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Quasi-isometric rigidity for the solvable Baumslag-Solitar groups, II.

Benson Farb and Lee Mosher

Revised, December 10, 1997

Introduction

Gromov’s Polynomial Growth Theorem [Gro81] characterizes the class of virtually nilpotent groups by their asymptotic geometry. Since Gromov’s theorem it has been a major open question (see, e.g. [GH91]) to find an appropriate generalization for solvable groups. This paper gives the first step in that direction.

One fundamental class of examples of finitely-generated solvable groups which are not virtually nilpotent are the solvable Baumslag-Solitar groups

BS(1, n) = ⟨a, b | bab^{−1} = a^{n}⟩

where n ≥ 2. Our main theorem characterizes the group BS(1, n) among all finitely-generated groups by its asymptotic geometry.

Theorem A (Quasi-isometric rigidity). Let G be any finitely generated group. If G is quasi-isometric to BS(1, n) for some n ≥ 2, then there is a short exact sequence

1 → N → G → Γ → 1

where N is finite and Γ is abstractly commensurable to BS(1, n).

In fact we will describe the precise class of quotient groups Γ which can arise, and will classify all torsion-free G; see section 5 in the outline below.

Theorem A complements the main theorem of [FM97], where it is shown that BS(1, n) is quasi-isometric to BS(1, m) if and only if they are abstractly commensurable, which happens if and only if m, n are positive integer powers of the same positive integer.

∗The first author is supported in part by NSF grant DMS 9604640, the second author by NSF grant DMS 9504946.

Theorem A says that every finitely generated group quasi-isometric to BS(1, n) can be obtained from BS(1, n) by first passing to some abstractly commensurable group and then to some finite extension. We describe this phenomenon by saying that the group BS(1, n) is quasi-isometrically rigid. This property is even stronger than what we know for nilpotent groups, for while Gromov’s theorem says that the class of nilpotent groups is a quasi-isometrically rigid class, outside of a few low-dimensional cases it is not known whether an individual nilpotent group must always be quasiisometrically rigid.

Comparison with lattices Recent work on lattices in semisimple Lie groups has established the quasi-isometric classification of all such lattices.

In the case of a nonuniform lattice Λ in a semisimple Lie group G ≠ SL(2, R), quasi-isometric rigidity of Λ follows from the deep fact that the quasi-isometry group QI(Λ) is the commensurator group of Λ in G, a countable group (see [Sch96b], [Sch96a], [FS96], [Esk96], or [Far96] for a survey).

In contrast, for uniform lattices Λ in the isometry group of X = H^{n} or CH^{n}, the quasi-isometry group QI(Λ) ≈ QI(X) is QC(∂X), the (Heisenberg) quasiconformal group of the sphere at infinity ∂X, an infinite dimensional group. In this situation it is the whole collection of lattices in Isom(X) which is quasi-isometrically rigid, not any individual lattice [Tuk86], [CC92], [Cho96].

For BS(1, n) something interesting happens. It exhibits both types of contrasting behavior just described: the group BS(1, n) is quasi-isometrically rigid and yet its quasi-isometry group is infinite-dimensional, as follows. Let Q_{n} be the metric space of “n-adic rational numbers”. Let Bilip(X) denote the group of bilipschitz homeomorphisms of the metric space X. In [FM97], Theorem 8.1 we showed:

QI(BS(1, n)) ≈ Bilip(R) × Bilip(Q_{n})

In proving Theorem A, this formula plays a role similar to that played by the formula QI(X) ≈ QC(∂X) when X = H^{n} or CH^{n}.

Outline of the paper

As we shall see, the “mixed behavior” of BS(1, n) allows for some analogies with proof techniques developed in the case of lattices. Some fundamentally new phenomena occur, however, and these require new methods. We point out in particular:

• The notion of quasisimilarity (a delicate though crucial variant of quasisymmetric map) and the corresponding notion of dilatation.

• The theory of biconvergence groups.

• A method which applies these dynamical properties to proving quasiisometric rigidity.

Section 1: Geometry and boundaries for BS(1, n). We review some results of [FM97]. We construct a metric 2-complex X_{n} on which BS(1, n) acts properly discontinuously and cocompactly by isometries, and we equip this complex with a boundary which is formed of two disjoint pieces: an upper boundary ∂^{u}X_{n} ≈ Q_{n}, which is the space of hyperbolic planes in X_{n}, and a lower boundary ∂_{ℓ}X_{n} ≈ R. We also review the fact that a quasiisometry of X_{n} induces bilipschitz homeomorphisms of the upper and lower boundaries (Proposition 1.2), giving the isomorphism

QI(BS(1, n)) ≈ Bilip(R) × Bilip(Q_{n})

Section 2: Representation into the quasi-isometry group. We recall a principle of Cannon and Cooper, that if a group G is quasi-isometric to a proper geodesic metric space X, then there is a “quasi-action” of G on X, and an induced representation G → QI(X). Combining this with the results of §1 we obtain the fact that if G is quasi-isometric to BS(1, n) then there is an induced action ρ = ρ_{ℓ} × ρ^{u}: G → Bilip(R) × Bilip(Q_{n}).

Important point: At this point one might try to make an analogy with [Tuk86] (see also [CC92]), and attempt to prove that a “uniformly quasiconformal” subgroup of Bilip(R) × Bilip(Q_{n}) is conjugate into some kind of “conformal” subgroup of Bilip(R) × Bilip(Q_{n}). However, there are serious difficulties with this approach (see comments in §6).

Section 3: Uniform quasisimilarity actions on R. Instead we consider the projected representation ρ_{l}: G → Bilip(R). We show in in Proposition 2.2 that ρ_{l}(G) is a group of quasisimilarities which is uniform with respect to a certain dilatation.

Inspired by Hinkkanen’s Theorem [Hin85], we prove in Theorem 3.2 that a uniform group of quasisimilarities of the real line is bilipschitz conjugate to an affine group (Theorem 3.2). Applying this to the representation ρ_{ℓ}: G → Bilip(R) we obtain a bilipschitz conjugate representation θ: G → Aff(R). It is crucial that this conjugacy is bilipschitz as opposed to just quasisymmetric. This is why we cannot use Hinkkanen’s original theorem.

Important point: For an arbitrary group quasi-isometric to a lattice in a semisimple Lie group, finding a representation into that Lie group usually finishes the proof of quasi-isometric rigidity. In the present case more work is required. One reason is that neither G nor even BS(1, n) is a lattice in Aff(R), although BS(1, n) is a nondiscrete subgroup of Aff(R). In fact Aff(R) is a nonunimodular Lie group and so does not admit any lattice. Hence we must find another way to prove that θ has finite kernel, and to analyze the image group θ(G).

Section 4: Biconvergence groups. We study the action of G on the boundary pair (∂_{ℓ}X_{n}, ∂^{u}X_{n}). Exploiting analogies with convergence groups, the dynamical behavior of this pair of actions is encoded in what we call a biconvergence group. Using this boundary dynamics, we show that the representation θ: G → Aff(R) is virtually faithful (Proposition 4.4), and that the group of affine stretch factors, or stretch group of θ(G) ⊂ Aff(R) is infinite cyclic (Proposition 4.5). The proof of Proposition 4.5 makes vital use of a bilipschitz conjugacy between ρ_{ℓ} and θ.

Section 5: Finishing the proof of Theorem A. We apply combinatorial group theory and quasi-isometry invariants, together with the results of §4, to identify the image group Γ = θ(G) ⊂ Aff(R). We show that Γ is the mapping torus of some injective, nonsurjective endomorphism ϕ: A → A where A is either the infinite cyclic group or the infinite dihedral group— that is, Γ has the presentation ⟨A, t | tat^{−1} = ϕ(t), ∀a ∈ A⟩. In particular Γ contains a subgroup of index ≤ 2 isomorphic to BS(1, m) for some integer m ≥ 2. Applying [FM97] it follows that BS(1, m) is abstractly commensurable to BS(1, n), and so the same is true of Γ. We also prove in Corollary 5.3 that if G is torsion free then G is isomorphic to BS(1, k) for some integer k with |k| ≥ 2.

Section 6: Final comments. We discuss the possibility of strengthening Theorem A.

Acknowledgements

We thank J. Sturm and M. Feighn for useful suggestions and discussions, and A. Hinkkanen for helpful correspondence. We also thank the members of the Workship on Hyperbolic and Automatic Groups at the University of Montreal where a preliminary version of this work was presented, especially S. Gersten for helpful criticism, and T. Delzant for suggesting the theorem of Bieri-Strebel which we use in §5.

1 Geometry and boundaries for BS(1, n)

In this section we briefly review the material from [FM97] which we will need.

1.1 Quasi-isometries

A (K, C)quasi-isometry between metric spaces is a map f: X → Y such that, for some constants K, C > 0:

• For all x_{1}, x_{2} ∈ X we have

\frac{1}{K} d_{X}(x_{1}, x_{2}) − C ≤ d_{Y} (f(x_{1}), f(x_{2})) ≤ Kd_{X}(x_{1}, x_{2}) + C

• The C-neighborhood of f(X) is all of Y.

A coarse inverse of a quasi-isometry f: X → Y is a quasi-isometry g: Y → X such that, for some constant C′ > 0, we have d(g ∘ f(x), x) < C′ and d(f ∘ g(y), y) < C′ for all x ∈ X and y ∈ Y. Every (K, C) quasi-isometry f: X → Y has a coarse inverse g: Y → X, namely, for each y ∈ Y define g(y) to be any point x ∈ X such that d(f(x), y) ≤ C.

A fundamental observation due to Effremovich-Milnor-Švarc states that if a group G acts properly discontinuously and cocompactly by isometries on a proper geodesic metric space X, then G is finitely generated, and X is quasi-isometric to G equipped with the word metric. Two finitely generated groups G_{1}, G_{2} are said to be (abstractly) commensurable if there are finite index subgroups H_{i} < G_{i} such that H_{1}, H_{2} are isomorphic to each other. It is easy to check that abstractly commensurable groups are quasi-isometric, and that if G is finitely generated and N is a finite normal subgroup then G is quasi-isometric to G/N.

Let QIMap(X) be the set of all quasi-isometries f: X → X, equipped with the binary operation of composition. Given f, g ∈ QIMap(X) and C ≥ 0, we write f ∼_{C} g if d(f(x), g(x)) < C for all x ∈ X. We write f ∼ g if there exists C ≥ 0 such that f ∼_{C} g; this is an equivalence relation on QIMap(X), known as Hausdorff equivalence or coarse equivalence. The set of equivalence classes is denoted QI(X). The operation of composition respects Hausdorff equivalence, in the sense that if f_{1} ∼ f_{2} and g_{1} ∼ g_{2} then f_{1} ∘ g_{1} ∼ f_{2} ∘ g_{2}. Composition therefore descends to a well-defined binary operation on QI(X). With respect to this operation, QI(X) is a group, whose identity element is the Hausdorff equivalence class of the identity map on X. Inverses exist in QI(X) because of the fact that every quasi-isometry has a coarse inverse.

1.2 The 2-complex X_{n}

Throughout this paper we use the upper half plane model of the hyperbolic plane

H^{2} = {(x, y) ∈ R^{2} | y > 0}

with metric ds^{2} = (dx^{2} + dy^{2})/y^{2}. Also, let T_{n} denote the unique homogeneous, directed tree such that each vertex has one incoming directed edge and n outgoing directed edges, with the geodesic metric that makes each edge of T_{n} isometric to the interval [0, log(n)].

In [FM97] we constructed a metric 2-complex X_{n} on which BS(1, n) acts properly discontinuously and cocompactly by isometries. The 2-complex X_{n} is homeomorphic to T_{n} × R. There is a geodesic metric on X_{n} with the following properties:

• For each directed edge E ⊂ T_{n}, the subset of X_{n} corresponding to E × R is isometric to the “horostrip of height log(n)”, namely

{(x, y) ∈ H^{2} | 1 ≤ y ≤ n}

• For each bi-infinite directed line L ⊂ T_{n}, the subset of X_{n} corresponding to L × R is isometric to H^{2}.

• For any two bi-infinite directed lines L ≠ L′ ⊂ T_{n}, the subset of X_{n} corresponding to (L ∩ L′) × R is isometric to a “horodisc exterior” {(x, y) ∈ H^{2} | y ≤ 1}.

