Geometric and algebraic aspects of spectrality in order unit spaces: a comparison
aa r X i v : . [ qu a n t - ph ] F e b Geometric and algebraic aspects of spectrality in orderunit spaces: a comparison
Anna Jenˇcov´a and Sylvia Pulmannov´a ∗ Abstract
Two approaches to spectral theory of order unit spaces are compared: the spectralduality of Alfsen and Shultz and the spectral compression bases due to Foulis. While theformer approach uses the geometric properties of an order unit space in duality with abase norm space, the latter notion is purely algebraic. It is shown that the Foulis approachis strictly more general and contains the Alfsen-Shultz approach as a special case. This isdemonstrated on two types of examples: the JB-algebras which are Foulis spectral if andonly if they are Rickart and the centrally symmetric state spaces, which may be Foulisspectral while not necessarily Alfsen-Shultz spectral.
The operational approach to foundations of quantum mechanics works with an order unit space A , in duality with a base normed space V , where the distinguished base of the positive conerepresents states of a physical system, whereas elements of the unit interval in A are interpretedas effects, or dichotomic measurements. This theory goes back to the works of Mackey [29]and Ludwig [28] and many others, see e.g. [26] for a historical account and further references.More recent approaches include the convex operational theories or general probabilistic theories(GPT) (e.g. [8, 9]), working mainly in finite dimension, see also [26] for an introduction andoverview or [33] for other approaches.An important property of any mathematical theory describing quantum mechanics is spec-trality, that is, existence of spectral resolutions of effects (and hence all elements in the space A )that allows an integral expression in terms some special elements called projections. Perhapsthe best well known extension of spectrality to order unit spaces is due to Alfsen and Shultz[5, 7], motivated by characterization of state spaces of operator algebras and JB-algebras. Thistheory is based on the notion of compressions on an order unit space A which employs a sepa-rating order and norm duality with a base norm space V . The compressions are in one-to-onecorrespondence with special elements of the unit interval in A called projective units and alsowith so-called projective faces of the set of states. The duality of A and V is called spectral ifthere are ”sufficiently many” compressions on A , it is also assumed that the order unit space ismonotone σ -complete [5] or that A = V ∗ [7]. Under these conditions, any element has a spectralresolution in terms of the projective units and there is a well defined functional calculus on A . ∗ Mathematical Institute, Slovak Academy of Sciences, ˇStef´anikova 49, SK-814 73 Bratislava, Slovakia;[email protected], [email protected]. Supported by the grant VEGA 2/0142/20 and the Slovak Re-search and Development Agency grant APVV-16-0073. A e.g. [20, 21, 31], also morerecently [22] or [24]. The relations of these various definitions are not clear and it is a naturalquestion whether the algebraic approach could lead to the same results as the Alfsen-Shultztheory (if similar conditions are assumed).The aim of the present paper is to compare the Alfsen-Shultz approach to that of [12].The latter notion of spectrality follows the works by Foulis [13, 14, 15, 16] in the more generalsetting of ordered groups with order unit. This definition also uses a notion of a compression,but since no duality is used, the definition is different from that of Alfsen-Shultz. In this case, A is spectral if there are enough compressions to form a spectral compression base: a set ofcompressions with given properties. It is proved in [12] that this implies existence of spectralresolutions for elements in A .In order to compare the two theories, we will assume that A is in a separating order andnorm duality with a base norm space V and that the (Foulis) compressions are continuous withrespect to this duality, note that we can always take V = A ∗ and then continuity always holds.It is noted in [12] that the spectral theory of Alfsen-Shultz is a special case of the spectraltheory of Foulis and examples are given showing that the latter notion works also in the casethat the order unit space is not σ -monotone complete. However, the exact relation of the twotheories is unknown. In particular, it is a question whether these two notions of spectralityare equivalent if A = V ∗ for some base norm space V . We prove that the Foulis spectrality isstrictly more general than Alfsen-Shultz spectrality, in the sense that if A = V ∗ , the Alfsen-Shultz spectrality implies Foulis spectrality, but we show examples that the opposite implicationdoes not hold. We compare the two notions of compressions and show that these are not thesame. We also analyze the case of JB-algebras and show that JB-algebras are Foulis spectralif and only if they are Rickart, but the Rickart JB-algebras are not Alfsen-Shultz spectral ingeneral, although the compressions are the same in both cases.The outline of the paper is as follows. In the next section, we introduce some basic definitionsrelated to order unit and base norm spaces and dualities between them. In Section 3 we describethe two basic notions of spectrality that we are going to compare. While the Alfsen-Shultztheory is developed in detail in [7] and is therefore introduced only briefly in Section 3.1, thetheory proposed in [12] is much less well known, with many proofs scattered about severalpapers and in different settings. For this reason, the exposition of this theory in Section 3.2 ismore detailed and quite self-contained, with many results reproved. In Section 4, we proceedto comparison of the two theories, here the main result is Theorem 4.7. The last two sectionscontain examples showing the difference between the two theories. In Section 5, we studythe Rickart JB-algebras in which case we may have A = V ∗ . We work with compressions inAlfsen-Shultz sense and show that they form a spectral compression base, while Alfsen-Shultzspectrality cannot hold. Section 6 is devoted to centrally symmetric theories, where A = V ∗ and the distinguished base in V is (isomorphic to) the unit ball in a (reflexive) Banach space( X, k · k ). We show that Alfsen-Shultz spectrality holds if and only if the space X is strictlyconvex and smooth, whereas smoothness is equivalent to Foulis spectrality. For more details for this section, we refer the reader to [4, 6, 7].Recall that an order unit space is an archimedean partially ordered real vector space with a2istinguished order unit. In the sequel, we denote the positive cone by A + := { a ∈ A : 0 ≤ a } and the order unit by 1, the order unit space will be denoted by the triple ( A, A + , A will be denoted by E := { e ∈ A : 0 ≤ e ≤ } . Note that E endowed with thepartial operation inherited from + in A is an effect algebra, see [11] for the definition and [10]for more information on effect algebras and related structures.The order unit norm k . k on A is defined by k a k = inf { λ ∈ R + : − λ ≤ a ≤ λ } , a ∈ A. A base norm space is a partially ordered normed real vector space V with a generatingcone V + and a distinguished base K of V + such that K lies in a hyperplane H V and co ( K ∪ − K ) is the unit ball in V . The base norm space will be denoted by ( V, K ). The normof V is the base norm with respect to K denoted by k · k K and is given by k v k K = inf { λ + µ : v = λx − µy, λ, µ ≥ , x, y ∈ K } , v ∈ V. An order unit space (
A, A + ,
1) and a base norm space (
V, K ) are in separating order andnorm duality if there is a duality h· , ·i : A × V → R such that(i) a ∈ A + ⇔ h a, ϕ i ≥ ϕ ∈ V + ,(ii) ϕ ∈ V + ⇔ h a, ϕ i ≥ a ∈ A + ,(iii) for all a ∈ A and ϕ ∈ V : k a k = sup k σ k K ≤ |h a, σ i| , k ϕ k K = sup k a k ≤ |h a, ϕ i| . In this case, k ϕ k K = h , ϕ i for ϕ ∈ V + and K = { ρ ∈ V + : h , ρ i = 1 } . Let (
A, A + ,
1) be an order unit space and let V := A ∗ , V + := { ϕ ∈ V, h a, ϕ i ≥ ∀ a ∈ A + } and K := { ρ ∈ V + : h , ρ i = 1 } . Then A and V are in separating order and normduality.Similarly, let ( V, K ) be a base norm space and let A := V ∗ , A + := { a ∈ A, h a, ϕ i ≥ , ∀ ϕ ∈ V + } and 1 ∈ A + be the functional determining the hyperplane H containing K . Then A and V are in separating order and norm duality. Let A = A sa be the self-adjoint part of the C*-algebra A . Let A + = A + = { b ∗ b, b ∈ A} be the positive cone in A and let 1 be the identity in A . Then ( A, A + ,
1) is anorder unit space. If V = A ∗ and K = S ( A ) is the set of states on A , then ( V, K ) is a base normspace in separating order and norm duality with A . If A is a von Neumann algebra, then thepredual V = A ∗ with the set K = S ∗ ( A ) of normal states is a base norm space in separatingorder and norm duality with A .More generally, we may take A to be a JB-algebra resp. a JBW-algebra and obtain similarresults.Let ( A, A + ,
1) and (
V, K ) be an order unit space and a base norm space in separating orderand norm duality. We denote by σ ( A, V ) be the locally convex topology in A or V , given bythe duality of A and V . Unless otherwise stated, we will always consider A and V endowedwith these topologies. 3ecall that a face of the positive cone A + (or similarly of V + ) is a hereditary subcone F ⊆ A + : a ∈ F and b ≤ a implies b ∈ F . The face of A + generated by a ∈ A + will be denotedas face A + ( a ) := { b ∈ A + : b ≤ λa, λ ≥ } . A subcone F ⊆ V + is a face of V + if and only if F ∩ K is a face of K : ρ ∈ F ∩ K and ρ = λσ + (1 − λ ) σ with λ ∈ (0 ,
1) and σ , σ ∈ K implies σ , σ ∈ F ∩ K .Let C ⊆ A , W ⊆ V be any subsets. We will use the notations C ◦ := { ϕ ∈ V : h a, ϕ i = 0 , ∀ a ∈ C } , W ◦ := { a ∈ A : h a, ϕ i = 0 , ∀ ϕ ∈ W } . Note that C ◦◦ is the closed linear span of C ∪ { } , similarly for W ◦◦ .If C ⊆ A + and W ⊆ V + , we will denote C • = { ϕ ∈ V + : h a, ϕ i = 0 , ∀ a ∈ C } , W • := { a ∈ A + : h a, ϕ i = 0 , ∀ ϕ ∈ W } . Note that C • is a face of V + , similarly for W • . A face F ⊆ V + is called exposed if there issome a ∈ A + such that F = { a } • and semiexposed if F = C • for some C ⊆ A + , equivalently, F •• = F . We have similar definitions for faces of V + . Note that a face F ⊆ V + is semiexposediff F ∩ K is semiexposed, which means that there is a subset of effects D ⊆ E such that F ∩ K = { ρ ∈ K : h a, ρ i = 1 , ∀ a ∈ D } , and F is exposed iff we may take a singleton D = { a } . In this section, we will recall the definitions and basic results of the two approaches to spectralityin order unit spaces that we are going to compare. Throughout the rest of the paper, we willwork with a pair (
A, V ) of an order unit space (
A, A + ,
1) and a base norm space (
V, K ) inseparating order and norm duality.
