THE UNCLASSIFIABILITY OF ERGODIC MEASURE PRESERVING TRANSFORMATIONS

Transcription

1 THE UNCLASSIFIABILITY OF ERGODIC MEASURE PRESERVING TRANSFORMATIONS

2 WHERE ARE WE?! Trees F Odometer Based F s F Di 1 (T 2, ) R Circular Systems

3 ! Trees F Odometer Based F s F We build a reduction Di 1 (T 2, ) R Circular Systems F : TREES! odometer based transformations such that F F is a reduction from ill-founded trees to circular systems T such that T = T 1.

4 PLAN Today we just talk about abstract ergodic MPT s. Tomorrow we ll finish by adding the conditions to make the reduction to circular systems.

5 WE BEGIN BY REPORTING ON THE RESULTS OF THIS PAPER

6 WE BEGIN BY REPORTING ON THE RESULTS OF THIS PAPER And then tomorrow we ll describe what more we have to do to take care of F.

7 MARGINALIA: POSITIVE ENTROPY The transformations T we build are zeroentropy. if we fix a Bernoulli shift B, then T is the Pinsker algebra of T B. if follows that T = T 1 if and only if T B = T 1 B.

8 MARGINALIA: CONTINUOS CONJUGACIES For the transformations T we build 1. T is not congruent to T 1 or 2. There is a G set of measure one L = L T on which T is continuous and a continuous isomorphism between T L and T 1 L 1.

9 MARGINALIA There is a continuous map S from the ergodic transformations on [0, 1] to the ergodic measure preserving R-flows on [0, 1] [0, 1] such that for all ergodic transformations S, T we have S = T i S(S) = R S(T ) where = R is isomorphism of R-flows. S is the ``suspension map

10 SUSPENSIONS Given an invertible measure preserving transformation T :[0, 1]! [0, 1] we consider the equivalence relation on [0, 1] R generated by setting (x, s) (T n (x),s n) foreachn. We can take [0, 1] [0, 1) as fundamental domain for ([0, 1] R)/ and the transformation S(T )(t, [(x, s)] )= [(x, s + t)] defines a measure preserving flow on [0, 1] [0, 1) with the product Lebesgue measure.

11 BACK TO THE PROOF THAT FOR ABSTRACT ERGODIC MPT S ISOMORPHISM IS NOT BOREL

12 RELATIVELY INDEPENDENT JOININGS Suppose that X and Y are ergodic and we have factor maps 1 : X! Z and 2 : Y! Z, where Z = (Z, D,,W). Then we get disintegrations hµ z : z 2 Zi of µ and h z : z 2 Zi of over.

13 RELATIVELY INDEPENDENT JOININGS The relatively independent joining of X and Y over Z is the unique measure on X Y such that for all A 2 B and B 2 C, Z (A B) = µ z (A z ) z (B z )d (z).

14 THE STRATEGY We are given a tree T = h n : n 2 Ni with each n 2 N <N and shorter sequences first. Fromh m : m apple ni we will build hw m : m apple ni.

15 CONTINUITY Let h n : n 2 Ni be an enumerate of N <N with initial segments coming before long sequences. By using our strategy we will have that if T \ { m : m apple n} = T 0 \ { m : m apple n}, then for all m apple n W m (T )=W m (T 0 ).

16 Paralleling the construction of the words W n,we construct groups of involutions G n s,n s, with the G n s having a direct limit G s as n tends to infinity. There are also homomorphisms n s+1 : G n s+1! G n s which converge to a homomorphism s+1 : G s+1! G s. The system (G s, s )hasaninverselimitdenoted by G, and it is non-trivial if and only if T has an infinite branch.

17 As s goes to infinity, the elements of the G s s give more and more precise information about an isomorphism of T with T 1 or a non-trivial isomorphism of T with itself. It will be relatively easy to see that an element of the inverse limit of the G s s are isomorphisms between T and T 1. The di cult part is showing that there are no other isomorphisms.

