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i∃x ∈ [x0 ]G ∃∗ g ∈ G (g · x ∈ Nλ(u,v) ∩ Ns(i ) ).

Granting such a sequence, let x = i s(i). Then ∀u ∈ P ∃v∃i∀i > i (Ns(i ) ∩ Nλ(u,v) = ∅). It follows that ∀u ∈ P ∃v (x is a limit point of Nλ(u,v) ). Since Nλ(u,v) is closed, we have that ∀u ∈ P ∃v (x ∈ Nλ(u,v) ), and hence x ∈ [x0 ]G . To define the sequence we use induction and simultaneously define a sequence c(i) of Borel codes for closed sets. To begin with let s(0) = ∅ and Bc(0) = X. In general suppose s(i) and c(i) have been defined and suppose Γ(s(i), c(i)) is the winning strategy for Player II in the game G(s(i), c(i)) defined in the preceding paragraph. Let Player I play ui and apply Γ(s(i), c(i)) to obtain s(i + 1)  s(i) as a part of Player II’s response. Let v be the other part of Player II’s moves. Then let Bc(i+1) = Bc ∩Nλ(ui ,v) . To verify that this is as required, let u ∈ P . Let ui = u and v be played by Player II according to the construction. Then for all i > i, Bc(i ) ⊆ Nλ(u,v) . Since Γ(s(i), c(i)) is winning for Player II, we have that ∃x ∈ [x0 ]G ∃∗ g ∈ G (g ·x ∈ Nλ(u,v) ∩Ns(i ) ). This finishes the proof of the theorem.

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Corollary 5.2.4 Let G be a Polish group, X a Borel G-space, Y a standard Borel space, and F a Borel equivalence relation on Y . Suppose f : X → Y is a Borel reduction X X to F such that [f (X)]F = Y . Then EG ∼B F . from EG Proof. The above proof gives rise to a Borel function g : Y → X such that for any y ∈ [f (X)]F , letting x = g(y), then f (x)F y. Note that the definition of x does not depend on the choice of x0 . It is easy to see that g is a reduction X from F to EG . Exercise 5.2.1 Show that the Borel reduction constructed in Corollary 5.2.4 is also faithful. Exercise 5.2.2 Let E, F be Borel orbit equivalence relations on Polish spaces X, Y , respectively. Show that if E ≤B F then there is an invariant Borel set Z ⊆ Y such that E ∼B F  Z.

5.3

Perfect set theorems for equivalence relations

In this and the next sections we study Borel reductions involving the identity relations. Recall from Proposition 5.1.12 that reduction from the identity relation on 2ω is equivalent to the existence of perfectly many equivalence classes. In this section we give various sufficient conditions for there being perfectly many equivalence classes for an equivalence relation on a Polish space. Again category arguments are playing the vital role in this study. Theorem 5.3.1 (Mycielski) Let X be a Polish space and E a meager equivalence relation on X. Then id(2ω ) c E. Proof. First note that X is perfect. In fact, otherwise X must contain at least one isolated point x, and any equivalence relation on X must contain the open set {(x, x)}, and therefore is nonmeager.

Let d < 1 be a compatible complete metric on X. Let E ⊆ n Fn with Fn ⊆ X 2 closed and nowhere dense for all n ∈ ω. We may assume that id(X) ⊆ F0 and Fn ⊆ Fn+1 for all n ∈ ω. We define a sequence (Us )s∈2<ω of open sets in X by induction on lh(s) such that, for all s ∈ 2<ω , (i) diam(Us ) ≤ 2−lh(s) ; (ii) Us 0 , Us 1 ⊆ Us ; Us 0 ∩ Us 1 = ∅;

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(iii) for all s, t ∈ 2<ω with lh(s) = lh(t) = n+1 and s = t, (Us ×Ut )∩Fn = ∅. To begin let U∅ = X. Assume now that for all s with lh(s) = n the set Us has been defined. We define all Us 0 and Us 1 . For notational convenience we enumerate all Us for lh(s) = n as U1 , . . . , Ui , . . . , Uk with 1 ≤ i ≤ k = 2n . Since X is perfect there are disjoint open subsets Vi , Vk+i ⊆ Ui for all 1 ≤ i ≤ k with diam(Vi ), diam(Vk+i ) ≤ 2−n−1 , and by regularity we may assume that Vi , Vk+i ⊆ Ui and Vi ∩Vk+i = ∅. Thus we have obtained a collection {Vi }1≤i≤2k of disjoint open sets satisfying (i) and (ii). To guarantee (iii) we make the following observation. Since Fn is nowhere dense, for any disjoint nonempty open sets W1 and W2 , W1 × W2 ⊆ Fn . It follows that there are nonempty W1 ⊆ W1 and W2 ⊆ W2 with (W1 × W2 ) ∩ Fn = ∅. Applying this observation successively to all pairs (i, j) with 1 ≤ i, j ≤ 2k and i = j, we obtain open sets Vi ⊆ Vi such that, for all such pairs (i, j), (Vi × Vj ) ∩ Fn = ∅. Now condition  (iii) is satisfied if we let Us 0 = Vi and Us 1 = Vk+i where Ui = Us . ω For u ∈ 2 we now let f (u) to be the unique element of the set n Uun . Then f is a continuous function from 2ω into X. To finish the proof of the theorem we only need to verify that (f (u), f (v)) ∈ E for u = v. For this let u, v ∈ 2ω with u = v. Let n be the least integer with u(n) = v(n). Then for all m > n, u  m = v  m. By (iii), (Uum × Uvm ) ∩ Fm = ∅. Thus (f (u), f (v)) ∈ Fm for all m > n. Therefore (f (u), f (v)) ∈ E as required. Corollary 5.3.2 Let X be a Polish space and E an equivalence relation on X. If there is a nonempty open set U ⊆ X such that E is meager on U × U , then id(2ω ) c E. Corollary 5.3.3 Let X be a Polish space and E an equivalence relation on X. Suppose E has the Baire property. If every E-equivalence class is meager, then there are perfectly many E-equivalence classes. Proof. The condition can be stated as ∀x ∀∗ y (x, y) ∈ E. By Kuratowski– Ulam Theorem 3.2.1 it follows that E is meager. Another corollary of Theorem 5.3.1 concerning the Gandy–Harrington topology is going to play an important role in the proof of Silver’s theorem later in this section. Corollary 5.3.4 Let τ be the Gandy–Harrington topology on the Baire space ω ω and E an equivalence relation on ω ω . If there is a Σ11 set V ⊆ ω ω such that E is τ × τ -meager in V × V , then there are perfectly many E-equivalence classes. Proof. Let Y = Xlow be the Polish space of low elements with the Gandy– Harrington topology (Theorem 1.8.5). By the Gandy Basis Theorem 1.8.4

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U = Y ∩ V is a nonempty open subset of Y . Now E is meager on U × U , and thus by Corollary 5.3.2 there is a continuous reduction f : 2ω → Y witnessing id(2ω )  E. Since τ is finer than the usual topology f as a function from 2ω into ω ω is still continuous. Thus there are perfectly many E-equivalence classes. The following theorem, often referred to as the Silver dichotomy theorem, is the first of a series of dichotomy theorems undoubtedly occupying the center of the invariant descriptive set theory of equivalence relations. It is also important because of its proof presented here, due to Harrington, which is an elegant example of the use of effective descriptive set theoretic methods in establishing an important result of classical character. Theorem 5.3.5 (Silver) Let X be a standard Borel space and E a coanalytic equivalence relation on X. Then either there are countably many E-equivalence classes or there are perfectly many E-equivalence classes. Proof. Without loss of generality we may assume X = ω ω . By relativization we also assume that E is Π11 . Let τ be the Gandy–Harrington topology on ω ω . First we define V = {x ∈ X : there is no Δ11 set U such that x ∈ U ⊆ [x]E }. We claim that if V = ∅ then there are only countably many E-equivalence classes. This is because, if V = ∅, then every E-equivalence class contains some nonempty Δ11 set. Since there are only countably many Δ11 sets, it follows that there are only countably many E-equivalence classes. For the rest of the proof we assume V = ∅. We claim that V ∈ Σ11 . Indeed, x ∈ V ⇐⇒ ∀U ∈ Δ11 ( x ∈ U → ∃y ∈ U (x, y) ∈ E ). With the coding of Δ11 sets given by Theorem 1.7.4 we get that x ∈ V ⇐⇒ ∀n [(n ∈ D ∧ x ∈ Pn+ ) → ∃y (y ∈ Pn− ∧ (x, y) ∈ E)]. This shows that V ∈ Σ11 . To finish the proof we claim that E is τ × τ -meager in V × V . Then by the above corollary there are perfectly many E-equivalence classes. We establish the claim in several steps. As the first step, we check that for every x ∈ V there is no Σ11 set U such that x ∈ U ⊆ [x]E . To see this, assume toward a contradiction that x ∈ V and U ∈ Σ11 and x ∈ U ⊆ [x]E . Then note that [x]E is Π11 since y ∈ [x]E ⇐⇒ ∀z(z ∈ U → (y, z) ∈ E). Now by the separation property for Σ11 sets (see Theorem 1.7.1 and the remarks following it) there is W ∈ Δ11 such that U ⊆ W ⊆ [x]E , thus x ∈ V .

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From this it follows immediately that every nonempty Σ11 subset U of V meets more than one E-equivalence class. Next we note that every set involved has the Baire property in the Gandy– Harrington topology or its products. Specifically, the equivalence relation E as a subset of ω ω × ω ω has the Baire property in τ × τ , and each equivalence class [x]E has the Baire property in τ . This is because, by a theorem of Nikodym (see the remarks preceeding Proposition 2.3.1), the collection of all sets with the Baire property in any topological space is closed under the Suslin operation. Now both E and [x]E are Π11 , and therefore coanalytic, and therefore their complements are results of the Suslin operation applied to sequences of closed sets in the usual topology of ω ω . Since the Gandy–Harrington topology is coarser, they continue to be the results of Suslin operation, and thus have the Baire property in the sense of τ . This allows us to use the Kuratowski– Ulam Theorem 3.2.1. Now to show that E is τ × τ -meager in V × V , by the Kuratowski–Ulam theorem it suffices to show that for all x ∈ V , [x]E is τ -meager in V . Thus it suffices to show that for all x ∈ V , [x]E is not τ -comeager in any U ⊆ V where U ∈ Σ11 . Again toward a final contradiction assume that for some x ∈ V and U ⊆ V , U ∈ Σ11 , [x]E is τ -comeager in U . Now by Louveau’s lemma Exercise 1.8.3, if [x]E is τ -comeager in U ∈ Σ11 , then [x]E × [x]E is comeager in U × U in the Gandy–Harrington topology of ω ω × ω ω . Since (U × U ) ∩ (ω ω × ω ω − E) is nonempty and Σ11 , we have that ([x]E × [x]E ) ∩ (U × U ) ∩ (ω ω × ω ω − E) = ∅, which is a contradiction. Exercise 5.3.1 Let X be a Polish space and E an equivalence relation on X with the Baire property. Suppose every E-equivalence class is countable. Show that there are perfectly many E-equivalence classes. Exercise 5.3.2 Let X be a Polish space, U ⊆ X open, and E an equivalence relation on X with the Baire property. Suppose τ is a strong Choquet topology on X finer than the Polish topology. Show that if E is τ × τ -meager on U × U , then there are perfectly many E-equivalence classes. Exercise 5.3.3 Let E be a Π11 equivalence relation on ω ω . Show that the following are equivalent: (i) There are only countably many E-equivalence classes. (ii) For every x ∈ ω ω there exists a Δ11 set U with x ∈ U ⊆ [x]E . (iii) For every x ∈ ω ω there exists a Σ11 set U with x ∈ U ⊆ [x]E . (iv) Every E-equivalence class is Π11 .

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Exercise 5.3.4 Let E be a Π11 equivalence relation on ω ω with only countably many classes. Show that E is a union of countably many Δ11 relations. In particular E is Borel.

5.4

Smooth equivalence relations

Definition 5.4.1 Let E be an equivalence relation on a standard Borel space X. We call E smooth or concretely classifiable if E ≤B id(2ω ). It is clear from the definition that smooth equivalence relations are Borel. Since any two uncountable standard Borel spaces are Borel isomorphic, E is smooth iff E ≤B id(X) for any uncountable Polish space X. Among the countable Polish spaces we consider ω with the discrete topology and let id(ω) denote its identity relation. The following is an immediate corollary of the Silver dichotomy theorem. Theorem 5.4.2 Let E be a smooth equivalence relation. Then exactly one of the following holds: (I) E ∼B id(2ω ); (II) E ≤B id(ω). Proof. By the Silver dichotomy theorem either there are perfectly many E-equivalence classes or else there are only countably many E-equivalence classes. In the first case we have both id(2ω ) ≤B E and, by smoothness, E ≤B id(2ω ). Hence E ∼B id(2ω ). For the second case let C0 , C1 , . . . enumerate all the E-equivalence classes. Then the function θ defined by θ(x) = n iff x ∈ Cn gives a Borel reduction from E to id(ω). The converse of this theorem is trivial. Easily we obtain a complete classification of smooth equivalence relations in terms of Borel bireducibility (see Exercise 5.4.2). Definition 5.4.3 Let E be an equivalence relation on a space X and S a family of subsets of X. We say that E is generated by S if for any x, y ∈ X, xEy ⇐⇒ for any S ∈ S, x ∈ S iff y ∈ S. In this situation S is also called a generating family or a separating family for E. If X is a Borel space then a generating family S is called Borel if every element of S is Borel.

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Proposition 5.4.4 Let E be an equivalence relation on a standard Borel space X. Then E is smooth iff there is a countable Borel generating family for E. Proof. Let S = {S0 , S1 , . . . , Sn , . . . } be a countable Borel generating family for E. Define θ : X → 2ω by θ(x)(n) = 1 ⇐⇒ x ∈ Sn . Then θ is a Borel reduction from E to id(2ω ). Conversely, suppose θ : X → 2ω is a Borel reduction. Then define Sn = {z ∈ 2ω : θ(z)(n) = 1}. We have that S = {S0 , S1 , . . . , Sn , . . . } is a countable Borel generating family for E. Corollary 5.4.5 Let E be a smooth equivalence relation on a Polish space X. Then there is a finer Polish topology σ on X such that E is closed as a subset of (X 2 , σ × σ). Proof. Let S0 , S1 , . . . , Sn , . . . enumerate a countable Borel generating family for E. By Theorem 4.2.3 there is a finer Polish topology σ on X such that each of Sn is σ-clopen. Now xEy ⇐⇒ ∀n ( x ∈ Sn ⇐⇒ y ∈ Sn ), and thus E is closed in (X 2 , σ × σ). Next we give some sufficient conditions for smoothness. First we show that closed equivalence relations are smooth. To do this we establish the following useful general lemma which is reminiscent of the Luzin separation theorem for Σ11 sets. Lemma 5.4.6 Let E be a Σ11 equivalence relation on a standard Borel space X. Let A, B ⊆ X be disjoint Σ11 E-invariant sets. Then there is an E-invariant Borel set C with A ⊆ C and B ∩ C = ∅. Proof. Let A0 = A and C0 be a Borel set separating A0 and B by the Luzin separation theorem. That is, A0 ⊆ C0 and B ∩ C0 = ∅. Let A1 = [C0 ]E . Then A1 is Σ11 and A1 ∩ B = ∅. By the separation theorem again let C1 be a Borel set separating A1 and B. Continue this way, and we obtain A = A0 ⊆ C0 ⊆ A1 ⊆ C1 ⊆ A2 ⊆ . . . Cn ⊆ An ⊆ . . . 1 where Cn is Borel

and An is E-invariant, Σ1 and disjoint from B. Let now C = n Cn = n An . Then C is invariant Borel, A ⊆ C, and B ∩ C = ∅ as required.

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Proposition 5.4.7 Let E be a closed equivalence relation on a Polish space X. Then E is smooth.

Proof. Since E is closed we can write X 2 − E = n Un × Vn where Un , Vn are basic open sets. Let An = [Un ]E and Bn = [Vn ]E . Then we have that An and Bn are disjoint Σ11 E-invariant sets. By Lemma 5.4.6 there is E-invariant Borel Cn such that An ⊆ Cn and Bn ∩ Cn = ∅. We claim that {Cn : n ∈ ω} is a generating family for E, and thus E is smooth. To see this, let x, y ∈ X be such that x ∈ Cn ⇐⇒ y ∈ Cn for all n ∈ ω. Assume (x, y) ∈ E. Then for some n ∈ ω we have that x ∈ Un and y ∈ Vn . However, this implies that x ∈ An ⊆ Cn and y ∈ Bn ⊆ X − Cn , a contradiction. In the next chapter we will show that any Borel equivalence relation whose equivalence classes are Gδ is smooth. For now we turn to the following easy fact. Proposition 5.4.8 Let E be an equivalence relation on a standard Borel space. If E has a Borel selector then E is smooth. Proof.

The selector function is a reduction.

Note that not every closed equivalence relation has a Borel selector (Exercise 5.4.6), hence the converse of the above proposition is not true. Next we give a characterization for the existence of Borel selectors for smooth equivalence relations. Definition 5.4.9 Let X be a standard Borel space and E an equivalence relation on X. We call E idealistic if there is an assignment C → IC that associates with each E-equivalence class C a σ-ideal IC of subsets of C such that (i) C ∈ IC ; (ii) for each Borel set A ⊆ X 2 the set AI defined by x ∈ AI ⇐⇒ {y ∈ [x] : (x, y) ∈ A} ∈ I[x] is Borel. Clause (ii) in the above definition is often referred to as Borelness of the map C → IC . The following proposition helps motivate the definition. Proposition 5.4.10 X is idealistic. Let G be a Polish group and X a Borel G-space. Then EG

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Let x ∈ X and C = [x]G . Then define IC on C by S ∈ IC ⇐⇒ {g ∈ G : g · x ∈ S} is meager in G.

It is easy to check that IC is a σ-ideal and C ∈ IC . To see that C → IC is Borel, let A ⊆ X 2 be a Borel set. Then x ∈ AI ⇐⇒ {y ∈ [x] : (x, y) ∈ A} ∈ I[x] ⇐⇒ {g ∈ G : (x, g · x) ∈ A} is meager in G ⇐⇒ ∀∗ g ∈ G (x, g · x) ∈ A, hence AI is Borel. Theorem 5.4.11 (Kechris) Let E be an equivalence relation on a standard Borel space X. Then E has a Borel selector iff E is smooth and idealistic.

Proof. (⇒) Let s be a Borel selector for E. We show that E is idealistic. For any x ∈ X and C = [x]E , define IC on C by S ∈ IC ⇐⇒ s(x) ∈ S. Then IC is a σ-ideal and C ∈ IC . To see that C → IC is Borel, let A ⊆ X 2 be a Borel set. Then x ∈ AI ⇐⇒ {y ∈ [x] : (x, y) ∈ A} ∈ I[x] ⇐⇒ (x, s(x)) ∈ A is Borel. (⇐) Suppose E is smooth and idealistic. Let f : X → 2ω witness that E ≤B 2ω . By Theorem 4.2.3 we may assume X is Polish and f is continuous. Let F be the graph of f , that is, (x, y) ∈ F ⇐⇒ f (x) = y. Then F ⊆ X × 2ω is closed. By the usual argument we can obtain a sequence

(Fs )s∈ω<ω of nonempty closed subsets of X × 2ω so that F∅ = F , Fs = n∈ω Fs n , and diam(Fs ) < 2−lh(s) . For each y ∈ 2ω and s ∈ ω <ω , let Fsy = {x ∈ X : (x, y) ∈ Fs }. Then for any E-equivalence class C = [x]E , if f (x) = y then F∅y = C and for any

<ω y s ∈ ω , Fs = n∈ω Fsy n . Now let C → IC witness that E is idealistic. Then F∅y ∈ IC , and moreover if Fsy ∈ IC then for some n ∈ ω, Fsy n ∈ IC . Thus if we define T y = {s ∈ ω <ω : Fsy ∈ IC }, then T y is a tree on ω. Let g(y) be the leftmost branch of T y . It is straightforward to check that g is y a Borel function. Furthermore, n∈ω Fg(y)n is a singleton, and we name its unique element h(y). It is clear that (h(y), y) ∈ F and therefore h(y) ∈ C. Let H be the set of all h(y) thus obtained. It is clear that H is a transversal for E on X. To finish the proof it suffices to show that H is Borel. For this, simply note that y ). x ∈ H ⇐⇒ ∀n ∈ ω (x ∈ Fg(y)n

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Corollary 5.4.12 (Burgess) X Let G be a Polish group and X a Borel G-space. If EG is smooth then it has a Borel selector. Exercise 5.4.1 Show that for any equivalence relation E on a Polish space, id(ω) ≤B E iff there are infinitely many E-equivalence classes. Exercise 5.4.2 Show that if an equivalence relation E is smooth then either E ∼B id(2ω ), or E ∼B id(ω), or for some finite n ∈ ω, E ∼B id(n), where id(n) denotes the identity relation on any finite set of size n. Exercise 5.4.3 Let E be an equivalence relation on a Polish space X. Show that E is smooth iff there is a finer Polish topology σ on X such that E is closed in (X 2 , σ × σ). Exercise 5.4.4 Show that an intersection of countably many smooth equivalence relations on a Polish space is smooth. Exercise 5.4.5 Let G be a compact Polish group and X a Borel G-space. X is smooth. Show that EG Exercise 5.4.6 Let F ⊆ ω ω × ω ω be closed so that {x : ∃y (x, y) ∈ F } is Σ11 but not Borel. Let X = F and define the equivalence relation E on X by (x1 , y1 )E(x2 , y2 ) ⇐⇒ x1 = x2 . Show that E is closed but has no Borel selector. Exercise 5.4.7 Show that the orbit equivalence relation induced by the conjugacy action of S∞ on itself is smooth. Exercise 5.4.8 Let G be a Polish group and X be a Polish G-space. Suppose X X is Gδ . Show that EG is smooth. (Hint: The map x → [x]G every orbit of EG is a Borel reduction.)

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Chapter 6 The Glimm–Effros Dichotomy

The Glimm–Effros dichotomy is undoubtedly the most important theoretical result that helped form the subject of invariant descriptive set theory. In the context of operator algebra Glimm and Effros have obtained the dichotomy for locally compact group actions and more generally Fσ orbit equivalence relations. Nowadays the terminology usually refers to the remarkable theorem for all Borel equivalence relations proved by Harrington, Kechris, and Louveau. The theorem and its proof were so influential that for a while the main activity of the field was to prove new dichotomy theorems. For orbit equivalence relations the Glimm–Effros dichotomy have been obtained by Solecki, Hjorth, Becker, and others. Another remarkable connection with other fields of mathematics is the study of hyperfinite equivalence relations. To this date there are still intriguing open problems around which exciting research is actively going on.

6.1

The equivalence relation E0

Definition 6.1.1 The equivalence relation E0 is the relation of eventual agreement on 2ω , that is, it is defined by xE0 y ⇐⇒ ∃m ∀n ≥ m x(n) = y(n). We also consider the eventual agreement relation on ω ω , denoted by E0 (ω). Proposition 6.1.2 E0 ∼B E0 (ω). Proof. The identity embedding from 2ω into ω ω is a reduction from E0 to E0 (ω). To define a reduction f to witness E0 (ω) ≤B E0 , fix a computable bijection ·, · : ω × ω → ω. Given x ∈ ω ω , let f (x) ∈ 2ω be defined by f (x)(n, k) = the (k + 1)-th least significant digit of the binary expansion of x(n).

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f is actually continuous, even computable. To see that it is a desired reduction, assume that xE0 (ω)y. Suppose x(n) = y(n) for n ≥ m. Then for all n ≥ m and k ∈ ω, f (x)(n, k) = f (y)(n, k). For n < m, since the binary expansions of x(n) and y(n) are of finite lengths, there is kn such that for all k ≥ kn , f (x)(n, k) = 0 = f (y)(n, k). Thus for any n, k ∈ ω, if f (x)(n, k) = f (y)(n, k) we must have that n < m and k < kn , and there are only finitely many such pairs n, k. This shows that f (x)E0 f (y). Conversely, if x E0 y, then the set A = {n ∈ ω : x(n) = y(n)} is infinite. For each n ∈ A, let kn be such that the (kn + 1)-th least significant digits of binary expansions of x(n) and y(n) differ. Then we have that f (x)(n, kn ) = f (y)(n, kn ) for all n ∈ A. This means that f (x) and f (y) are not E0 (ω)equivalent. Definition 6.1.3 The Vitali equivalence relation Ev is the equivalence relation on R defined by xEv y ⇐⇒ x − y ∈ Q. The terminology is motivated by the theorem of Vitali in real analysis that any transversal of Ev is not Lebesgue measurable. Proposition 6.1.4 E0 ∼B Ev . Proof.

We first show that Ev ≤B E0 . Given x ∈ R, express x as x = a0 +

a2 an a1 + + ··· + + ··· 2! 3! (n + 1)!

where a0 = x and for n > 0, an ∈ {0, 1, . . . , n}. Note that such an expression is uniquely determined. In fact, one can inductively define that an = (x − a0 −

an−1 a1 − ···− )(n + 1)!. 2! n!

This allows us to define a function f : R → ω ω by f (x)(n) = an . It is clear that f is Borel (but not continuous). To see that f is a reduction, let x, y ∈ R. First assume that xEv y. Without loss of generality assume x > y and let k for some k ∈ ω. r = x − y. Let m ∈ ω be the least such that r = (m + 1)! Then for all n > m, f (x)(n) = f (y)(n) from the definition of f . Conversely, assume f (x)E0 f (y). Then the expressions for x and y differ by only finitely many terms, and it is clear that x − y is rational. Thus we have shown that Ev ≤B E0 (ω). But since E0 (ω) ∼B E0 , we have that Ev ≤B E0 .

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To show the converse, that E0 ≤B Ev , we follow the same idea. Let x ∈ 2ω be given. We define g(x) = x(0) +

x(1) x(2) x(n) + + ··· + + ··· . 2! 3! (n + 1)!

The series is absolutely convergent, and thus g(x) is well-defined. It follows from the above argument that xE0 y iff g(x)− g(y) ∈ Q. Therefore g is a Borel reduction from E0 to Ev . Note that every E0 -equivalence class is countable. In fact, E0 can be viewed as an orbit equivalence relation of a countable group action, where the acting group is Z<ω 2 , the direct sum of infinitely many copies of the two-element group Z2 . For this reason we also refer to the E0 -equivalence classes as E0 -orbits. Also note that every E0 -orbit is dense. It follows that every E0 -orbit is not Gδ . In fact, a dense Gδ orbit would be comeager, whereas every E0 -orbit is countable, and therefore meager. These obvious facts distinguish E0 from all smooth equivalence relations. In the following we give two proofs of the nonsmoothness of E0 . Definition 6.1.5 Let X be a standard Borel space and E an equivalence relation on X. A σ-finite nonzero Borel measure μ on X is (E−)ergodic if for any E-invariant Borel set A ⊆ X, either μ(A) = 0 or μ(X − A) = 0. The measure μ is (E−)nonatomic if μ(C) = 0 for any E-equivalence class C. Proposition 6.1.6 Let E be an equivalence relation on a standard Borel space. If there is an E-nonatomic, E-ergodic measure on X, then E is not smooth. Proof. Let μ be E-nonatomic and E-ergodic. Assume E is smooth, and {Sn }n∈ω a countable Borel generating family for E. Since each Sn is Einvariant, either μ(Sn ) = 0 or μ(X − Sn ) = 0. Consider the set   C = {X − Sn : μ(Sn ) = 0} ∩ {Sn : μ(X − Sn ) = 0}. Then μ(X − C) = 0. But C is an equivalence class, hence μ(C) = 0. It follows that μ(X) = 0, a contradiction. Proposition 6.1.7 The usual product measure on 2ω is E0 -nonatomic and E0 -ergodic. Hence E0 is not smooth.

Proof. Let μ be the product measure on 2ω . Since μ is atomless, that is, μ({x}) = 0 for any x ∈ 2ω , and every E0 -orbit is countable, μ is E0 -nonatomic.

