The Unitary Representations of Z According to Shelah: “ I have always felt that examples usually just confuse you, having always specific properties that are traps as they do not hold in general.” Theorem (Folklore) The irreducible unitary representations of Z are ϕ z : Z → U 1 ( C ) = T = { c ∈ C : | c | = 1 } where z ∈ T and ϕ z ( k ) = z k . Observation The unitary representation Z � ℓ 2 ( Z ) has no Z -invariant 1-dimensional subspaces. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The Unitary Representations of Z The “multiplicity-free” unitary representations of Z can be parameterized by the probability measures µ on T so that the following are equivalent: (ii) the measures µ , ν have the same null sets; (i) the corresponding unitary representations ϕ µ , ϕ ν are unitarily equivalent. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The Unitary Representations of Z The “multiplicity-free” unitary representations of Z can be parameterized by the probability measures µ on T so that the following are equivalent: (ii) the measures µ , ν have the same null sets; (i) the corresponding unitary representations ϕ µ , ϕ ν are unitarily equivalent. Mackey (1955): it is not clear that these measure equivalence classes can be parameterized by the points of a Polish space. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The Unitary Representations of Z The “multiplicity-free” unitary representations of Z can be parameterized by the probability measures µ on T so that the following are equivalent: (ii) the measures µ , ν have the same null sets; (i) the corresponding unitary representations ϕ µ , ϕ ν are unitarily equivalent. Mackey (1955): it is not clear that these measure equivalence classes can be parameterized by the points of a Polish space. Definition A Polish space is a separable completely metrizable topological space. E.g. R , C , 2 N , N N ,... Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
What is a parameterization? Notation If E is an equivalence relation on the set X , then X / E denotes the set of E -equivalence classes. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
What is a parameterization? Notation If E is an equivalence relation on the set X , then X / E denotes the set of E -equivalence classes. A First Approximation If E is an equivalence relation on the set X, then a parameterization of X / E is an explicit map f : X → Z to a Polish space Z such that f ( x ) = f ( y ) ⇐ ⇒ x E y . Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
What is an explicit map? Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
What is an explicit map? Question Which functions f : R → R are explicit? Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
What is an explicit map? Question Which functions f : R → R are explicit? Church’s Thesis for Real Mathematics EXPLICIT = BOREL Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
What is an explicit map? Question Which functions f : R → R are explicit? Church’s Thesis for Real Mathematics EXPLICIT = BOREL Definition A function f : R → R is Borel if graph ( f ) is a Borel subset of R × R . Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
What is an explicit map? Question Which functions f : R → R are explicit? Church’s Thesis for Real Mathematics EXPLICIT = BOREL Definition A function f : R → R is Borel if graph ( f ) is a Borel subset of R × R . Equivalently, if f − 1 ( A ) is Borel for each Borel subset A ⊆ R . Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
What is a parameterization? Definition (Mackey) An equivalence relation E on a Polish space X is parameterizable or smooth if there exists a Borel map f : X → Z to a Polish space Z such that f ( x ) = f ( y ) ⇐ ⇒ x E y . Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
