spectral theory for ordered spaces - TAMU Math

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 partial 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. A huge thanks to my parents for their continuous support and encouragement. 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 noncommutative version. In this project, we build a spectral theory and functional calculus 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). A motivating example for this theory comes from the spectral theorem for 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 ordered 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, compatibility and also obtain certain lattice structures, under an assumption called standing hypothesis. Then, we construct an abstract notion of range projection 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 . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.2

Spectral Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Order Structure 2.1

2.2

15

Ordered Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.1.1

Order Unit Spaces . . . . . . . . . . . . . . . . . . . . . . . 17

2.1.2

Base Norm Spaces . . . . . . . . . . . . . . . . . . . . . . . 22

Duality between Order Unit Space and Base Norm Space . . . . . . 24

3 Spectral Theory for Ordered Spaces 3.1

3.2

30

Projections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 3.1.1

Tangent Spaces and Semi-exposed Faces . . . . . . . . . . . 31

3.1.2

Smooth Projections . . . . . . . . . . . . . . . . . . . . . . . 35

Compressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 3.2.1

3.3

7

Projective Units and Projective Faces . . . . . . . . . . . . . 47

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

Spaces in Spectral Duality . . . . . . . . . . . . . . . . . . . . . . . 91

3.7

Spectral Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 3.7.1

Functional Calculus . . . . . . . . . . . . . . . . . . . . . . . 111

4 Spectral Theory for Jordan Algebras 4.1

4.2

4.3

114

Order Unit Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . 115 4.1.1

Characterising Order Unit Algebras . . . . . . . . . . . . . . 122

4.1.2

Spectral Result . . . . . . . . . . . . . . . . . . . . . . . . . 127

Jordan Algberas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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

JBW algebras

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 generalisation 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. The objective 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 a spectral theory and functional calculus for order unit spaces, which generalised 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 a general spectral theory for suitable ordered spaces and understanding how this extends the commutative spectral theorem to its non-commutative version. In this project, we mainly work with an order unit space (A, A+ , e) which is in separating order and norm duality with a base norm space (V,V+ , K). On investigating 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, the set of compressions on A form a lattice. Using the properties of this lattice structure, we then construct order theoretic tools, called range projections, analogous to those in CR (X) spaces. Finally, we specialise to spaces in spectral duality, where these 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. Another 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 ) and learn how the 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 construct 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 representation 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 complete 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. Define a partial order on CR (X) as follows: f  g () f (x)  g(x), 8x 2 X Then, CR (X) forms a lattice with the following lattice operations 1 f _ g = (f + g + |f 2

1 g|), f ^ g = (f + g 2

|f

1

g|), for f, g 2 CR (X)

Here, |f | denotes the usual modulus function |f | : X ! R defined as: 8 < f (x) if f (x) 0 |f |(x) = : f (x) if f (x) < 0

Notation 1. Let f 2 CR (X) and let 0 denote the constant 0 function on X. Define f + = f _ 0 and f =

(f ^ 0).

Remark 1.1. If f 2 CR (X) , then f = f +

f ,

|f | = f + + f

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 CR (X) is 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 2 A, then the spectrum of a, denoted by (a), is extremally disconnected. Hence, the space CR ( (a)) is monotone complete. Remark 1.4. Every von-Neumann algebra is monotone complete. 1

If f 2 CR (X) , then |f | also belongs to CR (X)

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. CR (X)

+

= {f 2 CR (X) | f

Notation 3. Let E ✓ X. Then E (x)

E

=

0}

: X ! R is defined as 8 0} is both closed and open in X. Further, 0 on E and a  0 on X r E.

1. a

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 2 X | a(s) > 0}. Then Y = E. For n = 1, 2, 3, . . . define fn : R ! [0, 1] as

8 0, x0 > > > > < 1 fn (x) = nx, 0  x  n > > > 1 > : 1, x n 7

1 Spectral Theory for Function Spaces Then, {fn

a}1 n=1 is an increasing sequence in CR (X), bounded above by the

constant function 1. As CR (X) is monotone complete, {fn

a} ! b, for some

b 2 CR (X). Claim. b =

E

s 2 Y =) 9 n 2 N such that a(s)

1 . n

This implies (fn

a)(s) = 1 and

hence, b(s) = 1. This is true for all s 2 Y . So, b(Y ) = 1. And as b is continuous, we have b(E) = b(Y ) = 1 Next, fix t 2 X r E. Choose

2

c 2 CR (X) such that c(X) ✓ [0, 1], c(t) =

0, c(E) = 1. If s 2 E, then (fn a)(s)  1 = c(s), 8n. Thus, b(s)  c(s), 8s 2 E. If s 2 / E, then (fn a)(s) = 0, 8n. Thus, 0 = b(s)  c(s), 8s 2 / E. Hence, b  c. In particular, 0  b(t)  c(t) = 0. So, b(t) = 0. Since t 2 X r E was arbitrary, we have b(X r E) = 0 Hence, b =

E.

As, b is continuous, E = b 1 (1) and E c = b 1 (0) are closed subsets of X. Hence, E is both open and closed in X. 1. As a > 0 on Y , we have a

0 on E = Y . If s 2 / E, then s 2 / Y =) a(s)  0.

Thus, a  0 on X r E. Next, let F be a closed subset of X such that a  0 on X r F . Now, s 2 Y =) a(s) > 0 =) s 2 / X r F =) s 2 F . Hence, Y ✓ F =) Y = E ✓ F . Thus, E is the smallest closed subset of X such that a  0 on X r E. 2 Compact 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 For s 2 E,

E (s)a

+

Ea

+

= a+ .

(s) = 1.a+ (s) = a+ (s).

For s 2 / E, a(s)  0 =) a+ (s) = 0. Thus, Hence,

Ea

+

E (s)a

+

(s) = 0.0 = 0 = a+ (s).

(s) = a+ (s), 8s 2 X.

Next, assume F is a closed subset of X such that s 2 Y , a(s) = a+ (s) > 0. This implies

F (s)

+ Fa

= a+ . Then, for

= 1 =) s 2 F . Hence,

Y ✓ F =) Y = E ✓ F . 3. Note that +

a (s) =

8 < a(s), :

s2E

0, otherwise

and for the {fn } defined previously,

(fn a)(s) = fn (a(s)) =

Thus, fn a 2 CR (X)+ and (fn na+ =) fn b=

E.

8 > > > >
> > 1 > : 1, a(s) n + a)(s)  na (s), 8s 2 X. Hence, fn a  na(s),

0  a(s) 

a 2 face (a+ ). Further, we have shown that {fn

Therefore,

E

a} %

is the supremum in CR (X) of the increasing sequence

+ {fn a}1 n=1 in face (a ).

Definition 1.6. An element p of CR (X) is called a projection if p = p2 . Notation 5. For a projection p in CR (X), denote p0 = 1

p, where 1 is the

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 2 CR (X), then p = and open subset E of X. 9

E

for some closed

1 Spectral Theory for Function Spaces This is because if s 2 X, then p(s) = p(s).p(s) =) p(s) = 0 or p(s) = 1. Therefore, p(X) = {0, 1}. Define E = {s 2 X | p(s) = 1}. Then p =

E.

Also,

as p is a continuous function on X, the sets E = p 1 (1) and X r E = p 1 (0) are closed subsets of X. Hence, E is both open and closed in X. Definition 1.8. Let a 2 CR (X)+ . Define r(a) to be the least projection p in CR (X) such that pa = a. We call r(a) to be the range projection of a. Proposition 1.5 implies the existence of range projections for positive elements of CR (X) (a

0 () a+ = a). Infact, for a 2 CR (X)+ , we have r(a) =

E

where E = {s 2 X | a(s) > 0}. Remark 1.9. If p, q are two projections in CR (X) of the form

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 CR (X) and p↵ % p 2 CR (X), then p is also a projection in CR (X). Similarly, if {p↵ } is a decreasing net of projections in CR (X) and p↵ & p 2 CR (X), then p is a projection. Proof. First consider {p↵ } % p. Note that 0  p↵  1, 8↵ =) 0  p  1. Thus, p2  p. Next, 0  p↵  p =) p2↵  p2 . But p2↵ = p↵ ,

8↵. Thus,

p = sup↵ p↵ = sup↵ p2↵  p2 . Therefore, p2 = p. Now, let {p↵ } & p. Then {1

p↵ } % (1

p). Then, as above, (1

p) is a

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 2 CR (X) and {p↵ } be a net of projections in CR (X). If p↵ % p and p↵ a

0, 8↵, then pa

Proof. Consider a = a+

0.

a and p↵ a = (p↵ a)+

Claim. p↵ a = (p↵ a) 10

(p↵ a) .

1 Spectral Theory for Function Spaces Let s 2 X. As p↵ is positive and takes values only 0 or 1, we have p↵ (s)a (s) = 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. 0, we get p↵ a = (p↵ a)+ and (p↵ a) = 0. Note that p↵ a

Now as p↵ a

8↵. Hence,

pa in CR (X), because a is positive and bounded. But p↵ a = 0, 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 2 CR (X) and pa  p, then p  1

r (a

2 R. If p is a projection in CR (X) such that 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 =) p0 b+ = b+ . Hence, r(b+ )  p0 (by proposition 1.5 (2)). Therefore, 1)+ .

p1

r (a

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 2 CR (X). Then, there is a unique family {e } in CR (X) such that (i) e a  e ,

e0 a

(ii) e = 0 for




eµ = e ,

e0 , 8 2 R kak,

> kak

e = 1 for

kak, define k k = max1in (

i

i

Then, lim ks

1

E

i

< e

kak i 1

).

1)+ and E = {s 2 X | a(s) > }. Then e0 =

r (a

1)+ =

e = r (a

0

ak = 0

k k!0

Proof. Define e = 1

. . . n } with Pn 1 ) and s = i=1 i (e 1

and e = 1

=

E

Ec .

(i) By proposition 1.5, we know that a

1

1  0 on E c

0 on E , a

e = 1 on E c

e = 0 on E ,

Using the above and the fact that e is positive, we get that e (a

(1.2.1) (1.2.2) 1)  0

on X. Thus, e a e ,

8

Similarly, e0 = 0 on E c ,

e0 = 1 on E c

(1.2.3)

Combining 1.2.3 with 1.2.1, we get e0 a (ii) When Ec

(iii)

8

> kak, we have E = {s 2 X | a(s) >

> kak} = ; =) e =

kak, we have E = {s 2 X | a(s)

kak > } = X =) e =

= 1.

When Ec

e0 ,


} ◆ {s 2 X | a(s) > µ} =) E ◆ Eµ =) E c ✓ Eµc =)

Ec



c Eµ

=) e  eµ . 12

1 Spectral Theory for Function Spaces (iv) Fix

2 R. From (ii), (iii), we know that eµ = 1, 8µ > kak and eµ

e , 8µ > . Take µ0 > kak. Then {eµ }µ0

µ>

is a decreasing net of pro-

jections in CR (X), bounded below by e . By monotone completeness and lemma 1.10, {eµ }µ0 µ0 > & p , for some projection p in CR (X). Infact, V p = µ> eµ and thus p

e

Let E = {s 2 X | p(s) = 0}. Note that p(s) = 1 =) eµ (s) = 1, 8µ > . Hence, E c ✓ Eµc , 8µ > . Thus, by (1.2.1), a  µ1 on E c , 8µ > . This implies a  1 on E c . Thus, pa  p ( p = 1 on E c ). Now by lemma 1.12, we get pe So, p = e . Uniqueness: Let {f }

2R

be another family of projections in CR (X) satisfying (i), (ii),

(iii) and (iv). In particular, f a  f , 8 . Thus, by lemma 1.12, we get f e , 8 For each

(1.2.4)

2 R, define E = {s 2 X | e (s) = 0} and F = {s 2 X | f (s) =

0}. By (1.2.4), we get E ✓ F . Also, by (i),

For

a

1 on E , a  1 on E c

a

µ1 on Fµ , a  µ1 on Fµc

< µ and x 2 Fµ , we have a(x)

Fµ ✓ E , 8µ > . But e = e  fµ , 8µ > . Then, e  Hence, f = e , 8 2 R.

and fµ =

Ec

V

µ>

µ > Fµc .

=)

x 2 E . Thus,

So, by remark 1.9, we get

fµ = f . Thus, e  f , 8 .

13

1 Spectral Theory for Function Spaces Now, let with

0


kak. Define Ei = {s 2 X | e i (s) = 1}, i = 1, 2 . . . n.

n

Then by (ii) and (iii), we see that ; = E0 ✓ E1 ✓ E2 ✓ . . . ✓ En = X. Define P s = ni=1 i (e i e i 1 ). Then x 2 Ei r Ei 1 =) s (x) = i . Also, as a  i1

on E i and a

i 11

So,we get 0  s

a

i

i 1

on E c i 1 , we have i

i 1

i 1

on Ei r Ei

 k k, 8x 2 X. Therefore, ks

a

1

i

on Ei \Eic

1

= Ei rEi 1 .

which implies ks (x)

ak  k k. Hence, s

a(x)k 

! a in norm

as k k ! 0. Corollary 1.14. Let X be a compact Hausdor↵ space such that CR (X) is monotone complete. Then, the linear span of projections is dense in CR (X). Proof. Given a 2 CR (X), let {e }

be the unique family of spectral units asso⇥ ⇤ ciated with a. Let n be the regular partition of (kak + 1), (kak + 1) of norm P 2(kak+1) . Define s n = ni=0 i (e i e i 1 ). Then, 8n, s n belongs to linear span n of projections and s

n

2R

! a in norm as n ! 1, by theorem 1.13.

To summarise, we have obtained a spectral result for monotone complete CR (X) space, using order theoretic ideas. The space (CR (X) , ) is a particular example 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 a larger class of order unit spaces.

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 a base norm space and dual of a base norm space is an order unit space. 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 2 V + , 8 v, w 2 V + (ii)

v 2 V +, 8 v 2 V +,

0

V+ is said to be proper if V+ \ -V+ = {0} . V+ is said to be generating if for each v 2 V, 9 v1 , v2 2 V+ such that v = v1

v2 .

Definition 2.2. A relation  defined on V is said to be a vector order on V if 15

2 Order Structure (i) v  v, 8v 2 V (ii) u  v , v  w =) u  w (iii) u  v =) u + w  v + w, 8w 2 V (iv) u  v =)

u  v, 8

0

Definition 2.3. A real vector space with a vector order is called an ordered 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 2V+

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 2 V | 0 0 v} is a cone in V. And the order on V, obtained from U , is the same as 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. This order is obtained from the cone R2+ := {(a, b) 2 R2 | 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) 2 V+ , which implies (v

u) 2

v = u. Hence,  is anti-symmetric. 16

V+ . This implies v

u) 2 V+ and u = 0 =)

2 Order Structure (() Let v 2 V+ \ - V+ . Then, 0  v and 0 

v =) v  0. As  is

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. A positive element e of an ordered vector space V is said to be an order unit if for each a 2 V, there exists

0 such that

ea e

(2.1.1)

Remark 2.7. Existence of order unit implies V+ is generating because for any given v 2 V, we have v = v1

v2 where v1 =

e+v , 2

v2 =

e v 2

2 V+ ).

Example 2.8. Consider V = l1 (the space of all real bounded sequences), with ordering {ai }

0 () ai

0, 8i. Then e = (1, 1, 1, . . . ) is an order unit for V.

Definition 2.9. The order unit e is said to be Archimedian if for each a 2 V, we have na  e, for n = 1, 2, 3 . . . =) a  0

(2.1.2)

Equivalently, {v + ke 2 V+ , 8k > 0} =) v 2 V+ . One can show that an ordered vector space V, having an Archimedian order unit e, admits a norm pe , defined as follows: pe (a) = inf{

0|

e  a  e}

(2.1.3)

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 2 V , (i) pe (e) = 1 (ii)

pe (v)e  v  pe (v)e

(iii)

e  v  e () pe (v)  1

(iv) 0  u  v =) pe (u)  pe (v) Proof.

(i) 1e ± e = 2e, 0 2 V + . Hence, pe (e)  1. Suppose (1

for some

0, then it implies

)e ± e 2 V +

e 2 V + . Since V + is proper,

must be 0.

Hence, pe (e) = 1. (ii) By definition of infrimum, given ✏ > 0, 9 0 < v 2 V + . Adding the positive element (✏

< ✏ such that (pe (v) + )e ±

)e, we get pe (v)e+ e+(✏

)e±v 2

V + =) (pe (v) + ✏)e ± v 2 V + . So, 8 ✏ > 0, ✏e + (pe (v)e ± v) 2 V + . Hence, by Archimedian property, pe (v)e ± v 2 V + . (iii) Assume then

e  v  e. Then, by definition, pe (v)  1. Conversely, if pe (v)  1,

e

pe (v)e  v  pe (v)e  e.

(iv) Assume 0  u  v. By (ii), pe (v)e ± v 2 V + ; =) pe (v)e ± (v V+

=)

0  (v

u)  pe (v)e

u + u) 2

u. Also, pe (v)e + u 2 V + . Hence,

pe (u)  pe (v).

Proposition 2.11. Let e be an Archimedian order unit for (V, V + ). Assume V+ is proper. Define pe (v) = inf {k Proof. Clearly, pe (v)

0|

ke  v  ke}. Then, pe is a norm on V.

0, 8v 2 V and pe (0) = 0.

(i) Suppose pe (v) = 0 for some v 2 V. This implies ke ± v 2 V+ , 8k > 0. Therefore, by Archimedian property, v  0 and v get v = 0. So, pe (v) (ii) Let

2 R. If

pe ( v) = inf {k

0. As V+ is proper, we

0, 8v 2V and pe (v) = 0 () v = 0.

= 0, then clearly pe ( v) = pe (v). Suppose 0|

ke  v  ke} = inf {k 18

6= 0. Then,

0 | | |( |ke| ± v) 2 V + } =

2 Order Structure inf {k

0 | ( |ke| ± v) 2 V + } = | | inf {k 0

where k 0 =

k . | |

0 | (k 0 e ± v) 2 V + } = | |pe (v)

So, pe ( v) = | |pe (v), 8 2 R, v 2 V.

(iii) Take v1 , v2 2 V. Let a = pe (v1 ) and b = pe (v2 ). By 2.10-(ii), we have (a + b)e ± (v1 + v2 ) = (ae ± v1 ) + (be ± v2 ) 2 V + . So, a + b

inf {k

0|

ke ± (v1 + v2 ) 2 V + }. Hence pe (v1 + v2 )  pe (v1 ) + pe (v2 ). 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) = pe0 (v), 8v 2 V , then e = e0 (follows from remark 2.10). Definition 2.13. An ordered normed linear space (A, k.k) is said be an order unit space if the norm on A can be obtained from an Archimedian order unit e, as follows: kak = inf{

0|

e  a  e}

(2.1.4)

This e is called the distinguished order unit of A. Further, this norm satisfies the property that

kake  a  kake.

Example 2.14. Let CR (X) be the set of continuous real valued functions on a compact Hausdor↵ space X. Define an ordering on CR (X) as f

0 () f (x)

0, 8x 2 X. Let e denote the constant function 1 on X. Then, CR (X) endowed 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 {vn }1 n=1 be a sequence in A and let v 2 A such that pe (v

0 as n ! 1. For each k 2 N, 9 nk such that pe (v

vn ) < k1 , 8n

vn ) ! nk . Then,

0  vnk  k1 e + v, 8 k 2 N. Hence, by Archimedian property, v 2 A+ . Thus, A+ is closed w.r.t pe . 19

2 Order Structure Theorem 2.16. Let (A, A+ , k.k) be an ordered normed linear space and let e 2 A+ . Then, A is an order unit space with distinguished order unit e () A+ is closed in A (with respect to k.k) and the following holds for each a 2 A: kak  1 ()

eae

(2.1.5)

Proof. If e is an Archimedian order unit for A, then denote kake = inf {

0|

e  a  e}, for each a 2 A. ()) Suppose A is an order unit space and e is its distinguished Archimedian order unit. Then k.k = k.ke . By proposition 2.15, A+ is closed in A and by remark 2.10-(iii), (2.1.5) holds. (() Assume A+ is closed in A and (2.1.5) holds. Take a 2 A. If a = 0, then e  a  e clearly. If a 6= 0, define a0 = (2.1.5), we have

e  a0  e; =)

a . kak

Then ka0 k = 1. Hence, by

kake  a  kake. Thus, e satisfies

the order unit property (2.1.1) and, hence, is an order unit for A. Next, we show that e is Archimedian. Let a 2 A such that na  e for all n 2 N, i.e, a

e n

for all n 2 N. Note that {a

e } n

k.k

! a as n ! 1. But a

e n

0

for each n and -A+ is closed, by assumption. Hence, a 2 -A+ ; =) a  0. Thus, e is Archimedian. Now, we will show that for each a 2 A, we have kak = kake . This is clear if a = 0. So, assume a 6= 0. By (2.1.5), we have

kake  a  kake which implies kake  kak. Suppose kake < kak,

then k(kak 1 a)ke < 1. Thus, 9 0 < e =)

e

kak

1a

 e =) k

contradiction to the fact that

a k kak

< 1 such that

e  kak 1 a 

a  1 =) 1 = k kak k  , which is a

< 1. Hence, kake = kak.

We close the section on order unit spaces, with some results about the dual space. 20

2 Order Structure Let (A, A+ , e) be an order unit space. Let A0 denote the set of all linear functionals on A. Define A0+ = {f 2 A0 | f (v)

0, 8v 2 A+ }. Then A0+ is a cone

in A0 . Proposition 2.17. Let (A, A+ , e) be an order unit space and let f 2 A0 . Then, f 2 A0+ () f is bounded and kf k = f (e). Proof. Let us denote the order unit norm on A by pe . Then kf k = sup {|f (x)| | x 2 A, pe (x)  1}. ()) Assume f 2 A0+ . Take v 2 A with pe (v)  1. Then,

e  v  e. Now, f is

positive =) f ( e)  f (v)  f (e). Hence, |f (v)|  f (e). This shows that kf k  f (e). Hence, f is bounded. Now, pe (e) = 1 =) f (e)  kf k. So, kf k = f (e). (() Assume f is bounded with kf k = f (e). Take v 2 A+ . If v = 0, then f (v) = 0. If v 6= 0, put u =

v . pe (v)

v 2 A+ =) u

0. Hence,

Therefore, f (e

u)  kf kke

f (u)

0 and therefore f (v)

Now, pe (u) = 1 and so

e  u  e. Also,

e  u  e and 0  e u  e. So, pe (e u)  1. uk  f (e). So, f (e)

f (u)  f (e). Hence,

0. Thus, f 2 A0+ .

Definition 2.18. A linear functional ⇢ on an order unit space (A, e) is called a state if ⇢ is positive and ⇢(e) = 1. The set of all states on A is called the state 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 A⇤1 be the closed unit ball of A⇤ . Then, by Banach-Alaoglu theorem (A.1), A⇤1 is w⇤ -compact. As K is a w⇤ -closed subset of A⇤1 , K is also w⇤ -compact. Proposition 2.20. If a is an element of an order unit space A, with state space K, then a 2 A+ () ⇢(a) 21

0, 8⇢ 2 K

(2.1.6)

2 Order Structure and kak = sup {|⇢(a)| | ⇢ 2 K} Proof. If a 2 A+ , then clearly ⇢(a)

(2.1.7)

0, 8⇢ 2 K. Now, assume ⇢(a)

0, 8⇢ 2 K.

Suppose a 2 / A+ . Since A+ is closed, then by Hahn-Banach Separation theorem (A.3), there exists f 2 A⇤ and ↵ 2 R such that f (a) < ↵ and f (b)

↵ for all

b 2 A+ . Since A+ is a cone, we can choose the separating real number ↵ to be zero. Thus f is a positive linear functional, so ⇢ :=

f kf k

is a state on A with ⇢(a) < 0,

which is a contradiction. Hence, a 2 A+ . To prove (2.1.7), define ⇢ 2 K, we have

= sup {|⇢(a)| | ⇢ 2 K}. As |⇢(a)|  k⇢kkak for each

 kak. Suppose

either ( e + a) 2 / A+ or ( e

< kak. By the definition of order unit norm,

a) 2 / A+ . If suppose ( e + a) 2 / A+ , then by (2.1.6),

9 ⇢ 2 K such that ⇢( e + a) < 0 which implies ⇢(a) < contradiction. Similarly, if ( e

 0, i.e. |⇢(a)| < , a

a) 2 / A+ , we get a contradiction. So,

= kak.

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. A non-empty convex set K ✓ V+ \{0} is said to be a base for V+ if for each v 2 V + \{0}, 9! k 2 K,

> 0 such that v = k.

Example 2.22. If (A, A+ , e) is an order unit space, then the state space of A is a base for A⇤ . Remark 2.23. If K is base for V+ , then V+ = { k | k 2 K, Remark 2.24. If V+ has a base K, then V+ is proper. 22

0}

2 Order Structure Proof. Let ±v 2 V + \{0} with v = Now, 0 = v + ( v) =

1 k1

+

2 k2

1 k1 ,

=(

v=

1

+

some k 2 K (since K is convex). But, (

1

2 )(

+

2 k2 1 1+ 2

2)

where

k1 +

2 1+ 2

1,

2

> 0, k1 , k2 2 K.

k2 ) = (

1

+

2 )k,

for

> 0 =) k = 0. This contradicts

that 0 2 / K. Therefore, v must be 0. From remark 2.23 and 2.24, we see that if K is a base for V+ , then each v 2 V can be represented as v = putting

=

1

+

2

1 k1

2 k2 ,

for some

1,

2

0, k1 , k2 2 K. Now

and k = k1 + k2 , we find that k ± v 2 V + .

Proposition 2.25. Let (V, V + ) be an ordered vector space. Assume V + is generating and let K be a base for V + . Define, for each v 2 V, kvkK = inf {

0 | k ± v 2 V + , for some k 2 K}

(2.1.8)

Then, k.kK is a semi-norm on V. Proof. Clearly, kvkK

0, 8v 2 V and k0kK = 0.

(i) Let ↵ 2 R. If ↵ = 0, then clearly k↵vkK = |↵|kvkK . Now, assume ↵ 6= 0. Then, k↵vkK = inf {

0 | k ± ↵v 2 V + , for some k 2 K}

= inf {

0 | |↵|( |↵| k ± v) 2 V + , for some k 2 K}

= inf {

0|

|↵|

k ± v 2 V + , for some k 2 K}

= inf { 0 |↵|

0|

0

k ± v 2 V + , for some k 2 K} where

0

0|

0

k ± v 2 V + , for some k 2 K}

= |↵| inf {

0

=

|↵|

= |↵|kvkK . (ii) Let v, w 2 V. Let k

kvkK and l

kwkK . Then, 9 b1 , b2 2 K such that (kb1 ±

v) 2 V + and (lb2 ± w) 2 V + . So, (kb1 + lb2 ) ± (v + w) 2 V + . Also, k l k l ( k+l b1 + k+l b2 ) 2 K, since K is convex. Hence, (k+l)( k+l b1 + k+l b2 )±(v+w) 2

V + . This implies kv + wkK  k + l. Taking infrimum over k and l, we get, kv + wkK  kvkK + kwkK . So, k.kK is a semi-norm on K. 23

2 Order Structure Definition 2.26. An ordered normed vector space (V, k.k) with a generating cone 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+ . In this case, K = {v 2 V + | kvk = 1} and is called the distinguished base of V. Remark 2.27. The closed unit ball B of a base norm space (V, V+ , K) is of the form B = co(K [

K).

Example 2.28. Consider the vector space (Rn , k.k1 ), with positive cone C = {(a1 , a2 . . . an ) 2 Rn | ai

0, 8 i}. Define ei to be the vector whose ith component

is 1 and all other components are 0. Then K := co(e1 , e2 . . . en ) is a base for C and (Rn , C, K) is a base norm space.

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

: V x W ! K be a

bilinear form such that (i)

(v, w) = 0, 8v 2 V =) w = 0

(ii)

(v, w) = 0, 8w 2 W =) v = 0

Then, (V, W,

) is called a Dual Pair.

Notation 8. Let E be a normed linear space. Then E 0 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, hV, V ⇤ i form a dual pair under the bilinear map h, i : V X V ⇤ ! R defined as hv, f i = f (v). 24

2 Order Structure Let (V, W, h, i) be a dual pair. If (V, V+ ) is an ordered vector space, then W+ := {w 2 W | hv, wi

0, 8v 2 V + } defines a cone in W.

