aa r X i v : . [ m a t h . R A ] F e b GRAPH PRODUCTS OF LEFT ABUNDANT MONOIDS
YANG DANDAN AND VICTORIA GOULD
Abstract.
Graph products of monoids provide a common framework for direct and freeproducts, and graph monoids (also known as free partially commutative monoids). Ifthe monoids in question are groups, then any graph product is, of course, a group. Formonoids that are not groups, regularity is perhaps the first and most important algebraicproperty that one considers; however, graph products of regular monoids are not in generalregular. We show that a graph product of regular monoids satisfies the related weakercondition of being abundant. More generally, we show that the classes of left abundantand left Fountain monoids are closed under graph product. The notions of abundancyand Fountainicity and their one-sided versions arise from many sources, for example, thatof abundancy from projectivity of monogenic acts, and that of Fountainicity (also knownas weak abundancy ) from connections with ordered categories. As a very special casewe obtain the earlier result of Fountain and Kambites that the graph product of rightcancellative monoids is right cancellative. To achieve our aims we show that elementsin (arbitrary) graph products have a unique Foata normal form, and give some usefulreduction results; these may equally well be applied to groups as to the broader case ofmonoids. Introduction
Graph products arise from many sources and provide an important and wide rangingconstruction. They are defined by presentations, where the edges of a simple, non-directedgraph determine commutativity of elements associated with the vertices. Further detailsare given in Section 2.
Graph products of monoids are defined in the same way as graphproducts of groups , a notion introduced by Green in her thesis [25], and generalise at oneand the same time free products, restricted direct products, free (commutative) monoidsand graph monoids . The latter are graph products of free monogenic monoids, and wereintroduced by Cartier and Foata [6] to study combinatorial problems for rearrangementsof words; they have been extensively studied by mathematicians and computer scientists,having applications to the study of concurrent processes [12, 13]. Graph monoids are alsoknown as free partially commutative monoids, right-angle Artin monoids and trace monoids(sometimes with the condition the underlying graph is finite); corresponding terminology Date : February 15, 2021.2010
Mathematics Subject Classification.
Primary: 20M05, 20M10 Secondary: 20F05.
Key words and phrases. monoid, graph product, regularity, abundancy. The existing terminology is a little unfortunate. Graph monoids are a strict subclass of the class ofgraph products of monoids. Note also that graph groups should not be confused with the fundamentalgroups of graphs of groups. applies in the case for groups. Graph groups were first defined by Baudisch [4]; for a recentsurvey see [16] and for the analogous notion for inverse semigroups see [10, 14].Although mentioned in [25] and in other earlier works focussing on groups, graph prod-ucts of monoids per se were first defined in [8], and have subsequently been studied invarious contexts, e.g. [8, 19]. Much of the existing work in graph products of monoids, andgroups, has been to show that various properties are preserved under graph product, seee.g. [28, 15, 9, 32]. These properties are often of algorithmic type, for example, automatic-ity [28, 9]. In a different direction, articles such as [2, 3, 24] consider algebraic conditions.Of particular interest to us here is that Fountain and Kambites [24] show that a graphproduct of right cancellative monoids is right cancellative.A monoid M is regular if for any a ∈ M there is a b ∈ M such that a = aba ; note that ab, ba are, respectively, idempotent left and right identities for a . From an algebraic pointof view, regularity is often the first property to look for in a monoid. Yet, it is easy to seethat only in very special cases will a graph product of regular monoids be regular.The aim of this paper is easy to state. We consider two properties that each providea natural weakening of regularity, and show that the classes of monoids satisfying theseproperties are closed under graph product. In general, the properties we consider providethe natural framework to study classes of monoids that need not be regular, but whichhave behaviour strongly influenced by idempotent elements. We first prove: Theorem 5.22
The graph product of left abundant monoids is left abundant .A monoid M is left abundant if every principal left ideal is projective (so that sometimesa left abundant monoid is called left PP [20]). This property may handily be expressedby saying that every R ∗ -class of M contains an idempotent. We define the relation R ∗ inSection 2; it suffices to say here that R ∗ contains Green’s relation R , whence it followsimmediately that regular semigroups are left abundant. We note that a monoid is a single R ∗ -class if and only if it is right cancellative. Certainly then such monoids are abundant.The above mentioned result of [24] for right cancellative monoids easily follows.Our second main result is: Theorem 6.10
The graph product of left Fountain monoids is left Fountain .One way to define a left Fountain (also known as weakly left abundant , or left semiabun-dant ) monoid M is to say that every e R -class of M must contain an idempotent; we givefurther details in Section 2. Here e R is a relation containing R ∗ , whence it is clear that leftabundant monoids are left Fountain. As for left abundancy, there is a natural approachto left Fountainicity using principal one-sided ideals. Again as for left abundancy, suchsemigroups arise independently from a number of sources. They (and their two-sided ver-sions) appear in the work of de Barros [11], in that of Ehresmann on certain small orderedcategories [17] and in the thesis of El Qallali [18]. A systematic study of such semigroupswas initiated by Lawson, who establishes in [33] the connection with Ehresmann’s work.A useful source for the genesis of these ideas is Holling’s survey [29]. We note here that RAPH PRODUCTS OF LEFT ABUNDANT MONOIDS 3 the class of left Fountain monoids contains a number of important subclasses: we havementioned left abundant, but we also have left ample and left restriction [29]. The studyof left abundant monoids, left Fountain monoids, their two-sided versions, and monoids inrelated classes, continues to provide one focus in algebraic semigroup theory. Some resultsshow similarities with the structure of regular and inverse monoids [26, 23], whereas othersillustrate significantly different behaviour [31, 37, 5].In order to prove Theorems 5.22 and 6.10 we have considerable work to do to get a gripon normal forms of elements of graph products. Essentially, the difficulty in the transitionfrom graph monoids to graph products of monoids lies in the fact that for the broaderconcept the group of units of the monoids in question need not be trivial. Some of ourtechniques and results concerning normal forms and reduction of products of words maybe of independent interest. In particular, in Proposition 3.17, we establish that elementsin graph products of monoids have a left Foata normal; previously this was an importanttool in the study of graph monoids, and the same holds here.The structure of this paper is as follows. In Section 2 we give the necessary definitionsand gather together the results we need from the literature. In Section 3 we begin ouranalysis of the form of words, and how these behave with respect to products. We establishthe left Foata normal form for elements of graph products, not relying on any assumptionof cancellativity. In the next two sections we build a suite of techniques that allow us tosimplify the words we need to consider when determining the relation R ∗ , these then enableus eventually to prove Theorem 5.22. In Section 6 we use the earlier techniques, togetherwith a further analysis of words, to establish Theorem 6.10. There is a correspondingnotion of graph product for semigroups ; the behaviour of the resulting semigroup is similarto that of a graph monoid and hence sheds some of the technical difficulties we encounterin graph products of monoids. We apply our results to the semigroup case in Section 7,and mention a number of other applications. We finish with some open questions.2. Preliminaries
We outline the notions required to read this article. For further details, we recommendthe classic texts [7] and [30].2.1.
Presentations and graph products of monoids.
We begin with an account ofthe notion on which this article is based: that of graph product of monoids. They aredetermined by monoid presentations. Let X be a set. The free monoid X ∗ on X consistsof all words over X with operation of juxtaposition. We denote a non-empty word by x ◦ · · · ◦ x n where x i ∈ X for 1 ≤ i ≤ n ; we also use ◦ for juxtaposition of words. Theempty word is denoted by ǫ and is the identity of X ∗ . Throughout, our convention is thatif we say x ◦ · · · ◦ x n ∈ X ∗ , then we mean that x i ∈ X for all 1 ≤ i ≤ n , unless we explicitlysay otherwise. We denote by x r by the word x n ◦ · · · ◦ x ∈ X ∗ for x = x ◦ · · · ◦ x n .A monoid presentation h X | R i , where X is a set and R ⊆ X ∗ × X ∗ , determines themonoid X ∗ /R ♯ , where R ♯ is the congruence on X ∗ generated by R . In the usual way, weidentify ( u, v ) ∈ R with the formal equality u = v in a presentation h X | R i . Y. DANDAN AND V. GOULD
We now define graph products of monoids [25, 8]. Let Γ = Γ(
V, E ) be a simple, undi-rected, graph with no loops. Here V is a non-empty set of vertices and E ⊆ V is the setof edges of Γ, where V is the set of 2-element subsets of V . We think of { α, β } ∈ E asjoining the vertices α, β ∈ V . For notational reasons we denote an edge { α, β } as ( α, β ) or( β, α ); since our graph is undirected we are identifying ( α, β ) with ( β, α ). Definition 2.1.
Let Γ = Γ(
V, E ) be a graph and let M = { M α : α ∈ V } be a set ofmutually disjoint monoids. We write 1 α for the identity of M α and put I = { α : α ∈ V } .The graph product G P = G P (Γ , M ) of M with respect to Γ is defined by the presentation G P = h X | R i where X = S α ∈ V M α and R = R id ∪ R v ∪ R e are given by: R id = { α = ǫ : α ∈ V } ,R v = { x ◦ y = xy : x, y ∈ M α , α ∈ V } ,R e = { x ◦ y = y ◦ x : x ∈ M α , y ∈ M β , ( α, β ) ∈ E ) } . The monoids M α in Definition 2.1 are known as vertex monoids . Throughout we assume | V | ≥
2, as otherwise
G P is isomorphic to the single vertex monoid. We denote the R ♯ -class of w ∈ X ∗ in G P by [ w ]. It is worth noting that there are various different ways toset up graph products, which all yield equivalent constructions. In particular, if one startswith monoids that are groups, the process above yields the graph product of groups.The main focus of this article is on monoids, although we briefy visit graph products of semigroups in Section 7. Free products of semigroups, and a discussion of their universalproperties, may be found in [7, 30]. Free products of monoids may be viewed as a specialcase of an amalgamated free product of semigroups; this is commented on explicitly in [30,p. 266]. Here we remark that a free product of monoids is a graph product for a graphΓ( V, ∅ ).We now touch on the other extreme where E = V . Let M = { M α : α ∈ V } be asabove. The restricted direct product (or direct sum ) Σ α ∈ V M α of M is defined byΣ α ∈ V M α = { f ∈ Π α ∈ V M α : αf = 1 v for only finitely many v ∈ V } . Clearly Σ α ∈ V M α is a submonoid of Π α ∈ V M α and Σ α ∈ V M α = Π α ∈ V M α if and only if V isfinite. It is easy to see that a restricted direct product of monoids is a graph product fora graph Γ( V, V ).Graph products of monoids behave beautifully with respect to certain substructures, aswe now demonstrate. To do so we need some terminology. Definition 2.2.
Let
G P = G P (Γ , M ). Let s : X → V be a map defined by s ( a ) = α if a ∈ M α . The support s ( x ) of x = x ◦ · · · ◦ x n ∈ X ∗ is defined by s ( x ) = { s ( x i ) : 1 ≤ i ≤ n } . In particular, s ( ǫ ) = ∅ . RAPH PRODUCTS OF LEFT ABUNDANT MONOIDS 5
Notice that when s ( x ) is a singleton, we simply drop braces around it. Below we use [ , ]for the equivalence class of a word under two different relations, so the reader should bearin mind the context in each case. Proposition 2.3.
Let V ′ ⊆ V and let Γ ′ = Γ( V ′ , E ′ ) be the resulting full subgraph of Γ .Let G P ′ be the corresponding graph product of the monoids M ′ = { M α : α ∈ V ′ } . Then G P ′ is a retract of G P .Proof.
Let η := η V,V ′ : X ∗ → G P ′ be the morphism extending the map defined on X by xη = (cid:26) [ x ] s ( x ) ∈ V ′ [ ǫ ] else.We show that R ♯ ⊆ ker η .First, for any α ∈ V , whether or not α ∈ V ′ , we have 1 α η = [ ǫ ] = ǫη so that R id ⊆ ker η .To see that R v ⊆ ker η , let α ∈ V and let u, v ∈ M α . If α V ′ , then( u ◦ v ) η = ( uη )( vη ) = [ ǫ ][ ǫ ] = [ ǫ ] = ( uv ) η. If α ∈ V ′ , then ( u ◦ v ) η = ( uη )( vη ) = [ u ][ v ] = [ v ◦ v ] = [ uv ] = ( uv ) η. Now consider u ∈ M α , v ∈ M β with ( α, β ) ∈ E . If neither α nor β is in V ′ , then( u ◦ v ) η = ( uη )( vη ) = [ ǫ ][ ǫ ] = ( vη )( uη ) = ( v ◦ u ) η. If α, β ∈ V ′ with ( α, β ) ∈ E , then, as Γ ′ is a full subgraph of Γ, we have ( α, β ) ∈ E ′ , sothat ( u ◦ v ) η = ( uη )( vη ) = [ u ][ v ] = [ u ◦ v ] = [ v ◦ u ] = [ v ][ u ] = ( vη )( uη ) = ( v ◦ u ) η. If α ∈ V ′ but β V ′ then( u ◦ v ) η = ( uη )( vη ) = [ u ][ ǫ ] = [ ǫ ][ u ] = ( vη )( uη ) = ( v ◦ u ) η and dually if α V ′ but β ∈ V ′ . Thus R e ⊆ ker η .It follows that R ♯ ⊆ ker η and so η := η V,V ′ : G P → G P ′ given by [ w ] η = wη is a welldefined morphism.It is easy to see that ι := ι V ′ ,V : G P ′ → G P such that [ w ] ι = [ w ] is well defined, andby considering ιη it is clear that ι is an embedding. It is then immediate that and ηι is aretraction of G P onto a submonoid
G P ′ ι . (cid:3) We identify
G P ′ with its image under ι and regard G P ′ as a submonoid of G P . Remark 2.4.
Let α ∈ V . By taking V ′ = { α } in Proposition 2.3, we immediately seethat M α is naturally embedded in G P via ι α : M α → G P , where for x ∈ M α we have xι α = [ x ]. Proposition 2.5.
A graph product
G P = G P (Γ , M ) is a direct limit of the graph prod-ucts corresponding to the finite full subgraphs of Γ . Y. DANDAN AND V. GOULD
Proof.
The finite full subgraphs of Γ are partially ordered by inclusion, and form a directedset under union. It is routine to see that the direct limit of the graph products
G P ′ ,corresponding to finite full subgraphs with vertex set V ′ ⊆ V and embeddings ι V ′ ,V ′′ where V ′ ⊆ V ′′ , is isomorphic to G P . (cid:3) We end this subsection by remarking that there are universal approaches to describegraph products of monoids as indicated in [24, Proposition 1.6], in the same way as thereare for direct and free products.2.2.
Regular, abundant and Fountain monoids.