18 This is done by giving a complete analysis of the joinings between T and T 1. We find a canonical sequence of factors K s of K(T )correspondingto some equivalence relations Q s. These factors have K as their inverse limit. In passing from W s to W s+1 we will have randomly concatenated words from W s. This will allow us to show that any joining of T with T 1 is either an independent joining over a isomorphism of K s with K 1 arising from s an element of G s or comes from some element of the inverse limit of the G s s.

19 FIX A TREE T

20 Define maps M : N! N by setting M(s) =n i n is the least number such that n 2 T and lh( n )=s. Dually we define a function s : N! N by setting s(n)tobethelengthofthelongest sequence m such that m 2 T and m apple n.

21 A USEFUL POT (THEOREM) Let be a compact group and g 2 be such that {g n : n 2 Z} is dense in. Let T g be the Haar measure preserving transformation given by translating by g. If is an ergodic self-joining of T g, then is supported on the graph of the translation by some h 2,i.e.isoftheform (A B) =µ(a \ h 1 B) where µ is the Haar measure. Moreover each h 2 determines an invertible self joining.

22 GROUPS OF INVOLUTIONS If G = P i2i (Z 2) i and we have a distinguished basis B = {r i : i 2 I} for G then there is a well defined notion of parity

23 GROUPS OF INVOLUTIONS If G = P i2i (Z 2) i and we have a distinguished basis B = {r i : i 2 I} for G then there is a well defined notion of parity Parity is preserved under homomorphisms in the sense that if G has distinguished basis B and H has distinguished basis C and : H! G is a homomorphism sending C to B then sends even elements to even elements and odd elements to odd elements.

24 GROUPS OF INVOLUTIONS Hence if we are given an inverse limit system of groups of involutions {G s : s 2 S} over a linearly ordered set (S, < S )withmaps{ t,s : s< S t} where each G s is a group of involutions with a distinguished set of generators and G =limg s is the inverse limit, then the elements of G have a well defined parity.

25 To each level s of T we can associate a group of involutions G s by taking the sum of copies of Z 2 indexed by the nodes on T at level s. We will view the nodes of T at level s as the distinguished generators of G s = P 2T,lh( )=s (Z 2). If s<tare levels of T then we get a canonical homomorphism t,s : G t! G s that sends a generator of G t at level t to the unique generator initial segment of. of G s that is an

26 CONNECTING THE GROUPS TO ILL-FOUNDED TREES Let G 1 be the inverse limit of hg s, t,s : s<t< 1i. Let s : G 1! G s be the projection map to G s (T ).

27 CONNECTING THE GROUPS TO ILL-FOUNDED TREES Let G 1 be the inverse limit of hg s, t,s : s<t< 1i. Let s : G 1! G s be the projection map to G s (T ). G 1 has a non-identity element of odd parity i T is ill-founded. G 1 has a non-identity element of even parity i T has at least two infinite branches.

28 SPECIFICATIONS We will describe our construction step-by-step to satisfy some specifications. We add these specifications one at a time.

29 THE FIRST THREE Specifications E1-E3 from the paper simply say that hw n i n is a construction sequence, s n is a power of 2, s n s n+1 and each n-word occurs equally often in each n +1-word.

30 PRODUCT EQUIVALENCE RELATIONS Given an equivalence relation Q on X we can define the product equivalence relation n Q on n X by setting i for all i, x i x 0 i. x 0 x 1...x n 1 x 0 0x x 0 n 1

31 EQUIVALENCE RELATIONS We will have equivalence relations Q n s M(s). They will satisfy: for n Q4 Q 0 0 is the trivial equivalence relation on with just one class and Suppose that n = M(s). Then any two words in the same Q n s equivalence class agree on an initial segment of length at least (1 n )q n. Q5. For n M(s)+1,Q n s is the product equivalence relation of Q M(s) s. Hence we can view W n /Q n s as sequences of elements of W M(s) /Q M(s) s and similarly for rev(w n )/Q n s. It follows that Q n 0 is the equivalence relation on W n which has one equivalence class.

32 EQUIVALENCE RELATIONS It follows that each Q n s class has d-diameter at most n. We will have axioms that say that the Q n s - classes are far apart in d.