Remark 2.31. If V+ is generating, then W+ is proper. Definition 2.32. Let (V, V+ ), (W, W+ ) be two ordered vector spaces over R. V, W are said to be in separating order duality if there exists a bilinear form h, i: V x W ! R such that (i) (V, W, h, i) forms a dual pair (ii) V + = {v 2 V | hv, wi (iii) W + = {w 2 W | hv, wi

0, 8w 2 W + } 0, 8v 2 V + }

Definition 2.33. Let (V, k.k1 ), (W, k.k2 ) be two normed vector spaces over R. V, W are said to be in norm duality if there exists a bilinear form h, i: V x W ! R such that (i) (V, W, h, i) forms a dual pair (ii) For each v 2 V, kvk1 = sup {|hv, wi| | kwk2  1} (iii) For each w 2 W, kwk2 = sup {|hv, wi| | kvk1  1} Suppose (V,W,

) form a dual pair. For each w 2 W, define fw : V ! K as

fw (v) = (v, w). Then fw is a linear functional on V. Hence, W can be identified as a subset of V 0 , via the mapping

: W ! V 0 given by

(w) 7! fw . Similarly,

V can be identified as a subset of W 0 . Notation 9. (V, V + , k.k) denotes an ordered vector space V, with norm k.k and positive cone V+ . Proposition 2.34. Let (V, V+ , k.k1 ) be an ordered normed linear space. Assume V+ is closed in V, with respect to the norm induced topology. Then, V is in separating order and norm duality with (V ⇤ , (V ⇤ )+ , k.k2 ), where (V ⇤ )+ = {f 2 V ⇤ |f (v)

0, 8v 2 V + } and kf k2 = sup {|f (v)| | kvk1  1} for each f 2 V ⇤ . 25

2 Order Structure Proof. Consider the bilinear map h, i : V X V ⇤

! R defined as hv, f i = f (v).

Then hV, V ⇤ i form a dual pair under h, i because if f (v) = 0, 8v 2 V, then f = 0. Conversely, if f (v) = 0, 8f 2 V ⇤ and v 6= 0, then by Hahn-Banach Theorem, there exists f 2 V ⇤ such that f (v) = kvk = 6 0, which is a contradiction. Thus, v = 0. Next, we prove separating order duality. If v 2 V + , then f (v) Conversely, assume f (v)

0, 8f 2 V ⇤ + .

0, 8f 2 V ⇤ + . Suppose v 2 / V + . Since V + is closed,

then by Hahn-Banach Separation theorem (A.3), there exists g 2 V ⇤ and ↵ 2 R such that g(v) < ↵ and g(u)

↵ for all u 2 V + . Since V+ is a cone, we can

choose the separating real number ↵ to be zero. Thus, g is a positive linear functional. This contradicts the fact that f (v) So, V + = {v 2 V | f (v) V ⇤ |f (v)

0, 8f 2 V ⇤ + . Hence, v 2 V + .

0, 8f 2 V ⇤ + }. And, by definition, (V ⇤ )+ = {f 2

0, 8v 2 V + }. Therefore, hV, V ⇤ i are in separating order duality.

Now, we prove norm duality. Let v 2 V. Clearly, kvk1

sup {|f (v)| | kf k2 

1}, because |f (v)|  kf k2 kvk1 . And by Hahn-Banach theorem, there exists f 2 V ⇤ such that kf k2  1 and f (v) = kvk1 . Thus, kvk1 = sup {|f (v)| | kf k2  1}. And, by definition, kf k2 = sup {|f (v)| | kvk1  1} for each f 2 V ⇤ . Hence, hV, V ⇤ i are in norm duality. 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 2 V ⇤ + | kf k = f (e) = 1}. Then (V ⇤ , V ⇤ + , S(V )) is a base norm space and h(V, V + , e), (V ⇤ , V ⇤ + , S(V ))i are in separating order and norm duality. Proof. Let the order unit norm on V be denoted by pe . It is easy to see that S(V ) is a base for V ⇤ + . And as S(V ) is w⇤ -compact, the semi-norm induced by S(V ) becomes a norm on V ⇤ . Let the dual norm on V ⇤ be denoted by kf k = sup {|f (v)| | pe (v)  1}. 26

2 Order Structure Clearly, V ⇤ + is closed in V ⇤ , with respect to k.k. Let us denote the norm on V ⇤ , induced by the base S(V ), as kf kB = inf { 0 | g ± f 2 (V ⇤ )+ , for some g 2 S(V )} . We need to prove that kf k = kf kB , for each f 2 V ⇤ . This is clearly true for f = 0. So, assume f 6= 0. For simplicity, let us denote S(V ) by S. (i) Let v 2 V such that pe (v)  1. Then,

e  v  e. This implies 0  e

e and 0  e + v  2e. Suppose g ± f 2 (V ⇤ )+ for some g 2 S, ( g + f )(e ( g

v) = g(e)

f )(e + v) = g(e) + g(v)

Adding both, we get 2 by

g(v) + f (e)

v, we get

inf {

0

f (v)

0

0; =)

f ( v). This implies

f . kf k

> 0. Then,

f (v). Similarly, replacing v |f (v)|. So,

0 | g ± f 2 (V ⇤ )+ , for some g 2 S}

Hence, kf kB (ii) Let h =

2f (v)

f (e)

f (v)

v

sup {|f (v)| | pe (v)  1}.

kf k Then, h 2 V1⇤ = co(S [ S). So, h = ↵g1

g1 , g2 2 S, ↵ 2 [0, 1]. Take g = ↵g1 + (1

(1

↵)g2 , for some

↵)g2 2 S (since S is convex). Now

g ± h = 2↵g1 , 2(1 ↵)g2 2 (V ⇤ )+ . This implies g ± h 2 V ⇤ + =) kf kg ± f 2 (V ⇤ )+ =) kf k

kf kB .

Thus, kf kB = kf k, for all f 2 V ⇤ . Hence, (V ⇤ , V ⇤ + , S(V )) is a base norm space. Now, by proposition 2.34, it follows that the order unit space (V, V + , e) is in 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

eK (k) = 1, 8 k 2 K and then extended to V by linearity. Then (V ⇤ , V ⇤ + , eK ) is an order unit space with order unit eK and h(V, V + , K), (V ⇤ , V ⇤ + , eK )i are in separating order and norm duality. 27

2 Order Structure Proof. Let the base norm on V be denoted by k.kK . And let the dual norm on V ⇤ be denoted by kf k = sup {|f (v)| | kvkK  1}. Claim. eK is an Archimedian order unit for V⇤ . Let v 2 V+ . Then 9 k 2 K, 0. Hence, eK (v)

0 such that v = k. So, eK (v) = eK (k) =

0, 8v 2 V+ . So, eK 2 (V ⇤ )+ . Now, let f 2 (V ⇤ )+ . Define

= sup {|f (k)| | k 2 K}. Then, 8↵ ↵( ± f (k))

0, k 2 K, we have ( eK ± f )(↵k) =

0. Therefore, ( eK ± f ) 2 (V ⇤ )+ . Hence eK is an order unit for V ⇤ .

Next, we will show that eK is Archimedian. Suppose f 2 V ⇤ such that f  1 e , n K

8n 2 N. Then, for each k 2 K, we have f (k) 

1 , n

8n which implies

f (k)  0. Thus, f (v)  0, 8v 2 V + and so f  0. Hence, proved. Let the order unit norm on V ⇤ be denoted by kf ke = inf {

0 | eK ± f 2

(V ⇤ )+ }. Now, we will show that kf k = kf ke , for all f 2 V ⇤ . (i) Let v 2 V+ . So, v = k, for some (kf keK ± f )( k) = which implies kf k

[kf k ± f (k)]

0. Therefore, (kf keK ± f ) 2 (V ⇤ )+

kf ke

(ii) Suppose eK ±f 2 (V ⇤ )+ , for some which implies

0, k 2 K. Now, (kf keK ± f )(v) =

0. Then, ( eK ±f )(k)

0, 8k 2 K,

|f (k)|, 8k 2 K. Consider, v 2 V such that kvkK  1.

Since V1 = co(K [ -K), there exists k1 , k2 2 K such that v = ↵(k1 ) ↵)(k2 ), for some ↵ 2 [0, 1]. Now, |f (v)|  ↵|f (k1 )| + (1 sup {|f (k)| | k 2 K}  . This implies, inf { kf ke

0 | eK ± f 2 (V ⇤ )+ }

(1

↵)|f (k2 )| 

sup {|f (v)| | kvkK  1}. So,

sup {|f (v)| | kvkK  1} which implies

kf k.

Hence, kf k = kf ke . So, (V ⇤ , V ⇤ + , eK ) is an order unit space. Now, by proposition 2.34, it follows that the base norm space (V, V + , K) is in 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) he, vi = kvk, 8v 2 V+ (ii) kv1 + v2 k = kv1 k + kv2 k, 8v1 , v2 2 V+ (iii) K = {v 2 V + | he, vi = 1} (iv) If T is a positive linear map from A to A, then hT e, vi = kT ⇤ vk, 8v 2 V+ Here T ⇤ is the adjoint map on V, satisfying hT a, vi = ha, T ⇤ vi, for each a 2 A, v 2 V . Proof.

(i) Let v 2 V+ and let a 2 A such that kak  1. This implies

ea

e; =) h e, vi  ha, vi  he, vi. So, |ha, vi|  he, vi. Also, kek = 1. Hence, he, vi  sup {|ha, vi| | a 2 A, kak  1}  he, vi. So, kvk = he, vi. (ii) By (i), kv1 + v2 k = he, v1 + v2 i = he, v1 i + he, v2 i = kv1 k + kv2 k. (iii) By definition, K = { v 2 V + | kvk = 1}. Therefore, by (i), K = { v 2 V + | he, vi = 1}. (iv) {T is positive and v 2 V + } =) T ⇤ v 2 V + . Hence, by (i), hT e, vi = he, T ⇤ vi = kT ⇤ vk.

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 affine functions on the compact convex set K. The main ingredients of spectral theory are projections and their orthogonality, as observed in the motivating 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. The image of the order unit under a compression gives rise to projective unit, which generalise the idea of projections in CR (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. We use properties of this lattice structure 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 nonempty convex subset F of C is said to be a face of C if the following condition is satisfied: x, y 2 C and ↵x + (1

↵)y 2 F for some ↵ 2 (0, 1) =) x, y 2 F

Remark 3.2. F is a face of X + () F is a subcone of X + and for every given x 2 F, 0  y  x =) y 2 F Proof. Given: F ✓ X + , F 6= ;. ()) Let F be a face of X + . Let x 2 F . Now, 12 (0) + 12 (2x) = x 2 F . Hence, 0 2 F. Claim.

x 2 F, 8

First assume

0

1. Then, 1 ( x) + (1

Next, consider 0 

1

)0 = x 2 F . Therefore, x 2 F .

 1. Then, x = x+(1

x 2 F . So, x 2 F, 8

)0 (convex combination) =)

0, x 2 F . Hence, the claim.

Now, consider x, y 2 F . This implies 12 x + 12 y 2 F . Then, by the above claim, 2( x+y ) 2 F ; =) x + y 2 F . Hence, F is a subcone of X + . Now, let 2

31

3 Spectral Theory for Ordered Spaces y) 2 X + and 12 (x

x 2 F and 0  y  x. Then, y, (x

y) + 12 y = 12 x 2 F .

This implies y 2 F . (() Let F be a subcone of X + and assume that for every given x 2 F, we have 0  y  x =) y 2 F . As F is a subcone, it is a convex set. And if x, y 2 X + such that ↵x + (1

↵)y = z 2 F for some ↵ 2 (0, 1), then

0  ↵x, (1

↵)y 2 F which implies x, y 2 F ( as F is

↵)y  z. So, ↵x, (1

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, denoted by FaceC (x), is the smallest face of C, containing x. By remark 3.2, FaceC (x) = {b 2 C | b  kx, for some k

0}

Definition 3.4. A point x 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 6= x, z 6= x and

0 < ↵ < 1. Equivalently, x 2 X is an extreme point of a convex set K if FaceK (x) = {x}. The set of all extreme points of K is called the extreme boundary of K and we will denote it by @e K. Definition 3.5. H ✓ X is called a hyperplane of X if H = f

1

(↵) for some

non-zero linear functional f on X, ↵ 2 R. (i.e. H has codimension 1) If H is a hyperplane of an ordered vector space (X, X+ ), then it splits X into two half spaces, namely U1 = {a 2 X | a where H = f

1

f

1

(↵)} and U2 = {a 2 X | a  f

1

(↵)}

(↵).

Definition 3.6. Let (X , ) be an ordered vector space with a topology and let C be a convex subset of X. Then H is called a supporting hyperplane of C if H is a hyperplane of X such that C is contained in one of the half spaces of H and @C \H = 6 ;. Here, @C is the boundary of C (@C := C\C ) with respect to the topology on X. 32

3 Spectral Theory for Ordered Spaces Hereafter h(X, X + ), (Y, Y + )i is a pair of positively generated ordered vector spaces in separating order duality under a bilinear form . The topology on X, Y is the weak topology defined by the given duality, i.e. for a net {x↵ } 2 X, x↵ ! x () hx↵ , yi ! hx, yi, for each y 2 Y 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 C at F and is denoted by TanC F. If there is no supporting hyperplane of C which contains F, then TanC F = X , by convention. Definition 3.8. A face F of a convex set C ✓ X is semi-exposed if there exists a collection H of closed supporting hyperplanes of C containing F such that F = T C \ ( H2H H ). F is said to be exposed if H can be chosen to consist of a single hyperplane, i.e. F = C \ H, for some closed supporting hyperplane H of C. Remark 3.9. F is semi-exposed () F = C \ TanC F Definition 3.10. Let B ✓ X. The annihilator of B in the space Y is denoted by B := {y 2 Y | hx, yi = 0, 8x 2 B} and positive anhillator of B = B• = B \ Y+ = {y 2 Y + | hx, yi = 0, 8x 2 B}. Proposition 3.11. Let B ✓ X+ .Then H ✓ X is a closed supporting hyperplane of X + , containing B () H = y 1 (0) for some y 2 B • . Proof. Recall that y 1 (0) = {x 2 X | hx, yi = 0} (() Let H = y 1 (0) for some y 2 B • . As y is a continous linear functional on X, y 1 (0) is a closed hyperplane of X. Since y 2 Y+ , X + ✓ {x 2 X | hx, yi

0}. So, X+ is contained in one of the half spaces of H. Note that

0 2 X + \ (X + )C = @X + . Also, 0 2 H. Hence, 0 2 @X + \ H. So, H is a supporting hyperplane of X + . As y 2 B • , we have hx, yi = 0, 8x 2 B. Therefore, B ✓ H. 33

3 Spectral Theory for Ordered Spaces ()) Suppose H ✓ X is a closed supporting hyperplane of X + , containing B. Let H=f

1

(t) for some non-zero linear functional f on X, t 2 R. Since H is

closed, f is continuous 1 . By Hahn-Banach Theorem, the weak dual of X is Y. So, 9y 2 Y such that f (y) = hx, yi, 8x 2 X. Therefore, H = y 1 (t). As H is supporting, without loss of generality, we may assume X + ✓ {x 2 X | hx, yi

t}. So, 0 = h0, yi

t. This implies 0

Now, x0 2 @X + = X + \(X + )

t. Let x0 2 @X + \ H.

+ =) 9 {xn }1 n=1 2 X such that xn ! x0

as n ! 1. So, t  limn!1 h2xn , yi = 2limn!1 hxn , yi = 2hx0 , yi = 2t. So, t  2t; hence t

0. So, t = 0. This implies H = y 1 (0). Now,

B ✓ H = y 1 (0) =) hx, yi = 0, 8x 2 B. This implies y 2 B . If necessary, replacing y by

y, we have X + ✓ {x 2 X | hx, yi

0} which implies y

0.

Therefore, y 2 B • . Remark 3.12. If B ✓ X, then B =

T

x2B

x 1 (0).

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 Proof.

(i) B = {y 2 Y | hx, yi = 0, 8x 2 B} =

T

x2B

x 1 (0). Therefore B

is the intersection of a collection of closed supporting hyperplanes of Y + containing B • and B • = Y + \ B . Hence, B • is semi-exposed. (ii) By 3.11, TanX + B =

T

y2B •

y 1 (0) = B •

(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 + , containing B. 1

a property of topological vector space

34

3 Spectral Theory for Ordered Spaces This implies S = X + \ H, where H is the intersection of a collection of closed supporting hyperplanes of X + containing S. Since B ✓ S, H is the intersection of some closed supporting hyperplanes of X + containing B. So, T T 1 1 + y (0) ✓ H. This implies ✓ H \ X + ; hence • y2B y2B • y (0) \ X B • • ✓ S.

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. A linear map P: X ! X such that P2 = P and P(X+ ) ✓ X+ is called a positive projection on P. Notation 10. ker P = {x 2 X | P (x) = 0} and im P = {x 2 X | P x = x}. And we define ker+ P = ker P \ X+ and im+ P = im P \ 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 2 X. Then x = (x P x)+P x. This implies, Rx = R(x P x)+R(P x). Now, (x

P x) 2 ker P ✓ ker R and P x 2 im P = im R. Therefore, Rx = 0 + P x;

=) Rx = P x, 8x 2 X. Hence, P = R. Remark 3.16. If R : X

! X is a positive projection on X and R(X + ) := {Rx |

x 2 X + }, then im+ R = R(X+ ), i.e. R(X) \ X + = R(X + ). This is because if v 2 im+ R, then v

0 and v = Rv which implies v 2 R(X + ).

Conversely, if v 2 R(X + ), then v = Ru for some u 2 X + . Since R is positive, we have v = Ru

0. Hence, v 2 im+ R. 35

3 Spectral Theory for Ordered Spaces Remark 3.17. ker+ P is a face of X + . 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 hP x, yi = hx, P ⇤ yi, 8x 2 X, y 2 Y is called the

(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 2 (kerP ) . For any x 2 X, we have x hx

P x 2 ker P. This implies

P x, yi = 0. So, hx, yi = hP x, yi = hx, P ⇤ yi, 8x 2 X. So, y = P ⇤ y.

Hence, y 2 im P⇤ . (◆) Let y = P ⇤ y and let x 2 ker P. Now hx, yi = hx, P ⇤ yi = hP x, yi = h0, yi = 0. Hence, y 2 (kerP ) . 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 ) . Since P is continuous, ker P is a closed convex set of 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+ Q and ker+ 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 () P Q = QP = 0, P + Q = I. Proof. ()) Assume ker P = im Q and ker Q = im P. Let x 2 X. Then P Q(x) = P (Qx) = 0 as Qx 2 im Q = ker P . Hence, P Q = 0. Similarly, QP = 0. Now, (x x

P x) + P x = x, 8x 2 X. (x

P x) 2 ker P = im Q. Therefore,

P x = Qy for some y 2 X. Applying Q on both sides, Qx

Q2 y =) Qx + 0 = Qy = x

QP x =

P x =) P x + Qx = x. Since this is true

8x 2 X, we have P + Q = I. (() Assume P Q = QP = 0 and P + Q = I. Claim. im Q = ker P (✓) Let y 2 im Q =) y = Qx for some x 2 X. Then, P y = P (Qx) = P Q(x) = 0 =) y 2 ker P . Hence, im Q ✓ ker P . (◆) Let y 2 ker P . P + Q = I =) P y + Qy = y =) 0 + Qy = y =) y 2 im Q. Hence, im Q ◆ ker P . So, ker P = im Q. Similarly, ker Q = im P. Hence, P,Q are strongly complementary.

Remark 3.24. P admits a strong complement () I 3.23 with Q = I-P. 37

P

0. This follows from

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. P is said to be bicomplemented 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. A smooth projection is uniquely determined by its positive kernel and positive 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 2 im+ P } = { a - b | a,b 2 im+ R } = im R. So, we have ker P ⇢ ker R and im P = im R. By 3.15, P = R. Smoothness of P ensures that ker+ P and im+ P dualize properly under positive annihilators. Proposition 3.29. Let P be a continous positive projection on X with dual projection 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⇤ ) = (im P) =) (kerP ⇤ )

(imP ) . Since ker P⇤ is closed convex, ker P⇤ = (kerP ⇤ )

= (imP )

= =)

ker+ P ⇤ = (imP )• = (im+ P )• , as im P is positively generated. (ii) im P⇤ = (ker P) ⇢ (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 )• ) = (ker P) = im P⇤ =) (ker+ P)• ⇢ im+ P⇤ . The other way containment is satisfied by (ii). Hence, (ker+ P)• = im+ P⇤ . ()) (ker+ P )• = im+ P ⇤ (kerP )

=)

(ker+ P )•

kerP =) (ker+ P )•

= (im+ P ⇤ )

= (imP ⇤ )

=

kerP =) P is smooth.

•

(iv) By 3.13, (im+ P )• is the smallest semi-exposed face of X + containing im+ P. •

Therefore, im+ P is semi-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+ R and 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 complementary 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 semiexposed 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, E is a projection. (ii) Clearly, im(P+Q) ⇢ im P + im Q. Let x 2 im P \ im Q. This implies x = Px = P(Qx) = PQ(x) = 0. Therefore, the sum is direct and im(P+Q) ⇢ im P

im Q. Let u 2 im P

im Q. This implies u = Px1 + Qx2 for some

x1 , x2 2 X. Eu = (P+Q)(Px1 + Qx2 ) = PPx1 + PQx2 + QPx1 + QQx2 = Px1 + Qx2 = u. This implies u 2 im(E). Therefore, im P im(P+Q) =) im(P+Q) = im P

im Q ⇢

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 2 ker P. P,Q are bicomplementary =) P, Q is smooth. So, x 2 ker P = (ker+ P )• . Note that if y 2 Y+ , then T ⇤ P ⇤ y 2 (ker+ P )• , since if a 2 ker+ P = im+ Q ⇢ im Q ⇢ im T, then ha, T ⇤ P ⇤ yi = hT a, P ⇤ yi = ha, P ⇤ yi = hP a, yi = h0, yi = 0. Now, since x 2 (ker+ P )• and T ⇤ P ⇤ y 2 (ker+ P )• , hP T x, yi = hx, T ⇤ P ⇤ yi = 0, 8 y 2 Y+ . This implies hP T x, yi = 0, 8 y 2 Y as Y is positively generated. So, PTx = 0 =) x 2 ker PT. Claim: ker E = ker P \ ker Q Clearly, ker E

ker P \ ker Q. Let x 2 ker E =)

P x + Qx =) P x =

0 = Ex =

Qx =) Px = PPx = -PQx = 0 =) x 2 ker

P. Similarly, x 2 ker Q. Therefore, ker E ⇢ ker P \ ker Q i.e. ker(P+Q) ⇢ 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) be a pair of order unit space and base norm space 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. A weakly continous positive projection P on A or V is said to be normalised if kP k  1 ( () P = 0 or kP k = 1) 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) < P e, v > = kP ⇤ vk, 8 v 2 V+ (iii) If P has a complement Q, then P is normalised () Q is normalised Proof.

(i) kP ⇤ k = kP k. Hence, P ⇤ is normalised () P is normalised.

To Prove: P is normalised () Pe  e ()) Suppose P is normalised. Now, kek = 1 and kP k  1 =) kP ek  1 =) P e  e. (() Assume P e  e. Take a 2 A 3 kak  1. Then P e  P a  P e =)

e

e  a  e =)

P e  P a  P e  e =) kP ak 

1 =) kP k  1. (ii) Let v 2 V + . Then, < P e, v >=< e, P ⇤ v >= kP ⇤ vk (from 1.8). (iii) e - Pe 2 ker+ P = im+ Q. Hence, Q(e-Pe) = e - Pe =) e = Qe + Pe =) Qe  e =) Q is normalised (by (i)).

Proposition 3.34. If P, Q are weakly continuous positive projections on A, then Pe + Qe = e () kP ⇤ v + Q⇤ vk = kvk, 8 v 2 V+ () (P⇤ + Q⇤ )(K) ⇢ K. Proof. We know, kvk = < e, v >, 8 v 2 V+ 1. (i) ) (ii) Assume Pe + Qe = e. Let v 2 V+ =) (P ⇤ v + Q⇤ v) 2 V + . kP ⇤ v + Q⇤ vk = < e, P ⇤ v + Q⇤ v > = < e, P ⇤ v > + < e, Q⇤ v > = < P e, v > + < Qe, v > = < P e + Qe, v > = < e, v > = kvk. 42

3 Spectral Theory for Ordered Spaces 2. (ii) ) (iii) Assume kP ⇤ v + Q⇤ vk = kvk, 8 v 2 V+ . K = {v 2 V + | kvk = 1}. Let k 2 K ⇢ V+ . Then k(P ⇤ + Q⇤ )kk = kkk = 1. Also, since P⇤ and Q⇤ are positive, k(P ⇤ + Q⇤ )kk 2 V+ . Therefore, (P ⇤ + Q⇤ )k 2 K. 3. (iii) ) (i) Assume (P⇤ + Q⇤ )(K) ⇢ K. Let k 2 K. Let (P ⇤ + Q⇤ )k = k 0 for some k’ 2 K. < (P + Q)e, k > = < e, (P + Q)⇤ k > = < e, (P ⇤ + Q⇤ )k > = < e, k 0 > = kk 0 k = 1 = kkk = < e, k >. Since V+ = [

0

K, < (P + Q)e, v > =

< e, v >, 8 v 2 V+ =) < (P + Q)e, v > = < e, v >, 8 v 2 V (as V+ generates V). So, e = (P+Q)e = Pe + Qe.

Remark 3.35. In particular, if P,Q are normalised complementary projections on A, then ker+ P = im+ Q. So, (e - Pe) 2 ker+ P =) Q(e - Pe) = e - Pe. Hence, e = Pe + Qe. So, they satisy 3.34. 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 2 V+ : kP ⇤ vk = kvk =) P ⇤ v = v 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 2 (ker+ P )• . Since e - Pe 2 ker+ P,

= 0 =) < e, v > = < P e, v > = < e, P ⇤ v > =) kvk =

kP ⇤ vk =) v = P⇤ v (as P⇤ is neutral) =) (ker+ P )• ✓ im+ P ⇤ . Therefore, by 3.29, P is smooth. (ii) (a) Assume P,Q are smooth =) P⇤ , Q⇤ are complementary. Let v 2 V+ such that kvk = kP ⇤ vk =) < P e, v > = < e, v >. Since P is normalised, Pe + Qe = e. Therefore, < Qe, v > = 0 =) kQ⇤ vk = 0 =) Q⇤ v = 0 =) v 2 ker+ Q⇤ = im+ P ⇤ . So, P ⇤ v = v. Hence, P ⇤ is neutral. (b) ()) P,Q are bicomplementary =) P,Q are smooth =) P ⇤ , Q⇤ are neutral. (() P ⇤ , Q⇤ are neutral =) P,Q are smooth. Also they are complementary. 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, kP ⇤ v + Q⇤ vk = kvk, 8 v 2 V+ , (P⇤ + Q⇤ )(K) ⇢ K. : V ! V and

0

linear projections with dual maps P =

⇤

Theorem 3.40. Let

: V ! V be weakly continous positive : A ! A and P’ =

0⇤

P, P’ are complementary compressions () the following holds: 44

: A ! A. Then

3 Spectral Theory for Ordered Spaces (i) k vk + k 0 vk = kvk, 8 v 2 V + (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 k vk + k 0 vk = kvk, 8 v 2 V + . Finally, P, P’ are compressions =) they are smooth =) (() (a)

and

0

,

0

are neutral.

are neutral =) P,P’ are smooth.

(b) Let k 2 K ⇢ V+ . Then, k kk + k 0 kk = kkk = 1. =) k kk  1. Let v 2 V 3 kvk  1, Since Co(K [ -K) = closed unit ball of V, 9 k1 , k2 2 K 3 v = k1 k

k1

(1

(1

)k2 for some

) k2 k  k k1 k + (1

2 [0, 1]. Now, k vk =

)k k2 k  1. =) k k  1.

