  # Amenable actions of the infinite permutation group Lecture IV - PowerPoint PPT Presentation

## Amenable actions of the infinite permutation group Lecture IV Juris Stepr ans York University Young Set Theorists Meeting March 2011, Bonn Juris Stepr ans Amenable actions Questions The main question left open is the following

1. Amenable actions of the infinite permutation group — Lecture IV Juris Stepr¯ ans York University Young Set Theorists Meeting — March 2011, Bonn Juris Stepr¯ ans Amenable actions

2. Questions The main question left open is the following which, of course, does not assume any extra set theory beyond Choice. Question Is there an amenable subgroup of S ( ω ) whose natural action on ω has a unique mean? But there are a number of intermediate questions that may be of interest as well. Juris Stepr¯ ans Amenable actions

3. Question Is there a model where p � = u yet the Key Hypothesis still holds? Recall that the Key Hypothesis is the following: There is a generating set { G ξ } ξ ∈ κ for an ultrafilter on ω such that there exist infinite A ξ ⊆ ω satisfying: A ξ ⊆ ∗ G η for each η ≤ ξ A ξ ∩ A η is finite if ξ � = η . Is there such a model where no ultrafilter is generated by a tower? Juris Stepr¯ ans Amenable actions

4. Question Is there a model where the Key Hypothesis fails yet there is still an amenable subgroup of S ( ω ) whose natural action on ω has a unique mean? Juris Stepr¯ ans Amenable actions

5. Foreman has shown that assuming Martin’s Axiom every amenable group of size less than c has 2 c invariant means. A weaker version of the main question is: Question Is there an amenable subgroup of S ( ω ) whose natural action on ω has less than 2 c invariant means? Juris Stepr¯ ans Amenable actions

6. Until this point only discrete groups have been discussed. Pestov asked whether an amenable topological group acting on B can have a unique invariant mean. A modification of Foreman’s construction provides a positive answer. Let { ξ n } ∞ n =0 enumerate all finite one-to-one functions from N to N . For any partial partial involution σ let σ denote the permutation that agrees with σ on its domain and is the identity elsewhere, Construct by induction on n sets { A j n } ∞ j =0 and involutions σ j n : A j n → A j +1 such that n 1 N \ � n � J j =0 A j i is infinite for each J i =0 2 A j n ∩ A k n = ∅ of j � = k . 3 If A n = � ∞ j =1 A j n then A n ∩ A j m is finite for m < n and all j . n = B j yields that (1), 4 If there are { B j } ∞ j =0 such that setting A j (2) and (3) hold and if for some j there is an involution θ : B j → B j +1 such that ξ n ⊆ θ then ξ n ⊆ σ j n . Juris Stepr¯ ans Amenable actions

7. Let A be the group generated by { σ j n } n , j ∈ ω . Extend the ideal generated by { A j n } n , j ∈ ω to a maximal ideal J and for each X ∈ J choose A X such that A X ∩ X = ∅ and A X ∩ A j n is � � A j n \ X | n , j ∈ ω finite for all n and j . This is possible because generates a proper ideal on the complement of X . Let { A j X } ∞ j =1 partition A X into infinite sets and choose θ j X : A j X → A j +1 to be X j involutions. Let G be the group generated by A ∪ { θ X } X ∈J , j ∈ ω . Let G inherit the topology of pointwise convergence from the full symmetric group. Juris Stepr¯ ans Amenable actions

8. A is dense in G . A is locally finite and hence amenable. Hence G is also amenable. The natural action of G on N is uniquely amenable. Juris Stepr¯ ans Amenable actions

9. Given that all the arguments to this point have relied on group being locally finite in order to claim it is amenable on can ask whether the action of a non-locally finite amenable subgroup of S ( ω ) can have a unique mean. To eliminate trivialities, on can ask the following: Question Is there an amenable subgroup G ⊆ S ( ω ) whose action has a unique mean and such that every non-identity element of G has infinite order? Consistently? Juris Stepr¯ ans Amenable actions

10. Define an indexed set { σ i } i ∈ n of permutations of ω to be growing by induction on n . Let F ( σ ) denote the fixed points of a permutation σ . If n = 0 then { σ i } i ∈ 0 is growing. If n > 0 then { σ i } i ∈ n is growing if { σ i } i ∈ n − 1 is growing for each k ∈ ω \ F ( σ n − 1 ) the orbit of k under σ n − 1 is infinite σ n − 1 has infinitely many infinite orbits F ( σ n − 1 ) is infinite F ( σ j ) ⊇ ∗ F ( σ n − 1 ) for j ∈ n if O is an infinite orbit of F ( σ n − 1 ) and j ∈ n − 1 then | O \ F ( σ j ) | ≤ 1 Juris Stepr¯ ans Amenable actions

11. Example A growing family { σ i } i ∈ n on Z n +2 is provided by defining σ i ∈ S ( Z n +2 ) by � ( z 0 , . . . z n +1 ) ( ∃ j > i + 1) z j � = 0 σ i ( z 0 , . . . z n +1 ) = ( z 0 , . . . z i , z i +1 + 1 , 0 . . . , 0) otherwise In this example F ( σ j ) ⊇ F ( σ n − 1 ) for j ∈ n rather than just F ( σ j ) ⊇ ∗ F ( σ n − 1 ). Juris Stepr¯ ans Amenable actions

