Non-commutative Stone duality: inverse semigroups, topological groupoids and C*-algebras
aa r X i v : . [ m a t h . C T ] M a r NON-COMMUTATIVE STONE DUALITY:INVERSE SEMIGROUPS, TOPOLOGICAL GROUPOIDS AND C ∗ -ALGEBRAS M. V. LAWSON
Abstract.
We study a non-commutative generalization of Stone duality thatconnects a class of inverse semigroups, called Boolean inverse ∧ -semigroups,with a class of topological groupoids, called Hausdorff Boolean groupoids.Much of the paper is given over to showing that Boolean inverse ∧ -semigroupsarise as completions of inverse semigroups we call pre-Boolean. An inverse ∧ -semigroup is pre-Boolean if and only if every tight filter is an ultrafilter, wherethe definition of a tight filter is obtained by combining work of both Exeland Lenz. A simple necessary condition for a semigroup to be pre-Boolean isderived and a variety of examples of inverse semigroups are shown to satisfyit. Thus the polycyclic inverse monoids, and certain Rees matrix semigroupsover the polycyclics, are pre-Boolean and it is proved that the groups of unitsof their completions are precisely the Thompson-Higman groups G n,r . Theinverse semigroups arising from suitable directed graphs are also pre-Booleanand the topological groupoids arising from these graph inverse semigroups un-der our non-commutative Stone duality are the groupoids that arise from theCuntz-Krieger C ∗ -algebras. Introduction
From the appearance of Renault’s seminal monograph [55] and the work ofKumjian [25] to the more recent book by Paterson [49] it has been known thatthree areas of mathematicsinverse semigroups, topological groupoids, C ∗ -algebrasare closely related to each other. In the literature, most attention has focusedon the relationship between topological groupoids and C ∗ -algebras whereas thegoal of this paper is to shift attention to that between inverse semigroups andtopological groupoids. We prove three main theorems in this paper and exploretheir applications. In the remainder of this introduction, we explain what thesethree theorems are and touch on the kinds of applications we deal with.In this paper, classical Boolean algebras will be termed unital Boolean algebras whereas generalized Boolean algebras will be called simply Boolean algebras . Thus aBoolean algebra is a distributive lattice B , without necessarily having a top element,in which for each pair of elements a and b there exists a, perforce unique, element b \ a satisfying a ∨ b = a ∨ ( b \ a ) and a ∧ ( b \ a ) = 0. Equivalently, it is a lattice in which Mathematics Subject Classification.
Primary: 20M18; Secondary: 46L05, 06E15.
Key words and phrases.
Stone duality, graph algebras, Thompson-Higman groups.The author’s research was partially supported by an EPSRC grant (EP/F004184, EP/F014945,EP/F005881). He would particularly like to thank Daniel Lenz for many discussions on the subjectof this paper which uses in a crucial way some of his ideas, and also Ganna Kudryavtseva, StuartMargolis, Pedro Resende, Philip Scott and Benjamin Steinberg. each principal order ideal is a unital Boolean algebra. Homomorphisms of Booleanalgebras are lattice homomorphisms preserving the bottom element. Observe thatin a non-unital Boolean algebra, the joins of all finite subsets exist, including theemptyset whose join is 0, and that the meets exist of all finite nonempty subsets;for the emptyset to have a meet the algebra has to have a 1. A homomorphism ofBoolean algebras is called proper if every element in the codomain lies beneath anelement of the image.A
Boolean space is a Hausdorff topological space with a basis of compact-opensets. A continuous function between topological spaces is said to be proper if theinverse image of every compact set is also a compact set. For background resultsfrom topology needed in this paper, see [58, 59]. The following theorem is ultimatelydue to Marshall H. Stone [9].
Theorem 1.1 (Stone duality) . The category of Boolean algebras and proper homo-morphisms is dual to the category of Boolean spaces and proper continuous func-tions.
Let ( E, ≤ ) be a poset with zero. If e ∈ E such that f ≤ e implies that either f = e or f = 0 then we say that e is 0 -minimal . If X ⊆ E define X ↑ = { y ∈ E : ∃ x ∈ X ; x ≤ y } and X ↓ = { y ∈ E : ∃ x ∈ X ; y ≤ x } . If X = { x } we write x ↑ and x ↓ , respectively. If X = X ↓ we say that X is orderideal ; if X is finite then X ↓ is said to be a finitely generated order ideal. If X = X ↑ we say that X is closed . The set X is said to be (down) directed if for all x, y ∈ X there exists z ∈ X such that z ≤ x, y . Observe that if E is a meet semilattice thena closed set F is down directed precisely when x, y ∈ F implies that x ∧ y ∈ F . Asubset A ⊆ X is called a filter if it is directed and closed and does not contain zero.An ultrafilter is a maximal filter.Let S be an inverse semigroup. In what follows, the only order used in connec-tion with inverse semigroups will be the natural partial order. The semilattice ofidempotents of S is denoted by E ( S ). An inverse semigroup is said to be an inverse ∧ -semigroup if each pair of elements has a meet [44, 45]. In an inverse semigroup,we often write d ( s ) = s − s and r ( s ) = ss − . We refer the reader to [27] for anyunproved assertions about inverse semigroups. Remark 1.2.
Except where stated otherwise, all inverse semigroups in this paperwill be inverse ∧ -semigroups with zero. This is not a necessary condition to developour theory but it simplifies the mathematics — witness Lemma 2.6(2) — and issufficient for the examples we have in mind. The general theory is developed, froma different perspective, in the preprint [39].We say that elements s and t are compatible , denoted s ∼ t , if both s − t and st − are idempotents. A subset of S is compatible if each pair of elements in the subsetare compatible. If s − t = 0 = st − then s and t are said to be orthogonal . If a pairof elements are bounded above then they are easily seen to be compatible. It followsthat for a pair of elements to be eligible to have a join they must be compatible.Furthermore, if s and t are compatible then s ∧ t exists and d ( s ∧ t ) = d ( s ) d ( t ),and dually. ON-COMMUTATIVE STONE DUALITY 3
An inverse semigroup is said to be distributive if it satisfies two conditions. First,we require that every finite non-empty compatible subset has a join. Second, if { a , . . . , a m } is a non-empty finite compatible subset and if a ∈ S is any elementthen both W mi =1 aa i and W mi =1 a i a exist and we have the following two equalities a m _ i =1 a i ! = m _ i =1 aa i and m _ i =1 a i ! a = m _ i =1 a i a. A distributive inverse semigroup is said to be
Boolean if its semilattice of idempo-tents is a Boolean algebra. Distributive inverse ∧ -semigroups and Boolean inverse ∧ -semigroups will be the main classes of inverse semigroup considered in this paper.A morphism θ : S → T of Boolean inverse ∧ -semigroups is a homomorphism ofinverse ∧ -semigroups with the property that the restriction θ | E ( S ) : E ( S ) → E ( T )is a homomorphism of Boolean algebras. A morphism is said to be proper if theinverse images of ultrafilters in T are ultrafilters in S .Throughout this paper categories, apart from categories of structures, will besmall and objects replaced by identities. If a category is denoted by C then its setof identities will be denoted by C o . The elements of a category are called arrows .Each arrow a has a domain , denoted by d ( a ), and a codomain denoted by r ( a ),both of these are identities and a = a d ( a ) = r ( a ) a . A pair of elements ( a, b ) in acategory is composable if d ( a ) = r ( b ). The set of composable pairs in a category C is denoted by C ∗ C . Given an identity e the set of all arrows that begin and end at e forms a monoid called the local monoid at e . An arrow a is invertible if there isan arrow a − such that a − a = d ( a ) and aa − = r ( a ). A category in which everyarrow is invertible is called a groupoid .Let G be a groupoid. Denote the multiplication map by m and the inversionmap by i . A topological groupoid is a groupoid G which is also a topological spacein which both the multiplication and inversion maps are continuous. A topologicalgroupoid G is said to be open if the map d : G → G o is open. If d is a localhomeomorphism then G it is said to be ´etale . Since local homeomorphisms are openmaps it follows that every ´etale groupoid is an open groupoid. A local bisection ina groupoid G is a subset A such that A − A, AA − ⊆ G o . A Boolean groupoid isan ´etale topological groupoid with a basis of compact-open bisections whose spaceof identities is a Boolean space. We shall be interested in this paper in
HausdorffBoolean groupoids . A functor θ : G → H between groupoids is said to be a covering functor if it is star-injective , meaning that if θ ( g ) = θ ( g ′ ) and d ( g ) = d ( g ′ ) then g = g ′ , and star-surjective , meaning that if d ( h ) = θ ( e ), where e is an identity, thenthere exists g ∈ G such that d ( g ) = e and θ ( g ) = h .The key concept on which this paper is based is the Lenz arrow relation → [46].This concept is also implicit in Exel’s paper [12] since it is used to define the notionof a cover. Let a, b ∈ S . We define a → b iff for each non-zero element x ≤ a , wehave that x ∧ b = 0. Observe that a ≤ b ⇒ a → b . We write a ↔ b iff a → b and b → a . More generally, if a, a , . . . , a m ∈ S then we define a → { a , . . . , a m } iff foreach non-zero element x ≤ a we have that x ∧ a i = 0 for some i . Finally, we write { a , . . . , a m } → { b , . . . , b n } iff a i → ( b , . . . , b n ) for 1 ≤ i ≤ m , and we write { a , . . . , a m } ↔ { b , . . . , b n } I follow Ehresmann’s notation and write ‘o’ for ‘object’.
M. V. LAWSON iff both { a , . . . , a m ) → ( b , . . . , b n } and { b , . . . , b n } → { a , . . . , a m } . A subset Z ⊆ A is said to be a cover of A , denoted A → Z , if for each a ∈ A there exists z ∈ Z such that a ∧ z = 0. A special case of this definition is the following. A finitesubset A ⊆ a ↓ is said to be a cover of a if a → A . A homomorphism θ : S → T from an inverse ∧ -semigroup S to a distributive inverse semigroup T is said to bea cover-to-join map if for each element s ∈ S and each finite cover A of s we havethat θ ( s ) = W θ ( A ).A homomorphism θ : S → T between inverse semigroups with zero is said to be0 -restricted if θ ( s ) = 0 implies that s = 0. Congruences that are 0-restricted ariseas kernels of such homomorphisms. The first of our three main theorems may nowbe stated. It is a wide-ranging generalization of [33]. Completion Theorem
Let S be an inverse ∧ -semigroup. Then there is a dis-tributive inverse ∧ -semigroup D ( S ) and a -restricted cover-to-join homomorphism δ : S → D ( S ) having the following universal property: for every cover-to-join map θ : S → T to an arbitrary distributive inverse semigroup there is a unique join-preserving homomorphism ¯ θ : D ( S ) → T such that ¯ θδ = θ .We explain the intuitive idea behind this theorem as follows. Our goal is to con-struct from an inverse semigroup S the most general distributive inverse semigroup T generated by S subject to the condition that elements of S which ‘morally havethe same join’ should be identified in T . The precise meaning of ‘morally have thesame join’ is encoded by the notion of cover. Thus if { a , . . . , a m } ⊆ a ↓ is a coverof a then in our completion T we shall require that the join of the images of the a i be equal to the image of a . The theorem says that we can indeed find such acompletion of S . We call D ( S ) the distributive completion of S .Our second main theorem is as follows; it is clearly a generalization of Theo-rem 1.1. Duality Theorem
There is a duality between the category of Boolean inverse ∧ -semigroups and their proper morphisms and the category of Hausdorff Booleangroupoids and the proper continuous covering functors between them. Remark 1.3.
The monoid version of the above theorem was proved in [37] wherethe Hausdorff Boolean groupoids in that case have a compact space of identities.I shall now explain how these two theorems are related to each other. Obviousexamples of Boolean inverse ∧ -semigroups are the symmetric inverse monoids butit is hard to think of other examples which are not just Boolean algebras. Thisraises the obvious question of finding natural examples of such semigroups. TheCompletion Theorem gets us part of the way but only yields distributive inversesemigroups. This motivates the following definition. An inverse ∧ -semigroup S is said to be pre-Boolean if its distributive completion D ( S ) is actually Boolean.This definition does not of course solve anything: it simply changes the question.However, it turns out that pre-Boolean inverse semigroups are common in mathe-matics and are related to both group theory — specifically Thompson-Higman type The term ‘cover’ in this context is sanctioned by its use in frame theory. In the preprint [39],we show that this notion of cover is a special case of that of a coverage which in turn is closelyrelated to the notion of a Grothendieck topology.
ON-COMMUTATIVE STONE DUALITY 5 groups — and the theory of C ∗ -algebras — notably graph C ∗ -algebras. These arediscussed in Section 4 of this paper.All of this raises the question of how we can identify pre-Boolean inverse ∧ -semigroups. This is the subject of our third main theorem. To state it we needsome notation and definitions. Let E be a meet semilattice with zero. Let X, Y ⊆ E be finite subsets. Define X ∧ = { e ∈ E : e ≤ x, ∀ x ∈ X } , the set of all elements beneath every element of X , and define Y ⊥ = { e ∈ E : e ∧ y = 0 , ∀ y ∈ Y } , the set of all elements orthogonal to every element in Y . If X were non-empty,we could replace X by its meet, but it is convenient to have this extra flexibility.Furthermore, if X were empty we could only replace X by a single element if thesemilattice had a top which we do not want to assume. If the set X ∧ ∩ Y ⊥ doesnot consist solely of the zero we shall write X ∧ ∩ Y ⊥ = 0.A filter F ⊆ E is said to be tight if for each a ∈ F and each finite cover { a , . . . , a m } of a we have that a i ∈ F for some i . We shall prove later that everyultrafilter is tight. Remark 1.4.
Our definition of tight filter arises from the work of both Exel [12]and Lenz [46]. We shall say more about this definition at various places in thispaper.A meet semilattice with zero E satisfies the trapping condition if for all 0 = y < x the set x ↓ ∩ y ⊥ has a finite cover. The semilattice E is said to be 0 -disjunctive iffor each 0 = f ∈ E and e such that 0 = e < f , there exists 0 = e ′ < f such that e ∧ e ′ = 0. f ✁✁✁✁✁✁✁✁ ❃❃❃❃❃❃❃ e ❃❃❃❃❃❃❃❃ e ′ ⑧⑧⑧⑧⑧⑧⑧⑧ Booleanization Theorem (1)
An inverse ∧ -semigroup is pre-Boolean if and only if its semilattice of idem-potents is pre-Boolean. (2) Let E be a meet semilattice with zero. Then E is pre-Boolean if and onlyif every tight filter of E is an ultrafilter. (3) Let E be -disjunctive meet semilattice with zero. Then E is pre-Booleanif and only if the trapping condition holds. The examples of pre-Boolean inverse semigroups that we shall investigate inSection 4 will satisfy condition (3) above.The three theorems above will be proved in Section 2. All are new thoughthey have varying pedigrees. The Duality Theorem is a direct generalization of[37] which is nothing other than the monoid version of this theorem. Note thatwe have had to make some terminological changes from [37] since the theory hasoutgrown the framework in which it was originally conceived. The Booleanization
M. V. LAWSON
Theorem shows how the work of Exel [12] and Lenz [46] are related. It completesthe preliminary results of [38]. The Completion Theorem is new.2.
Proofs of the main theorems
In this section, we shall prove the three theorems discussed in the Introduction.To do this, we also need to prove a technical result, called the Comparison Theorem.2.1.
The completion theorem.
We shall construct the semigroup D ( S ) in threesteps. Step 1
We begin by slightly extending some results from Section 5 of [46].
Lemma 2.1.