We will denote the set of idempo-tents of a monoid M by E ( M ). We recall that Green’s relations R is defined on M by therule a R b if and only if aM = bM . Equivalently, a = bt and b = as for some s, t ∈ M ,thus, R is a relation of mutual divisibility. The relation L is defined dually. It is easy tosee that M is regular if and only if every a ∈ M is R -related to an idempotent, and so fromconsiderations of duality, if and only if every a ∈ M is L -related to an idempotent. Graphproducts do not behave well with regard to regularity. Let M and N be regular monoidscontaining elements m, n respectively which do not have one-sided inverses. Then [ m ◦ n ]is not regular in the graph product G P (Γ , M ) where Γ = ( { , } , ∅ ) and M = { M , M } (that is, in the free product). See [8] for a discussion of regularity in graph products. Wetherefore consider relations larger than R and L and ask whether they contain idempotents.The relation R ∗ on a monoid M was first defined in [35, 36]. For elements a, b ∈ M we have a R ∗ b if and only if a R b in some over-monoid N of M . Equivalently, for any x, y ∈ M we have xa = ya if and only if xb = yb. Thus, R ∗ is a relation of mutual cancellativity. A third equivalent condition is that theprincipal left ideals M a and
M b are isomorphic under a left ideal isomorphism where a b [21]. It is easy to see that R ⊆ R ∗ with equality if M is regular. The relation L ∗ is theleft-right dual of R ∗ . Definition 2.6.
A monoid M is left abundant if every element in M is R ∗ -related to anidempotent. The notion of right abundant is defined dually, and M is abundant if it is bothleft and right abundant.Examples of (left) abundant monoids abound; regular monoids are, of course, abundant;for a favourite non-regular example take the monoid M n ( Z ) of n × n integer matrices undermatrix multiplication [22]. Remark 2.7.
It is easy to see that for a ∈ M and e ∈ E ( M ) we have that a R ∗ e if andonly if ea = a and for any x, y ∈ Mxa = ya ⇒ xe = ye. A monoid M is right cancellative if for all a, b, c ∈ M , from ac = bc we deduce that a = b ; left cancellative is dual and M is cancellative if it is right and left cancellative. Itis easy to see that M is right cancellative if and only if it is a single R ∗ -class. Thus, aright cancellative monoid is left abundant. A right cancellative monoid has no non-identity RAPH PRODUCTS OF LEFT ABUNDANT MONOIDS 7 idempotents, and need not be left cancellative. It follows that left abundancy does notimply right abundancy, which contrasts with the case for regularity.The relation e R arose from many sources, as indicated in the Introduction. It extendsthe relation R ∗ and coincides with it in the case where the monoid is left abundant. Forelements a, b of a monoid M we have that a e R b if and only if ea = a ⇔ eb = b for all e ∈ E ( M ) . The relation e L is defined dually. Definition 2.8.
A monoid M is left Fountain if every element in M is e R -related to anidempotent. The notion of right Fountain is defined dually, and M is Fountain if it is bothleft and right Fountain.
Remark 2.9.
Similarly to Remark 2.7, it is easy to see that for a ∈ M and e ∈ E ( M ) wehave that a e R e if and only if ea = a and for any f ∈ E ( M ) f a = a ⇒ f e = e. Formerly, left Fountain was referred to as weakly left abundant , but in view of the per-ceived significance the notion was renamed by Margolis and Steinberg in [34]. It is easy tosee that M is left Fountain if and only if for any a ∈ M the intersection of the principal,idempotent generated, right ideals containing a is principal and idempotent generated. Asfor abundancy, there are many natural examples of (non-abundant) (left) Fountain semi-groups. These include finite monoids such that every principal (left) ideal has at mostone idempotent generator, for instance, any finite monoid with commuting idempotents[34]. For some recent examples of Fountain monoids, consisting of semigroups of tropicalmatrices, see [27]. Remark 2.10.
The relation R on a monoid M is easily seen to be a left congruence, forany a, b, c ∈ M , if a R b then ca R cb . Similarly, R ∗ is a left congruence. The same isnot true, in general, for e R , even for some quite natural monoids (see, for example, [27,Proposition 6.10]). Thus we do not assume that e R is a left congruence in our calculations.3. (left) Foata normal forms Throughout we let
G P = G P (Γ , M ) and follow the notation as established in Section 2.We show that elements in G P may be written in a normal form we refer to as left Foatanormal form. Such normal forms were previously known for elements of graph monoids,that is, where all the vertex monoids are free monogenic. The existing proofs rely oncancellativity, which is not available to us. Moreover, the presence of units in our vertexmonoids provides an added complication.
Definition 3.1.
Let x ◦ · · · ◦ x n ∈ X ∗ . A reduction step is one of:(id) x ◦ · · · ◦ x n → x ◦ · · · x i − ◦ x i +1 ◦ · · · ◦ x n where x i ∈ I ;(v) x ◦ · · · ◦ x n → x ◦ · · · x i − ◦ x i x i +1 ◦ x i +2 ◦ · · · ◦ x n where x i , x i +1 ∈ M α for some α ∈ V . Y. DANDAN AND V. GOULD A shuffle is a step:(e) x ◦ · · · ◦ x n → x ◦ · · · ◦ x i − ◦ x i +1 ◦ x i ◦ x i +2 ◦ · · · ◦ x n where ( s ( x i ) , s ( x i +1 )) ∈ E . Definition 3.2.
Two words in X ∗ are shuffle equivalent if one can be obtained from theother by applying relations in R e , or, equivalently, by shuffle steps. Definition 3.3.
A word x = x ◦ · · · ◦ x n ∈ X ∗ is pre-reduced if it is not possible to applya reduction step to x .A word x = x ◦ · · · ◦ x n ∈ X ∗ is reduced if for all 1 ≤ i ≤ n , x i I , and for all1 ≤ i < j ≤ n with s ( x i ) = s ( x j ), there exists some i < k < j with ( s ( x i ) , s ( x k )) E .We denote by K the set of reduced words in X ∗ .If x = x ◦· · ·◦ x n ∈ X ∗ is reduced, then any factor x i ◦ x i +1 ◦· · ·◦ x j is reduced. A reducedword is pre-reduced, but the converse is not necessarily true. For example, x ◦ x ◦ x where s ( x ) = s ( x ) = α , s ( x ) = β , ( α, β ) ∈ E and no x i is an identity, is pre-reduced,but not reduced. Notice that ǫ is always reduced. The following remarks are clear fromDefinition 3.3. Remark 3.4.
A word is reduced if and only if any word shuffle equivalent is pre-reduced.In particular, any word shuffle equivalent to a reduced word is reduced.
Remark 3.5.
Let x = x ◦ · · · ◦ x n , y = y = y ◦ · · · ◦ y n ∈ X ∗ be such that x i , y i I and s ( x i ) = s ( y i ) for all 1 ≤ i ≤ n . If one of x, x r , y, y r is reduced, then so are all four.We will frequently concatenate reduced words in X ∗ , wanting to know if the product isreduced. The next remark is useful in this regard. Remark 3.6.
Let x = x ◦ · · · ◦ x m , y = y ◦ · · · ◦ y n ∈ X ∗ be reduced. Then x ◦ y is not reduced exactly if there exists i, j with 1 ≤ i ≤ m, ≤ j ≤ n such that s ( x i ) = s ( y j ) andfor all h, k with i < h ≤ m, ≤ k < j we have ( s ( x i ) , s ( z )) ∈ E where z = x h or z = y k .The lemma below is standard but it is worth making explicit. Lemma 3.7.
Let w ∈ X ∗ . Applying reduction steps and shuffles leads in a finite numberof steps to a reduced word w with [ w ] = [ w ] .Proof. Note that applying reduction steps to w reduces its length. There are finitely manywords shuffle equivalent to w . Either these are all pre-reduced, and we let w = w , or wecan apply a reduction step to some w ′ shuffle equivalent to w . Continue applying reductionsteps to w ′ until we arrive at a pre-reduced word w . Notice that | w | < | w | . Repeat thisprocess, obtaining a finite list of words w = w , w , w , . . . , w m where all words shuffleequivalent w m are pre-reduced. By Remark 3.4, w m is reduced; let w = w m . (cid:3) The next result is fundamental to our arguments. As commented in [24], it is essentiallythe monoid version of a result of Green [25] which can be deduced from [8, Theorem 6.1].However, we note that [25] and [8] deal only with the case of a finite graph. Here we givethe general result, calling upon Proposition 2.3.
RAPH PRODUCTS OF LEFT ABUNDANT MONOIDS 9
Proposition 3.8.
Every element of
G P is represented by a reduced word. Two reducedwords represent the same element of
G P if and only if they are shuffle equivalent. Anelement x ∈ [ w ] is of minimal length if and only if it is reduced.Proof. We have already shown the first part.For the second, it is clear that if two reduced forms are shuffle equivalent then theyrepresent the same element of
G P . Conversely, suppose that w, w ′ ∈ X ∗ are reduced formsand [ w ] = [ w ′ ] in G P . Let V ′ = s ( w ) ∪ s ( w ′ ) and let Γ ′ = ( V ′ , E ′ ) be the corresponding fullsubgraph. Let X ′ = S α ∈ V ′ M α and let G P ′ be the corresponding graph product. Clearly, w, w ′ ∈ ( X ′ ) ∗ are pre-reduced and from Proposition 2.3, [ w ] = [ w ′ ] in G P ′ . We deducefrom [8, Theorem 6.1] that w and w ′ are shuffle equivalent in G P ′ and hence clearly shuffleequivalent in G P .For the final point, it is clear that a word w ∈ X ∗ such that | w | is minimal in [ w ] isa reduced form. For the converse, suppose that x ∈ X ∗ is a reduced form and [ x ] = [ y ].Choosing y as in Lemma 3.7 we have that [ x ] = [ y ] = [ y ] where y is reduced and | y | ≥ | y | .By the above x, y are shuffle equivalent and hence | x | = | y | ≤ | y | . (cid:3) Definition 3.9. If x ∈ X ∗ and [ x ] = [ w ] for a reduced word w ∈ X ∗ , then we say that w is a reduced form of x .Notice that:(1) The equality [ x ] = [ y ] where x = x ◦ · · · ◦ x n and y = y ◦ · · · ◦ y m , does not, ingeneral, imply that s ( x ) = s ( y ). However, if both x and y are reduced, we musthave m = n and s ( x ) = s ( y ), by Proposition 3.8.(2) If x ◦· · ·◦ x n is reduced and s ( x ◦· · ·◦ x n ) is a complete subgraph, then s ( x i ) = s ( x j )for all 1 ≤ i < j ≤ n , again by Proposition 3.8.We now show that, starting with a reduced word x ∈ X ∗ , and multiplying by a singleletter p from X , we have a narrow range of possibilities for any reduced form of the product p ◦ x . Lemma 3.10.
Let p ∈ X , where p I , and let x = x ◦ · · · ◦ x n ∈ X ∗ be reduced. Thenone of the following occurs: (i) p ◦ x ◦ · · · ◦ x n is reduced; (ii) there exists ≤ k ≤ n such that s ( x k ) = s ( p ) and ( s ( p ) , s ( x l )) ∈ E for all ≤ l ≤ k − , and p ◦ x ◦ · · · ◦ x n reduces to (1) px k ◦ x ◦ · · · ◦ x k − ◦ x k +1 ◦ · · · ◦ x n and also to (2) x ◦ · · · ◦ x k − ◦ px k ◦ x k +1 ◦ · · · ◦ x n . Further, in Case (ii)(a) if px k is not an identity then (1) and (2) are reduced; (b) if px k is an identity then p ◦ x ◦ · · · ◦ x n reduces to the reduced word (3) x ◦ · · · ◦ x k − ◦ x k +1 ◦ · · · ◦ x n . Consequently, if α ∈ s ( x ) and q ∈ X with s ( q ) = α , then α must be in the support ofany reduced form of q ◦ x .Proof. Suppose that p ◦ x is not reduced. Then, by the definition of reduced, k as definedin the statement must exist. Clearly, for p ◦ x , we may shuffle x k and glue it to p to obtain px k ◦ x ◦ · · · ◦ x k − ◦ x k +1 ◦ · · · ◦ x n which is shuffle equivalent to x ◦ · · · ◦ x k − ◦ px k ◦ x k +1 ◦ · · · ◦ x n . If px k is not an identity, then these words are reduced, by Remark 3.5.If px k is an identity then p ◦ x reduces to x ◦ · · · x k − ◦ x k +1 ◦ · · · ◦ x n , which is reduced, since it is a right factor of the word x k ◦ x ◦ · · · x k − ◦ x k +1 ◦ · · · ◦ x n ,which is shuffle equivalent to the reduced word x .The final statement is clear if q ∈ I ; if q / ∈ I it follows by examining the cases above. (cid:3) Corollary 3.11.
Let x, y ∈ X ∗ where y is reduced. If α ∈ s ( y ) but α s ( x ) , then α mustbe in the support of any reduced form of x ◦ y .Proof. Let x = x ◦ · · · ◦ x m and proceed by induction on m . If m = 1 then the result istrue by Lemma 3.10. Suppose therefore that m ≥ m −
1. Let z ◦ · · · ◦ z k be a reduced form of x ◦ · · · ◦ x m ◦ y . Then α is in the support of z ◦ · · · ◦ z k byassumption, and so α is in the support of the reduced form of x ◦ z ◦ · · · ◦ z k and hence x ◦ y , again by Lemma 3.10. (cid:3) We will make extensive use of Corollary 3.11 to find reduced forms of products of re-duced words. The expression of elements in
G P using reduced forms has a very usefulcancellation-type property, as we now explain. First, another technical result using astrategy that will be key in this paper. Recall from Definition 3.3 that K = { w ∈ X ∗ : w is reduced } . Lemma 3.12.
Let α ∈ V and define maps θ α : K −→ G P and η α : K −→ G P where for each x = x ◦ · · · ◦ x n ∈ K , xθ α = (cid:26) [ x i ( α ) ] α ∈ s ( x )[ x ] else and xη α = (cid:26) [ x i ( α ) ] α ∈ s ( x )[ ǫ ] else.Here i ( α ) is smallest i such that s ( x i ) = α and x i ( α ) is obtained by deleting x i ( α ) from x .Then θ α and η α are constant on R ♯ -classes, that is, they extend to maps θ α : G P −→ G P and η α : G P −→ G P given by [ w ] θ α = w ′ θ α and [ w ] η α = w ′ η α where w ′ is any reduced form of w . RAPH PRODUCTS OF LEFT ABUNDANT MONOIDS 11
Proof.
Let [ p ] = [ q ] where both p, q ∈ K are reduced. We need show pθ α = qθ α and pη α = qη α . By Proposition 3.8, p and q are shuffle equivalent; by finite induction we canassume that q is obtained from p by exactly one shuffle.Let p = x ◦ · · · ◦ x j − ◦ x j ◦ x j +1 ◦ x j +2 ◦ · · · ◦ x n and q = x ◦ · · · ◦ x j − ◦ x j +1 ◦ x j ◦ x j +2 ◦ · · · ◦ x n . If α s ( p ) (and so α s ( q )), then pθ α = [ p ] = [ q ] = qθ α and pη α = [ ǫ ] = qη α . Suppose now that α ∈ s ( p ). Considering p , pick the smallest k such that s ( x k ) = α . If1 ≤ k ≤ j − j + 2 ≤ k ≤ n , then, clearly, pθ α = qθ α and pη α = qη α . If k = j , thensince ( s ( x j ) , s ( x j +1 )) ∈ E we have s ( x j ) = s ( x j +1 ); it follows that pη α = qη α = [ x j ] and pθ α = qθ α = [ x ◦ · · · ◦ x j − ◦ x j +1 ◦ x j +2 ◦ · · · ◦ x n ]. Similarly if k = j + 1. (cid:3) We use the θ α maps to prove our first cancellation-type result. Lemma 3.13.
