The quasi-isometry classification of rank one lattices

Richard Schwartz

doi:10.1007/BF02698639

Schwartz, Richard

2007-08-30 00:00:00

THE QUASI-ISOMETRY CLASSIFICATION OF RANK ONE LATTICES by RICr~ARD EVAN SCHWARTZ* ABSTRACT. Let X be a symmetric space---other than the hyperbolic plane--of strictly negative sectional curvature. Let G be the isometry group of X. We show that any quasi-isometry between non-unlform lattices in G is equivalent to (the restriction of) a group element of G which commensurates one lattice to the other. This result has the following corollaries: I. Two non-uniform lattices in G are quasi-isometric if and only if they are commensurable. 2. Let I ~ be a finitely generated group which is quasi-isometric to a non-uniform lattice in G. Then 1' is a finite extension of a non-uniform lattice in G. 3. A non-uniform lattice in G is arithmetic if and only if it has infinite index in its quasi-isometry group. !. Introduction A quasi-isometry between metric spaces is a map which distorts distances by a uni- formly bounded factor, above a given scale. (See w 2.1 for a precise definition.) Quasi- isometrics ignore the local structure of metric spaces, but capture a great deal of their large scale geometry. A finitely generated group G has a natural word metric, which makes it into a path metric space. Different finite generating sets produce quasi-isometric spaces. In other words, quasi-isometric properties of the metric space associated to the group only depend on the group itself. There has been much interest recently in understanding these quasi- isometric properties. (See [Grl] for a detailed survey.) Lattices in Lie groups provide a concrete and interesting family of finitely gene- rated groups. A uniform (i.e. co-compact) lattice in a Lie group is always quasi-isometric to the group itself (equipped with a left-invariant metric). In particular, two uniform lattices in the same Lie group are quasi-isometric to each other. In contrast, much less is known about quasi-isometries between non-uniform lattices. S. Gersten posed the natural question: When are two non-uniform lattices quasi-isometric to each other? The purpose of this paper is to answer Gersten's question, completely, in the case of rank one semi-simple IAe groups. Such groups agree, up to index 2, with isometry groups * Supported by an NSF Postdoctoral Fellowship. 134 RICHARD EVAN SCHWARTZ of negatively curved symmetric spaces. (See w 2.3 for a list.) To save words, we shall enlarge our Lie groups so that they precisely coincide with isometry groups of negatively curved symmetric spaces. We will simply call these groups rank one Lie groups. We will call their lattices rank one lattices. The most familiar examples of non-uniform rank one lattices are fundamental groups of finite volume, non-compact, hyperbolic Riemann surfaces. Such lattices are finitely generated free groups. Every two such are quasi-isometric to each other. In all other cases, we uncover a rigidity phenomenon which is related to, and in some sense stronger than, Mostow rigidity. 1 .t. Statement of Results Let G be a Lie group. Given two lattices A1, As C G, we say that an element 7 e G commensurates A 1 to A s if 7 o A 1 o 7-1 n A s has finite index in A s. In particular, the group of elements which commensurate A to itself is called the commensurator of A. Given two lattices A1 and As, an isometry which commensurates A 1 to A s induces, by restriction, a quasi-isometry between A a and A s. We shall let Isom(I-I ~) be the isometry group of the hyperbolic plane. Theorem 1.1 (Main Theorem). -- Let G 4: Isom(H 2) be a rank one Lie group, and let A 1 and A s be two non-uniform lattices in G. Any quasi-isometry between A 1 and A s is equivalent to (the restriction of) an element of G which commensurates A 1 to A s. Roughly speaking, two quasi-isometrics between metric spaces are said to be equivalent if one can be obtained from the other by a uniformly bounded modification. (See w 2.1 for a precise definition.) The group of quasi-isometrics, modulo equivalence, of a metric space M to itself is called the quasi-isometry group of M. The Main Theorem immediately implies: Corollary 1.2. -- Let A C G be a non-uniform lattice in a rank one Lie group G 4: Isom(I-I~). Then the commensurator of A and the quasi-isometry group of A are canonically isomorphic. Here is the complete quasi-isometry classification of rank one lattices. Corollary 1.3. -- Suppose, for j = 1, 2, that A s is a lattice in the rank one Lie group Gj. Then A 1 and A 2 are quasi-isometric if and only if GI = G~ and exactly one of the following sta- tements holds: 1. A~ is a non-uniform lattice in Isom(I-I~). 2. A s is a uniform lattice. 3. A s is a non-uniform lattice in Gj + Isom(I-I~). Furthermore A 1 and A 2 are commensurable. In a rather tautological way, the Main Theorem gives the following extremely general rigidity result. THE QUASI-ISOMETRY CLASSIFICATION OF RANK ONE LATTICES 135 Corollary 1.4. --Let G 4: Isom(I-I 2) be a rank one Lie group. Suppose that r is an arbitrary finitely generated group, quasi-isometric to a non-uniform lattice A of G. Then I' is a finite extension of a non-uniform lattice A' of G. Furthermore A and A' are commensurable. Finally, combining the Main Theorem with Margulis' well-known characte- rization of arithmeticity we obtain: Corollary 1.5. -- Let G 4: Isom(H 2) be a rank one Lie group, and let A be a non-uniform lattice in G. Then A is arithmetic if and only if it has infinite index in its quasi-isometry group. This last corollary is philosophical in nature. It says that arithmeticity, which is a number theoretic concept, is actually implicit in the group structure. 1.2. Outline of the Proof Let X 4: H ~ be a negatively curved symmetric space. (See w 2.3 for a list.) A neutered space is defined to be the closure, in X, of the complement of a disjoint union of horoballs. The neutered space is equipped with the path metric induced from the Riemannian metric on X. This path metric is called the neutered metric. Let A denote the isometry group of ~. We say that D. is equivariant if the quotient space ~/A is compact (1). Step 1. -- Introducing Neutered Spaces We will show in w 2 (the standard fact) that an equivariant neutered space is quasi- isometric to its isometry group. Since any non-uniform lattice can be realized--up to finite index--as the group of isometrics of an equivariant neutered space, we can ignore the groups themselves, and work with neutered spaces. Step 2.- Horospheres Quasi-Preserved We will show that the image of a quasi-isometric embedding of a horosphere into a neutered space must stay close to some (unique) boundary horosphere of that neutered space. In particular, any quasi-isometry between neutered spaces must take boundary horospheres to boundary horospheres. This is done in w 3 and w 4. Step 3. -- Ambient Extension Let ~1, ~2 C X be neutered spaces, and let q : ~1 --~ ~ be a quasi-isometry (rela- tive to the two neutered metrics). From Step 2, q pairs up the boundary components of ~1 with those of ~. In w 5, we extend q to a map ~ : X -+ X. It turns out that this extension is a quasi-isometry of X, which "remembers" the horoballs used to define ~j. We will abbreviate this by saying that ~ is adapted to the pair (~a, ~2). (t) Equivariant neutered spaces are called invariant cores in [Grl]. Technically speaking, we are only concerned with results about equivariant neutered spaces, since these arise naturally in connection with lattices. However, many of the steps in our argument do not require the assumption of equivariance, and the logic of the argument is clarified by the use of more general terminology. 136 RICHARD EVAN SCHWARTZ Step 4. -- Geometric Limits We now introduce the assumption that fl~ is an equivariant neutered space. Let h = O~ be the boundary extension of ~ to 0X. It is known that h is quasiconformal, and almost everywhere differentiable. (In the real hyperbolic case, this is due to Mostow [M2] ; the general formulation we need is due to Pansu [P].) We will work in stereographic coordinates, in which OX- oo is a Heisenberg group. (Euclidean space in the real- hyperbolic case.) By "zooming-in" towards a generic point x of differentiability of h, we produce a new quasi-isometry q' : X -+ X having the following properties: 1. q' is adapted to a new pair (t)~, ~) of equivariant neutered spaces. 2. h' -- Oq' is a nilpotent group automorphism of 0X -- oo. (Linear transformation in the real case.) 3. h' = dh(x), the linear differential at x. In w 6, we make this construction in the real hyperbolic case. In w 8, we work out the general case. Step 5. -- Inverted Linear Maps Suppose q' is the quasi-isometry of X produced in Step 5. Using a trick involving inversion, we show that h' is in fact a Heisenberg similarity. (Ordinary similarity in the real-hyperbolic case.) This is to say that dh(x) is a similarity. Since x is generic, the original map k is 1-quasi-conformal, and hence is the restriction of an isometry of X. We work out the real hyperbolic case in w 7, and the complex hyperbolic case in w 8. The qua- ternionic (and Cayley) cases have similar proofs, and also follow directly from [P, Th. 1]. Step 6. -- The Commensurator Let ~ : X ~ X be a quasi-isometry adapted to the pair (f~l, f~2) of equivariant neutered spaces. We know from Steps 4-5 that ~ is equivalent to an isometry q.. In w 9, we use a trick, similar to the one developed in Step 5, to show that q. commensurates the isometry group of f~l to that of f~2. In w 10, we recall the correspondence between f~j and A j, and see that q. commensurates A1 to A~. Remark. -- Our quasi-isometry from one lattice to another is not a priori assumed to (virtually) conjugate one lattice to the other. This situation is in marked contrast to the situation in Mostow rigidity. Accordingly, the details of Steps w 4-6 are quite different (in places) from those usually associated with Mostow rigidity [M1]. 