Descriptive Properties of Measure Preserving. Actions and the Associated Unitary Representations. Tzer-jen Wei. Thesis by

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1 Descriptive Properties of Measure Preserving Actions and the Associated Unitary Representations Thesis by Tzer-jen Wei In Partial Fulllment of the Requirements for the Degree of Doctor of Philosophy California Institue of Technology Pasadena, California 2005 (Defended May 12, 2005)

2 ii c 2005 Tzer-jen Wei All Rights Reserved

3 iii Acknowledgements I would like to thank my advisor, Alekos Kechris, for his instruction, patience, and support in my graduate studies and while I was preparing this work. Thanks also to John Clemens, Christian Rosendal, Orestis Raptis, Todor Tsankov for listening to my presentations and providing useful feedback. I am grateful to the Department of Mathematics at Caltech for its support during my graduate study program and for providing a great environment for research. Finally, I would like to thank my parents and my sister for their encouragement and personal support.

4 iv Abstract Let be a countable group and X a Borel -space with invariant Borel probability measure µ. Let E = E be the countable equivalence relation dened by xey γ (γ x = y). This thesis consists of two independent parts, Chapter 2 and Chapter 3: In Chapter 2, we study the descriptive complexity of the full group [E]. The main result of this chapter is i) If E is not smooth, then [E] is Π 0 3-complete; ii) If E is smooth, then [E] is closed. We also study the descriptive complexity of N[E], the normalizer of [E]. It turns out that N[E] has the same complexity as [E], i.e., N[E] is Π 0 3-complete i E is not smooth and is closed if E is smooth. In Chapter 3, we study descriptive properties of π X, the Koopman unitary representation associated with the action. Consider the induced Polish action on L 2 (X), i.e., γ f = π X (γ)(f). Denote by E L2 (X) the induced countable Borel equivalence relation on L 2 (X), i.e., fe L2 (X) g γ (g = γ f).

5 v Let act on the measure algebra of µ, MALG µ, by γ A = γ(a) and on Aut(X, µ) by γ T = π X (γ)t. We have the induced countable equivalence relations E MALGµ and E Aut(X,µ) respectively. We relate the descriptive complexity of E L2 (X) to that of E X. We show that the smoothness of EL2 (X) is equivalent to the smoothness of E MALGµ and the compressibility of the nonconstant part of E L2 (X) is equivalent to the compressibility of E MALGµ\{X, }. We also connect the smoothness and compressibility of E L2 (X) to mixing properties of the action of on X. Finally, we will show that the amenability of E X implies a certain weak containment property of πx.

6 vi Contents Acknowledgements iii Abstract iv Chapter 1. Introduction 1 Chapter 2. Descriptive Complexity of Full Groups Full groups and their normalizers Upper bound of the complexity of [E] and N(E) Smooth equivalence relations and the closure of [E] The Descriptive Complexity of [E] and N(E) N(E) as a Polishable group 18 Chapter 3. Descriptive Properties of Measure Preserving Actions and the Associated Unitary Representations Introduction Smoothness of E L2 (X,µ) Compressibility Some Embedding and Containment Results 60 Bibliography 74

7 1 CHAPTER 1 Introduction Let be a countable group and X a standard Borel -space with an invariant (nonatomic) Borel probability measure µ. induces an equivalence relation E X on X, which is dened by xe X y γ (γ x = y). By a theorem of Feldman-Moore (see [KM], Theorem 1.3), every countable Borel equivalence relation on a standard Borel space X is induced by some Borel action of some countable group. Denote by Aut(X, µ) the group of µ-measure preserving automorphisms of X (modulo null sets). For each T Aut(X, µ), we can dene a corresponding unitary operator U T U(L 2 (X)), U T (f) = f T 1. And by identifying T and U T, we can view Aut(X, µ) as a subgroup of U(L 2 (X)). The weak topology of Aut(X, µ) is the subspace topology of the weak topology on U(L 2 (X)). Aut(X, µ) is a closed subspace of U(L 2 (X) in the weak topology, hence Polish. In Chapter 2, we will study the descriptive complexity of an important invariant of E, namely the full group of E. The full group of E, denoted by [E], is dened by [E] = {T Aut(X, µ) : xet x a.e.}. The main result of this chapter is i) If E is not smooth, then [E] is Π 0 3-complete. ii) If E is smooth, then [E] is closed. And the same result holds for N[E]:

8 2 i) If E is not smooth, then N[E] is Π 0 3-complete. ii) If E is smooth, then N[E] is closed. Dene π X : Aut(X, µ) U(L 2 (X)), γ T, where T is the element in Aut(X, µ) such that T (x) = γ x for µ-almost every x X. Or equivalently T (f) = U T (f) = f(γ 1 ) for every f L 2 (X). Clearly π X is a homomorphism of into U(L 2 (X)), i.e., π X is a unitary representation of on the Hilbert space L 2 (X). The unitary representation π X induces a natural Polish action on L 2 (X), i.e., γ f = π X (γ)(f). Denote by E L2 (X) the induced countable Borel equivalence relation on L 2 (X), i.e., fe L2 (X) g γ (g = γ f). In Chapter 3, we will study relations between E X and EL2 (X). We will obtain characterizations of the smoothness and compressibility of E L2 (X) and reducibility results. Denote by MALG µ the measure algebra of µ (see Section 3.1.2). Let act on MALG µ by γ A = γ(a). Similarly, we have the induced equivalence relation E MALGµ dened by AE MALGµ B γ (B = γ A).

9 3 We will show that E L2 (X) is smooth i E MALGµ is smooth and the nonconstant part of E L2 (X) is compressible i E MALGµ\{X, } is compressible. These descriptive properties of E L2 (X) also have connections with mixing properties of the action: i) If action is mildly mixing, then E L2 (X) is smooth. ii) The nonconstant part of E L2 (X) is compressible i the action is weakly mixing. Furthermore, we will show that smoothness is related to rigid factors and compressibility is related to isometric factors. At the end of this chapter, we will study some embedding properties, containment properties, and their applications. Denote by λ the regular unitary representation of and by λ /x the quasi-regular unitary representation on λ /x. If the action is amenable, then π X λ (see [Kuhn]). We show that if E X is amenable, then π X X λ / x dµ(x).

10 4 CHAPTER 2 Descriptive Complexity of Full Groups 2.1. Full groups and their normalizers Let µ be a Borel probability measure dened on a standard Borel space X. Recall that Aut(X, µ) denotes the group of all measure preserving Borel isomorphisms (modulo null sets) on X. There are two frequently used topologies dened on Aut(X, µ), namely the uniform topology and the weak topology. Let's use B(X) to denote the set of Borel subsets of X and A = {A n } an algebra generating B(X). The uniform topology has as basis the sets of the form V T,ɛ = {S Aut(X, µ) : sup{µ(s(a) T (A)) : A B(X)} < ɛ}. It has two compatible metrics: d 1 (S, T ) = µ{x S(x) T (x)} and d 2 (S, T ) = sup{µ(s(a) T (A)) A B(X)} = sup{µ(s(a n ) T (A n ))}. n N Aut(X, µ) is not separable in the uniform topology.

