Representations of positive projections 1 Introduction - Mathematics ...

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$MATH 5b; Solutions to Homework Set 1 January 2009 1. As K is a ...$

Now we are in a position to apply Lemma 2.14, from which it follows that m has a unique extension 0 m : K 0 m ! A 2 with the property that m (f ) (o) ;! 0 (f) inA 2 whenever f 2 K 0 m and ff g is a net in Km such that f (o) ;! f in A 1. It is straightforward to verify that K 0 m is an f-subalgebra of A 1 and that 0 m is an f-algebra homomorphism. Hence the pair (K 0 m 0 m) satis es (i) and (ii) above. To show that condition (iii) is satis ed as well, takeany f 2 K 0 m and let ff g be a net in Km such that f (o) ;! f in A1. By the order continuity of P1 of follows that P1f ;! P1f and so m (P1f ) (o) ;! 0 m (P1f). On the other hand, since mf (o) ;! 0 and P2 is order continuous we nd that mf m (P1f )=P2 ( mf ) (o) ;! P2 ( 0 mf) consequently 0 m (P 1f) =P 2 ( 0 mf). This shows that 0 m P 1 = P 2 (o) 0 mand so (K 0 m 0 m) 2K. Since (K 0 m 0 m) is a maximal element ofK, we may conclude that K 0 m = Km. Therefore, Km is order closed in A 1. By Lemma 6.3, the order closure of K (CB1 E 1) is equal to A 1, so Km = A 1. Hence = m is the desired f-algebra homomorphism. Finally, suppose that and are surjective isomorphisms. Then we can apply the above construction to ;1 and ;1 to obtain a corresponding falgebra homomorphism 1 : A 2 ! A 1. Then it is clear that 1 ( f) =f for all f 2 K (CB1 E 1). De ning M = ff 2 A 1 : 1 ( f) =fg it follows that M is an order closed f-subalgebra of A 1 with K (CB1 E 1) M. Again using Lemma 6.3 we conclude that M = A 1, so 1 ( f) = f for all f 2 A 1. Similarly we see that ( 1g) = g for all g 2 A 2, hence is a surjective isomorphism. By this the proof of the proposition is complete. 7 The construction of special subalgebras In this section we will discuss in some more detail the structure of positive projections onto f-subalgebras. The construction presented in this section are inspired by the results in [10], Section 14. We start by listing the hypotheses which will be assumed throughout this section: 48
A is a Dedekind complete f-algebra with unit element 1 2 A B is an f-subalgebra of A with 1 2 B P : A ! A is a strictly positive order continuous projection onto B A is nowhere full with respect to B. As before, for any non-empty subset D CA we will denote by S(D) the complete Boolean subalgebra of CA generated by D, i.e., S(D) is the smallest complete Boolean subalgebra of CA containing the set D. The main objective in the present section is to show that given any sequence fang 1 n=1 in CA, there exists a complete Boolean subalgebra E of CA such that an 2 S(CB [E) for all n and E S. Here S = SP is de ned by (28). It follows from Lemma 6.2 that the inclusion E Sis equivalenttosaying that there exists a completely additive strictly positive measure : E ! [0 1] such that P (e) = (e)1 for all e 2E. The proof will be divided in a number of lemmas. Lemma 7.1 Let a p 2CA be given. Then there exist elements Hk (p a) 2CA (k =0 1 2 3) suchthat (i). H 0 (p a) ::: H 3 (p a) are mutually disjoint (ii). H 0 (p a)+H 1 (p a) =ap (iii). H 2 (p a)+H 3 (p a) =(1;a) p (iv). P 3 k=0 Hk (p a) =p 1 (v). P (Hk (p a)) P (p) for k =01 2 3. 2 Proof. Since 0 that there exists H0 (p a) 2CA such that 1P (ap) P (ap) in B, it follows from Theorem 5.4 2 H0 (p a) ap and P (H0 (p a)) = 1 P (ap): 2 De ne H 1 (p a) = ap ; H 0 (p a). Similarly there exists H 2 (p a) 2 CA such that H2 (p a) (1 ; a) p and P (H2 (p a)) = 1 P ((1 ; a) p): 2 De ne H 3 (p a) =(1;a) p ; H 2 (p a). It is clear that the elements Hk (p a) satisfy all the requirements. 49
Page 1 and 2: Representations of positive project
Page 3 and 4: for all f 2 bL. We stress the fact
Page 5 and 6: that K itself is Dedekind ( -) comp
Page 7 and 8: in C (w) and so it follows that f +
Page 9 and 10: Pe (e) =e and Pe (f) =0whenever f 2
Page 11 and 12: (i). L is locally order separable (
Page 13 and 14: this example also shows that the an
Page 15 and 16: 3 Representations Many of the famil
Page 17 and 18: Lemma 3.3 Given a Dedekind -complet
Page 19 and 20: observations that Ker ( ) = bL \ M(
Page 21 and 22: It is readily veri ed that the exis
Page 23 and 24: of 1 by CA = CA (1). Note that CA i
Page 25 and 26: for all c 2 C, bytheaveraging prope
Page 27 and 28: and, using that Qf is 1-periodic, Z
Page 29 and 30: As we have seen in Example 2.12, a
Page 31 and 32: Proposition 4.12 Let A beaDedekind
Page 33 and 34: Then there exists a representation
Page 35 and 36: can adjust the values of the functi
Page 37 and 38: so which implies that u P (p) ; 1 1
Page 39 and 40: Proof. First observe that'jK 2 K s
Page 41 and 42: Proof. Let P : Z(L) ! Z(L) be the c
Page 43 and 44: (ii). If e 1e 2 2F such that e 1 ^
Page 45 and 46: Boolean subalgebra such that CAi =
Page 47: theorem that B 1 K 1 and 1 (g) = (g
Page 51 and 52: and (34) now follows by induction o
Page 53 and 54: and since P is strictly positive it
Page 55 and 56: Lemma 7.7 If 2 D and 0 < 1, then [q
Page 57 and 58: Since q( ) 2 CB for all 0 1, it is
Page 59 and 60: (ii). there exists a strictly posit
Page 61 and 62: y Proposition 4.2 there exists a st
Page 63 and 64: S (CB [E ). Moreover, there exists
Page 65 and 66: Proposition 8.6 Assume that A, B an
Page 67 and 68: Since B is Dedekind complete, it is
Page 69 and 70: E is a Dedekind complete f-algebra
Page 71 and 72: Then there exists a - nite measure
Page 73: [10] Dorothy Maharam, The represent

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