Let [ x ] = [ y ] where x = x ◦ · · · ◦ x n and y = y ◦ · · · ◦ y n are reduced and let ≤ m ≤ n . Then [ x ◦· · ·◦ x m ] = [ y ◦· · ·◦ y m ] if and only if [ x m +1 ◦· · ·◦ x n ] = [ y m +1 ◦· · ·◦ y n ] .Proof. Suppose that [ x ◦ · · · ◦ x m ] = [ y ◦ · · · ◦ y m ]. Since [ x ◦ · · · ◦ x n ] = [ y ◦ · · · ◦ y n ], wehave [ x ◦ · · · x m ◦ x m +1 ◦ · · · ◦ x n ] = [ x ◦ · · · x m ◦ y m +1 ◦ · · · ◦ y n ] . As x ◦ · · · x m ◦ x m +1 ◦ · · · ◦ x n is reduced and x ◦ · · · x m ◦ y m +1 ◦ · · · ◦ y n has the same length,we deduce that x ◦ · · · x m ◦ y m +1 ◦ · · · ◦ y n is reduced, by Proposition 3.8. Let s ( x r ) = α r for all 1 ≤ r ≤ m . Then, observing that any right factor of a reduced word is reduced,[ x ◦ · · · ◦ x m ◦ x m +1 ◦ · · · ◦ x n ] θ α · · · θ α m = [ x ◦ · · · ◦ x m ◦ y m +1 ◦ · · · ◦ y n ] θ α · · · θ α m by Lemma 3.12, which gives [ x m +1 ◦ · · · ◦ x n ] = [ y m +1 ◦ · · · ◦ y n ] as desired.The remainder of the lemma follows dually. (cid:3) Definition 3.14.
A word w ∈ X ∗ is a complete block if it is reduced, and s ( w ) forms acomplete subgraph of Γ = Γ( V, E ).We now show that any reduced word in X ∗ may be shuffled into a word that is a productof complete blocks. Definition 3.15.
Let w ∈ X ∗ . Then w is a left Foata normal form with block length k and blocks w i ∈ X ∗ , 1 ≤ i ≤ k , if:(i) w = w ◦ · · · ◦ w k ∈ X ∗ is a reduced word;(ii) s ( w i ) is a complete subgraph for all 1 ≤ i ≤ k ;(iii) for any 1 ≤ i < k and α ∈ s ( w i +1 ), there is some β ∈ s ( w i ) such that ( α, β ) E .If [ x ] = [ w ] where w is a left Foata normal form, then we may say w is a left Foatanormal form of x . Remark 3.16. (i) The empty word ǫ is a left Foata normal form with block length 0. (ii)A complete block is precisely a word in left Foata normal form with block length 1. (iii) If w = w ◦ · · · ◦ w k ∈ X ∗ is in left Foata normal form with blocks w i , 1 ≤ i ≤ k , then for any1 ≤ j ≤ j ′ ≤ k we have w j ◦ w j +1 ◦ · · · ◦ w j ′ is also in left Foata normal form, with blocks w h , j ≤ h ≤ j ′ . Proposition 3.17.
Every element in
G P may be represented by a left Foata normal form.Proof.
We know that any element of
G P may be represented by a reduced word. Takea reduced word w = y and let w be chosen such that w ◦ y is shuffle equivalent to w for some y , s ( w ) is complete, and | w | is maximum with respect to these constraints.Assume that w , y , w , y , . . . , w k , y k have been chosen such that for each 1 ≤ j ≤ k wehave y j − is shuffle equivalent to w j ◦ y j , s ( w j ) is complete, and | w j | is maximum withrespect to these constraints. Clearly this process must end after a finite number of stepswith y k = ǫ .For any 1 ≤ j ≤ k we have by finite induction that y j − is shuffle equivalent to w j ◦ w j +1 ◦ · · · ◦ w k and, in particular, w is shuffle equivalent to w ◦ · · · ◦ w k . We now claimthat w ◦ · · · ◦ w k is a left Foata normal form with blocks w i for 1 ≤ i ≤ k . Certainly (i)and (ii) of Definition 3.15 hold. To see that (iii) holds, suppose that 1 ≤ i < k and let α ∈ s ( w i +1 ); say w i +1 = p ◦ a ◦ q where a ∈ X and s ( a ) = α . Suppose for contradiction thatfor all β ∈ s ( w i ) we have ( α, β ) ∈ E . Since y i − is shuffle equivalent to w i ◦ w i +1 ◦ · · · ◦ w k we would have y i − being shuffle equivalent to w i ◦ a ◦ y ′ i +1 for some y ′ i +1 , where s ( w i ◦ a )is complete and | w i ◦ a | > | w i | , a contradiction. (cid:3) We are now in a position to prove the main result of this section, which tells us that theleft Foata normal form of an element of any
G P is essentially unique. In the result belowthe maps ϕ α and υ α are the left right duals of θ α and η α from Proposition 3.12. Theorem 3.18.
Let w ∈ X ∗ and let w ◦ w ◦ · · · ◦ w k and w ′ ◦ w ′ ◦ · · · ◦ w ′ h be left Foatanormal forms of w with blocks w i , w ′ j for ≤ i ≤ k, ≤ j ≤ h . Then k = h and [ w i ] = [ w ′ i ] for ≤ i ≤ k .Proof. We first make the following general observation. Let x = x ◦ · · · ◦ x n and z = z ◦ · · · ◦ z n be reduced forms of w . Pick α ∈ s ( x )(= s ( z )). Let i be least such that s ( x i ) = α and j be least such that s ( z j ) = α . Since x and z are shuffle equivalent, x i = z j .Suppose that there exists some 1 ≤ i ′ < i such that s ( x i ′ ) = β with ( β, α ) E ; note thatby minimality of i we have β = α . Then, again as x and z are shuffle equivalent, thereexists some 1 ≤ j ′ < j such that s ( z j ′ ) = β and z j ′ = x i ′ .Let p = w ◦ · · · ◦ w k and p ′ = w ′ ◦ · · · ◦ w ′ h ; by Remark 3.16 p and p ′ are also in leftFoata normal form. We claim that s ( w ) = s ( w ′ ). Expressing as products of letters, let w = a ◦ · · · a r , p = b ◦ · · · ◦ b m , w ′ = u ◦ · · · ◦ u t and p ′ = v ◦ · · · ◦ v n . Suppose that there exists some δ ∈ s ( w ) but not in s ( w ′ ), so that δ ∈ s ( p ′ ). Let i be least such that s ( a i ) = δ and let j be least such that s ( v j ) = δ . By definition of leftFoata normal form, either (i) v j is in the first block w ′ of p ′ , in which case there existssome 1 ≤ t ′ ≤ t with ( s ( u t ′ ) , δ ) E , or (ii) v j is in a subsequent component of p ′ in which RAPH PRODUCTS OF LEFT ABUNDANT MONOIDS 13 case certainly there exists 1 ≤ j ′ < j with ( s ( v j ′ ) , δ ) E . Let γ = s ( u t ′ ) (in Case (i)) and γ = s ( v j ′ ) (in Case (ii)). In either case we have γ = δ and ( δ, γ ) / ∈ E . By the observationabove there must be some i ′ with 1 ≤ i ′ < i such that s ( a i ′ ) = γ . This is impossiblesince s ( w ) is a complete subgraph. Together with the converse argument we deduce that s ( w ) = s ( w ′ ).We now show that [ w ] = [ w ′ ] and [ p ] = [ p ′ ]. Let s ( w ) = { α , · · · , α r } . It then followsfrom Lemma 3.12 that[ p ] = [ w ◦ p ] ϕ α · · · ϕ α r = [ w ′ ◦ p ′ ] ϕ α · · · ϕ α r = [ p ′ ]and [ w ] = [ w ◦ p ] υ α · · · [ w ◦ p ] υ α r = [ w ′ ◦ p ′ ] υ α · · · [ w ′ ◦ p ′ ] υ α r = [ w ′ ]as required.Noticing that | p | < | w ◦ p | , the result now follows by induction. (cid:3) Clearly, we may define the notion of a right Foata normal form of an element in X ∗ , andthe dual arguments to those for left Foata normal form hold.4. Towards a characterization of R ∗ We continue to consider a fixed, but arbitrary, graph product of monoids
G P . We nowshow how we can use the left Foata normal forms developed in Section 3 to describe therelation R ∗ in G P . We will build on this in Section 5 to show that if each vertex monoidis abundant, then so is
G P .The next lemma can be deduced from [8, Proposition 7.1], together with our Propo-sition 2.3 and Remark 2.4. Note that if x = x ◦ · · · ◦ x n is reduced, then in Costa’sterminology, the x i are components . Lemma 4.1.
Let x = x ◦ · · · ◦ x n ∈ X ∗ be reduced. Then the following are equivalent: (1) [ x ] is left invertible in G P ; (2) [ x i ] is left invertible in G P for ≤ i ≤ n ; (3) x i ∈ M s ( x i ) is left invertible in M s ( x i ) for ≤ i ≤ n .Moreover, if any of the above conditions hold, then any left inverse of [ x ] has the form [ y ] where y = y n ◦ · · · ◦ y and y i is a left inverse of x i for ≤ i ≤ n . The arguments in the next lemma essentially rely on the following simple observations. If x = x ◦ · · · ◦ x n ∈ X ∗ is shuffle equivalent to y = x j ◦ · · · ◦ x j n , then for any 1 ≤ i < k ≤ n with j i > j k we have ( s ( x j i ) , s ( x j k )) ∈ E . Consequently, if we can shuffle x to a word x ′ ◦ x ′′ , where x ′ has length m , then we can shuffle x ′ to a word x i ◦ x i ◦ · · · ◦ x i m where i < i < · · · < i m and x ′′ to the word obtained from x by deleting the letters x i , · · · , x i m .Moreover, for any 1 ≤ ℓ ≤ m we can shuffle x to x ◦ x · · · ◦ x i ℓ − ◦ x i ℓ ◦ x i ℓ +1 ◦ · · · ◦ x i m ◦ z where z is x i ℓ +1 ◦ x i ℓ +2 ◦ · · · ◦ x n with x i ℓ +1 , · · · , x i m deleted. Lemma 4.2.
Let u ∈ X ∗ . Then: (1) [ u ] = [ a ][ x ] where a ◦ x is reduced, [ a ] is left invertible, and | a | is maximum withrespect to these constraints; (2) if [ a ][ x ] = [ b ][ y ] where (in addition) b ◦ y is reduced, [ b ] is left invertible, and | b | = | a | ,then [ a ] = [ b ] and [ x ] = [ y ] ; (3) with [ u ] = [ a ][ x ] as in (1), x has a left Foata normal form x ◦ · · · ◦ x m with blocks x i , ≤ i ≤ m , such that x contains no left invertible letters.Proof. We begin by finding a reduced form p = p ◦ · · · ◦ p n for u . By shuffling p wemay find a and x as in (1). By Lemma 4.1 and the above remark we may assume that a = p i ◦ · · · ◦ p i k where i < i < · · · < i k , with p i h is left invertible for all 1 ≤ h ≤ k .Suppose now that b, y are as given; again we may assume that b = p j ◦ · · · ◦ p j k where j < j < · · · < j k . If i < j then we notice that we can shuffle p to p i ◦ p ◦ · · · ◦ p j − ◦ p j ◦ p j +1 ◦ · · · ◦ p n and then to p i ◦ p j ◦ p j ◦ · · · ◦ p j k ◦ y ′ where y ′ is y with p i deleted. But,this contradicts the maximality of | a | . With the dual argument we obtain that i = j .Suppose for finite induction that i s = j s for 1 ≤ s < k , and that i s +1 < j s +1 . Thensimilarly to the preceding argument we have that p shuffles to p i ◦ p i ◦ · · · ◦ p i s ◦ p i s +1 ◦ z where z is p with p i , p i , · · · , p i s +1 deleted. But then we can shuffle z to obtain a word p j s +1 ◦ p j s +2 ◦ · · · ◦ p j k ◦ w where w is p with p i , p i , · · · , p i s +1 , p j s +1 , p j s +2 , · · · , p j k deleted.Again, this contradicts the maximality of | a | . We deduce that i s = j s for 1 ≤ s ≤ k andhence [ a ] = [ b ]. Clearly then [ x ] = [ y ] follows.Suppose now that [ u ] = [ a ][ x ] as in (1), and shuffle x to left Foata normal form x ◦· · ·◦ x m ,where the x i are the blocks for 1 ≤ i ≤ m . Clearly, since s ( x ) is complete, x cannotcontain any left invertible letters, else this would contradict the maximality of | a | . (cid:3) To simplify the description of R ∗ on G P we now present two technical lemmas.
Lemma 4.3.
Let x = x ◦ · · · ◦ x n ∈ X ∗ and ( α, β ) E . Suppose that x l is non-leftinvertible with s ( x l ) = β , for some ≤ l ≤ n , and s ( x k ) is neither α nor β for all l < k ≤ n . Let z = z ◦ · · · ◦ z m ∈ X ∗ be any reduced form of x ◦ · · · ◦ x n . Then β ∈ s ( z ) and if j is greatest such that ≤ j ≤ m with s ( z j ) = β , then z j is non left invertible, and s ( z t ) = α for all j < t ≤ m .Proof. We begin by observing that if we can find one reduced form of x with the requiredproperty, then all reduced forms will have the required property.We proceed by induction on n . If n = 1 = l the result is clear, since x = x is the onlyreduced form of x . Suppose now that n > n .Let w = x ◦ · · · ◦ x l − , w = x l +1 ◦ · · · ◦ x n and let w ′ , w ′ ∈ X ∗ be reduced suchthat [ w ] = [ w ′ ] and [ w ] = [ w ′ ]. Certainly α, β / ∈ s ( w ′ ). Let w ′ = u ◦ · · · ◦ u h and w ′ = v ◦ · · · ◦ v r . If w ′ ◦ x l ◦ w ′ is a reduced form, then we are done.Suppose therefore that w ′ ◦ x l ◦ w ′ is not a reduced form, and consider first w ′ ◦ x l = u ◦ · · · ◦ u h ◦ x l . If w ′ ◦ x l is not a reduced form then, from Remark 3.6, there exists some t with 1 ≤ t ≤ h with s ( u t ) = β and ( s ( u k ) , β ) ∈ E for all t < k ≤ h . By shuffling w ′ , without loss ofgenerality we can assume that t = h . Let p = u h x l and notice that as x l is not left RAPH PRODUCTS OF LEFT ABUNDANT MONOIDS 15 invertible, then neither is p , and certainly p = ǫ . Then y = u ◦ · · · ◦ u h − ◦ p ◦ v ◦ · · · ◦ v r has length strictly less than n , s ( p ) = β , p is not left invertible, and α, β / ∈ s ( v ◦ · · · ◦ v r ).On the other hand, if w ′ ◦ x l is a reduced form, then again by Remark 3.6, and makinguse of the fact β / ∈ s ( w ′ ), we may assume that s ( u h ) = s ( v ) and ( β, s ( u h )) ∈ E . Then y = u ◦ · · · ◦ u h − ◦ u h v ◦ x l ◦ v ◦ · · · ◦ v r has length strictly less than n , and α, β / ∈ s ( v ◦ · · · ◦ v r ).In each case we have found a word y with [ y ] = [ x ] to which we can apply the inductionhypothesis. The result follows. (cid:3) Lemma 4.4.
