On the First Betti Number of Compact Quotient Spaces of Higher-Dimensional Symmetric Spaces

Annals of Mathematics

On the First Betti Number of Compact Quotient Spaces of Higher-Dimensional SymmetricSpacesAuthor(s): Yozo MatsushimaSource: Annals of Mathematics, Second Series, Vol. 75, No. 2 (Mar., 1962), pp. 312-330Published by: Annals of MathematicsStable URL: http://www.jstor.org/stable/1970176 .

Accessed: 20/11/2014 20:54

Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp

.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact [email protected].

.

Annals of Mathematics is collaborating with JSTOR to digitize, preserve and extend access to Annals ofMathematics.

http://www.jstor.org

This content downloaded from 157.89.65.129 on Thu, 20 Nov 2014 20:54:47 PMAll use subject to JSTOR Terms and Conditions

http://www.jstor.org/action/showPublisher?publisherCode=annals

http://www.jstor.org/stable/1970176?origin=JSTOR-pdf

http://www.jstor.org/page/info/about/policies/terms.jsp


ANNALS OF MATHEMATICS

Vol. 75, No. 2, March, 1962 Printed in Japan

ON THE FIRST BETTI NUMBER OF COMPACT QUOTIENT SPACES OF HIGHER-DIMENSIONAL SYMMETRIC SPACES

BY Yozo MATSUSHIMA

(Received July 19, 1961)

Recently several important results about the discrete subgroups of semi-simple Lie groups were obtained by A.Selberg, A.Borel, E.Calabi, E.Vesentini and A.Weil. It was found that, compared with the classical case of fuchsian groups, certain differences appear in the case of higher- dimensional simple Lie groups.

We shall study in this paper the following two problems: A. Let G be a semi-simple Lie groups, all of whose simple factors are

non-compact and of dimension > 3. Let F be a discrete subgroup of G with compact quotient space G/il. Is it always true that the first Betti number of the manifold G/F vanishes?

B. Let M be a simply connected symmetric space, all of whose irreducible factors are non-compact, non-euclidean, and of dimension > 2. Let F be a properly discontinuous group of isometries of M with compact quotient space r\M, and let r' be the commutator group of r. Is it always true that the quotient group r/r' is finite?

To treat these problems, we associate with each simple, non-compact Lie group G a quadratic form Hg(t) on the vector space of the twice contravariant symmetric tensors at the origin of the symmetric space of G, which, as in [3] and [7], is defined by the curvature tensor and also by a certain constant depending on the Lie algebra g of G. We shall prove that A is true if the quadratic form of each simple factor of G is positive definite (Theorem 1). We shall show also that B is true if the quadratic form of the largest connected group of isometries of each irreducible factor of Mis positive definite (Theorem 2). We consider also the case, where, for some irreducible factors, the associated quadratic forms are non-negative, but are not positive definite. We shall show that B is true even in this case, if we impose a certain condition on r (Theorem 3). Thus our Problems A and B are transformed into an algebraic problem. We consider then the case where G is locally isomorphic with the automorphism group of an irreducible bounded symmetric domain. Using a recent result of A. Borel [1], we shall show that the quadratic form of G is always non- negative and that it is positive definite, except in the case where G is locally isomorphic with the automorphism group of the unit open ball in Ctn (m > 1). Combined with Theorem 2, we get the result that B is true

312



QUOTIENT SPACES 313

if M is a bounded symmetric domain, none of whose irreducible factors is analytically equivalent with the unit open ball in Cm (m > 1). Thus, the unit open ball in Cm is exceptional in our result. But it is not yet known, whether or not there exists a discontinuous group F on the unit open ball in Cm with m > 1, for which the quotient group P/t' is infinite. We see also by direct calculation (which is omitted in this paper) that the quadratic form associated with a space of constant negative curvature is non- negative, but it is not positive definite.

1. Let G be a semi-simple Lie group, all of whose simple factors are non-compact. Let g be the Lie algebra of all right invariant vector fields on G. Let K be an analytic subgroup of G whose image in the adjoint group ad G is a maximal compact subgroup of ad G. Let f be the subalgebra of g corresponding to K and m be the orthogonal complement of f in g with respect to the Killing form cp(X, Y) of g. Then

(1.1) g = m + e, [m,m]ce, [e,m] c m, [,fc] c.

We know that the restriction of qp onto m (resp. f) defines a positive (resp. negative) definite bilinear form on m (resp. f). Hence we can choose a base {X1, ** A, X7} of m and a base {X7+1, ..., Xj} of f such that

(1.2) q(X, Xj) = ij, - 1 i,j< r,

(1.3) P(X<, X) = -8" , r + 1 _ a,fS _ n .

In the following, Latin indices i, j, k, ... will range from 1 to r, while Greek indices a, ,S, a, ... will range from r + 1 to n and the indices X, q, , ... from 1 to n. Let

(1.4) [XX, XJ = CELIV X .

