Modular Abbott Algebras

Abstract

This note adds to the investigation of Abbott algebras in relation to quantum logics (see the references below). We introduce a variety of modular Abbott algebras and show that they are isomorphic to the variety of modular quantum logics. We extend this isomorphism for the varieties endowed with a symmetric difference.

1. Introduction

Abbott algebras came into existence in an effort to shed light on certain questions regarding generalized mathematical logic. It was observed in [1] and more thoroughly studied in [2] that the theory of Abbott algebras with 0 can be built up in parallel with the theory of quantum logics (the theory of orthomodular lattices). This provides a possible link between formal mathematical logic and quantum logical structures. A certain novelty presented here is the introduction of modularity and symmetric difference in Abbott algebras.

Since the class of modular quantum logics constitutes an important part of theoretical physics (for example, the quantum logic $L\left({\mathbb{R}}^n\right)$ of linear subspaces of ${\mathbb{R}}^n$ or the logic $L\left({\mathrm{II}}_1\right)$ of projections in type-${\mathrm{II}}_1$ von Neumann algebras are modular [3]), it is desirable to consider an appropriate notion of modularity for Abbott algebras. The objective of this note is to initiate such an investigation.

Although we tried to present a self-contained text, we briefly recall the basic properties of the rather unusual calculus of Abbott algebras (the contents of this note slightly overlap those of [2,4]). It should be noted that readers acquainted with [2] are likely to find the text easier to follow.

2. Results

Let us begin by introducing a specific axiom for an algebra with one binary and one nullary operation.

Definition 1. 

Let $A(·,0)$ be an algebra where $·:A\times A\to A$ is a binary operation, and $0\in A$ is a nullary operation. Let us write $(a,b\in A)$

$$p(a,b)=\left(\left((a,0)·(b,0)\right)·(b,0)\right)·0.$$

Let us say that $A(·,0)$ has the modular property if

$$p\left(\left((a·c)·c\right),\left((a·b)·b\right)\right)=\left(a·p\left(c,\left((a·b)·b\right)\right)\right)·p\left(c,\left((a·b)·b\right)\right).$$

This condition will play a role in our study of Abbott algebras that follows (we omit the sign · in places, writing $ab$ instead of $a·b$). Obviously, this axiom is not as transparent as one would like to have, but, with advances in computer-assisted proofs in quantum theories (see [5,6], etc.), this matter could be simplified.

Definition 2. 

Let $A(·,0)$ be an algebra with a binary and a unary operation: · and 0. Then $A(·,0)$ is said to be an Abbott algebra, provided that the following conditions are satisfied $(a,b,c\in A)$:

1. 

$\left(ab\right)a=a$;

2. 

$\left(ab\right)b=\left(ba\right)a$;

3. 

$a\left(\left(ba\right)c\right)=ac$;

4. 

$0a=bb$.

We adopt the definition from [4], adding only the nullary operation 0 to adapt it for applications in quantum theory. It may be noted that the axioms of Abbott algebras are a kind of Lindenbaum–Tarski algebra of generalized propositional calculus (see also [4,7]). The rules of inference then reduce to

• $(a\to b)\to a=a$;
• $\left((a\to b)\to b\right)=(b\to a)\to a$;
• $a\to \left((b\to a)\to c\right)=a\to c$.

These equations (together with the falsity symbol) motivate the axioms of Abbott algebras.

Let us exhibit some examples of Abbott algebras. An obvious example is Boolean algebra. If $(X,\mathcal{B})$ is its Stone set-representation, then $0=\varnothing$, and for $A,B\in \mathcal{B}$, we set $AB=X\setminus (A\setminus B)$. Another natural example is a horizontal sum of Boolean algebras [8]. This is a disjoint union of Boolean algebras identified on $0s$ and $1s$; see a simple example of this in Figure 1.

The operation is $ab={(a\vee b)}^{\prime }\vee b$ (it will be shown later that this operation converts each orthomodular lattice into an Abbott algebra). Several other examples will be exhibited in the following text.

Let us first briefly review the basic properties of Abbott algebras (see [1,2,4]).

Proposition 1. 