Let x = x ◦ · · · ◦ x n be a left Foata normal form with blocks x i , ≤ i ≤ n ,such that x contains no left invertible letters. Let u ∈ X ∗ and let z be a reduced form of u ◦ x . Then z ◦ x ◦ · · · ◦ x n is a reduced form of u ◦ x .Proof. Certainly [ u ◦ x ] = [ z ◦ x ◦ · · · ◦ x n ]. Let z = z ◦ · · · ◦ z m . As both z and x ◦ · · · ◦ x n are reduced, if z ◦ x ◦ · · · ◦ x n is not reduced, then by Remark 3.6 we can shuffle a letter z k of z to the end of z and a letter a of x ◦ · · · ◦ x n to the start of x ◦ · · · ◦ x n where s ( z k ) = s ( a ) = α say. We may assume that k = m and as x ◦ · · · ◦ x n is a left Foatanormal form, that a is a letter of x , and then that it is the first letter of x . Since x is inleft Foata normal form, it follows that α s ( x ) and there exists a (unique) b ∈ x suchthat ( α, s ( b )) E . Let s ( b ) = β ; recall that b is non-left invertible. It then follows fromLemma 4.3 that β ∈ s ( z ) and if t is greatest such that 1 ≤ t ≤ m with s ( z t ) = β , then s ( z h ) = α for all t < h ≤ m . This contradicts the fact s ( z m ) = α .We deduce that z ◦ x ◦ · · · ◦ x n is a reduced form of u ◦ x , as required. (cid:3) We can now get our first handle on the consideration of the R ∗ -class of an element of G P in the general case. Subsequently, we will focus on the case where the vertex monoidsare abundant.
Proposition 4.5. (1)
Let x = x ◦ · · · ◦ x n be a left Foata normal form with blocks x i , ≤ i ≤ n , such that x contains no left invertible letters. Then [ x ] R ∗ [ x ] . (2) Let p ∈ X ∗ . Then [ p ] = [ a ][ x ] where a ◦ x is reduced, the letters of a are all leftinvertible, | a | is maximum with respect to these constraints and x is a left Foatanormal form x as in (1). Further, [ p ] R ∗ [ a ][ x ] .Proof. (1) Let [ p ] , [ q ] ∈ G P . Clearly it suffices to show that if [ p ][ x ] = [ q ][ x ], then [ p ][ x ] =[ q ][ x ]. Suppose therefore that [ p ][ x ] = [ q ][ x ] and let ( p ◦ x ) ′ and ( q ◦ x ) ′ be reduced formsof p ◦ x and q ◦ x , respectively. By Lemma 4.4, ( p ◦ x ) ′ ◦ x ◦· · ·◦ x n and ( q ◦ x ) ′ ◦ x ◦· · ·◦ x n are reduced forms of p ◦ x ◦ · · · ◦ x n and q ◦ x ◦ · · · ◦ x n , respectively. It then follows fromLemma 3.13 that [( p ◦ x ) ′ ] = [( q ◦ x ) ′ ] and so [ p ][ x ] = [ q ][ x ].(2) This existence of a and x is guaranteed by Lemma 4.2, and then the result followsfrom (1) and the fact that R ∗ is a left congruence. (cid:3) Graph products of left abundant monoids are left abundant
The aim of this section is to prove the claim of the heading; this will involve us in somecombinatorial intricacies. It might be helpful to the reader if we outline our strategy here.Proposition 4.5 is our first step in describing R ∗ in G P . In Proposition 5.20 we show thatif z = z ◦ · · · ◦ z n ∈ X ∗ is a complete block, then [ z ] R ∗ [ z ′ ] where z ′ = z ′ ◦ · · · ◦ z ′ n ischosen such that z ′ i ∈ M s ( z i ) and z i R ∗ z ′ i in M s ( x i ) for all 1 ≤ i ≤ n . In particular, if each M i is left abundant, then for any idempotents z + i with z i R ∗ z + i in M s ( z i ) , we have that[ z ] is R ∗ -related to the idempotent [ z + ] where z + = z +1 ◦ · · · ◦ z + n . Proposition 4.5 tells usthat for p ∈ X ∗ we can write [ p ] = [ a ][ x ] where a ◦ x is reduced, the letters of a are allleft invertible, and x is a left Foata normal form, the first block of which contains no leftinvertible letters. Moreover, calling this first block z we have that [ p ] R ∗ [ a ][ z ]. As R ∗ isa left congruence, [ p ] R ∗ [ a ][ z + ] and then as [ a ] has a left inverse [ a ′ ] (so that [ a ′ ] R [ ǫ ]) wehave [ p ] R ∗ [ a ][ z + ][ a ′ ]. The fact that [ a ][ z + ][ a ′ ] is idempotent is easily seen.To arrive at Proposition 5.20 we cannot escape a very careful analysis of products [ x ][ z ]in G P (remember, we are considering equations of the form [ x ][ z ] = [ y ][ z ]). To this endwe find a new factorisation of elements in G P that allows us to cancel and replace a finalterm in equalities. This we achieve in Lemma 5.19.To arrive at Lemma 5.19 we now define the notions of α -absorbing, α -good and subse-quently a stronger version of being α -good that we call α -amenable, where α ∈ V . We showin Proposition 5.14 that in an α -amenable word, the inner factor reduces to a word whichdoes not have α in its support. This enables us to pin down exactly which letters we canmove to the right of a word (see Definition 5.16) and hence we arrive at the factorisationof Lemma 5.19.First, we need to recall the description of idempotents in G P from [8].
Definition 5.1.
We say that an idempotent of
G P is in standard form if it is written as[ u ] where u = b ◦ e ◦ b ′ ∈ X ∗ is reduced, b = b ◦ · · · ◦ b n , e = e ◦ · · · ◦ e m , b ′ = b ′ n ◦ · · · ◦ b ′ where b ′ i b i is an identity for 1 ≤ i ≤ n , s ( e ) is complete and e i = e i for 1 ≤ i ≤ m .Note that [ u ] is idempotent for any word u of the form in Definition 5.1. Lemma 5.2. [8, Theorem 14.2]
Any idempotent in
G P can be written in standard form.
Definition 5.3.
Let α ∈ V . A word x ∈ X ∗ is said to be α - absorbing if α is not in thesupport of any reduced form of x . Definition 5.4.
Let α ∈ V . A word x ∈ X ∗ is said to be α - good if for all β in the supportof any reduced form of x , we have β = α or ( β, α ) ∈ E .We remark that in Definitions 5.3 and 5.4, α may not be in the support of x . If forany β in the support of x , we have β = α or ( β, α ) ∈ E , then certainly x is α -good, butthe converse need not be true. If [ x ] = [ y ], or x, y are reduced and s ( x ) = s ( y ), then x is α -good (resp. α -absorbing) if and only if y is α -good (resp. α -absorbing). Further, as s ( ǫ ) = ∅ , we have that ǫ is both α -good and α -absorbing, and hence so is 1 β for all β ∈ V . RAPH PRODUCTS OF LEFT ABUNDANT MONOIDS 17
Finally, if w ∈ X + and s ( w ) = { α } for some α ∈ V , then w is α -good. By Remark 3.5 wehave: Lemma 5.5.
Let x = x ◦ · · · ◦ x n , y = y ◦ · · · ◦ y n ∈ X ∗ be such that x i , y i I and s ( x i ) = s ( y i ) for all ≤ i ≤ n . If one of x, x r , y, y r is a reduced word that is α -good, thenso are all four. The next lemma is crucial in allowing us to deduce the α -goodness (or otherwise) of aword in terms of its factors. Lemma 5.6.
Let α ∈ V and x = x ◦ · · · ◦ x n ∈ X ∗ . (i) If x k ◦ · · · ◦ x n is α -good for some ≤ k ≤ n , then x ◦ · · · ◦ x n is α -good if and onlyif x ◦ · · · ◦ x k − is α -good. (ii) If x ◦ · · · ◦ x k − is α -good for some ≤ k ≤ n + 1 , then x ◦ · · · ◦ x n is α -good if andonly if x k ◦ · · · ◦ x n is α -good.Proof. Suppose that x k ◦ · · · ◦ x n is α -good.If x ◦ · · · ◦ x k − is α -good, then from Remark 3.6 and comments above it is clear that x ◦ · · · ◦ x n is α -good.Conversely, suppose that x ◦ · · · ◦ x n is α -good but x ◦ · · · ◦ x k − is not α -good. Let u ◦ · · · ◦ u m be a reduced form of x ◦ · · · ◦ x k − . Then, by Definition 5.4, there exists some1 ≤ t ≤ m such that s ( u t ) = β with β = α and ( β, α ) E . As x ◦ · · · ◦ x n is α -good, β isnot in the support of the reduced form of x ◦ · · · ◦ x n . Let v ◦ · · · ◦ v l be a reduced form of x k ◦ · · · ◦ x n . As x k ◦ · · · ◦ x n is α -good, β is not in the support of v ◦ · · · ◦ v l . Now considerthe word ( u ◦ · · · ◦ u m ) ◦ ( v ◦ · · · ◦ v l ). Of course,[( u ◦ · · · ◦ u m ) ◦ ( v ◦ · · · ◦ v l )] = [ x ◦ · · · ◦ x n ] . By the dual of Corollary 3.11, β lies in the support of the reduced form of ( u ◦ · · · ◦ u t ) ◦ ( v ◦ · · · ◦ v l ), and hence that of x ◦ · · · ◦ x n , contradiction.The proof of (ii) is the dual of (i). (cid:3) Corollary 5.7.
Let x ∈ X ∗ and let z, z ′ , t ∈ X where s ( z ) = s ( z ′ ) = α, s ( t ) = β and ( α, β ) ∈ E . The the following are equivalent: (1) x is α -good; (2) z ◦ z ′ ◦ x is α -good; (3) z ′ ◦ x is α -good; (4) zz ′ ◦ x is α -good; (5) z ◦ t ◦ x is α -good.Proof. From the remarks following Definition 5.4, z, z ◦ z ′ , zz ′ and z ◦ t are α -good. Theresult follows by Lemma 5.6. (cid:3) Our next definition is more subtle, but crucial for subsequent analysis of products in
G P . Definition 5.8.
Let α ∈ V and x = x ◦ · · · ◦ x n ∈ X ∗ be α -good. Then x is said to be α -amenable if one of the following holds: (i) n ≤ n > α s ( x ◦ · · · ◦ x n − ), or α ∈ s ( x ◦ · · · ◦ x n − ) and for all x k with2 ≤ k ≤ n − s ( x k ) = α , the word x k ◦ · · · ◦ x n is not α -good.It might help to bear in mind that x k ◦ · · · ◦ x n is not α -good if and only if there existssome β = α in the support of a reduced form, such that ( α, β ) / ∈ E . Notice that ǫ is α -amenable for any α ∈ V .As we remarked earlier, for x, y ∈ X ∗ , if [ x ] = [ y ], then x is α -good if and only if y is α -good. One might ask: Is it always true that x is α -amenable if and only if y is α -amenable? The answer is no, as illustrated by the following easy example. Let α, β, γ bedistinct elements of V with ( α, β ) , ( α, γ ) ∈ E and a ∈ M α , b ∈ M β and c ∈ M γ non-identityelements. The word a ◦ b ◦ c is reduced, α -amenable (by virtue of α / ∈ s ( b )). On the otherhand it shuffles to b ◦ a ◦ c which is α -good but not α -amenable (as s ( a ) = α and ( α, γ ) ∈ E ).On the positive side, we have the following result. Lemma 5.9.
Let α ∈ V and x = x ◦ · · · ◦ x n ∈ X ∗ be α -amenable. Let y be any wordobtained by applying reduction steps and shuffles to x ◦ · · · ◦ x n − . Then x ◦ y ◦ x n is also α -amenable.Proof. Clearly the result is true for n ≤ x ◦ · · · ◦ x n − = ǫ and there are no stepsto apply.Assume now that n >
2. Since x is α -good, so is any word in the same equivalence class,so that x ◦ y ◦ x n is also α -good. To show x ◦ y ◦ x n is α -amenable, it is sufficient toconsider the case where y is obtained from p = x ◦ · · · ◦ x n − in a single step.Clearly, if α / ∈ s ( p ), then we are done; suppose therefore that α ∈ s ( p ). We consider thefollowing cases.Case (1): s ( x j ) = β and s ( x j +1 ) = γ with ( β, γ ) ∈ E , where 2 ≤ j < n −
1. We showthat the word x ′ = x ◦ x ◦ · · · ◦ x j − ◦ x j +1 ◦ x j ◦ x j +2 ◦ · · · ◦ x n − ◦ x n is α -amenable. Clearly, we are fine in the case where neither β nor γ equals α . If β = α (and so γ = α ), then, by Definition 5.8, x j ◦ x j +1 ◦ x j +2 ◦ · · · ◦ x n − ◦ x n is not α -good. But,on the other hand, as x j ◦ x j +1 is α -good, x j +2 ◦ · · · ◦ x n is not α -good by Lemma 5.6, andhence, again by Lemma 5.6, x j ◦ x j +2 ◦ · · · ◦ x n is not α -good. For any k with 2 ≤ k ≤ n − k = j, j + 1 with s ( x k ) = α , it is clear that the factor x k ◦ · · · ◦ x n of x ′ is not α -goodby the assumption that x is α -amenable. Similarly if γ = α .Case (2): s ( x j ) = s ( x j +1 ) = β where 2 ≤ j < n −
1. We show that the word x ′′ = x ◦ x ◦ · · · ◦ x j − ◦ x j x j +1 ◦ x j +2 ◦ · · · ◦ x n − ◦ x n is α -amenable. As in Case (1) it is enough to show that if β = α then x j x j +1 ◦ x j +2 ◦ · · · ◦ x n is not α -good. To this end, if β = α , then as x j ◦ x j +1 and x j x j +1 are α -good but x j ◦ x j +1 ◦ x j +2 ◦· · ·◦ x n is not α -good, we deduce from Corollary 5.7 that x j x j +1 ◦ x j +2 ◦· · ·◦ x n is not α -good. RAPH PRODUCTS OF LEFT ABUNDANT MONOIDS 19
Case (3): s ( x j ) = β and x j = 1 β . An essentially vacuous argument easily gives that theword x ′′′ = x ◦ x ◦ · · · ◦ x j − ◦ x j +1 ◦ x j +2 · · · ◦ x n − ◦ x n is α -amenable. (cid:3) The next corollary is immediate from Lemmas 3.7 and 5.9.
Corollary 5.10.
Let α ∈ V and x = x ◦ · · · ◦ x n ∈ X ∗ be α -amenable. Let y be a reducedform of x ◦ · · · ◦ x n − . Then x ◦ y ◦ x n is also α -amenable. Lemma 5.11.