11 The weak topology has as sub-basis the sets of the form 5 W T,An,ε = {S µ(s(a n ) T (A n )) < ε} and a complete and compatible metric: ρ(s, T ) = 2 n (µ(s(a n ) T (A n )) + µ(s 1 (A n ) T 1 (A n ))). Aut(X, µ) is a Polish group in the weak topology. Now consider a countable Borel equivalence relation E dened on a standard Borel space X. By Theorem 1.3 in [KM], E is induced by Borel action of a countable group G acting on X, i.e., E = E G, xey g G(g x = y). We call a Borel measure µ on X E-invariant if µ is G-invariant for every countable Borel group G such that E G = E. Dene for µ that which is E invariant [E] = {T Aut(X, µ) : xet x a.e.} and let N(E) Aut(X, µ) be the normalizer of [E], i.e., N(E) = {T Aut(X, µ) : T 1 [E]T = [E]}. The goal of this chapter is to determine the descriptive complexity of the full groups of countable Borel equivalence relations and their normalizers in the weak topology.

12 Upper bound of the complexity of [E] and N(E) Let {f n } [E] be a Cauchy sequence in the uniform topology and assume n m > n d 1 (f n, f m ) < 2 n. For each n N, let Y n = {x m > n(f m (x) = f n (x))}. We have µ(y n ) > 1 2 n, and dene f : X X by f(x) = f n (x) if x Y n. Clearly f is well-dened µ-almost everywhere and is in [E]. Since d 1 (f, f n ) < 2 n, we have f = lim f n. Therefore [E] is complete in the uniform topology In general, [E] is not closed in the weak topology. For example, consider the Vitali equivalence relation E 0 on (X, µ) = ([0, 1], m), xe 0 y i 2 n (x y) N for some n N. Since E 0 is ergodic, for all measurable subsets A, B X, T [E 0 ] (T (A) = T (B) i µ(a) = µ(b). So clearly [E 0 ] = Aut(X, µ) in the weak topology. But [E 0 ] Aut(X, µ) (for example x x + π mod 1 is in Aut(X, µ)\[e 0 ]), thus [E 0 ] is not closed in the weak topology. More generally, let E be an ergodic countable Borel equivalence relation. Then [E] = Aut(X, µ) by the same reason above. We may assume (X, µ) = ([0, 1], m) and consider f r : x x + r mod 1. We have f r Aut(X, µ) and d 1 (f r1, f r2 ) = 1 if r 1 r 2, therefore Aut(X, µ) is not separable in the uniform topology. However, by the following proposition, we can show that [E] is separable in the uniform topology, hence [E] Aut(X, µ) = [E] in the weak topology. Proposition If E is a countable Borel equivalence relation on a standard Borel space X with invariant probability measure µ, then [E] is separable in the uniform topology.

13 7 Proof. Let T, S : X X be Borel maps. Consider the equivalence relation T S µ x(t (x) = S(x)). Denote by [T ] the equivalence class of T. When T is an automorphism, it is customary to write T instead of [T ] if there is no danger of confusion. To avoid the confusion with the [ ] notation of full groups, we will write the equivalence of T as T instead [T ] for all Borel maps too. So when we write T is Borel (modulo null), we mean the equivalence class of T. Let A = {T T : X X is Borel (modulo null).}. We can view (Aut(X, µ), d 1 ) as a metric subspace of (A, d 1 ). We only need to show that there exists a countable subset S A, such that [E] S. Construct S by the following steps. First, let {P m } be a sequence of nite Borel partitions of X where P m = {A m n }, such that for any nite Borel partition {X n } and ε > 0, there is a P m = {A m n } with n (µ(a m n X n ) < ε). Then since E is countable, we may assume X is a countable Borel G-space and E = E G. Let {g i } be a nite subset of G and {g i } P m, dene S {gi },m by S {gn},m A m n = g n A m n. Let S = {S {gn},m} {gn} <,{g n} G. Then clearly S A and is countable.

14 We only need to show [E] S, that is 8 ε > 0 T [E] S S (d 1 (S, T ) < ε). Since T [E], we can nd a nite subset {g n } n N G and a nite partition {X n } n N such that T X n = g n X n, if n < N, and µ(x N ) < ε. Fix an m so that 2 µ(a m n X n ) < ε 2N. Clearly d 1 (T, S {gn},m) < µ(x N ) + µ(a m n X n ) < ε. Therefore, [E] is separable in the uniform topology. The following proposition gives an upper bound on the Borel complexity of [E]. Proposition If E is a countable Borel equivalence relation on a standard Borel space X with invariant probability measure µ, then [E] is Π 0 3 in the weak topology. Proof. Since [E] is separable in the uniform topology by the above proposition, we can x a countable dense subset {T n } n N [E] in the uniform topology. And since [E] is closed in the uniform topology, we can write [E] = {S : d 2 (S, T n ) < 2 m }. m=1 n=0 Note that {S : d 2 (S, T n ) < 2 m } is an open ball in the d 2 metric. Let A = {A n } be an algebra generating B(X). We can write an open ball in the d 2 metric as

15 9 {S : d 2 (S, T ) < r} = = {S : sup{µ(sa n T A n )} r 2 m } m=1 m=1 n=0 n {S : µ(sa n T A n ) r 2 m }, which is clearly Σ 0 2 in the weak topology. Hence [E] is Π 0 3. We can use this to show that N(E) is in Π 0 3 in the weak topology. Proposition If E is a countable Borel equivalence relation on a standard Borel space X with invariant probability measure µ, and if [E] Π 0 3 in the weak topology, then N(E) Π 0 3 in the weak topology too. In particular, by Proposition 2.2.2, N(E) Π 0 3 in the weak topology. Proof. Let G be a countable group of Borel automorphisms generating E. First note that if T Aut(X, µ) and g G (T gt 1 [E]), then if xey, y = g(x) for some g G, T (y) = T (g(x)) = T gt 1 T (x). Since T gt 1 [E], there is a conull subset Y X, such that x, y Y (xey T (y)et (x)). Similarly if g G (T 1 gt [E]) then there is a conull subset Y X, x, y Y (T (x)et (y) xey). So if g G (T gt 1 [E]) and g G (T 1 gt [E]), then T N(E).

16 On the other hand, if T N[E], then 10 T 1 [E]T = [E] = T [E]T 1, so g G [E], T gt 1 [E], and T 1 gt [E]. Therefore, N(E) = g G{T T gt 1 [E]} {T T 1 gt [E]}. Clearly T T gt 1 and T T 1 gt (for any g G) are continuous maps from Aut(X, µ) to itself, and reduce {T T gt 1 [E]} and {T T 1 gt [E]} to [E], respectively. Therefore N(E) is Π Smooth equivalence relations and the closure of [E] To determine the exact complexity of [E], the simplest case is when E is µ- smooth, i.e., E Y = F Y, where F is smooth and Y X is µ-conull. We can even slightly loosen our conditions. Consider a (not necessarily countable) Borel equivalence relation F on X, andconsider µ a (not necessarily F -invariant) Borel probability measure on X. We dene [F ] = {T Aut(X, µ) µx (T (x)f x)} (which extends the concept of the full group to general Borel equivalence relations). We have the following simple proposition:

17 11 Proposition Let F be a Borel equivalence relation on a standard Borel space X with a Borel probability measure µ. If F is smooth, then [F ] is closed in the weak topology of Aut(X, µ). Proof. F is smooth, hence F B ([0, 1]). Let f : X [0, 1] be a Borel function such that xf y i f(x) = f(y). Then the assignment T f T is a continuous map from Aut(X, µ) to L 2 (X). Note that T [F ] i ft = f (modulo null sets), hence [F ] is closed. Notice that if two Borel equivalence relations F 1 and F 2 agree a.e. in the sense that F 1 Y = F 2 Y for some co-null subset Y X, then [F 1 ] = [F 2 ]. We have: Corollary Let F be a Borel equivalence relation on a standard Borel space X with a Borel probability measure µ. If F is µ-smooth, then [F ] is closed in the weak topology of Aut(X, µ). It remains to nd the complexity of E when it is not µ-smooth. Lemma If E is a countable Borel equivalence relation on a standard Borel space X with invariant probability measure µ, then [E] is closed or Σ 0 2 -hard. Proof. If [E] is not closed, then T [E]\[E]. If [E] is Π 0 2, then since [E] is dense in [E], [E] is comeager in [E]. Hence the coset [E]T is also comeager in [E], but [E] ([E]T ) =, a contradiction. Therefore [E] is not Π 0 2, so it is Σ 0 2 -hard (see [Kechris 2], Theorem 22.10). It makes sense to see what [E] is. Recall the uniform ergodic decomposition for invariant measures of E. Denote by P (X) the set of probability measures on X, by I E P (X) the set of E-invariant Borel probability measures on X and by

18 12 EI E P (X) the set of E-invariant ergodic Borel probability measures on X. We have (see [KM], Theorem 3.3) Theorem (Farrell, Varadarajan) Let E be a countable Borel equivalence relation on a standard Borel space X. Assume I E 0. Then there is a unique (up to null sets) Borel surjection π : X EI E such that (1) π(x) = π(y) if xey; (2) If X e = {x : π(x) = e}, for e EI E, then e(x e ) = 1; (3) For any µ I E, µ = π(x)dµ(x). So we can write EI E = {e x }, where e x = π(x). From now on, we will use the above notations: π, X e to denote the unique ergodic decomposition of (X, E) and F to denote the Borel equivalence relation on X, which is dened by xf y i π(x) = π(y). Since F is smooth, [F ] is closed by and clearly [E] [F ]. Furthermore, we can show that [F ] is the closure of [E]. Theorem Let E be a countable Borel equivalence relation on a standard Borel space X with invariant probability measure µ. Given S [F ] and a Borel set A, then there is a T [E], such that T (A) = S(A). Proof. Let Y be an F - invariant Borel set. Then T (A Y ) = T (A) T (Y ) = T (A) Y, hence µ(a Y ) = µ(t (A) Y ). Consider the set Y = {x : e x (A) > e x (T (A))}. If µ(y ) > 0, then µ(a Y ) > µ(t (A) Y ). But Y is F -invariant, so this contradicts that µ(a Y ) = µ(t (A) Y ). So µ(y )=0. We may assume therefore that x X, e x (A) = e x (T (A)) > 0.

19 13 By a well-known lemma (see [Kechris 3], p.117, Lemma 4.50), there are disjoint E-invariant sets P,Q, and R, such that [A] [T (A)] = P Q R, and A P T (A) P, T (A) Q A Q, A R T (A) R. But P, Q are also F -invariant, µ(a P ) = µ(t (A) P ), µ(a Q) = µ(t (A) Q), so µ(p ) = µ(q) = 0, A T (A). Corollary [F ] is the closure of [E]. Proof. We only need to show that [E] is dense in [F ]. Recall that weak topology of Aut(X, µ) has as basis the sets of the form U S,P,ε = A P{T : µ(t (A) S(A)) < ε}, where S Aut(X, µ), P is a nite Borel partition of X, and ε > 0 is a real. Fix an arbitrary S [F ], a nite Borel partition P = {A i } and an ε > 0. By Theorem 2.3.5, there is a T i [E] such that T i (A i ) = S(A i ) for each A i P. Let T = (T i A i ). Since S is an automorphism, {S(A i )} is a Borel partition. Thus T Aut(X, µ). Since T (x) = T i (x)ex for some i, T [E]. Finally, µ(t (A i ) S(A i )) = µ(t i (A i ) S(A i )) = 0 < ε. Therefore, [E] = [F ]. For a countable Borel E, we can divide E into the periodic part and the aperiodic part, E = E aperiodic E periodic. The periodic part is smooth, E = F in this part. So we only need to deal with the case that E is aperiodic. We have the following isomorphism theorem for (X, F, µ):

20 14 Lemma Let E be a countable Borel equivalence relation on a standard Borel space X with a nonatomic invariant probability measure µ. If E is aperiodic, then (X, µ, F ) = (EI E [0, 1], πµ m, I). Proof. By the isomorphism theorem of standard Borel spaces with nonatomic probability measures (see [Kechris 2], Theorem 17.41), we can assume (X, µ) = ([0, 1], m). Dene f : X EI E [0, 1] x (π(x), π(x)([0, x])) and g : EI E [0, 1] X (e, r) inf{y e([0, y]) r}. Clearly, f and g are Borel, and since gf(x) = g(π(x), π(x)([0, x])) = inf{y π(x)([0, y]) π(x)([0, x])} = inf{y π(x)([y, x]) = 0}.

21 15 Let A = {x gf(x) x}, we have A = {x y < x π(x)([y, x]) = 0} = n N{x π(x)([x 2 n, x]) = 0} Also let A n = {x π(x)([x 2 n, x]) = 0}. It is easy to see that e(a n ) = e(a n X e ) = 0 for all e EI, therefore µ(a n ) = 0, hence µ( A n ) = µ(a) = 0. That is gf(x) = x almost everywhere. It is easy to check that fµ = πµ m and note that f(x e ) = {e} [0, 1], hence (X, µ, F ) = (EI E [0, 1], πµ m, I) The Descriptive Complexity of [E] and N(E) By we know that [E] is either closed or Σ 0 2-hard. In fact, we can show that in the case that [E] is not closed, [E] is not only Σ 0 2-hard but also Π 0 3-complete: Proposition If E is a countable Borel equivalence relation on a standard Borel space X with invariant probability measure µ, then [E] is closed or Π 0 3 -complete. Proof. Assume [E] is not closed. We claim that we can nd a Borel partition {Y i } i N of X, such that µ(y i ) > 0 and [E Y i ] are not closed (hence Σ 0 2-hard by 2.3.3) in Aut(Y i, µ Y i ). Granting this, since [E Y i ] is Σ 0 2-hard, we have Q 2 c [E Y i ] (where Q 2 = {x C : m n > m (x(n) = 0)} is a Σ 0 2-complete set see [Kechris 2], p.179).

