Infinitely supported Liouville measures of Schreier graphs
aa r X i v : . [ m a t h . G R ] A ug INFINITELY SUPPORTED LIOUVILLE MEASURES OFSCHREIER GRAPHS
KATE JUSCHENKO AND TIANYI ZHENG
Abstract.
We provide equivalent conditions for Liouville property of actionsof groups. As an application, we show that there is a Liouville measure for theaction of the Thompson group F on dyadic rationals. This result should becompared with a recent result of Kaimanovich, where he shows that the actionof the Thompson group F on dyadic rationals is not Liouville for all finitelysupported measures. As another application we show that there is a Liouvillemeasure for lamplighter actions. This gives more examples of non-amenableLiouville actions. introduction Let a discrete group G act on a countable set X , denoted by G y X , and let µ be a probability measure on G . A measure µ on G is non-degenerate if supp µ generate the group G . We denote by P µ the transition matrix on X induced by µ ,that is P µ ( x, y ) = X g ∈ G { g · x = y } µ ( g ) . For the simplicity of the notations we write µ · x = P µ ( x, · ) . A function f : X → R is P µ -harmonic if f ( x ) = P y ∈ X f ( y ) P µ ( x, y ) , and ( X, P µ ) is Liouville if all bounded P µ -harmonic functions are constant. The action G y X is µ -Liouville if ( X, P µ ) isLiouville, and if this is the case we say µ is a Liouville measure for the action. Wecall an action Liouville if there is a measure µ on G which makes it µ -Liouville.Note that, one can make several definitions for Liouville actions by adapting thedefinition of Liouville measures on Cayley graphs. While many of the definitionsare equivalent for Cayley graphs, they are not equivalent when we pass to actions.The main reason of our current definition is a recent approach to amenability de-veloped by Kaimanovich in [5]. In order to show that a group is not amenable itis sufficient to find an action which does not admit any non-degenerate Liouvillemeasure. Indeed, this will insure that there is no non-denegerate Liouville measureon the group itself, thus, by renowned result of Kaimanovich and Vershik, [6], thegroup is not amenable. The problem of amenability of Thompson’s group F canbe approached with this technique. In particular, Kaimanovich showed that everyfinitely supported non-degenerate measure on Thompson group F is not Liouvillefor it’s action on dyadic rationals. In this paper we show that this action admitsinfinitely supported Liouville measure.In fact, our methods are more general, we give a criteria for a measure to be Li-ouville for the action. As another application of this method we show that certainSchreier graphs of lamplighter group are non-amenable but µ -Liouville for some Date : August 12, 2016. nfinitely supported measure µ . We note that examples of non-amenable graphswith Liouville measures of finite support were previously have been known, see forexample [2] and references therein. Acknowledgements:
We are grateful to Kaimanovich for several interestingand motivating discussions. We also thank Nico Matte Bon and Omer Tamusz forremarks on earlier versions of the paper.2.
Liouville measures on Schreier graphs
We fix a finite generating set S on G . Consider an action of G on a countableset X . Let k·k be the ℓ norm with respect to counting measure on X , that is k f k = P x ∈ X | f ( x ) | . Lemma 1.
Suppose P µ is irreducible and lazy such that P µ ( x, x ) ≥ for all x ∈ X .If ( X, P µ ) is Liouville, then for any two points x, y ∈ X , lim n →∞ (cid:13)(cid:13)(cid:13) µ ( n ) · x − µ ( n ) · y (cid:13)(cid:13)(cid:13) = 0 . Proof.