33 EQUIVALENCE RELATIONS Q6. Q n s+1 refines Q n s and each Q n s class contains 2 k(n) many Q n s+1 classes for some number k(n)! 1 as n!1.

34 GROUP DEFINITIONS Let G n 0 be the trivial group and For s>0, G n s = X {(Z 2 ) : 2 { m : m apple n}, lh( ) =s}. Then G n+1 factors naturally as G s n+1 s H where H is either trivial or Z 2. = G n s G s is the the direct limit of the G n s.

35 SOME NUMERICS We will have decreasing summable sequences n and s such that 2 n s 2 nm n 1 n k n s 2!1as n!1. n 1 etc. etc.

36 SUBORDINATE GROUP ACTIONS Suppose: 1. Q and R are equivalence relations on a set X with R refining Q. 2. G and H are groups with G acting on X/Q and H acting on X/R 3. : H! G is a homomorphism. Then we will say that the H action is subordinate to the G action if for all x 2 X, whenever [x] R [x] Q we have h [x] R (h) [x] Q.

37 DIAGONAL AND SKEW DIAGONAL ACTIONS If G acts on X then there is a canonical diagonal action of G on n X defined by g(x 0 x 1...x n 1 )= gx 0 gx 1...gx n 1. If G is a group of involutions with a distinguished collection of free generators, then we define the skew diagonal action on n X by setting g(x 0 x 1 x 2...x n 1 )=gx n 1 gx n 2...gx 0 for g acanonicalgenerator. in the skew diagonal action, elements of G with odd parity reverse the orders of the x i s, while elements of even parity preserve the order.

38 GROUP ACTION SPECIFICATIONS A7. G n s acts freely on W n /Q n s [ rev(w n /Q n s )and the G n s action is subordinate to the G n s 1 action via the natural homomorphism s,s 1 from G n s to G n s 1. A8. The canonical generators of Gs M(s) send elements of W M(s) /Q M(s) s to elements of rev(w M(s) )/Q M(s) and vice versa by reversing the words. s )

39 GROUP ACTION SPECIFICATIONS A9. If M(s) apple m, and we view G m+1 s = G m s H then the action of G m s on W m /Q m s [rev(w m /Q m s ) is extended to an action on W m+1 /Q m+1 s [ rev(w m+1 /Q m+1 s )bytheskewdiagonalaction. If H is non-trivial then its canonical generator maps W m+1 /Q m+1 s to rev(w m+1 /Q m+1 s ).

40 GROUP ACTION SPECIFICATIONS A9. If M(s) apple m, and we view G m+1 s = G m s H then the action of G m s on W m /Q m s [rev(w m /Q m s ) is extended to an action on W m+1 /Q m+1 s [ rev(w m+1 /Q m+1 s )bytheskewdiagonalaction. If H is non-trivial then its canonical generator maps W m+1 /Q m+1 s to rev(w m+1 /Q m+1 s ). We will refer to the words in some W n as having even parity and the words in some rev(w n )as having odd parity.

41 THE CANONICAL FACTORS The odometer factor K 0 factors K s determined by a sequence of sub- -algebras of B(K)

42 DEFINING K s For each s, Q M(s) s is an equivalence relation on W M(s). Enumerate the classes {c j : j<q M(s) s }. A typical x 2 K gives a well-defined bi-infinite sequence of such classes. This gives a shift invariant map s from K to {0, 1,...,Q M(s) s 1} Z.

43 DEFINING K s For each s, Q M(s) s is an equivalence relation on W M(s). Enumerate the classes {c j : j<q M(s) s }. A typical x 2 K gives a well-defined bi-infinite sequence of such classes. This gives a shift invariant map s from K to {0, 1,...,Q M(s) s 1} Z. Let K s be the space {0, 1,...,Q M(s) s 1} Z O T. The map s determines a continuous shift invariant map s : K! K s defined by letting the first coordinate of s (x) be the Z-sequence h s (x)(k) :k 2 Zi and the second coordinate be 0 (x)

44 THE SUB SIGMA ALGEBRAS Let H s be the sub- algebras of B(K) determined by s s(k s ). If q = q M(s+1) /q M(s) then equivalence relation Q M(s+1) refines s+1 q Q M(s) s, hence H s+1 refines H s. The factor maps can be computed explicitly and are continuous.

