# Flows on Classes of Regular Semigroups and Cauchy Categories

Authors: Suha Ahmed Wazzan
Journal: Journal of Mathematics
 (2019)

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## Abstract

We consider the structure of the flow monoid for some classes of regular semigroups (which are special case of flows on categories) and for Cauchy categories. In detail, we characterize flows for Rees matrix semigroups, rectangular bands, and full transformation semigroups and also describe the Cauchy categories for some classes of regular semigroups such as completely simple semigroups, Brandt semigroups, and rectangular bands. In fact, we obtain a general structure for the flow monoids on Cauchy categories.


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## Body

## 1. Introduction and Preliminaries

The term flow monoid first arose in unpublished typescript of Chase; in this manuscript, Chase determines the structure of the flow monoid (named it in his paperincidence monoid) and its group of units in general category theory which has categories with fixed vertices set X. In [1], Gilbert adopted the termflow monoid which connects the concepts with vector fields. Gilbert was solely interested in studying flow monoid in the theory of regular semigroup. He depicted the structure of the flow monoid in accordance to the Green relations on the semigroup. The aim of this paper is to describe flow monoids for some classes of regular semigroups. We characterize flows for Rees matrix semigroups, rectangular bands, and full transformation semigroups. We describe the Cauchy categories for some classes of regular semigroups such as completely simple semigroups, Brandt semigroups, and rectangular bands. We find the general structure of the flow monoids on Cauchy categories.A directed graph consists of two sets: a vertex set and an arrow set and two mappings:dom:E→V and the codomain operation cod:E→V. There are many possible formulations of the definition ofcategories (see, for instance, [2–4]). As in [5], we shall regard acategory as a directed graph with extra structure in the mean of abstract algebra. We start with a directed graph with the termobjectsused for the vertices, and the termarrowsused for directed edges, and which has supplementary operations:(C1)
Identity, which appoints to each objecta an arrow ida=1a:a↦a;(C2)
Composition, which designates each two arrowsf and g satisfying domg=codf with an arrow gf:domf↦codg. Composition satisfies two conditions:
(i)
Associativity: for objects and arrows such thata↦fb↦gc↦kd, we have the equality (1)kgf=kgf.(ii)
Identity: for each arrowf:a↦b, we have the composition with identity arrows 1a and 1b:f1a=f and 1bf=f.In general, a category contains objects setX and arrows set A between objects; arrows can sometimes be composed, whenever they fit together withsource andtarget maps σ,τ:A→X, and an identity map e:X→A such that xeσ=x=xeτ. Whenever aτ=bσ, every object has an identity arrow and so the composition of arrows is associative.According to [1], a flow is a set-theoratic section to the source map. A flow on a category C with vertex set X is a function φ:X→C that is a section to the source map: that is, for all x∈X,xφσ=x, let ΦC denote the set of all flows on C; then ΦC is a monoid, with composition ∗ defined by (2)xφ∗ψ=xφxφτψand identity e:x↦x.In Section3, we characterize the structure of the flow monoids for some regular semigroups such as full transformation semigroups, Rees matrix semigroups, and rectangular band.Cauchy category of a semigroupS is a small category whose vertices set is the set ES of idempotents of S. An arrow from f to e is a triple e,s,f where s∈eSf. Composition of arrows is given by (3)e,s,ff,t,g=e,st,g.The identity ate is e,e,e by [6]. Moreover, again by the same reference, the Cauchy category known asKaroubi envlopekS (or Cauchy completion or idempotent splitting) is a full subcategory of CS which is named as theSchützenberger category of S whose vertices are elements of the semigroup S. TheSchützenberger categoryCS of a semigroup S is defined as follows. The vertices set of CS is S. Also an arrow from t to s is a triple s,u,t such that u∈sS1∩S1s. The composition of arrows is given by (4)s,xt,tt,ty,r=s,xty,r,where x,y∈S1 and the identity at s is s,s,s.In Section4, we describe Cauchy categories for some small regular semigroups such as rectangular bands, Brandt semigroups, and completely simple semigroups and then find the general structure for the flow monoid on Cauchy categories. We actually prove that, in the Cauchy category ϱS, we have exactly one arrow from e to f (for all e,f∈ES) if and only if S is a rectangular band.


