Representations of positive projections 1 Introduction - Mathematics ...

'Pi (f) 1Ai for all f 2 Ai (Ei) (see (29)). Since 2 ( (e)) = 1 (e) for all
e 2E 1 it follows that 'P2 ( (f)) = 'P1 (f) for all f 2 A 1 (E 1). We de ne
by
0
0 : K (CB1 E 1) ! K (CB2 E 2)
nX
i=1
qifi
!
=
nX
i=1
(qi) (fi) (31)
for all fqig n
i=1 in CB1 and all ffig n
i=1 in A 1 (E 1). First we observe that 0 is
well de ned. Indeed, if f = P n
i=1 qifi has standard form f = P m
j=1 pjgj, then
it is easy to see that
nX
i=1
(qi) (fi) =
mX
j=1
(pj) (gj).
Since the standard form of an element inK (CB1 E1) is unique, it follows that
0 is well de ned by (31). Since and are f-algebra homomorphisms it
is now also clear that 0 is an f-algebra homomorphism.
Take f 2 K (CB1 E1) and write f = Pn i=1 qifi with fqig n
i=1 in CB1
ffig
and
n
i=1 in A1 (E1). Then
P 1f =
nX
i=1
P 1 (qifi) =
nX
i=1
qiP 1 (fi) =
nX
i=1
'P1 (fi) qi
and so P 1 [K (CB1 E 1)] K (CB1 E 1). Moreover, using that 'P2 ( (fi)) =
'P1 (fi), we nd that
0 (P 1f) =
=
nX
i=1
nX
i=1
'P1 (fi) (qi) =
(qi) P 2 ( (fi)) =
= P 2
nX
i=1
nX
i=1
nX
i=1
(qi) (fi)
'P2 ( (fi)) (qi)
P 2 ( (qi) (fi))
!
= P 2 ( 0f) :
This shows that 0 P 1 = P 2 0.
Let K 1 be the 1A1-uniform closure of K (CB1 E 1)inA 1. Then 0 extends
uniquely to an f-algebra homomorphism 1 : K 1 ! A 2. Since CB1 K 1 and
1 (q) = 0 (q) = (q) for all q 2CB1, it follows from Freudenthal's spectral
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