We first introduce the theory of spectral duality due to Alfsen and Shultz. We refer to [5, 7]for details.Let J : A → A be a positive idempotent linear map, continuous in the σ ( A, V )-topology.We will say that J is normalized if k J k ≤
1, equivalently p := J (1) ≤
1. Let us denoteKer + J := { a ∈ A + : J ( a ) = 0 } Im + J := { a ∈ A + : J ( a ) = a } . (1)We say that two such maps J and J ′ are complementary ifKer + J = Im + J ′ and Ker + J ′ = Im + J. In this case we say that J ′ is a complement of J .Notice that J has an adjoint map J ∗ : V → V , which is positive and idempotent as welland we have similar definitions of Ker + ( J ∗ ), Im + ( J ∗ ) and complementarity. We will say thatthe maps J and J ′ are bicomplementary if they are complementary and ( J ′ ) ∗ is a complementof J ∗ . In this case, we say that J (or J ′ ) is bicomplemented . Note that then J has a uniquecomplement J ′ [7, Corollary 7.11]. 4 .1 Definition. A compression on A is a bicomplemented normalized σ ( A, V )-continuouspositive idempotent linear map A → A .If J is a compression, the element p = J (1) is called a projective unit . If p is a projectiveunit, then there is a unique compression J such that J (1) = p and then J ′ (1) = 1 − p for thecomplementary compression. This implies that any compression has a unique complement. [7, Def. 8.42] We say that the spaces A and V are in spectral duality if A = V ∗ and for every a ∈ A , there is a least compression J such that J ( a ) ≥ a and J ( a ) ≥ a ∈ A and J is a compression as in the above definition with complement J ′ ,then we have a = J ( a ) + J ′ ( a ) = a + − a − , a + := J ( a ) , a − := − J ′ ( a ) ≥ . In general, for a ∈ A , b, c ∈ A + , we say that a = b − c is an orthogonal decomposition of a ifthere is a compression J such that J ( b ) = b , J ( c ) = 0.There are several equivalent characterizations of spectral duality. Here an important role isplayed by the following assumption: [7] Every exposed face F ⊆ K is projective, which means that itis determined by a projective unit. Since K is a base of V + , this means that for every a ∈ A + there is a projective unit p such that { a } • = { p } • . By [7, Thms. 8.52 and 8.55], A and V are in spectral duality iff the Standing Hypothesisholds and every element a ∈ A has a unique orthogonal decomposition: a = a + − a − .The importance of the notion of spectral duality is the fact that it allows us to define spectralresolutions of elements in A , as we briefly describe next. For a ∈ A , let e λ := 1 − J λ,a (1) , where J λ,a is the compression corresponding to a − λ (by Definition 3.2). The family { e λ } λ ∈ R has exactly the properties of the spectral resolution of a self-adjoint element in a von Neumannalgebra. In particular, we have a = Z λde λ , where the integral is defined as the norm-limit of approximating Riemann-Stieltjes sums. Thefamily is uniquely determined by a and called the spectral resolution of a .The following is a prototypical example of a spectral duality. Let A = M sa be the self-adjoint part of a von Neumann algebra M . For anyprojection p ∈ A , the map U p : a pap, a ∈ A is a compression on A and p = U p (1). Moreover, any compression on A is necessarily of thisform for some projection p , [7, Theorem 7.23]. In this way, projective units generalize theprojections in von Neumann algebras. Furthermore, A = V ∗ where V is the hermitian partof the predual of M and A and V are in spectral duality. For a ∈ A , let a = a + − a − bethe decomposition into the positive and negative part and let p be the support projection of a + , then U p ( a ) = a + ≥ a ≤ a + = U p ( a ) and U p is the compression corresponding to a inDefinition 3.2. It is easily seen that in this case, { e λ } for a ∈ A is the usual spectral resolutionfor a . Similar results hold in the case of a general JBW algebra.5 .2 Compression bases and spectrality In this section, we discuss the notion of spectrality in an order unit space along the lines of[12]. This notion is based on the works [13, 14, 15, 16] on compressions on unital orderedabelian groups, that were applied to order unit spaces in [12]. We stress that the originaldefinitions are of algebraic nature and no duality with a base norm space is assumed. Sincewe want to compare this to the theory of Alfsen and Shultz, we have to admit such dualityhere and the maps are assumed continuous. Note that we can always put V = A ∗ , then all theinvolved mappings are automatically continuous (with respect to the duality of A and V ). Todistinguish the maps used here from the compressions in the previous paragraph, we use theterm F-compressions.The following definition is based on [12, Definition 1.1 and Lemma 1.1]. Let J : A → A be a σ ( A, V )-continuous positive linear mapping. Then J isan F-compression with focus p on A if for all e ∈ E ,(F1) J (1) = p ∈ E (that is, J is normalized),(F2) e ≤ p = ⇒ J ( e ) = e ,(F3) J ( e ) = 0 = ⇒ e ≤ − p .If J satisfies (F1) and (F2) but not necessarily (F3), we say that J is a retraction .It is easily seen from this definition that any retraction is idempotent. Let J be an F-compression and let Ker + J and Im + J be as in (1). The following equalities follow easily fromthe properties (F2) and (F3):Ker + ( J ) = { a ∈ A + , ∃ t > , a ≤ t (1 − p ) } = face A + (1 − p ) (2)Im + ( J ) = face A + ( p ) , (3)We next collect some easy observations on F-compressions and their foci for later use. Let J and J ′ be F-compressions, J (1) = p and J ′ (1) = p ′ . Then J and J ′ arecomplementary if and only if p ′ = 1 − p .Proof. Let J and J ′ be complementary. Then 1 − p ∈ Ker + J = Im + J ′ and p ∈ Im + J =Ker + J ′ , so that J ′ ( p ) = 0 and 1 − p = J ′ (1 − p ) = J ′ (1) = p ′ . The converse is clear from theequalities (2) and (3). Let p be the focus of a retraction J . Then p is a principal element in E : for any a, b ∈ E such that a + b ≤ and a, b ≤ p we have a + b ≤ p . This property is equivalent to thefollowing facial property (cf. [7, Def. 8.45]): face A + ( p ) ∩ E = [0 , p ] . If J is an F-compression, the same is true for − p .Proof. Let a, b ∈ [0 , p ] be such that a + b ≤
1. Then by property (F2) we have a = J ( a ), b = J ( b ), so that a + b = J ( a + b ) ≤ J (1) = p. q ∈ E has the facial property and let a, b ≤ q , a + b ∈ E . Then a + b ≤ q , so that a + b ∈ face A + ( q ) ∩ E = [0 , q ]. Conversely, assume that q is principal, andlet 0 ≤ a ≤ λq for some λ >
0. If λ ≤ λ >
1. Then(1 /λ ) a ≤ q and a = (1 /λ ) a + (1 − /λ ) a . We can decompose1 − /λ = X i β i , i = 1 , . . . , n, ≤ β i ≤ /λ, ∀ i. Then β i a ≤ (1 /λ ) a ≤ q for all i and since q is principal, we must have a = (1 /λ ) a + P i β i a ≤ q .The last statement follows easily by (2) and the property (F3). For p ∈ E , consider the following statements:(a) p is principal;(b) p is extremal in E ;(c) p is sharp : if a ∈ E is such that a ≤ q and a ≤ − q , then a = 0 .Then (a) = ⇒ (b) = ⇒ (c).Proof. Assume (a) and let a, b ∈ E , λ ∈ (0 ,
1) be such that λa + (1 − λ ) b = p . By the facialproperty of p , we must have a, b ≤ p , but this is possible only if a = b = p , so that p is extremalin E . Assume (b) and let a ∈ E , a ≤ p , a ≤ − p . Then p ± a ∈ E and p = (( p − a ) + ( p + a )),so that p − a = p + a = p and a = 0. Hence p is sharp.We say that the elements a, b ∈ E are Mackey compatible , if there are some elements c, a , b ∈ E such that c + a + b ≤ a = c + a , b = c + b . A sub-effect algebra P in E is normal if for all d, e, f ∈ E such that d + e + f ≤ d + e, d + f ∈ P we have d ∈ P [16, Definition 1]. Note that this implies that elements of P are compatible in E iff they arecompatible in P . [15, Definition 2] A compression base for A is a family ( J p ) p ∈ P of F- compres-sions on A , indexed by their own foci, such that P is a normal subalgebra of E and whenever p, q, r ∈ P and p + q + r ≤
1, then J p + r ◦ J q + r = J r . Elements of P will be called projections . Let X be a compact Hausdorff space and let A = C ( X, R ) be the set of contin-uous functions X → R . With the usual ordering of functions and 1 = 1 X the constant unit, A is an order unit space and the order unit norm coincides with the maximum norm on C ( X, R ).Since A is the self-adjoint part of the C*-algebra C ( X, C ), we know by [16] (see also Theorem5.10 below) that the retractions on A are precisely of the form U p ( f ) = pf , f ∈ A , where p ∈ A is a projection, which means that p is the characteristic function of a clopen subset of X . Bya similar reasoning as in the proof of Corollary 5.11, we have that ( U p ) p ∈P ( X ) is a compressionbase, here P ( X ) is the set of all projections in A .We will encounter this example repeatedly below and we will always assume that C ( X, R )is endowed with the compression base { U p } p ∈P ( X ) .7n the sequel, we fix a compression base ( J p ) p ∈ P for A . Note that since P is a subalgebra,any J p has a (fixed) complementary F-compression J − p . Observe also that using the fact thatall projections are principal elements, it is easily seen that if p, q ∈ P and p + q ≤
1, then p + q = p ∨ q in P , so that P with the orthocomplementation p − p is an orthomodularposet (OMP) [30, 23]. It follows that if p, q ∈ P are Mackey compatible, then p ∧ q and p ∨ q exists in P . It can be shown as in [15, Thm. 2.5] that P is a regular OMP, which meansthat any pairwise Mackey compatible subset in P is contained in a Boolean subalgebra of P ([23]). A maximal subset of pairwise Mackey compatible elements is called a block of P . Everyelement p ∈ P is contained in some block B of P . The following notion for p ∈ P and a ∈ A was introduced in [12, Def. 1.5], see also [7, Def.7.41] for projective units p . For p ∈ P and a ∈ A , we say that a is compatible with p if a = J p ( a ) + J − p ( a ) . The set of all a compatible with p ∈ P will be denoted by C ( p ). For a subset Q ⊆ P , we denote C ( Q ) := T p ∈ Q C ( p ). If B ⊆ P is a block of P , the set C ( B ) is called a C-block of A .The following facts are easily checked (see also [12, Lemma 1.3]). Let a ∈ A , p ∈ P . Then(i) If J p ( a ) ≤ a , then a ∈ C ( p ) .(ii) If a ∈ A + , then a ∈ C ( p ) if and only if J p ( a ) ≤ a .(iii) If a ∈ E , then a ∈ C ( p ) if and only if a and p are Mackey compatible.