---
abstract: 'We show that finite-dimensional order unit spaces equipped with a continuous sequential product as defined by Gudder and Greechie are homogeneous and self-dual. As a consequence of the Koecher-Vinberg theorem these spaces therefore correspond to Euclidean Jordan algebras. We remark on the significance of this result in the context of reconstructions of quantum theory. In particular, we show that sequential product spaces that have locally tomographic tensor products, i.e. their vector space tensor products are also sequential product spaces, must be C$^*$ algebras. Finally we remark on a couple of ways these results can be extended to the infinite-dimensional setting of JB- and JBW-algebras and how changing the axioms of the sequential product might lead to a new characterisation of homogeneous cones.'
author:
- |
    John van de Wetering\
    Radboud University Nijmegen, Netherlands\
    `john@vdwetering.name`
bibliography:
- '../bibliography.bib'
title: Sequential Product Spaces are Jordan Algebras
---

Introduction
============

The set of observables of a quantum system can be represented as the space of self-adjoint operators on a complex Hilbert space $B(H)^{\text{sa}}$. This space has a variety of algebra-like structures that can be associated to it, the most well-known of which is the *Jordan product* $a*b := \frac12(ab+ba)$. In the 30’s Jordan, von Neumann and Wigner hoped to find generalisations of the quantum mechanical formalism by considering general spaces equipped with an axiomatisation of this algebraic structure. They however found that the resulting *Euclidean Jordan algebras* (EJAs) have a strikingly simple classification [@jordan1993algebraic], and hence that this algebraic approach does not allow you to go far beyond quantum theory. The significance of EJAs was further established by the Koecher-Vinberg theorem that states that any homogeneous and self-dual ordered vector space is a Euclidean Jordan algebra [@koecher1957positivitatsbereiche]. These two results, the Koecher-Vinberg theorem and the classification by Jordan, von Neumann and Wigner, lie at the heart of many *reconstructions of quantum theory* where intuitively sensible axioms from which quantum theory can be derived are sought [@hardy2011reformulating; @wetering2018reconstruction; @selby2018reconstructing; @barnum2014higher; @gunson1967algebraic] (although it should be noted that these theorems are not used directly in all approaches [@tull2016reconstruction; @hohn2017toolbox; @chiribella2011informational]).

Other algebraic structures on $B(H)^{sa}$ studied in an axiomatic way are the *quadratic Jordan algebras* that axiomatize the map $(a,b)\mapsto aba$ or the more general *triple product* $(a,b,c)\mapsto \frac{1}{2}(abc+cba)$. The definitions of the Jordan product and the triple product don’t have a particularly compelling physical motivation: the product does not correspond to any type of physical process. In this paper we will look at a different structure that *does* follow naturally from physical processes. Given two positive operators $a$ and $b$ we define their *sequential product* as $a{\,\&\,}b := \sqrt{a}b\sqrt{a}$. When $a$ and $b$ represent *effects*, i.e. possible outcomes in a measurement, then the sequential product models the act of first getting the outcome $a$ and then the outcome $b$, hence the name *sequential* product (although this composition is also known as a *Lüders process*). It is important to note that this product is only defined for positive operators (since otherwise the square root wouldn’t be defined), and that this operation is not bilinear, associative or commutative. Gudder and Greechie introduced the concept of a *sequential effect algebra* to study the sequential product in a more abstract setting [@gudder2002sequential]. While they studied the structure of the sequential product on the very general structure of *effect algebras*, we will restrict ourselves to the more concrete setting of *order unit spaces*:

An *order unit space* $(V, \leq, 1)$ is an ordered real vector space with the additional property that $1$ is a *strong Archimedean unit*:

1.  Strong unit: For all $a\in V$ we can find $n\in {\mathbb{N}}$ such that $-n1\leq a \leq n$.

2.  Archimedean: For $a\in V$ when $a\leq \frac1n 1$ for all $n\in{\mathbb{N}}_{>0}$, then $a\leq 0$.

We call the elements $a\in V$ with $0\leq a \leq 1$ the *effects* of $V$ which we will denote by $[0,1]_V$. The *states* of $V$ are positive linear maps $\omega:V\rightarrow {\mathbb{R}}$ such that $\omega(1) = 1$.

Ordered vector spaces with a strong unit represent the most general kinds of systems allowed in causal *generalised probabilistic theories* [@barrett2007information] and hence they form a suitable background to studying models of general physical theories. The Archimedity condition states intuitively that there are no effects that cannot be distinguished by a state. More precisely, order unit spaces are precisely the ordered vector spaces where the states order-separate the elements: if $\omega(v)\leq \omega(w)$ for all states $\omega$ then $v\leq w$.

Note that an order unit space has a norm induced by the order in the following way: ${\norm{a} := \inf\{r\in {\mathbb{R}}_{\geq 0}~;~ -r1\leq a\leq r1\}}$. Whenever we refer to continuity in the context of order unit spaces it should be understood to refer to this norm.

The object of study in this paper is an order unit space with an operation modelled after the sequential product on $B(H)^{\text{sa}}$. To be specific:

\[def:seqprod\] Let $(V, \leq, 1, \&)$ be an order unit space equipped with a binary operation\
${\&:[0,1]_V\times [0,1]_V \rightarrow [0,1]_V}$. We write $a{\,\lvert\,}b$ and say $a$ and $b$ are *compatible* when $a{\,\&\,}b = b{\,\&\,}a$. We call $V$ a *sequential product space* and $\&$ a *sequential product* when $\&$ satisfies the following properties for all $a,b,c \in [0,1]_V$:

1.  \[ax:add\] Additivity: $a{\,\&\,}(b+c) = a{\,\&\,}b+ a{\,\&\,}c$.

2.  \[ax:cont\] Continuity: The map $a\mapsto a{\,\&\,}b$ is continuous in the norm.

3.  \[ax:unit\] Unitality: $1{\,\&\,}a = a$.

4.  \[ax:orth\] Compatibility of orthogonal effects: If $a{\,\&\,}b = 0$ then also $b{\,\&\,}a =0$.

5.  \[ax:assoc\] Associativity of compatible effects: If $a{\,\lvert\,}b$ then $a{\,\&\,}(b{\,\&\,}c) = (a{\,\&\,}b){\,\&\,}c$.

6.  \[ax:compadd\] Additivity of compatible effects: If $a{\,\lvert\,}b$ then $a {\,\lvert\,}1-b$, and if also $a{\,\lvert\,}c$ then $a{\,\lvert\,}(b+c)$.

7.  \[ax:compmult\] Multiplicativity of compatible effects: If $a{\,\lvert\,}b$ and $a{\,\lvert\,}c$ then $a{\,\lvert\,}(b{\,\&\,}c)$.

The properties we require of $\&$ are the same as that of a sequential product in a sequential effect algebra [@gudder2002sequential] except for condition \[ax:cont\] which is new. It should be noted that the standard sequential product $a{\,\&\,}b = \sqrt{a}b\sqrt{a}$ on $B(H)^{\text{sa}}$ is not fully characterised by these axioms, as there are multiple binary operations that satisfy these axioms [@weihua2009uniqueness]. It is possible however to characterise the standard sequential product using related sets of axioms [@gudder2008characterization; @westerbaan2016universal; @wetering2018characterisation]. It has been established in the authors previous work [@wetering2018characterisation] that Euclidean Jordan algebras allow a binary operation satisfying these properties and hence are examples of sequential product spaces. The main purpose of this paper is to establish the converse:

Let $V$ be a finite-dimensional sequential product space, then $V$ is order-isomorphic to a Euclidean Jordan algebra.

Since EJAs are very well understood and in particular classified we can use this theorem to prove additional results. In particular, using the classification of Jordan algebras and a property called *local tomography* we can infer when sequential product spaces are C$^*$-algebras.

Let $V$ and $W$ be finite-dimensional sequential product spaces. We say that they have a *locally tomographic composite* when their vector space tensor product $V\otimes W$ is also a sequential product space with $(a_1\otimes b_1){\,\&\,}(a_2\otimes b_2) = (a_1{\,\&\,}a_2)\otimes (b_1{\,\&\,}b_2)$ for all effects $a_i$ in $V$ and $b_i$ in $W$.

In the context of generalised probabilistic theories, the property of local tomography states that local measurements on each of the subsystems is enough to fully characterise bipartite states. It is a property that holds for regular quantum theory, but fails for, for instance, quantum theory over the real numbers.

Let $V$ be a finite-dimensional sequential product space that has a locally tomographic composite with itself, then there exists a C$^*$-algebra $A$ such that $V$ is isomorphic to $A^{\text{sa}}$ as a Jordan algebra.

Note that C$^*$-algebras being singled out among all the EJAs by local tomography is not surprising as similar results were obtained in [@barnum2014local; @selby2018reconstructing; @masanes2014entanglement], but in combination with the results regarding the sequential product it does give a novel understanding of the mathematical structure of quantum theory:

> A causal probabilistic physical theory that satisfies local tomography and that has a well-behaved notion of sequential measurement must be modelled by C$^*$-algebras.

There have been quite a few characterisations of quantum theory using operational axioms [@barnum2014higher; @selby2018reconstructing; @masanes2014entanglement; @chiribella2011informational; @tull2016reconstruction; @wetering2018reconstruction; @wilce2016royal; @hohn2017toolbox; @hardy2001quantum], but the one presented above is different in a couple of ways. First of all, other characterisations and reconstructions have their axioms refer to a multitude of structures, like the existence of certain systems, transformations and pure states, instead of focusing on a single aspect, which this characterisation does with regards to sequential measurement. Second, all reconstructions of quantum theory that the author is aware of have axioms ensuring the existence of suitable reversible (i.e. invertible) dynamics in the theory. In contrast, this characterisation of C$^*$-algebras doesn’t directly say anything about the existence of reversible maps.

In addition to the above theorems, we will also establish a couple of infinite-dimensional versions of the first theorem relating infinite-dimensional sequential product spaces to JB- and JBW-algebras. We refer to section \[sec:infiniterank\] for the details.

The structure of the paper is as follows: the main theorem that sequential product spaces are EJAs will be proved using the Koecher-Vinberg theorem which requires us to show that the space is both *homogeneous* and *self-dual*. In section \[sec:prelim\] we will cover known results originally presented in [@gudder2002sequential; @wetering2018characterisation] regarding sequential product spaces, culminating in a proof of a spectral theorem and a proof of the homogeneity of the space. Then in section \[sec:selfdual\] we will prove self-duality of the space, using results regarding lattices of projections of Alfsen and Schultz [@alfsen2012state; @alfsen2012geometry] and a characterisation result concerning low rank homogeneous spaces of Ito and Louren[ç]{}o [@ito2017p]. At this point sequential product spaces have been established to be EJAs, but only in a rather indirect way. In section \[sec:jordanproduct\] we directly construct the Jordan product using the sequential product. In section \[sec:loctom\] we show how the additional requirement of local tomography forces the sequential product space to be a C$^*$-algebra. Section \[sec:infiniterank\] discusses infinite dimensional generalisations of the main theorem, while in section \[sec:axioms\] we discuss how changing the axioms of a sequential product impacts the results of this paper.

*Acknowledgements.*—The author would like to thank Bas and Bram Westerbaan for all the useful and insightful conversations regarding effect algebras and order unit spaces and Alex Kolmus and Ema Alsina for suggestions to improve the manuscript. This work is supported by the ERC under the European Union’s Seventh Framework Programme (FP7/2007-2013) / ERC grant n$^\text{o}$ 320571.

