6
REPRESENTATION THEORY AND NUMERICAL AF-INVARIANTS
so that
and
(1.25) Xi(x)
1/2
1 ifxi = i,
0 otherwise.
Conversely, if \i, H, x \— 7 Y (a;) and x i— » U (x) satisfy all the conditions in the
initial part of the theorem, the formulae (1.19)-(1.24) define a nondegenerate rep-
resentation of Od on Ti.
Remark 1.3. At the outset the formula (1.21) does not make sense, since U (x)
is an operator on TL (#), while £ (a (x)) G Ti (a (x)). Here we have actually made
an identification of the Hilbert spaces Ti (x) over each orbit of a. The Hilbert
spaces over each a-invariant set have constant dimension /j,-almost everywhere by
the argument after (1.53) below. Hence if we define
fi(n) = {x e ft I dim (H (x)) = n}
for n = 1,2,. ..,^o, then the sets Q(n) are /i-measurable and 7-, as well as cr^-,
invariant up to sets of measure zero. If Hn is the Hilbert space of dimension n for
n 1,..., No, then we may identify H (x) with Hn for x G fJ(n)
a n
d
w e n a v
e the
decomposition
H = n(x) d/x(a;) = 0 / W(x) d/i(x) = 0 H
n
® L
2
( ( ]
( n )
, d / i ) .
^
Q
n = l ^
n
( ^ )
n
= l
Hence we may view 17 (x) as a unitary operator on Hn for all x G fi(n). Since the
f2(n) are a- and ovinvariant, the formula (1.21) is meaningful, and expressions like
the one on the third line of (1.42) make sense since £ (x) and rj(a(x)) lie in the
same Hilbert space. The direct sum above is a decomposition of the representation
of Od- See also Remark 1.5, and see the book [71, Section 2.5.3] for more details.
Before proving this theorem, let us consider the intertwiner space between two
representations Si i— Si and Si ^ Ti. Recall that an operator T intertwines these
representations if and only if it intertwines the operators Si, Tf.
(1.26) TSt = TzT, z = l,...,d.
"Only if" is obvious. As for "if", note that if T satisfies (1.26), then
d
(1.27) T^T =
J2TtTSJSJ
d
3 = 1
= TS;.
Theorem 1.4. Let Si, Si be representations of Od on separable Hilbert spaces
(1.28) H= / H{x) dn(x)
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