Invariant eigendistributions on a semisimple lie algebra

INVARIANT EIGENDISTRIBUTIONS

ON A SEMISIMPLE LIE ALGEBRA

by H A R I S H - C H A N D R A

w I. I N T R O D U C T I O N

Let g be a semisimple Lie algebra over R and T an invariant distribution on g which is an eigendistribution of all invariant differential operators on g with constant coefficients. Then the first result of this paper (Theorem i) asserts that T is a locally summable function F which is analytic on the regular set g' of g (cf. Lemma i of [3(g)]). The second result (Theorem 5) can be stated as follows. Let D be an invariant analytic

differential operator on g such that D r = o for every invariant C ~~ function f on g.

Then D S = o for any invariant distribution S on g (cf. [3(g), Lemma 3])- This will be needed in the next paper of this series, in order to lift the first result, from g to the corresponding group G (see [3(g), Theorem i]), by means of the exponential

mapping. Proof of Theorem I proceeds by induction on dim g. In w 2 we show that there

exists an analytic function F on g' such that T = F on g'. Moreover we verify that F

is locally summable on g and therefore it defines a distribution Tr. on g. Thus it remains to prove that 0 = T - - T r is actually zero. The results of w 3 enable us to reduce this to the verification of the fact that no semisimple element H of g lies in Supp 0. I f H + o , this follows easily from [3 (i), Theorem 2] and the induction hypothesis. Hence we conclude (see Corollary I of Lemma 8) that Supp 0 c~/" where Jff is the set of all nilpotent elements ofg. Let to be the Killing form ofg. Then ~(to)T=-- cT (ceC).

Since T = 0 q - T F , we get

(a(~o)--c)0 =J

where J = - - ( 0 ( c 0 ) - - c ) T F. By making use of [3 (J), Theorem 4], one proves that J = o and therefore it follows from [3 (h), Theorem 5] that 0 = o.

In w 8 we study the function F in greater detail. Let a be a Cartan subalgebra of g and rc a the product of all the positive roots of (g, a). Define g , (H)=r~a(H)F(H)

for H c a ' = a o g ' . Then we show that 0(r~")g, can be extended to a continuous function h, on a and if b is another Cartan subalgebra of g, then h , = h b on arab

(Theorem 3). These results will be used in subsequent papers for a detailed study

of the irreducible characters of a semisimple Lie group. In w io we apply Theorem 3 to give a new and simpler proof of the main result of [3 (e)].

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The rest of this paper is devoted to the proof of the second result mentioned at the beginning. It depends, in an essential way, on Theorem I and the theory of Fourier transforms for distributions. However, since the given distribution S is not assumed to be tempered, one has to construct a method of reducing the problem to the tempered case. This is done by means of Lemma 29. The last seven sections (w167 I5-2I ) are devoted to the proof of this lemma.

This work was partially supported by grants from the Sloan and the National Science Foundations.

w 2. B E H A V I O U R OF T O N T H E R E G U L A R SET

We use the terminology of [3 (h)] and [3 (i)]. Let g be a reductive Lie algebra over R and g' the set of all regular elements of g. Let I(g,) denote the subalgebra of all invariants in S(gc) (see [3 (i), w 9]). Fix a Euclidean measure dX on g.

Lemma 1. - - Let T be a distribution on an open subset ~ of g. Assume that :

I) T is locally invariant ; ~) There exists an ideal lI in I(g~) such that dim(X(g,)/lI)<oe and O (u)T =o for uElI.

Then there exists an analytic function F on f~ '= kQng' such that

T ( f ) =ffFdX for all fEC~(f~ ' ) .

Fix a point H0ef~'. It is obviously enough to show that T coincides with an analytic function around H 0. Fix a connected Lie group G with Lie algebra g and let [? and A be the centralizers of H 0 in g and G respectively. Then D is a Cartan subalgebra of g and A is the corresponding Cartan subgroup of G [3(J), Lemma 8]. Let x-+x* denote the natural projection of G on G*=G/A. As usual we define x * H = x H (xeG, H~t)). T h e n i f n = d i m g and t)'=I)r~g', the mapping c? : (x*, H) ~ x * H has rank n everywhere on G*• I)' [3(i), Lemma I5]. Therefore we can select open connected neighborhoods Go and I)0 of i and H o in G and I?' respectively such that f~0=GoIk0cf~ and q~ is univalent on G o • Then f~o is open in g and q~ defines an

analytic diffeomorphism ?0 of G O x t) 0 onto ~o. Fix a Euclidean measure dH on I) and let *T denote the distribution on Do which

corresponds to T under Lemma 17 of [3(i)]. As usual let = denote the product of all the positive roots of (g, I)). Then if ~ = ~ , we conclude from Theorem 2 of [3(i)] that O(u~),=o for u~11. Let 1I~ denote the image of 1I under the homomorphism p ~ p ~ of I(gc) into S([L). Put ~=S( tL) I I~ . Since dim(I(g~)/11)<oo, it follows from Lemma 19 of [3(i)] that dim(S(I?,)/~)<oo. Moreover it is obvious that 3(v),-----o for ve~. Therefore from the corollary of [3(b), Lemma 27], we get the following result.

Lemma 2. - - We can choose linear functions X~ and polynomial functions Pi on Do ( I ~ i ( r)

such that

~(~) = f ~ g d g (~C7(~0) )

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where

This shows that

g(H) = ~<~< pi(H)eXd m

T(f=) = (-e C2(G0 • t)0))

in the notation of [3(i), Lemma i7]. Since % is an analytic diffeomorphism, we can now define an analytic function F

on ~o as follows:

F(x*H)=g(H)r:( i ' i ) -1 (x*eG0, HeDo ).

Then if ~eC~(Go• we have

ff FdX= f : H)F(xH)dxdH= f }=rc-lgdH= T(f=).

Since the mapping ~--*f~ of C~(G0• Do) into C~~ is surjective [3(h), Theorem ,], this implies that T = F on f~0 and so Lemma I is proved.

Lemrna 3. - - The function F of Lemrna 1 is locally sumrnable on fL Let l = rank g and t an indeterminate. We denote by ~(X) (Xegc) the coefficient

o f t l in d e t ( t - - a d X). Then we know (see [3(J), Corollary 2 of Lemma 3o]) that [~ [-1/2 is locally summable on g. Since the singular set of g is of measure zero, it would be enough to show that there exists a neighborhood V (in fl) of any given point Xoef~ , such that [ ~qI*'2{F[ is bounded on V n f~'.

Fix X 0 in f~ and a positive-definite quadratic form Q on g. For C>o, let f~ be the set of all X e g such that Q ( X - - X 0 ) < z 2. Then f~cf~ if s is sufficiently small. Put

t ( x ) = ( Q ( x - - Xo)--

Then p is a polynomial function on g. Let g" be the set of all points X e g where p ( X ) # o. By a theorem of Whitney [4, Theorem 4, P- 547] g" has only a finite number of connected components. It is obvious that any connected component of f~: = ~ n g ' is also a connected component of g". Hence f~'~ has only a finite number of connected

components (*). So it would be enough to show that I nl*Z=lFI remains bounded on a connected component f~0 of ~ ' .

We now fix an element H0sf~ ~ and use the notation of the proof of Lemma I. In particular q~ is the mapping (x*, H)-+x*H of G*x I)' into g'. Let U denote the connected component of (I*, H0) in 9-*(f2~ We claim that 9(U)=f~~ Since q~ is everywhere regular, 9(U) is open in f~0. Therefore since f,0 is connected, it would be enough to show that 9(U) is closed in f~0. So let (xk, I-Ik) (k> I) be a sequence in U suchthat X~=xT~ H k converges to some point Xef~ ~ Then B(I-Ik)=~(x k I-Ik) ~ ( X ) # o .

(I) This proof was pointed out to me by A. BoreI.

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Moreover Xkef~~ Since f~ is a bounded set in g, we can conclude from Lemma 2 3

of [3(J)] that H k remains bounded. Hence by selecting a subsequence, we can arrange that H k converges to some H'eI) . But then ~ ( H ' ) = ~ ( X ) + o and therefore H'et? ' .

Hence [3(J), Lemma 8] A is the centralizer of H ' in G and therefore [3(J), Lemma 7] xk* remains within a compact subset of G*. So again by selecting a subsequence we can

assume that xk*--->x* for some x*eG*. Then (xk* , Hk) --> (x*, H') in G* • I]'. Since X~--->X, it follows that x * H ' = Xef~ ~ and therefore (x*, H') et~-t(f~~ But U, being a connected component of q~-l(fp), isclosedin q)-l(~0). Hence (x*,I-I')eU and X=x*H'eq~(U) . This proves that q~(U) is closed in f~0 and therefore q~(U)= f~0.

Now choose Go, I)0 as in the proof of Lemma I. We may assume that Go* • t?0 c U. Moreover we recall (see Lemma 2) that g is defined and analytic on [). Consider the function v:(x*, H)---> F ( x * H ) - - n ( H ) - t g ( H ) on U. It is obviously analytic and it vanishes identically on G~ • I]0. Therefore, since U is connected, v = o. This shows that

I ~(x*H)lI/~l F(x*H)I = lg(H) I

for (x*, H ) e U . However q~(U)= f~0 is contained in the bounded set f~. Therefore

if V is the projection of U on I?, it follows from [3(j) , Lemma 23] that V is bounded. Hence g is bounded on V and therefore [~}]t/~]F] is bounded on ? ( U ) = f ~ ~ This

proves Lemma 3. Corollary. - - Let p~I(gc). Then 0(p)F is also locally summable on fL Since F is analytic and T = F on f~', it is clear that 0 ( p ) T = 0(p)F on f~'.

However the distribution O(p)T obviously also satisfies all the conditions of Lemma i.

Therefore our assertion follows by applying Lemma 3 to (O(p)T, ~(p)F) in place of (T, F). (P being a locally summable function on ~2, define the distribution T~, on f~ by

T r = ffCdX ( f e C 7 (a)) .

We intend to show (under some mild extra conditions) that T = T F. Let ~a be the set of all points X e ~ such that T coincides around X with an analytic

function. Clearly f~a is open and there exists an analytic function F a on ~2 a such that T = Fa o n ~-~a" Moreover f2D f~' from Lemma i and therefore F a = F on f~'. But

then, since the singular set of g has measure zero, it is obvious that TF~ = T F. Hence

we shall write F instead of F~. We say that an element H E g is of compact type if I) ad H is semisimple and 2) the

derived algebra of the centralizer 9 of H in g is compact. (It follows from I) that 3

is reductive in g and therefore [9, 3] is semisimple.) Lemrna ~. - - Every element of f~ of compact type lies in f~a" Fix an element H 0 in f2 of compact type and let 9 denote the centralizer of H o

in g. Then it is clear that 3 satisfies the conditions of [3(i), w 2]. Define ~ and 9' as in [3(i), w 2]. Let :~ be the analytic subgroup of G corresponding to ~ and x-+x* the natural mapping of G on G* = G/E. Since 9 is reductive, E is unimodular and therefore

there exists an invariant measure dx* on G*. Select open neighborhoods G O and ~o

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of i and H o in G and 3' respectively such that 3o~~ and G O is connected. Let G o

denote the image of G o in G*. Then if G o and 3o are sufficiently small, the following

conditions hold (see [3(e), pp. 654-655]). I) There exists an analytic mapping + of G o into G such that (+ (x*) )* ---- x* (x*sG;)

and + is regular on G o . 2) The mapping qo :(x*, Z) -+t~(x*)Z of G0• into ~ is univalent.

Put g)o = q~(Go • go)- Then f~o is open in fl and ~0 is an analytic diffeomorphism of G o • 3o

onto fl o. Moreover since 31=[3, 3] is compact, 30o=~ez30 is open. Hence by

replacing 3o by 30o, we can assume that ~ = 3o. Let aT be the distribution on 3o which corresponds to T under Lemma 17 of [3(i)].

Since ]~[t/z is an analytic function on 30, a=[~l~/Zar is also a distribution on 3o. Moreover since ~2>o on 3o it follows from Theorem 2 of [3(i)] that O(u3)a = o for ueR.

Let lI, denote the image of R in I(3,) under the mapping P-+P3 of I (gJ into I(3~). Put ~ 3 = I ( 3 J R ,. Then ~(v)~-=o for ve~3 and it follows from Lemma 19 of [3(i)] that dim(I(3~)/~) < oo.

Let q be the center and 3~ the derived algebra of 3. We identify 3~ with its dual under the Killing form ~ of 31- Select a base H1, . . . , H, for q over R and put

r = H~ + . . . q- H ~ - - co 1 .

Then o n ~.

that

Hence

r and since 3~ is compact, Q=0(o~) is an elliptic differential operator Let N = d i m ( I ( 3 ~ ) / $ ). Then we can choose complex numbers ca, . . . , cs such

oN+qco~-i+...+cse~.

(V? N + q D N- ' + . . �9 +cN)~ = o.

This shows that ~ satisfies an elliptic differential equation with constant coefficients.

Therefore there exists an analytic function g on 30 such that

~,(~) = f ~gdZ ( ~ c 7 ( ~ 0 ) ) .

Since ~ is invariant under E, it follows from [3(i), Lemma I7] that g is locally invariant (with respect to ]). Therefore since E is connected and 3o=30, it follows that g

is invariant under ~. Now consider the analytic function F 0 on fl 0 defined by

Fo(~(x*, Z ) ) = l K(Z)l-1/Zg(Z) (x*eG;, Ze3o )

Then if ~eC~(G0• 3o), we have (see [3(i), w 7])

fAFodX= f o~(x : Z)Fo(xZ)dxdZ.

However if x~Go, it is clear that x=~(x*)~ where GeE. Therefore

F0(~Z) = Fo(~(x*, ~Z)) = I~(Z)[-~;~g(Z) (Z~30)

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since ~ and g are invariant under ~ This shows that

fArodX=f (x : Z)[ (Z)l-l/ g(Z)dxdZ

= f ~, ] ~ [- i/~gdZ = c; T (~,) = T (f~,)

from Lemma 17 of [3(i)]. Hence T = F o on f~o and this proves that Houri, .

w 3. SOME P R O P E R T I E S OF COMPLETELY INVARIANT SETS

We keep to the above notation. An element H~g is called semisimple if ad H is semisimple. Moreover X~g is called nilpotent if X~g 1 = [g, g] and ad Xis nilpotent. It is obvious that if X is both semisimple and nilpotent then X = o.

Lemma 5. - - Any element Y ~ g can be written uniquely in the form Y = H + X where H

is a semisimple and X a nilpotent element of g and [H, X] = o. Let r be the center of g. Since 9 = c + 91, the lemma follows from well-known

facts about semisimple Lie algebras (see Bourbaki [2, p. 79]). H and X respectively

are called the semisimple and the nilpotent components of Y. Lemma 6. - - Let 3 be a subalgebra of g which is reductive in g. An element Z of 3 is

semisimple (or nilpotent) in 5, i f and only i f the same holds in 9.

Let c~ be the center of ,3. Since ~ is reductive in g, every element of c~ is semisimple

in g. The lemma follows easily from this (see [2, p. 79])- Corollary. - - Let Z E 5. Then the semisimple component of Z in 3 is the same as in g.

Similarly for the nilpotent component.

This is obvious from Lemma 5. Lemma 7. - - Let U 1 be a neighborhood of zero in 91 and X a nilpotent element of g. Then

we can choose x~G such that x X e U 1. We may assume that X + o. Then by the Jacobson-Morosow theorem [3(h),

Lemma 24], we can choose H~91 sucht hat [H, X] = 2 X. Put a t = exp( - - tH) cG (tEN). Then a t X = e - Z t X and therefore a tX~U 1 if t is positive and sufficiently large.

Corollary. - - Let H denote the semisimple component of an element Z~ 9. Then (1)

HeCI(Z~ Let X be the nilpotent component of Z so that Z = H + X . Consider the

centralizer 3 of H in 9. Then 3 is reductive in 9 and X e 3. Hence X is nilpotent in 3 (Lemma 6). Let E be the analytic subgroup of G corresponding to 5. Then by

Lemma 7, applied to 3, we have

H ~ H + Cl(X ~') = CI(Z ~') c e l (Z~

Let f~ be a subset of 9. We say that fi is completely invariant if it has the following

property: C being any compact subset of f~, CI(C~)cfL

(1) CIS denotes the closure of S.

