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SPECTRAL THEORY FOR ORDERED

SPACES

A Thesis Submittedin Partial Fulfilment of the Requirements

for the Degree of

Master of Science

by

G. PRIYANGA

to the

School of Mathematical Sciences

National Institute of Science Education and Research

Bhubaneswar

16-05-2017

DECLARATION

I hereby declare that I am the sole author of this thesis, submitted in par-

tial fulfillment of the requirements for a postgraduate degree from the National

Institute of Science Education and Research (NISER), Bhubaneswar. I authorize

NISER to lend this thesis to other institutions or individuals for the purpose of

scholarly research.

Signature of the Student

Date: 16th May, 2017

The thesis work reported in the thesis entitled Spectral Theory for Ordered

Spaces was carried out under my supervision, in the school of Mathematical

Sciences at NISER, Bhubaneswar, India.

Signature of the thesis supervisor

School: Mathematical Sciences

Date: 16th May, 2017

ii

ACKNOWLEDGEMENTS

I would like to express my deepest gratitude to my thesis supervisor Dr. Anil

Karn for his invaluable support, guidance and patience during the two year project

period. I am particularly thankful to him for nurturing my interest in functional

analysis and operator algebras. Without this project and his excellent courses, I

might have never appreciated analysis so much.

I express my warm gratitude to my teachers at NISER and summer project

supervisors, for teaching me all the mathematics that I used in this thesis work.

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 non-

commutative version. In this project, we build a spectral theory and functional cal-

culus for order unit spaces and explore how this generalised spectral theory fits in

the scheme of commutative case (function spaces) as well as the non-commutative

case (Jordan Algebras).

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 or-

dered spaces. So, we begin with an order unit space A, which is in separating

order and norm duality with a base norm space V. Here, we define maps called

compressions, which give rise to projective units in A. Using projective units and

projective faces, we develop various notions like comparability, orthogonality, com-

patibility and also obtain certain lattice structures, under an assumption called

standing hypothesis. Then, 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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

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

2 Order Structure 15

2.1 Ordered Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 15

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

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

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

3 Spectral Theory for Ordered Spaces 30

3.1 Projections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

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

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

3.2 Compressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

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

3.3 Relation between Compressions . . . . . . . . . . . . . . . . . . . . 54

3.3.1 Comparability . . . . . . . . . . . . . . . . . . . . . . . . . . 54

3.3.2 Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . 55

3.3.3 Compatibility . . . . . . . . . . . . . . . . . . . . . . . . . . 57

v

CONTENTS

3.4 The Lattice of Compressions . . . . . . . . . . . . . . . . . . . . . . 65

3.4.1 The lattice of compressions when A = V ⇤ . . . . . . . . . . . 81

3.5 Range Projections . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

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

3.7 Spectral Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

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

4 Spectral Theory for Jordan Algebras 114

4.1 Order Unit Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . 115

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

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

4.2 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

4.3 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 generali-

sation of integration theory. More precisely, as the lebesgue integral of a function

is approximated by a sequence of simple functions, every self-adjoint element of a

von-Neumann algebra can be written as a limit of finite linear sum over orthogonal

projections. This idea is known as Spectral Decomposition Theory. 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 inves-

tigating the spectral theorem for monotone complete CR(X) spaces, which is one

of the primary motivating example for this theory, one learns that projections and

their orthogonality is the fundamental object involved in developing the spectral

result. In order to construct these notions in a more general setting, we define

maps, called compressions, on the order unit space A. The image of the order unit

e under these compressions (known as projective unit) behave like projections in

A. We then explore various properties of the projective units and projective faces

and develop notions of comparability, orthogonality and compatibility between

compressions.

We further find that under an assumption called standing hypothesis, 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. An-

other concrete example of this theory appears in the non-commutative framework

of Jordan algebras. So, the last part of this thesis is devoted to understanding

2

CONTENTS

the spectral theory for JBW-algebras. Here, we study about JB-algebras through

order unit algebras, see examples of concrete compression (Up) 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 con-

struct abstract range projections and obtain a spectral theorem for JBW-algebras,

derived from the spectral theorem for monotone complete CR(X) . The topics on

JB-algebra and JBW-algbera are presented from chapters 1, 2 of [3], while the

section on order unit algebra is based on chapter 1 of [2].

Finally, a caveat: any errors found in this thesis are entirely my own!

4

Chapter 1

Spectral Theory for FunctionSpaces

We begin with the study of a spectral theory which is valid for a class of spaces

of the form CR(X). The spectral theory for function spaces gives an example of

spectral decomposition in the commutative case and serves as a motivation for the

general theory. So, this chapter is devoted to an order theoretic understanding

of the spectral theorem for monotone complete CR(X) spaces. The vector lattice

CR(X) is not monotone complete in general. However, certain important repre-

sentation theorems for abstract algebras (such as the commutative von Neumann

algebras and the normed Jordan algebras known as JBW-algebras) give rise to

compact Hausdor↵ spaces X, for which CR(X) is monotone complete.

In the following pages, we will show that every element of a monotone com-

plete CR(X) space can be approximated in norm by a linear span of projections.

Further, the family of projections associated with a given element is unique and

is characterised by certain order properties (Theorem 1.13). The objective of the

project is to use order theoretic ideas and generalise this spectral theorem to a

larger class of order unit spaces.

Let X be a compact Hausdor↵ space. Let CR(X) denote the set of all continuous

5

1 Spectral Theory for Function Spaces

real valued functions on X. 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 =1

2(f + g + |f � g|), f ^ g =

1

2(f + g � |f � g|), for f, g 2 CR(X)

Here, |f | denotes the usual modulus function |f | : X �! R defined as:

|f |(x) =8<

:

f(x) if f(x) � 0

�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.

1If f 2 CR(X) , then |f | also belongs to CR(X)

6


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 � 0}

Notation 3. Let E ✓ X. Then �E : X �! R is defined as

�E(x) =

8<

:

1 if x 2 E

0 otherwise

Notation 4. Let f 2 CR(X) . The face generated by f+ in CR(X)+, denoted by

face (f+), is the set {g 2 CR(X) | 0 g �f+, for some � � 0}.

1.1 Range Projections

Our first proposition proves the existence of range projections (defined later).

Proposition 1.5. Let X be a compact Hausdor↵ space and assume that CR(X) is

monotone complete. Then for each a 2 CR(X), the set E = {s 2 X | a(s) > 0} is

both closed and open in X. Further,

1. a � 0 on E and a 0 on X r E.

Also, E is the smallest closed subset of X such that a 0 on X r E.

2. E is the smallest closed subset of X for which �Ea+ = a+ (pointwise prod-

uct).

3. �E is the supremum in CR(X) of an increasing sequence in face (a+).

Proof. Let Y = {s 2 X | a(s) > 0}. Then Y = E. For n = 1, 2, 3, . . . define

fn : R �! [0, 1] as

fn(x) =

8>>>><

>>>>:

0, x 0

nx, 0 x 1

n

1, x � 1

n

7


Then, {fn � a}1n=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) � 1n. 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 Ec = 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.

2Compact Hausdor↵ spaces are normal. Hence, it is possible to separate a point and a closedset by a continuous function.

8


2. We will prove that �Ea+ = a+.

For s 2 E, �E(s)a+(s) = 1.a+(s) = a+(s).

For s /2 E, a(s) 0 =) a+(s) = 0. Thus, �E(s)a+(s) = 0.0 = 0 = a+(s).

Hence, �Ea+(s) = a+(s), 8s 2 X.

Next, assume F is a closed subset of X such that �Fa+ = a+. Then, for

s 2 Y , a(s) = a+(s) > 0. This implies �F (s) = 1 =) s 2 F . Hence,

Y ✓ F =) Y = E ✓ F .

3. Note that

a+(s) =

8<

:

a(s), s 2 E

0, otherwise

and for the {fn} defined previously,

(fn � a)(s) = fn(a(s)) =

8>>>><

>>>>:

0, a(s) 0 (i.e. s /2 E)

na(s), 0 a(s) 1

n

1, a(s) � 1

n

Thus, fn � a 2 CR(X)+ and (fn � a)(s) na+(s), 8s 2 X. Hence, fn � a na+ =) fn � a 2 face (a+). Further, we have shown that {fn � a} %b = �E. Therefore, �E is the supremum in CR(X) of the increasing sequence

{fn � a}1n=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 = �E for some closed

and open subset E of X.

9


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 � 0.

Proof. Consider a = a+ � a� and p↵a = (p↵a)+ � (p↵a)�.

Claim. p↵a� = (p↵a)�

10


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.

Now as p↵a � 0, we get p↵a = (p↵a)+ and (p↵a)� = 0. Note that p↵a� �!pa� in CR(X), because a� is positive and bounded. But p↵a� = 0, 8↵. Hence,

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 � 2 R. If p is a projection in CR(X) such that

pa �p, then p 1� r�(a� �1)+

�.

Proof. Note that pa �p =) p(a� �1) 0. Let b = a� �1. Then pb 0 =)(pb)+ = 0. But (pb)+ = pb+ (as proved in the claim of the previous lemma). Thus,

pb+ = 0 =) p0b+ = b+. Hence, r(b+) p0 (by proposition 1.5 (2)). Therefore,

p 1� r�(a� �1)+

�.

1.2 Spectral Theorem

We are now ready to prove our main spectral result for monotone complete CR(X)

spaces.

Theorem 1.13. Let X be a compact Hausdor↵ space such that CR(X) is monotone

complete and let a 2 CR(X). Then, there is a unique family {e�}�2R of projections

in CR(X) such that

(i) e�a �e�, e0�a � �e0�, 8� 2 R

(ii) e� = 0 for � < �kak, e� = 1 for � > kak

(iii) e� eµ for � < µ

(iv)V

µ>� eµ = e�, � 2 R

11


The family {e�} is given by e� = 1� r�(a� �1)+

�.

Further, for each increasing finite sequence � = {�0,�1 . . .�n} with �0 < �kakand �n > kak, define k�k = max1in(�i � �i�1) and s� =

Pni=1 �i(e�i � e�i�1).

Then,

limk�k!0

ks� � ak = 0

Proof. Define e� = 1 � r�(a � �1)+

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

1� e� = r�(a� �1)+

�= �E�

and e� = 1� �E�= �Ec

�.

(i) By proposition 1.5, we know that

a� �1 � 0 on E�, a� �1 0 on Ec� (1.2.1)

e� = 0 on E�, e� = 1 on Ec� (1.2.2)

Using the above and the fact that e� is positive, we get that e�(a� �1) 0

on X. Thus,

e�a �e�, 8�

Similarly,

e0� = 0 on Ec�, e0� = 1 on Ec

� (1.2.3)

Combining 1.2.3 with 1.2.1, we get

e0�a � �e0�, 8�

(ii) When � > kak, we have E� = {s 2 X | a(s) > � > kak} = ; =) e� =

�Ec�= 1.

When � < �kak, we have E� = {s 2 X | a(s) � �kak > �} = X =) e� =

�Ec�= 0.

(iii) � < µ =) {s 2 X | a(s) > �} ◆ {s 2 X | a(s) > µ} =) E� ◆ Eµ =)Ec

� ✓ Ecµ =) �Ec

� �Ec

µ=) e� eµ.

12


(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,

p =V

µ>� eµ and thus

p � e�

Let E = {s 2 X | p(s) = 0}. Note that p(s) = 1 =) eµ(s) = 1, 8µ > �.

Hence, Ec ✓ Ecµ, 8µ > �. Thus, by (1.2.1), a µ1 on Ec, 8µ > �. This

implies a �1 on Ec. Thus, pa p� ( p = 1 on Ec). 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� (1.2.4)

For each � 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),

a � �1 on E�, a �1 on Ec�

a � µ1 on Fµ, a µ1 on F cµ

For � < µ and x 2 Fµ, we have a(x) � µ > � =) x 2 E�. Thus,

Fµ ✓ E�, 8µ > �. But e� = �Ec�and fµ = �F c

µ. So, by remark 1.9, we get

e� fµ, 8µ > �. Then, e� V

µ>� fµ = f�. Thus, e� f�, 8�.

Hence, f� = e�, 8� 2 R.

13


Now, let � = {�0,�1 . . .�n} be a finite increasing sequence of real numbers

with �0 < �kak and �n > kak. Define Ei = {s 2 X | e�i(s) = 1}, i = 1, 2 . . . n.

Then by (ii) and (iii), we see that ; = E0 ✓ E1 ✓ E2 ✓ . . . ✓ En = X. Define

s� =Pn

i=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 on Ec

�i�1, we have �i�1 a �i on Ei\Ec

i�1 = EirEi�1.

So,we get 0 s� � a �i � �i�1 on Ei r Ei�1 which implies ks�(x) � a(x)k �i � �i�1 k�k, 8x 2 X. Therefore, ks� � ak k�k. Hence, s� �! a in norm

as k�k ! 0.

Corollary 1.14. Let X be a compact Hausdor↵ space such that CR(X) is mono-

tone complete. Then, the linear span of projections is dense in CR(X).

Proof. Given a 2 CR(X), let {e�}�2R be the unique family of spectral units asso-

ciated with a. Let �n be the regular partition of⇥� (kak+ 1), (kak+ 1)

⇤of norm

2(kak+1)n

. Define s�n =Pn

i=0 �i(e�i � e�i�1). Then, 8n, s�n belongs to linear span

of projections and s�n �! 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 2 V +

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) 2 V+ and

(u � v) 2 V+ , which implies (v � u) 2 �V+ . This implies v � u = 0 =)v = u. Hence, is anti-symmetric.

16

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+v2 , v2 =

�e�v2 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 � �)e ± e 2 V +

for some � � 0, then it implies ��e 2 V +. Since V + is proper, � must be 0.

Hence, pe(e) = 1.

(ii) By definition of infrimum, given ✏ > 0, 9 0 < � < ✏ such that (pe(v) + �)e±v 2 V +. Adding the positive element (✏��)e, we get pe(v)e+�e+(✏��)e±v 2V + =) (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 �e v e. Then, by definition, pe(v) 1. Conversely, if pe(v) 1,

then �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 � u + u) 2V + =) 0 (v � u) pe(v)e � 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 � 0 | �ke v ke}. Then, pe is a norm on V.

Proof. Clearly, pe(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 � 0. As V+ is proper, we

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

(ii) Let � 2 R. If � = 0, then clearly pe(�v) = �pe(v). Suppose � 6= 0. Then,

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

18

2 Order Structure

inf {k � 0 | ( ke|�| ± v) 2 V +} = |�| inf {k0 � 0 | (k0e ± v) 2 V +} = |�|pe(v)where k0 = k

|�| . 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}1n=1 be a sequence in A+ and let v 2 A such that pe(v � vn) �!0 as n ! 1. For each k 2 N, 9 nk such that pe(v � vn) < 1

k, 8n � nk. Then,

0 vnk 1

ke + 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 2A+. 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 = akak . Then ka0k = 1. Hence, by

(2.1.5), we have �e a0 e; =) �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 enfor 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�1a)ke < 1. Thus, 9 0 < � < 1 such that ��e kak�1a �e =) �e kak�1a

� e =) k a

�kakk 1 =) 1 = k akakk �, which is a

contradiction to the fact that � < 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 kfk = f(e).

Proof. Let us denote the order unit norm on A by pe. Then kfk = 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

kfk f(e). Hence, f is bounded. Now, pe(e) = 1 =) f(e) kfk. So,

kfk = f(e).

(() Assume f is bounded with kfk = f(e). Take v 2 A+. If v = 0, then

f(v) = 0. If v 6= 0, put u = vpe(v)

. Now, pe(u) = 1 and so �e u e. Also,

v 2 A+ =) u � 0. Hence, �e u e and 0 e�u e. So, pe(e�u) 1.

Therefore, f(e � u) kfkke � uk f(e). So, f(e) � f(u) f(e). Hence,

f(u) � 0 and therefore f(v) � 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) � 0, 8⇢ 2 K (2.1.6)

21

2 Order Structure

and

kak = sup {|⇢(a)| | ⇢ 2 K} (2.1.7)

Proof. If a 2 A+, then clearly ⇢(a) � 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 ⇢ := fkfk is a state on A with ⇢(a) < 0,

which is a contradiction. Hence, a 2 A+.

To prove (2.1.7), define � = sup {|⇢(a)| | ⇢ 2 K}. As |⇢(a)| k⇢kkak for each⇢ 2 K, we have � kak. Suppose � < kak. By the definition of order unit norm,

either (�e+ a) /2 A+ or (�e� 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) < �� 0, i.e. |⇢(a)| < �, a

contradiction. Similarly, if (�e � 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,� � 0}

Remark 2.24. If V+ has a base K, then V+ is proper.

22

2 Order Structure

Proof. Let ±v 2 V +\{0} with v = �1k1, �v = �2k2 where �1,�2 > 0, k1, k2 2 K.

Now, 0 = v+ (�v) = �1k1 + �2k2 = (�1 + �2)(�1

�1+�2k1 +

�2�1+�2

k2) = (�1 + �2)k, for

some k 2 K (since K is convex). But, (�1 + �2) > 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 = �1k1 � �2k2, for some �1,�2 � 0, k1, k2 2 K. Now

putting � = �1 + �2 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 gen-

erating 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 | �0k ± v 2 V +, for some k 2 K} where �0 = �

|↵|

= |↵| inf {�0 � 0 | �0k ± v 2 V +, for some k 2 K}= |↵|kvkK .

(ii) Let v, w 2V. 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,

( kk+l

b1+l

k+lb2) 2 K, since K is convex. Hence, (k+l)( k

k+lb1+

lk+l

b2)±(v+w) 2V +. 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 BaseNorm 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, fi = 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 � 0, 8w 2 W+}(iii) W+ = {w 2 W | hv, wi � 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 andpositive 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 2V ⇤|f(v) � 0, 8v 2 V +} and kfk2 = 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, fi = 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) � 0, 8f 2 V ⇤+.

Conversely, assume f(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) � 0, 8f 2 V ⇤+. Hence, v 2 V +.

So, V + = {v 2 V | f(v) � 0, 8f 2 V ⇤+}. And, by definition, (V ⇤)+ = {f 2V ⇤|f(v) � 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)| | kfk2 1}, because |f(v)| kfk2kvk1. And by Hahn-Banach theorem, there exists f 2V ⇤ such that kfk2 1 and f(v) = kvk1. Thus, kvk1 = sup {|f(v)| | kfk2 1}.And, by definition, kfk2 = 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 ⇤+ | kfk = 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 kfk = 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 kfkB = inf {� �

0 | �g ± f 2 (V ⇤)+, for some g 2 S(V )} .

We need to prove that kfk = kfkB, 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� v e and 0 e+ v 2e. Suppose �g± f 2 (V ⇤)+ for some g 2 S,� > 0. Then,

(�g + f)(e� v) = �g(e)� �g(v) + f(e)� f(v) � 0

(�g � f)(e+ v) = �g(e) + �g(v)� f(e)� f(v) � 0

Adding both, we get 2� � 2f(v) � 0; =) � � f(v). Similarly, replacing v

by �v, we get � � f(�v). This implies � � |f(v)|. So,inf {� � 0 | �g ± f 2 (V ⇤)+, for some g 2 S} � sup {|f(v)| | pe(v) 1}.Hence, kfkB � kfk

(ii) Let h = fkfk . Then, h 2 V ⇤

1 = co(S [�S). So, h = ↵g1� (1�↵)g2, for some

g1, g2 2 S,↵ 2 [0, 1]. Take g = ↵g1 + (1� ↵)g2 2 S (since S is convex). Now

g±h = 2↵g1, 2(1�↵)g2 2 (V ⇤)+. This implies g±h 2 V ⇤+ =) kfkg±f 2(V ⇤)+ =) kfk � kfkB.

Thus, kfkB = kfk, 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 kfk = sup {|f(v)| | kvkK 1}.

Claim. eK is an Archimedian order unit for V⇤.

Let v 2 V+ . Then 9 k 2 K,� � 0 such that v = �k. So, eK(v) = �eK(k) =

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

� = sup {|f(k)| | k 2 K}. Then, 8↵ � 0, k 2 K, we have (�eK ± f)(↵k) =

↵(�± 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 1neK , 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 kfke = inf {� � 0 | �eK ± f 2(V ⇤)+}. Now, we will show that kfk = kfke, for all f 2 V ⇤.