So, kP k = kP ⇤ k = k k  1 . Hence, P is normalised. Similarly, P’ is normalised. (c) 8 v 2 V + , k vk + k 0 vk = kvk . This implies k vk = kvk () k 0 vk = 0 =) ker+

= im+ 0 . Similarly, ker+

tary =)

0

0

v = v () = 0 and

0

0

v = 0 (as = im+ . So,

is neutral). Therefore ,

0

are complemen-

= 0. Dualising gives PP’ = 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 2 V + | + kvk =< 1, v >= 1}. Define A+ 1 := {a 2 A | kak = 1}.

Proposition 3.41. If P is a compression on A, then + ⇤ (i) Pe is the greatest element of A+ 1 which belongs to (ker P )

45

•

3 Spectral Theory for Ordered Spaces (ii) im+ P ⇤ = {v 2 V + |< P e, v >=< e, 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) P is normalised =) 0  P e  e =) P e 2 A+ 1 . Let v 2 ker P . •

Then 0 = < e, P ⇤ v > = < P e, v >. Hence, Pe 2 (ker+ P ⇤ ) . Let a 2 A+ 1 3 •

•

a 2 (ker+ P ⇤ ) . Then, kak  1 =) 0  a  e. a 2 (ker+ P ⇤ ) = im+ P (by 2.9) P positive =) 0  a = P a  P e. Hence, a  Pe. So, Pe is the largest such element. (ii) P is a compression =) P ⇤ is neutral. So, < P e, v >=< e, v > () < e, P ⇤ v >=< e, v > () kP ⇤ vk = kvk () P ⇤ v = v. So, im+ P ⇤ = {v 2 V + | P ⇤ v = v} = {v 2 V + |< P e, v >=< e, v >} (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), Pe determines im+ P ⇤ and Re determines im+ R⇤ . Since Pe = Re, im+ P ⇤ = im+ R⇤ . So, we have ker+ P ⇤ = ker+ R⇤ and im+ P ⇤ = im+ R⇤ . Since P⇤ 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, Q is determined as in a) and then P is determined as the unique complement of Q. (c) Treating P = (P ⇤ )⇤ , in a),b), we see that each of ker+ 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 A+ 1 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 ⇤ . P is linear =) 0. Thus, im+ P ⇤ ◆ [

F.

0

(✓) Let v 2 im+ P ⇤ ⇢ V + . Then v = k for some P ⇤ v = P ⇤ k If

= 0, then v 2 [

K \ im P ⇤ = F. So, v 2 [

F ⇢ im+ P ⇤ , for all

0

0

F . If

0 and k 2 K. k = v = 6= 0, then k = P ⇤ k =) k 2

F.

Remark 3.44. Each projective unit p is associated with a unique compression P Proof. p = Pe determines im+ 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) 8 a 2 A+ , a 2 im+ P () a  kakp () a 2 face(p) ()

a 2 p•• (ii) Pa Proof.

0 () < a, v >

0, 8 v 2 F

(i) Let a 2 A+ 1. (a) a  kake for all a in A. If a 2 im+ P , then a = Pa  kakP e = kakp. 47

3 Spectral Theory for Ordered Spaces (b) face(p) = {a 2 A | a  kp, for some k

0}. So, a  kakp =) a 2

f ace(p). (c) p•• is the smallest semi-exposed face of A+ containing p =) a 2 f ace(p) ⇢ p•• . (d) P is a compression =) im+ P is a semi-exposed face of A+ =) •

•

(im+ P )• = im+ P . Therefore, a 2 p•• ⇢ (im+ P )• = im+ P . (ii) Let a 2 A. ()) If < a, v > 3.6, < a, v >

0, 8 v 2 F, then < a, v >

0, 8 v 2 F for some

0, 8 v 2 im+ P ⇤ . Therefore, < P a, v >=< a, P ⇤ v >

(() Let v 2 F ⇢ im+ P ⇤ . < a, v >=< a, P ⇤ v >=< P a, v > a

0. By 0.

0. Hence,

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 2 K | P⇤ v = v } = { v 2 K | < p, v > = 1 } = { v 2 K | < p0 , v > =0} F’ = { v 2 K | P’ ⇤ v = v } = { v 2 K | < p0 , v > = 1 } = { v 2 K | < p, v > =0} (iii) For a 2 A, < P a, v >=< a, v >, 8 v 2 F and < P a, v >= 0, 8 v 2 F’

48

3 Spectral Theory for Ordered Spaces (iv) p = _ { a 2 A+ 1 |< a, v >= 0, 8 v 2 F’}, p = ^ { a 2 A+ 1 |< a, v >= 1, 8 v 2 F}. Hence, p is the unique element of A+ 1 which is 0 on F’ and 1 on F. Proof.

(i) Follows from Remark 8 and 3.2.

(ii) (a) F = K \ imP ⇤ = { v 2 K | P⇤ v = v }. (b) Since K ⇢ V+ , F = K \ imP ⇤ = K \ im+ P ⇤ . By 3.5(ii), F = { v 2 K | < p, e >=< P e, v >=< e, v >= kvk = 1 }. (c) Now replacing p by e - p’, we have F = { v 2 K | < e < p0 , v >= 1 } = { v 2 K | < e

p0 , v >= 1

p0 , v >=< e, v >

< p0 , v >= 1 } = { v

2 K | < p0 , v >= 0 } Similarly, for F’. (iii) (a) v 2 F ⇢ imP ⇤ =) P⇤ v= v. Therefore, < a, v >=< a, P ⇤ v > = < P a, v >, for all a 2 A. (b) If v 2 F’,then v 2 im+ (P 0 )⇤ = ker+ P ⇤ =) < P a, v >=< a, P ⇤ v > = 0 for all a 2 A. (iv) (a) If v 2 F 0 ⇢ im+ (P 0 )⇤ = ker+ P ⇤ . By 3.5(i), p is the largest element in + + ⇤ A+ 1 which vanishes on ker P . So, p is the largest element in A1 which

vanishes on F’. Therefore, p = _ { a 2 A+ 1 |< a, v >= 0, 8 v 2 F’} (b) e - p = p’ = _ { a 2 A+ 1 |< a, v >= 0, 8 v 2 F}. So, p = e - p’ = e - _ { a 2 A+ 1 |< a, v >= 0, 8 v 2 F} = ^ { e - a | a 2 A+ 1 , < a, v >= 0, 8 v 2 F} =^{b|e

b 2 A+ 1 ,< e

b, v >= 0, 8 v 2 F}

+ + = ^ { b 2 A+ 1 |< e b, v >= 0, 8 v 2 F} (since b 2 A1 () e b 2 A1 )

49

3 Spectral Theory for Ordered Spaces = ^ { b 2 A+ 1 | 1 = kvk =< e, v >=< b, v >, 8 v 2 F} = ^ { b 2 A+ 1 |< b, v >= 1, 8 v 2 F}. (c) Now, if t 2 A+ 1 such that t is 0 on F’, then p

t.

Also, if t is 1 on F, then p  t. Hence, p = t.

Theorem 3.48. Let P be a compression on A with associated projective unit p, projective face F. Then 1. [ p, p] = {a 2 A |

p  a  p} = A1 \ imP

2. Co(F [ -F) = V1 \ imP ⇤ 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 2 f ace(p) =)

a + p 2 im+ P (by 3.9(i)). Now, a+p = P(a+p) = Pa + PPe = Pa + p. This implies a = Pa =) a 2 im+ P . Also a  p =) kak  kpk  + 1.This implies a 2 A+ 1 \ im P . + (◆) Let a 2 A+ 1 \ im P =)

e  a  e =)

p  P a = a  p.

2. (✓) F ⇢ V1 \ imP ⇤ =) Co(F [ -F) ⇢ V1 \ imP ⇤ . (◆) Let a 2 V1 \ imP ⇤ . WLOG, we may assume kak = 1(else replace a by

a ). kak

V1 = Co(K [

K) =) a = ↵s

s,t in K and ↵ 2 [0, 1]. a = P ⇤ a = ↵P ⇤ (t) 1 = kak  ↵kP ⇤ (t)k

(1

(1 (1

↵)t, for some ↵)P ⇤ (s). Hence,

↵)kP ⇤ (s)k  1. Therefore, kP ⇤ (t)k =

kP ⇤ (s)k = 1 =) P ⇤ (s), P ⇤ (t) 2 imP ⇤ \ K = F . So, a 2 Co(F [ F ).

50

3 Spectral Theory for Ordered Spaces 3. Clearly, im+ P is a cone in im+ P since, if v1 , v2 2 im+ P , then v1 = P v1 0, v2 = P v2

0 =) v1 + v2 = P v1 + P v2 = P (v1 + v2 )

v1 + v2 2 im+ P . Similarly, v 2 im+ P , 8 v 2 im+ P,

0. Hence,

0.

0 3 e ± v 2 A+ . This

Let v 2 A. Since e is an order unit for A, 9

implies P ( e±v) 2 im+ P =) ( P e±P v) 2 im+ P =) ( p±v) 2 im+ P . Hence, p is an order unit for im P. So, (imP, im+ P, p) is an order unit space. 4. As in the proof of (3), im+ P ⇤ is a cone in imP ⇤ . Claim: F is a base for im P ⇤ . Consider f1 , f2 2 F and 1 f1

+

2 f2

1,

2

2 [0, 1] 3

2 K. Also P ⇤ ( 1 f1 +

f1 , f2 2 imP ⇤ . Hence,

1 f1

+

2 f2

2 f2 )

1

+

2

= 1. Since K is convex,

= ( 1 f1 +

2 f2 )

2 imP ⇤ since

2 K \ imP ⇤ = F . Thus, F is convex.

Let v 2 imP ⇤ \{0}. Since K is a base for V, 9

> 0, k 2 K 3 v = k. Now,

k = v = P ⇤ v = P ⇤ k =) k = P ⇤ k =) k 2 F . Hence, v = f where f = k 2 F. So, F is a base for im P⇤ and (im P⇤ , im+ P ⇤ , F ) is a base norm space.

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 ) be in separating order and norm duality. Let denote the restriction of 0 on {(im P) X (im P⇤ )}. (i) Dual pair: Let a 2 im P. Suppose < a, v >= 0 8 v 2 imP ⇤ . This implies < a, u >0 =< P a, u >0 =< a, P ⇤ u >0 =< a, P ⇤ u >= 0 8 u 2 V. But < a, u >0 = 0 8 u 2 V 51

3 Spectral Theory for Ordered Spaces =) a = 0. So, a = 0. Similarly,we can show, < a, v >= 0 8 a 2 im P⇤ =) v = 0. (ii) Separating order duality: To prove: im P+ = {a 2 imP |< a, v >

0 8 v 2 im+ P ⇤ }

(✓) If a 2 im P+ and v 2 im+ P ⇤ , then < a, v >=< a, v >0 (◆) Suppose a 2 imP 3< a, v > a, P ⇤ v >0 =< a, P ⇤ v >

0.

0 8 v 2 im+ P ⇤ . < a, v >0 =< P a, v >0 =


0 8 a 2 im+ P }.

(iii) Norm Duality: Denote: kakp = inf { > 0 | p ± a 2 im+ P }, for all a 2 im P; kake = inf { > 0 | e ± a 2 A+ }, for all a 2 A; kvkF = inf { > 0 | f ± v 2 im+ P ⇤ , for some f 2 F }, 8 v 2 im P⇤ ; kvkK = inf { > 0 | k ± v 2 V + , for some k 2 K}, 8 v 2 V. Claim: kakp = kake 8 a 2 im P. Let a 2 im P =) Pa = a. First, kakp p ± a 2 im+ P and e So, kakp e ± a 2 A+ . Hence, kakp e±a 2 A+ =)

kake . Second,for all

p.

> 0 3

p±a 2 imP + . So, kakp  kake . Thus, kakp = kake .

Claim: kvkF = kvkK 8 v 2 im P⇤ . Let v 2 im P⇤ =) P⇤ v = v. First, if f ± v 2 im+ P ⇤ , for some f 2 F and

> 0, then since F ⇢ K and f ± v 2 V + , kvkK  kvkF . Second,

if k ± v 2 V + , for some k 2 K and

> 0, then P ⇤ k ± v 2 im+ P ⇤ .

Now, kP ⇤ kkK  kP kkkkK  1. By 3.2.1, P ⇤ k 2 Co(F [ implies P ⇤ k = ↵f1 P ⇤k

(1

F ). This

↵)f2 for some f1 , f2 2 F and ↵ 2 [0, 1].

0 =) ↵ 6= 0. Therefore, P ⇤ k  ↵f1 and P ⇤ k  ↵f1  f1 52

3 Spectral Theory for Ordered Spaces as ↵ 2 (0, 1]. Now, P ⇤ k ± v 2 im+ P ⇤ =)

f1 ± v 2 im+ P ⇤ . Hence,

kvkF  kvkK . Thus kvkF = kvkK . To Prove: kvkF = kvkdual 8 v 2 im P⇤ We know, kvkF = kvkK = inf { > 0 | k ± v 2 V + for some k 2 K} = sup{| < a, v > | | a 2 A, kake  1} = sup{| < a, P ⇤ v > | | a 2 A, kake  1} = sup{| < P a, v > | | a 2 A, kake  1} = sup{| < b, v > | | b 2 imP, kbke  1}

2

= sup{| < b, v > | | b 2 imP, kbkp  1} = kvkdual . To Prove: kakp = kakdual 8 a 2 im P We know, kakp = kake = inf { > 0 | e ± a 2 A+ } = sup{| < a, v > | | v 2 V, kvkK  1} = sup{| < P a, v > | | v 2 V, kvkK  1} = sup{| < a, P ⇤ v > | | v 2 V, kvkK  1} = sup{| < a, u > | | u 2 imP ⇤ , kukK  1}

3

= sup{| < a, u > | | u 2 imP ⇤ , kukF  1} = kakdual . Hence, the two spaces are in separating order and norm duality. 2

sup{| < P a, v > | | a 2 A, kake  1}  sup{| < b, v > | | b 2 imP, kbke  1} and sup{| < b, v > | | b 2 imP, kbke  1}  sup{| < a, v > | | a 2 A, kake  1} = sup{| < P a, v > | | a 2 A, kake  1}. 3 sup{| < a, P ⇤ v > | | v 2 V, kvkK  1}  sup{| < a, u > | | u 2 imP ⇤ , kukK  1} and sup{| < a, u > | | u 2 imP ⇤ , kukK  1}  sup{| < a, v > | | v 2 V, kvkK  1} = sup{| < a, P ⇤ v > | | v 2 V, kvkK  1}.

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 2 X. Then Px 2 imP ⇢ imQ =) Q(P x) = P x =) QP = P . 2. (ii) =) (iii) Assume QP = P, then p = P e = QP e. Since P e  e and Q is positive, Q(P e)  Qe = q =) p  q. 3. (iii) =) (iv) Assume p  q. Let v 2 F. Then by, 3.10 (ii), 1 =< p, v >< q, v >< e, v >= kvk = 1. Now, since v 2 K and < q, v >= 1, by 3.10(ii), v 2 G. This implies F ✓ G. 54

3 Spectral Theory for Ordered Spaces 4. (iv) =) (v) Assume F ⇢ G =) [ imP ⇤ = im+ P ⇤

0

F ⇢ [

0

G =) im+ P ⇤ ⇢ im+ Q⇤ . Now,

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 2 Ker Q. Then Px = PQx = P0 = 0. Hence, x 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 Proposition 3.53. P

on the set of compressions on A is a partial ordering. Q () Q’

P’. So, the map P 7! P’ is an order

reversing map for this relation.

3.3.2

Orthogonality

Definition 3.54. P is said to be orthogonal to Q (denoted by P ? Q) if PQ = 0. (We also write p ? q and F ? 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 2 im P. Then Qx = Q(Px) = 0x = 0 =) imP ⇢ kerQ. So, im+ P ⇢ ker+ Q = im+ Q0 =) imP ⇢ imQ0 =) P

Q0 .

2. (ii) =) (iii), (iii) =) (iv) P

Q0 =) imP ⇢ imQ0 . Then by 3.11, p  q 0 which further implies F ⇢

G’. 3. (iv) =) (v) F ⇢ G0 =) P Q0 = 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 () 9 a 2 A+ 1 3 < a, v > = 1, 8 v 2 F and < a, v > = 0, 8 v 2 G Proof. Let p, q be the projective units associated with F, G respectively. ()) F ? G =) p  q 0 =) p  e

q =) p + q  e. By 3.10, v 2 F =)

< p, v >= 1. Also, if v 2 G, then 1 =< q, v >< p, v > + < q, v >=< p+q, v >< e, v >= kvk = 1. This implies < p, v >= 0, 8 v 2 G. Therefore, p is the required element a 2 A+ 1. 56

3 Spectral Theory for Ordered Spaces (() Assume 9 a 2 A+ 1 3 < a, v > = 1, 8 v 2 F and < a, v > = 0, 8 v 2 G. This implies 1 = kvk =< e, v >=< e e

3.3.3

a

q i.e. a  e

a, v >, 8 v 2 G. By 3.10(vi), a

p and

q = q 0 . So, p  a  q 0 . So, by 3.13, F ? G.

Compatibility

Definition 3.57. Let P , Q be two compressions on A. P is said to be compatible with Q if P Q = QP . P is said to be compatible with an element a 2 A if a = P a + P 0 a. So, if a 2 A is compatible with P , then a is also compatible with P 0 . Notation 11. P is compatible with Q is denoted by P ⇠ Q and P ⇠ a denotes that P is compatible with the element a 2 A. Lemma 3.58. If a 2 A+ , then a = Pa + P’a () Pa  a. Proof. ()) a (() a

P a = P 0a

0 =) a

P a.

P a 2 kerP . Further, P a  a =) a

P a 2 ker+ P = im+ P 0 . But P’P

= 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, by 3.3.1, 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, P q 0  q 0 =) P q  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 suffices to show (i) () (ii) ()) If PQ = QP, then Pq = PQe = QPe = Qp  Qe = q. So P q  q. (() Assume Pq  q. Let a 2 A+ . Since 0  a  kake, then 0  Qa  kakq =) 0  P Qa  kakP q  kakq = kakQe. Hence, 0  Q0 P Qa  kakQ0 Qe = 0. This implies Q’PQa = 0, 8 a 2 A+ . Therefore, P Qa 2 ker+ Q0 = im+ Q =) QP Qa = P Qa, 8 a 2 A+ and since A+ generates A, QPQa = PQa, 8 a 2 A. Hence, QPQ = PQ. Now P q  q =) P q 0  q 0 . So replacing q by q’ in the above arguments, we have QPQ’a = 0, 8 a 2 A+ =) QPQ’a = 0, 8 a 2 A. So, QPQ’ = 0. Dualizing gives Q0⇤ P ⇤ Q⇤ = 0. So, 8 v 2 V + , P ⇤ Q⇤ v 2 ker+ Q0⇤ = im+ Q⇤ . Therefore, P ⇤ Q⇤ v = Q⇤ P ⇤ Q⇤ v, 8 v 2 V + and hence for all v 2 V. Thus, P ⇤ Q⇤ = Q⇤ P ⇤ Q⇤ . Dualizing gives QP = QPQ. Thus, PQ = QPQ = QP.

Remark 3.61. P is compatible with Q () P is compatible with Q’. 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 , and SP = im(P + P 0 ). Proof. a, b 2 SP =) a = P a + P 0 a, b = P b + P 0 b =) (a + b) = (P a + P 0 a) + (P b + P 0 b) = (P a + P b) + (P 0 a + P 0 b) = P (a + b) + P 0 (a + b) =) (a + b) 2 SP . And, a 2 SP =) a = P a+P 0 a =)

a = P a+ P 0 a = P ( a)+P 0 ( a) 8 2 R.

Hence, a 2 SP 8 2 R. Therefore, SP is a linear subspace of A.

58

3 Spectral Theory for Ordered Spaces Now, (P + P 0 )2 = P 2 + P P 0 + P 0 P + P 02 = P + P 0 ( as P P 0 = P 0 P = 0). Therefore, (P + P 0 ) is a projection on A. Let a 2 A. a 2 im(P + P 0 ) () a = (P + P 0 )a () a = P a + P 0 a () a 2 SP . Hence, SP = im(P + P 0 ). Therefore, to show the compatibility of an element a 2 A, with P , it is sufficient to show compatibility of a + e with P , for any

2 R (since e 2 SP ).

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 =) P Q = P and QP = P

=) P Q = 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 2 A+ . If a = 0 on F 0 , then a is compatible with P (Pa = a, P 0 a = 0). Proof. Suppose < a, v > = 0 8v 2 F 0 . Now, F 0 = K \ im(P 0 )⇤ ✓ im+ (P 0 )⇤ = ker+ P ⇤ . a 2 (F 0 )• ◆ (ker+ P ⇤ )• = im+ P . Therefore, P a = a and 0 = (P 0 P )a = P 0 (P a) = P 0 a. Hence, a = a + 0 = P a + P 0 a =) a is compatible with P . Remark 3.66. Converse of the above is not true i.e. If a 2 A such that a is compatible with P , then a need not be 0 on F 0 . Example: e 2 SP but < e, f 0 > = kf k = 1 6= 0 8f 0 2 F 0 . Proposition 3.67. Let a 2 A. Then Pa is the unique element in A compatible with P which is equal to a on F and 0 on F 0 . 59

3 Spectral Theory for Ordered Spaces Proof. P (P a) + P 0 (P a) = P a + 0 = P a. Hence, P a is compatible with P . Let f 2 F ✓ im P ⇤ . Then, < P a, f > = < a, P ⇤ f > = < a, f >. Hence, P a = a on F . Let f 0 2 F 0 ✓ im+ P 0 ⇤ = ker+ P ⇤ . Then, < P a, f 0 > = < a, P ⇤ f 0 > = < a, 0 > = 0. Hence, P a = 0 on F 0 . Now, suppose b 2 A such that b is compatible with P and b = a on F , b = 0 on F 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 = 0 on F 0 =) b = 0 on im+ P 0 ⇤ . Let v 2 V + . < b, v > = < P b + P 0 b, v > = < b, P ⇤ v > + < b, P 0 ⇤ v > = < a, P ⇤ v > + < 0, P 0 ⇤ v > = < P a, v >. Hence, b = P a, thus proving the uniqueness. Proposition 3.68. Let a 2 A. Then (P a + P 0 a) is the unique element in A, compatible with P, which is equal to a on Co(F [ F 0 ). Proof. First, P (P a+P 0 a)+P 0 (P a+P 0 a) = P 2 a+P P 0 a+P 0 P a+P 0 2 a = P a+P 0 a. Hence, (P a + P a) is compatible with P . Let f 2 F ✓ im+ P ⇤ = ker+ P 0 ⇤ , f 0 2 F 0 ✓ im+ P 0 ⇤ = ker+ P ⇤ , ↵ 2 [0, 1]. Now, < P a + P 0 a, ↵f + (1 + (1

↵) < a, P ⇤ f 0 > + (1

= < a, ↵f + (1

↵)f 0 > =

↵ < a, P ⇤ f >

+ ↵ < a, P 0 ⇤ f >

↵) < a, P 0 ⇤ f 0 > = ↵ < a, f > + (1

↵) < a, f 0 >

↵)f 0 >. Hence, (P a + P 0 a) = a on Co(F [ F 0 ).

Suppose b 2 A such that b is compatible with P and b = a on Co(F [ F 0 ). Since, im+ P ⇤ = [

0

F and im+ P 0 ⇤ = [

0

F 0 , b = a on Co(F [ F 0 ) =) b =

a on im+ P ⇤ and im+ P 0 ⇤ . Now let v 2 V + . < b, v > = < P b + P 0 b, v >= = < b, P ⇤ v > + < b, P 0 ⇤ v > = < a, P ⇤ v > + < a, P 0 ⇤ v > = < P a + P 0 a, v >. So, b = P a + P 0 a on V + =) b = P a + P 0 a on V (as V + generates V). Hence, b = (P a + P 0 a), thus proving the uniqueness. Remark 3.69. An element a 2 A, compatible with P , is completely determined by its values on F and F 0 .

60

3 Spectral Theory for Ordered Spaces Proof. Suppose b 2 A such that b is compatible with P and b = a on F and F 0 . Then by 3.68, b = P a + P 0 a = a (as a is compatible with P ). Hence, b = a. Proposition 3.70. Let a 2 A+ . Then 1. P a = 2. P a =

W V

{b 2 A+ | b = 0 on F 0 , b  a on F } {b 2 A+ | b = 0 on F 0 , b

a on F }

Proof. Assume a 2 A+ . Then P a = 0 on F 0 and P a = a on F . Hence b can be chosen to be P a in (1), (2) above. 1. Let b 2 A+ such that b = 0 on F 0 and b  a on F . By 3.65, b is compatible with P . Also, b  a on F

=) b  a on im+ P ⇤ = [

0

F . Similarly,

b = 0 on F 0 =) b = 0 on im+ P 0 ⇤ . Now consider v 2 V + . < b, v > = < P b + P 0 b, v > = < b, P ⇤ v > + < b, P 0 ⇤ v >  < a, P ⇤ v > + 0 = < P a, v > W =) b  P a on V + . Hence, b  P a. So, P a = {b 2 A+ | b = 0 on F 0 , b  a on F }

2. Imitating the proof for (1), with  replaced by b = 0 on F 0 , b

, we get P a =

a on F }

V

{b 2 A+ |

The above proposition says that P a 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) kT k  1 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 ) is also positive. (ii) Let a 2 A such that kak  1. Then,

e  a  e. T = P + P 0 is

positive. So, (P +P 0 )( e)  (P +P 0 )a  (P +P 0 )(e) =)

e  Ta  e

( as p + p0 = e). Hence, kT ak  1. Thus, kT k  1. (iii) P (P + P 0 ) = P 2 + P P 0 = P and (P + P 0 )P = P 2 + P 0 P = P . So, P T = T P . Similarly, T P 0 = P 0 T . (() Given that im T = im (P + P 0 ). (i) Suppose T is positive. Then T = P + P 0 by ??. (ii) Suppose kT k  1. Let a 2 A+ . We may assume kak  1 (else replace a by

a ). kak

Now, 0  a  e =)

e =) ke e =) ee

ak  1. So, kT (e

e  Te

e 

a  0 =) 0  e

a)k  1 =)

e  T (e

a  a) 

T a  e. But T e = (P + P 0 )e = p + p0 = e. Hence,

T a  e =) 0  T a. So, T a

0 8a 2 A+ . Hence, T is

positive. Thus, T = P + P 0 by (i). (iii) Denote E = P + P 0 . T commutes with P, P 0 =) T commutes with E. Let a 2 A. T E(a) = T (Ea) = Ea (as Ea 2 im E = im T ). So, T E(a) = E(a) 8a 2 A

=)

T E = E. Similarly, ET = T . So,

T = ET = T E = 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 2 K. We know E ⇤ (K) ✓ K by ??, im+ P ⇤ = S 0 ⇤ ⇤ 0⇤ 0 0 F . So, E k = P k + P k = ↵f + f for some ↵,

S

0

F and im+ P 0 ⇤ =

0, f 2 F, f 0 2 F 0 .

↵ This implies (↵+ )( ↵+ f + ↵+ f 0 ) = E ⇤ k 2 K (↵+ 6= 0 else E ⇤ k = 0 2 K which ↵ is a contradiction). Now, F, F 0 ✓ K and K is convex. So, ( ↵+ f+

↵+

f 0 ) 2 K.

Therefore, E ⇤ k = (↵ + )k 0 2 K for some k 0 2 K. But K is a base and the representation of each element is unique =) (↵ + ) = 1. So, Hence, E ⇤ k = ↵f + (1 E ⇤ (K) ✓ Co(F [ F 0 ).

= 1

↵.

↵)f 0 2 Co(F [ F 0 ). Since k is arbitrary, we have

4

Now, consider v 2 Co(F [ F 0 ). Let v = ↵f + (1 F, f 0 2 F 0 . E ⇤ v = ↵P ⇤ f + (1

↵)P ⇤ f 0 + ↵P 0 ⇤ f + (1

↵)f 0 for some ↵ 2 [0, 1], f 2 ↵)P 0 ⇤ f 0 = ↵f + 0 + 0 + (1

↵)f 0 = v. Hence, E ⇤ fixes Co(F [ F 0 ) pointwise. Now, to show uniqueness of E ⇤ , assume there exists a weakly continuous projection 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 = im E ⇤ (◆) T fixes elements of Co(F [ F 0 ) =) T fixes F, F 0 pointwise =) T fixes im+ P ⇤ and im+ P 0 ⇤ . Since im P ⇤ and im P 0 ⇤ are positively generated, T also fixes im P ⇤ and im P 0 ⇤ pointwise. Now, consider u 2 im E ⇤ . T u = T (E ⇤ u) = 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, im T ◆ im E ⇤ . (✓) Let v 2 im+ T . This implies T v = v = k = v = T v = T ( k) = T k = (↵f + (1

k for some

0, k 2 K. i.e.