45 AN IMPORTANT PROPOSITION -algebra in- B(K) isthesmallestinvariant cluding S s2n H s. For a set of L of measure one, if x 6= y belonging to L, there is an open set S in the topology generated by the open sets in the K s s such that x 2 S and y/2 S. For all s 1, H s is a strict subalgebra of H s+1. Moreover, if h x : x 2 K s i is the disintegration of s+1 over s, then for s -a.e. x, x is non-atomic.

46 LET S PROVE ONE OF THESE B(K) isthesmallestinvariant S s2n H s. -algebra including

47 B(K) S isthesmallestinvariant s2n H s. -algebra including hc j, 0i is the equivalence class hc j i sitting at the position 0 in K s. ` It su ces to show that if u 2 W m then the basic open set hui K is arbitrarily well approximated in measure by elements of S s2n H s. We will use specification Q4. Fix > 0. Choose an n>msuch that n = M(s)forsomes and large enough that n + q m q n <. Let G K be the collection of x such that x(0) is not among the last ( n + q m q n )q n letters of the word in W n sitting at the n-block of x containing 0. By the ergodic theorem, the measure of G is at least 1. Let x 2 G \hui and w 2 W n sit at the n- block B =[ k, q n k)ofxcontaining 0. If c j is the Q M(s) s class of w then by specification Q4, sh k (x) 2 s 1 is a union of shifts of sets of the form 1 (hc j, 0i) sh k hui. Hence G \hui s (hc j, 0i). a

48 UNDERSTANDING JOININGS Let K 0 be K or K 1,andK 0 s the corresponding factor. Let h s : s 2 Ni be a sequence with each s ajoiningofk s with K 0 s. We will say that the sequence is coherent if s is the projection of s+1 to a joining of K s with K 0 s via s+1,s s+1,s. Coherent graph joinings yield a graph joining of K with K 0.

49 BUILDING GRAPH JOININGS If g 2 G n s then g gives a permutation of powers the equivalence classes that determine K s. In particular g gives an invertible graph joining of K s with K ±1 (depending on the s parity of g) andthisgraphjoiningprojects to the identity map on the odometer. If s 0,s(g s 0)=g s, then the g s action is subordinate to the g s 0 joinings cohere. action hence the graph

50 BUILDING GRAPH JOININGS If g 2 G n s then g gives a permutation of powers the equivalence classes that determine K s. In particular g gives an invertible graph joining of K s with K ±1 s (depending on the parity of g) andthisgraphjoiningprojects to the identity map on the odometer. If s 0,s(g s 0)=g s, then the g s action is subordinate to the g s 0 action hence the graph joinings cohere. In particular, if g 2 G 1 6=0,the s (g) s cohere to give a graph joining of K with K ±1 (depending on the parity of g).

51 WHERE ARE WE? Suppose we have built K with these properties. Then if T has an infinite branch then G 1 has an odd element so K = K 1.

52 WHERE ARE WE? Suppose we have built K with these properties. Then if T has an infinite branch then G 1 has an odd element so K = K 1. If T has two infinite branches then G 1 has an even element so there is a non-trivial graph joining of K with itself (i.e. an element of the centralizer not in the powers of the shift.)

53 NOW THE HARD PART

54 NOW THE HARD PART: We need to see that if T is isomorphic to its inverse then there is a branch through the tree.

55 JOINING SPECIFICATIONS The basic observation is that if u, v 2 n and each pair (a, b) 2 occurs the same number of times in (u, v), then d(u, v) =1 1/ and this can be (approximately) arranged by taking u and v to be random.

56 J10 Let u 0 and v 0 be elements of W n [ rev(w n ). Let 1 apple k < (1 n )(q n+1 /q n ). Then for each pair u, v 2 W n+1 [ rev(w n+1 )suchthatu 0 has the same parity as u and v 0 has the same parity as v, let r(u 0,v 0 )bethenumberofoccurrencesof(u 0,v 0 ) in (sh kq n (u),v)ontheiroverlap.then r(u 0,v 0 ) (q n /q n+1 ) k 1 s 2 n < n.