(i) Let v 2 V+ . So, v = �k, for some � � 0, k 2 K. Now, (kfkeK ± f)(v) =

(kfkeK ± f)(�k) = �[kfk ± f(k)] � 0. Therefore, (kfkeK ± f) 2 (V ⇤)+

which implies kfk � kfke

(ii) Suppose �eK±f 2 (V ⇤)+, for some � � 0. Then, (�eK±f)(k) � 0, 8k 2 K,

which implies � � |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) � (1 �↵)(k2), for some ↵ 2 [0, 1]. Now, |f(v)| ↵|f(k1)| + (1 � ↵)|f(k2)| sup {|f(k)| | k 2 K} �. This implies, � � sup {|f(v)| | kvkK 1}. So,

inf {� � 0 | �eK ± f 2 (V ⇤)+} � sup {|f(v)| | kvkK 1} which implies

kfke � kfk.

Hence, kfk = kfke. 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 + v2k = kv1k+ kv2k, 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 hTe, vi = kT ⇤vk, 8v 2 V+

Here T ⇤ is the adjoint map on V, satisfying hTa, 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 + v2k = he, v1 + v2i = he, v1i+ he, v2i = kv1k+ kv2k.

(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), hTe, vi =he, T ⇤vi = kT ⇤vk.

29

Chapter 3

Spectral Theory for OrderedSpaces

In this chapter, we develop a spectral theory and functional calculus for an order

unit space A, which is in separating order and norm duality with a base norm

space V, whose distinguished base is K. So, A may also be considered as the set

of continuous a�ne functions on the compact convex set K. The main ingredients

of spectral theory are projections and their orthogonality, as observed in the mo-

tivating example of spectral theory for function spaces. Here, we develop these

ideas in a more general setup, namely for order unit spaces. We fit this scheme

both in the commutative case (function spaces) as well as non-commutative case

(Jordan algebras).

To construct projections in A (not on A), we define certain smooth maps,

called compressions. 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 non-

empty 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, 1

2(0) +12(2x) = x 2 F . Hence,

0 2 F .

Claim. �x 2 F, 8� � 0

First assume � � 1. Then, 1�(�x) + (1 � 1

�)0 = x 2 F . Therefore, �x 2 F .

Next, consider 0 � 1. Then, �x = �x+(1��)0 (convex combination) =)�x 2 F . So, �x 2 F, 8� � 0, x 2 F . Hence, the claim.

Now, consider x, y 2 F . This implies 12x + 1

2y 2 F . Then, by the above

claim, 2(x+y2 ) 2 F ; =) x+ y 2 F . Hence, F is a subcone of X+. Now, let

31


x 2 F and 0 y x. Then, y, (x� y) 2 X+ and 12(x� y) + 1

2y = 12x 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 z. So, ↵x, (1� ↵)y 2 F which implies x, y 2 F ( as F is

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 @eK.

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)

IfH is a hyperplane of an ordered vector space (X, X+), then it splits X into two

half spaces, namely U1 = {a 2 X | a � f�1(↵)} and U2 = {a 2 X | a f�1(↵)}where H = 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 His 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


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 TanCF. If there is no supporting hyperplane of C which

contains F, then TanCF = 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 =

C \ (T

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 \ TanCF

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


()) 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 � t. Let x0 2 @X+ \ H.

Now, x0 2 @X+ = X+\(X+)� =) 9 {xn}1n=1 2 X+ such that xn ! x0

as n ! 1. So, t limn!1h2xn, yi = 2limn!1hxn, 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.

1a property of topological vector space

34


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

y2B• y�1(0) ✓ H. This impliesT

y2B• y�1(0) \ X+ ✓ H \ X+; hence

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 | Px = x}. Andwe 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�Px)+Px. This implies, Rx = R(x�Px)+R(Px).

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

=) Rx = Px, 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


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 hPx, 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 � Px 2 ker P. This implies

hx � Px, yi = 0. So, hx, yi = hPx, 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 = hPx, 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 Tan(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+ \ Tan(ker+P ). By (i), X+ \ Tan(ker+P ) ✓ X+ \ ker P =

ker+P . So, ker+P = X+ \ Tan(ker+P ). Hence, it is semi-exposed.

36


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 () PQ = QP = 0, P +Q = I.

Proof. ()) Assume ker P = im Q and ker Q = im P. Let x 2 X. Then PQ(x) =

P (Qx) = 0 as Qx 2 im Q = ker P . Hence, PQ = 0. Similarly, QP = 0.

Now, (x� Px) + Px = x, 8x 2 X. (x� Px) 2 ker P = im Q. Therefore,

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

Q2y =) Qx + 0 = Qy = x � Px =) Px + Qx = x. Since this is true

8x 2 X, we have P +Q = I.

(() Assume PQ = QP = 0 and P +Q = I.

Claim. im Q = ker P

(✓) Let y 2 im Q =) y = Qx for some x 2 X. Then, Py = P (Qx) =

PQ(x) = 0 =) y 2 ker P . Hence, im Q ✓ ker P .

(◆) Let y 2 ker P . P + Q = I =) Py + 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 comple-

mentary.

Remark 3.24. P admits a strong complement () I � P � 0. This follows from

3.23 with Q = I-P.

37


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 pro-

jection P⇤ on Y. Then

(i) (im+P)• = ker+P⇤

(ii) (ker+P)• � im+P⇤

(iii) (ker+P)• = im+P⇤ () P is smooth

38


(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 ⇤ =) (ker+P )•� = (im+P ⇤)� = (imP ⇤)� =

(kerP �)� � kerP =) (ker+P )•� � 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


(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 com-

plementary and smooth (replacing P,Q by P ⇤, Q⇤ in (a)) =) P,Q are

bicomplementary.

(c) Let P,Q be bicomplementary =) im+P ⇤ = ker+Q⇤ which is a semi-

exposed face of X+. By 3.29 - (iv), P is smooth. Similarly, Q is smooth.

Theorem 3.31. Let P,Q be bicomplementary continuous positive projections on

X. Then E = P + Q is the unique continuous positive projection onto the subspace

im(P+Q)= im(P) � im(Q).

Proof. Let E = P + Q. P, Q are complementary =) PQ = QP = 0.

(i) Since P, Q are continuous, positive and linear, E is a positive continuous

linear map. E2 = (P+Q)(P+Q) = P2 + PQ + QP + Q2 = P2 + Q2 = P +

Q = E. Therefore, 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 Q ⇢im(P+Q) =) im(P+Q) = im P � im Q.

(iii) Let T be another continuous positive projection on X such that im T = im

P � im Q = im E. Now, im T = im E =) ET = T.

40


Claim: ker P ⇢ ker PT

Let x 2 ker P. P,Q are bicomplementary =) P, Q is smooth. So, x 2ker 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 = hTa, P ⇤yi= ha, P ⇤yi = hPa, yi = h0, yi = 0. Now, since x 2 (ker+P )•� and

T ⇤P ⇤y 2 (ker+P )•, hPTx, yi = hx, T ⇤P ⇤yi = 0, 8 y 2 Y+. This

implies hPTx, 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 =) 0 = Ex =

Px+Qx =) Px = �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 kPk 1 (() P = 0 or kPk = 1)

41


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) < Pe, 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 = kPk. Hence, P ⇤ is normalised () P is normalised.

To Prove: P is normalised () Pe e

()) Suppose P is normalised. Now, kek = 1 and kPk 1 =) kPek 1 =) Pe e.

(() Assume Pe e. Take a 2 A 3 kak 1. Then �e a e =)�Pe Pa Pe =) �e �Pe Pa Pe e =) kPak 1 =) kPk 1.

(ii) Let v 2 V +. Then, < Pe, 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 > = < Pe, v > + < Qe, v >

= < Pe+Qe, v > = < e, v > = kvk.

42


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 = k0 for some k’ 2K. < (P + Q)e, k > = < e, (P + Q)⇤k > = < e, (P ⇤ + Q⇤)k > = < e, k0 >

= kk0k = 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


Proof. (i) Assume P ⇤ is neutral. Let v 2 (ker+P )•. Since e - Pe 2 ker+P,

< e � Pe, v > = 0 =) < e, v > = < Pe, 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 2V+ such that kvk = kP ⇤vk =) < Pe, 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 comple-

mentary. By 2.10(iii), P,Q are bicomplementary.

Definition 3.38. Let A be an order unit space, which is in separating order and

norm duality, with a base norm space V. A bicomplemented weakly continuous

normalised positive projection on A is called a compression.

Proposition 3.39. If P is a compression, then

(i) P is smooth

(ii) P ⇤ is neutral

(iii) Q is a compression, where Q is the unique complement of P and

(iv) Pe + Qe = e, kP ⇤v +Q⇤vk = kvk, 8 v 2 V+, (P⇤ + Q⇤)(K) ⇢ K.

Theorem 3.40. Let � : V ! V and �0 : V ! V be weakly continous positive

linear projections with dual maps P = �⇤ : A ! A and P’ = �0⇤ : A ! A. Then

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

44


(i) k�vk+ k�0vk = 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�0vk = kvk, 8 v 2 V +. Finally, P,

P’ are compressions =) they are smooth =) �,�0 are neutral.

(() (a) � and �0 are neutral =) P,P’ are smooth.

(b) Let k 2 K ⇢ V+. Then, k�kk + k�0kk = kkk = 1. =) k�kk 1.

Let v 2 V 3 kvk 1, Since Co(K [ -K) = closed unit ball of V, 9k1, k2 2 K 3 v = �k1 � (1 � �)k2 for some � 2 [0, 1]. Now, k�vk =

k��k1 � (1 � �)�k2k �k�k1k + (1 � �)k�k2k 1. =) k�k 1.

So, kPk = kP ⇤k = k�k 1 . Hence, P is normalised. Similarly, P’ is

normalised.

(c) 8 v 2 V +, k�vk + k�0vk = kvk . This implies k�vk = kvk ()k�0vk = 0 =) �v = v () �0v = 0 (as � is neutral). Therefore

ker+� = im+�0. Similarly, ker+�0 = im+�. So, �,�0 are complemen-

tary =) ��0 = 0 and �0� = 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


(ii) im+P ⇤ = {v 2 V + |< Pe, 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 Pe e =) Pe 2 A+1 . Let v 2 ker+P ⇤.

Then 0 = < e, P ⇤v > = < Pe, 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 = Pa Pe. Hence, a Pe. So, Pe is the largest

such element.

(ii) P is a compression =) P ⇤ is neutral. So, < Pe, v >=< e, v > ()< e, P ⇤v >=< e, v >() kP ⇤vk = kvk () P ⇤v = v. So, im+P ⇤ = {v 2V + | P ⇤v = v} = {v 2 V + |< Pe, 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


3.2.1 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 =) �F ⇢ im+P ⇤, for all

� � 0. Thus, im+P ⇤ ◆ [��0�F .

(✓) Let v 2 im+P ⇤ ⇢ V +. Then v = �k for some � � 0 and k 2 K. �k = v =

P ⇤v = �P ⇤k If � = 0, then v 2 [��0�F . If � 6= 0, then k = P ⇤k =) k 2K \ im P ⇤ = F. So, v 2 [��0�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 � 0 () < a, v >� 0, 8 v 2 F

Proof. (i) Let a 2 A+1 .

(a) a kake for all a in A. If a 2 im+P , then a = Pa kakPe = kakp.

47


(b) face(p) = {a 2 A | a kp, for some k � 0}. So, a kakp =) a 2face(p).

(c) p•• is the smallest semi-exposed face of A+ containing p =) a 2face(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 >� 0, 8 v 2 F, then < a, v >� 0, 8 v 2 �F for some � � 0. By

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

(() Let v 2 F ⇢ im+P ⇤. < a, v >=< a, P ⇤v >=< Pa, v >� 0. Hence,

a � 0 on F.

Let (P, F, p) denote a compression P on A , with associated projective face F

and projective unit p. Each of P, p, F determines the others.

Proposition 3.47. If P’ is the unique compression complementary to P, with

corresponding projective face F’ and projective unit p’ ,then

(i) p + p’ = e

(ii) F = { v 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, < Pa, v >=< a, v >, 8 v 2 F and < Pa, v >= 0, 8 v 2 F’

48


(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 >=< Pe, v >=< e, v >= kvk = 1 }.

(c) Now replacing p by e - p’, we have F = { v 2 K | < e�p0, v >=< e, v >

� < p0, v >= 1 } = { v 2 K | < e� p0, v >= 1� < 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 > =

< Pa, v >, for all a 2 A.

(b) If v 2 F’,then v 2 im+(P 0)⇤ = ker+P ⇤ =) < Pa, 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 A+

1 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 A+

1 () e�b 2 A+1 )

49


= ^ { 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 face(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 Pa = 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 akak). V1 = Co(K [ �K) =) a = ↵s � (1 � ↵)t, for some

s,t in K and ↵ 2 [0, 1]. a = P ⇤a = ↵P ⇤(t) � (1 � ↵)P ⇤(s). Hence,

1 = kak ↵kP ⇤(t)k � (1 � ↵)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. Clearly, im+P is a cone in im+P since, if v1, v2 2 im+P , then v1 = Pv1 �0, v2 = Pv2 � 0 =) v1 + v2 = Pv1 + Pv2 = P (v1 + v2) � 0. Hence,

v1 + v2 2 im+P . Similarly, �v 2 im+P , 8 v 2 im+P,� � 0.

Let v 2 A. Since e is an order unit for A, 9 � � 0 3 �e ± v 2 A+. This

implies P (�e±v) 2 im+P =) (�Pe±Pv) 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,�2 2 [0, 1] 3 �1 + �2 = 1. Since K is convex,

�1f1 + �2f2 2 K. Also P ⇤(�1f1 + �2f2) = (�1f1 + �2f2) 2 imP ⇤ since

f1, f2 2 imP ⇤. Hence, �1f1 + �2f2 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 du-

ality. 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=<

Pa, u >0=< a, P ⇤u >0=< a, P ⇤u >= 0 8 u 2 V. But < a, u >0= 0 8 u 2 V

51


=) 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� 0.

(◆) Suppose a 2 imP 3< a, v >� 0 8 v 2 im+P ⇤. < a, v >0=< Pa, v >0=<

a, P ⇤v >0=< a, P ⇤v >� 0 8 v 2 V . Hence, a � 0. So, a 2 im+P .

Similarly, we can show, im+P ⇤ = {v 2 imP ⇤ |< a, v >� 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, kakpp ± a 2 im+P and e � p.

So, kakpe ± a 2 A+. Hence, kakp � kake. Second,for all � > 0 3�e±a 2 A+ =) �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 kPkkkkK 1. By 3.2.1, P ⇤k 2 Co(F [ �F ). This

implies P ⇤k = ↵f1 � (1 � ↵)f2 for some f1, f2 2 F and ↵ 2 [0, 1].

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

52


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{| < Pa, 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{| < Pa, 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.2sup{| < Pa, 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{| < Pa, v > | |a 2 A, kake 1}.

3sup{| < a, P ⇤v > | | v 2 V, kvkK 1} sup{| < a, u > | | u 2 imP ⇤, kukK 1} andsup{| < 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


3.3 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(Px) =

Px =) QP = P .

2. (ii) =) (iii)

Assume QP = P, then p = Pe = QPe. Since Pe e and Q is positive,

Q(Pe) 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.

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4. (iv) =) (v)

Assume F ⇢ G =) [��0�F ⇢ [��0�G =) im+P ⇤ ⇢ im+Q⇤. Now,

imP ⇤ = im+P ⇤ � im+P ⇤ and imQ⇤ = im+Q⇤ � im+Q⇤. So, im+P ⇤ ⇢im+Q⇤ =) imP ⇤ ⇢ imQ⇤ =) Q⇤P ⇤ = P ⇤(as in (i)). Dualizing gives PQ

= P.

5. (v) =) (vi)

Assume PQ = P. Let x 2 Ker Q. Then Px = PQx = P0 = 0. Hence, x 2Ker P. This implies ker Q ✓ ker P.

6. (vi) =) (vii)

Assume Ker Q ⇢ Ker P. This implies Ker+Q ⇢ Ker+P =) im+Q0 ⇢im+P 0 =) imQ0 ⇢ imP 0.

7. (vii) =) (i)

Assume im Q’ ⇢ im P’. Using (i) =) (vii), with Q’, P’ in place of P, Q

respectively, we get im P ⇢ im Q.

Remark 3.52. The relation � on the set of compressions on A is a partial ordering.

Proposition 3.53. P � Q () Q’ � P’. So, the map P 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


(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 q0 which further implies F ⇢G’.

3. (iv) =) (v)

F ⇢ G0 =) PQ0 = P (by 3.11). So, PQ = PQ’Q = 0.

4. (v) =) (i)

Using (i) =) (v) with P and Q interchanged, we get QP = 0.

Proposition 3.56. Let (A, A+, e), (V, V+, K) be a pair of order unit space and

base norm space in separating order and norm duality. Assume A = V⇤ and let

F,G be projective faces of K. Then

F ? G () 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 q0 =) 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


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

e� a � q i.e. a e� q = q0. So, p a q0. So, by 3.13, F ? G.

3.3.3 Compatibility

Definition 3.57. Let P , Q be two compressions on A. P is said to be compatible

with Q if PQ = QP . P is said to be compatible with an element a 2 A if

a = Pa+ P 0a.

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� Pa = P 0a � 0 =) a � Pa.

(() a� Pa 2 kerP . Further, Pa a =) a� Pa 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, Pq0 q0 =) Pq q.

Proposition 3.60. PQ = QP () Pq q () Qp p

57


Proof. Since (i) is symmetric in P and Q, it su�ces to show (i) () (ii)

()) If PQ = QP, then Pq = PQe = QPe = Qp Qe = q. So Pq q.

(() Assume Pq q. Let a 2 A+. Since 0 a kake, then 0 Qa kakq =)0 PQa kakPq kakq = kakQe. Hence, 0 Q0PQa kakQ0Qe = 0.

This implies Q’PQa = 0, 8 a 2 A+. Therefore, PQa 2 ker+Q0 = im+Q =)QPQa = PQa, 8 a 2 A+ and since A+ generates A, QPQa = PQa, 8 a 2A. Hence, QPQ = PQ.

Now Pq q =) Pq0 q0. 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 = Pa + P 0a, b = Pb + P 0b =) (a + b) = (Pa + P 0a) +

(Pb+ P 0b) = (Pa+ Pb) + (P 0a+ P 0b) = P (a+ b) + P 0(a+ b) =) (a+ b) 2 SP .

And, a 2 SP =) a = Pa+P 0a =) �a = �Pa+�P 0a = P (�a)+P 0(�a) 8� 2 R.

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

58


Now, (P + P 0)2 = P 2 + PP 0 + P 0P + P 02 = P + P 0 ( as PP 0 = P 0P = 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 = Pa+ P 0a () a 2 SP . Hence, SP = im(P + P 0).

Therefore, to show the compatibility of an element a 2 A, with P , it is su�cient

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 =) PQ = P and QP = P =) PQ = QP . Hence,

P and Q are compatible .

2. P ? Q =) P � Q0 =) P and Q0 are compatible . Hence, P and Q are

compatible by 3.62.

Proposition 3.65. Let a 2 A+. If a = 0 on F 0, then a is compatible with P (Pa

= a, P 0a = 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, Pa = a and 0 = (P 0P )a =

P 0(Pa) = P 0a. Hence, a = a+ 0 = Pa+ P 0a =) a is compatible with P .

Remark 3.66. Converse of the above is not true i.e. If a 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 > =

kfk = 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


Proof. P (Pa) + P 0(Pa) = Pa+ 0 = Pa. Hence, Pa is compatible with P .

Let f 2 F ✓ im P ⇤. Then, < Pa, f > = < a, P ⇤f > = < a, f >. Hence,

Pa = a on F . Let f 0 2 F 0 ✓ im+P 0⇤ = ker+P ⇤. Then, < Pa, f 0 > = < a, P ⇤f 0 >

= < a, 0 > = 0. Hence, Pa = 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 > = < Pb + P 0b, v > = < b, P ⇤v > + < b, P 0⇤v > = < a, P ⇤v > + <

0, P 0⇤v > = < Pa, v >. Hence, b = Pa, thus proving the uniqueness.

Proposition 3.68. Let a 2 A. Then (Pa + P 0a) is the unique element in A,

compatible with P, which is equal to a on Co(F [ F 0).

Proof. First, P (Pa+P 0a)+P 0(Pa+P 0a) = P 2a+PP 0a+P 0Pa+P 02a = Pa+P 0a.