1.3. Suggested Itinerary It should be possible for the reader to read only portions of the paper and still come away with an understanding of the main ideas. Here are some suggestions for the order in which to read the paper: 1. To see the basic skeleton (and prettiest part) of the paper, without getting bogged down in the details of Step 2 and Step 3, read w 2. I-2.4, w 5.1, w 6, w 7, w 9 and w 10 1-10.2. THE QUASI-ISOMETRY CLASSIFICATION OF RANK ONE LATTICES 2. To understand Step 3, use Step 2 as a black box. Read w 2, w 3.1-3.2 and w 5. 3. To understand Step 2, read w 2 and w 3, using w 4 as a black box. 4. Read w 4 last. It helps to draw a lot of pictures here. In general--and especially for Step 2--the reader should first restrict his attention to real hyperbolic space. The general case is conceptually the same as this case, but requires tedious background results (w 2.5-2.7 and w 4.1) on rank one geometry. 1.4. Acknowledgements It almost goes without saying that this paper would have been impossible without the beautiful ideas of Mostow. I would also like to thank: 1. Benson Farb, for originally telling me about Gersten's question, for numerous conver- sations about geometric group theory, and for his enthusiasm about this work. 2. Peter Doyle, for a great deal of moral support. 3. Steve Gersten, for posing such an interesting question. 4. Pierre Pansu, for his theory of Carnot differentiability. 5. Mladen Bestvina, Martin Bridgeman, Jeremy Kahn, Misha Kapovich, Geoff Mess, Bob Miner, and Pierre Pansu, for helpful and interesting conversations on a variety of subjects related to this work. 1.5. Dedication I dedicate this paper to my wife, Brienne Elisabeth, on the occasion of our wedding, May 13, 1995. 2. Background 2.1. Quasi-isometrics Let (M, d) be a metric space, with metric d. A subset N C M is said to be a K-net if every point of M is within K of some point of N. A K-quasi-isometric embedding of (M, d) into (M', d') is a map q : N --> M' such that: 1. N is a K-net in M. 2. d'(q(x), q(y)) ~ [d(x,y)lK -- K, Kd(x,y) + K], for x,y ~ N. The map q is said to be a K-quasi-isometry if the set N' = q(N) is a K-net in M'. In this case, the two metric spaces (M, d) and (M', d') are said to be K-quasi-isometric. When the choice of K is not important, we will drop K from the terminology. Two quasi-isometric embeddings (or quasi-isometrics) qa, q2 :M ---> M' are said to be equivalent if there are constants C 1 and C2 having the following properties: 1. Every point of N1 is within C1 of N2, and vice versa. 2. If xj ~ Nj are such that d(xx, x~) ~< C1, then d'(ql(xl), q2(x~.)) <<. C~. 18 138 RICHARD EVAN SCHWARTZ It is routine to verify that the above relation is an equivalence relation, and that, modulo this relation, the quasi-isometries of M form a group. We call this group the quasi- isometry group of M. A K-quasi-geodesic (segment) in M is a K-quasi-isometric embedding of (a segment of) R into M. If M happens to be a path metric space, then every such segment is equi- valent to a K-bi-Lipschitz segment. The uniformity of the equivalence only depends on K. For more information about quasi-isometries, see [Grl] and [El]. 2.2. The Word Metric Let G be a finitely generated group. A finite generating set S C G is said to be symmetric provided that g ~ S if and only if g- 1 ~ S. The word metric on G (resp. S) is defined as follows: The distance from gl, g, ~ G is defined to be the minimum number of generators needed to generate the element gl g2-1. It is easy to see that this makes G into a path metric space. Two different finite generating sets $1 and $2 produce Lipschitz equivalent (and in particular, quasi-isometric) metric spaces. 2.3. Rank One Symmetric Spaces Here is the list of symmetric spaces which have (strictly) negative sectional curvature: 1. Real hyperbolic n-space, I-I". 2. Complex hyperbolic n-space, CH". 3. Quaternionic hyperbolic n-space, OH". 4. The Cayley plane. These spaces are also known as rank one symmetric spaces. For basic information about these spaces, see [C], [Grl], [Go], [E2], or [T]. We will let X denote a rank one symmetric space. We will never take X ----- I-~ ---- CH 1, unless we say so explicitly. Also, to simplify the exposition, we will omit the description of the Cayley plane. Readers who are familiar with (and interested in) this one exceptional case can easily adapt all our arguments to fit it. Let gx be the symmetric Riemannian metric on X. In the usual way, gx induces a path metric on X. We will denote the path metric by d x. 2.4. Neutered Spaces Let X be a rank one symmetric space. A horobaU of X is defined to be the limit of unboundedly large metric balls, provided that this limit exists, and is not all of X. The isometry group of X transitively permutes the horoballs. A horosphere is the boundary of a horoball. We define a neutered space ~ to be the closure, in X, of the complement of a non- empty disjoint union V of horoballs, equipped with the path metric. We will call this THE QUASI-ISOMETRY CLASSIFICATION OF RANK ONE LATTICES metric the neutered metric, and denote it by d n. We will say that the horoballs of V are koroballs of fL This is a slight abuse of language, because only the bounding horospheres actually belong to YL The metrics dx]n and da are not Lipschitz equivalent. However, they are Lipschitz equivalent below any given scale. The Lipschitz constant only depends on the scale, and not on the neutered space. We say that f~ is equivariant if it admits a co-compact group of isometrics. These isometrics are necessarily restrictions of isometrics of X. Lemma 2.1. -- Suppose A is a non-uniform lattice of isometries of X. Then there is an equiraviant neutered space f~ C X such that: 1. A has finite index in the isometry group of ~. 2. A is canonically quasi-isometric to fL Proof. -- We remove disjoint horoball neighborhoods of the quotient X/A, and then lift to X. These neighborhoods are covered by a disjoint union of horoballs. The closure, f~ C X, of the complement of these horoballs, is an equivariant neutered space. A well-known criterion of Milnor-Svarc says that, since A acts virtually freely, with compact quotient, on the path space ~, the two spaces A and ~ are quasi-isometric. The canonical quasi-isometry is given by mapping k E A to the point ~,(x), for some pre-chosen point x e f~. The equivalence class of this quasi-isometry is independent of the choice of x. [] To avoid certain trivialities, we will assume that any neutered space we consider has at least 3 horospheres. Certainly, this condition is fulfilled for equivariant neutered spaces. 2.5. Rank One Geometry: Horospheres Let a C X be a horosphere. We will let d, denote the path metric on a induced from the Riemannian metric gx]o- As is well known, (a, do) is isometric to a Euclidean space, when X = H". Below, we will describe the geometry of a, when X ---- FH". Here, we will take F to be either the complex numbers C, or the quaternions O. Let T(a) denote the tangent bundle of a, considered as a sub-bundle of the tangent bundle to FH". There is a canonical codimension dim(F) -- 1 distribution D(~) C T(~) defined by the maximal F-linear subspaces in T(a). This distribution is totally non- integrable, in the sense that any two points p, q e a can be joined by a curve which is integral to D(~). Furthermore, D(a) is totally symmetric, in the sense that the stabilizer of a in Isom(X) acts transitively on pairs (p, v), where p e a, and v ~ D~(a). There are explicit "coordinates" for both a and D(~), which we now describe. Let ~ denote the componentwise conjugate of ~, in F u. Let Im(~) denote the imaginary 140 RICHARD EVAN SCHWARTZ part of ~; Im(~) has exactly dim(F) -- 1 components. The product ~1 ~ will be the usual componentwise multiplication. G(F, n) is defined to be the semidirect product of F" and Im(F). The multipli- cation law is: (~1, vt) -b (~, v~) ----- (~ -b ~2, vx + v2 --k Im(~ ~2)). The horosphere (~, do) is isometric to G(F, n), equipped with a left invariant Riemannian metric do. The restriction of do to the tangent space to F" � { 0 } is the same as the Euclidean metric (~1, ~ ) ----Re(~l ~). The precise choice of d o depends on the nor- malization of the metric on FI'I "+ 1. There is a codimension dim(F) -- 1 distribution: D(F, n) C T(G(F, n)). It is defined to be left invariant, and to agree with the tangent space to F" X { 0 } at the point (0, 0). Any isometry which identifies a with G(F, n) identifies the distribu- tion D(o) with D(F, n). 