Geometry of Nonpositively Curved Manifolds (Chicago Lectures in Mathematics)

y in the cone topology. Then Td(x, y):5 lim inf -. Td(x,,, PROOF. The first six assertions of this result are found in lemma 4.10 of [BGS]. Our assertion (5) is a restatement of (iv) of that result. Assertion (8) is proved in section 4.9 of [BGS].

We prove (7). Let p be a point of F, and let x and y be arbitrary points of F(x). Then p(x, y) = Td(x, y) by (3.1.2) and assertion (2) of this result. 0 Tits geodesics

Given distinct points x and y of M(x) one may ask about the existence and uniqueness of geodesics in M(x) in the sense of (3.3.5) that join x to y. The next result is lemma 4.11 of [BGS]. 3.4.4. PROPOSITION. Let M be any simply connected manifold, and let x and y be arbitrary distinct points of M(x). If Td(x, y) < x, then there exists

a minimal Tits geodesic in M(x) that joins x to y, and this geodesic is unique up to parameterization if Td(x, y) = 4(x, y) < ir.

Convexity of (M (x), Td) Let M be any simply connected manifold, and let x, y, and z be distinct points in M(x) whose Tits distances are all pairwise smaller

3.5.

than ir. By the preceding result there are unique Tits geodesics joining the points x, y, and z, and we obtain a geodesic triangle A in M(x) as indicated in figure 3.1. Let x, y, and z be points in a standard 2-sphere S2(1) of radius I and Gaussian curvature K = 1 such that Td(x, z) =

4(z,1), Td(x, y) =
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u,

OcS2(1),K31

Ac M(oo)

Figure 1

3.5.1. PROPOSITION. Let M be any simply connected manifold. Let the geodesic triangles A c M(c) and a c S2(1) be as above, and let u, and u, and u2 and u2 be corresponding pairs of points in A and 0 (see fig. 3.1 for an example). Then

Td(u,,u2) 5 4(u,,it2). This result is lemma 4.12 of [BGS].

Convex subsets of M(c) 3.5.2. DEFINITION. Let M be any simply connected space. A subset A CM(cc) is said to be convex if for any two points a,, a2 E A there exists a

minimal Tits geodesic in M(x) that joins a, to a2 and is contained in A. Because of the comparison result above the space M(o) with the Tits

metric Td inherits some of the convexity properties of a standard sphere with curvature 1. One result in this direction is the following. 3.5.3. PROPOSITION. Let M be any simply connected manifold, and let A c M(oo) be any convex subset that is closed in the cone topology of M(00). Then:

(1) For any point x e M(oo) there exists a point x* EA such that Td(x, x*) = Td(x, A) = inf(Td(x, y): y E A). (2) If Td(x, A) < it/2, then the point x* in assertion (1) is unique.

PROOF. Assertion (1) follows from the compactness of A in the cone topology and the semicontinuity of Td with respect to the cone topology

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as expressed in (8) of proposition (3.4.3) above. Actually, (1) holds for any subset A of M(x) that is closed in the cone topology. Next, suppose

that Td(x, A) < IT/2 and that x; and x* are distinct points of A such that Td(x, xi) = Td(x, x*) = Td(x, A). By the convexity of A there exists a Tits geodesic o,: [0,1] --IA such that o(0) =x* and o-() =x*. If x* is the midpoint o(1/2), then x; E A and Tcl(x, x*) < max{Td(x, xi ), Td(x, x2 )) = Td(x, A) by comparison with the standard unit sphere S2(1), using proposition (3.5.1). This contradiction proves (2).

0

Small convex subsets and the center of gravity 3.5.4. DEFINITION. Let M be any simply connected manifold. A subset

A c M(x) is said to be small if there exists x E M(x) and a number 0 with 0 < a < a/2 such that Td(x, a) < a for every a E A. 3.5.5. PROPOSITION. Let Al he any simply connected manifold, and let A C M(x) be any set whose diameter = sup(Td(x, y): x E A, y E A) < Tr/2. Then A is small.

PROOF. Let B be the closed convex hull of A in M(x), and let B* = (x e B: B c B,,,2(x)), where (y E M(x): Td(x, y) < it/2). By (3.4.3) and (3.5.1) we have that A C B*, B* is convex, and B*

is closed in the cone topology of M(x). Hence B = B* and B has diameter < 7r/2. It follows that B is small by lemmas 4 and 5 in section

B of the appendix of [BGS] entitled "Tits metric and the action of isometries at infinity." We give a very brief sketch of the proof.

First, for any point p of M we have 4y (x, y) < Td(x, y) < it/2 for all x, y e B. By the geometry of the unit sphere in R", it = dim M, there

exists a constant a = a(n) < 7r/2 such that <,,(x, y) < a for all p E M and all x, y E B. For each point p E M let V(p) be the center of gravity (see below) of the set BP = (y,,s(0): x E B) c SP M. The unit vector field

V is continuous, and any integral curve of V is divergent in M. Let X* E M(x) be any point such that o- (t,,) ->x* as n -> + x for some sequence (t") - +x and some integral curve v of V. Then convexity properties of M show that Td(x*, y) < a < 7r/2 for any point

yeB. 0

We now define the center of gravity of a set A c M(x). 3.5.6. DEFINITION. Let X f-

any simply connected manifold, and let

A c M(x) be any subset. A point x in M(x) is said to be a center of gravity of A if the function r: M(x) - [0, x) U (x) given by

r(z) = sup[Td(z, a): a EA} assumes a global minimum at x.

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In words, a center of gravity of a subset A c M(x) is the center of a

closed metric ball in M(c) of smallest radius that contains A. The motivation for this definition comes from the proof in (1.4.6) of the Cartan fixed point theorem. By mimicking that proof and using the comparison theorem (3.5.1) above we obtain the following. 3.5.7. PROPOSITION. Let M be any simply connected manifold, and let

A c M(x) be any small subset of M(x). Then A has a unique center of gravity a* E M(x).

PROOF. Consider the function r: M - [0, x) u (x) defined in (3.5.6). By hypothesis there exists a point x (=- M(x) and a number a- with 0 < a <

1r/2 such that Td(x, a) < Q for all a E A. Hence r(x) < a and r* _ inf(r(z): z E M(-)} < a-. We show first that there exists a point a* E M(x) with r(a*) = r*. Let

(a*) c M(x) be a sequence such that r(a*) -9 r* as n -> x. Since M(x) is compact with respect to the cone topology we may assume, by passing

to a subsequence if necessary, that (a*) converges to a point a* in M(x). If a E A is given arbitrarily, then by proposition (3.4.3) we have Td(a*, a) s lim inf,, Td(a*, a) < lim inf x r(a*) = r*. Hence r(a*) < r* since a E A was arbitrary, and equality follows by the defini-

tion of r*. By definition, the point a* is a center of gravity of A. We now show the uniqueness of a*. Suppose that a' a* is another point of M(x) with r(a') = r(a*) = r* < it/2. Since

Td(a',a*)
a*. Let z = o,012) be the midpoint of a. If a EA is given arbitrarily, then by spherical trigonometry, proposition (3.5.1), and the fact that r* < 7r/2 there exists a positive number e such that Td(a, z) < max(Td(a, a'),Td(a, a*)} - e S max{r(a*), r(a')) - e = r* - e. Since a E A was arbitrary it follows that r(z) S r* - e < r*, contradicting the definition of r*. Therefore, the center of gravity in M(-) of A is unique.

We now obtain the following analogue of the Cartan fixed point theorem.

3.5.8. COROLLARY. Let M be any simply connected manifold, and let A c M(x) be a small convex subset that is closed with respect to the cone topology on M(oc). Let G c 1(M) be a group of isometrics that leaves A invariant. Then G has a fixed point in A.

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PROOF. Let r: M(x) -p [0, x) U (cx) be the function defined in (3.5.6).

Then r(4)z) = r(z) for all z E M(x) and all 0 E G. Hence G fixes the unique point a* of M(x) where r assumes its minimum value on M(x). If a' E A is the unique point closest to a* then G fixes a'. 3.5.9.

Examples of convex subsets of M(co)

Let M be any simply connected manifold of nonpositive curvature in the first two examples that follow. (1) (Metric balls) For each point x E M(x) and each positive number

r let T,(x) = {y e M(x): Td(x, y)
Then T,(x) is a convex subset of M(x) for all x E M(x) and each number r with 0 < r:5 7r/2. (2) (Fixed point sets) Let ¢ E I(M) be a parabolic isometry (see (1.9.1), and let

Mi(x) = (x E M(x): 4)(x) =x). Then ft (x) is a convex subset that is closed in the cone topology of M(x). Moreover, there exists a point x E Mi(x) such that Fix(4)) c {y E M(x): Td(x, y) < 7r/2). See proposition (4.1.1) below.

REMARK. Examples show that the set M,00 is not necessarily a small subset in the sense of (3.5.4). (3) (Weyl chambers or faces) For each point x E M(x), where M is a symmetric space of noncompact type and rank k >_ 2, we define F(X) _ (y E M(x): (GG)p = (G,,)o),

where for any z e M(x), G, = (g E G: g(z) = z) and (G,)o denotes the connected component of GZ that contains the identity. We shall prove in proposition (3.6.28) below that for each point x E M(x) the set '(x) is a convex subset of M(x) such that Td(y 1, Y2):5 it/2 for all y y2 E '(x). Moreover, the set W(x) is homeomorphic to an open subset of an (e(x) - 1)-dimensional sphere, where e(x) z 1 is defined in (2.21.7). If e(x) = 1 (i.e., x is maximally singular), then '(x) = (x), while if e(x) = k = rank M, then x r= R(te), and F(x) is a Weyl chamber in M(m) (see (2.17.20)).

(4) Let M be a symmetric space of noncompact type and rank k z 2. Let F be a k-flat in M. If A is a convex subset of F(x), then its closure A in the Td-topology is convex. We prove (4) since we shall need this result later. Let A and F be as above, and let p be a point of F. Then Td(x, y) = p (x, y) for x, y EA by (3.1.2) and assertion (2) of (3.4.3). Hence the map v - y,,(x) is an

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isometry of-(SPF,
Examples of Tits geometries (1) Euclidean spaces. If M is the Euclidean space 18" with its

3.6.

canonical flat metric, then the Tits metric on M(x) is just the standard

angle metric on the unit sphere S"-' c R", which can be identified with Rl'(x).

(2) Visibility manifolds. If M satisfies the Visibility axiom, then (x, y) = it and Td(x, y) = +x for any two distinct points x, y E M(x). This follows immediately from the definition of the Tits metric Td and the fact that any two distinct points in M(x) can be joined by a geodesic of M in the sense of (1.11.1). (3) Product manifolds. Let M=M, X M2 be a Riemannian product manifold. Let A; = M;(x) c M(x) for i = 1, 2. Then:

(a) 4(x,,x2)=Td(x,,x2)= it/2 for all x, EA, and x2EA2. (b) Let x be any point in M(x) - (A I U A,}. Then there exist points x, E A, and x, E A 2 such that x is an interior point of the unique Tits geodesic from x, to x2.

These properties are routine to prove, and a proof is therefore omitted. The existence of subsets A, and A2 in M(x) with the properties (a) and (b) above is a characterization of product manifolds M. See proposition (5.6.1) below. See Section 4.9 of [BGS] for a discussion of the Tits geometry on the product manifold M = H 2 x R, where H 2 is the hyperbolic plane.

Symmetric spaces of noncompact type Let M he a symmetric space of noncompact type. If M has rank 1, then its sectional curvature is strictly negative and M satisfies the Visibility axiom. Since this case has already been discussed in example (2) above, we assume that M has rank k >: 2. The Tits geometry in this case is interesting, and we present some of the most basic features in the remainder of this chapter.

Tits geodesics in M(oo) Let x and y be any two distinct points in M(x), and let F be a k-flat in M such that x and y both lie in F(x). This can always be done by

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proposition (2.21.14). Fix a point p E F, and let E, = y;,(0). Let E, be that unit vector in TPF orthogonal to E, such that y,,},(0) = cos(6)E, + sin(O)E, for some 0:5 0:5 it. If v(t) = cos(t)E, + sin(t)E2 for 0:5; t < it, then

(T (t) = x-( (x)

is a Tits geodesic in M(x) from x = o(0) to y = u(0). Moreover, this observation and proposition (3.4.4) yield the following. 3.6.1. PROPOSITION. Let M be a symmetric space of noncompact type and

rank k > 2. Then Td(x, y) 5 it for all points x, y E M(=). In particular, there is always a minimal Tits geodesic in M(me) that joins x to y, and if Td(x, y) < it then this geodesic is unique. If x and y are points in M(-) with Td(x, y) = it, then there exists a geodesic y of M such that y(x) = x and y(- x) =Y. If Td(x, y) = IT then the minimal Tits geodesic in M(oo) from x to y is definitely not unique. If F is a k-flat with x, y E F(cc), then x can be joined toy in the sense of (1.11.1) by a geodesic y of M that lies in F. Moreover, for every (flat) 2-dimensional subspace F* of F that contains y there are two distinct Tits geodesics (semicircles) in F*(-) that join x to y. The next result shows that all Tits geodesics from x to y have this form. 3.6.2. PROPOSITION. Let M be a symmetric space of noncompact type and rank k >- 2. Let o': [0,11--> M(me) be a minimal Tits geodesic from x = o- (0) toy = Then there exists a k-flat F in M such that (7[0, 1] e F(x).

PROOF. If Td(x, y) < it then the Tits geodesic from x to y is unique and lies in F(-) for any k-flat F such that x and y both lie in F(-). We assume therefore that Td(x, y) = it, and it follows that 4 (x, y) = it by proposition (3.4.3). Let y be any geodesic of M that joins x to y in the sense of (1.11.1), and let p = y(0). If z = o-(1/2) E M(x), then Td(x, z) = Td(y, z) = it/2. From the fact that 7r=

4 (x, z) + 4(z, y) = Td(x, z) + Td(z, y) = it

we

conclude

that

P(x,z)=4(x,z)=it/2 and
spanned by w and y'(0) is zero by proposition (3.1.2). The proof of (3.6.2) will now be completed by the following. 3.6.3. LEMMA. Let M be a symmetric space of noncompact type and rank k > 2. Let y be a geodesic of M, and let w be a unit vector tangent to M at y(0) such that w and y'(0) are noncollinear and span a_2-plane with zero

sectional curvature. Then there exists a k -flat F in M such that y is contained in F and w is tangent to F.