↵)f 0 ) for some ↵ 2 [0, 1], f 2

Alternately, as E ⇤ k 2 K, we have 1 = kE ⇤ kk = < e, E ⇤ k > = < e, ↵f + f 0 > = ↵ < e, f > + < e, f 0 > = ↵kf k + kf 0 k = ↵ + . 4

63

3 Spectral Theory for Ordered Spaces F, f 0 2 F 0 . If

= 0, then Ev = E0 = 0 = v. Suppose

k = T k = ↵f + (1

6= 0, then we have

↵)f 0 2 Co(F [ F 0 ). Since E fixes Co(F [ F 0 ), Ek =

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 = im E ⇤ = im (P ⇤ + P 0 ⇤ ). Thus, T = P ⇤ + P 0 ⇤ , proving the uniqueness. Central Compressions Definition 3.74. A compression P is said to be central if it is compatible with all a 2 A. An element a 2 A is said to be central if it is compatible with all 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) P a  a, 8a 2 A (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) P a  a =) I

P

0. By 3.24, P is strongly complemented.

(ii =) iii) Assume P is strongly complemented. Let P 0 = I P P e = e P 0 e  e =) P is normalised. Also, P 0⇤ = I

P⇤

0. Then,

0 =) P ⇤ , P 0 ⇤

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) Since P 0 = I

P, a = P a + P 0 a, 8a 2 A. So, P is a central

compression. (iv =) i) Let a 2 A+ . P is central =) a = P a + P 0 a =) P a  a.

3.4

The Lattice of Compressions

In this section, we will consider < (A, A+ , e), (V, V + , K) > - a pair of order unit space and base norm space under separating order and norm duality under an additional assumption called the standing hypothesis. We will see that in this case, the compressions have nice lattice properties. Definition 3.76. Let F ✓ K. Then F is said to be an exposed face of K if 9 a 2 A+ such that F = K \ a 1 (0) = {w 2 K| < a, w > = 0}. F is said to be a semi-exposed face of K if 9 a family {a↵ } 2 A+ such that F = {w 2 K| < a↵ , w > = 0, 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) denotes a compression P on A with associated projective face F and projective unit p. Its unique complementary compression is denoted by (P 0 , F 0 , p0 ). C is the set of all compressions on A, F is the set of all projective faces of V and P is the set of all projective units of A. 5

(F = K \ p0

1

(0))

65

3 Spectral Theory for Ordered Spaces Each of these sets have an ordering (C, ), (F, ✓), (P, ) and a notion of complementation (P, F, p) ! (P 0 , F 0 , p0 ). Also, for any two compressions (P1 , F1 , p1 ), (P2 , F2 , p2 ) on A, we have P1

P2 () F1 ✓ F2 () p1  p2

(3.4.1)

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 1

: C ! P given by P

! Pe = p

2

: C ! F given by P

! K \ im P ⇤ = F

3

: P ! F given by p ! {w 2 K| < p, w > = 1} = F .

Definition 3.80. A partially order set (L, ) is said to be a lattice if every two elements x, y 2 L have a greatest lower bound (x^y) and least upper bound (x_y) in L. Remark 3.81. Let (X, ), (Y, ) be two partially ordered sets and

: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 x1 , x2 2 X, y1 = Then, by order isomorphism, x1 x1 ^ x2

x1 , x2 =)

(x1 ^ x2 ) 

If y3 2 Y such that y3  x1 ^ x2 . Applying

x2 ()

(x1 ) 

(x1 ), y2 =

(x2 ).

(x2 ) () y1  y2 . Now,

(x1 ), (x2 ).

(x1 ), (x2 ) =)

on both sides, we get y3 

(x1 ) ^ (x2 ). 66

1

y3  x1 , x2 =)

(x1 ^ x2 ). Hence,

1

y3

(x1 ^ x2 ) =

3 Spectral Theory for Ordered Spaces Similarly,

(x1 _ x2 ) =

(x1 ) _ (x2 ). So, Y is a lattice and the bijective map

preserves the lattice structure, thus is a lattice isomorphism. Proposition 3.82. Assume < A, V > satisfy the standing hypothesis. Then C, F, P are lattices, isomorphic to each other, and the lattice operations are given by the following equations for each pair F, G 2 F: F _ G = (F 0 \ G0 )0

F ^ G = F \ G,

(3.4.2)

Proof. Let F, G 2 F with associated projective units p, q respectively. Clearly, F \ G will be the greatest lower bound for F, G, if it belongs to F. Let a = 1 0 (p 2

1 + q 0 ) 2 A+ 1 . Then x 2 K \ a (0) () x 2 K, < a, x > = 0 () x 2

K, < 12 (p0 + q 0 ), x > = 0 () x 2 K, < p0 , x > + < q 0 , x > = 0 () x 2 K, < p0 , x > = 0, < q 0 , x > = 0 () x 2 F and x 2 G () x 2 F \ G. So, F \ G = K \ a 1 (0). Hence, F \ G is an exposed face of K and thus projective, by the standing hypothesis. So, F \ G 2 F and F \ G = F ^ G. Now, consider F 0 , G0 2 F. Then, F 0 \ G0 ✓ F 0 , G0 . Complementation is order reversing. So, (F 0 \ G0 )0 ◆ F 00 , G00 = F, G. If J 2 F such that J ◆ F, G, then J 0 ✓ F 0 , G0 =) J 0 ✓ F 0 \G0 =) J = J 00 ◆ (F 0 \G0 )0 . So, F _G = (F 0 \G0 )0 2 F. Hence, F is a lattice. From 3.79, 3.81, it follows that C, P are also lattices and the maps

1,

2,

3

are lattice isomorphisms. Thus, C, F, P are isomorphic lattices. Remark 3.83. From 3.82, we observe that 8 F, G 2 F, (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 compressions (P, F, p), (Q, G, q) on A. Let a 2 A+ such that a  p, a  q. Then a  p ^ q. 67

3 Spectral Theory for Ordered Spaces Proof. The result is trivially true when a 2 P. But we are saying it even for those a2 / P. Let H = K \ a 1 (0). Then H is an exposed face of K and therefore is a projective face, by the standing hypothesis. Let f 0 2 F 0 . Then 0  < a, f 0 >  < p, f 0 > = 0 =) < a, f 0 > = 0 =) a = 0 on F 0 . Similarly, a = 0 on G0 . This implies F 0 , G0 ✓ H =) F 0 _ G0 = (F ^ G)0 ✓ H. So, a = 0 on (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 P Q = QP = P ^ Q. Proof. Define r = P q = P (Qe) = Q(P )e = Qp 2 A+ 1 . Then r = P q  P e = p =) r  p. Similarly, r  q. So, r  p ^ q. Now let s 2 A+ such that s  p, q =) s 2 im+ P and s 2 im+ Q (by ??). This implies s = P s  P q = r. In particular, taking s = p ^ q, we get p ^ q  r. So r = p ^ q = P e ^ Qe = (P ^ Q)e

(3.4.4)

Now let a 2 A+ 1 . Then P Qa  P Qe = r = (P ^ Q)e. P Q is a normalised projection, so kP Qak  1. So, by ??, P Qa 2 im+ (P ^ Q) =) P Qa = (P ^ Q)(P Q)a = 6 (P Q)(P ^ Q)a = 7 (P ^ Q)a. Therefore, P Q = P ^ Q on A+ 1 . Hence, P Q = P ^ Q. Remark 3.86. In the above proof, 3.4.4 tells us that P Qe = (P ^Q)e. But from this we directly can’t conclude that P Q = P ^ Q because in general P Q need not be a compression! (For example, in C ⇤ algebra, P is a projection () P = P ⇤ = P 2 . (P Q)⇤ = Q⇤ P ⇤ = QP . So, if P Q is a projection, then (P Q)2 = P Q =) QP = P Q. Conversely, if P Q = QP , then (P Q)2 = P Q, thus P Q is a projection. Hence, 6 7

P ^ Q P, Q =) P ^ Q is compatible with P, Q =) P ^ Q is compatible with P Q im+ (P ^ Q) ✓ im+ P, im+ Q

68

3 Spectral Theory for Ordered Spaces compatibility is a necessary and sufficient condition for P Q 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 2 A. Then, P ^ Q, P _ Q are also compatible with a. Proof. First assume a 2 A+ . P, Q compatible with a =) P a, Qa  a. So, (P ^ Q)a = (P Q)a = P (Qa)  P a  a =) P ^ Q ⇠ a. Next, P ⇠ Q =) P 0 ⇠ Q0 . Also, P, Q ⇠ a =) P 0 , Q0 ⇠ a. Then as above, P 0 ^ Q0 is compatible with a =) (P 0 ^ Q0 )0 = P _ Q is compatible with a. Now consider any a 2 A compatible with both P, Q. We have a = a1 where a1 = 12 (kake + a), a2 = 12 (kake

a2

a) 2 A+ . Then P a1 + P 0 a1 = 12 (kak(p +

p0 ) + a) = 12 (kake + a) = a1 =) P ⇠ a1 . Similarly, it can shown that P ⇠ a2 , Q ⇠ a1 and Q ⇠ a2 . Then P ^ Q, P _ Q ⇠ a1 , a2 =) P ^ Q, P _ Q ⇠ a. Proposition 3.88. Assume < A, V > satisfy the standing hypothesis. Let P, Q be orthogonal compressions, both compatible with an element a 2 A. Then P _ Q is compatible with a and (P _ Q)a = P a + Qa. Proof. P ? Q =) P ⇠ Q (by 3.64) and we are given P, Q ⇠ a. So, from 3.87 we have P _ Q ⇠ a. Then, a = (P _ Q)a + (P _ Q)0 a = (P _ Q)a + (P 0 ^ Q0 )a = (P _ Q)a + (P 0 Q0 )a. Now, P, Q ⇠ a =) a = Qa + Q0 a, a = P a + P 0 a. So, P 0 Q0 a = P 0 (a

Qa) = P 0 a

P 0 Qa = 8 P 0 a

this above, we get a = (P _ Q)a + (a

Pa

Qa = (a

P a)

Qa. Substituting

Qa) =) (P _ Q)a = P a + Qa.

˙ 2+ ˙ . . . Pn in place of P1 _P2 _. . . Pn when P1 , P2 , . . . Pn Notation 13. We write P1 +P are mutually orthogonal compressions. So by definition, ˙ 2+ ˙ . . . Pn () P = P1 _ P2 _ . . . Pn and Pi ? Pj for i 6= j P = P1 +P

8

P ? Q =) Q  P 0 =) im+ Q ✓ im+ P 0 =) P 0 Q = Q

69

(3.4.5)

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. Since Q 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˙ 3+ ˙ . . . Pn ). onal compressions, then P1 ? (P2 +P Proposition 3.91. Assume < A, V > satisfy the standing hypothesis. Let P1 , P2 . . . Pn be mutually orthogonal compressions, all compatible with an element a 2 A. Then, ˙ 2+ ˙ . . . Pn is compatible with a and (P1 +P ˙ 2+ ˙ . . . Pn )a = P1 a + P2 a + . . . Pn a. P1 +P 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

˙ 2+ ˙ . . . Pn 1 )a = P1 a + P2 a + . . . Pn 1 a. 1. i.e. (P1 +P

˙ 2 +P ˙ 3+ ˙ . . . Pn 1 ). Thus by 3.88 and Now, when n = k, by 3.90, Pn ? (P1 +P ˙ 2+ ˙ . . . Pn )a = Pn a + (P1 +P ˙ 2+ ˙ . . . Pn 1 )a = Pn a + induction hypothesis, (P1 +P P1 a + P2 a + . . . Pn 1 a. Proposition 3.92. Assume < A, V > satisfy the standing hypothesis. If p1 , p2 , . . . pn are projective units, then the following are equivalent: 1.

Pn

i=1

pi  e

2. pi ? pj for i 6= j 3.

Wn

i=1

pi =

Pn

i=1

pi

Proof. Let Pi be the compression associated with pi . (1) =) (2). Let i 6= j. Then pi + pj  e =) pi  e

pj =) pi ? pj .

Wn Wn Wn ˙ ˙ (2) =) (3). i=1 pi = i=1 Pi e = ( i=1 Pi )e = (P1 +P2 + . . . Pn )e = Pn Wn Pn i=1 Pi e (since each Pi ⇠ e). Thus i=1 pi = i=1 pi . 70

3 Spectral Theory for Ordered Spaces Wn

(3) =) (1). pi  e 8 i =)

i=1

pi  e =)

Pn

i=1

pi  e

Remark 3.93. From 3.92, it follows that p1 + p2 + . . . pn = p for some projective Wn unit p () i=1 pi = p and pi ? pj for i 6= j. So, if P1 , P2 , . . . Pn and P are compressions corresponding to p1 , p2 . . . pn and p respectively, then ˙ 2+ ˙ . . . Pn p = p1 + p2 . . . pn () P = P1 +P

(3.4.6)

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

q is the projective unit associated with P ^ Q0 = P Q0 . Further, R = P ^ Q0

p

˙ and if a 2 A such that a ⇠ P, Q is the unique compression such that P = Q+R then Ra = P a Proof. Q

Qa.

P =) Q ⇠ P =) Q0 ⇠ P . Then by 3.85, P ^ Q0 = P Q0 is a

compression. P Q0 e = P (e

P q = 9p

Qe) = P e

q.

˙ Now let R = P ^ Q0 = P Q0 . Then by 3.4.6, P = Q+R () p = q + r which is true since we have shown r = p

q. Now, if S is any other compression such

˙ then p = q + Se =) Se = p that P = Q+S,

q = r = Re =) S = R. So, R is

˙ the unique compression such that P = Q+R. Next, let a 2 A such that a ⇠ P, Q =) a ⇠ P, Q0 =) a ⇠ P ^ Q0 = R (by ˙ 3.87). Now, P = Q+R =) P a = Qa + Ra =) Ra = P a Remark 3.95. Let P1 Proof. P1

Q1

Q02

Q1 , P2

Qa.

Q2 , Q1 ? Q2 =) P1 ? P2 .

P20 =) P1 ? P2 .

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

9

˙ P ⇠ Q () P = (P ^ Q)+(P ^ Q0 ). q 2 im+ Q ✓ im+ P

71

3 Spectral Theory for Ordered Spaces Proof. First note that P ^ Q

Q, P ^ Q0

Q0 and Q ? Q0 . So, by 3.95,

(P ^ Q) ? (P ^ Q0 ) and hence, the orthogonal sum makes sense. Now, to prove ˙ ^ q 0 ). the result, it is sufficient to prove P ⇠ Q () p = (p ^ q)+(p ()) P ⇠ Q =) P ⇠ Q0 and so P ^ Q = P Q, P ^ Q0 = P Q0 . Now (p ^ q) + (p ^ q 0 ) = P Qe + P Q0 e = P (q + q 0 ) = P e = p. (() p^q  q =) p^q 2 im+ Q (by ??) =) Q(p^q) = p^q and p^q 0  q 0 =) p^q 0 2 im+ Q0 = ker+ Q =) Q(p^q 0 ) = 0. Applying Q to p = (p^q)+(p^q 0 ), we get Qp = Q(p ^ q) + Q(p ^ q 0 ) = 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 () 9 R, S, T 2 C such that S ? T, P = R+S, Q = R+T 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, ˙ ˙ and Q = by 3.95 S ? T and by 3.96, P = (P ^ Q)+(P ^ Q0 ) = R+S ˙ 0 ^ Q) = R+T ˙ . (P ^ Q)+(P ˙ ˙ . Then, (() Assume 9 R, S, T 2 C such that S ? T, P = R+S, Q = R+T R

P, Q. Hence P ^ Q.

R 72

(3.4.7)

3 Spectral Theory for Ordered Spaces S 0 =) R _ T

Also, R ? S, T ? S =) R, T S

Q0 . And S

S 0 =)

S _ R = P . This implies P ^ Q0

S So, P = R _ S

S 0 =) Q

(3.4.8)

(P ^ Q) _ (P ^ Q0 ) which is a sufficient condition for

compatiblity by 3.96. Hence, P ⇠ Q. ˙ ˙ and let Uniqueness: Assume 9 R, S, T 2 C such that S ? T, P = R+S, Q = R+T r = Re, s = Se and t = T e. Then we have, S ? R, T which implies S ? ˙ = Q. This implies R+T Qs = (QS)e = 0 Also, R

(3.4.9)

Q =) r 2 im+ Q, which implies 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, P e = Re + Se =) s = p

r = p

(p ^ q) = p ^ q 0 (because

p ⇠ q =) p = (p ^ q) + (p ^ q 0 )). 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 < A, V > satisfy the standing hypothesis. Let P, Q be two compatible compressions on A. Then ˙ ˙ ˙ 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 ) and Q = (P ^ Q) _ (P 0 ^ Q). =) P _ Q = (P ^ Q) _ (P ^ Q0 ) _ (P 0 ^ Q) ˙ ˙ 0 ^ Q) = (P ^ Q)+(P ˙ 0 ^ Q) +(P ˙ = (P ^ Q)+(P ^ Q0 )+(P ^ Q0 ) ˙ 0 ^ Q) = P +(P

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. This implies p = (p ^ q) + (p ^ q 0 ) and q = (p ^ q) + (p0 ^ q). Hence, p + q = (p ^ q) + (p ^ q 0 ) + (p ^ q) + (p0 ^ q) = (p ^ q) + (p _ q)

(by 3.96)

Sublattices of C Definition 3.101. A lattice L with greatest element I and least element 0, is said to be an orthocomplemented lattice if there exists a map x ! x0 in L satisfying the following properties: (i) x00 = x (ii) x  y =) y 0  x0 (iii) x ^ x0 = 0, x _ x0 = I Definition 3.102. An orthomodular lattice is an orthocomplemented lattice (L,  , 0, I,0 ) satisfying the orthomodular law i.e x  y =) y = (y ^x0 )_y, 8 x, y 2 L. Theorem 3.103. The lattice of compressions (C, )

as well as the isomorphic

lattices of projective units (P, ), projective faces (F, ✓)

is orthomodular.

Proof. We have already seen that (C, ) is a lattice. Let O, I denote the zero map and identity map on A respectively. Then O, I 2 C and im+ I = A+ ◆ 74

3 Spectral Theory for Ordered Spaces im+ P ◆ 0 = im+ O, 8 P 2 C. This implies O, I are the least and greatest elements in C respectively. Now, let (P, F, p), (Q, G, q) be any two compressions on A. (i) P 00 = P () (P e)00 = P e and this is true because (P e)00 = (e e (ii) Q P0

(e

P e)0 =

P e) = P e.

P =) q  p =)

q

p =) e

q

e

p =) q 0

p0 =)

Q0

(iii) P ? P 0 =) P ⇠ P 0 =) P ^ P 0 = P P 0 = O. Now, P _ P 0 = (P ^ P 0 )0 = O0 = 10 I. (iv) By 3.99, Q

˙ P =) P = Q+(P ^ Q0 ) = Q _ (P ^ Q0 ).

Definition 3.104. A Boolean algebra B is an orthocomplemented lattice satisfying the distributive law i.e. x ^ (y _ z) = (x ^ y) _ (x ^ z) and x _ (y ^ z) = (x _ y) ^ (x _ z), 8 x, y, z 2 B. Remark 3.105. Let L be an orthocomplemented lattice and x, y, z 2 L. Then 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. 10

ker+ 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. Because, if P, Q 2 B such that Q  P , then (P ^ Q0 ) _ Q = (P _ Q) ^ (Q0 _ Q) = P ^ I = P (using the distributive law). So, B satisfies the orthomodular condition. Definition 3.107. A non-empty subset of an orthocomplemented lattice is called a complemented sublattice if it is closed under complementation and closed under the lattice operations (^, _). Remark 3.108. If J is a sublattice of L, then P ^ P 0 = 0, P _ P 0 = I ( where P 2 J ) implies that 0, I 2 J . So, J has greatest and least elements, which are the same as the greatest and least elements of L. Hence, J itself is an orthocomplemented lattice. Theorem 3.109. Let < A, V > satisfy the standing hypothesis. Then, L ✓ C is a Boolean Algebra (under the lattice operations induced from C) () L is a complemented sublattice of C and each pair P, Q 2 L is compatible . Proof. Let P, Q, R 2 L with associated projective units p, q, r respectively. ()) Every boolean algebra is closed under complementation, join and meet. This implies L is a complemented sublattice of C. Now, for P, Q 2 L, using distributive law P = P ^ (Q _ Q0 ) = (P ^ Q) _ (P ^ Q0 ) =) P ⇠ Q by 3.96. (() Given L is a complemented sublattice of C under the lattice operations induced from C. Thus, L itself is an orthocomplemented lattice. To prove the distributive law, consider P, Q, R 2 L. Then, P, Q, R are mutually

76

3 Spectral Theory for Ordered Spaces compatible . P ^ (Q _ R) e = =

˙ ^ Q0 ) e (using 3.99) P (Q+R 0 ˙ P (Q+RQ ) e

(R ⇠ Q0 )

0 ˙ = P (Q+RQ )e

= P Qe + RQ0 e

( by 3.88 )

= P Qe + P RQ0 e So, we have p ^ (q _ r) = (p ^ q) + (p ^ q 0 ^ r) = (p ^ q) _ (p ^ q 0 ^ r) (by 3.92)  (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, L is a distributive orthocomplemented lattice, thus a boolean algebra.

Theorem 3.110. Assume < A, V > satisfy the standing hypothesis. Let (P0 , F0 , p0 ) be a compression on A. Then < P0 (A), P0⇤ (V ) > satisfy the standing hypothesis. Moreover, Compressions of P0 (a) = {P restricted to P0 | P 2 C, P

P0 },

Projective faces of P0 (a) = {F 2 F | F ✓ F0 } and Projective units of P0 (a) = {p 2 P | p  p0 }. Proof. We will first show that if P 2 C such that P

P0 , then P restricted to

P0 (A) is a compression on P0 (A). Let Pˆ denote the restriction of P to P0 (A). 77

3 Spectral Theory for Ordered Spaces P

P0 =) P, P 0 , P0 , P00 are mutually compatible . Now let x 2 P0 (A),

say x = P0 a for some a 2 A. Then Pˆ x0 = P P0 a = P0 P a 2 P0 (A). So, P0 (A) is 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 kxkP0 (A) = kxkA , 8 x 2 P0 (A), we have kPˆ k =

sup {kPˆ akP0 (A) | kakP0 (A)  1, a 2 P0 (A)}

=

sup {kP akA | kakA  1, a 2 P0 (A)}



sup {kP akA | kakA  1, a 2 A}

= kP k  1 Therefore, Pˆ is normalised. So, Pˆ is a weakly continous normalised positive projection on P0 (A). ˆ denote the restriction of Now, let Q = P0 P 0 = P 0 P0 = P0 ^ P 0  P0 and let Q ˆ is also a weakly continous normalised positive projection on Q to P0 (A). Then, Q P0 (A), like we showed for Pˆ . ˆ Claim 1: Pˆ 0 (the complement of Pˆ in P0 (A)) = Q Let x 2 P0 (A), x

0. Then

x 2 ker+ Pˆ () Pˆ x = 0 () P x = 0 () x 2 ker+ P = im+ P 0 () ˆ So, ker+ Pˆ = P 0 x = x () P 0 P0 x = x () Qx = x () x 2 im+ Q. ˆ im+ Q. Similarly, x 2 im+ Pˆ () Pˆ x = x () P x = x () x 2 im+ P = ˆ = 0 () x 2 ker+ Q ˆ ker+ P 0 () P 0 x = 0 () P 0 P0 x = 0 () Qx ˆ This shows that im+ Pˆ = ker+ Q. ˆ are complementary projections on P0 (A). So, Pˆ , Q Claim 2: Pˆ ⇤ = P ⇤ |P0⇤ (V ) = P ⇤ restricted to P0 (V ) First note that P P0 = P0 P =) P0⇤ P ⇤ = P ⇤ P0⇤ . From this it follows that P0⇤ (V ) is invariant under P ⇤ and so, P ⇤ is a projection on P0⇤ (V ). Now, let 78

3 Spectral Theory for Ordered Spaces x 2 P0 (A), y 2 P0⇤ (V ). Then < x, Pˆ ⇤ y > = < Pˆ x, y > = < P x, y > = < x, P ⇤ y >. This implies Pˆ ⇤ y = P ⇤ |P0⇤ (V ) y. Hence, the claim. ˆ ⇤ = (Pˆ ⇤ )0 Claim 3: Q ˆ ⇤ = Q⇤ |P ⇤ (V ) = (P0 P 0 )⇤ |P ⇤ (V ) = P 0⇤ P0⇤ |P ⇤ (V ) = P 0⇤ |P ⇤ (V ) . ThereLHS = (Q) 0 0 0 0 fore it is sufficient to prove (Pˆ ⇤ )0 = P 0⇤ |P0⇤ (V ) . Let x 2 P0⇤ (V ), x

0.

x 2 ker+ Pˆ ⇤ () Pˆ ⇤ x = 0 () P ⇤ x = 0 () x 2 ker+ P ⇤ = im+ (P ⇤ )0 = im+ (P 0 )⇤ . This implies ker+ Pˆ ⇤ = im+ P 0⇤ |P0⇤ (V ) . Conversely, if x 2 im+ (Pˆ )⇤ () Pˆ ⇤ x = x () P ⇤ x = x () x 2 im+ P ⇤ = ker+ (P 0 )⇤ = ker+ (P ⇤ )0 . This implies im+ Pˆ ⇤ = ker+ P 0⇤ |P0⇤ (V ) . ˆ ⇤ are complementary. Hence, Pˆ ⇤ , Q From the claims, we see that Pˆ is bicomplementary. Hence, Pˆ = P |P0 (A) is a compression on P0 (A).

Next, we will prove that < P0 (A), P0⇤ (V ) > satisfy the standing hypothesis. Let F be an exposed face of F0 . This implies 9 a 2 im+ P0 such that a = 0 on F and a > 0 on F0 \F . Define b = a + p00 . Then, on K, a, p00

0 =) b

0 on K.

On F0 , a > 0, p00 = 0 =) b > 0 on F0 . And on F , a = 0, p00 = 0 =) b = 0 on F . Therefore, F = {w 2 K | < b, w > = 0} = K \ b 1 (0). This implies F is an exposed face of A =) F is a projective face of V , say associated with a compression P . Now, F ✓ F0 =) P

P0 . Then, P restricted to P0 (A) is a

compression on P0 (A) with projective face F . Thus, every exposed face of F0 is a projective face of P0 (A). Hence, the standing hypothesis is satisfied.

Now, we will show that if (P, F, p) is some compression of P0 (A), then P = R|P0 (A) for some R 2 C such that R

P0 . 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 (P0 (A), p0 ). This implies F = {w 2 F0 |< p0 Claim: F = {w 2 K |< e (✓) Let w 2 F

p, w > = 0}.

p, w > = 0}

=) P0⇤ w = w and < p0 , w > = < p, w >. This implies

< P0 e, w > = < e, P0⇤ w > = < e, w > =) < p, w > = < e, w >. So, w 2 RHS. (◆) Let w 2 K and < e, w > = < p, w >. Then < p0 , w > = < p, w > () < p0 , w > = < e, w > () < e, P0⇤ w > = < e, w > () kP0⇤ wk = kwk () P0⇤ w = w () w 2 F0 . Now, we have p  p0  e (because p is a projective unit and p0 is the order unit in the space P0 (A)). Thus, < p0 , w >  < p, w >  < e, w > = < p, w >. Therefore, equality holds everywhere and we get < p0 , w > = < p, w > . This implies w 2 F . From the claim, it follows that F is an exposed face of K =) F is projective face of V R

=) 9 R 2 C with associated projective face F . But F ✓ F0 =)

P0 =) R|P0 (A) is a compression on P0 (A) with associated projective face F .

Now, P and R|P0 (A) are two compressions on P0 (A) with the same projective face. 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)) () P a  a, 8a 2 A+ (or Pˆ (A)+ ). (() If P is central for A, then P a  a for all a 2 Pˆ (A)+ ✓ A+ . Thus, P is central for Pˆ (A)+ . 80

3 Spectral Theory for Ordered Spaces ()) Assume P is central for Pˆ (A) =) P b  b, 8 b 2 im+ Pˆ . In particular, P (Pˆ a)  Pˆ a, 8 a 2 A+ . Now, P

Pˆ =) P ⇠ Pˆ and P = Pˆ P = P Pˆ .

Then, P a = P Pˆ a  Pˆ a  a, 8 a 2 A+ . Thus, P is central for A.

3.4.1

The lattice of compressions when A = V ⇤

In this section, we will continue to study < A, V > a pair of order unit space and base norm space under separating order and norm duality, satisfying the standing hypothesis, with the additional assumption that A = V ⇤ . In this case, A can be identified with Ab (K), the space of all bounded affine functions on K, by an order preserving isomorphism (a ! a ˆ). Remark 3.112. Under this mapping a 7! a ˆ, {an } ! a weakly in A () {ˆ an } ! a ˆ in Ab (K) pointwise Proof. {an } ! a weakly () limn!1 < an , v > = < a, v >, 8 v 2 V () limn!1 < an , k > = < a, k >, 8 k 2 K () limn!1 a ˆn (k) = a ˆ(k), 8 k 2 K () {ˆ an } ! a ˆ pointwise in Ab (K) Definition 3.113. A lattice L is called a complete lattice if every bounded subset of L has a greatest lower bound and least upper bound. ( In orthocomplemented lattice, every subset is bounded by 0 and I.) Definition 3.114. A lattice is said to be monotone complete if every bounded monotone net has a limit. Remark 3.115. Assume A = V ⇤ . Then A is monotone complete.

81

3 Spectral Theory for Ordered Spaces Proof. Let {a } be a bounded increasing net in A, say a  a0 , 8

2 ⇤. Fix

k 2 K. Then {< a , k >} is an increasing net of real numbers, bounded above by < a0 , k >. Since R is monotone complete, this net has a limit, say lim < a , k > = ↵(k), where ↵ is some real number dependent on k. Now, this map ↵ : K ! R is affine because lim < a , tk1 + (1 a , k2 > = t↵(k1 ) + (1

t)k2 > = lim t < a , k1 > + lim (1

t)
} = ↵(k) = < a, k > =) < a , k >  < a, k > , 8

2 ⇤ =) a  a, 8

Next, let b

a , 8

2 ⇤.

2 ⇤. Then < b, k >

sup < a , k > = < a, k >, 8 k 2 K =) b

< a , k >, 8 k 2 K =) < b, k > a.

Hence, sup{a } = a 2 A. Similarly, we can show that infrimum exists for a bounded decreasing sequence in A. So, A is monotone complete. Thus, every increasing net {a↵ } 2 A which is bounded above, has a least upper bound a 2 A, denoted as a↵ % a. In this case, a is also the weak limit of {a↵ }. Similarly, every decreasing net {a↵ } 2 A which is bounded below, has a greatest lower bound a 2 A, denoted as a↵ & a. In this case, a is also the weak limit of {a↵ }. Remark 3.116. If a↵ % a (or a↵ & a ) and P is a compression on A, then by weak continuity, P a↵ % P a. (or P a↵ & P a respectively). Proof. Note that a↵ % a () a↵

! a weakly. Since P is positive, P a↵ is

increasing net and by weak continuity and P a↵

! P a weakly. Hence, P a↵ %

P a. 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 2 A, then a 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 2 A. Let F↵ be the projective face corresponding to p↵ . Define G = {w 2 K | < a, w > = 0} = K \a 1 (0). Then, G is an exposed face of K and hence a projective face of V , by the standing hypothesis. Let (P, G0 , p) denote the compression complementary to G. Then F↵0 = {w 2 K | < p↵ , w > = 0} ✓

11

G, 8 ↵. This implies 8 ↵, G0 ✓

F↵ =) p  p↵ =) p  a. Now, p is the largest element in A+ 1 which vanishes on G. This implies p

a. Therefore, a = p 2 P.

Similarly, for an increasing net p↵ % a, consider the decreasing net (e (e

a). Then, (e

a) 2 P as above, which in turn implies that (e

p↵ ) &

a)0 = a 2 P.

Proposition 3.118. Assume < A, V > satisfy the standing hypothesis and A = V ⇤ . Then C, F, P are complete lattices and the lattice operations are given by the following equations for each family {F↵ } ✓ F: ^

F↵ =

↵

\ ↵

F↵ ,

_

F↵ =

↵

\

F↵0

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 ⇤ , every semi-exposed face of K 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. 11

0  < a, w >  < p↵ , w > = 0

83

3 Spectral Theory for Ordered Spaces Bicommutant Definition 3.121. Assume < A, V > satisfy the standing hypothesis and A = V ⇤ . Let a 2 A. Then the C -Bicommutant of a is defined as the set of all compressions on A, which are compatible with a as well as compatible with all compressions compatible with a. C-Bicommutant of a = {P 2 C | P ⇠ a, P ⇠ Q whenever Q ⇠ a} The corresponding set of projective units and projective faces is called P-Bicommutant of a and F- Bicommutant of a respectively. Proposition 3.122. Assume < A, V > satisfy the standing hypothesis and A = V ⇤ . Then the C -Bicommutant of a is a complete boolean algebra. Orthogonal decomposition in V Definition 3.123. Two elements ⇢, as ⇢ ? , if k⇢

2 V + are said to be orthogonal, denoted

k = k⇢k + k k.

Theorem 3.124. Assume < A, V > satisfy the standing hypothesis and A = V ⇤ . Then every w 2 V admits an orthogonal decomposition w = ⇢ 0, ⇢ ? . This decomposition is unique and is given by ⇢ = P ⇤ w,

, where ⇢, = 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 2 V is called central if (P ⇤ + P 0⇤ )w = w for all compressions P . Lemma 3.126. Assume < A, V > satisfy the standing hypothesis and A = V ⇤ . Let w be a central element of V and w = ⇢ Then ⇢,

be its unique orthogonal decomposition.

are also central.

84

3 Spectral Theory for Ordered Spaces Definition 3.127. An ordered vector space (V, ) which is a lattice under this order  (i.e. closed under ^, _) is called a vector lattice. Remark 3.128. Let V be a vector lattice. Then the following holds 8, u, v, w 2 V . (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 A+ 1 . The following is a partial converse to this result, under our additional assumptions. 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 A+ 1. Proposition 3.131. Assume < A, V > satisfy the standing hypothesis and A = V ⇤ . If p 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 = pˆ), where pˆ is the only point of K where p takes the value 1. And the map p 7! pˆ is a one-one map from the set of minimal points of P onto the set of exposed points of K.

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 2 K, there is a least central projection c(w) such that < c(w), w > = 1; the associated compression is the least central compression P such that P ⇤ w = w; the associated projective face is the smallest split face which contains w. Definition 3.133. Assume < A, V > satisfy the standing hypothesis and A = V ⇤ . For each w 2 K, the least central projective unit which takes the value 1 at w is called the central support or central carrier of w. It is denoted by c(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 2 A+ are orthogonal (a ? b) if there exists a compression (P, F, p) such that P a = a, P 0 b = b. Definition 3.135. An equality a = b

c is called an orthogonal decomposition of

an element of a 2 A if b, c 2 A+ and b ? c. Lemma 3.136. Assume < A, V > satisfy the standing hypothesis and A = V ⇤ . Let b, c 2 A+ and (P, F, p) be a compression on A. Then the following are equivalent conditions for orthogonality of b and c. (i) P b = b and P 0 c = c (ii) P 0 b = 0 and P c = 0 (iii) b = 0 on F 0 and c = 0 on F . Remark 3.137. Let a 2 A+ and P 2 C. Then (i) P a ? P 0 a (ii) If a = b

c is an orthogonal decomposition of a such that P b = b, P 0 c = c,

then P a = b, P 0 a = c. 86

3 Spectral Theory for Ordered Spaces Hence, a = P a + P 0 a 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 2 A+ , there exists a least projective unit p such that a 2 f ace(p). Further, p is the unique element of P such that {w 2 K | < a, w > = 0} = {w 2 K | < p, w > = 0} Proof. If a = 0, then we can take p = 0 =) a 2 face(p). And clearly, 0 will be the least such projective unit. So, assume a 6= 0. Let a1 =

a kak

2 A+ 1 . Let

G = K \ a 1 (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 2 K |< p, w > = 0} and p is the greatest element in A+ 1 such that p = 0 on G. Thus, a1  p =) a  kakp =) a 2 face(p). To prove that p is the least such projective unit, assume q 2 P such that a  q for some

0. Then, F 0 = K \ q 1 (0) ✓ K \ a 1 (0) = G. This implies

G0 ✓ F =) p  q. Hence, p is the least projective unit such that a 2 face(p). As projective units are uniquely determined by projective faces (p

! G0 ), p is

the unique element in P such that K \ a 1 (0) = G = {w 2 K | < p, w >= 0}. Definition 3.139. Assume < A, V > satisfy the standing hypothesis and A = V ⇤ . For each a 2 A+ , define r(a) to be the least projective unit p in A such that a 2 face(p). We call r(a) to be the range projection of a. 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 2 A+ , then r(a) is characterised by each of the following statements: 1. r(a) is the only projective unit such that < a, w > = 0 () < r(a), w > = 0, where w 2 K. 2. r(a) is the least projective unit p such that a  kakp. 3. The compression associated with r(a) is the least compression P such that P a = a. 4. The complement of the compression associated with r(a) is the greatest compression P 0 such that P 0 a = 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, proposition 3.138, and the fact that a 2 face r(a)

() a 2 im+ P (prop. ??).

To show (5), recall that the semi-exposed face generated by a in A+ = {a}•• . Now, for any s 2 A+ , s 2 {a}•• () < s, w > = 0,

8w 2 {a}•

(3.5.1)

Let v 2 V + . Then < a, v > = 0 () < r(a), v > = 0. So, < r(a), u > = 0, 8u 2 {a}• . Hence, by eqn. 3.5.1, r(a) 2 {a}•• . Thus, r(a) belongs to the semi exposed face generated by a in A+ . Let q be a projective unit with associated projective face H and let G be the projective face associated with r(a). Then q 2 {a}•• =) {w 2 K |< a, w > = 0} ✓ {w 2 K |< q, w > = 0} (by 3.5.1) =) G0 ✓ H 0 =) r(a)0  q 0 =) q  r(a). Thus, r(a) is the greatest projective unit contained in the semi-exposed face of A+ , generated by a. 88

3 Spectral Theory for Ordered Spaces Remark 3.141. If a, b 2 A+ such that a  b, then r(a)  r(b). This is because a  b  kbkr(b) =) a 2 face r(b) . Hence, r(a)  r(b). Proposition 3.142. Assume < A, V > satisfy the standing hypothesis and A = V ⇤ . Let a 2 A+ and P be a compression on A with projective unit p. Then r(P a) = r(a) _ p0 ^ p Proof. Let Q, R be the compressions associated with r(a) and r(P a) respectively. Now, a 2 face(p) =) a 2 im+ Q ✓ im+ (Q _ P 0 ). Thus, (Q _ P 0 )a = a. Also, P 0  Q _ P 0 =) P, P 0 ⇠ (Q _ P 0 ). Hence, (Q _ P 0 )(P a) = P (Q _ P 0 )a = P a. So, (Q _ P 0 ) fixes P a. By proposition 3.140 (3), R  (Q _ P 0 ). So, r(P a)  r(a) _ p0

(3.5.2)

Similarly, P fixes P a =) R  P (proposition 3.140 (3)). Therefore, r(P a)  p

(3.5.3)

r(P a)  r(a) _ p0 ^ p

(3.5.4)

From 3.5.2, 3.5.3, we have

Next, R  P =) R0 ? P . Thus, (R0 ^P )a = R0 P a = 0. Therefore, a 2 ker+ (R0 ^ P ) = im+ (R _ P 0 ). Hence, R _ P 0 fixes a =) Q  R _ P 0 . So, r(a)  r(P a) _ p0 . This implies r(a) _ p0  r(P a) _ p0 . Also, note that r(P a) _ p0 = r(P a) + p0 (as R ? P 0 ). Therefore, r(a) _ p0

p0  r(P a). Thus, by proposition ??, r(a) _ p0 ^ p  r(P a)

(3.5.5)

Hence, by eqn. 3.5.3 and 3.5.5, we have r(P a) = r(a) _ p0 ^ p.

Proposition 3.143. Assume < A, V > satisfy the standing hypothesis and A = V ⇤ . Let a, b 2 A+ . Then 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) and r(b) respectively. Claim: r(a) ? r(b) () a ? b ()) Assume r(a) ? r(b). Then Rb Ra = 0. This implies Rb a = Rb (Ra a) = 0 =) Rb0 a = a. And Rb b = b ( proposition 3.140 (3)). Thus, a ? b. (() Let a ? b. This implies 9 P 2 P such that P a = a,

P 0 b = b. By

proposition 3.140 (3), Ra  P and Rb  P 0 . Hence, by lemma ??, Ra ? Rb . 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  kakr(a) and b  kbkr(b). Hence, a + b  kakr(a) + kbkr(b) 

12

ka +

bk r(a) + r(b) . Thus by proposition 3.140 (2), 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, by proposition ??, 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 2 A+ and Q be a compression compatible with a. Then r(Q(a)) = Q(r(a)) 12

0  a, b  a + b =) kak, kbk  ka + bk

90

3 Spectral Theory for Ordered Spaces Proof. Qa is orthogonal to Q0 a. So, by proposition 4.1.6, r(Qa) + r(Q0 a) = r(Qa + Q0 a) = r(a) Let q be the projective unit associated with Q. Then, Qa  kakQe =) Q(a) 2 face(q). Hence, r(Qa)  q. So, r(Qa) 2 im+ Q (prop ??). Similarly, r(Q0 a) 2 im+ Q0 = ker+ Q. Thus, Q(r(a)) = Q r(Qa) + r(Q0 a) = Q(r(Qa)) + 0 = r(Qa). Hence, the result. Proposition 3.145. Assume < A, V > satisfy the standing hypothesis and A = V ⇤ . If a 2 A+ , then r(a) is in the P-bicommutant of a. Proof. Let P be the compression associated with r(a). Then P a = a and P 0 a = 0 (prop. 3.140). Thus, P a + P 0 a = 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) by prop ??. Therefore, r(a) is in P bicommutant 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, in addition, each a 2 A admits a least compression P such that P a

a and P a

0, then we say

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 2 A and P be a compression. Then Pa

a () P is compatible with a and P 0 a  0

Proof. First, assume that P ⇠ a and P 0 a  0. Then, a = P a + P 0 a =) P a = a

P 0a

a. 91

3 Spectral Theory for Ordered Spaces Next, assume P a Pa

0. Thus, P (P a a) = P 2 a P a = 0 =)

a. Then P a a

a 2 ker+ P = im+ P 0 . So, P a

a = P 0 (P a

a) =

P 0 a =)

P 0a

0 and

a = P a + P 0 a. Hence, P 0 a  0 and P ⇠ a. Lemma 3.148. Let A, V be an order unit space and base norm space in separating order and norm duality. Let a 2 A and P be a compression wth associated projective face F . Then the following are equivalent: (i) P a

a, P a

0, P 0 a  0.

(ii) P ⇠ a, P a (iii) F ⇠ a, a

0.

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, P a 0 8w 2 K () a

0 ()

hP a, wi

0, 8w 2 K ()

ha, P ⇤ wi

0 on F . Similarly, P 0 a  0 () a  0 on F 0 .

So, A, V are in spectral duality () for each a 2 A, there exists a least compression P compatible with a such that either (i) or (ii) holds, or there exists a least projective face F satisfying (iii). Notation 14. let b 2 A+ 1 . Then we denote [0, b] = {a 2 A | 0  a  b} and face (p) = {a 2 A | 0  a  b, for some

0}.

Definition 3.149. Let A, V be in spectral duality. An element b of A+ 1 is said to have facial property if [0, b] = A+ 1 \ face (b). Remark 3.150. Every projective unit satisfies the facial property. This is because, for a projective unit p, a  kakp () a 2 face (p) (by prop ??). So, {a 2 A | a  kakp} \ A+ = {a 2 face (p)} \ A+ This implies face (p) \ A+ 1 = {a 2 A | 0  a  p} = [0, p] 92

3 Spectral Theory for Ordered Spaces Proposition 3.151. Let A, V be in spectral duality. If b 2 A+ 1 has facial property, then b is a projective unit. Proof. Define a = 2b

e. By spectral duality, 9 P 2 P such that P a

0. By lemma 3.147, P ⇠ a, P 0 a  0. Then, b =

Pa 1 (P e 2

a+e 2

a and

= 12 (P a + P 0 a) +

+ P 0 e) = P ( a+e ) + P 0 ( a+e ) = P b + P 0 b. Thus, 2 2 P ⇠ b and b

Let p = P e and p0 = P 0 e. Now, P (2b 4.3.12, we get 2b

2P b

Pb

(3.6.1)

e) = P a

0 =) 2P b

p. By eqn

p =) p 2 face (b). Hence, by facial property of b, we

get pb Note that 0  b  e =)

e  a  e. Now, a  P a =) 2b

P a  P e = p =) 2b p  e. Also, 2b p = b + (b p) p  2b =) 2b

Next, 2b

p 2 face (b). So, 2b

facial property of b, we get 2b

e  P (2b

e) =

0. Thus, 0  2b p  e.

p 2 face (b) \ A+ 1 . Hence, by

p  b which implies b  p. Therefore, b = p.

Lemma 3.152. Let A, V be in spectral duality. If P, Q are compatible compressions, then P Q = QP = P ^ Q Proof. Define r = P Qe. Then, we have r  p, r  q. Note that p, q p^q

0 =)

0. Now, if s 2 A+ such that s  p, s  q, then s 2 im+ P, im+ Q (by

prop. ??). This implies s = P s  P q = r. So, r r is the greatest lower bound of (p, q) in A+

13

s whenever 0  s  p, q. Thus,

.

Claim. r has facial property. 13

Without 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 2 A+ 1 \ face (r). Since r  p, we have a 2 face (p). This implies a 2 im+ P =) a = P a  P e = p. Similarly, a  q. Thus, a  p ^ q = r. So, a 2 [0, r]. Hence, the claim. Now, by proposition 3.151, r is a projective unit. And as r = p ^ q, we see that precisely r = (P ^ Q)e. Now, we will show that P Q = P ^ Q

14

. Let

+ a 2 A+ 1 . Then P Qa  P Qe = r = (P ^ Q)e. So, P Qa 2 im (P ^ Q) ✓

im+ P, im+ Q (by prop ??). Also, P, Q, P ^ Q are mutually compatible. Hence, P Qa = (P ^ Q)(P Qa) = P (P ^ Q)(Qa) = P Q(P ^ Q)a = (P ^ Q)a. Thus, P Q = P ^ Q on A+ 1 =) P Q = P ^ Q.

Currently, C, P, F are order isomorphic sets. We don’t know if they are lattices. However, under spectral duality, we have shown that if P ⇠ Q, then P ^ Q exists and P ^ Q = P Q. Further, P ⇠ Q ⇠ P 0 ⇠ Q0

=) P 0 ^ Q0 = P 0 Q0 . 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 2 A. Then, P ^ Q, P _ Q are also compatible with a. Proof. First assume a 2 A+ . P, Q ⇠ a =) P a, Qa  a. So, by lemma 3.152, (P ^ Q)a = (P Q)a = P (Qa)  P a  a =) P ^ Q ⇠ a. Next, P ⇠ Q =) P 0 ⇠ Q0 . Also, P, Q ⇠ a =) P 0 , Q0 ⇠ a. Then as above, P 0 ^ Q0 ⇠ a which further implies (P 0 ^ Q0 )0 = P _ Q is compatible with a. Now consider any a 2 A compatible with both P, Q. We have a = a1 a2 where a1 = 21 (kake + a), a2 = 12 (kake

a) 2 A+ . Then P a1 + P 0 a1 = 12 (kak(p + p0 ) + a) =

14

P Qe = r = (P ^ Q)e does not imply P Q = P ^ Q because we do not know if P Q is a compression.

94

3 Spectral Theory for Ordered Spaces 1 (kake 2

+ a) = a1

=)

P ⇠ a1 . Similarly, it can shown that P ⇠ a2 , Q ⇠

a1 and Q ⇠ a2 . Then P ^ Q, P _ Q ⇠ a1 , a2 =) P ^ Q, P _ Q ⇠ a (by ??). Proposition 3.154. Let A, V be in spectral duality. Let P, Q be orthogonal compressions, both compatible with an element a 2 A. Then P _ Q is compatible with a and (P _ Q)a = P a + Qa. Proof. P ? Q =) P ⇠ Q and we are given P, Q ⇠ a. So, from lemma 3.153, we have P _ Q ⇠ a. Then, a = (P _ Q)a + (P _ Q)0 a = (P _ Q)a + (P 0 ^ Q0 )a = (P _ Q)a + (P 0 Q0 )a. Now, P, Q ⇠ a =) a = Qa + Q0 a, a = P a + P 0 a. So, P 0 Q0 a = P 0 (a

Qa) = P 0 a

P 0 Qa = 15 P 0 a

this above, we get a = (P _ Q)a + (a

Pa

Qa = (a

P a)

Qa. Substituting

Qa) =) (P _ Q)a = P a + Qa.

Remark 3.155. We had seen that if A = V ⇤ , then A 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 {pn } is a decreasing sequence of projective units, then the element inf n pn of A is a projective unit, hence it is the greatest lower bound of {pn } among the projective units. Similarly, if {pn } is an increasing sequence of projective units, then the element supn pn of A is a projective unit, and is the least upper bound of {pn } among the projective units. Proof. As A is monontone complete and the decreasing sequence of projective units {pn } is bounded below by 0, there exists b 2 A such that b = inf n pn . As, 0  pn  e, 8n, we have b 2 A+ 1 . We will prove that b has facial property. Let a 2 + A+ 1 \ face (b). Now, b  pn =) face (b) ✓ face (pn ). So, a 2 A1 \ face (pn ) =

[0, pn ]. This implies a  pn , 8n. Thus, a  inf n pn = b =) a 2 [0, b]. Now, by proposition 3.151, b is a projective unit and it is the greatest lower bound of {pn } in P. 15

P ? Q =) Q  P 0 =) im+ Q ✓ im+ P 0 =) P 0 Q = Q

95

3 Spectral Theory for Ordered Spaces For an increasing sequence of projective units {qn }, look at the corresponding decreasing sequence of complement projective units {qn0 }. Then {qn0 }

! r for

some projective unit r 2 A (as in the above case) which implies {qn } ! r0 , which is also a projective unit. Lemma 3.157. Assume A, V are in spectral duality. Let a 2 A+ and let such that 0


2

>

3

. . . ! 0, i.e. limn

n

= 0. By lemma 3.157, we can construct

Q0 ⌫ Q1 ⌫ Q2 ⌫ Q3 ⌫ . . . ⌫ Qn . . . such that Qi ⇠ a, Qi a 

i qi ,

Q0i a

0 i qi ,

for i = 1, 2, 3 . . .

where the notation for the compressions are (Qn , Gn , qn ), for each n. By lemma 3.156, {qn } & q , for some projective unit q 2 A. Let Q be the compression and G be the projective face associated with the projective q. Then, q = inf qn , G = n

\ n

97

Gn Q =

^ n

Qn

3 Spectral Theory for Ordered Spaces For a given i, have

i qi

i hQi e, ki

Qi a =) h i qi hQi a, ki =)

As G ✓ Gn , 8n, we have

i he, f i

ne

0, 8k 2 K. Thus, 8k 2 K, we

Qi a, ki

ha, f i, 8f 2 Gi =)

ie

a on G, 8n. Thus, 0 = limn

G =) a  0 on G. But a 2 A+ =) a

a on Gi . ne

a on

0 on G. Hence, a = 0 on G. Therefore,

G ✓ F. Conversely, if w 2 F , then 0 

0 n hqn , wi

 hQ0n a, wi  ha, wi = 0. This

implies hqn0 , wi = 0, 8n. Now, qn & q =) qn0 % q 0 =) hq 0 , wi = 0 =) w 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 hypothesis, are now valid under the spectral duality condition. In particular, we note that C, P and F are complete lattices, when A, V are in spectral duality. Now we will look at some equivalent ways of defining spectral duality. Lemma 3.159. Assume hA, V i are in separating order and norm duality such that they satisfy the standing hypothesis and A = V ⇤ . Let a 2 A and (P, F, p) be a compression on A such that P ⇠ a, a

0 on F, a  0 on F 0

Then G = {w 2 K | hP a, wi = 0} is a projective face and its complementary compression (P˜ , F˜ , p˜) satisfies F˜ ✓ F, P˜ a = P a and P˜ ⇠ a, a > 0 on F˜ , a  0 on F˜ 0 Proof. First note that, a

0 on F =) P a

0;

a  0 on F 0 =) P 0 a  0

98

3 Spectral Theory for Ordered Spaces Define G = {w 2 K | hP a, wi = 0}. Then G is an exposed face of K and hence projective, by theorem 3.158. By proposition ??, P a = 0 on F 0 which implies F 0 ✓ G =) G0 ✓ F . If w 2 G, then ha, wi = hP a, wi + hP 0 a, wi = 0 + hP 0 a, wi  0 If w 2 G0 ✓ F , then ha, wi = hP a, wi + hP 0 a, wi = hP a, wi + 0

0 (P 0 a = 0 on F )

And hP a, wi = 0 =) w 2 G. But, G \ G0 = ;. Thus, hP a, wi > 0 on G0 . Hence, a  0 on G and a > 0 on G0 . Let Q be the compression associated with the projective face G. Then, by proposition ??, P a = 0 on G =) Q0 (P a) = P a, Q(P a) = 0 P 0 a = 0 on G0 ✓ F =) Q(P 0 a) = P 0 a, Q0 (P 0 a) = 0 Therefore, Qa + Q0 a = Q(P a + P 0 a) + Q0 (P a + P 0 a) = P 0 a + P a = a =) Q ⇠ a, Q0 ⇠ a. Now, let (P˜ , F˜ , p˜) = (Q0 , G0 , Q0 e). Then, we have F˜ = G0 ✓ F and P˜

P.

Also, P˜ a = P˜ (P a + P 0 a) = Q0 P a + Q0 P 0 a = P a + 0 = P a.

Lemma 3.160. Assume that hA, V i are a pair of order unit space and base norm 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 2 A, there exists a unique projective face F compatible with a 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

0 on F, a  0 on F 0 . Then, by lemma 3.159, there

exists projective face F˜ such that F˜ ✓ F, a ⇠ F˜ , a > 0 on F˜ , a  0 on F˜ 0 . Note that F ✓ F˜ , by minimality of F . Hence, F = F˜ . Next, we will show uniqueness of F˜ . Suppose there exists a projective face F1 such that a ⇠ F1 , a

0 on F1 , a  0 on F10 . Again, by minimality of F ,

we get F ✓ F1 . By orthomodular lattice structure of F under spectral duality, F ✓ F1 =) F1 = F _ (F1 ^ F 0 ). Now, a > 0 on F1 and a  0 on F 0 =) F1 \ F 0 = ; =) F1 ^ F 0 = 0. Thus, F1 = F _ 0 = F . Hence F is the unique projective face satisfying the required properties.

(2) =) (1) We will first show that when (2) holds, the standing hypothesis is satisfied. Fix a 2 A+ and let G = {w 2 K | ha, wi = 0}. For this a, 9 ! F 2 F such that a ⇠ F, a > 0 on F, a  0 on F 0 . We will show that G0 = F . a 2 A+ =) a

0 on K ◆ F 0 . Thus, a = 0 on F 0 . This implies F 0 ✓ G.

