Global local covers
John L. Rhodes, Benjamin Steinberg, J.C. Birget
(
2.iv.2019)
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.
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 SA and MA 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 SA or (S,A).
We consider also the category AA 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 AA 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
AA, which we also denote by AA; 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 AA. For clarity, let us define the right
Cayley graph ΓA(S,σ) for a semigroup S and a map
σ:A→S that maps A onto a generating set of
S: ΓA(S,σ) is a directed labeled rooted graph, with
vertex set SI (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⋅σ(a)):s∈SI,a∈A}
(where ⋅ is the product in S). The directed edge
(s,a,s⋅σ(a)) is labeled by a. We often write Γ(SA)
instead of ΓA(S,σ).
For automata A1,A2 in AA we write
A1↞A2 iff there exists a surjective
AA-morphism from A2 onto A1.
The relation ↞ is a pre-order on AA.
Because of the role of the start state, a surjective morphism is unique if
it exists (for a given pair A1,A2); hence, this pre-order
is a partial order. We also write A2↠A1
for A1↞A2.
For automata A1,A2 in AA we define the
direct product in the category AA.
For i=1,2, let Ai=(Qi,Ii,⋅i), where
⋅i:Qi×A→Qi is the next-state function. Then
A1×AA2 \ =\
\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∈A∗, a∈A, we define
(I1⋅1w, I2⋅2w)⋅a \ =\
(I1⋅1w⋅1a, I2⋅2w⋅2a).
For two automaton morphisms φi:Ai→B in the
category AA we can define the direct product
φ1×Aφ2:A1×AA2→B
in AA by
(I1⋅1w,I2⋅2w)φ1×Aφ2
= I⋅Bw (for all w∈A∗).
This is well-defined: If (I1⋅1u, I2⋅2u)
= (I1⋅1v, I2⋅2v) then
Ii⋅iu=Ii⋅iv (for i=1,2), hence
(Ii⋅iu)φi=(Ii⋅iv)φi. Moreover,
(Ii⋅iu)φi=(Ii)φi⋅Bu
=I⋅Bu; similarly,
(Ii⋅iv)φi=I⋅Bv. Thus,
I⋅Bu=I⋅Bv.
We show next that ↞ is a lattice order.
Lemma 2.1
(1) The join A1∨A2 in AA is
isomorphic to the direct product A1×AA2.
In the case of right Cayley graphs of semigroups in SA, the
join is (up to isomorphism) the direct product in SA.
(2) The meet is determined by the join as
A1∧A2 = ⋁
{A:A1↠A
and A2↠A} (up to isomorphism).
Proof.
(1) The projections from A1×AA2 onto A1,
respectively A2, show that A1×AA2 is an
upper bound on A1 and A2 for the order ↞.
To show that it is the least upper bound, we consider the commutative
diagrams in the category AA that defines the direct product,
with arrows
A1↞X↠A2,
X↠A1×AA2, and the
projections A1↞A1×AA2
↠A2, for any object X in AA.
The join is defined by exactly the same diagrams, so the join is the direct
product.
For Cayley graphs ΓA(S1,σ1) and ΓA(S2,σ2),
viewed as A-automata, the direct product
ΓA(S1,σ1)×AΓA(S2,σ2) has the set
{(σ1(w),σ2(w)):w∈A∗} as its states (vertices);
here we extended σ1 and σ2 to homomorphisms from A∗ onto
S1 or S2. Hence ΓA(S1,σ1)×AΓA(S2,σ2), with start state (I1,I2), is the Cayley graph of
(S1)A×A(S2)A (semigroup direct product in the category
SA).
(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.
□
In the partial order ↞ we define an interval
[A1,A2] \ =\ {A∈AA:
A1↞A↞A2}.
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 A in AA and for two states
q1,q2 of A, we say that q2 is reachable from q1
(and we write q2≤Rq1) iff q2=q1⋅w
for some w∈A∗. This relation is a pre-order, and we write
q2≡Rq1 iff (q2≤Rq1 and
q1≤Rq2). An ≡R equivalence
class is called a reachability class, or an R-class;
this is the set of vertices of a strongly connected component of the
underlying digraph of the automaton. For two states q1,q2 we will
also use the notation q2=≡Rq1 as a shorthand
for (q2≡Rq1 and q2=q1).