## 2. Flows on Regular Semigroups

LetS be aregular semigroup, so for all a∈S there exists b∈S such that aba=a and bab=b. We call b aninverse for a. An element a may have many inverses, and the set of inverses of a is denoted by Va. At the same time, anidempotent in S is an element e∈S such that e2=e. If b∈Va, then ab and ba are idempotents. Let ES be the set of idempotents. In regular semigroup, ES need not be a subsemigroup since the product of two idempotents need not be an idempotent element. If S is regular and ES is a subsemigroup of S, then S is calledorthodox (cf. [7]).Aflowα on S is a pair of functions α1,α2:ES→S with the properties(5)for  alle∈ES,the  elementsα2e∈Vα1e;and(6)e=α1eα2e.Due to [1], by letting ΦS as the set of all flows on S, one can compose flows in the following way: if α and β are flows on S, then a pair of mappings γ1,γ2:ES→S is defined as follows: (7)γ1e=α1eβ1α2eα1eandγ2e=β2eα2eα1eα2e.Thus we have the following first lemma of this section.Lemma 1.
The pair of mappingsγ1,γ2 is a flow on S.Proof.
For alle∈ES and γ2e∈Vγ1e, we have to prove that (8)γ1eγ2eγ1e=γ1eandγ2eγ1eγ2e=γ2e.
Now(9)γ1eγ2eγ1e=α1eβ1α2eα1eβ2eα2eα1e︸α2eα1eβ1α2eα1e=α1eα2eα1eα2e︸α1eβ1α2eα1eβis  a  flow  onS=α1eα2eα1e︸β1α2eα1eαis  a  flow  onS=α1eβ1α2eα1eαis  a  flow  onS=γ1e.
Similarly, we obtainγ2eγ1eγ2e=γ2e. Additionally, we also have (10)γ1eγ2e=α1eβ1α2eα1eβ2eα2eα1e︸α2e=α1eα2e︸α1eα2e︸βis  a  flow  onS=e·e=e2=eαis  a  flow  onS.
Hence the pair of mappingsγ1,γ2 is a flow on S.Define the flowγ=γ1,γ2 to be the product of the flows α=α1,α2 and β=β1,β2 in ΦS, and write γ=α∗β. Thus we get the following proposition.Proposition 2.
Φ S , ∗ is a monoid which is called the flow monoid of S.Proof.
To show thatΦS is a semigroup, we first need to prove the operation ∗ is associative. Let a,b,c be flows on S, and so each of those is a pair of mapping from ES to S. To prove a∗b∗c=a∗b∗c, let p=a∗b∗c and q=a∗b∗c. We then have pairs of maps p1,p2 and q1,q2 again from ES to S. So we have to show p1=q1 and p2=q2. Now (11)p1e=a∗b1ec1a∗b2ea∗b1e=a1eb1a2ea1ec1(b2a2ea1ea2ea1eb1a2ea1e.
Sincep is a flow, then a2ea1e=b1a2ea1eb2a2ea1e, and so in p1 we have (12)c1b2a2ea1ea2ea1eb1a2ea1e=c1b2a2ea1eb1a2ea1eb2a2ea1eb1a2ea1e=c1(b2a2ea1eb1a2ea1esincep1fistheinverseforp2fforallf∈Swhich implies that(13)p1e=a1eb1a2ea1ec1b2a2ea1eb1a2ea1ewhereas q1e=a1eb∗c1a2ea1e. Thus(14)q1e=a1eb1a2ea1ec1b2a2ea1eb1a2ea1e.Therefore (13) and (14) together imply that p1=q1. Similarly, we get p2=q2, and hence ∗ is an associative operation. Now the trivial flow ε on S is the pair of mappings id,id, where id is the identity map on ES. So, for all e∈ES, we have εie=e, where i=1,2. Then ϵ is the identity element in ΦS since (15)ε=id,id:ES→Ssuchthatε1e=e,ε2e=e,∀e∈ES.
As the next step, for alla∈ΦS, we have to show ε∗a=a∗ε=a, where a is a flow on S. In other words, ε∗a1=a1 and ε∗a2=a2. Now, from the equality ε∗a1=p1, we get (16)p1e=ε1ea1ε2eε1e=ea1e·e=ea1e2=ea1e=a1ea2ea1eby6=a1eby5.
Hence we getp1e=a1e which implies that ε∗a1=a1. With a similar calculation, from the equation ε∗a2=p2, we obtain p2e=a2e which gives ε∗a2=a2.
Thereforeε is the identity element in ΦS, and so it becomes a monoid.