By (1.1), among the structure constants c, only the c a,

, Cj, c; can be # 0. We shall write c~ij instead of c~-. From the invariance of the Kill- ing form under the adjoint representation follows that

(1.5) cia= - = =

Moreover, we get from (1.2);

(1.6) r

CwikCk = -sitj

For any X, Ye m, we denote by R(X, Y) the endomorphism of the vector space m defined by

R(X, Y).Z = [[Y, X], Z]

for all Z e m. It can be interpreted as the riemannian curvature tensor



314 YOZ6 MATSUSHIMA

for the riemannian symmetric space of G. We have

R(Xk, X)*X; = j RkX

where

(1.7) RikA= Riik= E- =r+1C~iJC~kh

Then we get from (1.6)

Ri j = Er=,Rik jk = 2s k 2

this means that the Ricci curvature tensor Rij is proportional to the metric tensor gii = Isl

Now let ()A = 1, ... , n) be the right invariant 1-forms on G such that i(X,)rox = soX(X,,) = 18. We have the Maurer-Cartan equations

(1.8) d(o% = I CVxO?a)9 A Wx. 2 /L, X-1 1

2. Let P be a discrete subgroup of G with compact quotient space G/r. Since the vector fields XA. are right invariant, we can project them onto G/P. To simplify the notations, we denote by the same letter XX the vector field on G/P obtained by projecting XA. onto G/P. Analogously we may consider 49 as a 1-form on G/P. Now let f be a C--function on G/r. We have then

(2.1) df = E1X fqOX.

We define the riemannian metric ds2 on G/r by putting

(2.2) ds2 = En= The corresponding volume element dv is given by

dv = oil A *.. AOin.

Let so be a p-form on G/r. We can express it uniquely in the form

= I a., Aa... A * ... AO)xP

where the a~l ..P are skew-symmetric in the indices. If so n= ,x(lO is a 1-form, we get from (1.8) and (2.1):

Id~o)= iE,,LaxIJ(ox A (oil (2.3) 2 - u

a,, = X -g Xxx - SECAJgv



QUOTIENT SPACES 315

Since the vector fields X1, ***, Xn form an orthonormal base of the tangent vector space at each point of G/F, the usual inner product of two p-forms (o, r is given by

p!) Y)) =I - Ex Xzal.. b,,l.,pdv ,

where ax1...xP and b,1...Ap denoting the components of (o and (). Let so be a 1-form and f be a C--function on G/P. Then

(df, oj) = X0(Xxf)gX dv .

Since, for any C--function h, we have (see [7])

(2.4) Gr X dv= 0,

we get (df, oi) = (f, - E X~g.). Therefore,

(2.5) 8(0= En Xxg

where 8 denotes the operator of codifferentiation.

3. Now let so =n ,g(X be a harmonic 1-form on the riemannian manifold G/F. Since we have do = 0 and 8o = 0, we get from (2.3) and (2.5):

(3.1) Xxg. - X"gx = E= cA9V

(3.2) = 0

Denoting by 0(X) the operator of Lie derivation, let

(3.3) (XA)o) = 1a

Since O(XA) = d i(XA) + i(XA) d and do = 0, we have O(X)Go = d(i(Xx)o)) = dgx =ES(Xtgx)o *

Hence (3.4) if,, X = Xxxg (1 _ X, ,u n)X

We can write (3.1) in the form

(3.5) - f = pgvv

Further, Xxf99 - Xsf~w = [Xx, X'jga = n Thus

(3.6) Xxf,,>- Xjx1f =P.1

Now we see easily that the Lie derivatives O(X,,,)gA,, of the metric tensor g,,, are all equal to zero (r + 1 ? a ? n). This means that the vector fields X,,, are the Killing vector fields (= infinitesimal motions) of the riemannian manifold G/F. We know that the Lie derivative of a harmonic



316 YOZo MATSUSHIMA

form with respect to a Killing vector field vanishes (see [5]). Therefore O(X<I5)Go = 0 and this implies f X. = = 0 (1 < X < n, r + 1 < a ? n). Hence the g, are constant.

LEMMA 1. g =0 (r + 1 < a < n)

fij Li (1 < ,i-j< r) . PROOF. By (3.5), fij -fji'= , r+1casijga and hence fij fj is constant.

By (2.4), we get:

T cWigma* v(Gfr) = (fi - fji)dv = 0,

v(G/r) denoting the total volume of G/F. This implies that Yclijgl5 = 0 and hence fij = fji (1 _ i , j < r). Let Y= EgXf. Then, [Y, Xj =

TjT,5 C~kig.)Xk = 0 (1 ? i < r). We know that the tangential representation X - admX of f in m defined by admX(Z) = [X, Z] (X e I, Z e m) is faithful. Therefore we have Y= 0 and hence g 0 = 0.

Now from (3.6) we obtain:

(3.7) Xifjk Xgfik = En=r+1C&Qijfck , (1 _ i, j, k < r).

