Topological full groups of line-like minimal group actions are amenable
aa r X i v : . [ m a t h . G R ] J a n Topological full groups of line-like minimal groupactions are amenable
N´ora Gabriella Sz˝oke ∗ Abstract
We consider a finitely generated group acting minimally on a compactspace by homeomorphsims, and assume that the Schreier graph of at leastone orbit is quasi-isometric to a line. We show that the topological full groupof such an action is amenable.
Consider a group G and a compact Hausdorff topological space Σ. A group action G y Σ by homeomorphisms is called minimal if Σ has no proper G -invariant closedsubset. The topological full group [[ G y Σ]] is the group of all homeomorphisms ofΣ that are piecewise given by the action of elements of G , where each piece is openin Σ.The notion of topological full groups was first introduced for Z -actions by Gior-dano, Putnam and Skau [2]. Among others, Matui and Nekrashevych investigatedthese groups ([11], [12], [13], [15]). Their results show that the derived subgroupof the topological full group is often simple, and in many cases it is also finitelygenerated.In their groundbreaking paper [7], Juschenko and Monod developed a strategy forproving the amenability of topological full groups. They show that the topologicalfull group of a minimal Cantor Z -action is amenable. Combined with the results ofMatui, their paper provides the first examples of finitely generated infinite simpleamenable groups. A natural question arises: how far can we extend their technique?Several directions were investigated in [5], [8], [6], and by the author of the presentpaper in [16].The goal of this paper is to further stretch the Juschenko-Monod result in acertain direction. Namely, we consider a minimal action of a finitely generatedgroup such that there exists an orbit that is quasi-isometric to a line, and show thatthe topological full group of such an action is amenable. This is a generalization ofTheorem A in [16], where the group was virtually cyclic. Theorem 1.1.
Let G be a finitely generated group acting minimally on a compactHausdorff topological space Σ by homeomorphisms. Assume that there exists a G -orbit X ⊆ Σ , such that the Schreier graph of the action of G on X is quasi-isometricto Z . Then the topological full group [[ G y Σ]] is amenable. ∗ Institut Fourier, Universit´e Grenoble Alpes, France. Email: [email protected] supported by the Swiss National Science Foundation, Early Postdoc.Mobility fellowshipno. P2ELP2 184531.
1n order to illustrate the interest of our result, let us mention how to recovera result of Matte Bon about the Grigorchuk group. Let G be the first Grigorchukgroup (defined in [3]), which is usually defined as a transformation group of thebinary rooted tree. Its action on the boundary of the tree - a Cantor set - is knownto be minimal and its Schreier graphs are quasi-isometric to lines, as seen in [4].Theorem 1.1 can be applied to deduce the following. Corollary 1.2 (Matte Bon, [10]) . The topological full group of the Grigorchuk groupacting on the boundary of the rooted binary tree is amenable.
This result was first proved by Matte Bon, who showed that the Grigorchukgroup can be embedded in the topological full group of a minimal Cantor Z -action([10]).There are more groups to which our Theorem 1.1 can be applied, for instancethe groups defined by Nekrashevych in [14]. Let a be an involution on a Cantorspace Σ. We say that a finite group A of homeomorphisms of Σ is a fragmentation of a if for all h ∈ A and all x ∈ Σ, we have h ( x ) = x or h ( x ) = a ( x ) and forevery x ∈ Σ there exists h ∈ A such that h ( x ) = a ( x ). In [14] it is shown that fora fragmentation A, B of a minimal action of the dihedral group D ∞ = h a, b i , theaction of the topological full group G = h A, B i is minimal. It is not difficult to seethat the associated Schreier graphs are quasi-isometric to lines. Therefore, we getthe following result of Nekrashevych as a corollary of Theorem 1.1. Corollary 1.3 (Nekrashevych, [14]) . For any fragmentation
A, B of a minimalaction of the dihedral group D ∞ = h a, b i , the topological full group of G = h A, B i isamenable. Acknowledgements.
It was Nicol´as Matte Bon who asked me whether a similarresult in my thesis could be true in this more general setting, I would like to thankhim for this question. Furthermore, I am very grateful to Fran¸cois Dahmani for ournumerous discussions and his comments on a previous version of the paper. If G = (V( G ) , E( G )) is a connected graph, then we can think of G as a metricspace. The distance d : V( G ) × V( G ) → N is defined to be the length of the shortestpath between two vertices.Let G , G be two connected graphs with distance functions d , d respectively.Recall that the map f : V( G ) → V( G ) is a quasi-isometry if there exist constants α ≥ β ≥ γ ≥ u, v ∈ V( G ) we have α − d ( u, v ) − β ≤ d ( f ( u ) , f ( v )) ≤ α d ( u, v ) + β.
