SPECTRAL THEORY FOR ORDERED SPACES
A Thesis Submitted in Partial Fulfilment of the Requirements for the Degree of Master of Science
by G. PRIYANGA
to the School of Mathematical Sciences National Institute of Science Education and Research
Bhubaneswar 16-05-2017 DECLARATION
I hereby declare that I am the sole author of this thesis, submitted in par- tial fulfillment of the requirements for a postgraduate degree from the National
Institute of Science Education and Research (NISER), Bhubaneswar. I authorize NISER to lend this thesis to other institutions or individuals for the purpose of scholarly research.
Signature of the Student Date: 16th May, 2017
The thesis work reported in the thesis entitled Spectral Theory for Ordered
Spaces was carried out under my supervision, in the school of Mathematical Sciences at NISER, Bhubaneswar, India.
Signature of the thesis supervisor
School: Mathematical Sciences Date: 16th May, 2017
ii ACKNOWLEDGEMENTS
I would like to express my deepest gratitude to my thesis supervisor Dr. Anil Karn for his invaluable support, guidance and patience during the two year project period. I am particularly thankful to him for nurturing my interest in functional analysis and operator algebras. Without this project and his excellent courses, I might have never appreciated analysis so much.
I express my warm gratitude to my teachers at NISER and summer project supervisors, for teaching me all the mathematics that I used in this thesis work. Ahugethankstomyparentsfortheircontinuoussupportandencouragement.I am also indebted to my classmates who request the authorities and extend the deadline for report submission, each time!
iii ABSTRACT
This thesis presents an order theoretic study of spectral theory, developed by Alfsen and Shultz, extending the commutative spectral theorem to its non- commutative version. In this project, we build a spectral theory and functional cal- culus for order unit spaces and explore how this generalised spectral theory fits in the scheme of commutative case (function spaces) as well as the non-commutative case (Jordan Algebras). Amotivatingexampleforthistheorycomesfromthespectraltheoremfor monotone complete CR(X) spaces, which we study first. The goal is to extend the order theoretic ideas involved here, to a more general setting, namely to or- dered spaces. So, we begin with an order unit space A, which is in separating order and norm duality with a base norm space V. Here, we define maps called compressions, which give rise to projective units in A. Using projective units and projective faces, we develop various notions like comparability, orthogonality, com- patibility and also obtain certain lattice structures, under an assumption called standing hypothesis.Then,weconstructanabstractnotionofrangeprojection in A, which leads to a spectral decomposition result and functional calculus, for spaces satisfying a spectral duality condition. In the last section, we focus onto the spectral theorem for JBW-algebras and understand how the spectral theory works in non-commutative framework. Along the way, we also see concrete example of the order theoretic constructions developed above.
iv Contents
Introduction 1
1 Spectral Theory for Function Spaces 5
1.1 Range Projections ...... 7 1.2 Spectral Theorem ...... 11
2 Order Structure 15
2.1 Ordered Vector Spaces ...... 15 2.1.1 OrderUnitSpaces ...... 17 2.1.2 BaseNormSpaces ...... 22
2.2 Duality between Order Unit Space and Base Norm Space ...... 24
3 Spectral Theory for Ordered Spaces 30 3.1 Projections ...... 31
3.1.1 TangentSpacesandSemi-exposedFaces ...... 31 3.1.2 Smooth Projections ...... 35 3.2 Compressions ...... 41
3.2.1 ProjectiveUnitsandProjectiveFaces ...... 47 3.3 Relation between Compressions ...... 54
3.3.1 Comparability ...... 54 3.3.2 Orthogonality ...... 55 3.3.3 Compatibility ...... 57
v CONTENTS
3.4 The Lattice of Compressions ...... 65
3.4.1 The lattice of compressions when A = V ⇤ ...... 81 3.5 Range Projections ...... 87 3.6 SpacesinSpectralDuality ...... 91
3.7 Spectral Theorem ...... 107 3.7.1 Functional Calculus ...... 111
4 Spectral Theory for Jordan Algebras 114
4.1 OrderUnitAlgebras ...... 115 4.1.1 CharacterisingOrderUnitAlgebras ...... 122 4.1.2 Spectral Result ...... 127
4.2 JordanAlgberas...... 129 4.2.1 The Continuous Functional Calculus ...... 136 4.2.2 Triple Product ...... 140
4.2.3 Projections and Compressions in JB-algebras...... 145 4.2.4 Orthogonality ...... 150 4.2.5 Commutativity ...... 151
4.3 JBWalgebras ...... 158 4.3.1 Range Projections ...... 166 4.3.2 Spectral Resolutions ...... 172
5 Appendix 176
vi Introduction
The theory of von-Neumann algebras can be seen as a non-commutative generali- sation of integration theory. More precisely, as the lebesgue integral of a function is approximated by a sequence of simple functions, every self-adjoint element of a von-Neumann algebra can be written as a limit of finite linear sum over orthogonal projections. This idea is known as Spectral Decomposition Theory.Theobjective of this thesis work is to understand the order theoretic aspect of spectral theory, in the commutative case as well as non-commutative case.
The generalised spectral theory for ordered spaces, presented in this thesis, was originally developed by Erik M. Alfsen and Frederic W. Shultz, in the late 20th century. Motivated by structures in quantum mechanics, Alfsen and Shultz were interested in characterising the state space of operator algebras, particularly
Jordan algebras and C⇤-algebras. As part of this larger project, they established aspectraltheoryandfunctionalcalculusfororderunitspaces,whichgeneralised the corresponding results for von-Neumann algebras and JBW-algebras. This generalised theory seems to have applications in quantum mechanics. For instance, in the standard algebra model of quantum mechanical measurement, observables are represented by self-adjoint elements of a von Neumann algebra M and the states are represented by the elements in the normal state space K of M. Under the new generalisation, the basic concepts of this theory can be studied in a broader order-theoretic context: by representing states as elements in the distinguished base K of a base norm space V and observables by elements in the
1 CONTENTS
order unit space A = V ⇤. However, in my thesis work, the interest is not in state space characterisation or investigating its applications in physics. The focus of this project is developing ageneralspectraltheoryforsuitableorderedspacesandunderstandinghowthis extends the commutative spectral theorem to its non-commutative version. In this project, we mainly work with an order unit space (A, A+, e)whichisin separating order and norm duality with a base norm space (V,V+,K).Oninves- tigating the spectral theorem for monotone complete CR(X) spaces, which is one of the primary motivating example for this theory, one learns that projections and their orthogonality is the fundamental object involved in developing the spectral result. In order to construct these notions in a more general setting, we define maps, called compressions, on the order unit space A. The image of the order unit e under these compressions (known as projective unit) behave like projections in
A. We then explore various properties of the projective units and projective faces and develop notions of comparability, orthogonality and compatibility between compressions.
We further find that under an assumption called standing hypothesis,thesetof compressions on A form a lattice. Using the properties of this lattice structure, we then construct order theoretic tools, called range projections,analogoustothose in CR(X) spaces. Finally, we specialise to spaces in spectral duality,wherethese range projections lead to a spectral decomposition theory and functional calculus on A.
So, we obtain an abstract spectral theorem for order unit spaces. But how do these order theoretic constructions look in particular cases? A commutative example of this theory is seen in the case of monotone complete CR(X) spaces. An- other concrete example of this theory appears in the non-commutative framework of Jordan algebras. So, the last part of this thesis is devoted to understanding
2 CONTENTS the spectral theory for JBW-algebras. Here, we study about JB-algebras through order unit algebras, see examples of concrete compression (Up)andlearnhowthe spectral theory works in a non-commutative setting. In summary, we develop an order theoretic model of spectral theory for order unit spaces and explore how this generalised spectral theory fits in the scheme of commutative case (function spaces) as well as the non-commutative case (Jordan Algebras).
We now briefly describe the contents of each chapter.
Chapter 1 presents the order theoretic study of spectral theorem for monotone complete CR(X) spaces (continuous functions on a compact Hausdor↵space X). The spectral result and range projections constructed in this chapter are again used in Chapter 4, while developing the spectral theory for JBW-algebras. This material is mostly based on Alfsen and Shultz’s first book [2] (chapter 1).
Chapter 2 introduces order structure on vector spaces. Here, we explore vector order, cones, order unit space and base norm space. This is followed by a discussion on duality: dual pair, order duality, norm duality and then we show that order unit spaces and base norm spaces are dual to each other.
Chapter 3 focusses on the construction and development of the main spectral theorem. It begins with definitions and results on tangent spaces, semi-exposed faces and smooth projections in cones. This is followed by a discussion on general compressions, associated projective units, projective faces and relations between compressions: comparability, orthogonality and compatibility. Then, we study lattice structure on compressions, construction of abstract range projections and characterisation of spectral duality. We conclude by presenting the generalised spectral decomposition result and functional calculus for order unit spaces. This
3 CONTENTS content is largely based on Alfsen and Shultz’s second book [3] (chapters 7,8).
The last chapter is devoted to the spectral theory for JBW-algebras. Starting from the theory of order unit algebras, we obtain a continuous functional calculus on JB-algebras. Then, we look at triple products, construct concrete compressions and develop notions of orthogonality and commutativity in JB-algebra . Finally, we enter into JBW-algebras: basic definitions and relevant topologies. Here, we con- struct abstract range projections and obtain a spectral theorem for JBW-algebras, derived from the spectral theorem for monotone complete CR(X) . The topics on JB-algebra and JBW-algbera are presented from chapters 1, 2 of [3], while the section on order unit algebra is based on chapter 1 of [2].
Finally, a caveat: any errors found in this thesis are entirely my own!
4 Chapter 1
Spectral Theory for Function Spaces
We begin with the study of a spectral theory which is valid for a class of spaces of the form CR(X). The spectral theory for function spaces gives an example of spectral decomposition in the commutative case and serves as a motivation for the general theory. So, this chapter is devoted to an order theoretic understanding of the spectral theorem for monotone complete CR(X) spaces. The vector lattice
CR(X) is not monotone complete in general. However, certain important repre- sentation theorems for abstract algebras (such as the commutative von Neumann algebras and the normed Jordan algebras known as JBW-algebras) give rise to compact Hausdor↵spaces X, for which CR(X) is monotone complete. In the following pages, we will show that every element of a monotone com- plete CR(X) space can be approximated in norm by a linear span of projections. Further, the family of projections associated with a given element is unique and is characterised by certain order properties (Theorem 1.13). The objective of the project is to use order theoretic ideas and generalise this spectral theorem to a larger class of order unit spaces.
Let X be a compact Hausdor↵space. Let CR(X) denote the set of all continuous
5 1 Spectral Theory for Function Spaces
real valued functions on X.DefineapartialorderonCR(X) as follows:
f g f(x) g(x), x X () 8 2
1 Then, CR(X) forms a lattice with the following lattice operations
1 1 f g = (f + g + f g ),f g = (f + g f g ), for f,g C (X) _ 2 | | ^ 2 | | 2 R
Here, f denotes the usual modulus function f : X R defined as: | | | | ! f(x)iff(x) 0 f (x)= | | 8 f(x)iff(x) < 0 < Notation 1. Let f C (X) and: let 0 denote the constant 0 function on X. 2 R + Define f = f 0andf = (f 0). _ ^
+ + Remark 1.1. If f C (X) , then f = f f , f = f + f 2 R | |
Definition 1.2. A lattice L is said to be monotone complete if every bounded increasing (decreasing) net has a least upper bound (greatest lower bound) in L.
Example 1.3. If X = 1, 2 ...100 with the discrete topology, then C (X)is { } R monotone complete.
Throughout this chapter, we will assume that X is a compact Hausdor↵space and CR(X) is monotone complete. Infact, CR(X) is monotone complete precisely when X is extremally disconnected (i.e. the closure of every open set is open). One of the motivation to study such spaces, comes from the theory of von-Neumann algebra. For example, if A is a von-Neumann algebra and a A, then the spectrum 2 of a,denotedby (a), is extremally disconnected. Hence, the space CR( (a)) is monotone complete.
Remark 1.4. Every von-Neumann algebra is monotone complete.
1If f C (X) , then f also belongs to C (X) 2 R | | R
6 1 Spectral Theory for Function Spaces
Our goal is to develop a spectral result for monotone complete CR(X) spaces. We begin with some notations and definitions.
+ Notation 2. C (X) = f C (X) f 0 R { 2 R | } Notation 3. Let E X.Then E : X R is defined as ✓ ! 1ifx E (x)= 2 E 8 < 0otherwise Notation 4. Let f C (X) . The face: generated by f + in C (X)+, denoted by 2 R R face (f +), is the set g C (X) 0 g f +, for some 0 . { 2 R | } 1.1 Range Projections
Our first proposition proves the existence of range projections (defined later).
Proposition 1.5. Let X be a compact Hausdor↵space and assume that CR(X) is monotone complete. Then for each a C (X), the set E = s X a(s) > 0 is 2 R { 2 | } both closed and open in X. Further,
1. a 0 on E and a 0 on X r E. Also, E is the smallest closed subset of X such that a 0 on X r E.
+ + 2. E is the smallest closed subset of X for which Ea = a (pointwise prod- uct).
+ 3. E is the supremum in CR(X) of an increasing sequence in face (a ).
Proof. Let Y = s X a(s) > 0 .ThenY = E.Forn =1, 2, 3,... define { 2 | } fn : R [0, 1] as ! 0,x0 1 8 nx, 0 x fn(x)=> > n <> 1 1,x n > > :> 7 1 Spectral Theory for Function Spaces
Then, f a 1 is an increasing sequence in C (X), bounded above by the { n }n=1 R constant function 1. As C (X)ismonotonecomplete, f a b,forsome R { n } ! b C (X). 2 R
Claim. b = E
1 s Y = n N such that a(s) .Thisimplies(fn a)(s)=1and 2 )9 2 n hence, b(s)=1.Thisistrueforalls Y .So,b(Y ) = 1. And as b is continuous, 2 we have
b(E)=b(Y )=1
Next, fix t X r E.Choose2 c C (X)suchthatc(X) [0, 1],c(t)= 2 2 R ✓ 0,c(E)=1. If s E,then(f a)(s) 1=c(s), n.Thus,b(s) c(s), s E. 2 n 8 8 2 If s/E,then(f a)(s)=0, n.Thus,0=b(s) c(s), s/E. 2 n 8 8 2 Hence, b c.Inparticular,0 b(t) c(t)=0.So,b(t)=0.Sincet X r E 2 was arbitrary, we have
b(X r E)=0
Hence, b = E.
1 c 1 As, b is continuous, E = b (1) and E = b (0) are closed subsets of X. Hence, E is both open and closed in X.
1. As a>0onY ,wehavea 0onE = Y .Ifs/E,thens/Y = a(s) 0. 2 2 ) Thus, a 0onX r E. Next, let F be a closed subset of X such that a 0onX r F . Now, s Y = a(s) > 0= s/X r F = s F . Hence, Y F = 2 ) ) 2 ) 2 ✓ ) Y = E F .Thus,E is the smallest closed subset of X such that a 0on ✓ X r E. 2Compact Hausdor↵spaces are normal. Hence, it is possible to separate a point and a closed set by a continuous function.
8 1 Spectral Theory for Function Spaces
+ + 2. We will prove that Ea = a . For s E, (s)a+(s)=1.a+(s)=a+(s). 2 E For s/E, a(s) 0= a+(s)=0.Thus, (s)a+(s)=0.0=0=a+(s). 2 ) E Hence, a+(s)=a+(s), s X. E 8 2 + + Next, assume F is a closed subset of X such that F a = a .Then,for s Y , a(s)=a+(s) > 0. This implies (s)=1 = s F . Hence, 2 F ) 2 Y F = Y = E F . ✓ ) ✓
3. Note that a(s),sE a+(s)= 2 8 < 0, otherwise and for the f defined previously, { n} :
0,a(s) 0(i.e.s/E) 2 1 8 na(s), 0 a(s) (fn a)(s)=fn(a(s)) = > > n <> 1 1,a(s) n > Thus, f a C (X)+ and (f a>)(s) na+(s), s X. Hence, f a n 2 R n : 8 2 n na+ = f a face (a+). Further, we have shown that f a ) n 2 { n }%
b = E.Therefore, E is the supremum in CR(X)oftheincreasingsequence
+ f a 1 in face (a ). { n }n=1
2 Definition 1.6. An element p of CR(X)iscalledaprojectionifp = p .
Notation 5. For a pro jection p in C (X), denote p0 =1 p,where1isthe R function which takes the constant value 1 on X. Note that p0 is also a projection in CR(X).
Remark 1.7. Note that if p is a projection C (X), then p = for some closed 2 R E and open subset E of X.
9 1 Spectral Theory for Function Spaces
This is because if s X,thenp(s)=p(s).p(s)= p(s)=0orp(s)=1. 2 ) Therefore, p(X)= 0, 1 .DefineE = s X p(s)=1 .Thenp = . Also, { } { 2 | } E 1 1 as p is a continuous function on X,thesetsE = p (1) and X r E = p (0) are closed subsets of X. Hence, E is both open and closed in X.
Definition 1.8. Let a C (X)+.Definer(a)tobetheleastprojectionp in 2 R
CR(X)suchthatpa = a.Wecallr(a)tobetherange projection of a.
Proposition 1.5 implies the existence of range projections for positive elements of C (X)(a 0 a+ = a). Infact, for a C (X)+,wehaver(a)= R () 2 R E where E = s X a(s) > 0 . { 2 | }
Remark 1.9. If p, q are two projections in CR(X)oftheform E, F respectively (for some clopen subsets E, F of X), then p q E F . () ✓
Lemma 1.10. Let X be a compact Hausdor↵space and assume that CR(X) is monotone complete. If p is an increasing net of projections in C (X) and { ↵} R p p C (X), then p is also a projection in C (X). Similarly, if p is a ↵ % 2 R R { ↵} decreasing net of projections in C (X) and p p C (X), then p is a projection. R ↵ & 2 R
Proof. First consider p p. Note that 0 p 1, ↵ = 0 p 1. { ↵}% ↵ 8 ) Thus, p2 p. Next, 0 p p = p2 p2.Butp2 = p , ↵.Thus, ↵ ) ↵ ↵ ↵ 8 p =sup p =sup p2 p2.Therefore,p2 = p. ↵ ↵ ↵ ↵ Now, let p p.Then 1 p (1 p). Then, as above, (1 p)isa { ↵}& { ↵}% projection and hence p is also a projection.
Lemma 1.11. Let X be a compact Hausdor↵space and assume that CR(X) is monotone complete. Let a C (X) and p be a net of projections in C (X).If 2 R { ↵} R p p and p a 0, ↵, then pa 0. ↵ % ↵ 8
+ + Proof. Consider a = a a and p a =(p a) (p a) . ↵ ↵ ↵
Claim. p↵a =(p↵a)
10 1 Spectral Theory for Function Spaces
Let s X. As p is positive and takes values only 0 or 1, we have p (s)a (s)= 2 ↵ ↵ p (s)max a(s), 0 =max a(s)p (s), 0.p (s) =max a(s)p (s), 0 =(p a) (s). ↵ { } { ↵ ↵ } { ↵ } ↵ Hence, the claim.
+ Now as p a 0, we get p a =(p a) and (p a) = 0. Note that p a ↵ ↵ ↵ ↵ ↵ ! pa in C (X), because a is positive and bounded. But p a =0, ↵. Hence, R ↵ 8 + + + pa =0.So,pa = p(a a )=pa pa = pa 0. Thus, pa 0.
Lemma 1.12. Let X be a compact Hausdor↵space such that CR(X) is monotone complete. Let a C (X) and R.Ifp is a projection in C (X) such that 2 R 2 R pa p, then p 1 r (a 1)+ . Proof. Note that pa p = p(a 1) 0. Let b = a 1. Then pb 0= ) ) (pb)+ =0.But(pb)+ = pb+ (as proved in the claim of the previous lemma). Thus,
+ + + + pb =0 = p0b = b . Hence, r(b ) p0 (by proposition 1.5 (2)). Therefore, ) p 1 r (a 1)+ . 1.2 Spectral Theorem
We are now ready to prove our main spectral result for monotone complete CR(X) spaces.
Theorem 1.13. Let X be a compact Hausdor↵space such that CR(X) is monotone complete and let a CR(X). Then, there is a unique family e R of projections 2 { } 2 in CR(X) such that
(i) e a e ,e0 a e0 , R 8 2
(ii) e =0for < a ,e=1for > a k k k k
(iii) e e for <µ µ
(iv) eµ = e , R µ> 2 V 11 1 Spectral Theory for Function Spaces
The family e is given by e =1 r (a 1)+ . { } Further, for each increasing finite sequence = , ... with < a { 0 1 n} 0 k k n and n > a , define = max1 i n( i i 1) and s = i=1 i(e i e i 1 ). k k k k Then, P
lim s a =0 0 k k k k!
+ Proof. Define e =1 r (a 1) and E = s X a(s) > .Thene0 = { 2 | } + 1 e = r (a 1) = E and e =1 E = Ec . (i) By proposition 1.5, we know that
a 1 0onE ,a 1 0onEc (1.2.1) c e =0onE ,e =1onE (1.2.2)
Using the above and the fact that e is positive, we get that e (a 1) 0 on X.Thus, e a e , 8 Similarly,
c c e 0 =0onE ,e 0 =1onE (1.2.3)
Combining 1.2.3 with 1.2.1, we get
e0 a e0 , 8
(ii) When > a ,wehaveE = s X a(s) > > a = = e = k k { 2 | k k} ; ) c =1. E When < a ,wehaveE = s X a(s) a > = X = e = k k { 2 | k k } ) c =0. E
(iii) <µ = s X a(s) > s X a(s) >µ = E E = ){ 2 | }◆{ 2 | } ) ◆ µ ) c c E E = Ec Ec = e eµ. ✓ µ ) µ )
12 1 Spectral Theory for Function Spaces
(iv) Fix R.From(ii),(iii),weknowthateµ =1, µ> a and eµ 2 8 k k
e , µ> .Takeµ0 > a .Then eµ µ0 µ> is a decreasing net of pro- 8 k k { }
jections in CR(X), bounded below by e .Bymonotonecompletenessand
lemma 1.10, eµ µ0 µ0> p ,forsomeprojectionp in CR(X). Infact, { } &
p = µ> eµ and thus V p e Let E = s X p(s)=0 . Note that p(s)=1 = e (s)=1, µ> . { 2 | } ) µ 8 Hence, Ec Ec , µ> .Thus,by(1.2.1),a µ1onEc, µ> .This ✓ µ 8 8 implies a 1onEc.Thus,pa p ( p =1onEc). Now by lemma 1.12, we get p e
So, p = e .
Uniqueness:
Let f R be another family of projections in CR(X)satisfying(i),(ii), { } 2 (iii) and (iv). In particular, f a f , .Thus,bylemma1.12,weget 8
f e , (1.2.4) 8
For each R,defineE = s X e (s)=0 and F = s X f (s)= 2 { 2 | } { 2 | 0 .By(1.2.4),wegetE F . Also, by (i), } ✓
a 1onE ,a 1onEc a µ1onF ,a µ1onF c µ µ
For <µand x F ,wehavea(x) µ> = x E .Thus, 2 µ ) 2
Fµ E , µ> .Bute = Ec and fµ = F c .So,byremark1.9,weget ✓ 8 µ e f , µ> .Then,e f = f .Thus,e f , . µ 8 µ> µ 8 V Hence, f = e , R. 8 2
13 1 Spectral Theory for Function Spaces
Now, let = , ... be a finite increasing sequence of real numbers { 0 1 n} with < a and > a .DefineE = s X e (s)=1 ,i=1, 2 ...n. 0 k k n k k i { 2 | i } Then by (ii) and (iii), we see that = E E E ... E = X.Define ; 0 ✓ 1 ✓ 2 ✓ ✓ n n s = i=1 i(e i e i 1 ). Then x Ei r Ei 1 = s (x)= i. Also, as a 2 ) P c c i1onE i and a i 11onE i 1 ,wehave i 1 a i on Ei Ei 1 = EirEi 1. \ So,we get 0 s a i i 1 on Ei r Ei 1 which implies s (x) a(x) k k
i i 1 , x X.Therefore, s a . Hence, s a in norm k k 8 2 k kk k ! as 0. k k!
Corollary 1.14. Let X be a compact Hausdor↵space such that CR(X) is mono- tone complete. Then, the linear span of projections is dense in CR(X).
Proof. Given a CR(X), let e R be the unique family of spectral units asso- 2 { } 2 ciated with a.Let be the regular partition of ( a +1), ( a +1) of norm n k k k k 2( a +1) n k k ⇥ ⇤ n .Defines n = i=0 i(e i e i 1 ). Then, n, s n belongs to linear span 8 of projections and s P a in norm as n ,bytheorem1.13. n ! !1
To summarise, we have obtained a spectral result for monotone complete CR(X) space, using order theoretic ideas. The space (C (X) , )isaparticularexample R of an ordered vector space. More precisely, CR(X) is an order unit space. In the subsequent chapters, we will attempt to generalise the above spectral theorem to alargerclassoforderunitspaces.
14 Chapter 2
Order Structure
In this chapter, we introduce order structure on vector spaces. We begin with general definitions and properties of vector order and then shift our focus onto particular types of ordered spaces: namely order unit space and base norm space. The main objective of this chapter is to prove that dual of an order unit space is abasenormspaceanddualofabasenormspaceisanorderunitspace. As mentioned before, the goal of the project is to develop a spectral theory and functional calculus, for suitable order unit spaces. Showing that an order unit space is in separating order and norm duality with a base norm space, will provide us with a platform, where we can begin building the spectral theory.
2.1 Ordered Vector Spaces
Let V be a vector space over R and let V+ be a subset of V.
Definition 2.1. V+ is said to be a cone in V if: (i) v + w V +, v, w V + 2 8 2 (ii) v V +, v V +, 0 2 8 2 V+ is said to be proper if V+ -V+ = 0 .V+ is said to be generating if for \ { } each v V, v ,v V+ such that v = v v . 2 9 1 2 2 1 2
Definition 2.2. Arelation defined on V is said to be a vector order on V if
15 2 Order Structure
(i) v v, v V 8 2 (ii) u v , v w = u w ) (iii) u v = u + w v + w, w V ) 8 2 (iv) u v = u v, 0 ) 8
Definition 2.3. Arealvectorspacewithavectororderiscalledanordered vector space.
There is a correspondence between cones and vector orders on a given vector space. 1. Let V+ be a cone in V. Define a relation on V as
v w w v V + () 2
Then, defines a vector order on V, called the order obtained from V+ .
2. Conversely, if 0 is a vector order on V, then U := v V 0 0 v is a cone { 2 | } in V. And the order on V, obtained from U,isthesameas 0. Hence, each cone V+ in V is associated with a unique vector order on V.
Example 2.4. R2 is an ordered vector space, with vector order given as (a, b) (c, d)i↵a c, b d.ThisorderisobtainedfromtheconeR2+ := (a, b) R2 { 2 | a 0,b 0 }
Notation 6. Hereafter, (V, V+ ) denotes ordered vector space V, with positive cone V+ and vector order . Remark 2.5. V+ is proper is anti-symmetric. ()
Proof. Recall is said to be antisymmetric if a b, b a = a = b. { } ) ( ) Assume V+ -V+ = 0 . Now, if u v and v u,then(v u) V+ and ) \ { } 2 (u v) V+ ,whichimplies(v u) V+ .Thisimpliesv u =0 = 2 2 ) v = u. Hence, is anti-symmetric. 16 2 Order Structure
( ) Let v V+ -V+ . Then, 0 v and 0 v = v 0. As is ( 2 \ ) anti-symmetric, we have v =0.
Now, we will look at two special types of ordered vector spaces, namely order unit spaces and base norm spaces.
2.1.1 Order Unit Spaces
Throughout this section, (V, V+ ) is an ordered vector space and we assume V+ is proper.
Definition 2.6. Apositiveelemente of an ordered vector space V is said to be an order unit if for each a V, there exists 0suchthat 2 e a e (2.1.1) Remark 2.7. Existence of order unit implies V+ is generating because for any given
e+v e v + v V, we have v = v v where v = ,v = V ). 2 1 2 1 2 2 2 2
Example 2.8. Consider V = l1 (the space of all real bounded sequences), with ordering a 0 a 0, i.Thene =(1, 1, 1,...) is an order unit for V. { i} () i 8 Definition 2.9. The order unit e is said to be Archimedian if for each a V, we 2 have na e, for n =1, 2, 3 ... = a 0(2.1.2) ) Equivalently, v + ke V+ , k>0 = v V+ . { 2 8 } ) 2 One can show that an ordered vector space V, having an Archimedian order unit e,admitsanormpe,definedasfollows:
p (a)=inf 0 e a e (2.1.3) e { | }
Before we prove that pe is a norm on V, we first note some of its properties, in the form of the following remark.
17 2 Order Structure
Remark 2.10. For all u, v V , 2
(i) pe(e)=1 (ii) p (v)e v p (v)e e e (iii) e v e p (v) 1 () e (iv) 0 u v = p (u) p (v) ) e e
Proof. (i) 1e e =2e, 0 V +. Hence, p (e) 1. Suppose (1 )e e V + ± 2 e ± 2 for some 0, then it implies e V +.SinceV + is proper, must be 0. 2
Hence, pe(e)=1. (ii) By definition of infrimum, given ✏>0, 0 < <✏such that (p (v)+ )e 9 e ± v V +. Adding the positive element (✏ )e,wegetp (v)e+ e+(✏ )e v 2 e ± 2 V + = (p (v)+✏)e v V +.So, ✏>0, ✏e +(p (v)e v) V +. Hence, ) e ± 2 8 e ± 2 by Archimedian property, p (v)e v V +. e ± 2 (iii) Assume e v e.Then,bydefinition,p (v) 1. Conversely, if p (v) 1, e e then e p (v)e v p (v)e e. e e (iv) Assume 0 u v.By(ii),p (v)e v V +;= p (v)e (v u + u) e ± 2 ) e ± 2 V + = 0 (v u) p (v)e u. Also, p (v)e + u V +. Hence, ) e e 2 p (u) p (v). e e
Proposition 2.11. Let e be an Archimedian order unit for (V,V +). Assume V+ is proper. Define p (v) = inf k 0 ke v ke . Then, p is a norm on V. e { | } e
Proof. Clearly, p (v) 0, v Vandp (0) = 0. e 8 2 e (i) Suppose p (v)=0forsomev V. This implies ke v V+ , k>0. e 2 ± 2 8 Therefore, by Archimedian property, v 0andv 0. As V+ is proper, we get v =0.So,p (v) 0, v Vandp (v)=0 v =0. e 8 2 e () (ii) Let R.If =0,thenclearlype( v)= pe(v). Suppose =0.Then, 2 6 ke + pe( v) = inf k 0 ke v ke =inf k 0 ( v) V = { | } { || | | | ± 2 }
18 2 Order Structure
ke + + inf k 0 ( v) V = inf k0 0 (k0e v) V = pe(v) { | | | ± 2 } | | { | ± 2 } | | k where k0 = .So,pe( v)= pe(v), R,v V. | | | | 8 2 2 (iii) Take v ,v V. Let a = p (v ) and b = p (v ). By 2.10-(ii), we have 1 2 2 e 1 e 2 (a + b)e (v + v )=(ae v )+(be v ) V +.So,a + b inf k 0 ± 1 2 ± 1 ± 2 2 { | ke (v + v ) V + . Hence p (v + v ) p (v )+p (v ). ± 1 2 2 } e 1 2 e 1 e 2
Thus, pe is a norm on V.
Remark 2.12. If e and e0 are two Archimedian order units of V, such that pe(v)= p (v), v V ,thene = e0 (follows from remark 2.10). e0 8 2
Definition 2.13. An ordered normed linear space (A, . )issaidbeanorder k k unit space if the norm on A can be obtained from an Archimedian order unit e, as follows: a =inf 0 e a e (2.1.4) k k { | } This e is called the distinguished order unit of A. Further, this norm satisfies the property that a e a a e. k k k k
Example 2.14. Let CR(X) be the set of continuous real valued functions on a compact Hausdor↵space X. Define an ordering on C (X) as f 0 f(x) R () 0, x X.Lete denote the constant function 1 on X. Then, C (X) endowed 8 2 R with the sup norm, is an order unit space with distinguished order unit e.
Proposition 2.15. Let e be an Archimedian order unit for an ordered vector space
(A, A+ ). Then, A+ is closed in A, with respect to the topology induced from the norm pe. In particular, the positive cone in an order unit space is norm closed.
+ Proof. Let v 1 be a sequence in A and let v Asuchthatp (v v ) { n}n=1 2 e n ! 1 0asn .Foreachk N, nk such that pe(v vn) < , n nk.Then, !1 2 9 k 8 1 + + 0 vnk e + v, k N. Hence, by Archimedian property, v A .Thus,A k 8 2 2 is closed w.r.t pe.
19 2 Order Structure
Theorem 2.16. Let (A, A+ , . ) be an ordered normed linear space and let e k k 2 A+. Then, A is an order unit space with distinguished order unit e A+ is () closed in A (with respect to . ) and the following holds for each a A: k k 2
a 1 e a e (2.1.5) k k ()
Proof. If e is an Archimedian order unit for A, then denote a =inf 0 k ke { | e a e ,foreacha A. } 2
( ) Suppose A is an order unit space and e is its distinguished Archimedian order ) unit. Then . = . .Byproposition2.15,A+ is closed in A and by remark k k k ke 2.10-(iii), (2.1.5) holds.
( ) Assume A+ is closed in A and (2.1.5) holds. Take a A. If a =0,then ( 2 a e a e clearly. If a =0,definea0 = a .Then a0 = 1. Hence, by 6 k k k k (2.1.5), we have e a0 e;= a e a a e.Thus,e satisfies ) kk kk the order unit property (2.1.1) and, hence, is an order unit for A. Next, we
show that e is Archimedian. Let a Asuchthatna e for all n N,i.e, 2 2 . a e for all n N. Note that a e k k a as n .Buta e 0 n 2 { n } ! !1 n for each n and -A+ is closed, by assumption. Hence, a -A+ ;= a 0. 2 ) Thus, e is Archimedian. Now, we will show that for each a A, we have 2 a = a .Thisisclearifa =0.So,assumea =0.By(2.1.5),we k k k ke 6 have a e a a e which implies a a .Suppose a < a , k k kk k ke kk k ke k k 1 1 then ( a a) < 1. Thus, 0 < <1suchthat e a a k k k ke 9 kk 1 a a a a k k e = e e = a 1= 1= a ,whichisa ) )kk k k ) k k k k contradiction to the fact that <1. Hence, a = a . k ke k k
We close the section on order unit spaces, with some results about the dual space.
20 2 Order Structure
+ Let (A, A , e)beanorderunitspace.LetA0 denote the set of all linear
+ + + functionals on A. Define A0 = f A0 f(v) 0, v A .ThenA0 is a cone { 2 | 8 2 } in A0.
+ Proposition 2.17. Let (A, A , e) be an order unit space and let f A0. Then, 2 + f A0 f is bounded and f = f(e). 2 () k k Proof. Let us denote the order unit norm on A by p .Then f =sup f(x) e k k {| || x A, p (x) 1 . 2 e } + ( ) Assume f A0 .Takev A with p (v) 1. Then, e v e. Now, f is ) 2 2 e positive = f( e) f(v) f(e). Hence, f(v) f(e). This shows that ) | | f f(e). Hence, f is bounded. Now, p (e)=1 = f(e) f .So, k k e ) k k f = f(e). k k ( ) Assume f is bounded with f = f(e). Take v A+.Ifv =0,then ( k k 2 v f(v)=0.Ifv =0,putu = . Now, pe(u)=1andso e u e. Also, 6 pe(v) v A+ = u 0. Hence, e u e and 0 e u e.So,p (e u) 1. 2 ) e Therefore, f(e u) f e u f(e). So, f(e) f(u) f(e). Hence, k kk k + f(u) 0andthereforef(v) 0. Thus, f A0 . 2
Definition 2.18. Alinearfunctional⇢ on an order unit space (A, e)iscalleda state if ⇢ is positive and ⇢(e)=1.ThesetofallstatesonAiscalledthestate space of A and is denoted by K.
Remark 2.19. Let A⇤ denote the set of all continuous linear functionals on A and let A1⇤ be the closed unit ball of A⇤. Then, by Banach-Alaoglu theorem (A.1), A1⇤ is w⇤-compact. As K is a w⇤-closed subset of A1⇤,Kisalsow⇤-compact.
Proposition 2.20. If a is an element of an order unit space A, with state space K, then
a A+ ⇢(a) 0, ⇢ K (2.1.6) 2 () 8 2 21 2 Order Structure and
a = sup ⇢(a) ⇢ K (2.1.7) k k {| || 2 } Proof. If a A+,thenclearly⇢(a) 0, ⇢ K. Now, assume ⇢(a) 0, ⇢ K. 2 8 2 8 2 Suppose a/A+.SinceA+ is closed, then by Hahn-Banach Separation theorem 2 (A.3), there exists f A⇤ and ↵ R such that f(a) <↵and f(b) ↵ for all 2 2 b A+.SinceA+ is a cone, we can choose the separating real number ↵ to be zero. 2 f Thus f is a positive linear functional, so ⇢ := f is a state on A with ⇢(a) < 0, k k which is a contradiction. Hence, a A+. 2 To prove (2.1.7), define =sup ⇢(a) ⇢ K . As ⇢(a) ⇢ a for each {| || 2 } | |k kk k ⇢ K,wehave a .Suppose < a .Bythedefinitionoforderunitnorm, 2 k k k k either ( e + a) / A+ or ( e a) / A+.Ifsuppose( e + a) / A+,thenby(2.1.6), 2 2 2 ⇢ K such that ⇢( e + a) < 0whichimplies⇢(a) < 0, i.e. ⇢(a) < ,a 9 2 | | contradiction. Similarly, if ( e a) / A+,wegetacontradiction.So, = a . 2 k k This proves (2.1.7).
2.1.2 Base Norm Spaces
We begin with base norm spaces. Throughout this section, we assume (V, V+)is an ordered vector space and V+ is generating.
Notation 7. Let E be a real normed linear space. Then we denote the set of all continuous linear functionals on E, by E⇤.
Definition 2.21. Anon-emptyconvexsetK V+ 0 is said to be a base for ✓ \{ } V+ if for each v V + 0 , ! k K, > 0suchthatv = k. 2 \{ } 9 2 Example 2.22. If (A, A+,e)isanorderunitspace,thenthestatespaceofAis abaseforA⇤.
Remark 2.23. If K is base for V+ ,thenV+ = k k K, 0 { | 2 } Remark 2.24. If V+ has a base K, then V+ is proper.
