Uniform undistortion from barycentres, and applications to hierarchically hyperbolic groups

Definition 2.1

Let us say that a metric space ( X , d ) has barycentres if for each n there is a map b n : X n → X such that b n is

• idempotent: for every x ∈ X , we have b n ( x n ) = x , where x n denotes the tuple ( x , … , x ) ∈ X n ;

• symmetric: b n is invariant under permutation of co-ordinates;

• Isom X -invariant: for every g ∈ Isom X ,we have g b n ( x 1 , … , x n ) = b n ( g x 1 , … , g x n ) ;

• 1 n -Lipschitz: d ( b n ( x 1 , … , x n ) , b n ( y 1 , … , y n ) ) ≤ 1 n ∑ i = 1 n d ( x i , y i ) .

Definition 2.2

Let X be a metric space. A bicombing σ on X is a choice of path σ x y = σ ( x , y ) from x to y for every x , y ∈ X .

• If every σ ( x , y ) is a geodesic, then σ is a geodesic bicombing.

• If σ ( x , y ) = σ ( y , x ) for all x , y ∈ X , then σ is reversible.

• We say that a geodesic bicombing σ is conical if the following holds for all x , y , x ′ , y ′ ∈ X :

  d ( σ x y ( t ) , σ x ′ y ′ ( t ) ) ≤ ( 1 − t ) d ( x , x ′ ) + t d ( y , y ′ ) .

Definition 2.3

A metric space X is injective if for every metric space B and every subset A ⊆ B , if f : A → X is 1-Lipschitz, then there exists a 1-Lipschitz map f ˆ : B → X with f ˆ ∣ A = f .

Lemma 2.4

Injective metric spaces have barycentres.

Proof

Let ( X , d ) be an injective space. The map x ↦ d ( x , ⋅ ) defines an isometric embedding of X into R X , so we can view X as a subset. We now recall a construction from [39, §1]. First, let P X ⊆ R X be the set of functions f such that f ( x ) + f ( y ) ≥ d ( x , y ) for all x , y ∈ X . Observe that the map ( R X ) n → R X defined by ( f 1 , … , f n ) ↦ 1 n ∑ i = 1 n f i sends tuples of functions in P X to functions in P X .

Let T X ⊆ P X be the set of f such that

for all x ∈ X . Observe that X , regarded above as a subset of R X , is contained in T X . Indeed, for each z ∈ X , we have d ( z , x ) = sup y ∈ X { d ( x , y ) − d ( z , y ) } by the triangle inequality. On the other hand, as noted in [39], injectivity of X implies that X → T X is surjective, and we can identify X with T X .

Each isometry Ψ : X → X extends to a linear isomorphism f ↦ f Ψ − 1 , which is an isometry on each component of R X and which preserves P X and T X . Dress defines a 1-Lipschitz retraction p : P X → T X = X in [39, §1.9], which, by construction, is Isom ( X ) -equivariant (refer also [71, Prop. 3.7.(2)]). (From Definition 2.3, one could construct a 1-Lipschitz retraction R X → X directly, but we use the map p to ensure equivariance.)

Given points x 1 , … , x n in X , consider their affine barycentre 1 n ∑ i = 1 n d ( x i , ⋅ ) ∈ P X . The maps defined by b n : ( x 1 , … , x n ) ↦ p ( 1 n ∑ i = 1 n d ( x i , ⋅ ) ) satisfy the requirements of Definition 2.1, and hence provide barycentres for X .□

Lemma 2.5

If X has barycentres, then every pair of points is joined by an isometric image of a dense subset of an interval. If X is complete, then it has an Isom X -invariant, reversible, conical bicombing; and every space with an Isom -invariant conical bicombing has barycentres.

