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
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