2.6. Rank One Geometry: C-C Metric The Carnot-Caratheodory distance between two points p, q E a is defined to be the infimal do-arc-length of any (piecewise smooth) path which joins p to q and which remains integral to the distribution D(o). We will denote this metric by d~, and call it the C-C metric for brevity. The C-C metric is defined on G(F, n) in an entirely ana- logous way. Any isometry which identifies o with G(F, n) is also an isometry in the respective C-C metrics. Lemma 2.2. -- Let ~ > 0 be fixed. Then there is a constant K~ having the following property: do(p, q) >>. r => do(P, q) <~ d" (p, q) <<. K, do(P, q); K~ only depends on ~ and X. Proof. -- The first inequality is true by definition, and independent of ~. Consider the second inequality. If do(p, q) e [~, 2el, then by compactness and non-integrability, there is a constant K, such that p and q can be joined by a (piecewise) smooth integral path having arc length at most K, times do(p, q). Now, suppose that do(p, q)> 2~. Let y be the shortest path--not necessarily integral--which connects them. We can subdivide y into intervals having length between e and 2~, and then replace each of these intervals by piecewise smooth integral paths having arc-length at most K~ times as long. The concatenation of these paths is integral and has the desired arc-length. [] For more details on the C-C metric, see [Grl], [Gr2], or [P]. THE QUASI-ISOMETRY CLASSIFICATION OF RANK ONE LATTICES 2.7. Rank One Geometry: Projection Let a C X be a horosphere. We let ho denote the horoball which is bounded by or, we let bo C 0X denote the accumulation point of a and call bo the basepoint of or. Given any point x C X -- ko we define ~o(X) = xb. n ~. (Here xbo is the geodesic connecting x and bo.) On a, we define no to be the identity map. We call n the korospherical projection onto or. We will persistently use the convention that x is disjoint from ko. Under this convention, ~o is distance non-increasing. We say that two horospheres are parallel if they share a common base-point. Let 6t denote the horosphere of X which is parallel to a, contained in h~ and exactly t units t r from ~. Let ~o : ~ -+ be the restriction of ~o, to ~. For j = 1, 2, let (Mj, d~) be metric spaces. A bijection f: M 1 -+ M s is called a K-similarity if d2(f(x),f(y)) = Kd~(x,y) for all x,y e M~. Lemma 2.3. -- The projection ~o : ~ ---> r is an exp(--kt)-similarity, relative to the two C-C metrics. The constant k > 0 only depends on X. Proof. -- Let G denote the stabilizer of a. Then G is also the stabilizer of ~. Further- more, n commutes with G, and takes D(a) to D(at). Since G acts transitively on D(~) and D(at) , we see that n is an ~(t)-similarity when restricted to any linear subspace of D(~). It now follows from the definitions that ~ is an ~(t)-similarity relative to the t is part two C-C metrics. The equality aq(t) = exp(-- kt) follows from the fact that no of a one-parameter group of maps. [] If ~ and a t are both identified with G(F, n), then ~ = D -*t, where D'(~, v) ---- (exp(r) ~, exp(2r) v). The map (~, v) -+ ~ gives a natural fibration G(F, n) --~ F". The fiber is a Euclidean space of dimension dim(F) -- 1. The subgroup consisting of elements of the form (0, v) is central in G(F, n), and acts by translations along the fibers. The map D" obviously preserves this fibration structure, and induces a similarity of F" relative to the metric ( ~1, ~2 ) = Re(~l ~). 3. Detecting Boundary Components 3.1. Metric Space Axioms In this chapter, we shall be concerned with a metric space M which satisfies the following two axioms: Axiom 1. -- There is a constant K 0 having the following property: For every point p E M, there is a K0-quasi-geodesic in M which contains p. 142 RICHARD EVAN SCHWARTZ Axiom 2. -- There is a constant K 1 and functions ~, ~ : R + -+ R + having the following properties: If ql, q2 ~ M are points that avoid p ~ M by at least a(n) units, then there are points q~ and q~, and a Kl-quasi-geodesic segment y'C M, such that: 1. y' connects q~ and q~. 2. y' avoids the ball of radius n about p. 3. q~ is within ~(n) of qj. Here are some examples: I. Above dimension one, Euclidean spaces satisfy both axioms. 2. In w 3.7 we will see that a horosphere of a rank one symmetric space (other than I-1 "z) satisfies both axioms. 3. Axiom 2 fails for all simply connected manifolds with pinched negative curvature. 3.2. O.uasi-Flat Lemma Given a subset S C X, let T,(S) denote the r-tubular neighborhood of S in X. The goal of this chapter is to prove: Lemma 3.1. (Quasi-Flat Lemma). -- Suppose that: 1. X ~= H a is a rank one symmetric space. 2. ~ C X is a neutered space. 3. M is a metric space satisfying Axioms 1 and 2. 4. q : M -+ ~ is a K-quasi-isometric embedding. Then there is a constant K' and a (unique) horosphere r Of 2 such that q(M) C Tx,(cr ). The constant K' only depends on K and on M. Here is an overview of the proof (3) of the Quasi-Flat Lemma. Let q : N -+ f2 be a K-quasi-isometric embedding, defined relative to a net N C M. From Axiom 1, the set q(N) has at least one accumulation point on 0X. Let L denote the set of these accumulation points. We will show, successively, that: 1. L must be a single point. 2. L must be the basepoint of a horosphere of fL 3. q(N) must stay within K' of the horosphere based at L. The main technical tool for our analysis is the Rising Lemma, which we will state here, but prove in w 4. 8.8. The Rising Lemma We will use the notation established in w 2.7. Given any horosphere a C X, and a subset S C X -- ho, we let [ S Io denote the do-diameter of z~o(S) C ~r. (s) Steve Gersten has independently given a proof of the Quasi-Flat Lemma for the case where M is Euclidean space, and C2 C H n is an equivarient neutered space. Gersten's proof involves the asymptotic cone construction. THE QUASI-ISOMETRY CLASSIFICATION OF RANK ONE LATTICES 143 Suppose that fl is a neutered space. We say that S is visible on ~, with respect to ~, if ,~o(s) n n, o. Here, ~ is not assumed to be a horosphere of ~. Lemma 3.2 (Rising Lemma). -- Let ~ C X be a neutered space. Let ~ be an arbitrary horosphere, not necessarily belonging to ~. Let constants ~, K > 0 be given. Then there is a constant d = d(~}, K) having the following property. Suppose that 1. y is a K-bi-Lipschitz segment. 2. Ivlo> . 3. y is visible on ~. Then dx(P, ~) <<. d for some p ~ T. 3.4. Multiple Limit Points Suppose that L contains (at least) two limit points ~ and.~. Since we are assuming that ~ has at least 3 horospheres, we can choose one of them, ~, whose basepoint is neither k- nor ft. Recall that q is defined relative to a net N C M. Let { x, }, {y, } E N be sequences of points such that q(x,) ---> Yand q(y,) -+~. We will choose an extremely large number do, whose value is, as yet, undetermined. Let 0 C N be any chosen origin. Since M satisfies t I Axiom 2, we can find points x, and y, such that 1. There is a uniform bound ~(d0) from x', to x,. 2. There is a uniform bound ~(d0) from y', to y,. 3. There is a Kl-quasi-geodesic segment ~, connecting x', to y',. 4. 0, avoids 0 by d o units, independent of n. Clearly dn(q(x,), q(x',)) is uniformly bounded. Since dx~< dn, we have that dx(q(x,) , q(x',)) is uniformly bounded. It follows that q(x',) ---> ~. Likewise, q(y',) --->.~. Let y, = q(~,). Since ~ is a path metric space, we can assume that T, is (uniformly) bi-Lipschitz. Let B C ~ denote the ball of do-radius 1 about the point ~o(~). Let Y = ~-I(B). Let 8, = Y, n Y. It is easy to see that rc~(q(x',)) -+ ~o(~). Likewise, =o(q(y',)) -+ ~o(.~). Hence, there is a positive constant a such that, once n is large enough, [ 8, I~ >/ a. Let p = q(0) C ft. Let A(p) denote the ball of d~-radius r about p. For any fixed value of r, we can guarantee that y, does not intersect A(p) by initially choosing d o large enough. The sets { A(p) n Y } exhaust Y. Therefore, the distance from 8, to can be made arbitrarily large, by choosing the initial constant do large enough. Since 8, is visible on ~, we get a contradiction to the Rising Lemma. 144 RICHARD EVAN SCHWARTZ 3. ft. Location of the Limit Point From the previous section, we know that L cannot be more than one point. We will show in this section that L is the basepoint of a horosphere of ~. Since M satisfies Axiom 1, it contains at least one quasi-geodesic 0. The image y = q(~) C X is a quasi- geodesic which limits, on both ends, to L. We may assume that y is a bi-Lipschitz curve, since ~ is a path metric space. Choose any point L' C 0X distinct from L. Let l be the geodesic in X whose two endpoints are L' and L. Assume now that L is not the basepoint of a horosphere of ~. Then we can find a sequence of points Pl,P~,-.-C l such that 1. p, belongs to the interior of ~. 2. p, ~L. Let ~, be the horosphere based at L' and containing p,. Note that e, need not belong to ~. We will let d, be the induced path metric on ~,. Let B, C e, denote the ball of d,-radius 1 about p~. Let Y, = ~-I(B,). (Here r: = ~:o.) It is easy to see that the sets Y, are nested, and (,) fi Y. = O. Let x be any point of y. Let y~ and y~ be the two infinite half-rays of y divided by x. From (.), both ya and y~ must intersect OYn, for all n > n 0. Note, also, that yJ n Y. is visible from an, by choice of p.. Applying the Rising Lemma, we conclude there is a uniform constant d having the following property: There is a point x~ C yJ n Y, which is at distance at most d from ~.. Since the points x n and x, ~ belong to Yn, and are both close to a., they are uni- formly close to each other, independent of n. However, the length of the segment of y connecting x, 1 to x 2 n tends to co with n. This contradicts the fact that y is bi-Lipschitz. 3.6. Distance to a Horosphere We know that L is the basepoint ofa horosphere n of~. We will now show that q(N) remains within a small tubular neighborhood of a. Since M satisfies Axiom 1, we just have to show that any (say) K"-bi-Lipschitz curve in ~, which limits to L on both ends, remains within K' of ~. Let y be such a curve. Let p ~ y be any point, which divides y into two infinite rays, y~ and y~. Similar to the previous section, the Rising Lemma says that there are points x j C yJ which are close to each other, and close to a. The bi-Lipschitz nature of y bounds the length of the portion of y connecting x ~ to x 2, independent ofp. Since p lies on this (short) segment joining xl to x~, we see that p is also close to n. 3.7. Examples Let X + I-~ be a symmetric space. Let a C X be a horosphere. Recall from w 2.5 that a is isometric to the nilpotent group G(F, n). In this section, we will show that THE QUASI-ISOMETRY CLASSIFICATION OF RANK ONE LATTICES G(F, n) satisfies Axioms 1 and 2. Axiom 1 is obvious, by homogeneity. We concentrate on Axiom 2. Since the left invarlant metric on G(F, n) is quasi-isometric to the C-C metric, we will work with the C-C metric, which is more symmetric. Recall that there is a fibration p : G(F, n) ~ F", and that C-C similarities of G(F, n) cover Euclidean similarities of F ". From this, it is easy to see: Lemma 3.3. -- Let ~ > 0 be fixed. If two points ql, q~ e G(F, n) are suffciently far away, and y is a length minimizing geodesic segment which connects them, then the curvature of p (y) is bounded above by ~. Axiom 2 essentially follows from the existence of a central direction. Let p e G(F, n). Let B denote the ball of radius N aboutp. Given any real number r, let g, = (0, r) E G(F, n). We choose r sufficiently large so that g,(B) is disjoint from B. By homogeneity, this choice does not depend on p, but only on N. Since g, is central, d(x, gr(x)) <<. r', for all x E G(F, n). Here r' only depends on r, and r in turn only depends on N. Suppose that two points ql, q2 are at distance r"/?romp. Let T be a length-minimizing curve connecting ql to q2. We choose r" so large that the projection P(Y) has very small curvature. More precisely, we choose r" so that y c~ p-l(p(B)) consists of disjoint seg- ments Y1, Y,, 9 9 9 having the following properties: 1. At most one yj intersects B. 2. The distance from y~ to yj is at least 100r. (We picture y as a portion of a helix, and 9-1(p(B)) as an infinite solid tube, intersecting this helix transversally.) Our two conditions imply that gs(Y) is disjoint from B for some s ~< r. Also, d(q~, g,(qj)) <~ r'. We get Axiom 2 if we set q~ ---- g,(qj); e(N) = r"; ~(N) = r'; K1 ----- 1. 