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Lemma 8. - - Let f~ be a completely invariant subset of g and Z an element in s Then

i f H is the semisimple component of Z, H e f t .

This is obvious from the corollary of Lemma 7- Let ,4 ~ be the set of all nilpotent elements of g. Corollary 1. - - Let S be the set of all semisimple elements of f~ and �9 an invariant subset

of f~ which is closed in f~. Then �9 r~ S = o implies that @ = o. Similarly �9 n S C{o}

implies that Ocf~r~A/'.

For suppose ZeO. Then if H is the semisimple component of Z, HeCl(ZC' )ca . Since @ is invariant and closed in s it follows that H e q ) n S . The two statements

of the corollary are now obvious. Corollary 2. - - Let a o be an open and invariant subset of a . Assume that S c a 0.

Then f2 o = s

This follows from Corollary t by taking �9 to be the complement of f~0 in a . Let r be the center of g. Fix an open and completely invariant subset a of g

and a point X o = C 0 + Z o (Coco , Z0egi) in a . Select a relatively compact and open neighborhood c o of C o in c such that C l ( % ) - k z 0 c a .

Lemma 9. - - Let a I be the set of all Z e gl such that Z + C1% c f~. Then f t l is an open

and completely invariant neighborhood of Z 0 in gl.

It is obvious that a l is an open neighborhood of Z 0 in gl. Fix a compact set Q

in a 1. Then Clc 0 - f -Q is a compact subset of a and therefore

Cl(CI% + Q)~ = Clc o + CI (Q a) c a ,

since a is completely invariant. This shows that C I (Q ~ r and therefore a l is also

completely invariant. Lemma 10. - - The following three conditions on a are equivalent :

2) o e a ; 3) ~4r c f~. Let X e ~ n ~ " . By Lemma 8, o e a . Hence i) implies 2). Now assume o e a .

T h e n i f X e J f , it follows from Lemma 7 that X~ef~ forsome xeG. S incea is invar ian t ,

this means that X e ~ . Therefore 2) implies 3). It is obvious that 3) implies I).

w 4. THE M A I N P A R T OF THE P R O O F OF T H E O R E M l

We shall now begin the proof of the following theorem (cf. [3(g), Lemma I]). Theorem 1. - - Let g be a reductive Lie algebra over R , a an open and completely invariant

subset of g and T a distribution on ~. Assume that :

i) T is invariant ;

2) There exists an ideal lI in I(gc) such that dim(I(gc)/lI)<oo and 0(u)T = o for ue l i .

Then T is a locally summable function on f~ which is analytic on a ' = s

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We use induction on dim g. Let F be the analytic function on tl ' corresponding to Lemma I. Then by Lemma 3, F is locally summable on tl and we have to show that T = T F.

Let c be the center and gl the derived algebra of g. First assume that c4~{o}.

F i x a p o i n t X o = C o + Z 0 (C0er , Zocgl) int l . We have to prove that T = T F a r o u n d X o. Select an open and relatively compact neighborhood % of Co in c such that (C1r + Z 0 c fl. Let tll be the set of all elements Zeg 1 such that ClCo+ZCtl . Then by Lemma 9, tll is also completely invariant.

Fix Euclidean measures dC and dZ on c and g~ respectively such that dX = dCdZ for X = C + Z (Cec, Zegl) and, for any ~eC~(r consider the distribution 0~ on tll given by

0~(~) = T(~ x $) (~zC7 (tlx)).

Then if G~ is the analytic subgroup of G corresponding to gl, it is clear that O~ is invariant under G 1. Moreover I(gc)=S(%)I(gm~ ) since g = c + g 1. Put l I l= l In I (gxc ) . Then it is obvious that

dim(I (gl~)/Ul) ~ dim (I (go)/U) < oo

and O(u)0~=o for uelI1. Therefore, since dim g i < d i m g, it follows by the induction hypothesis that 0~ coincides on tll with a locally summable function g~. Put tl~ = ~1 n gs where fl~ is the set of those elements of gl which are regular in gl. Since g ' = c + g~, it is clear that c0+f~[c~ ' . Moreover since T = T F on tl', it follows that

0~(~) = T(~ • ~) =- TF(~ • ~) = f ~ ( c ) ~(Z)F(C + Z)dCdZ

for ~eC~(tl~). Since g~ is analytic on ~1~ (by the induction hypothesis), it is clear from the above relation that

g~(Z) = f e ( C ) F ( C + Z)dC (z~a l ) .

But since g~ and F are locally summable on tll and f~ respectively, we can now conclude

that

T(~ • ~) = 0~(~) = f ~(Z)e(C)F(C + Z)dCdZ •

for ~eCy(aa). This proves (see [3(h), Lemma 3]) that T = T r on c 0 + t l 1. So now we can assume that c = { o } and therefore g is semisimple. Fix a

semisimple element H o 4: o in ~1. We shall first prove that T = Tr around H 0. Let 3 be the centralizer of H o in g and E the analytic subgroup of G corresponding to 3. Define ~ and 3' as in [3(i), w 2]. Then ~(H0)+ o. Let f~ be the set of all Z~gof~ such that I (z)l>l (H0)l/2. Then n~ is an open neighborhood of H 0 in ( . Moreover since ~ is invariant under ~, it follows easily that f~ is completely invariant in ~. Let ~ be the distribution on f~ corresponding to T under [3(i), Lemma 17] with G O = G and 30=tl~. Then by Corollary I of [3(i), Lemma 17], aT is invariant under E. Now ~ > o on ~ . Hence a=l~l~/~a~ is also an invariant distribution on tl~ and

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it follows from T h e o r e m e of [3(i)] that O(us )o=o for ue l I . Let lI~ denote the image

of 11[ under the homomorph i sm P~Ps of I(gc) into I(3c). Then if ~ = I ( 3 c ) ~ I s , it is clear from L e m m a 19 of [3( i ) ] that d im(I (3~) /~)<oo . O n t h e other hand dim 3 < d i m g since g is semisimple and H0:t: o. Therefore the induct ion hypothesis is applicable to (~, f2s, ~3) in place of (T, f~, 1I). Let f2'~ be the set of all points in f2 s which are regular in 3- T h e n cr coincides with a locally summable function g on f2~ which

is analytic on f2' s. This shows that

T (A) )

in the nota t ion of [3(i), L e m m a I7]. O n the other hand since f2sc3' , it is clear that f~;cf2'. Moreover T = T F on f2'. Therefore

T(f~) = Tv(f~ ) =fo~(x : z ) r ( x Z ) d x d Z

for ~e C~ (G • f2'~). However T is invar iant and therefore the same holds for F. Hence

T ( A ) = f ~ ( Z ) F ( Z ) d Z .

This proves that g(Z)=[K(Z)I~/~F(Z) for Zef2; . N o w fix ~eC2~ Then

T ( f~ )= f ~ ]:l-1/2g dz = fn'~ ~ 1: t-t/Zg dg

= f o • ~'s "(x : Z)F (xZ)dxdZ = Tv(f~ )

a round H o. from Corollary 2 of [3(h), Th eo rem I]. This proves that T = T v Put 0 = T - - T F. Then 0 is an invariant distr ibution on f2.

Lemma 11. - - Let . ~ be the set of all nilpotent elements of g. Then

Supp 0 c~AZ c~ fL

I t follows from the above proof that no semisimple element of f2, other than zero,

can lie in Supp 0. Therefore our assertion follows immedia te ly by taking r = Supp 0

in Corollary r of L e m m a 8. As usual we identify gc with its dual under the Killing form r of g. Lemma 12. - - Assume that there exists a complex number c and an integer r ) o such

that (O(o~) - -c ) rT=o . Then T = T v. We shall prove this by induct ion on r. I f r - - o then T = o and our s ta tement

is true. So assume that r > i. Put T O =(O(o~)--c)T. Then T O satisfies all the conditions of Theo rem I and (~(o~)-- c) r - iT0 = o. Moreover since T = F on f~' and F is analytic on f2', it is obvious that T0=(O(o~) - -c )F on f2'. Therefore it follows by the induct ion

hypothesis that T O = Tr. Hence

and therefore

where F0=(0 (c0 ) - - c )F (see also the corollary of L e m m a 3).

(~(r (0 + TF) = Tr, ~

(0(~o)-- c)0 = T~(~o)v-- c~(co) T v.

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Lemma 1 3 . - To(~)p~--(~o)TF=o. Assuming this for a moment, we shall complete the proof of Lemma 12. For

then we have (0(o~)--c)0----o and therefore we conclude from [3(h), Theorem 5] that 0 = o . Hence T = T F.

The proof of Lemma 13 is based on Theorem 4 of [3(J)] and requires some preparation. Select a system of generators (1) (Pl, "' ' ,Pro) for the algebra I(gc) over C.

Lemma 14. - - Fix X0eg and for any r let Ux~162 ) denote the set of all X~g such

that ( i< i_<m) . Then Ux~ ) is open and completely invariant.

Ux~ ) is obviously open. Let C be a compact subset of Ux.(r ). Then it is clear that we can choose a ( o < a < s ) such that

sup IP,(X)--p , (Xo)l<a (i < i < m ) . X E C

Since Pi is invariant, it is obvious that [p~(Y)--p~(go)[<a for any YeCI(C G) and therefore Ux.(~) is completely invariant.

Now put J0=Te(~)v--0(co)TF and fix XoEfL We have to prove that J 0 = o around X o. Define f ~ ( e ) = ~ n U x ~ 1 6 2 ) for s>o . Then f2(r is an open and completely invariant neighborhood of X 0. We shall now use the notation of [3(J), Theorem 4]. Put (I)i(s)=I)~og2(r and (I),= 90 ( I ) i ( r I f HE*i , it is clear that p ( g ) = p ( X 0 )

for PeI(gc). Hence it follows from Chevalley's theorem [3(c), Lemma 9] that (I) i is a finite set. For each HE(I), choose two open convex neighborhoods UH, V H of H inD~such that C1UHCVHC(I)~(I) and V i t n V a , = O for H 4 : H ' (H, H ' ~ ) . Then C1V H is compact (see the proof of Lemma 2 3 of [3(J)]). Put

U,= 13 U~, V~= 13 V H H ~ i H e @ ,

and select ,r such that ~ i t = I on U H (H~(I)~). Define

0r i ~ ~ ~X H �9 HE~,

Let F~ denote the restriction o f F on f~'n[?~=~nI)~. Fix i and let Pc be the set of all complex positive roots of I)i. Let Q be a connected component of D~(S) and Q1 the set consisting of all regular and semiregular points of Q . I f ~ is a root of (g, t),) which vanishes at some point H 0 in Q i , then it is clear that ~ is compact and therefore H 0 is of compact type in g. Obviously Q1 is open in I)~. Therefore by Lemma 4, F, can be extended to an analytic function on Q l n f ~ which we again denote by F i.

Now fix He(1), and consider Q l n V t i . Then Q l n V H is connected (see the corollary of [3(J), Lemma i9] ). Also VHCf~ and therefore F~ is analytic on the connected set Q ~ n V H. Hence by Lemma 2, there exists an analytic function hH on I?~

such that r~,F~----h n on QlnVI~. Then eHr:~F~=:~ht~ on Q l O ~ and therefore

a,r:iFi = Y, 0~ H h a HEq) i

(1) Since g is semisimple, it follows from the theory of invaxiants that 1(%) is finitely generated.

618

I N V A R I A N T E I G E N D I S T R I B U T I O N S O N A S E M I S I M P L E L I E A L G E B R A I5

on QlOY2. Put g~ .... 0~i~F~. Then the above result shows that g( is of class Coo on C l ( Q 1 ) = C l ( Q ).

Choose , > o so small that q)/r ( ~ i < r ) . Then from Corollary i of [3(J), Lemma 30] we can choose numbers c~ (i < i < r ) such that

f f u d X = Y~ c,f~ +t.,~.,r~,u,d,H ( / ~ C : (a)) 1 < / <

for any invariant and locally summable function u on f2. (Here u~ is the restriction of u on f~nIki.) Now suppose f eC~( f~ ( , ) ) . Since ~(~) is completely invariant, it follows from [3(J), Lemma 22] that

Hence

Supp d/t,,cf~(, ) n b, = O,(~) cU, .

l < i < r J '

Now take u = F . Then C~sR,i~dz~Ui=q*R, ig~---gi (say). On the other hand, by the corollary of Lemma 3, we can also take u = ~(co)F. Then it follows from [3(c), Lemma 3] that u~=~-tO(o~i)(~F~) on I?~g~ and therefore

Ci~R,~O~irciu ~ = O(coi)g i

on UinI?~. Therefore

f (F 0(o~)f-- 0(co) F . f ) d X J0 (f)

= Y, f (0(~i)+t,~'g~--+t,~" 0(c@gi)d~U l < i < r bi

from [3(d), Theorem 3]. Now define J as in [3(J), Theorem 4], corresponding to the above functions gi (I < i < r). Then the above result shows that J =J0 on f2(~). Since g~ is obviously of class C OO on the closure of each connected component of D~(S), Theorem 4 of [3(J)] is applicable. Fix an open and relatively compact neighborhood V of X 0 in f~(-*). Then since ~(r is completely invariant, CI(V ~) cD(r Let 5 ~ denote the set of all semiregular elements of f~(r of noncompact type. Then in order to prove that J 0 = o on V, it is enough, from [3(J), Theorem 4], to verify that S u p p J 0 0 5 0 = o . However J0--(o(co)--c)0 and so it follows from Lemma i i that

Supp J0 c Supp 0 c X n fL

Since zero is the only semisimple element in dg', it is clear that M/'n 50 c{o}. Therefore we may assume that 50 contains zero. But then it follows from [3(J), w 4] that g is isomorphic to the three dimensional noncompact semisimple algebra I of [3(J), w 2]. We shall consider this case in detail in the next section.

619

I6 H A R I S H - C H A N D R A

w 5. S O M E C O M P U T A T I O N S O N I

So we now assume that f l = l and oef2. Then we have to show that J 0 = o

around zero. Hence we take X 0 = o (see w 4)- Then it follows from L e m m a IO

that Mzc~)(,). Now dg" is also the singular set of g in the present case. Therefore

SuppJC~4/'. However J = J 0 on ~2(,) and so it is obvious that J = J 0 on fl. Lemma 15. - - W e can choose complex numbers a, a + and a - such that

J ( f ) = af(o) + a + c + ( f ) + a - c - ( f ) (f~C~ ~ (g))

in the notation o f [3(J), L e m m a 34]. This is obvious from [3( j ) , Lemmas 2, 3 and 26].

C o r o l l a r y . - - ~oJ = o .

Since o~ = o on dg., this is an immedia te consequence of L e m m a i5.

We have seen i n w S u p p J o C ~ . Fix an element X 4 : o i n ~ . We shall

first prove that J0-- o around X. By theJacobson-Morosow theorem [3(h), L e m m a 24] ,

we can select H, Y in g such that

[H, X] = 2X, [H, Y] = - 2Y, [X, Y] = H.

Then ~x ----- R X is the centralizer of X in g and gx = IX, g] = R H + R X . Take U -- R Y

and V = R H + R Y so that g = U + gx = 3x + V . We now use the notat ion of [3(h), w 7]- Then 4 r o ~ ( X + t Y ) = 8 t and

rx+ty(Y2| = H2- -2Y, r x + t y ( g | = 2 ( X Y - - Y - - t Y 2 )

for tER. Hence

This means that 4A(8(o~)) ----- 3 D + 2tD 2

on U ' in the notat ion of [3 (h), w 8]. (Here D = d/dt.) On the other hand (8(o))--c)0 = J 0

and 0 and Jo are both invar iant distributions. Hence

( A - - c ) ~ 0 = %

in the notat ion of [3(h), Theorem 3] where A=A(~(o~)) . Since Supp 0 c~V, we can regard % as a distr ibution on an open neighborhood U 0

of the origin in R and assume that Supp o0C{o ) (see [3(h), L e m m a 23] ). I f * 0 = o ,

it follows from [3 (h), Theorem 2] that 0 = o a round X and therefore the same holds

for Jo. Hence we may assume that o c Su p p o 0. Then (see [3(h), L e m m a 20])

% = Y~ akDk8 O<k<m

620

INVARIANT EIGENDISTRIBUTIONS ON A SEMISIMPLE LIE ALGEBRA i7

where 8 denotes the Dirac distribution ~-7 ~ (o) (~ e C~ (Uo)) and a k are complex numbers

(%4:0). Now coJ---coJo=o o n ~ . Since ~o (X +t Y )=8 t , it follows that t a jo=o o n U 0. Hence

akt(3D + 2tD2--4c)Dk8 = o . 0<k<m

But it is easy to verify that tD k 8 = - - kD k- 18,

t2Dk8 -~ k(k--- I)Dk-2~ (k 2 o)

where D~8 should be interpreted to mean zero if ~<o. Therefore

0<~< mak{ (k -~ I) (")k + I)Dk~ N t- 4ckDk-~8} = o.

