Global local covers

TL;DR

This paper systematically constructs specific covers of finite semigroups to facilitate future research on their computational complexity.

Contribution

It introduces a new systematic method for constructing covers of finite semigroups, advancing the theoretical framework for complexity analysis.

Findings

Provides a systematic construction method

Lays groundwork for complexity analysis of semigroups

Enhances understanding of finite semigroup covers

Abstract

This paper gives a systematic construction of certain covers of finite semigroups. These covers will be used in future work on the complexity of finite semigroups.

Full text

1 Introduction

This paper is an incomplete description of the important glc cover for finite semigroups, which will play a vital role in the future proposed proof of the decidability of c. (See see e.g. [3], [11] for the definition of the complexity c of finite semigroups.)

In the following, certain surmorphisms between semigroups, or between automata, will be called covers. Covers are surmorphisms, but they are obtained by constructing a pre-image in a “systematic way”. Expansions (as in [1, 2]) are examples of covers; in general, covers are a weaker form of expansions.

2 A direct construction of the glc-cover

2.1 Lattices of automata

We consider the categories ${\mathcal{S}}_{A}$ and ${\mathcal{M}}_{A}$ of finite semigroups, respectively monoids, over a chosen generating set $A$; the morphisms are the usual semigroup (or monoid) morphisms that commute with the map that sends $A$ to a generating set of the semigroup (or monoid). A semigroup in that category will usually be denoted by $S_{A}$ or $(S,A)$.

We consider also the category ${\mathcal{A}}_{A}$ of finite automata with alphabet $A$. Here, the objects are all deterministic finite automata with input alphabet $A$, with a chosen start state i, that are trim (i.e., all states are reachable from i), and such that i cannot be returned to. The morphisms in ${\mathcal{A}}_{A}$ are all state maps that commute with the next-state function and preserve the start state. Usually, we work in the category of isomorphism classes of ${\mathcal{A}}_{A}$, which we also denote by ${\mathcal{A}}_{A}$; but because of the role of the start state this makes no difference.

For any semigroup $S$ generated by $A$, the right Cayley graph of $S$ belongs to ${\mathcal{A}}_{A}$. For clarity, let us define the right Cayley graph $\Gamma_{A}(S,\sigma)$ for a semigroup $S$ and a map $\sigma:A\to S$ that maps $A$ onto a generating set of $S$: $\Gamma_{A}(S,\sigma)$ is a directed labeled rooted graph, with vertex set $S^{I}$ (i.e., $S$ with a new identity element $I$ added, even in case $S$ already had an identity), with root $I$, and with edge set $\{(s,a,s\cdot\sigma(a)):s\in S^{I},\,a\in A\}$ (where $\cdot$ is the product in $S$). The directed edge $(s,a,s\cdot\sigma(a))$ is labeled by $a$. We often write $\Gamma(S_{A})$ instead of $\Gamma_{A}(S,\sigma)$.

For automata ${\bf A}_{1},{\bf A}_{2}$ in ${\mathcal{A}}_{A}$ we write ${\bf A}_{1}\twoheadleftarrow{\bf A}_{2}$ iff there exists a surjective ${\mathcal{A}}_{A}$-morphism from ${\bf A}_{2}$ onto ${\bf A}_{1}$. The relation $\twoheadleftarrow$ is a pre-order on ${\mathcal{A}}_{A}$. Because of the role of the start state, a surjective morphism is unique if it exists (for a given pair ${\bf A}_{1},{\bf A}_{2}$); hence, this pre-order is a partial order. We also write ${\bf A}_{2}\twoheadrightarrow{\bf A}_{1}$ for ${\bf A}_{1}\twoheadleftarrow{\bf A}_{2}$.

For automata ${\bf A}_{1},{\bf A}_{2}$ in ${\mathcal{A}}_{A}$ we define the direct product in the category ${\mathcal{A}}_{A}$. For $i=1,2$, let ${\bf A}_{i}=(Q_{i},I_{i},\cdot_{i})$, where $\cdot_{i}:Q_{i}\times A\to Q_{i}$ is the next-state function. Then

${\bf A}_{1}\times_{A}{\bf A}_{2}$ \ =\ \big{(}\{(I_{1}\cdot_{1}w,\ I_{2}\cdot_{2}w)\in Q_{1}\times Q_{2}: w\in A^{*}\},\ (I_{1},I_{2}),\ \cdot\big{)},

where for all $w\in A^{*}$, $a\in A$, we define  $(I_{1}\cdot_{1}w,\ I_{2}\cdot_{2}w)\cdot a$ \ =\ $(I_{1}\cdot_{1}w\cdot_{1}a,\ I_{2}\cdot_{2}w\cdot_{2}a)$.

For two automaton morphisms $\varphi_{i}:{\bf A}_{i}\to{\bf B}$ in the category ${\mathcal{A}}_{A}$ we can define the direct product $\varphi_{1}\times_{A}\varphi_{2}:{\bf A}_{1}\times_{A}{\bf A}_{2}\to{\bf B}$ in ${\mathcal{A}}_{A}$ by  $(I_{1}\cdot_{1}w,I_{2}\cdot_{2}w)\varphi_{1}\times_{A}\varphi_{2}$ $\ =\ I\cdot_{{}_{\bf B}}w$  (for all $w\in A^{*}$). This is well-defined: If  $(I_{1}\cdot_{1}u,\ I_{2}\cdot_{2}u)$ $=$ $(I_{1}\cdot_{1}v,\ I_{2}\cdot_{2}v)$  then  $I_{i}\cdot_{i}u=I_{i}\cdot_{i}v$  (for $i=1,2$), hence $(I_{i}\cdot_{i}u)\varphi_{i}=(I_{i}\cdot_{i}v)\varphi_{i}$. Moreover, $(I_{i}\cdot_{i}u)\varphi_{i}=(I_{i})\varphi_{i}\cdot_{{}_{\bf B}}u$ $=I\cdot_{{}_{\bf B}}u$; similarly, $(I_{i}\cdot_{i}v)\varphi_{i}=I\cdot_{{}_{\bf B}}v$. Thus, $I\cdot_{{}_{\bf B}}u=I\cdot_{{}_{\bf B}}v$.

We show next that $\twoheadleftarrow$ is a lattice order.

Lemma 2.1

(1) The join ${\bf A}_{1}\vee{\bf A}_{2}$ in ${\mathcal{A}}_{A}$ is isomorphic to the direct product ${\bf A}_{1}\times_{A}{\bf A}_{2}$. In the case of right Cayley graphs of semigroups in ${\mathcal{S}}_{A}$, the join is (up to isomorphism) the direct product in ${\mathcal{S}}_{A}$.

(2) The meet is determined by the join as  ${\bf A_{1}}\wedge{\bf A}_{2}\ =\ \bigvee$ $\{{\bf A}:{\bf A}_{1}\twoheadrightarrow{\bf A}$ and ${\bf A}_{2}\twoheadrightarrow{\bf A}\}$  (up to isomorphism).

Proof. (1) The projections from ${\bf A}_{1}\times_{A}{\bf A}_{2}$ onto ${\bf A_{1}}$, respectively ${\bf A}_{2}$, show that ${\bf A}_{1}\times_{A}{\bf A}_{2}$ is an upper bound on ${\bf A_{1}}$ and ${\bf A}_{2}$ for the order $\twoheadleftarrow$. To show that it is the least upper bound, we consider the commutative diagrams in the category ${\mathcal{A}}_{A}$ that defines the direct product, with arrows ${\bf A}_{1}\twoheadleftarrow{\bf X}\twoheadrightarrow{\bf A}_{2}$, ${\bf X}\twoheadrightarrow{\bf A}_{1}\times_{A}{\bf A}_{2}$, and the projections ${\bf A_{1}}\twoheadleftarrow{\bf A}_{1}\times_{A}{\bf A}_{2}$ $\twoheadrightarrow{\bf A}_{2}$, for any object ${\bf X}$ in ${\mathcal{A}}_{A}$. The join is defined by exactly the same diagrams, so the join is the direct product.

For Cayley graphs $\Gamma_{A}(S_{1},\sigma_{1})$ and $\Gamma_{A}(S_{2},\sigma_{2})$, viewed as $A$-automata, the direct product $\Gamma_{A}(S_{1},\sigma_{1})\times_{A}\Gamma_{A}(S_{2},\sigma_{2})$ has the set $\{(\sigma_{1}(w),\sigma_{2}(w)):w\in A^{*}\}$ as its states (vertices); here we extended $\sigma_{1}$ and $\sigma_{2}$ to homomorphisms from $A^{*}$ onto $S_{1}$ or $S_{2}$. Hence $\Gamma_{A}(S_{1},\sigma_{1})\times_{A}\Gamma_{A}(S_{2},\sigma_{2})$, with start state $(I_{1},I_{2})$, is the Cayley graph of $(S_{1})_{A}\times_{A}(S_{2})_{A}$ (semigroup direct product in the category ${\mathcal{S}}_{A}$).

(2) This follows from the general fact that if a partial order has finite joins and a global minimum, and if every principal ideal is finite, then this partial order is a lattice. This finiteness comes from the fact that here all automata are finite.    $\Box$

In the partial order $\twoheadleftarrow$ we define an interval  $[{\bf A}_{1},{\bf A}_{2}]$ \ =\ $\{{\bf A}\in{\mathcal{A}}_{A}:$ ${\bf A}_{1}\twoheadleftarrow{\bf A}\twoheadleftarrow{\bf A}_{2}\}$. This interval is a finite lattice with a global maximum and a global minimum. Finiteness comes from the fact that here all automata are finite.

For an automaton ${\bf A}$ in ${\mathcal{A}}_{A}$ and for two states $q_{1},q_{2}$ of ${\bf A}$, we say that $q_{2}$ is reachable from $q_{1}$ (and we write $q_{2}\leq_{\mathcal{R}}q_{1}$) iff $q_{2}=q_{1}\cdot w$ for some $w\in A^{*}$. This relation is a pre-order, and we write $q_{2}\equiv_{\mathcal{R}}q_{1}$ iff $(q_{2}\leq_{\mathcal{R}}q_{1}$ and $q_{1}\leq_{\mathcal{R}}q_{2})$. An $\equiv_{\mathcal{R}}$ equivalence class is called a reachability class, or an $\mathcal{R}$-class; this is the set of vertices of a strongly connected component of the underlying digraph of the automaton. For two states $q_{1},q_{2}$ we will also use the notation $q_{2}\neq\equiv_{\mathcal{R}}q_{1}$ as a shorthand for $(q_{2}\equiv_{\mathcal{R}}q_{1}$ and $q_{2}\neq q_{1})$.

An automaton morphism $\varphi:{\bf A}_{1}\to{\bf A}_{2}$ is said to be 1:1${\mathcal{R}}$ iff the restriction of $\varphi$ to any reachability class of ${\bf A}_{1}$ is injective.

Lemma 2.2

If two morphisms $\varphi_{1}:{\bf A}_{1}\to{\bf B}$ and $\varphi_{2}:{\bf A}_{2}\to{\bf B}$ are 1:1${\mathcal{R}}$ then the direct product morphism $\varphi_{1}\times_{A}\varphi_{2}:{\bf A}_{1}\times_{A}{\bf A}_{2}\to{\bf B}$ in ${\mathcal{A}}_{A}$ is also 1:1${\mathcal{R}}$.

Proof. The state set of ${\bf A}_{1}\times_{A}{\bf A}_{2}$ is $\{(I_{1}\cdot_{1}w,I_{2}\cdot_{2}w):w\in A^{*}\}$. The morphism $\varphi_{1}\times_{A}\varphi_{2}$ is defined by  $\varphi_{1}\times_{A}\varphi_{2}(I_{1}\cdot_{1}w,I_{2}\cdot_{2}w)$ $=I\cdot_{{}_{\bf B}}w$ $=\varphi_{1}(I_{1}\cdot_{1}w)=\varphi_{2}(I_{2}\cdot_{2}w)$. The last two equalities hold because $\varphi_{1}$ and $\varphi_{2}$ are morphisms; in particular, $\varphi_{1}(I_{1})=I=\varphi_{2}(I_{2})$. If  $(I_{1}\cdot_{1}u,I_{2}\cdot_{2}u)$ $\neq\equiv_{\mathcal{R}}$ $(I_{1}\cdot_{1}v,I_{2}\cdot_{2}v)$  then  $I_{1}\cdot_{1}u\equiv_{\mathcal{R}}I_{1}\cdot_{1}v$  and  $I_{2}\cdot_{2}u\equiv_{\mathcal{R}}I_{2}\cdot_{2}v$; moreover,  $I_{1}\cdot_{1}u\neq I_{1}\cdot_{1}v$  or  $I_{2}\cdot_{2}u\neq I_{2}\cdot_{2}v$. If  $I_{1}\cdot_{1}u\neq I_{1}\cdot_{2}v$  then the 1:1${\mathcal{R}}$ property of $\varphi_{1}$ implies  $\varphi_{1}(I_{1}\cdot_{1}u)\neq\varphi_{1}(I_{1}\cdot_{1}v)$, hence  $\varphi_{1}\times_{A}\varphi_{2}(I_{1}\cdot_{1}u,I_{2}\cdot_{2}u)$ $\neq$ $\varphi_{1}\times_{A}\varphi_{2}(I_{1}\cdot_{1}v,I_{2}\cdot_{2}v)$. If  $I_{2}\cdot_{2}u\neq I_{2}\cdot_{2}v$  the same conclusion holds. Thus, $\varphi_{1}\times_{A}\varphi_{2}$ is 1:1${\mathcal{R}}$.    $\Box$

For a finite semigroup $S$ and $x\in S$, let $x^{\omega}$ be the idempotent in $\{x^{n}:n>0\}$. The algebraic rank of $x$ is defined to be the length of a longest strict $\mathcal{J}$-chain of regular $\mathcal{J}$-classes, ascending from $x^{\omega}$ (see [8]). The rank of $x$ is a non-negative integer (possibly 0); we denote it by ${\sf rank}_{S}(x)$ or ${\sf rank}(x)$. Formally,

${\sf rank}_{S}(x)\ =\ {\sf max}\{r\ :\ x^{\omega}<_{\mathcal{J}}J_{1}$ $<_{\mathcal{J}}\ \ldots\ <_{\mathcal{J}}J_{r},$  where $J_{1},\ldots,J_{r}$ are regular $\mathcal{J}$-classes of $S\}$.

In other words, ${\sf rank}_{S}(x)$ is the regular $\mathcal{J}$-depth of the idempotent generated by $x$.

We also define the rank with respect to automata, i.e., we extend ${\sf rank}_{S}:S\to{\mathbb{N}}$ to a word-rank function ${\sf rank}_{\bf A}:A^{*}\to{\mathbb{N}}$   (where ${\mathbb{N}}=$ $\{0,1,\ldots\}$ denotes the natural integers). For an automaton A over $A$ and $w\in A^{*}$, let $[w]$ be the image of $w$ in the syntactic monoid $S$ of A. Then ${\sf rank}_{\bf A}(w)$ is defined to be ${\sf rank}_{S}([w])$.

Definition 2.3

(Rank condition).*  An automaton morphism $\varphi:{\bf A}\twoheadrightarrow{\bf B}$ in ${\mathcal{A}}_{A}$ satisfies the rank condition iff the following holds. For all $q\in Q_{\bf A}$ and all $\alpha,\beta\in A^{+}$ such that $q\cdot_{{}_{\bf A}}\alpha=q$ and $(q)\varphi\cdot_{{}_{\bf B}}\beta=(q)\varphi$ and ${\sf rank}_{\bf B}(\alpha)\geq{\sf rank}_{\bf B}(\beta)$, we have:  $q\cdot_{{}_{\bf A}}\beta=q$.*

The formal statement of the rank condition on $\varphi$ is as follows.