57 J10 Let u 0 and v 0 be elements of W n [ rev(w n ). Let 1 apple k < (1 n )(q n+1 /q n ). Then for each pair u, v 2 W n+1 [ rev(w n+1 )suchthatu 0 has the same parity as u and v 0 has the same parity as v, let r(u 0,v 0 )bethenumberofoccurrencesof(u 0,v 0 ) in (sh kq n (u),v)ontheiroverlap.then r(u 0,v 0 ) (q n /q n+1 ) k 1 s 2 n < n. This is saying that two words that are relatively shifted a small amount are separated by about 1 s n+1

58 J11 Suppose that u 2 W m+1 and v 2 W m+1 [rev(w m+1 ). We let s = s(u, v)bethemaximalisuch that there is a g 2 G m i unique such g 2 G m s will call g(u, v). such that g[u] i =[v] i. Then there is a with this property, which we The action of g(u, v) introducesacommonpattern between u and v. Specification J11 will say that relative to this pattern all occurrences of pairs (u 0,v 0 )occurrandomly.

59 J11 Write u = u 0 u 1...u (km 1) with u i 2 W m and v = v 0 v 1...v (km 1) with v i 2 Wm 0 where Wm 0 is either W m or rev(w m )dependingontheparityofg. Let Q m s be the number of Q m s classes. If C is a Q m s class, then a portion of approximately 1/Q m s of the u i s in u come from C. At the same locations in v we have occurrences of v i s from gc. Specification J11 says that in these locations all pairs from C gc occur with about the same frequency, C 2.

60 J11 Suppose that u 2 W m+1 and v 2 W m+1 [rev(w m+1 ). Let (u 0,v 0 ) 2 W m Wm 0 be such that g[u 0 ] s =[v 0 ] s. Let r(u 0,v 0 )bethecardinalityof{i : (u i,v i ) = (u 0,v 0 )}. Then: r(u 0,v 0 ) k m 1 Q m s 1 C m s 2 < n k m.

61 J11 RESTATEMENT r(u 0,v 0 ) k m 1 Q m s 1 C m s 2 < n.

62 J11 IN ENGLISH An element g 2 G s aligns Q s equivalence classes. But if there is no g 0 2 G s+1 aligning the Q s+1 classes then the words that occur in g-aligned classes do so randomly.

63 J11 IN ENGLISH Either: g can be continued to a g 0 2 G s+1 (so s 0,s(g 0 )=g) or The words are random conditioned on the map g from Q s equivalence classes.

64 THE CUNNING PLAN

65 JOINING SPECIFICATIONS The basic observation is that if u, v 2 n and each pair (a, b) 2 occurs the same number of times in (u, v), then d(u, v) =1 1/ and this can be (approximately) arranged by taking u and v to be random.

66 J11 IN ENGLISH An element g 2 G s aligns Q s equivalence classes. But if there is no g 0 2 G s+1 aligning the Q s+1 classes then the words that occur in g-aligned classes do so randomly.

67 J11 IN ENGLISH Either: g can be continued to a g 0 2 G s+1 (so s 0,s(g 0 )=g) or The words are random conditioned on the map g from Q s equivalence classes.

68 THE PLAN Anything resembling a conjugacy must be a finite shift of the odometer. Approximations to joinings given by stationary codes determine nodes through the tree: either they give a partial branch determined by a g 2 G s or they are relatively independent joinings over some H s.

69 SYNCHRONOUS JOININGS Suppose that is an ergodic joining between K and K ±1 Suppose further that H 0 H 0 is supported on the graph of some g 6= j for any j 2 Z. Then must be the relatively independent joining of K and K ±1 over the graph joining given by g.

70 SYNCHRONOUS JOININGS Suppose that is an ergodic joining between K and K ±1 Suppose further that H 0 H 0 is supported on the graph of some g 6= j for any j 2 Z. Then must be the relatively independent joining of K and K ±1 over the graph joining given by g. Note that if g = j, then composing g with sh j gives a synchronous or anti-synchronous joining.