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PROOF. Let p = y(0), and let t1= f + p be the Cartan decomposition

determined by p. Choose X, W E p so that dp(X) = v = y'(0) and dp(W) = w, where dp: p -> TPM is the canonical isomorphism of (2.3.6). By (2.14.3) the curvature transformation Rx = (ad X)2 when restricted

to p corresponds to the curvature transformation R,.: TP M - TP M by

the relation dp(Rx(U)) = -R,.(u) = -R(u,v)v, where U E p, and u = dp(U) E TPM are arbitrary. By hypothesis 0 = (R,.(w), w), and since R,, is symmetric and negative semidefinite it follows that R,.(w) = 0. There-

fore Rx(W) = (ad X)2(W) = 0. Since ad X: g - ; is symmetric on g relative to the inner product Op (see (2.7.1)) it follows that ad X is diagonalizable and hence kernel(ad X) = kernel((ad X)'). In particular, [X, W ] = ad X(W) = 0. If a c p is any maximal abelian subspace that

contains both X and W and if F = exp(a )(p), then F is a k-flat in M that contains y and w is tangent to F at p. Proposition (3.6.2) implies that any Tits geodesic in M(x) is piecewise contained in closed Weyl chambers. More precisely, we have the following.

3.6.4. PROPOSITION. Let M be a symmetric space of noncompact type and rank k _> 2. Let o : [a, b] - M(x) be a Tits geodesic. Then there exist points

XI,...,xN inR(x)andapartition a=t( <11 < .
v[t;_,,t;] cC'(x;)

for 1 si :!g N,

where '(x;) denotes the closure in the cone topology of M(x) of the Weyl chamber 1(x; ).

PROOF. It clearly suffices to prove this result in the case that Q is a minimal Tits geodesic. If o-: [a, b] -- M(x) is a minimal Tits geodesic, then b] c F(x) for some k-flat F in M by (3.6.2). The result now follows immediately from propositions (2.12.2) and (2.17.21), and the discussion of Weyl chambers in (2.12).

The groups K0(p, x) and the manifolds F(p, x) 3.6.5. DEFINITIONS. Let M be a symmetric space of noncompact type and rank k > 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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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

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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 <_ i < N). By hypothesis, dim W >_ 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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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) _ <(x, y) = Td(x, y) < e by proposition (3.1.2) since the geodesic rays yyr[0, x) and yny[0, x) lie in F and hence span a flat triangular sector in F. By the choice of a and lemma (2.20.17) it follows that ct c 1; x and hence (G.) c (GX),,. Next, if y is any point in Ts(x), then by (3.6.10) we may choose E (G,)0 such that 46(y) e T,(x) n F(m). It follows from the case just y,

considered that ¢ (G,,)0 4 ' _ (G,,(y)) c (G,)0, and hence (G,) c (Gx)0 since 46 E (Gx)o.

O

This result has several useful corollaries. 3.6.12. COROLLARY. Let M be a symmetric space of noncompact type and

rank k >- 2. Let x E M(cc) be any point. Let y he any geodesic of M that belongs to x, and let p = y(0). There exists a positive number e0 such that if 0 < e < eo then

TE(x) _ (y c- F(y)W: <,,(x, y) < e}. In particular, T,(x) with the cone topology of M(x) is homeomorphic to an open (r - 1)-ball, where r = r(x) is the rank of x as defined in (2.21.2) and r is also the dimension of F(y ).

For the next result we point out that the cone and Td topologies coincide on T(x) for x E R(x) and a sufficiently small; if y is any geodesic representing x, then F(y) is a k-flat and the assertion now follows from (3.6.12) and assertion (7) of (3.4.3).

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3.6.13. COROLLARY. Let M be a symmetric space of noncompact type and

rank k >- 2. Let x E R(oo) a M(c) be any regular point at infinity. There exists co > 0 such that if 0 < e < co then

TT(x) cW(x), the Weyl chamber at infinity determined by x, and hence Te(x) is homeomorphic to an open (k - 1)-ball. 3.6.14. COROLLARY. Let M be a symmetric space of noncompact type and rank k > 2. Let F be a k -fiat in M, and let x * be any point in M(") - F(00).

Let y* E F(co) be a point that is Td-closest to x*; that is, Td(x*, y*) 5 Td(x*, y) for all y E F(c). Then y* is a singular point at infinity; that is, y* E M(oo) - Moo), and hence s(y*) >- I in the notation of (2.21.7). We now prove the three corollaries. PROOF OF COROLLARY (3.6.12). By (3.6.11) we may choose 0 < eo < IT

such that if y E T6 (x), then (G,)(, c (Gx)o. We assert that e() has the properties described in the statement of the corollary. Let p, x, and y be as in the statement of the corollary. We show first that if 0 < e < co and y E TE(x), then y (=- F(yXc) and

in some k-flat F of M. Hence y E F(yXoo) since F c F(y). Finally,

0 the point yyy(t) lies in F(y), and hence the segment ypy[O,t] is contained in a flat strip that has y = yPx as one of its edges. By (3.6.3) the geodesics yPx and y, are contained in some k-flat of M. By (3.1.2)

and assertion (2) of (3.4.3) we see that Td(x, y) =

The map y - ypy(0) is a homeomorphism of TT(x) onto the a-ball

centered at y' (0) in the (r - 1)-sphere of unit vectors at p that are tangent to F(-y), where r is the dimension of F(y). It remains only to prove that r = r(x), the rank of x as defined in (2.21.2). Let g = I + p be the Cartan decomposition determined by p, and let X be that vector in p such that dp(X) = y, (0). By (2.11.4), F(y) = F(yyx) = exp(Z(X) n pX p), and hence

dim F(y) = dim[Z(X) n p ]. Since ad X is diagonalizable on g and (ad X )l leaves invariant both f and p it follows that Z(X) n p = p n kernel(ad X) = p n

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kernel((ad X)2) = kernel(RX), where RX is the restriction of (ad X)2 to P. If v = dp(X) = yx(0), then by (2.14.3) dp: p -, TPM maps the kernel of RX isomorphically onto the kernel of the curvature operator R,,: TPM -> TPM. From the definition of r(x) and the discussion above we obtain r(x) = dim kernel(R,.) =dim kernel(RX )

= dim{Z(X) n t,} = dim F(y). PROOF OF COROLLARY (3.6.13). Again, by (3.6.11) we may choose 0 <

e0 < 7r such that if y e 7 (x), then (Gy)0 c (Gx)0. By hypothesis, x E R(x), and by proposition (2.17.19), we conclude that (G,,)o = (Gx)0. Hence Ta(x) c F(x) by the definition of '(x), (2.17.20), for every e with 0 < e < co. The fact that TT(x) is a (k - 1)-sphere is a consequence of (3.6.12).

PROOF OF COROLLARY (3.6.14). We use the notation in the statement of

(3.6.14). Note that the existence of a point y* E F(x) closest to x* in the Tits metric is the first part of proposition (3.5.3). By proposition (3.6.1) there exists a minimal Tits geodesic a-: [0.11- M(x) such that 0(0) =x* and o(1) = y* E F(x). If y* were a regular point at infinity, then by (3.6.13) we could find positive numbers e and 6 such that a (t) E Ts(y*) c'(y*) for 1 - 6 < t 5 1. However, M(y*) is a subset of F(x) by proposition (2.17.21) and definition (2.12.1), and the points o-(t)

for 1 - 8:!g t < 1 would therefore lie in F(x) and be closer to x* than y*. This contradiction shows that y* is a singular point at infinity. Next, we show how to determine the degree of singularity s(x) as

defined in (2.21.7) from the metric balls Te(x) c M(x) if e > 0 is sufficiently small. We begin by defining a partial ordering on the points

of M(x). 3.6.15. DEFINITION. Let M be a symmetric space of noncompact type and

rank k > 2. Let G =10(M). For points x, y c- M(x) we set

x
if

(Gx)o -2 (Gy)0,

X
if

(G,)0

(Gy)(),

where for z EM(x) we set G, = (g E G: gz = z) and (G, )o = the connected component of GZ that contains the identity. 3.6.16. DEFINITION. Let i% be a symmetric space of noncompact type and

rank k >: 2. A chain in a subset A of M(x) is a finite sequence <x,,. The chain (x0, x,,..., xn) {x0, x,,..., x,,} eA with x0 <x, <x2 < is said to start at x0. The integer n is said to be the length of the chain.

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3.6.17. PROPOSITION. Let M be a symmetric space of noncompact type and rank k > 2. Let 0 < s(x) <- k - 1 denote the degree of singularity of x as defined in (2.21.7). Then for any positive number a the integer s(x) is the length of a maximal chain in Te(x) that starts at x.

PROOF. If x E R(x), the set of regular points at infinity, then the parabolic subgroup G, is minimal by proposition (2.17.19). Hence there

are no chains in M(c-) that start at x. On the other hand, s(x) = 0 by proposition (2.21.8), which- proves the result in this case. We may therefore assume that x c- M(--) - R(x). The proof now is an immediate consequence of the following two results. 3.6.18. LEMMA. Let M be a symmetric space of noncompact type and rank k >: 2. Let x and y be points of M(-) such that x < y. Then s(y) < s(x). In

particular, any chain in M(x) that starts at x has length at most s(x). 3.6.19. LEMMA. Let M be a symmetric space of noncompact type and rank k >- 2. Then for any point x E M(x) - R(x) and for any e > 0 there exists a

pointy E T6(x) such that x < y and s(y) = s(x) - 1. PROOF OF LEMMA (3.6.18). Let x and y be points of M(x) such that x
s(y)=N
O

PROOF OF LEMMA (3.6.19). Let x r=- M(x) - Woo) be given arbitrarily.

Fix a point p r= M, and let g = f + p be the corresponding Cartan decomposition. Let X E p be that element such that dp(X) = yP,J0), and let a c p be a maximal abelian subspace that contains X. Let A c a* be the set of roots determined by a, and let q = cl0 + Eo E R be the corresponding root space decomposition. Let (, ) denote the inner product on p induced from the inner product on Tt, M by the isomorphism dp: p -> TP M. Let < and 1111 denote the corresponding angles and lengths of vectors in p.

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Let {a,, a,,..., as) be a maximal linearly independent subset of Ax = (a E A: a(X) = 0). By hypothesis s >- I and by proposition (2.21.8)

s=s(x)
set .V = a if s = 1.) Since (a1,..., as) is a linearly independent subset (YE.': as(Y) _ of A it follows that asl; * 0. Therefore, the set 0) is a hyperplane in -cV, and X E.'. Now let 0 < e < Tr be given. Choose 0 < 6 < e so that if Y E a is a unit vector with <(X, Y) < 6, then a(X) a (Y) > 0 for every a E A with a (X) 0 0. Choose a unit vector Y in .W -.W' so that -(X, Y) < 6 and as(Y) > 0. This can be done since .Q -" is an open dense subset of .W. Let y,, be the geodesic in M given by yy(t) = e' '(p) for every t E R, and let y = yy(ce) E M(c). We assert that y E Te(x), x < y, and s(y) = s(x) - 1. To see that y E T,(x) it suffices to observe that Td(x, y) = 4(x, y) =

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 <- y} 3.6.20. DEFINITION. Let M be a symmetric space of noncompact type and rank k;-> 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)
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192

We define AX(,) _ {a E A: a(X(t)) = 0}, AX(,)= {a E A: a(X(t)) > (a E A: a(X(t)) < 0} for 0< t < 1. By (2.17.13) or (2.20.17) we know that the following statements are equivalent for a

0), and

number t E [0, 1]:

o.*(t) -
AX(1)c Ay,

AY.

Since a *(0) = x s y by hypothesis we obtain (i) AXC AY and AXC AY. We note that

(ii) a(X(t)) = (1 - t)a(X) + ta(Y) = a(X) + t{a(Y) - a(X)) for

all aEAand all tEIt

To show that (GQ(,))o = (GG)o for all t > 0 it suffices by the conditions

above to show (a) AX(,)c AY for all t > 0 and (b) AYc AX(,) for all t > 0. Fix t > 0. If a(X(t)) < 0 for some a E A, then a(X) < 0 by (i)

and (ii). If a(X) < 0, then a(Y) < 0 by (i). If a(X) = 0, then 0 > a(X(t)) = ta(Y) by (ii), and we also conclude that a(Y) < 0. This proves (a). Now suppose that a(Y) < 0. Then a(X) S 0 by (i), and a(X(t)) 0. By the previous paragraph we see that x E F(y). Hence Td(x, y) :!!g it/2 by (1) of (3.6.28) below. This completes the proof of (3). We prove (4). We use the notation and results of (3). Since x s y we have

(i) AXc AY and AXc AY, and from (i) we conclude that (ii) AX

As,. with equality if and only if AX= Ay and AX= AY.

Finally, if equality holds in (ii), then (Gx)o = (GY) by (2.20.17), and 9'(x) = '(y) contains the unique minimal Tits geodesic from x to y by (3).

It follows from (ii) above and (3) of (2.21.3) that r(y) < r(x), and if r(y) = r(x) then equality holds in (ii). Similarly, it follows from (ii) above and (3) of (2.21.8) that s(y) < s(x), and if s(y) = s(x), then equality holds in (ii). One can also see that if s(y) = s(x), then equality holds in (ii) by (3.6.18) or the proof of that result. This completes the proof of (4).