Since fj-Xgl = . 0, we get from (3.5):

(3.8) f= =1Cg, (1 <jr, r + 1 a < n) . From (3.7) and (3.8) follows that

(3.9) Xjfjk - Xjfik = , RijkhgA

Now, let

(3.10) = 2i,,,k=1(XfJk

- Xgfik)2

LEMMA 2*.

)dv j \

Rkh= j

Rijkhffhfjkdv

PROOF. By (3.9)

2 ijk h(Xifjk - Xfk)Rijkhgh 2

By () wRi jkh(Xifejk)h

BY (2.4) We get * The idea of evaluating the integral of the function 'D comes from E. Calabi (cf. [7]); I

became aware of it in reading the manuscript of A. Weil [7]. I should like to express here my thanks to Professor A. Weil, who permitted me to read his manuscript before publication.



QUOTIENT SPACES 317

YGl dv = i, JkIhFRikfthffjkdv

LEMMA 3.

EklXkfk=( = -kg, (1 ?< i < r) 2

PROOF. By (3.2), (3.7) and (3.8) we have

, J = A kl lXk fk XJ fkk + C=r+1Cvkifack

Er = 1 Lk,h ,Wl=r+1Cwki a51k9A

= k=18ihgh 2g 2 2

LEMMA 4.

8 g2dv 2 2 f fdv1 < i < r).

PROOF. By Lemma 3, g2 =-2 r=(Xjfj)g. Hence by (2.4)

gidv = 2 15 fii.Xjgidv = 2 fG dv.

4. In the following we shall evaluate the integral of 1 in a different way by using Lemma 4. We get from (3.7) and (3.8)

X jfk - XJfik = Toah CajCahkgh

Hence

(4.1) 2'I = kEWh I CwjjCwhkC,9ijC,1k hgl

= Eas .khI(i, jCai;C~ii)Ca~hkCj3kghgI

Let

*(X, Y) = tr (admX. adm Y), (X, Y r),

where X - admX denotes the tangential representation of e. Then

*(Xa', X,) = ,'CmjCqi

= T.1Caijcji= -T,.1CaijCqij

Thus we get

(4.2) *(XI, X0) = -z 1Jc=Caijcij From (4.1) and (4.2) follows:

2F = -Ek h lEa,,(Xa9 Xj)CwhkCj91kgh9.

Now, let + = A + t1 + f+ fq 9



318 YOzO MATSUSHIMA

where 3 denotes the center of f and fl, I.., e% denote the simple ideals of f. We see easily that 8, l, * * *, Cq are orthogonal to each other with respect to the Killing form qp of g. Therefore we can choose a base {X7+1, **, X1} of f such that p(X,,,, X,) = - 1 and that each X,,. belongs to i or to a simple ideal C8. This being done, we show that +(X,,, X,) = 0 for a # S. For this purpose, let X be the Killing form of the Lie algebra C. Then p(X, Y) = *(X, Y) + x(X, Y), X, Ye C. Since p(X,,, Xq) = 0, we get +(X?, X,3) = -X(XX, X,) for a # R. If one of X~. and X,3 belong to 3, or if X,. and X,3 belong to different simple ideals of C, we have X(X., XS) = 0 and hence J(XXI5, X,) = 0. Hence, let Xl5, X,3 e Cs. Let XZ be the Killing form of f.. Then xs is equal to the restriction of X onto f.. Since t8 is simple, a bilinear form on t8 which is invariant under the adjoint representation of C8 is a scalar multiple of X8. Hence there exists a real number c # 0 such that cp(X, Y) = cZ8(X, Y) = cZ(X, Y) for any X, Y e t8. Since Xl5, X,, e f, cX(Xe,,, X,) = p(X&, X) = 0 and hence #(X,, Xq) = 0.

We get then

24) = Sk h I 1aE1(X"9 X")cfhkCa5kg9gA.

Now, if Xl. e j, we have x(X,,, X5) = 0 and hence J(XXI5 X"') = p(XXI5 X"')= -1. The restriction of f and q onto r8 are negative definite bilinear forms

on r8 which are invariant under the adjoint representation of C8. There- fore, there exists a positive real number a8 such that

*(X, Y) = a8,9(X, Y), X, Ye C8.

Then

*(X, XI S) = -a, X""ser8.

