Representations of positive projections 1 Introduction - Mathematics ...

subalgebra of M (X ), it follows that we may assume in this situation, that
bA is an f-subalgebra of M (X ) and that is an f-algebra homomorphism.
If (X ) is a representation space for (A 1), then we will simply say that
(X ) is a representation space for A.
Although the result of the following lemma is known (the result is even
true under some weaker assumptions see [3], Theorem 6.1), we indicate the
simple proof for the convenience of the reader.
Lemma 4.1 Let A be an Archimedean f-algebra with unit element 1 and
suppose that B is an f-subalgebra of A with 1 2 B. If P is a positive
projection in A onto B (i.e., Ran(P ) = B), then P (ba) = bP (a) for all
b 2 B and a 2 A (i.e., the projection P is so-called averaging).
Proof. Fix some 0 a 2 A. First we assume in addition that 0 a n1
for some n 2 N. De ne the linear operator Ta : B ! B by Ta(b) = P (ba)
for all b 2 B. Then 0 Ta(b) nb for all 0 b 2 B. Hence Ta 2 Z(B),
the center of B. Since B is an f-algebra with unit element every operator in
Z(B) is given by multiplication by some element ofB, so there exists b 1 2 B
such that Ta(b) = bb 1 for all b 2 B. Taking b = 1 we see that b 1 = P (a).
It follows that P (ba) = Ta(b) = bP (a). If 0 a 2 A is arbitrary, then
we can use that a ^ n1 "n a (a 2 -uniformly) and the result now follows via
approximation (see e.g. [16], Theorem 142.7).
The following proposition will play an important role in the sequel. As
usual, positive projection P : A ! A will be called strictly positive if Pa > 0
whenever 0 < a 2 A. The result of the following proposition, which is of
interest in its own right, will play a crucial role in the proof of Lemma 8.2.
Proposition 4.2 Let A be aDedekind complete f-algebra with unit element
1, which is a strong unit as well. Let P : A ! A be a strictly positive order
continuous projection onto the f-subalgebra B, with 1 2 B. Suppose that C
is a complete f-subalgebra ofA such that B C. Then there exists a unique
positive projection Q : A ! A onto C such that P = P Q. Moreover, this
projection Q is strictly positive and order continuous.
by
Proof. Given 0 a 2 A de ne the positive linear operator Ta : C ! B
Ta(c) =P (ac)
for all c 2 C. Since C is a regular Riesz subspace of A and P is order
continuous, it follows that Ta is order continuous as well. Furthermore, if
b 2 B then
Ta (bc) =P (bac) =bP (ac) =bTa (c)
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