22 2 Order Structure
Proof. Let v V + 0 with v = k , v = k where , > 0, k ,k K. ± 2 \{ } 1 1 2 2 1 2 1 2 2 1 2 Now, 0 = v +( v)= 1k1 + 2k2 =( 1 + 2)( k1 + k2)=( 1 + 2)k, for 1+ 2 1+ 2 some k K(sinceKisconvex).But,( + ) > 0= k =0.Thiscontradicts 2 1 2 ) that 0 / K. Therefore, v must be 0. 2 From remark 2.23 and 2.24, we see that if K is a base for V+ ,theneachv V 2 can be represented as v = k k ,forsome , 0,k,k K. Now 1 1 2 2 1 2 1 2 2 putting = + and k = k + k , we find that k v V +. 1 2 1 2 ± 2 Proposition 2.25. Let (V, V +) be an ordered vector space. Assume V + is gen- erating and let K be a base for V +. Define, for each v V, 2
v = inf 0 k v V +, for some k K (2.1.8) k kK { | ± 2 2 }
Then, . is a semi-norm on V. k kK Proof. Clearly, v 0, v Vand 0 =0. k kK 8 2 k kK
(i) Let ↵ R.If↵ =0,thenclearly ↵v K = ↵ v K . Now, assume ↵ =0. 2 k k | |k k 6 Then, ↵v =inf 0 k ↵v V +, for some k K k kK { | ± 2 2 } + =inf 0 ↵ ( ↵ k v) V , for some k K { || | | | ± 2 2 } + =inf 0 ↵ k v V , for some k K { | | | ± 2 2 } + =inf 0 ↵ 0 0k v V , for some k K where 0 = ↵ { | | | ± 2 2 } | | + = ↵ inf 0 0 0k v V , for some k K | | { | ± 2 2 } = ↵ v . | |k kK (ii) Let v, w V. Let k v and l w .Then, b ,b K such that (kb 2 k kK k kK 9 1 2 2 1± v) V + and (lb w) V +.So,(kb + lb ) (v + w) V +. Also, 2 2 ± 2 1 2 ± 2 ( k b + l b ) K, since K is convex. Hence, (k+l)( k b + l b ) (v+w) k+l 1 k+l 2 2 k+l 1 k+l 2 ± 2 V +.Thisimplies v + w k + l.Takinginfrimumoverk and l,weget, k kK v + w v + w . k kK k kK k kK So, . is a semi-norm on K. k kK 23 2 Order Structure
Definition 2.26. An ordered normed vector space (V, . ) with a generating cone k k V+ is said to be a base norm space if V+ is closed in V and the norm on V can be obtained as in (2.1.8) from a base K of V+ .Inthiscase,K = v V + v =1 { 2 |k k } and is called the distinguished base of V.
Remark 2.27. The closed unit ball B of a base norm space (V, V+,K)isofthe form B = co(K K). [
n Example 2.28. Consider the vector space (R , . 1), with positive cone C = k k n th (a1,a2 ...an) R ai 0, i .Defineei to be the vector whose i component { 2 | 8 } is 1 and all other components are 0. Then K := co(e1,e2 ...en)isabaseforC and (Rn,C,K)isabasenormspace.
2.2 Duality between Order Unit Space and Base Norm Space
In this section, we discuss order and norm duality and prove that order unit spaces and base norm spaces are dual to each other.
Definition 2.29. Let V, W be vector spaces over R.Let :VxW K be a ! bilinear form such that (i) (v, w)=0, v V = w =0 8 2 ) (ii) (v, w)=0, w W = v =0 8 2 ) Then, (V, W, ) is called a Dual Pair.
Notation 8. Let E be a normed linear space. Then E0 denotes the set of all linear functionals on E and E⇤ denotes the set of all bounded (or norm continuous) linear functionals on E.
Example 2.30. Let V be a real vector space. Then, V,V ⇤ form a dual pair h i under the bilinear map , : V X V ⇤ R defined as v, f = f(v). h i ! h i
24 2 Order Structure
Let (V, W, , ) be a dual pair. If (V, V+)isanorderedvectorspace,thenW+ h i := w W v, w 0, v V + defines a cone in W. { 2 |h i 8 2 } Remark 2.31. If V+ is generating, then W+ is proper.
Definition 2.32. Let (V, V+), (W, W+)betwoorderedvectorspacesoverR. V, W are said to be in separating order duality if there exists a bilinear form
, :VxW R such that h i ! (i) (V, W, , )formsadualpair h i (ii) V + = v V v, w 0, w W + { 2 |h i 8 2 } (iii) W + = w W v, w 0, v V + { 2 |h i 8 2 }
Definition 2.33. Let (V, . 1), (W, . 2)betwonormedvectorspacesoverR. k k k k V, W are said to be in norm duality if there exists a bilinear form , :VxW h i R such that ! (i) (V, W, , )formsadualpair h i (ii) For each v V, v =sup v, w w 1 2 k k1 {|h i||k k2 } (iii) For each w W, w =sup v, w v 1 2 k k2 {|h i||k k1 }
Suppose (V,W, ) form a dual pair. For each w W, define fw : V K as 2 ! fw(v)= (v, w). Then fw is a linear functional on V. Hence, W can be identified as a subset of V 0,viathemapping :W V 0 given by (w) f .Similarly, ! 7! w VcanbeidentifiedasasubsetofW 0.
Notation 9. (V,V +, . ) denotes an ordered vector space V, with norm . and k k k k positive cone V+ .
Proposition 2.34. Let (V, V+, . ) be an ordered normed linear space. Assume k k1 V+ is closed in V, with respect to the norm induced topology. Then, V is in
+ + separating order and norm duality with (V ⇤, (V ⇤) , . ), where (V ⇤) = f k k2 { 2 + V ⇤ f(v) 0, v V and f = sup f(v) v 1 for each f V ⇤. | 8 2 } k k2 {| ||k k1 } 2
25 2 Order Structure
Proof. Consider the bilinear map , : V X V ⇤ R defined as v, f = f(v). h i ! h i Then V,V ⇤ form a dual pair under , because if f(v)=0, v V, then f =0. h i h i 8 2 Conversely, if f(v)=0, f V ⇤ and v = 0, then by Hahn-Banach Theorem, there 8 2 6 exists f V ⇤ such that f(v)= v =0,whichisacontradiction.Thus,v =0. 2 k k6 + + Next, we prove separating order duality. If v V ,thenf(v) 0, f V ⇤ . 2 8 2 + + + Conversely, assume f(v) 0, f V ⇤ .Supposev/V .SinceV is closed, 8 2 2 then by Hahn-Banach Separation theorem (A.3), there exists g V ⇤ and ↵ R 2 2 such that g(v) <↵and g(u) ↵ for all u V +.SinceV+ is a cone, we can 2 choose the separating real number ↵ to be zero. Thus, g is a positive linear
+ + functional. This contradicts the fact that f(v) 0, f V ⇤ . Hence, v V . 8 2 2 + + + So, V = v V f(v) 0, f V ⇤ . And, by definition, (V ⇤) = f { 2 | 8 2 } { 2 + V ⇤ f(v) 0, v V .Therefore, V,V ⇤ are in separating order duality. | 8 2 } h i Now, we prove norm duality. Let v V. Clearly, v sup f(v) f 2 k k1 {| ||k k2 1 ,because f(v) f v . And by Hahn-Banach theorem, there exists f } | |k k2k k1 2 V ⇤ such that f 1andf(v)= v .Thus, v =sup f(v) f 1 . k k2 k k1 k k1 {| ||k k2 } And, by definition, f =sup f(v) v 1 for each f V ⇤. Hence, k k2 {| ||kk1 } 2 V,V ⇤ are in norm duality. h i
We will now prove that an order unit space is in separating order and norm duality, with a base norm space.
Theorem 2.35. Let (V, V+, e) be an order unit space. Define S(V) = f { + + V ⇤ f = f(e)=1. Then (V ⇤,V⇤ ,S(V )) is a base norm space and 2 |kk } + + (V,V ,e), (V ⇤,V⇤ ,S(V )) are in separating order and norm duality. h i
Proof. Let the order unit norm on V be denoted by pe.ItiseasytoseethatS(V )
+ is a base for V ⇤ . And as S(V )isw⇤-compact, the semi-norm induced by S(V ) becomes a norm on V ⇤.
Let the dual norm on V ⇤ be denoted by f =sup f(v) p (v) 1 . k k {| || e }
26 2 Order Structure
+ Clearly, V ⇤ is closed in V ⇤,withrespectto . . k k Let us denote the norm on V ⇤,inducedbythebaseS(V ), as f =inf k kB { + 0 g f (V ⇤) ,forsomeg S(V ) . | ± 2 2 } We need to prove that f = f ,foreachf V ⇤.Thisisclearlytruefor k k k kB 2 f =0.So,assumef =0.Forsimplicity,letusdenoteS(V )byS. 6 (i) Let v V such that p (v) 1. Then, e v e.Thisimplies0 e v 2 e + e and 0 e + v 2e.Suppose g f (V ⇤) for some g S, > 0. Then, ± 2 2 ( g + f)(e v)= g(e) g(v)+f(e) f(v) 0 ( g f)(e + v)= g(e)+ g(v) f(e) f(v) 0 Adding both, we get 2 2f(v) 0; = f(v). Similarly, replacing v ) by v,weget f( v). This implies f(v) .So, | | + inf 0 g f (V ⇤) ,forsomeg S sup f(v) p (v) 1 . { | ± 2 2 } {| || e } Hence, f f k kB k k
f (ii) Let h = f . Then, h V1⇤ = co(S S). So, h = ↵g1 (1 ↵)g2,forsome k k 2 [ g ,g S,↵ [0, 1]. Take g = ↵g +(1 ↵)g S (since S is convex). Now 1 2 2 2 1 2 2 + + g h =2↵g ,2(1 ↵)g (V ⇤) .Thisimpliesg h V ⇤ = f g f ± 1 2 2 ± 2 )kk ± 2 + (V ⇤) = f f . )kk k kB
+ Thus, f = f ,forallf V ⇤. Hence, (V ⇤,V⇤ ,S(V )) is a base norm space. k kB k k 2 Now, by proposition 2.34, it follows that the order unit space (V,V +,e)isin
+ separating order and norm duality with the base norm space (V ⇤,V⇤ ,S(V )).
Next, we show that a base norm space is in separating order and norm duality, with an order unit space.
+ Theorem 2.36. Let (V, V , K) be a base norm space. Define eK :V R as ! + e (k) = 1, k K and then extended to V by linearity. Then (V ⇤,V⇤ ,e ) K 8 2 K + + is an order unit space with order unit e and (V,V ,K), (V ⇤,V⇤ ,e ) are in K h K i separating order and norm duality.
27 2 Order Structure
Proof. Let the base norm on V be denoted by . . And let the dual norm on V ⇤ k kK be denoted by f =sup f(v) v 1 . k k {| ||k kK }
Claim. eK is an Archimedian order unit for V⇤. Let v V+ .Then k K, 0suchthatv = k.So,e (v)= e (k)= 2 9 2 K K + + + 0. Hence, e (v) 0, v V .So,e (V ⇤) . Now, let f (V ⇤) .Define K 8 2 K 2 2 =sup f(k) k K . Then, ↵ 0,k K,wehave( e f)(↵k)= {| || 2 } 8 2 K ± + ↵( f(k)) 0. Therefore, ( e f) (V ⇤) . Hence e is an order unit for V ⇤. ± K ± 2 K Next, we will show that e is Archimedian. Suppose f V ⇤ such that f K 2 1 1 eK , n N.Then,foreachk K,wehavef(k) , n which implies n 8 2 2 n 8 f(k) 0. Thus, f(v) 0, v V + and so f 0. Hence, proved. 8 2
Let the order unit norm on V ⇤ be denoted by f =inf 0 e f k ke { | K ± 2 + (V ⇤) . Now, we will show that f = f ,forallf V ⇤. } k k k ke 2 (i) Let v V+ .So,v = k,forsome 0,k K. Now, ( f e f)(v)= 2 2 k k K ± + ( f e f)( k)= [ f f(k)] 0. Therefore, ( f e f) (V ⇤) k k K ± k k± k k K ± 2 which implies f f k k k ke
+ (ii) Suppose e f (V ⇤) ,forsome 0. Then, ( e f)(k) 0, k K, K ± 2 K ± 8 2 which implies f(k) , k K.Consider,v V such that v 1. | | 8 2 2 k kK Since V =co(K -K), there exists k ,k K such that v = ↵(k ) (1 1 [ 1 2 2 1 ↵)(k ), for some ↵ [0, 1]. Now, f(v) ↵ f(k ) +(1 ↵) f(k ) 2 2 | | | 1 | | 2 | sup f(k) k K . This implies, sup f(v) v 1 .So, {| || 2 } {| ||k kK } + inf 0 e f (V ⇤) sup f(v) v 1 which implies { | K ± 2 } {| ||kkK } f f . k ke k k
+ Hence, f = f .So,(V ⇤,V⇤ ,e )isanorderunitspace. k k k ke K Now, by proposition 2.34, it follows that the base norm space (V,V +,K)isin
+ separating order and norm duality with the order unit space (V ⇤,V⇤ ,eK ).
28 2 Order Structure
So, we have shown that dual of an order unit space is a base norm space and vice-versa. The following are some observations, arising from this duality.
Proposition 2.37. Let (A, A+, e), (V, V+, K) be a pair of order unit space and base norm space in separating order and norm duality under a bilinear form <, >. Then
(i) e, v = v , v V+ h i k k 8 2
(ii) v + v = v + v , v ,v V+ k 1 2k k 1k k 2k 8 1 2 2
(iii) K = v V + e, v =1 { 2 |h i }
+ (iv) If T is a positive linear map from A to A, then Te,v = T ⇤v , v V h i k k 8 2 Here T ⇤ is the adjoint map on V, satisfying T a, v = a, T ⇤v , for each h i h i a A, v V . 2 2
Proof. (i) Let v V+ and let a A such that a 1. This implies e a 2 2 k k e;= e, v a, v e, v .So, a, v e, v . Also, e = 1. Hence, )h ih ih i |h i| h i k k e, v sup a, v a A, a 1 e, v .So, v = e, v . h i {|h i| | 2 k k }h i k k h i
(ii) By (i), v + v = e, v + v = e, v + e, v = v + v . k 1 2k h 1 2i h 1i h 2i k 1k k 2k
(iii) By definition, K = v V + v =1 .Therefore,by(i),K= v V + { 2 |k k } { 2 | e, v =1 . h i }
+ + (iv) Tispositiveandv V = T ⇤v V . Hence, by (i), Te,v = { 2 } ) 2 h i e, T ⇤v = T ⇤v . h i k k
29 Chapter 3
Spectral Theory for Ordered Spaces
In this chapter, we develop a spectral theory and functional calculus for an order unit space A, which is in separating order and norm duality with a base norm space V, whose distinguished base is K. So, A may also be considered as the set of continuous a ne functions on the compact convex set K. The main ingredients of spectral theory are projections and their orthogonality, as observed in the mo- tivating example of spectral theory for function spaces. Here, we develop these ideas in a more general setup, namely for order unit spaces. We fit this scheme both in the commutative case (function spaces) as well as non-commutative case (Jordan algebras). To construct projections in A (not on A), we define certain smooth maps, called compressions.Theimageoftheorderunitunderacompressiongivesrise to projective unit,whichgeneralisetheideaofprojectionsinCR(X) spaces and JB algebras. Then, we study various properties and characterisations of projective units and projective faces and develop notions like comparability, orthogonality and compatibility on compressions. After this, we show that the set of compressions on A form a lattice, under an assumption called standing hypothesis.Weusepropertiesofthislatticestructure for spaces in spectral duality and construct order theoretic objects called range
30 3 Spectral Theory for Ordered Spaces projections. These range projections then give rise to a spectral decomposition theory and functional calculus on A; and characterise the spectral family associated with each element in A.
3.1 Projections
In this section, we study smooth projections on ordered vector spaces. To begin this, we require a few geometric objects, discussed below.
3.1.1 Tangent Spaces and Semi-exposed Faces
Throughout this section, X is a real vector space.
Definition 3.1. Let C be a cone of an ordered vector space (X, X+). A non- empty convex subset F of C is said to be a face of C if the following condition is satisfied: x, y C and ↵x +(1 ↵)y F for some ↵ (0, 1) = x, y F 2 2 2 ) 2
Remark 3.2. FisafaceofX+ FisasubconeofX+ and for every given () x F, 0 y x = y F 2 ) 2
Proof. Given: F X+,F = . ✓ 6 ; ( ) Let F be a face of X+.Letx F . Now, 1 (0) + 1 (2x)=x F . Hence, ) 2 2 2 2 0 F . 2 Claim. x F, 0 2 8 First assume 1. Then, 1 ( x)+(1 1 )0 = x F .Therefore, x F . 2 2 Next, consider 0 1. Then, x = x+(1 )0 (convex combination) = ) x F .So, x F, 0,x F . Hence, the claim. 2 2 8 2 Now, consider x, y F .Thisimplies1 x + 1 y F .Then,bytheabove 2 2 2 2 claim, 2( x+y ) F ;= x + y F . Hence, F is a subcone of X+. Now, let 2 2 ) 2
31 3 Spectral Theory for Ordered Spaces
x F and 0 y x.Then,y, (x y) X+ and 1 (x y)+ 1 y = 1 x F . 2 2 2 2 2 2 This implies y F . 2 ( ) Let F be a subcone of X+ and assume that for every given x F, we have ( 2 0 y x = y F . As F is a subcone, it is a convex set. And ) 2 if x, y X+ such that ↵x +(1 ↵)y = z F for some ↵ (0, 1), then 2 2 2 0 ↵x, (1 ↵)y z.So,↵x, (1 ↵)y F which implies x, y F (asFis 2 2 a subcone). Hence, F is a face of X+.
Definition 3.3. Let C be a cone in X (with associated ordering ). Face of C generated by x,denotedbyFaceC (x), is the smallest face of C, containing x. By remark 3.2, Face (x)= b C b kx, for some k 0 C { 2 | } Definition 3.4. Apointx in a convex set K X is said to be an extreme point ✓ if there is no convex combination x = ↵y +(1 ↵)z with y = x, z = x and 6 6 0 <↵<1. Equivalently, x X is an extreme point of a convex set K if Face (x) 2 K = x . The set of all extreme points of K is called the extreme boundary of K and { } we will denote it by @eK.
1 Definition 3.5. Xiscalledahyperplane of X if = f (↵)forsome H✓ H non-zero linear functional f on X, ↵ R.(i.e. has codimension 1) 2 H If is a hyperplane of an ordered vector space (X, X+), then it splits X into two H 1 1 half spaces, namely U = a X a f (↵) and U = a X a f (↵) 1 { 2 | } 2 { 2 | } 1 where = f (↵). H Definition 3.6. Let (X , )beanorderedvectorspacewithatopologyandlet C be a convex subset of X. Then is called a supporting hyperplane of C if H H is a hyperplane of X such that C is contained in one of the half spaces of and H @C = . Here, @C is the boundary of C (@C:=C C )withrespecttothe \H6 ; \ topology on X.
32 3 Spectral Theory for Ordered Spaces
Hereafter (X, X+), (Y,Y +) is a pair of positively generated ordered vector h i spaces in separating order duality under a bilinear form <, >. The topology on X, Yistheweaktopologydefinedbythegivenduality,i.e.foranet x X, { ↵}2
x x x ,y x, y , for each y Y ↵ ! () h ↵ i !h i 2
Definition 3.7. If F is a subset of a convex set C X, then the intersection of all ✓ closed supporting hyperplanes of C which contain F, is called the tangent space of
CatFandisdenotedbyTanC F. If there is no supporting hyperplane of C which contains F, then TanC F=X,byconvention.
Definition 3.8. AfaceFofaconvexsetC Xissemi-exposed if there exists a ✓ collection of closed supporting hyperplanes of C containing F such that F = H
C ( H H ). F is said to be exposed if can be chosen to consist of a single \ 2H H hyperplane,T i.e. F = C H,forsomeclosedsupportinghyperplaneH of C. \ Remark 3.9. Fissemi-exposed F=C Tan F () \ C Definition 3.10. Let B X. The annihilator of B in the space Y is denoted by ✓ + B := y Y x, y =0, x B and positive anhillator of B = B• =B Y { 2 |h i 8 2 } \ = y Y + x, y =0, x B . { 2 |h i 8 2 } Proposition 3.11. Let B X+.Then X is a closed supporting hyperplane ✓ H✓ + 1 of X , containing B = y (0) for some y B•. ()H 2
1 Proof. Recall that y (0) = x X x, y =0 { 2 |h i }
1 ( ) Let = y (0) for some y B•. As y is a continous linear functional ( H 2 1 + + on X, y (0) is a closed hyperplane of X. Since y Y , X x X 2 ✓{ 2 | x, y 0 .So,X+ is contained in one of the half spaces of . Note that h i } H 0 X+ (X+)C = @X+. Also, 0 . Hence, 0 @X+ .So, is 2 \ 2H 2 \H H + asupportinghyperplaneofX . As y B•,wehave x, y =0, x B. 2 h i 8 2 Therefore, B . ✓H 33 3 Spectral Theory for Ordered Spaces
( ) Suppose XisaclosedsupportinghyperplaneofX+,containingB.Let ) H✓ 1 = f (t) for some non-zero linear functional f on X, t R.Since is H 2 H closed, f is continuous 1. By Hahn-Banach Theorem, the weak dual of X is
1 Y. So, y Y such that f(y)= x, y , x X. Therefore, = y (t). As 9 2 h i 8 2 H is supporting, without loss of generality, we may assume X+ x X H ✓{ 2 | x, y t .So,0=0,y t. This implies 0 t.Letx @X+ . h i } h i 0 2 \H + + + + Now, x @X = X (X ) = x 1 X such that x x 0 2 \ )9{n}n=1 2 n ! 0
as n .So,t limn 2xn,y =2limn xn,y =2x0,y =2t. !1 !1h i !1h i h i 1 So, t 2t;hencet 0. So, t =0.Thisimplies = y (0). Now, H 1 B = y (0) = x, y =0, x B. This implies y B .Ifnecessary, ✓H )h i 8 2 2 replacing y by y,wehaveX+ x X x, y 0 which implies y 0. ✓{ 2 |h i } Therefore, y B•. 2
1 Remark 3.12. If B X, then B = x B x (0). ✓ 2 T Proposition 3.13. Let B X+. Then ✓
+ (i) B• is a semi-exposed face of Y
+ (ii) B• is the tangent space to X at B
+ (iii) B•• is the smallest semi-exposed face of X containing B
1 Proof. (i) B = y Y x, y =0, x B = x B x (0). Therefore B { 2 |h i 8 2 } 2 is the intersection of a collection of closed supportingT hyperplanes of Y +
+ containing B• and B• = Y B . Hence, B• is semi-exposed. \
1 (ii) By 3.11, TanX+ B = y B y (0) = B• 2 • T + (iii) We have B B•• and B•• is a semi-exposed face of X (replacing B by ✓ + B• in part (i)). Let S be another semi-exposed face of X ,containingB.
1a property of topological vector space
34 3 Spectral Theory for Ordered Spaces
This implies S = X+ ,where is the intersection of a collection of \H H closed supporting hyperplanes of X+ containing S. Since B S, is the ✓ H intersection of some closed supporting hyperplanes of X+ containing B. So,
1 1 + + y B y (0) .Thisimplies y B y (0) X X ;hence 2 • ✓H 2 • \ ✓H\ BT•• S. T ✓
3.1.2 Smooth Projections
Now, we will start exploring some properties of a positive projection map P, defined on an ordered vector space (X, X+), which is in separating order duality with
(Y,Y +). Also, recall that by continuous projection, we refer to the weak topology on X and Y, arising from this duality.
Definition 3.14. AlinearmapP:X XsuchthatP2 =PandP(X+) X+ ! ✓ is called a positive projection on P.
Notation 10. ker P = x X P (x)=0 and im P = x X Px = x . And { 2 | } { 2 | } we define ker+P=kerP X+ and im+P=imP X+ \ \ Remark 3.15. If P, R are two projections such that ker P ker R and im P = im ✓ R, then P = R.
Proof. Let x X. Then x =(x Px)+Px.Thisimplies,Rx = R(x Px)+R(Px). 2 Now, (x Px) ker P ker R and Px im P = im R. Therefore, Rx =0+Px; 2 ✓ 2 = Rx = Px, x X. Hence, P = R. ) 8 2
Remark 3.16. If R : X X is a positive projection on X and R(X+):= Rx ! { | x X+ ,thenim+R=R(X+), i.e. R(X) X+ = R(X+). 2 } \ This is because if v im+R,thenv 0andv = Rv which implies v R(X+). 2 2 Conversely, if v R(X+), then v = Ru for some u X+.SinceR is positive, we 2 2 have v = Ru 0. Hence, v im+R. 2 35 3 Spectral Theory for Ordered Spaces
Remark 3.17. ker+PisafaceofX+. And since P is positive, im P = im+P- im+P, i.e. im P is a positively generated linear subspace of X.
Definition 3.18. If P is a (continuous) positive projection on X, then the adjoint map P ⇤ : Y Y satisfying Px,y = x, P ⇤y , x X, y Y is called the ! h i h i 8 2 2 (continuous) dual projection of P in Y.
Proposition 3.19. (ker P) = im P⇤
Proof. First note that both (ker P) and im P⇤ are subsets of Y.
( ) Let y (kerP) .Foranyx X, we have x Px ker P. This implies ✓ 2 2 2 x Px,y =0. So, x, y = Px,y = x, P ⇤y , x X. So, y = P ⇤y. h i h i h i h i 8 2 Hence, y im P⇤. 2 ( ) Let y = P ⇤y and let x ker P. Now x, y = x, P ⇤y = Px,y = 0,y =0. ◆ 2 h i h i h i h i Hence, y (kerP) . 2
Proposition 3.20. If P is a continuous positive projection on X, then
+ + (i) Tan(ker P) = (ker P )• kerP ✓
(ii) ker+P is a semi-exposed face of X+
+ + Proof. (i) Taking B = ker P in 3.13, we get T an(ker P )=(ker P )• . Now,
+ + + ker P ker P = (ker P )• (kerP)• = (ker P )• (ker P)• . ✓ ) ◆ ) ✓ + + Also (ker P) = im P ⇤ = im P ⇤ im P ⇤ =(ker P)• (ker P)• = ) (ker P)• =(ker P) .SincePiscontinuous,kerPisaclosedconvexsetof
+ X. Hence, by Bipolar theorem (A.4) ker P = (kerP) . So, Tan(ker P) =
+ (ker P )• (ker P)• =(ker P) = ker P. ✓
(ii) ker+P X+ T an(ker+P ). By (i), X+ T an(ker+P ) X+ ker P = ✓ \ \ ✓ \ ker+P .So,ker+P = X+ T an(ker+P ). Hence, it is semi-exposed. \
36 3 Spectral Theory for Ordered Spaces
Complement of a Projection
Let P and Q be two continuous positive projections on X.
Definition 3.21. P, Q are said to be complementary if ker+P=im+Qandker+Q
=im+P. Q is said to be a complement of P and vice-versa. P,Q are said to be strongly complementary if ker P = im Q and ker Q = im P.
Remark 3.22. P,Q are complementary = PQ = QP = 0. ) Remark 3.23. P, Q are strongly complementary PQ = QP =0,P+ Q = I. () Proof. ( ) Assume ker P = im Q and ker Q = im P. Let x X.ThenPQ(x)= ) 2 P (Qx)=0asQx im Q =kerP . Hence, PQ =0.Similarly,QP =0. 2 Now, (x Px)+Px = x, x X.(x Px) ker P =imQ.Therefore, 8 2 2 x Px = Qy for some y X. Applying Q on both sides, Qx QP x = 2 Q2y = Qx +0=Qy = x Px = Px+ Qx = x.Sincethisistrue ) ) x X, we have P + Q = I. 8 2
( ) Assume PQ = QP =0andP + Q = I. ( Claim. im Q = ker P
( ) Let y im Q = y = Qx for some x X.Then,Py = P (Qx)= ✓ 2 ) 2 PQ(x)=0 = y ker P . Hence, im Q ker P . ) 2 ✓ ( ) Let y ker P . P + Q = I = Py + Qy = y = 0+Qy = y = ◆ 2 ) ) ) y im Q. Hence, im Q ker P . 2 ◆
So, ker P = im Q. Similarly, ker Q = im P. Hence, P,Q are strongly comple- mentary.
Remark 3.24. Padmitsastrongcomplement I P 0. This follows from () 3.23 with Q = I-P.
37 3 Spectral Theory for Ordered Spaces
Remark 3.25. The complement of P may not be unique, but strong complement of P is unique, if it exists.
Definition 3.26. Pissaidtobebicomplemented if there exists a continuous positive projection R on X such that P, R are complementary and P⇤,R⇤ are complementary.
We now define what a smooth projection is.
Definition 3.27. Let P be a continous positive projection on X. P is called smooth if Tan(ker+P) = ker P.
Asmoothprojectionisuniquelydeterminedbyitspositivekernelandpositive image.
Proposition 3.28. Let P, R be continous positive projections on X such that ker+P = ker+R and im+P = im+R. If P is smooth, then P = R.
Proof. ker+P=ker+R= Tan(ker+P) = Tan(ker+R). P is smooth = ker P ) ) =Tan(ker+P) = Tan(ker+R) ker R. And im+P=im+R= im P = a-b ⇢ ) { | a,b im+P = a-b a,b im+R =imR.So,wehavekerP ker R and 2 } { | 2 } ⇢ im P = im R. By 3.15, P = R.
Smoothness of P ensures that ker+Pandim+Pdualizeproperlyunderpositive annihilators.
Proposition 3.29. Let P be a continous positive projection on X with dual pro- jection P⇤ on Y. Then
+ + (i) (im P)• = ker P⇤
+ + (ii) (ker P)• im P⇤
+ + (iii) (ker P)• = im P⇤ P is smooth () 38 3 Spectral Theory for Ordered Spaces
+ + (iv) im P is semi-exposed face of X P⇤ is smooth ()
Proof. (i) Replacing P by P⇤ in 3.19, (ker P⇤) =(imP)= (kerP ⇤) = ) (imP ) .SincekerP⇤ is closed convex, ker P⇤ =(kerP ⇤) =(imP ) = ) + + ker P ⇤ =(imP )• =(im P )•, as im P is positively generated.
+ + + + (ii) im P⇤ =(kerP) (ker P) = im P⇤ X (ker P) X = ⇢ ) \ ⇢ \ ) + + im P ⇤ (ker P)•. ⇢
+ + + (iii) ( ) Assume P is smooth. So, (ker P )• = ker P. Now,(ker P )• (ker P )• ( ⇢ + + + =((ker P )• ) =(kerP) =imP⇤ = (ker P)• im P⇤.Theother ) ⇢ + + way containment is satisfied by (ii). Hence, (ker P)• =im P⇤.
+ + + + ( ) (ker P )• = im P ⇤ = (ker P )• =(im P ⇤) =(imP ⇤) = ) ) + (kerP ) kerP = (ker P )• kerP = Pissmooth. ) )
+ + + (iv) By 3.13, (im P )•• is the smallest semi-exposed face of X containing im P.
+ + + Therefore, im Pissemi-exposed im P=(im P )••.By(i)and(iii), () + + + (im P )•• =(ker P ⇤)• = im P P⇤ is smooth . ()
Proposition 3.30. Let P be a continuous positive projection with a complement
Q. Then
(i) P⇤ is smooth
(ii) Q is smooth = Q is the unique complement of P )
(iii) P,Q are smooth P⇤,Q⇤ are complementary smooth projections () () P,Q are bicomplementary
+ + + Proof. (i) im P=ker Q is a semi-exposed face of X = P⇤ is smooth (By ) 3.29).
39 3 Spectral Theory for Ordered Spaces
(ii) Let R be another complement of P. This implies im+Q=ker+P=im+Rand
ker+Q=im+P=ker+R. Since Q is smooth, by 3.28, Q = R.
(iii) (a) Assume P,Q are smooth. Since P, Q are complemented = P⇤,Q⇤ ) + + + + are smooth (by (i)). Now, ker P ⇤ =(im P )• =(ker Q)• = im Q⇤.
+ + Similarly, ker Q⇤ = im P ⇤. Hence, P⇤ and Q⇤ are complementary.
(b) Let P⇤,Q⇤ be smooth complementary projections = P,Q are com- ) plementary and smooth (replacing P,Q by P ⇤,Q⇤ in (a)) = P,Q are ) bicomplementary.
+ + (c) Let P,Q be bicomplementary = im P ⇤ = ker Q⇤ which is a semi- ) exposed face of X+. By 3.29 - (iv), P is smooth. Similarly, Q is smooth.
Theorem 3.31. Let P,Q be bicomplementary continuous positive projections on X. Then E = P + Q is the unique continuous positive projection onto the subspace im(P+Q)= im(P) im(Q). Proof. Let E = P + Q. P, Q are complementary = PQ = QP = 0. ) (i) Since P, Q are continuous, positive and linear, E is a positive continuous
linear map. E2 =(P+Q)(P+Q)=P2 +PQ+QP+Q2 =P2 +Q2 =P+ Q=E.Therefore,Eisaprojection. (ii) Clearly, im(P+Q) im P + im Q. Let x im P im Q. This implies x = ⇢ 2 \ Px = P(Qx) = PQ(x) = 0. Therefore, the sum is direct and im(P+Q) ⇢ im P im Q. Let u im P im Q. This implies u = Px +Qx for some 2 1 2 x ,x X. Eu = (P+Q)(Px +Qx )=PPx +PQx +QPx +QQx 1 2 2 1 2 1 2 1 2 =Px +Qx =u.Thisimpliesu im(E). Therefore, im P im Q 1 2 2 ⇢ im(P+Q) = im(P+Q) = im P im Q. ) (iii) Let T be another continuous positive projection on X such that im T = im
P im Q = im E. Now, im T = im E = ET = T. ) 40 3 Spectral Theory for Ordered Spaces
Claim: ker P ker PT ⇢ Let x ker P. P,Q are bicomplementary = P, Q is smooth. So, x 2 ) 2 + + + ker P = (ker P )• . Note that if y Y ,thenT ⇤P ⇤y (ker P )•,since 2 2 + + if a ker P=im Q im Q im T, then a, T ⇤P ⇤y = T a, P ⇤y 2 ⇢ ⇢ h i h i + = a, P ⇤y = P a, y = 0,y = 0. Now, since x (ker P )• and h i h i h i 2 + + T ⇤P ⇤y (ker P )•, PTx,y = x, T ⇤P ⇤y =0, y Y .This 2 h i h i 8 2 implies PTx,y =0, y YasYispositivelygenerated.So,PTx= h i 8 2 0= x ker PT. ) 2 Claim: ker E = ker P ker Q \ Clearly, ker E ker P ker Q. Let x ker E = 0=Ex = \ 2 ) Px+ Qx = Px = Qx = Px = PPx = -PQx = 0 = x ker ) ) ) 2 P. Similarly, x ker Q. Therefore, ker E ker P ker Q i.e. ker(P+Q) 2 ⇢ \ ker P ker Q. Hence, the claim. ⇢ \ We have, ker P ker PT and similarly ker Q ker QT. Now, ker E = ker ⇢ ⇢ P ker Q ker PT ker QT = ker(PT + QT) = ker (ET) = ker T. So, \ ⇢ \ ker E ker T and im E = im T. Therefore, by 3.15, E = T. ⇢
3.2 Compressions
Let (A, A+, e), (V, V+,K)beapairoforderunitspaceandbasenormspace in separating order and norm duality. In this section, the word continuous would imply continuity with respect to the norm topology and weakly continuous would mean continuity with respect to the weak topology given by the duality <, >.
Definition 3.32. AweaklycontinouspositiveprojectionPonAorVissaidto be normalised if P 1( P=0or P =1) k k () k k
41 3 Spectral Theory for Ordered Spaces
Proposition 3.33. Let P be a weakly continuous positive projection on A with dual projection P⇤ on V. Then
(i) P⇤ is normalised P is normalised Pe e () ()
+ (ii)
(iii) If P has a complement Q, then P is normalised Q is normalised ()
Proof. (i) P ⇤ = P . Hence, P ⇤ is normalised Pisnormalised. k k k k ()
To Prove: P is normalised Pe e () ( ) Suppose P is normalised. Now, e =1and P 1= Pe ) k k k k )k k 1= Pe e. ) ( ) Assume Pe e.Takea A a 1. Then e a e = ( 2 3k k ) Pe Pa Pe = e Pe Pa Pe e = Pa ) )k k 1= P 1. )k k
+ (ii) Let v V .Then,
(iii) e-Pe ker+ P=im+Q. Hence, Q(e-Pe) = e - Pe = e=Qe+Pe= 2 ) ) Qe e= Qisnormalised(by(i)). )
Proposition 3.34. If P, Q are weakly continuous positive projections on A, then
+ Pe + Qe = e P ⇤v + Q⇤v = v , v V (P⇤ +Q⇤)(K) K. () k k k k 8 2 () ⇢
Proof. We know, v =
1. (i) (ii) ) + + Assume Pe + Qe = e. Let v V = (P ⇤v + Q⇤v) V . P ⇤v + Q⇤v 2 ) 2 k k =
2. (ii) (iii) ) + + Assume P ⇤v + Q⇤v = v , v V .K= v V v =1 .Letk K k k k k 8 2 { 2 |k k } 2 + V .Then (P ⇤ + Q⇤)k = k = 1. Also, since P⇤ and Q⇤ are positive, ⇢ k k k k + (P ⇤ + Q⇤)k V .Therefore,(P ⇤ + Q⇤)k K. k k2 2
3. (iii) (i) )
Assume (P⇤ +Q⇤)(K) K. Let k K. Let (P ⇤ + Q⇤)k = k0 for some k’ ⇢ 2 2 K. < (P + Q)e, k > =
+ = k0 =1= k =
Remark 3.35. In particular, if P,Q are normalised complementary projections on A, then ker+P=im+Q. So, (e - Pe) ker+ P= Q(e - Pe) = e - Pe. Hence, e 2 ) =Pe+Qe.So,theysatisy3.34.
Let P be a normalised weakly continuous positive projection on A with dual projection P⇤ on V.
Definition 3.36. P⇤ is said to be neutral if the following implication holds for v
+ V : P ⇤v = v = P ⇤v = v 2 k k k k )
Proposition 3.37. Let P be a normalised weakly continuous positive projection on A. Then
(i) P⇤ is neutral = P is smooth )
(ii) If P has a complement Q, then
(a) P,Q are smooth = P⇤ is neutral ) (b) P,Q are bicomplementary P⇤,Q⇤ are neutral () 43 3 Spectral Theory for Ordered Spaces
+ + Proof. (i) Assume P ⇤ is neutral. Let v (ker P )•.Sincee-Pe ker P, 2 2
(ii) (a) Assume P,Q are smooth = P⇤,Q⇤ are complementary. Let v ) 2 + V such that v = P ⇤v =
(b) ( )P,Qarebicomplementary= P,Q are smooth = P ⇤, Q⇤ are ) ) ) neutral.
( ) P ⇤,Q⇤ are neutral = P,Q are smooth. Also they are comple- ( ) mentary. By 2.10(iii), P,Q are bicomplementary.
Definition 3.38. Let A be an order unit space, which is in separating order and norm duality, with a base norm space V. A bicomplemented weakly continuous normalised positive projection on A is called a compression.
