Kadison's antilattice theorem for a synaptic algebra

TL;DR

This paper proves that in a synaptic algebra with a complete orthomodular lattice of projections, the algebra is a factor if and only if it is an antilattice, extending Kadison's results on operator algebras.

Contribution

It establishes a new equivalence between being a factor and an antilattice in synaptic algebras with complete projection lattices, generalizing Kadison's earlier work.

Findings

A synaptic algebra is a factor iff it is an antilattice when the projection lattice is complete.

Generalization of Kadison's results on infima and suprema in operator algebras.

Provides new insights into the structure of synaptic algebras and their projection lattices.

Abstract

We prove that if A is a synaptic algebra and the orthomodular lattice P of projections in A is complete, then A is a factor iff A is an antilattice. We also generalize several other results of R. Kadison pertaining to infima and suprema in operator algebras.

Equations

$$a\odot b:=\frac{1}{2}(ab+ba)=\frac{1}{2}((a+b{)}^2-a^2-b^2)\in A,$$

$$(1-q_{\lambda })(a-\lambda )\le 0\le q_{\lambda }(a-\lambda ).$$

$$\text{if }a\in A^{+}\text{ and }a\le p,q\text{ then }a\le g.$$

$$\alpha :=-\frac{5}{4},\beta :=-\frac{3}{4},\gamma :=1,\text{ and }k=2s-{\alpha }^2\in A.$$

$${\alpha }^2-{\beta }^2=1,(\alpha +\beta )\gamma =-2,\text{ and }{\beta }^2+{\gamma }^2=\frac{25}{16}.$$

$$sp=p^{\perp }s,ps=s{p}^{\perp },sp+ps=sp+s{p}^{\perp }=s(p+p^{\perp })=s1=s$$

$$\text{and }s{p}^{\perp }+p^{\perp }s=ps+sp=s.$$

$$0\le d^2={\alpha }^{2}p+\alpha \gamma (ps+sp)+{\gamma }^2+\beta \gamma (s{p}^{\perp }+p^{\perp }s)+{\beta }^2{p}^{\perp }=$$

$${\alpha }^{2}p+{\beta }^2(1-p)+\alpha \gamma s+{\gamma }^2+\beta \gamma s=({\alpha }^2-{\beta }^2)p+(\alpha +\beta )\gamma s+{\beta }^2+{\gamma }^2$$

$$=p-2s+\frac{25}{16}=p-k,$$

Full text

1 Introduction

A synaptic algebra [3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 21] is a generalization of the self-adjoint part of several structures based on operator algebras. For instance, although a synaptic algebra $A$ need not be norm complete (i.e., Banach), $A$ is isomorphic to the self-adjoint part of a Rickart C*∗-algebra if and only if it is Banach [9, Theorem 5.3]. Also, $A$ is isomorphic to the self-adjoint part of an AW∗*-algebra iff it is Banach and its projection lattice is complete [9, Theorem 8.5]. Numerous additional examples of synaptic algebras can be found in the references cited above.

In [18], Richard Kadison calls the self-adjoint part $\mathfrak{S}$ of an operator algebra an antilattice if and only if, whenever two elements of $\mathfrak{S}$ have an infimum in $\mathfrak{S}$, then the elements are comparable (i.e., one is less than or equal to the other). As Kadison remarks [18, p. 505], “A moments thought shows that this is as strongly nonlattice as a partially ordered vector space can be.” He shows that, in important cases, the condition that $\mathfrak{S}$ is an antilattice is equivalent to the condition that the operator algebra in question is a factor (i.e., its center consists only of scalars). Our main theorem in this paper (Theorem 4.7) is a version of Kadison’s result for a synaptic algebra in which the projections form a complete lattice.

2 Some properties of a synaptic algebra

Axioms for a synaptic algebra can be found in [3, §1] and will not be repeated here. Rather, we shall briefly sketch some of the important features of a synaptic algebra that we shall need below. Readers who are informed about operator algebras will be familiar with many of these features—details can be found in the references given in Section 1, especially in [3]. We use ‘iff’ as an abbreviation for ‘if and only if,’ the notation $:=$ means ‘equals by definition,’ and ${\mathbb{R}}$ is the ordered field of real numbers.

In what follows, we assume that $A$ is a synaptic algebra [3, Definition 1.1]. Thus, associated with $A$ is a real or complex associative unital algebra $R$ with unit $1$ called the enveloping algebra of $A$ such that (1) $1\in A$, (2) $A$ is a real linear subspace of $R$, (3) the linear space $A$ is a partially ordered order-unit normed space with order unit $1$ and positive cone $A^{+}:=\{a\in A:0\leq a\}$, and (4) the order-unit norm of $a\in A$ is denoted and defined by $\|a\|:=\inf\{0<\lambda\in{\mathbb{R}}:-\lambda 1\leq a\leq\lambda 1\}$ [1, pp. 67–69]. In important examples, the enveloping algebra $R$ is a unital (concrete or abstract) operator algebra with an adjoint mapping $a\mapsto a^{\ast}$, $A$ is the self-adjoint part of $R$, and $A^{+}=\{rr^{\ast}:r\in R\}$. We shall assume that $A$ is nontrivial, i.e., $A\not=\{0\}$. Then $1\not=0$, which enables us, as usual, to identify each scalar $\lambda\in{\mathbb{R}}$ with the element $\lambda 1\in A$.

Let $M\subseteq A$. Then $M$ is understood to be partially ordered under the restriction of the partial order $\leq$ on $A$. If $m,n\in M$, then the infimum (greatest lower bound) and the supremum (least upper bound) of $m$ and $n$ in $M$—if they exist—are written as $m\wedge_{M}n$ and $m\vee_{M}n$, respectively. An involution on $M$ is a mapping $m\mapsto m^{\prime}$ that is order reversing (i.e., $m\leq n\Rightarrow n^{\prime}\leq m^{\prime}$) and of order two (i.e., $(m^{\prime})^{\prime}=m$). Such an involution provides a “duality” between existing infima and suprema in $M$ according to $m\wedge_{M}n=(m^{\prime}\vee_{M}n^{\prime})^{\prime}$ and $m\vee_{M}n=(m^{\prime}\wedge_{M}n^{\prime})^{\prime}$. Note that the mapping $a\mapsto-a$ is an involution on $A$ itself.

Let $a,b\in A$. Then the product $ab$ is understood to be calculated in $R$ and may or may not belong to $A$; however, it is assumed that $0\leq a^{2}\in A$. Consequently, the Jordan product

[TABLE]

so $A$ is a real unital special Jordan algebra under $\odot$ [20]. If $a$ commutes with $b$, i.e., $ab=ba$ in $R$, we write $aCb$. Evidently, if $aCb$, then $ab=a\odot b\in A$. It can be shown that $ab=0$ iff $ba=0$. Also, if $a,b\in A^{+}$ and $aCb$, then $ab\in A^{+}$. As a consequence, if $a,b,c\in A$, $a\leq b$, $0\leq c$, $cCa$, and $cCb$, then $ca\leq cb$; indeed, by the hypotheses, $0\leq c,b-a$ and $cC(b-a)$, whence $0\leq cb-ca$, i.e., $ca\leq cb$.

Suppose that $a,b\in A$ and put $c:=2(a\odot b)$. Then $aba=a\odot c-a^{2}\odot b\in A$ and the mapping $b\mapsto aba$ is called the quadratic mapping on $A$ determined by $a$. It can be shown that the quadratic mapping $b\mapsto aba$ is both linear and order preserving on $A$.

An element $e\in A$ such that $0\leq e\leq 1$ is called an effect, and it can be shown that $e\in A$ is an effect iff $e^{2}\leq e$. The subset $E:=\{e\in A:0\leq e\leq 1\}$ of $A$ forms a convex effect algebra [17] under the partially defined binary operation obtained by restriction to $E$ of the addition operation $+$ on $A$. The mapping $e\mapsto e^{\perp}:=1-e$, called the orthosupplementation on $E$, is an involution on $E$.

An idempotent element $p=p^{2}$ in $A$ is called a projection, and the set $P:=\{p\in A:p=p^{2}\}$ of all projections in $A$ is a subset of $E$; in fact, $P$ is precisely the set of all extreme points of the convex set $E$. Under the restriction to $P$ of the partial order on $A$, $P$ forms an orthomodular lattice (OML) [3, §5], [2, 19], and the orthocomplementation $p\mapsto p^{\perp}:=1-p$ on the OML $P$ is the restriction to $P$ of the orthosupplementation on $E$. The orthocomplementation mapping is an involution on $P$, whence we have an infimum-supremum duality on the lattice $P$. If $e\in E$ and $p\in P$, it can be shown that $e\leq p\Leftrightarrow e=ep\Leftrightarrow e=pe$ and $p\leq e\Leftrightarrow p=pe\Leftrightarrow p=ep$.