$(\forall q\in Q_{\bf A})(\forall\alpha,\beta\in A^{+}):$

$[\,q\cdot_{{}_{\bf A}}\alpha=q$   $\&$   $(q)\varphi\cdot_{{}_{\bf B}}\beta=(q)\varphi$   $\&$   ${\sf rank}_{\bf B}(\alpha)\geq{\sf rank}_{\bf B}(\beta)\,]$   implies   $q\cdot_{{}_{\bf A}}\beta=q$.

Yet another way to say this: If $\alpha$ fixes $q$ in A, and $\beta$ fixes $(q)\varphi$ in $({\bf A})\varphi$, and ${\sf rank}_{({\bf A})\varphi}(\alpha)$ $\geq$ ${\sf rank}_{({\bf A})\varphi}(\beta)$, then $\beta$ also fixes $q$ in A.

The following is a better statement of the rank condition:

$(\forall q\in Q_{\bf A})(\forall\beta\in A^{+}):$

if   $[\,(q)\varphi\cdot_{{}_{\bf B}}\beta=(q)\varphi$   $\&$   $(\exists\alpha\in A^{+})$ $[\,q\cdot_{{}_{\bf A}}\alpha=q$  $\&$  ${\sf rank}_{\bf B}(\alpha)\geq{\sf rank}_{\bf B}(\beta)\,]\,]$   then   $q\cdot_{{}_{\bf A}}\beta=q$.

In words:  If $\beta$ fixes $(q)\varphi$ in $({\bf A})\varphi$, and there is $\alpha$ that fixes $q$ in A such that ${\sf rank}_{({\bf A})\varphi}(\alpha)$ $\geq$ ${\sf rank}_{({\bf A})\varphi}(\beta)$, then $\beta$ also fixes $q$ in A.

Lemma 2.4

If two morphisms $\varphi_{1}:{\bf A}_{1}\to{\bf B}$ and $\varphi_{2}:{\bf A}_{2}\to{\bf B}$ satisfy the rank condition, then the direct product morphism $\varphi_{1}\times_{A}\varphi_{2}:{\bf A}_{1}\times_{A}{\bf A}_{2}\to{\bf B}$ in ${\mathcal{A}}_{A}$ also satisfies the rank condition.

Proof. Assume that  $(I_{1}\cdot_{1}w,\ I_{2}\cdot_{2}w)\cdot\alpha$ $=$ $(I_{1}\cdot_{1}w,\ I_{2}\cdot_{2}w)$  in ${\bf A}_{1}\times_{A}{\bf A}_{2}$, assume that  $I\cdot_{{}_{\bf B}}w\cdot_{{}_{\bf B}}\beta=I\cdot_{{}_{\bf B}}w$  in B, and assume that ${\sf rank}_{\bf B}(\alpha)\geq{\sf rank}_{\bf B}(\beta)$. The equality in ${\bf A}_{1}\times_{A}{\bf A}_{2}$ implies that for $i=1,2:$  $I_{i}\cdot_{i}w\cdot_{i}\alpha=I_{i}\cdot_{i}w$ and $\varphi_{i}(I_{i})\cdot_{{}_{\bf B}}w\cdot_{{}_{\bf B}}\beta$ $=$ $\varphi_{i}(I_{i})\cdot_{{}_{\bf B}}w$. Hence by the rank condition for $\varphi_{i}$ we have $I_{i}\cdot_{i}w\cdot_{i}\beta=I_{i}\cdot_{i}w$. So, $\beta$ fixes $(I_{1}\cdot_{1}w,\ I_{2}\cdot_{2}w)$ in ${\bf A}_{1}\times_{A}{\bf A}_{2}$.    $\Box$

2.2 Definition of the glc-cover

Let cov be any cover in the category ${\mathcal{A}}_{A}$. For an automaton A in ${\mathcal{A}}_{A}$ we consider the interval $[{\bf A},{\bf A}^{\sf cov}]$. This is a finite lattice, as we saw.

Definition 2.5

The glc-cover ${\bf A}^{\sf cov\,glc}$ is defined by the following join in the category ${\cal A}_{A}$:

{\bf A}^{\sf cov\,glc}\ =\* $\bigvee\,\{{\bf X}\in[{\bf A},{\bf A}^{\sf cov}]:$ there exists a morphism ${\bf A}\twoheadleftarrow{\bf X}$ in ${\mathcal{A}}_{A}$ that is 1:1${\mathcal{R}}$ and that*

satisfies the rank condition}.

More generally, if ${\bf A}$ and ${\bf B}$ are $A$-automata such that ${\bf A}\twoheadleftarrow{\bf B}$, we can define the glc-cover of the interval $[{\bf A},{\bf B}]$ by

[{\bf A},{\bf B}]^{\sf glc}\ =\ $\bigvee\,\{{\bf X}\in[{\bf A},{\bf B}]:$ there exists a morphism ${\bf A}\twoheadleftarrow{\bf X}$ in ${\mathcal{A}}_{A}$ that is 1:1${\mathcal{R}}$ and that

satisfies the rank condition}*. *

By the previous Lemmas, and since $[{\bf A},{\bf A}^{\sf cov}]$ and $[{\bf A},{\bf B}]$ are finite complete lattices, these joins exist. The name “glc” stands for global-local-cover.

Sometimes we use the glc-cover without the rank condition, i.e., with the 1:1${\mathcal{R}}$ requirement alone; we will make that clear in the context.

When the automata A and B in $[{\bf A},{\bf B}]^{\sf glc}$ are right Cayley graphs it makes sense to define the cover $[{\bf A},{\bf B}]^{\sf glc}$ either as above (letting X range over $A$-automata), or just in terms of right Cayley graphs (i.e., letting X range over right Cayley graphs only). We will prove next that the resulting cover is the same in either approach.

Let A and B be right Cayley graphs of $A$-generated semigroups, and suppose C is any $A$-automaton such that ${\bf B}\twoheadrightarrow{\bf C}$ $\stackrel{{\scriptstyle\varphi}}{{\twoheadrightarrow}}{\bf A}$, where $\varphi$ is 1:1${\mathcal{R}}$ (with or without the rank condition). Let $S({\bf C})$ be the $A$-generated syntactic semigroup of C, and let $\Gamma(S({\bf C}))=\Gamma_{A}(S({\bf C}),\sigma)$ be the right Cayley graph of $S({\bf C})$, where $\sigma$ embeds $A$ into a generating set of $S({\bf C})$. Let $\eta:\Gamma(S({\bf C}))\twoheadrightarrow{\bf C}$ be the canonical surmorphism, mapping semigroup elements to states.

Lemma 2.6

If  ${\bf B}\twoheadrightarrow\Gamma(S({\bf C}))$ $\stackrel{{\scriptstyle\eta}}{{\twoheadrightarrow}}{\bf C}$ $\stackrel{{\scriptstyle\varphi}}{{\twoheadrightarrow}}{\bf A}$, where A and B are Cayley graphs, and if $\varphi$ is 1:1${\mathcal{R}}$, then $\eta\circ\varphi:\Gamma(S({\bf C}))\twoheadrightarrow{\bf A}$ is also 1:1${\mathcal{R}}$. If, in addition, $\varphi$ satisfies the rank condition then $\eta\circ\varphi$ also satisfies the rank condition.

Proof. If for $s_{1},s_{2}\in S({\bf C})$ we have $s\neq s_{2}$, $s_{1}\equiv_{\mathcal{R}}s_{2}$, then by faithful action of $S({\bf C})$ in C there exists a state $q$ of C with $q\cdot s_{1}\neq q\cdot s_{2}$. Hence since $\varphi$ is 1:1${\mathcal{R}}$, $(qs_{1})\varphi\neq(qs_{2})\varphi$. Hence, since A is a Cayley graph, $I_{\bf A}\cdot s_{1}\neq I_{\bf A}\cdot s_{2}$, so $(s_{1})\eta\circ\varphi$ $=$ $I_{\bf A}\cdot s_{1}$ $\neq$ $I_{\bf A}\cdot s_{2}=(s_{2})\eta\circ\varphi$, i.e., $\eta\circ\varphi$ is 1:1${\mathcal{R}}$.

Suppose $\varphi$ satisfies the rank condition. We want to show for all $s\in S({\bf C})$ and $\alpha,\beta\in A^{+}:$  if   $s\cdot_{{}_{S}}\alpha=s$ and  $(s)\eta\varphi\cdot_{{}_{\bf A}}\beta=(s)\eta\varphi$ and ${\sf rank}_{\bf B}(\alpha)\geq{\sf rank}_{\bf B}(\beta)$, then $s\cdot_{{}_{S}}\beta=s$. Here $\cdot_{{}_{S}}$ is the next-state function in $\Gamma(S({\bf C}))$. By the rank condition for $\varphi$, we have  $(s)\eta\cdot_{{}_{\bf C}}\beta=(s)\eta$. Hence, since $\eta$ is bijective on the semigroups, $s\cdot_{{}_{\bf C}}\beta=s$.    $\Box$

Proposition 2.7

When A and B are $A$-generated Cayley graphs then the cover $[{\bf A},{\bf B}]^{\sf glc}$ is the same in the category of $A$-automata as in the category of $A$-generated Cayley graphs.

Proof. Since an $A$-generated Cayley graph is an $A$-automaton, the $A$-automata cover (which is the join over automata) maps onto the Cayley graph cover.

By the previous lemma, the Cayley graph of the automaton cover maps onto A by a 1:1${\mathcal{R}}$ morphism with rank condition. Hence the Cayley graph is equal to the cover, by maximality of the cover.    $\Box$

2.3 Examples of glc-covers for 1:1${\mathcal{R}}$ morphisms

(1)  Let V be a pseudovariety of semigroups, and let $S_{A}$ be an $A$-generated semigroup. Let $S_{A}^{\bf V}$ denote the maximum image of $S_{A}$ in V; i.e.,

S_{A}^{\bf V}\ =\ $\bigvee\,\{Y\in{\bf V}:$ there exists a morphism $S_{A}\twoheadrightarrow Y$ in ${\mathcal{S}}_{A}\}$,

where $\vee$ is the join in the category of $A$-generated semigroups; this join is finite since $S_{A}$ is finite.

Let R be the pseudovariety of $\mathcal{R}$-trivial semigroups. For any $A$-generated semigroup $S_{A}$, $S_{A}^{\bf R}$ denotes thus the maximum $\mathcal{R}$-trivial image. References for this and the next few definitions are [11] and [8].

Let {\scriptsize\sfm}⃝ be the Maltsev product. The Maltsev kernel of a morphism $\psi:S\twoheadrightarrow T$ is, by definition, the pseudovariety generated by the set of semigroups $\{(e)\psi^{-1}:e=e^{2}\in T\}$. We consider only the case where V is locally finite (i.e., every finitely generated semigroup in V is finite). For example, the pseudo-variety $\langle T\rangle$ generated by any finite semigroup $T$ is locally finite.

Then V {\scriptsize\sfm}⃝ (.) is an expansion of $A$-generated semigroups, where V {\scriptsize\sfm}⃝ $S_{A}$ is the unique functorially largest $A$-generated semigroup that maps onto $S_{A}$ by a surmorphism whose Maltsev kernel is in V:

V {\scriptsize\sfm}⃝ $S_{A}$ \ =\ $\bigvee\,\{X:$ there exists $\varphi:X\twoheadrightarrow S_{A}$ in ${\cal S}_{A}$ such that the Maltsev kernel of $\varphi$ is

contained in V}.

Since V is locally finite, this join is a finite set, so V {\scriptsize\sfm}⃝ $S_{A}$ is finite (by Brown’s theorem; see [8], [11]).

Proposition 2.8

(Example 2.8)*  Let $\Gamma(S_{A})=[\Gamma(T_{A}),\Gamma(U_{A})]^{\sf glc}$ be the glc-cover of $A$-generated right Cayley graphs with respect to the 1:1${\mathcal{R}}$ property; here we temporarily drop the rank condition. Hence, $S_{A}$ is the unique maximum $A$-generated semigroup such that*

$\Gamma(U_{A})\stackrel{{\scriptstyle\varphi_{1}}}{{\twoheadrightarrow}}\Gamma(S_{A})$* $\stackrel{{\scriptstyle\varphi_{2}}}{{\twoheadrightarrow}}\Gamma(T_{A})$,*

where $\varphi_{2}$ is 1:1${\mathcal{R}}$. Then:

(a)*  The Maltsev kernel of $\varphi_{2}$ (and in fact, of any 1:1*${\mathcal{R}}$* morphism) is contained in R.*

(b)*  If $T_{A}\in{\bf R}$ and $U_{A}=T_{A}^{\sf cov}$ for some cover   $(.)^{\sf cov}$, then*

$[\Gamma(T_{A}),\Gamma(T_{A}^{\sf cov})]^{\sf glc}$* \ =\ $T_{A}^{{\sf cov}\,{\bf R}}$.*

(c)*  For any $A$-generated semigroup $W_{A}$ we consider the expansion $(.)_{A}^{\sf exp}=\langle W_{A}\rangle$ {\scriptsize\sfm}⃝ $({\bf.})$, where $\langle W_{A}\rangle$ is the pseudovariety generated by $W_{A}$. Let ${\bf 1}_{A}$ be the one-element semigroup in the category of $A$-generated semigroups.*

Then ${\bf 1}_{A}^{\sf exp}=W_{A}$, and   $[\Gamma({\bf 1}_{A}),\Gamma(W_{A})]^{\sf glc}=W_{A}^{\bf R}$.

Thus, for any $A$-generated $R\in{\bf R}$, $\langle R\rangle$ can be the Maltsev kernel.

Proof. (a)  If  $\varphi_{2}:\Gamma(S_{A})\twoheadrightarrow\Gamma(T_{A})$  is 1:1${\mathcal{R}}$ then the corresponding map $S_{A}\twoheadrightarrow T_{A}$ (which we also denote by $\varphi_{2}$) is also 1:1${\mathcal{R}}$. Then for any idempotent $e=e^{2}\in T_{A}$, $(e)\varphi_{2}^{-1}$ is an ${\mathcal{R}}$-trivial semigroup. Indeed, if $x_{1}\equiv_{\mathcal{R}}x_{2}$ in $(e)\varphi_{2}^{-1}$ then $(x_{1})\varphi_{2}=e=(x_{2})\varphi_{2}$, hence the 1:1${\mathcal{R}}$-property implies $x_{1}=x_{2}$.

(b)  When $\varphi_{2}:S_{A}\twoheadrightarrow T_{A}$ is 1:1${\mathcal{R}}$, then $S_{A}\in{\bf R}$ iff $T_{A}\in{\bf R}$. Indeed, if $x_{1}\equiv_{\mathcal{R}}x_{2}$ in $S_{A}$ then $(x_{1})\varphi_{2}\equiv_{\mathcal{R}}(x_{2})\varphi_{2}$ in $T_{A}$, hence $(x_{1})\varphi_{2}=(x_{2})\varphi_{2}$ (since $T_{A}\in{\bf R}$), hence $x_{1}=x_{2}$ (since $\varphi_{2}$ is 1:1${\mathcal{R}}$); the converse is obvious.

