Quasiisometries between negatively curved Hadamard manifolds

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1 Quasiisometries between negatively curved Hadamard manifolds Xiangdong Xie Department of Mathematics, Virginia Tech, Blacksburg, VA Abstract. Let H 1, H 2 be the universal covers of two compact Riemannian manifolds (of dimension 4) with negative sectional curvature. Then every quasiisometry between them lies at a finite distance from a bilipschitz homeomorphism. As a consequence, every self quasiconformal map of a Heisenberg group (equipped with the Carnot metric and viewed as the ideal boundary of complex hyperbolic space) of dimension 5 extends to a self quasiconformal map of the complex hyperbolic space. Mathematics Subject Classification (2000). 53C20, 20F65, 30C65. Key words. Quasisymmetric extension, Hadamard manifolds, quasiisometry, bilipschitz. 1 Introduction A classical result in quasiconformal analysis says that every quasisymmetric map from R n to itself extends to a quasiconformal map of the upper half space R n+1 + = {(x 1,, x n+1 ) : x n+1 > 0}. Here R n is identified with the subset {(x 1,, x n+1 ) : x n+1 = 0} of R n+1. This result was first proved for n = 1 by L. Ahlfors and A. Beurling [BA], then for n = 2 by L. Ahlfors [A], for n = 3 by L. Carleson [C], and finally for all n by Tukia and Vaisala [TV]. Recall that R n+1 + is the upper half space model for the real hyperbolic space and Ṙ n = R n { } is its ideal boundary. The quasiconformal extension to R n+1 + turns out to be a bilipschitz homeomorphism in the hyperbolic metric. On the other hand, every self quasiisometry of the hyperbolic space R n+1 + induces a self quasisymmetric map of Ṙn (equipped with the spherical metric), and conversely every self quasisymmetric map of Ṙ n is induced by a self quasiisometry of R n+1 + (see [BS] for more 1

2 2 Xiangdong Xie general statements). Hence what Tukia-Vaisala and others have proved is this: every self quasiisometry of the hyperbolic space R n+1 + lies at a finite distance from a bilipschitz homeomorphism. The main result of this paper generalizes this statement to negatively curved Hadamard manifolds. A Hadamard manifold is a simply connected complete Riemannian manifold with nonpositive sectional curvature. Theorem 1.1. Let H 1, H 2 be the universal covers of two compact Riemannian manifolds (of dimension 4) with negative sectional curvature. Then every quasiisometry between them lies at a finite distance from a bilipschitz homeomorphism. The restriction on dimension is due to the following facts: a 4-dimensional topological manifold might have more than one lipschitz structures or none [DS], while every topological n-manifold with n 4 has a unique lipschitz structure [S]. The unit ball BC n in Cn is a model for the complex hyperbolic space, which is a negatively curved Hadamard manifold. The unit sphere in C n has a sub-riemannian structure coming from the complex structure of C n. In this case the natural metric on the unit sphere is the associated Carnot metric. Theorem 1.1 applied to the case H 1 = H 2 = BC n yields a generalization of the classical quasiconformal extension result to the complex hyperbolic case: Corollary 1.2. Every self quasisymmetric map of the unit sphere (equipped with the Carnot metric) in C n (n 2) extends to a self quasiconformal map of the complex hyperbolic space B n C. There are two general questions related to the results in this paper. It seems that solutions to both questions require the use of controlled topology (for example the theorem of Chapman-Ferry [CF] on small homotopy equivalences or its variants). The first general question asks when a power quasisymmetric map between the boundaries of two Euclidean domains extend to a quasisymmetric map between the domains. See Section 7 for a more precise formulation. The second general question asks when a quasiisometry lies at a finite distance from a bilipschitz homeomorphism. Also see Section 7 for a more precise formulation. For example, a natural question is whether the main theorem in this paper holds for all Hadamard manifolds, in particular, whether every self quasiisometry of R n lies at a finite distance from a self bilipschitz homeomorphism of R n. The example of Dranishnikov-Ferry-Weinberger [DFW] suggests that the question is very subtle: they constructed two uniformly contractible Riemannian manifolds and a quasiisometry between them that is not at a finite distance from any homeomorphism. The two manifolds they constructed are homeomorphic to the Euclidean space, but the metrics are somehow exotic. On the positive side, Whyte [W] showed that every quasiisometry between two uniformly discrete non-amenable spaces of bounded geometry (for example, finitely generated nonamenable groups with the word metric) lies at a finite distance from a bilipschitz map.

3 Quasiisometries between Hadamard manifolds 3 Our proof follows the strategy of Tukia-Vaisala [TV]. In particular, we use the boundary map induced by the quasiisometry. This means that our proof can not be generalized to nonpositively curved spaces like R n : a quasiisometry between CAT(0) spaces in general does not induce a boundary map. In Section 2 we review some basics about various maps and negatively curved spaces. In Section 3 we replace the quasiisometry with a homeomorphism F, constructed using the boundary map of the quasiisometry. In general the map F is not bilipschitz, but has very good compactness property: both F and F 1 are uniformly continuous. This is established in Sections 4 and 5. In Section 6 we modify F to obtain a bilipschitz map. The arguments in Section 6 are essentially due to Tukia-Vaisala [TV]. In the last Section we formulate some open questions. Acknowledgment. The author particularly thanks the Department of Mathematics at Virginia Tech for its generous support: the teaching load of one class per year is really a gift. The author would also like to thank Bruce Kleiner for drawing his attention to the paper of Block-Weinberger [BW]. 2 Preliminaries Various maps A bijection between two metric spaces f : X Y is L-bilipschitz (L 1) if d(x, y)/l d(f(x), f(y)) L d(x, y) for all x, y X. An embedding of metric spaces f : X Y is locally bilipschitz, if each point x X has a neighborhood U such that f U is bilipschitz; we say f is locally L-bilipschitz if f U is L-bilipschitz. Let L 1 and A 0 be constants. A (not necessarily continuous) map f : X Y is an (L, A)-quasiisometry if the following holds: (1) d(x, y)/l A d(f(x), f(y)) L d(x, y) + A for all x, y X; (2) For any z Y, there is some x X with d(z, f(x)) A. If f : I Y is a map defined on an interval I R and satisfies condition (1) above, then we say f is an (L, A)-quasigeodesic. It is clear that a bilipschitz map is a quasiisometry, but the converse is not true. Let η : [0, ) [0, ) be a homeomorphism. A homeomorphism between metric spaces f : X Y is η-quasisymmetric if for all distinct triples x, y, z X, we have ( ) d(f(x), f(y)) d(x, y) d(f(x), f(z)) η. d(x, z) A homeomorphism f : X Y is quasisymmetric if it is η-quasisymmetric for some η. A quasisymmetric map is called a power quasisymmetric map if it is η-quasisymmetric and η has the form η(t) = Ct α, where C > 0 and α > 0 are constants. Recall that