Proposition 3.39. If P is a compression, then
(i) P is smooth
(ii) P ⇤ is neutral (iii) Q is a compression, where Q is the unique complement of P and
+ (iv) Pe + Qe = e, P ⇤v + Q⇤v = v , v V , (P⇤ +Q⇤)(K) K. k k k k 8 2 ⇢
Theorem 3.40. Let :V V and 0 : V V be weakly continous positive ! ! linear projections with dual maps P = ⇤ : A A and P’ = 0⇤ : A A. Then ! ! P, P’ are complementary compressions the following holds: ()
44 3 Spectral Theory for Ordered Spaces
+ (i) v + 0v = v , v V k k k k k k 8 2 (ii) and 0 are neutral
+ + + + (iii) ker P im P 0 and ker P 0 im P ⇢ ⇢
+ + + Proof. ( ) If P,P’ are complementary, then ker P im P 0 and ker P 0 ) ⇢ ⇢ + + im P . Also, they satisfy 3.2, so v + 0v = v , v V . Finally, P, k k k k k k 8 2 P’ are compressions = they are smooth = , 0 are neutral. ) )
( ) (a) and 0 are neutral = P,P’ are smooth. ( ) + (b) Let k K V .Then, k + 0k = k =1. = k 1. 2 ⇢ k k k k k k )k k Let v V v 1, Since Co(K -K) = closed unit ball of V, 2 3kk [ 9 k ,k K v = k (1 )k for some [0, 1]. Now, v = 1 2 2 3 1 2 2 k k k (1 ) k k +(1 ) k 1. = 1. k 1 2k k 1k k 2k )kk So, P = P ⇤ = 1 . Hence, P is normalised. Similarly, P’ is k k k k k k normalised.
+ (c) v V , v + 0v = v .Thisimplies v = v 8 2 k k k k k k k k k k() 0v =0 = v = v 0v =0(as isneutral).Therefore k k ) () + + + + ker =im 0.Similarly,ker 0 = im . So, , 0 are complemen-
tary = 0 =0and 0 =0.DualisinggivesPP’=P’P=0. ) So, im+P ker+P’ and im+P ker+P’ . Along with (iii), this implies ✓ ✓ + + + + ker P =im P 0 and ker P 0 =im P . Hence, P,P’ are complementary.
So P, P’ are complementary compressions.
The distinguished base K of the base norm space V is given by K = v V + { 2 | v =< 1,v >=1 .DefineA+ := a A+ a =1 . k k } 1 { 2 |k k }
Proposition 3.41. If P is a compression on A, then
+ + (i) Pe is the greatest element of A1 which belongs to (ker P ⇤)•
45 3 Spectral Theory for Ordered Spaces
+ + (ii) im P ⇤ = v V
+ + + + + (iii) ker P and im P are semi-exposed faces of A , ker P ⇤ and im P ⇤ are semi-exposed faces of V +
+ + + + (iv) Each of ker P , im P , ker P ⇤ and im P ⇤ determine P
+ + Proof. (i) Pisnormalised= 0 Pe e = Pe A .Letv ker P ⇤. ) ) 2 1 2 + + Then 0 =
(ii) Pisacompression= P ⇤ is neutral. So,
+ + + (iii) Let Q be the unique complement of P. Then im P = ker Q and im P ⇤ =
+ ker Q⇤. So, by 3.13, they are all semi-exposed faces.
(iv) Let Q be the unique complement of P.
+ + (a) Suppose R is another compression such that ker P ⇤ =ker R⇤.By(i),
+ + ker P ⇤ determines Pe. Similarly, ker R⇤ determines Re. Then Pe = Re
+ + + as ker P ⇤ =ker R⇤.By(ii),Pedeterminesim P ⇤ and Re determines
+ + + + + im R⇤.SincePe=Re,im P ⇤ =im R⇤.So,wehaveker P ⇤ =ker R⇤
+ + and im P ⇤ =im R⇤.SinceP⇤ is smooth, by 3.28, P⇤ =R⇤.Dualizing
+ gives P = R. Hence, ker P ⇤ determines P.
+ + + (b) When im P ⇤ is given, im P ⇤ = ker Q⇤.So,Qisdeterminedasina) and then P is determined as the unique complement of Q.
+ + (c) Treating P = (P ⇤)⇤,ina),b),weseethateachofker P and im P uniquely determines P.
46 3 Spectral Theory for Ordered Spaces
3.2.1 Projective Units and Projective Faces
Let P be a compression on A.
+ Definition 3.42. The element p = Pe of A1 is called a projective unit of A
(associated with P) and F = K imP⇤ is called projective face (associated with \ P).
+ Proposition 3.43. im P ⇤ = 0 F [
+ + Proof. ( ) By definition, F im P ⇤.Pislinear= F im P ⇤,forall ◆ ⇢ ) ⇢ + 0. Thus, im P ⇤ 0 F . ◆[
+ + ( ) Let v im P ⇤ V .Thenv= kforsome 0andk K. k = v = ✓ 2 ⇢ 2
P ⇤v = P ⇤k If =0,thenv 0 F .If =0,thenk=P ⇤k= k 2[ 6 ) 2
K im P ⇤ = F. So, v 0 F . \ 2[
Remark 3.44. Each projective unit p is associated with a unique compression P
+ Proof. p=Pedeterminesim P ⇤ (by 3.5 (ii)) which in turn determines P uniquely (by 3.5(iv)).
Remark 3.45. Each projective face F is associated with a unique compression P
+ Proof. By 3.6, F determines im P ⇤ which in turn determines P
Remark 3.46. (i) a A+,a im+P a a p a face(p) 8 2 2 () k k () 2 () a p•• 2 (ii) Pa 0 < a, v > 0, v F () 8 2
Proof. (i) Let a A+. 2 1 (a) a a e for all a in A. If a im+P ,thena=Pa a Pe = a p. k k 2 k k k k 47 3 Spectral Theory for Ordered Spaces
(b) face(p) = a A a kp,forsomek 0 .So,a a p = a { 2 | } k k ) 2 face(p).
+ (c) p•• is the smallest semi-exposed face of A containing p = a ) 2 face(p) p••. ⇢ (d) Pisacompression= im+P is a semi-exposed face of A+ = ) ) + + + + (im P )•• = im P .Therefore,a p•• (im P )•• = im P . 2 ⇢
(ii) Let a A. 2
( ) If < a, v > 0, v F, then < a, v > 0, v F for some 0. By ) 8 2 8 2 + 3.6, < a, v > 0, v im P ⇤.Therefore,< P a, v >=< a, P ⇤v> 0. 8 2 + ( ) Let v F im P ⇤. < a, v >=< a, P ⇤v>=< P a, v > 0. Hence, ( 2 ⇢ a 0 on F.
Let (P, F, p) denote a compression P on A , with associated projective face F and projective unit p. Each of P, p, F determines the others.
Proposition 3.47. If P’ is the unique compression complementary to P, with corresponding projective face F’ and projective unit p’ ,then
(i) p + p’ = e
(ii) F= v K P⇤v=v = v K
F’ = v K P’ ⇤v=v = v K
(iii) For a A, < P a, v >=< a, v >, v F and < P a, v >=0, v F’ 2 8 2 8 2
48 3 Spectral Theory for Ordered Spaces
(iv) p= a A+ < a, v >=0, v F’ , _{ 2 1 | 8 2 } p= a A+ < a, v >=1, v F . ^{ 2 1 | 8 2 } + Hence, p is the unique element of A1 which is 0 on F’ and 1 on F.
Proof. (i) Follows from Remark 8 and 3.2.
(ii) (a) F=K imP ⇤ = v K P⇤v=v . \ { 2 | } + + (b) Since K V ,F=K imP ⇤ =K im P ⇤.By3.5(ii),F= v K ⇢ \ \ { 2
(c) Now replacing p by e - p’, we have F = v K
(iii) (a) v F imP ⇤ = P⇤v= v. Therefore, < a, v >=< a, P ⇤v>= 2 ⇢ ) < P a, v >,foralla A. 2 + + (b) If v F’,then v im (P 0)⇤ = ker P ⇤ = < P a, v >=< a, P ⇤v>= 2 2 ) 0foralla A. 2
+ + (iv) (a) If v F 0 im (P 0)⇤ = ker P ⇤.By3.5(i),pisthelargestelementin 2 ⇢ + + + A1 which vanishes on ker P ⇤.So,pisthelargestelementinA1 which vanishes on F’. Therefore, p = a A+ < a, v >=0, v F’ _{ 2 1 | 8 2 } (b) e-p=p’= a A+ < a, v >=0, v F . _{ 2 1 | 8 2 } So, p = e - p’
=e- a A+ < a, v >=0, v F _{ 2 1 | 8 2 } = e-a a A+, < a, v >=0, v F ^{ | 2 1 8 2 } = b e b A+,
= b A+ 1= v =
Theorem 3.48. Let P be a compression on A with associated projective unit p, projective face F. Then
1. [ p, p] = a A p a p = A imP { 2 | } 1 \
2. Co(F -F) = V imP ⇤ [ 1 \
3. P(A) = im P, is an order unit space with distinguished order unit p
4. P⇤(V) = im P⇤, is a base norm space with distinguished base F
Proof. 1. ( ) Let p a p = 0 a + p 2p = a + p face(p)= ✓ ) ) 2 ) a + p im+P (by 3.9(i)). Now, a+p = P(a+p) = Pa + PPe = Pa + 2 p. This implies a = Pa = a im+P . Also a p = a p ) 2 )kkk k 1.This implies a A+ im+P . 2 1 \ ( ) Let a A+ im+P = e a e = p Pa = a p. ◆ 2 1 \ ) )
2. ( ) F V imP ⇤ = Co(F -F) V imP ⇤. ✓ ⇢ 1 \ ) [ ⇢ 1 \
( ) Let a V imP ⇤. WLOG, we may assume a =1(elsereplace ◆ 2 1 \ k k a a by a ). V1 = Co(K K)= a = ↵s (1 ↵)t,forsome k k [ ) s,t in K and ↵ [0, 1]. a = P ⇤a = ↵P ⇤(t) (1 ↵)P ⇤(s). Hence, 2 1= a ↵ P ⇤(t) (1 ↵) P ⇤(s) 1. Therefore, P ⇤(t) = k k k k k k k k P ⇤(s) =1 = P ⇤(s),P⇤(t) imP ⇤ K = F .So,a Co(F F ). k k ) 2 \ 2 [
50 3 Spectral Theory for Ordered Spaces
3. Clearly, im+P is a cone in im+P since, if v ,v im+P ,thenv = Pv 1 2 2 1 1 0,v = Pv 0= v + v = Pv + Pv = P (v + v ) 0. Hence, 2 2 ) 1 2 1 2 1 2 v + v im+P .Similarly, v im+P , v im+P, 0. 1 2 2 2 8 2 Let v A. Since e is an order unit for A, 0 e v A+.This 2 9 3 ± 2 implies P ( e v) im+P = ( P e Pv) im+P = ( p v) im+P . ± 2 ) ± 2 ) ± 2 Hence, p is an order unit for im P.
So, (imP, im+P, p)isanorderunitspace.
+ 4. As in the proof of (3), im P ⇤ is a cone in imP ⇤.
Claim: F is a base for im P ⇤.
Consider f ,f F and , [0, 1] + =1.SinceKisconvex, 1 2 2 1 2 2 3 1 2 f + f K. Also P ⇤( f + f )=( f + f ) imP ⇤ since 1 1 2 2 2 1 1 2 2 1 1 2 2 2 f ,f imP ⇤. Hence, f + f K imP ⇤ = F .Thus,Fisconvex. 1 2 2 1 1 2 2 2 \
Let v imP ⇤ 0 . Since K is a base for V, >0,k K v = k.Now, 2 \{ } 9 2 3 k = v = P ⇤v = P ⇤k = k = P ⇤k = k F . Hence, v = f where ) ) 2 f = k F . 2 + So, F is a base for im P⇤ and (im P⇤, im P ⇤,F)isabasenormspace.
+ + Theorem 3.49. ((imP, im P, p), (imP ⇤, im P ⇤,F)) are in separating order and norm duality under the ordering, norm and bilinear form <, > relativised from A and V
+ + Proof. Assume ((A, A ,e), (V,V ,K),<,>0)beinseparatingorderandnormdu- ality. Let <, > denote the restriction of <, >0 on (im P) X (im P⇤) . { } (i) Dual pair:
Let a im P. Suppose < a, v >=0 v imP ⇤.Thisimplies< a, u >0=< 2 8 2 P a, u >0=< a, P ⇤u>0=< a, P ⇤u>=0 u V. But < a, u >0=0 u V 8 2 8 2 51 3 Spectral Theory for Ordered Spaces
= a=0.So,a=0.Similarly,wecanshow,< a, v >=0 a im P⇤ = ) 8 2 ) v=0.
(ii) Separating order duality:
+ + To prove: im P = a imP < a, v > 0 v im P ⇤ { 2 | 8 2 }
+ + ( ) If a im P and v im P ⇤,then< a, v >=< a, v >0 0. ✓ 2 2 + ( ) Suppose a imP < a, v > 0 v im P ⇤. < a, v >0=< P a, v >0=< ◆ 2 3 8 2 + a, P ⇤v>0=< a, P ⇤v> 0 v V . Hence, a 0. So, a im P . 8 2 2
+ + Similarly, we can show, im P ⇤ = v imP ⇤ < a, v > 0 a im P . { 2 | 8 2 }
(iii) Norm Duality:
Denote: a = inf >0 p a im+P ,foralla im P; k kp { | ± 2 } 2 a = inf >0 e a A+ ,foralla A; k ke { | ± 2 } 2 + v = inf >0 f v im P ⇤, for some f F , v im P⇤; k kF { | ± 2 2 } 8 2 v = inf >0 k v V +, for some k K , v V. k kK { | ± 2 2 } 8 2
Claim: a = a a im P. k kp k ke 8 2 Let a im P = Pa = a. First, a p a im+P and e p. 2 ) k kp ± 2 So, a e a A+. Hence, a a .Second,forall >0 k kp ± 2 k kp kke 3 e a A+ = p a imP +.So, a a .Thus, a = a . ± 2 ) ± 2 k kp k ke k kp k ke
Claim: v = v v im P⇤. k kF k kK 8 2 + Let v im P⇤ = P⇤v = v. First, if f v im P ⇤,forsomef F 2 ) ± 2 2 and >0, then since F Kand f v V +, v v .Second, ⇢ ± 2 k kK k kF + + if k v V ,forsomek K and >0, then P ⇤k v im P ⇤. ± 2 2 ± 2 Now, P ⇤k P k 1. By 3.2.1, P ⇤k Co(F F ). This k kK k kk kK 2 [ implies P ⇤k = ↵f (1 ↵)f for some f ,f F and ↵ [0, 1]. 1 2 1 2 2 2 P ⇤k 0= ↵ =0.Therefore,P ⇤k ↵f and P ⇤k ↵f f ) 6 1 1 1 52 3 Spectral Theory for Ordered Spaces
+ + as ↵ (0, 1]. Now, P ⇤k v im P ⇤ = f v im P ⇤. Hence, 2 ± 2 ) 1 ± 2 v v .Thus v = v . k kF k kK k kF k kK
To Prove: v = v v im P⇤ k kF k kdual 8 2 We know, v = v k kF k kK = inf >0 k v V + for some k K { | ± 2 2 } = sup < a, v > a A, a 1 {| || 2 k ke }
= sup < a, P ⇤v> a A, a 1 {| || 2 k ke } = sup < P a, v > a A, a 1 {| || 2 k ke } = sup
= sup < a, P ⇤v> v V, v 1 {| || 2 k kK } 3 = sup < a, u > u imP ⇤, u 1 {| || 2 k kK }
= sup < a, u > u imP ⇤, u 1 {| || 2 k kF } = a . k kdual
Hence, the two spaces are in separating order and norm duality.
2 sup
53 3 Spectral Theory for Ordered Spaces
3.3 Relation between Compressions
Let (P, F, p) and (Q, G, q) denote compressions on A with their corresponding projective faces and projective units. Let (P’, F’,p’) and (Q’, G’, q’) denote their respective complementary compressions.
3.3.1 Comparability
Definition 3.50. We write P Q and F G if im P im Q. ⇢
Proposition 3.51. The following are equivalent:
(i) im P im Q ⇢ (ii) QP = P (iii) p q (iv) F G ⇢ (v) PQ = P (vi) ker P ker Q ⇢ (vii) im Q’ im P’ ⇢
Proof. 1. (i) = (ii) ) Assume im P im Q. Let x X. Then Px imP imQ = Q(Px)= ⇢ 2 2 ⇢ ) Px = QP = P . )
2. (ii) = (iii) ) Assume QP = P, then p = Pe = QP e.SincePe e and Q is positive, Q(Pe) Qe = q = p q. )
3. (iii) = (iv) ) Assume p q.Letv F. Then by, 3.10 (ii), 1 =
4. (iv) = (v) ) + + Assume F G = 0 F 0 G = im P ⇤ im Q⇤. Now, ⇢ )[ ⇢[ ) ⇢ + + + + + imP ⇤ = im P ⇤ im P ⇤ and imQ⇤ = im Q⇤ im Q⇤.So,im P ⇤ ⇢ + im Q⇤ = imP ⇤ imQ⇤ = Q⇤P ⇤ = P ⇤(as in (i)). Dualizing gives PQ ) ⇢ ) =P.
5. (v) = (vi) ) Assume PQ = P. Let x Ker Q. Then Px = PQx = P0 = 0. Hence, x 2 2 Ker P. This implies ker Q ker P. ✓
6. (vi) = (vii) ) + + + Assume Ker Q Ker P. This implies Ker Q Ker P = im Q0 ⇢ ⇢ ) ⇢ + im P 0 = imQ0 imP 0. ) ⇢
7. (vii) = (i) ) Assume im Q’ im P’. Using (i) = (vii), with Q’, P’ in place of P, Q ⇢ ) respectively, we get im P im Q. ⇢
Remark 3.52. The relation on the set of compressions on A is a partial ordering.
Proposition 3.53. P Q Q’ P’. So, the map P P’ is an order () 7! reversing map for this relation.
3.3.2 Orthogonality
Definition 3.54. Pissaidtobeorthogonal to Q (denoted by P Q) if PQ = 0. ? (We also write p qandF G) ? ?
Proposition 3.55. The following are equivalent: (i) QP = 0
55 3 Spectral Theory for Ordered Spaces
(ii) P Q’ (iii) p q’ (iv) F G’ ⇢ (v) PQ = 0
Proof. 1. (i) = (ii) ) Let QP = 0. Let x im P. Then Qx = Q(Px) = 0x = 0 = imP kerQ. 2 ) ⇢ + + + So, im P ker Q = im Q0 = imP imQ0 = P Q0. ⇢ ) ⇢ )
2. (ii) = (iii), (iii) = (iv) ) )
P Q0 = imP imQ0.Thenby3.11,p q0 which further implies F ) ⇢ ⇢ G’.
3. (iv) = (v) )
F G0 = PQ0 = P (by 3.11). So, PQ = PQ’Q = 0. ⇢ )
4. (v) = (i) ) Using (i) = (v) with P and Q interchanged, we get QP = 0. )
Proposition 3.56. Let (A, A+, e), (V, V+, K) be a pair of order unit space and base norm space in separating order and norm duality. Assume A = V⇤ and let F,G be projective faces of K. Then F G a A+ < a, v > = 1, v F and < a, v > = 0, v G ? () 9 2 1 3 8 2 8 2
Proof. Let p, q be the projective units associated with F, G respectively.
( ) F G= p q0 = p e q = p + q e.By3.10,v F = ) ? ) ) ) 2 )
( ) Assume a A+ < a, v > =1, v Fand< a, v > =0, v G. This ( 9 2 1 3 8 2 8 2 implies 1 = v =
3.3.3 Compatibility
Definition 3.57. Let P , Q be two compressions on A. P is said to be compatible with Q if PQ = QP . P is said to be compatible with an element a Aif 2 a = Pa+ P 0a.
So, if a AiscompatiblewithP ,thena is also compatible with P 0. 2 Notation 11. P is compatible with Q is denoted by P Q and P a denotes ⇠ ⇠ that P is compatible with the element a A. 2
Lemma 3.58. If a A+, then a = Pa + P’a Pa a. 2 ()
Proof. ( ) a Pa = P 0a 0= a P a. ) )
+ + ( ) a Pa kerP.Further,Pa a = a Pa ker P = im P 0. But P’P ( 2 ) 2 = 0. So, a - Pa = P’(a - Pa) = P’a + 0. Hence, a = Pa + P’a.
Lemma 3.59. If P is a compression on A and r is some projective unit of A, then Pr r Pr’ r’ ()
Proof. Let P’ be the complementary compression of P. Assume Pq q. Therefore, by3.3.1, Pq+P’q=q. Now, Pe+P’e=e. q’=e-q=(Pe+P’e)-(Pq+P’q) = P(e-q) + P’(e-q) = Pq’ + P’q’. Hence, by 3.3.1, Pq’ q’. Similarly, Pq0 q0 = Pq q. )
Proposition 3.60. PQ = QP Pq q Qp p () ()
57 3 Spectral Theory for Ordered Spaces
Proof. Since (i) is symmetric in P and Q, it su ces to show (i) (ii) ()
( ) If PQ = QP, then Pq = PQe = QPe = Qp Qe = q. So Pq q. )
( ) Assume Pq q. Let a A+.Since0 a a e,then0 Qa a q = ( 2 k k k k ) 0 PQa a Pq a q = a Qe. Hence, 0 Q0PQa a Q0Qe =0. k k k k k k k k + + + This implies Q’PQa = 0, a A .Therefore,PQa ker Q0 = im Q = 8 2 2 ) QP Qa = PQa, a A+ and since A+ generates A, QPQa = PQa, a 8 2 8 2 A. Hence, QPQ = PQ.
Now Pq q = Pq0 q0.Soreplacingqbyq’intheabovearguments, ) we have QPQ’a = 0, a A+ = QPQ’a = 0, a A. So, QPQ’ = 0. 8 2 ) 8 2 + + + Dualizing gives Q0⇤P ⇤Q⇤ =0.So, v V , P ⇤Q⇤v ker Q0⇤ = im Q⇤. 8 2 2 + Therefore, P ⇤Q⇤v = Q⇤P ⇤Q⇤v, v V and hence for all v V. Thus, 8 2 2 P ⇤Q⇤ = Q⇤P ⇤Q⇤.DualizinggivesQP=QPQ.
Thus, PQ = QPQ = QP.
Remark 3.61. PiscompatiblewithQ PiscompatiblewithQ’. ()
Remark 3.62. P is compatible with Q P is compatible with Q0 Q0 is () () compatible with P Q0 is compatible with P 0. () Remark 3.63. The set of all elements in A, compatible with a compression P forms a linear subspace of A , denoted by SP ,andSP =im(P + P 0).
Proof. a, b S = a = Pa+ P 0a, b = Pb+ P 0b = (a + b)=(Pa+ P 0a)+ 2 P ) ) (Pb+ P 0b)=(Pa+ Pb)+(P 0a + P 0b)=P (a + b)+P 0(a + b)= (a + b) S . ) 2 P And, a SP = a = Pa+P 0a = a = P a+ P 0a = P ( a)+P 0( a) R. 2 ) ) 8 2 Hence, a SP R.Therefore,SP is a linear subspace of A. 2 8 2
58 3 Spectral Theory for Ordered Spaces
2 2 2 Now, (P + P 0) = P + PP0 + P 0P + P 0 = P + P 0 (asPP0 = P 0P =0).
Therefore, (P + P 0) is a projection on A. Let a A. a im(P + P 0) a = 2 2 () (P + P 0)a a = Pa+ P 0a a S . Hence, S =im(P + P 0). () () 2 P P
Therefore, to show the compatibility of an element a A, with P ,itissu cient 2 to show compatibility of a + e with P ,forany R (since e SP ). 2 2
Proposition 3.64. Let P, Q be two compressions on A. Then
1. P Q = P and Q are compatible )
2. P Q = P and Q are compatible ? )
Proof. 1. P Q = PQ = P and QP = P = PQ = QP . Hence, ) ) P and Q are compatible .
2. P Q = P Q0 = P and Q0 are compatible . Hence, P and Q are ? ) ) compatible by 3.62.
+ Proposition 3.65. Let a A . If a = 0 on F 0, then a is compatible with P (Pa 2 = a, P 0a = 0).
+ Proof. Suppose < a, v > =0 v F 0. Now, F 0 = K im(P 0)⇤ im (P 0)⇤ = 8 2 \ ✓ + + + ker P ⇤. a (F 0)• (ker P ⇤)• =im P .Therefore,Pa = a and 0 = (P 0P )a = 2 ◆ P 0(Pa)=P 0a. Hence, a = a +0=Pa+ P 0a = a is compatible with P . )
Remark 3.66. Converse of the above is not true i.e. If a A such that a is 2 compatible with P ,thena need not be 0 on F 0.Example:e S but
Proposition 3.67. Let a A. Then Pa is the unique element in A compatible 2 with P which is equal to a on F and 0 on F 0.
59 3 Spectral Theory for Ordered Spaces
Proof. P (Pa)+P 0(Pa)=Pa+0=Pa. Hence, Pa is compatible with P .
Let f F im P ⇤.Then,< P a, f > = < a, P ⇤f>= < a, f >. Hence, 2 ✓ + + Pa = a on F .Letf 0 F 0 im P 0⇤ =kerP ⇤.Then,< P a, f 0 > = < a, P ⇤f 0 > 2 ✓ = < a, 0 > = 0. Hence, Pa =0onF 0. Now, suppose b A such that b is compatible with P and b = a on F , 2 + + b =0onF 0. Now, im P ⇤ = 0 F and im P 0⇤ = 0 F 0.So,b = a on [ [ + + + F = b = a on im P ⇤ and b =0onF 0 = b =0onim P 0⇤.Letv V . ) ) 2
0,P0⇤v>= < P a, v >. Hence, b = Pa,thusprovingtheuniqueness.
Proposition 3.68. Let a A. Then (Pa + P 0a) is the unique element in A, 2 compatible with P, which is equal to a on Co(F F 0). [
2 2 Proof. First, P (Pa+P 0a)+P 0(Pa+P 0a)=P a+PP0a+P 0Pa+P 0 a = Pa+P 0a. Hence, (Pa+ Pa)iscompatiblewithP .
+ + + + Let f F im P ⇤ =kerP 0⇤,f0 F 0 im P 0⇤ =kerP ⇤,↵ [0, 1]. 2 ✓ 2 ✓ 2 Now,
+ + So, b = Pa+ P 0a on V = b = Pa+ P 0a on V (as V generates V). Hence, ) b =(Pa+ P 0a), thus proving the uniqueness.
Remark 3.69. An element a A, compatible with P ,iscompletelydetermined 2 by its values on F and F 0.
60 3 Spectral Theory for Ordered Spaces
Proof. Suppose b A such that b is compatible with P and b = a on F and F 0. 2 Then by 3.68, b = Pa+ P 0a = a (as a is compatible with P ). Hence, b = a.
Proposition 3.70. Let a A+. Then 2
+ 1. Pa = b A b =0on F 0,b a on F { 2 | } W + 2. Pa = b A b =0on F 0,b a on F { 2 | } V + Proof. Assume a A .ThenPa =0onF 0 and Pa = a on F . Hence b can be 2 chosen to be Pa in (1), (2) above.
+ 1. Let b A such that b =0onF 0 and b a on F .By3.65,b is compatible 2 + with P . Also, b a on F = b a on im P ⇤ = 0 F . Similarly, ) [ + + b =0onF 0 = b =0onim P 0⇤. Now consider v V .
2. Imitating the proof for (1), with replaced by ,wegetPa = b A+ { 2 | b =0onF 0,b a on F V }
The above proposition says that Pa is the best choice of an element in A+ which is 0 on F 0 and closest to a on F.
Proposition 3.71. Let P be a compression on A and T : A A be a weakly ! continuous projection of A onto the subspace im(P + P 0). Then
T = P + P 0 T satisfies one of the following conditions: ()
(i) T is positive
(ii) T 1 k k 61 3 Spectral Theory for Ordered Spaces
(iii) T commutes with P and P 0.
Proof. ( ) Suppose T = P + P 0. )
(i) P, P 0 are positive = (P + P 0)isalsopositive. )
(ii) Let a A such that a 1. Then, e a e. T = P + P 0 is 2 k k positive. So, (P +P 0)( e) (P +P 0)a (P +P 0)(e)= e Ta e ) (asp + p0 = e). Hence, Ta 1. Thus, T 1. k k k k 2 2 (iii) P (P + P 0)=P + PP0 = P and (P + P 0)P = P + P 0P = P .So,
PT = TP.Similarly,TP0 = P 0T .
( ) Given that im T =im(P + P 0). (
(i) Suppose T is positive. Then T = P + P 0 by ??.
(ii) Suppose T 1. Let a A+.Wemayassume a 1(elsereplace k k 2 k k a a by a ). Now, 0 a e = e a 0= 0 e a k k ) ) e = e a 1. So, T (e a) 1= e T (e a) )k k k k ) e = e Te Ta e.ButTe =(P + P 0)e = p + p0 = e. Hence, ) e e Ta e = 0 Ta.So,Ta 0 a A+. Hence, T is ) 8 2 positive. Thus, T = P + P 0 by (i).
(iii) Denote E = P + P 0. T commutes with P, P 0 = T commutes with ) E.Leta A. TE(a)=T (Ea)=Ea (as Ea im E =imT ). So, 2 2 TE(a)=E(a) a A = TE = E. Similarly, ET = T .So, 8 2 ) T = ET = TE = E = P + P 0.
Remark 3.72. Proposition 3.71 says that P + P 0 is the unique weakly continous positive projection of A onto the subspace of all elements compatible with P .
62 3 Spectral Theory for Ordered Spaces
Proposition 3.73. Let P be a compression on A. Define E⇤ = P ⇤ +P 0⇤. Then E⇤
is the unique weakly continuous projection on V which maps K onto Co(F F 0) [ and leaves this set point-wise fixed.
+ + Proof. Let k K.WeknowE⇤(K) K by ??,im P ⇤ = 0 F and im P 0⇤ = 2 ✓
0 F 0.So,E⇤k = P ⇤k + P 0⇤k = ↵f + f0 for some ↵,S 0,f F, f0 F 0. 2 2 ↵ ThisS implies (↵+ )( f + f 0)=E⇤k K (↵+ =0elseE⇤k =0 K which ↵+ ↵+ 2 6 2 ↵ is a contradiction). Now, F, F 0 K and K is convex. So, ( f + f 0) K. ✓ ↵+ ↵+ 2 Therefore, E⇤k =(↵ + )k0 K for some k0 K.ButK is a base and the 2 2 representation of each element is unique = (↵ + )=1.So, =1 ↵. ) Hence, E⇤k = ↵f +(1 ↵)f 0 Co(F F 0). Since k is arbitrary, we have 2 [ 4 E⇤(K) Co(F F 0). ✓ [ Now, consider v Co(F F 0). Let v = ↵f +(1 ↵)f 0 for some ↵ [0, 1],f 2 [ 2 2 F, f0 F 0. E⇤v = ↵P ⇤f +(1 ↵)P ⇤f 0 + ↵P 0⇤f +(1 ↵)P 0⇤f 0 = ↵f +0+0+(1 2 ↵)f 0 = v. Hence, E⇤ fixes Co(F F 0) pointwise. [ Now, to show uniqueness of E⇤,assumethereexistsaweaklycontinuouspro-
jection on V, say T : V V such that T (K) K and T fixes Co(F F 0) ! ✓ [ pointwise. T (K) K = T (V +) V +. Hence, T is positive. ✓ ) ✓
Claim: im T =imE⇤
( ) T fixes elements of Co(F F 0)= TfixesF, F 0 pointwise = Tfixes ◆ [ ) ) + + im P ⇤ and im P 0⇤.SinceimP ⇤ and im P 0⇤ are positively generated, T also
fixes im P ⇤ and im P 0⇤ pointwise. Now, consider u im E⇤. Tu = T (E⇤u)= 2 T (P ⇤u + P 0⇤u)=T (P ⇤u)+T (P 0⇤u)=P ⇤u + P 0⇤u = E⇤u = u.Thus,u 2 im T .So,imT im E⇤. ◆
( ) Let v im+T . This implies Tv = v = k for some 0,k K.i.e. ✓ 2 2 k = v = Tv = T ( k)= T k = (↵f +(1 ↵)f 0)forsome↵ [0, 1],f 2 2 4 Alternately, as E⇤k K,wehave1= E⇤k =
F, f0 F 0.If =0,thenEv=E0=0=v.Suppose =0,thenwehave 2 6 k = Tk = ↵f +(1 ↵)f 0 Co(F F 0). Since E fixes Co(F F 0),Ek= 2 [ [ + k = Ev = v = im T im E⇤ = im T im E⇤ (as im T is ) ) ✓ ) ✓ positively generated).
So, T is a positive projection on V and im T =imE⇤ =im(P ⇤ + P 0⇤). Thus,
T = P ⇤ + P 0⇤,provingtheuniqueness.
Central Compressions
Definition 3.74. AcompressionP is said to be central if it is compatible with all a A. An element a A is said to be central if it is compatible with all 2 2 compressions P on A.
Proposition 3.75 (Equivalent conditions for a central compression). Let P :
A A be a weakly continuous positive projection on A. Then the following are ! equivalent:
(i) Pa a, a A 8 2
(ii) P is strongly complemented
(iii) P is a compression such that P 0 = I P
(iv) P is a central compression
Proof. Given that P is a weakly continuous positive projection on A.
(i = ii) Pa a = I P 0. By 3.24, P is strongly complemented. ) )
(ii = iii) Assume P is strongly complemented. Let P 0 = I P 0. Then, ) Pe = e P 0e e = P is normalised. Also, P 0⇤ = I P ⇤ 0= P ⇤,P0⇤ ) ) are also strongly complemented with norm 1. So, P is a bicomplemented normalised weakly continuous positive projection and hence a compression.
64 3 Spectral Theory for Ordered Spaces
(iii = iv)SinceP 0 = I P, a = Pa+ P 0a, a A.So,P is a central ) 8 2 compression.
+ (iv = i)Leta A . P is central = a = Pa+ P 0a = Pa a. ) 2 ) )
3.4 The Lattice of Compressions
In this section, we will consider < (A, A+,e), (V,V +,K) > -apairoforderunit space and base norm space under separating order and norm duality under an additional assumption called the standing hypothesis.Wewillseethatinthis case, the compressions have nice lattice properties.
Definition 3.76. Let F K.ThenF is said to be an exposed face of K if ✓ + 1 a A such that F = K a (0) = w K < a, w > =0. 9 2 \ { 2 | } F is said to be a semi-exposed face of K if afamily a A+ such that F = 9 { ↵}2 w K =0, ↵ . { 2 | ↵ 8 }
Definition 3.77. The pair < A, V > is said to satisfy the standing hypothesis if each exposed face of K is projective.
Remark 3.78. Since every projective face of V is an exposed face 5 of K,the standing hypothesis implies that if F is a face of K,then F is exposed F is projective. () Notation 12. (P, F, p)denotesacompressionP on A with associated projective face F and projective unit p.Itsuniquecomplementarycompressionisdenoted by (P 0,F0,p0). is the set of all compressions on A, is the set of all projective faces of V C F and is the set of all projective units of A. P 5 1 (F = K p0 (0)) \
65 3 Spectral Theory for Ordered Spaces
Each of these sets have an ordering ( , ), ( , ), ( , )andanotionofcom- C F ✓ P plementation (P, F, p) (P 0,F0,p0). !
Also, for any two compressions (P1,F1,p1), (P2,F2,p2)onA,wehave
P P F F p p (3.4.1) 1 2 () 1 ✓ 2 () 1 2
Projective units and projective faces are uniquely associated with compressions and this correspondence respects the complementation. So, we have the following result:
Remark 3.79. The natural maps 1, 2, 3 associating compressions, projective faces and projective units are complementation preserving order isomorphisms, where : given by P Pe = p 1 C !P ! : given by P K im P ⇤ = F 2 C !F ! \ : given by p w K
Definition 3.80. Apartiallyorderset( , )issaidtobealattice if every two L elements x, y have a greatest lower bound (x y)andleastupperbound(x y) 2L ^ _ in . L
Remark 3.81. Let (X, ), (Y, )betwopartiallyorderedsetsand :X Y ! be an order isomorphism. Then, X is a lattice Y is a lattice and in this () case, becomes a lattice isomorphism.
Proof. Let us assume X is a lattice. Let x ,x X, y = (x ),y = (x ). 1 2 2 1 1 2 2 Then, by order isomorphism, x x (x ) (x ) y y . Now, 1 2 () 1 2 () 1 2 x x x ,x = (x x ) (x ), (x ). 1 ^ 2 1 2 ) 1 ^ 2 1 2 1 1 If y Y such that y (x ), (x )= y x ,x = y 3 2 3 1 2 ) 3 1 2 ) 3 x x . Applying on both sides, we get y (x x ). Hence, (x x )= 1 ^ 2 3 1 ^ 2 1 ^ 2 (x ) (x ). 1 ^ 2
66 3 Spectral Theory for Ordered Spaces
Similarly, (x x )= (x ) (x ). So, Y is a lattice and the bijective map 1 _ 2 1 _ 2 preservesthelatticestructure,thusisalatticeisomorphism.
Proposition 3.82. Assume < A, V > satisfy the standing hypothesis. Then , , are lattices, isomorphic to each other, and the lattice operations are given C F P by the following equations for each pair F, G : 2F
F G = F G, F G =(F 0 G0)0 (3.4.2) ^ \ _ \
Proof. Let F, G with associated projective units p, q respectively. Clearly, 2F F G will be the greatest lower bound for F, G,ifitbelongsto .Leta = \ F 1 + 1 (p0 + q0) A .Thenx K a (0) x K, < a, x > =0 x 2 2 1 2 \ () 2 () 2 1 K, < (p0 + q0),x > =0 x K, < p0,x > +
(F G)0 = F 0 G0, (F G)0 = F 0 G0 (3.4.3) ^ _ _ ^ which is similar to De-Morgan’s laws!
