2. SOME BASIC FACTS ABOUT Aut(X, µ) 7

Then S ∈ Aut(X, µ) and

SUS−1

= V .

Now given aperiodic T, W and 0, find periodic U, V of large enough

period N such that δu(T, U) , δu(W, V ) and find S ∈ Aut(X, µ) with

SUS−1

= V . Then as δu is 2-sided invariant,

δu(STS−1,W

) ≤

δu(STS−1,SUS−1)

+ δu(V, W ) 2,

so we are done.

For the last assertion, note that if T is not aperiodic, then for some n =

0, 1,δu(T

n,

1) ≤ and so

δu((STS−1)n,

1) ≤ for every S ∈ Aut(X, µ).

✷

On the other hand every conjugacy class is small in the sense of category.

In fact one has a stronger conclusion concerning unitary (or spectral) equiv-

alence classes. Consider Aut(X, µ) as a closed subgroup of

U(L2(X,

µ)) via

the identification T → UT . Then we say that T, S ∈ Aut(X, µ) are unitarily

equivalent or spectrally equivalent if they are conjugate in

U(L2(X,

µ)). This

is clearly a coarser equivalence relation than conjugacy in Aut(X, µ). One

now has:

Theorem 2.5 (Rokhlin). Every unitary equivalence class in Aut(X, µ) is

meager in the weak topology. In particular, this is true for every conjugacy

class in Aut(X, µ).

Proof (Hjorth). We work below in the weak topology. Fix a set A ⊆ X

with µ(A) =

1

2

. We claim that the open set

{(S, T) :

∃n[µ(Sn(A)∆A)

1

400

and µ(T

n(A)∆A)

1

9

]}

which is equal to

{(S, T) : ∃n[ US

n(χA)

− χA

1

20

and UT

n(χA)

− χA

1

3

]}

is dense in Aut(X,

µ)2.

To see this, notice that by 2.4 the set of mixing

T ∈ Aut(X, µ) is dense in Aut(X, µ). So, given any (S0,T0) ∈ Aut(X,

µ)2,

we can find mixing T as close as we want to T0. Then µ(T

n(A)

\ A) →

µ(A)µ(X \ A) =

1

4

, so for some N and all n ≥ N, µ(T

n(A)∆A)

1

9

. On the

other hand, by 2.1 we can find periodic S of period ≥ N as close as we want

to S0. For such S, if S has period n,

µ(Sn(A)∆A)

= 0, and we are done.

(Notice that it would be enough here to take T weak mixing, since then there

would be a set A ⊆ N of density 1 such that µ(T

n(A)∆A))

1

9

, ∀n ∈ A

(see, e.g., Bergelson-Gorodnik [BGo], 1.1).)

Now, by 2.4, if a unitary equivalence class is not meager it will have to be

comeager. So assume the unitary equivalence class of some T0 is comeager,

towards a contradiction. Then for comeager many T ∈ Aut(X, µ), there is

U ∈ U(L2(X, µ)) with UUT U −1 = UT0 . By the Jankov-von Neumann Se-

lection Theorem, we can find a Borel function f : Aut(X, µ) →

U(L2(X,

µ))

so that, letting fT = f(T),

∀∗T(fT

UT fT

−1

= UT0 ),