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2. Let o : [a, b] - M(x) be a Tits geodesic. Then there exist points
XI,...,xN inR(x)andapartition a=t( 2. Given a point p r= M and a point x E M(x) we define
K(p,x)= {0EG=1(,(If):Op =p and Ox=x), K0(p, x) = the connected component of K(p, x) containing the identity,
F(p, x) = (q
=q for all
.
EKO(p,x)).
It will be useful to establish some facts about K0(p, x) and F(p, x). We retain the same notation for the next three results.
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Geometry of Nonpositively Curved Manifolds
3.6.6. PROPOSITION. The group K0( p, x) is a compact subgroup of G =
10(M), and its Lie algebra is Z(X) n f, where g = f + p is the Cartan decomposition determined by p and X e p is that element such that dp(X) = yPx(O).
3.6.7. PROPOSITION. The set F(p, x) is a complete, totally geodesic submanifold of M for every choice of p e M and x E M(cc). Let g = f + p be the Cartan decomposition determined by p. Then: (1) If p* e p is the Lie triple system such that F(p, x) = exp(p*Xp), then
p*={Yep:Z(Y)nf:)Z(X)nf), where X E p is that element such that dp(X) = yp'x(0). (2) z(X)fl p * = EX, the intersection of all maximal abelian subspaces
of p that contain X. In particular, Ex is a maximal abelian subspace of p * and the rank of F(p, x) in the sense of (1.12) equals e(x) as defined in (2.21.7). (3) E(ypx) = F(p, x) n F(ypx) in the notation of (2.11.3) and (2.20.11).
The proof of these two results are elementary but will be omitted here. See the appendix for the proofs. REMARK. The space E(yIj,x) can be a proper submanifold of F(p, x), even in the case that M is an irreducible symmetric space. For any symmetric space M of noncompact type it follows from assertions (1)
and (2) above or from the proof of assertion (3) of (3.6.7) in the appendix that E(y X) = F(p, x) if and only if p c Z(X ). In particular,
if E(ypx)=F(p,xc, then [X,Y]=0 for all Yep*. For purposes of illustration we consider the case that M is the irreducible symmetric space M = SUn, Il)/SO(n, R) discussed in (2.13). If p = I SO(n, R), then by (2.13) we have cl = _:1(n, U8), f = (skew sym-
metric n x n real matrices), and p= (symmetric n x n real matrices with trace zero). Let X e p be a unit element that has at least two eigenvalues of multiplicity 1, and let x = yx(oo) a M(co), where yx(t) =
erx(p) for all t e R. We assert that E(ypx) is a proper submanifold of F(p, x). It will then follow that E(ygx) is a proper submanifold of
F(q, x) for all q e M since Gx acts transitively on M by (2.17.1); specifically, we use the easily established facts that K0(gp, x) = gK0(p,x)g-1, F(gp, x) = gF(p, x) and E() =gE(y) for all gCGx. Let X e p be a unit element that has at least two eigenvalues of multiplicity
1, and let W be the subspace of W spanned by all
eigenvectors of X whose corresponding eigenvalues have multiplicity 1. If W t{8", then we may write Ifs" as an orthogonal direct sum l>8" = V, ® ... ®VN ®W for some integer N z 1, where (1) dim V >_ 2 for 1:5 i < N and (2) there exist distinct real numbers (A,: 1 < i < N) such that
Tits Geometries
185
X = A; Id on V, for I < i _< N. By the definition of W it follows that
X(W) C W, and if W Rn, then the eigenvalues of X on W have multiplicity 1 and are distinct from the elements of (A;: 1 _ 2 and hence there exists an element Y E p such that Y(W) c W, [ X, Y ] 0 on W, and Y is a multiple of the identity on each V, for 1 _< i:5 N in the case that W # R'. It follows immediately
V. for 1 _i:Nand f=0on that Z(Y)nf;;? W}. Hence YE p" as defined in assertion (1) of (3.6.7). However, [X,Y] * 0 by the definition of Y, and it follows that E(yp,,) is a proper
submanifold of F(p, x) by the discussion at the beginning of this remark.
Using K (p, x) we can reduce the study of F(y,, X x) to the study of F(x) if F is any k-flat containing ypx. The proof of the next result can also be found in the appendix. 3.6.8. PROPOSITION. Let F be a k -fiat in M that contains the geodesic y, . Then: (1) If y (=- F(yp., X m) is any point, then there exists 46 E K0( p, x) such that 4)(y) E F(x). Hence, F(yp,, )(x) = K0(p, x X F(m)). (2) If F, and F, are any two k -flats of M that contain y,,, , then there x) such that 4i(F,) = F,. exists ¢ E
Metric balls in the Tits geometry of M(c) 3.6.9. DEFINITION. Let M be a symmetric space of noncompact type and rank k z 2. Given x E M(x) and a positive number R we define
TR(x) = (y E M(x): Td(x, y) < R).
The topological properties of the metric balls TR(x) reflect the geometry of M and M(x) to a considerable extent as we shall see. We first reduce the study of TR(x) to the study of TR(x) n F(x), where F is any k-flat in M such that x c- F(x). The proof of the next proposition follows from (2.21.14), (2.17.1), and (3.6.8).
3.6.10. PROPOSITION. Let M be a symmetric space of noncompact type and rank k >_ 2. Let a point x E M(x) and a positive number R be given. Let F be any k -flat in M such that x E F(m). Then TR(x) = (GX)O(TR(x) r) Moo)),
where Gx = (0 E G: Ox = x) and (G, )o is the connected component of G, that contains the identity.
Next, we look at the properties of metric balls 7 (x), where e > 0 is small.
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Geometry of Nonpositively Curved Manifolds
3.6.11. PROPOSMON. Let M be a symmetric space of noncompact type and rank k > 2. For any point x E M(x) there exists a positive number e such that if y E Te(x), then (GG)0 g (Gx)0.
PROOF. Let F be any k-flat of i%i with x E F(x), and let p be any point of F. Let g = f + p be the Cartan decomposition determined by p, and let a c p be the maximal abelian subspace such that exp(a X p) = F. Let 0 = 00 + E. EAU,, be the root space decomposition of q determined by
a. If q, denotes the Lie algebra of (G,)0, then from proposition (2.17.13) we see that
iia
r - RU+ a(X)_>0
Let Q be the inner product on p pulled back from Tp M by the isomorphism dp: p -- Te M. Let 4 (,) denote the corresponding angle measurement in p. Now choose X E a so that dp(X) = y,,,(0) Choose 0 < e < IT so that if YE a and 1:(X, Y) < e, then a(X) a(Y)> 0 for all a c= A such that a(X) 0 0. We show that if y E T(x), then (G,.) c (G.),,. We first consider the case that y is a point in Ta(x) n F(x). If YE a is that element such that dp(Y) = y;,y(0), then 1(X, Y) = o (x, y) _ - I and by proposition (2.21.8)
s=s(x) 0) is a subset of Ay',
and since as(X) = 0 and as(Y) > 0 it follows that AX is a proper subset of A. Hence (Gy)0 c (Gx)0 and x < y by lemma (2.20.17). Finally, s(y) >- s - 1 = s(x) - 1 since a;(Y) = 0 for 1 < i < s - 1. On the other hand s(y) <s(x) by the previous result, lemma (3.6.18). It follows
that s(y) = s(x) -1. The sets Wx = {y E 4(co): x 2. For any point x E M(oo) we define Wx = (y E M(00): X:5 Y), where x < y means that (Gx )o Q (Gy )0 as in (3.6.15).
We summarize some properties of Wx in the next result. 3.6.21. PROPOSITION. Let M be a symmetric space of noncompact type and rank k z 2. Let x be any point in M(me). Then: (1) W, is an open subset of M(e) in the Td-topology (but never in the cone topology). (2) Wx c F(yXoo) for every geodesic y that represents x. (3) If y is any point of Wx distinct from x, then Td(x, y):!-, it/2. If a: [0, 1] - M(oc) is the unique minimal Tits geodesic from x = a(0) toy = a(1), then (G(,))fl = (GG)o for all t > 0. (4)
If y is any point of Wx distinct from x, then s(y) < s(x) and r(y) < r(x), where r and s are the rank and degree of singularity functions defined in (2.21.2) and (2.21.7). If s(y) = s(x) or r(y)
Tits Geometries
= r(x), then
191
'(x) contains the unique minimal Tits geodesic
from x to y. (5)
Let p be any point of M, and let z E R(x) n F(yp., Xx) be any point with x E '(z). Then WX = Ko( p, x){WX n 7(z)}
= K (p, x){WX n F(yyX)(x)},
where &(p, x) is the group defined in (3.6.5),
'(z) = {y E
M(x): (G,. )o = (G2 )O) is the Weyl chamber in M(x) determined by
z (see (2.17.20)), and W (z) denotes the closure of F (z) in the cone topology.
REMARK. If z E R(x) and F is any k-flat with z e F(x), then W(z) C F(x) by (2.12.1) and (2.17.21). Hence the closures of V(z) with respect to the cone topology and Td-topology coincide. See also remark (3) following (3.6.24).
PROOF. (1) The fact that WX is open in the Td-topology follows from proposition (3.6.11) and the fact that W. c WX if y E WX. The set Wr is never open in the cone topology of M(x) by corollary (3.6.13) and the next assertion of this result. (2) If y e WX, then y, c F(ypy) c F(y,X) for every point p E M by proposition (2.20.16). It follows that y E F(y XXx) for any point p E M. (3) Let y E WX be given, and let a : [0,1 J"-* M(x) be a minimal Tits
goedesic such that o-(0) =x and Q(1) = y. Such a geodesic o always exists by proposition (3.6.1). By proposition (3.6.2) there exists a k-flat F
in M such that u[0,1] c F(x). Let p be a point of F, and let g = f + p be the corresponding Cartan decomposition. Let a c p be the maximal abelian subspace of p such that F = exp(aXp). Let X and Y be those unit vectors in a such that dp(X) = y;,(0) and dp(Y) = yp'y(0). We show first that Td(x, y) < ir. Suppose that this is not the case. Then Td(x, y) = 7r by proposition (3.6.1), and 4P (x, y) = 4(x, y) = it since x and y both lie in F(x). It follows that -X = Y. Let A C a* be the set of roots determined by a, and let g = 90 + Ea E g,, be the corresponding root space decomposition. Let g x and g y denote the Lie algebras of (GY)o and (GG),,. If a E A is a root such that a(Y) > 0, then a(X) = - a(Y) < 0. Hence fl a c lI y but cl,,g r by proposition (2.17.13). This contradicts the fact that g y c q, since y r=- W. Therefore
Td(x,y)0 so that if 0 < s< co, then Ac AX(7 If a(Y)>O and a(X) = 0 for some a E A, then a(X(e)) > 0 for all e > 0, and hence AycAX(e)
forall0<e<eo
since AXnAYCA+X(C) and Ay=A+u{AxnAY} by(*). X
It remains only to prove that A .(e) c Ay for all 0 < e < co. Let
a E AX (8) be given arbitrarily. If a(Y) = 0, then a(X) = 0 by (* ), and
hence a(X(e)) = 0, contradicting the assumption that a E AX(
).
If
Geometry of Nonpositively Curved Manifolds
196
a(Y) < 0, then a(X) < 0 by (*), and hence a(X(e)) < 0, again contradicting the fact that a e AX(.) . The only remaining possibility is that a E AY, which proves that AX(e)c Ay
We prove (5) - (2). Let x E s°(y) be given, and choose a sequence {yn} c'(y) such that Td(x, 0. By (3.6.11) we know that x :!5;y,, for sufficiently large n, and hence x< y since (GY)o = (Gy) for all n. We prove (2) - (1). Assume x < y, and let x* E be given. Then x* : 2. Let x E M(oo) be arbitrary. Then
?(x) = n E(ypx)(x), pEM
where WW denotes the closure of 9-'(x) in the cone topology of M(x).
Further properties of the sets V(x) are given in the next result. 3.6.28. PROPOSITION. Let M be a symmetric space of noncompact type and rank k >: 2. Let x be any point of M(oo). Then:
(1) '(x) is a convex subset of M(c) with respect to Tits geodesics in M(me), and Td(y, z) < it/2 for all points y, z E F(x). (2) There exists a positive number e such that Tt,(x) n E(yp,Xco) c W(x) for any point p E M.
(3) For any point p E M the set If(x) is an open subset of E(yPxXc) with respect to the induced cone topology on E(ypX Xco). Moreover, if T*: F(x) --> SpE(ypx) is the map given by
T*(y) = ypy'(0),
then Td(y, z) = p(T*(y), T*(z)) for all points y, z E W(x), and T* is a homeomorphism onto an open spherically convex subset of a hemisphere in the (e(x) - 1)-dimensional unit sphere S,,E(ypx). (4) Let o : [0, 1] -R-00 be any minimal Tits geodesic with o (0) = x. (a) If s(o(t)) = s(x), where s is the degree of singularity function defined in (2.21.7), then o([0,1]) e fi(x). (b) If r(o(t)) r(x), where r is the rank function defined in (2.21.2),
then o([0,1]) c '(x). PROOF. We recall that e(x) is defined in (2.21.7) above. We prove (1). The convexity of '(x) is an immediate consequence of assertion (3) of
Tits Geometries
197
proposition (3.6.21). The proof that '(x) has diameter < it/2 has several steps.
We first reduce to the case that x e R(x). If x E M(oo) is any point, then we may choose z E R(x) with x< z; for example, use the methods of (2.20.17). By (3.6.26) it follows that W (x) c W(z ), and hence it suffices to prove that 7(z), the Td-closure of W(z) in M(oo), has
diameter < a/2. Let x E R(-) be any point, and let y, and Y2 be points in F(x). Let
F be a k-flat in M with x e F(x). Let p be a point of F, and let q = f + p be the corresponding Cartan decomposition. Let a be the maximal abelian subspace of p such that exp(a X p) = F. Let A c a* be the corresponding set of roots. Let Y1, Y2, and X be those unit vectors
in p such that yY(x) = y; for i = 1, 2 and yx(c) = x; we adopt the notation of (2.20.17). We know that X E a by the fact that p E F and x E F(x). Moreover Y E '(X) for i = 1, 2 by (3.6.31) below. Note that X is a regular vector of a since yx(x) =x E Moo) by hypothesis. Since p(Y1,YZ) = p(y,, y2) = Td(y,, y,) by (3.1.2) and (2) of (3.4.3) it suffices to prove the following.
LEMMA A. Let X be a regular rector, and let Y, and Y, be arbitrary vectors in the Weyl chamber
'( X ). Then 4 (Y, , Y2) < 7r/2.
PROOF. We continue to use the notation of (2.20.17). Let k = rank M = dim a >_ 2. If A = ak) denotes the set of simple or indecompos-
able roots in A', then (a;, a, ) < 0 for all 1 < i, j < k with i # j; see (2.9.4) through (2.9.6) for a definition and properties of simple roots. The proof of lemma A will follow from the following.
ak) be a basis of RA such that (a;, a,) < 0 for all 1 < i, j < k with i * j. Let (a*,..., aR } be the basis of RA defined by LEMMA B. Let
(a*, a1)=5+1 for all i, 1. Then (a*,a*)z0 for all 1 0 for all r since Y, E W(X) and a,(X) > 0 for all r since a,. E A c AX. Hence y;, = a,(Y) > 0 and (Y,,Y2> _ E;-, Y),y2s( a*, as) z 0 by lemma B. We conclude that 4(Y1,Y2)
it/2. We now prove lemma B by induction on k [GI]. Let k = 2. If A and B are 2 x 2 matrices given by A,1 = (a,, a1) and Bi, = (a7, a* ), then it
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Geometry of Nonpositivcly Curved Manifolds
is routine to show that B = A - '. Hence if i # j, then Bit = - (1 /D)Ail, where D = det(A). Note that D = I a, A a21` > 0 and Ail < 0 if i < j by the hypothesis on A. Hence Bit 0 if i o j. We now consider the general case for k >- 3 and assume that the result has been proved for k - 1. We wish to prove that (a*, a*) >- 0 if i : j, and without loss of generality it suffices to do this in the case i = 1,
j=2. If V=akl=(vERk:(t',ak)=0), then (a*,...,a'_,)cV. For 1 < i < k - 1, let a, denote the orthogonal projection of a; on V. Assuming without loss of generality that ak is a unit vector we have
ai=ai - a,,ak ak. aj _ ,) c V and (a; , a*) = Sil for I < i, j < k - 1. For i *j we compute (a;, a;> = (ai, a1) - (a,, ak) (aj, ad !5;0 by It is clear that
the hypothesis that < a,, a,) -< 0 if 1 < r, s < k with r * s. We now apply the induction hypothesis to the dual bases (a.., ak_,) and
(a*,...,
in V; ff8k-' to conclude that (aa* ) >_ 0 for 1 0 such that
if y E T,(x) then (G,.)(, c (G.,)o. Let p be any point of M, and let yE
n E(y,,, Xx) be given. Since (G,.) is transitive on M and fixes
both y and x it follows that y e Ta(x) n E(yyxX x) for every point q E M. (Hence T(x) n E(yyxXx) is independent of q.) Therefore yq,, c E(ygx) for every point q E M, and it follows by propositions (2.20.14) and (2.20.16) that (GY)o (GX),,. This proves that TT(x) n E(yP,)(x) e
'(x).
We prove (3). Fix a point p e M. The set ?'(x) is a subset of
E(yP.,Xx) by corollary (3.6.27). If y E W(x), then E(yp,) = E(yyx) by proposition (2.20.16), and by (2) of this result there exists e > 0 such that TE(y) n E(yyxXx) = Te(y) n E(yp)Xc) c '(y) = is'(x). Since E(yyx) is a flat Euclidean space it follows from proposition (3.1.2) that Tt,(y) n E(y, Xx) = (z E E(yPxXx): 4P(z, y) < e). Hence f(x) is an open subset of E(yP,Xx) relative to the topology induced from the cone topology of M(x). The remaining assertions of (3) follow from assertion (1) or proposition (3.6.31) below.
We prove (4). Let a: [0,1] -> M(x) be a minimal Tits geodesic with a(0) = x. To prove either (a) or (b) it suffices by the connectedness of [0, 1] to show that for every to E [0, 1] there exists e > 0 such that
Ma(t)) = '(a(to)) if t e [0,1] and It -tol < e. Let to E [0,1] be given. By (3.6.11) there exists e > 0 such that if t e [0,1] and It - tol < e, then
a(t(,) < a(t). By hypothesis, s(a(t)) = s(x) in (a) or r(a(t)) = r(x) in (b) for all t E [0, 1]. If It - tol < e, then it follows from (4) of (3.6.21) that
in either case (a) or (b). The proof of (4)
F(a(t)) = complete.
0
is
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199
REMARK. Recall that e(x) = 1 if and only if s(x) = k - 1 if and only if x
is a maximally singular point at infinity by proposition (2.21.8) and definition (2.21.9). In this case '(x) is the single point {x). Coxeter complexes in a
We now give a slightly altered version of the usual aebraic description of the Tits building determined by M or G =10(M). This description is the algebraic dual of the description of the Tits building in M(x). See also [Most, pp. 13-14]. Let g = f + p be a Cartan decomposition determined by some point
p E M, and let a c p be any maximal abelian subspace of p. Let F = exp(a)(p) be the corresponding k-flat in M. We give an algebraic
description of the chambers and faces of the apartment AF, G) _ ((Gx),,: x e F(m)}.
For any nonzero vector X E a it follows routinely from lemma (2.20.9) that Ex, the intersection of all maximal abelian subspaces of p that contain X, is also a finite intersection of root hyperplanes as = (Z E a: a(Z) = 0): Namely, Ex =
n as Cr E.1% x
where Ax = (a E A: a(X) = 0). One may show that there are (k) sets E. of codimension i in a. 3.6.29. DEFINITION. Let f, p, and a be as above. For each nonzero X E a we define
all aEA - Ax). If X is a regular element of a, then Ax is empty and W(X) is a usual Weyl chamber of a as defined in (2.8.2). If X is a singular element of a, then Y(X) is called a face or Weyl face. 3.6.30. DEFINITION. Let f, p, and a be as above. Let X and Y be any elements of a. We say that W(X) is a face of '(Y) if '(X) c W(Y ).
The next result relates these two definitions to the corresponding definitions (3.6.24) and (3.6.25).
3.6.31. PROPOSITION. Let M be a symmetric space of noncompact type and rank k z 2. Let x e M(a) and p E M be given arbitrarily. Let g = f + p
be the Cartan decomposition determined by p, and let X E p be that element such that dp(X) = yx(0). Let a c p be any maximal abelian subspace that contains X. For any Z E p let y, denote the geodesic given
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Geometry of Nonpositively Curved Manifolds
by yz(t) = e`{z/uzu)( p). Then
W(x) _ {yy(x): Z E W(X)). PROOF. If Z E V(X ), then Az
Ax since Z E Ex= n. E A,
RR. How-
ever, A' g A; and AXc AZ be the definition of W(X), which implies that Az c Ax. Hence Az = Ax, and this implies that AX= AT and AX= Az. By lemma (2.20.17) we conclude that yz(x) E '(x).
Conversely, let z e '(x) be given, and let Z E p be that element such that dp(Z) = y, (0). By corollary (3.6.27) and proposition (2.20.12) it follows that Z E Ex, and by lemma (2.20.17) we see that A X = A'Z9 AX= Az, and A x = Az. Hence Z E 9'(X). 0 COROLLARY. Let M be a symmetric space of noncompact type and rank k 2-- 2. Let p be a point of M, and let tl = f + p be the Cartan decomposition determined by p. Let X be any unit vector in p, and let x = yX(x) E M(x). Let K = (g E G: g(p) = p) be the maximal compact subgroup whose Lie algebra is f. (1) Let ¢ be an element of K such that Ad(4,) leaves '(X) invariant. Then Ad(4,XY) = Y for all Y E Ex. In particular, Ad(4,XY) = Y for all Y E W(X ). that leaves invariant i_5'(x) e (2) Let g be an element of G M(). Then g(y) = y for all y c- K(x).
PROOF. (1) Let a be a maximal abelian subspace of p that contains X, and let A e a* denote the set of roots determined by a. If we let ?,(X) denote the unit vectors in '(X), then by (1) of (3.6.28) and its proof we conclude that 9,(X) is a compact convex subset of the unit sphere in a
with angular diameter < a/2. The transformation Ad(4,) is a linear isometry of p with its canonical inner product by (5) of (2.7.1), and hence Ad(4) leaves W',(X) and W,(X) invariant since by hypothesis Ad(4,) leaves W(X) invariant. It follows from (2) of (1.15.3) that Ad(4,) fixes the soul s(X) E ',(X) = Int(&,(X )) c a c p. By (2.20.18) we conclude that Ad(4,) fixes every vector Y E E,, x,.
It remains only to prove that Ex = E,, x,. We note that F(s(X )) _ '(X) and the sets ?(s(X )) and ?;"(X) are open subsets of E,, x, and Ex respectively. A vector space is the linear span of any open subset, which proves that Ex = E,(x and completes the proof of (1).
(2) Let g e=_ G be an element such that g('(x)) c '(x), and let y e F(x) be given. By (3) of (1.13.14) there exists an element 0 E K such that g(y) = 4,(y) E F(x). We must show that 0 fixes y. Let Y be
the unit vector in p such that y = yy(x). If Z is the unit vector
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201
Both Y Ad(4XY), then standard arguments show that y7(x) and Z lie in 9'(X) by (3.6.31), and it follows that Ad(4)(W(X )) _ Ad(o)(W(Y)) = K'(AdXOXY)) = K'(Z) = K'(X). By (1) it follows that Ad(4) fixes every vector in W(X ). In particular, Z = Ad(¢XY) = Y, which implies that 4(y) = y and completes the proof of (2). O
'(x) and the strata M, (x) (see 2.21.9)) 3.6.32. PROPOSITION. Let X1- be a symmetric space of noncompact type
and rank k >: 2. For each integer j with 0:5j S k - 1 the connected components of Mi(x) in the Td-topology are the sets K'(x), x C Mi(x).
PRooF. Each set K'(x), x E M(x), is convex with respect to Tits geodesics by (1) of (3.6.28). Let A c M1(x) be open and connected in the Td-topology. It suffices to show that A c K'(x) for some x C M!(00). Let x CA be given. By the connectedness of A it suffices to show that
there exists e > 0 such that if y E T,.(x) n A, then W(y) = W(x). By (3.6.11) we may choose e > 0 such that if Td(x, y) < e, then x 5 y. It follows from (4) of (3.6.21) that if y e then s(y) < s(x), where s is the degree of singularity function. If y E TT(x) n A c Mi(x), then s(y) = s(x) =j, and hence by the equality assertion in (4) of (3.6.21) we conclude that W(x) = W(y). 0 For each x e M(x) we define for W(x) a center of gravity CG(x) E
W(x), and we show that CG: Mi(x) -> M(x) is continuous for each integer j with 0 < j < k - 1. Using the sets W(x) we also define some functions S and rad: M(x) - (0, x), and we show that these functions are continuous on each stratum M,(-)3.6.33. DEFINITION. Let M be a _symmetric space of noncompact type and rank k >- 2. For each point x E M(x) we set
S(x) = inf(Td(x, y): y E V(x) - '(x)), rad(x) = sup(S(y): y E G'(x)).
The function S(x) is always positive by assertion (3) of proposition (3.6.28) unless £'(x) = {x} and gives the distance from a point x to the boundary of '(x). The function rad(x) is also positive unless W(x) = W. Note that _ 2. For each point x e Moo) there is a unique point x* _
CG(x) E '(x) such that rad(x) = S(x*). The function CG: M1(c) Mj(oo) is continuous in the cone topology for each integer j with 0:!5;
j 5 k_ 1. Moreover, CG(ox) = 0 CG(x) for all 4)E G and all x E M(c). We now begin the proof of propositions (3.6.34) and (3.6.35). a symmetric space of noncompact type and rank k >_ 2. Let G =10(M). Let x E M(me) be given, and let j = s(x), the degree of singularity of x (see (2.21.7)). Then for each positive number a and each neighborhood U c G of the identity the set 3.6.36. LEMMA. Let A f-
{4)(y): 4 E U, Y E W'(x)) is a neighborhood of x in M,.(°) with respect to the cone topology, where
Fe(x)={yEV(x):Td(x,y)<s}. To prove this we need the following result whose proof is given in the appendix. SUBLEMMA. Let M and G be as above. Let x E M(oc) be given. For each positive number e and each neighborhood U c G of the identity, the set
W(U, e, x) = {4)(y): 0 E U, y E M(x), Td(x, y) < e} is a neighborhood of x in M(x) with respect to the cone topology.
Assuming the sublemma we prove the lemma. Let x E M(me) be given, and let j = s(x). Let a positive number e and a neighborhood U c G of the identity be given. If W = M1(oo) n W(U, e, x), then W is a
neighborhood of x in M,(oc) by the sublemma. We show that W = using the notation in the statement of the lemma. Clearly, U- FU) c W by the definitions. Conversely, let z E W be given, and choose 0 in U and y e M(oo) such that Td(x, y) < e and (k(y) =z. It suffices to prove that y E W(x). By proposition (3.6.11) there exists a positive number co such that if Td(x, y) < co, then (G,)0 c (GX)o. Without loss of generality we may
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assume that 0 < e < co. Fix a point p c- M, and let c1= f + p be the corresponding Cartan decomposition. Let X and Y be those elements in p such that dp(X) = ypx(0) and dp(Y) = y y(0). Since (Gy )0 c (GG)o it follows from (2.17.16) that F(y7 y) c F(y,,x) and Z(Y) c Z(X). Hence [X,Y] = 0, and we may choose a maximal abelian subspace a c p that contains both X and Y. Let A c_ a* denote the set of roots determined by a, and let g = 0 + Ea E ,, q. be the corresponding root space decom-
position. By lemma (2.20.17) it follows that AXc Ay, AXc Ay, and hence Ay c AX. By hypothesis, s(y) = s(¢y) = s(z) = j = s(x). Let a,} be a maximal linearly independent set of roots in Ay. If Ay
were a proper subset of Ax, then {al,..., a, a} would be a larger linearly independent subset of Ax for any a E Ax - Ay, but by assertion (3) of proposition (2.21.8) this would contradict the fact that s(y) =Ax). Hence Ay = Ax, and this forces AX= Ay and AX= A. By lemma (2.20.17) we conclude that y E si(x), which completes the proof of the lemma. We now prove proposition (3.6.34). The fact that the functions S and rad are G-invariant is obvious from the definitions. It follows easily
from proposition (3.6.31) that S and rad are continuous on W(x). Combining this fact with lemma (3.6.36) we will obtain the continuity of
8 and rad on M1(-). We complete the proof only for S since the other proof is similar.
Let j be any integer with 0 !5j < k - 1. Let x e M,(m) and e > 0 be given. Choose 71 > 0 such that if y E 9'(x) and Td(x, y) < 77, then 16(y) - 6(x)I < e. Let U c G be any neighborhood of the identity, and let W = U i;,(x) c M1(cc). By lemma (3.6.36), W is a neighborhood of x in M1(oo). Given z E W choose y E W,,r(x) and 46 E U such that z = 4'(y).
Then 6(z) = S(y), and hence IS(z) - S(x)l < e for all z E W by the choice of -q. PROOF OF PROPOSITION (3.6.35). By (3.6.31), assertion (1) of (3.6.28), and
assertion (1) of (1.15.3) it follows that the center of gravity function CG: M(x) - M(oo) is well defined. It is clear from the definition that CG(4z) = 0-CG(z) for any 4 E G and any point z E M(me). Now let j be any integer with 0:< j :!g k - 1, and let x be any point in M,(o). It
follows from the definition that CG is constant on '(x). Now let O c M1(-) be any neighborhood of x* = CG(x), and let U c G be a neighborhood of the identity in G such that U(x*) c O. Then W= U-W(x) is a neighborhood of x in M,(oc) by lemma (3.6.36). Given z e W we choose 0 E U and y e so (x) such that z = ¢(y). Then CG(z) = CG(cby) = 0-CG(y) = 46 CG(x) = 4,(x*) E 0 by the choice of U. This proves that CG(W) c O.
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Incidence relations in M(c) 3.6.37. DEFINITION. Points x and y in M(x) are incident if they lie in some common face or chamber in M(x); that is, there exists z e M(x) such
that x E W(z) and y E'(z ). We recall from (3.6.26) that x e W(z) if and only if x _ x and z > y.
More generally, we can consider upper bounds and least upper bounds for arbitrary subsets of M(x). 3.6.38. DEFINMON. A subset S c M(x) is said to have an upper bound if there exists a point z E M(x) with z >_ x for every x E S. A point z E M(x) is a least upper bound for a set S c M(x) if z is an upper bound for S and Z:5 z* for any upper bound z* of S. We let lub(S) denote the set of least
upper bounds of S. Clearly, W(z,) _ W(z,) for any two upper bounds z,, z, E lub(S). We shall see later, in (3.6.43), that any subset S of i%(x) that has an upper bound has a least upper bound. However, we begin by considering sets S with two elements. 3.6.39. PROPOSITION. Let x and y be points in M(x) that are incident. Then:
(1) The function p - 4p (x, y) is constant in M and equals a < ir. In particular, Td(x, y) = a < IT. (2) If z is any interior point of the unique minimizing Tits geodesic from
xtoy,then g.=gxngy. PROOF. (1) By (3.6.37) there exists a point z* in M(x) such that z* zx and z * >-y. Hence, fi,. c g x n q Y by the definition of the partial order-
ing. It follows that the function p -'
_ 2.
3.7.1. DEFINMON. A splice in M(me) is a subset of M(me) of the fore S = F,(-) n F2(c), where F, and F2 are arbitrary k-flats in M, k = rank M.
Splices in M(ac) will play an important role in the proof of the Mostow rigidity theorem in the higher rank case. They are a boundary sphere version of the splices in M defined by Mostow in [Mos2, p. 56]. We discuss the relationship between splices in M(me) and splices in M below in (8.5.1).
We begin with some basic properties and examples of splices. 3.7.2. PROPOSITION. Let S g,400 be a splice in M(oo). Then: (1) The Tits topology and cone topology agree on S, and S is closed in these topologies.
(2) If x c- S, then F(x) c S.
(3) There exists a finite set {x,,...,xN} of points in S such that
S= U",W(x;). REMARK. We shall see below in (3.7.5) that a closed chamber or face '(x) is a splice for every x E M(oo).
PROOF. Let F be a k-flat in M such that S c F(m) and let p be a point of F. If x and y are any two points of S, then 4n(x, y) = Td(x, y) by (3.1.2) and (2) of (3.4.3). Therefore the cone and Td topologies coincide on S and more generally on F(oo). If we write S = F,(x) n F2(cc) for suitable k-flats F, and F2 in M, then S is closed in these topologies since both FPO and F200 are closed in these topologies.
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We prove (2). Let F, and F2 be k-flats in Al such that S = F,(oo) n F2(x). Let x be any point of S. By (3.6.26), '(x) = (Y E M(oo): Y:5 x}. It now follows immediately from (2) of (3.6.42) that T(x) c F,(oo) n F2(oo) = S.
We prove (3). Let F be any k-flat in M with S C F(oo). By (1) of xN) c S such that if x e S then gx = gx for some i with 1 0 and exp is one-to-one on W. This shows that (e'XFXc) n F(oo) c S = S(X, F) for all t E (0, e). 3.7.5. PROPOSITION. If S ='(x) for some point x E M(oo), then S is a splice in M(me).
PROOF. Let F be a k-flat with x E F(oo), and let p be a point of F. Let g = f + p be the Cartan decomposition determined by p, and let a c p
be the abelian subspace of p such that F = exp(a)(p). Let A g a* be the corresponding set of roots. We adopt the notation of (2.17.13) and (2.20.17).
Let X E a be the unit vector with yX(oO) =x, where yx(t) = (e"X p).
Let Z be a nonzero vector in g such that Zo is arbitrary and Zr is nonzero for a e A if and only if a E Ax U AX; here Z. denotes the component of Z in g. and Zo denotes the component of Z in go relative to the root space decomposition g = go + E. 1. A g Q. We assert that ig'(x) = S(Z, F), which will complete the proof since S(Z, F) is a splice in M(me) by (3.7.4).
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We note first that Z E 0 r by the definition of Z and (2.17.13). Hence
x E S(Z, F) and W(x) c S(Z, F) by (3.7.2) and (3.7.4). It remains to prove that S(Z, F) c F(x). Let y E S(Z, F) be given, and let YE a be the unit vector such that yy(x) = y. We know that Z E R by the y,
definition of S(Z, F), and by (2.17.13) or (2.20.17) we conclude that if Z. * 0 for some a E A, then a(Y) z 0. Hence by the definition of Z we have Ax U A X c Ay U A Y or, equivalently, AX D A. It follows from
(2.20.17) that y 5 x and hence y E '(x) by (3.6.26). This shows that S(Z, F) c lq(x). Irreducible splices Following Mostow [Mos2, §15] we introduce the notion of irreducible splice.
3.7.6. DEFINITION. A splice S in M(cc) is irreducible if whenever S is a finite union of splices S;, I 5 i:5 N, it follows that S = S;, for some i.
3.7.7. PROPOSITION. A splice S is irreducible if and only if S = '(x) for some x E M(c). PROOF. Suppose first that S is an irreducible splice. By (3) of (3.7.2) we
know that there exists a finite set (x1,...,xN) in S such that S = U, I W (x; ). By (3.7.5) each of the sets W (x,) is a splice in M(oo), and
hence S = '(x;) for some i by the irreducibility of S. Conversely, let S = W(x) for some x e M(me) and suppose that S = U. I Si, where each Si is a splice in M(oo). Choose an integer i such that x E S. Then S = W(x) C- Si by (2) of (3.7.2), and it follows that S = S,. Therefore S is irreducible. Splice mappings
Let M and M* denote two symmetric spaces of noncompact type and rank z 2, and let 51' and .51* denote the set of all splices in M(cc) and M *(oo) respectively.
>Y* will be called strongly order preserving if i/i(S) = U. , ii(S,) whenever S = U" I S; for splices S. S. E 3.7.8. DEFINITION. A function
REMARK. We shall see below in (8.5.3) that the pseudoisometries
0: M M * that arise in the proof of the Mostow rigidity theorem induce strongly order preserving maps ry: We recall from remark (1) following (3.6.24) that .97M is the quotient space M(x)/ - , where x - y for points x, y E M(oo) if (G,,)0 = (GG)0. A
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function :.9 f -'9M* is order preserving if fi([x]) 5 t ([y]) whenever [xl 5 [y]. 3.7.9. PROPOSITION. Let qi:.5"-+.9 and 4*:S-"* --.Y be inverse functions that are both strongly order preserving. Then iy and 1i * induce inverse order preserving bijections of the Tits buildings, t : 9M _ 5M * and *: .9M *
respectively.
For the proof of (3.7.9) we need the following.
3.7.10. LEMMA. Let ifr:be a strongly order preserving function. If S, c S, for splices S, , S, E.9, then ir(S,) c 41(S')' PROOF. Let S, and S2 be splices in M(x) with S, c S2. Without loss of generality we assume that S, # S2. By (3.7.2) we can find a finite set (x1, ... , xn+) c S, such that S1 = UN , W(x; ). If F is a k-flat of M such that S2 c F(oc), then by (1) of (2.17.22) there are only finitely many
distinct Weyl chambers or faces ('(x), x E F(x)}. Hence by (3.7.2) there exists a finite set (xN+ 1, ..., xN+,) c F(oo) such that S2 = UN I'W(x;). Since +li is strongly order preserving we conclude that I,J(S1) = U"_, +/(W(X;))C UN ,'y(W(x;))= .#(S2).
O
PROOF OF (3.7.9). We show first that 0 and i/i* map irreducible splices into irreducible splices. Let S be an irreducible splice, and suppose that a[i(S) = S* = UN , S*, where S* ES9* for each i. Then S = p*(S*) _ UN , 4i*(S*), and it follows that S = lp*(S*) for some i by the irreducibility of S. We conclude that S* = 4i(S) = Si*, which shows that S* is irreducible.
If x E M(oo) is given arbitrarily, then i/('(x)) = iV(x*) for some x* E M(c) by the previous paragraph since the irreducible splices are the closed chambers or faces by (3.7.7). We define +/ :9M -> 91if * by 4i([x]) = [x*], where O(F(x)) = W(x*). Note that if SB(x*) _ S-(y*) for points x* and y* in M_ *(oo), then 2(x*) = °(y*) by (3.6.26). It is now routine to show that , is well defined and that l' *: 9M * -* 9M is the
inverse of eji. If [x] 5 [ y] for points x, y E M(CC), then x :5y and '(x)cF(y) by (3.6.26). Hence ¢(F'(x))c i/i(F'(y)) by (3.7.10), which proves that +I([x]) 5 i([y]). Similarly, it follows that i,*:9M* --*9'M is order preserving.
0
Classification of splices We now use the root space decomposition of g to classify all possible splices in M(oc). This section is not needed for the proof of the Mostow
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rigidity theorem and may be omitted by a reader interested primarily in that result.
Let p be a point of M, and let g = t + p be the Cartan decomposition determined by p. Let F be a k-flat in M, with k = rank M, such that p E F. By (2) of (2.10) we may write F = exp(a X p) for some maximal abelian subspace a of p. Let fp: a --), F be the diffeomorphism
defined by fp(A) = e'(p). By (2) of (2.10) the map fp is an isometry with respect to the inner product on a defined in (2.3.7). The fact that fp: a -* p is an isometry immediately implies the following.
3.7.11. LEMMA. If C is a convex subset of a, then fp(C) is a convex subset
of F. Moreover, if there exists an r-dimensional subspace a' such that C e a' and C has nonempty interior in a', then fp(C) c fp(a') = F*, an r fiat in F, and ff(C) has nonempty interior in F*.
In the sequel we will need two elementary results about convex subsets in R". We state them now and prove them at the end of this section on the classification of splices in M(cc). 3.7.12. PROPOSITION. Let C be a closed convex subset of R", where n >- 1.
Then there exists an integer r with 0 < r < n such that C is contained in an r -flat F and C has nonempty interior in F. The r -flat F with these properties is unique.
An r-flat is the translate of an r-dimensional subspace of R". 3.7.13. PROPOSITION. Let C be a closed convex subset of B8" with nonempty interior. Let the boundary of C be contained in a finite union of hyperplanes (= (n -1)-flats) (H1,..., HN }. Then there exists a subset S c { 1, ... , N) such that C is the intersection of closed half-spaces determined by the hyperplanes (Hi: i e S). ROOT SPACE CONSTRUCTION OF SPLICES IN M(me). We begin with an
elementary construction and show eventually that all splices arise in this fashion. 3.7.14. NOTATION. For the remainder of this discussion of splices we let
p, 1, p, a, and fp: a ---> F = exp(aX p) be as above. Let g = go +
E. e A g a be the root space decomposition determined by a, where
A c a* denotes the finite set of roots. Fix a nonzero element A e a, and let AA = (a e A: a(A) = 0) and AA = (a E A: a(A) > 0) (cf. (2.20.17)). Let EA = na E A,, aa, where as = (B E a: a(B) = 0) (cf. discussion preceding (3.6.29)). The sets {aa: a E A) are the root hyperplanes
in a.
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For a subset A, c A; we consider the closed convex subset EA(A1) _ (B E EA: a(B) z 0 for all a E A,). We note that A lies in the interior of EA(A,) regarded as a subset of a' = EA. Hence by (3.7.11), C(A, A) = fp(EA(A 1)) is a convex subset with nonempty interior in the flat F* = fp(EA) c F. 3.7.15. PROPOSITION. S(A, A,) = C(A, A, Xx) is a splice in F(oo).
We define C(A, A,)(oo) = C(A, A,) n Af(x), where the closure is in the cone topology of M U M(x). Since EA(A,) is a convex subset of a that contains the origin it is easy to show that C(A, A,X-) = (yR(x): B is a unit vector in EA(A,)}, where y8(t) = e`R(p) for all t E R.
PROOF. Let X be an element of g, and write X = X + Ea E Xa, where X. E go and Xa E ga for all a c- A. We restrict X so that X,,,:# 0 if and only if a C I = AA U A, in the notation above. We show
that S(A, A,) = S(X, F) _ (x e F(x): X e g), where F= exp(a)(p). It then follows from (3.7.4) that S(A, A,) is a splice in M(x). Let X be an element of g with Xa # 0 if and only if a E I = AA U A,. We show first that S(A, A,) c S(X, F). Let x E S(A, A,) c F(x) be given, and let B be a unit vector in EA(A,) such that x = y8(x). By the definition of EA(A1) we know that a(B) ? 0 for all a E Y. It follows from the definition of X and (2.17.13) that X E + Ea E s g a c g o+ Ea(8) 2 o g a = gx Hence S( A, A,) c S(X, F) since x C S(A, A,) was arbitrary.
Conversely, let x C S(X, F) be given, and let B be the unit vector in a such that x = yR(x). If a E I = AA U A,, then Xa # 0 by the choice
of X and hence a(B) >- 0 since by hypothesis X C gx = go + EacR>> o g a If of E AA then - a e AA, and we conclude that a(B) = 0 since a(B) >- 0 and (- a )(B) >- 0. It follows that B E EA = n,,,, A a,, and, moreover, B E EA(A, ). We conclude that x = ya(x) E S(A, A,), which proves that S(X, F) c S(A, A,), and this completes the proof of the proposition.
We now apply the construction above in some cases of particular interest. We shall see later that any splice in M(x) arises as in (4). Proofs are given after the statement of the examples. 3.7.16. EXAMPLES. (1) Let Y. c A be any subset consisting of r linearly
independent roots, where 1 < r< k - 1 and k = dim a = rank M. Let az = na E x aa, and let F* = exp(aXp) = fp(a2), an r-flat in F. Then S = F*(x) is a splice.
If r = k - 1, then S consists of two points {x, y} such that (a) x = y(x) and y = y( - x) for some geodesic y of M and (b) x and y are
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maximally singular in M(x); that is, s(x) = s(y) = k - 1, where s(x) is the degree of singularity defined in (2.21.7). (2) Let x E M(x) be a maximally singular point. Then {x} is a splice
in M(x). (3) Let x E M(x) be a maximally singular point. Let p e M be given arbitrarily, and let y = yx( - x). Then S = (x, y} is a splice in M(). (4) Let I c A be any subset such that a,= na E s aR is nonempty, where aQ = (A E a: a(A) z 0) for a E A. If Cz = ff(a?) and S = C_(x), then S is a splice in M(x). PROOF. (1) Let A be an element of a such that AA consists of those
roots in a that are linear combinations of roots in E. Then a 2 = n., as = na E.\ ^ as = EA. If A,gA' is the empty set, then S= %
S(A, A,) in the notation above and S is a splice in M(x) by (3.7.15). If r = k - 1, then aI is a 1-dimensional subspace of a, and S = {x, y},
where x = yx(x), y = yx(- x), yx(t) = e`x(p), and X is a unit vector spanning a, . Clearly, a,= Ex= na E ,, aa, which by the discussion preceding (3.6.29) is the intersection of all maximal abelian subspaces of
p that contain X. The fact that x and y are maximally singular points of M(x) is now a consequence of the next result. LEMMA. Let p E M be given, and let it = f + p be the Cartan decomposition determined by p. Let x E M(x) be given, and let X be the unit vector in
p such that x = yx(x), where yx(t) = e"(p). Let Ex denote the intersection of all maximal abelian subspaces of p that contain X. 77ten x is a maximally singular point of M(x) if and only if Ex is 1-dimensional.
PROOF. This follows immediately from (2.20.12) and assertions (1) and (2) of (2.21.8).
(2) Let x E M(x) be a maximally singular point. Let p, f, p, and X E p be as in the lemma above, and let a be any maximal abelian subspace of p that contains X. By the lemma above in (1), Ex is 1-dimensional, and by (3.6.29) it follows that '(X) consists of all positive multiples of X. In particular, X is the only unit vector in i'(X ). By (3.6.31) we conclude that '(x) = (x), and hence (x) = F, (x) is a splice in M(x) by (3.7.5).
REMARK. If (x) is a point in M(x) that is also a splice, then '(x) {x} by (2) of (3.7.2). Hence x is a maximally singular point of M(x) as one may easily see by examing the argument above. (3) Let x E M(x) be a maximally singular point, and let p, f, p and
X E p be as in the lemma in (1). Let y = y1x(-x). Let a be any maximal abelian subspace of p that contains X. We know that Ex is 1-dimensional by the lemma in (1). By (2) of (2.20.9) it follows that there
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exists a subset I of k - 1 linearly independent roots in A C a* such that Ex= ay = n.. r a.. If F* = exp(a1Xp) and S = (x, y), then S is a splice in M(c) by (1) above.
(4) The set a; is a closed convex subset of a that contains the origin. By (3.7.12) there exists a unique r-dimensional subspace a' in a such that a i c a' and a E has nonempty interior, which we denote by
Int(a'), in a'. Let A. _ {a (E A: arc aa), and define a° _ (A E Int(aI):A0- a'na,, for any aEA-A.).Let A, A0. We will show that A, c A ; and a i = EA(A,) for every A E a?, which is a dense open subset of a? in the topology of a'. It then follows that
S = S(A, A,) for any A E a 0, and hence S is a splice in M(oo) by (3.7.15). We show further that a' = ao = fla E A aa. We observe that a+ is not contained in as if a E A - Ao, and hence a' n as is a hyperplane in a' for every a E A - A0. Since Int(a is a dense open subset of a+ we conclude that
(i) a? is open in a' and dense in a'. Our next goal is to show that a'=EA(A,) for every A E a°. Given
A E a o we know that a(A) * 0 for all a E A, since A, c A- A 0. However, a(A) z 0 for all a E A, c I since A E a+. We conclude that (ii) A, c AA for every A E a'
Y.
Note that AA c Ao for all A E a( since a(A) # 0 if a 0- A0. Conversely, if a is any element of A0, then a(A) = 0 for any A E ag since a c a j c a,,,. This proves that (iii) AA = A. for all A E ao,.
We fix A E a z and let B E 4 be given arbitrarily. By (ii) we know that a(B)> 0 for every of E A,. By (iii) we see that AA = AB = A0 and hence a(B) = 0 for all a E AA. Hence B E EA and B E EA(A 1) since
a(B) >- 0 for all a r= A, A. We conclude that a° c EA(A1) since B E a o was arbitrary. Since EA(A) is closed in a and a? is dense in a i by (i) we have shown that (iv)
If B is any element of EA(A1), then a(B) = 0 for all a E AA = A0 and a(B) >- O for all a e A, = - A0. Hence B E aQ for every a E I or equivalently, BEE a'. Since B E EA(A 1) was arbitrary we conclude that EA(A,) c a, and equality holds by (4). Hence S = S(A, A,) is a splice.
Although we do not use this fact, we now identify the subspace a' as the subspace ao = I la E n aa. By the uniqueness assertion in (3.7.12) it
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suffices to show (a) a F c a o and (b) a' has nonempty interior in a o. Assertion (a) follows immediately from the definitions of A0 and ao. To
prove (b) we show that a' contains a neighborhood in ao of any
element A of a'. By (ii) above we know that a(A) > 0 for all a E A, = I - A0. If 0 is a sufficiently small neighborhood of A in ao,
then a(A*) > 0 for all A* e 0 and all a E A1. On the other hand
if A*EOcao=naeAoaa, then a(A*)=0 for all aEAo. Hence a(A*) >_ 0 for all A* E O and all a E A, u A0 2Y., which proves that
Oca-.
El
Now we show that all splices occur as in the examples described above. We continue to use the notation of (3.7.14). 3.7.17. PROPOSITION. Let S be a splice in F(oo), where F = exp(a X p). Then S arises as in (4) of (3.7.16); that is, S = C1(oo), where I is a suitable subset of A c a* such that a E= n a c- f a a is nonemply.
PROOF. We define C(p, S) = Us E s yps[0, °°). (See (8.5.1) below.) We show the following. (1) LEMMA. C(p, S) is a closed convex subset of F.
PROOF OF THE LEMMA. If q, and q2 are distinct points of F with , M with a (0) = p and a (R) c F. Hence it suffices to consider the case that 0 and i = 1, 2. It follows from (3.1.2) and (2) of (3.4.3) that Td(x,, x2) = 4p (xI, x2) = < (q1, q2) < ir. Hence by (3.4.4) there exists a unique Tits geodesic a: [0, 11 -> M(oo) with Q(0) =x1 and 0'(1) =x2.