But since the distributions Dk8 (k>o) on U 0 are linearly independent, we conclude that ( m + i ) ( 2 m + I)am=O. However this is impossible since m > o and am4-o. This contradiction shows that a0 = o and therefore 0 = J 0 = o around X. This proves that

Supp0C{o} and SuppJoC{O }. Now J = J o on fL Hence it follows (see [3(e), p. 685] ) that a+- - - - a -=o in

Lemma 15 and therefore J 0 = J = a 8 0 on f~. Here 30 is the Dirac distribution f-->f(o) ( f e C [ ( g ) ) on g. But since Supp 0C{o}, we conclude from [3(h), Lemma ~o] that 0=0 (p )S o where psS(g~). On the other hand (o(o~)- -c )0=Jo=aSo on ~.

Therefore (o ) - - c )p=a again from [3(h), Lemma 2o]. Since o~ is homogeneous of degree 2, this is possible only if p = a = o . Therefore 0 = J 0 = o and so Lemma 13 is now proved.

w 6. C O M P L E T I O N OF T H E P R O O F OF T H E O R E M x

It remains to complete the proof of Theorem I in case g is semisimple. Let 3; be the vector space of all distributions on ~ of the form O(p)T (p~I(gc)). Then it is clear that

dim 3;< dim(I(g~)/U)< oo

and every element of 3; satisfies all the conditions of Theorem i. The mapping S-+O(o~)S (Se3;) is obviously an endomorphism of 3;. Hence we can choose a base Tj. (I < j < N) for 3; over C with the following property. There exist complex numbers c i and integers r~._>o such that

(O(r = o (i < j < N).

Then Lemma i2 is applicable to T i. Let F i be the analytic function on f~' such that T i = F i on f~' (Lemma I). Then F l is locally summable on ~ (Lemma 3) and

T / = T F i (Lemmai2) . Since ( T ] ) ( I < j < N ) is a base for 3;, T = ~ a I T ! for some a]~C.

Then if F = ~.a]Fi, it is obvious that T = T F. This proves Theorem I. 1

621 3

i8 H A R I S H - C H A N D R A

w 4]"

Theorem 1. Hence DF is locally summable on ~ and DT----TDF.

Corollary. - - Let D* denote, as usual, the adjoint o f D. Then

f f D F d X = f D * f . F d X

This is merely a restatement of the relation D T = TDF.

w 7. SOME CONSEQUENCES OF THEOREM x

We shall now derive some consequences of Theorem i. Define ,~(gc) as in [3(i), We keep to the notation of Theorem I.

Lemma 16. - - Fix De~(g~). Then the distribution D T also satisfies the conditions of

(fecy(n)).

Since the distribution D T is obviously invariant, it is enough to verify that the dimension of the space of all distributions of the form #(p)(DT) (peI(gc)) is finite. This requires some preparation.

Let us now use the notation of [3(i), w 3]. For any peS(E) , let rp and dp denote the endomorphisms D-+Do~(p) and (1) D ~ { ~ ( p ) , D} (De~)(E)) respectively of~)(E).

Lemma 17. - - Fix peS(E) . Then for every D e ~ ( E ) we can choose an integer N > o

such that d i D = o. Let A be the set of all peS(E) for which the lemma holds. We claim that A is

a subalgebra of S(E). Observe that dp, rp, dq, rq (p, qeS(E)) all commute with each other

and dpq = dpdq + rpdq + dprq.

Now fix p, q in A and D e ~ ( E ) and choose an integer N_>o such that d~D =dND = o . Then it is obvious that (dp+dq)2ND=o and

d ~ D = (d, dq + r, dq + dprq)3ND = o.

This shows that p + q and pq are both in A and therefore A is a subalgebra. On the other hand if peP(E) , qeS(E) and X e E , it is obvious that

(pO(q) ) = ( d ,p) e (q) = o

if N>d~ This shows that E c A and therefore A = S ( E ) .

We now return to the proof of Lemma i6. Let 32 denote the space of all distributions of the form O(p)T (peI(gc)). Then dim 32<oe. Since the algebra I(g~) is abelian, we can choose a base T1, . . . , T m for 32 over C and homomorphisms Z1, �9 �9 Zm of I(g~) into C such that

(e(p)--z,(p))mTi = o ( i < i< m).

Since T is a linear combination of T~, it would be enough to prove Lemma I6 under

the additional assumption that

(0(p)--z(p))mT = o (peI(gc))

(1) As usual {Da, D 2 } = D l o D 2 - - D 2 o D 1 for two differential operators D1, D 2

622

I N V A R I A N T E I G E N D I S T R I B U T I O N S ON A S E M I S I M P L E L I E A L G E B R A x 9

for some integer m_>o and some homomorphism ) of I(g~) into C. Now fix psi(go) and choose N > o so large that d~D = o (in the notation of Lemma 17 with E = g~). Then

( O ( p ) - - x ( p ) ) z~ +rooD = (dp + rp - - ~.(p) ) z~ +roD

0<k<Z~+m'~k ~'p x(P)) N

where c~s+m stands for the usual binomial coefficient. Now consider ~ k

((r ,--x(p)) ~ + m-~dpkD)T= (d~D)((~(p)--x(p)) ~ +m-kT).

I f k > N , d ~ D = o and if k < N , (~ (p ) - -x (p ) )N+m-kT=o . Hence

(~(p)--z(p)) ~ +m(DT) = o.

Choose Pl, .- . ,P~ in I(g,) such that I(fl,)=e[p~, ...,Pt]. Then we can choose an integer M > o such that

(~(p~)--?((p,))~tDT = o ( I < i < l).

But this implies that the space of all distributions of the form 0(p)DT (peI(gc)) has dimension at most M z. This proves that Theorem i is applicable to DT.

It is obvious that D T = DF on ~2'. Hence by applying Theorem i to D T we

conclude that DF is locally summable on f~ and D T = TDF.

w 8. FURTHER PROPERTIES OF F

Fix a Cartan subalgebra D of g and let us use the notation of [3(J), w 4]. Define

the analytic function g on D' n f~ by

g(H) = ~(H)F(H) (HeI) ' n ~).

Theorem 2. - - g can be extended to an analytic funct ion on I~'(R)c~ fL Fix an element H 0 e b ' ( R ) n f L It is enough to show that there exists an analytic

function gl on an open neighborhood U of H 0 in I)'(R)nf~ such that gl ~ g on U n [?'. First assume that H 0 is semiregular. Let ~ be the unique positive root of [? which vanishes

at H 0. Then clearly ~ is imaginary. I f ~ is compact, the required result follows immediately from Lemma 4- Hence we may assume that ~ is singular. Define a and b

as in [3(j) , w 7] corresponding to H 0. Then it follows from [3(J), Lemmas I2 and 13] that we may assume that t ) = b.

Let 3 be the centralizer of H 0 in g. Define ~ and 3' as usual [3(i), w167 ~, 7]-

We now use the notation of [3(J), w 7]. Let 30 be the set of all Ze~nf~ such that I (Z)l>l (H0)l/2. Then 30 is an open neighborhood of H o in 3' which is completely invariant (with respect to ~). Now ~ is the center of 3- Fix an open and convex

neighborhood a0 of H o in a such that C1% is compact and contained in 3o. Let f2 l

denote the set of all Z e[ such that C1% + Z c 30. Then by Lemma 9, ~I is a completely

invariant and open neighborhood of zero in 1.

623

20 H A R I S H - C H A N D R A

Now apply Lemma 17 of [3(i)] with G 0 = G and put T~=[~[I/2aT. Then it follows from Theorem 2 of [3(i)] that ~(u~)T 3 = o (uelI). Put $ =I(3~)1I~ where ~I 3 is the image of lI in I(3~) under the mapping p~pa (p~I(g~)). Then by [3(i), Lemma I9], dim (I(8~)/2~)<o~. Let ~ denote the Killing form ofl . Then, if we identify I with its dual under o~t, we have r Hence we can choose complex numbers q , . . . , c, such that

E c,~o,-~$ 0 < k < r

where c 0 = I. This proves that

Now fix yeC~(%) and let v v

E 6a(co~)r-kT~=o. 0 < k < r

denote the distribution

% :f--> T ~ ( y x f ) (fEC~(f~0)

is invariant under L and it is clear from the above relation that

E ckO(co )r-k = o. O < k < r ":

I(L)=C[col] , Theorem i and Lemma 16 are both applicable to (I, f~l, %) in

on D. 1. Obviously %

Since place of (g, f/, T). Let f~{ be the set of those points of f~ which are regular in I. Fix a Euclidean measure d[ on I. Then we can choose an analytic function q% on ~( which is locally summable on f~l and such that

( f ) = ff ,dr ( fe C 2 (f~O ). Ty

Hence it follows from Lemma I6 that

f { O(~,)kf. q~v-- f" a(~,)k %r} dl = o

for k > o and feC~(f~l). For any ~>o, let f~(r denote the set of all Ze l with ]~(Z) [<8r 2. I f a is sufficiently small, it is obvious that tH' and t ( X ' - - Y ' ) both lie in ~I whenever }t]<~ (teR). Since f~l is completely invariant under L, we can conclude (see [3(e), p. 68I]) that Clf~l(~)cf~ ,. It follows from Lemma 2 that there exist three analytic functions gv, g+, g~- on R such that

g~(t) = t%t(tH' )

gv + (0) = O~v(O(X'--Y'))

gv-(O) = Oq%(O(X'--Y'))

(o<O<,)

(--, <0<o).

Now define the distributions T k (k>o) on 1 as in Corollary i of [3(J), Lemma 35] with (g, g+, g-) replaced by (g-r, g+, g~-)" Then it follows from [3(e), Lemma i6]

and [3(c), Theorem I] that

Tk(f) = c f { O(co,)kf. 9v--fO(col) k ~%}dl = o

624

INVARIANT EIGENDISTRIBUTIONS ON A SEMISIMPLE LIE ALGEBRA 2 !

for feC2(f~ , ( r (Here c is a positive constant.) Therefore we conclude from the corollaries of [3(J), L e m m a 35] that g+ =g~- and

( - ~ )~( d ~ + ~ g~/ dt ~ + % = ( d ~ + ~ ~+ ,-'.~+ ~, ( k >_o) g7 I uu ) o

where the subscript o denotes the value at zero. O n the other hand let F~ denote the restriction of F to ]0. T h e n by Corollary 2

of [3(h), Theorem ~], F~ is locally summable on ]0 and since F is obviously invar iant under G, we have

T(f~) = f f ~ F d X = f ~rf lZ (~e C: (S0))

in the notat ion of [3(i), L e m m a i7]. This proves that aT-=-F ~ and therefore T~=t~lX/eF~. Now a : a + R H ' and b : t ) = a + R ( X ' - - Y ' ) . Let -~ and X be the unique positive roots of a and b respectively which vanish at H 0. We may assume that z ( H ' ) = 2 , X ( X ' - Y ' ) = - - 2 ( - - I ) ~/'~ and the positive roots of a go into positive roots

of b under the au tomorphism ,~ of [3(J), w 7]. Put is clear that

It(H) 11/2= [ ~$(H) I

I t (H) 11/,~ = j ~bx(H) j

Let I denote the open interval ( - - r r in

rr~=-~-lr~ ", r~x~=X-lrc ~. T h e n it

( H e a ) ,

(Heb).

R. Put a(~)=~o~-IH' and b(*)=*o+I(X'--Y'). T h e n a(r and b(*) are both connected sets. Since a( ,)C3o, it is obvious that no positive root of (g, a) other than v can vanish anywhere on a(r Hence [r~,"(H)I/~,"(H) is a continuous function on a(r But since its fourth power

is I (see [3(J), L e m m a 9]), it must be a constant. Put c- - In~(H0) l/r~(H0). Since b a v r~x ----- (r%) and H o remains fixed under v, it is clear that

c = ] ~ ( H 0 ) 1 / ~ ( H 0 ) .

Hence we conclude by a similar a rgument that

This shows that t I ~ (H + tH') [~/2 = 2 -~ c7: ~ -~ tH ') (1 t[ < ~)

01 ~(H § 0 ( X ' - - Y')) 1'/2 ---= 2 -1 ( - - ~)'/2cr~(H @ 0 ( X ' - - Y ' ) ) ([0I<~)

for H c ~ o. Now put ga(H) = r~a(H)F(H) (He a ' n g/),

gb(H) = n ~(H)F(H) (H e b ' n ~) .

and fix a Eucl idean measure da on a such that d a d I is equal to the Eucl idean measure dZ

on 3 used above. Since T a = [tll/eFs, it is obvious that

~ ( Y ) = f y ( H ) t t ( H + Y)['/2F(H + Y ) d , (YeggS).

625

Hence

g~(t) = 2-acfg"(H § tH')T(H)da

g+ (0) = 2-t (-- ~ )~l~ c f gb(H + 0(X'--Y'))T(H)da

g~-(0) = 2-~(-- ,)l/~cfg~(H § 0(X'--Y'))T(H)da

(o< t<~) ,

(o<0<~),

(--~<0<o).

On the other hand i f J is the open interval (o, e) in R, it is clear that e 0 ~ J H ' are connected sets contained in a ' n ~ . Let a • denote the connected component of a 'n containing % • Similarly let b • be the connected component of b' n ~ containing

a0 :~J (X ' - -Y ' ) . Then by Lemmas I and 2, there exist analytic functions g~• and gb on a and b respectively such that ~ a g~=g~_ ~ ~ b + g = g + on a +, on a - , g = g + on

and g b = g ~ on b- . It is then obvious that

2-1C f a gv(t) = j g + ( H + tH')T(H)da ( teR) ,

(0) = 2 - 1 ( - I)l/2cfgb, ( H § 0(X ' - -Y ' ) )T(H)da (0eR). g~

On the other hand we have seen above that g+ = g v for every u Therefore it is clear that g~+=g~_. This shows that g_=gb_=gb+ on b( r Since b(r is a neighborhood of H o in b, our assertion is proved in this case.

Moreover since

(d~k + l gvl dt~k + ')o = ( - - ' )k(d2k + lg+ /dOS + l)o (k~o)

and v ( H ' ) = ( I I ) I / ~ ( X ' - - Y ' ) , we find in the same way that

g~(H; 0(H')~+~)=g~(H; ~(~(H')) ~k+~) for He~.

We now use the notation of [3(J), w 8]. Lemma 18. - - Let s, be the Weyl reflexion in a corresponding to z. Then ( g ~

I f D is an element in ~ ( a~) such that D ~, = - - D , then Dg" can be extended to a continuous function

on 0(~) and (') g~(H; D ) = g b ( H ; D ~)

for HEa 0. Since "~ is real we know from [3(J), Lemma 6] that s~EW~. Therefore since F

is invariant under G, it is obvious that (gQ)~=__ga and hence (Dga )~=Dg a. This implies that (Dg+) ~ = Dg a _ and therefore Dg~ = Dg~ on a. It is now clear that Dg a can be extended to a continuous function on a(r So it remains to show that

g~_(H; D ) = g ~ ( H ; D ~)

for Heo0. Since ~(a~)=~(a~)~3(CH') and since s, leaves a pointwise fixed, it is sufficient to consider the case when D=AoxIO(H ' ) i. Here Ae~(a~) and i + j is odd.

(1) ga(H ; D) denotes, as usual, the value of the continuous function Dg a a t H. Similarly in other cases.