Lemma 3.84. Assume < A, V > satisfy the standing hypothesis. Consider com- pressions (P, F, p), (Q, G, q) on A. Let a A+ such that a p, a q. Then 2 a p q. ^ 67 3 Spectral Theory for Ordered Spaces
Proof. The result is trivially true when a .Butwearesayingitevenforthose 2P 1 a/ .LetH = K a (0). Then H is an exposed face of K and therefore is a 2P \ projective face, by the standing hypothesis. Let f 0 F 0.Then0 < a, f 0 > < 2 p, f 0 > =0 = < a, f 0 > =0 = a =0onF 0. Similarly, a =0onG0. This ) ) implies F 0,G0 H = F 0 G0 =(F G)0 H.So,a =0on(F G)0.By ✓ ) _ ^ ✓ ^ lattice isomorphism, the compression and projective unit associated with F G ^ are P Q and p q respectively. So, by 3.65, a =(P Q)a (P Q)e = p q. ^ ^ ^ ^ ^ Hence, a p q. ^ Proposition 3.85. Assume < A, V > satisfy the standing hypothesis. Let (P, F, p), (Q, G, q) be two compatible compressions. Then PQ = QP = P Q. ^ Proof. Define r = Pq = P (Qe)=Q(P )e = Qp A+.Thenr = Pq Pe = 2 1 p = r p.Similarly,r q.So,r p q. ) ^ Now let s A+ such that s p, q = s im+P and s im+Q (by ??). 2 ) 2 2 This implies s = Ps Pq = r.Inparticular,takings = p q,wegetp q r. ^ ^ So
r = p q = Pe Qe =(P Q)e (3.4.4) ^ ^ ^ Now let a A+.ThenPQa PQe = r =(P Q)e. PQ is a normalised 2 1 ^ projection, so PQa 1. So, by ??, PQa im+(P Q)= PQa =(P k k 2 ^ ) ^ Q)(PQ)a = 6 (PQ)(P Q)a = 7(P Q)a. Therefore, PQ = P Q on A+. Hence, ^ ^ ^ 1 PQ = P Q. ^ Remark 3.86. In the above proof, 3.4.4 tells us that PQe =(P Q)e.Butfromthis ^ we directly can’t conclude that PQ = P Q because in general PQ need not be a ^ 2 compression! (For example, in C⇤ algebra, P is a projection P = P ⇤ = P . () 2 (PQ)⇤ = Q⇤P ⇤ = QP . So, if PQ is a projection, then (PQ) = PQ = QP = ) PQ.Conversely,ifPQ = QP ,then(PQ)2 = PQ,thusPQ is a projection. Hence,
6P Q P, Q = P Q is compatible with P, Q = P Q is compatible with PQ ^ ) ^ ) ^ 7im+(P Q) im+P, im+Q ^ ✓ 68 3 Spectral Theory for Ordered Spaces compatibility is a necessary and su cient condition for PQ to be a projection, in a C⇤ algebra.) The above proposition tells us that under the standing hypothesis, product of compatible compressions is also a compression.
Proposition 3.87. Assume < A, V > satisfy the standing hypothesis. Let P, Q be mutually compatible compressions, both compatible with an element a A. 2 Then, P Q, P Q are also compatible with a. ^ _ Proof. First assume a A+. P, Q compatible with a = P a, Qa a.So, 2 ) (P Q)a =(PQ)a = P (Qa) Pa a = P Q a. Next, P Q = P 0 ^ ) ^ ⇠ ⇠ ) ⇠ Q0. Also, P, Q a = P 0,Q0 a.Thenasabove,P 0 Q0 is compatible with ⇠ ) ⇠ ^ a = (P 0 Q0)0 = P Q is compatible with a. ) ^ _ Now consider any a A compatible with both P, Q.Wehavea = a a 2 1 2 1 1 + 1 where a = ( a e + a),a = ( a e a) A .ThenPa + P 0a = ( a (p + 1 2 k k 2 2 k k 2 1 1 2 k k 1 p0)+a)= ( a e + a)=a = P a .Similarly,itcanshownthatP 2 k k 1 ) ⇠ 1 ⇠ a ,Q a and Q a . Then P Q, P Q a ,a = P Q, P Q a. 2 ⇠ 1 ⇠ 2 ^ _ ⇠ 1 2 ) ^ _ ⇠ Proposition 3.88. Assume < A, V > satisfy the standing hypothesis. Let P, Q be orthogonal compressions, both compatible with an element a A. Then P Q 2 _ is compatible with a and (P Q)a = Pa+ Qa. _ Proof. P Q = P Q (by 3.64) and we are given P, Q a.So,from3.87 ? ) ⇠ ⇠ we have P Q a.Then,a =(P Q)a +(P Q)0a =(P Q)a +(P 0 Q0)a = _ ⇠ _ _ _ ^ (P Q)a +(P 0Q0)a. Now, P, Q a = a = Qa + Q0a, a = Pa + P 0a.So, _ ⇠ ) 8 P 0Q0a = P 0(a Qa)=P 0a P 0Qa = P 0a Qa =(a Pa) Qa.Substituting this above, we get a =(P Q)a +(a Pa Qa)= (P Q)a = Pa+ Qa. _ ) _ Notation 13. We write P +˙ P +˙ ...P in place of P P ...P when P ,P ,...P 1 2 n 1_ 2_ n 1 2 n are mutually orthogonal compressions. So by definition,
P = P +˙ P +˙ ...P P = P P ...P and P P for i = j (3.4.5) 1 2 n () 1 _ 2 _ n i ? j 6 8 + + P Q = Q P 0 = im Q im P 0 = P 0Q = Q ? ) ) ✓ ) 69 3 Spectral Theory for Ordered Spaces
Lemma 3.89. Assume < A, V > satisfy the standing hypothesis. Let P, Q, R be mutually orthogonal compressions. Then P is orthogonal to Q+˙ R.
Proof. P Q = P Q0 and P R = P R0 = P Q0 R0 =(Q ? ) ? ) ) ^ _ R)0 = P Q R.SinceQ and R are orthogonal, we can write P Q+˙ R. ) ? _ ?
Remark 3.90. By induction, we can prove that if P1,P2 ...Pn are mutually orthog- onal compressions, then P (P +˙ P +˙ ...P ). 1 ? 2 3 n
Proposition 3.91. Assume < A, V > satisfy the standing hypothesis. Let P1,P2 ...Pn be mutually orthogonal compressions, all compatible with an element a A. Then, 2
P1+˙ P2+˙ ...Pn is compatible with a and (P1+˙ P2+˙ ...Pn)a = P1a + P2a + ...Pna.
Proof. The proof goes by induction on n. The case n=2 was shown in 3.88. Assume the result is true for n = k 1. i.e. (P1+˙ P2+˙ ...Pn 1)a = P1a + P2a + ...Pn 1a.
Now, when n = k,by3.90,Pn (P1+˙ P2+˙ P3+˙ ...Pn 1). Thus by 3.88 and ? induction hypothesis, (P1+˙ P2+˙ ...Pn)a = Pna +(P1+˙ P2+˙ ...Pn 1)a = Pna +
P1a + P2a + ...Pn 1a.
Proposition 3.92. Assume < A, V > satisfy the standing hypothesis. If p1,p2,...pn are projective units, then the following are equivalent:
1. n p e i=1 i P 2. p p for i = j i ? j 6
n n 3. i=1 pi = i=1 pi W P Proof. Let Pi be the compression associated with pi.
(1) = (2). Let i = j.Thenp + p e = p e p = p p . ) 6 i j ) i j ) i ? j
(2) = (3). n p = n P e =( n P )e =(P +˙ P +˙ ...P )e = ) i=1 i i=1 i i=1 i 1 2 n n P e (since eachW P e).W Thus n pW= n p . i=1 i i ⇠ i=1 i i=1 i P W P 70 3 Spectral Theory for Ordered Spaces
(3) = (1). p e i = n p e = n p e ) i 8 ) i=1 i ) i=1 i W P
Remark 3.93. From 3.92, it follows that p1 + p2 + ...pn = p for some projective unit p n p = p and p p for i = j.So,ifP ,P ,...P and P are () i=1 i i ? j 6 1 2 n compressions correspondingW to p1,p2 ...pn and p respectively, then
p = p + p ...p P = P +˙ P +˙ ...P (3.4.6) 1 2 n () 1 2 n
Lemma 3.94. Assume < A, V > satisfy the standing hypothesis. If P, Q are compressions such that P Q and p, q are their associated projective units, then p q is the projective unit associated with P Q0 = PQ0. Further, R = P Q0 ^ ^ is the unique compression such that P = Q+˙ R and if a A such that a P, Q 2 ⇠ then Ra = Pa Qa.
Proof. Q P = Q P = Q0 P .Thenby3.85,P Q0 = PQ0 is a ) ⇠ ) ⇠ ^ 9 compression. PQ0e = P (e Qe)=Pe Pq = p q. Now let R = P Q0 = PQ0.Thenby3.4.6,P = Q+˙ R p = q + r which ^ () is true since we have shown r = p q. Now, if S is any other compression such that P = Q+˙ S,thenp = q + Se = Se = p q = r = Re = S = R.So,R is ) ) the unique compression such that P = Q+˙ R.
Next, let a A such that a P, Q = a P, Q0 = a P Q0 = R (by 2 ⇠ ) ⇠ ) ⇠ ^ 3.87). Now, P = Q+˙ R = Pa = Qa + Ra = Ra = Pa Qa. ) ) Remark 3.95. Let P Q ,P Q ,Q Q = P P . 1 1 2 2 1 ? 2 ) 1 ? 2
Proof. P Q Q0 P 0 = P P . 1 1 2 2 ) 1 ? 2 Lemma 3.96. Assume < A, V > satisfy the standing hypothesis. Let P, Q be two compressions on A with associated projective units p, q. Then
P Q P =(P Q)+(˙ P Q0). ⇠ () ^ ^ 9q im+Q im+P 2 ✓ 71 3 Spectral Theory for Ordered Spaces
Proof. First note that P Q Q, P Q0 Q0 and Q Q0.So,by3.95, ^ ^ ? (P Q) (P Q0) and hence, the orthogonal sum makes sense. Now, to prove ^ ? ^ the result, it is su cient to prove P Q p =(p q)+(˙ p q0). ⇠ () ^ ^
( ) P Q = P Q0 and so P Q = PQ, P Q0 = PQ0. Now (p q)+ ) ⇠ ) ⇠ ^ ^ ^ (p q0)=PQe+ PQ0e = P (q + q0)=Pe = p. ^
+ ( ) p q q = p q im Q (by ??)= Q(p q)=p q and p q0 q0 = ( ^ ) ^ 2 ) ^ ^ ^ ) + + p q0 im Q0 =ker Q= Q(p q0) = 0. Applying Q to p =(p q)+(p q0), ^ 2 ) ^ ^ ^ we get Qp = Q(p q)+Q(p q0)=p q +0 p = Qp p = Q P . ^ ^ ^ ) ) ⇠
Remark 3.97. Note that we always have P Q, P Q0 P .So,(P Q) (P Q0) ^ ^ ^ _ ^ P trivially. Hence, the above the result can be simplified into P Q P ⇠ () (P Q)+(˙ P Q0). ^ ^
Theorem 3.98. Assume < A, V > satisfy the standing hypothesis. Let P, Q be two compressions on A. Then P Q R, S, T such that S T, P = R+˙ S, Q = R+˙ T . ⇠ () 9 2C ? If such a decomposition exists, then it is unique and infact, R = P Q, S = ^ P Q0 and T = Q P 0. ^ ^
Proof. Let p, q be the projective units corresponding to P, Q respectively.
( ) Assume P Q and let R = P Q, S = P Q0,T= Q P 0.Then, ) ⇠ ^ ^ ^ by 3.95 S T and by 3.96, P =(P Q)+(˙ P Q0)=R+˙ S and Q = ? ^ ^ (P Q)+(˙ P 0 Q)=R+˙ T . ^ ^
( ) Assume R, S, T such that S T, P = R+˙ S, Q = R+˙ T .Then, ( 9 2C ? R P, Q. Hence R P Q. (3.4.7) ^
72 3 Spectral Theory for Ordered Spaces
Also, R S, T S = R, T S0 = R T S0 = Q S0 = ? ? ) ) _ ) ) S Q0. And S S R = P . This implies _
S P Q0 (3.4.8) ^
So, P = R S (P Q) (P Q0) which is a su cient condition for _ ^ _ ^ compatiblity by 3.96. Hence, P Q. ⇠
Uniqueness: Assume R, S, T such that S T, P = R+˙ S, Q = R+˙ T and let 9 2C ? r = Re, s = Se and t = Te.Thenwehave,S R, T which implies S ? ? R+˙ T = Q.Thisimplies Qs =(QS)e =0 (3.4.9)
Also, R Q = r im+Q,whichimplies ) 2
Qr = r (3.4.10)
So, r = r +0=Qr + Qs = Q(r + s)=Qp = QP e =(Q P )e = q p. ^ ^ Hence, r = p q. ^ Now, Pe = Re + Se = s = p r = p (p q)=p q0 (because ) ^ ^ p q = p =(p q)+(p q0)). Similarly, t = q p0. ⇠ ) ^ ^ ^ So, any decomposition has to be of the form R = P Q, S = P Q0 and T = ^ ^ Q P 0. ^
Corollary 3.99. Assume
P Q =(P Q)+(˙ P Q0)+(˙ Q P 0)=P +(˙ Q P 0) _ ^ ^ ^ ^
Proof. First note that P P 0,Q Q0 = by 3.95 (P Q), (P Q0)and(P 0 Q) ? ? ) ^ ^ ^
73 3 Spectral Theory for Ordered Spaces
are mutually orthogonal. P =(P Q) (P Q0)andQ =(P Q) (P 0 Q). ^ _ ^ ^ _ ^ = P Q =(P Q) (P Q0) (P 0 Q) ) _ ^ _ ^ _ ^
=(P Q)+(˙ P Q0)+(˙ P 0 Q)= (P Q)+(˙ P 0 Q) +(˙ P Q0) ^ ^ ^ ^ ^ ^ = P +(˙ P 0 Q) ^
Corollary 3.100. Assume < A, V > satisfy the standing hypothesis. If p,q are compatible projective units, then p + q =(p q)+(p q) ^ _
Proof. Given p q.Thisimpliesp =(p q)+(p q0)andq =(p q)+(p0 q). ⇠ ^ ^ ^ ^ Hence, p + q =(p q)+(p q0)+(p q)+(p0 q) ^ ^ ^ ^ =(p q)+(p q)(by3.96) ^ _
Sublattices of C Definition 3.101. Alattice with greatest element I and least element 0, is said L to be an orthocomplemented lattice if there exists a map x x0 in satisfying ! L the following properties:
(i) x00 = x
(ii) x y = y0 x0 ) (iii) x x0 =0,x x0 = I ^ _
Definition 3.102. An orthomodular lattice is an orthocomplemented lattice ( , L , 0,I,0 )satisfyingtheorthomodularlawi.ex y = y =(y x0) y, x, y . ) ^ _ 8 2L
Theorem 3.103. The lattice of compressions ( , ) as well as the isomorphic C lattices of projective units ( , ), projective faces ( , ) is orthomodular. P F ✓ Proof. We have already seen that ( , )isalattice.LetO, I denote the zero C map and identity map on A respectively. Then O, I and im+I = A+ 2C ◆
74 3 Spectral Theory for Ordered Spaces im+P 0=im+O, P .ThisimpliesO, I are the least and greatest ◆ 8 2C elements in respectively. Now, let (P, F, p), (Q, G, q)beanytwocompressions C on A.
(i) P 00 = P (Pe)00 = Pe and this is true because (Pe)00 =(e Pe)0 = () e (e Pe)=Pe.
(ii) Q P = q p = q p = e q e p = q0 p0 = ) ) ) ) ) P 0 Q0
(iii) P P 0 = P P 0 = P P 0 = PP0 = O. Now, P P 0 =(P P 0)0 = ? ) ⇠ ) ^ _ ^ 10 O0 = I.
(iv) By 3.99, Q P = P = Q+(˙ P Q0)=Q (P Q0). ) ^ _ ^
Definition 3.104. A Boolean algebra is an orthocomplemented lattice satisfying B the distributive law i.e. x (y z)=(x y) (x z)andx (y z)=(x y) (x z), x, y, z . ^ _ ^ _ ^ _ ^ _ ^ _ 8 2B
Remark 3.105. Let be an orthocomplemented lattice and x, y, z .Then L 2L x (y z)=(x y) (x z) x (y z)=(x y) (x z). ^ _ ^ _ ^ () _ ^ _ ^ _
Proof. First assume x (y z)=(x y) (x z). Then ^ _ ^ _ ^ (x y) (x z)= (x y) x (x y) z _ ^ _ _ ^ _ _ ^ = x (x z) (y z) _ ^ _ ^ = x (x z) (y z ) _ ^ _ ^ = x (y z) _ ^ Hence, the dual distributive law is obtained.
Similarly, we can prove the other way implication also. 10ker+O = A+ = im+I, im+O =0=ker+I
75 3 Spectral Theory for Ordered Spaces
Remark 3.106. Boolean Algebras are special cases of orthomodular lattices. Be- cause, if P, Q such that Q P ,then(P Q0) Q =(P Q) (Q0 Q)= 2B ^ _ _ ^ _ P I = P (using the distributive law). So, satisfies the orthomodular condition. ^ B
Definition 3.107. Anon-emptysubsetofanorthocomplementedlatticeiscalled a complemented sublattice if it is closed under complementation and closed under the lattice operations ( , ). ^ _
Remark 3.108. If is a sublattice of ,thenP P 0 =0,P P 0 = I (whereP J L ^ _ 2 )impliesthat0,I .So, has greatest and least elements, which are the J 2J J same as the greatest and least elements of . Hence, itself is an orthocomple- L J mented lattice.
Theorem 3.109. Let < A, V > satisfy the standing hypothesis. Then, is a Boolean Algebra (under the lattice operations induced from ) L✓C C () is a complemented sublattice of and each pair P, Q is compatible . L C 2L
Proof. Let P, Q, R with associated projective units p, q, r respectively. 2L
( ) Every boolean algebra is closed under complementation, join and meet. This ) implies is a complemented sublattice of . Now, for P, Q ,using L C 2L distributive law P = P (Q Q0)=(P Q) (P Q0)= P Q by 3.96. ^ _ ^ _ ^ ) ⇠
( ) Given is a complemented sublattice of under the lattice operations in- ( L C duced from .Thus, itself is an orthocomplemented lattice. To prove C L the distributive law, consider P, Q, R .Then,P, Q, R are mutually 2L
76 3 Spectral Theory for Ordered Spaces
compatible .
P (Q R) e = P (Q+˙ R Q0) e (using 3.99) ^ _ ^ = P (Q+˙ RQ0) e (R Q0) ⇠ = P (Q+˙ RQ0)e = P Qe + RQ0e (by3.88) = PQe+ PRQ0e So, we have
p (q r)=(p q)+(p q0 r) ^ _ ^ ^ ^
=(p q) (p q0 r)(by3.92) ^ _ ^ ^ (p q) (p r) ^ _ ^ p (q r) p (q r) ^ _ _ ^ _ = p (q r) ^ _ Hence, equality holds everywhere and we have p (q r)=(p q) (p r)= ^ _ ^ _ ^ ) P (Q R)=(P Q) (P R). By 3.105, the dual distributive law also ^ _ ^ _ ^ holds. Hence, is a distributive orthocomplemented lattice, thus a boolean L algebra.
Theorem 3.110. Assume < A, V > satisfy the standing hypothesis. Let (P0,F0,p0) be a compression on A.
Then
Proof. We will first show that if P such that P P ,thenP restricted to 2C 0
P0(A)isacompressiononP0(A). Let Pˆ denote the restriction of P to P0(A).
77 3 Spectral Theory for Ordered Spaces
P P = P, P 0,P,P0 are mutually compatible . Now let x P (A), 0 ) 0 0 2 0 say x = P a for some a A.ThenPxˆ = PP a = P Pa P (A). So, P (A)is 0 2 0 0 0 2 0 0
invariant under Pˆ. This implies Pˆ is a projection on P0(A). Since P is positive
and weakly continous on A, Pˆ is positive and weakly continous on P0(A). Next, since x = x , x P (A), we have k kP0(A) k kA 8 2 0 Pˆ =sup Paˆ a 1,a P (A) k k {k kP0(A) |k kP0(A) 2 0 } =sup Pa a 1,a P (A) {k kA |k kA 2 0 } sup Pa a 1,a A {k kA |k kA 2 } = P 1 k k Therefore, Pˆ is normalised. So, Pˆ is a weakly continous normalised positive pro-
jection on P0(A).
Now, let Q = P P 0 = P 0P = P P 0 P and let Qˆ denote the restriction of 0 0 0 ^ 0
Q to P0(A). Then, Qˆ is also a weakly continous normalised positive projection on
P0(A), like we showed for Pˆ.
Claim 1: Pˆ0 (the complement of Pˆ in P0(A)) = Qˆ
Let x P (A),x 0. Then 2 0 + + + x ker Pˆ Pxˆ =0 Px =0 x ker P = im P 0 2 () () () 2 () + + P 0x = x P 0P x = x Qx = x x im Qˆ.So,ker Pˆ = () 0 () () 2 im+Qˆ.
Similarly, x im+Pˆ Pxˆ = x Px = x x im+P = 2 () () () 2 + + ker P 0 P 0x =0 P 0P x =0 Qxˆ =0 x ker Qˆ () () 0 () () 2 This shows that im+Pˆ = ker+Qˆ.
So, P,ˆ Qˆ are complementary projections on P0(A).
Claim 2: Pˆ⇤ = P ⇤ P (V ) = P ⇤ restricted to P0(V ) | 0⇤
First note that PP = P P = P ⇤P ⇤ = P ⇤P ⇤. From this it follows that 0 0 ) 0 0
P0⇤(V )isinvariantunderP ⇤ and so, P ⇤ is a projection on P0⇤(V ). Now, let
78 3 Spectral Theory for Ordered Spaces
x P (A),y P ⇤(V ). Then
x, P ⇤y>.ThisimpliesPˆ⇤y = P ⇤ P (V )y. Hence, the claim. | 0⇤
Claim 3: Qˆ⇤ =(Pˆ⇤)0
LHS = (Qˆ)⇤ = Q⇤ P (V ) =(P0P 0)⇤ P (V ) = P 0⇤P ⇤ P (V ) = P 0⇤ P (V ).There- | 0⇤ | 0⇤ 0 | 0⇤ | 0⇤
fore it is su cient to prove (Pˆ⇤)0 = P 0⇤ P (V ). | 0⇤
Let x P ⇤(V ),x 0. 2 0 + + + x ker Pˆ⇤ Pˆ⇤x =0 P ⇤x =0 x ker P ⇤ = im (P ⇤)0 = 2 () () () 2 + + + im (P 0)⇤. This implies ker Pˆ⇤ = im P 0⇤ P (V ). | 0⇤ + Conversely, if x im (Pˆ)⇤ Pˆ⇤x = x P ⇤x = x x 2 () () () 2 + + + + + im P ⇤ = ker (P 0)⇤ = ker (P ⇤)0.Thisimpliesim Pˆ⇤ = ker P 0⇤ P (V ). | 0⇤
Hence, Pˆ⇤, Qˆ⇤ are complementary.
From the claims, we see that Pˆ is bicomplementary. Hence, Pˆ = P is a |P0(A)
compression on P0(A).
Next, we will prove that
compression on P0(A)withprojectivefaceF .Thus,everyexposedfaceofF0 is a
projective face of P0(A). Hence, the standing hypothesis is satisfied.
Now, we will show that if (P, F, p)issomecompressionofP (A), then P = R 0 |P0(A) for some R such that R P . 2C 0 79 3 Spectral Theory for Ordered Spaces
We know that F is the projective face corresponding to projective unit p0 in the order unit space (P (A),p ). This implies F = w F
=0. 0 0 { 2 0 | 0 } Claim: F = w K
( ) Let w F = P ⇤w = w and
( ) Let w K and
p, w >.Therefore,equalityholdseverywhereandweget
From the claim, it follows that F is an exposed face of K = F is projective ) face of V = R with associated projective face F .ButF F = )9 2C ✓ 0 ) R P = R is a compression on P (A) with associated projective face F . 0 ) |P0(A) 0 Now, P and R are two compressions on P (A)withthesameprojectiveface. |P0(A) 0 Hence, P = R . |P0(A)
Corollary 3.111. Assume < A, V > satisfy the standing hypothesis. Let Pˆ be a central compression on A and let P be any other compression on A such that P Pˆ. Then P is central for A P is central for Pˆ(A). ()
Proof. By 3.75, P is central for A (or Pˆ(A)) Pa a, a A+ (or Pˆ(A)+). () 8 2
( ) If P is central for A,thenPa a for all a Pˆ(A)+ A+.Thus,P is ( 2 ✓ central for Pˆ(A)+.
80 3 Spectral Theory for Ordered Spaces
( ) Assume P is central for Pˆ(A)= Pb b, b im+Pˆ.Inparticular, ) ) 8 2 P (Paˆ ) Pˆ a, a A+. Now, P Pˆ = P Pˆ and P = PPˆ = P Pˆ. 8 2 ) ⇠ Then, Pa = P Paˆ Paˆ a, a A+.Thus,P is central for A. 8 2
3.4.1 The lattice of compressions when A = V ⇤
In this section, we will continue to study < A, V > apairoforderunitspaceand base norm space under separating order and norm duality, satisfying the standing hypothesis, with the additional assumption that A = V ⇤.Inthiscase,A can be identified with Ab(K), the space of all bounded a ne functions on K,byanorder preserving isomorphism (a aˆ). ! Remark 3.112. Under this mapping a aˆ, 7!
a a weakly in A aˆ aˆ in A (K)pointwise { n} ! (){ n} ! b
Proof. a a weakly { n} !
limn
limn
limn aˆn(k)=ˆa(k), k K () !1 8 2 aˆ aˆ pointwise in A (K) (){ n} ! b
Definition 3.113. Alattice is called a complete lattice if every bounded subset L of has a greatest lower bound and least upper bound. ( In orthocomplemented L lattice, every subset is bounded by 0 and I.)
Definition 3.114. Alatticeissaidtobemonotone complete if every bounded monotone net has a limit.
Remark 3.115. Assume A = V ⇤.ThenA is monotone complete.
81 3 Spectral Theory for Ordered Spaces
Proof. Let a be a bounded increasing net in A,saya a , ⇤. Fix { } 0 8 2 k K.Then is an increasing net of real numbers, bounded above by 2 { }
Thus, every increasing net a A which is bounded above, has a least upper { ↵}2 bound a A,denotedasa a.Inthiscase,a is also the weak limit of a . 2 ↵ % { ↵} Similarly, every decreasing net a A which is bounded below, has a greatest { ↵}2 lower bound a A,denotedasa a.Inthiscase,a is also the weak limit of 2 ↵ & a . { ↵} Remark 3.116. If a a (or a a )andP is a compression on A,thenbyweak ↵ % ↵ & continuity, Pa Pa.(orPa Pa respectively). ↵ % ↵ &
Proof. Note that a a a a weakly. Since P is positive, Pa is ↵ % () ↵ ! ↵ increasing net and by weak continuity and Pa Pa weakly. Hence, Pa ↵ ! ↵ % Pa.
Lemma 3.117. Assume < A, V > satisfy the standing hypothesis and A = V ⇤.If
82 3 Spectral Theory for Ordered Spaces
p is a decreasing (increasing) net of projective units and p a A, then a { ↵} ↵ & 2 is a projective unit. So, a is also the greatest lower bound (least upper bound) of p in P. { ↵}
Proof. Let p be a decreasing net of projective units such that p a A.Let { ↵} ↵ & 2 F be the projective face corresponding to p .DefineG = w K < a, w > = ↵ ↵ { 2 | 1 0 = K a (0). Then, G is an exposed face of K and hence a projective face of V , } \ by the standing hypothesis. Let (P, G0,p)denotethecompressioncomplementary
11 to G. Then F 0 = w K
=0 G, ↵.Thisimplies ↵,G0 ↵ { 2 | ↵ }✓ 8 8 ✓ F = p p = p a. Now, p is the largest element in A+ which vanishes ↵ ) ↵ ) 1 on G. This implies p a.Therefore,a = p . 2P Similarly, for an increasing net p a,considerthedecreasingnet(e p ) ↵ % ↵ & (e a). Then, (e a) as above, which in turn implies that (e a)0 = a . 2P 2P
Proposition 3.118. Assume < A, V > satisfy the standing hypothesis and A =
V ⇤. Then , , are complete lattices and the lattice operations are given by the C F P following equations for each family F : { ↵}✓F
0 F↵ = F↵, F↵ = F↵0 (3.4.11) ↵ ↵ ↵ ↵ ^ \ _ \ Corollary 3.119. Assume < A, V > satisfy the standing hypothesis and A = V ⇤. Then each semi-exposed face of K is projective.
Remark 3.120. From the above result it follows that under the standing hypothesis and A = V ⇤,everysemi-exposedfaceofK is exposed.
Now, we present a few miscellaneous topics, which were studied during the course of this project. Some of them are not directly needed for proving the main spectral theorem. Hence, the proofs of these have been omitted here.