We observe that Q[0,1] c S. We may write _S = F,(oo) n F2(z) for suitable k-flats F, and F2 since S is a splice in M(me). By the discussion
preceding (3.6.1) we see that v [0,1 ] c F,(x) for i = 1, 2, and hence
o[0,1]cS. Now let Q: [0, 1] -' F be the geodesic segment such that X6(0) = qI and j3(1) = q2. Let x(t) be the unique point in F(x) such that /3(t) is an interior point of the geodesic ray yp,,(,). By the discussion preceding (3.6.1) it follows that x(t) is a reparametrization of the Tits geodesic or
from x, to x2. Hence x[0, 1 ] = 0[0,1] c S, and we conclude that 0[0,1] c C(p, S). This proves that C(p, S) is convex, and it is routine to prove that C(p, S) is a closed subset of F since S is a closed subset of M(oc) in the cone topology. This completes the proof of (1). 0 By (3.7.2) there is a finite set (x 1, ... , X N) c S such that N
(2)
S = U F(x,) S F(x). =1
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232
Let X, be the unit vector in a such that x,
where yx(t) =
e`x'(p). It follows from (3.6.31) that W (x,) = {yy(x): Y e'(X, ), 1Y1 = 1),
where W(X,) is the closed Weyl chamber in a determined by X, as defined in (3.6.29). If W(a, S) = fp'(C( p, S)), where fn: a -+ F is the isometry of (2.10.2), then from (1), (2), and the discussion above we obtain N
C (a , S) = U '(X,) is a closed convex subset of a.
(3)
i=1
By (3.7.12) there exists a subspace a' of a such that CO, S) c a' and CO, S) has nonempty interior, Int(C(a,S)), in a'. Next we assert the following: (4)
Let 1={i:15i5Nand Ex =a'}. ThenC(a,S)= UW(X,). iE(
It suffices to prove that Int(C(a, S)) c U, E IW(X) since Int(C(a, S)) is a dense subset of CO, S) (cf. proof of (3.7.13) below). Let X E-= Int(C(a, S)) be given and choose r > 0 such that Int(C(a, S)) Q B,(X), the open ball in a' with center X and radius r. If 0 < r' < r, then by (3) and the Baire category theorem, B,.(X) n W(X,) contains an open subset of a' for some integer i with 1 < i 5 N. Hence there exists an integer i with 1 5 i 5 N and a sequence (r,,) of positive numbers such
that r -> 0 and Br( X) n W(X,) contains an open subset of a' for every n. It follows that X E '(X, ). Moreover, by (3.6.29) and the uniqueness assertion in (3.7.12) we conclude that Ex, = a' since F(X,) is a closed convex set with nonempty interior in both EX and a'. This proves (4). Next we establish that (5)
if Ao = (a e A: a' c c1,, }, then either a' = a and A. is empty or
a' is a proper subspace of a and a' = n a0. aE AO
If a' is a proper subspace of a, then by (4) and the discussion preceding (3.6.29) we see that AX a AO for all i e I since a' = EX = (la I AX aQ. Hence a ' c (la a A9 ao c (la E A a.. = EX = a' for every i e 1, which proves (5). Next we show that (6)
there exists a subset Al c A - AO such that CO, S)
= a' n I ( a+),awhere a+ ` a eA ,
=(ACa:a(A)>-0)).
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Note that if a E A - A,,, then a' is not contained in a,, and hence as =a' n as is a hyperplane in a'. If i E 1, then AX +,U A. c A - A0 since X; E EX = a'. Moreover, 9-'(X;) = EX n (fl a E ., , a
n, 1E AX (a' n au) by (3.6.29). If A* = U ; E , (AX,U AX,) c A - A0, then n a,)) 9 (U,, E A (a' n as )). by (4) we have C(a, S) = U; E , ( n , , , = - I It follows that dC(a, S) c (Ua E
a finite union of hyperplanes in
a', since d(a' n a+) = a', where d denotes the boundary in a'. If
aEA*, then -aEA* and aa ={AEa:a(A)50}=(a_,Y. The closed half spaces in a' bounded by a,, are a' n aQ and a' n aR = a'
n From (3.7.13) we conclude that there exists a subset A, c A* such that co, s) = na E ,(a' n aa) = a' n (na E 1. a'), which proves (6).
We conclude the proof of (3.7.17). If a E A,,, then -a e A(, since as = a. a, and moreover a,, = aQ n (a a )'. From (5) we conclude that
a' = na E A, aQ if a' is a proper subspace of a, and it now follows from (6) that C(a, s) = na E as , where I = A U A,. This proves that the splice S arises as in (4) of (3.7.16).
PROOFS OF (3.7.12) AND (3.7.13). We conclude the discussion of the classification of splices in M(x) by proving the two elementary convexity results above. Although both results seem virtually obvious we could find no explicit reference in the literature that contained either result. PROOF OF PROPOSITION (3.7.12). Let C be a closed convex subset of FR"
that contains at least two points. Define r = dim C = max{p >- 1: there exists a p-flat F in fly" such that C n F contains an open subset of F}. We leave as an exercise the assertion that an open subset of an r-flat cannot be contained in a p-flat, where p < r. Hence C is contained in at most one r-flat, where r = dim C. Let F be an r-flat in F" such that C n F contains an open subset 0 of F. If there exists a point x E C - F, then the cone C(x, 0) consisting of all segments from x to points of 0 is a subset of C that contains an
open subset of the (r + 1)-flat containing x and F. This implies that dimC > r + 1, a contradiction that proves C c F. Since 0 c C n F= C lies in the interior of C with respect to F the proof is complete. PROOF OF PROPOSITION (3.7.13). We proceed in several steps. Let C c FR"
and the hyperplanes (H,,..., HJ,) of FR" be as in the statement of (3.7.13).
For a convex set C c FR" let dC and Int(C) denote the boundary and
interior of C respectively. Let xy denote the line segment between points x, y E FR", and let Int(xy) denote the interior of xy.
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LEMMA 1. Let C be a closed convex subset of l
whose interior is
nonempty. Let x E C and y E Int(C) be given. Then:
(a) lnt(xy) c Int(C). (b) Int(C) is an open, dense, convex subset of C.
PROOF. Let 0 c Int(C) be an open neighborhood of y. The cone C(x, 0) consisting of all segments between x and points of 0 is contained in C and contains a neighborhood of any point of z in Int(xy). This proves (a), and (b) follows from (a).
0
COROLLARY. Let x c- R" - C be fixed, and define Q: Int(C) -> aC by Q(y) = xy n dC. Then Q is well defined and continuous. We omit the proof, which is a routine consequence of (a) of lemma 1.
LEMMA 2. Let C be a closed convex subset of R" whose interior is nonempty. Let P be a hyperplane in l" such that P n dC contains an open subset of P. Then C lies in one of the closed half-spaces of R" determined by P.
PROOF. Let P, and P2 denote the open half-spaces determined by P.
Suppose that the lemma is false and there are points z, and z, in P, n C and P2 n C respectively. Without loss of generality we may assume that z, E Int(C) for i = 1, 2 by (b) of lemma 1. If z3 = z, z2 n P,
then z3 E Int(C) by lemma 1. Let 0 be an open subset of P with D c P n aC, and let x be a point of 0. Then Int(xz3) c Int(C) by lemma 1 since z3 E Int(C). However, there exists y E Int(xz3) n 0 since 0 is open in P and xz3 c P. We conclude that y E aC n Int(C) _ (0), which is impossible.
HN) be as in the statement of (3.7.13). Let LEMMA 3. Let C and x E l " - C and y E Int(C) be given, and let Q: Int(C) -> dC be defined by Q(z) = xz n aC for z E Int(C). Let S c {1,..., N) be the subset such that
Q(y)EH, if and on lyifiES.Then there exists iESsuch that xEInt(H,') and C c H,-, where H,+ and H,- denote the closed half-spaces in 1k" determined by H,.
PROOF. Let P be the hyperplane containing y that is orthogonal to xy. For e > 0 let De(y) = (y* E P: J y* -yl < e). By the hypothesis of (3.7.13)
there exists S > 0 such that if z E dC n BS(Q(y)) then z E U,= s H,, where Bs(Q(y)) = {6 E Ps": I - Q(y)I < 8}. Choose e > 0 so that (a) Q(DE(y)) c B5(Q(y)) n dC c U, E s H, and (b) Q is one-to-one on DD(y). Condition (a) can be satisfied by the continuity of Q. Condition (b) can be satisfied since xy meets DD(y) c P orthogonally and hence xz n D,,(y) is a single point for all z in D,.(y) if e > 0 is sufficiently small.
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235
For i e S c {1,..., N) define A; _ {y* E DD(y): Q(y*) E H;}. By (a) and (b) above it follows that DE(y) = U;,= s A;, and hence some set A;
contains an open set O; of P by the Baire category theorem. Since Q: O; -> H; is a continuous, one-to-one function and since H; and P are both hyperplanes in R' it follows that Q(O;) is an open subset of H, by the invariance of domain (cf. chap. 1 of [Spi, vol. fl). Since Q(O;) c H; n dC it follows from lemma 2 that C lies in one of the closed half-spaces, _ say H; , determined by H;. The point x cannot lie in H; for then Int(xy) c Int(H,-) by lemma 1, and this would imply that Q(y) = xy n dC E Int(H), contradicting the
fact that Q(y) E Hi by lemma 3. The same argument shows that x cannot lie in Int(H, ), and therefore x c- Int(H, ), which completes the proof of lemma 3. We complete the proof of (3.7.13). Let C* denote the intersection of all closed half-spaces that contain C and are determined by the hyper-
H"). By lemma 3, C* is a nonempty set and clearly planes C C C*. If x is any point in R" - C, then by lemma 3, x E 1W' - C* and we conclude that C* C C. Therefore C* = C, and the proof of (3.7.13) is complete. 3.8.
The Furstenberg boundary
We now describe briefly a boundary of a symmetric space of noncompact type, due to H. Furstenberg, that plays a key role in, among other things, the proof of the Mostow and Gromov rigidity theorems discussed later in chapters 8 and 9. There are several equivalent definitions of the Furstenberg boundary. The most natural of these in our setting is the definition found in [Mos2, pp. 31-341; see also [I]. Let M be a symmetric space of noncompact type and rank k >_ 1. If k = 1, then the boundary FM(x) that we define coincides with the usual boundary M(x) so we assume that k >_ 2. Given a regular unit vector v j=- Sp M we recall from definition (2.12.4) that a Weyl chamber W(y) c M of the second type is given by
W(v) = {exp(ty*): v* E g'(v),t > 0), where 1;'(v) c S,,M denotes the Weyl chamber of the first type determined by v as defined in (2.12.1). The point p E M is called the vertex of the Weyl chamber W(v). We now define an equivalence relation of the Weyl chambers W(v), v a regular vector of SM, which is the exact analogue of the asymptotic
Geometry of Nonpositively Curved Manifolds
236
equivalence relation on unit vectors of SM. See lemma 4.1 of [Most, p. 331.
3.8.1. DEFINITION. For regular vectors v1, v, E SM we say that W(v1) is equivalent or asymptotic to W(v2) if
Hd(W(vl),W(v,)) <x, where Hd denotes Hausdorff distance (see (1.2.12)). We let W(vXx) denote the equivalence class of W(v), and we let FM(x) denote the set of equivalence classes W(vXx), v E SM a regular vector. The set FM(x) is called the Furstenberg boundary or maximal boundary of M. 3.8.2. PROPOSITION. Let M be a symmetric space of noncompact type and
rank k z 2. Let x E FM(x) be given arbitrarily. Then there exists a regular that for any point p E M the Weyl chamber unit vector v E A fW(v(p)) is the unique Weyl chamber with vertex p that represents x.
PROOF. Let v E SM be any regular unit vector such that W(vXoc) =x. We show that such a vector v satisfies the conditions of the proposition. If p E M is any point of M, then W(v(p)Xx) =x by assertion (2) of proposition (2.12.5). We complete the proof by showing that if v* is any
other regular unit vector at p with x = W(v*Xc ), then W(v(p)) = W(v*). If v* is such a vector, then there exists a positive number A such that Hd(W(v(p)), W(v*)) =A. For every positive integer n let p E W(v(p)) be a point such that d(p, y,,.(n)) _ 2. Let p be any point of M. Then the map v - W(v X x) defines a bijection of the Weyl chambers W(v) of regular unit vectors v in SP M onto the Furstenberg boundary FM(x).
Next, we define the action of G =1o(M) on FM(x). 3.8.4. DEFINITION. Let g E G and x E FM(x) be given arbitrarily. We define
g(x) = W(dg(v))(x) = (gW(v))(x), where t' e Al is any regular unit vector such that W(v Xx) = x.
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237
The action of G on FM is well defined by (3.8.2). For each point p E M we define a metric p on FM(x) and we let p denote the corresponding topology on FM(x). We shall see that the topology.,7p- does not depend on the point p. We define a function S: FM(x) -* R(x) as follows: Given x E FM(x) we choose a regular vector v E SM such that W(tv Xx) = x and we let
x* = y,.(x) E R(x). We then define the function S(x) = CG(x*), the center of gravity of V(x*). The function S is well defined by proposition (3.8.2), proposition (2.17.21), and the definitions of W(c) and CG given in (2.12.4) and (3.6.35). It is easy to see that S is one-to-one. 3.8.5. DEFINITION. Let M be a symmetric space of noncompact type and rank k >_ 2. For any point p E M we define a metric p in FM(x) by
I,(x,,x2) = `rp(S(x,),S(xz)), where -
_ 2. Let G =10(M ). Let S: FM(x) - R(x) and T : R(x) - FM(x) be the functions defined above. Then:
(1) S(gr)
for all g E G and all x E FM(-). (2) For each point p of M the function S is one-to-one and continuous with respect to the topology on FM(x) and the cone topology on R(x). (3) T S is the identity map on F1l%f(x). (4) S c T = CG, where CG is the center of gravity map of (3.6.35).
The proof of this result is routine. The next result shows that the map T: R(x) -> FM(x) is a G-equivariant fibration with fiber 0'(x) = T-'(Tx).
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Geometry of Nonpositively Curved Manifolds
3.8.7. PROPOSITION. Let M and T be as above. Then:
(1) The map T: R(a) - FM(x) is sugective and T(z,) = T(z2) for points z,, z2 E R(x) if and only if '(z,) = W'(z2). (2) T(gz) = g T(z) for all g E G for all z E R(-). (3) Let p be any point on M. Then a set 0 c FM(x) is open with respect to the topology p if and only if T-'(O) is open in R(x) with respect to the cone topology. Hence the topology 5'= p does not depend on the point p.
(4) T: R(x) - FM@) is an open map. PROOF. Assertions (1) and (2) follow routinely from the definition of T and the definition of the action of G on FM(x). We prove (3). First, we
fix a point p E M and let 0 c FM(x) be open relative to the topology
p. We prove that T-'(0) c R(x) is open in the cone topology. Let x* E T-'(O) be given arbitrarily, and let x = T(x*) E O. Choose e> 0 such that if y E FM(x) and
R(x) with respect to -9p, we can choose a neighborhood V c FM(x) of x in the topology such that
S(V) c W. By the first two parts of this result and by assertion (3) of (3.8.6) we have (x*))= T(U(x*)) c T(O*) = O. Hence 0 is open with respect to p. We prove (4). Let 0 be an open subset of R(x). We must show that
T-'(T(O)) is open in R(x). Let Y E T-'(T(O)) be given. By (1) of (3.8.7) it follows that T-'(T(O)) = Ux E n W(x), and hence y E W (x) for
some x GO. Since 0 is open in the cone topology there exists a neighborhood U of the identity in G such that U(x) c O. Hence U W(x) = Ug E u g ' W(x) = Ug
u
W(gx) c U., - 0 W(x) = T-'(T(O)).
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However,
239
U-W(x) is a neighborhood of y in Moo) by (3.6.36).
This proves that T-'(T(O)) is open in R(te).
0
We continue the discussion of the Furstenberg boundary FM(x) with two useful corollaries of the result just proved. 3.8.8. COROLLARY. Let M be a symmetric space of noncompact type and rank k >_ 2. Let G =10(M ), and let FM(oo) be the Furstenberg boundary of M. Let K be any maximal compact subgroup of G. Then: (1) G and K act transitively on FM(a). Hence FM(oc) is compact.
(2) The stabilizer in G of a point in FM(x) is a minimal parabolic subgroup of G. More precisely, given x E FM(x) let x* E R(x) be any point such that T(x*) = x, where T: R(x) - FM(oc) is the map defined after (3.8.5). Then Gx = Gx.; that is, (g E G: gx = x) = (g E G: gx* = x*). An analogous statement holds for the stabilizer in K of a point in FM(a). 3.8.9. COROLLARY. Let M be a symmetric space of noncompact type and rank k >_ 2. Let G =10(M). For any point x E R(a) let Ax: G/Gx FM (x) and Bx: K/Kx -* FM(c) be the surjective maps given by
T(gx),
T(gx).
Then:
(1) Ax and Bx are homeomorphisms with respect to the quotient topologies on the coset spaces G/Gx, K/Kx, and the topology 9r =s on FM(oo) defined by (3.8.5). (2) A1(gz) for any g E G and any z r=- G/Gx. B,,(gz) _ g- Bx(z) for any g E K and any z E K/Kx.
Corollary (3.8.9) follows immediately from (3.8.8). We prove (3.8.8). Let x, and x2 be any two points in FM(=), and choose x*, x* E R(oo) so that T(x;) = xt and T(x*) = x2. By assertion (2) of (2.17.24) it follows
that g(x*) E '(x*) for some g e G. Hence g(x,) = gT(x*) = T(gx*) _ T(x*) =x, by proposition (3.8.7). Therefore G acts transitively on FM(oe). By (3) of (1.13.14), G(x) = K(x) for every x r= M(me), and hence
K acts transitively on FM(cc). It follows that FM(x) is compact since it is an orbit of the compact group K.
Next, let x be an arbitrary point of FM(oo), and let x* E R(x) be any point such that T(x*) =x. If g(x*) =x*, then g(x) =gT(x*) = T(gx*) = T(x*) =x. Conversely, if g(x) =x, then T(gx*) =gT(X*) = g(x) =x = T(x*), and hence g(x*) E W(x*) by assertion (1) of (3.8.7). We conclude that g(x*) =x* by (1) of (2.17.25).
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Geometry of Nonpositively Curved Manifolds
3.8.10. SMOOTH MANIFOLD STRUCTURE ON FM(x). Using (3.8.9) we can
make FM(x) into a smooth manifold in a natural way. The space FM(x) is compact by (3.8.8). By (3.8.9), FM(x) is homeomorphic to the coset space G/GX under the map Ax for any point x E R(x). Now give
FM(x) the differentiable structure that makes AX a diffeomorphism. This differentiable structure is independent of the choice of x in R(x) since the parabolic subgroups (GX: x r= R(O) are all conjugate by elements of G by (2.17.23).
THE ACTION OF G =1o(M) ON FM(x). We have already seen by the
remarks following (3.8.5) that the action z: G x FM(x) -+ FM(x) is continuous. In fact, it is easy to show that this action is differentiable with respect to the manifold structure just defined on Moo). Our goal in this section is to show that the natural Lie topology of G agrees with the topology it inherits from the imbedding by i into the homeomorphism group of FM(x) with the compact open topology. 3.8.11. NOTATION. Let X(FM(x)) denote the set of homeomorphisms
of FM(x). R'(FM(x)) becomes a group under composition, and we equipZ(FM(x)) with the compact-open topology. Let r: G ->Z(FM(x)) be the homomorphism that defines the action of G on FM(x); that is, r(g) EX(FM(x)) is defined by (3.8.4) or, equivalently, r(gXTx) = T(gx), where T: R(x) -+ FM(oo) is the projection of (3.8.7). The homomorphism r: G -- X(FM(x)) has trivial kernel (i.e., G acts effectively on FM(-)) by (2) of (3.8.8), (5) of (1.9.4), and the fact that R(x) is dense in M(x) with respect to the cone topology.
By the Lie topology on G =10(M) we mean the compact-open topology in G regarded as a transformation group on M. For later use in the proof of the Mostow rigidity theorem we shall need the following result, which seems to be surprisingly difficult to prove. See the article [GP] for an enlightening discussion of the subtle properties of various topologies on transformation groups. In particular note corollary (5.15) and the word of caution following definition (5.9). We are grateful to R. Palais for the reference to [GP] and for many helpful comments. 3.8.12. PROPOSITION. Let G* = r(G) cY(FM(x)) and equip G* with the topology induced from the compact-open topology of T(FM(-)). If G is given the Lie topology, then the isomorphism r: G -+ G* is also a homeomorphism.
PROOF. The action f : G x FM(x) -* FM(x) is continuous relative to the Lie topology on G, and hence r: G -+ G* is continuous by definition of
the compact-open topology on G*. We have already seen that r is
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241
one-to-one so it suffices to prove that T is an open map. Since FM(c) is a compact metric space, relative to the metric 4,, of (3.8.5) for example, it is routine to show that the compact-open topology on G* equals the
uniform topology on G*, in which basic open sets have the form eE(f) _ (g e G*:
_ 2. Let G = I0(M) and let G be given the usual Lie topology. Given a point p E M and a neighborhood U C_ G of the identity there exists a number
e > 0 and a compact subset C c R(x) such that if
0 and q e M be given. To show that xo = xq it suffices to show that 'q (xo, xq) < 2e. Since Moo) is dense in M(x) with respect to the cone topology we can choose xq* E R(cc) such that 4q (xq, x9) < C. For sufficiently large n, x* E K,, and hence (1) g, (xq*) -- xq as n -> oo
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Geometry of Nonpositively Curved Manifolds
by the choice of (go}. By the angle sum law for geodesic triangles (see (1.4.5)) we observe that g-'(q)
(q,x4)+ 4Cq(gn'(q),x9) g(x) and g(x*) (1) and (2) above we obtain (3)
_ 2, and let 4':9M . +.9M * be a Tits isomorphism. Then 4)
induces a bijection a: FM(x) - F14 *00. If ¢ is equivariant with respect to subgroups r and r* and an isomorphism 0: IF -)- r*, then the bijection a: FM(x) - FM *(x) is also equivariant with respect to 0; that is, a ° T(y) = T*(9y)° a for all elements y E r. PROOF. As above, let 9'0 M and 170-M * denote the collections of maximal elements in .YM and 9M* with respect to the Tits partial orderings.
Let p: 9 M --+ FM (x) and p * : 9 M * _ FM *(x) denote the bijections defined in (3.9.4). Note that 4)(.9M) =.90M* since both 4) and (4))-'
preserve the partial ordering. If we define a: FM(x) -> FM*(x) by a = p* o 4) ° p', then a is a well-defined bijection. If 4) is 9-equivariant, then a routine calculation shows that a is also 8-equivariant by the equivariance properties of p and p* as expressed in (3.9.4). O The next result, due to Tits, is essential to the proof of the Mostow rigidity theorem in the higher rank case. 3.9.7. THEOREM (TITS). Let M and M * be symmetric spaces of noncom-
pact type and rank >_ 2 that have no de Rharn factors of rank 1. Let G =10(M) and G * = I,Z(M * ). Let ¢: 9M -+ 9M * be a Tits isomorphism, and let a: FM(x) -_ FM*(x) be the corresponding bijection. Then
(1) 46 ° µ(G)-W-' = µ*(G*); (2) a 0 T(G)° a' = T*(G*).
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PROOF. Assertion (1) is a reformulation of corollary (16.2) of [Mos2, p. 123], which itself follows from the work of Tits on buildings. Assertion
(2) follows from (1), the definition a = p* o 4) c p', and the equivariance properties of p and p* as expressed in (3.9.4). 3.9.8. COROLLARY. Let M, Z *,
and a be as in (3.9.7). Then there
exists an isomorphism 0: G - G* such that
(1) ¢o µ(g)=µ*(Bg)e ¢ for all gEG; (2) a oT(g) = T*(Og)a a for all g E G. Moreover, if a : FM(c) -> FM *(oc) is a homeomorphism, then 0: G - G* is continuous (and hence analytic).
PROOF. Given g E G there exists an element g* E G* such that 0 - µ(g-(0)- = A*(g*) by (1) of (3.9.7). The element g* is uniquely I
determined by g since G* acts effectively on Al" by (3.9.5). It is routine to check that the map g - g* is an isomorphism of G onto G*. This proves (1). Assertion (2) follows from (1) and the definition of a. We now show that if the bijection a: FM(-) -> FM*(oc) is a homeomorphism, then the isomorphism 0: G -i G* is continuous. We let Z(FM(c)) and ar(FM*(-)) denote the groups of homeomorphisms of FM(oo) and FM *(oc) equipped with the compact-open topologies. Let T: G --* '(FM(oo)) and T*: G* - X(FM*(c)) denote the injective homomorphisms defined by the actions of G and G* on FM(oo) and FM*(oo) respectively (see (3.8.11)). If we equip G and G* with the standard Lie topologies, then the maps T: G -3Z(FM(c)) and T*: G* -+XI(FM*(oo)) are homeomorphic imbeddings by (3.8.12). If we define T,:Z(FM(r)) -,A'(FM*(x)) by T0(f) = a of o a - ', then To is a homeomorphism since a is a homeomorphism by hypothesis. Assertion (2) implies that To G T = T* c 9. Since To is a homeomorphism and T and T* are homeomorphic imbeddings it follows immediately that 0 is continuous. 3.9.9. REMARK. An isomorphism 8: G -+ G* need not be continuous. For example, let G = G* = SL(n, C), and let 6: C -p C be any discontinuous field automorphism (such exist). Then 6 induces a discontinuous automorphism 1: G -* G in an obvious fashion: (j(g)),J = 6(g, j) for
15i,j:n.
3.9.10. Rescaling M by constants Let M be a symmetric space of noncompact type and rank z 2, and let go denote the left-invariant Riemannian inner product. For simplic-
ity we assume that M is irreducible. By (2.3.11) each point p of M determines an inner product gr, on p that arises from the Killing form B on ¢ and the Cartan decomposition g = f + p determined by p.
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Clearly go is determined by its values on p as in (2.3.7). Since M is irreducible it follows from the discussion in (2.3) and from theorem (8.2.9) of [W51 that go = APgp for some positive constant AP.
Conversely, if go is multiplied by an arbitrary positive constant c, then the manifold (M, cg(,) continues to be symmetric of noncompact type and rank k z 2. The geodesics of (M, g0) and (M, cgo) are the same, though with different speeds, and angles between geodesics remain the same. It follows that the boundary spheres (M, g0Xo) and (M, cgoXx) may be identified in a natural way. Moreover, the Tits metrics remain the same with respect to these identifications by (3.1.2) and (2) of (3.4.3). In the sequel we identify the boundaries (M, goXx) and (M, cg0Xx)
together with their Tits metrics, and we omit the definition of any explicit map identifying these boundaries.
3.9.11.
Continuous isomorphisms induce isometries
Next we show that a continuous isomorphism 9: G -> G* induces an isometry f: M - M* (after rescaling the metric of M by constants) such
that 0(g) =f c g o f-' for all g E G. Conversely, we note that for any isometry f: M -> M* the map g -f o g o f-' is a continuous isomorphism of G onto G*. PROPOSITION. Let M and M * be symmetric spaces of noncompact type and rank >_ 2. Let G =10(M) and G* = Io(M*), and let 9: G G* be a continuous isomorphism. Then, after multiplying the metric of M by suitable positive constants on de Rham factors, there exists a unique isometry f: M --> M* such that f o g = 0(g)- f for all g E G.
PROOF. The basic idea of the proof is that there is a non-to-one correspondence between maximal compact subgroups of G and points of M, with a similar statement holding for G* and M*. A continuous isomorphism of G onto G* defines a bijection between maximal compact subgroups of G and G*. One then has to show that the induced map from M to M* is an isometry after rescaling the metrics on M and M* appropriately by constant multiples. We make the outline above precise. First, we describe the one-to-one correspondence between points of M and maximal compact subgroups
of G =10(M). Given a point p of M the subgroup K = Gp = (g c G: gp = p) is a maximal compact subgroup of G by (1) of (1.13.14). Conversely, if K is a maximal compact subgroup of G, then K fixes a unique point p of M by (4) of (1.13.14). Now let M and M* be two symmetric spaces of noncompact type and rank >_ 2, and let G =10(M) and G* = I0(M*) be their connected
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isometry groups. Let 0: G -* G* be a continuous isomorphism. It is well
known that 0 must in fact be real analytic since G and G* are Lie groups. (See, e.g., theorem 2.6 of [Hell, p. 107].) Since 0 maps maximal compact subgroups of G onto maximal compact subgroups of G* we may define a bijection f: M - M* by the requirement (1) 8(GP) = G7P for every point p E M,
where as above GP and Gfp denote the maximal compact subgroups of G and G* that fix p and fp. One may reformulate (1) as
(1') fog=6(g)of for all To see the equivalence of (1) and (1') note that for any points p e M and g e G we have Gf gp) = 8(Ggp) = 0(gGpg-') = 0(g)8(Gp)8(g)-' = 8(g)Gfp0(g)-' = GG(g)fp. It follows that f(gp) = 8(gXfp), which is equivalent to (1'), since a maximal compact subgroup of G* fixes a unique point of M*. Assertion (1') says that the bijection f: M-+M* is equivariant with respect to 0. Conversely, if (1') holds, then it is straightforward to prove that (1) holds. This proves that f is uniquely determined by (1').
Let M and M* be equipped with the G-invariant canonical inner products of (2.3.11) that are defined by the Killing forms B and B* of g and 0. By (2.3.9) the canonical inner products restricted to irreducible de Rham factors are constant positive multiples of the original locally
symmetric metrics. Hence it suffices to show that f: M -+ M* is an isometry. We begin by showing the following. (2) If g = t ® p is the Cartan decomposition determined by a point p
of M, then g * = do(f) ®d 8(p) is the Cartan decomposition determined by f(p). We note that de(p) is the orthogonal complement in g* of d8(t) relative to the Killing form B * since B(X, Y) = B*(d 8(X ), MY)) for all X, Y E g. To prove (2) we need only show that do(f) is the Lie algebra of Gjp = 8(GP). But this is obvious since t is the Lie algebra
of G. Now let p and q be distinct points of M. We show that d(fp, fq) _ d(p, q), which proves that f is a differentiable isometry. Let g = t ® p be the Cartan decomposition determined by p. By (2.4) there exists
X E p such that y(t) = e'X(p) is a geodesic of M with y(0) = p and y(1) = q. From (2.3.11) we know that d(p, q) = B(X, X)''2. From (1') above we obtain f(y(t)) = (f a e") (p) = 8(er 'X fp) =e1ae(X)(fp) It follows from (2), (2.3.11), and (2.4) that f(y(t)) is a geodesic of M* with
speed B*(d 8(X ), d0(X ))''2 = B(X, X)'i2. Since f(y(0)) = f(p) and f(y(1)) = f(q) we conclude that d(fp, fq) = B(X, X )'I' = d(p, q). 0
Tits Geometries
3.10.
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Boundary rigidity
Let P M --+ M * be an isometry between two symmetric spaces of noncompact type and rank >_ 2. If 46: M(x) -+ M*(x) denotes the bijection induced by f, then 46 is a homeomorphism with respect to the cone topologies and an isometry with respect to the Tits metric Td. The main goal of this section is to prove the converse of this result for symmetric spaces of noncompact type that have no rank-1 factors. 3.10.1. THEOREM. Let M and M * be symmetric spaces of noncompact type and rank >_ 2. Let M have no rank-I factors, and let 4): M(x) M*(x) be a bijection that is a homeomorphism with respect to the cone topologies and an isometry with respect to the Tits metrics Td. Then, after multiplying the metric of M by positive constants on de Rham factors, there exists a unique isometry f: M -4 M* such that f induces the map 4): M(x) -> M*(x). REMARK. If M = M, X M, is the Riemannian product of two symmetric spaces of rank 1, then Td(x;, y;) = IT for any two distinct points x;, y; E
M;(x) for i = 1, 2. (This follows from the fact that Td(x,, x,) = it/2 for any two points x, E M,(x), x2 (=- M,(").) In view of this one may show that the result just stated fails if M is allowed to have rank-1 factors. We omit further details. As an immediate consequence of (3.10.1) we obtain the following. 3.10.2. COROLLARY. Let M be a symmetric space of noncompact type and
rank k >_ 2 that has no rank-1 factors. Let 0: M(x) -- M(x) be a bijection
that is a homeomorphism with respect to the cone topologies and an isometry with respect to the Tits metric Td. Then there is a unique isometry
f: M -> M such that f induces the map ¢: M(x) -, M(x). REMARK. Note that unlike (3.10.1) no rescaling of the metric of M is necessary in (3.10.2) since M is isometric to itself. The necessity of rescaling in (3.10.1) is explained in (3.9.10). We shall also give two further applications of (3.10.1). In (3.10.3) we extend (3.10.1) to the case that only one of the manifolds M and M* is assumed to be a higher rank symmetric space with no rank-1 factors. In (3.10.4) we weaken the hypothesis of (3.10.1) on the map 4): M(x) M*(x) by assuming that 0 is a homeomorophsim with respect to the cone and the Td topologies. We conclude that M and M* are isometric after rescaling, but we are no longer able to say that 46 is induced by an isometry f : M -> M* since 46 is not assumed to be an isometry with respect to the Tits metrics Td.
We are now ready to begin the proof of (3.10.1). First we use Q,: M(x) -> M*(x) to construct an isometry f: M -+ M* after rescaling the metric of M by positive constants on de Rham factors. Then we
Geometry of Nonpositively Curved Manifolds
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show that the map from M(x) to M *(x) induced by f equals 0. Finally, we show that f is uniquely determined by 0. Let 0: M(x) - M*(x) be a bijection that is a homeomorphism with
respect to the cone topologies and an isometry relative to the Tits metrics Td. We define a Tits isomorphism 4 :.9M - M* given by 0([x]) = [4(x)], where x is any point of M(x) and [ ]: M(x) -67M denotes the projection; the bijection 0 is well defined and preserves the Tits partial ordering by (3.9.2) since 4) preserves the metrics Td. Note that 4 (F(x)) = 9(4)x) for all x E M(x) (cf. (3.6.24)) since 0 preserves the Tits partial ordering. Moreover, .0 maps R(x) onto R*(x) since the regular points are the maximal points with respect to the Tits partial ordering and ¢ preserves this partial ordering. The bijection 4):9M -* 9?M* induces a bijection a: FM(x) -* FM *(x) by (3.9.6), and one may verify that a o T = T* o 0: R(x) -> FM *(x), using the notation of (3.8.7).
Hence a is a homeomorphism since T and T* are continuous open maps and ¢: R(x) - R*(x) is a homeomorphism in the cone topologies.
Let G and G* denote the isometry groups 10(M) and 10(M*) respectively. By (3.9.8) there exists a continuous and hence analytic isomorphism 0: G --, G*. By (3.9.11) we can multiply the metric of M by
positive constants on de Rham factors and construct an isometry f: M -* M* such that f c g = 0(g)o f for all g E G. The next stage of the proof is to show that the map F: M(x) -. M*(x) induced by f: M --> M* equals 46: M(x) - M*(x). This will require two lemmas.
LEMMA A. Let M be a symmetric space of noncompact type and rank k z 2. Let 4: M(x) -+ M(x) be a bijection that is a homeomorphism with respect to the cone topologies and an isometry with respect to the Tits metric
Td. Suppose that M*(x). If f: M --> M* is
another isometry that induces the map 0, then h =f` o f is an isometry of M whose induced map on M(oo) is the identity. It follows that h
is a Clifford translation of M by (5) of (1.9.4). Since M has no Euclidean de Rham factor it follows from (3) of (1.9.4) that h = e and
f =1. This completes the proof of (3.10.1) except for the proofs of lemmas A and B. We divide the proof of lemma A into two steps. STEP 1. If x e R(oo), then ql(x) E Moo) and
We begin by showing that
(* *)
qi(x) (=- F(yyx)(ao)
for any point p E M.
Fix a point p E M. Since x E R(oo) the geodesic yPx is regular by (2.17.17) and F(ypx) is a k-flat in M by (2.11.1). By hypothesis, q q (x, ift) = Td(x, ilix) s ir. If Td(x, qix) = Tr, then the 1-dimensional foliation q -> ygx(0) is parallel in M by proposition 2.2 of [EC], and M admits a nontrivial Euclidean de Rham factor, which is impossible since M is symmetric of noncompact type. Hence it follows from (3.1.2) that yPx[0, oo) and yy, (x)(0, ao) bound a flat triangular sector A in M. By (3.6.3)
we know that 0 is contained in some k-flat F, and since yyx[O, x) c 0 c F it follows that F = F(y, ). This proves (* * ).
By the equivalence of (2) and (3) in (3.6.26) and (* *) above we conclude that 4i(x) <x with respect to the Tits partial ordering. However, +y(x) is maximal with respect to the partial ordering since x E R(x) is maximal with respect to the partial ordering and 4i preserves the partial ordering. It follows that x 5 t #(x) and g x = g 4,(x), which proves step 1. STEP 2. Let F be any k -flat in M. Then
fixes every point in F(oo).
Step 2 will complete the proof of lemma A since every point x of M(me) lies in F(ao) for a suitable k-flat of M.
Let F be a k-flat in M, and let p be any point of F. Let g = t + p be
the Cartan decomposition determined by p, and let a c p be the maximal abelian subspace such that exp(a )(p) = F. If X is a unit vector of p we recall that yx(oo) denotes the asymptote class in M(,) of the geodesic yy(t) = erx(p)
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Let p denote the unit sphere in p, and define a continuous map /J*: 3 - pI by +i:2. The next result appears in appendix 4 of [BGS, 224-2291. Using harmonic maps as a tool this result and its generalization to the case where one of the manifolds has arbitrary nonpositive sectional curvature have recently been proved by Mok, Sui, and Yeung in [MSY]. COROLLARY. Let M be a symmetric space of noncompact type with rank >- 2 and with no rank-1 factors. Let M* be an arbitrary complete, simply connected manifold with sectional curvature K< 0. Let 3.10.3.
45: M(oo) -> M*(c) be a bijection that is a homeomorphism with respect to
the cone topologies and an isometry with respect to the Tits metrics Td. Then, after multiplying the metric of M by positive constants on de Rham factors of M, there exists a unique isometry f : M - M * such that f induces the map 0: M(me) -> M*(oo).
PROOF. By (3.10.1) it suffices to prove that ll%1 * is a symmetric space of
noncompact type with rank z 2.
We consider first the case that M is irreducible. We begin by showing that M* is also irreducible, and we then apply (5.4.4) below to
conclude that M* is symmetric. Suppose that M* is reducible, and write M* is a Riemannian product M; x M; . For i = 1, 2 the sets A* = M* (x) satisfy the two properties listed in the statement of (5.6.2)
as the reader may verify without difficulty from the discussion of product manifolds beginning at (3.6.57). If A. = 4-'(A*) c M(o) for i = 1, 2, then the sets A, and A 2 also satisfy the two properties of (5.6.2) since 0': M*(oc) -* M(oo) is an isometry with respect to the Tits metrics Td. By (5.6.2) we conclude that M is a Riemannian product M, X MZ such that M;(oo) = A, for i = 1, 2. This contradicts the assumption that M is irreducible, and we conclude that M * is also irreducible. Let S = M(me) - R(m) denote the singular points in M(me), and let
S* = 4(S) cM*(oc). The set S* is a proper subset of M(me) that is closed with repect to the cone topology since the set S has these properties in M(ao). To show that M* is a symmetric space of noncompact type with rank > 2 it suffices by (5.4.4) below to show that S* is involutive; that is, yp.x.( - cc) = Sp.(x*) E S* for all points p* E M* and X* ES
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Let p* E M* and x* E S* be given, and let y* = yP.x.(-oo). Then Td(x*, y*) z
- lr. However, Td(x,, x2) < ,r for any two points x,, x2 E =M(00) if M is symmetric of noncompact type
with rank z 2 by (3.6.1). Hence Td(x, y) = Td(x*, y*) = a, and by (3.6.1) we conclude that y = y,,,(- cc) for some point p r =M. By hypoth-
esis the point x = 4-'(x*) lies in the set of singular points S since x* E S* = O (S). The set S of singular points is involutive by example (2)
of (5.3), and hence y = y,, ( - oo) = SP(x) E S. Therefore y* _ 4(y) E ¢(S) = S*. This shows that S* is an involutive subset of M(x), and it follows by (5.4.4) below that M* is a symmetric space of noncompact type with rank > 2. We have shown that M* is symmetric of noncompact type with rank
2 if M is irreducible. Next we consider the case that M is a Riemannian product M, x ... X Mr, where r;-> 2 and each M, is irreducible for 1 < i s r. We show by induction on r that M * can be written as a Riemannian product M; x x M* such that 4(M,(oo)) = M*(co) for 1 < i < r. It will then follow from the work above that each factor M,* is symmetric, and hence M* is symmetric. We use here the fact that
the restriction to M,(cc) of the Tits metric on M(x) is the same as the Tits metric of M,(cc). To check this it suffices by (3.6.1) and (2) of (3.4.3)
to consider points x, and yi in M,(-) with Td(x,,y,) < 7r. In this case there exists a unique Tits geodesic a in M(c) that joins x, to y, by (3.4.4). Since v must lie in M,(oo) by (3.4.4) it follows that Td,H(x,, y,) _ Tdg(xi, y,).
Write M as a Riemannian product M, X . xM, and consider first the case r = 2. For i = 1, 2 let A, = M(oo) and A* = O(A,) c M*(-). The sets A, and A2 satisfy the two properties listed in the statement of (5.6.2), and hence A; and A also satisfy these properties in M *(oo). By (5.6.2) it follows that M* is za Riemannian product M* X M2 such that M* (oo) = A* for i = 1, 2. This completes the case r = 2. Next let r >- 3 be any integer and write M = M, X M2, where M2 is the Riemannnian product M2 x Mr. By the work above we can write M* as a Riemannian product M* X MZ such that 4,(M,()) = M* (oo) and 4(M2(w)) _ WOO. We now restrict 46 to M2(oa) and apply the induction hypothesis
It follows that M* is a Riemannian product X Mr such that 4(Mi(c)) = M* (oo) for 1 < i _< r. This completes the proof of (3.10.3) as explained above. to
M; x
In the next result we relax the restriction that 0: M(me) -, M*(-) be an isometry with respect to the Tits metrics, and we require instead that di preserve the Td-topology. This weakens the conclusion slightly since
0 will not be induced by an isometry f: M -- M* unless 0 does preserve the Tits metrics.
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3.10.4. PROPOSITION. Let M and M* be symmetric spaces of noncompact type with rank ? 2 and with no rank-1 factors. Let 4): M(oo) --* M*(oo) be a bijection that is a homeomorphism with respect to both the cone and Td topologies. Then, after rescaling the metric of M by positive constants on de
Rham factors of M, there exists an isometry f: M - M*.
PROOF. We may assume that M is irreducible. The passage to the case
that M is reducible is then carried out exactly as in the proof of the previous result. The map 0 preserves the Tits partial orderings by the discussion in (3.9.2), and hence 0 induces a Tits isomorphism :3' --> Y1%%1 * given by 4)([x])
= [ 4)(x)] for all x c- M(oo). Note that 4)(R(oo)) _
R*(oo) since the regular points in M(co) and M*(c) are maximal with
respect to the partial orderings, which 0 preserves. The bijection ¢:YM ->9M* induces a bijection a: FM(co) - FM*(oc) by (3.9.6), and
a is a homeomorphism for the same reasons given in the second paragraph of the proof of (3.10.1). The assertion of (3.10.4) now follows from (3.9.8) and (3.9.11). 0
3.11.
Rigidity for irreducible quotients of reducible symmetric spaces
Let M and M* be symmetric spaces of noncompact type of rank z 2 that are nontrivial Riemannian products. Let r c_ G = 10(M) and r* c G* =10(M *) be lattices that are algebraically irreducible; that is, they have no subgroups of finite index that are direct products. The main result of this section, theorem (3.11.7) below, is that if ¢:YM --+YM* is a Tits isomorphism that is equivariant with respect to an isomorphism 9: t --* r*, then 0 extends to a continuous isomorphism of G onto G*. It then follows by (3.9.11) that after normalizing the metric of M by positive constants on de Rham factors there exists an isometry f: M M* such that fog = 9(g) for all g e G. See (1) of the remark following (3.6.24) for a definition of YM.
Before proving the result just described we need to examine the geometry of M(m) as begun in propositions (3.6.57) through (3.6.61) in
the case that M is a product manifold. If M = M, x ... x Mk we let G =10(M) and G; =10(M; ), 1 < i:5 k. Note that G = G, x ... x Gk3.11.1. PROPOSITION. Let 11%1 be a symmetric space of noncompact type
that is a nontrivial Riemannian product M, x ... X Mk, k >_ 2. Then there exists a natural homeomorphism *: Moo) -, FM,(ao) X ... x FMk(co) with respect to the topology of (3.8) defined for FM(c3) and the corresponding product topology for FM,(c) x . . . X FMk(x). Moreover, 41 is equivariant
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257
with respect to the action of G; that is, 41 o T(g) = A(g)o i/i for all g E G, where
A(g)(T,(x,),...,Tk(xk)) = (T,(g,x,),...,Tk(gkxk))
forg=(91,...,gk)EG=G, x ... xGk. PROOF. We begin by explaining the notation of the last sentence in more detail. Let T: R(x) - FM(x) and T,: R,(x) --a FM,(x) denote the projections defined following (3.8.5), where R(x) and R;(x) denote the regular points of M(x) and Mi(x), 1 5 i:5 k. The points x,, 1 5 i:5 k, above are arbitrary points of R,(x). Now let 0: R(x) -, R,(x) x x Rk(x) denote the surjective, continuous open map of (3.6.60). Let 4),: R(x) a R,(x) be the projections defined in (3.6.61) so that 4)(x) _ (4) (x),..., 4. (x)) for every x E R(oc). Define 0: FM(x) - FM,(x) x x FMk(x) by 41(Tx) = (41(Tx),..., 4rk(TX)), where 4,: FM(x) --a FM,(x) is given by 4r,(Tx) = T,(4),x)
for 1 FM *(oc) be the bijection of (3.9.6) induced by 4). Let /y: FM(c) -> FA(o) and 0,*: FM *(oo) - FM* (r) be the maps defined by 41 and 41* as in the proof of (3.11.1). Then for each i with 1 _< i 5 k there exists a bijection i9,: FM;(oo) -* EM* (w) such that f3; o iJr; = qr;* o a. If a is a homeomorphism, then each f3; is a homeomorphism.
PROOF. (1) For 1 S i 5 k we define A* _ (x* E M*(c): [x*] E (91N;)}, where [ ] denotes the projection of M(oo) onto 9M and also of WOO
The sets (A, = onto 9M*, and where 9M; = {[x] E9M: x cMi(x), 1 < i < k) satisfy the hypotheses of (3.6.62), and hence {A 1 5 i 5 k) satisfy the hypotheses also since 4):9M _,.?M* is a Tits isomorphism. By (3.6.65) we can write M* as a Riemannian product M;* x ... x Mk*., where M;* (c) = A* for 1 5 i:5 k. Finally,
([x*]
E9M*:x* EM*(Z)=A*)=4)(9M;)for 1 _ 2. Let M be a Riemannian product M* X M2*, and let F C- G =10(M) be an irreducible lattice. Let 0: F -- I'* be an isomorphism onto a lattice F* C G* = I (M*). Let 0:-q'Af _ .?M* be a Tits isomorphism that is equivariant with resect to 0 (cf. (3.9.5)-(3.9.6)), and assume that the induced bijection a: FM(.c) -+ FM*(-) of (3.9.6) is a hotneomorphism. Under the hypotheses above 0 extends to a continuous (hence analytic) isomorphism of G onto G*. Moreover, after multiplying the metric of M by positive constants on de Rham factors, there exists an isometry f : M -p M
such that fog=0(g)of forallgEG. PROOF. By (3.11.3) we can write NI * as a Riemannian product M* _
M* X M* in such a way that 4(5M,) =9M;* for i = 1,2. Write Gi =10(M) and G; =10(M;*) for i = 1, 2, and let p;: G --- G. and P, : G* - G* denote the projection homomorphisms. (Note that G = G, x G2 and G; x Gz.) Let r, =p,(F) C G; and ri* =p*(r*) c G* for i = 1,2. The proof of (3.11.7) falls into two parts, which we now describe briefly. For an element y of 1' we write y = (y,, 72), where y; =p;(y) E I; C G; for i = 1, 2. Similarly, we write y* = (y*, y2) for an element y of r*. By (b) of (3.11.6) the isomorphism 0: 1' - 17* induces isomorphisms 9;: F, --> l,* for i = 1,2, such that 8(y) = y;*, where 0(y) = y*. The fact that I; is topologically dense in G; by (a) of (3.11.6) allows us
to extend each
9;
to an isomorphism of G; onto G* for i = 1, 2
(lemma A).
In the second part of the proof we use lemma A to extend 0 to an
isomorphism of G onto G* such that a -r(g) = r*(9g)- a for all g e G, where a: FM(S) --' FM*(--) is the homeomorphism of (3.9.6) and
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T and T* denote the actions of G and G* on FM(x) and FM*(x) (lemma B). The proof of (3.11.7) is now essentially complete. By the proof of (3.9.8) we conclude that 0 is continuous, hence analytic. The construction of the 0-equivariant isometry f: M -* M* then follows from (3.9.11).
Before beginning the proof we establish some notation. 3.11.8. NOTATION. (a) From (2) of (3.11.3) there exist homeomorphisms /3;: FM;(x) -+ FM;* (x) such that (3, o +i; = 4i,* o a for i = 1, 2. Here
a: FM -' FM*(x) is the bijection of (3.9.6) induced by /. By hypothesis,
a is assumed to be a homeomorphism. Moreover, 41=01,1412): FM(x) FM,(x) x FM2(x) and /i* = (0*, GZ ): FM*(x) -FM*(xx FM*(a) are the homeomorphisms constructed in (3.11.1).
