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,