Thus we get finally

21 =EX e3Ekh l1CahkCgkg9Agg

(4.3) + :=a Ca~xae9 9 + TS=1 ZxefaS~ E 7c,lc^clgg h,

5. In this section, we suppose that g is simple and that the center 3 of f is # (0). Then we know that dim j = 1. We can suppose j = {Xr~i}. Let

a.= Min (a, *.. , a8)

Then, since 1h,^,CakCakgh9Z

= E7(Ch

) a O. we get from (1.6) and (4.3)



QUOTIENT SPACES 319

21D ?> k h l Cr+1 hk Cr+l7kggl + aEa,>r+lEk h, 1 CcAkCacgk9gg1

= a = ++1 khl Cac1k Cackg9Ag9 + (1 a)k ,hI Cr+1 k Cr+1jkg,'g

-

h + (1 a)- (ZACr+1 k h

here the equality holds if the derived algebra [f, f] is simple. Now, since g is simple, we know that the tangential representation of

f is irreducible. Hence, for each X e I, admX has only two purely imagi- nary eigenvalues ri and -ri, r being real. Therefore we can find X0 E 3 such that (ad. X0)2 = -1, 1 denoting the identity endomorphism of m. Putting J= ad. X0, we have p(J. X, Y) +p9(X, J. Y) = 0 for any X, Y e m. Replacing Y by J. Y, we find that 9(J. X, J. Y) = p(X, Y), which means that J is an orthogonal transformation of m. Then we can find an orthonormal base X1, ** , Xr of m by which ad, X0 is represented by the matrix of the form

0 1 0

-1 0

0 1 0

-1 0

Then J(X0, X0) = p(X, X0) =-r. Since dim j = 1, Xr+j is a scalar multiple of X0 and since I(Xr+i, Xr+i) -1, we find that Xr+i=(1/1Vr)XO. Therefore the matrix (Cr+1 Ak)1,k=l...r is the form

0 1 0

-1 0 1

0 0 1 -1 0

Then

hCr+1 Ak9) = (1/r)Y = g.

Thus we get

(5.1) 2the e a + - (1 a)m = (ghi

here the equality holds, if [f, f] is simple. If f = 8 (this is the case if and



320 YOZO MATSUSHIMA

only if dim g = 3), we get directly from (4.3) and (1.6)

(5.2) 2(D = - lrg2 (dim g = 3) .

6. If g is simple and 3 # (0), the symmetric space of G is an irreducible symmetric bounded domain. From the classification of such domains, it is known that, except in the case of classical domain of type 1m , m' > 2), f has only one simple ideal. Suppose first that f has only one simple ideal f1 = [f, f] and we show that

(6.1) a dim f - 1 d 2 i)

In fact, a = a, = *-(X,, X,) for any X e f1 and *(Xr+1, Xr+1) = -1. Then E*(Xb, X,) = -1 +>E,>r+l(XW, X) =-1-a(dim f- 1). On the other hand, we get from (4.2) and (1.6):

E (X X) =(cij)2 =- = dim m. 2 2

From these (6.1) follows. Suppose now that the corresponding irreducible symmetric domain is of

type 1m, (m, m' > 2). In this case, the Lie algebra g is isomorphic with the Lie algebra of all (m + m') x (m + m') complex matrices of trace 0 which leave the hermitian form

XlXl + * +XmXm - Xm+iXm+l * - Xm+mlXm+mi

invariant. Then [f, f] is isomorphic with the Lie algebra Aou(m) + Aou(m'), iu(m) (resp. Aou(m')) denoting the Lie algebra of the special unitary group SU(m) (resp. SU(m')). By an easy direct calculation, we see that

a - Min (m, m') m + m'

Put

b(g) = -a + 1 (1- a), r = dimm. 2 r

Then we can state:

LEMMA 6. If g is simple and if the center of f is different from (0), we have:

(6.2) iF _ -b(g)E h=12



QUOTIENT SPACES 321

where

2 1 if dim g = 3; 2

b(g) = (dim m-2)2 + 4(dim f - 1) if [f, f] is simple 4 dim m(dim f - 1)

2mn(2 + 1 if [f, f] -u(m) + Bu(m'), m _ m' 2M'(m + m')

in (6.2) the equality holds, if dim g = 3 or if [f, f] is simple.

7. We consider now the case, where g is a simple Lie algebra for which the center of f reduces to (0). We shall use the notation introduced in ?4. We have then

4 =1 as AxlE , CmhkCaZkghgZ

Let a = Min (a*, aq)

We get then

2n a3 a3ccggg= 24) - a1:0r+1Ek ~hZfihkgz 2 Ehlh

If X,,. e f8, we have *(X., XW) = -as, and hence

:r(XWX) = Eq= a,dimf8.

On the other hand, we have already shown in ? 6 that X, X) = - 1 dim m. In particular, if f is simple, we get a = (dim m/2 dim f). Thus we can state:

LEMMA 7. Let g be a simple Lie algebra for which the center of f reduces to (0). Then

(7.1) 4) > 2-b(g)j: =1 gh

where 2b(g) = Min (a1, *** aq). If f is simple, we have

b(g) dim m 4 dim f

and the equality holds in (7.1).

8. We define for each simple non-compact Lie algebra g a quadratic form Hg(t) on the linear space of twice contravariant symmetric tensors e constructed over the vector space m by putting



322 YOZO MATSUSHIMA

(8.1) Ug(4) = b(g)Efr(, ) + Aik X

THEOREM 1. Let G be a semi-simple Lie group, all of whose simple factors are non-compact. Suppose that, for each simple factor Gi of G, the quadratic form Hgjf) is positive definite. Let P be a discrete subgroup of G with compact quotient space G/P. Then the first Betti number of G/F is equal to 0.