Hence, (Pa+ Pa) 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, < Pa + P 0a,↵f + (1 � ↵)f 0 > = ↵ < a, P ⇤f > + ↵ < a, P 0⇤f >

+ (1� ↵) < a, P ⇤f 0 > + (1� ↵) < a, P 0⇤f 0 > = ↵ < a, f > + (1� ↵) < a, f 0 >

= < a,↵f + (1� ↵)f 0 >. Hence, (Pa+ P 0a) = 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 > = < Pb + P 0b, v >= = <

b, P ⇤v > + < b, P 0⇤v > = < a, P ⇤v > + < a, P 0⇤v > = < Pa + P 0a, v >.

So, b = Pa + P 0a on V + =) b = Pa + P 0a on V (as V + generates V). Hence,

b = (Pa+ P 0a), thus proving the uniqueness.

Remark 3.69. An element a 2 A, compatible with P , is completely determined

by its values on F and F 0.

60


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 = Pa+ P 0a = a (as a is compatible with P ). Hence, b = a.

Proposition 3.70. Let a 2 A+. Then

1. Pa =W{b 2 A+ | b = 0 on F 0, b a on F}

2. Pa =V{b 2 A+ | b = 0 on F 0, b � a on F}

Proof. Assume a 2 A+. Then Pa = 0 on F 0 and Pa = a on F . Hence b can be

chosen to be Pa 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 > = <

Pb + P 0b, v > = < b, P ⇤v > + < b, P 0⇤v > < a, P ⇤v > + 0 = < Pa, v >

=) b Pa on V +. Hence, b Pa. So, Pa =W{b 2 A+ | b = 0 on F 0, b

a on F}

2. Imitating the proof for (1), with replaced by �, we get Pa =V{b 2 A+ |

b = 0 on F 0, b � a on F}

The above proposition says that Pa is the best choice of an element in A+

which is 0 on F 0 and closest to a on F.

Proposition 3.71. Let P be a compression on A and T : A �! A be a weakly

continuous projection of A onto the subspace im(P + P 0). Then

T = P + P 0 () T satisfies one of the following conditions:

(i) T is positive

(ii) kTk 1

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(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, kTak 1. Thus, kTk 1.

(iii) P (P + P 0) = P 2 + PP 0 = P and (P + P 0)P = P 2 + P 0P = P . So,

PT = TP . Similarly, TP 0 = P 0T .

(() Given that im T = im (P + P 0).

(i) Suppose T is positive. Then T = P + P 0 by ??.

(ii) Suppose kTk 1. Let a 2 A+. We may assume kak 1 (else replace

a by akak). Now, 0 a e =) �e �a 0 =) 0 e � a

e =) ke � ak 1. So, kT (e � a)k 1 =) �e T (e � a) e =) �e Te � Ta e. But Te = (P + P 0)e = p + p0 = e. Hence,

�e e � Ta e =) 0 Ta. So, Ta � 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. TE(a) = T (Ea) = Ea (as Ea 2 im E = im T ). So,

TE(a) = E(a) 8a 2 A =) TE = E. Similarly, ET = T . So,

T = ET = TE = E = P + P 0.

Remark 3.72. Proposition 3.71 says that P + P 0 is the unique weakly continous

positive projection of A onto the subspace of all elements compatible with P .

62


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 �F and im+P 0⇤ =S

��0 �F0. So, E⇤k = P ⇤k + P 0⇤k = ↵f + �f 0 for some ↵, � � 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 = (↵ + �)k0 2 K for some k0 2 K. But K is a base and the

representation of each element is unique =) (↵ + �) = 1. So, � = 1 � ↵.

Hence, E⇤k = ↵f + (1 � ↵)f 0 2 Co(F [ F 0). Since k is arbitrary, we have

E⇤(K) ✓ Co(F [ F 0). 4

Now, consider v 2 Co(F [ F 0). Let v = ↵f + (1� ↵)f 0 for some ↵ 2 [0, 1], f 2F, f 0 2 F 0. E⇤v = ↵P ⇤f +(1�↵)P ⇤f 0+↵P 0⇤f +(1�↵)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 pro-

jection on V, say T : V �! V such that T (K) ✓ K and T fixes Co(F [ F 0)

pointwise. T (K) ✓ K =) T (V +) ✓ V +. Hence, T is positive.

Claim: im T = 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⇤. Tu = 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 2im T . So, im T ◆ im E⇤.

(✓) Let v 2 im+T . This implies Tv = v = �k for some � � 0, k 2 K. i.e.

�k = v = Tv = T (�k) = �Tk = �(↵f + (1 � ↵)f 0) for some ↵ 2 [0, 1], f 24Alternately, as E⇤k 2 K, we have 1 = kE⇤kk = < e,E⇤k > = < e,↵f + �f 0 > = ↵ <

e, f > + � < e, f 0 > = ↵kfk+ �kf 0k = ↵+ �.

63


F, f 0 2 F 0. If � = 0, then Ev = E0 = 0 = v. Suppose � 6= 0, then we have

k = Tk = ↵f + (1 � ↵)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) Pa 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) Pa a =) I � P � 0. By 3.24, P is strongly complemented.

(ii =) iii) Assume P is strongly complemented. Let P 0 = I�P � 0. Then,

Pe = e�P 0e e =) P is normalised. Also, P 0⇤ = I�P ⇤ � 0 =) P ⇤, P 0⇤

are also strongly complemented with norm 1. So, P is a bicomplemented

normalised weakly continuous positive projection and hence a compression.

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(iii =) iv) Since P 0 = I � P, a = Pa + P 0a, 8a 2 A. So, P is a central

compression.

(iv =) i) Let a 2 A+. P is central =) a = Pa+ P 0a =) Pa 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 ofK 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


Each of these sets have an ordering (C,�), (F ,✓), (P ,) and a notion of com-

plementation (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 = (x1), y2 = (x2).

Then, by order isomorphism, x1 � x2 () (x1) (x2) () y1 y2. Now,

x1 ^ x2 � x1, x2 =) (x1 ^ x2) (x1), (x2).

If y3 2 Y such that y3 (x1), (x2) =) �1y3 x1, x2 =) �1y3 �x1 ^ x2. Applying on both sides, we get y3 (x1 ^ x2). Hence, (x1 ^ x2) =

(x1) ^ (x2).

66


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 \G, F _G = (F 0 \G0)0 (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 =

12(p

0 + q0) 2 A+1 . Then x 2 K \ a�1(0) () x 2 K, < a, x > = 0 () x 2

K, < 12(p

0 + q0), x > = 0 () x 2 K, < p0, x > + < q0, x > = 0 () x 2K, < p0, x > = 0, < q0, 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 com-

pressions (P, F, p), (Q,G, q) on A. Let a 2 A+ such that a p, a q. Then

a p ^ q.

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Proof. The result is trivially true when a 2 P . But we are saying it even for those

a /2 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 PQ = QP = P ^Q.

Proof. Define r = Pq = P (Qe) = Q(P )e = Qp 2 A+1 . Then r = Pq Pe =

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 = Ps Pq = r. In particular, taking s = p ^ q, we get p ^ q r.

So

r = p ^ q = Pe ^Qe = (P ^Q)e (3.4.4)

Now let a 2 A+1 . Then PQa PQe = r = (P ^ Q)e. PQ is a normalised

projection, so kPQak 1. So, by ??, PQa 2 im+(P ^ Q) =) PQa = (P ^Q)(PQ)a = 6 (PQ)(P ^Q)a = 7(P ^Q)a. Therefore, PQ = P ^Q on A+

1 . Hence,

PQ = P ^Q.

Remark 3.86. In the above proof, 3.4.4 tells us that PQe = (P^Q)e. But from this

we directly can’t conclude that PQ = P ^Q because in general PQ need not be a

compression! (For example, in C⇤ algebra, P is a projection () P = P ⇤ = P 2.

(PQ)⇤ = Q⇤P ⇤ = QP . So, if PQ is a projection, then (PQ)2 = PQ =) QP =

PQ. Conversely, if PQ = QP , then (PQ)2 = PQ, thus PQ is a projection. Hence,

6P ^Q � P, Q =) P ^Q is compatible with P, Q =) P ^Q is compatible with PQ7im+(P ^Q) ✓ im+P, im+Q

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compatibility is a necessary and su�cient condition for PQ to be a projection, in

a C⇤ algebra.) The above proposition tells us that under the standing hypothesis,

product of compatible compressions is also a compression.

Proposition 3.87. Assume < A, V > satisfy the standing hypothesis. Let P,Q

be mutually compatible compressions, both compatible with an element a 2 A.

Then, P ^Q, P _Q are also compatible with a.

Proof. First assume a 2 A+. P, Q compatible with a =) Pa,Qa a. So,

(P ^Q)a = (PQ)a = P (Qa) Pa a =) P ^Q ⇠ a. Next, P ⇠ Q =) P 0 ⇠Q0. Also, P, Q ⇠ a =) P 0, Q0 ⇠ a. 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 � a2

where a1 = 12(kake + a), a2 = 1

2(kake � a) 2 A+. Then Pa1 + P 0a1 = 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 _Qis compatible with a and (P _Q)a = Pa+Qa.

Proof. P ? Q =) P ⇠ Q (by 3.64) and we are given P, Q ⇠ a. So, from 3.87

we have P _Q ⇠ a. Then, a = (P _Q)a+ (P _Q)0a = (P _Q)a+ (P 0 ^Q0)a =

(P _ Q)a + (P 0Q0)a. Now, P,Q ⇠ a =) a = Qa + Q0a, a = Pa + P 0a. So,

P 0Q0a = P 0(a � Qa) = P 0a � P 0Qa = 8P 0a � Qa = (a � Pa) � Qa. Substituting

this above, we get a = (P _Q)a+ (a� Pa�Qa) =) (P _Q)a = Pa+Qa.

Notation 13. We write P1+P2+ . . . Pn in place of P1_P2_. . . Pn when P1, P2, . . . Pn

are mutually orthogonal compressions. So by definition,

P = P1+P2+ . . . Pn () P = P1 _ P2 _ . . . Pn and Pi ? Pj for i 6= j (3.4.5)8P ? Q =) Q P 0 =) im+Q ✓ im+P 0 =) P 0Q = Q

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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-

onal compressions, then P1 ? (P2+P3+ . . . Pn).

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,

P1+P2+ . . . Pn is compatible with a and (P1+P2+ . . . Pn)a = P1a+ P2a+ . . . Pna.

Proof. The proof goes by induction on n. The case n=2 was shown in 3.88. Assume

the result is true for n = k � 1. i.e. (P1+P2+ . . . Pn�1)a = P1a+ P2a+ . . . Pn�1a.

Now, when n = k, by 3.90, Pn ? (P1+P2+P3+ . . . Pn�1). Thus by 3.88 and

induction hypothesis, (P1+P2+ . . . Pn)a = Pna + (P1+P2+ . . . Pn�1)a = Pna +

P1a+ P2a+ . . . Pn�1a.

Proposition 3.92. Assume < A, V > satisfy the standing hypothesis. If p1, p2, . . . pn

are projective units, then the following are equivalent:

1.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.

(2) =) (3).Wn

i=1 pi =Wn

i=1 Pie = (Wn

i=1 Pi)e = (P1+P2+ . . . Pn)e =Pn

i=1 Pie (since each Pi ⇠ e). ThusWn

i=1 pi =Pn

i=1 pi.

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(3) =) (1). pi e 8 i =) Wni=1 pi e =) Pn

i=1 pi e

Remark 3.93. From 3.92, it follows that p1 + p2 + . . . pn = p for some projective

unit p () Wni=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

p = p1 + p2 . . . pn () P = P1+P2+ . . . Pn (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

p � q is the projective unit associated with P ^ Q0 = PQ0. Further, R = P ^ Q0

is the unique compression such that P = Q+R and if a 2 A such that a ⇠ P, Q

then Ra = Pa�Qa.

Proof. Q � P =) Q ⇠ P =) Q0 ⇠ P . Then by 3.85, P ^ Q0 = PQ0 is a

compression. PQ0e = P (e�Qe) = Pe� Pq = 9p� q.

Now let R = P ^Q0 = PQ0. 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

that P = Q+S, then p = q + Se =) Se = p� 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 =) Pa = Qa+Ra =) Ra = Pa�Qa.

Remark 3.95. Let P1 � Q1, P2 � Q2, Q1 ? Q2 =) P1 ? P2.

Proof. P1 � Q1 � Q02 � P 0

2 =) 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

P ⇠ Q () P = (P ^Q)+(P ^Q0).9q 2 im+Q ✓ im+P

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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

the result, it is su�cient to prove P ⇠ Q () p = (p ^ q)+(p ^ q0).

()) P ⇠ Q =) P ⇠ Q0 and so P ^ Q = PQ, P ^ Q0 = PQ0. Now (p ^ q) +

(p ^ q0) = PQe+ PQ0e = P (q + q0) = Pe = p.

(() p^q q =) p^q 2 im+Q (by ??) =) Q(p^q) = p^q and p^q0 q0 =)p^q0 2 im+Q0 = ker+Q =) Q(p^q0) = 0. ApplyingQ to p = (p^q)+(p^q0),we get Qp = Q(p^ q) +Q(p^ q0) = p^ q+0 p =) Qp p =) Q ⇠ P .

Remark 3.97. Note that we always have P^Q, P^Q0 � P . So, (P^Q)_(P^Q0) �P trivially. Hence, the above the result can be simplified into P ⇠ Q () P �(P ^Q)+(P ^Q0).

Theorem 3.98. Assume < A, V > satisfy the standing hypothesis. Let P, Q be

two compressions on A. Then

P ⇠ Q () 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,

by 3.95 S ? T and by 3.96, P = (P ^ Q)+(P ^ Q0) = R+S and Q =

(P ^Q)+(P 0 ^Q) = R+T .

(() Assume 9 R, S, T 2 C such that S ? T, P = R+S, Q = R+T . Then,

R � P, Q. Hence

R � P ^Q. (3.4.7)

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Also, R ? S, T ? S =) R, T � S 0 =) R _ T � S 0 =) Q � S 0 =)S � Q0. And S � S _R = P . This implies

S � P ^Q0 (3.4.8)

So, P = R _ S � (P ^ Q) _ (P ^ Q0) which is a su�cient condition for

compatiblity by 3.96. Hence, P ⇠ Q.

Uniqueness: Assume 9 R, S, T 2 C such that S ? T, P = R+S, Q = R+T and let

r = Re, s = Se and t = Te. Then we have, S ? R, T which implies S ?R+T = Q. This implies

Qs = (QS)e = 0 (3.4.9)

Also, R � Q =) r 2 im+Q, which implies

Qr = r (3.4.10)

So, r = r + 0 = Qr + Qs = Q(r + s) = Qp = QPe = (Q ^ P )e = q ^ p.

Hence, r = p ^ q.

Now, Pe = Re + Se =) s = p � r = p � (p ^ q) = p ^ q0 (because

p ⇠ q =) p = (p ^ q) + (p ^ q0)). Similarly, t = q ^ p0.

So, any decomposition has to be of the form R = P ^Q, S = P ^Q0 and T =

Q ^ P 0.

Corollary 3.99. Assume < 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


are mutually orthogonal. P = (P ^Q) _ (P ^Q0) and Q = (P ^Q) _ (P 0 ^Q).

=) P _Q = (P ^Q) _ (P ^Q0) _ (P 0 ^Q)

= (P ^Q)+(P ^Q0)+(P 0 ^Q) =�(P ^Q)+(P 0 ^Q)

�+(P ^Q0)

= P +(P 0 ^Q)

Corollary 3.100. Assume < A, V > satisfy the standing hypothesis. If p,q are

compatible projective units, then p+ q = (p ^ q) + (p _ q)

Proof. Given p ⇠ q. This implies p = (p^ q)+ (p^ q0) and q = (p^ q)+ (p0 ^ q).Hence, p+ q = (p ^ q) + (p ^ q0) + (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 =) y0 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+ ◆

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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 () (Pe)00 = Pe and this is true because (Pe)00 = (e � Pe)0 =

e� (e� Pe) = Pe.

(ii) Q � P =) q p =) �q � �p =) e � q � e � p =) q0 � p0 =)P 0 � Q0

(iii) P ? P 0 =) P ⇠ P 0 =) P ^ P 0 = PP 0 = O. Now, P _ P 0 = (P ^ P 0)0 =

O0 = 10I.

(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. Thenx ^ (y _ z) = (x ^ y) _ (x ^ z) () x _ (y ^ z) = (x _ y) ^ (x _ z).

Proof. First assume x ^ (y _ z) = (x ^ y) _ (x ^ z). Then

(x _ y) ^ (x _ z) =�(x _ y) ^ x

� _ �(x _ y) ^ z�

= x _ �(x ^ z) _ (y ^ z)�

=�x _ (x ^ z)

� _ (y ^ z)

= x _ (y ^ z)

Hence, the dual distributive law is obtained.

Similarly, we can prove the other way implication also.10ker+O = A+ = im+I, im+O = 0 = ker+I

75


Remark 3.106. Boolean Algebras are special cases of orthomodular lattices. Be-

cause, 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 2J ) 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 orthocomple-

mented 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, usingdistributive 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 in-

duced 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


compatible .�P ^ (Q _R)

�e =

�P (Q+R ^Q0)

�e (using 3.99)

=�P (Q+RQ0)

�e (R ⇠ Q0)

= P�(Q+RQ0)e

�

= P�Qe+RQ0e

�( by 3.88 )

= PQe+ PRQ0e

So, we have

p ^ (q _ r) = (p ^ q) + (p ^ q0 ^ r)

= (p ^ q) _ (p ^ q0 ^ 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), P ⇤0 (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 P denote the restriction of P to P0(A).

77


P � P0 =) P, P 0, P0, P 00 are mutually compatible . Now let x 2 P0(A),

say x = P0a for some a 2 A. Then P x0 = PP0a = P0Pa 2 P0(A). So, P0(A) is

invariant under P . This implies P is a projection on P0(A). Since P is positive

and weakly continous on A, P is positive and weakly continous on P0(A).

Next, since kxkP0(A) = kxkA, 8 x 2 P0(A), we have

kPk = sup {kP akP0(A) | kakP0(A) 1, a 2 P0(A)}

= sup {kPakA | kakA 1, a 2 P0(A)}

sup {kPakA | kakA 1, a 2 A}

= kPk 1

Therefore, P is normalised. So, P is a weakly continous normalised positive pro-

jection on P0(A).

Now, let Q = P0P 0 = P 0P0 = P0 ^ P 0 P0 and let Q denote the restriction of

Q to P0(A). Then, Q is also a weakly continous normalised positive projection on

P0(A), like we showed for P .

Claim 1: P 0 (the complement of P in P0(A)) = Q

Let x 2 P0(A), x � 0. Then

x 2 ker+P () P x = 0 () Px = 0 () x 2 ker+P = im+P 0 ()P 0x = x () P 0P0x = x () Qx = x () x 2 im+Q. So, ker+P =

im+Q.

Similarly, x 2 im+P () P x = x () Px = x () x 2 im+P =

ker+P 0 () P 0x = 0 () P 0P0x = 0 () Qx = 0 () x 2 ker+Q

This shows that im+P = ker+Q.

So, P , Q are complementary projections on P0(A).

Claim 2: P ⇤ = P ⇤|P ⇤0 (V ) = P ⇤ restricted to P0(V )

First note that PP0 = P0P =) P ⇤0P

⇤ = P ⇤P ⇤0 . From this it follows that

P ⇤0 (V ) is invariant under P ⇤ and so, P ⇤ is a projection on P ⇤

0 (V ). Now, let

78


x 2 P0(A), y 2 P ⇤0 (V ). Then < x, P ⇤y > = < Px, y > = < Px, y > = <

x, P ⇤y >. This implies P ⇤y = P ⇤|P ⇤0 (V )y. Hence, the claim.

Claim 3: Q⇤ = (P ⇤)0

LHS = (Q)⇤ = Q⇤|P ⇤0 (V ) = (P0P 0)⇤|P ⇤

0 (V ) = P 0⇤P ⇤0 |P ⇤

0 (V ) = P 0⇤|P ⇤0 (V ). There-

fore it is su�cient to prove (P ⇤)0 = P 0⇤|P ⇤0 (V ).