## 3. The Structure of Flow Monoids on Classes of Regular Semigroups

In this section, by describing the construction of the flow monoid on a regular semigroup (which is a special case of a flow on a category), we will investigate the special classes of regular semigroups and flow monoids more precisely.In the following example, by taking into account a regular semigroupS=ℸX where X=1,2 (in fact, by [8], S is the full transformation semigroup on X), we will compute the number of elements in the flow monoid ΦS.Example 1.
Recall that the full transformation semigroupℸX on a set X consists of all mappings from X to itself under the operation composition of mappings. It is known that ℸx=nn when X=n.
By taking any elementf∈ℸx, we will first show that ℸx is actually a regular semigroup. To do that we need to find a g∈ℸx such that f=fgf and g=gfg. Since f is a map from X to itself, we can define it as fay=y where y∈imf and ay∈X. Also define g:X→X by (17)gy=ayify∈imfyify∉imf.Then (18)fgfx=fgyify=fx=fay=y=fxwhich implies fgf=f. Similarly we can get gfg=g. Hence ℸx is a regular semigroup.
Now we compute the number of elements in the flow monoidΦℸ1,2. In fact, ℸ1,2 is defined by (19)ℸ1,2=1212︸1,1222︸e2,1211︸e1︸idempotents,1221︸τ︸invertable.Recall that a∈Vb⇔aba=a,bab=b. In Table 1, we find all inverses.
It is clear from the table thatV1=1, Vτ=τ, and Ve1=Ve2=e1,e1. Now, let (20)φi:Eℸ1,2→ℸ1,2,i=1,2φ=φ1,φ2∈Φℸ1,2ifφ1xφ2x=xandφ2x∈Vφ1x
We can get a (bad) estimate of the size ofΦS here: number of maps ES→S×S without any restriction is 163-much too big. Let us look in term at the possibilities for φ1,φ2:ES→S for which element of ES.
Forx=1, we have two possibilities 1φ1x=φ2x=1 or 2φ1x=φ2x=τ.
Forx=e1, we have six possibilities for choosing two elements in S whose product is e1: (21)φ1x=1,φ2x=e1φ1x=τ,φ2x=e2φ1x=e1,φ2x=1φ1x=e1,φ2x=τ3φ1x=e1,φ2x=e14φ1x=e1,φ2x=e2
Forx=e2, as for e1, we can only choose (22)5φ1x=e2,φ2x=e26φ1x=e2,φ2x=e1
Therefore, we get eight flows onS as shown in Table 2.
A sample of the composition can be considered asφ,ψ∈ΦS,φ11=τ=φ21,  φ1e1=e1,φ2e1=e2,φ1e2=e2=φ2e2, and ψ11=τ=ψ21,ψ1e1=e1,ψ2e1=e2,ψ1e2=e2=ψ2e2. Then φ∗ψ∈ΦS satisfies (23)φ∗ψ1x=φ1xψ1φ2xφ1x,φ∗ψ2x=ψ2φ2xφ1xφ2x.For the casesx=1, we have φ∗ψ11=τψ11=τ2=1 and φ∗ψ2x=1 (same as φ∗ψ1)x=e1, we have φ∗ψ1e1=e1ψ1e2e1=e1ψ1e2=e1e2=e1and(24)φ∗ψ2e1=ψ2e2e1φ2e1=ψ2e2φ2e1=e1e2=e1x=e2, we have φ∗ψ1e2=e2ψ1e2=e2e2=e2and(25)φ∗ψ2e2=ψ2e2φ2e2=e1e2=e1.