2. For every w ∈ V( G ) there exists u ∈ V( G ) such that d ( f ( u ) , w ) ≤ γ .Two graphs are quasi-isometric if there exists a quasi-isometry between them.2 .2 Group actions and graphs As a convention, throughout the paper we always consider groups acting fromthe left.Let G be a group acting on a set X . The piecewise group PW( G y X ) ofthe action is defined as follows. A bijection ϕ : X → X is a piecewise G map , i.e.,an element of the piecewise group, iff there exists a finite subset S ⊂ G such that ϕ ( x ) ∈ S · x for every x ∈ X . In other words, we cut the space X into finitely manypieces, and act on each of them with a group element. It is clear that the piecewise G maps form a group.If the group G acts on a compact space by homeomorphisms, then the topologicalfull group of this action is always a subgroup of its piecewise group. Indeed, by thecompactness of the space a partition into open subsets is necessarily finite.Let G be a finitely generated group with a symmetric generating set S , andassume that G acts on a set X . Recall that the Schreier graph of this actionSch(
G, X, S ) is defined to be the graph with vertex set X and edge set { ( x, sx ) : x ∈ X, s ∈ S } . Sometimes it is also called the graph of the action G y X .Note that the Cayley graph of G is the Schreier graph of its action on itself by(left) multiplication. Definition 2.1. If G is a finitely generated group with a fixed symmetric generatingset S , then the length of a group element g ∈ G is defined aslen( g ) = min { n ∈ N : g = s s . . . s n with s , s , . . . , s n ∈ S } . In other words, the length of an element is its distance from the identity elementin the Cayley graph.
Definition 2.2.
For any graph G = ( V, E ) with distance function d and a number n ∈ N we define the n -ball around a point p ∈ V to be B n ( p ) = { q ∈ V : d( p, q ) ≤ n } . For a set W ∈ V , the n -neighborhood of W isΓ n ( W ) = { q ∈ V : d( q, W ) ≤ n } . If a group G acts on the graph G , then for a set of elements D ⊆ G the D -neighborhood of a point p ∈ V is D · p = { d · p : d ∈ D } , and the D -neighborhood ofa set W ⊆ V is D · W = [ q ∈ W D · q. Keep in mind that when G is a Schreier graph of a G -action, and G = h S i ,then the n -ball around a point is exactly the S n -neighborhood of that point and the n -neighborhood of a set is equal to its S n -neighborhood. A group action G y X is amenable if there exists a G -invariant mean on X . Inthe proof of our result we will use a stronger property, the extensive amenability ofan action. 3 efinition 2.3. For a set X , let us denote the set of all finite subsets of X by P f ( X ). Note that this set becomes an abelian group with the symmetric difference.If a group G acts on X , it gives rise to a G -action on P f ( X ).We say that the action G y X is extensively amenable if there exists a G -invariant mean on P f ( X ) that gives full weight to the collection of sets containingany given finite subset of X .Extensively amenable actions were first used (without a name) in [7]. Thename was given in [6] and this concept turned out to be very useful for provingthe amenability of topological full groups, see [7], [8], [6], or [16]. For a detailedintroduction to extensive amenability, see Chapter 11 of [1]. The following twostatements about extensive amenability will be among the core ingredients in ourproof. Proposition 2.4 (Proposition 3.6 in [16]) . Let G be a group acting on a set X .Assume that for all finitely generated subgroups H ≤ G and all H -orbits Y ⊆ X theSchreier graph of the action H y Y is recurrent. Then the action of the piecewisegroup PW( G y X ) on X is extensively amenable. Proposition 2.5 (Remark 1.5 in [6]) . Let G y X be an extensively amenableaction. Assume that there exists an embedding G ֒ → P f ( X ) , g ( c g , g ) such thatthe subgroup { g ∈ G : c g = ∅ } ≤ G is amenable. Then G itself is also amenable. In Proposition 2.5, such a map c : G → P f ( X ), g c g is called a cocycle withamenable kernel . Thus, we can rephrase the statement of the proposition as follows:If G y X is an extensively amenable action, and there exists a cocycle on G withamenable kernel, then G is amenable. Let us denote the orbit of p by X , i.e., X = G · p . In this section we consider a finitely generated group G = h S i acting minimallyon a compact space Σ satisfying the assumption in Theorem 1.1. Let X be an orbitsuch that the Schreier graph Sch( G, S, X ) is quasi-isometric to Z . X The set X is dense in Σ, since the action G y Σ is minimal. Consider therestricted action of G on X . We can define the embedding ε X : [[ G y Σ]] ֒ −→ PW( G y X ); ϕ ϕ (cid:12)(cid:12) X . Since X is dense in Σ, the ϕ -action on X determines the ϕ -action on Σ, so the map ε X is injective. Definition 3.1.
Let d denote the distance function on the graph X . Let f : X → Z be the quasi-isometry between X and Z . By definition, there exist constants α ≥ β ≥ γ ≥ x, y ∈ X we have α − · d( x, y ) − β ≤ | f ( x ) − f ( y ) | ≤ α · d( x, y ) + β,
2. for every n ∈ Z there is x ∈ X such that | f ( x ) − n | ≤ γ . Lemma 3.2.
For every n ∈ Z , the set f − ( n ) ⊆ X is finite.Proof. If f − ( n ) = ∅ , then it is finite. Now assume that it is non-empty. Let x ∈ f − ( n ), then by Definition 3.1, for any y ∈ f − ( n ) we have α − d( x, y ) − β ≤ | f ( x ) − f ( y ) | = 0d( x, y ) ≤ αβ. Hence, f − ( n ) is contained in the ball of radius αβ around x . This ball is a finiteset since the graph X is locally finite, so f − ( n ) is also finite.The following two propositions are well-known for any graph that is quasi-isometric to Z , but we include their proofs for completeness. Proposition 3.3.
There exists a bi-infinite geodesic in X . Lemma 3.4.