Let x 2 P ⇤0 (V ), x � 0.

x 2 ker+P ⇤ () P ⇤x = 0 () P ⇤x = 0 () x 2 ker+P ⇤ = im+(P ⇤)0 =

im+(P 0)⇤. This implies ker+P ⇤ = im+P 0⇤|P ⇤0 (V ).

Conversely, if x 2 im+(P )⇤ () P ⇤x = x () P ⇤x = x () x 2im+P ⇤ = ker+(P 0)⇤ = ker+(P ⇤)0. This implies im+P ⇤ = ker+P 0⇤|P ⇤

0 (V ).

Hence, P ⇤, Q⇤ are complementary.

From the claims, we see that P is bicomplementary. Hence, P = P |P0(A) is a

compression on P0(A).

Next, we will prove that < P0(A), P ⇤0 (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


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 � p, w > = 0}.Claim: F = {w 2 K |< e� p, w > = 0}

(✓) Let w 2 F =) P ⇤0w = w and < p0, w > = < p,w >. This implies

< P0e, w > = < e, P ⇤0w > = < 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, P ⇤0w > = < e,w > () kP ⇤

0wk = kwk ()P ⇤0w = 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 =) 9 R 2 C with associated projective face F . But F ✓ F0 =)R � 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 P be

a central compression on A and let P be any other compression on A such that

P � P . Then P is central for A () P is central for P (A).

Proof. By 3.75, P is central for A (or P (A)) () Pa a, 8a 2 A+ (or P (A)+).

(() If P is central for A, then Pa a for all a 2 P (A)+ ✓ A+. Thus, P is

central for P (A)+.

80


()) Assume P is central for P (A) =) Pb b, 8 b 2 im+P . In particular,

P (P a) P a, 8 a 2 A+. Now, P � P =) P ⇠ P and P = PP = PP .

Then, Pa = PPa P 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 a�ne functions on K, by an order

preserving isomorphism (a �! a).

Remark 3.112. Under this mapping a 7! a,

{an} �! a weakly in A () {an} �! a 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 an(k) = a(k), 8 k 2 K

() {an} �! a 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


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 a�ne because lim� < a�, tk1 + (1� t)k2 > = lim� t < a�, k1 > + lim�(1� t) <

a�, k2 > = t↵(k1) + (1 � t)↵(k2), for t 2 [0, 1]. Also, supk2K{↵(k)} supk2K <

a0, k > ka0k =) ↵ is bounded. Hence, ↵ 2 Ab(K). Let ↵ correspond to the

element a 2 A under the isomorphism between Ab(K) and A.

Claim: sup{a�} = a

First, note that sup{< a�, k >} = ↵(k) = < a, k > =) < a�, k > < a, k >

, 8 � 2 ⇤ =) a� a, 8 � 2 ⇤.Next, let b � a�, 8 � 2 ⇤. Then < b, k > � < a�, k >, 8 k 2 K =) < b, k > �sup� < a�, k > = < a, k >, 8 k 2 K =) b � 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, Pa↵ % Pa. (or Pa↵ & Pa respectively).

Proof. Note that a↵ % a () a↵ �! a weakly. Since P is positive, Pa↵ is

increasing net and by weak continuity and Pa↵ �! Pa weakly. Hence, Pa↵ %Pa.

Lemma 3.117. Assume < A, V > satisfy the standing hypothesis and A = V ⇤. If

82


{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} ✓ 11G, 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� p↵)&(e� a). Then, (e� a) 2 P as above, which in turn implies that (e� 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.

110 < a,w > < p↵, w > = 0

83


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.


V ⇤. Then the C -Bicommutant of a is a complete boolean algebra.

Orthogonal decomposition in V

Definition 3.123. Two elements ⇢, � 2 V + are said to be orthogonal, denoted

as ⇢ ? �, if k⇢� �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 = ⇢ � �, where ⇢, � �0, ⇢ ? �. This decomposition is unique and is given by ⇢ = P ⇤w, � = P 0 ⇤ w for

some compression P of A.

Central elements


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 = ⇢�� be its unique orthogonal decomposition.

Then ⇢, � are also central.

84


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)


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.


V ⇤. Then the projective units of A are w⇤-dense in the set of extreme points of

A+1 .


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 = p), where p is the

only point of K where p takes the value 1. And the map p 7! p is a one-one map

from the set of minimal points of P onto the set of exposed points of K.

85


Central support


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.


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+


We will say that two elements a, b 2 A+ are orthogonal (a ? b) if there exists a

compression (P, F, p) such that Pa = a, P 0b = 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) Pb = b and P 0c = c

(ii) P 0b = 0 and Pc = 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) Pa ? P 0a

(ii) If a = b� c is an orthogonal decomposition of a such that Pb = b, P 0c = c,

then Pa = b, P 0a = c.

86


Hence, a = Pa+ P 0a and thus a is compatible with P .

3.5 Range Projections

In this section, we use our knowledge of compressions to develop a concept called

range projection, which will be a fundamental tool for the spectral theorem.


V ⇤. Then, for each a 2 A+, there exists a least projective unit p such that a 2face(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 = akak 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}.


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



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

Pa = a.

4. The complement of the compression associated with r(a) is the greatest com-

pression P 0 such that P 0a = 0.

5. r(a) is the greatest projective unit contained in the semi-exposed face of A+

generated by a.

Proof. Let P be the compression associated with r(a).

Then (1), (2), (3), (4) directly follow from the definition of range projections, propo-

sition 3.138, and the fact that a 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 q0 =) q r(a). Thus, r(a) is the greatest projective unit contained in the semi-exposed face

of A+, generated by a.

88


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).


V ⇤. Let a 2 A+ and P be a compression on A with projective unit p. Then

r(Pa) =�r(a) _ p0

� ^ p

Proof. Let Q, R be the compressions associated with r(a) and r(Pa) 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)(Pa) = P (Q _ P 0)a = Pa.

So, (Q _ P 0) fixes Pa. By proposition 3.140 (3), R (Q _ P 0). So,

r(Pa) r(a) _ p0 (3.5.2)

Similarly, P fixes Pa =) R P (proposition 3.140 (3)). Therefore,

r(Pa) p (3.5.3)

From 3.5.2, 3.5.3, we have

r(Pa) �r(a) _ p0

� ^ p (3.5.4)

Next, R P =) R0 ? P . Thus, (R0^P )a = R0Pa = 0. Therefore, a 2 ker+(R0^P ) = im+(R_P 0). Hence, R_P 0 fixes a =) Q R_P 0. So, r(a) r(Pa)_ p0.This implies r(a) _ p0 r(Pa) _ p0. Also, note that r(Pa) _ p0 = r(Pa) + p0 (as

R ? P 0). Therefore,�r(a) _ p0

�� p0 r(Pa). Thus, by proposition ??,

�r(a) _ p0

� ^ p r(Pa) (3.5.5)

Hence, by eqn. 3.5.3 and 3.5.5, we have r(Pa) =�r(a) _ p0

� ^ p.


V ⇤. Let a, b 2 A+. Then

r(a) ? r(b) () a ? b () r(a+ b) = r(a) + r(b)

89


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 RbRa = 0. This implies Rba = Rb(Raa) = 0 =)R0

ba = a. And Rbb = b ( proposition 3.140 (3)). Thus, a ? b.

(() Let a ? b. This implies 9 P 2 P such that Pa = a, P 0b = 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) 12ka +

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.


V ⇤. Let a 2 A+ and Q be a compression compatible with a. Then

r(Q(a)) = Q(r(a))120 a, b a+ b =) kak, kbk ka+ bk

90


Proof. Qa is orthogonal to Q0a. So, by proposition 4.1.6,

r(Qa) + r(Q0a) = r(Qa+Q0a) = r(a)

Let q be the projective unit associated with Q. Then, Qa kakQe =) Q(a) 2face(q). Hence, r(Qa) q. So, r(Qa) 2 im+Q (prop ??). Similarly, r(Q0a) 2im+Q0 = ker+Q. Thus, Q(r(a)) = Q

�r(Qa) + r(Q0a)

�= Q(r(Qa)) + 0 = r(Qa).

Hence, the result.


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 Pa = a and P 0a = 0

(prop. 3.140). Thus, Pa+ P 0a = a which implies r(a) ⇠ a.

Next, let Q be any other compression compatible with a. Then, Qa a (prop

??). By remark 3.141, r(Qa) r(a). Now, r(Q(a)) = Q(r(a)) (prop 4.1.7). Hence,

Q(r(a)) r(a). Thus, Q ⇠ r(a) 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 Pa � a and Pa � 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 0a 0

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

a� P 0a � a.

91


Next, assume Pa � a. Then Pa�a � 0. Thus, P (Pa�a) = P 2a�Pa = 0 =)Pa� a 2 ker+P = im+P 0. So, Pa� a = P 0(Pa� a) = �P 0a =) �P 0a � 0 and

a = Pa+ P 0a. Hence, P 0a 0 and P ⇠ a.

Lemma 3.148. Let A, V be an order unit space and base norm space in separat-

ing order and norm duality. Let a 2 A and P be a compression wth associated

projective face F . Then the following are equivalent:

(i) Pa � a, Pa � 0.

(ii) P ⇠ a, Pa � 0, P 0a 0.

(iii) F ⇠ a, a � 0 on F , a 0 on F 0

Proof. (i) () (ii) follows from lemma 3.147. For (ii) () (iii), recall that a ⇠P () a ⇠ F . Also, Pa � 0 () hPa,wi � 0, 8w 2 K () ha, P ⇤wi �0 8w 2 K () a � 0 on F . Similarly, P 0a 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


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 Pa � a and

Pa � 0. By lemma 3.147, P ⇠ a, P 0a 0. Then, b = a+e2 = 1

2(Pa + P 0a) +

12(Pe+ P 0e) = P (a+e

2 ) + P 0(a+e2 ) = Pb+ P 0b. Thus,

P ⇠ b and b � Pb (3.6.1)

Let p = Pe and p0 = P 0e. Now, P (2b � e) = Pa � 0 =) 2Pb � p. By eqn

4.3.12, we get 2b � 2Pb � 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 Pa =) 2b� e P (2b� e) =

Pa Pe = p =) 2b�p e. Also, 2b�p = b+(b�p) � 0. Thus, 0 2b�p e.

Next, 2b� p 2b =) 2b� p 2 face (b). So, 2b� p 2 face (b) \ A+1 . Hence, by

facial property of b, we get 2b� p b which implies b p. Therefore, b = p.

Lemma 3.152. Let A, V be in spectral duality. If P,Q are compatible compres-

sions, then

PQ = QP = P ^Q

Proof. Define r = PQe. Then, we have r p, r q. Note that p, q � 0 =)p ^ q � 0. Now, if s 2 A+ such that s p, s q, then s 2 im+P, im+Q (by

prop. ??). This implies s = Ps Pq = r. So, r � s whenever 0 s p, q. Thus,

r is the greatest lower bound of (p, q) in A+ 13.

Claim. r has facial property.

13Without the standing hypothesis, we don’t yet know whether join of projective units is aprojective unit. So we can’t conclude that r = (P ^Q)e

93


Let a 2 A+1 \ face (r). Since r p, we have a 2 face (p). This implies

a 2 im+P =) a = Pa Pe = 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 PQ = P ^ Q 14. Let

a 2 A+1 . Then PQa PQe = r = (P ^ Q)e. So, PQa 2 im+(P ^ Q) ✓

im+P, im+Q (by prop ??). Also, P,Q, P ^ Q are mutually compatible. Hence,

PQa = (P ^ Q)(PQa) = P (P ^ Q)(Qa) = PQ(P ^ Q)a = (P ^ Q)a. Thus,

PQ = P ^Q on A+1 =) PQ = 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 = PQ. Further, P ⇠ Q ⇠ P 0 ⇠ Q0 =) P 0 ^ Q0 = P 0Q0. So,

P _Q = (P 0 ^Q0)0 also exists, if P and Q are compatible.

Also, P ? Q =) P ⇠ Q. Therefore, P _Q exists for orthogonal compressions

P and Q.

Lemma 3.153. Let A, V be in spectral duality. Let P,Q be mutually compatible

compressions, both compatible with an element a 2 A. Then, P ^ Q, P _ Q are

also compatible with a.

Proof. First assume a 2 A+. P,Q ⇠ a =) Pa, Qa a. So, by lemma 3.152,

(P ^Q)a = (PQ)a = P (Qa) Pa a =) P ^Q ⇠ a. Next, P ⇠ Q =) P 0 ⇠Q0. Also, P, Q ⇠ a =) P 0, Q0 ⇠ a. 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 wherea1 =

12(kake+a), a2 =

12(kake�a) 2 A+. Then Pa1+P 0a1 = 1

2(kak(p+p0)+a) =

14PQe = r = (P ^ Q)e does not imply PQ = P ^ Q because we do not know if PQ is acompression.

94


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 (by ??).

Proposition 3.154. Let A, V be in spectral duality. Let P,Q be orthogonal com-

pressions, both compatible with an element a 2 A. Then P _Q is compatible with

a and (P _Q)a = Pa+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)0a = (P _Q)a+ (P 0 ^Q0)a =

(P _ Q)a + (P 0Q0)a. Now, P,Q ⇠ a =) a = Qa + Q0a, a = Pa + P 0a. So,

P 0Q0a = P 0(a�Qa) = P 0a� P 0Qa = 15P 0a�Qa = (a� Pa)�Qa. Substituting

this above, we get a = (P _Q)a+ (a� Pa�Qa) =) (P _Q)a = Pa+Qa.

Remark 3.155. We had seen that if A = V ⇤, 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 se-

quence of projective units, then the element infn 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 = infn 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 A+

1 \ face (pn) =

[0, pn]. This implies a pn, 8n. Thus, a infn 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 .

15P ? Q =) Q P 0 =) im+Q ✓ im+P 0 =) P 0Q = Q

95


For an increasing sequence of projective units {qn}, look at the corresponding

decreasing sequence of complement projective units {q0n}. Then {q0n} �! 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 �1,�2 2 R

such that 0 < �1 < �2. If Q2 is a compression compatible with a, with associated

projective unit q2, such that

Q2a �2q2, Q02a � �2q

02 (3.6.2)

then there exists a compression Q1 compatible with a, with associated projective

unit q1, such that Q1 � Q2 and

Q1a �1q1, Q01a � �1q

01 (3.6.3)

Proof. Define b = Q2a 2 A+. Let g = b� �1e. By spectral duality, 9 P 2 C, withassociated projective unit p, such that P ⇠ g, Pg � 0, P 0g 0, i.e.

Pb � �1p, P 0b �1p0, P ⇠ g (3.6.4)

Let Q = P 0, with associated projective unit q = p0. Then

Q0b � �1q0, Qb �1q, Q0 ⇠ g

Now, Q0 is compatible with b, so we have Q0b b =) 0 �1q0 Q0b b.

Applying Q02 to this inequality, we get 0 �1Q0

2q0 Q0

2Q0b Q0

2b = Q02(Q2a) = 0.

Thus, Q02q

0 = 0 =) q0 2 ker+Q02 = im+Q2. Therefore,

Q2q0 = q0 =) Q2 ⇠ Q0 =) Q0Q2 = Q2Q

0 = Q0 ^Q2 (3.6.5)

Thus, Q2Q0 is a compression on A (in general, product of two compressions is not

a compression!) So, Q2q0 = q0 =) Q2Q0e = Q0e. This implies

Q0Q2 = Q2Q0 = Q0 and Q0 � Q2 (3.6.6)

96


Define Q1 = QQ2 = Q ^ Q2. Q0a = Q0Q2a = Q0b b a =) Q0 ⇠ a. Now,

Q ⇠ Q2, Q ⇠ a,Q2 ⇠ a. So, by lemma 3.153, Q1 = Q ^Q2 ⇠ a.

Next, applying Q1 on the inequality 0 Qb �1q, we get Q1(Qb) �1Q1(Qe).

But Q1(Qb) = Q(Q1b) = Q1b = Q1(Q2a) = Q2(Q1a) = Q1a. Also, �1Q1(Qe) =

�1Q(Q1e) = �1Q1e = �1q1. So, we get Q1a �1q1.

Eqn 3.6.6 gives Q0 ? Q02 and (Q1)0 = (Q ^ Q2)0 = Q0 _ Q0

2. As Q,Q2 ⇠ a, by

proposition 3.154, Q01a = Q0a+Q0

2a. Similarly, Q01e = Q0e+Q0

2e =) q01 = q0+q02.

Also, Q0a = Q0Q2a = Q0b. Thus, Q01a = Q0

1b +Q02a � �1q0 + �2q02 � �1(q0 + q02) =

�1q01. Hence, we get Q01a � �1q01.

Now we will show that under the spectral duality condition, the standing hy-

pothesis is satisfied.

Theorem 3.158. If A, V are in spectral duality, then every exposed face of K is

projective.

Proof. Let F be an exposed face of K. Then 9 a 2 A+1 such that F = {w 2 K |

ha, wi = 0}. Take �0 = 1 and Q0 = I. This implies Q0 ⇠ a, Q0a �0e, Q00a �

�00. Choose a decreasing sequence of positive real numbers converging to 0, say

�0 > �1 > �2 > �3 . . . �! 0, i.e. limn �n = 0. By lemma 3.157, we can construct

Q0 ⌫ Q1 ⌫ Q2 ⌫ Q3 ⌫ . . . ⌫ Qn . . . such that

Qi ⇠ a, Qia �iqi, Q0ia � �iq

0i, 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 = infnqn, G =

\

n

Gn Q =^

n

Qn

97


For a given i, �iqi � Qia =) h�iqi � Qia, ki � 0, 8k 2 K. Thus, 8k 2 K, we

have �ihQie, ki � hQia, ki =) �ihe, fi � ha, fi, 8f 2 Gi =) �ie � a on Gi.

As G ✓ Gn, 8n, we have �ne � a on G, 8n. Thus, 0 = limn �ne � a on

G =) a 0 on G. But a 2 A+ =) a � 0 on G. Hence, a = 0 on G. Therefore,

G ✓ F .

Conversely, if w 2 F , then 0 �nhq0n, wi hQ0na, wi ha, wi = 0. This

implies hq0n, wi = 0, 8n. Now, qn & q =) q0n % q0 =) hq0, wi = 0 =) w 2G =) F ✓ G.

Thus, F = G. Hence, F is a projective face.

So, all the results that we had shown previously, assuming the standing hy-

pothesis, are now valid under the spectral duality condition. In particular, we note

that 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 | hPa,wi = 0} is a projective face and its complementary

compression (P , F , p) satisfies F ✓ F, P a = Pa and

P ⇠ a, a > 0 on F , a 0 on F 0

Proof. First note that,

a � 0 on F =) Pa � 0; a 0 on F 0 =) P 0a 0

98


Define G = {w 2 K | hPa,wi = 0}. Then G is an exposed face of K and hence

projective, by theorem 3.158. By proposition ??, Pa = 0 on F 0 which implies

F 0 ✓ G =) G0 ✓ F .

If w 2 G, then

ha, wi = hPa,wi+ hP 0a, wi = 0 + hP 0a, wi 0

If w 2 G0 ✓ F , then

ha, wi = hPa,wi+ hP 0a, wi = hPa,wi+ 0 � 0 (P 0a = 0 on F )

And hPa,wi = 0 =) w 2 G. But, G \G0 = ;. Thus, hPa,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 ??,

Pa = 0 on G =) Q0(Pa) = Pa, Q(Pa) = 0

P 0a = 0 on G0 ✓ F =) Q(P 0a) = P 0a, Q0(P 0a) = 0

Therefore, Qa + Q0a = Q(Pa + P 0a) + Q0(Pa + P 0a) = P 0a + Pa = a =) Q ⇠a, Q0 ⇠ a.

Now, let (P , F , p) = (Q0, G0, Q0e). Then, we have F = G0 ✓ F and P � P .

Also, P a = P (Pa+ P 0a) = Q0Pa+Q0P 0a = Pa+ 0 = Pa.

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


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 F 01. 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 =hPa + P 0a, 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


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 F1, a 0 on F 0

1. Then, by lemma 3.159, 9 F1 2 F such that F ✓ F1, 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 decomposi-

tion.

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, Pa � 0, P 0a 0.

Then a = b � c, with b = Pa � 0, c = �P 0a � 0, is an orthogonal

decomposition of a.