12. Claim If { σ i } i ∈ n is growing then the group generated by { σ i } i ∈ n is solvable and, hence, amenable. Before examining the proof of the claim, suppose that { A ξ } ξ ∈ κ is a ⊆ ∗ -tower generating a maximal ideal on ω . Without loss of generality, it may be assumed that A ξ +1 \ A ξ is infinite and A ξ +1 ⊃ A ξ for all ξ . Given ξ let { a n , m } n ∈ ω, m ∈ Z \{ 0 } enumerate A ξ +1 \ A ξ and let { a n } n ∈ ω enumerate A ξ . Define the permutation θ ξ by  m if m / ∈ A ξ +1     a n , k +1 if m = a n , k and k � = − 1  θ ξ ( m ) = a n if m = a n , − 1     a n , 0 if m = a n  Juris Stepr¯ ans Amenable actions

13. Observe that if ξ 1 < ξ 2 < . . . < ξ n then { θ ξ i } n i =1 is a growing family of permutations. The same argument as in Foreman’s construction shows that the action of the subgroup of S ( ω ) generated by { θ ξ } ξ ∈ κ has a unique invariant mean. Moreover, if every finite subset of { θ ξ } ξ ∈ κ generates a solvable group then the group is locally solvable, hence locally amenable and hence amenable by the Følner equivalence. Juris Stepr¯ ans Amenable actions

14. To sketch the main idea of the proof that growing families are solvable consider the example of the σ i ∈ S ( Z n +2 ) defined by � ( z 0 , . . . z n +1 ) ( ∃ j > i + 1) z j � = 0 σ i ( z 0 , . . . z n +1 ) = ( z 0 , . . . z i , z i +1 + 1 , 0 . . . , 0) otherwise and recall that this would be isomorphic to the general case except that F ( σ j ) ⊇ F ( σ n − 1 ) for j ∈ n rather than F ( σ j ) ⊇ ∗ F ( σ n − 1 ). Let G n be the subgroup of S ( Z n +2 ) generated by { σ i } i ∈ n . Let F n be the free group using the letters { σ i , σ − 1 } i ∈ n and for i w ∈ F n let σ ( w ) ∈ S ( Z n +2 ) be the corresponding permutation. Let I n ⊆ F n be the subgroup of all words w such that σ ( w ) is the identity. Of course, G n is isomorphic to F n / I n . Juris Stepr¯ ans Amenable actions

15. Given a family { w t ∈ F n | t : k → 2 } define [ w t ] t ∈ 2 k by induction on k . Let [ w t ] t ∈ 2 1 denote the usual commutator [ w 0 , w 1 ] = w 0 w 1 w − 1 1 w − 1 1 and for k > 1 let [ w t ] t ∈ 2 k denote [ w 0 t ] t ∈ 2 k − 1 , [ w 1 � � t ] t ∈ 2 k − 1 where w i t = w i ⌢ t . It will be shown that [ w t ] t ∈ 2 n ∈ I n for every family { w t } t ∈ 2 n ⊆ F n . From this it follows that the derived series of G n has length n + 1 and, hence, G n is solvable. Juris Stepr¯ ans Amenable actions

16. First some notation is needed. Given a word w ∈ F n and j ∈ n Let w / j be the word obtained from w by deleting all instances of σ j and σ − 1 in w . j Let e j ( w ) be the sum of the exponents of σ j occuring in w . Let w be the element of G n corresponding to w . The following claim is the key. Claim If w ∈ F n and e n − 1 ( w ) = 0 then for each j ∈ Z there is w [ j ] ∈ F n − 1 such that w ( j ,� x ) = ( j , w j ( � x )) . Juris Stepr¯ ans Amenable actions

17. To prove this proceed by induction on the length L of w . Let w = σ ℓ ( L ) σ ℓ ( L − 1) . . . σ ℓ (1) and suppose first that ℓ (1) � = n − 1. In this case, if j � = 0 then w ( j ,� x ) = σ ℓ ( L ) σ ℓ ( L − 1) . . . σ ℓ (2) ( j ,� x ) and so it is possible to set w [ j ] = σ ℓ ( L ) σ ℓ ( L − 1) . . . σ ℓ (2) [ j ] by the induction hypothesis. On the other hand, if j = 0 then σ ℓ (1) (0 ,� x ) = (0 , σ ℓ (1) ( � x )) and it is possible to set � � w [ j ] = σ ℓ ( L ) σ ℓ ( L − 1) . . . σ ℓ (2) [ j ] σ ℓ (1) Juris Stepr¯ ans Amenable actions

18. In the case that ℓ (1) = n − 1 let J be the least integer such that e n − 1 ( σ ℓ ( J ) σ ℓ ( J − 1) . . . σ ℓ (1) ) = 0 and let { x i } t i =1 be an increasing enumeration of � ℓ ( k + 1) � = n − 1 and e n − 1 ( σ ℓ ( k ) σ ℓ ( k − 1) . . . σ ℓ (1) ) = − j � � � k < J and let w ∗ = σ ℓ ( x t +1) σ ℓ ( x t − 1 +1) . . . σ ℓ ( x 1 +1) and let w ∗ � � w [ j ] = σ ℓ ( L ) σ ℓ ( L − 1) . . . σ ℓ ( J +1) [ j ] Juris Stepr¯ ans Amenable actions

19. To prove the main claim proceed by induction on n noting that G 1 is abelian to begin. For the general case, let { w t } t ∈ 2 n ⊆ F n and suppose that n > 1. Observe that if t : n − 1 → 2 then e j ([ w t ⌢ 0 , w t ⌢ 1 ]) = 0 and, in particular, e n − 1 ([ w t ⌢ 0 , w t ⌢ 1 ]) = 0. For each such t and m ∈ Z let z m t = [ w t ⌢ 0 , w t ⌢ 1 ][ m ]. The induction hypothesis yields that [ z m t ] t ∈ 2 n − 1 ∈ I n − 1 for each m . It follows that [ w t ] t ∈ 2 n ∈ I n because z m t ( m ,� x ) = ( m , [ w t ⌢ 0 , w t ⌢ 1 ]( � x )) Juris Stepr¯ ans Amenable actions

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