Let $p,q\in P$. Then $p\leq q$ iff $p=pq$ iff $p=qp$. We shall write the infimum and the supremum of $p$ and $q$ in the lattice $P$ as $p\wedge q$ and $p\vee q$, respectively, (without subscripts on $\wedge$ and $\vee$). If $pCq$, then $p\wedge q=pq=qp$ and $p\vee q=p+q-pq$. The projections $p$ and $q$ are said to be orthogonal, in symbols $p\perp q$, iff $p\leq q^{\perp}$, or equivalently iff $q\leq p^{\perp}$. Note that $p\perp q$ iff $pq=0$ iff $qp=0$ iff $p+q=p\vee q$. In particular, for the orthocomplement $p^{\perp}$ of $p$ in $P$, we have $p\perp p^{\perp}$, $pCp^{\perp}$, $p\wedge p^{\perp}=pp^{\perp}=0$ and $p\vee p^{\perp}=p+p^{\perp}=1$.

If $M\subseteq A$, then $C(M):=\{a\in A:aCm\text{\ for all\ }m\in M\}$ is called the commutant of $M$ and $CC(M):=C(C(M))$ is called the bicommutant of $M$. Evidently, $C(M)$ is a linear subspace (or a vector subspace) of $A$. The subset $M$ of $A$ is said to be commutative iff $mCn$ for all $m,n\in M$, i.e., iff $M\subseteq C(M)$. If $M$ is commutative, then so is $CC(M)$. If $a\in A$, then $C(a):=C(\{a\})$ and $CC(a):=CC(\{a\})$.

If $a\in A^{+}$, then there exists a unique $a^{1/2}\in A^{+}$, called the square root of $a$, such that $(a^{1/2})^{2}=a$; moreover, $a^{1/2}\in CC(a)$. The absolute value of $a\in A$ is denoted and defined by $|a|:=(a^{2})^{1/2}$ and the positive and negative parts of $a$ are denoted and defined by $a^{+}:=\frac{1}{2}(|a|+a)$ and $a^{-}:=\frac{1}{2}(|a|-a)$, respectively. Then $0\leq a^{+},a^{-}\in CC(a)$, $a=a^{+}-a^{-}$, $|a|=a^{+}+a^{-}$, and $a^{+}a^{-}=0$. It turns out that an element $a\in A$ has an inverse $a^{-1}$ in $A$ such that $aa^{-1}=a^{-1}a=1$ iff there exists $0<\epsilon\in{\mathbb{R}}$ such that $\epsilon\leq|a|$.

If $a\in A$, there exists a unique projection $a^{\text{\rm o}}\in P$, called the carrier of $a$, such that, for all $b\in A$, $ab=0\Leftrightarrow a^{\text{\rm o}}b=0$. (Some authors would refer to $a^{\text{\rm o}}$ as the support of $a$.) It turns out that $a^{\text{\rm o}}\in CC(a)$, $a^{\text{\rm o}}$ is the smallest projection $p\in P$ such that $a=ap$, and if $e\in E$, then $e^{\text{\rm o}}$ is the smallest projection $p\in P$ such that $e\leq p$. See [3, Theorem 2.10] for additional properties of the carrier.

2.1 Lemma**.**

If $p,q\in P$, then $p\wedge q=p\wedge_{E}q$ and $p\vee q=p\vee_{E}q$.

Proof.

We have $p,q,p\wedge q\in P\subseteq E$ and $p\wedge q\leq p,q$. Suppose $e\in E$ with $e\leq p,q$. Then $e^{\text{\rm o}}\leq p,q$, whence $e\leq e^{\text{\rm o}}\leq p\wedge q$, and therefore $p\wedge q=p\wedge_{E}q$. By duality, $p\vee q=p\vee_{E}q$. ∎

In view of Lemma 2.1, no confusion will result if we use the same notation $e\wedge f$ and $e\vee f$ (without subscripts) for existing infima and suprema of effects as we do for infima and suprema of projections. (Actually, the question of just when two effects have an infimum in $E$ is important, but not easy to resolve [16].)

There is a very satisfactory spectral theory for $A$ based on the following notions [3, §8]: Let $a\in A$. The spectral resolution of $a$ is the one-parameter family of projections $\{p_{\lambda}:\lambda\in{\mathbb{R}}\}$ given by $p_{\lambda}:=((a-\lambda)^{+})^{\text{\rm o}})^{\perp}\in CC(a)$ for all $\lambda\in{\mathbb{R}}$. (Recall that $\lambda$ is identified with $\lambda 1\in A$.) The spectral lower and upper bounds for $a$ are defined by $L:=\sup\{\lambda\in{\mathbb{R}}:\lambda\leq a\}$ and $U:=\inf\{\lambda\in{\mathbb{R}}:a\leq\lambda\}$, respectively. By [4, Theorem 3.1], we have $-\infty<L\leq a\leq U<\infty$ and $\|a\|=\max\{|L|,|U|\}$.