Suppose a = b1 � c1 is another decomposition of a, with b1 = P1a, c1 =

�P 01a for some compression (P1, F1, p1) compatible with a (prop. ??). Then,

P1a � 0 =) a � 0 on F1; P 01a 0 =) a 0 on F1. Now, by lemma

3.159, 9 F 2 F such that F ✓ F, P a = P1a 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 = P1a = P a =

Pa = b and c1 = a�b1 = a�b = c. Hence, a = b�c is the unique orthogonaldecomposition of a.

101


(() Given a 2 A, let a = b � c be the unique orthogonal decomposition of a,

with b = Pa, c = �P 0a, a ⇠ P for some compression (P, F, p) on A. Now,

Pa � 0 =) a � 0 on F and P 0a 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 = Pa and F = G0

where G = {w 2 K | hPa,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 F1, a 0 on F 0

1. Then b1 = P1a � 0, c1 = �P 01a � 0 and a = b1�c1 is an orthogonal

decomposition of a. By assumption, a has a unique orthogonal decomposi-

tion. Thus, b1 = P1a = Pa and c1 = �P 01a = �P 0a . By lemma ??,

b1 = 0 on F 01. Thus, F 0

1 ✓ G = {w 2 K | hP1a = Pa,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+ = Pa and a� = �P 0a 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


1. F is the unique projective face compatible with a such that a > 0 on F and

a 0 on F 0.

2. F is the least projective face compatible with a such that a � 0 on F and

a 0 on F 0.

3. P is the least compression such that Pa � a and Pa � 0.

Proof. Given a 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+ = Pa and a� = �P 0a.

Let H be the projective face associated with the projective unit r(a+).

Claim. F = H

From lemma 3.159, we know that F 0 = G = {w 2 K | hPa,wi = 0} = {w 2K | 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) (3.6.9)

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


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�1P ⇤w 2 F ; � = (1� µ)�1Q⇤w 2 G.

µ = hp, wi = he, P ⇤wi =) 1 = he, µ�1P ⇤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 = hPa + Qa,wi = hPa,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) =) 0 R�b

104


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, R

0�a = R0R0

�a �R0r0�(a) = �R0R0�(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 ? Q0a, we have r�(Qa) + r�(Q0a) = r�(Qa + Q0a) = r�(a), by

proposition 4.1.6. By lemma 3.169 and 4.1.7, r�(Qa) r(Qa) = Q(r(a)) Qe =

105


q. So, r�(Qa) 2 face (q). Then, by lemma ??, r�(Qa) 2 im+Q. Similarly,

r�(Q0a) 2 im+Q0 = ker+Q. Hence, Q(r�(a)) = Q�r�(Qa) + r�(Q0a)

�= 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) 2P�bicommutant of a, for all � 2 R.

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. � < 0

ka��ek�r(a��e)� � a��e � ��e. Thus, e 1��ka��ek�r(a��e)� =) e 2

face (r(a� �e)). Hence, by facial property of projective units, e r(a� �e) e.

Thus, r(a� �e) = e =) r�(a) ⇠ Q.

Case. � = 0. Then, r�(a) = r(a) 2 P-bicommutant of a. So, r�(a) ⇠ Q.

Case. � > 0

By lemma 3.170, 3.171, Q(r�(a)) + Q0(r�(a)) = r�(Qa) + r�(Q0a) = r�(Qa +

Q0a) = r�(a). Thus, r�(a) ⇠ Q.

So, we have proved the result for a � 0. Now, let a 2 A be arbitary. Define

b = a+ kake 2 A+ and µ = �+ kak. Then Q ⇠ b and Q ⇠ rµ(b) (like a � 0). But

rµ(b) = r((µ� be)+) = r((a� �e)+) = r�(a). Hence, Q ⇠ r�(a).

Corollary 3.173. Assume A, V are in spectral duality. Let a 2 A, � 2 R and F

be a compression compatible with a such that a � µe on F , where µ > �. Then,

106


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 � < µ, then rµ(a) r�(a).

3.7 Spectral Theorem

We are now ready to prove our spectral theorem, which generalises the correspond-

ing theorem for von Neumann algebras and JBW algebras.

Theorem 3.175. Assume A, V are in spectral duality and let a 2 A. Then,

there is a unique family {e�}�2R of projective units with associated compressions

(U�, G�, e�) on A such that

(i) e� is compatible with a, for each � 2 R

(ii) U�a �e�, U 0�a � �e0�, 8� 2 R

(iii) e� = 0 for � < �kak, e� = e for � > kak

(iv) e� eµ for � < µ

(v)V

µ>� eµ = e�, for each � 2 R

The family {e�} is given by e� = e�r�(a��e)+� and each e� is in the P�bicommutant

of a.

Proof. Define e� = e � r�(a � �e)+

�= e � r�(a) = r0�(a). Then U� = R0

�, G� =

F 0� 8�.

(i) By proposition 3.172, we have r�, r0� 2 P�bicommutant of a, for each �.

Thus, each e� is in the P�bicommutant of a.

107


(ii) Holds by proposition 3.167.

(iii) If � < �kak, then �e �kake a =) a� �e > 0. So, e r(a� �e) e.

Thus, r(a� �e) = e =) e� = 0.

If � > kak, then a� �e = (a�kake)� (��kak)e 0� (��kak)e 0. So,

(a� �e)+ = 0 =) r�(a) = 0 =) e� = e.

(iv) Let � < µ. Then by result 3.173, Fµ ✓ F� =) F 0� ✓ F 0

µ =) e� eµ.

(v) We will showV

µ>� F0µ = F 0

�, for each � 2 R. Fix � 2 R and µ0 > kak. Then,by (iv), {F 0

µ}µ>�µ=µ0

is a bounded decreasing net of projective faces in K. By

lemma 3.156, this net converges to some projective face G =V

µ>� F0µ =

Tµ>� F

0µ. By proposition ??, G ⇠ a. As F 0

µ ◆ F 0�, 8µ > �, we have

G ◆ F 0� =) G0 ✓ F�. This implies

a > �e on G0

Let w 2 G ✓ F 0µ, 8µ > � =) ha, wi µ, 8µ > � =) ha, wi �. This

implies

a �e on G

Thus, by result 3.167, we have G = F 0�. Hence,

Vµ>� F

0µ = F 0

�.

Uniqueness:

Let {f�}�2R 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 � �r�(a); R0�a �r0�(a)

So, R� � T 0� =) F� ✓ H 0

� =) F 0� ✓ H�.

108


Next, H� =V

µ>� Hµ. T 0µa = a on H 0

µ and T 0µa � µf 0

µ = µT 0µe. Thus,

a � µe on H 0µ. By lemma 3.173, H 0

µ ✓ F�, 8µ > � =) Hµ ✓ F 0�, 8µ >

� =) Tµ>� Hµ = H� ✓ F 0

�.

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�}�2Rbe the spectral resolution of a. For each real increasing finite sequence � = {�0,�1 . . .�n}with �0 < �kak and �n > kak, define k�k = max1in(�i � �i�1). Then the Rie-

mann sums s� :=Pn

i=1 �i�1(e�i � e�i�1), converge in norm to a when k�k ! 0,

i.e.

limk�k!0

ks� � ak = 0

Proof. Given ✏ > 0, choose finite increasing sequence � = {�0,�1 . . .�n} with

�0 < �kak, �n > kak and �i��i�1 < ✏, 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 = Uµa�U 0

�a.

Note that e� eµ =) U�, U 0�, Uµ, U 0

µ are mutually compatible compressions.

Now, Uµa µeµ =) U 0�(Uµa) µU 0

�eµ =) (U 0� ^ Uµ)a µU 0

�Uµe =)(U 0

� ^ Uµ)a µ(e0� ^ eµ). Similarly, U 0�a � �e0� =) UµU 0

�a � �Uµe0� =)(Uµ ^ U 0

�)a � �UµU 0�e =) (Uµ ^ U 0

�)a � �(eµ ^ e0�).

Therefore,

�(eµ � e�) (Uµ ^ U 0�)a µ(eµ � e�) (3.7.1)

109


Let pi = e�i � e�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) Pia �i(e�i � e�i�1), i 2 {1, 2 . . . n} (3.7.2)

Now,Pn

i=0 pi =Pn

i=0(e�i � e�i�1) = en � e0 = e. This implies pi ? pj, for i 6= j

andPn

i=1 Pi = I, a ⇠ Pi, 8i. Hence, by prop ??,Pn

i=1 Pia = a. Summing over n

in (3.7.2), we get

nX

i

�i�1(e�i � e�i�1) a nX

i

�i(e�i � e�i�1)

=) 0 a� s� nX

i

(�i � �i�1)(e�i � e�i�1) k�knX

i

(e�i � e�i�1) = k�ke

=) ka� s�k k�k. Hence, limk�k!0 ks� � ak = 0.

Corollary 3.178. Assume A, V are in spectral duality. Then each a 2 A can be

approximated in norm by linear combinationsPn

i=1 �i�1pi of mutually orthogonal

projections pi in the P�bicommutant of a.

Proof. Let {e�}�2R be the spectral resolution of a and � = {�0,�1 . . .�n} be a

finite increasing sequence of real numbers with �0 < �kak, �n > kak. Putting

pi = e�i � e�i�1 , for i = 1, 2 . . . n, we see that

nX

i=0

pi =nX

i=0

(e�i � e�i�1) = en � e0 = e

By lemma ??, pi ? pj, for i 6= j. Moreover, e�i and e�i�1 belong to P-bicommutant

of a implies pi = e�i � e�i�1 = e�i ^ e0�i�1also belongs to P-bicommutant of a, by

lemma 3.153. By proposition 3.177, the Riemann sums s� =Pn

i �i�1(e�i�e�i�1) =Pn

i=1 �i�1pi converge in norm to a.

110


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

a

f(x)dg(x) = limkPk!0

nX

i=1

f(yi)[g(xi)� g(xi�1)], where yi 2 [xi, xi�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�}�2Rbe 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

⌦ Zf(�)de�,�

↵=

Zf(�)dhe�,�i, for all � 2 V + (3.7.3)

The integral on the right hand side is defined in the sense of Riemann Stieltjes

integral of f with respect to the increasing function � 7! he�,�i.

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,�) =nX

i=1

f(⌘i)he�i � e�i�1 ,�i, L(P,�) =nX

i=1

f(⇠i)he�i � e�i�1 ,�i

111


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 con-

tinous), given ✏ > 0, 9 � > 0 such that |f(x)� f(y)| < ✏ whenever |x� y| < �. So,

for a partition P with kPk �, we have

kU(P,�)� L(P,�)k = knX

i=1

(f(⌘i)� f(⇠i))he�i � e�i�1 ,�ik (3.7.4)

< ✏knX

i=1

he�i � e�i�1 ,�ik (3.7.5)

= ✏he,�i = ✏k�k (3.7.6)

Hence, for each � 2 V +, the Riemann sum

Zf(�)dhe�,�i = lim

kPk!0

nX

i=1

f(�i)he�i � 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)

to be pointwise bounded onK (Rdhe�,�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�}�2Rbe 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) =

Zf(�)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


1. kf(a)k kfk1

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,

Zdhe�, wi = lim

kPk!0

nX

i=1

he�i � e�i�1 , wi = limkPk!0

he, wi = 1

1. kf(a)k = supw2K |hf(a), wi| supw2K | R f(�)dhe�, wi|

supw2K

Z|f(�)|dhe�, wi kfk1 sup

w2K

Zdhe�, wi = kfk1

2. Let v 2 V +. Then ↵hf(a), vi+�hg(a), vi = R↵f(�)dhe�, vi +

R�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 =) Rf(�)dhe�, vi

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

f(a), vi � 0, 8v 2 V + =) f(a) g(a).

4. Fix v 2 V +. By monotone convergence theorem,

fn �! f =)Z

fn(�)dhe�, vi �!Z

f(�)dhe�, vi =) hfn(a), vi �! hf(a), vi

Then, hfn(a), vi �! hf(a), vi, 8v 2 V =) fn(a) �! f(a) in weak*

topology.

113

Chapter 4

Spectral Theory for JordanAlgebras

In the previous chapters, we have developed an abstract spectral theory, for a

general order unit space satisfying certain conditions. Now, we present a concrete

example of these order theoretic constructions and spectral decomposition result,

in the case of Jordan algebras.

In this chapter, we specialise the spectral theory for JBW-algebras. We try to

understand JB algebras, through a larger class of ordered algebras, namely order

unit algebras, which have both algebraic structure as well as order unit space struc-

ture. We show that JB algebras are locally isomorphic to CR(X) spaces and use this

correspondence to develop notions like orthogonality and range projections. Then,

the spectral theorem for JBW algebra follows from the spectral theory for function

spaces, using the local isomorphism with monotone complete CR(X) spaces.

Jordan algebras are commutative algebras by definition. However, they are lo-

cally embedded in an associative parent algebra, which is highly non-commutative,

in general. Hence, this chapter will give us a glimpse of the spectral theorem in a

non-commutative framework.

114

4 Spectral Theory for Jordan Algebras

4.1 Order Unit Algebras

We now begin with the study of order unit algebras. Order unit algebras are order

unit spaces with a normed algebra structure. These spaces contain JB algebras, as

a special case, namely every commutative order unit algebra is a JB-algebra (and

vice-versa). We will show that order unit algebras are locally isomorphic to CR(X)

spaces and later use this result to obtain a spectral result for JBW algebras.

We start with some basic definitions. Throughout this section, A denotes an

order unit algebra with order unit e.

Definition 4.1 (Algebra). An algebra A is a real vector space, with a bilin-

ear product (not necessarily associative or commutative). The algebra is said to

normed if it is equipped with a norm such that for each a, b 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 = aman, 8 m,n 2N, 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


(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 � : A1 �! A2 which preserves the algebra structure,

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.

Proof. Using Taylor series expansion of the function f(x) = (1� x)12 , about 0, we

get

(1� x)12 =

1X

n=0

�nxn where �n =

(�1)n(12)(12 � 1) . . . (12 � (n� 1))

n!(4.1.2)

This series is absolutely and uniformly convergent for |x| 1 (A.9) and absolute

convergence of the series at x = 1 givesP1

n=0 |�n| <1. Now, fix x 2 [0, 1]. Then

(1�x) = (1�x) 12 (1�x) 1

2 =1X

n=0

�nxn

1X

n=0

�nxn =

1X

n=0

�nxn where �n =

nX

k=0

�k�n�k

(4.1.3)

Note that the product series in (4.1.3) is convergent because each of the seriesP1

n=0 �nxn is absolutely convergent in [0, 1]. Also, see that �0 = 1, �1 = �1, �n =

0, 8 n > 1. Next, given a 2 A, satisfying k1� ak 1, define b = 1� a 2 A. Then,

kbk 1. Define s =P1

n=0 �nbn. Note that this series is absolutely convergent 1

and as A is complete, the series converges in A. Now, s2 =P1

n �nbn = 1� b = a.

Therefore, s2 = a.1P1

n=0 |�nbn| P1

n=0 |�n||b|n P1

n=0 |�n| <1

116


Proposition 4.8. If A is an order unit algebra, then A2 = A+.

Proof. By definition of order unit algebra , A2 ✓ A+ . Conversely, let a 2 A+ .

Denote ↵ = kak. Then 0 ↵�1a e. This implies 0 e�↵�1a e which in turn

implies ke � ↵�1ak 1. Now, by proposition 4.7, 9s 2 A such that s2 = ↵�1a.

Then, (p↵s)2 = a. Thus, a 2 A2 implying A+ ✓ A 2.

Remark 4.9. As A+ is closed under addition, lemma 4.8 tells us that in an order

unit algebra, sum of squares is again a square.

Definition 4.10 (Pure States). A linear functional ⇢ on an order unit space (A,

e) is called a state if it is positive and ⇢(e) = 1. The set of all states on A is called

the state space of A. An extreme point of the state space is called a pure state.

Definition 4.11 (Multiplicative States). A linear functional ⇢ on an algebra A is

said to be multiplicative if ⇢(xy) = ⇢(x)⇢(y), 8 x, y 2 A.

Next, we will show a bijective correspondence between the set of pure states

and multiplicative states on an order unit algebra.

Our first lemma is a version of the Cauchy-Schwarz inequality.

Lemma 4.12. Let A be an order unit space, which is also an algebra such that

the distinguished order unit (e) is a multiplicative identity for A. If ⇢ is a state on

A which is positive on squares, then for each a 2 A,

⇢(a)2 ⇢(a2) (4.1.4)

And if ab = ba 2 A, then

⇢(ab)2 ⇢(a2)⇢(b2) (4.1.5)

Proof. Define a product h, i on A as ha, bi = ⇢(12(ab + ba)), 8 a, b 2 A. As, ⇢ is

positive on squares, we see that ha, ai � 0, 8a 2 A. Also, by linearity of ⇢ and

bilinearity of the product on A, it is clear that h, i is a bilinear map. Thus, h, i is

117


a semi-inner product on A. Now, by the general Cauchy-Schwarz inequality (A.6),

we get that |ha, bi|2 ha, aihb, bi, for each a, b 2 A. In particular, when ab = ba,

we get (4.1.5). And, putting b = e in (4.1.5) gives (4.1.4).

Proposition 4.13. Let A be an order unit space, which is also an algebra such

that the distinguished order unit (e) is a multiplicative identity for A. Further,

assume a2 2 A+ , for each a 2 A. Then:

(i) Each multiplicative state on A is pure.

(ii) If ab 2 A+ for each pair a, b 2 A+ , then each pure state is multiplicative.

Proof. Let S denote the state space of A.

(i) Let ⇢ be a multiplicative state on A such that ⇢ = 12⇢1+

12⇢2 for some ⇢1, ⇢2 2

S. Note that ⇢, ⇢1, ⇢2 satisfy the hypothesis of proposition 4.12. Then, for

each a 2 A, we have:

⇢(a2) =1

2⇢1(a

2) +1

2⇢2(a

2) � 1

2⇢1(a)

2 +1

2⇢2(a)

2 (4.1.6)

As ⇢ is multiplicative, we also have

⇢(a2) = ⇢(a)2 =1

4(⇢1(a)

2 + 2⇢1(a)⇢2(a) + ⇢2(a)2) (4.1.7)

Subtracting (4.1.7) from (4.1.6), we get

0 � ⇢1(a)2 + ⇢2(a)

2 � 2⇢1(a)⇢2(a) = (⇢1(a)� ⇢2(a))2 � 0 (4.1.8)

Hence, for each a 2 A, we get ⇢1(a) = ⇢2(a). This implies ⇢ = ⇢1 = ⇢2.

Thus, ⇢ is pure.

(ii) Assume A+ is closed under multiplication. Let ⇢ be a pure state on A. Take

0 a e. Then, e�a 2 A+ . Now, for each x 2 A+ , we have ax, (e�a)x 2A+ . This implies 0 (e � a)x =) ax ex. Define a linear functional

⇢a : A �! R given by s 7! ⇢(as). Also, ax ex = x =) ⇢a(x) ⇢(x), for

each x 2 A+ . Hence, 0 ⇢a ⇢.

118


Claim. 0 ⇢a ⇢ =) 9 � > 0 such that ⇢a = �⇢.

Since ⇢a is a positive linear functional on A, ⇢a = �⇢1 for some � > 0, ⇢1 2S. Infact, � = k⇢ak = ⇢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��. Thus, ⇢e�a = (1��)⇢2.Now, ⇢a + ⇢e�a = ⇢ =) �⇢1 + (1 � �)⇢2 = ⇢. As ⇢ is a pure state on A,

⇢ = ⇢1 = ⇢2. Hence, ⇢a = �⇢.

Let x 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 the-

orem for Ordered Algebras.

Theorem 4.15. Suppose A is a complete order unit space, with state space K,

which is also an algebra satisfying the following:

(i) the order unit (e) is a multiplicative identity for A

(ii) a2 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 @eK, is a w⇤-compact set consisting

of all multiplicative states, and the map a 7! a|@eK is an isometric, order and

algebra isomorphism of A onto CR(@eK).

119


Proof. By proposition 4.13, we see that @eK is precisely the set of multiplicative

states on A.

• w*-compactness of @eK

As K is a w*-compact subset of A⇤, it is su�cient to show that @eK is w*-

closed in K. Let ⇢nw⇤�! ⇢, where ⇢ 2 K, ⇢n 2 @eK, 8n. Then, ⇢n(a) �!

⇢(a), 8a 2A. Take a, b 2A. ⇢(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 im-

plies ⇢ 2 @eK. Thus, @eK is a w*-closed subset of K and hence, w*-compact.