Preliminaries {#sec:prelim}
=============

As mentioned in the introduction, our main goal is to show that a sequential product space is homogeneous and self-dual, let us start therefore with the definition of these properties.

Let $V$ be an order unit space. An *order isomorphism* is a linear map $\Phi: V\rightarrow V$ such that $\Phi(a)\geq 0 \iff a\geq 0$ for all $a\in V$. Denote the interior of the positive cone of $V$ by $C$, i.e. $a\in C \iff \exists \epsilon\in{\mathbb{R}}_{>0}: \epsilon 1 \leq a$. We call $V$ *homogeneous* when for all $a,b \in C$ there exists an order isomorphism $\Phi$ such that $\Phi(a) = b$.

Let $V$ be an order unit space. We call $V$ *self-dual* when there exists an inner product $\inn{\cdot,\cdot}$ such that for all $a\in V$: $a\geq 0$ if and only if $\inn{a,b} \geq 0$ for all $b\geq 0$.

To give a complete picture of the theory of sequential product spaces we will repeat some of the known basic results regarding sequential products and sequential product spaces that can be found in for instance [@gudder2002sequential; @gudder2005uniqueness; @wetering2018characterisation]. This section on preliminaries will end with the existence proofs of spectral decompositions of effects and the corollary of homogeneity that follows from it originally shown in [@wetering2018characterisation].

Unless otherwise stated, we will let $V$ denote a finite-dimensional sequential product space, ${E=[0,1]_V}$ its set of effects and $\&: E\times E\rightarrow E$ a sequential product. For $a\in E$ we let $a^\perp=1-a$ denote its *complement* which by virtue of $a$ lying in the unit interval of $V$ is also an effect.

\[prop:basic\] Let $a,b,c \in E$.

1.  \[prop:unitzero\] $a{\,\&\,}0 = 0{\,\&\,}a = 0$ and $a{\,\&\,}1=1{\,\&\,}a = a$.

2.  \[prop:decreasing\] $a{\,\&\,}b \leq a$.

3.  \[prop:orderpreserve\] If $a\leq b$ then $c{\,\&\,}a\leq c{\,\&\,}b$.

Originally proved in Ref. [@gudder2002sequential].

1.  We of course have $a{\,\lvert\,}a$ and by \[ax:compadd\] we have $a{\,\lvert\,}a^\perp$. Using \[ax:compadd\] again we then see that $a{\,\lvert\,}a+a^\perp = 1$ so that by \[ax:unit\] $1{\,\&\,}a = a{\,\&\,}1 = a$. Using \[ax:compadd\] again we also have $a{\,\lvert\,}1^\perp = 0$ so that it remains to show that $a{\,\&\,}0 = 0$. By \[ax:add\] we get $a{\,\&\,}0 = a{\,\&\,}(0+0) = a{\,\&\,}0 + a{\,\&\,}0$ so that indeed $a{\,\&\,}0 = 0$.

2.  By the previous point and \[ax:add\] $a = a{\,\&\,}1 = a{\,\&\,}(b+(1-b)) = a{\,\&\,}b + a{\,\&\,}(1-b)$ so that indeed $a{\,\&\,}b \leq a$, as $a{\,\&\,}(1-b) \geq 0$.

3.  Since $a\leq b$ we have $b-a \geq 0$ so that using \[ax:add\] we have $c{\,\&\,}b = c{\,\&\,}(b-a + a) = c{\,\&\,}(b-a) + c{\,\&\,}a$, from which we derive $c{\,\&\,}(b-a) = c{\,\&\,}b - c{\,\&\,}a$. Since the lefthandside is greater than zero, the righthandside must be as well.

\[prop:linearity\] Let $a,b\in E$ and let $q$ be any rational number between zero and one, and $\lambda$ any real number between zero and one.

1.  $a{\,\&\,}(qb) = q(a{\,\&\,}b)$.

2.  $a{\,\&\,}(\lambda b) = \lambda(a{\,\&\,}b)$.

3.  $(\lambda a){\,\&\,}b = a{\,\&\,}(\lambda b) = \lambda(a{\,\&\,}b)$.

4.  \[prop:commumult\] If $a{\,\lvert\,}b$ then $a{\,\lvert\,}\lambda b$.

Originally proved in Ref. [@wetering2018characterisation].

1.  Of course $a{\,\&\,}b = a{\,\&\,}(n \frac1n b) = n a{\,\&\,}(\frac1n b)$ by \[ax:add\]. Dividing by $n$ gives $a{\,\&\,}(\frac1n b) = \frac1n (a{\,\&\,}b)$. By summing this equation multiple times we see that we get $a{\,\&\,}(q b) = q(a{\,\&\,}b)$ for any rational $0\leq q\leq 1$.

2.  Let $q_i$ be an increasing sequence of positive rational numbers that converges to $\lambda$. Using the norm of the order unit space we compute $$\begin{aligned}
            \norm{\lambda (a{\,\&\,}b) - a{\,\&\,}(\lambda b)} &= \norm{(\lambda - q_i)(a{\,\&\,}b) + q_i(a{\,\&\,}b) - a{\,\&\,}(\lambda b)} \\
            &= \norm{(\lambda-q_i)(a{\,\&\,}b) - a{\,\&\,}((\lambda - q_i)b)}.
            \end{aligned}$$ Because $(\lambda - q_i)b\leq (\lambda-q_i)\norm{b}1$ and using proposition \[prop:orderpreserve\] we have $\norm{a{\,\&\,}((\lambda - q_i)b)} \leq \norm{a}\norm{(\lambda -q_i)b} = (\lambda - q_i)\norm{a}\norm{b}$. Then $\norm{\lambda (a{\,\&\,}b) - a{\,\&\,}(\lambda b)} \leq 2(\lambda - q_i)\norm{a}\norm{b}$. This expression indeed goes to zero as $i$ increases so that indeed $\lambda (a{\,\&\,}b) = a{\,\&\,}(\lambda b)$.

3.  Clearly $\frac{1}{n}a{\,\lvert\,}\frac{1}{n}a$ so that by \[ax:compadd\] $\frac{1}{n}a{\,\lvert\,}a$. In the same way we also get $qa{\,\lvert\,}a$ and $qa^\perp {\,\lvert\,}a^\perp$ for any rational $0\leq q\leq 1$. Using the rule $a{\,\lvert\,}b \implies a{\,\lvert\,}b^\perp$ from \[ax:compadd\] we then also get $qa^\perp {\,\lvert\,}a$ so that $a{\,\lvert\,}(qa+qa^\perp)=q1$. Then indeed $(q1){\,\&\,}a = a{\,\&\,}(q1) = q(a{\,\&\,}1) = qa$ so that also $(qa){\,\&\,}b = (a{\,\&\,}(q1)){\,\&\,}b = a{\,\&\,}((q1){\,\&\,}b)) = a{\,\&\,}qb = q(a{\,\&\,}b)$. Now let $\lambda\in[0,1]$ be a real number and let $q_i$ be a sequence of rational numbers converging to $\lambda$ so that also $q_i a \rightarrow \lambda a$ and $q_i(a{\,\&\,}b) \rightarrow \lambda(a{\,\&\,}b)$. Then $q_i(a{\,\&\,}b) = (q_ia){\,\&\,}b \rightarrow (\lambda a){\,\&\,}b$ by \[ax:cont\]. We conclude that $(\lambda a){\,\&\,}b = \lambda(a{\,\&\,}b) = a{\,\&\,}(\lambda b)$.

4.  Suppose $a{\,\lvert\,}b$, then using the previous point $a{\,\&\,}(\lambda b) = \lambda (a{\,\&\,}b) = \lambda (b{\,\&\,}a) = (\lambda b){\,\&\,}a$.

As a result of this proposition, the *left-multiplication map* $L_a:E\rightarrow E$ for $a\in E$ given by $L_a(b) = a{\,\&\,}b$ can be extended by linearity to the entirety of $V$ by $L_a(b-c) = L_a(b) - L_a(c)$. Similarly we can define the sequential product for any element in the positive cone of $V$ (not necessarily below the identity) by rescaling: $a{\,\&\,}b := \norm{a} ((\frac{1}{\norm{a}} a){\,\&\,}b)$.

An effect $p\in E$ is called *sharp* when the only effect below both $p$ and $p^\perp$ is the zero effect, i.e when the following implication holds: $b\leq p$ and $b\leq p^\perp$ implies $b=0$.

When $V=B(H)^{\text{sa}}$ the sharp effects are precisely the projections. This should be clear considering the following proposition:

\[prop:sharpness\] Let $a\in E$ be an effect, $a$ is sharp if and only if $a{\,\&\,}a^\perp = 0$ if and only if $a{\,\&\,}a = a$.

Originally proved in Ref. [@gudder2002sequential].

The equivalence of $a{\,\&\,}a^\perp =0$ and $a{\,\&\,}a = a$ follows straightforwardly by \[ax:add\] and proposition \[prop:unitzero\]: $a = a{\,\&\,}1 = a{\,\&\,}(a+a^\perp) = a{\,\&\,}a + a{\,\&\,}a^\perp$.

So let us assume $a$ is sharp. By \[ax:compadd\] we have $a{\,\lvert\,}a^\perp$ so that $a{\,\&\,}a^\perp = a^\perp {\,\&\,}a$. By \[prop:decreasing\] we have $a{\,\&\,}a^\perp \leq a$ and similarly $a^\perp {\,\&\,}a \leq a^\perp$, the expression $a{\,\&\,}a^\perp$ is therefore below both $a$ and $a^\perp$ and by sharpness of $a$ has to be zero. Now suppose $a{\,\&\,}a^\perp = 0$ and let $b\leq a$ and $b\leq a^\perp$. If $b\leq a^\perp$ then by \[prop:decreasing\] we get $a{\,\&\,}b \leq a{\,\&\,}a^\perp =0$, and similarly we get $a^\perp {\,\&\,}b = 0$. By \[ax:orth\] we have $b{\,\lvert\,}a$ and $b{\,\lvert\,}a^\perp$ so that $b = b{\,\&\,}1 = b{\,\&\,}(a+a^\perp) = b{\,\&\,}a + b{\,\&\,}a^\perp = a{\,\&\,}b + a^\perp {\,\&\,}b = 0 + 0 = 0$.

We call two sharp effects $p$ and $q$ *orthogonal* when $p{\,\&\,}q = 0$.

Of course by \[ax:orth\] orthogonality is a symmetric relation, and we note that therefore orthogonal effects are also compatible.

Let $a\in E$ be an effect. We define the powers of $a$ inductively to be $a^0 := 1$ and $a^n := a{\,\&\,}a^{n-1}$. We define the *classical algebra of $a$* to be the linear space $C(a)$ spanned by all the powers of $a$ and $a^\perp$.

\[prop:polyspace\] Let $a\in E$ be an effect. $C(a)$ is a commutative sequential product space.

$C(a)$ inherits the order structure from $V$ in the obvious way. Of course $a{\,\lvert\,}a^\perp$ and because of \[ax:compmult\] we have $a^n {\,\lvert\,}(a^\perp)^m$ for all $n$ and $m$. Because of \[ax:compadd\] and proposition \[prop:commumult\] linear combinations of compatible effects are also compatible and hence all effects of $C(a)$ are compatible.