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To see that μ is also E0 -ergodic, let A ⊆ 2ω be an E0 -invariant Borel set. Suppose μ(A) > 0. Let  > 0. Then there exists a basic open set Ns such that μ(A ∩ Ns )/μ(Ns ) > 1 − . We claim then that for any t ∈ 2<ω with lh(t) = lh(s), μ(A ∩ Nt )/μ(Nt ) > 1 − . In fact, let ϕs,t : 2ω → 2ω be given by ⎧ if s ⊆ x and t ⊆ x, ⎨ x, ϕs,t (x) = t y, if x = s y, ⎩  s y, if x = t y. Then ϕs,t (Ns ) = Nt and ϕs,t (A ∩ Ns ) = A ∩ Nt , and moreover ϕs,t preserves the measure μ. It follows that μ(A) > 1 −  for any  > 0, and therefore indeed μ(A) = 1. Definition 6.1.8 Let X be a Polish space and E an equivalence relation on X. We say that E is generically ergodic if every E-invariant Borel set is either meager or comeager. Proposition 6.1.9 Let G be a group of homeomorphisms on a Polish space X. Then the following are equivalent: (i) EG is generically ergodic; (ii) Every nonempty EG -invariant open set is dense; (iii) There is an invariant dense Gδ set Y ⊆ X all of whose orbits are dense in X; (iv) There exists a dense orbit; (v) Every invariant set with the Baire property is meager or comeager. Proof. The implications (i)⇒(ii), (iii)⇒(iv), and (v)⇒(i) are obvious. The implication (iv)⇒(v) follows from Exercise 3.3.2. It remains to show (ii)⇒(iii). Let {Un }n∈ω be a countable base for the topology of X. Every saturation [Un ]G is EG -invariant open, and therefore dense. Let Y = n [Un ]G . Then Y is an EG -invariant dense Gδ set. If x ∈ Y , then [x]G ⊆ [Un ]G and therefore [x]G ∩ Un = ∅. Thus every orbit in Y is dense. Now we have an analog of Proposition 6.1.6 for generically ergodic equivalence relations, whose proof we leave as an exercise (Exercise 6.1.4). Proposition 6.1.10 Let E be an equivalence relation on a Polish space X. If E is generically ergodic and has no comeager orbits, then E is not smooth.

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Exercise 6.1.1 Define the tail equivalence relation Et on 2ω by xEt y ⇐⇒ ∃n∃m∀k ( x(n + k) = y(m + k) ). Show that Et is not smooth. Exercise 6.1.2 Show that the Lebesgue measure on R is Ev -nonatomic and Ev -ergodic. Exercise 6.1.3 Show that if E0 ≤B E then there is an E-nonatomic, Eergodic measure. Exercise 6.1.4 Prove Proposition 6.1.10. Exercise 6.1.5 Show that both E0 and Ev are generically ergodic. Exercise 6.1.6 Let X, Y be Polish spaces and G, H be groups of homeoX morphisms on X, Y respectively. Show that if EG is generically ergodic and X Y Z EG ≤c EH then there is a closed invariant subspace Z of Y so that EH is generically ergodic. Exercise 6.1.7 Let G be a countably infinite group acting by shift on 2G . Let EG be the induced orbit equivalence relation. Show that EG is generically ergodic.

6.2

Orbit equivalence relations embedding E0

In this section we give a construction of an embedding of E0 into a given equivalence relation. In particular we will derive the original Glimm–Effros dichotomy theorems. Most theorems about E0 -embeddability require that the equivalence relation is dense and meager. However, we first note that more needs to be assumed. Take the example of the following equivalence relation E on R: xEy ⇐⇒ x = y ∨ (x, y ∈ Q). Then E is dense and meager as a subset of R2 , but E is smooth. The following theorem assumes a condition weaker than a continuous action of a Polish group. Theorem 6.2.1 (Becker–Kechris) Let X be a Polish space and G a group of homeomorphisms of X. Suppose EG is meager and that there is a dense orbit. Then E0 c EG .

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Proof. Let D be a dense orbit and {Wn }n∈ω be a decreasing sequence of dense open subsets of X 2 with EG ∩ n Wn = ∅. Without loss of generality assume that W0 ∩ {(x, x) : x ∈ X} = ∅. We construct a sequence {Us }s∈2<ω of open sets and elements gs,t ∈ G for s, t ∈ 2<ω and lh(s) = lh(t), such that, for all s, t, u ∈ 2<ω with lh(s) = lh(t) = lh(u): (a) U∅ = X; diam(Us ) < 2−lh(s) ; Us 0 , Us 1 ⊆ Us ; Us 0 ∩ Us 1 = ∅; (b) If lh(s) = lh(t) = n and s(n − 1) = t(n − 1), then Us × Ut ⊆ Wn ; −1 (c) gs,t · Us = Ut ; gs,s = 1G ; gt,s = gs,t ; gs,u = gt,u gs,t ;

(d) If n ≤ lh(s) = lh(t) is the largest with s(n − 1) = t(n − 1), then gs,t = gsn,tn . Granting the construction, we may continuously embed E0 into E G as follows. For x ∈ 2ω , we let f (x) be the unique element of the singleton n Uxn . Then f is a continuous embedding of 2ω into X. We claim that f is also a reduction from E0 into EG . Suppose first x  E 0 y. Then for infinitely many n, x(n − 1) = y(n − 1). By (c), Uxn × Uyn ⊆ Wn for infinitely many n, and therefore (f (x), f (y)) ∈ n Wn ⊆ X 2 − EG . On the other hand, suppose xE0 y and n is the largest such that x(n − 1) = y(n − 1). Let g = gxn,yn . Then by (d) gxm,ym = g for any m ≥ n. Since g ·Uxm = Uym for all m ≥ n, we have that g · f (x) = f (y), and thus f (x)EG f (y). The construction is by induction on lh(s). We start with lh(s) = 1. At this stage we need to find disjoint open sets U0 , U1 and g0,1 ∈ G such that diam(U0 ), diam(U1 ) < 1/2, g0,1 · U0 = U1 , and U0 × U1 ⊆ W1 . To do this we first let V0 , V1 ⊆ X be open sets such that V0 × V1 ⊆ W1 . Without loss of generality we may assume diam(V0 ), diam(V1 ) < 1/2. By the density of D we can find x0 ∈ D ∩ V0 and x1 ∈ D ∩ V1 . Let g0,1 be an arbitrary g ∈ G such that g ·x0 = x1 . Let now U0 = g −1 (g ·V0 ∩V1 ) and U1 = g ·U0 . Since U0 ⊆ V0 and U1 ⊆ V1 , we have the required properties. In the inductive step assume Us and gs,t have been defined for all s, t with lh(s) = lh(t) ≤ n. We need to define Us i and gs i,t j for all s, t with lh(s) = lh(t) = n and i, j ∈ {0, 1}. First note that it is required by the condition (d) that gs i,t i = gs,t for all s, t with lh(s) = lh(t) = n and i ∈ {0, 1}. Therefore in view of (c) all gs i,t j will be determined once we have defined g0 0,0 1 , where 0 here stands for the element s with lh(s) = n and s(k) = 0 for all k < n. Similarly, all Us i will be determined once we have defined U0 0 . As in the base step we start with choosing disjoint open sets V0 0 and V0 1 , both of diameter < 2−n−1 , whose closures are contained in U0 , and so that V0 0 × V0 1 ⊆ Wn+1 . The last requirement can be fulfilled since Wn+1 is dense open. For any other s with lh(s) = n, we can then let Vs 0 = g0,s · V0 0 and Vs 1 = g0,s · V0 1 . Note that Vs 0 and Vs 1 are disjoint, open, and their closures are contained in Us . By shrinking we may assume also that each of

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them has diameter < 2−n−1 . Note that in the shrinking process we always maintain that g0,s · V0 i = Vs i . We next go through a finite number of steps to shrink the sets Vs i so as to guarantee that Vs 0 × Vt 1 ⊆ Wn+1 . Again in this shrinking process we always maintain that g0,s ·V0i = Vs i . To illustrate the argument we consider V0 0 and V0 1 . Since Wn+1 is dense open, it meets the set V0 0 × V0 1 . Thus we could shrink both of them so as to guarantee that V0 0 × V0 1 ⊆ Wn+1 . Remember this could result in other sets being shrunken too. In general, as we go through all possible pairs Vs 0 and Vt 1 , each of the sets in the collection could be shrunken, but the requirements fulfilled in the earlier process are maintained. Thus at the end of this shrinking process, we obtained open sets satisfying the properties prescribed by (a) and (b). Now by the density of D we may find x0 0 ∈ V0 0 and x0 1 ∈ V0 1 . Let g0 0,0 1 be an arbitrary g ∈ G such that g · x0 0 = x0 1 . Finally let U0 0 = g −1 (g · V0 0 ∩ V0 1 ). As remarked above, all other sets Us i and elements gs i,t j are determined. This finishes the proof of the theorem. Corollary 6.2.2 Let G be a Polish group and X a Polish G-space. Suppose there is no Gδ X orbit. Then E0 c EG . Proof. Let x ∈ X be arbitrary and consider Y = [x]G . Then Y is a Polish Y X G-space and EG c EG via the identity embedding. In Y there is obviously a dense orbit. Since there is no Gδ orbit in X, the same is true for Y . It follows from Effros’ theorem that every orbit is meager in itself, and in Y is analytic, it has the Baire property. particular meager in Y . Since EG Y Y By the Kuratowski–Ulam theorem EG is meager. Thus E0 c EG and so X E0 c EG . The corollary is very useful since it explicitly relates the descriptive complexity of orbits to the complexity of the entire equivalence relation. It is usually easier to investigate the complexity of a particular orbit. For instance, in a Baire space a dense Fσ set with a dense complement is not Gδ , since otherwise both the set and its complement would be dense Gδ , hence having a nonempty intersection. Using this it is easy to show that E0 and Ev have no Gδ orbits. The following theorem establishes a strong dichotomy if the Polish action meets some strong requirement. From it we can deduce the original Glimm– Effros dichotomy theorems.

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Theorem 6.2.3 Let G be a Polish group and X a Polish G-space. Suppose every Gδ orbit is X . In particular, either also Fσ . Then either every orbit is Gδ or else E0 c EG X X EG is smooth or else E0 c EG . X Proof. Suppose E0 c EG . We show that every orbit is Gδ . Let x ∈ X. Consider Y = {y ∈ X : [y]G = [x]G }.

It is easy to see that Y is invariant. We also note that Y is Gδ . To see this, let U be a countable base for the topology of X. Then  ∅)∧ y ∈ Y ⇐⇒ ∀U ∈ U ( U ∩ [x]G = ∅ ⇒ U ∩ [y]G = ∀U ∈ U ( U ∩ [y]G = ∅ ⇒ U ∩ [x]G =  ∅)  ∅ ⇒ y ∈ [U ]G )∧ ⇐⇒ ∀U ∈ U ( U ∩ [x]G = ∀U ∈ U ( U ∩ [x]G = ∅ ⇒ y ∈ [U ]G ). Since [U ]G is an open set, the first condition is Gδ and the second is closed. It follows that Y is Polish, and in fact a Polish G-space with the inherited action Y X Y c EG via the identity embedding, we have that E0 c EG . of G. Since EG Now by Corollary 6.2.2 there is a Gδ orbit in Y . Let [y]G ⊆ Y be Gδ . It is also true that every orbit of Y is dense. Hence Y cannot contain more than one orbit. In fact, by our assumption Y − [y]G is also Gδ , hence if it were nonempty then it would be dense Gδ and meeting [y]G , a contradiction. It follows that Y = [x]G , and hence [x]G is Gδ . X X Finally by Exercise 5.4.8 if every orbit of EG is Gδ then EG is smooth. Corollary 6.2.4 (Effros) X is Fσ . Then Let G be a Polish group and X a Polish G-space. Suppose EG X X either EG is smooth or else E0 c EG .

Proof. Since the equivalence relation is Fσ every orbit is Fσ too; in particular, every Gδ orbit is also Fσ . Corollary 6.2.4 is the original Glimm–Effros dichotomy theorem, proved by Effros, strengthening earlier work of Glimm for locally compact Polish group actions (see Exercise 6.2.2). Exercise 6.2.1 Consider the following equivalence relation E on Rω : xEy ⇐⇒ x − y ∈ 1 ⇐⇒

∞  n=0

Show that E0 c E.

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Exercise 6.2.2 (Glimm) Let G be a locally compact Polish group and X a X X is smooth or else E0 c EG . Polish G-space. Then either EG Exercise 6.2.3 Let G be a Polish group and X a Polish G-space. Define an equivalence relation F on X by xF y ⇐⇒ [x]G = [y]G . X ⊆ F. (a) Verify that F is an equivalence relation and show that EG

(b) Show that F is Gδ . X (c) Show that xF y ⇐⇒ for all EG -invariant open (or closed) sets C, x ∈ C iff y ∈ C.

6.3

The Harrington–Kechris–Louveau theorem

In this section we prove the Glimm–Effros dichotomy theorem for Borel equivalence relations [64]. This is again a theorem of classical character for which the only known proof uses effectively descriptive set theory. Theorem 6.3.1 (Harrington–Kechris–Louveau) Let E be a Borel equivalence relation on a Polish space X. Then either E is smooth or else E0 c E. The real theorem being proved is the following effective version of the Glimm–Effros dichotomy. Theorem 6.3.1 follows immediately by relativization. Theorem 6.3.2 Let E be a Δ11 equivalence relation on ω ω . Then either there is a Δ11 generating family for E or else E0 c E. The rest of this section is devoted to a proof of Theorem 6.3.2. We leave further remarks and corollaries to the next section. First we state an effective version of Lemma 5.4.6. The proof is left as an exercise (Exercise 6.3.1). Lemma 6.3.3 Let E be a Σ11 equivalence relation on ω ω . Let A, B be disjoint Σ11 E-invariant sets. Then there is a Δ11 E-invariant set C with A ⊆ C and B ∩ C = ∅. Next for any Σ11 equivalence relation E on ω ω we define a derived equivalence relation E ∗ by xE ∗ y ⇐⇒ for all Δ11 E-invariant sets C, x ∈ C iff y ∈ C. Then the following lemma collects some basic properties of E ∗ .

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Lemma 6.3.4 E ∗ ⊇ E is a Σ11 equivalence relation with a Δ11 generating family. Proof. It follows straightforward from the definition of E ∗ that E ⊆ E ∗ , E ∗ is an equivalence relation, and that E ∗ has a Δ11 generating family, that is, the family of all Δ11 E-invariant sets. We show that E ∗ is Σ11 . Using the coding in Theorem 1.7.4 we have xE ∗ y ⇐⇒ ∀n { ∃w∃z (w ∈ Pn− ∧ z ∈ Pn+ ∧ wEz ) ∨ [ ( x ∈ Pn+ ⇒ y ∈ Pn− ) ∧ ( y ∈ Pn+ ⇒ x ∈ Pn− ) ]} . Thus E ∗ is Σ11 . To consider further properties of E ∗ we take into account the Gandy– Harrington topologies. Let τ be the Gandy–Harrington topology on ω ω . Let τ 2 = τ ×τ be the product topology on (ω ω )2 and τ2 be the Gandy–Harrington topology on (ω ω )2 . Then τ 2 ⊆ τ2 . Lemma 6.3.5 E ∗ is Gδ in τ 2 . Proof.

Let C be the (countable) family of all Δ11 E-invariant sets. Then E∗ =



{(C × C) ∪ [(ω ω − C) × (ω ω − C)]}.

C∈C

Since (ω ω , τ ) is a second countable Baire space, it follows that E ∗ is Baire. Lemma 6.3.6 E ∗ is the closure of E in τ 2 . Proof. We first show that E is dense in E ∗ . For this we show that, for any Σ11 A, B ⊆ ω ω , if (A × B) ∩ E = ∅, then (A × B) ∩ E ∗ = ∅. Assume A, B are Σ11 sets with (A × B) ∩ E = ∅. Then [A]E ∩ [B]E = ∅. Since both [A]E and [B]E are Σ11 , by Lemma 6.3.3 there is a Δ11 E-invariant set C with A ⊆ C and B ∩ C = ∅. Now for any x ∈ A and y ∈ B, x ∈ C but y ∈ C, hence x E ∗ y. This means (A × B) ∩ E ∗ = ∅ as required. Next we show that E ∗ is closed in τ 2 . For this let (x, y) ∈ E ∗ , then there is a Δ11 E-invariant set C with x ∈ C and y ∈ C. Let A = C and B = ω ω − C. Then both A and B are Σ11 sets, and (A × B) ∩ E ∗ = ∅. This shows that (ω ω )2 − E ∗ is open in τ 2 , and therefore E ∗ is closed.

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If E ∗ = E then E has a Δ11 generating family and the first alternative of the conclusion of Theorem 6.3.2 is obtained. From now on we assume E ∗ = E. Consider the set X = {x ∈ ω ω : [x]E ∗ = [x]E }. Then X is nonempty. If in addition E is Δ11 then X is Σ11 , since x ∈ X ⇐⇒ ∃y ( yE ∗ x ∧ y E  x ). From now on we also assume that E is Δ11 . We are now ready to take category arguments into account. Lemma 6.3.7 If A, B ⊆ X are nonempty Σ11 sets such that E is comeager in (A × B) ∩ E ∗ , then (A × A) ∩ E ∗ ⊆ E. Proof. Consider Y = {(x, y, z) ∈ A × A × B : xE ∗ yE ∗ z} with the τ2 × τ topology. Similar to Lemma 6.3.5 we have that  {(C × C × C) ∪ [(ω ω − C) × (ω ω − C) × (ω ω − C)]}, Y = (A × A × B) ∩ C∈C

where C is the countable family of all Δ11 E-invariant sets. Thus Y is Gδ in τ 2 × τ as well as in τ2 × τ . Let Y1 = {(x, y, z) ∈ Y : xEz} and Y2 = {(x, y, z) ∈ Y : yEz}. Since E is comeager in (A × B) ∩ E ∗ , both Y1 and Y2 are comeager in Y . Now suppose (A × A) ∩ E ∗ ⊆ E. Consider Y3 = {(x, y, z) ∈ Y : x  E y}. Then Y3 is nonempty. Since E is Δ11 , Y3 is Σ11 and open in τ2 × τ . It follows that Y1 ∩ Y2 ∩ Y3 = ∅. But if (x, y, z) ∈ Y1 ∩ Y2 ∩ Y3 , then xEz, yEz, and xE  y, a contradiction. Lemma 6.3.8 E is dense and meager in X 2 ∩ E ∗ in τ 2 . Proof. Since X 2 is Σ11 , and thus open in τ 2 , the density is immediate from Lemma 6.3.6. Assume toward a contradiction that E is nonmeager in X 2 ∩E ∗ . Then there exist nonempty Σ11 sets A, B ⊆ X such that E is comeager in (A × B) ∩ E ∗ . By the preceding lemma we have that (A × A) ∩ E ∗ ⊆ E. Therefore in fact ([A]E × [A]E ) ∩ E ∗ ⊆ E. We claim that [A]E = [A]E ∗ . To see this, assume that [A]E = [A]E ∗ . Then the set Z = {x ∈ [A]E ∗ : ∃y ∈  y )} is nonempty, Σ11 , and (Z×[A]E )∩E ∗ = ∅. By the density [A]E ( xE ∗ y ∧ x E of E we have that (Z × [A]E ) ∩ E = ∅. But if z ∈ Z, x ∈ [A]E , and xEz then  y; since (x, y) ∈ ([A]E × [A]E ) ∩ E ∗ ⊆ E, there is y ∈ [A]E with zE ∗ y and z E we have xEy, a contradiction. Finally we claim that for any x ∈ A, [x]E = [x]E ∗ , a contradiction to the assumption that x ∈ X. So let x ∈ A and yE ∗ x. Then y ∈ [A]E ∗ = [A]E , and hence (x, y) ∈ ([A]E × [A]E ) ∩ E ∗ ⊆ E. This means that xEy and y ∈ [x]E .

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We are finally ready to embed E0 into E. Let us comment on the main ideas before giving the construction itself. As usual we will construct a sequence {Us }s∈2<ω of open sets and define the embedding from 2ω by letting f (z) to be the unique element of the singleton n Uzn . However, since we are working with the Gandy–Harrington topology, the strong Choquet property is used to guarantee the nonemptiness of the intersection n Uzn in place of complete metrizability. For this we need to play the strong Choquet game and let Us be coming from a winning strategy for Player II. Thus for each s ∈ 2<ω we need to find xs and Vs as Player I’s move. To make sure we are creating a reduction of E0 into E a scheme similar to the proof of Theorem 6.2.1 is used. To ensure that z E 0 y ⇒ f (z) E f (y) we ask that Us × Ut be contained in some dense open set avoiding the meager relation E if s, t have the same length but their last entries differ. For the other implication, namely zE0 y ⇒ f (z)Ef (y), the difficulty now is that no group action is present to provide the uniform linkage we had in the proof of Theorem 6.2.1. The alternative is also making use of the strong Choquet property, this time on the space of pairs, to guarantee that appropriate linkage exists among the elements xs . When zE0 y, the pairs (xzn , xyn ) will converge to a unique linked pair, forcing that f (z)Ef (y). To deal with the combinatorics associated with the linkage of elements xs , we introduce the following notation. For s, t ∈ 2<ω with lh(s) = lh(t) and k ∈ ω, define a relation Rk by sRk t ⇐⇒ ∀i < k s(i) = t(i) = 0 ∧ s(k) = t(k) ∧ ∀k < i < lh(s) s(i) = t(i). Note that if k = k  then Rk ∩ Rk = ∅. If sRk t then (s u)Rk (t u) for any u ∈ 2<ω . Also if sRk t and k < n < lh(s) then (s  n) Rk (t  n). Let also R = k Rk . The following lemma collects some other basic properties of the relation R. The proof is an easy induction on n ∈ ω, which we leave as an exercise (Exercise 6.3.3). Lemma 6.3.9 For each n ∈ ω the following are true:

(i) R ∩ (2n × 2n ) ⊆ k
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Proof. Fix winning strategies for Player II in the strong Choquet games played on (X, τ ) and ((X × X) ∩ E, τ2 ) respectively. Since E is meager in (X × X) ∩ E ∗ in τ 2 we can find a decreasing sequence of dense open subsets of (X × X) ∩ E ∗ in τ 2 such that E ∩ n Wn = ∅. Without loss of generality we may assume that W0 ∩ {(x, x) : x ∈ X} = ∅. We define sequences {xs }s∈2<ω of elements of X, {Vs }s∈2<ω , {Us }s∈2<ω of τ -open subsets of X, and {Fs,t }sRt , {Es,t }sRt of τ2 -open subsets of (X ×X)∩E such that, for any s, t ∈ 2<ω with lh(s) = lh(t), the following requirements are fulfilled: (i) V∅ = U∅ = X; (ii) The following is a play in the strong Choquet game on X according to Player II’s winning strategy: I

xs1 , Vs1

II

···

xs2 , Vs2 Us1

xs , Vs ···

Us2

Us

In particular, xs ∈ Us ⊆ Vs ; (iii) diam(Vs ) ≤ 2−lh(s) (in the usual topology of ω ω ); (iv) If lh(s) = lh(t) = n and s(n − 1) = t(n − 1) then Vs × Vt ⊆ Wn ; (v) If lh(s) = lh(t) = n and sRk t then the following is a play in the strong Choquet game on (X ×X)∩E according to Player II’s winning strategy: I

···

(xsk , xtk ), Fsk,tk

II

Esk,tk

(xs , xt ), Fs,t ···

Es,t

(vi) diam(Fs,t ) ≤ 2−lh(s) for sRt (in the usual topology of (ω ω )2 ). The construction is by induction on lh(s). We start with lh(s) = 1. At this stage we need Σ11 sets V0 , V1 ⊆ X of diameter ≤ 1/2 such that V0 × V1 ⊆ W1 and elements x0 ∈ V0 , x1 ∈ V1 so that (x0 , x1 ) ∈ E. Since W1 is dense open in (X × X) ∩ E ∗ in τ 2 such V0 , V1 exist; and since E is dense in (X × X) ∩ E ∗ , E ∩ (V0 × V1 ) = ∅. Let F0,1 be any Σ11 subset of (X × X) ∩ E containing (x0 , x1 ) and of diameter < 1/2. The sets U0 , U1 and E0,1 are obtained by playing the strong Choquet games. This finishes the construction for lh(s) = 1. In the inductive step assume all xs , Vs , Us , Fs,t , and Es,t have been defined for lh(s) ≤ n and sRt. We need to define xs i , Vs i , Fs i,t i for lh(s) = n, sRt, and i ∈ {0, 1}, as well as F0 0,0 1 . Once these are defined the sets Us i , Es i,t i , and E0 0,0 1 are obtained by playing the strong Choquet games.

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To satisfy the requirements we need at least to maintain, for all s, t with lh(s) = lh(t) = n and sRt, that xs 0 , xs 1 ∈ Us , xt 0 , xt 1 ∈ Ut , and (xs 0 , xt 0 ), (xs 1 , xt 1 ) ∈ Es,t . We consider the set of all such tuples Y = {((ys 0 )s∈2n , (ys 1 )s∈2n ) : ∀s, t ∈ 2n [ys i ∈ Us ∧ (ys i , yt i ) ∈ E ∗ ∧ (sRt ⇒ (ys i , yt i ) ∈ Es,t )], i = 0, 1}. n

n

As a subset of X 2 × X 2 the set Y is obviously of the form Z × Z for n n some Z ⊆ X 2 open in the Gandy–Harrington topology on X 2 . And it is nonempty since (xs )s∈2n ∈ Z and therefore ((xs )s∈2n , (xs )s∈2n ) ∈ Y . To further satisfy requirement (iv) we fix any s0 , t0 ∈ 2n . Since Wn+1 is dense open in (X × X) ∩ E ∗ , the set Ys0 ,t0 = {((ys 0 )s∈2n , (ys 1 )s∈2n ) ∈ Y : (ys 0 , yt 1 ) ∈ Wn+1 } 0

0

is dense open in Y . It follows that the finite intersection s,t∈2n Ys,t is still dense open, and thus there are Σ11 sets Vs 0 , Vt 1 such that    Vs 0 × Vt 1 ⊆ Ys,t . s

t

s,t

With possible shrinking we may require that diam(Vs 0 ), diam(Vt 1 ) ≤ 2−lh(s)−1 . Thus (iii) and (iv) are satisfied. Furthermore since E is dense in (X × X) ∩ E ∗ , the set Y  = {((ys 0 )s∈2n , (ys 1 )s∈2n ) ∈ Y : (y0 0 , y0 1 ) ∈ E} is dense in Y . Let ((xs 0 )s∈2n , (xs 1 )s∈2n ) be any element of the nonempty intersection   Vs 0 × Vt 1 . Y∩ s

t

We have defined the required elements (xu )u∈2n+1 satisfying (ii). Finally for each pair (s, t) with sRt and i ∈ {0, 1} let Fs i,t i be any Σ11 subset of (X × X) ∩ E containing (xs 0 , xt 1 ) and with diameter ≤ 2−lh(s)−1 . Let also F0 0,0 1 be any Σ11 subset of (X × X) ∩ E containing (x0 0 , x0 1 ) and with diameter ≤ 2−lh(s)−1 . We have thus finished the definition of all elements and sets involved, and conditions (i) through (vi) are all satisfied. It follows from (ii) that for any z ∈ 2ω the objects xzn , Vzn , Uzn form a play in the strong Choquet game on X according to Player II’s winning strategy. Thus n Uzn is nonempty, and by (iii), is a singleton. We define f (z) to be this unique element. This defines a continuous embedding of 2ω into X. It remains to verify that f is a reduction from E0 to E. First suppose z E  0 y. Then for infinitely many n, z(n − 1) = y(n − 1), and by (iv), for infinitely many n, Uzn ×Uyn ⊆ Wn . It follows that (f (z), f (y)) ∈ W , and hence (f (z), f (y) ∈ E. n n

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For the other direction we show that for any s, t with lh(s) = lh(t) = n and z ∈ 2ω , f (s 0 z)Ef (t 1 z). This is by induction on n. For n = 0 we note that (0 z  k)R1 (1 z  k) for any k ∈ ω. Thus the objects (x0 zk , x1 zk ), F0 zk,1 zk , E0 zk,1 zk form a play in the strong Choquet game on (X × X) ∩ E according to Player II’s winning strategy. It follows from (v) and (vi) that k E0 zk,1 zk is a singleton. By our construction this unique pair must be (f (0 z), f (1 z)). Therefore f (0 z)Ef (1 z). For the inductive step when n > 0, consider the elements f (0 0 z) and f (0 1 z). By the inductive hypothesis f (0 0 z)Ef (s 0 z) and f (0 1 z)Ef (t 1 z). Thus it suffices to show that f (0 0 z)Ef (0 1 z). However this is similar to the argument for the case n = 0. For any k ∈ ω let uk = 0 0 z  k and vk = 0 1 z  k. By our construction the objects (xuk , xvk ), Fuk ,vk , Euk ,vk form a play in the strong Choquet game on (X × X) ∩ E according to Player II’s winning strategy. It follows again from (v) and (vi) that k Euk ,vk is a singleton and its unique element must be (f (0 0 z), f (0 1 z)). Therefore f (0 0 z)Ef (0 1 z). Exercise 6.3.1 Prove Lemma 6.3.3. (Hint: Δ11 is closed under effective countable unions but not arbitrary countable unions.) Exercise 6.3.2 Show that if E is a Σ11 equivalence relation on ω ω then E ∗ is clopen in the Gandy–Harrington topology τ2 . Exercise 6.3.3 Prove Lemma 6.3.9.

