intertwining operators II, Bull. A.M.S. 82 (1976), 271-273. [4] A. KNAPP and E. M. STEIN, (a) Intertwining operators for semi-simple Lie groups, Ann.

of Math. 93 (1971), 489-578. (b) Singular integrals and the principal series III, IV, Proc. Nat. Acad. Sci. USA 71

(1974), 4622-4624; 72 (1975), 2459-2461. [5] R. A. KUNZE and E. M. STEIN, Uniformly bounded representations I, II, III, Amer. J.

Math. 82 (1960), 1-62; 83 (1961), 723-786; 89 (1967), 385-442. (6] F. BRUHAT, Sur les representations induites des groupes de Lie, Bull. Soc. Math. France

84 (1956), 97-205. (7] S. G. GINDIKIN and F. I. KARPELEVIC, Plancherel measure of Riemannian symmetric spaces

of nonpositive curvature, Sov. Math. 3 (1962), 962-965. (8] G. SHIFFMAN, Integrales d'entrelacement et fonctions de Whittaker, Bull. Soc. Math. France

99 (1971), 3-72. (9] P. C. TROMBI, On Harish-Chandra's theory of the Eisenstein integral for real semisimple

Lie groups, preprint. [10] N. WALLACH, Cyclic vectors and irreducibility for principal series representations, Trans.

A.M.S. 158 (1971), 107-112. [11] D. P. ZELOBENKO, The analysis of irreducibility in the class of elementary representations

of a complex semi-simple Lie group, Math.-USSR-Izvestija 2 (1968), 105-128.

(Received February 6, 1976)



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