We are now in a position to use the seminal representation theorem from Kadison:

[@kadison1951representation] Let $V$ be a complete order unit space with a bilinear operation $\circ$ that preserves positivity: $a\circ b \geq 0$ when $a,b\geq 0$. Then there exists a compact Haussdorff space $X$ such that $V\cong C(X)$, the space of continuous real-valued functions on $X$. This isomorphism is both an order- and algebra-isomorphism.

Let $a\in E$ be an effect. $C(a)\cong {\mathbb{R}}^n$ for some $n\in {\mathbb{N}}$.

The sequential product is linear in the second argument. Since $C(a)$ is a commutative sequential product space by proposition \[prop:polyspace\], its product is also linear in the first argument, and hence this operation is bilinear. It obviously preserves positivity by definition, so that Kadisons theorem applies and $C(a) \cong C(X)$. Since $V$ is finite-dimensional, $C(a)$ has to be so as well, and hence $C(X)$ is finite-dimensional. Of course $C(X)$ is finite-dimensional only when $X$ is finite so that $X$ is necessarily discrete. We conclude that indeed $C(X) \cong {\mathbb{R}}^n$.

Let $a\in E$ be an effect. There exists a set of orthogonal sharp effects $p_i$ compatible with $a$ and positive scalars $\lambda_i$ such that $a = \sum_i \lambda_i p_i$.

By the previous proposition $C(a) \cong {\mathbb{R}}^n$ and this space is obviously spanned by orthogonal sharp effects, hence we can find the desired $p_i$ and $\lambda_i$. By construction $p_i\in C(a)$ so that they are compatible with $a$.

We will refer to a decomposition of $a$ in the above sense as a *spectral decomposition* of $a$. The existence of these decomposition is already enough to show that the space must be homogeneous:

\[prop:homogen\] Let $C$ denote the cone of *strictly positive* elements in $V$, i.e the elements $v\in V$ such that $\exists \epsilon>0$ with $\epsilon 1\leq v$. The cone $C$ is homogeneous, i.e. for every $v,w\in C$ there exists an order isomorphism $\Phi: V\rightarrow V$ such that $\Phi(v)=w$.

Originally proved in Ref. [@wetering2018characterisation].

For an arbitrary positive element $a$ we can find a spectral decomposition $a=\sum_i \lambda_i p_i$ such that $\lambda_i > 0$, i.e. we don’t write the zero ‘eigenvalues’. It is then straightforward to check that $a$ lies in the interior of the positive cone if and only if $\sum_i p_i=1$. In that case we define its *inverse* $a^{-1}=\sum_i \lambda_i^{-1} p_i$. Since the $p_i$ are all compatible and $a$ and $a^{-1}$ are linear combinations of these effects, they are also compatible and we calculate $a{\,\&\,}a^{-1} = \sum_{i,j} \lambda_i \lambda_j^{-1} p_i {\,\&\,}p_j = \sum_i \lambda_i \lambda_i^{-1} p_i = \sum_i p_i = 1$ so that $a^{-1}$ is indeed the inverse of $a$ with respect to the sequential product. The multiplication map $L_a(b) := a{\,\&\,}b$ is positive and has a positive inverse $L_{a^{-1}}$ due to \[ax:assoc\]: $a^{-1}{\,\&\,}(a{\,\&\,}b) = (a^{-1}{\,\&\,}a){\,\&\,}b = 1{\,\&\,}b = b$. The map $L_a$ is therefore an order isomorphism when $a$ is strictly positive. Now, for $a$ and $b$ strictly positive and hence invertible, define $\Phi: V\rightarrow V$ by $\Phi = L_bL_{a^{-1}}$. As this is a composition of order isomorphisms, it is also an order isomorphism and of course $\Phi(a) = b{\,\&\,}(a^{-1}{\,\&\,}a) = b{\,\&\,}1 = b$ as desired.

Proof of self-duality {#sec:selfdual}
=====================

With homogeneity of $V$ now established, we set our sights on proving self-duality. We will do this by first showing that the sharp effects form an atomic lattice with the *covering property* as defined in [@alfsen2012state]. The covering property states that for every sharp effect $p$ there is a unique number $r$ called the *rank* of $p$ such that we can write $p=\sum_{i=1}^r p_i$ where the $p_i$ are atomic and orthogonal. Using this definition we can define the rank of a space as equal to the rank of the unit effect. The existence of well-defined ranks of sharp effects allows us to reduce the question of self-duality to that of self-duality in rank 2 spaces. This problem is in turn solved by appealing to the classification result of Ref. [@ito2017p] that homogeneous spaces of rank 2 are always self-dual.

The lattice of sharp effects
----------------------------

\[prop:sharpprop\] Let $a\in E$ be any effect and let $p\in E$ be sharp.

1.  \[prop:belowsharp\] $a\leq p$ if and only if $p{\,\&\,}a = a{\,\&\,}p = a$ if and only if $p^\perp {\,\&\,}a = 0$.

2.  \[prop:abovesharp\] $p\leq a$ if and only if $p{\,\&\,}a = a{\,\&\,}p = p$.

 

1.  Suppose $a\leq p$ with $p$ sharp. Then $p^\perp {\,\&\,}a \leq p^\perp {\,\&\,}p = 0$ by proposition \[prop:sharpness\] and \[prop:decreasing\]. Hence $a{\,\lvert\,}p^\perp$ and $a{\,\lvert\,}p$ so that $a = a{\,\&\,}(p+p^\perp) = a{\,\&\,}p + a{\,\&\,}p^\perp = a{\,\&\,}p = p{\,\&\,}a$. For the other direction we note that $a=p{\,\&\,}a \leq p$ by \[prop:decreasing\].

2.  Suppose $p\leq a$ with $p$ sharp, then $a^\perp \leq p^\perp$ with $p^\perp$ sharp so that by the previous point $a{\,\lvert\,}p$ and $p{\,\&\,}a^\perp = 0$ so that $p=p{\,\&\,}(a+a^\perp) = p{\,\&\,}a$.

For an effect $a\in E$ we let $\ceil{a}$ denote the smallest sharp element above $a$ (when it exists) and $\floor{a}$ the largest sharp element below $a$ (when it exists).

The ceiling and the floor exist for any $a$. Moreover, writing $a=\sum_i \lambda_i p_i$ with $1\geq \lambda_i> 0$ and the $p_i$ sharp and orthogonal, then $\ceil{a} = \sum_i p_i$ and $\floor{a}=\ceil{a^\perp}^\perp$.

Write $a=\sum_i \lambda_i p_i$. Of course $\sum_i p_i$ is an upper bound of $a$. Suppose $a\leq r$ for some sharp $r$. Then $\lambda_i p_i \leq r$, so by proposition \[prop:belowsharp\] $r{\,\&\,}(\lambda_i p_i) = \lambda_i p_i$. Using linearity we can rewrite this expression to $\lambda_i (r{\,\&\,}p_i) = \lambda_i p_i$ so that $r{\,\&\,}p_i = p_i$. So again by \[prop:belowsharp\] $p_i\leq r$ and $p_i{\,\lvert\,}r$ from which we get $r{\,\&\,}\sum_i p_i = \sum_i r{\,\&\,}p_i = \sum_i p_i$ so that also $\sum_i p_i\leq r$ which proves that it is the least upper bound. The other statement now follows because $a\leq b \iff b^\perp \leq a^\perp$.

As a corollary of the above we also see that $\ceil{\lambda a} = \ceil{a}$ when $1\geq \lambda > 0$ and that $a$ is sharp if and only if $\ceil{a}=a$ or $\floor{a}=a$. We also note that $a\leq b$ implies that $\ceil{a}\leq \ceil{b}$.

\[prop:lattice\] The sharp effects form a lattice: for two sharp effects $q$ and $p$, their least upper bound $p\vee q$ and greatest lower bound $p\wedge q$ exist. Furthermore the following relation holds between them: $(p\vee q)^\perp = p^\perp \wedge q^\perp$.

We claim that $p\vee q = \ceil{\frac12(p+q)}$. Note that $p\leq p+q$ and thus that $\frac12 p \leq \frac12(p+q)$ so that $p = \ceil{p} = \ceil{\frac12p} \leq \ceil{\frac12(p+q)}$. Similarly we also have $q\leq \ceil{\frac12(p+q)}$ and thus this is an upper bound. Suppose now that $p\leq a$ and $q\leq a$ for some sharp $a$. Then of course also $\frac12(p+q)\leq \frac12(a+a) = a$. Taking the ceiling on both sides then shows that indeed $p\vee q = \ceil{\frac12(p+q)}$.

To find $p\wedge q$ we note that $(\cdot)^\perp$ is an order-antiautomorphism, and thus that it interchanges joins with meets: $(p\vee q)^\perp = p^\perp\wedge q^\perp$.

\[prop:sharplattice\] Let $a\in E$ be any effect and let $p\in E$ be sharp.

1.  \[prop:meetsharp\] $p{\,\&\,}a = 0$ if and only if $p+a\leq 1$ in which case it is the least upper bound of the two. When $p{\,\&\,}a = 0$ their sum $p+a$ is sharp if and only if $a$ is also sharp.

2.  \[prop:joinsharp\] If both $a$ and $p$ are sharp and $a$ and $p$ are compatible then $p{\,\&\,}a$ is sharp and equal to their join: $p\wedge a = p{\,\&\,}a$.

Originally proved in Ref. [@gudder2002sequential].

1.  $p{\,\&\,}a = 0$ if and only if $p^\perp{\,\&\,}a = a$ which by proposition \[prop:abovesharp\] is true if and only if $a\leq p^\perp = 1-p$ so that indeed $p+a\leq 1$. That $p+a$ is an upper bound of $p$ and $a$ is obvious. Suppose now that $b$ is also an upper bound so that $p\leq b$ and $a\leq b$. We then calculate using proposition \[prop:abovesharp\] $p = p{\,\&\,}b = p{\,\&\,}(b-a + a) = p{\,\&\,}(b-a) + p{\,\&\,}a = p{\,\&\,}(b-a)$ so that $p\leq b-a$ again by \[prop:abovesharp\]. This gives $p+a \leq b$ so that $p+a$ is indeed the least upper bound.

    Now to show $p+a$ is sharp if and only if bot $p$ and $a$ are sharp: since $p{\,\&\,}a = 0$ we have $p{\,\lvert\,}a$ and thus also $p{\,\lvert\,}p+a$ and $a{\,\lvert\,}p+a$ by \[ax:compadd\]. We calculate $(p+a){\,\&\,}(p+a) = p{\,\&\,}p + 2 p{\,\&\,}a + a{\,\&\,}a = p + a{\,\&\,}a = (p+a) + (a-a{\,\&\,}a)$. We therefore have $(p+a){\,\&\,}(p+a) = p+a$ if and only if $a-a{\,\&\,}a = 0$ which proves the result by proposition \[prop:sharpness\].