22 16 Dene F : i N Aut(Y i, (µ Y i )/µ(y i )) Aut(X, µ), (f i ) i N f = f i, i.e, f Y i = f i. It is easy to see that F is continuous and (f i ) i N [E Y i ] i(f i [E Y i ]) i((( f i ) Y i [E Y i ]) f [E] where f i Aut(Y i, (µ Y i )/µ(y i ). Therefore i N [E Y i] c [E], hence P 3 = {x 2 N N : m (x m Q 2 )} = Q N 2 c [E Y i ] c [E]. i N Since P 3 is Π 0 3-complete (see [Kechris 2], p.179), [E] is Π 0 3-complete. It remains to show that the claim is true. Let X 0 = X. We will dene X i and Y i = X i \X i+1 inductively. Suppose we already have X i such that µ(x i ) > 0 and [E X i ] is not closed (which is clearly true in the base case that i = 0). We can therefore x a T [E X i ]\[E X i ], and the non-null Borel subset A = X i \{x X i T (x)ex}. Then we simply dene X i+1 and Y i+1 to be any Borel partition of X i such that both X i+1 A and Y i+1 A are not null. That is X i+1 Y i+1 = X i and 0 < µ(x i+1 A) < µ(a), 0 < µ(y i+1 A) < µ(a). By Theorem 2.3.5, there is an S [E X i ] such that S(X i+1 ) = T (X i+1 ). Since S 1 T (x)et (x) Ex

23 17 for almost every x A X i and S 1 T [E X i ], we have S 1 T X i+1 [E X i+1 ]\[E X i ], hence [E X i+1 ] is not closed. Similarly, [E Y i+1 ] is not closed. And since X i = X i+1 Y i+1 and X = X 0, {Y i } is a Borel partition of X. This completes the proof. We are ready to prove our main theorem. Theorem Let E be a countable Borel equivalence relation on a standard Borel space X with invariant probability measure µ. If E is µ-smooth, then [E], N(E) are closed. If E is not µ-smooth, then [E] and N(E) are Π 0 3 -complete. Proof. If E is µ-smooth, the result follows from and For non µ-smooth E, since the periodic part is smooth, we can assume E is aperiodic and, furthermore, by 2.3.7, we can assume that (X, µ, F ) = (EI E [0, 1], πµ m, I). Dene h s : (e, r) (e, r + s mod 1). We have h s [ I] = [F ] and d 0 (h s, h t ) = 1 δ s,t. So [F ] is not separable in the uniform topology, while [E] is, hence [E] [F ], [E] is not closed, and [E] is Π 0 3-complete from To determine the complexity of N[E], by and we can nd a Borel subset Y X, S [E], such that S 2 = 1 and {Y, S(Y )} is a partition of X (modulo null sets). For example, Y = g({(e, r) r < 1}), S = g (λ(e, r).(e, r + 1 mod 1)) f with 2 2

24 18 notations as in Lemma Dene h : Aut(Y, 2µ Y ) Aut(X, µ), T (x) if x Y h(t )(x) = x if x S(Y ). It is easy to check that h is continuous. If T [E Y ], then h(t ) [E] N(E). Conversely, if h(t ) [E], then there is a null subset N X, such that x, y X\N (xey h(t )xeh(t )y). We may also assume S(N) = N, otherwise, just replace N with N S(N). If x Y \N, then Sx S(Y )\N X\N. We have then T (x) = h(t )(x)eh(t )S(x) = S(x)E(x)) for all x Y, hence T [E Y ]. Therefore T [E Y ] i h(t ) N([E]), so [E Y ] c N[E]. Since E Y is not smooth almost everywhere, [E Y ] is Π 0 3-complete. Combining with 2.2.3, N(E) is also Π 0 3-complete N(E) as a Polishable group Let E be a countable Borel equivalence relation on X and µ an E-invariant ergodic Borel probability measure. Consider [E] as a standard Borel subgroup of Aut(X, µ) in the weak topology. Since [E] is a Polish group in the uniform topology and the identity map is a Borel isomorphism between the weak topology and the uniform topology, [E] is Polishable. N(E) is also a subgroup of Aut(X, µ). But N(E) is in general not separable, hence it is not Polish in the uniform topology. However, we can dene a new topology (see [HO], p.91) in the normalizer group N(E) by saying

25 19 that T n T in N(E) i T n converge to T weakly and T n ST 1 n converges to T ST 1 uniformly for all S [E]. It is easy to check that N(E) is a topological group with this topology. In fact N(E) is a Polish group in this topology (see [HO], Lemma 53). It is easy to check that the identity map is a Borel isomorphism between the weak topology and this topology, so N(E) is also Polishable.

26 20 CHAPTER 3 Descriptive Properties of Measure Preserving Actions and the Associated Unitary Representations 3.1. Introduction Measure Preserving Actions and Unitary Representations. Let be a countable group and X a standard Borel -space with an invariant (nonatomic) Borel probability measure µ. induces an equivalence relation E X on X, which is dened by xe X y γ (γ x = y). On the other hand, by a theorem of Feldman-Moore (see [KM], Theorem 1.3), every countable Borel equivalence relation on a standard Borel space X is induced by some Borel action of some countable group. Denote by Aut(X, µ) the group of µ-measure preserving automorphisms of X. For each T Aut(X, µ), we can dene a corresponding unitary operator U T U(L 2 (X)), U T (f) = f T 1. And by identifying T and U T, we can view Aut(X, µ) as a subgroup of U(L 2 (X)). The weak topology of Aut(X, µ) is the subspace topology of the weak topology dened on U(L 2 (X)). Aut(X, µ) is a closed subspace of U(L 2 (X) in the weak topology, hence it is Polish. Dene π X : Aut(X, µ) U(L 2 (X)), γ T, where T is the element in Aut(X, µ) such that T (x) = γ x for µ-almost every x X. Or equivalently T (f) =

27 21 U T (f) = f(γ 1 ) for every f L 2 (X). Clearly π X is a homomorphism of into U(L 2 (X)), i.e., π X is a unitary representation of on the Hilbert space L 2 (X). The unitary representation π X induces a natural Polish action on L 2 (X), i.e., γ f = π X (γ)(f). Denote by E L2 (X) the induced countable Borel equivalence relation on L 2 (X), i.e., fe L2 (X) g γ (g = γ f). If there is no danger of confusion, we will set E L2 (X) = E L2 (X). In this chapter, we will study relations between E X and (X) EL2. We will obtain some characterizations of the smoothness and compressibility of E L2 (X) and some reducibility results The action on MALG µ and Aut(X, µ). Let X be a standard Borel space with Borel probability measure µ. Denote by MEAS µ the σ-algebra of measurable sets and for A, B MEAS µ, let A = µ B µ(a B) = 0 and denote by [A] the equivalence class of A. Let MALG µ = {[A] : A MEAS µ }. Let [A] [B] = [A B] and δ([a], [B]) = µ(a B). Then (MALG µ, ) is an abelian Polish group with invariant metric δ. For every measure preserving automorphism T Aut(X, µ), [A] [T (A)] is a measure algebra automorphism of MALG µ. Conversely, every measure algebra automorphism of MALG µ is of the form [A] [T (A)] for some T Aut(X, µ). Therefore, we can canonically identify Aut(X, µ) and the group of measure algebra preserving automorphisms of MALG µ (see [Kechris 2], p.118). If µ is continuous, MALG µ is independent of µ and called the Lebesgue Measure Algebra. Consider a countable group and X a Borel -space with invariant Borel probability measure µ. There is a action on MALG µ dened by γ [A] = [γ(a)]. This