6.4

Consequences of the Glimm–Effros dichotomy

In this section we derive some consequences of the Glimm–Effros dichotomy for Borel equivalence relations. The main results of this section are due to Harrington, Kechris, and Louveau [64]. We first recall the following definition. Definition 6.4.1 Let X, Y be standard Borel spaces. A subset A ⊆ X is called universally measurable if it is μ-measurable for any σ-finite Borel measure μ on X. A function f : X → Y is universally measurable if it is μ-measurable for any σ-finite Borel measure μ on X. A theorem of Luzin states that every analytic set is universally measurable (see Reference [97] Theorem 21.10). Let σ(Σ11 ) be the σ-algebra generated by the analytic sets. Then it follows that every set in σ(Σ11 ) is universally measurable. If X and Y are standard Borel spaces, then a function f : X → Y

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is σ(Σ11 )-measurable if for any Borel set B ⊆ Y , f −1 (B) is a σ(Σ11 ) subset of X. We also recall the following definition. Definition 6.4.2 Let X, Y be sets and P ⊆ X × Y . Let projX (P ) = {x ∈ X : ∃y ∈ Y (x, y) ∈ P }. A uniformization of P is a function f : projX (P ) → Y such that for all x ∈ projX (P ), (x, f (x)) ∈ P . We will use the following theorem known as the Jankov–von Neumann uniformization theorem. Theorem 6.4.3 (Jankov–von Neumann) Let X, Y be standard Borel spaces and P ⊆ X × Y an analytic set. Then there is a uniformization of P that is σ(Σ11 )-measurable. For a proof of this theorem see Reference [97] Theorem 18.1. We are now ready to derive some consequences of the Glimm–Effros dichotomy. Theorem 6.4.4 Let X be a standard Borel space and E a Borel equivalence relation on X. Then the following are equivalent: (i) E is smooth; (ii) There is a compatible Polish topology τ on X such that E is closed in (X 2 , τ 2 ); (iii) There is a compatible Polish topology τ on X such that E is Gδ in (X 2 , τ 2 ); (iv) There is a compatible Polish topology τ on X such that every Eequivalence class is Gδ ; (v) There is a countable generating family for E consisting of analytic sets; (vi) There is a countable generating family for E consisting of universally measurable sets; (vii) There is a universally measurable reduction from E to id(2ω ); (viii) E has a σ(Σ11 )-measurable selector; (ix) E has a universally measurable selector.

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Proof. The implication (i)⇒(ii) was proved in Corollary 5.4.5. The implications (ii)⇒(iii)⇒(iv), (v)⇒(vi), and (viii)⇒(ix) are obvious. The equivalence between (vi) and (vii) follows from the proof of Proposition 5.4.4. Also (ix)⇒(vi) by a similar argument. We prove the other implications. For (iv)⇒(v), let {Un }n∈ω enumerate a countable base for (X, τ ). We claim that {[Un ]E }n∈ω is a countable generating family for E consisting of analytic sets. It is clear that each [Un ]E is analytic. To see that they form a generating family for E, let x, y ∈ X be such that x ∈ [Un ]E ⇐⇒ y ∈ [Un ]E for all n ∈ ω. Since for any n ∈ ω, [x]E ∩ Un = ∅ ⇐⇒ [x]E ∩ Un = ∅ ⇐⇒ x ∈ [Un ]E , we have that for all n ∈ ω, [x]E ∩ Un = ∅ ⇐⇒ [y]E ∩ Un = ∅, which in turn implies that [x]E = [y]E . Now [x]E , [y]E are both dense Gδ subsets of [x]E , thus [x]E ∩ [y]E = ∅, and therefore in fact xEy. Next we show that (vi)⇒(i) from the Glimm–Effros dichotomy. Let {An }n∈ω be a countable generating family for E where each An is universally measurable. Assume that E is not smooth. Then there is a Borel embedding θ : 2ω → X of E0 into E. It is easy to verify that the family {θ−1 (An )}n∈ω is a generating family for E0 . Note also that each θ−1 (An ) is universally measurable. To see this let μ be a σ-finite Borel measure on 2ω . Then θμ defined by θμ (A) = μ(θ−1 (A)) is a σ-finite Borel measure on X. Since each An is universally measurable, An is θμ -measurable, and hence θ−1 (An ) is μmeasurable. To summarize, we now have a countable separating family for E0 consisting of universally measurable sets. By the proofs of Propositions 6.1.6 and 6.1.7 show that E0 does not admit such families. Finally we show that (i)⇒(viii). For this let ϕ : X → 2ω be a Borel reduction of E to id(2ω ). Consider A = {(z, x) ∈ 2ω × X : z = ϕ(x)}. Then A is Borel, in particular Σ11 , and by the Jankov–von Neumann uniformization theorem, allows a σ(Σ11 )-measurable uniformization. Let ψ : ϕ(X) → X be a σ(Σ11 )-measurable uniformization of A. Then ψ ◦ ϕ is σ(Σ11 )-measurable. But ψ ◦ ϕ is clearly a selector for E. Recall that smooth equivalence relations need not have Borel selectors. In contrast, clauses (viii) and (ix) provide characterizations for smoothness in terms of σ(Σ11 )-measurable and universally measurable selectors. Also for Borel equivalence relations the existence of Borel selectors and of Borel transversals are equivalent (see the remarks preceding Proposition 3.4.6), therefore smooth equivalence relations need not have Borel transversals either. On the flip side we also get characterizations for E0 -embeddability.

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Theorem 6.4.5 Let X be a Polish space and E a Borel equivalence relation on X. Then the following are equivalent: (a) E0 c E; (b) E0 ≤B E; (c) There is a universally measurable reduction from E0 to E; (d) There is an E-nonatomic, E-ergodic probability Borel measure on X. Proof. The implications (a)⇒(b)⇒(c) are obvious. To see that (c)⇒(a), suppose there is a universally measurable reduction θ from E0 to E. By the proofs of Propositions 6.1.6 and 6.1.7 we know that there are no countable generating families of E0 consisting of universally measurable sets. However, if (a) fails then by the Glimm–Effros dichotomy we have that E is smooth and thus admitting a countable Borel generating family {An }n∈ω . It follows that {θ−1 (An )}n∈ω is a generating family for E0 consisting of universally measurable sets, a contradiction. A similar argument shows that (d)⇒(a). We finally show that (a)⇒(d). Let μ be the product measure on 2ω and θ : 2ω → X be a continuous embedding of E0 into E. Then the measure θμ defined by θμ (A) = μ(θ−1 (A)) is a probability Borel measure on X. It is E-nonatomic because for any x ∈ X, θμ ([x]E ) = μ(θ−1 ([x]E )) = 0 since θ−1 ([x]E ) is contained in at most one E0 -equivalence class. It is also E-ergodic since for any invariant Borel set A ⊆ X, θ−1 (A) is also invariant Borel. Exercise 6.4.1 Give a direct proof that an equivalence relation E on a Polish space X with every E-equivalence class Gδ admits a σ(Σ11 )-measurable selector. Exercise 6.4.2 Show that a Borel equivalence relation with a Σ11 transversal is smooth. Exercise 6.4.3 Let G be a Polish group and X a Polish G-space. Suppose X is Fσ . Show that the following are equivalent: EG X (1) EG is smooth;

(2) Every orbit is Gδ ; X (3) EG is Gδ ; X . (4) There is a Borel selector for EG

Exercise 6.4.4 Let G be a Polish group and H a Borel subgroup of G. Let E be the coset equivalence relation on G defined by xEy ⇐⇒ xH = yH. Show that E is smooth iff H is closed.

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The Glimm–Effros Dichotomy

6.5

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Actions of cli Polish groups

In this section we prove a theorem of Becker [5] that establishes the Glimm– Effros dichotomy for orbit equivalence relations induced by Borel actions of cli Polish groups. The main theorem of this section is the following. Theorem 6.5.1 (Becker) Let G be a cli Polish group and X a Borel G-space. Then there is a Polish topology on X such that the action is continuous and every Gδ orbit is closed. By Theorem 6.2.3, the Glimm–Effros dichotomy follows immediately. Theorem 6.5.2 (Becker) X Let G be a cli Polish group and X a Borel G-space. Then either EG is smooth X . or else E0 B EG The rest of the section is devoted to a proof of Theorem 6.5.1. We fix a cli Polish group G and a compatible complete left-invariant metric dG on G. Also fix a countable dense subgroup G0 of G and a countable base U for the topology of G such that U is closed under left-translates by elements of G0 , that is, for g ∈ G0 and U ∈ U, gU ∈ U. Without loss of generality assume that G ∈ U and ∅ ∈ U. For U, V ∈ U, define G0 (U, V ) = {g ∈ G0 : gU ∩ V = ∅}. Then note that G0 (U, V ) is nonempty. Lemma 6.5.3

For any open U  ⊆ U , {gU  : g ∈ G, gU  ∩ V = ∅} ⊆ G0 (U, V )U . In particular, V ⊆ G0 (U, V )U . Proof. Let g ∈ G and h, k ∈ U  with gh ∈ V . We need to show that gk ∈ G0 (U, V )U . By the continuity of the group operations there are open sets U0 , U1 , U2 , N such that g ∈ U0 , 1G ∈ N , h ∈ U1 , k ∈ U2 , N U1 ⊆ U  , N U2 ⊆ U , and U0 N U1 ⊆ V . Note that gN −1 ∩ U0 is nonempty open. We let g0 ∈ G0 ∩ gN −1 ∩ U0 . Then g0 h ∈ U0 N U1 ⊆ V , and therefore g0 ∈ G0 (U, V ). Also gk = g0 (g0−1 g)k, where (g0−1 g)k ∈ N U2 ⊆ U . Thus gk ∈ G0 (U, V )U . This proves the first half of the lemma. The second half follows since V ⊆ {gU : g ∈ G, gU ∩ V = ∅}. Next we consider the action of G on X and find a Polish topology on X by the method of Section 4.4. Lemma 6.5.4 There is a countable collection B of Borel sets on X such that

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(i) the topology τ generated by B is Polish, (ii) the action of G on (X, τ ) is continuous, and (iii) for all A, B ∈ B and U, V ∈ U, the set [A − G0 (U, V ) · B] G is τ -open. Proof. Let B0 be a countable base for a Polish topology on X generating the given Borel structure. Let A be a countable Boolean algebra of Borel subsets of X with the following closure properties: (a) B0 ⊆ A; (b) For all A ∈ A and U, V ∈ U, G0 (U, V ) · A ∈ A; (c) For all A ∈ A and U ∈ U, A U ∈ A; (d) The topology generated by A is Polish. This can be easily done by the remarks preceding Lemma 4.4.5. Now let τ be the topology on X generated by B = A U . By the lemmas in Section 4.4 (X, τ ) becomes a Polish G-space. To check (iii) note that B ⊆ A, and for any A, B ∈ A and U, V ∈ U, [A − G0 (U, V ) · B] G ∈ A U by condition (b) and the fact that A is a Boolean algebra. We fix a countable base B and the Polish topology τ on X with the properties described above, and investigate the Gδ orbits. For this we also fix x ∈ X so that [x]G is Gδ . By Effros’ Theorem 3.2.4 the canonical map g → g · x is open from G onto [x]G . It follows that for any open U ⊆ G, U ·x is a relatively open subset of [x]G . We define for open U ⊆ G,  Ω(U ) = {A ⊆ X : A ∈ τ, A ∩ [x]G ⊆ U · x}. It is clear that Ω(U ) is the largest open subset O of X such that O∩[x]G = U ·x. It is also easy to see that U ⊆ V implies Ω(U ) ⊆ Ω(V ), and for any g ∈ G, g · Ω(U ) = Ω(gU ). Note as well that for any sequence (Un ) of open sets in G, n Ω(Un ) ⊆ Ω( n Un ). We also define, for U, V ∈ U,  Λ(U, V ) = [Ω(V ) − {Ω(hU ) : h ∈ G0 (U, V )}] G . Then Λ(U, V ) is an invariant Borel subset of X. The main issue is whether it is empty. We first show that it is empty on the orbit [x]G . Lemma 6.5.5 For any U, V ∈ U, Λ(U, V ) ∩ [x]G = ∅. Proof. Assume Λ(U, V ) ∩ [x]G =  ∅. Since Λ(U, V ) is invariant, we have x ∈ Λ(U, V ). By definition there is g ∈ G such that g · x ∈ Ω(V ) but for all

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h ∈ G0 (U, V ), g · x ∈ Ω(hU ). From g · x ∈ Ω(V ) we get that g · x ⊆ V · x. Thus we may assume without loss of generality that g ∈ V . But by Lemma 6.5.3 g ∈ G0 (U, V )U . Thus for some h ∈ G0 (U, V ), g ∈ hU , and g · x ∈ hU · x ⊆ Ω(hU ), a contradiction. The following lemma is one of two key lemmas for the proof of Theorem 6.5.1. Lemma 6.5.6 Suppose y ∈ Λ(U, V ). Then there are A, B ∈ B such that [y]G ⊆ [A − G0 (U, V ) · B] G and [x]G ∩ [A − G0 (U, V ) · B] G = ∅. Proof. Since y ∈ Λ(U, V ) there is a nonmeager set N0 ⊆ G such that for all g ∈ N0 ,  g · y ∈ Ω(V ) − {Ω(hU ) : h ∈ G0 (U, V )}. Thus for all g ∈ N0 , g ·y ∈ Ω(V ) but for all h ∈ G0 (U, V ), g ·y ∈ Ω(hU ). Since Ω(V ) is open in X and B is a countable base, Ω(V ) = {A : A ∈ B ∧ A ⊆ Ω(V )}. It follows that there is A ∈ B and a nonmeager set N1 ⊆ N0 such that for all g ∈ N1 , g · y ∈ A ⊆ Ω(V ). Let B ∈ B be such that B ⊆ Ω(U ) and B ∩ U · x = ∅. Then B ∩ [x]G ⊆ U · x. It follows that for h ∈ G0 (U, V ), h · B ∩ [x]G ⊆ h · (U · x) = (hU ) · x, and thus h · B ⊆ Ω(hU ). This implies that for all g ∈ N1 , g · y ∈ h · B. We have shown that y ∈ [A − G0 (U, V ) · B] G . It remains to show [x]G ∩ [A − G0 (U, V ) · B] G = ∅. Assume this is not the case. Then there is k ∈ G such that k · x ∈ A − G0 (U, V ) · B. Since A ⊆ Ω(V ) we may assume without loss of generality that k ∈ V . Since B ⊆ Ω(U ) is open there is an open U  ⊆ U such that B ∩ [x]G = U  · x. U  = ∅ since B ∩ U · x = ∅. Note that G0 (U  , V ) ⊆ G0 (U, V ), and hence k · x ∈ G0 (U  , V ) · B. However, by Lemma 6.5.3 V ⊆ G0 (U  , V )U  . Thus k · x ∈ G0 (U  , V ) · (U  · x) ⊆ G0 (U  , V ) · B, a contradiction. Note that so far we have not used the assumption that G is a cli Polish group. The next key lemma is the only place where we essentially use this assumption. Lemma 6.5.7 If y ∈ [x]G − [x]G , then there are U, V ∈ U such that y ∈ Λ(U, V ). Proof. Suppose y ∈ [x]G but assume there are no U, V ∈ U such that y ∈ Λ(U, V ). We construct sequences (Un )n∈ω of elements of U and (gn )n∈ω of elements of G such that, for all n ∈ ω, (i) diam(Un ) < 2−n , (ii) Un ∩ Un+1 = ∅,

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(iii) dG (gn , gn+1 ) < 2−n , (iv) gn · y ∈ Ω(Un ). The construction is by induction on n. To begin with, let U0 = G and g0 = 1G . Since Ω(G) = X we have that g0 · y ∈ Ω(U0 ). In general, suppose Un and gn have been defined to satisfy the requirements. We now define Un+1 ∈ U and gn+1 ∈ G. Let V = Un and U ∈ U such that diam(U ) < 2−(n+1) . Since y ∈ Λ(U, V ) and Λ(U, V ) is invariant, we have that gn · y ∈ Λ(U, V ). Thus there is a comeager set C ⊆ G such that for all g ∈ C,  g · (gn · y) ∈ Ω(Un ) − {Ω(hU ) : h ∈ G0 (U, Un )}. Let

W = {g ∈ G : g · (gn · y) ∈ Ω(Un ) ∧ dG (ggn , gn ) < 2−n }.

By (iv) gn · y ∈ Ω(Un ), and since Ω(Un ) is open in X, W is nonempty open. Let g ∈ C ∩ W and gn+1 = ggn . Then dG (gn , gn+1 ) = dG (gn , ggn ) < 2−n . Now gn+1 · y = g · (gn · y) ∈ Ω(Un ) since g ∈ W , so by g ∈ C we obtain an h ∈ G0 (U, Un ) such that gn+1 · y = g · (gn · y) ∈ Ω(hU ). Let Un+1 = hU . Since diam(U ) < 2−(n+1) , the left invariance gives that diam(Un+1 ) = diam(hU ) < 2−(n+1) . Also Un ∩ Un+1 = Un ∩ hU = ∅ since h ∈ G0 (U, Un ). Lastly, gn+1 · y = g · (gn · y) ∈ Ω(hU ) = Ω(Un+1 ). Thus the construction is finished with all the requirements fulfilled. By the construction we have that the sequence (gn ) is dG -Cauchy, and therefore by the completeness of dG there is g∞ ∈ G such that gn → g∞ as n → ∞. Let y∞ = g∞ · y. Let dX be a complete metric on X compatible with τ . Let (Mn )n∈ω be a countable open nbhd base of y∞ with gn · y ∈ Mn . We claim that Un · x ∩ Mn = ∅ for all n ∈ ω. To see this fix an n and let A = Ω(Un ) ∩ Mn . Then A is open and gn · y ∈ A by (iv). Since y ∈ [x]G , we get that gn · y ∈ [x]G and thus A ∩ [x]G = ∅. It follows that A ∩ [x]G = ∅, and therefore Un · x ∩ Mn = ∅. For each n ∈ ω let xn ∈ Un · x ∩ Mn and hn ∈ Un be such that hn · x = xn . By the above construction (hn ) is a dG -Cauchy sequence, and therefore there is h∞ ∈ G such that hn → h∞ as n → ∞. By the continuity of the action we thus have that xn = hn · x → h∞ · x as n → ∞. On the other hand, xn → y∞ as (Mn ) is a nbhd base for y∞ . Thus y∞ = h∞ · x, and y ∈ [x]G . We are now ready to give the proof of Theorem 6.5.1. Proof. Following the notation of this section let x ∈ X be such that [x]G is Gδ . We show that [x]G is closed. Assume that y ∈ [x]G −[x]G . By Lemma 6.5.7 there are U, V ∈ U such that y ∈ Λ(U, V ). And so by Lemma 6.5.6 there are A, B ∈ B such that, letting D = [A − G0 (U, V ) · B] G , we have [y]G ⊆ D and [x]G ∩ D = ∅. By Lemma 6.5.4 D is open, and so y ∈ [x]G , contradicting our assumption.

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Becker’s theorem implies earlier results of Sami (for abelian Polish groups [134]; see Exercises 9.5.3 through 9.5.5), Hjorth–Solecki (for nilpotent Polish groups [83]), and Solecki (for Polish groups with two-sided invariant metrics [83]). The following exercise problems follow the notation of this section, where G is an arbitrary Polish group. Exercise 6.5.1 Show that for any g ∈ G and open U ⊆ G, g ·Ω(U ) = Ω(gU ). Exercise 6.5.2 Show that for any A ∈ B, [A]G is clopen. Exercise 6.5.3 Define an equivalence relation ∼ on X by x ∼ y iff [x]G = [y]G . Show that every ∼-equivalence class is closed.

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Chapter 7 Countable Borel Equivalence Relations

The topic of Borel equivalence relations is so vast that it certainly cannot fit into two chapters of any book. In this and the next chapters we just give an introduction of the topic and leave further explorations of the literature to the interested reader. There are many theoretical results about Borel equivalence relations, and at the same time we have also identified many classification problems and equivalence relations from other fields of mathematics that turn out to be Borel equivalence relations on standard Borel spaces. In this chapter we will discuss countable Borel equivalence relations. For reasons that will become clear the study of countable Borel equivalence relations have become intertwined with the study of countable group theory and ergodic theory. Thus most of the important results about countable Borel equivalence relations cannot be proved using set theoretic or general topological tools alone. Our objective in this chapter is to provide a self-contained inroad into the subject.

7.1

Generalities of countable Borel equivalence relations

Definition 7.1.1 Let X be a standard Borel space. An equivalence relation E on X is called finite if every E-equivalence class is finite. E is called countable if every E-equivalence class is countable. Any finite Borel equivalence relation is smooth (see Exercise 7.1.1). The following theorem of Luzin–Novikov on Borel uniformizations will be useful in proving theorems about countable Borel equivalence relations. The proof can be found in Reference [97] Theorem 18.10. Theorem 7.1.2 (Luzin–Novikov) Let X, Y be standard Borel spaces and let P ⊆ X × Y be Borel. Suppose every section Px = {y ∈ Y : (x, y) ∈ P } is countable. Then projX (P ) = {x ∈ X : ∃y (x, y) ∈ P } is Borel, and P has a Borel uniformization, that is, there is a Borel function f : projX (P ) → Y such that (x, f (x)) ∈ P for any x ∈ projX (P ).

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Moreover, P can be written as n Pn where each Pn is the graph of some Borel function fn : An → Y for An Borel, that is, Pn = {(x, fn (x)) : x ∈ An }. This uniformization theorem greatly simplifies the study of reductions among countable Borel equivalence relations, as the following lemma shows. Lemma 7.1.3 Let E be a countable Borel equivalence relation on a standard Borel space X, Y a standard Borel space, and f : X → Y a Borel function such that f (x) = f (y) implies xEy. Then f (X) is Borel in Y and there is a Borel function g : f (X) → X such that f ◦ g = id. Proof. Consider P = {(z, x) ∈ Y ×X : f (x) = z}. Then P is a Borel subset of Y × X with each Pz countable. By the Luzin–Novikov uniformization theorem f (X) = projY (P ) is Borel. Let g : f (X) → X be a Borel uniformization function. Then f ◦ g = id. The following is another application of the Luzin–Novikov uniformization theorem. The theorem states that all countable Borel equivalence relations are orbit equivalence relations of countable group actions. Theorem 7.1.4 (Feldman–Moore) Let E be a countable Borel equivalence relation on a standard Borel space X. Then there is a countable group G and a Borel action of G on X such that X . E = EG Proof. Consider E as a Borel subset of X × X. Since E is countable, each section Ex is

countable. By the Luzin–Novikov uniformization theorem we can write E as n Pn where each Pn is the graph of a Borel function fn : An → X for An Borel. Without loss of generality we may assume that Pn ∩ Pm = ∅ for n = m. Let Δ = {(x, x) : x ∈ X}. First note that X − Δ can be partitioned into countably many sets of the form A × B, where A, B are disjoint Borel subsets of X (see Exercise 7.1.3). We write X − Δ = k Rk , where {Rk }k∈ω is such a partition. Next for n, m, k ∈ ω we let En,m,k = {(x, y) ∈ E : (x, y) ∈ Pn , (y, x) ∈ Pm , x = y, (x, y) ∈ Rk }. Then {En,m,k } is a partition of E − Δ. And since each Pn is the graph of a Borel function, so is each En,m,k . We let hn,m,k : An,m,k → Bn,m,k be a Borel isomorphism between Borel subsets An,m,k and Bn,m,k such that En,m,k = {(x, hn,m,k (x)) : x ∈ An,m,k }. Define a Borel automorphism gn,m,k of X by ⎧ ⎨ hn,m,k (x), if x ∈ An,m,k , (x), if x ∈ Bn,m,k , gn,m,k (x) = h−1 ⎩ n,m,k x, otherwise.

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Let G be the group of Borel automorphisms generated by gn,m,k for n, m, k ∈ ω. Then xEy ⇐⇒ ∃n, m, k (gn,m,k (x) = y) ∨ (x = y) ⇐⇒ ∃g ∈ G(g · x = y). X Thus E = EG .

The following lemma is another basic and useful result about countable Borel equivalence relations. It is often referred to as the marker lemma. Lemma 7.1.5 (Slaman–Steel) Let E be a countable Borel equivalence relation on a standard Borel space X. Let A = {x ∈ X : [x]E is finite} and B = X − A. Then there are Borel sets B ⊇ S0 ⊇ S1 ⊇ S2 ⊇ . . . such that [Sn ]E = B for all n, and n Sn = ∅. Proof. Without loss of generality we may assume that X = 2ω . Let G be X . Enumerate a countable group acting on X in a Borel manner with E = EG G as {gn }n∈ω . For any E-equivalence class C, let C be the closure of C and xC be the lexicographically least element of C. Then map y → x[y]E is Borel, since for any s ∈ 2<ω , x[y]E ∈ Ns ⇐⇒ ∃n ( gn · y ∈ Ns ) ∧ ∀s
(i) ω ω − Δ = {Ns × Nt : sRt}. (ii) If sRt and s Rt but (s, t) = (s , t ), then (Ns × Nt ) ∩ (Ns × Nt ) = ∅.

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Exercise 7.1.4 Show that the sets Sn defined in the proof of Lemma 7.1.5 are as required. Exercise 7.1.5 Show that every countable Borel equivalence relation is idealistic. Exercise 7.1.6 Show that the following are equivalent for a countable Borel equivalence relation E: (i) E is smooth; (ii) E has a Borel transversal; (iii) E has a Borel selector. Exercise 7.1.7 Show that for every countable Borel equivalence relation E

there exist finite equivalence relations En such that E = n En . Exercise 7.1.8 Let X, Y be standard Borel spaces and E, F countable Borel equivalence relations on X, Y , respectively. Show that if E ≤B F then there is an F -invariant Borel subset B of Y such that E ∼B F  B.

7.2

Hyperfinite equivalence relations

Definition 7.2.1 Let X be a standard Borel space. A Borel equivalence relation E on X is hyperfinite if there are finite Borel equivalence relations En , n ∈ ω, with En ⊆ En+1 for all n, such that E = n En . By definition every hyperfinite equivalence relation is a countable Borel equivalence relation. Note that the equivalence relation E0 is hyperfinite. In fact, we can define, for n ≥ 1, equivalence relations En on 2ω by xEn y ⇐⇒ ∀m ≥ n x(m) = y(m).

Then each En is apparently Borel and finite, En ⊆ En+1 , and E0 = n≥1 En . The theorems proved in this section will show that E0 is a typical hyperfinite equivalence relation in a strong sense. The following proposition collects some basic closure properties of hyperfinite equivalence relations. Proposition 7.2.2 Let E and F be countable Borel equivalence relations on standard Borel spaces X and Y , respectively. Then the following hold:

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(1) If X = Y , F is hyperfinite and E ⊆ F , then E is hyperfinite. (2) If E is hyperfinite and A ⊆ X is Borel, then E  A is hyperfinite. (3) If A ⊆ X is Borel, [A]E = X, and E  A is hyperfinite, then E is hyperfinite. (4) If E ≤B F and F is hyperfinite, then E is hyperfinite. (5) If both E and F are hyperfinite, then E × F is hyperfinite. Proof. Properties (1), (2) and (5) follow easily

from the definition. We only show (3) and (4). For (3) suppose E  A = n Fn , where Fn are finite Borel X equivalence relations and Fn ⊆ Fn+1 for all n. By Theorem 7.1.4 E = EG for some countable group G acting in a Borel manner on X. Enumerate G by {gm }m∈ω . For each x ∈ X, define m(x) to be the least m such that gm ·x ∈ A. By the assumption [A]E = X, m(x) is well defined for all x ∈ X. We then define xEn y ⇐⇒ ( m(x), m(y) < n ∧ gm(x) · x Fn gm(y) · y ) ∨ x = y.

Then En are finite Borel equivalence relations, En ⊆ En+1 , and E = n En . Thus E is hyperfinite. For (4) we let f : X → Y be a Borel reduction of E to F . Then f (x) = f (y) implies xEy. By Lemma 7.1.3 there is a Borel function g : f (X) → X so that f ◦g = id. Now g is a Borel injection, thus g(f (X)) is Borel. Let A = g(f (X)). Then [A]E = X. Moreover E  A is Borel isomorphic (via g −1 ) to F  f (X). Therefore by (2) F  f (X) is hyperfinite. It follows that E  A is hyperfinite and by (3) E is hyperfinite. The following theorem gives various useful characterizations of hyperfiniteness. Especially of interest is the characterization in terms of Borel reducibility to E0 . The theorem was based on earlier work by Weiss and by Slaman–Steel, and appeared in this form in Reference [30]. Theorem 7.2.3 (Dougherty–Jackson–Kechris) Let X be a standard Borel space and E a countable Borel equivalence relation. Then the following are equivalent: (i) E ≤B E0 . (ii) E is hyperfinite.