Let G be a locally finite graph, i.e., the degree of every vertex is finite.The following are equivalent for a vertex v ∈ G .1. For every n ∈ N , there exists a geodesic of length n with midpoint v .2. There exists a bi-infinite geodesic through the vertex v .Proof. The implication 2 ⇒ v that satisfies the first statement.Let us construct the rooted tree T as follows. The vertices of T are the finitegeodesics in G of even length with midpoint v . Two such geodesics are connected in T if their length difference is exactly 2 and the shorter one is a subset of the longerone. The root is the “geodesic” of length zero consisting only of the point v , and the n -th level of the tree consists of the geodesics of length 2 n . By the local finitenessof G , the rooted tree T is also locally finite.By K˝onig’s lemma, there exists an infinite ray in T from the root, say { v } = ℓ , ℓ , ℓ , ℓ , . . . , where the length of ℓ i is 2 i and ℓ i ⊆ ℓ i +1 for every i ∈ N . Then theunion S i ∈ N ℓ i = ℓ ⊆ G is a bi-infinite geodesic in G . This proves the implication1 ⇒ Proof of Proposition 3.3.
Let f : X → Z be the quasi-isometry with constants α , β , γ , and let us define B = f − ([0 , α + β ]). By Lemma 3.2, the set f − ( z ) ⊆ X is finitefor every z ∈ Z . Consequently, B is finite.Let B i = { x ∈ B | x is the midpoint of a length 2 i geodesic in X } for all i ∈ N . We have B i +1 ⊆ B i for every i .We show that B i = ∅ for every i ∈ N . Consider a fixed i ∈ N and take k ∈ N such that k ≥ α · i + β . Note that this choice ensures that | f ( x ) − f ( y ) | > k implies d( x, y ) > i for some x, y ∈ X by Definition 3.1. Take x , x ∈ X such that5 ( x ) < − k and f ( x ) > α + β + k . (It is possible to find such points by the secondstatement of Definition 3.1.) Then d( x , B ) > i and d( x , B ) > i both hold.Let [ x , x ] be a shortest path between the two points, i.e., a geodesic. Notethat if u, v ∈ X are neighbors, then | f ( u ) − f ( v ) | ≤ α + β . Indeed, this holds byDefinition 3.1: | f ( u ) − f ( v ) | ≤ α d( u, v ) + β = α + β . Hence, by walking along thepath [ x , x ], the f -image changes by at most α + β at every step. On the otherhand, we know that f ( x ) < f ( x ) > α + β , so there must be a point on thispath y ∈ [ x , x ], such that 0 ≤ f ( y ) ≤ α + β . This implies that y ∈ B , so we haved( x , y ) > i and d( y, x ) > i . Therefore, there exists a geodesic of length 2 i withmidpoint y , and hence B i = ∅ .We proved that B = B ⊇ B ⊇ B ⊇ . . . is a decreasing sequence of non-empty finite sets. Therefore, their intersection is alsonon-empty. Take a point x ∈ T i ∈ N B i . Then for every i ∈ N , there exists a geodesicof length 2 i in X with midpoint x . Hence, by Lemma 3.4, there is a bi-infinitegeodesic through the point x . Proposition 3.5.
There exists a constant m ∈ N such that X is contained in the m -neighborhood of any bi-infinite geodesic in X .Proof. Let f : X → Z be the quasi-isometry with constants α , β , γ as in Definition3.1. Let m = α + 2 αβ . Let ℓ be a bi-infinite geodesic in X , we would like to showthat X is contained in the m -neighborhood of ℓ .First, note that if u, v ∈ X such that d( u, v ) > m = α + 2 αβ , then we have α − d( u, v ) − β ≤ | f ( u ) − f ( v ) | α − m − β < | f ( u ) − f ( v ) | α + β < | f ( u ) − f ( v ) | (1)Take an arbitrary x ∈ X , and suppose for contradiction that d( x, ℓ ) > m = α + 2 αβ . This implies that for every y ∈ ℓ , we have α + β < | f ( x ) − f ( y ) | by (1).Note that if u, v ∈ X are neighbors, then | f ( u ) − f ( v ) | ≤ α d( u, v ) + β = α + β .Therefore, as we walk along the geodesic ℓ , the f -image cannot jump over the value f ( x ), since the distance of f ( ℓ ) from f ( x ) is more than α + β . Hence, f ( ℓ ) must becontained in a half-line, either ( −∞ , f ( x ) − α − β ) or ( f ( x ) + α + β, + ∞ ).We will show that f ( ℓ ) cannot be contained in a half-line. Without loss ofgenerality, suppose that f ( ℓ ) ⊆ ( N, + ∞ ), such that N = min f ( ℓ ) ≥ f ( x ) + α + β ,and take x ∈ ℓ so that f ( x ) = N . Let I = B m ( x ) ∩ ℓ be a geodesic segment oflength 2 m on ℓ around x .Take y , y ∈ ℓ \ I be in different components of ℓ \ I such that f ( y ) ≤ f ( y ).Consider [ x , y ], which denotes a shortest path between the two points in X , in thiscase we may take the path that is contained in ℓ . We know that | f ( u ) − f ( v ) | ≤ α + β if u and v are neighbors, hence if we “walk” along the path [ x , y ], the f -imagechanges by at most α + β in each step. Since N = f ( x ) ≤ f ( y ) ≤ f ( y ), wecan find y ∈ [ x , y ], such that | f ( y ) − f ( y ) | ≤ α + β . On the other hand,d( y , y ) ≥ d( y , x ) > m = α + 2 αβ , which is a contradiction by (1).Therefore, for every x ∈ X , we have d( x, ℓ ) ≤ m , so X is contained in the m -neighborhood of ℓ . 6 .2 Definition of the cocycle Definition 3.6.