28 22 action is continuous. The map [A] χ A is a continuous -space embedding of MALG µ into L 2 (X, µ). Assume now that acts on Aut(X, µ) by left translation and Aut(X, µ) acts on L 2 (X) by T f = U T (f). Let f L 2 (X) and denote by Aut(X, µ) f the stabilizer of f, i.e., Aut(X, µ) f = {T Aut(X, µ) : T f = f}. If Aut(X, µ) f = {1}, then T T f is a continuous -space embedding of Aut(X, µ) into L 2 (X). Many such f exist, for example any f L 2 (X) that is an injection of X into C has a trivial Aut(X, µ)-stabilizer Borel Reducibility. Suppose we have Borel equivalence relations E, F on X, Y respectively. If there is a Borel map α : X Y such that x 1 Ex 2 α(x 1 )F α(x 2 ), for all x 1, x 2 X, we say E is Borel reducible to F. Put E B F if E is Borel reducible to F. When α is a reduction and also an injection, E is said to be Borel embedded in F, in symbols E B F. If, moreover, the image of α, α(x), is F -invariant, E is said to be Borel invariantly embedded in F or E i B F. B stands for Borel in the above symbols. We can replace Borel by another class of maps, say in a class A, and generalize the concept and notations to A- reducible, A, etc. For example, for the class of continuous maps, we will denote continuous reducible, embedded, invariantly embedded by c, c, i c, respectively, where c stands for continuous.

29 Smoothness of E L2 (X,µ) Smooth equivalence relations are the simplest relations in the reduction hierarchy. We rst review some basic properties of smooth relations and show there are simple characterizations of the smoothness of E Aut(X,µ) and E MALGµ. Then Theorem shows that the smoothness of E L2 (X) is equivalent to the smoothness of E MALGµ. It turns out the smoothness of these equivalence relations has interesting connections with the concept of rigid factors, which we will discuss in In 3.2.4, we will discuss some relations between mixing properties and smoothness. In the rest of this section, we will develop some techniques to deal with smoothness in some special kinds of -spaces by using the Peter-Weyl theorem, which is sometimes easier than using Theorem directly General Facts on Smooth Relations. Recall that an equivalence relation E on X is (Borel) smooth if E B (Y ), where (Y ) is the equivalence relation dened on some Polish space Y by y 1 (Y )y 2 y 1 = y 2. A selector is a map s : X X such that xey s(x) = s(y). A transversal for E is a set T X that meets each E-class at exactly one point. Having a Borel selector is equivalent to having a Borel transversal and implies smoothness. The converse is not true in general. But in the case that the E is generated by a discrete Borel group action, i.e., E = E for some countable group, the smoothness of E implies the existence of Borel selectors for E (see [KM], Proposition 6.4). Moreover, we have: Theorem Let X be a Borel -space, where is a countable group. Then the following are equivalent:

30 24 i) E is smooth; ii) E has a Borel selector; iii) X/ = X/E is standard Borel (in its quotient σ-algebra); iv) There is a Polish topology T on X compatible with its Borel structure such that E is closed in the product topology (X, T ) (X, T ); v) There is a countable sequence (A n ) of Borel E -invariant subsets of X separating E i.e. [ n(x A n y A n )] [xey]; vi) E 0 cannot be Borel embedded in E X, where E 0 is the equivalence relation on the Cantor space C = 2 N dened by se 0 t n N(s(n) = t(n)). Proposition Let E be a countable Borel equivalence relation on a uncountable Polish space X. i) If every equivalence class of E is G δ, then E is smooth. ii) If E admits a nonatomic ergodic measure, then E is not smooth. iii) If E is generically ergodic (i.e., every invariant Borel subset is either meager or comeager and every class is meager), then E is not smooth. Let X be a Polish metric space. In this case, we have a converse to Proposition (i). Proposition Let X be a Polish metric space, a countable group acting on X by homeomorphisms. The following are equivalent: i) E is smooth; ii) Every orbit is discrete; iii) x X, x is an isolated point in x; iv) The closure of each orbit is not perfect.

31 25 Proof. iii) ii): If γ i x γ x for some x X γ i, γ, then (γ 1 γ i ) x x. So these two conditions are clearly equivalent. ii) i ): Proposition (i). i) iv): Proof by contradiction. Suppose x is perfect for some x X. Note that x is dense in x which is equivalent to the condition that E ( x) is generically ergodic. By Proposition (iii), E ( x) is not smooth. Therefore E is not smooth. iv) iii): If x is a limit point in x, then γ (γ x is a limit point in x). Therefore x is perfect. Assume now acts on (X, µ) by measure preserving automorphisms. Recall that MALG µ and Aut(X, µ) can be embedded into L 2 (X) as -spaces. We have Corollary E MALGµ is smooth i for every Borel subset A X, there is an r > 0 such that µ(γ(a) A) / (0, r) for every γ. E Aut(X,µ) is smooth i π X ()is discrete in the weak topology The Characterization of Smoothness of E L2 (X). Clearly the smoothness of E L2 (X) implies the smoothness of E MALGµ. The converse is also true. Theorem E MALGµ is smooth i E L2 (X) is smooth. Before we prove the theorem, it would be rst convenient to prove the following lemma: Lemma Let X be a standard Borel space with Borel probability measure µ and g, f 0, f 1, f 2 L 2 (X, µ). If f i µ = f j µ for all i, j, then the following are equivalent:

32 26 i) f 1 i (A) g 1 (A) for every analytic A C; ii) f 1 i (A) g 1 (A) for every Borel A C; iii) f 1 i (A) g 1 (A) for every basic open set A C; iv) f i g. Proof. Obviously, i) ii) iii). Put f ω = g. (ii) i)) Assume ii) is true. We may assume f i is Borel for all i ω. Let B i X\f 1 i (A) be a Borel set such that B i = X\f 1 (A) (modulo null). So f i (B i ) is analytic and f i (B i ) A =. Let B = i ω f i(b i ). By the separation theorem (see [Kechris 2] 28.B), there is a Borel set A, such that A A B. We have i f 1 i (A) f 1 i (A ) f 1 i ( B) f 1 i ( B) = X\B i. Since X\B i = f 1 (A) (modulo null), we have f 1 i (A) = f 1 i (A ) for all i ω. Therefore f 1 i (A) = f 1 i (A ) g 1 (A ) = g 1 (A). (iii) ii)) It is easy to check that the convergence property is preserved under complement and nite union, that is [f 1 i (A) g 1 (A)] [f 1 ( A) g 1 ( A)] i and [ k<n (f 1 i (A k ) g 1 (A k ))] [f 1 ( A k ) f 1 ( A k )]. k<n k<n So we only need to check that the convergence property is preserved under countable union. Suppose now f 1 i (A k ) g 1 (A k ) for all k N. We may assume (A k ) is increasing and put A = k N A k. For any ε > 0, we can nd an n large enough so

