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Banach-Tarski Paradox Amenability Orbit Equivalence
Banach-Tarski Paradox and Amenability
Konrad Wrobel
Texas A&M UniversityDepartment of Mathematics
GIG’EM 2018
March 4, 2018
Banach-Tarski Paradox Amenability Orbit Equivalence
Banach-Tarski
What is Banach Tarski
Start with a sphere
Partition it into finitely many sets
Rotate and translate these sets
Get two spheres of the same volume as the first
Banach-Tarski Paradox Amenability Orbit Equivalence
Some Quick Observations/History
Let S2 be the sphere. Let (Ai )ni=1 be a partition of S that allows
the Banach-Tarski Paradox to occur.
At least one of these subsets Ai has to be nonmeasurable.
The standard tool we use for constructing nonmeasurable setsis the Axiom of Choice(although we don’t require the fullpower of AC).
This was one of the reasons Axiom of Choice wascontroversial in the early 20th century.
Banach-Tarski Paradox Amenability Orbit Equivalence
Some Quick Observations/History
Let S2 be the sphere. Let (Ai )ni=1 be a partition of S that allows
the Banach-Tarski Paradox to occur.
At least one of these subsets Ai has to be nonmeasurable.
The standard tool we use for constructing nonmeasurable setsis the Axiom of Choice(although we don’t require the fullpower of AC).
This was one of the reasons Axiom of Choice wascontroversial in the early 20th century.
Banach-Tarski Paradox Amenability Orbit Equivalence
Some Quick Observations/History
Let S2 be the sphere. Let (Ai )ni=1 be a partition of S that allows
the Banach-Tarski Paradox to occur.
At least one of these subsets Ai has to be nonmeasurable.
The standard tool we use for constructing nonmeasurable setsis the Axiom of Choice(although we don’t require the fullpower of AC).
This was one of the reasons Axiom of Choice wascontroversial in the early 20th century.
Banach-Tarski Paradox Amenability Orbit Equivalence
Construction
Let G be a group and A an arbitrary set.Fix a group action G y A.
Paradoxicality
The set A is G -paradoxical if there exist disjoint subsetsA1, ...,An,B1, ...,Bm of G and elements a1, ..., an, b1, ..., bm of Gsuch that A =
⋃i ai · Ai =
⋃j bj · Bj .
This should look similar to the statement of Banach-Tarski, sinceBanach-Tarski can be restated as the sphere beingSO(3)-paradoxical.
Banach-Tarski Paradox Amenability Orbit Equivalence
Construction
Paradoxicality of F2 = 〈a, b〉For a generator g ∈ {a, a−1, b, b−1}, set S(g) to be the set of allwords that begin with the letter g .
Notice that a · S(a−1) = {e} t S(a−1) t S(b) t S(b−1)
So F2 is F2-paradoxical.
Banach-Tarski Paradox Amenability Orbit Equivalence
Construction
Paradoxicality of F2 = 〈a, b〉For a generator g ∈ {a, a−1, b, b−1}, set S(g) to be the set of allwords that begin with the letter g .Notice that a · S(a−1) = {e} t S(a−1) t S(b) t S(b−1)
So F2 is F2-paradoxical.
Banach-Tarski Paradox Amenability Orbit Equivalence
Construction
Paradoxicality of F2 = 〈a, b〉For a generator g ∈ {a, a−1, b, b−1}, set S(g) to be the set of allwords that begin with the letter g .Notice that a · S(a−1) = {e} t S(a−1) t S(b) t S(b−1)
So F2 is F2-paradoxical.
Banach-Tarski Paradox Amenability Orbit Equivalence
Construction
An action of F2 on the sphere S2
Consider the matrices
A =
13
2√2
3 0
−2√2
313 0
0 0 1
,B =
1 0 0
0 13
2√2
3
0 −√23
13
.
These are elements of SO(3) that generate a copy of F2.
Now we have an action of F2 on the sphere S2 by rotations. Eachelement g ∈ F2 stabilizes only two points on the sphere.
So, in fact, we have a free action F 2 y S2 \ C where C is acountable set.
Banach-Tarski Paradox Amenability Orbit Equivalence
Construction
An action of F2 on the sphere S2
Consider the matrices
A =
13
2√2
3 0
−2√2
313 0
0 0 1
,B =
1 0 0
0 13
2√2
3
0 −√23
13
.