Since P µ is irreducible, it is sufficient to prove the claim for y = s · x with s ∈ supp ( µ ) . Let ( W n ) be a left random walk on G with step distribution µ .By Corollary 14.13 of [7], ( X, P µ ) is Liouville if and only if the invariant σ -field I is trivial. Since P µ is lazy, by Theorem 14.18 [7], the completion of the tail σ -fieldcoincides with the completion of the invariant σ -field. Thus Liouville property of ( X, P µ ) implies that the tail σ -field T is trivial. Therefore, P ( W · x = y | W n · x ) → P ( W · x = y ) almost surely when n → ∞ . Since P ( W · x = y | W n · x = y n ) = P µ ( x, y ) P n − µ ( y , y n ) P nµ ( x, y n ) , we have for y = s · x , lim n →∞ P n − µ ( s · x, W n · x ) P nµ ( x, W n · x ) = 1 a.s Then it implies for any ǫ > , lim n →∞ µ ( n ) ( g : (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) P n − µ ( s · x, g · x ) P nµ ( x, g · x ) − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ ǫ )! = 0 . It follows that lim n →∞ (cid:13)(cid:13) P n − µ ( s · x, · ) − P nµ ( x, · ) (cid:13)(cid:13) = 0 . Finally, by Theorem 14.16 in [7], laziness of P µ guarentees that (cid:13)(cid:13) P n − µ ( s · x, · ) − P nµ ( s · x, · ) (cid:13)(cid:13) → when n → ∞ . (cid:3) In the other direction, we can build a measure µ on G such that P µ is Liouvillefrom pieces that have good coupling properties. emma 2. Suppose there exists an increasing sequence of finite subsets ( K n ) exhausting X and a sequence ( ǫ n ) decreasing to such that for each n , there existsa probability measure ν n of finite support on G such that for any x, y ∈ K n suchthat y = s · x for some s ∈ S , we have k ν n · x − ν n · y k < ǫ n . Then there exists a non-degenerate probability measure µ on G such that ( X, P µ ) is Liouville. Proof.
Our proof is reminiscent to the proof of Theorem 4.3 in [6]. The measure µ is obtained as a convex combination of a subsequence of ( ν n ) , µ = ∞ X j =0 c j ζ j , ζ j = ν n j To make µ non-degenerate, we take ν to be uniform on S ∪ S − .First note that to show P µ is Liouville, it suffices to show for any x, y ∈ X connected by an edge, y = s · x for some s ∈ S , we have lim inf m →∞ (cid:13)(cid:13)(cid:13) µ ( m ) · x − µ ( m ) · y (cid:13)(cid:13)(cid:13) = 0 . (This implies that for any bounded P µ -harmonic function h , h ( x ) = h ( y ) for neigh-boring points, thus the function must be constant.)Since each ν n is assumed to be of finite support on G , let B ( e, r n ) be a ball largeenough on G such that support ν n ⊂ B ( e, R n ) . Let r n be the largest ball such that B ( o, r n ) ⊆ K n , we have r n → ∞ since ( K n ) exhausts X . Fix a consequence ofweights ( c j ) , select ( n j ) inductively as follows: let m j be the smallest integer suchthat ( c + . . . + c j − ) m j ≤ /j , take n j to be the least integer such that m j R n j − ≤ r n j and m j R n j − ǫ n j ≤ /j. For m -th convolution power of µ , µ ( m ) = X k c k . . . c k m ζ k m ∗ . . . ∗ ζ k . Consider two parts, µ ( m )1 consists these terms with max ≤ i ≤ m k i ≥ j , and µ ( m )2 = µ ( m ) − µ ( m )1 . For m = m j , the total mass of the first part is (cid:13)(cid:13)(cid:13) µ ( m j )1 (cid:13)(cid:13)(cid:13) = ( c + . . . + c j − ) m j . For each term in the second part, let i = i ( k ) be the lowest index such that k i ≥ j . Starting at two neighboring points x, y , consider the distribution induced by ζ k i ∗ . . . ∗ ζ k on these two points. Since for i < k , k i < j , it follows that thesupport of ζ k i − ∗ . . . ∗ ζ k · x and ζ k i − ∗ . . . ∗ ζ k · y are contained in the ball B X (cid:0) o, d ( o, x ) + 1 + ( m j − R n j − − (cid:1) . By the choice of the ( ζ n ) , we have for k i ≥ j , (cid:13)(cid:13) ζ k i ∗ ζ k i − ∗ . . . ∗ ζ k · x − ζ k i ∗ ζ k i − ∗ . . . ∗ ζ k · y (cid:13)(cid:13) ≤ m j − R n j − − ǫ n j . Combine the two parts, we have (cid:13)(cid:13)(cid:13) µ ( m j ) · x − µ ( m j ) · y (cid:13)(cid:13)(cid:13) ≤ ( c + . . . + c j − ) m j + 2( m − R ℓ m − ǫ ℓ m ≤ /j (cid:3) . Applications to the Thompson group F
We denote by F the Thompson group F . In [5] and [8], authors show that theSchreier graph of the action of F on the orbit of / is not Liouville with respect toany measure of finite support. Here we show that there are measures with infinitesupport that make this action Liouville. In fact, one can even choose symmetricones. Theorem 3.