33 27 that (f 0 µ)(a\a n ) + (gµ)(a\a n ) < ε. Combine with the condition f i µ = f j µ for all i, j < ω, so we actually have (f i µ)(a\a n ) < ε for all i ω. We can also nd m large enough so that µ(f 1 i (A n ) g 1 (A n )) < ε for all i > m. Therefore, for every i > m µ(f 1 i (A) g 1 (A)) µ(f 1 i (A)\f 1 i (A n )) + µ(f 1 i (A n ) g 1 (A n )) +µ(g 1 (A)\g 1 (A n )) µ(f 1 i (A\A n )) + µ(f 1 i (A n ) g 1 (A n )) +µ(g 1 (A\A n )) 3ε. So f 1 i (A) g 1 (A). (ii) iv)) By simple function approximation, we can nd Borel sets A k C so that and f 0 a k χ f 1 g a k χ f 1 0 (A k) 0 (A k) < ε < ε

34 28 for some a k A k. We can use nitely many A k and assume them to be bounded. Then, because of the condition f i µ = f j µ for all i, j < ω, we have f 1 i (A k ) f i (x) a k 2 dµ(x) = = = s a k 2 d(f i µ)(s) s A k s a k 2 d(f 0 µ)(s) s A k f 0 (x) a k 2 dµ(x). f 1 0 (A k) So we actually have f i a k χ f 1 i (A k ) < ε for all i ω. Since f 1 i (A k ) g 1 (A k ), we can nd m large enough so that ak 2 µ(f 1 i (A k ) g 1 (A k )) < ε 2 for all i > m. Therefore, for every i > m f i g f i a k χ f 1 i (A k ) + a k χ g 1 (A k ) a k χ f 1 < 2ε + ak χ g 1 (A k )\f 1 + g a k χ g 1 (A k ) i (a k ) i (A k ) + ak χ f 1 i (A k )\g 1 (a k ) = 2ε + < 3ε. ak 2 µ(f 1 i (A k ) g 1 (A k )) We have f i g.

35 29 ((iv) (iii)) Since f i g, we have f i µ gµ, hence f i µ = gµ for all i. Let A = B a,r = {s C : s a < r} be an open ball with arbitrary radius r and center a. If f 1 0 (A) is null, then g 1 (A) is null. So assume µ(f 1 0 (A)) > 0. Toward a contradiction, assume f 1 i for all i. Since µ(f 1 i (A)) = µ(g 1 (A)), we have (A) g 1 (A). Pick t > 0 so that µ(f 1 (A) g 1 (A)) > t i µ(f 1 i (A)\g 1 (A)) = µ(g 1 (A)\f 1 i (A)) > t 2. Let A s = B a,r s and x an s such that µ(a\a s ) < t. Then we have 4 µ(f 1 i (A s )\g 1 (A)) µ(f 1 i (A)\g 1 (A)) µ(a\a s ) > t 4. Finally we have f i g i 2 f 1 i (A ε)\g 1 (A) f i g 2 dµ > s2 t 4, contradicting the assumption f i g. Remark. So if f i µ = f j µ for all i, j, then (f i ) converge i f 1 i (A) converge for all basic open sets A, as the limit g in the above lemma can be easily constructed by simple function approximation. We are ready to prove the theorem: Proof. We need to show that if E L2 (X) is not smooth, then E MALGµ is not smooth. Assume E L2 (X) is not smooth. By Proposition 3.2.3, there is an f L 2 (X)

36 30 such that f is a limit point of ( f)\{f}. We can nd a countable sequence {γ i } \ f such that γ i f f. By Lemma 3.2.6, we may assume f(x) C [0, 1]. Otherwise, replace f by α f where α is a Borel isomorphism of C to the Cantor set C. In order to prove this by contradiction, assume E MALGµ is smooth. So for every Borel subset A X, there is an r A > 0 such that µ(γ(a) A) / (0, r A ) for every γ. And since µ(γ i (f 1 (B)) B) 0 for every Borel B C, there is a number m B N such that µ(γ mb (f 1 (B)) B) r f 1 (B) and µ(γ i (f 1 (B)) B) = 0 for all i > m B. Let S = {s C : µ(f 1 (s)) > 0}. Suppose S =. Let P n = {S s,n : s C}, where S s,n = {t C : t n = s n}. Fix an s C, n N f 1 (S s,n ) = f 1 (s), lim n µ(f 1 (S s,n )) = 0. Dene s i, n i inductively. To simplify the notations, let A i = S si,n i, r i = r Ai, m i = m Ssi,n i. m i. Find s 0, n 0 such that 0 < µ(f 1 (S s0,n 0 )) < 1. Let A 0 = f 1 (S s0,n 0 ). Suppose we have s i, n i. Find s i+1, n i+1 > n i such that A i+1 = f 1 (S si+1 ), 0 < µ(a i+1 ) < r i 6 and m i+1 > We can always nd such s i+1, n i+1 because n(#{s P n : µ(f 1 (S)) > r} = 0) for every r > 0. And #{S P n : µ(f 1 (S)) 0} > 0. Let A n = A 0 A 1... A n 1, and [A] = lim n [A n] in MALG µ.

37 and 31 Clearly γ mi A A. Let T n = A\A n. Since r i < 2(A i ), we have µ(a i+1 ) < µ(a i) 3 µ(t i ) µ(a j ) < 3 2 µ(a i). j=i Therefore µ((γ mi A) A) = µ((γ mi (A i )) (γ mi (T i+1 )) A i T i+1 ) µ(γ mi (A i )) A i ) µ(γ mi (T i+1 )) µ(t i+1 ) r i 2µ(T i+1 ) > r i 3µ(A i+1 ) > 0, contradicting the assumption that E MEASµ is smooth. Suppose S, then S is countable. Let f = f 1 + f 2, where f 1 (X) S and f 2 (X) C\S. Note that γ i f 1 f 1 and γ i f 2 f 2 because S and C\S are Borel. If i(γ i f 2 f 2 ), then replacing f by f 2, we are back to the case that S =. So we may assume f = f 1. If S is nite, let m = max s S {m {s} }. Thus γ i f = f for all i > m, contradicting the assumption γ i f f for all i. If S is countably innite, dene s i S inductively. To simplify the notation, let A i = f 1 (s i ), r i = r Ai, m i = m {si }.

38 32 Pick s 0 such that 0 < µ(a 0 ) = µ(f 1 (s 0 )) < 1. Suppose we have s i. Pick s i+1 such that and m i+1 > m i. 0 < µ(a i+1 ) < r i 6 Let A = i=0 A i, T i = j=i A i. We have γ mi A A, and since r i < 2(A i ), we have µ(a i+1 ) < µ(a i) 3 and µ(t i ) < 3 2 µ(a i). Therefore, µ((γ mi A) A) = µ((γ mi (A i )) (γ mi (T i+1 )) A i T i+1 ) µ(γ mi (A i )) A i ) µ(γ mi (T i+1 )) µ(t i+1 ) r i 2µ(T i+1 ) > r i 3µ(A i+1 ) > 0, contradicting the assumption that E MEASµ is smooth. So E MEASµ is not smooth, when E L2 (X) is not smooth Rigid factors and smoothness of E L2 (X). The smoothness of E Aut(X,µ) does not imply the smoothness of E L2 (X,µ). We will study some connections between rigid factors, mixing properties, and the smoothness of E L2 (X,µ). Denition A action on (X, µ) is faithful i π X : Aut(X, µ) is injective. The following notion of rigid factor is from [SW].