Next, let w 2 G and P be the compression associated with F . Then, ha, wi = hP a + P 0 a, wi = ha, P ⇤ wi + ha, P 0⇤ wi. Now, a = 0 on G ◆ F 0 =) ha, wi = 0, ha, P 0⇤ wi = 0. Thus, ha, P ⇤ wi = 0. But a > 0 on F . This implies P ⇤ w = 0 =) w 2 ker+ P ⇤ = im+ (P ⇤ )0 , i.e. w 2 F 0 =) G ✓ F 0 . 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 2 A, the projective face F satisfying (2) is the least projective face 100

3 Spectral Theory for Ordered Spaces in K such that a ⇠ F, a

0 on F, a  0 on F 0 .

Suppose there exists a projective face F1 such that a ⇠ F1 , a 0 on F˜10 . Then, by lemma 3.159, 9 F1 2 F such that F˜ ✓ F1 ,

0 on F1 , a  a ⇠ F˜ ,

a >

0 on F˜ , a  0 on F˜ 0 . By uniquness of F mentioned in (2), we get F = F˜ ✓ F1 . Hence, the claim. So, we have shown that for each a 2 A, there is a least projective face F in K such that a ⇠ F, a

0 on F, a  0 on F 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 2 A has a unique orthogonal decomposition. Proof. ()) Assume A, V are in spectral duality and let a 2 A. By spectral duality (lemma 3.160), 9 ! F 2 F such that a ⇠ F, a > 0 on F, a  0 on F 0 . If P is the compression associated with F , then we have a ⇠ P, P a Then a = b

c, with b = P a

0, c =

P 0a

0, P 0 a  0.

0, is an orthogonal

decomposition of a. Suppose a = b1

c1 is another decomposition of a, with b1 = P1 a, c1 =

P10 a for some compression (P1 , F1 , p1 ) compatible with a (prop. ??). Then, P1 a

0 =) a

0 on F1 ; P10 a  0 =) a  0 on F1 . Now, by lemma

3.159, 9 F˜ 2 F such that F˜ ✓ F, P˜ a = P1 a and P˜ ⇠ a, a > 0 on F˜ , a  0 on F˜ 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 = P1 a = P˜ a = P a = b and c1 = a b1 = a b = c. Hence, a = b c is the unique orthogonal decomposition of a. 101

3 Spectral Theory for Ordered Spaces (() Given a 2 A, let a = b with b = P a, c = Pa

0 =) a

c be the unique orthogonal decomposition of a,

P 0 a, a ⇠ P for some compression (P, F, p) on A. Now, 0 on F and P 0 a  0 =) a  0 on F 0 . By lemma 3.159,

9 F˜ ✓ F such that a ⇠ F˜ , a > 0 on F˜ , a  0 on F˜ 0 , P˜ a = P a and F˜ = G0 where G = {w 2 K | hP a, wi = 0}. Claim. F˜ is the least projective face satisfying a ⇠ F, a

0 on F˜ , a 

0 on F˜ 0 Suppose (P1 , F1 , p1 ) is a compression on A such that a ⇠ F1 , a 0 on F10 . Then b1 = P1 a

0, c1 =

P10 a

0 on F1 , a 

0 and a = b1 c1 is an orthogonal

decomposition of a. By assumption, a has a unique orthogonal decomposition. Thus, b1 = P1 a = P a and c1 =

P10 a =

P 0 a . By lemma ??,

b1 = 0 on F10 . Thus, F10 ✓ G = {w 2 K | hP1 a = P a, wi = 0}. This implies F˜ = G0 ✓ F1 . Hence, the claim. So, for each a 2 A, we have found a least projective face F such that a ⇠ F, a

0 on F˜ , a  0 on F˜ 0 . Hence, A, V are in spectral duality.

Definition 3.162. Assume A, V are in spectral duality. For each a 2 A, we will 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+ = P a and a =

P 0 a where P is the compression associated

with the unique projective face F satisfying a ⇠ F, a > 0 on F, a  0 on F 0 . Proposition 3.164. Assume A, V are in spectral duality. Let a 2 A and r(a+ ) be 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 P a

a and P a

0.

Proof. Given a 2 A, by spectral duality, there exists unique compression (P, F, p) such that a ⇠ F, a > 0 on F, a  0 on F 0 . By theorem 3.161, the unique orthogonal decomposition of a = a+

a is given by a+ = P a and a =

P 0 a.

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 2 K | hP a, wi = 0} = {w 2 K | ha+ , wi = 0}. But, range projections have the unique property that {w 2 K | ha+ , wi = 0} = {w 2 K | hr(a+ ), wi = 0} = H 0 . Hence, F 0 = H 0 =) F = H. Now, by lemma 3.160, the properties (1), (2) and (3) follow. Definition 3.165. Fix

2 R, a 2 A. Define r (a) = r (a

e)+ . The

corresponding compression is denoted by (R , F , r (a)). Remark 3.166. By proposition 3.164, we have r (a) ⇠ (a

e) and so r (a) ⇠ a.

Corollary 3.167. R , F , r (a) is the unique compression compatible with a such that a > e on 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)

Lemma 3.168. Assume A, V are in spectral duality and let a 2 A,

(3.6.9) 2 R. If F, G

are two projective faces compatible with a such that F ? G and a > e on F [ G, then a > e on F _ G. 103

3 Spectral Theory for Ordered Spaces Proof. We are given (P, F, p) ? (Q, G, q). So, p _ q = p + q. Take w 2 F _ G. Then hp, wi + hq, wi = hp _ q, wi = 1. Let µ = hp, wi and 1

µ = hq, wi.

µ = 1 =) hp, wi = 1 =) w 2 F ✓ F [ G. Therefore, ha, wi > . µ = 0 =) hq, wi = 1 =) w 2 G ✓ F [ G. Therefore, ha, wi > . Now, assume 0 < µ < 1. Define ⇢ = µ 1 P ⇤ w 2 F ; µ = hp, wi = he, P ⇤ wi

=)

= (1

µ) 1 Q⇤ w 2 G.

1 = he, µ 1 P ⇤ wi. Then he, ⇢i = 1. Similarly,

he, i = 1. Now, w 2 F _ G =) (P _ Q)⇤ w = w. Thus, ha, wi = ha, (P _ Q)⇤ wi = h(P _ Q)a, wi = hP a + Qa, wi = hP a, wi + hQa, wi = ha, P ⇤ wi + ha, Q⇤ wi = µha, ⇢i + (1

µ)ha, i > µhe, ⇢i + (1

µ)he, i = (µ + (1

µ)) = .

Hence, a > e on each w 2 F _ G. Lemma 3.169. Assume A, V are in spectral duality. If a 2 A+ and

> 0, then

r (a)  r(a). Proof. We know that r (a) ⇠ a and r(a) 2 P bicommutant of a. Thus, r (a) ⇠ 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 2 F 0 , ha, wi = hr(a), wi = 0. For each w 2 F , ha, wi > he, wi =

> 0.

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 2 A+ and

> 0.

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) and r (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  kbkr(b) 104

=)

0  R b 

3 Spectral Theory for Ordered Spaces kbkR (r(b)) = 0. Thus, R b = 0  b =) F ⇠ b. Hence, F ⇠ (a + b). Similarly, G ⇠ (a + b). As, F ? G and both are compatible with a + b, by lemma 3.154, we get F _ G ⇠ (a + b) By (3.6.8), we have a > e on F and b > e on G . Thus, a + b >

on F [ G .

Hence, by lemma 3.168, a + b > e on F _ G

(3.6.10)

Claim. a  r(a) on F 0 and b  r(b) on G0 r (a)  r(a) =) R0 ⇠ R , R0 . So, R0 a = R0 R0 a  R0 r0 (a) = R0 R0 (e) = R0 R0 (e) =

R0 r(a). Therefore, R0 ( r(a)

a)

0 =) a 

r(a) on F 0 .

Similarly, b  r(b) on G0 . Hence, the claim. Now, r(a) ? r(b) =) r(a) + r(b)  e. So, on F 0 \ G0 = (F _ G )0 , we have 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) must be the projective unit of F _ 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 2 A+ and (Q, G, q) is a compression compatible with a, then for each

> 0,

r (Qa) = Q(r (a)) Proof. As Qa ? Q0 a, we have r (Qa) + r (Q0 a) = r (Qa + Q0 a) = 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) 2 face (q). Then, by lemma ??, r (Qa) 2 im+ Q. Similarly, r (Q0 a) 2 im+ Q0 = ker+ Q. Hence, Q(r (a)) = Q r (Qa) + r (Q0 a) = r (Qa).

We know that if a

0 and

= 0, then r (a) = r(a) and hence belongs to

(P )-bicommutant of a. The following result shows that this is true for arbitary a 2 A and

2 R.

Proposition 3.172. Let A, V be in spectral duality. If a 2 A, then r (a) 2 2 R.

P bicommutant of a, for all Proof. Fix

2 R. By remark 4.2.23, r (a) ⇠ a. Let Q be a compression compati-

ble with a. First, assume that a 2 A+ . Case. ka

0

By lemma 3.170, 3.171, Q(r (a)) + Q0 (r (a)) = r (Qa) + r (Q0 a) = r (Qa + Q0 a) = r (a). Thus, r (a) ⇠ Q. So, we have proved the result for a b = a + kake 2 A+ and µ = rµ (b) = r((µ

be)+ ) = r((a

0. Now, let a 2 A be arbitary. Define

+ kak. Then Q ⇠ b and Q ⇠ rµ (b) (like a

0). But

e)+ ) = r (a). Hence, Q ⇠ r (a).

Corollary 3.173. Assume A, V are in spectral duality. Let a 2 A, be a compression compatible with a such that a 106

2 R and F

µe on F , where µ > . Then,

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 2 A and

3.7

< µ, then rµ (a)  r (a).

Spectral Theorem

We are now ready to prove our spectral theorem, which generalises the corresponding theorem for von Neumann algebras and JBW algebras. Theorem 3.175. Assume A, V are in spectral duality and let a 2 A. Then, there is a unique family {e }

2R

of projective units with associated compressions

(U , G , e ) on A such that 2R

(i) e is compatible with a, for each (ii) U a  e , (iii) e = 0 for (iv) e  eµ for (v)

V

µ>

e0 , 8 2 R

U0 a
kak

kak, then a

(a

kake  a =) a

e = (a

kake)

(

kak)e  0

(

kak)e  0. So,

e)+ = 0 =) r (a) = 0 =) e = e. < µ. Then by result 3.173, Fµ ✓ F

(v) We will show

V

µ>

Fµ0 = F 0 , for each

=) F 0 ✓ Fµ0 =) e  eµ .

2 R. Fix

2 R and µ0 > kak. Then,

by (iv), {Fµ0 }µ> µ=µ0 is a bounded decreasing net of projective faces in K. By V lemma 3.156, this net converges to some projective face G = µ> Fµ0 = T 0 0 0 8µ > , we have µ> Fµ . By proposition ??, G ⇠ a. As Fµ ◆ F , G ◆ F 0 =) G0 ✓ F . This implies

a > e on G0 Let w 2 G ✓ Fµ0 , 8µ >

=) ha, wi  µ, 8µ >

=) ha, wi  . This

implies a  e on G Thus, by result 3.167, we have G = F 0 . Hence, Uniqueness: Let {f }

2R

V

µ>

Fµ0 = F 0 .

be arbitrary projective units, with associated compressions (T , H , F ),

satisfying (i), (ii), (iii), (iv) and (v). We will show that H = F 0 , 8 where 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 So, R

r (a); R0 a  r0 (a)

T 0 =) F ✓ H 0 =) F 0 ✓ H . 108

3 Spectral Theory for Ordered Spaces

a

V

Hµ . Tµ0 a = a on Hµ0 and Tµ0 a

µfµ0 = µTµ0 e. Thus,

µe on Hµ0 . By lemma 3.173, Hµ0 ✓ F , 8µ > T =) µ> Hµ = H ✓ F 0 .

=) Hµ ✓ F 0 , 8µ >

Next, H =

µ>

So, H = F 0 , 8 .

Definition 3.176. If A, V are in spectral duality and a 2 A, then the unique family {e }

2R

of projective units, given in theorem 3.175, is called the spectral

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 2 A. Let {e } = { 0,

be the spectral resolution of a. For each real increasing finite sequence

2R 1

...

kak and n > kak, define k k = max1in ( i i 1 ). Then the RieP mann sums s := ni=1 i 1 (e i e i 1 ), converge in norm to a when k k ! 0, with

0


0, choose finite increasing sequence 0


kak and

i

i 1

= { 0,

1

...

n}

with

< ✏, 8i 2 {1, 2 . . . n}. Let U be the projective

face associated with e . By theorem 3.175 (iv), we know that e  eµ whenever < µ. So, for

< µ, we have e ? e0µ =) eµ

e = eµ ^ e0 and (Uµ ^ U 0 )a = Uµ a

U 0 a.

Note that e  eµ =) U , U 0 , Uµ , Uµ0 are mutually compatible compressions. Now, Uµ a  µeµ

=) U 0 (Uµ a)  µU 0 eµ

(U 0 ^ Uµ )a  µ(e0 ^ eµ ). Similarly, U 0 a (Uµ ^ U 0 )a

Uµ U 0 e =) (Uµ ^ U 0 )a

=) (U 0 ^ Uµ )a  µU 0 Uµ e =) e0

=) Uµ U 0 a

Uµ e0

=)

(eµ ^ e0 ).

Therefore, (eµ

e )  (Uµ ^ U 0 )a  µ(eµ 109

e )

(3.7.1)

n}

3 Spectral Theory for Ordered Spaces Let pi = e

e

i

i 1

, for i = 1, 2 . . . n. Then Pi := U i ^ U

i 1

is compatible with a

and satisfies i 1 (e

i

e

i 1

)  Pi a 

i (e

e

i

i 1

), i 2 {1, 2 . . . n}

(3.7.2)

P P Now, ni=0 pi = ni=0 (e i e i 1 ) = en e0 = e. This implies pi ? pj , for i 6= j P P and ni=1 Pi = I, a ⇠ Pi , 8i. Hence, by prop ??, ni=1 Pi a = a. Summing over n in (3.7.2), we get

n X

i 1 (e

i

e

i 1

i

=) 0  a =) ka

s 

n X

(

i

)a

1 )(e

i

e

i

i

s k  k k. Hence, limk

n X

k!0

ks

i (e

e

i

i 1

)

i

i

)k k 1

ak = 0.

n X

(e

e

i

i 1

i

) = k ke

Corollary 3.178. Assume A, V are in spectral duality. Then each a 2 A can be P approximated in norm by linear combinations ni=1 i 1 pi of mutually orthogonal projections pi in the P bicommutant of a. Proof. Let {e }

be the spectral resolution of a and

2R

finite increasing sequence of real numbers with pi = e

i

e

i 1

0


kak. Putting

, for i = 1, 2 . . . n, we see that n X

pi =

i=0

n X

(e

i

e

i 1

) = en

e0 = e

i=0

By lemma ??, pi ? pj , for i 6= j. Moreover, e i and e = e i ^ e0 i

i 1

belong to P-bicommutant

also belongs to P-bicommutant of a, by P lemma 3.153. By proposition 3.177, the Riemann sums s = ni i 1 (e i e i 1 ) = Pn i=1 i 1 pi converge in norm to a. of a implies pi = e

i

e

i 1

1

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] and P = {x0 , x1 . . . xn } be a partition of [a, b]. Let g be an increasing real valued function on [a, b]. Then, the Riemann - Stieltjies integral of f with respect to g is defined as Z

b

f (x)dg(x) = lim

kP k!0

a

n X

f (yi )[g(xi )

where yi 2 [xi , xi 1 ]

g(xi 1 )],

i=1

whenever this limit exists. If we define Riemann-Stieltjies integrals with respect to {e }

2R

as the norm

limit of the approximating Riemann sums, then we can restate the result of theorem 3.177 as a=

Z

de

Definition 3.180. Assume A, V are in spectral duality. Let a 2 A and {e }

2R

be the spectral resolution of a. For a continous function f defined on [ kak, kak], we can define the integral with respect to {e }

2R ,

as the weak integral in A = V ⇤ ,

determined by the equation ⌦

Z

↵

f ( )de ,

=

Z

f ( )dhe , i,

for all

2V+

(3.7.3)

The integral on the right hand side is defined in the sense of Riemann Stieltjes 7! he , i.

integral of f with respect to the increasing function

Remark 3.181. Note that if f is a continuous function on [ kak, kak], then we are guaranteed that the integral on RHS of (3.7.3) exists. Fix

2 V + and let P = {x0 , x1 . . . xn } be a partition of [ kak, kak]. Define

U (P, ) =

n X i=1

f (⌘i )he

i

e

i

, i, L(P, ) = 1

111

n X i=1

f (⇠i )he

i

e

i 1

, i

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 [ kak, kak] (uniformly continous), given ✏ > 0, 9

> 0 such that |f (x)

f (y)| < ✏ whenever |x

y| < . So,

for a partition P with kP k  , we have kU (P, )

L(P, )k = k

n X

(f (⌘i )

i=1 n X

< ✏k

i=1

he

i

f (⇠i ))he e

i 1

i

e

i 1

, ik

, ik

(3.7.5)

= ✏he, i = ✏k k

(3.7.6)

2 V + , the Riemann sum

Hence, for each Z

(3.7.4)

f ( )dhe , i = lim

kP k!0

n X

f ( i )he

i

i=1

e

i 1

, i

exists

The map Ta (f ) : V + ! R given by 7!

Z

f ( )dhe , i

is a positive linear functional on V + . As f is bounded on [ kak, kak], we get Ta (f ) R to be pointwise bounded on K ( dhe , i = k k). Thus, Ta (f ) 2 Ab (K) ⇠ = V⇤ =A (proposition ??).

Using this notion, we define a spectral calculus for an element a 2 A. Definition 3.182. Assume A, V are in spectral duality. Let a 2 A and {e }

2R

be the spectral resolution of a. For each continuous function f on [ kak, kak], assign the element f (a) = Ta (f ) 2 A, denoted by f (a) =

Z

f ( )de

Proposition 3.183. If A, V are in spectral duality and a 2 A, then the spectral calculus for a satisfies the following (where f , g, fn are continuous functions on [ kak, kak] and ↵,

are real numbers): 112

3 Spectral Theory for Ordered Spaces 1. kf (a)k  kf k1 2. (↵f + g)(a) = ↵f (a) + g(a) 3. f  g =) f (a)  g(a) 4. If {fn } is a bounded sequence and fn ! f pointwise, then fn (a) ! f (a) in the weak* topology of A = V ⇤ . Proof. First note that for w 2 K, Z n X dhe , wi = lim he kP k!0

i

e

i 1

i=1

, wi = lim he, wi = 1 kP k!0

R 1. kf (a)k = supw2K |hf (a), wi|  supw2K | f ( )dhe , wi| Z Z  sup |f ( )|dhe , wi  kf k1 sup dhe , wi = kf k1 w2K

w2K

R R 2. Let v 2 V + . Then ↵hf (a), vi+ hg(a), vi = ↵f ( )dhe , vi + g( )dhe , vi = R ↵f ( ) + g( ) dhe , vi = h(↵f + g)(a), vi. So, ⌦

↵ ⌦ ↵ ↵f (a) + g(a), v = (↵f + g)(a), v 8v 2 V +

(3.7.7)

As V + generates V , (3.7.7) is true for all v 2 V . Thus, (↵f + g)(a) = ↵f (a) + g(a). 3. f  g =) f (a), vi

R

f ( )dhe , vi 

R

g( )dhe , vi for each v 2 V + . Thus, hg(a)

0, 8v 2 V + =) f (a)  g(a).

4. Fix v 2 V + . By monotone convergence theorem, Z Z fn ! f =) fn ( )dhe , vi ! f ( )dhe , vi =) hfn (a), vi ! hf (a), vi Then, hfn (a), vi

! hf (a), vi, 8v 2 V

topology.

113

=)

fn (a)

! f (a) in weak*

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 structure. 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 locally 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 bilinear 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 2 A, we have: kabk  kakkbk

(4.1.1)

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 = am an , 8 m, n 2 N, a 2 A. So, the n-th power of an element a, denoted by an , 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) is a multiplicative identity (iv) a2 2 A+ , for each a 2 A 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 7! 12 (ab + ba). Definition 4.6. Two order unit algebras A1 and A2 are said to be isomorphic if there exists a bijection

! A2 which preserves the algebra structure,

: A1

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 2 A satisfies the inequality k1 ak  1, then 9 s 2 A such that s2 = a. 1

Proof. Using Taylor series expansion of the function f (x) = (1

x) 2 , about 0, we

get (1

1

x) 2 =

1 X

n where nx

n

=

( 1)n ( 12 )( 12

n=0

1) . . . ( 12 n!

(n

1))

(4.1.2)

This series is absolutely and uniformly convergent for |x|  1 (A.9) and absolute P convergence of the series at x = 1 gives 1 n=0 | n | < 1. Now, fix x 2 [0, 1]. Then 1 2

1 2

(1 x) = (1 x) (1 x) =

1 X

nx

n

n=0

1 X

nx

n=0

n

=

1 X n=0

nx

n

where

n

=

n X

k n k

k=0

(4.1.3)

Note that the product series in (4.1.3) is convergent because each of the series P1 n 1, n = n=0 n x is absolutely convergent in [0, 1]. Also, see that 0 = 1, 1 = 0, 8 n > 1. Next, given a 2 A, satisfying k1 ak  1, define b = 1 a 2 A. Then, P n 1 kbk  1. Define s = 1 n=0 n b . Note that this series is absolutely convergent P n and as A is complete, the series converges in A. Now, s2 = 1 b = a. n nb = 1 Therefore, s2 = a. 1

P1

n=0

|

nb

n

|

P1

n=0

|

n ||b|

n



P1

n=0

|

n|

0 such that ⇢a = ⇢.

Since ⇢a is a positive linear functional on A, ⇢a = ⇢1 for some S. Infact,

> 0, ⇢1 2

= k⇢a k = ⇢a (e) = ⇢(ae) = ⇢(a). Also note that 0  ⇢(a) 

⇢(e) = 1 =) 0 

 1. Similarly, ⇢e

a

= µ⇢2 for some µ > 0, ⇢2 2 S.

Further, µ = ⇢e a (e) = ⇢(e ae) = 1 ⇢a (e) = 1 Now, ⇢a + ⇢e

a

= ⇢ =)

⇢1 + (1

. Thus, ⇢e

a

= (1

)⇢2 .

)⇢2 = ⇢. As ⇢ is a pure state on A,

⇢ = ⇢1 = ⇢2 . Hence, ⇢a = ⇢. Let x 2 A. Then, ⇢(a)⇢(x) = ⇢(ae)⇢(x) = ⇢a (e)⇢(x) = ⇢(e)⇢(x) = ⇢(x) = ⇢a (x) = ⇢(ax). Thus, ⇢(a)⇢(x) = ⇢(ax) for all x 2 A, a 2 A+ . Now, as A is positively generated, ⇢(ax) = ⇢(a)⇢(x) for all a, x 2 A. Hence, ⇢ is 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 theorem 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 2 A+ , for each a 2 A (iii) ab 2 A+ , for each pair a, b 2 A+ . Then, the set of pure states on A, denoted by @e K, is a w⇤ -compact set consisting of all multiplicative states, and the map a 7! a ˆ|@e K is an isometric, order and algebra isomorphism of A onto CR (@e K).

119

4 Spectral Theory for Jordan Algebras Proof. By proposition 4.13, we see that @e K is precisely the set of multiplicative states on A. • w*-compactness of @e K As K is a w*-compact subset of A⇤ , it is sufficient to show that @e K is w*closed in K. Let ⇢n

w⇤

! ⇢, where ⇢ 2 K, ⇢n 2 @e K, 8n. Then, ⇢n (a) !

⇢(a), 8a 2 A. Take a, b 2 A. ⇢(ab) = limn!1 ⇢n (ab) = limn!1 ⇢n (a)⇢n (b) = limn!1 ⇢n (a) limn!1 ⇢n (b) = ⇢(a)⇢(b). Hence, ⇢ is multiplicative which implies ⇢ 2 @e K. Thus, @e K is a w*-closed subset of K and hence, w*-compact. • Algebra homomorphism Consider the map

: A ! CR (@e K) given by a 7! a ˆ|@e K . This map is well

defined because if a 2 A and ⇢n

w⇤

! ⇢ in @e K, then ⇢n (a) ! ⇢(a) which

implies a ˆ(⇢n ) ! a ˆ(⇢). Thus, a ˆ|@e K is continous. Clearly,

ˆ is linear. For each ⇢ 2 @e K and a, b 2 A, we have ab(⇢) = ⇢(ab) =

⇢(a)⇢(b) = a ˆ(⇢)ˆb(⇢). Thus, •

is an algebra homomorphism. A.5

is an isometry Let a 2 A. Then, kak = sup⇢2K |⇢(a)| = sup⇢2@e K |⇢(a)| = k (a)k.

•

is one-one Suppose a ˆ(⇢) = ˆb(⇢), 8 ⇢ 2 @e K. By Krein-Milman Theorem (A.5), K = Co(@e K [

@e K). As a ˆ, ˆb are continuous linear funtionals on K, it follows

that a ˆ(⇢) = ˆb(⇢), 8⇢ 2 K. So, ⇢(a) = ⇢(b), 8⇢ 2 A⇤ . By Hahn-Banach separation theorem (A.3), a = b. •

is onto We will prove this using Stone Weierstrass theorem (A.7). As A is complete and

is an isometry, it follows that (A) is closed in CR (@e K).

(A)

contains the constant function eˆ. Also, if ⇢1 6= ⇢2 in @e K, then 9 a 2 120

4 Spectral Theory for Jordan Algebras A such that ⇢1 (a) 6= ⇢2 (a) which implies

(a) ⇢1 6= (a) ⇢2 . Hence,

(A)

separates points in @e K. Thus, by Stone - Wierestrass theorem (A.7), we get that

(A) = (A) = CR (@e K). Therefore,

is onto.

• Order isomorphism Let a1  a2 . Then a2

a1

0. As elements of @e K are positive, ⇢(a2 )

⇢(a1 ), 8⇢ 2 @e K. Hence, a ˆ 2 | @e K Conversely, if f

Hence, Hence,

is positive.

1

0 in CR (@e K), then, f 2 2 CR (@e K) and is positive. 1

is an algebra isomorphism, 9 a 2 A such that a =

Since, a2 =

a ˆ1 |@e K . So,

1

(f ). As A2 ✓ A+ , we get

1

(f )

0 in A. So,

1

1

(f 2 ). Then,

is positive.

is an order isomorphism. is an order and algebra isomorphism of A onto CR (@e K).

Theorem 4.16. If A is an order unit algebra, then the following are equivalent: 1. A is associative and commutative 2. ab 2 A+ , for each pair a, b 2 A+ . 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 2 A+ . Then, by proposition 4.7, a = s2 , b = t2 for some s, t 2 A. As A is associative and commutative, ab = (ss)(tt) = (st)(st) 2 A2 = A+ . Hence, A+ is closed under multiplication. (2 ) 3) Clear from Theorem 4.15. (3 ) 1) As CR (X) is an associative and commutative space, so is A.