4 4 Xiangdong Xie a quasisymmetry between connected metric spaces is a power quasisymmetry (see Theorem 6.14 in [V]). Gromov hyperbolic spaces Let X be a geodesic metric space. We assume X is proper, that is, all closed balls in X are compact. Let δ 0. We say X is δ-hyperbolic, if for any x, y, z X, and any geodesics xy, yz, zx between them, yz is contained in the δ-neighborhood of xy xz. A metric space is Gromov hyperbolic if it is δ-hyperbolic for some δ. A Gromov hyperbolic geodesic space X has an ideal boundary X: by definition, X is the set of equivalence classes of geodesic rays in X, where two rays are equivalent if the Hausdorff distance between them is finite. There is a natural topology on X := X X, in which X is compact and X is an open dense subset of X. Let X and Y be Gromov hyperbolic spaces. Then every quasiisometry f : X Y induces a boundary map f : X Y, which is a homeomorphism. Moreover, f is a power quasisymmetry with respect to the so-called visual metrics on the Gromov boundary; conversely, if X and Y satisfy some mild conditions, then every power quasisymmetry X Y is induced by a quasiisometry. See [BS] for more details. See also Lemma 3.1 for a statement in the Hadamard manifold case. Let X be a δ-hyperbolic geodesic space, x, y, z X, and xy, yz, zx geodesics between them. We say xy yz zx is a triangle. Let C 0. We say a point w X is a C-quasicenter of xy yz zx if d(w, xy), d(w, xz), d(w, yz) C. There is a constant C = C (δ, C) with the following property: for any triangle in X and any its two C-quasicenters w 1, w 2, the inequality d(w 1, w 2 ) C holds. Quasigeodesics in Gromov hyperbolic spaces have the so-called stability property. For any δ 0, any L 1, A 0, there is a constant C = C(L, A, δ) with the following property: for any δ-hyperbolic space X, any two (L, A)-quasigeodesics γ 1 : I 1 X, γ 2 : I 2 X with the same endpoints, the inequality HD(γ 1 (I 1 ), γ 2 (I 2 )) C holds. Here we use the notation HD d (A, B) to denote the Hausdorff distance between two subsets A, B X of a metric space (X, d); we often write HD(A, B) if the metric in question is clear. Hyperbolic trigonometry For λ > 0, let H 2 ( λ 2 ) be the hyperbolic plane with constant sectional curvature λ 2. We abbreviate H 2 ( 1) by H 2. Let be a triangle in H 2 ( λ 2 ). Denote the three angles by A, B, C and the lengths of their opposite sides by a, b, c. If the angle C is a right angle, then ([G], p.24) cosh(λa) = cos A sin B. Hadamard manifolds Let H be a Hadamard manifold. A classical theorem of Hadamard says that for every x H, the exponential map exp x : T x H H is a diffeomorphism from the

5 Quasiisometries between Hadamard manifolds 5 tangent space T x H onto H. The ideal boundary H of H is defined in the same way as for Gromov hyperbolic spaces. There is the so-called cone topology on H := H H, in which H is homeomorphic to a sphere and H is homeomorphic to a closed ball. The distance function on H is convex: for any two geodesics c 1, c 2 : R H, the function f(t) := d(c 1 (t), c 2 (t)) is convex. In particular, if c 1 ( ) = c 2 ( ), then d(c 1 (t), c 2 (t)) is decreasing. For each x H, let S x T x H be the unit tangent sphere of H at x. There is a map L x : S x H, where for each v S x, L x (v) is the equivalence class containing the ray starting at x with initial direction v. The map L x is a homeomorphism. It follows that for each equator (intersection of S x with a hyperplane of T x H) in S x, its image under L x is a codimension 1 sphere separating H into two balls. For x H and y, z H\{x}, the angle x (y, z) is defined to be the angle in the tangent space T x H between the initial directions of xy and xz. This is a continuous function on y, z H\{x}: if {y i } i=1, {z i } i=1 H\{x} are two sequences with y i y and z i z, then x (y i, z i ) x (y, z). From now on, we shall assume that the sectional curvature of H satisfies λ 2 K 1, where λ 1 is some fixed constant. Then H is a δ 0 -hyperbolic space, where δ 0 = log 3. See [CDP], p.12. There is a family of visual metrics on H: there exists a universal constant C 0, such that for each x H, there is a metric d x on H satisfying 1 C 0 e d(x,ξη) d x (ξ, η) C 0 e d(x,ξη) for all ξ η H. See [B] Section 2.5. Here ξη is the geodesic connecting ξ and η. Let ξ 1 ξ 2 H and set [ξ 1, ξ 2 ] = ξ 1 ξ 2 {ξ 1, ξ 2 }. The orthogonal projection P ξ1 ξ 2 : H [ξ 1, ξ 2 ] is defined as follows: for x / [ξ 1, ξ 2 ], P ξ1 ξ 2 (x) is the unique point w on ξ 1 ξ 2 such that wx is perpendicular to ξ 1 ξ 2, and for x [ξ 1, ξ 2 ], P ξ1 ξ 2 (x) = x. For x, y, z H, let (x, y, z) be the triangle with vertices x, y, z. A triangle (x λ, y λ, z λ ) in H 2 ( λ 2 ) is a comparison triangle of (x, y, z) if d(x, y) = d(x λ, y λ ), d(y, z) = d(y λ, z λ ) and d(z, x) = d(z λ, x λ ). Comparison triangle always exists and is unique up to isometry. The comparison angle x (y, z) is defined to be xλ (y λ, z λ ). For p xy and q xz, points p λ x λ y λ and q λ x λ z λ are said to correspond to p and q if d(p, x) = d(p λ, x λ ) and d(q, x) = d(q λ, x λ ). Let (x, y, z) be a triangle in H. Let (x λ, y λ, z λ ) and (x 1, y 1, z 1 ) respectively be the comparison triangles of (x, y, z) in H 2 ( λ 2 ) and H 2. Then we have the following: (1) xλ (y λ, z λ ) x (y, z) x1 (y 1, z 1 ); (2) For any p xy, q xz, if p λ x λ y λ, q λ x λ z λ and p 1 x 1 y 1, q 1 x 1 z 1 correspond to p, q, then d(p λ, q λ ) d(p, q) d(p 1, q 1 ). See [BH] p.161, p and [CE] p.42 for more details.

6 6 Xiangdong Xie Lemma 2.1. Given any ɛ > 0, there is a constant ɛ 1 = ɛ 1(ɛ) > 0 with the following property: for any three points x, y, z H 2, if d(x, y) ɛ and x (y, z), y (x, z) π/4, then x (y, z) + y (x, z) π ɛ 1. Proof. Let m be the midpoint of xy and x mx and y my with d(m, x ) = d(m, y ) = ɛ/2. Let S be the unique circle in H 2 satisfying the following: (1) the point z and the center p of S lie at the same side of xy; (2) S is tangent to xy at the midpoint m of xy; (3) x (p, y) = π/8. Then the ball B inside S is contained in the triangle (x, y, z), due to our assumption on the angles x (y, z), y (x, z). Hence the area A of (x, y, z) is at least the area of B, which is a constant depending only on ɛ. Now the lemma follows from the Gauss-Bonnett formula: x (y, z) + y (x, z) + z (x, y) = π A. Lemma 2.2. Let c i : [0, ) H (i = 1, 2) be two equivalent rays. Set x = c 1 (0), y = c 2 (0) and ξ = c 1 ( ). Suppose d(x, y) ɛ and x (y, ξ), y (x, ξ) 3π/8. Then x (y, ξ) + y (x, ξ) π ɛ 1, where ɛ 1 is the constant in Lemma 2.1. Proof. For any 0 < η < π/8, pick a point z yξ with x (y, z) x (y, ξ) η. Consider the comparison triangle of (x, y, z) in H 2. Since H has sectional curvature K 1, we have x (y, z) x (y, z) x (y, ξ) η π/4 and y (x, z) y (x, z) 3π/8. It now follows from Lemma 2.1 and the assumption d(x, y) ɛ that x (y, z) + y (x, z) π ɛ 1. Hence x (y, ξ) + y (x, ξ) π ɛ 1 + η. The lemma follows by letting η 0. Lemma 2.3. Given any ɛ > 0, ɛ 1 > 0, there is some ɛ 1 = ɛ 1(ɛ, ɛ 1) > 0 with the following property: for any x, y, z H, if d(x, z) ɛ and d(x, y) ɛ 1, then z (x, y) ɛ 1/10. Proof. The statement clearly holds for H 2. The statement for H follows by comparison. Lemma 2.4. There is an absolute constant C 1 with the following property: for any x H and y, z H\{x}, if x (y, z) π/2, then d(x, yz) C 1 ; furthermore, HD(yz, xy xz) C 1. Proof. It suffices to prove the first claim for y, z H. It clearly holds for H 2, and the general case follows by comparison. The second claim then follows from the convexity of distance function.