39 33 Denition Let X be a -space with invariant probability measure µ. The rigid factor is the set R(, X, µ) = {f L 1 (X, µ, S 1 ) : lim inf γ γ f f L 1 = 0}. The action on X is said to be rigid i the rigid factor is L 1 (X, µ, S 1 ). The action on X is said to have no rigid factor if R(, X, µ) contains only constant functions. There are many equivalent denitions of mildly mixing. For an action of countable group, no rigid factor is equivalent to mildly mixing. Assume now acts on (X, µ) faithfully and µ is a -invariant Borel probability measure. We have: Proposition E Aut(X,µ) is smooth i the action on X is not rigid. Proof. Let f be a Borel isomorphism of X to [0, 1]. Put d(t, S) = T f S f for all S, T Aut(X, µ) so that d is a (left -invariant) metric of the weak topology. Suppose E Aut(X,µ) is not smooth, then lim inf γ d(γ, 1) = 0. Since L 2 (X, µ) is a Polish Aut(X, µ)-space, the map γ γ f is continuous for the weak topology restrict on for every f L 2 (X, µ). We have lim inf γ γ f f = 0 for all f L 2 (X, µ). Assume now the action on X is rigid. We have lim inf γ γ f f = 0. Since the action is faithful, f is trivial and π X () is discrete. In fact, the smoothness of E L2 (X) is strictly between X having no rigid factors and being nonrigid. Let

40 34 (NM) the action on X is mildly mixing. (NRF) the action on X has no rigid factors. (LS) E L2 (X,µ) (MS) E MALGµ (AS) E Aut(X,µ) is smooth. is smooth. is smooth. (NR) the action on X is not rigid. Then we have the following picture: (MM) (NRF) (LS) (MS) (AS) (NR) In 3.2.4, we will show (MM) (LS) (Proposition (iii)) and give an example to show (AS) (LS). Example will show (LS) (NM) Mixing Properties and Smoothness of E L2 (X). Consider the countable Borel -space X with invariant probability measure µ and recall the induced

41 -action on L 2 (X) by unitary operators. Denote by L 2 0(X) the set 35 {f L 2 (X) : f, 1 = 0}. Then L 2 0(X) is an invariant subspace of L 2 (X). Therefore, we have a action on L 2 0(X) and π X 0 := π X L 2 (X), the subrepresentation of π X on L 2 0(X). The mixing properties, including (strong) mixing, mild mixing, weak mixing, ergodicity can be read from the action on L 2 0(X). Let us review these properties from the strongest one to the weakest one: 1) (Strongly) Mixing: An action is strongly mixing if and only if for any Borel A, B X, lim µ(γ A B) = µ(a)µ(b). γ Considering the action on L 2 0(X), this is equivalent to f, g L 2 0(X)( lim γ γ f, g = 0). In the language of unitary representation theory, it is equivalent to saying that π X 0 is a c 0 -representation. 2) Mildly mixing: The action on X is mildly mixing i lim inf γ µ((γ A) A) > 0 for any Borel subset A X that is neither null or conull.

42 36 It is equivalent to that for all f L 2 0(X)\{0}, lim sup γ f, f < f 2. γ 3) Weakly mixing: This is equivalent to saying that the action on L 2 0(X) has no nite dimensional invariant subspace. Or in the language of unitary representation theory, π X 0 has no nite dimensional subrepresentation. 4) Ergodic: The action is ergodic i every -invariant Borel subset of X is either µ-null or µ-conull. This is equivalent to saying that f L 2 0(X)\{0}( f ), where f = {γ : γ f = f}, the stabilizer of f. Or in the language of unitary representations, π X 0 does not contain the trivial one-dimensional representation. Proposition i) E L2 (X) is smooth i E L2 0 (X) is smooth. ii) If E L2 (X) is smooth, then π X () Aut(X, µ) is discrete in the weak topology. iii) If E X H is ergodic for every innite subgroup H, then the action on X is mildly mixing i E L2 0 (X) is smooth. In particular mildly mixing implies the smoothness of E L2 0 (X). iv) Assume E H is ergodic for every subgroup H such that [π X () : π X (H)] <. If E L2 0 (X) is smooth, then the action on X is weakly mixing. Proof. i) It is easy to check that E L2 (X) = E L2 0 (X) (C).

43 37 Therefore E L2 (X) is smooth i E L2 0 (X) is smooth. ii) This follows directly from Corollary and E Aut(X,µ) i c E L2 (X,µ). iii) Suppose the action on X is mildly mixing. Let A MALG µ If µ(a) = 0 or µ(a) = 1, γ [µ((γ A) A) = 0]. Assume A is neither null or conull. Since is mildly mixing, R = lim inf γ µ((γ A) A) > 0. So there S = {γ : 0 < µ((γ A) A) < R} is a nite set. Let r = min γ S {µ((γ A) A))}. We have µ((γ A) A) / (0, min(r, R)) for all γ. Therefore, by Theorem and Corollary 3.2.4, E L2 (X) is smooth. On the other hand, assume E L2 (X) is smooth. Suppose is not mildly mixing on X. Then there is an A MALG µ such that lim inf γ µ((γ A) A) = 0 and 0 < µ(a) < 1. Therefore, S r = {γ : µ((γ A) A) < r} is innite for every r > 0. By Theorem and Corollary 3.2.4, there is an R > 0 such that µ(γ(a) A) / (0, R) for all γ. So S R = {γ : µ((γ A) A) < r} = {γ : µ((γ A) A) = 0}. Let H be the subgroup of generated by S R. E X H is not ergodic because A is H-invariant. This contradicts the condition that E X H is ergodic, when H is innite. So the action on X is mildly mixing. iv) Proof by contradiction. Suppose E L2 (X) is smooth and there is a nite dimensional -invariant subspace V L 2 0(X). Pick an arbitrary f V. Since E L2 (X) is smooth, f V is discrete.

44 38 Therefore f is nite. Thus π X ()/π X ( f ) <. But f xes f L 2 0(X), contradicting the hypothesis that E f is ergodic. Example Let X, Y be faithful Borel -spaces with invariant probability measures µ, ν respectively. Assume the action on X is mildly mixing, and E L2 (Y,ν) not smooth. Consider acting on X Y by the diagonal action. π X Y () is discrete is in Aut(X Y, µ ν), hence E Aut(X Y,µ ν) is smooth. In fact, pick any A X that is neither null or conull. There exists an r > 0 such that (µ ν)((γ (A Y ) (A Y )) < r for almost every γ. But E L2 (X Y ) is clearly nonsmooth because we can nd a B Y and a sequence of γ i so that γ i (X B) X B and γ i (X B) X B for all i. For example, let = S < and G = 2 2 N. Consider the action on G dened by γ ((a g ), (b i )) = ((a γ 1 g), (b γ 1 (i)). Then the action on 2 is mixing, so E Aut(G,µ) is smooth. But E L2 (G) is not smooth because E L2 (2 N ) is not smooth The Peter-Weyl Theorem. We are going to develop several other techniques to determine the smoothness and nonsmoothness of E L2 (X). In some situation, it is easier to use them than directly check the conditions in Theorem Most of these techniques involve the Peter-Weyl theorem.