Let x = x ◦ · · · ◦ x m , y = y ◦ · · · y n ∈ X ∗ be reduced words. If s ( x m ) = α but α s ( y ) and there exists β ∈ s ( y ) with ( β, α ) E , then β must be in the support ofthe reduced form of x ◦ y .Proof. We proceed by induction on n . If n = 1, then x ◦ y = x ◦ · · · ◦ x m ◦ y . We musthave s ( y ) = β so that x ◦ y is clearly reduced by Remark 3.6. Suppose now that n > y of length strictly less than n .Clearly, the result is true if ( x ◦ · · · ◦ x m ) ◦ ( y ◦ · · · y n ) is reduced. If not, by Remark 3.6,there exists some 1 ≤ k ≤ m, ≤ j ≤ n such that s ( x k ) = s ( y j ) and ( s ( y j ) , s ( z )) ∈ E for any z = x h or z = y t with k + 1 ≤ h ≤ m, ≤ t ≤ j −
1. Let y ′ = y ◦ · · · ◦ y j − ◦ y j +1 ◦ · · · ◦ y n ; notice that y shuffles to y j ◦ y ′ , so that y ′ is a reduced form. Further, let p = x ◦ · · · ◦ x k − ◦ x k y j ◦ x k +1 ◦ · · · ◦ x m . Let x ′ = p if x k y j is not an identity and otherwiselet x ′ = x ◦ · · · ◦ x k − ◦ x k +1 ◦ · · · ◦ x m ; in either case, x ′ is a reduced form. Now consider x ′ ◦ y ′ . Clearly, [ x ◦ y ] = [ x ′ ◦ y ′ ]. As α s ( y ), we have α / ∈ s ( y ′ ) and s ( x k ) = s ( y j ) = α , sothat k = m . Moreover, as s ( x m ) = α and ( β, α ) E , we have s ( y j ) = β , and so β ∈ s ( y ′ ).By induction, β is in the support of any reduced form of x ′ ◦ y ′ , and hence in that of x ◦ y . (cid:3) Lemma 5.12.
Let α ∈ V and x = x ◦ · · · ◦ x n ∈ X ∗ be α -amenable with s ( x n ) = α . Thenthe word x ′ = x ◦ · · · ◦ x n − is α -absorbing.Proof. If n ≤
2, then we may take x ◦ · · · ◦ x n − as ǫ , which is certainly α -absorbing.Assume now that n >
2. Let y = y ◦ · · · ◦ y m be a reduced form of x ′ . By Corollary 5.10, x ◦ y ◦ x n is α -amenable. We claim that x ′ is α -absorbing. To prove this, we assume thecontrary, so that T = ∅ where T = { k : 1 ≤ k ≤ m, s ( y k ) = α } . Let l and l ′ be the least and greatest elements of T , respectively. Since x ◦ y ◦ x n is α -amenable, we have that for any k ∈ T the word y k ◦ y k +1 ◦ · · · ◦ y m ◦ x n is not α -good.We consider the following cases.Case (1): x ◦ y is a reduced form. It follows that x ◦ y ◦ · · · ◦ y l ′ is also a reduced form.Let z be a reduced form of y l ′ +1 ◦ · · · ◦ y m ◦ x n . As commented, α -amenability gives usthat y l ′ ◦ y l ′ +1 ◦ · · · ◦ y m ◦ x n is not α -good. We deduce y l ′ +1 ◦ · · · ◦ y m ◦ x n is not α -goodby Corollary 5.7, hence neither is z . Thus there exist β ∈ s ( z ) such that β = α and( β, α ) E . Further, as s ( x n ) = α , we have α s ( y l ′ +1 ◦ · · · ◦ y m ◦ x n ) and so α s ( z ). By Lemma 5.11, β is in the support of the reduced form of x ◦ y ◦ · · · ◦ y l ′ ◦ z , but ( β, α ) E ,implying that x ◦ y ◦ · · · ◦ y l ′ ◦ z is not α -good, and hence neither is x , a contradiction.Case (2): x ◦ y is not a reduced form and s ( x ) = α . By Remark 3.6, ( β, α ) ∈ E for all β ∈ s ( y ◦ · · · ◦ y l − ), and so[ x ] = [ x ◦ y ◦ x n ] = [ x ◦ y ◦ · · · ◦ y m ◦ x n ] = [ y ◦ · · · ◦ y l − ◦ x y l ◦ y l +1 ◦ · · · ◦ y m ◦ x n ] . Notice that y ◦ · · · ◦ y l − ◦ x y l is α -good. By α -amenability y l ◦ y l +1 ◦ · · · ◦ y m ◦ x n is not α -good. As y l is α -good, we deduce that y l +1 ◦ · · · ◦ y m ◦ x n is not α -good by Corollary 5.7,so that y ◦ · · · y l − ◦ x y l ◦ y l +1 ◦ · · · ◦ y m ◦ x n is not α -good, a contradiction.Case (3): x ◦ y is not a reduced form and s ( x ) = α . Then, by Remark 3.6, thereexists some 1 ≤ j ≤ m , j / ∈ T , such that s ( y j ) = s ( x ) and ( s ( x ) , s ( y k )) ∈ E for all1 ≤ k ≤ j −
1. We consider two sub-cases.Case (3)(a): j < l ′ . Let w = y ◦ · · · ◦ y j − ◦ x y j ◦ y j +1 ◦ · · · ◦ y l ′ . Let w ′ = w if x y j is not an identity, and otherwise let w ′ = y ◦ · · · ◦ y j − ◦ y j +1 ◦ · · · ◦ y l ′ , so that w ′ is areduced form of w . Let z be a reduced form of y l ′ +1 ◦ · · · ◦ y m ◦ x n . Then [ w ′ ◦ z ] = [ x ].Since y l ′ ◦ y l ′ +1 ◦ · · · ◦ y m ◦ x n is not α -good, we deduce y l ′ +1 ◦ · · · ◦ y m ◦ x n is not α -goodby Corollary 5.7, so that neither is z . Hence there exists β ∈ s ( z ) such that β = α and( β, α ) E . Further, as s ( x n ) = α , we have α s ( y l ′ +1 ◦ · · · ◦ y m ◦ x n ) and so α s ( z ). Itthen follows from Lemma 5.11 that β is in the support of the reduced form of w ′ ◦ z . But,( β, α ) E , implying that w ′ ◦ z and hence x is not α -good, a contradiction.Case (3)(b): j > l ′ . Notice first that [ y l ′ +1 ◦ · · · ◦ y j − ◦ x y j ◦ y j +1 ◦ · · · ◦ y m ◦ x n ] = [ w ]where w = x y j ◦ y l ′ +1 ◦ · · · ◦ y j − ◦ y j +1 ◦ · · · ◦ y m ◦ x n . We claim that w is not α -good.As s ( y j ) = s ( x ) and ( s ( x ) , s ( y l ′ )) ∈ E , we have ( s ( y j ) , α ) ∈ E , so that y j is α -good.By α -amenability, y l ′ ◦ y l ′ +1 ◦ · · · ◦ y m ◦ x n is not α -good and so y l ′ +1 ◦ · · · ◦ y m ◦ x n isnot α -good by Corollary 5.7. As [ y l ′ +1 ◦ · · · ◦ y j − ◦ y j ◦ y j +1 ◦ · · · ◦ y m ◦ x n ] = [ w ′ ] where w ′ = y j ◦ y l ′ +1 ◦ · · · ◦ y j − ◦ y j +1 ◦ · · · ◦ y m ◦ x n we deduce that w ′ is not α -good and so y l ′ +1 ◦ · · · ◦ y j − ◦ y j +1 ◦ · · · ◦ y m ◦ x n is not α -good by Corollary 5.7; similarly, as x y j is α -good, we deduce x y j ◦ y l ′ +1 ◦ · · ·◦ y j − ◦ y j +1 ◦ · · · y m ◦ x n is not α -good. Let z be a reducedform of x y j ◦ y l ′ +1 ◦ · · · ◦ y j − ◦ y j +1 ◦ · · · y m ◦ x n and notice α / ∈ s ( z ). As z is not α -good,there is β ∈ s ( z ) such that β = α and ( β, α ) E . Consider the word v = y ◦ · · · ◦ y l ′ ◦ z .Clearly [ x ] = [ v ]. By Lemma 5.11, β is in the support of the reduced form of v and hencethat of x . But ( β, α ) E , contradicting x being α -good.We conclude that x ◦ · · · ◦ x n − is α -absorbing, thus completing the proof. (cid:3) Corollary 5.13.
Let α ∈ V and x = x ◦ · · · ◦ x n ∈ X ∗ be α -amenable with s ( x n ) = α ,and let β ∈ V . Then x ◦ · · · ◦ x n − ◦ β is also α -amenable.Proof. Certainly 1 β is α -good, as its unique reduced form is ǫ . Since s ( x n ) = α and x is α -good, two applications of Corollary 5.7 give that x ◦ · · · ◦ x n − ◦ β is α -good. Supposethat n ≥ s ( x k ) = α where 2 ≤ k ≤ n −
1. By α -amenability, x k ◦ · · · ◦ x n is not α -good, but as x n is α -good, two applications of Corollary 5.7 give that x k ◦ · · · ◦ x n − ◦ β is not α -good. Therefore x ◦ · · · ◦ x n − ◦ β is α -amenable. (cid:3) We have been working towards the following:
RAPH PRODUCTS OF LEFT ABUNDANT MONOIDS 21
Proposition 5.14.
Let α ∈ V and x = x ◦ · · · ◦ x n ∈ X ∗ be α -amenable. Then the factor x ◦ · · · ◦ x n − is α -absorbing.Proof. The result is true when s ( x n ) = α , by Lemma 5.12. Suppose that s ( x n ) = α . ByCorollary 5.13, x ◦ · · · ◦ x n − ◦ β is α -amenable, for any β ∈ V . Since | V | ≥
2, taking β = α Lemma 5.12 tells us that x ◦ · · · ◦ x n − is α -absorbing. (cid:3) Corollary 5.15.
Let α ∈ V and x = x ◦ · · · ◦ x n ∈ X ∗ be α -amenable. (i) If s ( x ) = s ( x n ) = α , then for all β in the support of the reduced form of x ◦ · · ·◦ x n − we have β = α and ( α, β ) ∈ E . (ii) If s ( x ) = α, s ( x n ) = α , then for all β in the support of the reduced form of x ◦· · ·◦ x n we have β = α and ( α, β ) ∈ E .Proof. Clearly we may assume that n >
2. Let y be a reduced form of x ◦ · · · ◦ x n − , sothat α / ∈ s ( y ) by Proposition 5.14.(i) Clearly, the result is true when y = ǫ , so we assume that y = ǫ . Let w be a reducedform of x ◦ y . It follows from Corollary 3.11 that s ( y ) ⊆ s ( w ). Further, by the dual ofCorollary 3.11, s ( y ) is contained in the support of the reduced form of w ◦ x n . As x is α -good, so are x ◦ y ◦ x n and w ◦ x n , implying ( α, β ) ∈ E for all β ∈ s ( y ).(ii) Let w be a reduced form such that [ w ] = [ y ◦ x n ] = [ x ◦ · · · ◦ x n ]. Since s ( x n ) = α ,we deduce that α / ∈ s ( w ). Let v be a reduced form of x ◦ w . Since x is α -good and[ v ] = [ x ◦ w ] = [ x ], we have that v is α -good, so that β = α or ( β, α ) ∈ E for all β ∈ s ( v ).Further, as s ( x ) = α but α s ( w ), we have s ( w ) ⊆ s ( v ) by Corollary 3.11, so that( β, α ) ∈ E for all β ∈ s ( w ). (cid:3) In what follows we use the foregoing analysis to allow us to factorise elements of
G P ina way that will enable us to achieve the aim of this section. First, another definition.
Definition 5.16.
Let x = x ◦ · · · ◦ x n ∈ X ∗ and α ∈ V . We define a set N α ( x ) = { k ∈ { , · · · , n } : s ( x k ) = α and x k ◦ · · · ◦ x n is α -good } . We will show that for a word x as in Lemma 5.16 we can move the letters indexedby elements of N α ( x ) to the right of x (maintaining their order). Where convenient, insituations where the enumeration of indices is particularly involved, and where there is nodanger of ambiguity, we may identify N α ( x ) with { x k : k ∈ N α ( x ) } .Notice that N α ( x ) may be empty and, in particular, N α ( ǫ ) = ∅ . Further, s ( x n ) = α if andonly if n ∈ N α ( x ). If l, k ∈ N α ( x ) with l < k , there may exist some l < j < k with s ( x j ) = α such that j N α ( x ). For example, suppose that n = 6, s ( x ) = s ( x ) = s ( x ) = s ( x ) = α ,and s ( x ) = s ( x ) = β where α = β , ( α, β ) E , x x = 1 α , x , x / ∈ I and x x = 1 β .Then N α ( x ) = { , } . This also provides an example of an α -amenable word. Lemma 5.17.
Let α ∈ V and x = x ◦ · · · ◦ x n ∈ X ∗ with s ( x n ) = α . Write N α ( x ) = { l , · · · , l r : 1 ≤ l < · · · < l r = n } . Then [ x ] = [ x ′ ][ x l ◦ · · · ◦ x l r ] where x ′ is the word obtained from x ◦ · · · ◦ x n by deleting the letters x l , · · · , x l r .Further, if z is a word obtained from x by replacing x l , · · · , x l r by letters z l , · · · , z l r ∈ M α , respectively, we have [ z ] = [ x ′ ][ z l ◦ · · · ◦ z l r ] . Proof.
Let 1 ≤ k ≤ r − x l k ◦ · · · ◦ x l k +1 is α -good. Wenow claim that x l k ◦ · · · ◦ x l k +1 is α -amenable.Clearly, x l k ◦ · · · ◦ x l k +1 is α -amenable if either l k +1 = l k + 1 or l k +1 > l k + 1 and thereexists no l k < j < l k +1 such that s ( x j ) = α . Suppose now that there exists l k < j < l k +1 such that s ( x j ) = α . Since j N α ( x ), the word x j ◦ · · · ◦ x n is not α -good. On the otherhand, we know x l k +1 +1 ◦ · · · ◦ x n is α -good, giving that x j ◦ · · · ◦ x l k +1 is not α -good byLemma 5.6, and hence x l k ◦ · · · ◦ x l k +1 is α -amenable.For each k in the range above let w k be a reduced form of x l k +1 ◦ · · · ◦ x l k +1 − . ByCorollary 5.15, since x l k ◦ · · · ◦ x l k +1 is α -amenable, for any β ∈ s ( w k ) we have β = α and( β, α ) ∈ E . Further,[ x ] = [ x ◦ · · · ◦ x l r ] = [ x ◦ · · · ◦ x l ◦ w ◦ x l ◦ · · · ◦ x l r − ◦ w r − ◦ x l r ] , so that [ x ] = [ y ′ ◦ x l ◦ · · · ◦ x l r ] = [ y ′ ][ x l ◦ · · · ◦ x l r ] = [ x ′ ][ x l ◦ · · · ◦ x l r ]where y ′ is the word obtained form x ◦ · · · ◦ x l ◦ w ◦ x l ◦ · · · ◦ x l r − ◦ w r − ◦ x l r by deleting x l , · · · , x l r and x ′ is the word obtained from x ◦ · · · ◦ x n by deleting x l , · · · , x l r .Suppose now that z is a word obtained from x by replacing x l , · · · , x l r by letters z l , · · · , z l r ∈ M α , respectively. Since [ y ′ ] = [ x ′ ] we have[ z ] = [ y ′ ◦ z l ◦ · · · ◦ z l r ] = [ y ′ ][ z l ◦ · · · ◦ z l r ] = [ x ′ ][ z l ◦ · · · ◦ z l r ] . (cid:3) We now remove the restriction that s ( x n ) = α in Lemma 5.17. Lemma 5.18.
