Representations of positive projections 1 Introduction - Mathematics ...

Given a Dedekind complete order separable f-algebra A in which the
unit element 1 is a strong order unit and a strictly positive order continuous
projection P : A ! A onto the f-subalgebra B with 1 2 B, letA = A (p 1)
A (p 2) be the decomposition of Proposition 4.12. As shown in the above
example, it is easy to construct a representation for P p1 on A (p 1). The main
e ort in the remainder of the present paper is to construct a representation
for the projection P p2 on the nowhere B-full part A (p 2).
5 Maharam's theorem for positive projections
In this section we will discuss a generalization of a result of D. Maharam
( [11], Lemma 3) to positive projections, which will play an important role
in the sequel. Actually, the main theorem in the present section can be
considered as a vector valued version of Maharam's lemma. As before, A
will denote a Dedekind complete f-algebra with unit element 1 and B will
be an f-subalgebra with 1 2 B. Furthermore we will assume that P : A ! A
is a positive order continuous projection onto B. As observed in Lemma 3.6,
this implies that B is a complete f-subalgebra of A. The proof of the main
result in the present section is divided intoanumber of lemmas.
Lemma 5.1 Assume that A is nowhere full with respect to B. If 0 6= e 2CA,
then there exists p1 2CA such that 0 6= p1 e and P (p1) P (e).
Proof. Since e 6= 0, the element e is not B-full and so there exists
p 2CA(e) such that p =2 eCB. Let q = e ; p and put u = p:q. First observe
that u 6= 0. Indeed, suppose that u =0,then pq =0,which implies that
so p 2 eCB, a contradiction.
ep =(p + q) p = pp = p
Let e 1 2CB be the component of p in the band
1
2
n P (p) ; 1
2 P (e) +o d
B. Then e1 P (p) ; 1 +
P (e) =0andso from Lemma 4.1 it follows that
2
P (e 1p) =e 1P (p)
1
P (e) :
2
If e 1 6= 0,then, by Lemma 4.4, e 1p 6= 0and so we can take p 1 = e 1p in this
case.
Now assume that e 1 =0. Then p 2
in
n P (p) ; 1
2 P (e) +o dd
and since 0 <
u p, it follows that u P (p) ; 1 +
P (e) > 0 and u P (p) ; 2 1
2P (e) ; = 0,
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