(iv) If q ∈ P , then q ∈ C ( p ) if and only if p ∈ C ( q ) if and only if J p J q = J q J p = J p ∧ q . We denote the set of all projections compatible with a ∈ A by P C ( a ) and for B ⊆ A , P C ( B ) := T a ∈ B P C ( a ). For B ⊆ A we have:(i) If p , . . . , p n ∈ P C ( B ) are such that P i p i ≤ , then J P i p i ( a ) = X i J p i ( a ) ∀ a ∈ B. (ii) For compatible p, q ∈ P C ( B ) , we have p ∨ q, p ∧ q ∈ P C ( B ) .Consequently, P C ( B ) is a normal subalgebra in E and a regular OMP. roof. Let p, q ∈ P C ( a ), p + q ≤ r = 1 − p − q . Then since p ≤ p + q and 1 − p = r + q ,we have for a ∈ BJ p + q ( a ) = J p + q ( J p ( a ) + J − p ( a )) = J p ( a ) + J p + q J r + q ( a ) = J p ( a ) + J q ( a ) , By induction, this proves (i).Assume that p, q ∈ P C ( B ) are compatible and put p ′ = 1 − p , q ′ = 1 − q . Then for all a ∈ Ba = J p ( a ) + J p ′ ( a ) = J p ( J q ( a ) + J q ′ ( a )) + J p ′ ( J q ( a ) + J q ′ ( a ))= J p ∧ q ( a ) + J p ∧ q ′ ( a ) + J p ′ ∧ q ( a ) + J p ′ ∧ q ′ ( a ) . Since 1 − p ∧ q = p ∧ q ′ + q ′ ∧ p + q ′ ∧ p ′ , all the involved elements are compatible and we obtainby Lemma 3.12 J − p ∧ q ( a ) = J − p ∧ q ( J p ∧ q ′ ( a ) + J p ′ ∧ q ( a ) + J p ′ ∧ q ′ ( a ))= J p ∧ q ′ ( a ) + J p ′ ∧ q ( a ) + J p ′ ∧ q ′ ( a )so that a = J p ∧ q ( a ) + J − p ∧ q ( a ) . Hence p ∧ q ∈ P C ( B ), (ii) follows from p ∨ q = 1 − p ′ ∧ q ′ . The last statement follows by [23,Cor. 2.1.14]. For Q ⊆ P , C = C ( Q ) is a norm-closed subspace and an order unit space, withorder unit 1. Let ˜ J p := J p | C , then { ˜ J p } p ∈ P ∩ C is a compression base in C . If Q is a block, then P ∩ C = Q .Proof. Let q ∈ P , then C ( q ) = { a ∈ A, a = J q ( a ) + J − q ( a ) } = J − (0), where J = id − J q − J − q is a bounded linear map. Hence C ( q ) is a norm-closed subspace and since 1 ∈ C ( q ), it is anorder unit space with order unit 1. Further, let p ∈ C ( q ) ∩ P , then since J p J q = J q J p , we havefor any a ∈ C ( q ) J p ( a ) = J p ( J q ( a ) + J − q ( a )) = J q ( J p ( a )) + J − q ( J p ( a )) , so that J p ( C ( q )) ⊆ C ( q ). Since C = ∩ q ∈ Q C ( q ), we see that C is a norm closed subspace in A and an order unit space with order unit 1. Moreover, for p ∈ C ∩ P , ˜ J p ( C ) ⊆ C and itis straightforward that ˜ J p is an F-compression. By Lemma 3.13, P ∩ C ( Q ) = P C ( Q ) is anormal subalgebra in E and hence also in E ∩ C . It is also easily checked that { ˜ J p } p ∈ C ∩ P is acompression base in C .If Q is a block, then clearly Q ⊆ C ( Q ) ∩ P , the other inclusion follows by maximality of Q .The following subset can be seen as the bicommutant of a in P : P ( a ) := P C ( P C ( a ) ∪ { a } ) , i.e. the set of all projections compatible with a and with all projections compatible with a .Note that all elements in P ( a ) are pairwise compatible, so that by Lemma 3.13, P ( a ) is aboolean subalgebra in P . If q ∈ P , then P ( q ) = { , , q, − q } .9 .2.2 Projection cover property3.15 Definition. We say that the compression base ( J p ) p ∈ P (or A ) has the projection coverproperty if for every effect e ∈ E there is an element p ∈ P such that for q ∈ P , we have e ≤ q iff p ≤ q . Such an element p is necessarily unique and is called the projection cover of e ,denoted by e .In this section, we assume that ( J p ) p ∈ P has the projection cover property. Assume that ( J p ) p ∈ P has the projection cover property. For a ∈ E , a ∈ P ( a ) .Proof. Let q ∈ P C ( a ), then a = b + c , with b = J q ( a ), c = J − q ( a ). Since b ≤ q , c ≤ − q ,we have b ≤ q and c ≤ − q , so that b , c ∈ P C ( q ). We will show that a = b + c = b ∨ c ∈ P C ( q ). Indeed, since a = b + c ≤ b + c , we have a ≤ b + c . On the other hand, b, c ≤ a ≤ a , so that b , c ≤ a and b + c = b ∨ c ≤ a . The fact that a ∈ P C ( a ) is clearfrom a ≤ a .The following theorem was proved in [15, Theorem 6.4 (iii)] in a more general context. Wegive a simpler proof here. Suppose that ( J p ) p ∈ P has the projection cover property. Then(i) P is an orthomodular lattice (OML);(ii) P is sup/inf closed in E , in the sense that for every subset M ⊆ P , whenever the supre-mum ∨ M exists in E , then ∨ M ∈ P , and similarly for the infimum ∧ M .Proof. Since, as noted above, P is an OMP, we only need to prove that it is a lattice, in fact,it is enough to show that any two elements p, q ∈ P have a supremum in P . Let us denote r λ := ( λp + (1 − λ ) q ) for some λ ∈ (0 , λp ≤ r λ , we have p ≤ r λ by the facialproperty, similarly q ≤ r λ . Let r ∈ P be such that p, q ≤ r , then λp + (1 − λ ) q ≤ r , so that r λ ≤ r (since r λ is the projection cover). Hence we may put p ∨ q = r λ , for any λ ∈ (0 , M ⊆ P and ∧ M =: a exists in E . For all b ∈ M , a ≤ b = ⇒ a ≤ b , hence a ≤ ∧ M = a , which entails a = a ∈ P . [12, Def. 1.6] We say that the compression base ( J p ) p ∈ P in A has the com-parability property if, for every a ∈ A , P ± ( a ) = ∅ , where P ± ( a ) := { p ∈ P ( a ) and J − p ( a ) ≤ ≤ J p ( a ) } . Let p ∈ P ± ( a ) and put b := J p ( a ), c := − J − p ( a ), then we have a = b − c, b, c ∈ A + , J p ( b ) = b, J p ( c ) = 0 . (4)Any decomposition of a of the form (4) for some p ∈ P is called a P - orthogonal decomposition of a . 10 .19 Proposition. If P ± ( a ) = ∅ , then a ∈ A has a unique P -orthogonal decomposition, whichwill be written as a = a + − a − . Moreover we have a + , a − , | a | := a + + a − ∈ C ( P C ( a )) .Proof. Let p ∈ P ± ( a ), we have seen that a has a P -orthogonal decomposition a = J p ( a ) − ( − J − p ( a )). To prove uniqueness, let a = b − c with b, c ∈ A + and J q ( b ) = b , J q ( c ) = 0 forsome q ∈ P . Then a ∈ C ( q ) and hence p and q are compatible, by definition of P ± ( a ). Since J − p ( a ) ≤ J q is positive, we have J q J − p ( a ) ≤
0, similarly, we also have J − p J q ( a ) ≥ q and 1 − p are compatible, so that0 ≤ J − p J q ( a ) = J q J − p ( a ) ≤ , hence J − p J q ( a ) = 0 and J p J − q ( a ) = 0 follows by a similar argument. We obtain J p ( a ) = J p J q ( a ) = J q J p ( a ) = J q ( a ) , this proves that the P -orthogonal decomposition is unique. For the last statement, let q ∈ P C ( a ), then a + = J p ( a ) = J p ( J q ( a ) + J − q ( a )) = J q ( a + ) + J − q ( a + ) , so that a + ∈ C ( P C ( a )). The proof for a − and | a | is similar.We will assume below that ( J p ) p ∈ P is a compression base with the comparability property.We will show that in this case any a ∈ A is contained in some C-block of A and that theC-blocks are isomorphic to (sub)spaces of functions as in Example 3.10. Moreover, we willextend the notion of compatibility to all pairs of elements in A . We start by showing that inthis case all sharp elements are projections (cf. [15, 31]). Suppose that ( J p ) p ∈ P has the comparability property and let q ∈ E . Then q ∈ P if and only if q is sharp.Proof. We know by Lemmas 3.7 and 3.8 that all elements in P are sharp. Conversely, let p ∈ E be sharp and let a = 2 p −
1. Let q ∈ P ± ( a ), then J q ( a ) ≥
0, so that J q (1 − p ) ≤ J q ( p ). On theother hand, since a ∈ C ( q ), it is easily seen that q is compatible with p , so that J q ( p ) ≤ p andalso J q (1 − p ) ≤ − p . Hence J q (1 − p ) ≤ p, − p and we must have J q (1 − p ) = 0. It followsthat a + = J q ( a ) = J q (1 − − p )) = q. Similarly, we obtain that a − = 1 − q , so that a = a + − a − = 2 q − p = q ∈ P . (cf. [15, Thm. 3.7]) Assume that ( J p ) p ∈ P has the comparability property. Let a ∈ A and let λ ≤ λ ≤ . . . be a sequence. Then there are projections q i ∈ P ± ( a − λ i ) suchthat q i ≥ q i +1 , i = 1 , , . . . .Proof. We take any q ∈ P ± ( a − λ ) and construct the rest of the sequence by induction. Soassume that for some k ≥ q ≥ · · · ≥ q k with the requiredproperties. Choose some q ∈ P ± ( a − λ k +1 ) and put q k +1 := q ∧ q k . Clearly, q k +1 ∈ P ( a ) = P ( a − λ k +1 ) and J q k +1 ( a − λ k +1 ) = J q k J q ( a − λ k +1 ) ≥ . q ′ = 1 − q , q ′ k = 1 − q k . Since q, q k are elements of the boolean algebra P ( a ) = P ( a − λ k +1 ),so are q ′ , q ′ k and we have 1 − q ∧ q k = q ′ ∧ q k + q ∧ q ′ k + q ′ ∧ q ′ k . By Lemma 3.13, J − q ∧ q k ( a − λ k +1 ) = ( J q ′ ∧ q k + J q ∧ q ′ k + J q ′ ∧ q ′ k )( a − λ k +1 )and we have J q ′ ∧ q k ( a − λ k +1 ) = J q k J q ′ ( a − λ k +1 ) ≤ ,J q ∧ q ′ k ( a − λ k +1 ) ≤ J q ∧ q ′ k ( a − λ k ) = J q J q ′ k ( a − λ k ) ≤ J q ′ ∧ q ′ k ( a − λ k +1 ) ≤ J q ′ J q ′ k ( a − λ k ) ≤ . Hence J − q k +1 ( a − λ k +1 ≤ q k +1 ∈ P ± ( a − λ k +1 Assume that ( J p ) p ∈ P has the comparability property. Then any a ∈ A is inthe norm-closed linear span of P ( a ) : a ∈ span( P ( a )) , the same is true for a + , a − , | a | .Proof. For ǫ >
0, let us choose a sequence −k a k =: λ < λ < · · · < λ n < λ n +1 := k a k , λ i +1 − λ i ≤ ǫ, i = 0 . . . , n. By Lemma 3.21, there is a sequence q i ∈ P ± ( a − λ i ), i = 0 , . . . n + 1 such that q i ≥ q i +1 , notethat we may put q := 1 and q n +1 := 0. Put p i := q i − − q i = q i − ∧ q ′ i , i = 1 , . . . , n + 1 . Then p i ∈ P ( a ), P i p i = 1, so by Lemma 3.13, a = P i J p i ( a ). We also have λ i − p i ≤ J p i ( a ) ≤ λ i p i , i = 1 , . . . , n + 1Indeed, J p i ( a − λ i ) = J q i − J q ′ i ( a − λ i ) ≤
0, so that J p i ( a ) ≤ λ i p i . The other inequality followsfrom J p i ( a − λ i − ) = J q ′ i J q i − ( a − λ i − ) ≥
0. For any choice of ξ i ∈ ( λ i − , λ i ), we obtain − ǫp i ≤ J p i ( a ) − ξ i p i ≤ ǫp i , i = 1 , . . . , n + 1 , so that, summing up over i , we obtain − ǫ ≤ a − P i ξ i p i ≤ ǫ which means that k a − P i ξ i p i k ≤ ǫ .Let p ∈ P ± ( a ), so that a + = J p ( a ), and let a n ∈ span( P ( a )) be such that a n → a in norm,then also J p ( a n ) ∈ span( P ( a )) and J p ( a n ) → J p ( a ) = a + . The proof for a − and | a | is similar. Assume that ( J p ) p ∈ P has the comparability property.(i) For any a ∈ A and p ∈ P , a ∈ C ( p ) if and only if P ( a ) ⊆ C ( p ) .(ii) For any a ∈ A , there is some block B ⊆ P such that a ∈ C ( B ) .(iii) For any block B ⊆ P , C ( B ) = span( B ) . roof. For (i), let a ∈ A and p ∈ P . It is clear that if a ∈ C ( p ) then P ( a ) ⊆ C ( p ). Conversely,assume that P ( a ) ⊆ C ( p ), then q = J p ( q ) + J − p ( q ) for all q ∈ P ( a ), hence for all elements ofspan( P ( a )), the statement (i) now follows from Theorem 3.22.Since P ( a ) is a boolean subalgebra in P , it is contained in some block B ⊆ P . Then P ( a ) ⊆ C ( B ) and by Lemma 3.14 also span( P ( a )) ⊆ C ( B ), this shows (ii).For (iii) let a ∈ C ( B ), then B ⊆ P C ( a ), so that P ( a ) ⊆ P C ( B ) = B , the last equalityfollows by maximality of B . We obtain a ∈ span( P ( a )) ⊆ span( B ) , so that C ( B ) ⊆ span( B ). The converse inclusion is obvious.We will now look at the structure of the C-blocks of A in the comparability case. Assume that ( J p ) p ∈ P has the comparability property and let B be a block of P .Then C ( B ) is an order unit space and ( ˜ J p ) p ∈ B where ˜ J p = J p | C ( B ) is a compression base in C ( B ) with the comparability property.Proof. By Lemma 3.14, C ( B ) is an order unit space and ( ˜ J p ) p ∈ B is a compression base in C ( B ).By Corollary 3.23 (i), we see that a ∈ C ( B ) if and only if P ( a ) ⊆ C ( B ), where P ( a ) is computedwith respect to all of P . By maximality, this means that P ( a ) ⊆ B . On the other hand, sinceall elements in C ( B ) are compatible with all projections in B , the set P B ( a ), computed in C ( B )with respect to B , is all of B . Hence ∅ 6 = P ± ( a ) ⊆ B has the required properties also withrespect to B . It follows that ( ˜ J p ) p ∈ B has the comparability property. Let { J p } p ∈ P be a compression base in A with the comparability property. Let B ⊆ P be a block. Then there is a totally disconnected compact Hausdorff space X such that(i) B is isomorphic (as a Boolean algebra) to the Boolean algebra P ( X ) of all clopen subsetsin X .(ii) C ( B ) is isomorphic (as an order unit space) to a norm-dense order unit subspace in C ( X, R ) .