Let α ∈ V and x = x ◦ · · · ◦ x n ∈ X ∗ . Write N α ( x ) = { l , · · · , l r : 1 ≤ l < · · · < l r ≤ n } . Then [ x ] = [ x ′ ][ x l ◦ · · · ◦ x l r ] where x ′ is the word obtained from x ◦ · · · ◦ x n by deleting the letters x l , · · · , x l r .Further, if z is a word obtained from x by replacing x l , · · · , x l r by letters z l , · · · , z l r ∈ M α , respectively, we have [ z ] = [ x ′ ][ z l ◦ · · · ◦ z l r ] . Proof.
We are done with the case where s ( x n ) = α , by Lemma 5.17. Suppose now that s ( x n ) = α , and so l r = n . Let p = x ◦ · · · ◦ x l r . Applications of Lemma 5.6 that are nowstandard yield N α ( p ) = { l , · · · , l r } . By Lemma 5.17, [ p ] = [ p ′ ◦ x l ◦ · · · ◦ x l r ] where p ′ isthe word obtained from p by deleting letters x l , · · · , x l r . We now have[ x ] = [ p ◦ x l r +1 ◦ · · · ◦ x n ] = [ p ′ ◦ x l ◦ · · · ◦ x l r ◦ x l r +1 ◦ · · · ◦ x n ] . RAPH PRODUCTS OF LEFT ABUNDANT MONOIDS 23
To show the required result, we now consider the α -good word x l r ◦ · · · ◦ x n . We nowclaim that it is α -amenable. Clearly, we are done with the cases where either n = l r + 1or n > l r + 1 and there exists no l r < j < n such that s ( x j ) = α . Suppose therefore thatthere exists l r < j < n such that s ( x j ) = α . As j N α ( x ), we have that x j ◦ · · · ◦ x n isnot α -good, and so x l r ◦ · · · ◦ x n is α -amenable. Let q be a reduced form of x l r +1 ◦ · · · ◦ x n .Since x l r ◦ · · · ◦ x n is α -amenable and s ( x n ) = α , we have that β = α and ( α, β ) ∈ E forall β ∈ s ( q ) by Corollary 5.15. Therefore,[ x ] = [ p ′ ◦ x l ◦ · · · ◦ x l r ◦ x l r +1 ◦ · · · ◦ x n ] = [ p ′ ◦ x l ◦ · · · ◦ x l r ◦ q ] = [ p ′ ◦ q ◦ x l ◦ · · · ◦ x l r ] . Since [ p ′ ◦ q ] = [ x ′ ], we have [ x ] = [ x ′ ][ x l ◦ · · · ◦ x l r ] . Suppose now that z is a word obtained from x by replacing x l , · · · , x l r by letters z l , · · · , z l r from M α , respectively. Clearly, z = z ′ ◦ x l r +1 ◦ · · · ◦ x n where z ′ is the word obtained from p by replacing x l , · · · , x l r by z l , · · · , z l r ∈ M α . We have shown that N α ( p ) = { l , · · · , l r } and so from Lemma 5.17 we have [ z ′ ] = [ p ′ ◦ z l ◦ · · · ◦ z l r ] . Then[ z ] = [ z ′ ◦ x l r +1 ◦ · · · ◦ x n ] = [ p ′ ◦ z l ◦ · · · ◦ z l r ◦ q ] = [ p ′ ◦ q ◦ z l ◦ · · · ◦ z l r ] = [ x ′ ][ z l ◦ · · · ◦ z l r ] . (cid:3) The reader should note that we are not claiming that the maps φ α and ψ α in Lemma 5.19are morphisms. Lemma 5.19.
Let α ∈ V . Then the maps φ α : X ∗ −→ G P and ψ α : X ∗ −→ G P defined by xφ α = [ x l ◦ · · · ◦ x l r ] and xψ α = [ x m ◦ · · · ◦ x m t ] where x = x ◦ · · · ◦ x n , with N α ( x ) = { l , · · · , l r } , ≤ l < · · · < l r ≤ n and { m , · · · , m t } = { , · · · , n }\ N α ( x ) , ≤ m < · · · < m t ≤ n, induce maps φ α : G P −→ G P and ψ α : G P −→ G P defined by [ x ] φ α = xφ α and [ x ] ψ α = xψ α . Further, [ x ] = ( xψ α )( xφ α ) .Proof. To show that φ α and ψ α are well defined we need to show that R ♯ ⊆ ker φ α and R ♯ ⊆ ker ψ α . Let L be the binary relation on X ∗ defined by L = { ( y ◦ a ◦ z, y ◦ b ◦ z ) : y, z ∈ X ∗ , ( a, b ) ∈ R } . Since R ♯ is the transitive closure of L , and ker φ α and ker φ α are, of course, equivalencerelations, it suffices to show that L ⊆ ker φ α and L ⊆ ker ψ α . This can be seen in a routinemanner by using Corollary 5.7 and considering ( a, b ) ∈ R id , R v and R e in turn. It follows from Lemma 5.18 that [ x ] = ( xψ α )( xφ α ). (cid:3) Proposition 5.20.
Let z = z ◦ · · · ◦ z n ∈ X ∗ such that s ( z ) is a complete subgraph suchthat s ( z j ) = s ( z k ) for any ≤ j < k ≤ n . Suppose that z k R ∗ z ′ k in M s ( z k ) for ≤ k ≤ n and put z ′ = z ′ ◦ · · · ◦ z ′ n . Then [ z ] R ∗ [ z ′ ] in G P .Proof.
Let x = x ◦ · · · ◦ x m , y = y ◦ · · · ◦ y h ∈ X ∗ be such that [ x ][ z ] = [ y ][ z ]. We proceedby induction on n to show [ x ][ z ′ ] = [ y ][ z ′ ]. Clearly, the result is true when n = | z | = 0,i.e. z = ǫ = z ′ . Suppose now that n > z with | z | < n .Let s ( z ) = α . Then s ( z k ) = α and ( α, s ( z k )) ∈ E for all 1 < k ≤ n , so that certainly z is α -good. Suppose that N α ( x ◦ z ) = { r , · · · , r l } and N i ( y ◦ z ) = { d , · · · , d t } where r < · · · < r l and d < · · · < d t . We have r l = m + 1 and d t = h + 1. By Lemma 5.18,[ x ◦ z ] = [ x ′ ◦ z ◦ · · · ◦ z n ][ x r ◦ · · · ◦ x r l − ◦ z ]and [ y ◦ z ] = [ y ′ ◦ z ◦ · · · ◦ z n ][ y d ◦ · · · ◦ y d t − ◦ z ] . By replacing the ( m + 1)-th place letter z by z ′ in x ◦ z , we have[ x ◦ z ′ ◦ · · · ◦ z n ] = [ x ′ ◦ z ◦ · · · ◦ z n ][ x r ◦ · · · ◦ x r l − ◦ z ′ ]by Lemma 5.18. Similarly,[ y ◦ z ′ ◦ · · · ◦ z n ] = [ y ′ ◦ z ◦ · · · ◦ z n ][ y d ◦ · · · ◦ y d t − ◦ z ′ ] . On the other hand, by applying the maps φ α and ψ α to each side of [ x ◦ z ] = [ y ◦ z ], wehave[ x ′ ◦ z ◦ · · · ◦ z n ] = [ y ′ ◦ z ◦ · · · ◦ z n ] , and [ x r ◦ · · · ◦ x r l − ◦ z ] = [ y d ◦ · · · ◦ y d t − ◦ z ] . Using Remark 2.4, the latter gives x r · · · x r l − z = y d · · · y d t − z . As z R ∗ z ′ in M α , wehave x r · · · x r l − z ′ = y d · · · y d t − z ′ so that [ x r ◦ · · · ◦ x r l − ◦ z ′ ] = [ y d ◦ · · · ◦ y d t − ◦ z ′ ]. Therefore,[ x ◦ z ′ ◦ · · · ◦ z n ] = [ y ◦ z ′ ◦ · · · ◦ z n ]and so [ x ◦ z ′ ][ z ◦ · · · ◦ z n ] = [ y ◦ z ′ ][ z ◦ · · · ◦ z n ] . Our inductive assumption now gives[ x ][ z ′ ◦ z ′ ◦ · · · ◦ z ′ n ] = [ x ◦ z ′ ][ z ′ ◦ · · · ◦ z ′ n ] = [ y ◦ z ′ ][ z ′ ◦ · · · ◦ z ′ n ] = [ y ][ z ′ ◦ z ′ ◦ · · · ◦ z ′ n ] . The result follows by induction. (cid:3)
RAPH PRODUCTS OF LEFT ABUNDANT MONOIDS 25
Proposition 5.21.
Let u ∈ X ∗ and let [ u ] = [ a ][ v ] where a, v ∈ X ∗ are such that all letterscontained in a are left invertible, and v = v ◦· · ·◦ v m is a left Foata normal form with blocks v k , ≤ k ≤ n , such that v contains no left invertible letters. Let v = z ◦ · · · ◦ z s ∈ X ∗ .Suppose that for each ≤ j ≤ s an idempotent z + j ∈ M s ( z j ) is chosen such that z + j R ∗ z j in M s ( z j ) , and put v +1 = z +1 ◦ · · · ◦ z + s . Let [ a ′ ] be a left inverse of [ a ] in G P . Then [ u ] R ∗ [ a ][ v +1 ][ a ′ ] and [ a ][ v +1 ][ a ′ ] is idempotent.Proof. Under the conditions of the hypothesis, it follows from (1) of Proposition 4.5 that[ v ] R ∗ [ v ] and then from Proposition 5.20 that [ v ] R ∗ [ v +1 ]. Since [ a ′ ][ a ] = [ ǫ ], we have[ a ′ ] R [ ǫ ] and so certainly [ a ′ ] R ∗ [ ǫ ]. Then[ u ] = [ a ][ v ] R ∗ [ a ][ v +1 ] R ∗ [ a ][ v +1 ][ a ′ ] , using the fact that R ∗ is a left congruence. Further, [ a ][ v +1 ][ a ′ ] is idempotent by Lemma5.2. (cid:3) The main result of our paper now follows.
Theorem 5.22.
The graph product
G P = G P (Γ , M ) of left abundant monoids M = { M α : α ∈ V } with respect to Γ is left abundant.Proof. Let [ u ] ∈ G P . By Lemma 4.2 we are guaranteed a decomposition of u as inProposition 5.21. The result now follows from the assumption that each vertex monoid isleft abundant. (cid:3) Of course, the left-right dual of Theorem 5.22 holds, and hence one may also deducethat the graph product of abundant monoids is abundant. A consequence is worth statingseparately.
Corollary 5.23.
The graph product
G P = G P (Γ , M ) of regular monoids M = { M α : α ∈ V } with respect to Γ is abundant. Graph Products of left Fountain Monoids are left Fountain
We now discuss the left Fountainicity of the graph product
G P = G P (Γ , M ) of leftFountain monoids M = { M α : α ∈ V } with respect to Γ.Our strategy is as follows. We know from Lemma 4.2 that any element of G P hasreduced form a ◦ x where the letters of a are all left invertible, x = x ◦ · · · ◦ x n is a leftFoata normal form with blocks x i , 1 ≤ i ≤ n , such that x contains no left invertible letters.From Proposition 4.5 we then have [ a ◦ x ] R ∗ [ a ◦ x ] and so certainly [ a ◦ x ] e R [ a ◦ x ]. Wetake an idempotent of G P in standard form u and examine the reduction processes forthe word u ◦ a ◦ x in the case [ u ◦ a ◦ x ] = [ a ◦ x ]. This eventually enables us to showthat [ u ◦ a ◦ x ] = [ a ◦ x ] if and only if [ u ◦ a ◦ x ] = [ a ◦ x ] where x is obtained from x by replacing each letter by an idempotent in the same e R -class in the relevant vertexmonoid. Hence [ a ◦ x ] e R [ a ◦ x ] but then with [ a ′ ] being a left inverse for [ a ] we arrive at[ a ◦ x ] e R [ a ◦ x ◦ a ′ ]. The latter element is clearly idempotent. To proceed, we rely on the analysis of α -good suffices of words provided in Section 5. Inaddition, we need some further analysis of the way in which the product of two reducedwords reduces in G P .It is worth remarking that if every vertex monoid has the property that left invertibleelements are also right invertible, the our arguments would need to be less delicate. Since,in that case, [ u ◦ a ◦ x ] = [ a ◦ x ] if and only if [ a ′ ◦ u ◦ a ◦ x ] = [ x ], and the fact that s ( x ) is complete then makes the subsequent analysis somewhat easier.Lemma 3.10 shows the different ways in which multiplying a reduced word by p ∈ X \ I leads to a reduced word. In some cases, we need to delete a letter of I , that is, use Step(id) of Definition 3.1; in other cases, we need only Steps (v) and (e). This leads to thefollowing notion. Definition 6.1.
Let x = x ◦ · · · ◦ x n , y = y ◦ · · · ◦ y m ∈ X ∗ be reduced words. We saythat x ◦ y is S -reducible if in reducing x ◦ y to a reduced form we only use Steps (v) and(e) in Definition 3.1.We use the term ‘ S -reducible’ since using Steps (v) and (e) would be allowed in thecorresponding notion of a semigroup graph product: see Section 7. Lemma 6.2.
Let x = x ◦ · · · ◦ x n , y = y ◦ · · · ◦ y m ∈ X ∗ be reduced words. Suppose that x ◦ y is S -reducible. Then x ◦ y shuffles to p ◦ · · · ◦ p n ◦ y ′ and has reduced form q ◦ · · · ◦ q n ◦ y ′ where for all ≤ j ≤ n , q j = x j = p j or p j = x j ◦ y r j and q j = x j y r j for some distinctindices r j ∈ { , · · · , m } , and y ′ ∈ X ∗ is the word obtained from y by deleting the letters y r j .Proof. We use induction on the length n of x . Clearly, the result is true for n = 1 byLemma 3.10. Suppose that n > x of lengthstrictly less than n . Let x ′ = x ◦ · · · ◦ x n . Clearly, x ′ ◦ y is also S -reducible, and so x ′ ◦ y shuffles to u = p ◦ · · · ◦ p n ◦ y ′ and has a reduced form u = q ◦ · · · ◦ q n ◦ y ′ where for all 2 ≤ j ≤ n , q j = x j = p j or p j = x j ◦ y r j and q j = x j y r j for some distinctindices r j ∈ { , . . . , m } , and y ′ is the word obtained from y by deleting the letters y r j .Now consider the words w = x ◦ u and w = x ◦ u . If w is reduced then we are done, with p = q = x . Suppose therefore that w is notreduced. Since s ( q j ) = s ( x j ) for all 2 ≤ j ≤ n , the word x ◦ q ◦ · · · ◦ q n is reduced byRemark 3.5. So, there must exist some letter y t in y ′ with s ( x ) = s ( y t ) that can be shuffled RAPH PRODUCTS OF LEFT ABUNDANT MONOIDS 27 to the front of both u and u . Clearly t is distinct from any existing r j ; we put r = t . As x ◦ y is S -reducible, x y r is not an the identity. Therefore, w shuffles to p ◦ p ◦ · · · ◦ p n ◦ y ′′ and, from Lemma 3.10, has reduced form q ◦ q ◦ · · · ◦ q n ◦ y ′′ where p = x ◦ y r and q = x y r and y ′′ is the word obtained by deleting y r from y ′ . (cid:3) Corollary 6.3.