These are elements of SO(3) that generate a copy of F2.
Now we have an action of F2 on the sphere S2 by rotations. Eachelement g ∈ F2 stabilizes only two points on the sphere.
So, in fact, we have a free action F 2 y S2 \ C where C is acountable set.
Banach-Tarski Paradox Amenability Orbit Equivalence
Construction
An action of F2 on the sphere S2
Consider the matrices
A =
13
2√2
3 0
−2√2
313 0
0 0 1
,B =
1 0 0
0 13
2√2
3
0 −√23
13
.
These are elements of SO(3) that generate a copy of F2.
Now we have an action of F2 on the sphere S2 by rotations. Eachelement g ∈ F2 stabilizes only two points on the sphere.
So, in fact, we have a free action F 2 y S2 \ C where C is acountable set.
Banach-Tarski Paradox Amenability Orbit Equivalence
Construction
Pushing forward the paradoxicality of F 2
Recall F2 = S(a) t a · S(a−1) = S(b) t b · S(b−1).
Use Axiom of Choice to choose a single point in every orbit.Think of this set of points as the identity of F2.
Now simultaneously find sets T (a),T (a−1),T (b),T (b−1)analogously to S(a), ...
S2 \ C = T (a) t a · T (a−1) = T (b) t b · T (b−1).
So, S2 \ C is, in fact, F 2-paradoxical.
Banach-Tarski Paradox Amenability Orbit Equivalence
Definitions
Amenability
A countable discrete group G is amenable if there exists asequence of nonempty finite sets (Fn)n∈N such that for every g ∈ G
|Fn∆(g · Fn)||Fn|
→ 0 as n→∞
Such a sequence is called a Følner sequence.
Examples
finite groups
Zn
abelian groups
solvable groups
Banach-Tarski Paradox Amenability Orbit Equivalence
Some implications
G amenable=⇒ ∃ left-invariant finitely-additive probability measure on G
Proof. Take a Følner sequence (Fn) for G and fix an ultrafilter Uon N. Let A ⊆ G . Define m(A) := limU
|A∩Fn||Fn| .
G amenable =⇒ G is not paradoxicalProof. Take a left-invariant finitely additive probability measure mon G . If G were paradoxical, thenG =
⊔Ai t
⊔Bj =
⊔ai · Ai =
⊔bj · Bj . But we now get that
1 = m(G ) = m(G ) + m(G ) = 2 due to the properties of m.
Other directions are also true but not as easy.
Banach-Tarski Paradox Amenability Orbit Equivalence
Some implications
G amenable=⇒ ∃ left-invariant finitely-additive probability measure on GProof. Take a Følner sequence (Fn) for G and fix an ultrafilter Uon N. Let A ⊆ G . Define m(A) := limU
|A∩Fn||Fn| .
G amenable =⇒ G is not paradoxicalProof. Take a left-invariant finitely additive probability measure mon G . If G were paradoxical, thenG =
⊔Ai t
⊔Bj =
⊔ai · Ai =
⊔bj · Bj . But we now get that
1 = m(G ) = m(G ) + m(G ) = 2 due to the properties of m.
Other directions are also true but not as easy.
Banach-Tarski Paradox Amenability Orbit Equivalence
Some implications
G amenable=⇒ ∃ left-invariant finitely-additive probability measure on GProof. Take a Følner sequence (Fn) for G and fix an ultrafilter Uon N. Let A ⊆ G . Define m(A) := limU
|A∩Fn||Fn| .
G amenable =⇒ G is not paradoxical
Proof. Take a left-invariant finitely additive probability measure mon G . If G were paradoxical, thenG =
⊔Ai t
⊔Bj =
⊔ai · Ai =
⊔bj · Bj . But we now get that
1 = m(G ) = m(G ) + m(G ) = 2 due to the properties of m.
Other directions are also true but not as easy.
Banach-Tarski Paradox Amenability Orbit Equivalence
Some implications
G amenable=⇒ ∃ left-invariant finitely-additive probability measure on GProof. Take a Følner sequence (Fn) for G and fix an ultrafilter Uon N. Let A ⊆ G . Define m(A) := limU
|A∩Fn||Fn| .