45 39 Recall the Peter-Weyl theorem from unitary representation theory (see [ Folland, Kechris 1]). Theorem (Peter-Weyl) Let G be a compact Polish group. Then (i) Every irreducible unitary representation of G is nite dimensional; (ii) Ĝ is countable; (iii) Every unitary representation of G is a direct sum of irreducible unitary representations. Consider a compact Polish group G with the (normalized) Haar measure µ. For each irreducible unitary representation π of G, denote by ˆπ the isomorphism class of π and by H π the Hilbert space of it. Also denote by Ĝ the dual of G, which is the countable set {ˆπ : π is an irreducible unitary representation of G}. Denote by ρ G : G U(L 2 (G)) the right regular representation of G, which is the unitary representation dened by (ρ G (g))(f(h)) = f(hg) for all g, h G and f L 2 (G). Fix a representative π for each ˆπ Ĝ and an orthonormal basis {eπ i } 1 i dπ, where d π = dim(h π ). Let π ij (g) = π(g)e π j, e π i be the matrix coecients of π in this basis. π ij L 2 (G) and denote by E π the linear span of {π ij }. Clearly this space is independent of the choice of π, thus we can write Eˆπ = E π. Theorem (Peter-Weyl) Let G be a compact Polish group. Then

46 40 (i) L 2 (G) = ˆπ Ĝ Eˆπ. (ii) { d π π ij } 1 i,j dπ is an orthonormal basis for Eˆπ, so dim(eˆπ ) = d 2 π. (iii) For i = 1,..., d π, the subspace Eˆπ,i of Eˆπ spanned by the ith row of the matrix ( d π π ij ) is invariant under the right regular representation, and the subrepresentation of ρ G determined by Eˆπ,i is isomorphic to π. Dene the character χ π by χ π (g) = trace(π(g)). Trace is also independent of the choice of the basis and the representative π of each isomorphic class, so we put χˆπ = χ π. {χˆπ : ˆπ Ĝ} is an orthonormal set in L2 (G) (see [Folland], 5.23). Suppose is a countable group acting by (topological group) automorphisms on G. Clearly preserves µ. There is a natural action on Ĝ. For a γ, dene (γ π)(g) = π(γ 1 g). Since π is irreducible, γ π is also irreducible. And again, γ π is independent of the choice of π in each isomorphic class. So we can dene γ ˆπ = γ π. Also note that (γ χˆπ )(g) = χˆπ (γ 1 g) = trace(π(γ 1 g)) = trace((γ π)(g)) = χ γ π (g). So γ χˆπ = χ γ ˆπ, nally γ Eˆπ = E γ ˆπ Some Characterizations of Smoothness. Corollary ) Assume X = G is a compact Polish group and acts on G by automorphisms. If E L2 (G) is smooth, then ˆπ π is nite for every irreducible unitary representation π of X. Or equivalently, ˆπ / π is nite for every irreducible unitary representation π of X.

47 41 2) Assume acts on a compact Polish space X by isometries and X has a Borel probability measure that is invariant under any isometry. Then E L2 (X) is smooth i E is uniformly periodic µ-a.e., i.e., N < µ x X( [x] E < N). 3) Assume acts on X by topological group automorphisms where X = G is a connected semisimple Lie group with an invariant Borel probability measure µ. Then E L2 (X) is smooth i E X is uniformly periodic µ-a.e. Proof. 1) Recall that ˆπ is the stabilizer of ˆπ, i.e., ˆπ = {γ : γ π = ˆπ}. Hence ˆπ π = {γ π γ ˆπ } = {γ π γ π = ˆπ}. By the Peter-Weyl Theorem, the subrepresentation of ρ G determined by Eˆπ,i is isomorphic to π. If ˆπ π is innite, then ˆπ π ij is innite for some j. But Eˆπ,i is nite dimensional, so ˆπ π ij, which is contained in the unit sphere of Eˆπ,i, is compact. Therefore ˆπ π ij has at least one limit point and therefore is not closed. By Proposition (iii), E L2 (G) is not smooth. 2) ( ) Since γ N! f = f, E L2 (G) is smooth. ( ) Since Iso(X) is compact, by the Peter-Weyl theorem, π X is the direct product of irreducible nite dimensional unitary representations, say L 2 (X) = i N V i, and V i are nite dimensional π X invariant subspaces.

48 42 For every f L 2 (X), we can write f = f i where f i V i. Suppose f is innite. If f i is innite for some i, then E L2 (X) is not smooth. Assume now f i is nite for every i. Since i( f i ) is nite and f is innite, i<m f i is innite for every M <. Therefore, we can nd a sequence g M i<m f i f\f, and g M f, hence E L2 (X) is not smooth. So if E L2 (X) is smooth, then every orbit of f L 2 (X) is nite. Therefore, we must have a nite upper bound of -orbit on X µ-a.e. 3) Since Aut(G) is compact, this is from the proof of 2) Stabilizers. Consider the action on L 2 (X). We can also describe the stabilizer of f L 2 (X) in terms of the action on X and its full group. Denition Let F be a (not necessarily countable) Borel equivalence relation dened on X and µ a (not necessarily F -invariant) Borel probability measure on X. Denote by [F ] = {T Aut(X, µ) µ x (T (x)ex)} the full group of F. This is a straightforward generalization of the usual concept of full group, which is usually dened in the case that µ is E-invariant and in the context that E is countable. We have the following simple proposition: Proposition Let F be a Borel equivalence relation on a standard Borel space X with a Borel probability measure µ. If F is smooth, then [F ] is closed in the weak topology of Aut(X, µ). Proof. F is smooth, hence F B ([0, 1]). Let f : X [0, 1] be a Borel function such that xf y i f(x) = f(y). Then the assignment T f T is a continuous map

49 43 from Aut(X, µ) to L 2 (X). Note that T [F ] i f T = f (modulo null sets), hence [F ] is closed. Recall the uniform ergodic decomposition for invariant measures of E. Denote by P (X) the set of probability measures on X, by I E P (X) the set of E-invariant Borel probability measures on X, and by EI E P (X) the set of E-invariant ergodic Borel probability measures on X. We have (see [KM], Theorem 3.3) Theorem (Farrell, Varadarajan) Let E be a countable Borel equivalence relation on a standard Borel space X. Assume I E 0. Then there is a unique (up to null sets) Borel surjection π : X EI E such that (1) π(x) = π(y) if xey; (2) If X e = {x : π(x) = e}, for e EI E, then e(x e ) = 1; (3) For any µ I E, µ = π(x)dµ(x). So we can write EI E = {e x }, where e x = π(x). Until the end of this subsection, we will use the above notations: π, X e to denote the unique ergodic decomposition of (X, E) and F (F Xif E = EX ) to denote the Borel equivalence relation on X, which is dened by xf y i π(x) = π(y). Since F is smooth, [F ] is closed by Proposition and clearly [E] [F ]. Furthermore, we can show that [F ] is the closure of [E]. Theorem Let E be a countable Borel equivalence relation on a standard Borel space X with invariant probability measure µ. Given S [F ] and a Borel set A, then there is a T [E], such that T (A) = S(A). Proof. Let Y be an F -invariant Borel set. Then T (A Y ) = T (A) T (Y ) = T (A) Y, hence µ(a Y ) = µ(t (A) Y ).