To understand the full structure, we would need to complete the8×8 table for the composition in ΦS. Note that we have the trivial flow ε, ε1x=x,ε2x=x, and φ∗ε=ε∗φ=φ for all φ∈ΦS.Table 1
Cayley table ofℸ1,2.
ℸ1,2 1 τ e 1 e 2  1 1 τ e 1 e 2  τ τ 1 e 2 e 1  e1 e 1 e 1 e 1 e 1  e2 e 2 e 2 e 2 e 2Table 2
Flows onℸ1,2.
ϕ β γ σ ψ τ δ α 1 1 1 1 1 2 2 2 2  e1 3 4 3 4 3 4 3 4  e2 5 5 6 6 5 5 6 6In [1, Proposition 3.3], Gilbert characterized that a simplicity groupoid arises from the Nambooripad construction. In the following example, we will investigate the flow monoid on a rectangular band.Example 2.
Recall that a semigroupS is a band if every element of S is an idempotent and a rectangular band if aba=a for all a,b in S. Define a,bc,d=a,d∈X×X=S. Then S is regular and for all a,b∈S, Va,b=S, and ES=S. For flow on S, we have (26)φ1xφ2x=xandφ2x∈Vφ1xwhich is trivially true. So, for a flow is now a pair of maps [i=1,2]φi:S→S such that φ1xφ2x=x, let us suppose (27)φ1a,b=φ11a,b,φ12a,b∈Sφ2a,b=φ21a,b,φ22a,b∈S.So φ1a,bφ2a,b=φ11a,b,φ22a,b=a,b which gives φ11a,b=a,φ22a,b=b. Therefore (28)φ1a,b=a,φ12a,bandφ2a,b=φ21a,b,b,where φ12,φ21:X×X→X. So we get a flow whenever we choose any pair of mappings f,g:X→X and the flow is given by φ1a,b=a,fa,b and φ2a,b=ga,b,b, or could combine f,g into single mappings X×X→X×X and so a mapping S→S. Now suppose pairs of functions f,g:X×X→X determine the flow φ a pair of functions φi:S→S, for i=1,2, such that (29)φ1a,b=a,fa,bandφ2a,b=ga,b,band h,k:X×X→X determines the flow ψ a pair of functions ψi:S→S, for i=1,2, such that (30)ψ1a,b=a,ha,bandψ2a,b=ka,b,b.Then the corresponding functions φ∗ψ:X×X→X are defined by (31)φ∗ψ1a,b=φ1a,bψ1φ2a,bφ1a,b=a,fa,bψ1(ga,b,ba,fa,b=a,fa,b(ga,b,hga,b,fa,b=a,hga,b,fa,b.Similarly, we obtain φ∗ψ2a,b=kga,b,fa,b,b. Hence φ∗ψ is determined by a single function: X×X→X×X with the rule (32)a,b→kga,b,fa,b,hga,b,fa,b.LetS be a semigroup, I and ∧ nonempty sets, and P a matrix indexed by I and ∧ with entries pi,λ taken from S. Then the Rees matrix semigroup ([7]) MS;I;∧;P is the set I×S×∧ together with the multiplication (33)i,s,λj,t,μ=i,spλ,jt,μ.In the next example we describe flows on Rees matrix semigroups.Example 3.
