# Characterizations of classes of countable Boolean inverse monoids

Mark V. Lawson  Mark V. Lawson, Department of Mathematics and the Maxwell
Institute for Mathematical Sciences, Heriot-Watt University, Riccarton,
Edinburgh EH14 4AS, UNITED KINGDOM<EMAIL_ADDRESS>and  Philip Scott
Philip Scott, Department of Mathematics and Statistics, University of Ottawa,
STEM Complex, Ottawa, Ontario K1N 6N5, CANADA<EMAIL_ADDRESS>This paper
is dedicated to the memory of Pieter Hofstra, our friend and colleague.

###### Abstract.

A countable Boolean inverse monoid that can be written as an increasing union
of finite Boolean inverse monoids (suitably embedded) is said to be of finite
type. Borrowing terminology from $C^{\ast}$-algebra theory, we say that such a
Boolean inverse monoid is AF (approximately finite) if the finite Boolean
inverse monoids above are isomorphic to finite direct products of finite
symmetric inverse monoids, and we say that it is UHF (uniformly hyperfinite)
if the finite Boolean inverse monoids are in fact isomorphic to finite
symmetric inverse monoids. We characterize abstractly the Boolean inverse
monoids of finite type and those which are AF, and by using MV-algebras, we
also characterize the UHF monoids.

##  1\. Introduction

The theory of finite Boolean inverse monoids generalizes that of finite
Boolean algebras by replacing the finite sets of the theory of finite Boolean
algebras by the finite groupoids of the theory of finite Boolean inverse
monoids. The theory of finite Boolean inverse monoids is summarized in [22].
The goal of our paper is to generalize that theory to the countably infinite
case. To do this, we shall describe first what we need from the finite case.
Any undefined terms used below will be defined in later sections.

Let $S$ be a finite Boolean inverse monoid and denote by $\mathsf{at}(S)$ the
set of atoms of $S$ equipped with the restricted product. Then
$\mathsf{at}(S)$ is a groupoid. More generally, if $G$ is any groupoid denote
by $\mathsf{K}(G)$ the set of all local bisections of $G$. We now have the
following results:

  * •

By [22, Theorem 4.5], we have that $S\cong\mathsf{K}(\mathsf{at}(S))$.

  * •

By [22, Lemma 4.6], $S$ is fundamental if and only if $\mathsf{at}(S)$ is
principal.

  * •

By [22, Lemma 4.6], $S$ is $0$-simplifying if and only if $\mathsf{at}(S)$ is
connected.

We see that the groupoid structure of $\mathsf{at}(S)$ determines the
structure of $S$. If $G$ is a group, we denote by $G^{0}$ the group $G$ with a
zero adjoined, and we denote by $R_{n}(G^{0})$ the Boolean inverse monoid of
all $n\times n$ rook matrices over $G^{0}$. If $n$ is a non-zero natural
number, we denote by $\mathscr{I}_{n}$ the finite symmetric inverse monoid on
$n$ letters. The finite Boolean inverse monoids have more concrete
descriptions as follows:

  * •

By [22, Theorem 4.7], given a finite Boolean inverse monoid $S$ there are
finite groups $G_{1},\ldots,G_{r}$ such that $S\cong
R_{n_{1}}(G_{1}^{0})\times\ldots R_{n_{r}}(G_{r}^{0})$.

  * •

By [19, Theorem 4.18], the finite fundamental Boolean inverse monoids are
isomorphic to the finite direct products
$\mathscr{I}_{n_{1}}\times\ldots\times\mathscr{I}_{n_{r}}$.

  * •

By [19, Theorem 4.18], the finite simple Boolean inverse monoids are
isomorphic to the symmetric inverse monoids $\mathscr{I}_{n}$.

In this paper, we shall be interested in countably infinite Boolean inverse
meet-monoids $S$ which can be written in the form
$S=\bigcup_{i=1}^{\infty}S_{i}$ where $S_{1}\subseteq S_{2}\subseteq
S_{3}\subseteq\ldots$ and where each $S_{i}$ is a subalgebra111This term will
be defined later. It is much stronger than merely being an inverse submonoid.
of $S$. We shall characterize $S$ in three different situations:

  1. (1)

The $S_{i}$ are finite Boolean inverse monoids; we say that $S$ is of finite
type. They are characterized in Theorem 4.1.

  2. (2)

The $S_{i}$ are finite fundamental Boolean inverse monoids; we say that $S$ is
an AF monoid. These were the subject of [23]. They are characterized in
Theorem 4.3.

  3. (3)

The $S_{i}$ are finite simple Boolean inverse monoids; we say that $S$ is a
UHF monoid. These are a special case of (2). They are characterized in Theorem
6.33.

Cases (2) and (3) are clearly motivated by the theory of $C^{\ast}$-algebras.

We refer the reader to [17] for classical inverse semigroup theory, in
particular the theory of the natural partial order; to [38] for general
results on Boolean inverse semigroups; and to [2] for MV-algebras. In writing
this paper, we have found the thesis [33] very useful.

##  2\. Background on inverse semigroups

In this section, we shall review the important definitions we shall need from
inverse semigroup theory. See [14] for general semigroup theory and [17] for
inverse semigroup theory.

###  2.1. General

Our inverse semigroups will have a zero and homomorphisms will be required to
map zero to zero. In addition, if our semigroups are monoids then
homomorphisms will be required to map identities to identities. We shall use
Green’s relations when convenient and any basic semigroup theory needed. If
$S$ is a semigroup then a subset $I$ is an ideal if $SI,IS\subseteq I$; our
ideals will always contain at least the zero and so are non-empty. If $X$ is a
non-empty set then $\mathscr{I}(X)$ denotes the symmetric inverse monoid on
$X$; that is, the inverse monoid of all partial bijections of the set $X$. If
$X$ is a finite set with $n$ elements then we denote the corresponding
symmetric inverse monoid by $\mathscr{I}_{n}$. If $S$ is an inverse semigroup,
we call the operation $a\mapsto bab^{-1}$ conjugation. We denote the set of
idempotents of $S$ by $\mathsf{E}(S)$. If $X\subseteq S$ define
$\mathsf{E}(X)=X\cap\mathsf{E}(S)$. If $a$ and $b$ are elements of an inverse
semigroup, we say that $a$ and $b$ are compatible, written $a\sim b$, if both
$a^{-1}b$ and $ab^{-1}$ are idempotents. Being compatible is a necessary
condition for $a$ and $b$ to have an upper bound.

If $a\in S$, define $\mathbf{d}(a)=a^{-1}a$, called the domain of $a$, and
$\mathbf{r}(a)=aa^{-1}$, called the range of $a$; we often write
$e\stackrel{{\scriptstyle a}}{{\rightarrow}}f$ to mean that $\mathbf{d}(a)=e$
and $\mathbf{r}(a)=f$. In an inverse semigroup, this is equivalent to
$e\,\mathscr{D}\,f$. An inverse semigroup is said to be Dedekind finite222The
term ‘completely semisimple’ is usually applied in this case, but the word
‘semsimple’ is overloaded in mathematics which is why we have introduced this
term. if $e\,\mathscr{D}\,f\leq e$ implies that $e=f$. In any semigroup, we
always have that $\mathscr{D}\subseteq\mathscr{J}$, but in a Dedekind finite
inverse semigroup, we have that $\mathscr{D}=\mathscr{J}$. An inverse monoid
is said to be directly finite if for any idempotent $e$ we have that
$e\,\mathscr{D}\,1$ implies that $e=1$. It is immediate that Dedekind finite
inverse monoids are directly finite. If $S$ is an inverse semigroup and $e$ is
an idempotent then $eSe$ is an inverse monoid; we say that $eSe$ is a local
monoid of $S$.

An inverse semigroup is said to be $0$-disjunctive if for all idempotents $e$
and $f$ such that $0\neq e<f$ there exists $0\neq e^{\prime}\leq f$ such that
$ee^{\prime}=0$. The following definitions can be found in [14] and are
standard. An inverse semigroup in which every idempotent is central is said to
be Clifford. An inverse semigroup is said to be fundamental if the only
elements commuting with all idempotents are themselves idempotents. On every
inverse semigroup, we may define the maximum idempotent-separating congruence
$\mu$. This has the property that $S/\mu$ is fundamental. See [17, Section
5.2].

######  Lemma 2.1.

Let $S$ be an inverse semigroup. Let $e$ and $f$ be idempotents. Then
$e\,\mathscr{D}\,f$ if and only if $\mu(e)\,\mathscr{D}\,\mu(f)$.

###### Proof.

Only one direction needs proving. Suppose that $\mu(e)\,\mathscr{D}\,\mu(f)$.
Then there is an element $s\in S$ such that $\mu(e)=\mu(s^{-1}s)$ and
$\mu(f)=\mu(ss^{-1})$. But $\mu$ is idempotent-separating. Thus $e=s^{-1}s$
and $f=ss^{-1}$. It follows that $e\,\mathscr{D}\,f$. ∎

If $\mathbf{d}(a)=\mathbf{r}(b)$ then $ab=a\cdot b$ is called a restricted
product. The inverse semigroup with respect to the restricted product is a
groupoid; see [13] for the necessary background on groupoids. We shall use the
term principal groupoid for the groupoid obtained from an equivalence
relation. We shall also have to refer to connected groupoids; a groupoid $G$
is connected if for any identities $e$ and $f$ there is an element $g$ of the
groupoid such that $e=g^{-1}g$ and $f=gg^{-1}$. Every product in an inverse
semigroup can be written as a restricted product since $ab=(ae)\cdot(eb)$
where $e=\mathbf{d}(a)\mathbf{r}(b)$; observe that $ae\leq a$ and $eb\leq b$.

Let $S$ be an inverse monoid. Denote its group of units by $\mathsf{U}(S)$;
the elements of $\mathsf{U}(S)$ will be called units or invertible elements.
The group of units of $S$, $\mathsf{U}(S)$, acts on the semilattice of
idempotents $\mathsf{E}(S)$ by $e\mapsto geg^{-1}$. We call this the natural
action of $\mathsf{U}(S)$ on $\mathsf{E}(S)$.

###  2.2. Factorizable inverse monoids

An inverse monoid $S$ is said to be factorizable if every element is below a
unit. It was proved in [5], that $\mathscr{I}(X)$ is factorizable if and only
if $X$ is finite, and that factorizable inverse monoids are directly finite.
The key feature of factorizable inverse monoids is that they can be
constructed from groups and semilattices in a simple way. The following is
well-known but we include a proof because of its importance.

######  Lemma 2.2.

Let $S$ be an inverse monoid.

  1. (1)

Let $G\leq\mathsf{U}(S)$ be a subgroup and let $E\subseteq\mathsf{E}(S)$ be a
subsemilattice that contains the identity. Then $T=GE$ is an inverse submonoid
of $S$ if and only if $E$ is closed under the natural action by $G$.

  2. (2)

If $G\leq\mathsf{U}(S)$ is a subgroup and $E\subseteq\mathsf{E}(S)$ is a
subsemilattice that contains the identity such that $E$ is closed under the
natural action by $G$ then $T=GE$ is a factorizable inverse monoid with group
of units $G$ and semilattice of idempotents $E$.

  3. (3)

If $T$ is a factorizable inverse submonoid of $S$ then $T=GE$ where $G$ is the
group of units of $T$ and $E$ is the semilattice of idempotents of $T$ (and is
closed under the natural action by $G$.)

###### Proof.

(1) Suppose first that $E$ is closed under the natural action by $G$; this
means that if $e\in E$ and $g\in G$ then $geg^{-1}\in E$. Let $ge$ and $hf$ be
elements of $T$. Then $(ge)(hf)=gh(h^{-1}eh)f$. But $E$ is closed under the
natural action by $G$ and so $(ge)(hf)\in T$. Thus $T$ is closed under
multiplication. Let $ge\in T$. Then $(ge)^{-1}=g^{-1}(geg^{-1})$. But $E$ is
closed under the natural action by $G$ and so $(ge)^{-1}\in T$. We have
therefore proved that $T$ is a submonoid of $S$. We now prove the converse: if
$T$ is an inverse submonoid then $E$ is closed under the natural action by
$G$. Let $g\in G$ and $e\in E$. We prove that $geg^{-1}\in E$. Put $s=ge\in
T$. Then $ss^{-1}\in T$, since $T$ is an inverse submonoid of $S$, from which
it follows that $geg^{-1}\in E$.

(2) Put $T=GE$ which by part (1) is an inverse submonoid of $S$. Let $ge\in T$
and suppose that $\mathbf{d}(ge)=1=\mathbf{r}(ge)$. From $\mathbf{d}(ge)=1$ we
deduce that $e=1$. It follows that the group of units of $T$ is $G$. Let $ge$
be an idempotent of $T$. Then $gege=ge$ and $ge=(ge)^{-1}=eg^{-1}$. Now
$ege=e$ and so $e(geg^{-1})=eg^{-1}=ge$. It follows that $ge$ is the product
of $e$ and $geg^{-1}$, both elements of $E$. Thus, since $E$ is a semilattice,
it follows that $ge\in E$. Observe that $ge\leq g$ and so $T$ is factorizable.

(3) Let $G$ be the group of units of $T$ and let $E$ be its semilattice of
idempotents. Clearly $GE\subseteq T$. On the other hand we have equality since
every element of $T$ is beneath an element of $G$. ∎

###  2.3. Meet semigroups

An inverse semigroup $S$ is called a meet-semigroup if each pair of elements
has a meet. There is a different way of thinking about meet-semigroups that
will be important. Let $S$ be an inverse semigroup. A function $\phi\colon
S\rightarrow\mathsf{E}(S)$ is called a fixed-point operator if it satisfies
the following two conditions:

(FPO1):

$s\geq\phi(s)$.

(FPO2):

If $s\geq e$ where $e$ is any idempotent then $\phi(s)\geq e$.

Thus, $\phi(s)$ is the largest idempotent below $s$. The following are proved
in [25] or are immediate from the definition.

######  Proposition 2.3.

  1. (1)

An inverse semigroup $S$ is an inverse meet-semigroup if and only if it
possesses a fixed-point operator.

  2. (2)

In an inverse meet-semigroup, we have that $\phi(s)=s\wedge s^{-1}s$. In the
monoid case, $\phi(s)=s\wedge 1$.

  3. (3)

If $\phi$ is the fixed-point operator then $a\wedge b=\phi(ab^{-1})b$.

  4. (4)

$\phi(s)=\phi(s^{-1})$.

  5. (5)

$\phi(s)=s\phi(s)=\phi(s)s$.

  6. (6)

In an inverse meet-semigroup, we have that $\phi(se)=\phi(s)e$ when $e$ is an
idempotent.

  7. (7)

If $g$ and $h$ are units then $\phi(hgh^{-1})=h\phi(g)h^{-1}$.

The following is proved as [17, Proposition 1.4.19].

######  Lemma 2.4.

Let $S$ be an inverse semigroup. Then for any $a,b,c\in S$ such that $a\wedge
b$ exists then both $ac\wedge bc$ and $ca\wedge cb$ exist and $(a\wedge
b)c=ac\wedge bc$ and $c(a\wedge b)=ca\wedge cb$.

######  Lemma 2.5.

Let $T$ be a submonoid of the inverse meet-monoid $S$. Let $\phi$ be the
fixed-point operator for $S$. Then $T$ is closed under binary meets if and
only if for each $t\in T$ we have that $\phi(t)\in T$.

###### Proof.

Suppose first that $T$ is closed under all binary meets in $S$. Let $t\in T$.
We claim that $\phi(t)\in T$. By Proposition 2.3, we have that
$\phi(t)=t\wedge 1$. By assumption, $S$ contains the identity of $S$ (since it
is a submonoid) and it is closed under binary meets. Thus $\phi(t)\in T$.
Suppose, now, that for each $t\in T$ we have that $\phi(t)\in T$. Let
$t_{1},t_{2}\in T$. Then, again by Proposition 2.3, we have that $t_{1}\wedge
t_{2}=\phi(t_{1}t_{2}^{-1})t_{2}$. Since $T$ is, in particular, an inverse
submonoid of $S$, we have that $t_{1}t_{2}^{-1}\in T$. Thus, by assumption,
$\phi(t_{1}t_{2}^{-1})\in T$. Whence, $t_{1}\wedge t_{2}\in T$. ∎

######  Lemma 2.6.

Let $S$ be an inverse meet-monoid and let $F$ be an inverse submonoid of $S$
which is also factorizable. Then $F$ is closed under binary meets if and only
if $\phi(g)\in F$ for every unit $g$ in $F$.

###### Proof.

Denote by $\phi$ the fixed-point operator of $S$. Suppose first that
$\phi(g)\in F$ for every unit $g$ in $F$. By assumption, typical elements of
$F$ have the form $ge$ and $hf$ where $g$ and $h$ are units and $e$ and $f$
are idempotents. Observe that

$\phi(ge(hf)^{-1})=\phi(gefh^{-1})=\phi(gh^{-1}(hefh^{-1}))=\phi(gh^{-1})(hefh^{-1}),$

where we have used the fact that $hefh^{-1}$ is an idempotent and Proposition
2.3. By assumption, $\phi(gh^{-1}),hefh^{-1}\in F$. Thus $ge\wedge hf\in F$ by
Proposition 2.3. By Lemma 2.5, the proof of the converse is immediate. ∎

###  2.4. Boolean algebras and posets

In this paper, generalized Boolean algebras are referred to as Boolean
algebras and Boolean algebras will be called unital Boolean algebras. See
[35]. If $e$ and $f$ are elements of a Boolean algebra we write $e\perp f$ if
$ef=fe=0$ and say that $e$ and $f$ are orthogonal. In the case where $e\perp
f$ the join $e\vee f$ is often written $e\oplus f$. In a Boolean algebra, if
$f\leq e$ then denote by $e\setminus f$ the unique element such that
$e=(e\setminus f)\vee f$ and $f\perp(e\setminus f)$. In a unital Boolean
algebra, the complement of $e$ is denoted by $\bar{e}$. Observe that in a
unital Boolean algebra, $ef=0$ if and only if $e\leq\bar{f}$. The two-element
Boolean algebra will be denoted by $\mathbf{2}$.

If $P$ is a poset and $a\in P$, we write $a^{\downarrow}$ for the set $\\{b\in
P\colon b\leq a\\}$ and $a^{\uparrow}$ for the set $\\{b\in P\colon a\leq
b\\}$. A subset $Q$ of $P$ is called an order-ideal if $q\in Q$ and $p\leq q$
implies that $p\in Q$. An order-preserving map between posets will be called
isotone. An order-isomorphism between two posets is an isotone bijection whose
inverse is also isotone. Let $P$ be a poset with zero. A non-zero element
$a\in P$ is called an atom if $b\leq a$ implies that $b=a$ or $b=0$.

The following definition will play a key role in this paper.

Definition. Let $S$ be a group (respectively, a Boolean algebra, respectively
an MV-algebra). Let $X=\\{s_{1},\ldots,s_{n}\\}$ be a finite subset of $S$.
Then $\langle s_{1},\ldots,s_{n}\rangle$ denotes the subgroup (respectively,
Boolean subalgebra, respectively MV-subalgebra) of $S$ generated by $X$. A
structure is locally finite if every finitely generated substructure is
finite.

We follow [3] in our definition of a locally finite MV-algebra. The following
is proved in [9, Chapter 11].

######  Lemma 2.7.

Unital Boolean algebras are locally finite

It will be useful to describe the elements of the Boolean subalgebra of the
unital Boolean algebra $B$ generated by the elements
$X=\\{p_{1},\ldots,p_{n}\\}$. A product $x_{1}\ldots x_{n}$ where each $x_{i}$
is either $p_{i}$ or $\overline{p_{i}}$ is called a minterm [29]. The Boolean
subalgebra generated by $X$ consists of all finite joins of minterms.

###  2.5. Some directed colimits of inverse monoids

We begin with a notion that is familiar in category theory. Let $S_{e}$ be a
set of objects of some category indexed by a set $\\{e\in I\\}$. We suppose
that there are some connecting arrows $\phi^{e}_{f}\colon S_{e}\rightarrow
S_{f}$. Let $T$ be any object in the catgeory. A set of arrows $\tau_{e}\colon
S_{e}\rightarrow T$ in the category is called a cocone with base the $S_{e}$
and target the structure $T$ if $\tau_{f}\phi^{e}_{f}=\tau_{e}$ for all $e$
and $f$ and connecting arrows $\phi_{f}^{e}$.

We shall use the theory of directed colimits throughout this paper. Directed
colimits exist in any variety of algebras [4, Proposition III.1.3]; so, in
particular, directed colimits of MV-algebras and of Boolean inverse semigroups
(see below) exist; observe that, despite appearances, Boolean inverse
semigroups are definable by a finite set of identities in an appropriate
language [38, Section 3.2]. We shall only be interested in directed colimits
of systems of the form
$S_{1}\stackrel{{\scriptstyle\theta_{1}}}{{\rightarrow}}S_{2}\stackrel{{\scriptstyle\theta_{2}}}{{\rightarrow}}S_{3}\stackrel{{\scriptstyle\theta_{3}}}{{\rightarrow}}\ldots$
where the maps $\theta_{i}$ are injections. The directed colimit of this
system is denoted $\varinjlim S_{i}$. Define $\theta_{i}^{i}$ to be the
identity function on $S_{i}$ and, if $j>i$, define
$\theta_{j}^{i}=\theta_{j-1}\ldots\theta_{i}$. The functions $\theta^{i}_{j}$
are injections. Denote the maps from $S_{i}$ to $\varinjlim S_{i}$ by
$\tau_{i}$; in the cases that interest us in this paper, these maps are
injective. Furthermore, $\varinjlim S_{i}$ is the union of these images. In
the cases that interest us, if we have a structure $S$ written as a union
$S=\bigcup_{i=1}S_{i}$ of substructures $S_{i}$ where $S_{1}\subseteq
S_{2}\subseteq S_{3}\subseteq\ldots$ then $S$ is the directed colimit of
$S_{1}\rightarrow S_{2}\rightarrow S_{3}\rightarrow\ldots$ via the natural
inclusion maps; and if
$S_{1}\stackrel{{\scriptstyle\theta_{1}}}{{\rightarrow}}S_{2}\stackrel{{\scriptstyle\theta_{2}}}{{\rightarrow}}S_{3}\stackrel{{\scriptstyle\theta_{3}}}{{\rightarrow}}\ldots$,
where the maps $\theta_{i}$ are injections, has colimit $S$ then
$S=\bigcup_{i=1}S_{i}^{\prime}$ where each $S_{i}^{\prime}$ is the isomorphic
image of $S_{i}$ in $S$. We shall use the following result in this paper. The
proof is routine.

######  Lemma 2.8.

Let
$S_{1}\stackrel{{\scriptstyle\theta_{1}}}{{\rightarrow}}S_{2}\stackrel{{\scriptstyle\theta_{2}}}{{\rightarrow}}\ldots$
be a sequence of Boolean inverse monoids and morphisms and let
$T_{1}\stackrel{{\scriptstyle\phi_{1}}}{{\rightarrow}}T_{2}\stackrel{{\scriptstyle\phi_{2}}}{{\rightarrow}}\ldots$
be a sequence of Boolean inverse monoids and morphisms. Suppose that
$\alpha_{i}\colon S_{i}\rightarrow T_{i}$ is an isomorphism such that
$\phi_{i}\alpha_{i}=\alpha_{i+1}\theta_{i}$. Then $\varinjlim
S_{i}\cong\varinjlim T_{i}$.

We shall now expand upon some topics discussed after [23, Lemma 3.9].
Specifically, we shall show that certain directed colimits of inverse monoids
— in particular, the ones of interest to us in this paper — can be constructed
purely in terms of semigroup theory. This is based on two constructions. Here
is our first construction. Let $(P,\leq)$ be a meet-semilattice. For each
$e\in P$ let $S_{e}$ be an inverse monoid. If $e\geq f$ we assume that there
is a homomorphism of inverse monoids (called a connecting homomorphism)
$\phi^{e}_{f}\colon S_{e}\rightarrow S_{f}$ such that $\phi_{e}^{e}$ is the
identity function on $S_{e}$ and if $e\geq f\geq g$ then
$\phi_{g}^{e}=\phi_{g}^{f}\phi_{f}^{e}$. We say that the $S_{e}$, together
with the homomorphisms, form a strong semilattice of inverse semigroups over
$P$. We shall use the notation $(P,S_{e},\phi_{f}^{e})$ to denote the
ingredients for such a strong semilattice.333This is the classical semigroup
theory terminology, but we could also say that we have a presheaf of inverse
semigroups over the semilattice $P$. If $e\in P$ denote the identity in
$S_{e}$ by $1_{e}$. The following is folklore, and easy to prove.

######  Lemma 2.9.

Let $(P,S_{e},\phi_{f}^{e})$ be a strong semilattice of inverse monoids.

  1. (1)

Put $S=\sqcup_{e\in P}S_{e}$, the disjoint union of the monoids. Define a
binary operation $\circ$ on $S$ by $s\circ t=\phi_{e\wedge
f}^{e}(s)\phi_{e\wedge f}^{f}(t)$, where $s\in S_{e}$ and $t\in S_{f}$. Then
$(S,\circ)$ is an inverse semigroup. We denote this semigroup by $S^{\circ}$.
Observe that $\mathsf{E}(S^{\circ})=\bigcup_{e\in P}\mathsf{E}(S_{e})$, and
that if $a\in S_{e}$ then the unique inverse of $a$ in $S^{\circ}$ is
$a^{-1}$, the inverse of $a$ in $S_{e}$.

  2. (2)

Put $\mathscr{E}=\\{1_{e}\colon e\in P\\}$. Then $\mathscr{E}$ is a set of
central idempotents in $S$ with the property that for each $s\in S$ there
exists $e\in\mathscr{E}$ such that $se=s$.

  3. (3)

If $P$ has a top element $t$ then $1_{t}$ is the identity of $S$.

We now come to our second construction.

######  Lemma 2.10.

Let $S$ be an inverse semigroups and let $\mathscr{E}$ be a set of central
idempotents such that for each $s\in S$ there exists $e\in\mathscr{E}$ such
that $se=s$ and such that $\mathscr{E}$ is closed under products. Define the
binary relation $a\equiv b$ if and only if $ae=be$ for some $e\in\mathscr{E}$.
Then $\equiv$ is a congruence on $S$. Denote the $\equiv$-congruence class
containing $a$ by $[a]$. We have that $S/\equiv$ is an inverse semigroup, and
$\equiv$ is the smallest congruence on $S$ which identifies all the elements
of $\mathscr{E}$.

We now combine the above two constructions, of Lemma 2.9 and Lemma 2.10, to
construct colimits of inverse monoids over meet-semilattices. In this special
case, colimits will be built using only ideas from semigroup theory. In what
follows, we shall always assume that $P$ has a top element.

Let $(P,S_{e},\phi_{f}^{e})$ be a strong semilattice of inverse monoids. We
shall construct $\varinjlim S_{e}$. First, form the inverse semigroup
$(S,\circ)$. Then, on this inverse semigroup, define
$\mathscr{E}=\\{1_{e}\colon e\in P\\}$. We may therefore define the congruence
$\equiv$ on $S$ to get the quotient inverse monoid $T=S/\equiv$. Observe that
$a\equiv b$ if and only if $\phi_{g}^{e}(a)=\phi_{g}^{f}(b)$ where $a\in
S_{e}$, $b\in S_{f}$ and $g\leq e\wedge f$ is some element of $P$. We shall
prove that $T=\varinjlim S_{e}$.

We begin by showing that there is a cocone with base $S_{e}$ and target $T$.
Define $\tau_{e}\colon S_{e}\rightarrow T$ by $\tau_{e}(a)=[a]$. Suppose that
$\phi_{f}^{e}\colon S_{e}\rightarrow S_{f}$. We calculate

$\tau_{f}(\phi_{f}^{e}(a))=[\phi_{f}^{e}(a)]=[a]=\tau_{e}(a),$

where we have used the fact that $a\equiv\phi_{f}^{e}(a)$. The following
result shows that cocones can be replaced by homomorphisms.

######  Lemma 2.11.

There is a bijective correspondence between monoid homomorphisms on
$S^{\circ}$ which identify all elements of $\mathscr{E}$ and cocones with base
$S_{e}$.

###### Proof.

Let $\theta\colon S^{\circ}\rightarrow T^{\prime}$ be any homomorphism which
identifies all the elements of $\mathscr{E}$. Define $\tau_{e}\colon
S_{e}\rightarrow T^{\prime}$ by $\tau_{e}(a)=\theta(a)$. We prove that this
defines a cocone with base $S_{e}$ and target $T^{\prime}$. Let $a\in S_{e}$.
Then

$\tau_{f}(\phi_{f}^{e}(a))=\tau_{f}(a\circ f)=\theta(a\circ
f)=\theta(a)=\tau_{e}(a),$

where we have used the fact that $\theta$ is a homomomorphism from $S^{\circ}$
that identifies all the elements of $\mathscr{E}$. It is clear that each
$\tau_{e}$ is a homomorphism.

We now go in the opposite direction. Let $\tau_{e}\colon S_{e}\rightarrow
T^{\prime}$ be a cocone with base $S_{e}$ and target $T^{\prime}$. Define
$\theta\colon S^{\circ}\rightarrow T^{\prime}$ by $\theta(a)=\tau_{e}(a)$
where $a\in S_{e}$. We now show that $\theta$ is a homomorphism. We calculate

$\theta(a\circ b)=\theta(\phi_{e\wedge f}^{e}(a)\phi_{e\wedge
f}^{f}(b))=\tau_{e\wedge f}(\phi_{e\wedge f}^{e}(a)\phi_{e\wedge f}^{f}(b)).$

But, using the fact that we have a cocone, we have that this is equal to
$\tau_{e}(a)\tau_{f}(b)=\theta(a)\theta(b)$. Finally, we show that $\theta$
identifies all the elements of $\mathscr{E}$. Let $1_{e}\in\mathscr{E}$. Then
$\theta(1_{e})=\tau_{e}(1_{e})$ which is the identity of $T^{\prime}$.

We have defined two processes above. We now show that they are mutually
inverse. Let $\theta\colon S^{\circ}\rightarrow T^{\prime}$ be any
homomorphism which identifies all the elements of $\mathscr{E}$. The
corresponding cocone is defined by $\tau_{e}\colon S_{e}\rightarrow
T^{\prime}$ by $\tau_{e}(a)=\theta(a)$. The homomomorphism $\theta^{\prime}$
associated with this cocone is $\theta^{\prime}(a)=\tau_{e}(a)=\theta(a)$. It
follows that $\theta^{\prime}=\theta$. Now, let $\tau_{e}\colon
S_{e}\rightarrow T^{\prime}$ be a cocone with base $S_{e}$ and target
$T^{\prime}$. The corresponding homomorphism is defined by
$\theta(a)=\tau_{e}(a)$ where $a\in S_{e}$. The cocone with base $S_{e}$ and
target $T^{\prime}$ is defined by
$\tau_{e}^{\prime}(a)=\theta(a)=\tau_{e}(a)$. It follows that
$\tau_{e}^{\prime}=\tau_{e}$. ∎

If $\theta\colon S\rightarrow T$ is a semigroup homomorphism then
$\mbox{ker}(\theta)$ denotes the congruence defined on $S$ by
$(a,b)\in\mbox{ker}(\theta)$ if and only if $\theta(a)=\theta(b)$. The
following theorem combines Lemma 2.11 with [14, bottom of page 23]. The proof
is now routine.

######  Lemma 2.12.

Let $\tau_{e}^{1}\colon S_{e}\rightarrow T_{1}$ be a cocone with base $S_{e}$
and target $T_{1}$ and let $\tau_{e}^{2}\colon S_{e}\rightarrow T_{2}$ be a
cocone with base $S_{e}$ and target $T_{2}$. Let $\theta_{1}$ be the
homomorphism with domain $S^{\circ}$ corresponding to the cocone
$\tau_{e}^{1}$, and let $\theta_{2}$ be the homomorphism with domain
$S^{\circ}$ corresponding to the cocone $\tau_{e}^{2}$, as described by Lemma
2.11. Then there is a unique monoid homomorphism $\alpha\colon
T_{1}\rightarrow T_{2}$ such that $\alpha\tau_{e}^{1}=\tau_{e}^{2}$ if and
only if $\mbox{\rm ker}(\theta_{1})\subseteq\mbox{\rm ker}(\theta_{2})$.

It follows from Lemma 2.11 and Lemma 2.12 that we can construct the colimit of
inverse monoids over a meet-semilattice using ideas only from semigroup
theory. In this paper, we are only interested in the meet semilattice
$\mathbb{N}^{\ast}=\mathbb{N}\setminus\\{0\\}$ with the meet given by $i\wedge
j=\mbox{\rm max}\\{i,j\\}$. The connecting homomorphisms are the maps
$\theta_{j}^{i}$ as defined earlier in this section.

The following result is important since it tells us that we shall be working
with meet-monoids in this paper.

######  Lemma 2.13.

Let the $(P,S_{e},\phi_{f}^{e})$ be a strong semilattice of inverse meet-
monoids where the connecting homomorphisms preserve meets. Then their colimit
is a meet-monoid.

###### Proof.

Denote the colimit of the $S_{e}$ by $T$. It is constructed as the inverse
monoid $S^{\circ}/\equiv$. We prove that this is a meet-monoid. Let $a\in
S_{e}$, $b\in S_{f}$. Then $\phi_{e\wedge f}^{e}(a),\phi_{e\wedge f}^{f}(b)\in
S_{e\wedge f}$. We may therefore form $\phi_{e\wedge
f}^{e}(a)\wedge\phi_{e\wedge f}^{f}(b)$ in $S_{e\wedge f}$. Define

$[a]\cap[b]=[\phi_{e\wedge f}^{e}(a)\wedge\phi_{e\wedge f}^{f}(b)],$

an element of $T$. We shall prove that in the inverse semigroup $T$ we have
that $[a]\wedge[b]=[a]\cap[b]$. To that end, let $[c]\leq[a],[b]$ where $c\in
S_{g}$. It follows that $[c]\leq[\phi_{e\wedge f}^{e}(a)],[\phi_{e\wedge
f}^{f}(b)]$. We therefore have that $c\equiv\phi_{e\wedge f\wedge
g}^{e}(a)\phi_{e\wedge f\wedge g}^{g}(c^{-1}c)$ and $c\equiv\phi_{e\wedge
f\wedge g}^{f}(b)\phi_{e\wedge f\wedge g}^{g}(c^{-1}c)$. We can find an
idempotent $1_{h}\in\mathscr{E}$ such that

$c\circ 1_{h}=\phi_{e\wedge f\wedge g}^{e}(a)\phi_{e\wedge f\wedge
g}^{g}(c^{-1}c)\circ 1_{h}=\phi_{e\wedge f\wedge g}^{f}(b)\phi_{e\wedge
f\wedge g}^{g}(c^{-1}c)\circ 1_{h}.$

It follows that

$c\circ 1_{h}=\phi_{e\wedge f\wedge g}^{e}(a)\phi_{e\wedge f\wedge
g}^{g}(c^{-1}c)\circ 1_{h}\wedge\phi_{e\wedge f\wedge g}^{f}(b)\phi_{e\wedge
f\wedge g}^{g}(c^{-1}c)\circ 1_{h}$

since we are assuming that each of our semigroups $S_{e}$ is a meet-monoid. By
Lemma 2.4, we have that

$c\circ 1_{h}=(\phi_{e\wedge f\wedge g}^{e}(a)\wedge\phi_{e\wedge f\wedge
g}^{f}(b))\phi_{e\wedge f\wedge g}^{g}(c^{-1}c)\circ 1_{h}.$

It follows that

$[c]=[(\phi_{e\wedge f\wedge g}^{e}(a)\wedge\phi_{e\wedge f\wedge
g}^{f}(b))\phi_{e\wedge f\wedge g}^{g}(c^{-1}c)].$

Thus $[c]\leq[a]\cap[b]$ where we have used the fact that the connecting
homomorphisms preserve meets. ∎

##  3\. Background on Boolean inverse monoids

In this section, we shall revise the important definitions around Boolean
inverse monoids. For more on such monoids, see [38].

###  3.1. General

An inverse semigroup is said to be distributive if every compatible pair of
elements has a join and multiplication distributes over such joins. A
distributive inverse semigroup is said to be Boolean if its semilattice of
idempotents is a Boolean algebra. The symmetric inverse monoids are always
Boolean. A morphism of Boolean inverse monoids is a homomorphism that
preserves the $0$ and $1$ and maps finite joins to finite joins. A meet-
morphism between Boolean inverse meet-monoids is a morphism of Boolean inverse
monoids that also preserves binary meets. The finite fundamental Boolean
inverse monoids are isomorphic to a finite direct product of finite symmetric
inverse monoids; we say that such Boolean inverse monoids are matricial. If
$a$ is an element of a Boolean inverse semigroup, define
$\mathbf{e}(a)=\mathbf{d}(a)\vee\mathbf{r}(a)$, called the extent of $a$.
Observe that $a\in\mathbf{e}(a)S\mathbf{e}(a)$. It follows that
$S=\bigcup_{e\in\mathsf{E}(S)}eSe$. The proof of the following is
straightforward.

######  Lemma 3.1.

Let $S$ be a Boolean inverse semigroup. If $e$ is an idempotent then $eSe$ is
a Boolean inverse monoid.

We say that $a$ and $b$ are orthogonal, written $a\perp b$, if
$a^{-1}b=0=ab^{-1}$. If an orthogonal pair of elements has a join then we
speak of orthogonal joins. The join of $a$ and $b$ is written $a\vee b$ and if
they are orthogonal their orthogonal join is written $a\oplus b$.444The
context will determine what is meant by the symbol $\oplus$ since we shall
also use it to denote the binary operation in an MV-algebra. Let $S$ be a
Boolean inverse semigroup. If $a\leq b$ define $b\setminus
a=b(\mathbf{d}(b)\setminus\mathbf{d}(a))$ where
$\mathbf{d}(b)\setminus\mathbf{d}(a)$ is defined in the Boolean algebra of
$S$. Let $S$ be a Boolean inverse monoid with group of units $\mathsf{U}(S)$
and Boolean algebra of idempotents $\mathsf{E}(S)$. The natural action of the
group $\mathsf{U}(S)$ on the unital Boolean algebra $\mathsf{E}(S)$ is given
by $g\cdot e=geg^{-1}$. The following are all easy to prove:

  1. (1)

$g\cdot(e\vee f)=g\cdot e\vee g\cdot f$.

  2. (2)

$g\cdot(e\wedge f)=g\cdot e\wedge g\cdot f$.

  3. (3)

$g\cdot\bar{e}=\overline{(g\cdot e)}$.

It follows that the natural action of the group of units of a Boolean inverse
monoid preserves the Boolean algebra structure of the Boolean algebra of
idempotents. The following was proved as [21, Proposition 3.1].

######  Lemma 3.2.

Let $S$ be a Boolean inverse monoid. Then $S$ is fundamental if and only if
the natural action of $\mathsf{U}(S)$ on $\mathsf{E}(S)$ is faithful.

######  Lemma 3.3.

A commutative fundamental Boolean inverse semigroup is a Boolean algebra.

###### Proof.

We prove that every element is an idempotent. Let $a$ be an arbitrary element
and let $e$ be any idempotent. Then, by commutativity, $ae=ea$. But we are
assuming that the semigroup is fundamental. Thus $a$ is an idempotent. ∎

Let $S$ be a finite Boolean inverse monoid containing elements $a$ and $b$.
The set of elements beneath $a$ and $b$ is a finite compatible set and so has
a join. This will be precisely the meet of $a$ and $b$. We have therefore
proved the following.

######  Lemma 3.4.

Every finite Boolean inverse monoid has meets and any morphism between finite
Boolean inverse monoids preserves those meets.

By Lemma 3.4, it follows that morphisms between finite Boolean inverse monoids
are automatically meet-morphisms.

###  3.2. Rook matrices

We shall need a way of constructing Boolean inverse semigroups. The following
is a matrix technique, the details of which can be found in [16, 38], but we
will outline it here now. Let $S$ be a Boolean inverse semigroup. Let $m$ and
$n$ be ordinals; we shall only need the cases where $m$ and $n$ are both
finite or $m=n=\omega$. An $m\times n$ matrix over $S$ is said to be a rook
matrix if it satisfies the following two conditions:

  1. (1)

In each column and each row only a finite number of elements are non-zero.

  2. (2)

If $a$ and $b$ are in the same row but distinct columns then $a^{-1}b=0$; if
$a$ and $b$ are in the same column but distinct rows then $ab^{-1}=0$.

We denote by $B_{n}(S)$ the set of all $n\times n$ rook matrices over $S$ and
by $R_{n}(S)$ the set of all $n\times n$ rook matrices over $S$ in which each
rook matrix contains only a finite number of non-zero entries. Multiplication
of rook matrices is defined in the usual way except that addition is replaced
by union (which makes sense because of the conditions we have imposed on the
rows and columns). If $A$ is a rook matrix then $A^{\ast}$ is defined to be
the transpose of $A$ with the inverse taken of all entries. It can be checked
that $A^{\ast}$ is a rook matrix such that $A=AA^{\ast}A$ and
$A^{\ast}=A^{\ast}AA^{\ast}$. If $S$ is a Boolean inverse semigroup then both
$B_{n}(S)$ and $R_{n}(S)$ are Boolean inverse semigroups. If $S$ is a Boolean
inverse monoid then $B_{\omega}(S)$ is a Boolean inverse monoid but
$R_{\omega}(S)$ is a Boolean inverse semigroup. On a point of notation, we
write $R_{n}$ instead of $R_{n}(\mathbf{2})$. Observe that
$R_{n}\cong\mathscr{I}_{n}$; we shall return to the nature of this isomorphism
later.

###  3.3. Subalgebras

Let $B$ be a unital Boolean algebra. We say that $B^{\prime}$ is a subalgebra
of $B$ if $B^{\prime}\subseteq B$ contains the 0 and 1 and is closed under
joins, meets and complements. Now, let $S$ and $T$ be Boolean inverse monoids.
Each morphism $\theta\colon S\rightarrow T$ induces a homomorphism of Boolean
algebras
$(\theta\mid\mathsf{E}(S))\colon\mathsf{E}(S)\rightarrow\mathsf{E}(S)$.

Definition. Let $S$ be a Boolean inverse monoid. An inverse submonoid $T$ of
$S$ will be called a subalgebra if the following two conditions hold:

  1. (1)

If $a,b\in T$ and $a\sim b$ then $a\vee b\in T$, where $a\vee b$ means the
join calculated in $S$.

  2. (2)

If $e\in\mathsf{E}(T)$ then $\bar{e}\in\mathsf{E}(T)$.

Wehrung [38, Definition 3.1.7] uses the terminology ‘additive inverse
subsemigroup’. The justification for our use of the term ‘subalgebra’ is based
on [38, Corollary 3.2.7].

######  Remark 3.5.

Observe that if $S$ is a Boolean inverse monoid and $T$ is a subalgebra of $S$
then units of $T$ are units of $S$; this is immediate since $S$ has the same
identity as any of its subalgebras.

######  Lemma 3.6.

  1. (1)

Let $S$ and $T$ be Boolean inverse monoids. Let $\theta\colon S\rightarrow T$
be a morphism which is also an embedding. Then the image of $S$ is a
subalgebra of $T$.

  2. (2)

If $S$ is a subalgebra of $T$ then the natural embedding of $S$ into $T$ is a
morphism.

###### Proof.

(1) Let $\theta\colon S\rightarrow T$ be an embedding and morphism of Boolean
inverse monoids. We prove that $\theta(S)$ is a subalgebra of $T$. Let
$\theta(a)\sim\theta(b)$ in $\theta(S)$. Then $a\sim b$ in $S$. Thus
$\theta(a\vee b)=\theta(a)\vee\theta(b)$ in $T$. It follows that joins in
$\theta(S)$ are the same as joins in $T$. Now, let $\theta(e)$ be an
idempotent in $\theta(S)$. Then $1=e\vee\bar{e}$ and $0=e\bar{e}$ in $S$. It
follows that $\overline{\theta(e)}=\theta(\bar{e})$.

(2) Straightforward from the definitions. ∎

The definition above has to be refined in the case where $S$ is a meet-monoid.
We then say that $T$ is a subalgebra of $S$ if $T$ is closed under meets
formed in $S$. We can modify Lemma 3.6 to accomodate meet-monoids: we replace
morphisms by meet-morphisms.

###  3.4. Additive ideals

Let $S$ be a Boolean inverse semigroup. A subset $I\subseteq S$, where $0\in
I$, is called an additive ideal if $I$ is a semigroup ideal of $S$ and $I$ is
closed under binary compatible joins. The following is a related notion. Let
$e$ and $f$ be non-zero idempotents. A finite set $X$ is said to be a pencil
from $f$ to $e$, denoted by $f\preceq e$, if $f=\bigvee\mathbf{d}(X)$ and
$\bigvee\mathbf{r}(X)\leq e$. If $S$ is a Boolean inverse monoid then both
$\\{0\\}$ and $S$ are additive ideals and if these are the only additive
ideals we say that $S$ is $0$-simplifying. A Boolean inverse semigroup which
is both $0$-simplifying and fundamental is called simple [19, Theorem 4.18];
we shall also require that $S\neq\\{0\\}$. The simple Boolean inverse monoids
are additive congruence-free. The finite simple Boolean inverse monoids are
precisly the finite symmetric inverse monoids. Let $S$ be a distributive
inverse monoid. If $X\subseteq S$ define $X^{\vee}$ to be the set of all joins
(in $S$) of finite compatible subsets of $X$.

######  Lemma 3.7.

Let $S$ be a Boolean inverse monoid. Let $T$ be an inverse submonoid of $S$
such that $\mathsf{E}(T)$ is a Boolean subalgebra of $\mathsf{E}(S)$. Then
$T^{\vee}$ is a subalgebra of $S$. In addition, if $T$ is finite then
$T^{\vee}$ is finite.

###### Proof.

Observe that since $T$ is an inverse submonoid of $S$ so too is $T^{\vee}$.
The fact that $T^{\vee}$ is distributive is inherited from $S$. The
idempotents of $T^{\vee}$ are the elements of the form
$e=\bigvee_{i=1}^{m}e_{i}$ where the $e_{i}\in\mathsf{E}(T)$. It follows that
the idempotents of $T^{\vee}$ are precisely the elements of $\mathsf{E}(T)$.
But $\mathsf{E}(T)$ is a Boolean algebra. It follows that $T^{\vee}$ is a
Boolean inverse monoid. It is therefore a subalgebra by construction. The fact
that $T^{\vee}$ is finite if $T$ is finite follows from the fact that the
number of finite compatible subsets of $T$ is finite. ∎

There is an important corollary to the above lemma.

######  Corollary 3.8.

Let $S$ be a Boolean inverse meet-monoid. Let $F$ be an inverse submonoid of
$S$ such that $\phi(g)\in F$ for each unit $g$ of $F$ where $\phi$ is the
fixed-point operator for $S$. Then $F^{\vee}$ is closed under all binary meets
in $S$.

###### Proof.

By Lemma 2.5, $F$ is closed under binary meets. The fact that $F^{\vee}$ is
closed under binary meets follows by [20, Corollary 2.6]. ∎

######  Lemma 3.9.

Let $S$ be a distributive inverse semigroup. If $I$ is a non-empty semigroup
ideal of $S$ then $I^{\vee}$ is an additive ideal of $S$.

###### Proof.

A typical element of $I^{\vee}$ is $\bigvee_{i=1}^{m}a_{i}$ where $a_{i}\in
I$. If $s\in S$ then
$s\left(\bigvee_{i=1}^{m}a_{i}\right)=\bigvee_{i=1}^{m}sa_{i}$, and dually. ∎

We shall now connect additive ideals of $S$ with certain subsets of
$\mathsf{E}(S)$. Let $S$ be a Boolean inverse semigroup. A subset
$F\subseteq\mathsf{E}(S)$, where $0\in F$, is called an additive ideal if it
is an order-ideal closed under conjugation by elements of $S$ and closed under
finite compatible joins. The proofs of the following are all straightforward.

######  Lemma 3.10.

Let $S$ be a Boolean inverse semigroup.

  1. (1)

Let $I$ be an additive ideal of $S$. Then $\mathsf{E}(I)$ is an additive ideal
of $\mathsf{E}(S)$.

  2. (2)

If $I\subseteq J$, where $I$ and $J$ are additive ideals, then
$\mathsf{E}(I)\subseteq\mathsf{E}(J)$.

  3. (3)

If $\mathsf{E}(I)=\mathsf{E}(J)$ then $I=J$.

######  Lemma 3.11.

Let $F$ be an additive ideal of $\mathsf{E}(S)$, where $S$ is a Boolean
inverse semigroup. If $e\in F$ and $f\preceq e$ then $f\in F$.

###### Proof.

By assumption, there is a finite set $X=\\{x_{1},\ldots,x_{m}\\}$ such that
$f=\bigvee_{i=1}^{m}\mathbf{d}(x_{i})$ and
$\bigvee_{i=1}^{m}\mathbf{r}(x_{i})\leq e$. Because $F$ is an order-ideal, we
have that $\bigvee_{i=1}^{m}\mathbf{r}(x_{i})\in F$. Thus
$\mathbf{r}(x_{i})\in F$ for $1\leq i\leq m$. Because $F$ is closed under
conjugation, we have that $\mathbf{d}(x_{i})\in F$ for $1\leq i\leq m$. We now
use the fact that $F$ is closed under finite compatible joins to get that
$f\in F$. ∎

The following is the key result we shall need later.

######  Lemma 3.12.

Let $S$ be a Boolean inverse semigroup. Then there is an order-isomorphism
between the poset of additive ideals of $S$ and the poset of additive ideals
of $\mathsf{E}(S)$.

###### Proof.

If $I$ is an additive ideal of $S$ then $\mathsf{E}(I)$ is an additive ideal
of $\mathsf{E}(S)$ by Lemma 3.10. If $F$ is an additive ideal of
$\mathsf{E}(S)$ then $(SFS)^{\vee}$ is an additive ideal of $S$ by Lemma 3.9.
We prove that $\mathsf{E}((SFS)^{\vee})=F$. Clearly,
$F\subseteq\mathsf{E}((SFS)^{\vee})$. Let $e$ be an idempotent of
$(SFS)^{\vee}$. Then $e$ is a finite join of idempotents $f$ in $SFS$. If
$f\in\mathsf{E}(SFS)$. Then $f\preceq e$ where $e\in F$. By Lemma 3.11, we
have that $f\in F$. But $F$ is closed under finite joins. Thus $e\in F$. It is
now routine to check that the maps: $I\mapsto\mathsf{E}(I)$ and
$F\mapsto(SFS)^{\vee}$ are mutually inverse isotone maps. ∎

###  3.5. Factorizable Boolean inverse monoids

Let $S$ be a Boolean inverse monoid. We say that a Boolean inverse monoid has
$\mathscr{D}$-complementation if $e\,\mathscr{D}\,f$ implies that
$\bar{e}\,\mathscr{D}\,\bar{f}$ for any idempotents $e,f$. The following was
proved as [23, Proposition 2.7].

######  Proposition 3.13.

Let $S$ be a Boolean inverse monoid. Then $S$ is factorizable if and only if
$S$ has $\mathscr{D}$-complementation.

######  Proposition 3.14.

Let $S$ be a Boolean inverse monoid. Then $S$ is factorizable if and only if
$S/\mu$ is factorizable.

###### Proof.

Only one direction needs proving. Let $e\,\mathscr{D}\,f$ in $S$ where $e$ and
$f$ are idempotents. Then $\mu(e)\,\mathscr{D}\,\mu(f)$. By assumption,
$S/\mu$ is factorizable. Thus
$\overline{\mu(e)}\,\mathscr{D}\,\overline{\mu(f)}$ by Proposition 3.13. But
$\mu$ induces an isomorphism of the Boolean algebras of idempotents. Thus
$\mu(\bar{e})\,\mathscr{D}\,\mu(\bar{f})$. It follows that
$\bar{e}\,\mathscr{D}\,\bar{f}$ in $S$ by Lemma 2.1. Thus $S$ is factorizable.
∎

######  Proposition 3.15.

All finite Boolean inverse monoids are factorizable.

###### Proof.

Let $S$ be a finite Boolean inverse monoid. Then $S/\mu$ is a finite
fundamental Boolean inverse monoid. This is factorizable since it is a finite
direct product of finite symmetric inverse monoids each of which is
factorizable. The result now follows by Proposition 3.14. ∎

The following lemma deals with what we might call ‘normalization’.

######  Lemma 3.16.

Let $S$ be a Boolean inverse monoid. Let $G\leq\mathsf{U}(S)$ be a finite
subgroup and let $E\subseteq\mathsf{E}(S)$ be a finite Boolean subalgebra.
Then there is a finite Boolean subalgebra $E^{\prime}\subseteq\mathsf{E}(S)$
such that $E\subseteq E^{\prime}$ and $E^{\prime}$ is closed under the natural
action of $G$.

###### Proof.

Let $E^{\prime}$ be the Boolean subalgebra of $\mathsf{E}(S)$ generated by the
finite set $F=\\{geg^{-1}\colon g\in G,e\in E\\}$. Observe that the set $F$ is
closed under complements since $g\bar{e}g^{-1}=\overline{geg^{-1}}$. Each
element of $E^{\prime}$ can be written as a join of terms each of which is a
meet of elements of $F$. It is immediate that $E^{\prime}$ is closed under the
natural action by $G$. ∎

The definition of factorizability presupposes the existence of an identity,
but we can also make a definition that does not require the existence of such
an identity. We say that a Boolean inverse semigroup $S$ is
$\mathscr{D}$-cancellative if $e_{1}\oplus f_{1}\,\mathscr{D}\,e_{2}\oplus
f_{2}$ and $e_{1}\,\mathscr{D}\,e_{2}$ implies that
$f_{1}\,\mathscr{D}\,f_{2}$.

######  Lemma 3.17.

Let $S$ be a Boolean inverse monoid. Then $S$ is $\mathscr{D}$-cancellative if
and only if $S$ has $\mathscr{D}$-complementation.

###### Proof.

Suppose that $S$ is $\mathscr{D}$-cancellative. We prove that $S$ has
$\mathscr{D}$-complementation. Suppose that $e\,\mathscr{D}\,f$. Then
$e\vee\bar{e}=1\,\mathscr{D}\,1=f\vee\bar{f}$. We now use the fact that $S$ is
$\mathscr{D}$-cancellative to deduce that $\bar{e}\,\mathscr{D}\,\bar{f}$.

Now, we assume that $S$ has $\mathscr{D}$-complementation. Suppose that
$e_{1}\oplus f_{1}\,\mathscr{D}\,e_{2}\oplus f_{2}$ where
$e_{1}\,\mathscr{D}\,e_{2}$. By assumption, $f_{1}\leq\overline{e_{1}}$ and
$f_{2}\leq\overline{e_{2}}$. By $\mathscr{D}$-complementation, we have that
$\overline{e_{1}}\,\overline{f_{1}}\,\mathscr{D}\,\overline{e_{2}}\,\overline{f_{2}}$.
It follows that
$e_{1}\vee\overline{e_{1}}\,\overline{f_{1}}\,\mathscr{D}\,e_{2}\vee\overline{e_{2}}\,\overline{f_{2}}$.
Now apply $\mathscr{D}$-complementation again to get
$\overline{e_{1}}(e_{1}\vee f_{1})\,\mathscr{D}\,\overline{e_{2}}(e_{2}\vee
f_{2})$. It follows that $f_{1}\,\mathscr{D}\,f_{2}$. ∎

######  Lemma 3.18.

Let $S$ be a Boolean inverse semigroup. If it is $\mathscr{D}$-cancellative
then $S$ is Dedekind finite.

###### Proof.

Suppose that $e\,\mathscr{D}\,f\leq e$. Then $e=f\oplus(e\setminus f)$. We
therefore have that $f\oplus(e\setminus f)\,\mathscr{D}\,f\oplus 0$. By
$\mathscr{D}$-cancellation, we deduce that $(e\setminus f)\,\mathscr{D}\,0$
and so $e\setminus f=0$ yielding $e=f$. ∎

###  3.6. Basic Boolean inverse monoids

A non-zero element $a$ of an inverse semigroup is said to be an infinitesimal
if $a^{2}=0$ [20]. Clearly, there are no idempotent infinitesimals and an
infinitesimal cannot be above a non-zero idempotent. The following is proved
as [20, Lemma 2.18(2)].

######  Lemma 3.19.

In an inverse semigroup, the following are equivalent:

  1. (1)

$a$ is an infinitesimal.

  2. (2)

$\mathbf{d}(a)\perp\mathbf{r}(a)$.

  3. (3)

$a\perp a^{-1}$.

The proof of the following is straightforward.

######  Lemma 3.20.

Let $S$ be an inverse semigroup. Any element beneath an infinitesimal is
either zero or an infinitesimal.

The proof of the following is straightforward.

######  Lemma 3.21.

Let $S$ be an inverse semigroup and let $x$ be any element.

  1. (1)

If $e$ is an idempotent then $xex^{-1}$ is an idempotent.

  2. (2)

If $a$ is an infinitesimal then $xax^{-1}$ is either zero or an infinitesimal.

The following is a slight generalization of [21, Proposition 2.6].

######  Proposition 3.22.

Let $S$ be an inverse semigroup whose semilattice of idempotents is
$0$-disjunctive. Then $S$ is Clifford if and only if $S$ contains no
infinitesimals.

###### Proof.

Assume, first, that the semigroup is Clifford. Let $a$ be an infinitesimal. By
Lemma 3.19, $a^{-1}aaa^{-1}=0$. In a Clifford semigroup, $a^{-1}a=aa^{-1}$
[14, Section IV.1]. Thus $a^{-1}a=0$ from which it follows that $a=0$, a
contradiction.

Assume, now, that there are no infinitesimals. Let $a\in S$. If
$a^{-1}aaa^{-1}=0$ then $a$ would be an infinitesimal. Thus
$e=\mathbf{d}(a)\mathbf{r}(a)\neq 0$. If $e<\mathbf{d}(a)$ then there is a
non-zero idempotent $f$ such that $f\leq\mathbf{d}(a)$ where $fe=0$. Put
$x=af$. Then $\mathbf{d}(x)=f\neq 0$. It follows that $x\neq 0$. But
$x^{2}=afaf=a\mathbf{d}(a)f\mathbf{r}(a)f=aefa=0$ and so $af$ is an
infinitesimal, which is a contradiction. Thus $e=\mathbf{d}(a)$ and so
$\mathbf{d}(a)\leq\mathbf{r}(a)$. By symmetry, we get that
$\mathbf{d}(a)=\mathbf{r}(a)$ and so $S$ is a Clifford semigroup. ∎

The following is proved as [24, Proposition 5.5].

######  Proposition 3.23.

Let $S$ be a Boolean inverse monoid. Then $S$ is fundamental if and only if
each non-idempotent element of $S$ is above an infinitesimal.

Recall that an involution in a group is a non-identity element $g$ such that
$g^{2}=1$. In Boolean inverse monoids, there is a close association between
infinitesimals and involutions. In the following, recall that
$\mathbf{e}(s)=s^{-1}s\vee ss^{-1}$.

######  Lemma 3.24.

Let $S$ be a Boolean inverse semigroup. Let $a$ be an infinitesimal where
$a\in eSe$. Then $a^{-1}\vee a\vee(e\setminus\mathbf{e}(a))$ is an involution
in the group of units of $eSe$.

###### Proof.

Because $a$ is an infinitesimal, $a\perp a^{-1}$ by Lemma 3.19. Thus $a\vee
a^{-1}$ is defined and $a\vee a^{-1}\in eSe$. Clearly, $\mathbf{e}(a)\leq e$
and $\mathbf{e}(a\vee a^{-1})=\mathbf{e}(a)$. Thus the element $g=a^{-1}\vee
a\vee(e\setminus\mathbf{e}(a))$ is well-defined. It is routine to check that
$\mathbf{d}(g)=e=\mathbf{r}(a)$ and that $g^{2}=e$. ∎

The following is now immediate from Lemma 3.24. It enables us to construct
involutions from infinitesimals.

######  Corollary 3.25.

Let $S$ be a Boolean inverse monoid. If $a$ is any infinitesimal then $g=a\vee
a^{-1}\vee\overline{(\mathbf{e}(a))}$ is an involution.

The following definition and an explanation of where it came from was first
given in [20].

Definition. A Boolean inverse semigroup is said to be basic if each non-zero
element can be written as a join of an idempotent (possibly $0$) and a finite
number of infinitesimals.

The proof of the following is straightforward.

######  Lemma 3.26.

The finite product of basic Boolean inverse semigroups is a basic Boolean
inverse semigroup.

Let $S$ be a Boolean inverse monoid with group of units $\mathsf{U}(S)$. Then
$S$ is said to be piecewise factorizable if for each $a\in S$ we can write
$a=\bigvee_{i=1}^{n}g_{i}e_{i}$ where the $g_{i}$ are units and the $e_{i}$
are idempotents. The following is proved as [20, Lemma 4.30]. We give proofs
anyway.

######  Proposition 3.27.

Let $S$ be a basic Boolean inverse monoid. Then $S$ is:

  1. (1)

A meet-monoid.

  2. (2)

Piecewise factorizable.

  3. (3)

Fundamental.

###### Proof.

(1) This follows by Proposition 2.3. (2) This follows by Corollary 3.25. (3)
This follows by Proposition 3.23. ∎

######  Proposition 3.28.

Let $S$ be a finite Boolean inverse monoid. Then $S$ is fundamental if and
only if $S$ is basic.

###### Proof.

By Proposition 3.27 we know that being basic implies that the semigroup is
fundamental We prove the converse. Because $S$ is fundamental, by Proposition
3.23, each non-idempotent element is above an infinitesimal. Since binary
meets exist in finite Boolean inverse monoids, there is a fixed-point operator
$\phi\colon S\rightarrow\mathsf{E}(S)$ by Proposition 2.3. Let $a$ be any non-
idempotent element of $S$. Then the only idempotent beneath
$a\setminus\phi(a)$ is $0$. By assumption, there is an infinitesimal
$a_{1}\leq a\setminus\phi(a)$. If $(a\setminus\phi(a))\setminus a_{1}=0$ then
we are done otherwise $(a\setminus\phi(a)\setminus a_{1}<a\setminus\phi(a)$
and the process can be repeated. Because of finiteness, we deduce that we can
write $a\setminus\phi(a)$ as a join of a finite number of infinitesimals. We
have therefore shown $a$ can be written as the join of a finite number of
infinitesimals and an idempotent. We have therefore proved that $S$ is basic.
∎

The following corollary is immediate from Proposition 3.28 but is also clear
from the properties of finite symmetric inverse monoids and an application of
Lemma 3.26; it is the motivation for defining basic Boolean inverse monoids in
the first place.

######  Corollary 3.29.

Finite direct products of finite symmetric inverse monoids are basic.

##  4\. Characterization theorems

By Lemma 2.13, the directed colimit of inverse meet-monoids is an inverse
meet-monoid. By Lemma 3.4, every finite Boolean inverse monoid is a meet-
monoid and morphisms between finite Boolean inverse monoids automatically
preserve meets. These results imply that in this section we should focus on
Boolean inverse meet-monoids. We begin by characterizing the Boolean inverse
monoids of finite type.

######  Theorem 4.1 (Boolean inverse monoids of finite type).

Let $S$ be a countable Boolean inverse meet-monoid. Then $S$ is of finite type
if and only if $S$ is factorizable and its group of units is locally finite.

###### Proof.

We prove the easy direction first. Suppose that $S$ is of finite type. Then
$S=\bigcup_{i=1}^{\infty}S_{i}$ where $S_{1}\subseteq S_{2}\subseteq
S_{3}\subseteq\ldots$ and where each $S_{i}$ is a finite subalgebra of $S$.
Let $a\in S$. Then $a\in S_{i}$ for some $i$. But $S_{i}$ is factorizable by
Proposition 3.15. Thus $a\leq g$ where $g$ is a unit of $S_{i}$. It follows
that $g$ is a unit of $S$ since $S_{i}$ is a subalgebra of $S$. Thus $S$ is
factorizable. Now, let $X$ be any finite subset of the group of units of $S$.
Then $X\subseteq S_{i}$ for some $i$ where $S_{i}$ is a subalgebra of $S$. It
follows that $X$ are units of $S_{i}$ which is finite. Hence, the subgroup
generated by $X$ is finite. We have therefore proved that the group of units
of $S$ is locally finite.

We now prove the converse. Let $S$ be a countable Boolean inverse meet-monoid
which is factorizable and whose group of units is locally finite. We prove
that we can write $S=\bigcup_{i=1}^{\infty}S_{i}$ where $S_{1}\subseteq
S_{2}\subseteq\ldots$ and the $S_{i}$ are finite subalgebras.

Since $S$ is countable both the group $\mathsf{U}(S)$ and the Boolean algebra
$\mathsf{E}(S)$ are countable. Let $\mathsf{U}(S)=\\{g_{1}=1,g_{2},\ldots\\}$
and $\mathsf{E}(S)=\\{e_{1}=1,e_{2}\ldots\\}$. Define $G_{i}=\langle
g_{1},\ldots,g_{i}\rangle$ and $E_{i}=\langle e_{1},\ldots,e_{i}\rangle$.
Since $S$ is locally finite the groups $G_{i}$ are all finite. The Boolean
algebras $E_{i}$ are finite since they are finitely generated by Lemma 2.7. By
construction $G_{1}\subseteq G_{2}\subseteq\ldots$ and $E_{1}\subseteq
E_{2}\subseteq\ldots$. As a first step, we may enlarge each $E_{i}$ by
including $\phi(g)$ for each $g\in G_{i}$ and then taking the Boolean algebra
generated. We shall assume this done in what follows. By Lemma 3.16, we may
enlarge each Boolean algebra $E_{i}$ to a Boolean algebra $E_{i}^{\prime}$ in
such a way that $E_{i}^{\prime}$ is closed under the natural action by $G_{i}$
and such that $E_{1}^{\prime}\subseteq E_{2}^{\prime}\subseteq\ldots$: to do
this, first enlarge $E_{1}$ to $E_{1}^{\prime}$ and then enlarge $\langle
E_{1}^{\prime}\cup E_{2}\rangle$ to get $E_{2}^{\prime}$ and then enlarge
$\langle E_{2}^{\prime}\cup E_{3}\rangle$ to get $E_{3}^{\prime}$ and so on.
We relabel $E_{i}^{\prime}$ as $E_{i}$. We may therefore assume that each
$E_{i}$ is closed under the natural action of $G_{i}$. We now apply parts (1)
and (2) of Lemma 2.2 and Lemma 2.5, to get an increasing sequence
$F_{i}=G_{i}E_{i}$ of factorizable inverse meet-submonoids $F_{1}\subseteq
F_{2}\subseteq F_{3}\subseteq\ldots$. We now apply Lemma 3.7 and Corollary 3.8
to get an increasing sequence $F_{1}^{\vee}\subseteq F_{2}^{\vee}\subseteq
F_{3}^{\vee}\subseteq\ldots$ of finite subalgebras. Now, let $a\in S$ be
arbitrary. Then $a=ge$ for some unit $g$ and idempotent $e$ by
factorizability. It follows that $a\in F_{i}$ for some $i$ and so $a\in
F_{i}^{\vee}$. ∎

We now turn to the characterization of AF inverse monoids. We need the
following result first.

######  Proposition 4.2.

Let $S$ be a countable, basic Boolean inverse monoid whose group of units is
locally finite. Then $S$ is factorizable.

###### Proof.

We shall use the fact that $S$ has all binary meets by Proposition 3.27. By
Proposition 3.27, the monoid $S$ is piecewise factorizable. Let
$a=\bigvee_{i=1}^{n}g_{i}e_{i}$ where the $g_{i}$ are units and the $e_{i}$
are idempotents. It is enough to prove that $a\leq g$ where $g$ is a unit. Put
$G=\langle g_{1},\ldots,g_{n}\rangle$ which is finite since the group of units
of $S$ is locally finite. Let $\mathscr{E}=\\{e_{1},\ldots,e_{n}\\}$. Let $E$
be the Boolean algebra generated by $\mathscr{E}$ together with the
idempotents $\phi(g)$ where $g\in G$ which is also finite by Lemma 2.7. By
Lemma 3.16, there is a finite Boolean subalgebra $B$ of $\mathsf{E}(S)$
invariant under the natural action by $G$. Put $T=GB$. By Lemma 2.2 and Lemma
2.6, $T$ is a factorizable inverse meet-monoid with semilattice of idempotents
$B$. By Lemma 3.7 and Corollary 3.8, $T^{\vee}$ is a finite Boolean inverse
monoid and subalgebra of $S$. It follows by Proposition 3.15, that $T^{\vee}$
is factorizable. By construction, $a\in T^{\vee}$. The result follows. ∎

We now come to our second main theorem.

######  Theorem 4.3 (AF monoids).

Let $S$ be a countable Boolean inverse meet-monoid. Then $S$ is AF if and only
if $S$ is basic and its group of units is locally finite.

###### Proof.

We prove the easy direction first. Let $S$ be an AF inverse monoid. Then
$S=\bigcup_{i=1}^{\infty}S_{i}$ where each $S_{i}$ is a matricial subalgebra
of $S$; that is, each $S_{i}$ is a finite fundamental Boolean inverse monoid.
By Proposition 3.28, each $S_{i}$ is basic and so, since $S_{i}$ is a
subalgebra of $S$, it follows that $S$ is basic. The fact that the group of
units of $S$ is locally finite follows by the same argument as that used in
the proof of Theorem 4.1.

We now prove the converse. Let $S$ be a countable basic Boolean inverse monoid
whose group of units is locally finite. We prove that we can write
$S=\bigcup_{i=1}^{\infty}S_{i}$ where $S_{1}\subseteq S_{2}\subseteq\ldots$
and the $S_{i}$ are matricial subalgebras. By Proposition 4.2, we know that
$S$ is factorizable. Thus by Theorem 4.1, we can write
$S=\bigcup_{i=0}^{\infty}S_{i}$ where the $S_{i}$ are finite Boolean inverse
monoids each of which is a subalgebra of $S$. We shall have proved our result
if we prove that every finite subalgebra (which is a Boolean inverse monoid)
can be embedded into a finite fundamental subalgebra. Let $T$ be a finite
subalgebra of $S$. We shall show that there is a finite fundamental subalgebra
$T^{\prime}$ of $S$ such that $T\subseteq T^{\prime}$. This will prove the
result. Since $T$ is a finite Boolean inverse monoid we know that is
factorizable. We can write $T=\mathsf{U}(T)\mathsf{E}(T)$. Since $S$ is basic,
for each $g\in\mathsf{U}(T)$, we may write $g=a_{1}\vee\ldots\vee a_{m}\vee e$
where the $a_{1},\ldots,a_{m}$ are infinitesimals and $e$ is an idempotent.
Put $\iota(g)=\\{\mathbf{d}(a_{1}),\ldots,\mathbf{d}(a_{m}),e\\}$ and call
this a signature set of idempotents of $g$; observe that $\phi(g)=e$. Put
$\mathsf{E}(S)^{\prime}$ equal to the set $\mathsf{E}(S)$ together with the
sets $\iota(g)$ for each of the $g\in\mathsf{U}(S)$, and then adjoin all the
conjugates of these idempotents by elements of $\mathsf{U}(S)$. Now, define
$B$ to be the finite Boolean algebra generated by these idempotents. By
construction $T_{1}=\mathsf{U}(T)B$ is a factorizable inverse meet-submonoid
of $S$ with $B$ as its Boolean algebra of idempotents. Put
$T^{\prime}=T_{1}^{\vee}$. This is a finite subalgebra of $S$ that contains
$T$. We have used both Lemma 2.6 and Corollary 3.8 above. It remains to show
that $T^{\prime}$ is fundamental. We shall show that $T^{\prime}$ is basic and
use Proposition 3.28. Each element of $a\in T^{\prime}$ is of the form
$a=\bigvee_{i=1}^{m}g_{i}e_{i}$ where $g_{i}\in\mathsf{U}(T)$ and $e_{i}\in
B$. If we can show that each $g_{i}e_{i}$ is a join of infinitesimals and an
idempotent then the element $a$ will also be a join of infinitesimals and an
idempotent. Let $g\in\mathsf{U}(T)$. By assumption $\iota(g)\subseteq B$. It
follows that in $T^{\prime}$ we can write $g=a_{1}\vee\ldots\vee a_{m}\vee e$
where $a_{1},\ldots,a_{m}\in\mathsf{U}(S)B$ are inifinitesimals and $e\in B$.
If $f\in B$ then $gf=a_{1}f\vee\ldots\vee a_{m}f\vee ef$. If we weed out the
products which are zero, we get a representation of $gf$ as a sum of
infinitesimals and an idempotent, using Lemma 3.20. We have therefore proved
that $T^{\prime}$ is basic and so it is fundamental by Proposition 3.27. ∎

##  5\. Finite simple Boolean inverse monoids

In the next section, we shall characterize the UHF monoids. Since these are
constructed from increasing unions of finite simple Boolean inverse monoids,
we shall begin by describing their properties. Recall that every finite simple
Boolean inverse monoid is isomorphic to a finite symmetric inverse monoid by
[19, Theorem 4.18]. However, in this section, we shall be interested in the
nature of those isomorphisms not only their mere existence.

###  5.1. Normalized invariant means

Let $S$ be a Boolean inverse monoid. The following definitions are taken from
[16]. A function $\nu\colon\mathsf{E}(S)\rightarrow[0,\infty)$ is called an
invariant mean if for any $s\in S$ we have that $\nu(s^{-1}s)=\nu(ss^{-1})$
and if $e$ and $f$ are orthogonal idempotents then $\nu(e\vee
f)=\nu(e)+\nu(f)$. We say that $\nu$ is normalized if $\nu(1)=1$. In the case
of normalized invariant means, we have that
$\nu\colon\mathsf{E}(S)\rightarrow[0,1]$ by [16, Lemma 2.1(3)]. The proof of
the following can be found at [16, Lemma 2.17(1)], but we give a direct proof.

######  Lemma 5.1.

Finite symmetric inverse monoids $\mathscr{I}_{n}$ have exactly one normalized
invariant mean.

###### Proof.

Each non-zero idempotent is the join of a finite number of orthogonal atomic
idempotents and, since $\mathscr{I}_{n}$ is $0$-simplifying, any two atomic
idempotents are $\mathscr{D}$-related. Let the atomic idempotents be
$e_{1},\ldots,e_{n}$. Then $1=e_{1}\oplus\ldots\oplus e_{n}$. It follows that
if $\nu$ is a normalized invariant mean then $\nu(e)=\frac{1}{n}$ where $e$ is
any atomic idempotent. Thus if a normalized invariant mean exists it is
unique. We now prove existence. If $1_{A}$ is any idempotent in
$\mathscr{I}_{n}$, where $A$ is some subset of $\\{1,2,3,\ldots,n\\}$, define
$\nu(1_{A})=\frac{\left|A\right|}{n}$. Then $\nu$ is a normalized invariant
mean. ∎

We denote the unique normalized invariant mean on the finite simple Boolean
inverse monoid $S$ by $\nu_{S}$.

######  Lemma 5.2.

Let $S$ be a finite simple Boolean inverse monoid. Then for
$e,f\in\mathsf{E}(S)$ we have that $e\,\mathscr{D}\,f$ if and only if
$\nu_{S}(e)=\nu_{S}(f)$.

###### Proof.

Only one direction needs proving. Suppose that $\nu_{S}(e)=\nu_{S}(f)$.
Observe that $e=1_{A}$ and $f=1_{B}$ and, by assumption, the cardinalities of
$A$ and $B$ are the same. It follows that $e\,\mathscr{D}\,f$. ∎

The proof of Lemma 5.3 is immediate by the proof of Lemma 5.2.

######  Lemma 5.3.

Let $S$ be a finite simple Boolean inverse monoid. Then $S/\mathscr{J}$ is
linearly ordered.

######  Lemma 5.4.

Let $S$ and $T$ be finite simple Boolean inverse monoids. Let $\theta\colon
S\rightarrow T$ be a morphism. Then for all $e\in\mathsf{E}(S)$ we have that
$\nu_{S}(e)=\nu_{T}(\theta(e))$.

###### Proof.

Observe first that $\nu_{T}\theta\colon\mathsf{E}(S)\rightarrow[0,\infty)$. It
is easy to check that this function satisfies the conditions to be a
normalized invariant mean. By the uniqueness guaranteed by Lemma 5.1, it
follows that $\nu_{T}\theta$ restricted to the idempotents of $S$ is just
$\nu_{S}$. ∎

###  5.2. Morphisms between finite simple Boolean inverse monoids

We shall need a concrete description of finite simple Boolean inverse monoids
as finite symmetric inverse monoids. The following is immediate by [23, Lemma
2.5], but we give a direct proof.

######  Lemma 5.5.

Let $\theta\colon S\rightarrow T$ be a morphism between finite simple Boolean
inverse monoids (recall that: $0\neq 1$). Then $\theta$ is injective.

###### Proof.

Observe first that because $S$ is $0$-simplifying the set of all elements that
are mapped to zero is either $\\{0\\}$ or $S$ itself. However, identities are
mapped to identities and we are assuming that $0\neq 1$. It follows that
$\theta(x)=0$ if and only if $x=0$. Suppose, now, that $\theta(a)=\theta(b)$.
By Lemma 3.4, finite Boolean inverse monoids are meet-monoids and $\theta$
preserves meets. Put $c=a\wedge b$. Then $\theta(a)=\theta(c)$. But $c\leq a$,
so we may write $a=c\vee(a\setminus c)$ where $c\perp a\setminus c$. It
follows that $\theta(a)=\theta(c)\vee(\theta(a)\setminus\theta(c))$ and
$\theta(c)\perp\theta(a)\setminus\theta(c)$. Thus $\theta(a\setminus c)=0$ and
so $a\setminus c=0$. It follows that $a=a\wedge b$. By symmetry, $b=a\wedge
b$. We have therefore shown that $a=b$, as required. ∎

######  Remark 5.6.

Our finite symmetric inverse monoids will also be on a non-empty set of
letters so the condition in Lemma 5.5 that $0\neq 1$ will always be satisfied.

######  Proposition 5.7.

Let $S$ be a finite simple Boolean inverse monoid. Then $S$ is isomorphic to
the symmetric inverse monoid $\mathscr{I}(\mathscr{E}(S))$ where
$\mathscr{E}(S)$ is the set of atomic idempotents of $S$.

###### Proof.

Recall from the Introduction that if $S$ is a finite Boolean inverse monoid
with groupoid of atoms $G$ then $S$ is isomorphic to $\mathsf{K}(G)$. Denote
by $\mathscr{E}(S)$ the set of atomic idempotents of $S$. Let $a\in S$. Then
for each atom $e\leq\mathbf{d}(a)$ we have that $aea^{-1}$ is an atom. Thus
$a$ induces a partial map $f_{a}$ on $\mathscr{E}(S)$. Define
$S\rightarrow\mathcal{I}(\mathscr{E}(S))$ by $a\mapsto f_{a}$. It is easy to
check that this is a bijection from the set of atoms beneath $\mathbf{d}(a)$
to the set of atoms beneath $\mathbf{r}(a)$. The fact that $S$ is fundamental
shows that this is injective. We now show that
$S\rightarrow\mathcal{I}(\mathscr{E}(S))$ is surjective. Let $f\colon
E\rightarrow F$ be an element of $\mathcal{I}(\mathscr{E}(S))$. Let $e\in E$.
Then $f(e)\in F$. By assumption, $e\,\mathscr{D}\,f(e)$ and there is a unique
element that witnesses this. Thus there is an element $a_{f,e}$ such that
$\stackrel{{\scriptstyle a_{f,e}}}{{e\longrightarrow f}}$. Define $a$ to be
the (othogonal) join of the elements $a_{f,e}$. Then $f_{a}=f$. We have
therefore defined a bijection. It remains to show that this is a homomorphism.
Let $e$ be an atom less than or equal to $\mathbf{d}(a)\mathbf{r}(b)$. Then
$i=b^{-1}eb$ is an atom less than or equal to $\mathbf{d}(ab)$ and all such
atoms are of this form. Observe that
$(f_{a}f_{b})(i)=a(bib^{-1})a^{-1}=ab(i)(ab)^{-1}=f_{ab}(i)$. ∎

We now describe the isomorphisms between finite simple Boolean inverse monoids
and finite symmetric inverse monoids.

######  Lemma 5.8.

Let $S$ and $T$ be finite simple Boolean inverse monoids.

  1. (1)

Let $\alpha,\beta\colon S\rightarrow T$ be isomorphisms which agree on the
atomic idempotents of $S$. Then $\alpha=\beta$.

  2. (2)

Let $e_{1},\ldots,e_{n}$ be a particular listing of the atomic idempotents of
$S$ and let $f_{1},\ldots,f_{n}$ be a particular listing of the atomic
idempotents of $T$. Then there is a unique isomorphism from $S$ to $T$ that
maps $e_{i}$ to $f_{i}$ where $1\leq i\leq n$.

###### Proof.

(1) Let $x$ be an atom of $S$. Then $\mathbf{r}(x)$ and $\mathbf{d}(x)$ are
atomic idempotents. Put $e=\alpha(\mathbf{d}(x))=\beta(\mathbf{d}(x))$ and
$f=\alpha(\mathbf{r}(x))=\beta(\mathbf{r}(x))$. Both $e$ and $f$ are atomic
idempotents since $\alpha$ and $\beta$ are isomorphisms. The elements
$\alpha(x)$ and $\beta(x)$ both begin at $e$ and end at $f$. By simplicity
$\alpha(x)=\beta(x)$. Thus $\alpha$ and $\beta$ agree on all the atoms of $S$.
However, each element of $S$ is a finite join of atoms. Thus $\alpha=\beta$.

(2) Define $\theta(e_{i})=f_{i}$. Let $e_{i}\stackrel{{\scriptstyle
x}}{{\rightarrow}}e_{j}$ be the unique element. Define $\theta(x)$ to be the
unique element with domain $f_{i}$ and codomain $f_{j}$. We can now define
$\theta$ on all elements of $S$ since every non-zero element of $S$ is a
finite join of atoms. Define $\theta(0)=0$. It is clearly a bijection. It
remains to show that it is a homomorphism. Observe that $\theta$ induces a
functor from the groupoid $\mathsf{at}(S)$ to the groupoid $\mathsf{at}(T)$.
Now, observe that if $x$ and $y$ are atoms either $xy=0$ or $xy$ is an atom
where $\mathbf{d}(x)=\mathbf{r}(y)$. It can now be checked that $\theta$ is a
homomorphism. ∎

We shall now use rook matrices. Let $R_{n}$ be the set of $n\times n$ rook
matrices over the set $\\{0,1\\}$, regarded as the $2$-element unital Boolean
algebra. They consist of all $n\times n$ matrices with at most one non-zero
entry in each row and column. The atomic idempotents of $R_{n}$ are the rook
matrices $E_{i}$ which are zero everywhere except that the element $(i,i)$ is
$1$. In what follows, we use the word permutation to mean a listing of
elements without repetition.

######  Lemma 5.9.

Let $S$ be a finite simple Boolean inverse monoid with $n$ atomic idempotents
listed thus: $e_{1},\ldots,e_{n}$. Then there is an isomorphism from $S$ to
$R_{n}$. Each such isomorphism is determined by a permutation of the atomic
idempotents of $S$.

###### Proof.

By Proposition 5.7, there is an isomorphism between $S$ and
$\mathscr{I}(\mathscr{E}(S))$ where $\mathscr{E}(S)$ is the set of atomic
idempotents of $S$. Suppose that $\mathscr{E}(S)$ consists of $n$ elements
which we list as follows: $e_{1},\ldots,e_{n}$. With respect to this order,
define an isomorphism from $\mathscr{I}(\mathscr{E}(S))$ to $R_{n}$ which
sends, in particular, the partial bijection with domain $\\{e_{i}\\}$ and
range $\\{e_{j}\\}$ to the $n\times n$ rook matrix that has a $1$ in row $j$
and column $i$ and $0$s everywhere else. In this way, we obtain an isomorphism
from $S$ to $R_{n}$ The proof of the final statement follows by part (2) of
Lemma 5.8. ∎

Let $S$ be a Boolean inverse monoid. Let $g$ be a unit. Then the function
$s\mapsto gsg^{-1}$ is an automorphism of $S$ called an inner automorphism; we
say that this inner automorphism is induced by $g$. Suppose now that $S$ is
fundamental. Then the function $s\mapsto gsg^{-1}$ is the identity function if
and only if $g$ is the identity; this follows by Lemma 3.2. In this case, two
inner automorphisms are the same if and only if they are induced by the same
unit.

######  Lemma 5.10.

Each automorphism of $R_{n}$ is induced by a permutation of the atomic
idempotents of $R_{n}$. In particular, every automorphism of $R_{n}$ is an
inner automorphism.

###### Proof.

Every automorphism of $R_{n}$ is determined by its effect on the atomic
idempotents of $R_{n}$; in fact, every automorphism permutes the atomic
idempotents. To prove the result, we simply need to show that every
permutation of the atomic idempotents can be implemented by an inner
automorphism. Denote the atomic idempotents of $R_{n}$ by
$E_{1},\ldots,E_{n}$. Let $\sigma$ be any permutation, in the usual sense, of
the set $\\{1,2,\ldots,n\\}$. Consider now the list
$E_{\sigma(1)},\ldots,E_{\sigma(n)}$. We shall prove that there is an inner
automorphism $\theta$ such that $E_{i}\mapsto E_{\sigma(i)}$. Let $A$ be the
permutation matrix that arises when $\sigma$ is used to permute the rows of
the $n\times n$ identity matrix $I$. We claim that
$A^{-1}E_{i}A=E_{\sigma(i)}$. We know that $A^{-1}E_{i}A$ has to be one of the
atomic idempotents. Thus we need only show that
$(A^{-1}E_{i}A)_{\sigma(i),\sigma(i)}=1$. In fact,
$(A^{-1}E_{i}A)_{\sigma(i),\sigma(i)}=A_{i,\sigma(i)}$ but
$A_{i,\sigma(i)}=1$. ∎

We shall now prove that morphisms between finite simple Boolean inverse
monoids have a normal form. Let $A,B\in R_{m}$. Then $A\sim B$ if and only if
the sum $A+B$ — in the sense of pointwise Boolean joins — is a rook matrix in
which case $A\vee B=A+B$. Let $A$ be an $m\times m$ rook matrix and $B$ an
$n\times n$ rook matrix. We denote by $A\oplus B$ the $(m+n)\times(m+n)$ rook
matrix

$\left(\begin{array}[]{cc}A&0\\\ 0&B\end{array}\right).$

Let $s$ be any positive natural number. Then we may iterate the above
construction

$sA=\overbrace{A\oplus\ldots\oplus A}^{s\,\text{\small times}}.$

Let $n=sm$. Then we can define a map $\sigma_{s}\colon R_{m}\rightarrow R_{n}$
by $A\mapsto sA$. This is a semigroup homomorphism that maps the zero to the
zero and the identity to the identity. It is now routine to check that
$\sigma$ is a morphism of Boolean inverse monoids. We call this a standard
morphism.

The following is just a variant of [23, Lemma 3.1], but we give the proof
anyway.

######  Lemma 5.11.

Let $S$ and $T$ be finite simple Boolean inverse monoids. Let the number of
atomic idempotents in $S$ be $m$ and let the number of atomic idempotents on
$T$ be $n$. If there is a morphism from $S$ to $T$ then $m\mid n$.

###### Proof.

Let the atomic idempotents in $S$ be $e_{1},\ldots,e_{m}$. Then
$1=e_{1}\oplus\ldots\oplus e_{m}$. Suppose there is a morphism $\theta\colon
S\rightarrow T$. Then $1=\theta(e_{1})\oplus\ldots\theta(e_{m})$, where now
$1$ is the identity in $T$. Since $S$ is simple, the idempotents
$e_{1},\ldots,e_{m}$ are all $\mathscr{D}$-related. It follows that the
idempotents $\theta(e_{1}),\ldots,\theta(e_{m})$ are all
$\mathscr{D}$-related. Let $e$ and $f$ be idempotents in a finite Boolean
inverse monoid such that $e\,\mathscr{D}\,f$. Then the number of idempotent
atoms below $e$ is the same as the number of idempotent atoms below $f$. Let
the number of idempotent atoms beneath $\theta(e_{1})$ be $s$. Then the number
of idempotent atoms beneath each of the idempotents $\theta(e_{i})$ is also
$s$. Since the total number of idempotent atoms in $T$ is $n$ it follows that
$n=sm$. We have therefore proved that $m\mid n$. ∎

We shall now connect the above result with standard morphisms. We shall use
Proposition 5.7. Let $S$ be a finite simple Boolean inverse monoid with $m$
atomic idempotents. Linearly order this set. We may therefore identify this
set with the linearly ordered set $1,2,\ldots,m$. On the basis of this linear
order, we can define an isomorphism $\alpha\colon S\rightarrow R_{m}$. The
element $s\in S$ induces a partial bijection $f$ of the set $1,2,\ldots,m$
which leads to the rook matrix $A$ where $A$ has a 1 in the row $i$ and column
$j$ if and only if $i=f(j)$.

The following is a corrected version of [23, Lemma 3.2]; we have to say more
precisely how we shall order the atoms below each idempotent $\theta(e_{i})$
than is said in [23]. It tells us that morphisms between finite simple Boolean
inverse monoids are always standard morphisms in disguise.

######  Lemma 5.12 (Morphisms between simple semigroups).

Let $S$ and $T$ be finite simple Boolean inverse monoids. Let the number of
atomic idempotents in $S$ be $m$ and let the number of atomic idempotents in
$T$ be $n$ and suppose that $n=sm$. Let $\theta\colon S\rightarrow T$ be a
morphism and let $\alpha\colon S\rightarrow R_{m}$ be any isomorphism. Then
there is an isomorphism $\beta\colon T\rightarrow R_{n}$ and a standard
morphism $\sigma\colon R_{m}\rightarrow R_{n}$ such that
$\sigma=\beta\theta\alpha^{-1}$.

###### Proof.

The reader will need to refer back to the proof of Lemma 5.11. Let the linear
ordering of the atomic idempotents of $S$ that determines the isomorphism
$\alpha$ be $e_{1},\ldots,e_{m}$. Denote the set of atomic idempotents in $T$
below $\theta(e_{i})$ by $A_{i}$. Then, the set of atomic idempotents of $T$
is partitioned by the set $\\{A_{1},\ldots,A_{m}\\}$ and each set $A_{i}$ has
$s$ elements. We shall linearly order the atomic idempotents of $T$ by first
linearly ordering each of the sets $A_{i}$. Inside $S$, for each $1\leq i\leq
m$, there is exactly one element $\stackrel{{\scriptstyle
x_{i}}}{{e_{1}\longrightarrow e_{i}}}$. In particular, $x_{1}=e$. We have that
$\stackrel{{\scriptstyle\theta(x_{i})}}{{\theta(e_{1})\longrightarrow\theta(e_{i})}}$.
Linearly order the elements of $A_{1}$ arbitrarily. Now linearly order each of
the sets $A_{i}$ in such a way that $\theta(x_{i})$ induces an order-
isomorphism from $A_{1}$ to $A_{i}$. We have therefore linearly ordered each
of the sets $A_{1},\ldots,A_{m}$.

We can now linearly order the set of atomic idempotents of $T$. We do this by
choosing the first elements from each of the sets $A_{1},\ldots,A_{m}$ in turn
and writing them down in order; then choosing the second elements in turn and
writing them down next, and so on. This gives us our linear ordering of the
atomic idempotents of $T$. This, in turn, induces an isomorphism $\beta\colon
T\rightarrow R_{n}$.

We shall now prove that $\beta\theta\alpha^{-1}=\sigma$. To do this, we
observe that the atoms of $S$ are the elements of the form $x_{i}x_{j}^{-1}$
and that each element of $S$ is a join of some of the atoms. Therefore, to
prove the claim, it will be enough to calculate what $\beta\theta\alpha^{-1}$
does to each element $x_{i}$. Observe that $\alpha(x_{i})=A\in R_{m}$ where
$A$ has a $1$ in row $i$ and column $1$ and zeros everywhere else. We now
calculate $\beta(\theta(x_{i}))$. Observe that
$\stackrel{{\scriptstyle\theta(x_{i})}}{{\theta(e_{1})\rightarrow\theta(e_{i})}}$.
A routine calculation shows that
$\beta(\theta(x_{i}))=sA=\sigma(\alpha(x_{i}))$. ∎

We may now phrase what we have found in categorical terms. The following
theorem is immediate by the results we have proved above.

######  Theorem 5.13.

The category of rook matrices and standard morphisms is a subcategory of the
category of finite simple Boolean inverse monoids and their morphisms. Let $S$
and $T$ be any finite simple Boolean inverse monoids, let $\theta\colon
S\rightarrow T$ be any morphism and let $\alpha\colon S\rightarrow R_{m}$ be
any isomorphism. Then there is an isomorphism $\beta\colon T\rightarrow R_{n}$
such that $\beta\theta\alpha^{-1}$ is a standard morphism. In particular,
every morphism between finite simple Boolean inverse monoids is isomorphic to
a standard morphism.

##  6\. UHF monoids

In this section, we shall characterize the UHF monoids. We could try to obtain
such a characterization along the lines of [36], but we shall use MV-algebras
instead.

###  6.1. MV-algebras

We refer the reader to [2] for background information on MV-algebras. An MV-
algebra $(A,\oplus,\neg,0)$ is a commutative monoid $(A,\oplus,0)$555Our use
of the symbol $\oplus$ is not to be confused by our use of this symbol to
denote the orthogonal join of two elements of a Boolean inverse monoid.
together with a unary operation $\neg$ that satisfies the following three
axioms; in what follows, we define $1=\neg 0$:

  1. (1)

$\neg\neg x=x$.

  2. (2)

$x\oplus 1=1$.

  3. (3)

$\neg(\neg x\oplus y)\oplus y=\neg(\neg y\oplus x)\oplus x$.

If $x$ is an element of an MV-algebra and $n$ is a natural number we define

$nx=\overbrace{x\oplus\ldots\oplus x}^{n\,\text{\small times}}.$

By [2, Lemma 1.1.2], we may define a partial order on an MV-algebra by $x\leq
y$ if $y=z\oplus x$ for some $z$. The unit interval $[0,1]$ becomes an MV-
algebra with respect to the following definitions:

$x\oplus y=\mbox{min}(x+y,1)\,\text{and}\,\neg x=1-x.$

The partial order is the usual partial order. Boolean algebras are MV-
algebras; define an idempotent in an MV-algebra to be an element $x$ such that
$x\oplus x=x$. Then by [2, Theorem 1.5.3], the idempotent elements of an MV-
algebra form a Boolean algebra. It follows that Boolean algebras are the MV-
algebras in which every element is an idempotent. An ideal $I$ of an MV-
algebra $A$ is a subset that contains $0$, is an order-ideal and is closed
under $\oplus$. For $n\geq 2$, define $\L_{n}$ to be the MV-algebra with
cardinality $n$ that contains the elements $\frac{r}{n-1}$ where $0\leq r\leq
n-1$ and which is an MV-subalgebra of $[0,1]$. These are precisely the finite
simple MV-algebras. More generally, we have the following by [2, Theorem
3.5.1]:

######  Proposition 6.1.

The simple MV-algebras are precisely those which can be embedded in the MV-
algebra $[0,1]$.

By a rational MV-algebra we mean an MV-subalgebra of $\mathbb{Q}\cap[0,1]$. By
[3, Theorem 5.1], these are precisely the simple, locally finite MV-algebras.
The finite rational MV-algebras are precisely those of the form $\L_{n}$ by
[2, Corollary 3.5.4].

We can construct all MV-algebras as follows. Let $G$ be a lattice-ordered
abelian group (usually called an $l$-group). An order-unit in $G$ is positive
element $u$ such that for every positive element $x$ there is a natural number
$n$ such that $x\leq nu$. Given such an element $u$, define $[0,u]$ to be all
the elements $x$ of $G$ such that $0\leq x\leq u$. Define $x\oplus
y=u\wedge(x+y)$ and $\neg x=u-x$. Then $([0,u],\oplus,\neg,0)$ is an MV-
algebra which we denote by $\Gamma(G,u)$, see [2, Proposition 2.1.2]. In fact,
every MV-algebra arises in this way, and $\Gamma$ is an equivalence of
categories between the category of $l$-groups with distinguished order-units
and the category of MV-algebras. See [2, Chapter 7] and [30]. Observe that the
MV-algebra $\mathbb{Q}\cap[0,1]=\Gamma(\mathbb{Q},1)$. More generally, the
rational MV-algebras are all obtained in this way from the subgroups of the
rationals that contain $1$; such subgroups are classified by the supernatural
numbers (see below).

######  Lemma 6.2.

There is a homomorphism from $\L_{m}$ to $\L_{n}$ if and only if
$(m-1)\mid(n-1)$. If there is such a homorphism then it is unique and it is
completely determined by where the atom of $\L_{n}$ is sent.

###### Proof.

Let $\theta\colon\L_{m}\rightarrow\L_{n}$ be a homomorphism. Since $\L_{m}$ is
simple, $\theta$ is an injection by [2, Lemma 1.2.3(iii)]. In [2, Remark, page
71], it is stated that $\L_{m}$ is an MV-subalgebra of $\L_{n}$ if and only if
$(m-1)\mid(n-1)$. The proof of our first claim now follows. If $a$ is the atom
of $\L_{m}$ then every element of $\L_{m}$ is of the form $pa$ for some
natural number $p$; specifically, the elements of $\L_{m}$ are:
$0a,1a,\ldots,(m-1)a=1$. It follows that any homomorphism with domain $\L_{m}$
is determined by what it does to this atom. If $\theta$ is such a homomorphism
then $(m-1)\theta(a)=1$. If $\phi$ is any such homomorphism then we also have
that $(m-1)\phi(a)=1$. We deduce that in $\L_{n}$ we have the equality
$(m-1)\theta(a)=(m-1)\phi(a)$ and so $\theta(a)=\phi(a)$. ∎

###  6.2. Supernatural numbers

Put $\mathbb{N}^{\ast}=\mathbb{N}\setminus\\{0\\}$. We order
$\mathbb{N}^{\ast}$ via divisibility $\mid$. In this way, $\mathbb{N}^{\ast}$
becomes a distributive lattice where $m\wedge n=\gcd{(m,n)}$ and $m\vee
n=\operatorname{lcm}{(m,n)}$. This much is elementary; see [32]. Recall that
an ideal in a lattice is an order-ideal which is closed under binary joins.
Denote the set of all ideals in $\mathbb{N}^{\ast}$ by
$\mbox{Id}(\mathbb{N}^{\ast})$. We say that an ideal $I$ of
$\mathbb{N}^{\ast}$ is bounded if there is an element $n\in I$ such that $n$
does not properly divide any element in $I$. We call $n$ a bound for $I$.

######  Lemma 6.3.

The bounded ideals of $\mathbb{N}^{\ast}$ have the form $n^{\downarrow}$ for
some non-zero natural number $n$.

###### Proof.

It is clear that $n^{\downarrow}$ is a bounded ideal with bound $n$. Let $I$
be a bounded ideal with bound $n$. We prove that $I=n^{\downarrow}$. One
inclusion is immediate. Let $m\in I$. We prove that $m\mid n$. By assumption,
$m\vee n\in I$. But $n\mid m\vee n$. It follows that $m\vee n=n$. Thus $m\mid
n$, as claimed. ∎

Thus every bounded ideal in $\mathbb{N}^{\ast}$ is of the form
$n^{\downarrow}$ for some $n\in\mathbb{N}^{\ast}$. An ideal which is not
bounded is said to be unbounded. We shall now describe the unbounded ideals
using the concept of a division sequence. Let
$\sigma=s_{1},s_{2},s_{3},\ldots$ where $s_{1}\mid s_{2}$ and $s_{2}\mid
s_{3}$, …, but where only proper divisions occur. We shall call this a
division sequence. Each $s_{i}$ is called a term of $\sigma$. Define
$I_{\sigma}$ to be all those elements of $\mathbb{N}^{\ast}$ that divide some
term of $\sigma$.

######  Lemma 6.4.

Let $\sigma$ be a division sequence. Then $I_{\sigma}$ is an ideal.

###### Proof.

Clearly, $\bigcup_{i=1}^{\infty}s_{i}^{\downarrow}\subseteq I_{\sigma}$. Let
$m,n\in I_{\sigma}$. Then $m\mid s_{i}$ and $n\mid s_{j}$ for some $i$ and
$j$. Without loss of generality, we may assume that $s_{i}\mid s_{j}$. Thus,
by assumption, $m,n\mid s_{j}$. It follows that $m\vee n\mid s_{j}$. Thus
$m\vee n\in I_{\sigma}$. ∎

We now prove the converse to Lemma 6.4.

######  Lemma 6.5.

Let $I$ be an unbounded ideal. Then there is a division sequence $\sigma$ such
that $I=I_{\sigma}$.

###### Proof.

Let $I$ be any unbounded ideal. By assumption, it is countably infinite. Let
$I\setminus\\{1\\}=\\{a_{1},a_{2},a_{3},\ldots\\}$. Put $s_{1}=a_{1}$,
$s_{2}=s_{1}\vee a_{2}$, $s_{3}=s_{2}\vee a_{3}$, …. Define
$\sigma=s_{1},s_{2},s_{3}\ldots$. By construction, $\sigma$ is a division
sequence. We prove that $I=I_{\sigma}$. Let $a_{i}\in I$. Then $a_{i}\mid
s_{i}$. It follows that $I\subseteq I_{\sigma}$. The reverse inclusion holds
since $I$ is closed under binary joins. ∎

It is clear that many division sequences determine one and the same ideal. We
now determine when division sequences determine the same ideal. Let
$\sigma=s_{1},s_{2},s_{3}\ldots$ and $\tau=t_{1},t_{2},t_{3}\ldots$ be
division sequences. We say that they are interleaved if and only if two
conditions hold: for each $i$ there exists a $j$ such that $s_{i}\mid t_{j}$
and for each $j$ there exists an $i$ such that $t_{j}\mid s_{i}$.

######  Lemma 6.6.

Let $\sigma$ and $\tau$ be division sequences. Then $I_{\sigma}=I_{\tau}$ if
and only if $\sigma$ and $\tau$ are interleaved.

###### Proof.

Suppose first that $I_{\sigma}=I_{\tau}$. Let $s_{i}$ be any term of $\sigma$.
Then $s_{i}\in I_{\sigma}$. By assumption, $s_{i}\in I_{\tau}$. Therefore
there is a term $t_{j}$ of $\tau$ such that $s_{i}\mid t_{j}$. By symmetry, we
have proved that $\sigma$ and $\tau$ are interleaved. Conversely, suppose that
$\sigma$ and $\tau$ are interleaved. Let $a\in I_{\sigma}$. Then $a\mid s_{i}$
for some $i$. By assumption, $s_{i}\mid t_{j}$ for some $j$. It follows that
$a\mid t_{j}$. Thus $a\in I_{\tau}$. We have proved that $I_{\sigma}\subseteq
I_{\tau}$. The reverse inclusion follows by symmetry. ∎

A supernatural number is a function $\nu\colon
P\rightarrow\mathbb{N}\cup\\{\infty\\}$ where $P$ is the set of primes. From
[3], there is an order-isomorphism between the set of ideals of
$\mathbb{N}^{\ast}$ and the set of supernatural numbers $\mathbb{S}$. We now
describe this order-isomorphism. List the primes in order
$p_{1},p_{2},p_{3},\ldots$. Let $I$ be any ideal of $\mathbb{N}^{\ast}$.
Denote the corresponding supernatural number by $\nu(I)$; informally, we can
regard this as a formal infinite product
$\prod_{i=1}^{\infty}p_{i}^{\nu(p_{i})}$ but formally we usually just focus on
the infinite sequence $\nu(p_{1}),\nu(p_{2}),\nu(p_{3}),\ldots$. For each
prime $p$, there are two possibilities: either there is a largest power of $p$
that belongs to $I$ or there is no bound to the powers of $p$ in $I$; in the
former case, define $\nu(p)$ be the largest power of $p$ that belongs to $I$
in the latter case define $\nu(p)=\infty$. The reason why
$\mbox{Id}(\mathbb{N}^{\ast})\cong\mathbb{S}$, induced by $\nu$, is that an
ideal is determined by the prime powers it contains since if $m\in I$, where
$m\neq 1$, and if $m=q_{1}^{s_{1}}\ldots q_{k}^{s_{k}}$ is the prime
factorization of $m$ then $q_{1}^{s_{1}},\ldots,q_{k}^{s_{k}}\in I$, since $I$
is an order-deal, and $m=q_{1}^{s_{1}}\vee\ldots\vee q_{k}^{s_{k}}$. If
$\sigma$ is a division sequence denote the corresponding supernatural number
by $\mathbf{n}_{\sigma}$. With this in mind, we can rephrase Lemma 6.6 as
follows.

######  Corollary 6.7.

Let $\sigma$ and $\tau$ be two division sequences. Then $\sigma$ and $\tau$
are interleaved if and only if $\mathbf{n}_{\sigma}=\mathbf{n}_{\tau}$.

Elements of $\mathbb{N}^{\ast}$ we shall call finite supernatural numbers. A
supernatural number that is not finite is said to be infinite. We shall assume
that our supernatural numbers are infinite.

Supernatural numbers completely describe the subgroups of $\mathbb{Q}$ that
contain $1$; see [34, Proposition 7.4.3]. This means that supernatural numbers
parametrize the rational MV-algebras. We now sketch out how. Let $\mathbf{n}$
be a supernatural number regarded as an infinite formal product of prime
powers. Denote by $\L_{\mathbf{n}+1}$ the set consisting of $0$ and $1$ and of
all positive fractions $\frac{p}{q}$ such that $0\leq\frac{p}{q}\leq 1$
satisfying two conditions: (1) $p\wedge q=1$ and (2) $q$ is a product of
primes each of which divides some term of $\mathbf{n}$. The peculiar notation
arises for the following reason. Consider the finite supernatural number $n$.
Then, by definition, $\L_{n+1}$ consists of $0$ and $1$ and of all positive
fractions $\frac{p}{q}$ such that $0\leq\frac{p}{q}\leq 1$ satisfying two
conditions: (1) $p\wedge q=1$ and (2) $q$ is a product of primes each of which
divides $n$ — this means precisely that $q$ divides $n$. We may therefore
identify the elements of $\L_{n+1}$ with the MV-algebra of all fractions
$\frac{p}{n}$ where $0\leq p\leq n$. The cardinality of this set is $n+1$.

###  6.3. The partial type monoid (or type interval) of a Boolean inverse
semigroup

Let $S$ be a Boolean inverse semigroup. Recall that
$\mathsf{E}(S)/\mathscr{D}$ is equipped with a partially defined addition
$[e]+[f]=[e^{\prime}\vee f^{\prime}]$ where $e\,\mathscr{D}\,e^{\prime}$ and
$f\,\mathscr{D}\,f^{\prime}$ and $e^{\prime}f^{\prime}=0$; here $[e]$ denotes
the set of all idempotents $e_{1}$ such that $e_{1}\,\mathscr{D}\,e$. Put
$\mathbf{0}=[0]$ (and $\mathbf{1}=[1]$ in the monoid case). Put
$\mathsf{Int}(S)=\mathsf{E}(S)/\mathscr{D}$ with the above partial addition.
This structure is called the partial type monoid; Wehrung [38, Definition
4.1.3] calls this the type interval. By [16], this is a partial commutative
monoid satisfying the refinement property. We say that a partial commutative
monoid is conical if $\exists e+f=0$ implies that $e=f=0$. The following is
stated in [23] but because of its importance we state it again and give the
(simple) proof.

######  Lemma 6.8.

Let $S$ be a Boolean inverse semigroup. Then $\mathsf{Int}(S)$ is conical.

###### Proof.

Suppose that $\exists[e]+[f]=[0]$. Then there exist
$e^{\prime}\,\mathscr{D}\,e$ and $f^{\prime}\,\mathscr{D}\,f$ such that
$e^{\prime}f^{\prime}=0$ and $e^{\prime}\vee f^{\prime}=0$. Thus
$e^{\prime}=f^{\prime}=0$ and so $e=f=0$. ∎

######  Proposition 6.9.

Let $S$ be a Boolean inverse semigroup. The partial type monoid of $S$ is
cancellative if and only if $S$ is $\mathscr{D}$-cancellative.

###### Proof.

We prove first that $S$ being $\mathscr{D}$-cancellative is equivalent to the
partial type monoid being cancellative. Suppose first that the partial type
monoid is cancellative. We prove that $S$ is $\mathscr{D}$-cancellative.
Suppose that $e_{1}\oplus e_{2}\,\mathscr{D}\,f_{1}\oplus f_{2}$ and
$e_{1}\,\mathscr{D}\,f_{1}$. Then $[e_{1}]+[e_{2}]=[f_{1}]+[f_{2}]$ and
$[e_{1}]=[f_{1}]$. Since the partial type monoid is cancellative, we deduce
that $[e_{2}]=[f_{2}]$ and so $e_{2}\,\mathscr{D}\,f_{2}$. Thus $S$ is
$\mathscr{D}$-cancellative. If, now, $S$ is $\mathscr{D}$-cancellative then it
is routine to check that the partial type monoid is cancellative. ∎

We shall show that for a special class of Boolean inverse monoids (the Foulis
monoids below), the partial type monoid is naturally an MV-algebra. We shall
not do this directly, instead we shall use special kinds of effect algebras
which are MV-algebras in disguise. A partial commutative monoid is called a
generalized effect algebra if is conical and cancellative. In a generalized
effect algebra, the preorder $\leq$ is actually a partial order. A generalized
effect algebra with a maximum element is called an effect algebra [31, 23].
Effect algebras were introduced by Foulis and Bennet [8] as an abstraction of
the algebraic structure of self-adjoint operators with spectrum in $[0,1]$
(quantum effects). For more on effect algebras, see [1, 8, 7] and [6]. Denote
the partial addition on an effect algebra by $+$. We may define an order
relation on an effect algebra by $x\leq y$ if and only if there exists $z$
such that $x+z=y$. An effect algebra is called an MV-effect algebra if it is a
lattice with respect to this partial order and satisfies the Riesz
decomposition property; see, [26, Section 1.4]. The key point is that there is
a connection between MV-effect algebras and MV-algebras; see [26, Theorem
1.4.3]. In an MV-algebra, define $p\perp q$ if and only if $p\leq\neg q$. Now,
define a partial addition by $a+b=a\oplus b$ if and only if $a\perp b$ and
undefined otherwise. Then we get an MV-effect algebra. Conversely, given an
MV-effect algebra, we can define the binary operation of an MV-algebra by
$a\oplus b=a+(\neg a\wedge b)$.

Now back to Boolean inverse semigroups $S$. We have that $\mathsf{Int}(S)$ is
a generalized effect algebra if and only if $S$ is $\mathscr{D}$-cancellative.
The element $[e]$ is maximum in $\mathsf{Int}(S)$ (assuming that $S$ is
$\mathscr{D}$-cancellative) if and only if $S=SeS$. We have therefore proved
the following.

######  Proposition 6.10.

Let $S$ be a Boolean inverse semigroup. Then $\mathsf{Int}(S)$ is an effect
algebra if and only if $S$ is $\mathscr{D}$-cancellative and $S=SeS$ for some
idempotent $e$.

By Lemma 3.1, if $e$ is an idempotent of a Boolean inverse semigroup $S$ then
$eSe$ is a Boolean inverse monoid with identity $e$. It follows by [38,
Theorem 4.2.2], that if we are interested in the effect algebras that arise
from Boolean inverse semigroups, then we may restrict our attention to Boolean
inverse monoids which are $\mathscr{D}$-cancellative.

We now have a different proof of [23, Theorem 2.13].

######  Proposition 6.11.

Let $S$ be a Boolean inverse monoid. Then the partial type monoid of $S$ is an
effect algebra if and only if $S$ is factorizable.

###### Proof.

Suppose first that the partial type monoid of $S$ is an effect algebra. Then
$S$ is $\mathscr{D}$-cancellative by Proposition 6.10. By Lemma 3.17, it
follows that $\mathscr{D}$-complementation holds. Thus by Proposition 3.13, it
follows that $S$ is factorizable. Conversely, suppose that $S$ is
factorizable. Then by Proposition 3.13, $S$ satisfies
$\mathscr{D}$-complementation. By Lemma 3.17, $S$ is
$\mathscr{D}$-cancellative. Finally, by Proposition 6.10, we deduce that the
partial type monoid is an effect algebra. ∎

The above result tells us when the partial type monoid of a Boolean inverse
monoid is an effect algebra. Our goal is to determine for which Boolean
inverse monoids the partial type monoid is in fact an MV-effect algebra.

###  6.4. The type monoid of a Boolean inverse semigroup

The partial type monoid can be embedded in a commutative monoid in two
(equivalent) ways. One can either adopt the approach described in [38,
Definition 4.1.3] or the equivalent approach using $R_{\omega}(S)$ in [16]. In
either case, we denote the resulting monoid by $\mathsf{Typ}(S)$ and call it
the type monoid of $S$; the binary operation in $\mathsf{Typ}(S)$ is also
denoted by $+$. There is a map
$\tau\colon\mathsf{E}(S)\rightarrow\mathsf{Typ}(S)$. The type monoid is always
a refinement monoid. The following is [38, Theorem 5.1.11].

######  Theorem 6.12.

Let $S$ and $T$ be AF inverse monoids. Then there is an isomorphism
$(\mathsf{Typ}(S),\mathbf{1})\cong(\mathsf{Typ}(T),\mathbf{1})$ (that is, an
isomorphism of commutative monoids that maps distinguished element to
distinguished element) if and only if there is an isomorphism $S\cong T$.

###  6.5. Foulis monoids

We shall be interested in those Boolean inverse monoids where the partial type
monoid can be endowed with the structure of an MV-algebra. We shall go via MV-
effect algebras. We say that a Boolean inverse monoid $S$ satisfies the
lattice condition if $S/\mathscr{J}$ is a lattice under subset-
inclusion.666The reader should note that this is an ordering on the principal
ideals of the inverse semigroup and is therefore not the same as the natural
partial order; for example, in the symmetric inverse monoid $\mathscr{I}_{2}$
the idempotents $e=1_{\\{1\\}}$ and $f=1_{\\{2\\}}$ are such that $e\wedge
f=0$ but they generate the same principal ideal since $e\,\mathscr{D}\,f$. If
$SiS=SeS\cap SfS$ define $[e]\wedge[f]=[i]$ in the partial type monoid, which
is well-defined if $\mathscr{D}=\mathscr{J}$. A Foulis monoid is defined to be
a factorizable Boolean inverse monoid satisfying the lattice condition.777This
is different from the way the term was used in [23] where the ‘lattice
condition’ was omitted. Let $S$ be a Foulis monoid. Put $\L(S)$ equal to the
partial type monoid as a set. By [7, 5.1 Definition], [7, 5.5 Theorem] and [1,
Lemma 2.7], if $S$ is a factorizable Boolean inverse monoid satisfying the
lattice condition, then $(\L(S),\oplus,\neg,\mathbf{0},\mathbf{1})$ is an MV-
algebra when we define $\neg[e]=[\bar{e}]$ and

$[e]\oplus[f]=[e]+([\bar{e}]\wedge[f]).$

We say that the MV-algebra $A$ is coordinatized by the Foulis monoid $S$ if
$A\cong\L(S)$. It is known that every MV-algebra is coordinatized by some
Boolean inverse monoid; the case of countable MV-algebras was established in
[23] and the general case was proved in [38, Theorem 5.2.10].

From [11, Chapter IV, Section 1], each commutative monoid $M$ has a universal
group $\mathsf{G}(M)$. This is an abelian group equipped with a mononoid
homomorphism $\gamma\colon M\rightarrow\mathsf{G}(M)$ which is universal. This
is often known as the Grothendieck construction. If $M$ is cancellative then
$\gamma$ is injective. Recall that we say that a monoid is conical if it has a
trivial group of units. If $M$ is cancellative and conical then
$\mathsf{G}(M)$ is a partially ordered abelian group. Recall that in the
conical case, we can define a partial order $\leq$ by $x\leq y$ if and only if
$x+z=y$ for some $z$. If $M$ is a lattice with respect to this order then
$\mathsf{G}(S)$ is a lattice-ordered abelian group (or, $l$-group). We now
apply the above results to $\mathsf{Int}(S)$ and $\mathsf{Typ}(S)$ where $S$
is a Boolean inverse monoid. The type monoid $\mathsf{Typ}(S)$ is always
conical: by [38, Corollary 4.1.4], $\mathsf{Typ}(S)$ is a conical refinement
monoid and $\mathsf{Int}(S)$ is an order-ideal. By Proposition 6.11, [38,
Lemma 3.6] and [38, Corollary 2.7.4], if $S$ is factorizable then
$\mathsf{Typ}(S)$ is cancellative. From [23, Proposition 2.12], we have that
$e\preceq f$ if and only if $[e]\leq[f]$ in $\mathsf{Int}(S)$. If $S$
satisfies the lattice condition, then $\mathsf{Int}(S)$ is a lattice; it
follows by [38, Lemma 3.14], that $\mathsf{G}(\mathsf{Typ}(S))$ is an
$l$-group with positive cone $\mathsf{Typ}(S)$. In this case,
$[\mathbf{0},\mathbf{1}]=\mathsf{Int}(S)$. We summarise what we have found as
follows.

######  Proposition 6.13.

Let $S$ be a Foulis monoid. Then $\mathsf{G}(\mathsf{Typ}(S))$ is an
$l$-group, $\mathbf{1}$ is an order-unit, the positive cone of this $l$-group
is $\mathsf{Typ}(S)$, and $[\mathbf{0},\mathbf{1}]=\mathsf{Int}(S)$.

We now calculate the MV-algebra that underlies a Foulis monoid in the light of
Proposition 6.13. To make our calculations easier, we shall assume that
$S/\mathscr{J}$ is linearly ordered. This implies that the associated
$l$-group is linearly ordered. The underlying set of this MV-algebra is
$\mathsf{Int}(S)$. The sum is given by
$[e]\oplus[f]=\mathbf{1}\wedge([e]\dotplus[f])$, where the $\dotplus$ refers
to the addition in the group $\mathsf{G}(\mathsf{Typ}(S))$. We prove first
that $[e]+[f]$ is defined in $\mathsf{Int}(S)$ if and only if $SfS\subseteq
S\bar{e}S$. Suppose first that $[e]+[f]$ is defined. Then
$e\,\mathscr{D}\,e_{1}$, $f\,\mathscr{D}\,f_{1}$ and $e_{1}\perp f_{1}$ for
some idempotents $e_{1}$ and $f_{1}$. By definition $[e]+[f]=[e_{1}\vee
f_{1}]$. It follows that $SfS\subseteq S\bar{e}S$. Conversely, suppose that
$SfS\subseteq S\bar{e}S$. Then it is almost immediate that $[e]+[f]$ is
defined. Observe that in the case where $SfS\subseteq S\bar{e}S$ then
$[e]+[f]=[e]\oplus[f]$. If this does not hold then $S\bar{e}S\subseteq SfS$
from which it follows that $[e]\oplus[f]=[1]$. However, if
$[e]\dotplus[f]\nleq[1]$ then $[1]\leq[e]\dotplus[f]$. It follows in this case
that $[1]\wedge([e]\dotplus[f])=[1]$. We have therefore proved the following;
it is a refinement of [23, Proposition 4.5].

######  Proposition 6.14.

Let $S$ be a Foulis monoid in which $S/\mathscr{J}$ is linearly ordered. Then
$\L(S)=\Gamma(\mathsf{G}(\mathsf{Typ}(S)),\mathbf{1})$.

We can now refine Theorem 6.12. This will be crucially important for our
characterization of UHF monoids later.

######  Theorem 6.15.

Let $S$ and $T$ be Foulis monoids which are also AF monoids where both
$S/\mathscr{J}$ and $T/\mathscr{J}$ are linearly ordered. Then
$\L(S)\cong\L(T)$ as MV-algebras if and only if $S\cong T$.

###### Proof.

Only one direction needs proving. Suppose that $\L(S)\cong\L(T)$. Then by
Proposition 6.14, we have that
$\Gamma(\mathsf{G}(\mathsf{Typ}(S)),\mathbf{1})\cong\Gamma(\mathsf{G}(\mathsf{Typ}(S)),\mathbf{1})$.
By [30], the isomorphism is an isomorphism of $l$-groups that maps
distinguished order-units to distinguished order-units. It follows that the
positive cones of $\mathsf{G}(\mathsf{Typ}(S))$ and
$\mathsf{G}(\mathsf{Typ}(T))$ are isomorphic and so the type monoids of $S$
and $T$, respectively, are isomorphic. The result now follows from Theorem
6.12. ∎

######  Proposition 6.16.

Let $S$ be a Foulis monoid. Then there is an order-isomorphism between the
additive ideals of $\mathsf{E}(S)$ and the MV-algebra ideals of $\L(S)$.

###### Proof.

Let $F$ be an additive ideal of $\mathsf{E}(S)$. Put $[F]=\\{[e]\colon e\in
F\\}$. Then $[F]$ is an ideal in $\L(S)$. If $F_{1}\subseteq F_{2}$ then
$[F_{1}]\subseteq[F_{2}]$ with equality if and only if $F_{1}=F_{2}$. Thus we
have a map given by $F\mapsto[F]$.

Now let $\mathcal{I}\subseteq\L(S)$ be an ideal. Define
$\mathsf{E}(\mathcal{I})=\\{e\in\mathsf{E}(S)\colon[e]\in\mathcal{I}\\}$. Then
$\mathsf{E}(\mathcal{I})$ is an additive ideal of $\mathsf{E}(S)$. There are
three conditions to check each of which is easy: it is an order-ideal; it is
closed under orthogonal joins and so it is closed under arbitrary joins; it is
closed under the $\mathscr{D}$-relation. Suppose that
$\mathcal{I}_{1}\subseteq\mathcal{I}_{2}$. Then
$\mathsf{E}(\mathcal{I}_{1})\subseteq\mathsf{E}(\mathcal{I}_{2})$ with
equality if and only if $\mathcal{I}_{1}=\mathcal{I}_{2}$. Thus we have a map
given by $\mathcal{I}\mapsto\mathsf{E}(\mathcal{I})$

We now show that the above maps are mutually inverse. Let $F$ be an additive
ideal in $\mathsf{E}(S)$. Then $[F]$ is an ideal in $\L(S)$. Thus
$\mathsf{E}([F])$ is an additive ideal in $\mathsf{E}(S)$. Observe that
$F\subseteq\mathsf{E}([F])$. Let $f\in\mathsf{E}([F])$. Then $[f]\in[F]$. Thus
$[f]=[e]$ where $e\in F$. But $E$ is closed under the $\mathscr{D}$-relation
and so $f\in F$. We have proved that $F=\mathsf{E}([F])$.

Let $\mathcal{I}$ be an ideal in $\L(S)$. Then $\mathsf{E}(\mathcal{I})$ is an
additive ideal in $\mathsf{E}(S)$. Thus $[\mathsf{E}(\mathcal{I})]$ is an
ideal in $\L(S)$. Let $[e]\in\mathcal{I}$. Then $e\in\mathsf{E}(\mathcal{I})$.
Thus $[e]\in[\mathsf{E}(\mathcal{I})]$. It follows that
$\mathcal{I}\subseteq[\mathsf{E}(\mathcal{I})]$. Let
$[e]\in[\mathsf{E}(\mathcal{I})]$. Then $e\in\mathsf{E}(\mathcal{I})$. Thus
$[e]\in\mathcal{I}$. We have therefore proved that
$\mathcal{I}=[\mathsf{E}(\mathcal{I})]$. ∎

The following is immediate by Proposition 6.16 and Lemma 3.12.

######  Corollary 6.17.

Let $S$ be a Foulis monoid. Then there is an order-isomorphism between the
additive ideals of $S$ and the MV-algebra ideals of $\L(S)$.

We have the following by Corollary 6.17. It is the one of the keys to our
characterization theorem of UHF monoids.

######  Corollary 6.18.

Let $S$ be a Foulis monoid. Then $S$ is $0$-simplifying if and only if $\L(S)$
is simple.

It is of interest to determine for which Foulis monoids the associated MV-
algebra is actually a Boolean algebra.

######  Lemma 6.19.

Let $S$ be a Foulis monoid. Then $[e]\in\L(S)$ is an idempotent in the MV-
algebra if and only if $SeS\cap S\bar{e}S=\\{0\\}$.

###### Proof.

Suppose that $SeS\cap S\bar{e}S=\\{0\\}$ for each idempotent $e$ in $S$. Then
$[\bar{e}]\wedge[e]=[0]$. Thus $[e]\oplus[e]=[e]$, and so $[e]$ is an
idempotent. We now prove the converse. Suppose that $[e]$ is an idempotent and
that $SeS\cap S\bar{e}S=SfS$. Then $[e]\oplus[e]=[e]+[f]=[e\vee
f^{\prime}]=[e]$ where $f^{\prime}\mathscr{D}f$ and $ef^{\prime}=0$. Observe
that $e\vee f^{\prime}\,\mathscr{D}\,e$. But, by Dedekind finiteness, we have
that $e=e\vee f^{\prime}$ and so $f^{\prime}\leq e$. But $f^{\prime}e=0$. It
follows that $f^{\prime}=0$ and so $f=0$, as required. ∎

We now have the following result.

######  Proposition 6.20.

Let $S$ be a Foulis monoid. Then $\L(S)$ is a Boolean algebra if and only if
$S$ is Clifford.

###### Proof.

Suppose that $S$ is Clifford. Then $SeS\cap S\bar{e}S=Se\cap S\bar{e}=\\{0\\}$
and so by Lemma 6.19 every element of $\L(S)$ is an idempotent and so $\L(S)$
is a Boolean algebra. Suppose that $\L(S)$ is a Boolean algebra. Then every
element of $\L(S)$ is an idempotent. Thus for every idempotent $e$ of $S$ we
have that $SeS\cap S\bar{e}S=\\{0\\}$. Let $a$ be any element of $S$. Then
$ea\bar{e}=0=\bar{e}ae$ since $SeS\cap S\bar{e}S=\\{0\\}$. But $ea\bar{e}=0$
implies $ea=ea1=ea(e\vee\bar{e})=eae$ and $\bar{e}ae=0$ implies
$ae=1ae=(e\vee\bar{e})ae=eae$. It follows that $ae=ea$ and so every idempotent
is central. ∎

The following result tells us that when it comes to studying the MV-algebras
associated with a Boolean inverse monoid, we can always assume that the monoid
is fundamental.

######  Lemma 6.21.

Let $S$ be a Foulis monoid. Then $S/\mu$ is a Foulis monoid and
$\L(S)\cong\L(S/\mu)$ as MV-algebras.

###### Proof.

Let $S$ be a Foulis monoid. We prove that $S/\mu$ is also a Foulis monoid.
Observe that $S/\mu$ is a factorizable Boolean inverse monoid by Proposition
3.13 and Proposition 3.14. It therefore only remains to check that the poset
of principal ideals of $S/\mu$ forms a lattice. We shall in fact prove that it
is isomorphic with the poset of principal ideals of $S$. Put $T=S/\mu$. Let
$SeS$ be a principal ideal in $S$ where $e$ is an idempotent. We shall map it
to the principal ideal $T\mu(e)T$. This is an isotone map. We shall now prove
that it is an order-isomorphism. Observe that $T\mu(e)T=T\mu(f)T$ if and only
if $\mu(e)\,\mathscr{D}\,\mu(f)$ if and only if $e\,\mathscr{D}\,f$, where we
have used Lemma 2.1 and Lemma 3.18. It follows that the map $SeS\mapsto
T\mu(e)T$ is a bijection. (This map is well-defined. Observe that $SeS=SfS$ if
and only if $e\,\mathscr{J}\,f$. But, by assumption, $S$ is a Foulis monoid.
It follows that $e\,\mathscr{D}\,f$ since $S$ is Dedekind finite. Thus
$\mu(e)\,\mathscr{D}\,\mu(f)$ and so $T\mu(e)T=T\nu(f)T$.) We need to prove
that it is an order-isomorphism. Suppose that $T\mu(e)T\subseteq T\mu(f)T$.
Then $\mu(e)\,\mathscr{D}\,\mu(e^{\prime})\leq\mu(f)$ for some idempotent
$e^{\prime}$. We deduce that $e\,\mathscr{D}e^{\prime}$ and
$\mu(e)=\mu(e^{\prime}f)$ by Lemma 2.1 where we have used the fact that
$\mu(e^{\prime})\leq\mu(f)$ and so $\mu(e^{\prime})=\mu(e^{\prime}f)$. But the
fact that $\mu$ is idempotent-separating together with Lemma 2.1 yields
$e\,\mathscr{D}\,e^{\prime}\leq f$. Thus $SeS\subseteq SfS$.

It remains to prove that $\L(S)\cong\L(S/\mu)$ as MV-algebras. Define the map
from $\L(S)$ to $\L(S/\mu)$ by $[e]\mapsto[e]_{\mu}$ where $[e]$ is the
idempotent $\mathscr{D}$-class in $S$ and $[\mu(e)]_{\mu}$ is the idempotent
$\mathscr{D}$-class in $S/\mu$. Since $e\,\mathscr{D}\,f$ implies
$\mu(e)\,\mathscr{D}\,\mu(f)$ this map is well-defined. In fact, by Lemma 2.1,
it is a bijection that induces an order-isomorphism. Observe that the natural
map associated with $\mu$ induces an order-isomorphism between $\mathsf{E}(S)$
and $\mathsf{E}(S/\mu)$. Thus these Boolean algebras are isomorphic. It
follows that, $[e]+[f]$ exists if and only if $[\mu(e)]_{\mu}+[\mu(f)]_{\mu}$
exists; and $[e]+[f]$ maps to $[\mu(e)]_{\mu}+[\mu(f)]_{\mu}$. The fact that
$\L(S)\cong\L(S/\mu)$ as MV-algebras now follows. ∎

###  6.6. Characterization of UHF monoids

We begin by establishing some properties of UHF monoids.

######  Lemma 6.22.

Let $S=\bigcup_{i=1}^{\infty}S_{i}$ be a UHF monoid. Then $S$ is
$0$-simplifying.

###### Proof.

Let $I$ be a non-zero additive ideal of $S$. We prove that $I=S$. Let $a\in I$
be any non-zero element of $I$. Then $a\in S_{i}$ for some $i$. It follows
that $S_{i}\cap I\neq\\{0\\}$. By assumption, $S_{i}$ is $0$-simplifying. It
follows that $S_{i}\subseteq I$. We can now see that $I\cap S_{j}\neq\\{0\\}$
for all $j\geq i$. In every case $S_{j}\subseteq I$. We deduce that $I=S$. ∎

Unfortunately, the counterexample on page 157 of [38, Example 5.1.9] shows
that $0$-simplifying AF-inverse monoids need not be UHF.

######  Lemma 6.23.

Let $S=\bigcup_{i=1}^{\infty}S_{i}$ be a UHF monoid. Then $S/\mathscr{J}$ is
linearly ordered.

###### Proof.

Let $a$ and $b$ be non-zero elements of $S$. Then $a,b\in S_{i}$ for some $i$.
But $S_{i}$ is a finite symmetric inverse monoid. On $S_{i}$ the $\mathscr{J}$
relation induces a linear order by Lemma 5.3. Without loss of generality, we
may assume that $a=xby$ where $x,y\in S_{i}$. It follows that $J_{a}\leq
J_{b}$ in $S$, as required. ∎

######  Lemma 6.24.

Let $S=\bigcup_{i=1}^{\infty}S_{i}$ be a UHF monoid. Then $\L(S)$ is a locally
finite MV-algebra.

###### Proof.

We shall use the fact that $S$ is equipped with a (unique) normalized
invariant mean $\nu$ [16, Theorem 2.2]. We shall prove that $\L(S)$ is an
increasing union of finite MV-algebras each of which is linearly-ordered. Fix
$i$. Define $\theta\colon\mathsf{E}(S_{i})/\mathscr{D}_{i}\rightarrow\L(S)$ by
$\theta([e]_{i})=[e]$. We prove first that $\theta$ is injective. Suppose that
$\theta([e]_{i})=\theta([f]_{i})$ where $e,f\in S_{i}$. Then $[e]=[f]$ and so
$e\,\mathscr{D}\,f$. It follows that $\nu(e)=\nu(f)$ and so
$\nu_{i}(e)=\nu_{i}(f)$, where we have used the fact that on a simple finite
Boolean inverse monoid the normalized invariant mean is unique. From this, we
get that $e\,\mathscr{D}_{i}\,f$ and so $[e]_{i}=[f]_{i}$. We have therefore
proved that $\theta$ is injective. We now prove that $\theta$ is a morphism of
MV-algebras. Let $e,f\in\mathsf{E}(S_{i})$. Since $S_{i}$ is isomorphic to a
finite symmetric inverse monoid, there are therefore two possibilities: either
$S_{i}\bar{e}S_{i}\subseteq S_{i}fS_{i}$ or $S_{i}fS_{i}\subseteq
S_{i}\bar{e}S_{i}$. In the first case, $[e]_{i}\oplus[f]_{i}=[1]_{i}$, whereas
in the second case $[e]_{i}\oplus[f]_{i}=[e\oplus f_{1}]_{i}$ where
$f\,\mathscr{D}_{i}\,f_{1}\leq\bar{e}$. In the first case, we have that
$S\bar{e}S\subseteq SfS$ and so $[e]\oplus[f]=[1]$. In the second case,
$SfS\subseteq S\bar{e}S$ giving $f\,\mathscr{D}\,f_{1}\leq\bar{e}$. and so
$[e]\oplus[f]=[e\oplus f_{1}]$. It now follows that $\theta$ is an isomorphism
onto its image. ∎

######  Lemma 6.25.

Let $S$ be a Boolean inverse monoid that satisfies the following four
conditions:

  1. (1)

$S$ is an AF monoid.

  2. (2)

$S/\mathscr{J}$ is linearly ordered.

  3. (3)

$S$ is $0$-simplifying

  4. (4)

$\L(S)$ is a locally finite MV-algebra.

Then $\L(S)$ is a rational MV-algebra.

###### Proof.

We begin by showing that our conditions make sense. Observe first that since
$S$ is AF, we know that $S$ is factorizable by Proposition 4.2. We are given
that $S/\mathscr{J}$ is linearly ordered which, in particular, implies that
$S$ satisfies the lattice condition. It follows that $S$ is a Foulis monoid
and so $\L(S)$ can be regarded as an MV-algebra. Thus our conditions make
sense. Now, $S$ is $0$-simplifying and so by Corollary 6.18 it follows that
$\L(S)$ is simple. By Proposition 6.1, we deduce that $\L(S)$ can be embedded
in the MV-algebra $[0,1]$. Finally, the fact that $\L(S)$ is locally finite
implies that $\L(S)$ is a rational MV-algebra by [3, Theorem 5.1]. ∎

If we combine Lemma 6.22, Lemma 6.23, Lemma 6.24 and Lemma 6.25, we obtain the
following.

######  Proposition 6.26.

Let $S$ be a UHF monoid. Then $S$ is an AF monoid, it is $0$-simplifying, the
set $S/\mathscr{J}$ is linearly ordered, and $\L(S)$ is a rational MV-algebra.

We now wish to prove two statements:

  1. (1)

If a Boolean inverse monoid $S$ is AF, $0$-simplifying, the set
$S/\mathscr{J}$ is linearly ordered, and $\L(S)$ is a rational MV-algebra then
$S$ is a UHF monoid

  2. (2)

Every rational MV-algebra is coordinatized by a UHF monoid.

Let $S=\bigcup_{i=1}^{\infty}S_{i}$ be a UHF monoid. Then, since $S_{i}$ is a
finite simple Boolean inverse monoid, we have that
$S_{i}\cong\mathscr{I}_{s_{i}}$ for some $s_{i}$. By Lemma 5.11, Lemma 5.12
and Lemma 3.6, we have that $s_{i}\mid s_{i+1}$. We may therefore associate
with $S$ a division sequence $\sigma_{S}$.

######  Proposition 6.27.

Let $S=\bigcup_{i=1}^{\infty}S_{i}$ and $T=\bigcup_{j=1}^{\infty}T_{j}$ be UHF
monoids. Then $S\cong T$ if and only if the division sequences $\sigma_{S}$
and $\sigma_{T}$ are interleaved.

###### Proof.

Suppose first that $S\cong T$ by means of the isomorphism $\theta\colon
S\rightarrow T$. Observe that $\theta(S_{i})$ is isomorphic to $S_{i}$ but is
a subalgebra of $T$. It follows that $\theta(S_{i})\subseteq T_{j}$, a
subalgebra, for some $j$. The fact that $\sigma_{S}$ and $\sigma_{T}$ are
interleaved now follows.

We now prove the converse. Without loss of generality, we can assume that the
terms of $\sigma_{S}$ and $\sigma_{T}$ are strictly increasing; if there is
any ‘stuttering’ then we can simply omit the corresponding $S_{i}$’s and
$T_{j}$’s. Suppose that $\sigma_{S}$ and $\sigma_{T}$ are interleaved. We
prove that $S\cong T$. Observe first that by Lemma 5.12 and Lemma 5.11 that we
may replace $S_{1}\rightarrow S_{2}\rightarrow S_{3}\rightarrow\ldots$ by
$R_{m_{1}}\stackrel{{\scriptstyle\sigma}}{{\rightarrow}}R_{m_{2}}\stackrel{{\scriptstyle\sigma}}{{\rightarrow}}R_{m_{3}}\stackrel{{\scriptstyle\sigma}}{{\rightarrow}}\ldots$
where the $\sigma$’s are standard maps; observe that the composition of
standard maps is a standard map. The colimits of these two diagrams will be
isomorphic. We may in a similar way replace $T_{1}\rightarrow T_{2}\rightarrow
T_{3}\rightarrow\ldots$ by
$R_{n_{1}}\stackrel{{\scriptstyle\sigma}}{{\rightarrow}}R_{n_{2}}\stackrel{{\scriptstyle\sigma}}{{\rightarrow}}R_{n_{3}}\stackrel{{\scriptstyle\sigma}}{{\rightarrow}}\ldots$.
By assumption, $\sigma_{S}=m_{1},m_{2},m_{3},\ldots$ and
$\sigma_{T}=n_{1},n_{2},n_{3},\ldots$ are interleaved. We now define two
functions $f,g\colon\mathbb{N}^{\ast}\rightarrow\mathbb{N}^{\ast}$. We define
$f$ first. The idea is that given $m_{i}$, we know that it divides at least
one term of $\sigma_{T}$ by the interleaving property; thus by the well-
ordering property, we know that there is a smallest $f(i)$ such that
$m_{i}\mid n_{f(i)}$; However, we need to ensure that $f$ is an increasing
function. We know that $S_{1}\cong R_{m_{1}}$. Define $f(1)$ such that
$m_{1}\mid n_{f(1)}$ and where $f(1)$ is as small as possible. Now suppose
that $i>1$. Given such an $i$ let $f(i)$ be as small as possible such that
$m_{i}\mid n_{f(i)}$ and $f(i)>f(i-1)$ The definition of $g$ is analogous. We
therefore have a sequence of embeddings by standard maps:
$R_{m_{1}}\rightarrow R_{m_{gf(1)}}\rightarrow\ldots$ and
$R_{n_{f(1)}}\rightarrow R_{n_{fgf(1)}}\rightarrow\ldots$ and embeddings
$R_{m_{1}}\rightarrow R_{n_{f(1)}}$, $R_{n_{f(1)}}\rightarrow R_{m_{gf(1)}}$,
$\ldots$ with the diagrams commuting since all inclusions are standard maps.
It follows by [34, Exercise 6.8] interpreted in the category of Boolean
inverse monoids, that the corresponding Boolean inverse monoids (as colimits)
are isomorphic. ∎

In the light of Proposition 6.27 and Corollary 6.7, we have the following.

######  Corollary 6.28.

Let $S=\bigcup_{i=1}^{\infty}S_{i}$ and $T=\bigcup_{j=1}^{\infty}T_{j}$ be UHF
monoids. Then $S\cong T$ if and only if we have
$\mathbf{n}_{\sigma_{S}}=\mathbf{n}_{\sigma_{T}}$.

By Corollary 6.28, the following now makes sense.

Definition. Given a supernatural number $\mathbf{n}$, define
$\mathscr{I}_{\mathbf{n}}$ to be isomorphic to the colimit $S$ of
$S_{1}\rightarrow S_{2}\rightarrow\ldots$ by means of embeddings such that the
associated division sequence gives rise to the supernatural number
$\mathbf{n}$. The Boolean inverse monoids $\mathscr{I}_{\mathbf{n}}$ are
Foulis monoids by construction. We shall call them generalized symmetric
inverse monoids.

Finite symmetric inverse monoid $\mathscr{I}_{n}$ are Foulis monoids: they are
factorizable and their set of principal ideals is linearly ordered by Lemma
5.3. By [23, Theorem 2.14], we have that $\L(\mathscr{I}_{n})\cong\L_{n+1}$.

######  Lemma 6.29.

Let $S$ and $T$ be Foulis monoids such that $S/\mathscr{J}$ and
$T/\mathscr{J}$ are both linearly ordered.

  1. (1)

Then every $\theta\colon S\rightarrow T$ morphism induces an MV-algebra
homomorphism $\theta^{\ast}\colon\L(S)\rightarrow\L(T)$ given by
$\theta^{\ast}([e])=[\theta(e)]$.

  2. (2)

Suppose now that $S$ and $T$ are finite simple Boolean inverse monoids. Then
$\theta^{\ast}$ is injective.

###### Proof.

(1) The proof of this is straightforward when we use the fact that
$\theta(\bar{e})=\overline{\theta(e)}$. (2) We use Lemma 5.2 and Lemma 5.4.
Suppose that $\theta^{\ast}([e])=\theta^{\ast}([f])$. Then
$\theta(e)\,\mathscr{D}\,\theta(f)$. Thus
$\nu_{T}(\theta(e))=\nu_{T}(\theta(f))$. It follows that
$\nu_{S}(e)=\nu_{S}(f)$. We deduce that $e\,\mathscr{D}\,f$ and so $[e]=[f]$.
It follows that $\theta^{\ast}$ is injective. ∎

The proof of the following is immediate by Lemma 6.29 and Lemma 6.2.

######  Lemma 6.30.

Let $\alpha\colon\L_{m+1}\rightarrow\L_{n+1}$ be an MV-algebra homomorphism
where $n=mp$. Then the standard homomorphism $\sigma_{p}\colon
R_{m}\rightarrow R_{n}$ given by $A\mapsto pA$ is such that
$\sigma_{p}^{\ast}=\alpha$ (where we use the notation of Lemma 6.29).

We now come to a key result.

######  Proposition 6.31.

Every rational MV-algebra can be coordinatized by a UHF monoid; in particular,
it can be coordinatized by a generalized symmetric inverse monoid.

###### Proof.

Let $A$ be a rational MV-algebra. Then $A$ is described by the supernatural
number $\mathbf{n}$. By assumption, $A=\bigcup_{i=1}^{\infty}A_{i}$ where each
$A_{i}$ is a finite simple MV-algebra. Let $A_{i}\cong\L_{n_{i}}$. Then
$\L(R_{n_{i}-1})\cong\L_{n_{i}}$. Put $m_{i}=n_{i}-1$. Then we have a sequence
of embeddings $R_{m_{1}}\rightarrow R_{m_{2}}\rightarrow R_{m_{3}}\rightarrow$
by means of standard maps. Let the colimit of this sequence be $S$. Then
$\L(S)\cong A$ by [26, Theorem 4.4.1] and Lemma 6.30. ∎

The following is immediate by Proposition 6.31.

######  Corollary 6.32.

Let $\mathbf{n}$ be any supernatural number. Then
$\L(\mathscr{I}_{\mathbf{n}})\cong\L_{\mathbf{n}+1}$.

We now come to our third main theorem.

######  Theorem 6.33 (UHF monoid).

Let $S$ be an AF monoid. Then $S$ is UHF if and only of $S$ is
$0$-simplifying, $S/\mathscr{J}$ is linearly ordered and $\L(S)$ is a rational
MV-algebra.

###### Proof.

Let $S$ be a UHF monoid. Then by Lemma 6.22, Lemma 6.23, Lemma 6.24 and Lemma
6.25, we see that $S$ is $0$-simplifying, $S/\mathscr{J}$ is linearly ordered
and $\L(S)$ is a rational MV-algebra. We prove the converse. By Lemma 6.25,
$\L(S)$ is a rational MV-algebra. Thus it is isomorphic to $\L_{\mathbf{n}+1}$
for some supernatural number $\mathbf{n}$. But
$\L(\mathscr{I}_{\mathbf{n}})\cong\L_{\mathbf{n}+1}$ by Corollary 6.32. It
follows by Proposition 6.31, that $\L(S)$ is isomorphic to
$\L(\mathscr{I}_{\mathbf{n}})$. But both $S$ and $\mathscr{I}_{\mathbf{n}}$
are AF Foulis monoids and the lattices of principal ideals are linearly
ordered in each case. Thus $S\cong\mathscr{I}_{\mathbf{n}}$ by Theorem 6.15.
It follows that $S$ is a UHF monoid (since it is isomorphic to one). ∎

Our classification of UHF monoids follows the same lines as that of Glimm [10]
in our use of supernatural numbers. Given the supernatural number
$\mathbf{n}$, we may define the generalized symmetric inverse monoid
$\mathscr{I}_{\mathbf{n}}$. This is the prototype of all UHF monoids. An
explicit coordinatization of $\mathbb{Q}\cap[0,1]$ can be found in [27,
Theorem 3.4].

###  6.7. Concluding remarks

The authors are grateful to Dr Ganna Kudryavtseva for her suggestion to
include the following discussion.

Observe first that the groups studied in [15] are precisely the groups of
units of our UHF monoids; this follows from [23, Lemma 3.11] and the fact that
the ‘diagonal embeddings’ of [15] are restrictions to the groups of units of
our standard maps. We shall now justify this latter claim. The paper [15] uses
permutations but it is more transparent to use rook matrices. Define $U_{m}$
to be the group of units of $R_{m}$; thus, $U_{m}$ consists of all $m\times m$
matrices over $\\{0,1\\}$ where there is exactly one $1$ in each row and each
column: in other words, permutation matrices. If $m\mid n$ where $n=pm$ then
we can define an injection from $U_{m}$ to $U_{n}$ by $A\mapsto pA$. This is
exactly a diagonal embedding in the sense of [15] and is precisely a
restriction of a standard map between to finite symmetric inverse monoids
(regarded as rook matrices).

We now apply non-commutative Stone duality [18] (and subsequent papers). By
[20, Proposition 4.31], suppose that $S$ is a Boolean inverse meet-monoid.
Then $S$ is basic precisely when its associated Boolean groupoid is
principal.888The condition that $S$ be a meet-monoid is missing from the
statement of the proposition but should be added. UHF monoids are simple: that
is, they are $0$-simplifying and fundamental (since they are AF and so basic);
this follows by Proposition 3.27. These monoids are also meet-monoids by
Proposition 3.27. Recall that we call the (unique) countable atomless Boolean
algebra the Tarski algebra. A countable Boolean inverse meet-monoid which has
a Tarski algebra of idempotents is called a Tarski monoid. By what we term the
Dichotomy Theorem [20, Proposition 4.4], a simple countable Boolean inverse
meet-monoid is either finite, and so isomorphic to a finite symmetric inverse
monoid, or a Tarski monoid. It follows that an infinite UHF monoid is a Tarski
monoid. By [21, Theorem 2.10] — which is an interpretation of work by Matui
[28, Theorem 3.9] using non-commutative Stone duality — two infinite UHF
monoids are isomorphic if and only if their groups of units are isomorphic.
However, as we explained above, these are just the groups studied in [15].
This shows that, as we would expect, there is a close connection between our
work on UHF monoids and the groups defined in [15]. However, we do not know of
an abstract characterization of the groups in [15].

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