An automaton morphism φ:A1→A2 is said to be
1:1R iff the restriction of φ to any reachability
class of A1 is injective.
Lemma 2.2
If two morphisms φ1:A1→B and
φ2:A2→B are 1:1R then the direct
product morphism
φ1×Aφ2:A1×AA2→B
in AA is also 1:1R.
Proof. The state set of A1×AA2 is
{(I1⋅1w,I2⋅2w):w∈A∗}.
The morphism φ1×Aφ2 is defined by
φ1×Aφ2(I1⋅1w,I2⋅2w)
=I⋅Bw
=φ1(I1⋅1w)=φ2(I2⋅2w). The last two
equalities hold because φ1 and φ2 are morphisms; in
particular, φ1(I1)=I=φ2(I2). If (I1⋅1u,I2⋅2u) =≡R
(I1⋅1v,I2⋅2v) then
I1⋅1u≡RI1⋅1v and I2⋅2u≡RI2⋅2v; moreover,
I1⋅1u=I1⋅1v or I2⋅2u=I2⋅2v.
If I1⋅1u=I1⋅2v then the 1:1R
property of φ1 implies
φ1(I1⋅1u)=φ1(I1⋅1v),
hence φ1×Aφ2(I1⋅1u,I2⋅2u)
= φ1×Aφ2(I1⋅1v,I2⋅2v).
If I2⋅2u=I2⋅2v the same conclusion holds.
Thus, φ1×Aφ2 is 1:1R.
□
For a finite semigroup S and x∈S, let xω be the idempotent
in {xn:n>0}. The algebraic rank of x is defined to be the
length of a longest strict J-chain of regular
J-classes, ascending from xω (see [8]).
The rank of x is a non-negative integer (possibly 0); we denote it by
rankS(x) or rank(x). Formally,
rankS(x) = max{r : xω<JJ1
<J … <JJr, where
J1,…,Jr are regular J-classes of S}.
In other words, rankS(x) is the regular
J-depth of the idempotent generated by x.
We also define the rank with respect to automata, i.e., we extend
rankS:S→N to a word-rank function
rankA:A∗→N (where N=
{0,1,…} denotes the natural integers).
For an automaton A over A and w∈A∗, let [w] be the image
of w in the syntactic monoid S of A. Then rankA(w)
is defined to be rankS([w]).
Definition 2.3
(Rank condition).* An automaton morphism φ:A↠B in
AA satisfies the rank condition iff the following holds.
For all q∈QA and all α,β∈A+ such that
q⋅Aα=q and
(q)φ⋅Bβ=(q)φ and
rankB(α)≥rankB(β),
we have: q⋅Aβ=q.*
The formal statement of the rank condition on φ is as follows.
(∀q∈QA)(∀α,β∈A+):
[q⋅Aα=q &
(q)φ⋅Bβ=(q)φ &
rankB(α)≥rankB(β)]
implies q⋅Aβ=q.
Yet another way to say this: If α fixes q in A, and β
fixes (q)φ in (A)φ, and
rank(A)φ(α) ≥
rank(A)φ(β), then
β also fixes q in A.
The following is a better statement of the rank condition:
(∀q∈QA)(∀β∈A+):
if [(q)φ⋅Bβ=(q)φ & (∃α∈A+)
[q⋅Aα=q & rankB(α)≥rankB(β)]]
then q⋅Aβ=q.
In words: If β fixes (q)φ in (A)φ,
and there is α that fixes q in A such that
rank(A)φ(α) ≥
rank(A)φ(β), then
β also fixes q in A.
Lemma 2.4
If two morphisms φ1:A1→B and
φ2:A2→B satisfy the rank condition,
then the direct product morphism
φ1×Aφ2:A1×AA2→B
in AA also satisfies the rank condition.
Proof. Assume that (I1⋅1w, I2⋅2w)⋅α =
(I1⋅1w, I2⋅2w) in
A1×AA2, assume that
I⋅Bw⋅Bβ=I⋅Bw
in B, and assume that
rankB(α)≥rankB(β).