(iii) If A is norm-complete, then C ( B ) ≃ C ( X, R ) (as an order unit space).Proof. By Corollary 3.23, C ( B ) is the norm-closed linear span of B . Let X be the Stone spaceof the block B of P , so that X is a totally disconnected compact Hausdorff space such that(i) is satisfied. Let φ : B → P ( X ) be the boolean algebra isomorphism. Since B is a Booleanalgebra, any element a ∈ span( B ) can be (uniquely) written as a = P ki =1 a i p i , where p i ∈ B and P i p i = 1. In this case, a ∈ A + iff a i ≥ i and k a k = max i | a i | . Similarly, let F ( X, R ) be the set of all simple functions in C ( X, R ). Since the clopen sets separate points of X , F ( X, R ) is norm-dense in C ( X, R ) by Stone-Weierstrass theorem. Any f ∈ F ( X, R ) can bewritten as f = P i a i χ X i , where { X i } ki =1 is a decomposition of X and χ X i is the characteristicfunction of X i . So the map φ extends to a bijection φ : span( B ) → F ( X, R ), given as φ ( X i a i p i ) = X i a i χ φ ( p i ) . a = P i a i p i , b = P i b i q i , where p i , q i ∈ B , P i p i = P j q j = 1. Then p i = ∨ j ( p i ∧ q j ) = P j p i ∧ q j , q j = P i p i ∧ q j and we have a + b = X i,j ( a i + b j ) p i ∧ q j , p i ∧ q j ∈ B, X i,j p i ∧ q j = 1 . Hence φ ( a + b ) = X i,j ( a i + b j ) χ φ ( p i ∧ q j ) = X i,j ( a i + b j ) χ φ ( p i ) χ φ ( q j ) = φ ( a ) + φ ( b ) , so that φ is a linear isomorphism of span( B ) onto F ( X, R ). It is also clear that φ is a unitalorder and norm isomorphism. Extending φ to span( B ), we obtain (ii), (iii) is clear.In the comparability case, we can extend the notion of compatibility to all pairs of elementsin A . Let ( J p ) p ∈ P be a compression base in A with the comparability property. Wesay that a, b ∈ A are compatible if p and q are compatible for all p ∈ P ( a ) and q ∈ P ( b ).Note that by Corollary 3.23 (i), a and b are compatible iff a ∈ C ( P ( b )) or, equivalently, b ∈ C ( P ( a )). This also shows that if a or b is in P , we obtain the same notion of compatibilityas in Definition 3.11. Let ( J p ) p ∈ P be a compression base in A with the comparability property.Then:(i) Two elements in A are compatible if and only if they are in the same C-block.(ii) The C-blocks of A are precisely the maximal sets of mutually compatible elements.Proof. By Corollary 3.23 (i) and the definition, we see that a and b are compatible iff P ( a ) and P ( b ) are contained in the same block B , which is equivalent to a, b ∈ C ( B ). This proves (i).To prove (ii), note that by (i), the elements on any C-block are mutually compatible. On theother hand, if c ∈ A is compatible with all elements of C ( B ), then clearly c ∈ C ( B ). Let A = M sa be the self-adjoint part of a von Neumann algebra M . For anyprojection p ∈ M , let U p : A → A be defined as U p ( a ) = pap , a ∈ A . Then ( U p ) p ∈P isa compression base, where P is the set of all projections in M . It is easily seen that thiscompression base has the comparability property.The elements a, b ∈ A are compatible in the sense of Definition 3.26 if and only if theycommute. Blocks of P are the maximal sets of mutually commuting projections in M and theC-blocks are precisely the maximal abelian von Neumann subalgebras in M . [12, Def. 1.7] The compression base ( J p ) p ∈ P in an order unit space is spectral if it has both the projection cover and the comparability property. A spectral order unit space is an order unit space with a spectral compression base.14ecall that we say that P is monotone σ -complete if suprema (infima) of ascending (de-scending) sequences in P exist and belong to P . Assume that P is monotone σ -complete. Then ( J p ) p ∈ P is spectral if and onlyif it has the comparability property.Proof. Assume that P is monotone σ -complete and let ( J p ) p ∈ P have the comparability property.We will show that ( J p ) p ∈ P also has the projection cover property. So let a ∈ E . By consideringthe sequence λ n = 1 − /n and 1 − a in Lemma 3.21, we obtain a descending sequence q n ≥ q n +1 such that q n ∈ P ± (1 /n − a ). Then J q n (1 /n − a ) ≥
0, so that J q n ( a ) ≤ /nq n . Put q = ∧ n q n ,then J q ( a ) = J q J q n ( a ) ≤ /nJ q ( q n ) = 1 /nq, ∀ n ∈ N , it follows that J q ( a ) = 0 and a ≤ − q . Assume that p ∈ P is such that a ≤ p , then a ∈ C ( p )so that by Corollary 3.23 (i), q n ∈ C ( p ) for all n . Since J q ′ n ( a ) ≥ /nq ′ n , we obtain1 /nJ p ′ ( q ′ n ) ≤ J p ′ J q ′ n ( a ) = J q ′ n J p ′ ( a ) = 0 , hence J p ′ ( q ′ n ) = 0, so that q ′ n ≤ p for all n . It follows that 1 − q = ∨ n q ′ n ≤ p , so that 1 − q isthe projection cover of a . This shows that ( J p ) p ∈ P is spectral. Let X be a compact Hausdorff space. The following are equivalent.(i) C ( X, R ) is spectral.(ii) C ( X, R ) is monotone σ -complete.(iii) X is basically disconnected.(iv) The Boolean algebra P ( X ) is monotone σ -complete.Proof. The equivalences (i) - (iii) were proved in [17, Thm. 4.8]. The equivalence (iii) ⇐⇒ (iv) is well known. (cf. [12, Ex. 1.7]) Let X be a totally disconnected compact Hausdorff spacewhich is not basically disconnected and let A = F ( X, R ), with the compression base ( J p ) p ∈P ( X ) ,where J p = U p | A , see Example 3.10. Then P ( X ) is not monotone σ -complete, but it is easilychecked that the compression base is spectral.As we will see next, the property that A is not norm complete was crucial in the aboveexample. Assume that A is norm complete and let ( J p ) p ∈ P be a compression base withthe comparability property. The following are equivalent.(i) ( J p ) p ∈ P is spectral.(ii) For any block B of P , ( ˜ J p = J p | C ( B ) ) p ∈ B is a spectral compression base in C ( B ) .(iii) Any C-block in A is monotone σ -complete. iv) Any C-block in A is isomorphic to C ( X, R ) for some basically disconnected compact Haus-dorff space X .(iv) P is monotone σ -complete.Proof. Assume that (i) holds and let B be a block of P , then by Corollary 3.23 (i), P ( a ) ⊆ B if and only if a ∈ C ( B ). By Lemma 3.16, a ∈ P ( a ) ⊆ B , so that ( ˜ J p ) p ∈ B has the projectioncover property. We have the comparability property by Lemma 3.24. This proves (ii).Assume (ii). By Theorem 3.25 (iii), any C-block C ( B ) is isomorphic to C ( X, R ), where X is the Stone space of B . It is clear that in this isomorphism, ( ˜ J p ) p ∈ B is identified withthe compression base ( U p ) p ∈P ( X ) , see Example 3.10. The equivalence of (ii), (iii) and (iv) nowfollows by Lemma 3.31.Assume (iii), we will prove that then every descending sequence in P has an infimum in P .We use a method similar to the one used in [32, Lemma 1.3]. So let ( p n ) be such a sequenceand let L ⊆ P be the set of all lower bounds of ( p n ) in P . Notice that L is upward directed.Indeed, let p, q ∈ L and let a = ( p + q ). Then a is compatible with each p n , so that there isa block B in P such that ( p n ) ∪ { a } ⊆ C ( B ). Since C ( B ) is spectral, a has a projection cover r = a ∈ B . Since a ≤ p n , we have r ≤ p n for all n , so that r ∈ L . Also p, q ≤ a ≤ r , wehave p, q ≤ r by the facial property of r . Hence L is upward directed.Further, let L be an increasing chain in L , then again L ∪ ( p n ) is contained in some block B of P . Since B is monotone σ -complete, ( p n ) has an infimum p in B . Clearly, p ∈ L and p ≥ q for any q ∈ L . It follows that every increasing chain in L has an upper bound, soby Zorn’s Lemma, L has a maximal element, but since L is upward directed, L has a greatestelement, which is the greatest lower bound for ( p n ). This proves (iv).The implication (iv) = ⇒ (i) is proved in Theorem 3.30.In the rest of this section we assume that ( J p ) p ∈ P is a spectral compression base in A . Insuch a case, there is a unique mapping ∗ : A → P , called the Rickart mapping , such that forall a ∈ A and p ∈ P p ≤ a ∗ ⇔ a ∈ C ( p ) and J p ( a ) = 0 . Indeed, put a ∗ := 1 − ( k a k − | a | ) , a ∈ A. If p ≤ a ∗ , then ( k a k − | a | ) ≤ − p . It follows that a + , a − ≤ k a k (1 − p ) and hence J p ( a ) = 0, J − p ( a ) = a , so that a ∈ C ( p ). For the converse, assume that a ∈ C ( p ) and J p ( a ) = 0. Let q ∈ P ± ( a ), then q ∈ C ( p ) and 0 = J p ( a ) = J p ∧ q ( a ) + J p ∧ q ′ ( a ) , since 0 ≤ J p ∧ q ( a ) ≤ q and 0 ≤ − J p ∧ q ′ ( a ) ≤ − q , we must have J p ∧ q ( a ) = J p ∧ q ′ ( a ) = 0. Hence J p ( a + ) = J p ( a − ) = J p ( | a | ) = 0, so that | a | ∈ Ker + J p . By (2) and the facial property, we see that | a | ≤ k a k (1 − p ), which entails that k a k − | a | ≤ − p . This implies that p ≤ − ( k a k − | a | ) = a ∗ .Uniqueness is clear.We note that if an order unit space has the comparability property, then the projectioncover property is equivalent to the existence of the Rickart mapping [12, Thm 2.1][15, Thm.6.5].Some important properties of the Rickart mapping are collected in the following lemma.16 .34 Lemma. [12, Lemma 2.1] (i) For ≤ a ≤ b , we have b ∗ ≤ a ∗ ;(ii) For a ∈ A , a ∗ ∈ P ( a ) ;(iii) For a ∈ A , a ∗∗ := ( a ∗ ) ∗ = 1 − a ∗ is determined by the condition J p ( a ) = a ⇔ a ∗∗ ≤ p ; (iv) For e ∈ E , e ∗∗ is the projection cover of e . For a ∈ A , ( a + ) ∗∗ is the least element in P ± ( a ) .Proof. Let p = ( a + ) ∗∗ and q ∈ P ± ( a ), so that a + = J q ( a ), a − = − J − q ( a ). Then J q ( a + ) = a + ,so that p ≤ q . Hence 1 − q ≤ − p , so that J p ( a − ) = 0. It follows that J p ( a ) = J p ( a + ) = a + ≥ J − p ( a ) = J − p ( − a − ) = − a − ≤
0. In particular, p ∈ P C ( a ). We next show that p ∈ P C ( P C ( a )).Assume that q ∈ P C ( a ) and let r ∈ P ± ( a ). Then r is compatible with q and r, q ∈ P C ( a ),so that by Lemma 3.13, we obtain a = ( J q ∧ r + J q ′ ∧ r + J q ∧ r ′ + J q ′ ∧ r ′ )( a ) , so that a + = J r ( a ) = J q ∧ r ( a ) + J q ′ ∧ r ( a ) = J q ( a + ) + J q ′ ( a + ) . It follows that
P C ( a ) ⊆ P C ( a + ). Using Lemma 3.34, we obtain p ∈ P ( a + ) ⊆ P C ( P C ( a + )) ⊆ P C ( P C ( a )) . Now we see that p ∈ P ± ( a ) and we have already shown that p ≤ q for all q ∈ P ± ( a ).Similarly as in the case of spectral duality, we may now define the spectral resolution of a as a family ( p a,λ ) λ ∈ R ⊆ P defined by [12, Def. 3.2] p a,λ := (( a − λ ) + ) ∗ . Note that by Lemma 3.34, p a,λ ∈ P ( a ) for all λ ∈ R .Let λ ≤ · · · ≤ λ n be any sequence. Using Proposition 3.35 and the construction in theproof of Lemma 3.21, we see that q i := (( a − λ i ) + ) ∗∗ is a descending sequence in P , which byLemma 3.34 (iii) gives an ascending sequence of spectral projections p a,λ ≤ · · · ≤ p a,λ n . Proceeding as in the proof of Theorem 3.22, we can see that every a ∈ A can be written as anorm-convergent integral of Riemann-Stieltjes type a = Z U a L a − λdp ( λ ) , where the spectral lower and upper bounds for a are defined by L a := sup { λ ∈ R : λ ≤ a } and U a := inf { λ ∈ R : a ≤ λ } , respectively.Properties of spectral resolutions are described in [12, Section 3]. In particular, we have If p ∈ P , then a ∈ C ( p ) ⇔ p a,λ ∈ C ( p ) for all λ ∈ R . There exists an ascending sequence a ≤ a ≤ · · · in C ( P C ( a )) such thateach a n is a finite linear combination of projections in the family ( p a,λ ) λ ∈ R and k a − a n k → . A comparison of the two spectrality theories
In this section we compare the spectral theory in order unit spaces presented in [12] with thetheory of Alfsen and Shultz [7]. We show in more details that the Alfsen and Shultz theorymay be considered as a special case of the theory presented in [12].