Let f : X → Z be the quasi-isometry from Definition 3.1. Let usdefine Y = f − ( N ) ⊆ X. For a subgraph H of X let us denote by ∂H the vertices on the boundary of H ,i.e., let ∂H = { x ∈ H : there exists y ∈ X \ H such that ( x, y ) ∈ E ( X ) } ⊆ H ⊆ X. Lemma 3.7.
The set Y is infinite and ∂Y is finite.Proof. By the second requirement in Definition 3.1, we have that f − ( I ) = ∅ forevery interval I of length at least 2 γ . Since N contains infinitely many pairwisedisjoint intervals of length 2 γ , the preimage f − ( N ) = Y is infinite.For x ∈ H and y ∈ X \ H we have ( x, y ) ∈ E ( X ) if and only if d( x, y ) = 1. ByDefinition 3.1, we have α − − β = α − d( x, y ) − β ≤ | f ( x ) − f ( y ) | ≤ α d( x, y ) + β = α + β. Therefore, since f ( x ) ∈ N and f ( y ) ∈ Z \ N , we must have 0 ≤ f ( x ) ≤ α + β −
1, so ∂Y ⊆ f − ([0 , α + β − ∂Y isalso finite. Lemma 3.8.
For every group element g ∈ G , the set gY \ Y is finite.Proof. Notice that for all x ∈ X we have d( x, gx ) ≤ len( g ). Therefore, the set gY \ Y is contained in the len( g )-neighborhood of ∂Y . Since ∂Y is finite by Lemma3.7, the len( g )-neighborhood is also a finite set by the local finiteness of X . Hence, gY \ Y is finite. Proposition 3.9.
For every piecewise map ϕ ∈ PW( G y X ) , the set Y △ ϕ ( Y ) isfinite.Proof. There exists a finite set T ⊆ G such that for every x ∈ X we have ϕ ( x ) ∈ T · x .Hence, we have the inclusion ϕ ( Y ) \ Y ⊆ [ t ∈ T tY ! \ Y = [ t ∈ T ( tY \ Y ) . By Lemma 3.8, tY \ Y is finite for all t ∈ T , so ϕ ( Y ) \ Y is also finite. The sameargument works for ϕ − ( Y ) \ Y , and hence ϕ ( ϕ − ( Y ) \ Y ) = Y \ ϕ ( Y ) is finite aswell. This implies that the set Y △ ϕ ( Y ) = ( Y \ ϕ ( Y )) ∪ ( ϕ ( Y ) \ Y )is also finite, finishing the proof. Definition 3.10.
For ϕ ∈ PW( G y X ) let us define c ϕ = Y △ ϕ ( Y ) ∈ P f ( X ) . emark . We defined the map c : PW( G y X ) → P f ( X ). This gives riseto the cocycle c : [[ G y Σ]] → P f ( X ). We would like to show that its kernel { ϕ ∈ [[ G y Σ]] : c ϕ = ∅ } is amenable in order to use this cocycle in Proposition2.5. Note that ker c = { ϕ ∈ [[ G y Σ]] : c ϕ = ∅ } = { ϕ ∈ [[ G y Σ]] : Y △ ϕ ( Y ) = ∅ } = { ϕ ∈ [[ G y Σ]] : ϕ ( Y ) = Y } = [[ G y Σ]] Y . (2)Hence, the kernel of c is exactly the stabilizer of the set Y in the topological fullgroup [[ G y Σ]]. In the next sections we prove that this stabilizer is amenable.
Definition 3.12.
Let G be a group acting on the space Σ. Let D ⊂ G be a finiteset containing the identity element. For an element ϕ ∈ PW( G y Σ) and for twopoints q , q ∈ Σ, we say that the ϕ -action is the same on the D -neighborhood of q and q , if the D -neighborhoods of q and q are isomorphic, and for every d ∈ D , ϕ acts by the same element of G on d · q and on d · q , i.e., there exists g ∈ G suchthat ϕ ( d · q ) = gd · q and ϕ ( d · q ) = gd · q . Lemma 3.13.
Let G = h S i be a group acting minimally on the compact space Σ with a finite symmetric generating set S , and take an arbitrary point q ∈ X .For every finite subset F ⊂ [[ G y Σ]] and every n ∈ N , there exists r = r ( q, F, n ) ∈ N so that for every y ∈ X there exists z ∈ B r ( x ) such that for all ϕ ∈ F , the ϕ -action is the same on the S n -neighborhood of q and z .Proof. Let us fix the elements ϕ , . . . , ϕ k ∈ [[ G y σ ]] and a number n ∈ N .Choose a finite partition P of Σ such that every ϕ i is acting with one elementof G when restricted to any element of P . Then there exists an open neighborhood V of q such that the sets g · V for g ∈ S n are pairwise disjoint, and every g · V iscontained in some element of P . Since V is open and non-empty, the union [ g ∈ G g · V = [ j ≥ [ g ∈ S j g · V is non-empty, open and G -invariant, so by minimality we have [ j ≥ [ g ∈ S j g · V = Σ . Due to the compactness of Σ, already a finite union must cover it, so there exists j ∈ N such that [ g ∈ S j g · V = Σ . Let r = j . Now let y ∈ X = G · q be an arbitrary point. Then y = h · q for some h ∈ G . We have Σ = h − · Σ = [ g ∈ S r h − g · V,
8o there exists ˆ g ∈ S r such that q ∈ h − ˆ g · V . This means that ˆ g − h · q ∈ V . Let z = ˆ g − h · q , and note that z = ˆ g − h · q ∈ B r ( h · q ) = B r ( y ). On the other hand, q and z are both in V , so for every g ∈ S n , the points g · q and g · z are in the sameelement of the partition P , so every ϕ i acts with the same element of G on them.Therefore, for all i = 1 , . . . , k , the ϕ i -action is the same on the S n -neighborhood of q and z .This proves the statement of the lemma for r . Lemma 3.14.