The equality in A1×AA2 implies that for i=1,2:
Ii⋅iw⋅iα=Ii⋅iw and
φi(Ii)⋅Bw⋅Bβ =
φi(Ii)⋅Bw. Hence by the rank condition for
φi we have Ii⋅iw⋅iβ=Ii⋅iw. So,
β fixes (I1⋅1w, I2⋅2w) in
A1×AA2.
□
2.2 Definition of the glc-cover
Let cov be any cover in the category AA.
For an automaton A in AA we consider the interval
[A,Acov]. This is a finite lattice, as we saw.
Definition 2.5
The glc-cover Acovglc is defined by the following
join in the category AA:
{\bf A}^{\sf cov\,glc}\ =\*
⋁{X∈[A,Acov]: there exists a
morphism A↞X in AA that is
1:1R and that*
satisfies the rank condition}.
More generally, if A and B are A-automata such that
A↞B, we can define the glc-cover
of the interval [A,B] by
[{\bf A},{\bf B}]^{\sf glc}\ =\
⋁{X∈[A,B]:
there exists a morphism A↞X in
AA that is 1:1R and that
satisfies the rank condition}*.
*
By the previous Lemmas, and since [A,Acov]
and [A,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 1:1R requirement alone;
we will make that clear in the context.
When the automata A and B in [A,B]glc
are right Cayley graphs it makes sense to define the cover
[A,B]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
B↠C
↠φA, where φ is
1:1R (with or without the rank condition). Let S(C)
be the A-generated syntactic semigroup of C, and let
Γ(S(C))=ΓA(S(C),σ) be the right Cayley graph
of S(C), where σ embeds A into a generating set of
S(C). Let η:Γ(S(C))↠C be
the canonical surmorphism, mapping semigroup elements to states.
Lemma 2.6
If B↠Γ(S(C))
↠ηC
↠φA, where A and
B are Cayley graphs, and if φ is 1:1R, then
η∘φ:Γ(S(C))↠A is
also 1:1R.
If, in addition, φ satisfies the rank condition then
η∘φ also satisfies the rank condition.
Proof. If for s1,s2∈S(C) we have s=s2,
s1≡Rs2, then by faithful action of S(C)
in C there exists a state q of C with
q⋅s1=q⋅s2.
Hence since φ is 1:1R,
(qs1)φ=(qs2)φ.
Hence, since A is a Cayley graph,
IA⋅s1=IA⋅s2, so
(s1)η∘φ = IA⋅s1 =
IA⋅s2=(s2)η∘φ, i.e.,
η∘φ is 1:1R.
Suppose φ satisfies the rank condition. We want to show for all
s∈S(C) and α,β∈A+: if
s⋅Sα=s and
(s)ηφ⋅Aβ=(s)ηφ and
rankB(α)≥rankB(β),
then s⋅Sβ=s. Here ⋅S is the next-state function
in Γ(S(C)).
By the rank condition for φ, we have
(s)η⋅Cβ=(s)η. Hence, since η is
bijective on the semigroups, s⋅Cβ=s.
□
Proposition 2.7
When A and B are A-generated Cayley graphs then the cover
[A,B]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 1:1R morphism with rank condition. Hence
the Cayley graph is equal to the cover, by maximality of the cover.
□
2.3 Examples of glc-covers for 1:1R morphisms
(1) Let V be a pseudovariety of semigroups,
and let SA be an A-generated semigroup. Let SAV denote
the maximum image of SA in V; i.e.,
S_{A}^{\bf V}\ =\
⋁{Y∈V: there exists a morphism
SA↠Y in SA},
where ∨ is the join in the category of A-generated
semigroups; this join is finite since SA is finite.
Let R be the pseudovariety of R-trivial semigroups.
For any A-generated semigroup SA, SAR denotes thus the
maximum 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 ψ:S↠T is, by definition, the
pseudovariety generated by the set of semigroups
{(e)ψ−1:e=e2∈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 ⟨T⟩ generated by any finite semigroup T is locally finite.
Then V {\scriptsize\sfm}⃝ (.) is an expansion of A-generated semigroups,
where V {\scriptsize\sfm}⃝ SA is the unique functorially largest
A-generated semigroup that maps onto SA by a surmorphism whose
Maltsev kernel is in V:
V {\scriptsize\sfm}⃝ SA \ =\
⋁{X: there exists φ:X↠SA in
SA such that the Maltsev kernel of φ is
contained in V}.
Since V is locally finite, this join is a finite set, so
V {\scriptsize\sfm}⃝ SA is finite (by Brown’s theorem; see [8],
[11]).
Proposition 2.8
(Example 2.8)*
Let Γ(SA)=[Γ(TA),Γ(UA)]glc be the
glc-cover of A-generated right Cayley graphs with respect to the
1:1R property; here we temporarily drop the rank condition.
Hence, SA is the unique maximum A-generated semigroup such that*
Γ(UA)↠φ1Γ(SA)*
↠φ2Γ(TA),*
where φ2 is 1:1R. Then:
(a)* The Maltsev kernel of φ2 (and in fact, of
any 1:1*R* morphism) is contained in R.*
(b)* If TA∈R and UA=TAcov
for some cover (.)cov, then*
[Γ(TA),Γ(TAcov)]glc* \ =\
TAcovR.*
(c)* For any A-generated semigroup WA we consider
the expansion (.)Aexp=⟨WA⟩ {\scriptsize\sfm}⃝ (.),
where ⟨WA⟩ is the pseudovariety generated by WA.
Let 1A be the one-element semigroup in the category of
A-generated semigroups.*
Then 1Aexp=WA, and
[Γ(1A),Γ(WA)]glc=WAR.
Thus, for any A-generated R∈R, ⟨R⟩ can be
the Maltsev kernel.
Proof. (a) If φ2:Γ(SA)↠Γ(TA) is
1:1R then the corresponding map SA↠TA
(which we also denote by φ2) is also 1:1R.
Then for any idempotent e=e2∈TA, (e)φ2−1 is an
R-trivial semigroup. Indeed, if x1≡Rx2
in (e)φ2−1 then (x1)φ2=e=(x2)φ2, hence
the 1:1R-property implies x1=x2.
(b) When
φ2:SA↠TA is 1:1R, then
SA∈R iff TA∈R.
Indeed, if x1≡Rx2 in SA then
(x1)φ2≡R(x2)φ2 in TA, hence
(x1)φ2=(x2)φ2 (since TA∈R), hence
x1=x2 (since φ2 is 1:1R);
the converse is obvious.
Since [Γ(TA),Γ(TAexp)]glc
↠Γ(TA) is 1:1R and since
TA∈R, it follows that
[Γ(TA),Γ(TAexp)]glc is in R.
Since the glc-cover is maximal with respect to the
1:1R-property, it follows that the glc-cover is
TAexpR (by the definition of (.)R).
(c) Obviously, the map SA↠1A is
1:1R iff SA∈R.
Hence, by the definition of the expansion (.)exp =
⟨WA⟩ {\scriptsize\sfm}⃝ (.) we have
1Aexp=WA. And by the definition of (.)R,
[Γ(1A),Γ(WA)]glc=WAR.
Now let WA (=R) be any A-generated semigroup in R. Then
WA=WAR, so in that case, [Γ(1A),Γ(WA)]glc=WA (by (c)).
And the Maltsev kernel is ⟨WA⟩ (which is
⟨R⟩).
□
Remark. From Example 2.8 we see that
[ΓA(T),ΓA(Tcov)]glc can be:
(case 1) ΓA(T), or
(case 2) ΓA(Tcov), or
(case 3) strictly between the bounds of the interval.
Let TA=1A and
(.)Acov=⟨WA⟩ {\scriptsize\sfm}⃝ (.). Then
1Acov=WA, and by Example 2.8.(c),
[ΓA(T),ΓA(Tcov)]glc =
[1A,1Acov]glc=WAR.
(1) When WA is a non-trivial A-generated group, then
WAR=1A. So we get (case 1).
(2) When 1A=WA∈R (e.g., WA 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 WA∈R and WAR=1; e.g.,
let WA=Z2×AAℓ, where Aℓ is the left-zero
semigroup generated by A. Then WA↠Aℓ by
projection; then [1A,1Acov]glc=WAR
is strictly between 1A and 1Acov=WA.