We start by describing more properties of F-compressions that can be obtained taking the σ ( A, V )-continuity into account.Let J : A → A be an F-compression with focus p and let J ∗ : V → V be its adjoint: J ∗ : V → V, J ∗ ϕ = ϕ ◦ J, ϕ ∈ V. It is easy to see that J ∗ is σ ( A, V )-continuous, positive and idempotent. Moreover, we haveKer + J ∗ = { p } • , Im + J ∗ ⊆ { − p } • = (Im + J ∗ ) •• . (5)Indeed, we haveKer + J ∗ = { ϕ ∈ V + : J ∗ ϕ = 0 } = { ϕ ∈ V + : h J ( a ) , ϕ i = 0 , ∀ a ∈ A + } = (Im + J ) • = { p } • and similarlyKer + J = { a ∈ A + : J ( a ) = 0 } = { a ∈ A + : h J ( a ) , ϕ i = 0 , ∀ ϕ ∈ V + } = { a ∈ A + : h a, J ∗ ϕ i = 0 , ∀ ϕ ∈ V + } = (Im + J ∗ ) • , so that Im + J ∗ ⊆ (Im + J ∗ ) •• = (Ker + J ) • = { − p } • . Let p be the focus of an F-compression. Then { − p } •• ∩ E = [0 , − p ] . Proof.
As we have seen,face A + (1 − p ) = Ker + J = (Im + J ∗ ) • = { − p } •• . The result follows by Lemma 3.7.For an F-compression J with focus p , we denote K p := { − p } • ∩ K = { ρ ∈ K, h p, ρ i = 1 } , S ( J ) := Im( J ∗ ) ∩ K. Then S ( J ) is a base of the subcone Im + J ∗ ⊆ V + and by (5), S ( J ) ⊆ K p = S ( J ) •• ∩ K. (6)18 .2 Lemma. Let J be an F-compression. Then J is uniquely determined by its focus p andthe subset S ( J ) .Proof. Any idempotent map J is uniquely determined by its image and kernel. Since Im( J ) ispositively generated and Im + J = face A + ( p ), we see thatIm( J ) = { a ∈ A : ∃ λ ≥ , − λp ≤ a ≤ λp } is the order ideal in A generated by p . Further, it is easy to see thatKer( J ) = Im( J ∗ ) ◦ = (Im + J ∗ ) ◦ = S ( J ) ◦ , since Im( J ∗ ) is positively generated as well and S ( J ) is a base of Im + J ∗ .Following [7], we introduce an important property. A continuous positive idempotent map J : A → A is smooth if Ker( J ) = Tan A + (Ker + J ) , where Tan A + (Ker + J ) is the tangent space of A + at Ker + J . If J is an F-compression with focus p , then we have (see [7, Prop. 7.3])Tan A + (Ker + J ) = Tan A + (1 − p ) = { − p } •◦ . We say that the dual map J ∗ : V → V is neutral if ρ ∈ K p = ⇒ h J ( a ) , ρ i = h a, ρ i , ∀ a ∈ A. Let J be an F-compression with focus p . The following are equivalent.(i) J is smooth;(ii) S ( J ) = K p ;(iii) Im + J ∗ = { − p } • ;(iv) Im + J ∗ is a semi-exposed face of V + ;(v) J ∗ is neutral.Proof. It is clear that (ii) is equivalent to (iii), the equivalence of (iii) and (iv) is clear from(5). The fact that (i) is equivalent to (iv) and (v) was proved in [7, Chap. 7] for more generalmaps, we give an easy proof for F-compressions. Note that (v) can be reformulated as: ϕ ∈ V + , h − p, ϕ i = 0 = ⇒ J ∗ ϕ = ϕ, in other words, { − p } • ⊆ Im + J ∗ . Again by (5), this is equivalent to (iii). Observe that { − p } •◦ ⊆ (Im + J ∗ ) ◦ = { a ∈ A : h J ( a ) , ϕ i = 0 , ∀ ϕ ∈ V + } = Ker( J ) (7)and (i) means that equality holds in (7). It is now clear (iii) implies (i). Conversely, assume(i), then { − p } • ⊆ { − p } •◦◦ ∩ V + = (Im + J ∗ ) ◦◦ ∩ V + = Im( J ∗ ) ∩ V + = Im + J ∗ , here we used the fact that Im( J ∗ ) is a closed linear subspace generated by Im + J ∗ , so that(Im + J ∗ ) ◦◦ = Im( J ∗ ). The proof now follows by (5).19 .4 Lemma. Let p ∈ E be the focus of a smooth F-compression J . Then J is the uniqueF-compression with focus p .Proof. Let ˜ J be an F-compression with focus p . Then Im( ˜ J ) = Im( J ) andKer( J ) = K ◦ p ⊆ S ( ˜ J ) ◦ = Ker( ˜ J ) . But then for any a ∈ A , ˜ J ( a ) = ˜ J ( J ( a ) + a − J ( a )) = ˜ J ( J ( a )) = J ( a ) . We can now compare compressions (Definition 3.1) and F-compressions (Definition 3.5).
Any compression is an F-compression. An F-compression is a compression ifand only if it is smooth and has a smooth complementary F-compression.Proof.
Let J : A → A be a compression. Since J is a normalized idempotent σ ( A, V )-continuouslinear mapping, we only need to check the properties (F2) and (F3) of Definition 3.5.Let J ′ be the complement of J , then J ′ (1) = 1 − p , this can be seen similarly as in Lemma3.6. To show (F2), let 0 ≤ e ≤ p , then J ′ ( e ) ≤ J ′ ( p ) = 0, so e ∈ Ker + J ′ = Im + J , whence J ( e ) = e . For (F3), observe that J ( e ) = 0 implies e ∈ Ker + J = Im + J ′ . Hence J ′ ( e ) = e ≤ e = J ′ ( e ) ≤ J ′ (1) = 1 − p .By Definition 3.1, a continuous positive idempotent map J is a compression if and only if it isbicomplemented, which means that there is a complementary continuous positive idempotentmap J ′ such that the dual maps J ∗ and ( J ′ ) ∗ are also complementary. In this case, J ′ is acompression as well. In particular, J and J ′ are complementary F-compressions such that by(5) { p } • = Ker + J ∗ = Im + ( J ′ ) ∗ , { − p } • = Ker + ( J ′ ) ∗ = Im + J ∗ . By Lemma 4.3, this means that both J and J ′ are smooth (cf. [7, Prop. 7.20]). Recall that the focus of a compression is called a projective unit. Let P denote the set of allprojective units in A and for each p ∈ P , let J p be the unique compression with focus p . Theset J = ( J p ) p ∈P of compressions has a natural ordering given as J p (cid:22) J q if J q J p = J p and themap J p p is an order isomorphism onto the set P with the ordering induced from A , [7,Prop. 7.32].In general, it is not clear whether ( J p ) p ∈P is a compression base in A . Therefore, althoughall J p are F-compressions, we cannot directly use the properties of the map p J p proved forcompression bases. We will next show that it is the case if the Standing Hypothesis 3.3 is true,recall that this means that for any a ∈ A + there is a projective unit denoted by r ( a ) such thatthe exposed face of K (or, equivalently, of V + ) determined by a is also determined by r ( a ),that is { a } • = { r ( a ) } • . It is proved in [7, Prop. 8.1] that under this assumption, P is an orthomodular lattice and bythe above order isomorphism, J (with complementation) is an OML as well, with J p ∨ J q = J p ∨ q and J p ∧ q = J p ∧ J q . Moreover, two compressions J p and J q commute if and only if J p ( q ) ≤ q and then J p J q = J p ∧ J q = J p ∧ q , [7, Thm. 8.3].20 .6 Theorem. If the Standing Hypothesis is satisfied, then { J p } p ∈P is a compression base in A , with the projection cover property.Proof. We first prove that P is a normal subalgebra in E . If p + q ∈ E , then p ≤ − q and p + q = p ∨ q = ( J p ∨ J q )(1) ∈ P , so that P is a subalgebra. Further, let e, f, d ∈ E , e + d + f ≤ p := e + d, q := f + d ∈ P . Then f ≤ − p, d ≤ p implies that J p ( q ) = J p ( f ) + J p ( d ) = d ≤ q so that J p and J q commute and we obtain d = J p J q (1) = J q ∧ J p (1) ∈ P . Similarly, if p, q, r ∈ P and p + q + r ≤
1, then J p + r J q + r = J q + r J p + r = J r . Hence ( J p ) p ∈P is acompression base. Let now a ∈ E and let p = r ( a ). This implies that a ∈ { a } •• ∩ E = { p } •• ∩ E = [0 , p ] , hence a ≤ p , here we have used Lemma 4.1. If a ≤ q ∈ P , then clearly { q } • ⊆ { a } • = { p } • , sothat p ∈ { p } •• ∩ E ⊆ { q } •• ∩ E = [0 , q ] , hence p ≤ q . It follows that p = a . (Also [7, Prop. 8.24]: p is the least projective unit suchthat a ∈ face A + ( p )). Assume that A = V ∗ . Then the following are equivalent:(i) A and V are in spectral duality;(ii) ( J p ) p ∈P is a spectral compression base;(iii) There is a spectral compression base ( J p ) p ∈ P such that all J p are smooth.Proof. By [7, Thm. 8.52,Thm. 8.55], A and V are in spectral duality if and only if the StandingHypothesis holds and every element a ∈ A has a unique orthogonal decomposition a = a + − a − .So assume (i) spectral duality. We already know by Theorem 4.6 that ( J p ) p ∈P is a compressionbase with the projection cover property. To show comparability, let a ∈ A and let a = a + − a − be the unique orthogonal decomposition. By [7, Thm. 8.57], J r with r = r ( a + ) is the leastcompression such that J r ( a ) ≥ J r ( a ) ≥ a , which implies that J − r ( a ) ≤
0. Moreover, by[7, Thm. 8.62], r ∈ C ( P C ( a )). Hence r ∈ P ± ( a ) and ( J p ) p ∈P has the comparability property,so (ii) is true.The implication (ii) = ⇒ (iii) follows by Theorem 4.5, so assume (iii). Then by Theorem4.5, all J p are compressions and so P ⊇ P . Moreover, by Lemma 3.20, all sharp elements arein P , so that P = P and ( J p ) p ∈ P is the set of all compressions. Let a ∈ A . By Proposition3.35, r = ( a + ) ∗∗ is the least element in P ± ( a ) and since p J p is an order isomorphism, J r isthe least compression J such that J ′ ( a ) ≤ ≤ J ( a ), equivalently, J ( a ) ≥ a and J ( a ) ≥