Let $A(·,0)$ be an Abbott algebra, and let $a,b,c\in A$. Then

1. 

$aa=1$ (1 is a constant in $A(·,0)$);

2. 

$1a=a$;

3. 

$a1=1$;

4. 

$ab=ba\Rightarrow a=b$;

5. 

$a\left(ba\right)=1$;

6. 

$ab=1\Rightarrow a\left(bc\right)=ac$;

7. 

$ab=1\Rightarrow \left(bc\right)\left(ac\right)=1$.

Proof. 

Conditions 1–4 of Proposition 1 are direct consequences of the axioms of Definition 2. According to Definition 2, 3, it follows that by substituting $c=ba$ we obtain

$$a\left(\left(ba\right)\left(ba\right)\right)=a\left(ba\right).$$

Therefore, we derive

$$1=\left(a1\right)=a\left(\left(ba\right)\left(ba\right)\right)=a\left(ba\right).$$

This verifies the condition in Proposition 1, $5.$ Further, making use of Definition 2, 2 and Definition 2, 3, it follows that if $ab=1$, then $b=\left(ba\right)a$, and therefore

$$a\left(bc\right)=a\left(\left(\left(ba\right)a\right)c\right)=a\left(\left(ba\right)c\right)=ac.$$

This proves Proposition 1, $6.$ Finally, to check Proposition 1, 7, since $ab=1$, we infer that

$$\left(bc\right)\left(ac\right)=\left(bc\right)\left(a\left(bc\right)\right)=1.$$

□

Definition 3. 

Let $A(·,0)$ be an Abbott algebra. Let us call $A(·,0)$ a mod-Abbott algebra if $A(·,0)$ has the modular property of Definition 1.

In our first result, let us recall that by quantum logic, we mean an orthomodular lattice (see, e.g., [8,9,10,11]). Recalled briefly, by orthomodular lattice, we mean a lattice $A(\le {,}^{\prime },0,1)$ with respect to ≤, where ′: $A\to A$ is an orthocomplementation operation ($a\le b\Rightarrow b^{\prime }\le a^{\prime }, a^{″}=a, a\vee a^{\prime }=1, a\wedge a^{\prime }=0$, and the orthomodular law $(a\le b)\Rightarrow b=a\vee (a^{\prime }\wedge b)$ is valid.

As generally assumed, quantum logics are often interpreted as event structures of quantum experiments (see, e.g., [9,10,11]).

Theorem 1. 

Let $A(·,0)$ be an Abbott algebra with modular property. Let us set $a^{\prime }=a0$ and write $a\le b$ precisely when $ab=1$. Then ${A(\le , }^{\prime },0,1)$ becomes a modular quantum logic. Vice versa, if ${A(\le , }^{\prime },0,1)$ is a modular quantum logic, then $A(·,0)$ with the operation $ab={(a\vee b)}^{\prime }$ becomes an Abbott algebra.

Proof. 

It is easy to check that, if we set $a\le b$ precisely when $ab=1$, it follows that ≤ is a partial ordering (one uses the axioms of Definition 2 and Proposition 1, 7). Moreover, A becomes a lattice with respect to ≤ when we define $a\vee b=\left(ab\right)b$. We immediately see that $a\le a\vee b$ and $b\le a\vee b$. Moreover, if $a\le c$ and $b\le c$, then $ac=1$ and $bc=1$. By $ac=1$ and Proposition 1, 7, we infer that $\left(cb\right)\left(ab\right)=1$ (this remedies [4]). It gives us

$$\left(\left(ab\right)b\right)\left(\left(cb\right)b\right)=1$$

and therefore $a\vee b=\left(ab\right)b\le \left(cb\right)b=\left(bc\right)c\le 1c=c$. Hence $a\vee b\le c$, and therefore A is a lattice. Further, let us observe that if we define $a^{\prime }=a0$, then A becomes an orthocomplemented lattice. We see that

$${\left(a^{\prime }\right)}^{\prime }=\left(a0\right)0=a\vee 0=a.$$

Moreover, let $a\le b$. We have to check that $b^{\prime }\le a^{\prime }$. In other words, we must show that condition $a\le b$ implies $\left(b0\right)\le \left(a0\right)$. Indeed, if $\left(ab\right)=1$, then $\left(b0\right)\left(a0\right)=1$ (Proposition 1, 7), and this means that $b^{\prime }\le a^{\prime }$. Suppose finally that $A(·,0)$ has the modular property. We want to show that when $A(·,0)$ is viewed as a lattice (with respect to ≤ defined above), then $A(\le )$ fulfills the modular law.