4. The Rising Lemma In this chapter, we will prove the Rising Lemma, which was stated in w 3.3. 4.1. Geodesics and Horospheres Let X be a rank-one symmetric space. In this section, we give some estimates concerning the interaction of geodesics and horoballs in X. A theme implicit in our discussion below, and worth making explicit here, is that horoballs in X are convex with respect to d x. Here is a piece of notation we will use often: Let [y ['a denote the d~ diameter of y. Here, d~ is the C-C metric on a. We say that y and ~r are tangent if they intersect 19 146 RICHARD EVAN SCHWARTZ in exactly one point. In this case, all other points of y are disjoint from the horoball ho. The following lemma is well known: Lemma 4.1. -- Let y be a geodesic and ~ a horosphere. Suppose that y is tangent to a. Then K-'4 [Vlo-~ K, where K on/y depends on X. Proof. -- Compactness and equivariance. [] Lemma 4.9.. __ Let y be a geodesic and ~ a horospkere. Suppose that dx(v, ,,) n. Then ] V ]~ ~< K exp(-- n/K), where K only depends on X. Proof. -- Combine Lemma 4.1 and Lemma 2.3. [] Lemma 4.3. -- Let a be a horosphere. Let y' be a geodesic segment joining two points xl, xz ~ ~. Let y', C y' denote the points which are at least s units away from both xt and x~. Then dx(y',, a) >/s -- K, provided that y', is nonempty. Here, the constant K only depends on X. Proof. -- Let ,' be the horosphere parallel to a and tangent to y' at a single point, 4. Let ~ denote the geodesic segment connecting x~ to no,(x~). Let y'~ be the portion of y' connecting xj to 4. Using Lemma 4.1, and the usual comparison theorems, y'j remains t t within the K-tubular neighborhood of ~j. Hence, Ys moves away from ~ at the same linear rate that ~'j does, up to the constant K. [] The following Lemma is obvious for real hyperbolic space, and actually a bit surprising in the general case. Lemma 4.4. -- Let ~1 and ~ be two horospheres. Suppose that x~,yj C ,j satisfy a, (xl,yl) <. w + Then dol(xx,yl) <~ W" do,(x2,y2), where W" only depends W and on X. Proof. -- To avoid trivialities, we will assume that dx(x ~,yj) >t 1. Below, the positive constants Kx, Ks, ... have the desired independence. Let Y5 be the geodesic segment THE QUASI-ISOMETRY CLASSIFICATION OF RANK ONE LATTICES 147 in X which connects x 5 to y~. Let o'j denote the horosphere parallel to o~ and tangent to 7'~. From Lemma 4.1, l/K1 ~< [ 7j ]~j ~< K1. (The first inequality follows from the assumption that dx(xj,y~) >>. 1.) From this, it follows that 2dx(e~, o'~) ~< dx(x~,yj) <~ 2dx(~, e'j) q- K 1. Therefore dx(oa, a~) ~< dx(~, a'~) + K,. From Lemma 2.3, we see that (*) d'(xl,yl) K. By assumption, d~(x~,yj)t> 1. Now apply Lemma 2.2 to (.). [] Let o be a horosphere. Let ho be the closed horoball bounded by o. Lemma 4.5. -- Let ~ be a horosphere. Suppose that S C X -- h a is closed, and %(S) is compact. Let Z be a horoball which intersects S but does not contain ~o(S). Then OZ n ~ has diameter at most Ka, independent of Z and ~. Proof. -- If this was false, let Zl, Z,, ... be a sequence forming a counterexample. Since none of these horoballs contains all of ~o(S), and since this set is compact, no sub- sequence of these horoballs can converge to all of X. Furthermore, since Z, intersects and S, no subsequence can converge to the empty set. Hence, some subsequence converges to a horoball Zoo of X. The intersection Z~ n ~ has infinite diameter. This is impossible unless Z~o = a. This contradicts the fact that Z, always intersects S. [] Let a t denote the horosphere parallel to a, disjoint from ha, and exactly t units away from o. The following result it obvious in the real hyperbolic case, but requires work in general: Lemma 4.6. -- Let Z be a horoball. Then diamx(0Z n ~) ~ Ka =~ I 0Z n ~, 1'o ~< Ks exp(-- t/Ks), where K, only depend on X and on K 1. Proof. -- Let -: --= OZ. By moving o parallel to itself by at most K 1 units, we can assume that ha and h, are disjoint. This move only changes the constants in the estimates by a uniformly bounded factor. Let y be a geodesic joining two points xl, x2 ~ "~ n 0,. 148 RICHARD EVAN SCHWARTZ Sub-Lemma 4.7. -- We have dx(y, ~) >1 t[2 -- K. Proof. -- The endpoints of y are t units away from a. If our sub-lemma is false, then there is a point p ~ y for which dx(P, ~) <~ t/2 -- Rt, for unboundedly large R,. By convexity, y C hr. Since ks and a are disjoint, (,) dx( p, -:) ~< t/2 -- R t . On the other hand, since y comes within t/2 -- R t of ~, its length must be at least t. (Recall the location of the endpoints). This says--in the notation of Lemma 4.3--that Yt/2 is nonempty. Lemma 4.3 now contradicts Equation (,) for large R t . [] It follows from Lemma 4.2 that d~,(x,y) ~< K~. exp(-- t/K2). But x 1 and x2 were arbitrary points in ,v n a,. [] 4.2. Shading Let ~ be a horosphere. We say that a point x r h, is s-shaded with respect to q~ if r%(x) r f~, and d~(%(x), 0f~ n ~)/> s. Lemma 4.8. -- Suppose that 1. xe~. 2. dx(x, < w. 3. x is not s-shaded with respect to ~. Then da(x, ~) <~ W'(x q- 1), where the constant W' only depends on X and on W. Proof. -- The case where r%(x) ~f~ is trivial. So, we will assume that there is a horoball Z of f~ which contains %(x). By hypothesis, there is a point y C OZ n ~ such that d,(y, %(x)) <<. s. Since x r Z, there is a point z e 0Z such that 7%(z) = %(x), and dx(x, z) + dx(z, = w. By the triangle inequality, we have dx(y , z) <<. dx(y , 7%(x)) + W. From Lemma 4.4, there is a path on 0Z which connects y to z, having arc-length at most W"(s + 1). Hence, da(3, z) ~< W"(s + 1). Also, dn(x , z) is uniformly bounded, in terms of W. The result now follows from the triangle inequality, and an appropriate choice of W'. [] THE QUASI-ISOMETRY CLASSIFICATION OF RANK ONE LATTICES 149 Let ~ be any (compact) curve in ~. Let O a ~ and 0 2 ~ denote the two endpoints of 0c. We say that ~ is W-controlled by the horosphere ~ provided that 1. ~ is disjoint from h~. 2. For any p e ~, dx(P, q~) >i W. 3. dx(O ~ ~, r <<, 2W. We say that ~ is X-fractionally shaded with respect to r provided that at least one endpoint of ~ is X]~ ], shaded with respect to q~. Lemma 4.9. -- Suppose oc is a K-bi-Lipschitz segment. Then there are positive constants X, W, L having the following property: If ~ is W-controlled by ~ and I ~ ]~ >>" L, then o~ is X-fractionally shaded with respect to ~. These three constants only depend on K and on X. Proof. -- For the present, the three constants W, L and X will be undetermined. We will let A(,) denote the arc-length. Let A = [ a ], 1> L. For points at least W away from % the projection n, descreases distances exponentially. This is to say that A(~) /> exp(k' W) A. The constant k' only depends on the symmetric space X. We set 0j = 0j ~. Suppose that 0t is not X-fractionally shaded with respect to ~. Then, according to Lemma 4.8, there are paths ~j C D connecting 0 5 to points p~ C ? such that A(~) ~< W'(XA + 1). Since the projection onto r is distance non-increasing, we have, by the triangle inequality, d,(px,p,) <~ W'(kA + 1) + h + W'(XA + 1). Let y be the geodesic segment in X connecting Pl to p,. Every time 7 intersects a horoball Z of D, we replace y r~ Z by the shortest path on 8Z having the same end- points as y r~ OZ. Call the resulting path 8. By Lemma 4.4, A(8) <~ K 0 d,(pl,P2). Here K o is a universal constant, only depending on X. The path ~ = ~1 u 8 u ~ C ~ is a rectifiable curve connecting the endpoints of 0~. Combining the previous equations, we get A(~)~< K 1W'xA+K 1W'+K1A. The constant K 1 is a universal constant, depending only on X. For whatever value of W we choose, we wiLl take L> W', X< W---;" 150 RICHARD EVAN SCHWARTZ With these choices, we get A(~) ~< 3K 1 A ~< 3Ka exp(-- k' W) A(~). For sufficiently large W, this contradicts the bi-Lipschitz constant of 0~. [2 4.8. Main Construction We will use the notation from the Rising Lemma. Let p ~ ~, be a point such that no(p) ~. Let B denote the ball of do-radius B about zoo(p). Let Y = z~-I(B). Let 8 be the component of y n Y which contains p. Choose two points xo,y ~ ~ such that I. l{xo,y}lo > 12. 2. x 0 is a point of ~ which minimizes the dx-distance to ~. We delete portions of 8 so that x 0 and y are the endpoints. We orient 8 so that it pro- gresses from x 0 to y. By construction, x 0 is a point of 8 which is dx-closest to a. Let N denote this distance. Choose the constants W, L and X as in Lemma 4.9. Forj = 1, 2, ..., we inductively define xj to be the last point along ~ which is at most jW + N units away from a. (It is possible that Xl =y.) Let 8 5 denote the segment of ~ connecting xs_ 1 to xj. By cons- truction, 8j is W-controlled by the horosphere ~0j = ~(j--2)W+N" We insist that N >/ 2W + 1. This guarantees that ?