(iii) E = n En , where En are finite Borel equivalence relations, En ⊆ En+1 , and every En -equivalence class has at most n elements. (iv) There is a Borel assignment C →
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(v) There is a Borel automorphism T of X such that xEy Z T n (x) = y.

⇐⇒ ∃n ∈

(vi) There is a Borel action of the additive group Z on X such that E = EZX . Proof. The implication (i)⇒(ii) follows immediately from Proposition 7.2.2 (4). The implication (iii)⇒(ii) and the equivalence of (iv), (v), and (vi) are obvious. We show the other implications. (ii)⇒(iii): Let E = n Fn , where Fn are finite Borel equivalence relations and Fn ⊆ Fn+1 . Without loss of generality we may assume F1 = id(X). We define a sequence of finite Borel equivalence relations En , n ≥ 1, such that En ⊆ En+1 and each En -equivalence class has at most n elements. Define E1 = id(X) = F1 . For n > 1, inductively define the following sequence of sets: Xn = {x ∈ X : [x]Fn has at most n elements}, Xn−1 = {x ∈ Xn : [x]Fn−1 has at most n elements}, · · · · · · Xi = {x ∈ i<j≤n Xj : [x]Fi has at most n elements}, ······ X2 = {x ∈ Xn ∪ · · · ∪ X3 : [x]F2 has at most n elements}, X1 = X −

1<j≤n

Xj .

Note that the definition of X1 can be formulated in the same manner as those of the other sets. Since F1 ⊆ F2 ⊆ · · · ⊆ Fn , each Xi is Fj -invariant if j ≤ i. Let En = (Fn  Xn ) ∪ (Fn−1  Xn−1 ) ∪ · · · ∪ (F2  X2 ) ∪ (F1  X1 ). Then En is a finite Borel equivalence relation with each En -equivalence class containing at most n elements. To see that En ⊆ En+1 let Yn+1 , Yn , . . . , Y2 , Y1 be the sequence of sets defined in the definition of En+1 . Suppose xEn y. Then for some i ≤ n, x, y ∈ Xi and xFi y. Since Xi ⊆ Yn+1 ∪ Yn ∪ · · · ∪ Yi and each of Yn+1 , Yn , . . . , Yi is Fi -invariant, there is j ≥ i such that x, y ∈ Yj and xFj y, and thus xEn+1 y. This shows that En ⊆ En+1 . It is easy to see that E = n En .

(ii)⇒(iv): Let E = n En where En are finite Borel equivalence relations and En ⊆ En+1 . Fix a Borel linear order < on X. We define
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It follows from this construction that if C is an En -equivalence class and D is an En+1 -equivalence class with C ⊆ D, then
from the property that Sn ⊇ Sn+1 , and E  X2 = n En since n Sn = ∅. (vi)⇒(i): This would make the above proof of (iv)⇒(ii) appear redundant, but in fact we are going to give a proof based on the above one. First by Theorem 3.3.4 the orbit equivalence relation induced by the Z-action on 2ω×Z is a universal equivalence relation for all Z-orbit equivalence relations. Denote this equivalence relation by EZ . It is thus sufficient to show that EZ ≤B E0 . For convenience we use the variation of E0 on the Baire space ω ω and continue to denote EZ by E. By Exercise 7.2.1 it again suffices to find a Borel partition {Xn }n∈ω so that E  Xn ≤B E0 for all n ∈ ω. We repeat the above proof of (iv)⇒(ii) to define the invariant Borel sets X0 , X1 , and X2 . It suffices to show that E  X2 ≤B E0 . And we also have the finite equivalence relations En defined. Now for each x ∈ X2 and n ∈ ω we let πn (x) = {g ∈ Z : g · xEn x}. We note that πn (x) is a finite set of consecutive integers and for xEn y, πn (x) · x = πn (y) · y. Let m = |πn (x)| and gn (x) = inf πn (x). Then define an n × m matrix Mn (x) = (ai,j )i
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have that Mn (x) = Mn (y). Moreover, Mn−1 (x) is a submatrix of Mn (x), and its relative position in Mn (x) can be determined by gn−1 (x) − gn (x). It is also true that if xEn−1 y then Mn−1 (x) = Mn−1 (y), Mn (x) = Mn (y), and gn−1 (x) − gn (x) = gn−1 (y) − gn (y). Let cn (x) be an integer coding the matrix Mn,m (x) and the integer gn−1 (x) − gn (x). Then we have that if xEn−1 y, then rn (x) = rn (y). We claim that xEy iff (cn (x))E0 (cn (y)). If xEy then for some n we have that xEn y for all n > n. By the construction we have that πn (x) · x = πn (y) · y and eventually cn (x) = cn (y). Thus (cn (x))E0 (cn (y)). Conversely, from the sequence (cn (x)) we can determine x ∈ 2ω×Z up to a shift by an element of Z. This is because, from the information of the sequence Mn (x) and gn−1 (x) − gn (x) we may recover x if g0 (x) is given. Thus if (cn (x)) = (cn (y)) then x and y are in the same orbit of the Z-action, and so xEy. Example 7.2.4 Let X be a standard Borel space and let Z act on X Z by (n · x)(m) = x(m − n). Then the orbit equivalence relation is hyperfinite. In view of the Glimm–Effros dichotomy, we have the following immediate corollary of the theorem. Corollary 7.2.5 Let X be a standard Borel space and E a hyperfinite equivalence relation on X. Then either E is smooth or else E ∼B E0 . Thus E0 is a typical nonsmooth hyperfinite equivalence relation and in fact the unique one up to Borel bireducibility. The theorem also motivates the following definitions. Definition 7.2.6 Let X be a standard Borel space. An equivalence relation E is essentially hyperfinite if E ≤B E0 . E is essentially countable if there is a countable Borel equivalence relation F such that E ≤B F . However, no good characterizations of essential hyperfiniteness are known. Exercise 7.2.1 Let E be a countable Borel equivalence relation on a standard Borel space X. Suppose {Xn }n∈ω is a partition of X into disjoint Borel sets and E  Xn ≤B E0 for all n ∈ ω. Show that E ≤B E0 . Exercise 7.2.2 (Jackson) Let E, F be countable Borel equivalence relations on a standard Borel space X. Suppose E ⊆ F , E is hyperfinite, and every F -equivalence class contains only finitely many E-equivalence classes. Show that F is hyperfinite.

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Countable Borel Equivalence Relations

7.3

165

Universal countable Borel equivalence relations

In this section we investigate countable Borel equivalence relations that are most complicated in the Borel reducibility hierarchy. Such equivalence relations are called universal countable Borel equivalence relations. All results in this section are due to Dougherty, Jackson, and Kechris [30]. By the Feldman–Moore theorem (Theorem 7.1.4) every countable Borel equivalence relation is the orbit equivalence relation of a Borel action of a countable group. If G is a fixed countable group, then among all Borel actions of G one is identified as universal by Theorem 3.3.4. The action is the shift action of G on the power of the Effros Borel space F (G)ω . Here, however, G has the discrete topology, and hence every subset of G is closed. It is thus equivalent to consider the shift action of G on the space (2G )ω or 2G×ω defined by (g · f )(h, n) = f (g −1 h, n) for g ∈ G, f ∈ 2G×ω , h ∈ G, and n ∈ ω. Note that this action is continuous. We have the following immediate corollary. Proposition 7.3.1 Let G be a countable group and Let X be the Polish G-space 2G×ω . Then X is a universal G-orbit equivalence X is a universal Borel G-space and EG relation. Also by Theorem 3.5.2 and a trivial analog of it for quotient groups we have the following extension result. Proposition 7.3.2 Xi for i = 1, 2. If Let G1 , G2 be countable groups, Xi = 2Gi ×ω , and Ei = EG i G1 ≤w G2 , then E1 B E2 . Thus the Borel complexity of the orbit equivalence relations of the shift actions reflect the complexity of the structures of the acting groups. It turns out that this connection between the group structures and the dynamics they produce is much stronger than we would suspect. This strong bondage is now refered to as the rigidity or superrigidity of the groups. It is therefore not surprising that ergodic theory, especially the superrigidity theory, is playing a central role in the study of countable Borel equivalence relations nowadays. Here we only provide an inroad to the study by focusing on the universal countable Borel equivalence relations. For 1 < n < ω let Fn be the free group with n generators. Let Fω be the free group with countably infinitely many generators. It is an easy fact of group theory that Fω is a surjectively universal countable group, that is, every countable group is a quotient group of Fω . Also Fω can be embedded as a subgroup of F2 . This implies that each of Fn , for 1 < n ≤ ω, is a

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weakly universal countable group. Therefore we have the following examples of universal countable Borel equivalence relations. Proposition 7.3.3 Let 1 < n ≤ ω, X = 2Fn ×ω , and E = EFXn . Then E is a universal countable Borel equivalence relation. In search for a simpler representation of the universal countable Borel equivalence relation, we prove the following sharper results. Proposition 7.3.4 Let G1 , G2 be countable groups and Y a Polish space. For i = 1, 2 let Xi = Xi . If G1 ≤w G2 then E1 B E2 . Y Gi with the shift action of Gi and Ei = EG i Proof. X2 by

First suppose G1 ≤ G2 . Let y0 ∈ Y . Define an embedding ϕ : X1 →  ϕ(f )(g) =

f (g), if g ∈ G1 , if g ∈ G1 . y0 ,

Then for g1 ∈ G1 , ϕ(g1 · f ) = g1 · ϕ(f ). Thus f E1 f  implies ϕ(f )E2 ϕ(f  ). Conversely, suppose g ·ϕ(f ) = ϕ(f  ) for some g ∈ G2 . If g ∈ G1 then g ·f = f  . Otherwise it must be the case that both f and f  take constant value of y0 , and thus f = f  . Next suppose G1 is a quotient of G2 , and let π : G2 → G1 be an onto homomorphism. Then define ψ : X1 → X2 by ψ(f )(g) = f (π(g)). Then for any g2 ∈ G2 , g2 ·ψ(f ) = ψ(π(g2 )·f ). It follows easily that E1 B E2 . The general case G1 ≤w G2 is a composition of the above cases considered, and we are done by the transitivity of B . In several steps we will build up a reduction of the shift action of F2 on 2F2 ×ω to simply its shift action on 2F2 . Proposition 7.3.5 X Let G be a countable group, X = 2G×ω , and E = EG . Let Y = 3G×Z with the shift action of G × Z and let F be the orbit equivalence relation. Then E B F . Proof. Fix a bijection of ω with Z−{0}. We can then view X as 2G×(Z−{0}) since they are Borel isomorphic as Borel G-spaces. Define an embedding ϕ : 2G×(Z−{0}) → Y by  f (h, n), if n = 0, ϕ(f )(h, n) = 2, if n = 0.

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Then for any g ∈ G, ϕ(g · f ) = (g, 0) · ϕ(f ). Conversely, if ϕ(f  ) = (g, m) · ϕ(f ), then m = 0 and f  = g · f . This is because, suppose m = 0, then ϕ(f )(h, 0) = 2 for all h ∈ G and therefore ϕ(f  )(h, m) = ((g, m)·ϕ(f ))(h, m) = ϕ(f )(g −1 h, 0) = 2 for all h ∈ G, but this is a contradiction since ϕ(f  )(h, m) = f  (h, m) ∈ {0, 1} by definition. Proposition 7.3.6 X Let G be a countable group, X = 3G , and E = EG . Let F be the orbit equivalence relation of the shift action of G × Z2 on 2G×Z2 . Then E B F . Proof. by

View 0 as encoded by 00, 1 by 01, and 2 by 11. Define ϕ : X → 2G×Z2  ϕ(f )(h, i) =

0, if f (h) = 0 or if f (h) = 1 and i = 0, 1, if f (h) = 1 and i = 1 or if f (h) = 2.

Then for any g ∈ G, ϕ(g · f ) = (g, 0) · ϕ(f ). Conversely, if ϕ(f  ) = (g, i)ϕ(f ), then we claim that ϕ(f  ) = (g, 0) · ϕ(f ) and f  = g · f . Suppose ϕ(f  ) = (g, 1) · ϕ(f ), then ϕ(f  ) = (1G , 1) · ϕ(g · f ), and thus without loss of generality we may assume ϕ(f  ) = (1G , 1) · ϕ(f ). It is a property of the coding used that if ϕ(f  )(h, 0) = 1 then ϕ(f  )(h, 1) = 1 as well. Thus by our assumption if ϕ(f )(h, 1) = ϕ(f  )(h, 0) = 1 then ϕ(f )(h, 0) = ϕ(f  )(h, 1) = 1. Similarly we also have that if ϕ(f )(h, 1) = 0 then ϕ(f )(h, 0) = 0. It follows that for any h ∈ G, ϕ(f )(h, 0) = ϕ(f )(h, 1) = ϕ(f  )(h, 0) = ϕ(f  )(h, 1). So f = f  . Notation 7.3.7 Let E∞ denote the orbit equivalence relation induced by the shift action of F2 on 2F2 . Theorem 7.3.8 E∞ is a universal countable Borel equivalence relation. Proof. By Proposition 7.3.3 the orbit equivalence relation of F2 shift action on 2F2 ×ω is already a universal countable Borel equivalence relation. By Proposition 7.3.5 it is Borel embeddable into orbit equivalence relation of the shift action of F2 × Z on 3F2 ×Z , and by Proposition 7.3.6 the latter orbit equivalence relation is in turn Borel embeddable into that of the shift action of F2 × Z × Z2 on 2F2 ×Z×Z2 . The group F2 × Z × Z2 is a quotient of Fω , and thus by Proposition 7.3.4, the last equivalence relation is Borel embeddable into that of the shift action of Fω on 2Fω . This shows that the shift action of Fω on 2Fω is a universal countable Borel equivalence relation. Now since Fω is a subgroup of F2 , by Proposition 7.3.4 again E∞ is also universal. In the next section we will show that E0
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complicated (see Reference [1] by Adams and Kechris). Exercise 7.3.1 Give a direct proof of Proposition 7.3.1. Exercise 7.3.2 Prove Proposition 7.3.2. Exercise 7.3.3 Show that for any n ≥ 2 the shift action of Fn on 2Fn gives rise to a universal countable Borel equivalence relation. Consider the shift action of Fn on 2Fn . Let Fn be the free part of this action, that is, Fn = {x ∈ 2Fn : ∀g ∈ Fn ( g = 1G ⇒ g · x = x ) }. Exercise 7.3.4 Show that Fn is an invariant dense Gδ subset of 2Fn . Exercise 7.3.5 Let μn be the usual product measure on 2Fn . Show that μn (Fn ) = 1. Exercise 7.3.6 Let En be the orbit equivalence relation on Fn of the induced Fn action. Show that En B E2 for all n ≥ 2.

7.4

Amenable groups and amenable equivalence relations

In this section we show that the universal countable Borel equivalence relation E∞ defined in the preceding section is not hyperfinite. To do this we discuss the concept of amenability for countable groups and their orbit equivalence relations. The study of amenable groups and amenable equivalence relations is interesting in its own right and is a well-established subject in ergodic theory. We first give the definitions. Definition 7.4.1 Let X be a set. A finitely additive probability measure on X is a map μ : P (X) → [0, 1], where P (X) = {A : A ⊆ X} is the power set of X, such that μ(∅) = 0, μ(X) = 1, and μ(A ∪ B) = μ(A) + μ(B) if A ∩ B = ∅. If G is a group, a finitely additive probability measure μ on G is leftinvariant if for all g ∈ G and A ⊆ G, μ(gA) = μ(A). A countable group G is amenable if there is a left-invariant finitely additive probability measure μ on G.

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Definition 7.4.2 A countable group G satisfies the Reiter condition if for all  > 0 and  g1 , . . . , gn ∈ G there is f : G → R≥0 such that g∈G f (g) = 1 and for any  1 ≤ i ≤ n, g∈G |f (g) − f (gi−1 g)| < . G satisfies the Følner condition if for all  > 0 and g1 , . . . , gn ∈ G there is a finite F ⊆ G such that for any 1 ≤ i ≤ n, |gi F F | < |F |. It is known that the Reiter condition and the Følner condition are equivalent to amenability for countable groups. Here, to keep our exposition selfcontained, we only prove the following weaker theorem. Theorem 7.4.3 Let G be a countable group. If G satisfies the Reiter condition then it also satisfies the Følner condition. If G satisfies the Følner condition then it is amenable. Proof. First we assume that G satisfies the Reitercondition. Let  > 0 and g1 , . . . , gn ∈ G. Let f : G → R≥0 be such that g∈G f (g) = 1 and for  any 1 ≤ i ≤ n, g∈G |f (g) − f (gi−1 g)| < /n. For each 0 < r ≤ 1 we let Fr = {g ∈ G : f (g) ≥ r} and χr be the characteristic function of Fr . Since  1 g∈G f (g) = 1 the set Fr is finite for any r > 0. Note that f (g) = 0 χr (g)dr,   |Fr | = g∈G χr (g), and |Fr gi Fr | = |Fr gi−1 Fr | = g∈G |χr (g)− χr (gi−1 g)| for any 1 ≤ i ≤ n. Then for any 1 ≤ i ≤ n, ! 1 ! 1 |Fr gi Fr |dr = |χr (g) − χr (gi−1 g)|dr 0

=

0 g∈G

! g∈G

<

1

0

|χr (g) − χr (gi−1 g)|dr =

|f (g) − f (gi−1 g)|

g∈G

! !     1  1  = f (g) = χr (g)dr = χr (g)dr n n n n 0 0 g∈G

=



 n

!

It follows that

1

g∈G

g∈G

|Fr |dr.

0

! 0

n 1 i=1

! |Fr gi Fr |dr <  0

1

|Fr |dr,

n and thus there is 0 < r ≤ 1 such that i=1 |Fr gi Fr | < |Fr |. This Fr witnesses the Følner condition for G. Now for the second part of the theorem assume that G satisfies the Følner condition. Enumerate G as g0 , g1 , . . . , gn , . . . and for each n ∈ ω let Fn ⊆ G be

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a finite subset of G such that |Fn gi Fn | < 2−n |Fn |. Let U be a nonprincipal ultrafilter on ω. For A ⊆ G, define μ(A) = lim

n∈U

|A ∩ Fn | . |Fn |

Then μ is a finitely additive probability measure on G. It remains to check that μ is left-invariant. For this let g ∈ G and A ⊆ G. Then | |gA ∩ Fn | − |A ∩ Fn | | |Fn | " " " |A ∩ g −1 Fn | − |A ∩ Fn | " = lim n∈U |Fn | " −1 " "g Fn Fn " ≤ lim = 0. n∈U |Fn |

|μ(gA) − μ(A)| = lim

n∈U

Thus G is amenable. It is easy to see that Z satisfies the Reiter condition. In fact for any integer N > 1 let fN : Z → R≥0 be defined as  fN (n) =

1/N, if 0 < n ≤ N , 0, otherwise.

Then for any finite collection of elements of Z one of the functions fN witnesses the Reiter condition. It follows that Z is amenable. Proposition 7.4.4 F2 is not amenable. Proof. Assume that μ is a left-invariant finitely additive probability measure on F2 . Let a and b be the generators of F2 . For each s ∈ {a, b, a−1 , b−1 } let A(s) be the set of all elements of F2 whose reduced form starts with s. Then A(a), A(b), A(a−1 ), A(b−1 ) form a partition of F2 . Also we have that A(a) = a(A(b) ∪ A(b−1 ) ∪ A(a)). Since μ is left-invariant, we have that μ(A(a)) = μ(A(b) ∪ A(b−1 ) ∪ A(a)) = μ(A(b)) + μ(A(b−1 )) + μ(A(a)). It follows that μ(A(b)) = μ(A(b−1 )) = 0. Similarly we also obtain that μ(A(a)) = μ(A(a−1 )) = 0. But then μ(F2 ) = 0, a contradiction. We now turn to equivalence relations. The following definition of amenable equivalence relations is inspired by the Reiter condition.

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Definition 7.4.5 Let X be a standard Borel space, μ a probability Borel measure on X, and E a countable equivalence relation. We say that (X, E, μ) is amenable, or E is μ-amenable, if there is a sequence of Borel functions ϕn : E → R≥0 such that (a) for any x ∈ X,

 yEx

ϕn (x, y) = 1, and

(b) there is a Borel E-invariant set A ⊆ X with μ(A) = 1 such that for all (x, x ) ∈ E ∩ (A × A), 

|ϕn (x, y) − ϕn (x , y)| → 0 as n → ∞.

yEx

Proposition 7.4.6 Let G be a countable group, X a Borel G-space, and μ a Borel probability X measure on X. If G satisfies the Reiter condition, then EG is μ-amenable. Proof. Enumerate . . . and for each n ≥ 1 let fn : G → G as g0 , g1 , . . . , gn , R≥0 be such that g∈G f (g) = 1 and g∈G |f (g) − f (gi−1 g)| < 1/n for all i < n. Then for x, y ∈ X let  ϕn (x, y) = fn (g). g·y=x



  Then for every x ∈ X, yEx ϕn (x, y) = g∈G fn (g) = 1. If xEx and  x = h · x , then for any yEx,    fn (g) − fn (g  ) = fn (g) − fn (h−1 g), ϕn (x, y) − ϕn (x , y) = g·y=x

and



g ·y=x

|ϕn (x, y) − ϕn (x , y)| ≤

yEx



g·y=x

|fn (g) − fn (h−1 g)| → 0

g∈G

X is μ-amenable. as n → ∞. Thus EG

In particular, every hyperfinite equivalence relation is μ-amenable for any Borel probability measure μ. We will show below that the amenability of an orbit equivalence relation can be pulled back to conclude the amenability of the group. For this we need the notion of G-invariance for measures, as follows.

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Definition 7.4.7 Let G be a countable group, X a Borel G-space, and μ a Borel probability measure on X. We say that μ is G-invariant if for all g ∈ G and Borel S ⊆ X, μ(S) = μ(g · S). Theorem 7.4.8 Let G be a countable group, X a Borel G-space, and μ a Borel probability X is measure on X. Suppose the action is free and μ is G-invariant. If EG μ-amenable, then G is amenable. Proof. It suffices to show that G satisfies the Reiter condition. Let ϕn be X . Define Borel functions witnessing the μ-amenability of EG ! fn (g) = ϕn (x, g · x)dμ(x).   Then g∈G fn (g) = g∈G ϕn (x, g · x)dμ(x) = 1. In addition, for every h ∈ G, by the assumptions we have  |fn (g) − fn (h−1 g)| g∈G ! !  = | ϕn (x, g · x)dμ(x) − ϕn (x, h−1 g · x)dμ(x)| g∈G ! !  = | ϕn (x, g · x)dμ(x) − ϕn (h · x, g · x)dμ(x)| g∈G !  ≤ |ϕn (x, g · x) − ϕ(h · x, g · x)|dμ(x) g∈G !  = |ϕn (x, g · x) − ϕn (h · x, g · x)|dμ(x) ! g∈G  = |ϕn (x, y) − ϕn (h · x, y)|dμ(x) → 0 yEx

as > 0 and g1 , . . . , gm ∈ G, then for large enough n we have  n → ∞. Now if  −1 |f (g) − f (g g)| <  for all 1 ≤ i ≤ m. This shows that G satisfies n n i g∈G the Reiter condition, and is therefore amenable. The usual product measure on 2G is a canonical example of a G-invariant measure (Exercise 7.4.5). To apply the above theorem we check below that its free part has full measure. Lemma 7.4.9 Let G be a countably infinite group, X = 2G with the shift action of G, and μ the usual product measure on X. Let F be the free part of X. Then μ(F ) = 1.

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Proof. We show that for any g ∈ G, the set {x ∈ 2G : g · x = x} has measure 0. Consider two cases. Case 1: g is of infinite order, that is, for any n = m ∈ Z, g n = g m . In this case note that if g · x = x then for all n ∈ Z, we have x(g 2n ) = x(g 2n+1 ). Thus # $ μ ({x : g · x = x}) ≤ μ {x : ∀n ∈ Z x(g 2n ) = x(g 2n+1 )} =

 n∈Z

# $ 1 = 0. μ {x : x(g 2n ) = x(g 2n+1 )} = 2 n∈Z

Case 2: g has finite order, that is, for some n ∈ Z, g n = 1G . In this case let H = g ≤ G. Then H is a finite subgroup of G. Since G is infinite, there are infinitely many right-cosets of H in G. Let g0 , g1 , . . . be chosen from infinitely many distinct cosets of H in G. Note that if g · x = x then for all n ∈ ω, we have x(gn ) = x(ggn ). Moreover, for distinct n, m ∈ ω, {gn , ggn } ∩ {gm , ggm } = ∅. Thus similar to Case 1 we have μ ({x : g · x = x}) ≤ μ ({x : ∀n ∈ ω x(gn ) = x(ggn )}) =



μ ({x : x(gn ) = x(ggn )}) =

n∈ω

1 = 0. 2

n∈Z

We are finally ready to prove the main theorem of this section. Theorem 7.4.10 E∞ is not hyperfinite. In particular, E0
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Exercise 7.4.4 Prove that if G and H both satisfy the Reiter condition then so does G × H. Exercise 7.4.5 Let G be a countable group, X = 2G with the shift action of G, and μ the usual product measure on X. Show that μ is G-invariant. Exercise 7.4.6 Let F2 be the free part of 2F2 with the shift action of F2 . Let E∞T denote the equivalence relation E∞ restricted to F2 . Show that E∞T is not hyperfinite.

7.5

Actions of locally compact Polish groups

In this section we prove a theorem of Kechris [92] that any orbit equivalence relation of an action of a locally compact Polish group is essentially countable, that is, Borel reducible to a countable Borel equivalence relation. We will use the following theorem, which is a generalization of the Luzin–Novikov uniformization theorem (Theorem 7.1.2) to sets with Kσ sections. Recall that a subset of topological space is Kσ if it is a union of countably many compact sets. Theorem 7.5.1 (Arsenin–Kunugui) Let Y be a Polish space, X a standard Borel space, and P ⊆ X × Y Borel. Suppose each section Px is Kσ . Then projX (P ) is Borel and P has a Borel uniformization. A proof can be found in Reference [97] Theorem 35.46. Also recall that a complete section of an equivalence relation E on X is a subset Y of X which meets every E-equivalence class. We say that a complete section Y is countable if Y ∩ [x]E is countable for every x ∈ X. Theorem 7.5.2 (Kechris) Let G be a locally compact Polish group and X a Borel G-space. Then there X . is a Borel countable complete section of EG Proof. In view of Theorem 4.4.6 we may assume X is a Polish G-space without loss of generality. Let d be a compatible metric on X. Let K0 be a compact nbhd of 1G and K a compact symmetric nbhd (that is, K = K −1 ) of 1G with K 2 ⊆ K0 . Let N = Int(K). Consider the following relation defined on X: R(x, y) ⇐⇒ ∃g ∈ K (g · x = y).

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R is symmetric and reflexive, but not necessarily an equivalence relation on X. However, we will construct a sequence of Borel sets Xn ⊆ X with X = n Xn such that R  Xn is an equivalence relation. For  > 0 put A = {x ∈ X : ∀g ∈ K0 ( d(x, g · x) ≤  ⇒ g ∈ N Gx ) }. Recall that Gx is the stabilizer of x. We check that each A is Borel. Note that x ∈ A ⇐⇒ ∃g P (x, g), where P (x, g) ⇐⇒ g ∈ K0 − N Gx and d(x, g · x) ≤ . By the Arsenin–Kunugui uniformization theorem it suffices to show that P is Borel and that each section Px is Kσ . It is clear that every section Px is Kσ since G as a locally compact Polish group is Kσ and each Px is closed in G. To see that P is Borel, note that the conditions g ∈ K0 and d(x, g · x) ≤  are closed. Hence it is enough to see that g ∈ N Gx is a Borel condition. For this we note that g ∈ N Gx ⇐⇒ ∃h ∈ Gx ∃k ∈ N ( g = hk ) ⇐⇒ ∃h∃k Q(g, x, h, k), where Q(g, x, h, k) ⇐⇒ h · x = x ∧ k ∈ N ∧ g = hk. Q is obviously Borel, with sections Qg,x Fσ in G × G and hence Kσ . Thus again by the Arsenin–Kunugui theorem we have that g ∈ N Gx is Borel, which in turn implies that A is Borel.