For every piecewise map ϕ ∈ PW( G y X ) , there exists a number d ϕ ∈ N , such that for every x ∈ X , d( x, ϕ ( x )) ≤ d ϕ .Proof. There exists a finite set T ⊆ G such that for every x ∈ X , we have ϕ ( x ) ∈ T · x . The statement of the lemma holds for d ϕ = max { len( t ) : t ∈ T } . Definition 3.15.
Let m ∈ N be the constant from Proposition 3.5. Let us fix abi-infinite geodesic ℓ in X (it exists by Lemma 3.4), and a point p ∈ ℓ .Let us denote the two ends of ℓ by + ∞ and −∞ . For a set A ⊆ X we willsay that + ∞ ∈ A , if there exists a point x ∈ ℓ such that [ x, + ∞ ] ⊆ A , where[ x, + ∞ ] ⊆ ℓ denotes the half-line from x towards + ∞ . Similarly, −∞ ∈ A if thereexists x ∈ ℓ such that [ x, −∞ ] ⊆ A .Let R ∈ N be such that the R -ball around the point p contains both ∂Y and ∂Y c (such a radius exists since ∂Y is finite by Lemma 3.7, and hence ∂Y c is alsofinite).For a piecewise map ϕ ∈ PW( G y X ) let us define the number N ϕ = 6 m + R + 2 d ϕ . Lemma 3.16.
If a set A ⊆ X and its complement A c are both infinite, but itsboundary ∂A is finite, then it contains exactly one end of ℓ .Proof. It is enough to prove that if A is infinite and ∂A is finite, then it contains atleast one end of ℓ . Indeed, we can apply this statement to both A and A c – since ∂A c is also finite – to prove the statement of the lemma.Suppose for contradiction that A ⊆ X is an infinite set with finite boundarysuch that + ∞ , −∞ / ∈ A . Since its boundary is finite, there exists a ball B r ( x ) withfinite radius such that ∂A ⊆ B r ( x ). By the definition of the boundary, a connectedcomponent of the set ℓ \ B r ( x ) must entirely belong either to A or to A c . Since B r ( x )is finite, there exists a connected component of ℓ \ B r ( x ) containing + ∞ , and thereis one (possibly the same) containing −∞ . Since we assumed that + ∞ , −∞ / ∈ A ,we have + ∞ , −∞ ∈ A c .Now consider an arbitrary point y ∈ X \ B r + m ( x ), where m is the constant fromProposition 3.5. By Proposition 3.5, there exists a point ˆ y ∈ ℓ such that d( y, ˆ y ) ≤ m .Since ˆ y is connected to either + ∞ or −∞ outside of B r ( x ) (and ∂A ⊆ B r ( x )), wemust have ˆ y ∈ A c . We have y ∈ X \ B r ( x ), but we can say even more: there isa path of length at most m connecting y and ˆ y that lies outside of the ball B r ( x ).Since y is connected to ˆ y outside of B r ( x ), it must also belong to A c .Therefore, X \ B r + m ( x ) ⊆ A c , and hence A ⊆ B r + m ( x ). This contradicts theassumption that A is infinite, so we must have + ∞ ∈ A or −∞ ∈ A . Corollary 3.17.
The set Y contains exactly one end of ℓ , we can assume that + ∞ ∈ Y , but −∞ / ∈ Y . emma 3.18. Let us fix a finite subset F ⊆ [[ G y Σ]] Y of the stabilizer of Y anda number n > max { N ϕ : ϕ ∈ F } . Assume that there is a point z ∈ X such thatthe ϕ -action is the same on the S n -neighborhood of p and of z for every ϕ ∈ F .Then there exists a set Y z ⊆ X , such that ∂Y z ⊆ B R ( z ) , + ∞ ∈ Y z , −∞ ∈ Y cz and F ⊆ [[ G y Σ]] Y z .Proof. Since the ϕ -action is the same on the S n -neighborhood of p and z for every ϕ ∈ F , there exists a bijection h : S n p −→ S n z, such that for every x ∈ S n p = B n ( p ), ϕ acts by the same group element on thepoints x and h ( x ). Let us define B + = h ( S n p ∩ Y ) and B − = h ( S n p ∩ Y c ), and let A + = { x ∈ X : there exists a path from x to B + that does not intersect B − } ,A − = { x ∈ X : there exists a path from x to B − that does not intersect B + } . We will show that setting Y z = A + or Y z = A − satisfies the statement of the lemma. Claim 3.19.