~ is disjoint from ho, and at least one unit away from a. Below, the constants K1, K2, ... have the desired dependence. Recall that d~ is the C-C metric on a. Let [ S ['o denote the d~ diameter of%(S). Note that [ S [o ~< [ S ]'o- Sub-Lemma 4.10. -- We have [ ~j 1'~ ~< Kx exp(-- N/Kx) exp(--j/K1). Proof. -- xf L, then, by compactness, 18 lLi < K2. The bound now follows immediately from Lemma 2.3. Suppose that ] 8j ]~./> L. Then ~j is X-fractionally shaded, from Lemma 4.9. Let Z be the horoball of f2 with respect to which ~j is X-frac- tionally shaded. The shading condition says: I x-ll 0z n Lemma 2.2 says that I P Let S C Y denote the set of points which are at least one unit away from h a. Since Z shades ~5 with respect to ?j, we see that Z intersects S. Also, Z cannot contain B = %(S), because p ~ B c~ ~. Lemma 4.5 says therefore that diam(0Z r~ a) ~< K 4. THE QUASI-ISOMETRY CLASSIFICATION OF RANK ONE LATTICES From Lemma 4.6 we conclude that [ OZ n % ]'o ~< K~ exp(-- N/K~) exp(--j/K~). The desired bound now follows from (,) and Lemma 2.3. [] Summing terms from both cases over all j, we see that For sufficiently large N, this contradicts the choice of x 0 and y. 5. Ambient Extension 5.1. Overview Suppose that ~ : X -+ X is a quasi-isometry, defined relative to nets N1 and N2- We say that ~ is adapted to the pair (~1, f12) provided that: 1. No two distinct points of N~ are within one unit of each other. 2. q is a bi-Lipschitz bijecfion between N1 and N,. 3. Ifx EN~ c~j, then ~(x) ENi+ 1 r~+ 1. 4. Let ~ C 0~ s be a horosphere. Then N~ c~ a~ is a K-net of ~j, where the constant K does not depend on ~5" 5. For each horosphere ajC a~, there is a horosphere aj+lC 0~+1 such that n = N +a n To simplify notation, we have taken the indices mod 2 above, and also blurred the distinction between ~ and ~-1. Hopefully, this does not cause any confusion. The goal of this chapter is to prove the following: Lemma 5.1 (Ambient Extension). -- Suppose that q : f~l --+ f~2 is a quasi-isometry. Then there is a quasi-isometry q:X-+ X such that 1. -~ is adapted to (f~, f~). 2. ~ la i # equivalent to q, relative to d~i. Remark. -- The map ~ is a quasi-isometry relative to d x. The restriction qlaj is being considered as a quasi-isometry relative to daj. In fact, since ~ is adapted to (f~l, f~2), this restricted map is a quasi-isometry relative to both dni and dxlaj. 5.2. Cleaning Up We can add points to Na so that N~ n al is a uniform net of a~, for every horosphere (~1C ~2 x. Given a horosphere e~ C f~, there is, by the Quasi-Flat Lemma, a unique horosphere (~ such that q(N~ n ~) remains within a uniformly thin tubular neigh- borhood of as. We define a new quasi-isometry q', as follows: For each point x e ex, let = 152 RICHARD EVAN SCHWARTZ Doing this for every horosphere of f~l, we obtain a quasi-isometry q' which is equivalent to q, and which has the property that q'(al n N1) s a2. We set q = q'. Lemma 5.2. -- The image q(al t~ N1) is a uniform net of a~. Proof. -- The map q-1 is well defined, up to bounded modification. Applying the Quasi-Flat Lemma to q-~, we see that every point near or, is near a point of the form y = q(x), where x is near e h. Our lemma now follows from the triangle inequality. [] Since q' is a quasi-isometry, there is a constant X 0 having the following property: If x,y ~ N~ are at least X 0 apart, then q'(x) and q'(y) are at least one unit apart. By thinning the net N 1 as necessary, we obtain nets N' 1 and N' 2 = q'(N~) having the following properties: 1. No two distinct points of N'j are within 1 unit of each other. 2. q' is a bi-Lipschitz bijection from N~ to N~. 3. q'(N~ -- 0f2j) = N'~+~ -- 0f~j+ 1. 4. Let ~ C 0f~ be a horosphere. Then N~. ~ a~ is a uniform net of ~5" such that 5. For each horosphere ~ C 0~, there is a horosphere ~i+ 1 C 0ffj+ 1 = N' i+1 ~ 1~.~ + 1 " 5.3. Main Construction For the rest of w 5, we adopt the notation of w 2.7. We now construct certain special nets which extend N' 1 and N~. For simplicity, we will drop subscripts. Let x ~ N'. If x ~ N' -- Off, let S~ = { x }. If x C Off, then x belongs to a horosphere C Off. Let S, be the infinite ray joining x to the basepoint bo C 0X. Define N= U S,. z~EN' Clearly l~I is a net of X. For each point x e N~, there is an obvious isometric bijection from S, to Sr The union of these isometric bijecfions gives a bijection Lemma 5.3. -- Let C 2> 0 be given. Then there exists a constant C' having the following property: If x, y ~ ~1 satisfy dx(x,y ) < C, then dx(~(x), ~(y)) < C'. Proof. -- Below, the constants Ca, C,, ... depend on (C, q', N[, Ns For each point p e X, let S(x) denote the minimum distance from p to ~. For positive k, the con- nected components of the level set S-l(k) are horospheres. Say that two points p, q e ~l a THE QUASI-ISOMETRY CLASSIFICATION OF RANK ONE LATTICES 153 are horizontal if they belong to the same connected level set of 8. Say that two points of lq 1 are vertical if they both belong to the same set S,. Now suppose that dx(x,y ) ~ C. It is clear from the construction of ~ that there are points x =Po,P,,P2,P3 =yC Iq a such that 1. P0 and Pl are vertical. 2. Px and pz are horizontal. 3. P2 and P3 are vertical. 4. dx(pj,pj+~) <~ C~. By construction, -< Cl. ax(O(p0), -< G, For the points Pl,P2, there are two cases: Case 1. -- Ifpl,P~ ~f]l, then dnx(px,p, ) ~< C,, since d x and d o are Lipschitz equi- valent below any given scale. Since q is a quasi-isometry relative to dni , we have that dn,(~(pl), ~(p,)) ~< Cz. Since d x In, ~< dt~2, the same bound holds for dx(~(pl), q(P2)). Case 2. -- Suppose that Pl, P2 r t21. Then there is a horosphere cr 1 C OO x and a number d such that Pl, P2 e crx d. Let a2 C Of 2~ be the horosphere which is paired to ax, via q. From Lemma 2.2, the map q' s'~ c~ o~ : N~ o eq -+ N., o er 2 I ' ' is uniformly bi-Lipschitz relative to the C-C metrics on these two horospheres. From Lemma 2.3, therefore, Here di' is the C-C metric on Crj.d (Recall that ad is the horosphere parallel to ~, contained in ho, and d units away from a.) Since dx(pl,p2) ~< C1, it follows from compactness that dl(Pl, P2) ~< C5. Finally, O(p,)). Putting everything together gives dx(~(p,), q(P2))~< C8. The triangle inequality completes the proof. [] Lemma 5.3 also applies to the inverse map ~-1. Since X is a path metric space, it follows that ~ is a quasi-isometry. To finish the proof of the Ambient Extension Lemma, we thin out the nets lqj appropriately. 20 154 RICHARD EVAN SCHWARTZ 6. Geometric Limits: Real Case Let q:X-+ X be a quasi-isometry, adapted to the pair (f~l, f~2) of neutered spaces. From this point onward, we will assume that ~1 and f~, are equivariant neutered spaces. It is the goal of w 6-w 8 to prove Lemma 6.1 (Rigidity Lemma). -- Suppose that q is a quasi-isometry of X which is adapted to the pair (f~x, f~2) of equivariant neutered spaces. Then q is equivalent to an isometry of X. In this chapter, and the next, we will work out the real hyperbolic case. In w 8, we will make the modifications needed for the complex case. The other cases follow immediately from [P, Th. 1]. 6.1. Q uasiconformal Extension Let H = H" be real hyperbolic space, for some n >/ 3. We will use the upper half- space model for H, and set E = OH -- ~. Finally, we will let T,(E) denote the tangent space to E at x. It is well known that q has a quasi-conformal extension h = Oq. (See [M2], or [T, Ch. 5].) We normalize so that h(~) = oo. It is also known that 1. h is a.e. differentiable (3) on E, and this differential is a.e. nonsingular [M2, Th. 9.1]. Let dh(x) denote the linear differential at x. 2. If dh(x) is a similarity for almost all x, then h is a conformal map. This is to say that q is equivalent to an isometry of H [M2, Lemma 12.2]. This chapter is devoted to proving: Lemma 6.2 (Real Case). -- Let q be a quasi-isometry of H which is adapted to the pair (f~l, f~z) of equivariant neutered spaces. Suppose that h = Oq fixes oo. Let x C E be a generic point of differentiabitity for h. Then there are isometric copies f~'j of O~, and a quasi-isometry q' : X -+ X such that 1. q' is adapted to the pair (f~'l, f~). 2. h' = Oq' is a real linear transformation of E. 8. h' = dh(x), under the canonical identification of E and T,(E). 6.2. Hausdorff Topology Let M be a metric space. The Hausdorff distance between two compact subsets K1, Kg. C M is defined to be the minimum value 8 = 8(K1, K2) such that every point of K~ is within 8 of a point of Kj+ 1. (Indices are taken mod 2.) A sequence of closed (s) For the purist, an additional trick can make our proof work under the easier assumption that h is just ACL. THE Q UASI-ISOMETRY CLASSIFICATION OF RANK ONE LATTICES 155 subsets $1, S~, ... C M is said to converge to S C M in the Hausdorfftopology if, for every compact set K C M, the sequence { 8(S, n K, S n K)} converges to 0. We say that a net N C It is sparse provided that no two distinct points of N are within 1 unit of each other. We say that a quasi-isometry q : It ~ It is sparse if it is a bi-Lipschitz bijecfion between sparse nets Nx and N~. Let F(q)C I-I � I-I denote the graph of q. Let q" be a sparse quasi-isometry defined relative to sparse nets N~ and N~. We say that qt, q~, ... converges to a sparse quasi-isometry q, defined relative to nets Ns, provided that 1. N~ converges to Nj in the Hausdorff topology. 