Proof of Lemma 2.5

Given x , y ∈ X , the point b 2 ( x , y ) has d ( x , b 2 ( x , y ) ) = d ( y , b 2 ( x , y ) ) = 1 2 d ( x , y ) . Iterating, we obtain an isometrically embedded dyadic interval from x to y , as in the first assertion. If X is complete, then we obtain a uniquely defined geodesic from x to y by taking limits of the dyadic interval. By the properties of barycentres, these geodesics form a reversible, Isom X -invariant, conical bicombing. The fact that every space with an Isom -invariant conical bicombing has barycentres is [36, Thm 2.1].□

Remark 2.6

If X is complete, then Lemma 2.5 implies that it has a reversible conical bicombing. According to [36, Prop. 2.4], the barycentres can then be perturbed so that they additionally satisfy b n m ( x 1 m , … , x n m ) = b n ( x 1 , … , x n ) for every n , m , x 1 , … , x n , where z m denotes the tuple ( z , … , z ) ∈ X m . We could therefore have assumed this stronger property to begin with in most situations. Moreover, observe that the barycentres on X naturally extend to its metric completion.

Lemma 2.7

If G acts on a metric space X with barycentres, then ∣ g ∣ = τ X ( g ) for all g ∈ G .

Proof

Fix x ∈ X , and let x n = b n ( x , g x , … , g n − 1 x ) . We compute

Hence, τ X ( g ) ≤ ∣ g ∣ ≤ d ( x n , g x n ) → τ X ( g ) .□

Definition 2.8

Let G be a group acting on a metric space X . The action is said to be proper, if for every ε > 0 there exists N such that

for all x ∈ X . The action is uniformly proper if for every ε > 0 , there exists N such that

for all x ∈ X . The action is acylindrical if for every ε > 0 , there exist R , N such that if d ( x , y ) > R , then

Proposition 2.9

Let G act on a metric space X with barycentres. If the action is:

• acylindrical, then there exists δ > 0 such that τ X ( g ) > δ for every g ∈ G whose action is not elliptic;

• uniformly proper, then τ X ( g ) > δ for every infinite-order g ∈ G ;

• proper and cobounded, then G has finitely many conjugacy classes of finite subgroups.

Proof

Supposing that the action is acylindrical, let R and N be such that if d ( x , y ) > R , then ∣ { g ∈ G : d ( x , g x ) , d ( y , g y ) ≤ 1 } ∣ ≤ N . Suppose that g ∈ G is not elliptic. By Lemma 2.7 there is some x ∈ X such that d ( x , g x ) ≤ τ X ( g ) + 1 2 N . As g is not elliptic, there is some n such that d ( x , g n x ) > R . If τ X ( g ) ≤ 1 2 N , then

for all i ≥ 0 . Since i ∈ { 0 , … , N } is a contradiction, the first statement is proved.

If the action is proper, then no infinite-order element is elliptic, and if the action is uniformly proper, then it is acylindrical.

Finally, suppose the action is proper and cobounded. Let x ∈ X . If F = { 1 , f 2 , … , f n } is a finite subgroup of G , then b n ( x , f 2 ⋅ x , … , f n ⋅ x ) is fixed by F . Using that finite subgroups have fixed points, a standard argument shows that G has finitely many conjugacy classes of finite subgroups; refer [22, I.8.5], for instance.□

Lemma 2.10

M ε ( g ) is closed undertaking barycentres and is set wise stabilised by the action of the centraliser of g.

Proof

If { x 1 , … , x n } ⊆ M ε ( g ) , then

If h commutes with g and x ∈ M ε ( g ) , then d ( h x , g h x ) = d ( h x , h g x ) ≤ τ X ( g ) + ε .□

Definition 2.11

For q ≥ 0 , a q -quasiaxis of an isometry g of a metric space X is a ⟨ g ⟩ -invariant subset A g ⊆ X admitting a q -coarsely surjective ( 1 + q , q ) -quasi-isometry R → A g .

Proposition 2.12

Let G act on a metric space X with barycentres. For every ε > 0 , every g ∈ G with τ X ( g ) > 0 has a ε -quasiaxis.

Hence, if G acts on X acylindrically, then for all ε > 0 , every non-elliptic g ∈ G has an ε -quasiaxis. If the action is uniformly proper, then this holds for all g ∈ G of infinite order.