2.  Suppose both $p$ and $a$ are sharp and that $p{\,\lvert\,}a$. We calculate: $$(p{\,\&\,}a){\,\&\,}(p{\,\&\,}a) = (p{\,\&\,}a){\,\&\,}(a{\,\&\,}p) =p{\,\&\,}(a{\,\&\,}(a{\,\&\,}p)) = p{\,\&\,}(a{\,\&\,}p) = p{\,\&\,}(p{\,\&\,}a) = p{\,\&\,}a$$ where we have used that $a{\,\lvert\,}a{\,\&\,}p$ and $p{\,\lvert\,}a{\,\&\,}p$ by \[ax:compmult\]. Hence $p{\,\&\,}a$ is sharp. It is a lower bound of $p$ and $a$ by \[prop:decreasing\]. Suppose $b\leq p, a$ is also a lower bound. We calculate $p{\,\&\,}a = p{\,\&\,}(a-b + b) = p{\,\&\,}(a-b) + p{\,\&\,}b = p{\,\&\,}(a-b) + b \geq b$, where we have used that $p{\,\&\,}b = b$ as a consequence of proposition \[prop:belowsharp\].

\[lem:ceilzero\] Let $a,b\in E$. If $b{\,\&\,}a = 0$ then $b{\,\&\,}\ceil{a} = 0$.

Write $a= \sum_i \lambda_i p_i$. If $b{\,\&\,}a = 0 = \sum_i \lambda_i b{\,\&\,}p_i$, then we must have $b{\,\&\,}p_i=0$ for all $p_i$. Since $\ceil{a}=\sum_i p_i$ the claim follows.

\[lem:ceilceil\] Let $p\in E$ be sharp and $a\in E$ arbitrary, then $\ceil{p{\,\&\,}a} = \ceil{p{\,\&\,}\ceil{a}}$.

First of all $p{\,\&\,}a \leq p{\,\&\,}\ceil{a}$ so that $\ceil{p{\,\&\,}a}\leq \ceil{p{\,\&\,}\ceil{a}}$. It suffices therefore to prove the other inequality. Because $p{\,\&\,}a \leq p{\,\&\,}1 = p$ we also have $\ceil{p{\,\&\,}a}\leq \ceil{p}=p$ implying $\ceil{p{\,\&\,}a}^\perp{\,\&\,}p =0$ so that $\ceil{p{\,\&\,}a}^\perp$ and $p$ are compatible. Now because $p{\,\&\,}a \leq \ceil{p{\,\&\,}a}$ we can use proposition \[prop:belowsharp\] to write $0=\ceil{p{\,\&\,}a}^\perp {\,\&\,}(p{\,\&\,}a) = (\ceil{p{\,\&\,}a}^\perp{\,\&\,}p){\,\&\,}a = (\ceil{p{\,\&\,}a}^\perp {\,\&\,}p){\,\&\,}\ceil{a} = \ceil{p{\,\&\,}a}^\perp{\,\&\,}(p{\,\&\,}\ceil{a})$ where we have used lemma \[lem:ceilzero\] to replace $a$ with $\ceil{a}$. Since $\ceil{p{\,\&\,}a}^\perp{\,\&\,}(p{\,\&\,}\ceil{a}) = 0$ we use \[prop:belowsharp\] again to conclude $p{\,\&\,}\ceil{a}\leq \ceil{p{\,\&\,}a}$ so that indeed $\ceil{p{\,\&\,}\ceil{a}}\leq \ceil{p{\,\&\,}a}$.

Atomic effects {#sec:atomeffect}
--------------

We call a nonzero sharp effect $p\in E$ *atomic* if for all $a\in E$ with $a\leq p$ we have $a=\lambda p$ for some $\lambda\in[0,1]$.

Every sharp effect can be written as a sum of orthogonal atomic effects.

Let $p$ be sharp and suppose it is not atomic, then we can find $0\leq a\leq p$ such that $a\neq \lambda p$ for any $\lambda\in [0,1]$. Write $a= \sum_i \lambda_i q_i$ where the $q_i$ are sharp and orthogonal. Then $\lambda_i q_i\leq p$ and thus also $\ceil{\lambda_i q_i} = q_i\leq \ceil{p}=p$. If all the $q_i$ are equal to $p$, then $a$ is a multiple of $p$, so at least one of the $q_i$ is strictly smaller than $p$. We can repeat this process for $q_i$ and $p-q_i$, getting a sequence of nonzero orthogonal sharp effects that sum up to $p$. Since the space is finite-dimensional and orthogonal effects are linearly independent this process must stop after a finite amount of steps in which case we are left with atomic effects.

\[cor:spectralatomic\] Every $a\in V$ can be written as $a=\sum_i \lambda_i p_i$ where the $p_i$ are orthogonal sharp atomic effects.

For every $a\in V$ we can find a spectral decomposition in terms of orthogonal sharp effects. The previous proposition shows that these sharp effects can be further decomposed into atomic effects.

Recall that the norm in an order unit space is defined as ${\norm{a}:=\inf\{r~;~-r 1 \leq a \leq r1\}}$.

\[lem:atomicnorm\] A non-zero effect $p$ is atomic if and only if we have $p{\,\&\,}a = \norm{p{\,\&\,}a}p$ for all $a\in E$.

First we establish that the norm of any non-zero sharp effect is equal to $1$. Let $q$ be sharp. We see that $q = q{\,\&\,}q \leq q {\,\&\,}(\norm{q} 1) = \norm{q} q{\,\&\,}1 = \norm{q} q$ so that $\norm{q}\geq 1$. But since $q\leq 1$ we also have $\norm{q}\leq 1$.

Suppose $p$ is atomic. Because $0\leq p{\,\&\,}a \leq p$ we must have $p{\,\&\,}a = \lambda p$ for some $0\leq \lambda \leq 1$ so that $\norm{p{\,\&\,}a} = \lambda \norm{p} = \lambda$ because $p$ is sharp.

For the other direction first note that $p= p{\,\&\,}\ceil{p} = \norm{p{\,\&\,}\ceil{p}} p = \norm{p}p$ so that necessarily $\norm{p}=1$ (since $p\neq 0$). By writing $p$ as a linear combination of sharp effects we see that then also $\norm{p^2}=1$. Now $p{\,\&\,}p = \norm{p^2} p = p$ so that $p$ is sharp. Let $q\leq p$ be non-zero sharp. Then $q=p{\,\&\,}q = \norm{p{\,\&\,}q}p=p$ (using again that $\norm{q}=1$ because $q$ is sharp) so there are no non-trivial sharp effects below $p$. Now if $a=\sum_i \lambda_i q_i$ lies below $p$ we see that $\lambda_iq_i\leq p$ so that $\ceil{\lambda_i q_i} = q_i\leq \ceil{p}=p$ so that $q_i=p$ and thus $a=\lambda p$. Since all $a\in E$ can be written in this way we conclude that this holds for all $a\leq p$, so that $p$ is indeed atomic.

The set of atomic effects is closed in the norm topology.

Let $p_n\rightarrow p$ be a norm converging set of atomic effects $p_n$. We need to show that $p$ is also atomic. As a result of the previous lemma we have $p_n{\,\&\,}a = \norm{p_n {\,\&\,}a} p_n$ for all effects $a$. By continuity of $\&$ (i.e. axiom \[ax:cont\]) we have $p_n{\,\&\,}a \rightarrow p{\,\&\,}a$ so that $p{\,\&\,}a = \lim p_n {\,\&\,}a = \lim \norm{p_n{\,\&\,}a}p_n = \norm{p{\,\&\,}a} p$. Using the previous lemma again we conclude that $p$ is indeed atomic.

\[prop:atompreservation\] Let $a\in E$ be arbitrary and $p\in E$ be atomic, then $a{\,\&\,}p$ is proportional to an atomic effect.

The property that $0\leq a\leq p \implies a=\lambda p$ is determined by the order, so any order isomorphism preserves it. If $a$ is invertible then $L_a: V\rightarrow V$ given by $L_a(b):= a{\,\&\,}b$ is an order isomorphism, so that $L_a(p)$ must be proportional to an atomic effect. For non-invertible $a$ we write $a_n = a+\frac{1}{n}$, so that $a_n$ is invertible and the sequence $a_n$ converges to $a$. Let $q_n = (a_n{\,\&\,}p)/\norm{a_n {\,\&\,}p}$, then all the $q_n$ are atomic. By the continuity condition \[ax:cont\] we get $a_n{\,\&\,}p \rightarrow a{\,\&\,}p$ so that also $\norm{a_n {\,\&\,}p}\rightarrow \norm{a{\,\&\,}p}$. The sequence $q_n$ is therefore also convergent and since the set of atomic effects is closed by the previous corollary we conclude that $q_n\rightarrow q=(a{\,\&\,}p)/\norm{a{\,\&\,}p}$ is atomic.

The Covering Property {#sec:coverprop}
---------------------

At this point we know that the set of sharp effects forms an atomic lattice, but in fact it has the much stronger *covering property* that allows us to attach a *rank* to each sharp effect: the amount of atomic effects needed to make the effect. To show this we need some results from Alfsen and Schultz [@alfsen2012geometry] that unfortunately were proven in a slightly different setting. We will repeat these results with very similar proofs, but adapted to work in the setting of sequential product spaces.

\[lem:wedgeminus\] [@alfsen2012geometry Lemma 8.9] Let $q$ and $p$ be sharp with $q\leq p$, then $p-q = p\wedge q^\perp$.

Let $q$ and $p$ be sharp with $q\leq p$, then $p-q$ is sharp and $p{\,\lvert\,}q$, so also $p{\,\lvert\,}q^\perp$ so that $p-q = p{\,\&\,}q^\perp = p\wedge q^\perp$ by \[prop:joinsharp\].

[@alfsen2012geometry Theorem 8.32]: Let $p$ be sharp and $a$ arbitrary, then $\ceil{p{\,\&\,}a} = (\ceil{a}\vee p^\perp)\wedge p$.

Because $\ceil{p{\,\&\,}a} = \ceil{p{\,\&\,}\ceil{a}}$ by lemma \[lem:ceilceil\] it suffices to prove this for sharp $a$. We prove the equality by showing that an inequality holds in both directions.

Since $p^\perp \leq a\vee p^\perp$ we have $p^\perp {\,\lvert\,}(a\vee p^\perp)$ by \[prop:belowsharp\] so that in turn $p{\,\lvert\,}(a\vee p^\perp)$ by \[ax:compadd\]. We proceed by using \[ax:assoc\]: $(a\vee p^\perp){\,\&\,}(p{\,\&\,}a) = ((a\vee p^\perp){\,\&\,}p){\,\&\,}a = p{\,\&\,}((a\vee p^\perp){\,\&\,}a) = p{\,\&\,}a$ because $a\vee p^\perp \geq a$. Therefore $p{\,\&\,}a \leq a\vee p^\perp$ which implies that $\ceil{p{\,\&\,}a} \leq a\vee p^\perp$. Since also $p{\,\&\,}a\leq p$ and therefore $\ceil{p{\,\&\,}a}\leq p$ we conclude that $\ceil{p{\,\&\,}a}\leq (a\vee p^\perp)\wedge p = (\ceil{a}\vee p^\perp)\wedge p$.