• Algebra homomorphism

Consider the map � : A �! CR(@eK) given by a 7! a|@eK . This map is well

defined because if a 2 A and ⇢nw⇤�! ⇢ in @eK, then ⇢n(a) �! ⇢(a) which

implies a(⇢n) �! a(⇢). Thus, a|@eK is continous.

Clearly, � is linear. For each ⇢ 2 @eK and a, b 2 A, we have ab(⇢) = ⇢(ab) =

⇢(a)⇢(b) = a(⇢)b(⇢). Thus, � is an algebra homomorphism.

• � is an isometry Let a 2 A. Then, kak = sup⇢2K |⇢(a)| A.5= sup⇢2@eK |⇢(a)| =

k�(a)k.

• � is one-one

Suppose a(⇢) = b(⇢), 8 ⇢ 2 @eK. By Krein-Milman Theorem (A.5), K =

Co(@eK [ �@eK). As a, b are continuous linear funtionals on K, it follows

that a(⇢) = b(⇢), 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 com-

plete and � is an isometry, it follows that �(A) is closed in CR(@eK). �(A)

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

120


A such that ⇢1(a) 6= ⇢2(a) which implies �(a)�⇢1� 6= �(a)�⇢2

�. Hence, �(A)

separates points in @eK. Thus, by Stone - Wierestrass theorem (A.7), we

get that �(A) = �(A) = CR(@eK). Therefore, � is onto.

• Order isomorphism

Let a1 a2. Then a2 � a1 � 0. As elements of @eK are positive, ⇢(a2) �⇢(a1), 8⇢ 2 @eK. Hence, a2|@eK � a1|@eK . So, � is positive.

Conversely, if f � 0 in CR(@eK), then, f12 2 CR(@eK) and is positive.

Since, � is an algebra isomorphism, 9 a 2 A such that a = ��1(f12 ). Then,

a2 = ��1(f). As A2 ✓ A+ , we get ��1(f) � 0 in A. So, ��1 is positive.

Hence, � is an order isomorphism.

Hence, � is an order and algebra isomorphism of A onto CR(@eK).

Theorem 4.16. If A is an order unit algebra, then the following are equivalent:

1. A is associative and commutative

2. ab 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.

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Let a be an element in a power associative normed algebra A, with multiplica-

tive identity 1. Then the norm closure of all polynomials in a and 1, denoted by

C(a, 1), is the least norm closed subalgebra in A, containing a and 1. Note that

C(a, 1) is also associative and commutative (A.8). If A is an order unit algebra,

then C(a, 1) is also an order unit algebra (for the ordering, order unit, norm and

product inherited from A). This is referred to as the order unit subalgebra generated

by a.

Corollary 4.17. Let a be an element of an order unit algebra (A, e). Then,

C(a, e) ⇠= CR(X) for some compact Hausdor↵ space X.

Proof. C(a, e) 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 mul-

tiplicative identity e. Assume A satisfies the following condition:

ka2k ka2 + b2k, 8a, b 2 A (4.1.9)

Then the following are equivalent:

(i) a 2 A2

(ii) k↵e� ak kak, for all ↵ � kak(iii) k↵e� ak kak, for one ↵ � kak

Proof. Clearly (ii) ) (iii).

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(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 = t2 for some t 2A. Then, s2+ t2 = a+(e�a) = e. So, ke�ak = kt2k ks2+ t2k = kek = 1.

Therefore, we have shown that kak 1 =) ke� ak 1. Now, for arbitary

a 2 A, take ↵ � kak. Then k a↵k 1 which implies ke � a

↵k 1 by the

previous case. Hence, k↵e� ak ↵.

(iii ) i) Suppose k↵e � ak kak, for some ↵ � kak. Then ke � ↵�1ak 1. By

proposition 4.7, ↵�1a 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 � ak ↵, k�e � bk �.

Therefore, k(↵ + �)e � (a + b)k k↵e � ak + k�e � bk ↵ + �. Again, by the

equivalence of (iii) and (i) in lemma 4.18, we have a + b 2 A2. So, A2 is closed

under addition. And clearly if a = s2 2 A2 and � � 0, then �a = (p�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, k12(a + b)k, k12(a � b)k 1.

So, �e 12(a+ b), 12(a� b) e. By (ii), we get

0 1

4(a+ b)2 e, �e �1

4(a� b)2 0 (4.1.10)

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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 2 A0. Now for arbitary a, b 2 A0, we have k akakk, k b

kbkk 1. Hence, by previous

case, k abkakkbkk 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 (4.1.11)

Proof. First, assume A is an order unit algebra . Let �e a e. Then, kak 1.

This implies ka2k kak2 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) ka2k = kak2

(iii) ka2k ka2 + b2k

Moreover, the ordering of A is uniquely determined by the norm and the identity

element e.2If a1 b1, a2 b2, then a1 + a2 b1 + a2 b1 + b2

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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 kank �! kak as n!1.

Now, kkake�ak = kkake�a+kanke�kanke�an+ank (kak�kank)kek+kan�ak+kkanke�ank (kak�kank)kek+kan�ak+kank �! 0+0+kakas n ! 1. Hence, kkake � 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 = ka2k ka2 + b2k = 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 () �e a e.

()) Let kak 1. Then ke � (e � a)k 1. By proposition 4.7, we see

that e � a 2 A2 = A+ This implies e � a � 0 =) a e. Similarly,

k � ak 1 =) �a e =) �e a. Hence, �e a e.

(() Given�e a e, we get e�a, e+a 2 A+ = A2. Let � = ke�ak. Then,by lemma 4.18, we get ke��1(e�a)k 1. Now consider ��1(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��1(e� a)k 1

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holds in C(a, e) as well. Now, by proposition 4.7, ��1(e�a) has a squareroot 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 = s2t2 = (e� a)(e+ a) = e� a2. Thus,

(e� a2) 2 A2 = A+ . So, we have shown

� e a e =) 0 a2 e (4.1.12)

Therefore, kak2 = ka2k 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 =) ka2k ka2 + b2k for each a, b 2 A. Thus,

(iii) holds. Next we prove (ii). Suppose a 2 A such that ka2k 1. Then,

0 a2 e. Now,

a =1

2

�a2 + e� (a� e)2

� 1

2(a2 + e) e

a =1

2

�(a+ e)2 � a2 � e

� � 1

2(�a2 � e) � �e

Therefore, �e a e which in turn implies kak 1. So, we have shown that

ka2k 1 =) kak 1. Therefore, in general, ka2k � =) kak p�.Taking � = ka2k, we get kak ka2k 1

2 . Hence, kak2 ka2k. The other

inequality is clear because A is a normed algebra . So, ka2k = 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.

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Theorem 4.23. A commutative and associative real Banach algebra A is isomet-

rically isomorphic to CR(X) i↵ the following holds for each a, b 2 A:

(i) kabk kakkbk(ii) ka2k = kak2


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 com-

plete and allows us to use the previous spectral result (theorem 1.13). Further, we

assume that multiplication is separately w*-continuous in A. Under these assump-

tions, we get a least w*-closed subalgebra W (a, e) of A containing a given element

a and e. This subalgebra is the w*-closure of all polynomials in a and e, and it

127


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�nX

i=1

�ipik ✏ (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 a�ne, so is b. Also,

kbk = supk2K |b(k)| 3 supk2K |c(k)| kck. Therefore, b is an a�ne bounded

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

3b↵ c, implies that for each k 2 K, b↵(k) c(k). This is true forall ↵. Hence, lim↵ b↵(k) c(k)

128


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

a � b = 1

2(ab+ ba) (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

a � b = 1

2(ab+ ba)

Also, any subspace of A closed under � forms a Jordan algebra .

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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 asso-

ciative.

Lemma 4.29. Let A be an algebra over R, equipped with a product that is com-

mutative and bilinear, but not necessarily associative. Then x � (x � x2) = x2 �x2 holds for all x 2 A ()

Xxi(xj(xkxl)) =

X(xixj)(xkxl) (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 2A and define x = �1x1+�2x2+�3x3+�4x4 where �t 2 R, 1 t 4. Substituting

this in (4.2.3), we get

4X

i=1

4X

j=1

4X

k=1

4X

l=1

�i�j�k�l xi(xj(xkxl)) =4X

i=1

4X

j=1

4X

k=1

4X

l=1

�i�j�k�l (xixj)(xkxl)

(4.2.5)

130


Let � : A �! R be a linear functional. Then applying � to both sides of (4.2.9)

leads to two polynomials with real coe�cients which are equal for all values of the

variables �1,�2,�3,�4. Therefore, the coe�cients of the corresponding variables

must be same. In particular, equating the coe�cient of �1�2�3�4, we get

��X

xi(xj(xkxl))�= �

�X(xixj)(xkxl)

�(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 argu-

ment. The complete proof may be found in [1].

Corollary 4.31. Every Jordan algebra over R is power associative.

The motivation to single out Jordan algebras that act like self-adjoint operators

on a Hilbert space, leads to the following definition:

Definition 4.32. A JB-algebra A is a Jordan algebra over R with identity element

1, equipped with a complete norm satisfying the following conditions, for each

a, b 2 A:

(i) ka � bk kakkbk(ii) ka2k = kak2


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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 � b =

12(ab+ ba).

A subalgebra of a Jordan algebra which is closed under the Jordan product is

said to be a Jordan subalgebra.

Definition 4.34. Let A be a JB-algebra . A JB-subalgebra of A is a norm closed

linear subspace of A, closed under the Jordan product.

Definition 4.35. A JC-algebra is a JB-algebra A which is isomorphic to a norm

closed Jordan subalgebra of B(H)sa.

Next, we show that a JB-algebra is a commutative order unit algebra .

Proposition 4.36. Every JB-algebra A is a commutative order unit algebra , with

given norm and positive cone A2 = {a2|a 2 A}.

Proof. Since A is a JB-algebra , it is a real Banach space, equipped with a com-

mutative bilinear product, has an identity element e and satisfies the three norm

conditions. Also, by corollary 4.31, A is power associative. Now, by theorem 4.22,

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.

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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 polyno-

mials 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


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(a

2b+ ba2) = 12(a+ a) = a.

(() If a � b = e, then

ab+ ba = 2e (4.2.9)

And if a2 � b = a, then

a2b+ ba2 = 2a (4.2.10)

Now, a(ab+ ba) = 2ae = 2a = 2ea = (ab+ ba)a. So, a2b+ aba = aba+ ba2.

So, we get

ab2 = ba2 (4.2.11)

Putting (4.2.11) in (4.2.10), we get 2a2b = 2a =) a2b = a =) ba2b = ba.

Similarly, ba2 = a =) ba2b = ab. Hence, ba = ba2b = ab. 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.

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We will now show that in the context of JB-algebra , the two notions of invert-

ibility coincide.

Proposition 4.42. For elements a, b in a JB-algebra A, the following are equiv-

alent:

(i) b is the inverse of a in the Banach algebra C(a,e)

(ii) b is the Jordan inverse of a in A

Proof. Note that, in order to use lemma 4.40 for A, we need A to be embedded

inside an associative algebra 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 � : B �! CR(X) be the isometric isomorphism mentioned above. Now,

a � b = e =) �(a)�(b) = 1, i.e. �(a) is invertible in CR(X) . So, 0 /2

135


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 � 7! ��1.

Then, f 2 CR(Y ) and can be approximated by polynomials pn (by Stone-

Weierstrass theorem). Let pn �! f in CR(Y ). Then, pn � �(a) �! f ��(a) in CR(X) . But f � �(a) = (�(a))�1 = �(b). This means that �(b)

can be approximated in CR(X) , by polynomials in �(a). By the isometric

isomorphism, this implies that b can be approximated in B, by polynomials

in a. Thus, b 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. As � preserves invertibility,

�(a) = �(a).

Claim 4.45.1. �(a) = Range (a)

� 2 �(a) () �1 � a is not invertible in CR(X) () 9 x 2 X such that (�1 �a)(x) = 0 () a(x) = �1 () � 2 Range a. Hence, the claim.

Now, a is a continuous function on a compact space X. Hence range(a) is

compact and non-empty. Thus �(a) = �(a) is non-empty and compact.

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Claim 4.45.2. a : X �! �(a) defined as x 7! a(x) is a homeomorphism.

This map is surjective because we have just shown that �(a) = �(a) = Range (a).

To prove injectivity, assume that a(x) = a(y) for some x, y 2 X. Then (a(x))n =

(a(y))n for all n 2 N. Thus, p(x) = p(y) for all polynomials p in a. Now, P (a, e)

is dense in C(a, e) and the isometry � preserves density. Hence, P (a, 1) is dense

in CR(X) . Hence, f(x) = f(y), for all f 2 CR(X) Now, X is a compact, Haus-

dor↵ 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 is a continuous function from X to R. Hence, a 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 Tf = f � (a)�1. Then the map = T � � is an

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

: C(a, e)��! CR(X)

T�! CR(�(a)) (4.2.12)

Note that (e) = T (1) = 1 and takes squares to squares. Thus, is a unital

order isomorphism. Also, (a) = T (a) = a� (a)�1 = I (identity function on �(a)).

Further, is an isometry because kbk = inf{� > 0 | ��e b �e} = inf{� >

0 | (��e) (b) (�e)} = inf{� > 0 | ��1 (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 func-

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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 func-

tional calculus is an isometric order isomorphism from CR(�(a)) onto C(a,e). Fur-

thermore,

1. f(a) has its usual meaning when f is a polynomial.

2. �(f(a)) = f(�(a)), for f 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 �(I) = a. So, if f(x) =Pk

n=0 �nxn, then

�(f) =kX

n=0

�n�(xn) =

kX

n=0

�n

��(x)

�n=

kX

n=0

�n��I(x)

�n=

kX

n=0

�nan = f(a)

(4.2.13)

2. Unital isomorphism preserves invertibility. Thus, �(f(a)) = �(f) = Range

(f) = f(�(a)).

3. Note that �(a)g�! g(�(a)) = �(g(a))

f�! 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 �nxn. Then, (f � g)(x) = f(g(x)) =Pk

n=1 �n

�g(x)

�n=

f(g(a)). Thus, (f � g)(a) = f(g(a)).

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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 � g �! f � g in CR(�(a)). Since

� is an isometry, (pn � g)(a) �! (f � g)(a) in C(a, e) as n ! 1. But,

(pn � 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 � 0 () �(a) ✓ [0,1)

Also, kak = sup{|�| | � 2 �(a)}.

Proof. a � 0 () I(x) � 0, 8 x 2 �(a) () x � 0, 8 x 2 �(a) () �(a) ✓[0,1). Also, kak = kIk = sup{|�| | � 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.

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4.2.2 Triple Product

Definition 4.50 (Triple Product). In any Jordan algebra, we define the triple

product {abc} as:

{abc} = a � (b � c) + c � (b � a)� b � (a � c) (4.2.14)

The following are some properties of this triple product, which can easily be

verified:

(i) {abc} is linear in each factor.

(ii) {abc} = {cba}(iii) In a special Jordan algebra , {abc} = 1

2(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)� (a2 � b) (4.2.15)

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Thus, for a special Jordan algebra , Ua(b) = aba.

Remark 4.54 (Identities). Using Macdonald’s theorem, one can prove that the

following identities hold in any Jordan algebra :

{{aba}c{aba}} = {a{b{aca}b}a} (4.2.16)

{bab}2 = {b{ab2a}b} (4.2.17)

(a � b)2 = 1

4(2a � {bab}+ {ab2a}+ {ba2b}) (4.2.18)

{b{bab}b} = {b2ab2} (4.2.19)

{(e� b){bab}(e� b)} = {(b� b2)a(b� b2)} (4.2.20)

{ab2a}2 = {a{b{ba2b}b}a} (4.2.21)

If p2 = p, then p � a =1

2(a+ {pap}� {p0ap0}) (4.2.22)

Using (4.2.17), one can see that

U{aba} = UaUbUa (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.

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()) Suppose a is invertible in A, with c = a�1 in C(a, e). Now, C(a, e) is as-

sociative implies {ac2a} = a2c2 = e. This implies U{ac2a} = Ue = Identity

function on A. So, I = U{ac2a} = UaUc2Ua (by (4.2.23)). Then, UaUc2 is the

left inverse of Ua and Uc2Ua is the right inverse of Ua. 4 Therefore, Ua is

invertible, say (Ua)�1 = T . Now, {aca} = a2 � c = a � (a � c) = a. Therefore,

Ua = U{aca} = UaUcUa

Now, I = TUa = TUaUcUa = UcUa and I = UaT = UaUcUaT = UaUc.

Hence, UcUa = UaUc = 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 kUabnk �! 0. If Ua had a bounded

inverse, then applying it to the sequence {Uabn} would show bn �! 0, which

contradicts that kbnk = 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�1b�1a�1}.

Proof. If a, b are invertible elements in A, then by lemma 4.55, Ua, Ub have bounded

inverses. Now, U{aba} = UaUbUa implies U{aba} also has a bounded inverse. Then,

by lemma 4.55, {aba} is invertible in A. Using the fact that Ux(x�1) = x�x�x�1 =

x, for each invertible x 2 A, we get: (aba)�1 = U{aba}�1{aba} = (U{aba})�1{aba} =

(UaUbUa)�1{aba} =�(Ua)�1(Ub)�1(Ua)�1

�{aba} = Ua�1Ub�1Ua�1{aba} = Ua�1Ub�1Ua�1Ua(b) =

Ua�1Ub�1(b) = Ua�1(b�1) = {a�1b�1a�1}. So, {aba}�1 = {a�1b�1a�1}.

Theorem 4.57. Let A be a JB-algebra . For each a 2 A, the map Ua is positive

and has norm kak2.4For associative algebras, left inverse is same as right inverse

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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 � 0, then

b is invertible () b � �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 � > 0 such that a � �e. Now, look at b 2 B(a, �2 ) \A�1 = {c 2 A�1 | ka� ck < �

2}. So, ka� bk < �2 =) ��

2 e b� a �2e =) 0

�2e a� �

2e b a+ �2e. Thus, 0 �

2e b. So, by (4.2.24), b 2 A0. Therefore,

A0 is open.

Claim 4.57.2. A0 is closed in A�1.

A+ is closed in A. So, A0 = A+ \ A�1 is closed in A�1.

Claim 4.57.3. A0 is connected.

Let a1, a2 2 A0. Then, 9 �1,�2 > 0 such that a1 � �1e, a2 � �2e. This implies

↵a1 + (1� ↵)a2 � (↵�1 + (1� ↵)�2)e. By (4.2.24), ↵a1 + (1� ↵)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,

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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,

UcUa(b) = {c{aba}c} = {c{ac2a}c} = {cac}2 � 0 (4.2.25)

Now, by (4.2.25), Ua(b) = Uc�1(UcUab) = 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, kUak = kUa(e)k = ka2k = kak2.

Lemma 4.58. If a, b are positive elements of JB-algebra A, then

{aba} = 0 () {bab} = 0 (4.2.26)

Also,

{aba} = 0 =) a � b = 0 (4.2.27)

Proof. Given a, b 2 A+ .

1. Assume {aba} = 0. As b � 0, by functional calculus, 0 b2 kbkb.Applying the positive operator Ua, we get 0 Uab2 kbkUab, which implies

0 {ab2a} kbk{aba} = 0. Thus,

{aba} = 0 =) {ab2a} = 0 (4.2.28)

Now, {bab}2 = {b{ab2a}b} = 0. Thus, {bab}2 = 0 which implies k{bab}k2 =k{bab}2k = 0. Hence, {bab} = 0.

2. Assume {aba} = 0. Then, also {bab} = 0, by (1). Now, by (4.2.28), {ab2a} =

0 and {ba2b} = 0. By (4.2.18), (a�b)2 = 14(2a�{bab}+{ab2a}+{ba2b}) = 0.

Thus, ka � bk2 = k(a � b)2k = 0. Hence, (a � b) = 0.

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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)� (p � a), 8a 2 A (4.2.29)

We denote the projection e� 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

kUpk 1, U2p = Up, UpUe�p = 0 (4.2.31)

If a � 0, then

Up(a) = 0 () Ue�p(a) = a (4.2.32)

Proof. Note that if p 6= 0, then kpk = kp2k = kpk2 =) kpk = 1. So, by theorem

4.57, kUpk = kpk2 = 1.

Using the identity (4.2.19): we get {p{pap}p} = {p2ap2}. So, Up(Up(a)) =

Up{pap} = {p{pap}p} = {p2ap2} = {pap} = Up(a). Therefore, (Up)2 = Up.