71 WHY? J10 Let u 0 and v 0 be elements of W n [ rev(w n ). Let 1 apple k < (1 n )(q n+1 /q n ). Then for each pair u, v 2 W n+1 [ rev(w n+1 )suchthat u 0 has the same parity as u and v 0 has the same parity as v, let r(u 0,v 0 )bethenumber of occurrences of (u 0,v 0 )in(sh kq n (u),v)on their overlap. Then r(u 0,v 0 ) (q n /q n+1 ) k 1 s 2 n < n.

72 WHY? With any slippage words occur approximately independently (in the probabilistic sense). if H 0 H 0 is not a j then g has to shift more and more at larger and large scales. asymptotically the occurrences of words becomes precisely independent, relative to the graph joining j.

73 THE KRONECKER FACTOR The Kronecker Factor L is defined to be the largest compact factor of K. Any automorphism of L can be lifted to a joining of K with K 0 using the relatively independent product joining. If L strictly extends O then there is a t 2 L that doesn t map to any j in the odometer. This gives an automorphism of L and its lift to is NOT the relatively independent joining of K with K 0 over O.

74 So the odometer is the Kronecker factor

75 UPSHOT Any joining J of K with K ±1 is either arelativelyindependentjoiningovero or projects to a finite j on the odometer. In the second case J (1, sh j )projectstothe identity on the odometer. It follows that if J is an isomorphism then we have an isomorphism that it is identity on the odometer; e.g. it is synchronous.

76 JOININGS OF LARGER K s For g 2 G s, let I g = {(x, y) :g s (x) = 0 s(y)} K K 0 T g = I g \{(x, y) :fornog 2 G s+1 is g s+1 (x) = 0 s+1(y)}.

77 Let s, t 2 N and g 2 G s,h2 G t be di erent. Then: Each I g is a closed set, as is S g2g s I g, T g is a G set, T g \ T h = ;, If is the relatively independent joining of K with K 0 over the graph joining g of K s with K 0 s then (T g )=1, If g 1 2 G 1 (T ), is the graph joining of K with K 0 determined by g 1 and g s = s (g 1 ), then T s2n I g s is measure one for.

78 THE TWO ALTERNATIVES Let I be the collection of joinings of K with K 0 of the following form: 1. the relatively independent joining of K with K 0 over the graph joining g of K s with K 0 s for some g 2 G s, 2. the graph joining of K with K 0 determined by some g 2 G 1 (T ). Then any two distinct members of I are mutually singular measures.

79 FINISHING THE THEOREM

80 FINISHING THE THEOREM?? Let be an ergodic joining of K with K 0. Then exactly one of the following holds: 1. For some s, some j 2 Z and some g 2 G s, is the relatively independent joining of K with K 0 over the graph joining g (1, sh j ) of K s K 0 s. 2. There is a g 2 G 1 (T )andaj 2 Z such that (1, sh j )isthegraphjoiningofkwith K 0 determined by g.

81 Fix an ergodic. If is not a relatively independent joining over O, then we can assume that is synchronous. Assume is not a graph joining arising from some g 2 G 1. Let (x, y) be generic for. Then there is a maximal s such that g s (x) = s (y). For this s and g we apply specification J11, to see that words are independent, conditioned on the g-mapping of the classes. This is a restatement of being a relatively independent joining of K with K 0 over K s.

82 SOME INFORMAL COMMENTS

83 WHAT YOU CAN SEE WITH A STATIONARY CODE

84 Note that patterns of lengths of words can be seen, and you can see words that are lined up and so on.

85 If we are given a conjugacy : K! K 0 then can be approximated by a sequence of codes i : i 2 Ni. We apply them to generic points (x, (x)) 2 K K 0. At some point the codes code more accurately than the radius of the s-equivalence classes and thus can be error-corrected back to maps from equivalence classes to equivalence classes. This error correction takes the patters of s- classes in x to the pattern of s-classes in (x). The only way it can do that is if the pattern comes from some g s 2 G s. Thus a conjugacy gives us a coherent sequence hg s : s 2 Ni in the inverse limit, hence the tree is ill-founded.

