A maximal Boolean sublattice that is not the range of a Banaschewski function

TL;DR

The paper constructs a specific countable bounded sublattice of a vector space's subspace lattice, demonstrating that not all maximal Boolean sublattices are ranges of Banaschewski functions, thus solving an open problem.

Contribution

It provides a counterexample of a maximal Boolean sublattice not arising from a Banaschewski function, addressing a question posed by Wehrung.

Findings

Constructed a countable bounded sublattice with two non-isomorphic maximal Boolean sublattices.

One sublattice is the range of a Banaschewski function, the other is not.

Resolved an open problem in lattice theory regarding Banaschewski functions.

Abstract

We construct a countable bounded sublattice of the lattice of all subspaces of a vector space with two non-isomorphic maximal Boolean sublattice. We represent one of them as the range of a Banschewski function and we prove that this is not the case of the other. Hereby we solve a problem of F. Wehrung.

Equations

$$a\wedge b=a\wedge c=b\wedge c$$

$$\mu ⟨a,b,c⟩=(a\wedge b)\vee (a\wedge c)\vee (b\wedge c)$$

$$\overline{⟨a,b,c⟩}=⟨a\vee \mu ⟨a,b,c⟩,b\vee \mu ⟨a,b,c⟩,c\vee \mu ⟨a,b,c⟩⟩.$$

$$\begin{array}[]{ll}a\vee\mu{\left\langle a,b,c\right\rangle}&=a\vee(b\wedge c),\\ b\vee\mu{\left\langle a,b,c\right\rangle}&=b\vee(a\wedge c),\\ c\vee\mu{\left\langle a,b,c\right\rangle}&=c\vee(a\wedge b).\end{array}$$

$$⟨a,b,c⟩\vee ⟨a^{\prime },b^{\prime },c^{\prime }⟩=\overline{⟨a\vee a^{\prime },b\vee b^{\prime },c\vee c^{\prime }⟩}$$

$$⟨a,b,c⟩\wedge ⟨a^{\prime },b^{\prime },c^{\prime }⟩=⟨a\wedge a^{\prime },b\wedge b^{\prime },c\wedge c^{\prime }⟩.$$

$$\mathcal{F}(\kappa ):=\{X\subseteq \kappa \mid X\text{ is finite or }\kappa \setminus X\text{ is finite}\}.$$

$$\mathbf{T}=\{⟨A,B,C⟩\in \mathcal{F}(\kappa {)}^3\mid C\setminus \mu ⟨A,B,C⟩\text{ is finite}\}.$$

$$\mu ⟨A\cup A^{\prime },B\cup B^{\prime },C\cup C^{\prime }⟩\supseteq \mu ⟨A,B,C⟩\cup \mu ⟨A^{\prime },B^{\prime },C^{\prime }⟩$$

$$\displaystyle\big{(}C\cup C^{\prime}\big{)}\setminus\mu{\left\langle A\cup A^{\prime},B\cup B^{\prime},C\cup C^{\prime}\right\rangle}$$

$$\subseteq$$

$$\big{(}C\cup\mu{\left\langle A,B,C\right\rangle}\big{)}\setminus\mu{\overline{\left\langle A,B,C\right\rangle}}\subseteq C\setminus\mu{\left\langle A,B,C\right\rangle},$$

$$\mu ⟨A\cap A^{\prime },B\cap B^{\prime },C\cap C^{\prime }⟩=\mu ⟨A,B,C⟩\cap \mu ⟨A^{\prime },B^{\prime },C^{\prime }⟩.$$

$$\displaystyle\big{(}C\cap C^{\prime}\big{)}\setminus\mu{\left\langle A\cap A^{\prime},B\cap B^{\prime},C\cap C^{\prime}\right\rangle}$$

$$=$$

$$\subseteq$$

$$|\mathcal{F}(\kappa )|\le |\mathbf{S}|\le |\mathcal{F}(\kappa {)}^3|.$$

$$f⟨A,B,C⟩:=⟨\kappa \setminus A,\kappa \setminus (B\cup C),\kappa \setminus (A\cup B\cup C)⟩,\text{for all}⟨A,B,C⟩\in \mathbf{S},$$

$$A^{\prime }\cap B^{\prime }=\mu ⟨A^{\prime },B^{\prime },A^{\prime }\cap B^{\prime }⟩=\mu f⟨A,B,C⟩.$$

$$1_{\mathbf{S}}=⟨\kappa ,\kappa ,\kappa ⟩=⟨A,B,C⟩\oplus f⟨A,B,C⟩,\text{for all}⟨A,B,C⟩\in \mathbf{S}.$$

$$⟨A,B,C⟩\wedge f⟨A,B,C⟩=⟨\varnothing ,\varnothing ,\varnothing ⟩=0_{\mathbf{S}}.$$

$$\kappa =\mu ⟨A\cup (\kappa \setminus A),B\cup (\kappa \setminus (B\cup C)),C\cup (\kappa \setminus (A\cup B\cup C))⟩.$$

$$\mathbf{E}=\{⟨A,B,A\cap B⟩\mid A,B\in \mathcal{F}(\kappa )\}$$

$$⟨A,B,A\cap B⟩\mapsto ⟨A,B⟩$$

$$\mathbf{E}\subseteq \{⟨A,B,C⟩\in \mathbf{S}\mid C=A\cap B\}=\{⟨A^{\prime },B^{\prime },A^{\prime }\cap B^{\prime }⟩\mid A^{\prime },B^{\prime }\in \mathcal{F}(\kappa )\}.$$

$$g⟨A,C⟩\wedge g⟨A^{\prime },C^{\prime }⟩=g⟨A\cap A^{\prime },C\cap C^{\prime }⟩$$

$$g⟨A,C⟩\vee g⟨A^{\prime },C^{\prime }⟩=g⟨A\cup A^{\prime },C\cup C^{\prime }⟩$$

$$\mathbf{A}=\{⟨A,C⟩\in \mathcal{F}(\kappa )\mid A\sim C\}$$

$$\mathbf{B}=\{⟨A,B,C⟩\in \mathbf{S}\mid A\sim C\text{ and }B\subseteq A\}.$$

$$\mu ⟨A\cup A^{\prime },B\cup B^{\prime },C\cup C^{\prime }⟩\subseteq B\cup B^{\prime }\cup C\cup C^{\prime }$$

$$\varnothing =F\cap A=F\cap B\cap A=F\cap B\cap C=F\cap C.$$

Full text

The first author was partially supported by the project SVV-2015-260227 of Charles University in Prague. The second author was partially supported by the Grant Agency of the Czech Republic under the grant no. GACR 14-15479S

1. Introduction

In [14] Friedrich Wehrung defined a Banaschewski function on a bounded complemented lattice ${\boldsymbol{L}}$ as an antitone (i.e., order-reversing) map sending each element of ${\boldsymbol{L}}$ to its complement, being motivated by the earlier result of Bernhard Banaschewski that such a function exists on the lattice of all subspaces of a vector space [1]. Wehrung extended Banaschewski’s result by proving that every countable complemented modular lattice has a Banaschewski function with a Boolean range and that all the possible ranges of Banaschewski functions on ${\boldsymbol{L}}$ are isomorphic [14, Corollary 4.8].

Still in [14] Wehrung defined a ring-theoretical analogue of Banaschewski function that, for a von Neuman regular ring $R$, is closely connected to the lattice-theoretical Banaschewski function on the lattice ${\boldsymbol{L}}(R)$ of all finitely generated right ideals of $R$. He made use of these ideas to construct a unit-regular ring $S$ (in fact of bounded index $3$) of size $\aleph_{1}$ with no Banaschewski function [15].

Furthermore in [14] Wehrung defined notions of a Banaschewski measure and a Banaschewski trace on sectionally complemented modular lattices and he proved that a sectionally complemented lattice which is either modular with a large $4$-frame or Arguesian with a large $3$-frame is coordinatizable (i.e. isomorphic to ${\boldsymbol{L}}(R)$ for a possibly non-unital von Neumann regular ring $R$) if and only if it has a Banaschewski trace. Applying this results, he constructed a non-coordinatizable sectionally complemented modular lattice, of size $\aleph_{1}$, with a large $4$-frame [14, Theorem 7.5].

The aim of our paper is to solve the second problem from [14]:

Problem** (Problem 2 from [14]).**

Is every maximal Boolean sublattice of an at most countable complemented modular lattice ${\boldsymbol{L}}$ the range of some Banaschewski function on ${\boldsymbol{L}}$? Are any two such Boolean sublattices isomorphic?

We construct a countable complemented modular lattice ${\boldsymbol{S}}$ with two non-isomorphic maximal Boolean sublattices ${\boldsymbol{B}}$ and ${\boldsymbol{E}}$. We represent ${\boldsymbol{E}}$ as the range of a Banaschewski function on ${\boldsymbol{S}}$ and we prove that ${\boldsymbol{B}}$ is not the range of any Banaschewski function. Finally we represent the lattice ${\boldsymbol{S}}$ as a bounded sublattice of the subspace-lattice of a vector space.

2. Basic concepts

We start with recalling same basic notions as well as the precise definition of the Banaschewski function adopted from [14]. Next we outline the Schmidt’s ${\boldsymbol{M}}_{3}[{\boldsymbol{L}}]$ construction, which we then apply to define the bounded modular lattice ${\boldsymbol{S}}$ containing a pair of non-isomorphic maximal Boolean sublattices.

2.1. Some standard notions, notation, and the Banaschewski function

A lattice ${\boldsymbol{L}}$ is bounded if it has both the least element and the greatest element, denoted by $0_{\boldsymbol{L}}$ and $1_{\boldsymbol{L}}$, respectively. A bounded sublattice of a bounded lattice is its sublattice containing the bounds. Given elements $a,b,c$ of a lattice ${\boldsymbol{L}}$ with zero, we will use the notation $c=a\oplus b$ when $a\wedge b=0_{\boldsymbol{L}}$ and $a\vee b=c$. A complement of an element $a$ of a bounded lattice ${\boldsymbol{L}}$ is an element $a^{\prime}$ of ${\boldsymbol{L}}$ such that $a\oplus a^{\prime}=1_{\boldsymbol{L}}$. A lattice ${\boldsymbol{L}}$ is said to be complemented provided that it is bounded and each element of ${\boldsymbol{L}}$ has a (not necessarily unique) complement. A lattice ${\boldsymbol{L}}$ is relatively complemented if each of its interval is complemented. Note that a relatively complemented lattice is not necessarily bounded.

We say that a lattice ${\boldsymbol{L}}$ is uniquely complemented if it is bounded and each element of ${\boldsymbol{L}}$ has a unique complement. By a Boolean lattice we mean a lattice reduct of a Boolean algebra, that is, a distributive uniquely complemented lattice. For the clarity, let us recall the formal definition of the Banaschewski function [14, Definition 3.1]:

Definition 2.1**.**

Let ${\boldsymbol{L}}$ be a bounded lattice. A Banaschewski function on ${\boldsymbol{L}}$ is a map $f\colon{\boldsymbol{L}}\to{\boldsymbol{L}}$ such that both

1. (1)

$x\leq y$ implies $f(x)\geq f(y)$, for all $x,y\in{\boldsymbol{L}}$, and 2. (2)

$f(x)\oplus x=1_{{\boldsymbol{L}}}$ for all $x\in{\boldsymbol{L}}$,

hold true.

2.2. The ${\boldsymbol{M}}_{3}[{\boldsymbol{L}}]$-construction.

Let ${\boldsymbol{L}}$ be a lattice. We will call a triple $\left\langle a,b,c\right\rangle\in{\boldsymbol{L}}^{3}$ balanced, if it satisfies

[TABLE]

and we denote by ${\boldsymbol{M}}_{3}[{\boldsymbol{L}}]$ the set of all balanced triples. It is readily seen that ${\boldsymbol{M}}_{3}[{\boldsymbol{L}}]$ is a meet-subsemilattice of the cartesian product ${\boldsymbol{L}}^{3}$. However, it is not necessarily a join-subsemilattice, for one easily observes that the join of balanced triples may not be balanced. The ${\boldsymbol{M}}_{3}[{\boldsymbol{L}}]$-construction was introduced by E. T. Schmidt [12, 13] for a bounded distributive lattices ${\boldsymbol{L}}$. He proved [13, Lemma 1] that in this case ${\boldsymbol{M}}_{3}[{\boldsymbol{L}}]$ is a bounded modular lattice and that it is a congruence-preserving extension of the distributive lattice ${\boldsymbol{L}}$. This result was later extended by Grätzer and Schmidt in various directions [2, 3]. In particular, in [2] they proved that every lattice with a non-trivial distributive interval has a proper congruence-preserving extension. This was further improved by Grätzer and Wehrung in [7], where they introduced a modification of the ${\boldsymbol{M}}_{3}[{\boldsymbol{L}}]$-construction, called ${\boldsymbol{M}}_{3}\langle{\boldsymbol{L}}\rangle$-construction. Using this new idea they proved that every non-trivial lattice admits a proper congruence-preserving extension.

The lattice constructions ${\boldsymbol{M}}_{3}[{\boldsymbol{L}}]$ and ${\boldsymbol{M}}_{3}\langle{\boldsymbol{L}}\rangle$ appeared in the series of papers by Grätzer and Wehrung [4, 5, 6, 7, 8, 9, 10] dealing with semilattice tensor product and its related structures, namely the box product and the lattice tensor product [6, Definition 2.1 and definition 3.3]. Indeed, ${\boldsymbol{M}}_{3}\boxtimes{\boldsymbol{L}}\simeq{\boldsymbol{M}}_{3}\langle{\boldsymbol{L}}\rangle$ for every lattice ${\boldsymbol{L}}$ and ${\boldsymbol{M}}_{3}\otimes{\boldsymbol{L}}\simeq{\boldsymbol{M}}_{3}[{\boldsymbol{L}}]$ whenever ${\boldsymbol{L}}$ has zero and ${\boldsymbol{M}}_{3}\otimes{\boldsymbol{L}}$ is a lattice (see [10, Theorem 6.5] and [5, Corollary 6.3]). In particular, the latter is satisfied when the lattice ${\boldsymbol{L}}$ is modular with zero. Note also, that if ${\boldsymbol{L}}$ is a bounded distributive lattice both the constructions ${\boldsymbol{M}}_{3}[{\boldsymbol{L}}]$ and ${\boldsymbol{M}}_{3}\langle{\boldsymbol{L}}\rangle$ coincide. In our paper we get by with this simple case.

Let ${\boldsymbol{L}}$ be a distributive lattice. Given a triple $\left\langle a,b,c\right\rangle\in{\boldsymbol{L}}^{3}$, we define

[TABLE]

and we set

[TABLE]

Using the distributivity of ${\boldsymbol{L}}$ one easily sees that $\overline{\left\langle a,b,c\right\rangle}$ is the least balanced triple $\geq\left\langle a,b,c\right\rangle$ in ${\boldsymbol{L}}^{3}$ and that the map $\overline{\left\langle-\right\rangle}\colon{\boldsymbol{L}}^{3}\to{\boldsymbol{L}}^{3}$ determines a closure operator on the lattice ${\boldsymbol{L}}^{3}$ (see [14, Lemma 2.3] for a refinement of this observation). It is also clear that

[TABLE]

A triple $\left\langle a,b,c\right\rangle\in{\boldsymbol{L}}^{3}$ is closed with respect to the closure operator if and only if it is balanced. Therefore the set of all balanced triples, denoted by ${\boldsymbol{M}}_{3}[{\boldsymbol{L}}]$, forms a lattice [14, Lemma 2.1], where

[TABLE]

and

[TABLE]

By [5, Lemma 2.9] the lattice ${\boldsymbol{M}}_{3}[{\boldsymbol{L}}]$ is modular if and only if the lattice ${\boldsymbol{L}}$ is distributive. The “if” part of the equivalence is included in the above mentioned [13, Lemma 1].

3. The lattice

Fix an infinite cardinal $\kappa$. As it is customary, we identify $\kappa$ with the set of all ordinals of cardinality less than $\kappa$. Let us denote by $\mathcal{P}(\kappa)$ the Boolean lattice of all subsets of $\kappa$ and set

[TABLE]

It is well-known that $\mathcal{F}(\kappa)$ is a bounded Boolean sublattice of $\mathcal{P}(\kappa)$. Next, let us define

[TABLE]

Lemma 3.1**.**

The set ${\boldsymbol{T}}$ forms a bounded join-subsemilattice of $\mathcal{F}(\kappa)^{3}$.

Proof.

Being a lattice polynomial, the map $\mu\colon\mathcal{P}(\kappa)^{3}\to\mathcal{P}(\kappa)$ is monotone. It follows that for all $\left\langle A,B,C\right\rangle$, $\left\langle A^{\prime},B^{\prime},C^{\prime}\right\rangle\in\mathcal{P}(\kappa)$, the inclusion

[TABLE]

holds, whence also

[TABLE]

Thus if both $C\setminus\mu{\left\langle A,B,C\right\rangle}$ and $C^{\prime}\setminus\mu{\left\langle A^{\prime},B^{\prime},C^{\prime}\right\rangle}$ are finite, then $(C\cup C^{\prime})\setminus\mu{\left\langle A\cup A^{\prime},B\cup B^{\prime},C\cup C^{\prime}\right\rangle}$ is finite as well. It follows that ${\boldsymbol{T}}$ is join-subsemilattice of $\mathcal{F}(\kappa)^{3}$. Finally, it is clear that both $0_{\mathcal{F}(\kappa)^{3}}=\left\langle\emptyset,\emptyset,\emptyset\right\rangle$ and $1_{\mathcal{F}(\kappa)^{3}}=\left\langle\kappa,\kappa,\kappa\right\rangle$ belong to ${\boldsymbol{T}}$. ∎

Let ${\boldsymbol{S}}:={\boldsymbol{T}}\cap{\boldsymbol{M}}_{3}[\mathcal{F}(\kappa)]$ denote the set of all balanced triples from ${\boldsymbol{T}}$.

Lemma 3.2**.**

The join-semilattice ${\boldsymbol{T}}$ is closed under the $\overline{\left\langle-\right\rangle}$ operation.

Proof.

Let $\left\langle A,B,C\right\rangle\in{\boldsymbol{T}}$. Since $\mathcal{F}(\kappa)$ is a lattice, we have that all $A\cup\mu{\left\langle A,B,C\right\rangle}$, $B\cup\mu{\left\langle A,B,C\right\rangle}$ and $C\cup\mu{\left\langle A,B,C\right\rangle}$ belong to $\mathcal{F}(\kappa)$. Since the map $\mu\colon\mathcal{P}(\kappa)^{3}\to\mathcal{P}(\kappa)$ is monotone, the inclusion $\mu{\left\langle A,B,C\right\rangle}\subseteq\mu{\overline{\left\langle A,B,C\right\rangle}}$ holds. It follows that

[TABLE]

which is finite due to $\left\langle A,B,C\right\rangle$ being an element of ${\boldsymbol{T}}$. ∎

Lemma 3.3**.**

The set ${\boldsymbol{S}}$ forms a bounded sublattice of the lattice ${\boldsymbol{M}}_{3}[\mathcal{F}(\kappa)]$.

Proof.

Applying Lemmas 3.1 and 3.2, we deduce that ${\boldsymbol{S}}$ is a bounded join-subsemilattice of ${\boldsymbol{M}}_{3}[\mathcal{F}(\kappa)]$. Therefore, it suffices to verify that ${\boldsymbol{S}}$ is a meet-subsemilattice of $\mathcal{F}(\kappa)^{3}$. It is easy to observe that if at least one of $\left\langle A,B,C\right\rangle,\left\langle A^{\prime},B^{\prime},C^{\prime}\right\rangle\in\mathcal{P}(\kappa)^{3}$ is balanced, then

[TABLE]

¿From this we we get that if $\left\langle A,B,C\right\rangle,\left\langle A^{\prime},B^{\prime},C^{\prime}\right\rangle\in{\boldsymbol{S}}$, then

[TABLE]

so the set $(C\cap C^{\prime})\setminus\mu{\left\langle A\cap A^{\prime}.B\cap B^{\prime},C\cap C^{\prime}\right\rangle}$ is finite. This concludes the proof. ∎

As discussed in Section 2, since the lattice $\mathcal{F}(\kappa)$ is distributive, the lattice ${\boldsymbol{M}}_{3}[\mathcal{F}(\kappa)]$ is modular. Observe that the mapping $A\mapsto\left\langle A,A,A\right\rangle$ embeds $\mathcal{F}(\kappa)$ into ${\boldsymbol{S}}$, from which we deduce that

[TABLE]

Since the size of both $\mathcal{F}(\kappa)$ and $\mathcal{F}(\kappa)^{3}$ is $\kappa$, we get that $|{\boldsymbol{S}}|=\kappa$. Let us sum up these observations in the following corollary to Lemma 3.3.

Corollary 3.4**.**

For $\kappa$ countable infinite, ${\boldsymbol{S}}$ forms a countable bounded modular lattice.

Remark 3.5*.*

Note that unlike ${\boldsymbol{S}}$, the lattice ${\boldsymbol{T}}$ is not a meet-subsemillatice of $\mathcal{F}(\kappa)^{3}$. Indeed, both $\left\langle\kappa,\emptyset,\kappa\right\rangle,\left\langle\emptyset,\kappa,\kappa\right\rangle\in{\boldsymbol{T}}$ while $\left\langle\kappa,\emptyset,\kappa\right\rangle\wedge\left\langle\emptyset,\kappa,\kappa\right\rangle=\left\langle\emptyset,\emptyset,\kappa\right\rangle\notin{\boldsymbol{T}}$.

4. A Banaschewski function on ${\boldsymbol{S}}$

In this section we define a Banaschewski function $f\colon{\boldsymbol{S}}\to{\boldsymbol{S}}$ and describe, element-wise, its range.

Lemma 4.1**.**

The map $f\colon{\boldsymbol{S}}\to{\boldsymbol{S}}$ defined by

[TABLE]

is a Banaschewski function on ${\boldsymbol{S}}$.

Proof.

First we prove that ${\boldsymbol{S}}$ contains the range of the map $f$. Observe that if we put $A^{\prime}:=\kappa\setminus{A}$ and $B^{\prime}:=\kappa\setminus{(B\cup C)}$, then $f\left\langle A,B,C\right\rangle=\left\langle A^{\prime},B^{\prime},A^{\prime}\cap B^{\prime}\right\rangle$. Since $\mathcal{F}(\kappa)$ is a Boolean lattice, the sets $A^{\prime},B^{\prime}$ and $A^{\prime}\cap B^{\prime}$ all belong to $\mathcal{F}(\kappa)$. Furthermore, we have that

[TABLE]

In particular, $A^{\prime}\cap B^{\prime}\setminus\mu{f\left\langle A,B,C\right\rangle}=\emptyset$, whence $f\left\langle A,B,C\right\rangle\in{\boldsymbol{S}}$.

It is clear from (4.1) that the map $f$ is antitone. Finally, we check that

[TABLE]

It follows immediately from the definition of $f$ that

[TABLE]

To prove that $\left\langle A,B,C\right\rangle\vee f\left\langle A,B,C\right\rangle=1_{\boldsymbol{S}}$, let us verify that

[TABLE]

Note that each element of $\kappa$ that is not contained in $C$ belongs to $B\cup(\kappa\setminus{(B\cup C)})$. Together with $A\cup(\kappa\setminus{A})=\kappa$, we get that (4.2) holds, which concludes the proof. ∎

Lemma 4.2**.**

Let ${\boldsymbol{E}}$ denote the range of the Banaschewski function $f\colon{\boldsymbol{S}}\to{\boldsymbol{S}}$. Then

[TABLE]

and the mapping

[TABLE]

determines an isomorphism from ${\boldsymbol{E}}$ onto the Boolean lattice $\mathcal{F}(\kappa)\times\mathcal{F}(\kappa)$.

Proof.

While proving Lemma 4.1, we have observed that

[TABLE]

A straightforward computation gives that $f(f\left\langle A^{\prime},B^{\prime},A^{\prime}\cap B^{\prime}\right\rangle)=\left\langle A^{\prime},B^{\prime},A^{\prime}\cap B^{\prime}\right\rangle$, so the lattice ${\boldsymbol{E}}$ is equal to the right-hand side of (4.4). Finally, it is readily seen that the correspondence (4.3) determines an isomorphism ${\boldsymbol{E}}\to\mathcal{F}(\kappa)\times\mathcal{F}(\kappa)$. ∎

It was noted in [14] that if the range of a Banaschewski function on a lattice ${\boldsymbol{L}}$ is Boolean, then it is a maximal Boolean sublattice of ${\boldsymbol{L}}$. Thus we derive from Theorem 4.2 that ${\boldsymbol{E}}$ is a maximal Boolean sublattice of ${\boldsymbol{B}}$.

5. The counter-example

In the present section, we construct another maximal Boolean sublattice ${\boldsymbol{B}}$ of the lattice ${\boldsymbol{S}}$. We show that the lattices ${\boldsymbol{B}}$ and ${\boldsymbol{S}}$ are not isomorphic and we prove directly that the lattice ${\boldsymbol{B}}$ is not the range of any Banaschewski function on ${\boldsymbol{S}}$.

Lemma 5.1**.**

The assignment $\left\langle A,C\right\rangle\mapsto g\left\langle A,C\right\rangle:=\left\langle A,A\cap C,C\right\rangle$ defines a bounded lattice embedding $g\colon\mathcal{F}(\kappa)\times\mathcal{F}(\kappa)\to{\boldsymbol{M}}_{3}[\mathcal{F}(\kappa)]$. In particular, the range of $g$ is a bounded Boolean sublattice of ${\boldsymbol{M}}_{3}[\mathcal{F}(\kappa)]$ isomorphic to $\mathcal{F}(\kappa)\times\mathcal{F}(\kappa)$.

Proof.

It is clear from the definition of the map $g$ that it is injective and that its range is included in ${\boldsymbol{M}}_{3}[\mathcal{F}(\kappa)]$. Further, for any $A,A^{\prime},C,C^{\prime}\subseteq\kappa$, the equality

[TABLE]

holds by (2.3), while

[TABLE]

can be easily deduced from (2.1) and (2.2). Finally, observe that $g\left\langle\kappa,\kappa\right\rangle=\left\langle\kappa,\kappa,\kappa\right\rangle$ and $g\left\langle\emptyset,\emptyset\right\rangle=\left\langle\emptyset,\emptyset,\emptyset\right\rangle$, which concludes the proof. ∎

For any $A,C\in\mathcal{F}(\kappa)$, we say that $\left\langle A,C\right\rangle$ is finite if both $A$ and $C$ are finite, and we say that $\left\langle A,C\right\rangle$ is co-finite if both $\kappa\setminus{A}$ and $\kappa\setminus{C}$ are finite. Let us write $A\sim C$ if $\left\langle A,C\right\rangle$ is either finite or co-finite. Note that there are pairs $A,C\in\mathcal{F}(\kappa)$ such that $\left\langle A,C\right\rangle$ is neither finite nor co-finite; namely, $A\sim C$ if and only if the symmetric difference $(A\setminus C)\cup(C\setminus A)$ is finite.

Lemma 5.2**.**

The set

[TABLE]

form a bounded Boolean sublattice of $\mathcal{F}(\kappa)\times\mathcal{F}(\kappa)$.

Proof.

Let $\left\langle A,C\right\rangle$, $\left\langle A^{\prime},C^{\prime}\right\rangle$ be a pair of elements from ${\boldsymbol{A}}$. If at least one of them is finite, then $\left\langle A\cap A^{\prime},C\cap C^{\prime}\right\rangle$ is clearly finite as well. If both $\left\langle A,C\right\rangle$ and $\left\langle A^{\prime},C^{\prime}\right\rangle$ are co-finite, then so is $\left\langle A\cap A^{\prime},C\cap C^{\prime}\right\rangle$. In either case, $\left\langle A\cap A^{\prime},C\cap C^{\prime}\right\rangle\in{\boldsymbol{A}}$.

If at least one of the pairs $\left\langle A,C\right\rangle,\left\langle A^{\prime},C^{\prime}\right\rangle$ is co-finite, then $\left\langle A\cup A^{\prime},C\cup C^{\prime}\right\rangle$ is co-finite, while if both $\left\langle A,C\right\rangle$ and $\left\langle A^{\prime},C^{\prime}\right\rangle$ are finite, then so is $\left\langle A\cup A^{\prime},C\cup C^{\prime}\right\rangle$. In particular, $\left\langle A\cup A^{\prime},C\cup C^{\prime}\right\rangle\in{\boldsymbol{A}}$ whenever $\left\langle A,C\right\rangle,\left\langle A^{\prime},C^{\prime}\right\rangle\in{\boldsymbol{A}}$.

We have shown that ${\boldsymbol{A}}$ is a sublattice of $\mathcal{F}(\kappa)\times\mathcal{F}(\kappa)$. To complete the proof, observe that $\left\langle\emptyset,\emptyset\right\rangle$ is finite and $\left\langle\kappa,\kappa\right\rangle$ is co-finite and that the unique complement in $\mathcal{F}(\kappa)\times\mathcal{F}(\kappa)$ of each $\left\langle A,C\right\rangle\in{\boldsymbol{A}}$, namely $\left\langle\kappa\setminus{A},\kappa\setminus{C}\right\rangle$ belongs to ${\boldsymbol{A}}$. ∎

Lemma 5.3**.**

The $g$-image ${\boldsymbol{B}}=g({\boldsymbol{A}})$ of ${\boldsymbol{A}}$ is a bounded Boolean sublattice of ${\boldsymbol{S}}$.

Proof.

Due to Lemma 5.1 and Lemma 5.2, ${\boldsymbol{B}}$ is a bounded Boolean sublattice of ${\boldsymbol{M}}_{3}[\mathcal{F}(\kappa)]$. Thus in view of Lemma 3.3, it suffices to verify that ${\boldsymbol{B}}\subseteq{\boldsymbol{S}}$, that is, that $C\setminus(A\cap C)$ is finite for every $\left\langle A,C\right\rangle\in{\boldsymbol{A}}$. This is clear when $\left\langle A,C\right\rangle$ is finite. If $\left\langle A,C\right\rangle$ is co-finite, then $C\setminus(A\cap C)=C\setminus A\subseteq\kappa\setminus A$ is finite and we are done. ∎

Observe that if $\left\langle A,B,C\right\rangle$ is a balanced triple then $B\subseteq A$ if and only if $B=A\cap B=A\cap C$. It follows that

[TABLE]

Lemma 5.4**.**

Let $\left\langle A,B,C\right\rangle\in{\boldsymbol{S}}\setminus{\boldsymbol{B}}$ and let $\left\langle A^{\prime},B^{\prime},C^{\prime}\right\rangle$ be a complement of $\left\langle A,B,C\right\rangle$ in ${\boldsymbol{S}}$. If $B\subseteq A$, then $B^{\prime}\not\subseteq A^{\prime}$.

Proof.

Since $\left\langle A,B,C\right\rangle\not\in{\boldsymbol{B}}$ and $B\subseteq A$, it follows from (5.2) that $A\nsim C$. Hence exactly one of the two sets $A,C$ is finite. From $B\subseteq A$ and $C\setminus B$ being finite we conclude that $C$ and $\kappa\setminus A$ are finite. It follows that the set $B=B\cap C$ is finite as well.

Suppose now that $B^{\prime}\subseteq A^{\prime}$. Since $\left\langle A,B,C\right\rangle\wedge\left\langle A^{\prime},B^{\prime},C^{\prime}\right\rangle=0_{{\boldsymbol{S}}}$, we have that $A\cap A^{\prime}=\emptyset$, whence the set $A^{\prime}$ $\subseteq\kappa\setminus A$ is finite. A fortiori, the set $B^{\prime}$ is also finite due to the assumption that $B^{\prime}\subseteq A^{\prime}$. As $C^{\prime}\setminus B^{\prime}=C^{\prime}\setminus(B^{\prime}\cap A^{\prime})=C^{\prime}\setminus\mu{\left\langle A^{\prime},B^{\prime},C^{\prime}\right\rangle}$ is also finite, we conclude that so is $C^{\prime}$. But then

[TABLE]

is a finite set, which contradicts the assumption that $\left\langle A,B,C\right\rangle\vee\left\langle A^{\prime},B^{\prime},C^{\prime}\right\rangle=\left\langle\kappa,\kappa,\kappa\right\rangle=1_{\boldsymbol{S}}$. ∎

Corollary 5.5**.**

Every complemented bounded sublattice ${\boldsymbol{C}}$ of ${\boldsymbol{S}}$ such that ${\boldsymbol{B}}\subsetneq{\boldsymbol{C}}$ contains an element $\left\langle A,B,C\right\rangle$ with $B\not\subseteq A$.

Proof.

Let $\left\langle A,B,C\right\rangle\in{\boldsymbol{C}}\setminus{\boldsymbol{B}}$ and let $\left\langle A^{\prime},B^{\prime},C^{\prime}\right\rangle$ be its complement in ${\boldsymbol{C}}$. Applying Lemma 5.4, we get that either $B\not\subseteq A$ or $B^{\prime}\not\subseteq A^{\prime}$. ∎

Proposition 5.6**.**

The lattice ${\boldsymbol{B}}$ is a maximal Boolean sublattice of ${\boldsymbol{S}}$.

Proof.

Let ${\boldsymbol{C}}$ be a complemented bounded sublattice of ${\boldsymbol{S}}$ satisfying ${\boldsymbol{B}}\subsetneq{\boldsymbol{C}}$. There is $\left\langle A,B,C\right\rangle\in{\boldsymbol{C}}$ with $B\not\subseteq A$ by Corollary 5.5. We can pick a finite nonempty $F\subseteq(B\setminus A)$. Since the triple $\left\langle A,B,C\right\rangle$ is balanced,

[TABLE]

Now observe that both $g\left\langle F,\emptyset\right\rangle$ and $g\left\langle\emptyset,F\right\rangle$ are in ${\boldsymbol{B}}$. Applying (5.1) and (5.3), we get that

[TABLE]

while

[TABLE]

It follows from (5.4) and (5.5) that the lattice ${\boldsymbol{C}}$ is not distributive, a fortiori it is not Boolean. ∎

Proposition 5.7**.**

The sublattice ${\boldsymbol{B}}$ of ${\boldsymbol{S}}$ is not the range of any Banaschewski function on ${\boldsymbol{S}}$.

Proof.

The range of a Banaschewski function on ${\boldsymbol{S}}$ must contain a complement of each element of ${\boldsymbol{S}}$. We show that no complement of $\left\langle\kappa,\emptyset,\emptyset\right\rangle$ in ${\boldsymbol{S}}$ belongs to ${\boldsymbol{B}}$.

Suppose the contrary, that is, that there is $\left\langle A,B,C\right\rangle=g\left\langle A,C\right\rangle\in{\boldsymbol{B}}$ satisfying $\left\langle\kappa,\emptyset,\emptyset\right\rangle\oplus\left\langle A,B,C\right\rangle=1_{{\boldsymbol{S}}}$. Then $A=A\cap\kappa=\emptyset$, and by (5.2) also $B=\emptyset$. Then from $B=\emptyset$ and $\left\langle\kappa,\emptyset,\emptyset\right\rangle\vee\left\langle A,B,C\right\rangle=1_{\boldsymbol{S}}$, one infers that $C=\kappa$. It follows that $\left\langle A,B,C\right\rangle\notin{\boldsymbol{S}}$; indeed, $C\setminus\mu{\left\langle A,B,C\right\rangle}=C\setminus\emptyset=\kappa$ is not finite. Thus $\left\langle A,B,C\right\rangle\not\in{\boldsymbol{B}}$, which is a contradiction. ∎

Remark 5.8*.*

Note that for the particular case of $\kappa=\aleph_{0}$, the assertion of Proposition 5.7 follows from Proposition 5.9 together with [14, Corollary 4.8], which states that the ranges of two Boolean Banaschewski functions on a countable complemented modular lattice are isomorphic.

Proposition 5.9**.**

The lattices ${\boldsymbol{B}}$ and ${\boldsymbol{E}}$ are not isomorphic.

Proof.

In ${\boldsymbol{B}}$, every finite element $g\left\langle A,C\right\rangle$ is a join of a finite set of atoms, namely

[TABLE]

and, dually, every co-finite element is a meet of a finite set of co-atoms. On the other hand, there are elements in $\mathcal{F}(\kappa)\times\mathcal{F}(\kappa)$ that are neither finite joins of atoms nor finite meets of co-atoms. Recall that in Lemma 4.2, we have observed that the lattice ${\boldsymbol{E}}$ is isomorphic to $\mathcal{F}(\kappa)\times\mathcal{F}(\kappa)$. Therefore the lattices ${\boldsymbol{B}}$ and ${\boldsymbol{E}}$ are not isomorphic. ∎

6. Representing ${\boldsymbol{S}}$ in a subspace-lattice

Although the construction in the three previous sections was performed for an infinite cardinal $\kappa$, the results of the present section on embedding the lattice ${\boldsymbol{M}}_{3}[\mathcal{P}(\kappa)]$ into $\mathrm{Sub}({\boldsymbol{V}})$ (namely Theorem 6.4) work just as well for $\kappa$ finite. In particular, Proposition 6.5 (an enhancement of [5, Lemma 2.9]) holds for lattices of any cardinality.

Let $\mathbb{F}$ be an arbitrary field and let $\boldsymbol{V}$ denote the vector space over the field $\mathbb{F}$ presented by generators $x_{\alpha},y_{\alpha},z_{\alpha}$, $\alpha\in\kappa$, and relations $x_{\alpha}+y_{\alpha}+z_{\alpha}=0$. For a subset $X$ of the vector space $\boldsymbol{V}$ we denote by $\mathrm{Span}(X)$ the subspace of $\boldsymbol{V}$ generated by $X$. Given subspaces of $\boldsymbol{V}$, say $\boldsymbol{X}$ and $\boldsymbol{Y}$, we will use the notation $\boldsymbol{X}+\boldsymbol{Y}=\mathrm{Span}(\boldsymbol{X}\cup\boldsymbol{Y})$. Let $\mathrm{Sub}(\boldsymbol{V})$ denote the lattice of all subspaces of the vector space $\boldsymbol{V}$.

For all $A,B,C\subseteq\kappa$ we put $\boldsymbol{X}_{A}=\mathrm{Span}(\{x_{\alpha}\mid\alpha\in A\})$, $\boldsymbol{Y}_{B}=\mathrm{Span}(\{y_{\beta}\mid\beta\in B\})$, and $\boldsymbol{Z}_{C}=\mathrm{Span}(\{z_{\gamma}\mid\gamma\in C\})$.

We define a map $F\colon\mathcal{P}(\kappa)^{3}\to\mathrm{Sub}(\boldsymbol{V})$ by the correspondence

[TABLE]

Each of the sets $\{x_{\alpha}\mid\alpha\in\kappa\}$, $\{y_{\beta}\mid\beta\in\kappa\}$, and $\{z_{\gamma}\mid\gamma\in\kappa\}$ is clearly linearly independent. It follows that $\boldsymbol{X}_{A\cup A^{\prime}}=\boldsymbol{X}_{A}+\boldsymbol{X}_{A^{\prime}}$ for all $A,A^{\prime}\subseteq\kappa$ and, similarly, $\boldsymbol{Y}_{B\cup B^{\prime}}=\boldsymbol{Y}_{B}+\boldsymbol{Y}_{B^{\prime}}$ and $\boldsymbol{Z}_{C\cup C^{\prime}}=\boldsymbol{Z}_{C}+\boldsymbol{Z}_{C^{\prime}}$ for all $B,B^{\prime},C,C^{\prime}\subseteq\kappa$. A straightforward computation gives the following lemma:

Lemma 6.1**.**

The map $F\colon\mathcal{P}(\kappa)^{3}\to\mathrm{Sub}(\boldsymbol{V})$ is a bounded join-homomorphism.

Proof.

Clearly $F\left\langle\emptyset,\emptyset,\emptyset\right\rangle=\boldsymbol{0}$ and $F\left\langle\kappa,\kappa,\kappa\right\rangle=\boldsymbol{V}$. Following the definitions, we compute $F(\left\langle A,B,C\right\rangle)+F(\left\langle A^{\prime},B^{\prime},C^{\prime}\right\rangle)=\boldsymbol{X}_{A}+\boldsymbol{Y}_{B}+\boldsymbol{Z}_{C}+\boldsymbol{X}_{A^{\prime}}+\boldsymbol{Y}_{B^{\prime}}+\boldsymbol{Z}_{C^{\prime}}=\boldsymbol{X}_{A\cup A^{\prime}}+\boldsymbol{Y}_{B\cup B^{\prime}}+\boldsymbol{Z}_{C\cup C^{\prime}}=F(\left\langle A\cup A^{\prime},B\cup B^{\prime},C\cup C^{\prime}\right\rangle).$ ∎

Let $G\colon\mathrm{Sub}(\boldsymbol{V})\to\mathcal{P}(\kappa)^{3}$ be a map defined by

[TABLE]

for all $\boldsymbol{W}\in\mathrm{Sub}(\boldsymbol{V})$.

It is straightforward that $G$ is a bounded meet-homomorphism and that it is the right adjoint of $F$ (i.e., replacing the lattice $\mathrm{Sub}(\boldsymbol{V})$ with its dual, the maps $F$ and $G$ form a Galois correspondence [11]). Indeed, one readily sees that

[TABLE]

The maps $F$ and $G$ induce a closure operator $GF$ on $\mathcal{P}(\kappa)^{3}$.

Lemma 6.2**.**

The composition $GF\colon\mathcal{P}(\kappa)^{3}\to\mathcal{P}(\kappa)^{3}$ is precisely the closure operator $\overline{\left\langle-\right\rangle}$ on $\mathcal{P}(\kappa)^{3}$ defined by (2.1).

Proof.

We shall prove that $GF\left\langle A,B,C\right\rangle=\overline{\left\langle A,B,C\right\rangle}$ for every $\left\langle A,B,C\right\rangle\in\mathcal{P}(\kappa)^{3}$. By symmetry, it suffices to prove that

[TABLE]

Let $\alpha\in A\cup(B\cap C)$. If $\alpha\in A$, then $x_{\alpha}\in F\left\langle A,B,C\right\rangle$ by the definition (6.1), while if $\alpha\in B\cap C$, then $x_{\alpha}=-y_{\alpha}-z_{\alpha}\in F\left\langle A,B,C\right\rangle$ by (6.1) and the defining relations of $\boldsymbol{V}$. It follows that $A\cup(B\cap C)\subseteq\{\alpha\in\kappa\mid x_{\alpha}\in F\left\langle A,B,C\right\rangle\}$.

In order to prove the opposite inclusion, take any $\xi\in\kappa\setminus A$ satisfying $x_{\xi}\in F\left\langle A,B,C\right\rangle$; if there is one, there is nothing to prove. We need to show that then $\xi\in B\cap C$. Certainly

[TABLE]

for suitable $a_{\alpha}$, $b_{\beta}$, and $c_{\gamma}\in\mathbb{F}$ such that all but finitely many of them are zero. We set $a_{\alpha}=0$ for $\alpha\notin A$, $b_{\beta}=0$ for $\beta\notin B$, and $c_{\gamma}=0$ for $\gamma\notin C$. Since $z_{\gamma}+x_{\gamma}+y_{\gamma}=0$ for every $\gamma\in\kappa$, it follows from (6.2) that

[TABLE]

It easily follows from the defining relations of $\boldsymbol{V}$ that $\{x_{\alpha},y_{\alpha}\mid\alpha\in\kappa\}$ forms a basis of $\boldsymbol{V}$. Thus, applying (6.3) we get that

[TABLE]

Since by our assumption $\xi\notin A$, we get from (6.2) that $a_{\xi}=0$. Substituting to (6.4) we get that $b_{\xi}=c_{\xi}=-1$, hence $\xi\in B\cap C$. This concludes the proof that $A\cup(B\cap C)\supseteq\set{\alpha\in\kappa}{x_{\alpha}\in F\left\langle A,B,C\right\rangle}$. ∎

The next lemma shows that $F\restriction{\boldsymbol{M}}_{3}[\mathcal{P}(\kappa)]$ preserves meets. Note that with Lemma 6.1, this means that $F\restriction{\boldsymbol{M}}_{3}[\mathcal{P}(\kappa)]$ is a lattice embedding of ${\boldsymbol{M}}_{3}[\mathcal{P}(\kappa)]$ into the lattice $\mathrm{Sub}(\boldsymbol{V})$.

Lemma 6.3**.**

Let $\left\langle A,B,C\right\rangle,\left\langle A^{\prime},B^{\prime},C^{\prime}\right\rangle\in{\boldsymbol{M}}_{3}[\mathcal{P}(\kappa)]$ be balanced triples. Then

[TABLE]

Proof.

Since, by Lemma 6.1, $F$ is a join-homomorphism, it is monotone, whence $F\left\langle A\cap A^{\prime},B\cap B^{\prime},C\cap C^{\prime}\right\rangle\subseteq F\left\langle A,B,C\right\rangle\cap F\left\langle A^{\prime},B^{\prime},C^{\prime}\right\rangle$. Thus it remains to prove the opposite inclusion.

Let $v\in F\left\langle A,B,C\right\rangle\cap F\left\langle A^{\prime},B^{\prime},C^{\prime}\right\rangle$ be a non-zero vector. Then $v$ can be expressed as

[TABLE]

Consider such an expression of $v$ with

[TABLE]

minimal possible. Put $a_{\alpha}=0$ for $\alpha\notin A$, $b_{\beta}=0$ for $\beta\notin B$, and $c_{\gamma}=0$ for $\gamma\notin C$. By symmetry, we can assume that $a_{\alpha}\neq 0$ for some $\alpha\in A$. Suppose for a contradiction that $\alpha\notin A^{\prime}$. Since the triple $\left\langle A^{\prime},B^{\prime},C^{\prime}\right\rangle$ is balanced, $B^{\prime}\cap C^{\prime}\subseteq A^{\prime}$, whence $\alpha\notin B^{\prime}\cap C^{\prime}$. Without loss of generality we can assume that $\alpha\notin B^{\prime}$. If all $a_{\alpha},b_{\alpha}$, and $c_{\alpha}$ were non-zero, we could replace $c_{\alpha}z_{\alpha}$ with $-c_{\alpha}x_{\alpha}-c_{\alpha}y_{\alpha}$ and reduce the value of the expression in (6.6) which is assumed minimal possible. Thus either $b_{\alpha}=0$ or $c_{\alpha}=0$ (recall that we assume that $a_{\alpha}\neq 0$). We will deal with these two cases separately. If $b_{\alpha}=0$, then the equality

[TABLE]

must hold true. Since $x_{\alpha}$ and $z_{\alpha}$ are linearly independent, it follows from (6.7) that $a_{\alpha}=0$ which contradicts our choice of $\alpha$. The remaining case is when $c_{\alpha}=0$. Under this assumption we have that

[TABLE]

It follows that

[TABLE]

Since $x_{\alpha}$ and $y_{\alpha}$ are linearly independent, we infer from (6.8) that $a_{\alpha}=-c_{\alpha}^{\prime}=b_{\alpha}$. Then we could reduce the value of (6.6) by replacing $a_{\alpha}x_{\alpha}+b_{\alpha}y_{\alpha}$ with $c_{\alpha}^{\prime}z_{\alpha}$ in (6.5). This contradicts the minimality of (6.6). ∎

Combining Lemma 6.1, Lemma 6.2, and Lemma 6.3, we conclude:

Theorem 6.4**.**

The restrictions $F\restriction{\boldsymbol{M}}_{3}[\mathcal{P}(\kappa)]\colon{\boldsymbol{M}}_{3}[\mathcal{P}(\kappa)]\to\mathrm{Sub}(\boldsymbol{V})$ and, a fortiory, $F\restriction{\boldsymbol{S}}\colon{\boldsymbol{S}}\to\mathrm{Sub}(\boldsymbol{V})$ are bounded lattice embeddings. In particular, the lattice ${\boldsymbol{S}}$ is isomorphic to a bounded sublattice of the subspace-lattice of a vector space.

It is well-known that a distributive lattice ${\boldsymbol{L}}$ embeds (via a bounds-preserving lattice embedding) into the lattice $\mathcal{P}(\kappa)$, where $\kappa$ is the cardinality of the set of all maximal ideals of ${\boldsymbol{L}}$. Such embedding induces an embedding ${\boldsymbol{M}}_{3}[{\boldsymbol{L}}]\hookrightarrow{\boldsymbol{M}}_{3}[\mathcal{P}(\kappa)]$ (cf. Lemma 3.3). By Theorem 6.4, the lattice ${\boldsymbol{M}}_{3}[\mathcal{P}(\kappa)]$ embeds into the lattice $\mathrm{Sub}(\boldsymbol{V})$ for a suitable vector space $\boldsymbol{V}$ (note again that we now also admit finite $\kappa$). Since the lattice $\mathrm{Sub}({\boldsymbol{V}})$ is Arguesian, so are ${\boldsymbol{M}}_{3}[\mathcal{P}(\kappa)]$ and ${\boldsymbol{M}}_{3}[{\boldsymbol{L}}]$.

On the other hand, [5, Lemma 2.9] states that a lattice $L$ is distributive if and only if ${\boldsymbol{M}}_{3}[{\boldsymbol{L}}]$ is modular. Hence, if ${\boldsymbol{M}}_{3}[{\boldsymbol{L}}]$ is modular, it follows that ${\boldsymbol{L}}$ is distributive, and, by the above argument, ${\boldsymbol{M}}_{3}[{\boldsymbol{L}}]$ is even Arguesian. We have thus proven the following strengthening of [5, Lemma 2.9]:

Proposition 6.5**.**

Let $L$ be a lattice. Then $L$ is distributive iff the lattice ${\boldsymbol{M}}_{3}[{\boldsymbol{L}}]$ is modular iff ${\boldsymbol{M}}_{3}[{\boldsymbol{L}}]$ is Arguesian. If this is the case, then ${\boldsymbol{M}}_{3}[{\boldsymbol{L}}]$ can be embedded into the lattice of all subspaces of a suitable vector space over any given field.