(5) Let p be any point of M, and let z be any point of R(o) n F(ypxXc) such that x E '(z). By proposition (2.17.21) and definition (2.12.1) we see that '(z) c F(yyXXoc). If y E WX n F(y,,xXoc) and 0 E

K0(p, x) are arbitrary, then (G,( Y))(, = 4(G).4' c 4 (GX)o - 0-' (Gx)o, and hence Ko(p, x){Wx n F(ypxXo)} c W. It remains only to prove that WX c K0(p, x)(Wx n W(z)), and this proof will be given in the appendix. This completes the proof of (3.6.21). 0

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193

Tits buildings in M(ao) We now give a brief description of the Tits building associated to a symmetric space of noncompact type and rank k z 2. See also [Most, pp. 120-124] and, of course, [T2]. Tits shows that M is determined by its Tits building together with the partial ordering defined above in (3.6.15). See corollary 16.2 of [Mos2, p. 1231 for a precise statement. We shall discuss Tits buildings again later in the chapter, and in chapter 8 when

we discuss the Mostow rigidity theorem and its generalization by Gromov.

As usual, let M denote a symmetric space of noncompact type and rank k >_ 2. Let G =I(M), a semisimple Lie group with trivial center and no compact factors. 3.6.22. DEFINITION. Let 9(G) = {(GX)(,: x E M(x)). 9(G) is called the Tits building associated to M or to G. 3.6.23. DEFINITION. Let F be any k -flat in M. Let AF, G) = ((GX)0: x E

F(x)). The set 9(F, G) is called an apartment in 9(G). By proposition (2.17.22) each apartment 9(F, G) is a finite set. The minimal parabolic subgroups (GX ),,, x c- R(x) n F(c), are called cham-

bers in 9(F,G). A parabolic subgroup (GX),, with x E F(x) - R(x) is called a face in 9(F, G). 3.6.24. DEFINITION. Let M he a symmetric space of noncompact type and rank k >- 2. Let x E M(x) be any point. We define

'(x) = {y e M(x): (Gx)o =

If x E R(x), then the set g'(x) is precisely the Weyl chamber in M(x) determined by x (see (2.17.20)). If x E M(x) - R(w), then we call W(x) a face or Weyl face. REMARKS. (1) We recall from (3.6.15) that by definition x 5 y if (G.,)o

(GY)0. We introduce the equivalence relation - in M(x) defining x -y if '(x) _'(y) for points x, y E M(x). Equivalently, x - y if and only if (GX) = (Gy)(, if and only if x 5y and y Sx. We let .Y7M denote the quotient space M(x)/ - and let [ ]: M(x) -*.fTM denote the projection map. The quotient space .9M can obviously be identified with the Tits building 9(G) by means of the map [x] - (GX)o, where x is an arbitrary point of M(x). If we define g[x] = [&x] for g e G, then this action of G on 9M is well defined and corresponds to the action of G on 9(G) by conjugation.

We may extend the Tits partial ordering in M(x) to 9M as follows: define [x]5 [y] for points x, y E M(co) if and only if x s y in M(oo). It is easy to see that this partial ordering in 9M is well defined.

194

Geometry of Nonpositively Curved Manifolds

(2) If z (=- R(x) is any point, then the closed Weyl chamber 1(z) c M(x) is a fundamental domain for the action of G =10(M) in M() by propositions (2.17.24) and (2.17.25). Hence many geometric properties of the Tits metric Td on M(x) can be studied by considering a single closed Weyl chamber W(z).

(3) If x is any point of M(x), then the closure of '(x) in the Td-topology equals the closure of '(x) in the cone topology. If a sequence (x,,) c M(c) converges to a point x E M(x) in the Td-topology,

then it converges to x in the cone topology also since 4p (xx) < Td(x,,, x) for any point p E M by the definition of the Tits metric Td. Hence the Td-closure of W(x) is contained in the cone closure of F(x). Conversely, if a point y lies in the cone closure of '(x), then (GY)o

(G.,)o since G acts continuously on M(c) with respect to the cone topology. Let (x,,) c F(x) be a sequence that converges to y in the cone topology. Since (G.,)o acts transitively on M by (2.17.1) and (GG)0 = (Gx)o c (G,,)o for every n, it follows that for each n the function

p - 4o (x,,, y) is constant in M. Therefore Td(x,,, y) _y (x,,, ),) for every n and every p c- M by (3.1.2) and (2) of (3.4.3). We conclude that converges to y in the Td-topology, which proves that the Td-closure and the cone closure of W(x) are equal. 3.6.25. DEFINITION. Let M be a symmetric space of noncompact type and

rank k z 2. Let x and y be arbitrary points of M(a). Then '(x) is said to

be a face of '(y) if '(x) c W(y ), where W(y) denotes the closure of ma(y) in the cone or Td topology of M(x).

The justification for this definition is given in proposition (3.6.31) below and in the discussion of Coxeter complexes in a that precedes that result. 3.6.26. PROPOSITION. Let 11%1 be a symmetric space of noncompact type

and rank k;-> 2. Let x and y be points of M(x). Then the following conditions are equivalent.

(1) F(x) is a face of F(y).

(2) x Sy

(i.e., (GG)0 ? (G,,)(,).

(3) If F is any k -flat in M with y E F(x), then x c- F(x). (4) If y is any geodesic of M that represents y, then x E E(y)(x), where

E(y) is the intersection of all k -flats in M that contain y (see (2.20.11)).

(5) xEW(y). PROOF. We prove the equivalences by showing that (1) - (2), (2) - (4), (4)

(3), (3) - (2), (2) - (5), (5) - (2), and (2) - (1). We prove (1) - (2). Let F be any k-flat in M such that y E F(oc). If

z is any point in F(y) and p is any point of F, then y,, cE(yp,)_

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195

E(yy),) c F by proposition (2.20.16) since ypy c F. Hence z E F(m), and it follows that x E F(y) c F(-). By remark (3) following (3.6.24) we see that TT(x) n F(y) is nonempty for every positive number e. If e > 0 is sufficiently small and z is any point in T8(x) n W(y), then (Gy)p = (G2)0 C_ (Gx)o by proposition (3.6.11). Hence x _< y.

We prove (2) - (4). Suppose that (G,,)0 c (G.,)0, and let y be any

geodesic of M that represents y. If p = y(0), then yyx c E(yyx) c E(yy y) = E(y) by proposition (2.20.16), and hence x E E(y X°°).

We prove (4) - (3). Let F be any k-flat in M with y e F(oo), and let

p be any point of F. Then yyy c F, and hence E(yyy) c F, by the definition of E(yyy) in (2.20.11). It follows by the hypothesis of (4) that x E E(yyyXoo) c F(co).

We prove (3) - (2). Let p be any point of M, and let F be any k-flat in M that contains yy y. By the hypothesis of (3) we see that x (=- F(oo) since y E F(oo). Hence yyx c F since p E F, and it follows that yyx c E(yyy) since F was arbitrary. We conclude from proposition (2.20.14) and (2.20.16) that (G.,)0 2 (GG)e.

We prove (2) - (5). Let p be any point of M, and let g = f + p be the corresponding Cartan decomposition. Let X, Y r= p be those elements such that dp(X) = y,,,(0) and dp(Y) = ypy(0). Since (G,,)o c (G.,)0 by hypothesis it follows from proposition (2.17.16) that F( yyy) c F(yyx)

and Z(Y) c Z(X). Hence [X, Y] = 0 since YE Z(Y). Let a c p be a maximal abelian subspace that contains both X and Y, and let A c a * be the corresponding set of roots. For each e > 0 let X(e) = X + eY, and let x(e) E M(oc) be the asymptote class of the

geodesic y(t) = e`zt"(p), where Z(e) =X(e)/IIX(e)Il. It suffices to prove that AX(C)= AY for all small e> 0, using the notation of lemma (2.20.17), for it will then follow by that result that x(e) E F(y) for all small e > 0 and hence x E '(y), completing the proof of (2) - (5). Since (G)),) c (G.,) it follows by (2.20.17) that we have the inclusion relations

(*)

AXc A*y,

AX- g; Ay,

Ay c Ax.

Choose s0>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

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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*
and this completes the proof of (2) - (1)

since x* E W(x) was

arbitrary.

The equivalence of conditions (4) and (5) of the result just proved may be restated as follows. 3.6.27. COROLLARY. Let M be a symmetric space of noncompact type and rank k >: 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

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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<1,j
Y = E;=, y;, a*. Note that a,(Y) > 0 for i= 1,2 and 1:5r< k; recall that a,(Y,) - a,(X) > 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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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 <1, j <

k - 1. In particular, (a , a2) >- 0. We prove (2). By proposition (3.6.11) we can choose e > 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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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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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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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. Let G=1004 ). Then for each integer j with 0 < j < k - 1

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the functions S and rad: M(oo) - (0, oo) are G-invariant and continuous in the cone topology.

We defer the proof momentarily in order to define the center of gravity function.

3.6.35. PROPOSITION. Let M be a symmetric space of noncompact type

and rank k >_ 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 -'

y, and let

F be a k-flat in M such that z* e F(x). By (2) of (3.6.42) below it

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follows that x and y also lie in F(oo). 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 such that exp(a X p) = F, and let A c a* be the corresponding set of roots. We adopt the notation of (2.20.17).

Let X, Y, and Z* be the unit vectors in a such that yX(cc) =x, yr(") =y, and y,. (x) = z*. By (2.20.17) we have (a) AXC AZ. and AXC A;. ,

(b) Ayc Az. and Ayc AZ. ZFor t e [0,1} define X(t) = (1 - t)X + tY. We assert that
To prove (c) let a r= AXC AZ.. Then a(Y) >_ 0 since Ay C Az. by (b). Hence a(X(t)) > 0 for 0 < t < 1, which proves the first assertion of (c). The second assertion of (c) follows similarly. If X*(t) =X(t)/I X(t)I and a.*(t) = yx.(,)(x), then by the discussion

prior to (3.6.1) o,* is a reparametrization of the unique geodesic 0 from x to y. The uniqueness of a follows from the fact that Td(x, y) < Tr. If z is an interior point of v, then z = a.*(t) for some t E (0, 1), and it follows from (c) above and (2.20.17) that x< z and y < z. Hence .1 Y.

To prove the reverse inclusion we consider an arbitrary nonzero element A of R x n 4l Y. Then for any t e R, e`A fixes both x and y and

hence fixes any point of the unique geodesic a from x to y. In particular, e`A fixes the given interior point z of a for all t, and we conclude that A E q.. Hence it. 2 cl x n g y, which completes the proof of (2).

We now extend the previous result to show that any two incident points belong to a smallest face or chamber. 3.6.40. PROPOSITION. Let x, y, and z be arbitrary points in M(oc). Then the following are equivalent.

,=i1Z. (2) z E lub{x, y}. (3) There exists a unique minimizing Tits geodesic xy from x toy whose interior Int(xy) is contained in W(z).

(1) clxr

(

3.6.41. COROLLARY. If x and y are incident points in M(x), then the intersection of all chambers or faces '(z*) that contain (x, y) is itself a chamber or face K(z ), where z E lub(x, y) is any interior point of the unique Tits geodesic from x to y.

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Geometry of Nonpositively Curved Manifolds

The corollary is an immediate consequence of (3.6.26), (3.6.39), and (3.6.40). We begin the proof of (3.6.40). We prove the assertions in the cyclic order (1) - (3), (3) (2), and (2) - (1). We prove (1) (3). If g x n g = q then z z x and z >_ y and x and y are incident. By (3.6.39) there exists a unique minimizing Tits geodesic o, from x to y, and if z * is an interior point of v, then q 2. = g x n q Y = qz. Hence z* E W(z), which proves (3). We prove (3) (2). If (3) holds, then x c= '(z) and y c- K(z) or equivalently x 5 z and y 5 z by (3.6.26). Hence z is an upper bound for {x, y}. Now let z* be any upper bound for {x, y}. It follows from (3.6.26) that both x and y lie in W(z*) since x:5 z* and y _
any interior point of xy then z' E W(z*) and z'
We prove (2) - (1). From (2) we know that x:5 z and y:5 z or, equivalently, that both x and y lie in '(z) by (3.6.26). By (3.6.39) we know that gx n q,, = gZ., where z' is an interior point of the unique Tits geodesic xy from x to y. The point z' is an upper bound for (x, y) since

cjz,cgx and gZ.cgY. Hence z_z' or ij,_ q,. =q,r)q,, by (2). On

the other hand, q: c q x n qy since z is an upper bound for (x, y). We conclude that gZ = qX n gy which proves that (2) - (1) and completes the proof of (3.6.40).

It will be useful to generalize the results above to arbitrary subsets S of M(x) that have an upper bound.

3.6.42. LEMMA. Let S c M(x) be a set with an upper bound z E M(x). Then:

(1) For any two points x, y E S the function p - 4: (x, y) is constant in M and equals a < ir. In particular, Td(x, y) = a. (2) If F is any k -fiat in M with z (=- F(x), then S c F(x).

PROOF. (1) If x and y are points of S, then x 5 z and y 5 z. Hence x and y are incident and both lie in '(z) by (3.6.26). Assertion (1) now follows from assertion (1) of (3.6.39).

(2) Let F be a k-flat in M such that x E F(x), and let p be a point of F. Let q = f + p be the Cartan decomposition determined by p, and let a c p be the maximal abelian subspace such that exp(a)(p) = F. Let x

be any point of S, and let X be the unit vector in p such

that yy(x) = x in the notation of (2.20.17). Let Z be the unit vector in a such that yz(x) =z. We have x E'(z) by (3.6.26) since x s z. Hence X E '(Z)c a by (3.6.29) and (3.6.31). It follows that x e F(x) since

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3.6.43. PROPOSrnoN. Let S c M(x) be a set with an upper bound. Then S has a least upper bound. Moreover, the following are equivalent for a point z E M(oo).

(1) z E lub(S).