Let S be an inverse semigroup. (1) The relation → is reflexive and transitive. (2) a → b if and only if a − → b − . (3) If a → b and bc = 0 then ac = 0 . (4) The relation → is left and right compatible with the multiplication in thesense that if a → b and ac, bc = 0 (respectively, ca, cb = 0 ) then ac → bc (respectively, ca → cb .) (5) If u → s and u → t then u → s ∧ t .Proof. (1) Clearly → is reflexive. To prove transitivity, suppose that a → b and b → c . Let 0 = x ≤ a . Then x ∧ b = 0. But also x ∧ b ≤ b and so x ∧ b ∧ c = 0.Hence x ∧ c = 0, as required.(2) Straightforward.(3) Suppose that a → b and bc = 0 but that ac = 0. Then acc − = 0 and acc − ≤ a . Thus by assumption b ∧ acc − = 0. But b ∧ acc − ≤ acc − and so( b ∧ acc − ) cc − = b ∧ acc − . Thus ( b ∧ acc − ) c = 0 and so bc ∧ ac = 0, whichcontradicts the fact that bc = 0. We have proved that bc = 0 implies that ac = 0.(4) Suppose that a → b and ac, bc = 0. We prove that ac → bc . Let 0 = x ≤ ac .Then x = xx − ac and so xc − c = x . Thus xc − = 0 and xc − ≤ acc − ≤ a . Itfollows that xc − ∧ b = 0. Now xc − ∧ b ≤ xc − and so ( xc − ∧ b ) cc − = xc − ∧ b .It follows that ( xc − ∧ b ) c = 0. But ( xc − ∧ b ) c = xc − c ∧ bc ≤ x ∧ bc . Hence x ∧ bc = 0, as required.(5) Let 0 = v ≤ u . Then since u → s we have that 0 = v ∧ s . But v ∧ s ≤ u alsoand so since u → t we have that 0 = v ∧ s ∧ t . We have proved that u → s ∧ t . (cid:3) It follows from the lemma above that ↔ is a 0-restricted congruence on S . Wedenote the ↔ -equivalence class containing s by s , the quotient semigroup by S and the natural map S → S by λ . If ↔ is just equality, we say that the inversesemigroup is separative . Lemma 2.2.
We have that a → b if and only if λ ( a ) ≤ λ ( b ) . In particular, asemigroup is separative if and only if a ≤ b ⇔ a → b .Proof. Suppose that a → b . We prove that λ ( a ) ≤ λ ( b ). Thus we need to provethat λ ( a ) = λ ( ba − a ); that is, a ↔ ba − a . From a → b we get that a − → b − .Thus by compatibility, we have that a = aa − a → ba − a . It remains to showthat ba − a → a . From a → b we get that a − → b − . Thus by compatibility ba − a → bb − a . But bb − a ≤ a implies that bb − a → a . Thus ba − a → a . ON-COMMUTATIVE STONE DUALITY 7
Conversely, suppose that λ ( a ) ≤ λ ( b ). Then λ ( a ) = λ ( ba − a ) which implies that a ↔ ba − a . However, a → ba − a and ba − a ≤ b implies that a → b .We now prove the final claim. Suppose that a ≤ b ⇔ a → b . Then a ↔ b iff a ≤ b and b ≤ a giving a = b . Thus the condition implies separativity. Conversely,suppose that the semigroup is separative and that a → b . Then λ ( a ) ≤ λ ( b ). Thus λ ( a ) = λ ( ba − a ). But under the assumption of separativity this implies a = ba − a and so a ≤ b , as required. (cid:3) The semigroup S is an inverse ∧ -semigroup. The same is true of the quotientsemigroup S . Lemma 2.3.
The semigroup S has all finite non-empty meets.Proof. Let s , t ∈ S . We shall prove that s ∧ t exists and that λ ( s ) ∧ λ ( t ) = λ ( s ∧ t ).The map λ is a homomorphism of inverse semigroups and so λ ( s ∧ t ) ≤ λ ( s ) , λ ( t ).Suppose that λ ( u ) ≤ λ ( s ) , λ ( t ). Then by Lemma 2.2, we have that u → s and u → t . By Lemma 2.1, we have that u → s ∧ t . Thus λ ( u ) ≤ λ ( s ∧ t ). We havetherefore proved that λ ( s ∧ t ) = λ ( s ) ∧ λ ( t ). (cid:3) For inverse ∧ -semigroups, we are interested not merely in congruences but inthose congruences that also preserve the meet structure. Such congruences were firstdescribed by Leech in [44]. In this paper, we shall refer to them as ∧ -congruences . Acongruence ρ on an inverse semigroup S will be called separative if S/ρ is separative.
Proposition 2.4. ↔ is the smallest -restricted, separative, ∧ -congruence on S .Proof. The fact that ↔ is a 0-restricted congruence follows by Lemma 2.1 and itfollows easily from the fact that ↔ is 0-restricted that it is separative. In Lemma 2.3,we proved that it is a ∧ -congruence.Now let ρ be any 0-restricted, separative ∧ -congruence on S . Let a ↔ b . Supposethere exists 0 = ρ ( x ) ≤ ρ ( a ) such that ρ ( x ) ∧ ρ ( b ) = 0. Then ρ ( x ∧ b ) = 0 andso since ρ is 0-restricted we have that x ∧ b = 0. Now x ∧ a ≤ a and x ∧ a = 0since ρ ( x ) ≤ ρ ( a ). Thus by assumption x ∧ a ∧ b = 0. It follows that x ∧ b = 0which is a contradiction. Thus ρ ( x ) ∧ ρ ( b ) = 0. It follows that we have proved that ρ ( a ) → ρ ( b ). By a symmetrical argument we get that ρ ( a ) ↔ ρ ( b ). By assumption ρ ( a ) = ρ ( b ), as required. (cid:3) Step 2
The second step is a simple modification of a construction due to Boris Scheindescribed in Section 1.4 of [27] and applies to any inverse semigroup. An orderideal in an inverse semigroup is said to be compatible if it is a compatible subset. Ifthe order ideal is finitely generated then this is equivalent to requiring that its setof generators form a compatible subset. We denote by FC ( S ) the set of all finitelygenerated compatible order ideals of S . We define ι : S → FC ( S ) to be the map s s ↓ . We have the following finitary version of Theorems 1.4.23 and 1.4.24 of[27]. Proposition 2.5.
Let S be an inverse semigroup with zero. Then FC ( S ) is adistributive inverse semigroup. If θ : S → T is any homomorphism to a dis-tributive inverse semigroup then there is a unique join-preserving homomorphism θ ∗ : FC ( S ) → T such that θ ∗ ι = θ . Step 3
M. V. LAWSON
The third step is more involved. It uses in a crucial way results to be foundin [46] and [42] as described in [37]. We begin by constructing the groupoid G ( S )from the ultrafilters in the inverse semigroup S . The proofs of the following maybe found in [12], Lemma 3.2, and [37], Lemma 2.5. Lemma 2.6.
Let S be an inverse ∧ -semigroup. (1) Every non-zero element of S is contained in an ultrafilter. (2) Let F be a filter in S . Then F is an ultrafilter if and only if F containsevery element b ∈ S such that b ∧ c = 0 for all c ∈ F . The proofs of (1) and (2) below are Lemmas 2.6 and 2.7 of [37].
Lemma 2.7.
Let S be an inverse semigroup. (1) If A is a filter then A = AA − A . (2) If A and B are filters then ( AB ) ↑ is the smallest filter containing AB . If A and B are filters we define A · B = ( AB ) ↑ . Filters have extra structure which makes them behave in a way analogous to cosetsin group theory. See Lemmas 2.8, 2.9 and 2.10 of [37] for the proofs of (1),(2) and(3) respectively, below.
Lemma 2.8.
Let S be an inverse semigroup. (1) Let S be a filter. Then B = A − · A is a filter and inverse subsemigroup of S and A = ( aB ) ↑ for any a ∈ A . (2) Let A be a filter. Then A = A · A iff A is an inverse subsemigroup iff A contains an idempotent. (3) If A and B are filters such that A ∩ B = ∅ and A − · A = B − · B (respectively, A · A − = B · B − ) then A = B . Denote by G ( S ) the set of all ultrafilters of the inverse semigroup S . If we restrictthe definition of A · B to the case where A − · A = B − · B then G ( S ) becomes agroupoid.We denote the set of all local bisections of the groupoid G by Bi ( G ). This is aninverse semigroup under subset multiplication. For a proof see [49], for example. Lemma 2.9.
The semigroup Bi ( G ) is a Boolean inverse ∧ -semigroup where thenatural partial order is given by inclusion and the idempotents are the subsets of G o . For each a ∈ S , define V a to be the set of all ultrafilters in S that contain a . PutΩ = Ω( S ) = { V a : a ∈ S } . Lemma 2.10.
With the above notation we have the following. (1) V a ∩ V b = V a ∧ b . (2) V a is a local bisection in the groupoid G ( S ) . (3) V − a = V a − . (4) V a V b = V ab .Proof. (1) We deal with a special case first. Suppose that V a ∩ V b = ∅ . Then a ∧ b = 0because if not there would be an ultrafilter containing a ∧ b and so an ultrafiltercontaining both a and b which contradicts the assumption that the intersection is ON-COMMUTATIVE STONE DUALITY 9 empty. But if a ∧ b = 0 then V a ∧ b = ∅ . Now suppose that a ∧ b = 0. Then V a ∧ b = ∅ .On the other hand we must have V a ∩ V b = ∅ because if not we would be able toshow that a ∧ b = 0.Now we can prove the general case. Suppose that V a ∩ V b = ∅ . Let A ∈ V a ∩ V b .Then a, b ∈ A . But A is an ultrafilter and so a ∧ b ∈ A . It follows that A ∈ V a ∧ b .Now suppose that a ∧ b = 0. Any ultrafilter containing a ∧ b must contain both a and b .(2) Let A, B ∈ V a such that A − · A = B − · B . By assumption A ∩ B = ∅ . Thusby Lemma 2.8(3), we have that A = B . A similar argument works dually.(3) This follows from the observation that A is an ultrafilter iff A − is an ultra-filter.(4) Let A ∈ V a and B ∈ V b . Then a ∈ A and b ∈ B so ab ∈ A · B and A · B ∈ V ab .Now let C ∈ V ab . Thus ab ∈ C . Put H = C − · C so that C = ( abH ) ↑ .We have that ( ab ) − ab ∈ H but b − a − ab ≤ b − b and so b − b ∈ H . It followsthat B = ( bH ) ↑ is a well-defined ultrafilter containing b . Thus B ∈ V b . Observethat B · B − = ( bHb − ) ↑ . Now ( ab ) − ab ∈ H and so bb − a − abb − ∈ ( bHb − ) ↑ .Thus a − abb − ∈ ( bHb − ) ↑ from which it follows that a − a ∈ ( bHb − ) ↑ . It followsthat A = ( a ( bHb − ) ↑ ) ↑ is a well-defined ultrafilter containing a . Observe that A − · A = B · B − and that A · B = C . (cid:3) The following result is in part from [46] and links the Lenz arrow with ultrafilters.
Lemma 2.11.
Let S be an inverse semigroup. (1) Let a, b ∈ S . Then V a ⊆ V b if and only if a → b . (2) Let a, a , . . . , a m ∈ S . Then a → ( a , . . . , a m ) if and only if V a ⊆ S mi =1 V a i . (3) Let a , . . . , a m , b , . . . , b n ∈ S . Then ( a , . . . , a m ) ↔ ( b , . . . , b n ) if and onlyif S mi =1 V a i = S mj =1 V b j . (4) Let a ∈ S be such that every ultrafilter in V a is idempotent. Then a ↔ a ∧ a − a .Proof. We prove (1); the proof of (2) is an immediate generalization, and (3) isimmediate from (2). Suppose that V a ⊆ V b . Let 0 = x ≤ a . By Lemma 2.6(1),there is an ultrafilter F containing x . But then a ∈ F . By assumption b ∈ F . But x, b ∈ F , where F is an ultrafilter, implies that x ∧ b = 0.To prove the converse, suppose that for each non-zero element x ≤ a , we havethat x ∧ b = 0. Let F ∈ V a . Thus a ∈ F . Suppose that b / ∈ F . Then byLemma 2.6(2), there exists y ∈ F such that b ∧ y = 0. Now a, y ∈ F implies that x = a ∧ y = 0 and by construction x ≤ a . Thus by our assumption, x ∧ b = 0. But x ∧ b = a ∧ y ∧ b ≤ b ∧ y = 0, which is a contradiction. Thus b ∈ F and so V a ⊆ V b .(4) Put e a = a ∧ a − a . Suppose that every ultrafilter in V a is idempotent. ByLemma 2.8(2), this is equivalent to saying that every such ultrafilter is an inversesubsemigroup. It follows that if F ∈ V a then from a ∈ F we get that a − ∈ F and so a − a ∈ F . Hence a ∧ a − a ∈ F . We have show that V a ⊆ V e a . On the other hand, e a ≤ a and so V e a ⊆ V a . It follows that V a = V e a and so a ↔ a ∧ a − a = e a . (cid:3) Lemma 2.12.
Let A = { a , . . . , a n } ↓ be a finitely generated compatible order ideal.Then [ a ∈ A V a = n [ i =1 V a i . Proof.
Only one direction needs proving. Let F ∈ S a ∈ A V a . Then F ∈ V a for some a ∈ A . But a ≤ a i for some i and so F ∈ V a i , as required. (cid:3) Lemma 2.13. If a and b are compatible then V a ∪ V b is a bisection.Proof. Suppose that F ∈ V a and G ∈ V b and that F − · F = G − · G . Then F · G − isdefined and contains the idempotent ab − . It follows by Lemma 2.8(2) that F · G − is an idempotent ultrafilter and so is an identity in the groupoid. Thus F = G , asrequired. The dual result follows by symmetry. (cid:3) Let S be an inverse ∧ -semigroup. Then G ( S ) is the groupoid of ultrafilters of S .Observe that here this groupoid is considered without a topology. By Lemma 2.9, Bi ( G ( S )) is the Boolean inverse ∧ -semigroup of all local bisections of G ( S ). ByLemmas 2.12 and 2.13, the map β : FC ( S ) → Bi ( G ( S )) defined by β ( A ) = n [ i =1 V a i where A = { a , . . . , a n } ↓ is well-defined. If A = { a , . . . , a m } ↓ and B = { b , . . . , b n } ↓ are elements of FC ( S ) define A ≡ B ⇔ ( a , . . . , a m ) ↔ ( b , . . . , b n ) . Lemma 2.14.
The map β is a homomorphism. In addition, if A = { a , . . . , a m } ↓ and B = { b , . . . , b n } ↓ then β ( A ) = β ( B ) ⇔ A ≡ B. Proof.
Let A = { a , . . . , a m } ↓ and B = { b , . . . , b n } ↓ . Then β ( A ) = m [ i =1 V a i and β ( B ) = n [ j =1 V b j . Multiplying the two unions together and then using the fact that Bi ( G ( S )) is aBoolean inverse ∧ -semigroup by Lemma 2.9 together with Lemma 2.10 we get that β ( A ) β ( B ) = [ i,j V a i b j . But it is easy to check that AB = { a i b j : 1 ≤ i ≤ m, ≤ j ≤ n } ↓ and so β ( A ) β ( B ) = β ( AB ). The final claim follows by Lemma 2.11(3). (cid:3) It follows that the kernel of the homomorphism β is the congruence ≡ . Wedenote the ≡ -class containing A by [ A ]. We denote the natural map from FC ( S ) to FC ( S ) / ≡ by ξ . We denote by β ′ the unique map such that β ′ ξ = β . The map β ′ is, of course, injective and the image of β ′ is the same as the image of β . S ι / / FC ( S ) ξ / / / / β (cid:15) (cid:15) FC / ≡ kK β ′ y y sssssssss Bi ( G ( S )) Restriction to the case where S is separative ON-COMMUTATIVE STONE DUALITY 11
At this point, we shall restrict the class of inverse semigroups we consider. Weshall, however, return to the general case later.
