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Group Colorings and Bernoulli Subflows Su Gao University of North Texas Steve Jackson University of North Texas Brandon Seward* University of Michigan RTG Logic and Dynamics Conference June 48, 2012 Brandon Seward () Group Colorings and

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Group Colorings and Bernoulli Subflows

Su Gao University of North Texas Steve Jackson University of North Texas Brandon Seward* University of Michigan RTG Logic and Dynamics Conference June 4–8, 2012

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 1 / 16

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Bernoulli Flows

For G a countable group and k ≥ 2 an integer, we define the Bernoulli flow kG := set of all functions G → {0, . . . , k − 1} =

g∈G

{0, 1, . . . , k − 1}.

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 2 / 16

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Bernoulli Flows

For G a countable group and k ≥ 2 an integer, we define the Bernoulli flow kG := set of all functions G → {0, . . . , k − 1} =

g∈G

{0, 1, . . . , k − 1}. We give kG the product topology.

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 2 / 16

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Bernoulli Flows

For G a countable group and k ≥ 2 an integer, we define the Bernoulli flow kG := set of all functions G → {0, . . . , k − 1} =

g∈G

{0, 1, . . . , k − 1}. We give kG the product topology. G naturally acts on kG: (g · x)(h) = x(g−1h).

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 2 / 16

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Bernoulli Flows

For G a countable group and k ≥ 2 an integer, we define the Bernoulli flow kG := set of all functions G → {0, . . . , k − 1} =

g∈G

{0, 1, . . . , k − 1}. We give kG the product topology. G naturally acts on kG: (g · x)(h) = x(g−1h). kG is compact and G acts continuously.

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 2 / 16

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Bernoulli Flows

For G a countable group and k ≥ 2 an integer, we define the Bernoulli flow kG := set of all functions G → {0, . . . , k − 1} =

g∈G

{0, 1, . . . , k − 1}. We give kG the product topology. G naturally acts on kG: (g · x)(h) = x(g−1h). kG is compact and G acts continuously. For µ a probability measure on {0, 1, . . . , k − 1}, we call the product measure µG a Bernoulli measure.

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 2 / 16

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Bernoulli Flows

For G a countable group and k ≥ 2 an integer, we define the Bernoulli flow kG := set of all functions G → {0, . . . , k − 1} =

g∈G

{0, 1, . . . , k − 1}. We give kG the product topology. G naturally acts on kG: (g · x)(h) = x(g−1h). kG is compact and G acts continuously. For µ a probability measure on {0, 1, . . . , k − 1}, we call the product measure µG a Bernoulli measure. Under a Bernoulli measure, the action is measure preserving and ergodic.

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 2 / 16

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Bernoulli Flows

For G a countable group and k ≥ 2 an integer, we define the Bernoulli flow kG := set of all functions G → {0, . . . , k − 1} =

g∈G

{0, 1, . . . , k − 1}. We give kG the product topology. G naturally acts on kG: (g · x)(h) = x(g−1h). kG is compact and G acts continuously. For µ a probability measure on {0, 1, . . . , k − 1}, we call the product measure µG a Bernoulli measure. Under a Bernoulli measure, the action is measure preserving and ergodic. A point x ∈ kG is aperiodic (or free) if its stabilizer is trivial. We let F(kG) denote the free part of the action, i.e. the set of all aperiodic (free) points of kG.

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 2 / 16

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Hyperfinite Equivalence Relations

Recall that a countable Borel equivalence relation E is hyperfinite if it can be expressed as the increasing union of finite Borel equivlance relations.

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 3 / 16

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Hyperfinite Equivalence Relations

Recall that a countable Borel equivalence relation E is hyperfinite if it can be expressed as the increasing union of finite Borel equivlance relations. A long standing open problem asks if every Borel action of an amenable group induces a hyperfinite equivalence relation (via its orbits).

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 3 / 16

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Hyperfinite Equivalence Relations

Recall that a countable Borel equivalence relation E is hyperfinite if it can be expressed as the increasing union of finite Borel equivlance relations. A long standing open problem asks if every Borel action of an amenable group induces a hyperfinite equivalence relation (via its orbits). This is known to hold if G is abelian (Gao–Jackson, 2007) or finitely generated and virtually nilpotent (Jackson–Kechris–Louveau, 2002).

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 3 / 16

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Hyperfinite Equivalence Relations

Recall that a countable Borel equivalence relation E is hyperfinite if it can be expressed as the increasing union of finite Borel equivlance relations. A long standing open problem asks if every Borel action of an amenable group induces a hyperfinite equivalence relation (via its orbits). This is known to hold if G is abelian (Gao–Jackson, 2007) or finitely generated and virtually nilpotent (Jackson–Kechris–Louveau, 2002). Progress on this problem hinges on finding constructions for better marker sets and better marker regions.