Next we claim that X = n≥1 A1/n . Toward a contradiction assume x ∈ A1/n for any n ≥ 1. We obtain a sequence gn ∈ K0 such that d(x, gn ·x) ≤ 1/n but gn ∈ N Gx . By compactness of K0 there is an accumulation point g∞ ∈ K0 of gn . Then g∞ · x = x but g∞ ∈ N Gx since N Gx is open. This is a contradiction since 1G ∈ N . Next we note that if B ⊆ A has diameter ≤  then R  B is an equivalence relation. If suffices to check that R  B is transitive. For this let x, y, z ∈ B with R(x, y) and R(y, z). So there are g, h ∈ K with g · x = y and h · y = z. Since K 2 ⊆ K0 , hg ∈ K0 and d(x, hg · x) = d(x, z) ≤ diam(B) ≤ . Thus hg ∈ N Gx . This means that there is k ∈ N ⊆ K with z = k · x, and hence R(x, z). For each n ≥ 1 let {Bn,m : m ∈ ω} be a countable open cover of X with diam(Bn,m ) ≤ 1/n. Then the collection {A1/n ∩ Bn,m : n ≥ 1, m ∈ ω} is a countable Borel cover of X so that the restriction of R on each set is an equivalence relation. Let Xn enumerate this collection of Borel sets.

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We claim that R  Xn is smooth. To see this note that R is closed in X ×X. Let τ be a Polish topology on X extending the topology induced by d so that Xn is clopen in (X, τ ). Then R is still closed in τ × τ . Now R  Xn is a closed equivalence relation on the Polish space (Xn , τ  Xn ), therefore it is smooth. X By Proposition 5.4.10 EG is idealistic, that is, there is a Borel map C → IC X assigning to each E -equivalence class C a σ-ideal IC so that C ∈ IC G

. Now X since X = n Xn , for each EG equivalence class C we also have C = n (C ∩ Xn ). It follows that there is an n ∈ ω such that C ∩ Xn ∈ IC . Let n(C) be the least such n for the class C. Put  Y0 = [x]G ∩ Xn([x]G ) . x∈X

Then Y0 is Borel since for C = [x]G , x ∈ Y0 ⇐⇒ ∃n [ ∀m < n ( C ∩ Xm ∈ IC ) ∧ C ∩ Xn ∈ IC ∧ x ∈ Xn ]. X It is clear that Y0 is a complete section for EG . Let Y0,n = Y0 ∩ Xn . Then {Y0,n : n ∈ ω} is a partition of Y0 . Since R  Xn is a smooth equivalence relation, it follows that R  Y0,n is also a smooth equivalence relation, and X therefore so is R  Y0 . It is clear that each EG  Y0 equivalence class in Y0 contains only countably many R  Y0 -classes. Define Y ⊆ Y0 by

x ∈ Y ⇐⇒ x ∈ Y0 ∧ [x]RY0 ∈ I[x]G . X . We still have that R  Y is Then again Y is a Borel complete section of EG smooth. Now R  Y is idealistic, witnessed by the assignment

[x]RY −→ I[x]G  [x]RY . By Theorem 5.4.11 R  Y has a Borel selector as well as a Borel transversal. X , A is a complete Let A be a Borel transversal of R  Y on Y . Since R ⊆ EG X X section of EG . Each EG  Y class in Y contains only countably many R  Y X , as required. classes. Thus A is a Borel countable complete section for EG Corollary 7.5.3 X Let G be a locally compact Polish group and X a Borel G-space. Then EG X is essentially countable, that is, EG ≤B E∞ . X Proof. Let A ⊆ X be a Borel countable complete section of EG . Let X F = EG  A. Then F is a countable Borel equivalence relation on the standard Borel space A. By Theorem 7.1.2 there is a Borel function f : X → A X X x for all x ∈ X. Then f witnesses that EG ≤B F . so that f (x)EG

Exercise 7.5.1 Show that the following are equivalent for a Borel equivalence relation E:

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(a) E is essentially countable; (b) there is a Borel countable complete section of E;

(c) E = n En where each En is smooth and idealistic;

(d) E = n En where each En has a Borel selector. Exercise 7.5.2 Let n ≥ 1. Consider the space F (Rn ) of all closed subsets of Rn . If A, B ∈ F (Rn ) we say that A is congruent to B, denoted A ∼ B, if there is an isometry from A onto B. Show that the congruence relation on F (Rn ) is essentially countable.

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Chapter 8 Borel Equivalence Relations

In this chapter we continue our introduction of Borel equivalence relations. The primary classification here is orbit equivalence relations versus equivalence relations not reducible to orbit equivalence relations. Again we only give a self-contained account of the basic facts. The more interested reader should go beyond this book and find a vast literature about the subject. Along with many theorems there are also a large number of open problems. Many of the problems are fundamental, the answers of which might potentially redefine the subject.

8.1

Hypersmooth equivalence relations

The concept of a hypersmooth equivalence relation is a natural generalization of hyperfiniteness. Definition 8.1.1 Let X be a standard Borel space. An equivalence relation E on X is hypersmooth if E = n En , where En are smooth equivalence relations and En ⊆ En+1 for all n. Similar to the equivalence relation E0 we define a canonical hypersmooth equivalence relation E1 . Definition 8.1.2 The equivalence relation E1 is defined on 2ω×ω as xE1 y ⇐⇒ ∃m ∀n ≥ m ∀k x(n, k) = y(n, k). It is easy to check that E1 is hypersmooth. We can also consider the eventual agreement for sequences of actual real numbers. Let E1 (R) denote the equivalence relation on Rω defined by (xn )E1 (yn ) ⇐⇒ ∃m ∀n ≥ m xn = yn . It is easy to see that E1 (R) is also hypersmooth and that E1 ∼B E1 (R). For notational simplicity we also write x = (xn ) for x ∈ 2ω×ω and xn ∈ 2ω , where x(n, k) = xn (k) for all n, k ∈ ω.

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We have the following closure properties for hypersmooth equivalence relations. The proof is left to the reader (Exercise 8.1.1). Proposition 8.1.3 Let X, Y be standard Borel spaces and E, F equivalence relations on X, Y , respectively. Then the following are true: (1) If F is hypersmooth and E ≤B F , then E is hypersmooth. (2) If E is hypersmooth and A ⊆ X is Borel, then E  A is hypersmooth. (3) If E and F are hypersmooth, then so is E × F . The following characterization of hypersmoothness is analogous to the one on hyperfiniteness. Proposition 8.1.4 Let X be a standard Borel space and E an equivalence relation on X. Then the following are equivalent: (i) E is hypersmooth. (ii) E ≤B E1 . (iii) E B E1 . Proof. By Proposition 8.1.3 (1) we have that (iii)⇒(ii)⇒(i). It suffices to

show (i)⇒(iii). For this let E be hypersmooth. Let E = n Fn where Fn is smooth and Fn ⊆ Fn+1 for all n ∈ ω. Without loss of generality assume that F0 = id(X). Let fn : X → 2ω witness that Fn ≤B id(2ω ). Then in particular f0 is a Borel injection from X into 2ω . Define f : X → 2ω×ω by f (x)(n, k) = fn (x)(k). Then f is a Borel injection and xEy ⇐⇒ f (x)E1 f (y), as required. Hyperfinite equivalence relations are of course hypersmooth. Conversely, we have the following theorem. Theorem 8.1.5 (Dougherty–Jackson–Kechris) Let X be a standard Borel space and E a countable Borel equivalence relation on X. If E is hypersmooth, then it is hyperfinite.

Proof. Let E = n Fn , where each Fn is smooth, Fn ⊆ Fn+1 for all n ∈ ω. Without loss of generality assume F0 = id(X). Each Fn is a countable equivalence relation. Hence by the Feldman–Moore Theorem 7.1.4 there are countX . By Proposiable groups Gn with Borel actions on X so that Fn = EG n tion 5.4.10 and Theorem 5.4.11 there are Borel selectors sn for Fn , that is,

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sn (x)Fn x and sn (x) = sn (y) for all x, y ∈ X with xEn y. Fix an enumeration for Gn as {gn,k }k∈ω . We define a sequence of relations Rn as follows: xRn y ⇐⇒ ∃m ≤ n ( xFm y∧ ∃k0 , . . . , km ≤ n [x = g0,k0 s0 g1,k1 s1 . . . gm,km sm (x)]∧ ∃l0 , . . . , lm ≤ n [y = g0,l0 s0 g1,l1 s1 . . . gm,lm sm (y)]). Then Rn ⊆ Fn and Rn ⊆ Rn+1 . We claim that E = n Rn . Suppose xEy. Then there is m ∈ ω such that xFm y. For every j ≤ m, there is gj,kj ∈ G with x = gj,kj sj (x). Similarly, for each j ≤ m there is gj,lj ∈ G with y = gj,lj sj (y). Then we have that x = g0,j0 s0 g1,j1 s1 . . . gm,km sm (x) and y = g0,l0 s0 g1,l1 s1 . . . gm,lm sm (y). Let n = max{m, k0 , k1 , . . . , km , l0 , . . . , lm }. Then xRn y. Next we check that Rn is an equivalence relation. Clearly Rn is reflexive and symmetric. We only show its transitivity. For this let xRn yRn z. Let m, k0 , . . . , km , l0 , . . . , lm ≤ n witness that xRn y and, without loss of generality, let p ≤ m and h0 , . . . , hp ≤ n be such that yEp z and g0,h0 s0 g1,h1 s1 . . . gp,hp sp (z) = z. If m = p then there is nothing to prove. We assume p < m. Then we have that xFm yFp z and thus xFm z. Note that sm (y) = sm (z) since yFp z and p < m. Let v = gp+1,lp+1 sp+1 . . . gm,lm sm (y) = gp+1,lp+1 sp+1 . . . gm,lm sm (z). Then y = g0,l0 s0 g1,l1 s1 . . . gp,lp sp (v) and hence vFp y. Since yFp z it follows that vFp z, and thus sp (v) = sp (z). Therefore z = g0,h0 s0 g1,h1 s1 . . . gp,hp sp (z) = g0,h0 s0 g1,h1 s1 . . . gp,hp sp (v) = g0,h0 s0 g1,h1 s1 . . . gp,hp sp gp+1,lp+1 sp+1 . . . gm,lm sm (y) = g0,h0 s0 g1,h1 s1 . . . gp,hp sp gp+1,lp+1 sp+1 . . . gm,lm sm (z). This shows that xRn z, and thus Rn is an equivalence relation. Finally note that Rn is finite, since sm (x) = sm (y) for any yFm x. We have thus obtained

an increasing sequence of finite Borel equivalence relations Rn with E = n Rn , and therefore E is hyperfinite. In the rest of this section we analyze the equivalence relation E1 and show that it is not essentially countable. This implies in particular that it is not essentially hyperfinite, and hence E0
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Theorem 8.1.6 (Kechris–Louveau) Let X be a standard Borel space, F a countable Borel equivalence relation on X, and f : 2ω×ω → X a Borel map such that xE1 y ⇒ f (x)F f (y). Then there are x, y ∈ 2ω×ω such that (a) for n = m, xn = xm and yn = ym , (b) for n, m ∈ ω, xn = ym , and (c) f (x) = f (y). In particular, E1 ≤B F . Proof. Let C = n Un ⊆ 2ω×ω where Un ⊇ Un+1 are dense open such that f  C is continuous. We construct sequences (ln ), (kn ), (pn ), (qn ), (xn ), (yn ) such that the following conditions are met: (i) ln , kn ∈ ω, ln < ln+1 , kn < kn+1 for all n ∈ ω; (ii) pn , qn ∈ 2ln ×kn ; (iii) Npn , Nqn ⊆ Un , where Ns = {x ∈ 2ω×ω : s ⊆ x} for any s ∈ 2l×k with l, k ∈ ω; (iv) xn , yn ∈ 2ln ×ω with pn ⊆ xn ⊆ xn+1 , qn ⊆ yn ⊆ yn+1 ; for i = j < ln , (xn )i = (xn )j , (yn )i = (yn )j ; for any i, j < ln , (xn )i = (xn )j ; (v) ∀∗ z ∈ 2ω×ω (xn  z ∈ C ∧ yn  z ∈ C); (vi) ∃∗ z ∈ 2ω×ω f (xn  z) = f (yn  z).

Granting this construction, we can let x = n xn and y = n yn . Then x and y satisfy conditions (a) and (b) of the theorem, and by (iii) x, y ∈ C. Let zn ∈ 2ω×ω be such that xn  zn , yn  zn ∈ C and f (xn  zn ) = f (yn  zn ). Such zn exists by (v) and (vi). Then xn  zn → x and yn  zn → y as n → ∞, and by the continuity of f on C we get that f (x) = f (y). To show that the construction is possible, we need to use the following claim. Claim. Let p ∈ 2l×k and ϕ : D → X, where D ⊆ Np is comeager in Np , such that for x, y ∈ D, xE1 y ⇒ ϕ(x)F ϕ(y). Then there is p ≥ p and r both in 2<ω×<ω such that ∀∗ u ∈ Np ∀∗ v ∈ Np ∀∗ z ∈ Nr ϕ(u z) = ϕ(v  z). To prove the claim, consider a family of functions defined as follows. For x ∈ 2ω×ω define ψx : 2l×ω → X by letting ψx (u) = ϕ(u x) wherever ϕ is defined. For a comeager set of x in 2ω×ω , ψx is well defined on a comeager set of u in Np . Since for u, v ∈ 2l×ω , ux E1 v x for any x ∈ 2ω×ω , we also get that ψx (u) = ϕ(u x)F ϕ(v  x) = ψx (v).

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Therefore, whenever ψx is defined, its range is a countable set. Thus there are px ⊇ p in 2l×<ω such that ψx is constant on a comeager subset of Npx in 2l×ω . A similar argument shows that there is r ∈ 2<ω×<ω such that for a comeager set of x in Nr the map x → px is constant. Let p be this constant value of px for comeager many x ∈ Nr . Thus ∀∗ x ∈ Nr ∀∗ u ∈ Np ∀∗ v ∈ Np ϕ(u x) = ϕ(v  x). The claim follows by the Kuratowski–Ulam theorem. We now turn to the construction by induction on n. Assume that ln , kn , pn , qn , xn , and yn have been constructed to satisfy (i) through (vi). By (v) and (vi) there is r ∈ 2<ω×<ω such that $ # ∀∗ z ∈ Nr xn  z, yn  z ∈ C ∧ f (xn  z) = f (yn  z) . 



Without loss of generality assume r ∈ 2l ×k for some l > 0 and k  > kn .  Also by extending we may assume that if we let s = xn  (ln × k  ) r then ∗ l ×ω ∗ ω×ω ∀ z∈2 Ns ⊆ Un+1 . Then it follows that ∀ u ∈ Nr ∩ 2 f (xn  u z) = f (yn  u z). Now define ϕ(x) = f (xn  x) = f (yn  x) for x = u z where u ∈ Nr and z ∈ 2ω×ω . Then on a comeager set of Nr ϕ is well defined. It certainly satisfies that xE1 y ⇒ ϕ(x)F ϕ(y) since f satisfies it. Thus by the above claim we can find extensions r ⊇ r such that ∀∗ u ∈ Nr ∀∗ v ∈ Nr ∀∗ z ∈ 2ω×ω : f (xn  u z) = f (yn  v  z). Now it is easy to extend r to pn+1 and qn+1 so that for any u ⊇ pn+1 and v ⊇ qn+1 , condition (iv) holds for u, v. Let xn+1 ⊇ pn+1 and yn+1 ⊇ qn+1 be witnesses in the appropriate comeager set. Then (v) and (vi) are satisfied. This completes the construction as well as the proof of the theorem. Kechris and Louveau [102] have shown a dichotomy theorem that any hypersmooth equivalence relation is either essentially hyperfinite or else Borel bireducible with E1 . The proof uses the Gandy–Harrington topology as in the proof of the Glimm–Effros dichotomy theorem. This penetrating theorem completely determines the Borel reducibility structure of hypersmooth equivalence relations. A by-product of their proof is that E1 is not Borel reducible to any idealistic Borel equivalence relation. A further result of theirs states X (even if it that E1 is not Borel reducible to any orbit equivalence relation EG is non-Borel). These results suggest the possibility that idealistic equivalence relations coincide with the orbit equivalence relations. We will give a proof of this last result, that E1 is not Borel reducible to any orbit equivalence relation, later in Section 10.6 (Theorem 10.6.1) as an application of the techniques developed in Chapter 10.

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Exercise 8.1.1 Prove Proposition 8.1.3. Exercise 8.1.2 Recall that the tail equivalence relation Et on 2ω is defined by xEt y ⇐⇒ ∃n∃m∀k ( x(n + k) = y(m + k) ). Show that Et is hyperfinite but not smooth. Thus E ∼B E0 . Exercise 8.1.3 Show that a hypersmooth equivalence relation is essentially countable iff it is idealistic. (Hint: Use Exercise 7.5.1.)

8.2

Borel orbit equivalence relations

In Section 3.4 we already analyzed some conditions guaranteeing Borelness of orbit equivalence relations. For instance, if a Borel action of a Polish group is free then the orbit equivalence relation is Borel. We now give some more characterizations of Borelness of orbit equivalence relations. Recall the following notation about group actions. Let G be a Polish group and X a Borel G-space. For x ∈ X the stabilizer of x is the closed subgroup Gx = {g ∈ G : g · x = x}. For x, y ∈ X we denote Gx,y = {g ∈ G : g · x = y}. Then Gx,y is a left coset of Gx since for any g, h ∈ Gx,y , h−1 g ∈ Gx . We view Gx and Gx,y as elements of the Effros Borel space F (G) of all closed subsets of G. Theorem 8.2.1 (Becker–Kechris) Let G be a Polish group and X a Borel G-space. Then the following are equivalent: X (i) EG is Borel.

(ii) The map x → Gx from X into F (G) is Borel. (iii) The map (x, y) → Gx,y from X × X into F (G) is Borel.

X y Proof. We show that (i)⇒(ii)⇒(iii)⇒(i). (iii)⇒(i) is obvious since xEG iff Gx,y = ∅. We next show (ii)⇒(iii). For this let U ⊆ G be nonempty open. Then

Gx,y ∩ U = ∅ ⇐⇒ ∃g ∈ G (g · x = y ∧ Gx ∩ g −1 U = ∅) ⇐⇒ Gx,y = ∅ ∧ ∀g ∈ G (g · x = y ⇒ Gx ∩ g −1 U = ∅)

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The computation shows that the condition is both Σ11 and Π11 , and is therefore Borel. X It remains to show (i)⇒(ii). For this we assume EG is Borel. We use the X ⊆ X × X, to obtain uniform version of Theorem 4.4.6, Exercise 4.4.4, for EG a Borel set B ⊆ X ×X ×ω with the following properties. For y ∈ X and n ∈ ω let By,n = {x ∈ X : (x, y, n) ∈ B} and let σy be the topology generated by the countable collection {By,n : n ∈ ω}. Then for each y ∈ X, By,0 = [y]G , and (X, σy ) is a Polish G-space. We may assume that {By,n : n ∈ ω} is a countable base for σy . We are now ready to show that the map x → Gx from X into F (G) is Borel. For this we fix a countable base U = {Um } of nonempty open sets for −1 Uk : the topology of G. It is easy to see that the countable collection {Um m, k ∈ ω} is still a base for the topology of G. Moreover, −1 Uk = ∅ ⇐⇒ Um Gx ∩ Uk = ∅. Gx ∩ Um

Thus, in order to finish the proof, it suffices to show that for any nonempty open U, V ⊆ G, the set {x ∈ X : U Gx ∩ V = ∅} is Borel. For this we claim that Um V → x ∈ Bx,n ). U Gx ∩ V = ∅ ⇐⇒ ∃m ∀n (Um ⊆ U ∧ x ∈ Bx,n

Suppose U Gx ∩ V = ∅ and let g0 ∈ Gx and h0 ∈ U with h0 g0 ∈ V . Let Um ⊆ U with h0 ∈ Um and Um g0 ⊆ V . Now for any A ⊆ X, x ∈ A Um ⇐⇒ ∃∗ h ∈ Um (h · x ∈ A) ⇐⇒ ∃∗ h ∈ Um g0 ⊆ V (h · x ∈ A) =⇒ x ∈ A V . Thus the direction (⇒) follows with A = Bn,x for any n ∈ ω. For the direction (⇐) assume U Gx ∩V = ∅. Then U ·x∩V ·x = ∅. Consider the topology σx on X. (X, σx ) is a Polish G-space and [x]G is σx -open. By Effros’ Theorem 3.2.4 the map gGx → g · x is a homeomorphism from G/Gx onto [x]G . It follows that U · x = U Gx · x is σx -open. Since {Bx,n : n ∈ ω} is a countable base for V σx , there exists n ∈ ω such that x ∈ Bx,n ⊆ U · x. It follows that x ∈ Bn,x . This completes the proof of the claim. The following is another useful characterization of Borelness of orbit equivalence relations. Theorem 8.2.2 (Sami) X Let G be a Polish group and X a Polish G-space. Then EG is Borel iff there 0 is α < ω1 such that every G-orbit is Πα . X X Proof. If EG is Borel then there is α < ω1 such that EG is a Π0α subset of X X × X. Then every G-orbit is a section of EG , and hence is Π0α . Conversely,

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consider for any α < ω1 the equivalence relation Eα defined by xEα y ⇐⇒ for all Π0α invariant subsets A ⊆ X, x ∈ A iff y ∈ A. X If every G-orbit is Π0α then Eα = EG . Hence it suffices to show that Eα is 1 Π1 for all α < ω1 . For this we use the fact that there is a Π0α subset U of ω ω × X which is universal for Π0α subsets of X (see Exercise 1.5.6). Consider the action of G on ω ω × X defined by

g · (z, x) = (z, g · x). Then the action is still continuous. Now let T = U ∗G ⊆ ω ω × X. Then for any z ∈ ω ω and x ∈ X, x ∈ Tz ⇐⇒ ∀∗ g ∈ G g ·(z, x) ∈ U ⇐⇒ ∀∗ g ∈ G g ·x ∈ Uz ⇐⇒ x ∈ (Uz )∗G . We claim that {Tz : z ∈ ω ω } consists of all Π0α invariant subsets of X. Since Tz = (Uz )∗G and Uz ∈ Π0α , Tz is Π0α invariant. Conversely, if A is Π0α invariant, then there is z ∈ ω ω with Uz = A, and furthermore, Tz = (Uz )∗G = A∗G = A. This finishes the proof of the claim. We thus have that xEα y ⇐⇒ ∀z ∈ ω ω (x ∈ Tz ↔ y ∈ Tz ), and hence Eα is Π11 . The assumption of a Polish group action is crucial in the theorem. It is easy to construct non-Borel equivalence relations with all equivalence classes closed (see Exercise 7.1.2). We give some examples of Borel orbit equivalence relations. By Corollary 7.5.3 any locally compact Polish group action generates an essentially countable equivalence relation, and thus a Borel one. In particular, the following scheme will yield many examples of Borel orbit equivalence relations with potentially distinct complexity in the Borel reducibility hierarchy. Consider an arbitrary uncountable Polish group H. Let G be an arbitrary countable subgroup of H. Then the coset equivalence relation H/G is a countable equivalence relation. Note that G is regarded to have the discrete topology, rather than the subspace topology inherited from H. However, the Borel structures on G generated by these topologies are the same. The following concept is an abstraction of this circumstance. Definition 8.2.3 Let H be a Polish group. A Polishable subgroup G of H is a Borel subgroup of H which admits a Polish topology τ generating the Borel structure on G inherited from that of H, and such that (G, τ ) is a topological group. If H is a Polish group and G is a Polishable subgroup of H, then the translation action of G on H is continuous and the coset equivalence relation

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H/G is obviously a Borel orbit equivalence relation. The seemingly technical requirement of Polishability is actually fulfilled by many examples other than countable subgroups. For example, the classical Banach spaces p (1 ≤ p < ∞) and c0 are Polishable subgroups of the additive group Rω . The Borel reducibility relation among these examples of Borel orbit equivalence relations are extremely complicated. In fact many questions about them remain open. We will discuss them again in the next few sections and the rest of the book. Exercise 8.2.1 Let G be a Polish group. Let S(G) be the set of all closed subgroups of G and C(G) be the set of all (left) cosets of closed subgroups of G. Show that both S(G) and C(G) are Borel subsets of F (G), and hence are standard Borel spaces. Exercise 8.2.2 (Dixmier) Show that for any Polish group G there is a Borel set T ⊆ S(G) × G such that for any H ∈ S(G), TH is a transversal for the left cosets of H. Exercise 8.2.3 Let G be a Polish group and X a Borel G-space. Show that X EG is Borel iff there is a Borel function g : X × X → G such that for all X xEG y, g(x, y) · x = y. Exercise 8.2.4 Show that p (1 ≤ p < ∞) and c0 are Polishable subgroups of Rω .

8.3

A jump operator for Borel equivalence relations

In this section we discuss an operation that will increase the Borel complexity of any nontrivial Borel equivalence relation. Definition 8.3.1 Let E be a Borel equivalence relation on a standard Borel space X. The Friedman–Stanley jump (or simply jump) of E, denoted by E + , is the equivalence relation on X ω defined by (xn )E + (yn ) ⇐⇒ {[xn ]E : n ∈ ω} = {[yn ]E : n ∈ ω}. E + is obviously an equivalence relation. To see that it is Borel, just note that (xn )E + (yn ) ⇐⇒ ∀n∃m(xn Eym ) ∧ ∀m∃n(xn Eym ). Thus in general if E is Σ0α for α < ω1 , then E + is Π0α+1 . It is easy to see that E ≤B E + for any equivalence relation E. We will show later in this

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section that E
Let C ⊆ 2ω × ω ω be Π01 such that x ∈ A ⇐⇒ ∃y ∈ ω ω (x, y) ∈ C.

Let π : 2ω × ω ω → 2ω be the projection to the first coordinate. Then for any z ∈ C, π(z) ∈ A, and in particular, π(z) ∈ LO. Assume toward a contradiction that the described Δ11 set B exists. Then for any u, v ∈ C, if π(u) ∈ WO and π(v) ∈ WO then (π(u), π(v)) ∈ B, and if π(u) ∈ WO and π(v) ∈ WO then (π(u), π(v)) ∈ B. For each Π11 set P we consider a game GP as follows. First let C P ⊆ ω ω ×ω ω be Π01 such that x ∈ P ⇐⇒ ∃y ∈ ω ω (x, y) ∈ C P .

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Then let T P be a tree on ω × ω such that [T P ] = C P . For each x ∈ ω ω , let TxP be the tree on ω defined by s ∈ TxP ⇐⇒ (x  lh(s), s) ∈ T P . Then we have that x ∈ P ⇐⇒ ∀y ∈ ω ω (x, y) ∈ [T P ] ⇐⇒ TxP is well-founded. The game GP is now played as follows: I n II



a1 , u1 , ϕ1 a2 , u2 , ϕ2

where (i) n ∈ ω,  ∈ {0, 1}, a1 , a2 ∈ 2ω ; (ii) if  = 0 then n ∈ a1 but n ∈ a2 ; if  = 1 then n ∈ a2 but n ∈ a1 ; (iii) u1 , u2 ∈ C (thus π(u1 ), π(u2 ) ∈ A ⊆ LO); (iv) for i = 1, 2, ϕi : TaPi → ω is strictly (reverse) order-preserving: if s, t ∈ TaPi and s  t then ϕi (t) <π(ui ) ϕi (s). Then elements a1 , u1 , ϕ1 , a2 , u2 , ϕ2 are regarded as coded by subsets of ω and are understood to be played alternatively by the two players just as in the standard game. Then Player I wins the play iff (π(u1 ), π(u2 )) ∈ B. For each fixed Π11 set P the above game can be equivalently played on ω as a standard game where the payoff set is Borel. Thus by Borel determinacy the game GP is determined. We make some observations about the game GP where P is a singleton. For this we fix an arbitrary Π11 singleton a and replace all notation with superscript P by superscript a. We claim that Player II has a winning strategy in the game Ga . Since Ga is determined, it is enough to show that Player I does not have a winning strategy in Ga . Toward a contradiction assume he does. Consider the following play of the game. At the beginning Player I chooses n ∈ ω. Let Player II play  according to whether n ∈ a. If n ∈ a let  = 1; otherwise let  = 0. Then let Player II play a2 = a. Now Taa is well-founded, hence there exist well-ordering x of ω and strictly order-preserving map ϕ : Taa → ω with ϕ(t) <x ϕ(s) for all s  t in Taa . Let u2 ∈ C be such that π(u2 ) = x and ϕ2 = ϕ. Then the rules (i) through (iv) of the game are obeyed. Let Player I play according to his winning strategy. Then (i) through (iv) hold and (π(u1 ), π(u2 )) ∈ B. Since π(u2 ) ∈ WO the separation property of B requires that π(u1 ) ∈ WO. Thus we have that ϕ1 : Taa1 → ω is a strictly order-preserving map from Taa1 into a well-order coded by π(u1 ). This implies that Taa1 is well-founded, and thus a1 ∈ {a} and in fact a1 = a = a2 . However, by (ii) n ∈ a2 iff n ∈ a1 , and thus a1 = a2 , a contradiction.