(2) gz= nxesgx PROOF. The set n, E s g,r is nonempty since it contains g z. for any upper bound z * of S. We show first that nx E s q x = q z for some z r= M(x). Choose points x,, ... , xN E S such that nx E s g x = n N i . x, We prove by induction on N that nl , , g x, = g . for some z e M(x). If N = 2, we are done by (3.6.39), so we assume that N >_ 3. By (3.6.39) there exists z* E M(x) such that gx. = gx, n gx`. Any upper bound z* for the subset {x,, x,, ... , XN) of S is also an upper bound for the set (x*,x3,...,xN), and hence by applying the induction hypothesis to the set (x*, x3,... _xN) we conclude that fl" , c = gx n (fl" 3 9x) = 9z for some z E M(x). We show that (2) - (1). If g z = fl x E s g x for some z E M(x), then g z c (q x or, equivalently, z >_ x for all x E S. Thus z is an upper bound for S. If z* is any upper bound for S, then gz c nx, s gx = (jz and hence z* z z. Therefore z E lub(S). Finally, we prove (1) - (2) of (3.6.43). Let z be a point of M(x) such

that z E lub(S). The earlier part of this proof shows that there exists z' E M(x) such that g z. = nx E s g x and z' E lub(S). Since z and z' are both least upper bounds of S we conclude that W(z) = '(z') and hence

gz=g, =

proves (1)=* (2).

O

Relationship between the geometry of M and the Tits partial ordering in M(x) In the remainder of (3.6) we show that several geometric properties of M can be recovered from properties of the Tits partial ordering in M(co). These properties include the degree of singularity of a point x E M(), the rank of M, and the de Rham decomposition of M. Before we consider these properties we show that the Tits metric and the Tits topology (the .topology of the Tits metric) determine the Tits partial ordering in M(x). Although this is logically redundant we include an argument that the Tits metric determines the Tits partial ordering since this argument is simpler than the corresponding argument for the Tits topology.

Tits metric determines Tits partial ordering We show how one may recover the Tits partial ordering from the Tits metric on M(x). See also appendix 5 of [BGS] for some discussion of

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the relationship between the Tits metric and the Tits partial ordering in the case that M has rank 2. I am grateful to Jens Heber who suggested the approach used here. 3.6.44. DEFINITION. An r-spherical cap of radius a in M(x) is a subset S of M(x) equipped with the Tits metric Td such that there exists an isometry e}, where S' denotes the stanf : (S, Td) -> S'(e, x(,) = {x e S': 4(x, dard r-sphere of radius I and 4 denotes the standard angle metric on Sr.

A closed r-spherical cap of radius e in M(x) is defined in similar fashion using weak inequality in the definition above.

The center of an r-spherical cap S of radius e < it is the unique point x* in S such that S = {x (=- S: Td(x, x*) < e}. If f: (S, Td) -* Sr(__, x0) is any isometry, then x* =f- '(x,,) since x0 is the unique center of Sr(__, xo ). A closed r-spherical cap of radius e = it is called an r-sphere. If F is an (r + 1)-flat in M, then F(x) is an r-sphere in M(x). One of the key results of this section (corollary (3.6.49)) is a proof of the converse for (k - 1)-spheres, where k is the rank of M: every (k - 1)-sphere in M(x) equals F(x) for some k-flat f of M. In the discussion below M or M * will always denote a symmetric space of noncompact type and rank k > 2. We begin with some elementary facts about spherical caps. 3.6.45. LEMMA. Let S c M(x) be a subset that is both an r-spherical cap of radius e and an r*-spherical cap of radius a*. Then r = r* and e = e*.

PROOF. Let f:(S, isometries, and define g =f

f-

Td) :

S'*(e', x0*) be

Sr(__, xo) -> S''(e *, x* ). The map g

is a bijection that preserves the angle metrics, and hence g is a homeomorphism, which shows that r = r*. Moreover, g is the restriction of an element of 0(r), the group of differentiable isometries of Sr, since the domain of g contains an open subset of Sr. It follows that

e=e*. o 3.6.46. LEMMA. Let A and B be r-spherical caps of radius a such that A C_ B. Then A = B.

PROOF. As above let

and f*:(B,Td)->S'(e,.r0)

be isometries, and define g =f f- ': S'(e, x0) -, S'(e, x* ). The map g preserves angle metrics and is the restriction of an element of O(r) as

above. It follows that g is surjective since S'(e, x*) and its subset g(S'(e,x0)) are closed subsets of S' with the same volume. The surjectivity of g implies that A = B. 0

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The next result shows that the rank of M can be detected from information provided by the Tits metric.

3.6.47. LEMMA. Let S be an r-spherical cap of radius e, where r is maximal. Then r = k - 1, and S n R(x) is dense in S in the Td-topology. PROOF. A point x in S will be called locally maximal in S if there exists

S > 0 such that [x] = [y] for all y c- S with Td(x,y) < S. We recall by (3.6.11) that there exists S > 0 such that x < y for all y E M(x) with Td(x, Y) < S. Recall also by (3.6.17) that every chain in the Tits partial ordering has length at most k - 1. It follows easily from these two facts that So = (x c= S: x is locally maximal in S) is a dense subset of S in the Td-topology.

We show that So c R(x) and r = k - 1, which will complete the proof. Let x E S. and p c- M be given, and let y be the geodesic y,,1. Let E(y)

denote the intersection of all k-flats in M that contain y. If s(x) denotes the degree of singularity of x, then by (2.21.8), E(y) is an r*-flat, where r* = e(x) = k - s(x). Moreover, by (2) of (2.21.8) we see that r* < k with equality if and only if x c= R(x).

By the definition of So there exists 8 > 0 such that if y E S with Td(x, y) <8 then [x] _ [y]. By (4) of (3.6.26) it follows that TT(x) n S c E(y)(x), where TS(x) _ (y E M(x): Td(x, y) < 8). However, E(y)(o) is

an (r* - 1)-sphere, and TS(x) n S is an r-spherical cap of radius S. Hence r:5 r* - 1 < k - 1. Since every k-flat F determines a (k - 1)sphere it follows from the maximality of r that r = r* - 1 = k - 1, and hence x E R(x) since r* = k. 3.6.48. LEMMA. Let S be a (k - 1)-sphere in M(x), and let x E S n R(x). Let yo be the unique point in S with Td(xo, yo) = 7r. Then: (1) There exists a geodesic y of M with y(x) = xo and y(- x) = y0. (2) There exists a unique k-flat F in M that contains y, and F(x) = S.

PROOF. (1) The existence of y follows from (3.6.1). The geodesic y is regular by (2.17.17) since x0 = y(x) E R(x), and hence there is a unique k-flat that contains y by (2.11.1). By applying (3.6.46) with e= 7r it suffices to show that S c F(x) since both S and F(x) are (k - 1)-spheres. Since F is the unique k-flat containing y it follows from (3.6.2) and its [0, 1] --. M(x) proof that F(x) contains any minimal Tits geodesic such that Q(0) =xo and =yo. Every point of the (k - 1)-sphere S lies on such a minimal Tits geodesic o,, and hence S c F(a).

3.6.49. COROLLARY. Every (k - 1)-sphere S in A (x) has the form S = F(x) for some k -flat in M.

PROOF. If S is a (k - 1)-sphere in M(x), then S n R(x) is nonempty by (3.6.47). The result follows immediately from (3.6.48).

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The next result has an analogue for closed Weyl chambers as we shall see below in the corollary to (3.6.55). 3.6.50. COROLLARY. Let x and y be arbitrary points of *c). Then x <- y if and only if every (k - 1)-sphere S in M(cc) that contains y also contains x.

PROOF. In view (3.6.49) this is a restatement of (3) of (3.6.26). 3.6.51. COROLLARY. Let ¢: M(x) -* M *(co) be a bijection that preserves Tits distance. Then 0 preserves the Tits partial order.

PROOF. This is an immediate consequence of (3.6.50) since 0 sends (k - 1)-spheres in M(x) into (k - 1)-spheres in M*(x). We conclude with a description in terms of (k - 1)-spheres in M(x) of a closed Weyl chamber or face W(x) in M(x). 3.6.52. PROPOSITION. Let x be any point of M(x), and let Sx denote the intersection of all (k - 1)-spheres in M(x) that contain x. Then Sx = (y e M(x): y <x} = W(x), the closure of 1(x) in the Td-topology. Moreover, for small positive numbers e, T,(x) n Sx = T,(x) n W(x) is an r-spherical cap of radius e, where r = e(x) - 1.

PROOF. By (3.6.49) every (k - 1)-sphere in M(x) equals F(m) for some k-flat F in M. From the equivalence of (2), (3) and (5) of (3.6.26) we conclude that

(1) Sx={yEM(x):y<x}=W(x). By (3.6.11) we may let e be a positive number such that x 5y for all y E TT(x). We show that T(x) n Sx = T6(x) n W(x) is an r-spherical cap of radius e, where r = e(x) - 1. Let y be any geodesic with y(w) = x. We assert that (2) T8(x) n Sx = T8(x) n '(x) = T(x) n E(y)(x). Assume for the moment that (2) has been proved. The manifold E(y) is flat and totally geodesic with dimension r = e(x) by (1) of (2.21.8), and hence Te(x) n E(yXco) is an (e(x) - 1)-spherical cap of radius e. This completes the proof of (3.6.52). We prove (2). By the definition of Sx we obtain

(2a) Sx = fl

,+i E(ypx X x) S E(y Xx).

If y is any point of Sx, then y :::-:x by (1) above. On the other hand, if y E TR(x) then x _<;y by the choice of e. We conclude that

(2b) T8(x)nSxg'(x).

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From (1), (2a), and (2b) we obtain

(2e) T8(x) n S, = TT(x) n r(x) e TE(x) n E(yX0).

However, Te(x) n E(yxco) S Te(x) n W(x) by (2) of (3.6.28) and its proof. This completes the proof of (2).

Tits topology determines Tits partial ordering By working somewhat harder than in the previous section one may

recover the Tits partial ordering from the Tits topology alone. We outline this process.

STEP 1. A point x E M(oo) lies in R(te) if and only if x has a compact neighborhood in M(co) with respect to the Td-topology. STEP 2. Let x (=- Moo) be given. Then W(x) is the connected component of Moo) that contains x with respect to the Td-topology. 3.6.53. DEFINITION. A closed Weyl chamber in M(c) is a set of the form

'(x), where x (=- Moo) and '(x) denotes the closure of F(x) in M(me) in the Td-topology.

STEP 3. For each x E M(me) the set '(x) is the intersection of all closed Weyl chambers containing x. In particular, for arbitrary points x, y (-= M(me) one concludes from (3.6.26) that x < y if and only if every closed Weyl chamber that contains y also contains x. We now fill in the details of this outline. 3.6.54. PROPOSITION. Let M be a symmetric space of noncompact type and rank k > 2. A point x e-= M(co) lies in Moo) if and only if x has a compact neighborhood in M(cc) with respect to the Td-topology.

PROOF. Let x r=- Moo) be given. Let p be any point of M, and let F be the k-flat containing yP.,(Il8). By corollaries (3.6.12) and (3.6.13) there exists e > 0 such that Be(x) = {y E M(a): Td(x, y) s e} g W(x) n F(x). The set Be(x) is clearly a closed neighborhood of x in the Td-topology, and it suffices to show that B3(x) is sequentially compact with respect to Td. Let (yn) c BB(x) be any sequence, and let (ynk) be a subsequence converging to a point y E M(oo) with respect to the sphere topology. By

(7) of (3.4.3) the sphere and Td-topologies coincide in F(x). Hence (y,,,) -+y E B8(x) in the Td-topology, and B8(x) is sequentially compact with respect to Td.

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Conversely, let x E M(x) - R(x) be given. We show that x has no compact neighborhood in M(x) in the Td-topology, which will complete

the proof of (3.6.53). It suffices to show that for every e > 0 the set Br(x) = (y E M(x): Td(x, y) < e} contains a sequence (z,,) with no subsequence convergent in the Td-topology. Let e > 0 be given. Without loss of generality we may assume by (3.6.11) that x s y for all y E BE(x), making e smaller if necessary. Let

p be any point in M, and let F be any k-flat containing the geodesic yPJR). Choose z E F(x) r) R(x) such that Td(x, z) = < (x, z) < e. By the choice of e and the fact that x e R(x) it follows that x < z and hence G1 G.. Let X be a unit vector in O s - g .. We For every positive integer n define g = exp(-LX) and z assert that {z,,} is a sequence in BF(x) that has no subsequence converging in the Td-topology. We observe that

for every n, follows from (2.17.26) that c Be(x). Since g - 1 it and hence has a subsez e F(x) for all sufficiently large n. Suppose that z* in the Td-topology. that converges to a point quence (z*) Then z = z* since z - z in the sphere topology and p (z*, z*) s Td(z*, z*) - 0. On the other hand, by corollaries (3.6.12) and (3.6.13) we can choose S > 0 such that TS(z) C W(z) n F(m) since z E R(x) ( F(x). This implies that Td(z**, z*) = Td(z*, z) >- S for large n since z* e- F(x) for large n. This contradiction shows that {z,} has no Tdconvergent subsequence and completes the proof of (3.6.54). 0 The result just proved completes step 1 of the outline above. Step 2 says that for x r= R(x) the Weyl chamber ?(x) is the connected component of R(x) that contains x; this is a special case of (3.6.32).

To complete step 3 we need the following result. Compare (3.6.53). 3.6.55. PROPOSITION. Let M be a symmetric space of noncompact type and rank k >t 2. If x is any point of M(x), then W(x) is the intersection of all closed Weyl chambers in M(x) that contain x.

As a consequence of this result we obtain the following analogue of (3.6.26) and (3.6.50).

COROLLARY. Let M be a symmetric space of noncompact type and rank k >- 2. If x and y are points in M(x), then x < y if and only if every closed Weyl chamber in M(x) that contains y also contains x. PROOF OF THE COROLLARY. Let x s y. If y E W(z) for some z E R(x),

then x E F(y)9 '(z) by the equivalence of (1), (2), and (5) of (3.6.26).

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Hence every closed Weyl chamber that contains y contains x. Conversely, if every closed Weyl chamber that contains y also contains x, then by (3.6.55) it follows that W(x) c'(y). Hence x 5 y by the equivalence of (1) and (2) in (3.6.26).

