5. TURBULENCE OF CONJUGACY 37

For each finite measure σ on T consider the Hilbert space

L2(T,σ)

and

the unitary operator Vσ(f)(z) = zf(z). Suppose now that Λ ⊆ T is an

independent set, i.e., if z1,...,zk ∈ Λ are distinct, n1,...,nk ∈ Z, and

z1

n1

· · · zk

nk

= 1, then n1 = · · · = nk = 0. If a symmetric σ is supported

by Λ ∪

¯

Λ, then the n-fold convolution

σ∗n

= σ ∗ · · · ∗ σ is supported by

(Λ ∪

¯

Λ)

n

(= {z1 . . . zn : zi ∈ (Λ ∪

¯

Λ) }). Moreover if σ is non-atomic, then

actually

σ∗n

is supported by (Λ ∪

¯

Λ)

∗n

= {z1 . . . zn : zi ∈ (Λ ∪

¯

Λ) , zi =

zk, ¯k, z if i = k} and these sets are pairwise disjoint, so σ =

σ∗1,σ∗2,...

have

pairwise disjoint supports. Now the spectral theory of Gaussian shifts shows

that the unitary operator Uσ on HC

:n:,n

≥ 1, is isomorphic to the unitary

operator Vσ∗n on

L2(T,σ∗n);

see Cornfeld-Fomin-Sinai [CFS], Chapter 14,

§4, Theorem 1. If σ is also a probability measure, let σ∞ =

∑∞

n=1

1

2n

σ∗n.

Then

L2(T,σ∞)

=

n≥1

Kn, where Kn = {f ∈

L2(T,σ∞)

: f is 0 off

Λ∗n},

each Kn is invariant under Vσ∞ and the map f ∈ Kn →

√1

2n

f ∈

L2(T,σ∗n)

is an isomorphism of Vσ∞ |Kn with Vσ∗n . It follows that

Uσ|L0(RZ,µϕ(σ))2

is isomorphic to Vσ∞ on

L2(T,σ∞).

In the language of spectral theory this

says that the maximal spectral type of the operator Uσ

0

=

Uσ|L0(RZ,µϕ(σ))2

is equal to σ∞ and it has multiplicity 1. Since, by the spectral theory,

two unitary operators are isomorphic iff their maximal spectral types are

equivalent and they have the same multiplicity function, we conclude that

whenever σ, τ are non-atomic symmetric probability Borel measures on T

each supported by a set of the form Λ ∪

¯

Λ, Λ independent, and Uσ

0,Uτ 0

are

the unitary operators associated to the respective Gaussian shifts (restricted

to the orthogonal of the constant functions), then

σ∞ ∼ τ∞ ⇔ Uσ

0

∼

= Uτ

0,

where

∼

= denotes isomorphism of the unitary operators.

But one can now verify that if σ, τ are supported by the same set Λ ∪

¯

Λ,

where Λ is an independent set, then

σ ∼ τ ⇔ σ∞ ∼ τ∞.

Indeed, σ ∼ τ ⇒ σ∞ ∼ τ∞ holds simply because σ τ ⇒ σ ∗ ρ τ ∗ ρ,

for any ρ. Conversely, assume that σ∞ ∼ τ∞. Say σ, τ are both supported

by Λ ∪

¯

Λ. Suppose now σ(A) = 0 but τ(A) 0, towards a contradiction.

Replacing A by A ∩ (Λ ∪

¯

Λ), we can assume that A ⊆ (Λ ∪

¯

Λ). Since

σ∞ ∼ τ∞ τ, we have σ∞(A) 0, so as σ(A) =

0,σ∗n(A)

0 for some

n 1. Thus A ∩ (Λ ∪

¯

Λ)

∗n

= ∅ and therefore (Λ ∪

¯

Λ) ∩ (Λ ∪

¯

Λ)

∗n

= ∅, a

contradiction. Thus if σ, τ are supported by Λ∪

¯

Λ for the same independent

set Λ, then

σ ∼ τ ⇔ Uσ

0

∼

=

Uτ

0.

Note also that

Uσ

0

∼

= Uτ

0

⇔ Uσ

∼

= Uτ ,

so we finally have: