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)
n
(= {z1 . . . zn : zi
¯
Λ) }). Moreover if σ is non-atomic, then
actually
σ∗n
is supported by
¯
Λ)
∗n
= {z1 . . . zn : zi
¯
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 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
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
0,Uτ 0
are
the unitary operators associated to the respective Gaussian shifts (restricted
to the orthogonal of the constant functions), then
σ∞ τ∞
0

=
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), we can assume that A
¯
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)
∗n
= and therefore
¯
Λ)
¯
Λ)
∗n
= ∅, a
contradiction. Thus if σ, τ are supported by Λ∪
¯
Λ for the same independent
set Λ, then
σ τ
0

=

0.
Note also that

0

=
0


= ,
so we finally have:
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