PROOF OF PROPOSITION (3.6.55). Let x E M(x) be any point, and let 6'(z ), z E R(co), be a closed Weyl chamber with x E W(z). then F(x)c W(z) by the equivalence of (1) and (5) of (3.6.26), and it follows

that 9'(x) is contained in the intersection of all closed Weyl chambers that contain x. Conversely, let y be a point of this intersection. We must show that y E'(x) or, equivalently, that y <x by (3.6.26). Let F be any k-flat of M with x E F(x). By (3.6.31) we may choose z E R(x) n F(x) such that x E ?;'(z). By the hypothesis on y and (3) of

(3.6.28) we conclude that y E'(z) c F(x). Hence y < x by (3) of (3.6.26).

Tits partial ordering determines de Rham decomposition Some geometric properties of a symmetric space M of noncompact type and rank k z 2 can be obtained from information provided by the Tits partial ordering. 3.6.56. PROPOSITION. The Tits partial ordering determines the rank of M and the integers s(x), and e(x) for a point x E M(x).

PROOF. We recall from (2.21.7) that if x is a point in M(cc) and y is any geodesic of M that belongs to x, then s(x) and e(x) are respectively the

dimensions of the noncompact type symmetric space factor and the Euclidean factor of F(y), the totally geodesic submanifold of M that is the union of all geodesics in M that are parallel to y. By lemma (3.6.18) and proposition (3.6.17), s(x) is the length of a maximal chain that begins at x, where chain is defined in (3.6.16). Hence s(x) is determined by the partial ordering. By (2.21.8) the integer s(x) has k - 1 as its largest value, where k = rank M. Hence the rank of M is determined by the partial ordering since it is one more than the

length of the longest chain in M(x). Finally, e(x) is determined by the partial ordering since s(x) + e(x) = k for every x e M(x) by (2.21.8).

The main effort of this section will be spent proving that the Tits partial ordering in M(x) determines the de Rham decomposition of M. The precise result is stated below as proposition (3.6.65). To motivate this result and to show the necessity of the conditions that occur in its statement we begin with a discussion of geodesics in a product manifold M and stability groups G, where G =10(M) and x e M(x).

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Let M be a Riemannian product M = M, x ... X Mk, where M and each factor M, is a symmetric space of noncompact type. If G = Io(M) x Gkand G; = J( f ), then G = G, x 3.6.57. PROPOSITION. Let y(t) be a unit speed geodesic in M. (1) There exist unit speed geodesics y; and numbers a; >_ 0 for 1 S i < k

such that E;`_ a? = 1 and

y(t) _ (y,(a,t),...,Yk(akt)) for all t (=- R. The geodesic y; is uniquely determined whenever a; > 0. (2) Let x = y(oc) and define x; = y,(oc) E M;(oc) whenever a; > 0. Then

Gz= n Gs,=H, x ... xHk, a,> 0

where

H;=G;=10(M;) =(G;)x

if a;=0, if a;>0.

REMARK. We recall that Mi(n) can be regarded as a subset of P (c) in a natural way so that G,,makes sense. PROOF. Let a;: M -* M; be the natural projection for each i,and define y*(t) = (vi o yXt) for a given unit speed geodesic y(t) in M. If V and V denote the Levi-Civita connections in M and M;, then 0 = V,.y' _ Ek= , V ,y*', and hence each curve y;* is a geodesic in M; with speed a; z 0. Clearly, E;`,. , a,2 = 1, and y* has a unique unit speed reparametrization y; if a; > 0. This proves (1). ¢k) be an element of G = G, x ... x We prove (2). Let 0 Gk. From (1) we obtain

(a) (0 ° yXt) _ ((4)I ° YlXa,t),...,ON °YkX akt)) for all t E R. From this it follows that

(b) d((4); c y;Xa;t), y;(a;t)) _ 0, for some C > 0. Hence from (b) we see that 0 E Gs if and only if ¢ E GX and ¢; E (G,),, whenever a; > 0. If a; > 0 it is clear that GG. = H, x X Hk, where

H1= G1 =10(Mj) if i:# i and H; = (G1). The assertions of (2) now follow.

3.6.58. COROLLARY. Let y(t) be a unit speed geodesic in M, and write y(t) = (y,(a, t ), ... , yk (ak t )) as above. Let Moo), and R; (w) denote re-

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spectively the regular points in M(x) and Mi(x). Let x = y(x) E M(x). Then x E R(x) if and only if

(1) a.>0 for every i, 1
We shall need the following result for the proof of this corollary. 3.6.59. LEMMA. Let S be a subset of { 1, ... , k}. For each integer i E S let x; E Mi(x) be assigned arbitrarily. Then there exists a point x E M(x) such

thatGx= fiEsG, =H, x

ifieS.

XHk,where Hi=(Gi),' ifiESand Hi =Gi

PROOF OF LEMMA (3.6.59). Let S and {xi} be given as above. Choose positive numbers a,, i E S, such that Ei E s a, = 1 . Let p = (p 1, ... , Pd

be any point_ of M, and for each i E S let yi(t) be the unit speed geodesic of M; such that 7ri(y(t)) = yi(ait) if i E S and iri(y(t)) =pi if i 44 S, where Tr,: M -+ M; is the projection. If x = y(x), then it is routine to check that x satisfies the assertions of the lemma. PROOF OF COROLLARY (3.6.58). We suppose first that x E R(x). Then x

is maximal in the Tits partial ordering of M(x) or, equivalently, Gx is minimal among stability subgroups {Gy: y c= M(x)). It now follows from

(2) of (3.6.57) and from lemma (3.6.59) that a, > 0 and x, E Ri(x) for every i. Conversely, suppose that a; > 0 and x; E R1(x) for every i. Then G, is minimal among the stability subgroups {Gy: y c=- M(x)} by (2) of (3.6.57) and lemma (3.6.59). Hence x E R(x).

The next result is convenient to prove now although it will not be used until later and does not fit exactly into the theme of this section. 3.6.60. COROLLARY. Let M = M, x ... x Mk be a Riemannian product of symmetric spaces M; of noncompact type. Then there exists a (naturally defined) surjective, continuous, open map (A: R(x) -, R1(x) x - x Rk(x),

where R(x) and Ri(x) denote the regular points in M(x) and Mi(x) and Ri(oo) x x Rk(x) is given the product topology. Moreover, if 4)(x) _ 4)(y) then Gx = Gy.

PROOF. Let x r= R(x) be given, and let y(t) be a unit speed geodesic of M such that y(x) = x. As in (1) of (3.6.57) we write y(t) = (y1(a1t)...... k(akt)) where a; > 0 for every i by (3.6.58) and yi is a

unit speed geodesic of M. for every i. Let x; = yi(x) E Mi(x). Then x; E Ri(x) for every i by (3.6.58). We define

¢(x) _ (x1,..., xk) E R10) x ... X Rk(x).

It is routine to check that ¢ does not depend on the choice of the geodesic y that represents x (in fact, even the numbers a; above are

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independent of the choice of y). It is also routine to check that 0 is open and continuous. The map 46 is surjective by (3.6.59). The final assertion of (3.6.60) follows from the definition of 0 and (2) of (3.6.57).

The next result gives a useful sharpening of (2) of (3.6.57) for points x c- R(x). 3.6.61. PROPOSITION. Let 4,;: R(x) -' R;(x) be defined by 4,; = pi o 46,

where ¢: R(x) - R,(x) x

x Rk(x) is the map defined in (3.6.60) and X Rk(x) -+ R;(x) is the projection for 1 5 i:5 k. Then: for every xER(x). (2) If g x = nka , q y for x E R(x) and y; E Mi(x), then y; E R,(x) and

p;: R,(x) X (1) Gx

n=,Go(x)

(I 4,x) = gy for every i.

PROOF. Assertion (1) follows from the definitions of 0 and 0,, and from (2) of (3.6.57). We prove (2). By (3.6.59) we may choose y E M(x) such that G).= n;`= , G,. Hence ci v = w- , g y, = s x by the hypothesis of (2). We conclude that (G.,)o = (GG,)0, and hence y E R(x) since x e Moo). It follows from (2) of (3.6.58) that y; E R(x) for 1 < i < k. It follows from (3) of (2.17.25) that Gx = G,, and hence Gs. = Gy, for 1 S i:5 k by (2) of (3.6.57). We conclude that g x = q y, for 1 5 i s k. THE SPLITTING CRITERION. We now present a result whose converse gives a description of the de Rham decomposition of M in terms of the

Tits partial ordering on M(x). 3.6.62. PROPOSITION. Let M = Ml X ... X Mk be a Riemannian yroduct of symmetric spaces M; of noncompact type. Let A; = M;(x) c M(x) for

1 5 i 5 k. Then the subsets {A,,..., Ad of M(x) satisfy the following properties: (1) The sets {A;} are disjoint. (2) If x, is any point of A; and if y is any point of R-00 with y 5 xi, then

y EA;. (3) Let S be any subset of { 1, ... , k) with at least two elements. For each

i r=- S choose x; E A, arbitrarily. Then there exists a point z in M(x) - (Uk_ , A;) that is an upper bound for the set {xi: i E S}. (4) Let z be a n y point o f M(x) - (U _ , A; ). Then there exists a subset S of { 1, ... , k} with at least two elements and points x; E A; for each i E S such that z is a least upper bound for the set (x;: i (=- S}.

(a) If z E=-R(x) is an upper bound for a set (x;: i e S) as in (3), then z must necessarily lie in M(x) - (Uk=, Ai) if properties (1) and (2) above hold. This will be shown in the proof of (3.6.62). In 3.6.63.

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view of (3.6.43) we could also have required z to be a least upper bound of (x1: i E S) in (3). (b) The result above and its converse may be regarded as a reworking

of the discussion of semisimple groups of the form G = GI x .. x Gk found in the first part of section 18 of [Mos2]. PROOF OF PROPOSITION (3.6.62). The elements of -A, are asymptote

classes of geodesics of M tangent to the ith factor M. of M. Hence the sets (A;} are disjoint. We prove (2). Let x; (=- A; be any point for I < i < k, and let y E M(x) be a point such that y <_xi. Let y(t) be a unit speed geodesic such that y(x) =y. By (1) of (3.6.57) we may write y(t) = (y,(a, t ), ... , yk(ak 0, where y, is a unit speed geodesic of M1 whenever a1 > 0, and E; _ , a1Z = 1 with a1 >- 0 for all I. From (2) of (3.6.57) we see that Gx = HI X X Hk, where H1 = G1 =10(M1) if j # i and H; _ (G,),,. Since y -x; we know that (G).)), Z) (G.,)0, and hence a1 = 0 for j

i by (2) of (3.6.57).

This shows that y E M,(x) =A i. We prove (3). Let S be any subset of { 1, ... , k} with at least two elements, and let x, be an arbitrarily given point of A. for each i E S. By (3.6.59) we can find an element z EM(x) such that G,= fl ; E . G,Y,, and hence 0: = n;,_ s 0 z, . By (3.6.43) it follows that z is the least upper bound of (x,: i E S). It remains only to show that z E=-M(x) - (Uk_ I A;)

(cf. remark (a) of (3.6.63)). By hypothesis we can find at least two distinct integers in S. Suppose that z lies in A, for some 1 <- r:5 k and choose an integer i in S distinct from r. Since x; < z it would follow from (2) that x; lies in A,, contradicting the disjointness of A; and A,.

Therefore z E M(x) - U. I A;. We prove (4). Let z E M(x) - (Uk_ I A;) be given arbitrarily, and let

y(t) be a unit speed geodesic of M such that y(a) = z. Let y(t) be written as in (1) of (3.6.57), and let S be the subset of { 1, ... , k) for which the numbers a; in (1) of (3.6.57) are positive. The set S has at least two elements since z does not lie in any of the sets A. = M;(x). By (2) of (3.6.57) there exists a point x; E M1(x) =A, for every i E S such

that G _ n,,-, G, . Hence c, = n,,= stt.r, and z is the least upper bound of the set (xi: i E S) by (3.6.43). This completes the proof of (4). 0 We now show that the converse of (3.6.62) describes the de Rham decomposition of M. We proceed by induction on k beginning with the case k = 2, which we restate for concreteness. This result is the ana-

logue of (5.6.2) below. Note that in the case k = 2 we can make condition (4) of (3.6.64) slightly weaker than the condition (4) of (3.6.62).

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3.6.64. PROPOSITION. Let M be a symmetric space of noncompact type and rank r z 2. Assume that M(cc) admits two subsets A, and A2 with the following properties:

(1) A, and A, are disjoint. (2) If x; is any point of A; for i = 1, 2 and if y is any point of M(oo) such that y - x;, then y E A;. (3) If x, E A, and x2 E A, are arbitrarily given points, then there exists a point z E M(oo) - (A, U A 2) such that z is an upper bound for {x,, x2}_

(4) If z E M(me) - (A, U A 2) is given arbitrarily, then there exist points x, E A, and x2 E A 2 such that z is an upper bound for {x,, x2}. Under these conditions M is isometric to a Riemannian product M ='Q-1 X M2, where M1(m) = A. for i = 1, 2.

PROOF. We shall apply proposition (5.6.1) below, and hence we need to establish the following two properties:

(a) Given a point p e M and a point x; E A; for i = 1, 2, the point yi = Sp(x;) = yp,, ,(- °°) E A;.

(b) Td(x,, x2) < 7r/2 for every x, E A,, x2 A2. The map SP that appears in (1) is the geodesic symmetry at p (cf. the discussion following (1.7.9)), and (a) asserts that the sets A, and A2 are involutive in the sense of (1.7.10). We begin the proof of (a) above.

LEMMA A. Let x and y be points of M(oo) such that 4p (x, y) = IT for some point p of M(oo). Let z be a point of M(c) such that g x n g y c g z c

a.. Then gZ=gx. PROOF. Let p c- M be a point such that 4p (x, y) _ IT, and let F be a

k-flat in M that contains the geodesic ypz(R). 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. By hypothesis, g, c g x or equivalently z z x, and hence x E'(z) by (3.6.26). Let X and Z be the unit vectors in p such that yX(oo) =x and yz(oo) = z in the notation of (2.20.17), which we adopt for the rest of the proof. We know that Z E a since z E F(oo), and it follows from (3.6.29) and (3.6.31) that X E

'(Z)ca since xEV(z). If Y is the unit vector in p with yy(x) = y, then Y = -X E a since 4p (x, y) = it by hypothesis. By (2.17.13) we know that gz = go +

E.(X)Z09. and gy=go+Ea(y)aoga. Hence (t) 9.1, ngy=90+EQEAX g.-

By hypothesis, g X n g, c g:. If a E AX, then g,, c g s n g c g, and hence a(Z) z 0 by (2.17.13). Similarly, if a E AX, then (- a) E AX and

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219

(-aXZ)> 0 by the argument above. This proves that (ii) AX c AZ.