Let S be a separative semigroup.From Lemmas 2.2 and 2.11(1), this means that for non-zero elements a, b ∈ S wehave that a = b if and only if a ↔ b if and only if V a = V b . Define D ( S ) = FC ( S ) / ≡ and δ ( s ) = [ s ↓ ] . Thus δ = ξι . We shall write δ S rather than δ when there is a danger of ambiguity.The inverse semigroup D ( S ) is embedded in Bi ( G ( S )) by β ′ . We now describe theimage of β ′ . Lemma 2.15.
Let S be separative. (1) S mi =1 V a i is a bisection in Bi ( G ( S )) if and only if { a , . . . , a m } is a compat-ible subset of S . (2) The image of β consists precisely of all bisections in Bi ( G ( S )) of the form S mi =1 V a i where { a , . . . , a m } is a compatible subset of S .Proof. (1) It follows by Lemma 2.13 that if { a , . . . , a m } is a compatible subsetof S then S mi =1 V a i is a bisection in Bi ( G ( S )). We prove the converse. Supposethat S mi =1 V a i is a bisection in Bi ( G ( S )). Then using Lemmas 2.9 and 2.10, thesets V a − i a j and V a i a − j are either empty if a − i a j = 0 or a i a − j = 0 or consist ofidentities in the groupoid that is, by Lemma 2.8(2), ultrafilters that are also inversesubsemigroups. It follows by Lemma 2.11(4) that if V x consists of only idempotentultrafilters then x ↔ x ∧ x − x . But S is separative and so x is an idempotent. Hencein all cases a − i a j and a i a − j are idempotents and so a i and a j are compatible.(2) This is now immediate. (cid:3) Proposition 2.16.
Let S be separative inverse ∧ -semigroup. Let θ : S → T be acover-to-join homomorphism to the distributive inverse semigroup T . Then thereexists a unique join-preserving homomorphism ¯ θ : D ( S ) → T such that ¯ θδ = θ .Proof. By Proposition 2.5, there exists a unique join-preserving homomorphism θ ∗ : FC ( S ) → T such that θ ∗ ι = θ . By definition θ ∗ ( { a , . . . , a m } ↓ ) = m _ i =1 θ ( a i ) . We now prove that if ( a , . . . , a m ) ↔ ( b , . . . , b n ) then W i θ ( a i ) = W j θ ( b j ). Bydefinition, for each i we have that a i → ( b , . . . , b n ). It is easy to check that { a i ∧ b , . . . , a i ∧ b n } ⊆ a ↓ i is a cover. But θ is a cover-to-join map and so for each i we have that θ ( a i ) = W j θ ( a i ∧ b j ). It follows that m _ i =1 θ ( a i ) = _ i,j θ ( a i ∧ b j ) . But a similar result holds by symmetry for W j θ ( b j ) and we have proved our claim.It follows that we may define ¯ θ ([ { a , . . . , a m } ↓ ]) = W i θ ( a i ) and so have a well-defined function ¯ θ : D ( S ) → T such that ¯ θξ = θ ∗ . Thus ¯ θδ = θ . It is clear thatit is a homomorphism and that it is join-preserving. Uniqueness follows almostimmediately. (cid:3) The definition of D ( S ) for arbitrary S will be given below. Proof of the Completion TheoremWe now return to the general case. Let S be an arbitrary inversesemigroup. Put S = S/ ↔ , a separative semigroup by Proposition 2.4. Recallthat the ↔ -equivalence class containing s is denoted by s . Define D ( S ) = FC ( S ) / ≡ and δ ( s ) = [ s ↓ ] . Observe that δ is 0-restrictive since it is essentially the map s V s and byLemma 2.6(1) this is non-empty if s is non-zero. Furthermore, D ( S ) is an inverse ∧ -semigroup essentially by Lemma 2.10. Proof.
Let θ : S → T be a cover-to-join map to a distributive inverse semigroup T .The first step in the proof is to show that if a, b ∈ S are non-zero elements suchthat a ↔ b then θ ( a ) = θ ( b ). Observe first that because 0 = a ≤ a we have that a ∧ b = 0. We prove now that a → a ∧ b . The fact that also b → a ∧ b will followby symmetry. Let 0 = x ≤ a . Then x ∧ b = 0. But x = x ∧ a and so x ∧ a ∧ b = 0,as required. Hence { a ∧ b } is a cover of both a and b . But θ is a cover-to-join mapand so θ ( a ) = θ ( a ∧ b ) = θ ( b ), as required. It follows that there is a homomorphism φ : S → T such that φλ = θ .It remains to show that φ is a cover-to-join map. Let { a , . . . , a n } ⊆ a ↓ be acover. Then by Lemma 2.2, a i → a for each i in the semigroup S . We claim that { a ∧ a i : 1 ≤ i ≤ n } ⊆ a ↓ is a cover. Let 0 = x ≤ a . Then = x ≤ a in S using thefact that ↔ is 0-restricted. It follows that x ∧ a i = . However, by Proposition 2.4the congruence ↔ is a ∧ -congruence. This gives us that x ∧ a i = 0. It follows that x ∧ a ∧ a i = 0. Since θ is a cover-to-join map, we have that θ ( a ) = n _ i =1 θ ( a ∧ a i ) . However, a i ≤ a and so a ∧ a i = a i . It follows that a ∧ a i ↔ a i . Hence θ ( a i ∧ a ) = θ ( a i ). So we have that θ ( a ) = n _ i =1 θ ( a i ) . But φλ = θ and so θ ( a ) = n _ i =1 θ ( a i )as required.The homomorphism φ : S → T is a cover-to-join map to a finitely completedistributive inverse semigroup. Thus by Proposition 2.16, there is a unique join-preserving homomorphism ¯ φ : D ( S ) → T such that ¯ φδ S = φ . S λ / / θ (cid:30) (cid:30) ❃❃❃❃❃❃❃❃ S φ (cid:15) (cid:15) δ S / / D ( S ) ¯ φ } } ④④④④④④④④ T The map δ : S → D ( S ) is just δ S λ and we shall rename ¯ φ as ¯ θ . We then have that¯ θδ = θ . Uniqueness is almost immediate. (cid:3) This is the definition of D ( S ) for arbitrary S not just those S which are separative. ON-COMMUTATIVE STONE DUALITY 13
The Booleanization Theorem.
In the previous section, we constructed adistributive inverse ∧ -semigroup from every inverse ∧ -semigroup S . The goal of thissection is to find conditions on S that imply that D ( S ) is Boolean. That is: when isan inverse ∧ -semigroup pre-Boolean? This section is in three parts correspondingto the three statements in the Booleanization Theorem given in the Introduction. Part 1
We show first that the problem of determining whether an inverse ∧ -semigroupis pre-Boolean can be answered by looking only at its semilattice of idempotents. Inwhat follows we shall denote the set of ultrafilters in E ( S ) containing the element e by U e to avoid ambiguity.The inverse semigroup D ( S ) is isomorphic to the inverse subsemigroup Bi ( G ( S ))that consists of all local bisections of the form S mi =1 V a i where a i ∈ S . The proofof the following is immediate. Lemma 2.17.
The semilattice E ( D ( S )) consists of all those bisections of the form S mi =1 V a i where each V a i consists of only idempotent ultrafilters. The partial order-ing is subset inclusion. The following result is a version of Proposition 2.13 [37]. It enables us to comparethe set of ultrafilters in S with the set of ultrafilters in E ( S ). Lemma 2.18.
Let S be an inverse ∧ -semigroup. (1) If A is an idempotent filter in S then E ( A ) is a filter in E ( S ) and A = E ( A ) ↑ . (2) If F is a filter (respectively, an ultrafilter) in E ( S ) then F ↑ is an idempotentfilter (respectively, ultrafilter) in S such that E ( F ↑ ) = F . (3) A is an idempotent ultrafilter in S iff E ( A ) is an ultrafilter in E ( S ) . (4) The function ǫ : G ( S ) o → G ( E ( S )) defined by ǫ ( A ) = E ( A ) is a bijection. (5) Let V a consist entirely of idempotent ultrafilters. Then ǫ : V a → U e a is abijection where e a = a ∧ a − a .Proof. (1) Let A be an idempotent filter in S . By Lemma 2.8(2), it follows that E ( A ) is non-empty. Since E ( A ) ⊆ A we have that E ( A ) ↑ ⊆ A . Let a ∈ A . ByLemma 2.8(2), we know that idempotent filters are inverse subsemigroups. Thus a − a ∈ A . But A is a filter and so e = a ∧ a − a ∈ A , an idempotent. But e ≤ a byconstruction. It follows that a ∈ E ( S ) ↑ and so A = E ( A ) ↑ .(2) The fact that F is a filter in E ( S ) implies that F ↑ is a filter in S . This filtercontains idempotents by construction and so be Lemma 2.8(2), it is an idempotentfilter. We have that F ⊆ F ↑ and so F ⊆ E ( F ↑ ). Let e ∈ E ( F ↑ ). Then f ≤ e forsome f ∈ F . But F is a filter in E ( S ) and so f ∈ F .Suppose now that F is an ultrafilter in E ( S ). Let F ↑ ⊆ A where A is a filter in S . Clearly, A is an idempotent filter and so A = E ( A ) ↑ . But F ⊆ E ( A ) and E ( A )is a filter and so F = E ( A ). It follows that F ↑ = A and so F ↑ is an ultrafilter in S .(3) Suppose that A is an idempotent ultrafilter in S . Let E ( A ) ⊆ F where F is a filter in E ( S ). Then A = E ( A ) ↑ ⊆ F ↑ . But A is an ultrafilter and F ↑ is afilter and so A = F ↑ . Hence E ( A ) = F and so E ( A ) is an ultrafilter. The conversefollows by (2).(4) The fact that we have a bijection follows by the results above.(5) The proof of this is straightforward. (cid:3) Proof of the Booleanization Theorem (1)
We prove that if S is an inverse ∧ -semigroup then E ( D ( S )) is isomorphic to D ( E ( S )). Proof.
By Lemma 2.17 we may identity a typical element of E ( D ( S )) with a bisec-tion of the form S mi =1 V a i where each V a i consists of only idempotent ultrafilters.Define a function E : E ( D ( S )) → D ( E ( S ))by E ( [ i V a i ) = [ i U e ai . To prove that this map is well-defined we need to prove the following. Suppose that( a , . . . , a m ) ↔ ( b , . . . , b n )and that a i ↔ e a i and b j ↔ e b j . Then( e a , . . . , e a m ) ↔ ( e b , . . . , e b n ) . This is straightforward. It follows by Lemma 2.18 that this sets up an order iso-morphism between E ( D ( S )) and D ( E ( S )). (cid:3) Part 2
In this section, we shall obtain necessary and sufficient conditions on an inverse ∧ -semigroup that it be pre-Boolean. We showed above that we may restrict attentionto semilattices so from now on E will denote a meet semilattice with zero. Wedenote by F ( E ) the set of all filters on E and by U ( E ) the set of all ultrafilters on E . Clearly U ( E ) ⊆ F ( E ). Lemma 2.19.
Let S be an inverse semigroup. (1) Every ultrafilter is a tight filter. (2) A is is a tight filter if and only if ( A − A ) ↑ is a tight filter. (3) Let A be an inverse subsemigroup. Then A is a tight filter in S if and onlyif E ( A ) is a tight filter in E ( S ) . (4) Every tight filter in S is an ultrafilter if and only if every tight filter in E ( S ) is an ultrafilter.Proof. (1) Let F be an ultrafilter. Let a ∈ F and suppose that { a , . . . , a m } is acover of a . Suppose that { a , . . . , a m } ∩ F = ∅ . Then by Lemma 2.6(2), we mayfind an element f ∈ F such that f ∧ a i = 0 for all i . Now f ∧ a ∈ F . By assumptionthere exists a i such that f ∧ a ∧ a i = 0. Thus f ∧ a i = 0, which is a contradiction.Therefore we must have that a i ∈ F for some i , and so F is a tight filter.(2) Let A be a tight filter. We prove that H = ( A − A ) ↑ is a tight filter. Let x ∈ H and let { x , . . . , x m } be a cover of x . By definition a − a ≤ x for some a , a ∈ A . Let a ∈ A be arbitrary. Then aa − a ≤ ax . But aa − a ∈ A andso ax ∈ A . We claim that { ax , . . . , ax m } is a cover of ax . Let 0 = y ≤ ax .Observe that aa − y = y . It follows that a − y = 0. But a − y ≤ aa − x ≤ x . Thus a − y ∧ x i = 0 for some i . But a ( a − y ∧ x i ) = aa − y ∧ ax i and since a − y ∧ x i ≤ a − y we have that a − a ( a − y ∧ x i ) = a − y ∧ x i . It follows that aa − y ∧ ax i = y ∧ ax i = 0.By assumption, A is a tight filter and so ax j ∈ A for some j . Thus a − ax j ∈ A − A and so x j ∈ H , as required. ON-COMMUTATIVE STONE DUALITY 15
Suppose now that H = ( A − A ) ↑ is a tight filter. We prove that A is a tightfilter. Let x ∈ A and let { x , . . . , x m } be a cover of x . Let a ∈ A be arbitrary. Thenwe may prove as above that { a − x , . . . , a − x m } is a cover of a − x . It follows that a − x j ∈ H for some j . Thus x j ∈ A , as required.(3) It is immediate that E ( A ) is a tight filter in E ( S ) if A be a tight filter andinverse subsemigroup in S . We prove the converse. Let a ∈ A and let { a , . . . , a m } be a cover of a . Then d ( a ) ∈ A , since A is an inverse subsemigroup and it is easyto check that { d ( a ) , . . . , d ( a m ) } is a cover of d ( a ). By assumption, d ( a i ) ∈ E ( A )for some i . But a, d ( a i ) ∈ A implies that a i = a d ( a i ) ∈ A since A is an inversesubsemigroup. We have therefore proved that A is a tight filter.(4) Suppose that every prime filter in S is an ultrafilter. Let F be a tight filterin E ( S ). Then F ↑ is a tight filter in S by (3) above. By assumption F ↑ is anultrafilter and so by Lemma 2.18, F is an ultrafilter in E ( S ).Suppose now that every tight filter in E ( S ) is an ultrafilter. Let A be a tightfilter in S . Then H = ( A − A ) ↑ is a prime filter and inverse subsemigroup of S by(2) above. Thus by (3) above E ( H ) is a tight filter in E ( S ). It follows that E ( H )is an ultrafilter in E ( S ). Thus by Lemma 2.18, H is an ultrafilter in S . It followsby Proposition 2.13 of [37] that A is an ultrafilter in S . (cid:3) We denote the set of tight filters in E by T ( E ). We therefore have that U ( E ) ⊆ T ( E ) ⊆ F ( E ) . Remark 2.20.
In Definition 2.6 of [12], Exel defines a tight representation tobe a homomorphism β : E → B to a unital boolean algebra B satisfying certainconditions. From Definition 4.5 of [12], a filter F is defined to be tight if itscharacteristic function χ F : E → , to the two-element boolean algebra, is tight.For such homomorphisms, condition (i) of Proposition 2.7 of [12] holds. It is noweasy to check from this that F is tight in the sense of Exel [12] if and only if it istight in the sense of this paper.The following property gives the essence of tight filters. Lemma 2.21.