A real number $\rho$ belongs to the resolvent set of $a$ iff there is an open interval $I$ in ${\mathbb{R}}$ such that $\rho\in I$ and $p_{\lambda}=p_{\rho}$ for all $\lambda\in I$. The spectrum of $a$, in symbols $\operatorname{spec}(a)$, which is defined to be the complement in ${\mathbb{R}}$ of the resolvent set of $a$, is a nonempty closed and bounded subset of ${\mathbb{R}}$ with all of the expected properties.

2.2 Theorem**.**

Let $a\in A$ with $0<a$ and let $\{p_{\lambda}:\lambda\in{\mathbb{R}}\}$ be the spectral resolution of $a$. For $\lambda\in{\mathbb{R}}$, define $q_{\lambda}:=1-p_{\lambda}=((a-\lambda)^{+})^{\text{\rm o}}$. Then: (i)* $q_{\lambda}\in CC(a)\subseteq C(a)$. (ii) If $\lambda<0$, then $q_{\lambda}=1$. (iii) $0<q_{0}=a^{\text{\rm o}}$. (iv) If $0<\lambda<\|a\|$, then $0<q_{\lambda}$ and $\lambda q_{\lambda}\leq q_{\lambda}a=aq_{\lambda}\leq a$. (v) If $\lambda\geq\|a\|$, then $q_{\lambda}=0$.*

Proof.

Part (i) follows from [3, Theorem 8.4 (i)]. Let $L$ and $U$ be the lower and upper spectral bounds for $a$. Since $0<a$, we have $0\leq L$, whence if $\lambda<0$, then $\lambda<L$, and (ii) then follows from [3, Theorem 8.4 (vi)]. Since $0<a$, we have $a=a^{+}$, whence $q_{0}=((a-0)^{+})^{\text{\rm o}}=a^{\text{\rm o}}\geq 0$. Also, $a^{\text{\rm o}}\not=0$, else $a=0$, contradicting $0<a$, and we have (iii).

As $0\leq L\leq U$ and $\|a\|=\max\{|L|,|U|\}$, we have $U=\|a\|$. Suppose $0<\lambda<\|A\|=U$. Then by [3, Theorem 8.4 (ii)],

[TABLE]

Thus, if $q_{\lambda}=0$, then $a\leq\lambda$, so $-\lambda\leq a\leq\lambda$, whence $\|a\|\leq\lambda$, contradicting $\lambda<\|a\|$. Therefore, $0<q_{\lambda}$, and by (1), we have $\lambda q_{\lambda}\leq q_{\lambda}a=aq_{\lambda}$. Since $(1-q_{\lambda})Ca$ and $0\leq 1-q_{\lambda},a$, it follows that $0\leq(1-q_{\lambda})a$, whence we also have $q_{\lambda}a\leq a$, and therefore (iv) holds. Finally, (v) follows from [3, Theorem 8.4 (v)]. ∎

2.3 Corollary**.**

If $0<a\in A$, then there exists $0<\lambda\in{\mathbb{R}}$ and $0<p\in P$ such that $\lambda p\leq a$.

Proof.

Choose $\lambda\in{\mathbb{R}}$ with $0<\lambda<\|a\|$, and in Theorem 2.2 let $p=q_{\lambda}$. ∎

An element $s\in A$ such that $s^{2}=1$ is called a symmetry, and two elements $a,b\in A$ are said to be exchanged by the symmetry $s$ iff $sas=b$ (or, equivalently, iff $sbs=a$) [7]. An element $t\in A$ is called a partial symmetry iff $t^{2}=p\in P$, and $a$ and $b$ are exchanged by $t$ iff $tat=b$ and $tbt=a$. There is a bijective correspondence $s\leftrightarrow p$ between symmetries $s\in A$ and projections $p\in P$ given by $s=2p-1$ and $p=\frac{1}{2}(s+1)$.

2.4 Lemma**.**

Let $s\in A$ be a symmetry, $s\not=-1$, and $\lambda,\mu\in{\mathbb{R}}$. Then $\lambda s\leq\mu\Rightarrow\lambda\leq\mu$.

Proof.

Assume the hypotheses. Suppose $\lambda s\leq\mu$, and let $p=\frac{1}{2}(s+1)$ be the projection corresponding to $s$. Then $s=2p-1=p-p^{\perp}$, and since $s\not=-1$, we have $0<p$. Then $\lambda p-\lambda p^{\perp}\leq\mu$, so $\lambda p=(\lambda p-\lambda p^{\perp})p\leq\mu p$, i.e., $0\leq(\mu-\lambda)p$. If $\mu-\lambda<0$, than $(\mu-\lambda)p<0$, contradicting $0\leq(\mu-\lambda)p$, whence $0\leq\mu-\lambda$, so $\lambda\leq\mu$. ∎

A subset $S\subseteq A$ is called a sub-synaptic algebra of $A$ iff $S$ is a linear subspace of $A$, $1\in S$, and $S$ is closed under the formation of squares, square roots, carriers, and inverses, in which case $S$ is a synaptic algebra in its own right. For instance, if $M\subseteq A$, then $C(M)$ is a sub-synaptic algebra of $A$. In particular, if $M$ is a commutative subset of $A$, then $CC(M)$ is a commutative sub-synaptic algebra of $A$.

If $p\in P$, then $pAp:=\{pap:a\in A\}=\{a\in A:a=pa=ap\}$ is a linear subspace of $A$ that is closed under the formation of squares, square roots, carriers, and inverses; it is a synaptic algebra in its own right with $p$ as its unit element and with $\{prp:r\in R\}$ as its enveloping algebra. The OML of projections in $pAp$ is the interval $P[0,p]:=\{q\in P:q\leq p\}$. If $t$ is a partial symmetry in $A$ with $t^{2}=p$, then $t$ is a symmetry in $pAp$.

The center of $A$ is the commutative sub-synaptic algebra $C(A)$. If $C(A)={\mathbb{R}}$, then $A$ is called a factor. For instance, the self-adjoint part ${\mathcal{B}}^{sa}({\mathfrak{H}})$ of the unital C*∗*-algebra of all bounded linear operators on a Hilbert space ${\mathfrak{H}}$ is a factor.

2.5 Theorem**.**

The synaptic algebra $A$ is a factor iff the only projections in $C(A)$ are [math] and $1$.

Proof.

If $C(A)={\mathbb{R}}$, then clearly the only projections in $C(A)$ are [math] and $1$. Conversely, suppose that the only projections in $C(A)$ are [math] and $1$ and let $a\in C(A)$. By [3, Theorem 8.10], the spectral resolution $\{p_{\lambda}:\lambda\in{\mathbb{R}}\}$ of $a$ is contained in $C(A)\cap P$, whence $p_{\lambda}\in\{0,1\}$ for all $\lambda\in{\mathbb{R}}$. Therefore, by [3, Theorem 8.4 (iii) and (vii)], there exists $\alpha\in{\mathbb{R}}$ such that $p_{\lambda}=0$ for $\lambda<\alpha$ and $p_{\lambda}=1$ for $\alpha\leq\lambda$, and it follows that $\operatorname{spec}(a)=\{\alpha\}$. Consequently, by [3, Theorem 8.9], $a=\alpha 1=\alpha$. ∎

3 Two commuting projections

3.1 Lemma**.**

[18, Lemma 2]* If $p,q\in P$, then $p\wedge q$ is also the infimum of $p$ and $q$ in $A^{+}$.*

Proof.

Clearly, $p\wedge q\in A^{+}$. Suppose that $a\in A^{+}$ and $a\leq p,q$. Then $0\leq a\leq p\leq 1$, so $a\in E$, and since $p\wedge q$ is also the infimum of $p$ and $q$ in $E$ (Lemma 2.1), we have $a\leq p\wedge q$. Therefore, $p\wedge q$ is the infimum of $p$ and $q$ in $A^{+}$. ∎

3.2 Lemma**.**

(Cf. [18, Proof of Theorem 1])* Suppose that $V$ is a vector subspace of $A$; $p,q\in P\cap V$; $p\wedge q\in V$; there exists $w\in P\cap V$ with $p,q\leq w$; and $p\wedge_{V}q$ exists. Put $p_{1}:=p-p\wedge_{V}q$ and $q_{1}:=q-p\wedge_{V}q$. Then*:**

1. (i)

$p\wedge_{V}q=p\wedge q\in P\cap V$. 2. (ii)

$p_{1},q_{1}\in P\cap V$* with $p_{1}\wedge_{V}q_{1}=0$.* 3. (iii)

$p_{1}q_{1}=q_{1}p_{1}=0$. 4. (iv)

$p$* commutes with $q$.*

Proof.