Since $[\Gamma(T_{A}),\Gamma(T_{A}^{\sf exp})]^{\sf glc}$ $\twoheadrightarrow\Gamma(T_{A})$ is 1:1${\mathcal{R}}$ and since $T_{A}\in{\bf R}$, it follows that $[\Gamma(T_{A}),\Gamma(T_{A}^{\sf exp})]^{\sf glc}$ is in R. Since the glc-cover is maximal with respect to the 1:1${\mathcal{R}}$-property, it follows that the glc-cover is $T_{A}^{{\sf exp}\,{\bf R}}$ (by the definition of $(.)^{\bf R}$).

(c)  Obviously, the map $S_{A}\twoheadrightarrow{\bf 1}_{A}$ is 1:1${\mathcal{R}}$ iff $S_{A}\in{\bf R}$. Hence, by the definition of the expansion $(.)^{\sf exp}$ $=$ $\langle W_{A}\rangle$ {\scriptsize\sfm}⃝ $({\bf.})$ we have ${\bf 1}_{A}^{\sf exp}=W_{A}$. And by the definition of $(.)^{\bf R}$,  $[\Gamma({\bf 1}_{A}),\Gamma(W_{A})]^{\sf glc}=W_{A}^{\bf R}$.

Now let $W_{A}$ ($=R$) be any $A$-generated semigroup in R. Then $W_{A}=W_{A}^{\bf R}$, so in that case,  $[\Gamma({\bf 1}_{A}),\Gamma(W_{A})]^{\sf glc}=W_{A}$  (by (c)).

And the Maltsev kernel is $\langle W_{A}\rangle$ (which is $\langle R\rangle$).    $\Box$

Remark. From Example 2.8 we see that $[\Gamma_{A}(T),\Gamma_{A}(T^{\sf cov})]^{\sf glc}$ can be:

(case 1)  $\Gamma_{A}(T)$, or

(case 2)  $\Gamma_{A}(T^{\sf cov})$, or

(case 3)  strictly between the bounds of the interval.

Let $T_{A}={\bf 1}_{A}$ and $(.)_{A}^{\sf cov}=\langle W_{A}\rangle$ {\scriptsize\sfm}⃝ $({\bf.})$. Then ${\bf 1}_{A}^{\sf cov}=W_{A}$, and by Example 2.8.(c), $[\Gamma_{A}(T),\Gamma_{A}(T^{\sf cov})]^{\sf glc}$ $=$ $[{\bf 1}_{A},{\bf 1}_{A}^{\sf cov}]^{\sf glc}=W_{A}^{\bf R}$.

(1) When $W_{A}$ is a non-trivial $A$-generated group, then $W_{A}^{\bf R}={\bf 1}_{A}$. So we get (case 1).

(2) When ${\bf 1}_{A}\neq W_{A}\in{\bf R}$ (e.g., $W_{A}$ is the left-zero semigroup generated by $A$) then by the Example 2.8.(b), we are in (case 2) and not at the same time in (case 1).

(3) Let $W_{A}\not\in{\bf R}$ and $W_{A}^{\bf R}\neq{\bf 1}$; e.g., let $W_{A}={\mathbb{Z}}_{2}\times_{A}A^{\ell}$, where $A^{\ell}$ is the left-zero semigroup generated by $A$. Then $W_{A}\twoheadrightarrow A^{\ell}$ by projection; then $[{\bf 1}_{A},{\bf 1}_{A}^{\sf cov}]^{\sf glc}=W_{A}^{\bf R}$ is strictly between ${\bf 1}_{A}$ and ${\bf 1}_{A}^{\sf cov}=W_{A}$. This illustrates (case 3).

(2)  Let us consider the reverse of the delay pseudovariety ${\bf D}=[xy^{\omega}=y^{\omega}]$. So ${\bf D}^{\rm rev}=[x^{\omega}y=x^{\omega}]$, corresponding to the limit of the set of equations $\{x_{1}\ldots x_{k}x_{k+1}=x_{1}\ldots x_{k}:k\geq 1\}$. It is known that $S\in{\bf D}^{\rm rev}$ iff $S$ is a nilpotent extension of a left-zero semigroup; such a semigroup consist of a minimal ideal which is an $\cal L$-class $L$ of left-zeros, and every element of the semigroup satisfies $(\exists k)\ x^{k}\in L$. The left-zero semigroups form the pseudovariety LZ $=$ $[xy=x]$. For fixed $k>0$, let ${\bf D}_{k}^{\rm rev}=[x_{1}\ldots x_{k}x_{k+1}=x_{1}\ldots x_{k}]$ be the reverse of the delay-$k$ pseudovariety; this pseudovariety is locally finite. We consider the expansion $(.)^{\sf exp}={\bf D}_{k}^{\rm rev}{\footnotesize{\scriptsize\sfm}⃝\ }(.)$. Then we have for every $A$-generated semigroup $T_{A}$:

${\bf D}_{k}^{\rm rev}{\footnotesize{\scriptsize\sfm}⃝\ }T_{A}$*   $=$   $T_{A}^{\sf exp}\ \twoheadrightarrow\ T_{A}$  is injective on every regular ${\mathcal{R}}$-class. *

Indeed, let us apply Rees’ Theorem to a regular ${\mathcal{R}}$-class $\{a\}\times G\times B$ of $T_{A}^{\sf exp}$. If the map identifies two elements in this ${\mathcal{R}}$-class then it identifies two $\cal L$-classes, hence it identifies to ${\mathcal{L}}$-equivalent idempotents. So in that case there are two idempotents $e\neq f$. Thus, $\{e,f\}$ is a right-zero semigroup, but a non-trivial right-zero semigroup cannot be in ${\bf D}_{k}^{\rm rev}$.

Notation: A morphism that is injective on every regular $\cal R$-class is said to be 1:1reg$\cal R$.

Digression:   Example of a morphism $\varphi:S_{A}\twoheadrightarrow T_{A}$ that is 1:1reg$\cal R$**, but not 1:1$\cal R$ **

Let $S=\{\i,x,\overline{\i},\overline{x},0\}$,  $T=\{\i,x,y,0\}$,  $\varphi=\{(\i,\i),(x,x),(\overline{\i},y),(\overline{x},y),(0,0)\}$, and $A$ is a two-element set embedded injectively into $\{x,\overline{\i}\}$ and $\{x,y\}$. The multiplication tables of $S$ and $T$ are:

[TABLE]

[TABLE]

The structure of these semigroups can be understood as follows: $S$ has three $\cal J$-classes, namely the group ${\mathbb{Z}}_{2}=\{\i,x\}$ at the top, a null $\cal R$-class $\{\overline{\i},\overline{x}\}$ in the middle, and a zero 0. The group ${\mathbb{Z}}_{2}$ acts on the $\cal R$-class in the obvious way on the left and on the right (see the multiplication table), and “null” means that products in the middle $\cal R$-class are 0. The semigroup $T$ differs from $S$ only in that its middle null $\cal J$-class is a singleton. The action of ${\mathbb{Z}}_{2}$ and the null property of the middle $\cal J$-class make it easy to check associativity.

It is clear that $\varphi$ is injective on regular $\cal R$-classes (namely ${\mathbb{Z}}_{2}$ and $\{0\}$), and not injective on the null middle $\cal R$-class.    [End, digression]

We continue to use glc-covers based on 1:1$\cal R$ morphisms, ignoring the rank condition. We also consider 1:1reg$\cal R$ morphisms, and we distinguish the two by the notation $(.)^{{\sf glc}\,1:1{\cal R}}$, respectively $(.)^{{\sf glc}\,1:1{\rm reg}{\cal R}}$. We will use the following.

Lemma 2.9

Let $\alpha,\beta$ be surmorphisms of $A$-generated semigroups. Then $\alpha\circ\beta$ is 1:1${\cal R}$ (or 1:1reg$\cal R$) iff each of $\alpha$ and $\beta$ is 1:1${\cal R}$ (respectively 1:1reg$\cal R$).

Proof. The right-to-left implication is obvious. Assume now that $\beta\circ\alpha$ is 1:1$\cal R$, from which it immediately follows that $\alpha$ is 1:1$\cal R$. For $\beta$, let $R$ be an $\cal R$-class of ${\sf dom}(\beta)$, and let $y_{1},y_{2}\in R$ with $y_{1}\neq y_{2}$. Let $x_{1},x_{2}\in{\sf dom}(\alpha)$ be such that $x_{1}\equiv_{\cal R}x_{2}$ and $y_{1}=(x_{1})\alpha$ and $y_{2}=(x_{2})\alpha$; $x_{1},x_{2}$ exist since $\alpha$ is surjective, and since the inverse image of an $\cal R$-class is a union of $\cal R$-classes. Then $(y_{1})\beta=(x_{1})\alpha\beta\neq(x_{2})\alpha\beta=(y_{2})\beta$, where “$\neq$” holds because $(.)\alpha\beta$ is 1:1$\cal R$. Hence $(y_{1})\beta\neq(y_{2})\beta$, i.e., $\beta$ is 1:1$\cal R$. The same proof (when $R$ is regular) works for 1:1reg$\cal R$, using the fact that the inverse image of a regular $\cal R$-class contains a regular $\cal R$-class.    $\Box$

We consider the expansion defined by   $T_{A}^{\sf exp}$ \ =\ ${\bf D}_{k}^{\rm rev}$ {\scriptsize\sfm}⃝ $T_{A}\ \ \twoheadrightarrow\ \ T_{A}$. Then

T_{A}^{\sf exp}\ \ \twoheadrightarrow\ \ $[T_{A},T_{A}^{\sf exp}]^{{\sf glc}\ 1:1{\rm reg}{\cal R}}$

is an isomorphism, since $T_{A}^{\sf exp}\twoheadrightarrow T_{A}$ is 1:1reg$\cal R$. So we have

$T_{A}^{\sf exp}$ \ =\ $[T_{A},T_{A}^{\sf exp}]^{{\sf glc}\ 1:1{\rm reg}{\cal R}}$ \ \ \ \stackrel{{\scriptstyle\varphi_{1}}}{{\twoheadrightarrow}}\ \ \ $[T_{A},T_{A}^{\sf exp}]^{{\sf glc}\ 1:1{\cal R}}$ $\ \ \ \stackrel{{\scriptstyle\varphi_{2}}}{{\twoheadrightarrow}}\ \ \ T_{A}$,

where $\varphi_{1}$ is not an isomorphism (in general), because 1:1reg$\cal R$ does not imply 1:1$\cal R$. The following is straightforward.

Proposition 2.10

Let $(.)^{\sf E}$ be an expansion such that $S_{A}^{\sf\ E}\twoheadrightarrow S_{A}$ is 1:1$\cal R$ for all $S_{A}$. If the semigroups $U_{A}$ and $T_{A}$ satisfy $U_{A}^{\sf\ E}=U_{A}$ and $T_{A}^{\sf\ E}=T_{A}$, then   $[T_{A},U_{A}]^{{\sf glc}\ {\sf E}}=[T_{A},U_{A}]^{\sf glc}$.    $\Box$

For example, the right Rhodes expansion $(.)^{\wedge R}$ has the 1:1$\cal R$ property used above.

2.4 The ${\mathcal{R}}$- and ${\mathcal{L}}$-expansions of an

automaton

The ${\mathcal{R}}$-expansion:   We generalize the right Rhodes ${\wedge_{\mathcal{R}}}$-expansion to the automaton category ${\mathcal{A}}_{A}$. In the special case of the right Cayley graph automaton $(S^{I},I,\cdot)$ of a semigroup $S_{A}$ over the alphabet $A$ we recover the known Rhodes expansion $S_{A}^{\wedge_{\mathcal{R}}}$.

The definition proceeds in two steps. Let ${\bf A}=(Q,i,\cdot)$ be an automaton in ${\mathcal{A}}_{A}$ with start state $i$. We assume that the start state $i$ is not reachable from any other state.

(1) Let Q^{\wedge_{\mathcal{R}}}=\ $\{(i>_{\mathcal{R}}q_{1}>_{\mathcal{R}}\ \ldots\ >_{\mathcal{R}}q_{k}):$ $k\geq 0\ {\rm and}\ q_{1},\ldots,q_{k}\in Q\}$  be the set of strict reachability chains, starting at the start state. From this we build an automaton with state set $Q^{\wedge_{\mathcal{R}}}$, start state $(i)$, and next-state function $\bullet$; the latter is defined as follows. For any $a\in A$ and ${\bf q}=$ $(i>_{\mathcal{R}}q_{1}>_{\mathcal{R}}\ \ldots\ >_{\mathcal{R}}q_{k-1}$ $>_{\mathcal{R}}q_{k})$ $\in Q^{\wedge_{\mathcal{R}}}$,

{\bf q}\bullet a\ =\ $(i>_{\mathcal{R}}q_{1}>_{\mathcal{R}}\ \ldots\ >_{\mathcal{R}}q_{k-1}$ $>_{\mathcal{R}}q_{k}>_{\mathcal{R}}q_{k}\cdot a)$,   if  $q_{k}>_{\mathcal{R}}q_{k}\cdot a$;   and

{\bf q}\bullet a\ =\ $(i>_{\mathcal{R}}q_{1}>_{\mathcal{R}}\ \ldots\ >_{\mathcal{R}}q_{k-1}$ $>_{\mathcal{R}}q_{k}\cdot a)$,   if  $q_{k}\equiv_{\mathcal{R}}q_{k}\cdot a$.

(2) Let $Q_{A}^{\wedge_{\mathcal{R}}}\subseteq Q^{\wedge_{\mathcal{R}}}$ be the set of states of the above automaton that are reachable from the start state $(i)$; i.e., $Q_{A}^{\wedge_{\mathcal{R}}}=\{(i)\bullet w\ :\ w\in A^{*}\}$. We define ${\bf A}_{A}^{\wedge_{\mathcal{R}}}$ to be the sub-automaton $(Q_{A}^{\wedge_{\mathcal{R}}},\ (i),\ \bullet)$ of the above automaton.

The canonical natural transformation $\eta$ of the expansion is defined for each A by $\eta_{\bf A}:{\bf A}_{A}^{\wedge_{\mathcal{R}}}\to{\bf A}$, where  (i>_{\mathcal{R}}q_{1}>_{\mathcal{R}}\ \ldots\ $>_{\mathcal{R}}q_{k})\eta_{\bf A}\ =\ q_{k}$. One easily checks that this is an automaton morphism in the category ${\mathcal{A}}_{A}$.

Any automaton morphism $\varphi:{\bf A}\to{\bf B}$ in ${\mathcal{A}}_{A}$ can be expanded to $\varphi_{A}^{\wedge_{\mathcal{R}}}:$ ${\bf A}_{A}^{\wedge_{\mathcal{R}}}\to{\bf B}_{A}^{\wedge_{\mathcal{R}}}$ defined by

(i_{\bf A}>_{\mathcal{R}}q_{1}>_{\mathcal{R}}\ \ldots\ $>_{\mathcal{R}}q_{k})\varphi_{A}^{\wedge_{\mathcal{R}}}$  $=$  {\sf red}\big{(}i_{\bf B}>_{\mathcal{R}}(q_{1})\varphi \geq_{\mathcal{R}}\ \ldots\ \geq_{\mathcal{R}}(q_{k})\varphi\big{)}.

Here, the “reduction operation” red has the effect of replacing every maximal $\equiv_{\mathcal{R}}$-chain by its rightmost element. More formally, ${\sf red}({\bf q})={\bf q}$ if q is a strict chain; and

${\sf red}(\ \ldots\ \geq_{\mathcal{R}}q_{i-1}\geq_{\mathcal{R}}$ $q_{i}\equiv_{\mathcal{R}}q_{i+1}\geq_{\mathcal{R}}\ \ldots\ )$  $=$  ${\sf red}(\ \ldots\ \geq_{\mathcal{R}}q_{i-1}\geq_{\mathcal{R}}$ $q_{i+1}\geq_{\mathcal{R}}\ \ldots\ )$.

It is easy to check that $\varphi_{A}^{\wedge_{\mathcal{R}}}$ is an automaton morphism.

In summary, $(.)_{A}^{\wedge_{\mathcal{R}}}$ is an expansion in the category of automata ${\mathcal{A}}_{A}$, according to the categorial definition of expansions.

The analogy between the Rhodes $\wedge_{\mathcal{R}}$-expansion and the automata $\wedge_{\mathcal{R}}$-expansion goes beyond the similarity of the definitions. Let $S_{A}$ be the syntactic monoid of an $A$-automaton ${\bf A}=(Q,i,\cdot)$; hence $(Q,S_{A})$ is a faithful right action.

Proposition 2.11

The Rhodes expansion $S_{A}^{\wedge_{\mathcal{R}}}$ acts on the set $Q_{A}^{\wedge_{\mathcal{R}}}$ by the following right action.

For ${\bf q}=$ $(i>_{\mathcal{R}}q_{1}>_{\mathcal{R}}\ \ldots\ >_{\mathcal{R}}q_{k})$ $\in Q_{A}^{\wedge_{\mathcal{R}}}$ and ${\bf s}=$ $(1>_{\mathcal{R}}s_{1}>_{\mathcal{R}}>_{\mathcal{R}}s_{2}>_{\mathcal{R}}$ $\ \ldots\ >_{\mathcal{R}}s_{h})\in S_{A}^{\wedge_{\mathcal{R}}}$ this action is defined by:

{\bf q}\bullet{\bf s}\ =\ ${\sf red}(i>_{\mathcal{R}}q_{1}>_{\mathcal{R}}\ \ldots\ >_{\mathcal{R}}$ $q_{k}\geq_{\mathcal{R}}q_{k}\cdot s_{1}\geq_{\mathcal{R}}q_{k}\cdot s_{2}$ $\geq_{\mathcal{R}}\ \ldots\ \geq_{\mathcal{R}}q_{k}\cdot s_{h})$.

This action yields a homomorphism from $S_{A}^{\wedge_{\mathcal{R}}}$ onto the syntactic monoid of ${\bf A}_{A}^{\wedge_{\mathcal{R}}}$. The action is not necessarily faithful, i.e., this homomorphism is not necessarily an isomorphism. Just as for the Rhodes expansion for the semigroup category ${\mathcal{S}}_{A}$, we have in the automaton category ${\mathcal{A}}_{A}$:

Proposition 2.12

For every automaton A in ${\mathcal{A}}_{A}:$  $({\bf A}_{A}^{\wedge_{\mathcal{R}}})_{A}^{\wedge_{\mathcal{R}}}$ $={\bf A}_{A}^{\wedge_{\mathcal{R}}}$.

Proof. The proof is similar to the case of $S_{A}^{\wedge_{\mathcal{R}}}$.    $\Box$

Proposition 2.13

The reachability order among the states of ${\bf A}_{A}^{\wedge_{\mathcal{R}}}$ is unambiguous.

Proof. The proof is similar to the case of $S_{A}^{\wedge_{\mathcal{R}}}$.    $\Box$

Proposition 2.14

For any automaton A, the canonical map $\eta_{\bf A}:{\bf A}_{A}^{\wedge_{\mathcal{R}}}\to{\bf A}$ is fully 1:1${\mathcal{R}}$ and satisfies the rank condition.

Proof.  [1:1${\mathcal{R}}$]  If ${\bf p}\equiv{\bf q}\in$ $A^{\wedge_{\mathcal{R}}}$ then p and q have the form ${\bf p}=(r_{1}>\ldots>r_{k-1}>p)$, respectively ${\bf q}=(r_{1}>\ldots>r_{k-1}>q)$. Hence, when ${\bf p}\equiv{\bf q}$ we have:  ${\bf p}\neq{\bf q}$  iff  $\eta({\bf p})\neq\eta({\bf q})$.

[fully 1:1${\mathcal{R}}$]  Let $p=\eta({\bf p})$ and suppose $p\equiv q$ in A. We need to show that there is q such that $\eta({\bf q})=q$ and ${\bf p}\equiv{\bf q}$. Since $p=\eta({\bf p})$, the chain p has the form $(r_{1}>\ldots>r_{k-1}>p)$ for some $r_{1},\ldots,r_{k-1}\in Q$. Since $p\equiv q$, the chain $(r_{1}>\ldots>r_{k-1}>q)$ (let’s call it q) exists in $Q^{\wedge_{\mathcal{R}}}$. To show that ${\bf q}\in Q_{A}^{\wedge_{\mathcal{R}}}$ and that ${\bf p}\equiv{\bf q}$ in $A^{\wedge_{\mathcal{R}}}$, suppose $q=p\cdot a_{1}\cdot\ldots\cdot a_{n}$ for some $a_{1},\ldots,a_{n}\in A$. Then  ${\bf p}\cdot a_{1}\cdot\ldots\cdot a_{n}$ $=$ ${\sf red}(r_{1}>\ldots>r_{k-1}>p\geq p\cdot a_{1}\geq\ \ldots\geq$ $p\cdot a_{1}\ldots a_{n-1}\geq p\cdot a_{1}\ldots a_{n-1}a_{n}=q)$ $=$ $(r_{1}>\ldots>r_{k-1}>q)$, since $p\equiv q$. Thus, ${\bf q}\in Q_{A}^{\wedge_{\mathcal{R}}}$ and ${\bf p}\equiv{\bf q}$.

[Rank condition]  Suppose there is a morphism $\psi:{\bf A}\to{\bf T}$. Suppose ${\bf q}\cdot\alpha={\bf q}$ in ${\bf A}_{A}^{\wedge_{\mathcal{R}}}$, and $\eta({\bf q})\cdot\beta=\eta({\bf q})$ in A, and ${\sf rank}_{\bf T}(\alpha)\geq{\sf rank}_{\bf T}(\beta)$. We need to show that ${\bf q}\cdot\beta={\bf q}$ in ${\bf A}_{A}^{\wedge_{\mathcal{R}}}$. One easily verifies that $\eta({\bf q})\cdot\beta=\eta({\bf q})$ by itself already implies ${\bf q}\cdot\beta={\bf q}$ (without any further hypotheses).    $\Box$

Proposition 2.15

For any automaton A in ${\mathcal{A}}_{A}$ we have (up to isomorphism):

If  ${\bf A}_{A}^{{\sf cov}\ \wedge_{\mathcal{R}}}={\bf A}_{A}^{\sf cov}$  then  ${\bf A}_{A}^{{\sf cov\,glc}\ \wedge_{\mathcal{R}}}$ $=$ ${\bf A}_{A}^{\sf cov\,glc}$.

Proof. Since ${\bf A}_{A}^{{\sf cov}\ \wedge_{\mathcal{R}}}={\bf A}_{A}^{\sf cov}$, the interval  $[{\bf A}_{A}^{\wedge_{\mathcal{R}}},$ ${\bf A}_{A}^{{\sf cov}\ \wedge_{\mathcal{R}}}]$  is contained in $[{\bf A},\ {\bf A}_{A}^{\sf cov}]$, with upper boundaries equal. The canonical map   ${\bf A}_{A}^{{\sf cov\,glc}\ \wedge_{\mathcal{R}}}$ $\twoheadrightarrow$ ${\bf A}_{A}^{\sf cov\,glc}$   is 1:1${\mathcal{R}}$ and satisfies the rank condition (by the previous Prop.). Moreover, by definition of $(.)^{\sf glc}$, ${\bf A}_{A}^{\sf cov\,glc}$ is maximal in the order $\twoheadrightarrow$ for maps that are 1:1${\mathcal{R}}$ and that satisfy the rank condition. The Proposition follows.    $\Box$

The ${\mathcal{L}}$-expansion:  For any right action $(Q\cdot,S)$ (or any automaton A with syntactic monoid $S$), we define a left action of $S$ on ${\mathcal{P}}(Q)$ by

$s\star X=Xs^{-1}$

for all $s\in S$ and $X\subseteq Q$. Recall the standard notation $Xs^{-1}=\{q\in Q:q\cdot s\in X\}$, for any $s\in S$ and $X\subseteq Q$. For a singleton $\{q\}$ we also write $s\star q$ and $(q)s^{-1}$ for $s\star\{q\}$, respectively $\{q\}s^{-1}$.

This is indeed a left action: $(s_{1}s_{2})\star X$ $=$ $X(s_{1}s_{2})^{-1}$ $=$ $(Xs_{2}^{-1})s_{1}^{-1}$ $=$ $s_{1}\star(s_{2}\star X)$. One can prove fairly easily that this is a faithful left action of $S$: If $s_{1}\neq s_{2}$ then, since the given right action is faithful, there is $q\in Q$ with $q\cdot s_{1}=p_{1}$ $\neq$ $p_{2}=q\cdot s_{2}$; then $q\in(p_{1})s_{1}^{-1}$ but $q\not\in(p_{1})s_{2}^{-1}$.

This action can be restricted to a faithful action on the set $\{(q)s^{-1}:\ q\in Q,\ s\in S^{1}\}\ \subseteq\ {\mathcal{P}}(Q)$.

The above construction can be applied to automata. Suppose an automaton ${\bf A}=(Q,i,\cdot)$ has a final state $f$, which is reachable from every state in $Q$, i.e., $(\forall q\in Q)(\exists w\in A^{*})[\,q\cdot w=f\,]$. Then  ${\bf A}^{\sf left}=(\{(f)s^{-1}:s\in S^{1}\},\ \{f\},\ \star)$  is a left automaton with the same syntactic semigroup as A.

For an automaton ${\bf A}=(Q,i,\cdot)$ and its left version ${\bf A}^{\sf left}=$ $(Q^{\sf left}=\{(f)s^{-1}:s\in S^{1}\},\ \{f\},\ \star)$, we can construct a left expansion ${\bf A}_{A}^{\wedge_{\mathcal{L}}}$; this generalizes the left Rhodes expansion $S_{A}^{\wedge_{\mathcal{L}}}$ of a semigroup, and yields a left action (not necessarily faithful) of $S_{A}^{\wedge_{\mathcal{L}}}$. The state set of ${\bf A}_{A}^{\wedge_{\mathcal{L}}}$ is

$\{(P_{k}<\ \ldots\ <P_{1}<\{f\}):k\geq 0,\ {\rm and}$ $P_{k},\ldots,P_{1}\in Q^{\sf left}\}$,

where, for any $P_{2},P_{1}\in Q^{\sf left}$, we define

$P_{2}\leq P_{1}$  iff  $(\exists s\in S^{1})$ $P_{2}=s\star P_{1}=P_{1}s^{-1}$.

The action is defined by

a\star(P_{k}<\ \ldots\ <P_{1}<\{f\})\ =\ ${\sf red}(a\star P_{k}\leq P_{k}<\ \ldots\ <P_{1}<\{f\})$.

The left action of $S_{A}^{\wedge_{\mathcal{L}}}$ is

$(s_{m}<_{\mathcal{L}}\ldots<_{\mathcal{L}}s_{1})$ $\star$ (P_{k}<\ \ldots\ <P_{1}<\{f\})\ =\

${\sf red}(s_{m}\star P_{k}\leq\ldots\leq$ $s_{1}\star P_{k}\leq P_{k}<\ \ldots\ <P_{1}<\{f\})$.

Here, the effect of the reduction operation red is to take the left-most element of an $\equiv$-chain.

When we start with a left action, we can define a corresponding right action in a similar way. This way we can iterate $(.)_{A}^{\wedge_{\mathcal{L}}}$ and $(.)_{A}^{\wedge_{\mathcal{R}}}$.

Proposition 2.16

When the automaton expansion $(.)_{A}^{\wedge_{\mathcal{L}}}$ is applied to the right-regular representation $(S^{1},1,\cdot)$ over the alphabet $A$, the classical Rhodes expansion $S_{A}^{\wedge_{\mathcal{L}}}$ is obtained.

Proof. We saw that $S_{A}^{\wedge_{\mathcal{L}}}$ acts on the state set of $(S^{1},1,\cdot)_{A}^{\wedge_{\mathcal{L}}}$; we want to show that this action is faithful: If ${\bf s,t}\in S_{A}^{\wedge_{\mathcal{L}}}$ act in same way on all states of $(S^{1},1,\cdot)_{A}^{\wedge_{\mathcal{L}}}$ we want to show that ${\bf s}={\bf t}$.

Let ${\bf s}=$ $(s_{m}<_{\mathcal{L}}\ \ldots\ <_{\mathcal{L}}s_{1})$ and ${\bf t}=$ $(t_{n}<_{\mathcal{L}}\ \ldots\ <_{\mathcal{L}}t_{1})$. We will prove by induction that ${\bf s}={\bf t}$. We will use the following (for any $x,y\in S^{1}$):

(l)                                   If  $y<_{\mathcal{L}}x$  then  $(x)y^{-1}=\varnothing$.

Indeed, if there exists $t\in(x)y^{-1}$ then $ty=x$, which contradicts $y<_{\mathcal{L}}x$. Moreover,

(i)                                   $x=y$  iff  $1\in(x)y^{-1}$.

For the base-case of the induction we show that $s_{n}=t_{m}$ follows from  ${\bf s}\star(\{s_{m}\}))={\bf t}\star(\{s_{m}\})$. Indeed, ${\bf s}\star(\{s_{m}\})=((s_{m})s_{m}^{-1}<\ \ldots\ )$  $=$  ${\bf t}\star(\{s_{m}\})$  $=$  $((s_{m})t_{n}^{-1}<\ \ldots\ )$. Since $1\in(s_{m})s_{m}^{-1}=(s_{m})t_{n}^{-1}$, it follows by (i) that $s_{m}=t_{n}$.

For the inductive step, let us assume that $s_{m}=t_{n}$, \ \ldots\, $s_{m-i+1}=t_{n-i+1}$, for some $i>1$; we want to show that $s_{m-i}=t_{n-i}$. We have

${\bf s}\star(\{s_{m-i}\})$  $=$  ${\sf red}((s_{m-i})s_{m}^{-1}\leq\ \ldots\ \leq(s_{m-i})s_{m-i+1}^{-1}$ $\leq(s_{m-i})s_{m-i}^{-1}\leq\ \ldots\ )$  $=$

{\bf t}\star(\{s_{m-i}\})\ =\ {\sf red}((s_{m-i})s_{m}^{-1}\leq\ \ldots\ $\leq(s_{m-i})s_{m-i+1}^{-1}\leq(s_{m-i})t_{m-i}^{-1}\leq\ \ldots\ )$.

By property (l) this becomes

${\bf s}\star(\{s_{m-i}\})\ =\ (\varnothing<(s_{m-i})s_{m-i}^{-1}<$ $\ \ldots\ )$  $=$  {\bf t}\star(\{s_{m-i}\})\ =\ $(\varnothing<(s_{m-i})t_{m-i}^{-1}\leq\ \ldots\ )$.

Also, $1\in(s_{m-i})s_{m-i}^{-1}=(s_{m-i})t_{m-i}^{-1}$, hence $s_{m-i}=t_{m-i}$.

By induction we conclude that  ${\bf t}=$ $(s_{m}<_{\mathcal{L}}\ \ldots\ <_{\mathcal{L}}s_{1}<_{\mathcal{L}}$ $t_{n-m}<_{\mathcal{L}}\ \ldots\ <_{\mathcal{L}}t_{1})$, and $m\leq n$. Moreover, by letting s and t act on the states $(\{t_{n}\}),\ \ldots\ ,(\{t_{1}\})$, we similarly find that $n\leq m$. Hence $n=m$, and ${\bf s}={\bf t}$.    $\Box$

The following can be further developed.

Variations on the definition: Consider automata with start state and final state. Redefine the right and left expansions so that the resulting automata have a start state and a final state.

More generally, take automata with a set $I$ of initial states and a set $F$ of final states. Then ${\bf A}^{\sf left}$ will have $Q^{\sf left}=\{(f)s^{-1}:\,s\in S^{1},\,f\in F,\,(\exists t\in S^{1})[(f)s^{-1}t^{-1}\cap I\neq\varnothing]\}$, and it will also have a set of initial states $I^{\sf left}=\{\{f\}:f\in F\}$ and a set of final states $F^{\sf left}=$ $\{P\in Q^{\sf left}:P\cap I\neq\varnothing\}$. And both ${\bf A}_{A}^{\wedge_{\mathcal{L}}}$ and ${\bf A}_{A}^{\wedge_{\mathcal{R}}}$ will have a set of initial states and a set of final states.

Now we can iterate the left and right expansions and obtain two-sided unambiguity.

3 A bottom-up inductive construction of the glc cover

3.1 Rank functions

The rank functions that we defined earlier will now be extended to edges and to walks of an $A$-automaton, within a reachability class. Let ${\bf A}=(Q,\cdot)$ be a finite automaton, let $\psi:{\bf A}\to{\bf B}$ be an $A$-automaton morphism. An edge is of the form $(q,\,a,\,q\cdot a)$, where $q\in Q,a\in A$; we will also simply write $(q,a)$. We say that an edge is in a reachability class iff $q\equiv_{\mathcal{R}}q\cdot a$.

More generally, a walk of length $n$ is of the form $(q,\,a_{1},\,q_{1},\,a_{2},\,q_{2},\ \ldots\ ,a_{i},\,q_{i},\,a_{i+1},$ $\ \ldots\ ,a_{n},\,q_{n})$, where $q,q_{1},\ldots,q_{n}\in Q$ and $a_{1},a_{2},\ldots,a_{n}\in A$, with $q_{i}=qa_{1}\ldots a_{i}$ for $i=1,2,\ldots n$. We will also simply denote a walk by $(q,\,a_{1}\ldots a_{n})$. A walk $(q,\,a_{1}\ldots a_{n})$ is said to be within a reachability class   iff   $q\cdot a_{1}\ldots a_{n}\equiv_{\mathcal{R}}q$.

Definition 3.1

When $q\equiv_{\mathcal{R}}q\cdot a$ we define the edge-rank of $(q,a)$ by

${\sf rank}_{\bf B}(q,a)$* \ =\ ${\sf min}\{{\sf rank}_{\bf B}(ax):\ x\in A^{*},\ q\cdot ax=q\}$.*

For a path $(q,w)$ within a reachability class we define the path-rank by

${\sf rank}_{\bf B}(q,w)$* \ =\ ${\sf min}\{{\sf rank}_{\bf B}(wx):\ x\in A^{*},\ q\cdot wx=q\}$.*

Notation:  For an $A$-automaton ${\bf A}=(Q,\cdot)$ and a word $w\in A^{*}$, $[w]$ denotes the element represented by $w$ in the syntactic monoid of A. This means that $[w]$ is the congruence class of $w$ for the monoid congruence $\equiv_{\bf A}$ on $A^{*}$ defined by   $x\equiv_{\bf A}y$   iff   $(\forall q\in Q)[\,q\cdot x=q\cdot y\,]$.

Definition 3.2

An $A$-automaton A has idempotent stabilizers iff for every state $q$ and every word $w\in A^{*}:$  $q\cdot w=q$ implies $[w]^{2}=[w]$.

Definition 3.3

An $A$-automaton A has ${\cal R}$-trivial stabilizers iff for every state $q$ and every word $w\in A^{*}:$  $q\cdot w=q$ implies that the $\equiv_{\mathcal{R}}$-class of $[w]$ is a singleton (in the syntactic monoid of A).

Proposition 3.4

(Invariance of rank under conjugation)*   Let ${\bf A}=(Q,\cdot)$ be an $A$-automaton with idempotent stabilizers, and suppose $(q,w)$ is a closed walk, i.e., $q\cdot w=q$, $w\in A^{*}$, $q\in Q$. If $w$ is factored as $w=xy$, then  ${\sf rank}_{\bf B}(xy)={\sf rank}_{\bf B}(yx)$.*

Proof. Since $qw=q$ and stabilizers are idempotents, we have $[w]^{2}=[w]$. Also, $qw=q$ implies $qwx=qx$, i.e., $qxyx=qx$ (since $w=xy$). So, $yx$ stabilizes $qx$, hence $[yx]^{2}=[yx]$. Moreover, $[xy]\equiv_{\mathcal{J}}[yx]$; indeed, $[xy]=[xyxy]\leq_{\mathcal{J}}[yx]$, and $[yx]=[yxyx]\leq_{\mathcal{J}}[xy]$. Now, since $[xy]$ and $[yx]$ are $\equiv_{\mathcal{J}}$-related idempotents, they have the same algebraic rank.    $\Box$

For Prop. 3.6 we will need an expansion with special properties; we first make sure that such an expansion is available. Let $(.)_{A}^{\wedge_{\cal L}}$ be the Rhodes expansion, let $(.)^{\sf RB}$ be the rectangular-bands expansion RB {\scriptsize\sfm}⃝ $(.)$, and let $(.)^{\sf IS}$ be the expansion from [7]. For properties of the Rhodes expansion, see [10] Appendix A.IV, [2], and Tilson’s chapter XII in [3]. We will use the fact that  $(.)^{{\sf RB}\,\wedge_{\cal L}}=(.)^{\sf RB}$, i.e., $(.)^{\sf RB}$ is stable under $(.)^{\wedge_{\cal L}}$.

Proposition 3.5

For any $A$-generated semigroup $S_{A}$, the expansions $S_{A}^{{\sf IS}\,\wedge_{\cal L}}$ and $S_{A}^{\sf IS\,RB}$ in the category of $A$-generated semigroups have the following properties: All right-stabilizers are $\cal R$-trivial bands, and the $\cal L$-order in each right-stabilizer is unambiguous.

Proof. In [7] it is proved that in $S_{A}^{\sf IS}$ the right-stabilizers are $\cal R$-trivial bands. For any $A$-generated semigroup $T_{A}$ the natural maps $\eta_{\cal L}:T_{A}^{\wedge_{\cal L}}\to T_{A}$ and $\eta_{\sf RB}:T_{A}^{\sf RB}\to T_{A}$ have the property that the inverse image $\eta^{-1}(B)$ of any band $B\subseteq T_{A}$ is a band (where $\eta$ stands for either $\eta_{\cal L}$ or $\eta_{\sf RB}$). And any stabilizer $\Sigma^{\prime}\subseteq T_{A}^{\wedge_{\cal L}}$ is contained in $\eta^{-1}\eta(\Sigma^{\prime})$, and $\eta(\Sigma^{\prime})$ is contained in a stabilizer $\Sigma$ in $T_{A}$; so any stabilizer $\Sigma^{\prime}\subseteq T_{A}^{\wedge_{\cal L}}$ is contained in $\eta^{-1}(\Sigma)$ where $\Sigma$ is a stabilizer of $T_{A}$ . Hence any stabilizer in $S_{A}^{{\sf IS}\,\wedge_{\cal L}}$ or in $S_{A}^{\sf IS\,RB}$ is a band.

Moreover, for any $T_{A}$, every right-stabilizer in $T_{A}^{\wedge_{\cal L}}$ is $\cal R$-trivial and the $\cal L$-order of $T_{A}^{\wedge_{\cal L}}$ is unambiguous. Since $T_{A}^{\sf RB}=T_{A}^{{\sf RB}\,\wedge_{\cal L}}$, every right-stabilizer is $\cal R$-trivial there too, and the $\cal L$-order is unambiguous.    $\Box$

Proposition 3.6

(Path-rank vs. edge-rank).*  Let A be an $A$-automaton whose stabilizers are $\cal R$-trivial bands with unambiguous $\cal L$-order. Let $(q_{0},\,a_{1}\ldots a_{n})$ be a walk within an ${\mathcal{R}}$-class, i.e., $q_{0}a_{1}\ldots a_{n}\equiv_{{}_{\mathcal{R}}}q_{0}$. Let $q_{i}$ denote $q_{0}a_{1}\ldots a_{i}$ for $i=1,\ldots,n$. Then the rank of the path is equal to the maximum of the edge-ranks, i.e.,*

(1)*               {\sf rank}_{\bf B}(q_{0},\,a_{1}\ldots a_{n})\ =\ ${\sf max}\{\,{\sf rank}_{\bf B}(q_{i-1},a_{i}):i=1,\ldots,n\}$.*

If the path $(q_{0},\,a_{1}\ldots a_{n})$ is closed (i.e., $q_{0}a_{1}\ldots a_{n}=q_{0}$) then the rank of the path is equal to the rank of the labeling word, i.e.,

(2)*                {\sf rank}_{\bf B}(q_{0},\,a_{1}\ldots a_{n})\ =\ ${\sf rank}_{\bf B}(a_{1}\ldots a_{n})$.*

(This Proposition will also be called the “Sausage Lemma” since a path between several reachability classes can be pictured as a string of sausages.)

Proof. In the proof we drop the ubiquitous subscript B of rank.

[(1) $\geq$]  By the definition of path-rank there exists $\beta\in A^{*}$ such that ${\sf rank}(q_{0},\,a_{1}\ldots a_{n})$ $=$ ${\sf rank}(a_{1}\ldots a_{n}\beta)$ and $q_{0}a_{1}\ldots a_{n}\beta=q_{0}$. The word $\beta$ is the “$x$” in the definition of path-rank for which the min is achieved. Then we have for any $j=1,\ldots,n$:

{\sf rank}(q_{0},\,a_{1}\ldots a_{n})\ =\ {\sf rank}(a_{1}\ldots a_{n}\,\beta)\

$=\ {\sf rank}(a_{j}\,a_{j+1}\ldots a_{n}\beta a_{1}\ldots a_{j-1})$             (by Prop. 3.4)

$\geq\ {\sf min}\{{\sf rank}(a_{j}x):$ $x\in A^{*},\ q_{j-1}a_{j}x=q_{j-1}\}$             (since $q_{j-1}\cdot a_{j}\ldots a_{n}\beta a_{1}\ldots a_{j-1}$ $=q_{j-1}$)

$=\ {\sf rank}(q_{j-1},a_{j})$             (by the definition of edge-rank).

Therefore, ${\sf rank}(q_{0},a_{1}\ldots a_{n})\geq{\sf rank}(q_{j-1},a_{j})$ for every $j=1,\ldots,n$; hence, ${\sf rank}(q_{0},a_{1}\ldots a_{n})\geq{\sf max}\{{\sf rank}(q_{j-1},a_{j}):$ $j=1,\ldots,n\}$.

[(1) $\leq$]  We use induction on the path-length $n$. For $n=1$, the two sides of (1) are identical. For $n>1$, assume   {\sf rank}(q_{0},a_{1}\ldots a_{n-1})\ \leq\ ${\sf max}\{{\sf rank}(q_{j-1},a_{j}):j=1,\ldots,n-1\}$; equivalently, {\sf rank}(q_{0},a_{1}\ldots a_{n-1})\ \leq\ ${\sf rank}(q_{j-1},a_{j})$ for some $j=1,\ldots,n-1$. We want to prove that ${\sf rank}(q_{0},a_{1}\ldots a_{n-1}a_{n})\leq$ ${\sf rank}(q_{i-1},a_{i})$ for some $i=1,\ldots,n-1,n$.

Let $\beta\in A^{*}$ be such that   $q_{0}a_{1}\ldots a_{n-1}\beta=q_{0}$   and   ${\sf rank}(q_{0},a_{1}\ldots a_{n-1})={\sf rank}(a_{1}\ldots a_{n-1}\beta)$, i.e., the min in the definition of path-rank is reached when $x$ is $\beta$. Similarly, let $\beta^{\prime}\in A^{*}$ be such that   $q_{n-1}a_{n}\beta^{\prime}=q_{n}$   and   ${\sf rank}(q_{n-1},a_{n})={\sf rank}(a_{n}\beta^{\prime})$. Then each of the following is a closed path:

$(q_{n-1},\ a_{n}\beta^{\prime})$,   $(q_{n-1},\ \beta a_{1}\ldots a_{n-1})$,   $(q_{0},\ a_{1}\ldots a_{n-1}\beta)$ ,  and  $(q_{0},\ a_{1}\ldots a_{n-1}a_{n}\beta^{\prime}\beta)$.

It follows that  $[a_{n}\beta^{\prime}]$,  $[\beta a_{1}\ldots a_{n-1}]$,  $[a_{1}\ldots a_{n-1}\beta]$,  and  $[a_{1}\ldots a_{n-1}a_{n}\beta^{\prime}\beta]$  belong to stabilizers, hence they are idempotents.

By the definition of rank (using min), ${\sf rank}(q_{0},\,a_{1}\ldots a_{n-1}a_{n})\leq$ ${\sf rank}(a_{1}\ldots a_{n-1}a_{n}\beta^{\prime}\beta)$, and by Prop. 3.4,   {\sf rank}(a_{1}\ldots a_{n-1}a_{n}\beta^{\prime}\beta)\ =\ ${\sf rank}(\beta a_{1}\ldots a_{n-1}a_{n}\beta^{\prime})$.

Case 1:  $[\beta a_{1}\ldots a_{n-1}]$ $\equiv_{\cal R}$ $[\beta a_{1}\ldots a_{n-1}a_{n}\beta^{\prime}]$

Then by $\cal R$-triviality of stabilizers, $[\beta a_{1}\ldots a_{n-1}a_{n}\beta^{\prime}]=[\beta a_{1}\ldots a_{n-1}]$, and these are idempotents. Hence, ${\sf rank}(q_{0},\,a_{1}\ldots a_{n-1}a_{n})\leq$ ${\sf rank}([\beta a_{1}\ldots a_{n-1}a_{n}\beta^{\prime}])=$ ${\sf rank}([\beta a_{1}\ldots a_{n-1}])=$ ${\sf rank}(q_{0},\,a_{1}\ldots a_{n-1})\leq{\sf rank}(q_{j-1},a_{j})$ for some $j$ (the latter by the induction hypothesis).

Case 2:  $[\beta a_{1}\ldots a_{n-1}]$ $>_{\cal R}$ $[\beta a_{1}\ldots a_{n-1}a_{n}\beta^{\prime}]$ $\leq_{\cal L}$ $[a_{n}\beta^{\prime}]$.

Case 2.1:  $[\beta a_{1}\ldots a_{n-1}a_{n}\beta^{\prime}]\equiv_{\cal L}[a_{n}\beta^{\prime}]$

Then, since these are idempotents, ${\sf rank}(q_{0},\,a_{1}\ldots a_{n-1}a_{n})\leq$ ${\sf rank}([\beta a_{1}\ldots a_{n-1}a_{n}\beta^{\prime}])=$ ${\sf rank}([a_{n}\beta^{\prime}])={\sf rank}(q_{n-1},a_{n})$.

Case 2.2:  $[\beta a_{1}\ldots a_{n-1}]$ $>_{\cal R}$ $[\beta a_{1}\ldots a_{n-1}a_{n}\beta^{\prime}]<_{\cal L}[a_{n}\beta^{\prime}]$.

Let $e_{1}=[\beta a_{1}\ldots a_{n-1}]$,  $e_{2}=[a_{n}\beta^{\prime}]$; then $e_{1}e_{2}=[\beta a_{1}\ldots a_{n-1}a_{n}\beta^{\prime}]$, and $e_{1}e_{2}=(e_{1}e_{2})^{2}$, and $e_{1}>_{\cal R}e_{1}e_{2}<_{\cal L}e_{2}$. Since stabilizers are $\cal R$-trivial bands, $e_{1}e_{2}=e_{1}e_{2}e_{1}$. Similarly, $e_{2}e_{1}=e_{2}e_{1}e_{2}$.

Thus, $e_{1}>_{\cal L}e_{2}e_{1}=e_{2}e_{1}e_{2}<_{\cal L}e_{2}$, which by unambiguity of the $\cal L$-order implies that either $e_{1}\leq_{\cal L}e_{2}$, i.e., $e_{1}=e_{1}e_{2}$, or $e_{2}\leq_{\cal L}e_{1}$, i.e., $e_{2}=e_{2}e_{1}$. But $e_{1}>_{\cal R}e_{1}e_{2}$ contradicts $e_{1}=e_{1}e_{2}$. And $e_{1}e_{2}<_{\cal L}e_{2}$ contradicts $e_{2}=e_{2}e_{1}$ (since $e_{1}e_{2}=e_{1}e_{2}e_{1}\equiv_{\cal L}e_{2}e_{1}$). Thus, when the syntactic monoid of the automaton A has unambiguous $\cal L$-order case 2.2 is impossible.

Proof of (2):  By definition, ${\sf rank}(q_{0},a_{1}\ldots a_{n})=$ ${\sf min}\{{\sf rank}(a_{1}\ldots a_{n}x):q_{0}a_{1}\ldots a_{n}x=q_{0}\}$. We assumed that $a_{1}\ldots a_{n}$ stabilizes $q_{0}$. By the properties of stabilizers, $[a_{1}\ldots a_{n}]$ and $[a_{1}\ldots a_{n}x]$ are idempotents. Since $a_{1}\ldots a_{n}x\leq_{\cal J}a_{1}\ldots a_{n}$ and since these are idempotents, we have ${\sf rank}(a_{1}\ldots a_{n}x)\geq{\sf rank}(a_{1}\ldots a_{n})$ for any $[x]$, with equality when $x$ is empty. Hence, ${\sf min}\{{\sf rank}(a_{1}\ldots a_{n}x):q_{0}a_{1}\ldots a_{n}x=q_{0}\}$ $={\sf rank}(a_{1}\ldots a_{n})$.    $\Box$

For the rest of this subsection let $T_{A}$ be an $A$-generated semigroup. Every $\cal R$-class $R$ of $T_{A}$ will be viewed as a deterministic partial finite automaton (with state set $R$ and alphabet $A$). This is a strongly connected component of the right Cayley graph $\Gamma(T_{A})$. We call this the $\cal R$-class automaton of $R$.

We assume that the semigroup $T_{A}$ satisfies the assumptions of Prop. 3.6, i.e., that its stabilizers are $\cal R$-trivial bands with unambiguous $\cal L$-order.

We consider the algebraic rank of elements of $T_{A}$ with respect to the identity map on $T_{A}$, denoted by ${\sf rank}(.)$. We also consider the edge-rank function, as in Definition 3.1, with respect to the identity map on $T_{A}$. We also call this rank function ${\sf rank}(.)$. So for an edge $(q,a)$ with $q,qa\in R$ we have: ${\sf rank}(q,a)={\sf min}\{{\sf rank}(ax):x\in A^{*},\ qax=a\}$; here, ${\sf rank}(ax)$ is the algebraic rank.

Definition 3.7

For the $\cal R$-class $R$ of an $A$-automaton we define

{\sf rank}(R)\ =\* ${\sf max}\{{\sf rank}(q,a)\,:\,q,\,qa\in R,\,a\in A\}$.*

For $r\in R$ and $j=0,1,\ldots$ we define

{\sf paths}(r,j)\ =\* $\{w\in A^{*}:\,r\cdot w\in R$ and every edge $e$ of the path $(r,w)$ in the automaton $R$*

has rank ${\sf rank}(e)\leq j\}$,

$P(r,j)$* ($\subseteq R$) is the set of vertices of the paths in ${\sf paths}(r,j)$,*

$E(r,j)$* is the set of edges of the paths in ${\sf paths}(r,j)$.*

Proposition 3.8

For every $\cal R$-class $R$ of an $A$-automaton and every fixed $j_{o}=0,1,\ldots$ we have: The set of sets $\{P(r,j_{o}):r\in R\}$ is a partition of $R$.

Proof. Since $r\in P(r,j_{o})$ the sets $P(r,j_{o})$ are obviously non-empty and $R=\,\bigcup_{r\in R}P(r,j_{o})$. Let us prove that non-equal sets that overlap are equal. If $r\in P(r_{i},j_{o})$, then there is a path in $R$ from $r_{i}$ to $r$, labelled by some $\alpha\in A^{+}$, with maximum edge rank $\leq j_{o}$. Then by Prop. 3.6 (Sausage Lemma), this path can be extended to a cycle containing $r_{i}$ with no increase in maximum edge rank. Hence, if $r\in P(r_{1},j_{o})\cup P(r_{2},j_{o})$ then there are paths from $r_{1}$ to $r_{2}$ and from $r_{2}$ to $r_{1}$ of maximum edge rank $\leq j_{o}$; so $P(r_{1},j_{o})=P(r_{2},j_{o})$.    $\Box$

Examples of $P(.,.)$ and ${\sf paths}(.,.)$:

(1)  Suppose the state diagram of $R$, with ranks above and below the edges, is

1 2 1 q1 -----> q2 -----> q3 -----> q4 <----- <----- <----- 1 1 1

Then the partition for $j=0$ is $\,\{\{q_{1}\},\{q_{2}\},\{q_{3}\},\{q_{4}\}\}$, for $j=1$ it is $\,\{\{q_{1},q_{2}\},\{q_{3},q_{4}\}\}$, and for $j=2$ it is $\,\{\{q_{1},q_{2},q_{3},q_{4}\}\}$.

(2)  Suppose the state diagram of $R$ consists of the states $q_{1}$ and $q_{2}$, with edges $(q_{1},a\,(1),q_{2})$,   $(q_{1},b\,(2),q_{2})$, and $(q_{2},c\,(1),q_{1})$, with ranks in parentheses. Then ${\sf paths}(q_{1},1)$ consists of all paths from $q_{1}$ with label in $(ac)^{*}\cup(ac)^{*}a$. And $P(q_{1},1)=P(q_{2},1)=P(q_{1},2)=P(q_{2},2)=\{\{q_{1},q_{2}\}\}$.

[End, Examples.]

We are now going back to the covers  U_{A}=T_{A}^{\sf cov\ IS\ RB}\ $\stackrel{{\scriptstyle\varphi_{1}}}{{\twoheadrightarrow}}\ [T_{A},U_{A}]^{\sf glc}$ $\ \stackrel{{\scriptstyle\varphi_{2}}}{{\twoheadrightarrow}}\ T_{A}$, where $\varphi_{2}$ is 1:1$\cal R$ and satisfies the rank condition. In this setting we will also write UP for $U_{A}$, MIDDLE for $[T_{A},U_{A}]^{\sf glc}$, and DOWN for $T_{A}$. Let $\tilde{R}$ be an $\cal R$-class of MIDDLE, let $R$ be the $\cal R$-class of DOWN such that $\varphi_{2}:\tilde{R}\ \hookrightarrow\ R$. To assign a rank to an edge $\tilde{r}\stackrel{{\scriptstyle a}}{{\longrightarrow}}\tilde{r}a$ of $\tilde{R}$, we apply $\varphi_{2}$, yielding $(\tilde{r})\varphi_{2}$ $\stackrel{{\scriptstyle a}}{{\longrightarrow}}(\tilde{r}a)\varphi_{2}$, and take the rank of this arrow in $R$. Recall that ${\sf rank}(\tilde{R})$ is the max rank over all idempotents in $E(\tilde{R})$ ($=E(R)$). Notice that by Prop. 3.6, ${\sf rank}(R)$ is the same as algebraic rank of all idempotents that fix any element of $R$ under right-multiplication. We also have $\tilde{n}={\sf rank}(\tilde{R})\leq{\sf rank}(R)=n$.

For an $\cal R$-class $C$ of a semigroup, viewed as an $A$-automaton, we denote the edge set of $C$ by $E(C)$. As above, $E(r,j)$ is the set of edges of ${\sf paths}(r,j)$ in the $\cal R$-class automaton $R$.

Proposition 3.9

For $\tilde{R}$ and $R$ as above and for any $\tilde{r}\in\tilde{R}$ and $r=(\tilde{r})\varphi_{2}\in R$ we have:

$E(\tilde{R})\ =\ E(r,\tilde{n})\ \subseteq\ E(R)$.

Proof. Let $\alpha\in A^{*}$ be the label of a cycle in $\tilde{R}$, starting at $\tilde{r}$, that traverses every edge of $\tilde{R}$. Then each path in ${\sf paths}(r,\tilde{n})$ can be extended to a cycle that returns to $r$, and that uses only edges of edge-rank $\leq\tilde{n}$ (by Prop. 3.6, the “Sausage Lemma”). Now the rank condition implies the result as follows: We have $\tilde{r}\alpha=\tilde{r}$, and $\alpha$ has at least one edge of rank $\tilde{n}$, and passes through all vertices of $\tilde{R}$. If $r\beta=r$ and ${\sf rank}(\beta)\leq{\sf rank}(\alpha)=\tilde{n}$ then $\tilde{r}\beta=\tilde{r}$, by the rank condition. So, $E(\tilde{R})=E(r,\tilde{n})\subseteq E(R)$.    $\Box$

Corollary 3.10

If ${\sf rank}(\tilde{R})={\sf rank}(R)$ then the restriction of $\varphi_{2}$ to $\tilde{R}$ is an isomorphism from $\tilde{R}$ to $R$.     $\Box$

We will give a formula for ${\sf rank}(R)$ in terms of idempotents. The following Lemma will be used. For idempotents $e,f$ we write   $e\geq f$  iff  $f=ef=fe$; this is called the idempotent order, or equivalently, the $\cal H$-order between idempotents; it is a partial order. We write $e>f$ iff $e\geq f$ and $e\neq f$.

Lemma 3.11

(Rhodes 1966).*  Let $J_{1}$ and $J_{2}$ be regular $\cal J$-classes of a finite semigroup, such that $J_{1}>_{\cal J}J_{2}$. Then for every idempotent $e\in J_{1}$ there exists an idempotent $f\in J_{2}$ such that $e>f$.*

Proof. See the proof of Prop. 3.1 in [9].     $\Box$

Proposition 3.12

Suppose $R$ is a regular $\cal R$-class of $T_{A}$ ($={\sf DOWN}$) such that the right-stabilizer of any element of $R$ is an $\cal R$-trivial band with unambiguous $\cal L$-order. Then ${\sf rank}(R)=n$ implies that there exist $r\in R$, and idempotents $e_{0},e_{1},\ \ldots\ ,e_{n}$, such that

$e_{0}>e_{1}>\ \ldots\ >e_{n}\equiv_{\cal L}r$* ,*

where “$>$” is the strict idempotent order. Hence the stabilizer $(r){\sf St}$ contains the idempotent-order chain $e_{0}>e_{1}>\ \ldots\ >e_{n}$.

Proof. By definition, ${\sf rank}(R)$ is the maximum of the algebraic ranks of all idempotents that fix any element of $R$ under right multiplication. So if ${\sf rank}(R)=n$, there exists $e=e^{2}\in R$ with ${\sf rank}(e)=n$, such that  $e\equiv_{\cal J}e^{\prime}_{n}<_{\cal J}\ \ldots\ <_{\cal J}e^{\prime}_{1}<_{\cal J}$ $e^{\prime}_{0}$   (a $\cal J$-chain of idempotents). Applying Lemma 3.12 $n$ times yields $e\equiv_{\cal J}$ $e_{n}<\ \ldots\ <e_{1}<e_{0}$, a chain in the idempotent order. Hence there exists $r\in R$ with $e\equiv_{\cal R}r$ and $r\equiv_{\cal L}e_{n}$. So, $(r){\sf St}=(e_{n}){\sf St}$, which contains $e_{n}<\ \ldots\ <e_{1}<e_{0}$.         $\Box$

Examples.

The following examples show that the stabilizers of the elements $r\in R$ are not all isomorphic. And when ${\sf rank}(R)=n$, some of these stabilizers contains an idempotent chain $e_{n}<\ \ldots\ <e_{1}<e_{0}$, and some stabilizers do not.

(a)  Consider the semigroup $T=\{f,a,b\}$ with the multiplication table

[TABLE]

Then the rank of the ${\cal R}$-class $\{a,b\}$ is 1, $(a){\sf St}=\{a\}$, and $(b){\sf St}=\{b,f\}$ (with $b<f$). So $(b){\sf St}$ contains a chain of idempotents of length 1 (counting the inequalities in the chain), while $(a){\sf St}$ does not.

(b)  Suppose the following diagram is part of the Cayley graph of a semigroup $S_{A}$. The vertices are $\{q_{1},q_{2},q_{3},q_{4}\}$ and the edges are

$(q_{1},a,q_{2})$,   $(q_{2},b,q_{1})$,   both with rank 1;

$(q_{2},c,q_{3})$,   $(q_{3},d,q_{2})$,   both with rank 2;

$(q_{3},e,q_{4})$,   $(q_{4},f,q_{3})$,   both with rank 3.

Then under the hypotheses of Prop. 3.6, the ranks of the right-stabilizers are as follows.

$(q_{1}){\sf St}\,$ has elements of ranks 1, 2, 3;

$(q_{4}){\sf St}\,$ has rank 3 only;

$(q_{2}){\sf St}\,$ has ranks 1, 2, 3.;

$(q_{3}){\sf St}\,$ has ranks 2 and 3.

This is proved as follows.

First, if a word $\ell_{j}\in\{a,b,c,d,e,f\}^{+}$ (for $j=1,2$) fixes $q\in\{q_{1},q_{2},q_{3},q_{4}\}$, i.e., $q\ell_{j}=q$, then $\ell_{j}$ is an idempotent (by the hypotheses of Prop. 3.6); moreover, $\ell_{2}\ell_{1}=\ell_{2}$ iff $\ell_{2}\leq_{\cal L}\ell_{1}$. Also, the $\cal L$-order of $(q){\sf St}$ is the $\cal L$-order of the whole semigroup since $\ell_{2},\ell_{1}$ are idempotents. Since $(q){\sf St}$ has unambiguous $\cal L$-order, either $\ell_{2}\leq_{\cal L}\ell_{1}$ or $\ell_{1}\leq_{\cal L}\ell_{2}$; i.e., either $\ell_{2}\ell_{1}=\ell_{2}$ or $\ell_{1}\ell_{2}=\ell_{1}$. The rank condition determines the direction of the $\cal L$-order: ${\sf rank}(\ell_{2})\geq{\sf rank}(\ell_{2})$ implies $\ell_{2}\ell_{1}=\ell_{2}$, etc., as is easy to see.

Second, we have already proved the conjugation property of loops (Prop. 3.4), i.e., the rank of a loop does not depend on the chosen starting point on the loop. We will denote conjugation by $\sim$.

From these two observations we now prove the rank properties of the stabilizers. To see that ${\sf St}(q_{1})$ has ranks 1, 2, 3, we consider the loops $(q_{1},ab)$,  $(q_{1},acdb)$,  $(q_{1},acefdb)$ with start point $q_{1}$; these loops have rank respectively 1, 2, 3. For example, to see that $acefdb$ has rank 3, we observe that $(q_{1},acefdb)\sim(q_{3},efdbac)$, and $(q_{3},dbac)\sim(q_{2},cdba)$; moreover, $cd$ has rank 2, and $ba$ has rank 1. So, ${\sf rank}(cdba)={\sf rank}(cd)$, and $dbac\sim cdba$, $cd\sim dc$; so ${\sf rank}(dbac)={\sf rank}(dc)=2$, Since ${\sf rank}(ef)=3$ and ${\sf rank}(dbac)=2$, we have ${\sf rank}(efdbac)={\sf rank}((ef)(dbac))={\sf rank}(ef)=3$. So, ${\sf rank}(acefdb)=3$.

3.2 Bottom-up construction

We construct the cover ${\bf A}^{\sf cov\,glc}$ in a different way than before, using “bottom-up” induction. We will later prove that under certain conditions this bottom-up construction yields the same cover ${\bf A}^{\sf cov\,glc}$ as defined earlier. However, in the meanwhile we need to distinguish the two, and we will denote the cover resulting from the bottom-up construction by ${\bf A}^{\sf cov\,GLC}$.

We start with a cover morphism $\varphi:{\bf A}^{\sf cov}\to{\bf A}$ and the interval $[{\bf A},{\bf A}^{\sf cov}]$, which is a finite lattice. Since ${\bf A}^{\sf cov\,GLC}$ is intended to belong to the interval $[{\bf A},{\bf A}^{\sf cov}]$, we want to construct morphisms $\varphi_{1}:{\bf A}^{\sf cov}\to{\bf A}^{\sf cov\,GLC}$ and $\varphi_{2}:{\bf A}^{\sf cov\,GLC}\to{\bf A}$ such that $(.)\varphi=(.)\varphi_{1}\circ\varphi_{2}$, and such that $\varphi_{2}$ is 1:1${\mathcal{R}}$ and satisfies the rank condition. Moreover, $\varphi_{2}$ should be maximal (i.e., ${\sf mod}\varphi_{1}$ should be maximally fine) with respect to these properties. Hence the congruence ${\sf mod}\varphi_{1}$ is a refinement of the congruence ${\sf mod}\varphi$ on the state set $Q^{\sf cov}$ of ${\bf A}^{\sf cov}$, and every ${\sf mod}\varphi_{1}$ class is mapped to one element of $Q$ by $\varphi$ (where $Q$ is the state-set of A). We will write states of ${\bf A}^{\sf cov}$ with an overline to make them recognizable. For a state ${\overline{q}}$ of ${\bf A}^{\sf cov}$, we denote the ${\sf mod}\varphi_{1}$ congruence class of ${\overline{q}}$ by $[{\overline{q}}]_{{\sf mod}\varphi_{1}}$ or more briefly by $[{\overline{q}}]_{\varphi_{1}}$.

We will construct ${\bf A}^{\sf cov\,GLC}$ by defining the congruence ${\sf mod}\varphi_{1}$ on $Q^{\sf cov}$, and by defining $\varphi_{2}$ on each constructed congruence class. The construction of $[{\overline{q}}]_{\varphi_{1}}$ proceeds by induction on the directed path-length from ${\overline{\i}}$ to ${\overline{q}}$ in ${\bf A}^{\sf cov}$, where ${\overline{\i}}$ is the start state of ${\bf A}^{\sf cov}$. In this induction we assume that the $\cal R$-order of each ${\bf A}^{\sf cov}$ is a tree; this will hold if $(.)^{\sf cov}$ is closed under the right Rhodes expansion $(.)^{\wedge_{\cal R}}$.

First, the congruence class $[{\overline{\i}}]_{\varphi_{1}}$ is $\{{\overline{\i}}\}$, since the ${\sf mod}\varphi$ congruence class of ${\overline{\i}}$ is just $\{{\overline{\i}}\}$ (assuming that ${\overline{\i}}$ is not reachable from any state).

For the inductive step we assume that for every state $\overline{q}_{1}$ with a certain directed path-length from ${\overline{\i}}$, a congruence class $[{\overline{q}}_{1}]_{\varphi_{1}}$ $=\{{\overline{q}}_{1},\ \ldots\ ,{\overline{q}}_{\ell}\}$ has been constructed. Our goal is to construct the congruence class $[{\overline{q}}_{1}\cdot a]_{\varphi_{1}}$ for each $a\in A$ such that the directed path-length from ${\overline{\i}}$ to $\overline{q}_{1}\cdot a$ is larger than the directed path-length from ${\overline{\i}}$ to $\overline{q}_{1}$.

Note that $[{\overline{q}}_{j}\cdot a]_{\varphi_{1}}$ $=$ $[{\overline{q}}_{1}\cdot a]_{\varphi_{1}}$, for $j=1,\dots,\ell$; hence $\{{\overline{q}}_{1}\cdot a,\ \ldots\ ,{\overline{q}}_{\ell}\cdot a\}$ $\subseteq$ $[{\overline{q}}_{1}\cdot a]_{\varphi_{1}}$. Also, since ${\overline{q}}_{j}$ is ${\sf mod}\varphi_{1}$-equivalent to ${\overline{q}}_{1}$ and since ${\sf mod}\varphi_{1}$ refines ${\sf mod}\varphi$, we have for all $j$:  $({\overline{q}}_{j})\varphi=({\overline{q}}_{1})\varphi$  and  $({\overline{q}}_{j}\cdot a)\varphi=({\overline{q}}_{1}\cdot a)\varphi$.

In A we consider the $\equiv_{\mathcal{R}}$-class $R=[({\overline{q}}_{1}a)\varphi]_{\equiv_{\mathcal{R}}}$, and the corresponding $\equiv_{\mathcal{R}}$-class $\overline{R}$ $=$ $[\overline{q}_{1}a]_{\equiv_{\mathcal{R}}}$ in ${\bf A}^{\sf cov}$. Since $({\overline{q}}_{j})\varphi=({\overline{q}}_{1})\varphi$, we have $R=[({\overline{q}}_{1}a)\varphi]_{\equiv_{\mathcal{R}}}=$ $[({\overline{q}}_{j}a)\varphi]_{\equiv_{\mathcal{R}}}$ for $j=1,\ldots,\ell$. Recall that in ${\bf A}^{\sf cov}$ we define ${\sf rank}(\overline{R})$ to be the maximum of the ranks of all the edges in $\overline{R}$, and that the rank of an edge $(q_{1},a,q_{2})$ in $\overline{R}$ is defined to be the rank of the image edge $((q_{1})\varphi,a,(q_{2})\varphi)$ in $R$. Let $E(S)$ be the set of edges between states in a set $S$. Below, ${\sf paths}(.)$ is taken in the $\cal R$-class $R$.

We now proceed in a number of stages.

Stage 1:  Let

$n_{1}\ =\ {\sf max}\{\,{\sf rank}(\overline{e})\ :\ \overline{e}\,\in\,$ $\bigcup_{1\leq j\leq\ell}$ E([{\overline{q}}_{j}a]_{\equiv_{\mathcal{R}}})\ \subseteq\ $E({\bf A}^{\sf cov})\,\}$   (where ${\sf max}(\varnothing)=-1$),

S_{1}\ =\ {\sf paths}\big{(}(\overline{q}_{1}a)\varphi,\ n_{1}\big{)} $\ \cup\ \{\varepsilon\}$,

$\overline{P}_{1}\ =\ \{\overline{q}_{j}at\in Q^{\sf cov}\ :\ t\in S_{1},$ $\ 1\leq j\leq\ell\}$.

Stage $i+1$:  Assuming $S_{h},\,{\overline{P}}_{h}$ have been defined for $1\leq h\leq i$, let

$n_{i+1}\ =\ {\sf max}\{\,{\sf rank}(\overline{e})\ :\ \overline{e}\,$ is an edge of an $\cal R$-class of $\overline{P}_{i}\,\}$,

S_{i+1}\ =\ {\sf paths}\big{(}(\overline{q}_{1}a)\varphi,\ n_{i+1}\big{)} $\ \cup\ \ \{\varepsilon\}$,

${\overline{P}}_{i+1}\ =\ \bigcup\,\{[\overline{q}_{j}at]_{\equiv_{\cal R}}\ :$ \ \overline{q}_{j}at\in Q^{\sf cov},\ $t\in S_{i+1},\ 1\leq j\leq\ell\}$,

where $[\overline{q}_{j}at]_{\equiv_{\cal R}}$ is the $\cal R$-class of $\overline{q}_{j}at$ in ${\bf A}^{\sf cov}$.

Stage End:  Continuing in this way we construct chains

n_{1}\leq n_{2}\leq\ \ldots\ \leq n_{i}\leq\ \ldots\ $\leq n_{\infty}={\sf max}\{n_{i}:i=1,2,\ \ldots\ \}$,

$S_{1}\subseteq S_{2}\subseteq\ \ldots\ \subseteq S_{i}\subseteq$ $\ \ldots\ \subseteq\ S_{\infty}=\bigcup_{i}S_{i}\ \subseteq A^{*}$,  and

$\overline{P}_{1}\subseteq\overline{P}_{2}\subseteq\ \ldots\ \subseteq\overline{P}_{i}$ \subseteq\ \ldots\ \subseteq\ $\overline{P}_{\infty}=\bigcup_{i}{\overline{P}}_{i}\ \subseteq Q^{\sf cov}$.

These sequences are actually finite. Indeed, ${\overline{P}}_{i}\subseteq Q^{\sf cov}$, which is a fixed finite set. Hence, $n_{i}$ is bounded since it is defined in terms of ${\overline{P}}_{i-1}$. Hence the sequence $S_{i}$ is of bounded, being defined in terms of $n_{i}$.

To define $\varphi_{1}$, $\varphi_{2}$, and ${\bf A}^{\sf cov\,GLC}$, we want to construct the ${\sf mod}\varphi_{1}$ congruence class $[\overline{q}_{j}a]_{\varphi_{1}}$ so that $\overline{q}_{j}a\stackrel{{\scriptstyle\varphi_{1}}}{{\longmapsto}}[\overline{q}_{j}a]_{\varphi_{1}}$ $\stackrel{{\scriptstyle\varphi_{2}}}{{\longmapsto}}(\overline{q}_{1}a)\varphi$, where $\varphi_{1}\varphi_{2}=\varphi$, and where $\overline{q}_{j}a$ and $(\overline{q}_{j}a)\varphi=(\overline{q}_{1}a)\varphi$ are known. And we want $\varphi_{2}$ to be injective on each $\cal R$-class. So $([(\overline{q}_{1}a)\varphi]_{\equiv_{\cal R}})\varphi^{-1}$ is a union of $\cal R$-classes of ${\bf A}^{\sf cov}$, each of which maps into $[(\overline{q}_{1}a)\varphi]_{\equiv_{\cal R}}$. By construction, $\overline{P}_{\infty}$ is a union of $\cal R$-classes of ${\bf A}^{\sf cov}$.

Therefore, for $\overline{q}_{j}a\in\overline{P}_{\infty}$ we define $[\overline{q}_{j}a]_{\varphi_{1}}$ to be  $(\overline{q}_{1}a)\varphi\varphi^{-1}\ \cap\ \overline{P}_{\infty}$. More generally, for any $\overline{q}\in\overline{P}_{\infty}$ we define $[\overline{q}]_{\varphi_{1}}$ to be

$(\overline{q})\varphi\varphi^{-1}\ \cap\ \overline{P}_{\infty}$.

Then $\varphi_{2}$ is defined by

$[\overline{q}]_{\varphi_{1}}$ \ \longmapsto\ $([\overline{q}]_{\varphi_{1}})\varphi\ \in Q$

where $Q$ is the state set of A. This completes the inductive step of the construction of ${\bf A}^{\sf cov\,GLC}$.

3.3 Proof of correctness

Proposition 3.13

Suppose that A is the Cayley graph of an $A$-monoid $M_{A}$ such that the stabilizers of ${\bf A}^{\sf cov}$ are $\cal R$-trivial bands with unambiguous $\cal L$-order, and such that the $\cal R$-order of $M_{A}$ is a tree. Then ${\sf glc}={\sf GLC}$.

Proof.  By construction $\varphi$, $\varphi_{1}$, $\varphi_{2}$ are $A$-automaton morphisms satisfying $\varphi=\varphi_{1}\circ\varphi_{2}$. We want to show that $\varphi_{2}:{\bf A}^{\sf cov\,GLC}\to{\bf A}$ is 1:1${\mathcal{R}}$ and obeys the rank condition; and we want to show that $\varphi_{2}$ is maximal among all the right-factors of $\varphi$ that are 1:1${\mathcal{R}}$ and that satisfy the rank condition.

Proof that $\varphi_{2}$ is 1:1${\mathcal{R}}$ and satisfies the rank condition

Let ${\overline{q}}\in Q^{\sf cov}$. If $(\overline{q}a)\varphi$ is in the $\cal R$-class $R$, and $(\overline{q}a)\varphi_{1}$ is in the $\cal R$-class $\tilde{R}$, then by construction (see also Prop. 3.9), E\big{(}(\overline{q}a)\varphi,\,n_{\infty}\big{)}\subseteq E(R); recall that $E(r,j)$ denotes the set of edges of ${\sf paths}(r,j)$. Then, clearly, $\varphi_{2}$ is 1:1$\cal R$ and satisfies the rank condition, since it embeds the edges of {\sf paths}\big{(}(\overline{q}a)\varphi,\,n_{\infty}\big{)} into $E(R)$.

Proof of maximality of $(.)^{\sf cov\,GLC}$ with respect to the 1:1${\mathcal{R}}$ property and the rank condition

We want to show that $\varphi_{2}$ is the maximal right-factor of $\varphi$ that is 1:1${\mathcal{R}}$ and satisfies the rank condition. Let B be an $A$-automaton, and let $\theta_{1}:{\bf A}^{\sf cov}\twoheadrightarrow{\bf B}$ and $\theta_{2}:{\bf B}\twoheadrightarrow{\bf A}$ be automaton morphisms, where $\theta_{2}$ is 1:1${\mathcal{R}}$ and satisfies the rank condition, and such that $(.)\varphi=(.)\theta_{1}\theta_{2}$. Hence the congruence ${\sf mod}{\theta_{1}}$ refines ${\sf mod}{\varphi}$. We want to show that the congruence ${\sf mod}{\varphi_{1}}$ refines ${\sf mod}{\theta_{1}}$.

Claim.  For any ${\overline{q}}\in A^{\sf cov}$,  $[\overline{q}]_{\varphi_{1}}\subseteq[\overline{q}]_{\theta_{1}}$.

Proof of the Claim: We follow the inductive construction of ${\sf mod}{\varphi_{1}}$. First, ${\overline{q}}\in[\overline{q}]_{\theta_{1}}$, obviously.

We will use $E(V)$ to denote the set of edges between vertices in $V$ (including loops). We have:  $[{\overline{q}}]_{\equiv_{\mathcal{R}}}\subseteq$ $\bigcup_{{\overline{p}}\in[{\overline{q}}]_{\theta_{1}}}[{\overline{p}}]_{\equiv_{\mathcal{R}}}$ $=$ $[[{\overline{q}}]_{\theta_{1}}]_{\equiv_{\mathcal{R}}}$; hence, since $\theta_{2}$ is 1:1$\mathcal{R}$, we obtain  $E_{1}^{o}\subseteq$ E\big{(}\big{(}[[{\overline{q}}]_{\theta_{1}}]_{\equiv_{\mathcal{R}}}\big{)}\varphi\big{)}. Since $\theta_{2}$ satisfies the rank condition we also have $E_{1}\subseteq$ E\big{(}\big{(}[[{\overline{q}}]_{\theta_{1}}]_{\equiv_{\mathcal{R}}}\big{)}\varphi\big{)}. Thus, ${\overline{P}}_{1}\subseteq[{\overline{q}}]_{\theta_{1}}$.