2. F(q") converges to F(q) in the Hausdorff topology. We will make use of several compactness results: Lemma 6 9 8. -- Let { q" } be a sequence of sparse K-quasi-isometries of I-I. Let h" = Oq" be the extension of q". Suppose that 1. h"(0) = 0. = F.. 3. h" converges uniformly on compacta to a homeomorphism h : E ~ E. q" converge on a subsequence to a sparse quasi-isometry q. Furthermore Oq --- h. Then the maps Proof. -- Let 0 be any chosen origin of hyperbolic space. We will first show that the set (q,(0)} is bounded 9 Let ~1 and Y2 be two distinct geodesics through 0. Then the quasi-geodesics q"(y~) remain within uniformly thin tubular neighborhoods of geo- desics 8~ and 8~. (This is a standard fact of hyperbolic geometry.) The endpoints of 8~ and 8~ converge to four distinct points of OH. Furthermore, the point q"(0) must lie close to both 8~ and 8~. This implies that { q"(0)} is bounded 9 Statement i now follows from a routine diagonalization argument. By thinning out the sequence, we can assume that q" converges to a quasi- isometry q~. Let h ~~ -~ Oq ~~ Let p be any point in E. Let y be any geodesic, one of whose endpoints is p. Let 8" denote the geodesic whose tubular neighborhood contains q"(y). Then 8" converges to some geodesic 8~o, in the Hausdorff topology. Hence the endpoints of S" converge to those of 8 ~176 This means that h"(p) converges to h ~~ (p). Hence h(p) = h ~176 (p). Since p is arbitrary, we get Statement 2. [] Lemma 6 9 4. -- Let I" be a sequence of hyperbolic isometrics. Let ~ be an equivariant neu- tered space. Let ~2" ---- I"(~2). Suppose that n ~" is nonempty. Then, on a subsequence, these neutered spaces converge to an isometric copy ~' of ~2. Proof. -- Let K" C ~" denote a compact fundamental domain for fl", modulo its isometry group. Since n ~" is nonempty, we can choose K" so that n K" is nonempty. 156 RICHARD EVAN SCHWARTZ Note that K" has uniformly bounded diameter. Hence, we can choose isometrics j1, j2, ... such that 1. J"(n) = I"(n). 2. The sequence { JJ } lies in a compact subset of the isometry group of H. From these two statements, the claims of the Lemma are obvious. [] 6.3. Taking a Derivative Suppose that f: E -~ E is a homeomorphism which is differentiable at the origin. Suppose also that the differential if(O) is a nonsingular linear transformation of the tangent space To(E ). Let D ~ denote the dilation v ~ exp(n) v. Consider the sequence of maps f. = i)- ofo D-- It is a standard fact from several variable calculus that f" converges, uniformly on compacta, to a linear transformation f~, and that f~ equals if(O), under the canonical identification of T0(E ) with E. We will call this the differentiability principle, and will use it below. 6.4. Zooming In We will use the notation established above. Suppose that q is a hyperbolic K-quasi- isometry satisfying the hypothesis of Lemma 6.2. By translation we can assume that the point x is the origin, 0 C E. By further translation, we can assume that h(0) = 0. Since x is generic, we can assume that 0 is not the basepoint of a horosphere of ~1- Since q is adapted to the pair (~1, ~2), we see that 0 is not the basepoint of a horosphere of ~ either. Recall that D' is the dilation by exp(r) discussed in the preceding section. Let T' denote the hyperbolic isometry which extends D'. Consider the following sequence of objects: 1. q" =T'oqoT -r. 3. h" = Oq'. From the differentiability principle, the maps k" converges to the nonsingular linear map h' = dh(O), as r -+ oo. Let l be the geodesic connecting 0 to oo. Since 0 is not the basepoint of a horosphere of f~l, we can find points Pl,P2,... e l n ~1 which converge to 0. By choosing r appropriately, we extract a subsequence rx, r2, ... such THE QUASI-ISOMETRY CLASSIFICATION OF RANK ONE LATTICES 157 that T',(p,) = Pl- By Lemma 6.4, T',(~I) converges to a neutered space ~ isometric to ~1. Every quasi-isometry in sight is sparse. By Lemma 6.3, the maps q' converge, on a thinner subsequence, to a quasi-isometry q' with 0q'= h'. Since the maps q'. converge, the points qr,(pl) remain in a compact subset. Hence, for some thinner sub- sequence (labelled the same way) there is some point p~ which belongs to every neutered space D.~, ----q",(~)~,). By Lemma 6.4 there is a thinner subsequence on which these neutered spaces converge to a limit ~ which is isometric to ~2. Note that q" is adapted to the pair (~, ~). Since everything in sight converges, q' is adapted to the pair (~i, ~.~). This establishes Lemma 6.2. 7. Inversion Trick: Real Case The goal of this chapter is to prove the Rigidity Lemma stated in w 6, in the real hyperbolic case. Using the facts about quasi-conformal maps of the Euclidean space listed in w 7.1, and Lemma 6.2, we just have to prove: Lemma 7.1 (Real Case). -- Suppose that q is a quasi-isometry of H which is adapted to a pair (~1, ~2) of equivariant neutered spaces. Suppose also that h = Oq is a real linear trans- formation when restricted to F.. Then h is a similarity on E. The remainder of this chapter is devoted to proving this result. The technique is somewhat roundabout, since the map q is not assumed to conjugate (or virtually conjugate) the isometry group of ~1 to that of ~2- 7.1. Inverted Linear Maps Let T : E ---> E be a real linear transformation. Let I denote inversion in the unit sphere of E. Technically, I is well-defined only on the one-point compactification E u oo. Alternatively, I is well-defined and conformal on F. -- { 0 }. We will call the map I o T o I an inverted linear map. There are two possibilities. If T is a similarity, then so is I o T o I. However, if T is not a similarity, then I o T o I is quite strange. The key to our proof of the Lemma 7.1 is a careful analysis of the map I o T o I, when T is not a similarity. For notational conve- nience, we will set T = I o T o I. We will say that a dilation of F.. is any map of the form v ~-* kv. Here, k is a scalar, and v is a vector. Lemma 7.2. -- The transformation T__ commutes with the one-parameter subgroup of dila- tions. Furthermore, T is bi-Lipschitz on E- { 0 }. Proof. -- Let D" denote the dilation by exp(n). Then, we have D" o I = I o D -". Also, D" commutes with T. These two facts imply that T commutes with dilations. 158 RICHARD EVAN SCHWARTZ Now, __T is clearly (say) K-bi-Lipschitz when restricted to a thin annulus containing the unit sphere. Since T commutes with dilations, it must in fact be K-bi-Lipschitz on the image of this annulus under an arbitrary dilation. [] 7.~.. Images of Orbits In this section, we will use the vector space structure of E. A translation of !~. is a map of the formf,(x) -- x + v, where v C !~. is a vector. Let A be a co-compact lattice of translations of E. Once and for all, we fix a basis { vx, ..., v d } for A. Here d is the dimension of 1~.. Let G be an orbit of A. We will say that a cell of G is a collection of 2 d vertices made up of the form: g-~-r - ... -~-zaV d. Here g e G, and z~ can either be 0 or 1. The orbit G is made up of a countable union of cells, all of which are translation equivalent. Each cell consists of the vertices of a parallelepiped. Say that two ceils of G are adjacent if they have nonempty intersection. Clearly, two ceils can be joined by a sequence of adjacent cells. We say that two compact sets S~, S~ C E are z-translation equivalent if there is a translation f, such thatf~(S1) and S~ are z-close in the Hausdorff topology. Let B, C E denote the n-ball about 0. Lemma 7.3. -- Let r > 0 be fixed. Then for sufficiently large n, the following is true: If C1 and C2 are adjacent cells of G, and disjoint from B,, then T(C1) and T(C2) are ~-translation equivalent. The constant n only depends on T, and on ~. Proof. -- Let Y be any ray emanating from 0. Let 0 be a number significantly larger than the diameter of a cell. Let A denote the ball of radius p centered on the point of Y which is exp(r) units away from 0. Note that the ball A' _-- D-'(A) is centered on a point of the unit sphere, and has radius tending to zero as r tends to oo. From Lemma 7.2, we know what __T commutes with dilations. Hence, in particular ' = n'o_Tl o D-'. Note that T is differentiable on the unit sphere. Hence, using the differentlability prin- ciple of w 6.4, we see that, pointwise, T is within r of an affine map, when restricted to A, provided that r is sufficiently large. (It is worth emphasizing that the diameter of A is large, and independent of r.) By compactness, the choice of r can be made inde- pendent of the ray y. [] Lemma 7.4. -- Suppose that T is not a similarity. Then, for any n the foUowing is true: There are cells C1 and C~ of G, which are disjoint from B,, such that _T(Cx) and _T_(C~) are not translation equivalent. THE OUASI-ISOMETRY CLASSIFICATION OF RANK ONE LATTICES 159 Proof. -- If this was false, then there would be a value of n having the following property: The images of all cells avoiding B, would be translation equivalent. Let L denote the union of these cells. (Note that L is a countable set of points.) From the reasoning in Lemma 7.3, it follows that T is affine when restricted to L. Since T__ commutes with dilations, __T is affine when restricted to the set L ~ = D-"(L). The set L" becomes arbitrarily dense as n -+ oo. Taking a limit, we see that T is affine. This is only possible if T is a similarity. [] We now come to the main fact about the map T. Lemma 7.5. ~ Suppose that T__ is not a similarity. Suppose that A 1 and As are two co-compact lattices of translations Of E. Let G be an orbit Of A1. Then __T(G) cannot be contained in a finite union of orbits of A s. Proof. -- We will assume the contrary, and derive a contradiction. By Lemma 7.2, T__ is bi-Lipschitz. Hence, there is a bound, above and below, on the size ofT(C), where C is a cell of G. Hence, ff T(G) was contained in a finite union of A2-orbits, there would be only a finite set of possible shapes for the image T(C). But this contradicts Lemma 7.3 and Lemma 7.4. [] 7.3. Packing Contradiction We will now apply the above theory to prove Lemma 7.1. Suppose that q and k = Oq satisfy the hypotheses of Lemma 7.1. By translating, we can assume that 0 is the basepoint of a horosphere of ~1. Since q is adapted to the pair (~a, ~s), and h is real linear, 0 is also the basepoint of a horosphere of ~s- Let J be the isometry of H which extends inversion. Consider the following objects: 1. = 2. q=JoqoJ. 