(b) Let /3= (/3,,/32): FM,(x) x FM2(x) - FM*(x) x FM* (x). The maps /3, are homeomorphisms for i = 1, 2 by (2) of (3.11.3) since a is a homeomorphism. Hence /3 is a homeomorphism. (c) For elements g E G,_and g; E G1, let T(g) and T,(g,) denote the
actions of g and g, on FM(x) and FM,(x) for i = 1,2. Define r*(g*) and T,*(g*) similarly for elements g* E G* and g* E G. (d) For elements g E G and g; E G;, let µ(g) and µ;(g;) denote the actions of g and g; on .9M(x) and .7M;(-) for i = 1, 2. Define µ*(g*) and µ*(g*) similarly for elements g* E G* and g* E G*.
(e) For an element g E G let A(g) denote the action of g on FM 00 X FM ,(x) given by A(g) = (T,(gl ), r2(g2)), where g = (g,, g2) E G = G, X G2. Define A*(g*) similarly for g* E G*. We now begin the proof of (3.11.7).
LEMMA A. Let y be an element of I' c G, and let y* = 8(y) E r' c G*. Then:
(1) a -r(Y) = T*(y*)o a.
(2) If Y;=p;(Y)EF1 and y*=p*(y*)EI',*, then /3;or,(y;)_ T*(y,*)0 /3; for i = 1,2.
is defined by Q;(f) _ 13, o f o /3, i, then Q;(T;(G,)) = T;* (G*) for i = 1,2. In particular,
(3) If Q,: Z(FM;(x)) - e`Z'(FM;*(o=)) there exist isomorphisms 0,: G.
G* for i = 1, 2 such that
9(y1 , Y2) = (81(y1), 92(72)) for every Y = (Y11 Y2) E r.
PROOF OF LEMMA A. (1) By hypothesis :5M _45M* is equivariant
with respect to 0; that is, 0 o µ(y) = µ*(y*)o 4 for every element yE I', where y* = 9(y) (=- F. It follows routinely from the definition of
a: FM(x)- FM*(x) in (3.9.6) that a cT(y) = T*(y*)o a. (2) From the equivariance assertion 41 -r(g) = A(g)e 4 in (3.11.1) it follows from the definitions of in (3.11.1) and A(g) =
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(T,(9,), T2(g, )) in (e) of (3.11.8) that (a) / i ; oT(g) = T,(g,)o +/r, for i = 1, 2,
where g = (91, 92) E G = G, X G2. Similarly, ii,* 0T*(g*) = T*(g*)o q,,*
for i = 1,2,where g*=(g;,g;)EG*=G; xG2. From (a) of (3.11.8) we have (b) J3, o 41, = qi,* o a for i=1,2. From (a) and
(b) we compute
iy,* u a o T(y) = p, U 41, o T(y) _
(3, o T,(y,) o 41, for i = 1, 2. On the other hand, from (a) and (b) and (1) above we compute r/i,* a r(y) _ r,* oT*(y*) a = T*(y*)a $, i;. This proves that
)3 o7,('Y,)°4, =T,*(y,*)°R,°4G,
fori= 1, 2.
Since t/i,: FM(x) --> FM,(-) is surjective for i = 1,2 we obtain assertion (2) of lemma A. We prove (3). Since l: c G = G, X G2 is irreducible by hypothesis it follows from the definition of irreducibility in (3.11.4) that F* = 9(r) is irreducible in G* = G i x G? . By (3.11.6) we conclude that F, = p,(F) is dense in G, and r,* = p* (I,,*) is dense in G* for i = 1, 2 with respect to the Lie topologies of G, and G,*. By (3.8.12) we know that for i = 1, 2 the maps r,: G; -Z(FM,(x)) and r,*: G* - A (FM,*(x)) are homeomor-
phic imbeddings onto their images r,(G,) and r*(G*) with respect to the Lie topologies in G, and G* and the compact-open topologies on Z(FM,(x)) and Z(FM,*(x)). Hence
(a) r,(F) is dense in r,(G,) and T*(F,*) is dense in r,*(G*) for
i=1,2.
If Q,:'(FM,(x)) _ r(FM,*(x)) is the homeomorphism defined in (3) of lemma A for i = 1, 2, then we may restate (2) of lemma A as
(b) Q,(T,(y,))=r*(y*)for i=1,2and every y=(y1,y2)EFcG= G, xG2. Hence Q,(T,(F, )) = r* (I;*) for i = 1, 2. Passing to the closures of these sets in the compact-open topologies of G; and G* and using (a) and (b) we obtain (c) Q,(T,(G,)) = r,*(G*) for i = 1, 2.
This is the first assertion of (3) of lemma A. Given an element g, E G, there exists an element g* E G* such that
Q,(r;(g,)) = T,*(g*) by (c) above. The element g* is uniquely determined by g, since G* acts effectively on FM,*(x) by (3.9.5). It is easy to check that the map B,: g; --> g* is a homomorphism of G, onto G* and 0, has trivial kernel since G, acts effectively on FM,(x). Hence 0,: G, -->
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G* is an isomorphism for i = 1,2, and the assertion that 0(y1, y2) _ (81(y1), 02(y2 )) for every y = (y1, y2) E r is equivalent to (b) above. This completes the proof of lemma A.
We now use lemma A to extend the isomorphism o: r - r* to an isomorphism of G onto G*. LEMMA B. The isomorphism o: r - r* extends to an isomorphism of G onto G* such that
ao-r(g)=T*(8g)oa for all PROOF OF LEMMA B. We break the proof of lemma B into two steps. STEP 1. Let /3 = (/31, J32): FM1(c) X FM2(o) -* FM* (x) x FM2 (oo) be the homeomorphism discussed in (a) and (b) of (3.11.8). Then for every g E G
there exists a unique element g* E G* such that /3 o A(g) = A*(g*)o 16, where A(g) is the action of g defined in (e) of (3.11.8).
Let g = (g1, 92) E G = G1 X G2 be given. By (3) of lemma A there exist elements g* E G* such that Q;(T;(g,)) = T* (g*) or, equivalently,
fori=l,2. (*} A oTi (g) =r,*(g*)o$, From (e) in (3.11.8) and the equation of (*) we compute J3 ° A(g) = /3 -(TI (gl ), T2(g2 )) = ( N1 ° 'rl (gl ), 132 -'r2(92))
_ (TI (gl )o '1,T2 (g2 )o 162) = A*(g*)o J
The element g* is unique since G* acts effectively on FA*(-) for
i=1,2.
STEP 2 (Conclusion). The map g --> g* of step I is easily seen from the assertion of step 1 to be a surjective homomorphism of G onto G*, and we denote this homomorphism also by 0. Comparing (*) of step 1 with
(2) of lemma A we see that 0 extends our original isomorphism 0: IF -> r*. The assertion of step 1 shows that 0 has trivial kernel since G. acts effectively on FM,(-) for i = 1, 2.
It remains only to show that a -r(g) = T*(g*)o a for all g e G. From (3.11.1) we recall that
(a) 41 a r(g) = A(g)o ,/ for all g E G, and i* o r*(g*) = A*(g*)o r/r*
for all g*EG*. From (2) of (3.11.3) we know that J3; o tlr, _ +/,* o a for i = 1, 2. Since Q=(13 ,12), 4=(+/11,42), and 1/,* =(1/ii ,1/rZ ) we obtain (b) 0 o
* o a.
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Let g E G be given. From (a), (b), step 1, and (3.11.1) we obtain
W*Da°T(g)=R °4°T(g)=0 °A(g)o /i=A*(g*)°fioi/i = A*(g*)° 4,*° a=rA*°T*(g*)° a. Hence a ° T (g) = T *(g *) ° a since * is one-to-one. This completes the proof of lemma B.
The isomorphism 0: G - G* just constructed is continuous by the proof of (3.9.8). From (3.9.11) we conclude that after multiplying the metric of M* by positive constants on de Rham factors there exists an isometry f: M -> M* such that f ° g = 0(g) for all g c- G. This completes the proof of (3.11.7). 3.12. The Karpelevic boundary of ) Let M be a symmetric space of noncompact type and rank k >_ 2. In [Ka] Karpelevic has defined a boundary for M that is more refined than M(o) with respect to the properties of singular points at infinity. A point x r= M(co) - R(oo) determines a whole complex in the Karpelevic boundary, while a point x E R(-) determines a single point in the Karpelevic
boundary. We shall not go into detail here about the Karpelevic boundary, and we refer those interested to section 13 of [Ka]. Instead we give a somewhat vague flavor of the nature of this boundary. Fix a point p EM, a symmetric space of noncompact type and rank k >_ 2. For any x r= M(co) we decompose F(ypx) as a Riemannian product E(ypx) X F,.(ypx) according to (2.20.10) and (2.20.13), where E(ypx) is the
intersection of all k-flats in M that contain ypx and F,(ypx) is a symmetric space of noncompact type (possibly empty) whose rank is less than the rank of M. Consider the set of finite sequences (x1, x2, ... , xN) where (1) x, E M(c) for every 1 < i < N, (2) ypx,(R) S n k _ , FS(ypxx) for every 1 < i < N - 1. We define two sequences x = (xl, x2,..., xN) and y = (y1, y2,..., yM) to
be equivalent if N=M, x, = y1, and x,= 4(y,) for some 0(-=G.,, = Gy, and all i >: 2. (This removes the dependence of x = (x1, x2,..., xN) upon the base point p.) The set of equivalence classes of sequences x =
(x1,...,xN) is the Karpelevic boundary of M, denoted KM(oo). If x, E R(x), then FS(ypx) is empty and x, determines a single point in KM(-). With a suitable topology the space M U KM(oo) becomes a
compactification of M and G =l()(M) acts by homeomorphisms on KM(-). Pointwise description of KM(cr) Fix a point p E M, and let c = f + p, be the Cartan decomposition determined by p. We now describe the Karpelevic boundary KM(oo) by
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algebraic objects in V. We then illustrate these objects for the symmetric space M,, = SUn, R)/SO(n, IR) discussed in (2.13). PROPOSITION. Let (x1,...,xN) be any finite sequence of points in M(x), and let {X,,..., XN} be those unit vectors in p such that ypX(t) = e`X'(p)
for all t c- R and 1 < i:5 N. Then {x1,.. . , x,,) satisfies (2) above and defines a point in KM(-) if and only if (a) [X;, XJ} =0 for I S i, j 5 N and
(b) X; E E' for all i >- 2 and 1 < j < i - 1, where E
is the intersection of all abelian subspaces of p that contain X, (cf. (2.20.8)).
REMARK. By definition, X E EX for any element X of p, and hence the sequences (X,,..., XN) in p that satisfy (a) and (b) are orthonormal and
span an abelian subspace of V. It follows that N< rank M. PROOF OF THE PROPOSITION. From (2.1 1.4) we recall that F(yp.,) =
exp(Z(X;) n p). Hence by (2.20.3) and (1) of (2.20.10) it follows that F,(ypz) = exp(Z(X;) n p n EX) for 15 i < N. It follows that FJyp.,)=exp(p,Xp),whereppn((1kv,Z(Xk))n(ni E,r,} (1 If {x,,..., xN} satisfies (2), then p; 0 (0) for 1 < i 5 N and X,, , E p; for
1 5 i 5 N - 1. Hence (a) and (b) hold by the definition of p;. Conversely, if (X,,..., XN) are unit vectors in p that satisfy (a) and (b), then X;+, E p; for I < i < N - I and ypz,+ (R) c exp(p;Xp) = flk=1 EXAMPLE. As in
(2.13) we let M = M = SL(n, R)/SO(n, R), p =
PSL(n, I8), fi = (n x n real matrices of trace 1 SO(n, R), G zero), f = (skew symmetric n x n real matrices), and p = (symmetric n x n real matrices of trace zero).
ASSERTION. Let (X,,..., XN) be a finite sequence of unit vectors in p. Then (X,,..., XN) satisfies (a) and (b) above if and only if the following conditions are satisfied: (i) X; X1 - X1 X; = 0 for 1 < i, j S N.
(ii) For any integers i, j with 1 < i < j < N it is true that X1 leaves invariant any eigenspace V of X; and trace (X1 I v) = 0.
Condition (i) is obviously a restatement of (a) above, while (ii) is a restatement of (b) as one can see immediately from the discussion just prior to (2.20.10).
Appendix We present the proofs of the unproved results from chapter 3. PROOF OF (3.6.6). Let X E p be the element such that ypx(t) = e`x(p) for all t e R. If K = Gp, the maximal compact subgroup of G that fixes
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p, then by the definition of K(p, x), the remarks preceding (2.17.5), and the proof of (5) of (2.17.5) we conclude that K(p, x) = K n G1 = K n ZX, where Z, = {g e G: ge`X = e`Xg for all t c= R). Hence K0(p, x) and Ks = K n Zr have the same Lie algebra fs = Z(X) n f by (3) and (4) of (2.17.13). The group K0(p, x) is a closed subgroup of K and hence is compact. PROOF OF (3.6.7). If M,, is the set of points in M fixed by an element 0 of G, then MM is also the minimum locus of the displacement function
d.. Hence M. is a complete, totally geodesic submanifold of M by (1) of (4.1.4), and it follows that F(p, x) _ fl4 E Mm is also a complete, totally geodesic submanifold of M. (1) Let q be any point of F(p, x) distinct from p, and let y(t) be the K0(p. X)
unit speed geodesic of M such that y(O) = p and y(to) = q, where to = d(p, q) > 0. Let Y E p be the unique element such that y(t) = exp(tY)(p) for all t E R. Since F(p, x) is complete and totally geodesic it contains a geodesic o : R - F(p, x) that joins p to q. Since o is also a geodesic of M we have o = y, and hence exp(tY) E F(p, x) for all tEI}B.
Let y = y(o), where y(t) = exp(tYXp) for t ca R. If 6 is any element of Z(X) n f, which is the Lie algebra of K0(p, x) by (3.6.6), then for any t E tR, exp(tf) fixes every point of F(p, x) and in particular every
point of y(R). Hence, for every t E ilL exp(t f) fixes p = y(O) and y = y(w), which means that exp(tf) E K (p, y). Hence CE Z(Y) n f, by (3.6.6), and this proves that Y E p * since was an arbitrary element of Z(X) n f. Hence q = exp(t0Y X p) E exp(p *)(p), and this shows that F(p, x) c exp(p*)(p) since q (=- F(p, x) was arbitrary. Conversely, let Y be any unit vector in p* c p, and let y = yy(oo), where yy is the unit speed geodesic of M given by yy(t) = exp(tYXp). We see that Z(X) n f c Z(Y) n f by the definition of p and hence K0(p, x) c Ko(p, y) by (3.6.6). If 46 is any element of K0(p, x), then 0 fixes p = yy(0) and y = yy(Oo) and hence every point of yy(F ). It follows that yy(OB) c F(p, x) since ¢ E K0(p, x) was arbitrary, and hence
exp(p *)(p) c F(p, x) since Y E p * was arbitrary. The proof of (1) is complete. (2) We shall need the following.
LEMMA. Let p be a point of M, and let g = f + p be the Cartan decomposition determined by p. Let 0 = 9p: q -> q be the Cartan involution defined
in (2.3.2) whose + I and - I eigenspaces are f and p respectively. Let X be a nonzero element of p, and let a e p be a maximal abelian subspace that contains X. Let g = qo + E. E A g,, be the root space decomposition determined by a. For a E A let f. = (I + 0)(g.) c f and 1). = (I - 0Xg0)cp (cf. (2.14).
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Assertions:
(1) Z(X) = go + Ea(X)-o ga
(2) Z(X)np=a+Ea(x)=opa (3) Z(X)nf=fo+Ea(x)_ofog where fo=goof. PROOF OF THE LEMMA. The first assertion is (3) of (2.17.13). Assertions
(2) and (3) follow immediately from (1), the definitions of fa and pa, and the facts that
Z(X) n p = (i - 8)(Z(X)), Z(X) n f = (I + 9)(Z(X)), and a=gong=(I-OXgo). We are now ready to complete the proof of (2). Let Y E Z(X) n p be given, and let a be a maximal abelian subspace of p that contains both X and Y. By (4) of (2.14.2), f = to + E. F , fa (direct sum), where to = go n L By hypothesis YE p*, and hence Z(Y) n f DZ(X) n f. By (3) of the lemma above we conclude that if a(X) = 0 for some a r= A, then a(Y) = 0. Hence Z(Y) ;? Z(X) by (1) of the lemma above, and we conclude by (2.20.9) that Y E Ex.
Conversely, if Y E Ex, then Z(Y) Z(X) by (2.20.9). Hence Z(Y) n f Q Z(X) n f, and we conclude that Y E p *. Since Ex is an abelian subspace of p that contains Y it follows that YE Z(X) n p*, which completes the proof of (2). (3) Identifying TP M with p c g we have Tp F(ypx) = Z(X) n p by (2.11.4) and TP F(p, x) = p * by (1) of this result. Hence, by (2) of this result, (2.20.12), and the fact that both F(ypx) and F(p, x) are complete totally geodesic submanifolds we see that
Tp{F(ypx) n F(p, x)} = T,F(ypx)nTTF(p,x) = (Z(X) n p) n p*
=Z(X) n p* =Ex=T,(E(ypx)). Since F(ypx) n F(p, x) and E(ypx) are complete, totally geodesic submanifolds of M with the same tangent space at p it follows that they are equal, which proves (3). PROOF OF (3.6.8). Let g = f + p be the Cartan Decomposition determined by p, and let X E p be the unit vector such that ypx(t) = e`x(p) for all t r= R. If F is any k-flat in M such that p C F and x E F(°), then ypx(f{8) cF, and hence by (1) of (2.10), F = exp(aXp), where a is a maximal abelian subspace of p that contains X. Conversely, if a is a maximal abelian subspace of p that contains X, then x E F(x), where F = exp(a)(p). Nye prove (1). Let y be any point of F(ypxXo), and let F be a k-flat in M with p E F and x E F(co). Let a be a maximal abelian subspace of
p such that X E a and F = exp(a X p). Let Y E p be the unit vector
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such that yp,,(t) = e"(p) for all t E R. Then Y E Z(X) n p by (2.11.4). If 0 is any element of K0(p, x) c K, then ((P O ypyX t) _ (cae` 46-' X p) _
t e l since 0(p) =p. Hence 46(y) A-) if Ad(4i)Y e a. The proof of (1) is now an immediate consequence of the next
eiAdem>r(p) for all
result. LEMMA. Let g = f + p be the Cartan decomposition determined by p E M. Let X be a unit vector in p, and let a be a maximal abelian subspace of p that contains X. Let x = y(x) E M(x), where y(t) = e`x(p). Then for any YE Z(X) n p, there exists 0 E K0(p, x) such that Ad(4))(Y) E a.
PROOF OF THE LEMMA. From the bracket relations of (2.3.5) and the
fact that Z(X) is a subalgebra of g we obtain
(a) [Z(X)nf, Z(X) n p I g Z(X) n p, [Z(X) n p, Z(X) n p] c Z(X) n f,
[Z(X)nf, Z(X)nf]cZ(X)nf.
By (3.6.6) we know that Z(X) n f is the Lie algebra of K0(p, x). If Y E Z(X) n p and f E Z(X) n f are arbitrary elements, then Ad(e£XY) = ead £(Y) E Z(X) n p by (a). Moreover, Ad(e4XX) ead f(X) = X. Since exp(Z(X) n f) generates K0(p, x) (in fact, it equals Ko(p, x)) we obtain the following:
(b) Ad(¢ X X) = X and Ad(4) leaves Z(X) n p invariant for all c= Ko(p, x)
Now let Y E Z(X) n p be given, and let A be a regular element of a. We define a function f: K0(p, x) -> R by f(4)) = B(Ad(4)XY), A) where B denotes the Killing form of g. The function f has critical points since K0(p, x) is compact. We shall show that Ad(4)XY) E a if 4)
is a critical point of f, and this will complete the proof of the lemma. The proof is essentially the same as the proof of (1) of (2.8.3). Let 0 E K0(p, x) be a critical point of f, and let f be an arbitrary
element of Z(X) n f, the Lie algebra of K0(p, x). Define gf(t) = f(4)e`f) = B(Ad(¢)Ad(e`4 XY), A) = B(e`ad 4(Y),Ad(4))-'A). In the last equality we use the fact that B is invariant under Lie algebra automorphisms of g by (1.13.8). Let A* denote Ad(4))-'A. Since di is a critical point we use (2) of (1.13.8) to conclude that 0 =gM) = B(ad (Y), A*) = B(- ad Y(e ), A*) = B(6, ad Y(A*)) = B(,~, [Y, A*D. Since A E a c
Z(X) n p we see that A* = Ad(4)) -')(A) E Z(X) n p by (b) above, and hence [Y, A*] E Z(X) n f by (a). Since B is negative definite on f by the discussion preceding (2.7.1) (see also lemma (8.2.1) of [W5]) and since BQ, [Y, A*]) = 0 for all 6 E Z(X) n f it follows that [Y, A*] = 0. Hence [Ad(4))Y, A] = Ad(4)XY, A*] = 0 and Ad(4)XY) E p n Z(A). The
fact that A is regular together with (2.8.1) and (3) of (2.17.13) implies
that Z(A) = go. Hence Ad(4)XY) E p n Z(A) = p n go = a, which completes the proof of the lemma.
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We prove (2). Let F, and F2 be any two k-flats of M that contain yp,, and let y be a regular point of F,(x). Note that F;(x) c since F; c F(yp,) for i = 1, 2. By (1) there exists 0 E K0( p, x) such that y* = ¢(y) E F2(0. The geodesic y y. = 0 v ypy, is regular since y is a regular point of F,(x), and ypy. is contained in both 4,(F,) and F2 since y,, c F,. Hence 4,(F,) = F2 by (2.11.1). 0 PROOF OF (3.6.21).
(6) Let x E'(z ), where z E R(x) n F(y, Xx), and
let y E: W be given. To prove that W, c K0( p, x){W n''(z )) it suffices to find cb E K0(p, x) such that Q,(y) E W, n W(z ).
We outline the proof. Let A, X, and Y be unit vectors in p such that z = yA(-), x = yX(x), and y = yy(x), where y,,(t) = e'A(p), yX(t) = e'X(p), and yy(t) = e'''(p) for all i E R. Note that Y E Z(X) n p by (2.11.4). Define f: K0(p, x) --> R by F(¢) = B(Ad(4,)Y, A), where B
is the Killing form of q. (We considered this function earlier in the proof of (3.6.8).) We show that 4,(y) E W, n B(z) if 0 is a local
maximum for f. Clearly f has a local maximum since K (p, x) is compact.
Let 46 be a local maximum for the function f: K (p, x)
(l8 defined
above. Let a be the unique maximal abelian subspace of p that contains the regular vector A. Since 0 is a critical point for f it follows from the proof of (3.6.8) above in this appendix that Ad(4,XY) E a.
Let a' = Ad(¢)- '(a) and A* = Ad(4,)-'(A) E a'. Since x c= J(z) we conclude by (3.6.31) that X E F(A) c a. It follows that X E a' since Ad(4,)X = X by (b) of the proof of (3.6.8). If z* = ¢ -' (z) = yA.(x), then x E 0-'(S-W) = 2(z*). Hence x< z* by the equivalence of (2) and (5) of (3.6.26).
Let q = g + Ev E , q a be the root space decomposition deter-
mined by a'. For an element 6 E a' we recall the notation AF = (a E A: a(f) = 0) and A+ = {a E A: a(e) > 0}. Since x:5 z* we obtain from (2.20.17)
(a) AXc A ;. and AXc A,-,. By hypothesis, y E W or equivalently x < y and hence from (2.20.17) we obtain
(b) AXC Ay and AXC Ay. From (a) and (h) we obtain
(c) a(Y)a(A*) > 0 if a(X) 0 0, a E A. Clearly ¢ leaves W, invariant since 4,(x) =x. To prove that 4(y) E
W, n f(z) it suffices to prove that y E W, n ¢-'('(z)) = W, which by (3.6.26) is equivalent to the inequalities x < y < z*. We are
Geometry of Nonpositively Curved Manifolds
270
given that x 0, and U c G be given as in
the statement of the sublemma. Without loss of generality assume e < ir. Let X be a unit vector in p with respect to the canonical inner product Q of (2.3.7) such that x = yx(oo), where yX(t) = e'X(p), for all tEOB.
We break the proof of the sublemma into several steps that we now outline. STEP 1. Define a function f: f x {Z(X) n p) - p by f(f, Y) Ad(exp f )(Y) = ead t(Y). We show that f has maximal rank at (0, X).
_
STEP 2. Choose S > 0 with the following properties.
M f has maximal rank at each point of Us = (Q, Y) E f x (Z(X) n p):
IfI<s,IY-XI<s}. (2) If f E f and 1 I < 5, then exp(t) E U. (3) If Y E Z(X) n p with I Y - X I < 3 then 0 with the following property: if Y is an element of p (X, Y) < o-, then YE f(US).
w i t h IYI = 1 and
We remark that f : US --> p is an open map by (1) of step 2 and the constant rank theorem. Hence the existence of o> 0 follows immediately from step 2 and the fact that X =f(0, X) E f(UU). STEP 4. Let 0 = {x* E M(oo):
=o fa = Z(X) n f by (iii). This completes the proof of (iv) and hence of step 1. PROOF OF STEP 4. Let x* be a point of M(oo) such that
y (x,
x*) < Cr
where a> 0 is chosen as in step 3. Let X* be the unit vector in p such that yX.(x) = x*, where yX.(t) = et' (p). We are given that 0 is chosen as in step 2. We note that I
IYI = IX*I = 1 by (2.3.8).
If 0 = exp(C ), then 0 E K n U by (2) of step 2. If yy(t) = e"(p), then di(yy(t))=(cfiety0 1Xp)=etAa(4 Y(p)=etX'(p)= yx.(t) for all t E R. Hence x* = f(y), where y = yy(x). To show that x* E W(U, e, x) and complete the proof of the sublemma it suffices to show that Td(x, y) < e. Since YE Z(X) n p and IY-XI <S it follows that 4(X,Y)<eby (3)of step 2.Since [X,Y]=0 there exists a maximal abelian subspace a of of p that contains both X and Y. Let F = exp(a)(p) be the corresponding k-flat in M, and note
that F(x) contains x = yx(x) and y = yy(x) since F contains the geodesics yx and yy. In particular, etAa(m)x(p) is a geodesic with initial velocity Ad(O)X. On the other hand, erAd(m)x(p) =
(4)e'xc,-')(p) = 4)e1x(p)
= e'x(p)
for all t since t -> e'x(p) is a geodesic that lies in Md,. Therefore X=Ad(4))X since both elements lie in p and are the initial velocity of the same geodesic. Hyperbolic elements 4.1.7. PROPOSITION. Let di a 1,(M) be hyperbolic, where M is a symmetric space of noncompact type and rank k >_ 2. Let Mm = (p e M: dm(p) S
dd,(q) for all q e M}. Let y be any geodesic of M that is translated by 0. Then:
(1) M4, is the union of all geodesics parallel in M to y, and Md, is a complete, totally geodesic submanifold of M. (2) M,s is invariant under Z(4)), and Z(4)) acts transitively on Md,.
PROOF. Let p E Md, be given, and let y(t) be the unit speed geodesic such that y(O) = p and y(w) = 4)(p), where w = d(p, 4)p) > 0 is the minimum value of d., on M. Let g = I +!p be the Cartan decomposition determined by the point p = y(0), and let X e p be that element such
that y(t) = e'x(p) for all t r= R. If h = e'"x and e = h -'¢, then by corollary (2.19.19) we have
0=eh = he with h hyperbolic and e elliptic with fixed point p. Hence e = 1 and 0 = h = e'"x by the uniqueness of the Jordan decomposition (see (2.19.23)). In particular, exp(Y) E Z(4)) if Y e p and [ X, Y ] = 0. Now let p' = {Yep : [ X, Y ] = 0). We assert that Md, = F(y) =
exp(p*Xp), where F(y) denotes the union of all geodesics in M that
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are parallel to y and exp(p*) = (e1: Ye p*}. This will show that M. is a complete, totally geodesic submanifold of M by proposition (2.6.1) since p * is a Lie triple system (cf. (2) of (2.6.3)). Moreover, it will then follow
that Z(4) acts transitively on M, since exp(p *) c Z(d). The fact just mentioned implies that exp(p *X p) c M , since Al is invariant under Z(¢). Note that M,, c F(y) by proposition 6.7 of [EO] and proposition (1.9.2) of this paper.
It suffices to prove that F(y) = exp(p*)(p), but this is (2.11.4).
o
PROOF OF PROPOSITION (4.1.4). Let 0 = eh = he be the Jordan decom-
position of a semisimple (axial) element O-E G. In assertion (1) of the proposition it suffices to prove that MM = M, n Mh. The fact that Mm is a complete, totally geodesic submanifold of M is then a consequence of the fact that Me and Mh are complete, totally geodesic submanifolds by propositions (4.1.6) and (4.1.7). We saw above in the proof of proposition (4.1.3) that Me n R. is nonempty and invariant under gyp. If p E Me n M,, is an arbitrary point, then by proposition (1.9.2) the geodesic yp joining p to 4(p) = h(p) is translated by h and fixed pointwise by e.
Hence 0 = he translates yp, and it follows from (1.9.2) that p E M. Hence Me n M,, c M,. We prove that Mm c Me n Mh. Let p E M. be given, and let p* be the orthogonal projection of p onto the closed convex subset Me n Mh. By the lemma in the proof of (4.1.3), p* lies in M. If yp (t) is the unit
speed geodesic joining p* to 4)(p*), then yp is fixed pointwise by e, and hence yp (t+w)=(4-yp X0 =(h0yF Xt) by (1.9.2), where w>0 is the minimum value of both d4, and dh. If yy(t) denotes the unit speed geodesic joining p to 4)(p), then by (1-9.2),(46 c y,X t) = yP(t + w) for all t e OB and hence yp is parallel in M to yP . By (1) of proposition (4.1.7) the geodesic yp lies in Mh, which proves that MM c Mh. To prove that
Mo c Mr observe that h translates a geodesic o, through p E Mh by (1.9.2) and o, must be parallel to yP and hence to yp. Therefore o, = yp.
Since 0 and h leave yp invariant it follows that e = h -'4) leaves yp invariant, and hence e fixes some point q of yp by the lemma. The fact that e commutes with 4) implies that e fixes the points 4)"(q) in yp for every integer n. Therefore ¢ fixes every point of yp, and in particular, p lies in Me, which proves that M6 c Me n Mh. This proves (1). Clearly, Z(4)) leaves R. invariant, and the fact that Z(4)) = Z(e) n Z(h) follows from theorem (2.19.23). To show that Z(¢) acts transitively on MM we fix a point p E Mm, and we let g = f + p be the Cartan decomposition determined by p. Let p * = (Y E p: elY(p) E M,, for all
t (=- R). Since M,, c Me and M,, c Mh it follows from the proofs of propositions (4.1.6) and (4.1.7) that exp(p *) c Z(e) and exp(p *) 9 Z(h).
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283
Hence exp(p*) c Z(0), and since M,, = exp(p*X p) it follows that Z(4)) acts transitively on M., which proves (2). PROOF OF PROPOSITION (4.1.5). We shall need the following result,
which is lemma 12.3 of [Mos2). See also the lemma in the proof of (2.19.26) in the chapter 2 appendix. LEMMA. Let s = eh = he denote the semisimple component of 0. Then for
every number a > 0 there exists a number c = c(a) > 0 such that if di(p) < a for some point p e M, then d(p, Ms) < c.
We now proceed to the proof of the proposition. Clearly, Mr(cc) n c M,,(cc). Now let x E Md,(cc) be given arbitrarily. For any point p E M the convex function (d4, o ypx): ll -- U8 is bounded above for t >_ 0 and hence is nonincreasing since .0 fixes x. If we set a = I + di(p), then by the lemma above there exists c > 0 such that d(ypxt, M) < c for all t z 0. Hence x e Ms(cc), and it follows that s fixes x and (d3 ° yp1Xt) is nonincreasing in /.-If we choose p to be any point of M, then it follows that ypx[O, x) c MS. By proposition (4.1.4), M, = Mh(cO) n
Me n Mh, which proves that e fixes y,,X(t) for each t >_ 0 and (dh o yn1Xt)
is constant on [0, cc). Therefore both e and h fix x, and it follows that u = h -' a -'c) fixes x.
Unipotent elements u
We now consider the set A,00 in the case that 45 e G =10(M) is a unipotent element. The story here is both more interesting and more complicated than for elliptic or hyperbolic elements. We begin by recalling the definition of a unipotent element 0 of G. Given points p E M and x E M(me) we let c = f + p denote the Cartan decomposition determined by p, and we choose X e p so that ypx(t) = e`x(p) for all t E R. The horospherical subgroup Nx was defined in (2.17.4) to be Nx = (4)E Gx: e-'x4e`X -' 1 as t - +co), and this definition is independent of the choice of p and the corresponding Cartan decomposition by (2.17.7). In (2.19.21) we defined 0 to be unipotent if 45 E Nx for some x e M(-), and by (5) of (2.19.18), 0 is unipotent if and only if Ad(4) is a unipotent element of Ad(G) c SL(g). The horospherical subgroup in which a given unipotent element 0 lies is in general not unique as we shall see in the next result. 4.1.8. PROPOSITION. Let M be a symmetric space of noncompact type,
and rank k>_ 2. Let 0 E G =10(M) be a unipotent element, and define M, (oc)=(xCMOO: ¢ENx}. Then: (1) M,*, (cc) c M4(oc). Moreover, d d(y t) --> 0, as t --* + cc for any x E M (oc) and for any geodesic y of M that belongs to x. Conversely, if
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Geometry of Nonpositively Curved Manifolds
x e fi(oo) is a point such that d,(y t) -* 0 as t --> x for some geodesic y that belongs to x, then x E (x). (2) No two points of M,0* (x) can be joined by a geodesic of M; that is, given distinct points x, y E M,,* (x) there exists no geodesic y of M such that (y(oo), y( - x)} = (x, y). (3) M,,* (x) is open in the Td-topology, and Mm (x) rl R(x) is nonempty. (4) If x and y are any two distinct points of M, (x), then Td(x, y) < IT and MVx) contains the unique Tits geodesic in M(x) from x to y. PROOF. (1) Clearly, M,*, (x) c Mm(x) since by definition N,, c GX = (g E
G: gx = x}. Now let x E MM (x), and let y be any geodesic of M with y(x) = x. Let p = y(0), and let g = f + p be the Cartan decomposition determined by p. Choose X E p so that y(t) = e`x(p) for all t E R. By the definition of NX in (2.17.4) and by proposition (2.17.7) it follows that
e-Ixderx -, 1 as t --* +x. Hence d,,(yt) = d(e`xp, 4,e`xp) = d(p, e"4e`xp) - 0
as t->+x. Next let x (=- M(x) be a point such that d,(yt) -> 0 as t - x for some geodesic y that belongs to x. Let p, f, p, and X have the meaning of the previous paragraph. It follows immediately from the hypothesis that 46 fixes x, and hence by (2.17.3) there exists an element Q,* E G such that e-`x4e`x - 4,* as t - x. The elements Ad(e-1x0e1x) = e-i ad XAd(4,)e' ad x are unipotent elements of Ad(G) by (2.19.27) and the hypothesis that 0 is a unipotent element of G (cf. (2.19.21)). Hence
Ad(4,*) is a unipotent element of Ad(G) since a- tad xAd(M)et ad x
,
-
Ad(4,*) as t -' x. On the other hand, ¢*(p) =p since d(p, 4,*p) _ d,6(yt) = 0. Hence lim, -= d(p, e-`x4e`xp) = lim, -% d,(e`xp) = lim, Ad(4,*) is an elliptic element of Ad(G) by (1) of (2.19.18). We conclude
that Ad(4*) = Id since Ad(¢*) is both semisimple and unipotent, and hence 4,* is the identity in G since G has trivial center. We have shown that 0 E Nx, which completes the proof of (1). (2) Suppose there did exist a geodesic y of M such that y(x) and y(-x) were both contained in Mi(x). It would follow from (1) that d4,(yt) - 0 as t -> +oo and as t -- -x. Hence dm o y would be bounded
on R and must be constant by the convexity of this function. We conclude that d4, C y = 0 or 4, fixes every point of y, contradicting the fact that d4, assumes no minimum value. (3) If x e M (x) is given, then by proposition (3.6.11), there exists e > 0 such that if Td(x, y) < e, then (G, )o (Gy )o, and hence NX c Ny by (2.20.16). It follows that the open e-ball around x in the Td-metric is contained in MM (x) (see also corollary (3.6.12)). By lemma (3.6.19) we
Action of Isometries on M(me)
can find y E Moo) with (Gr)o y E M, (x) n R(te).
285
(G,)0. Hence NX c N. by (2.20.16) and
(4) Let x, y (-=MVc) be arbitrary distinct points. By proposition (2.21.14) there exists a point p E M such that the geodesics yPX(I18), yp,,(R) lie in some k-flat F c M, where k = rank M. Let q = f + p be
the Cartan decomposition determined by p, and let a c p be the maximal abelian subspace such that F = exp(a X p). Choose X, Y E a such that yPx(t) = e`X(p) and yPy(t) = e`Y(p) for all t E R. By (2) it
cannot happen that X = - Y, and hence
4P(x, y) < Tr. However,
Td(x, y) = .,(x, y) by proposition (3.1.2), and hence there exists a unique minimizing Tits geodesic o * in M(c) from x to y by proposition (3.4.4). In fact, the discussion preceding (3.6.1) shows that if v(t) is the initial velocity of the geodesic s - eSX(`)(p), where 0:!g 1< 1 and X(t) = (1 - t)X + tY/IKI - t)X + tYll, then a (t) = y,,(,)(oo) is a reparameterization of o,*.
We show that o[0,1) c M (x). Fix a number i E (0,1). For each positive number s let o be the geodesic segment in F joining ypx(s) to yp,,(s), and let y,.(,)(gs) be the intersection point of os and y,.(,)(0,00). By (1) we know that d,,(yPss) -> 0 and d,y(yyys) --, 0 as s -+ cc and hence d,,(y,,(Jgs) < max(d,,(yp1s), d,(yPys)) --* 0 as s --> x by the convexity of d. and the fact that y,.(Jgs) is an interior point of o . By Euclidean
geometry g(s) -
as s --> , and hence d,(y,.(,)s) - 0 as s - x by the
convexity of d.. We conclude that ¢ E proof of (4). 0
by (1), which completes the
4.1.9. REMARK. The hypothesis that 0 be unipotent is essential for the last assertion in (1) of (4.1.8) as the next example shows. 4.1.10. EXAMPLE. We show there exists a nonunipotent element 46c=
G =10(M) and a point x E M(me) fixed by 0 such that 0 e NX but do(yt) -> 0 as t --* + oc for all geodesics y belonging to x.
Let F be a k-flat in M, where k = rank M, and let p be a point of M. Let c = f + p be the Cartan decomposition determined by p, and let
a c p be the maximal abelian subspace such that F = exp(a )(p). Let 0 = 00 + E« E A 9 « be the root space decomposition determined by a. By (3) of proposition (2.7.3) we may write 00 = f0 + a, where f = 90 n f and a = 00 n p. If f0 has positive dimension, then any nonzero element
Z of f commutes with all elements of a by the definition of g0. If 460 =eZ, then 00 fixes p in M and Ad(4)0) fixes every element X of a.
Geometrically, this means that 00 fixes every point of the k-flat F=exp(aXp). Now let y(t) be a regular geodesic of M such that y(O) =p and y lies in F. Let X E a be the regular element such that y(t) = e'X(p) for
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Geometry of Nonpositively Curved Manifolds
t (-= It Let x = y(x), and fix an arbitrary element 0 of N, even 0 = 1. If 41 = 4) o, then 4)= lim, .,, e-'Xqiesx since 0 E Nx and 0o all
commutes with es' for all s. Hence ili does not lie in Nx. On the other hand, if y is any geodesic of M with y*(x) =x, then y*(t) = ( o yXt) for some E Gx since Gx acts transitively on M. By (4) of (2.17.5) we Nx and C, E Zx, using the notation of may write C _ C1 - CZ, where y) (t), then of(t) is parallel to y(t) by (1.11.5) that result. If Q(t) and the definition of Z. Hence Q(t) lies in F for all i by (2.11.4) since y(t) is a regular geodesic of M. We conclude that d(y*t, o t) _ d(C, a(t), Qt) = ds1(ot) -> 0 as t -* x by (1) above. Hence
lim d,`(y*t) = lim dg,(crt) = 0
t-+%
I -+z
since d,1, s do + do0, lim, . x do(o t) = 0 by (1), and 0o fixes every point
of F. 4.1.11. FURTHER QUESTIONS AND PROBLEMS. Although propositions
(4.1.1) and (4.1.8) give some information about the nature of the fixed point set Mi(x) c M(x), for c¢ unipotent, the answers to many other elementary questions are unknown. For example: (1) What are the relationships between the following subsets X CM,,(x)? (a) X = M4,(x);
(b) X = the set of centers for gravity for M4,(x) = CG(4)) = (x E Md,(x): r(x) s r(y) for all y E Mi(x)}, where r(x) = sup{Td(x, z): z E Mi(x)); (c) X = the set of accumulation points in M(x) of (4)"(p): p E M,
(2) Can one give an explicit geometric construction of the distinguished center of gravity of M4,(x) when 0 is unipotent? (3) One can define a distinguished center for gravity for Mo(x) by the same procedure that is used to define a distinguished center of gravity for M,(x). Are these two centers of gravity the same? If not, can one give an explicit geometric construction of the distinguished center of gravity for Mm (x)?
(4) Given a parabolic element 4)E G =10(M) how many distinct minimal horospherical subgroups Nx exist that contain 0? What features do these minimal containing groups Nx have in com-
mon? For example, is s(x), the degree of singularity of x (cf. (2.21.7)), the same for all such groups?
(5) If a unipotent element 0 leaves invariant a proper, complete totally geodesic submanifold M* of M, then it is true that Mi(x) c M *(x)? (The set M *(x) may be regarded as a subset of M(x).)
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287
If the answer to (5) is affirmative, then by using induction on the dimension of M one might be able to describe M, (c) for a unipotent element ¢ E G =10(M). In this case, one would have to consider the
more general case of a set M4,(c) in a symmetric space M whose Euclidean de Rham factor is possibly nontrivial. An affirmative answer to (5) would also lead to a simple geometric proof of the lemma in the proof of (4.1.5) setting M* = MS, where s = eh = he is the semisimple Jordan component of an arbitrary element 45 E G. 4.1.12. TOTALLY GEODESIC SUBMANIFOLDS OF CONSTANT DISPLACEMENT.
The existence of complete, totally geodesic submanifolds M* of M on which the displacement function dm is constant for a given unipotent element 0 E G can be investigated via Cartan decompositions of 0.
Note, however, that 0 cannot leave M* invariant; in this case 0 would translate the geodesic y,4,,, for all p EM*, which implies that d. has a minimum in M by proposition 4.2 of [BO] and contradicts the fact that 0 is unipotent.
Given a point p E M let 0 = f + p be the corresponding Cartan decomposition. If p 0 = (X E p : Ad(4)X X) = X}, then p m is a Lie triple system and hence it follows from proposition (2.6.1) that F4,(p) = exp(p,X p) is a complete, totally geodesic submanifold of M. The fact that ¢e'X = e'X4) for all X E p and all t E R implies that the displace-
ment function dm is constant on F,(p); that is, the restriction of ¢ to F4(p) is a Clifford translation on Fo(p). Of course, if one chooses the point p incorrectly, then the dimension of p m may be zero, in which case F,(p) is the point p. The example below will show that p m is often nonzero. Now suppose that M* is any complete, totally geodesic submanifold of M on which d., is constant. If y is any geodesic of M*, then 0 fixes
both y(w) and y(- o) and it follows that ¢ leaves invariant the complete, totally geodesic submanifold F(y), the union of all geodesics
in M that are parallel to y. In particular, for each point p E M * the element ¢ leaves invariant the complete, totally geodesic submanifold MP consisting of the intersection of all F(y) such that y lies in M* and y(0) = p.
4.1.13. EXAMPLE. An r-flat M* and a unipotent element 0 of G = 10(M) such that d4, is-constant on M*.
Fix a k-flat F of M and a point p e F. Let g= f + p be the Cartan decomposition induced by p, and let a c p be the maximal abelian subspace such that F = exp(a X p). Let g = g + E« e A l; « be the root space decomposition determined by ad(a ). Let y be a unit speed geodesic of F with y(O) = p, and let x = y(oo). If X E p is the unique U
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Geometry of Nonpositively Curved Manifolds
vector such that y(t) = e"(p) for all t e 68, then nx, the Lie algebra of Nx, is given by Ea(x > 0 q,,, according to proposition (2.17.13).
CASE 1. Let a E A be any root such that a(X) > 0. Let Z be any nonzero element of g a, and let ca = eZ E Nx. Let as = (Y E a: a(Y) _ 0), a hyperplane in a, and let Fa = exp(a a X p), a hyperplane in F. If a(Y)Z = 0 since Z C g a. It folad Y(Z) Y E aa, then ad Z(Y) lows that Ad(¢) = ead Z fixes every YC aa, and hence d4, is constant on Fa c F.
CASE 2 (General case). Let S c A . = (a e A: a(X) > 0) be any subset, and let as denote the span of the root vectors Ha, a C S (cf. (2.9.1)). If
a5 is a proper subspace of a, then a(Y) = 0 for all Y E as and all a e S, where as denotes the orthogonal complement in a of as relative to the Killing form B of g restricted to a. It follows that if 0 = eZ for some Z E Ea E s 0 a, then Ad(¢) fixes every vector in as . Hence 0 lies in Nx and d4, is constant on the subspace FS = exp(a')(p) of F. 4.2. Action on M(oo) of minimal parabolic subgroups Throughout (4.2) we assume that 11%1 is a symmetric space of noncom-
pact type and rank k z 2. Let G denote 10(M). For any x c- M(°) we recall that Gx = {g e G: g(x) =x}. The parabolic subgroup Gx is minimal if and only if x e R(te), the set of regular points in M(co). 4.2.1. THEOREM. Let x (=- Moo) and z E M(c) be given arbitrarily. Let F be any k -flat in M such that x E Ac). Then the orbit Gx(z) intersects F(x) in exactly one point.
The proof of this result will be broken up into two lemmas.
4.2.la. LEMMA. Let y be a regular geodesic of M, and let 6 E G be an element that fixes both y(x) and y( - co). Then a leaves invariant the k -flat F = F(y) and the displacement function d£: q -b d(q, q) is constant on F.
PROOF OF (4.2.1a). Clearly, 6(y) is parallel toy since f fixes y(cc) and y( -oc). Hence 6(y) is a regular geodesic of M that is contained in the
k-flats F and f(F). It follows that (F) = F since a regular geodesic of M lies in exactly one k-flat of M. If x = y('), then '(x) is an open subset of F(°) by (3.6.11) and (3) of (3.6.26). Recall that the cone topology and the Td-topology coincide on F(cc) by (7) of (3.4.3). Since f fixes x, 4 fixes every point of W(x) by (2.17.25). It now follows from the Euclidean geometry of F that the
restriction of a to F is a translation of F; that is, dt is constant on F.
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289
4.2.1b. LEMMA. Let x (=- R(x) be given, and let F be a k -flat in M with x E F(x). Then for any point z E M(x) the set Gx(z) n F(x) is nonempty.
PROOF OF (4.2.1b). Let x, z, and F be as above. Fix a point p E F, and
O E G =10(M) be any axial element such that (4 c y Xt) =
let
yvx(t + w) for all t c- R and some w > 0. We will show that any cluster point of {0-"(z): n (=- Z') lies in G2(z) n F(x). By lemma 2.8 of [BBE] we see that all cluster points of the sequence
(¢-"(z): n c= Z') lie in F(x) since F = F(yyx) is the unique k-flat that contains y, and is also the union of all geodesics in M that are parallel to ypx. Let (nk) -a +x be a divergent sequence of positive integers such that zk =-"k(z) converges to a point z* E F(x) as k -> +x. We now proceed as follows to show that z* lies in Gx(z) n F(x): (1) We will show that there exists q e M such that z E F(ygxXo). (2) We construct a sequence ( k) c Gx such that (a) k(p) is a bounded sequence in M; (b) if p = yqx, where q is the point from (1), and if Y'k = Sk 0-"k, then Pik( p) = p for every k. Assuming for the moment that (1) and (2) have been carried out we complete the proof of the lemma. Let q, p, 6k, and rfik be as in (1) and (2). Note that O k fixes both p(x) and p( - x) for every k since 4ik( p) = p. The geodesic p is a regular geodesic of M since x = p(x) is a regular point at infinity. It follows from lemma (4.2.1a) that qik fixes every point of F( pXx) and in particular
(*)
tIik (z) = z
for all k
by (1). By (2a) and proposition (1.2.3) there exists a divergent sequence (mk) c 71+ such that { Finally, since z* = limk
-
converges to an element f * E Gx as n - +x. 4-11(z) and 4ik = CkO-"k we conclude that
*(z*) = lim
k-x
(Z) =z
by (*). This shows that z (=- Gx(z*) and hence z* E G1(z) n F(x), which completes the proof of the lemma.
Assertion (1) follows immediately from (2.21.14). We prove (2). Define p = yqx and Pk = 4-"k( p) for every integer k. Note that 4-"k(q) lies on pk and pk(x) =x for every k. For a fixed integer k the convex function s - d(ypxs, pk) is nonincreasing and it follows that d(p, pk) 5
d(ypx(-nkw),pk)=d(¢-"k(p),Pk) 0 choose p k E pk so that d(p, pk) = d(P, Pk)
_ 2, and let G = I0( Ab. Let x and y be any points of M(x) that can be joined by a geodesic y of M. Let F be any k-flat that contains y, and let p be any point of y. Choose (4,)^! I c K(p, F) to be a complete set of coset representatives for the finite group K(p, F)/K( p, F)*. Then N
N
G= U GX-46,-G, = U i=I
.=1
We shall need the following result.
LEMMA. Let x and y be points in R(x) such that W(x) = in(y). Then GX=Gy. PROOF OF THE LEMMA. By definition W(x) = W(y) if and only if (G') _ (GG),,. Now apply (3) of (2.17.25).