Let X = { x , . . . , x m } and Y = { y , . . . , y n } be finite subsets of E with X non-empty and let F be a tight filter such that X ⊆ F and Y ∩ F = ∅ . Then (1) X ∧ ∩ Y ⊥ = 0 . (2) If Z is any finite cover of X ∧ ∩ Y ⊥ then Z ∩ F = ∅ .Proof. (1) Suppose that X ∧ ∩ Y ⊥ = { } . Then ∅ is a finite cover. Put x = V i x i ∈ F . We claim that { x ∧ y , . . . , x ∧ y n } is a cover of x . Let 0 = y ≤ x . Then byassumption y ∧ y j = 0 for some j . But y = y ∧ x which proves the claim. However, F is a tight filter and so x ∧ y k ∈ F for some k . But this implies that y k ∈ F , whichis a contradiction.(2) From (1) above, any cover Z is non-empty. Put x = V i x i ∈ F . Let { z , . . . , z p } be a cover of { x } ↓ ∩ { y , . . . , y n } ⊥ . We claim that Z ′ = { z , . . . , z p , x ∧ y , . . . , x ∧ y n } is a cover of x . Let 0 = y ≤ x . Suppose that y is orthogonal to allthe y i . Then y ∧ z k = 0 for some k . It follows that Z ′ ∩ F = ∅ . But clearly thisintersection cannot contain any of the x ∧ y j and so must in fact be a non-emptyintersection with Z . (cid:3) For each e ∈ E , define U e = { F ∈ F ( E ) : e ∈ F } . Observe that U = ∅ . Given e and a finite set { e , . . . , e n } , possibly empty, define U e : e ,...,e n = U e ∩ U ce ∩ . . . ∩ U ce n , the set of all filters that contain e and omit all of e , . . . , e n , where we write X c todenote the set-theoretic complement of the set X . Observe that U e : e ,...,e n = U e : e ∧ e,...,e n ∧ e so that we may assume e , . . . , e n ≤ e if it simplifies calculations. Lemma 2.22.
The sets above form the basis for a topology on F ( E ) which isHausdorff.Proof. Let A ∈ U e : e ,...,e m ∩ U f : f ,...,f n . Consider U = U e ∧ f : e ,...,e m ,f ,...,f n . Clearly A ∈ U and U ⊆ U e : e ,...,e m ∩ U f : f ,...,f n . It follows that the sets do form a basis.Let
A, B ∈ F ( E ) where A = B . If A \ B and B \ A , with the set-theoreticmeanings, are both non-empty, then choose e ∈ A \ B and f ∈ B \ A . We have that A ∈ U e : e ∧ f and B ∈ U f : e ∧ f and U e : e ∧ f ∩ U f : e ∧ f = ∅ . If, say, A is properly containedin B then choose e ∈ A and f ∈ A \ B . Observe that B ∈ U f and A ∈ U e : e ∧ f and U f and A ∩ U e : e ∧ f = ∅ . Thus the topology is Hausdorff. (cid:3) The following was proved in [46] as Proposition 4.7 using functional analysis.We give a direct elementary proof.
Lemma 2.23.
The sets U e : e ,...,e n are compact-openProof. We begin with a general construction. Put = { , } , the unital Booleanalgebra with two elements. Let E be the product space. Each element is a functionfrom E to { , } . By Tychonoff’s theorem it is a compact space. A subbase for thistopology is given by subsets of the form T e and T ce , where T e = { θ : E → : θ ( e ) = 1 } . It follows that these sets are clopen and so the sets T e ∩ T ce ∩ . . . ∩ T ce n are compact-open.Next we look at the subspace of E that consists of all semigroup homomorphisms θ : E → { , } . These are functions θ : E → such that θ ( e ∧ f ) = θ ( e ) ∧ θ ( f ). Afunction φ fails to be a homomorphism if and only if θ ( e ∧ f ) = θ ( e ) θ ( f ) for some e, f ∈ E if and only if θ belongs to the union of sets of the form U e ∩ U f ∩ U ce ∧ f or U ce ∩ U f ∩ U e ∧ f or U e ∩ U cf ∩ U e ∧ f or U ce ∩ U cf ∩ U e ∧ f for some e, f ∈ E . It follows that the set of functions which are not semigrouphomomorphisms is open and so the set of homomorphisms is closed. Thus thesubspace ¯ E of E that consists of all homomorphisms f : E → { , } is the closedsubset of all semigroup homomorphisms intersected with the closed set T whichforces φ (0) = 0. This is therefore a closed subset.Finally, we remove from this space the homomorphism that sends everything tozero. We therefore obtain the locally compact space ˆ E of all non-zero homomor-phisms from E to { , } . ON-COMMUTATIVE STONE DUALITY 17
There is a bijection between F ( E ) and ˆ E which takes a filter F to its characteristicfunction χ F , and which associates with each non-zero homomorphism the set of allelements of E that map to 1. This bijection is actually a homeomorphism.We now prove the lemma. We may assume that e = 0 since no filter contains 0.This set is mapped by our bijection above to the set T e ∩ T ce ∩ . . . ∩ T ce n intersectedwith the closed subset ¯ E of homomorphisms. It is therefore a closed subset of E and so is itself compact. (cid:3) We summarize what we have found in the following.
Proposition 2.24.
The topological space F ( E ) of all filters on E is Hausdorff witha basis of compact-open subsets. The following is the analogue of Theorem 4.4 of [12].
Proposition 2.25.
The closure of the set U ( E ) of ultrafilters in the space of allfilters F ( E ) is precisely the set of tight filters P ( E ) .Proof. We show first that T ( E ) is a closed subset of F ( E ). Let F be a filter whichis not tight. Then by definition there is an element f ∈ F and a cover { e , . . . , e m } of f such that F ∩ { e , . . . , e m } = ∅ . Thus F ∈ U f : e ,...,e m an open set. Let G ∈ U f : e ,...,e m be an ultrafilter. Then since { e , . . . , e m } is a cover of f we musthave that e i ∈ G for some i which is a contradiction.Let F be a tight filter on E . Let F ∈ U be any open subset of F ( E ). Then thereis an open subset F ∈ V ⊆ U such that V contains an ultrafilter. To see why, wefirst observe that from the definition of the topology, we may find an open set V ofthe form U e : e ,...,e m containing F . Thus e ∈ F and { e , . . . , e m } ∩ F = ∅ . Since F is a tight filter, there is a non-zero z ∈ e ↓ ∩ { e , . . . , e m } by Lemma 2.21. Let G beany ultrafilter containing z ; such exists by Lemma 2.6(1). Observe that z ≤ e andso e ∈ G , and that z ∧ e i = 0 and so e i / ∈ G . It follows that G ∈ U e : e ,...,e m . (cid:3) Define V e : e ,...,e n = U e : e ,...,e n ∩ U ( E ) . This defines the subspace topology on U ( E ). Lemma 2.26.
The above topology has { V e : e ∈ E } as a basis.Proof. Let F ∈ V e : e ,...,e n be an ultrafilter. Then e ∈ F and { e , . . . , e n } ∩ F = ∅ .Thus by Lemma 2.21, there exists f ∈ F ∩ e ↓ ∩ { e , . . . , e n } ⊥ . Clearly F ∈ V f . Let G ∈ V f be any ultrafilter. Then f ∈ G and f ≤ e implies that e ∈ G and if e i ∈ G then f ∧ e i = 0. Thus in fact G ∈ V e : e ,...,e n . Hence F ∈ V f ⊆ V e : e ,...,e n . Finally,observe that V = ∅ and that V e ∧ f = V e ∩ V f . (cid:3) The following result could be proved by using Proposition 6.3 of [46] and Propo-sition 2.24, but given its importance, we prefer to give a direct proof.
Proposition 2.27.
The sets V e : e ,...,e n are all compact in F ( E ) if and only if everytight filter is an ultrafilter.Proof. Suppose that every tight filter is an ultrafilter. Then U ( E ) is a closed subsetof F ( E ) by Proposition 2.25. By Proposition 2.24, F ( E ) is a Hausdorff space. ByLemma 2.23, the sets U e : e ,...,e n are compact-open. Therefore they are closed. Thusthe set V e : e ,...,e n ⊆ U e : e ,...,e n is closed. But a closed subset of a compact space isitself compact. Thus the subsets V e : e ,...,e n are compact. Suppose now that all sets V e : e ,...,e n are compact in F ( U ). We prove that everytight filter is an ultrafilter. Let F be a tight filter in F ( E ). Let e ∈ F so that F ∈ U e .Let O be any open set containing F . Then O ∩ U e is an open set containing F .Since F is tight we have that O ∩ U e contains an ultrafilter by Proposition 2.25.It follows that O ∩ V e is non-empty. The set V e is compact in the Hausdorff space F ( U ) and so is closed. However, we have shown that F is a limit point for V e andso must belong to V e and is therefore an ultrafilter, as required. (cid:3) Proof of the Booleanization Theorem (2)
Recall that the elements of D ( E ) are the sets of the form S mi =1 V e i and that thepartial ordering is subset inclusion. We shall prove that D ( E ) is a Boolean algebraif and only if every tight filter in E is an ultrafilter. Proof.
Suppose first that every tight filter is an ultrafilter. Let A ⊆ B where A = S mi =1 V e i and B = S nj =1 V f i . The set B/A is a finite union of sets of the form V e /V f . Observe that V e /V f = V e : e ∧ f . Under our assumption, V e : e ∧ f is compactand so by Lemma 2.26, it is a finite union of sets of the form V g . It follows that B/A is a finite union of sets of the form V g . Hence B/A ∈ D ( E ) and so, since D ( E )is already a distributive lattice, it is a Boolean algebra.To prove the converse, assume that D ( E ) is a Boolean algebra. In our proofbelow, we use the fact that in a Boolean algebra, not necessarily unital, everyultrafilters are the same as prime filters and that maps to are determined byultrafilters. We now prove that every tight filter is an ultrafilter. Let F be atight filter in E . Then the characteristic function χ F : E → is a cover-to-joinmap. Thus by the universal property of the map δ : E → D ( E ) there is a uniquehomomorphism χ G : D ( E ) → such that χ G δ = χ F . The filter G is an ultrafilterbecause D ( E ) is a boolean algebra. Since the diagram of maps commutes we alsohave that F = δ − ( G ). We shall prove that F is an ultrafilter. Suppose not. Thenby Lemma 2.6(2), there is an e ∈ E such that e has a non-zero intersection withevery element of F but e / ∈ F . It follows that δ ( e ) / ∈ G . But δ is 0-restricted and so δ ( e ) = 0. Now G is an ultrafilter and so there exists g ∈ G such that δ ( e ) ∧ g = 0.Each non-zero element of D ( S ) is a join of a finite number of elements in the imageof δ . Thus g = W mi =1 δ ( e i ). But G is an ultrafilter and so a prime filter. It followsthat δ ( e i ) ∈ G for some i . Also δ ( e ) ∧ δ ( e i ) = 0. Again using the fact that δ is0-restricted, we have that e ∧ e i = 0. But e i ∈ F . This is a contradiction. It followsthat F is a tight filter. (cid:3) One obvious question is when the Boolean completion of a pre-Boolean semilat-tice is unital. A finite set of idempotents { e , . . . , e n } in a semilattice E is said tobe essential if it is a cover of E \ { } . We shall say that a pre-Boolean semilattice E is compactable if it has at least one essential set of idempotents. Theorem 2.28.
Let E be a pre-Boolean semilattice. Then D ( E ) is unital if andonly if E is compactable, and the essential sets of idempotents are precisely the setsof idempotents mapped to the identity of D ( E ) .Proof. Suppose that D ( E ) is unital. Then D ( E ) = S ni =1 V e i for some finite setof idempotents e , . . . , e n . Let e ∈ E be an arbitrary non-zero element. Then byLemma 2.6(1), there is an ultrafilter F containing e . By assumption, e i ∈ F forsome i . In particular, e ∧ e i = 0. Thus { e , . . . , e n } is a finite cover. ON-COMMUTATIVE STONE DUALITY 19
Conversely, suppose that { e , . . . , e n } is an essential set of idempotents. Let F be any ultrafilter. Suppose that e i / ∈ F for all i . Then for each i there exists f i ∈ F such that f i ∧ e i = 0 by Lemma 2.6(2). Put f = ∧ ni =1 f i . Then f ∈ F and so is non-zero. But f ∧ e i = 0 for all i . This contradicts our assumption that { e , . . . , e n } is a finite cover. Thus we must have e i ∈ F for some i . It follows that D ( E ) = S ni =1 V e and so D ( E ) is unital. (cid:3) Part 3Lemma 2.29.
Let X and Y be finite subsets of E with X non-empty. Put e = V X and let Y = { e , . . . , e n } . Assume that X ∧ ∩ Y ⊥ is non-zero. Then Z ⊆ X ∧ ∩ Y ⊥ is a finite cover if and only if [ z ∈ Z V z = V e : e ,...,e n . Proof.
Suppose that [ z ∈ Z V z = V e : e ,...,e n . We prove that Z ⊆ X ∩ Y ⊥ is a cover. Let x ∈ X ∧ ∩ Y ⊥ . Then x ≤ e and x ∧ e i = 0for all i . Let F be an ultrafilter containing x . Then F contains e and omits all the e i . Thus F ∈ V e : e ,...,e n . By assumption z ∈ F for some z ∈ Z . But then z ∧ x = 0,as required.Conversely, suppose that Z ⊆ X ∧ ∩ Y ⊥ is a finite cover. Let F ∈ V e : e ,...,e n .Then F contains e and omits all the e i . Let f ∈ F such that f ∧ e i = 0 for all i .Suppose that Z ∩ F = ∅ . Let f ′ ∈ F be such that z ∧ f ′ = 0 for all z ∈ Z . Then0 = e ∧ f ∧ f ′ ∈ X ∧ ∩ Y ⊥ . Thus there is z ∈ Z such that z ∧ e ∧ f = 0 which is acontradiction. Thus Z ∩ F = ∅ . (cid:3) Let E be a 0-disjunctive semilattice. We now prove that E is pre-Boolean if andonly if it satisfies the trapping condition. Proof of the Booleanization Theorem (3)
Proof.
Suppose that E is 0-disjunctive and pre-Boolean. Let 0 = f < e . Because E is 0-disjunctive, the set e ↓ ∩ f ⊥ = { } . The set V e : f is non-empty and by assumptioncompact and so by Lemma 2.29, e ↓ ∩ f ⊥ has a finite cover.Suppose now that the trapping condition holds. We prove that every tight filteris an ultrafilter. Let F be a tight filter that is not an ultrafilter. Then we can findan ultrafilter G such that F ⊂ G . Let g ∈ G \ F and let f ∈ F be arbitrary. Then g ′ = g ∧ f ∈ G . We shall prove that F tight implies that g ′ ∈ F which implies g ∈ F , a contradiction. We have that 0 = g ′ < f . Since E is 0-disjunctive, the set f ↓ ∩ ( g ′ ) ⊥ = 0. Let { f , . . . , f n } be a cover. Suppose that f i ∈ F . Then f i , g ′ ∈ G and so f i ∧ g ′ = 0. But this contradicts the fact that f i ∧ g ′ = 0. It follows that f ∈ F and { f , . . . , f n } ∩ F = ∅ . Observe that g ′ ∈ f ↓ ∩ { f , . . . , f n } ⊥ . Let i be any nonzero element of f ↓ ∩ { f , . . . , f n } ⊥ . If i ∧ g ′ = 0 then i ∧ f i = 0 forsome i which is a contradiction. It follows that i ∧ g ′ = 0. Thus { g ′ } is a cover of f ↓ ∩ { f , . . . , f n } ⊥ . By Lemma 2.21, it follows that g ′ ∈ F giving our contradiction.It follows that F is an ultrafilter. (cid:3) The semilattices arising above can be characterized in a more direct way. Wesay that a semilattice ( E, ∧ ) is densely embedded in a Boolean algebra B if the ∧ -operation on E is precisely the restriction of the ∧ -operation in B and if everynon-zero element of B is a finite join of elements of E . Proposition 2.30.
A semilattice is -disjunctive and satisfies the trapping condi-tion if and only if it can be densely embedded in a Boolean algebra.Proof. Suppose that our semilattice E can be densely embedded in Boolean algebra B . We prove that it is 0-disjunctive and that the trapping condition holds. Let0 = f < e in E . Then e ↓ ∩ f ⊥ = ( e \ f ) ↓ . By denseness, we may write e \ f = W ni =1 e i where e i ∈ E . In particular, this shows that E is 0-disjunctive. Put Z = { e , . . . , e n } . It is easy to check that this is a cover of e ↓ ∩ f ⊥ . For the converse,we apply part (3) of the Booleanization Theorem and observe by Lemma 2.2 that0-disjunctive semilattices are separative. (cid:3) The Comparison Theorem.
The main theorem this section shows how toconstruct a Hausdorff Boolean groupoid directly from a pre-Boolean inverse ∧ -semigroup.With each pre-Boolean inverse semigroup S we may associate a Boolean inversesemigroup D ( S ). Our proof of this in Section 2.1 used the groupoid of ultrafiltersof S . We show now how to regard this as a Hausdorff Boolean groupoid G ( S )and establish the exact connection between it and the Boolean inverse ∧ -semigroup D ( S ).Let G be a topological groupoid. We make the following definitions: Ω( G ) is theset of all open subsets of G ; Bi ( G ) is the the inverse semigroup of all local bisectionsof G ; oBi ( G ) = Bi ( G ) ∩ Ω( G ) is the set of all open local bisections of G ; B ( G ) isthe set of all compact-open local bisections of G . The proofs of the following canbe found in [56]. Lemma 2.31.
Let G be a topological groupoid. (1) G is open if and only if the multiplication map m is open if and only if Ω( G ) is a semigroup under the pointwise product. (2) If G is open then oBi ( G ) is an inverse semigroup. (3) G is ´etale if and only if it is open and G o is open in G . (4) If G is ´etale then Ω( G ) is a monoid and oBi ( G ) is an inverse monoid. (5) If G o is open in G then the open bisections form a basis for the topology on G . (6) If G is ´etale then oBi ( G ) is an inverse monoid and a basis for the topology Ω( G ) . Lemma 2.32.
Let G be a Hausdorff, ´etale topological groupoid. (1) G ∗ G is a closed subspace of G × G . (2) The product of two compact-open bisections is a compact-open bisection. (3) G has a basis of compact-open bisections if and only if G o has a compact-open basis. (4) If G is a Hausdorff Boolean groupoid then B ( G ) is a Boolean inverse ∧ -semigroup.Proof. (1) This is well-known but we give the proof for completeness’ sake. Put∆( G o ) = { ( e, e ) : e ∈ G o } . Now G is hausdorff and G o is an open subset of G byLemma 2.36. Thus G o is hausdorff. By Theorem 13.7 [59], it follows that ∆( G o ) isa closed subspace of G o × G o . But G ∗ G = ( d , r ) − (∆( G o )) and the map ( d , r ) iscontinuous. Thus G ∗ G is a closed subsapce as claimed. ON-COMMUTATIVE STONE DUALITY 21 (2) Let A and B be compact-open bisections. It only remains to prove that AB is compact. By Theorem D of Section 26 of [58], both A and B are closed subsetsof G . It follows that A × B is a closed subset of G × G . By (1), we therefore havethat A ∗ B = ( a × B ) ∩ ( G ∗ G ) is a closed subset. But A × B is compact and soby Theorem A of Section 21 of [58] the set A ∗ B is compact. By Theorem B ofSection 21 of [58], the continuous image of a compact space is compact. Thus AB is compact.(3) Suppose that G has a basis of compact-open bisections. We prove that G o has a basis of compact-open sets. Consider the set of all compact-open bisectionsthat are subsets of G o . We prove that they form a basis for the subspace topologyon G o . Every open set U of G o has the form U = G o ∩ V where V is an open setin G . But G o is an open subspace of G and so U is also an open subset of G . Byassumption, we may write U = S i B i where B i are compact-open bisections. But B i ⊆ U ⊆ G o and so each B i is a compact-open bisection contained in G o . It isclear that the B i are open in G o and also compact.Suppose that G o has a basis of compact-open sets. We prove that G has a basisof compact-open bisections. By assumption, G has a basis of open bisections. Let A be such an open bisection. Then A − A is an open subset of G o . By assumptionwe may write A − A = S i E i where the E i are compact-open subsets of G o . Put A i = AE i . This is an open bisection satisfying A − i A i = E i . We shall have provedthe result if we can prove that every open bisection A such that A − A is compact-open is itself compact-open. Suppose that A = S j B j where the B j are openbisections. Then A − A = S j B − j B j . By assumption A − A is compact-open andso we may find a finite subset of the j such that A − A = S nj =1 B − j B j . But then A = S nj =1 AB − j B j = S nj =1 B j . Thus if A is an open bisection such that A − A iscompact-open then A is compact-open.(4) It only remains to observe that if A and B are both compact-open bisectionsthen A ∩ B being closed and a subset of A is also compact and so compact-openand a bisection, and if A is compatible with B then A ∪ B is a bisection, open andcompact. (cid:3) Recall that if S be an inverse ∧ -semigroup with zero, then we defined Ω = { V a : a ∈ S } . By Lemma 2.10(1), the set Ω is a basis for a topology on the groupoid G ( S ). Proposition 2.33.
Let S be an inverse ∧ -semigroup with zero. Then with thetopology above G ( S ) is a Hausdorff ´etale topological groupoid whose topological spaceof identities is homeomorphic to the topological space constructed from E ( S ) .Proof. We show first that the topology is Hausdorff. Let
A, B ∈ G ( S ) be distinctelements. Then there exists a ∈ A such that a / ∈ B otherwise we would have A ⊆ B and so A = B since both A and B are ultrafilters. But b / ∈ B implies that thereexists b ∈ B such that a ∧ b = 0 by Lemma 2.6(2). Thus V a ∩ V b = ∅ and A ∈ V a and B ∈ V b .That inversion is continuous follows by Lemma 2.10(3).Multiplication is continuous. Let G ( S ) ∗ G ( S ) denoted the subset of G ( S ) × G ( S )consisting of pairs ( A, B ) such that A − · A = B · B − . We prove that the map µ : G ( S ) ∗ G ( S ) → G ( S ) given by ( A, B ) A · B is continuous. To do this we prove that µ − ( V a ) = [ = bc ≤ a V b × V c ∩ ( G ( S ) ∗ G ( S )) . Let (
B, C ) ∈ µ − ( V a ). Then A = B · C is an ultrafilter containing a . Then a ∈ ( BC ) ↑ and so bc ≤ a for some b ∈ B and c ∈ C . Thus B ∈ V b , C ∈ V c and0 = bc ≤ a . We have proved that the lefthand side is contained in the righthandside. To prove the reverse inclusion, suppose that 0 = bc ≤ a and B ∈ V b , C ∈ K c and the product B · C exists. Then B · C is an ultrafilter containing a and so B · C ∈ K a .The map d is a local homeomorphism. We are required to show that d : G ( S ) → G ( S ) o is a local homeomorphism. To do this it is enough to prove that the map d : V s → V s − s given by A A − · A is a homeomorphism. It is bijective byLemma 2.8(3). It is continuous because inversion and multiplication are continuous.It remains to show that it is open. A base of open subsets of V s is given by the sets V s ∩ V t ; that is, the sets V s ∧ t . It follows that a base of open subsets of V s is givenby the sets V t where t ≤ s . But it follows that V t − t is an open set in V s − s . G ( S ) o is homeomorphic to the space constructed from E ( S ). In Lemma 2.18(4),we defined a bijection ǫ : G ( S ) o → G ( E ( S )) by ǫ ( A ) = E ( A ). We prove that thisis a homeomorphism. We need some notation. If e ∈ E ( S ) is an idempotent wedenote the set of ultrafilters in E ( S ) containing e by V Ee . We prove that the map ǫ is continuous and open. By definition ǫ − ( V Ee ) = { A ∈ G ( S ) o : e ∈ E ( A ) } . But the set on the righthand side is just V e because a filter contains an idempotentif and only if it is an inverse submonoid if and only if it is an idempotent filter.It follows that ǫ is continuous. To show that ǫ is open we calculate ǫ ( V a ∩ G ( S ) o ).The elements of V a ∩ G ( S ) o are the idempotent ultrafilters that contain a . Theytherefore contain a − a and so also the idempotent e = a ∧ a − a . But if F ⊆ E ( S )is an ultrafilter containing e then F ↑ is an idempotent ultrafilter in S containing a .Thus ǫ ( V a ∩ G ( S ) o ) = V Ee . (cid:3) The main theorem of this section now follows.
Theorem 2.34 (Comparison theorem) . Let S be a pre-Boolean inverse ∧ -semigroup.Then G ( S ) is a Hausdorff Boolean groupoid, and D ( S ) is isomorphic to B ( G ( S )) .Proof. The fact that G ( S ) is a Hausdorff Boolean groupoid follows from Propo-sition 2.33 and the fact that the semilattice of idempotents of S is a pre-Booleansemilattice whose associated topological space is Boolean and homeomorphic to thespace of identities of G ( S ). After Lemma 2.14, we defined what we can now see isa map β ′ : D ( S ) → B ( G ( S )). We proved there that this was an injective homomor-phism. By Lemma 2.15, this homomorphism is surjective because every element of B ( G ( S )), being compact-open, is a finite union of elements of the form V s where s ∈ S since these form a compact-open basis. (cid:3) The Duality Theorem.
We have seen how to complete pre-Boolean inverse ∧ -semigroups to Boolean inverse ∧ -semigroups. We have also seen how to constructBoolean inverse ∧ -semigroups from Hausdorff Boolean groupoids. We now complete ON-COMMUTATIVE STONE DUALITY 23 the circle of ideas by proving the Duality Theorem. The following was proved asLemma 2.17 of [37].
Lemma 2.35.
Let α : G → H be a covering functor between groupoids. Then α − : Bi ( H ) → Bi ( G ) is a morphism of inverse ∧ -semigroups and defines a con-travariant functor from the category of groupoids and their covering functors to thecategory of Boolean inverse ∧ -semigroups and their morphisms. Let G be a Hausdorff Boolean groupoid. For each g ∈ G define F g = { A ∈ B ( G ) : g ∈ A } . The following was proved as Lemma 2.19 of [37]; although the definition of Booleangroupoid used in that paper is more restricted than the meaning used here, theproofs are equally valid for our more general case.
Lemma 2.36.
With the above definition we have the following. (1) F g is an ultrafilter and every ultrafilter in B ( G ) is of this form. (2) If gh exists in the groupoid G then F g · F h = F gh . (3) F − g = F g − . (4) F g = F h if and only if g = h .Proof. (1) Let g ∈ G . It is immediate that F g is a filter so it only remains to checkthat it is an ultrafilter. We use Lemma 2.6(2). Let A ∈ B ( G ) be a compact-openbisection with the property that A ∩ B = ∅ for each B ∈ B ( G ). We shall prove that g belongs to the closure of A ; but in a Hausdorff space compact sets are closed andso this will imply that g ∈ A . Let O be any open set containing g . By definitionthere is a compact-open bisection C such that g ∈ C ⊆ O . But C is a compact-openbisection containing g and so C ∈ F g . By assumption C ∩ A = ∅ . We have provedthat every open set containing g contains elements of A . It follows that g belongsto the closure of A , as required.Now let F be any ultrafilter in B ( G ). We shall prove that F ⊆ F g for some g ∈ G from which the claim will follow. Let A ∈ F be any compact-open bisectionbelonging to F . Consider the set F ′ = { A ∩ B : b ∈ F } , a subset of F because F isa filter. In addition, this is a family of closed subsets of G , since compact subsetsof hausdorff spaces are closed, with the property that any finite intersection is non-empty, because F is a filter. But each element of F ′ is a subset of the compact set A . It follows that the set F ′ has a non-empty intersection. Let g belong to thisintersection. Then g belongs to every element of F and so F ⊆ F g , as required.The proofs of (2) and (3) are straightforward, and (4) follows from the fact thatthe groupoid G is Hausdorff. (cid:3) Proposition 2.37.
Define B to take the Hausdorff Boolean groupoid G to theBoolean inverse ∧ -semigroup B ( G ) and the proper continuous covering functor α : G → H to the function α − : B ( H ) → B ( G ) . Then B is a contravariant functorfrom the category of Hausdorff Boolean groupoids to the category of Boolean inverse ∧ -semigroups.Proof. Let α : G → H be a proper continuous covering functor between two Booleangroupoids. By Lemma 2.35 and the fact that α is proper, we have that α − is a ∧ -homomorphism from B ( H ) to B ( G ). It remains to prove that α − pulls ultra-filters in B ( G ) back to ultrafilters in B ( H ). Let F be an ultrafilter in B ( G ). ByLemma 2.36(1), there exists g ∈ G such that F = F g . Put h = α ( g ). Then G = F h is an ultrafilter in H . The inverse image of F under α − consists of all thosecompact-open local bisections B such that α − ( B ) ∈ F . But α − ( B ) ∈ F if andonly if g ∈ α − ( B ) if and only if α ( g ) ∈ B if and only if B ∈ G , as required. (cid:3) The following was proved as Proposition 2.15 of [37].
Lemma 2.38.
Let θ : S → T be a morphism of Boolean inverse ∧ -semigroups.Then θ − : G ( T ) → G ( S ) is a covering functor. Proposition 2.39.
Define G to take the Boolean inverse ∧ -semigroup S to theHausdorff Boolean groupoid G ( S ) and the proper morphism θ : S → T betweenBoolean inverse ∧ -semigroups to the function θ − : G ( T ) → G ( S ) . Then G definesa contravariant functor from the category of Boolean inverse ∧ -semigroups to thecategory of Hausdorff Boolean groupoids.Proof. Let θ : S → T be a proper morphism between Boolean inverse ∧ -semigroups.By assumption, θ − : G ( T ) → G ( S ) is a well-defined function. Put φ = θ − . ByLemma 2.38, it follows that φ is a covering functor. It remains to prove that φ iscontinuous and proper. The basic open sets of G ( S ) are of the form V s where s ∈ S .Put t = θ ( s ). Then F belongs to the inverse image of V s under φ if and only if θ − ( F ) ∈ V s if and only if s ∈ θ − ( F ) if and only if θ ( s ) ∈ F if and only if F ∈ V t ,as required.It remains to show that θ − is proper. Let X be a compact subset of G ( S ).Clearly X ⊆ S s ∈ S V s . But X is compact and so we can find a finite number ofelements of S such that X ⊆ S ni =1 V s i . It follows that φ − ( X ) ⊆ S ni =1 V θ ( s i ) . Now X is a compact subset of a Hausdorff space and so is closed. It follows that φ − ( X )is closed. But S ni =1 V θ ( s i ) is a finite union of compact subsets and so is compact.But a closed subset of a compact spaces is compact. It follows that φ − ( X ) iscompact, as claimed. (cid:3) Proof of the Duality Theorem
Proof.
Let S be a Boolean inverse ∧ -semigroup. Then G ( S ) is a Boolean groupoidand B ( G ( S )) is a Boolean inverse semigroup. By Theorem 2.34, the ComparisonTheorem, D ( S ) is isomorphic to B ( G ( S )). We therefore need to show that S isisomorphic to D ( S ). It is enough to show that if θ : S → T is a cover-to-joinmap to a distributive inverse semigroup then θ is automatically a join-preservinghomomorphism. The isomorphism will then follow from the universal property of D ( S ). Let s = W ni =1 s i in S . Then the s i are pairwise compatible. We prove that { s , . . . , s n } is a cover of s . Let 0 = a ≤ s . Then a = sa − a . It follows that a = W ni =1 s i a − a . Since a is non-zero one of the s i a − a must be non-zero. Thusthe idempotent s − i s i a − a is non-zero. But a and s i are both in s ↓ and so they arecompatible. It follows also that s i ∧ a = s − i s i a − a . Thus s i ∧ a = 0, as required.It follows that θ ( s ) = W ni =1 θ ( s i ), as required.Let G be a Hausdorff Boolean groupoid. Then B ( G ) is a Boolean inverse semi-group and G ( B ( G )) is a Hausdorff Boolean groupoid. The function g
7→ F g is abijective functor by Lemma 2.36. We need to show that this map is continuousand open. Let U be a compact-open bisection of G . The image of U under thismap is {F g : g ∈ U } . Now U is an element of the inverse semigroup B ( G )). Withinthis inverse semigroup, we may consider all the ultrafilters that contain U as an ON-COMMUTATIVE STONE DUALITY 25 element. That is, the set V U . By Lemma 2.36, ultrafilters in B ( G )) all have theform F h for h ∈ G . Now the ultrafilter F h contains U if and only if h ∈ U . Hence V U = {F g : g ∈ U } is the image of U under our map and so is an open set.Finally, we prove that our map is continuous. A basic compact-open subset of G ( B ( G )) is of the form V U where U is a compact-open bisection of G . The inverseimage of V U under our map is U . (cid:3) Basic properties
In this section, we shall prove some results that will be used to help us analysethe examples in Section 4.3.1.
Unambiguous E ∗ -unitary inverse semigroups. The following lemma willsimplify checking when a map is a cover-to-join map. Note that D is one of Green’srelations. Lemma 3.1.
Let S be an inverse ∧ -semigroup and let θ : S → T be a homomor-phism to a distributive inverse semigroup. (1) The map θ is a cover-to-join map if and only if the map θ | E ( S ) : E ( S ) → E ( T ) is a cover-to-join map. (2) Let { f i : i ∈ I } be an idempotent transversal of the non-zero D -classes of S . Then θ is a cover-to-join map if and only if it is a cover-to-join mapfor the distinguished family of idempotents.Proof. (1) Only one direction needs proving: we prove that if θ | E ( S ) : E ( S ) → E ( T ) is a cover-to-join map then θ is a cover-to-join map. Let { a , . . . , a m } be acover of a . We prove that { d ( a ) , . . . , d ( a m ) } is a cover of d ( a ). Let 0 = e ≤ d ( a ).Put a ′ = ae . Then d ( a ′ ) = e and so 0 = a ′ ≤ a . Thus by assumption a i ∧ a ′ = 0for some i . But a i and a ′ are bounded above by a and so are compatible. Bycompatibility, we get that d ( a i ∧ a ′ ) = d ( a i ) d ( a ′ ) = e = 0, as required. By theassumption on θ it follows that θ ( d ( a )) = _ i θ ( d ( a i )) . Multiply this equality on the left by θ ( a ). The lefthand-side becomes θ ( a ) and therighthand side becomes W i θ ( a i ), as required.(2) Let e be an arbitrary idempotent such that e D f where f belongs to ourtransversal. Let { e , . . . , e m } be a cover of e , where all elements are idempotents.Let a be any element of S such that a − a = e and aa − = f . Put f i = ae i a − allbounded above by f . We claim that { f , . . . , f m } is a cover of f . Let 0 = j ≤ f .Then 0 = a − ja ≤ e . By assumption there exists e i such that e i ∧ a − ja = 0.Observe that e ( e i ∧ a − ja ) = e i ∧ a − ja . It follows that a ( e i ∧ a − ja ) a − = 0.Thus a ( e i ∧ a − ja ) a − = j ∧ f i = 0, as required, where we use the fact thatmultiplication always distributes over finite meets by Proposition 1.4.19 of [27].Thus { f , . . . , f m } is a cover of f . It follows that θ ( f ) = W i θ ( f i ). Now multiplythis equation on the left by θ ( a ) − and on the right by θ ( a ). This then gives us θ ( e ) = W i θ ( e i ), as required. (cid:3) We now locate an important class of inverse ∧ -semigroups. An inverse semigroupwith zero is is said to be E ∗ -unitary if 0 = e ≤ s where e is an idempotent impliesthat s is an idempotent. The following is Remark 2.3 of [46] which is worth repeatingsince it was a surprise to many people not least the author. Lemma 3.2. If S is an E ∗ -unitary inverse semigroup then S is an inverse ∧ -semigroup. There are many naturally occurring examples of E ∗ -unitary inverse semigroupsand it is a condition that is easy to verify. In particular, both the graph inversesemigroups and the tiling semigroups are E ∗ -unitary. All the inverse semigroups inSection 4 will be E ∗ -unitary.A poset with zero X is said to be unambiguous if for all x, y ∈ X if there exists0 = z ≤ x, y the either x ≤ y or y ≤ x . An inverse semigroup with zero S will besaid to be unambiguous if its semilattice of idempotents is unambiguous. That is,for any two non-zero idempotents e and f if ef = 0 then e and f are comparable.Let E be a meet semilattice with zero. Given e, f ∈ E we say that e is directlyabove f and that f is directly below e if e > f and there is no g ∈ E such that e > g > f . For each e ∈ E define ˆ e to be the set of elements of E that aredirectly below e . The semilattice E is said to be pseudofinite if for each e ∈ E theset ˆ e is finite and whenever e > f there exists g ∈ ˆ e such that e > g > f . Aninverse semigroup is pseudofinite if its semilattice of idempotents is pseudofinite.The meet semilattice E is said to satisfy the Dedekind height condition if for allnon-zero elements e the set (cid:12)(cid:12) e ↑ (cid:12)(cid:12) < ∞ . An inverse semigroup is said to satisfy thiscondition if its semilattice of idempotents does. Proposition 3.3.
Let E be an unambiguous semilattice with zero which is pseu-dofinite and satisfies the Dedekind finiteness condition. Then E is pre-Boolean.Proof. We claim first that if 0 = f < e and if V e : f = ∅ then we may find idempotents e , . . . , e m ≤ e such that V e : f = m [ i =1 V e i . For each non-zero element e , the set ˆ e is a finite orthogonal set by unambiguity.Let 0 = f < e . If V e : f = ∅ , then V e : f may be written as a finite union of setsof the form V e : j where j < i is immediately below e . By assumption, the set f ↑ isfinite. There is therefore an element f < g ≤ e such that g is immediately above f . Observe that V e : f = V g : f ∪ V e : g . The argument is now repeated with V e : g and by induction we have proved the claim.Let 0 = f < e be such that ˆ e = { f, e , . . . , e m } . Then if V e : f = ∅ then V e : f = m [ i =1 V e i . Let F ∈ V e : f . Then e ∈ F and f / ∈ F . By Lemma 2.6(2), there exists g ∈ F suchthat g ∧ f = 0. Since g ∧ e ∈ F we may in fact assume that g ≤ e . Now g ≤ e implies that either g ≤ f or g ≤ e i for some i . We cannot have the former andso we must have the latter. It follows that e i ∈ F and so F ∈ V e i . To prove the Strictly speaking ‘unambiguous except at zero’ but that is too much of a mouthful.
ON-COMMUTATIVE STONE DUALITY 27 reverse inclusion, let F ∈ V e i . Then F ∈ V e . But we cannot have f ∈ F because f ∧ e i = 0. It follows that F ∈ V e : f .We have therefore proved that if V e : f = ∅ then V e : f = m [ i =1 V e i . The claim is proved when we observe that the equality holds when the e i arereplaced by e i ∧ e .We now prove that every tight filter is an ultrafilter which implies that E ispre-Boolean by part (2) of the Booleanization Theorem. Let P be a tight filter andsuppose that it is not an ultrafilter. Then there is an ultrafilter Q such that P ⊂ Q .Let f ∈ Q \ P and let e ∈ P . Now e ∧ f ∈ Q \ P . It follows that we may choose f ∈ Q \ P such that f ≤ e . Now P is a tight filter that contains e and omits f .It follows by Lemma 2.21(1), that there exists 0 = g ≤ e such that g ∧ f = 0. ByLemma 2.6(1), there is an ultrafilter containing g that must therefore omit f . Butthen we have shown that V e : f = ∅ . By what we proved above we have that V e : f = m [ i =1 V e i for some idempotents e , . . . , e m . It follows that V e = m [ i =1 V e i ! ∪ V f . Thus by Lemma 2.29, we have that e → ( e , . . . , e m , f ) . Now e ∈ P and P is a tight filter. It follows that either f ∈ P or e i ∈ P for some i .By construction, f / ∈ P . It follows that e i ∈ P for some i . It follows that f, e i ∈ Q .But Q is an ultrafilter containing e i . It follows that F ∈ V e : f and so f / ∈ Q . Wehave therefore arrived at a contradiction. It follows that every tight filter is anultrafilter. (cid:3) Concrete examples of semilattices satisfying the conditions of the above propo-sition may easily be constructed. Let G be a directed graph and let G ∗ denote thefree category generated by G . Denote by E the set of all elements of G ∗ togetherwith a zero element. Define e ≤ f if and only if e = f g for some element g ∈ G ∗ ;in other words, the prefix ordering. With respect to this order, E becomes a semi-lattice which is unambiguous and satisfies the Dedekind finiteness condition. Thefollowing are easy to check. Lemma 3.4.
With the above definitions, we have the following. (1)
The semilattice E has no -minimal idempotents if and only if the in-degreeof each vertex is at least 1. (2) The semilattice E is -disjunctive if and only if the in-degree of each vertexis either zero or at least 2. (3) The semilattice E is pseudofinite if and only if the in-degree of each vertexis finite. The group of units of D ( S ) . Let S be a pre-Boolean inverse ∧ -semigroup.In this section, we shall be interested in calculating the group of units of D ( S ) inthe case where it is unital. By Theorem 2.28, this occurs precisely when E ( S ) iscompactable. We shall focus on the case where S is unambiguous and E ∗ -unitary.The proofs of the following can be found in [19] or easily proved directly. Lemma 3.5. (1)
Let S be an unambiguous inverse semigroup. Then ( S, ≤ ) is an unambigu-ous poset if and only if S is E ∗ -unitary. (2) Let S be an unambiguous E ∗ -unitary inverse semigroup. Then S is separ-ative if and only if the semilattice of idempotents E ( S ) is -disjunctive. Lemma 3.6.
Let S be an unambiguous E ∗ -unitary inverse semigroup. Given afinitely generated order ideal { s , . . . , s m } ↓ , there exists an orthogonal set { t , . . . , t n } such that { s , . . . , s m } ↓ = { t , . . . , t n } ↓ .Proof. Starting with i = 1 compare s i with s j where j > i . If they are orthogonalcontinue; if s i < s j then discard s i and increase i by 1; if s j < s i then discard s j and continue comparing. (cid:3) The following is now immediate.
Corollary 3.7.
Let S be an unambiguous E ∗ -unitary inverse semigroup. Thenevery element of D ( S ) is an orthogonal join of elements from δ ( S ) . A homomorphism of inverse semigroups is said to be idempotent-pure if theinverse images of idempotents consist only of idempotents.
Lemma 3.8.
Let S be E ∗ -unitary. Then the congruence ≡ defined on FC ( S ) isidempotent-pure.Proof. Suppose that ( a , . . . , a m ) ↔ ( e , . . . , e n ) where the e j are idempotents.Then the a i are idempotents. This follows because for each i there exists j suchthat a i ∧ e j = 0. But a i ∧ e j is a non-zero idempotent beneath a i and so a i is alsoan idempotent. (cid:3) A finite compatible subset A ⊆ S is said to be essential when both d ( A ) = { a − a : a ∈ A } and r ( A ) = { aa − : a ∈ A } are essential sets of idempotents; thatis, both are covers of E ( S ) \ { } .Let T be an inverse semigroup with zero. An idempotent e in T is said to be essential if for each non-zero idempotent f ∈ S we have that f ∧ e = 0. This saysthat { e } is a cover of E ( T ). We denote by T e the set of all elements s ∈ T such that d ( s ) and r ( s ) are essential idempotents. By Lemma 4.2 of [32], T e is an inversesubsemigroup of T . Lemma 3.9.
Let S be separative. An idempotent { e , . . . , e n } ↓ in FC ( S ) is essentialif and only if the set { e , . . . , e n } is essential.Proof. Suppose that { e , . . . , e n } ↓ is an essential idempotent. Let f ∈ E ( S ) be anynon-zero idempotent. Then f ↓ { e , . . . , e n } ↓ is a non-zero idempotent. Thus forsome i we have that f e i is non-zero, as required.Suppose that { e , . . . , e n } is a cover of E ( S ). Then it is easy to check that { e , . . . , e n } ↓ is an essential idempotent. (cid:3) ON-COMMUTATIVE STONE DUALITY 29
On an inverse semigroup S define s σ t if and only if there exists u ≤ s, t . Then σ is a congruence and S/σ is a group. It is, in fact, the minimum group congruence .See Section 2.4 of [27].
Theorem 3.10.
Let S be E ∗ -unitary, unambiguous, compactable and separative.Then the group of units of D ( S ) is isomorphic to the group FC ( S ) e /σ .Proof. If A ∈ FC ( S ) then we denote the ≡ -class containing A by [ A ]. Since A is afinitely generated order ideal, we have that A = X ↓ where X is a finite compatiblesubset of S . By definition, the element [ A ] is in the group of units of D ( S ) if andonly if [ A ] − [ A ] and [ A ][ A ] − are both the identity element of D ( S ). But A is acompatible order ideal and so we have that A − A = { d ( a ) : a ∈ A } = d ( A ) andsimilarly for AA − . Thus both [ d ( A )] and [ r ( A )] are equal to the identity andso both d ( X ) and r ( X ) are essential sets of idempotents by Theorem 2.28. Bydefinition, it follows that X is an essential subset of S . By Lemma 3.6, we mayassume that the set X is orthogonal.Let [ A ] = [ X ↓ ] and [ B ] = [ Y ↓ ] be two invertible elements of D ( S ) where both X and Y are finite, orthogonal essential sets. We shall prove that A ≡ B if and onlyif A σ B . First we need an auxiliary result.Define { a , . . . , a m } (cid:22) { b , . . . , b n } if and only if { a , . . . , a m } ↓ ⊆ { b , . . . , b n } ↓ and { b , . . . , b n } → { a , . . . , a m } . Now observe that if { a , . . . , a m } ↔ { b , . . . , b n } then { a , . . . , a m } → { a i ∧ b j : 1 ≤ i ≤ m, ≤ j ≤ n } and { b , . . . , b n } → { a i ∧ b j : 1 ≤ i ≤ m, ≤ j ≤ n } . It follows that if { a , . . . , a m } ↔ { b , . . . , b n } then there is { c , . . . , c p } such that { c , . . . , c p } (cid:22) { a , . . . , a m } and { c , . . . , c p } (cid:22) { b , . . . , b n } . On the other hand, if { a , . . . , a m } (cid:22) { b , . . . , b n } then in fact { a , . . . , a m } ↔ { b , . . . , b n } . Thus { a , . . . , a m } ↔ { b , . . . , b n } if and only if there is { c , . . . , c p } such that { c , . . . , c p } (cid:22) { a , . . . , a m } and { c , . . . , c p } (cid:22) { b , . . . , b n } . It is immediate by the calculation above that if A ≡ B then A σ B . To provethe converse suppose that
A σ B . Then there is an essential set C such that C ⊆ A, B . We prove that if C ⊆ A and both C and A are essential then C ≡ A . Let A = { a , . . . , a m } and C = { c , . . . , c p } . It is enough to prove that ( a , . . . , a m ) → ( c , . . . , c p ). Let 0 = x ≤ a i . Then 0 = d ( x ) ≤ d ( a i ). Thus d ( x ) ∧ d ( c j ) = 0 forsome j . Now c j ≤ a k for some k . But d ( x ) ≤ d ( a i ) and d ( c j ) ≤ d ( a k ). Thus0 = d ( x ) ∧ d ( c j ) ≤ d ( a i ) ∧ d ( a k ). But a i and a k are assumed orthogonal if theyare not equal. Thus a i = a k . We deduce that we must have x, c j ≤ a i . But then x and c j are compatible and so d ( x ∧ c j ) = d ( x ) ∧ d ( c j ). It follows that x ∧ c j = 0,as required. (cid:3) When is D ( S ) congruence-free? On an inverse semigroup S define s µ t ifand only if ses − = tet − for all idempotents e ∈ S . Then µ is a congruence andthe natural homomorphism from S to S/µ is injective when restricted to the semi-lattice of idempotents. Such homomorphisms are said to be idempotent-separating and the congruence µ is thus idempotent separating. It is, in fact, the maximumidempotent-separating congruence on S . Semigroups for which µ is equality are called fundamental . See Section 5.2 of [27]. An inverse semigroup with zero is saidto be 0 -simple if it has no, non-trivial ideals. The following is proved in [52]. Lemma 3.11.