The Rees matrix semigroupMS;I;∧;P is regular since, for all elements i,a,λ in this semigroup, there exist j,b,μ such that(34)i,a,λj,b,μi,a,λ=i,a,λand(35)j,b,μi,a,λj,b,μ=j,b,μ.From (34), we get (36)a=apλjbpμia→1=pλjbpμi→b=pλj-1a-1pμi-1,and from (35), we have (37)b=bpμiapλjb→1=pμiapλjb→b=pλj-1a-1pμi-1.So Vi,a,λ=j,b,μ:b=pλj-1a-1pμi-1. Hence once we have selected j and μ, b is uniquely determined.
For any elemente=i,a,λ∈MS;I;∧;P is an idempotent if (38)i,a,λi,a,λ=i,a,λ→i,apλia,λ=i,a,λ→a=pλi-1.So EMS;I;∧;P=i,pλi-1,λ:i∈I,λ∈∧. Therefore we can think of EMS;I;∧;P as being identified (once we are given P) with I×∧.
For eachi,μ∈I×∧, a flow φ∈ΦMS;I;∧;P determines a pair of inverse elements i,a,λ,j,b,μ (in which b=pλj-1a-1pμi-1) such that (39)i,a,λ,j,b,μ=i,apλjb,μ=i,pμi-1,μ.Once i is determined, we have free choice for a,λ,j and μ is determined, and so b is determined. Hence flow is a function φ:I×∧→S×∧×I. We can now think of a flow φ as two functions: (40)δφ:I×∧→Si,μ↦aπφ:I×∧→I×∧i,μ↦j,λ.
We determine the composition of functions that corresponds to the composition of flows. Take two flowsφ1 and φ2 with corresponding functions (41)δ1,δ2′:I×∧→Sandπ1,π2′:I×∧→I×∧.We have (42)φ1∗φ2i,μ=φ1i,μ·φ2τφ1i,μ=φ1i,μ·φ2π1i,μ.Now φ1i,μ=i,δ1i,μ,λ1,j1,b1,μ, where π1i,μ=j,λ1 and (43)φ2π1i,μ=φ2j1λ1=j1,δ2j1,λ1,λ2,j2,b2,λ1,where π2j1,λ1=j2,λ2. So (44)φ1∗φ2i,μ=i,δ1i,μ,λ1j1,δ2j1,λ1,λ2,j2,b2,λ1j1,b1,μ=i,δ1i,μpλ1j1δ2j1,λ1,λ2,j2,b2pλ1j1b1,μ.The functions δ and π corresponding to φ1∗φ2 are (45)δi,μ=δ1i,μpλ1j1δ2j1,λ1=δ1i,μpλ1j1δ2π1i,μ=δ1i,μpπ1¯i,μδ2π1i,μπ1¯i,μ=λ1,j1-switchingorderofcomponentsπi,μ=j2,λ2=π2j1,λ1=π2π1i,μThus π=π2∘π1, and so, for any flow monoid ΦC on any category C, we get a monoid homomorphism XistheobjC)(46)π:ΦC→EndXopsuch that elements pλ1j1 occur to reflect rule for multiplication in an H-class of MS;I;∧;P.LetS=MH;I;∧;P be a Rees matrix group. Consider the subset Hiλ=i,h,λ:h∈H of S. In fact Hiλ is a subgroup of S which is isomorphic to H. Consider (47)i,h,λi,h′,λ=i,hpλih′,λ∈HiλThus we have the following lemma and result.Lemma 4.
LetS be a group, and fix an element p∈S. Define a new binary operation ∗p on S as follows: for all a,b∈S, a∗pb=apb∈S. Then S,∗p is a group isomorphic to H.Corollary 5.
S = M H ; I ; ∧ ; P contains many copies of the group H, all isomorphic to H one copy Hiλ for each i,λ∈I×∧.


## 4. Flows on Cauchy Categories

In this section we will investigate another category that can be built from a semigroup. No special conditions are required: any semigroupS gives us its Cauchy category ϱS.The Cauchy category is defined as follows: the set of vertices (or objects) isES. Given e,f∈ES the set of arrows in ϱS from e to f is (48)ϱe,f=e,s,f:es=s=sf.as in Figure 1(a).Figure 1
(a) An arrow inϱS. (b) Composition of arrows in ϱS. (c) Category with one vertex. (d) Typical arrow in Rees matrix semigroup.
(a) (b) (c) (d)The identity arrow ate∈ES is e,e,e and the composition of arrows is shown in Figure 1(b). (49)e,s,ff,t,g=e,st,g.Lemma 6.
In the Cauchy categoryϱS, e,s,f∈ρe,f if and only if s∈eSf. Hence ϱe,f is always nonempty (so that ϱ is a connected category).Proof.
Suppose thate,s,f∈ϱe,f; then es=s=sf. Hence s=esf⇒s∈eSf, where s∈S.
Conversely, ifs∈eSf, then s=eaf where a∈S. We want to show that es=s=sf. Since (50)es=eeaf=e2af=eaf=seis  an  idempotent;sf=eaff=eaf2=eaf=sfis  an  idempotent.
Hencees=s=sf; therefore e,s,f∈ϱe,f.
SinceeSf≠ϕ, and if e=f, the identity arrow at e∈ES is e,e,e∈ϱe,f such that ee=e2=e. Therefore ϱe,f is always nonempty.If we thought ofϱ as a category with one vertex, then we have the following. If G is a group then ϱG is equal to G. We have one object (vertex ∗) which is not an element of G and one arrow for each element of G, with the arrow 1∈G being the identity arrow at ∗. Hence we have one arrow for each element of G; i.e., ϱG=G. See Figure 1(c).Lemma 7.
LetM be a monoid with identity 1; then {1,m,1:m∈M} is a submonoid of ϱM and isomorphic to M. Hence M is a submonoid of ΦϱM.Proof.
LetM¯=1,m,1:m∈M be a submonoid of ϱM. It is easy to check that the operation is associative since associativity will hold automatically in a category. Also 1,1,1∈M¯ is the identity arrow. To show that M¯≃M, suppose f:M→M¯ defined by fm=1,m,1 again it is easy to check that f is one-one correspondence and it is a morphism. In order to prove that M is a submonoid of ΦϱM, given m∈M, define a flow φm on ϱM as follows: (51)φm1=1,m,1andφme=e,e,e,e≠1.
Thenφm:m∈M is isomorphic to M; this gives us a copy of M inside ΦϱM. Hence M is a submonoid of ΦϱM.In the following we describe the Cauchy categories of some small semigroups given in [7].(i) Cauchy category of a rectangular bandS: we write S=A×B using aba=a for all a,b∈S. If e,s,f∈ϱe,f then s∈eSf: so s=etf for some t∈S and e,f∈ES. But it follows that (52)s=etf=eetff=eetffefsincefef=f=eetfef=e2f=efThe set of vertices (or objects) ofϱS is ES=S, and arrow from idempotent a,b to idempotent c,d is defined by element x,y∈S such that a,b︸e,x,y︸s,c,d︸f∈ϱS where es=a,bx,y=a,y=x,y so x=a and sf=x,yc,d=x,d=x,y so y=d. Hence e,ef,f is the only arrow connecting e,f.If we have in the Cauchy category of a regular semigroupS which has exactly one arrow from any two idempotents vertices, assume that etf=ef for all e,f in the set of idempotents ES and t∈S. If we put t=fe then etf=efef=ef2=ef∈ES. Hence ES is a subsemigroup of S.Now sinceS is regular let a∈S and b∈Va; then ab,ba∈ES,e=ab,f=ba and t=a; we want to prove that every element of S is an idempotent. Since etf∈S is an arbitrary element of S, (53)etf2=etfetf=abab︸aab︸aba=ababa=aba=a;and (54)etf=ababa=aba=aHence every element ofS is an idempotent; therefore S is a band; our assumption on S tells us that for all a,b,c∈S, abc=ac, if c=a, then aba=a2=a. Therefore S is a rectangular band.We prove the coming result.Proposition 8.
LetS be a regular semigroup. Then in the Cauchy category ϱS we have exactly one arrow from e to f (for all e,f∈ES) if and only if S is a rectangular band.Take the semigroupS with ES=φ (for example, infinite monogenic semigroup). Define S0=S∪0;0s=s0=0, and 00=0 for all s∈S; then ES0={0} and S0 satisfies our assumption: etf=ef. Since we must have e=0=f, S0 is not regular (0 is the only regular element). Can we find a semigroup S that does not have a zero, is not regular, and satisfies for all e,f∈ES,t∈S such that etf=ef? Consider T=x,x2,x3=x2 (which is a monogenic subsemigroup) is not a band since x≠x2 and not regular since x is not a regular element, but satisfies xyz=xz for all x,y,z since (55)Define S=N×T with multiplication m,yn,z=n,yz, which is a semigroup such that no element acts as 0 (if n,z∈S behave like a zero then z=x2.n,zm,y=m,z≠n,z if m≠n) moreover n,x is not regular for any n∈N: we compute n,xm,yn,x=n,x2≠n,x for all y. But S satisfies the low etf=ef since m1,y1m2,y2m3,y3=m3,y1y2y3=m3,y1y3=m1y1m3,y3 (remember that y1y2y3=y1y3 in T). Hence if T satisfies abc=ac as a low, then S satisfies same low. As a semigroup, if T is a rectangular band, then so is S.(i)
We describe Cauchy categoriesϱS where S is completely simple semigroup which is the Rees matrix semigroup S=MI,∧,P,H where i,a,λj,b,μ=i,apλjb,μ. The vertex set is ES=e:e=i,a,λ,a=pλi-1. The arrows between the vertices e=i,a,λ and f=j,b,μ are l=e,s,f where s=i,c,μ such that es=s,sf=s. See Figure 1(d).Check the composition of arrows: So lete=i,a,λ,a=pλi-1,f=j,b,μ,b=pμj-1 and g=k,c,γ, c=pγk-1 as drawn in Figure 2(a). If s=i,d,μ and t=j,h,γ, then (56)est=i,a,λi,d,μj,h,γ=i,a,λi,dpμjh,γ=i,apλidpμjh,γ=i,dpμjh,γ=stand(57)stg=i,dpμjh,γk,pγk-1,γ=i,dpμjhpγkpγk-1,γ=st.Identity ate∈ES is e,e,e.(ii)
We describe Cauchy categoriesϱS where S is a Brandt semigroup B2. The general construction of Brandt semigroups proceeds as follows. Let X be a finite set with n elements. The Brandt semigroup Bn is the set Bn=X×X∪{0} with multiplication given by (58)a,bc,d=a,difb=c0otherwise,soBn has n2+1 elements, and n+1 idempotents: (59)EBn=a,a:a∈X∪0.Arrows from idempotent e=a,a to idempotent f=b,b are a triple e,s,f such that S=x,y such that (es=s=sf), es=a,ax,y=a,y if a=x and sf=x,yb,b=x,b if b=y. So if e,f≠0 then e=a,a,f=b,b, then a,b∈ϱS but if a≠b, then ef=0, and so we get two arrows from e to f as shown in Figure 2(b).(60)e,0,fande,a,b,fMoreover, 0∈ES and 0 is a vertex of ϱS. Then 0·s=0 so s=0 and then sf=0f=0.Figure 2
(a) Composition of arrows in Rees matrix semigroup. (b) Arrows in Brandt semigroup.
(a) (b)NowBn has n+1 idempotents. If n+1≥3 there are at least two nonzero idempotents and e,f≠0; then ϱS has two arrows. If n=1, B1 has only two elements B1=a,a,0 which is certainly a band; but since a,a0a,a=0≠a,a it is not a rectangular band. See Figure 3.Figure 3
Cauchy categoryϱB2.Our task here is to describe the flow monoid on a Cauchy category. We describe it in terms of functions defined onES. A flow ϕ on ϱS assigns to each e∈ES an arrow e,s,f where e,f∈ES and s∈eSf. So we can first select the function πϕ:e↦f, and then having chosen πϕ we have a second function δϕ say, with δϕ:ES→S satisfying δϕe∈eSπϕe for all e. The pair πϕ,δϕ then determines the flow ϕ. If ϕ1 and ϕ2 are two flows determined by the pairs π1,δ1 and π2,δ2, then we compute the composition of flows ϕ1∗ϕ2 as (61)ϕ1∗ϕ2x=ϕ1x·ϕ2π1x=x,δ1x,π1xπ1x,δ2π1x,π2π1x=(x,δ1xδ2π1x,π2π1x.For future work from the point of semigroup theory, it is quite natural to consider partial flows that are only defined on a subset ofX. This is similar idea of considering the partial transformation semigroup ℘X of a set X instead of the full transformation semigroup IX. Partial flow then is a function from a subset U⊆X assigning to each u∈U an arrow ϕu starting at u; composition of ϕ defined on U and ψ defined on V is defined on the subset W=u∈U:ϕu ends at an element of V and then ϕ∗ψu=ϕu·ψτϕu; we need to allow the empty flow which acts as a zero; ΦC comes equipped with a monoid homomorphism πϕ:ΦC→℘X (partial transformation semigroup) such that πϕ:u↦τϕu. We will try to set up the theory of partial flows on category, generalizing [1, Sections 2 and 4]. Another future work for flows monoid is to present a Gröbner-Shirshov basis for each example in this paper. We may refer to [9–13] for more details.



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