Representations of positive projections 1 Introduction - Mathematics ...

More documents

Recommendations

Info

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

Since B(p 0) is Dedekind complete, B(p 0) is an ideal in B(p 0) u and so 0 P (p 0) ;1 P (f) 2 B(p 0). Hence 0 P p0(f) 2 B p0 A(p 0). This shows that P p0 is a well de ned positive linear mapping with its range contained in B p0. If f 2 B p0 then, by de nition, f = p 0g for some g 2 B. Hence P p0(f) =p 0P (p 0) ;1 P (p 0g) =p 0P (p 0) ;1 P (p 0)g = p 0p 0g = p 0g = f which shows that P p0 is a projection onto B p0. It is clear that P p0 is order continuous, so it remains to prove that P p0 is strictly positive. Suppose that 0 f 2 A(p 0) such that P p0(f) = 0. Since 0 P (p 0) ;1 P (f) 2 B(p 0), it follows from Lemma 4.4 that P (p 0) ;1 P (f) = 0 and so p 0P (f) = P (p 0)P (p 0) ;1 P (f) =0. Since 0 P (f) np 0 for some n 2 N, this implies that P (f) = 0 and so f =0. In the above situation we will say that P p0 is the restriction of P to A(p 0) (or, the restriction to p 0). We endthis section with a technical result, which shows how representations of restrictions of a positive projection P can be glued together to obtain a representation of P (as de ned in De nition 3.7). We assumethatA is a Dedekind complete f-algebra in which the unit element 1 is a strong order unit and that P : A ! A is a strictly positive order continuous projection onto the f-subalgebra B with 1 2 B. Suppose that (X ) is a representation space for B with representation homomorphism B : Mb (X ) ! B. If 0 < p 0 2 CA, then (X ) is also a representation space for B p0, the representation given by the homomorphism p0 B : Mb (X ) ! B p0, (17) de ned by p0 B (f) =p0 B (f) for all f 2 Mb (X ). In this situation we will p0 call B the induced representation of B p0. Lemma 4.15 Let B : Mb (X ) ! B be arepresentation of B. We assume furthermore that: (i). fpjg is an at most countable disjoint system in CA such that P j pj = 1 (ii). For every j there exists a representation space (X Yj j) for P pj and A (pj) such that the corresponding representation j : Mb (X Yj j) ! A (pj) is compatible with the induced representation pj B of B pj. 32
Then there exists a representation space (X Y ) for P and A such that the corresponding representation :Mb (X Y ) ! A is compatible with B. Proof. We will denote ( j Fj) = (X Yj j). By assumption, there exists a - nite measure j on j and for each j =12::: there is an operator Rj : Mb ( j Fj) ! Mb ( j Fj), given by Rjf (x y) = Z Yj mj (x z) f (x z) d j (z) for all (x y) 2 j and all f 2 Mb ( j Fj), with 0 mj R 2 M ( j Fj) and Yj m (x z) d j (z) =1for all x 2 X, such that j Rj = P pj j. We de ne Y = F j Yj (i.e., Y is the disjoint union of the Yj, j =12:::), = L j j and = L j j. Then (Y ) is a - nite measure space. Denoting ( F) =(X Y ), we identify Mb ( j Fj) with the ideal in Mb ( F) consisting of all functions of the form f1 j with f 2 Mb ; ( F). If 0 f 2 Mb ( F), then j f1 j is a disjoint order bounded sequence in A, so we can de ne (f) = X ; f1 j . j It is easily veri ed that this de nes a -order continuous surjective f-algebra homomorphism :Mb ( F) ! A, which is compatible with B, as all j are compatible with pj ; B . Note that f1 j = pj (f) for all f 2 Mb ; ( F). For f 2 Mb ( F) we will from now on consider the function Rj f1 j as an element of Mb ( F).The commutation relation j Rj = P pj j then corresponds to ; 1 jRj f1 j = P pj j ; f1 j = P pj [pj (f)] (18) for all f 2 Mb ( F). For j =12::: wechoose 0 uj 2 Mb (X ) such that B (uj) =P (pj). Take f 2 Mb ( F) and put g = (f). De ne hj 2 Mb (X ) by ; hj = ujRj f1 j . ; ; Since 0 P (pj) pj, Rj f1 j 2 B and hj = P (pj) Rj f1 j , it follows that hj 2 B (pj). Furthermore, using (18) and the de nition of P pj (see Lemma 4.14), we nd pjhj = ; ; ; 1 j P (pj) Rj f1 j = P (pj) 1 jRj f1 j = P (pj) P pj (pjg) =P (pj) pjP (pj) ;1 P (pjg) 33 = pjpjP (pjg) =pjP (pjg).
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: Proposition 4.12 Let A beaDedekind
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 and 48: theorem that B 1 K 1 and 1 (g) = (g
Page 49 and 50: A is a Dedekind complete f-algebra
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

Representations of positive projections 1 Introduction - Mathematics ...

Create successful ePaper yourself

Delete template?

Save as template?