86 SO WHY AREN T WE DONE?

87 WE HAVE A BUNCH OF AXIOMS ABOUT WORDS.

88 WE HAVE A BUNCH OF AXIOMS ABOUT WORDS. HOW DO WE BUILD A CONSTRUCTION SEQUENCE SATISFYING THEM?

89 THE WORD SPECIFICATIONS Specifications E1-E3 from the paper simply say that hw n i n is a construction sequence, s n is a power of 2, s n s n+1 and each n-word occurs equally often in each n +1-word. These are easy to satisfy.

90 Q4. Suppose that n = M(s). Then any two words in the same Q n s equivalence class agree on an initial segment of length at least (1 n )l n. Q5. For n M(s)+1,Q n is the product equivalence relation of Q M(s) s. Hence we can view s W n /Q n as sequences of elements of W s M(s) /Qs M(s) and similarly for rev(w n )/Q n. s Q6. Q n s+1 refines Q n s and each Q n s class contains 2 k(n) many Q n s+1 classes.

91 A7. G n s acts freely on W n /Q n s [ rev(w n /Q n s )and the G n s action is subordinate to the G n s 1 action via the natural homomorphism s,s 1 from G n s to G n s 1. A8. The canonical generators of G M(s) s send ele- to elements of rev(w M(s) )/Q M(s) s ments of W M(s) /Q M(s) s and vice versa. A9. If M(s) apple m, and we view G m+1 s = G m s H then the action of G m s on W m /Q m s [rev(w m /Q m s ) is extended to an action on W m+1 /Q m+1 s [ rev(w m+1 /Q m+1 s )bytheskewdiagonalaction. If H is non-trivial then its canonical generator maps W m+1 /Q m+1 s to rev(w m+1 /Q m+1 s ).

92 THE ISSUE: THE JOINING AXIOMS

93 J10. Let u and v be elements of W m+1 [rev(w m+1 ). Let 1 apple k<(1 m+1 )(k n ). Then for each pair u 0,v 0 2 W m [ rev(w m ) such that u 0 has the same parity as u and v 0 has the same parity as v, let r(u 0,v 0 )bethenumberofoccurrences of (u 0,v 0 )in(sh kq m (u),v)ontheir overlap. Then r(u 0,v 0 ) (k n ) k 1 W 2 m < m+1. J11. Suppose that u 2 W m+1 and v 2 W m+1 [ rev(w m+1 ). We let s = s(u, v) bethemaximal i such that there is a g 2 G m i such that g[u] i =[v] i. Let g = g(u, v) betheuniqueg with this property and (u 0,v 0 ) 2 W m Wm 0 be such that g[u 0 ] s =[v 0 ] s. Let r(u 0,v 0 )be the number of occurrences of (u 0,v 0 )in(u, v). Then: r(u 0,v 0 ) q m+1 q m 1 Q m s 1 C m s 2 < m+1(k n ).

94 J10 says that shifts of words are mutually random. J11 says that conditioned on the action of G s words are mutually random.

95 HOW DO YOU BUILD RANDOM WORDS?

96 HOW DO YOU BUILD RANDOM WORDS? You DONT!

97 Law of Large Numbers Suppose that {X i : i 2 N} is a sequence of independent identically distributed 2-value random variables defined on a measure space (X, B,µ) taking value 1 with probability p and 0 with probability 1 p. Let > 0. Then P ( 1 n n 1 X X i p ) <e n 2 /4 (1) i=0

98 J10 says that two words that are shifted so that the overlap is non-trivial are mutually random. To satisfy this: For each shift the Law of Large numbers says that the vast majority of sequences are mutually random for long words the percentage of words that are not mutually random decays exponentially the requirements we are putting on about shifts are only linear as a function of the length of the word. So: we lose the same proportion of words at each shift, that proportion of words goes to zero exponentially, but the number of shifts grows linearly... For long enough word-length there are large collections of words that satisfy J10.