PROOF. Let Go be a harmonic 1-form on the riemannian manifold G/P and let, as in the preceding sections, o= ng(X*. Let g = g1 +* +gP be the decomposition of the Lie algebra g into simple factors. Then, m= ml + * * + mP, mi = m n gi, and f = f1+ * * + fpqfi=f n g. If we divide up the ranges of the indices X, a, i accordingly, the c", are 0 unless all three indices belong to one and the same simple factor. We obtain from (4.3) and from the considerations in sections 6 and 7,

( >L=1 2b(gi)Exh e2ig

Integrating over G/F, we get by Lemma 4:

/rv Et=1b(gi) Exktat, G fkhdv + 5r Qdv where

Q = ?=l(b(j) + b(gj))Ex Exk eMjA -

On the other hand, we have by Lemma 2:

G~ ( dv = _f~jl,k,&=1Gir Rik7&fiifk7&dv

Here the Rik,,j are 0 unless all four indices belong to one and the same simple factor. Hence we obtain

?- I = Gir Hgl((f% k))dv + GI Qdv .

Since Hg,((fllk)) ? 0 by the assumption of the theorem, and since Q ?0 by the definition of Q, we must have

Hg,((f1k)) = 0 and Q = 0. But we

have assumed that HgjI) is positive definite. Then Hg,((fnk)) =0 (1 < i < p) implies that fhk =? for the indices h and k belonging to one and the same simple factor. On the other hand, Q = 0 implies fhk = 0 for the indices h and k belonging to the different simple factors. Thus fhk = 0 for any indices h, k. Then, by Lemma 4, we get gi = 0 (1 < i < r). On the other hand, we have already proved that ga. = 0 (r + 1 ? a ? n) (see Lemma 1). Thus a harmonic 1-form w on G/P always vanishes. Therefore the first



QUOTIENT SPACES 323

Betti number of G/P is equal to 0.

COROLLARY OF THEOREM 1. The notation and the assumptions being as in Theorem 1, let F' be the commutator group of F. Then the quotient group F/F' is finite.

In fact, let G be the universal covering group of G and let p be the natural projection of G onto G. Let P=p=-(F). Then P is a discrete subgroup of G with compact quotient G/F. Clearly the manifolds GuI and GIl are homeomorphic. By Theorem 1, the integral homology group H1(G/P, Z) is a finite group. Since P is isomorphic with the fundamental group of G/P, F/F' is isomorphic with H1(G/P, Z) and hence finite, F' denoting the commutator group of F. Let p-'(F') = A. Then F/' F-/A. Clearly A contains F' and hence FaiA is finite and so is P/I'.

9. THEOREM 2. Let M be a simply connected symmetric space, all of whose irreducible factors are non-compact and non-euclidean. Suppose that the quadratic form Hg(e) of the largest connected group of isometries of each irreducible factor of M is positive definite. Let F be a properly discontinuous group of isometries of M with compact quotient space J'\M and let F' be the commutator group of F. Then the quotient group F/F' is finite.

PROOF. Let I(M) be the group of isometries of M with compact open topology, and let G be the identity component of I(M). Then G is a semi- simple Lie group satisfying the conditions in Theorem 1. Let F. = F f G. Then F, is a normal subgroup of F and the quotient group F/FO is isomorphic with the image of F in I(M)/G. It is known that I(M)/G is a finite group. Hence F/FO is finite. This implies that FO\M is also compact. Since F0 is properly discontinuous on M, F0 is a discrete subgroup of G and we see easily that G/FO is compact. Then FO/F' is finite by Corollary of Theorem 1. Then F/F' is also finite and, since F' contains F', F/F' is finite.

COROLLARY. Let the notation and assumptions be as in Theorem 2. Suppose that F contains no element of finite order different from the identity. Then the first Betti number of the compact manifold F\M is equal to 0.

In fact, in this case F\M is a compact manifold and, since M is simply connected, F is isomorphic with the fundamental group of F\M. Then the first Betti number of F\M is equal to 0 by Theorem 2.

10. THEOREM 3. Let G1, *.., ,Gp be simple, non-compact Lie groups. Suppose that the quadratic forms Hg() (1 ? i ? s) are non-negative and are not positive definite and that Hg,() (s + 1? i ? p) are positive definite. Let G = G x ... x Gp and let F be a discrete subgroup of G with



324 YOZO MATSUSHIMA

compact quotient space G/I. Suppose that, for each subset {il, *..., it} (il < ... <it) of the set of indices {1, * * *, s}, the image of F in Gil x * * * x Gi, under the projection G Gil x ... x Git is not discrete. Then, the first Betti number of GIF is equal to 0. Moreover, the quotient group P/F' is finite, where F' denotes the commutator group of F.