G amenable =⇒ G is not paradoxicalProof. Take a left-invariant finitely additive probability measure mon G . If G were paradoxical, thenG =
⊔Ai t
⊔Bj =
⊔ai · Ai =
⊔bj · Bj . But we now get that
1 = m(G ) = m(G ) + m(G ) = 2 due to the properties of m.
Other directions are also true but not as easy.
Banach-Tarski Paradox Amenability Orbit Equivalence
Some implications
G amenable=⇒ ∃ left-invariant finitely-additive probability measure on GProof. Take a Følner sequence (Fn) for G and fix an ultrafilter Uon N. Let A ⊆ G . Define m(A) := limU
|A∩Fn||Fn| .
G amenable =⇒ G is not paradoxicalProof. Take a left-invariant finitely additive probability measure mon G . If G were paradoxical, thenG =
⊔Ai t
⊔Bj =
⊔ai · Ai =
⊔bj · Bj . But we now get that
1 = m(G ) = m(G ) + m(G ) = 2 due to the properties of m.
Other directions are also true but not as easy.
Banach-Tarski Paradox Amenability Orbit Equivalence
Ergodicity
Probability Measure Preserving
A group action on a standard measure space G y ([0, 1], λ) isprobability measure preserving(p.m.p.) if for every g ∈ G andmeasurable A ⊆ X , the measure λ(A) = λ(g · A).
Ergodic
A p.m.p. group action G y ([0, 1], λ) is ergodic if for everyG-invariant set A ⊆ X , λ(A) = 10 or λ(A) = 1.
Example Irrational translation(fix q ∈ R \Q) by Z,
n · x 7→ x + nq mod 1
Banach-Tarski Paradox Amenability Orbit Equivalence
Ergodicity
Probability Measure Preserving
A group action on a standard measure space G y ([0, 1], λ) isprobability measure preserving(p.m.p.) if for every g ∈ G andmeasurable A ⊆ X , the measure λ(A) = λ(g · A).
Ergodic
A p.m.p. group action G y ([0, 1], λ) is ergodic if for everyG-invariant set A ⊆ X , λ(A) = 10 or λ(A) = 1.
Example Irrational translation(fix q ∈ R \Q) by Z,
n · x 7→ x + nq mod 1
Banach-Tarski Paradox Amenability Orbit Equivalence
Orbit Equivalence
Orbit Equivalence
Two group actions on a standard measure space G ,H y ([0, 1], λ)are Orbit Equivalent if there is a measure space isomorphismf : ([0, 1], λ)→ ([0, 1], λ) such that
f (G · x) = H · f (x) for almost every x ∈ [0, 1]
Dye’s Theorem
If G ,H are amenable, then any 2 free p.m.p. ergodic actions on astandard measure space are orbit equivalent.
So, up to orbit equivalence, there is one action of amenable groups.
Banach-Tarski Paradox Amenability Orbit Equivalence
Orbit Equivalence
Orbit Equivalence
Two group actions on a standard measure space G ,H y ([0, 1], λ)are Orbit Equivalent if there is a measure space isomorphismf : ([0, 1], λ)→ ([0, 1], λ) such that
f (G · x) = H · f (x) for almost every x ∈ [0, 1]
Dye’s Theorem
If G ,H are amenable, then any 2 free p.m.p. ergodic actions on astandard measure space are orbit equivalent.
So, up to orbit equivalence, there is one action of amenable groups.
Banach-Tarski Paradox Amenability Orbit Equivalence
What about nonamenable groups?
Hjorth 2005
Every nonamenable group has at least 2 free p.m.p ergodicnon-orbit equivalent actions.
Epstein 2007
Every nonamenable group has at least c ≥ 2ℵ0 free p.m.p ergodicnon-orbit equivalent actions.
So, nonamenable groups have a lot of actions.
Banach-Tarski Paradox Amenability Orbit Equivalence
What about nonamenable groups?
Hjorth 2005
Every nonamenable group has at least 2 free p.m.p ergodicnon-orbit equivalent actions.
Epstein 2007
Every nonamenable group has at least c ≥ 2ℵ0 free p.m.p ergodicnon-orbit equivalent actions.
So, nonamenable groups have a lot of actions.