Now for the other direction: we obviously have $p^\perp{\,\&\,}(p{\,\&\,}a) = (p^\perp {\,\&\,}p){\,\&\,}a = 0$ by \[ax:compadd\] and \[ax:assoc\] so that by lemma \[lem:ceilzero\] $p^\perp{\,\&\,}\ceil{p{\,\&\,}a}=0$. Using \[prop:belowsharp\] we see then that $p{\,\lvert\,}\ceil{p{\,\&\,}a}^\perp$ and therefore by \[prop:joinsharp\] that $\ceil{p{\,\&\,}a}^\perp{\,\&\,}p = \ceil{p{\,\&\,}a}^\perp\wedge p$. Since $p{\,\&\,}a \leq \ceil{p{\,\&\,}a}$ we calculate using \[prop:belowsharp\]: $0=\ceil{p{\,\&\,}a}^\perp{\,\&\,}(p{\,\&\,}a) = (\ceil{p{\,\&\,}a}^\perp {\,\&\,}p){\,\&\,}a = (\ceil{p{\,\&\,}a}^\perp \wedge p){\,\&\,}a$ so that $a\leq (\ceil{p{\,\&\,}a}^\perp\wedge p)^\perp = \ceil{p{\,\&\,}a}\vee p^\perp$ by proposition \[prop:lattice\]. Then of course also $a\vee p^\perp \leq \ceil{p{\,\&\,}a}\vee p^\perp$ and by noting that $\ceil{p{\,\&\,}a}$ and $p^\perp$ are orthogonal and using \[prop:meetsharp\]: $\ceil{p{\,\&\,}a}\vee p^\perp = \ceil{p{\,\&\,}a}+p^\perp$. Bringing the $p^\perp$ to the other side and using lemma \[lem:wedgeminus\] (which applies because $p^\perp\leq a\vee p^\perp$) then gives $(a\vee p^\perp)\wedge p = a\vee p^\perp - p^\perp \leq \ceil{p{\,\&\,}a}$.

\[prop:coverprop\] [@alfsen2012geometry Proposition 9.7]: The lattice of sharp effects has the covering property: for $q$ atomic, the expression $(q\vee p)\wedge p^\perp = (q\vee p)-p$ is either zero or atomic. In other words: when $q$ does not lie below $p$ then there is no sharp effect lying strictly between $p$ and $q\vee p$.

By the previous lemma $(q\vee p)\wedge p^\perp = \ceil{p^\perp {\,\&\,}q}$. Since $p^\perp{\,\&\,}q$ is proportional to an atom by proposition \[prop:atompreservation\], it is either zero (when $q\leq p$) in which case we are done, or non-zero in which case $\ceil{p^\perp {\,\&\,}u}$ is an atom, which also proves the statement. The equality $(q\vee p)\wedge p^\perp = (q\vee p)-p$ follows directly from lemma \[lem:wedgeminus\].

The last observation is proven as follows. Suppose $p< r < q\vee p$. Subtract $p$ to get $0< r-p < q\vee p - p$. As $r-p$ is sharp and $q\vee p -p$ has been established to be atomic, this is not possible.

Let $p$ be sharp and let $p_i$ be a collection of atomic orthogonal effects such that $p=\sum_i^n p_i$. The minimal size of such a collection is called the *rank* of $p$. The rank of a sequential product space is defined to be the rank of the unit effect.

With the covering property proven we can finally prove the following ‘dimension’ theorem:

[@alfsen2012state Proposition 1.66] Write $p=\sum_i^n p_i$ where the $p_i$ are orthogonal and atomic, then $n={\text{rnk}\xspace}~ p$, i.e. all ways of writing $p$ as a sum of atomic effects require an equal amount of atomic effects. Furthermore, suppose $q\leq p$ then ${\text{rnk}\xspace}~ q \leq {\text{rnk}\xspace}~ p$ and if also ${\text{rnk}\xspace}~ q = {\text{rnk}\xspace}~ p$ then necessarily $q=p$.

Let $p^\prime = p_1\vee\ldots\vee p_{n-1}$. Then $p^\prime\vee p_n = p$ and by the covering property (proposition \[prop:coverprop\]) there is no sharp effect strictly between $p^\prime$ and $p$. Suppose now $q\leq p$ is atomic and suppose that $q$ is not below $p^\prime$. Then $p^\prime \vee q$ must be strictly greater than $p^\prime$, but since this must also lie below $p$ we conclude that $p^\prime \vee q = p$.

Let $p = \sum_j^r q_j = q_1\vee\ldots\vee q_r$ where $r:={\text{rnk}\xspace}~ p$ is the minimal amount of terms needed to write $p$ as a sum of atomic effects. We must then of course have $r\leq n$. Let $q=q_2\vee\ldots\vee q_r$, then $q$ must lie strictly below $p$ since $q_1\leq p$ but not $q_1\leq q$. It then follows that there must be a $p_i$ such that $p_i$ does not lie below $q$ as well, since otherwise $p = p_1\vee\ldots\vee p_n \leq q < p$. Without loss of generality let this $p_i$ be $p_1$. By the previous paragraph we must have $p_1\vee q = p_1\vee q_2\ldots\vee q_r = p$. This procedure can be repeated with $q_2,\ldots, q_r$ until we are left with the equation $p_1\vee \ldots\vee p_r = p$. Suppose $n>r$, then because $p_n$ is orthogonal to all the other $p_i$’s we have in particular $p_n\leq p_1^\perp \wedge \ldots \wedge p_r^\perp = (p_1\vee \ldots \vee p_r)^\perp = p^\perp$. Since also $p_n\leq p$ we get $p_n=0$ by sharpness which is a contradiction. We therefore have $n=r$.

Now suppose $q=\sum_j^s q_j\leq p=\sum_i^r p_i$. Where $s={\text{rnk}\xspace}~ q$. Since $p-q$ is sharp we can write $p-q = \sum_k^t v_k$. Then because $p = \sum_j^s q_j + \sum_k^t v_k$ we must by the previous points have $s+t = r$ so that indeed ${\text{rnk}\xspace}~ q\leq {\text{rnk}\xspace}~ p$. When these ranks are equal we must have $t=0$ so that indeed $p-q = 0$.

\[cor:rank\] Let $p\neq q$ be two atomic sharp effects and suppose $0\leq a\leq p\vee q$, then $a=\lambda_1 r_1 + \lambda_2 r_2$ where the $r_i$ are orthogonal and atomic and $r_1 + r_2 = p\vee q$.

By proposition \[prop:coverprop\] $(p\vee q) - p$ is atomic so that $p\vee q$ can be written as the sum of two atomic sharp effects so that indeed ${\text{rnk}\xspace}~ p\vee q = 2$. Suppose $0\leq a \leq p\vee q$. Let $a=\sum_i^n \lambda_i r_i$ be a spectral decomposition of $a$ with the $r_i$ orthogonal and atomic. Of course $\ceil{a} \leq p\vee q$ so that by the previous proposition we must have ${\text{rnk}\xspace}\ceil{a} \leq 2$. Since also by the previous proposition ${\text{rnk}\xspace}~\sum_i^n r_i = n$ we see that we must have $n=2$ and thus that $a$ is as desired.

Self-duality {#sec:subselfdual}
------------

The important concept of this section will be that of *strict convexity* of a cone, since this is related to a characterisation theorem for homogeneous spaces.

Let $C$ be a positive cone of an order unit space $V$. We call $F\subseteq C$ a *face* of $C$ if $F$ is a convex set such that whenever $\lambda a+\lambda^\perp b \in F$ with $0<\lambda<1$ then $a,b \in F$. The face $\{\lambda p~;~ \lambda \in {\mathbb{R}}_{\geq 0}\}$ of $C$ defined by an extreme point $p\in C$ is called an *extreme ray*. A face is called *proper* when it is non-empty and not equal to $C$. If the only proper faces of a cone are extreme rays the cone is *strictly convex*.

\[prop:itochar\] [@ito2017p] Let $V$ be a finite-dimensional ordered vector space with a strictly convex homogeneous positive cone, then $V$ is order isomorphic to a spin-factor, i.e. $V\cong H\oplus {\mathbb{R}}$ where $H$ is a real finite-dimensional Hilbert space with the order on $H\oplus {\mathbb{R}}$ given by $(v,t)\geq 0 \iff t\geq \norm{v}_2$.

\[def:orderideal\] Let $p$ and $q$ be two unequal atomic effects. We define the *order ideal* generated by $p$ and $q$ as $V_{p,q}:= \{v\in V~;~\exists n: -n~p\vee q \leq v \leq n ~p\vee q\}$.

$V_{p,q}$ is an order unit space with unit $p\vee q$. If we have $a,b\in [0,1]_{V_{p,q}}$ then $a{\,\&\,}b \leq a \leq p\vee q$ so that the sequential product restricts to this space. We therefore conclude by proposition \[prop:homogen\] that this space has a homogeneous positive cone.

Let $p$ and $q$ be two unequal atomic effects. The positive cone of $V_{p,q}$ is strictly convex.

Let $F$ be a proper face of the positive cone of $V_{p,q}$. Let $a\in F$ and write $a=\lambda (\lambda^{-1} a) + \lambda^\perp 0$, so that $\lambda^{-1} a \in F$. We see that $F$ is closed under positive scalar multiplication and thus that we can restrict ourselves to effects. Let $a\in F$ be an effect. By corollary \[cor:rank\] we can write $a=\lambda r + \mu r^\perp$ for some $\lambda,\mu\geq 0$ and $r$ atomic. Suppose both $\lambda,\mu >0$, then because $F$ is a face $r,r^\perp \in F$ so that $\frac{1}{2}1 = \frac{1}{2}(r+r^\perp)\in F$. But then since $1=s+s^\perp$ for any atomic $s$ we see that $F$ has to be the entire positive cone. We conclude that we must have had $a=\lambda r$ for some atomic $r$. If there were some other atomic $s \in F$, then we can consider $a = \frac{1}{2}(r + s)$. We know that $a$ can’t be atomic so we can write it as $a=\lambda r + \mu r^\perp$ with $\lambda, \mu > 0$ which is a contradiction. We conclude that $F$ is an extreme ray and thus that the positive cone of $V_{p,q}$ is strictly convex.

Let $p$ and $q$ be two different atomic effects, then $V_{p,q}$ is isomorphic to a spin-factor.

Follows directly from the previous lemma and proposition \[prop:itochar\].

Recall that a *state* on an order unit space is a positive linear map $\omega: V\rightarrow {\mathbb{R}}$ such that $\omega(1) = 1$. For an atomic effect $p$ in a spin-factor there exists a unique state $\omega_p$ such that $\omega_p(p)=1$. A spin-factor has *symmetry of transition probabilities* [@alfsen2012geometry]: $\omega_p(q)=\omega_q(p)$ for any two atomic effects $p$ and $q$. We can use the previous results to prove that symmetry of transition probabilities also holds for arbitrary (finite-dimensional) sequential product spaces.

\[prop:uniquestate\] Let $p,q \in E$ be atomic effects. There exist unique pure states $\omega_p$ and $\omega_q$ such that $\omega_p(p)=1$ and $\omega_q(q)=1$. Furthermore for these states we have $\omega_p(q)=\omega_q(p)$.

The states separate the points of an order unit space [@alfsen2012state Corollary 1.27] so that for $p$ we can find a state $\omega$ such that $\omega(p)\neq 0$. Define $\omega_p(a) := \omega(p{\,\&\,}a)/(\omega(p))$, then $\omega_p$ is a state and $\omega_p(p)=1$. Suppose there is another state $\omega^\prime$ such that $\omega^\prime(p)=1$. Let $q\neq p$ be any other atomic effect (if there is no atomic $q\neq p$ then $V\cong {\mathbb{R}}$ and we are already done) and look at the restrictions of the states $\omega_p$ and $\omega^\prime$ to the space $V_{p,q}$. These restriction maps are still states as $\omega_p(p\vee q)\geq \omega_p(p)=1$ (and similarly for $\omega^\prime$). Because states with this property are unique on spin-factors we can conclude that these restricted states are equal on this subspace and in particular $\omega_p(q)=\omega^\prime(q)$. Since $q$ was arbitrary and the atomic effects span $V$ we conclude that $\omega_p=\omega^\prime$ so that $\omega_p$ is indeed unique.