PROOF. We shall use the notation introduced in sections 1-8. Let o be a harmonic 1-form on G/F. Just as in the proof of Theorem 1, we see that

0-?i lrHgi((fk))dv + 5 Qdv .

Since Hg,((fhk)) ? 0 and Q _ 0, we must have Hg,((fgh)) = 0 and Q = 0. Q = 0 implies that X11gk = fnk = 0, if the indices h and k belong to different simple factors of g. Since Hg(e) (s + 1 ? i < p) are positive definite, Hg,((fnk)) = 0 implies that Xhgk = fAk = 0, if the indices h and k belong to one and the same simple factor gi with i > s. Hence, if the index k belongs to gi with i > s, then flk = 0 for any h. By Lemma 4, we obtain gk = 0. Since O(Xk)) = E~fxkoX, we have O(Xk) = 0. This means that the 1-form co is invariant by the left translations by elements of G8+1 x * x GP. Let wr be the projection of G onto G/F and let ') = ;ds. Then rj is a closed 1-form on G which is left invariant by G8+1 x .. x GP and right invariant by F. Moreover, we have gk = 0 for any index k belonging to the simple factor g, with i > s. Hence we can identify ') with a closed 1-form on the group G1 x ... x G8. Let A be the image of F in G1 x ... x G8 under the projection G1 x * * * x G. x * * * x GP )G1 x * x G,. Then ') is right invariant by A. Let A be the closure of A and let A0 be the identity component of the closed subgroup A of G1 x ... x G,. For any 8 e A, we have 8A&8-1 = A0. Since A0 c G1 x * * * x G,, this imples that rrA<rh = A. for any y e F. Then, by a theorem of Borel [2], A0 is a normal subgroup of G1 x ... x G,. We may suppose that A0 = G1 x ... x Gt. Suppose that t < s. Then A = G1 x ... x Gt x A1, where A1 is a discrete subgroup of Gt+1 x ... x G,. Clearly the image of F in Gt+l x ... x G, under the projection G -- Gt+l x ... x G, is contained in Al. Since A1 is discrete, the image of F is also discrete, contradicting our assumption. Hence t = s, and this shows that A is everywhere dense in G1 x ... x G,. Since e) is right invariant by A and A is everywhere dense, A2 is a right invariant form on G1 x ... x G,. Let Y1, . . ., Ym be a base of the Lie algebra g' of all right invariant vector fields on G, x ... x G,. Then, since e2 is right invariant, 7)(Yi) are constant. Moreover, we have 2(d7))( Yi, Yj) = Yi(7(Y)) - Yj(7(YJ)) - 7)([IYi, Yj]). This implies that ()([ IY, Yj]) = 0 1 < i, j < m. Since g' is semi-simple, we have [g', g'] = g'. Hence 7( Y) = 0 for any Ye g', and this implies e = 0; and therefore (o=0. Then the first



QUOTIENT SPACES 325

Betti number of G/r vanishes. We can prove then that r/F' is finite by the arguments used in the proof of Corollary of Theorem 1.

Let M be a simply connected symmetric space, all of whose irreducible factors are non-compact and non-euclidean and let

M= Ml x ...x Mp

be the decomposition of M into the product of irreducible factors. Let G (resp. GJ be the identity component of the group of isometries of M(resp. Mi). Then we have G = G1 x * x G,.

THEOREM 4. The notation being as above, suppose that the quadratic forms Hgj()(1 < i < s) are non-negative and are not positive definite and that Hg,()(s+ 1 < i ? p) are positive definite. Let F be a properly discontinuous group of isometries of M with compact quotient space F\M, and let F' be the commutator group F. Let Fo = F n G. Suppose that, for each subset {il, * *., it} (il< ... < it) of the set of indices {1,* - , s} the image of F, in G,, x ... x G,, under the projection G =G x ... x Gp Gi x *... x Gi, is not discrete. Then the quotient group F/F' is finite.

Theorem 4 may be deduced easily from Theorem 3, hence the proof is omitted.

11. We shall prove here the following theorem.

THEOREM 5. Let G be a simple non-compact Lie group, with center reduced to {e}, and K a maximal compact subgroup of G. Suppose that the center of K is not finite. Then the quadratic form Hg(4) is non-negative. Moreover, except in the case where G is locally isomorphic with the Lie group of all (m + 1) x (m + 1) complex unimodular matrices which leave invariant the hermitian form

xll + * * * + Xmm - xm+,m,+l (m > 1) the quadratic form Hg(4) is positive definite.

Let {X1, * * *, X7} be an arbitrary base of the vector space m. Let gij = q(X,, Xj) (1 < i , j < r) and let (gii) be the inverse matrix of (gj). We define a linear transformation P of the vector space m m of twice contravariant tensors by putting

P(X, 0 X3) = Sk hl R kijh(Xk (0 XJ)

where Rkjh=ih -M. If e = > t"Xi 0 Xj, the components of the

tensor P(4) are given by

P(shkh = Eg tR kijiti

Using the well-known identities



326 YOZO MATSUSHIMA

Rijkh= RkAtj = Rjik A RiAk

we see easily that Rkh - hk

We see from this identity that, if t is a symmetric tensor, so is P(4). Hence P induces a linear transformation P of the vector space m m of twice contravariant symmetric tensors. Let Xi *Xj = Xi 0 Xj + X3 0 Xl. Then

P(Xi * Xj) = E2 h R iijXk Xh .