A cocompact, properly discontinuous, free action of BS(1, n) on X_{n} is described in [FM97] by exhibiting X_{n} as the universal cover of a certain metric 2-complex C_{n} whose fundamental group is BS(1, n).

Various important features of the complex X_{n}, and the action of BS(1, n), are summarized in the following proposition. Given a continuous map between metric spaces f: A → B, an isometric section is a subset B′ ⊂ A such that f | B′ is an isometry onto B. The isometric sections of the height function h: H^{2} → R, defined by h(x, y) = log(y), are precisely the vertical lines in H^{2}, lines of the form x = (constant). Also define a height function h: T_{n} → R, by requiring h(v_{0}) = 0 for some chosen base vertex v_{0} ∈ T_{n}, and requiring that h takes each edge of T_{n} onto some interval of length log(n) in R by an orientation preserving isometry. The isometric sections of h: T_{n} → R are precisely the bi-infinite directed lines in T_{n} (called “coherent lines” in [FM97]).

Proposition 1.1. There exist actions of BS(1, n) on X_{n}, H^{2}, T_{n}, and R, and equivariant maps between these spaces as summarized in the following commutative diagram:

\xymatrix{ & X_{n} \ar[dl]_{p} \ar[dd]^{h} \ar[dr]^{q} \\ H^{2} \ar[dr]_{h} &  & T_{n} \ar[dl]^{h} \\  & R}

Moreover:

• The action on X_{n} is properly discontinuous, cocompact, and free.

• The function h: X_{n} → R is the fiber product of the height functions h: H^{2} → R and h: T_{n} → R. That is, the map X_{n} → H^{2} × T_{n} taking x to (p(x), q(x)) is an equivariant homeomorphism onto the subset of H^{2} × T_{n} consisting of all pairs (a, b) such that h(a) = h(b).

• The map q: X_{n} → T_{n} induces a 1–1 correspondence between vertical lines in T_{n} and isometric sections of the map p: X_{n} → H^{2}.

• The map p: X_{n} → H^{2} induces a 1–1 correspondence between vertical lines in H^{2} and isometric sections of the map q: X_{n} → T_{n}.

Isometric sections of p: X_{n} → H^{2} are called hyperbolic planes in X_{n}, and isometric sections of q: X_{n} → T_{n} are called trees in X_{n}. Since BS(1, n) clearly acts on the set of vertical lines in T_{n}, it follows that BS(1, n) acts on the set of hyperbolic planes in X_{n}. Similarly, BS(1, n) acts on the set of trees in X_{n}.

The proof of the above proposition can be gleaned from the information in [FM97]. Here is an alternative proof, which gives an interesting new construction of X_{n}.

Proof. Recall the presentation BS(1, n) = ⟨a, b | bab^{−1} = a^{n}⟩. Define the height action of BS(1, n) on R by

a · t = t, b · t = t + log(n)

To describe the actions of BS(1, n) on H^{2} and T_{n}, we first define affine actions on R and on the n-adic rational numbers Q_{n}, and then we describe how these induce the desired actions on H^{2} and T_{n}.

If R is a ring with unit, let Aff(R) be the group of all matrices of the form ( \begin{matrix}r & s \\ 0 & 1 \end{matrix} ), where r, s ∈ R and r is invertible; the group law is ordinary matrix multiplication. The group Aff(R) acts on R by fractional linear transformations: ( \begin{matrix}r & s \\ 0 & 1 \end{matrix} ) · x = rx + s for all x ∈ R. Note that if R is a commutative ring then Aff(R) is a solvable group.

If the integer n is invertible in the ring R, there is a representation BS(1, n) ↦ Aff(R) defined by

a ↦ (\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix})  b ↦ (\begin{matrix}n & 0 \\ 0 & 1 \end{matrix})
a · r = r + 1  b · r = nr

for all r ∈ R. We call this the affine action of BS(1, n) on the ring R.

As a special case we obtain an affine action of BS(1, n) on the real numbers. The action of Aff(R) on R extends to an isometric action on H^{2}, and by composition we obtain the desired action of BS(1, n) on H^{2}.

For another special case, let Q_{n} be the ring of n-adic rational numbers. This is the ring of all formal series ∑_{i∈Z} ζ_{i}n^{i} with ζ_{i} ∈ {0, …, n − 1} for all i, and ζ_{i} = 0 for i sufficiently close to −∞, with the obvious addition and multiplication. We write the series ∑_{i∈Z} ζ_{i}n^{i} more succinctly as (ζ_{i})_{i∈Z} or just (ζ_{i}) when i ∈ Z is understood. The ring of n-adic integers Z_{n} is the subring of all (ζ_{i}) ∈ Q_{n} with ζ_{i} = 0 for all i < 0. Note that the integer n is invertible in Q_{n}, and so the affine action of BS(1, n) on Q_{n} is defined.

The metric on Q_{n} is the usual n-adic metric, where the distance between (ζ_{i}) and (ζ′_{i}) in Q_{n} is n^{−k}, where k ∈ Z is the maximum integer such that ζ_{i} = ζ′_{i} for all i ≤ k.

The tree T_{n} may be identified with the Bruhat-Tits building of Q_{n}, and so the action of BS(1, n) on Q_{n} induces an action on T_{n}. To be explicit, for each truncated series η = (η_{i})_{i≤k}, where η_{i} ∈ {0, …, n − 1} for i ≤ k, and η_{i} = 0 for i sufficiently close to −∞, define a set

C_{η} = {(ζ_{i}) ∈ Q_{n} | ζ_{i} = η_{i}, ∀i ≤ k}

The set C_{η} is called a clone of Z_{n} in Q_{n}. The integer k is called the combinatorial height of the clone C_{η}, denoted h_{c}(C_{η}). The inclusion lattice on clones defines the directed tree T_{n} as follows. The vertices of T_{n} are the clones of Z_{n} in Q_{n}. There is a directed edge C_{η} → C_{η′} if and only if the following hold:

• C_{η} ⊃ C_{η′}.

• For all C_{η′′}, if C_{η} ⊃ C_{η′′} ⊃ C_{η′} then C_{η′′} = C_{η} or C_{η′′} = C_{η′}.

Note that h_{c}(C_{η′}) = h_{c}(C_{η}) + 1. Note also that T_{n} is a tree, because any two clones are either disjoint or one contains the other, and it is easy to check that each vertex has one incoming and n outgoing edges.

The action of Aff(Q_{n}) on Q_{n} takes clones to clones, preserving inclusion. The affine action of BS(1, n) on Q_{n} therefore induces a direction preserving action of BS(1, n) on the tree T_{n}.

Now we may define X_{n}, and the height function h: X_{n} → R, by applying the fiber product construction to the height functions h: H^{2} → R and h: T_{n} → R. Since the latter two height functions are BS(1, n) equivariant, we obtain an action of BS(1, n) on X_{n} so that h: X_{n} → R is equivariant.

To see that the action on X_{n} is properly discontinuous, note that the general element of BS(1, n) is a matrix of the form ( \begin{matrix}n^{i} & k/n^{j} \\ 0 & 1 \end{matrix} ), and if a 1–1 sequence of such matrices is bounded in Aff(R) then it is unbounded in Aff(Q_{n}). The rest of the proof follow easily. ♢

Remarks

1. This construction of X_{n} is equivalent to the construction given in [FM97].

2. The action of BS(1, n) on T_{n} is isomorphic to the action on the Bass-Serre tree of the HNN decomposition BS(1, n) ≈ Z∗_{ϕ} where ϕ: Z → Z is given by ϕ(k) = nk.

3. The action of BS(1, n) on H^{2} is a “laminable action” as described in [Mos96]. The action on T_{n} is also laminable: the decomposition of X_{n} into trees gives a T_{n}-lamination which resolves the indiscreteness of the action of BS(1, n) on T_{n}.

4. Let Z[1/n] be the ring of fractions obtained from Z by inverting n. Note that the affine representation BS(1, n) → Aff(Z[1/n]) is a group isomorphism. The natural inclusions from Aff(Z[1/n]) into Aff(R) and Aff(Q_{n}) then give the affine representations BS(1, n) → Aff(R) and BS(1, n) → Aff(Q_{n}).

1.3 The upper and lower boundaries of X_{n}

The boundary of the hyperbolic plane in the upper half plane model may be written as ∂H^{2} = R ∪ {+∞}. Using the affine action of BS(1, n) on R, Proposition 1.1 gives BS(1, n)-equivariant bijections

{trees in X_{n}} ≈ {vertical lines in H^{2}} ≈ R

The lower boundary ∂_{ℓ}X_{n} is defined to be any of these objects. Note that any two hyperbolic planes Q, Q′ ⊂ X_{n} intersect in a common horodisc exterior, and so ∂Q and ∂Q′ share a line at infinity which may be identified with ∂_{ℓ}X_{n}. This was how ∂_{ℓ}X_{n} was defined in [FM97].

As a dual picture we have the upper boundary ∂^{u}X_{n}, defined to be the set of isometrically embedded hyperbolic planes in X_{n}. The space of ends of the directed tree T_{n} can be written as Ends(T_{n}) = Q_{n} ∪ {−∞} where Q_{n} is naturally identified with the set of positively asymptotic ends and −∞ is the unique negatively asymptotic end. By Proposition 1.1 we have BS(1, n)-equivariant bijections

{hyperbolic planes in X_{n}} ≈ {vertical lines in T_{n}} ≈ Q_{n}

and so ∂^{u}X_{n} may be identified with any of these.

Remark. The set \overline{X}_{n} = X_{n} ∪∂_{ℓ}X_{n} ∪∂^{u}X_{n} may be given a topology so that X_{n} is dense and so that the action of BS(1, n) on X_{n} extends to a continuous action on \overline{X}_{n}. However \overline{X}_{n} is not compact, nor even locally compact— compare this with the compactifications of X_{n} described in [Bes95].

1.4 Metrics on ∂_{ℓ}X_{n} and ∂^{u}X_{n}

The metric on ∂_{ℓ}X_{n} = R is the usual metric, and the metric on ∂^{u}X_{n} = Q_{n} is the n-adic metric discussed above.

There are a few other ways of visualizing this metric. If (ζ_{i}), (ζ′_{i}) ∈ Q_{n} correspond to hyperbolic planes Q, Q′ ⊂ X_{n}, and if S is the common horodisc exterior Q ∩ Q′, then d(Q, Q′) = e^{−h(∂S)}. Equivalently, if L, L′ are the vertical lines in T_{n} corresponding to ζ, ζ′, and if v ∈ T_{n} is the finite endpoint of the ray L ∩ L′, then d(L, L′) = e^{−h(v)}.

Note that ∂^{u}X_{n} is a proper metric space, that is, closed balls are compact. The Hausdorff dimension of Q_{n} equals 1. The metric d is an ultrametric, also called a nonarchimedean metric, in other words a metric satisfying d(x, z) ≤ Max{d(x, y), d(y, z)} for any x, y, z ∈ ∂^{u}X_{n}. Distance in Q_{n} takes values in the discrete set {n^{k} | k ∈ Z}. For any ζ ∈ Q_{n} and any k ∈ Z, the closed ball around ζ of radius n^{−k} is precisely the clone of combinatorial height k that contains ζ.

1.5 The groups Aff(Q_{n}) and Sim(Q_{n})

Define a similarity of a metric space X to be a bijection ϕ: X → X such that, for some constant k > 0 we have d(ϕ(x), ϕ(y)) = k d(x, y) for all x, y ∈ X. The number k is called the stretch factor of ϕ, denoted Stretch(ϕ); we also say that ϕ is a k-similarity. The set of all similarities of X forms a group under composition, denoted Sim(X).

The groups Sim(R) and Aff(R) acting on R are obviously identical, but the situation is different in Q_{n}.

First note that there is a natural isomorphism between Sim(Q_{n}) and the group Aut(T_{n}) of direction preserving automorphisms of T_{n}. Since the closed balls in Q_{n} are precisely the clones, each similarity of Q_{n} takes clones to clones preserving inclusion. Every element of Aut(T_{n}) clearly arises in this manner.