$$a\le b\Rightarrow (a\vee c)\wedge b=a\vee (c\wedge b).$$

Our assumption is that

$$p\left(\left(ac\right)c,\left(ab\right)b\right)=\left(a\left(p\left(c,\left(ab\right)b\right)\right)\right)p\left(c,\left(ab\right)b\right),$$

where

$$p(r,s)=\left(\left(\left(r0\right)\left(s0\right)\right)\left(s0\right)\right)0,(a,b,c,r,s\in A).$$

Considering the left-hand side, we obtain the following by using de Morgan’s law:

$$\begin{array}{cc}\hfill p\left(\left(ac\right)c,\left(ab\right)b\right)& =p(a\vee c,a\vee b)\hfill \\ & =\left(\left((a\vee c)0\right)\vee \left((a\vee b)0\right)\right)0\hfill \\ & ={\left((a^{\prime }\wedge c^{\prime })\vee (a^{\prime }\wedge b^{\prime })\right)}^{\prime }\hfill \\ & =(a\vee c)\wedge (a\vee b).\hfill \end{array}$$

Regarding the right-hand side, we obtain

$$\begin{array}{cc}\hfill \left(a\left(p\left(c,\left(ab\right)b\right)\right)\right)p\left(c,\left(ab\right)b\right)& =a\vee p\left(c,a\vee b\right)\hfill \\ & =a\vee \left(\left(c0\right)\vee \left((a\vee b)0\right)0\right)\hfill \\ & =a\vee {\left(c^{\prime }\vee {(a\vee b)}^{\prime }\right)}^{\prime }\hfill \\ & =a\vee \left(c\wedge (a\vee b)\right).\hfill \end{array}$$

So we derived the classical modular identity

$$(a\vee c)\wedge (a\vee b)=a\vee \left(c\wedge (a\vee b)\right)$$

for any $a,b,c\in A$. Obviously, supposing that $a\le b$, we obtain

$$(a\vee c)\wedge b=a\vee (c\wedge b),$$

and this is the modular law.

In order to show the opposite implication, let us suppose that $A(\le {,}^{\prime },0,1)$ is a modular quantum logic. We want to show that if we set $ab={(a\vee b)}^{\prime }\vee b$, we will make $A(\le {,}^{\prime },0,1)$ an Abbott algebra. The only technically complicated verification is the validity of Property 3 of Definition 2. This property is of certain importance within orthomodular structures in its own right. Let us prove it by a Proposition 2, adding thus to the paper [4] (see also [2]).

Proposition 2. 

Let $A(\le {,}^{\prime },0,1)$ be an othomodular lattice. Let us set $ab={(a\vee b)}^{\prime }\vee b$. Then for each $a,b,c\in A$, we infer that the equality $a\left(\left(ba\right)c\right)=ac$ holds true.

Proof. 

We have the equalities

$$\begin{array}{cc}\hfill & a\left(\left(ba\right)c\right)=a\left(\left({(b\vee a)}^{\prime }\vee a\right)c\right)=a\left(\left((b^{\prime }\wedge a^{\prime })\vee a\right)c\right).\hfill \end{array}$$

Write $y=b^{\prime }\wedge a^{\prime }$.