Proof

As noted in Remark 2.6, the barycentre maps on X naturally extend to its completion, so there is no loss in assuming that X is complete. Let g ∈ G satisfy τ ≔ τ X ( g ) > 0 . Let ε > 0 . We are free to assume that ε < τ . Let x ∈ M ε ( g ) . Let I : [ 0 , d ( x , g x ) ] → X be the unit-speed geodesic from x to g x provided by the proof of Lemma 2.5. Note that a dense subset of I is constructed by repeatedly taking barycentres, so Lemma 2.10 implies that I ⊂ M ε ( g ) . Hence, the subset A g = ⋃ n ∈ Z g n I , which is stabilised by g , is also contained in M ε ( g ) . It remains to show that A g is an ε -quasiline.

Given t ∈ R , write t = n τ + r , where n ∈ Z and r ∈ [ 0 , τ ) , and let f ( t ) = g n I ( r ) , which is well-defined since τ ≤ ∣ I ∣ . This defines a map f : R → A g that is ε -coarsely onto. Let t 1 , t 2 ∈ R , and write t i = n i τ + r i for i ∈ { 1 , 2 } . If n 1 = n 2 , then by definition we have d ( f ( t 1 ) , f ( t 2 ) ) = ∣ t 1 − t 2 ∣ . Otherwise, we may assume that n 1 < n 2 , and we compute

We similarly obtain a lower bound as follows:

Combining these estimates, we see that f is a ( 1 + ε , ε ) -quasi-isometry. The statement about acylindrical and uniformly proper actions now follows using Proposition 2.9.□

Proposition 2.13

Let G act properly coboundedly on a metric space X with barycentres, and let ε > 0 . If H = ⟨ g 1 , … , g n ⟩ ≅ Z n is a free abelian subgroup of G, then there is an H-equivariant ( 1 + ε , ε ) -coarsely Lipschitz map ( R n , ℓ 1 ) → X .

Proof of Proposition 2.13

Let δ ∈ ( 0 , 1 ] be given by Proposition 2.9, and fix ε > 0 .

We first show that ⋂ i = 1 n M ε ( g i ) ≠ ∅ , arguing by induction on n . By Lemma 2.7, M ε ( g 1 ) ≠ ∅ . Fix j ∈ { 2 , … , n } , and assume by induction that there exists x ∈ ⋂ i = 1 j − 1 M ε ( g i ) . By Lemma 2.10, for every m , the point y m = b m ( x , g j x , … , g j m − 1 x ) lies in ⋂ i = 1 j − 1 M ε ( g i ) , using that [ g j , g i ] = 1 for all i . Moreover,

Thus, y m ∈ ⋂ i = 1 j M ε ( g i ) for sufficiently large m . So ⋂ i = 1 n M ε ( g i ) ≠ ∅ . Fix x ∈ ⋂ i = 1 n M ε ( g i ) .

Next let { e 1 , … , e n } be the standard basis of R n . For brevity, let τ i = τ X ( g i ) for 1 ≤ i ≤ n . Let D n ⊆ R n be the set of vectors of the form ∑ i = 1 n r i τ i e i , where r i is a dyadic rational. Define a map f : D n → X as follows.

Set f ( 0 ) = x . For i ≥ 1 , suppose that f has been defined on ( D i − 1 × { 0 } n − i + 1 ) ∩ ∏ s = 1 n [ 0 , τ s ) . Given an element p thereof, set f ( p + 1 2 τ i e i ) = b 2 ( f ( p ) , g i f ( p ) ) . We can define f ( q ) for every q ∈ D i × { 0 } n − i ∩ ∏ s = 1 n [ 0 , τ s ) by repeatedly taking barycentres in this way. Inductively, this defines f on D n ∩ ∏ s = 1 n [ 0 , τ s ) . Specifically, for any p ∈ D n − 1 × { 0 } , the image of the restriction of f to the set of dyadic rationals in { p } × [ 0 , τ n ) is an isometrically embedded (dense subset of an) interval of length d ( f ( p ) , g n f ( p ) ) from f ( p ) to g n f ( p ) .