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Let σ be a winning strategy of Player II in Ga . σ is coded by a real in ω . Let σ ∗ : ω → {0, 1} be such that σ ∗ (n) =  iff Player II responses with  according to σ when Player I’s first move is n. Apparently σ ∗ is recursive in σ. We note that σ ∗ (n) = 1 iff n ∈ a. To see this suppose n ∈ a but σ ∗ (n) = 0. Then let Player I plays a1 = a (note that (ii) is satisfied), u1 ∈ C with π(u1 ) ∈ WO, ϕ1 : Taa → ω strictly order-preserving into <π(u1 ) , where the choice of u1 and ϕ1 are similar to the argument in the preceding paragraph. Then since Player II wins we have (π(u1 ), π(u2 )) ∈ B, and by the separation property of B this requires that π(u2 ) ∈ WO. By the above argument again we have that a2 = a, a contradiction to (ii). The argument for the situation n ∈ a but σ ∗ (n) = 1 is similar. Finally note that for all Π11 sets P the payoff set for the game GP is Δ11 , hence Σ0α for some α < ω1 . Since there are only countably many Π11 sets there exists β < ω1 such that all payoff sets are Σ0β . Let U ⊆ ω × ω ω be universal for all Σ0β subsets of ω ω . Consider the set Q of all τ ∈ (ω ω )ω such that for any k ∈ ω, τ (k) ∈ ω ω is a winning strategy for either Player I or Player II in the game GUk . Then Q is nonempty Π11 . By Theorem 1.7.7 Q contains a Π11 singleton τ0 . We arrive at a final contradiction to Corollary 1.7.8 by showing that every Π11 singleton is recursive in τ0 . Let a be an arbitrary Π11 singleton. Let k ∈ ω be such that Uk is the payoff set of the game Ga . Then τ0 (k) is a winning strategy for Player II in Ga . Therefore n ∈ a iff τ0 (k)∗ (n) = 1. This shows that a is recursive in τ0 (k)∗ , which is in turn recursive in τ0 , and hence a is recursive in τ0 . ω

Recall that D ⊆ ω ω is Π11 -complete if D is Π11 and for all Π11 sets P ⊆ ω ω there is a continuous function f : ω ω → ω ω such that f −1 (D) = P . The set WO is Π11 -complete, and it is easy to see that a set D is Π11 -complete iff there is a continuous function f : LO → ω ω such that f −1 (D) = WO. With these concepts the above theorem can be generalized and relativized. Theorem 8.3.4 Let D ⊆ ω ω be Π11 -complete and A ⊆ ω ω be Σ11 with D ⊆ A. Then there is no Borel set B ⊆ ω ω ×ω ω such that D×(A−D) ⊆ B and (A−D)×D∩B = ∅. Proof. First note that the proof of the previous theorem can be relativized. Let f : LO → ω ω be continuous such that f −1 (D) = WO. Let A = f −1 (A). Then A ⊆ LO is Σ11 and WO ⊆ A since D ⊆ A. Assume that there is a Borel set B with the separation property. Let B  = {(u, v) ∈ LO × LO : (f (u), f (v)) ∈ B}. Then B  is Borel in LO × LO, and WO × (A − WO) ⊆ B  and (A − WO) × WO ∩ B  = ∅. Let x ∈ ω ω be such that A ∈ Σ11 (x) and B  ∈ Δ11 (x). This contradicts the previous theorem relativized to x. We are now ready to give Harrington’s proof of Friedman’s Borel diagonalization theorem for Borel equivalence relations.

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Theorem 8.3.5 (Friedman) Let E be a Borel equivalence relation on a standard Borel space X. Then there is no Borel function F : X ω → X such that, for all x = (xn ) and y = (yn ) in X ω , (a) if xE + y then F (x)EF (y), and (b) (F (x), xn ) ∈ E for all n ∈ ω.

Proof. Assume that the Borel function F exists with properties (a) and (b). Without loss of generality we may assume that X = ω ω . Let z0 ∈ ω ω be an arbitrarily fixed element. Let A ⊆ 2ω × (ω ω )ω be the set of all elements u = (x, f ) where x ∈ LO and f : ω → ω ω such that, for all n ∈ ω, (i) if n is <x -least then f (n) = z0 ; and (ii) if n is not <x -least, then letting m0 , m1 , . . . enumerate the set {m ∈ ω : m <x n} in the increasing order of natural numbers (possibly with repetitions), we have that f (n) = F (f (m0 ), f (m1 ), . . . ). Because of property (a) we can write f (n) = F (f  {m : m <x n}) without specifying the order in which the elements are enumerated. Note that A is Borel. Let π : 2ω × (ω ω )ω → 2ω be the projection to the first coordinate. Then WO ⊆ π(A) since for any x ∈ WO a function f satisfying (i) and (ii) can be defined by a transfinite induction on the ordinal length of <x . Define D ⊆ A by (x, f ) ∈ D ⇐⇒ (x, f ) ∈ A ∧ x ∈ WO. Then it is clear that D is a Π11 -complete subset of 2ω × (ω ω )ω . We are going to apply Theorem 8.3.4 to the sets D and A, regarding the underlying space as a homeomorphic copy of ω ω . For any u = (x, f ) and v = (y, g) in A, we define a binary relation Ru,v ⊆ ω × ω by nRu,v m ⇐⇒ f (n)Eg(m). Because of property (b) we have that if nRu,v m1 and nRu,v m2 then m1 = m2 . Similarly, if n1 Ru,v m and n2 Ru,v m then n1 = n2 . Thus Ru,v is in fact a partial bijection between two subsets of ω. Let ϕu,v denote the partial bijection given by Ru,v . Define Iu,v ⊆ dom(ϕu,v ) and Ju,v ⊆ range(ϕu,v ) by letting n ∈ Iu,v ⇐⇒ {k : k ≤x n} ⊆ dom(ϕu,v ) ∧ ϕu,v ({k : k ≤x n}) = {l : l ≤y ϕu,v (n)} ∧ ∀k, k  ≤x n (k <x k  → ϕu,v (k)
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Ju,v = ω). Note that B is Borel. We will arrive at a contradiction by checking that D × (A − D) ⊆ B and (A − D) × D ∩ B = ∅. For this let u = (x, f ) and v = (y, g) be in A. First suppose x ∈ WO and y ∈ WO. If Iu,v = ω, then since (Iu,v , <x ) is a well-order and ϕu,v is order-preserving, we get that (Ju,v ,
Proof. It suffices to show that E + ≤B E. Assume toward a contradiction that f : X ω → X is a Borel reduction from E + to E. We will consider the space (X ω )ω and the equivalence relation (E + )+ on it. For notational clarity we denote a typical element of X ω by x = (xn ), where each xn ∈ X, and a typical element of (X ω )ω by x = (xk ), where xk ∈ X ω . We will also omit the subscripts when we denote the equivalence class of an element. This will not cause any confusion. Let C = {x = (xn ) ∈ X ω : ∀n, m xn = xm }. Thus elements of C are constant sequences in X ω . We note that there is x ∈ C such that f (x) = x0 . Otherwise we would have that f (C) = X. This implies that, for every y ∈ X ω , there is some x ∈ C with f (x) = f (y), and thus xE + y since f is a reduction, and in fact yn Ex0 for all n. It follows that there is only one E-equivalence class on X, contradicting our assumption. We fix a z ∈ C with f (z) = z0 . For each x = (xk ) ∈ (X ω )ω let F (x) ∈ X ω be a canonical enumeration (possibly with repetitions) of the set {f (z), f (xk ) : ∀n [f (xk )] = [xkn ]}. Note that F : (X ω )ω → X ω is Borel. We claim that F is invariant, that is, if x(E + )+ y then F (x)E + F (y). For this suppose x(E + )+ y. Let k be such that [f (xk )] = [xkn ] for all n. Let l be such that xk E + y l . Since f is a reduction [f (y l )] = [f (xk )] whereas {[xkn ] : n ∈ ω} = {[ynl ] : n ∈ ω}. Hence [f (y l )] = [ynl ] for all n. By symmetry we have that F (x)E + F (y). Next we claim that for any x, (F (x), xk ) ∈ E + for all k. Assume that for some x and k, F (x)E + xk . Let A = {z, xp : ∀n [f (xp )] = [xpn ]}. We consider two cases. Case 1: there is y ∈ A with f (xk )Ef (y). Then in fact xk E + y,

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[f (xk )] = [f (y)], and {[yn ] : n ∈ ω} = {[xkn ] : n ∈ ω}. This implies that [f (xk )] = [xkn ] for all n, and therefore xk ∈ A. However since F (x)E + xk , [f (xk )] ∈ {[f (y)] : y ∈ A} = {[xkn ] : n ∈ ω}, a contradiction. Case 2: there is no y ∈ A with f (xk )Ef (y). In particular xk ∈ A. Thus for some n, [f (xk )] = [xkn ]. Again by F (x)E + xk it follows that [f (xk )] ∈ {[f (y)] : y ∈ A}, contradicting our case assumption. We have thus constructed a Borel diagonalizer of E + , contradicting Theorem 8.3.5. Exercise 8.3.1 Show that E ≤B E + for any equivalence relation E on a standard Borel space X. Exercise 8.3.2 Show that if E ≤B F then E + ≤B F + . Exercise 8.3.3 Let =+ denote id(ω ω )+ . Show that E∞ ≤B =+ and in fact + ∼B =+ . E∞ Let E be an equivalence relation on a standard Borel space X. Define an equivalence relation E ∗ on X ω by xE ∗ y ⇐⇒ xE + y ∧ ∀z ∈ X |{n ∈ ω : xn = z}| = |{n ∈ ω : yn = z}|. Exercise 8.3.4 Show that E ∗ ∼B E + . Exercise 8.3.5 Let G be a Polish group and X a Polish G-space. Show that X ∗ ) is the orbit equivalence relation induced by an action of G × S∞ on (EG ω X , where S∞ is the infinite permutation group.

8.4

Examples of Fσ equivalence relations

Recall from Theorem 6.4.4 that every Gδ equivalence relation is smooth. Thus from the point of view of descriptive complexity the simplest kind of Borel equivalence relations are Fσ ones. However, this is not to say that the Borel complexity of these equivalence relations in the Borel reducibility hierarchy is easy to describe. In this section we give an overview of some canonical examples of Fσ equivalence relations. We have introduced at least three concrete nonsmooth Fσ equivalence relations: E0 , E∞ , and E1 . Of course, any countable Borel equivalence relation is Fσ . Consider the following additional examples arising from classical Banach spaces.

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Recall that for a real number 1 ≤ p < ∞,   ∞  ω p p = x = (xn ) ∈ R : |xn | < ∞ , n=0

and

 ∞ =

x = (xn ) ∈ R

ω

: sup |xn | < ∞ . n∈ω

All of them are Fσ subgroups of the additive group Rω . Thus for 1 ≤ p ≤ ∞ the coset equivalence relation Rω /p defined by x Rω/p y ⇐⇒ x − y ∈ p is an Fσ equivalence relation on the Polish space Rω . For notational simplicity we also denote the equivalence relation by p . We first note that the equivalence relations p all have nontrivial Borel complexity. Lemma 8.4.1 For any 1 ≤ p ≤ ∞, E0 ≤B p . Proof. We use Theorem 6.2.1. For this note that each p is a group of homeomorphisms on Rω . It is also easy to see that p is dense, and thus the equivalence relation has a dense orbit. It remains to show that the equivalence relation is meager. By Kuratowski–Ulam it suffices to show that every orbit is meager. Since each orbit is a homeomorphic copy of p , it suffices to show that p is meager in Rω . We do this for p = ∞ first. For each N ∈ ω let AN = {x ∈ Rω : supn |xn | ≤ N }. Then AN is closed in Rω and ∞ = N ∈ω AN . We claim that AN is nowhere dense. Otherwise there is a basic open U ⊆ Rω with U ⊆ AN . Without loss of generality assume U = (a0 , b0 ) × (a1 , b1 ) × . . . (an , bn ) × Rω . Consider V ⊆ U defined by V = (a0 , b0 ) × (a1 , b1 ) × . . . (an , bn ) × (N, N + 1) × Rω . Then V ⊆ AN . However, for any x ∈ V we have that |xn+1 | > N and therefore supn |xn | > N , contradicting x ∈ AN . A similar argument shows that p is also meager for 1 ≤ p < ∞. Our next goal is to show that ∞ is the most complex among these equivalence relations. In fact, we show that ∞ is universal among all Kσ equivalence relations on Polish spaces. Recall that a subset of a topological space is Kσ if it is the union of countably many compact sets. If X is a Polish space, E a Kσ equivalence relation on X, and Δ = {(x, x) : x ∈ X}, then Δ is a closed subset of E which is homeomorphic to X, and it follows that X is Kσ .

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Theorem 8.4.2 (Rosendal) Let E be a Kσ equivalence relation on a Polish space X. Then E ≤B ∞ .

Proof.

Let (Xn ) be an increasing sequence of compact subsets of X with X = n Xn . Let Δ = {(x, x) : x ∈ X}. Let (Fn ) be an increasing sequence of compact subsets of X × X with E = n Fn . Without loss of generality we may assume Δ ∩ Xn2 ⊆ Fn . Define an increasing sequence (Rn ) of compact relations by induction as follows. Let R0 = F0 ∩ X02 . Assuming Rn has been defined, let Rn+1 = Rn ∪ (Rn ◦ Rn ) ∪ (Fn ∩ Xn2 ).

It is easy to see that each Rn is compact and that E = n Rn . Let {Um }m∈ω be a countable base for the topology of X. Fix a bijection ·, · : ω × ω → ω such that for all n, m ∈ ω, n, m ≥ n. We define a function ϕ : X → Rω by  k, if k ≤ n is the least such that ∃y ∈ Um (x, y) ∈ Rk , ϕ(x)(n, m) = n, if ∀y ∈ Um (x, y) ∈ Rn . Note that ϕ is a Borel map. To see this it suffices to check that for any m, n ∈ ω, the set {x ∈ X : ∃y ∈ Um (x, y) ∈ Rn } is Borel. We have that ∃y ∈ Um (x, y) ∈ Rn ⇐⇒ ∃l(Ul ⊆ Um ∧ ∃y ∈ Ul ∩ Xn (x, y) ∈ Rn ). The set {x : ∃y ∈ Ul ∩ Xn (x, y) ∈ Rn } is compact, and thus the desired set is in fact Fσ . We claim that ϕ is Borel reduction from E to ∞ . Suppose x, y ∈ X and xEy. Fix the least n0 ∈ ω with (x, y) ∈ Rn0 and (y, x) ∈ Rn0 . Then for any z ∈ X and k ∈ ω, if (x, z) ∈ Rk then (y, z) ∈ Rn0 ◦ Rk ⊆ Rmax{n0 ,k}+1 , and by symmetry, if (y, z) ∈ Rk then (y, z) ∈ Rmax{n0 ,k}+1 . It follows that if ϕ(x)(n, m) = k < n, then ϕ(y)(n, m) ≤ max{n0 , k} + 1, and therefore |ϕ(x) − ϕ(y)| ≤ n0 + 1. By symmetry if ϕ(x)(n, m) < n then |ϕ(x)(n, m) − ϕ(y)(n, m)| ≤ n0 +1. If both ϕ(x)(n, m) = n = ϕ(y)(n, m), then trivially |ϕ(x)(n, m)−ϕ(y)(n, m)| ≤ n0 +1. We have shown that ϕ(x)−ϕ(y) ∈ ∞ . Conversely, suppose ϕ(x) − ϕ(y) ∈ ∞ . Let k be large enough such that y ∈ Xk and |ϕ(x)(N ) − ϕ(y)(N )| ≤ k for all N . Fix an arbitrary m with y ∈ Um . Since Δ ∩ Xk2 ⊆ Fk , (y, y) ∈ Rk+1 and then ϕ(y)(2k + 2, m) ≤ k + 1. It follows that ϕ(x)(2k + 2, m) ≤ 2k + 1, and thus there is z ∈ Um with (x, z) ∈ Rk+1 . We have shown that, for any m with y ∈ Um there is z ∈ Um with (x, z) ∈ Rk+1 . Thus we have a sequence (zm ) with zm → y and (x, zm ) ∈ Rk+1 . Since Rk+1 is compact, we have that (x, y) ∈ Rk+1 , and therefore (x, y) ∈ E. It is easy to see that all of the equivalence relations E0 , E∞ , E1 , and p (1 ≤ p ≤ ∞) are Kσ (see Exercise 8.4.3). From this we immediately get that

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all of them are Borel reducible to ∞ . It is not known if ∞ is universal among all Fσ equivalence relations. Before closing this section we mention some results without proofs. Dougherty and Hjorth [29] have shown that the Borel reducibility order
∞ 

1/p |xn |

p

.

n=0

This makes the equivalence relation p an orbit equivalence relation of a Polish group action. In particular, all p (1 ≤ p < ∞) are idealistic. Since E1 is not Borel reducible to any orbit equivalence relations (Theorem 10.6.1), E1 ≤B p for 1 ≤ p < ∞. On the other hand we just showed that E1 ≤B ∞ , and so the equivalence relation ∞ is not an orbit equivalence relation of a Polish group action. This implies that topological group ∞ is not a Polishable subgroup of Rω , which can also be seen directly (see Lemma 9.3.3). For now we note that, with the usual norm x∞ = sup |xn |, n∈ω

the space ∞ is not separable. Exercise 8.4.1 Show that for any 1 ≤ p ≤ ∞, p is an Fσ subset of Rω . Exercise 8.4.2 Show that p is meager in Rω for 1 ≤ p < ∞. Exercise 8.4.3 Show that all equivalence relations E0 , E∞ , E1 , and p (1 ≤ p ≤ ∞) are Kσ .

8.5

Examples of Π03 equivalence relations

We define the power operator on equivalence relations. Definition 8.5.1 Let E be an equivalence relation on a standard Borel space X. The power of E is the equivalence relation E ω on X ω defined by xE ω y ⇐⇒ ∀n ∈ ω xn Eyn .

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Apparently E ≤B E ω and E ω ≤B F ω if E ≤B F . However, (E ω )ω ∼B E ω . Thus the power operator is not a jump operator. If E is Π0α for α < ω1 , then E ω is also Π0α . Thus the powers of all Fσ equivalence relations we considered in the last section are examples of Π03 equivalence relations, among which E0ω has the simplest Borel complexity. It is easy to see that if E is an orbit equivalence relation induced by an action of a Polish group G, then E ω is induced by an action of Gω . We show that E0ω is not Borel reducible to any Σ03 equivalence relation. It follows that the powers of all other equivalence relations have the same property. Theorem 8.5.2 Let E be a Σ03 equivalence relation on a Polish space X. Then E0ω ≤B E. Proof. Assume toward a contradiction that f : (2ω )ω → X is a Borel reduction of E0ω to E. Since f is Borel, and hence Baire measurable, there is a comeager set C ⊆ (2ω )ω such that f  C is continuous. Without loss of generality we may assume C is Gδ . Then E0ω ∩ C 2 is Σ03 , since (x, y) ∈ E0ω ∩ C 2 ⇐⇒ x, y ∈ C ∧ (f (x), f (y)) ∈ E. To arrive at a contradiction it suffices to establish the following claim: For every dense Gδ set C ⊆ (2ω )ω , E0ω ∩ C 2 is Π03 -complete. Fix any bijection ·, · : ω × ω → ω. Let B = {x ∈ 2ω : ∀i ∃j ∀k ≥ j x(i, k) = 1}. Then B is Π03 -complete. To see this, let A ⊆ 2ω be Π03 , say  A= Ai,j,k , i

j

k

where each Ai,j,k is clopen. Define ρ : 2ω → 2ω as follows. Given x ∈ 2ω , for each i, k ∈ ω let ax,i,k ∈ ω be the least integer j ≤ k, if one exists, such that ∀k  ≤ k (x ∈ Ai,j,k ). Let  1, if ax,i,k and ax,i,k−1 are both defined and are equal, ρ(x)(i, k) = 0, otherwise. It is easy to see that ρ is continuous and x ∈ A ⇐⇒ ρ(x) ∈ B. This shows that B is Π03 -complete. Write C = n Dn where each Dn is open dense in (2ω )ω . We next define two continuous functions π1 , π2 : 2ω → 2ω×ω so that x → (π1 (x), π2 (x)) reduces the set B to E0ω  C 2 . For this we define two sequences (us )s∈2<ω , (vs )s∈2<ω and a function p : ω → ω so that the following hold: (i) for any n ∈ ω, if n = i, k then i, p(n) + k < p(n + 1);

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(ii) for any n ∈ ω and lh(s) = n, us , vs ∈ 2p(n)×p(n) ; (iii) for s ⊆ t, us ⊆ ut and vs ⊆ vt ; (iv) if lh(s) = n then Nus , Nvs ⊆ Dn ; (v) for any n ∈ ω, if s(n) = 1 then us (j, l) = vs (j, l) for all (j, l) ∈ p(n + 1) × p(n + 1) − p(n) × p(n); (vi) for n = i, k, if s(n) = 0 then us (i, p(n) + k) = vs (i, p(n) + k) and us (j, l) = vs (j, l) for (j, l) ∈ p(n + 1) × p(n + 1) − p(n) × p(n) − {(i, p(n) + k)}. It follows easily from

the density of Dn that

such sequences can be constructed. We let π1 (x) = n uxn and π2 (x) = n vxn . Clearly π1 , π2 are continuous. By (iv) π1 (x), π2 (x) ∈ C for any x ∈ 2ω . If x ∈ B, then for each i let k(i) be such that ∀k ≥ k(i) x(i, k) = 1. Fix i ∈ ω. For any n ≥ i, k(i), if x(n) = 1 then by (v) for any p(n) ≤ j < p(n + 1) we have π1 (x)(i, j) = π2 (x)(i, j); if x(n) = 0 then n = i , k for some i = i, since otherwise k ≥ k(i) and we have x(i, k) = 1 by assumption. In this latter case by (vi) we still have that π1 (x)(i, j) = π2 (x)(i, j) if i < p(n + 1) and p(n) ≤ j < p(n + 1). Thus π1 (x)(i, k) = π2 (x)(i, k) for all k ≥ k0 . This shows that π1 (x) E0ω π2 (x). If x ∈ / B, then for some i we have that x(i, k) = 0 for infinitely many k. Fix such an i. Then π1 (x)(i, p(i, k) + k) = π2 (x)(i, p(i, k) + k) for infinitely many k. This shows that (π1 (x), π2 (x)) ∈ E0ω . We introduce some other Π03 equivalence relations. First consider the classical Banach space c0 = {x ∈ Rω : lim |xn | = 0} n

with the norm  · ∞ . c0 is a subgroup of Rω . We consider the coset ω equivalence relation R /c0 defined by Π03

x Rω/c0 y ⇐⇒ x − y ∈ c0 and still denote it by c0 . Then the equivalence relation c0 is Π03 . Lemma 8.5.3 E0ω ≤B c0 . Proof. Let ·, · : ω × ω → ω be a bijection such that n, m ≤ n , m  if n ≤ n and m ≤ m . Define ϕ : 2ω×ω → Rω by ϕ(x)(n, m) = 2−n x(n, m). Clearly ϕ is a Borel function. We check that it is a reduction from E0ω to c0 . Let xE0ω y, that is, ∀n∃m(n)∀m ≥ m(n) x(n, m) = y(n, m). Let  > 0. Fix

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n0 large enough such that 2−n0 < . Let m0 = max{m(i) : i < n0 }. We show that for all N ≥ n0 , m0 , |ϕ(x)(N ) − ϕ(y)(N )| < . Let N ≥ n0 , m0 . Say N = n, m. If n ≥ n0 then |ϕ(x)(N ) − ϕ(y)(N )| ≤ 2−n ≤ 2−n0 < . If n < n0 then m ≥ m(n) since N = n, m ≥ n0 , m0  ≥ n, m0  ≥ n, m(n). Thus x(n, m) = y(n, m) and ϕ(x)(N ) − ϕ(y)(N ) = 0. This shows that limN ϕ(x)(N ) − ϕ(y)(N ) = 0 and so ϕ(x) − ϕ(y) ∈ c0 . Conversely let (x, y) ∈ E0ω . Fix n such that (xn , yn ) ∈ E0 , that is, x(n, m) = y(n, m) for infinitely many m. Then |ϕ(x)(n, m) − ϕ(y)(n, m)| = 2−n for infinitely many m, and thus limN [ϕ(x)(N ) − ϕ(y)(N )] = 0. This shows that ϕ(x) − ϕ(y) ∈ c0 . Another class of examples of Π03 equivalence relations are the Friedman– Stanley jumps of Fσ equivalence relations. Among these the simplest is the jump of the identity relation on a Polish space, for instance, id(2ω )+ . For notational simplicity we denote this equivalence relation by =+ . Lemma 8.5.4 E0ω ≤B =+ . Proof. It is easy to see that E0+ ∼B =+ (see Exercise 8.3.3). Thus it suffices to show that E0ω ≤ E0+ . Fix a bijection ·, · : ω × ω → ω. Let τ : 2ω → 2ω be defined by  z(j), if i = 0, τ (z)(i, j) = 1, otherwise. For each n ∈ ω define a function ϕn : 2ω → 2ω by  τ (z)(j), if i = n, ϕn (z)(i, j) = 0, otherwise. Then for each n ∈ ω and z1 , z2 ∈ 2ω , we have that z1 E0 z2 iff ϕn (z1 )E0 ϕn (z2 ). Moreover, for n1 = n2 and any z1 , z2 ∈ 2ω , (ϕn1 (z1 ), ϕn2 (z2 )) ∈ E0 since for infinitely many m, ϕn1 (z1 )(n1 , m) = τ (z1 )(m) = 1, but for all m, ϕn2 (z2 )(n1 , m) = 0. We then define ρ : (2ω )ω → (2ω )ω by letting (ρ(x))n = ϕn (xn ). Suppose xE0ω y. Then for all n ∈ ω, xn Eyn , and therefore ϕn (xn )E0 ϕn (yn ). It follows that {[ϕn (xn )]E0 : n ∈ ω} = {[ϕn (yn )]E0 : n ∈ ω}, and thus ρ(x)E0+ ρ(y). Conversely, suppose (x, y) ∈ E0ω . Let n be such that (xn , yn ) ∈ E0 . Then (ϕn (xn ), ϕm (y)) ∈ E0 for all m ∈ ω. Thus [ϕn (xn )]E0 ∈ {[ϕn (yn )]E0 : n ∈ ω} and (ρ(x), ρ(y)) ∈ E0+ . The last equivalence relation we consider in this section is measure equivalence. Let M be the space of Borel probability measures on 2ω . Every μ ∈ M

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is completely determined by the sequence (μ(Ns ))s∈2<ω . Hence M can be viewed as a subspace of [0, 1]ω , and therefore is a Polish space. For μ, ν ∈ M , we say that μ is absolutely continuous with respect to ν, denoted μ  ν, if for any Borel set A ⊆ 2ω , ν(A) = 0 ⇒ μ(A) = 0. Two measures μ and ν are equivalent if both μ  ν and ν  μ. We use ≡m to denote the measure equivalence relation. It is easy to see that μ  ν ⇐⇒ ∀ > 0∃δ > 0 ∀ basic open U (ν(U ) < δ ⇒ μ(U ) < ). It follows that  is a Π03 relation and ≡m is a Π03 equivalence relation. Lemma 8.5.5 =+ ≤B ≡m . Proof. We give an informal sketch of the proof. The exact details of the definitions are left to the reader. Intuitively we will assign to each countable subset S of 2ω a Borel probability measure μS so that S = S  iff μS ≡m μS  . This is easy to do: let μS be any Borel probability measure supported on S. Then for any Borel set A ⊆ 2ω , μ(A) = 0 iff A ∩ S = ∅ iff ν(A) = 0. There are many important facts omitted in our discussion so far. Before closing this section we mention some known facts without proof. The Spectral Theorem in functional analysis essentially established that the measure equivalence relation is Borel bireducible to the orbit equivalence relation of the conjugacy action of the unitary group on itself. In particular this establishes that the measure equivalence is Borel bireducible with an orbit equivalence relation of a Polish group action. This implies (by Theorem 10.6.1) that E1 is not reducible to any of the equivalence relations we discussed in this section. Kakutani studied the measure equivalence relation on product measures. He essentially showed that 2 ≤B ≡m (see Reference [104]). It follows from Theorem 8.4.3 that p ≤B ≡m if 1 ≤ p ≤ 2. It is not known if p ≤B ≡m for any p > 2. Hjorth and Kechris [78] proved a dichotomy theorem that for any E ≤ E0ω , either E ∼B E0ω or E is hyperfinite. Exercise 8.5.1 Let C ⊆ Rω be comeager. Show that c0 ∩C 2 is Π03 -complete. Exercise 8.5.2 Show that (c0 )ω ∼B c0 . Exercise 8.5.3 Give an example of an equivalence relation E such that E +
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Chapter 9 Analytic Equivalence Relations

The study of general analytic equivalence relations is a very important and yet challenging subject. In this chapter we will discuss general dichotomy theorems for analytic equivalence relations, universal analytic equivalence relations, and other examples of analytic but non-Borel equivalence relations. Orbit equivalence relations of Polish group actions are in general analytic, and often non-Borel, and therefore provide many interesting examples and opportunities of applications of general theorems. The Borel reducibility hierarchy for analytic equivalence relations is more complicated than the one for Borel equivalence relations.