626

INVARIANT EIGENDISTRIBUTIONS ON A SEMISIMPLE LIE ALGEBRA ~3

Now A and z commute . Therefore, i f i > i, our assertion is obvious from the fact that

and X are both zero on ~. So we m a y assume that i = o so t h a t j is odd. I t is enough

to verify that g~_(H; ~ (H ' ) i )=g~(H; 0(~(H')) i) ( H ~ 0 )

since the required relation would then follow by applying the differential operator A

to this equation. However ~ b g = g + on b(r and so this follows from the result proved

above. Now we re turn to the proof of Theorem 2. Fix a point H0e~)'(R)n~2 and an

open convex neighborhood U of H 0 in b ' ( R ) n ~ . Let U1 be the set consisting of all regular and semiregular elements of U. Then U1 is open, and if ~ is a root of (g, I))

which vanishes at some point of U1, it is clear that ~ is imaginary. Hence it follows

from the above proof tha t there exists an analytic function gx on Ux such that g ~ = g

on Uln I ) ' . Now fix a connected component Us of U l n l ) ' = U n I ) ' . T h e n b y L e m m a 2

there exists an analytic function g2 on D such that g =g~ on Us. Since U1 is connected

(see the corollary of [3(J), L e m m a I9]), we conclude that gl =g~ and therefore g = g 2

on U r~ D'. Since g2 is analytic on D, we have shown that g can be extended to an analytic

function on U. Thus Theorem ~ is proved.

We denote the extended analytic function on D ' ( R ) n ~ again by g.

Lemma 19 (1). __ Let H o be a point in t) n ~ and D an element in 79 (De) such that D ~ = - - D

jbr every real root ~ o f (g, D) which vanishes at H 0. Then Dg can be extended to a continuous

function around H 0. Fix an open, convex and relatively compact neighborhood U of H 0 in ~ n [ .

By taking it sufficiently small we can arrange that no real root ~ of (g, I)) vanishes

anywhere on U unless ~(Ho) = o. Let U o be the set consisting of all regular and semiregular elements of U. Then, as before, U 0 is open and connected and it follows from Theorem 2

and L e m m a 18 tha t there exists a continuous function go on U 0 such that D g = g o on Uor~I3'(R ). The set U n I ) ' has only a finite number of connected components,

say U 1 , . . . , Ur. By L e m m a 2 we can choose an analytic function gi on D such

that g = g i on U i (~ < i < r). This shows that Dg is of class C ~ on C1U~ (see [3(J), w I4])" But C 1 U i = C I ( U i n U o ) and D g = g 0 on Uir~U o. Therefore go is also of class Coo

on C1U i ( I < i < r ) . Fix a Euclidean norm on I? and put

v(go) = sup Igo(H1 ; O(H~))]

where Hx, H 2 vary in U n D' and t) respectively under the sole restriction that I I H II< i.

Then it is obvious from what we have said above that v(go)<Oo. Moreover (see

[3(J), w Io]) [ g0(H~) --g0(H~) I < v (go) II H 1 - H211

for any two points H1, H 2 in U r~ b'. Obviously this means that Dg can be extended

to a continuous function on U.

(1) Cf. [3(J), Theorem I].

627

~4 H A R I S H - C H A N D R A

Corollary. - - Let D be an element of ~( tL) such that D ~ -~ - - D for every real root ~ of t).

Then Dg can be extended to a continuous function on I ) r ~ .

This is obvious from the above lemma. We denote the extended function again by Dg. Moreover g(H; D) (HEI)nf2) will stand for the value of Dg at H.

Put ~ = ] - I H ~ where ~ runs over all positive roots of (g, I?). Then ~ES(IL) ~ > 0

and m ~ = - - m for every root 0~. Hence 0(m)g is a continuous function on I)oYL Since the differential operator ~(m)07: is obviously independent of the choice of positive roots of I), it is clear that the function D(~)g also does not depend on this choice. Corresponding to any Cartan subalgebra a of g, we define ~", g" and D(m")g" in an analogous way.

Theorem 3. - - Let a and b be two Cartan subalgebras of g. Then

on ctr~bn~.

Before giving the proof we derive a consequence of this theorem. Let g~ denote the function g of Theorem 2 corresponding to the distribution T. For any DEg(g~), DT also fulfills the conditions of Theorem I (Lemma I6). Hence we can consider the corresponding function gl)T. It follows from [3 (i), Theorem I] that gDT ~---~g/~(D)gT" Therefore

O(~)gi)T = (~(~)o3g/~ (D))g

can also be extended to a continuous function on D nf l . Coro l lary . - (O(~a)oSo/,(D))g"=(D(~)o8g/~(D))g ~ on ar~bnf~ for any D ~ ( g c ) . This follows by applying Theorem 3 to DT instead of T. We shall prove Theorem 3 by induction on dim g. Fix a point H o e a n b o [ L

We have to show that g"(H0; O(~o))----g~(H0; ~(~)) . Let c be the center and gl the derived algebra of g and first suppose that r Let H0=C0- [ -H 1 where C0~r and H l e g 1. Then it is clear that H~ea~nb~ where [)l=I)mg 1 ( D = a or b). Choose an open and relatively compact neighborhood r of C o in c and let fl~ be the set of all Z egl such that CI% + Z cf l . Then (Lemma 9) fl~ is an open and completely invariant neighborhood of H~ in gl, if c o is sufficiently small. Fix a~C~~162 and consider the distribution

"G : f - + T ( e • ( f e C / ( ~ l ) )

on ~ . Put l I l = l l n I ( g x c ) . Then it is clear that

and 0(ul)1: ~ = o

by (gl, ~1, "~). thesis. Put

dim (I(glc)/lI1)_<dim (I(gc)/lI) <oo

for ulelI 1. Hence Theorem I also holds if we replace (g, f~, T) Since dim g l < d i m g, Theorem 3 applies to v, by the induction hypo-

g~(H) = f~(C)g~(C -}- H)dC

628

where dC is a Euclidean measure on c. Then we conclude that

g~(H; 0(~,~))=g~(H; 0(~,~))

for Hear~br~f~ 1. Since this is true for every ~eC~(c0), it is clear that ~( a)ga and ~(~)gb coincide around H 0 on ctnbr~f~.

So now we can assume that c = { o } and therefore g is semisimple. Then we identify g and t) with their respective duals by means of the Killing form (see [3 (i), w 6]) so that ~ = ~ b ( I ) = a or b). First assume that H 0 + o and let ,3 be the centralizer of H 0 in g. Then dim 3 < d i m g and we can identify g with its dual by means of the restriction (to 3) of the Killing form of g. Define ~ and 3' as in [3(i), w 2] and put ~ = ] ' r ~ f ~ . Then f~ is an open neighborhood of H 0 in g which is completely invariant (with respect to ~). Take G 0 = G and 3 0 = f ~ in Lemma 1 7 of [3(i)] and let a T denote the corresponding distribution on f~. Then

aT(~) = T(f~) =ff~FdX ( ~ C ~ ~ (G • f~)).

But since F is invariant under G, we conclude from Corollary 2 of [3(h), Theorem i] that the function F~ : Z --- F(Z) (ZerO3) is locally summable on f~ and

f f~FdX = f f~F, dZ. This shows that ~ = F~.

Let I ) = a or b. Then t)r Define q as in [3(i), w 2]. P~ being the set of all positive roots of (g, I)), let P~ and Pq~ denote the subsets of those 0~P ~ for which X~ lies in g~ and q~ respectively. Let r~ and rr~ be the products of all roots in P~ and P~ respectively. Then ~ = ~ and it is clear that ( rc~)~=rq ~ for all ~ P ~ . Hence, by Chevalley's theorem [3(c), Lemma 9], there exists an invariant polynomial function p on 3 such that p(H)=r~2(H ) for H~a . But

~(H) = ( - - I)q(Tv~(H)) ~ (H~ , )

where q = 2 - ~ dim q is the number of roots in P~. Therefore ~ = ( - - I ) q p z again by Chevalley's theorem. Let P0 denote the restriction o f p to D. Then since ~ coincides with (--I)~(r~) ~ on b, it is clear that p ~ = r where r 1 7 7

Now put T , = p a ~ . Then it follows from Theorem ~ of [3(i)] (see also w 4) that Theorem i still holds if we replace (g, ~, T) by (~, Y~, T~). Put

g~(H) = r~(H)p(H)F(H) (H ~ I)' r~ ~ ) .

Since dim g < d i m g, both Theorem 3 and its corollary are applicable to Ta. Moreover ~/~(O(p))=O(po) [3(c), Theorem I] and so we conclude that

on ~ n a o b . However %~p~=n~ and ~z~p~=s~ ~. Therefore g[=g" and g~=~g~ on ~ r~ a' and ~c~ b' respectively. So it follows that O(~z")g ~ = O(~)g ~ on Y~c~ a~ b. Since H0~)~r~nr~b , our assertion 'is proved in this case.

6'29

So it remains to consider the case when H 0 = o. Hence we may assume that o E~. For any Cartan subalgebra ~1 of g, put c(I?)=g~(o; ~(~)).

Lemma 20. - - Let Q and b be two Cartan subalgebras of g. Then, i f a n b # { o } , c(a)=c(b).

Since f~ is an open neighborhood of zero in g, we can choose H # o in a n b such that tHE a n b n fl for o < t < I. Then in view of what we have proved above, it is

clear that ga(tH; e(~za))=gb(tH; 0(=~)) ( o < t < i ) .

Making t tend to zero we get c (a )=c(b ) .

Lemma 21. - - Let t) be a Cartan subaIgebra of g. Then c(i))=c(t) ~) for any x in G.

Without loss of generality we may assume that 7:"= (=~)~. Then it Let a = I)*. is clear that

and therefore

g~(H *) ----- g~ (H) (Hef ln i ) ' )

g"(H~; ~(~z"))=g~(H; 0(~b))

for Hef~n~)'. We obtain the required result by making H tend to zero. Fix a Cartan involution 0 of g and let g = ~ - k P be the corresponding Cartan

decomposition. I f t) is a Cartan subalgebra of g which is stable under 0, we put

l+ (i)) : dim(i) np) , l_(i)) : dim(i)n ~).

Then l+(i))-t-I ( i ) )=d im I ) = l where / = r a n k g. Let l + = s u p l+(I)), l _ = s u p l_(i))

where I) runs over all Cartan subalgebras stable under 0. Fix two Cartan subalgebras t)+,

I)_, both stable under 0, such that l+ =l+(i)+), l _= l_ ( i )_ ) . Lemma 22. - - Let b be a Caftan subalgebra ofg which is stable unde~ O. Then c(t)) = c(~O+)

/ f l+(i))>o and c(i))=c(i)_) / f l (I))>o. Let K be the analytic subgroup of G corresponding to L Then I)+np and I)_n~

are maximal abelian subspaces of p and ~ respectively. Since any two maximal abelian subspaces of p (or ~) are conjugate under K, it is clear that we can choose kl, k~eK

such that

Then

and similarly

dim (t)k' n t)+) • dim(t) n p)kl = l+ (i))

dim (t)k' ~ I?_) > l_ (I?).

Hence our assertion follows from Lemmas 20 and 2I. Lemma 23. - - c(t)+ )=c( t )_ ) . We may obviously assume that l > I . I f l_(~)+)-[-l+(I)_)>I, our statement

follows from Lemma 22. Hence we may assume that D+cp and D_r Then I3+ is not fundamental in g and so there exists a positive real root ~ of (g, I)+) (see [3(d),

630

INVARIANT EIGENDISTRIBUTIONS ON A SEMISIMPLE LIE ALGEBRA o7

Lemma 33])- We assume, as we may (see [3(d), Lemma 46]), that 0 ( X ~ ) = - - X _ ~ and X~, X_~ are in g. Take H ' = a - 2 H ~ , X ' = a - t X ~ , Y ' = a - t X _ ~ where

a=(~(H~) /2) 1/2. Define the automorphism,~ of g, as in [3(J), w 7]. Then 0 ( X ' ) = - - Y ' and b = v ( ( I ) + ) ~ ) c ~ g = a ~ + R ( X ' - - Y ' ) is a Cartan subalgebra of g which is stable under 0. Here ~ is the hyperplane consisting of all points HED+ where 0~(H)=o.

Since b n p = ~ and br~ t = R ( X ' - - Y ' ) , it is obvious that l+ (b) = l - - i and l (b) = i. Hence, if l > 2 , it follows from Lemma 22 that c(O+)=c(b)=c(O_) . On the other hand

if l = I, zero is a semiregular element of g and our assertion follows immediately from

Lemmas I8 and 2~. We shall now finish the proof of Theorem 3. Choose x, y in G such that cff and b v

are stable under 0 (see [3(b), p. ioo]). Then it is clear from Lemmas 21, 22 and 23 ~hat c(a)=c(b) . The proof of Theorem 3 is now complete.

w 9. THE DIFFERENTIAL OPERATOR Vg AND THE FUNCTION V~F

Lemma 24. - - There exists a unique differential operator V~ on g' with the following two properties :

I) V~ is invariant under G. 2) Let D be a Cartan subalgebra of g. Then

f ( H ; V ~ ) = f ( H ; OC~)or~ ~)

for fEc*(g) and Moreover Vg is analytic.

Since two distinct Cartan subalgebras cannot have a regular element in common, the uniqueness is obvious. The existence is proved as follows. Fix a Cartan subalgebra a of g and define g, ~ (a') 0. Then go is an open subset of g. Let A be the Cartan subgroup

G* of G corresponding to a and x-+x* the natural projection of G on G/A. Then the mapping q~ : (x*, H) -+x*H of G* • a' onto g~ (in the notation of w 2) is everywhere

regular. Define W G = A / A where A is the normalizer of a in G. Then WQ operates

G* on and a as follows. Fix seW~ and choose y e a lying in the coset s. Then

sH = H y, x*s = (xy)*

for H E a and xeG. It is clear that the complete inverse image under q~ of a point x*HEg, (x*eG*, H e a ' ) consists of the elements (x's, s - l H ) (seW~), which are all distinct. Since q~ is locally an analytic diffeomorphism and since ~(w")or:" is obviously

invariant under WG, it is clear that there exists an analytic differential operator V on ga such that

f ( x ' H ; V ) = f ( x * : H; ~(~")or: a) (x*eG*, HEa ' )

for f s C ~ ( g , ) . Here f(x*: H ) = f ( x * H ) as usual. It is easy to verify that V satisfies

the two conditions of the lemma on ga-

631

Now select a maximal set I?1, . . . , I)r of Cartan subalgebras of g, no two of which

are conjugate under G. Put g~=(th') ~ and define a differential operator V~ on g~

as above corresponding to a=I)~. Since g' is the disjoint union of the open sets gl, . - . , gr, we can define Vo by setting Vg=Vi on gi ( I < i < r ) .

Lemma 25. - - For any De~(gc), (VgoD)F can be extended to a continuous function on fL

We shall use induction on dim g. In view of Lemma 16, it is enough to consider the case when D = I. Define r and gl as in w 4 and first assume that r Fix

a point X 0 = C 0-r Z 0 (C0e c, Z0egl) in YL We have to show that V~F can be extended to a continuous function around X 0. Select an open, connected and relatively compact

neighborhood c0 of C o in c such that (Clc0)§ Z0cfL Define f~l to be the set of all ZEg 1 such that C l c 0 + Z c f L Then, by Lemma 9, g~l is also open and completely invariant

in gl- Since S(c~)CI(g~), it is clear that

dim(S(co)/l ln S (c~))~dim(I (gc)/1I) < oo.

Let E be the space of all analytic functions Z on r such that 0 ( u ) z = o for uelInS(r Then (see the proof of Lemma t 3 of [3(c)]) dim E < m . Let ~(j ( I < j < N ) be a base for E over C. Fix ZEf2t '=f21ng ' . Then it is obvious that F ( Z + C ; 0 (u ) )=o for

uelInS(c~) and Ceco. Therefore

F(C § Z) = <~<N ;q(C)F~.(Z) (CEc0)

Since F is analytic on rE, it is obvious that F i (I < j < N ) are analytic where Fj(Z) eC. functions on f~.

Fix Euclidean measures dC and dZ on c and gl respectively such that dX-= dCdZ

for X = C + Z (Cec, Zegl) and for any eeC~(c0) define the distribution 0~ on f21 by

0c~(~ ) = T((xX ~) (~@C2(~'~1)).

Then, as we have seen in w 4, the induction hypothesis is applicable to (gl, 0~, f~t) in

place of (g, T, f~). Put

F=(Z)= z F (Z)fxj(C) (C)dC I</<N

Then V~F~ can be extended to a continuous function on f~l. Since this is true for every eaC2(%) , the same holds for V~Ff, I < j < N (see [3(e), Lemma 2o]). But it

is obvious that

F ( C d - Z ; Vs)= Z zi(C)Fi(Z; Vs,) I<:<N

Hence VgF extends to a continuous function on c0-~ f~1, which for Ceco and Ze~)~. proves our assertion.