PROOF OF PROPOSITION (4.2.11). We first construct a point x* E R(x)
such that x e c(x*). Let F be a k-flat in M with x E Roc). The set F(m) n Moo) is dense in F(oo), and if z E F(w) n R(x) is given arbitrarily,
then '(z) is the connected component of F(x) n R(x) that contains z by propositions (2.8.2), (2.12.2), and (2.17.21). It is now clear that we can
find x* E F(x) n R(x) with x r= 2(x*) since F(x)n R(x) has only finitely many connected components by (1) of (2.17.22).
Now let x, y, p, y, and F be as in the statement of the proposition. By the previous paragraph we may choose x* E F(x) n R(x) so that
x E W(x*). If y* = y, .(-x), then y E F(y*) by the lemma in the proof of (4.2.3). If c¢ E G. is given arbitrarily and if (x*) c F(x*) is a sequence converging to x, then 4(x) = lim 4(x*) = lim x* =x by the lemma above. Hence G. c G.,, and a similar argument shows that Gy. c G,,. The two assertions of the proposition now follow from the two versions of the Bruhat decomposition stated in proposition (4.2.9) and (4.2.10).
COROLLARY. Let M, G, x, y, F, and (0,) c K(p, F) be as in the statement of proposition (4.2.11). Then N
G(x) = U Gx(46,x),
i-l N
,/ G(y) = U GG(.Oiy). 1=1
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REMARK. If x and y are singular points at infinity that can be joined by
a geodesic y of M and if F is any k-flat that contains y, then either x or y may be fixed by elements in K(p, F) - K(p, F*), where p = y(0) (investigate, e.g., the case G = SL(n, R) using the discussion in (2.13)). Hence the number of Gx orbits in G(x) or G(y) is in general strictly smaller than the cardinality of the Weyl group. We have seen above in the corollaries to (4.2.9) and (4.2.10) that the number of Gx orbits in G(x) or G(y) is exactly the cardinality of the Weyl group if x and y lie in R(x). 4.2.12. PROPOSITION. Let M be a symmetric space of noncompact type and rank k > 2. Let G =10(M ). Let x and y be points of M(°°) that can be joined by a geodesic of M. Then: (1) Gx(y) is a dense open subset of G(y). (2) If x and y lie in R(te), then G,,(y') is a dense open subset of G(y') for each y' E `ma(y).
PROOF. To prove (1) we shall need two preliminary results. 4.2.12a. Lemma. Let x and y be any points of M(cc) that can be joined by a geodesic y of M. Then there exists a neighborhood 0 of the identity in G such that 0(y) S Gx(y), and x can be joined to every point y* E 0(y) by a geodesic of M.
PROOF. Let y(t) be a geodesic of M with y(cx) =x and y( -co) = y. Let g = f + p be the Cartan decomposition determined by p = y(0), and let
K = (g E G: gp = p). Let X E p be chosen so that y(t) = e"(p) for all t E R. Let a c p be a maximal abelian subspace containing X, and let g = go + E. F n g,, be the corresponding root space decomposition of g determined by ad(a).
Let g, =g0+Ea(x)>oga and g2=go+E,(x): 2. Let G =10(M ). Let x e R(x) be given. Then: (1) M(x) - "(x) = GG('(y)), where y c- R(x) is any point that can be joined to x by a geodesic of M. (2)
"(x) is a compact subset of M(x) that contains 1'(x) and has empty interior.
(3) Let F be any k -flat with x E F(x), and let y = ypx(- x), where p is any point of F. Then: (a) "(x) = Gx(F(x) - My));
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304
(b) PxF: M(c) -p F(oo) is continuous on M(oo) - W'(x), and PXF(M(00) - W" (x)) = F(Y); F(00) is not (c) PxF: M(oc)
GG(F(c) - W(y)).
continuous at any point of
ExAMPLE. Ai,, = SUn, R)/SO(n, R). For this example we give a more explicit description of the set W'(x) discussed in (4.2.15). For terminology and definitions see (2.13.8). ASSERTION. Let x and y be points of Mn(co). Let F(x) and F(z) denote the flags in R' determined by x and z. Then z (-= M (oo) - W'(x) if and only if
F(x) and F(z) are in opposition.
PROOF. Suppose first that z c M(oo) - Y"(x). Let p = I SO(n, R), and let y = ypx( - no). By (1) of (4.2.15) there exists g E Gx and y* E W(y) such that z = g(y*). By the discussion in (2.13.8) and (2.17.28) we know that F(x) and F(y) = F(y*) are in opposition. Hence from the discussion
in (2.13.8) we conclude that F(z) = F(g(y*)) =g(F(y*)) and
F(x) = F(g(x)) =g(F(x)) are in opposition. Conversely, suppose that F(x) and F(z) are in opposition, and let y = ypx(- x). By the discussion in (2.13.8) there exists an element g E SL(n,R) such that F(z) =g(F(y)) and F(x) =g(F(x)). In particular, g e Gx since F(x) = g(F(x)). By (2.17.28) we see that g-'(z) E W(y)
since F(g-'(z)) =g-'(F(z)) = F(y). Hence z E Gx(f(y)) = '(x) by (1) of (4.2.15).
We shall need the following result in the proof of proposition (4.2.15).
LEMMA. Let x E R(co) be fixed, and let Px: Moo) -. T(x) be the map given
by Px(z) = G(z) n '(x). Then Px is continuous. PROOF. The set G(z) n W(x) is a single point for any z E R(°) by (2) of c R(-) be any proposition (2.17.24). Let z E R(oo) be given, and let sequence that converges to z. To show that Px is continuous at z it will suffice since M(oo) is compact to show that there is a subsequence (z such that Px(z) as k -> + Let K c G =10(M) be any maximal compact subgroup. Since G(z) = K(z) by proposition (1.13.14) we
may choose 4n E K such that
E W(x) for every n.
Passing to a subsequence (nk) of 71 and using the compactness of K we converges to some element di E K as k -+ +x. may assume that Then O (Z) = limk 0,,(z,,,,) E'(x). Since G(z) c R(m) it follows that O(z) lies in W(x) and hence Px(z) = j(z) = limk. limk The proof of the lemma is complete.
-
Action of Isometrics on M(x)
305
PROOF OF PROPOSITION (4.2.15). Assertion (1) follows immediately from
the definition of g"(x) and from proposition (4.2.13). We prove (2). The fact that W'(x) is a compact subset of M(x) with empty interior follows from (1) and from (2) of proposition (4.2.14). Proposition (2.17.13) shows
that if y is a regular geodesic of M with endpoints x' = y(x) and y' = y(-x), then (Gx.)0 (G,.),,. This shows that '(x) c F"(x) and completes the proof of (2).
We prove (3), beginning with (3a). Let x, y, and F be as in the statement of (3). Let z E W'(x) be given, and let z* = PXF(z) = G,(z) n
F(x). If z* E Fe(y), then z e GG(K(y)) = M(x) - "'(x) by (1). Hence z* E F(x) - '(y) and z E G,r(F(x) - 9-'(y)), which shows that !V(x) c Gx(F(x) - '(y)). Conversely, if z E F(x) - W(y) and x' E W(x) c F(x), then < (z, x') < 7r for any point p E F since W(y) _ (y.(-): x' (-= '(x)) by the lemma in the proof of (4.2.3). Since Td(z, x') = 4 (z, x') for all x' E F(x) by (3.1.1), (3.1.2), and (2) of (3.4.3) it follows that z E "(x) and hence G1(F(x) - ma(y)) c "(x) since W'(x) is invariant under G. by (1).
We prove OR Note that '(y) c F(x) by (3) of (3.6.26). By (1) we have M() - W'(x) = Gx(W(y)) c R(x). Given z e M(x) - W'(x) we write z = qi(y') for suitable 41 E G, and y' E'(y) c F(x). It follows that PXF(z) = GG(z) n F(x) = y', which proves that PXF(M(a) - "'(x)) = W(Y)' Moreover, Y' = G(z) n W(y) = PP(z), where Py: R(x) -> W(y) is contin-
uous by the lemma above. The discussion shows that PxF = Py on M(a) - "'(x), which proves (3b). We prove (3c). By its definition the map PxF fixes every point of F(x). Given z E F(x) - W(y) it follows from (2) of (4.2.14) that there exists a sequence (zn) c G,(W(y)) such that Z. --> z as n -- + x. By (1) and (3b) we see that PXF(zn) E F'(y) for every n but
PPF(z) = z E F(x) - K(y). and PxF is Therefore P1F(Zn) does not converge to PXF(z) as n not continuous at z. Since z E F(x) - W(y) was arbitrary and PxF is constant on Gx orbits it follows that PxF is not continuous at any point of Gx(F(x) - '(y)), which proves (3c) and completes the proof of this proposition. We conclude (4.2) with the following useful result. 4.2.16. PROPOSITION. Let M be a symmetric space of noncompact type and rank k >: 2. Let G = !0(M). Let x E R(x) be fixed, Then: (1) For each point z E M(x) - FW'(x) there is a unique point P.,* (z) in '(x) to which z can be joined by a geodesic of M. (2) The map Px* : M(oo) - W'(x) - FS'(x) is continuous.
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Geometry of Nonpositively Curved Manifolds
PROOF. (1) By definition of
"(x) it follows that for every point
z E M(x) - "'(x) there exists some point x' E (x) to which z can be joined be a geodesic of M. Conversely, let x* E '(x) be any point to which z can be joined by a geodesic of M, and let p be any point of M. If z' = yyz(-x), then x* E GZ(z') n W(x) by (2) of proposition (2.21.13). That result also shows that Gz(z') = G(x'). It follows that x* =x' since
both x* and x' lie in GZ(z') n W(x) which is a single point by (2) of proposition (2.17.24). This proves (1).
(2) Fix p E M, and let SP: M(x) -> M(x) be the continuous map given by Sp(z) = y, (- x). The discussion above shows that P': M(x) W U) -> F'(x) is the composition PX o S,,, where PX: R(x) - W(x) is the
map defined by Px(z) = G(z) n I(x). By the lemma of the proof of (4.2.15) the map PX is continuous and hence Px is continuous. 4.3.
Action on M(x) of axial isometries +
In this section we shall assume, at least at the beginning, that Al is an arbitrary complete, simply connected manifold of nonpositive sectional curvature. However, the main results of interest occur in the case that M is a symmetric space of noncompact type and rank k z 2. We begin with a result that follows from lemma 2.8 of [BBEI, whose proof is valid without any hypotheses on the structure of 1(M). Before stating the result we introduce some notation. If M is any complete, simply connected manifold of nonpositive sectional curvature and if A c M is any noncompact subset, then we define
A(x) =X n M(x) = {x E M(x): p" -' x for some sequence (p,,) cA) . Here A denotes the closure of A in M * = M U M(x) with respect to the cone topology. 4.3.1. PROPOSITION. Let M be any complete, _simply connected manifold of nonpositive sectional curvature. Let 46 E I(M) be an axial isometry, and let y be a geodesic of M that is translated by 0. Let F(y) denote the union of all geodesics in M that are parallel to y. If z E M(x) is an arbitrary point, then all cluster points of the doubly infinite sequence {4"(z): n (=- Z} lie in F(y )(x).
PROOF. See lemma 2.8 of [BBEI.
Symmetric spaces of noncompact type If M is a symmetric space of noncompact type and if y is a regular geodesic of M, then we may greatly strengthen the result above.
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307
4.3.2. PROPOSITION. Let M be a symmetric space of noncompact ON and
rank k >- 2, and let 0 E G = 10(M) be an axial element that translates a regular geodesic y of M. Orient y so that (0 0 yX t) = y(t + w) for all
t E R and some to > 0. Let x = y(o) and y = y(- o). Let F = F(y ) denote the unique k -flat containing y. For any point z E 4f,(-) let z, = G,(z) n F(o) =: PyF(z) and z2 = Gx(z) n F(w) _: PXF(z) (cf. theorem (4.2.1)). Then: (1) F(c) = M4,(w) _ {x E M(oo): 4)(x) =x}. (2) 0"(z) --> z, and 0-"(z) -> z2 as n -> +
PROOF. Let 4), y, x, y, and F be as in the statement of the proposition, and let z E=M(cc) be given arbitrarily. By (4.3.1) all cluster points of the sequences {¢"(z): n E 71+} and (¢-"(z): n c= Z') lie in F(x). The proof
of lemma (4.2.1b) shows that every cluster point of the sequence W"(z): n E Z') lies in Gs(z) n F(x) and similarly, replacing 0 by 0', we conclude that every cluster point of the sequence {4)"(z): n E Z1 } lies in G5(z) n F(-). Since the sets Gx(z) n F(-) and GG(z) n F(x) are
single points, z2 and z, respectively, it follows that ai"(z) -'z, and
4-"(Z)->z2 as n-> +-. The proof of the proposition will be complete when we show that F(y) = M., the set of points where d4, has a minimum in M. It will then
follow by the discussion at the beginning of (4.1) that M,,(-) = (x E M(x): 4(x) =x) = (x E M(oo): p" - x for some (po) c M,} = F(y)(m). Let 0 = eh = he be the Jordan decomposition of 0 into a product of
commuting elliptic and hyperbolic elements, e and h. By proposition (4.1.4) M, = Me n Mh, and hence if p E M, is given arbitrarily, then by proposition (1.9.2) both cp and h translate the unit speed geodesic y* from p to 4)(p) = h(p). By proposition (4.1.7) it follows that Mh = F(y*)
and F(y*) = F(y) since the geodesics y and y* translated by 0 must be parallel. Since y * is a regular geodesic lying in M4, c Me we conclude
by lemma (4.2.1a) that e fixes every point of F(y*)=F(y). Hence Me
F(y), and it follows that M. = F(y).
O
Our next result, the main one of this section, shows that in the notation of the result just proved the convergence of 0" to PyF and of
47" to
P.,F
is uniform on compact subsets of M(co) - W'(y) and
M(o) - 8"(x) respectively. More precisely we have the following. 4.3.3. THEOREM. Let M be a symmetric space of noncompact type and rank k >- 2, and let 0 E lfl(M) be an axial element that translates a regular
geodesic y of M. Orient y so that (0 o yXt) = y(t + w) for all t c- R,
-
where w > 0 is the minimum value of d.6. Let x = y(oo) = limo ¢"(p), for p c= M, and let y = y(-oo) _ limo _x 0-"(p). Let F = F(y) denote the unique k -fiat containing y. Let U, Uy and Ox, Oy be open subsets of M(oo) that contain W'(x), W'(y) and
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Geometry of Nonpositively Curved Manifolds
W(x), W(y) respectively. Then there exists a positive integer N such that if n >- N, then
(1) ¢`(M(x) - Ux) c OY and (2) 4)"(M(x) - Uy) c Ox. PROOF. It suffices to prove (1) since (2) follows from (1) if one replaces
0 by ¢-'. The proof of (1) will follow from a series of lemmas that we state now and prove later. 4.3.3a. LEMMA. Let z E M(x) - "'(x) be given, and let {zk} c M(x) "(x) and (nk) c 71 + be sequences such that Zk - z, nk -' + x, and ai-"k (zk) --> * for some * (=- M(x). Then * E G(z) n F(x). 4.3.3b. LEMMA. Let z (=- M(x) - W U) be given, and let (Zk) c M(x) W'(x) be any sequence such that zk - z. Then there exists a point 6 * C-
AT(-) such that 0"A(zk) - * as k - + x for any divergent sequence {nk}
c71+.
4.3.3c. LEMMA. Let z E M(x) - FU) be given, and let y* = PXF(z) E F(x). Then for any sequence (zk) c M(x) - F"(x) with Zk - z we can find a divergent sequence (n k) c71+ such that 46 -"A(zk) --y* as k - +x. Deferring for the moment the proofs of the lemmas we complete the proof of the proposition by using a contradiction argument. Suppose that assertion (1) of the proposition is false for some open sets Ux and
Oy in M(x) that contain "'(x) and F(y). Hence we can choose sequences (zk) c M(x) - Ui and (nk} c 7L+ such that nk - + x as k - x but O-"kM(zk) E (x) - O,, for every k. Passing to a subsequence we
may assume that zk -' z e M(x) - Ux c M(x) - "'(x) and 0-""(zk) 6 * E M(x) - Oy as k -. + x. By lemmas (4.3.3b) and (4.3.3c) it follows that 6 * = P,,F(z) E F(x). However, by (3b) of proposition (4.2.15) it follows that PXF(z) E PXF(M(x) - 9'(X)) = sp(y), contradicting the fact that * E M(x) - Oy c M(x) - '(y). It follows that assertion (1) of the proposition is true. o We now prove the lemmas stated above. PROOF OF LEMMA (4.3.3a). Let z, Zk, nk, and * be as in the statement of the lemma. Let K be a maximal compact subgroup of G. The point 4-"k(zk) lies in G(zk) = K(zk) for every k by proposition (1.13.14). Since Zk -+ z and 4) ' (zk) - e * it follows from the compactness of K that f * E K(z) = G(z). It remains only to show that f * lies in F(oo).
We show that,(, )(f *, y) is constant in t, where y = y(- x). It then follows from [E13, p. 78] that the geodesic ray yP,.[0, x) is contained in a flat half-plane bounded by y for any point p of y, and hence * c
F(x) = F(yxx) since y is a regular geodesic of M. In fact, we shall
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309
show that a and Zk - z as k -> + cc there exists an integer k0 >- 0 such that if k z k(,, then (1) a-e-(N-nk)wandk>-k(,,then a-e 6* as k -, +x. Let p = y(0) E F. We
shall show that for any number a with 0 < a < Sa as k -> +x. The previous lemma (4.3.3a) shows that the points {a} lie in G(z) n F(x) for every a, but
this contradicts the fact that G(z) n F(x) is a finite set by (2) of proposition (2.17.22). This contradiction will complete the proof of the lemma.
Let a be given with 0 < a < 4P (1 , * By passing to subsequences we may assume that Mk < nk for every k. Since C ,S*)>a
P(4' m(Zk), it follows that
+x. The point S. lies in G(z) n F(x) by lemma (4.3.3a), and hence 4(a) = ba since 0 fixes all points of F(x) = F(y Xx) by lemma (4.2.1 a); recall that 0 translates the regular geodesic y. From (1) above we obtain
_ 0 such that r leaves F invariant and IF(C) = F for some compact subset C of F. See, for example, the lemma in (7.1) of [BGS] or theorem 1' of [GW]. Moreover, d4, is constant on F for every element 0 in r, and it follows that F(x) = L(1') c Mr(x). By inspection the group 1' is not admissible even though I' has fixed points in M(x). By theorem 1' of [GW] this discussion is also valid for solvable groups F c 1(M) that contain only axial or elliptic isometrics. See also theorem (10.3.6) below.
The situation is different if r c 1(M) is an abelian group that contains nonidentity parabolic isometrics.
4.4.2. PROPOSITION. Let r c I(t f) be an abelian group that contains a nonidentity parabolic element. Then 1' is an admissible group.
PROOF. Let 0 0 1 be a parabolic isometry of M that lies in r, and let x(¢) E M4,(x) be the distinguished center of gravity of 46 as defined above in (4.1). Since r is abelian it leaves invariant the set M,,(x) and
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hence I' fixes x(¢) by (2) of proposition (4.1.2). Hence x(4) E Mr(-) c A 7f,6(-). It follows from (1) of proposition (4.1.2) that Td(x(4), y) :!g ir/2 for all y E Mr(oo), which proves that r is admissible.
We may mimic some of the discussion of (4.1) and apply it to admissible_ groups
I' c I(M). In particular, we may define a set
CG(F) c Mr(oo) of centers of gravity of Mr(oo) and a distinguished center of gravity x(F) E CG(F). Using the same methods of proof we obtain the following analogue of theorem (4.1.1') and proposition (4.1.2). 4.4.3. PROPOSITION. Let 1' c I(M) be an admissible subgroup. Then: (1) Mr(oo) is closed in M(oo) with respect to the cone topology.
(2) If x and y are distinct points of Mr(-) that admit a unique Tits geodesic cr joining them, then a lies in Mr(co). (3) There exists a point x(F) in Mr(oo) such that (a) Td(x(I'), y) < -7r/2 for ally (=-,4 r(-) and (b) x(F) is fixed by any isometry of M that leaves Mr(co) invariant.
From this result we obtain some useful corollaries. 4.4.4. PROPOSITION. Let F and r* be subgroups of 1(A1) such that F is normal in r*. If r is admissible, then r* is admissible.
PROOF. Let x(F) be a point of tblr(oo) that satisfies the two conditions of proposition (4.4.3). Since F* normalizes IF it follows that F* leaves Mr(-) invariant, and hence r* fixes the point x(F) by (3b) of proposic Mr(°) since F* 2 F, and tion (4.4.3). Therefore x(F) (=Td(x(F), y) < it/2 for all y EM1.,(o) by (3a) of proposition (4.4.3). It follows that f* is admissible. 4.4.5. COROLLARY. Let r c 1(M) admit a normal abelian subgroup A such that A contains nonidentity parabolic elements. Then r is admissible.
PROOF. This is an immediate consequence of proposition (4.4.2) and (4.4.4).
REMARK. The corollary is a strengthened version of lemma 6 of appendix 3 of [BGS] in which the Tits metric in MA-) is discussed in detail. 4.4.6. PROPOSITION. Let M be an arbitrary complete, simply connected manifold of nonpositive sectional curvature whose Euclidean de Rham factor is trivial. Let r c I(M) be a subgroup with limit set L(1') = M(oo) such that IF admits a nonidentity normal abelian subgroup A. Then: (1) A contains only parabolic isometries of M. (2) 1' is an admissible subgroup of 1(M ).
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PROOF. Assertion (2) follows immediately from (1) by using corollary
(4.4.5). We prove (1) by contradiction. Let H cA be the subset of elements that are semisimple (i.e., axial or elliptic), and assume that H # (1). The set H is in fact a normal abelian subgroup of F by lemma 7.7 of [BGS]. By the lemma in (7.1) of [BGS] or by theorem 1' in [GW] there exists a complete, flat, totally geodesic submanifold F of M of
dimension r >_ 0 such that H leaves F invariant and H(C) = F for some compact subset C of F. Moreover, d. is constant on F for every 0 E H. See also theorem (10.3.6) below. The group H cannot have any fixed points in M; the set of points in M that are fixed by H is a closed convex subset of M that is invariant under t since H is normal in T. If H fixed a point of M, then it would follow that M is the fixed point set of H since L(t) = M(co), contradicting the assumption that H is not the identity. Therefore the dimension of the flat submanifold F c M described above is positive, and the set B = F(-) c M(x) is nonempty. The fact that H leaves F invariant and F/H is compact implies that L(H) = F(m) = B. Since H is a normal
subgroup of r it follows that r leaves B invariant. Therefore M is isometric to the Riemannian product of F with another manifold by proposition 2.2 of [Eli], which contradicts the hypothesis that M has no Euclidean de Rham factor and completes the proof of (1). The proof of the proposition is now complete. As a corollary of the result just proved we obtain the following. 4.4.7. PROPOSITION. Let M be a homogeneous space whose Euclidean de Rham factor is trivial. Then either M is a symmetric space of noncompact type or I(M) is an admissible subgroup and in particular I(M) fixes some point in M(c).
PROOF. If M is not a symmetric space of noncompact type, then I(lt1) is not a semisimple Lie group and hence by [Bou, parts I, III, section 9.8, remark 11 there exists a nonidentity normal abelian subgroup A of
1(M). Note that L(I(M)) = M(oo) since I(M) acts transitively on M. The fact that 1(M) is admissible now follows from proposition (4.4.6). REMARK. The argument found in the discussion of example (1.9.17) shows that if t c1(M) has a fixed point in M(oo), then I' cannot satisfy the duality condition (cf. (1.9.15)). From this observation and the preceding result we obtain the following. 4.4.8. COROLLARY. Let M be a homogeneous space that is not a symmetric space. Then 1(M) does not satisfy the duality condition. In particular, M admits no quotient manifolds with finite Riemannian volume.
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PROOF. Write M = M0 X M,, where Mo is the Euclidean de Rham
factor of M and M, is the product of all non-Euclidean de Rham factors of M. Since M is not a symmetric space it follows that M, is not a symmetric space and hence 1(M,) fixes a point in M,(-) by proposition (4.4.7). Therefore 1(M,) does not satisfy the duality condition by
the remark above, and it follows from (1.9.22) that 1(M) = I(MO) x 1(M,) does not satisfy the duality condition.
REMARK. This result was proved by a different method in [CE] as theorem 5.4. 4.5.
The set D(F) c M(x) determined by F c 1(4)
Let M be an arbitrary complete, simply connected manifold of nonpositive sectional curvature, and let F c I(M) be any subgroup. In this section we introduce the duality set D(F) c M(cc) for r and describe some of its properties. We recall from (1.9.12) that two points x, y CM(cc) are F-dual if there exists a sequence (o j c r such that O (p) - x
and 0, '(p) -y as n - co for some (hence any) point p of M. 4.5.1. DEFINITION. Let 1' c 1(M) be any group of isometries. Let
D(I') = {x e AL (cc): x is t-dual to ypx(-oo) for all p E M}. The set D(T) may of course be empty, although we shall see that it is
nonempty if L(r) = M(x). Clearly, r satisfies the duality condition if and only if D(F) = M(oo).
4.5.2. PROPOSITION. For any group F c I(M) the set D(F) a M(c) is a closed subset of the limit set L(F) with respect to the cone topology and is invariant under F. PROOF. D(F) is clearly a subset of L(F) (cf. (1.9.5) and (1.9.12)). Recall
that the geodesic symmetry SP: M -. M determined by a point p E M
extends to a continuous map Sp: M u M(x) = M * -M* given by Sp(x) = ypx( - cc) for alI x E M(x). If x E D(F) is any point, then by definition, x is F-dual to Sp(x) for all p E M, and hence fi(x) is F-dual
to cSp(x) = 4(40x) for all 0 E F and for all p c=- M by (2) of proposition (1.9.13). Therefore 4,(x) E D(F) and D(I') is I'-invariant. Now let {xn} c D(r) be a sequence converging to some point x E M(cc).
For any point p E M and all integers n the points x are r-dual to Sp(x ), and hence it follows without difficulty from the definition of F-duality that x is F-dual to Sp(x). Therefore D(F) is a closed subset of M(oo) with respect to the cone topology.
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4.5.3. PROPOSITION. Let r c I(M) be any group, and let x be a point of D(r). Choose p E M, and let y = ypx(- cD). Then the set of points in M(0) that are r-dual to x is precisely T (y ). PROOF. This follows immediately from (2) and (3) of proposition (1.9.13)
since x is F-dual to y by the definition of D(F).
Symmetric spaces of noncompact type If M is a symmetric space of noncompact type, then one can say
more about D(r), Fc1(M), as the results in the remainder of this section show. The proofs of these results are also valid for rank-I symmetric spaces, but we have preferred to restrict ourselves to the case
of rank k z 2.
The action of isometry groups on regular Weyl chambers W(x) ) Let r be a subgroup of G =10(M), where M is a symmetric space of noncompact type and rank k 2. Before investigating the properties of
D(F) we consider the action of t on a Weyl chamber '(x), where x E Moo).
4.5.4. PROPOSITION. Let M be a symmetric space of noncompact type and
rank k z 2, and let G =10(M ). Let r c G be any group, and let x and x * be points of Moo) such that '(x) = '(x * ). Then for every point z e F(x) there exists a unique point z * E'(z) n Fix * ). Before proving this result we shall need two lemmas. 4.5.4a. LEMMA. Let M be as above, and let x and x* be distinct points of R(co). Then '(x) = S"(x*) if and only if there exists a constant a > 0 such that p(x, x*) = a for every point p e M.
x*) = a for some a > 0 since PROOF. If F(x) = W(x*), then p (G,)0 = (Gx.)0 acts transitively on M by proposition (2.17.1). Con 0 such that oo. Then there exists a point z * E'(z) such that
z * as
n -> oo.
PROOF. By lemma (4.4.5a) there exists a > 0 such that P(x, x*) = a for all p EM. If {nk} c71+ is any_sequence such that nk --> +oo and converges to a point z* E M(me) as k - +c, then for any point p of M we have
4P(z,z*)= lim
lira
1(P)(x,x*)=a.
Therefore, all cluster points of the sequence (i (x*)) lie in G(x*) n'(z) by (4.5.4a) and (3) of (1.13.14). The assertion of the lemma follows since
G(x*) n '(z) is a single point by (2) of proposition (2.17.24). PROOF OF PROPOSMON (4.5.4). Let x and x* and z E T(x) be as in the statement of the proposition. By (2) of proposition (2.17.24) there exists
at most one point z* E F(x*) n '(z) c G(x*) n W(z). The existence of a point z* E F(x*) n w(z) follows immediately from lemma (4.5.4b). 11
The proposition just proved has some useful consequences. 4.5.5. PROPOSITION. Let M be a symmetric space of noncompact type and rank k >_ 2, and let G = la(M ). Let F c G be any group, and let F(x) be a
minimal set for some point x E Moo). Then I' (x *) is a minimal set for r for every point x* E '(x).
PROOF. Let x and r be as above, and let x* be a point of io°(x). If z* E F(x*) is given, then there exists a point z E F(x) n W(z*) by proposition (4.5.4). The fact that r(x) is a minimal set for r implies that x E F(z ). From (4.5.4) it follows that there exists a point x' E I'(z*) n t?°(x) since W(z) = F(z*). However, r(z*)c r(x*) since
z* E F(x*), and hence both x' and x* lie in G(x*) n '(x). Therefore x' =x* by (2) of proposition (2.17.24), and it follows that x* E I'(z*) for all z* E F(x*). This proves that I'(x*) is a minimal set for F. 4.5.6. PROPOSITION. Let M be a symmetric space of noncompact type and
rank k 2-.2, and let G = IO(M). Let F C- G be any group. If r(x) = G(x) for some x c- Moo), then F(x*) = G(x*) for every x* E '(x).
PROOF. Let F and x be as above, and let x* be any point of '(x). Let
z* be any point of G(x*), and let 41 E G be an element such that tfi(x*) = z*. Let z = i/i(x) E G(x) = F(x). By proposition (4.5.4) there
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317
exists a point z' e 1'(z) n r(x*). Note that z* = i/r(x*) lies in B(irx) = W(z) since x* lies in '(x). Therefore both z' and z* lie in G(x*) n
'(z), and hence z' =z*. We conclude that z* E r(x*) and hence I'(x*) = G(x*). The next result is a special case of (3.6.36) that we need later. For convenience we state the result here, and we give a short alternative proof that uses the methods of this chapter. 4.5.7. PROPOSITION. Let M be a symmetric space of noncompact type and
rank k z 2, and let G =10(M ). Let x e R(oc) be any point, and let U be any open subset of fi(x) equipped with the topology induced from the cone
topology of M(co). Then G(U) is an open subset of M(me) in the cone topology.
PROOF. Let x and U be as above. Let z E G(U) be any point, and write z = qi(x*) for some tji E G and some x* E U. Then U* = qr(U) is an
open subset of K(z) such that z E U*. If z' is any point of M(oo) to which z can be joined by a geodesic of M, then GZ.(U*) is an open subset of M(oo) that contains z by the invariance of domain argument used in the proof of (2) of proposition (4.2.14). Since GZ.(U*) c G(U) it follows that G(U) is an open subset of M(oo).
We now consider the structure of the set D(r) c M(w), where r is any subgroup of G = I0(M) and M is a symmetric space of noncompact type and rank k >_ 2. 4.5.8. PROPOSITION. Let Il%1 be a symmetric space of noncompact type and
rank k ;-> 2, and let G =10(M ). Let r c G be a group such that D(r) is nonempty. Let y e M(me) be a point such that y = y,, (- a) for some p E M and some x E D(r). Then:
(1) r(y)=G(y). (2) If OY. = (y* ,E G(y): r(y*) = G(y)), then Oyl contains a dense open subset of G(y).
PROOF. Let x, y, p. and r be as above. By the transitivity of Gx on X1-
it follows that G,,(y) is the set of points in M(co) to which x can be joined by a geodesic of M; that is, Gx(y) = (yqx(- °C): q E M). By proposition (4.2.12) GG(y) is a dense open subset of G(y), and hence by the definition of D(r) and (2) and (3) of proposition (1.9.13) it follows that
G(y) is the set of points in M(oo) that are r-dual to x. Finally, r(y) = G(y) by proposition (4.5.3), which proves (1).
We prove (2). Assertion (1) and the argument of the previous paragraph show that Gx(y) c 0yi., and this proves (2) since G1(y) is dense and open in G(y).
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4.5.9. PROPOSITION. Let M be a symmetric space of noncompact type and
rank k z 2, and let G = IO(M). Let r c G be any group such that D(r) is nonempty. Then for each x E D(r) we have
D(r) n G(x) = n
r(x').
x' E G(x)
Hence for each x E D(r) the set D(F) n G(x) is the unique minimal set for
r in G(x). PROOF. Recall that a closed subset X C- M(me) is said to be minimal for
r if r(z) = X for every z e X. Now let x E D(r), x* E D(r) n G(x), and x' E G(x) be given arbitrarily. It suffices to show that x* E r(x').
It will then follow that D(r) n G(x)c- nx.E G(x) r(x') and equality must hold since D(F) is a closed set invariant under r; clearly, nx' G(x) r(x') is a minimal set for F.
Fix a point p in M, and let y * = y,, .(- co) and y' = yPx.(- oo). We show first that G(y*) = G(y'). The group G =10(M) satisfies the duality condition, and hence y* is G-dual to every point in G(x*) = G(x) by (2) of proposition (1.9.13). In particular, y* is G-dual to x'. It follows from (3) of proposition (1.9.13) that y* e G(y') and hence G(y*) = G(y'). The fact that x* lies in D(r) means that x* is r-dual to every point in r(y*) = G(y*) by propositions (4.5.3) and (4.5.8). Hence x* is r-dual
to y' since G(y') = G(y*) by the work above, and it follows that x* E r(x') by (3) of proposition (1.9.13). 0 4.5.10. PROPOSITION. Let M be a symmetric space of noncompact type
and rank k z 2, and let G = 10(M). If r c G is any subgroup, then Moo) n Int(D(r)) is an involutive set that is invariant under G.
REMARK. It may be true that D(r) = M(o) (i.e., F satisfies the duality condition) whenever Int(D(r)) is nonempty, but we are unable to prove this.
We recall that a set X C- M(oo) is involutive if X is invariant under each geodesic symmetry SP: Moo) -. M(me) given by SP(x) = y, (- oo) (cf.
(1.7.10)). The notation Int(D(r)) denotes the interior of D(r) with respect to the cone topology of M(oo). We shall need the following result.
LEMMA. Let M and r be as above. If x E R(x) n Int(D(r)), then r(x)= G(x).
PROOF. Let x E R(oo) n Int(D(r)) and p E M be given. Let y = ypx(- x'). By proposition (2.21.13) (see also prop. (4.2.12)) we may
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319
choose a neighborhood Uy of y in G(y) such that x can be joined by a geodesic of M to all points y* in U,. Since x E D(I') we obtain (1) x is F-dual to all points y* E Uy.
By proposition (2.21.13) and the fact that x c- MOW)) we can choose an open neighborhood Ux of x in G(x) such that UX c G(x) n Int(D(I')) and y can be joined by a geodesic of M to all points x* E Ux. From the definition of Ux we obtain (2) y is F-dual to all points x* E Ux.
To show that F(x)=G(x) it suffices to prove that F(x) is an open subset of the connected set G(x). Let z E F(x) be given, and let w be a
point of M(x) to which z can be joined by a geodesic of M. Now y = ypx(- 00) is F-dual to x since x E D(F), and hence y is F-dual to z by (2) of proposition (1.9.13). By (3) of proposition (1.9.13) it follows
that y E I'(w)c G(w), and hence G(y) = G(w). Choose a sequence {4,,) c F such that 4 (w) -y as n - x. Since w E G(y) it follows that E UY for sufficiently large n, and hence x is F-dual to
for
large n by (1) above. Therefore x is r-dual to w and x E F(z) by (2) and (3) of proposition (1.9.13) since w can be joined to z by a geodesic
of M. Choose a sequence (,,,) c F such that
x as n - + x.
Since z E F(x) c G(x) it follows that On (z) E Ux for large n, and hence z E I/I,-, '(Ux). By (2) above y is r-dual to every point x* E Ux, which implies that Ux c F(x) by (3) of proposition (1.9.13). Therefore gl,-,'(Ux) is an open subset of G(x) that is contained in I'(x) and contains z for large n. It follows that IF(x) is an open subset of G(x) since z E F(x) was arbitrary, and we conclude that r(x) = G(x). PROOF OF PROPOSITION (4.5.10). Let x e R(x) n Int(D(F)) be given, and
let U be an open subset of '(x) such that x (=- U c R(x) n Int(D(F)). It follows from lemma (4.5.10a) that
G(U) = U G(x*) = U F(x*) c D(F). X*EU
X'EU
Hence qi(x) E G(U) c R(x) n D(F) for any 41 E G, and it follows that
+i(x) E R(x) n Int(D(F)) since G(U) is an open subset of M(x) by proposition (4.5.7). Therefore R(x) n Int(D(F)) is invariant under G. We prove that R(x) n Int(D(F)) is involutive. Let x and U retain the same meaning as above. Let p E M be any point, and let y = Sp(x) = ypx(- x). If U * = SP(U) = (ypx.(- x): x* E U), then U * is an open subset of '(y) by the lemma in the proof of (4.2.3). We show that U* C D(T). Given y* E U* we write y* = ypx.(- x) for
a suitable point x* E U. It follows that y* is T-dual to x* since
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U c Moo) n Int(D(I')), and moreover, we see that r(x*) = G(x*) by the
lemma above. Hence y* is r-dual to every point of G(x*) by (2) of proposition (1.9.13). By proposition (2.21.13) the set of points in M(oo)
to which y* can be joined by a geodesic of M is precisely Gy.(x*) c G(x*), and it now follows that y* E D(r). This proves that U* c D(T). Since U c D(F) it follows from the work above and (1) of proposition (4.5.8) that
G(y*) = r(y*) c D(r) for every y* E U*. Therefore
G(U*) c R(c) n D(F) since U * c f(y) c R(oo). It follows that y= SP(x) E R(oo) n Int(D(r)) since y lies in G(U*), which is an open subset
of M(x) by proposition (4.5.7). This shows that R(x) n Int(D(r)) is involutive since p E M and x (=- R(oo) n Int(D(F)) were arbitrary. The proof of the proposition is complete.
Next we relate the set D(r) c M(oo) to the nonwandering set fl(F) c SM that is defined in (1.9.10). 4.5.11. PROPOSITION. Let M be a symmetric space of noncompact type .
and rank k >_ 2, and let G =10(M). Let r c G be any group, and let c SM denote the dense open subset of regular vectors in SM (cf. (1.12.8)). Then a unit vector v of SM lies in 9P n Int(fi(r)) if and only if yjoo) E Moo) n Int(D(r)). As an immediate consequence we obtain the following.
4.5.12. COROLLARY. Let M, G, and r be as above, and suppose that
f(r) contains an open subset of SM. If v E fi(F) and v* E SM is asymptotic to v, then v* E fi(F). PROOF OF PROPOSITION (4.5.11). Let M, G, and I' be as in the state-
ment of the proposition. Suppose that y,,(oo) E R(oo) n WOW)). In particular, v lies in 5P. By the continuity of the map v* -> y,*0 we can find an open subset 0 c5P such that v E 0 and y,.(-) ,E Rte) n Int(D(F)) for all v* E O. It follows from proposition (1.9.14) and the definition of D(F) that 0 c si(r), and hence v E.f' n Int(fZ(F)). To prove the converse assertion of the proposition we will need the following.
LEMMA. Let M and G be as above, and let y be any geodesic of M. Let
x = y(c) and y = y(- -). If U c Gy is any open neighborhood of the identity, then U(x) is an open subset of G(x) that contains x.
PROOF OF THE LEMMA. Let M, G, x, y, and U be as above. Let x* = 4(x) E U(x) for some ¢ E U. Choose a neighborhood W c Gy of the identity such that W(x*) c U(x). Since y can be joined to x* by a geodesic of M it follows from the proof of lemma (4.2.12a) that we can find open neighborhoods 0, Oy, and Ox. of the identity in G, Gy, and G.,. respectively such that Oy c W and 0 = Oti, Ox.. Hence
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O(x*) = Oy(x*). By lemma (4.2.12b) the set O(x*) is an open subset of
G(x*) = G(x). Hence U(x) is open in G(x) since O(x*) = Oy(x*) c W(x*) c U(x).
We now complete the proof of the proposition. Let v ESP n Int(f(F))
be given, and let x = y,.00 and y = y,,( - co). We wish to show that x e R(te) fl Int(D(r)). Clearly both x and y lie in R(o) since v ESP. By hypothesis there exists an open set A c SM such that v (=- A c WI'). Let U c Gy be an open neighborhood of the identity such that A D dU(v) = {d4)(v): ¢ E U). Since A c fl(F) it follows from proposition (1.9.14) that y is T-dual to x* for each x* E U(x). By the lemma above, U(x) is an open subset of G(x). Since y can be joined to all points in U(x) c GG(x) we have shown the following:
(1) There exists an open neighborhood Ux of x in G(x) such that y can be joined to all points of UX and y is I'-dual to all points of Us.
Replacing v by - v and reversing the roles of x and y in the discussion above we obtain the following: (2) There exists an open neighborhood Uy of y in G(y) such that x can be joined to all points of Uy and x is I'-dual to all points of Uy.
It now follows exactly as in the proof of lemma (4.5.10a) that IF(x) = G(x) and r(y) = G(y). Hence x is F-dual to all points of G(y) = I'(y) by (2) of proposition (1.9.13). Since G,r(y) is precisely the set of points in M(me) to which x can be joined by a geodesic of M it follows that x E D(I'). We have shown that x = y,.(co) E R(x) n D(r) whenever v E.W fl Int(SZ(I')). If 0 is any open set of SM, then O* = (y,,(0: v E 0) is an
open subset of M(x) since for any point q in M the map z
yq,(0)
is a homeomorphism of M(me) onto the unit sphere in TQ M. Therefore (yt,(oo): v ESP fl Int(l(F))) is an open subset of M(O) that lies in Mao) n D(I-), which completes the proof.
Existence of r -periodic vectors Let M be a symmetric space of noncompact type and rank k >- 2, and let G =10(M). 4.5.13. DEFINITION. If t C G is any group, then we define a vector v E SM to be r-periodic if there exists o E T and to > 0 such that
(46 °y,,)(t)= y,.(t+w)
for all tER.
Let 51per(I') denote the set of I' periodic vectors in SM.
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The set flpc,(F) is clearly a subset of f2(IF) by proposition (1.9.14). 4.5.14. PROPOSITION. Let M be a symmetric space of noncompact type
and rank k 2. Let G =10(M ), and let I' c G be any subgroup. If v E f2(F) n- q, then there exists a sequence (v,,) a f per(F) n.-q such that
vn - v as n -4 +c. In particular, fl,er(r) is a dense subset of fl(I') if l(F) has nonempty interior. For the proof of this result we need the following. LEMMA. Let M and G be as above, and let y be any unit speed geodesic of
M with endpoints x = y(c) and y = y( - -) in M(cc). For any e > 0 there exists a neighborhood U of the identity in G such that if x * E U(x) and y * E U(y) are given arbitrarily, then there exists a geodesic y* of M such that d(y (0), y *) < e, y * (oo) = x * and y * (- cc) = y *.
PROOF OF THE LEMMA. Let y, x, and y be as above, and let e > 0 be given. Let p = y(0). Choose a neighborhood Vy, c Gy. of the identity such that d(p, 4)p) < e/2 for all 0 E Vy. By the lemma in the proof of (4.5.12) the set V,,(x) is an open subset of G(x) that contains x. Let V be an open neighborhood in G of the identity such that V(x) c Vy(x). Finally, let U be an open neighborhood in G of the identity such that U-' - U c V and d(p, Op) < e/2 for all ijr E U. We assert that U has the properties stated in the lemma. From the fact that V;, c Gy, it follows that for every point x* E Vy(x) there exists a geodesic y* with y*(oc) =x*, y*(-o) = y, and d(p, y*)
0 so that if -P(x, x*) < S and x* E G(x), then x* E 14(x). Choose a positive integer N such that if n z N, then (1) P(4 p*, x) < 5/2 for all p* E M with d(p, p*) s 1 and (2) 0. Then we can find sequences { fin) c 1(M), {vn} c C, and (tn) c R such that (a) to z n for every positive integer n, (b) d*((dcbn o g'"Xv,,),v,,) 5 1/n for every n, and (c) for vn E. and to (=- lJ with It,, - to I s e and d *(vn, v;,) < e we have (d-On o g',,Xv;,) # v;, for every n.
Passing to a subsequence we let {vn} converge to a point v e C. If p and p denote respectively the points of attachment in M for vn and v, then pn -+ p as n -' + oo. We shall prove that (1) On'(p)- woo) and 4n(p) y,.(-oo) as n oo and (2) It,, -d(pn, 4n'pn)I -' 0 as n -- co.
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325
Assuming for the moment that these facts have been established we complete the proof of the proposition. It follows from (1) and from the proof of proposition (4.5.14) that there exist sequences ( wn) c IIB * and
(vn) c. ' such that
(i) vn-'vas n-++x,
(ii) (On' a y, Xt) = yy.(t + wn) for all t r=- R and for all n, and (iii) I d(p, On `p) - wnI - 0 as n -* + x.
Since pn -* p as n -+ - it follows from (2) and (iii) that I to 0 as n --* +c. From (ii) we see that (don o g`".Xv,,) = v,, for every n, but this contradicts hypothesis (c) above for large n since both (vn) and (v,',} converge to v as n --> co. This contradiction completes the proof of the proposition.
We now prove assertions (1) and (2) above, starting with (1). By definition of the vectors v,, and (b) above we have d(4; pn, y,,.tn) _ d(p,,,(on °Y,, )(tn)) -> 0 as n - +a. Hence
asn -,oo since pn -> p, v'n -> v, and In - - as n -> oo. If v,* = (d4 a g")(vn) = (g`^ c dcnX vn) then y,,.(-tn) = 4n(pn) for every n. Since v,' - v as n ---, +oo by (b) above we conclude that On(p) --> y,.(-oo)
as n --' +°C,
which proves (1). To prove (2) we observe that for every n.
In
Similarly,
d(pn,On'pn) sd(pn,Y,,.tn)+d(y tn,On'Pn) We conclude that
I to - d(p,,, 0. 'Pn)I s
On'Pn)
for all n.
0 as n - + oo by the work above, we obtain (2), which completes the proof of the proposition. 0 Since d(4R ' pn,
4.6.
Groups I' C I( M) with LM = M(c)
If t is a subgroup of I(M) whose limit set L(F) is M(Qo), then we may improve the results of the previous section. 4.6.1. PROPOSITION. Let M be an arbitrary complete, simply connected
manifold of nonpositive sectional curvature, and let r CI(M) be a subgroup such that L(T') = M(oo). If A c M(oo) is a minimal set of r, then A C_ D(1' ). In particular, D(r) n r (x) is nonempty for every x E M(me).
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PROOF. Let M, F, and A be as above, and let x be any point of A. Let y = ypx(- x) for some point p E M. Since L(T) = M(x) there exists a sequence ( 4,,) c 1' such that y as n -> +x. Passing to a subse-
quence if necessary let 0,`(p) converge toapoint x* E M(x). The points x* and y are F-dual, and hence x* E f(x)cA by (3) of proposition (1.9.13). It follows that x E F(x*) since A was assumed to be a minimal set for I', and hence x is T-dual to y by (2) of proposition (1.9.13). It follows that x (=- D(I'). Since F(x) contains a I'-minimal set for every x E M(x) it follows that D(r) n r(x) is nonempty. 4.6.2. PROPOSITION. Let M be a symmetric space of noncompact type and
rank k > 2, and let G =10(M ). Let r c G be a subgroup such that L(t) = M(x). Then for every x E M(x) the set D(I') n G(x) _ nx' E G(x) F W) is the unique minimal set for r in G(x).
PROOF. Given x E M(x) there exists x* E D(F) n r(x) c D(F) n G(x) by the previous result. Now apply proposition (4.5.9).
O
4.6.3. PROPOSITION. Let M be a symmetric space of noncompact type and
rank k >- 2, and let G =10(M ). Let F C G be a subgroup such that L(t) = M(x) and D(F) has nonempty interior. Then D(F) = M(x), and r satisfies the duality condition.
As an immediate consequence of this result and proposition (4.5.11) we then obtain the following. 4.6.4. COROLLARY. Let M be a symmetric space of noncompact type and
rank k >- 2, and let G =10(M ). Let IF e G be a subgroup such that L(F) = M(x) and l(r) has nonempty interior in SM. Then (1(F) = SM, and r satisfies the duality condition. PROOF OF PROPOSITION (4.6.3). Since R(x) is a dense open subset of
M(x) and D(F) is a closed subset of M(x) it suffices to show that R(x) c D(F) under the hypotheses of the proposition. Let M, G, and r be as above, and let x be any point of R(x). Fix a point p of M. Since L0') = M(x) there exists a sequence -'x as c F such that
n - x. Passing to a subsequence let 0;'(p) converge to a point y' in M(x). If y = yx(- x), then y E R(x) and y' E F(y) c G(y) by proposition (1.9.13) since x is f-dual to y'. Now let z be a point of R(x) n Int(D(I')). By proposition (4.5.10) the orbit G(z) is contained in R(x) n Int(D(F)), and hence by proposition (2.17.24) there exists a point y* in
'(y') n Int(D(F)) n G(z). By the lemma in the proof of (4.5.10) we have I' y*)=G(y*), and hence from proposition (4.5.6) we conclude that r(y') = G(y') = G(y). It follows from (2) of proposition (1.9.13)
that x is F-dual to every point of G(y) since x is t-dual to y'. Therefore x E D(I') since Gx(y) is precisely the set of points in M(x) to
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327
which x can be joined by a geodesic of M. This concludes the proof since x E R(x) was arbitrarily chosen.