For any two atomic $p$ and $q$ we can look at their unique pure states $\omega_p$ and $\omega_q$ as restricted to $V_{p,q}$ for which we know that $\omega_p(q)=\omega_q(p)$ which finishes the proof.

\[prop:atomicprod\] Let $p$ and $q$ be atomic sharp effects. $p$ and $q$ are orthogonal, i.e. $p{\,\&\,}q = q{\,\&\,}p = 0$ if and only if $\omega_p(q) = \omega_q(p) = 0$. Furthermore, $p{\,\&\,}q = \omega_p(q) p$.

Note that if $q{\,\&\,}p =0$ then by proposition \[prop:meetsharp\] $q+p\leq 1$ so that $1=\omega(p)\leq\omega_p(q+p)\leq \omega(1)=1$ from which we conclude that $\omega_p(q)=0$. So if $p$ and $q$ are orthogonal then $\omega_p(q)=\omega_q(p)=0$.

For the converse we will show that $p{\,\&\,}q = \omega_p(q) p$, from which it directly follows that $\omega_p(q) = 0 \implies p{\,\&\,}q = 0$. Since $p$ is atomic we of course have $p{\,\&\,}q = \lambda p$ for some $\lambda \geq 0$. Let $\omega^\prime(a) = \omega_p(p{\,\&\,}a)$, then $\omega^\prime(p) = \omega_p(p{\,\&\,}p) = \omega_p(p) = 1$, so that by the uniqueness of $\omega_p$ we have $\omega^\prime = \omega_p$. We then see that $\omega_p(q) = \omega^\prime(q) = \omega_p(p{\,\&\,}q) = \omega_p(\lambda p) = \lambda \omega_p(p) = \lambda$.

There exists an inner product on $V$ such that the positive cone is self-dual with respect to this inner product.

For atomic $p$ and $q$ we let $\inn{p,q}:= \omega_p(q)=\omega_q(p)=\inn{q,p}$. We can then extend it by linearity to arbitrary $a=\sum_i \lambda_i p_i$ and $b=\sum_j \mu_j q_j$ by $\inn{a,b}:= \sum_{i,j} \lambda_i\mu_j \inn{p_i,q_i}$. This is well-defined since $\inn{a,b} = \sum_i \lambda_i \omega_{p_i}( \sum_j \mu_j q_j) = \sum_i \lambda_i \omega_{p_i}(b) = \sum_j \mu_j \omega_{q_j}(a)$ so that this is independent of the representation of $a$ and $b$ in as linear combinations of atomic effects. Now $\inn{a,a} = \sum_{i,j} \lambda_i \lambda_j \omega_{p_i}(p_j) = \sum_i \lambda_i^2$ since $p_i$ and $p_j$ are orthogonal when $i\neq j$ and $\omega_{p_i}(p_i)=1$. We conclude that $\inn{a,a}\geq 0$ and that it is only equal to zero when $a=0$ so that $\inn{\cdot,\cdot}$ indeed is an inner product.

If $a$ and $b$ are positive elements then we can write them as $a=\sum_i \lambda_i p_i$ and $b=\sum_j \mu_j q_j$ where all the $\lambda_i$ and $\mu_j$ are greater than zero. It then easily follows that $\inn{a,b}\geq 0$ because $\omega_{p_i}(q_j)\geq 0$. Conversely if we have $a=\sum_i \lambda_i p_i$ with $\lambda_i$ not necessarily positive with $\inn{a,b}\geq 0$ for all $b\geq 0$, then we can in particular take $b=p_j$ to see that $0\leq \inn{a,p_j} = \lambda_j$ from which we conclude that indeed $a\geq 0$.

A finite-dimensional sequential product space is isomorphic to a Euclidean Jordan algebra.

By proposition \[prop:homogen\] the space is homogeneous, and by the previous proposition it is self-dual. The Koecher-Vinberg theorem [@koecher1957positivitatsbereiche] states that any homogeneous self-dual ordered vector space is order-isomorphic to a Euclidean Jordan algebra.

The Jordan product from a sequential product {#sec:jordanproduct}
============================================

The Koecher-Vinberg theorem is a rather indirect way of establishing the Jordan algebra structure of the space. Since we don’t have just a homogeneous self-dual space, but we also have access to the sequential product we can in fact construct the Jordan product directly. That is what we will strive for in this section. We will use the construction of the Jordan product from the work of Alfsen and Schultz [@alfsen2012geometry], but then adapted to our setting.

We call a real vector space $V$ a *Jordan algebra* when it has a bilinear commutative operation $*$ that satisfies the *Jordan identity*: $a*(b*(a*a)) = (a*b)*(a*a)$. We call $V$ a *Euclidean* Jordan algebra when it is furthermore a Hilbert space with inner product $\inn{\cdot, \cdot}$ such that $\inn{a*b,c} = \inn{b,a*c}$.

Note: By writing the Jordan product operator $T_a(b) := a*b$ and using the commutativity of the product we can also write the Jordan identity as $T_aT_{a*a} = T_{a*a}T_a$.

Let $p$ be an atomic sharp effect and let $b\in V$ be arbitrary. We define their *Jordan product* as $p*b = \frac{1}{2}({\text{id}}+ L_p - L_{p^\perp}) b$.

\[lem:restrict\] Let $p$ and $q$ be atomic sharp effects, then $p^\perp {\,\&\,}q = p^\prime {\,\&\,}q$ where $p^\prime = p\vee q - p$.

First note that $p^\perp = 1-p = 1-p\vee q + p\vee q -p = (p\vee q)^\perp + p^\prime$ and hence that $p^\prime \leq p^\perp$ so that $p^\prime {\,\lvert\,}p^\perp$ by proposition \[prop:belowsharp\]. We then also have $p^\perp{\,\&\,}(p\vee q) = (p\vee q){\,\&\,}p^\perp = (p\vee q) {\,\&\,}((p\vee q)^\perp + p^\prime) = p^\prime$. Now using the fact that we are working with compatible effects and that $q\leq p\vee q$ we calculate $p^\perp {\,\&\,}q = p^\perp {\,\&\,}((p\vee q){\,\&\,}q) = (p^\perp {\,\&\,}(p\vee q)){\,\&\,}q = p^\prime {\,\&\,}q$.

\[lem:atomiccommute\] [@alfsen2012geometry Lemma 9.29]: Let $p$ and $q$ be sharp atomic effects.

1.  $p*q = q*p$.

2.  When $p{\,\&\,}q = 0$ we have $p*q=0$ and in that case for any $b\in V$: $p*(q*b)=q*(p*b)$.

3.  $p*p = p$.

 

1.  If $p=q$ this is trivial, so assume that $p\neq q$. Let us denote $p^\prime = p\vee q - p$. By proposition \[prop:coverprop\] $p^\prime$ is atomic. By proposition \[prop:atomicprod\] we have $p{\,\&\,}q = \omega_p(q) p = \inn{p,q} p$ and similarly $p^\prime {\,\&\,}q = \inn{p^\prime ,q} p^\prime$. Expanding the definition of $p*q$ and using lemma \[lem:restrict\] to write $p^\perp {\,\&\,}q = p^\prime {\,\&\,}q$ where $p^\prime = p\vee q - p$ we calculate $$\begin{aligned}
            2 (p*q) &= q + \inn{p,q}p - \inn{p^\prime, q} p^\prime \\
            &= q + \inn{p,q}p - \inn{p^\prime, q} (p\vee q - p) \\
            &= q + (\inn{p,q}p + \inn{p^\prime, q})p - \inn{p^\prime, q} (p\vee q) \\
            &= q + \inn{p\vee q,q}p + \inn{p\vee q - p, q} (p\vee q) \\
            &= q+p + (1-\inn{p,q})(p\vee q)
            \end{aligned}$$ which is indeed symmetric in $p$ and $q$.

2.  When $p{\,\&\,}q = 0$ we have $q\leq p^\perp$ so that $p^\perp {\,\&\,}q = q$ which indeed gives $p*q = {\frac12(q + p{\,\&\,}q - p^\perp {\,\&\,}q)} = \frac12(q-q) = 0$. For the second point we note that because $p{\,\&\,}q = 0$ we have $p{\,\lvert\,}q, q^\perp$ and $q{\,\lvert\,}p^\perp$, and hence that the maps $L_p, L_{p^\perp}, L_q$ and $L_{q^\perp}$ commute so that the maps $b\mapsto p*b$ and $b\mapsto q*b$ will commute as well.

3.  Follows immediately from $p{\,\&\,}p = p$ and $p^\perp {\,\&\,}p =0$.

As a result of this lemma we can extend the Jordan product by linearity to the entirety of the space.

Let $a,b\in V$ be arbitrary. Let $a=\sum_i \lambda_i p_i$ and $b=\sum_j \mu_j q_j$ be spectral decompositions with the $p_i$ and $q_j$ atomic. Define their Jordan product as $a*b = \sum_{i,j} \lambda_i \mu_j p_i*q_j$. We write $T_a:V\rightarrow V$ for the operator that sends $b$ to $a*b$.

\[prop:jordanproduct\] The Jordan product is well-defined, bilinear, commutative and furthermore

1.  If $a{\,\lvert\,}b$ then $T_aT_b = T_bT_a$.

2.  \[prop:jordancomm\] If $a{\,\lvert\,}b$ then $T_a b = a^+{\,\&\,}b - a^- {\,\&\,}b$ where $a^+$ and $a^-$ are the unique orthogonal positive elements such that $a=a^+ - a^-$.

We first note that we of course have $a*b = \sum_i \lambda_i p_i*b$ so that the definition is independent of how $b$ is represented as a sum of atomic sharp effects. By the previous lemma the product is commutative and therefore we see it is bilinear and well-defined.

1.  Suppose $a{\,\lvert\,}b$. By considering the classical algebra spanned by both $a$ and $b$ we can find an orthogonal set of atomic sharp effects $p_i$ such that $a=\sum_i \lambda_i p_i$ and $b=\sum_i \mu_i p_i$. Since $p_i*p_j=0$ for $i\neq j$ we have $p_i*(c*p_j) = p_j*(c*p_i)$ by lemma \[lem:atomiccommute\] so that we can then write $$T_bT_a c = b*(a*c) = \sum_{i,j} \mu_i\lambda_j p_i*(c*p_j) = \sum_{i,j} \mu_i \lambda_j p_j*(c*p_i) = a*(c*b) = T_aT_b c$$ which holds for all $c$. We conclude that $T_bT_a=T_aT_b$.

2.  If atomic $p_i$ commutes with $b$, then $p_i*b = \frac{1}{2}({\text{id}}+ U_{p_i} - U_{p_i^\perp})b = \frac{1}{2}(b + p_i{\,\&\,}b - p_i^\perp b) = \frac{1}{2}((p_i+p_i^\perp){\,\&\,}b + p_i{\,\&\,}b - p_i^\perp b) = p_i{\,\&\,}b$. So by writing $a=\sum_i \lambda_i p_i = \sum_{i, \lambda_i>0} \lambda p_i - \sum_{i, \lambda_i<0}\lvert \lambda_i\rvert p_i = a^+ - a^-$ with $p_i{\,\lvert\,}b$ the desired result follows by linearity.