So now we may assume that 9 is semisimple. Let f~0 be the set of all points X0af~ such that VgF can be extended to a continuous function around X 0. Clearly ~0 is an

open and invariant subset of f~. Therefore, in view of Corollary 2 of Lemma 8, it would be enough to show that every semisimple element of g~ lies in f~0.

632


Fix a semisimple element HoE~. First assume that H04:o. Let 3 be the centralizer of H o in g. Define q and ~ as in [3(i), w 2]. Then as we have seen during the proof of Theorem 3, there exists an invariant polynomial function p on 5 such that ~ = ( - - I ) q p 2 where q = (dim q) /2. Let I? be a Cartan subalgebra of 5. We identify fl, 3, I) with their respective duals by means of the Killing form of g. Define r:~ and % as in the proof of Theorem 3. Since ~ = ( - - I)qr~q 2 on ~), it follows from [3(c), Theorem I] that

~'3,( ~(p)op) = ~- l O(r~.)or:.

Hence if H E D o [2', we get

F(H; V0)=F(H ; a(rc)orc) = F(H; O(%)o%o~',/~(O(p)op)).

On the other hand define [2~, a r and F~ as before (see the proof of Theorem 3) and put T 3 = p q r. Then by [3(i), Theorem 2 and Lemma I9] , the induction hypothesis is applicable to (5, ~2,, T,) in place of (g, [2, T). On the other hand we have seen during the proof of Theorem 3 that *r ~ F3- Therefore T~ =pFa and so by the induction hypothesis (V~oO(p))(pF~) extends to a continuous function g~ on f2~.

Let E denote the analytic subgroup of G corresponding to 5 and x-->x* the natural projection of G on G*-----G/E. Select open connected neighborhoods G O and 30 of I and H o in G and f2~ respectively and let G O denote the image of G O in G*. Then if G o and 3o are sufficiently small, we can define +, q~ and [2 o as in the proof of Lemma 4. Define a function g on f~0 as follows:

g(~(x*, Z)) =g~(Z) ( ; ~ G o , Z~3o).

Since q~ is an analytic diffeomorphism of G o • 30 with [20, g is obviously continuous. Fix XE[20ng'. We claim that g ( X ) = F ( X ; Vo). Let X-~q0(x*, H) (x*eG0* , H~30 ). Then it is clear that g ( X ) = g ( H ) . Similarly, since V~F is invariant under G, it follows that F (X;Vo)=F(H;V0) . Hence it would be enough to show that g ( H ) = F ( H ; V~). Obviously H is regular in both g and 3. Let I) be the centralizer of H in 3. Then D is a Cartan subalgebra of 3 and H e b n f l ' . Therefore, as we have seen above,

F(H; V~) = F(H; ~(7~)o=aoS;/~(O(p)op)).

Put F~ = (O(p)op)F,. Since F~ is invariant under E and O(p)ope~(3,) , it follows from

[3(i), Lemma I4] that F'~(H') = F~(H'; 8~/~(~(p)op)) (n 'eb 'n[2~) ,

and therefore g,(H) ----- F'~(H ; Vs)=- Fs(H; ~(r~3)or~so3~/~(~(p)op) )

= F ( H ; Vg)

from the definition of V 3. This proves that V ~ F = g on ~0r~g' and therefore H0ef2 ~

So in order to complete the proof o fL em m a 25, we may assume that o~[2. Then, by Lemma io, J g ' c ~ and it follows from Corollary I of Lemma 8 that V~F can be

633

extended to a continuous function g on (1) ~ n % U . Hence it would be sufficient to

prove the following result.

Lemma 26. - - There exists a number c with following property. I f (Xk) (k > ~ ) is a sequence

in f2' which converges to some element X in J ' , then g(Xk)~c .

Define I)i and gi (I<i<_r) as in the proof of L e m m a 24 and c(I)i ) as in w 8.

Then c(Ih) . . . . . c(b~)=c (say) from the results of w 8. Since g' is the union of

gl, . . . , gr we can select, for each k, an index i k and elements xkeG , Hkel)} k such

that X k = x k H k. Since Xk--->X , it is clear (see the proof of [3(J), L e m m a 23]) that Hk--->o. Hence it follows from the definition o fV 9 and c that

g(Hk) = F(Hk; V~) -->c.

But since g-----VqF is invar iant under G, g(Xk)=g(Hk) and therefore g(Xk)->c.

w xo. A D I G R E S S I O N

We shall now apply Theorem 3 to give a new proof of the main result of [3(e)].

We keep to the notat ion of Theorem I. Lemma 27. - - Assume that F is locally constant on fY. Then T is locally constant on fL

Given any point HoEf~ , we have to show that T coincides with a constant

a round H 0. In view of Corollary 2 of L e m m a 8, it would be sufficient to consider the

case when H 0 is semisimple. However we first prove the following lemma.

Lemma 28. - - There exists a number a > o such that

= a

for every Cartan subalgebra I) of g.

Take T----- ~ and f 2 = g in Theorem ~. Then 0(~)g ~ = ~(~)rc ~ in the notat ion

of Theorem 3. But 0(~)r~ ~ is obviously a constant which we denote by a(t)). Since

zero belongs to every Car tan subalgebra D, it follows from Theorem 3 that a(I~) is actually

independent of D. Hence we m a y denote it by a. On the other hand we know (see [3(c), p. I io] ) tha t a(I))>o. This proves the l emma (2).

Let us now re turn to the proof of L e m m a 27. Since F is locally constant on fY,

it follows from L e m m a 28 that VgF-~aF. Therefore we conclude from L e m m a 25

that F can be extended to a continuous function on f2. Since T----TF, this proves

that T is locally constant on ~. Now we know that the distr ibution T ' of [3(d), L e m m a 3 o] is locally constant

on g' (see [3(d), p. 235]). Hence from L e m m a 27, it is a constant. This gives a new

proof of L e m m a 17 of [3(J)].

634

(1) e S denotes the complement of any set S. (2) I t is obviously possible to give a direct proof of L e m m a 98.

w II . P R O O F OF T H E O R E M 4

In order to prove that the irreducible unitary characters of G are actually

functions [3(g), Theorem 1], we have to develop a method of lifting our results from g to G. The remainder of this paper is devoted to this task.

We use the notation of [3(i), Theorem I]. Theorem 4. - - Let f~ be a completely invariant open set in g and T an invariant distribution

on f~. Let D be a diJ~erential operator in ~(g~) such that Dp = o for all peJ(g~). Then DT = o. We proceed by induction on dim g. Let c be the center and gl the derived algebra

o f g and first assume that r Then -~(g,)=~(c,)~(gl~) (see [3(0, w 3]). Hence

D = Y, ~D~ where ~e~(co), DCe,~(g~) and r . . . , ~ are linearly independent l < i < r

over C. Fix a point Xo~f~ and let X o = C 0 + Z o (Coco , Zoega). Define Co and ~a as in

the proof of Lemma 2 5. Since J (g , )=P(c , )J (g lo) , we conclude that

X (~.q)(Dip1) = o i

for all qeP(c~) and p leJ(g~) . that

Fix pleJ(gl~). Then it follows from the above result

(D~pl) ~ ~ -= o

in ~3(g,). Therefore we can conclude (see [3(i), w 3]) that D i p l = o ( i < i < r ) .

Now fix ~eC~~ Then if ~C~( f~ l ) , we have

where Ti(~)=-T(~i*~x ~). Since dim g l < d i m g, Theorem 4 holds for (f21, T~, Di) in place of (~, T, D) by the induction hypothesis. Hence D~T~=o. In view of [3(h),

Lemma 3] this shows that DT = o on Co + f~x and therefore X 0 q~ Supp DT. Since X o was an arbitrary point in ~, this proves that D T = o .

Hence we may now assume that c- -{o} and therefore g is semisimple. Let H o4 = o

be a semisimple element in fL We intend to show that H 0r DT. Let 3 be the centralizer of H 0 in g. Define ~ and 3' as usual (see [3(i), w 2]) and let f~3 be the set of all Zef~n 3 such that [~(Z) I>I~(H0)[/2. Then a, is open and completely

invariant in 3. Take G o = G and g 0 = f ~ in [3(i), Lemma 17] and let *T be the corresponding distribution on f~3" Let ~. be the analytic subgroup of G corresponding

to 3. Then a T is invariant under E (see Corollary I of [3(i), Lemma 17] ). Now it

follows from [3(i), Lemma IO and Corollary 2 of Lemma 2] that D 1 = ~m3~/3(D ) ~-~(3~), if m is a sufficiently large positive integer. Moreover

635

3~ H A R I S H - C H A N D R A

by Corollary 2 of [3(i), Lemma 17]- Fix a Cartan subalgebra I) of 9- Then

83/I)(D1) = ~l, S~:~(D)

from [3(i), Lemma i i ] where ~ is the restriction of ~ on I?. Moreover we know from [3(i), Theorem I] that 8~/~(D)=o. Therefore, by applying [3(i), Theorem I] to gc

instead of g,, we conclude that Dip 1 = o for all pl eJ(3c). Therefore, since dim 3 < dim g,

we conclude from the induction hypothesis that D la T = ~mai~ T = o. Since ~ is nowhere zero on f~a, this implies that apT = o and therefore DT = o around H 0.

In view of the above result, it follows from Corollary I of Lemma 8 that Supp D T c ~ n ~ / . Hence, in order to complete the proof of Theorem 4, we may assume that ~nM/'4=o. But then dg'cf~ from Lemma IO.

Lemma 29. - - We can select a function .)c~C~(g) such that:

I) f is invariant under G;

2) SuppfCf~; 3) f = I on some neighborhood of zero in g;

4) the distribution f T on g is tempered.

Notice that since SuppfCf~, the distribution f T : g - ~ T ( f g ) ( g r is well defined on g. The proof of this lemma is rather long and therefore, in order not to

interrupt our main line of argument, we shall postpone it until later (see w I9). We have seen above that Supp D T c J V . Choose an open neighborhood V of

zero in ~ such that f = I on V. Fix a point Xe~/ ' . Then, b y L e m m a 7, we can choose y e G such that y -~XEV. Now T a n d f T are both invariant distributions which obviously coincide on V. Hence they also coincide on V y. Therefore, in order to show

that D T = o around X, it would be sufficient to verify that D ( f T ) = o around X. This means that in order to complete the proof of Theorem 4, it is enough to prove that D ( f T ) = o. Therefore, replacing T b y f T , we may now assume that T is an invariant and tempered distribution on g. Moreover we know from the above proof that Supp DT cJg'.

Define the space c~(g) as in [3(c), p. 9 I] and for any feC~(g) define its Fourier

transform f by

f (Y)----fgexp((-- I)I/=B(Y, X ) ) f ( X ) d X (Y~g)

where d X is a fixed Euclidean measure on g and B(Y, X) = tr(ad Y ad X) (X, Y~g,) as usual. I f c is any tempered distribution on g, its Fourier transform S is also a tempered

distribution on g given by S ( f ) = a ( f ) (fec~(g)). Since f ~ f is a topological mapping

of ~f(g) onto itself, S = o implies that a = o. As usual we identify g~ with its dual under B and use the notation of [3(i),

Lemma 12]. Then the mapping 0~:A-~(A)*(Ae~(g~)) is an anti-automorphism

of ~)(g~) and therefore 0~ 2 is an automorphism. However it is easy to check that 0~ 2

leaves gc q-0(go) fixed pointwise and therefore it must be the identity. The relation

636

0c~A=A (Ae~)(gc)) implies that A*=(0cA) ̂ . On the other hand (~A)*=A from the definition of ~. Therefore

(Aa) ^ ( f ) ---- e(A*f) = a(((eA)f) ^)

= ~((eA)f) = ( s ( f e ~(g)),

for any tempered distribution ~. This proves that (Aa) ̂ = A ~ . Similarly, since B is

invariant under G, ( f f )^=( f )* for fe~f(g) and xeG. Therefore (d )^=(~ )x .

Thus T is an invariant distribution on g and ( D T ) ^ = f ) T . Fix a Cartan subalgebra I) of g. Then it follows from [3(i), Theorem I] and our hypothesis on D,

t that 3o/~(D)=o. Therefore we conclude from [3(i), Lemma I3] that De~(g~) and

l~p----o for every p e J ( g c ) = I(g~). So the above proof is also applicable to (I), ~') instead ^

of (D, T). Hence S u p p D T c ~ . Now put a = D T and fix an element peI(g~) such that p vanishes at zero. Then

it follows from Lemma 7 that p = o on M r. Hence (see [3(h), Lemma 2I ] ) we can choose an integer m ~ o such that pine = o around zero. Then, if we take f~ = g and

r in Corollary I of Lemma 8, we can conclude that p m a = o . Choose a

finite number of homogeneous elements Pz,. . . ,P~ of positive degrees in I(g~) such that I(g~)=C[pl, ...,pJ. Fix an integer m > o such that p~'*~=o ( I<i<l) . Then, if ~ is the ideal in I(gr generated by p~, . . . , p 2 , it is obvious that dim(I(g~)/~) < m z

and w = o for w ~ . On the other hand, by [3(i), Lemmas 12 and I3] , A--->A (Ae~)(ge))

is an automorphism of~)(g~) of order 4 which maps I(gc) onto O(I(g~)). Therefore ~ is an ideal in O(I(g,)) and ~ = ( w ) ^ = o for w ~ . Hence we conclude from Theorem i,

applied to ~ instead of T, that a is a locally summable function on g. But ~ = D T and therefore, as we have seen above, Supp ~ cdg'. Since JV is of measure zero in g, it follows that ~----o and therefore D T = ~----o. This proves Theorem 4-

w x2. ANALYTIC D I F F E R E N T I A L O P E R A T O R S

For applications we have to generalize Theorem 4 to the case when D is an analytic

differential operator on fL For this we need some preparation. Let E be a vector space over R of finite dimension, f~ a non-empty open subset

of E and ~),(f~ : E) the algebra of all differential operators on ~. Then any such

operator D can be written in the form

D =1 ~ < f ie(p,)

where f e C ~ ( ~ ) ) and pieS(E). For any Xef~,

expression of D at X (see [3(c), p. 90]) so that

D x = ~ f ( X ) 0 (p,).

D x denotes, as usual the local

637


Let ~(~2) be the algebra of all analytic functions on f~. Then ~'(f~) is a subalgebra of C~(f~). We denote by T)~(f~:E) the subalgebra of ~ | generated by ~ ' ( f~)u0(S(E)) . I f f~ is empty, we define ~)~(f~ :E)=~)~(f~ : E)={o} .

Let f~a be an open subset of fL Then we get a homomorphism

j : ~ (f~ : E) -+ ~ o (a~ : E )

as follows. I f D e ~ (~ : E), then j (D) is the restriction of D on f~ . It is clear that j maps ~a(f~ : E) into ~ ( f ~ : E). We say that an element De~oo(f/ : E) is analytic on f~l if j (D) e~a(f~ : E). In particular D is analytic if De~a(f~ : E).

w 13. EXTENSION OF SOME RESULTS TO ANALYTIC DIFFERENTIAL OPERATORS

Now let g be a reductive Lie algebra over R and g~ a non-empty open set in g. I f f~ is invariant, G operates on ~3oo(~ : g) (see [3(h), w 5]). We denote by ~| : g) the subalgebra consisting of all invariant elements and put .~a(f~ : g ) = ~o (f~ : g)n~a(f~ : g).

Fix ~ and define ~ and 3' as in [3(i), w167 2, 7]. Put f ~ = f ~ n 3 ' and for any D E ~ ( ~ : g) define an element A ( D ) s ~ o ( f ~ : 3) as follows. Fix ZEf~ and choose pz~S(gc) such that Dz=0(pz) . Then, corresponding to Corollary i of [3(i), Lemma 2], ez(pz)eS(3c). It follows from Corollary 2 of [3(i), Lemma 2] that there exists a unique element V e ~ ( f ~ : 3 ) such that Vz=0(ez(pz)) for ZEf~. We define A(D)----V. (In case ~ is empty, A ( D ) = o by definition.)