4.6.5. EXAMPLE (see also (1.9.17)). Let M and G be as above, and let
r = G, the parabolic subgroup determined by some point x r= M(x). Then L(r) = M(x) since Gx acts transitively on M. However, r does not satisfy the duality condition by the discussion in (1.9.17).
EXERCISE. Determine the set D(r) c M(x) in the example above. 4.7.
Groups IF c I(M) that satisfy the duality condition
We first give a characterization of groups r that satisfy the duality condition in the class of all groups r with L(r) = Mw4.7.1. PROPOSITION. Let M be an arbitrary complete, simply connected
manifold of nonpositive sectional curvature, and let rc1(M) be a subgroup such that L(r) = M(x). Then r satisfies the duality condition if and only if r(x) is a minimal set for r for each x E M(x).
PROOF. Let M and r be as above, and suppose first that F(x) is a minimal set for r for every x E M(x). Let y be an arbitrary geodesic of M with endpoints x = y(x) and y = y( - x). Fix a point p of M. Since x as L(r) = M(x) we can find a sequence {on} c r such that
n - x. Passing to a subsequence let 0;'(p) converge to a point y* E M(x) as n - x. By (3) of proposition (1.9.13) it follows that y* E r(y). By hypothesis, r(y) is a minimal set for r and hence y E r(y*). It follows from (2) of proposition (1.9.13) that x is r-dual to y since x is r-dual to y*. Therefore r satisfies the duality condition. Next, suppose that r satisfies the duality condition. Let x E M(x) and x* E r(x) be given. To show that r(x) is a minimal set for r it suffices to show that x E r(x*). Fix a point p in M, and let y = ypx(-x). The fact that x* E r(x) implies that y is r-dual to x* by (2) of (1.9.13). Hence if y* = ypx.(-x), then y E r(y*) by (3) of (1.9.13). Choose a If x = ypy. (-x), sequence { fin} c r such that y,, = 4n(y*) --*y as n then x,, -+ x = ypy( - x) as n - x. By the duality condition x is r-dual to y for every n, and hence x,, is r-dual to y* = 0 ' (yn) for every n by (2) of (1.9.13). Therefore x is r-dual to y* by (2) of (1.9.13) since x,, -x as n - x, and x E r(x*) by (3) of (1.9.13). Therefore F(x) is a minimal set for r for every x r= M(x). 4.7.2. COROLLARY. Let M be a symmetric space of noncompact type and
rank k 2, and let G =10(M ). Let r c G be a subgroup such that L(r) = M(x). Then r satisfies the duality condition if and only if r(x)= G(x) for every x r= M(x).
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PROOF. Let M, G, and F be as above, and suppose first that F(x) = G(x)
for every x E M(oo). By the definition of minimality it follows that F(x) = G(x) is a minimal set for r for every x E M(x), and hence r satisfies the duality condition by the previous result. Conversely, if r satisfies the duality condition, then by (4.7.1), F(x) is a minimal set for F for every x r= M(me). Since D(F) = M(x) it now follows from proposition (4.5.9) that F(x) = G(x) for every x C- M(-). 11
Groups r that act minimally on M(cc) 4.7.3. PROPOSITION. Let M be an arbitrary complete, simply connected manifold of nonpositive sectional curvature. Let F c I(M) be a subgroup that satisfies the duality condition. Then the following properties are equivalent :
(1) F acts minimally on M(c). (2) F(x) = M(oo) f o r some x E M(x). (3) There exists a vector v E SM such that {(d¢ c g`)(v): 0 E F, t (-= II2l) is a dense subset of SM.
REMARK. Here (g') denotes the geodesic flow in SM. In the case that r is a discrete group this result is contained in theorem (4.14) of [E31. PROOF OF PROPOSITION (4.7.3). We shall prove that (3) - (2), (2) - (1),
and (1) - (3). (3) - (2). Let v E SM be a vector satisfying the property of (3), and let x = yt,(co). If w = (do o g `X v) for some 0 E F and some t E R, then yµ,(-) _ fi(x). It is now clear that (3) - (2). The assertion (2) - (1) is an immediate consequence of proposition (4.7.1). (1)
(3). We shall need to introduce and discuss some new terminology before proving this assertion (cf. pp. 71-74 of [E3]). Let M be any complete, simply connected manifold of nonpositive sectional curvature, and let F c 1(M) be any group of isometries. Given a vector v c SM we define
P, (v) = (w e SM : for any neighborhoods 0 and U in SM of v and w there exist sequences
c f and {4n} c F
such that t -. +- and (do o g`^)(O) n U is nonempty for every n}.
We define the set Pr (v) similarly by requiring that t -' -
in the description above. The sets P1 (v) and PI=(v) are closed in SM and invariant under {g'}, and dF = (d4): 4) (.- F) for every vector v E SM.
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329
Note that v E fl(F) if and only if v E P, (v) if and only if v E PI =(v). The proof of proposition (3.7) of [E2] now yields the following (cf. also prop. (4.9) of [E3]). LEMMA. Let M and IF be as above, and let v and w he any vectors in SM. Then:
(1) w E Pit( v) if and only if y,.(x) and y,,(-x) are F-dual. (2) w E Pr (v) if and only if y,.(- x) and y,,,(x) are F-dual.
Given a group F c 1(M) we let Hr denote the group of diffeomor-
phisms of SM generated by d r and (g': t E R). Note that
H1. =
(do o g': 4 E IF and t e R} since do o g'=g'- do for all ¢ E F and all t E R. Therefore property (3) is equivalent to the condition that Hr have a dense orbit in SM, and by a standard argument the existence of
a dense orbit for Hr is equivalent to the condition that Hr(O) be a dense open subset of SM for every open subset 0 of SM. We prove the assertion (1) - (3), which will complete the proof of the proposition. By the discussion above condition (3) will hold if we show that P'(0 = SM for every vector v E SM. Let v c- SM be given,
and let x = y,,(-) E M(x). Since F satisfies the duality condition it follows that x is F-dual to y = y,,( -x). By (1) of this result and (2) of (1.9.13) it follows that x is F-dual to every point in F(y) = M(x). Hence Pr (v) = SM by the lemma above, which completes the proof. 4.7.4. PROPOSITION. Let M be a simply connected manifold of nonpositive
sectional curvature and rank 1 (see (1.12.3)). If F c 1(M) is a group that satisfies the duality condition, then IF acts minimally on M(x).
PROOF. This follows from lemma 2.3b of [E8] and the fact that if l f' has
rank 1, then M admits a geodesic y that bounds no flat half-plane in M.
REMARK. The converse of the previous result is also true. Let M have rank k > 2, and let IF c 1(M) be any subgroup that satisfies the duality
condition. Then r does not act minimally on M(x). We see this as follows. If M has a nontrivial Euclidean de Rham factor Rio, then F leaves invariant Mj(x), a proper, closed I'-invariant subset of M(x). If M(x) has no Euclidean de Rham factor, and M is a nontrivial Riemannian product Ml X M2, then IF has a finite index subgroup I`* that preserves this splitting by proposition (1.2.4). Hence I,* leaves invariant the proper, closed subsets M,(x) in M(x) for i = 1, 2, and it now follows from proposition (4.8) of [EC] that I' also fails to act minimally on
M(x). Finally, if M is irreducible, then M is a symmetric space of noncompact type by theorem (9.3.1) below. In this case there are many
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proper, closed subsets X of MA-) that are left invariant not only by r but also by I(M). For example, let X = M(oo) as defined in (2.21.9).
Symmetric spaces of noncompact type Density of T periodic vectors and F -periodic k-flats We conclude chapter 4 with a discussion of some density properties of subgroups r c 1(M) that satisfy the duality condition in a symmetric space M of noncompact type and rank k >_ 2. For notation see (4.5.13). The next result is an immediate corollary of proposition (4.5.14). 4.7.5. PROPOSITION. Let Al- be a symmetric space of noncompact type and
rank k >_ 2, and let G =10(M ). Let F c G be a subgroup that satisfies the duality condition. Then the subset SZpe,(r) of r -periodic vectors is a dense subset of SZ(r) = SM. We now define and discuss r-periodic k-flats, where k is the rank of M.
4.7.6. DEFINITION. Let M be a symmetric space of noncompact type and
rank k z 2, and let G =10(M ). Let r c G be any subgroup. If F c M is any k -flat, then define rF = (g r= r: g(F) = F). We say that a k -flat F is r -periodic if there exists a compact subset C of F such that F = 17,(C) U0E r,, 0(C).
Before stating the final result of this section we recall some definitions. Let M be any complete, simply connected manifold of nonpositive sectional curvature, and let r c1(M) be a subgroup. By (1.9.23) IF
is a discrete group if for any compact subset C of M there are only finitely many isometries 46 in r such that cb(C) n C is nonempty. We recall from (1.9.27) that r is a lattice if r is discrete and if the quotient space M/r has finite Riemannian volume; equivalently, there
exists a constant c > 0 such that if 0 c M is any open set with ,(O) n 0 empty for all 0 * 1 in r then vol(O) 5 c. A lattice r is called uniform (nonuniform) if the quotient space M/r is compact (noncompact). 4.7.7. PROPOSITION. (Density of r-periodic k-flats). Let M be a symmetric space of noncompact type and rank k z 2, and let G =10(M). Let F be
any k -flat of M. Let y be any geodesic of M that lies in F, and let v = y'(0). Let IF c G be a uniform lattice. Then there exist sequences {4,,} c r and {v,,} cR c SM such that:
(1) v -> v as n -> + oo, and ¢ translates the geodesic
for
every n.
(2) If F = F(,,) is the unique k -flat of M that contains y,,, then F. is r -periodic for every n.
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331
REMARK. This result is lemma 8.3 of [Mos2]. Prasad and Raghunathan in theorem 2.8 of [PR] showed that this result is valid for any lattice F in G, uniform or nonuniform; see also theorem 1.14 of [PR] and lemma 8.3' of [Mos2]. PROOF OF PROPOSITION (4.7.7). Since the subset
.
n TF of regular
vectors of SM that are tangent to F is dense in TF it suffices to prove the proposition in the case that v E.W. By (1.9.32), 11(F) = SM for any lattice F c G; that is, r satisfies the duality condition by (1.9.15). By proposition (4.5.14) we can choose sequences (on) c r and (vn) cE' such that assertion (1) is satisfied. It remains only to prove that Fn is F-periodic for every n, where yn = y,,. and F,, = F(y,). This fact is a consequence of the following somewhat more general result. 4.7.8. LEMMA. Let M and G be as above. Let F c 1(M) be a discrete group, and let 4) E t be an axial element that translates a regular geodesic y of M. Let F be the unique k -flat of M that contains y, where k is the rank of M, and let o- be any regular geodesic in F. Suppose there exist sequences
{tn} c R and-{ n} c r such that t,, - + x and (fin
is a bounded
sequence in M. Then by passing to a subsequence we have: (1) an = f, 1t;, commutes with 0 for every n. (2) an leaves F invariant, and the displacement function da, is constant on F for every n.
(3) For any point p of F there exists a constant A > 0 such that d(an(p), o,(tn)) 0 on F. It follows that d(qn, /3ngn) = d(a(tn),
w > 0 for all n. Since {qn}
is a bounded sequence in M and r is a discrete group it follows that there are only finitely many distinct elements fin. Passing to a subsequence we may assume that F'n = fit for every n. It now follows immediately that a,, = 6 16, commutes with 0 for every n, which proves (1). The fact that an commutes with 0 implies that an fixes the endpoints {y(oc), y( - oc)) of the regular geodesic y that is translated by
0. Assertion (2) now follows from lemma (4.2.1a) applied to each element an. Finally, if p is any point of F, then d(an(p), cr(tn)) = d(4n 1e1(p), cr(tn)) GF/Gby T(K(p, F)*) = 6GF for 6(=F K(p, F). It is routine to show that T is a well-defined group homomorphism. T is injective since K(p, F) n GF = (K n GF) n GF = K n GF = K(p, F)*. To show that T is surjective it suffices to note that GF = K(p, F) exp(a) by (5) and exp(a) c GF by (6). This completes the proof of (4.2.8).
5
A Splitting Criterion
5.1.
Isometries that preserve a Riemannian splitting
337
5.2.
Symmetry diffeomorphisms
338
5.3.
Involutive subsets of M(c)
338
5.4.
Holonomy
339
Dc Rham decomposition theorem 339 Theorem of Berger 340 Relationship to symmetry diffeomorphism group orbits on M(x) 340 Relationship to isometry group orbits on M(-) 340 Characterization of symmetric spaces of higher rank 340 5.5. Riemannian splitting criterion
5.6.
Applications of the splitting criterion
341
344
in this chapter we present a useful criterion for determining when a simply connected Riemannian manifold of nonpositive sectional curva-
ture splits as the Riemannian product of two manifolds of lower dimension. Our treatment is a modification of that found in [E15] and [El 6], where a more complete discussion can be found. In this and later sections we obtain applications of the splitting criterion. Let X = X, x X2 be a Riemannian product manifold, and let and
9 be the foliations of X induced by the factors X, and X2. From the discussion following (1.2.3) we recall the following. 5.1.
Isometries that preserve a Riemannian splitting
DEFINITION. An isometry ca of X is said to preserve the splitting if d 09(p) = l,((Ap) for i = 1, 2 and all points p E X.
If ¢ E I(X) preserves the splitting X = X, x X2, then ¢ permutes the leaves of the foliations .9, and and ¢ may be written 0 = ( p,(O), p2(0)) E 1(X,) x 1(X2) c I(X). If r cI(X) is a group of isometries that preserves the splitting, then the projections p,: F - 1(X,) for i = 1, 2 are homomorphisms.
Geometry of Nonpositively Curved Manifolds
338
5.2.
Symmetry diffeomorphisms
Let A%1 be a complete, simply connected manifold of nonpositive sectional curvature. For every point p E M there is a natural involutive diffeomorphism SP: M -> M given by SP = exp c S o expp , where S: TM - M is the map given by S(v) =_- v. Equivalently, SP(y(t)) = Y(- 1) if y is any unit speed geodesic of M with y(O) =p (see (1.3.1)). The diffeomorphism SP: M - M is called the geodesic symmetry at p. Clearly, (SP )2 = 1, SP fixes p, and dSP = -1 on Ti, M. DEFINITION. We define groups G* and Ge* in Diff(M) by (a) G* is the group of diffeomorphisms of M generated by the geodesic symmetries {SP: p (=- M); (b) Ge* is the subgroup of G* generated by those elements {SP o S9:
p,geM}, Since the geodesic symmetries SP, p c- M, are all involutions it 0 SP,,' where follows that G* consists of all finite products SP, ° SP2 0 the points p 1, ... , pN are chosen arbitrarily in M. Moreover, Ge* has o S,,N, index 2 in G* and consists of all finite products S,, o SP2 o where N is an even integer. The groups G* and Ge* are called the symmetry diffeomorphism
group and the even symmetry diffeomorphism group of M respectively. Note that M is a symmetric space if and only if G* c 1(M). If M is a Euclidean space, then Ge4 is the translation subgroup of 1(M). If M is symmetric of noncompact type (i.e., M has no Euclidean factor), then one may show that Ge =10(M ). A geodesic symmetry SP extends naturally to a homeomorphism of M(x) by defining SP(x) = y,x( - x) (cf. (1.7.10)). Hence we may regard
G* and Ge* as groups of homeomorphisms of M(x) with the cone topology.
5.3.
Involutive subsets of M(c)
A subset X of the boundary sphere M(x) is said to be involutive if X is invariant under all geodesic symmetries SP, p C -M (see (1.7.10)). A priori we do not assume that an involutive subset is closed in the cone topology, although this is often useful in applications. We list some useful examples of involutive subsets.
EXAMPLE 1. Let M = MI X M2 be a nontrivial Riemannian product manifold. Then the sets X, = M;(x) for i = 1, 2 are closed involutive subsets of M(x).
A Splitting Criterion
339
EXAMPLE 2. Let AYI be a symmetric space of noncompact type and rank
k Z 2. Let S(x) c M(x) denote the singular points of M(x); that is, S(x) = {y,.(oo): v e SM is not regular} (see also (2.17.17)). Then X is a closed involutive subset of M(x).
More generally, the singular points S(x) c M(x) of a symmetric space M of noncompact type may be further stratified into subsets M1(x) according to their degree of singularity (see (2.21.9). Each of these strata M,(x) is a closed involutive subset of M(x). For further examples and discussion of involutive subsets of M(x), see section 2 of [E15]. We also present further examples in the discussion that follows. 5.4.
Holonomy
Let X denote a complete Riemannian manifold. For each p E M let 4'P c O(TP X) denote the holonomy group at p, where O(TP X) denotes the orthogonal group of TPX. By definition, 4'P is the set of all parallel
translation operators PY, where y: [a, b] -> X is a piecewise smooth curve with y(a) = y(b) = p. For a detailed discussion of holonomy groups see [KN, pp. 179-193]. If [a, b] -X is a piecewise smooth path from p = o-(a) to q = o,(b), then P,,-' O 4'q ° P, = OP. Hence the holonomy groups at any two points are essentially the same. The group 4'P is a Lie subgroup of O(TP M) (cf. theorem 4.2 of [KN, p. 73]).
If X = M, a complete, simply connected manifold of nonpositive sectional curvature, then we shall see shortly that there is a close relationship between the action of the holonomy group 4'P on SP M, the
unit vectors of M at p, and the action of the even symmetry diffeomorphism group Ge on M(x). Before describing this relationship we discuss two important results about holonomy.
If X is any complete Riemannian manifold and if W c TPX is a subspace invariant under 4'P for some point p of X, then W defines a distribution ''' in TX as follows: given a point q c- X let W'(q) c TQ X
be that subspace obtained by parallel translation of W along any
piecewise smooth path y from p to q. The definition of /(q) is independent of the path y chosen since W is invariant under 4'P. 5.4.1. THEOREM. Let X be a complete, simply connected Riemannian manifold. Let W be a proper subspace of TP X invariant under 4'P for some
point p of X. Let W ' denote the orthogonal complement of W in T. X, and let 2V and ' denote the corresponding distributions in TX. Then: (1) Both V and 7'1 are involutive (integrable). (2) If X, and X2 denote the integral manifolds of 71' and W/' through a fixed point q of X, then X, and X2 are complete, totally geodesic
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340
submanifolds of X, and X is isometric to the Riemannian product X1 X X2.
For a proof see theorem 6.1 of [KN, pp. 187-191]. A complement to this splitting result is the following special case of a result of Marcel Berger [Be]. 5.4.2. THEOREM. Let X be a complete, irreducible Riemannian manifold,
and assume that the holonomy group fip restricted to the unit sphere SP X c TP X has more than one orbit for some point p of X. Then X is a Riemannian symmetric space.
REMARK. If X = M in the result above, then M must necessarily be symmetric of rank k Z 2 since each holonomy group fiP acts transitively
on SPM if M is symmetric of rank 1. In fact, 1p consists of the differential maps of elements of the maximal compact subgroup K of G = 10(M) consisting of those isometries of G that fix p. The next result, which is theorem A of [E16], relates the action of 4)P on SPM to the action of GG on M(me). We omit the proof. 5.4.3. THEOREM. Let M be a complete, simply connected Riemannian manifold of nonpositive sectional curvature. Let v E SPM be given and let w be any vector in the holonomy orbit c1,(v) c SP M. Let x = y,; (co) and y = yw(oo) be the points in M(me) determined by v and w. Then y e GG W.
The proof uses a lemma of W. Ballmann from [Ba2]; see [E16] for further details. As an immediate corollary of this result and the result of Berger preceding it we obtain theorem B of [E16]. 5.4.4. THEOREM. Let M be a complete, irreducible, simply connected Riemannian manifold of nonpositive sectional curvature. Let X be a proper closed subset of M(me) that is invariant under G,*. Then M is symmetric of noncompact type and rank k z 2.
PROOF. Fix a point p e M, and let AP = (v (=- SPM: V°) E X). Then AP is a proper closed subset of SPM, and AP is invariant under cP by (5.4.3). The result is now immediate from (5.4.2) and the remark that follows.
0
In general, it is not easy to study the action of GG on 141(oo) directly. The next result is frequently useful. 5.4.5. THEOREM. Let 141 be any complete, simply connected manifold of nonpositive sectional curvature. Let F c 1(M) be any subgroup that satisfies the duality condition, and let X c M(oo) be any closed subset that is invariant under I'. Then X is invariant under G,*.
A Splitting Criterion
341
PROOF. Let x E X and points p, q E M be given arbitrarily. It suffices
to show that (SPOSgxx)EF(x)cX. If y=Sq(x) and x*=SP(y)= (SP 0 SqXX), then y is r-dual to both x and x* (cf. (1.9)). It follows that X* E F(x) by (3) of (1.9.13). 11
5.5.
Riemannian splitting criterion
The next result (theorem A of [E15}) is basic for many later applications and can be used to prove most splitting theorems in the literature for nonpositively curved manifolds. See [E15] for a more complete discussion.
THEOREM. Let M be a complete, simply connected Riemannian manifold of nonpositive sectional curvature, and let X be a closed subset of M(30) that is invariant under GG . For each point p e M define AP = (v r= SM: y,.(oo) E X),
BP = closed convex hull of (yt.(R): v EAP), ,41 p) = subspace of TP M spanned by APP. Then:
(1) If BP is a proper subset of M for some point p E M, then Y(q) is a proper subspace of T. M for some point q E M. (2) If .N( p) is a proper subspace of TP M for some point p E M, then M splits as a Riemannian product M, X M2. Moreover: (a) X c M,(-) c M(oo). (b) T h e distribution q - . q ) is integrable, and the maximal integrable manifold through q = (q,, q2) is M, x (q2) = Bq. (c) If 0 E 1(M) leaves the set X invariant, then 46 preserves the splitting M = M, X M, . PROOF. For the proof of (1) we need two preliminary steps.
5.5.1. LEMMA. Let X * denote the smallest closed involutive subset of M(w) that contains X. Then X * = X U SP(X) for any point p E M.
PROOF. Clearly X * contains X U SP(X) for any point p E M since X * is involutive. Conversely, X U SP(X) is involutive for any point p of M since it is invariant under SP and Ge , which together generate G*. 5.5.2. LEMMA. Fix a point p E M, and let X * be as in (5.5.1). If q E BP and x E X * are arbitrary points, then ygx(R) c- Bp.
PROOF. Let q E BP and x e X* be given. Since X* = X U SP(X) by (5.5.1) it follows from the definition of BP that y,,.(l8) c BP. The geodesic
342
Geometry of Nonpositively Curved Manifolds
segments y,, from q to y,x(n) lie in BP for every n by the convexity of Bp. Moreover, yn(0) -* yyx(0) as n -+ x since ypx(n) ->x as n --* x. Hence ygx[0, x) c B,, since Bp is closed.
Now let y = Sq(x) = yqx(- x), and note that y e X* since X *
is
involutive. The argument above applied to y shows that yqx( - x, 01 _ ygy.[0,x)cBP.
We now prove (1). Let p E M be any point such that. Bp is a proper
closed convex subset of M. Let r be any point of M - Bp, and let q E Bo be the footpoint of r on Bp (cf. (1.6.3)). Let t! be any unit vector
in Aq, and let x = y,.(x) e X c X *. By the result just proved, the geodesic y,.(l>8) = ygx(1) lies in Bp, and hence q is also the footpoint of r on ygx(Q8). It follows that V(q, r) = yq,(0) is orthogonal to
yyx(0) = v. Therefore V(q, r) is orthogonal to AAq) since v E A. was arbitrary, and this shows that A(q) is a proper subspace of T. M.
We begin the proof of (2). We assume that . f(p) is a proper subspace of TPM for some point p of M. Without loss of generality we for all points q c= M. The may assume also that dim.N(p) < set A p c SP M that is defined in the statement of the theorem is invariant under the holonomy group (Pp by (5.4.3) since X is invariant under G,*. Hence . l p) is invariant under 4P as is the orthogonal complement '(p)1 c TP M. Let S and .9 1 denote the parallel distribu-
tions in TM determined by.N(p) and .0p)1 as described in (5.4). By the de Rham theorem (5.4.1) the distributions Sand S1 are integrable with totals geodesic leaves, and M is isometric to the Riemannian product M, X M2, where M, and M2 are the leaves of S and -9-' that pass through the point p. For each point q E M we let F2 denote the leaf of the foliation 1z that passes through q (note that M, = FP). We show that 5q) =Mq) and Fq = Bq for every q E M, which is equivalent to assertions (a) and (b) of (2). We proceed in several steps. STEP 1. X c A%1(00) c M(x).
(yp'x(0):xEX}=ApcMp)=5(p)=TPF,=TPM,. The result
fol-
lows since y,.(x) E M,(x) for every vector v E TPM,; each leaf of S is a complete, totally geodesic submanifold of M.
STEP 2. Bq c Fq and .N(q) c5q) for all q E M. By -step 1, X c M,(x) = Fp(x) = Fq(x) for any point q E M since Fq = M, x (q2), where q = (q,, q2) with q; E-=M; for i = 1, 2. The subman-
ifold Fq is complete and totally geodesic, hence also closed and convex. Since X c Fq(x) we see that Aq c TqFq =5q). Step 2 follows.
STEP 3. Aq) =A q) for all q c =M.
A Splitting Criterion
343
By the choice of p we know that .0 p) =,4(p) and dim.N(p) < for all q E M. The assertion follows from step 2 since dim 5q)= dim5(p)= dim.4(p) < dimMq). STEP 4. BP = Fp.
We know that BP c FP by step 2. Suppose that BP is a proper subset of Fp, and let r be a point of FP - BP. If q is the footpoint of r on Bp, then V(q, r) = y r(0) E TgFp n.N(q)' by the proof of (1) above. On the other hand, q E BP c FP and hence B. c Fp and M q) c Tq FP since FP is
a closed, convex submanifold of M with X c Fp(x) = M,(x). We conclude that A' q) is a proper subspace of TgFp, and hence dimA'(q) < dim FP = dimA'(p). This contradicts the choice of p. STEP 5. Bq = Fq for every point q C -M.
Fix a point q E M, and let fq: M -* Il be the convex function given by fq(r) = d(r, Bq). The function fq is not identically zero by step 2. We show first that fq(r) < d(p, q) for all points r in BP, where p E M
is the distinguished point chosen at the beginning of the proof. It suffices to prove that fq(ypxt) < d(p, q) for all x E X and all t E R by the definition of BP and the convexity of fq. In fact, we shall establish this inequality for all points x r= X *. Recall that X * = X U SP(X) = X U Sq(X) by (5.5.1). If X E X* = X U Sq(X) is any point, then ygx(R) c Bq by the defini-
tion of Bq. If g(t) = d(y,xt, ygxt), then g(t) is bounded above t > 0 and is convex by (4) of (1.6.6). Hence g(t) is nonincreasing in t, and it follows that fq(ypxt) s d(ypxt, ygxt) < d(p, q) for all t >- 0. If y= SP(x) E X*, then the argument above shows that fq < d(p, q) for all points in ypy[0, x) = ypx(- x, 0). This completes the proof that fq(r) < d(p, q) for all points r in Bp. Now let x E A(m) = Fq(x) = FP(x) be given. By step 4 it follows that ypx[0, oo) c FP = Bp. By the previous paragraph there exists a sequence fq(ypx(n)) < d(p, q) for every n. It (qn) c Bq such that d(y,x(n), follows that q,, -'x as n --> - by (3) of (1.4.4). The closed convex set B. contains the geodesic segments yn from q to qn for every n, and hence Bq contains the geodesic ray ygx[0, x) since q --+ x. Hence B. -2 Fq = Ux a M'(x)
ygx[0, x), and equality follows from step 2.
In steps 1-5 we have proved assertions (2a) and (2b) of the theorem. We now prove (2c). Let 0 E 1(M) be any isometry that leaves invariant the set X c M,(x). It follows from the definition of the distribution rr
that dOAp)=.'(4)p) for all points p of M; that is, 0 preserves the splitting M = M, X M,. This completes the proof of the theorem.
0
344
5.6.
Geometry of Nonpositively Curved Manifolds
Applications of the splitting criterion
We use the splitting theorem of (5.5) to obtain useful characterizations of product manifolds in terms of the Tits metric Td. The main result, proposition (5.6.2), was proved originally in [BGS, pp. 222-224]. 5.6.1. PROPOSITION. Let M be a complete, simply connected manifold of
nonpositive sectional curvature. Assume that M(me) admits disjoint nonempty involutive subsets AI and A2 such that Td(a1, a2) S 7r/2 for
all points a, E A, and a2 E A2. _ Then M splits as a Riemannian product M, X M2 such that A, c M,(cc) and A, c M2(oo). In particular, Td(a I, a2) _ it/2 for all points a, E A, and a2 E A2. REMARK. The sets AI and A2 may actually be proper subsets of M,(cc)
and M2(x) in the result above. For example, let M, and M2 be symmetric spaces of noncompact type and rank >- 2, and let A, and A2 be the singular points in M,00 and M2(w) respectively. PROOF OF PROPOSITION (5.6.1). For i = 1, 2 let B; denote the closure of
A; in the sphere topology of M(x). It follows immediately from the definition that the sets BI and B2 are also involutive subsets of M(x). We show that 4P(a , a2) = it/2 for all points p E M, a, ca B,, and a2 E B2. By the definition of B, and B2 it suffices to consider points
p E M, a, EA,, and a2 EA2. If hI =SP(a,) = y,.(-x), then b, EAI since A, is involutive. By (3.1.1) and assertion (2) of (3.4.3) we know that 4 (x, y) 5 Td(x, y) for all points x, y e M(c). Hence Tr= 4P(a,,b,) = 4.P(a,,a2) + 4,,(a,, b,)
:
4P(a, b1) _ ir. If b1 1 A 1, then Td(b1, a,) < it/2 for some point a, E A2 by hypothesis (2). From (1) we know that Td(a1, a2) = .7r/2, and hence Td(a1, b,) :!g Td(a,, a2) + Td(a2, b,) < 7r/2 + 7r/2 = 7r. This contradicts the observation above that Td(a 1, b 1) >: Tr and proves that b, E A 1. Hence A, is involutive, and by a similar argument A2 is involutive.
By (5.6.1), M admits a Riemannian splitting MI X M, such that A; C- M;(°) for i = 1, 2. It remains only to prove that A. = M;(c) for i = 1, 2. Let x1 E M10) be given. For any given point a2 E A2 c M2(3o) and any point p E M we see that 4P (x,, a,) = it/2. Hence Td(x,, a,) = it/2 for all a2 EA2 by (3.1.1) and (2) of (3.4.3). Since x1 '4 A2 it follows
from hypothesis (2) that x1 E A,; if this were not the case, then we would have Td(x, a2) < it/2 for some point a2 E A 2. Hence A I = M1(cc), and a similar argument shows that A2 = M,(00).
6
Isometries of IIg°
6.1.
Commutator estimates in End(C")
346
6.2.
Nilpotent groups and unipotent elements
349
6.3.
Displacement functions of Euclidean isometries
351
6.4.
Commutator estimates in I(R")
352
6.5.
Solvable subgroups of 1(R") Existence of f-compact flats for solvable subgroups r of 1(R") 354 Extension to almost solvable subgroups of 1(R") 358
354
6.6.
Discrete subgroups of I(R") A strengthened version of a Bieberbach theorem 358
358
In sections (6.1) through (6.4) of this chapter we present a number of useful but soporific technical results. We encourage the reader to skip these sections on a first reading and proceed directly to (6.5). 6.1.
Commutator estimates in End(C")
We are particularly interested in estimating the norm m(f) = I I - f I of a commutator f = [ A, B ] =ABA - t B - t in terms of the norms m(A) = II -AI and m(B) = II - B1, where A and B are C-linear transformations of C" that are close to the identity I. Our treatment is based on [Buser] and [R, pp. 143-144]. The main results of this part are (6.1.5) and (6.1.7).
Let End(C" = (C-linear maps f: C" - C"), and define End(R") similarly, where C and C" are replaced by R and ll". For A E End(C") or
End(") one defines (6.1.1)
JAI = sup{IAvI: v E C",11)1=1),
where IvI = (v,v)1"2 and (v,w) = E. I v.i , for elements v = (v1,...,v") and w = (w,, ..., w") E C".
Isometrics of I8"
347
One obtains routinely from the definitions (6.1.2)
IA + BI s JAI + IBI,
IABI < JAI-IBI
for all A, B E End(U) or End(R"). For A E End(C") or End(ll ") we define
m(A) =11-AI.
(6.1.3)
For elements in 0(n) = (A E End(R"): (Av, Aw) = (v, w) for all one can make stronger statements. Similar assertions hold for elements of U(n) = (A E End(C"): (Ate, Aw) = (v, w) for all v, w e 118")
L', W E C").
6.1.4. LEMMA. Let A and B be elements of End(R"). (1) If A e 0(n), then IAuI = lid for all v r= LW'. Hence IAI = 1.
(2) If A E 0(n), then (a) IABA -'I= JBI,
(b) m(ABA ') = m(B). (3) If A E O(n), then m(A) = m(A -' ). PROOF. Assertions (1) and (2a) follow routinely from the definitions. To
prove (2b) observe that I - ABA -' = A(1- B) A -' and then apply (2a). To prove (3) let v e U8" be arbitrary. Then KI -A)v1= I v -Avl =
IA-'v-vI=K1-A-')vl by(1). We now derive some useful commutator estimates. Compare [Buser] and [R, pp. 143-144]. 6.1.5. LEMMA. (1)
if A, B E O(n) are arbitrary, then m([ A, B ]) 5
2m(A)m(B). (2) Let A, B E End(C1) be elements with m(A) < 1, m(B) < 1. Then A and B are invertible, and
2m(A)m(B) m([ A, B]) -
2 by induction. 6.1.7. LEMMA. (1) Fix a number 77 < Z, and let U = (A E0(n): m(A) < n}. Then U is a Zassenhaus neighborhood for 0(n) and m( ) < Z(2,7)" for all integers n z 2 and all E U("). (2) Let U = (A E End(C"): m(A) < a}. Then U is a Zassenhaus neighborhood for End(En) and m( ) < (I )n-' for all integers n >: 2 and all s E U("). PROOF. Both assertions of (1) follow routinely by using (1) of (6.1.5) and
induction. To prove the assertions of (2) observe that if m(A) 0 so that Bee c oil , where for any positive number a, B. = (A E 0(n): d(1, A) < a) and O,I = (A E 0(n): m(A)
0, and choose a positive integer N such that No, > 1. Let S = (A,,..., AN) be any subset of 0(n) with N elements. For 1 2 be the largest integer such that S(k) * ( I), Note that S(k) c U(k) c U since U is a Zassenhaus neighborhood.
Let B 0I be an element of S(k). We wish to show that B is
unipotent, and we begin by showing that B has exactly one eigenvalue.
350
Geometry of Nonpositively Curved Manifolds
Suppose that B has r > 2 distinct eigenvalues A,, ... , A, in C. We may write C" as a direct sum of V,,..., V,, where (B - Ail) is nilpotent on I for 1 s i < r. The elements of S commute with B and hence leave each V, invariant since (1) = S(k+') = [S,S(k)]. We may assume without loss of generality that A, * 1 since B has at least two eigenvalues, and hence B E S(k) is not unipotent on V,. If m,(A) denotes KI - A)I v,I, then clearly, m,(A) 5 m(A) = II -AI < e for all A E=- S. Now we restrict S to act on V, and apply the induction
hypothesis to conclude that F restricted to V, is abelian; note that B e S(k) is not unipotent and not the identity on V,. It follows that S(') = I on V, for all j >- 2. However, the group r was assumed to be nonabelian on C", and hence B e S(k) for some k 2 since S(2)0(1). Therefore B restricted to V, is the identity, which contradicts the fact
-
that (B - A,1) is nilpotent on V, where A, # 1. We conclude that every B E S(k ) has exactly one eigenvalue. Let B 0 1 be an element of S(k), and let A be the unique eigenvalue of B. The elements of S(') have determinant 1 for all j >- 2, and hence I = det(B) = A". Since B e U and I A - I I < e by (*) above, it follows from the definition of e that A = 1. Therefore B is unipotent. 6.2.2. COROLLARY. For every integer n > I there exists a Zassenhaus neighborhood W of the identity in 0(n) such that if S is a subset of W that generates a nilpotent subgroup r of O(n), then r is abelian.
PROOF. Let U c End(C") be a neighborhood of the identity that satisfies the hypotheses of (6.2.1), and let W = U n 0(n). Let S be a subset of W that generates a nilpotent subgroup of 0(n). The only unipotent element of 0(n) is the identity, and hence by (6.2.1) we conclude that I' is abelian. REMARK. This corollary and proof are valid also for U(n). 6.2.3. COROLLARY. Let G be a Lie group with discrete center. Then there exists a Zassenhaus neighborhood W of the identity in G such that if IF is a discrete subgroup of G, then either the elements of S = F n W commute or
S generates a nonabelian nilpotent group Co and Ad(S(k)) consists of unipotent elements, where k > 2 is the largest integer such that S(k) * {1). (Ad: G -), GL(g) denotes the adjoint representation.)
PROOF. Let U c GL(g) be a Zassenhaus neighborhood of the identity that satisfies the conditions of (6.2.1), and let W c G be a Zassenhaus neighborhood of the identity in G such that Ad(W) c U. Making W still smaller we may assume that W n Z = (1), where Z denotes the center of G. We show that W satisfies the assertions of the corollary.
Isometrics of 1W'
351
If r c G is any discrete group and s= F n w, then S(k) c c n Wtk> = (1) for large k since W(k) - 1 as k - oo. Hence S generates a nilpotent subgroup F0 by (7.2.3) below. If S* = Ad(S) c U, then S* generates the nilpotent subgroup r* = Ad(I'o) c End(C"), identifying g with C" or 1' c C". By the choice of U and (6.2.1) either r* is abelian or (S*)(k) consists of unipotent elements where k >- 2 is the largest integer such that (S*)(k) 0 (1). If r* is abelian, then Ad(S(2)) = (S*)(2) = (1), and hence S(2) C W n Z = (1) since W is a Zassenhaus neighbor-
hood and Z is the kernel of Ad. In this case we conclude that r is abelian. If r* is nonabelian, then Ad(S(k)) = (S*)(k) consists of unipo-
tents if k z 2 is the largest integer such that
(S *)(k)
(1). If r is the
largest integer such that S(') # (1), then clearly r >: k. However, if r > k {1}, which implies that S(') C W n Z = {1}, a then Ad(S(')) = contradiction. Therefore r = k and the corollary is proved. REMARK. If G is a closed subgroup of GL(n, C), then we may omit Ad in the result above. We now begin the study of 1(118"), the group of isometrics of R'. 6.3.
Displacement functions of Euclidean isometries
Every isometry ¢ of R" can be written uniquely as (k = T c R, where R E O(n) and T is a translation of 118". We define maps t:1(18") -' R " and r: AR") -> O(n) as follows. 6.3.1. NOTATION. If 0= T I R E I(88"), then
(1) r(4)) = R E O(n), and (2) t(4)) E R" is that vector such that T(v) = v + t(4) for all v r= 68".
Note that r: AR') --' O(n) is a homomorphism whose kernel is the translation subgroup of 1(Q8").
For an isometry 46 of R" we consider the displacement function do: R" -* 68 given by dm(p) = d(p, ¢p) = I4)(p) - pl for all p E R'. We recall from (1.9.2) that if d,, assumes a minimum value w >- 0, then the
set F,, where do assumes its minimum is a closed, totally geodesic submanifold if w = 0 and is a closed, convex subset if w > 0. If w > 0, then furthermore F. is the union of all geodesics translated by 0.
For any isometry 46 of 1' the displacement function d, always assumes a minimum value. More precisely, we have the following. 6.3.2. PROPOSITION. Let ¢ E I(ll ") be any isometry, where n > 1 is any integer. Write ¢ = T c R, where R = r(4)) and T is translation by 6 = t(4)).
Let W = (I - RXR") c O8", and write _ 1 + 1 2, where 6, E W and 62 E Wl , the orthogonal complement of W. Let F. denote the subset of
352
Geometry of Nonpositively Curved Manifolds
lib" on which the displacement function dm assumes its minimum. Then F4,
is always nonempty and is a k -flat, where k z 0 is the dimension of Wo = kernel(I - R). Moreover:
(1) pER" lies in F4, if and only if (I-RXp)_61. Hence Fm=po+ Wo, where po is any point in F4,.
(2) 1621= W is the minimum value for dm on R". In particular, 0 has a fired point in R" if and only if 6 = t(4)) E W = (I - RXP").
PROOF. Recall that the critical points of a C" convex function f: X -- fl8
are the absolute minima for f, where X is any complete, Riemannian manifold. Regard F. as the locus of critical points for the C" function d 2: 68" - R. It is routine to calculate the critical points of d 2 and verify the assertions of the proposition. 6.4.
Commutator estimates in I(P")
In this part we apply the commutator estimates of (6.1) to 1(U8"). The main results are (6.4.6) and (6.4.7).
6.4.1. LEMMA. Let 01 and 02 be arbitrary elements of I(R"). Let
r: 1(R") -> 0(n) and t: AR") - R" be defined as in
(6.3.1). Let
[01, 4)2] ='010201 '02 1 Then: (1) r([4)1, 4)2]) _
(2) t([4)1, 02] = r(4))(1- r(4)2)r(4),)-'r(4)2)-'}1(4)2) +
(3) If 01 is a translation of R" by translation of R" by r(4)2)
b1=
t(4)1), then
4Y2 0102
is a
1.
PROOF. The proof is routine. In proving (2) it is helpful to observe that MOD 02 ]) is the image of the origin under 146,14621-
6.4.2. DEFINITION. For 46 E I(R") define m(4)) = m(r(4)) = II - r(4))I = sup{KI - r(O)X01: v e E8" with Iv I = 1).
6.4.3. LEMMA. Let ¢i and 46, be arbitrary elements of R". Then
lt([-01,42])Ism(4)1)It(4)2)1+m(4)2)lt(4)1)I.
PROOF. From (2) and (3) of (6.1.4) we have II - r(4)2)r(4i)-'r(4)2)-' I = 11 - r(4)1) 'I = 11 - r(4)1)I = m(4)1). Similarly 11- r(4)1)r()2)r(4)1)-' I = I I - r(02)1 = m(4)2). The lemma now follows immediately from (2) of (6.4.1), (6.1.2), and (1) of (6.1.4).
We introduce a left-invariant metric on AR").
Isometries of R"
353
6.4.4. DEFINITION. For arbitrary elements 0, ¢,, and 452 in 1(R") we define
(1) n(¢) = m(4)) + It(O)I and (2) d(4),, 02) = n(¢, '¢2). 6.4.5. PROPOSITION. Let ¢, 41, and 4' be arbitrary elements in 1(08"). Then
(1) n(¢)=n((P-), (2) n([46,,4121) 5 n(4))m(4)2) + n(412)m(¢1), and
(3) d is a metric on 1(08") and d(4)41i, 4662) = d(411, 4),).
PROOF. Assertion (1) follows from (3) of (6.1.4) and the fact that It(4))I = It(4-' )I. To prove (2) observe that
m([0>,'02]) :=m(r([4)1, 02])) =m([r(4,),r(412)]) < 2m(4),)m(4)2) by (1) of (6.4.1) and (1) of (6.1.5). Hence by (6.4.3) we obtain
n([4,,4)21) s 2m(4,)m(4)2) +m(4))11(4)2)1 +m(42)It(4i)I = m(4i){m(4)2) + 11(4)2)1) +m(412){m(41) + It(4l)l)
=m(01)n(42) +m(4)2)n(cb1). We prove (3). From (1) we obtain 4)2) = n(41 '02) = n(4 '41) =d(4)2,411).
Hence d is symmetric. It is obvious from the definition that d(4¢1, 0-02) = d(¢,, 02). Now let 0 d, 02, and ¢, be any elements of I(R"), and write R. = r(4),) and 6; = t(0j) for 1 5 i < 3. Observe that It(0j-'0j)1 =1t(0; '¢;)I =1,=; - 6;I for 1 < i, j:5 3. Observe also that 1Ri 'R3 = (1- R) 'R2) + Rj 'R2(1- R2'R3). Hence M(01-'03) = II R'R31-II-R;'R21+11-R;'R3I=m(41i'4)2)+m(41Z'¢3) by (6.1.2), (1) of (6.1.4), and the fact that r:1(08') -> O(n) is a homomorphism. Finally, d(4)1, 03) = m(4)1 '4)3) + It(41 '¢3)I 5 m(41j '02) + m(4z'4)3) + I f1 -X31
Sm(41 62)+161 -621+m(41263)+162-531
0 The next two results are the main results of this section. The first is an analogue for 1(08") of (6.1.7). 6.4.6. PROPOSITION. Let S = (4) , ... , 46N } be any finite subset of I(R') such that m(¢;) < for all i, 1 < i:5 N. Let 17 < i and M > 0 be constants
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Geometry of Nonpositively Curved Manifolds
such that m(4,) < 71 and It(0;)I <M for 1 S i , k >_ 2, by (2) of (6.4.5) and induction.
13
Solvable subgroups of I(P")
6.5.1. DEFINITION. Let r c I(R') be any subgroup, and let W c R" be any k -flat, where 0 _< k 5 n. The k -flat W is said to be F-periodic if
(a) r(W) c Wand (b) the quotient space W/F is compact; that is, there exists a compact subset C of W such that W = IF(C) = (di(x): ¢ E F, x E C). REMARK. In (b) the space W/F need not be Hausdorff.
The main result of this section is the following 6.5.2. PROPOSITION. Let F c I(R') be a solvable group. Then there exists a I' periodic k-flat W c 68" for some integer k z 0.
Isometries of Fr
355
In the result above, the case k = 0 is the case that F fixes a point of R". D. Gromoll and J. Wolf in [GW] have shown that this result remains true for a solvable group F c 1(M), where M is arbitrary, provided that the displacement function d4, assume a minimum value in M for every E F. Our proof of (6.5.2) is very similar to that of [GW].
We shall prove (6.5.2) first in the case that 1' is abelian and then reduce to this case. 6.5.3. LEMMA. Let I' c 1(R") be an abelian group. Then for some integer k with 0 < k < n there exists a r -periodic k -flat W such that the displacement function dm is constant on W and equal to its minimum value for every element 4' E F.
PROOF. Given an isometry 45 of 1(O.") we recall that F. denotes the
subset of R" on which the displacement function dm assumes its minimum value. In (6.3.2) we showed that F. is a k-flat in li" for some integer k >- 0 that depends on n. The first step in the proof of (6.5.3) is:
(*) Let S 9;1(R") be any finite set of commuting elements. Then F = n Fm is a nonempty k-flat for some integer k >- 0. .oES
We prove (*) by induction on the number N of elements in S. The result is true for N = 1 by (6.3.2). Now let N > 2 be given, and assume that (*) is true for sets S with at most N - 1 elements. Let S be a set n,.-,I Fthen C is of N commuting elements (00 ... , ¢N}. If c=
nonempty by the induction hypothesis and C is an r-flat for some integer r >- 0. Note that ON leaves C invariant since ON commutes with
each ¢; and hence leaves F4, invariant for 1 s i < N - 1. It follows by
the lemma in the proof of (4.1.3) that F= n " ,
F44, = C n F.,
is a
nonempty k-flat for some integer k >- 0. The proof of (*) is complete. Now let 91' denote the set of k-flats F of the form F= n , `_ , F4,for ON) c I'. The So is nonempty and contains a some finite set S = k-flat F0 of smallest dimension. We assert that F0 = n o E r Fm and IF
leaves F. invariant. To see this let 46 be any element of F. Since r is abelian it follows from (*) that F. n Fo is nonempty and belongs to Y. Hence F0 c F4, by the minimality of F0, and this shows that F0 = n 4E r Fm. If 4' and tG are any two elements of F, then 4'(F,) = Fy commute. Hence r leaves F0 invariant. Since F. c F4, since 45 and for every 46 E IF it follows that each displacement function d4,, 4' E r, is
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Geometry of Nonpositively Curved Manifolds
constant on F0 and equal to its minimum value. Hence the restriction of F to F. is a group of translations of F0. Finally, let po be any point of F0, and define W to be the closed convex hull of the orbit r(p0). It is easy to see that W is F-periodic, and in fact W = F0. We have proved (6.5.2) in the case that F is abelian. To reduce the general case to the abelian case we shall also need the following. 6.5.4. LEMMA. Let F* C:1(W) be a group that possesses a r * periodic k -fiat W for some integer k >- 0. Let ."denote the k-dimensionalfoliation of R" whose leaves are k -flats parallel to W, and let JV 1 denote the (n - k)dimensional foliation orthogonal to .3 Let R" = V, X V2 be the isometric
splitting such that the foliation of R" determined by V, X V2 are .3 and respectively. If r e 1(6i:") is any subgroup that normalizes r*, then r preserves this splitting of W. PROOF. Let 7 denote the collection of all k-flats in R" that are parallel to the r*-periodic k-flat W. If 0 is any element of F, then 4)r*4-, = r* and it follows that ¢(W) is also a r*-periodic k-flat. If C is a compact subset of W such that r*(C) = W, then 4)(C) is a
compact subset of 4(W) and r*(4(C)) = ¢(r*(C)) = ¢(W). Hence Hd(W, 4)(W)) = Hd(C, 4(C)) < oo, where Hd denotes Hausdorff distance (see (1.2)). It follows that for any element ¢ of r, the k-flat 4)(W) is parallel to W. We conclude that each 45 E r permutes the k-flats of Y, which are the leaves of the foliation .3 defined above. This proves the lemma. We are now ready to complete the proof of (6.5.2). We proceed by induction on n. Let n = 1. If F c I(R) contains a translation, then F is a r-periodic 1-flat. If r c I(R) does not contain a translation, then r has
at most two elements and fixes some point x of 08, a r-periodic zero-flat.
Now let n >- 2 be given, and assume that (6.5.2) holds in I1 for all k S n - 1. Define the derived series r(k) inductively by setting r(,, = r and r(k) = ["(k_ I), r(k_ I)] for all k >- 2. By hypothesis, r(k) = (1) for some
integer k z 2, and if k is the first such integer, then r* = r(k _ ,) is abelian and not the identity. By the proof of (6.5.3) we know that W * = fl m E r. F. is nonempty, where F4. is the set of points in R where d4,. assumes its minimum value. Note that IF leaves W* invariant
since r* is normal in r and y(F4,.) =
,) for all -j r= r, 4* E r*.