The group Aff(Q_{n}) acts by similarities on Q_{n}: given M = ( \begin{matrix}ζ & ζ′ \\ 0 & 1 \end{matrix} ) ∈ Aff(Q_{n}), if k is the least integer such that ζ_{k} ≠ 0, then Stretch(M) = n^{−k}. We therefore have a natural monomorphism Aff(Q_{n}) ↪ Sim(Q_{n}), but it is not surjective. For example, given x, y ∈ Q_{n} such that x − y is invertible in Q_{n}, the only affine transformation fixing x and y is the identity, but for any clone C not containing x or y the subgroup of Sim(Q_{n}) fixing x, y and preserving C acts transitively on C, as can be seen by constructing appropriate automorphisms of T_{n}.

1.6 Boundary maps induced by a quasi-isometry

In [FM97] we showed that any quasi-isometry f: X_{m} → X_{n} induces maps f^{u}: ∂^{u}X_{m} → ∂^{u}X_{n} and f_{ℓ}: ∂_{ℓ}X_{m} → ∂_{ℓ}X_{n} characterized as follows. For each hyperbolic plane Q ⊂ X_{m}, there is a unique hyperbolic plane f^{u}(Q) ⊂ X_{n} such that

d_{H}(f(Q), f^{u}(Q)) < C < ∞

where d_{H} denotes Haudorff distance, and C depends only on the quasiisometry constants of f. Also, for each hyperbolic plane Q ⊂ X_{m}, the closest point projection from f(Q) to f^{u}(Q) induces a map from the line at infinity of Q to the line at infinity of f^{u}(Q), which induces the desired map f_{ℓ}. One could also prove that for any tree τ ⊂ X_{m}, there is a unique tree f_{ℓ}(τ) ⊂ X_{n} such that

d_{H}(f(τ), f_{ℓ}(τ)) < C < ∞

It is proved in [FM97] that the maps f^{u} and f_{ℓ} are bilipschitz homeomorphisms. In fact what we proved is a little stronger, as seen in the next proposition.

Given a homeomorphism h: X → Y between metric spaces, and given k ≥ 1, we say that h is k-bilipschitz if

\frac{1}{k} d(x, y) ≤ d(f(x), f(y)) ≤ k d(x, y) for all x, y ∈ X

We find it useful to introduce a more precise notion, as follows. Given 0 < a < b, we say that h is [a, b]-bilipschitz if

a · d(x, y) ≤ d(h(x), h(y)) ≤ b · d(x, y) for all x, y ∈ X

We use the notation [a(h), b(h)] to denote the stretch interval of h, the smallest subinterval of R_{+} such that h is [a(h), b(h)]-bilipschitz.

For example, “k-bilipschitz” is equivalent to “[1/k, k]-bilipschitz”, and “k-similarity” is equivalent to “[k, k]-bilipschitz”.

The next proposition comes from [FM97] Proposition 5.3 and the following remark.

Proposition 1.2. For each K ≥ 1, C > 0 there exists L ≥ 1 such that if f: X_{m} → X_{n} is a (K, C)-quasi-isometry, then there exist 0 < a ≤ b such that b/a < L, the map f_{ℓ}: ∂_{ℓ}X_{m} → X_{n} is [a, b] bilipschitz, and the map f^{u}: ∂^{u}X_{m} → ∂^{u}X_{n} is [1/b, 1/a]-bilipschitz. ♢

Warning. This does not say that the bilipschitz constants of f^{u}, f_{ℓ} depend only on K, C. Even the natural isometric action of BS(1, n) on X_{n} does not induce uniformly bilipschitz actions on ∂_{ℓ}X_{n}, ∂^{u}X_{n}. For example the element b ∈ BS(1, n) acts on ∂_{ℓ}X_{n} ≈ R as a similarity with stretch factor n, and so the best bilipschitz constant of b^{i} acting on ∂_{ℓ}X_{n} is n^{i}, although b^{i} acts isometrically on X_{n}.

2 Representation into the quasi-isometry group

In this short section we use a (now standard) principle due essentially to Cannon-Cooper [CC92], also indicated by Gromov [Gro83], to prove that if a finitely generated group G is quasi-isometric to BS(1, n) then there is an induced representation G → QI(BS(1, n)) ≈ Bilip(R) × Bilip(Q_{n}). We also study the quality of this representation.

2.1 Quasi-isometries and quasi-actions

Given a proper, geodesic metric space X and a group G, a quasi-action of G on X is a map ψ: G → QIMap(X) such that, for some constants K ≥ 1 and C ≥ 0, we have:

• ψ(g) is a (K, C)-quasi-isometry, for all g ∈ G.

• ψ(Id) ∼_{C} Id_{X}.

• ψ(g) ∘ ψ(h) ∼_{C} ψ(gh), for all g, h ∈ G.

A quasi-action ψ: G → QIMap(X) evidently induces a homomorphism Ψ: G → QI(X), where Ψ(g) is the Hausdorff equivalence class of ψ(g).

A quasi-action ψ: G → QIMap(X) is cocompact if for some x ∈ X and C ≥ 0 the C-neighborhood of the set {ψ(g)(x) | g ∈ G} is X. Also, ψ is properly discontinuous if for each x, y ∈ X and C ≥ 0 the set {g ∈ G | d(ψ(g)(x), y) ≤ C)} is finite.

Proposition 2.1 (QI rigidity condition). Let X be a proper geodesic metric space, and Γ a finitely generated group. If f: Γ → X is a quasiisometry with coarse inverse f̄: X → Γ, and if L_{g}: Γ → Γ denotes left multiplication by g ∈ Γ, then ψ(g) = f ∘ L_{g} ∘ f̄ defines a properly discontinuous, cocompact quasi-action ψ: Γ → QIMap(X).

Proof. It is evident that ψ is a quasi-action. To see that ψ is cocompact, for any x ∈ X we have {L_{g}(f̄(x)) | g ∈ G} = G, and so {ψ(g)(x) | g ∈ G} = f(G), whose C neighborhood equals X.

To see that ψ is properly discontinuous, fix x, y ∈ X and C ≥ 0. Given g ∈ G, if d(ψ(g)(x), y) ≤ C then d(f̄∘ ψ(g)(x), f̄(y)) ≤ KC + C^{2}, and since f̄∘ ψ(g) = f̄∘ f ∘ L_{g} ∘ f̄ we also have d(f̄∘ ψ(g)(x), L_{g}(f̄(y))) ≤ C and so d(f̄(y), g · f̄(y)) ≤ KC + C^{2} + C. Since Γ is finitely generated, there are only finitely many g which satisfy this inequality. ♢ Remarks

1. If ψ: G → QIMap(X) is a properly discontinuous, cocompact quasiaction on a proper geodesic metric space X, does G have a true action on X by quasi-isometries?

2. The passage from a quasi-action ψ: G → QIMap(X) to its associated representation Ψ: G → QI(X) seems to involve a loss of information. The multiplicative constant can be recovered, because if f ∼ g and if f is a (K, C)-quasi-isometry then g is a (K, C′)-quasi-isometry for some C′. In particular, one can define the dilatation of an element of QI(X) as the infimum of K such that each representative f ∈ QIMap(X) is a (K, C)- quasi-isometry for some C.

It is unclear how to recover the additive constant in general, but for example if ψ: G → QI(H^{3}) = QC(S^{2}) has bounded dilatation then ψ(G) is induced by a uniformly quasi-isometric action of G on H^{3}; this follows from [Tuk86].

2.2 Consequences for BS(1, n)

By Proposition 2.1, if G is a finitely generated group quasi-isometric to BS(1, n) we obtain a quasi-action ψ: G → QIMap(X_{n}). Let ρ: G → QI(X_{n}) be the induced representation. Applying Proposition 1.2 to ψ(g) for each g ∈ G, we obtain:

Proposition 2.2. Let G be a finitely generated group quasi-isometric to BS(1, n). Then there is an induced representation ρ: G → QI(BS(1, n)) ≈ Bilip(R) × Bilip(Q_{n}). Furthermore, let ρ_{ℓ}: G → Bilip(R) and ρ^{u}: G → Bilip(Q_{n}) be the projected actions, and let [a_{ℓ}(g), b_{ℓ}(g)] and [a^{u}(g), b^{u}(g)] be the stretch intervals of ρ_{ℓ}(g), ρ^{u}(g) respectively. There exists L ≥ 1 such that for all g ∈ G we have

• b_{ℓ}(g)/a_{ℓ}(g) ≤ L.

• b^{u}(g)/a^{u}(g) ≤ L.

• [a_{ℓ}(g), b_{ℓ}(g)] · [a^{u}(g), b^{u}(g)] ⊂ [1/L, L].

The multiplication of intervals used in this proposition is defined as follows: if [a, b], [c, d] are subintervals of R_{+}, then

[a, b] · [c, d] = {xy | x ∈ [a, b] and y ∈ [c, d]}

3 Uniform quasisimilarity actions on R

An important part of our proof of Theorem A will be to understand uniform groups of quasisimilarities of the real line. This entire section is devoted to such a study, and is based on the following theorem of Hinkkanen (the terms are defined below). Let Aff_{+}(R) be the index 2 subgroup of Aff(R) which preserves orientation of R.

Theorem 3.1 (Hinkkanen’s Theorem). A uniform group of orientation preserving quasisymmetric homeomorphisms of R is quasisymmetrically conjugate to a subgroup of Aff_{+}(R).

We cannot use Hinkkanen’s Theorem directly because we will need to make use of a bilipschitz conjugacy, which is not generally produced by a uniformly quasisymmetric group. In Theorem 3.2 we recast Hinkkanen’s Theorem in the quasisimilarity setting. This setting has a pedagogical advantage as well—the technical details of the proof are simpler, and we believe that it is easier to see the geometric ideas underlying Hinkkanen’s proof.

3.1 A Hinkkanen Theorem for uniform quasisimilarity groups

Let X be a metric space.

Definition (quasisimilarity). A function f: X → X is a K-quasisimilarity if for each distinct triple x, y, z ∈ X we have

\frac{1}{K} ≤ \frac{d(f(z), f(x))/d(z, x)}{d(f(y), f(x))/d(y, x)} ≤ K  (∗)

An action of a group G on X is a uniform quasisimilarity action if there exists K ≥ 1 such that the action of each g ∈ G is a K-quasisimilarity of X.

Remarks

1. Note that a similarity is the same as a 1-quasisimilarity.

2. This property is called “quasiconformal” in the appendix to [FM97], because it implies that if S is a metric sphere in X then f(S) is nested between two metric spheres S_{1}, S_{2} such that radius(S_{2})/ radius(S_{1}) is bounded. However, even more is true: the ratios

\frac{radius(S_{2})}{radius(S)} and \frac{radius(S)}{radius(S_{1})}

lie in a fixed interval in R_{+}, independent of the original sphere S. For this reason it now seems more appropriate to us to refer to this property as “quasisimilarity”.

3. An orientation preserving homeomorphism f: R → R is said to be Kquasisymmetric if the above inequality (∗) holds whenever z − y = y − x. An orientation preserving K-quasisimilarity is therefore K-quasisymmetric. The converse is not true: K-quasisimilarities are bilipschitz and hence absolutely continuous; whereas there exist K-quasisymmetric maps which are not absolutely continuous.

As Cooper notes in [FM97], if f is a K-quasisimilarity then, fixing two points z, w ∈ X, we have

\frac{1}{K^{2}} ≤ \frac{d(f(x), f(y))/d(x, y)}{d(f(z), f(w))/d(z, w)} ≤ K^{2} for all x, y ∈ X

With L = d(f(z), f(w))/d(z, w) it follows that f is K^{2} · Max{L, 1/L}- bilipschitz. However, this conclusion throws some information away. What one really obtains from this argument is that

K^{−2}L ≤ \frac{d(f(x), f(y))}{d(x, y)} ≤ K^{2}L

and so f is [a, b]-bilipschitz with a = K^{−2}L and b = K^{2}L. Thus, if f is a Kquasisimilarity then f is [a, b]-bilipschitz for some a, b such that b/a ≤ K^{4}. Conversely, it is easy to see that if f: X → X is [a, b]-bilipschitz with b/a ≤ K, then f is a K-quasisimilarity. Note that there does not exist a constant C depending only on K such that every K-quasisimilarity is Cbilipschitz, however a map f: R → R is a K-quasisimilarity if and only if there exists a similarity g: R → R such that f ∘ g is \sqrt{(}K)-bilipschitz.