On the other hand, x:5 z since g z c g x, and hence from (2.20.17) we obtain

(iii) AXC AZ and AXC AZ.

The inclusions in (iii) imply that

(iv) AX Q Az with equality if and only if AX= AZ and AX= A. From (ii) and (iv) and the equality assertion in (iv) we see that AX= Az and hence g x = g . by (2.20.17). This completes the proof of lemma A.

The next result is property (a) stated at the beginning of the proof. We recall from the discussion after (1.7.9) that for any point p of M -the geodesic symmetry SP: M --> M extends to a homeomorphism of M(me) given by Sp(x) = ypx( - x) for x E M(oo).

LEMMA B. Let points x, c- A, and x2 E A , and a point p E M be given arbitrarily. Then y 1 = Sp(x,) E A, and Y2 = Sp(x2) E A2.

PROOF. We verify the property above for A, only. Let p e M and x, E A, be given, and let y1 = Sp(x,) = ypx,(- oo). We assume that y1 0 A

and obtain a contradiction. Assuming that y, 0- A, we show that y, iZ A2 . If y 1 E A 2 then by condition (3) in the statement of (3.6.64) we can find an upper bound

- (A1 UA2) for the set {x1, y1}. By (3.6.39) there exists a number a with 0 < a < it such that 4 (x,, y1) = a for all q E M, but

z E=-,Woo)

this contradicts the hypothesis that
of (3.6.64) that we can find points x E A, and x2 E A2 such that y, is an upper bound for {x*, x2 }. By (3.6.43) and condition (3) of (3.6.64) there exists a least upper bound z for (x1, x*) and g, = g x, n c. Since x1 and y, zx* it follows that gyp c gx. n gx2, and hence qx, n y1

c

n in.the

._y91.

proof of (3) of (3.6.62) we note that z E M(z) - (A, UA2). As If z e A 1 then x* E A, by condition (2) of (3.6.64) since x* s z, but this

contradicts the disjointness of A, and A2- Therefore z e A 1, and a similar argument with x1 shows that z 1Z A,. By (3.6.39) there exists a unique minimizing Tits geodesic x1x* from

to x* whose interior Int(x1x*) is contained in '(z). We may therefore find a point z' in Int(x1x*) c V(z) such that Td(x,, z') is

x1

arbitrarily small. By (3.6.11) we know that x1 5 z' if Td(x1, z')

is

sufficiently small or, equivalently, g2. c gx,. Hence we can find z' E '(z)

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220

By lemma A above we conclude such that gx n gyp c g2 = gZ. c that g x = g, since 4 (x1, y,) _ IT, and this implies that z E A, by condition (2) of (3.6.64 since z Sx,. We obtain a contradiction to the fact that z E M(-) - (A, U A,). We conclude that our original assumption y, A, is invalid, which completes the proof of lemma B.

We now establish the condition (b) stated at the beginning of the proof. LEMMA C.

Td(x,,x2) < it/2 for all x, EA1, x, EA,.

PROOF. Given x, E A, and x2 E A2 we know by condition (3) of (3.6.64)

that there exists an upper bound z for (x1, x2). Hence (x,, x2) c '(z) by (3.6.26), and it follows from (1) of (3.6.28) that Td(x,, x2) < Ir/2. We complete the proof of (3.6.64). The conditions (a) and (b) verified

above imply by (5.6.1) below that M splits as a Riemannian product M1 X M2 such that A. c M;(c) for i = 1, 2. It remains only to prove that A; = M;(00) for i = 1, 2. We do this only for A, since the proof for A 2 is similar.

Let z be any point of M0) and suppose that z A,. Since z e A2 c MZ(x) we know by (4) of (3.6.64) that there exist points x, E A, and x2 E A, such that z >- x, and z > x,. Therefore q-' c- q -r2 = g, ®(g 2)xZ since Gx2 = G, X (G2)x, by (3.6.57). Here g; is the Lie algebra of G; =10(M1) for i= 1, 2. However, g, =(GI),, ®g2 since G, = (G1)Z X G2 by (3.6.57) and the fact that z E M1(oo). Therefore g, Z g xZ since g 2 is

strictly larger than (2)x, by (2.17.13). This contradiction shows that A 1 = M1(cc).

We now prove the full converse of (3.6.62). 3.6.65. PROPOSITION. Let M be a symmetric space of noncompact type and rank r > 2. Suppose for some integer k >: 2 that M(cc) admits k subsets

{A,,..., Ak} that satisfy properties (1) through (4) of (3.6.62). Then M splits as a Riemannian product M =,4I X ... X Mk such that M;(oo) =A j

fort
We proceed by induction on k. The previous result is the case k = 2.

Now let k >: 3 and suppose that the result is true for k - 1 sets satisfying the four properties of (3.6.62). Define B1 = A, and B2 = (x E M(oo) - B1: for every x 1 E A 1 there exists an upper bound in M(cc) for the set (x, x,)}. We note that B2 W- 2 A. by condition (3) of (3.6.62).

We verify that B, and B, satisfy the four conditions of (3.6.64). It

will then follow from (3.6.64) that M = M, X M*, where M1(oo) = B1 and

M * = B2. We then apply the induction hypothesis to M * and the sets

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(A,,..., A,,) in M*(x) to conclude that M* -M2 x

x Mk, where M;(x) =A i for i > 2. This will complete the proof of (3.6.65). It is clear from the definitions that B, and B2 are disjoint subsets of M(x). We verify condition (2) of (3.6.64). Let x, be a point of B, =A,,

and let y be any point of M(x) with y < x,. Then y E A, = B, by hypothesis (2) of (3.6.62). Next let x2 be any point of B2, and let y be any point of M(x) such that y <x2. If x, E B, is any point, then by the definition of B2 there exists an upper bound z for the set {x,, x,). But z is also an upper bound for the set (x,, y) since y s x,, and hence y lies in B, since the point x, E B, was chosen arbitrarily. We verify condition (3) of (3.6.64). If x, E B, and x2 E B2 are given arbitrarily, then the set (x,, x2) has an upper bound z by the definition

of B,. Property (2) in (3.6.64) of the sets B, and B, implies that z E M(x) - (B, U B2) for any upper bound z of {x,, x2}; see the proof of lemma B of (3.6.64) or the proof of (3) in (3.6.62). The verification of condition (3) is complete. We verify condition (4) of (3.6.64) for the sets B, and B,. Recall that U;..2 Ak c B2. Let z E M(x) - (B, U B2) c M(x) - Uk- i Ak be given.

By hypothesis (4) of (3.6.62) we know that there exists a subset S c {1,..., k} with at least two elements and elements {x; EA;: i (-= S} such that z is a least upper bound for {x;: i E S}. By (3.6.43) we know that q, = ni E s q x, We outline the remainder of the proof. We show first

that the integer 1 belongs to S so that n; E s q x, = q x, n (n, # ; E s q x ) We then use (3.6.43) to show that (l, f ; E s q x, = q y.: for some element y2 E M(x). The final step is to show that y2 belongs to B2. We then have q., = n; E .s q x, = 9 x, (1 q y.,, which implies that z is an upper bound for (x,, y2) and verifies condition (4) of (3.6.64).

We suppose that 1 does not belong to S; that is, S c (2,..., k), and obtain a contradiction. Choose y, E B, =A , arbitrarily. Let S' = {1} U S c {1,..., k), and let S* = {y,, x;: i E S). By condition

(3) of (3.6.62) we know that S* has a least upper bound z*, and by (3.6.43) we see that

n niEsgx,=gv,nq: Hence

z* is an upper bound for {y,, z}, and we conclude that z E B2 since y, E B, was arbitrary. This contradicts the fact that z E M(x) - (B, U B2) and shows that 1 belongs to S. Since z is an upper bound for (x;: i E S, i # 1) it follows from (3.6.43)

that there exists y2 E M(x) such that q y Z = (1, * Esc; x . Hence q z = n; E s qx, = q x, n q y,z. We conclude that z is an upper bound for {x1, y2} (in fact a least upper bound by (3.6.43)). It remains only to prove that y2 E B2 to verify that condition (4) of

(3.6.64) is satisfied. Note that y2 a B, by properties (1) and (2) of (3.6.62) since y2 z x; for some i E S with i >- 2. Let y, E B, be given arbitrarily. We must show that (yI, y2) has an upper bound in M(x). If

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Geometry of Nonpositively Curved Manifolds

S* = (y,, x;: i E S, i # 1), then by condition (3) of (3.6.62) we know that S* has a least upper bound z* in M(x), and by (3.6.43) it follows that ^ q,. = f x E s. = c1,', ' f n, * i E s ti, - q y, n q y,. Hence z * is an upper bound for (Y I I y2) and we conclude that y2 E B,. We now apply (3.6.64) to the sets B, and B2-in M(x) to conclude

that M splits as a Riemannian product M, X M*, where B, = M,(x) and B2 = M*(x). We observed earlier that A; c B, = M*(x) for i < 2 < k. We now verify that the sets {A2,..., Ak) verify the four conditions of (3.6.62) for M*(x) and then apply the induction hypothesis of (3.6.65). Clearly, the sets (A;: 2:5 i < k} are disjoint by condition (1) of (3.6.62).

We prove (2) of (3.6.62) for M*(x). Let x; EAR be given for some integer i with 2 5 i < k, and let y be a point of M*(x) such that y <x; in the partial ordering of M*(x). This means that g* - g* , where g* is the Lie algebra of G* =10(M*). It follows immediately from (3.6.57) that gy= g, ®gy andgs = g, ®g*;. Hence gy Q gz, and y <x; in the partial ordering _of M(x). By condition (2) of (3.6.62) for the sets {A,,..., Ad in M(x) we conclude that y EA;. This proves condition (2) for the sets (A2,..., Ak) in M*(x). We verify condition (3) for the sets {A2,..., Ad in M*(x). Let S be a subset of (2,. .. , k) with at least two elements, and let elements {x, (=- A;, i E S) be given arbitrarily. By condition (3) of (3.6.62) and (3.6.43) there exists a least upper bound z for {x,: i E S) and g, = n;,=- s a X,. As explained in (a) of (3.6.63) it follows from condition (2) of (3.6.62) that z E M(x) - Uk_ , A,. It remains only to show that z e M*(x) and z is a least upper bound for (x;: i E S) with respect to the partial ordering of M*(x). For each i e S it follows from (3.6.57) and the fact that i Z 2 that GX = G, X H;, where G, =10(M1) and H; = G,*, G* =10(M*). Hence cX, = g, g*

for all i r= S, where g* is the Lie algebra of G*. It follows that A, = ni E s g.r, = R, ® (n;. s a*,). By (3.6.57) we conclude that z E M*(x) and g* = n;,=- s g*,. By (3.6.43), z is a least upper bound for {x;, i (=- S} with respect to the partial ordering in M*(x). This completes

the verification of (3) of (3.6.62) for the sets (A2,..., Ad in M*(x).

We verify (4) of (3.6.62) for the sets (A2,..., Ak) in M*(x). Let z E M*(x) - Uk_ 2 A. c M(x) -

Uk_ , A. be given. By hypothesis (4) of

(3.6.62) for the sets {A,,..., Ak} in M(cc) there exists a subset S of (1,... , k} with at least two elements and elements x; E A, for each i E S such that z is the least upper bound of {x;: i E S}. To verify (4) for the

sets {A,,..., Ad in M*(x) it suffices to show that S c {2,..., k} and z is a least upper bound of (x,: i E S) relative to the partial ordering of M *(x).

Since z is a least upper bound for {x;: i E S} relative to the partial

ordering of M(x) we know from (3.6.43) that g, = n;,: s g,r,. Since

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223

z E M *(co) we know from (3.6.57) applied to M = M, x M* that q, = g ®g z . We apply (3.6.57) again to conclude that x, c- M *(oo) for every i E S since g, c g,,for every i. This shows that 1,4 S since A, = M,(r),

which is disjoint from M*(c). Hence S c (2,..., k). Moreover, since x; E M*(oo) for every i it follows from (3.6.57) that gX. = g, g . Hence g,9g2=gZ=nrEsgx=gIED (ni.sg* By gZ=niEsg* since (3.6.43) we conclude that z is a least upper bound for {x;: i (=- S) relative

to the partial ordering in M *00. This completes the verification of (4) of (3.6.62) for the sets {A2,..., Ak) in M*(oo).

We now apply the induction hypothesis of (3.6.65) to the sets (A2".., Ak} in M *(cc) to conclude that M * is the Riemannian product

x Mk such that A. = M;(oc) for 2 < i < k. Hence M = M, X M* =M1 x X Mk and A, = M1(oo) for 1 < j < k. M2 x -

Splices at infinity In this section M will continue to denote a symmetric space of

3.7.

noncompact type and rank k >_ 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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Geometry of Nonpositively Curved Manifolds

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 <_i SN. If gx = gx, then W(x)_'(x,) and (2.17.22) there exists a finite set (XI,..

the assertion (3) now follows from (2).

We now present a basic example of a splice. We shall see below in (3.7.15), (3.7.16), and (3.7.17) that all splices are of this form. 3.7.3. NOTATION. Given a nonzero element X E g and a k-flat in M we define S(X, F) _ (x e F(a): X E cix}. 3.7.4. PROPOSITION. Let S(X, F) be nonemply for some nonzero element X of g and some k -flat F of M. Then there exists a positive number e such

that S(X, F) = F(-) fl (e'XF)(o) for all t with 0 < t < e. In particular, S(X, F) is a splice in M(oc).