Let 8~I~ denote the mapping D-+A(D) of ~),(f~ : g) into ~3~o(~)~ : 3) (cf. [3(i),

w 4]). Lemma 30. - - 3~1 ~ maps ~ ( f 2 : g) into ~ ( f ~ : 3). Moreover, i f f~ is invariant,

~/~ maps ~ (f~ :g) into ~ o ( f~ : 3). The first statement is obvious from Corollary 2 of [3(i), Lemma 2]. Moreover,

if f~ is invariant, then f~ is invariant in 3 and the second assertion follows from [3(i),

Lemma 3]. Let I? be a Cartan subalgebra of 3. Lemma 3 1 . - 3~(D)=S~(3~/~(D)) for De~)~(f~ : g). The proof of this is the same as that of [3(i), Lemma ~ ] . Lemma 82. - - Let f be a locally invariant C ~ function on an open subset f~o of fL Then

f ( Z ; D) = f ( Z ; 8~(D))

for Zef~0r~ 3' and De~) | This is proved in the same way as Lemma 14 of [3(i)]. Lemma 33. - - Let D be a:differential operator on an open subset t) of g. Then the following

two conditions on D are equivalent.

i) For every Cartan subalgebra b of g, 3g/~(D)=o. ~) I f ~o is an open subset of ~ and f a locally invariant C ~~ function on s then D f = o.

638


Assume i) holds and let f be a locally invariant C ~ function on Q0. Since ~ )0=~0ng ' is dense in f~0, it is enough to verify that D f = o on f~0. Fix Hoefl o and let D be the centralizer of H o in g. Then D is a Cartan subalgebra of g and since f is locally invariant, it follows from Lemma 32 that f ( H o ; D ) = f ( H o ; 8 ~ ( D ) ) = o . This proves that D f = o on Y2 o.

Conversely assume that 2) holds. Fix a Cartan subalgebra b of g and a point H0ef2r~D'. Let A be the Cartan subgroup of G corresponding to D. We now use the notation of the proof of Lemma i. Then q~ defines an analytic diffeormorphism of Go• with Y2 o. Fix ~C~176 and define feC~~ by the relation

f(x*H) = ~(H) (x'eGo* , H~[o). Then it is obvious tha t , f is locally invariant and therefore

o = f ( H ; D ) = f ( H ; ~/~(D))= ~(H; S~/~(D)) (HeDo)

Since ~ was an arbitrary element of C~(D0), this implies that S~/~(D)=o on Do. Hence in particular (3~/~(D))i~o=O. This shows that 2) implies i).

Corollary. - - Assume g~ is invariant. Then either one of the two conditions above is equi-

valent to the following. 3) For every invariant function f in C~176 D f = o . Obviously 2) implies 3). Now assume 3) holds. Fix a Cartan subalgebra D

of g and a point Hoef~nD'. Let D0 be an open neighborhood of H 0 in I ) 'n~ . We assume that Do is relatively compact in I)' and sD0r~[)0=o for s4: I in W a (see w 9 for the definition of Wo). Fix ~0eC~ (D0) and put

[3(H)----, ~wG }o(sH ) (H ~ D).

and the mapping q~ : G * x D ' ~ g is everywhere regular. Put The group W e operates on G* • D' on the right as follows:

(x*, H)s=(x*s, s - IH) (seWa)

Then ~ = ~ (seWa) go = v(G" x W) = (D') a.

in the notation of the proof of Lemma 24. Since no point of G* • D' is left fixed by s if s + I, it follows that q0 defines an analytic diffeomorphism of the quotient manifold (G*• with g~. Now define a function F on G*• by

F(x* : H ) = ~(H) (x*eG*, HeD' ).

Then F(x*s : s - l H ) = ~ ( s - l H ) = ~ ( H ) = F ( x * : H) and therefore F defines a C ~~ f u n c t i o n f on g0" Since t)0 is relatively compact in [~' it follows from [3(J), Lemma 7] that CI(D0 ~ cg~ and therefore we can e x t e n d f t o a C | function on 9 by defining it to be zero outside g0" Then it is clear that f is invariant and therefore D f = o on f~ by 3). But

f ( H ; D)~---f(H; 8~/~(D))= ~o(H; 8s (HeDo)

b e c a u s e f = ~ = ~ 0 o n Do" Since ~owas arbitrary in Cc| this shows that (8~F0(D))Ho=O. Therefore 3) implies I) and the corollary is proved.

639

w x 4. P R O O F OF T H E O R E M 5

We shall now prove the following generalization of Theorem 4- Theorem 5. - - Let ~ be a completely invariant open set in g and T an invariant distribution

on f2. Let D be an analytic and invariant differential operator on f~ such that D f = o for every

invariant C | function f on ~ . �9 Then DT = o. We again use induction on dim g. Define c and gl as in w 4 and fix a semisimple

element HoEf~ such that H06c. We shall prove that D T = o around H0. Let denote the centralizer of H0 in g and define ~ and ~' as in [3 (i), w 2]. Then ~s---- f~ n 3' is an open and completely invariant set in ~. Let c T and eDT be the distributions on f~, corresponding to T and DT respectively under [3(i), Lemma 17] with G 0 = G and 30=f2, . Then it would be enough to show that ,DT=O. However it is easy to prove (of. Corollary 2 of [3(i), Lemma 17] ) that ~DT = A~T where A = ~/~(D). Now ~T is an invariant distribution on f~s (see Corollary I of [3(i), Lemma 17] ) and A ~ a ( f l 3 : 3) by Lemma 3 o. Therefore since dim 3<d im g, it follows by induction hypothesis

that A e T = o (see Lemma 31 and the corollary of Lemma 33). Now fix C0ecr~fL We claim that T = o around C 0. Applying the translation

b y - - C O to the whole problem, we are reduced to the case when C o = o . Let

where Pl, . . . , P r are linearly independent homogeneous elements in S(flc) and al, . . . , a r are analytic functions on fL Let V be the subspace of S(gc) spanned by p~. ( i<_ i<r , xeG). Then obviously dim V < o o and we may, without loss of generality,

assume that (Pl, - - . ,Pr) is a base for V. Then

px = X c#(x)pj (x G)

where cj~ are analytic functions on G. Since D = D x, it follows that D~x----(Dx) * and

therefore

z a,(xX) a(p,) = X a(x) a(pr

This shows that

a,(xX) = 2 (xeG, X e ~ ) .

denote the subspace of ~(gc) spanned by elements of For any integer m, let ~,,, the form pS(q) where p and q are homogeneous elements in P(gc) and S(gc) respectively

and deg p - - d e g q = m. Then if A ~ , ~ and Q i s a homogeneous polynomial function on g, it is clear that A Q is homogeneous and

deg(AQ) = deg Q - b m.

6dO

I N V A R I A N T E I G E N D I S T R I B U T I O N S O N A S E M I S I M P L E LIE A L G E B R A 37

Choose an open and convex neighborhood f2 0 of zero in ~ such that each a~ (i Xoq,~(x ) ( x ~no)

where q,,i is a homogeneous polynomial function on g of degree v. our result above that

q. (xX) = <~< ~,j(x) q~;(X)

and therefore

~D = ~ q~,O(p~)

It is obvious from

(xeG, Xef~o)

lies in ~(g~). On the other h a n d it is clear that ~)(g~) is the direct sum of ~,~ for all rn ( - -oo<m<oo) and each ~3,, is stable under G. Let ~D,, denote the component of ~D in ~,, is this direct sum. Then it is clear that ~D,~e~(g,). Moreover ~D,~+ o implies that v = m + d e g p ~ forsomei . H ence i f rn0=supdegp~ , it follows that ~D,,-----o for , J>m+m0. Put i

D,, = ~>o~Dm.

Then hypothesis

Dme~(gc)n~)m. Moreover if p is a homogeneous element in J(gc), then by

o----Dp= ]~ Dmp

on f~o. Since D,,p is homogeneous of degree m + degp, it is clear that Drop = o. D ~ T = o by Theorem 4-

Now fix feC~(f~o). It is clear that for any peS(gc), the series

v > 0

Therefore

(~<i<r)

converge uniformly on any compact subset of f~0. Hence it follows without difficulty

that the series

Y~ D~,f m ~ - - m 9

(Here the star denotes adjoint, as usual.) This implies converges in C~~ to D*f. that the series

Y,, T ( D : f ) m ~> - - m 0

converges to T(DV). But T ( D U ) = o since DmT=o. Therefore T(D*f)=o. This means that DT-----o on fl 0.

The above proof shows that Supp DT contains no semisimple element of f~.

641

3 8 H A R I S H - C H A N D R A

Hence it follows from Corollary I of Lemma 8 that D T = o. This completes the proof

of Theorem 5. Remark. - - I do not know whether Theorem 5 continues to hold when the condition

of analyticity of D is dropped.

w x 5. S O M E P R E P A R A T I O N F O R T H E P R O O F O F L E M M A 0 9

Let G be a connected semisimple Lie group with a faithful finite-dimensional representation, g its Lie algebra over R, 0 a Cartan involution of g and g = f-}-p the

corresponding Cartan decomposition. Let ct be a maximal abelian subspace of p. We introduce an order in the space of all (real) linear functions on a and denote by Z

the set of all positive roots of (g, a) (see [3 ( f ) , P. 244]). Let a + be the set of all points HEa where ~ ( H ) > o for every 0r Put A = e x p a and A + = e x p ( a +) in G. The exponential mapping from a to A is bijective. We denote its inverse by log. Introduce a partial order in A as follows. Given two elements hi, h2 in A, we write

h i g h 2 if hlhz-lEA +. Let / = d i m a. Then we can choose a simple system of" roots

= a , - . . , =, in s (see [3(d), Lemma i]). Lemma 84. - - Fix some norm v on the finite-dimensional space g. Then for any number

a > o , wecanchoosetwonumbers b,c (b>a, c_.'>_i) withthefollowingproperty. Suppose Xeg, heA + and ,~(X)<a. Then there exist elements Xo~g and hoeA + such that

I) X h = X : ', v(Xo) <~b , I -~ho-~,h; 2) max exp ~(log ho) < c ( i +~(Xh)) z.

l < i < l

(In case l = o , max exp a~(log h0) should be taken to mean I.) We shall give

a proof of this lemma in w 2o. As usual put B(X, Y) ----- tr(ad Xad Y) (X, Yeg) . Then the quadratic form

I I x ll 2 = - - B ( X , 0(X)) (X g)

is positive-definite and defines the structure of a real Hilbert space on g. For any a>o ,

let ~ denote the set of all X e g with [ [X[ l<a and put f~a=(%) ~.

Lemma 35. - - Suppose a > b > o . Then Clf~bcf~ a. We shall give a proof of this in w 2I. Corollary. - - f~a is an open and completely invariant subset of g.

It is obvious that f~ is open and invariant. Fix X0eg~ ~. Then X 0 = Y0 ~~ where

[IYol[<a and x0~G. Choose b such that [IY01]<b<a. Then X0~f~ b and Cl(f~b) Cf~ a by Lemma 35. This shows that every point of f~a has an open invariant neighborhood whose closure (in g) is contained in f~. From this it is clear that ~2~ is completely

invariant. For any linear transformation T in g, let T* denote its adjoint (in the sense of

Hilbert-space theory). Put [1 x [I 2 -= tr (Ad(x)*ad(x)) (x~G).

642

I N V A R I A N T E I G E N D I S T R I B U T I O N S O N A S E M I S I M P L E LIE A L G E B R A 39

Choose a base (Hi, . . . , Hz) for a over R dual t o (~i, " " , eZ) SO that

e~(Hi) = 8~i (i < i, j < l)

and define m ( e ) = Z e ( H i ) for eeE. Since Hiea +, it is clear that m(e) is a positive integer.

Lemma 36. - - Given a>o , we can choose numbers b, c (b > a, c> i ) such that the following condition holds. For any Xe~2,, we can select x e G such that

I) lIX - ll b, 2) Ilxl[ c(i-+-[lXll)

where m = l max m(0~). e E Z

(If l = o then m = o by definition.) Choose b0, c o such that Lemma 34 holds for (b0, Co) instead of (b, c) with the norm (Z)=llZJI (z~g). Let K be the analytic subgroup of G corresponding to f. Then K is compact, G = K A + K and Ad(k) is unitary for keK. Hence if X = k l h k 2 (kl, k2~K, hsA+), it follows that Ilxll=llh[I. However Ad(h) is self-adjoint (t) and its eigenvalues are i and e •176 (e~Z). Since ~(log h ) > o, it is clear that

I I h I I < nl/2 max e ~(l~ h) - - ~EZ

where n = d i m g. But ~ = ~E mie i where mi=e(Hi ) are integers > o . Hence l ( i (

e(log h) <m(~)max e~(log h)<m o max a,(log h)

where m o = m a x m(~). Therefore

[[hlt<nl/2(max e~diogh)) m~

(This holds also if l = o . We define m 0 = o in that case.) Now fix Xef~, and choose Yag, y e 6 such that I ]g l l<a and X=YU. Let

y = k x h k 2 (k~, k~eK, heA+). Replacing (Y, y) by (yk,, k~h), we can assume that y = k l h .

Select Yoeg and ho~a + such that Yh=Y0h~ , ]lYol]__<bo, I-~ho-~h and

m a x eei(l~ Av IIYh]])/. i

This is possible from the definition of bo, c o. Then

11 holl <n'/%~"(~ + llYhll) ~.

Now put x = k x h o. Then X = Y Y = Y g and therefore ]lX~-'t]<b0. Moreover

II x l l=l l hoIl~n'/2Com~ I - t - I lxII) m.

Therefore we can take b = b o and c=nl/2c~ ~ in the lemma.

(1) T he facts stated here are all well known. T h e y can be found in [3(a)].

643

4o H A R I S H - C H A N D R A

w I6. SOME INEQUALITIES

For t ~ I , let G(t) denote the se tofa l l x e G with IIx[l~t. Then G(t) is obviously compact.

Lemma 87. - - Let ~ denote the Haar measure of G. Then there exists a number c > o and an integer M > o such that

~(G(t) ) <ct M

for t > I .

The statement is trivial if G is compact. Hence we may assume that l > I. Put

A+( t )=A+r~G(t ) . Then it is clear that G ( t ) = K A + ( t ) K and therefore (see [3(a), Lemma 22])

~x(G(t)) = f A§ dh

where dh is the (suitably normalized) Haar measure on A,

(h) = e I I z (e ~0~ h)__ e- =0og h))m~ (h s A) D

and m~ is the multiplicity of ~ (m~ is the dimension of the space 9~ consisting of all X e g

such that [H, X ] = ~(H)X for all H in a.) Put 2p = ~2 m~0t. Then it is obvious that " a C E

D (h)<_e 2~('~ h) (h ~ A +).

Now 2p= ~ mi~ ~ where m~ are positive integers. Put -~=~i(logh). Then 1 < i < ~

dh= q dv 1 . . . dv 1 where Cl is a positive constant and

e2p0oeh) = exp(mlv 1 + . . . + mv:z).

Now if heA+(t), we have

h [l<t and therefore o_<z~<log t. Hence

(a (t))_< fA+(,) h) dh<ct

where c=-q / (mlm 2 . . . mz) and M = m l + . . . + m t. Lemma 38. - - There exists a compact neighborhood U of i in G and two constants al, q > o

with the following propert.y. For any t > I , we can choose a finite set of points x i (i < i < N ( t ) ) in G such that

G(t)cy ,v; 2) I]x~ll<axt;

3) N(t)<--Ca tM. By a theorem of Borel [I, Theorem C], there exists a discrete subgroup F of G

such that F \ G is compact. Choose a compact neighborhood U of I in G such that

U-----U - t and G = F U . Put r(t) = P•(G(t)U).

I N V A R I A N T E I G E N D I S T R I B U T I O N S ON A S E M I S I M P L E LIE ALGEBRA 41

Select a compact neighborhood V = V - t of ~ in U such that V 2 n P = { ~ } (V2=VV). Then the union

,LI o-rv = r(t) V Y

is disjoint and

since

r (t) V c G (t) UV c G (t) U 2.

Choose a1>I so large that UZcG(al). Then

r(t) VcG(t)G(a ) cC(ta )

}[xy]J<_[]xl].lly]l (x, yeG) . Hence

~(F (t)V) < ~ (G (tal))<calMt M

from Lemma 37. But since the above union was disjoint,

= N(t) ~(V)

where N(t) is the number of elements in r(t). Hence N ( t ) < q t M where q = c a ~ / ~ ( V ) .