Now we define as usual the inner product (t, I) of two tensors e and e by putting

(t, a) = Ad ?e??j = Ad, jgi~gjtt{?St

s.t

Then we see easily that

(P(0),r ) = (ISP(j)), : teMm .

Thus P is symmetric. The quadratic form Hg(4) is written as follows:

Hg(t) = b(g)(&, t) + (P(4), t) e

The quadratic form Hg(e) is non-negative (resp. positive definite) if and only if b(g) is > (resp. >) than the absolute value of the minimal eigen- value of the symmetric linear transformation P.

G will denote here a simple non-compact, connected Lie group, with center reduced to {e}. Let

g = m +, [my ] c m, [m, m] c A,

as in section 1. Let gc and tc be the complexifications of g and f and Gc be the complex Lie group of Lie algebra gc and with center reduced to {e}. Let p = V-1 m. Then g, = ip + f is a real subalgebra of gc such that 9C = gC. The real simple Lie algebra g& is a compact real form of gc. Let G,, be the analytic subgroup of G6 corresponding to go. Then GJIK is a compact irreducible symmetric space. Let 0 be the isomorphism of the vector space m onto the vector space p defined by 0(X) = V-i X. We see that R(0(X), 0(Y))0(Z) = -0(R(X, Y).Z), (X, Y, Z e m). Thus the curvature tensors of G,/IK and GIK are opposite in sign. Moreover, if we denote by qAd the Killing form of the Lie algebra go, we have q'(X, Y) = -q'p(0(X), 0( Y)). Therefore, to find the minimal eigen-value of the linear transformation P, it is sufficient to find the maximal eigen-value of the corresponding linear transformation defined by the curvature of the compact irreducible symmetric space G,/IK.



QUOTIENT SPACES 327

From now on, G will be a simple compact connected Lie group, with center reduced to {e}, and K the identity component of the fixed point set of an involutive automorphism a of G. Let g = m + f, where m = {X e g I v(X) = - X}. Then, [m, m] c f and [m, fl c m. We shall asume that the center of K is not finite, Then it is one-dimensional. Let tj be a Cartan subalgebra of f. It is known that f is also a Cartan subalgebra of g. Let Y. be the set of all non-zero roots of gc with respect to the Cartan subalgebra bc. For each root a e Y., we can choose an element E,,, e gc with the following properties:

[H, En] = a(H)E, (He tc);

q'(E., E_.) = 1, [E., E-.] = H

where Ha. denotes the element of tc for which qp(H, Ha,) = a(H) for any He tc;

N., s N, _ ,

where N, , (a, f e X, a + f 8 0) is defined by

[Ea,, E] = Na,,pEa+p .

The subspace mc of qc is spanned by the En, such that E,,, to [1]. We call such roots a complementary roots. Now, we extend the linear transformation

P(Xi * Xi) = 5fkl hsgRXijSXk 1Xh

((gAs) being the inverse matrix of the matrix (gij) with glj = -q(Xi, Xj)) complex linearly to a linear transformation of mc mc. Choosing a suitable ordering of the roots as in [1], let T be the set of positive complementary roots and let

n+ = UEaYCE n = AdeCE-a'*

Then mic = + n- and the vector space mc. mc is identified with the direct

sum of the vector spaces n+tt+, t+ 3 ?i- and n- it-. Put

- a = a .

Then, for any a, I8 e P, g,, = - p(E,,, E,) = 0, Gu = -9(E-,,, E_) = 0 and go-= -q'(E., E-,3) Moreover, the components of the curvature tensor with respect to the base {E,,, E-,,,, a e P} satisfies the relations:

R- = Ri = 0 (a,ePi, jePU-T) .

Since gal = ga- = 0. g'a = 8,p, we have Ra = Rs= R . It follows then that the linear transformation P leaves invariant the subspaces



328 YOZO MATSUSHIMA

u+.n+Y, nr- -n and ut+ 0 n-. Let Q, Q and S be the linear transformations of these vector spaces induced by P. Since the eigen-values of P are real, and since the coefficients of the matrices (Ra$^8) and (Ra-_' ) of Q and Q are conjugate complex, it is sufficient for our purpose to consider only the eigen-values of Q and S. The linear transformation Q is defined by

Q(Ew E3) = ,y8 c Ry8Ey *Es,

and S is defined by

S(Ew (0 E) = R- 0 EEy .

Following the method of Borel [1], we find that the eigen-values of S are all non-positive, while the maximal eigen-value of Q is equal to (a1, a1) where a, is a root of maximal length. Thus we find that the maximal eigen-value of P is equal to Max,, (a, a).