9.1

The Burgess trichotomy theorem

The Silver dichotomy theorem for Π11 equivalence relations fails for Σ11 equivalence relations. For an example, consider the equivalence relation Eω1 defined on the standard Borel space LO by xEω1 y ⇐⇒ (x ∈ WO ∧ y ∈ WO) ∨ ∃ϕ : (ω, <x ) ∼ = (ω,
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countably many equivalence classes it is in fact Borel and smooth. Thus having ω1 many equivalence classes is typical of all counterexamples of the Silver dichotomy for Σ11 equivalence relations. To prove the Burgess trichotomy theorem we need to show another theorem of Burguess representing every Σ11 equivalence relation as a intersection of ω1 many Borel equivalence relations. We need also recall some basic notions about ordinals. The notation ω1 is used to represent the least uncountable ordinal as well as the set of all countable ordinals. We also use it to denote the least uncountable cardinal number. The space ω1 of all countable ordinals is endowed with the order topology, where subbasic open sets are of the form {α < ω1 : α < α0 } and {α < ω1 : α > α0 } for α0 < ω1 . A subset C of ω1 is closed if C is closed in the order topology of ω1 , and C is a club if it is closed and unbounded above. The intersection of countably many clubs in ω1 is still a club in ω1 . If f : ω1 → ω1 is a continuous function satisfying f (α) ≥ α for all α < ω1 , then there is a club C ⊆ ω1 such that f (α) = α for all α ∈ C. The following lemma is an immediate corollary of the boundedness principle Theorem 1.6.10. Lemma 9.1.1 Let X be a standard Borel space and f1 , . . . , fn , g1 , . . . , gm : X → LO be Borel −1 functions. Suppose f1−1 (WO) ∩ · · · ∩ fn−1 (WO) ⊆ g1−1 (WO) ∪ · · · ∪ gm (WO). Then there is a club C ⊆ ω1 such that for all α ∈ C, −1 (WOα ). f1−1 (WOα ) ∩ · · · ∩ fn−1 (WOα ) ⊆ g1−1 (WOα ) ∪ · · · ∪ gm

Proof. We define a continuous function α → β(α) as follows. For each α < ω1 , let β(α) be the least β ≥ α such that −1 (WOβ ). f1−1 (WOα ) ∩ · · · ∩ fn−1 (WOα ) ⊆ g1−1 (WOβ ) ∪ · · · ∪ gm

To see that this β(α) is well defined, note that by our assumption −1 (WO). f1−1 (WOα ) ∩ · · · ∩ fn−1 (WOα ) ⊆ g1−1 (WO) ∪ · · · ∪ gm

Let A denote the set on the left-hand side. Then A is a Borel subset of X. It −1 (WO)) is Σ11 . Since follows that A − (g2−1 (WO) ∪ · · · ∪ gm −1 (WO)) ⊆ g1−1 (WO), A − (g2−1 (WO) ∪ · · · ∪ gm

we have that −1 (WO))) ⊆ WO. g1 (A − (g2−1 (WO) ∪ · · · ∪ gm

The set on the left-hand side is Σ11 . Hence by the boundedness principle there is some β1 ≥ α such that −1 (WO)) ⊆ g1−1 (WOβ1 ). A − (g2−1 (WO) ∪ · · · ∪ gm

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Repeating the argument for g2 , . . . , gm , we obtain β2 , . . . , βm ≥ α similarly. Let β = max{β1 , . . . , βm }. Then we have that A ⊆ g1−1 (WOβ ) ∪ · · · ∪ −1 gm (WOβ ). This shows that β(α) is well defined. It is easy to see that α → β(α) is continuous. We can thus get a club C ⊆ ω1 such that for all α ∈ C, β(α) = α. Theorem 9.1.2 (Burgess) Let E be a Σ11 equivalence relation on a standard Borel X. Then there are Borel equivalence relations Eα on X for α < ω1 such that E = α<ω1 Eα .

Proof. Let f : X 2 → LO be a Borel function such that X 2 − E = f −1 (WO). 2 −1 For α < ω1 let A α = X −f (WOα ). Then each Aα is a Borel binary relation on X and E = α<ω1 Aα . We claim that there is a club C ⊆ ω1 such that for all α ∈ C, Aα is an equivalence relation. Clearly the theorem follows from this claim. It is easy to see that every Aα is reflexive. To prove the claim we first show that there is a club C ⊆ ω1 such that for all α ∈ C, Aα is symmetric. For this let f1 (x, y) = f (x, y) and g1 (x, y) = f (y, x) for all (x, y) ∈ X 2 . Then since E is symmetric, f1−1 (WO) = X 2 − E = g1−1 (WO). By applying Lemma 9.1.1 for both f1−1 (WO) ⊆ g1−1 (WO) and g1−1 (WO) ⊆ f1−1 (WO), we get a club C ⊆ ω1 such that for all α ∈ C, f1−1 (WOα ) = g1−1 (WOα ). It follows that Aα is symmetric for all α ∈ C. Next we show that there is a club C ⊆ ω1 such that for all α ∈ C, Aα is transitive. The claim then follows since the intersection of two clubs is still a club. For this define f1 , g1 , g2 : X 3 → LO by f1 (x, y, z) = f (x, z), g1 (x, y, z) = f (x, y), and g2 (x, y, z) = f (y, z). Then for any (x, y, z) ∈ X 3 , (x, y, z) ∈ f1−1 (WO) ⇐⇒ (x, z) ∈ E =⇒ (x, y) ∈ E ∨ (y, z) ∈ E ⇐⇒ (x, y, z) ∈ g1−1 (WO) ∪ g2−1 (WO). This means that f1−1 (WO) ⊆ g1−1 (WO) ∪ g2−1 (WO). Thus by Lemma 9.1.1 there is a club C ⊆ ω1 such that for all α ∈ C, f1−1 (WOα ) ⊆ g1−1 (WOα ) ∪ g2−1 (WOα ). It follows that for α ∈ C, Aα is transitive, since if (x, y), (y, z) ∈ Aα then (x, y, z) ∈ g1−1 (WOα )∪g2−1 (WOα ) and therefore (x, y, z) ∈ f1−1 (WOα ) and so (x, z) ∈ Aα . Recall from Definition 5.4.3 that a collection S of subsets of a standard Borel space X generates the equivalence relation xEy ⇐⇒ ∀S ∈ S (x ∈ S ↔ y ∈ S).

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Definition 9.1.3 An equivalence relation E on a standard Borel space X is ω1 -smooth if E is generated by a set S of ω1 many Borel subsets of X. Theorem 9.1.4 Let E be an ω1 -smooth equivalence relation on a Polish space X. Then either E has ≤ ω1 many equivalence classes or else E has perfectly many classes. Proof. We fix a complete metric for X for the entire proof. Let Bα , α < ω1 , witness that E is ω1 -smooth. Since both Bα and X − Bα are Suslin, we have sequences (Csα,0 )s∈ω<ω , (Csα,1 )s∈ω<ω of closed sets with diam(Csα,i ) < 2−lh(s) for i = 0, 1 and all α < ω1 , such that   α,0   α,1 Bα = Cxn , and X − Bα = Cxn . x∈ω ω n∈ω

x∈ω ω n∈ω

As usual we may assume that Ctα,i ⊆ Csα,i if s ⊆ t. For s ∈ ω <ω we let  α,i  Bsα,i = Cxn . x∈ω ω ,s⊆x n∈ω

Then Bsα,i ⊆ Csα,i . Assume that E has more than ω1 many equivalence classes. Let Z ⊆ X be a set of ω2 pairwise E-inequivalent elements. By induction on n ∈ ω, and for each s ∈ 2n , we construct a set Ds , an ordinal αs < ω1 , and σ(s, m) ∈ ω n for m < n so that the following conditions are satisfied: (i) σ(s, m) ⊆ σ(t, m) if s ⊆ t and m < lh(s); αsm ,s(m) for n > 0; (ii) Ds = m
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which is a finite union of sets of the form  α ,s(m) sm Bσ(s,m) ∩ Btαs ,0 k m m
for km ∈ ω and t ∈ ω n+1 . At least one of these sets contains ω2 many elements of Z. Thus there exists km for m < n and t ∈ ω n+1 such that if we define σ(s 0, m) = σ(s, m) km for m < n and σ(s 0, n) = t, then the corresponding Ds 0 contains ω2 many elements of Z. It is clear that all of (i) through (iii) are satisfied. The definitions of σ(s 1, m) for m ≤ n is similar by considering the set (Z ∩ Ds ) − Bαs . This finishes our construction. Now the sequence (Ds )s∈2ω gives rise to a Suslin set   A= Dxn x∈2ω n∈ω

via the Suslin operation. Since diam(Ds ) < 2−lh(s) we get that for any x ∈ 2ω , the set n∈ω Dxn is a singleton, say with element zx . Moreover, if x(m) = 0, then by our construction, for any n > m, α

,0

x(m+1) . Dxn ⊆ Cσ(xn,m)

It follows from (i) that   αx(m+1) ,0 zx ∈ Dxn ⊆ Cσ(xn,m) ⊆ Bαx(m+1) . n>m

n>m

Thus zx ∈ Bαx(m+1) . Similarly, if y(m) = 1, then zy ∈ Bαy(m+1) . These imply that if x, y ∈ 2ω with x = y, then there is α < ω1 such that either x ∈ Bα and y ∈ Bα , or x ∈ Bα and y ∈ Bα . In either case we have that x and y are E-inequivalent. Thus we have produced a Suslin set A of E-inequivalent elements containing uncountably many elements. By the perfect set theorem (Theorem 1.6.6) A contains a perfect set of E-inequivalent elements. Theorem 9.1.5 (Burgess) Let E be a Σ11 equivalence relation on a standard Borel space X. Then E has either perfectly many, countably many, or ω1 many equivalence classes.

Proof. By Theorem 9.1.2 E can be written as the intersection of ω1 many Borel equivalence relations. Let Eα , α < ω1 , be Borel equivalence relations such that E = α<ω1 Eα . Each Eα has either countably many or perfectly many equivalence classes by the Silver dichotomy theorem. If any of Eα has perfectly many equivalence classes, then so does E. So assume each Eα has only countably many equivalence classes, and let them be enumerated as Cα,n for n ∈ ω. Then we have that xEy ⇐⇒ ∀α∀n (x ∈ Cα,n ↔ y ∈ Cα,n ).

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This means that E is ω1 -smooth, and hence by Theorem 9.1.4, E has ≤ ω1 many equivalence classes. It follows that E has either perfectly many, countably many, or ω1 many equivalence classes. Exercise 9.1.1 Write Eω1 as an intersection of ω1 many Borel equivalence relations. Exercise 9.1.2 Recall that a quasiorder is a reflexive and transitive relation. Show that any Σ11 quasiorder on a standard Borel space X is the intersection of ω1 many Borel quasiorders. Exercise 9.1.3 Let X be a standard Borel space and E ⊆ F equivalence relations on X, where E is Σ11 and F is Π11 . Show that there is a Borel equivalence relation R such that E ⊆ R ⊆ F . Exercise 9.1.4 Let X be a standard Borel space and E = α<ω1 Eα , where Eα are Σ11 equivalence relations on X. Show that E has either perfectly many, countably many, or ω1 many equivalence classes. Exercise 9.1.5 Let X be a standard Borel space and E = α<ω1 Eα , where Eα are Π11 equivalence relations on X. Show that E has either perfectly many, countably many, or ω1 many equivalence classes. Exercise 9.1.6 For a standard Borel space X and a collection S of subsets of X, let E(S) denote the equivalence relation generated by S. Show the following corollary of the proof of Theorem 9.1.4: If S is a set of ω1 many Borel subsets of a standard Borel space X, then either E(S) has ≤ ω1 many equivalence classes or there is a countable T ⊆ S such that E(T ) has perfectly many equivalence classes.

9.2

Definable reductions among analytic equivalence relations

In this section we discuss definable reductions among Σ11 equivalence relations. We start with a discussion of Borel reducibility and show that it is sometimes inadequate to capture the intuitive reductions among general Σ11 equivalence relations. We will then introduce a more general notion of definable reducibility. The downside, however, of considering this broader notion is that a significant amount of metamathematics is needed to prove statements about it. Since these metamathematical techniques are beyond the scope of this book, we will only mention some results without proving them. On the

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other hand, we will introduce the very useful notion of Ulm classifiability and also briefly discuss universal analytic equivalence relations. Recall that the conclusion of the Silver dichotomy can be reformulated in terms of Borel reducibility as either E ≤B id(ω) or id(2ω ) ≤B E. For the Burgess trichotomy theorem, these Borel reducibility statements still correspond to the cases when E has either countably many or perfectly many equivalence classes. However, when E has ω1 many equivalence classes, there is no obvious way to characterize E in the Borel reducibility hierarchy. We elaborate some more below. Consider the equivalence relation Fω1 defined on 2ω by CK(x)

xFω1 y ⇐⇒ ω1

CK(y)

= ω1

.

CK(x)

Since {ω1 : x ∈ 2ω } is unbounded in ω1 , clearly Fω1 has ω1 many equivalence classes. Lemma 9.2.1 Fω1 is Σ11 equivalence relation with every equivalence class Borel. Proof.

CK(x)

Note that ω1

CK(y)

≤ ω1

iff

∀m∃n ({m}x ∈ WO → ∃ϕ : (ω, {m}x ) ∼ = (ω, {n}y ) ) , which implies that Fω1 is Σ11 . Now fix x ∈ 2ω and m, n ∈ ω with {m}x ∈ WO. Then we also have that (ω, {m}x ) ∼ = (ω, {n}y ) iff (a) {n}y ∈ WO, (b) for all strictly order-preserving ϕ from (ω, {n}y ) into (ω, {m}x ), ϕ is onto, and (c) for all strictly order-preserving ψ from (ω, {m}x) into (ω, {n}y ), ψ is onto. Thus if we let Bm,n = {y : (ω, {m}x ) ∼ = (ω, {n}y )}, then Bm,n is Δ11 , in x particular Borel, when {m} ∈ WO. Now 

  CK(x) CK(y) = y : ω1 = ω1 Bm,n , {m}x ∈WO n

hence the Fω1 -equivalence class of x is Borel. Now Fω1 is indeed a Σ11 equivalence relation and apparently has ω1 many equivalence classes. However, if we compare it to Eω1 defined in the preceding section via Borel reducibility, they turn out to be incomparable.

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Lemma 9.2.2 Eω1 ≤B Fω1 and Fω1 ≤B Eω1 . Proof. Note that if x ∈ WO then the Eω1 -equivalence class of x is not Borel. Now every Fω1 -equivalence class is Borel, which implies that any equivalence relation Borel reducible to Fω1 has the same property. Thus Eω1 ≤B Fω1 . On the other hand, assume f : 2ω → LO is a Borel reduction from Fω1 to Eω1 . There is at most one Fω1 -equivalence class C such that f (C) ⊆ LO − WO. Since C is Borel, it follows that f (2ω −C) is Σ11 in LO. Since f (2ω −C) ⊆ WO, by the boundedness principle there is α < ω1 such that f (2ω − C) ⊆ WOα . This is a contradiction to the assumption that f is a reduction since Eω1  WOα has only countably many equivalence classes. Nevertheless on an intuitive level we should be able to define a reduction from Fω1 to Eω1 , that is, given x ∈ 2ω to define Φ(x) ∈ WO so that its order CK(x) type is ω1 . A straightforward strategy to achieve this is the following. Consider the countable set {e ∈ ω : {e}x ∈ WO} and enumerate it as {en : n ∈ ω}. Then define ⎧ if n < m, ⎨ 1, if m < n, Φ(x)(n, k, m, l) = 0, ⎩ {en }(k, l), if m = n. It is easy to see that Φ is as required. From the discussions above Φ is not a Borel function. To allow reductions like Φ we need to consider classes of functions larger than that of Borel functions. This takes us to the realm of the projective hierarchy, and one natural class of functions to use is that of Δ12 functions. One can check that Φ is Δ12 . Two basic issues arise with the consideration of Δ12 reductions. The first is that the usual axioms of set theory no longer seem sufficient to determine the answers of many questions. Thus for many results axioms beyond ZFC are assumed and it is not clear that they can be removed from the assumptions. The second, a more subtle, issue is that, even for results which turn out to be provable within ZFC, the proof techniques go beyond the classical or effective descriptive set theory we have developed and been using in this book. The most powerful method employed in these proofs is that of forcing. Because of these issues we will not prove any result involving Δ12 reductions, even if some of the proofs are classical. Instead, we will mention some results without proof so as to provide the reader a brief overview of the subject. One of the common axioms beyond ZFC used in the results about Δ12 reductions is Σ11 determinacy. This is the statement that every game GA is determined if the payoff set A ⊆ ω ω is Σ11 . Burgess has shown under Σ11 determinacy that if an analytic equivalence relation E has ≤ ω1 -many equivalence classes, then E is Δ12 reducible to Eω1 . For this reason it is customary to denote Eω1 by id(ω1 ), so that the Burgess trichotomy theorem

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becomes a full analog to the Silver dichotomy. Hjorth and Kechris [76] have considered a generalization of the Glimm– Effros dichotomy for Σ11 equivalence relations. Here the analog of the smooth clause, or the concrete classifiability, is the following notion of Ulm classifiability, inspired by Ulm’s classification of countable abelian p-groups (which we will review more in Section 13.4). For any α < ω let 2α be the space of all transfinite sequences (xβ )β<α where xβ ∈ {0, 1} for all β < α. Elements of 2α code subsets of α. If α < β then we view 2α as a subset of 2β in the natural sense: for any x ∈ 2ω , regard x(ξ) = 0 for all α ≤ ξ < β. Let 2<ω1 be the increasing union of all 2α for α < ω1 . Elements of 2<ω1 are called Ulm invariants, since Ulm used such elements to classify countable abelian p-groups up to isomorphism (see Section 13.4). We provide a way to code Ulm invariants by reals, as follows. Given any (x, y) ∈ WO × 2ω , let α = ot(<x ). For each n ∈ ω with y(n) = 1, let <x,n be the restriction of <x on the set {m ∈ ω : m <x n}, and then let βn = ot(<x,n ). We then associate U (x, y) ∈ 2α by letting it code the set {βn : y(n) = 1}. Consider the equivalence relation EUlm on LO × 2ω defined by (x, y)EUlm (x , y  ) if U (x, y) and U (x , y  ) code the same set of ordinals. A somewhat tedious but straightforward calculation shows that EUlm is a Σ11 equivalence relation (Exercise 9.2.2). Definition 9.2.3 An equivalence relation E on a standard Borel space X is Ulm classifiable if it is Δ12 reducible to EUlm . The equivalence relation EUlm is customarily denoted by id(2<ω1 ) for the obvious reason that it can be viewed as the identity relation on 2<ω1 . In this notation an equivalence relation E is Ulm classifiable iff E is Δ12 reducible to id(2<ω1 ). Hjorth and Kechris [76] proved under Σ11 determinacy that if E is a Σ11 equivalence relation, then either E is Ulm classifiable or E0 ≤B E. Finally we note that there are equivalence relations universal for all analytic equivalence relations with respect to Borel reducibility. Proposition 9.2.4 There exists a universal Σ11 equivalence relation, that is, a Σ11 equivalence relation E such that for any Σ11 equivalence relation F , F ≤B E. Proof. Let U ⊆ ω ω × (ω ω )2 be a universal set for all Σ11 subsets of (ω ω )2 , that is, for any Σ11 A ⊆ (ω ω )2 there is x ∈ ω ω such that A = Ux . Define U ∗ by letting (x, y, z) ∈ U ∗ ⇐⇒ (x, y, z) ∈ U ∨ (x, z, y) ∈ U. Then U ∗ is still Σ11 . Let ·, · : ω ω × ω ω → ω ω be a homeomorphism. Then define an equivalence relation E on ω ω by x, yEx , y   ⇐⇒ x = x ∧ (y = y  ∨ (x, y, y  ) ∈ U ∗ ∨ ∃y0 = y, y1 , . . . , yn = y  ∀i < n (x, yi , yi+1 ) ∈ U ∗ ).

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Informally E is a direct sum of transitive closures of Ux∗ for all x ∈ ω ω . It is clear that E is a Σ11 equivalence relation. Suppose F is any Σ11 equivalence relation on ω ω . Then F ⊆ (ω ω )2 and so there is x ∈ ω ω such that F = Ux . We have that Ux∗ = Ux = F is a transitive relation. Define f : ω ω → ω ω by letting f (y) = x, y. Then for any y, y  ∈ ω ω , yF y  iff (y, y  ) ∈ Ux∗ = F iff (x, y, y  ) ∈ U ∗ iff x, yEx, y  . This shows that, in fact, F c E. Exercise 9.2.1 Show that the reduction Φ defined in this section is σ(Σ11 )measurable. Exercise 9.2.2 Show that EUlm is a Σ11 equivalence relation. Exercise 9.2.3 Show that there is a universal Σ11 equivalence relation.

9.3

Actions of standard Borel groups

Orbit equivalence relations of Polish group actions form an important class of analytic equivalence relations. More generally, one can consider actions in which the acting group is standard Borel and still get that the resulting orbit equivalence relation is analytic. Definition 9.3.1 A standard Borel group is a group G with a standard Borel structure B on G such that the group operations are Borel with respect to B. Borel subgroups of Polish groups are standard Borel groups with their induced Borel structure. In particular, all Polish groups are standard Borel groups with their Borel structure. The converse, as we will see below, is not true. Definition 9.3.2 A standard Borel group G is Polishable if there is a Polish topology τ inducing the Borel structure on G so that (G, τ ) is a topological group, that is, the group operations are continuous with respect to τ . Recall that in Definition 8.2.3 we defined Polishable subgroups of Polish groups. The above definition is a generalization of Definition 8.2.3. By Theorem 2.3.3, and more explicitly by Exercise 2.3.7, the topological group topology witnessing the Polishability of a standard Borel group is unique. Lemma 9.3.3 ∞ is a standard Borel group that is not Polishable.

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Proof. ∞ is an Fσ subgroup of the Polish group Rω , hence is a standard Borel group. Assume τ is a Polish group topology on ∞ witnessing its Polishability. For each N ∈ ω define  AN = (xn ) ∈ Rω : (xn )∞ = sup |xn | ≤ N . n

Then AN is a compact, hence closed subset of Rω and ∞ = N AN . It follows that AN is a Borel subset of (∞ , τ ), and in particular has the Baire property. By the Baire category theorem there is N ∈ ω such that AN is nonmeager in (∞ , τ ). By Theorem 2.3.2 AN − AN contains a τ -open nbhd of the identity element of ∞ . Note that AN − AN ⊆ A2N , and from the separability of τ there is a countable set C ⊆ ∞ such that  (x + A2N ). ∞ = x∈C

Let C = {xn : n ∈ ω}. We construct a y ∈ A5N such that y ∈ xn + A2N for any n ∈ ω, arriving at a contradiction. The construction of y = (yn ) is by diagonalization:  5N, if |(xn )n | ≤ 2N , yn = 0, if |(xn )n | > 2N . It is easy to see that |yn − (xn )n | > 2N , hence y ∈ xn + A2N for any n ∈ ω. There are many examples of non-Polishable standard Borel groups. Another such example can be found in Exercise 9.3.1. Here we mention a third one which is also useful in our discussion later this section. Recall from Section 2.6 that for any metric space (X, d) the Graev metric δ can be defined on the free group F(X) making it a topological group. Moreover if (X, d) is separable then F(X) is separable and its completion F(X) becomes a Polish group. If in addition (X, d) is compact, then by Exercise 2.6.3 F(X) becomes a Kσ subset of F(X). We summarize these simple observations in the following lemma. Lemma 9.3.4 If X is a compact Polish space then there is a Borel structure on F(X) making it a standard Borel group. Proof. Let d be any compatible metric on X and let δ be the Graev metric on F(X). Then F(X) becomes a Borel subgroup of the Polish group F(X) and therefore a standard Borel group. If X is not finite then it has an accumulation point, and from the definition of the Graev metric we can see that δ on F(X) is not complete. In fact by a similar argument as in the proof of Lemma 9.3.3 one can show that F(X) is

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not Polishable if X is any uncountable compact Polish space. We leave the details of these proofs to the reader (Exercises 9.3.2 and 9.3.3). We now turn to actions of standard Borel groups. Definition 9.3.5 If G is a standard Borel group, X a standard Borel space and a : G × X → X a Borel action, then we call X a standard Borel G-space. If G is a standard Borel group and X a standard Borel G-space, then the X is obviously Σ11 . The following result of Shelah orbit equivalence relation EG 1 shows that every Σ1 equivalence relation arises this way. Theorem 9.3.6 (Shelah) Let X be a standard Borel space and E a Σ11 equivalence relation on X. Then there is a standard Borel group G and a Borel action of G on X such that X . E = EG Proof. Let τ be a Polish topology on X giving rise to the standard Borel structure. Since E is Σ11 as a subset of (X, τ )2 , there is a closed set F ⊆ (X, τ )2 × ω ω such that xEy ⇐⇒ ∃z ∈ ω ω (x, y, z) ∈ F . As a closed subset of (X, τ )2 × ω ω , F is a Polish space. For each (x, y, z) ∈ F we define a Borel isomorphism π(x,y,z) : X → X by ⎧ ⎨ y, if w = x, π(x,y,z) (w) = x, if w = y, ⎩ u, otherwise. Now let Y be a compact Polish space with the same cardinality as F . For example, if F is uncountable then let Y = 2ω ; if F is countable then let Y be a closed countable subset of 2ω with |F | = |Y |. Then there is a Borel isomorphism ϕ between Y and F . We can therefore associate for any a ∈ Y a Borel isomorphism πϕ(a) . Let G = F(Y ) be the standard Borel group as given in Lemma 9.3.4. Then there is a natural action of G on X induced by a · w = πϕ(a) (w) for a ∈ Y and w ∈ X. It is straightforward to check that this action is Borel X and that E = EG . Exercise 9.3.1 Let T be the multiplicative group of the unit circle {eiθ : 0 ≤ θ < 2π}. Let H = Tω and G = {(xn ) ∈ H : ∃m ∀n ≥ m ( xn = 1 ) }. Show that

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(i) T and H are compact Polish groups; (ii) G is an Fσ subgroup of the Polish group H, hence is a standard Borel group; (iii) G is not Polishable; (iv) E1 is Borel reducible to the coset equivalence relation H/G. Exercise 9.3.2 Let X be a compact Polish space, d a compatible metric on X, and x an accumulation point in X. Show that there is a sequence (yn ) in X with yn → x as n → ∞ such that the following sequence (un ) in F(X) is Cauchy in the Graev metric δ but does not converge in F(X): u0 = e, un+1 = un yn x−1 . Thus δ is not complete on F(X). Exercise 9.3.3 Show that for any uncountable compact Polish space X the standard Borel group F(X) is not Polishable.

9.4

Wild Polish groups

For a general Σ11 equivalence relation the very first attempt to classify it is to decide whether it is Borel or non-Borel. In practice a good place to start is to examine its equivalence classes. Of course, if there is one non-Borel class then we conclude that the equivalence relation is non-Borel. However, for an orbit equivalence relation induced by a Polish group action this strategy will not work since every one of its orbits is Borel. Thus it is somewhat more challenging to decide if an orbit equivalence relation, being Σ11 in general, is Borel or not. In this section we investigate some non-Borel orbit equivalence relations. Theorem 8.2.2 provides a criterion: an orbit equivalence relation on a Polish space is non-Borel iff it has orbits of arbitrarily high complexity. This is useful in some context. For instance, if a locally compact Polish group acts continuously on a Polish space, then every orbit is Fσ and hence the equivalence relation is Borel (in fact Fσ ). However, it is not clear how to directly apply this theorem when the acting group is not locally compact. For instance, one of the simplest nonlocally compact Polish groups is Zω . More work is needed to decide if Zω can induce a non-Borel orbit equivalence relation. In fact, it was a documented question asked by Sami [134] whether any abelian group can induce a non-Borel orbit equivalence relation. This was answered positively by Solecki [140]. We introduce some terminology.