If W * has dimension < n - 1, then r admits a r-periodic k-flat W c- W* by the induction hypothesis. Hence we may assume that W* = R", which implies that the elements of r* are translations in R".
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357
If W1 is the closed convex hull in R" of any orbit of P, then W, is a t*-periodic k-flat for some integer k >_ 0. Let .T be the foliation of R' whose leaves are the k-flats parallel to W,. Let R" = V, X V. be the Riemannian splitting such that the foliation of R" determined by V, is .5r The group t preserves this splitting by (6.5.4) since t normalizes P. Let p,: r -b 1(V) denote the projection homomorphisms for i = 1, 2. We assert that (a) p2(I') is a solvable subgroup of 1(V2) and (b) p2(F*) = (1) in 1(V). The first assertion is obvious since IF is solvable. To prove the second assertion note that p2(I'*) fixes a point P2 E V21 where V, X (p2) = W, is the F*-periodic k-flat that defined the splitting R" = V, X V2. However, p2(]'*) consists of translations in V2 since F*
consists of translations in R". This proves (b). We now regard F* _ p,(f *) as a subgroup of 1(V1).
Since V2 has dimension less than n we may apply the induction hypothesis and conclude that there exists a p2(r)-periodic r-flat W2 c V2 for some integer r;-> 0. Let W = V, X W2 = W, X W2, a k-flat in W. We complete the proof of (6.5.2) by showing that w is I'-periodic. Clearly, W is invariant under r. Since W, = V1 X (P2) is a P -periodic flat there exists a compact set C, c V, such that p,(F*)(C1) = V1. Let C2 c W2 be a compact set such that p2(I'XC2) = W2. If C = C1 X C2, then clearly C is a compact subset of V, X W2, and
we assert that r(c) = V1 X W2 = W. Let q = (q1, q2) E W be given. Choose 4)2 E I' such that p2(42)(g2) E C2, and choose 4i, E r* such that p1(¢1)p1(OZXq,) E C1. If 4) = 4)1 ¢2 E I', then
4)(q) _ (P1(4)1)P1((P2)(q,),P2(4)2)(g2)) E C1 X C2 = C
since P2(4)1) = 1 by (b) above. This completes the proof of (6.5.2). REMARK. The assertion (6.5.3) for abelian groups IF c I(R") is stronger
than the assertion (6.5.2) for solvable groups I' c I(R"). This discrepancy cannot be avoided as the following example shows.
ExAMpL.E. A 2-step solvable group r c I(R') such that R3/I' is compact but R3/To - R is noncompact, where I'o denotes the translation subgroup of r. Write R3 = R2 X R and let I'a = (4) E I(R3): (k(x, t) = (T(x), t)
for all (x, t) E R 2 X R where T is a translation of R').
Let a E 1(R2) be any element such that r(a) E 0(2) has infinite order in 0(2). Define a E 1(R3) by
a(x,t)=(a(x),t+1)
for all (x,t)ER2xP.
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Geometry of Nonpositively Curved Manifolds
Let r c 1(083) denote the subgroup generated by F0 and a. It is easy to see that 083/I' is compact and [I', tl c r0 since r(r) is the infinite cyclic
group generated by r(a). Hence r is 2-step solvable. Since I70 is a normal subgroup of r it follows that for every /3 E t there exists an element y of 1'0 and an integer k such that /3(x, t) = (yaI(x), t + k) for all (x,t) E 082 x R. If /3 is a translation of 083, then ak is a translation of 08 2 and r(a )k = r( a k) = 1. Hence k = 0 and 6 = y E I70 since r(a )
has infinite order in 0(n). This shows that r. is the translation subgroup of r, and clearly 083/I'0 = R. We can make a slight but useful extension of (6.5.2). A group r is said to be almost solvable if r admits a solvable subgroup r* with finite index in T. 6.5.5. PROPOSITION. Let IF c 1(08") be an almost solvable subgroup. Then there exists a F -periodic k -flat W c P" for some integer k z 0.
PROOF. Let I'* be a solvable subgroup of F with finite index in r. By replacing r* by the intersection of all conjugate subgroups 4)I7*¢',
0 E t, we may also assume that r* is normal in F. By (6.5.2) there exists a I'*-periodic k-flat WC W. Let R" = V, X V2 be the splitting such that W = V, x q2 for some q2 E V2 and the k-flats parallel to W are the leaves of the foliation of R" determined by V1. By (6.5.4), t preserves this splitting 08" = V, x V2. Let p;: t be the projection homomorphisms for i = 1, 2. Note that p2(r*) fixes the point q2, where W= V, x (q2), since r* leaves W invariant. Hence the orbit p2(1'Xg2) is finite, and it follows from the Cartan fixed point theorem (see (1.4)) that P2(r) fixes a point qi E V2. We conclude that W* = V, x q2 is a I'-periodic k-flat in 08". 6.6.
Discrete subgroups of AR")
The main result of this section is a strengthened version of one of the Bieberbach theorems. 6.6.1. THEOREM. Let r c 1(08") be a discrete group. Then there exists a r -periodic k -flat W c R" for some integer k >_ 0. Moreover, there exists a
subgroup F0 of finite index in t such that the displacement function d. is constant on W and assumes its minimum value on W for each element 0 of F0. 6.6.2. COROLLARY. Let r c 1(W) be a discrete group that does not leave
invariant any k -flat W for k < n. Let r* denote the subgroup of t
Isometries of W
359
consisting of translations of R". Then r* has finite index in r and R,/r* is compact.
The corollary follows immediately from the theorem. The corollary itself is a slight generalization of one of the Bieberbach theorems, which in its usual form assumes that R"/r is compact. Our proof of (6.6.1) is very similar in spirit to the proof in [Au] of the Bieberbach theorem just
mentioned. For a completely elementary proof of the Bieberbach theorems see [Buser]. The statement of (6.6.1) looks somewhat peculiar, and in particular, it does not assert the existence of translations in r. In fact, the example below shows that an arbitrary discrete group r c I(R") may contain no
translations of R" whatsoever if R'/r is noncompact. EXAMPLE. Let k and n be any positive integers, and let {R1,..., R"} be
a set of commuting elements in O(k) that generate a free abelian subgroup of 0(k). Let {T1,...,T"} be generators for a lattice of linearly independent translations in R". If q = (q,, q2) E 138" X Rk is any point, then define 4;(q) = (T,.(q,), R;(q2)) for 1 < i < n. Then the set generates a discrete abelian group r c1(W x 338k), but IF contains no translations except the identity. The n-flat R" x {origin} is IF-compact.
Before proving (6.6.1) we need two preliminary results.
6.6.3. LEMMA. Let r c 1(W) be a discrete group. Then r admits a subgroup r* of finite index such that either r* or [r*, r*] consists of translations of R". In particular, r* is solvable of at most two steps.
6.6.4. LEMMA. Let M be arbitrary, and let r e I(M) be a discrete group. Let r0 be the subgroup of r consisting of Clifford translations (cf. (1.9)). If ro {1}, then there exists a finite index subgroup r* of r such that r* centralizes 170; that is, 46,, 0o = (bo * for all 46 * E r 4o e ro. PROOF OF LEMMA 6.6.3. Let r: 1(R") -> O(n) be the homomorphism of (6.3.1). We consider separately the cases
(1) r(r) is finite and (2) r(r) is infinite. E r: r(c) = 1}. Clearly, r* has finite In the first case we let r* index in r and consists of translations of R". Now suppose that r(r) is infinite, and let G = r(r), the closure in O(n) of r(r). The group G is a Lie group of positive dimension since O(n) is compact. Let Go denote the connected component of G that contains the identity, and define
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Geometry of Nonpositively Curved Manifolds
F* = {0 (-= F: r(4) E G0). The subgroup F* has finite index in F since Go has finite index in the compact group G. To prove that [F*, F*] consists of translations of R' it suffices to prove that G is abelian.
Let U be a neighborhood of the identity in 0(n) that satisfies the conditions of (6.2.2). Fix a number n with 0 < 71 < -',. Making U still smaller if necessary we may assume that 11-A I = m(A) < 71 for all A E U. Let W = U n Go. Since W is a neighborhood of the identity in Go, W generates Go and it suffices to prove that the elements of W commute.
Let g, and g2 be arbitrary elements of W. Since W c r(l:') = G we can find sequences {y"} c r and {Q,} c I' such that r(N) --+g, and r((T") - 92 as n -> oo. Choose a positive integer No such that r(y") and r(o,) lie in W for all n >_ No. It suffices to show that r(') and r(o ) commute for all n >- No. Fix n >_ N., and let S = {y", o"}. By the definition of W we know that m(y") < 71 < Z and m(o ,) < q < Z . It follows
from (6.4.6) that S(k) -* 1 as k --> . Therefore S(k) _ {1} for some positive integer k since S(k) c F for all k and I' is discrete. Therefore the group generated by S is nilpotent by (7.2.3) below, and it follows that the group generated by r(S) = (r(y,,, r(q")} is nilpotent since r is a homomorphism. By the choice of U and the fact that r(S) c W C U it follows from (6.2.2) that the elements of r(S) commute. 0
PROOF OF LEMMA (6.6.4). By (3) of (1.9.4) we see that 1'0 is a free abelian group of rank k >- 1. Let *1 ... I 4k} be a set of generators for I's, and let C; > 0 be the (constant) value of the displacement function d f, on M for 1 < i < k. For each y E 1' the element y¢; y' is a Clifford translation whose displacement function also has the constant value C;. By the discreteness of F each element O; can have only finitely many distinct conjugates in r, and it follows that the centralizer of (}ii in r is
a subgroup F, with finite index in I'. Hence r* = w_ , index subgroup of r that centralizes r.. 0
1';
is a finite
PROOF OF THEOREM (6.6.1). By (6.6.3), F contains a solvable subgroup
I`* of finite index, and hence by (6.5.6) there exists a F-periodic k-flat W c I8" for some integer k >- 0. Therefore it suffices to prove (6.6.1) in
the case that R"/F is compact. In this case we show that if Fo is the subgroup of r consisting of translations of R", then F' has finite index in r. Assuming that IR"/F is compact we proceed by induction on n. We leave the case n = 1 as an exercise. If F 0 is the translation subgroup we show first that F0 0 {1}. By (6.6.3) there exists a finite index subgroup F'* of r such that either (1) r* consists of translations of I8" or (2) [F*, t*] consists of translations of (W". Note that R,/F* is compact since F* has
Isometries of R"
361
finite index in r. It suffices to consider the case that r* is abelian, for otherwise 1 # [r*, r*] c ro. In this case we show r* c F0. Given an element 4 E r* we observe that the displacement function d , is con-
stant on r*-orbits in R" since r* is abelian. Hence d4, is bounded above in R" since it assumes all of its values on any compact set C c R"
such that r*(C) = fl". By (2) of (1.6.5) it follows that the convex function dm must be constant on R", and hence 0 is a translation of R". This proves that ro * (1). If r' denotes the subgroup of r that commutes with all elements in
the translation subgroup ro * (1), then r' has finite index in r by (6.6.4). Hence ro = ro n r, the translation subgroup of r', has finite index in ra and lies in the center of r. Moreover, R,/r' is compact since r' has finite index in r. For each point p E R" define WP to be the closed convex hull of the orbit ro(p). Each set WP is a k-flat in R", where k >_ I is the rank of the free abelian group ro, and the k-flats {WP: p E W"} are parallel. Moreover, WP/ro is compact for all p r= (' since rF consists of translations.
If WP = R' for some p E R", then we are done. We assume that k = dim WP < n and obtain a contradiction. Since ro is a normal sub-
group of r' it follows that O(W,) = WP for all 4, E r' and p E R". Hence there exists a r'-invariant splitting R" = V1 X V2 such that V, x (q2) = Wq for all points q = (q,, q2) E V, X V2 = W". Let p;: r' ->
1(l;) denote the projection homomorphism for i = 1,2. We note that Vl/p1(ro) is compact since Wp/ro is compact for all p E R". We assert that p2(r') is a discrete subgroup of 1(V2). Observe first that p2(F) = {1}; this follows since 170' leaves invariant Wq = V, x {q2) for every point q = (q1, q2) E V1 X V2 = R". Suppose now that (y") is a
sequence in r' with (p2(y")) converging to the identity in 1(V2). We show that only finitely many of the elements {p2(y")} are distinct and hence P2(,,) = 1 for all n z No. Let q = (q,, q2) be any point of R". Since V1/p1(ra) is compact we may choose a sequence {a"} c ro such that (p1(a")p1(y"Xq1)) is a bounded sequence in V,. If On = a" y" E r, then ;,(q) = (p,(a")p,(y")(q1), p2(a")p2(y)g2)) is a bounded sequence in R" since p2(a") = 1 for all n and p2(y" X q2) -' q2 as n -> co. By the discreteness of r there are only finitely many distinct elements {o"}, and hence there are only finitely many distinct elements (P2(0',,)) = (P2(%))' We have proved that p2(r') is discrete in 1(V2). Observe that p,(ro) lies in the center of p,(r') since rF lies in the center of r'. Hence p1(r') consists of translations in V1 by an argument earlier in the proof of (6.6.1); V,/pl(ro) is compact and dm. is constant on pl(r;,)-orbits for every 0' Ep,(r'). Note that V2/p2(r') is compact since 68"/r' is compact. By the induction hypothesis, p2(r')
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Geometry of Nonpositively Curved Manifolds
contains a nonidentity translation p2(4)') for some element 0' E T'. It follows that 46' is a translation of R" since p,()') is a translation of R". However, r is the translation subgroup of T', and hence p2(4)') E p2(r) = (1) by work above. This contradiction shows that the closed convex hull of FO'(q) equals R" for every n. Hence R"/I'o is compact. Since r c t' and R "/F' is compact it follows that r has finite index in r' (cf. (1.9.34)). This completes the proof of (6.6.1) since r' has finite index in r. O
7
Spaces with Euclidean Factors
7.1. Clifford translations
363
Criteria for the existence of Clifford translations in groups of isometries 363 7.2.
Discrete isometry groups in spaces with Euclidean factors Relationship between the center and the Clifford subgroup of
368
a lattice 370 Projections of lattices into isometry groups of non-Euclidean factors 372
73. Compact nonpositively curved manifolds whose fundamental groups have nontrivial center Relationship between nontrivial center and existence of Euclidean
372
factors 372 Canonical construction of such manifolds 372
In this chapter we consider isometry groups in a simply connected manifold M whose Euclidean de Rham factor has positive dimension. We are especially interested in the structure of lattices I' c 1(M) or more generally subgroups IF c 1(M) that satisfy the duality condition. The main results are theorems (7.2.1) and (7.3.3). 7.1.
Clifford translations
We recall from (1.9) that an isometry 0 of M is a Clifford translation if the displacement function d., is constant on M. In this case 0 acts as a translation on the Euclidean factor of M (cf. (1.9.4)). 7.1.1. DEFINITION. Let C(M) denote the group of Clifford translations of
M. If r c 1(M) is any subgroup let c(r) = r n c(m), the subgroup of Clifford translations in r.
We are interested in conditions on a group r c 1(M) that imply that c(r) * (1). The next result is a useful criterion. 7.1.2. LEMMA. Let M be arbitrary, and let r c 1(141) be a subgroup whose limit set L(I') equals M(x). If 0 E 1(M) commutes with every element of
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Geometry of Nonpositively Curved Manifolds
t, then 0 E OR). In particular, Z(I') c C(r), where Z(I') denotes the center of r. REMARK. If IF c 1(141) satisfies the duality condition, then L(t) = M(oo) as one may see directly from the definitions in (1.9.5) and (1.9.15). PROOF OF LEMMA (7.1.2). If 0 E I(141) commutes with every element of
I', then the displacement function d, is constant on the orbit r(p) for any point p of M. The closed convex hull of r(p) equals M since L(t) = M(me). The convex function d. equals d,,(p) on I'(p) and hence is bounded above by d ,,(p) on M. Therefore d4, is constant on M and 4) is a Clifford translation by (1.9.4).
0
The next result has a technical appearance but is the key to several later results. 7.1.3. PROPOSITION. Let M be arbitrary, and let N c I(141) be a nonidentity subgroup whose normalizer IF in I(M) satisfies the duality condition and whose centralizer Z in 1(M) is not the identity. Then either (1) N c C(M) or (2) there exists a nontrivial Riemannian splitting M = MI X M2 that is preserved by I' such that
(a) p,(N) c C(M) and (b) if p2(N)
{1}, then p2(N) has trivial centralizer in 1(M2).
Here pl_ and P2 denote the projection homomorphisms of IF into AM,) and 1(M2 ).
REMARK. Note that Z is a subgroup of T. This result is similar to the lemma in section 4 of [E15]. Note also that cases (1) and (2) are not mutually exclusive; for example, let M = M1 X M2 be a Riemannian product such that M2 has no Euclidean factor and let N = C(M1) X (1) = C(M). PROOF OF PROPOSITION (7.1.3). The proof requires several steps.
STEP 1. Z has no fixed points in M.
The group r normalizes the subgroup Z since IF normalizes N. Hence if Z fixes a point p E M, then Z fixes every point in the closed convex hull C of r(p). However, I' satisfies the duality condition, which implies that L(t) = Moo) and C = M. This contradicts the hypothesis
that Z*{1}. STEP 2. The limit set of Z is a nonempty involutive subset of MI(oo) (cf. (5.3)). Moreover, N fixes every point of L(Z).
We show first that L(Z) is nonempty. If L(Z) were empty, then Z would fix a point in M by (1) of (1.9.6), but this is ruled out by step 1. Next, let p e M and x E L(Z) be given. To show that L(Z) is involutive
Spaces with Euclidean Factors
365
we need to show that y = SP(x) = ypx( - oo) E L(Z). Choose a sequence x as n - oo. Passing to a subsequence we let (0,J C Z such that 4),-, '(p) converge to x* E L(Z). By (3) of (1.9.6), r leaves L(Z) invari-
ant since Z is a normal subgroup of F. By (3) of (1.9.13) we see that x* E F(y) since the points x and x* are Z-dual and hence F-dual. By (4.7.1) the set r(y) is a minimal set for r since r satisfies the duality
condition. Therefore Y E F(x*)cL(Z), which proves that L(Z) is involutive.
We prove that N fixes every point of L(Z). Let x r=- L(Z) and p E M
c Z such that as be given, and choose a sequence n --p oo. For any element 0 of N we have 4)(x) = lim -, 00,,(p) = limn -,o c (4)p) = x by (3) of (1.4.4) and the fact that N commutes with
Z. This completes step 2.
STEP 3 (Conclusion of the proof). Let X = L(Z) c M(oo), and for each point p e M let BP denote the closed convex hull of {yPx(R): x E X). Fix points p E M and x E X. If 45 E N is arbitrary, then by step 2, 0 fixes both x and y = y,x(- cc) E L(Z). The convex function d4, is bounded above on y,x(R) and hence constant on y,x(R). It follows that d4, < di(p)
on BP for all points p E M since the points p e M and x E X were arbitrary. We consider separately two cases.
Case 1. BP = M for some point p e M. Case 2. BP is a proper subset of M for all points p E M. In the first case it follows from (2) of (1.9.4) that N c C(M) since dm is bounded above on M for all 0 E N. Assume that the second case holds. By the theorem in (5.5) there exists a nontrivial Riemannian splitting M = M, X M2 such that for each point p E M the set BP is the integral manifold through p of the foliation in M determined by Ml.
Moreover, X = L(Z) c M,(oo) and r preserves the splitting since it leaves X invariant. Let p, and P2 denote the projection homomorphisms of r into 1(M,) and I(M2) respectively. A convexity argument above shows that d. s d,(p) on BP =MI X (P2) for each 0 E N and each point p = (p P2) E M, X M2 = M. This shows that p,(N) c C(MI) by (2) of (1.9.4) and proves (2a) of (7.1.3).
We show that N2 = p2(N) has trivial centralizer Z2 in 1(M2) if N2 * (1). Note that p2(F) normalizes p2(N) and P2(F) satisfies the duality condition in 1(M2) by (1.9.22). If N2 * (1) and Z2 # (1), then we may apply steps 1 and 2 of the argument above to conclude that L(Z2) is a nonempty subset of M2(oo). Note that (1) x Z2 c Z, the centralizer of N in 1(M), since N c p,(N) x p2(N) and Z2 centralizes p2(N) = N2. Hence L(Z2) c L(Z) = X c M,(co) by the construction of the splitting of M = M, X M2. We obtain a contradiction since M,(0 and M2(°) are
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disjoint subsets of M(oo), and this shows that Z2 = (1). This completes (2b) of (7.1.3) and concludes the proof of (7.1.3).
We now obtain some applications of (7.1.3). The next result
is
theorem 2.4 of [CE]. 7.1.4. PROPOSITION. Let N c I(M) be a nonidentity abelian group whose normalizer satisfies the duality condition. Then N c C(M).
PROOF. If t and Z denote the normalizer and centralizer of N in 1(M), then N c Z and hence Z is nonempty. We may assume that case (2) of (7.1.3) arises, for in case (1) we are done. Since p2(N) is abelian, using the notation in (7.1.3), it follows from (2b) of that result that p2(N) = (1). Hence N = p,(N) c C(Ml) c C(M) by (2a). The next result is theorem C of [E15]. 7.1.5. PROPOSITION. Let I' c 1(M) satisfy the duality condition, and suppose that the centralizer Z of f in 1(M) is not the identity. Then there exists a Riemannian splitting M = M, X M2 that is invariant under t and Z such that (1) M, is a Euclidean space of positive dimension,
(2) Z = C(M,) X (1), (3) p,(I') c C(M, ), and (4) p2(r) has trivial centralizer in 1070-
The maps p, and P2 are the projection homomorphisms into 1(M,) and 1(M2). The factor M2 may be absent. PROOF. We apply (7.1.3) to the group N = F. If I'* denotes the normalizer of IF in I(R' ), then F* satisfies the duality condition since r* : r. We suppose first that t c C(M). Note that L(I') = M(oo) since I' satisfies the duality condition, and hence Z c C(M) by (7.1.2). In this case we get M = M, and the M2 factor is absent.
Next we suppose that r is not contained in C(M). By (7.1.3) there exists a nontrivial Riemannian splitting M = M, x M2 that is preserved by T and Z such that p,(r) c_ C(M,) and p2(1') has trivial centralizer in 1(M2), where p, and P2 denote the projection homomorphisms into 1(M,) and I(M2 ). By (1.9.22), I', = p,(r) satisfies the duality condition and hence L(F,) = M,(r). By (1.9.4) it follows that M, is a Euclidean space of positive dimension since t, c C(M) . Next observe that p2(Z) = (1) since p2(Z) centralizes p2(r), which satisfies the duality condition in I(M2) and hence is nonempty. Hence Z cl(M,) x {1} c 1(M). Since Z = p,(Z) centralizes t, = p,(r) and L(I',) =A11(00) it follows from (7.1.2) that Z c C(M,) x {1}. Equality holds since C(M,) is
an abelian group and p,(r) c C(M, ). This completes the proof of (7.1.5).
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The first three assertions of the next result are contained in theorem 1 of [Sc2]. The formulation here, except for (5), is theorem D of [E15], whose proof is slightly different from that given here. 7.1.6. PROPOSITION. Let IF C I(M) satisfy the duality condition, and sup-
pose that r = A B, where A and B are commuting subgroups of r. Then there exists a r-invariant splitting Q' = MI X M2 X M3 such that: (1) M, is a Euclidean space of dimension r >_ 0 and p,(r) c C(M, ).
(2) C(MI) x Id X Id is the centralizer of r in 1(R). (3) (a) If c¢ E A, then 0 _ (01, 02, Id), where 0, e C(M,) and ¢2 E 1(M2 ).
(b) If 4 E B, then d _ (di,, Id, 03), where 0, c= C(M,) and 03 1(M3).
(4) if r is discrete, then p2(r) = p2(A) and p3(r) = p3(B) are discrete and have trivial centralizer in 1(M,) and 1(M3) respectively.
(5) If r is a lattice in 1(M) and if N = kernel(p,) n kernel(p3), then N is the center of IF and N is a lattice in 1(M1). Here p,: r -- 1(M;) denotes the projection homomorphism for i = 1, 2, 3. Any of the factors M, may be absent, and M = M, may also be true.
PROOF. Let Z denote the centralizer of r in 1(M). If Z = (Id), then we proceed as below and the result will be true with the M, factor missing. Assume that Z # (Id) and let M = Ml X M2 be a Riemannian splitting
invariant under IF and Z that satisfies the properties of (7.1.5). If r2 =p2(r), A2 =p2(A), and B, =p2(B), then F2 and the subgroups A2 and B2 are normal in F2 and commute with each other. If one of these subgroups, say A2, is the identity, then A cC(M1) x Id by (3) of (7.1.5) and the result is true without the factor M2. We assume that A, and B2 are both nonidentity subgroups of r2. By (1.9.22), F2 satisfies the duality condition. The group B2 does not lie in C(M,) since otherwise B2 would centralize r2, which contradicts (4) of (7.1.5). Applying (7.1.3) to the group N = B2 we obtain a r2-invariant Riemannian splitting M2 = Ma X M. such that (i) pa(B2) c C(Ma) and (ii) if ps(B2) # (Id), then p8(B2) has trivial centralizer in 1(M8). Here pa: F2 -* 1(Ma) and p : r2 -> I(M8) denote the projection homomorphisms. If pa(B2) * (Id), then pa(B2) is abelian by (i) and hence pa(B2) x Id centralizes F2 in 1(M2) since pa(r2) This contradicts condition (4) of the splitting M = M, x il%I2 from (7.1.5), and we conclude that B2 c Id x AM.). Next we observe that p8 (A,) = (Id)
by (ii) since p8(A,) commutes with pp(B2) = B2. Therefore A2 c I(Ma) X Id, and the splitting M = Mi X M. X M8 satisfies (3) of the proposition. Conditions (1) and (2) hold by the conditions of the splitting M = A X M2 determined by (7.1.5).
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We prove (4). From (3) it follows that p2(F) = p2(A), and from (1) and (3) we see that aya'' = p2(a)yp2(a)-' for all elements y r= I' and a EA. Hence p2(r) normalizes F. If p2(T) were not discrete, then the lemma in the proof of (3) of (1.9.35) shows that r would be normalized
by G = p2(F) and centralized by Go * (Id), which contradicts (2). Hence p2(F) is a discrete subgroup of 1(M2). From (3) it follows that p3(F) = p3(B), and a similar argument shows that p3(F) is a discrete subgroup of 1(M3). It follows from (2) that p2(F) and p3(F) have trivial centralizer in 1(M2) and 1(M3) respectively. We prove (5). Let F be a lattice in 1(M). If M2 = M2 X M3 and p2 :
F --> 1(M2) is the projection homomorphism, then pZ (t) 9p2(F) x p3(T) and we conclude that p2(F) is discrete since p2(F) and p3(F) are discrete by (4). If N = kernel(p2) n kernel(p3) = kernel(p2 ), then N is a lattice in 1(M,) by (2) of (1.9.35). Finally, N = C(M,) n t = center(F) by (1) and (2) of this result. 7.2.
Discrete isometry groups in spaces with Euclidean factors
The main result here is the following. 7.2.1. THEOREM. Let M have a Euclidean de Rham factor of dimension n >_ 1, and write M as a Riemannian product M = R" X M,, where M, has no Euclidean de Rham factor. Let F c I(M) be a lattice. Then:
(1) an is a lattice in I(IP"), and C(T) is a free abelian group of rank n. (2) F admits a finite index subgroup_ F with the following property: if r*
is a finite index subgroup of r, then C(I'*) = Z(I'*), a lattice in I(R') and a free abelian group of rank n. Here Z(F*) denotes the center of r *.
To prove the result above we first need the following result. 7.2.2. PROPOSITION. Let P,? be a Riemannian product M = Rn X M where n z 1, and let F c 1(M) be a discrete group that preserves this splitting. Let po: I' - I(R") and p,: F -> I(M,) denote the projection homomorphisms. Then either
(1) p,(F) is discrete in I(MI) or (2) if G = p,(I ), the closure of p,(C) in I(M,), then G is a Lie group of positive dimension and Go is solvable.
REMARK. In this result we allow M, to have a nontrivial Euclidean de
Rham factor. If M, has no Euclidean de Rham factor, then any subgroup r c I(M) preserves the splitting M = Rn X M1.
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PROOF. Assume that p,(I') is not a discrete subgroup of AM,). Then G = p,(f) is a Lie group of positive dimension. It remains to prove that Go is solvable. Let g denote the Lie algebra of G. We first construct special neighborhoods U, c Go and U0 c 1(98") of the identities. Fix a positive number 7 with q < i. If 0,7 = (A E 0(n):
m(A) =I I -< 71}, then On is a Zassenhaus neighborhood of the identity in 0(n) by (6.1.7). Choose a > 0 such that exp: B,. - 00 = exp(B,) is a diffeomorphism, where Ba = (X E g : I X I < a' } and exp: g -- G is the Lie group exponential map. We assume that g has been equipped with an inner product (, > and corresponding norm I I. Choose a,> 0 so small that U, = exp(B0) is a Zassenhaus neighborhood of the identity in Go. Let U0 c I(R') be a neighborhood of the identity such that r(Uo) c 0,,, where r:1(98") -- 0(n) is the homomorphism of (6.3.1). If ri < i is the positive constant chosen above we define S. = {y r= r: p,(y) E U1, m(po(y)) < rt}. We recall that m(po(y)) = m((r c p(, X-)) by definition. We shall show that S,, generates a solvable subgroup r, of r
such that p,(rn) U,. Since p,(r,,) is solvable and U, generates G it will follow that Go is solvable. We begin by listing two results that will be useful. 7.2.3. LEMMA. Let S be a subset of a group H, and define S(k) inductively
as in (6.1.6) by S' = S and S(k) = [S, S(k-')] = {[a, b] = aba-'b-': a E S, b (=- S(k-1)). If S(k) _ (1) for some positive integer k, then the subgroup of H generated by S is nilpotent. 7.2.4. LEMMA. A group H is solvable if every finitely generated subgroup of H is solvable.
The first of these results is lemma 8.17 of [R, p. 147) and the second is corollary 8.4 of [R, p. 142].
To show that S,, generates a solvable subgroup r',' of r we break the proof into two steps.
STEP 1. U, cp,(r7). STEP 2. Any finite subset S of S,, generates a nilpotent subgroup of r,,.
It will follow from these two steps and the second lemma above that r,, is solvable. PROOF OF STEP 1. By definition U, = exp(B0.), where BQ = {X E g:
I X1 < a } and o is some small positive number. Let g E U, be given and
write g = exp(X) for some X E B,,. It suffices to prove that for every e > 0 there exists t E [0,1 ] such that I t - 11 < e and exp(tX) E p, (I',, ).
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By (6.1.8) we may choose an integer N such that if {A,,..., AN} are any N distinct elements of 0(n), then m(A; 'A,) < 71 for some distinct integers i, j with 1 5 i, j < N. Now let e < 1 be any positive number.
Then (e/N )X E B0. and exp((e/N) X) E U, and we may choose a sequence {yk) c r such that p,(yk) -, exp((e/N)X) E U,; recall that p,(r) = G 2 Go 2 U,. Passing to a subsequence if necessary we let (r po)( yk) - a E O(n) as k -* cc. If we set a` =A; for 1 5 i 5 N, then by the discussion above m(a3) < i7 for some integer j with 1 < j 5 N. Now define o k = yj for every k. Then p,(o k) -+ exp(to X ), where to = je/N < e, and (r o poX o.k) - aJ as k --> +cc. Hence o-k lies in S,, for large k and exp(t0X) E p,(S,,)c p,(I' ). Since 0 < to 5 e we may choose an integer M such that t = Mt0 E [0,1i and It - 11:5 e. Hence exp(tX) _ exp(to X )m E p,(F,, ). This completes step 1.
Note that r. consists of all finite words in the elements of Sn and their inverses. If H is a finitely generated subgroup of I'.7, "then there is a finite subset S of S. such that the subgroup H" of r,, generated by S
contains the generators of H and hence H itself. To prove that r. is a solvable subgroup of r it therefore suffices by (7.2.4) to prove step 2. This will show that every finitely generated subgroup H of I',, is nilpotent. PROOF OF STEP 2. Let S S so
=
yN.} be a finite subset of S,,. Let cGo, and let
cI(R')
By (6.4.6), Sok) -' 1 as k -+ cc and S(k) c UU k' -- 1 as k -+ oo since U, is a
Zassenhaus neighborhood of the identity in Go. Therefore S(k) -. 1 as k - cc, and since S(k) lies in the discrete group t for all k it follows that
S'k) = (1) for some positive integer k. Hence the subgroup of F, generated by S is nilpotent by (7.2.3). This completes step 2.
Steps 1 and 2 show that r,, is a solvable subgroup of t and U, c p I (I',, ). Hence Go is solvable since U, generates Go. This completes the proof of (7.2.2). 0 As a corollary of (7.2.2) we obtain lemma A of [E12]. 7.2.5. PROPOSITION. Let I' c I(M) be a discrete groom that satisfies the duality condition. Write M as a Riemannian product M = R " X M,, where n >- 0 and M, has no Euclidean de Rham factor. Let po and p, denote the projection homomorphisms of 1(M) into I(R") and 1(M1), respectively. Then p,(F) is a discrete subgroup of 1(M, ).
PROOF. We suppose that p,(t) is not discrete and obtain a contradiction. If G = p,(F), then G is a Lie group of positive dimension and Go
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is solvable by (7.2.2). The last nontrivial group in the derived series of
Go is an abelian subgroup N of Go that
is left invariant by any automorphism of G(,. Inner automorphisms by elements of p,(r) leave
invariant Go and hence also N since G = p,(r). By (1.9.22), p,(r) satisfies the duality condition, and therefore the normalizer of N in 1(M,) satisfies the duality condition. By (7.1.4), N consists of Clifford translations of M,, and hence by (1.9.4) M, has a nontrivial Euclidean de Rham factor. This contradicts the hypothesis on M,, and it follows that p,(I') is discrete. PROOF OF THEOREM (7.2.1). Any isometry of M preserves the splitting
M = W" X M,, where R" is the Euclidean de Rham factor of M and n >_ 1. Let po and p, denote the projection homomorphisms from I(M) to 1(R') and 1(A%1,) respectively. By (7.2.5), p,(r) is discrete in I(11f,) since a lattice r c I(M) satisfies the duality condition by (1.9.32). Hence N = kernel(p,) is a lattice in 1(R") by (1.9.35). The Clifford subgroup
c(r) of r lies in N by (3) of (1.9.4), and c(r) has finite index in N by the Bieberbach theorem (6.6.2). Hence an is a lattice of translations in AR"), and as a consequence c(r) is a free abelian group of rank n. This proves (1) of (7.2.1).
By (6.6.4) there exists a finite index subgroup F of r such that I' centralizes C(r). Let f* be a subgroup of r with finite index in P. Since C(r*) c C(r) it follows that C(r*) c Z(r*), the center of r*. By (1.9.21), r* satisfies the duality condition since it has finite index in I'.
Hence Z(r*) c C(r*) by (7.1.2). Since f* has finite index in r it follows that C(r*) = F* n c(r) has finite index in C(r). Therefore C(r*) is a lattice of translations in I(!{8") by (1), and C(r*) = Z(r*) is a free abelian group of rank n. This proves (2) of (7.2.1) and completes the proof of (7.2.1). 7.2.6. PROPOSiTroN. Let r c 1(111) be a lattice, and let 11%1= M, X M2 be
a r-invariant Riemannian splitting. Let p;: IF -- I(M1) be the projection homomorphism for i = 1, 2, and let p2(F) be a discrete subgroup of 1(M2). Then either
(1) p,(r) is a discrete subgroup of 1(M,), or (2) M, has a nontrivial Euclidean de Rham factor. Moreover, if we write M, = M. X 11f,,, where M. is the Euclidean de Rham factor of M, and MQ is the Riemannian product of all other de Rham factors of M,, then (a) the Riemannian splitting M = Ma X Ms X M2 is r-invariant and
(b) if dim MR> 0, then ps(r) is a discrete subgroup of I(Mo), where po: r -' I(Ms) is the projection homomorphism.
REMARK. Note that there is no assumption on the existence or nonexistence of a Euclidean de Rham factor of 1112.
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PROOF. Assume that p,(F) is not a discrete subgroup of I(M,), and let G = p,(F). If N = kernel(p2), then by (1.9.35), N is a lattice in I(M1), and N is centralized by Go, the identity component of G. Since lattices satisfy the duality condition by (1.9.32) it follows that L(N) = A1-1(m) by
(1.9.16), and from (1.9.4) and (7.1.2) we conclude that M, has a nontrivial Euclidean de Rham factor. Write M1 = Ma X MM, where Ma is the Euclidean de Rham factor of
M, and Mif is the product of all other de Rham factors of M1. The splitting M1 = Ma X M,3 is invariant under 1(M,) and hence invariant under p,(F). This implies that the splitting M = Ma X MS X' f2 is invariant under I'. If ps: F -,1(Md is the projection homomorphism, then p,,(N) is a discrete subgroup of 1(Ms) by (7.2.5) since N is a -lattice in AM,). Hence p,(N) is a lattice in AM Ms) by (1.9.35). Since M, has no Euclidean de Rham factor it follows from the argument of the previous paragraph that p,,(I') is a discrete subgroup of 1(M,). O 7.3.
Compact nonpositively curved manifolds whose fundamental groups have nontrivial center
Our goal is to describe the structure of those manifolds defined in the title above. For motivation we begin with an application of (7.2.1). 7.3.1. PROPOSITION. Let M be a compact Riemannian manifold with sectional curvature K 5 0. Assume that the universal Riemannian cover Al
has a Euclidean de Rham factor of dimension n >_ 1. Then M admits a finite Riemannian covering M* such that the center of the fundamental group of M * is a free abelian group of rank n.
PROOF. Write A%I = R" X M,, Riemannian product, where M1 has no
Euclidean de Rham factor. Write M as a quotient manifold M/C, where r c 1(M) is a lattice. By (2) of (7.2.1) there exists a finite index
subgroup r* of t such that Z(I'*) = C(1'*), a free abelian group of rank n. If m* = M/r*, then M* is a finite Riemannian cover of M = M/F, and the center of the fundamental group of M* is isomorphic to Z(F*).
El
We now describe the main results of [E10] but omit the details of the proofs. We begin with a typical construction of a complete manifold M with finite volume and sectional curvature K:5 0 whose fundamental
group has nontrivial center. We refer to the example below as a canonical manifold with nontrivial center.
7.3.2. ExAMPLE. Given an integer r >_ 1 let Z be a lattice of translations in 1(R' ), and let T' denote the flat r-torus R "/Z. Now let I'a be
any lattice in I(M,) with trivial center, where Ml is complete and
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simply connected with sectional curvature K5 0. Let p: F0 T' be a homomorphism whose kernel contains no nonidentity elements of F0 with fixed points in M,. We let V. act by isometries on the Riemannian 4, 40(h)) for evproduct M* = T' X M, by defining 00(I;, h) _ h) E T' x M,. The quotient space M = ery 0o E I'o and every point
(T' x Ml)/I'o has finite volume and K 5 0. The center of the fundamental group of M is isomorphic to Z and hence is free abelian of rank r. We now restate the main result of [E10]. 7.3.3. THEOREM. Let M be a complete Riemannian manifold with finite volume and sectional curvature K :!g 0 whose fundamental group is finitely generated and has nontrivial center Z. Then:
(1) M is a canonical manifold with nontrivial center, and Z is a free abelian group of rank r, with 1 S r 5 n, where n is the dimension of the Euclidean de Rham factor of the universal Riemannian cover M. (2) There exists a finite Riemannian covering M * such that if M** is any finite Riemannian covering of M *, then M * * is a trivial n-torus bundle over a smooth orientable manifold M** with finite volume, sectional curvature K:5 0, and no local Euclidean de Rham factor. In particular, M * * is diffeomorphic to T" X M* * .
REMARK. The integer r in (1) above may be strictly smaller than the dimension n of the local Euclidean de Rham factor of M (cf. p. 24 of [E1OD. For example, if M is the 2-dimensional flat Klein bottle, then the center of the fundamental group of M is infinite cyclic. However, M admits a double covering by a flat 2-torus whose fundamental group is free abelian of rank 2. PROOF. This theorem follows from the main theorem of [E10], which considers only the case that M is compact. However, the proof of the main theorem of [E10] remains valid for finite volume manifolds with the properties stated in (7.3.3). For the convenience of the reader we give a detailed outline of the proof of (2) of (7.3.3) and indicate some changes and simplifications of the arguments of [E10]. The proof of (1) of (7.3.3) is exactly the same as the proof of (1) in the main theorem of [E10], and we omit it. We note that M has a nontrivial Euclidean de Rham factor by (7.1.2) and (1.9.4). The first step in the proof of (2) of (7.3.3) is the following. LEMMA A. Let M be the universal cover of M, and let r c I(M) be the deck group of the covering. Write M as a Riemannian product R" X M,, where 08" is the Euclidean de Rham factor of M and M, is the product of all other de Rham factors of M. Let po: F -' I(F ) and p,: I' -1(M,) denote the projection homonwrphisms. Then F admits a finite index
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subgroup F * with the following properties:
(1) I'o* = kernel(p 1) n I' * is a lattice in I(W) that consists of translations. Moreover, ro is both the Clifford subgroup C(I' *) and the
center Z(r*) of r*. (2) po(r*) consists of translations of R'.
(3) p,(r*) is a discrete subgroup of 1(M) that contains no elliptic elements. Hence the quotient space M, = M, /p,(r *) is a smooth manifold with finite volume, sectional curvature K5 0, and no local Euclidean de Rham factor.
PROOF. Let F be a finite index subgroup of IF with the properties of (2) in (7.2.1). Since F is a lattice in 1(M) it follows from (7.2.5) that p,(r)
is discrete, and hence p,(F) is a lattice in 1(M,) by (1.9.35). By the argument on page 26 of [E10] there exists a subgroup r* of r with finite index in F such that p,(r*) contains no elliptic elements. The elements of po(r*) commute with those of po(Z(r*)) = Z(r*) and Z(r*) = C(r*) by the choice of F. By (1) of (7.2.1) we know that C(r*)
is a lattice in I(R') consisting of translations, and we conclude that po(r*) consists of translations of R". We have proved (2) and (3). From (3) and (1.9.35) we know that ro = kernel(p,) n r* is a lattice in I(R"), and r,* consists of translations of 118" by (2).
In the sequel we use the splitting M = R" X M, from the lemma above for lemma 1 of [E10]. THE CANONICAL HOMOMORPHISM. Let M = R" X M,, F*, po: r -1(68"),
and p,: r -.I(M,) be as in lemma A. Let z* = Z(r*) = C(r*) _ kernel(p,) n r*. By (1) of lemma A the quotient manifold R'/Z* is a flat n-torus that we denote by T". Let P: 118" -> T" denote the covering homomorphism. If we set r; = p,(r*) c I(M, ), then we define a homo-
morphism p: r; - T" by
p(p,(y)) =P(po(y))
for all yE r*.
The homomorphism p is well defined since kernel(p,) n r* = Z*, the deck group of P: R" -> T". The next step in the proof of (2) of (7.3.3) is lemma 4 of [E10], which
we restate here. We refer the reader to [E10] for details of the proof. LEMMA B. Let r* be the finite index subgroup of r with properties (1), (2), and (3) of lemma A. Then there exists a C°° map F: M, --> T" = I8"/Z*
such that F(yq,) = p(y) F(q,) for all y E r; = p,(r*) and q, E M,. We are now ready to prove (2) of (7.3.3). We adopt the notation of lemma A, and we show that M* = M/r* satisfies (2). Let M** be any finite Riemannian covering of M*, and let r** be the finite index subgroup of r* such that m** = M/r**. Let r** =p,(r**) cp,(r*) =
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IT c 1(M,). First we replace M** by an isometric manifold N, = N/t**
with respect to a twisted isometric action of r,** on N = T" X M,. Given y, E r** and q,) E N we define y,(f, q,) _ (p(y,) - , y,(q, )), where p: r* -' T" is the homomorphism defined above. The action of r** on N is freely acting and properly discontinuous, and we let k*: N -* N, = N/I'** denote the corresponding covering map. We assert that
(1) N, = (T" x 1N,)/I'** is isometric to M** = M/F**.
To verify this let ir: R" X M, --+ M** be the universal cover and define F: M** - N, by F(ir(e, m,)) = k *(P( ), m,), where P: R" -> T" is the covering homomorphism. It is routine to show that F is a well-defined bijection, and F is an isometry since IT, P, and k* are local isometries. Next we show that
(2) N, is an n-torus bundle over M; * = M,/F**.
If k*: T" X M, - N, and irM, --> M** = M,/I'** are the covering maps, then define p: N, - M* by p(k*(, m, )) = Tr,(m, ). The projection p is well defined and in remark I following lemma 4 of [E101 it is
shown that p: N, - M** has the structure of an n-torus bundle over M**. i
To show that M** is a trivial n-torus bundle over M**, and hence diffeomorphic to T" X M**, it suffices by (1) and (2) to establish the following.
(3) There exists a C' section s: M* * - N, of the bundle p: N, --> M**.
Let F: M, -' T" be the C' map of lemma B and let 7r,: M, - M* be the covering projection. It is routine to verify that the map s: M** -.- N, given by s(1r,(m, )) = k *(F(m, ), m,) is a well-defined section.
8
Mostow Rigidity Theorem
U. Two statements of the theorem
376
8.2. Outline of the proof
381
Equality of ranks 381 Construction of equivariant pseudoisometries 381 Bijective correspondence of k-flats 382 Induced mappings on the spaces of pointed k-flats 382 Induced mappings on the space of splices 383 Induced mappings on Furstenberg boundaries 383 Completion of the proof 384
83. Convergence of totally geodesic submanifolds 8.4.
The space of pointed k-flats #9A topology on 6R9- 386 Convergence of sequences of pointed k-flats 387 Continuity of the natural action of isometrics on PY 388 Pseudoisometries induce continuous maps between spaces of pointed k-flats
8.5.
384 386
389
Induced mappings on the space of splices
The sets C(p, S)
391
391
Pseudoisometries induce mappings between spaces of splices 392 The mapping ¢: 392 Equivalent description of the induced mappings 392 Mappings induced by pseudoisometries are strongly order preserving 394 Pseudoisometries induce mappings between closed Weyl chambers in M(oc) 395 8.6.
8.1.
Induced mappings on Furstenberg boundaries Pseudoisometries induce homeomorphisms between Furstenberg boundaries 396
3%
Two statements of the theorem
8.1.1. THEOREM (Mostow). Let G and G* be connected, semisimple Lie groups with trivial center that have no compact normal subgroups other
than the identity. Let r and r* be discrete, irreducible, cocompact
Mostow Rigidity Theorem
377
subgroups of G and G* that admit no nonidentity elements of finite order.
Let G have real rank k z 2. Then any isomorphism o: r -' r* extends to an analytic isomorphism 0: G -+ G*.
REMARK. In Mostow's slightly more general formulation of this result
he omits the hypothesis that r and r* be irreducible and adds instead the hypothesis that G have no PSL(2, f8) factor that is closed modulo r. The hypothesis in the version above that r be irreducible implies by (3.11.6) that G has no factor of any sort that is closed modulo r. Margulis has generalized the Mostow rigidity theorem to a superrigidity theorem. See chapter 7 of [Mars] and [Z1). In chapter 9 of these
notes we shall discuss rigidity theorems for compact manifolds of nonpositive sectional curvature that are inspired by the Mostow rigidity
theorem. In particular, see the generalization by Gromov in theorem (9.4.3).
We recall that r is cocompact if the coset space r/G is compact. The group r is irreducible if no finite index subgroup is a direct product. Similarly, a compact locally symmetric space M is said to be irreducible if no finite Riemannian cover of M splits as a Riemannian product. If we write m = M/r, where M is symmetric of noncompact type and IF is a discrete subgroup of G =1(,(M), then M is irreducible if and only if r is irreducible by (10.3.9). We note that if r is any discrete cocompact subgroup in a connected semisimple group G with trivial center and no compact normal subgroups except the identity, then by lemma 8 of [Sell (see also [Se31) r admits a finite index subgroup that contains no elements of finite order except the identity.
We may reformulate the result above in differential geometric fashion. 8.1.1'. THEOREM (Mostow). Let M and M* be compact locally symmetric
spaces of noncompact type. Let M be irreducible of rank ;-> 2, and let 0: it 1(M) -* Tr I(M *) be an isomorphism. Then 0 is induced by an isomett f : M -+ M * after multiplying the metric of M by positive constants on local de Rham factors.
Some explanation of the statement "0 is induced by an isometry f " is required. Let M and M* denote the universal Riemannian coverings of M and M*, and let r and r* be discrete subgroups of G =10(M) and
G* = lo(M*) such that m = M/r and m* = M*/r*. We regard the groups r and r* as the fundamental groups of M and M*. An isometry f: M -> M* induces an analytic isomorphism 9: G G* such that 8(r) = r*, and this isomorphism 0 is uniquely determined up
to composition with inner automorphisms by elements of r or P.
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Geometry of Nonpositively Curved Manifolds
More precisely, let 7r: M - M and Ir*: M* M* denote the covering projections, and let F: M - M* be a lift of f; that is,
(a) 7r*oF=fo7r. The map F is an isometry and defines an isomorphism OF: G - G* given by
(b) OF(g) = F 0 g o F-' for all g E G.
It follows from (a) that OF(F) = P. If F': M - M* is another lift of f, then since Tr * C F = iT * o F' = f 0 IT, a routine connectedness argument
shows that there exist elements y E t and y* E r* such that
(c) Foy=F'=y*oF. It follows directly from the definition that 9F. =1.,. o OF = OF o I , where
17 and I., denote the inner automorphisms of G* and G by y and y. Observe that if F: M -> M* satisfies either equality of (c), then F' is a lift of the isometry F: M -> M*. DEFINITION. Given an isomorphism 0: F -* I'* we say that 0 is induced by
an isometry f : M -> M * if there exists a lift F: M - M * of f such that OF = 0 on F.