Combining these observations with Proposition 2.2, it follows that any group quasi-isometric to BS(1, n) has a uniform quasisimilarity action on ∂_{ℓ}X_{n} = R and on ∂^{u}X_{n} = Q_{n}.

Here is our quasisimilarity version of Hinkkanen’s Theorem:

Theorem 3.2 (Quasisimilarity Hinkkanen’s Theorem). Let ρ: G → Bilip(R) be a uniform quasisimilarity action of a group G on R. Then there exists ϕ ∈ Bilip(R) such that ϕ ∘ ρ(g) ∘ ϕ^{−1} ∈ Aff(R) for all g ∈ G. Moreover the bilipschitz constant for ϕ depends only on the quasisimilarity constant for ρ.

Remark. Hinkkanen’s paper considers only groups that preserve orientation on R, but his methods indicate an easy way to reduce the general case to the orientation preserving case. We use these methods at the end of our proof.

3.2 Proof of Theorem 3.2

The proof of this theorem is an adaptation of Hinkkanen’s proof for quasisymmetric maps. The first part is the following proposition which proves Theorem 3.2 in the special case that G = Z and G preserves orientation. This is analogous to §3 of [Hin85], but the details are quite different in the current quasisimilarity setting. After this proposition, the remainder of the proof is an almost verbatim quotation of Hinkkanen’s proof, and we will mention only the highlights.

Proposition 3.3. Let {f^{n} | n ∈ Z} be a nontrivial, orientation preserving, uniform quasisimilarity action of Z on R. Then one of two things happens:

• f has no fixed points, the action is uniformly bilipschitz, and f is bilipschitz conjugate to a translation.

• f has a single fixed point p, the action is not uniformly bilipschitz, and f is bilipschitz conjugate to multiplication M_{s}(x) = sx, for some s ∈ (0, ∞) − {1}.

In either case, there is a conjugating map whose bilipschitz constant depends only on the quasisimilarity constant for {f^{n}}.

Proof. Let K′ be a quasisimilarity constant for each f^{n}, and so the stretch interval [a(f^{n}), b(f^{n})] satisfies b(f^{n})/a(f^{n}) ≤ (K′)^{4} = K.

Case 1: f has no fixed points. It follows that f^{n} has no fixed points for n ≠ 0. The stretch interval [a(f^{n}), b(f^{n})] must contain 1, for otherwise f^{n} or f^{−n} would be a contraction mapping of R which always has a fixed point. Since b(f^{n})/a(f^{n}) ≤ K^{4} it follows that [a(f^{n}), b(f^{n})] ⊂ [1/K^{4}, K^{4}] and so f^{n} is K^{4}-bilipschitz for all n.

By replacing f with f^{−1}, if necessary, we may assume that f(x) > x for all x ∈ R. Let x_{n} = f^{n}(x_{0}) be the orbit of some point x_{0}, and so

… x_{−2} < x_{−1} < x_{0} < x_{1} < x_{2} < …

Let α = x_{1} − x_{0}. Define ϕ | [x_{0}, x_{1}] to be the translation x ↦ x − x_{0}. Extend ϕ to a homeomorphism of R as follows: for each n ∈ Z define ϕ | [x_{n}, x_{n+1}] by ϕ(x) = ϕ(f^{−n}(x)) + nα. Clearly ϕ is a homeomorphism of R conjugating f to the map x ↦ x+α. The maps ϕ | [x_{0}, x_{1}] and x ↦ x+α are 1-bilipschitz. Since f^{−n} is K-bilipschitz it follows that ϕ | [x_{n}, x_{n+1}] is K-bilipschitz.

The proof is completed by applying the following:

Rubber band principle. Given metric spaces X, Y and a collection of subsets A covering X, suppose that ϕ: X → Y is a homeomorphism such that:

1. For each A ∈ A the map ϕ | A: A → ϕ(A) is K-bilipschitz.

2. For any pair x, y ∈ X there is a sequence x = x_{0}, x_{1}, …, x_{m} = y such that d(x, y) = ∑_{i=1}^{m} d(x_{i−1}, x_{i}), and for each i = 1, …, m there exists A ∈ A with x_{i−1}, x_{i} ∈ A.

3. For any pair ξ, η ∈ Y, there is a sequence ξ = ξ_{0}, ξ_{1}, …, ξ_{m} = η such that d(ξ, η) = ∑_{i=1}^{m} d(ξ_{i−1}, ξ_{i}), and for each i = 1, …, m there exists A ∈ A with ξ_{i−1}, ξ_{i} ∈ ϕ(A).

Then ϕ is K-bilipschitz. ♢

Applying this principle to ϕ using A = {[x_{n}, x_{n+1}] | n ∈ Z}, it follows that ϕ is K-bilipshitz.

Case 2: f has at least one fixed point. In this case we will prove that f is bilipschitz conjugate to the multiplication map M_{s}(x) = sx, for some s ∈ (0, ∞) − {1}. Note that if s ≠ r then M_{s} and M_{r} are not bilipschitz conjugate, so we have to search carefully to find the correct value of s.

Step 1: f has a unique fixed point. Suppose f has more than one fixed point. Since f is not the identity map, the fixed point set is a proper, closed subset of R, and so the non-fixed point set is an open subset of R one of whose components is a finite interval (x, y). We have f(x) = x and f(y) = y, and replacing f by f^{−1} if necessary we have f(z) > z for all z ∈ (x, y). Since no point in (x, y) is fixed it follows that f^{n}(z) → y as n → ∞. Choose z = (x + y)/2, and so d(x, z)/d(z, y) = 1. By choosing n sufficiently large, the ratio

\frac{d(z, y) · d(f^{n}(x), f^{n}(z))}{d(x, z) · d(f^{n}(z), f^{n}(y))} = \frac{d(x, f^{n}(z))}{d(f^{n}(z), y)}

may be made larger than K′, violating the fact that f^{n} is a K′-quasisimilarity.

Let p be the unique fixed point of f.

Step 2: p is either attracting or repelling. Suppose not. Then either f(x) > x for all x ≠ p, or f(x) < x for all x ≠ p. We assume the former and derive a contradiction; the latter case is similar.

If x < p then f^{n}(x) → p as n → ∞, and if y > p then f^{n}(y) → +∞ as n → ∞. Taking x = p − 1 and y = p + 1, if n is sufficiently large we may make the ratio

\frac{d(x, p) · d(f^{n}(p), f^{n}(y))}{d(f^{n}(x), f^{n}(p)) · d(p, y)} = \frac{d(p, f^{n}(y))}{d(f^{n}(x), p)}

as large as we like, contradicting that {f^{n}} is a uniform quasisimilarity action.

Replacing f by f^{−1} if necessary, we will assume that p is repelling, so f^{n}(x) → +∞ if x > p, and f^{n}(x) → −∞ if x < p. Under this assumption the stretch factor s will be > 1.

Step 3: f^{n} is not uniformly bilipschitz. If f^{n} is L-bilipschitz for all n, then for any y > p consider the sequence f^{n}(y) as n → +∞. Then d(p, f^{n}(y)) = d(f^{n}(p), f^{n}(y)) ≤ L · d(p, y) and so f^{n}(y) is a bounded sequence. Since y < f(y) < f^{2}(y) < ⋯ it follows that f^{n}(y) converges to a fixed point of f distinct from p, a contradiction.

Step 4: Finding the expansion constant s. Using the assumptions that f preserves orientation and p is repelling, we show that there is a unique real number s > 1 such that s^{n} ∈ [a(f^{n}), b(f^{n})] for all n.

First we note an interesting property of any uniform quasisimilarity action of a group G on a metric space X. Let [a(g), b(g)] be the stretch interval for the action of g on X. Note that the map g ↦ [a(g), b(g)] satisfies the following properties:

• b(g)/a(g) ≤ K for all g ∈ G.

• a(Id) = b(Id) = 1.

• [a(gh), b(gh)] ⊂ [a(g)a(h), b(g)b(h)] for all g, h ∈ G.

• [a(g^{−1}), b(g^{−1})] = [b(g)^{−1}, a(g)^{−1}] for all g ∈ G.

Definition (uniform quasihomomorphism). Given a group G, a uniform quasihomomorphism from G to R_{+} is a map which associates to each g ∈ G an interval [a(g), b(g)] satisfying the properties listed above, for some constant K ≥ 1 independent of g.

Hence to each uniform quasisimilarity action of a group G on a metric space, there is an associated uniform quasihomomorphism from G to R_{+}.

The following lemma, applied to the uniform quasihomomorphism n ↦ [a(f^{n}), b(f^{n})], produces the expansion factor s:

Lemma 3.4. For any uniform quasihomomorphism n ↦ [a_{n}, b_{n}] from Z to R_{+}, there exists a unique s > 0 such that s^{n} ∈ [a_{n}, b_{n}].

Proof. Multiplying Z by −1 if necessary, and assuming that [a_{n}, b_{n}] is not uniformly bounded, we may assume that if n is sufficiently large then [a_{n}, b_{n}] ⊂ (1, ∞). In this case the s that we find will be larger than 1. Let K be a quasihomomorphism constant for n ↦ [a_{n}, b_{n}].

For all n ≥ 1 define the following subinterval of R_{+}:

I_{n} = [ \sqrt[n]{a_{n}}, \sqrt[n]{b_{n}}].

Given m, n ≥ 1 we have I_{mn} ⊂ I_{n}, because

[a_{mn}, b_{mn}] = [a\underbrace{n + ⋯ + n}_{m times}, b\underbrace{n + ⋯ + n}_{m times}] ⊂ [a_{n}^{m}, b_{n}^{m}]

In particular, I_{n} ⊂ I_{1} = [a_{1}, b_{1}] for all n ≥ 1. Also, the ratio of the upper and lower endpoints of I_{n} is \sqrt[n]{b_{n}/a_{n}} ≤ \sqrt[n]{K}, which approaches 1 as n → +∞. Since I_{n} is a subinterval of the fixed interval [a_{1}, b_{1}] it follows that Length(I_{n}) approaches 0 as n → +∞.

We have shown that the intervals I_{n} are nested with respect to the divisor lattice on the natural numbers. Since Length(I_{n}) → 0 it follows that ∩_{n}I_{n} is a singleton {s}. This proves the existence of a unique s > 1 such that s^{n} ∈ [a_{n}, b_{n}] for all n ∈ Z_{+}.

Since [a_{−n}, b_{−n}] = [b_{n}^{−1}, a_{n}^{−1}], it follows that s^{n} ∈ [a_{n}, b_{n}] for all n ∈ Z. ♢

Step 5: f is bilipschitz conjugate to M_{s}. Conjugating by a translation if necessary, we may assume that the fixed point of f is p = 0. We’ll define the conjugacy ϕ on [0, ∞), and prove it is bilipschitz there. The extension to (−∞, 0] is defined similarly, and the rubber band principle proves that ϕ is bilipschitz on R.

Let x_{0} = 1, x_{n} = f^{n}(x_{0}). Define ϕ | [x_{0}, x_{1}] to be the unique orientation preserving affine homeomorphism from [x_{0}, x_{1}] to [1, s]. This map has constant derivative (s − 1)/(x_{1} − x_{0}) = (s − 1)/(x_{1} − 1), and we must get a bound on this derivative depending only on K.

Lemma 3.5. With the notation above,

\frac{s − 1}{x_{1} − 1} ∈ [\frac{1}{K^{2}} , K^{2}]

Accepting this lemma for the moment, let us define the conjugacy ϕ on all of [0, ∞) and prove that it is bilipschitz there, with a constant depending only on K.

Define ϕ(0) = 0, and define ϕ | [x_{n}, x_{n+1}] by

ϕ(x) = ϕ(f^{−n}(x)) · s^{n}

Obviously ϕ is a homeomorphism of [0, +∞) conjugating f to M_{s}.

We prove that ϕ: [0, +∞) → [0, +∞) is bilipschitz by applying the rubber band principle to the family of sets

S_{n} = {0, x_{n}}, I_{n} = [x_{n}, x_{n+1}], n ∈ Z

Clearly this family of sets covers [0, +∞), and it satisfies properties (2) and (3) in the rubber band principle.