121

4 Spectral Theory for Jordan Algebras Let a be an element in a power associative normed algebra A, with multiplicative 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) is a commutative and associative order unit algebra ( with ordering, 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 multiplicative identity e. Assume A satisfies the following condition: ka2 k  ka2 + b2 k, 8a, b 2 A Then the following are equivalent: (i) a 2 A2 (ii) k↵e

ak  kak, for all ↵

(iii) k↵e

ak  kak, for one ↵

kak kak

Proof. Clearly (ii) ) (iii). 122

(4.1.9)

4 Spectral Theory for Jordan Algebras (i ) ii) Let a 2 A2 such that kak  1. Then, 9 s 2 A such that s2 = a. Now, ke

(e

a)k = kak  1. Hence, by proposition 4.7, e

A. Then, s2 + t2 = a + (e

ak = kt2 k  ks2 + t2 k = kek = 1.

a) = e. So, ke

Therefore, we have shown that kak  1 =) ke

ak  1. Now, for arbitary

kak. Then k ↵a k  1 which implies ke

a 2 A, take ↵

previous case. Hence, k↵e (iii ) i) Suppose k↵e

a = t2 for some t 2

a k ↵

 1 by the

ak  ↵.

ak  kak, for some ↵

kak. Then ke

↵ 1 ak  1. By

proposition 4.7, ↵ 1 a 2 A2 . Hence, a 2 A2 . 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 2 A2 . Denote kak = ↵, kbk = . Now, ka + bk  ↵ + . By the equivalence of (i) and (ii) in lemma 4.18, we have k↵e Therefore, k(↵ + )e

(a + b)k  k↵e

ak + k e

ak  ↵, k e

bk  .

bk  ↵ + . Again, by the

equivalence of (iii) and (i) in lemma 4.18, we have a + b 2 A2 . So, A2 is closed p under addition. And clearly if a = s2 2 A2 and 0, then a = ( s)2 2 A2 . 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 7! ab is a bilinear map from A0 x A0 into A such that for each a, b 2 A0 , (i) ab = ba (ii)

e  a  e =) 0  a2  e

Then, kabk  kakkbk for each a, b 2 A0 . Proof. Let a, b 2 A0 and assume kak, kbk  1. Then, k 12 (a + b)k, k 12 (a So,

e  12 (a + b), 12 (a

b)k  1.

b)  e. By (ii), we get

1 0  (a + b)2  e, 4

e 123

1 (a 4

b)2  0

(4.1.10)

4 Spectral Theory for Jordan Algebras Adding the two

2

and using the fact that ab = ba, we get

e  ab  e which

implies kabk  1. So, we have shown that kak, kbk  1 =) kabk  1 for all a b a, b 2 A0 . Now for arbitary a, b 2 A0 , we have k kak k, k kbk k  1. Hence, by previous ab case, k kakkbk k  1 =) kabk  kakkbk.

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, 8a 2 A Proof. First, assume A is an order unit algebra . Let This implies ka2 k  kak2  1 =)

(4.1.11)

e  a  e. Then, kak  1.

e  a2  e. Also, A2 = A+ . So, 0  a2  e.

Next, assume (4.1.11) holds for all a 2 A. Taking A0 as A in lemma 4.20, we get kabk  kakkbk for all a, b 2 A. Therefore, A is a normed algebra . Further, by (4.1.11), we see that 0  a2 for all a 2 A. Thus, A is an order unit algebra . 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 2 A, we have the following: (i) kabk  kakkbk (ii) ka2 k = kak2 (iii) ka2 k  ka2 + b2 k Moreover, the ordering of A is uniquely determined by the norm and the identity element e. 2

If a1  b1 , a2  b2 , then a1 + a2  b1 + a2  b1 + b2

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 2 A|kkake

ak  kak}.

Claim 4.22.1. A2 is norm closed. Let an ! a in A, where an 2 A2 , 8n. This implies kan k ! kak as n ! 1. Now, kkake ak = kkake a + kan ke kan ke an + an k  (kak kan k)kek + kan

ak + kkan ke

an k  (kak

as n ! 1. Hence, kkake

kan k)kek + kan

ak + kan k ! 0 + 0 + kak

ak  kak which implies a 2 A2 . Thus, A2 is

norm closed. Claim 4.22.2. A2 is proper. Suppose a2 , a2 2 A2 . Note that

a2 2 A2 =) 9 b 2 A such that b2 =

a2 . Hence, 0 = a2 + b2 . Now, kak2 = ka2 k  ka2 + b2 k = 0. This implies kak = 0 implying a = 0. Thus, A2 is proper. Now, define A+ = A2 . Then, A becomes an ordered vector space with positive cone A2 . Claim 4.22.3. For each a 2 A, kak  1 () ()) Let kak  1. Then ke that e k

(() Given

a)k  1. By proposition 4.7, we see

(e

a 2 A2 = A+ This implies e

ak  1 =)

e  a  e.

a  e =)

a

0 =) a  e. Similarly,

e  a. Hence,

e  a  e.

e  a  e, we get e a, e+a 2 A+ = A2 . Let

by lemma 4.18, we get ke

1

(e

= ke ak. Then, 1

a)k  1. Now consider

(e

a) as

an element of the associative and commutative algebra C(a, e). Since the norm on C(a, e) is inherited from A, it follows ke 125

1

(e

a)k  1

4 Spectral Theory for Jordan Algebras holds in C(a, e) as well. Now, by proposition 4.7,

1

(e a) has a square

root in C(a, e). Thus, 9 s 2 C(a, e) such that s2 = e

a. Similarly,

9 t 2 C(a, e) such that t2 = e + a. Now, because C(a, e) is associative and commutative, we get (st)2 = s2 t2 = (e (e

a)(e + a) = e

a2 . Thus,

a2 ) 2 A2 = A+ . So, we have shown e  a  e =) 0  a2  e

(4.1.12)

Therefore, kak2 = ka2 k  1 which implies kak  1. Now, by proposition 2.16, A is an order unit space with order unit e. A is 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, given norm and positive cone A2 . ()) Now, assume (A,e) is an order unit algebra with positive cone A2 . Since A is a normed algebra, (i) follows. Moreover, being an order unit space, we see that 0  a2  a2 + b2 =) ka2 k  ka2 + b2 k for each a, b 2 A. Thus, (iii) holds. Next we prove (ii). Suppose a 2 A such that ka2 k  1. Then, 0  a2  e. Now, 1 2 a +e 2

(a

1 (a + e)2 2

a2

a= a=

1 e)2  (a2 + e)  e 2 1 e ( a2 e) e 2

e  a  e which in turn implies kak  1. So, we have shown that p ka2 k  1 =) kak  1. Therefore, in general, ka2 k  =) kak  . Therefore,

Taking

1

= ka2 k, we get kak  ka2 k 2 . Hence, kak2  ka2 k. The other

inequality is clear because A is a normed algebra . So, ka2 k = kak2 for each a 2 A. Thus, (ii) holds. Note that A+ = A2 = {a 2 A|kkake

ak  kak} implies that A+ is uniquely

determined by the norm (k.k) and order unit e. 126

4 Spectral Theory for Jordan Algebras Theorem 4.23. A commutative and associative real Banach algebra A is isometrically isomorphic to CR (X) i↵ the following holds for each a, b 2 A: (i) kabk  kakkbk (ii) ka2 k = kak2 (iii) ka2 k  ka2 + b2 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 CR (X) , for some compact Hausdor↵ space X. Define A+ = A2 . Using the isometric isomorphism, one can see that A becomes an order unit algebra with given norm, positive cone A+ = A2 and distinguished order unit e, (where e 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) which is the dual of a base norm space V. Considering A to be a dual space makes it monotone complete 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 assumptions, we get a least w*-closed subalgebra W (a, e) of A containing a given element a and e. This subalgebra is the w*-closure of all polynomials in a and e, and it 127

4 Spectral Theory for Jordan Algebras is also associative (A.8) and commutative. Further, W (a, e) is norm closed and satisfies the three norm conditions (i), (ii), and (iii) of theorem 4.22. Therefore, by theorem 4.22, W (a, e) is itself an order unit algebra , with the norm, 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. Two projections p, 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 2 A, and each ✏ > 0, there exist mutually orthogonal projections 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 ka

n X i=1

i pi k

✏

(4.1.13)

Proof. Let K the distinguished base for the base norm space V. As W (a, e) is an associative and commutative order unit algebra , it follows from theorem 4.16 that W (a, e) is isometrically isomorphic to CR (X) for some compact Hausdor↵ space X. Claim 4.25.1. W (a, e) is monotone complete. Let {b↵ } be an increasing net in W (a, e) which is bounded above, say by some c 2 W (a, e). Define a map b : K

! R given by b(k) = lim↵ b↵ (k). This

limit exists because for each k 2 K, {b↵ (k)} is an increasing net of real numbers bounded above by c(k) and R is monotone complete. Hence, this net converges to some real number, denoted by b(k). Since all b↵ are affine, so is b. Also, kbk = supk2K |b(k)| 

3

supk2K |c(k)|  kck. Therefore, b is an affine bounded

function on K, i.e. b 2 Ab (K) ⇠ = V ⇤ = A. So, b can be identified with an element 3

c(k)

b↵  c, implies that for each k 2 K, b↵ (k)  c(k). This is true forall ↵. Hence, lim↵ b↵ (k) 

128

4 Spectral Theory for Jordan Algebras ˜b of A and by construction, b↵ w⇤ ! ˜b. As W (a, e) is w*-closed, ˜b 2 W (a, e). Thus, 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) 2

(4.2.1)

This product is bilinear and commutative but not associative. However, it satisfies a weakened form of associativity, which leads to the following abstraction. 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) for all a, b 2 A

(4.2.2)

Any commutative bilinear product on a real algebra A satisfying (4.2.2) is said to be a Jordan product. Example 4.27. If A is any associative algebra over R, then A becomes a Jordan algebra when equipped with the symmetrised product 1 a b = (ab + ba) 2 Also, any subspace of A closed under

forms a Jordan algebra . 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 A, such that the subspace is closed under the symmetrised product in A. If it cannot be so embedded, then the Jordan algebra is said to be exceptional. In any Jordan algebra , we define powers of an element recursively: x1 = x and xn = xn

1

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 associative. Lemma 4.29. Let A be an algebra over R, equipped with a product that is commutative and bilinear, but not necessarily associative. Then x

(x

x2 ) = x2

x2 holds for all x 2 A () X

xi (xj (xk xl )) =

X

(xi xj )(xk xl )

(4.2.4)

where x1 , x2 , x3 , x4 2 A and the summation is over distinct i,j,k,l with 1  i, j, k, l  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 2 A. Fix x1 , x2 , x3 , x4 2 A and define x =

1 x1

+

2 x2

+

3 x3

+

4 x4

where

t

2 R, 1  t  4. Substituting

this in (4.2.3), we get 4 X 4 X 4 X 4 X

i j k l

xi (xj (xk xl )) =

i=1 j=1 k=1 l=1

4 X 4 X 4 X 4 X

i j k l

(xi xj )(xk xl )

i=1 j=1 k=1 l=1

130

(4.2.5)

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 coefficients which are equal for all values of the variables

1,

2,

3,

4.

Therefore, the coefficients of the corresponding variables

must be same. In particular, equating the coefficient of X

xi (xj (xk xl )) =

X

1 2 3 4,

we get

(xi xj )(xk xl )

(4.2.6)

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 argument. 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 2 A: (i) ka bk  kakkbk (ii) ka2 k = kak2 (iii) ka2 k  ka2 + b2 k

131

4 Spectral Theory for Jordan Algebras Example 4.33. If A is a C*-algebra (with an identity), then the self-adjoint elements of A form a JB-algebra with respect to the Jordan product a 1 (ab 2

b =

+ 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 2 A}. Proof. Since A is a JB-algebra , it is a real Banach space, equipped with a commutative 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, A is a commutative order unit algebra with positive cone A2 . 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 2 A, 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 2 A. Note that A being a Jordan algebra, is power associative. Then, 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) denote the norm closed Jordan subalgebra generated by a and e. Let P (a, e) denote the set of all polynomials in a and e. Then, it easy to see that C(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) is also associative. 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 2 A. 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 by a

1

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 = 12 (ab + ba). Then for a, b 2 A: 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. Then a b = 12 (2e) = e. Similarly, a2 b = 12 (a2 b + ba2 ) = 12 (a + a) = a. (() If a b = e, then ab + ba = 2e

(4.2.9)

a2 b + ba2 = 2a

(4.2.10)

And if a2 b = a, then

Now, a(ab + ba) = 2ae = 2a = 2ea = (ab + ba)a. So, a2 b + aba = aba + ba2 . So, we get ab2 = ba2

(4.2.11)

Putting (4.2.11) in (4.2.10), we get 2a2 b = 2a =) a2 b = a =) ba2 b = ba. Similarly, ba2 = a =) ba2 b = ab. Hence, ba = ba2 b = ab. Putting this in (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, then a 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 invertibility coincide. Proposition 4.42. For elements a, b in a JB-algebra A, the following are equivalent: (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 A such that the Jordan product on A comes from the symmetrised product on A. (i ) ii) Suppose b = a

1

in C(a, e). As C(a, e) is associative, we have a 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 (B0 , +, ) be the subalgebra generated by a, b, e in A. By theorem 4.52, B0 is special and can be viewed as a subalgebra of an associative algebra (C, +, .) such that x y = 12 (x.y + y.x) for all x, y 2 B0 . Then, by lemma 4.40, a.b = b.a = e. Let C0 be the subalgebra generated by a, b, e in C. Then, (C0 , +, .) is an associative and commutative subalgebra of C and is in bijective correspondence (a 7! a, b 7! b, x

y 7! x.y) with (B0 , +, ). It is easy to check that this bijection is

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 2 C(a, e). Let a

: B ! CR (X) be the isometric isomorphism mentioned above. Now, b = e =)

(a) (b) = 1, i.e. 135

(a) is invertible in CR (X) . So, 0 2 /

4 Spectral Theory for Jordan Algebras Range( (a)). As X is compact and

(a) is continuous, Range( (a)) is a

compact subset of R, say Y. Now, define f : Y

! R given as

1

7!

.

Then, f 2 CR (Y ) and can be approximated by polynomials pn (by StoneWeierstrass theorem). Let pn (a) in CR (X) . But f

! f in CR (Y ). Then, pn

(a) = ( (a))

1

=

(a)

! f

(b). This means that

can be approximated in CR (X) , by polynomials in

(b)

(a). By the isometric

isomorphism, this implies that b can be approximated in B, by polynomials in a. Thus, b 2 P (a, e) = C(a, e). Hence, the claim. 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) is defined as (a) = { 2 R | e

a is not invertible}.

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 2 A. By theorem 4.38, there exists an isometric (order and algebra) isomorphism

: C(a, e) ! CR (X) . Let

(a) = a ˆ. As

preserves invertibility,

(a) = (ˆ a). Claim 4.45.1. 2 (ˆ a) ()

(ˆ a) = Range (ˆ a) 1

a ˆ is not invertible in CR (X) () 9 x 2 X such that ( 1

a ˆ)(x) = 0 () a ˆ(x) = 1 ()

2 Range a ˆ. Hence, the claim.

Now, a ˆ is a continuous function on a compact space X. Hence range(ˆ a) is compact and non-empty. Thus (ˆ a) = (a) is non-empty and compact. 136

4 Spectral Theory for Jordan Algebras Claim 4.45.2. a ˆ : X ! (a) defined as x 7! a ˆ(x) is a homeomorphism. This map is surjective because we have just shown that (a) = (ˆ a) = Range (ˆ a). To prove injectivity, assume that aˆ(x) = a ˆ(y) for some x, y 2 X. Then (ˆ a(x))n = (ˆ a(y))n for all n 2 N. Thus, p(x) = p(y) for all polynomials p in a ˆ. Now, P (a, e) is dense in C(a, e) and the isometry

preserves density. Hence, P (ˆ a, 1) is dense

in CR (X) . Hence, f (x) = f (y), for all f 2 CR (X) Now, X is a compact, Hausdor↵ space and hence normal. Therefore, distinct points in X can be separated by continuous functions. So, f (x) = f (y) for all f 2 CR (X) =) x = y. And we already know that a 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 : CR (X) ! CR ( (a)) given by T f = f

(ˆ a) 1 . Then the map

=T

is an

algebra isomorphism from C(a, e) onto CR ( (a)). T

: C(a, e) ! CR (X) ! CR ( (a)) Note that

(e) = T (1) = 1 and

order isomorphism. Also, Further, 0|

(

e) 

takes squares to squares. Thus,

(a) = T (ˆ a) = a ˆ (ˆ a)

1

( e)} = inf{ > 0 |

1

is a unital

= I (identity function on (a)).

is an isometry because kbk = inf{ > 0 | (b) 

(4.2.12)

e  b  e} = inf{ >

(b)  1} = k (b)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) to a. Since polynomials are dense in C(a, e), ⇥ is uniquely determined by its value at a. So, if ⇥ takes a to the identity map on (a), then ⇥ =

.

Definition 4.46. Let a be an element of a JB-algebra A. The continuous func137

4 Spectral Theory for Jordan Algebras tional calculus is the unique isomorphism f 7! f (a) from CR ( (a)) onto C(a,e), taking the identity function on (a) to a. Proposition 4.47. Let a be an element of a JB-algebra A. The continuous functional calculus is an isometric order isomorphism from CR ( (a)) onto C(a,e). Furthermore, 1. f (a) has its usual meaning when f is a polynomial. 2.

(f (a)) = f ( (a)), for f 2 CR ( (a))

3. (f

g)(a) = f (g(a)) for g 2 CR ( (a)) and f 2 CR ( (g(a)))

4. If f 2 CR ( (a)) and f (0) = 0, then f (a) is in the (non-unital) norm closed 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) and 1 denote the constant function 1 on (a). 1. Recall that (f ) =

k X

(I) = a. So, if f (x) = n

(x ) =

n

n=0

k X

n

Pk

(x)

n=0

n

=

n=0

nx

k X

n

, then

n

I(x)

n

n=0

=

k X

na

n

= f (a)

n=0

(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) 2 C(a, e) and f (g(a)) 2 C(g(a), e) ✓ C(a, e). Case (I). f is a polynomial. Let f (x) =

Pk

n=1

f (g(a)). Thus, (f

nx

n

. Then, (f

g)(a) = f (g(a)). 138

g)(x) = f (g(x)) =

Pk

n=1

n

g(x)

n

=

4 Spectral Theory for Jordan Algebras Case (II). f is an arbitary continuous function on (g(a)). P (a, e) is dense in C(a, e) and P (g(a), e) is dense in C(g(a), e). Recall that f 2 CR ( (g(a))) ⇠ = C(g(a), e) and the identity function on g(a), say Ig , maps to g(a) under this isomorphism. Now, (g(a)) is a compact Hausdor↵ space and hence f can be approximated by polynomials pn , in CR (g(a)). Now, pn

! f in CR ( (g(a))) implies pn

is an isometry, (pn (pn

g)(a)

! (f

g

!f

g in CR ( (a)). Since

g)(a) in C(a, e) as n ! 1. But,

g)(a) = pn (g(a)) by case I and pn (g(a)) ! f (g(a)) in CR ( (g(a))) as

n ! 1. Thus, (f

g)(a) = f (g(a)).

4. Let f 2 CR ( (a)). Then f can be approximated by polynomials pn in CR ( (a)). This implies pn (x) ! f (x) for each x 2 (a). Thus, qn (x) := pn (x)

pn (0) ! f (x)

f (0). Now f (0) = 0 =) qn ! f in CR ( (a)). As

qn have constant term 0, qn (a) is a polynomial in a, not involving e. Thus, qn (a) 2 C(a), 8 n. Hence, f (a) = limn qn (a) 2 C(a).

Corollary 4.48. Let a be an element of a JB-algebra A. Then, a Also, kak = sup{| | | Proof. a

0 ()

(a) ✓ [0, 1)

2 (a)}.

0 () I(x)

0, 8 x 2 (a) () x

[0, 1). Also, kak = kIk = sup{| | |

0, 8 x 2 (a) ()

(a) ✓

2 (a)}.

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

0 is equivalent whether a 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} = 12 (abc + cba) 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 = 12 (ab + ba). We will now explore the special case of the triple product {abc} with a = c. Definition 4.53. Let a 2 A. Define Ua : A ! A given as Ua (b) = {aba} = 2a (a b) 140

(a2 b)

(4.2.15)

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{ab2 a}b}

(4.2.17)

1 (a b)2 = (2a {bab} + {ab2 a} + {ba2 b}) 4

(4.2.18)

{b{bab}b} = {b2 ab2 }

(4.2.19)

{(e

b){bab}(e

b)} = {(b

b2 )a(b

b2 )}

{ab2 a}2 = {a{b{ba2 b}b}a} 1 If p2 = p, then p a = (a + {pap} 2

(4.2.20)

(4.2.21)

{p0 ap0 })

(4.2.22)

Using (4.2.17), one can see that U{aba} = Ua Ub Ua

(4.2.23)

Our next goal is to show that the maps Ua are positive for each a 2 A. Lemma 4.55. Let a be an element of a JB-algebra A. Then a is an invertible element i↵ Ua has a bounded inverse. In this case, (Ua )

1

= Ua 1 .

Proof. First note that the map Ua is bounded for each a 2 A. 141

4 Spectral Theory for Jordan Algebras ()) Suppose a is invertible in A, with c = a

1

in C(a, e). Now, C(a, e) is as-

sociative implies {ac2 a} = a2 c2 = e. This implies U{ac2 a} = Ue = Identity function on A. So, I = U{ac2 a} = Ua Uc2 Ua (by (4.2.23)). Then, Ua Uc2 is the left inverse of Ua and Uc2 Ua is the right inverse of Ua . invertible, say (Ua )

1

4

Therefore, Ua is

= T . Now, {aca} = a2 c = a (a c) = a. Therefore, Ua = U{aca} = Ua Uc Ua

Now, I = T Ua = T Ua Uc Ua = Uc Ua and I = Ua T = Ua Uc Ua T = Ua Uc . Hence, Uc Ua = Ua Uc = I Thus, (Ua )

1

= Uc = Ua 1 .

(() If a is not invertible in A, then by functional calculus, there are elements {bn } in C(a, e) which have norm 1 and satisfy kUa bn k ! 0. If Ua had a bounded inverse, then applying it to the sequence {Ua bn } would show bn ! 0, which contradicts that kbn k = 1, 8 n. Hence, a is invertible in A.

Lemma 4.56. If a, b are invertible elements of JB-algebra , then {aba} is invertible with inverse {a 1 b 1 a 1 }. Proof. If a, b are invertible elements in A, then by lemma 4.55, Ua , Ub have bounded inverses. Now, U{aba} = Ua Ub Ua implies U{aba} also has a bounded inverse. Then, by lemma 4.55, {aba} is invertible in A. Using the fact that Ux (x 1 ) = x x x x, for each invertible x 2 A, we get: (aba) (Ua Ub Ua ) 1 {aba} = (Ua ) 1 (Ub ) 1 (Ua )

1

1

1

=

= U{aba} 1 {aba} = (U{aba} ) 1 {aba} =

{aba} = Ua 1 Ub 1 Ua 1 {aba} = Ua 1 Ub 1 Ua 1 Ua (b) =

Ua 1 Ub 1 (b) = Ua 1 (b 1 ) = {a 1 b 1 a 1 }. So, {aba}

1

= {a 1 b 1 a 1 }.

Theorem 4.57. Let A be a JB-algebra . For each a 2 A, the map Ua is positive and has norm kak2 . 4

For associative algebras, left inverse is same as right inverse

142

4 Spectral Theory for Jordan Algebras Proof. Let A

1

denote the set of all invertible elements in A and let A0 = A 1 \A+ .

Using functional calculus, one can see that if b b is invertible () b

0, then

e, for some

>0

(4.2.24)

Claim 4.57.1. A0 is open in A 1 . Let a 2 A0 . By (4.2.24), 9 A 2

1

= {c 2 A

ea

2

1

| ka

e. Now, look at b 2 B(a, 2 ) \

> 0 such that a

ck < 2 }. So, ka

bk
0 such that a1

1 e,

a2

↵) 2 )e. By (4.2.24), ↵a1 + (1

2 e.

This implies

↵)a2 2 A0 . So, A0

is convex, hence connected. Claim 4.57.4. A0 is dense in A+ . Let a

0 and ✏ > 0 be given. Then, a + ✏e

✏e

0. Hence, by (4.2.24),

a + ✏e 2 A0 . And, a + ✏e ! a as ✏ ! 0. Hence, the claim. Now, we will prove that Ua is positive. Case (I). a 2 A

1

By functional calculus, (a 1 )2 2 A0 . Now, if b 2 A 1 , then Ua (b) = {aba} 2 A 1 , by lemma 4.56. Hence, Ua (A 1 ) ✓ A 1 . In particular, Ua (A0 ) ✓ A 1 . Next, A0 is connected and Ua is continuous implies that Ua (A0 ) is connected. Now, A

1

= A0 t (A0 )c (A0 is both open and closed in A 1 .) But, Ua (A0 )

is a connected subset of A 1 . Therefore, it is contained entirely in either A0 or (A0 )c . Note that 1 2 Ua (A0 ) \ A0 because Ua ((a 1 )2 ) = a2 (a 1 )2 = e = 1. Thus, 143

4 Spectral Theory for Jordan Algebras Ua (A0 ) ✓ A0 . Now, A0 is dense in A+ and Ua is continuous. Hence, Ua (A0 ) = Ua (A+ ) ✓ A0 = A+ . Thus, Ua is positive. Case (II). a is an arbitary element in A. Let b 2 A0 . By functional calculus, 9 c 2 A

1

such that c2 = b. Now,

Uc Ua (b) = {c{aba}c} = {c{ac2 a}c} = {cac}2

0

(4.2.25)

Now, by (4.2.25), Ua (b) = Uc 1 (Uc Ua b) = Uc 1 {cac}2 . By case I, Uc 1 {cac}2

0.

Hence, Ua (A0 ) ✓ A+ . Again, using density argument, Ua (A+ ) ✓ A+ . So, Ua is positive for each a 2 A. Now, since Ua is a positive operator on A, kUa k = kUa (e)k = ka2 k = kak2 . Lemma 4.58. If a, b are positive elements of JB-algebra A, then {aba} = 0 () {bab} = 0

(4.2.26)

{aba} = 0 =) a b = 0

(4.2.27)

Also,

Proof. Given a, b 2 A+ . 1. Assume {aba} = 0. As b

0, by functional calculus, 0  b2  kbkb.

Applying the positive operator Ua , we get 0  Ua b2  kbkUa b, which implies 0  {ab2 a}  kbk{aba} = 0. Thus, {aba} = 0 =) {ab2 a} = 0

(4.2.28)

Now, {bab}2 = {b{ab2 a}b} = 0. Thus, {bab}2 = 0 which implies k{bab}k2 = k{bab}2 k = 0. Hence, {bab} = 0. 2. Assume {aba} = 0. Then, also {bab} = 0, by (1). Now, by (4.2.28), {ab2 a} = 0 and {ba2 b} = 0. By (4.2.18), (a b)2 = 14 (2a {bab} + {ab2 a} + {ba2 b}) = 0. Thus, ka bk2 = k(a b)2 k = 0. Hence, (a b) = 0. 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 2 A satisfying p2 = p. A compression on A is the map Up : A ! A, when p is a projection. Up (a) = {pap} = 2p (p a) We denote the projection e

(p a), 8a 2 A

(4.2.29)

p by p0 .

A priori we do not know whether the maps Up 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 kUp k  1, If a

Up2 = Up ,

Up Ue

p

=0

(4.2.31)

0, then Up (a) = 0 () Ue p (a) = a

(4.2.32)

Proof. Note that if p 6= 0, then kpk = kp2 k = kpk2 =) kpk = 1. So, by theorem 4.57, kUp k = kpk2 = 1. Using the identity (4.2.19): we get {p{pap}p} = {p2 ap2 }. So, Up (Up (a)) = Up {pap} = {p{pap}p} = {p2 ap2 } = {pap} = Up (a). Therefore, (Up )2 = Up . Putting b = p in identity (4.2.20), we get Up Ue 145

p

= 0.

4 Spectral Theory for Jordan Algebras Next, let a

0. Assume Up0 a = a. Then applying Up on both sides, we get:

0 = Up a. Conversely, if Up a = 0, then {pap} = 0. By lemma 4.58, we have p a = 0. So, p0

p) a = a. Thus, Up0 a = 2p0

a = (e

(p0

a)

p0

a = 2p0

a

a = a.

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) = ⇢(T a) for

a 2 A and ⇢ 2 A⇤ . 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) kUp⇤ k  1 (ii) kUp⇤ k = 1 ()

(p) = 1 () Up⇤ ( ) =

(iii) Up⇤ = 0 () Up⇤0 = Proof. First we claim that if

is a state on A, then (p0 ) = 0 =) Up⇤ =

Proof: Suppose

(4.2.33)

(p0 ) = 0. Then, by Cauchy-Schwarz inequality, ( (p0

b))2 

((p0 )2 ) (b2 ) = 0. Thus, (p0 b) = 0, 8 b 2 A. This implies (b) = (p b) for all b 2 A. Hence, Up⇤ (a) = (Up (a)) = [2p (p a) p a] = [2p (p a)]

[p a] =

(a). This is true for all a 2 A. Hence, Up⇤ = . Now, we start proving the proposition. (i) kUp⇤ k = kUp k = 1. (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 Up⇤

is a positive linear functional on A. Hence, kUp⇤ k =

Up⇤ (e) = (Up e) = (p). Therefore, kUp⇤ k = 1 () 146

(p) = 1.