What is a parameterization? Definition (Mackey) An equivalence relation E on a Polish space X is parameterizable or smooth if there exists a Borel map f : X → Z to a Polish space Z such that f ( x ) = f ( y ) ⇐ ⇒ x E y . Theorem (Jordan 1870) The similarity relation on the space Mat n ( C ) of complex n × n matrices is smooth. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
What is a parameterization? Definition (Mackey) An equivalence relation E on a Polish space X is parameterizable or smooth if there exists a Borel map f : X → Z to a Polish space Z such that f ( x ) = f ( y ) ⇐ ⇒ x E y . Theorem (Jordan 1870) The similarity relation on the space Mat n ( C ) of complex n × n matrices is smooth. Theorem (Freedman 1966) If X is an uncountable Polish space, then the measure equivalence relation on the space M ( X ) of probability measures on X is not smooth. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
A “simple” non-smooth equivalence relation Definition E 0 is the equivalence relation on 2 N defined by: x E 0 y ⇐ ⇒ x n = y n for all but finitely many n . Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
A “simple” non-smooth equivalence relation Definition E 0 is the equivalence relation on 2 N defined by: x E 0 y ⇐ ⇒ x n = y n for all but finitely many n . Notation Let µ be the uniform product probability measure on 2 N . Theorem (Kolmogorov Zero-One Law 1933) If f : 2 N → [ 0 , 1 ] is a Borel map which is constant on E 0 -classes, then f is constant on a µ -measure 1 subset. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
Borel reductions Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
Borel reductions Definition (Friedman-Stanley 1989) Let E, F be equivalence relations on the Polish spaces X, Y. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
Borel reductions Definition (Friedman-Stanley 1989) Let E, F be equivalence relations on the Polish spaces X, Y. E ≤ B F if there exists a Borel map ϕ : X → Y such that x E y ⇐ ⇒ ϕ ( x ) F ϕ ( y ) . In this case, f is called a Borel reduction from E to F. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
Borel reductions Definition (Friedman-Stanley 1989) Let E, F be equivalence relations on the Polish spaces X, Y. E ≤ B F if there exists a Borel map ϕ : X → Y such that x E y ⇐ ⇒ ϕ ( x ) F ϕ ( y ) . In this case, f is called a Borel reduction from E to F. E ∼ B F if both E ≤ B F and F ≤ B E. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
Borel reductions Definition (Friedman-Stanley 1989) Let E, F be equivalence relations on the Polish spaces X, Y. E ≤ B F if there exists a Borel map ϕ : X → Y such that x E y ⇐ ⇒ ϕ ( x ) F ϕ ( y ) . In this case, f is called a Borel reduction from E to F. E ∼ B F if both E ≤ B F and F ≤ B E. E < B F if both E ≤ B F and E ≁ B F. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
Borel reductions Definition (Friedman-Stanley 1989) Let E, F be equivalence relations on the Polish spaces X, Y. E ≤ B F if there exists a Borel map ϕ : X → Y such that x E y ⇐ ⇒ ϕ ( x ) F ϕ ( y ) . In this case, f is called a Borel reduction from E to F. E ∼ B F if both E ≤ B F and F ≤ B E. E < B F if both E ≤ B F and E ≁ B F. Remark In particular, an equivalence relation E is smooth iff E is Borel reducible to the identity relation ∆ Z on some Polish space Z . Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The Glimm-Effos Dichotomy 1961/65 Theorem (Harrington-Kechris-Louveau 1990) If E is a Borel equivalence relation on the Polish space X, then exactly one of the following holds: (i) E is smooth; or (ii) E 0 ≤ B E. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The Glimm-Effos Dichotomy 1961/65 Theorem (Harrington-Kechris-Louveau 1990) If E is a Borel equivalence relation on the Polish space X, then exactly one of the following holds: (i) E is smooth; or (ii) E 0 ≤ B E. Definition An equivalence relation E on a Polish space X is Borel if E is a Borel subset of X × X. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The unitary equivalence relation Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The unitary equivalence relation Let G be a countably infinite group. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The unitary equivalence relation Let G be a countably infinite group. Let H be a fixed infinite dimensional separable Hilbert space and let U ( H ) be the corresponding unitary group. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The unitary equivalence relation Let G be a countably infinite group. Let H be a fixed infinite dimensional separable Hilbert space and let U ( H ) be the corresponding unitary group. Let Irr ∞ ( G ) be the Polish space of irreducible representations ϕ : G → U ( H ) . Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The unitary equivalence relation Let G be a countably infinite group. Let H be a fixed infinite dimensional separable Hilbert space and let U ( H ) be the corresponding unitary group. Let Irr ∞ ( G ) be the Polish space of irreducible representations ϕ : G → U ( H ) . Let ≈ G be the unitary equivalence relation defined on Irr ∞ ( G ) by ( ∃ A ∈ U ( H ) ) ( ∀ g ∈ G ) A ϕ ( g ) A − 1 = ψ ( g ) . ϕ ≈ G ψ ⇐ ⇒ Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The unitary equivalence relation Let G be a countably infinite group. Let H be a fixed infinite dimensional separable Hilbert space and let U ( H ) be the corresponding unitary group. Let Irr ∞ ( G ) be the Polish space of irreducible representations ϕ : G → U ( H ) . Let ≈ G be the unitary equivalence relation defined on Irr ∞ ( G ) by ( ∃ A ∈ U ( H ) ) ( ∀ g ∈ G ) A ϕ ( g ) A − 1 = ψ ( g ) . ϕ ≈ G ψ ⇐ ⇒ Question Is ≈ G a smooth equivalence relation? Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The Glimm-Thoma Theorem Theorem (Glimm & Thoma 1964) If G is a countable group, then the following conditions are equivalent: Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The Glimm-Thoma Theorem Theorem (Glimm & Thoma 1964) If G is a countable group, then the following conditions are equivalent: (i) G is not abelian-by-finite. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The Glimm-Thoma Theorem Theorem (Glimm & Thoma 1964) If G is a countable group, then the following conditions are equivalent: (i) G is not abelian-by-finite. (ii) G has an infinite dimensional irreducible representation. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The Glimm-Thoma Theorem Theorem (Glimm & Thoma 1964) If G is a countable group, then the following conditions are equivalent: (i) G is not abelian-by-finite. (ii) G has an infinite dimensional irreducible representation. (iii) The unitary equivalence relation ≈ G on the space Irr ∞ ( G ) of infinite dimensional irreducible unitary representations of G is not smooth. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The Glimm-Thoma Theorem Theorem (Glimm & Thoma 1964) If G is a countable group, then the following conditions are equivalent: (i) G is not abelian-by-finite. (ii) G has an infinite dimensional irreducible representation. (iii) The unitary equivalence relation ≈ G on the space Irr ∞ ( G ) of infinite dimensional irreducible unitary representations of G is not smooth. Question Does this mean that we should abandon all hope of finding a “satisfactory classification” for the irreducible unitary representations of these groups? Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
When it’s bad, it’s worse ... Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
When it’s bad, it’s worse ... Theorem (Hjorth 1997) If the countable group G is not abelian-by-finite, then the action of U ( H ) on Irr ∞ ( G ) is turbulent. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