We have A − = ( A + ) c .Proof. Since X is connected, every x ∈ X is connected to some point of B + ∪ B − = S n z = B n ( z ), and hence A + ∪ A − = X . Therefore, we have to prove A + ∩ A − = ∅ .Suppose for contradiction that there exists a point that can be connected toboth B + and B − without intersecting the other. This means that we can findpoints z + ∈ B + and z − ∈ B − that are connected by a path outside of B n ( z ), i.e.,there exists a path z + = x , x , x , . . . , x k − , x k = z − , such that x i ∈ X \ B n ( z ) for i = 1 , . . . , k − x ∈ X , we will denote its ‘projection’ to ℓ by ˆ x , i.e., the closest pointto x on ℓ . If there are several such points, let us choose the closest one to the end −∞ . By Proposition 3.5, for every x ∈ X , we have d( x, ℓ ) = d( x, ˆ x ) ≤ m .We know that d( z, x i ) ≥ n > N ϕ for i = 0 , , . . . , k . By the triangle inequality,we have 4 m + R + 2 d ϕ = N ϕ − m < n − m ≤ d(ˆ z, ˆ x i ) ≤ n + 2 m. (3)The two projections ˆ z + and ˆ z − are either separated by ˆ z on ℓ or they are on thesame side of it. In both cases, we get a contradiction:1. Suppose that ˆ z + and ˆ z − are separated by ˆ z on ℓ . Consider the points x i ofthe path connecting z + and z − , and their projections ˆ x i . Since ˆ z + = ˆ x andˆ z − = ˆ x k are separated by ˆ z , there exists i such that ˆ x i and ˆ x i +1 are alsoseparated by ˆ z on ℓ . For this i , we must have2(4 m + R + 2 d ϕ ) ≤ d(ˆ x i , ˆ x i +1 )by (3). On the other hand, d(ˆ x i , x i ) ≤ m , and d( x i +1 , ˆ x i +1 ) ≤ m , and henced(ˆ x i , ˆ x i +1 ) ≤ m + 1 , this gives a contradiction. 10. Suppose that ˆ z + and ˆ z − are on the same side of ˆ z on ℓ . Our goal is to find apath of length at most 12 m connecting z + and z − that lies in the ball B n ( z ).Since d( z + , z ) = n , we can choose y + such that d( z + , y + ) = 3 m and d( y + , z ) = n − m . Let [ z + , y + ] denote a shortest path between these two points in thegraph X . Clearly this path lies in B n ( z ). We define y − similarly for z − .Consider the projection ˆ y + . First, note that d( y + , ˆ y + ) ≤ m and d( y + , z ) = n − m , and hence [ y + , ˆ y + ] ⊆ B n ( z ). By the triangle inequality, we haved(ˆ z + , ˆ y + ) ≤ m , so ˆ y + cannot be separated from ˆ z + by ˆ z on ℓ . Similarly, thepoint ˆ y − is also on the same side of ˆ z and [ y − , ˆ y − ] ⊆ B n ( z ).Again by the triangle inequality, we have that n − m ≤ d(ˆ y + , ˆ z ) ≤ n − m,n − m ≤ d(ˆ y − , ˆ z ) ≤ n − m. Since ˆ y + and ˆ y − are not separated by ˆ z , we must have d(ˆ y + , ˆ y − ) ≤ m , anda shortest path connecting them lies on the geodesic ℓ , so it is contained in B n ( z ).Now look at the path P = [ z + , y + ] ∪ [ y + , ˆ y + ] ∪ [ˆ y + , ˆ y − ] ∪ [ˆ y − , y − ] ∪ [ y − , z − ] . We have seen that all sections of this path are contained in B n ( z ). Its lengthis at most 3 m + m + 4 m + m + 3 m = 12 m .Therefore, there exists a path P of length at most 12 m connecting z + with z − that lies in B n ( z ). Since d( z + , z ) = n > N ϕ = 6 m + R + 2 d ϕ , and alsod( z − , z ) > m + R + 2 d ϕ , we have that d( P, z ) > R + 2 d ϕ . Consequently,taking the h -preimage of the path P , we have that d( h − ( P ) , p ) > R + 2 d ϕ .Since ∂Y ⊆ B R ( p ) (by Definition 3.15), the path h − ( P ) cannot intersect theboundary of Y , so it must lie entirely in Y or in Y c . This contradicts theassumption that z + ∈ B + = h ( Y ∩ B n ( p )) and z − ∈ B − = h ( Y c ∩ B n ( p )).This concludes the proof of the fact that A + ∩ A − = ∅ . Claim 3.20.
We have ∂A + = h ( ∂Y ) (and similarly ∂A − = h ( ∂Y c ) ).Proof. Since ∂Y ⊆ B R ( p ) and R < n , we have h ( ∂Y ) ⊆ ∂A + . For the otherdirection, consider a point x ∈ ∂A + . Then x has a neighbor y ∈ A − . There are fourpossibilities.1. If x, y / ∈ B n ( z ), then there is a path from x (going through y ) to B − withouttouching B + , so x ∈ A − , this contradicts the fact that A + ∩ A − = ∅ .2. If x / ∈ B n ( z ), y ∈ B n ( z ), then x, y is a path from x to B − that does notintersect B + , and hence x ∈ A − . This is again a contradiction.3. If x ∈ B n ( z ), y / ∈ B n ( z ), then y, x is a path from y to B + without goingthrough B − , so y ∈ A + , which is also a contradiction.4. The only remaining possibility is x, y ∈ B n ( z ). In this case we have x ∈ B + , y ∈ B − , so h − ( x ) ∈ Y , h − ( y ) ∈ Y c , and hence x ∈ h ( ∂Y ).11e have proved the equality ∂A + = h ( ∂Y ). Similarly, one can prove that ∂A − = h ( ∂Y c ). Claim 3.21.