110 =0 ↵ 83 3 Spectral Theory for Ordered Spaces Bicommutant Definition 3.121. Assume < A, V > satisfy the standing hypothesis and A = V ⇤. Let a A.Thenthe -Bicommutant of a is defined as the set of all compressions 2 C on A,whicharecompatiblewitha as well as compatible with all compressions compatible with a. -Bicommutant of a = P P a, P Q whenever Q a C { 2C| ⇠ ⇠ ⇠ } The corresponding set of projective units and projective faces is called -Bicommutant P of a and - Bicommutant of a respectively. F Proposition 3.122. Assume < A, V > satisfy the standing hypothesis and A = V ⇤. Then the -Bicommutant of a is a complete boolean algebra. C Orthogonal decomposition in V Definition 3.123. Two elements ⇢, V + are said to be orthogonal,denoted 2 as ⇢ ,if ⇢ = ⇢ + . ? k k k k k k Theorem 3.124. Assume < A, V > satisfy the standing hypothesis and A = V ⇤. Then every w V admits an orthogonal decomposition w = ⇢ , where ⇢, 2 0,⇢ . This decomposition is unique and is given by ⇢ = P ⇤w, = P 0 w for ? ⇤ some compression P of A. Central elements Definition 3.125. Assume < A, V > satisfy the standing hypothesis and A = V ⇤. An element w V is called central if (P ⇤ + P 0⇤)w = w for all compressions P . 2 Lemma 3.126. Assume < A, V > satisfy the standing hypothesis and A = V ⇤. Let w be a central element of V and w = ⇢ be its unique orthogonal decomposition. Then ⇢, are also central. 84 3 Spectral Theory for Ordered Spaces Definition 3.127. An ordered vector space (V, )whichisalatticeunderthis order (i.e. closed under , )iscalledavector lattice. ^ _ Remark 3.128. Let V be a vector lattice. Then the following holds ,u,v,w V . 8 2 (i) (u + w) (u + v)=u +(v w) _ _ (ii) (u + w) (u + v)=u +(v w) ^ ^ (iii) ↵u ↵v = ↵(u v),↵ 0 _ _ (iv) ↵u ↵v = ↵(u v),↵ 0 ^ ^ (v) ↵u ↵v = ↵(u v),↵ 0 _ ^ (vi) ↵u ↵v = ↵(u v),↵ 0 ^ _ (vii) u v = (u v) _ ^ (viii) u v = (u v) ^ _ Proposition 3.129. Assume < A, V > satisfy the standing hypothesis and A = V ⇤. Then the set of all central elements in V is a vector lattice. + We know that each projective unit is an extreme point of A1 .Thefollowingis apartialconversetothisresult,underouradditionalassumptions. Proposition 3.130. Assume < A, V > satisfy the standing hypothesis and A = V ⇤. Then the projective units of A are w⇤-dense in the set of extreme points of + A1 . Proposition 3.131. Assume < A, V > satisfy the standing hypothesis and A = V ⇤.Ifp is a minimal (non-zero) projective unit, then the associated compression P has a one dimensional range (im P = Rp) Also, the associated projective face F is a singleton (F =ˆp), where pˆ is the only point of K where p takes the value 1. And the map p pˆ is a one-one map 7! from the set of minimal points of onto the set of exposed points of K. P 85 3 Spectral Theory for Ordered Spaces Central support Proposition 3.132. Assume < A, V > satisfy the standing hypothesis and A = V ⇤. For each w K, 2 there is a least central projection c(w) such that Definition 3.133. Assume < A, V > satisfy the standing hypothesis and A = V ⇤. For each w K, the least central projective unit which takes the value 1 at w is 2 called the central support or central carrier of w.Itisdenotedbyc(w). Orthogonality in A+ Definition 3.134. Assume < A, V > satisfy the standing hypothesis and A = V ⇤. We will say that two elements a, b A+ are orthogonal (a b)ifthereexistsa 2 ? compression (P, F, p)suchthatPa = a, P 0b = b. Definition 3.135. An equality a = b c is called an orthogonal decomposition of an element of a A if b, c A+ and b c. 2 2 ? Lemma 3.136. Assume < A, V > satisfy the standing hypothesis and A = V ⇤. Let b, c A+ and (P, F, p) be a compression on A. Then the following are equivalent 2 conditions for orthogonality of b and c. (i) Pb = b and P 0c = c (ii) P 0b =0and Pc =0 (iii) b =0on F 0 and c =0on F . Remark 3.137. Let a A+ and P .Then 2 2C (i) Pa P 0a ? (ii) If a = b c is an orthogonal decomposition of a such that Pb = b, P 0c = c, then Pa = b, P 0a = c. 86 3 Spectral Theory for Ordered Spaces Hence, a = Pa+ P 0a and thus a is compatible with P . 3.5 Range Projections In this section, we use our knowledge of compressions to develop a concept called range projection, which will be a fundamental tool for the spectral theorem. Proposition 3.138. Assume < A, V > satisfy the standing hypothesis and A = + V ⇤. Then, for each a A , there exists a least projective unit p such that a 2 2 face(p). Further, p is the unique element of such that w K < a, w > = P { 2 | 0 = w K Proof. If a =0,thenwecantakep =0 = a face(p). And clearly, 0 will ) 2 a + be the least such projective unit. So, assume a =0.Leta1 = a A1 .Let 6 k k 2 1 G = K a (0). Then G is an exposed face of K and hence a projective face by \ the standing hypothesis. Let p be the projective unit associated with the face G0. So, by proposition 4.101, G = w K Definition 3.139. Assume < A, V > satisfy the standing hypothesis and A = V ⇤. For each a A+,definer(a)tobetheleastprojectiveunitp in A such that 2 a face(p). We call r(a)tobetherange projection of a. 2 The above proposition shows the existence and uniqueness of range projections for elements of A+. 87 3 Spectral Theory for Ordered Spaces Proposition 3.140. Assume < A, V > satisfy the standing hypothesis and A = + V ⇤. Let a A , then r(a) is characterised by each of the following statements: 2 1. r(a) is the only projective unit such that < a, w > =0 2. r(a) is the least projective unit p such that a a p. k k 3. The compression associated with r(a) is the least compression P such that Pa = a. 4. The complement of the compression associated with r(a) is the greatest com- pression P 0 such that P 0a =0. 5. r(a) is the greatest projective unit contained in the semi-exposed face of A+ generated by a. Proof. Let P be the compression associated with r(a). Then (1), (2), (3), (4) directly follow from the definition of range projections, propo- sition 3.138, and the fact that a face r(a) a im+P (prop. ??). 2 () 2 + To show (5), recall that the semi-exposed face generated by a in A = a ••. { } Now, for any s A+, 2 s a •• 88 3 Spectral Theory for Ordered Spaces Remark 3.141. If a, b A+ such that a b,thenr(a) r(b). This is because 2 a b b r(b)= a face r(b) . Hence, r(a) r(b). k k ) 2 Proposition 3.142. Assume < A, V > satisfy the standing hypothesis and A = + V ⇤. Let a A and P be a compression on A with projective unit p. Then 2 r(Pa)= r(a) p0 p _ ^ Proof. Let Q, R be the compressions associated with r(a)andr(Pa)respectively. + + Now, a face(p)= a im Q im (Q P 0). Thus, (Q P 0)a = a. Also, 2 ) 2 ✓ _ _ P 0 Q P 0 = P, P 0 (Q P 0). Hence, (Q P 0)(Pa)=P (Q P 0)a = Pa. _ ) ⇠ _ _ _ So, (Q P 0)fixesPa.Byproposition3.140(3),R (Q P 0). So, _ _ r(Pa) r(a) p0 (3.5.2) _ Similarly, P fixes Pa = R P (proposition 3.140 (3)). Therefore, ) r(Pa) p (3.5.3) From 3.5.2, 3.5.3, we have r(Pa) r(a) p0 p (3.5.4) _ ^ + Next, R P = R0 P . Thus, (R0 P )a = R0Pa =0.Therefore,a ker (R0 ) ? ^ 2 ^ + P )=im (R P 0). Hence, R P 0 fixes a = Q R P 0.So,r(a) r(Pa) p0. _ _ ) _ _ This implies r(a) p0 r(Pa) p0. Also, note that r(Pa) p0 = r(Pa)+p0 (as _ _ _ R P 0). Therefore, r(a) p0 p0 r(Pa). Thus, by proposition ??, ? _ r(a) p0 p r(Pa)(3.5.5) _ ^ Hence, by eqn. 3.5.3 and 3.5.5, we have r(Pa)= r(a) p0 p. _ ^ Proposition 3.143. Assume < A, V > satisfy the standing hypothesis and A = + V ⇤. Let a, b A . Then 2 r(a) r(b) a b r(a + b)=r(a)+r(b) ? () ? () 89 3 Spectral Theory for Ordered Spaces Proof. Let Ra,Rb be the compressions associated with r(a)andr(b)respectively. Claim: r(a) r(b) a b ? () ? ( ) Assume r(a) r(b). Then R R =0.ThisimpliesR a = R (R a)=0 = ) ? b a b b a ) R0 a = a. And R b = b (proposition3.140(3)).Thus,a b. b b ? ( ) Let a b.Thisimplies P such that Pa = a, P 0b = b.By ( ? 9 2P proposition 3.140 (3), R P and R P 0. Hence, by lemma ??, R R . a b a ? b So, r(a) r(b). ? Claim: a b r(a + b)=r(a)+r(b) ? () ( ) Assume a b. This implies r(a) r(b) by the first claim. We know ) ? ? a a r(a)andb b r(b). Hence, a + b a r(a)+ b r(b) 12 a + k k kk k k k k k b r(a)+r(b) .Thusbyproposition3.140(2), k r(a + b) r(a)+r(b) Next, by remark 3.141, r(a),r(b) r(a + b). So, r(a) r(b) r(a + b). And _ r(a) r(b)= r(a) r(b)=r(a)+r(b)(proposition??). Thus, ? ) _ r(a)+r(b)=r(a) r(b) r(a + b) _ Hence, r(a)+r(b)=r(a + b). ( ) Assume r(a)+r(b)=r(a + b) e.Thus,byproposition??, r(a) r(b). ( ? And then by the first claim, we have a b. ? Proposition 3.144. Assume < A, V > satisfy the standing hypothesis and A = + V ⇤. Let a A and Q be a compression compatible with a. Then 2 r(Q(a)) = Q(r(a)) 120 a, b a + b = a , b a + b )kk k kk k 90 3 Spectral Theory for Ordered Spaces Proof. Qa is orthogonal to Q0a.So,byproposition4.1.6, r(Qa)+r(Q0a)=r(Qa + Q0a)=r(a) Let q be the projective unit associated with Q.Then,Qa a Qe = Q(a) k k ) 2 + face(q). Hence, r(Qa) q.So,r(Qa) im Q (prop ??). Similarly, r(Q0a) 2 2 + + im Q0 =ker Q.Thus,Q(r(a)) = Q r(Qa)+r(Q0a) = Q(r(Qa)) + 0 = r(Qa). Hence, the result. Proposition 3.145. Assume < A, V > satisfy the standing hypothesis and A = + V ⇤.Ifa A , then r(a) is in the -bicommutant of a. 2 P Proof. Let P be the compression associated with r(a). Then Pa = a and P 0a =0 (prop. 3.140). Thus, Pa+ P 0a = a which implies r(a) a. ⇠ Next, let Q be any other compression compatible with a.Then,Qa a (prop ??). By remark 3.141, r(Qa) r(a). Now, r(Q(a)) = Q(r(a)) (prop 4.1.7). Hence, Q(r(a)) r(a). Thus, Q r(a)byprop??.Therefore,r(a)isin bicommutant ⇠ P of a. 3.6 Spaces in Spectral Duality Definition 3.146. Assume that A, V are a pair of order unit space and base norm space under separating order and norm duality, with A = V ⇤.If,inaddition,each a A admits a least compression P such that Pa a and Pa 0, then we say 2 that A and V are in Spectral Duality. Lemma 3.147. Let A, V be an order unit space and base norm space in separating order and norm duality. Let a A and P be a compression. Then 2 Pa a P is compatible with a and P 0a 0 () Proof. First, assume that P a and P 0a 0. Then, a = Pa+ P 0a = Pa = ⇠ ) a P 0a a. 91 3 Spectral Theory for Ordered Spaces Next, assume Pa a.ThenPa a 0. Thus, P (Pa a)=P 2a Pa =0 = ) + + Pa a ker P =im P 0.So,Pa a = P 0(Pa a)= P 0a = P 0a 0and 2 ) a = Pa+ P 0a. Hence, P 0a 0andP a. ⇠ Lemma 3.148. Let A, V be an order unit space and base norm space in separat- ing order and norm duality. Let a A and P be a compression wth associated 2 projective face F . Then the following are equivalent: (i) Pa a, P a 0. (ii) P a, Pa 0, P 0a 0. ⇠ (iii) F a, a 0 on F , a 0 on F 0 ⇠ Proof. (i) (ii) follows from lemma 3.147. For (ii) (iii), recall that a () () ⇠ P a F . Also, Pa 0 P a, w 0, w K a, P ⇤w () ⇠ () h i 8 2 () h i 0 w K a 0onF .Similarly,P 0a 0 a 0onF 0. 8 2 () () So, A, V are in spectral duality for each a A,thereexistsaleast () 2 compression P compatible with a such that either (i) or (ii) holds, or there exists aleastprojectivefaceF satisfying (iii). Notation 14. let b A+.Thenwedenote[0,b]= a A 0 a b and 2 1 { 2 | } face (p)= a A 0 a b, for some 0 . { 2 | } + Definition 3.149. Let A, V be in spectral duality. An element b of A1 is said to have facial property if [0,b]=A+ face (b). 1 \ Remark 3.150. Every projective unit satisfies the facial property. This is because, for a projective unit p, a a p a face (p)(byprop??). So, a A k k () 2 { 2 | a a p A+ = a face (p) A+ This implies k k }\ { 2 }\ face (p) A+ = a A 0 a p =[0,p] \ 1 { 2 | } 92 3 Spectral Theory for Ordered Spaces Proposition 3.151. Let A, V be in spectral duality. If b A+ has facial property, 2 1 then b is a projective unit. Proof. Define a =2b e.Byspectralduality, P such that Pa a and 9 2P a+e 1 Pa 0. By lemma 3.147, P a, P 0a 0. Then, b = = (Pa + P 0a)+ ⇠ 2 2 1 a+e a+e 2 (Pe+ P 0e)=P ( 2 )+P 0( 2 )=Pb+ P 0b.Thus, P b and b Pb (3.6.1) ⇠ Let p = Pe and p0 = P 0e. Now, P (2b e)=Pa 0= 2Pb p.Byeqn ) 4.3.12, we get 2b 2Pb p = p face (b). Hence, by facial property of b,we ) 2 get p b Note that 0 b e = e a e. Now, a Pa = 2b e P (2b e)= ) ) Pa Pe = p = 2b p e. Also, 2b p = b+(b p) 0. Thus, 0 2b p e. ) Next, 2b p 2b = 2b p face (b). So, 2b p face (b) A+. Hence, by ) 2 2 \ 1 facial property of b,weget2b p b which implies b p.Therefore,b = p. Lemma 3.152. Let A, V be in spectral duality. If P, Q are compatible compres- sions, then PQ = QP = P Q ^ Proof. Define r = PQe.Then,wehaver p, r q. Note that p, q 0= ) p q 0. Now, if s A+ such that s p, s q,thens im+P, im+Q (by ^ 2 2 prop. ??). This implies s = Ps Pq = r.So,r s whenever 0 s p, q.Thus, r is the greatest lower bound of (p, q)inA+13. Claim. r has facial property. 13Without the standing hypothesis, we don’t yet know whether join of projective units is a projective unit. So we can’t conclude that r =(P Q)e ^ 93 3 Spectral Theory for Ordered Spaces Let a A+ face (r). Since r p,wehavea face (p). This implies 2 1 \ 2 a im+P = a = Pa Pe = p.Similarly,a q.Thus,a p q = r.So, 2 ) ^ a [0,r]. Hence, the claim. 2 Now, by proposition 3.151, r is a projective unit. And as r = p q,wesee ^ that precisely r =(P Q)e. Now, we will show that PQ = P Q 14.Let ^ ^ a A+.ThenPQa PQe = r =(P Q)e.So,PQa im+(P Q) 2 1 ^ 2 ^ ✓ im+P, im+Q (by prop ??). Also, P, Q, P Q are mutually compatible. Hence, ^ PQa =(P Q)(PQa)=P (P Q)(Qa)=PQ(P Q)a =(P Q)a.Thus, ^ ^ ^ ^ PQ = P Q on A+ = PQ = P Q. ^ 1 ) ^ Currently, , , are order isomorphic sets. We don’t know if they are lattices. C P F However, under spectral duality, we have shown that if P Q,thenP Q exists ⇠ ^ and P Q = PQ.Further,P Q P 0 Q0 = P 0 Q0 = P 0Q0.So, ^ ⇠ ⇠ ⇠ ) ^ P Q =(P 0 Q0)0 also exists, if P and Q are compatible. _ ^ Also, P Q = P Q.Therefore,P Q exists for orthogonal compressions ? ) ⇠ _ P and Q. Lemma 3.153. Let A, V be in spectral duality. Let P, Q be mutually compatible compressions, both compatible with an element a A. Then, P Q, P Q are 2 ^ _ also compatible with a. Proof. First assume a A+. P, Q a = P a, Qa a.So,bylemma3.152, 2 ⇠ ) (P Q)a =(PQ)a = P (Qa) Pa a = P Q a. Next, P Q = P 0 ^ ) ^ ⇠ ⇠ ) ⇠ Q0. Also, P, Q a = P 0,Q0 a.Thenasabove,P 0 Q0 a which further ⇠ ) ⇠ ^ ⇠ implies (P 0 Q0)0 = P Q is compatible with a. ^ _ Now consider any a A compatible with both P, Q.Wehavea = a a where 2 1 2 1 1 + 1 a = ( a e+a),a = ( a e a) A .ThenPa +P 0a = ( a (p+p0)+a)= 1 2 k k 2 2 k k 2 1 1 2 k k 14PQe = r =(P Q)e does not imply PQ = P Q because we do not know if PQ is a compression. ^ ^ 94 3 Spectral Theory for Ordered Spaces 1 ( a e + a)=a = P a .Similarly,itcanshownthatP a ,Q 2 k k 1 ) ⇠ 1 ⇠ 2 ⇠ a and Q a .ThenP Q, P Q a ,a = P Q, P Q a (by ??). 1 ⇠ 2 ^ _ ⇠ 1 2 ) ^ _ ⇠ Proposition 3.154. Let A, V be in spectral duality. Let P, Q be orthogonal com- pressions, both compatible with an element a A. Then P Q is compatible with 2 _ a and (P Q)a = Pa+ Qa. _ Proof. P Q = P Q and we are given P, Q a.So,fromlemma3.153, ? ) ⇠ ⇠ we have P Q a.Then,a =(P Q)a +(P Q)0a =(P Q)a +(P 0 Q0)a = _ ⇠ _ _ _ ^ (P Q)a +(P 0Q0)a. Now, P, Q a = a = Qa + Q0a, a = Pa + P 0a.So, _ ⇠ ) 15 P 0Q0a = P 0(a Qa)=P 0a P 0Qa = P 0a Qa =(a Pa) Qa.Substituting this above, we get a =(P Q)a +(a Pa Qa)= (P Q)a = Pa+ Qa. _ ) _ Remark 3.155. We had seen that if A = V ⇤,thenA is monotone complete. So, when A, V are in spectral duality, then A is monotone complete too. Lemma 3.156. Assume A, V are in spectral duality. If p is a decreasing se- { n} quence of projective units, then the element infn pn of A is a projective unit, hence it is the greatest lower bound of p among the projective units. { n} Similarly, if p is an increasing sequence of projective units, then the element { n} sup p of A is a projective unit, and is the least upper bound of p among the n n { n} projective units. Proof. As A is monontone complete and the decreasing sequence of projective units p is bounded below by 0, there exists b A such that b =inf p . As, { n} 2 n n 0 p e, n,wehaveb A+.Wewillprovethatb has facial property. Let a n 8 2 1 2 A+ face (b). Now, b p = face (b) face (p ). So, a A+ face (p )= 1 \ n ) ✓ n 2 1 \ n [0,p ]. This implies a p , n.Thus,a inf p = b = a [0,b]. Now, by n n 8 n n ) 2 proposition 3.151, b is a projective unit and it is the greatest lower bound of p { n} in . P 15 + + P Q = Q P 0 = im Q im P 0 = P 0Q = Q ? ) ) ✓ ) 95 3 Spectral Theory for Ordered Spaces For an increasing sequence of projective units q ,lookatthecorresponding { n} decreasing sequence of complement projective units q0 .Then q0 r for { n} { n} ! some projective unit r A (as in the above case) which implies q r0,which 2 { n} ! is also a projective unit. + Lemma 3.157. Assume A, V are in spectral duality. Let a A and let 1, 2 R 2 2 such that 0 < 1 < 2.IfQ2 is a compression compatible with a, with associated projective unit q2, such that Q a q ,Q0 a q0 (3.6.2) 2 2 2 2 2 2 then there exists a compression Q1 compatible with a, with associated projective unit q , such that Q Q and 1 1 2 Q a q ,Q0 a q0 (3.6.3) 1 1 1 1 1 1 Proof. Define b = Q a A+.Letg = b e.Byspectralduality, P ,with 2 2 1 9 2C associated projective unit p, such that P g, Pg 0,P0g 0, i.e. ⇠ Pb p, P 0b p0,P g (3.6.4) 1 1 ⇠ Let Q = P 0, with associated projective unit q = p0.Then Q0b q0,Qb q, Q0 g 1 1 ⇠ Now, Q0 is compatible with b,sowehaveQ0b b = 0 q0 Q0b b. ) 1 Applying Q0 to this inequality, we get 0 Q0 q0 Q0 Q0b Q0 b = Q0 (Q a)=0. 2 1 2 2 2 2 2 + + Thus, Q0 q0 =0 = q0 ker Q0 =imQ .Therefore, 2 ) 2 2 2 Q q0 = q0 = Q Q0 = Q0Q = Q Q0 = Q0 Q (3.6.5) 2 ) 2 ⇠ ) 2 2 ^ 2 Thus, Q2Q0 is a compression on A (in general, product of two compressions is not acompression!)So,Q q0 = q0 = Q Q0e = Q0e.Thisimplies 2 ) 2 Q0Q = Q Q0 = Q0 and Q0 Q (3.6.6) 2 2 2 96 3 Spectral Theory for Ordered Spaces Define Q = QQ = Q Q . Q0a = Q0Q a = Q0b b a = Q0 a. Now, 1 2 ^ 2 2 ) ⇠ Q Q ,Q a, Q a.So,bylemma3.153,Q = Q Q a. ⇠ 2 ⇠ 2 ⇠ 1 ^ 2 ⇠ Next, applying Q on the inequality 0 Qb q,wegetQ (Qb) Q (Qe). 1 1 1 1 1 But Q1(Qb)=Q(Q1b)=Q1b = Q1(Q2a)=Q2(Q1a)=Q1a. Also, 1Q1(Qe)= Q(Q e)= Q e = q .So,wegetQ a q . 1 1 1 1 1 1 1 1 1 Eqn 3.6.6 gives Q0 Q0 and (Q )0 =(Q Q )0 = Q0 Q0 . As Q, Q a,by ? 2 1 ^ 2 _ 2 2 ⇠ proposition 3.154, Q0 a = Q0a+Q0 a.Similarly,Q0 e = Q0e+Q0 e = q0 = q0 +q0 . 1 2 1 2 ) 1 2 Also, Q0a = Q0Q a = Q0b.Thus,Q0 a = Q0 b + Q0 a q0 + q0 (q0 + q0 )= 2 1 1 2 1 2 2 1 2 q0 . Hence, we get Q0 a q0 . 1 1 1 1 1 Now we will show that under the spectral duality condition, the standing hy- pothesis is satisfied. Theorem 3.158. If A, V are in spectral duality, then every exposed face of K is projective. Proof. Let F be an exposed face of K.Then a A+ such that F = w K 9 2 1 { 2 | a, w =0 .Take = 1 and Q =I.ThisimpliesQ a, Q a e, Q0 a h i } 0 0 0 ⇠ 0 0 0 00. Choose a decreasing sequence of positive real numbers converging to 0, say > > > ... 0, i.e. lim =0.Bylemma3.157,wecanconstruct 0 1 2 3 ! n n Q Q Q Q ... Q ... such that 0 ⌫ 1 ⌫ 2 ⌫ 3 ⌫ ⌫ n Q a, Q a q ,Q0 a q0, for i =1, 2, 3 ... i ⇠ i i i i i i where the notation for the compressions are (Qn,Gn,qn), for each n.Bylemma 3.156, q q , for some projective unit q A.LetQ be the compression and { n}& 2 G be the projective face associated with the projective q.Then, q =infqn,G= Gn Q = Qn n n n \ ^ 97 3 Spectral Theory for Ordered Spaces For a given i, q Q a = q Q a, k 0, k K.Thus, k K,we i i i )hi i i i 8 2 8 2 have Q e, k Q a, k = e, f a, f , f G = e a on G . ih i i h i i ) ih i h i 8 2 i ) i i As G G , n,wehave e a on G, n.Thus,0=lim e a on ✓ n 8 n 8 n n G = a 0onG.Buta A+ = a 0onG. Hence, a =0onG.Therefore, ) 2 ) G F . ✓ Conversely, if w F ,then0 q0 ,w Q0 a, w a, w =0.This 2 nh n ihn ih i implies q0 ,w =0, n. Now, q q = q0 q0 = q0,w =0 = w h n i 8 n & ) n % )h i ) 2 G = F G. ) ✓ Thus, F = G. Hence, F is a projective face. So, all the results that we had shown previously, assuming the standing hy- pothesis, are now valid under the spectral duality condition. In particular, we note that , and are complete lattices, when A, V are in spectral duality. C P F Now we will look at some equivalent ways of defining spectral duality. Lemma 3.159. Assume A, V are in separating order and norm duality such h i that they satisfy the standing hypothesis and A = V ⇤. Let a A and (P, F, p) be a 2 compression on A such that P a, a 0 on F, a 0 on F 0 ⇠ Then G = w K P a, w =0 is a projective face and its complementary { 2 |h i } compression (P,˜ F,˜ p˜) satisfies F˜ F, Pa˜ = Pa and ✓ P˜ a, a > 0 on F,˜ a 0 on F˜0 ⇠ Proof. First note that, a 0onF = Pa 0; a 0onF 0 = P 0a 0 ) ) 98 3 Spectral Theory for Ordered Spaces Define G = w K P a, w =0 .ThenG is an exposed face of K and hence { 2 |h i } projective, by theorem 3.158. By proposition ??, Pa =0onF 0 which implies F 0 G = G0 F . ✓ ) ✓ If w G,then 2 a, w = P a, w + P 0a, w =0+ P 0a, w 0 h i h i h i h i If w G0 F ,then 2 ✓ a, w = P a, w + P 0a, w = P a, w +0 0(P 0a =0onF ) h i h i h i h i And P a, w =0 = w G.But,G G0 = .Thus, P a, w > 0onG0. h i ) 2 \ ; h i Hence, a 0onG and a>0onG0. Let Q be the compression associated with the projective face G.Then,by proposition ??, Pa =0onG = Q0(Pa)=P a, Q(Pa)=0 ) P 0a =0onG0 F = Q(P 0a)=P 0a, Q0(P 0a)=0 ✓ ) Therefore, Qa + Q0a = Q(Pa+ P 0a)+Q0(Pa+ P 0a)=P 0a + Pa = a = Q ) ⇠ a, Q0 a. ⇠ Now, let (P,˜ F,˜ p˜)=(Q0,G0,Q0e). Then, we have F˜ = G0 F and P˜ P . ✓ Also, Pa˜ = P˜(Pa+ P 0a)=Q0Pa+ Q0P 0a = Pa+0=Pa. Lemma 3.160. Assume that A, V are a pair of order unit space and base norm h i space under separating order and norm duality, with A = V ⇤. Then, the following are equivalent: 1. A, V are in spectral duality. 2. For each a A, there exists a unique projective face F compatible with a 2 such that a>0 on F and a 0 on F 0. 99 3 Spectral Theory for Ordered Spaces If these conditions are satisfied, then the face F in (2) is also the least projective face compatible with a such that a 0 on F and a 0 on F 0. Proof. (1) = (2) ) Assume A, V are in spectral duality. By lemma 3.148, there exists a least projective face F such that a F, a 0onF, a 0onF 0. Then, by lemma 3.159, there ⇠ exists projective face F˜ such that F˜ F, a F,˜ a > 0onF,˜ a 0onF˜0. ✓ ⇠ Note that F F˜,byminimalityofF . Hence, F = F˜. ✓ Next, we will show uniqueness of F˜.Supposethereexistsaprojectiveface F such that a F ,a 0onF ,a 0onF 0. Again, by minimality of F , 1 ⇠ 1 1 1 we get F F .Byorthomodularlatticestructureof under spectral duality, ✓ 1 F F F = F = F (F F 0). Now, a>0onF and a 0onF 0 = ✓ 1 ) 1 _ 1 ^ 1 ) F F 0 = = F F 0 =0.Thus,F = F 0=F . Hence F is the unique 1 \ ; ) 1 ^ 1 _ projective face satisfying the required properties. (2) = (1) ) We will first show that when (2) holds, the standing hypothesis is satisfied. Fix a A+ and let G = w K a, w =0.Forthisa, ! F such that 2 { 2 |h i } 9 2F a F, a > 0onF, a 0onF 0.WewillshowthatG0 = F . ⇠ + a A = a 0onK F 0.Thus,a =0onF 0.ThisimpliesF 0 G. 2 ) ◆ ✓ Next, let w G and P be the compression associated with F .Then, a, w = 2 h i Pa + P 0a, w = a, P ⇤w + a, P 0⇤w . Now, a =0onG F 0 = a, w = h i h i h i ◆ )h i 0, a, P 0⇤w =0.Thus, a, P ⇤w =0.Buta>0onF .ThisimpliesP ⇤w = h i h i + + 0= w ker P ⇤ =im(P ⇤)0,i.e.w F 0 = G F 0. ) 2 2 ) ✓ So, G = F 0 and hence G is a projective face. In conclusion, we have shown that the standing hypothesis holds, when we assume (2). Claim. For a A,theprojectivefaceF satisfying (2) is the least projective face 2 100 3 Spectral Theory for Ordered Spaces in K such that a F, a 0onF, a 0onF 0. ⇠ Suppose there exists a projective face F such that a F ,a 0onF ,a 1 ⇠ 1 1 0onF˜0.Then,bylemma3.159, F such that F˜ F ,a F,˜ a > 1 9 1 2F ✓ 1 ⇠ 0onF,˜ a 0onF˜0.ByuniqunessofF mentioned in (2), we get F = F˜ F . ✓ 1 Hence, the claim. So, we have shown that for each a A,thereisaleastprojectivefaceF in K 2 such that a F, a 0onF, a 0onF 0. Hence, A, V are in spectral duality. ⇠ Theorem 3.161. Assume that the standing hypothesis holds and A = V ⇤. Then, A, V are in spectral duality Each a A has a unique orthogonal decomposi- () 2 tion. Proof. ( ) Assume A, V are in spectral duality and let a A.Byspectralduality ) 2 (lemma 3.160), ! F such that a F, a > 0onF, a 0onF 0.IfP is 9 2F ⇠ the compression associated with F ,thenwehavea P, Pa 0,P0a 0. ⇠ Then a = b c,withb = Pa 0,c= P 0a 0, is an orthogonal decomposition of a. Suppose a = b c is another decomposition of a,withb = P a, c = 1 1 1 1 1 P 0a for some compression (P ,F ,p )compatiblewitha (prop. ??). Then, 1 1 1 1 P a 0= a 0onF ; P 0a 0= a 0onF . Now, by lemma 1 ) 1 1 ) 1 3.159, F˜ such that F˜ F, Pa˜ = P a and 9 2F ✓ 1 P˜ a, a > 0onF,˜ a 0onF˜0 (3.6.7) ⇠ But, by lemma 3.160, F is the unique projective face in K satisfying the properties in eqn 3.6.7. Thus, F˜ = F, P˜ = P . Hence, b1 = P1a = Pa˜ = Pa = b and c = a b = a b = c. Hence, a = b c is the unique orthogonal 1 1 decomposition of a. 101 3 Spectral Theory for Ordered Spaces ( ) Given a A,leta = b c be the unique orthogonal decomposition of a, ( 2 with b = P a, c = P 0a, a P for some compression (P, F, p)onA. Now, ⇠ Pa 0= a 0onF and P 0a 0= a 0onF 0.Bylemma3.159, ) ) F˜ F such that a F,˜ a > 0onF,˜ a 0onF˜0, Pa˜ = Pa and F˜ = G0 9 ✓ ⇠ where G = w K P a, w =0 . { 2 |h i } Claim. F˜ is the least projective face satisfying a F, a 0onF,˜ a ⇠ 0onF˜0 Suppose (P ,F ,p )isacompressiononA such that a F ,a 0onF ,a 1 1 1 ⇠ 1 1 0onF 0.Thenb = P a 0,c = P 0a 0anda = b c is an orthogonal 1 1 1 1 1 1 1 decomposition of a. By assumption, a has a unique orthogonal decomposi- tion. Thus, b = P a = Pa and c = P 0a = P 0a .Bylemma??, 1 1 1 1 b =0onF 0.Thus,F 0 G = w K P a = P a, w =0 .Thisimplies 1 1 1 ✓ { 2 |h 1 i } F˜ = G0 F . Hence, the claim. ✓ 1 So, for each a A,wehavefoundaleastprojectivefaceF such that a 2 ⇠ F, a 0onF,˜ a 0onF˜0. Hence, A, V are in spectral duality. Definition 3.162. Assume A, V are in spectral duality. For each a A,wewill 2 + write a = b and a = c where a = b c is the unique orthogonal decomposition of a. Remark 3.163. By theorem 3.161, the unique orthogonal decomposition of a = + + a a is given by a = Paand a = P 0a where P is the compression associated with the unique projective face F satisfying a F, a > 0onF, a 0onF 0. ⇠ Proposition 3.164. Assume A, V are in spectral duality. Let a A and r(a+) be 2 the range projection of a+. Then the compression P, F, r(a+) has the following properties: 102 3 Spectral Theory for Ordered Spaces 1. F is the unique projective face compatible with a such that a>0 on F and a 0 on F 0. 2. F is the least projective face compatible with a such that a 0 on F and a 0 on F 0. 3. P is the least compression such that Pa a and Pa 0. Proof. Given a A,byspectralduality,thereexistsuniquecompression(P, F, p) 2 such that a F, a > 0onF, a 0onF 0.Bytheorem3.161,theunique ⇠ + + orthogonal decomposition of a = a a is given by a = Pa and a = P 0a. Let H be the projective face associated with the projective unit r(a+). Claim. F = H From lemma 3.159, we know that F 0 = G = w K P a, w =0 = w { 2 |h i } { 2 K a+,w =0 .But,rangeprojectionshavetheuniquepropertythat w K |h i } { 2 | + + a ,w =0 = w K r(a ),w =0 = H0. Hence, F 0 = H0 = F = H. h i } { 2 |h i } ) Now, by lemma 3.160, the properties (1), (2) and (3) follow. + Definition 3.165. Fix R,a A.Definer (a)=r (a e) .The 2 2 corresponding compression is denoted by (R ,F ,r (a)). Remark 3.166. By proposition 3.164, we have r (a) (a e)andsor (a) a. ⇠ ⇠ Corollary 3.167. R ,F ,r (a) is the unique compression compatible with a such that a> eon F ,a e on F 0 (3.6.8) Also, R is the least compression compatible with a satisfying R a r (a) R0 a r0 (a) (3.6.9) Lemma 3.168. Assume A, V are in spectral duality and let a A, R.IfF, G 2 2 are two projective faces compatible with a such that F G and a> eon F G, ? [ then a> eon F G. _ 103 3 Spectral Theory for Ordered Spaces Proof. We are given (P, F, p) (Q, G, q). So, p q = p + q.Takew F G. ? _ 2 _ Then p, w + q, w = p q, w =1.Letµ = p, w and 1 µ = q, w . h i h i h _ i h i h i µ =1 = p, w =1 = w F F G.Therefore, a, w > . )h i ) 2 ✓ [ h i µ =0 = q, w =1 = w G F G.Therefore, a, w > . )h i ) 2 ✓ [ h i 1 1 Now, assume 0 <µ<1. Define ⇢ = µ P ⇤w F ; =(1 µ) Q⇤w G. 2 2 1 µ = p, w = e, P ⇤w = 1= e, µ P ⇤w .Thene, ⇢ =1.Similarly, h i h i ) h i h i e, = 1. Now, w F G = (P Q)⇤w = w.Thus, a, w = a, (P Q)⇤w = h i 2 _ ) _ h i h _ i (P Q)a, w = Pa + Qa, w = P a, w + Qa, w = a, P ⇤w + a, Q⇤w = h _ i h i h i h i h i h i µ a, ⇢ +(1 µ) a, > µe, ⇢ + (1 µ) e, = (µ +(1 µ)) = . h i h i h i h i Hence, a> eon each w F G. 2 _ Lemma 3.169. Assume A, V are in spectral duality. If a A+ and >0, then 2 r (a) r(a). Proof. We know that r (a) a and r(a) bicommutant of a.Thus,r (a) ⇠ 2P ⇠ r(a). Let F ,F be the projective faces associated with r (a),r(a)respectively.So, we have F, F 0 F . Now, F 0 =(F 0 F ) (F 0 F 0 ). ⇠ ^ _ ^ For each w F 0, a, w = r(a),w =0. 2 h i h i For each w F , a, w > e, w = >0. 2 h i h i Thus, F 0 F = . Hence, F 0 =(F 0 F 0 ) F 0 .Thus,F F = r (a) \ ; ^ ✓ ✓ ) r(a). Lemma 3.170. Assume A, V are in spectral duality. Let a, b A+ and >0. 2 Then a b = r (a) r (b) and r (a + b)=r (a)+r (b) ? ) ? Proof. a b = r(a) r(b) (proposition 4.1.6). Now, by lemma 3.169, we have ? ) ? r (a) r(a)andr (b) r(b)= r (a) r (b). ) ? Let the notations for the compressions be (R0,F,r(a)), (R ,F ,r (a)), (S ,G ,r (b)). We already know that F a. Next, 0 b b r(b)= 0 R b ⇠ kk ) 104 3 Spectral Theory for Ordered Spaces b R (r(b)) = 0. Thus, R b =0 b = F b. Hence, F (a + b). Similarly, k k ) ⇠ ⇠ G (a + b). As, F G and both are compatible with a + b,bylemma3.154, ⇠ ? we get F G (a + b) _ ⇠ By (3.6.8), we have a> eon F and b> eon G .Thus,a + b> on F G . [ Hence, by lemma 3.168, a + b> eon F G (3.6.10) _ Claim. a r(a)onF 0 and b r(b)onG0 r (a) r(a)= R R ,R0 .So,R0 a = R R0 a R r0 (a)= R R0 (e)= ) 0 ⇠ 0 0 0 R0 R (e)= R0 r(a). Therefore, R0 ( r(a) a) 0= a r(a)onF 0 . 0 ) Similarly, b r(b)onG0 . Hence, the claim. Now, r(a) r(b)= r(a)+r(b) e.So,onF 0 G0 =(F G )0,wehave ? ) \ _ a + b (r(a)+r(b)) e. Hence, a + b e on (F G )0 (3.6.11) _ So, by result 3.167, we see that r (a + b)mustbetheprojectiveunitofF G . _ And r (a) r (b)= r (a) r (b)=r (a)+r (b). Thus, ? ) _ r (a + b)=r (a) r (b)=r (a)+r (b) _ Lemma 3.171. Assume A, V are in spectral duality. If a A+ and (Q, G, q) is a 2 compression compatible with a, then for each >0, r (Qa)=Q(r (a)) Proof. As Qa Q0a,wehaver (Qa)+r (Q0a)=r (Qa + Q0a)=r (a), by ? proposition 4.1.6. By lemma 3.169 and 4.1.7, r (Qa) r(Qa)=Q(r(a)) Qe = 105 3 Spectral Theory for Ordered Spaces q.So,r (Qa) face (q). Then, by lemma ??, r (Qa) im+Q.Similarly, 2 2 + + r (Q0a) im Q0 =kerQ. Hence, Q(r (a)) = Q r (Qa)+r (Q0a) = r (Qa). 2 We know that if a 0and =0,thenr (a)=r(a)andhencebelongsto (P )-bicommutant of a.Thefollowingresultshowsthatthisistrueforarbitary a A and R. 2 2 Proposition 3.172. Let A, V be in spectral duality. If a A, then r (a) 2 2 bicommutant of a, for all R. P 2 Proof. Fix R.Byremark4.2.23,r (a) a.LetQ be a compression compati- 2 ⇠ ble with a. First, assume that a A+. 2 Case. <0 1 a e r(a e) a e e.Thus,e a e r(a e) = e k k k k ) 2 face (r(a e)). Hence, by facial property of projective units, e r(a e) e. Thus, r(a e)=e = r (a) Q. ) ⇠ Case. =0.Then,r (a)=r(a) -bicommutant of a.So,r (a) Q. 2P ⇠ Case. >0 By lemma 3.170, 3.171, Q(r (a)) + Q0(r (a)) = r (Qa)+r (Q0a)=r (Qa + Q0a)=r (a). Thus, r (a) Q. ⇠ So, we have proved the result for a 0. Now, let a A be arbitary. Define 2 b = a + a e A+ and µ = + a .ThenQ b and Q r (b)(likea 0). But k k 2 k k ⇠ ⇠ µ r (b)=r((µ be)+)=r((a e)+)=r (a). Hence, Q r (a). µ ⇠ Corollary 3.173. Assume A, V are in spectral duality. Let a A, R and F 2 2 be a compression compatible with a such that a µe on F , where µ> . Then, 106 3 Spectral Theory for Ordered Spaces F F . ✓ Proof. By proposition 3.172, we know that F F . Now, F =(F F ) (F F 0 ). ⇠ ^ _ ^ But a µe on F and a e on F 0 = F F 0 = .Thus,F =(F F ) ) \ ; ^ ✓ F . Remark 3.174. In particular, if a A and <µ,thenr (a) r (a). 2 µ 3.7 Spectral Theorem We are now ready to prove our spectral theorem, which generalises the correspond- ing theorem for von Neumann algebras and JBW algebras. Theorem 3.175. Assume A, V are in spectral duality and let a A. Then, 2 there is a unique family e R of projective units with associated compressions { } 2 (U ,G ,e ) on A such that (i) e is compatible with a, for each R 2 (ii) U a e ,U0 a e0 , R 8 2 (iii) e =0for < a ,e= e for > a k k k k (iv) e e for <µ µ (v) eµ = e , for each R µ> 2 V The family e is given by e = e r (a e)+ and each e is in the bicommutant { } P of a. + Proof. Define e = e r (a e) = e r (a)=r0 (a). Then U = R0 ,G = F 0 . 8 (i) By proposition 3.172, we have r ,r0 bicommutant of a,foreach . 2P Thus, each e is in the bicommutant of a. P 107 3 Spectral Theory for Ordered Spaces (ii) Holds by proposition 3.167. (iii) If < a ,then e a e a = a e > 0. So, e r(a e) e. k k k k ) Thus, r(a e)=e = e =0. ) If > a ,thena e =(a a e) ( a )e 0 ( a )e 0. So, k k k k k k k k (a e)+ =0 = r (a)=0 = e = e. ) ) (iv) Let <µ.Thenbyresult3.173,F F = F 0 F 0 = e e . µ ✓ ) ✓ µ ) µ (v) We will show Fµ0 = F 0 ,foreach R. Fix R and µ0 > a .Then, µ> 2 2 k k µ> by (iv), F 0 V is a bounded decreasing net of projective faces in K.By { µ}µ=µ0 lemma 3.156, this net converges to some projective face G = µ> Fµ0 = F 0 .Byproposition??, G a. As F 0 F 0 , µ> V,wehave µ> µ ⇠ µ ◆ 8 GT F 0 = G0 F . This implies ◆ ) ✓ a> eon G0 Let w G F 0 , µ> = a, w µ, µ> = a, w .This 2 ✓ µ 8 )h i 8 )h i implies a e on G Thus, by result 3.167, we have G = F 0 . Hence, µ> Fµ0 = F 0 . V Uniqueness: Let f R be arbitrary projective units, with associated compressions (T ,H ,F ), { } 2 satisfying (i), (ii), (iii), (iv) and (v). We will show that H = F 0 , where 8 F is the projective face associated with the compression (R ,F ,r ). By result 3.167, R is the least compression compatible with a satisfying R a r (a); R0 a r0 (a) So, R T 0 = F H0 = F 0 H . ) ✓ ) ✓ 108 3 Spectral Theory for Ordered Spaces Next, H = H . T 0 a = a on H0 and T 0 a µf 0 = µT 0 e.Thus, µ> µ µ µ µ µ µ a µe on H0 V.Bylemma3.173,H0 F , µ> = H F 0 , µ> µ µ ✓ 8 ) µ ✓ 8 = H = H F 0 . ) µ> µ ✓ T So, H = F 0 , . 