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 3 / 16

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Hyperfinite Equivalence Relations

Recall that a countable Borel equivalence relation E is hyperfinite if it can be expressed as the increasing union of finite Borel equivlance relations. A long standing open problem asks if every Borel action of an amenable group induces a hyperfinite equivalence relation (via its orbits). This is known to hold if G is abelian (Gao–Jackson, 2007) or finitely generated and virtually nilpotent (Jackson–Kechris–Louveau, 2002). Progress on this problem hinges on finding constructions for better marker sets and better marker regions. A classical construction for marker sets comes from the following. Lemma (Slaman–Steel, 1988) There is a decreasing sequence (Sn) of Borel complete sections of F(kG) such that

n Sn = ∅.

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 3 / 16

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Motivating Question

A strengthening of the Slaman–Steel lemma could lead to improved understanding of hyperfinite equivalence relations.

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 4 / 16

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Motivating Question

A strengthening of the Slaman–Steel lemma could lead to improved understanding of hyperfinite equivalence relations. Question Does there exist a decreasing sequence (Sn) of complete sections of F(kG) such that each Sn is relatively closed in F(kG) and

n Sn = ∅?

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 4 / 16

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Motivating Question

A strengthening of the Slaman–Steel lemma could lead to improved understanding of hyperfinite equivalence relations. Question Does there exist a decreasing sequence (Sn) of complete sections of F(kG) such that each Sn is relatively closed in F(kG) and

n Sn = ∅?

If F(kG) where to contain a compact G-invariant subset, then by compactness the answer to the above question must be ’No.’

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 4 / 16

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Free Subflows

A subflow of a Bernoulli flow kG is a closed (hence compact) subset of kG which is invariant under the action of G. A subflow X ⊆ kG is free if X ⊆ F(kG).

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 5 / 16

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Free Subflows

A subflow of a Bernoulli flow kG is a closed (hence compact) subset of kG which is invariant under the action of G. A subflow X ⊆ kG is free if X ⊆ F(kG). A point x ∈ kG is a k-coloring if the closure of its orbit, [x], is a free subflow (equivalently [x] ⊆ F(kG)).

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 5 / 16

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Free Subflows

A subflow of a Bernoulli flow kG is a closed (hence compact) subset of kG which is invariant under the action of G. A subflow X ⊆ kG is free if X ⊆ F(kG). A point x ∈ kG is a k-coloring if the closure of its orbit, [x], is a free subflow (equivalently [x] ⊆ F(kG)). Notice that x is a k-coloring if and only if it is contained in some free

subflow. Thus, the set of all k-colorings coincides with the union of all free

subflows.

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 5 / 16

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Free Subflows

A subflow of a Bernoulli flow kG is a closed (hence compact) subset of kG which is invariant under the action of G. A subflow X ⊆ kG is free if X ⊆ F(kG). A point x ∈ kG is a k-coloring if the closure of its orbit, [x], is a free subflow (equivalently [x] ⊆ F(kG)). Notice that x is a k-coloring if and only if it is contained in some free

subflow. Thus, the set of all k-colorings coincides with the union of all free

subflows. When free subflows are present, the Slaman–Steel marker lemma cannot be strengthened in the manner discussed.

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 5 / 16

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A Natural Question

Question Does every Bernoulli flow contain a free subflow?

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 6 / 16

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A Natural Question

Question Does every Bernoulli flow contain a free subflow? It is well known that F(kG) is always comeager and always has full measure under every Bernoulli measure.

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 6 / 16

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A Natural Question

Question Does every Bernoulli flow contain a free subflow? It is well known that F(kG) is always comeager and always has full measure under every Bernoulli measure. Lemma (Gao–Jackson–S) The set of k-colorings (i.e. the union of all free subflows) is always meager and always has measure zero under every Bernoulli measure. Thus measure and Baire category arguments seem incapable of addressing the above question.

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 6 / 16

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A Natural Question

Question Does every Bernoulli flow contain a free subflow? It is well known that F(kG) is always comeager and always has full measure under every Bernoulli measure. Lemma (Gao–Jackson–S) The set of k-colorings (i.e. the union of all free subflows) is always meager and always has measure zero under every Bernoulli measure. Thus measure and Baire category arguments seem incapable of addressing the above question. If the answer is to be ‘yes,’ a constructive proof is needed.

Brandon Seward () Group Colorings and Bernoulli Subflows Logic & Dynamics Conf. 2012 6 / 16

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