3. h=JokoJ---IokoI. Note that q is a quasi-isometry adapted to the pair (_~1, ~z)- Note also that there is a horosphere % of ~j based at oo. Since Y~5 is an equivariant neutered space, there is a co-compact lattice of trans- lations Aj of E having the following property: The hyperbolic extension of any element of Aj is an isometry of--~5 which preserves %. For each point x ~ E, we definef~(x) to be the hyperbolic distance from the horo- sphere of~j based at x to %. If x is not the basepoint of a horosphere of ~j, we define fs(x) = oo. Note thatfj is Aj-equivariant. It follows easily from packing considerations that (.): the set Sj., =f~-l[0, r] is contained in a finite union of A~-orbits. 160 RICHARD EVAN SCHWARTZ Recall that h = Oq, where q is a hyperbolic quasi-isometry adapted to (_~x, ~2). This implies (**): for each r > 0, h(S1, ~,) C 82, . for some s which depends on r and the quasi-isometry constant of q. Now, let x be any basepoint of a horosphere of f~x, and let G be the orbit of x under A 1. From (,) and (**) we see that _h(G) is contained in a finite union of A~-orbits. This contradicts Lemma 7.5, unless h is a similarity. 8. Rigidity Lemma: Complex Case The purpose of this chapter is to prove the Rigidity Lemma in the case of complex hyperbolic space CH. The technique is exactly the same as that for the real case. The only difference is that the analytic underpinnings are less well known. Our source for information about quasi-conformal maps on the boundary of complex hyperbolic space is [P]. Another reference is [KR]. 8.1. Three Kinds of Automorphisms A horosphere of CH" +1 has the geometry of the Heisenberg group G(C, n), described in w 2. In this section, we will describe some of the automorphisms of the Heisenberg group. We distinguish three types, listed in order of generality. Heisenberg Dilations. -- A Heisenberg dilation is a map of the form D', where D'(~, v) = (exp(r) ~, exp(2r) v). Here r is a real number. Heisenberg Similarities. -- A Heisenberg similarity is a map of the form (~, v) ---> (T(~), det(r) I/" v). Here T is a similarity of C" relative to the inner product ( ~1, 42 ) = Re(41 ~2). Linear Contact Automorphisms. -- An LCA of G(C, n) is a smooth group automorphism wich preserves the (contact) distribution D(C, n). Such transformations have the form: (4, v) -+ (T(4), det(T) 1/" v). Here T = S 1 $2, where S 1 is a similarity of C", as described above, and S~ is a symplectic transformation ofC ". In other words, S~ preserves the symplectic form (41, ~2) -+ Im(~1 ~). 8.2. Stereographic Projection Let X = CH "+ 1. The sphere at infinity, 0X, can be considered the one point compactification ofG(C, n), as follows: Let oo C 0X be any point, and let ~ be a horosphere based at oo. For any point x e 0X -- o% we define p~ = no(x). THE QUASI-ISOMETRY CLASSIFICATION OF RANK ONE LATTICES 161 Suppose that 0 4= oo is another point of OX. We identify ~ to G(C, n) by an iso- metry I 0 which takes ?~(0) to the identity element (0, 0) e G(C, n). We define the stereo- graphic projection ~ = I0 o p~ : ~X - co ~ G(G, n) q~o is well defined up to post-composition with a Heisenberg similarity. For map h : 0X ~ 0X which fixes oo, we define h~, = q~oho~ -1. If T is an isometry of X which fixes 0 and 0% and h = 0T, then h, is a Heisenberg simi- laxity. IfT is a pure translation along the geodesic connecting 0 to o% then h, is a Heisenberg dilation. 8.8. Heisenberg Differentiability Let D" be the Heisenberg dilation defined above. Let f: G(C, n) ~ G(C, n) be a homeomorphism such that f(0, 0) = (0, 0). We say that f is tleisenberg differentiable at (0, 0) if the sequence D" ofo D-'; r~oo converges uniformly on compacta to an LCA. More generally, we say thatfis Heisenberg differentiable at some other point of g e G(a, n) if the map f(g)- i o f o g is Heisenberg differentiable at (0, 0). We will denote this differential by dr(g). By hypo- thesis, dr(g) is an LCA. Let q:X -+ X be a quasi-isometry. Composing by isometrics of CI-I "+ 1, we can assume that h = Oq fixes oo. Let q~ be the stereographic projection described above. We now list two facts about k. 1. h, is a.e. (resp. Haar measure) Heisenberg differentiable [P, Th. 5]. 2. If dh, is a.e. a Heisenberg similarity, then h is the boundary extension of an isometry of X [P, Prop. 11.5]. 8.4. Zooming In Suppose that T : X ~ X is an isometry which is pure translation along the geodesic whose endpoints are 0 and oo. Let h : OX ~ 0X be a map which fixes both 0 and oo. Then the stereographic projection ~o conjugates 0T o h o OT-1 to D'oh, o D-'. The constant r is essentially the translation length of T. 21 162 RICHARD EVAN SCHWARTZ All of the limiting/compactness arguments of w 6 work in the complex hyperbolic setting, mutatis mutandis. The conjugation above, together with these arguments, gives: Lemma 8.1 (Complex Case). -- Suppose that q : CH" -+ CH" is a quasi-isometry, adapted to a pair (~1, ~) of equivariant neutered spaces. Suppose h = Oq fixes oo. Let x e G(C, n) be a generic point of Iteisenberg differentiability for h,. Then there are isometric copies ~'~ of ~j, and a quasi-isometry q' : X~ X such that 1. q' is adapted to the pair (~, f~). 2. h'~ is an LCA. 8. h; = dh (x). 8.5. Inversion Let J : X -+ X be an isometric involution. (Unlike the real hyperbolic case, such an involution cannot have a codimension-one fixed point set.) Choose any point 0 e 0X, such that J(0) + 0, and let oo = oJ(0). We define tIeisenberg Inversion I : o(c, n) - (0, 0) G(C, n) - (0, 0) to be the composition I = ~o oJ o (9~o)-a. Lemma 8.2. -- If T is a Heisenberg similarity, then so is T = I o T o I. Proof. m Clearly, if T is an isometry of X which fixes 0 and 0% then J o T o J is also an isometry fixing 0 and oo. Our lemma now follows from the fact such isometrics, under stereographic projection, induce Heisenberg similarities. [] Lemma 8.3. -- Suppose T is an LCA. If T is not Heisenberg similarity, then T is not an LCA. Proof. J It follows from symmetry that I preserves V = C" � { 0 }. It follows from Lemma 8.2 that I v is the composition of a Euclidean similarity and an inversion. Also, T preserves V. The map Tic is a linear transformation which is not a similarity of V. Hence __T Iv cannot be a linear transformation. This implies that T is not an LCA. [] Apology. -- For the reader who is unwilling to use "symmetry" to see that I pre- serves V, here is a more ad-hoc line of reasoning. The manifold G(F, n) --(0, 0) is foliated by codimension-one hypersurfaces which are invariant under Heisenberg simi- larities. Call this foliation ~'. Lemma 8.2 implies that I preserves the leaves of o~-. Every leaf of o~-, except V, is a punctured rotationally symmetric paraboloid. If follows that T cannot take a leaf of ~- to a leaf of 5, unless this leaf is V. If T is an LCA, then it is certainly not a similarity. Hence T_T - preserves V. But then I preserves V as well. The following results have proofs exactly analogous to those in the real case. THE QUASI-ISOMETRY CLASSIFICATION OF RANK ONE LATTICES 163 Lemma 8.4. -- Suppose that T is an LCA. Then 1. T commutes with Iteisenberg dilations. 2. T is Heisenberg differentiable away from (0, 0). 8. T__ is bi-Lipschitz in the C-C metric. Exactly as in w 7, Lemma 8.3 and Lemma 8.4 imply: Lemma 8.5. -- Let A1, A 2 C G(C, n) be two co-compact discrete subgroups. If T is and LCA, but not a Heisenberg similarity, then T cannot take an orbit ofA 1 into a finite union of orbits of Az. 8.6. Packing Contradiction Let J be the isometric involution of X defined above. Let ~ = ~ be stereographic projection. Let I be Heisenberg inversion. We normalize so that 0 is the basepoint of a horosphere of ~j. We define ~j, q' and _h' as in w 7. Note, in particular, that t t _h, = Ioh, oI. There are co-compact lattices Aa C G(C, n) whose elements have the form 0T,, where T is an isometry of ~j preserving the relevant horosphere based at oo. Let x ~ G(C, n) be a point such that ?-l(x) is th e basepoint of a horosphere of _~1. Let G, denote the Al-orbit of x. The same argument as in w 7 says that h, takes G x into a finite union of Ag-orbits. Lemma 8.3 therefore implies that h', is a Heisenberg similarity. This suffices to prove the Rigidity Lemma in the complex case. 9. The Commensurator Let X 4= H 2 be any rank one symmetric space. Let q be a quasi-isometry of X adapted to the pair (ill, ~z), where fl~ is an equivariant neutered space. We know from the Rigidity Lemma of w 6 that q is equivalent to an isometry q,:X ~X. The goal of this chapter is to show that q. commensurates the isometry group of fix to that of fla. For the sake of exposition, we first sketch a proof in the simplest arithmedc case. Afterwards, we turn to the general case. 9.1. Special Case Suppose that X = H 3. Nonuniform arithmetic lattices in X are all commensurable with PSLa(0), where 0 is the ring of integers in an imaginary quadratic field F. We will consider exactly these lattices. 164 RICHARD EVAN SCHWARTZ Let g~ and q, be as above. If ~ is suitably normalized, then the union of base- points of horoballs of ~ coincides with Fj u oo. Hence, Oq, is a Mobius transformation which induces a bijection between F 1 t3 oo and F, u oo. From this it is easy to see that F a = F~, and Oq. e PGL2(F~). This last group is isomorphic to the commensurator of PSL~(~j). 9.9.. General Case We now give a general argument, which works in all cases. We begin by isolating the key feature of q,. Lemma 9.1. (Bounded Distance). -- Let ~1 be a horosphere of Ox. Tken the horosphere q.(ol) is parallel to some horosphere ~ of ~)2. Furthermore, dx(q.(ol), o~) ~< K, where K does not depend on the choice of ol. Proof. -- Since q is adapted to (~1, ~), the extension h = Oq induces a bijection between basepoints of horospheres of ~1 and basepoints of horospheres of ~. Hence q,(ol) is parallel to a horosphere o~ of ~2. Since q and q, are equivalent, q,(ol) and 02 are uniformly close. [] To say that q, commensurates the isometry group of D1 to that of ~2 is to say that the common isometry group of D~ and q,(~x) has finite index in the isometry group of~2. We will suppose that this is false, and derive a contradiction. By assumption, there is an infinite family of isometrics I, having the following properties: 1. I.