# An introduction to ergodic theory - Walters P.

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Diagramatically the basis has the form

/o = l

â– - tV/ÑŒ ^Tlfufu UTfu Urfit â€¢ â€¢ â€¢

..., Ð¸Ñ‚Ð³/2, Urfh fiy â€¢ â€¢ â–

We shall show in Â§4.9 that every Kolmogorov automorphism has countable Lebesgue spectrum. This implies that every Bernoulli shift has countable Lebesgue spectrum. We shall now indicate why the two-sdied (i,i)-shift T has countable Lebesgue spectrum. Here the state space, {0,1} consists of two points each with measure A basis for the L2-space of the state space consists of the constant function 1 and the map {0,1} -* Ð¡ given by Â£ -*â€¢ enâ€œ, t e {0,1}. The transformation T acts on the direct product space X â€” {0,1}Z equipped with the product measure m. Since L2{m) is the tensor product the L2-space of the state space there is a basis for L2(m) of the form {9n,.....n> Â£ yÂ«i < Â«2 â€¢ â€¢ â€¢ < Â»,} u {1} where

' Sfâ€ž,.....â€žr({*i}) = eKilx-+x^+-+x"r\

Note that UTgni.....â€žr = g 1+n,.....1+Ilr. It is now clear that we can rename the

basis so that it has the form {U\f\i ^1,Â«eZ}u {1}.

One can use this method to show directly that the two-sided (p0,..., Pfc-J-shift has countable Lebesgue spectrum. It is also easy to give a direct proof that an ergodic automorphism A of a compact abelian metric group G has countable Lebesgue spectrum. The elements of the character group G form an orthonormal basis. The ergodicity of A implies that if Ñƒ e G and ÑƒÑ„ 1 then the collection {A"y\n e Z} consists of district characters. (Here A is the dual automorphism to A.) Therefore A has countable Lebesgue spectrum once it is shown that there are infinitely many district sets of the form {.4"y [ Ð¸ g Z} in G. This can be done by a simple group theory argument (Halmos [1] p, 54).

The following results are elementary.

Theorem 2.11. Any two invertible measure-preserving transformations with countable Lebesgue spectrum are spectrally isomorphic.

Proof. Let (X^SS^m,) i = 1, 2 be a probability space and let Tt\Xi -*â€¢ X-, be an invertible measure-preserving transformation. Suppose L2(has a basis {/0} u {UrJilj ^ L ne Z} where f0 = I and L1(m2) has a basis {g0} u

66

2 Isomorphism, Conjugacy, and Spcctial Isomorphism

{Uâ€™r^Ai ^ 1.H6 Z} where <jp s 1. Define W:L2(m2) -*â– L2(m,) by W(cj0) = /Ð¾. WWti9j) = U"r,fj an<i extend by linearity. Then WUTl = UTiW and T, and T2 are spectrally isomorphic. â–¡

It follows from this and the discussion above that any two Bernoulli shifts are spectrally isomorphic. This was known in 1943 and only when entropy was introduced by A. N. Kolmogorov in 1958 was it shown that there are non-isomorphic Bernoulli shifts (see Chapter 4).

Theorem 2.12. If a measure-preserving transformation T of a probability space (X Ð¨, in) has countable Lebesgue spectrum it is strong-mixing.

Proof. Let {/0} Ð¸ l.neZ} be a basis of L2(m) where f0 = 1.

Then if j, q> 0 lim,^ (Ð© Ð¾ Ð©/j. UkTf) = (UnTfm, 1)(1, UkTQ V/c, n e Z, since both sides are zero unless 7 = 9 = 0 and then both sides equal one. Fix Ðº and q and consider

= {/ e L\m)-. lim (Urrf, UkTf4) = (f, 1)(1, U%fq)

l p-'Â«3

Then ffk q is a closed subspace of L2(m) and contains the basis by the above calculation.. Hence Ð¯Ð¡Ðº-Ñ‡ = L2(m). Fix / g L2(m) and let

set = \g e L2(m): lim (l/f/, g) = (/, 1)(1, g)

(. p-<*>

Then is a closed subspace of L2(m), contains the basis by the above, and therefore is equal to L2{m). Hence

lim (l/f/, g) = (/, 1)(1, g) V/, g e L\m). â–¡

p~* CO

Â§2.6 Spcctral Invariants

Definition 2.10. A property P of measure-preserving transformations is a

(isomorphism

conjugacy invariant if the following holds: spectral

f isomorphic

Given Tl has P and T2 is ^conjugate to T1

[spectrally isomorphic

then T2 Ms property P.

Remark. A spectral invariant is a conjugacy invariant, and a conjugacy invariant is an isomorphism invariant.

jj2.6 Spcctral Invariants

67

The following shows that the properties we have considered up to now are spectral invariants.

Theorem 2 13. The following are spectral invariants of measure-preserving transformations: (1) ergodicity, (2) weak-mixmrp, (3) strong-mixing.

Proof â€¢"

(1) We know T is ergodic iff {/ e L2(m): UTf = /} is a one-dimensional subspace, and the latter condition is preserved under spectral isomorphism.

(2) We know T is weak-mixing iff 1 is the only eigenvalue and T is ergodic, and this is preserved under spectral isomorphism.

(3) Suppose WUTl= UTlW and Tj is strong-mixing. We have to show that

(Ð©Ð³Ðš Ðº) -Â»(A, 1)(1, k) VA, Ðº e LW

Since this is true if A is constant or if Ðº is constant, it suffices to consider the cases when (A, 1) = 0 = (Ðº, 1). Since T, is ergodic then T2 is ergodic by (1) and since W sends the invariant functions for T2 onto those for 7\ W maps the subspace of constants in L2{m2) onto the subspace of constants in L2(m{). So (Wh, 1) = 0 = (1, Wk). Since W preserves the inner product,

{Wr2h, Ðº) = (WU"Tlh, Wk) = (UnTl Wh, Wk) - 0

since Tj is strong-mixing. Therefore T2 is strong-mixing. â–¡

This theorem allows us to easily display non-spectrally isomorphic transformations. For example a non-ergodic transformation (such as a rotation of Ðš by a root of unity) cannot be spectrally isomorphic to an ergodic transformation (such as a rotation of Ðš by a non root of unity). Also a rotation of a compact group, which is not weak-mixing, cannot be spectrally isomorphic to an ergodic automorphism of a compact group because such automorphisms are weak-mixing.

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