Note that

d(0, x_{n}) = d(f^{n}(0), f^{n}(1))
∈ [a(f^{n}), b(f^{n})] · d(0, 1)
= [a(f^{n}), b(f^{n})]

According to Step 4 this interval is a subset of [s^{n}/K, s^{n} · K]. Since

d(ϕ(0), ϕ(x_{n})) = d(0, s^{n}) = s^{n}

it follows that ϕ is K-bilipschitz on the set S_{n}.

To prove that ϕ is bilipschitz on I_{n}, we have the following decomposition of ϕ | I_{n}:

\xymatrix{I_{n} \ar[r]_{f^{−n}} \ar[rrr]^{ϕ} & [x_{0}, x_{1}] \ar[r]_{ϕ} & [1, s] \ar[r]_{y→ys^{n}} & [s^{n}, s^{n+1}]}

The first map is [a(f^{−n}), b(f^{−n})]-bilipschitz, and this interval is contained in [s^{−n}/K, s^{−n}K]. The second map is [1/K^{2}, K^{2}] bilipschitz by Lemma 3.5. The third map is [s^{n}, s^{n}]-bilipschitz. The composition is therefore [1/K^{3}, K^{3}]-bilipschitz. ♢

Proof of Lemma 3.5. The idea of the proof is that x_{1}/s = d(f(1), f(0))/s ∈ [a(f), b(f)] · d(0, 1)/s ⊂ [s/K, K] · 1/s = [1/K, K], and so x_{1}/s is bounded by a constant depending only on K. If s were large it would follow that (x_{1} − 1)/(s − 1) is bounded. However, since s may not be large, we must work a little harder to make use of the fact that x_{n}/s^{n} is also bounded, by a constant depending only on K.

It follows from Step 4 that for all n ∈ Z the map f^{n} is [s^{n}/K, s^{n}K]- bilipschitz. In particular,

d(x_{n+1}, x_{n}) = d(f^{n}(x_{1}), f^{n}(x_{0})) ∈ [1/K, K] · s^{n} · d(x_{1}, x_{0})

and so

\frac{d(x_{n+1}, x_{n})}{s^{n+1} − s^{n}} ∈ [1/K, K] · \frac{s^{n}}{s^{n+1} − s^{n}} · d(x_{1}, x_{0})

= [1/K, K] · \frac{d(x_{1}, x_{0})}{s − 1}

We also have

d(x_{n}, x_{0}) = d(x_{n}, x_{n−1}) + ⋯ + d(x_{1}, x_{0})

∈ [1/K, K] · \frac{d(x_{1}, x_{0})}{s − 1} · (s^{n} − s^{n−1}) + ⋯

+ [1/K, K] · \frac{d(x_{1}, x_{0})}{s − 1} · (s^{1} − s^{0})

= [1/K, K] · \frac{d(x_{1}, x_{0})}{s − 1} · (s^{n} − 1)

and so

\frac{d(x_{n}, x_{0})}{s^{n} − 1} ∈ [1/K, K] · \frac{d(x_{1}, x_{0})}{s − 1}

On the other hand, we have

x_{n} = x_{n} − 0 = f^{n}(x_{0}) − f^{n}(0) ∈ [1/K, K] · s^{n} · (x_{0} − 0)

and, since x_{0} = 1 it follows that

\frac{x_{n}}{s^{n}} ∈ [1/K, K]

As n → ∞, the ratio

\frac{d(x_{n}, x_{0})}{s^{n} − 1} / \frac{x_{n}}{s^{n}} = \frac{x_{n} − 1}{s^{n} − 1} / \frac{x_{n}}{s^{n}}

approaches 1. Hence, for any ϵ > 0, if n is sufficiently large we have

\frac{d(x_{n}, x_{0})}{s^{n} − 1} ∈ [1/(K + ϵ), K + ϵ]

We have shown that the two intervals

[1/K, K] · \frac{d(x_{1}, x_{0})}{s − 1}  and  [1/(K + ϵ), K + ϵ]

have nonempty intersection, both containing d(x_{n}, x_{0})/(s^{n} − 1), and it follows that

\frac{1}{K + ϵ} ≤ K \frac{d(x_{1}, x_{0})}{s − 1}  and  \frac{1}{K} · \frac{d(x_{1}, x_{0})}{s − 1} ≤ K + ϵ

and so

\frac{1}{K(K + ϵ)} ≤ \frac{d(x_{1}, x_{0})}{s − 1} ≤ K(K + ϵ)

Since ϵ > 0 is arbitrary, the lemma follows. ♢

Now we prove Theorem 3.2 in the special case that G preserves orientation of R, in which case we conjugate G into Aff_{+}(R). The proof proceeds along the lines laid out on p. 332 of [Hin85], with a few comments needed to translate the quasisymmetric setting to the quasisimilarity setting.

Let Homeo_{+}(R) be the topological group of orientation preserving homeomorphisms of R in the topology of uniform convergence on compact sets. Given a sequence g_{i} ∈ Homeo_{+}(R) of K-quasisimilarities of R, the following two statements are equivalent:

• The sequence g_{i} has a convergent subsequence in Homeo_{+}(R).

• There exist x_{1}, x_{2} ∈ R such that g_{i}(x_{1}) is bounded and |g_{i}(x_{2})−g_{i}(x_{1})| is bounded away from 0 and ∞.

In this case, the limiting homeomorphism is also a K-quasisimilarity. To prove that the second statement above implies the first, one uses the Ascoli-Arzela Theorem, observing that the second statement is equivalent to saying that {g_{i}} is uniformly bilipschitz and {g_{i}(x_{1})} is bounded for some x_{1}.

If G ⊂ Homeo_{+}(R) is a group of K-quasisimilarities of R, it follows that the closure \overline{G} in Homeo_{+}(X) is also a group of K-quasisimilarities. If \overline{G} is bilipschitz conjugate to a subgroup of Aff_{+}(R) then so is G. We may therefore assume for the rest of the proof that G is closed in Homeo_{+}(R).

Lemma 6 of [Hin85] says that if g, h ∈ G−{Id} have distinct fixed points then H = g ∘ h ∘ g^{−1} ∘ h^{−1} has no fixed points. Lemma 7 of [Hin85] says that if g, h ∈ G each have no fixed points and g ≠ h then g(x) ≠ h(x) for all x ∈ R. We need not reprove these lemmas, because in each case the hypothesis requires only that the group generated by g, h be uniformly quasisymmetric, which is true in the present situation where g, h generate a uniform quasisimilarity group. If one desires, the proofs of Lemmas 6 and 7 may by improved by using the quasisimilarity conjugacies of Proposition 3.3, instead of the quasisymmetric conjugacies of Hinkkanen’s paper.

Define G_{T} to be the set of all elements of G without fixed points. By the previous paragraph, G_{T} is a subgroup of G. By Proposition 3.3, G_{T} consists of those elements which are bilipschitz conjugate to translations.

Lemma 8 of [Hin85] proves that if G is not cyclic and G ≠ G_{T} ≠ {Id} then G_{T} is not cyclic. Lemma 8 also proves, from the fact that G is closed in Homeo(R), that G_{T} is also closed. Lemma 9 of [Hin85] proves that if G_{T} = Transl(R), the group of all translations of R, then G ⊂ Aff_{+}(R).

Lemma 10 of [Hin85] says, assuming G is closed in Homeo(R), that the theorem is true in two special cases: G_{T} = G; and G_{T} = {Id}. Hinkkanen’s proof gives a quasisymmetric conjugacy, but if one follows the proof through verbatim one sees in the present setting that the conjugating map is bilipschitz. The point is this. Suppose G = G_{T}. One constructs a strictly increasing sequence of cyclic subgroups G_{1} ⊃ G_{2} ⊃ ⋯ whose union is dense in G_{T}, and then one conjugates each G_{i} to a translation group by applying the cyclic version of the theorem, using a conjugating map f_{i}. Then one shows that there is a limiting map f = lim f_{i} which conjugates all of G_{T} to a translation group. In our present case, the maps f_{i} are provided by Proposition 3.3, and f_{i} has a bilipschitz constant depending only on the quasisimilarity constant of the subgroups G_{i}. Since each G_{i} is a subgroup of G_{T}, the quasisimilarity constants of G_{i} are uniformly bounded, and so the f_{i} are uniformly bilipschitz. The limiting map f is therefore bilipschitz.

The only remaining case is when G ≠ G_{T} ≠ {Id}, and in this case one applies Lemmas 8, 9, and 10 as on page 332 of [Hin85] to finish the proof in the case that G preserves orientation.

To complete the proof we consider the case that G does not preserve orientation. Let G_{0} be the orientation preserving subgroup of G, a normal subgroup of index 2. Applying the orientation preserving case of the theorem, we may do a bilipschitz conjugacy of G_{0} into Aff_{+}(R). Applying this conjugacy to all of G, we may assume that G_{0} ⊂ Aff_{+}(R). Choose a single element g ∈ G − G_{0}.

The most interesting subcase (and the only case we really care about) is when (G_{0})_{T} is a dense subgroup of Transl(R). We follow the proof of Lemma 9 of [Hin85]. There is a dense additive subgroup S ⊂ R, and an isomorphism ϕ: S → S, such that g(x) + s = g(x + ϕ(s)) for all x ∈ R and s ∈ S. Since g reverses orientation on R, it follows that ϕ reverses order on S, and so ϕ is continuous on S. The map ϕ therefore extends to a continuous isomorphism of the additive group R. It follows that ϕ(s) = cs for some c ≠ 0. We therefore have g(x) + s = g(x + cs) for all x, s ∈ R. Taking x = 0 and r = cs we have g(r) = c^{−1}r + g(0), and so g is affine.

If (G_{0})_{T} is not a dense subgroup of Transl(R), then as we have seen there are two further subcases. In one subcase G_{0} is a cyclic group of translations, and in the other case (G_{0})_{T} is trivial and G_{0} is a group of dilations with a global fixed point. In particular G_{0} is abelian and so G is virtually abelian (these subcases will therefore not arise in applications to groups quasi-isometric to BS(1, n), because of Proposition 4.4). The interested reader can show in either of these cases that one can do a further conjugation keeping G_{0} in Aff_{+}(R) and taking g to an orientation reversing affine transformation.

4 A virtually faithful affine action on R

Recall the similarity groups Sim(R) = Aff(R) and Sim(Q_{n}). There is an obvious inclusion Isom(X_{n}) ↪ Aff(R)×Sim(Q_{n}), whose image is the subgroup of pairs (f, g) ∈ Aff(R)×Sim(Q_{n}) such that Stretch(f)·Stretch(g) = 1. We have natural inclusions

BS(1, n) ↪ Isom(X_{n})

↪ Aff(R) × Sim(Q_{n}) ↪ Bilip(R) × Bilip(Q_{n}) ≈ QI(BS(1, n))

the final isomorphism being Theorem 7.1 in [FM97].

There is a split short exact sequence

1 → Isom(R) → Aff(R) → Stretch(R) → 1 where Isom(R) is the group of isometries of R, and Stretch(R) is the group of orientation preserving stretch homeomorphisms of R, otherwise known as SL(1, R). Note that there are canonical isomorphisms

Isom(R) ≈ R ⋊ (Z/2Z) and Stretch(R) ≈ R_{+} ⟶^{log} R

where Z/2Z acts on R by a reflection.

Given a subgroup Γ < Aff(R) we therefore obtain a commutative diagram

\xymatrix{1 \ar[r] & A(Γ) \ar@{^{(}->}[r] \ar@{^{(}->}[d] & Γ \ar[r] \ar@{^{(}->}[d] & Stretch(Γ) \ar[r] \ar@{^{(}->}[d] & 1 \\ 1 \ar[r] & Isom(R) \ar@{^{(}->}[r] & Aff(R) \ar[r] & Stretch(R) \ar[r] & 1}

The purpose of this section is to prove:

Theorem 4.1. Let G be a finitely generated group quasi-isometric to BS(1, n). There exists a representation θ: G → Aff(R) such that ker(θ) is finite. Moreover, setting Γ = θ(G), the group Stretch(Γ) is infinite cyclic.