There is a non-degenerate symmetric measure µ on Thompson group F such that the action on Orb (1 / is µ -Liouville. Proof.
By Lemma 2, we have to find a measure that approximates any finite subsetin
Orb (1 / . The Schreier graph of Orb (1 / was described by Savchuk, see [9].There are two parts of the graph: the one that corresponds to the binary tree andanother is rays attached to every node of the tree. These ray imitate positive partof the Cayley graph of Z , and we will call them hairs.Let K ⊂ Orb (1 / . Since F is strongly transitive (see [1]), we can find an element g that maps this set to the hair. We can assume that this set is mapped deepenough into the hair. One the hairs one of the generators, say g (in the notationsof Savchuk), act as Z . The set g ( K ) might not be connected, but it is clear that theuniform measure on { g k : k ∈ [ − n, n ] } will satisfy Lemma 2 for sufficiently large n ,therefore the uniform measure on { g k g : k ∈ [ − n, n ] } is Liouville. We can make itsymmetric by taking { g − g k g : k ∈ [ − n, n ] } (cid:3) Liouville non-amenable Schreier graphs and lamplighters
As another application of Lemma 2 we show that certain Schreier graphs oflamplighter group are non-amenable but µ -Liouville for some infinitely supportedmeasure µ . We note that examples of non-amenable graphs with Liouville measureswere previously discovered in [2]. There are many examples when a non-amenablegroup G admits an action on a set X such that the induced action of the lamplightergroup L X Z / Z ⋊ G on L X Z / is not amenable. In fact, if the action of G on X is not amenable then the action of L X Z / Z ⋊ G on L X Z / , see for example [3],[4]. However, this action always admits Liouville measure. Lemma 4.
Consider the semi-direct product G ⋉ A with G discrete and A amenable.Then there exists a non-degenerate probability measure µ on G ⋉ A such that theaction of G ⋉ A on A is µ -Liouville. Proof.
Fix a sequence of finite subsets ( K n ) that exhaust A . Since A is amenable,we can find a sequence of measures ( ν n ) on A such that sup x,y ∈ K n k ν n · x − ν n · y k ℓ ( A ) ≤ n . Regard ν n as a measure supported on { ( e G , a ) : a ∈ A } , then by Lemma 2 we canfind a non-degenerate measure µ on G ⋉ A such that P µ is Liouville. (cid:3) Let G be an amenable group. The every action of G admits a Liouville measure(possibly infinitely supported), moreover, the stabilizers of the action are amenable.hile the following questions should have a negative answer, we currently don’thave any examples to support it. Question 5.
Let G act transitively on a set X and assume that this action is µ -Liouville action of G on X for some measure µ on G such that Stab G ( x ) is abelianfor some (equivalently for all) x in X . Is G amenable? Question 6.
Let G act transitively on a set X and assume that this action is µ -Liouville action of G on X for some measure µ on G . Denote by X n the set ofall finite subsets of size n . Then G acts on X n , however, this action may not betransitive. Is it true that the action of G on orbit of x ∈ X n is Liouville for somemeasure? References [1]
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Kate Juschenko, Department of Mathematics, Northwestern University, 2033 Sheri-dan Road, Evanston, IL 60208, USA
E-mail address : [email protected] Tianyi Zheng, Department of Mathematics, UC San Diego, 9500 Gilman Dr, LaJolla, CA 92093, USA
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