Let x~(I < i < N ( t ) ) be all the elements of P(t). Since P ( t ) c G ( t ) U c G ( t a ~ ) , it follows that [Ix~ll<axt. Finally since G = F U , it is obvious that

G ( t ) r V .

w x 7. P R O O F O F L E M M A 39

Fix a number a > o and let ~ = ~ in the notation of Lemma 35. For o < s < t ,

let f~(s, t) denote the set of all X e n with s<] ]X]J<t . Also put ~2(t)=f~(o, t). Lemma 89. - - Let T be an invariant distribution on g. Then there exist elements Pl , �9 �9 Pr

in S(gc) and an integer , ~ o such that

] T ( f ) ] < (~ + t)" 1 5 < r supl ~(P~)f[

for all feC~~ and t>o. This requires some preparation. As before let co, ( t>o)

X e g with I l x l l< t . Lemma 40. - - Define b, r and m as in Lemma 36 and for any t > o

of all x e G with I lxl l<c(I + t ) m. Then ~]t~f~(t) .

This is obvious from Lemma 36. Define U and M as in Lemma 38 . Lemma 41. - - There exist two numbers q , c2> o with the following property.

we can choose a finite set F t of points in G such that :

I) GtCFtU; 2) rlxll_<cl(i+t) for .veNt; 3) [Ft]--<c~(I+t) me.

Here [Ft] denotes the number o f elements in F t.

be the set of all points

let G t denote the set

For any t>o,

645

6

This follows immediately from Lemma 38 if we note that G t = G ( t ' ) where t' = 4 I +t ) ~.

Now choose bl>b such that Cl(%)vc%~ and fix ~C~(%~) such that o < ~ o , put

xE F t

Since f~(t)cco~tc(~) F, and ~ = I on eo~, it is clear that qh~i on f~(t). Put ~=~x/q~t on ~(t) (xeFt).

Lemma 42. ~ Given pcS(g,) , we can choose a number c(p)>o and aninteger m(p)>o

such that sup[ O(p) e*l<_c(p) (I § t) "(p/

for xeF, and t>o . Let V be the subspace of S(g,) spanned by pv (y~G) and let Pl, " " ",Pr be a base

for V. Then

PV=l h < f~(Y)Pi

where al are analytic functions on G. that (see (1) [3(d), p. 2o3] )

la,(y)l~c'llyll ~ Then

Hence

We can choose c '>o and an integer ,~>o such

o(p)~ ~ = (e(p,-') ~)~ = x ~,(x -1) (o(p,)~),.

(yeG, I<i<r) .

sup l ~(P)~]--<c0 [1 x-t] 1~

where Co-=C'Z sup] 0(P3~I. I f x = k l h k 2 (kl, k~EK, hEa) , it is obvious that t lx l l = I lh l l . i

Moreover 0 is a unitary transformation of g and therefore since 0Ad(h)0-1=Ad(h-1) , it is clear that [ I h l l - - [ I h - l l l = l [ x - l l [ . This shows that Ilx-l[I----[lxll and therefore

sup] o(p)~X[<Co[[xl[~<coq~(i + t ) "~

since ]]x][<q(i+t)" for xeF,. So we can take c(p)=coq ~ and m(p)=mv. Corollary 1. - - sup [ e(p) % [ < c(p)c~(i § t) m(') + mu (t> o). This is obvious since [F~]<c~(i + t ) raM. Corollary 2 . - Given p~S(g~), we can choose c'(p) > o and an integer ~(p) > o such that

sup [ O(p)~l <c'(p)(i + t) ~(vl n(t)

for xeF t and t>o . Since ex= 0d/% and % ~ I on f~(t), this is an immediate consequence of Lemma 42

and Corollary i above.

(1) T h e proof of L e m m a 6 o f [3(d)] is clearly independent of the assumpt ion tha t rank g = rank t~ which was made at the beginning of w 3 of [3(d)].

646

I N V A R I A N T E I G E N D I S T R I B U T I O N S O N A S E M I S I M P L E LIE ALGEBRA

Now we come to the proof of Lemma 39. on g~(t), it is obvious that

and therefore

But T ( L ) = T ( ( L ) ),

Hence

Put f x=~, f (xeF, ) . Since

T ( f ) = Z T ( L ). x ~ F t

since T is invariant. Moreover

Suppf~ c S u p p f n Supp % c Supp ~ .

~r,-1 Supp(f,) CSupp ~ c % .

Since %, is relatively compact in g, we can select Pl, . . . ,p ,eS(g~)

[T(g ) [~ Y, sup[c~(p~)g[ l< /<r

Therefore [T(f,) I = l T((L) =') I < E sup l o(A)f~l[.

43

x@ F t

such that

(geC2 (%,)).

But s u p I o(p,)fF'[ = s u p [ o(pg)L[.

Let V be the subspace of S(g~) spanned by pcu (yEG, I o and an integer v->o such that l a~ ; (y) l_<c011yl l ~ for y e G (see [3(d), p. 3o2]). Then

E sup [O(p,)f: I<CollXll~r z sup l e(q;)f~l. l</<r l</<s

We can obviously select qkj, q'~i in S(gc) (I <k<u, I<j<s) such that

a(qi)([3y)= E a(qki)~.O(qki')y (I <j<s) l < k < u

for any two C ~~ functions [~ and y on g. Then since fx = =xf, we get

~.sup [ a(qi)f.] <~ . sup [ 0(q~j)~l I a(q;;)/I. ,~ n(t)

Therefore

IT(A)I<corI[x[[ ~ Z Z sup [a(qk~)%[.sup [a(q~j)fl. - - l < k o and an integer m3>o such that

I T(f~)}--<c~( I + t) m'z sup l e(qk/)ft k,i

647

4 4 H A R I S H - C H A N D R A

for xeFt, feC~'(f~(t)) and t>o . Since f = ~ F f ~ and [Ft]<c2(I+t) mM ( L e m m a 4 i ) , we conclude that E t

I T ( f ) I <Q(I + t)m, Z sup[ O (q~)fl k, i

where q = c2c 3 and m 4 = m 3 %-mM. Obviously this implies the statement of Lemma 39-

w x8. PROOF OF LEMMA 43

For any a > o define f~a as in Lemma 35. Lemma 43. - - Let T be an invariant distribution on g and fix a number a>o . Then we

can choose Pl, . . . , P r in S(gc) and an integer d > o such that

I T ( f ) [ < Z s u p ( x + l l x l [ ) a l f ( x ; , 9 ( p 3 ) l l a/4 ( teR). Put e k ( t ) = e ( t - - k ) for any integer k and let

-- ~ < k < o o

Fix t0eR and select an integer ko such that lto--kol<i/2. Then eko(t0)=I and therefore ~(to)>I. Moreover tor k unless I t o - - k l < 3 / 4 . Since the closed interval of length 3/2 with t o at its center, can contain at most two integral points, it is clear that

I (d'f~ /dt"), = ,. IX 2 sup ] (dm ~ /dt") l t

for any integer m>o. Therefore 1_<9_<2 everywhere and

sup [(d" ~/dt")l < 2 sup I(d" ~/dtm)[< oo. t t

Put Y~=~k[~. Then it is clear that

sup l(amyjat")l t

is finite and independent of k. We denote it by c,,. Since or if k4=o, it is clear that ~ = i around the origin. Hence

B(s)=Nlsfl~2)(seR) is a C ~ funct ion on R and

sup l(dmB/dsm)[= sup l(d/2tdt)r"~[. s l

Since ~ = I around zero, it is clear that

sup ] t-P ( d~N / dtq) l < oo t

648

I N V A R I A N T E I G E N D I S T R I B U T I O N S O N A S E M I S I M P L E LIE ALGEBRA 45

for two integers p and q (p>o , q > I ) . Hence it follows that

supl(d~B/dsm)]< oo. 8

Similarly if A~(s)=0~k( [ s 11/~) (se R), one sees that A k is a function of class C ~. if k>o , it follows in the same way that

supl(dmAJdsm) l = sup l(d/2tdt)m%l s t > O

5 sup I (d/2 (t -t- k)dt)'~= t 5 c'~ t>O

Moreover

where c~, is a positive number independent of k.

Now put g(X)=f~(llXll)=g([lxll" ), hk(X)= ~(l[Xll)=&([IXll ~) for X e g and k_>o. Since Q.: X-+l[X][ 2 is a quadratic form on g, it is obvious that g and h k are C ~~ functions on g.

Lemma 44. - - Let p be an element in S(gc) of degree <d. Then we can choose a number

cp > o such that

Ig(X; ~ ( P ) ) l < 9 ( I + I l X l I ) a , INKX; o ( p ) ) l < : c p ( I + l l x l D d

for X e g and k > o . One proves by an easy induction on d that there exist polynomial functions

ql (o<_j<d) on g of degrees < d such that

g(X; O(p)) = o<~<dqi(X ) (diB/dsi)s= Ilxll',

hk(X; ~(P)) = 0<~<dqi(X)(diAk/dsi)s= Nxlt'

for X e g and k > o . Our assertion now follows immediately from the facts proved above.

Put g k = h J g ( k>o) . Since g~I, gk is also of class C ~ Corollary. - - We can choose c~ > o such that

Igk(X; O(P))l<_c'p(t+ [IXll) a

for X e g and k > o . This is obvious from Lemma 44 if we take into account the fact that g > I. We now come to the proof o f L e m m a 43- Since 0ck(t)=o for k < o and t > o ,

it follows that )2, gk= Z. Fix feC~(~2~) and put fk=gkf" Then Y~ f k = f . It is clear k>0 k~.O

that if XeSuppgk, then [ [ [XL]--k[<3/4 . Therefore f~-----o if k is large. Hence

T ( f ) = ~ > o T ( ~ ) "

649

46 HARISH-CHANDRA

Define f~(s,t) and f~( t ) (o<s<t) for f~=~a as in the beginning of w 17 . Then SuppfkCf~(k+I). Therefore, by Lemma 39, we can choose P l , . . . , P r in S(g~) and an integer v > o such that

IT(L) l< (2 § sup I ~(p,)A[

for all feC2(f~) and all k_>o. Moreover since [[X[l_>k--~ if XeSuppfk , it follows that

(2 § [ O(p,)fkl<sup( 3 + IIXII) + IA(X; a(#,)) I.

Choose qi~., q~. in S(g0)(I<i<r , I < j < s ) such that

O(p~) (F~F2)= ~. O(q~)Fx. O(q~'~)Fz (i < i< r)

for any two C ~ functions F1, F. on g. Then

Ifk(X; a(p,))]_<~. [gk(X; 0(q~i))I If(X; 0(q,'j)) [

since fk ---- gkf. and a number

Therefore there exist, from tile corollary of Lemma 44, an integer d0_>o c > o such that

IA(X; O(Pi))I.~C( I - t - I lx ! l )~ 'Z l f (X; 0(q,~)) I

for all feCT(f~a) , k>o , Xeg and I ~ ~ (k + 2)-2< oo.

Then it follows that

IT(f) ] <k~>0[ T(fk)[_< Co~ j sup(i § [IX[l)d[ O(q~i)f[

This completes the proof of Lemma 43. for f C2(no).

w x 9. C O M P L E T I O N OF THE P R O O F OF L E M M A 29

As usual we identify g~ with its dual under the Killing form. Call an element peS(g~) real ifp(X) is real for Xeg. Then we can select P l , . . . , P r in I(g~) such that I)p~ is real and homogeneous of degree > I and 2) I (gc )=C[p l , . . . , p r ] . Put

q(X) = f i

Lemma 45.--Wecanchooseanumber 8>0 such that q(X)<~(Xeg) implies that X eO a. Suppose this is false. Then we can choose a sequence Xkeg (k> i) such that

q(Xk)-~o and Xkef~ a. Let Yk and Z k respectively be the semisimple and nilpotent

650

I N V A R I A N T E I G E N D I S T R I B U T I O N S O N A S E M I S I M P L E L I E A L G E B R A 47

components of X k (see w 3). Then YkeCI(X~) from the corollary of Lemma 7. Therefore q(Yk)=q(Xk). Since ~)a is open and invariant and XkCf~, it is clear that Ykq~f2a. Therefore q(Yk)=q(Xk)-+O and Yk~f~a.

Let Ill, �9 �9 -, Ilm be a maximal set of Cartan subalgebras of g, no two of which are conjugate under G. Yk, being semisimple, lies in some Cartan subalgebra of g which must be conjugate to Ill for some j. Hence we can choose xkeG and an index Jk such

Xk , Xk that Yk ~I)/" k By choosing a subsequence we may assume that H k = Y k E[? (k_~ I)

where I) is a fixed Cartan subalgebra of g. Then q(Hk)---- q(Yk) -+o and therefore it is obvious that p ( H k ) ~ o for any psI(g~) which is homogeneous of degree > I. N o w define qi (i_~j_~n) in I(g~) by

de t ( t - - ad X ) = t n + ~.~ qi(X)t n- i (Xeg), 1<.~ <.n

where t is an indeterminate. Then qi is homogeneous of positive degree and therefore qi(H~) -+o. However

act ( t - - ad H) = t*I~> ~ ( t - - a (H)2) (H ~ D)

where l-~ dim D and • runs over all positive roots of (g, D). Therefore 0c(Hk)-+o for every root e and hence Hk~o . But then [[S~l[<a if k is large and therefore Yk = xk-1Hke ~ , giving a contradiction with our earlier result. This proves Lemma 45.

Corollary 1. - - There exists a C ~~ function g on g such that :

i) g is invariant and SuppgCg~,; 2 ) g = I around zero;

3) for any peS(g~), we can choose cp, mp>_o such that

[g(X; 0(p))l< cp( + I[x[I) m,

Select a C ~ function F on R such that i) F ( t ) = F ( - - t ) , 2) F ( t ) = I if [ t1~8 /3

and F ( t ) = o if I t [>8/2 ( te l l ) . Put

g(X) = F(q(X))

I f XESuppg, it is clear that q (X)<8/2 and therefore Xef~a. Moreover if q(X)_<8/3. Fix p~:o in S(gc) and let d = d ~ Then it is clear that

(Xeg).

g(X) =

g(X; O(p) )=o<~i<~ (dJF /dti)t=q(x)pi(X ) (Xeg)

where Pi (o<_j<d) are suitable elements in S(gc). Henceg obviously satisfies condition 3). Corollary 2. - - Let T be an invariant distribution on f~a. Then g2T is a tempered distri-

bution on g. Put T k = gkT (k = I, 2). Then T k is an invariant distribution on g. We now apply

Lemma 43 to T 1. So we can choose an integer d > o and elements pieS(go) (i <i__<r) such that

] T l ( f ) ] < Y~ sup(i + [ [XI ] )e I f (X; 0(A)) ] l < i < r

651

for f~C~~ (n~). Therefore if fEC~(g) , we have

[ T , ( f ) l = [T~(fa)] < ~ sup(I -I-[I Xll) lf (x; ~(P,))I

in such a way that

(i_<i_<,) where f l = g f . Now select pq, q#eS(gc) (I o such that

lT2( f ) l<--cZsup(I+ IlXll)a+ lf(x; $,

for feC~~ This proves that T 2 is tempered.

We can now complete the proof of Lemma 29. Since f~ is an open neighborhood of zero, we can choose a > o such that Xef~ whenever l l X l l <a (X~0). Therefore

f ~ c ~ . Now take f = g 2 where g is defined as in Corollary I of Lemma 45. Then it follows from Corollary 2 above that f T is a tempered distribution on g. This proves

Lemma 29.

w 20. P R O O F OF L E M M A 34

We shall now begin the proof of Lemma 34. Since any two norms on g are equivalent, it is enough to consider the case when v(X)----[Ixll (Xeg) . The case 1----o being trivial, we assume l_> I and use induction. For any (real-valued) linear function k on a, let gx denote the space of all X~g such that [H, X]----X(H)X for all Hea .

We denote by E z the orthogonal projection of g on gx. Then gx-----{o} unless X----o or -t-a for some a~Z. Since ad H is self-adjoint for H e a (see [3(h), Lemma 27]), the spaces gx and g~ (k+ ~) are mutually orthogonal. Therefore if

E+---- 2] E~, E _ = Y, E_~,

it is clear that E+ -k Eo + E_ = I. Let S denote the set {I, 2, . �9 l} and for any subset Q o f S, let y~q denote the set

of all e e Z which are linear combinations of 0c i ( i~Q) . Define nq = ~ z Q g . _ and let gq

be the subalgebra of g generated by nq-}-O(rtq). Then O(gq)=gq and therefore gq ---- ~q + pq where fq = f n gq, pq ---- p n gq.