7 Quasiisometries between Hadamard manifolds 7 The following lemma says that for x H and ξ, η H, d x (ξ, η) is small if and only if x (ξ, η) is small. Lemma 2.5. For any ɛ > 0, there is some δ = δ(ɛ) > 0 with the following properties: for x H and ξ, η H, (1) if d x (ξ, η) < δ, then x (ξ, η) < ɛ; (2) if x (ξ, η) < δ, then d x (ξ, η) < ɛ. Proof. Let z ξη be the point such that xz is perpendicular to ξη. (1) Fix ɛ > 0. By Lemma 2.4, d(z, xξ), d(z, xη) C 1. There is some b = b(ɛ) > 0 such that x (z, ξ) < ɛ/2 whenever d(x, z) b. Similarly for x (z, η). Now the claim follows from d x (ξ, η) e d(x,ξη) /C 0 = e d(x,z) /C 0. (2) For n 1, let ξ n zξ, η n zη be points at distance n from z. Then x (ξ n, η n ) < 2 x (ξ, η) for sufficiently large n. Let ( x, ξ n, η n ) be a comparison triangle of (x, ξ n, η n ) in H 2 ( λ 2 ) and z n ξ n η n the point corresponding to z. Since H has curvature K λ 2, we have x (ξ n, η n ) x ( ξ n, η n ) and d(x, z) d( x, z n ). Notice that d( x, z n ) is bounded above by a number independent of n and d( x, η n ) as n. It follows that ηn ( x, ξ n ) 0. Similarly, ξn ( x, η n ) 0. Hence for sufficiently large n, the projection w of x on ξ n η n lies in the interior of ξ n η n, xw is perpendicular to ξ n η n and x (w, η n ) x ( ξ n, η n ) < 2 x (ξ, η). Also notice d( x, w) d( x, z n ) d(x, z) = d(x, ξη). We may assume x (ξ, η) π/4. Now we have cosh(λ d(x, ξη)) cosh(λ d( x, w)) = cos η n ( x, w) sin x (w, η n ) cos η n ( x, ξ n ) sin 2 x (ξ, η) 1 sin 2 x (ξ, η). Hence cosh(λ d(x, ξη)) 1 sin 2 x (ξ,η). Now the claim follows since d x(ξ, η) C 0 e d(x,ξη). Lemma 2.6. Given any ɛ > 0, there is some δ = δ(ɛ, λ) > 0 with the following property: for any three distinct points p, ξ, η H, any x pξ, (1) if d x (ξ, η) < δ, then d(x, pη) < ɛ; (2) if d(x, pη) < δ, then d x (ξ, η) < ɛ. Proof. (1) Fix ɛ > 0. Let ɛ 1 be the constant in Lemma 2.1. By Lemma 2.5, there is a constant δ = δ(ɛ 1/10) > 0 such that if d x (ξ, η) < δ, then x (ξ, η) < ɛ 1/10. Then x (p, η) π ɛ 1/10. Let z be the projection of x on pη. Then x (z, η), x (z, p) < π/2 and x (z, η) + x (z, p) x (p, η) π ɛ 1/10. It follows that x (z, η), x (z, p) π/2 ɛ 1/10. Now Lemma 2.2 applied to xp and zp implies that d(x, pη) = d(x, z) < ɛ. (2) Claim: for any ɛ > 0, there is some δ = δ(ɛ, λ) > 0 such that for x H, ξ, η H, if d(x, ξη) < δ, then x (ξ, η) > π ɛ. To see this, first notice that for any ɛ > 0, there is some δ = δ(ɛ, λ) > 0 with the following property: for any x, y, z H 2 ( λ 2 ), if d(x, y) = d(x, z) 10 and d(x, yz) δ, then x (y, z) > π ɛ.

8 8 Xiangdong Xie Now for x H, ξ, η H with d(x, ξη) < δ, choose y xξ and z xη such that d(x, y) = d(x, z) 10 and d(x, yz) < δ. Let (x λ, y λ, z λ ) be a comparison triangle of (x, y, z) in H 2 ( λ 2 ). Then d(x λ, y λ z λ ) d(x, yz) and x (y, z) xλ (y λ, z λ ). Since x (ξ, η) = x (y, z), the claim follows. Now fix ɛ > 0. By Lemma 2.5, there is some ɛ = ɛ (ɛ) > 0 such that if x (ξ, η) < ɛ, then d x (ξ, η) < ɛ. By the above claim, there is some δ = δ(ɛ, λ) > 0 such that if d(x, pη) < δ, then x (p, η) > π ɛ. Now assume d(x, pη) < δ. Then x (ξ, η) = π x (p, η) < ɛ and hence d x (ξ, η) < ɛ. Lemma 2.7. Given any ɛ > 0, there is some δ = δ(ɛ, λ) > 0 with the following property: for x, y H and ξ H, if x (y, ξ) < π/2, y (x, ξ) π/2 and d(x, y) δ, then x (y, ξ) > π/2 ɛ. Proof. Let z n yξ be the point at distance n from y. Then x (y, ξ) = lim n x (y, z n ). Consider a comparison triangle (x, y, z n) of (x, y, z n ) in H 2 ( λ 2 ). Then x (y, z n) x (y, z n ) π/2 and y (x, z n) y (x, z n ) = y (x, ξ) π/2. Let w be the projection of z n on x y. Now by hyperbolic trigonometry sin x (y, z n) = cos z n (w, x ) cosh(λ d(w, x )) Since z n (w, x ) 0 as n, we have cos z n (w, x ) cosh(λ d(x, y)). sin x (y, ξ) = lim sin x (y, z n ) lim inf sin x n n (y, z n) 1 cosh(λ d(x, y))). It follows that there is a function g(t) satisfying g(t) π/2 as t 0 such that x (y, ξ) g(d(x, y)). The lemma follows from this. 3 Constructing the map F Let H 1, H 2 be the universal covers of two compact Riemannian manifolds with negative sectional curvature, and f : H 1 H 2 a quasiisometry. In this section we construct a map F : H 1 H 2 which has the same boundary map as f. In general F is not a bilipschitz homeomorphism and we shall modify F in Section 6 to obtain a bilipschitz map. Notice that, after rescaling the metrics on H 1 and H 2, we may assume that their sectional curvature satisfies λ 2 K 1 for some constant λ 1. We shall assume this from now on.

9 Quasiisometries between Hadamard manifolds 9 The Hadamard manifolds H 1, H 2 are Gromov hyperbolic spaces, and f induces a boundary map f : H 1 H 2. We set ξ := f(ξ) for any ξ H 1. Fix a point p H 1. For each ξ H 1 \{p}, the map F shall send the geodesic line pξ H 1 into the geodesic line p ξ H 2. The map F depends on the point p. However, we shall suppress this information to simplify the notation. Let ξ H 1 \{p} and x pξ. Let F x S x H 1 be the set of unit tangent vectors at x that are perpendicular to pξ. Then F x is an equator in S x H 1. Set E x = L x (F x ) H 1, where L x : S x H 1 H 1 is the homeomorphism defined in Section 2. Then E x is a codimension 1 sphere in H 1 separating H 1 into two open balls that contain p and ξ respectively. We let F (x) be the point in P p ξ ( f(e x)) that is closest to p ; that is, if c : R H 2 is an arc-length parametrization of ξ p from ξ to p and t x := sup{t R : c(t) = P p ξ ( f(β)) for some β E x}, then F (x) = c(t x ). The compactness of f(e x ) and the continuity of the projection P p ξ imply that F (x) P p ξ ( f(e x)). Notice that f(e x ) separates H 2 into two open balls that contain p and ξ respectively. This property implies that F is injective along pξ. Since f is a homeomorphism, F is injective. In the next two sections we shall show that F has good compactness property: both F and F 1 are uniformly continuous. Below we first prove some preliminary results. Let f : H 1 H 2 be an (L, A)-quasiisometry. The point of the following lemma is that the control function η is independent of the base point x. Lemma 3.1. Let p, q, r H 1 be three distinct points. Set x = P pq (r) and x = P p q (r ). Then f : ( H 1, d x ) ( H 2, d x ) is an η-quasisymmetry, where η(t) = Ct 1 L and C depends only on L and A. Proof. Notice that xr is perpendicular to pq. By Lemma 2.4 there is an absolute constant C 1 with d(x, pr), d(x, qr) C 1. Similarly d(x, p r ), d(x, q r ) C 1. It follows that d(f(x), f(pr)), d(f(x), f(qr)) LC 1 + A. Since f is a (L, A)-quasiisometry, the images of pr and qr under f are (L, A)-quasigeodesics. By the stability of quasigeodesics, there is a constant C 2 = C 2 (L, A), such that HD(p r, f(pr)) C 2 and HD(q r, f(qr)) C 2. Hence d(f(x), p r ), d(f(x), q r ), d(f(x), p q ) C 2 + LC 1 + A. That is, f(x) is a C 3 -quasicenter of the three points p, q, r, where C 3 = C 2 +LC 1 +A. Since x is also a C 3 -quasicenter of p, q, r, there is a constant C 4 = C 4 (L, A) such that d(x, f(x)) C 4. Now for any ξ, η H 1, we have d(x, ξ η ) d(f(x), ξ η ) d(x, f(x)) d(f(x), f(ξη)) HD(ξ η, f(ξη)) C 4 d(f(x), f(ξη)) C 2 C 4 d(x, ξη)/l A C 2 C 4. It follows that d x (ξ, η ) C 0 e d(x,ξ η ) C 0 e A+C 2+C 4 (e d(x,ξη) ) 1 L C(dx (ξ, η)) 1 L,