Continuing along the inductive construction of ${\sf mod}{\varphi_{1}}$ we obtain ${\overline{P}}_{\infty}\subseteq[{\overline{q}}]_{\theta_{1}}$. Hence, $[\overline{q}]_{\varphi_{1}}\subseteq[\overline{q}]_{\theta_{1}}$. This proves the Claim.

Since the Claim holds for every state ${\overline{p}}$ in $[\overline{q}]_{\theta_{1}}$, we conclude that $[\overline{q}]_{\theta_{1}}$ is a union of congruence classes $[{\overline{p}}]_{\varphi_{1}}$.    $\Box$

4 The Key Lemma

We will use $\stackrel{{\scriptstyle+}}{{=}}$ to denote equality in $A^{+}$ or $A^{*}$, i.e., literal equality of words.

Definition 4.1

A cover $(.)^{\sf cov}$ of $A$-automata has the backwards-$k$ property (for $k\geq 1$)  iff  for every $A$-automaton ${\bf A}=(Q,\i,\cdot)$ and every $\alpha,\beta\in A^{+}$, the following holds. If $\alpha\beta=\alpha$ in the syntactic semigroup $S_{{}_{A}}^{\sf cov}$ of ${\bf A}_{{}_{A}}^{\sf cov}$, then $\alpha$ can be factored in $A^{+}$ as $\alpha\stackrel{{\scriptstyle+}}{{=}}\tilde{\alpha}\beta_{1}\ldots\beta_{k}$ (with $\tilde{\alpha},\beta_{1},\ \ldots,\beta_{k}\in A^{+}$) such that in the syntactic semigroup $S_{A}$ of A we have $\tilde{\alpha}=\alpha$ and $\beta_{1}=\beta_{2}=\ \ldots\ =\beta_{k}=\beta=\beta^{2}$.

Proposition 4.2

For $A$-generated semigroups, let $(.)_{A}^{\sf IS}$ be an expansion with $\cal R$-trivial idempotent stabilizers, let $(.)_{A}^{\wedge_{(k+1)}}$ be the Henckell $(k+1)$-factor expansion, and let $(.)_{A}^{\sf RB}$ be the rectangular-bands expansion. Then for any $A$-generated semigroup $S$, the triply expanded semigroups $S_{\,{}_{A}}^{{\sf IS}\,\wedge_{(k+1)}\,{\sf RB}}$ and $S_{\,{}_{A}}^{\wedge_{(k+1)}\,{\sf IS}\ {\sf RB}}$ have the backwards-$k$ property.

Proof. If $\alpha\beta=\alpha$ holds in $S_{\,{}_{A}}^{{\sf IS}\,\wedge_{(k+1)}\,{\sf RB}}$ or in $S_{\,{}_{A}}^{\wedge_{(k+1)}\,{\sf IS}\ {\sf RB}}$ then $\alpha\beta=\alpha$ also holds in $S_{\,{}_{A}}^{{\sf IS}\,\wedge_{(k+1)}}$, respectively $S_{\,{}_{A}}^{\wedge_{(k+1)}}$. The existence of a factorization $\alpha=\tilde{\alpha}\beta_{1}\ldots\beta_{k}$ in $S_{A}$ then follows from the basic properties of the expansion $(.)_{A}^{\wedge_{(k+1)}}$.    $\Box$

For an $A$-automaton A and an expansion $(.)^{\sf exp}$, let ${\bf A}^{\sf exp\,GLC}$ be the glc-cover in the interval $[{\bf A},\,{\bf A}^{\sf exp}]$, with start state $\tilde{\i}$ and state set $Q_{\sf glc}$. From ${\bf A}^{\sf exp\,GLC}$ and a string $s=a_{1}a_{2}\ \ldots\ a_{n}$, where $a_{i}\in A$, we define a string automaton ${\sf str}(s)$ as follows. In the state graph of ${\bf A}^{\sf exp\,GLC}$ we consider any walk

$\tilde{\i}\stackrel{{\scriptstyle a_{1}}}{{\longrightarrow}}u_{1}$ $\stackrel{{\scriptstyle*}}{{\longrightarrow}}v_{1}$ \stackrel{{\scriptstyle a_{2}}}{{\longrightarrow}}\ \ \ldots\ \ $\stackrel{{\scriptstyle a_{i}}}{{\longrightarrow}}u_{i}$ $\stackrel{{\scriptstyle*}}{{\longrightarrow}}v_{i}$ \stackrel{{\scriptstyle a_{i+1}}}{{\longrightarrow}}\ \ \ldots\ \ $\stackrel{{\scriptstyle a_{n}}}{{\longrightarrow}}u_{n}\in R_{n}$,

where $u_{1},v_{1},\ \ldots,\ u_{n-1},v_{n-1},u_{n}\in Q_{\sf glc}$ are such that $u_{i}\equiv_{\cal R}v_{i}$ for $i=1,\ \ldots,\ n-1$; there is no $v_{n}$. Let $R_{i}$ be the reachability class of $u_{i}$ (and of $v_{i}$) in ${\bf A}^{\sf exp\,GLC}$. We assume $R_{i}\neq R_{j}$ (hence they are vertex-disjoint) when $i\neq j$; hence $R_{i}>_{\cal R}R_{i+1}$ for all $i$. The state set of ${\sf str}(s)$ is  $\{\tilde{\i}\}\ \cup\ \bigcup_{1\leq i\leq n}R_{i}$, and the arrows of ${\sf str}(s)$ are just the ${\bf A}^{\sf exp\,GLC}$-arrows between states of ${\sf str}(s)$. We will use $\tilde{\i}$ as the start state, and $R_{n}$ as the set of accept states of ${\sf str}(s)$. Conversely, ${\sf str}(s)$ uniquely determines the above walk: $u_{i}$ is the entry point into $R_{i}$, and $v_{i}$ is the exit point.

Any $A$-generated monoid $M_{A}$ will be viewed as an $A$-automaton via its right Cayley graph.

Theorem 4.3

(“Key Lemma”).* Let $M_{A}$ be an $A$-generated monoid, viewed as an $A$-automaton A. We consider the $A$-generated expansion $M_{\,{}_{A}}^{\wedge_{(k+1)}\,{\sf IS}\ {\sf RB}}$ (for some $k>1$), viewed as an $A$-automaton ${\bf A}^{\sf exp}$. Let ${\bf A}^{\sf exp\,GLC}$ be the glc-cover of $A$-automata in the interval $[{\bf A},\,{\bf A}^{\sf exp}]$. Let ${\sf str}(a_{1}a_{2}\ldots a_{n})$ be a string automaton in ${\bf A}^{\sf exp\,GLC}$, with $\cal R$-classes $R_{i}$ for $i=1,\ \ldots,\ n$. Finally, let $r_{n}={\sf max}\{{\sf rank}_{\bf A}(e):e\in E(R_{n})\}$, where $E(R_{n})$ is the edge-set of the reachability class $R_{n}$.*

Then there exist $d_{1},\ \ldots,\ d_{n-1},d_{n}\in A^{*}$ such that $d_{i}$ labels a path $u_{i}\stackrel{{\scriptstyle d_{i}}}{{\longrightarrow}}v_{i}$ in $R_{i}$ (in ${\bf A}^{\sf exp\,GLC}$), $1\leq i\leq n-1$, and such that the word $w\stackrel{{\scriptstyle+}}{{=}}a_{1}d_{1}a_{2}\ \ldots\ a_{n-1}d_{n-1}a_{n}d_{n}$ satisfies:

(1)*   There is $\ell_{n}\in A^{*}$ such that in ${\bf A}^{\sf exp}$:  $w\,\ell_{n}=w$.*

(2)*   ${\sf rank}_{\bf A}(\ell_{n})\ =\ r_{n}\ >\ {\sf rank}_{\bf A}(d_{n})$.*

(3)*   (Backwards-$k$ property):  There exist $w^{\prime},t_{1},\ldots,\ t_{k}\in A^{+}$ (where $k$ is as in $\wedge_{(k+1)}$) such that $w\stackrel{{\scriptstyle+}}{{=}}w^{\prime}\,t_{k}\ \ldots\,t_{2}\,t_{1}$ in $A^{+}$, and in $M_{A}$:  $w=w^{\prime}$,  $t_{k}=\ \ldots\ =t_{1}=\ell_{n}$.*

*The path labeled by $t_{1}$, when read backwards from the state ${\tilde{q}}_{n}$ that $\ell_{n}$ loops on, eventually visits the source state $v_{n-1}$ of the edge labeled by $a_{n}$ in ${\sf str}(a_{1}a_{2}\ldots a_{n})$; i.e., $a_{n}d_{n}$ is a suffix of $t_{1}$, as seen in the next graph \ \ \ \ldots\ $\stackrel{{\scriptstyle*}}{{\longrightarrow}}v_{n-1}$ $\stackrel{{\scriptstyle a_{n}}}{{\longrightarrow}}u_{n}$ $\stackrel{{\scriptstyle d_{n}}}{{\longrightarrow}}{\tilde{q}}_{n}$ $\circlearrowleft\ell_{n}$   (in ${\bf A}^{\sf exp\,GLC}$). *

Proof. Let us pick $d_{1},\ \ldots,\ d_{n-1}$ arbitrarily so that $u_{i}\stackrel{{\scriptstyle d_{i}}}{{\longrightarrow}}v_{i}$ in $R_{i}$, for $1\leq i\leq n-1$. Reading $w_{1}\stackrel{{\scriptstyle+}}{{=}}a_{1}d_{1}a_{2}\ \ldots\ a_{n-1}d_{n-1}a_{n}$ from the start state $\overline{\i}$ in ${\bf A}^{\sf exp}$, we reach the state $\overline{\i}\cdot w_{1}$, which maps into $R_{n}$ by the map ${\bf A}^{\sf exp}\twoheadrightarrow{\bf A}^{\sf exp\,GLC}$. But $\overline{\i}\cdot w_{2}$ does not map into $R_{n}$, where $w_{2}=w_{1}a_{n}^{-1}$. (We use the notation $waa^{-1}=w$; i.e., $a^{-1}$ is the operation of removing the last letter $a$ from a word that ends in $a$, and $a^{-1}$ is undefined on words that do not end in $a$.)

Thus we can run Stage 1 of the Bottom-up construction on state ${\overline{q}}\cdot a$, where for ${\overline{q}}$ we take ${\overline{\i}}\cdot w_{2}$ in ${\bf A}^{\sf exp}$, and for $a$ we take $a_{n}$. Then we run the inductive Stage $i+1$ for increasing $i$ until we reach $n_{\infty}$.

If $n_{1}=n_{\infty}$ then as we enter the $\cal R$-class $[{\overline{q}}a]_{\equiv_{\cal R}}$ we have maximum edge-rank equal to $n_{\infty}$; we let $d_{n}$ be the empty word, and let $\ell_{n}$ be a loop at the entrance of $[{\overline{q}}a]_{\equiv_{\cal R}}$, passing through all the edges in $E([{\overline{q}}a]_{\equiv_{\cal R}})$.

If $n_{1}<n_{\infty}$ we go to the 2nd stage of the induction, i.e., we find an $\cal R$-class $\overline{R}$ of ${\bf A}^{\sf exp}$ with rank $n_{2}>n_{1}$. So we have a path  $\ldots\ \stackrel{{\scriptstyle d_{n}}}{{\longrightarrow}}{\overline{R}}_{2}$, and this path up to (but excluding) ${\overline{R}}_{2}$ has edges of rank $n_{1}$. Then we take $d_{n}$ as indicated, and we take $\ell_{n}$ to be a loop at the entrance of ${\overline{R}}_{2}$, passing through all the edges in $E({\overline{R}}_{2})$.

We continue with $n_{3}$, etc., until $n_{\infty}$ is first encountered. Now, item (3) follows from Prop. 4.2, applied to the $\ell_{n}$-loop in   $\ldots$ $\stackrel{{\scriptstyle d_{n}}}{{\longrightarrow}}\bullet\circlearrowleft\ell_{n}$.        $\Box$

The Key Lemma can be generalized by replacing $k$ by two parameters, $k_{1}$ and $k_{2}$. For this we use the expansion  $(.)_{\,{}_{A}}$ ${}^{\wedge_{k_{1}}\,{\sf RB}\ \wedge_{(k_{2}+1)}\ {\sf IS}\ {\sf RB}}$.

Theorem 4.4

(Key Lemma with $k_{1}$ and $k_{2}$).*  Under the assumptions of Theorem 4.3, except that the expansion is now based on  $(.)_{\,{}_{A}}$ ${}^{\wedge_{k_{1}}\,{\sf RB}\ \wedge_{(k_{2}+1)}\ {\sf IS}\ {\sf RB}}$, we have the same conclusions, except for changes in item (3):*

(3)*  (Backwards-$k_{1}$-$k_{2}$ property): There exist $w^{\prime},t_{1},\ldots,\ t_{k_{2}}\in A^{+}$ such that $w\stackrel{{\scriptstyle+}}{{=}}w^{\prime}\,t_{k_{2}}\ \ldots\,t_{2}\,t_{1}$ in $A^{+}$, and in $M_{A}^{\wedge_{k_{1}}\,{\sf RB}}$:  $w=w^{\prime}$,  $t_{k_{2}}=\ \ldots\ =t_{1}=\ell_{n}$.*

The path labeled by $t_{1}$, when read backwards from the state ${\tilde{q}}_{n}$ that $\ell_{n}$ loops on, eventually visits the source state $v_{n-1}$ of the edge labeled by $a_{n}$ in ${\sf str}(a_{1}a_{2}\ldots a_{n})$; i.e., $a_{n}d_{n}$ is a suffix of $t_{1}$, as seen in the next graph \ \ \ \ldots\ $\stackrel{{\scriptstyle*}}{{\longrightarrow}}v_{n-1}$ $\stackrel{{\scriptstyle a_{n}}}{{\longrightarrow}}u_{n}$ $\stackrel{{\scriptstyle d_{n}}}{{\longrightarrow}}{\tilde{q}}_{n}$ $\circlearrowleft\ell_{n}$   (in ${\bf A}^{\sf exp\,GLC}$).

Proof. The proof is the same as for Theorem 4.3, but we use the morphisms  $M_{A}^{\wedge_{k_{1}}\,{\sf RB}\ \wedge_{(k_{2}+1)}\ {\sf IS}\ {\sf RB}}$ $\twoheadrightarrow$ $M_{A}^{\wedge_{k_{1}}\,{\sf RB}}$ $\twoheadrightarrow$ $M_{A}^{\wedge_{k_{1}}}$ for item (3).    $\Box$

Notation: Let $({\sf GM},A)$ be the starting group mapping semigroup whose complexity we are trying to compute, let $M_{A}=({\sf GM},A)^{\sf IS\,RB}$, and let $(.)^{\sf exp}$ be $(.)^{\wedge_{k_{1}}\,{\sf RB}\ \wedge_{(k_{2}+1)}\ {\sf IS}\ {\sf RB}}$. Then the semigroup automaton ${\bf A}_{A}^{\sf exp\ GLC}$ will be called PreFF$(k_{1},k_{2})$ (for “pre-funny fractal”).

For any finite semigroup $S$, let $\omega(S)$ be the smallest positive integer $m$ such that for all $s\in S$, $s^{m}$ is an idempotent.

To be continued.