To begin the proof, by Proposition 2.2 we obtain a properly discontinuous, cocompact quasi-action ψ: G → QIMap(X_{n}), inducing a representation ρ: G → QI(X_{n}) ≈ Bilip(R)×Bilip(Q_{n}), with projected representations ρ_{ℓ}: G → Bilip(R) and ρ^{u}: G → Bilip(Q_{n}). The representations ρ_{ℓ} and ρ^{u} are uniform quasi-similarity actions.

Applying Theorem 3.2 to the representation ρ_{ℓ}, there exists a bilipschitz homeomorphism ϕ: R → R such that θ(g) = ϕ ∘ ρ_{ℓ}(g) ∘ ϕ^{−1} ∈ Aff(R) for all g ∈ G, and so we obtain a representation

θ: G → Aff(R)

We now need to use further properties of the situation to show that the representation θ has finite kernel and that the group θ(G) ⊂ Aff(R) has infinite cyclic stretch group.

4.1 Biconvergence groups

Recall that if Γ is a word hyperbolic group then the action of Γ on its boundary ∂Γ is a uniform convergence group action, which means that Γ acts properly discontinuously and cocompactly on the triple space

T (∂Γ, ∂Γ, ∂Γ) = {(x, y, z) ∈ ∂Γ | x ≠ y, y ≠ z, z ≠ x}

Furthermore if G is quasi-isometric to Γ then the induced action of G on ∂Γ is also a uniform convergence group action.

Despite the fact that the groups BS(1, n) are far from being word hyperbolic, the upper and lower boundaries interact in a way reminiscent of convergence group actions, motivating the following definition.

Definition (Biconvergence group). Suppose that X, Y are topological spaces. Define two triple spaces

T (X, X, Y) = {(x, y, ζ) ∈ X × X × Y | x ≠ y}

and

T (X, Y, Y) = {(x, η, ζ) ∈ X × Y × Y | η ≠ ζ}

A biconvergence action of a group G on the pair (X, Y) consists of an action of G on X and an action on Y, such that the induced diagonal actions of G on T (X, X, Y) and on T (X, Y, Y) are properly discontinuous. If furthermore the actions of G on T (X, X, Y) and on T (X, Y, Y) are cocompact, then we call this a uniform biconvergence action of G on (X, Y).

Example. BS(1, n) is a uniform biconvergence group on (R, Q_{n}). Note that BS(1, n) does not act properly discontinuously on the triple space T (R, R, R) = {(x, y, z) ∈ R × R × R | x ≠ y, y ≠ z, z ≠ x}, nor on T (Q_{n}, Q_{n}, Q_{n}). For example the action on R contains translations of R by any rational number of the form i/n^{j}.

Example. Let Sol be the unique 3-dimensional, solvable, non-nilpotent, connected Lie group which has a cocompact lattice. Sol can be identified with R^{3} in such a way that the left invariant metric is

ds^{2} = e^{−2t}dx^{2} + e^{2t}dy^{2} + dt^{2}

There is an “upper boundary” ∂^{u}Sol consisting of the set of all “right side up” hyperbolic planes of the form y =(constant), and we have an identification ∂^{u}Sol = R. There is also a “lower boundary” ∂_{ℓ}Sol consisting of all “upside down” hyperbolic planes x =(constant), also identified with R. If Γ is the fundamental group of a compact 3-manifold M fibering over S^{1} with torus fiber T^{2}, so that the monodromy map T^{2} → T^{2} is an Anosov homeomorphism, then M has a Riemannian metric locally modelled on Sol, and so Γ may be identified with a cocompact, discrete subgroup of Sol. Under these conditions, one can show that the induced pair of actions of Γ on (∂_{ℓ}Sol, ∂^{u}Sol) ≈ (R, R) is a biconvergence action.

Figure 1: The point π(x, y, ζ) is the barycenter of the triangle Δ(x, y, +∞) in the hyperbolic plane H ⊂ X_{n} correponding to ζ.

Proposition 4.2 (G is a biconvergence group). Let G be a finitely generated group which is quasi-isometric to BS(1, n). The induced action of G on (∂_{ℓ}X_{n}, ∂^{u}X_{n}) ≈ (R, Q_{n}) is a uniform biconvergence action.

Proof. We use the symbols ∂_{ℓ}, ∂^{u} as shorthand for ∂_{ℓ}X_{n}, ∂^{u}X_{n}. First we give a detailed proof that the action of G on T (∂_{ℓ}, ∂_{ℓ}, ∂^{u}) is properly discontinuous and cocompact. We then indicate the changes needed for T (∂_{ℓ}, ∂^{u}, ∂^{u}).

By Proposition 2.1 we have a (K, C)-quasi-action ψ: G → QIMap(X_{n}) which induces the given action of G on T (∂_{ℓ}, ∂_{ℓ}, ∂^{u}).

There is a map π: T (∂_{ℓ}, ∂_{ℓ}, ∂^{u}) → X_{n} defined as follows (see Figure 1). Given (x, y, ζ) ∈ T (∂_{ℓ}, ∂_{ℓ}, ∂^{u}), let H ⊂ X_{n} be the hyperbolic plane corresponding to ζ ∈ ∂^{u}. The boundary ∂H is identified with ∂_{ℓ} plus a point denoted +∞. Consider the ideal triangle Δ = Δ(x, y, +∞) ⊂ H. Define π(x, y, ζ) to be the barycenter of Δ, that is, the intersection of the perpendiculars from each vertex of Δ to the opposite side.

The map π is obviously continuous, proper, and equivariant with respect to each isometry ϕ of X_{n}, that is

π(ϕ(x, y, ζ)) = ϕ(π(x, y, ζ))

We claim that π is almost equivariant with respect to a quasi-isometry. That is, there exists C_{1} depending on K, C such that if f is a (K, C)-quasi-isometry of X_{n}, with induced boundary maps f_{ℓ}: R → R and f^{u}: Q_{n} → Q_{n}, then for all x, y ∈ R, ζ ∈ Q_{n} we have

d(π(f_{ℓ}(x), f_{ℓ}(y), f^{u}(ζ)), f(π(x, y, ζ))) ≤ C_{1}

To prove this claim, note first that d_{H}(f(ζ), f^{u}(ζ)) ≤ C_{2} for some constant C_{2} depending only on K, C. Composing f | ζ with the closest point projection from f(ζ) to f^{u}(ζ), we obtain a quasi-isometry from the hyperbolic plane ζ to the hyperbolic plane f^{u}(ζ). This quasi-isometry takes x, y, regarded as elements of the line at infinity of ζ, to f_{ℓ}(x), f_{ℓ}(y), regarded as elements of the line at infinity of f^{u}(ζ), and it takes the point at infinity of ζ to the point at infinity of f^{u}(ζ). The claim now follows from the fact that the barycenter map of H^{2} is almost equivariant with respect to quasi-isometries of H^{2}.

Suppose that the action of G on T (∂_{ℓ}, ∂_{ℓ}, ∂^{u}) is not properly discontinuous. Then there is a sequence of distinct elements g_{i} ∈ G and a point (x_{0}, y_{0}, ζ_{0}) ∈ T (∂_{ℓ}, ∂_{ℓ}, ∂^{u}) such that, setting (x_{i}, y_{i}, ζ_{i}) = g_{i}(x_{0}, y_{0}, ζ_{0}), the sequence (x_{i}, y_{i}, ζ_{i}) converges to some (x, y, ζ) ∈ T (∂_{ℓ}, ∂_{ℓ}, ∂^{u}). After chopping off an initial subsequence, it follows that ψ_{g_{i}}(π(x_{0}, y_{0}, ζ_{0})) stays in a bounded neighborhood of π(x, y, ζ) in X_{n}. However, this contradicts the fact that the quasi-action ψ is properly discontinuous.

Next we prove that the action of G on T (∂_{ℓ}, ∂_{ℓ}, ∂^{u}) is cocompact. Note that the map π: T (∂_{ℓ}, ∂_{ℓ}, ∂^{u}) → X_{n} is continuous and proper. By cocompactness of the quasi-action ψ: G → QIMap(X_{n}), there exists x_{0} ∈ X_{n} and C_{2} > 0 such that for each y ∈ X_{n} there exists g ∈ G with d(y, ψ_{g}(x_{0})) ≤ C_{2}.

We claim that the compact set π^{−1}(\overline{B}(x_{0}, KC_{2} + 2C + C_{1})) is a fundamental domain for the action of G on T (∂_{ℓ}, ∂_{ℓ}, ∂^{u}).

To prove this claim, consider any t ∈ T (∂_{ℓ}, ∂_{ℓ}, ∂^{u}), and choose g so that d(π(t), ψ_{g}(x_{0})) ≤ C_{2}. Applying ψ_{g^{−1}} we obtain

d(ψ_{g^{−1}}(π(t)), ψ_{g^{−1}}ψ_{g}(x_{0})) ≤ KC_{2} + C
d(ψ_{g^{−1}}(π(t)), x_{0})) ≤ KC_{2} + C + C

Also,

d(π(g^{−1} · t), ψ_{g^{−1}}(π(t))) ≤ C_{1}
d(π(g^{−1} · t), x_{0}) ≤ KC_{2} + 2C + C_{1}
g^{−1} · t ∈ π^{−1}(\overline{B}(x_{0}, KC_{2} + 2C + C_{1}))

Figure 2: The point κ(x, η, ζ) is the intersection of the three sides of the triangle Δ(−∞, η, ζ) in the tree τ ⊂ X_{n} correponding to x.

completing the proof of cocompactness.

To prove that the action of G on T (∂_{ℓ}, ∂^{u}, ∂^{u}) is properly discontinuous and cocompact, it suffices to describe a continuous, proper, almost Gequivariant map κ: T (∂_{ℓ}, ∂^{u}, ∂^{u}) → X_{n} (see Figure 2), and mimic the above proof. Given (x, η, ζ) ∈ T (∂_{ℓ}, ∂^{u}, ∂^{u}), the point x corresponds to some tree τ ⊂ X_{n}, the image of an isometric section of the projection map q: X_{n} → T_{n}. The set Ends(τ) is identified with ∂^{u} plus a point denoted −∞. The three points −∞, η, ζ ∈ Ends(τ) determine an ideal triangle Δ(−∞, η, ζ) ⊂ τ whose three sides intersect in a unique point which we take to be κ(x, η, ζ).

♢

We now give a property of uniform biconvergence groups which is similar in spirit to what holds for uniform convergence groups.

Proposition 4.3 (Contraction property). Let G be a finitely generated group quasi-isometric to X_{n}. Let (ρ_{ℓ}, ρ^{u}): G → (Bilip(R), Bilip(Q_{n})) be the uniform biconvergence action given by Proposition 4.2. For every compact subset K ⊂ Q_{n} and every open subset U ⊂ Q_{n}, there exists g ∈ G such that g · K ⊂ U.

Proof. The appropriate generalization of this proposition should hold for any uniform biconvergence group, but here we will make use of the fact that ρ^{u} is a uniform quasisimilarity action.

Choose x ∈ R and ζ ∈ Q_{n}, and choose a one-to-one convergent sequence ζ_{1}, ζ_{2}, … → ζ in Q_{n} such that ζ_{i} ≠ ζ for all i ∈ N. Choose a compact fundamental domain K ⊂ T (R, Q_{n}, Q_{n}) for the action of G. Choose g_{i} ∈ G so that g_{i}^{−1} · (x, ζ_{i}, ζ) ∈ K. Pass to a subsequence so that g_{i}^{−1} · (x, ζ_{i}, ζ) converges to (y, η, ω) ∈ K. Of course η ≠ ω, because (y, η, ω) ∈ T (R, Q_{n}, Q_{n}). Note that d(g_{i}^{−1} ·ζ_{i}, g_{i}^{−1} ·ζ) → d(η, ω) ≠ 0, while d(ζ_{i}, ζ) → 0. It follows that the stretch interval [a_{i}, b_{i}] for the action of g_{i} on Q_{n} converges to 0, and so for sufficiently large i the map g_{i} contracts distance uniformly in Q_{n}, with a contraction factor b_{i} that goes to 0 as i → ∞. On the other hand, we also know that d(ω, g_{i}^{−1} · ζ) converges to 0, and so

d(g_{i} · ω, ζ) = d(g_{i} · ω, g_{i} · (g_{i}^{−1} · ζ))
≤ b_{i} d(ω, g_{i}^{−1} · ζ)
→ 0 as i → ∞

It follows that the action of the sequence (g_{i}) on Q_{n} converges uniformly on compact sets to the constant map with value ζ. Choosing ζ to be any point in U, and choosing K ⊂ Q_{n} to be any compact set, the proposition follows. ♢

Remark. In the above proof, we know by Proposition 2.2 that the stretch intervals for the action of g_{i} on R and on Q_{n} are inversely related, and so the stretch interval for g_{i}^{−1} on R converges to 0. Since d(g_{i}^{−1}(x), y) → 0 it follows that g_{i}^{−1} converges uniformly on compact sets to the constant map with value y. In some sense the sequence (g_{i}) therefore has “source-sink” dynamics, with source y ∈ R and sink ζ ∈ Q_{n}. This gives another analogy between convergence groups and biconvergence groups.