Returning to the non-compact case, we find then that the minimal eigen-value Xi of the original P is given by

= - Max~,(a, a) .

The calculation of the values Xi and b(g) can be carried out for each type of a, as in the following table. (We remark that, under the assumption of Theorem 5, the symmetric space of G is an irreducible bounded symmetric domain. As for the classification of the irreducible symmetric domains, see Calabi-Vesentini [3]; we follow the notation in [3].) We find

Type of the bounded b(g) 21 b(g) + 2, domain of G

Is1 mrn'2+1 1 (m'-1)2 (m ml' > 1) 2m'(m + m') mr+m' 2m'(m + m')

IIr 1 + 3r-nm 1 1 _ r3m-5 (m _ 3) 4 4(m-1)2 2(mr-1) 4 4(m-1)2

IIm 1 3 + m 1 1 3m + 1 (m _ 2) 4 4(m +1)2 m + 1 4 4(m+ 1)2

IVm _ 1_ 1 1 1 (m ? 3) 2r r2 12 2m mn

V ~~~3 1 5 V ~~~~~16 12 48

VI 29 1 10 VI l ~~~162 18 | 1

that b(g) + Xi is always non-negative and that it is equal to 0 in the case of Type I.,, and II3. But we know that II3 I3.1. Thus b(g) + X1 0 only in the case of Type I.Jm(m _ 1). Therefore, the quadratic form Hg(t) is



QUOTIENT SPACES 329

always non-negative and it is positive definite, except in the case of Type LA. Theorem 5 is thus proved.

12. In the exceptional case stated in Theorem 5, the symmetric space of G is the irreducible symmetric domain of type Im, which is analytically equivalent to the unit open ball in Cm consisting of all (z1l , Zn) such that I z1 12 + * * + I Zm 12 < 1.

From Theorems 1 - 5, we get the following results.

THEOREM 6. Let G be a semi-simple Lie group, all of whose simple components are non-compact. Suppose that the center of a maximal compact subgroup of the adjoint group of each simple factor is not discrete. Suppose further that G has no simple factor which is locally isomorphic with the group of all (m + 1) x (m + 1) complex unimodular matrices which leave invariant the hermitian form

X1X1 + * * * + XmXm - Xm.+Xm,+l (m > 1).

Let F be a discrete subgroup of G with compact quotient space G/F. Then the first Betti number of G/F is equal to 0. Moreover, if F' denotes the commutator group of F, then the quotient group F/F' is finite.

THEOREM 7. Let D be a bounded symmetric domain, none of whose irreducible factor is analytically equivalent with the unit open ball in Cm (m > 1). Let F be a properly discontinuous group of analytic transformations of D with compact quotient space F\D. Let F' be the commutator group of F. Then the factor group F/F' is finite. In particular, if F contains no element of finite order different from the identity, the first Betti number of the compact complex manifold F\D is equal to 0.

Let D be a bounded symmetric domain, and let

D=Dlx...x DP

be the decomposition of D into the product of the irreducible factors. Let G(resp. Go) be the identity component of the group of automorphisms of D(resp. Di). Then we have G = G1 x * x GP.

THEOREM 8. The notation being as above, let D1, , D3 be those irreducible factors which are analytically equivalent with the unit open balls of complex vector spaces of respective dimensions. Let F be a properly discontinuous group of analytic transformations of D such that F\D is compact. Let F' be the commutator group of F and let rF= F fl nG.

Suppose that, for each subset {i1, * *, i},(i1 ..< ... < i) of the set of indices {1, *.. ,s}, the image of Fr in Gi1x * x Gi under the projection G = G, x * x GP Gil x * x Git is not discrete. Then the quotient group



330 YOZO MATSUSHIMA

F/I" is finite. In particular, if P contains no element of finite order different from the identity, then the first Betti number of the compact complex manifold F\D is equal to 0.

OSAKA UNIVERSITY

BIBLIOGRAPHY

1. A. BOREL, On the curvature tensor of the hermitian symmetric manifolds, Ann. of Math., 71 (1960), 508-521.

2. , Density properties for certain subgroups of semi-simple groups without compact components, Ann. of Math., 72 (1960), 179-188.

3. E. CALABI and E. VESENTINI, On compact, locally symmetric Kthler manifolds, Ann. of Math., 71 (1960), 472-507.

4. A. SELBERG, On discontinuous groups in higher-dimensional symmetric spaces, Proceed- ings of the Bombay Conference on Analytic functions 1960.

5. K. YANO and S. BOCHNER, Curvature and Betti numbers, Ann. of Math. Studies, No. 32, 1953.

6. A. WEIL, On discrete subgroups of Lie groups, Ann. of Math., 72 (1960), 369-384. 7. , On discrete subgroups of Lie groups, II, to appear in Ann. of Math.



Documents

On the First Betti Number of Compact Quotient Spaces of Higher-Dimensional Symmetric Spaces