Put $g:=p\wedge q\in P\cap V$ and $g_{V}:=p\wedge_{V}q\in V$. By Lemma 3.1, $g$ is the infimum of $p$ and $q$ in $A^{+}$; hence,

[TABLE]

Since $0\in V$ and $0\leq p,q$, it follows that $0\leq g_{V}$, hence by (1) with $a:=g_{V}$, we have $g_{V}\leq g$. Also, $g\in V$ and $g\leq p,q$, so $g\leq g_{V}$. Thus, $g_{V}=g$, and we have (i).

By (i), $g_{V}\in P\cap V$, and since $g_{V}\leq p$, it follows that $p_{1}=p-g_{V}\in P\cap V$. Likewise, $q_{1}=q-g_{V}\in P\cap V$. Evidently $0\leq p_{1},q_{1}\in V$. Suppose that $v\in V$ and $v\leq p_{1},q_{1}$. Then $v+g_{V}\in V$ with $v+g_{V}\leq p,q$, whence $v+g_{V}\leq g_{V}$, so $v\leq 0$, and we have $p_{1}\wedge_{V}q_{1}=0$. This completes the proof of (ii).

Since $p_{1}\leq p\leq w$ and $q_{1}\leq q\leq w$, we have $p_{1}-w,q_{1}-w\leq 0$, whence $p_{1}+q_{1}-w\leq p_{1},q_{1}$. Therefore, $p_{1}+q_{1}-w\leq p_{1}\wedge_{V}q_{1}=0$, whence $p_{1}\leq w-q_{1}$. As $w$ and $q_{1}$ are projections and $q_{1}\leq w$, it follows that $w-q_{1}$ is a projection, whereupon $p_{1}=p_{1}(w-q_{1})=p_{1}w-p_{1}q_{1}=p_{1}-p_{1}q_{1}$, so $p_{1}q_{1}=0$, and we have (iii).

By (i), we have $p_{1}=p-p\wedge q$, so $p\wedge q=p-p_{1}\leq 1-p_{1}$, and it follows that $(p\wedge q)Cp_{1}$. Likewise, $(p\wedge q)Cq_{1}$. By (iii), $p_{1}Cq_{1}$, and therefore $(p_{1}+p\wedge q)C(q_{1}+p\wedge q)$, i.e., $pCq$, and we have (iv). ∎

3.3 Theorem**.**

(Cf. [18, Theorem 1])* Two projections $p$ and $q$ in $A$ commute iff there exists a vector subspace $V$ of $A$ such that $p,\,q,\,p\wedge q\in V$, there is a projection $w\in V\cap P$ with $p,q\leq w$, and the infimum $p\wedge_{V}q$ in $V$ of the two projections exists.*

Proof.

In view of Lemma 3.2, we have only to prove that if $p$ and $q$ commute, then there is a vector subspace $V$ of $A$ with the indicated properties. It suffices to take $V:=CC(\{p,q\})$, noting that $V$ is commutative, and $a\in V\Rightarrow|a|,a^{\text{\rm o}}\in V$. Therefore, by [12, Theorem 5.11] $V$ is a lattice. ∎

3.4 Corollary**.**

Let $p,q\in P$. Then: (i)* If $p\wedge_{A}q$ exists, then $pq=qp$. (ii) If $p\wedge_{A}q=0$, then $pq=qp=0$.*

Proof.

Part (i) is an obvious consequence of Theorem 3.3. To prove (ii), assume that $p\wedge_{A}q=0$. Then in Lemma 3.2 with $V=A$, we have $p=p_{1}$ and $q=q_{1}$, whence $pq=0$ by part (iii) of the lemma. ∎

In the following theorem we generalize part (ii) of Corollary 3.4 to an arbitrary pair of elements $a,b\in A$. In [18, Corollary 8], Kadison obtains this result and its corollary for the special case in which $A$ is the self-adjoint part of a W*∗*-algebra.

3.5 Theorem**.**

If $a,b\in A$ and $a\wedge_{A}b=0$, then $ab=ba=0$.

Proof.

Assume the hypotheses. We have $0=a\wedge_{A}b\leq a,b$, i.e., $a,b\in A^{+}$. We shall use the notion of the generalized infimum of $a$ and $b$, defined and denoted by $a\sqcap b:=\frac{1}{2}(a+b-|a-b|)$ [12, §4]. By [12, Lemma 4.1 (i)], $a\sqcap b\leq a,b$, whence $a\sqcap b\leq a\wedge_{A}b=0$. Thus, since $a,b\in A^{+}$, $ab=ba=0$ by [12, Lemma 4.4 (ii)]. ∎

3.6 Corollary**.**

If $a,b\in A$, $c:=a\wedge_{A}b$ exists, $cCa$, and $cCb$, then $aCb$.