8 Definition 3.176. If A, V are in spectral duality and a A,thentheunique 2 family e R of projective units, given in theorem 3.175, is called the spectral { } 2 resolution of a and the projective units e in this family are called the spectral units of a. Proposition 3.177. Assume A, V are in spectral duality and a A. Let e R 2 { } 2 be the spectral resolution of a. For each real increasing finite sequence = , ... { 0 1 n} with 0 < a and n > a , define =max1 i n( i i 1). Then the Rie- k k k k k k n mann sums s := i=1 i 1(e i e i 1 ), converge in norm to a when 0, k k! i.e. P lim s a =0 0 k k k k! Proof. Given ✏>0, choose finite increasing sequence = , ... with { 0 1 n} 0 < a , n > a and i i 1 <✏, i 1, 2 ...n .LetU be the projective k k k k 8 2{ } face associated with e .Bytheorem3.175(iv),weknowthate e whenever µ <µ. So, for <µ,wehavee e0 = e e = e e0 and (U U 0 )a = U a U 0 a. ? µ ) µ µ ^ µ ^ µ Note that e e = U ,U0 ,U ,U0 are mutually compatible compressions. µ ) µ µ Now, U a µe = U 0 (U a) µU 0 e = (U 0 U )a µU 0 U e = µ µ ) µ µ ) ^ µ µ ) (U 0 U )a µ(e0 e ). Similarly, U 0 a e0 = U U 0 a U e0 = ^ µ ^ µ ) µ µ ) (U U 0 )a U U 0 e = (U U 0 )a (e e0 ). µ ^ µ ) µ ^ µ ^ Therefore, (e e ) (U U 0 )a µ(e e )(3.7.1) µ µ ^ µ 109 3 Spectral Theory for Ordered Spaces Let pi = e i e i 1 ,fori =1, 2 ...n.ThenPi := U i U i 1 is compatible with a ^ and satisfies i 1(e i e i 1 ) Pia i(e i e i 1 ),i 1, 2 ...n (3.7.2) 2{ } n n Now, i=0 pi = i=0(e i e i 1 )=en e0 = e. This implies pi pj, for i = j ? 6 and Pn P = I,Pa P , i.Hence,byprop??, n P a = a. Summing over n i=1 i ⇠ i 8 i=1 i in (3.7.2),P we get P n n i 1(e i e i 1 ) a i(e i e i 1 ) i i X X n n = 0 a s ( i i 1)(e i e i 1 ) (e i e i 1 )= e ) k k k k i i X X = a s . Hence, lim 0 s a =0. )k kk k k k! k k Corollary 3.178. Assume A, V are in spectral duality. Then each a A can be 2 n approximated in norm by linear combinations i 1pi of mutually orthogonal i=1 projections p in the bicommutant of a. P i P Proof. Let e R be the spectral resolution of a and = 0, 1 ... n be a { } 2 { } finite increasing sequence of real numbers with < a , > a .Putting 0 k k n k k pi = e i e i 1 ,fori =1, 2 ...n,weseethat n n pi = (e i e i 1 )=en e0 = e i=0 i=0 X X By lemma ??, pi pj, for i = j.Moreover,e i and e i 1 belong to -bicommutant ? 6 P of a implies pi = e i e i 1 = e i e 0 i 1 also belongs to -bicommutant of a,by ^ P n lemma 3.153. By proposition 3.177, the Riemann sums s = i i 1(e i e i 1 )= n i 1pi converge in norm to a. P i=1 P 110 3 Spectral Theory for Ordered Spaces 3.7.1 Functional Calculus Definition 3.179. Let f be a bounded real valued function defined on an interval [a, b]andP = x ,x ...x be a partition of [a, b]. Let g be an increasing real { 0 1 n} valued function on [a, b]. Then, the Riemann - Stieltjies integral of f with respect to g is defined as b n f(x)dg(x)= lim f(yi)[g(xi) g(xi 1)], where yi [xi,xi 1] P 0 2 a k k! i=1 Z X whenever this limit exists. If we define Riemann-Stieltjies integrals with respect to e R as the norm { } 2 limit of the approximating Riemann sums, then we can restate the result of theorem 3.177 as a = de Z Definition 3.180. Assume A, V are in spectral duality. Let a A and e R 2 { } 2 be the spectral resolution of a.Foracontinousfunctionf defined on [ a , a ], k k k k we can define the integral with respect to e R,astheweakintegralinA = V ⇤, { } 2 determined by the equation f( )de , = f( )d e , , for all V + (3.7.3) h i 2 Z Z ⌦ ↵ The integral on the right hand side is defined in the sense of Riemann Stieltjes integral of f with respect to the increasing function e , . 7! h i Remark 3.181. Note that if f is a continuous function on [ a , a ], then we are k k k k guaranteed that the integral on RHS of (3.7.3) exists. Fix V + and let P = x ,x ...x be a partition of [ a , a ]. Define 2 { 0 1 n} k k k k n n U(P, )= f(⌘i) e i e i 1 , ,L(P, )= f(⇠i) e i e i 1 , h i h i i=1 i=1 X X 111 3 Spectral Theory for Ordered Spaces where ⌘i is a point where f attains maximum in [xi,xi 1]and⇠i is a point where f attains minimum in [xi,xi 1]. As f is continous on [ a , a ](uniformlycon- k k k k tinous), given ✏>0, >0suchthat f(x) f(y) <✏whenever x y < .So, 9 | | | | for a partition P with P ,wehave k k n U(P, ) L(P, ) = (f(⌘i) f(⇠i)) e i e i 1 , (3.7.4) k k k h ik i=1 Xn <✏ e i e i 1 , (3.7.5) k h ik i=1 X = ✏ e, = ✏ (3.7.6) h i k k Hence, for each V +,theRiemannsum 2 n f( )d e , =lim f( i) e i e i 1 , exists h i P 0 h i k k! i=1 Z X + The map Ta(f):V R given by ! f( )d e , 7! h i Z is a positive linear functional on V +. As f is bounded on [ a , a ], we get T (f) k k k k a to be pointwise bounded on K ( d e , = ). Thus, T (f) A (K) = V ⇤ = A h i k k a 2 b ⇠ (proposition ??). R Using this notion, we define a spectral calculus for an element a A. 2 Definition 3.182. Assume A, V are in spectral duality. Let a A and e R 2 { } 2 be the spectral resolution of a.Foreachcontinuousfunctionf on [ a , a ], k k k k assign the element f(a)=T (f) A,denotedby a 2 f(a)= f( )de Z Proposition 3.183. If A, V are in spectral duality and a A, then the spectral 2 calculus for a satisfies the following (where f, g, fn are continuous functions on [ a , a ] and ↵, are real numbers): k k k k 112 3 Spectral Theory for Ordered Spaces 1. f(a) f k kk k1 2. (↵f + g)(a)=↵f(a)+ g(a) 3. f g = f(a) g(a) ) 4. If f is a bounded sequence and f f pointwise, then f (a) f(a) { n} n ! n ! in the weak* topology of A = V ⇤. Proof. First note that for w K, 2 n d e ,w =lim e i e i 1 ,w =lime, w =1 h i P 0 h i P 0h i k k! i=1 k k! Z X 1. f(a) =supw K f(a),w supw K f( )d e ,w k k 2 |h i| 2 | h i| R sup f( ) d e ,w f sup d e ,w = f w K | | h ik k1 w K h i k k1 2 Z 2 Z 2. Let v V +.Then↵ f(a),v + g(a),v = ↵f( )d e ,v + g( )d e ,v = 2 h i h i h i h i ↵f( )+ g( ) d e ,v = (↵f + g)(a)R,v .So, R h i h i R ↵f(a)+ g(a),v = (↵f + g)(a),v v V + (3.7.7) 8 2 ⌦ ↵ ⌦ ↵ As V + generates V ,(3.7.7)istrueforallv V . 2 Thus, (↵f + g)(a)=↵f(a)+ g(a). 3. f g = f( )d e ,v g( )d e ,v for each v V +.Thus, g(a) ) h i h i 2 h f(a),v 0,R v V + = fR(a) g(a). i 8 2 ) 4. Fix v V +.Bymonotoneconvergencetheorem, 2 f f = f ( )d e ,v f( )d e ,v = f (a),v f(a),v n ! ) n h i ! h i )hn i !h i Z Z Then, f (a),v f(a),v , v V = f (a) f(a)inweak* h n i !h i 8 2 ) n ! topology. 113 Chapter 4 Spectral Theory for Jordan Algebras In the previous chapters, we have developed an abstract spectral theory, for a general order unit space satisfying certain conditions. Now, we present a concrete example of these order theoretic constructions and spectral decomposition result, in the case of Jordan algebras. In this chapter, we specialise the spectral theory for JBW-algebras. We try to understand JB algebras, through a larger class of ordered algebras, namely order unit algebras, which have both algebraic structure as well as order unit space struc- ture. We show that JB algebras are locally isomorphic to CR(X) spaces and use this correspondence to develop notions like orthogonality and range projections. Then, the spectral theorem for JBW algebra follows from the spectral theory for function spaces, using the local isomorphism with monotone complete CR(X) spaces. Jordan algebras are commutative algebras by definition. However, they are lo- cally embedded in an associative parent algebra, which is highly non-commutative, in general. Hence, this chapter will give us a glimpse of the spectral theorem in a non-commutative framework. 114 4 Spectral Theory for Jordan Algebras 4.1 Order Unit Algebras We now begin with the study of order unit algebras. Order unit algebras are order unit spaces with a normed algebra structure. These spaces contain JB algebras, as a special case, namely every commutative order unit algebra is a JB-algebra (and vice-versa). We will show that order unit algebras are locally isomorphic to CR(X) spaces and later use this result to obtain a spectral result for JBW algebras. We start with some basic definitions. Throughout this section, A denotes an order unit algebra with order unit e. Definition 4.1 (Algebra). An algebra A is a real vector space, with a bilin- ear product (not necessarily associative or commutative). The algebra is said to normed if it is equipped with a norm such that for each a, b A, we have: 2 ab a b (4.1.1) k kk kk k Definition 4.2. An normed algebra which is complete with respect to the norm is called a Banach Algebra. Note that the algebras we are working with, are not associative in general. Definition 4.3. An algebra is said to be power associative if am+n = aman, m, n 8 2 N,a A. 2 So, the n-th power of an element a,denotedbyan, is well-defined in a power associative algebra . Definition 4.4 (Order Unit Algebra). An order unit space A, which is also a normed algebra (with respect to the order unit norm) is said to be an order unit algebra if it has the following properties: (i) A is power associative (ii) A is norm complete 115 4 Spectral Theory for Jordan Algebras (iii) the distinguished order unit (e)isamultiplicativeidentity (iv) a2 A+ ,foreacha A 2 2 Example 4.5. The real linear space of bounded self-adjoint linear operators on a complex Hilbert space H is an order unit algebra, under the symmetrized product a, b 1 (ab + ba). 7! 2 Definition 4.6. Two order unit algebras A1 and A2 are said to be isomorphic if there exists a bijection : A A which preserves the algebra structure, 1 ! 2 ordering, order unit and norm. We denote this by A1 ⇠= A2. Proposition 4.7. Let A be a power associative complete normed algebra, with multiplicative identity 1. If a A satisfies the inequality 1 a 1, then s A 2 k k 9 2 such that s2 = a. 1 Proof. Using Taylor series expansion of the function f(x)=(1 x) 2 , about 0, we get 1 1 1 1 n 1 n ( 1) ( 2 )( 2 1) ...( 2 (n 1)) (1 x) 2 = x where = (4.1.2) n n n! n=0 X This series is absolutely and uniformly convergent for x 1 (A.9) and absolute | | convergence of the series at x =1gives 1 < . Now, fix x [0, 1]. Then n=0 | n| 1 2 n 1 1 P 1 1 1 n n n (1 x)=(1 x) 2 (1 x) 2 = nx nx = nx where n = k n k n=0 n=0 n=0 k=0 X X X X (4.1.3) Note that the product series in (4.1.3) is convergent because each of the series n 1 x is absolutely convergent in [0, 1]. Also, see that =1, = 1, = n=0 n 0 1 n 0P, n>1. Next, given a A, satisfying 1 a 1, define b =1 a A. Then, 8 2 k k 2 n 1 b 1. Define s = 1 b . Note that this series is absolutely convergent k k n=0 n 2 n and as A is complete,P the series converges in A. Now, s = 1 b =1 b = a. n n Therefore, s2 = a. P 1 n n 1 b 1 b 1 < n=0 | n | n=0 | n|| | n=0 | n| 1 P P P 116 4 Spectral Theory for Jordan Algebras Proposition 4.8. If A is an order unit algebra, then A2 = A+. Proof. By definition of order unit algebra , A2 A+ .Conversely,leta A+ . ✓ 2 1 1 Denote ↵ = a . Then 0 ↵ a e. This implies 0 e ↵ a e which in turn k k 1 2 1 implies e ↵ a 1. Now, by proposition 4.7, s A such that s = ↵ a. k k 9 2 Then, (p↵s)2 = a.Thus,a A2 implying A+ A 2. 2 ✓ Remark 4.9. As A+ is closed under addition, lemma 4.8 tells us that in an order unit algebra, sum of squares is again a square. Definition 4.10 (Pure States). Alinearfunctional⇢ on an order unit space (A, e)iscalledastate if it is positive and ⇢(e)=1.ThesetofallstatesonAiscalled the state space of A. An extreme point of the state space is called a pure state. Definition 4.11 (Multiplicative States). Alinearfunctional⇢ on an algebra A is said to be multiplicative if ⇢(xy)=⇢(x)⇢(y), x, y A. 8 2 Next, we will show a bijective correspondence between the set of pure states and multiplicative states on an order unit algebra. Our first lemma is a version of the Cauchy-Schwarz inequality. Lemma 4.12. Let A be an order unit space, which is also an algebra such that the distinguished order unit (e) is a multiplicative identity for A. If ⇢ is a state on A which is positive on squares, then for each a A, 2 ⇢(a)2 ⇢(a2)(4.1.4) And if ab = ba A, then 2 ⇢(ab)2 ⇢(a2)⇢(b2)(4.1.5) Proof. Define a product , on A as a, b = ⇢( 1 (ab + ba)), a, b A. As, ⇢ is h i h i 2 8 2 positive on squares, we see that a, a 0, a A. Also, by linearity of ⇢ and h i 8 2 bilinearity of the product on A,itisclearthat , is a bilinear map. Thus, , is h i h i 117 4 Spectral Theory for Jordan Algebras a semi-inner product on A. Now, by the general Cauchy-Schwarz inequality (A.6), we get that a, b 2 a, a b, b ,foreacha, b A. In particular, when ab = ba, |h i| h ih i 2 we get (4.1.5). And, putting b = e in (4.1.5) gives (4.1.4). Proposition 4.13. Let A be an order unit space, which is also an algebra such that the distinguished order unit (e) is a multiplicative identity for A. Further, assume a2 A+ , for each a A. Then: 2 2 (i) Each multiplicative state on A is pure. (ii) If ab A+ for each pair a, b A+ , then each pure state is multiplicative. 2 2 Proof. Let S denote the state space of A. (i) Let ⇢ be a multiplicative state on A such that ⇢ = 1 ⇢ + 1 ⇢ for some ⇢ ,⇢ 2 1 2 2 1 2 2 S. Note that ⇢,⇢1,⇢2 satisfy the hypothesis of proposition 4.12. Then, for each a A, we have: 2 1 1 1 1 ⇢(a2)= ⇢ (a2)+ ⇢ (a2) ⇢ (a)2 + ⇢ (a)2 (4.1.6) 2 1 2 2 2 1 2 2 As ⇢ is multiplicative, we also have 1 ⇢(a2)=⇢(a)2 = (⇢ (a)2 +2⇢ (a)⇢ (a)+⇢ (a)2)(4.1.7) 4 1 1 2 2 Subtracting (4.1.7) from (4.1.6), we get 0 ⇢ (a)2 + ⇢ (a)2 2⇢ (a)⇢ (a)=(⇢ (a) ⇢ (a))2 0(4.1.8) 1 2 1 2 1 2 Hence, for each a A, we get ⇢ (a)=⇢ (a). This implies ⇢ = ⇢ = ⇢ . 2 1 2 1 2 Thus, ⇢ is pure. (ii) Assume A+ is closed under multiplication. Let ⇢ be a pure state on A. Take 0 a e.Then,e a A+ . Now, for each x A+ ,wehaveax, (e a)x 2 2 2 A+ .Thisimplies0 (e a)x = ax ex. Define a linear functional ) ⇢a : A R given by s ⇢(as). Also, ax ex = x = ⇢a(x) ⇢(x), for ! 7! ) each x A+ . Hence, 0 ⇢ ⇢. 2 a 118 4 Spectral Theory for Jordan Algebras Claim. 0 ⇢ ⇢ = >0suchthat⇢ = ⇢. a )9 a Since ⇢ is a positive linear functional on A, ⇢ = ⇢ for some >0,⇢ a a 1 1 2 S. Infact, = ⇢ = ⇢ (e)=⇢(ae)=⇢(a). Also note that 0 ⇢(a) k ak a ⇢(e)=1 = 0 1. Similarly, ⇢e a = µ⇢2 for some µ>0,⇢2 S. ) 2 Further, µ = ⇢e a(e)=⇢(e ae)=1 ⇢a(e)=1 .Thus,⇢e a =(1 )⇢2. Now, ⇢a + ⇢e a = ⇢ = ⇢1 +(1 )⇢2 = ⇢. As ⇢ is a pure state on A, ) ⇢ = ⇢1 = ⇢2. Hence, ⇢a = ⇢. Let x A. Then, ⇢(a)⇢(x)=⇢(ae)⇢(x)=⇢ (e)⇢(x)= ⇢(e)⇢(x)= ⇢(x)= 2 a ⇢ (x)=⇢(ax). Thus, ⇢(a)⇢(x)=⇢(ax)forallx A, a A+ . Now, as a 2 2 Aispositivelygenerated,⇢(ax)=⇢(a)⇢(x)foralla, x A. Hence, ⇢ is 2 multiplicative. Remark 4.14. From the above proposition, we see that in an order unit algebra, the set of pure states is same as the set of multiplicative states. Using this proposition, one can prove the following Stone’s representation the- orem for Ordered Algebras. Theorem 4.15. Suppose A is a complete order unit space, with state space K, which is also an algebra satisfying the following: (i) the order unit (e) is a multiplicative identity for A (ii) a2 A+ , for each a A 2 2 (iii) ab A+ , for each pair a, b A+ . 2 2 Then, the set of pure states on A, denoted by @eK, is a w⇤-compact set consisting of all multiplicative states, and the map a aˆ is an isometric, order and 7! |@eK algebra isomorphism of A onto CR(@eK). 119 4 Spectral Theory for Jordan Algebras Proof. By proposition 4.13, we see that @eK is precisely the set of multiplicative states on A. w*-compactness of @ K • e As K is a w*-compact subset of A⇤,itissu cienttoshowthat@eK is w*- w closed in K. Let ⇢ ⇤ ⇢,where⇢ K, ⇢ @ K, n.Then,⇢ (a) n ! 2 n 2 e 8 n ! ⇢(a), a A. Take a, b A. ⇢(ab)=limn ⇢n(ab) = limn ⇢n(a)⇢n(b) = 8 2 2 !1 !1 limn ⇢n(a)limn ⇢n(b)=⇢(a)⇢(b). Hence, ⇢ is multiplicative which im- !1 !1 plies ⇢ @ K.Thus,@ K is a w*-closed subset of K and hence, w*-compact. 2 e e Algebra homomorphism • Consider the map : A C (@ K)givenbya aˆ .Thismapiswell ! R e 7! |@eK defined because if a Aand⇢ w⇤ ⇢ in @ K,then⇢ (a) ⇢(a)which 2 n ! e n ! impliesa ˆ(⇢ ) aˆ(⇢). Thus,a ˆ is continous. n ! |@eK Clearly, is linear. For each ⇢ @ K and a, b A, we have abˆ (⇢)=⇢(ab)= 2 e 2 ⇢(a)⇢(b)=ˆa(⇢)ˆb(⇢). Thus, is an algebra homomorphism. A.5 isanisometry Let a A.Then, a =sup⇢ K ⇢(a) =sup⇢ @eK ⇢(a) = • 2 k k 2 | | 2 | | (a) . k k is one-one • Supposea ˆ(⇢)=ˆb(⇢), ⇢ @ K. By Krein-Milman Theorem (A.5), K = 8 2 e Co(@ K @ K). Asa, ˆ ˆb are continuous linear funtionals on K, it follows e [ e thata ˆ(⇢)=ˆb(⇢), ⇢ K. So, ⇢(a)=⇢(b), ⇢ A⇤. By Hahn-Banach 8 2 8 2 separation theorem (A.3), a = b. is onto • We will prove this using Stone Weierstrass theorem (A.7). As A is com- plete and is an isometry, it follows that (A)isclosedinCR(@eK). (A) contains the constant functione ˆ. Also, if ⇢ = ⇢ in @ K, then a 1 6 2 e 9 2 120 4 Spectral Theory for Jordan Algebras A such that ⇢ (a) = ⇢ (a)whichimplies (a) ⇢ = (a) ⇢ . Hence, (A) 1 6 2 1 6 2 separates points in @eK. Thus, by Stone - Wierestrass theorem (A.7), we get that (A)= (A)=CR(@eK). Therefore, is onto. Order isomorphism • Let a a .Thena a 0. As elements of @ K are positive, ⇢(a ) 1 2 2 1 e 2 ⇢(a ), ⇢ @ K. Hence,a ˆ aˆ .So, ispositive. 1 8 2 e 2|@eK 1|@eK 1 Conversely, if f 0inC (@ K), then, f 2 C (@ K)andispositive. R e 2 R e 1 1 Since, is an algebra isomorphism, a A such that a = (f 2 ). Then, 9 2 2 1 2 + 1 1 a = (f). As A A ,weget (f) 0 in A. So, is positive. ✓ Hence, is an order isomorphism. Hence, is an order and algebra isomorphism of A onto CR(@eK). Theorem 4.16. If A is an order unit algebra, then the following are equivalent: 1. A is associative and commutative 2. ab A+ , for each pair a, b A+ . 2 2 3. A ⇠= CR(X) for some compact Hausdor↵space X (here, ⇠= is an isometric order and algebra isomorphism) Proof. First recall that A+ =A2. (1 2) Let a, b A+ .Then,byproposition4.7,a = s2,b = t2 for some s, t A. ) 2 2 As A is associative and commutative, ab =(ss)(tt)=(st)(st) A2 =A+ . 2 Hence, A+ is closed under multiplication. (2 3) Clear from Theorem 4.15. ) (3 1) As C (X) is an associative and commutative space, so is A. ) R 121 4 Spectral Theory for Jordan Algebras Let a be an element in a power associative normed algebra A, with multiplica- tive identity 1. Then the norm closure of all polynomials in a and 1, denoted by C(a, 1), is the least norm closed subalgebra in A, containing a and 1. Note that C(a, 1) is also associative and commutative (A.8). If A is an order unit algebra, then C(a, 1) is also an order unit algebra (for the ordering, order unit, norm and product inherited from A). This is referred to as the order unit subalgebra generated by a. Corollary 4.17. Let a be an element of an order unit algebra (A, e). Then, C(a, e) ⇠= CR(X) for some compact Hausdor↵space X. Proof. C(a, e)isacommutativeandassociativeorderunitalgebra(withordering, norm, product and order unit inherited from A). Now, by theorem 4.16, the result follows. 4.1.1 Characterising Order Unit Algebras In this section, we formulate various characterisations of order unit algebras, which will later help us to relate them with JB algebras. Lemma 4.18. If A is a power associative complete normed real algebra with mul- tiplicative identity e. Assume A satisfies the following condition: a2 a2 + b2 , a, b A (4.1.9) k kk k 8 2 Then the following are equivalent: (i) a A2 2 (ii) ↵e a a , for all ↵ a k kk k k k (iii) ↵e a a , for one ↵ a k kk k k k Proof. Clearly (ii) (iii). ) 122 4 Spectral Theory for Jordan Algebras (i ii) Let a A2 such that a 1. Then, s A such that s2 = a. Now, ) 2 k k 9 2 e (e a) = a 1. Hence, by proposition 4.7, e a = t2 for some t k k k k 2 A. Then, s2 + t2 = a +(e a)=e.So, e a = t2 s2 + t2 = e =1. k k k kk k k k Therefore, we have shown that a 1= e a 1. Now, for arbitary k k )k k a A, take ↵ a .Then a 1whichimplies e a 1bythe 2 kk k ↵ k k ↵ k previous case. Hence, ↵e a ↵. k k 1 (iii i) Suppose ↵e a a , for some ↵ a .Then e ↵ a 1. By ) k kkk kk k k 1 2 2 proposition 4.7, ↵ a A . Hence, a A . 2 2 Lemma 4.19. If A is real algebra for which the statements (i), (ii), (iii) of lemma 4.18 are equivalent, then A2 is a cone. Proof. Let a, b A2.Denote a = ↵, b = . Now, a + b ↵ + .Bythe 2 k k k k k k equivalence of (i) and (ii) in lemma 4.18, we have ↵e a ↵, e b . k k k k Therefore, (↵ + )e (a + b) ↵e a + e b ↵ + . Again, by the k kk k k k equivalence of (iii) and (i) in lemma 4.18, we have a + b A2.So,A2 is closed 2 under addition. And clearly if a = s2 A2 and 0, then a =(p s)2 A2. 2 2 Hence, A2 is a cone. Lemma 4.20. Let A0 be a linear subspace of an order unit space (A, e) and suppose that a, b ab is a bilinear map from A x A into A such that for each 7! 0 0 a, b A , 2 0 (i) ab = ba (ii) e a e = 0 a2 e ) Then, ab a b for each a, b A . k kk kk k 2 0 Proof. Let a, b A and assume a , b 1. Then, 1 (a + b) , 1 (a b) 1. 2 0 k k k k k 2 k k 2 k So, e 1 (a + b), 1 (a b) e.By(ii),weget 2 2 1 1 0 (a + b)2 e, e (a b)2 0(4.1.10) 4 4 123 4 Spectral Theory for Jordan Algebras Adding the two 2 and using the fact that ab = ba,weget e ab e which implies ab 1. So, we have shown that a , b 1= ab 1forall k k k k k k )kk a b a, b A0. Now for arbitary a, b A0,wehave a , b 1. Hence, by previous 2 2 k k k k k k k k ab case, a b 1= ab a b . k k kk k k )k kk kk k Lemma 4.21. Suppose A is complete order unit space which is a power associative commutative algebra, where the distinguished order unit (e) acts as identity. Then A is an order unit algebra i↵the following implication holds: e a e = 0 a2 e, a A (4.1.11) ) 8 2 Proof. First, assume A is an order unit algebra . Let e a e.Then, a 1. k k This implies a2 a 2 1= e a2 e. Also, A2 =A+ .So,0 a2 e. k kk k ) Next, assume (4.1.11) holds for all a A. Taking A as A in lemma 4.20, we 2 0 get ab a b for all a, b A. Therefore, A is a normed algebra . Further, by k kk kk k 2 (4.1.11), we see that 0 a2 for all a A. Thus, A is an order unit algebra . 2 The following theorem gives a characterisation of order unit algebras in terms of norm. Theorem 4.22. Suppose A is real Banach space which is equipped with a power associative commutative bilinear product with identity element e. Then A is an order unit algebra, with distinguished order unit e, cone A+ = A2 and given norm i↵for each a, b A, we have the following: 2 (i) ab a b k kk kk k (ii) a2 = a 2 k k k k (iii) a2 a2 + b2 k kk k Moreover, the ordering of A is uniquely determined by the norm and the identity element e. 2If a b ,a b ,thena + a b + a b + b 1 1 2 2 1 2 1 2 1 2 124 4 Spectral Theory for Jordan Algebras Proof. Note that A is not an ordered space to begin with. ( ) Assume that A satisfies the three given norm conditions. (i) says that A ( is a normed algebra . (iii) says that A satisfies the hypothesis of lemma 4.18 and hence, by lemma 4.19, A2 is a cone in A. Infact, by lemma 4.18, A2 = a A a e a a . { 2 |kk k kk k} Claim 4.22.1. A2 is norm closed. Let a a in A, where a A2, n.Thisimplies a a as n . n ! n 2 8 k nk !k k !1 Now, a e a = a e a+ a e a e a +a ( a a ) e + kk k k kk k k nk k nk n nk k k k nk k k a a + a e a ( a a ) e + a a + a 0+0+ a k n k kk nk nk k k k nk k k k n k k nk ! k k as n . Hence, a e a a which implies a A2.Thus,A2 is !1 kk k kkk 2 norm closed. Claim 4.22.2. A2 is proper. Suppose a2, a2 A2. Note that a2 A2 = b A such that b2 = 2 2 )9 2 a2. Hence, 0 = a2 + b2. Now, a 2 = a2 a2 + b2 =0.Thisimplies k k k kk k a =0implyinga =0.Thus,A2 is proper. k k Now, define A+ =A2. Then, A becomes an ordered vector space with positive cone A2. Claim 4.22.3. For each a A, a 1 e a e. 2 k k () ( ) Let a 1. Then e (e a) 1. By proposition 4.7, we see ) k k k k that e a A2 =A+ This implies e a 0= a e.Similarly, 2 ) a 1= a e = e a. Hence, e a e. k k ) ) ( ) Given e a e,wegete a, e+a A+ =A2.Let = e a .Then, ( 2 k k 1 1 by lemma 4.18, we get e (e a) 1. Now consider (e a)as k k an element of the associative and commutative algebra C(a, e). Since 1 the norm on C(a, e) is inherited from A, it follows e (e a) 1 k k 125 4 Spectral Theory for Jordan Algebras 1 holds in C(a, e) as well. Now, by proposition 4.7, (e a)hasasquare root in C(a, e). Thus, s C(a, e)suchthats2 = e a.Similarly, 9 2 t C(a, e) such that t2 = e + a. Now, because C(a, e)isassociative 9 2 and commutative, we get (st)2 = s2t2 =(e a)(e + a)=e a2.Thus, (e a2) A2 =A+ .So,wehaveshown 2 e a e = 0 a2 e (4.1.12) ) Therefore, a 2 = a2 1whichimplies a 1. k k k k k k Now, by proposition 2.16, A is an order unit space with order unit e.Ais given to be complete, power associative and commutative. Hence, by (4.2.9) and lemma 4.21, it follows that A is an order unit algebra , with order unit e,givennormandpositiveconeA2. ( ) Now, assume (A,e) is an order unit algebra with positive cone A2.SinceA ) is a normed algebra, (i) follows. Moreover, being an order unit space, we see that 0 a2 a2 + b2 = a2 a2 + b2 for each a, b A. Thus, )k kk k 2 (iii) holds. Next we prove (ii). Suppose a Asuchthat a2 1. Then, 2 k k 0 a2 e. Now, 1 1 a = a2 + e (a e)2 (a2 + e) e 2 2 1 1 a = (a + e)2 a2 e ( a2 e) e 2 2 Therefore, e a e which in turn implies a 1. So, we have shown that k k a2 1= a 1. Therefore, in general, a2 = a p . k k )kk k k )kk 2 2 1 2 2 Taking = a ,weget a a 2 . Hence, a a .Theother k k k kk k k k k k inequality is clear because A is a normed algebra . So, a2 = a 2 for each k k k k a A. Thus, (ii) holds. 2 Note that A+ =A2 = a A a e a a implies that A+ is uniquely { 2 |kk k kkk} determined by the norm ( . )andorderunite. k k 126 4 Spectral Theory for Jordan Algebras Theorem 4.23. A commutative and associative real Banach algebra A is isomet- rically isomorphic to C (X) i↵the following holds for each a, b A: R 2 (i) ab a b k kk kk k (ii) a2 = a 2 k k k k (iii) a2 a2 + b2 k kk k Proof. First recall that CR(X) is itself an associative and commutative order unit algebra with the constant function 1 as the distinguished order unit. ( ) If A satisfies the three norm conditions, then by theorem 4.22, A is an order ( unit algebra. Further, A is given to be associative and commutative. Hence, by theorem 4.16, A is isometrically isomorphic to CR(X) , for some compact Hausdor↵space X. ( ) Let A be isometrically isomorphic to C (X) , for some compact Hausdor↵ ) R space X. Define A+ =A2. Using the isometric isomorphism, one can see that Abecomesanorderunitalgebrawithgivennorm,positiveconeA+ =A2 and distinguished order unit e,(wheree is the image of 1 under the given isomorphism). Now, by theorem 4.22, the three norm conditions hold in A. 4.1.2 Spectral Result We now give a general version of spectral theorem for certain order unit spaces. Here, we consider a commutative order unit algebra (A, e)whichisthedualof a base norm space V. Considering A to be a dual space makes it monotone com- plete and allows us to use the previous spectral result (theorem 1.13). Further, we assume that multiplication is separately w*-continuous in A. Under these assump- tions, we get a least w*-closed subalgebra W (a, e) of A containing a given element a and e.Thissubalgebraisthew*-closureofallpolynomialsina and e,andit 127 4 Spectral Theory for Jordan Algebras is also associative (A.8) and commutative. Further, W (a, e)isnormclosedand satisfies the three norm conditions (i), (ii), and (iii) of theorem 4.22. Therefore, by theorem 4.22, W (a, e)isitselfanorderunitalgebra,withthenorm,ordering, order unit and multiplication inherited from A. Definition 4.24. An element p of an order unit algebra A is called a projection if p2 = p.Twoprojectionsp, q are said to be orthogonal if pq = qp =0. Theorem 4.25. Let (A, e) be a commutative order unit algebra that is the dual of a base norm space V such that the multiplication in A is separately w⇤-continuous. Then for each a A, and each ✏>0, there exist mutually orthogonal projections 2 p1,p2,p3 ...pn in the w⇤-closed subalgebra W (a, e) generated by a and e, and there exist scalars 1, 2 ... n such that n a p ✏ (4.1.13) k i ik i=1 X Proof. Let K the distinguished base for the base norm space V. As W (a, e)isan associative and commutative order unit algebra , it follows from theorem 4.16 that W (a, e)isisometricallyisomorphictoCR(X) for some compact Hausdor↵space X. Claim 4.25.1. W (a, e)ismonotonecomplete. Let b be an increasing net in W (a, e)whichisboundedabove,saybysome { ↵} c W (a, e). Define a map b : K R given by b(k)=lim↵ b↵(k). This 2 ! limit exists because for each k K, b (k) is an increasing net of real numbers 2 { ↵ } bounded above by c(k)andR is monotone complete. Hence, this net converges to some real number, denoted by b(k). Since all b↵ are a ne, so is b. Also, 3 b =supk K b(k) supk K c(k) c .Therefore,b is an a ne bounded k k 2 | | 2 | |kk function on K, i.e. b A (K) = V ⇤ = A.So,b can be identified with an element 2 b ⇠ 3 b↵ c, implies that for each k K, b↵(k) c(k). This is true forall ↵. Hence, lim↵ b↵(k) c(k) 2 128 4 Spectral Theory for Jordan Algebras w ˜b of A and by construction, b ⇤ ˜b. As W (a, e)isw*-closed,˜b W (a, e). Thus, ↵ ! 2 W (a, e) is monotone complete. Hence, the claim. So, the space CR(X) isomorphic to W (a, e) is also monotone complete. Now, the result follows from the spectral theorem for monotone complete CR(X) spaces (Theorem 1.13). Now, we are ready to begin with Jordan algebras. 