4 Spectral Theory for Jordan Algebras (iii) Suppose Up⇤

= 0. Then (Up e) = 0 =)

(p) = 0 =) Up⇤0

=

(by

(4.2.33)). Conversely, suppose Up⇤0 = , then applying Up⇤ on both sides, we 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 bicomplemented 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] denotes their associated order interval, i.e. [a, b] = {x 2 A | a  x  b}. If a is any positive element of A, then f ace(a) denotes the face in A+ generated by a, i.e. f ace(a) = {y 2 A | 0  y  a, for some Lemma 4.64. Let p be a projection in a JB-algebra A. Then (i) f ace(p) = im+ Up (ii) f ace(p) \ [0, e] = [0, p]

147

0}

(4.2.34)

4 Spectral Theory for Jordan Algebras Proof.

(i) Let a 2 f ace(p). This implies 0  a 

implies Up0 0  Up0 a 

p for some

0. This

Up0 Up e = 0. So, Up0 a = 0. Hence, by (4.2.32),

Up a = a which implies a 2 im+ Up . Conversely, let a 2 im+ Up . Then a = Up a  Up (kake) = kakp which implies a 2 f ace(p). (ii) Let a 2 f ace(p) \ [0, e]. Then, by (i), a 2 im+ Up . Thus, a = Up a  Up e = p =) 0  a  p =) a 2 [0, p]. The other way inclusion is easy to see.

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] and let @e I denote the set of extreme points of I. (✓) Let a 2 @e I. Then, by functional calculus, 0  a2  a  e. Hence, a2 2 I. Now, 0  a  e =) 0  i(x)  1, for all x 2 (a). Now, 0  (1 + i(x))2 a2  e. Hence, 2a

for each x 2 (a) implies that 0  2a a = 12 (a2 + (2a

a2 2 I. Now,

a2 )) and a is an extreme point of I. Hence, a = a2 which

implies that a is a projection in A. (◆) Let p be a projection in A. If p = a, (1

)b for some a, b 2 I, then

a + (1

)b  p which implies that a, b 2 f ace(p) \ [0, e] = [0, p]. So,

p = a + (1

)b  p + (1

)p = p. Thus, a = b = p. So, p 2 @e I.

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. Recall that for each state ⇢ on a JB-algebra A, the map ha, bi = ⇢(a b) is a positive semi-definite bilinear form and thus we have the Cauchy-Schwarz inequality 1

1

|⇢(a b)|  ⇢(a2 ) 2 ⇢(b2 ) 2 1

And taking b = e, we get |⇢(a)|  ⇢(a2 ) 2 . 148

(4.2.35)

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. Up (A) = {pAp} is a JB-subalgebra with identity p. Proof. Since Up is a linear operator on A, its image is a subspace of A. Let b = {pap} 2 Up (A). Then b2 = {pap}2 = {p{ap2 a}p} 2 Up (A). So, Up (A) is closed under squares. Now, if a, b 2 Up (A), then a

b = 12 [(a + b)2

a2

b2 ] 2 Up (A).

Thus, Up (A) is closed under product. Claim 4.66.1. Up (A) = {a 2 A | Up a = a} Let a = {pbp} 2 Up (A). Then Up (a) = (Up )2 (b) = Up (b) = a. Hence, the claim. Note that Up is continuous on A. If an all n, then Up (an )

! a in A, where an 2 Up (A) for

! Up (a). But Up (an ) = an for all n, by the claim. Hence,

Up (a) = limn Up (an ) = limn an = a. Thus, a 2 Up (A) and hence Up (A) is norm closed. Claim 4.66.2. p is the identity for Up (A). Let a 2 {pAp}. Then, by the claim, a = Up a which implies Up0 a = 0

=)

{p0 ap0 } = 0 =) p0 a = 0 =) p a = a. So, p is the identity for Up (A). 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 = 0 for all a 2 Ap , b 2 A0p . Proof. Let a 2 Ap and b 2 Ap0 . As A is positively generated and the maps Up , Up0 are positive, it follows that Up (A) and Up0 (A) are also positively generated. Hence, it is sufficient to prove the result for 0  a, b. Note that since p is the identity for 149

4 Spectral Theory for Jordan Algebras Ap , p 0

a=a

a) = 0 for all a 2 Ap . Now, 0  b  kbke =) 0  b =

(p

Up0 b  Up0 kbke = kbkp0 . Applying Ua , we get 0  {aba}  kbk{ap0 a} = 0. So, we 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, denoted by a ? b, if {aba} = {bab} = 0. Proposition 4.70. Let A be a JB-algebra . Then each a 2 A can be uniquely 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 decomposition is called the orthogonal decomposition of a. Proof. Let X be a compact Haudsor↵ space and let f 2 CR (X) . Now, define f + , f : X ! R by f + (x) =

f (x) = Then, f + , f

8 < f (x) :

8
, t (constant function 1 n1 ) > > > n n > > > 1 > > (n 1)t, t0 (linear from n1 to 0) > > n > < hn (t) = 0, t=0 > > > > 1 > > (n 1)t, 0t (linear from 0 to n1 ) > > n > > > > n 1 1 : , t (constant function 1 n1 ) n n We claim that

|t2 (1

hn (t))| 

Fix t 2 [ 1, 1] and n 2 N. If |t|

1 whenever t 2 [ 1, 1] n 1 , n

then |t2 (1

case t = 0 is clear. Now let 0  t  0  hn (t)  1

1 n

So, 0  t2 (1

hn (t))  t2 

1 n

This gives 0

 t  0, we have |t2 (1

hn (t) 1 n2



1 . n

(4.2.40)

hn (t))| = |t2 n1 | 

1 . n

Then 0  (n

1 n

1 which implies 1

1)t  1

1 . n

The

n 1 . n

So,

hn (t)

1 . n

Similarly, it can be shown that when

hn (t))|  n1 . Hence, (4.2.40) holds.

Claim. For each a 2 J, limu2U ka2

{aua}k = 0

First assume kak  1. Then, (a) ✓ [ 1, 1]. Restricting hn to (a), we get

hn 2 CR ( (a)) ⇠ = C(a, e). Denote the image of hn under this isomorphism as hn (a). Note that for all t 2 (a), we have |hn (t)|  1

1 n

< 1. So, khn (a)k =

khn k  1. And hn (0) = 0 =) hn (a) 2 C(a) ✓ J, by proposition 4.47. Also, hn

0 =) hn (a) ka2

0. Hence, hn (a) 2 U . Now,

{ahn (a)a}k = 9 ka2

(a2 hn (a))k = ka2 (e

hn (a))k 

1 n

(4.2.41)

where the last inequality follows from functional calculus and (4.2.40). Now, given ✏ > 0, choose n such that 9

1 n

 ✏. If u

hn (a), then we have {ahn (a)a} 

hn (a) 2 C(a) implies that {ahn (a)a} = (a2 hn (a))

157

4 Spectral Theory for Jordan Algebras {aua}  {aea} = a2 . So, for all u 2 U with u ka2

{ahn (a)a}k 

1 n

hn (a), we have ka2

by (4.2.41). Hence, limu2U ka2

{aua}k 

{aua}k = 0, whenever

kak  1. As norm and triple product are linear, the claim holds for all a 2 J. Now, if u 2 U and a 2 J, thenka u)a}k (by lemma 4.78) = k(e limu2U ka

(a u)k2 = ka (e

u)kka2

{aua}k  ka2

(a u)k  limu2U ka2

u)k2  ke

ukk{a(e

{aua}k. Hence,

{aua}k = 0

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 ka + Jk = inf b2J ka + bk 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 symmetrised 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 2 K}. Then, V+ is cone in V and K is a

base for V+ . Further, as K is a w⇤ -compact subset of M ⇤ , we see that (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 ha, bi 7! ⇢(a

b) is a positive semi-definite bilinear form and hence, by Cauchy-Schwarz

inequality, for each a, b 2 M, we get: 1

1

⇢(a b)  (⇢(a2 )) 2 (⇢(b2 )) 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

w)

is induced by the semi-norms a 7! |⇢(a)|, for ⇢ 2 K. The subbasis elements

are: B(x, ⇢, r) = {y 2 M | |⇢(x

y)| < r}, where x 2 M, ⇢ 2 K, r > 0. Therefore,

w b↵ ! b () ⇢(b↵ ) ! ⇢(b), 8⇢ 2 K

(4.3.2)

Remark 4.87. Note that -weak limit of a given net is unique, if it exists, because if ⇢(a) = ⇢(b) for all ⇢ 2 K, then a = b ( M admits separating set of normal 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 s)

1

is induced by the semi-norms a 7! (⇢(a2 )) 2 for ⇢ 2 K. The subbasis elements

are given as B(x, ⇢, r) = {y 2 M | (⇢(x

1

y)2 ) 2 < r}, where x 2 M, ⇢ 2 K, r > 0.

Hence, s b↵ ! b () ⇢((b↵

b)2 ) ! 0, 8⇢ 2 K

(4.3.3)

Remark 4.89. Using Cauchy-Schwarz inequality, one can see that sum of -strong limits is -strong limit of sums, i.e.

10

s {a↵ ! a, b

over directed set (↵, )

s s ! b} =) (a↵ + b ) ! a + b10

159

(4.3.4)

4 Spectral Theory for Jordan Algebras Recall the linear maps Ua (b) = {aba} and Ta (b) = a b, defined on M, for each a 2 M. Proposition 4.90. For each a 2 M: 1. The maps Ta⇤ , Ua⇤ : M ⇤ ! M ⇤ take V into V. 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 2 M. It is easy to verify that 1 Ta = (Ue+a 2

Ua

I)

(4.3.5)

1. We will first show that Ua⇤ (V ) ✓ V . Case (I). a is invertible. If a is invertible, then Ua becomes an order isomorphism on M, with positive inverse map Ua b

1

(lemma 4.55). So, x  y () Ua x  Ua y. Let ⇢ 2 K. If

0, then (Ua⇤ ⇢)(b) = ⇢(Ua b)

0. Also, if b↵ % b in M, then Ua b↵ % Ua b,

since Ua is an order isomorphism. So, (Ua⇤ ⇢)(b↵ ) = ⇢(Ua b↵ ) % ⇢(Ua b) = Ua⇤ ⇢(b). Thus, Ua⇤ ⇢ is positive and normal for each ⇢ 2 K. Hence, Ua⇤ (K) ✓ V which implies Ua⇤ (V ) ✓ V . Case (II). a is an arbitary element of A. Choose

> kak. Then by functional calculus, e a and e+a are invertible,

and from the linearity of the triple product, we get 1 Ua = (U 2

e+a

160

+U

e a

2 2 I)

(4.3.6)

4 Spectral Theory for Jordan Algebras By case I, It follows that Ua⇤ maps V into V for every a 2 A. Now, by (4.3.5), we also see that Ta⇤ maps V into V, for all a 2 A. w 2. Let b↵ ! b. This implies ⇢(b↵ ) ! ⇢(b), 8⇢ 2 K =) ⇢(b↵ ) ! ⇢(b), 8⇢ 2

V . In particular, (Ua⇤ ⇢)(b↵ )

! (Ua⇤ ⇢)(b), 8 ⇢ 2 K, by (1). This implies,

w ⇢(Ua (b↵ )) ! ⇢(Ua (b)), 8⇢ 2 K =) Ua (b↵ ) ! Ua (b). Therefore, Ua is

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, 8 ⇢ 2 K which implies b2↵

w !

w 0 =) Ua (b2↵ ) ! 0. Now, as 0  a2  ka2 ke = kak2 e, we have {ab↵ a}2 =

{a{b↵ a2 b↵ }a} = Ua {b↵ a2 b↵ } = Ua Ub↵ (a2 )  Ua Ub↵ (kak2 e) = kak2 Ua (b2↵ ) = kak2 {ab2↵ a}

w ! 0. Thus, for all ⇢ 2 K, we got ⇢({ab↵ a}2 )

s implies {ab↵ a} ! 0. Now, if b↵

! 0 which

s s ! b, then Ua (b↵ ) ! Ua (b), by remark

4.89. Hence, Ua is continuous in -strong topology. And, by (4.3.5), we see that Ta is also continuous in -strong topology. s ! 0 and b

4. Consider two nets a↵

s ! 0 in M such that 9 N1 , N2 >

0 such that ka↵ k  N1 , kb k  N2 , 8↵, . For each a 2 M, by functional calculus, a4  ka2 ka2 . So, for ⇢ 2 K, we get ⇢(a4↵ )  ka2↵ k⇢(a2↵ ) ! 0. So, ⇢(a4↵ ) ! 0 for all ⇢ 2 K which implies a2↵

s ! 0. So, we have prove that

for a bounded net {a↵ }, s s a↵ ! 0 =) a2↵ ! 0

(4.3.7)

Also, note that for a, b 2 M, 1 a b = [(a + b)2 2

a2

b2 ]

(4.3.8)

s Then using (4.3.4), (4.3.7) and (4.3.8), we get that (a↵ b ) ! 0.

Now, consider two bounded nets a↵ s a↵ ! 0 and b b

s ! a and b

s ! b in M. Then, a

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

we get that a↵ b

a↵ ) b + (a

a↵ ) (b

b ) + a (b

b ) (4.3.9)

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. s 1. Suppose a↵ ! a. Then, by Cauchy-Swartz inequality, ⇢(a 1

a↵ ) 2 ) 2

! 0, for ⇢ 2 K. Thus, ⇢(a

a↵ )

a↵ )  (⇢(a

! 0, for all ⇢ 2 K. Hence,

w a↵ ! 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, by definition of normal state, ⇢ 2 K =) ⇢(a↵ ) ! ⇢(a). By functional calculus, a2  kaka, for each a 2 M. Now, ⇢((a a↵ )2 )  ka

a↵ k⇢(a

s a↵ ) ! 0. Hence, a↵ ! a.

3. Suppose {p↵ } is a net of projections in M. Then, p↵  e for each ↵. As M is monotone complete, p↵ % p, for some p 2 M. By (2), p↵ by proposition 4.90-(3), p↵ = p↵

s ! p. Now,

s p↵ ! p p. By remark 4.92, 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 s w M. Then, by proposition 4.91, a↵ ! a which inturn implies a↵ ! a. Now, as f

weakly continuous, we get f (a↵ ) ! f (a).

is

Remark 4.94. For a monotone net of elements {a↵ } and any element b in M, we have s a↵ % a =) U↵ (b) ! Ua (b)

(4.3.10)

s s Proof. Let a↵ % a. Then, by proposition 4.91, a↵ ! a and (a↵ a↵ ) ! (a a).

Then, by proposition 4.90-2, it follows that s Tb (a2↵ ) ! Tb (a2 ) s Next, Tb (a↵ ) ! Tb (a) implies s (a↵ Tb (a↵ )) ! (a Tb (a))

So, Ua↵ = 2a↵ (a↵ b)

s (a2↵ b) ! 2a (a b)

(a2 b) = Ua (b).

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, 9 b 2 M such that {b↵ } % b. Then, by proposition 4.91, {b↵ }

s ! b which in turn implies {b↵ }

w ! b. As B is

-weakly closed, b 2 B.

Hence {b↵ } % b in B. Thus, B is monotone complete. w Note that if {a↵ } is a net in B such that {a↵ } ! a in norm, then {a↵ } ! a.

Hence, a 2 B. Thus, B is norm closed. 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 2 B. Then, v

a ! a in norm and also v

a !v a

in -weak topology. But norm limit is also the -weak limit. Thus, v a = a, for all a 2 B. Hence, v is an identity for B. 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 Mp is

weakly closed. Note that by 4.66.1, Mp = {a 2 M | Up a = a}.

Now, if {a↵ } is a net in Mp such that a↵ a↵ = Up (a↵ )

w ! Up (a). By uniqueness of

w ! a in M, then proposition 4.90-(2),

-weak limits, Up (a) = a and thus,

a 2 Mp . Now, by proposition 4.95, it follows that Mp is a JBW-subalgebra. 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 = 0 for all x 2 Mp , y 2 Mp0 . Thus, Mp + Mp0 is a Jordan subalgebra containing Mp and Mp0 . Now, if b 2 Mp and c 2 Mp0 , then by lemma 4.61, we get (Up + Up0 )(b + c) = b + c. Thus, Mp + Mp0 = {a 2 M | (Up + Up0 )a = a} 164

(4.3.11)

4 Spectral Theory for Jordan Algebras We will now show that Mp + Mp0 is

-weakly closed. Let {a↵ } be a net in

w w Mp + Mp0 and let a↵ ! a in M. Then, a↵ = (Up + Up0 )(a↵ ) ! (Up + Up0 )(a). By

uniqueness of -weak limits, (Up + Up0 )(a) = a and thus, a 2 Mp + Mp0 . Hence, Mp + Mp0 is

weakly closed.

Thus, by proposition 4.95, it follows that Mp + Mp0 is a JBW subalgebra of M. Let W (a, e) denote the -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 = C is also associative (see A.8) . Now, by theorem 4.38, C is isometrically isomorphic to some CR (X) . Since C is monotone complete, so is CR (X) . Corollary 4.100. If p is a projection in M, which operator commutes with an element a 2 M, then p operator commutes with every element in W (a, e). Proof. Note that for x 2 M, 4.73

4.3.11

x operator commutes with p () x = Up x+Up0 x () x 2 (Mp +Mp0 ) (4.3.12) So, if a operator commutes with p, then a 2 (Mp + Mp0 ) and (Up + Up0 )e = e 2 Mp + Mp0 . As Mp + Mp0 is a -weakly closed JB subalgebra of M (corollary 4.98), it follows that W (a, e) ✓ Mp + Mp0 . Now, by (4.3.12), we see that p operator 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 2 M, 9! projection p 2 W (a, e) \ f ace(a+ ) ( -weak closure) such that Up a

0, Up a

a. Furthermore, p is the smallest

projection in M such that Up a+ = a+ . Proof. By theorem 4.99, we can identify W (a, e) with a monotone complete CR (X) : W (a, e) ! CR (X) . Let

. Let us call denote this isomorphism as

(a) = a ˆ.

Define E = {x 2 X | a ˆ(x) > 0}. By proposition 1.5, E is both closed and open in X,

E

2 CR (X) , a ˆ

0 on E, a ˆ  0 on X/E and

a) E (ˆ

+

= (ˆ a)+ . Let p =

1

(

E ).

Then, p 2 W (a, e) and satisfies p2 = p. Note that, as W (a, e) is commutative and p 2 W (a, e) it follows that Up (x) = {pxp} = p x, 8 x 2 W (a, e) So,

a) E (ˆ As a ˆ

p a

+

= (ˆ a)+ =) p a+ = a+ =) Up a+ = a+ . 0 on E and a ˆ  0 on X/E, it follows that

a and p a

0. Therefore, Up a

By proposition 1.5,

E

a and Up a

ˆ Ea

0 and

ˆ Ea

a ˆ. So,

0.

is the supremum of an increasing sequence in f ace((ˆ a)+ ).

By the order isomorphism

, it follows that p is the supremum (in W (a, e)) of an

increasing sequence {a↵ } in f ace(a+ ). If b is the supremum of {a↵ } in M, then w by proposition 4.91-(2),(3), a↵ ! b. As, W (a, e) is -weakly closed, b 2 W (a, e)

and so b = p. Thus, p belongs to the

-weak closure of f ace(a+ ). Hence, p 2

W (a, e) \ f ace(a+ ) ( -weak closure). To prove uniqueness, assume that there is another projection q 2 W (a, e) \ f ace(a+ ) ( -weak closure) such that Uq a 166

a and Uq a

0. Let

(q) =

F

for

4 Spectral Theory for Jordan Algebras some F ✓ X and denote G = X/F . Note that 0  Uq a = q Also, 0

a

Uq a =) a ˆ

ˆ Fa

= (1

a F )ˆ

=

ˆ Ga

a =)

 0. Thus, a ˆ

a ˆ  0 on G. By proposition 1.5, E ✓ F , which implies

E



F

ˆ Fa

0.

0 on F and

=) p  q. On

the other hand, Up a+ = a+ =) a+ 2 im+ Up = f ace(p). Thus, q 2 f ace(a+ ) ✓ f ace(p) ✓ im Up . So, q = Up q  Up e = p. Hence, p = q. Finally, if q is any projection in M such that Uq a+ = a+ , then a+ 2 im+ Uq = f ace(q). So, p 2 f ace(a+ ) ✓ f ace(q) ✓ im Uq . Then, as above, p  q. Hence, p is the least projection in M such that Up a+ = a+ . Definition 4.102. Let a be a positive element of a JBW-algebra M. The smallest projection p in M such that Up a = a is called the range projection of a and is denoted by r(a). Proposition 4.103. Let a 2 M + . The range projection r(a) satisfies the following properties: 1. r(a) 2 W (a, e) \ f ace(a) ( -weak closure) 2. a  kakr(a) 3. If

2 K, then (a) = 0 ()

(r(a)) = 0

4. a  b =) r(a)  r(b) 5. r(a)  p = p2 =) Up a = a Proof. Note that if a

0, then a+ = a.

1. Follows from by proposition 4.101. 2. As a  kake, applying Ur(a) on both sides, we get a = Ur(a) a  kakUr(a) e = kakr(a).

167

4 Spectral Theory for Jordan Algebras 3. Applying

2 K on both sides of (2), we get (r(a)) = 0 =)

(a) = 0.

Conversely, if (a) = 0, then (b) = 0, 8 b 2 f ace(a) (4.2.34). And as continuous in the -weak topology,

is

(b) = 0, 8 b in the -weak closure of

f ace(a). Thus, by (1), (r(a)) = 0. 4. Suppose a  b, then by (2) a  b  kbkr(b). So, a 2 f ace(r(b)) = im+ Ur(b) , Thus, Ur(b) a = a. Hence, by definition of range projection, r(a)  r(b). 5. Suppose p is a projection such that r(a)  p, then a 2 im+ Ur(a) = f ace(r(a)) ✓ f ace(p) = im+ Up . Thus, Up a = a.

Remark 4.104. If B is -weakly closed subalgebra of M, containing the identity of M and if 0  a 2 B, then W (a, e) ✓ B. So, r(a) 2 B. Thus, r(a) is also the least projection p in B such that Up a = a, i.e. the notions of range projections in B and M coincide. Proposition 4.105. Let a, b 2 M + . Then a ? b () r(a) ? r(b). Proof. Recall that two positive elements a, b are orthogonal if {aba} = {bab} = 0. (() If r(a) ? r(b), then by definition Ur(a) r(b) = {r(a) r(b) r(a)} = 0. So, by proposition 4.103-(2), we see that 0  Ur(a) b  kbkUr(a) r(b) = 0. This implies Ur(a) b = 0 which is equivalent to Ub (r(a)) = 0. Hence, 0  Ub a  kakUb (r(a)) = 0. So, Ub (a) = 0. Thus, a ? b. ()) If a ? b, then Ua b = 0. So, Ua annihilates f ace(b). As r(b) 2 f ace(b) ( 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 proposition we see that a ? b =) r(a) ? b. Hence, Ur(a) b = 0 which further implies Ur(a)0 b = b (4.2.32). So, a ? b =) {Ur(a) b = 0, Ur(a)0 b = 0}

(4.3.13)

Corollary 4.107. The normal states on M determine the norm and order on M, i.e. for each a 2 M , a

0 ()

0, 8

(a)

kak = sup{| (a)| |

2K

(4.3.14)

2 K}

(4.3.15)

Proof. We first prove (4.3.14). 0, 8 2 K. Let a = a+

(() Let a 2 M such that (a)

orthogonal decomposition of a. Suppose a

a be the unique

6= 0. Since M admits a sep-

arating set of normal states, 9 ⇢ 2 K such that ⇢(a ) 6= 0. Then, by ⇤ proposition 4.103-(3), ⇢(r(a )) 6= 0. Let ⌧ = Ur(a ) ⇢. By proposition

4.90-(1), ⌧ is a positive multiple of a normal state. Hence, ⌧ (a)

0 (by

⇤ hypothesis). Now, ⌧ (e) = (Ur(a ) ⇢)(e) = ⇢(Ur(a ) e) = ⇢(r(a )) 6= 0. Also,

we know that a+ ? a . So, by (4.3.13), we get Ur(a ) a+ = 0. Hence, ⇤ + ⌧ (a) = (Ur(a ) ⇢)(a) = ⇢(Ur(a ) (a

a )) = ⇢ Ur(a ) ( a ) =

which is a contradiction to the fact that ⌧ (a) implies a ()) Let

0. Thus, a

⇢(a ) < 0 = 0 which

0.

2 K. Then, a

0 =)

(a)

0 because normal states are positive

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 2 M, kak  1 ()

e  a  e () (e

a), (e + a)

169

0 () ⇢(e)

⇢(a), ⇢( a), 8 ⇢ 2

4 Spectral Theory for Jordan Algebras K () 1 = ⇢(e)

|⇢(a)|, 8 ⇢ 2 K. Hence, in general, kak 

() |⇢(a)|  , 8 ⇢ 2 K

(4.3.16)

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 equivalent: 1. p ? q 2. p q = 0 3. p  q 0 4. p + q  e 5. Up Uq = 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  q 0 . (3 , 4) p + q  e () p  e (3 ) 5) If p  e

q = q0.

q, then q  e

p. So, Up (q)  Up (e

p) = 0. Now, for a

0,

Up Uq (a)  Up Uq (kake) = kakUp q = 0. So, Up Uq (a) = 0, 8a 2 M + . As M is positively generated, it follows that Up Uq a = 0, 8a 2 M. (5 ) 1) If Up Uq = 0, then 0 = Up Uq (e) = Up q = {pqp}. Thus, p ? q.

Remark 4.110. The positive cone M + is -weakly closed. So, for each projection p 2 M, face(p) = im+ Up = im Up \ M + is

-weakly closed. If 0  a, then

r(a) 2 f ace(a) ( -weak closure) and a 2 f ace(r(a)). So, f ace(r(a)) = f ace(a) ( -weak closure)

(4.3.17)

Proposition 4.111. Let a 2 M and p be a projection in M. Then the following are equivalent: (i) Up a

a, Up a

0

(ii) Up a

0, Up0 a  0 and p operator commutes with a

For each a 2 M , r(a+ ) is the smallest projection satisfying (i) and (ii). Proof. Fix a 2 A. (i ) ii) Suppose p satisfies (i). Then, Up a Up0 (Up a

a) = Up a

a

a. Hence, Up0 a = Up a

0. As Up (Up a

a) = 0, we get

a. Thus, a = Up a + Up0 a which

implies p commutes with a, by proposition 4.73. Also, Up0 a = a Hence, (ii) holds. 171

Up a  0.

4 Spectral Theory for Jordan Algebras (ii ) i) Suppose (ii) holds. Then a = Up a + Up0 a. This implies a 0 =) Up a

Up a = Up0 a 

a.

Next, we will show that for each a 2 M , r(a+ ) is the smallest projection satisfying (i) and (ii). Note that r(a+ ) satisfies (i) by proposition 4.101. If p is any other projection satisfying (i) and (ii), then Up a

0 and Up (Up a) = Up a. Thus,

r(Up a)  p. Similarly, r( Up0 a)  p0 . Now, r(Up a) + r( Up0 a)  (p + p0 ) = e. Then, by proposition 4.109, r(Up a) ? r( Up0 a). By proposition 4.105, we get Up a ?

Up0 a. Also, by (ii), a commutes with p and hence a = Up a + Up0 a. So, a =

(Up a)

( Up0 a) is an orthogonal decomposition of a. By uniqueness of orthogonal

decomposition, it follows that Up a = a+ , Up0 a = follows that

a . Hence, by (4.2.32), it

a 2 Im+ Up0 = Ker+ Up . So, a+ = Up (a) = Up (a+

a ) = Up (a+ ).

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) is isometrically (order and algebra) isomorphic to some monotone complete CR (X) (Theorem 4.99). = { 0,

For each increasing finite sequence and

n

> kak, define k k = max1in (

1

...

n}

in R, with

0




eµ = e ,

kak, define s = ni=1 i (e i e i 1 ). Then, lim ks

k k!0

1

...

n}

with

0