We show the equivalence of the two statements of the theorem. Before proving-that (8.1.1) implies (8.1.1') we need some preliminary remarks. Let M be a symmetric space of noncompact type, and let K be a maximal compact subgroup of G =10(M). Then K = G,, = (_g E
G: g(p) =p) for some p E M by (2) of (1.13.14), and hence M is diffeomorphic to the coset space G/K since G acts transitively on M. Let I' be a discrete subgroup of G. Since K is compact it follows that r is cocompact in G (i.e., r \ G is compact) if and only if M/t = I' \ G/K is compact. By (2.1.1), G is a connected, semisimple Lie group with trivial center. We show that G has no compact normal subgroups except the identity. If N is a compact subgroup of G, then N has a fixed point
in M by (1.4.6). If in addition N is a normal subgroup of G, then the fixed point set of N in M is invariant under G. Therefore G has no compact normal subgroups except the identity since G acts transitively on M. We are now ready to prove that the first statement of the theorem implies the second. Let M and M* satisfy the hypotheses of the second statement, and let ir: M -* M and lr *: M * M * denote the universal Riemannian covers of M and M*. Let F cl(M) and r* cl(M*) be the deck groups of these coverings, and let an isomorphism 6: F --> r* be given. We now encounter a slight technical difficulty since the lattices r
and r* may not be contained in the connected groups 10(M) and
Mostow Rigidity Theorem
379
Io(M*) respectively. This delays an immediate application of the first statement and (3.9.11), which would otherwise yield the second statement. However, the delay is not serious. There exists a uniformly continuous map h: M - M* such that (1) h o ¢ = 0(¢) o h for all ¢ E 1'. The construction of h is elementary. See, for example, proposition 2.1 of [E9) for details. Since 10(M) and 10(M *) have finite index in 1(M )
and 1(M*) we can find a finite index subgroup r0 of r such that r0 c G =10(M) and I'o = 0(r0) c G* = I0(M*). By (8.1.1) the isomorphism 0:1'0 -* F extends to an analytic isomorphism 0: G - G*. Let M and M* have the canonical metrics defined in (2.3.11) by the Killing forms of q and q *. On each irreducible de Rham factor of M and M * the canonical metric is a constant positive multiple of the given locally symmetric metric by (2.3.9).
By (3.9.11) there exists an isometry F: M -. M * with respect to the canonical metrics such that
(2) Fog = 0(g)" F for all g E G. To derive the second statement of Mostow's theorem from the first it suffices to show that F c 0 = 0(4)) o F for all 0 E r, which we finally achieve in (5) below. By the discussion following (5) we then obtain an isometry f : M -* M * with respect to the canonical metrics that induces
the isomorphism 0: r - P. By (2) we already know that F o 0 _ 0(4))" F for all 0 e I70, but this is not enough. In the remainder of the proof we let ¢* denote 0(4)) for any element of F. We recall from (4.5.13) that a vector v e SM is F-periodic if there exist 46 E t and w > 0 such that (4) c y,,Xt) = y,.(t + w) for all t E R. Equivalently, v is r-periodic if there exist 0 E T and w > 0 such that
d4)(v)=g'v, where (g') denotes the geodesic flow. It is easy to see that v E SM is r-periodic if and only if der(v) is periodic in SM, where ir: M -> M is the universal cover. By (4.7.7) it follows that the rperiodic vectors are dense in SM. If we set Pr = {x E M(00): x = y,.(oo)
for some r-periodic vector v in SM), then Pr is dense in M(oo) with respect to the cone topology. If h and F are the maps of (1) and (2), then we assert that
(3) if x is a point of Pr and v is a I'-periodic vector in SM such that c y,,Xt). x = y,,(oo), then F(x) =
Let x and v be as above, and choose 4) E r and w > 0 such that d ¢(v) = g *'v. Without loss of generality we may assume that 0 E T0, replacing 0 by ON for a suitable positive integer N. Since 0 translates
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Geometry of Nonpositively Curved Manifolds
the geodesic y,, by w it follows from (1) that 0* = 0(0) translates the curve h o y,, by (o. However, 0* = F o ¢ o F-' by (2) since 46 E 1'0 c G,
and hence 0* translates the geodesic F(y,;) by w. Since the sets (F o y,.XR) and (h o y,.XR) have finite Hausdorff distance we conclude
from (3)of(1.4.4)that lim,.x(hoy,,xt)=lim,- xF(y,.t)=(F- y,Xoc)= F(x). This proves (3).
If x is any point of Pr, then choose 0 E 170 and v E SM such that y,.(cc) = x and 46 translates y,, by w. If a,1i E F is arbitrary, then 1i c 0 o 4-' translates 41 o y,, = yd,y(,,) by w and (+1 c y,.Xx) = 4i(x). From (1) and (3)
we obtain F(4x) = lim, -,(h o 4 o y,.xt) = lim, -r(4)* 0 h o y,,Xt) _ i/i*(Fx). We have proved the following.
(4) If x E Pr and 4) E F, then F(qix) = 4) *(Fx). With respect to the cone topology on M(me) the maps F, 4', and 0 * are homeomorphisms of M(me) and Pr is dense in M(cc). From (4) it follows that F c 4 C F-' = 0* on M(oo) for all iIie F. By (5) of (1.9.4) and the fact that M has no Euclidean factor we now obtain the following.
(5) F°4,0F-' =4)* on M for all 0E F. If we define f: M - M* by f(Trp) = a*(Fp) for all p E M, where ir: M -, M and lr*: M* - M* denote the universal Riemannian covers of M and M*, then by (5), f is a well-defined isometry that induces the isomorphism 0: r - t*. Hence the first version of the Mostow rigidity theorem implies the second. Before proving that the second version implies the first we need some further remarks. Let G be a connected semisimple group with trivial center and no compact normal subgroups except the identity. If K is a maximal compact subgroup of G, then the coset space M = G/K admits a G-invariant metric making M a symmetric space of noncompact type by theorem 1.1 of [Hell, pp. 214-215]. If an element g e G acts on the left as the identity transformation on G/K, then g lies in the intersection of all G-conjugates of K, which is a compact normal subgroup K* of G. By hypothesis K* = (1), and hence G acts effectively on G/K. By theorem 4.1 of [Hell, p. 207] it follows that G = 10(M). We are ready to prove that the second version of the rigidity theorem
implies the first. Let G, G*, r, and r* satisfy the hypotheses of the first
version, and let 0: F - I'* be an isomorphism. Let K and K* be maximal compact subgroups of G and G*, and let M = G/K and M* = G*/K* denote the corresponding symmetric spaces of noncompact type. The spaces M = M/I' and M* = M*/F* are smooth compact manifolds since r and r* have no elements of finite order. By
hypothesis the fundamental groups IF and r* of M and M* are isomorphic, and hence by the second version of the theorem there
Mostow Rigidity Theorem
381
exists, after rescaling the metric of M, an isometry F: M M* such that B(y) = BF(y) = F c y c F-' for all y r= F, where the isomorphism OF: G --* G* given by 9F(g) = F o g o F-' for g E G is analytic. There-
fore the second version of the rigidity theorem implies the first. We have also used the fact that if the metric of M or M* is rescaled by positive constants on de Rham factors, then the groups G =10(M) and G* =1o(M*) remain unchanged. 8.2.
Outline of the proof
We sketch a proof of (8.1.1). Let G and G* be as in the statement (8.1.1), and let K and K* be maximal compact subgroups of G and G*.
Let M and M* denote the coset spaces G/K and G*/K* equipped with left-invariant metrics making them into symmetric spaces of noncompact type. If k > 2 is the real rank of G = !0(M), then k is also the rank of M as a symmetric space. 8.2.1.
Equality of ranks
We observe first that M and M* have the same rank k z 2 under the hypothesis of (8.1.1). A k-flat is r -periodic if F/rF is compact, where rr = (y c- r: y(F) = F). If F is r-periodic, then by the Bieberbach theorem rF contains a free abelian subgroup of rank k. There exist
r-periodic flats F in M, and in fact, any flat F of M is a limit of a sequence (Fn) of r-periodic flats by lemma 8.3 of [Mos2) or (4.7.8). Conversely, if A is a free abelian subgroup of r with rank r >_ 2, then M admits an r-flat F such that the elements of A leave F invariant and F'/A is compact; see [GW], [LY], or (6.5.3). Since M admits no r-flats with rank r > k = rank M we obtain the following characterization of rank M: let IF c G = I0(M) be a discrete cocompact subgroup and let k = rank M. Then r admits a free abelian subgroup of rank k but no free abelian subgroup of rank r > k. Hence the isomorphism e: r -> r* ensures that M and M* have the same rank k z 2. 8.2.2.
Construction of equivariant pseudoisometries
The next step in the proof of (8.1.1) is the construction of a (k, b) pseudoisometry 0: M - M* that is equivariant with respect to 0 on r (cf. §9 of [Mos2]). This means that there exist positive constants k and b and a map ca: M -> M* such that
(1) d(4)p, ¢q):5 kd(p, q) for all p, q E M, (2) (1/k)d(p, q) :!g d(c)p, 4)q) for all p, q e M with d(p, q)
b, and
(3) O(yp) = 6(y)4)(p) for all p,q EM, for all yE r._ Similarly one constructs a (k*, b*) pseudoisometry 4)*: M* - M that is equivariant with respect to 0'. Moreover, for some positive constant A
Geometry of Nonpositively Curved Manifolds
382
one has the following further property_
(4) d(¢4*p*, p*) 5A for all p* E M*, and d(4) *4)p, p) 5A for all p E M.
For the convenience of the reader we outline the construction of Further details may be found in section 04f -1M* and 0*: M* 9 of [Mos2], section 2 of [E9], and [E13, pp. 53-541. We construct a r-invariant triangulation of M from a triangulation of M = M/r, and we then define 0 first on the vertices so that 0 is B-equivariant. Extend ¢ inductively over the m-skeleton of the triangulation, 1:5 m 5 dim M, so that ¢ is 9-equivariant at each stage. Now define *: M* -M in a similar fashion from a r*-invariant triangulation of M*. 8.2.3.
B{jective correspondence of k flats
Let M = G/K and M* = G*/K* be as above. Let r c G and r* = 8(r) c G* be isomorphic cocompact lattices, and let 0: M -+ M* be the B-equivariant (k, b) pseudoisometry described above. Then there exists a positive constant R with the following property: for each k-flat F in M there exists a unique k-flat F* in M* such that
HAW), F*) 5R,
(1)
where Hd denotes Hausdorff distance (cf. lemma 14.1 of [Mos2]). Mostow proves this first for r-periodic flats F (i.e., F/rF is compact, where rF = (y E r: y(F) = F)) and then uses the density of r-periodic flats in the space .9 of all k-flats in M. If F is a r-periodic flat in M, then it is easy to show that Hd((A(F), F*) 5 R for some k-flat F* in M*
and some positive constant R; simply let F* be the k-flat in M left invariant by O(FF). However, it requires more work to show that one may choose R to be independent of the r-periodic flat F. Similarly, let 0*: M* -* M be the 6-'-equivariant (k*, b*) pseudoisometry described earlier. Then there exists a positive constant R* such that for every k-flat F* in M* there exists a unique k-flat F in M such that (2)
Hd(4)*(F*), F) 5 R*.
By property (4) of (8.2.2) and (8.4.6) below it follows that the correspondences (1) and (2) of k-flats in M and M* are inverses. 8.2.4.
Induced mappings on the spaces of pointed k- lats
A pointed k-flat in M is a pair (p, F), where F is a k-flat in M and p is a point of F. Let .9.F and SAY' denote the spaces of pointed k-flats in M and M*. The space .9'.m admits a natural topology such that a
Mostow Rigidity Theorem
sequence {(p,,,
383
converges to (p, F) in A9?- if and only if
Hd(B(p,,, R') n F, B(p, R') n F) - 0 as n -> x for every positive constant R'. Here B(q, R) denotes the open ball in M with center q and radius R. The pseudoisometry 0: M -- M* induces a map 4i0: _40_9r09' as follows: dbo(p, F) =
(p*,
F*),
where
(a) F* is the unique k-flat in M* such that Hd(4(F), F*) s R, where R is the positive constant of (8.2.3), and (b) p* is the unique point on F* closest to d(p). We show that the map 4,,: is continuous and 0a 0 y = 6(y)° 0o for all yE F. 8.2.5. Induced mappings on the spaces of splices
Let Y and ..9'' denote the spaces of splices at infinity for M and M*
as described in (3.7). Let S be a splice in 5°, and let Fl and F2 be k-flats in M such that S = F,(x) n F20). Let F, and F2 be the k-flats in M * that correspond to F, and F2 as in (8.2.3). We define
i(S) =F, (x) nFZ (x). We show that i(S) does not depend on the choice of k-flats F, and F2. We also show that the function ¢: is strongly order preserving in the sense of (3.7.8). By (3.7.7) the irreducible splices of $° are those of the form F(x),
x E ,*x). By (3.6.26) the irreducible splices that are maximal with respect to set inclusion are those of the form F(x), x (-= R(oo), where R(x) denotes the regular points of M(x). Since 4 9 -* '* is strongly order preserving it follows from (3.7.10) and the proof of (3.7.9) that 4 maps maximal irreducible splices in .' to maximal irreducible splices in
.S*. In particular, if x is any point in Moo), then ('(x)) = W (x*) for some point x* E R*(x). 8.2.6. Induced mappings on Furstenberg boundaries
Let FM(x) and FM *(x) denote the Furstenberg boundaries of h%I and M* as defined in (3.8), and let T: R(x) - FM(x) and T*: R*(x) --. FM *(x) denote the continuous, open surjective maps defined following (3.8.5). Let I' c G = Io(M) and r* = 0(F) c G* = lo(M*) be isomorphic cocompact lattices, and let 04f --> M* and*: M* -M be the corresponding "inverse" 0 and 0-' equivariant maps of (8.2.2). By (8.2.5) the induced maps 4: S°- SP' are strongly order preserving. Hence by induces a Tits isomorehism By (3.7.9) the map (3.9.6), induces a bijection a: FM(x) --> FM*(x).
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Geometry of Nonpositively Curved Manifolds
We may summarize as follows: given a point x E R(oo) c M(me) choose
a point x* E R*(oo) c M*(oo) such that 4)(F(x) _ W(x*) (see (8.2.5)). Then the bijection a: FM(-) - FM*(co) is given by
a(Tx) = T*(x*).
Using the fact from (8.2.4) that 0.: .- 5r* is continuous we show that a is continuous. Hence a is a homeomorphism since both FM(°) and FM*(cc) are compact Hausdorff spaces. Moreover, a is O-equivariant since 4): M --p M* is O-equivariant.
8.2.7.
Completion of the proof
The proof falls naturally into two cases. CASE 1. M is irreducible with rank k >_ 2. CASE 2. M is reducible.
In case 1 the result follows immediately from (8.2.6), (3.9.8), and (3.9.11). The work of Tits on buildings is essential for this case; see corollary 16.2 of [Mos2] or (3.9.7). In case 2 the result follows immediately from (8.2.6) and (3.11.7). This case is actually the easier of the two since the theory of Tits buildings is not needed.
Convergence of totally geodesic submanifolds Before filling in the details of the outline in (8.2) we need some
8.3.
preliminary sections on the convergence of totally geodesic submanifolds and the space of pointed k-flats. The main result of this section is the following. PROposmoN. Let M be an arbitrary complete, simply connected manifold
with sectional curvature K _,! 0. Let k z 2 be an integer. Let {Fn} be a sequence of k-dimensional complete, totally geodesic submanifolds of M, and let (pa) c M be a bounded sequence of points in M such that p E F,, for every n. Then there exists a k-dimensional complete, totally geodesic submanifold F of M, a point p E F, and a subsequence (p,,, ) of (p,,) such that p,,, -' p as j oo and for every positive number R, Hd(B(R, p) rl F, B(R, p,,) r1 0 as j -> oc. If the submanifolds {Fn} are all flat, then F is flat. PROOF. Let {e (' ), ... , eki>) be an orthonormal basis of TP, F,. Passing to a
subsequence if necessary we let p1-> p and e;>> -+ e; E TP M for 1:5 i < k. The set (e,, ... , ek)_ is orthonormal by continuity. We set 1T=
span{e,,... , ek} c TP M and F = expP(ir ). We will show that p and F satisfy the assertions of the proposition.
Mostow Rigidity Theorem
385
Note that F is a complete, k-dimensional submanifold of M since expp: TPM -> M is a diffeomorphism. Define diffeomorphisms f1: F - Fj as follows. First, define linear isometries Ij: T,,F -> TP,Fj by k
k
If
a; e;
_
a; e;! .
Then define fj by requiring that fj = expp o Ii o expy' restricted to F. Note that fj(B(R, p) n F) = B(R, pj) n F for all R > 0. STEP 1. Given positive numbers e and R we can find a positive integer jo such that
d(q,fq)<e
forj>_joandgEB(R,p)nF.
Let d* be the Riemannian metric in SM induced from the Riemannian metric d in M. Given s > 0 and R > 0 choose S > 0 such that if v E SPM and w E SM with d*(v,w) < S, then d(exp(tv),exp(tw)) < e for all t E [0, R]. Choose jo ca 7L+ such that if j >_ jo, then d*(v, Ij(v)) < S for all v E SP F. This can be done since the linear isometry Ij: Tp F -> TP F
"converges to the identity" as j --> - from its definition.
Now let q E B(R, p) n F be given and write q = expp(av), where v E SPF and a = d(p, q) < R. If vj = 1j(v) E Sp,FI, then by definition fj(q) = expP(avj). By hypothesis, d*(v, v1) < S for j >_ jo, and hence d(q, f (q)) < e by the choice of S. STEP 2. For every R > 0 we have Hd (B(R, pj) n F, B(R, p) n F) -> 0 as
It follows from the definition of fj: F - F that fj(B(R, p) n F) _ B(R, pj) n Fj and fj(p) = p1. Given e > 0 and R > 0 choose jo E Z+ as in step 1. Then Hd(B(R, pj) n F, B(R, p) n F) < e for j >_ jo by the conclusion of step 1.
STEP 3. Let q E F be given, and let (qj) c M be a sequence such that qj E Fj for every j and q, -> q as j - oc. Let vj be a unit vector in Tg Fj for every j. Then any accumulation point v of (vj) is a unit vector in F.
We let vi --' v e Sq M, passing to a subsequence if necessary. To prove that v E SqF it suffices to prove that expq(ty) E F for all t E [0, 1]. Let t E [0, 1] be given. For large R > 0 we see that expg1(tvj) E B(R, p) n F. for all j since F. is complete and totally geodesic. By step
2 we can find qj(t) E B(R, p) n F such that d(expg(tvj), qj(t)) -0 as j -* cc. However,
expg(ty) as j --+ 00 by the continuity of the
exponential map and the fact that tvj -> tv. Hence expq(ty) _ limj
._
qj(1) E F.
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Geometry of Nonpositively Curved Manifolds
STEP 4. Let q E F and let v e SqF be given. Let q. = f1(q) E F,. Then there exists a sequence (v,) c SM such that v, - v and v1 E Sq,F1 for every j.
Choose R > d(p, q). Since Hd(B(R, p) n F,, B(R, p) n F) - 0 by step 2 we can find q, = fj(q) E B(R, p1) n F with q, -> q. Let Q; J), ... , k i )) be an orthonormal basis of Tq, F . Passing to a subsequence let e,(1
f i E Sq M for 1 w.
STEP 5. F is a totally geodesic submanifold.
Let q E F and v E SqF be given. It suffices to prove that expq(ty) E F for all t r= R. By step 4 we can find q, = ff(q) E F and v, e Sq,F such that qj - q and v, -> v as j -' -. Fix t E R. We know that expq (tv,) E F for every j since F is complete and totally geodesic. By step 2 there exists q1(t) E F such that d(expq(tv,), q,(t)) -> 0 as j - +oo. Hence expq(ty) = lim1-. expq(tv1) = lim1 _; , q.(t) E F. STEP 6. If the submanifolds F are flat for all j, then F is flat. Let q E F and orthonormal vectors v, w E Sq F be given. By step 4 we can find q; E Fj and unit vectors U1, w1 E Sq, F such that v1 - v, w1 -10 w,
and qj - q. If ir, and it denote the 2-planes spanned by (v,, w1) and {v, w} respectively, then K(ir) = lime the curvature tensor. 8.4.
K(ay) = 0 by the continuity of
The space of pointed k-flats .9F
Let M denote a symmetric space of noncompact type and rank k >_ 2. A pointed k -fiat in M is a pair (p, F), where F is a k-flat in M and p is
a point of F. Let AF denote the space of pointed k-flats in M. We define a natural topology on .9'.m
8.4.1. DEFINmON. Given positive numbers E and R and an element (p, F) of AF we let
N.,R(p,F) = (q, G) E.lay':d(p,q) g. It suffices to show that (g(p), g(F)). Let An =g g, Let R > 0 be given. Then
Hd(B(R,g,,(P"))ng,,(F,),B(R,g(P))ng(F)) p") nFn},g( B(R, p) n F))
= Hd(µ"(B(R, p") n Fn), B(R, p) n F) (p, F) as n - oo. Hence (g(p), g(F)) as n - cc since' R > 0 was arbitrary.
We next show that admissible pseudoisometries 041 -, M* between symmetric spaces of noncompact type and the same rank k >_ 2 induce
continuous maps 0o: A -AF* on the spaces of pointed k-flats.
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389
8.4.5. PROPOSITION. Let M and M * be symmetric spaces of noncompact type with the same rank k >_ 2. Let 0: M -> M* be a (k, b) pseudoisometry
for some positive constants k and b (cf. (8.2.2)) that has the following additional property: there exists R > 0 such that for any k-flat F in M there exists a unique k -flat F* in M* with Hd(cb(F), F*) 5 R. Then 0 induces a
continuous map 4 :AF- *9'
*.
PROOF. We define ¢o as follows. Given (p, F) we let F* be the unique k-flat in M* such that Hd(4)(F), F*) :5 R. We then define
00(p, F) _ (p*, F*),
where p* = it .(4)(p)), the unique point on F* closest to 4)(p). In th remainder of the proof we use (q*, F*) to denote 4)a(q, F) for eve (q,F)(=-A7. We need the following result. 8.4.6. LEMMA. Let M be a symmetric space of noncompact type and rank
k z 2. If F, and F2 are k -flats in M such that Hd(F,, F2) is finite, then F, = F2.
REMARK. This result indicates why there is a unique k-flat F* in Y-9-* such that Hd(4)(F), F*) < oo, where F E.9.9:
PROOF OF LEMMA (8.4.6). The function f: M -> R given by f(q) _ d(q, F2) is convex by (1) of (1.6.6) and is bounded above on F, by hypothesis. Restricting f to each geodesic of F, we conclude that f has a constant value c >_ 0 on F, by (1.6.5). If c = 0 we are done, so we assume that c > 0 and obtain a contradiction. By the sandwich lemma 2.1 of [Ell] there exists an isomorphic, totally geodesic imbedding F: F, x [0, c ] -+ M such that F(p, 0) = p for all p E F, and F(F1 x (c)) _ F2. In particular, F(t) = F(F1 x (t)) is a k-flat in M parallel to F, for every t E [0, c]. Hence, if y is any geodesic in F,, then F(F, x [0, c]) c_
F(y), the union of all geodesics in M parallel to y. It follows that F, = F(F1 x {0}) is a proper subset of F(y). However, if y is a regular geodesic of F,, then F, = F(y) by (2.11.1) and the remark following (2.11.3). This contradiction shows that F, = F2 and completes the proof of the lemma.
We now prove that the function 0o: 9a.4"-+A9 * defined above is continuous. Let (p, F) E.9'.9 be given and let ((p,,, EAg be a sequence converging to (p, F). We must show that
We break the proof into two steps.
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Geometry of Nonpositively Curved Manifolds
STEP 1. If (p', F') is an accumulation point in 9a3* of the sequence ((P*, F,,*)), then F' = F*.
REMARK. Since 4)(pp) - 4)(p) and d(p*, 4i(pp)) 5 R for every n it follows from the main result of (8.3) that every subsequence of ((p*, F,,*))
has a further subsequence that converges.
PROOF OF STEP 1. To prove that F' = F* it suffices to prove that Hd(F', F*) < cc by (8.4.6). By the definition of F*, Hd(4)(F), F*) 5 R, and hence it suffices to prove that Hd(4'(F), F'):5 R. Let (pp**, F,*) - (p', F'), passing to a subsequence if necessary. Let R' > R be given arbitrarily. It suffices to show (a) 4)(F) c TR.F' and (b) F' c TR.(4)(F)). We prove (a). Let q e F be given. By (2) of (8.4.3) we can find a sequence (qp) c M such that q,, -. q as n - cc and qp E F. for every n. Hence 4)(q,,) --* 4)(q) as n -), oo. Since F,*) oo, where p E M is fixed, then it would follow by (2) of (8.2.2) that d(4)(p), q') - + 00, which is not the case. Passing to a subsequence we let (r,,) converge to a point r of M. Then r E F by (1) of (8.4.3), and hence d(q', Or) = limp Or,,) s
R* < R'. This shows that F' c
and completes the proof of
step 1.
STEP 2. If (p', F') is an accumulation point in .9'.`m of the sequence {(P*, F* )), then p' = p* = 'rF.(43P) PROOF OF STEP 2. Let ((p*, F* )} -- (p', F'), passing to a subsequence if
necessary. Then F' = F* by step 1, and hence p* = 1rF. (4,pp) -' irF.(4)p) = p* by (3) of (8.4.3). Hence p' = p*.
We now complete the proof of (8.4.5). Steps 1 and 2 show that if (p', F') is any cluster point of {(p*, F,*)}, then (p', F') = ((p*, F*)).
Mostow Rigidity Theorem
391
However, every subsequence of ((p,*,, F.*)) has a further convergent subsequence by the main result of (8.3). Therefore (p.,*, F*) --> (p*, F*). O
8.5.
Induced mappings on the space of splices
We are now ready to begin the proof of theorem (8.1.1), and we adopt the notation used in its statement. The symmetric spaces M = G/K and M* = G*/K* associated to G and G* have the same rank k z 2 as explained in (8.2.1). We construct a B-equivariant (k, b) pseudoisometry 0: M -- M* and a 0-'-equivariant (k*, b*) pseudoisometry ¢*: M* -* M as in (8.2.2). The maps .0 and ¢* lead to inverse bijective correspondences of k-flats as explained in (8.2.3) and induce -->9. as we saw in continuous maps 4o: 5a.F-60-9or * and 0*: (8.4). Next we explain how 0 and ¢* induce strongly order preserving where 5" and .So denote the * and 4)*:5 * splice maps spaces of all splices in M and M* (see (3.7)). The definition given here is different from but equivalent to the definition given in (8.2.5). It follows routinely from the definitions that 0o and 4) are equivariant with respect to 9 on r since 0: M -+ M* has this property. 8.5.1. The sets C(p, S)
Let M be a symmetric space of noncompact type and rank k >_ 2. Let S c M(x) be a splice; that is, S = F,(x) n F2(x) for k-fiats F, and F2 in M. DEFINITION. For each point p E M we let
C(p, S) = U yp"[0,oo). XES
REmwxxs. (1) If S = F,(x) n F2(0 for k-flats F, and F2 and if p is any point of F2, then C(p, S) equals the splice Ft np F2 defined by Mostow in [Mos2, p. 56). In this case, C(p, S) is a closed convex subset of M. (2) Let v be a regular unit vector of SM, and let x = yi,(x) E R(x). The set S = '(x) is an irreducible splice that is maximal with respect to set inclusion by (3.6.26) and the discussion in (3.7). If p is the footpoint of v in M, then C(p, S) is the closure in M of the Weyl chamber W(v) defined in (3.8).
(3) If S is an arbitrary irreducible splice, then S = '(x) for some x E M(x) by (3.7.7). Hence Gs = (g e G: g(S) = S)
G1, and it follows
that Gs acts transitively on M since Gx acts transitively on M by (2.17.1). We conclude that (a) Gs acts transitively on the set {C(p, S): p e M) and
(b) the set C(p, S) is a closed convex subset of M for each point
pEM.
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Geometry of Nonpositively Curved Manifolds
Assertion (a) follows from the transitivity of Gs in M, while (b) follows from (a) and the first remark.
The remarks above will not be needed in the construction of the function 4):.m-->92*. However, we shall need the next result. PROPOSITION. Let S c M(O0) be a splice, and let p and q be any points of
M. Then Hd(C(p, S), C(q, S) < d(p, q).
PROOF. Let x be any point of S and let R > d(p, q). By (4) of (1.6.6) the function f(t) = d(yp,t, ygxt) is convex and f(t) is bounded above for t >_ 0 since ypx and yqx are asymptotic geodesics. Hence f(t) is nonin-
creasing and f(t) 5 f(0) = d(p, q) < R for
all t Z 0. In particular, ypx[0, oo) c TR(ygs[O, 00)) c TR(C(q, S)) since ygx[0, x0) c C(q, S). It follows
that C(p, S) c TR(C(q, S)) since x (=- S was arbitrary. By symmetry we
see that C(q, S) c TR(C(p, S)), and we conclude that Hd(C(p, S), C(q, S)) S R. 8.5.2.
The mapping
For an unbounded set A g M we define Am) = Z n M(00), where A
denotes the closure of A in M = M u M(c) with respect to the cone topology. The set A(00) is obviously closed in MOO and represents the set of accumulation points of A in M(00). We have already seen this
construction in the case that A = r(p), the orbit of a point p (-=M under a group r c1(M). In this case, the set A(00) is the limit set L(1') (cf. (1.9.5)).
Now let M and R* be symmetric spaces of noncompact type with
the same rank k z 2, and let 0: M -* M* be a O-equivariant (k, b) pseudoisometry as in the discussion at the beginning of (8.5). DEFINITION. Given a splice S in M(me) we define
i(S) = {4)(C(p,S)))(c0) cM*(00) for any point p of M.
The set i(S) is nonempty and independent of the point p chosen in M since 0 is a pseudoisometry and any two sets (C(p;, S): i = 1, 2) have finite Hausdorff distance by the proposition in (8.5.1). However, it is not obvious from the definition that i(S) is a splice in M*(00). The next result shows that 4)(S) is a splice and also yields the equivalence of the definition of 4)(S) given here with the definition given in (8.2.5). PROPOSITION. Let F, and F2 be k -flats in M such that S = F,(co) n F2(00)
is nonempty. Let F* and F2 be those unique k -flats in M such that Hd(4)(F), F,*) < oo for i = 1, 2 (cf. (8.2.3) and (8.4.6)). Then ¢(S) =
Mostow Rigidity Theorem
393
F; (oo) n FF (o0. In particular, (S) is a splice in M*(o). Moreover,
0oy=6(y)-,0 for all yEl'. LEMMA. Let F be a k -flat in M, and let p be a point of F. Let (po) and {qo} be divergent sequences in F such that d(p,,, qn)/d(p, po) --> 0 and 0 for every Then 4p. (Opo, d(po, qo)/d(p, qo) - 0 as n point p* E M*.
We postpone the proof of the lemma and complete the proof of the proposition. It follows easily from the definitions that 4) is equivariant with respect to 0 on r since 0: M --p M * has this property. We show 4)(S) be given. Fix a point next that (S) c Fi (oo) n FZ (oo). Let x p E F2. By the definition of 4)(S) we can find a divergent sequence ( pn) c C(p, S) c FF such that 4)(p,,) -* x* as n - co. By (8.2.3) there exists a positive number R such that Hd(4)(F,), F;*) < R for i = 1, 2.
Since p, e F2 for every n we can find points q* E F2* such that d(4)(pn), q*) < R for every n. Hence x* = limo - x q* E Fz (cc). If we F,* (00). start with a point p E F,, then a similar argument shows that x Hence ¢(S) c Fl (co) n F2 (oo) since x* E ¢(S) was arbitrary. Now let x* be any point of F*(c) n F2 (0. To prove the proposition
we need to show that x* E ¢(S). Fix a point p E F2. We construct a point x e S and a divergent sequence (qn) c yps[0, oo) c C(p, S) such that 4)(q,,) _i x* as n -o oo. This will show that x* E {4)(C(p, S)))(c) _ 4(S) and will complete the proof. We construct a point x c- S. Fix a point p* E F2*. Since yp.x.[O, 00) c FF
and Hd(4)(F2), F2*) < R we can find a divergent sequence (po) c F2 such that (a) d(4)(po), yp.x.(n)) < R for every n.
Let x E F2(oo) be a cluster point of (po), and let (pn) - x, passing to a
subsequence if necessary. We show that x lies in S. There exists a positive number r such that yp.x.[O, x) c T,.F* since x* E Fr (x); any number r> d(p*, Fr) will suffice by a standard convexity argument. Since Hd(4)(F,), F*):!-. R we can find a sequence (r,,) c F, such that (b)
yp.x.(n)) < R, = r + R for every n.
It follows from (a) and (b) that d(4)(rr ), ()(po )) 5 R2 = R, + R, and hence by (4) of (8.2.2) we conclude that d(p,,, rn) < R3 for all n and
-
some positive constant R3. Hence
x = lim pn = lim r E F,(oo) f R-.a
S
since { pn} c F2 and (r} c F,. Fix a point p E F2, and let qn be that point on ypx[O, oo) such that d(p, pn) = d(p, qn) --+ + oo as n -> oo. By hypothesis, { pn} - x, and by
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Geometry of Nonpositively Curved Manifolds
The points p, construction {qn} -* x. Hence 4p (p,,, qn) -* 0 as n p,,, and qn all lie in the flat F2, and by applying the Euclidean law of
cosines in F2 to the triangle 0 n with these points as vertices we conclude that d(pn, 9n)/d(p, Pn) = d(pn, qn)/d(p, qn) -* 0 as n -' -. By the lemma above we conclude that 4.p. (4)pn, 4qn) -* 0 for every point p* of M*. Hence x* = limn_x 4)(pn) = limn,. 4)(q,,) by (a) above. Since x E S and qn E ypJ0, oo) c C(p, S) for every n it follows that x* E 4)(S). Therefore F*(oc) n FZ (oo) c 4)(S) since the point x* was arbitrary. The proof of the proposition is complete except for the proof of the lemma.
We prove the lemma. Let F be a k-flat, and let p, p, and q,, be points of F as in the lemma. Let 9n be the vertex angle at ¢(p) of the geodesic triangle in M with vertices 4)(p), 0(pn), and 4)(q,,). Let a = d(4)p, 4p,), bn = d(4p, 4)q,,), and c,, = d(4p,,, 4)q,,). We first consider the case that c,, s R' for all n and some positive constant R'. In this case, it follows that zip. (4)pn, ¢qn) --"0 for every
point p* of M* by (3) of (1.4.4). It suffices therefore to consider the case that cn -* +cc as n -' o. Since 0: M -41* is a (k, b) pseudoisometry it follows from (8.2.2) that
(*)
cn/an Sk2d(p,q,,)/d(p,p,,), 5 k2 d(pn, qn)/d(p, q,,),
for sufficiently large n. Hence cn/an -+ 0 and cn/bn - 0 as n -+
by
the hypotheses of the lemma.
By the law of cosines, (1) of (1.4.4), we have cR z a,2, + b. 2anbn cos On or equivalently cos 0 Z (an + By 0 as n -+ oo and (a, + b,?)/2anbn >: 1 for every (*) above, n. It follows that cos 0,, -> 1 and hence (p)(41(pn), 4)(qn)) = On - 0 as n -> oc. This completes the proof of the lemma. 0
Next we show that the function 4)is strongly order preserving in the sense of (3.7.9). 8.5.3. PROPOSITION. Let S be a splice in M(oo) such that S = U"'_ 1 Si, where each Si is a splice in M(oo). Then 4)(S) = U" 1 (S,).
PROOF. Fix p E 14f. Clearly, C(p, S,) c C(p, S) for all i, and hence ¢(S,) c 4)(S) for all i by the definition of 0. It suffices to prove that (S) c UN 4)(S,). Let x* E 4)(S) be given. Let F1 and F2 be k-flats in 1
M such that S = F1(oo) n F2(o), and let p be a point of F2. Let { pn} c C(p, S) c F2 be a divergent sequence such that 4)(p,,) -+x*. Passing to a subsequence let pn -> x E S. Choose an integer i such that x e S,. Let qn be the point on ypx[0, co) such that d(p, pn) = d(p, for
Mostow Rigidity Theorem
395
every n. Since (p,,) and follows that 4P (p,,,
are sequences in F2 that converge to x it 0 as n -- -. By the Euclidean law of cosines in F2 applied to the triangles 0 with vertices p, p, and q it follows that d(p,,, q.)/d(p, q.)/d(p, qrt) , 0 as n -> x. Hence x* = by the lemma in (8.5.2) above. It follows that x* E 4(S) since c yp,,[0, x) c C(p, S,) for every n. This shows that O(S) c U" , O (S,) and completes the proof of proposition (8.5.3). ,1
8.5.4. LEMMA. The functions and *:Y* -.So induced by the pseudoisometries 0: M --+ M* and 0*: M* --* M are inverse functions.
PROOF. Let S 5 be an arbitrary splice in M(oo), and let F, and F2 be k-flats in M such that S = F,(°) n F2(x). By (8.2.3) and (8.4.6) there exist unique k-flats Fl and F2 in M* such that (a) Hd(O(F,), F,*) < - for i = 1, 2. By (l) and (4) of (8.2.2) it follows that
(b) Hd(4 *(F,* ), F;) < - for i = 1, 2.
If S* = i(S) C=Y*, then S* = F; (x) n F2*(-) by (a) and the main result of (8.5.2). Hence ¢*(S*) = F,(x) n F2(oo) = S by (b) and another application of (8.5.2). We conclude that d*t(S) = *(S*) = S for all splices S in Y. By symmetry or a similar argument we find that ¢4i*(S*) = S* for all splices S* in 5°*.
The proof of (3.7.9) shows that cb:.'-5° * maps irreducible splices into irreducible splices. Hence the proof also shows that for each point
x ,E M(oc) there exists a point x* E M*(c) such that (F(x)) = F(x*) since the irreducible splices of M(x) are those of the form (0'(x), x E M(oc)} by (3.7.7). If x lies in R(c), the set of regular points of M(oo), then the irreducible splice 9'(x) is maximal with respect to set inclusion is an irreducible splice that is maximal with by (3.6.26). Hence respect to set inclusion by (3.7.10). Another application of (3.6.26) shows that ¢('(x)) = W(x*), where x* ,E R*(oo). The discussion above has proved the following. 8.5.5. PROPOSITION. For each point x E M(oo) there exists a point x* E M *(c) such that
('(x)) _W(x*). If x lies in R(x), then x* lies in R*(c).
REMARK. If '(x;) _'(x2) for points xT , x2 E M*(oo), then and xZ <x; by (3.6.26) and hence '(xr) W(xZ
xi xi
Geometry of Nonpositively Curved Manifolds
396
8.6.
Induced mappings on Furstenberg boundaries
We now consider the bijection_:_q70 a : FM(x) --+ FM *(co) induced as in which itself is induced (3.9.6) by the Tits isomorphism as in (3.7.9) by the splice map c : Y -Y *. Our main result is that a is a homeomorphism. Before proving this we establish the following. 8.6.1. LEMMA. Let T: R(te) -> FM(-) and T*: R*(oo) -* FM*(oo) be the continuous, open, surjective mappings defined following (3.8.5). Given a point x e Woo) letx* E R*(oo) be a point such that cb(9(x)) = W(x*) as in (8.5.5). Then
a(Tx) = T*(x*). PROOF. Let x E A'(co) be given. By (8.5.5) there exists x* E M*(o) such
that 4,('(x)) _ 9(x*). By the proof of (3.7.9) the induced Tits isomorphism cb:YM _, 9M* is given by 4,([x]) = [x*], where [ ] denotes the projection of M(c) onto .9M and also the projection of M*(oo) onto 9M*; see remark (1) following (3.6.24). Let .9,M and 170 M* denote the
points in .9M and 5'M* that are maximal with respect to the partial ordering in .9M and Y74 * as defined in the same remark. By (3.9.4) FM(oo) and p*:9 M* - Moo) such there exist bijections that p([x]) = T(x) and p*([x*]) = T*(x*) for arbitrary points x r= R(te) and x* E R*(c). By (3.9.6), a satisfies the relation a o p = p* c 4i. Hence
from _the discussion above we obtain a(Tx) _ (a c pX[xl) = (p* - ¢X[x]) = p*([x*]) = T*(x*), where given x c R(x) we choose x* E R*(oo) so that 4,(W(x)) = W(x*).
We now arrive at the main result of this section. 8.6.2. PROPOSnoN. The bijection a: FM(co) -> FM*(oo) induced by 4, and 4, is a homeomorphism. Moreover, a o r(y) = T *(9y) a a for all y E 17, where T and r * denote the actions of G and G* on FM(oo) and FM *(co).
PROOF. By (8.6.1) a is equivariant with respect to 0 on IF, as stated above, since is equivariant with respect to 0 on F. Since FM(c) and FM*(-) are compact Hausdorff spaces it suffices to prove that a is continuous. We will use the continuity of to achieve this. Let T: R(x) - FM(oo) and T*: R*(co) -* FM*(oc) denote the natural
projections. Let T(x) be a point in FM(co), where x E R(-), and let c R(te) be a sequence such that T(x) n We must show that a(Tx) n --), co. Since G =10(M) acts transitively on FM(m) by (3.8.8) we can find a sequence c G such that g -' 1 as n - oo and g (Tx) = for
Mostow Rigidity Theorem
397
every n. Without loss of generality we may assume that x for every n since T(x) A& x). By (8.5.5) we may choose points x*, X* E R*(c) such that
(1) O('(x)) _ F(x*), ('(x* )) for every n. We may further assume in (1) that x* and x* are the centers of gravity of the sets '(x*) and W(x*) (cf. (3.6.35)). We now consider the irreducible splice S = F(x), where x (=- Moo) is as above, and write S = F(oc) n F'(oo) for suitable k-flats F and F' in
M. If S =
F
F
then S = F(co) n
and F*' be those k-flats in M* that lie at finite Hausdorff distance from O(F), 4(F), and ¢(F') respectively; as usual 0: M -. M* is our 9-equivariant (k, b) pseudoisometry. Let S* = 4)(S) and S* = 4)(S). Then from (1), the discussion above and the main result of (8.5.2) we obtain (2) S F(oo) n F(cc) for all n, S, = W(x*) = F,* (r) n F, "(oo) for all n, S = F(x) = F(oo) n F'(O, S* = W(x*) = F*(cc) n F*'(0. of (8.4.5). Fix We now introduce the continuous map 00:
points p E F and p' E F', and let p =
g (p') for every -s (p, F) and (p;,, (p', F') as n - oo in the space of pointed k-flats S'3 since g -- I as n -+ co. The map 4):M -* M* by (8.2.4) or (8.4.5), and this map is induces a map 4) n. Then (p,,,
continuous by (8.4.5). Choose points p* E F*, p,*,' C- F*', p* E F*, and
p*' E F*' such that
(p*, F*) for all n, (3) 'o(P,,, 00(p;,F:)=(p*',F,*') for all n,
40(p,F)=(p*,F*), 00(p" F') = (p*', F*'). By the continuity of 0o we obtain
(4) (p*,F,*)->(p*,F*) as n -,oo, (p*', F,*') --> (p *', F*') as n --> cc.
From (2) and (8.6.1) we know that a(Tx) = T*(x*) and T *(x* ). To prove the continuity of a it suffices to show that x* -> x* in R*(co) as n -, co. Let v* = yp.*x'.(0), which lies in T P. F* by (2). Let v* be
any unit vector that is a cluster point of the sequence of unit vectors and let z* = y,,.(oo). To show that x* -sx* as n -4 cc it suffices to show that z* =x* for any choice of cluster point v*. If v* is a cluster point of (v*), then since (p*, F,*) - (p*, F*) we conclude that v* E TP.F* by step 3 in the proof of the main result of (8.3). Let v* -> v * by passing to a subsequence. Then x* = y,.: (oo) y,,.(oo) = z*. By the hypothesis following (1) each point x* is the center
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of gravity of W(x* ). The centers of gravity of {W(x): x E R(te)} are G-invariant by (3.6.35), and each G-orbit in M(me) is compact by (3) of (1.13.14). Moreover, the centers of gravity of ('(x): x (=- R(te)) lie on a single G-orbit in R(te) since G acts transitively on FM(oo) and T-'(Tx) = '(x) for every x e R(te) by (3.8.7). Hence z* = limn -a x* lies in R(te), and z* is the center of gravity of W(z*). We note that z* = y,,.(w) lies in F*(x) since v* E TP.F*. Next we show that z* E F*'(0. Assuming this fact for the moment we complete the proof of (8.6.2). We note that z* E F*(c) n F*'(oo) = S* = '(x*) by (2) above. Hence z* sx* by (3.6.26). However, z* is maximal with
respect to the Tits partial ordering since z* lies in R*(c) by the previous paragraph. We conclude that (G,.)0 = (Gx.)0; that is, 9(z*) =
'(x*). It follows that x* =z* since both x* and z* are the centers of gravity of their respective Weyl chambers. This completes the proof of (8.6.2), assuming that z* e F*'(o).
We prove that z* E F*'(0. Let r > d(p*, p*') be given, where p* E F* and p*' E F*' are the points appearing in (3). To show that z * E F*'(c) it suffices to show that y,..[0,oc) c T,(F") since T,(F*'Xc) = F*'(oc) by (3) of (1.4.4). Choose r' with r > r' > d(p*, p*'). Then d(p*, AV) < r' for large n since p* -* p* and p*' -> p*' by (4). For each integer n let w,' E TP.. Fn*' be the unit vector such that (00) = y,,n (00) = x* .
Fix a positive number T, and set q*(T) = yw.(T) for each n. The function f (t) = d(y,.* t, y,,,^ t) is a nonincreasingn convex function for each n, and hence we obtain
(5a) d(y,.:(T),q*(T)) 0 as n -+ oo. In particular, d(y,,. (T ), F*') ..5 Ay,.-M, q*') < r' for large n by (5a) and the choice of q*'. Since y,,.(T) -- y,..(T) as n we conclude that d(y,,.(T), F*') s r', and hence
(5c) y,..(T) E Tr(F*')
since r' < r. We have proved that y,..[0, x) c T,(F*') since T > 0 was arbitrary. This shows that z* = y,,.(oo) E F*'(oc) and completes the proof of (8.6.2) as we indicated above.
9
Rigidity Theorems and Characterizations of Symmetric Spaces of Higher Rank
9.1.
Characterizations of symmetric spaces of higher rank Orbits of 1(M) in M(oo) 400 Orbits of the geodesic flow in SM 400
400
9.2.
Characterizations of symmetric spaces of arbitrary rank
400
Geometric splitting when 1(M) satisfies the duality condition 400 Reducibility of finite volume manifolds 403 Existence of irreducible lattices in products of simple groups 404 A lattice characterization of irreducible symmetric spaces 404 9.3.
Higher rank rigidity theorem Higher-rank rigidity for homogeneous spaces
9.4.
405 405
Applications of the higher rank rigidity theorem
408
Reducibility of finite volume manifolds, II 408 Density of periodic vectors 408 Gromov rigidity theorem 409 Uniqueness of higher rank locally symmetric metrics 410 Tits finiteness theorem 411 Topological rigidity 411
In this chapter we discuss various generalizations of the Mostow rigidity
theorem that lead to differential geometric characterizations of symmetric spaces of noncompact type and rank at least 2. See also section 8 of [EHS] for a survey without proofs of such results. We avoid altogether a discussion of rigidity theorems that generalize the rank-1 case of the Mostow rigidity theorem. The methods in this case are completely different, and many beautiful results have been obtained, some of which are described in section 8 of [EHS]. See also the discussion in the introduction. We begin with two characterizations of higher rank symmetric spaces
that are proved in [E16]. We then continue with several characteriza-
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400
tions of symmetric spaces of arbitrary rank from [Ell]. We conclude the chapter with a discussion of the higher rank rigidity theorem of Ballmann [Ba2] and Burns-Spatzier [BuS] together with a description of several rigidity theorems that follow as consequences. 9.1.
Characterizations of symmetric spaces of higher rank
The first result is theorem 4.1 of [E16]. 9.1.1. THEOREM. Let M be irreducible, and let F c 1(M) be a subgroup that satisfies the duality condition and leaves invariant a proper, closed subset X of M(oo). Then M is a symmetric space of noncompact type and rank at least 2.
PROOF. This is an immediate consequence of (5.4.4) and (5.4.5). The topology of M(x) is the cone topology. Next we present a special case of theorem 4.3 of [E16]. 9.1.2. THEOREM. Let M be a complete Riemannian manifold with sectional curvature K :!g 0 and finite volume, and assume that the universal cover M is irreducible. Then M is a symmetric space of noncompact type
and rank at least 2 if and only if the geodesic flow in the unit tangent bundle SM fails to have a dense orbit in SM.
PROOF. Let r c I(M) be the lattice such that M = M/f. Since
1'
satisfies the duality condition it follows from theorem 4.14 of [E3] that the geodesic flow in the unit tangent bundle SM has a dense orbit in SM if and only if F has a dense orbit in M(oo). Hence if the geodesic flow has no dense orbit in SM, then Al is symmetric of noncompact type and rank at least 2 by (9.1.1). Conversely, if M is symmetric of noncompact type and rank at least 2, then M(=) admits many proper, closed subsets X that are invariant under 1(M) and in particular under
F. For example, let X be the set of singular points of M(oc) (cf. (2.21.7)).
9.2.
Characterizations of symmetric spaces of arbitrary rank
We present some results that can either be found in [Ell] or follow from the work of [Ell]. The next result is contained in proposition 4.2 of [El 1], and the proof was later simplified in theorem E of [E151. 9.2.1. THEOREM. Let M have no Euclidean de Rham factor, and let G c 1(M) be a closed, connected Lie subgroup of positive dimension whose normalizer IF in I(M) satisfies the duality condition. Then there exists a r-invariant Riemannian splitting M = MI X M, such that (1) M1 is a symmetric space of noncompact type and
(2) G=10(M,)x{Id}.