4.2 Jordan Algberas The motivating example for a JB-algebra is the normed closed real linear space of self-adjoint operators on a Hilbert space, closed under the symmetrised product 1 a b = (ab + ba)(4.2.1) 2 This product is bilinear and commutative but not associative. However, it satisfies aweakenedformofassociativity,whichleadstothefollowingabstraction. Definition 4.26. A Jordan Algebra over R is a real vector space A equipped with a commutative bilinear product that satisfies the following identity, known as the Jordan condition: (a2 b) a = a2 (b a)foralla, b A (4.2.2) 2 Any commutative bilinear product on a real algebra A satisfying (4.2.2) is said to be a Jordan product. Example 4.27. If is any associative algebra over R,then becomes a Jordan A A algebra when equipped with the symmetrised product 1 a b = (ab + ba) 2 Also, any subspace of closed under forms a Jordan algebra . A 129 4 Spectral Theory for Jordan Algebras Definition 4.28. A Jordan algebra is said to be special if it isomorphic to a subspace of an associative algebra ,suchthatthesubspaceisclosedunderthe A symmetrised product in . If it cannot be so embedded, then the Jordan algebra A is said to be exceptional. In any Jordan algebra , we define powers of an element recursively: x1 = x and n n 1 x = x x. Note that by taking a = b = x in (4.2.2), we see that in any Jordan algebra, the following holds: x (x x2)=x2 x2 (4.2.3) The following lemma will help us show that Jordan algebras are power asso- ciative. Lemma 4.29. Let A be an algebra over R, equipped with a product that is com- mutative and bilinear, but not necessarily associative. Then x (x x2)=x2 x2 holds for all x A 2 () xi(xj(xkxl)) = (xixj)(xkxl)(4.2.4) X X where x ,x ,x ,x A and the summation is over distinct i,j,k,l with 1 i, j, k, l 1 2 3 4 2 4. Proof. Suppose (4.2.4) holds. Then substituting x = x1 = x2 = x3 = x4 gives x (x x2)=x2 x2. Conversely, assume x (x x2)=x2 x2 holds for all x A. Fix x ,x ,x ,x 2 1 2 3 4 2 A and define x = 1x1 + 2x2 + 3x3 + 4x4 where t R, 1 t 4. Substituting 2 this in (4.2.3), we get 4 4 4 4 4 4 4 4 i j k l xi(xj(xkxl)) = i j k l (xixj)(xkxl) i=1 j=1 k=1 l=1 i=1 j=1 k=1 l=1 X X X X X X X X (4.2.5) 130 4 Spectral Theory for Jordan Algebras Let : A R be a linear functional. Then applying to both sides of (4.2.9) ! leads to two polynomials with real coe cients which are equal for all values of the variables 1, 2, 3, 4.Therefore,thecoe cientsofthecorrespondingvariables must be same. In particular, equating the coe cient of 1 2 3 4,weget xi(xj(xkxl)) = (xixj)(xkxl) (4.2.6) X X where the summation is over distinct i, j, k, l with 1 i, j, k, l 4. But (4.2.6) is true for all linear functionals on A. Hence, by Hahn-Banach Separation theorem, (4.2.4) follows. Proposition 4.30. Let A be an algebra over R, equipped with a product that is commutative and bilinear, but not necessarily associative. Then A is power associative it satisfies the identity x (x x2)=x2 x2 or the linearised () version (4.2.4). The proof of this proposition is based on a lengthy, yet simple, inductive argu- ment. The complete proof may be found in [1]. Corollary 4.31. Every Jordan algebra over R is power associative. The motivation to single out Jordan algebras that act like self-adjoint operators on a Hilbert space, leads to the following definition: Definition 4.32. A JB-algebra A is a Jordan algebra over R with identity element 1, equipped with a complete norm satisfying the following conditions, for each a, b A: 2 (i) a b a b k kk kk k (ii) a2 = a 2 k k k k (iii) a2 a2 + b2 k kk k 131 4 Spectral Theory for Jordan Algebras Example 4.33. If is a C*-algebra (with an identity), then the self-adjoint A elements of form a JB-algebra with respect to the Jordan product a b = A 1 2 (ab + ba). A subalgebra of a Jordan algebra which is closed under the Jordan product is said to be a Jordan subalgebra. Definition 4.34. Let A be a JB-algebra . A JB-subalgebra of A is a norm closed linear subspace of A, closed under the Jordan product. Definition 4.35. A JC-algebra is a JB-algebra A which is isomorphic to a norm closed Jordan subalgebra of B(H)sa. Next, we show that a JB-algebra is a commutative order unit algebra . Proposition 4.36. Every JB-algebra A is a commutative order unit algebra , with given norm and positive cone A2 = a2 a A . { | 2 } Proof. Since A is a JB-algebra , it is a real Banach space, equipped with a com- mutative bilinear product, has an identity element e and satisfies the three norm conditions. Also, by corollary 4.31, A is power associative. Now, by theorem 4.22, AisacommutativeorderunitalgebrawithpositiveconeA2. Infact, the converse of this proposition is also true. Every commutative order unit algebra is a JB-algebra (although we don’t prove it in this project!) Theorem 4.37. If A is a JB-algebra , then A is a norm complete order unit space with the identity element as the distinguished order unit (e) and with order unit norm coinciding with the given one. Furthermore, for each a A, 2 e a e = 0 a2 e (4.2.7) ) Conversely, if A is a complete order unit space equipped with a Jordan product for which the distinguished order unit is the identity and satisfies (4.90), then A is JB-algebra with order unit norm. 132 4 Spectral Theory for Jordan Algebras Proof. First, assume that A is a JB-algebra . Then, by proposition 4.36, A is a commutative order unit algebra with the given norm, positive cone A2 and with the distinguished order unit being the identity of A. And, (4.90) follows from lemma 4.21. Next, assume that A is a complete order unit space equipped with a Jordan product for which the distinguished order unit is the identity and satisfies (4.90) for all a A. Note that A being a Jordan algebra, is power associative. Then, 2 by lemma 4.21, A is an order unit algebra and hence the 3 norm properties of JB-algebra are satisfied. Hence, it is a JB-algebra. For each element in a JB-algebra A, we let C(a, e)denotethenormclosed Jordan subalgebra generated by a and e.LetP (a, e)denotethesetofallpolyno- mials in a and e.Then,iteasytoseethatC(a, e)=P (a, e). As JB-algebras are power associative, P (a, e) is an associative Jordan subalgebra of A. Further, the submultiplicativity of norm in a JB-algebra implies that multiplication is jointly continuous. Hence, C(a, e)isalsoassociative. Theorem 4.38. If A is a JB-algebra and B is a norm closed associative subalgebra of A containing e, then B is isometrically (order and algebra ) isomorphic to CR(X) for some compact Hausdor↵space X. In particular, the result is true when B = C(a,e), for a A. 2 Proof. Under the given hypothesis, B itself becomes an associative JB-algebra, with the inherited norm. Then, by proposition 4.36, B is also a commutative order unit algebra. Now, B being an associative and commutative order unit algebra, is isometrically isomorphic to some CR(X) , by theorem 4.16. Invertibility in a Jordan Algebra Note that we generally require associativity to prove that inverse of an element in an algebra is unique. This leads to the following definition: 133 4 Spectral Theory for Jordan Algebras Definition 4.39. An element a of a JB-algebra is invertible if it is invertible in the associative Banach algebra C(a, e). Its inverse in this subalgebra is denoted 1 by a and is called the inverse of a. The next few results relate this notion to other concepts of invertibility. Lemma 4.40. Let A be an associative algebra with identity element e. Consider the symmetrised Jordan product a b = 1 (ab + ba). Then for a, b A: 2 2 ab = ba = e a2 b = a and a b = e (4.2.8) () Proof. First, recall that A is given to be associative. ( ) Assume ab = ba = e.Thena b = 1 (2e)=e. ( 2 Similarly, a2 b = 1 (a2b + ba2)= 1 (a + a)=a. 2 2 ( ) If a b = e,then ( ab + ba =2e (4.2.9) And if a2 b = a,then a2b + ba2 =2a (4.2.10) Now, a(ab + ba)=2ae =2a =2ea =(ab + ba)a.So,a2b + aba = aba + ba2. So, we get ab2 = ba2 (4.2.11) Putting (4.2.11) in (4.2.10), we get 2a2b =2a = a2b = a = ba2b = ba. ) ) Similarly, ba2 = a = ba2b = ab. Hence, ba = ba2b = ab.Puttingthisin ) (4.2.9), we get 2ab =2e = ab = e.Similarly,ba = e. ) Definition 4.41. If elements a, b in a Jordan algebra satisfy the condition a2 b = a and a b = e,thena and b are said to be Jordan invertible and b is said to be the Jordan inverse of a. 134 4 Spectral Theory for Jordan Algebras We will now show that in the context of JB-algebra , the two notions of invert- ibility coincide. Proposition 4.42. For elements a, b in a JB-algebra A, the following are equiv- alent: (i) b is the inverse of a in the Banach algebra C(a,e) (ii) b is the Jordan inverse of a in A Proof. Note that, in order to use lemma 4.40 for A, we need A to be embedded inside an associative algebra such that the Jordan product on A comes from the A symmetrised product on . A 1 (i ii) Suppose b = a in C(a, e). As C(a, e)isassociative,wehavea b = e and ) a2 b = a (a b)=a e = a. As the product in C(a, e) is inherited from A, these equalities hold in A as well. Thus, b is the Jordan inverse of a, in A. (ii i) Let b be the Jordan inverse of a in A. Let (B , +, )bethesubalgebra ) 0 generated by a, b, e in A. By theorem 4.52, B0 is special and can be viewed as asubalgebraofanassociativealgebra(C, +,.)suchthatx y = 1 (x.y + y.x) 2 for all x, y B .Then,bylemma4.40,a.b = b.a = e.LetC be the 2 0 0 subalgebra generated by a, b, e in C.Then,(C0, +,.)isanassociativeand commutative subalgebra of C and is in bijective correspondence (a a, b 7! 7! b, x y x.y)with(B , +, ). It is easy to check that this bijection is 7! 0 infact an algebra isomorphism. Therefore, B0 is also associative. And as multiplication is jointly continuous in A, the norm closure of B0 in A, say B, is an associative JB-subalgebra of A. Now, by theorem 4.38, B is isometrically isomorphic to some CR(X) . Claim 4.42.1. b C(a, e). 2 Let : B C (X) be the isometric isomorphism mentioned above. Now, ! R a b = e = (a) (b)=1,i.e. (a)isinvertibleinC(X) . So, 0 / ) R 2 135 4 Spectral Theory for Jordan Algebras Range( (a)). As X is compact and (a)iscontinuous,Range( (a)) is a 1 compact subset of R, say Y. Now, define f : Y R given as . ! 7! Then, f C (Y )andcanbeapproximatedbypolynomialsp (by Stone- 2 R n Weierstrass theorem). Let p f in C (Y ). Then, p (a) f n ! R n ! 1 (a)inC(X) . But f (a)=( (a)) = (b). This means that (b) R can be approximated in CR(X) , by polynomials in (a). By the isometric isomorphism, this implies that b can be approximated in B, by polynomials in a.Thus,b P (a, e)=C(a, e). Hence, the claim. 2 Therefore, b is the inverse of a in C(a, e). Remark 4.43. Using the associativity in C(a, e), one can now prove that Jordan inverse of an element, if it exists, is unique. 4.2.1 The Continuous Functional Calculus Definition 4.44. If a is an element of a JB-algebra A, then the spectrum of a, denoted by (a)isdefinedas (a)= R e a is not invertible . { 2 | } Proposition 4.45. For each element a in a JB-algebra A, (a) is non-empty and compact. Moreover, there is a unique isometric order isomorphism from CR( (a)) onto C(a,e), taking the identity function on (a) to a. Proof. Fix a A. By theorem 4.38, there exists an isometric (order and algebra) 2 isomorphism : C(a, e) C (X) . Let (a)=ˆa. As preserves invertibility, ! R (a)= (ˆa). Claim 4.45.1. (ˆa)=Range(ˆa) (ˆa) 1 aˆ is not invertible in C (X) x X such that ( 1 2 () R () 9 2 aˆ)(x)=0 aˆ(x)= 1 Rangea ˆ. Hence, the claim. () () 2 Now,a ˆ is a continuous function on a compact space X. Hence range(ˆa)is compact and non-empty. Thus (ˆa)= (a)isnon-emptyandcompact. 136 4 Spectral Theory for Jordan Algebras Claim 4.45.2. aˆ : X (a)definedasx aˆ(x)isahomeomorphism. ! 7! This map is surjective because we have just shown that (a)= (ˆa)= Range(ˆa). To prove injectivity, assume thata ˆ(x)=ˆa(y)forsomex, y X. Then (ˆa(x))n = 2 (ˆa(y))n for all n N.Thus,p(x)=p(y)forallpolynomialsp ina ˆ. Now, P (a, e) 2 is dense in C(a, e) and the isometry preserves density. Hence, P (ˆa, 1) is dense in C (X) . Hence, f(x)=f(y), for all f C (X) Now, X is a compact, Haus- R 2 R dor↵space and hence normal. Therefore, distinct points in X can be separated by continuous functions. So, f(x)=f(y)forallf C (X) = x = y. And 2 R ) we already know thata ˆ is a continuous function from X to R. Hence,a ˆ is a continuous bijection from a compact space to a Hausdor↵space. Thus, it is a homeomorphism. So, we have shown that X is homeomorphic to (a). Consider the map T : 1 C (X) C ( (a)) given by Tf = f (ˆa) .Thenthemap =T isan R ! R algebra isomorphism from C(a, e)ontoCR( (a)). :C(a, e) C (X) T C ( (a)) (4.2.12) ! R ! R Note that (e)=T (1) = 1 and takes squares to squares. Thus, is a unital 1 order isomorphism. Also, (a)=T (ˆa)=ˆa (ˆa) = I (identity function on (a)). Further, is an isometry because b =inf >0 e b e =inf > k k { | } { 0 ( e) (b) ( e) =inf >0 1 (b) 1 = (b) . | } { | } k k Claim 4.45.3. Uniqueness of Let ⇥is any other isometric order and algebra isomorphism from CR( (a)) onto C(a, e), that takes the identity function on (a)toa. Since polynomials are dense in C(a, e), ⇥is uniquely determined by its value at a.So,if⇥takesa to the identity map on (a), then ⇥= . Definition 4.46. Let a be an element of a JB-algebra A. The continuous func- 137 4 Spectral Theory for Jordan Algebras tional calculus is the unique isomorphism f f(a)fromC ( (a)) onto C(a,e), 7! R taking the identity function on (a)toa. Proposition 4.47. Let a be an element of a JB-algebra A. The continuous func- tional calculus is an isometric order isomorphism from CR( (a)) onto C(a,e). Fur- thermore, 1. f(a) has its usual meaning when f is a polynomial. 2. (f(a)) = f( (a)), for f C ( (a)) 2 R 3. (f g)(a)=f(g(a)) for g C ( (a)) and f C ( (g(a))) 2 R 2 R 4. If f C ( (a)) and f(0) = 0, then f(a) is in the (non-unital) norm closed 2 R subalgebra C(a), generated by a. Proof. From theorem 4.45, it is clear that the continuous functional calculus, say , is an isometric order isomorphism from CR( (a)) onto C(a, e). Let I denote the identity function on (a)and1denotetheconstantfunction1on (a). k n 1. Recall that (I)=a.So,iff(x)= n=0 nx ,then k k P k k n n n n (f)= n (x )= n (x) = n I(x) = na = f(a) n=0 n=0 n=0 n=0 X X X X (4.2.13) 2. Unital isomorphism preserves invertibility. Thus, (f(a)) = (f)=Range (f)=f( (a)). g f 3. Note that (a) g( (a)) = (g(a)) R implies that (f g)(a) C(a, e) ! ! 2 and f(g(a)) C(g(a),e) C(a, e). 2 ✓ Case (I). f is a polynomial. n Let f(x)= k xn.Then,(f g)(x)=f(g(x)) = k g(x) = n=1 n n=1 n f(g(a)). Thus,P (f g)(a)=f(g(a)). P 138 4 Spectral Theory for Jordan Algebras Case (II). f is an arbitary continuous function on (g(a)). P (a, e)isdenseinC(a, e)andP (g(a),e)isdenseinC(g(a),e). Recall that f C ( (g(a))) = C(g(a),e)andtheidentityfunctionong(a), say I , 2 R ⇠ g maps to g(a) under this isomorphism. Now, (g(a)) is a compact Hausdor↵ space and hence f can be approximated by polynomials pn,inCR(g(a)). Now, p f in C ( (g(a))) implies p g f g in C ( (a)). Since n ! R n ! R isanisometry,(p g)(a) (f g)(a)inC(a, e)asn .But, n ! !1 (p g)(a)=p (g(a)) by case I and p (g(a)) f(g(a)) in C ( (g(a))) as n n n ! R n .Thus,(f g)(a)=f(g(a)). !1 4. Let f C ( (a)). Then f can be approximated by polynomials p in 2 R n C ( (a)). This implies p (x) f(x)foreachx (a). Thus, q (x):= R n ! 2 n p (x) p (0) f(x) f(0). Now f(0) = 0 = q f in C ( (a)). As n n ! ) n ! R qn have constant term 0, qn(a)isapolynomialina,notinvolvinge.Thus, q (a) C(a), n. Hence, f(a)=lim q (a) C(a). n 2 8 n n 2 Corollary 4.48. Let a be an element of a JB-algebra A. Then, a 0 (a) [0, ) () ✓ 1 Also, a =sup (a) . k k {| || 2 } Proof. a 0 I(x) 0, x (a) x 0, x (a) (a) () 8 2 () 8 2 () ✓ [0, ). Also, a = I =sup (a) . 1 k k k k {| || 2 } Remark 4.49. If A is a JB-algebra and B is a norm closed Jordan subalgebra of A containing e, then B itself becomes a JB-algebra for the inherited norm. Now, if a belongs to B, then C(a, e) B and hence the spectrum of a is the same in A and ✓ B. Further, by corollary 4.48, a 0isequivalentwhethera is viewed as a member in A or B. So, the order on B viewed as a JB-algebra is that inherited from A. 139 4 Spectral Theory for Jordan Algebras 4.2.2 Triple Product Definition 4.50 (Triple Product). In any Jordan algebra, we define the triple product abc as: { } abc = a (b c)+c (b a) b (a c)(4.2.14) { } The following are some properties of this triple product, which can easily be verified: (i) abc is linear in each factor. { } (ii) abc = cba { } { } (iii) In a special Jordan algebra , abc = 1 (abc + cba) { } 2 The following are two famous results in Jordan algebras, which we state without proof. The interested reader may find the complete proofs in [5]. Theorem 4.51 (Macdonald). Any polynomial identity involving three or fewer variables (and possibly involving the identity 1) which is of degree atmost 1 in one of the variables and which holds for all special Jordan algebras will hold for all Jordan algebras. The following result can be derived from Macdonald’s theorem. Theorem 4.52 (Shirshov-Cohn). Any Jordan algebra generated by 2 elements (and 1) is special. This theorem implies that calculations in any Jordan algebra , involving just two elements can be done assuming that the Jordan algebra is a subalgebra of an associative algebra , with respect to the product a b = 1 (ab + ba). 2 We will now explore the special case of the triple product abc with a = c. { } Definition 4.53. Let a A. Define U : A A given as 2 a ! U (b)= aba =2a (a b) (a2 b)(4.2.15) a { } 140 4 Spectral Theory for Jordan Algebras Thus, for a special Jordan algebra , Ua(b)=aba. Remark 4.54 (Identities). Using Macdonald’s theorem, one can prove that the following identities hold in any Jordan algebra : aba c aba = a b aca b a (4.2.16) {{ } { }} { { { } } } bab 2 = b ab2a b (4.2.17) { } { { } } 1 (a b)2 = (2a bab + ab2a + ba2b )(4.2.18) 4 { } { } { } b bab b = b2ab2 (4.2.19) { { } } { } (e b) bab (e b) = (b b2)a(b b2) (4.2.20) { { } } { } ab2a 2 = a b ba2b b a (4.2.21) { } { { { } } } 2 1 If p = p, then p a = (a + pap p0ap0 )(4.2.22) 2 { } { } Using (4.2.17), one can see that U aba = UaUbUa (4.2.23) { } Our next goal is to show that the maps U are positive for each a A. a 2 Lemma 4.55. Let a be an element of a JB-algebra A. Then a is an invertible 1 1 element i↵ Ua has a bounded inverse. In this case, (Ua) = Ua . Proof. First note that the map U is bounded for each a A. a 2 141 4 Spectral Theory for Jordan Algebras 1 ( ) Suppose a is invertible in A, with c = a in C(a, e). Now, C(a, e)isas- ) 2 2 2 sociative implies ac a = a c = e.ThisimpliesU ac2a = Ue =Identity { } { } function on A. So, I = U ac2a = UaUc2 Ua (by (4.2.23)). Then, UaUc2 is the { } 4 left inverse of Ua and Uc2 Ua is the right inverse of Ua. Therefore, Ua is 1 2 invertible, say (U ) = T . Now, aca = a c = a (a c)=a.Therefore, a { } Ua = U aca = UaUcUa { } Now, I = TUa = TUaUcUa = UcUa and I = UaT = UaUcUaT = UaUc. 1 1 Hence, UcUa = UaUc = I Thus, (Ua) = Uc = Ua . ( ) If a is not invertible in A, then by functional calculus, there are elements b ( { n} in C(a, e)whichhavenorm1andsatisfy U b 0. If U had a bounded k a nk ! a inverse, then applying it to the sequence U b would show b 0, which { a n} n ! contradicts that b =1, n. Hence, a is invertible in A. k nk 8 Lemma 4.56. If a, b are invertible elements of JB-algebra , then aba is invertible { } 1 1 1 with inverse a b a . { } Proof. If a, b are invertible elements in A, then by lemma 4.55, Ua,Ub have bounded inverses. Now, U aba = UaUbUa implies U aba also has a bounded inverse. Then, { } { } 1 1 by lemma 4.55, aba is invertible in A. Using the fact that U (x )=x x x = { } x 1 1 x,foreachinvertiblex A, we get: (aba) = U aba 1 aba =(U aba ) aba = 2 { } { } { } { } 1 1 1 1 (U U U ) aba = (U ) (U ) (U ) aba = U 1 U 1 U 1 aba = U 1 U 1 U 1 U (b)= a b a { } a b a { } a b a { } a b a a 1 1 1 1 1 1 1 1 U 1 U 1 (b)=U 1 (b )= a b a .So, aba = a b a . a b a { } { } { } Theorem 4.57. Let A be a JB-algebra . For each a A, the map U is positive 2 a and has norm a 2. k k 4For associative algebras, left inverse is same as right inverse 142 4 Spectral Theory for Jordan Algebras 1 1 + Proof. Let A denote the set of all invertible elements in A and let A = A A . 0 \ Using functional calculus, one can see that if b 0, then b is invertible b e, for some >0(4.2.24) () 1 Claim 4.57.1. A0 is open in A . Let a A .By(4.2.24), >0suchthata e. Now, look at b B(a, ) 2 0 9 2 2 \ 1 1 A = c A a c < .So, a b < = e b a e = 0 { 2 |k k 2 } k k 2 ) 2 2 ) e a e b a + e.Thus,0 e b.So,by(4.2.24),b A .Therefore, 2 2 2 2 2 0 A0 is open. 1 Claim 4.57.2. A0 is closed in A . + + 1 1 A is closed in A. So, A = A A is closed in A . 0 \ Claim 4.57.3. A0 is connected. Let a ,a A .Then, , > 0 such that a e, a e.Thisimplies 1 2 2 0 9 1 2 1 1 2 2 ↵a +(1 ↵)a (↵ +(1 ↵) )e.By(4.2.24),↵a +(1 ↵)a A .So,A 1 2 1 2 1 2 2 0 0 is convex, hence connected. + Claim 4.57.4. A0 is dense in A . Let a 0and✏>0begiven.Then,a + ✏e ✏e 0. Hence, by (4.2.24), a + ✏e A . And, a + ✏e a as ✏ 0. Hence, the claim. 2 0 ! ! Now, we will prove that Ua is positive. 1 Case (I). a A 2 1 2 1 1 By functional calculus, (a ) A . Now, if b A ,thenU (b)= aba A , 2 0 2 a { }2 1 1 1 by lemma 4.56. Hence, U (A ) A .Inparticular,U (A ) A . a ✓ a 0 ✓ Next, A0 is connected and Ua is continuous implies that Ua(A0)isconnected. 1 c 1 Now, A = A (A ) (A is both open and closed in A .) But, U (A ) 0 t 0 0 a 0 1 is a connected subset of A .Therefore,itiscontainedentirelyineitherA0 or c 1 2 2 1 2 (A ) . Note that 1 U (A ) A because U ((a ) )=a (a ) = e =1.Thus, 0 2 a 0 \ 0 a 143 4 Spectral Theory for Jordan Algebras U (A ) A . a 0 ✓ 0 Now, A is dense in A+ and U is continuous. Hence, U (A )=U (A+) 0 a a 0 a ✓ + A0 = A .Thus,Ua is positive. Case (II). a is an arbitary element in A. 1 2 Let b A . By functional calculus, c A such that c = b. Now, 2 0 9 2 U U (b)= c aba c = c ac2a c = cac 2 0(4.2.25) c a { { } } { { } } { } 2 2 Now, by (4.2.25), U (b)=U 1 (U U b)=U 1 cac .BycaseI,U 1 cac 0. a c c a c { } c { } Hence, U (A ) A+. Again, using density argument, U (A+) A+. a 0 ✓ a ✓ So, U is positive for each a A. a 2 Now, since U is a positive operator on A, U = U (e) = a2 = a 2. a k ak k a k k k k k Lemma 4.58. If a, b are positive elements of JB-algebra A, then aba =0 bab =0 (4.2.26) { } (){ } Also, aba =0 = a b =0 (4.2.27) { } ) Proof. Given a, b A+ . 2 1. Assume aba = 0. As b 0, by functional calculus, 0 b2 b b. { } kk Applying the positive operator U ,weget0 U b2 b U b,whichimplies a a k k a 0 ab2a b aba =0.Thus, { }k k{ } aba =0 = ab2a =0 (4.2.28) { } ){ } Now, bab 2 = b ab2a b =0.Thus, bab 2 = 0 which implies bab 2 = { } { { } } { } k{ }k bab 2 = 0. Hence, bab =0. k{ } k { } 2. Assume aba =0.Then,also bab = 0, by (1). Now, by (4.2.28), ab2a = { } { } { } 0and ba2b =0.By(4.2.18),(a b)2 = 1 (2a bab + ab2a + ba2b )=0. { } 4 { } { } { } Thus, a b 2 = (a b)2 = 0. Hence, (a b)=0. k k k k 144 4 Spectral Theory for Jordan Algebras 4.2.3 Projections and Compressions in JB-algebras Definition 4.59. Let A be a JB-algebra. A projection in A is an element p A 2 satisfying p2 = p. A compression on A is the map U : A A,whenp is a projection. p ! U (a)= pap =2p (p a) (p a), a A (4.2.29) p { } 8 2 We denote the projection e p by p0. AprioriwedonotknowwhetherthemapsUp are compressions in the sense of chapter 3 (definition 3.38). But in later sections, we will prove that, when p is a projection in a JBW-algebra, the map Up is indeed a compression, consistent with the earlier abstract definition (3.38). Lemma 4.60. Let a be a positive element and p be a projection in a JB-algebra A. Then pap =0 p a =0 (4.2.30) { } () Proof. The result follows from lemma 4.58 and (4.2.29). Lemma 4.61. Let p be a projection in a JB-algebra A. Then 2 Up 1,Up = Up,UpUe p =0 (4.2.31) k k If a 0, then Up(a)=0 Ue p(a)=a (4.2.32) () Proof. Note that if p =0,then p = p2 = p 2 = p =1.So,bytheorem 6 k k k k k k )kk 4.57, U = p 2 =1. k pk k k Using the identity (4.2.19): we get p pap p = p2ap2 .So,U (U (a)) = { { } } { } p p U pap = p pap p = p2ap2 = pap = U (a). Therefore, (U )2 = U . p{ } { { } } { } { } p p p Putting b = p in identity (4.2.20), we get UpUe p =0. 145 4 Spectral Theory for Jordan Algebras Next, let a 0. Assume U a = a.ThenapplyingU on both sides, we get: p0 p 0=U a.Conversely,ifU a =0,then pap =0.Bylemma4.58,wehavep a =0. p p { } So, p0 a =(e p) a = a.Thus,U a =2p0 (p0 a) p0 a =2p0 a a = a. p0 Hence, the result. Definition 4.62 (Adjoint map). If T : A A is a continuous linear operator, ! then T ⇤ : A⇤ A⇤ denotes the adjoint map, defined by (T ⇤⇢)(a)=⇢(Ta)for ! a A and ⇢ A⇤. 2 2 Proposition 4.63. Let p be a projection in a JB-algebra A. Let p0 = e p and be a state on A. Then (i) U ⇤ 1 k p k (ii) U ⇤ =1 (p)=1 U ⇤( )= k p k () () p (iii) U ⇤ =0 U ⇤ = p () p0 Proof. First we claim that if is a state on A, then (p0)=0 = U ⇤ = (4.2.33) ) p 2 Proof: Suppose (p0) = 0. Then, by Cauchy-Schwarz inequality, ( (p0 b)) 2 2 ((p0) ) (b )=0.Thus, (p0 b)=0, b A.Thisimplies (b)= (p b)forall 8 2 b A. Hence, U ⇤ (a)= (U (a)) = [2p (p a) p a]= [2p (p a)] [p a]= 2 p p (a). This is true for all a A. Hence, U ⇤ = . 2 p Now, we start proving the proposition. (i) U ⇤ = U =1. k p k k pk (ii) First assume that (p)=1.Then (p0)=0.So,by(4.2.33),Up⇤ = . Conversely, if Up⇤ = ,then (p)= (Up(e)) = Up⇤ (e)= (e)=1. Also, note that U ⇤ is a positive linear functional on A. Hence, U ⇤ = p k p k U ⇤ (e)= (U e)= (p). Therefore, U ⇤ =1 (p)=1. p p k p k () 146 4 Spectral Theory for Jordan Algebras (iii) Suppose U ⇤ =0.Then (Upe)=0 = (p)=0 = U ⇤ = (by p ) ) p0 (4.2.33)). Conversely, suppose U ⇤ = , then applying U ⇤ on both sides, we p0 p get Up⇤ =0. We already know that the maps Up are positive on A. Now, lemma 4.61 tells us that Up is infact a normalised projection on A, with complementary projection + + + + Ue p (ker Up =im Ue p and im Up =ker Ue p by (4.2.32)). Also, Up⇤ and Ue⇤ p are complementary projections, by lemma 4.63. Thus, the maps Up are bicomple- mented normalised positive projections on A. In subsequent sections, we will prove that these maps are also continuous in some kind of weak topology, arising from the duality with a base norm space. This will show that Up is an example of the abstract compression, defined in chapter 3. Properties of Compressions The maps Up hold many properties, similar to the abstract compressions of Chapter 3. The following are some such results, which have been proved independently using the theory of JB algebras. Notation 15. If a, b are elements of a JB-algebra A, then [a, b]denotestheir associated order interval, i.e. [a, b]= x A a x b .Ifa is any positive { 2 | } element of A, then face(a)denotesthefaceinA+ generated by a,i.e. face(a)= y A 0 y a, for some 0 (4.2.34) { 2 | } Lemma 4.64. Let p be a projection in a JB-algebra A. Then + (i) face(p)=im Up (ii) face(p) [0,e]=[0,p] \ 147 4 Spectral Theory for Jordan Algebras Proof. (i) Let a face(p). This implies 0 a p for some 0. This 2 implies U 0 U a U U e =0.So,U a = 0. Hence, by (4.2.32), p0 p0 p0 p p0 U a = a which implies a im+U .Conversely,leta im+U .Then p 2 p 2 p a = U a U ( a e)= a p which implies a face(p). p p k k k k 2 (ii) Let a face(p) [0,e]. Then, by (i), a im+U .Thus,a = U a U e = 2 \ 2 p p p p = 0 a p = a [0,p]. The other way inclusion is easy to see. ) ) 2 Proposition 4.65. The extreme points of [0,e] (the positive part of the unit ball) of a JB-algebra A are the projections. Proof. Let I denote the set [0,e]andlet@eI denote the set of extreme points of I. ( ) Let a @ I. Then, by functional calculus, 0 a2 a e. Hence, a2 I. ✓ 2 e 2 Now, 0 a e = 0 i(x) 1, for all x (a). Now, 0 (1 + i(x))2 ) 2 for each x (a)impliesthat0 2a a2 e. Hence, 2a a2 I. Now, 2 2 a = 1 (a2 +(2a a2)) and a is an extreme point of I. Hence, a = a2 which 2 implies that a is a projection in A. ( ) Let p be a projection in A. If p = a +(1 )b for some a, b I,then ◆ 2 a, (1 )b p which implies that a, b face(p) [0,e]=[0,p]. So, 2 \ p = a +(1 )b p +(1 )p = p.Thus,a = b = p.So,p @ I. 2 e Let A⇤ denote the set of continuous linear functionals on A. The norm of each positive linear functional on A, is its value at e.Recallthatforeachstate⇢ on a JB-algebra A, the map a, b = ⇢(a b) is a positive semi-definite bilinear form h i and thus we have the Cauchy-Schwarz inequality 2 1 2 1 ⇢(a b) ⇢(a ) 2 ⇢(b ) 2 (4.2.35) | | 2 1 And taking b = e,weget ⇢(a) ⇢(a ) 2 . | | 148 4 Spectral Theory for Jordan Algebras Proposition 4.66. If p is a projection in a JB-algebra A, then the image of Up, i.e. U (A)= pAp is a JB-subalgebra with identity p. p { } Proof. Since Up is a linear operator on A, its image is a subspace of A. Let b = pap U (A). Then b2 = pap 2 = p ap2a p U (A). So, U (A)isclosed { }2 p { } { { } }2 p p under squares. Now, if a, b U (A), then a b = 1 [(a + b)2 a2 b2] U (A). 2 p 2 2 p Thus, Up(A)isclosedunderproduct. Claim 4.66.1. U (A)= a A U a = a p { 2 | p } Let a = pbp U (A). Then U (a)=(U )2(b)=U (b)=a. Hence, the claim. { }2 p p p p Note that U is continuous on A. If a a in A, where a U (A)for p n ! n 2 p all n,thenU (a ) U (a). But U (a )=a for all n, by the claim. Hence, p n ! p p n n U (a)=lim U (a )=lim a = a.Thus,a U (A)andhenceU (A)isnorm p n p n n n 2 p p closed. Claim 4.66.2. p is the identity for Up(A). Let a pAp .Then,bytheclaim,a = U a which implies U a =0 = 2{ } p p0 ) p0ap0 =0 = p0 a =0 = p a = a.So,p is the identity for U (A). { } ) ) p So, Up(A) is a norm closed Jordan subalgebra of A, and has identity element p. With respect to the inherited norm, Up(A) thus becomes a JB-subalgebra of A Definition 4.67. If p is a projection in a JB-algebra A, then Ap denotes the JB-subalgebra pAp . { } Proposition 4.68. If p is a projection in a JB-algebra A, then a b =0for all a A ,b A0 . 2 p 2 p Proof. Let a A and b A . As A is positively generated and the maps U ,U 2 p 2 p0 p p0 are positive, it follows that Up(A)andUp0 (A) are also positively generated. Hence, it is su cient to prove the result for 0 a, b. Note that since p is the identity for 149 4 Spectral Theory for Jordan Algebras A , p0 a = a (p a)=0foralla A . Now, 0 b b e = 0 b = p 2 p kk ) U b U b e = b p0. Applying U ,weget0 aba b ap0a =0.So,we p0 p0 k k k k a { }kk{ } get a b =0. 4.2.4 Orthogonality Definition 4.69. Let A be a JB-algebra . Two positive elements a, b in A are said to be orthogonal,denotedbya b,if aba = bab =0. ? { } { } Proposition 4.70. Let A be a JB-algebra . Then each a A can be uniquely 2 expressed as a di↵erence of two positive orthogonal elements: + + + a = a a , where a ,a 0,a a (4.2.36) ? + Both a and a will be in the norm closed subalgebra C(a, e). This unique decom- position is called the orthogonal decomposition of a. Proof. Let X be a compact Haudsor↵space and let f C (X) . Now, define 2 R + f ,f : X R by ! f(x)iff(x) 0 f +(x)= 8 < 0otherwise : 0iff(x) 0 f (x)= 8 f(x)otherwise < + + Then, f ,f are continuous positive functionals on X and f = f f . : + Moreover, this is the unique decomposition of f with the property that f ,f 0 + and f .f =0. We know that there exists an isometric isomorphism, C(a, e) ⇠= CR( (a)) + + with a I.Byfunctionalcalculus, a ,a C(a, e)suchthata = a 7! 9 2 + + 5 a ; a ,a 0anda a =0. Also, this decomposition is unique in C(a, e), 5 + + + + Since a ,a belong to the associative algebra C(a, e), a a =0 = a a a = 0. + ){ } Hence, a a . ? 150 4 Spectral Theory for Jordan Algebras by functional calculus. Now, we need to show that the orthogonal decomposition of a is unique in A. Suppose a = b c is another orthogonal decomposition of a in A. Then, b, c 0 and b c. ? Claim 4.70.1. bn cm =0, m, n 1 8 We will first show that 0 x, y, x y = xn y, n 1(4.2.37) ? ) ? 8 WLOG, assume that x , y 1. Then, by functional calculus, xn x, n N. k k k k 8 2 Now, fix n N.Then0 yxny yxy =0 = yxny =0 = xn y. 2 { }{ } ){ } ) ? Similarly, begin with xn and y,andproceedasabovetogetxn ym for every ? fixed m N.Bylinearityoftripleproduct,theresultistrueevenforany x , y . 2 k k k k Hence, the claim. Claim 4.70.2. C(b, c, e) is an associative JB-subalgebra of A. Since bn cm =0, m, n 1, the subalgebra P (b, c, e) of A, is associative and 8 commutative. As multiplication is jointly continuous in A, C(b, c, e)isalsoasso- ciative. By theorem 4.38, C(b, c, e) ⇠= CR(X) for some compact Hausdor↵space X. Now, + a = b c C(b, c, e). Thus, C(a, e) C(b, c, e). Thus, a = a a and a = b c 2 ✓ are two orthogonal decompositions of the image of a in CR(X) . By uniqueness + of orthogonal decomposition in CR(X) , we get a = b and a = c. Hence, the result. 4.2.5 Commutativity Next, we want to describe the Jordan analog of elements commuting in an asso- ciative algebra. Definition 4.71. Let A be a JB-algebra and let a A. Define T : A A given 2 a ! 151 4 Spectral Theory for Jordan Algebras by T (b)=a b.Twoelementsa, b Aaresaidtooperator commute in A if a 2 TaTb = TbTa. It is easy to see that in a special JB-algebra , elements commuting in the surrounding associative algebra, operator commute in the Jordan algebra. Lemma 4.72. Let d belong to a JB-algebra A and let p be a projection in A. Then p d = p = T T = T T . ) p d d p 2 2 Proof. Note that (a b) a = a (b a)= T 2 T (b)=T T 2 (b)= ) a a a a ) 6 [Ta2 ,Ta] = 0. Now, putting a = d + 1d + 2d in [Ta2 ,Ta] = 0 and equating the coe cients of 1 2 in the resulting polynomials gives [Td,Tp]+[Tp,Tp d]=0 (4.2.38) Putting p d = d gives [T ,T ] = 0. Hence, p, d operator commute. d p Proposition 4.73. Let A be a JB-algebra and let a A and p be a projection in 2 A. Then the following are equivalent: (i) a and p operator commute (ii) Tpa = Upa (iii) (Up + Up0 )a = a (iv) p and a are contained in an associative subalgebra Proof. Recall that U (b)=2p (p b) (p b)=2T T (b) T (b). p p p p (i ii) Assume T T = T T .Then,U (a)=U T (e)=T U (e)=T (p)= ) a p p a p p a a p a TaTp(e)=TpTa(e)=Tp(a). (ii iii) Let T a = U a.Byidentity(4.2.22),wehaveT = 1 (I + U U ). So, here ) p p p 2 p p0 1 we have U (a)=T (a)= (a + U a U a). This implies (U + U 0 )a = a. p p 2 p p0 p p 6adopt the procedure used in the proof of lemma 4.29 to justify why the coe cients can be equated 152 4 Spectral Theory for Jordan Algebras (iii i) Assume (U + U 0 )a = a.Letr = U a, s = U a.Thenr + s = a and ) p p p p0 r A ,s A . Now, p is the identity for A and p0 is the identity 2 p 2 p0 p for A .So,r p = r and s p0 = s. Now, by lemma 4.72, we have p0 TpTr = TrTp and Tp0 Ts = TsTp0 and this also implies TpTs = TsTp. Now, TpTa = Tp(Tr + Ts)=TpTr + TpTs = TrTp + TsTp =(Tr + Ts)Tp = TaTp.So, a, p operator commute. (iii iv) We know that, p is the identity for the JB-subalgebra A . Hence, p and ) p Up(a)generateanassociativesubalgebraofAp.Similarly,p0 and Up0 (a) generate an associative subalgebra of A .Forx A and y A0 ,wehave p0 2 p 2 p x y =0,byproposition4.68.Itfollowsthatp, U (a),p0 and U (a)generate p p0 an associative subalgebra B in A. Then, p and (Up + Up0 )a = a are contained in the associative subalgebra B. (iv iii) If p and a are contained in an associative subalgebra B of A, then U (a)= ) p p a.ThenadjoiningeandtakingthenormclosureofBgivesanassociative subalgebra of A, which also contains p0.So,herewegetU (a)=p0 a = p0 a (p a). And, hence (U + U 0 )a = a. p p Proposition 4.74. Let A be a JB-algebra and let p be a projection in A. Let a 0. Then a operator commutes with p U a a (4.2.39) () p Proof. If a 0andp is a projection which operator commutes with a,then U a + U a = a (proposition 4.73). So U a a. p p0 p Conversely, if a 0andU a a,thena U a is a positive element such that p p U (a U a)=0.By(4.2.32),U (a U a)=(a U a). Thus U a + U a = a and p p p0 p p p p0 so p and a operator commute by proposition 4.73. 