Putting b = p in identity (4.2.20), we get UpUe�p = 0.

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Next, let a � 0. Assume Up0a = a. Then applying Up on both sides, we get:

0 = Upa. Conversely, if Upa = 0, then {pap} = 0. By lemma 4.58, we have p�a = 0.

So, p0 � a = (e � p) � a = a. Thus, Up0a = 2p0 � (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) = ⇢(Ta) 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) kU⇤pk 1

(ii) kU⇤p�k = 1 () �(p) = 1 () U⇤

p (�) = �

(iii) U⇤p� = 0 () U⇤

p0� = �

Proof. First we claim that if � is a state on A, then

�(p0) = 0 =) U⇤p� = � (4.2.33)

Proof: Suppose �(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 allb 2 A. Hence, U⇤

p�(a) = �(Up(a)) = �[2p�(p�a)�p�a] = �[2p�(p�a)]��[p�a] =�(a). This is true for all a 2 A. Hence, U⇤

p� = �.

Now, we start proving the proposition.

(i) kU⇤pk = kUpk = 1.

(ii) First assume that �(p) = 1. Then �(p0) = 0. So, by (4.2.33), U⇤p� = �.

Conversely, if U⇤p� = �, then �(p) = �(Up(e)) = U⇤

p�(e) = �(e) = 1.

Also, note that U⇤p� is a positive linear functional on A. Hence, kU⇤

p�k =

U⇤p�(e) = �(Upe) = �(p). Therefore, kU⇤

p�k = 1 () �(p) = 1.

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(iii) Suppose U⇤p� = 0. Then �(Upe) = 0 =) �(p) = 0 =) U⇤

p0� = � (by

(4.2.33)). Conversely, suppose U⇤p0� = �, then applying U⇤

p on both sides, we

get U⇤p� = 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, U⇤p and U⇤

e�p

are complementary projections, by lemma 4.63. Thus, the maps Up are bicomple-

mented normalised positive projections on A. In subsequent sections, we will prove

that these maps are also continuous in some kind of weak topology, arising from

the duality with a base norm space. This will show that Up is an example of the

abstract compression, defined in chapter 3.

Properties of Compressions

The maps Up hold many properties, similar to the abstract compressions of Chapter

3. The following are some such results, which have been proved independently

using the theory of JB algebras.

Notation 15. If a, b are elements of a JB-algebra A, then [a, b] denotes their

associated order interval, i.e. [a, b] = {x 2 A | a x b}. If a is any positive

element of A, then face(a) denotes the face in A+ generated by a, i.e.

face(a) = {y 2 A | 0 y �a, for some � � 0} (4.2.34)

Lemma 4.64. Let p be a projection in a JB-algebra A. Then

(i) face(p) = im+Up

(ii) face(p) \ [0, e] = [0, p]

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Proof. (i) Let a 2 face(p). This implies 0 a �p for some � � 0. This

implies Up00 Up0a �Up0Upe = 0. So, Up0a = 0. Hence, by (4.2.32),

Upa = a which implies a 2 im+Up. Conversely, let a 2 im+Up. Then

a = Upa Up(kake) = kakp which implies a 2 face(p).

(ii) Let a 2 face(p) \ [0, e]. Then, by (i), a 2 im+Up. Thus, a = Upa Upe =

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 @eI denote the set of extreme points of I.

(✓) Let a 2 @eI. 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

for each x 2 �(a) implies that 0 2a � a2 e. Hence, 2a � a2 2 I. Now,

a = 12(a

2 + (2a � 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 face(p) \ [0, e] = [0, p]. So,

p = �a+ (1� �)b �p+ (1� �)p = p. Thus, a = b = p. So, p 2 @eI.

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

|⇢(a � b)| ⇢(a2)12⇢(b2)

12 (4.2.35)

And taking b = e, we get |⇢(a)| ⇢(a2)12 .

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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{ap2a}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 | Upa = 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 �! a in A, where an 2 Up(A) for

all n, then Up(an) �! 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 = Upa which implies Up0a = 0 =){p0ap0} = 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 su�cient to prove the result for 0 a, b. Note that since p is the identity for

149


Ap, p0 � a = a � (p � a) = 0 for all a 2 Ap. Now, 0 b kbke =) 0 b =

Up0b Up0kbke = kbkp0. Applying Ua, we get 0 {aba} kbk{ap0a} = 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 decom-

position 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) =

8<

:

f(x) if f(x) � 0

0 otherwise

f�(x) =

8<

:

0 if f(x) � 0

�f(x) otherwise

Then, f+, f� are continuous positive functionals on X and f = f+ � f�.

Moreover, this is the unique decomposition of f with the property that f+, f� � 0

and f+.f� = 0.

We know that there exists an isometric isomorphism, C(a, e) ⇠= CR(�(a))

with a 7! I. By functional calculus, 9 a+, a� 2 C(a, e) such that a = a+ �a�; a+, a� � 0 and a+ � a� = 05. Also, this decomposition is unique in C(a, e),

5Since a+, a� belong to the associative algebra C(a, e), a+ � a� = 0 =) {a+a�a+} = 0.Hence, a+ ? a�.

150


by functional calculus. Now, we need to show that the orthogonal decomposition

of a is unique in A.

Suppose a = b�c is another orthogonal decomposition of a in A. Then, b, c � 0

and b ? c.

Claim 4.70.1. bn � cm = 0, 8 m,n � 1

We will first show that

0 x, y, x ? y =) xn ? y, 8 n � 1 (4.2.37)

WLOG, assume that kxk, kyk 1. Then, by functional calculus, xn x, 8 n 2 N.

Now, fix n 2 N. Then 0 {yxny} {yxy} = 0 =) {yxny} = 0 =) xn ? y.

Similarly, begin with xn and y, and proceed as above to get xn ? ym for every

fixed m 2 N. By linearity of triple product, the result is true even for any kxk, kyk.Hence, the claim.

Claim 4.70.2. C(b, c, e) is an associative JB-subalgebra of A.

Since bn � cm = 0, 8 m,n � 1, the subalgebra P (b, c, e) of A, is associative and

commutative. As multiplication is jointly continuous in A, C(b, c, e) is also asso-

ciative.

By theorem 4.38, C(b, c, e) ⇠= CR(X) for some compact Hausdor↵ space X. Now,

a = b� c 2 C(b, c, e). Thus, C(a, e) ✓ C(b, c, e). Thus, a = a+ � a� and a = b� c

are two orthogonal decompositions of the image of a in CR(X) . By uniqueness

of orthogonal decomposition in CR(X) , we get a+ = b and a� = c. Hence, the

result.

4.2.5 Commutativity

Next, we want to describe the Jordan analog of elements commuting in an asso-

ciative algebra.

Definition 4.71. Let A be a JB-algebra and let a 2 A. Define Ta : A �! A given

151


by Ta(b) = a � b. Two elements a, b 2 A are said to operator commute in A if

TaTb = TbTa.

It is easy to see that in a special JB-algebra , elements commuting in the

surrounding associative algebra, operator commute in the Jordan algebra.

Lemma 4.72. Let d belong to a JB-algebra A and let p be a projection in A. Then

p � d = p =) TpTd = TdTp.

Proof. Note that (a2 � b) � a = a2 � (b � a) =) Ta2 � Ta(b) = Ta � Ta2(b) =)[Ta2 , Ta] = 0. Now, putting a = d + �1d + �2d in [Ta2 , Ta] = 0 and equating 6 the

coe�cients of �1�2 in the resulting polynomials gives

[Td, Tp] + [Tp, Tp�d] = 0 (4.2.38)

Putting p � d = d gives [Td, Tp] = 0. Hence, p, d operator commute.

Proposition 4.73. Let A be a JB-algebra and let a 2 A and p be a projection in

A. Then the following are equivalent:

(i) a and p operator commute

(ii) Tpa = Upa

(iii) (Up + U 0p)a = a

(iv) p and a are contained in an associative subalgebra

Proof. Recall that Up(b) = 2p � (p � b)� (p � b) = 2TpTp(b)� Tp(b).

(i ) ii) Assume TaTp = TpTa. Then, Up(a) = UpTa(e) = TaUp(e) = Ta(p) =

TaTp(e) = TpTa(e) = Tp(a).

(ii ) iii) Let Tpa = Upa. By identity (4.2.22), we have Tp =12(I +Up �Up0). So, here

we have Up(a) = Tp(a) =12(a+ Upa� Up0a). This implies (Up + U 0

p)a = a.

6adopt the procedure used in the proof of lemma 4.29 to justify why the coe�cients can beequated

152


(iii ) i) Assume (Up + U 0p)a = a. Let r = Upa, s = Up0a. Then r + s = a and

r 2 Ap, s 2 Ap0 . Now, p is the identity for Ap and p0 is the identity

for Ap0 . So, r � p = r and s � p0 = s. Now, by lemma 4.72, we have

TpTr = TrTp and Tp0Ts = TsTp0 and this also implies TpTs = TsTp. Now,

TpTa = Tp(Tr + Ts) = TpTr + TpTs = TrTp + TsTp = (Tr + Ts)Tp = TaTp. So,

a, p operator commute.

(iii ) iv) We know that, p is the identity for the JB-subalgebra Ap. Hence, p and

Up(a) generate an associative subalgebra of Ap. Similarly, p0 and Up0(a)

generate an associative subalgebra of Ap0 . For x 2 Ap and y 2 A0p, we have

x� y = 0, by proposition 4.68. It follows that p, Up(a), p0 and Up0(a) generate

an associative subalgebra B in A. Then, p and (Up +U 0p)a = a are contained

in the associative subalgebra B.

(iv ) iii) If p and a are contained in an associative subalgebra B of A, then Up(a) =

p�a. Then adjoining e and taking the norm closure of B gives an associative

subalgebra of A, which also contains p0. So, here we get Up0(a) = p0 � a =

a� (p � a). And, hence (Up + U 0p)a = a.

Proposition 4.74. Let A be a JB-algebra and let p be a projection in A. Let a � 0.

Then

a operator commutes with p () Upa a (4.2.39)

Proof. If a � 0 and p is a projection which operator commutes with a, then

Upa+ Up0a = a (proposition 4.73). So Upa � a.

Conversely, if a � 0 and Upa a, then a� Upa is a positive element such that

Up(a�Upa) = 0. By (4.2.32), Up0(a�Upa) = (a�Upa). Thus Upa+Up0a = a and

so p and a operator commute by proposition 4.73.

153


Ideals and Approximate Identities

While developing the spectral result for JBW-algebras, we somewhere use the

fact that every norm closed ideal in a JB-algebra has an increasing approximate

identity. The goal of this subsection is to prove the same.

Definition 4.75. An ideal of a Jordan algebra is a linear subspace closed under

multiplication by elements from the algebra .

Definition 4.76. A partially ordered set X is said to be directed upwards if for

each ↵, � 2 X, there exists � 2 X such that ↵, � �. A net is a set which has a

bijection with a directed set (directed upwards).

Definition 4.77. Let J be a norm closed ideal in a JB-algebra A. An increasing

net {v↵} in J is an increasing approximate identity for J if 0 v↵ e, 8↵ and

lim↵ kb � v↵ � bk = 0 for all b 2 J.

Lemma 4.78. If a, b are elements of a JB-algebra , then

1. k{ab2a}k = k{ba2b}k

2. If a � 0, then ka � bk2 kakk{bab}k

Proof. First recall identity (4.2.21), {ab2a}2 = {a{b{ba2b}b}a}.

1. k{ab2a}k2 = k{ab2a}2k = k{a{b{ba2b}b}a}k = kUaUb{ba2b}k kUaUbkk{ba2b}k= k(UaUb)(e)kk{ba2b}k = k{ab2a}kk{ba2b}k. So, we get k{ab2a}k k{ba2b}k.Similarly, we can show k{ba2b}k k{ab2a}k. Hence, k{ab2a}k = k{ba2b}k.

2. Assume a � 0. Let k(a � b)2k = k(a � b)k2 (4.2.18)= 1

4k2a � {bab} + {ab2a} +

{ba2b}k 14k2akk{bab}k+ 1

2k{ba2b}k (by (1)). So, k(a�b)k2 = 12kakk{bab}k+

12kUb(a2)k. By functional calculus, we get a2 kaka. Thus, 0 Ub(a2) kak{bab}. Since order unit norm respects the order for positive elements, we

get kUb(a2)k kakk{bab}k. So, k(a � b)2k 12kakk{bab}k+ 1

2kakk{bab}k =kakk{bab}k.

154


Lemma 4.79. If a, b are invertible elements of a JB-algebra A. Then 0 a b =) b�1 a�1.

Proof. b � 0 =) b�1 � 0, by functional calculus. Also, b�12 exists and (b�1)

12 =

(b12 )�1 � 0. Let us denote (b

12 )�1 by b�

12 . Now, 0 a b =) 0 U

b�12a

Ub�

12b. So, 0 {b� 1

2ab�12} {b� 1

2 bb�12} = e. By functional calculus 7 we get

0 e�1 {b� 12ab�

12}�1 = {b 1

2a�1b12} (by prop 4.56). Hence, 0 b�1 U

b�12e

Ub�

12{b 1

2a�1b12} =

�Ub�

12Ub12

�(a�1) =

�U(b

12 )�1

Ub12

�(a�1) =

�(U

b12)�1U

b12

�(a�1) =

a�1. Thus, 0 b�1 a�1.

Proposition 4.80. Every norm closed ideal J in a JB-algebra A contains an

increasing approximate identity.

Proof. Define U = {a 2 J+ | kak < 1}.

Claim. U is directed upwards.

Note that J+ is directed upwards because if a, b 2 J+, then a, b (a+b) 2 J+.

We will show that U is directed upwards, by constructing an order isomorphism

between U and J+.

Define f : [0, 1) �! [0,1) given by f(t) = t1�t

. Note that f is a continuous

function. Let a 2 U . Then a � 0 and kak < 1 which implies �(a) ✓ [0, 1).

Restricting f to �(a), we get f 2 CR(�(a)) ⇠= C(a, e). Denote the image of f ,

under this isomorphism, as f(a). f � 0 =) f(a) � 0. And f(0) = 0 =)f(a) 2 C(a) ✓ J , by proposition 4.47. Hence, f(a) 2 J+. Moreover, if i denotes

the identity function on �(a), then f = i1�i

= 11�i� 1 =) f(a) = (e� a)�1 � e.

7If x, y 2 CR(X) and 0 x, y are invertible, then 0 x�1, y�1. So, y x =) x�1.y x�1.x =) x�1.y 1 =) x�1.y.y�1 y�1 =) x�1 y�1.

155


Hence, we get a map 8

f : U �! J+ defined as f(a) = (e� a)�1 � e

Next, define g : [0,1) �! [0, 1) given by g(t) = t1+t

. Then g is a continuous

function. Let b 2 J+. Then b � 0 and �(b) ✓ [0,1). Restricting g to �(b), we

get g 2 CR(�(b)) ⇠= C(b, e). Denote the image of g, under this isomorphism, as

g(b). Now, g(0) = 0 =) g(b) 2 C(b) ✓ J , by proposition 4.47. Next, note that

g(x) < 1 for all x 2 �(b). As �(b) is a compact set, g attains its supremum and

hence kgk < 1. This implies kg(b)k < 1. Hence, g(b) 2 U . And, if i denotes the

identity function on �(b), then g = i1+i

= 1� 11+i

=) g(b) = e� (e+ b)�1. Hence,

we get a map

g : J+ �! U defined as g(b) = e� (e+ b)�1

Now, we will show that f is an order isomorphism between U and J+.

Let x, y 2 U such that x y. Then, (e� x) � (e� y). By lemma 4.79, we see

that (e�x)�1 (e�y)�1. Thus, f(x) f(y). Hence, f is an order preserving map.

Similarly, if x, y 2 J+ such that x y, then (e+ x) (e+ y) =) �(e+ x)�1 �(e+ y)�1 =) g(x) g(y). So, g is also an order preserving map.

Next, take a 2 U . Then, by proposition 4.47-(3), we get (g � f)(a) = g(f(a)) =

g�(e � a)�1 � e

�= e � �

e + (e � a)�1 � e��1

= a. Similarly, if b 2 J+, then by

proposition 4.47-(3), we get (f � g)(b) = f(g(b)) = f�e � (e + b)�1

�=�e � (e �

(e+ b)�1)��1� e = b. Hence, (f � g)(b) = b, (g � f)(a) = a, 8 a 2 U, b 2 J+. Thus,

f is an order isomorphism from U to J+, with (f)�1 = g.

Now, if x, y 2 U , then f(x), f(y) 2 J+. And as f(x), f(y) �f(x) + f(y)

� 2J+, we have g(f(x)), g(f(y)) g

�f(x) + f(y)

� 2 U , i.e. x, y g�f(x) + f(y)

� 2U . Thus, U is directed upwards.

8Note that 0 a kake =) (e� a) � (1�kak)e. As kak < 1, we have (e� a) is invertible,by (4.2.24).

156


Consider U as a net over itself. We will show that U is an increasing approxi-

mate identity for J . Clearly, U is an increasing net and 0 a e, for each a 2 U .

For each n 2 N, define hn : R �! R given by

hn(t) =

8>>>>>>>>>>>><

>>>>>>>>>>>>:

n� 1

n, t �1

n(constant function 1� 1

n)

�(n� 1)t,�1n t 0 (linear from �1

nto 0)

0, t = 0

(n� 1)t, 0 t 1

n(linear from 0 to 1

n)

n� 1

n, t � 1

n(constant function 1� 1

n)

We claim that

|t2(1� hn(t))| 1

nwhenever t 2 [�1, 1] (4.2.40)

Fix t 2 [�1, 1] and n 2 N. If |t| � 1n, then |t2(1 � hn(t))| = |t2 1

n| 1

n. The

case t = 0 is clear. Now let 0 t 1n. Then 0 (n � 1)t n�1

n. So,

0 hn(t) 1� 1nThis gives 0 � �hn(t) � 1

n� 1 which implies 1 � 1�hn(t) � 1

n.

So, 0 t2(1 � hn(t)) t2 1n2 1

n. Similarly, it can be shown that when

�1n t 0, we have |t2(1� hn(t))| 1

n. 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 � 1n< 1. So, khn(a)k =

khnk 1. And hn(0) = 0 =) hn(a) 2 C(a) ✓ J , by proposition 4.47. Also,

hn � 0 =) hn(a) � 0. Hence, hn(a) 2 U . Now,

ka2 � {ahn(a)a}k = 9ka2 � (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 1n ✏. If u � hn(a), then we have {ahn(a)a}

9hn(a) 2 C(a) implies that {ahn(a)a} = (a2 � hn(a))

157


{aua} {aea} = a2. So, for all u 2 U with u � hn(a), we have ka2 � {aua}k ka2 � {ahn(a)a}k 1

nby (4.2.41). Hence, limu2U ka2 � {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� (a � u)k2 = ka � (e� u)k2 ke� ukk{a(e�u)a}k (by lemma 4.78) = k(e� u)kka2 � {aua}k ka2 � {aua}k. Hence,

limu2Uka� (a � u)k limu2Uka2 � {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 = infb2J 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 sym-

metrised product is a JBW-algebra.

Throughout this section, M will denote a JBW-algebra.

Notation 16. The set of normal states on M will be denoted by K and V will

denote the linear span of K in M⇤.

158


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:

⇢(a � b) (⇢(a2))12 (⇢(b2))

12 (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 2M | |⇢(x� y)| < r}, where x 2M, ⇢ 2 K, r > 0. Therefore,

b↵�w�! 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) is induced by the semi-norms a 7! (⇢(a2))12 for ⇢ 2 K. The subbasis elements

are given as B(x, ⇢, r) = {y 2 M | (⇢(x� y)2)12 < r}, where x 2 M, ⇢ 2 K, r > 0.

Hence,

b↵�s�! 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.

{a↵ �s�! a, b��s�! b} =) (a↵ + b�)

�s�! a+ b10 (4.3.4)10over directed set (↵,�)

159


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 T ⇤a , U

⇤a : 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

Ta =1

2(Ue+a � Ua � I) (4.3.5)

1. We will first show that U⇤a (V ) ✓ V .

Case (I). a is invertible.