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We shall sketch a proof of this result below. First we derive a corollary, which is proposition 4.1 of [Ell]. 9.2.2. COROLLARY. Let M be any complete, simply connected Riemannian
manifold with sectional curvature K:5 0 whose isometry group I(M) satisfies the duality condition. Then there exist manifolds M0, M,, and M2, any two of which may have dimension zero, such that: (1) M is a Riemannian product M0 X M, X M2. (2) Mo is a Euclidean space. (3) M, is a symmetric space of noncompact type. (4) 1(M,) is discrete and satisfies the duality condition. PROOF OF COROLLARY (9.2.2). Write M as a Riemannian product M0 X
M*, where Mo is the Euclidean de Rham factor of M and M* is the Riemannian product of all other de Rham factors of M. Then I(M*) satisfies the duality condition by (1.9.22) since 1(M) =1(Mo) x I(M*). The desired result now follows by applying (9.2.1) to the group G = 10(M*) and using (1.9.22) to conclude (4). PROOF OF THEOREM (9.2.1). Full details may be found in the proof of
theorem E of [E15], but for the convenience of the reader we give an outline of the proof that conveys the main ideas. The proof rests on the following facts:
(i) G is a semisimple Lie group with trivial center and no compact normal subgroups except the identity. (ii) If L(G) = M(c), then M is a symmetric space of noncompact type and G =10(M ).
(iii) Let r denote the normalizer of G in 1(M), and let (4 E 1r: og =g(k for all g r= G}. Then F0 = G Zr(G) has finite index in F.
We sketch the proof of these facts. To begin the proof of (i) one observes that if G is not semisimple, then it contains a nontrivial abelian subgroup A that is invariant under all continuous automorphisms of G. Since r normalizes G it also normalizes A, and by (7.1.4), A must consist of Clifford translations. This contradicts the hypothesis that M has no Euclidean de Rham factor by (1.9.4). If Z denotes the center of G, then Z is normalized by r, and (7.1.4) implies again that Z consists of Clifford translations. Hence Z = (Id). The semisimple Lie algebra g is the direct sum of its finitely many simple ideals, which are
permuted by Ad(r). Let r* be a finite index subgroup of r such that Ad(F*) leaves invariant all simple ideals of g. If G admitted a compact normal subgroup N, then N would be normalized by I'* and Fix(N), the set of points in M fixed by N, would be invariant under r*. The set Fix(N) is closed and convex in M and is nonempty by the Cartan fixed
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point theorem. Since r* satisfies the duality condition it follows that Fix(N) = M, which implies that N = (Id). To prove (ii) one considers a maximal compact subgroup K of G and a point p E M fixed by K. The orbit G(p) in M is diffeomorphic to the coset space G/K, and hence G(p) with the induced Riemannian metric is a symmetric space of noncompact type by (i). Using the hypothesis that L(G) = M(oo) an ad hoc geometric argument now shows that G(p) = M, which proves (ii). To prove (iii) one again fixes a maximal compact subgroup K of G. The maximal compact subgroups of G are conjugate by elements of G
since (i) holds; see (1.13.14). If ¢ E r is any element, then 4)Kc-' is a maximal compact subgroup of G and hence equals gKg-' for some g E G. This proves that
(iii-a) t = G-Nr(K), where Nr(K) = (4) E T: (kK4-' = K).
Let g and f denote the Lie algebras of G and K, and let Atit r(g) _ {46 E Aut(g ): ¢(f) c f}. If p is the orthogonal complement in g of the subalgebra f relative to the nondegenerate Killing form B, then we recall from (2.6.4) that one may define a positive definite inner product
4) on g by setting 4) = B on p, (P _ -B on f, and making p and f orthogonal relative to 4). It is easy to show from (1) of (1.13.8) that the elements of Aut 1(g) preserve the inner product 4), and hence Aut t(g ) is a closed subgroup of the orthogonal group O(g, (b). We conclude that (iii-b) Aut }(g) is a compact subgroup of Aut(g ).
Now we define a homomorphism p: I F-+ Aut(g) by setting 46e'xO-' = e'P(m)X for all 4) E I', X E g, and t E R. If 0 e Nr(K) and X E f, then p(4)) e Aut,(g (g)and by restriction we obtain a homomorphism p: Nr(K) -' Aut t(g ). A routine argument using (iii-b) shows that No = Nr(K) f1 [G-Zr(G)] has finite index in Nr(K). Assertion (iii) now follows from (iii-a).
We use assertions (i)-(iii) to complete the proof of (9.2.1). If I'0 = G Zr(G), then FO has finite index in T by (iii) above, and hence t0 satisfies the duality condition by (1.9.21). Since M has no Euclidean de
Rham factor it follows from (7.1.6) that there exists a r0-invariant splitting M = M1 X M2 such that G c 1(M1) x (Id) and Zr(G) c (Id) x 1(M2). If denotes the projection homomorphism, then G = p,(F0) satisfies the duality condition by (1.9.22), and hence L(G) = M1(oo) by (1.9.16). By (ii) above we conclude that M, is symmetric of noncompact type and G =10(M1) x (Id). The leaves of the splitting M = M1 X M2 are left invariant by G and hence are permuted by the normalizer r of G. Therefore the splitting is r-invariant. 0
Characterizations of Symmetric Spaces of Higher Rank
403
We describe further useful consequences of (9.2.1). The next result is proposition 4.3 of [Ell).
9.2.3. PROPOSITION. Let M be irreducible, and let r c 1(M) be a subgroup that satisfies the duality condition. If r is not discrete, then M is symmetric of noncompact type and Go = 10(M), where G = r, the closure
of r in I(M). PROOF. This follows immediately from (9.2.1) since Go has positive dimension if r is not discrete and r normalizes G.. 0
As an immediate consequence of (9.2.3) or (9.2.2) we obtain the following.
9.2.4. PROPOSITION. Let M be irreducible, and let 1(M) satisfy the duality
condition. Then either 1(M) is discrete or M is a symmetric space of noncompact type.
The next result is a restatement of proposition 4.5 of [Eli]. 9.2.5. PROPOSITION. Let M be a complete Riemannian manifold with sectional curvature K< 0 and finite volume. Assume that the universal cover M has no Euclidean de Rham factor and that M splits as a nontrivial Riemannian product. Then either M is symmetric of noncompact type or M admits a finite Riemannian cover M * that splits as a nontrivial Riemannian product M, X M2*.
X Mk be the de Rham decomposition of M into irreducible factors. We know that k z 2 by hypothesis. If IF cI(M) is the lattice such that m = M/r, then by (1.2.4) there exists a finite PROOF. Let Al = Al, x
index subgroup r* that preserves the de Rham splitting of M. Let p,: r* -4 1(M;) denote the projection homomorphisms for I s i 5 k. If p,(r*) is not a discrete subgroup of 1(M,) for some i, then M, is a symmetric space of noncompact type by (9.2.3).
Now assume that M is not a symmetric space of noncompact type. We show that M has a finite covering by a nontrivial Riemannian product M* = M* X M2*. By the previous paragraph p,(r*) is discrete for some integer i, and we may write M = M* X M; , where MZ = M. and M; is the Riemannian product of the remaining de Rham factors M,, j i. The group r* preserves this splitting, and hence we obtain projection homomorphisms p,*: r* -* 1(M,*) for i = 1, 2. Since p; (r*) = p,(r*) is a discrete subgroup of I(M;) it follows from (7.2.6) that p; (r*) is a discrete subgroup of I(M!) since M has no Euclidean de Rham factor.
Let r, = kernel(p2 ), rZ = kernel(p* ),- and r** = r; x ]r2*. By (1.9.36) it follows that I,* is a lattice in I(M,) for i = 1,2, and r is a lattice in I(M) with finite index in r*. If M* = M/r** and M,* =
Geometry of Nonpositively Curved Manifolds
404
M;/lj* for i = 1, 2, then M* is a finite Riemannian covering of M = M/F and M* is isometric to the nontrivial Riemannian product M; X M*.
D
It has been known for a long time that certain reducible symmetric spaces M admit irreducible quotient manifolds of finite volume. For example, if M is the Riemannian product of k >: 2 hyperbolic planes
with K = - 1, then there are classical procedures for constructing irreducible, finite volume quotient manifolds, both compact and noncompact. A sketch of one such procedure may be found in [Shim, p. 64]. In general, a symmetric space M of noncompact type always admits quotient manifolds of finite volume, both compact and noncompact, by a result of A. Borel [Bo2]; see also [R]. However, if M is reducible, then M does not admit irreducible quotient manifolds of finite volume unless the de Rham factors of M are compatible in an appropriate way. This compatibility of the factors is made precise in the next result due to F. Johnson [J], which may be regarded as a complement to the previous result. For the definition of irreducible see (1.2.1) and (3.11.4). 9.2.6. THEOREM. Let G = G, x x G be a product of noncompact simple Lie groups G;, and let g be the Lie algebra of G. Then the following conditions are equivalent: (1) G contains an irreducible lattice r. (2) G is isomorphic as a Lie group to HR, the identity component of the group of real points of a cl-simple algebraic group H.
(3) The complexification q ® C is isomorphic to a ® . . 0 a for some simple Lie algebra a over C (G is called isotypic of type a). Moreover, any isotypic group G contains both cocompact and noncocompact irreducible lattices.
Recall that if M is a symmetric space of noncompact type, then by (2.1.1) G =10(M) is a semisimple Lie group with trivial center and no compact normal subgroups except the identity. A lattice 1' in G is cocompact if the coset space G/I' is compact (or equivalently if the quotient space M/I' is compact since M = G/K, where K is compact). If M is a Riemannian product M, x ... x Mk, then G = G, x ... X Gk, where G =10(M) and G; =10(M;) for i < i 5 k. The next result also complements the previous discussion. 9.2.7. PROPOStm'toN. Let M be irreducible. Then M is a symmetric space of noncompact type if and only if M admits a compact quotient manifold and a noncompact quotient manifold of finite volume.
PROOF. If i?1 is a symmetric space of noncompact type, then it admits both compact and noncompact finite volume quotient manifolds by the discussion above. Conversely, suppose that there are lattices r, and 172
Characterizations of Symmetric Spaces of Higher Rank
405
in 1(A!) for an arbitrary space M such that M/r, is compact and M/r, is noncompact with finite volume. If 1(M) were discrete, then by (1.9.34) it would follow that r,, r2, and r* = i', n I'2 have finite index in
1(M). Hence M* =M/r* would be a finite cover of the compact manifold M/r, and also a finite cover of the noncompact manifold M/1`2, which is impossible. Therefore 1(M) is not discrete, and M is a symmetric space of noncompact type by (9.2.3).
Higher rank rigidity theorem We refer the reader to (1.12) for the definition of the rank of a
9.3.
nonpositively curved manifold. In this section we give a more complete discussion of the result stated in (1.12.7).
9.3.1. THEOREM. Let hi be irreducible and have rank k >_ 2. If 1(M) satisfies the duality condition, then M is a symmetric space of rank k. 9.3.2. COROLLARY. Let M be a complete manifold with sectional curvature K 5 0, rank k >_ 2, and finite volume. If M is irreducible, then M is a symmetric space of noncompact type and rank k.
The first result in this direction was (9.3.2), which was proved by W. Ballmann [Ba2] and K. Burns and R. Spatzier [BuS] under the additional hypothesis that M have a lower bound for the sectional curva-
tures. This hypothesis is automatically satisfied of course if M is compact. The term "higher rank rigidity theorem" usually refers to (9.3.2). The more general version (9.3.1) is proved in [EH].
If M has finite volume, then m = M/r, where r c 1(M) is a lattice. Hence 1(M) satisfies the duality condition since r does, and (9.3.2) follows immediately from (9.3.1).
Theorem (9.3.1) has been generalized to homogeneous spaces by J. Heber [Heb]. 9.3.3. THEOREM. Let M be an irreducible homogeneous space of rank k z 2. Then M is a symmetric space of noncompact type and rank k.
Although (9.3.1) and (9.3.3) have a superficial similarity in their statements, the proofs of these results are necessarily different. If M is a homogeneous space such that 1(M) satisfies the duality condition, then M is a symmetric space by (1.9.19). Hence one cannot use (9.3.1) to prove (9.3.3).
For the convenience of the reader we sketch the proof of (9.3.1) as given in [EH] and simplify the proof of one part of it, namely, theorem 2.6 of [BBE], which is (1.12.11) in these notes. We begin by recalling some definitions and information from (1.12). For v e SM we define r(v) to be the dimension of the space of parallel
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Jacobi vector fields along the geodesic y and r(M) = inf{r(v): v E SM).
If r(v) = r(M) then v is called a regular vector, and we let .9/ denote the set of all regular vectors. The set 9P is nonempty and open in SM. For v E SM we let F(y,,) denote the union of all geodesics in M that are parallel to y,,. An r-flat in M is a complete, flat, imbedded totally geodesic r-dimensional submanifold of M. The first step in the proof of (9.3.1) is (1.12.11), which we restate here in slightly modified form. For our purposes it suffices to consider the case in which M is irreducible.
(1) PRoPOSmoN. Let M be irreducible, and let I(M) satisfy the duality condition. Then: (a) The set R of regular vectors is dense in SM.
(b) If u E9P, then F(y,,) is a k-flat, where k = r(M). (c) If v re SM is arbitrary, then the geodesic y,, is contained in at least one k -fiat and exactly one k -flat if v is regular.
PROOF. For each positive integer m let R. = {v c- SR: r(v) = m and r(w) = m for all w in some neighborhood of v in SM). A moment's thought shows that 5P* = U,,, m is a dense open subset of SM and Mk =,9P. It is proved in proposition 2.3 of [BBE] that F(y,.) is an m-flat
for every v Em and every positive integer m. We prove that .9/ =.9P*, which will prove (a) and (b) above. Assertion (c) will then follow from (a) and (b) and the observation that if F is any k-flat that contains a geodesic y,,, then F c F(y,.). To prove that ..9P =6P* it suffices to prove that r(v) = r(w) for any two vectors v, w E,9/*. Let vectors v, w E.9P* be given, and let x = y,,(oo).
If the orbit 10i Xx) is not dense in M(oo), then the set X = I(M)(x) is a proper closed subset of M(zo), and by (9.1.1) we conclude that M is a symmetric space of rank at least 2. In this case the proposition is known,
and hence it suffices to consider the case that I(MXx) is dense in M(co). (Actually this case does not arise since M is symmetric with rank at least 2 by (9.3.1).) Choose sequences c SM and c 1(M) such that v -, v and c5n(x) = y,, (co) for all n. Since v E.9* there exists an integer N such
that vN E,9 * and r(vN) = r(v). Since x = yw(-) it follows that wN = d'N(w) is asymptotic to v,,v and hence by (1.9.37) there exist sequences (41k) c 1(M), {vk } c SM, and (tk) c R such that tk vk -> vN, and (d qik o g'4 X vk)
- wN as k - x. Hence for sufficiently large k we have
r(w) = r(wN) = r((d+/ik o g"Xtrk )) = r(vk) = r(vN) = r(v) by (1.12.4) and the fact that WN and uN lie in .9p*. This proves (1). D
For the next step of the proof we use the Tits pseudometric Td on M(oo) as defined in (3.4). Recall that Td is the inner metric on M(°°) determined by the angle metric 4(x, y) = sup(p(x, y): p E M), where
Characterizations of Symmetric Spaces of Higher Rank
407
x and y are any two points in M(x). To know when a space M that we
are considering is symmetric of higher rank we use the following criterion, which is proposition 3.1 of [EH]. (2) PROPOSITION. Let M be irreducible and suppose that 1(M) satisfies the duality condition. Then M is isometric to a symmetric space of noncompact
type and rank at least 2 if there exist a constant a * > 0 and a point x E M(x) that satisfy the following conditions:
(a) Ba.(x) = (y E M(x): Td(x, y) < a*) contains a point other than x.
(b) If y, and y2 are any two points of B,,.(x), then 4q (y,, Y2) _ Td(y,, y2) for all points q E M.
The idea of the proof of this criterion is quite simple. Assume that
(a) and (b) hold for some a* > 0 and x E M(x). For any number Cr E (0, 7T] let Ma(a) = {x E M(x): there exists y E M(x) with 4q (x, y) = a for all points q E M). From the definition it follows immediately that Ma(x) is closed in M(x) and invariant under 1(M) for any a. Since (a) and (b) hold there exists a number p E (0, a*) such that Td(x, y) = p for some y e B,.(x) with y *x. The set MP(x) is nonempty since it- contains x and y, and we may now define f3 = sup( a E (0, ir): a(x) is nonempty). Since M is irreducible it is not flat, and
lemma 3.2 of [EH] shows that MR(x) is a nonempty proper subset of M(oo). We conclude that M is symmetric by (9.1.1).
The next stage of the proof of (9.3.1) is to establish lemma 4.3 of [EH]. We merely state this result here in (3) below and omit details of the proof, which is elementary. A vector v E SM is 1(M)-recurrent if there exist sequences {t,,} c !f$ and c 1(M) such that t - +x and (dc c g`-Xv) - v as n -- x. If F is a k-flat in M, where k = r(M ), then we recall that F(x) = (y(x): y
is a geodesic of M contained in F). We note that if p E F and x, y C M(x) are arbitrary points, then Td(x, y) =
q
(x, y) by (3.1.1),
(3.1.2), and (2) of (3.4.3).
Let M be irreducible, and let 1(M) satisfy the duality condition. Let v C SM be a regular, 1(M)-recurrent vector, and let x = y, (cc). Let F(y,.) denote the unique k-flat that contains y,,, where k = r(M ). Then there exists a positive number a with the following property: Let F be any k -fiat in M such that x E F(x). Then F(x) 2 Ba(x) = {y E M(me): Td(x, y) < a}. (3)
We now use (1), (2), and (3) to complete the proof of (9.3.1). Let x E M(x) and a > 0 be as in (3), and let p be any point of M. By (1) the geodesic is contained in some k-flat F of M. If z, and z2 are any two points of M(x) such that Td(x, z;) < a for i = 1, 2, then z; C F(x) for i = 1, 2 by (3). By a remark in the paragraph preceding (3) it follows that < (z,, z2) = Td(z,, z2) since p c- F. Since p E M was arbitrary we
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have shown that (2a) and (2b) hold for x and a* = a. Hence M is a symmetric space of noncompact type and rank at least 2 by (2). This completes the proof of (9.3.1). 9.4.
0
Applications of the higher rank rigidity theorem
9.4.1. THEOREM. Let M be a complete Riemannian manifold with sectional curvature K:5 0, finite volume, and no local Euclidean de Rham factor. Then either M is locally symmetric or M has rank 1 or M admits a
finite cover M * that splits as a nontrivial Riemannian product MS X M* X ... x M,*, where Ms is locally symmetric and M1* has rank 1 for
1Si5k.
In the result above the factor M* could be absent if k >_ 2, or we could have k = 1 if M, is present. Note that the spaces M,*, M*,..., Mk have finite volume since M* has finite volume. PROOF. We consider first the case that the universal Riemannian cover
M is irreducible. If M has rank 1, then M has rank 1. If M has rank k z 2, then M is a symmetric space of rank k by (9.3.2) and M is locally symmetric.
Next we consider the case that M is a nontrivial Riemannian product. If M admits no finite cover M* that splits as a nontrivial Riemannian product, then M is locally symmetric by (9.2.5). If M does admit a finite cover M* that splits as a nontrivial Riemannian product M* XM; x ... x M*, then by passing to a further finite cover if necessary we may assume that each M,* admits no finite cover that splits as a nontrivial Riemannian product. The discussion above now implies that each M* is locally symmetric or has rank 1. 0
A vector tangent to a periodic (= closed) geodesic of a Riemannian manifold X is called a periodic vector. 9.4.2. PROPOSITION. Let M be a complete Riemannian manifold with sectional curvature K:!5 0, finite volume, and finitely generated fundamental group. Then the periodic vectors in SM form a dense subset of SM.
REMARK. We do not know whether the condition on the fundamental
group is essential. It is unnecessary if the universal cover M has no Euclidean de Rham factor, as we show here. If M has a nontrivial Euclidean de Rham factor, then an extra technical argument is required (see [CEK]), and this argument requires that the fundamental group be finitely generated. We shall prove (9.4.2) only in the case that M has no Euclidean de Rham factor.
PROOF. Assume that M has no Euclidean de Rham factor. If M has rank 1, then the result follows from corollary 3.9 of [Bal]. If M is a
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symmetric space of rank k ;,f-- 2, then the result follows from lemma 8.3'
of [Mos2]. If the periodic vectors are dense in SM*, where M* is a Riemannian cover of M, then clearly the periodic vectors are dense in SM. The statement of (9.4.2) now follows from (9.4.1), the remarks above, and the next result.
0
LEMMA. Let X be a complete Riemannian manifold that is a nontrivial Riemannian product X) X X2. If the periodic vectors are dense in SX, for i = 1, 2, then the periodic vectors are dense in SX.
PROOF. Let v = v) + v2 E SX be given, where v, e TX, for i =1, 2 and Iv)I' + Iv212 = Iv12 = 1. We may assume that v) and v2 are both nonzero since such vectors v are dense in SX. For the geodesic flow (g') of any complete Riemannian manifold X we observe that
(1) g'(cv) = c(g"v) for all v E TX, t E R, and c > 0. Hence if v is a periodic vector with period (o (i.e., gwv = v), then cv is a periodic vector with period (1/c)w for any c > 0. It now follows from the hypotheses that there exist sequences { wn) and { w,*) c (0, oo), (vi")} C TX), and {vr")) c TX2 such that (2) g' 'v(") = vi") and gw^vZ") = V2(") for all n,
vi") - U)
and 027) -> v2 as n
Choose positive integers r,, and sn and a sequence (en) c (0, oo) such that (3) wn(1 + en)/wn = r,,/sn for all n, where en - 0 as n -> oc. If u(")' = vi ")/(1 + en), then from (1), (2), and (3) we obtain
(4) g
v (n)' = v (n)'
,(n) for all n, v 1n)' -> v i as and g r-wnv 2(") == v( 2
n
Let ,,)(t) and yrt2(t) be the geodesics in Xl and X2 with initial velocities v(,")' and v2") respectively. If v,, is the initial velocity of the geodesic NO = (y")(t), yn2(t)), then vn -* v by (2) and (4) and gr"w"vn = vn for all n by (4). 0
9.4.3. THEOREM (Gromov). Let M) and M2 be compact Riemannian manifolds with sectional curvature K:5 0 whose fundamental groups are isomorphic. Assume that M, has no finite cover that splits as a nontrivial Riemannian product and that the universal cover Ml is a symmetric space with rank k >_ 2 and no Euclidean de Rham factor. Then M, and M2 are isometric after multiplying the metric of Ml by suitable positive constants on local de Rham factors.
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Geometry of Nonpositively Curved Manifolds
REMARK. Gromov proved this result in theorem 14.2 of [BGS]. In the
special case that the universal cover M, is a nontrivial Riemannian product, this result was also proved in [E131.
PROOF. By the Mostow rigidity theorem (theorems 18.1 and 24.1 of [Mos2]) it suffices to prove that M2 is locally symmetric. We use the fact
that many geometric properties of a compact nonpositively curved manifold are determined by the algebraic structure of the fundamental group. This philosophy is the subject of chapter 10. We observe first that the universal cover M2 of M2 has no Euclidean de Rham factor. This follows from (10.3.10), which states that the rank of the Clifford subgroup of 7r,(M2 ), the unique maximal abelian normal subgroup of ir,(M2), equals the dimension of the Euclidean de Rham factor of M2. By hypothesis, M, has no Euclidean de Rham factor, and hence the same is true for M2.
Next we observe that M, and M, have the same rank k;-> 2. This follows from (10.3.11), which states that the rank of M, equals the algebraic rank of ir1(M,) for i = 1,2. The fundamental groups ir,(M,) and 7r1(M2) have trivial center by (7.1.2) and (1.9.4) since M1 and M, have no Euclidean de Rham factor. This implies that 7r,(M,) has no nontrivial direct product subgroup of finite index. If it did, then M would admit a finite cover M* that splits as a nontrivial Riemannian product by (10.3.9) below, which contradicts the hypothesis on M,. Hence Tr 1(M2) has no nontrivial direct product
subgroup of finite index, and it follows that M2 admits no finite Riemannian cover that splits as a nontrivial Riemannian product. If M2 is a nontrivial Riemannian product, then M, is locally symmetric by the remarks above and (9.2.5). If M2 is irreducible, then M2 has rank k > 2 by the discussion above and M2 is a symmetric space by (9.3.2). The proof of (9.4.3) is complete. As an immediate application of (9.4.3) we obtain the following. 9.4.4. THEOREM. Let (M, g) be a compact, locally symmetric Riemannian manifold with sectional curvature K< 0 and rank k >_ 2 whose universal cover M has no Euclidean de Rham factor. Assume that M admits no finite cover that splits as a nontrivial Riemannian product, and let g* be any Riemannian metric on M with sectional curvature K :::-t, 0. Then (M, g*) is
isometric to (M, g'), where g' is a Riemannian metric on M that is obtained by multiplying g by suitable positive constants on local de Rham factors of M.
REMARK. If g' is any metric obtained by rescaling the metric g on local
de Rham factors, then clearly (M, g') has nonpositive sectional curvature since (M, g) does. Theorem (9.4.4) says that this is the only way, up to isometric equivalence, to obtain nonpositively curved metrics g' on M.
Characterizations of Symmetric Spaces of Higher Rank
411
Tits finiteness theorem
The finiteness of the Tits metric Td on M(x) can be used to characterize symmetric spaces of higher rank. If M is a symmetric space
of noncompact type and rank k >- 2, then Td(x, y) < IT for all x, y E M(x) by (3.6.1). Conversely, we have the following. 9.4.5. THEOREM. Let M be irreducible and suppose that 1(171) satisfies the
duality condition. Assume furthermore that there exists an open set 0 of M(x) such that Td(x, y) is finite for any two points x, y E 0. Then M is a symmetric space of noncompact type and rank k >- 2. PROOF. By (9.3.1) it suffices to show that M has rank k > 2. We assume
that M has rank 1 and obtain a contradiction. We establish some terminology. We recall from (4.5.13) that a vector v (=- SM is 1(M )-periodic if there exists 0 E 1(M) and w > 0 such that (0 0 y,,Xt) = y,.(t + w) for all t E R. A geodesic y of M is said to bound
a flat half-plane in M if there exists an isometric totally geodesic imbedding f: R x [0,x) -> M with f(t,0) = y(t) for all t e R. Now let 0 c M(x) be an open set such that Td(x, y) is finite for all points x, y E 0. By (1.9.15) and by theorem 3.8 of [Bal] there exists a dense subset A of SM such that if v (=-A then tv is 1(M)-periodic and y,, bounds no flat half-plane in M. Choose a vector v E A such that x = y,.(x) e 0. By (2.2) and (2.3) of [Bal] it follows that if y E M(x) is any point distinct from x, then there exists a geodesic y of M such that y(x) = x and y(- oo) = y. Choose v E A such that x = ,,(ac) E 0 c M(x). We assert that Td(x, y) = + x for all y e M(x) with y # x, which will contradict the
hypothesis on 0. If y E M(x) is distinct from x, then Td(x, y) z -C(x, Y) = IT since Td is the inner metric obtained from c and since it z sup( 0, which is impossible for small t. Hence Td(x, y) = +x, which is contradicts our assumption that Td is finite on 0 and proves that M must have rank k z 2. O Topological rigidity
If M is compact, locally symmetric and irreducible with sectional
curvature K 5 0, rank at least 2, and no local Euclidean de Rham factor, then the next result shows that there are only two kinds of
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Geometry of Nonpositively Curved Manifolds
homotopy classes of continuous maps between M and another compact nonpositively curved manifold N whose dimension is no larger than that of M.
9.4.6. THEOREM. Let M and N be compact manifolds with sectional curvature K :g 0 and dim M z dim N. Let M be orientable, irreducible, and locally symmetric with rank k >_ 2 and no local Euclidean de Rham factor. Let f: M - N be a continuous map. Then either (1) f is homotopic to a constant map, or (2) dim M = dim N and after multiplying the metric of M by suitable positive constants on local de Rham factors there exists a Riemannian covering map g: M --> N of finite multiplicity such that g is homotopic to f.
We first show that maps f and g between nonpositively curved manifolds M and N are homotopic if and only if they induce the same homomorphisms between the fundamental groups of M and N. We
begin with a discussion of the homomorphism between ir,(M) and ir,(N) induced by a map f: M-> N. Let ir: M -* M and it *: N --s N be fixed universal covers of M and N with deck groups r c 1(M) and r * c I(N). If f : M -> N is any continuous map, then one can find a lift f: M --' N, that is, a map f: M -> N such that 9r* o f = f 0 IT. Any lift f induces a homomorphism 9: r --+ r*
such that f o 0 = 9(¢) o f for all cb E r. The lift f is unique up to precomposition with an element of r or postcomposition with an element of r*. It follows that the induced homomorphism 9: r --s r* is
unique up to precomposition by an inner automorphism of r or postcomposition by an inner automorphism of F. We say that two continuous maps f and g: M - N induce the same homomorphism between the fundamental groups ir,(M) and Tri(N) if they have lifts f and g:.4 --> N that induce the same homomorphism
9:r->r*.
LEMMA. Let M and N be complete Riemannian manifolds with sectional curvature K:5 0, and let f and g: M -s N be continuous maps. Then f and g are homotopic if and only if they induce the same homomorphism between
irl(M) and ir1(N). PROOF OF THE LEMMA. Suppose first that f and g are homotopic maps
between M and N, and let (f,) c C(M, N) be a homotopy with fo =f and fl =g. Let or: M ---> M and or*: N -> N be fixed universal covers of
M and N with deck groups r c1(M) and r* c1(N). By the homotopy lifting property_ of coverings there exists a homotopy (f,) c C(M, N) such that 7r * o f, = f, c ar for all t e [0,1]. The lifts f, induce homomor-
phisms 9,: r - r* with f, o 0 = 9,(¢)o f, for t E [0,1] and 45 E F. It
Characterizations of Symmetric Spaces of Higher Rank
413
follows that 0, is constant in t since r* is a discrete group, and hence the lifts fo and f1 of f and g induce the same homomorphism between
I' and r*. Conversely, suppose there exist a homomorphism o: r -> r* and lifts
f and off and g such that _f o 4)= 8(¢) o f and g o o= 0(4) o g for all ¢ r. For points n, n* E N let y,, . denote the unique geodesic in N with ynn.(0) = n and ynn.(1) = n*. For m e M and t c= [0,1] define h,(m) = yflm)RIm)(t). Then ho = f and h1 =g, and for 0 E IF, m E M, and t E [0,1 ] we have
(h, o cb)(m) = yj(om)8(o.)(1) = 0(4)(yI(,n)8(m)(t)) _
If we define h,(irm) = it*(h,(m)) for m E P. and t E [0, 1], then (h,) is a well-defined homotopy between h0 =f and h1 =g. This completes the proof of the lemma. 0 PROOF OF THEOREM (9.4.6). Let f : M - N be any continuous map, and
let f : M -+ N be a lift of f, where 7r: M --* M and lr*: N -* N are universal covers of M and N with deck groups F c 1(M) and r* c 1(N ). Let 8: r --+ r* be the induced homomorphism such that f o (k = 0(4)) o f for all ¢ E r. If F0 = IF n G, where G =10(M), then F0 has finite index in F since G has finite index in 1(M). The group ro is an irreducible, cocompact lattice in the semisimple Lie group G of real rank k;-> 2.
Moreover, G has trivial center and no compact normal subgroups except the identity; see (2.1.1). If A = kernel(8), then by the Margulis finiteness theorem (cf. theorem (8.1.2) of [Z4]) we know that either A n ro is finite or F0/A n ro is finite. It follows that either A is finite or
r/A is finite since ra has finite index in F. However, since r and r* are the fundamental groups of the smooth nonpositively curved manifolds M and N they contain no nontrivial subgroups of finite order by (1.5.1). Since 8(r) is a subgroup of r* isomorphic to r/A we conclude that
(*) either 8: r -+ r * is injective or 0(y) = Id for all y E r.
If 0(y) = Id for all y E r, then f: M -. N is homotopic to a constant map by the lemma above. We consider the case that 8: r -> r* is injective. Let r** = 8(r) c r*, and let N* = N/r**. We show first that N* is compact and dim M = dim N*. We recall from (1.5) that both M and N* are K(ir,1) spaces; that is, the kth homotopy group is zero for all k z 2. Hence the group homology of r and r** is isomorphic to the homology of the manifolds M and N*. See, for example, proposition 4.1
of [Br]. Since M is compact and orientable and 8: r -' r** is an isomorphism we conclude that 0 # Hn(M) - Hn(r) = Hn(r**) = Hn(N*), where n = dim M. Since dim N* 5 n by hypothesis we conclude that dim N * = n = dim M, and since H (N *)
0 we see that N * is compact.
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Geometry of Nonpositively Curved Manifolds
The compact manifolds M and N* have isomorphic fundamental groups, and by (9.4.3), N * is locally symmetric and isometric to M after
rescaling the metric of M by suitable constant positive multiples on local de Rham factors of M. Assume that the metric of M has been rescaled in this fashion. By the second version of the Mostow rigidity theorem (8.1.1') there exists an isometry F: M - N* and a lift F: M -' N such that f o 0 = 9(0) o F for all 0 E F. From the definition of N * it follows immediately that N* is a Riemannian covering of N. Since N* and N are compact the Riemannian covering map h: N* -> N has finite multiplicity. If p: N - N* is the universal covering map, then h o p = yr *: N -* N, and it follows that h = the identity on Al is a lift of h: N* -> N. If g = h o F: M -> N, then g is a Riemannian covering map of finite multiplicity and g = h c F: M --+N is a lift of g that induces the homomorphism 9:F - F. Hence f and g are homotopic maps by the lemma above.
0
10
Fundamental Group and Geometry
10.1.
Homotopy, homeomorphism, and diffeomorphism type
416
10.2.
Geometry determines fundamental group Cartan finite order lemma 416 Preissmann's theorem 416 Property (P) 416 A Bieberbach theorem 416
416
103. Fundamental group determines geometry Flat torus theorem 417 A converse to Bieberbach's theorem 418 An extension of Preissmann's theorem 418 Solvable fundamental group implies flat 418 Existence of F-periodic flats for solvable groups with semisimple elements 419 An algebraic splitting theorem 423 Dimension of the Euclidean de Rham factor 424
417
Algebraic rank of the fundamental group 424 The rank of a group 425 Visibility axiom and fundamental group 427 r-closed k-flats in simply connected analytic spaces 428
One of the principles of Riemannian geometry is that the geometric and topological properties of a Riemannian manifold X are often strongly
related, especially when X is compact. For a compact manifold of nonpositive sectional curvature the homotopy information is carried in
the fundamental group since the higher homotopy groups vanish. Therefore, it is reasonable to look for links between the algebraic structure of the fundamental group and the geometric properties of a compact nonpositively curved manifold. For a survey without proofs of results in this section, see also section 6 of [EHS].
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Geometry of Nonpositively Curved Manifolds
Homotopy, homeomorphism, and diffeomorphism type In a series of papers T. Farrell and L. Jones investigated the
10.1.
relationship between the homotopy, homeomorphism, and diffeomorphism types of compact nonpositively curved manifolds. The next result shows that the homotopy and homeomorphism types are identical. 10.1.1. THEOREM [FJ2]. Let M, and M2 be compact Riemannian manifolds with sectional curvature Ks 0 whose fundamental groups are isomorphic. Then M, and M2 are homeomorphic.
At roughly the same time, Farrell and Jones showed in a particularly striking way that the homeomorphism and diffeomorphism types can be different. 10.1.2. THEOREM [FJ 1 J. For every 5 > 0 and every integer n >- 5 there exist
compact Riemannian manifolds M and M,* such that (1) Mn and M,* are homeomorphic but not dif j`eomorphic and
(2) K= -1 in M,*, and -1-5SK__ 2. Then the universal cover M admits a
k flat F such that A leaves F invariant and the quotient space F/A is compact. In particular, M admits a totally geodesic, isometrically immersed flat k-torus T k = F/A.
We omit a proof of this result, which is a corollary of (10.3.6).
REMARK. If A is any finitely generated abelian subgroup of Tr,(M ), then A must be free abelian by (10.2.1).
From the result above we immediately obtain a converse to the Bieberbach theorem (10.2.4).
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Geometry of Nonpositively Curved Manifolds
10.3.2. COROLLARY. Let M be a compact Riemannian manifold with sectional curvature K 5 0. If 7r,(M) admits a normal abelian subgroup of finite index, then M is flat.
PROOF. Write M = M/1', where M is the universal covering of M and
t c 1(M) is the deck group of the covering. If r* is a normal abelian
subgroup of 1' with finite index in F, then M* = M/r* is a finite regular covering of M. Hence M* is compact and F* is finitely generated. By (10.2.1), F* is free abelian of rank k for some integer k >- 1,
and by (10.3.1) there exists a k-flat F in M such that r* leaves F invariant and F/r* is compact. If k < dim M then we obtain a contradiction to the fact that m* = M/r* is compact; if p E M - F is a point with d(p, F) > diam(M*), then d(yp, F) = d(p, F) > diam(M*) for all y r= r*. Hence M= F and M is flat. We can also use (10.3.1) to extend Preissmann's theorem as follows: 10.3.3. COROLLARY. Let M be a compact Riemannian manifold with K< 0 whose universal covering M satisfies the Visibility axiom (cf. (1.8)). Then M satisfies property (P) (cf. (10.2.3)).
PROOF. Write M =M/r, where F c 1(M) is the deck group, and let A be a nonidentity abelian subgroup of r. If A is not infinite cyclic, then A contains a free abelian subgroup A* of rank k >- 2, and by (10.3.1), M admits a k-flat F. However, the Euclidean geometry of the k-flat F is incompatible with the Visibility axiom in F and hence also in M. The next advances were by S.-T. Yau and D. Gromoll and J. Wolf. 10.3.4. THEOREM [Y1]. Let M be a compact Riemannian manifold with sectional curvature K :!g 0. If Ir1(M) is solvable, then M is flat.
REMARK. R. Zimmer has generalized this result by proving that if M is
a compact, Riemannian manifold with sectional curvature K:5 0 and amenable fundamental group, then M is flat; see theorem 3 of [Z3, p. 10121.
We omit a proof of (10.3.4), which is an immediate consequence of the next result and the argument used to prove (10.3.2). We prove a powerful result due to D. Gromoll and J. Wolf [GW] that
generalizes the Preissmann theorem and flat torus theorem stated earlier. See section 7 of [BGS] for related results. Before stating this result we introduce some terminology and recall some useful facts. Recall that a group r is solvable if the descending series (F1) defined by r0 = I' and I'; = [I; - 1, F;_ I ] for i >- 1 eventually terminates in the
identity. The first integer r for which descending series.
F', = (1)
is the length of the
Fundamental Group and Geometry
419
If C is any closed convex subset of M with positive dimension, then the elements of the group I(C) of isometries of C are differentiable on the interior of (cf. (1.6.1)). Following the definition in (2.19.21) for symmetric spaces we introduce the following. 10.3.5. DEFINITION. An element 4P of I(C) is semisimple if .0 is elliptic or axial; that is, the displacement function do has a minimum in C that is either zero or positive.
Note that parts (2) and (3) of (1.9.2) apply to semisimple elements of I(C). For elliptic elements 0 of 1(C) we must be content with a weaker version of (1) of (1.9.2), which says that the minimum locus of do (those points in C fixed by 0) is a closed convex subset of C. In keeping with (4.7.6) we say that a k-flat F c C is r -periodic for a
subgroup F c 1(C) if F is invariant under IF and the quotient space F/t is compact. 10.3.6. THEOREM (Gromoll-Wolf [GW]). Let C C M be a closed convex
subset, and let F c 1(C) be a solvable group whose elements are semisimple. Then for some integer k >_ 0 there exists a F -periodic k -flat F in C. By convention a point is a zero-flat and a geodesic of M defined on IF8 is a 1-flat. We prove the result above by induction on the length r of the descending series of IF. We begin with the case r = 1 when r is abelian.
LEMMA 1. Let r c I(C) be an abelian subgroup whose elements are semisimple. If Co c C denotes the minimum locus of the displacement function do for an element 4) of r, then Cr = noE r co is a closed, nonempty, convex subset of C.
PROOF. We prove this by induction on the dimension of C. If dim C = 0, then C is a point and the result is obvious. Suppose now that dim C = k
for some integer k > 1, and assume that the result is true for convex subsets with dimension < k - 1.
We first consider the case that IF contains an axial element 0. By (1.9.2) the subset Co of C is closed and convex and 0 translates a geodesic yx through every point x of C. by w = inf do > 0. If 4) is any element of IF, then 4)4) = 4)4) and it follows that 4) leaves Co invariant and t//(yx) = y4,(x) for all x r=- C4,. By (1.6.7) we conclude that C. is
isometric to C,* x y(lJ), where C,* is a closed convex subset of C and y(R) is one of the axes yx, x (=- Co. Moreover, each element 41 of F c 1(C; x y(R)) can be written 4)(p, y(t)) = (41(p), 412 y(t)) for p E CO* and t r=- ff8, where fir, E I(CS) and 412 E I(y(R)) translates y by some amount w4, # 0.
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Geometry of Nonpositively Curved Manifolds
Let p: IF --> 1(C) be the projection homomorphism given by p(4,) _
4, and let r, =p(t) c I(C, ). The group r, is also abelian with semisimple elements and dim C,* = dim C. - 1 < dim C - 1. It follows by induction that Cr, = fl,`, c r, C4,, is a nonempty subset of C,*,. Hence Cr = Cry x y(R) is nonempty since d1, assumes its minimum value at (p, y(t)) E CC x y(R) = C. for i, E r if and only if d,,' assumes its minimum value at p E C. Next we consider the case that t contains only elliptic elements. If 45 and 4, are any two elements of I', then ¢ leaves C,1, invariant since 04 = 4,4), and by (3) of (1.9.2) we conclude that C. n C,` is nonempty. If S is any finite subset of r, then Cs = fl, E s Co is nonempty by (3) of
(1.9.2) and induction on the cardinality of S. Now let S be a finite subset of r such that Cs has minimal dimension, and let a E r be any element # 1. By the remarks above Ca n Cs is nonempty and hence dim(Ca n Cs) = dim Cs for all a E r. It follows that Ca n Cs is an open subset of Cs, but if a fixes all points in an open subset of the convex set Cs, then a fixes all points of Cs. Hence Ca n Cs = Cs for all a E F or equivalently Cs = C,. = n s E r Co c C. LEMMA 2. Let F c 1(C) be abelian with semisimple elements, and let c,. = fl4, E C,, c C. Then there exists an integer k z 0 such that for every point x E C,. there exists a Fperiodic k -flat F, with x E FX c Cr. Moreover, any two k -flats Fx and FY for x, y e Cr are parallel.
PROOF. We again proceed by induction on the dimension of C. If dim C = 0, then C is a point and the result is obvious. Let C have dimension k >- 1, and suppose that the lemma holds for all closed, convex subsets of dimension < k - 1. If IF contains only elliptic elements, then i' fixes every point of C,. and lemma 1 holds with k = 0.
If t contains an axial element ¢, then 0 translates a geodesic yx through every point x of Cr by (1.9.2). Hence by (1.6.7), Cr is isometric
to Cr x y(R), where y is a geodesic in Cr translated by ¢ and C j is some closed convex subset of Cr. If 41 is any element of F, then d,, is constant on Cr and +/s leaves Cr invariant. Since 41 permutes the geodesics in Cr translated by 0 it follows from (1.6.7) that '/r(p, y(t)) _
(4,,(p), 4k2(yt)) for p E Cr and t e R, where Vi, E AM and 4/2 E 1(y(R)) translates y. Let IF, c I(Cr) be the image of F under the projection homomorphism that sends t# to 41,. Then r, is an abelian group with semisimple elements, and it is easy to see that C* = flu, e r, Co,. Since dim Cr = dim Cr - 1:5 dim C - 1, the induction hy-
pothesis implies that any point p E Ci is contained in a I',-periodic k-flat FP e C*r, where k >t 0 is an integer that does not depend on p. If
x = (p, y(t)) is any point in Cr x y(R) = Cr, then Fx = FP x y(R) is a (k + 1)-flat in C,.. By the induction hypothesis we also conclude that
Fundamental Group and Geometry
421
FP */r, is compact for all p E C* and hence F,/r is compact for all X = (p, y(t)) (=- Cr since r 2 (4) ), where (4)) denotes the infinite cyclic
group generated by ¢. Since Fs/r and Fy/r are compact for any two points x and y of C,. it follows that for some positive number c we have d(p, Fy) < c and d(q, F,):5 c for all points p E F, and q E Fy. Hence F, and FY are parallel (k + 1)-flats by (1.6.7), and the proof of lemma 2 is complete.
We are now ready to complete the proof of the theorem by induction
on the length r of the descending series of the solvable group r. If r = 1, then r is abelian and the result follows from lemma 2. Let r be a solvable subgroup of 1(C) with semisimple elements and length r >: 2 and suppose that the result is true for solvable groups r* of length 'Q
p fl{go.+g_Q), (2.14.2)
AF
space of pointed k-flats in M, where M is a symmetric space of noncompact type and rank k, (8.2.4) and C8-41 Q
q
point of M
Index of Notation
447
R
r
r(M) r(v)
r(x)
function from 1(R") to O(n, R) that assigns to each isometry V of R" the rotational part r(V) of cp, (6.3.1) rank of a complete, Riemannian manifold M with sectional curva-
ture K50, (1.12.3) rank of a unit vector v in a complete, Riemannian manifold M with sectional curvature K:5 0, (1.12.3) rank of x E M(°), where M is a symmetric space of noncompact type, (2.21.2)
r*(x)
r(y)
r(r) R(u,v) R(oo)
integer assigned to each point x (=- M(c), where M is a symmetric space of noncompact type, (2.21.4) rank of a unit speed geodesic y in a complete Riemannian manifold M with sectional curvature K 5 0, (1.12.3) integer determined by a group IF c 1(M), (10.3.13) skew symmetric curvature transformation of TPX determined by vectors u, v in TPX, (1.1) regular points in M(me), where M is a symmetric space of noncompact type, (2.17.17)
R,,
Rx
symmetric curvature transformation in TPX defined by R,,(u) _ R(u, v)v, where u and v are arbitrary vectors of TPX curvature transformation (ad X)' determined by an element X of p, (2.14.3)
rad(x)
sup(S(y):y6'(x)), where S(y)=inf{Td(y,z):zCs '(y)-{'(y)) and x, y, and z are points of M(-) and M is a symmetric space of noncompact type, (3.6.33)
rank(F) .9p
R
rank of a group I'c1(M), (10.3.13) set of regular unit vectors in SM or SM, (1.12.8) real numbers S
s(x)
degree of singularity of a point x E M(o) = the rank of FF(yyx) for any point p of M, where M is a symmetric space of noncompact
S
center of gravity map from FM(oo) to R(w), discussion preceding
S("`)
(3.8.5) [S, S(k-1)], where S = St's c End(C"), (6.1.6)
SP
geodesic symmetry of M or h1(oo) determined by a point p of M,
type, (2.21.7)
(1.7.10)
SL(n,R) elements of GL (n, R) with determinant 1 unit tangent bundle of a Riemannian manifold X SX S(X, F) (x E F(-): X E gX}, where F is a k-flat in a symmetric space M of noncompact type and rank k, G =10(M) and X E g; this is the splice in M(-) determined by X and F, (3.7.3) unit vectors in the tangent space TPX, where X is a Riemannian SPX manifold
448
Geometry of Nonpositively Curved Manifolds
Sx
(y (=- M(-): y sx), (3.6.52)
S.
reflection in a determined by a root a E A c a*, (2.9.2) unit speed geodesic in a Riemannian manifold involution of G = 1(M) determined by a point p of M, where M is a symmetric space of noncompact type, (2.3.1)
01
op
T t
function from I(R") to R" that assigns to each isometry qp of R" the
T
translational part t(cp) of gyp, (6.3.1) projection map from R(oo) onto FM(o), (3.8.5) and following
TTX TR(x)
tangent space at a point p of a manifold X
Tx
TX tx
homomorphism from Gx to G, where G =1 (M) and M is symmetric of noncompact type, (2.17.4) symmetric linear operator (ad X) -(ad Y) on g determined by commuting elements X, Y of p, (2.14) Tits metric on M(c), (3.4.1) tangent bundle of a manifold X Lie algebra homomorphism of gx tog determined by TX : Gx -, G,
9(G)
Tits building associated to G =10(M), where M is symmetric of
9(F,G) 9M
apartment in 9(G) determined by a k-flat F in M, (3.6.23) quotient space M(cc)/ - , where x - y if (Gx)0 _ (GG)0, discussion
9'M
elements of YM that are maximal with respect to the Tits partial
Txy
Td(,)
(y E M(oc) : Td(x, y) < R}, where x E M(oo) and R > 0, (3.6.9)
(2.17.9)
noncompact type, (3.6.22)
following (3.6.24)
ordering, (3.9.4) Bp
Cartan involution of g determined by a point p of a symmetric space M of noncompact type, where G = !0(M), (2.3.2) V
vol(F)
volume of the quotient space X/T, where f is a discrete subgroup of AX), (1.9.26) W
W
W(v) WF
W.
usually the Weyl group, (2.9.6) and (2.9.8) Weyl chamber in a k-flat F(y,.) determined by a regular unit vector v E SM, (2.12.4) quotient group GF/GF, the Weyl group determined by a k-flat F in a symmetric space M of noncompact type and rank k, (4.2.5) (y E M(A): x 5 y), (3.6.20)
Index of Notation
449
X
x
point of M(-)
X xsy
Riemannian manifold of arbitrary dimension (GY)0 c (G.,)0, the Tits partial ordering in M(oc), (3.6.15) Y
Y(t)
usually a Jacobi vector field on a geodesic in a Riemannian manifold
z Z(g)
(h E G =10(M) : gh = hg), where M is a symmetric space of noncompact type; more generally, the centralizer of g in any group that contains it
Z(X) Z(F) Z.
(Y E g : [ X, Y ] = 0), where X is an element of a Lie algebra g
,3 z
center of the group r, usually r c1(M) subgroup T,,(GX) c G., where Tx is the homomorphism from Gs to G, (2.17.4) Lie algebra of ZX
ISBN
Y