99 J11 says that conditioned on the maximal s for which there is a g s sending Q s -classes to Q s classes the words are random. There are only linearly many g s s so we can treat them separately. If the vast majority of words works for each g s then we intersect the collections of words. We use the Law of Large Numbers to substitute into equivalence classes with the frequencies required. (The actual Substitution Lemma is quite complicated to state.)

100 WE VE PROVED THAT ISOMORPHISM FOR ABSTRACT ERGODIC MPT S IS COMPLETE ANALYTIC

101 PAR CONTRA

102 Theorem: The collection of pairs (S, T )ofrank one ergodic transformations that are conjugate is aborelsubsetofmpt MPT.

103 RECALL Rank one transformations are those that can be built using a cut-and-stack construction with one tower at each stage.

104 WE USE THE FOLLOWING THEOREM OF J. KING Theorem: Suppose that T is an ergodic rank one transformation. Then the centralizer of T is the closure of {T n : n 2 Z} in MPT.

105 WE USE THE FOLLOWING THEOREM OF J. KING Theorem: Suppose that T is an ergodic rank one transformation. Then the centralizer of T is the closure of {T n : n 2 Z} in MPT. The closure will typically be more than the powers of T, since rigid transformations are a generic class.

106 THE PROOF The E ros Borel space of a Polish space is the collection F(X) ofclosedsubsetsofx with the -algebra generated by basic sets of the form: {F 2 F(X) :F \ U 6= ;}, where U X is open. It is a standard Borel space in the sense that there is a Polish topology (the Fell topology) on F(X) forwhichitisthecollectionof Borel sets.

107 FACTS: 1. (Dixmier) If G is a Polish group then there is a Borel T G F(G) suchthat (a) {x : T(x, H)} 6= ; i H is a closed subgroup of G, and (b) if T H = {x : T(x, H)} and H is a closed subgroup of G then T H is a transversal for G/H, the space of left cosets of H in G. 2. Suppose that X and Y are Polish spaces and B X Y is a Borel set such that for all x 2 X there is at most one y 2 Y with (x, y) 2 B. Then {x :(9y)(x, y) 2 B} is Borel.

108 Let R be the collection of ergodic rank one transformations. Define C : R! F(MPT) byt 7! C(T ). Then C is a Borel map.

109 Let R be the collection of ergodic rank one transformations. Define C : R! F(MPT) byt 7! C(T ). Then C is a Borel map. ` It su ces to see that the C-inverse of a basic set is Borel. Suppose that O = {F 2 F(MPT) : F \ U 6= ;} for some open set U in MPT. Then C 1 (O) ={T : C(T )\U 6= ;}. By King s theorem C(T ) \ U 6= ; i (9n)T n 2 U. The collection of T such that for some n, T n 2 U is an open set in MPT, hence C 1 (O) isopen. a

110 Theorem: The collection of pairs (S, T )ofrank one ergodic transformations that are conjugate is aborelsubsetofmpt MPT.

111 THE THEOREM Theorem: The collection of pairs (S, T )ofrank one ergodic transformations that are conjugate is aborelsubsetofmpt MPT. ` Let T C(T ) be the Borel transversal MPT/C(T ). Define B = {(S, T, g) :gtg 1 = S and g 2 T C(T ) } (R R) MPT. Then for each (S, T ) 2 R R there is at most one g with (S, T, g) 2 B. Hence {(S, T ):forsome g 2 MPT, (S, T, g) 2 B} is Borel. a

112 A PUZZLE Isomorphism on rank 1 transformations is Borel, but not reducible to an S 1 -action. It SHOULD have a structure theorem, but what can it be? A subquestion: what is its Borel complexity?

113 TOMORROW Trees F

114 We need to explain how to build the construction sequences in the reduction so that composing with F is a reduction. We know one direction already: If T is in the image of F and T = T 1, then there is an anti-synchronous isomorphism between T and T 1 This means that we have to add to the randomness assumptions so that if F(T ) = F(T ) 1, then there is an anti-synchronous isomorphism. Every isomorphism has a projection to a graph joining of S 1, and hence by the theorem is associated with a rotations we identify which rotations have to be in the closure of the powers of T. We call these central we randomize so that every isomorphism is associated with a central value.

115 END OF THURSDAY LECTURE