The sets A + and A − are both invariant under the action of F .Proof. It is enough to prove the F -invariance of A + , the other statement followsfrom this.Take a point x ∈ A + and let us fix ϕ ∈ F . We distinguish three cases.1. If x / ∈ B n ( z ), then d( x, A − ) > d ϕ , since ∂A + ⊆ B R ( z ) and n > R + d ϕ .We know that the distance of x and ϕ ( x ) is at most d ϕ , so we must have ϕ ( x ) ∈ A + . Similarly, we have ϕ − ( x ) ∈ A + .2. If x ∈ B n ( z ), but ϕ ( x ) / ∈ B n ( z ), then we must have d( x, X \ B n ( z )) ≤ d ϕ .Hence, d( x, B R ( z )) > d ϕ (since n > R + 2 d ϕ ), so we get d( x, A − ) > d ϕ again.Therefore, we have ϕ ( x ) ∈ A + and ϕ − ( x ) ∈ A + .3. If x, ϕ ( x ) ∈ B n ( z ), then we have ϕ ( h − ( x )) = h − ( ϕ ( x )) since the ϕ -action isthe same on the S n -neighborhood of p and of z . We know that h − ( x ) ∈ Y ,so ϕ ( h − ( x )) ∈ Y by the ϕ -invariance of Y . Hence, we have ϕ ( x ) ∈ h ( Y ∩ B n ( p )) ⊆ A + . Similarly, ϕ − ( x ) ∈ A + .Therefore, we have ϕ ∈ [[ G y Σ]] A + and hence also ϕ ∈ [[ G y Σ]] A − . Claim 3.22. A + and A − are both infinite.Proof. Suppose for contradiction that A + is finite. This implies that all infinitecomponents of ℓ \ B R ( z ) belong to A − (the number of such components is one ortwo). Hence, we have that ℓ ∩ ( B n + m ( z ) \ B n − m ( z )) ⊆ A − . Now consider any point x ∈ B n ( z ) \ B n − m ( z ). There exists a projection ˆ x , such that d( x, ˆ x ) ≤ m . Therefore,by the triangle inequality, we haveˆ x ∈ ℓ ∩ ( B n + m ( z ) \ B n − m ( z )) ⊆ A − . Furthermore, the shortest path connecting x to ˆ x lies outside of B R ( z ), so it cannotintersect ∂A + , and hence x ∈ A − .We proved that B n ( z ) \ B n − m ( z ) ⊆ A − . Therefore, B n ( z ) \ B n − m ( z ) ⊆ B − , so B n ( p ) \ B n − m ( p ) ⊆ Y c . Since ∂Y c ⊆ B R ( p ), this implies that X \ B n ( p ) ⊆ Y c , so Y is finite. This is a contradiction, hence A + is infinite. We can prove the same waythat A − is also infinite.We showed that A − = ( A + ) c , and that A + and A − are both infinite. Therefore,they both contain exactly one end of the geodesic ℓ by Lemma 3.16. Let us define Y z = ( A + if + ∞ ∈ A + ,A − if + ∞ ∈ A − . In both cases, we have + ∞ ∈ Y z , −∞ ∈ Y cz , ∂Y z ⊆ B R ( z ) (by Claim 3.20) and F ⊆ [[ G y Σ]] Y z (by Claim 3.21). This concludes the proof of the lemma.12 .4 Amenable kernel Proposition 3.23.
The stabilizer [[ G y Σ]] Y is locally finite.Proof. Consider a finite set F ⊆ [[ G y Σ]] Y , our goal is to prove that the subgroup h F i is also finite. Define N F = max { N ϕ : ϕ ∈ F } .Let n > N F . Let r = r ( p, F, n ) from Lemma 3.13 for the point p , the finite set F and the number n . Let y = z = p , and pick y i ∈ ℓ for all i ∈ Z \ { } such thatd( y i , y i +1 ) = 2 r + 2 n + 2 m + 2 and y i is closer to −∞ than y i +1 for every i ∈ Z . Nowfor every i ∈ Z \ { } let us use Lemma 3.13 for the point y i . Thus, we get the points z i ∈ B r ( y i ) (for all i ∈ Z ) such that for every ϕ ∈ F , the ϕ -action is the same onthe S n -neighborhood of p and z i . Note that due to the choice of the y i ’s, the n -ballsaround the points z i are pairwise disjoint.Let Y = Y and for every i ∈ Z \ { } let us use Lemma 3.18 for F , the point z i and the number n . For every i , there exists an infinite set Y i = Y z i ⊆ X , such thatwe have + ∞ ∈ Y i , −∞ ∈ Y ci , furthermore ∂Y i ⊆ B R ( z i ) and F ⊆ [[ G y Σ]] Y i . Claim 3.24.