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Definition 9.4.1 A Polish group G is tame if every G-orbit equivalence relation is Borel; it is wild if it is not tame. In this section we present a proof by  Solecki that Zω is wild. In fact, Solecki has considered groups of the form n Gn , where Gn are abelian, and has completely determined which of them are tame. Let X be the set of all trees on Z, that is, T ∈ X iff T ⊆ Z<ω is closed <ω under taking initial segments. X is a Borel subset of 2Z , hence is itself a ω standard Borel space. We define an action of Z on X as follows. For T ∈ X and σ ∈ Zω , let  σ·T = (σ  n) + (T ∩ Zn ). n∈ω

Clearly σ · T ∈ X and the action is Borel. Throughout this section we denote the orbit equivalence relation by E(Zω ), or E for simplicity. Our objective is to show that E is non-Borel. Recall that a tree is well-founded if it does not have any infinite branch, and is ill-founded otherwise. If T1 , T2 ∈ X, we let Φ(T1 , T2 ) = {t ∈ Z<ω : ∀m ≤ lh(t) (t  m) + (T1 ∩ Zm ) = T2 ∩ Zm }. Then it is easy to see that Φ(T1 , T2 ) ∈ X, and T1 ET2 iff [Φ(T1 , T2 )] = ∅ iff Φ(T1 , T2 ) is ill-founded. Note that for any n ∈ ω, Φ(T1 , T2 ) ∩ Zn is a coset of a subgroup of Zn . This motivates the following concepts. Definition 9.4.2 Let T ⊆ Z<ω be a tree on Z. T is a coset tree if for all n ∈ ω, T ∩ Zn is a coset of a subgroup of Zn . T is a group tree if for all n ∈ ω, T ∩ Zn is a subgroup of Zn . Note that in general if G is a group and C ⊆ G is nonempty, then C is a left-coset of a subgroup of G iff for all g1 , g2 , g3 ∈ C, g1 g2−1 g3 ∈ C. The same condition also characterizes the right-cosets in G. Thus even without commutativity we may address them simply as cosets. If C is a coset in G and g ∈ C, then g −1 C is a subgroup of G. In fact g −1 C is independent of the choice of g ∈ C, and it is the unique subgroup of G for which C is a left-coset. For an arbitrary C ⊆ G, we define ⎧ −1 ⎨ g C, if C is a nonempty coset in G and g ∈ C, γC = ⎩ ∅, otherwise. Then for any C ⊆ G, γC is a subgroup of G. Now if T is a coset tree on Z, we let  γ(T ∩ Zn ). γT = n∈ω

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Then γT is a group tree on Z. Moreover, Φ(γT, T ) = T . When a tree T on Z is well-founded we may speak of its rank, denoted by rk(T ), which is a countable ordinal. Here in order to study ill-founded trees, we define the following more general notion. For any tree T on Z, let define T  = {t ∈ T : ∃s ∈ T t  s}. Then T  ⊆ T . By transfinite induction T α for α < ω1 as follows: T 0 = T, T α+1 = (T α ) , and T λ = α<λ T α , if λ is limit. Since T is countable there is a countable ordinal β < ω1 such that T β = T β+1. Define the height of T , denoted ht(T ), to be the least ordinal β such that T β = T β+1 . For a well-founded tree T the height coincides with the rank of T , and if β = ht(T ) = rk(T ), then T β = ∅. Putting all these concepts together we can now reduce the problem to the construction of well-founded coset trees on Z of arbitrarily high rank. Lemma 9.4.3 If there are well-founded coset trees on Z of arbitrarily high rank then E(Zω ) is non-Borel. Proof. Assume toward a contradiction that E is Borel. Note that Φ is a Borel map from X 2 into X, and (T1 , T2 ) ∈ E ⇐⇒ Φ(T1 , T2 ) is well-founded. Thus Φ(X − E) is a Σ11 set of well-founded trees, and by the boundedness principle for Σ11 sets of well-founded trees (Theorem 1.6.11) there is α < ω1 such that rk(Φ(T1 , T2 )) < α for all (T1 , T2 ) ∈ E. Now if T is any well-founded coset tree with rk(T ) ≥ ω, γT is ill-founded, and since Φ(γT, T ) = T is wellfounded, (γT, T ) ∈ E. It follows that rk(T ) = rk(Φ(γT, T )) < α. This shows that all well-founded coset trees on Z have rank < max{ω, α}, contradicting our assumption. 2

Next we further reduce the problem to the construction of group trees on Z of arbitrarily high height. Note that there are no well-founded group trees of rank ≥ ω, since the element (0, 0, 0, . . . ) is always a branch of such a tree. So we do need to consider the concept of height when dealing with complex group trees. We also need to examine closely the tree nodes in the transfinite pruning process. For this define for any t ∈ T the rank of t in T by  the least β such that t ∈ T β , if such β exists, rkT (t) = ω1 , otherwise. For any T ∈ X, if α ≤ ht(T ) is a successor ordinal, then there is t ∈ T with rkT (t) = α. Lemma 9.4.4 Let T be a group tree with ht(T ) > ω. tn ∈ Zn such that for

Then there exist n all n ∈ ω, (tn+1  n) − tn ∈ T , and n tn + (T ∩ Z ) is a well-founded coset tree.

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Proof. First we note a general fact about countable groups. If G is a countable group and (Gn ) is a strictly decreasing sequence of subgroups of G, then there exist gn ∈ G such that for all n ∈ ω, gn+1 ∈ gn Gn , and n gn Gn = ∅. To see this, enumerate G as {hn : n ∈ ω} and inductively define gn+1 ∈ G such that gn+1 ∈ gn Gn and hn ∈ gn+1 Gn+1 . Now let T be a group tree with ht(T ) > ω. Let s0 ∈ T be such that rkT (s0 ) = ω + 1. Let k0 = lh(s0 ) + 1. Then there exists a sequence (un )n∈ω such that for all n ∈ ω, lh(un ) = k0 , s ⊆ un , and rkT (un ) < rkT (un+1 ) < ω. Let ln = k0 + rkT (un ) + 1 and πn be the projection of Zln onto the first k0 coordinates. Then πn is a group homomorphism from Zln to Zk0 , and therefore πn (T ∩ Zln ) is a subgroup of T ∩ Zk0 . Note that un ∈ πn (T ∩ Zln ) but un ∈ πm (T ∩ Zlm ) for all m > n. Thus if we let Gn = πn (T ∩ Zln ), then (Gn ) is a strictly decreasing sequence of subgroups of T ∩ Zk0 . By the general fact above we may let vn ∈ T ∩ Zk0 be such that vn+1 ∈ vn + Gn and n (vn + Gn ) = ∅. Next we inductively define a sequence (wn ) such that wn ∈ Zln , πn (wn ) = vn , and (wn+1  ln ) − wn ∈ T . To begin with, let w0 ∈ Zl0 be any extension of v0 . In general, assume wn has been defined to satisfy the inductive hypothesis. Since vn+1 ∈ vn + πn (T ∩ Zln ), there is w ∈ T ∩ Zln such that πn (w) = vn+1 − vn = vn+1 − πn (wn ). Thus vn+1 = πn (w + wn ). Let wn+1 ∈ Zln+1 be any extension of w + wn . Then πn+1 (wn+1 ) = πn (w + wn ) = vn+1 and (wn+1  ln ) − wn = w ∈ T . Finally we define tn ∈ Zn . For any n ∈ ω let m ∈ ω be the least such that n ≤ lm . Then let tn = wm  n. We verify that the tn have the required properties. To see that (tn+1  n) − tn ∈ T , it is without loss of generality to assume n = lm for some m ∈ ω. Then tn = wm but tn+1 = wm+1  (n + 1). Therefore, (tn+1  n) − tn = (wm+1  lm ) − wm ∈ T . Now let S = n tn + (T ∩ Zω ). It follows that S is a coset tree. It remains to verify that S is well-founded. We claim that for any s ∈ S ∩ Zk0 , rkS (s) < ω. For this let s ∈ S ∩ Zk0 . Since n vn + πn (T ∩ Zln ) = ∅, we have that s ∈ vn + πn (T ∩ Zln ) for some n ∈ ω. Assume s ∈ S ∩ Zln with s ⊆ s . Then s = πn (s ) ∈ πn (tln )+ πn (T ∩Zln ) = πn (wn )+ πn (T ∩Zln ) = vn + πn (T ∩Zln ), a contradiction. This shows that rkS (s) < ln , and therefore rkS (s) < ω. It follows that rk(S) < ω + k0 , and hence S is well-founded. Lemma 9.4.5 Let T be a group tree, α < ω1 , and tn ∈ Zn satisfy (tn+1  n) − tn ∈ T α . Let S = n tn + (T ∩ Zn ). Then S is a coset tree, and for all β ≤ α, S β = n tn + (T α ∩ Zn ). Proof. It is easy to see that if T is a coset (group) tree then T  is a coset (respectively group) tree. It follows that for any β < ω1 , T β is a coset (group)

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tree if T is. We only check that S  = n tn + (T  ∩ Zn ). The lemma then follows by an easy induction. Let s ∈ S  ∩Zn . Then there are t ∈ T ∩Zn , s ∈ S ∩Zn+1 , and t ∈ T ∩Zn+1 such that s = tn + t, s ⊆ s , and s = tn+1 + t . Since (tn+1  n) − tn ∈ T  there is also u ∈ T ∩ Zn+1 with (tn+1  n) ⊆ tn ⊆ u. Then t = s−tn = (s  n)−tn = (tn+1  n)+(t  n)−tn = u  n+t  n = (u+t )  n.

Since T is a group tree u + t ∈ T , and so S  ⊆ n tn + (T  ∩Zn ). The converse is obvious. Lemma 9.4.6 For any α < ω1 if there is a group tree of height > α + ω then there is a well-founded coset tree S of rank ≥ α. Proof. Let T be a group tree of height > α + ω. Then T α = ∅. Apply α n that (tn+1  n) − tn ∈ T α Lemma

9.4.4 to αobtainω tn ∈ T ∩ Z for n ∈ ω such and n tn + (T ∩ Z ) is well-founded. Let S = n tn + (T ∩ Zn ). Then by Lemma 9.4.5 S is a coset tree with S α = ∅ well-founded. Thus S is a well-founded coset tree with rk(S) ≥ α. Theorem 9.4.7 (Makkai–Solecki) There are group trees on Z of arbitrarily high height. Proof. Let φ : Z2 → Z be the homomorphism φ(a, b) = a+b. (G 0n ), (G1n ) Let 0 be strictly decreasing sequences of subgroups of Z such that n Gn = n G1n = {0} and φ(G0n × G1n ) = Z for all n ∈ ω. For example, we may take G0n = 2n Z and G1n = 3n Z. By transfinite induction on α < ω1 we define a group tree Tα such that (a) if α = β + 1 is a successor, then (Tα )β ∩ Z = Z and (Tα )a is well-founded if a = 0; (b) if α is a limit, then for any β < α there is n ∈ ω such that G0n × G1n ⊆ (Tα )β ∩ Z2 , and if t ∈ Z2 and t = 0 0, then (Tα )t is well-founded. Granting this construction it is clear that ht(Tα ) ≥ α. Let T0 = {0n : n ∈ ω} and T1 = T0 ∪ Z. If α = β + 1 and β is a successor, let Tα = {∅} ∪ Z ∪ {t(0) t : t ∈ Tβ , lh(t) ≥ 1}. If α = β + 1 and β is a limit, let Tα = the tree generated by {φ(t(0), t(1)) t : t ∈ Tβ , lh(t) ≥ 2}.

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With these definitions (a) is satisfied. Suppose α is a limit. To satisfy (b) we construct two group trees S0 and S1 and an increasing sequence of successor ordinals βn → α such that (i) G0n ⊆ (S0 )βn and G1n ⊆ (S1 )βn for all n ∈ ω; (ii) (S0 )a and (S1 )a are well-founded if a = 0. If S0 and S1 are defined as above, then let Tα = {t : t  {2k : k ∈ ω} ∈ S0 and t  {2k + 1 : k ∈ ω} ∈ S1 } . Then it is easy to see that (b) is satisfied. Fix any increasing sequence of successor ordinals βn → α. Apparently the constructions of S0 and S1 are parallel. We only construct S0 and then the construction of S1 will be similar. Note that Tβn have already been constructed to satisfy (a). S0 will be an amalgamation of the Tβn as follows. Let {In }n∈ω be a partition of ω − {0} into infinitely many infinite subsets. For each n ∈ ω define a tree Un by s ∈ Un ⇐⇒ s = ∅ ∨ [ s(0) ∈ G0n ∧ s  In ∈ Tβn ∧ ∀i < lh(s)(i = min In ⇒ s(i) = s(0) ∧ i ∈ In ⇒ s(i) = 0) ]. Then Un is a group tree. Let S0 =

   Un ∩ Zk . k

n

Then S0 is a group tree by definition. To see that G0n ⊆ (S0 )βn let a ∈ G0n . Let s ∈ Zmin In +1 be such that s(0) = s(min In ) = a and s(i) = 0 for all 0 < i < min In . Then s ∈ Un ⊆ S0 . Since (Tβn )βn −1 ∩ Z = Z, s ∈ (Un )βn −1 ⊆ (S0 )βn −1 . Thus a ∈ (S0 )βn and (i) is satisfied. To show (ii) let a = 0 and toward a contradiction let σ ∈ Zω be an infinite branch of (S0 )a . Then σ(0) = a. Let n ∈ ω be large enough such that a ∈ G0n . We have that σ(min Ii ) = 0 for all i < n. Otherwise if σ(min Ii ) = 0 for i < n, then by our construction of S0 , σ  Ii is an infinite branch of Tβi with (σ  Ii )(0) = σ(min Ii ) = 0, violating our inductive assumption (a) for Tβi . Let k > max{min Ii : i < n}

and s = σ  k ∈ S0 . Then by the definition of S0 we have that s ∈  m Um ∩ Zk . Since Um are group trees we may assume s = sm1 + sm2 + · · · + sml + sml+1 + · · · + smp with m1 < m2 < ml < n ≤ ml+1 < · · · < mp , where smj ∈ Umj for j = 1, . . . , p. It follows that a = sm1 (min Im1 )+· · ·+sml (min Iml )+sml+1 (min Iml+1 )+· · ·+smp (min Imp ). From the above argument we get that smj (min Imj ) = 0 for j = 1, . . . , l. However, smj (min Imj ) ∈ G0n for j = l + 1, . . . , p. Hence a ∈ G0n , a contradiction. This shows that S0 has the required properties.

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Corollary 9.4.8 (Solecki) E(Zω ) is non-Borel and Zω is wild. Exercise 9.4.1 Let G be a group and C ⊆ G is nonempty. Show that the following are equivalent: (i) C is a left-coset of some subgroup of G. (ii) C is a right-coset of some subgroup of G. (iii) For all g1 , g2 , g3 ∈ C, g1 g2−1 g3 ∈ C. Exercise 9.4.2 Show that if T is a coset tree then Φ(γT, T ) = T . Give a counterexample where T is not a coset tree. Exercise 9.4.3 Show that if T is a coset tree then (γT ) = γ(T  ). Exercise 9.4.4 Let T be a group tree and tn ∈ Zn such that (tn+1  n)−tn ∈ T ∩ Zn . Show that n tn + (T ∩ Zn ) is a coset tree.

9.5

The topological Vaught conjecture

It is not known if the Silver dichotomy holds for orbit equivalence relations of Polish group actions. This is one of the central open problems in the entire invariant descriptive set theory. Proposition 9.5.1 Let G be a Polish group. Then the following statements are equivalent: (a) For any Polish G-space X, either there are countably many orbits or there are perfectly many orbits. (b) For any Polish G-space X and invariant Borel set B ⊆ X, either B contains countably many orbits or B contains perfectly many orbits. (c) For any Borel G-space X, either there are countably many orbits or there are perfectly many orbits.

Proof. The implications (c)⇒(b)⇒(a) are obvious. (a)⇒(c) follows from Theorem 4.4.6. For any Polish group G, the topological Vaught conjecture for G refers to any of the above equivalent statements. We denote it as TVC(G). Then the topological Vaught conjecture is the statement

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The topological Vaught conjecture was first formulated by Miller. But its counterpart in countable model theory goes back to Vaught. We will elaborate on the connections in Chapter 11. The topological Vaught conjecture for any tame Polish group is true by the Silver dichotomy. In particular, TVC(G) holds for any locally compact Polish group G. Sami [134] proved the topological Vaught conjecture for all abelian Polish groups. Hjorth and Solecki [83] proved it for all nilpotent Polish groups. Solecki [83] proved it for all Polish groups with two-sided invariant metrics. The strongest result up to date is the following theorem of Becker. Theorem 9.5.2 (Becker) TVC(G) holds for all cli Polish groups, that is, Polish groups with complete left-invariant metrics. Proof.

This is a corollary of Theorem 6.5.1.

By Theorem 2.2.11 the class of cli Polish groups is closed under closed subgroups, quotient groups, and group extensions. Also it is easy to see that it is closed under taking countable products. We next consider the abstract closure properties of the class of all Polish groups satisfying the topological Vaught conjecture. Theorem 9.5.3 Let C be the class of Polish groups G for which TVC(G) holds. Then (i) if G ∈ C and H is a closed subgroup of G, then H ∈ C; (ii) if G ∈ C and H is a closed normal subgroup of G, then G/H ∈ C; (iii) if G ∈ C and G is a closed subgroup of H where H/G is countable, then H ∈ C. Proof. For (i) we let X be a Polish H-space. By Theorem 3.5.2 there is a Polish G-space Y such that X is a closed subset of Y and that every G-orbit contains a unique H-orbit. It follows that if Y has only countably many G-orbits then X has only countably many G-orbits. Assume Y has Y perfectly many G-orbits, and let Z ⊆ Y be a perfect set of pairwise EG Y 1 inequivalent elements. Then the relation R = EG  (Z × X) is Σ1 and for any Y y ∈ Z there is x ∈ X such that yEG x by Theorem 3.5.2. By the Jankov–von Neumann uniformization theorem R has a σ(Σ11 )-measurable uniformization, that is, there is a σ(Σ11 )-measurable function f : Z → X such that for any y ∈ Z, yRf (y). Since σ(Σ11 )-measurable functions are Baire measurable (by the remarks preceding Proposition 2.3.1), it follows that there is a comeager set C ⊆ Z such that f  C is continuous. Without loss of generality assume

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C is Gδ , hence an uncountable Polish space. Then f : C → X is a continuous X Y reduction from id(C) to EH , since for any y1 = y2 ∈ C, (y1 , y2 ) ∈ EG , and Y X thus (f (y1 ), f (y2 )) ∈ EG and therefore (f (y1 ), f (y2 )) ∈ EH . This shows that X has perfectly many G-orbits. (ii) is obvious since every Polish G/H-space is also a G-space. For (iii) we consider a Polish H-space X. X is also a Polish G-space with the restricted action of G, and so either there are only countably many G-orbits or there are perfectly many G-orbits. The assumption implies that every H-orbit contains countably many G-orbits. Thus in the case there are only countably many G-orbits there are only countably many H-orbits as well. In the second X -inequivalent elements. Then case, assume Y ⊆ X is a perfect subset of EG X 1 E = EG  Y is a Σ1 equivalence relation on Y with every equivalence class countable. Thus E has the Baire property. By Theorem 5.3.1, and more explicitly Exercise 5.3.1, there are perfectly many E-equivalence classes in Y , X -inequivalent elements in X. which gives a perfect set of EH It follows that the topological Vaught conjecture is equivalent to TVC(G) for any universal Polish group G. By the results of Section 2.5 it is equivalent to TVC(Iso(U)) or to TVC(H(IN )). In the rest of this section we consider a more general notion. Definition 9.5.4 Let X be a standard Borel space and E an equivalence relation on X. The topological Vaught conjecture for (X, E), denoted TVC(X, E), is the statement every invariant Borel subset of X either contains only countably many E-equivalence classes or perfectly many E-equivalence classes. Recall from Section 5.2 that a faithful Borel reduction from an equivalence relation E on X to F on Y is a Borel reduction f : X → Y witnessing E ≤B F and such that for any invariant Borel subset A ⊆ X, [f (A)]F is Borel. The following proposition shows that TVC is closed under faithful Borel reductions. Theorem 9.5.5 Let X, Y be standard Borel spaces and E, F be Σ11 equivalence relations on X, Y , respectively. If E ≤f B F , then TVC(Y, F ) implies TVC(X, E). Proof. Without loss of generality assume that X and Y are Polish spaces. Let f be a faithful Borel reduction from E to F . Let A ⊆ X be invariant Borel. Then [f (A)]F is invariant Borel. Suppose TVC(Y, F ). If [f (A)]F contains only countably many F -classes, then A contains only countably many E-classes. Otherwise, suppose [f (A)]F contains perfectly many F -classes, and let P ⊆ [f (A)]F be a perfect subset of F -inequivalent elements. Consider the

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set D = {(y, x) ∈ P × X : f (x)F y}. and for all y ∈ P there is x ∈ X with (y, x) ∈ D. By the Then D is Jankov–von Neumann uniformization theorem D has a σ(Σ11 )-measurable uniformization g. Thus g : P → X is Baire measurable and for y1 = y2 ∈ P , (g(y1 ), g(y2 )) ∈ E. Let now C ⊆ P be a dense Gδ subset such that g  C is continuous. Then g  C : C → X and g(C) is a perfect subset of X with E-inequivalent elements. Σ11

Thus to prove TVC(G) for a Polish group it suffices to consider a Polish G-space to which every other Polish G-space is faithfully Borel reducible. Exercise 9.5.1 Show that if H is a countable group then TVC(G) holds iff TVC(G × H) holds. For any standard Borel space X and equivalence relation E on X the Glimm–Effros dichotomy for (X, E), denoted GED(X, E), is the statement for any invariant Borel subset A ⊆ X, either E  A is smooth or E0 c E  A. Exercise 9.5.2 Let X, Y be standard Borel spaces and E, F be Σ11 equivalence relations on X, Y , respectively. Show that if E ≤f B F , then GED(Y, F ) implies GED(X, E). The following exercise problems outline a proof of the topological Vaught conjecture for abelian Polish groups by Sami. Assume G is an abelian Polish group and X a Polish G-space. Define an equivalence relation E on X by xEy ⇐⇒ Gx = Gy , where Gx = {g ∈ G : g · x = x} is the stabilizer group of x ∈ X. X ⊆ E. Exercise 9.5.3 Show that E is a Π11 equivalence relation and EG Deduce that if there are perfectly many E-equivalence classes then there are perfectly many G-orbits.

Exercise 9.5.4 Assume there are only countably many E-equivalence classes. Show that each E-equivalence class is Borel and that for each E-equivalence X  C is Borel. (Hint: Use Theorem 8.2.1.) class C the relation EG Exercise 9.5.5 Show that either there are only countably many G-orbits in X or there are perfectly many G-orbits.

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Chapter 10 Turbulent Actions of Polish Groups

In this chapter we turn back to Polish group actions and orbit equivalence relations. The Borel reducibility hierarchy of analytic equivalence relations has taken shape with the results we have established in the previous chapters. However, it is also noticeable that the results were obtained with a variety of machineries ranging from Baire category methods, measure theoretic methods, group theory, and descriptive set theory. There are still many pairs of equivalence relations we have introduced so far but have not mentioned their reducibility relation, either because it is still an open problem or because its known proof goes beyond the limitation of length or scope of this book. It is thus clear that a sweeping method to prove reducibility or, more challengingly, nonreducibility would be much desirable. The greatest success up to date is Hjorth’s theory of turbulence, which we present in this chapter. It will turn out that orbit equivalence relations from turbulent actions are not reducible to any S∞ -orbit equivalence relation. This powerful theorem has many applications in the classification problems of mathematics, and its potential is still being discovered in current research. The results in this entire chapter are due to Hjorth [70], unless otherwise indicated. Expositions of further results of the theory of turbulence can be found in References [70] and [101].

10.1

Homomorphisms and generic ergodicity

In this section we review the Baire category method for showing nonsmoothness of orbit equivalence relations and introduce some new concepts. Let G be a Polish group and X a Polish G-space. Recall from Section 6.1 X is generically ergodic iff every G-invariant Borel set is either meager that EG X or comeager. We showed in Proposition 6.1.9 that EG is generically ergodic X iff it has a dense orbit, and in Proposition 6.1.10 that if EG is generically X ergodic and has no comeager orbits, then EG is not smooth. The proof of Proposition 6.1.10 (similar to that of Proposition 6.1.6) is by contradiction. We now turn this proof into a positive statement by working a little harder.

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Proposition 10.1.1 Let G be a Polish group, X a Polish G-space, and f : X → 2ω a Baire X y. Suppose measurable G-invariant map, that is, f (x) = f (y) whenever xEG X EG is generically ergodic. Then there is a comeager set C ⊆ X on which f is constant. Proof. First by Proposition 6.1.9 (iii) there is an invariant dense Gδ set Y ⊆ X such that every orbit in Y is dense. Now Y is still a Polish G-space. Thus it suffices to find a comeager set C ⊆ Y on which f is constant. Let C ⊆ Y be comeager such that f  C is continuous. By the continuity of the action the map x → g · x is a homeomorphism of X onto X for all g ∈ G. Thus we have that ∀g ∈ G ∀∗ x ∈ Y g · x ∈ C. Then by the Kuratowski–Ulam theorem, ∀∗ x ∈ Y ∀∗ g ∈ G g · x ∈ C. Therefore we may find and fix an x ∈ Y such that ∀∗ g ∈ G g · x ∈ C. We claim that [x]G ∩ C is dense. For this let y ∈ Y be arbitrary and U ⊆ Y be open with y ∈ U . Since [x]G is dense the set {g ∈ G : g · x ∈ U } is nonempty. But it is also open, so there is g ∈ G such that g · x ∈ U ∩ C. This shows that [x]G ∩ C ∩ U = ∅. Now f  C takes the constant value f (x). To see this, let z ∈ C. By the density of [x]G ∩ C there is a sequence (xn )n∈ω in [x]G ∩ C such that xn → z as n → ∞. However, f is continuous on C, and therefore f (xn ) → f (z) as n → ∞. But f (xn ) = f (x) by the invariance assumption, so f (z) = f (x). In view of its proof the hypotheses of the proposition can be weakened to assume a Baire measurable G-invariant map f defined on a comeager set. Proposition 10.1.2 Let G be a Polish group, X a Polish G-space, C ⊆ X a comeager set, and X is generically f : C → 2ω a Baire measurable G-invariant map. Suppose EG  ergodic. Then there is a comeager set C ⊆ X on which f is constant. Proof. As in the above proof there is an invariant comeager set C0 such that every orbit in C0 is dense. Now C ∩ C0 is still comeager, we may let C  ⊆ C ∩ C0 be comeager such that f  C  is continuous. The same proof as above gives that f is constant on C  . A number of things can be extracted from these propositions and their proofs. We start with some definitions. Definition 10.1.3 Let X, Y be sets and E, F equivalence relations on X, Y , respectively. (1) A map f : X → Y is a homomorphism from E to F if for any x1 , x2 ∈ X, x1 Ex2 ⇒ f (x1 )F f (x2 ).

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(2) If X is a Polish space, then a generic homomorphism from E to F is a map f from a comeager subset C0 of X to Y so that for some comeager set C1 ⊆ C0 f is a homomorphism from E  C1 to F . Definition 10.1.4 Let X, Y be Polish spaces and E, F equivalence relations on X, Y , respectively. We say that E is generically F -ergodic if for every Baire measurable generic homomorphism f from E to F there is a comeager set C ⊆ X such that f (x1 )F f (x2 ) for any x1 , x2 ∈ C. With the new terminology Proposition 10.1.2 can be succinctly stated as Any generically ergodic orbit equivalence relation is generically id(2ω )-ergodic. It turns out that these notions are actually equivalent. Lemma 10.1.5 X is generically Let G be a Polish group and X a Polish G-space. Then EG ergodic iff it is generically id(2ω )-ergodic. X Proof. Suppose EG is generically id(2ω )-ergodic. Let {Un }n∈ω be a countable base for X consisting of nonempty open sets. Define f : X → 2ω by

f (x)(n) = 1 ⇐⇒ ∃g ∈ G g · x ∈ Un ⇐⇒ x ∈ [Un ]G . Since [Un ]G is open for each n ∈ ω, f is a Borel map. It is clear that f is X