An inverse semigroup with zero S is congruence-free if and only ifit is fundamental, -simple and E ( S ) is -disjunctive. Lemma 3.12.
Let S be an unambiguous E ∗ -unitrary pre-Boolean inverse semi-group. If S is fundamental and E ( S ) is -disjunctive then D ( S ) is fundamentaland E ( D ( S )) is -disjunctive.Proof. By Lemma 3.5(2), the semigroup S is separative. It follows that δ : S → D ( S ) is an embedding. We shall identify S with its image. Put T = D ( S ). Eachelement of T can be written as a join of a finite number of elements of S . We mayassume that this join is orthogonal by Corollary 3.7.Let t be a non-idempotent element of T . By assumption, t is a join of a finitenumber of elements s , . . . , s n . Not all of these elements can be idempotent elsetheir join would be idempotent. Without loss of generality assume that s is notan idempotent. By assumption S is fundamental and there exists an idempotent e ∈ S such that s e = es . We claim that et = te . Suppose that et = te .Then W i es i = W j s j e . Multiply this equation on the right by d ( s ). Then we get es = s e , a contradiction. It follows that et = te and so t is not in Z ( E ( T )).Thus the centralizer of the idempotents just consists of idempotents and so D ( S )is fundamental.It is immediate that E ( D ( S )) is 0-disjunctive. (cid:3) We shall say that an inverse semigroup S is sufficiently branching if for every non-zero idempotent e and every n ≥ e , . . . e n ≤ e which are pairwise orthogonal. The proof of the following is straightforward. Lemma 3.13.
Let E be a semilattice which is -disjunctive and contains no -minimal idempotents. Then E is sufficiently branching. Lemma 3.14.
Let S be an unambiguous E ∗ -unitary pre-Boolean inverse semigroupwith a -disjunctive semilattice of idempotents which is sufficiently branching. If S is -simple then D ( S ) is -simple.Proof. By Lemma 3.5, the semigroup S is separative. We may therefore identifyit with its image in D ( S ). Put T = D ( S ). Let e and f be arbitrary non-zeroidempotents in T . Then we can write e = W ni =1 e i . By assumption we can find n orthogonal idempotents f i ≤ f . We now use the fact that S is 0-simple to find n idempotents g i and elements s i such that e i s i −→ g i ≤ f i . Put s = W ni =1 s i . Then e s −→ W ni =1 f i ≤ f . (cid:3) We may sum up what we have found in the following.
Theorem 3.15.
Let S be an unambiguous E ∗ -unitary pre-Boolean inverse semi-group having no -minimal idempotents. Then if S is congruence-free so too is D ( S ) .Proof. By Lemma 3.11, E ( S ) is 0-disjunctive and so by Lemma 3.5(2), S is sep-arative. By Lemma 3.13, E ( S ) is sufficiently branching. By Lemmas 3.11 andLemma 3.14, D ( S ) is 0-simple. By Lemma 3.12, D ( S ) is fundamental and E ( D ( S ))is 0-disjunctive. It follows by Lemma 3.11, that D ( S ) is congruence-free. (cid:3) ON-COMMUTATIVE STONE DUALITY 31 Examples
In this section, we shall describe some examples of pre-Boolean inverse ∧ -semigroupsand their Boolean completions. In particular, we show that the Thompson-Higmangroups arise from the theory described in this paper, where the elements of thegroups are obtained by ‘glueing together’ elements of a suitable inverse semigroup.4.1. The polycyclic monoids and the Thompson groups G n, . In this sectionwe shall work with the polycyclic inverse monoids and show that work by the authorin [32, 33, 36] can be viewed as a special case of the general theory of this paper.Let A = { a , . . . , a n } be an alphabet with n letters. A string in A ∗ , the freemonoid generated by A n , will be called positive . The empty string is denoted ε . If u = vw are strings, then v is called a prefix of u , and a proper prefix if w is not theempty string. A pair of elements of A ∗ n is said to be prefix-comparable if one is aprefix of the other. If x and y are prefix-comparable we define x ∧ y = (cid:26) x if y is a prefix of xy if x is a prefix of y The polycyclic monoid P n , where n ≥
2, is defined as a monoid with zero by thefollowing presentation P n = h a , . . . , a n , a − , . . . , a − n : a − i a i = 1 and a − i a j = 0 , i = j i . Every non-zero element of P n is of the form yx − where x, y ∈ A ∗ . Identify theidentity with εε − . The product of two elements yx − and vu − is zero unless x and v are prefix-comparable. If they are prefix-comparable then yx − · vu − = (cid:26) yzu − if v = xz for some string zy ( uz ) − if x = vz for some string z The polyclic monoid P n is an inverse monoid with zero: the inverse of xy − is yx − ; the non-zero idempotents in P n are the elements of the form xx − ; thenatural partial order is given by yx − ≤ vu − iff ( y, x ) = ( v, u ) p for some positivestring p .Polycyclic inverse monoids are unambiguous, E ∗ -unitary and congruence-free.By Lemma 3.5(2) they are therefore separative. The semilattice of idempotentsof P n is the regular n -tree and so by Proposition 3.3 polycyclic monoids are alsopre-Boolean. An inverse semigroup is combinatorial if s − s = t − t and ss − = tt − implies that s = t . An inverse semigroup with zero S is said to be 0 -bisimple if forany two non-zero idempotents e and f there exists an element s such that e = ss − and f = ss − . Proposition 4.1.
The polycyclic inverse monoids are combinatorial, -bisimple,unambiguous, E ∗ -unitary, congruence-free, pre-Boolean and separative. It follows by the Completion Theorem that P n has a Boolean completion whichwe denote by C n and call the Cuntz inverse monoid . By Theorem 3.15, the Cuntzmonoid is itself congruence-free. To determine the group of units of C n , we useTheorem 3.10 which makes a direct connection with the calculations carried out in[33]. We therefore have the following. This term is frequently used, particularly by those working in C ∗ -algebra theory, for thepolycyclic monoids themselves. It seems appropriate to make a distinction as we have done here. Theorem 4.2.
The Boolean completion of the polycyclic inverse monoid P n is theCuntz inverse monoid C n . The Cuntz inverse monoid is congruence-free and itsgroup of units is the Thompson group G n, . By Lemma 3.1 and the theory developed in [33], a homomorphism θ : P n → T where T is a distributive inverse monoid is a cover-to-join map if and only if W ni =1 θ ( a i a − i ) is the identity of T . In fact, the covers of the identity of P n cor-respond bijectively to the maximal prefix codes in the free monoid on n generators,and all maximal prefix codes may be constructed from the simplest one { a , . . . , a n } using order-theoretic and algebraic operations [32, 33]. It follows that we may de-scribe C n in the following way: • It is distributive. • It contains a copy of P n and every element of C n is the join of a finitesubset of P n . • W ni =1 a i a − i . • It is the freest inverse semigroup satisfying the above conditions.There is a further consequence of this characterization. The symmetric inversemonoids I ( X ) are distributive inverse monoids. A map from P n to I ( X ) whichis a cover-to-join map is precisely what we have called a strong representation of P n in the papers [18, 36]. They lead to isomorphic representations of C n inside I ( X ). The theory of such strong representations is, as shown in [18], preciselywhat the monograph [3] by Bratteli and Jorgensen is about. In addition, strongrepresentations of P n lead to isomorphic representations of the Thompson group G n, . An example of such a representation motivated by linear logic is describedin Section 9.3 of [27].4.2. The polycyclic monoids and the Thompson-Higman groups G n,r . Weshall now show how to obtain the remaining Thompson groups from a generalizationof the polycyclic inverse monoids.An inverse semigroup S is said to have maximal idempotents if for each non-zeroidempotent e there is an idempotent e ◦ such that e ≤ e ◦ where e ◦ is a maximalidempotent such that if e ≤ i ◦ , j ◦ then i ◦ = j ◦ . The set of idempotents { e ◦ : e ∈ E ( S ) ∗ } is called the set of maximal idempotents . The proof of the following isimmediate. Lemma 4.3.
Let S be an inverse semigroup with a finite number of maximal idem-potents F . Then F is a cover of E ( S ) \ { } . We shall need the Rees matrix construction in a very simple form. Let M beany inverse monoid with zero, and let r be any (finite) cardinal. Define B r ( M ) asfollows. The non-zero elements are triples ( i, m, j ) where 1 ≤ i, j ≤ r and m ∈ M is non-zero, together with a zero . The product( i, m, j )( k, n, l ) = ( i, mn, l )if j = k and mn is non-zero; otherwise it is . Observe that B ( M ) is isomorphicto M . The proofs of the following are routine. Lemma 4.4.
The semigroup B r ( M ) is inverse with a semilattice of idempotentswhich is a -direct union of r copies of the semilattice of S . In addition, we havethe following: (1) If M is -bisimple then so too is B r ( M ) . ON-COMMUTATIVE STONE DUALITY 33 (2) If M is unambiguous then so too is B r ( M ) . (3) If M is E ∗ -unitary then so too is B r ( M ) . (4) The semigroup B r ( M ) has r maximal idempotents. Let n ≥ r ≥ P n,r = B r ( P n )and call it the extended polycyclic inverse semigroup with parameters ( n, r ).By Lemmas 4.3 and 4.4 together with Proposition 3.3 and Theorem 3.15, wehave the following. Theorem 4.5.
The extended polycyclic inverse semigroup P n,r is a compactablepre-Boolean inverse semigroup. Its Boolean completion, denoted by C n,r , is acongruence-free Boolean inverse ∧ -monoid. Our main theorem is the following.
Theorem 4.6.
The group of units of C n,r is the Higman-Thompson group G n,r .Proof. We shall need some definitions. If X is a set on which the monoid S acts onthe right then X is called a right S -act . Our actions will always be on the right. Subacts are defined in the obvious way and cyclic subacts have the form xS where x ∈ X . Subacts with the property that their intersection with any non-emptysubact is also non-empty are said to be essential . The act X is said to be finitelygenerated if X = S ni =1 x i S for some elements x i ∈ X . If X is any set and S isany monoid then S acts on the set X × S by right multiplication on the secondcomponent, and this determines what is called a free S -act . (Right) homomorphisms between right S -acts are defined in the obvious way.We begin by analysing the construction of the Higman-Thompson groups G n,r by Scott [57]. Let X = { x , . . . , x r } and A = { a , . . . , a n } where we assume that X ∩ A = ∅ . We denote by A ∗ the free monoid on A . Scott considers the set XA ∗ on which A ∗ acts on the right in the obvious way. The elements of X are simplyplaying the role of indices, so we could equally well write XA ∗ as X × A ∗ with theaction given by ( x, w ) a = ( x, wa ). Thus the starting point for Scott’s work is freeactions of free monoids. Because X has r elements, it is a free r -generated action.Scott now goes on to consider subsets, subspaces in Scott’s terminology, of thefree action which are closed under the action of the free monoid. These are subacts.She singles out those subacts which are ‘cofinite’ and ‘inescapable’. Birget [2] notedthat these two conditions together translate into ‘finitely generated’ and ‘essential’.Thus we are interested in those finitely generated subacts which are essential. Weconsider the set of all isomorphisms between the finitely generated essential subacts.These form an inverse monoid. In fact, an inverse monoid with a special property:each element sits beneath a unique maximal element. Such inverse monoids aresaid to be F -inverse. It is this property that enables us to define a group. Let G n,r be the set of maximal isomorphisms between the finitely generated essentialsubacts. If f and g are two such maps, their composition f g will not in generalbe a maximal map but will sit beneath a unique such map. We define this to bethe product of f and g , and in this way G n,r is a group. Alternatively, this groupis also the maximum group image of the inverse semigroup of isomorphisms. ThusThompson’s group G n,r is constructed from the free r -generated action of the freemonoid on n generators by considering the isomorphisms between essential finitelygenerated subacts. We now link this definition of the Thompson group G n,r to the inverse semigroup P n,r .We describe first how to construct all 0-bisimple inverse semigroups. Let X be aset on which the monoid S acts on the right. We say that ( X, S ) is an
RP-system if S is left cancellative, the intersection of any two cyclic subacts of X is eitherempty or again a cyclic subact, and if xs = x ′ s then x = x ′ . Define B ( X, S ) tobe the set of all right S -isomorphisms between the cyclic subacts together withthe empty function. Then B ( X, S ) is a 0-bisimple inverse semigroup and every0-bisimple inverse semigroup arises in this way. This is a classical theorem ofsemigroup theory which is described in a wider context with references in [28]. Aleft cancellative monoid satisfies
Clifford’s condition if the intersection of any twoprincipal right ideals is either empty or again a principal right ideal. If S is such amonoid then S acts on itself on the right and B ( S, S ) is a 0-bisimple inverse monoid ,and every 0-bisimple inverse monoid is isomorphic to one constructed in this way.Let S be a left cancellative monoid satisfying Clifford’s condition, whose associated0-bisimple inverse monoid is B ( S ) = B ( S, S ). Let ( X × S, S ) be a free S -system.Then it is an easy exercise to check that B ( X × S, S ) is isomorphic to B | X | ( B ( S )).It follows from the above discussion that the extended polycyclic semigroup P n,r isconstructed from the free system ( { , . . . , r } × A ∗ n , A ∗ n ).Consider now the RP -system ( X, S ) where X is finitely generated. Then theinverse monoid D e ( B ( X, S )) is isomorphic to the inverse monoid of isomorphismsbetween the finitely generated essential subacts of X . It follows that when X = { , . . . , r } × A ∗ n and S = A ∗ n then D e ( B ( X, S )) /σ is the Thompson group G n,r .Thus by Theorem 3.10, the group of units of C n,r is also the Thompson group G n,r . (cid:3) Self-similar groups.
The theory of self-similar groups has been developed byGrigorchuk and his school [47]. However, in [34], the author showed that they hadin fact first been defined by Perrot [51] as part of a generalization of the pioneeringwork by Rees [54]. Define a
Perrot semigroup to be an inverse semigroup thatis unambiguous and has the Dedekind height property. Then self-similar groupactions correspond to 0-bisimple Perrot monoids. They can be regarded informallyas polycyclic monoids which have acquired a group of units. Those 0-bisimple Perrotmonoids which are also E ∗ -unitary are pre-Boolean. This corresponds to the group G acting on the free monoid A ∗ n in such a way that the functions φ x : G x → G areall injective, where φ x ( g ) = g | x . The self-similar action is faithful if and only ifthe inverse monoid is fundamental. It follows that in this case the inverse monoidis congruence-free. In this way, certain kinds of self-similar group actions can beused to construct congruence-free Boolean inverse monoids generalizing the Cuntzinverse monoids. In fact, arbitrary self-similar group actions are pre-Boolean [39].4.4. Graph inverse semigroups.