0. Thisimplies (i) (see Definition 3.2).
Let A and V be in spectral duality. Then F-compressions coincide with com-pressions.Proof. Let A and V be in spectral duality. We already know that every compression is anF-compression and that ( J p ) p ∈P is a spectral compression base. By Lemma 3.20, every sharpelement in E is in P and therefore is a projective unit. Since the focus of an F-compression issharp, it is a projective unit. The statement follows by Lemma 4.4.21 Spectrality for JB-algebras
We describe an important example of (spectral) compression bases on JB-algebras. An overallreference we use for JB-algebras is [7, Chap. 1]. [7, Definition 1.1] A
Jordan algebra A over R is a vector space equipped witha commutative bilinear product ◦ that satisfies the identity( a ◦ b ) ◦ a = a ◦ ( b ◦ a ) for all a, b ∈ A. (8) [7, Definition 1.5] A JB-algebra is a Jordan algebra A over R with identityelement 1 equipped with a complete norm satisfying the following requirements for a, b ∈ A : k a ◦ b k ≤ k a kk b k , k a k = k a k , k a k ≤ k a + b k . The self-adjoint part A sa of a C*-algebra A is a JB-algebra with Jordan product a ◦ b = ( ab + ba ). Any norm closed Jordan subalgebra of B ( H ) sa is a JB-algebra.Let A be a JB-algebra. The positive cone in A is defined as A + = A = { a : a ∈ A } . Then (
A, A + ,
1) is a norm-complete order unit space and its order unit norm k·k coincides withthe norm k · k in A , [7, Theorem 1.11]. An element p ∈ A is called a projection if p = p . The setof all projections in A will be denoted by P . Clearly, we then have p ∈ A + and (1 − p ) = 1 − p ,so that 1 − p ∈ P and p ≤ A is defined as { abc } := ( a ◦ b ) ◦ c + ( c ◦ b ) ◦ a − ( a ◦ c ) ◦ b ) , a, b, c ∈ A. If a = c we have { aba } = 2( a ◦ b ) ◦ a − a ◦ b. [7, Lemmas 1.26 and 1.37] If a, b ∈ A + , then (i) and (ii) are equivalent and imply (iii) , where (i) { aba } = 0 , (ii) { bab } = 0 , (iii) a ◦ b = 0 .If a is a projection, then all (i) - (iii) are equivalent. For any a, c ∈ A , we define a linear map U a,c : A → A by U a,c ( b ) = { abc } , b ∈ A. If a = c , we put U a ( b ) := U a,a ( b ). For any a ∈ A , the map U a is positive [7, Theorem 1.25]. If p ∈ P , we get U p ( a ) = 2 p ◦ ( p ◦ a ) − p ◦ a, a ∈ A. .5 Lemma. Let a ∈ E , p ∈ P . Then a ≤ p if and only if a ◦ p = a if and only if U p ( a ) = a ifand only if U − p ( a ) = 0 .Proof. Using Lemma 5.4, we obtain the following equivalences: a ◦ p = a ⇐⇒ a ◦ (1 − p ) = 0 ⇐⇒ U − p ( a ) = 0 . On the other hand, it is clear that a ◦ p = a implies U p ( a ) = a and since U p is positive and U p (1) = p , we obtain that a = U p ( a ) ≤ U p (1) = p . From a ≤ p , it similarly follows that0 ≤ U − p ( a ) ≤ U − p ( p ) = 0, which implies a ◦ p = a , closing the chain of implications.Elements a, b ∈ A + are orthogonal ( a ⊥ b ) if { aba } = { bab } = 0. Note that if p, q ∈ P , then p ⊥ q if and only if p ≤ − q (by Lemma 5.5). By [7, Proposition 1.28], each a ∈ A can beexpressed uniquely as a difference of positive orthogonal elements: a = a + − a − , a + , a − ∈ A + , a + ⊥ a − . (9)Both a + , a − are contained in the norm-closed subalgebra A ( a, ⊆ A generated by a and 1.The decomposition in (9) is called the orthogonal decomposition of a .From now on, we will consider A in the separating order and norm duality with V = A ∗ . For any p ∈ P , U p is a compression on A with focus p .Proof. We will first show that U p is an F-compression. As we have seen, U p is linear and positiveand it is clear that U p (1) = p ≤
1, so (F1) holds. The conditions (F2) and (F3) follow fromLemma 5.5. Let U ∗ p : V → V be the dual map. By [7, Prop. 1.41], U ∗ p is neutral, hence U p issmooth (Lemma 4.3). Since U − p has the same properties and is complementary to U p (Lemma3.6), the proof is finished by Theorem 4.5.We have the following characterization of projections. For p ∈ E , the following are equivalent.(i) p is a projection;(ii) p is principal;(iii) p is sharp.Proof. Assume that p ∈ P , then by Proposition 5.6, p is the focus of the compression U p . Hence p is principal, so (i) implies (ii). The implication (ii) = ⇒ (iii) is by Lemma 3.8. Finally, let p be sharp. Since p ∈ E , we have by the functional calculus [7, Prop. 1.21] that p ≤ p . Itfollows that 0 ≤ p − p = (1 − p ) − (1 − p ) ≤ p, − p , so that p − p = 0 and p ∈ P .It follows by Proposition 5.6, Lemmas 3.8 and 4.4 that any F-compression on A is of theform U p for some projection p . We will next show that also any retraction on A is of this form.By [16], this means that the additive group of A is a c ompressible group, which will show that( U p ) p ∈ P is a compression base. We will need some preparation first.Let T a : A → A be the map obtained by Jordan multiplication by a . We say that a, b ∈ A operator commute if T a T b = T b T a . This will be denoted by a ↔ b . The next result shows thatif b or a is a projection, then this relation is the same as compatibility.23 .8 Theorem. [7, Proposition 1.47] Let A be a JB-algebra, a ∈ A , and let p be a projection in A . The following are equivalent: (i) a and p operator commute; (ii) T p a = U p a ; (iii) ( U p + U − p )( a ) = a (that is, a ∈ C ( p ) ); (iv) a and p are contained in an associative subalgebra in A . It also follows that any element in the subalgebra A ( a,
1) generated by a and 1 commuteswith any projection that commutes with a , that is, A ( a, ⊆ C ( P C ( a )) . (10)Indeed, let a ∈ C ( p ), then a and p are contained in an associative subalgebra A ⊆ A , whichclearly also contains A ( a, A ( a, ⊆ C ( p ).We will also need the following Peirce decomposition of elements in A . [3, Thm. 1.4] Let A be a Jordan algebra with unit , p ∈ A be a projection.Then A decomposes into the direct sum A = U p ( A ) ⊕ U p, − p ( A ) ⊕ U − p ( A ) , where U p ( A ) = { x ∈ A : p ◦ x = x } ,U p, − p ( A ) = { x ∈ A : p ◦ x = 12 x } ,U − p ( A ) = { x ∈ A : p ◦ x = 0 } . If x ∈ U p ( A ) or x ∈ U − p ( A ) then x ∈ C ( p ) ([3, Prop. 1.5]). The proof of the following result is a modification of the proof of [14, Theorem 4.5].
Let A be a JB-algebra and let J : A → A be a retraction with focus p = J (1) .Then p ∈ P and J = U p .Proof. As the focus of a retraction, p is principal and hence p ∈ P by Lemma 5.7. Let p ′ = 1 − p .If e ∈ A , 0 ≤ e ≤ ≤ { p ′ ep ′ } ≤ p ′ , whence0 ≤ e ≤ ⇒ J ( { p ′ ep ′ } ) = 0 . If 0 ≤ e ≤ ≤ pep ≤ p , whence0 ≤ e ≤ ⇒ J ( { pep } ) = { pep } . We claim that for a ∈ A , J ( { pap } ) = { pap } = U p ( a ). Indeed, this is easily seen to be true forall positive elements by normalization. Since we can always write a = x − y with x, y ∈ A + , wehave J ( { pap } ) = { pap } for all A . By the same argument, J ( { p ′ ap ′ } ) = 0 for a ∈ A . For theprojection p we have the Peirce decomposition as in Theorem 5.9: A = U p ( A ) ⊕ U p, − p ( A ) ⊕ U − p ( A ) . a ∈ A , then a = a + a + a where a ∈ U p ( A ), a ∈ U p, − p ( A ), a ∈ U − p ( A ). The we have a = { pa p } , a ◦ p = 12 a , a = { p ′ a p ′ } . For n ∈ Z , consider 0 ≤ ( a − np ′ ) = a − na ◦ p ′ + n p ′ = a − na + n p ′ . Applying J and taking into account that J ( p ′ ) = 0 we get nJ ( a ) ≤ J ( a ) , ∀ n ∈ Z . For positive n we obtain J ( a ) ≤
0, for n negative we obtain − J ( a ) ≤
0, hence J ( a ) = 0.Consequently, J ( a ) = J ( a ) + J ( a ) + J ( a ) = { pa p } = { pap } . The set { U p } p ∈ P is a compression base in A .Proof. The statement is a special case of [16, Theorem 2.3]. We rewrite the proof in the presentsituation for the convenience of the reader.Assume that p, q ∈ P , p + q ≤
1, so that p ≤ − q . It follows that p ◦ (1 − q ) = p andhence p ◦ q = 0. We have ( p + q ) = p + q + 2 p ◦ q = p + q , so that p + q ∈ P and P is asubalgebra in E . Let us prove that P is normal. So let e, f, d ∈ E , e + f + d ≤ p := e + d ∈ P and q := f + d ∈ P . Put J := U p ◦ U q , then J : A → A is a positive linearmap and since f ≤ − p and d ≤ p , we have J (1) = U p ( q ) = U p ( f + d ) = U p ( d ) = d ∈ E. Further, let 0 ≤ a ≤ d , then since a ≤ p and a ≤ q , we have J ( a ) = U p ( U q ( a )) = U p ( a ) = a. Hence J is a retraction and by Theorem 5.10, d ∈ P and J = U d . This also shows that for p, q, r ∈ P , p + q + r ≤
1, we have U p + r ◦ U q + r = U r .In [3] the following symbols are defined for a subset S of A : S ⊥ = { a ∈ A : U a ( x ) = 0 , ∀ x ∈ S } , ⊥ S = { x ∈ A : U a ( x ) = 0 , ∀ a ∈ S } , ⊥ S + = ⊥ S ∩ A + . The following notion of a Rickart Jordan algebra was introduced by Ayupov and Arzikulov. [2, Definition], [2, Theorem 1.6, Theorem 1.7] A Jordan algebra A is Rickart if one of the following equivalent statements is true25A1) For every element x ∈ A + there is a projection p ∈ A such that { x } ⊥ = U p ( A )(A2) For every element x ∈ A there is a projection p ∈ A such that ⊥ { x } + = U p ( A ) + . It was proved in [2] that the self-adjoint part of a C*-algebra A is Rickart JB-algebra iff A is a Rickart C*-algebra. An element p ∈ P is called a carrier of a ∈ A if it is the smallest projectionsuch that p ◦ a = a . The carrier of a will be denoted by s ( a ).The following result is straightforward from Lemma 5.5. Let a ∈ E . Then s ( a ) is the smallest projection such that a ≤ s ( a ) . [3, Lemma 2.7] Let a ∈ A + , p ∈ P . Then p = s ( a ) if ⊥ { a } + = U − p ( A ) + .Proof. Note that using Lemma 5.5, we have for q ∈ P and a ∈ A + that q ◦ a = a iff (1 − q ) ◦ a = 0iff 1 − q ∈ ⊥ { a } + . It follows that 1 − s ( a ) is the largest projection in ⊥ { a } + . If p ∈ P is suchthat ⊥ { a } + = U − p ( A ) + , then 1 − p = U − p (1 − p ) ∈ ⊥ { a } + and for any projection r ∈ ⊥ { a } + we have U − p ( r ) = r , so that r ≤ − p by Lemma 5.5. Hence p = s ( a ). Let A be a Rickart JB-algebra. For a, b ∈ A + then (i) { aba } = 0 ⇔ { s ( a ) bs ( a ) } = 0 .If, in addition, a ↔ b , then (ii) a ◦ b = 0 = ⇒ s ( a ) ↔ b and s ( a ) ◦ b = 0 .Proof. For (i), note that { aba } = 0 ⇒ b ∈ ⊥ { a } + = U − s ( a ) ( A ) + , hence