4.2 Two applications of the biconvergence property

Our first application shows that the representation θ: G → Aff(R) is virtually faithful.

Proposition 4.4 (θ is virtually faithful). The homomorphisms ρ_{ℓ}: G → Bilip(R) and θ: G → Aff(R) each have finite kernel.

Proof. Since ρ_{ℓ} and θ are conjugate in Bilip(R), they have the same kernel. So we need only prove that ρ_{ℓ} has finite kernel.

Pick a fixed clone C ⊂ Q_{n}. Let A = 0 × 1 × C, a compact subset of T (R, R, Q_{n}). Note that C is also an open subset of Q_{n}.

Suppose the proposition is false. Then there exists an arbitrarily large subset {g_{1}, …, g_{k}} ⊂ G acting trivially on R. We shall construct g ∈ G, depending on g_{1}, …, g_{k}, such that (gg_{i}g^{−1}) · A ∩ A ≠ ∅ in T (R, R, Q_{n}), for i = 1, …, k. Since k is arbitrarily large, this contradicts proper discontinuity of the action of G on T (R, R, Q_{n}) (Proposition 4.2).

Pick any point ζ_{0} ∈ Q_{n}. Let ζ_{i} = g_{i} · ζ_{0}. Let K be the smallest clone containing ζ_{0}, ζ_{1}, …, ζ_{k}, in particular K is compact. Furthermore, (g_{i} ·K)∩ K ≠ ∅ for i = 1, …, k.

By Proposition 4.3 applied to the compact set K and the open set C, there exists g ∈ G such that g · K ⊂ C in Q_{n}. Thus, (gg_{i}g^{−1}) · C ∩ C ≠ ∅ in Q_{n}, for each i = 1, …, k. Since gg_{i}g^{−1} acts as the identity on R it follows that (gg_{i}g^{−1}) · A ∩ A ≠ ∅ in T (R, R, Q_{n}), contradicting proper discontinuity as indicated above. ♢

Our next application of the biconvergence property shows that the stretch group of θ(G) ⊂ Aff(R) is infinite cyclic.

Proposition 4.5 (stretch group is cyclic). Let Γ = θ(G). The group Stretch(Γ) is infinite cyclic.

Proof. Recall that the representation θ: G → Aff(R) is obtained from ρ_{ℓ}: G → Bilip(R) by a bilipschitz conjugacy: there exists a bilipschitz map ϕ: R → R such that

θ(g) = ϕ ∘ ρ_{ℓ}(g) ∘ ϕ^{−1}

for all g ∈ G.

Let s(g) = Stretch(θ(g)). Let [a_{ℓ}(g), b_{ℓ}(g)] be the stretch interval of ρ_{ℓ}(g) acting on R, and let [a^{u}(g), b^{u}(g)] be the stretch interval of ρ^{u}(g) acting on Q_{n}. We know by Proposition 2.2 that there exists L_{1} ≥ 1 such that

[a^{u}(g), b^{u}(g)] · [a_{ℓ}(g), b_{ℓ}(g)] ⊂ [\frac{1}{L_{1}} , L_{1}]

We also know since the conjugating map ϕ is bilipschitz that

\frac{[a_{ℓ}(g), b_{ℓ}(g)]}{s(g)} ⊂ [\frac{1}{L_{2}} , L_{2}]  (∗)

where L_{2} is the square of the bilipschitz constant of ϕ. Therefore, setting L_{3} = L_{1}L_{2} we have

[a^{u}(g), b^{u}(g)] · s(g) ⊂ [\frac{1}{L_{3}} , L_{3}]  (∗∗)

Suppose the proposition is false. Then there exists a sequence g_{i} ∈ G such that s(g_{i}) ≠ 1 but s(g_{i}) → 1 as i → ∞. Since s(g_{i}) ≠ 1 it follows that θ(g_{i}) has a unique fixed point in R, and so ρ_{ℓ}(g_{i}) has a unique fixed point in R.

We claim also that ρ^{u}(g_{i}) has a unique fixed point in Q_{n}. To see this, since s(g_{i}) ≠ 1 we may choose a sufficiently high power g_{i}^{k} so that s(g_{i}^{k}) ∉ [\frac{1}{L_{3}} , L_{3}]. It follows that

1 ∉ [a^{u}(g_{i}^{k}), b^{u}(g_{i}^{k})]

and therefore either ρ^{u}(g_{i}^{k}) or ρ^{u}(g_{i}^{−k}) is a contraction mapping of Q_{n} and so has a unique fixed point, and so ρ^{u}(g_{i}) has a unique fixed point.

Let x_{i} ∈ R be the unique fixed point of ρ_{ℓ}(g_{i}), and let ζ_{i} ∈ Q_{n} be the unique fixed point of ρ^{u}(g_{i}). For the rest of the proof we use only the actions ρ_{ℓ}: G → Bilip(R) and ρ^{u}: G → Bilip(Q_{n}), and so we may unambiguously use the “dot” notation for these group actions.

Since the action of G on T (R, R, Q_{n}) is cocompact, we may pick a compact fundamental domain A.

Pick any point w ∈ R distinct from all the points x_{i}. Choose h_{i} ∈ G so that h_{i}(x_{i}, w, ζ_{i}) ∈ A. Replacing g_{i} by its conjugate h_{i}g_{i}h_{i}^{−1}, and replacing x_{i}, w, ζ_{i} by their images under h_{i}, we may assume that (x_{i}, w_{i}, ζ_{i}) ∈ A for some sequence w_{i}; since s is invariant under conjugation we still have s(g_{i}) → 1. By compactness of A we may pass to a subsequence so that x_{i} → x in R and ζ_{i} → ζ in Q_{n}.

Since s(g_{i}) → 1, using (∗) and (∗∗) it follows that for all sufficiently large i we have

[a_{ℓ}(g_{i}), b_{ℓ}(g_{i})] ⊂ [\frac{1}{L}, L]

and

[a^{u}(g_{i}), b^{u}(g_{i})] ⊂ [\frac{1}{L}, L]

where L = 2 Max(L_{2}, L_{3}).

Choose a point y ≠ x in R. We will contradict proper discontinuity of the action of G on T (R, R, Q_{n}) by showing that (g_{i} · x, g_{i} · y, g_{i} · ζ) stays in a compact subset of T (R, R, Q_{n}).

We know that g_{i} · x → x in R, because

d(x, g_{i} · x) ≤ d(x, x_{i}) + d(x_{i}, g_{i} · x)

= d(x, x_{i}) + d(g_{i} · x_{i}, g_{i} · x)

∈ d(x, x_{i}) + d(x, x_{i}) · [\frac{1}{L}, L]

= d(x, x_{i})[1 + \frac{1}{L}, 1 + L]

→ 0 as i → ∞

because d(x, x_{i}) → 0. The same argument shows that g_{i} · ζ → ζ in Q_{n}.

To complete the proof, we will show that g_{i} · y stays in a compact subset of R disjoint from x, from which it follows that (g_{i} · x, g_{i} · y, g_{i} · ζ) stays in a compact subset of triple space, a contradiction as noted above.

We have

d(x_{i}, g_{i} · y) = d(g_{i} · x_{i}, g_{i} · y)

⊂ [\frac{1}{L}, L] · d(x_{i}, y)

⊂ [\frac{1}{2L}, 2L] · d(x, y)

for sufficiently large i, because of the fact that d(x_{i}, y) → d(x, y). Since d(x_{i}, x) → 0 it follows that

d(x, g_{i} · y) ∈ [\frac{1}{3L}, 3L] · d(x, y)

for sufficiently large i. The set of points in R whose distance from x is in the interval [1/3L, 3L] · d(x, y) is a compact subset of R disjoint from x, and this compact subset contains g_{i} · y for all sufficiently large i, as desired. ♢

Remark. The inequality (∗) is precisely where we need Theorem 3.2, which gives a bilipschitz conjugacy between ρ_{ℓ} and θ, instead of the quasisymmetric conjugacy provided by Hinkkanen’s original theorem.

5 Finishing the proof of Theorem A

Suppose G is quasi-isometric to BS(1, n). By Theorem 4.1 we have a quotient group Γ = G/N, where N is a finite normal subgroup of G, and we have a diagram of short exact sequences

\xymatrix{1 \ar[r] & A = A(Γ) \ar@{^{(}->}[r] \ar@{^{(}->}[d] & Γ \ar[r] \ar@{^{(}->}[d] & Stretch(Γ) = ⟨t⟩ \ar[r] \ar@{^{(}->}[d] & 1 \\ 1 \ar[r] & Isom(R) \ar@{^{(}->}[r] & Aff(R) \ar[r] & Stretch(R) \ar[r] & 1}

where ⟨t⟩ is infinite cyclic. Choosing a splitting, we may regard t as an element of Aff(R), and replacing t with t^{−1} if necessary we may assume Stretch(t) > 1.

We have the following additional properties of Γ:

• Γ is solvable, because Aff(R) is solvable.

• Γ has a torsion free subgroup of index at most 2, namely

Γ_{+} = Γ ∩ Aff_{+}(R)

• Γ is quasi-isometric to BS(1, n).

Since finite presentability is a quasi-isometry invariant ([GH91], Proposition 10.18) we also have:

• Γ is finitely presented.

Note that there are finitely generated subgroups of Aff(R) which are not finitely presented; once such group is described in [Str84].

We may now quote the following theorem of Bieri-Strebel, taken from [Str84] (see also [BS78]):

Theorem 5.1 (Bieri-Strebel). Let Γ be a finitely presented solvable group, and suppose that Γ has an HNN presentation of the form

Γ = A∗_{Z} = ⟨A, t | tA_{1}t^{−1} = A_{2}⟩

where A_{1}, A_{2} are subgroups of A. Then there is another HNN presentation

Γ = B∗_{Z} = ⟨B, t | tBt^{−1} = B′⟩

where B is a finitely generated subgroup of A and B′ is a subgroup of B.

For completeness, here is a quick proof suggested to us by T. Delzant.

Proof. Let K be a finite complex with fundamental group Γ and universal covering map p: K̃ → K. Let T be a Bass-Serre tree for the given HNN decomposition of Γ. Let f: K̃ → T be a Γ-equivariant map, transverse to the midpoint of each edge of T. Pulling back those midpoints gives a Γequivariant 1-complex L̃ ⊂ K̃. Let L = p(L̃), a 1-complex in K. Construct a new Γ-tree T′ whose edges correspond to components of L̃ and whose vertices correspond to components of K̃ − L̃. The vertex and edge stabilizers of T′ are subgroups of vertex and edge stabilizers of T. Moreover, if v is a vertex of T′ corresponding to a component C̃ of K̃ − L̃, and if C = p(C̃), then Stab(v) is isomorphic to the image of the inclusion induced map π_{1}(C) → π_{1}(K), and so Stab(v) is finitely generated. Passing to the quotient graph of groups obtained from the action of Γ on T′, and collapsing subgraphs in the appropriate manner, one obtains an HNN decomposition Γ = ⟨B, t′ | t′B_{1}t′^{−1} = B_{2}⟩ where B is a finitely generated subgroup of A, B_{1}, B_{2} < B, and t′ is conjugate into the infinite cyclic subgroup ⟨t⟩. If both B_{1} and B_{2} are proper subgroups of B, then Γ has a free subgroup of rank ≥ 2; this follows from the normal forms theorem for HNN decompositions, or from a ping-pong argument on the Bass-Serre tree. Since Γ is solvable it has no free subgroups of rank ≥ 2, and so one of B_{1}, B_{2} equals B. ♢

Applying the above theorem to Γ, we have a finitely generated subgroup B ⊂ A(Γ), and an HNN decomposition

Γ = ⟨B, t | tbt^{−1} = ϕ(b), ∀b ∈ B⟩

where ϕ: B → B is an injective endomorphism, and B is a finitely generated subgroup of Isom(R).