Graph inverse semigroups are constructed asa special case of a general procedure for constructing inverse semigroups from leftcancellative categories [28, 31] originating in the work of Leech [43].A
Leech category is a left cancellative category in which any pair of arrows witha common range that can be completed to a commutative square have a pullback.A principal right ideal in a category C is a subset of the form aC where a ∈ C .A category C is said to be right rigid of aC ∩ bC = ∅ implies that aC ⊆ bC or bC ⊆ aC ; this terminology is derived from Cohn [5]. A left Rees category is a ON-COMMUTATIVE STONE DUALITY 35 left cancellative, right rigid category in which each principal right ideal is properlycontained in only finitely many distinct principal right ideals. Left Rees categoriesare Leech categories.With each Leech category C we may associate an inverse semigroup S ( C ) asfollows; all proofs may be found in [31]. Put U = { ( a, b ) ∈ C × C : d ( a ) = d ( b ) } . Define a relation ∼ on U as follows( a, b ) ∼ ( a ′ , b ′ ) ⇔ ( a, b ) = ( a ′ , b ′ ) u for some invertible element u ∈ C . This is an equivalence relation on U and wedenote the equivalence class containing ( a, b ) by [ a, b ]. The product [ a, b ][ c, d ] isdefined as follows: if there are no elements x and y such that bx = cy then theproduct is defined to be zero; if such elements exist choose such a pair that is apullback. The product is then defined to be [ ax, dy ]. Observe that [ a, b ] − = [ b, a ].Thus [ a, b ] L [ c, d ] if and only if [ b, b ] = [ d, d ] and [ a, b ] R [ c, d ] if and only if [ a, a ] =[ c, c ]. The non-zero idempotents of the inverse semigroup S ( C ) are the elements ofthe form [ a, a ]. Define [ a, a ] ◦ = [ r ( a ) , r ( a )]. The natural partial order is given by[ a, b ] ≤ [ c, d ] if and only if ( a, b ) = ( c, d ) p for some arrow p . Proposition 4.7.
Let C be a Leech category. (1) S ( C ) is an inverse semigroup with maximal idempotents and each D -classcontains a maximal idempotent. (2) If the groupoid of invertible elements in C is trivial then S ( C ) is combina-torial and each D -contains exactly one maximal idempotent. (3) If C is a left Rees category then S ( C ) is a Perrot semigroup (Section 4.3). In the case where C has no non-trivial invertible elements, an equivalence class[ a, b ] consists of the singleton ( a, b ). In this case, we denote the equivalence classby ab − .Let G be a directed graph and denote by G ∗ the free category on G . Lemma 4.8.
Free categories are left Rees categories with trivial groupoids of in-vertible elements.
Given a directed graph G , we define the graph inverse semigroup P G to be thesemigroup with zero defined by the above construction. The free category has nonon-trivial invertible elements and so each equivalence class is denoted by xy − .Thus the non-zero elements of P G are of the form uv − where u, v are paths in G with common domain. With elements in this form, multiplication works in thefollowing way: xy − · uv − = xzv − if u = yz for some path zx ( vz ) − if y = uz for some path z xy − and uv − be non-zero elements of P G . Then xy − ≤ uv − ⇔ ∃ p ∈ G ∗ such that x = up and y = vp. If xy − ≤ uv − or uv − ≤ xy − then we say xy − and uv − are comparable .The following is proved in [19] and is in fact a characterization of graph inversesemigroups. Proposition 4.9.
A graph inverse semigroup is a combinatorial Perrot semigroupwith maximal idempotents such that each D -class contains a unique maximal idem-potent. It is in addition E ∗ -unitary. The key result is the following and follows from our theory and an analysis thatgeneralizes the polycyclic case.
Theorem 4.10.
Graph inverse semigroups P G over graphs G in which each vertexhas finite in-degree are pre-Boolean. We call the Boolean completions of graph inverse semigroups
Cuntz-Krieger in-verse semigroups CK G . We may describe CK G in the following way: • It is distributive. • It contains a copy of P G and every element of CK G is the join of a finitesubset of P G . • e = W f ′ ∈ ˆ e f ′ for each maximal idempotent e of P G . • It is the freest inverse semigroup satisfying the above conditions.A direct construction of this semigroup is given in [18]. The Cuntz-Kriegerinverse semigroups constructed from finite directed graphs are monoids. Theirgroups of units are analogues of the Thompson groups G n, .5. A characterization of finite symmetric inverse monoids
In classical Stone duality, there is an order isomorphism between the ideals ofthe Boolean algebra and the open subsets of the associated Boolean space. Wegeneralize this result to our setting by reformulating the results of Section 7 of [46].We shall apply this result to bring out the analogy between the finite symmetricinverse monoids I ( X ), where X has n elements, and the finite-dimensional C ∗ -algebras M n ( C ).In a groupoid G , define g L h if g − g = h − h , and define g R h if gg − = hh − .Put D = L ◦ R = R ◦ L , an equivalence relation. The D -classes are precisely the connected components of the groupoid G . A subset of a groupoid G is said to be invariant if it is a union of D -classes.In an inverse semigroup S an ideal I ⊆ S is determined by the idempotents itcontains E ( I ), since a ∈ I if and only if aa − ∈ I if and only if a − a ∈ I . Theset E ( I ) is an order ideal of E ( S ) and is self-conjugate in the sense that e ∈ E ( I )implies that ses − ∈ E ( I ) for all s ∈ S . An ideal T of S is said to be tightly closed if s , . . . , s n ∈ T and { s , . . . , s n } a cover of s implies that s ∈ T . Self-conjugateorder ideals of E ( S ) are the same thing as the invariant order ideals of E ( S ) usedin [46]; the simple proof is left to the reader. It follows that we may rephrasethe correspondence found in Lemma 7.7 of [46] in terms of ideals of the inversesemigroup rather than as order ideals of its semilattice of idempotents. We givethe proof from this point of view. Theorem 5.1.
Let S be a pre-Boolean inverse semigroup and G ( S ) its associatedtopological groupoid. Then there is an order isomorphism between the set of tightlyclosed ideals of S and the set of open invariant subsets of G ( S ) .Proof. Let T be a tightly closed ideal of S . Define O ( T ) = [ t ∈ T V t . ON-COMMUTATIVE STONE DUALITY 37
Let O be an open invariant subset of G ( S ). Define C ( O ) = { s ∈ S : V s ⊆ O } . Observe that both of these functions are order-preserving.The set O ( T ) is open by construction. We prove that it is invariant. Let F D G ∈ O ( T ). By definition, there is an ultrafilter A such that F − · F = A · A − and G − · G = A − · A . By definition, G ∈ V s for some s ∈ T . Thus s ∈ G . It followsthat s − s ∈ G − · G = A − · A . Let a ∈ A be arbitrary. Then b = as − s ∈ A and b ∈ I since I is an ideal. Now bb − ∈ A · A − = F − · F . Let t ∈ F be arbitrary.Then c = tbb − ∈ F and c ∈ I since I is an ideal. Then F ∈ V c where c ∈ I . Itfollows that F ∈ O ( I ), as required.We prove that I = C ( O ) = { s ∈ : V s ⊆ O } is a tightly closed ideal. Let s ∈ I and t ∈ S . We prove that st ∈ I ; the dual result follows by symmetry. Byassumption, V s ⊆ O . We prove that V st ⊆ O . Let F ∈ V st . Then st ∈ F . Thus st ( st ) − ∈ F · F − . But st ( st ) − ≤ ss − and so ss − ∈ F · F − . It follows that A = ( F · F − s ) ↑ is an ultrafilter containing s and so A ∈ O . But A R F and so F ∈ O since O is an invariant subset. We have therefore shown that I is an ideal.The proof that it is tightly closed is immediate by Lemma 2.11(2).Finally, it remains to show what happens when we iterate our two functions.First, for every tighty closed ideal I we have that I = CO ( I ). The proof ofthis uses the fact that the sets V s are compact, the definition of a tightly closedideal and Lemma 2.11(2). Second, for every open invariant subset O we have that O = OC ( O ). The proof of this is routine. (cid:3) Let S be an inverse ∧ -semigroup. If e and f are non-zero idempotents define e (cid:22) f if and only if there are elements x , . . . , x m such that d ( x ) , . . . , d ( x m ) ≤ f and e → { r ( x ) , . . . , r ( x n ) } . Lemma 5.2.
With the above definition, (cid:22) is a preorder on E ( S ) \ { } .Proof. We need only prove that (cid:22) is transitive. Let e (cid:22) f and f (cid:22) g . By definitionthere are elements x , . . . , x m and y , . . . , y n such that e → { r ( x ) , . . . , r ( x m ) } and d ( x i ) ≤ f for 1 ≤ i ≤ m , and f → { r ( y ) , . . . , r ( y n ) } and d ( y j ) ≤ f for 1 ≤ j ≤ n .It can be checked that e → { r ( x i y j ) : 1 ≤ i ≤ m, ≤ j ≤ n } and d ( x i y j ) ≤ g forall 1 ≤ i ≤ m, ≤ j ≤ n . This shows that e (cid:22) g . (cid:3) Define ≡ on E ( S ) \ { } by e ≡ f if and only if e (cid:22) f and f (cid:22) e . We say thatan inverse ∧ -semigroup is 0 -simplifying if it has no non-trivial tightly closed ideals.The following is immediate from Lemma 7.8 of [46] and our discussion about idealsabove. Proposition 5.3.
An inverse ∧ -semigroup is -simplifying if and only if ≡ is auniversal equivalence. Our use of the term 0-simplifying was motivated by Kumjian’s use of the term simplification . Clearly, every 0-simple semigroup is 0-simplifying. We now describean example which shows that the converse is not true.
Example 5.4.
Let I ( X ) be a finite symmetric inverse monoid on the set { , . . . , n } .This semigroup is not 0-simple when n ≥ g D f (cid:22) e then g (cid:22) e . Let e be the partial identity on the set { , . . . , r } where 1 ≤ r ≤ n − identity on the whole of X . We show that e ≡
1. First e (cid:22)
1. This can be seen bydefining the r partial bijections f i ( i ) = i . Second we prove that 1 (cid:22) e . This canbe seen by defining the n partial bijections f i (1) = i for 1 ≤ i ≤ n . The proof isconcluded by observing that every non-zero idempotent whose rank is r is D -relatedto e .Alternatively, we may use Theorem 5.1 and the fact that the groupoid associatedwith I ( X ) is just the connected groupoid X × X with the discrete topology.A groupoid is said to be principal if all its local groups are trivial. Such groupoidsare essentially equivalence relations. We shall now investigate the question of whichBoolean inverse ∧ -semigroups S have the property that G ( S ) are principal. Suchgroupoids are interesting from the C ∗ -algebra perspective [55].We recall first a construction to be found in [52]. Let S be an arbitrary inversesemigroup and let F ⊂ E ( S ) be a subsemigroup. Put F c = { s ∈ S : s − s, ss − ∈ F, and sF s − , s − F s ⊆ F } . Then the following is easy to verify.
Lemma 5.5.
With the above definition, we have the following: (1) F c is an inverse subsemigroup of S whose semilattice of idempotents is F . (2) If T is any inverse subsemigroup of S whose semilattice of idempotents is F then T ⊆ F c . (3) Let F be a filter in E ( S ) . Then F ↑ ⊆ F c . We can now prove our first characterization.
Lemma 5.6.
Let S be a pre-Boolean inverse ∧ -semigroup. Then G ( S ) is principalif and only if F ↑ = F c for each ultrafilter F ⊆ E ( S ) .Proof. Suppose first that the condition holds. We prove that G ( S ) is principal.Let H be an ultrafilter and inverse subsemigroup of S . Then F = E ( H ) is anultrafilter in E ( S ) such that H = F ↑ . Suppose that A is an ultrafilter such that A − · A = H = A · A − . we may write A = ( aH ) ↑ = ( Ha ) ↑ where a ∈ A is arbitrary.Clearly a − a, aa − ∈ F . Also aHa − ⊆ H and a − Ha ⊆ H . It follows that a ∈ F c .By assumption e ≤ a where e ∈ F . Thus A ⊆ F ↑ . But A is an ultrafilter and so A = F ↑ = H , as required.We now assume that G ( S ) is principal and prove the condition. Let F be anarbitrary ultrafilter in E ( S ). Then H = F ↑ is an ultrafilter in S . Suppose that a ∈ F c . Then in particular a − a, aa − ∈ F . Put A = ( aH ) ↑ . Then A is anultrafilfter in S . Observe that A − · A = H . Also A · A − = ( aHa − ) ↑ . But a ∈ F c and so ( aHa − ) ↑ = H . By assumption A = H . Thus a ∈ F ↑ , as required. (cid:3) We have the following useful consequence of the above result.
Lemma 5.7.
Let S be a separative or E ∗ -unitary pre-Boolean inverse ∧ -semigroup.If G ( S ) is principal then S is fundamental.Proof. Let aµa − a . Let H be any ultrafilter containing a − a . It is necessarily aninverse subsemigroup. Put F = E ( H ) so that H = F ↑ . We claim that a ∈ F c . Byassumption a − a = aa − . Also if f ∈ F then af a − = f a − a and a − f a = f a − a .It follows that a ∈ F c and so, by assumption, there exists e ∈ F such that e ≤ a .In particular, a ∈ H . We have shown that V a − a ⊆ V a . ON-COMMUTATIVE STONE DUALITY 39
Let A be any ultrafilter containing a . Then A = ( aA − · A ) ↑ . But a − a ∈ A − · A and A − · A is an ultrafilter containing a − a . It follows by our result above that a ∈ A − · A and so a − ∈ A − · A . Thus aa − = a − a ∈ A .We have proved that a ↔ a − a .If S is E ∗ -unitary let F be any ultrafilter containing a . Then it must also contain a − a . It follows that the idempotent a ∧ a − a is non-zero. But it lies beneath a and so a is an idempotent and so a = a − a . It follows that S is fundamental.If S is separative, we have that a = a − a . It follows that S is fundamental. (cid:3) The converse of the above lemma is not true because by Theorem 4.2 the Cuntzinverse monoid C n is congruence-free and so fundamental, but its groupoid G ( C n )is not principal, because there are idempotent ultrafilters that can be constructedfrom right-infinite periodic strings [55]. The best we can do at the moment is thefollowing which is enough for our purposes. Lemma 5.8.
Let S be a finite Boolean inverse ∧ -semigroup. Then S is fundamentalif and only if G ( S ) is principal.Proof. Boolean inverse ∧ -semigroups are separative, so by Lemma 5.7 we have onlyone direction to prove. Thus let S be a fundamental finite Boolean inverse ∧ -semigroup. By finiteness, every ultrafilter in S is principal and is generated byan element immediately above zero. It follows that the groupoid of ultrafilters of S is isomorphic to the groupoid of 0-minimal elements M of S . The set M = M ∪ { } is an ideal of S . Since S is fundamental so too is M . But M is aprimitive inverse semigroup. Such semigroups are fundamental precisely when theyare combinatorial; see Exercises II.3.10(i) of [52]. It follows that M is a principalgroupoid. (cid:3) We now have a characterization of the finite symmetric inverse monoids.
Theorem 5.9.
The only finite Boolean inverse ∧ -monoids that are fundamentaland -simplifying are the finite symmetric inverse monoids.Proof. By Example 5.4, the finite symmetric inverse monoids are 0-simplifying andit is well-known, and can be verified directly, that they are inverse ∧ -semigroups andfundamental. It remains to prove the converse. Let S be a finite Boolean inverse ∧ -monoid that is fundamental and 0-simplifying. By Lemma 5.8, the groupoid G ( S ) is principal. By finiteness, the topology is discrete. But by Theorem 5.1, thegroupoid G ( S ) can have no non-trivial open invariant subsets and so this groupoidmust consist of just one component. It follows that G ( S ) is isomorphic to a groupoidof the form X × X where X is a finite set. The inverse monoid S is isomorphic tothe inverse semigroup of compact-open bisections of G ( S ) and so is isomorphic tothe inverse monoid I ( X ). (cid:3) It is worth observing that in Boolean inverse ∧ -semigroups, the tightly closedideals are precisely those ideals that are also closed under joins whenever they exist.5.1. Concluding remarks.
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Department of Mathematics and the Maxwell Institute for Mathematical Sciences,Heriot-Watt University, Riccarton, Edinburgh EH14 4AS, Scotland
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