PROOF. Let F be a k-flat in M, and let X be any nonzero element of g such that S = S(X, F) = {x E F(z): X E gx} is nonempty. Clearly, S c_ (e'XFXc) f F(-) for all t E R. By (2.17.26) we can find a neighborhood

W of the origin in g such that (a) exp is one-to-one on W and (b) if x E (gFX-) n F(x) for some g e exp(W), then g = exp(Z) for some Z E W n g x and g(x) = x. Let e be a positive number such that tX E W for 0 < t < e. If x is a point in (e" F)(-) n F(c) for some t e (0, e), then e'X = ez for some Z E W n gx by the choice of W. Hence tX = Z and X E g x since t > 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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225

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 <,, (q,, q2) < v. Let x, and x2 be points of S such that q, = yux(t;) for some t, > 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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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

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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))
p

of W(v(p)) there exists a unit vector v,, E'(v(p)) such that yt, for some t,, > 0. By the law of cosines, assertion (3) of corollary (1.4.4), it follows that v -1, v * as n --> +x, which implies that v* is tangent at p to the k-flat F = F(y,,(P)) by the definition of '(v(p)) in (2.12.1). By assertion (1) of proposition (2.12.3), W(v*) is an open subset of SPF that contains v*, and hence v,, E W(v*) for large n. It follows that F(t:*) W(v(p)), and hence W(v*) = W(v(p)). The following is an immediate consequence of the result just proved. 3.8.3. COROLLARY. Let 11 be a symmetric space of noncompact type and rank k >_ 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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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 -

z,, z, E R(x).

From the fact that the action G X R(x) -+ R(x) is continuous with respect to the cone topology it is easy to see that the action G X FM(x) -* FM(x) is continuous with respect to the topology .r on FM(x) defined by

FM(x). Given z E R(x) and P E M we define T(z) = W(y1,2(0))(x) E FM (x).

It follows from proposition (3.8.2) that T does not depend on the point p in M. Moreover T is obviously surjective. 3.8.6. PROPOSITION. Let M be a symmetric space of noncompact type and

rank k >_ 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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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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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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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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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*:

If x is an element of G or G*, then L, the left multiplication by x, is a homeomorphism of G or G*. Moreover, T o L8 = L,(g) o T for all g e G. Hence to show that T is an open map it suffices to show that for every neighborhood U of the identity in G in the Lie topology of G there exists a positive number e such that ®e = (g E G: 4,, (x, T(g)x) < e

for all x E Moo)) C U. This is an immediate consequence of the following two results.

LEMMA A. Let M be a symmetric space of noncompact type and rank k >_ 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

LEMMA B. Let A. and G be as in lemma A. Let a positive number e, a point p EM, and a compact subset C c R(cc) be given. There exists a positive number S such that if g E ®s = (g E G:

PROOF OF LEMMA A. We suppose that the lemma is false for some symmetric space M, some point p E M, and some neighborhood U of the identity in G. Write R(te) = U'_ , K,,, where each K,, is compact and K,, S K,, , for every n. For each positive integer n choose g,, e G - U such that 4P (x, g,, x) < 1 In for every x E K,,. STEP 1. The sequence

is bounded in M.

Suppose that this is not the case, and let g,,(p) - x(, E M(oo) and g '(p) -yo E MA-) by passing to a suitable subsequence. Given a point q E M we let xq = ygya(- -) = Sq(yo), where Sq is the geodesic symmeis-the try at q. We shall show that xo = xq for all q e and this will prove that M has a nontrivial Euclidean de Rham factor by proposition 2.2 of [EC], which is impossible.

Let e > 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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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)
_ 4q(yo,xq) < Hence we obtain (2)

q (yp,

gn'(q)) + 4q (xq*, xq)

We know that go(q) -'x0, g-'(q) -3 yo by (3) of (1.4.4) and go(xq) ->xq by (1). Hence from (2) we obtain (3) 4q (XO, Xq) < 4q (xq, xq* ) < e.

Finally, 4q(xo,xq)<4q(xo,xq*)+
STEP 2. (Completion of the proof). Fix a point p of M. It follows from step I and (1.2.3) that g converges to some element g of G, passing to a subsequence if necessary. If x E Moo), then x E K for sufficiently large n and hence 4P (x, gx) = limo . x

To prove lemma B we shall need a preliminary result about the center of gravity CG(x) of a Weyl chamber F(x) (cf. (3.6.35)).

SUBLEMMA. For every point p E M, compact subset C of R(-), and positive number a there exists a positive number S = S(p, C, e) with the following property: If x E C and g E C are elements such that 4: (CG(x), gCG(x)) < 6, then

point p E M, some compact subset C c R(ao), and some positive number

e. Then for any positive integer n we can find elements xo E C and gn E G such that (1)
1/n for every n;

(2) - a for every n.

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We may assume without loss of generality that each element g lies

in a fixed maximal compact subgroup K of G. To see this we set x* =

and we recall that

(a) (Gxn)0 = (Gx^ )p since (b) G = by (2.19.2).

Now write g = k 4,,, where k E K and 4 E (G., )o = (Gx. )o. Then g, (x,,) = k (x*) for all n, so (1) and (2) hold if g is replaced by k E K. By the preceding paragraph we may assume that g,, lies in K for every n. Let g,, -' g E K, passing to a subsequence if necessary. Let x -j x E C c R(x) by passing to a further subsequence. Then CG(x)

the continuity of the map CG (cf. (3.6.35)). It n Hence from follows that g,,(x,,) -->g(x) and g(x*) (1) and (2) above we obtain (3)

_ e.

By (3) it follows that g(x*) =x* and hence g(z) = z for all z E W(x by (1) of (2.17.25). However, since x* = CG(x) it follows that W(x) 1; '(x*) and by the remark above we conclude that gx =x, which contradicts (4). This completes the proof of the sublemma. PROOF OF LEMMA B. Let a point p r=- M, a compact subset C of R(x),

and a positive number a be given. Let S = S(p, C, e) > 0 be chosen as in the sublemma above. We show that 6 satisfies the conditions of lemma B. Let g E 6S = {g E G:

p (x*, T(g)x*) < S for all x* E FM(x)) and x E C be given arbitrarily. Let x* = T(x), where T: R(x) - FM(x) is

the open, continuous surjective map defined in the discussion following (3.8.5). From the discussion preceding (3.8.5) and (4) of (3.8.6) we recall the map S: FM(x) -+ R(x) given by S(Tx) = CG(x) for every x E R(x). Since g E 6S we conclude from (3.8.5), (3.8.6), and (3.8.7) that

S> 4P(x*,T(g)x*) = 4P(Tx,T(g)Tx)=
Hence

3.9. Tits isomorphisms and induced mappings We recall from the remarks following (3.6.24) that if M is a symmet-

ric space of noncompact type and rank k z 2, then 9M, the Tits

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associated to M, is the quotient space 911%1= M(oo)/ - , where x -y for points x, y c= M(oo) if (Gx)o = (Gy)0 or equivalently if W(x) = W(y). We let [ ]: M(co) -9M denote the projection map. Recall also that [x] <_ [y] if x < y for points x, y E M(c). 3.9.1. DEFINITION. Let M and M* be symmetric spaces of noncompact type and rank >_ 2. A function 4':9M -+9M* is order preserving if

¢([x]) < 4'([y]) whenever [x]:5 [y] for points [x] and [y] in 9M. A function 4:9M -9M* is a Tits isomorphism if 0 is a bijection such that 4) and (c))-' are order preserving. 3.9.2. EXAMPLES. (1) If 4': M(co) -' M*(co) is an order preserving bijec-

tion (i.e., 4)(x) s ¢(y) whenever x < y for points x, y E M(-)), then 4 induces a Tits isomorphism 4):9M -9M* given by 4([x]) = [4)(x)] for a point x E M(x). In particular, if 0: M(oo) - M*(-) is a homeomorphism with respect to the Td-topologies, then 0 is order preserving by the outline preceding the statement of (3.6.54). (2) Let ..59 and _91* denote the splices at infinity for M and M* (see (3.7)). Suppose there exist inverse functions i/i: Y-91* and 4i*: S°* -S° that are both strongly order preserving (cf. (3.7.8)). Then by proposition (3.7.9), 4) and 4i* induce inverse Tits isomorphisms between 9M and

9M*. The splice example above arises in the proof of the Mostow rigidity theorem in the higher rank case. See (8.5) below. 3.9.3. NOTATION. Let M be a symmetric space of noncompact type and

rank k z 2, and let G =10(M). Given an element g E G, we let µ(g) and r(g) denote the actions of g on 9M and FM(-) given by

µ(g)([x]) = [gx],

r(g)(Tx) = T(gx),

where x E M(oo) or Moo) and T is defined following (3.8.5). We now identify FM(x) with a subset of 9M in a way that is equivariant with respect to G. 3.9.4. PROPOSITION. Let To M denote the subset of 9M consisting of those elements of 9M that are maximal with respect to the partial ordering. Then there exists a bijection p: TO M -4 FM(oo) such that p o µ(g) = r(g) o p for

every gEG. PROOF. It is easy to see that [ ](R(-)) = 7-OM since R(co) is the set of points in M(x) that are maximal with respect to the partial ordering. If x is any point of R(-), then F(x) is the fiber of [ ] over [x] and W(x) is

also the fiber of T over T(x) by (1) of (3.8.7). Hence if we define FM(oo) by p([x]) = T(x), then p is well defined. The map p is

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injective by (1) of (3.8.7) and is surjective since both [ ] and T are surjective. The equivariance property of p follows from the corresponding equivariance properties of [ ] and T (cf. (2) of (3.8.7)). 3.9.5. COROLLARY. G =10(M) acts effectively on both _c770 and FAI-00.

PROOF. If T(g) is the identity on FM(x) for some g E G, then we observed earlier in the discussion preceding (3.8.12) that g = e by (2) of (3.8.8), (5) of (1.9.4), and the fact that R(x) is dense in M(x) in the cone topology. Hence G acts effectively on FM(x). If µ(g) is the identity on 9M, then T(g) is the identity on FM(x) by the equivariance of p:90M

- FM(x). Hence G acts effectively on 9M (and even effectively on

9M). Equivariance properties of Tits isomorphisms Let M and M * be symmetric spaces of noncompact type and rank

2. Let r and r* be subgroups of G =10(M) and G* =10(M*) respectively, and let 9: r -* r* be an isomorphism. A Tits isomorphism 4:9M -), 9M * is said to be 9-equivariant if 4 ° µ(y) = µ*(Oy) ° 4) for all elements y(=- r. 3.9.6. PROPOSITION. Let Mand M * be symmetric spaces of noncompact type and rank >_ 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

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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

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250

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 <, (x, 4W = Td(x, i/ix) for all points p E M and x E M(). Then ip is the identity on M(x). LEMMA B. Let F: M(x) - M*(x) denote the map induced by the isometry

f: M + M*. Then g

g

for all x e M(x).

The proofs of lemmas A and B are lengthy, and we postpone them

for the moment to complete the proof of (3.10.1). The map 4i= F- c 0: M(x) -) M(x) is a homeomorphism in the cone topologies and an isometry with respect to the Tits metric Td since the maps F and 0 have this property. We show that

(*)

4p(x, 4ix) = Td(x, 4ix)

for all p E M, for all x E M(x).

If x* and y* are points of M* with gA. = q,., then p* -,
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251

(3.1.2) and (2) of (3.4.3). This observation and lemma B show that if x is any point of M(oc), then Td(x, 4'x) = Td(Fx, Ox) =
p* E M*. However, <,P. (Ox, Fx) = 4 Oft, x), where p = f-'(p*) since f: M - M * is an isometry. This proves (*). From (*) and lemma A we conclude that 41 = F-' C ¢ is the identity and hence F = 4). It remains only to show that f: M -R' * is the unique isometry that induces the given map ¢: M(ao) --> 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<(yX(x))

We show that 4* is the identity on a, = a n p, and by the definition of 4r* this will imply that 4 fixes every point of F(x). We show first that 4r* leaves a, invariant, which by the definition of /i * is equivalent to showing that 41 leaves F(m) invariant. If x is any point of F(cc) n R(te), then W(x) c F(o) by (3.6.26) and W(x) = W(41x) by step 1. Hence 41 leaves F(x) n R(oo) invariant, and it follows that 4 leaves F(m) invariant since F(co) is closed in the cone topology and F(oo) n R(te) is a dense subset of F(cc).

We have shown that 0* leaves a, invariant. By step 1 4i* maps regular vectors of p to regular vectors of p. We show next that the restriction of 41* to a, is the restriction of a linear isometry of a that leaves invariant every Weyl chamber of a. Let X and Y be unit vectors in a,, and let x = yX(-) and y = yy(oc) be the corresponding points in M(°). Then x, y. +y(x), and P(y) lie in F(cc) since 41 leaves F(oo) invariant. We assert that

(1) 4(,P*X,+/*Y)= 4

Mix,r/'y)=Td(qix,+l
The first equality follows from the definition of i/r* and the identifi-

cation of a and Tp F via (2.3.7). The second equality follows from (3.1.2), (2) of (3.4.3), and the fact that yp,,(X)[O, oo) and y (y)[O, oo) span a

flat triangular sector in F since /i(x) and O(y) lie in F(oo). The third equality follows from the hypothesis that ip preserves Td. Similarly, we have

(2) <(X, Y) = -*:p (x, y) = Td(x, y).

Comparing (1) and (2) we see that <(X, Y) = 4(/I *X, 41 *Y) if X and Y are unit vectors. By its definition * carries unit vectors in p to unit

vectors in p. It follows that /r* leaves the unit sphere a, invariant and preserves the angle metric on a,. Therefore ir* is the restriction of a linear isometry of a, which henceforth we also denote by r/r*. If A is a regular unit vector of a, then iP*(io°(A)) = W(A) by (3.6.31) since +/r(F(a)) = F(lJra) = F(a) by step 1, where a = yA(oc) E R(te). Hence a/i* fixes the algebraic centroid c(A) of io'(A); see (2.9.15).