For every i ∈ Z , the set Y i contains Y i +1 .Proof. Let us denote by v the midpoint between y i and y i +1 on ℓ . Note that wehave d( v, B n ( z i )) ≥ m + 1 and d( v, B n ( z i +1 )) ≥ m + 1 by the choice of the distancebetween y i and y i +1 .Therefore, v ∈ Y i since + ∞ ∈ Y i and the half line [ v, + ∞ ] does not intersect ∂Y i ⊆ B n ( z i ). Similarly, we have v ∈ Y ci +1 since −∞ ∈ Y ci +1 and [ v, −∞ ] does notintersect ∂Y ci +1 ⊆ B n ( z i +1 ).Suppose for contradiction that there exists a point x ∈ Y i +1 \ Y i . Let ˆ x be theclosest point to x on ℓ , and let [ x, ˆ x ] denote a shortest path between them. We haved(ˆ x, x ) ≤ m by Proposition 3.5. There are two possibilites.1. If ˆ x ∈ [ v, + ∞ ], then the path [ x, ˆ x ] ∪ [ˆ x, v ] does not intersect B n ( z i ) sinced(ˆ x, B n ( z i )) ≥ m + 1. However, x / ∈ Y i and v ∈ Y i , so any path between thetwo must intersect ∂Y i ⊆ B n ( z i ). Hence, we get a contradiction.2. If ˆ x ∈ [ −∞ , v ], then we can use a similar argument: We have x ∈ Y i +1 but v / ∈ Y i +1 , so any path between them must intersect ∂Y i +1 ⊆ B n ( z i +1 ). However,the path [ x, ˆ x ] ∪ [ˆ x, v ] does not intersect B n ( z i +1 ), leading to a contradiction.We get a contradiction in both cases, so such a point x cannot exist. This provesthat Y i +1 ⊆ Y i . Claim 3.25.
For every i ∈ Z , the set Y i \ Y i +1 is a finite set. Moreover, there is auniform bound on the cardinality of the sets Y i \ Y i +1 .Proof. Consider the m -neighborhood of the segment [ y i − , y i +2 ] ⊂ ℓ , denoted by B m ([ y i − , y i +2 ]). First of all, the size of these sets has a uniform bound, since thegraph X is regular, and hence the size of the m -neighborhood of a set of 3(2 r + 2 n +2 m + 2) + 1 points is uniformly bounded.We show that Y i \ Y i +1 ⊆ B m ([ y i − , y i +2 ]). Take any point x ∈ Y i \ Y i +1 , letˆ x denote its projection to ℓ . Suppose for contradiction that ˆ x ∈ [ y i +2 , + ∞ ], then[ˆ x, + ∞ ] does not intersect ∂Y i +1 since ∂Y i +1 ⊆ B n ( z i +1 ) ⊆ B n + r ( y i +1 ). Hence, wehave ˆ x ∈ Y i +1 , and also x ∈ Y i +1 , since [ x, ˆ x ] cannot intersect ∂Y i +1 either. Thiscontradicts the face that x ∈ Y i \ Y i +1 . Therefore, we must have ˆ x ∈ [ −∞ , y i +2 ].13imilarly, one can prove that if ˆ x ∈ [ −∞ , y i − ], then x / ∈ Y i , leading to a contradic-tion again. This proves that ˆ x ∈ [ y i − , y i +2 ]. Recall that d( x, ˆ x ) ≤ m , and hence x ∈ B m ([ y i − , y i +2 ]).We proved that Y i \ Y i +1 ⊆ B m ([ y i − , y i +2 ]), this shows that the set Y i \ Y i +1 isfinite for every i ∈ Z , and that there is a uniform bound on their cardinalities.Notice that due to the F -invariance of every Y i , the sets Y i \ Y i +1 are also F -invariant for all i ∈ Z . Their union is the whole graph X , therefore, we can embed h F i into the direct product of the finite symmetric groups on the sets Y i \ Y i +1 .By Claim 3.25, the size of these finite symmetric groups is uniformly bounded, andhence their direct product is locally finite.Since h F i can be embedded into a locally finite group, and is finitely generated,it must be finite. This concludes the proof of the proposition.Finally, we are ready to prove our main theorem. Proof of Theorem 1.1.
Consider a minimal action G y Σ of the finitely generatedgroup G , such that there exists an orbit X which is quasi-isometric to Z .First, we show that the action [[ G y Σ]] y X is extensively amenable. TheSchreier graph of the action G y X is quasi-isometric to Z , and hence it is recur-rent. As a corollary of Rayleigh’s monotonicity principle, all connected subgraphsof a recurrent graph are also recurrent (for a proof see [9], Chapter 2). Thus, byProposition 2.4, the action PW( G y X ) y X is extensively amenable. It followseasily from the definition of extensive amenability that the action of any subgroupof PW( G y X ) on X is also extensively amenable. Therefore, [[ G y Σ]] y X isextensively amenable, as desired.Next, we apply Proposition 2.5 for [[ G y Σ]] and X , with the cocycle c definedin Definition 3.10. By (2) in Remark 3.11, we have ker c = [[ G y Σ]] Y . ByProposition 3.23, this stabilizer [[ G y Σ]] Y is locally finite, and hence it is amenable.Therefore, the conditions of Proposition 2.5 are satisfied: the action [[ G y Σ]] y X is extensively amenable, and the kernel of the cocycle c is amenable. This provesthat the topological full group [[ G y Σ]] is also amenable.
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