Proof.

Assume the hypotheses. Then $0\leq a-c,b-c$, and if $k\in A$ with $k\leq a-c,b-c$, it follows that $k+c\leq a,b$, whence $k+c\leq c$, so $k\leq 0$. Therefore $(a-c)\wedge_{A}(b-c)=0$, so by Theorem 3.5, $(a-c)(b-c)=(b-c)(a-c)=0$. Consequently, $ab=cb+ac-c^{2}=bc+ca-c^{2}=ba$. ∎

4 The antilattice theorem

In this section we prove our main theorem (Theorem 4.7) giving necessary and sufficient conditions for a synaptic algebra $A$ with a complete projection lattice $P$ to be an antilattice. In particular, Theorem 4.7 shows that a synaptic algebra with a complete projection lattice is an antilattice iff it is a factor. To begin with, we have the following.

4.1 Theorem**.**

If $A$ is an antilattice, then $A$ is a factor.

Proof.

Suppose that $A$ is an antilattice. It will be sufficient to show that the only projections in $C(A)$ are [math] and $1$ (Theorem 2.5). Let $p\in C(A)\cap P$. We claim that $p\wedge_{A}p^{\perp}$ exists and equals [math]. Obviously, $0\leq p,p^{\perp}$. Suppose $a\in A$ and $a\leq p,p^{\perp}$. As $p\in C(A)$, it follows that $a$ commutes with both $p$ and $p^{\perp}$, whence $pa\leq pp^{\perp}=0$ and $p^{\perp}a\leq p^{\perp}p=0$, so $a=(p+p^{\perp})a=pa+p^{\perp}a\leq 0$. Therefore, $p\wedge_{A}p^{\perp}=0$. Consequently, if $A$ is an antilattice and $p\in C(A)\cap P$, then either $p\leq p^{\perp}$ or $p^{\perp}\leq p$, i.e., either $p=0$ or $p=1$. ∎

4.2 Lemma**.**

If $P$ is a complete OML, $A$ is a factor, and $0<p,q\in P$ with $p\perp q$, then there exists a symmetry $t\in A$ such that $tpt\leq q$ or $tqt\leq p$. Moreover, $t\not=\pm 1$.

Proof.

Assume the hypotheses. According to [7, Lemma 8.5], There is a central projection $h\in C(A)\cap P$ and a symmetry $t\in A$ such that $ph$ and a subprojection of $q$ are exchanged by $t$ and $q(1-h)$ and a subprojection of $p$ are exchanged by $t$. Since $A$ is a factor, the central projection $h$ is either [math] or $1$ (Theorem 2.5), whence $tpt\leq q$ or $tqt\leq p$. If $t=\pm 1$, then $tpt=p$ and $tqt=q$, so $p\leq q$ or $q\leq p$, and since $p\perp q$, $p=0$ or $q=0$ contradicting $0<p,q$. ∎

4.3 Lemma**.**

Let $a,b\in A$ with $a\wedge_{A}b=0$ and suppose that $p,q\in P$, $0<\lambda,\mu\in{\mathbb{R}}$, $\lambda p\leq a$ and $\mu q\leq b$. Then $p\wedge_{A}q=pq=qp=0$.

Proof.

Assume the hypotheses and let $\kappa:=\min\{\lambda,\mu\}$. Suppose that $g\in A$ and $g\leq p,q$. Then $\kappa g\leq\kappa p\leq a$ and $\kappa g\leq\kappa q\leq b$, whence $\kappa g\leq 0$, and therefore $g\leq 0$. Thus $p\wedge_{A}q=0$, and by Corollary 3.4 (ii), $pq=qp=0$. ∎

4.4 Theorem**.**

Suppose that $A$ is not an antilattice. Then there are projections $p,q\in P$ with $p\perp q$, $0<p,q$, and $p\wedge_{A}q=pq=qp=0$.

Proof.

Since $A$ is not an antilattice, there exist $c,d\in A$ such that $c\not\leq d$, $d\not\leq c$, and $c\wedge_{A}d$ exists in $A$. Put $a:=c-c\wedge_{A}d$ and $b:=d-c\wedge_{A}d$. Obviously, $0\leq a,b$, and since $c\not\leq d$ and $d\not\leq c$, we have $0<a,b$. Suppose $k\in A$ with $k\leq a,b$. Then $k+c\wedge_{A}d\leq c,d$, whence $k+c\wedge_{A}d\leq c\wedge_{A}d$, and it follows that $k\leq 0$. Therefore, $a\wedge_{A}b=0$.