If a is invertible, then Ua becomes an order isomorphism on M, with positive

inverse map Ua�1 (lemma 4.55). So, x y () Uax Uay. Let ⇢ 2 K. If

b � 0, then (U⇤a⇢)(b) = ⇢(Uab) � 0. Also, if b↵ % b in M, then Uab↵ % Uab,

since Ua is an order isomorphism. So, (U⇤a⇢)(b↵) = ⇢(Uab↵) % ⇢(Uab) =

U⇤a⇢(b). Thus, U⇤

a⇢ is positive and normal for each ⇢ 2 K. Hence, U⇤a (K) ✓

V which implies U⇤a (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

Ua =1

2(U�e+a + U�e�a � 2�2I) (4.3.6)

160


By case I, It follows that U⇤a maps V into V for every a 2 A.

Now, by (4.3.5), we also see that T ⇤a maps V into V, for all a 2 A.

2. Let b↵�w�! b. This implies ⇢(b↵) �! ⇢(b), 8⇢ 2 K =) ⇢(b↵) �! ⇢(b), 8⇢ 2

V . In particular, (U⇤a⇢)(b↵) �! (U⇤

a⇢)(b), 8 ⇢ 2 K, by (1). This implies,

⇢(Ua(b↵)) �! ⇢(Ua(b)), 8⇢ 2 K =) Ua(b↵)�w�! 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�!0 =) Ua(b2↵)

�w�! 0. Now, as 0 a2 ka2ke = kak2e, we have {ab↵a}2 ={a{b↵a2b↵}a} = Ua{b↵a2b↵} = UaUb↵(a

2) UaUb↵(kak2e) = kak2Ua(b2↵) =

kak2{ab2↵a} �w�! 0. Thus, for all ⇢ 2 K, we got ⇢({ab↵a}2) �! 0 which

implies {ab↵a} �s�! 0. Now, if b↵�s�! b, then Ua(b↵)

�s�! 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.

4. Consider two nets a↵�s�! 0 and b�

�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 ka2ka2. 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↵},

a↵�s�! 0 =) a2↵

�s�! 0 (4.3.7)

Also, note that for a, b 2 M,

a � b = 1

2[(a+ b)2 � a2 � b2] (4.3.8)

Then using (4.3.4), (4.3.7) and (4.3.8), we get that (a↵ � b�) �s�! 0.

Now, consider two bounded nets a↵�s�! a and b�

�s�! b in M. Then, a �a↵

�s�! 0 and b�b��s�! 0 in M and the nets are bounded. Applying the joint

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continuity of multiplication on bounded sets at 0 and separate continuity of

multiplication (3), in the following:

(a � b)� (a↵ � b�) = (a� a↵) � b+ (a� a↵) � (b� b�) + a � (b� b�) (4.3.9)

we get that a↵ � b� �s�! a � b in M.

The following are some basic results about convergences in these topologies.

Proposition 4.91. Let M be a JBW-algebra. Then:

1. �-strong convergence implies �-weak convergence.

2. A bounded monotone net in M converges �-strongly.

3. A monotone net of projections converges �-strongly to a projection.

Proof. Let {a↵} be a net in M.

1. Suppose a↵�s�! a. Then, by Cauchy-Swartz inequality, ⇢(a � a↵) (⇢(a �

a↵)2)12 �! 0, for ⇢ 2 K. Thus, ⇢(a � a↵) �! 0, for all ⇢ 2 K. Hence,

a↵�w�! a.

Remark 4.92. We know that �-weak limit of a net is unique, if it exists. And

the proposition above says that strong convergence implies weak convergence.

Hence, �-strong limit of a net is unique, if it exists.

2. Suppose a↵ % a. Then, by definition of normal state, ⇢ 2 K =) ⇢(a↵) �!⇢(a). By functional calculus, a2 kaka, for each a 2M. Now, ⇢((a�a↵)2) ka� a↵k⇢(a� a↵) �! 0. Hence, a↵

�s�! 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↵�s�! p. Now,

by proposition 4.90-(3), p↵ = p↵ � p↵ �s�! p � p. By remark 4.92, p = p � p.Hence, p is a projection in M.

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Corollary 4.93. Every �-weakly continuous linear functional on M is normal.

Proof. Let f be a �-weakly continuous linear functional on M and let a↵ % a in

M. Then, by proposition 4.91, a↵�s�! a which inturn implies a↵

�w�! a. Now, as f

is ��weakly continuous, we get f(a↵) �! f(a).

Remark 4.94. For a monotone net of elements {a↵} and any element b in M, we

have

a↵ % a =) U↵(b)�s�! Ua(b) (4.3.10)

Proof. Let a↵ % a. Then, by proposition 4.91, a↵�s�! a and (a↵ � a↵) �s�! (a � a).

Then, by proposition 4.90-2, it follows that

Tb(a2↵)

�s�! Tb(a2)

Next, Tb(a↵)�s�! Tb(a) implies

(a↵ � Tb(a↵))�s�! (a � Tb(a))

So, Ua↵ = 2a↵ � (a↵ � b)� (a2↵ � b) �s�! 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.

Note that if {a↵} is a net in B such that {a↵} �! a in norm, then {a↵} �w�! a.

Hence, a 2 B. Thus, B is norm closed.

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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 � ain �-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 2M | Upa = a}.Now, if {a↵} is a net in Mp such that a↵

�w�! a in M, then proposition 4.90-(2),

a↵ = Up(a↵)�w�! Up(a). By uniqueness of �-weak limits, Up(a) = a and thus,

a 2Mp. 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 + U 0p) = 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 2Mp, y 2Mp0 . Thus, Mp+Mp0

is a Jordan subalgebra containing Mp and Mp0 . Now, if b 2Mp and c 2Mp0 , then

by lemma 4.61, we get (Up + Up0)(b+ c) = b+ c. Thus,

Mp +Mp0 = {a 2M | (Up + Up0)a = a} (4.3.11)

164


We will now show that Mp + Mp0 is �-weakly closed. Let {a↵} be a net in

Mp +Mp0 and let a↵�w�! a in M. Then, a↵ = (Up +Up0)(a↵)

�w�! (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,

x operator commutes with p4.73() x = Upx+Up0x

4.3.11() x 2 (Mp+Mp0) (4.3.12)

So, if a operator commutes with p, then a 2 (Mp + Mp0) and (Up + Up0)e = e 2Mp +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).

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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) \ face(a+)

(�-weak closure) such that Upa � 0, Upa � a. Furthermore, p is the smallest

projection in M such that Upa+ = a+.

Proof. By theorem 4.99, we can identify W (a, e) with a monotone complete CR(X)

. Let us call denote this isomorphism as � : W (a, e) �! CR(X) . Let �(a) = a.

Define E = {x 2 X | a(x) > 0}. By proposition 1.5, E is both closed and open in

X, �E 2 CR(X) , a � 0 on E, a 0 on X/E and �E(a)+ = (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, �E(a)+ = (a)+ =) p � a+ = a+ =) Upa+ = a+.

As a � 0 on E and a 0 on X/E, it follows that �E a � 0 and �E a � a. So,

p � a � a and p � a � 0. Therefore, Upa � a and Upa � 0.

By proposition 1.5, �E is the supremum of an increasing sequence in face((a)+).

By the order isomorphism �, it follows that p is the supremum (in W (a, e)) of an

increasing sequence {a↵} in face(a+). If b is the supremum of {a↵} in M, then

by proposition 4.91-(2),(3), a↵�w�! b. As, W (a, e) is �-weakly closed, b 2 W (a, e)

and so b = p. Thus, p belongs to the �-weak closure of face(a+). Hence, p 2W (a, e) \ face(a+) (�-weak closure).

To prove uniqueness, assume that there is another projection q 2 W (a, e) \face(a+) (�-weak closure) such that Uqa � a and Uqa � 0. Let �(q) = �F for

166


some F ✓ X and denote G = X/F . Note that 0 Uqa = q � a =) �F a � 0.

Also, 0 � a � Uqa =) a � �F a = (1 � �F )a = �Ga 0. Thus, a � 0 on F and

a 0 on G. By proposition 1.5, E ✓ F , which implies �E �F =) p q. On

the other hand, Upa+ = a+ =) a+ 2 im+Up = face(p). Thus, q 2 face(a+) ✓face(p) ✓ im Up. So, q = Upq Upe = p. Hence, p = q.

Finally, if q is any projection in M such that Uqa+ = a+, then a+ 2 im+Uq =

face(q). So, p 2 face(a+) ✓ face(q) ✓ im Uq. Then, as above, p q. Hence, p is

the least projection in M such that Upa+ = a+.

Definition 4.102. Let a be a positive element of a JBW-algebra M. The smallest

projection p in M such that Upa = a is called the range projection of a and is

denoted by r(a).

Proposition 4.103. Let a 2M+. The range projection r(a) satisfies the following

properties:

1. r(a) 2 W (a, e) \ face(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 =) Upa = 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


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 face(a) (4.2.34). And as � is

continuous in the �-weak topology, �(b) = 0, 8 b in the �-weak closure of

face(a). Thus, by (1), �(r(a)) = 0.

4. Suppose a b, then by (2) a b kbkr(b). So, a 2 face(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) = face(r(a)) ✓face(p) = im+Up. Thus, Upa = 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 Upa = a, i.e. the notions of range projections in B and

M coincide.

Proposition 4.105. Let a, b 2M+. 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 Uba kakUb(r(a)) = 0. So, Ub(a) = 0. Thus, a ? b.

()) If a ? b, then Uab = 0. So, Ua annihilates face(b). As r(b) 2 face(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


Remark 4.106. If a, b are two positive elements in M, then from the above propo-

sition we see that a ? b =) r(a) ? b. Hence, Ur(a)b = 0 which further implies

Ur(a)0b = b (4.2.32). So,

a ? b =) {Ur(a)b = 0, Ur(a)0b = 0} (4.3.13)

Corollary 4.107. The normal states on M determine the norm and order on M,

i.e. for each a 2M ,

a � 0 () �(a) � 0, 8 � 2 K (4.3.14)

kak = sup{|�(a)| | � 2 K} (4.3.15)

Proof. We first prove (4.3.14).

(() Let a 2 M such that �(a) � 0, 8� 2 K. Let a = a+ � a� be the unique

orthogonal decomposition of a. Suppose a� 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 ⌧ = U⇤r(a�)⇢. By proposition

4.90-(1), ⌧ is a positive multiple of a normal state. Hence, ⌧(a) � 0 (by

hypothesis). Now, ⌧(e) = (U⇤r(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) = (U⇤r(a�)⇢)(a) = ⇢(Ur(a�)(a+ � a�)) = ⇢

�Ur(a�)(�a�)

�= �⇢(a�) < 0

which is a contradiction to the fact that ⌧(a) � 0. Thus, a� = 0 which

implies a � 0.

()) Let � 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) � 0 () ⇢(e) � ⇢(a), ⇢(�a), 8 ⇢ 2

169


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 equiva-

lent:

1. p ? q

2. p � q = 0

3. p q0

4. p+ q e

5. UpUq = 0

Proof. Recall that p ? q () {pqp} = {qpq} = 0.

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(1) 2) Follows from proposition 4.58.

(2) 3) If p � q = 0, then (p + q)2 = (p + q). Hence, p + q is a projection and so

p+ q e =) p q0.

(3, 4) p+ q e () p e� q = q0.

(3) 5) If p e � q, then q e � p. So, Up(q) Up(e � p) = 0. Now, for a � 0,

UpUq(a) UpUq(kake) = kakUpq = 0. So, UpUq(a) = 0, 8a 2 M+. As M is

positively generated, it follows that UpUqa = 0, 8a 2 M.

(5) 1) If UpUq = 0, then 0 = UpUq(e) = Upq = {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 face(a) (�-weak closure) and a 2 face(r(a)). So,

face(r(a)) = face(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) Upa � a, Upa � 0

(ii) Upa � 0, Up0a 0 and p operator commutes with a

For each a 2M , r(a+) is the smallest projection satisfying (i) and (ii).

Proof. Fix a 2 A.

(i) ii) Suppose p satisfies (i). Then, Upa � a � 0. As Up(Upa � a) = 0, we get

Up0(Upa� a) = Upa� a. Hence, Up0a = Upa� a. Thus, a = Upa+Up0a which

implies p commutes with a, by proposition 4.73. Also, Up0a = a � Upa 0.

Hence, (ii) holds.

171


(ii) i) Suppose (ii) holds. Then a = Upa + Up0a. This implies a � Upa = Up0a 0 =) Upa � 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 Upa � 0 and Up(Upa) = Upa. Thus,

r(Upa) p. Similarly, r(�Up0a) p0. Now, r(Upa) + r(�Up0a) (p + p0) = e.

Then, by proposition 4.109, r(Upa) ? r(�Up0a). By proposition 4.105, we get

Upa ? �Up0a. Also, by (ii), a commutes with p and hence a = Upa+Up0a. So, a =

(Upa)� (�Up0a) is an orthogonal decomposition of a. By uniqueness of orthogonal

decomposition, it follows that Upa = a+, Up0a = �a�. Hence, by (4.2.32), it

follows that �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).

For each increasing finite sequence � = {�0,�1 . . .�n} in R, with �0 < �kakand �n > kak, define k�k = max1in(�i � �i�1).

Theorem 4.112 (Spectral Theorem). Let M be a JBW-algebra. Let a 2 M. Then,

there is a unique family {e�}�2R of projections in M such that

(i) Each e� operator commutes with a

(ii) Ue�a �e�, Ue0�a � �e0�, 8� 2 R

(iii) e� = 0 for � < �kak, e� = e for � > kak

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(iv) e� eµ for � < µ

(v)V

µ>� eµ = e�, � 2 R (greatest lower bound in M)

The family {e�} is given by e� = e� r�(a� �e)+

�and is contained in W (a, e).

Further, for each increasing finite sequence � = {�0,�1 . . .�n} with �0 < �kakand �n > kak, define s� =

Pni=1 �i(e�i � e�i�1). Then,

limk�k!0

ks� � ak = 0

Proof. By theorem 4.99, W (a, e) is isometrically (order and algebra) isomorphic

to some monotone complete CR(X) . Note that if p is a projection in M which

belongs to W (a, e), then Upa = p � a. Now, for each � 2 R, (a � �e)+ 2 W (a, e)

and hence e� := (e� r((a��e)+)) is a projection in W (a, e). Thus, Ue�a = e� � a.The range projection of (a � �e)+ calculated in W (a, e) or M coincide (remark

4.104) and also coincide with the notion of range projection in CR(X) . Using the

isomorphism between W (a, e) and CR(X) , these correspondences and theorem

1.13, we see that e� defined as e � r((a � �e)+) satisfies (ii), (iii) and (iv). Also,

decreasing nets in M converge ��weakly to their greatest lower bounds. So these

greatest lower bounds are the same whether interpreted in W (a, e) or M. Hence,

(v) is also satisfied, by theorem 1.13. Since each e� belongs to the associative

algebra W (a, e), it follows from proposition 4.73, that e� operator commutes with

a. Hence, (i) is also satisfied. And by theorem 1.13, the family {e�} is the unique

family of projections in W (a, e) satisfying the properties (i) through (v).

We need to prove the uniqueness of this family in M. Suppose {f�} is any other

family of projections in M satisfying (i) - (v). If � µ, then f� fµ; so f� 2face(fµ) = im+Ufµ ; therefore Ufµf� = f�. Thus, f� and fµ operator commute, by

proposition 4.74. So, a and all of {f�}mutually operator commute. By proposition

4.73, we see that the subalgebra generated by a and {f�} is associative and hence

its ��weak closure is isomorphic to some monotone complete CR(X) , by theorem

173


4.99. Now, by uniqueness of spectral resolutions in CR(X) , it follows that f� =

e� r((a� �e)+). Hence, f� = e�, 8�.

We would like to note here that this spectral theorem also follows from the

generalised spectral theorem (3.175), developed in chapter 3. To see this, recall

that the JBW-algebra M is an order unit space, which is separating order and

norm duality with the base norm space V. Further, it is a well known fact that for

a JBW-algebra M, the predual of M is the set of all normal states on M (A.10).

So, M = V ⇤. And, proposition 4.111 states that M satisfies the spectral duality

condition (3.146). So, M satisfies the hypothesis of 3.175, and hence the result

follows.

Conclusion

Using an order theoretic approach, we have developed a spectral theorem and

functional calculus for suitable ordered spaces A. 11 Further, we have seen two

concrete examples of this theory: one in commutative setting (function spaces),

namely spectral theory for monotone complete CR(X) space and another in non-

commutative setting (Jordan algebras), namely spectral theory for JBW-algebras.

Hence, this spectral theorem generalises the spectral theory for von-Neumann al-

gebras and JBW-algebras, to a larger class of order unit spaces.

11A must be in separating order and norm duality with base norm space V, such that A = V⇤

and A satisfies the spectral duality condition.

174

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Basic theory, Orientation and C⇤- products. Birkhauser, 2001.

[3] Erik M. Alfsen and Frederic W. Shultz. Geometry of State Spaces of Operator

Algebras. Birkhauser, 2003.

[4] L. Asimow and A. J. Ellis. Convexity theory and its applications in Functional

Analysis. Academic Press, 11 December, 1976.

[5] H. Hanche-Olsen and E. Stormer. Jordan Operator Algebras. Pitmann Pub-

lisher, 1984.

[6] Graham Jameson. Ordered linear Spaces. Springer, 1970 edition, July 1, 1970.

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175

Chapter 5

Appendix

Theorem A.1 (Banach-Alaoglu). Let V be a normed linear space. Then B⇤ = {f 2 V ⇤ | kfk = 1 } is weak⇤ compact.

Theorem A.2. If X is a locally convex space, then X⇤ separates points in X.

Theorem A.3. Let X be a real locally convex space and let C ✓ X be a closed

(convex) cone. If x0 2 X and x0 /2 C, then there exists ! 2 X⇤ such that !(x) � 0

for all x 2 C and !(x0) < 0.

Theorem A.4. Let X be a vector space. Then C ✓ X is a closed convex cone in

X () C = C��.

Theorem A.5 (Krein-Milman Theorem). Let K be a compact convex set of a

locally convex space V and let @eK denote the extreme points of K. Then, @e(K) 6= ;and K = co(@eK).

Theorem A.6 (Cauchy-Schwarz Inequality). If h, i is a semi-inner product on

complex or real vector space X, then

|ha, bi|2 ha, aihb, bi, 8 a, b 2 X (A.0.1)

Theorem A.7 (Stone Weierstrass Theorem). Suppose X is a compact Hausdor↵

space and A is a subalgebra of CR(X) , which contains a non-zero constant function.

Then A is dense in CR(X) i↵ it separates points in X.

176

5 Appendix

Theorem A.8. Let (A,+, �) be a commutative algebra (not necessarily associa-

tive), endowed with a topology ⌧ such that multiplication is separately continuous

in A, with respect to ⌧ , i.e. if x↵ �! x in A and y 2 A, then (lim↵ x↵) � y =

lim↵(x↵ � y) Now, if B is an associative subalgebra of A, then the closure of B, in

A, is also associative.

Proof. Let {ax}, {by} and {cz} be nets in B such that ax �! a, by �! b, cz �! c

in A. We need to prove that a � (b � c) = (a � b) � c.a � (b � c) = (lim

xax) � (b � c)

= limx(ax � (b � c))

= limx[ax � {(lim

yby) � c}]

= limx[ax � {lim

y(by � c)}]

= limx[lim

y{ax � (by � c)}]

= limx[lim

y{ax � (by � (lim

zcz) )}]

= limx[lim

y{ax � (lim

z(by � cz) )}]

= limx[lim

y{lim

z(ax � (by � cz) )}]

= limx[lim

y{lim

z( (ax � by) � cz)}]

= limx[lim

y{( (ax � by) � lim

zcz) )}]

= limx[lim

y{(ax � by) � c}]

= limx[{lim

y(ax � by)} � c]

= limx[{ax � (lim

yby)} � c]

= limx[{ax � b} � c]

= [limx{ax � b}] � c

= [{limx

ax} � b] � c

= [a � b] � c

177

5 Appendix

Theorem A.9. The series

(1� x)12 =

1X

n=0

�nxn where �n =

(�1)n(12)(12 � 1) . . . (12 � (n� 1))

n!(A.0.2)

is absolutely and uniformly convergent for |x| 1.

Theorem A.10 (Predual of JBW-algebra). A JB-algebra M is JBW-algbera i↵ it

is a Banach dual. In that case, the predual M⇤ is unique and and consists of the

normal linear functionals on M.

178

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