Then we have

$$\begin{array}{cc}\hfill & a\left(\left(ba\right)c\right)\hfill \\ \hfill & =a\left((y\vee a)c\right)\hfill \\ \hfill & =a\left({\left((y\vee a)\vee c\right)}^{\prime }\vee c\right)\hfill \\ \hfill & ={\left(a\vee \left({(y\vee a)}^{\prime }\wedge c^{\prime }\right)\vee c\right)}^{\prime }\vee \left(\left({(y\vee a)}^{\prime }\wedge c^{\prime }\right)\vee c\right)\hfill \\ \hfill & =\left((a^{\prime }\wedge c^{\prime })\wedge {\left({(y\vee a)}^{\prime }\wedge c^{\prime }\right)}^{\prime }\right)\vee \left((y^{\prime }\wedge a^{\prime }\wedge c^{\prime })\vee c\right)\hfill \\ \hfill & =\left((a^{\prime }\wedge c^{\prime })\wedge (y\vee a\vee c)\vee (y^{\prime }\wedge a^{\prime }\wedge c^{\prime })\right)\vee c.\hfill \end{array}$$

So, we obtained

$$\begin{array}{cc}\hfill & a\left(\left(ba\right)c\right)\hfill \\ \hfill & =\left((a^{\prime }\wedge c^{\prime })\wedge \left(\left((b^{\prime }\wedge a^{\prime })\vee a\right)\vee c\right)\right)\vee \left(\left({(b^{\prime }\wedge a^{\prime })}^{\prime }\wedge a^{\prime }\right)\wedge c^{\prime }\right)\vee c.\hfill \end{array}$$

Observing that $a^{\prime }\wedge a^{\prime }\ge \left({(b^{\prime }\wedge a^{\prime })}^{\prime }\wedge (a^{\prime }\wedge c^{\prime })\right)$, we use the orthomodular law to obtain

$$\left((a^{\prime }\wedge c^{\prime })\wedge \left(\left((b^{\prime }\wedge a^{\prime })\vee a\right)\vee c\right)\right)\vee \left(\left({(b^{\prime }\wedge a^{\prime })}^{\prime }\wedge a^{\prime }\right)\wedge c^{\prime }\right)=a^{\prime }\wedge c^{\prime }$$

As a result, we have

$$\begin{array}{cc}\hfill & a\left(\left(ba\right)c\right)=(a^{\prime }\wedge c^{\prime })\vee c=ac\hfill \end{array}$$

This completes the proof. □

The previous result can be strengthened as follows:

Theorem 2. 

The variety of mod-Abbott algebras and the variety of modular quantum logics are isomorphic.

Proof. 

Let $A(·,0)$ be an Abbott algebra with modular property. According to the previous theorem, we can assign a modular quantum logic ${A(\le ,}^{\prime },0,1)$ to $A(·,0)$, whose underlying set is A; the relation $a\le b$ is true precisely when $ab=1$ and $a^{\prime }=a0$. In the opposite direction, if ${A(\le ,}^{\prime },0,1)$ is a modular quantum logic, we assign $A(·,0)$ to it with $ab={(a\vee b)}^{\prime }\vee b$. The only technical property of the proof is the fact that the operation $ab$ defined in the above-stated manner fulfills the condition $a\left(\left(ba\right)c\right)=ac$ of Definition 2, 3. This was achieved in general in Proposition 2 (in the modular setup, Proposition 2 is even easier to check). As known, the modular law implies the modular identity. One uses the modular law starting with the inequalities

$$a\le (a\vee c)\wedge (a\vee b)\le (a\vee b),$$

see, e.g., [12]. This implies that $A(·,0)$ has the modular property.

Regarding morphisms, if $f:A(·,0)\to B(·,0)$ is a morphism in Abbott algebras, then $f:A(\wedge ,\vee ,0,1)\to B(\wedge ,\vee ,0,1)$ is a morphism in orthomodular lattices. Indeed, if

$$f(a\vee b)=f\left(\left(ab\right)b\right)=\left(f\left(a\right)f\left(b\right)\right),f\left(b\right)=f\left(a\right)\vee f\left(b\right).$$

Further,

$$f\left(a^{\prime }\right)=f\left(a0\right)=f\left(a\right)f\left(0\right)=f\left(a\right)0=f{\left(a\right)}^{\prime }.$$

Conversely, if $g:A(·,0)\to B(·,0)$ is a morphism of orthomodular lattices, then $g:A(·,0)\to B(·,0)$ is a morphism of Abbott algebras. Indeed,

$$g\left(ab\right)=g\left({(a\vee b)}^{\prime }\vee b\right)={\left(g\left(a\right)\vee g\left(b\right)\right)}^{\prime }\vee g\left(b\right)=g\left(a\right)g\left(b\right).$$

The proof is complete. □

With the interplay of Theorem 2.3 of [2], we can extend the previous result to mod-Abbott algebras with an XOR operation. Let us recall another notion of quantum logics (see [13]).

Definition 4. 

Let us consider the tuple $(L,\wedge ,\vee ,0,1{,}^{\prime },\Delta )$, where $(L,\wedge ,\vee ,0,1{,}^{\prime })$ is an orthomodular lattice, and the binary operation Δ fulfills the following properties $(a,b,c\in L)$:

1. 

$a\Delta (b\Delta c)=(a\Delta b)\Delta c$;

2. 

$1\Delta a=a^{\prime },a\Delta 1=a^{\prime }$;

3. 

$a\Delta b\le a\vee b$.

Then $(L,\wedge ,\vee ,0,1{,}^{\prime },\Delta )$ is said to be a quantum logic with a difference operation.

The investigation of when quantum logic can be endowed with a symmetric difference operation $\Delta$ has been presented in [13,14]. The recent intense research on an abstract symmetric difference can be seen in [5,15,16,17,18,19]. The intrinsic structure is then considerably richer.

In the next notion, we will introduce the axioms of a symmetric difference operation in Abbott algebras. It models a kind of logical XOR operation.

Definition 5. 

Let $A(.,0,\Delta )$ be an Abbott algebra with binary operation $\Delta :A\times A\to A$ that satisfies the following conditions $(a,b,c\in A)$:

1. 

$(a\Delta b)\Delta c=a\Delta (b\Delta c)$;

2. 

$bb\Delta a=a0$,$a\Delta bb=a0$;

3. 

$(a\Delta b)\left(\left(ab\right)b\right)=aa$.

Then $A(.,0,\Delta )$ is said to be an XOR Abbott algebra.

Theorem 3. 

The variety of XOR mod-Abbott algebras is equivalent to the variety of modular quantum logics with a symmetric difference.

Proof. 

The properties of $\Delta$ are easy to verify:

• Obvious;
• $1\Delta a=\left(bb\right)\Delta a=a0=a^{\prime }$, $a\Delta 1=a\Delta \left(bb\right)=a0=a^{\prime }$;
• $(a\Delta b)\left(\left(ab\right)b\right)=aa\Rightarrow (a\Delta b)\left(\left(ab\right)b\right)=1\Rightarrow (a\Delta b)\le (a\vee b)$.

□

Let us present two natural examples of XOR Abbott algebras with (or without) the modular condition.

Example 1. 

Let $S=\{1,2,3,4\}$. Let A be the collection of all subsets of S with an even cardinality. Define the operation $ab={(a\wedge b)}^{\prime }\vee b$, where the supremum ∨ is introduced with respect to the inclusion ordering. Then $A(.,\varnothing )$ is an XOR Abbott algebra with a modular condition. This algebra generates a variety of XOR mod-Abbott algebras (in particular, the Cartesian products seem relevant to the application in quantum theories).

Example 2. 

Let B be a proper Boolean subalgebra of a Boolean algebra, C. For each $c\in C\setminus B$, let us take the four-element Boolean subalgebra of C, $B_c=\{0,1,c,c^{\prime }\}$. Let us take the horizontal sum H of B and all algebras $B_c,c\in C\setminus B$ (see [8]). Then H with the operation $ab={(a\vee b)}^{\prime }\vee b$ and with a naturally defined Δ inherited from C forms an XOR Abbott algebra. Moreover, H has the modular condition exactly when $\mathrm{card}B\le 4$ (if $\mathrm{card}B>4$, then H obviously contains a five-element non-modular lattice, disproving the modularity of H).

Let H be a horizontal sum of four-element Boolean algebras. So $H={MO}_k$. Then H is a modular quantum logic and can be converted to a modular quantum logic with a symmetric difference exactly when $k=2^{n-1}$ or when k is infinite (see [13]). It follows that, although modular quantum logics with a symmetric difference (and therefore XOR mod-Abbott algebras) can be viewed as “almost Boolean”, they still allow for an arbitrarily high degree of non-compatibility and therefore remain relevant to quantum theories.