A Coloring Property for Countable Groups - Personal.Psu.Edu

Motivation I: Hyperfiniteness Proofs Motivation II: Dynamical Systems A Coloring Property Orthogonality and Minimality If There Is Time...

A Coloring Property for Countable Groups Su Gao*, Steve Jackson, and Brandon Seward Department of Mathematics University of North Texas

AMS-ASL Special Session on Logic and Dynamical Systems 6 January 2009 Washington, DC

Su Gao*, Steve Jackson, and Brandon Seward

A Coloring Property for Countable Groups


Motivation I: Hyperfiniteness Proofs Let G be a countable group with a Borel action on a Polish space X.




Motivation I: Hyperfiniteness Proofs Let G be a countable group with a Borel action on a Polish space X . The orbit equivalence relation is EG = {(x, y ) ∈ X × X : y = g · x for some g ∈ G }




Motivation I: Hyperfiniteness Proofs Let G be a countable group with a Borel action on a Polish space X . The orbit equivalence relation is EG = {(x, y ) ∈ X × X : y = g · x for some g ∈ G } EG is hyperfinite if EG = n Fn where F0 ⊆ F1 ⊆ F2 ⊆ · · · ⊆ Fn ⊆ . . . . . . are finite Borel equivalence relations, i.e., each Fn -equivalence class is finite.




A canonical example of hyperfinite equivalence relations

• •

X = 2ω xE0 y ⇐⇒ ∃n ∀m ≥ n x(m) = y (m)




A canonical example of hyperfinite equivalence relations

X = 2ω xE0 y ⇐⇒ ∃n ∀m ≥ n x(m) = y (m) E0 is hyperfinite, since E0 = n Fn , where • •

xFn y ⇐⇒ ∀m ≥ n x(m) = y (m) and each Fn is finite Borel.




Theorem (Dougherty-Jackson-Kechris, based on earlier work of Weiss, Slaman-Steel) TFAE: (i) E is hyperfinite; (ii) E is the orbit equivalence relation of a Borel Z-action. (iii) E ≤B E0 , i.e., there is a Borel (injective) function f : X → 2ω such that, for all x, y ∈ X , xEy ⇐⇒ f (x)E0 f (y ).




Theorem (Dougherty-Jackson-Kechris, based on earlier work of Weiss, Slaman-Steel) TFAE: (i) E is hyperfinite; (ii) E is the orbit equivalence relation of a Borel Z-action. (iii) E ≤B E0 , i.e., there is a Borel (injective) function f : X → 2ω such that, for all x, y ∈ X , xEy ⇐⇒ f (x)E0 f (y ). In particular, consider the shift action of Z on 2Z : (n · x)(k) = x(k − n) The orbit equivalence relation “embeds” into E0 . Su Gao*, Steve Jackson, and Brandon Seward



Slaman-Steel Lemma Let G be a countable group with a free Borel action on a Polish space X . Then there is an infinite decreasing sequence of Borel complete sections (i.e. sets meeting every equivalence class) S0 ⊇ S1 ⊇ S2 ⊇ · · · ⊇ Sn ⊇ . . . . . . such that

n

Sn = ∅.




Slaman-Steel Lemma Let G be a countable group with a free Borel action on a Polish space X . Then there is an infinite decreasing sequence of Borel complete sections (i.e. sets meeting every equivalence class) S0 ⊇ S1 ⊇ S2 ⊇ · · · ⊇ Sn ⊇ . . . . . . such that

n

Sn = ∅. An orbit in the Z-order

· · · · · · · · · · · · · · marker points are in red




Theorem (a) (Boykin-Jackson) There is a continuous embedding from (the free part of) 2Z into E0 . (b) (Gao-Jackson) For any 1 ≤ n < ∞ there is a continuous n embedding from (the free part of) 2Z into E0 .




Theorem (a) (Boykin-Jackson) There is a continuous embedding from (the free part of) 2Z into E0 . (b) (Gao-Jackson) For any 1 ≤ n < ∞ there is a continuous n embedding from (the free part of) 2Z into E0 . Observation If the Slaman-Steel lemma can be improved to obtain clopen marker sets in the free part of 2Z , then there is a continuous embedding of the free part of 2Z into E0 .




Question Is there a sequence of clopen complete sections of the free part of 2Z S0 ⊇ S1 ⊇ S2 ⊇ · · · ⊇ Sn ⊇ . . . . . . such that n Sn = ∅?




Question Is there a sequence of clopen complete sections of the free part of 2Z S0 ⊇ S1 ⊇ S2 ⊇ · · · ⊇ Sn ⊇ . . . . . . such that n Sn = ∅? Available techniques had given Fact There is a decreasing sequence (S n ) of clopen complete sections of the free part of 2Z such that n Sn meets each orbit at at most one point.




Answer No, because there are closed invariant subsets in the free part of 2Z .




Answer No, because there are closed invariant subsets in the free part of 2Z . Question What about countable groups other than Z?




Answer No, because there are closed invariant subsets in the free part of 2Z . Question What about countable groups other than Z? No for solvable groups and free groups.




Answer No, because there are closed invariant subsets in the free part of 2Z . Question What about countable groups other than Z? No for solvable groups and free groups. No for arbitrary countable groups.




Theorem (Gao-Jackson-Seward) Let G be any countably infinite group. Consider the action of G on 2G by shift: g · A = {gh : h ∈ A}. Let F (G ) be the free part of 2G . Then there is a closed invariant subset in F (G ).




Motivation II: Dyamical Systems

Let G be a countable group. Bernoulli G -flow: the G -space 2G subflow: closed invariant subset of 2G free subflow: closed invariant subset of F (G ), the free part of 2G




Motivation II: Dyamical Systems

Let G be a countable group. Bernoulli G -flow: the G -space 2G subflow: closed invariant subset of 2G free subflow: closed invariant subset of F (G ), the free part of 2G Question (Glasner-Uspenskij) Is there a free subflow of the Bernoulli G -flow for any countable group G ?




Partial Answers (Glasner-Uspenskij) Yes for (1) abelian groups (2) residually finite groups (3) S