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Classification problems in ergodic theory by William Parry

By William Parry

The isomorphism challenge of ergodic conception has been widely studied considering the fact that Kolmogorov's advent of entropy into the topic and particularly considering that Ornstein's answer for Bernoulli approaches. a lot of this learn has been within the summary measure-theoretic surroundings of natural ergodic thought. besides the fact that, there was transforming into curiosity in isomorphisms of a extra restrictive and maybe extra real looking nature which realize and appreciate the country constitution of methods in a variety of methods. those notes supply an account of a few fresh advancements during this course. a distinct characteristic is the common use of the data functionality as an invariant in various distinctive isomorphism difficulties. teachers and postgraduates in arithmetic and learn staff in verbal exchange engineering will locate this booklet of use and curiosity.

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It is useful to note that d(( t, e) = sup { d( a, 0) : a C a is a finite partition) since (X, a, m) is Lebesgue. 46(i) -(iii) show that d is nearly a metric - it is not symmetric. To get a metric we define D(a, C) = max {d(a, C) , d(C, (t) ) for sub-Q-algebras a, C C 63. 47. Theorem [P7]. Let a, e be sub-Q-algebras. Then d(a, e) < 2 iff I((t IC) is finite on a set of positive measure. Proof. Since d((t "C , e) = d(( t, e) and I(a " C I C) = I(a I C), we may assume a C a. Suppose d((I, C) < 2. Pick E > 0 small enough to have d(a, C) < 2 - 2s.

Proposition. e. f= 1). We may now prove with little effort the following interesting 44. Proposition. For a Markov chain T the following are equivalent (i) 92(T) =0 (ii) IT is cohomologous to a constant (iii) T is of maximal type. Proof. We easily see from 22 that (i) and (ii) are equivalent. We show (ii) and (iii) are equivalent. If T is defined by P(i, j) = A(i, j) r(j) where A is a Pr(i) 0-1 matrix, A its maximum eigenvalue and r a corresponding eigenvector, then I,1, is cohomologous to the function equal on [i, ji to -log P(i, j) = log Is - log r(j) + log r(i) i.

By definition a C T a C T' d C .. and n 0Tna generates (B so L2 (X, (1, m) C L2 (X, T( I, m) C L2(X, T2(1 ,m) C.. Proof. and n 0 L2 (X, Tn a, m) is dense in L2 (X, 63 , m) . Denote by U the unitary operator defined by T, Uf = f o T for f E L2 (X, (B, M). Put V = L2 (X, Ta, m) G L2 (X, a , m) and note that for n E Z Un(L2 (X, d, m)) = L2 (X, T -n(,, m) and Un(V) =L2(X, T-n+la m) OL2(X, T-na M) We have L2 (X, 63, m) = L2 (X, a, m) ®® U T1V n=0 since the subspace on the oo right is closed and contains each L2 (X, Tn (t, m) , n > 0.

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