EQUIVALENCE

Computing

f(Tfl M)

• (T M) = T

2

(T (a)))

a2 ax a2

f(T (a)) 0,00,1 (u))

a 1 a2

f(T (a)))+a (a),T (a))

2 2

a2(a3,Ta (u)))+f(u)

T ^ Oo) = T (T f(a))(a))) = T (#((0))

al al al al

and T1 is a factor of T .

Note: ty is clearly nonsingu'lar as f is integer valued.

We argue further, if i| were not 1-1 a.e., then on a set of positive

measure we would have

1 1

and

,/ ,\ * / i\ a-, (a),o)f)

al al 1

as w,a)f must be on the same orbit. Thus

f

, v f (a)f)+a (o)^')

a1 a

and as T is aperiodic a.e., f(u) = f(oof) + a..(u),wf) and a^Ga,*!)1) = 0

and T is not aperidoic a.e. Thus d is invertible a.e. so T and

a2 °1

T are isomorphic.

a2