10 10 Xiangdong Xie where C = C 1+ 1 L 0 e A+C 2+C 4 depends only on L and A. For any x H 1, we let q x H 1 be the unique point such that x pq x. Lemma 3.2. There are two absolute constants B 2 B 1 > 0 such that B 1 d x (ξ, q x ) B 2 for all x H 1 and all ξ E x. Proof. Let x H 1 and ξ E x. Then xξ is perpendicular to pq x. By Lemma 2.4 d(x, ξq x ) C 1. The lemma now follows from e d(x,ξ 1ξ 2 ) /C 0 d x (ξ 1, ξ 2 ) C 0 e d(x,ξ 1ξ 2 ). From now on we set x := F (x) for x H 1. Lemma 3.3. There are two constants B 4 B 3 > 0 that depend only on L and A such that B 3 d x (q x, β ) B 4 for all x H 1 and all β E x. Proof. There is some ξ E x such that ξ E x. By Lemma 3.2, we have B 1 d x (q x, ξ ) B 2. On the other hand, Lemma 3.1 says that f : ( H 1, d x ) ( H 2, d x ) is an η-quasisymmetry. For any β E x, we have d x (q x, β ) d x (q x, ξ ) η ( ) dx (q x, β) η d x (q x, ξ) ( B2 Hence d x (q x, β ) η( B 2 B 1 )d x (q x, ξ ) B 2 η( B 2 B 1 ). Similarly by considering d x (q x,ξ ) we d x (q x,β ) obtain d x (q x, β ) B 1. η( B 2 ) B 1 B 1 ). Lemma 3.4. There is a constant B 5 = B 5 (L, A) with the following property: for any x H 1, the projection P p q x ( f(e x)) of f(e x ) on the geodesic p q x is a closed segment with length at most B 5. Proof. P p q x ( f(e x)) is a closed segment since the projection is continuous and f(e x ) is compact and connected. By the definition of x we know that x is the endpoint of the segment that is closer to p. Let η f(e x ). Denote by z the projection of η on p q x and w the projection of x on q xη. Since B 3 d x (q x, η ) C 0 e d(x,q xη ), we have d(x, w) C for some C = C(L, A). Lemma 2.4 applied to z, p, η implies that d(x, p η ) C 1. It follows that w is a (C + C 1 )-quasicenter of p, η, q x. Lemma 2.4 also implies that z is a C 1 -quasicenter of p, η, q x. Hence d(z, w) C for some C = C (L, A). Now d(x, z) d(x, w) + d(w, z) C + C.

11 Quasiisometries between Hadamard manifolds 11 4 F is uniformly continuous In this Section we prove that F is uniformly continuous. It follows that F is a homeomorphism. For any ɛ > 0, we need to find δ > 0 such that d(f (x), F (y)) < ɛ for all x, y H 1 satisfying d(x, y) < δ. This is achieved in Lemmas Recall our notation: q = f(q) for q H 1 and x = F (x) for x H 1. For simplicity, for α, β > 0, the notation β = β(α) means that the number β depends on α and possibly L, A, λ, but nothing else. Lemma 4.1. Given any ɛ > 0, there exists ɛ 1 = ɛ 1 (ɛ) > 0 with the following property: for any x, y H 1, if d x (q x, q y) < ɛ 1 and HD dx ( f(e x ), f(e y )) < ɛ 1, then d(x, y ) < ɛ. Proof. Fix ɛ > 0. Claim: there exists δ 1 = δ 1 (ɛ) > 0 with the following property: for any x, y H 1, if d x (q x, q y) < δ 1 and HD dx ( f(e x ), f(e y )) < δ 1, then for any z P p q y ( f(e y)), there is some w P p q x ( f(e x)) such that d(z, w) < ɛ/2. Assuming the claim, we first finish the proof of the lemma. Set ɛ 1 = δ 1 /(C 2 0e B 5+ɛ ). Let x, y H 1 and assume d x (q x, q y) < ɛ 1, HD dx ( f(e x ), f(e y )) < ɛ 1. By Lemma 3.4, P p q x ( f(e x)) has length at most B 5. It follows from the claim that d(x, y ) < ɛ/2 + B 5. The inequality e d(z,ξη) /C 0 d z (ξ, η) C 0 e d(z,ξη) now implies d y (q x, q y) < δ 1 and HD dy ( f(e x ), f(e y )) < δ 1. Now switching the role of x and y and applying the claim, we see that the Hausdorff distance between P p q x ( f(e x)) and P p q y ( f(e y)) is < ɛ/2. Now let c 1 : R H 2 and c 2 : R H 2 be parametrizations of q xp and q yp respectively such that c 1 ( ) = c 2 ( ) = p and c 1 (t) and c 2 (t) are on the same horosphere centered at p. There are t 1, t 2 R with x = c 1 (t 1 ) and y = c 2 (t 2 ). We may assume t 2 t 1. Pick w P p q x ( f(e x)) such that d(y, w) < ɛ/2. By the definition of x we have w = c 1 (t 0 ) for some t 0 t 1. Since c 1 (t 2 ) and c 2 (t 2 ) = y are on the same horosphere centered at p and c 1 (t 2 ) is the point on the horosphere that is closest to w, we have d(w, c 1 (t 2 )) d(w, y ) < ɛ/2. Consequently d(x, w) < ɛ/2 and d(x, y ) < ɛ. We next prove the claim. Below we shall define ρ i = ρ i (ɛ) > 0 (i = 1, 2, 3, 4). Set δ 1 = min{ρ 1, ρ 2, ρ 3, ρ 4 }. Assume d x (q x, q y) < δ 1 and HD dx ( f(e x ), f(e y )) < δ 1. Let z P p q y ( f(e y)). Let ξ f(e y ) such that its projection on p q y is z. Pick some η f(e x ) with d x (η, ξ ) < δ 1. Let w and w respectively be the projections of η and ξ on p q x. We shall prove d(w, w ) < ɛ/4 and d(w, z) < ɛ/4. We first show that there exists ρ 1 = ρ 1 (ɛ) > 0 with the following property: for any u, v H 1, any ξ f(e v ), and any η f(e u ), if d u (ξ, η ) < ρ 1, then d(w, w ) < ɛ/4, where w and w are respectively the projections of η and ξ on p q u. Let ɛ 1 = ɛ 1(ɛ/4) be given by Lemma 2.2. Since P p q u ( f(e u)) has length at most B 5, Lemma 2.5 and the property of visual metrics imply that there is some ρ 1 =

12 12 Xiangdong Xie ρ 1 (ɛ 1) > 0 such that for any s P p q u ( f(e u)) and ξ 1, ξ 2 H 2, if d u (ξ 1, ξ 2 ) < ρ 1, then s (ξ 1, ξ 2 ) < ɛ 1/10. Let ξ f(e v ), η f(e u ), and assume d u (ξ, η ) < ρ 1. Let w and w respectively be the projections of η and ξ on p q u. Then w (ξ, η ) < ɛ 1/10. Hence w (w, ξ ) > π/2 ɛ 1/10. Since w (w, ξ ) = π/2, Lemma 2.2 implies d(w, w ) < ɛ/4. It remains to prove d(w, z) < ɛ/4. Let ɛ 1 be as above and ɛ 1 = ɛ 1(ɛ/4, ɛ 1) be given by Lemma 2.3. Let c 1, c 2 and t 1, t 2 be as above, and t 0 R such that c 1 ([t 0, t 1 ]) = P p q x ( f(e x)). Then t 1 t 0 B 5. Lemma 2.6 implies that there is some ρ 2 = ρ 2 (ɛ 1) > 0 such that if d x (q x, q y) < ρ 2, then d(c 1 (t), c 2 (t)) < ɛ 1 for all t [t 0 ɛ, t 1 + ɛ]. Notice w = c 1 (t ) for some t [t 0 ɛ/4, t 1 + ɛ/4]. Also z = c 2 (t ) for some t R. We first assume t t 0 ɛ. The choice of ρ 2 and the convexity of distance function imply d(z, c 1 (t )) < ɛ 1 and d(w, c 2 (t )) < ɛ 1. Suppose d(z, w ) ɛ/4. Then Lemma 2.3 implies w (c 1 (t ), z) < ɛ 1/10. Hence w (z, ξ ) π/2 ɛ 1/10. Similarly, z (w, ξ ) π/2 ɛ 1/10. It follows that w (z, ξ )+ z (w, ξ ) π ɛ 1/5, contradicting Lemma 2.2. Now assume t t 0 ɛ. Let ρ 3 = B 3 /4, where B 3 is the constant in Lemma 3.3. Then d x (q y, ξ ) d x (q x, η ) d x (q x, q y) d x (η, ξ ) B 3 /2. Hence d(x, ξ q y) D for some constant D = D(L, A). Noting d(w, x ) d(w, w) + d(w, x ) ɛ/4 + B 5, we have d(w, ξ q y) D + B 5 + ɛ/4. Since d(w, c 2 (t )) ɛ 1 and c 2 (t ) zp, Lemma 2.4 implies that w is a D -quasicenter of p, ξ, q y, where D = D + B 5 + ɛ/4 + C 1 + ɛ 1. Meanwhile, z is also a C 1 -quasicenter of p, ξ, q y. It follows that d(z, w ) D for some D = D (ɛ). Lemma 2.6 implies that there is some ρ 4 = ρ 4 (ɛ) > 0 such that if d x (q x, q y) < ρ 4, then d(c 1 (t), c 2 (t)) < ɛ 1 for all t [t 0 D ɛ, t 1 + D + ɛ]. Then the argument in the preceding paragraph shows d(w, z) < ɛ/4. Lemma 4.2. Given any ɛ 1 > 0, there is some ɛ 2 = ɛ 2 (ɛ 1 ) > 0 with the following property: for any x, y H 1, if d x (q x, q y ) < ɛ 2 and HD dx (E x, E y ) < ɛ 2, then d x (q x, q y) < ɛ 1 and HD dx ( f(e x ), f(e y )) < ɛ 1. Proof. For ɛ 1 > 0, let ɛ 2 be determined by η(ɛ 2 /B 1 )B 4 = ɛ 1 /2. Then ɛ 2 = ɛ 2 (ɛ 1 ). Now suppose d x (q x, q y ) < ɛ 2. By Lemma 3.1 f : ( H 1, d x ) ( H 2, d x ) is an η-quasisymmetry. For any ξ E x, we have d x (q y, q x) ( ) ( ) d x (ξ, q x) η dx (q y, q x ) ɛ2 η. d x (ξ, q x ) B 1 It follows that d x (q y, q x) η( ɛ 2 B 1 )d x (ξ, q x) B 4 η( ɛ 2 B 1 ) = ɛ 1 /2 < ɛ 1. inequality is proved similarly. The second

13 Quasiisometries between Hadamard manifolds 13 The following lemma follows from Lemma 2.6 (2). Lemma 4.3. Given any ɛ 2 > 0, there is some δ = δ(ɛ 2 ) > 0 with the following property: for any x, y H 1, if d(x, y) < δ, then d x (q x, q y ) < ɛ 2. Lemma 4.4. Given any ɛ 2 > 0, there is some δ = δ(ɛ 2 ) > 0 with the following property: for any x, y H 1, if d(x, y) < δ, then HD dx (E x, E y ) < ɛ 2. Proof. Fix ɛ 2 > 0. Set ɛ 2 = ɛ 2. We shall show that there is some δ = δ(ɛ ec 2), 1 > δ > with the following property: for any x, y H 1, if d(x, y) < δ, then for any ξ E y, there is some η E x with d x (η, ξ) < ɛ 2. By symmetry, for any β 1 E x, there is some β 2 E y with d y (β 1, β 2 ) < ɛ 2. Since δ < 1, the property of visual metrics implies that d x (β 1, β 2 ) < ɛ 2. By Lemma 2.5, there is some δ 1 = δ 1 (ɛ 2) > 0 such that if x H 1 and ξ, η H 1 with x (ξ, η) < δ 1, then d x (ξ, η) < ɛ 2. By Lemma 2.7, there is some δ 2 = δ 2 (δ 1 ) > 0 with the following property: for any y 1, y 2 H 1 and ξ H 1, if d(y 1, y 2 ) < δ 2 and y1 (y 2, ξ) π/2, y2 (y 1, ξ) < π/2, then y2 (y 1, ξ) > π/2 δ 1. Let ξ E y and w its projection on pq x. Claim: there is some δ = δ(δ 2 ) > 0 such that d(x, w) < δ 2 whenever d(x, y) < δ. We first finish the proof assuming the claim. We may assume w x, otherwise ξ E x and we may choose η = ξ. Then w (x, ξ) = π/2 and x (w, ξ) < π/2. The choice of δ 2 and the claim imply x (w, ξ) > π/2 δ 1. Let v T x H 1 be a vector perpendicular to pq x such that the angle between v and the direction of xξ is < δ 1. The geodesic ray starting from x in the direction of v determines a point η E x. Now the choice of δ 1 implies d x (η, ξ) < ɛ 2. We next prove the claim. Below we shall define δ = δ (δ 2 ) > 0, δ = δ (δ 2 ) > 0 and δ = δ (L, A) > 0. Set δ = min{δ, δ, δ, 1/2}. By Lemma 2.2, there is a constant ɛ 1 = ɛ 1(δ 2 /2) > 0 with the following property: for any y 1, y 2, y 3 H 1, if d(y 1, y 2 ) δ 2 /2, and y1 (y 2, y 3 ), y2 (y 1, y 3 ) 3π/8, then y1 (y 2, y 3 ) + y2 (y 1, y 3 ) π ɛ 1. Let ɛ 1 = ɛ 1(δ 2 /2, ɛ 1) be the constant in Lemma 2.3. Set δ = min{ɛ 1, δ 2 /10}. By Lemma 4.3, there is some δ = δ (L, A) > 0 such that d y (q x, q y ) < B 3 /2 whenever d(x, y) < δ. Assume d(x, y) < δ. To prove d(x, w) < δ 2 it suffices to show d(y, w) < δ 2 /2. We suppose d(y, w) δ 2 /2 and will get a contradiction. We use the argument in the proof of Lemma 4.1. There are two cases, depending on whether w lies between x and p on pq x. If w lies between x and p, then d(w, yp) < δ, and the argument shows w (y, ξ), y (w, ξ) π/2 ɛ 1/10, contradicting Lemma 2.2. If x lies between w and p, then a similar argument as in the proof of Lemma 4.1 shows that d(w, y) D for some constant D = D(L, A). Then by Lemma 2.6 there is some δ = δ (ɛ 1) > 0 such that d(y, pq x ), d(w, pq y ) < ɛ 1 whenever d(x, y) < δ. Then one gets a contradiction as above.