(~2) = t)~. 2. I,~(q.(~l) ) # I,(q,(~l) ) ff m # n. Let oo be any point of OX. We normalize so that oo is the basepoint of a horo- sphere co~ of~. Let ~0 = q.(~l), and let ~, = I,(~0). Modulo the isometry group of ~o, there are only finitely many equivalent horospheres. Let q~, denote the horosphere of ~. based at oo. By taking a subsequence, we can assume that the horospheres I$1(q~.) of ~0 are all equivalent. By replacing (I) o by (I) 1 if necessary, we can assume that the horo- spheres I$1(~,) are in fact all equivalent to ~o. This is to say that there is an isometry J, of X such that 1. J,(~0) -- r 2. J, fixes oo. Let Y, denote the group of isometrics of X which fix oo. Let A 0 denote the isometry subgroup of (I) 0 that fixes oo. Clearly A 0 C Z. We form the coset space Z/A 0 ---- g" as follows: We identify elements of the form S and S o ~, where ~, e A 0. The space g" is THE QUASI-ISOMETRY CLASSIFICATION OF RANK ONE LATTICES 165 topologically the product of a line with a compact nilpotent orbifold. Let ,1 C g" denote the image of {J, } in the quotient space. Since ~, # ~,,, the set ar is a infinite. We now contradict this. Observe that = I~I(o,)). These last two horospheres are parallel horospheres, belonging respectively to q.(f2~) and f~,. It follows from Lemma 9.1 that (1) dx(%, c02) ~< K 1. From (1) it follows that J is precompact in W.. To show that J is finite, then, we just have to show it is a discrete set. Suppose ar is not a discrete set. Let x C 0X be the basepoint of a horosphere q~g of ~o. Let G, denote the orbit of x under A 0. Define Y ---- IJ J,,(G.). IfJ, is indiscrete then so is Y C 0X. Lety ~ Y be a point. For some n, there is a horosphere qY, of~, based aty. Clearly, ~v is parallel to a horosphere cog of f~z. The same reasoning as above shows that (2) dx(~*,, o~) ~< Kz. By construction, (3) ax(q?., ~,) = ax(~, ~o) = Ka. Putting (1), (2) and (3) together, we see that Sincey was arbitrary, we see that every point of Y is the basepoint of a horosphere of f~ 2 which is within K, of the horosphere c%. Packing considerations say that Y is therefore a discrete set. 10. Main Theorem and Corollaries 10.1. The Main Theorem Suppose that A1, A 2 C G are two non-uniform rank one lattices. Let X be the rank one symmetric space associated to G. As usual, we assume X # H ~. Suppose that q0 : A1 ~ A2 is a quasi-isometry. 166 RICHARD EVAN SCHWARTZ From Lemma 2.1, A~ is canonically quasi-isometric to an equivariant neutered space ~5- Furthermore, A~ has finite index in the isometry group of~. Under the iden- tification of Aj and ~j, the quasi-isometry q0 induces a quasi-isometry ql : ~1 -+ ~. From the Ambient Extension Lemma, qx is equivalent to (the restriction of) a quasi-isometry q3:X~X which is adapted to the pair (~i, ~)- From the Rigidity Lemma of w 7, q3 is equivalent to an isometry q4 e G. From w 9, q4 commensurates the isometry group of Di to that of D2. Hence q4 commensurates A 1 to A s. All of this proves that q0 is equivalent to (the restriction of) an element of G which commensurates A 1 to A s. 10.2. Canonical Isomorphism Let A C G be a non-uniform rank one lattice. Let X be the associated rank one symmetric space. Every commensurator of A induces, by restriction and bounded modi- fication, a self-quasi-isometry of A. Thus there is a canonical homomorphism from the commensurator of A into the quasi-isometry group of A. Since different commensurators have different actions on 0X, this homomorphism is injective. Our Main Theorem says that this injection is in fact a surjection. Hence, the commensurator of A is canonically isomorphic to the quasi-isometry group of A. 10.3. Classification In this section, we will give the quasi-isometry classification of rank one lattices. The quasi-isometry classification of uniform lattices is well known. We will concentrate on the non-uniform case. Suppose that Aj is a non-uniform rank one lattice, acting on the rank one symmetric space X~. Suppose that q:A 1 --> A 2 is a quasi-isometry. If X 1 = H ~, then X 2 = H% This follows from the fact that non-uniform lattices in Xj have infinitely many ends iff Xj : Ht Suppose that X 5 4= H% We will now give three separate, but sketchy proofs, that X 1 : X~. 1. IfA 1 and A 2 are quasi-isometric, then they have isomorphic commensurators. Standard Lie group theory now implies that X 1 = X 2. 2. The Ambient Extension Lemma can be generalized (a bit) to show that q extends to a quasi-isometry between Xx and X~. It is (fairly) well known that this implies X 1 : X 2. 3. By the Quasi-Flat Lemma of w 3, q induces a quasi-isometry between a horosphere of X 1 and a horosphere of X 2. It is (fairly) well known that this implies X x = X~. THE QUASI-ISOMETRY CLASSIFICATION OF RANK ONE LATTICES We now reach the interesting part of the classification. If A 1 and A s are two non- uniform lattices acting on X, then any quasi-isometry between them commensurates A, to As, by the Main Theorem. This implies that the two lattices are commensurable. 10.4. Quasi-Isometric Rigidity Suppose that P is a finitely generated group quasi-isometric to a non-uniform lattice A. We will show that F is the finite extension of a non-uniform lattice A', and that A' is commensurable with A. Let f~ be an equivariant neutered space quasi-isometric to A. Let q: F --->~ be a quasi-isometry. Let q-1 be a (near) inverse of q. Note that each element y ~ F acts isometrically on F. Let I v be this isometry. Let O(y) =qoI voq-1. Then do(y) is a uniform quasi-isometry of f/. Let W(y) denote the isometry of X whose restriction to t2 is equivalent to dO(y). The constants in all of our arguments only depend on the pair (K(q), f~) where K(q) is the quasi-isometry constant of q. Hence, there is a constant K' such that dO (y) and W(y)In differ pointwise by at most K', independent of y. Now, W gives a representation of F into the commensurator of A. In steps, we will characterize W. Finite Kernel. -- Let e be any element of P. All but fnitely many elements of F move the point e ~ F more than N units away from itselE Hence, all but finitely many dO(y) move the point q(e) more than N units away from itselE From the uniformity of the distance between dO and W, we see that only fiuitely many ~F(y) can fix q(e). This is to say that W has finite kernal (4). Discreteness. -- If the image of W was not discrete, then there would be an infinite sequence of elements W(y1) , ~'(Ys) --- converging to the identity. This is ruled out by an argument just like the one given for the finiteness of the kernal. Cofinite Volume. -- Suppose that g = ~F(y). From Lemma 9.1, and the uniformity mentioned above, there is a uniform constant K" having the following property: Every horosphere of g(f2) is within K" of the corresponding horosphere of f~. It follows that ~'= U ~(Y)(~) is a neutered space on which iF(F) acts isometrically. Since F acts transitively on itself, the quotient fl'/tF has finite diameter, and hence is compact. This says that W acts on X with finite volume quotient. (4) I would like to thank B. Farb for supplying this argument. 168 RICHARD EVAN SCHWARTZ Thus, we see that P is the finite extension of a non-uniform lattice A'. From the classification above, we see that A' is commensurable with A. 10.5. Arithmeticity Let A C G 4: PSL2(R) be a non-uniform rank one lattice. It is a result of Margulis that A is arithmetic if and only if it has infinite index in its commensutaror. (See [Z, Ch. 6] for details.) By our Main Theorem, the commensurator of A is isomorphic to the quasi-isometry group of A. Thus, A is arithmetic if and only if A has infinite index in its quasi-isometry group. REFERENCES [C] K. CORLETTE, Hansdorff Dimension of Limits Sets I, Invent. Math., 102 (1990), 521-541. [El] D.B.A. Epstein et al., Word Processing in Groups, Jones and Bartlett, 1992. [E2] D.B.A. Epstein, Analytical and Geometric Aspects of Hyperbolic Space, LMS Lecture Notes, series 111, Cam- bridge University Press, 1984. [Go] W. GOLDMAN, Complex Hyperbolic Space, notes available from the author. [Grl] M. GRo~ov, Asymptotic Invariants of Infinite Groups, LMS Lecture Notes Series, 1994. [Gr2] M. GROMOV, Carnvt-Caratheodorr Spaces Seen from Within, IHES, preprint, 1994. [KR] A. KORA~CYI and H. M. REi~nn, Foundations for the Theory of O.uasi-Conformal Mappings of the Heisenberg Group, Preprint, 1991. [M1] G.D. MOSTOW, Strong Rigidity of Locally Symmetric Spaces, Annals of Math. Studies, No. 78, Princeton Uni- versity Press, 1973. [M2] G.D. MOSTOW, Quasiconformal Mappings in n-Space and the Rigidity of Hyperbolic Space Forms, Publ. Math. IHES, 34 (1968), 53-104. [P] P. PANSU, M~triques de Carnot-Caratheodory et Q uasi-Isom6tries des Espaces Sym6triques de Rang Un., Annals of Math., 129 (1989), 1-60. [T] W. THURSTON, The Geometry and Topology of Three-Manifolds, Princeton University Lecture Notes, 1978. [Z] R. ZIMsmR, Ergodic Theory and Semi-Simple Lie Groups, Birkhauser, Boston, 1984. Dept. of Mathematics University of Maryland College Park, MD 20742-0001 USA Manuscrit refu le 19 juillet 1994.

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G. D. Mostow (1968)
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Analytical and geometric aspects of hyperbolic space : Warwick and Durham 1984

Publisher: Springer Journals
Subject: Mathematics; Mathematics, general; Algebra; Analysis; Geometry; Number Theory
ISSN: 0073-8301
eISSN: 1618-1913
DOI: 10.1007/BF02698639
Publisher site: See Article on Publisher Site

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The quasi-isometry classification of rank one lattices

The quasi-isometry classification of rank one lattices

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The quasi-isometry classification of rank one lattices

The quasi-isometry classification of rank one lattices

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