Lemma 46. - - flq # sem@imple. Let < X , Y > = - - B ( X , 0(Y)) (X, Yeg) denote the scalar product in the Hilbert

space g and, for any linear function ?, on a, l e t H x denote the element in a such that

652

I N V A R I A N T E I G E N D I S T R I B U T I O N S O N A S E M I S I M P L E L I E A L G E B R A 49

< H , Hx>=X(H) for all Hea. We know (see [3(d), Lemma 3]) that if Keg x and IlXl/= I, then [O(X), X]----Hx.

First we claim that gq is reductive in g. Let U be any subspace of g such that [9q, U]CU. Since gq = 0(9q) , ad gq is a self-adjoint family of transformations in 9 [3(h), Lemma 27]. Hence if V is the orthogonal complement of U in g, V is stable under ad gq. This proves our assertion. Therefore g~ = [flq, gq] is semisimple. Now fix x~Y,Q and XEg~ with [[Xl[= I . Then [0(X), X]=S~Egq and therefore [H~, X]=0c(H~)X~g~.

C t i Since a(H~)=][H~]12>o, this proves that g~ gO" However gq is obviously stable under 0 and so we conclude that n q + 0(nq)cg~. But, in view of the definition of gQ, this implies that g~ = gq. This proves that gq is semisimple.

Let FQ denote the orthogonal projection of g on gq. We have seen above that

H~ar~gq for ~eZq. Put aq---- ~ RH~ and let bq denote the orthogonal complement

of aq in a. Then a q = ~ RH~cgq. a~q

Lemma 47. - - aq = angq. Moreover Fq commutes with 0 and Ex for any linear fimction X On l l .

LetHebq. Then a i (H)=<H~r H > = o (ieQ) and therefore ~ ( H ) = o for 0~eZq.

Hence H commutes with rtq + 0(rtq) and therefore also with gq. Since gq is semisimple, it follows that gqnbq={o}. Therefore since aqcgq, it is obvious that a o g q = a q .

Let mQ be the set of all Xsgq such that [H, X]egq for all Hea. Then mq is a subalgebra of gQ which contains nq+0(rtq). Hence mq=gq. Therefore gq is stable under ad H (H~a) and this implies that EzgqCg q for any linear function X on a. This shows that FQ commutes with E z. Similarly since gq is stable under 0, Fq commutes with 0.

Corollary. - - aq is maximal abelian in pq and aq = Fq a. Since go-l-rtq@O(llq) is a subalgebra of g, it must contain gq. Therefore

X = E o X + 2~ E ~ X + 2~ E_~X (Xegq). cc~Zq c~Q

Now suppose X~Oq and it commutes with aq. Then

o = [ H , X ] = Y~ ~(H)E~X-- Z a(H)E_~X (HeaQ)

and therefore ~(H)E• for Heaq and a~Zq. But H~eaq for a~Zq and

~(H~)--[[H~][S>o. Hence E• o (~eZq) and therefore X~-- EoXeg o. This means that Xegonp=a since a is :maximal abelian in p. But then X~angq=aq. This proves that aq is maximal abclian in pq.

Since Fq commutes with 0 and E o and aCp, it is clear that FqaCpqng 0. But

since aq is maximal abelian in pq, Pqt~go----aq. This proves that Fqa =a a- Let lq denote the number of elements in Q. Then dim aq---- lq. Let Gq and AQ

be the analytic subgroups of G corresponding to gq and aq respectively. If Q~e S, Lemma 34 holds for (gq, aQ) instead of (g, a) by the induction hypothesis. Let A~ be the

653

5o H A R I S H - C H A N D R A

set of all h~AQ such that o~i(log h)_> o ( ieQ). Then we obviously have the following result.

Lemma 48. - - Assume that Q + S. Then there exist numbers bq, % > 1 with the following

properties. Suppose Xegq, llXll< i and h e A ~ . Then we can choose Xocg q, h0eA ~ such

that :

I) x h = x 0 h~ IlX011_<4, o<_o:i(logho)<_ei(logh) ( ieQ), 2) max exp(ei(log ho))<%(I + [1X)~ 4. i~q - - -

Let A+(Q) be the set of all h~A + such that aj(log h)----o (jq~Q). For any heA,

define

hq ---- exp(i~Q~i(log h)H~).

Then ~i(log h)=~i(log hq) (i~Q,) and therefore log h-- log hq commutes with gq so that x h = x hQ (X~gq). Moreover if hEA +, it is clear that I-~,hq-.<h and hq~A+(Q).

Corollary. - - Suppose XE%, [IX[]_<~ and heA+(Q). Then we can choose X0~g Q and h0eA+(Q) such that

I) x h = X o h', I lXoll<4, 1..~ho..~h, 2) max exp o~g(log ho)_< Cq(I +[]Xoh"[t) 4. ~<~<~ Put h '=exp (~q~( log h)FqH~). Then h'~A~ and (h ' )q=h from the corollary

of Lemma 47. Hence Xh '= X h. Choose hoaA ~ and X0~gq such that the conditions of Lemma 4 8 hold for (X, h', X0, h'0) in place of (X, h, X0, h0). Then if we put ho = (h0) Q all the conditions of the corollary are fulfilled.

For any iES and ZEg, define

~(i: z ) = max IIE~ZII a ~ E a(gi)* o

and let Q(Z) be the set of all ieS for which ~t(i : Z)_>I. of S, let Z~ denote the complement of Z 0 in I2.

Lemma 49. - - Let Z be an element of g. Then I[E~ZIII for some acE. We have to show that ~aZq(z). such that 0~(Hi)4:o. Then

a(i : Z)> Itg~Zll_>~

Moreover for any subset Q

~(r~qiz/)'. Fix i~S

and therefore ieQ(Z) . Since this holds for every i for which e(H~)+o, it is clear

that ~eEqcz/. Put F ~ = i - -Fq for any subset Q ofS. Fix Xeg and heA + and assume that

Iil11_<I. Put Q 0 = e ( x h) and let s denote the number of elements in Z. Lemma 50. - - [[F~oXa[l<l+S 1/2 for any a e A such that I ~ a . ~ h .

Let X be a linear function on a such that gx4={o}. Then

ExX" = e zO~ a) Ex X.

654

INVARIANT EIGENDISTRIBUTIONS ON A SEMISIMPLE LIE ALGEBRA 5I

Now a~- I and therefore X(log a ) ~ o if X<o. Therefore since F' _ Q~ commutes with E 0 and E , it is obvious that

l[(Eo + E_)F~oXa I[_<ll(E0 + E_)xa l [~ II x 11 < x.

On the other hand a ( loga)<~( logh) (a~Y,) since a-<h. Therefore

from Lemma 49. Since r ~ X ~ o + E ) F ~ X ~

our assertion is now obvious. Lemma 51. - - For any (1) Q < S, select bq and cq corresponding to Lemma 48 and define

b o = I + s t/~ + max b., c o = max %. Q<S ~ q < s

Let X~g, hEA + and suppose that I tXI I<I and Q(Xh)4=S. Then we can choose XoE 9 and ho~A + such that

x) x~=X~o o, x < h o < h , l lXol /<bo; 2) m a x exp ~,(log ho)<_Co(~ + IIX~'ll ~-~

1<~<~

Put Q----O..(Xh). Then X h = F Q X h + F ~ X ~. But since FQ commutes with Ad(h) (Lemma 47), we have

FQX h = (FqX)h= X~ q

where XQ=FQX. Since Q < S , we can apply the corollary of Lemma 48 to (XQ, hQ). Hence we can choose X l e g Q and hoeA+(Q) such that:

I) xhQ-~-X h., l IXl l t_~bq, x -~ho-~hQ;

2) max exp ~( logho)<cQ(i+l lXhol l )Z~. 1<i<1

Then X h = X h. -~ F ~ X h = (X 1 j - F~ Xh') h.

Since I -< h o-4 hQ-< h, it follows that I -< h 2-< h. Put

X 0 = X 1 + FQ X h'.

where h2 =: hh~ ~.

Then

from Lemma 5 o.

Therefore

11 X.ll~ll Kill + I[ F~Xh'll_<bq-? I -t-st'2<_bo Moreover

Xho_-- X~._= FQX h .

H e n c e

II x ~ ~ IlXhll = II xoholl ,

m a x exp ct i (log ho) ~ c 0 ( I q- 11 xh' I1) ~ < Co(I + II Xoh' I I) ~-' I < i < /

and so the lemma is proved.

(1) Q < S means that Qis a subset orS and Q4:S.

655

Put c=2tc0 and b = b o. to verify the following result.

Lemma 5 2 . - Let X~g and hCA + and suppose I [ X [ ] ~ I .

Xor fl and h0~A + such that: ~) X~=Xho ~ tlX0ll__<b, I< :h0 -<h ; 2) ~<,<tmax exp 0:,(log ho)_<c(i -~-llX~.llL

I f Q ( X h ) < S , our s ta tement follows immedia te ly f rom L e m m a 51.

assume that ~z(i : X h ) > I

Let [2 be the set of all a~A + such that I) I ~,a-<h and 2) Obviously [2 is a compac t set containing h. Put

f (a )= 5<,~(i : X")

T h e n in order to prove L e m m a 34, itis obviously enough

Then we can choose

So we may

(i_<i<t).

ot(i : X a ) 2 I/2 (I~i_~</).

( ae~) .

T h e n f i s a continuous function on [2 which must take its m i n i m u m at some point a0~tl.

First suppose a o = i. T h e n i ~ and therefore

~t(i : X ) > I/2

Now fix i~S and choose e~Y~ such that e(H~)4:o and I[E~X][_>I/2.

I[ E+ Xn[t > e ~a~ h) [I E~XI[ > 2 - r e ~(l~ h).

Therefore

( i< i< l ) .

T h e n

max e~# ~ < 2 I I E+ xhl[ < 2 I I Xhll.

Since b > I and c = 2 % 0 > 2 , we can take X 0 = X and ho=h in this case. So now assume that a04= I. T h e n we claim that ~x( i :X a~ I/2 for some i.

For otherwise suppose &(i : X a') > I/2 for every i. Choose j such that ej(log %) 4: o. Put a, = a0(exp(--~Hi)) where ~ is a small positive number . I f r is sufficiently small,

it is clear tha t a~[2 . Hence f(a,)>f(ao). O n the other hand since

it is clear that w(i : X%) < ~(i : X a~

for every i. Moreover e-~(I~J) I/2,

it is obvious that ~ ( j : X % ) < ~ ( j : Xa~

But this implies that f(a,) <f(ao) and so we get a contradiction. Hence ~(i : X a') = I ]2 for some i and therefore Q ( X a ~ But then by L e m m a 51 we can choose Xoeg

and a t~A + such that X~o=X~) ', [[Xo[]<bo, I.<al.<ao and

max exp e, ilog al) <c0(I + tlX;'lf) ' -1. l<i<t

656

INVARIANT EIGENDISTRIBUTIONS ON A SEMISIMPLE LIE ALGEBRA

Now put h0=ha0-1al.

and therefore

Then x h = ( x a ' ) ha`-I = k~=O(Yax'~ha*-l] ---- ~=oYh"

11X~ll 2 II E~X~' I[ exp e(log(hao-1)) Fix ieS. Then since ~( i :X"~ we can select ~eX IIE~X~ Therefore since i-<ao-<h , we have

I I X~ll _> 2- ' exp ai (log (ha o- 1)). On the other hand

e%(l~ I -1-11Xg'll) z-~ =Co(~ + II x~ ~-1.

Therefore since ho=hao-laD we get

e~'(l~ h')~ 2c0 II Xh II (~ + II x~' II)/-L

But since i -<ao~,h , we have (see the proof of Lemma 5 o)

II E+ X~176 < [I E+ X~ll < l[ X~ll and

Therefore

and hence

Since

11 (Eo-r IlXll~ i,

llX~~ I +llX~ll

e=,(a~ h~ tl xh I1(~-r + II Xhll) t.

IlXoll<bo=b, Lemma 51 (and therefore also Lemma 34) is proved.

53

(~eX).

such that 0~(Hi) # 0 and

w 2I. P R O O F OF L E M M A 35

We have still to prove Lemma 35. Fix a>b>o and let x~ and X~ ( i > I ) be two sequences in G and g respectively such that ]lX~l[<b and x~X~ converges to some Y~9. We have to prove that Y~f2 a. Let xi=kih~k ~ (ki, k~EK;h~A+). Replacing (xi, Xi) by (k~h i, k~X~) we may assume that xi=kih i. Moreover by selecting a subsequence we can arrange that ki-+k and X~---~X (k~K, X~g). Then by replacing (xi, Xi, Y) by (k-lxi, X~, k- lY) , we are reduced to the case when k = i. Now

x ? - x~,= ( i - Ad(k; 1))X?. xi hi X/hi~Y. Since X~--~Y and ki-+I, it is clear that ] I X / - - X i [[-+o. Hence

By selecting a subsequence we can obviously arrange that the following condition holds. There exists a subset Q of S such that %.(log h i ) ~ t i ( t i eR ) for j ~ Q and ~i(loghi)-++oo for j C Q ( I < j < l ) as i -+~ . Then it is clear that

hi E _ a X i ~ e-~(l~ hi)E_~Xi--.->o t for cceZq. Put

E = E 0 + ~] (E=+E_=)

657

a n d

T h e n i t is c l e a r t h a t

O n t h e o t h e r h a n d i f

w e h a v e

h = e x p ( i ~ Q t iHj) .

EX/h<+ E X h .

I = E + E + + Z E_~,.

T h e r e f o r e s ince E_~X/hi- -> o ( ~ e E ~ ) , w e c o n c l u d e t h a t

(E + E ~ _ ) x h ' - + Y .

T h e r e f o r e Y = E Y + E + Y a n d E Y = E X h. N o w se lec t H e a

for j E Q a n d 0r for j 6 Q , ( I ~ j < / ) . T h e n ~ ( H ) > o for

A d ( e x p ( - - t H ) ) E ~ Y - + o

as t--->+oo. P u t Yt = ( h e x p t H ) -1 . T h e n

yVt = E X + (E+ Y) Yt--+ E X

as t - + + o o . S ince ] [ E X l l < l ] X [ l < b , i t fo l lows t h a t l l g Y ' l ] < a

l a r g e a n d pos i t i ve . T h e r e f o r e Y e ~ a a n d th is p r o v e s L e m m a 35-

such t h a t %.(H) = o !

~ e E q a n d t he r e fo re

i f t is su f f i c i en t ly

T h e I n s t i t u t e for A d v a n c e d S t u d y ,

P r i n c e t o n , N e w J e r s e y .

REFERENCES

[I] A. BOREL, Compact Clifford-Klein forms of symmetric spaces, Topology, 2 (I963) , IIioi22. [2] N. BOURB~a, Groupes et algObres de Lie, chapitre I er : << Alg~bres de Lie >>, I96O , Hermann, Paris. [3 ] H A R I S H - C H A N D R A :

a) Representations of semisimple Lie groups VI, Arner. 3our. Math., "/8 (I956), 564-628. b) The characters of semisimple Lie groups, Trans. Amer. Math. Soc., 83 (I956), 98-163 . c) Differential operators on a semisimple Lie algebra, Amer. oTour. Math., 79 (I957) , 87-i2o. d) Fourier transforms on a semisimple Lie algebra, I, Arner. Jour. Math., 79 (1957) , I93-257. e) Fourier transforms on a semisimple Lie algebra, II, Amer. oTour. Math., 79 (I957) , 653-686. f ) Spherical functions on semisimple Lie groups, I, Amer. oTour. Math., 80 (x958), 24i-3IO. g) Invariant eigendistributions on semisimple Lie groups, Bull. Amer. Math. Sot., 69 (I963) , II7-123. h) Invariant distributions on Lie algebras, Amer. Jour. Math., 86 (I964) , 27I-3o9 . i) Invariant differential operators and distributions on a semisimple Lie algebra, Amer. aTour. Math., 86 (1964),

534-564. j ) Some results on an invariant integral on a semisimple Lie algebra, Ann. of Math., 80 (I964), 551-593 .

[4] H. WHITNEY, Elementary structure of real algebraic varieties, Ann. of Math., 66 (i957) , 545-556.

Manuscr i t refu le 8 avril 1 9 6 4 .

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Invariant eigendistributions on a semisimple lie algebra

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