14 14 Xiangdong Xie 5 F 1 is uniformly continuous In this Section we prove that F 1 is also uniformly continuous. Notice that metric spheres in H i are separating. Since F is a homeomorphism, the following proposition implies that F 1 is uniformly continuous: Proposition 5.1. For any 0 < ɛ < 1, there exists a constant δ = δ(ɛ) > 0 with the following property: for any x H 1, d(f (x), F (S(x, ɛ))) δ. Here S(x, ɛ) denotes the metric sphere in H 1 with center x and radius ɛ. Proposition 5.1 follows from Lemmas Lemma 5.2. Given any ɛ > 0, there is some δ = δ (ɛ) > 0 with the following property: for any x, y H 1 and p H 1, if y xp and d(x, y) ɛ/2, then d(x, y ) > 2δ. Proof. Assume y xp and d(x, y) ɛ/2. We first show that there is some δ 1 = δ 1 (ɛ) > 0 such that d x (ξ 1, ξ 2 ) δ 1 for all ξ 1 E y and ξ 2 E x. Let ɛ 1 = ɛ 1(ɛ/2) be the constant in Lemma 2.2. Fix any ξ 1 E y, ξ 2 E x. The assumption d(x, y) ɛ/2 and Lemma 2.2 imply x (y, ξ 1 ) π/2 ɛ 1. Hence x (ξ 1, ξ 2 ) ɛ 1. Lemma 2.5(1) implies that there is some δ 1 = δ 1 (ɛ 1) > 0 such that d x (ξ 1, ξ 2 ) δ 1. Since f : ( H 1, d x ) ( H 2, d x ) is an η-quasisymmetric map, for any ξ 1 E y, ξ 2 E x, we have B 3 d x (ξ 1, ξ 2) d ( ) ( ) x (q x, ξ 2) d x (ξ 1, ξ 2) η dx (q x, ξ 2 ) B2 η. d x (ξ 1, ξ 2 ) It follows that d x ( f(e x ), f(e y )) δ 2 = δ 2 (ɛ), where δ 2 = B 3 (η( B 2 δ 1 )) 1. By Lemma 2.5(2), there is a constant δ 3 = δ 3 (δ 2 /ec0) 2 > 0: for any z H 2 and η 1, η 2 H 2, if z (η 1, η 2 ) < δ 3, then d z (η 1, η 2 ) < δ 2 ec 2 0 δ 1. Let δ 4 = δ(δ 3, λ) be the constant in Lemma 2.7. We shall show that δ := 1 2 min{δ 4, 1} has the required property. Suppose d(x, y ) 2δ δ 4. Fix some η f(e x ) such that the projection of η on p q x is x. Lemma 2.7 applied to x, y and η implies y (x, η ) > π/2 δ 3. Denote by v 1 T y H 2 the initial direction of y η at y. Let v 2 be the projection of v 1 on the hyperplane of T y H 2 orthogonal to the direction of p q x, and v 3 the unit vector in the direction of v 2. Then the angle between v 1 and v 3 is less than δ 3. The geodesic in the sphere S y H 2 from v 1 to v 3 gives rise to a continuous path γ in H 2 from η f(e x ) to a point β in E y. Since the angle between v 1 and v 3 is less than δ 3, for any ξ γ, we have y (η, ξ) < δ 3. Now by the choice of δ 3 we have d y (ξ, η ) < δ 2 /(ec 2 0) for all ξ γ. Since d(x, y ) 1, the property of visual metric implies d x (ξ, η ) < δ 2. Since η f(e x ) and d x ( f(e x ), f(e y )) δ 2, the path γ must lie in the component B of H 2 \ f(e y ) that contains f(e x ). In particular, β B. Notice B = f(b), where B is the component of H 1 \E y that contains E x.

15 Quasiisometries between Hadamard manifolds 15 Hence β = ξ 3 for some ξ 3 B. There is some z yq x, z y such that ξ 3 E z. Since F is a homeomorphism, z y q x, z y. It follows that the projection of β on p q x lies in z q x, contradicting β E y. Lemma 5.3. There is a constant ɛ 2 = ɛ 2 (ɛ) > 0 with the following property: for any x, y H 1, any p H 1, if d(x, y) = ɛ and d(x, pq y ) < ɛ 2, then d(x, y ) δ, where δ is the constant in Lemma 5.2. Proof. Since F is uniformly continuous, there exists some ɛ 1 = ɛ 1 (δ ) > 0 such that d(y 1, y 2) < δ for all y 1, y 2 H 1 satisfying d(y 1, y 2 ) < ɛ 1, where δ is the constant in Lemma 5.2. Set ɛ 2 = min{ɛ/10, ɛ 1 }. Now let x, y H 1 and p H 1 with d(x, y) = ɛ and d(x, pq y ) < ɛ 2. Let z pq y with d(x, z) < ɛ 2. Then d(x, z ) < δ. On the other hand, d(y, z) ɛ/2. Lemma 5.2 implies d(z, y ) 2δ. It follows that d(x, y ) δ. Lemma 5.4. There is a constant δ = δ (ɛ) with the following property: for any x, y H 1, any p H 1, if d(x, y) = ɛ and d(x, pq y ) ɛ 2, then d(x, y ) δ, where ɛ 2 is the constant in Lemma 5.3. Proof. Since d(x, pq y ) ɛ 2, Lemma 2.6(1) implies d x (q x, q y ) δ 2, where δ 2 depends only on ɛ 2. Then the argument in the second paragraph of the proof of Lemma 5.2 shows that d x (q x, q y) δ 3, where δ 3 depends only on ɛ. Now Lemma 2.6(2) implies that d(x, p q y) δ for some constant δ = δ (δ 3 ). Since y p q y, the lemma follows. Now Proposition 5.1 holds with δ = min{δ, δ }. 6 Modifying F In this Section we modify F to obtain a bilipschitz homeomorphism. The argument is a minor modification of that in Section 7 of [TV2]. We include it here mainly for completeness. Hidden behind this is Sullivan s theory of Lipschitz structures. It is used in the proof of Lemma 3.9 in [TV2], which we record as Lemma 6.4 below. Let f : X Y and g : X Y be two maps between metric spaces. The distance between f and g is d(f, g) := sup{d(f(x), g(x)) : x X}. We say f and g lie at finite distance from each other if d(f, g) <. For any subset A X, we also denote d(f, g; A) := d(f A, g A ). Let f : H 1 H 2 be a quasiisometry and F : H 1 H 2 the homeomorphism constructed in Section 3. Proposition 6.1. Suppose H 1 and H 2 are not 4-dimensional. Then for any ɛ > 0, there is a bilipschitz homeomorphism G : H 1 H 2 such that d(f, G) ɛ.