Recall that Γ_{+} = Γ ∩ Aff_{+}(R), and let B_{+} = B ∩ Isom_{+}(R). The group B_{+} is a finitely generated subgroup of Isom_{+}(R) = Transl(R), and so B_{+} is free abelian of some finite rank r ≥ 1. The indices [Γ : Γ_{+}], [B : B_{+}] are both ≤ 2, and so we fall into three cases:

Case 1. [Γ : Γ_{+}] = [B : B_{+}] = 1.

Case 2. [Γ : Γ_{+}] = 2 and [B : B_{+}] = 1.

Case 3. [Γ : Γ_{+}] = [B : B_{+}] = 2.

In case (1) we obviously have t ∈ Aff_{+}(R), that is, t preserves orientation of R. In case (3), if t ∈ Aff(R) − Aff_{+}(R), in other words if t reverses orientation of R, then we can replace t by its product with an element of B − B_{+}, a reflection of R; and hence we may assume in case (3) that t ∈ Aff_{+}(R). In case (2) we necessarily have t ∈ Aff(R) − Aff_{+}(R). Setting s = t^{2} and ψ = ϕ^{2} in case (2), or s = t and ψ = ϕ in cases (1) and (3), we have an HNN decomposition

Γ_{+} = ⟨B_{+}, s | sbs^{−1} = ψ(b), ∀b ∈ B_{+}⟩

where s ∈ Aff_{+}(R). Since B_{+} ≈ Z^{r} and ψ(B_{+}) ≈ B_{+}, it follows that ψ(B_{+}) has finite index in B_{+}. Let I = [B_{+} : ψ(B_{+})].

Next we extract some homological information about Γ_{+} (see [Bro82]). Recall that if K is a finitely generated, virtually torsion free group, the virtual cohomological dimension of K, denoted vcd(K), is the cohomological dimension of any finite index, torsion free subgroup K′ < K; by Serre’s Theorem ([Bro82] Theorem VIII.3.1) this number is independent of the choice of K′.

Lemma 5.2. For any injective homomorphism ψ: Z^{r} → Z^{r}, the HNN group K = ⟨Z^{r}, t | tbt^{−1} = ψ(b), ∀b ∈ Z^{r}⟩ has virtual cohomological dimension equal to r + 1.

Proof. We first construct a compact Eilenberg-Maclane space for K by mimicking the construction in [FM97] of an Eilenberg-Maclane space for BS(1, n).

Extend the endomorphism ψ: Z^{r} → Z^{r} to a linear isomorphism ψ: R^{r} → R^{r} which commutes with the action of Z^{r}. Let T^{r} = R^{r}/Z^{r} be the rdimensional torus. The map ψ descends to a self covering map Ψ: T^{r} → T^{r}. Let I be the index [Z^{r} : ψ(Z^{r})], and so I is the determinant of ψ acting on R^{r}, and I is the degree of the covering map Ψ. Let C be the mapping torus of Ψ,

C = T^{r} × [0, 1] / (x, 0) ∼ (Ψ(x), 1)

Note that π_{1}(C) ≈ K. Let X be the universal covering space of C. Let T_{I} be the tree constructed in §1, a homogeneous directed tree with one incoming edge and I outgoing edges at each vertex. By construction, the space X fibers over T_{I} with fiber R^{r}. Since T_{I} is contractible we have a homeomorphism

X ≈ R^{r} × T_{I}

In particular X is contractible, and so C is a compact Eilenberg-Maclane space for the group K.

Since K is torsion free we have vcd(K) = cd(K). By [Bro82] corollary VIII.7.6, the number cd(K) is the maximum dimension k such that H_{c}^{k}(X; Z) ≠ 0, where H_{c}^{k} denotes cohomology with compact supports. By [Spa81] (p. 341, corollary 9 and p. 360, exercise 6), there is a Künneth formula

H_{c}^{k}(X; Z) ≈ ∑_{i=0}^{k} H_{c}^{i}(R^{r}; Z) ⊗ H_{c}^{k−i}(T_{I}; Z)

from which it follows that

H_{c}^{k}(X; Z) ≈ \begin{cases}0 & if k ≠ r + 1 \\ H_{c}^{1}(T_{I}; Z) & if k = r + 1 \end{cases}

Since T_{I} has infinitely many ends, it follows that H_{c}^{1}(T_{I}; Z) is an infinite rank abelian group, and so vcd(K) = r + 1. ♢

Remark. This proof gives extra information about the group K, namely a complete computation of the cohomology groups H_{c}^{k}(X; Z) ≈ H^{k}(K; ZK), which by [Ger93] are quasi-isometry invariants of K. Note in particular that K is a duality group in the sense of Bieri and Eckmann (see [Bro82] §VIII), which means that, setting D = H^{r+1}(K; ZK) ≈ H_{c}^{r+1}(X; Z), we have

H^{i}(K, M) ≈ H_{r+1−i}(K, D ⊗ M) for all K-modules M.

Now we complete the proof of Theorem A.

Applying Lemma 5.2 we have the following computations of virtual cohomological dimensions:

vcd(Γ_{+}) = r + 1 and vcd(BS(1, n)) = 2

By a theorem of Gersten [Ger93] and Block-Weinberger [BW97], if two virtually torsion free groups have finite Eilenberg-Maclane spaces, and if the groups are quasi-isometric, then they have the same vcd. It follows that vcd(Γ_{+}) = vcd(BS(1, n)), and so r = 1 and B_{+} is an infinite cyclic group B_{+} = ⟨b⟩. The endomorphism ψ: Z → Z preserves orientation, and so ψ(b) = b^{m} for some m ≥ 1. If m = 1 then Γ_{+} ≈ Z^{2}, but Z^{2} is not quasiisometric to BS(1, n)—for example, Z^{2} has quadratic growth but BS(1, n) has exponential growth. It follows that m ≥ 2, and Γ_{+} is isomorphic to BS(1, m). Applying the main result of [FM97] it follows that Γ_{+} is abstractly commensurable to BS(1, n), in fact m, n are positive powers of the same positive integer.

As for the group Γ itself, we have already shown that Γ has a subgroup Γ_{+} isomorphic to BS(1, m) of index ≤ 2, and BS(1, m) is abstractly commensurable to BS(1, n); therefore Γ is abstractly commensurable with BS(1, n). This completes the proof of Theorem A. ♢

Here is a corollary:

Corollary 5.3. Let G be a finitely generated, torsion free group which is quasi-isometric to BS(1, n) for some n ≥ 2. Then G ≈ BS(1, k) for some integer k with |k| ≥ 2, such that BS(1, n) and BS(1, k) are abstractly commensurable.

Proof. Apply Theorem A to get a short exact sequence

1 → N → G → Γ → 1

with N finite. Since G is torsion free it follows that N = {Id} and G ≈ Γ. In the proof of Theorem A, it also follows that B = B_{+} is infinite cyclic, B = ⟨b⟩. Thus we fall into case 1 or 2 in the proof of Theorem A. The injective endomorphism ϕ: B → B must have the form ϕ(b) = b^{k} for some integer k ≠ 0. If k = ±1 then Γ has polynomial growth, a contradiction as above. Therefore, |k| ≥ 2 and G ≈ BS(1, k). Since BS(1, k) contains BS(1, k^{2}) with index 2, and since k^{2} ≥ 2, the main result of [FM97] shows that BS(1, k^{2}) is abstractly commensurable to BS(1, n), and so G ≈ BS(1, k) is abstractly commensurable to BS(1, n). ♢

In the proof of Theorem A, recall the three cases in the analysis of the group Γ. As the proof of the corollary shows, in case 1 we have Γ ≈ BS(1, k) for some k ≥ 2, and in case 2 we have Γ ≈ BS(1, k) for some k ≤ −2.

In case 3, B is the infinite dihedral group B = ⟨a, r | r^{2} = 1, rar = a^{−1}⟩. To enumerate the possibilities for Γ, it suffices to enumerate the injective, nonsurjective, orientation preserving endomorphisms ϕ: B → B (orientation preserving means that ϕ fixes each of the two ends of the group B). Fix the representation B → Aff(R) given by a·x = x+2, r·x = −x. The reflections of B may be enumerated as r_{i} = a^{−i}r, i ∈ Z, and r_{i} is the reflection about the point i ∈ R. Up to conjugation by an automorphism of B, there are two infinite sequences of endomorphisms ϕ, depending on whether ϕ fixes some reflection, as follows.

Case 3.i. If ϕ fixes a reflection then, replacing r by the fixed reflection, there exists an integer m ≥ 2 such that ϕ(a) = a^{m}, ϕ(r) = r. We may represent ϕ in Aff_{+}(R) as the expansion with fixed point 0 and stretch m, that is, ϕ(x) = mx. We therefore have

Γ = B∗_{ϕ} = ⟨a, r, t | r^{2} = 1, rar^{−1} = a^{−1}, tat^{−1} = a^{m}, trt^{−1} = r⟩ Case 3.ii. If ϕ does not fix a reflection then, up to a replacement of r,

there is an odd integer m = 2k + 1 ≥ 3 such that ϕ(a) = a^{m} and ϕ(r) = a^{−k}r. We may represent ϕ in Aff_{+}(R) as the expansion with fixed point −1/2 and stretch m, that is ϕ(x) = m(x + 1/2) − 1/2. It is clear from this description that ϕ indeed fixes no reflections; one can also compute directly that ϕ(r_{i}) = r_{2ki+k+i}, and clearly there is no i ∈ Z such that 2ki + k + i = i. We have:

Γ = B∗_{ϕ} = ⟨a, r, t | r^{2} = 1, rar^{−1} = a^{−1}, tat^{−1} = a^{m}, trt^{−1} = a^{−k}r⟩

6 Final comments: The Sullivan-Tukia Theorem

The Sullivan-Tukia Theorem says that a uniformly quasiconformal subgroup of QC(S^{2}) ≈ QI(H^{3}) is quasiconformally conjugate into the conformal group Conf(S^{2}) ≈ Isom(H^{3}).

Here is one possible way to formulate an analogue of this theorem for BS(1, n). The analogue of the group QC(S^{2}) ≈ QI(H^{3}) is QI(BS(1, n)) ≈ QI(X_{n}) ≈ Bilip(R)×Bilip(Q_{n}). The analogue of a uniformly quasiconformal subgroup of QC(S^{2}) is a “bounded” subgroup G < QI(X_{n}), which means that for some K ≥ 1, each g ∈ G is represented by a (K, C)-quasi-isometry of X_{n} (where C may depend on g). Equivalently, there is a constant K′ ≥ 1 such that for each g ∈ G, letting ρ_{ℓ}(g) ∈ Bilip(R), ρ^{u}(g) ∈ Bilip(Q_{n}) be the projections, we have

[a(ρ_{ℓ}(g)), b(ρ_{ℓ}(g))] · [a(ρ^{u}(g)), b(ρ^{u}(g))] ⊂ [1/K′, K′]

What is the analogue of Conf(S^{2})? One guess is Isom(X_{n}). However, the Sullivan-Tukia Theorem is false in this case, because if n = m^{2} then BS(1, m) is a bounded subgroup of QI(X_{n}), but there does not exist an isometric action of BS(1, m) on X_{n}.

A more reaonable guess is that if n ≥ 2 is an integer which is not a proper power, then every bounded subgroup of QI(X_{n}) is conjugate to a subgroup of Isom(X_{n}). One possible approach is to prove a version of Theorem 3.2 for Q_{n}, namely that any uniform quasisimilarity group action on Q_{n} is bilipschitz conjugate to a similarity action; for this to be true it is necessary that n not be a proper power. However, we do not know how to prove this.

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Benson Farb: Department of Mathematics, Univeristy of Chicago 5734 University Ave. Chicago, Il 60637 E-mail: farb@math.uchicago.edu

Lee Mosher: Department of Mathematics, Rutgers University, Newark Newark, NJ 07102 E-mail: mosher@andromeda.rutgers.edu