Let $V$ be a finite-dimensional sequential product space, then it is a Euclidean Jordan algebra with the Jordan product as defined above.

We have already established that the product is bilinear and commutative. Note that since $a{\,\lvert\,}a$ we get $a*a = a^+{\,\&\,}a - a^-{\,\&\,}a = (a^+)^2 + (a^-)^2 = a^2$ using proposition \[prop:jordancomm\]. Because of course $a{\,\lvert\,}a^2$ we get $T_aT_{a*a} = T_{a*a}T_a$ as a consequence of the previous proposition so that the Jordan identity holds. Since $a*a = a^2\geq 0$ we also see that the algebra is *formally real*: if $\sum_i a_i*a_i = 0$ then for all $i$: $a_i=0$. It is a well-known result (see for instance [@faraut1994analysis Proposition VIII.4.2]) that if a Jordan product is formally real, that the algebra is Euclidean, with the product being symmetric with regards to the (essentially unique) self-dual inner product.

Note: It is also possible to show in a more direct manner that the Jordan product is symmetric with respect to the inner product. We sketch here how to do so. First it must be established that $L_a$ for invertible $a$ commute with their adjoints $L_a^*$ by exploiting the fact that $\Theta = L_{a^{-1}}L_a^*$ must be a unital order-isomorphism necessarily satisfying $\Theta^{-1} = \Theta^*$. Because the mapping $a\mapsto L_a$ is continuous, the result extends to all $a$ and hence it holds in particular for $L_p$ with $p$ sharp. It is a standard result that an idempotent map that commutes with its adjoint is in fact self-adjoint. Since the Jordan product is defined as a linear combination of product maps of sharp effects this indeed establishes the desired result.

Local Tomography and C\*-algebras {#sec:loctom}
=================================

In this section we will let $V$ and $W$ be finite-dimensional sequential product spaces, and hence by the previous sections Euclidean Jordan algebras. For the duration of this section we will assume that their linear algebraic tensor product $V\otimes W$ is also a sequential product space and that the sequential product satisfies $$(a_1\otimes b_1){\,\&\,}(a_2\otimes b_2) = (a_1{\,\&\,}a_2) \otimes (b_1{\,\&\,}b_2).$$

Note that by definition of the tensor product any element of $V\otimes W$ can be written as $\sum_i \lambda_i a_i\otimes b_i$ where $a_i\in V$ and $b_i\in W$.

\[prop:atomictensor\] Let $p\in V$ and $q\in W$ be atomic, then $p\otimes q\in V\otimes W$ is also atomic.

Because $(p\otimes q){\,\&\,}(p\otimes q) = (p{\,\&\,}p)\otimes (q{\,\&\,}q) = p\otimes q$ it is sharp. Let $c = \sum_i \lambda_i a_i\otimes b_i$ be an arbitrary element of $V\otimes W$, then using lemma \[lem:atomicnorm\] $(p\otimes q){\,\&\,}c = \sum_i \lambda_i (p{\,\&\,}a_i)\otimes (q{\,\&\,}b_i) = \sum_i \lambda_i \norm{p{\,\&\,}a_i}\norm{q{\,\&\,}b_i} (p\otimes q) = \mu (p\otimes q)$ for some $\mu\in {\mathbb{R}}$. Since $c$ was arbitrary we conclude that $p\otimes q$ is atomic as a result of lemma \[lem:atomicnorm\].

Let $V$ be a sequential product space with effects $E$. We call $c\in V$ *classical* when it is compatible with all other effects: $a{\,\lvert\,}c$ for all $a\in E$. We will call a classical effect *minimal* when there is no non-zero classical effect strictly below it.

\[prop:classicaltensor\] Let $c\in V$ and $d\in W$ be classical, then $c\otimes d$ is classical in $V\otimes W$.

$c{\,\lvert\,}a$ for all $a \in V$ and $d{\,\lvert\,}b$ for all $b\in W$, therefore $c\otimes d{\,\lvert\,}a\otimes b$ and the same holds for linear combinations of these elements which span the entirety of $V\otimes W$.

We call an EJA *simple* if it contains no non-trivial classical effects. We can write any EJA uniquely as $E_1\oplus\ldots \oplus E_k$ where the $E_i$ are simple EJAs, which we will refer to as the *summands* of the EJA. An EJA with $k$ summands has exactly $k$ minimal classical effects, corresponding to the units of each of the summands. Each other sharp classical effect is a sum of these minimal ones.

\[lem:atomicoverlap\] Let $p$ and $q$ be atomic effects in an EJA. If $q{\,\&\,}p \neq 0$, then $p$ and $q$ belong to the same simple summand.

Suppose $c$ is a sharp classical effect. Then $c{\,\&\,}p = p{\,\&\,}c = \lambda p$, for some $\lambda\in[0,1]$. But also $\lambda p = c{\,\&\,}p=(c^2){\,\&\,}p = c{\,\&\,}(c{\,\&\,}p) = \lambda (c{\,\&\,}p) = \lambda^2 p$, so that $\lambda = 1$ or $\lambda = 0$. So either $p\leq c$ or $p$ and $c$ are orthogonal. Suppose $p\leq c$, and that $q{\,\&\,}p \neq 0$, then $0\neq q{\,\&\,}p \leq q{\,\&\,}c =\lambda q$, so that $\lambda\neq 0$ and thus $q\leq c$. If we let $c$ be the identity of the summand that $p$ belongs to we see that the desired property follows.

Let $p_1,\ldots,p_r$ be a maximal collection of orthogonal non-zero atomic effects in a simple EJA, then there exists an atomic effect $q$ such that $q{\,\&\,}p_i\neq 0$ for all $i$.

We do this by case distinction using the classification of simple EJAs [@jordan1993algebraic]. Either the space is a spin-factor, in which case a maximal collection is always given by a sharp atomic $p$ and its complement $p^\perp$. Any $q\neq p,p^\perp$ cannot be orthogonal to them, because the space is of rank 2, so that indeed $q{\,\&\,}p\neq 0$ and $q{\,\&\,}p^\perp \neq 0$.

If the space is not a spin-factor, then it must be of the form $B(H)^{\text{sa}}$ for a real, complex, quaternionic or octonion finite-dimensional Hilbert space $H$ (in the case of the octonions we must have $\dim H=3$). For such a space the atomic idempotents correspond to unit vectors of the underlying Hilbert space: $p_i = \lvert v_i\rangle\langle v_i \rvert$ where $v_i\in H$ is some unit vector. We can then take $q=\lvert w\rangle\langle w \rvert$ with $w=\frac{1}{\sqrt{r}}\sum_{i=1}^r v_i$. It should then be clear that $\inn{q,p_i}\neq 0$ and hence $q{\,\&\,}p_i \neq 0$.

\[prop:tensordirectsum\] Let $V = E_1\oplus\ldots \oplus E_m$ and $W = F_1\oplus \ldots \oplus F_n$ with the $E_i$ and $F_j$ being simple EJAs. Let $1\leq k \leq n$ and $1\leq l\leq m$. Let $p_1,\ldots, p_r$ be a maximal collection of orthogonal non-zero atomic effects in $E_k$, and let $q_1,\ldots, q_s$ be such a maximal collection in $F_l$. Then $(p_i\otimes q_j)_{i=1,j=1}^{r,s}$ belong to the same simple summand in $V\otimes W$ and they form a maximal collection of orthogonal non-zero atomic effects in this summand.

We let $p$ be atomic such that $p{\,\&\,}p_i\neq 0$ for all $i$ which exists by the previous lemma and similarly we let $q$ be atomic such that $q{\,\&\,}q_j \neq 0$ for all $j$. By proposition \[prop:atomictensor\] $p\otimes q$ and $p_i\otimes q_j$ will be atomic for all $i$ and $j$. By construction we of course have $0\neq (p{\,\&\,}p_i)\otimes (q{\,\&\,}q_j)=(p\otimes q){\,\&\,}(p_i\otimes q_j)$ and by lemma \[lem:atomicoverlap\] the $p_i\otimes q_j$ must then belong to the summand of $p\otimes q$ for all $i$ and $j$.

Since $\sum_i p_i=1_{E_k}$, this sum is a classical effect. The same holds for $\sum_j q_j=1_{F_l}$. Their tensor product $1_{E_k}\otimes 1_{F_l} = \sum_{i,j} p_i\otimes q_j$ is then also classical by proposition \[prop:classicaltensor\]. Since the only nonzero classical effect in a simple summand is the identity this expresion must be equal to the identity of this summand. As a result the set $(p_i\otimes q_j)_{i,j}$ is indeed maximal.

Using this proposition we conclude that for each of the summands $E$ of $V$ and $F$ of $W$ there must exist a summand in $V\otimes W$ which has rank ${\text{rnk}\xspace}~ E~ {\text{rnk}\xspace}~ F$. Because the tensor product map is obviously injective this factor must have dimension at least $\dim E~ \dim F$, and then because of local tomography the dimension must be strictly equal. Now let $V$ be a sequential product space for which the tensor product $V\otimes V$ exists so that the above must in particular be true when $E=F$, i.e. if $E$ is a simple factor of $V$ then there must exist a simple factor with rank $({\text{rnk}\xspace}E)^2$ and dimension $(\dim E)^2$.

\[prop:simpleEJAsquare\] Let $E$ be a simple Euclidean Jordan algebra of rank $r> 1$ and dimension $N$. There exists a simple Euclidean Jordan algebra of rank $r^2$ and dimension $N^2$ if and only if $E=B(H)^{\text{sa}}$ where $H$ is a complex finite-dimensional Hilbert space.

If $E=B(H)^{\text{sa}}$ is a complex matrix algebra the property is obviously true by considering $B(H\otimes H)^{\text{sa}}$ as the simple EJA of rank $r^2$ and dimension $N^2$. We simply check that every other simple EJA is not a possibility. If $E= B(H)^{\text{sa}}$ where $H$ is the 3-dimensional octonion Hilbert space, then $r=3$ and $N=27$. The highest dimensional simple EJA of rank $9$ is the quaternionic system which has dimension $9*(2*9-1)=153<27^2 = 729$ so that this is not possible. If $E=B(H)^{\text{sa}}$ with $H$ quaternionic, then $N = r(2r-1)$. The highest dimensional simple EJA of rank $r^2$ is also quaternionic so that its dimension is $r^2(2r^2-1)$. It is easy to check that $N^2 = r^2(2r-1)^2 > r^2(2r^2-1)$ when $r>1$ so that again, it cannot be this space. If $E=B(H)^{\text{sa}}$ where $H$ is real, then by dimension counting we can again see that there does not exist an EJA with rank $r^2$ and dimension $N^2$. A spin factor always has rank 2. The rank 4 EJAs have dimension 10, 16 and 28. The only one of these which is a square is 16. The 4 dimensional spin-factor corresponds to the qubit which is indeed $B(H)^{\text{sa}}$ with $H$ a 2-dimensional complex Hilbert space.

Suppose $V$ is a finite-dimensional sequential product space for which the linear algebraic tensor product space $V\otimes V$ is also a sequential product space with product satisfying $(a\otimes b){\,\&\,}(c\otimes d) = (a{\,\&\,}c)\otimes (b{\,\&\,}d)$ for all $a,b,c,d \in V$, then there exists a C$^*$-algebra $A$ such that $V\cong A^{\text{sa}}$.

