4
REPRESENTATION THEORY AND NUMERICAL AF-INVARIANTS
(Conditions like (1.8) and (1.9) make sense in this setting since ai and a are local
homeomorphisms, and thus map measurable sets into measurable sets.) Note that
(1.10) a-l(E) = \Jcrl(E)
i
for all sets E by the chain: x G |Jf o~i (E) = x = iy for some i = 1,..., d,
y G E = a(x) e E = x e
a"1
(E). Note also that if /i is both a- and
cr^-quasi-invariant, we have the connection
dpfajy))
=
dii{jj{y))
=
/ d / z o a V
1
between the Radon-Nikodym derivatives.
Note also that the two quasi-invariance conditions (1.5) and (1.6) together
imply the d equivalences
(1.11) fj,(E) = 0 ^= /x (7*£) = 0, i = 1,..., d,
for all Borel sets E C O , and that (1.11) implies the cr-invariance (1.5). (When
referring to (1.6) (or (1.11)) in the following we mean "(1.6) (or (1.11)) for all
i = 1,..., d"). Let us prove this.
(1.5) fc (1.6) =s (1.11): If (1.5) and (1.6) hold and p(E) = 0, it follows from
(1.5) that ^(a^iE)) = 0 and hence from (1.10) that fi(ai(E)) = 0 for all i.
Conversely, if \i {&i (E)) = 0 for some z, then since E = a^a^ it follows from
(1.6) that fj,(E) = 0 .
(1.11) (1.5): Assume that (1.11) holds and that \i (E) = 0. Then fi fa (E)) =
0 for all i by (1.11) and hence \x (a'1 (E)) = 0 by (1.10).
The condition (1.11) does not, however, imply cr^-quasi-invariance (1.6), by the
following example: d 2, n (5-measure on (1,1,1,...). Then (1.11) holds for
all E, but (1.6) fails for E {(2,1,1,1,...)} and i = 2. In this case /i is cr-quasi-
invariant and G\-quasi-invariant, but not a2-quasi-invariant, so (1.5) does not imply
(1.6). More interestingly, the converse implication is always valid:
Proposition 1.1. If fi is a probability measure on Q and /x is ai-quasi-invariant
for i 1,..., d, then \i is a-quasi-invariant.
Proof Put pi (y) = d(V Since the maps ai are injective and have disjoint
ranges, there is actually one function G such that
(1.12) Gfa(y)) = Pi(y).
One now proves as in (1.38) below (the tacit assumption there that /i is cr-quasi-
invariant is not needed for this) that
(1-13) / g (x) dfji (x)= [ R (g) (x) dfi (x),
where
(1-14) R(9){x)= £ G(y)g(y).
y
cr(y)=x
Note that the Ruelle operator R has the property
(1.15) R(foa) = R(l)-f,
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