When it’s bad, it’s worse ... Theorem (Hjorth 1997) If the countable group G is not abelian-by-finite, then the action of U ( H ) on Irr ∞ ( G ) is turbulent. Remarks This is a much more serious obstruction to the existence of a “satisfactory classification” of the irreducible unitary representations of G . Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
When it’s bad, it’s worse ... Theorem (Hjorth 1997) If the countable group G is not abelian-by-finite, then the action of U ( H ) on Irr ∞ ( G ) is turbulent. Remarks This is a much more serious obstruction to the existence of a “satisfactory classification” of the irreducible unitary representations of G . But hopefully this is not the end of the story ... Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
An aside: Classifying homeomorphisms of [ 0 , 1 ] Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
An aside: Classifying homeomorphisms of [ 0 , 1 ] Definition Hom + ([ 0 , 1 ]) is the group of homeomorphisms ϕ : [ 0 , 1 ] → [ 0 , 1 ] satisfying ϕ ( 0 ) = 0 and ϕ ( 1 ) = 1 . Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
An aside: Classifying homeomorphisms of [ 0 , 1 ] Definition Hom + ([ 0 , 1 ]) is the group of homeomorphisms ϕ : [ 0 , 1 ] → [ 0 , 1 ] satisfying ϕ ( 0 ) = 0 and ϕ ( 1 ) = 1 . Two maps ϕ , ψ ∈ Hom + ([ 0 , 1 ]) are conjugate if there exists θ ∈ Hom + ([ 0 , 1 ]) such that ψ = θ ◦ ϕ ◦ θ − 1 . θ [ 0 , 1 ] − − − − → [ 0 , 1 ] ϕ � ψ � θ [ 0 , 1 ] − − − − → [ 0 , 1 ] Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
An aside: Classifying homeomorphisms of [ 0 , 1 ] Definition Hom + ([ 0 , 1 ]) is the group of homeomorphisms ϕ : [ 0 , 1 ] → [ 0 , 1 ] satisfying ϕ ( 0 ) = 0 and ϕ ( 1 ) = 1 . Two maps ϕ , ψ ∈ Hom + ([ 0 , 1 ]) are conjugate if there exists θ ∈ Hom + ([ 0 , 1 ]) such that ψ = θ ◦ ϕ ◦ θ − 1 . θ [ 0 , 1 ] − − − − → [ 0 , 1 ] ϕ � ψ � θ [ 0 , 1 ] − − − − → [ 0 , 1 ] Problem Classify the elements of Hom + ([ 0 , 1 ]) up to conjugacy by “discrete invariants”. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The Bump Structure Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The Bump Structure Definition A bump of ϕ ∈ Hom + ([ 0 , 1 ]) is a maximal open interval I ⊂ [ 0 , 1 ] such that one of the following conditions hold: (a) ϕ ( x ) > x for all x ∈ I. (b) ϕ ( x ) < x for all x ∈ I. (c) ϕ ( x ) = x for all x ∈ I. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The Bump Structure Definition � � A bump of ϕ ∈ Hom + ([ 0 , 1 ]) is a � maximal open interval I ⊂ [ 0 , 1 ] � such that one of the following � � conditions hold: � � (a) ϕ ( x ) > x for all x ∈ I. � � (b) ϕ ( x ) < x for all x ∈ I. (c) ϕ ( x ) = x for all x ∈ I. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The Bump Structure Definition � � A bump of ϕ ∈ Hom + ([ 0 , 1 ]) is a � maximal open interval I ⊂ [ 0 , 1 ] � such that one of the following � � conditions hold: � � (a) ϕ ( x ) > x for all x ∈ I. � � (b) ϕ ( x ) < x for all x ∈ I. I 1 (c) ϕ ( x ) = x for all x ∈ I. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The Bump Structure Definition � � A bump of ϕ ∈ Hom + ([ 0 , 1 ]) is a � maximal open interval I ⊂ [ 0 , 1 ] � such that one of the following � � conditions hold: � � (a) ϕ ( x ) > x for all x ∈ I. � � (b) ϕ ( x ) < x for all x ∈ I. I 1 I 2 (c) ϕ ( x ) = x for all x ∈ I. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The Bump Structure Definition � � A bump of ϕ ∈ Hom + ([ 0 , 1 ]) is a � maximal open interval I ⊂ [ 0 , 1 ] � such that one of the following � � conditions hold: � � (a) ϕ ( x ) > x for all x ∈ I. � � (b) ϕ ( x ) < x for all x ∈ I. I 1 I 2 I 3 (c) ϕ ( x ) = x for all x ∈ I. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The Bump Structure Definition � � A bump of ϕ ∈ Hom + ([ 0 , 1 ]) is a � maximal open interval I ⊂ [ 0 , 1 ] � such that one of the following � � conditions hold: � � (a) ϕ ( x ) > x for all x ∈ I. � � (b) ϕ ( x ) < x for all x ∈ I. I 1 I 2 I 3 I 4 (c) ϕ ( x ) = x for all x ∈ I. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The Bump Structure Definition � � A bump of ϕ ∈ Hom + ([ 0 , 1 ]) is a � maximal open interval I ⊂ [ 0 , 1 ] � such that one of the following � � conditions hold: � � (a) ϕ ( x ) > x for all x ∈ I. � � (b) ϕ ( x ) < x for all x ∈ I. I 1 I 2 I 3 I 4 I 5 (c) ϕ ( x ) = x for all x ∈ I. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
The Bump Structure Definition � � A bump of ϕ ∈ Hom + ([ 0 , 1 ]) is a � maximal open interval I ⊂ [ 0 , 1 ] � such that one of the following � � conditions hold: � � (a) ϕ ( x ) > x for all x ∈ I. � � (b) ϕ ( x ) < x for all x ∈ I. I 1 I 2 I 3 I 4 I 5 (c) ϕ ( x ) = x for all x ∈ I. Theorem (Folklore) Two maps ϕ , ψ ∈ Hom + ([ 0 , 1 ]) are conjugate iff the corresponding colored linear orders are isomorphic. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
Classifying homeomorphisms of [ 0 , 1 ] 2 Definition Hom + ([ 0 , 1 ] 2 ) is the group of homeomorphisms ϕ : [ 0 , 1 ] 2 → [ 0 , 1 ] 2 satisfying ϕ ( v ) = v for each v ∈ { ( 0 , 0 ) , ( 0 , 1 ) , ( 1 , 0 ) , ( 1 , 1 ) } . Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
Classifying homeomorphisms of [ 0 , 1 ] 2 Definition Hom + ([ 0 , 1 ] 2 ) is the group of homeomorphisms ϕ : [ 0 , 1 ] 2 → [ 0 , 1 ] 2 satisfying ϕ ( v ) = v for each v ∈ { ( 0 , 0 ) , ( 0 , 1 ) , ( 1 , 0 ) , ( 1 , 1 ) } . Problem Classify the elements of Hom + ([ 0 , 1 ] 2 ) up to conjugacy by “discrete invariants”. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
Classifying homeomorphisms of [ 0 , 1 ] 2 Definition Hom + ([ 0 , 1 ] 2 ) is the group of homeomorphisms ϕ : [ 0 , 1 ] 2 → [ 0 , 1 ] 2 satisfying ϕ ( v ) = v for each v ∈ { ( 0 , 0 ) , ( 0 , 1 ) , ( 1 , 0 ) , ( 1 , 1 ) } . Problem Classify the elements of Hom + ([ 0 , 1 ] 2 ) up to conjugacy by “discrete invariants”. Perhaps there exists a “blister structure”? Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
Classifying homeomorphisms of [ 0 , 1 ] 2 Definition Hom + ([ 0 , 1 ] 2 ) is the group of homeomorphisms ϕ : [ 0 , 1 ] 2 → [ 0 , 1 ] 2 satisfying ϕ ( v ) = v for each v ∈ { ( 0 , 0 ) , ( 0 , 1 ) , ( 1 , 0 ) , ( 1 , 1 ) } . Problem Classify the elements of Hom + ([ 0 , 1 ] 2 ) up to conjugacy by “discrete invariants”. Perhaps there exists a “blister structure”? Theorem (Hjorth 1997) The conjugacy relation on Hom + ([ 0 , 1 ] 2 ) does not admit classification by “discrete invariants” ... Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
Classifying homeomorphisms of [ 0 , 1 ] 2 Definition Hom + ([ 0 , 1 ] 2 ) is the group of homeomorphisms ϕ : [ 0 , 1 ] 2 → [ 0 , 1 ] 2 satisfying ϕ ( v ) = v for each v ∈ { ( 0 , 0 ) , ( 0 , 1 ) , ( 1 , 0 ) , ( 1 , 1 ) } . Problem Classify the elements of Hom + ([ 0 , 1 ] 2 ) up to conjugacy by “discrete invariants”. Perhaps there exists a “blister structure”? Theorem (Hjorth 1997) The conjugacy relation on Hom + ([ 0 , 1 ] 2 ) does not admit classification by “discrete invariants” ... because the conjugacy relation is turbulent. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
When it’s bad, it’s worse ... Theorem (Hjorth 1997) If the countable group G is not abelian-by-finite , then the action of U ( H ) on Irr ∞ ( G ) is turbulent. Remarks This rules out the existence of a “satisfactory classification” of the irreducible unitary representations of G . But hopefully this is not the end of the story ... Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