b = U − s ( a ) ( b ), whichimplies U s ( a ) ( b ) = 0, i.e., { s ( a ) bs ( a ) } = 0. If { s ( a ) bs ( a ) } = 0, then { bs ( a ) b } = 0 by Lemma5.4, and a ≤ k a k s ( a ) implies { bab } ≤ k a k{ bs ( a ) b } = 0 which yields { aba } = 0. For (ii), notethat if a ↔ b , then a ◦ b = 0 implies { aba } = 0. Indeed, we have { aba } = 2( a ◦ b ) ◦ a − a ◦ b = 2 T a T b ( a ) − T b T a ( a ) = ( a ◦ b ) ◦ a = 0 . This implies by (i) that { s ( a ) bs ( a ) } = 0, and this in turn implies by Lemma 5.4 that s ( a ) ◦ b = 0and by Theorem 5.8 that s ( a ) ↔ b .We are now ready to prove the main result of this section. Let A be a JB-algebra. Then A is a Rickart if and only if A is a spectral.Proof. The family ( U p ) p ∈ P is a compression base by Corollary 5.11. If A is Rickart, then itfollows by Lemma 5.15 that any a ∈ A + has a carrier s ( a ) (this was proved in [3, Theorem 2.1,Theorem 2.8] for all a ∈ A ). By Lemma 5.14, if a ∈ E , then s ( a ) is a projection cover of a .Hence ( U p ) p ∈ P has the projection cover property. We have to prove the comparability property.For a ∈ E , then by Lemma 3.16, s ( a ) = a ∈ P ( a ). By normalization, we have s ( a ) ∈ P ( a )for all a ∈ A + . Let now a ∈ A , then the elements a + , a − are contained in the subalgebra26 ( a, a + , a − ∈ C ( P C ( a )) by (10). Hence P C ( a ) ⊆ P C ( { a + , a − } ) ⊆ P C ( a + ) sothat s ( a + ) ∈ P ( a + ) ⊆ C ( P C ( a + )) ⊆ C ( P C ( a )) . Applying Lemma 5.16 subsequently to a + , a − , and then to s ( a + ) , a − instead of a, b , weobtain first that a − ∈ C ( s ( a + )) and { s ( a + ) a − s ( a + ) } = 0, and then that s ( a + ) ↔ s ( a − ) and { s ( a + ) s ( a − ) s ( a + ) } = 0. Since a + , a − ∈ C ( s ( a + )), we have a ∈ C ( s ( a + )).Put p := s ( a + ), then we proved that p ∈ P ( a ). From a = a + − a − we have U p ( a ) = U p ( a + ) − U p ( a − ) = a + − a + ≥ , and U − p ( a ) = U − p ( a + ) − U − p ( a − ) = 0 − a − = − a − ≤ . Hence p ∈ P ± ( a ) and the comparability property is satisfied.Conversely, assume that A is spectral. Since the sharp elements are precisely the projections,we have by Lemma 3.20 and Theorem 5.10 that ( U p ) p ∈ P is a spectral compression base. Wewill check that the condition (A2) of Definition 5.12 is satisfied, so that A is Rickart. Let a ∈ A and and let C be a C-block of A such that a ∈ C (Corollary 3.23 (ii)). Then by (10) we have A ( a, ⊆ C . By Theorem 3.33 (iv), there is a basically disconnected compact Hausdorff space X such that C ≃ C ( X, R ). We will denote the elements of C and their representing functionsin C ( X, R ) by the same symbol. Let Y := { x ∈ X : a ( x ) = 0 } , then the support of a is theclosure ¯ Y , which is clopen in X .Let p be the projection in C corresponding to ¯ Y , then observe that p = a ∗∗ . Indeed, wehave U p ( a ) = ˜ U p ( a ) = pa = a , hence a ∗∗ ≤ p by Lemma 3.34. Conversely, a ∗∗ ∈ P ( a ) ⊆ C , sothat a = U a ∗∗ ( a ) = ˜ U a ∗∗ ( a ) = a ∗∗ a . Since ¯ Y is the support of a , this implies that p ≤ a ∗∗ . Notethat it also follows that a ∗∗ = ( a ) ∗∗ .We will next prove that ⊥ { a } + = U − p ( A ) + , so that (A2) holds. This amounts to provingthat { b ∈ A + : U a ( b ) = 0 } = Ker + U p . Assume first that b ∈ Ker + U p . Since a = U p ( a ), we have by [7, Eq. (1.16)] U a ( b ) = U { pap } ( b ) = U p U a U p ( b ) = 0 . Conversely, let b ∈ A + and let U a ( b ) = 0. Then we also have U a ( b ) = U a U a ( b ) = 0, so that a ⊥ b . Let c := a − b, then by the uniqueness of the orthogonal decomposition, we get a = c + and b = c − . Con-sequently, if now C ′ is a block of A containing c , then also a , b ∈ C ′ . Considering again therepresenting functions in C ( X ′ , R ) for a basically disconnected compact Hausdorff space X ′ ,we see that U a ( b ) = 0 ⇐⇒ U b ( a ) = b a ⇐⇒ ba = 0 . It follows that b must be 0 on the support of a , and we have seen above that the projectioncorresponding to the support is ( a ) ∗∗ = a ∗∗ = p . This implies that U p ( b ) = 0, so that we musthave b ∈ Ker + U p . 27 Centrally symmetric state spaces
In this section, we obtain a class of order unit spaces such that A = V ∗ for a base norm space V , where a spectral compression base exists but the duality of A and V is not spectral.Let ( X, k · k ) be a normed space and let B be its closed unit ball. We will construct a dualpair ( A, V ) with A = V ∗ such that the state space K ∼ = B , such state spaces are sometimescalled centrally symmetric (cf. [27]).Put V := R × X and V + := { ( α, x ) , k x k ≤ α } , K := { (1 , x ) , x ∈ B } . It is easy to see that V + is a generating cone: indeed, any element ( α, x ) ∈ V can be writtenas ( α, x ) = (max { α, k x k} , x ) − (max {k x k − α, } , ∈ V + − V + . Moreover, K is a base of V + and is located on the hyperplane H := { ( α, x ) ∈ V, α = 1 } 6∋ . Note that co ( K ∪ − K ) = { ( α, x ) , | α | ≤ , x ∈ B } . This set is clearly radially compact, that is, B ∩ L is a closed segment for every line L through the origin of V . By [4, Prop. II.1.12], ( V, K )is a base normed space and co ( K ∪ − K ) is the unit ball of the base norm k · k K . It follows that k ( α, x ) k K = max {| α | , k x k} . Let A = V ∗ , then A is isomorphic to the space A b ( K ) of bounded affine functions K → R ([7]). With the cone A + ∼ = A b ( K ) + of positive functions on K and with 1 ∼ = 1 K the constantunit functional, ( A, A + ,
1) is an order unit space, in separating order and norm duality with V .It is easily seen that we may put A = R × X ∗ with the cone A + = { ( a , y ) , k y k ∗ ≤ a } and the unit 1 = (1 , k ( a , y ) k = k y k ∗ + | a | and the unit interval E has the form E = { ( a , y ) , k y k ∗ ≤ min { a , − a }} , note that this implies ( a , y ) ∈ E = ⇒ a ∈ [0 ,
1] and k y k ∗ ≤ / . (11)We next characterize some special elements and subsets of E and show their relation to thestructure of the dual unit ball, which will be denoted by B ∗ . Let a ∈ E , a = 0 , a = 1 . The following are equivalent.(i) a is sharp;(ii) k a k = k − a k = 1 ; iii) a = (1 / , y ) with k y k ∗ = 1 / .Proof. Assume that k a k <
1, then there is some s ∈ (0 ,
1) such that (1 + s ) a ≤
1. But then sa ≤ a, sa ≤ − a, so that a is not sharp. Since a is sharp iff 1 − a is sharp, we obtain that (i) implies (ii).Assume the equality k a k = k − a k = 1, that is k y k ∗ + a = 1 = k − y k ∗ + 1 − a Hence a = 1 − a = 1 / k y k ∗ and (ii) implies (iii).To show that (iii) implies (i), let a = (1 / , y ) with k y k ∗ = 1 / b = ( b , w ) ∈ E , b ≤ a . Then (1 / − b , y − w ) ∈ A + , so that k y − w k ∗ ≤ / − b . We obtain1 / − b ≤ / − k w k ∗ ≤ |k y k ∗ − k w k ∗ | ≤ k y − w k ∗ ≤ / − b . This entails that b = k w k ∗ and 1 / − k w k ∗ = k y − w k ∗ . If also b ≤ − a we similarly obtain1 / − k w k ∗ = k y + w k ∗ . But then1 / k y k ∗ ≤ / k y − w k ∗ + 1 / k y + w k ∗ = 1 / − k w k ∗ ≤ / , which implies that b = k w k ∗ = 0. Hence b = 0 and a is sharp. A subset F ⊆ E is a face of E not containing 0 or 1 if and only if there is a face F of / B ∗ such that F = { (1 / , y ) , y ∈ F } . In particular, a ∈ E is an extreme point of E other than 0 or 1 if and only if a = (1 / , y ) forsome extreme point y of / B ∗ .Proof. Let F be of the given form and let a = ( a , y ) , b = ( b , w ) ∈ E be such that λa +(1 − λ ) b ∈ F . Then λa + (1 − λ ) b = 1 / λy + (1 − λ ) w ∈ F . Since by (11) k y k ∗ , k w k ∗ ≤ / y, w ∈ F as well. In particular, k y k ∗ = k w k ∗ = 1 /
2, this and (11) imply that a , b = 1 /
2. This shows that F is a face of E , obviously not containing 0 or 1.Conversely, let F be a face of E , 0 , / ∈ F , and let a = ( a , y ) ∈ F . Note that if ta ≤ t >
1, then a = t − ta + (1 − t − )0, so that 0 ∈ F , similarly, k − a k < ∈ F , So wemust have k a k = k − a k = 1 and by Lemma 6.1, this implies that a = k y k ∗ = 1 /
2. Let F := { y ∈ / B ∗ , (1 / , y ) ∈ F } , then it is easy to see that F is a face of 1 / B ∗ . The last statement is now obvious. Any extremal element p ∈ E is one dimensional, that is, ≤ a ≤ p implies a = tp for some t ∈ [0 , .Proof. By Lemma 6.2 an extremal element in E is of the form p = (1 / , y ) for some y extremalin 1 / B ∗ . Let 0 ≤ b = ( b , w ) ≤ p . Since k y k ∗ = 1 /
2, we obtain similarly as in the proof ofLemma 6.1 that we must have k w k ∗ = b and k y − w k ∗ = 1 / − k w k ∗ = k y k ∗ − k w k ∗ . Put λ := 2 k y − w k ∗ = 1 − k w k ∗ , then we get y = λ ( 12 k y − w k ∗ ( y − w )) + (1 − λ )( 12 k w k ∗ w ) . As y is extremal, we get w = 2 k w k ∗ y = 2 b y , so that b = ( b , b y ) = 2 b p , with 2 b ∈ [0 , A . For this, we need tointroduce the following notations for x ∈ X and y ∈ X ∗ : ∂ x := { y ∈ B ∗ , h y, x i = k x k} , ∂ ∗ y := { x ∈ B, h y, x i = k y k ∗ } . Then ∂ x , ∂ ∗ y are faces of the respective unit ball and ∂ x = ∅ . An element p ∈ E is the focus of a retraction if and only if p = (1 / , y ) ,where y is an extremal element in / B ∗ attaining its norm on B (i.e. ∂ ∗ y = ∅ ). A map J : A → A is a retraction with focus p if an only if J ( a ) = ( a + h x, w i ) p, a = ( a , w ) ∈ A for some x ∈ ∂ ∗ y . Moreover(i) The map J is an F-compression if and only if ∂ x is a singleton, that is, ∂ x = { y } .(ii) The map J is a compression if and only if ∂ ∗ y = { x } and ∂ x = { y } .Proof. Assume that J : A → A is a retraction with focus p . By Lemma 3.8, p must be extremal,so that p = (1 / , y ) for an extremal element y ∈ / B ∗ . Note that 1 − p = (1 / , − y ), so that,by symmetry of B ∗ , 1 − p is an extremal element in E as well. It follows that both p and 1 − p are one dimensional. By (3), we obtain Im + J = R + p and since Im( J ) is positively generated,Im( J ) = R p . Consequently, J must be of the form J ( a ) = h a, ρ i p, a ∈ A, for some linear functional ρ on A . Since J is continuous, positive and J (1) = p , we must have ρ ∈ K and since J is idempotent, h p, ρ i = 1, so that ρ ∈ K p = { (1 , x ) , x ∈ B, h (1 / , y ) , (1 , x ) i = 1 } = { (1 , x ) , x ∈ ∂ ∗ y } . Let 0 ≤ a ≤ p , then by Corollary 6.3, a = tp for some t ∈ [0 ,