Let α ∈ V and let x, y ∈ X ∗ be reduced words such that x is not α -goodbut x ◦ y is α -good. Then x ◦ y is not S -reducible.Proof. Let x, y be as given. If x ◦ y is S -reducible, then s ( x ) is a subset of the supportof the reduced form of x ◦ y , by Lemma 6.2. Since x is not α -good, neither is x ◦ y , acontradiction. (cid:3) In what follows, we use u = b ◦ e ◦ b ′ to denote a standard form of an idempotent[ u ] ∈ G P , as described in Definition 5.1. We use a ◦ x to denote a word in X ∗ satisfyingthe following conditions:(a) a = a ◦ · · · ◦ a l is a reduced word such that all letters in a are left invertible;(b) x = x ◦ · · · ◦ x k such that s ( x ) is complete and s ( x j ) = s ( x t ) for all 1 ≤ j < t ≤ k ;(c) there exists no j with 1 ≤ j ≤ l such that ( s ( a j ) , s ( a t )) ∈ E for all j + 1 ≤ t ≤ l and s ( a j ) ∈ s ( x ).The reader by now might think we should assume a ◦ x is reduced and no letter in x isleft invertible. However, we need this rather looser set up. The reason for this will becomeapparent later, when we apply Lemma 6.8 iteratively in Corollary 6.9. Lemma 6.4.
Let a ◦ x be defined as above. Then (i) for any y = y ◦ · · · ◦ y k ∈ X ∗ such that s ( y j ) = s ( x j ) for all ≤ j ≤ k , a ◦ y is of thesame form as a ◦ x ; (ii) a ◦ x ′ is a reduced form of a ◦ x , where x ′ is the word obtained from x by deleting allletters in x which are identities; (iii) for each α ∈ s ( x ) , N α ( a ◦ x ) contain the unique letter x j in x such that s ( x j ) = α .Proof. (i) and (ii) are clear.(iii) Let α ∈ s ( x ) and let j be the unique index guaranteed by (b) such that s ( x j ) = α .Since s ( x ) is complete, x j ∈ N α ( a ◦ x ). Suppose (with some abuse of notation) that a h ∈ N α ( a ◦ x ). Then s ( a h ) = α and a h ◦ · · · ◦ a l ◦ x is α -good, hence so is its reduced form a h ◦ · · · ◦ a l ◦ x ′ . Let h ≤ t ≤ l be the largest such that s ( a t ) = α . Then ( s ( a t ) , s ( a r )) ∈ E for all t + 1 ≤ r ≤ l , contradicting (c). Thus, N α ( a ◦ x ) = { x j } . (cid:3) In Corollary 6.6, and Lemmas 6.7 and 6.8 let a ◦ x and u = b ◦ e ◦ b ′ be defined as abovesuch that [ u ][ a ◦ x ] = [ a ◦ x ], and let w = u ◦ a ◦ x . Lemma 6.5.
Suppose that u is α -good. For any j ∈ { , · · · , n } we have b ′ j ∈ N α ( u ) if andonly if b j ∈ N α ( u ) . Proof.
Using Lemma 5.6, Corollary 5.7 and Lemma 5.5, the following are equivalent b ′ j ∈ N α ( u ) b ′ j ◦ · · · ◦ b ′ is α -good b j ◦ · · · ◦ b is α -good b ◦ · · · ◦ b j is α -good b j +1 ◦ · · · ◦ b n ◦ e ◦ b ′ is α -good b j ◦ b j +1 ◦ · · · ◦ b n ◦ e ◦ b ′ is α -good b j ∈ N α ( u ) . (cid:3) We can now make progress in the case where s ( x ) = α and a ◦ x is α -good. Corollary 6.6.
Suppose that s ( x ) = α and a ◦ x is α -good. Then for any j ∈ { , · · · , n } we have b ′ j ∈ N α ( w ) if and only if b j ∈ N α ( w ) .Proof. Since a ◦ x is α -good, so is u ◦ a ◦ x and hence from Lemma 5.6 so is u . Moreover(with substantial abuse of notation), z ∈ N α ( u ) if and only if z ∈ N α ( w ), for any letter z of u . The result follows from Lemma 6.5. (cid:3) Without the assumption that a ◦ x is α -good, our analysis of the elements of N α ( w )becomes more delicate. We remark that in what follows, we could replace the suffix a ◦ x of w by any word v and the same argument would apply to u ◦ v as it does to w . Lemma 6.7.
Let α ∈ V . If b ′ j / ∈ N α ( w ) for all ≤ j ≤ n , then b j / ∈ N α ( w ) for all ≤ j ≤ n .Proof. If α / ∈ s ( b ) there is nothing to show. Otherwise, let h be greatest such that s ( b ′ h ) = α ,so that v = b ′ h ◦ b ′ h − ◦ · · · ◦ b ′ ◦ a ◦ x is not α -good. Suppose that there exist some b j ∈ N α ( w ), so that z = b j ◦ · · · ◦ b n ◦ e ◦ b ′ n ◦ · · · ◦ b ′ ◦ a ◦ x is α -good. Notice that j ≤ h .Suppose for contradiction that b ′ ◦ a ◦ x is not α -good. Then neither is e ◦ b ′ ◦ a ◦ x .To see this, let y = y ◦ · · · ◦ y r be a reduced form of b ′ ◦ a ◦ x , so that y is not α -good.Notice that a product pq of two elements p, q in the same vertex monoid with at least oneof p, q being a non-identity idempotent cannot be the identity, so that using Lemma 3.10iteratively we see that e ◦ y is S -reducible. It follows from Lemma 6.2 that e ◦ y reduces to q ◦ · · · ◦ q m ◦ y ′ where for all 1 ≤ t ≤ m , q t = e t or q t = e t y r t for some distinct indices r t , and y ′ is the wordobtained from y by deleting the letters y r t . Clearly, s ( y ) ⊆ s ( q ◦ · · · ◦ q m ◦ y ′ ), implyingthat q ◦ · · · ◦ q m ◦ y ′ is not α -good, and hence neither is e ◦ b ′ ◦ a ◦ x .By assumption, z ′ = b j ◦ · · · ◦ b n ◦ q ◦ · · · ◦ q m ◦ y ′ RAPH PRODUCTS OF LEFT ABUNDANT MONOIDS 29 is α -good. We next claim that it is a reduced form. Since s ( q t ) = s ( e t ) for 1 ≤ t ≤ m and b j ◦· · ·◦ b n ◦ e ◦· · ·◦ e m is a reduced form, we deduce that b j ◦· · ·◦ b n ◦ q ◦· · ·◦ q m is also reducedby Remark 3.5. Further, it is impossible to shuffle some b t ( j ≤ t ≤ n ) in z ′ and glue it tosome letter in y ′ , as this would imply that in the reduced form b j ◦ · · · ◦ b n ◦ e ◦ b ′ n ◦ · · · ◦ b ′ j we may shuffle b t and glue it to b ′ t , contradicting the fact b ◦ e ◦ b ′ is reduced. Thus z ′ isindeed reduced. Since q ◦ · · · ◦ q m ◦ y ′ is not α -good, neither is z ′ , contradicting the factthat [ z ] = [ z ′ ] and b j ∈ N α ( w ).We have shown that b ′ ◦ a ◦ x must be α -good. Since v is not α -good, there exists β = α inthe support of the reduced form of v such that ( α, β ) E . On the other hand, b ′ ◦ a ◦ x andhence b ′ n ◦ · · · ◦ b ′ h +1 ◦ v are α -good, Corollary 3.11 forces there to be some l with h < l ≤ n such that s ( b ′ l ) = β . Since s ( b ′ l ) = s ( b l ) and h ≥ j , and z ′ is a reduced form, we have that z is not α -good, which again contradicts our initial assumption that b j ∈ N α ( w ). (cid:3) We can now show that, given [ u ◦ a ◦ x ] = [ a ◦ x ], we can replace a letter of x by anycorresponding element in the same e R -class in the relevant vertex monoid. Note that itmay be we replace a letter not in I by a letter in I . It is for this reason that our set-upfor a ◦ x is so delicate. Lemma 6.8.
Let s ( x ) = α and let ˜ x = x ′ ◦ x ◦ · · · ◦ x k where x ′ ∈ M α is chosen so that x e R x ′ in M α . Then [ u ][ a ◦ x ] = [ a ◦ x ] implies that [ u ][ a ◦ ˜ x ] = [ a ◦ ˜ x ] . Proof. If a ◦ x is α -good, then by Corollary 6.6 and Lemma 6.4 (iii) N α ( w ) = { b t , · · · , b t r , e h , b − t r , · · · , b − t , x } or N α ( w ) = { b t , · · · , b t r , b − t r , · · · , b − t , x } for some 0 ≤ r ≤ n and 1 ≤ t < · · · < t r ≤ n and 1 ≤ h ≤ m . Whether or not a ◦ x is α -good, in the case where b ′ j / ∈ N α ( u ◦ a ◦ x ) for all 1 ≤ j ≤ n , we have that b j N α ( u ◦ a ◦ x )for all 1 ≤ j ≤ n , by Lemma 6.7, so that N α ( u ◦ a ◦ x ) equals either { e h , x } or { x } forsome 1 ≤ h ≤ m .In either of these two special cases, let f be the idempotent b t · · · b t r e h b − t r · · · b − t or b t · · · b t r b − t r · · · b − t ; note that we could have f = ǫ . Then by Lemma 5.18,[ u ][ a ◦ x ] = [ u ][ a ◦ x ◦ · · · ◦ x k ] = [ w ′ ][ f ◦ x ] , or [ x ] if f = ǫ, where w ′ is the word obtained from w by deleting all letters in N α ( w ) . By replacing thefirst letter x of x by x ′ in u ◦ a ◦ x , we have[ u ][ a ◦ x ′ ◦ x ◦ · · · ◦ x k ] = [ w ′ ][ f ◦ x ′ ] , or [ x ] , again by Lemma 5.18.On the other hand, by applying the maps φ α and ψ α to [ u ][ a ◦ x ] and [ a ◦ x ], we have[ w ′ ] = [( a ◦ x ) ′ ] and [ f ◦ x ] = [ x ] (if f = ǫ ) where ( a ◦ x ) ′ is the word obtained from a ◦ x by deleting the first letter x of x . The latter gives f x = x in M α (if f = ǫ ). If f ∈ M α is idempotent, then given x e R x ′ in M α , we have f x ′ = x ′ . Therefore[ u ][ a ◦ x ′ ◦ x ◦ · · · ◦ x k ] = [( a ◦ x ) ′ ][ x ′ ] = [ a ◦ x ′ ◦ x ◦ · · · ◦ x k ]so that [ u ][ a ◦ ˜ x ] = [ a ◦ ˜ x ] . We now proceed by induction on the length of u . If | u | = 1, then u = e for some non-identity idempotent e from a vertex monoid. Clearly b ′ j / ∈ N α ( w ) for all j ∈ { , . . . , n } sothat if [ u ][ a ◦ x ] = [ a ◦ x ], then [ u ][ a ◦ ˜ x ] = [ a ◦ ˜ x ], by the above.Suppose now that 1 < | u | and the result is true for all idempotents having length lessthan u , when written in standard form. By the above we only need to consider the casewhere a ◦ x is not α -good and there exists some b ′ j ∈ N α ( u ◦ a ◦ x ). We pick j to be smallestsuch index. Then b ′ j − ◦ · · · ◦ b ′ ◦ a ◦ x is α -good. Since x is α -good we have b ′ j − ◦ · · · ◦ b ′ ◦ a is α -good and since a ◦ x is not α -good we also have that a is not α -good. We see fromCorollary 6.3 that b ′ j − ◦ · · · ◦ b ′ ◦ a is not S -reducible. There must therefore be a smallest t such that b ′ t ◦ · · · ◦ b ′ ◦ a is S -reducible, but b ′ t +1 ◦ b ′ t ◦ · · · ◦ b ′ ◦ a is not. By Lemma 6.2,we know b ′ t ◦ · · · ◦ b ′ ◦ a shuffles to some p t ◦ · · · ◦ p ◦ a ′ and reduces to a reduced form q t ◦ · · · ◦ q ◦ a ′ where for all 1 ≤ r ≤ t we have q r = b ′ r = p t or p t = b ′ r ◦ a r j and q r = b ′ r a r j , for somedistinct indices r j ∈ { , . . . , l } , and a ′ is the word obtained from a by deleting the letters a r j .Now consider the reduced form of b ′ t +1 ◦ q t ◦ · · · ◦ q ◦ a ′ or, equivalently, b ′ t +1 ◦ p t ◦ · · · ◦ p ◦ a ′ . Since s ( q r ) = s ( b ′ r ) = s ( p r ) for all 1 ≤ r ≤ t and b ′ t +1 ◦ b ′ t ◦ · · · ◦ b ′ is a reduced form, we havethat b ′ t +1 ◦ q t ◦ · · · ◦ q is a reduced form. As b ′ t +1 ◦ b t ◦ · · · ◦ b ◦ a and hence b ′ t +1 ◦ q t ◦ · · · ◦ q ◦ a ′ is not S -reducible, there must be a letter a r t +1 in a ′ such that s ( b ′ t +1 ) = s ( a r t +1 ), b ′ t +1 a r t +1 is an identity and such that we must be able to shuffle a r t +1 to the front of q t ◦ · · · ◦ q ◦ a ′ .Note that we can therefore also shuffle a r t +1 to the front of p t ◦ · · · ◦ p ◦ a ′ and hence tothe front of a , and b ′ t +1 to the right of p t ◦ · · · ◦ p and hence to the right of b ′ t ◦ · · · ◦ b ′ . Wecan therefore assume that t + 1 = 1 = r t +1 so that b ′ a is an identity.We now have[ u ◦ a ◦ x ] = [ b ◦ · · · ◦ b n ◦ e ◦ b ′ n ◦ · · · ◦ b ′ ◦ a · · · ◦ a l ◦ x ] = [ a ◦ x ]so that multiplying by [ b ′ ] on the left we have(4) [ b ◦ · · · ◦ b n ◦ e ◦ b ′ n ◦ · · · ◦ b ′ ][ a · · · ◦ a l ◦ x ] = [ a ◦ · · · ◦ a l ◦ x ] . We note that a · · · ◦ a l ◦ x is a word of the correct form for us to apply our inductiveassumption, which gives us that RAPH PRODUCTS OF LEFT ABUNDANT MONOIDS 31 (5) [ b ◦ · · · ◦ b n ◦ e ◦ b ′ n ◦ · · · ◦ b ′ ][ a · · · ◦ a l ◦ ˜ x ] = [ a ◦ · · · ◦ a l ◦ ˜ x ] . Now multiplying Equation (5) by [ b ] on the left and re-instating b ′ ◦ a we obtain[ u ◦ a ◦ ˜ x ] = [ b ◦ a ◦ · · · ◦ a l ◦ ˜ x ] . But multiplying Equation (4) by [ b ] on the left and re-instating b ′ ◦ a we also obtain[ u ◦ a ◦ x ] = [ a ◦ x ] = [ b ◦ a ◦ · · · ◦ a l ◦ x ] . Let x ′ be the word obtained from x by deleting letters which are identities. Then[ u ◦ a ◦ x ′ ] = [ a ◦ x ′ ] = [ b ◦ a ◦ · · · ◦ a l ◦ x ′ ] . Since a ◦ x ′ is a reduced form by Lemma 6.4 (ii) and | b ◦ a ◦ · · · ◦ a l ◦ x ′ | = | a ◦ x ′ | , wededuce that b ◦ a ◦ · · · ◦ a l ◦ x ′ is a reduced form, so that [ a ] = [ b ◦ a ◦ · · · ◦ a l ] by Lemma3.13. Therefore, [ u ◦ a ◦ ˜ x ] = [ a ◦ ˜ x ] . (cid:3) Corollary 6.9.