By Corollary 2.3, there exist projections $0<p,q$ and real numbers $0<\lambda,\mu$ such that $\lambda p\leq a$ and $\mu q\leq b$. Then by Lemma 4.3, $p\wedge_{A}q=pq=qp=0$, and since $pq=0$, we have $p\perp q$. ∎

4.5 Lemma**.**

Suppose that $p\in P$, $p\not=0,1$, and there exists a symmetry $s\in A$ that exchanges $p$ and $p^{\perp}$. Then there exists $k\in A$ with $k\leq p,p^{\perp}$ but $k\not\leq 0$.

Proof.

By the hypotheses, we have $0<p,p^{\perp}\in P$ and there is a symmetry $s\in A$ such that $sps=p^{\perp}$. Clearly, $s\not=\pm 1$. Put

[TABLE]

Note that $k=2s-\frac{25}{16}$, $sks=k$, and

[TABLE]

We have

[TABLE]

[TABLE]

Put $d:=\alpha p+\gamma s+\beta p^{\perp}$. Then by (2) and (3),

[TABLE]

[TABLE]

[TABLE]

whence $0\leq p-k$, so $k\leq p$. Therefore, $0\leq s(p-k)s=sps-k=p^{\perp}-k$, so $k\leq p^{\perp}$. But $2\not\leq\frac{25}{16}$, whence by Lemma 2.4 $2s\not\leq\frac{25}{16}$ and therefore $k=2s-\frac{25}{16}\not\leq 0$. ∎

4.6 Theorem**.**

Suppose that whenever $0<p,q\in P$ with $p\perp q$, there exists a symmetry $t$ in $A$ such that $tpt\leq q$ or $tqt\leq p$. Then $A$ is an antilattice.

Proof.

Assume the hypothesis. Aiming for a contradiction, we assume that $A$ is not an antilattice. By Lemma 4.4, there are projections $0<p,q\in P$ with $p\perp q$ and $p\wedge_{A}q=pq=qp=0$. Thus, by hypothesis, there exists a symmetry $t$ in $A$ such that $tpt\leq q$ or $tqt\leq p$. By relabeling if necessary, we can and do assume that $tpt\leq q$. Thus, $0<tpt\in P$ and $tpt\leq q\leq p^{\perp}$ so $tpt\perp p$. Therefore, $p(tpt)=(tpt)p=0$ and $p\vee tpt=p+tpt$. If $a\in A$ and $a\leq p,tpt$, then $a\leq p,q$, and it follows that $a\leq 0$; hence $p\wedge_{A}tpt=0$.

Now we are going to drop down to the synaptic algebra $A_{1}\subseteq A$ defined by $A_{1}:=(p+tpt)A(p+tpt)$ in which $u:=p+tpt$ is the unit element. The projection lattice of $A_{1}$ is the interval $P[0,u]=\{q\in P:q\leq u\}$ in $P$ and we have $0<p,tpt\in P[0,u]$ with $p(tpt)=(tpt)p=0$ and $p+tpt=p\vee tpt=u$, whence $tpt$ is the orthocomplement of $p$ in $A_{1}$. Clearly, $p\wedge_{A_{1}}tpt=0$. Put $s:=tp+pt$. Then $s$ is a partial symmetry in $A$ with $s^{2}=p+tpt$, and $s(p+tpt)=(p+ptp)s=s$, so $s$ is a symmetry in $A_{1}$. Moreover, $sps=tpt$. Applying Lemma 4.5 to the synaptic algebra $A_{1}$, we find that there exists $k\in A_{1}$ with $k\leq p,tpt$ but $k\not\leq 0$, contradicting $p\wedge_{A_{1}}tpt=0$. ∎

4.7 Theorem**.**

Suppose that the OML $P$ of projections in $A$ is complete. Then the following conditions are mutually equivalent:

1. (i)

$A$* is an antilattice.* 2. (ii)

$A$* is a factor.* 3. (iii)

If $0<p,q\in P$ with $p\perp q$, then there exists a symmetry $t$ in $A$ such that $tpt\leq q$ or $tqt\leq p$.

Proof.

Theorem 4.1 shows that (i) $\Rightarrow$ (ii), Lemma 4.2 shows that (ii) $\Rightarrow$ (iii), and Theorem 4.6 shows that (iii) $\Rightarrow$ (i). ∎