When it’s bad, it’s worse ... Theorem (Hjorth 1997) If the countable group G is not abelian-by-finite , then the action of U ( H ) on Irr ∞ ( G ) is turbulent. Open Question (Thomas 2011) Do there exist countable groups G , H such that (i) G , H are not abelian-by-finite; and (ii) ≈ G , ≈ H are not Borel bireducible? Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
When it’s bad, it’s worse ... Theorem (Hjorth 1997) If the countable group G is not abelian-by-finite , then the action of U ( H ) on Irr ∞ ( G ) is turbulent. Open Question (Thomas 2011 & Effros 2007) Do there exist countable groups G , H such that (i) G , H are not abelian-by-finite; and (ii) ≈ G , ≈ H are not Borel bireducible? Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
When it’s bad, it’s worse ... Theorem (Hjorth 1997) If the countable group G is not abelian-by-finite , then the action of U ( H ) on Irr ∞ ( G ) is turbulent. Open Question (Thomas 2011 & Effros 2007 & Dixmier 1967) Do there exist countable groups G , H such that (i) G , H are not abelian-by-finite; and (ii) ≈ G , ≈ H are not Borel bireducible? Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
Dixmier’s Question Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
Dixmier’s Question Question (Dixmier 1967) Do there exist countable groups G , H such that (i) G , H are not abelian-by-finite; and (ii) the unitary duals Irr ∞ ( G ) / ≈ G and Irr ∞ ( H ) / ≈ H are not Borel isomorphic? Definition If E, F are equivalence relations on the Polish spaces X, Y, then X / E, Y / F are Borel isomorphic if there exist Borel maps ϕ : X → Y and ψ : Y → X which induce mutually inverse bijections ϕ , ˆ ˆ ψ between X / E and Y / F. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
Borel bireducibility vs Borel isomorphism Theorem (Motto Ros 2012) If E, F are Borel orbit equivalence relations of actions of Polish groups on the Polish spaces X, Y, then the following are equivalent: E and F are Borel bireducible. X / E and Y / F are Borel isomorphic. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
Borel bireducibility vs Borel isomorphism Theorem (Motto Ros 2012) If E, F are Borel orbit equivalence relations of actions of Polish groups on the Polish spaces X, Y, then the following are equivalent: E and F are Borel bireducible. X / E and Y / F are Borel isomorphic. Corollary If G, H are countable groups, then the following are equivalent: The unitary equivalence relations ≈ G and ≈ H are Borel bireducible. The unitary duals Irr ∞ ( G ) / ≈ G and Irr ∞ ( H ) / ≈ H are Borel isomorphic. Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
Four test cases Question What can we say about the relative complexity of the unitary duals of the following groups? Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
Four test cases Question What can we say about the relative complexity of the unitary duals of the following groups? The free group F ∞ on infinitely many generators Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
Four test cases Question What can we say about the relative complexity of the unitary duals of the following groups? The free group F ∞ on infinitely many generators SL ( 3 , Z ) Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
Four test cases Question What can we say about the relative complexity of the unitary duals of the following groups? The free group F ∞ on infinitely many generators SL ( 3 , Z ) The infinite alternating group Alt ( N ) Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
Four test cases Question What can we say about the relative complexity of the unitary duals of the following groups? The free group F ∞ on infinitely many generators SL ( 3 , Z ) The infinite alternating group Alt ( N ) The direct sum Σ of infinitely copies of Sym ( 3 ) Simon Thomas (Rutgers) Lezione Lagrangiana 26th May 2016
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