153 4 Spectral Theory for Jordan Algebras Ideals and Approximate Identities While developing the spectral result for JBW-algebras, we somewhere use the fact that every norm closed ideal in a JB-algebra has an increasing approximate identity. The goal of this subsection is to prove the same. Definition 4.75. An ideal of a Jordan algebra is a linear subspace closed under multiplication by elements from the algebra . Definition 4.76. ApartiallyorderedsetXissaidtobedirected upwards if for each ↵, X, there exists Xsuchthat↵, .Anetisasetwhichhasa 2 2 bijection with a directed set (directed upwards). Definition 4.77. Let J be a norm closed ideal in a JB-algebra A. An increasing net v in J is an increasing approximate identity for J if 0 v e, ↵ and { ↵} ↵ 8 lim b v b =0forallb J. ↵ k ↵ k 2 Lemma 4.78. If a, b are elements of a JB-algebra , then 1. ab2a = ba2b k{ }k k{ }k 2. If a 0, then a b 2 a bab k k k kk{ }k Proof. First recall identity (4.2.21), ab2a 2 = a b ba2b b a . { } { { { } } } 1. ab2a 2 = ab2a 2 = a b ba2b b a = U U ba2b U U ba2b k{ }k k{ } k k{ { { } } }k k a b{ }k k a bkk{ }k = (U U )(e) ba2b = ab2a ba2b .So,weget ab2a ba2b . k a b kk{ }k k{ }kk{ }k k{ }k k{ }k Similarly, we can show ba2b ab2a . Hence, ab2a = ba2b . k{ }k k{ }k k{ }k k{ }k (4.2.18) 2. Assume a 0. Let (a b)2 = (a b) 2 = 1 2a bab + ab2a + k k k k 4 k { } { } ba2b 1 2a bab + 1 ba2b (by (1)). So, (a b) 2 = 1 a bab + { }k 4 k kk{ }k 2 k{ }k k k 2 k kk{ }k 1 U (a2) .Byfunctionalcalculus,wegeta2 a a.Thus,0 U (a2) 2 k b k k k b a bab . Since order unit norm respects the order for positive elements, we k k{ } get U (a2) a bab .So, (a b)2 1 a bab + 1 a bab = k b kk kk{ }k k k 2 k kk{ }k 2 k kk{ }k a bab . k kk{ }k 154 4 Spectral Theory for Jordan Algebras Lemma 4.79. If a, b are invertible elements of a JB-algebra A. Then 0 a 1 1 b = b a . ) 1 1 1 1 Proof. b 0= b 0, by functional calculus. Also, b 2 exists and (b ) 2 = ) 1 1 1 1 1 (b 2 ) 0. Let us denote (b 2 ) by b 2 . Now, 0 a b = 0 U 1 a ) b 2 1 1 1 1 7 U 1 b.So,0 b 2 ab 2 b 2 bb 2 = e.Byfunctionalcalculus we get b 2 { }{ } 1 1 1 1 1 1 1 1 0 e b 2 ab 2 = b 2 a b 2 (by prop 4.56). Hence, 0 b U 1 e { } { } b 2 1 1 1 1 1 1 1 1 2 2 1 1 1 1 1 1 U b a b = U U (a )= U 1 U (a )= (U ) U (a )= b 2 { } b 2 b 2 (b 2 ) b 2 b 2 b 2 1 1 1 a .Thus,0 b a . Proposition 4.80. Every norm closed ideal J in a JB-algebra A contains an increasing approximate identity. Proof. Define U = a J + a < 1 . { 2 |k k } Claim. U is directed upwards. Note that J + is directed upwards because if a, b J +,thena, b (a+b) J +. 2 2 We will show that U is directed upwards, by constructing an order isomorphism between U and J +. t Define f :[0, 1) [0, )givenbyf(t)= 1 t . Note that f is a continuous ! 1 function. Let a U.Thena 0and a < 1whichimplies (a) [0, 1). 2 k k ✓ Restricting f to (a), we get f C ( (a)) = C(a, e). Denote the image of f, 2 R ⇠ under this isomorphism, as f(a). f 0= f(a) 0. And f(0) = 0 = ) ) f(a) C(a) J, by proposition 4.47. Hence, f(a) J +.Moreover,ifi denotes 2 ✓ 2 i 1 1 the identity function on (a), then f = 1 i = 1 i 1= f(a)=(e a) e. ) 7 1 1 1 If x, y CR(X) and 0 x, y are invertible, then 0 x ,y . So, y x = x .y 1 2 1 1 1 1 1 1 ) x .x = x .y 1= x .y.y y = x y . ) ) ) 155 4 Spectral Theory for Jordan Algebras Hence, we get a map 8 + 1 f˜ : U J defined as f˜(a)=(e a) e ! Next, define g :[0, ) [0, 1) given by g(t)= t .Theng is a continuous 1 ! 1+t function. Let b J +.Thenb 0and (b) [0, ). Restricting g to (b), we 2 ✓ 1 get g C ( (b)) = C(b, e). Denote the image of g,underthisisomorphism,as 2 R ⇠ g(b). Now, g(0) = 0 = g(b) C(b) J, by proposition 4.47. Next, note that ) 2 ✓ g(x) < 1forallx (b). As (b)isacompactset,g attains its supremum and 2 hence g < 1. This implies g(b) < 1. Hence, g(b) U. And, if i denotes the k k k k 2 i 1 1 identity function on (b), then g = =1 = g(b)=e (e+b) . Hence, 1+i 1+i ) we get a map + 1 g˜ : J U defined asg ˜(b)=e (e + b) ! Now, we will show that f˜ is an order isomorphism between U and J +. Let x, y U such that x y.Then,(e x) (e y). By lemma 4.79, we see 2 1 1 that (e x) (e y) .Thus,f˜(x) f˜(y). Hence, f˜is an order preserving map. + 1 Similarly, if x, y J such that x y,then(e + x) (e + y)= (e + x) 2 ) 1 (e + y) = g˜(x) g˜(y). So,g ˜ is also an order preserving map. ) Next, take a U.Then,byproposition4.47-(3),weget(˜g f˜)(a)=˜g(f˜(a)) = 2 1 1 1 + g˜ (e a) e = e e +(e a) e = a.Similarly,ifb J ,thenby 2 1 proposition 4.47-(3), we get (f˜ g˜)(b)=f˜ (˜g(b)) = f˜ e (e + b) = e (e 1 1 + (e + b) ) e = b. Hence, (f˜ g˜)(b)=b, (˜g f˜)(a)= a, a U, b J .Thus, 8 2 2 + 1 f˜ is an order isomorphism from U to J ,with(f˜) =˜g. Now, if x, y U,thenf˜(x), f˜(y) J +. And as f˜(x), f˜(y) f˜(x)+f˜(y) 2 2 2 J +, we haveg ˜(f˜(x)), g˜(f˜(y)) g˜ f˜(x)+f˜(y) U,i.e.x, y g˜ f˜(x)+f˜(y) 2 2 U.Thus,U is directed upwards. 8Note that 0 a a e = (e a) (1 a )e. As a < 1, we have (e a)isinvertible, by (4.2.24). k k ) k k k k 156 4 Spectral Theory for Jordan Algebras Consider U as a net over itself. We will show that U is an increasing approxi- mate identity for J.Clearly,U is an increasing net and 0 a e,foreacha U. 2 For each n N,definehn : R R given by 2 ! n 1 1 ,t (constant function 1 1 ) n n n 8 1 1 (n 1)t, t 0(linearfrom to 0) > n > n > hn(t)=> 0,t=0 > > < 1 1 (n 1)t, 0 t (linear from 0 to n ) > n > n 1 1 > ,t(constant function 1 1 ) > n > n n > We claim that :> 1 t2(1 h (t)) whenever t [ 1, 1] (4.2.40) | n |n 2 1 2 2 1 1 Fix t [ 1, 1] and n N.Ift ,then t (1 hn(t)) = t .The 2 2 | | n | | | n | n 1 n 1 case t = 0 is clear. Now let 0 t .Then0 (n 1)t .So, n n 0 h (t) 1 1 This gives 0 h (t) 1 1whichimplies1 1 h (t) 1 . n n n n n n So, 0 t2(1 h (t)) t2 1 1 . Similarly, it can be shown that when n n2 n 1 2 1 t 0, we have t (1 h (t)) . Hence, (4.2.40) holds. n | n | n 2 Claim. For each a J,limu U a aua =0 2 2 k { }k First assume a 1. Then, (a) [ 1, 1]. Restricting h to (a), we get k k ✓ n h C ( (a)) = C(a, e). Denote the image of h under this isomorphism as n 2 R ⇠ n h (a). Note that for all t (a), we have h (t) 1 1 < 1. So, h (a) = n 2 | n | n k n k h 1. And h (0) = 0 = h (a) C(a) J, by proposition 4.47. Also, k nk n ) n 2 ✓ h 0= h (a) 0. Hence, h (a) U. Now, n ) n n 2 1 a2 ah (a)a = 9 a2 (a2 h (a)) = a2 (e h (a)) (4.2.41) k { n }k k n k k n kn where the last inequality follows from functional calculus and (4.2.40). Now, given ✏>0, choose n such that 1 ✏.Ifu h (a), then we have ah (a)a n n { n } 9h (a) C(a) implies that ah (a)a =(a2 h (a)) n 2 { n } n 157 4 Spectral Theory for Jordan Algebras aua aea = a2.So,forallu U with u h (a), we have a2 aua { }{ } 2 n k { }k 2 1 2 a ahn(a)a by (4.2.41). Hence, limu U a aua =0,whenever k { }k n 2 k { }k a 1. As norm and triple product are linear, the claim holds for all a J. k k 2 Now, if u U and a J,then a (a u) 2 = a (e u) 2 e u a(e 2 2 k k k k k kk{ u)a (by lemma 4.78) = (e u) a2 aua a2 aua . Hence, }k k kk { }k k { }k 2 limu U a (a u) limu U a aua =0 2 k k 2 k { }k Thus, U is an increasing approximate identity for J. Remark 4.81. If J is a norm closed ideal in a JB-algebra A, then the quotient space A/J is a Jordan algebra and a Banach space with the usual quotient norm a + J =infb J a + b k k 2 k k We now have all the required machinery to venture into JBW-algebras. 4.3 JBW algebras Definition 4.82. An ordered Banach space A is said to be monotone complete if every increasing net b in A, which is bounded above, has a least upper bound in { ↵} A. (We write b b for such a net.) A bounded linear functional on a monotone ↵ % complete space A is said to be normal if whenever b b,then (b ) (b). ↵ % ↵ ! Definition 4.83. A JBW-algebra is a JB-algebra that is monotone complete and admits a separating set of normal states. Example 4.84. The self-adjoint part of a von Neumann algebra with the sym- metrised product is a JBW-algebra. Throughout this section, M will denote a JBW-algebra. Notation 16. The set of normal states on M will be denoted by K and V will denote the linear span of K in M⇤. 158 4 Spectral Theory for Jordan Algebras Remark 4.85. Define V + = k 0,k K .Then,V+ is cone in V and K is a { | 2 } + + base for V .Further,asKisaw⇤-compact subset of M ⇤,weseethat(V,V ,K) is a base norm space, with distinguished base K. We note for future reference, that if ⇢ is a normal state on M, then a, b h i 7! ⇢(a b) is a positive semi-definite bilinear form and hence, by Cauchy-Schwarz inequality, for each a, b M, we get: 2 2 1 2 1 ⇢(a b) (⇢(a )) 2 (⇢(b )) 2 (4.3.1) Definition 4.86 ( -weak topology). The topology on M defined by the duality of M and K, is called the -weak topology. Thus, the -weak topology (denoted by )isinducedbythesemi-normsa ⇢(a) ,for⇢ K. The subbasis elements w 7! | | 2 are: B(x,⇢, r)= y M ⇢(x y) b w b ⇢(b ) ⇢(b), ⇢ K (4.3.2) ↵ ! () ↵ ! 8 2 Remark 4.87. Note that -weak limit of a given net is unique, if it exists, because if ⇢(a)=⇢(b)forall⇢ K,thena = b (Madmitsseparatingsetofnormal 2 states!) And by linearity of elements in K, we also see that sum of -weak limits is -weak limit of sums. Definition 4.88 ( -strong topology). The -strong topology on M (denoted by 2 1 )isinducedbythesemi-normsa (⇢(a )) 2 for ⇢ K. The subbasis elements s 7! 2 2 1 are given as B(x,⇢, r)= y M (⇢(x y) ) 2 a s a, b s b = (a + b ) s a + b10 (4.3.4) { ↵ ! ! } ) ↵ ! 10over directed set (↵, ) 159 4 Spectral Theory for Jordan Algebras Recall the linear maps U (b)= aba and T (b)=a b,definedonM,foreach a { } a a M. 2 Proposition 4.90. For each a M: 2 1. The maps T ⇤,U⇤ : M ⇤ M ⇤ take V into V. a a ! 2. Ua,Ta is continuous with respect to the -weak topology on M. 3. Ua,Ta is continuous with respect to the -strong topology on M. 4. Jordan multiplication is jointly continuous on bounded sets, with respect to the -strong topology. Proof. Fix a M. It is easy to verify that 2 1 T = (U U I)(4.3.5) a 2 e+a a 1. We will first show that U ⇤(V ) V . a ✓ Case (I). a is invertible. If a is invertible, then Ua becomes an order isomorphism on M, with positive inverse map U 1 (lemma 4.55). So, x y U x U y.Let⇢ K.If a () a a 2 b 0, then (U ⇤⇢)(b)=⇢(U b) 0. Also, if b b in M, then U b U b, a a ↵ % a ↵ % a since U is an order isomorphism. So, (U ⇤⇢)(b )=⇢(U b ) ⇢(U b)= a a ↵ a ↵ % a U ⇤⇢(b). Thus, U ⇤⇢ is positive and normal for each ⇢ K. Hence, U ⇤(K) a a 2 a ✓ V which implies U ⇤(V ) V . a ✓ Case (II). a is an arbitary element of A. Choose > a .Thenbyfunctionalcalculus, e a and e+a are invertible, k k and from the linearity of the triple product, we get 1 2 Ua = (U e+a + U e a 2 I)(4.3.6) 2 160 4 Spectral Theory for Jordan Algebras By case I, It follows that U ⇤ maps V into V for every a A. a 2 Now, by (4.3.5), we also see that T ⇤ maps V into V, for all a A. a 2 2. Let b w b.Thisimplies⇢(b ) ⇢(b), ⇢ K = ⇢(b ) ⇢(b), ⇢ ↵ ! ↵ ! 8 2 ) ↵ ! 8 2 V .Inparticular,(U ⇤⇢)(b ) (U ⇤⇢)(b), ⇢ K,by(1).Thisimplies, a ↵ ! a 8 2 ⇢(U (b )) ⇢(U (b)), ⇢ K = U (b ) w U (b). Therefore, U is a ↵ ! a 8 2 ) a ↵ ! a a continuous in -weak topology. Now, using (4.3.5), we see that Ta is also continuous in -weak topology. 3. Let b s 0. This implies ⇢(b2 ) 0, ⇢ K which implies b2 w ↵ ! ↵ ! 8 2 ↵ ! 0= U (b2 ) w 0. Now, as 0 a2 a2 e = a 2e,wehave ab a 2 = ) a ↵ ! k k k k { ↵ } a b a2b a = U b a2b = U U (a2) U U ( a 2e)= a 2U (b2 )= { { ↵ ↵} } a{ ↵ ↵} a b↵ a b↵ k k k k a ↵ a 2 ab2 a w 0. Thus, for all ⇢ K,wegot⇢( ab a 2) 0which k k { ↵ } ! 2 { ↵ } ! implies ab a s 0. Now, if b s b, then U (b ) s U (b), by remark { ↵ } ! ↵ ! a ↵ ! a 4.89. Hence, Ua is continuous in -strong topology. And, by (4.3.5), we see that Ta is also continuous in -strong topology. 4. Consider two nets a s 0andb s 0inMsuchthat N ,N > ↵ ! ! 9 1 2 0 such that a N , b N , ↵, .Foreacha M, by functional k ↵k 1 k k 2 8 2 calculus, a4 a2 a2.So,for⇢ K,weget⇢(a4 ) a2 ⇢(a2 ) 0. So, k k 2 ↵ k ↵k ↵ ! ⇢(a4 ) 0forall⇢ K which implies a2 s 0. So, we have prove that ↵ ! 2 ↵ ! for a bounded net a , { ↵} a s 0= a2 s 0(4.3.7) ↵ ! ) ↵ ! Also, note that for a, b M, 2 1 a b = [(a + b)2 a2 b2](4.3.8) 2 Then using (4.3.4), (4.3.7) and (4.3.8), we get that (a b ) s 0. ↵ ! Now, consider two bounded nets a s a and b s b in M. Then, a ↵ ! ! a s 0andb b s 0 in M and the nets are bounded. Applying the joint ↵ ! ! 161 4 Spectral Theory for Jordan Algebras continuity of multiplication on bounded sets at 0 and separate continuity of multiplication (3), in the following: (a b) (a b )=(a a ) b +(a a ) (b b )+a (b b )(4.3.9) ↵ ↵ ↵ we get that a b s a b in M. ↵ ! The following are some basic results about convergences in these topologies. Proposition 4.91. Let M be a JBW-algebra. Then: 1. -strong convergence implies -weak convergence. 2. A bounded monotone net in M converges -strongly. 3. A monotone net of projections converges -strongly to a projection. Proof. Let a be a net in M. { ↵} 1. Suppose a s a.Then,byCauchy-Swartzinequality,⇢(a a ) (⇢(a ↵ ! ↵ 2 1 a ) ) 2 0, for ⇢ K. Thus, ⇢(a a ) 0, for all ⇢ K. Hence, ↵ ! 2 ↵ ! 2 a w a. ↵ ! Remark 4.92. We know that -weak limit of a net is unique, if it exists. And the proposition above says that strong convergence implies weak convergence. Hence, -strong limit of a net is unique, if it exists. 2. Suppose a a.Then,bydefinitionofnormalstate,⇢ K= ⇢(a ) ↵ % 2 ) ↵ ! ⇢(a). By functional calculus, a2 a a,foreacha M. Now, ⇢((a a )2) k k 2 ↵ a a ⇢(a a ) 0. Hence, a s a. k ↵k ↵ ! ↵ ! 3. Suppose p is a net of projections in M. Then, p e for each ↵. As M { ↵} ↵ is monotone complete, p p,forsomep M. By (2), p s p. Now, ↵ % 2 ↵ ! by proposition 4.90-(3), p = p p s p p.Byremark4.92,p = p p. ↵ ↵ ↵ ! Hence, p is a projection in M. 162 4 Spectral Theory for Jordan Algebras Corollary 4.93. Every -weakly continuous linear functional on M is normal. Proof. Let f be a -weakly continuous linear functional on M and let a a in ↵ % M. Then, by proposition 4.91, a s a which inturn implies a w a. Now, as f ↵ ! ↵ ! is weakly continuous, we get f(a ) f(a). ↵ ! Remark 4.94. For a monotone net of elements a and any element b in M, we { ↵} have a a = U (b) s U (b)(4.3.10) ↵ % ) ↵ ! a Proof. Let a a.Then,byproposition4.91,a s a and (a a ) s (a a). ↵ % ↵ ! ↵ ↵ ! Then, by proposition 4.90-2, it follows that T (a2 ) s T (a2) b ↵ ! b Next, T (a ) s T (a)implies b ↵ ! b (a T (a )) s (a T (a)) ↵ b ↵ ! b So, U =2a (a b) (a2 b) s 2a (a b) (a2 b)=U (b). a↵ ↵ ↵ ↵ ! a Proposition 4.95. Let B be a -weakly closed Jordan subalgebra of M. Then B has an identity and B is a JBW-algebra. Proof. B is given to be a Jordan subalgebra of M. We will first show that B is monotone complete. Let b be an increasing bounded net in B. As M is { ↵} monotone complete, b M such that b b.Then,byproposition4.91, 9 2 { ↵}% b s b which in turn implies b w b. As B is -weakly closed, b B. { ↵} ! { ↵} ! 2 Hence b b in B. Thus, B is monotone complete. { ↵}% Note that if a is a net in B such that a a in norm, then a w a. { ↵} { ↵} ! { ↵} ! Hence, a B.Thus,Bisnormclosed. 2 163 4 Spectral Theory for Jordan Algebras Let C be the linear span of B and e in M. Then, C is a norm closed Jordan subalgebra in M, and hence a JB-subalgebra of M. Note that B is a norm closed ideal in C. By proposition 4.80, B has an increasing approximate identity v . { } Let v v in M. Let a B.Then,v a a in norm and also v a v a % 2 ! ! in -weak topology. But norm limit is also the -weak limit. Thus, v a = a,for all a B. Hence, v is an identity for B. 2 And finally the normal states in M (which are not zero on v), when restricted to B, give a separating set of normal positive functionals on B. Hence, B becomes a JBW algebra, with the inherited norm. Definition 4.96. If M is a JBW-algebra, a JBW-subalgebra is a -weakly closed Jordan subalgebra N of M. By proposition 4.95, N itself is a JBW-algebra. The following are some corollaries of proposition 4.95. Proposition 4.97. If p is projection in M, then Mp = Up(M) is a JBW-subalgebra of M, with identity p. Proof. By proposition 4.66, Mp = Up(M) is a JB-subalgebra with identity p.We claim that M is weakly closed. Note that by 4.66.1, M = a M U a = a . p p { 2 | p } Now, if a is a net in M such that a w a in M, then proposition 4.90-(2), { ↵} p ↵ ! a = U (a ) w U (a). By uniqueness of -weak limits, U (a)=a and thus, ↵ p ↵ ! p p a M . Now, by proposition 4.95, it follows that M is a JBW-subalgebra. 2 p p Corollary 4.98. If p is a projection in M, then Im(Up + Up0 )=Mp + Mp0 is a JBW-subalgebra of M. Proof. By proposition 4.97, we know that Mp and Mp0 are JBW-subalgebras. And by proposition 4.68, we see that x y =0forallx M ,y M .Thus,M +M 2 p 2 p0 p p0 is a Jordan subalgebra containing M and M . Now, if b M and c M ,then p p0 2 p 2 p0 by lemma 4.61, we get (Up + Up0 )(b + c)=b + c.Thus, M + M = a M (U + U )a = a (4.3.11) p p0 { 2 | p p0 } 164 4 Spectral Theory for Jordan Algebras We will now show that M + M is -weakly closed. Let a be a net in p p0 { ↵} M + M and let a w a in M. Then, a =(U + U )(a ) w (U + U )(a). By p p0 ↵ ! ↵ p p0 ↵ ! p p0 uniqueness of -weak limits, (U + U )(a)=a and thus, a M + M . Hence, p p0 2 p p0 M + M is weakly closed. p p0 Thus, by proposition 4.95, it follows that Mp + Mp0 is a JBW subalgebra of M. Let W (a, e)denotethe -weakly closed subalgebra generated by a and e in M. The following is the main theorem that allows us to find a correspondence between JBW-algebras and monotone complete CR(X) spaces. Theorem 4.99. If B is an associative Jordan subalgebra of M, then the -weak closure of B is an associative JBW-algebra, which is isomorphic to a monotone complete CR(X) . In particular, this holds for W (a, e). Proof. By proposition 4.95, the -weak closure of B, say C, is a JBW subalgebra of M. Moreover, C is also associative because Using associativity of B and separate continuity of Jordan multiplication, it follows that B =Cisalsoassociative(see A.8) . Now, by theorem 4.38, C is isometrically isomorphic to some CR(X) . Since Cismonotonecomplete,soisCR(X) . Corollary 4.100. If p is a projection in M, which operator commutes with an element a M, then p operator commutes with every element in W (a, e). 2 Proof. Note that for x M, 2 x operator commutes with p 4.73 x = U x+U x 4.3.11 x (M +M )(4.3.12) () p p0 () 2 p p0 So, if a operator commutes with p,thena (M + M )and(U + U )e = e 2 p p0 p p0 2 Mp + Mp0 . As Mp + Mp0 is a -weakly closed JB subalgebra of M (corollary 4.98), it follows that W (a, e) M + M . Now, by (4.3.12), we see that p operator ✓ p p0 commutes with every element in W (a, e). 165 4 Spectral Theory for Jordan Algebras 4.3.1 Range Projections The following proposition and lemmas talk about the existence and properties of range projections in M. These notions coincide with our earlier definitions of range projections (definition1.8, definition 3.139). Proposition 4.101. For every a M, ! projection p W (a, e) face(a+) 2 9 2 \ ( -weak closure) such that U a 0,Ua a. Furthermore, p is the smallest p p + + projection in M such that Upa = a . Proof. By theorem 4.99, we can identify W (a, e)withamonotonecompleteCR(X) .Letuscalldenotethisisomorphismas :W (a, e) C (X) . Let (a)=ˆa. ! R Define E = x X aˆ(x) > 0 .Byproposition1.5,Eisbothclosedandopenin { 2 | } + + 1 X, C (X) ,a ˆ 0onE,ˆa 0 on X/E and (ˆa) =(ˆa) .Letp = ( ). E 2 R E E Then, p W (a, e)andsatisfiesp2 = p. Note that, as W (a, e)iscommutativeand 2 p W (a, e)itfollowsthat 2 U (x)= pxp = p x, x W (a, e) p { } 8 2 So, (ˆa)+ =(ˆa)+ = p a+ = a+ = U a+ = a+. E ) ) p Asa ˆ 0onEandˆa 0 on X/E, it follows that aˆ 0and aˆ aˆ.So, E E p a a and p a 0. Therefore, U a a and U a 0. p p + By proposition 1.5, E is the supremum of an increasing sequence in face((ˆa) ). By the order isomorphism , it follows that p is the supremum (in W (a, e)) of an increasing sequence a in face(a+). If b is the supremum of a in M, then { ↵} { ↵} by proposition 4.91-(2),(3), a w b. As, W (a, e)is -weakly closed, b W (a, e) ↵ ! 2 and so b = p. Thus, p belongs to the -weak closure of face(a+). Hence, p 2 W (a, e) face(a+)( -weak closure). \ To prove uniqueness, assume that there is another projection q W (a, e) 2 \ face(a+)( -weak closure) such that U a a and U a 0. Let (q)= for q q F 166 4 Spectral Theory for Jordan Algebras some F X and denote G = X/F . Note that 0 U a = q a = aˆ 0. ✓ q ) F Also, 0 a U a = aˆ aˆ =(1 )ˆa = aˆ 0. Thus,a ˆ 0onFand q ) F F G aˆ 0onG.Byproposition1.5,E F ,whichimplies = p q.On ✓ E F ) the other hand, U a+ = a+ = a+ im+U = face(p). Thus, q face(a+) p ) 2 p 2 ✓ face(p) im U .So,q = U q U e = p. Hence, p = q. ✓ p p p Finally, if q is any projection in M such that U a+ = a+,thena+ im+U = q 2 q face(q). So, p face(a+) face(q) im U .Then,asabove,p q. Hence, p is 2 ✓ ✓ q + + the least projection in M such that Upa = a . Definition 4.102. Let a be a positive element of a JBW-algebra M. The smallest projection p in M such that Upa = a is called the range projection of a and is denoted by r(a). Proposition 4.103. Let a M +. The range projection r(a) satisfies the following 2 properties: 1. r(a) W (a, e) face(a) ( -weak closure) 2 \ 2. a a r(a) k k 3. If K, then (a)=0 (r(a)) = 0 2 () 4. a b = r(a) r(b) ) 5. r(a) p = p2 = U a = a ) p Proof. Note that if a 0, then a+ = a. 1. Follows from by proposition 4.101. 2. As a a e,applyingU on both sides, we get a = U a a U e = k k r(a) r(a) k k r(a) a r(a). k k 167 4 Spectral Theory for Jordan Algebras 3. Applying Konbothsidesof(2),weget (r(a)) = 0 = (a)=0. 2 ) Conversely, if (a)=0,then (b)=0, b face(a) (4.2.34). And as is 8 2 continuous in the -weak topology, (b)=0, b in the -weak closure of 8 face(a). Thus, by (1), (r(a)) = 0. 4. Suppose a b,thenby(2)a b b r(b). So, a face(r(b)) = im+U , kk 2 r(b) Thus, U a = a. Hence, by definition of range projection, r(a) r(b). r(b) 5. Suppose p is a projection such that r(a) p,thena im+U = face(r(a)) 2 r(a) ✓ + face(p)=im Up.Thus,Upa = a. Remark 4.104. If B is -weakly closed subalgebra of M, containing the identity of Mandif0 a B, then W (a, e) B.So,r(a) B. Thus, r(a)isalsotheleast 2 ✓ 2 projection p in B such that Upa = a,i.e.thenotionsofrangeprojectionsinBand Mcoincide. Proposition 4.105. Let a, b M +. Then a b r(a) r(b). 2 ? () ? Proof. Recall that two positive elements a, b are orthogonal if aba = bab =0. { } { } ( ) If r(a) r(b), then by definition U r(b)= r(a) r(b) r(a) =0.So, ( ? r(a) { } by proposition 4.103-(2), we see that 0 U b b U r(b)=0.This r(a) kk r(a) implies U b =0whichisequivalenttoU (r(a)) = 0. Hence, 0 U a r(a) b b a U (r(a)) = 0. So, U (a)=0.Thus,a b. k k b b ? ( ) If a b,thenU b =0.So,U annihilates face(b). As r(b) face(b)( - ) ? a a 2 weak closure) and Ua is -weakly continuous, it follows that Ua(r(b)) = 0. So, we have proved that a b = a r(b). Applying the same logic ? ) ? again, we get r(b) a = r(b) r(a). ? ) ? 168 4 Spectral Theory for Jordan Algebras Remark 4.106. If a, b are two positive elements in M, then from the above propo- sition we see that a b = r(a) b. Hence, U b =0whichfurtherimplies ? ) ? r(a) Ur(a)0 b = b (4.2.32). So, a b = U b =0,U b =0 (4.3.13) ? ){ r(a) r(a)0 } Corollary 4.107. The normal states on M determine the norm and order on M, i.e. for each a M, 2 a 0 (a) 0, K (4.3.14) () 8 2 a =sup (a) K (4.3.15) k k {| || 2 } Proof. We first prove (4.3.14). + ( ) Let a M such that (a) 0, K.Leta = a a be the unique ( 2 8 2 orthogonal decomposition of a.Supposea =0.SinceMadmitsasep- 6 arating set of normal states, ⇢ K such that ⇢(a ) =0.Then,by 9 2 6 proposition 4.103-(3), ⇢(r(a )) =0.Let⌧ = U ⇤ ⇢.Byproposition 6 r(a ) 4.90-(1), ⌧ is a positive multiple of a normal state. Hence, ⌧(a) 0(by hypothesis). Now, ⌧(e)=(U ⇤ ⇢)(e)=⇢(Ur(a )e)=⇢(r(a )) = 0. Also, r(a ) 6 + + we know that a a .So,by(4.3.13),wegetU a = 0. Hence, ? r(a ) + ⌧(a)=(U ⇤ ⇢)(a)=⇢(Ur(a )(a a )) = ⇢ Ur(a )( a ) = ⇢(a ) < 0 r(a ) which is a contradiction to the fact that ⌧(a) 0. Thus, a =0which implies a 0. ( ) Let K. Then, a 0= (a) 0becausenormalstatesarepositive ) 2 ) functionals on M. We next prove (4.3.15). Note that the norm on M is the order unit norm and the ordering is determined by the normal states (4.3.14). So, for a M, a 2 k k 1 e a e (e a), (e + a) 0 ⇢(e) ⇢(a),⇢( a), ⇢ () () () 8 2 169 4 Spectral Theory for Jordan Algebras K 1=⇢(e) ⇢(a) , ⇢ K. Hence, in general, () | | 8 2 a ⇢(a) , ⇢ K (4.3.16) k k ()| | 8 2 Hence, the result. Finally, we are ready to show that Up are examples of the abstract compressions defined in chapter 3. Proposition 4.108. Let p be a projection in a JBW-algebra M. Then the map Up defined on M is a compression, in the sense of definition 3.38. Proof. First note that V is a base norm space with distinguished base K. From corollary 4.107, we infer that M and V are in separating order and norm duality; and the -weak topology on M is just the weak topology, arising from the duality between M and V. From results 4.57, 4.61 and 4.63, we know that the map Up is bicomplemented, normalised and positive projection, whose complement is Ue p. Thus, Up is a bicomplemented normalised weakly continuous positive projection on the order unit space M, and hence a compression. The following results show that M satisfies conditions of orthogonality and spectral duality. Proposition 4.109. Let p, q be projections in M. Then the following are equiva- lent: 1. p q ? 2. p q =0 3. p q0 4. p + q e 5. UpUq =0 Proof. Recall that p q pqp = qpq =0. ? (){ } { } 170 4 Spectral Theory for Jordan Algebras (1 2) Follows from proposition 4.58. ) (2 3) If p q =0,then(p + q)2 =(p + q). Hence, p + q is a projection and so ) p + q e = p q0. ) (3 4) p + q e p e q = q0. , () (3 5) If p e q,thenq e p.So,U (q) U (e p) = 0. Now, for a 0, ) p p U U (a) U U ( a e)= a U q =0.So,U U (a)=0, a M +. As M is p q p q k k k k p p q 8 2 positively generated, it follows that U U a =0, a M. p q 8 2 (5 1) If U U =0,then0=U U (e)=U q = pqp .Thus,p q. ) p q p q p { } ? Remark 4.110. The positive cone M + is -weakly closed. So, for each projection p M, face(p)=im+U =imU M + is -weakly closed. If 0 a,then 2 p p \ r(a) face(a)( -weak closure) and a face(r(a)). So, 2 2 face(r(a)) = face(a)( -weak closure) (4.3.17) Proposition 4.111. Let a M and p be a projection in M. Then the following 2 are equivalent: (i) U a a, U a 0 p p (ii) U a 0,U a 0 and p operator commutes with a p p0 For each a M,r(a+) is the smallest projection satisfying (i) and (ii). 2 Proof. Fix a A. 2 (i ii) Suppose p satisfies (i). Then, U a a 0. As U (U a a)=0,weget ) p p p U (U a a)=U a a. Hence, U a = U a a.Thus,a = U a + U a which p0 p p p0 p p p0 implies p commutes with a, by proposition 4.73. Also, U a = a U a 0. p0 p Hence, (ii) holds. 171 4 Spectral Theory for Jordan Algebras (ii i) Suppose (ii) holds. Then a = U a + U a.Thisimpliesa U a = U a ) p p0 p p0 0= U a a. ) p Next, we will show that for each a M,r(a+)isthesmallestprojection 2 satisfying (i) and (ii). Note that r(a+)satisfies(i)byproposition4.101.Ifp is any other projection satisfying (i) and (ii), then U a 0andU (U a)=U a.Thus, p p p p r(U a) p.Similarly,r( U a) p0. Now, r(U a)+r( U a) (p + p0)=e. p p0 p p0 Then, by proposition 4.109, r(U a) r( U a). By proposition 4.105, we get p ? p0 U a U a. Also, by (ii), a commutes with p and hence a = U a + U a.So,a = p ? p0 p p0 (U a) ( U a)isanorthogonaldecompositionofa.Byuniquenessoforthogonal p p0 + decomposition, it follows that U a = a ,Ua = a . Hence, by (4.2.32), it p p0 + + + + + follows that a Im U = Ker U .So,a = U (a)=U (a a )=U (a ). 2 p0 p p p p Thus, r(a) p. 4.3.2 Spectral Resolutions We are now ready to state the spectral theorem. The spectral result follows directly from Theorem 1.13, since we have shown that W (a, e)isisometrically(orderandalgebra)isomorphictosomemonotone complete CR(X) (Theorem 4.99). For each increasing finite sequence = 0, 1 ... n in R,with 0 < a { } k k and n > a ,define =max1 i n( i i 1). k k k k Theorem 4.112 (Spectral Theorem). Let M be a JBW-algebra. Let a M. Then, 2 there is a unique family e R of projections in M such that { } 2 (i) Each e operator commutes with a (ii) Ue a e ,Ue0 a e0 , R 8 2 (iii) e =0for < a ,e= e for > a k k k k 172 4 Spectral Theory for Jordan Algebras (iv) e e for <µ µ (v) eµ = e , R (greatest lower bound in M) µ> 2 V The family e is given by e = e r (a e)+ and is contained in W (a, e). { } Further, for each increasing finite sequence = , ... with < a { 0 1 n} 0 k k n and n > a , define s = i=1 i(e i e i 1 ). Then, k k P lim s a =0 0 k k k k! Proof. By theorem 4.99, W (a, e)isisometrically(orderandalgebra)isomorphic to some monotone complete CR(X) . Note that if p is a projection in M which + belongs to W (a, e), then Upa = p a. Now, for each R,(a e) W (a, e) 2 2 and hence e := (e r((a e)+)) is a projection in W (a, e). Thus, U a = e a. e The range projection of (a e)+ calculated in W (a, e)orMcoincide(remark 4.104) and also coincide with the notion of range projection in CR(X) . Using the isomorphism between W (a, e)andCR(X) , these correspondences and theorem 1.13, we see that e defined as e r((a e)+) satisfies (ii), (iii) and (iv). Also, decreasing nets in M converge weakly to their greatest lower bounds. So these greatest lower bounds are the same whether interpreted in W (a, e) or M. Hence, (v)isalsosatisfied,bytheorem1.13.Sinceeache belongs to the associative algebra W (a, e), it follows from proposition 4.73, that e operator commutes with a. Hence, (i) is also satisfied. And by theorem 1.13, the family e is the unique { } family of projections in W (a, e)satisfyingtheproperties(i)through(v). We need to prove the uniqueness of this family in M. Suppose f is any other { } family of projections in M satisfying (i) - (v). If µ,thenf f ;sof µ 2 + face(fµ)=im Ufµ ;thereforeUfµ f = f .Thus,f and fµ operator commute, by proposition 4.74. So, a and all of f mutually operator commute. By proposition { } 4.73, we see that the subalgebra generated by a and f is associative and hence { } its weak closure is isomorphic to some monotone complete C (X) , by theorem R 173 4 Spectral Theory for Jordan Algebras 4.99. Now, by uniqueness of spectral resolutions in CR(X) , it follows that f = e r((a e)+). Hence, f = e , . 8 We would like to note here that this spectral theorem also follows from the generalised spectral theorem (3.175), developed in chapter 3. To see this, recall that the JBW-algebra M is an order unit space, which is separating order and norm duality with the base norm space V. Further, it is a well known fact that for a JBW-algebra M, the predual of M is the set of all normal states on M (A.10). So, M = V ⇤. And, proposition 4.111 states that M satisfies the spectral duality condition (3.146). So, M satisfies the hypothesis of 3.175, and hence the result follows. Conclusion Using an order theoretic approach, we have developed a spectral theorem and functional calculus for suitable ordered spaces A. 11 Further, we have seen two concrete examples of this theory: one in commutative setting (function spaces), namely spectral theory for monotone complete CR(X) space and another in non- commutative setting (Jordan algebras), namely spectral theory for JBW-algebras. Hence, this spectral theorem generalises the spectral theory for von-Neumann al- gebras and JBW-algebras, to a larger class of order unit spaces. 11 A must be in separating order and norm duality with base norm space V, such that A = V⇤ and A satisfies the spectral duality condition. 174 References [1] Erik M. Alfsen. Compact Convex sets and Boundary Integrals.Springer-Verlag Berlin Heidelberg, 1 edition, 1971. [2] Erik M. Alfsen and Frederic W. Shultz. State Spaces of Operator Algebras - Basic theory, Orientation and C⇤- products.Birkhauser,2001. [3] Erik M. Alfsen and Frederic W. Shultz. Geometry of State Spaces of Operator Algebras.Birkhauser,2003. [4] L. Asimow and A. J. Ellis. Convexity theory and its applications in Functional Analysis. Academic Press, 11 December, 1976. [5] H. Hanche-Olsen and E. Stormer. Jordan Operator Algebras.PitmannPub- lisher, 1984. [6] Graham Jameson. Ordered linear Spaces. Springer, 1970 edition, July 1, 1970. [7] Walter Rudin. Functional Analysis. Tata McGraw-Hill, 2 edition, 1991. [8] Lieven Vandenberghe and Stephen P. Boyd. Convex Optimisation.Cambridge University Press, 1 edition, 8 March, 2004. 175 Chapter 5 Appendix Theorem A.1 (Banach-Alaoglu). Let V be a normed linear space. Then B⇤ = { f V ⇤ f =1 is weak⇤ compact. 2 |k k } Theorem A.2. If X is a locally convex space, then X⇤ separates points in X. Theorem A.3. Let X be a real locally convex space and let C X be a closed ✓ (convex) cone. If x X and x / C, then there exists ! X⇤ such that !(x) 0 0 2 0 2 2 for all x C and !(x ) < 0. 2 0 Theorem A.4. Let X be a vector space. Then C X is a closed convex cone in ✓ X C = C . () Theorem A.5 (Krein-Milman Theorem). Let K be a compact convex set of a locally convex space V and let @ K denote the extreme points of K. Then, @ (K) = e e 6 ; and K = co(@eK). Theorem A.6 (Cauchy-Schwarz Inequality). If , is a semi-inner product on h i complex or real vector space X, then a, b 2 a, a b, b , a, b X (A.0.1) |h i| h ih i 8 2 Theorem A.7 (Stone Weierstrass Theorem). Suppose X is a compact Hausdor↵ space and A is a subalgebra of CR(X) , which contains a non-zero constant function. Then A is dense in CR(X) i↵it separates points in X. 176 5 Appendix Theorem A.8. Let (A, +, ) be a commutative algebra (not necessarily associa- tive), endowed with a topology ⌧ such that multiplication is separately continuous in A, with respect to ⌧, i.e. if x x in A and y A, then (lim x ) y = ↵ ! 2 ↵ ↵ lim (x y) Now, if B is an associative subalgebra of A, then the closure of B, in ↵ ↵ A, is also associative. Proof. Let a , b and c be nets in B such that a a, b b, c c { x} { y} { z} x ! y ! z ! in A. We need to prove that a (b c)=(a b) c. a (b c)=(limax) (b c) x =lim(ax (b c)) x =lim[ax (lim by) c ] x { y } =lim[ax lim(by c) ] x { y } =lim[lim ax (by c) ] x y { } =lim[lim ax (by (lim cz)) ] x y { z } =lim[lim ax (lim(by cz)) ] x y { z } =lim[lim lim(ax (by cz)) ] x y { z } =lim[lim lim((ax by) cz) ] x y { z } =lim[lim ((ax by) lim cz)) ] x y { z } =lim[lim (ax by) c ] x y { } =lim[ lim(ax by) c] x { y } =lim[ ax (lim by) c] x { y } =lim[ ax b c] x { } =[limax b ] c x { } =[lim ax b] c { x } =[a b] c 177 5 Appendix Theorem A.9. The series 1 1 1 1 n 1 n ( 1) ( 2 )( 2 1) ...( 2 (n 1)) (1 x) 2 = x where = (A.0.2) n n n! n=0 X is absolutely and uniformly convergent for x 1. | | Theorem A.10 (Predual of JBW-algebra). A JB-algebra M is JBW-algbera i↵it is a Banach dual. In that case, the predual M is unique and and consists of the ⇤ normal linear functionals on M. 178