Proposition (2.9.15) also shows that the centroids of the Weyl chambers of a are a spanning set for a. Therefore 4r* is the identity on a since it is linear and fixes each element of a spanning set of a. It follows that 41 fixes every point of F(oo), which completes step 2 and also lemma A as indicated above. O

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We begin the proof of lemma B, which we divide into three steps. We

have shown earlier in the proof of (3.10.1) that 4: M(x) _1 M*(x) induces a continuous isomorphism of G = 10(M) onto G* =10(M*) and

induces an isometry f: M - M* (after rescaling) such that f ° g = B(g)o f for all g E G. STEP 1. Let g E G and x c=- M(x) be given. Then

(a) g E G., if and only if 8(g) E GF(x); (b) Z E gx if and only if d0(Z) E cF(X).

Fix a point p e M, and let g = f + p be the Cartan decomposition determined by p. Let p* denote f (p), and let g * = f * + p * denote the

Cartan decomposition determined by p*. If K = GP and K* = Gp , then it follows from the construction of f: M - M* in (3.9.11) that 0(K) = K*, dO(f) = 1*, and dO(p) = p*. Given a vector X E g we let X* denote d6(X) E g *. Let a point x e M(x) be given, and let X E p be the unique unit vector such that yx(x) = x, where yx(t) = e'X(p), which equals the geodesic y, (t) for all t by (2.4.1). Then the geodesic yx.(t) = e`X'(p*) represents F(x) E M*(x) since f(e'Xp) = 8(e'X)f(p) = e`X'(p*).

-,

e-'Xge`X Now let g e G be given. Then by (2.17.3), g e Gx a lim, exists « lim, x e-`X 9(g)e`X' exists,** 8(g) E GFThis completes (a) of step 1. To prove (b) of step 1 we observe that Z E g x e'Z E Gx for all t E R p 0(e'Z) = e'Z' E G*(x) for all t E R «Z* E F(x) The second equivalence follows from (a). The proof of (b) is complete.

STEP 2. Let x E ll%1(x) and Z E g be given. If [x] _ [e`Z(x)] for tI < e for some e > 0, then Z E g x.

Recall that [ ]: M(x) ->.91b1 is the projection map. The assertion for [x] = [e`Z(x)] for ItI < e means that (Gx)o = (Ge:,-, ) = ItI < e. Hence e'Z normalizes (Gx)o for Id I < e. It follows from (2) of (2.17.25) that e`Z E Gx for I t I < e, and hence Z E g x.

STEP 3 (Conclusion of the proof). If 4:911%1-+9M* denotes the Tits isomorphism induced by the given map 46: M_(x) - M*(-), then by (1) of (3.9.8) we know that 4) c µ(g) = µ*(8g)o ¢ for all g E G,'where µ

and µ* denote the actions of G and G* on 9M and 9M*. Since fi([x]) = [ 4,(x)] for X E M(x) by the definition of 4) we conclude that

(*)

[¢(gx)] = 10(9)0(x)]

for all g E G, for all x E M(00).

Now let Z* E gF(x) be given. Then Z = (d8)- I (Z*) E gx by (b) of step 1. Hence [4)(x)] = [4(e`Zx)] = [e'Z'(4x)] for all t e O by (*). From

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Geometry of Nonpositively Curved Manifolds

step 2 we conclude that Z* E 9This shows that cF(

)

c OOcx

Conversely, if Z* E then [¢(x)] = [e`z'(4x)] = [4'(e`Zx)] for all t e l' by (*) where Z = (d B-' X Z*). Hence [x] _ [erzx] for all t E R since (A4' preserves the Tits partial ordering. By step 2 we conclude

that Z e gx, and hence Z* = d9(Z) E 9F(x) by (b) of step 1. This proves that q *( , ) c qF(x) and completes the proof of lemma B. 0 Next we extend (3.10.1) to the case where only one of the manifolds M and M * is assumed to be symmetric of noncompact type with rank >: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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255

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 <_ i<_ k.

If Tx = Ty for points x, y E R(x), then 0 x = g y, by (1) of (3.8.7), and hence fi co = 0 m (r) for 1 < i 5 k by (3.6.61). Hence ,/i is well defined by (1) of (3.8.7). Conversely, if rlr(Tx) = .r(Ty) for points x, y E R(x), then g O c x) = g ,,(Y) for 1 5 i 5 k by (1) of (3.8.7) and hence q x = cl , by (1) of (3.6.61). It follows that Tx = Ty by (1) of (3.8.7), which proves that +/r is one-to-one. The map 41 is surjective since the maps gyp, T, and T, are surjective. It is routine to show that i/r is a homeomorphism, and we omit the details. Similarly, the equivariance assertion ijr -r(g) = A(g)o 41 follows from the definitions in a straightforward manner.

3.11.2. REMARK. The corresponding product decomposition 9M

9MI x x9Mk for the Tits buildings is not valid. However, each 3M, can be identified naturally with a subset of 9M; each M;(x) can be identified naturally with a subset of M(°°) and GX = Gy for points x, y E A(m) if and only if (G,)X = (G,),, by (3.6.57).

On the other hand .9 M, the set of maximal points in .9M, is bijectively equivalent to 9QM, x X.9 Mk by (3.11.1) since the pairs M and FM(x) and 9o M, and FM,(x) are bijectively equivalent by (3.9.4).

3.11.3. PROPOSITION. Let M and M* be symmetric spaces of noncompact

type and rank z 2, and let M be a Riemannian product M, x

X M, .

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Let 0: 9M --> 9M * be a Tits isomorphism. Then: X Mk (1) M can be decomposed into a Riemannian product M* X in such a way that 4)(9M;) = 9M* for 1 < i _< k. (2) Let a: Moo) -> 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 <_i
tive, continuous open maps of (3.6.61) for I s i s k. If 4)([x]) = [x*] for x c= Moo), x* E R*@), then 4)([ 4;(x)]) = [c)*(x*)] for 1 <_ i < k.

PROOF OF THE LEMMA. If x and y are points of M(c) we recall from (1) of the remark following (3.6.24) that [x] _< [Y] in 9101 if and only if x 5 y

in M(ao). Now let x E R(te) be given arbitrarily. By (1) of (3.6.61) we know that g x = f;`_ , g and by (3.6.43) we see that x E lub{¢,(x),..., 4)k(x)). Hence (a) [x] E lub{[4)i(x)],...,[4k(x)]).

If fi([x]) _ [x*], where x* E R*(x), then since 4:97151 _.9.f* is a Tits isomorphism we obtain

(b) [x*] E lub{4)([,(x)]),..., b([4)k(x)])}.

For 1 < i 5 k the point [ 4);(x)] is maximal with respect to the partial ordering in 911%1; since 4);(x) E R;(oo). Hence 4)([4;(x)]) is maximal with

respect to the partial ordering in 4)(9M,) =.9M;*, and we can find a point y* E R*(oo) such that [y* ] = 4)([4;(x)]). From (b) we can see that x* E lub{y*,... , yk }, and hence by (3.6.43) we obtain (c)

c .= nk_,gy*.

On the other hand, from (1) of (3.6.61) we obtain

(d) gx. = nk-, ¢01`(x6)

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259

Comparing (c) and (d) and using (2) of (3.6.61) we obtain g 0P.) = 9y*' for 1 5 i:5 k. We conclude that [4*(x*)] = [ y* ] = 4([¢;(x)]) for every i. We prove (2) of (3.11.3). For an integer i with 1 5 i 5 k we let x and y be points of R,(a). From (2) of (3.6.57) we see that Gx = Gy if and only if (G), = (G;)y, and hence by (3) of (2.17.25) and (1) of (3.8.7) [x] = [y] if and only if T(x) = T,(y), where Ti: R;(cc) -* FM;(cc) is the projection. Similarly, [x*] = [ y*] for points x*, y* e R*(x) if and only if T*(x*) = T*(y*). We define 6;: FM;(z) -- FM*(x) as follows: for x E R;(cc) we let

f3,(Tx) = T*(x*) where x* E R*(c) is chosen so that fi([x]) = [x*]. By the remarks above and the surjectivity of Ti and Ti* it follows immediately that 6; is a well-defined bijection. We prove that Pi c ali; = 4i;* o a. Let x E Woo) be given. By definition

a(Tx) = T*(x*), where x* E R*(oo) and 4([x]) = [x*]. Hence by the definition of 41i*: FM *(co) --> FM;* (c) in (3.11.1) we have (,i;* o a )(Tx) = 4ii*(T*x*) = T*(Oi*(x*)).

On the other hand, 4i;(Tx) = T(4;(x)) and hence (f3; c i/r;)(Tx) = a T X Oi(x)) = Ti* (y* ), where y* E R* (x) is chosen so that ¢([0;(x)]) = [y* ]. However, 4([ca;(x)]) = [¢*(x*)] by the lemma above and hence [y* ] _ [4*(x*)]. Thus Ti* Ti* _ (+/;* c a)(Tx) by the previous paragraph. Finally, if a: Woo) -+ FM *(oo) is a homeomorphism, then it follows that f3;: FM;(o) -- FM;* (ao) is a homeomorphism for each i since i6; cry; = 4r;* c a and the maps 4k; and 4i;* are continuous, open, and surjective. This completes the proof of (3.11.3). O

Irreducible lattices in product manifolds 3.11.4. DEFINITION. Let M be a complete, simply connected manifold with

sectional curvature K:5 0. A lattice r c 1(M) is said to be reducible if r admits a finite index subgroup F0 that is a nontrivial direct prod-

uct ro = r, x r2. If a lattice r is not reducible, then r is said to be irreducible.

3.11.5. REMARK. If IF is a reducible lattice with trivial center and with no elements of finite order, then it follows from [GW] that the smooth

manifold M = M/1' has a finite cover M * that splits as a Riemannian product M* x M? . See also [LY] and (10.3.9) below. If M is a Rieman-

nian product M* x M2 with no Euclidean factor, then a lattice r c 1(M) must be reducible unless M is a symmetric space of noncompact type by proposition 4.5 of [Ell]. Not all reducible symmetric spaces M = M* x M2 of noncompact type admit irreducible lattices r c 1(M). See (9.2.6) below for a necessary and sufficient condition.

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For the purposes of this section we need only the following fact about irreducible lattices.

3.11.6. PRoPosrrtoN. Let M be a symmetric space of noncompact type and rank k z 2 that is a Riemannian product M, X M2. Write G = G, x G2, where G =10(M) and G; = 10(M,) for i = 1, 2. Let p;: G - G; denote the

projection homomorphism for i = 1, 2. If r c G is an irreducible lattice then: (a) F, = p;(F) is dense in G; for i = 1, 2 with respect to the Lie topology of G1;

(b) p,: r --* F, has trivial kernel for i = 1, 2.

PROOF. See corollary 5.21 of [R] or proposition 4.7 of [Ell]. We now reach the main result of this chapter, which is essentially a special case of theorem 18.1 of [Mos2]. This result differs from (3.9.8) in

that M is allowed to have rank-I de Rham factors. 3.11.7. THEOREM. Let M and M* be symmetric spaces of noncompact type of rank >_ 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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261

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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263

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

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given that x
(d) AycAA. and AYCAA..

Note that a(A*) * 0 for all a e A since A* is regular. Hence (d) is equivalent to the single condition a(Y)a(A*) z 0 for all a E A. In view of (c) it therefore suffices to prove that

(e) a(Y)a(A*) >- 0 if a(X) = 0, a e A.

Up to this point we have only used the fact that 0 E Ko(p, x) is a p, x) -* R. To prove (e) we must also critical point of the function f: use the fact that 0 is a local maximum of f. By (3.6.6) and the lemma in the proof of (2) of (3.6.7) we have (f) to +

En(x)=

fq = Z(X) n f, the Lie algebra of K0(p, x),

where to = f n go, fo = (1 + OX cIn) for all a E A, and 0 = BP is the Cartan involution of g determined by p as in (2.3.2). Now let TyA. _ ad Y o ad A*: 0 -+ g, where Y and A* =Ad(d)-' (A) are the elements of a' defined above. Note that ad Y ° ad A* - ad A* c ad Y = ad([Y, A*])

= 0. It follows from (5) of (2.14.2) that TyA. = 0 on to and TyA. = a(Y)a(A*)Id on fry for every a e A. From (f) we see that (e) is equivalent to the statement that TyA. is positive semidefinite on

Z(X)nf. Let f be an arbitrary element of Z(X) n f and proceed as in the proof of (3.6.8) by defining

& (t)=f(4e')=B(Ad((k)Ad(e'1)(Y),A)=B(e'a"(Y),A*) where f has a local maximum at ¢ E KO(p, x). We recall that B is negative definite on f by the discussion preceding (2.7.1). We know that g"(0) 5 0 for all E Z(X) n t since 0 is a local maximum for f. Hence

to prove that TyA. is positive semidefinite on Z(X) n f it suffices to prove (g) g-(0) = B(TyA.( ), 0) for all 6 E Z(X) n f.

We calculate g" (t) = B((ad 6 )2e'ad F(Y), A*) and hence by (2) of (1.13.8) we have

g(0) = B((ad 6)2(Y), A*) = - B(ad e(Y),ad f (A*))

=B(adY(f ),ad C(A*)) = -B(6,adYoad f(A*)). Since Y and A* lie in a' we know that ad Y(A*) = 0 and hence ad Y -ad C(A*) = ad

oad Y(A*) + ad([Y, f])(A*)

= -adA*([Y,6])= -(adA*oadY)(g)= -TyA.(6)

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since ad A* and ad Y commute. The proof of (g) is now complete, and this also completes the proof of (6) of (3.6.21). 0 SUBLEMMA IN PROOF OF (3.6.36). Let A ;f'

G be as above. Let x (=- M(oo)

be given. For each positive number a and each neighborhood U c G of the identity we set

W(U,e,x) = (4(y):0EE U, yEM(ao) andTd(x,y) <e}. Then W(U, e, x) is a neighborhood of x in M(-) with respect to the cone topology.

PROOF. Fix a point p E M, and let g = f + p be the Cartan decomposition determined by p. Let x E M(oo), e > 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 <(X, Y) < e. STEP 3. Choose o > 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):