1] so that J ( a ) = tJ ( p ) = tp = a. Hence a map of this form is indeed a retraction. This proves the first part of the statement.A retraction is an F-compression if and only if it satisfies (F3) of Definition 3.5. So let a = ( a , w ) ∈ E , then clearly J ( a ) = 0 if and only if a + h x, w i = 0, which means that − a − w ∈ ∂ x . So (F3) is true if and only if such a is in [0 , − p ], but since 1 − p is one-dimensional, this means that ( a , w ) = t (1 − p ) = ( t/ , − ty ), which implies − a − w = 2 y . Sincewe always have 2 y ∈ ∂ x , this proves (i).By the symmetry of the unit ball, we see from (i) that if J is an F-compression, then it hasa complementary F-compression J ′ with focus 1 − p . Indeed, if x ∈ ∂ ∗ y and ∂ x = { y } , thenclearly − x ∈ ∂ ∗− y and ∂ − x = {− y } . By Theorem 4.5, J is a compression if and only if both J and J ′ are smooth. By Lemma 4.3, J is smooth if and only if K p = S ( J ) = Im( J ∗ ) ∩ K. Since J ∗ ϕ = h p, ϕ i ρ , we see that S ( J ) = { ω ∈ K : J ∗ ω = ω } = { ω ∈ K : h p, ω i ρ = ω } = { ρ } . This means that K p and consequently also ∂ ∗ y must be a singleton, and ∂ ∗ y = { x } . If this istrue, J is smooth and again by symmetry, J ′ is smooth as well. This proves (ii).30 normed space X is called strictly convex if its closed unit ball B is strictly convex, thatis, every boundary point of B is an extreme point of B . A normed space is called smooth ifany nonzero element attains its norm at a unique point of the dual unit ball. If X ∗ is strictlyconvex (smooth), then X is smooth (strictly convex). If X is reflexive, then these propertiesare mutually dual. More precisely, X is smooth iff X ∗ is strictly convex iff ∂ x is a singleton forall 0 ≤ x ∈ X . For more details, see e.g. [25, § Let ( X, k · k ) be a Banach space and let the pair ( A, V ) be constructed as above.Then there is a spectral compression base ( J p ) p ∈ P on A if and only if X is reflexive and smooth.Proof. Let ( J p ) p ∈ P be a spectral compression base. By Lemma 3.20, all sharp elements are in P .By Lemma 6.1 and Proposition 6.4, this means that ∂ ∗ y = ∅ for all y ∈ X ∗ with k y k ∗ = 1 /
2, andhence for all nonzero y in X ∗ . By James’s theorem, this means that X is reflexive. Moreover,since for any y with k y k ∗ = 1 / { y } = ∂ x is a face of B ∗ , we see that X ∗ must be strictlyconvex, so that X is smooth.Conversely, if X is a reflexive smooth Banach space, then ∂ ∗ y = ∅ for all y = 0 and ∂ x is a singleton for all x = 0. Therefore, any p = (1 / , y ) with k y k ∗ = 1 / J p of the form J p ( a ) = ( a + h x p , w i ) p, a = ( a , w ) , for some choice of x p ∈ ∂ ∗ y .Let P = { (1 / , y ) , k y k ∗ = 1 / } ∪ { , } be the set of all sharp elements in E . Since B ∗ is strictly convex, all p ∈ P , p = 0 , P is a normal subalgebra in E and that ( J p ) p ∈ P is a compression basewith the projection cover property (we put J = id and J = 0). To prove comparability, let a = ( a , w ) ∈ A . If ± a ∈ A + , then clearly 0 or 1 is in P ± ( a ). So assume that ± a A + , whichamounts to k w k ∗ > | a | . Put y := (2 k w k ∗ ) − w , then p = (1 / , y ) ∈ P and we have( a , w ) = ( k w k ∗ + a ) p − ( k w k ∗ − a )(1 − p ) , hence p ∈ P ± ( a ). So ( J p ) p ∈ P is a spectral compression base. Let ( X, k · k ) be a Banach space and let the pair ( A, V ) be constructed as above.Then ( A, V ) is in spectral duality if and only if X is reflexive, smooth and strictly convex.Proof. Assume that (
A, V ) are is spectral duality. Let P be the set of all projective units andlet J p be the corresponding compressions. By Theorem 4.7, ( J p ) p ∈P is a spectral compressionbase, so that by Theorem 6.5, X is reflexive and smooth. As before, all sharp elements arein P , so that (1 / , y ) with k y k ∗ = 1 / ∂ ∗ y must be asingleton, which implies that X ∗ is smooth. Hence X is strictly convex.Conversely, let X be a strictly convex and smooth reflexive Banach space. Then we mayconstruct a spectral compression base ( J p ) p ∈ P as in the proof of Theorem 6.5, where P is theset of all sharp elements. Since ∂ ∗ y is a singleton for all y = 0, it follows by Proposition 6.4 thatall J p are compressions. The proof now follows by Theorem 4.7.31 Conclusions
We studied in some detail the notion of spectrality in order unit spaces in the sense of Foulis[12]. This notion is based on a particular set of idempotent mappings, called a compressionbase. Spectrality is characterized by two properties of the compression base: projection coverproperty and comparability. While the projection cover property can be trivial in some cases(consider for example the compression base consisting only of the zero and identity maps), weproved that the comparability property is already rather strong. If comparability holds, theorder unit space is covered by subsets called C-blocks that admit a functional representation.Moreover, this property is equivalent to spectrality if the set P of projections is monotone σ -complete, in particular, in the finite dimensional case.We compared this notion of spectrality with the spectral theory of Alfsen and Shultz [7],where it is assumed that the order unit space is the dual of a Banach space. We proved that thistheory is a special case of the Foulis spectrality and found conditions under which they coincide,as well as examples when they are different. We studied the spectral theory for JB-algebrasand proved that Rickart JB-algebras are exactly those that are spectral in the Foulis sense. Inthis case, the compression base is formed by compressions in the sense of Alfsen-Shultz, butthe Alfsen-Shultz theory is not applicable since a Rickart JB-algebra does not have a predualin general.There is also a more general version of Alfsen-Shultz theory [5], where A is only assumed tobe monotone σ -complete. For C*-algebras, it is known that a C*-algebra is Rickart if and onlyif the self-adjoint part of any maximal abelian subalgebra is monotone σ -complete [32], butit is not known if this implies monotone σ -completeness for the self-adjoint part of the wholealgebra. A similar characterization of Rickart JB-algebras is unknown. Here the maximalabelian subalgebras should be replaced by maximal associative subalgebras. It is an interestingquestion whether the maximal associative subalgebras of Rickart JB-algebras coincide with theC-blocks. References [1] M. C. Abbati and A. Mania, A spectral theory for order unit spaces,
Ann. Inst.H. Poincar´e Sect. A (N.S.) (1981), 259-285.[2] Sh. A. Ayupov, F.N. Arzikulov: Jordan counterparts of Rickart and Baer *-algebras, Uzbek. Math. J. (2013) ??-21.[3] F.N. Arzikulov: On abstract JW-algebras, Sibirsk. Math. Zh. (in Russian) , No1.20-27[4] E.M. Alfsen: Compact Convex Sets and Boundary Integrals , Springer, New York1971.[5] E.M. Alfsen, F.W. Shultz: Non-commutative spectral theory for affine functionspaces on convex sets,
Mem. Amer. Math. Soc. (1976) No. 172.[6] E. M. Alfsen and F. W. Shultz, State spaces of operator algebras: basic theory,orientations and C*-products , Mathematics: Theory & Applications, Birkh¨auserBoston, 200l. 327] E. M. Alfsen, F.W. Shultz:
Geometry of State Spaces of Operator Algebras ,Birkh¨auser, Boston-Basel-Berlin 2003.[8] J. Barrett, Information processing in generalized probabilistic theories,
Phys. Rev.A (2007), 032304[9] G. M. D’Ariano, G. Chiribella, and P. Perinotti, Quantum theory from first prin-ciples: an informational approach , Cambridge University Press (2017).[10] A. Dvureˇcenskij, S. Pulmannov´a:
New Trends in Quantum Structures , Kluwer,Academic, Dordrecht, 2000.[11] D.J. Foulis, M.K. Bennett: Effect algebra and unsharp quantum logics,
Found.Phys. (1994) 1331-1352.[12] D.J. Foulis, S. Pulmannov´a: Spectral resolutions in an order unit space, Rep. Math.Phys, (2008) 323-344.[13] D.J. Foulis: Compressible groups, Math. Slovaca (5) (2003) 433-455.[14] D.J. Foulis: Compressions on partially ordered abelian groups, Proc. Amer. Math.Soc. (2004) 3581-3587.[15] D.J. Foulis: Compressible groups with general comparability,
Math. Slovaca (4)(2005) 409-429.[16] D.J. Foulis: Compression bases in unital groups, Int. J. Theoret. Phys. (12)(2005) 2153-2160.[17] D.J. Foulis, S. Pulmannov´a: Monotone σ -complete RC-groups, J. London Math.Soc. (2) (2006) 1325-1346.[18] S.P. Gudder, S. Pulmannov´a: Representation theorem for convex effect algebra, Comment. Math. Univ. Carolinae (4) (1998) 645-659.[19] S. Gudder, S. Pulmannov´a, E. Beltrametti, S. Bugajski: Convex and linear effectalgebras, Rep. Math. Phys. (1999) 359-379.[20] S. Gudder: Compressible effect algebras, Rep. Math. Phys. (2004) 93-114.[21] S. Gudder: Compression bases in effect algebras, Demonstratio Math. (2006)43-58.[22] S. Gudder: Convex and sequential effect algebras, arXiv:1802.01265, 2018.[23] J. Harding, Regularity in quantum logic, Int. J. Theor. Phys. (1998), 1173–1212[24] A. Jenˇcov´a, M. Pl´avala; On the properties of spectral effect algebras, Quantum (2019), 148[25] G. K¨othe, Topological Vector Spaces I , Springer, Berlin, Heidelberg, 1983.[26] L. Lami: Non-classical correlations in quantum mechanics and beyond, PhD thesis,Universitat Autonoma de Barcelona (2018), arXiv:1803.029023327] L. Lami, C. Palazuelos, A. Winter: Ultimate data hiding in quantum mechanicsand beyond, Commun. Math. Phys. (2018), 661-708[28] G. Ludwig. Versuch einer axiomatischen Grundlegung der Quantenmechanik undallgemeinerer physikalischer Theorien, Z. Phys., (1964), 233–260[29] G. Mackey:
Mathematical Foundations of Quantum Mechanics . Benjamin, 1963.[30] P.Pt´ak, S. Pulmannov´a:
Orthomodular Structures as Quantum Logics , Kluwer,Dordrecht and VEDA, Bratislava 1991.[31] S. Pulmannov´a: Effect algebras with compressions,
Rep. Math. Phys. (2006)301-324.[32] K. Saitˆo, J.D.M. Wright: On defining AW*-algebras and Rickart C*-algebras, TheQuarterly Journal of Mathematics66