16 16 Xiangdong Xie A decomposition of H 1 Recall that H 1 is the universal cover of a compact Riemannian manifold H 1 /G 1 with sectional curvature λ 2 K 1, where G 1 acts on H 1 as the group of deck transformations. Fix a compact connected fundamental domain D 1 H 1 for the action of G 1 on H 1. Then H 1 is covered by {g(d 1 ) : g G 1 }. Let Γ be the incidence graph of this covering: the vertex set of Γ is G 1 ; two vertices g 1 and g 2 are connected by an edge if g 1 (D 1 ) g 2 (D 1 ). The action of G 1 on H 1 induces an action of G 1 on Γ and this action is transitive on the vertices. In particular, there is some positive integer m such that each vertex has valence m. Recall that a graph is finitely colorable if there is an integer N 1 and a map φ : V {1, 2,, N} defined on the vertex set V of the graph, such that φ(v 1 ) φ(v 2 ) whenever v 1 and v 2 are connected by an edge. One may assume φ is surjective. The following well-known (and easy to prove) lemma implies that the graph Γ defined above is finitely colorable. Lemma 6.2. Let Γ be a graph. If there is some integer m such that each vertex has valence at most m, then Γ is finitely colorable. Hence there is a surjective map φ : G 1 {1, 2,, N} such that g 1 and g 2 have different colors (i.e., φ(g 1 ) φ(g 2 )) whenever they are joined by an edge. For 1 i N, set K i = {g(d 1 ) : φ(g) = i} and K i = K 1 K i. Notice that each family K i consists of disjoint translates of D 1, and the N families K 1,, K N cover H 1. Since the action of G 1 on H 1 is proper and cocompact, there is a constant b 1 > 0 such that d(g(d 1 ), h(d 1 )) b 1 whenever g(d 1 ) h(d 1 ) =. A solid family of maps The notion of solid family is defined and discussed in [TV2]. It is closely related to the approximation of embeddings by locally bilipschitz maps. Let X and Y be metric spaces. A family F of embeddings f : X Y is said to be solid if its closure is a compact family of embeddings. If Y = R n and X R n is either open or compact, then F is solid if and only if the following three conditions hold (see p. 315 of [TV2]): (1) For every x 0 X, the set {f(x 0 ) : f F} is bounded; (2) For every x 0 X and ɛ > 0, there is a neighborhood U of x 0 such that d(f(x), f(x 0 )) < ɛ whenever x U and f F; (3) For every x 0 X and every neighborhood U of x 0, there is ɛ > 0 such that d(f(x), f(x 0 )) ɛ whenever x X\U and f F. We next construct a solid family from the map F. Let f : H 1 H 2 be a quasiisometry and F : H 1 H 2 the homeomorphism constructed in Section 3. Let i = 1 or 2. Then H i is the universal cover of a compact

17 Quasiisometries between Hadamard manifolds 17 Riemannian manifold H i /G i whose sectional curvature satisfies λ 2 K 1. Fix a compact connected fundamental domain D i for G i and an interior point x i of D i. Let d i be the diameter of D i. There is a constant b i > 0 such that d(g(d i ), h(d i )) b i whenever g(d i ) h(d i ) =. Let A = N b1 /2(D 1 ) and A = N 3b1 /8(D 1 ), where for t > 0 and a subset Z X of a metric space X, N t (Z) := {x X : d(x, Z) < t} denotes the open t-neighborhood of Z. Set B = {g(d 1 ) : g(d 1 ) D 1 }. Then B is compact connected and A B. Since F is uniformly continuous, there exists some constant α > 0 such that the diameter of F (g 1 (B)) is at most α for all g 1 G 1. The exponential map h i := exp xi : T xi H i H i is a diffeomorphism. Denote U = h 1 1 (A) and U = h 1 1 (A ). Then there is some M i 1 such that h 1 : U A is a M 1 -bilipschitz map and the restriction of h 2 on the closed ball B(o, α+d 2 +1) T x2 H 2 is M 2 -bilipschitz. For every g 1 G 1, we fix some g 1 G 2 such that g 1(F (g 1 (x 1 ))) D 2. For each Q = g 1 (D 1 ), let F Q : U T x2 H 2 be defined by F Q := h 1 2 g 1 F g 1 h 1. Notice that g 1(F (g 1 (B))) B(x 2, α + d 2 ) and F Q (U) B(o, α + d 2 ) T x2 H 2. Set Lemma 6.3. The family F is solid. F = {F Q : Q = g 1 (D 1 ), g 1 G 1 }. Proof. It suffices to verify the three conditions above. (1) holds because F Q (U) B(o, α + d 2 ) for all F Q F. (2) is true since F is uniformly continuous, G 1, G 2 act as isometries and h 1 U, h 2 B(o,α+d2 ) are fixed bilipschitz maps. Similarly (3) follows from the fact that F 1 is uniformly continuous. Proof of Proposition 6.1 The following result is key in the proof. Recall that for n 4, every topological n- manifold has a unique lipschitz structure and homeomorphisms between n-manifolds can be approximated by locally bilipschitz homeomorphisms, see [S] or Section 4 of [TV2]. Lemma 6.4. (Lemma 3.9 in [TV2]) Let U, U, V, W be open sets in R n such that W V U, U U, W U V, and U is compact. Let F be a solid family of embeddings g : U R n. For n = 4, we also assume that the members of F can be approximated by locally bilipschitz embeddings. Let ɛ > 0, L 1. There exist δ > 0 and L 1 with the following properties:

18 18 Xiangdong Xie Let h : V R n be a locally bilipschitz embedding, and g F be such that d(g, h; V ) δ. Then there is a locally bilipschitz embedding h : U R n such that (1) d(h, g; U ) ɛ; (2) h = h in W U ; (3) h U is L -bilipschitz if h is locally L-bilipschitz. Here δ depends only on τ = (U, U, V, W, F, ɛ) and L depends only on τ and L. Let N, K i, Ki (1 i N) and U, U be as above. For Q KN and 1 i N, let Q i = N 2 i 1 b 1 (Q). Set V i = {Q i : Q Ki }, W i = {Q i+1 : Q Ki }. We also let V 0 =, W 0 =. For Q = g 1 (D 1 ) K i and 0 < t b 1 /2, let Note W Q (t) V Q (t) U. V Q (t) = h 1 1 (N t (D 1 ) g 1 1 (V i 1 )), W Q (t) = h 1 1 (N t (D 1 ) g 1 1 (W i 1 )). Since G 1 acts transitively on the family K N, the choice of b 1 implies the following: for any 0 < t b 1 /2, there is a finite family S(t) such that V Q (t) and W Q (t) belong to S(t) for every Q K N. We apply Lemma 6.4 with U, U above, V = V Q (b 1 /2), W = W Q (b 1 /2), and F = {F Q }. Since the family S(b 1 /2) is finite, we obtain: Lemma 6.5. Let ɛ > 0 and L 1. Then there are positive numbers δ = δ(ɛ) ɛ and L = L (ɛ, L) L with the following property: Let Q K N, h : V Q(b 1 /2) T x2 H 2 be a locally L-bilipschitz embedding, and g F be such that d(g, h; V Q (b 1 /2)) δ. Then there is a locally bilipschitz embedding h : U T x2 H 2 such that (1) d(h, g; U ) ɛ; (2) h = h in W Q (3b 1 /8); (3) h U is L -bilipschitz. Since F 1 is uniformly continuous, there is a number q > 0 such that d(f (x), F (y)) q whenever x, y H 1 with d(x, y) b 1 /8. Define δ N δ N 1 δ 0 > 0 by δ N = min{q/3, ɛ, 1} and δ j 1 = δ(δ j /M 2 )/M 2, where δ( ) is the function in Lemma 6.5. We also define numbers L 0 L N by L 0 = 1 and L j = M 1 M 2 L (δ j /M 2, M 1 M 2 L j 1 ), where L (, ) is the function in Lemma 6.5. Observe that the sequences (δ 0,, δ N ) and (L 0,, L N ) depend only on ɛ. We show by induction that the following lemma is true for every integer j [0, N]: Lemma 6.6. Statement S(j) : There is an embedding F j : V j H 2 with the following property: (1) d(f j, F ; V j ) δ j ; (2) F j (Q j ) F (N 3b1 /8(Q)) for every Q K j; (3) F j is locally L j -bilipschitz.