Given p ∈ D n , we can write

where each p i ∈ [ 0 , 1 ) is a dyadic rational and each a i ∈ Z . We define [ p ] = ( p i τ i ) i = 1 n . Then, we let f ( p ) = g 1 a 1 … g n a n f ( [ p ] ) . Letting H act on D n by declaring g i to be a unit translation by τ i e i , observe that f is well-defined and H -equivariant by construction.

We now check that if f is coarsely Lipschitz. For convenience, let C = D n ∩ ∏ i = 1 n [ 0 , τ i ) , and let C ¯ be its closure in D n . Note that taking this closure in D n has the effect of adding the endpoint of τ i if it is a dyadic rational, and leaving the interval half-open otherwise. Note that C contains exactly one point in each H -orbit.

We first show that f is ( 1 + ε ⁄ δ ) -Lipschitz on C ¯ . If p , q ∈ C ¯ , we can write p = ∑ i = 1 n p i τ i e i and q = ∑ i = 1 n q i τ i e i . By construction,

Here we used that τ i ≥ δ > 0 for all i .

We next show that f is ( 1 + ε ⁄ δ ) -Lipschitz on D n . Let p , q ∈ D n be given. Let γ be an ℓ 1 -metric geodesic in R n from p to q such that γ ∩ D n is dense in γ . Then, γ = γ 1 … γ m , where each γ j is an ℓ 1 -metric geodesic whose intersection with D n lies in some H -translate of C ¯ . Since f is ( 1 + ε ⁄ δ ) -Lipschitz on C ¯ , it follows from equivariance that f is ( 1 + ε ⁄ δ ) -Lipschitz on each H -translate of C ¯ . Hence, letting p = p 0 , … , p m − 1 be the initial points of γ 1 , … , γ m and p m = q the terminal point of γ m , we have d ( f ( p i ) , f ( p i + 1 ) ) ≤ ( 1 + ε ⁄ δ ) ‖ p i − p i + 1 ‖ 1 for each i . We conclude that

as required.

Finally, we note that since D n is dense in R n , we can extend f : D n → X equivariantly to a map f : R n → X that is ( 1 + ε ⁄ δ , ε ⁄ δ ) -coarsely Lipschitz. Since this argument holds for all ε , it holds, in particular, for ε δ , which concludes the proof.□

Lemma 2.14

Let H = ⟨ g 1 , … , g n ⟩ ≅ Z n be a free abelian group acting on a metric space X. For every T , δ > 0 , there exists δ ′ = δ ′ ( n , δ , T ) > 0 such that the following holds. If every h ∈ H has τ X ( h ) > δ and max { τ X ( g i ) } ≤ T , then in fact, every h ∈ H has τ X ( h ) ≥ δ ′ d H ( 1 , h ) .

Proof

In the terminology of [26, Thm 4.2], τ X defines a Z -norm on H . Consider the group embedding H → R n given by g i ↦ e i , where e i is the standard basis vectors. The Z -norm τ X extends to a norm N on R n . By linearity, N ( x ) ≥ r ‖ x ‖ 1 for all x ∈ R n , where r = inf { N ( z ) : ‖ z ‖ 1 = 1 } . Let us find a lower bound for r .

Let x ∈ R n have ‖ x ‖ 1 = 1 . By an application of the pigeonhole principle, there is a constant M = M ( n , δ , T ) , a natural number q ≤ M , and integers p 1 , … , p n such that ∣ q x i − p i ∣ ≤ 1 2 n δ T (e.g. [63, Thm 201]). Following [89], let p = ( p 1 , … , p n ) , so that ‖ q x − p ‖ 1 ≤ 1 2 δ T . Because N is a norm, every point z with ‖ z ‖ 1 = 1 has N ( z ) ≤ max { N ( e i ) } ≤ T , which shows that N ( q x − p ) ≤ δ 2 . As ‖ q x ‖ 1 ≥ ‖ x ‖ 1 = 1 , the vector p must be nonzero; hence, N ( p ) > δ , and therefore, N ( q x ) > δ 2 . We have shown that r > δ 2 M . If h ∈ H , then ‖ h ‖ 1 = d H ( 1 , h ) , so we can compute