As established, $V$ is a Euclidean Jordan algebra. As a result of proposition \[prop:tensordirectsum\] for each summand of $V = E_1\oplus\ldots\oplus E_n$ there must exist a simple EJA of rank $({\text{rnk}\xspace}E_i)^2$ and dimension $(\dim E_i)^2$. By proposition \[prop:simpleEJAsquare\] this is only possible if $E_i=B(H)^{\text{sa}}$ where $H$ is a complex Hilbert space. Therefore $V$ is a direct sum of complex matrix algebras which means it is the set of self-adjoint elements of a C$^*$-algebra.

Infinite-dimensional sequential product spaces {#sec:infiniterank}
==============================================

In order to state the following infinite-dimensional generalisations of the main theorem we must first give an appropriate definition of infinite-dimensional Jordan algebras.

We call an order unit space $V$ a *JB-algebra* when it is complete in its norm topology and it has a Jordan product $*$ such that $\norm{a*a} = \norm{a}^2$ and $\norm{a*b} \leq \norm{a}\norm{b}$ for all $a,b \in V$.

Let $V$ be an order unit space. A subset $S\subseteq V$ is called *bounded* when there exists $r\in {\mathbb{R}}$ such that $\norm{a}\leq r$ for all $a\in S$. It is called *directed* when for all $s_1,s_2 \in S$ we can find $s\in S$ such that $s_1,s_2 \leq s$. We call $V$ *bounded directed complete* when for any bounded directed set $S$ we can find a least upper bound, i.e. an element $t\in V$ such that for all $s\leq t$ for all $s\in S$.

We call a state $\omega: V\rightarrow {\mathbb{R}}$ on an order unit space *normal* when it preserves suprema of bounded directed sets: $\omega(\bigvee S) = \bigvee \omega(S)$. We say that $V$ has *enough normal states* when the normal states order-separate the elements, i.e. when $\omega(v) \leq \omega(w)$ for all normal states $\omega$ implies that $v\leq w$.

A *JBW-algebra* is a JB-algebra that is bounded directed complete and has enough normal states.

When thinking about these algebras it is helpful to keep in mind the following analogy: JB-algebras are to C$^*$-algebras as JBW-algebras are to von Neumann algebras.

As shown in the authors previous work [@wetering2018characterisation], we can still derive a spectral decomposition theorem and homogeneity for an infinite-dimensional bounded direct complete sequential product space when we also require the sequential product to be normal. In order to use homogeneity as in proposition \[prop:itochar\] however we need atomic effects. In the main theorem, the finite-dimensionality of the space ensured the existence of atomic effects, but in infinite dimension, atomic effects don’t have to exist. For example, if $V$ would be the space of self-adjoint elements of a type II von Neumann algebra then it does not contain a single non-zero atomic effect. We see therefore that if we want to work with atomic effects, that we must require them explicitly:

Let $V$ be a sequential product space. We call $V$ *atomic* when below every sharp effect we can find a non-zero atomic sharp effect.

Unfortunately, at this point we still cannot use the characterisation of rank 2 spaces from [@ito2017p] as this classification only holds for finite-dimensional spaces. It is currently unclear whether the generalisation of this theorem to infinite-dimensional spaces holds, although it seems reasonable. At this point we therefore have to add one more condition:

We say that an atomic sequential product space $V$ has *finite bits* when the spaces $V_{p,q}$ (see definition \[def:orderideal\]) generated by sharp atomic effects $p$ and $q$ are finite-dimensional.

Note that there do exist sequential product spaces that do not have finite bits, namely spin-factors of infinite-dimension. But the author is not aware of any space of rank greater than 2 that does not contain a spin-factor as a subsystem, that doesn’t have finite bits.

Let $V$ be an atomic bounded directed complete sequential product space with finite bits where the sequential product is normal, then $V$ is a JB-algebra.

As shown in [@wetering2018characterisation] a directed complete sequential product space with normal sequential product is homogeneous. The necessary propositions presented in sections \[sec:atomeffect\], \[sec:coverprop\] and \[sec:subselfdual\] still hold and since the spaces $V_{p,q}$ are required to be finite-dimensional, the characterisation theorem of [@ito2017p] applies so that we indeed have symmetry of transition probabilities. Since the atomic effects lie dense in the space (see also [@wetering2018characterisation]) we have then satisfied all the conditions of [@alfsen2012geometry Theorem 9.38] so that $V$ is indeed a JB-algebra.

This theorem can with a few extra requirements be lifted to a more specific statement:

Let $V$ be an atomic bounded directed complete sequential product space with finite bits and a normal sequential product and enough normal states, then $V$ is a JBW-algebra. If furthermore $V$ is of infinite rank and the only classical effects are $0$ and $1$ then $V$ is isomorphic to $B(H)^{\text{sa}}$ where $H$ is an infinite-dimensional real, complex or quaternionic Hilbert space.

It has already been established that a $V$ with these properties is a JB-algebra. It is also a JBW-algebra, since we explicitly require bounded directed completeness and enough normal states. Because $V$ contains no non-trivial classical effects it must be a factor. A JBW factor of infinite rank such that there is an atomic effect below each sharp effect is isomorphic to $B(H)^{\text{sa}}$ where $H$ is a real, complex or quaternionic Hilbert space [@hanche1984jordan Theorem 7.5.11].

The statement of this theorem in its current form is quite hampered. It seems reasonable that this theorem should be able to be generalised. In particular, the requirement of enough normal states is probably unnecessary: there is a theorem stating that type I AW$^*$-algebra factors correspond exactly to type I von Neumann algebra factors. A similar theorem seems reasonable to hold for type I JBW-algebras and the ‘JBAW’-algebras. Furthermore, it seems reasonable that the characterisation of rank 2 homogeneous spaces of [@ito2017p] holds for infinite-dimensional spaces, in which case the condition on finite bits can be dropped.

A much larger class of spaces could be covered if the atomicity requirement could somehow be dropped. In fact, we conjecture the following:

> **Conjecture:** Let $V$ be a bounded directed complete sequential product space with a normal sequential product and enough normal states, then $V$ is a JBW-algebra.

It is currently not clear whether the tools we have now are adequate to prove this statement. We will however show that using an additional assumption on the sharp effects we can prove a similar theorem:

Let $V$ be a bounded directed complete sequential product space with a normal sequential product and enough normal states, such that for all sharp effects $p$ and $q$ the relation $ p{\,\&\,}q + p^\perp{\,\&\,}q^\perp = q{\,\&\,}p + q^\perp {\,\&\,}p^\perp$ holds, then $V$ is a JBW-algebra.

When $V$ allows spectral decompositions, is bounded directed complete and has enough normal states, a sufficient condition for it being a JBW-algebra is that the relation defined above for sharp effects holds true for so called *compressions* (Corollary 9.45 and Theorem 9.43 of [@alfsen2012geometry]). As shown in [@wetering2018characterisation] we know that $V$ has spectral decompositions. It is however a priori not clear whether the sequential product map of a sharp effect $L_p(a) = p{\,\&\,}a$ is indeed a compression as defined by Alfsen and Schultz. It is true though that these maps are *complemented*, so that the proof of the relevant theorem in [@alfsen2012geometry] still carries trough when the compressions are replaced with the maps $L_p$.

Minimality of axioms {#sec:axioms}
====================

In this section we will discuss the minimality of the conditions and the axioms needed to show that finite-dimensional sequential product spaces are Euclidean Jordan algebras. For easy reference we copy definition \[def:seqprod\] here:

A map ${\&:[0,1]_V\times [0,1]_V \rightarrow [0,1]_V}$ is called a sequential product when it satisfies the following properties for all $a,b,c \in [0,1]_V$:

1.  Additivity: $a{\,\&\,}(b+c) = a{\,\&\,}b+ a{\,\&\,}c$.

2.  Continuity: The map $a\mapsto a{\,\&\,}b$ is continuous in the norm.

3.  Unitality: $1{\,\&\,}a = a$.

4.  Compatibility of orthogonal effects: If $a{\,\&\,}b = 0$ then also $b{\,\&\,}a =0$.

5.  Associativity of compatible effects: If $a{\,\lvert\,}b$ then $a{\,\&\,}(b{\,\&\,}c) = (a{\,\&\,}b){\,\&\,}c$.

6.  Additivity of compatible effects: If $a{\,\lvert\,}b$ then $a {\,\lvert\,}1-b$, and if also $a{\,\lvert\,}c$ then $a{\,\lvert\,}(b+c)$.

7.  Multiplicativity of compatible effects: If $a{\,\lvert\,}b$ and $a{\,\lvert\,}c$ then $a{\,\lvert\,}(b{\,\&\,}c)$.

First of all, for the proofs of the main theorems, axiom \[ax:compmult\] is actually not needed since the following weaker version is sufficient:

Suppose $a{\,\lvert\,}b,c$ and that $b{\,\lvert\,}c$, then $a{\,\lvert\,}(b{\,\&\,}c)$.

Using axiom \[ax:assoc\] repeatedly: $a{\,\&\,}(b{\,\&\,}c) = (a{\,\&\,}b){\,\&\,}c = (b{\,\&\,}a){\,\&\,}c = b{\,\&\,}(a{\,\&\,}c) = b{\,\&\,}(c{\,\&\,}a) = (b{\,\&\,}c){\,\&\,}a$.

The reason we included axiom \[ax:compmult\] is because it is part of the definition of a sequential effect algebra, and because when defining the classical algebra of an element when working in infinite dimension, it is needed to show that the algebra is closed under multiplication.

Of the other axioms, the ones that seem less essential are \[ax:orth\] and \[ax:cont\], so it would be interesting to see what can be done without them.

To define and study the classical algebra of an effect, \[ax:orth\] is not needed and \[ax:cont\] is only needed to show that $(\lambda 1){\,\&\,}a = \lambda a$. The spectral theorem and the homogeneity of the space can thus be proven without using these axioms if \[ax:unit\] is changed to $(\lambda 1){\,\&\,}a = \lambda a$. When restricting to rank 2 spaces, axioms \[ax:add\], \[ax:unit\], \[ax:assoc\] and \[ax:compadd\] are then sufficient to prove that the space is a Euclidean Jordan algebra (or specifically, a spin-factor). Since the spectral theorem is also what is needed to show that $L_a$ for $a$ invertible is an order-isomorphism, it should be clear that on an EJA this restricted set of axioms already greatly reduces the possible sequential-product-like maps.

Note also that using the T-algebra formalism of Vinberg [@vinberg1967theory] it is possible to find an associative binary operation (see the beginning of section 4 of [@chua2003relating] for this operation) for the positive elements in any finite-dimensional homogeneous space that satisfies axioms \[ax:add\], \[ax:cont\], \[ax:unit\], \[ax:assoc\] (and by associativity also \[ax:compmult\]) but this product does not satisfy axioms \[ax:orth\] and \[ax:compadd\]. This actually leads to an interesting observation: if either the proof of homogeneity in section \[sec:prelim\] can be shown to hold without use of axiom \[ax:compadd\] or if a binary product on homogeneous spaces can be found that also satisfies \[ax:compadd\], then this would give us a new characterisation of homogeneous spaces. In particular, in the second case, it would show that \[ax:orth\] is the key to establishing self-duality.

When one considers more general ordered vector spaces than order unit spaces, one can find non-trivial totally ordered vector spaces that allow a commutative bilinear product, and hence a sequential product [@basmaster]. These spaces are pathological in the sense that they have ‘infinitesimal’ effects, i.e. effects than cannot be distinguished using states.