19 Quasiisometries between Hadamard manifolds 19 Proof. Since V 0 =, S(0) is true. Suppose that S(j 1) is true. Thus we have an embedding F j 1 : V j 1 H 2. We define F j (x) = F j 1 (x) for x W j 1. Recall that for each g 1 G 1 we fixed some g 1 G 2 such that g 1(F (g 1 (x 1 ))) D 2. Let Q = g 1 (D 1 ) K j. Then F Q = h 1 2 g 1 F g 1 h 1. Set h Q = h 1 2 g 1 F j 1 g 1 h 1 VQ (b 1 /2). S(j 1) implies that image(h Q ) B(o, α+d 2 +1) T x2 H 2, h Q is locally M 1 M 2 L j 1 - bilipschitz, and d(h Q, F Q ; V Q (b 1 /2)) M 2 δ j 1. Hence we can apply Lemma 6.5 with g = F Q, h = h Q, ɛ = δ j /M 2 and L = M 1 M 2 L j 1. We obtain a locally bilipschitz embedding h Q : U T x 2 H 2 such that (a) d(h Q, F Q; U ) δ j /M 2, (b) h Q = h Q on W Q (3b 1 /8), (c) h Q U is L (δ j /M 2, M 1 M 2 L j 1 )-bilipschitz. Setting F j = (g 1) 1 h 2 h Q h 1 1 g1 1 in Q j we obtain a well-defined map F j : V j H 2. We show that F j satisfies the conditions (1), (2), (3) and that F j is injective. Let Q K j. If Q K j 1, (1) follows from S(j 1). If Q K j, (a) implies h Q (U ) B(o, α + d 2 + 1) T x2 H 2 and hence d(f j, F ; Q j ) M 2 d(h Q, F Q ; U ) δ j. To prove (2), let again Q K j. If Q K j 1, (2) follows from S(j 1). Suppose Q K j. Since δ j < q, the choice of q and S(j) (1) imply (2). If Q Kj 1, F j is locally L j -bilipschitz in Q j by S(j 1) (3). If Q K j, then (c) implies F j Qj is L j -bilipschitz. Hence F j is a locally L j -bilipschitz immersion. We finally show that F j is injective. We know that F j Qj is injective for every Q Kj. Moreover, if Q, R Kj and Q R =, then (2) implies that F j (Q j ) F j (R j ) =. Hence it suffices to show that F j (x) F j (y) when j 2, x y, x Q j and y R j where Q K j, R Kj 1 and Q R. The equality F j = (g 1) 1 h 2 h Q h 1 1 g1 1 is valid in N 3b1 /8(Q) W j 1. Hence we may assume that y / N 3b1 /8(Q). By the choice of q, we have d(f (x), F (y)) q. By (1) we obtain d(f j (x), F j (y)) d(f (x), F (y)) d(f j (x), F (x)) d(f j (y), F (y)) q 2δ j q/3 > 0. Now notice V N = H 1. Hence F N : H 1 H 2 is an embedding with d(f N, F ) δ N ɛ. It follows that F N is a homeomorphism. By (3), F N is locally L N -bilipschitz. Since H 1 and H 2 are geodesic metric spaces, F N is L N -bilipschitz. This completes the proof of Proposition 6.1.

20 20 Xiangdong Xie Proof of Theorem 1.1 Let f : H 1 H 2 be a quasiisometry, F the map constructed in Section 3 and G the map in Proposition 6.1. Since G is bilipschitz and d(f, G) <, F is also a quasiisometry. From the construction of F one sees easily that F = f. We shall prove d(f, F ) <. Let L 1 and A 0 be such that both f and F are (L, A)-quasiisometries. Let x H 1. Pick p, q, r H 1 such that x pq and that xr is perpendicular to pq. By Lemma 2.4 d(x, pr), d(x, qr) C 1. So d(f(x), f(pr)), d(f(x), f(qr)) L C 1 + A. Since f(pr), f(qr) are (L, A)-quasigeodesics, the stability of quasigeodesics yields HD(f(pr), p r ) C, HD(f(qr), q r ) C and HD(f(pq), p q ) C, where C depends only on L and A. It follows that d(f(x), p r ) C + L C 1 + A, d(f(x), q r ) C + L C 1 + A and d(f(x), p q ) C. In other words, f(x) is a (C + L C 1 + A)- quasicenter of p, q, r. Similarly F (x) is also a (C + L C 1 + A)-quasicenter of p, q, r. It follows that d(f (x), f(x)) D for some D depending only on L and A. This is true for every x H 1. Hence d(f, f) D. Proof of Corollary 1.2 Let h : BC n Bn C (n 2) be a quasisymmetric map, where BC n is equipped with the Carnot metric. Then there is a quasiisometry f : BC n Bn C with f = h (see [BS]). By Theorem 1.1, there is a bilipschitz homeomorphism G : BC n Bn C with d(g, f) <. G is clearly a quasiconformal map in the complex hyperbolic metric. The fact d(g, f) < implies that G and f have the same boundary map, which is h. 7 Open questions In this Section we present several questions related to the result in this paper. The first natural question is the following. Question 7.1. Let H 1 and H 2 be two Hadamard n-manifolds (whose sectional curvatures are bounded from below) with n 4, and f : H 1 H 2 a quasiisometry. Is f always a finite distance from a bilipschitz homeomorphism? Notice that a Hadamard manifold has bounded geometry if the sectional curvature is bounded from below. Recall that a subset A X of a metric space X is a separated net if there are constants a, b > 0 such that d(x, y) a for distinct x, y A and d(x, A) b for all x X. Observe that the restriction of a quasiisometry f : X Y to a suitable separated net is a bilipschitz embedding of the net into Y. Hence an affirmative answer to the following problem implies the positive answer to Question 7.1. Question 7.2. Let H 1 and H 2 be two Hadamard n-manifolds (whose sectional curvatures are bounded from below) with n 4, A H 1 a separated net, and f : A H 2

21 Quasiisometries between Hadamard manifolds 21 a bilipschitz embedding. Does f always extend to a bilipschitz homeomorphism from H 1 to H 2? In the case H 1 = H 2 = R n, Question 7.2 has been asked by Alestalo-Trotsenko- Vaisala [ATV]. Hadamard manifolds have no topology: they are contractible. But Question 7.1 can also be asked for more general manifolds. For example one can consider quasiisometries between noncompact hyperbolic surfaces. Question 7.3. Let X and Y be two open complete Riemannian n-manifolds (n 4). Suppose X and Y are Gromov hyperbolic and have bounded geometry. Let f : X Y be a quasiisometry. Suppose there is a homeomorphism F : X Y such that F X = f. Is f always at a finite distance from a bilipschitz homeomorphism? Here X and Y are Gromov compactifications of X and Y respectively. For an arbitrary quasiisometry f : X Y, the boundary map f in general does not have a homeomorphic extension, let alone a bilipschitz extension. Every quasisymmetric map f : X Y between two metric spaces extends to a quasisymmetric map between their completions. Hence quasisymmetric maps between Euclidean domains extend to quasisymmetric maps between their closures. A basic question is to what extent the converse is true, that is, under what conditions, a quasisymmetric map between the boundaries of two domains extends to a quasisymmetric map between the domains? The theorem of Tukia-Vaisala shows that it is the case when the domains are balls in R n. How about more general domains? A necessary condition is that the quasisymmetric map between the boundaries must be power quasisymmetric: quasisymmetric maps between connected metric spaces are power quasisymmetric, and domains are connected. Again, in general a quasisymmetric map between the boundaries may not have a homeomorphic extension to the closures. Question 7.4. Let Ω 1, Ω 2 R n be two domains, and f : Ω 1 Ω 2 a power quasisymmetric map. Suppose there is a homeomorphism F : Ω 1 Ω 2 such that F Ω1 = f. Does f extend to a quasisymmetric map from Ω 1 to Ω 2? One may have to restrict attention to the so-called uniform domains. Uniform domains are considered nice domains in many analysis problems. And they are Gromov hyperbolic in the quasihyperbolic metric. See [BHK] for more details. References [A] L. Ahlfors, Extension of quasiconformal mappings from two to three dimensions, Proc. Nat. Acad. Sci. U.S.A