Suppose that for each ≤ j ≤ k we have x ′ j ∈ M s ( x j ) such that x j e R x ′ j in M s ( x j ) . Let ¯ x = x ′ ◦ x ′ ◦ · · · ◦ x ′ k . Then [ a ◦ x ] e R [ a ◦ ¯ x ] . Proof.
Suppose that [ u ][ a ◦ x ] = [ a ◦ x ] . By Lemma 6.8, we have[ u ][ a ◦ x ′ ◦ x ◦ · · · ◦ x k ] = [ a ◦ x ′ ◦ x ◦ · · · ◦ x k ] . Clearly, we may shuffle x ′ to the back of x ′ ◦ x ◦ · · · ◦ x k and note that, by Lemma 6.4 (i), a ◦ x ◦ · · · ◦ x k ◦ x ′ is of the correct form to apply Lemma 6.8. By repeating this process,and reshuffling, we obtain [ u ][ a ◦ ¯ x ] = [ a ◦ ¯ x ].Since a ◦ ¯ x is of the same form as a ◦ x , we may show that [ u ][ a ◦ ¯ x ] = [ a ◦ ¯ x ] implies[ u ][ a ◦ x ] = [ a ◦ x ] by exactly the same arguments as above. Therefore, [ a ◦ x ] e R [ a ◦ ¯ x ]. (cid:3) We can now prove our second main result.
Theorem 6.10.
The graph product
G P = G P (Γ , M ) of left Fountain monoids M = { M α : α ∈ V } with respect to Γ is a left Fountain monoid.Proof. Let [ w ] ∈ G P . From Proposition 4.5 we may write [ w ] = [ a ][ v ], where all letterscontained in a are left invertible, a ◦ v is a reduced form, and v = v ◦ · · · ◦ v m is a left Foatanormal form with blocks v i , 1 ≤ i ≤ m , such that v contains no left invertible letters; weprefer to use v here since for convenience in this section we have been using x to denotea single block. Suppose that v = x ◦ · · · ◦ x k = x and for each j ∈ { , . . . , k } choose anidempotent x + j ∈ M s ( x j ) such that x j e R x + j in M s ( x j ) . Let v +1 = x +1 ◦ · · · ◦ x + k = ¯ x . Let [ a ′ ] be a left inverse for [ a ]. Using the fact that R and R ∗ are left congruences contained in e R , Proposition 4.5 and Corollary 6.9 give us that[ a ][ v ] e R [ a ][ v ] e R [ a ][ v +1 ] e R [ a ][ v +1 ][ a ′ ] , the final step following from the fact [ a ′ ], being right invertible, is R -related to the identityof G P . We have earlier seen that [ a ][ v +1 ][ a ′ ] is an idempotent, so that G P is indeed a leftFountain monoid. (cid:3)
Of course, the left-right dual of Theorem 6.10 holds, and hence one may also deducethat the graph product of Fountain monoids is Fountain.7.
Applications and open questions
The aim of this section is to explore some applications of Theorems 5.22 and 6.10.Further, we will discuss some open problems related to this work.We make the following observation before re-obtaining one of the main results of [24]. If M is a right cancellative monoid with identity 1 and b ∈ M is a left inverse of a ∈ M , then1 a = a a ( ba ) = ( ab ) a , giving 1 = ab , so that b is also a right inverse of a , and hence aninverse. Corollary 7.1. [24, Theorem 1.5]
The graph product
G P = G P (Γ , M ) of right cancella-tive monoids M = { M α : α ∈ V } with respect to Γ is right cancellative.Proof. In Proposition 5.21 we take z + j as the identity of the vertex monoid M s ( z j ) for each1 ≤ j ≤ s . By Lemma 4.1, bearing in mind [ a ] is a reduced form, we have that [ a ′ ] as aproduct of left inverses (hence two-sided inverses) of the letters in a . Then[ u ] R ∗ [ a ][ v +1 ][ a ′ ] = [ a ][ ǫ ][ a ′ ] = [ ǫ ] , and it follows from the comment after Remark 2.7 that G P is right cancellative. (cid:3)
Of course, the corresponding result is true for graph products of left cancellative, andcancellative, monoids.We now turn our attention to graph products of semigroups [1]. This is an essentiallydifferent construction to that for monoids, since semigroups are algebras with a differentsignature from that for monoids. The combinatorics of graph products of semigroups aresignificantly easier to handle than graph products of monoids; they behave in a way moreakin to graph monoids, where the only unit in any vertex monoid is the identity.As in the case for monoids, graph products of semigroups are given by a presentation.The difference here is that a presentation denotes a quotient of a free semigroup X + on aset X , where X + = X ∗ \ { ǫ } is the set of non-empty words on X under juxtaposition. Stillwith Γ = Γ( V, E ), let S = { S α : α ∈ V } be a set of semigroups, called vertex semigroups ,such that S β ∩ S γ = ∅ for all β = γ ∈ V . Definition 7.2.
The graph product
G PS = G PS (Γ , S ) of S with respect to Γ is definedby the presentation G PS = h X | R s i RAPH PRODUCTS OF LEFT ABUNDANT MONOIDS 33 where X = S α ∈ V S α and R s = R v ∪ R e , with R v and R e as in Definition 2.1.As before, identifying a relation in R s with a pair in X + × X + , we have G PS = X + / ( R s ) ♯ where ( R s ) ♯ is the congruence on X + generated by R s .Note that, in Definition 7.2, even if S α and S β are monoids for some α, β ∈ V , we do notidentify their identities in G PS . We denote the ( R s ) ♯ -class of x ◦ · · · ◦ x n ∈ X + in G PS by ⌊ x ◦· · ·◦ x n ⌋ . As we remarked in Section 1, graph products of semigroups do not possessthe complexities existing for monoid (or, indeed, group) graph products. Essentially, thisis because (with obvious notation), for words x, y ∈ X + we have s ( x ) ⊆ s ( w ) for any word w such that ⌊ w ⌋ = ⌊ xy ⌋ or ⌊ yx ⌋ . Moreover, if x is of minimal length in its ( R s ) ♯ -class,then | x | ≤ | w | . Details will appear in [1]. However, the following result will enable us touse results for graph products of monoids to deduce corresponding results for semigroups. Proposition 7.3.
Let
G PS be the graph product of semigroups S = { S α : α ∈ V } withrespect to Γ = Γ(
V, E ) . For each α ∈ V let M α be the semigroup S α with an identity α adjoined whether or not S α is a monoid and put M = { M α : α ∈ V } .Let G P be the graph product of monoids M with respect to Γ . Then the map θ : G PS −→ G P : ⌊ x ◦ . . . x n ⌋ 7→ [ x ◦ . . . ◦ x n ] is a (semigroup) embedding.Proof. For clarity here we take Y = S v ∈ V S v and X = S v ∈ V M v . Let a semigroup morphism κ : Y + → G P be defined by its action on generators as yκ = [ y ] for all y ∈ Y . We have (with slight abuseof notation) R s ⊆ R , and it follows that κ induces the semigroup morphism θ as given.We now show that θ is one-one. Let G PS be the monoid obtained from G PS byadjoining an identity 1. We define a monoid morphism ξ : X ∗ −→ G PS by its action on generators as xξ = (cid:26) ⌊ x ⌋ x ∈ Y x = 1 α for some α ∈ V. We claim that R ♯ ⊆ ker ξ .Let u, v ∈ M α for some α ∈ V . If u, v ∈ S α , then( u ◦ v ) ξ = ( uξ )( vξ ) = ⌊ u ⌋⌊ v ⌋ = ⌊ u ◦ v ⌋ = ⌊ uv ⌋ = ( uv ) ξ. If u = 1 α and v ∈ S α , then( u ◦ v ) ξ = ( uξ )( vξ ) = 1 ⌊ v ⌋ = ⌊ v ⌋ = vξ = ( uv ) ξ and dually if u ∈ S α and v = 1 α . If u = v = 1 α , then( u ◦ v ) ξ = ( uξ )( vξ ) = 11 = 1 = ( uv ) ξ. Now consider u ∈ M α , v ∈ M β with ( α, β ) ∈ E . If u = 1 α and v = 1 β , then( u ◦ v ) ξ = (1 α ◦ β ) ξ = (1 α ξ )(1 β ξ ) = 11 = (1 β ξ )(1 α ξ ) = (1 β ◦ α ) ξ = ( v ◦ u ) ξ. If u = 1 α and v ∈ S β , then ( u ◦ v ) ξ = 1 ⌊ v ⌋ = ⌊ v ⌋ v ◦ u ) ξ and dually if u ∈ S α and v = 1 β . If u ∈ S α and v ∈ S β , then( u ◦ v ) ξ = ⌊ u ⌋⌊ v ⌋ = ⌊ u ◦ v ⌋ = ⌊ v ◦ u ⌋ = ⌊ v ⌋⌊ u ⌋ = ⌊ v ◦ u ⌋ ξ. Finally, for α ∈ V , we have 1 α ξ = 1 = ǫξ .We have shown that R ⊆ ker ξ . It follows that R ♯ ⊆ ker ξ and hence ξ : G P −→ G PS , [ w ] wξ is a well defined morphism. Further, for any ⌊ w ⌋ ∈ G PS , we have ⌊ w ⌋ θξ = [ w ] ξ = wξ = ⌊ w ⌋ so that θξ = 1 G PS , and hence θ is an embedding. (cid:3) The result below will appear in [1].
Corollary 7.4.
The graph product
G PS of left abundant semigroups S = { S α : α ∈ V } with respect to Γ is left abundant.Proof. Let Y = S α ∈ V S α and X = S α ∈ V M α , where M α = S α ∪ { α } as in Proposition 7.3.Since each S α is left abundant, it is easy to check that the same is true of each M α , and,moreover, if u, v ∈ S α then u R ∗ v in S α if and only if u R ∗ v in M α .It follows from Proposition 7.3 that G PS is isomorphic to a subsemigroup N of G P ,where N = { [ x ◦ · · · ◦ x n ] : x i ∈ Y, ≤ i ≤ n } and ϕ : G PS −→ N , ⌊ x ◦ · · · ◦ x n ⌋ 7→ [ x ◦ · · · ◦ x n ]is an isomorphism.Let x = x ◦ · · · ◦ x n ∈ Y + and let v = v ◦ · · · ◦ v m ∈ X ∗ be a left Foata normal formof x with blocks v i , 1 ≤ i ≤ m . Since the only left or right invertible element of anyvertex monoid M α is 1 α , we have that v ∈ Y + and v contains no left invertible letters.Choosing v +1 ∈ Y + as in Proposition 5.21 and noticing that a = ǫ in that result, we havethat [ x ] = [ v ] R ∗ [ v +1 ] in G P and hence in N . It follows that ⌊ x ⌋ R ∗ ⌊ v +1 ⌋ in G PS . (cid:3) The proof of the following result is similar to that of Corollary 7.4.
Corollary 7.5.
The graph product
G PS of left Fountain semigroups S = { S α : α ∈ V } is a left Fountain semigroup. Of course, the right (two-sided) versions of Corollaries 7.4 and 7.5 also hold.We remarked in Section 2 that free products and restricted direct products of monoidscan be regarded as special cases of graph products of monoids. We therefore have thefollowing result.
RAPH PRODUCTS OF LEFT ABUNDANT MONOIDS 35
Corollary 7.6.
The free product
F PM and the restricted direct product Σ α ∈ V M α of leftabundant monoids (resp. left Fountain monoids) M = { M α : α ∈ V } are left abundant(resp. left Fountain). Remark 7.7.
The corresponding statement to that of Corollary 7.6 is true for semigroupsand in the right/two-sided case for both monoids and semigroups.We finish this paper by posing the following open problems. Let M be a monoid. Wehave commented that the relations R and R ∗ are left congruences on M but, in general,this need not be true of e R . Since e R being a left congruence is an important property inmany structural results for left Fountain monoids and semigroups we first pose: Question 7.8.
Let
G P = G P (Γ , M ) be a graph product of monoids M = { M α : α ∈ V } with respect to Γ, where e R is a left congruence on each M α . Is e R a left congruence on G P ?The above could first be asked in the corresponding case for semigroups, and startingwith the vertex semigroup being left Fountain.A monoid is inverse if it is regular and its idempotents commute. Inverse monoids form avariety not of monoids but of unary monoids , that is, monoids equipped with an additionalunary operation. In this case the unary operation is given by a a − , where a − is theunique element such that a = aa − a and a − = a − aa − . The notion of a graph productof inverse monoids (see [10, 14], at least for the case where the vertex monoids are free)is analogous to that for monoids and semigroups, and is obtained as a quotient of a freeinverse monoid, by relations given as for R ; from its very construction, it is inverse. Amonoid is left adequate if it is left abundant and its idempotents commute. These are thefirst non-regular analogues of inverse monoids, and form quasivarieties of unary monoids.Here the unary operation is a a + where a + is the unique idempotent in the R ∗ -classof a . We therefore ask the following question, which can be interpreted in more than oneway. Of course, one could also begin with the semigroup case. Question 7.9.
Is the graph product of left adequate monoids left adequate?Finally, we would hope that using left Foata normal forms and other reduction tech-niques developed in this article we could both find new approaches to old results (such ascalculating centralizers in graph products of groups [3]) and extend these to the monoidcase. For example, we ask:
Question 7.10.
Determine centralisers in graph products of monoids.
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School of Mathematics and Statistics, Xidian University, Xi’an 710071, P. R. China
Email address : [email protected] Department of Mathematics, University of York, Heslington, York, YO10 5DD, UK
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