This illustrates (case 3).
(2) Let us consider the reverse of the delay
pseudovariety D=[xyω=yω].
So Drev=[xωy=xω], corresponding to
the limit of the set of equations
{x1…xkxk+1=x1…xk:k≥1}.
It is known that S∈Drev iff S is a nilpotent extension
of a left-zero semigroup; such a semigroup consist of a minimal ideal which
is an L-class L of left-zeros, and every element of the semigroup
satisfies (∃k) xk∈L. The left-zero semigroups form the
pseudovariety LZ = [xy=x]. For fixed k>0, let
Dkrev=[x1…xkxk+1=x1…xk] be the
reverse of the delay-k pseudovariety; this pseudovariety is locally finite.
We consider the expansion (.)exp=Dkrev\sfm⃝ (.).
Then we have for every A-generated semigroup TA:
Dkrev\sfm⃝ TA* =
TAexp ↠ TA is injective on every
regular R-class.
*
Indeed, let us apply Rees’ Theorem to a regular
R-class {a}×G×B of TAexp. If the
map identifies two elements in this R-class then it identifies
two L-classes, hence it identifies to L-equivalent
idempotents. So in that case there are two idempotents e=f. Thus,
{e,f} is a right-zero semigroup, but a non-trivial right-zero
semigroup cannot be in Dkrev.
Notation: A morphism that is injective on every regular
R-class is said to be 1:1regR.
Digression: Example of a morphism
φ:SA↠TA that is
1:1regR**, but not 1:1R **
Let S={\i,x,\i,x,0}, T={\i,x,y,0},
φ={(\i,\i),(x,x),(\i,y),(x,y),(0,0)}, and
A is a two-element set embedded injectively into {x,\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 J-classes, namely the group
Z2={\i,x} at the top, a null R-class
{\i,x} in the middle, and a zero 0. The group
Z2 acts on the 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 R-class are 0. The semigroup
T differs from S only in that its middle null J-class is a
singleton. The action of Z2 and the null property of the
middle J-class make it easy to check associativity.
It is clear that φ is injective on regular R-classes (namely
Z2 and {0}), and not injective on the null middle
R-class.
[End, digression]
We continue to use glc-covers based on 1:1R morphisms,
ignoring the rank condition. We also consider 1:1regR morphisms, and
we distinguish the two by the notation (.)glc1:1R,
respectively (.)glc1:1regR.
We will use the following.
Lemma 2.9
Let α,β be surmorphisms of A-generated semigroups. Then
α∘β is 1:1R (or 1:1regR)
iff each of α and β is 1:1R (respectively
1:1regR).
Proof. The right-to-left implication is obvious. Assume now that
β∘α is 1:1R, from which it immediately follows that
α is 1:1R. For β, let R be an R-class of
dom(β), and let y1,y2∈R with y1=y2. Let
x1,x2∈dom(α) be such that x1≡Rx2
and y1=(x1)α and y2=(x2)α; x1,x2 exist since
α is surjective, and since the inverse image of
an R-class is a union of R-classes. Then
(y1)β=(x1)αβ=(x2)αβ=(y2)β,
where “=” holds because (.)αβ is 1:1R.
Hence (y1)β=(y2)β, i.e., β is 1:1R.
The same proof (when R is regular) works for 1:1regR, using
the fact that the inverse image of a regular R-class contains a
regular R-class.
□
We consider the expansion defined by TAexp
\ =\ Dkrev {\scriptsize\sfm}⃝ TA ↠ TA.
Then
T_{A}^{\sf exp}\ \ \twoheadrightarrow\ \
[TA,TAexp]glc 1:1regR
is an isomorphism, since TAexp↠TA
is 1:1regR. So we have
TAexp \ =\
[TA,TAexp]glc 1:1regR
\ \ \ \stackrel{{\scriptstyle\varphi_{1}}}{{\twoheadrightarrow}}\ \ \
[TA,TAexp]glc 1:1R
↠φ2 TA,
where φ1 is not an isomorphism (in general), because
1:1regR does not imply 1:1R.
The following is straightforward.
Proposition 2.10
Let (.)E be an expansion such that
SA E↠SA is 1:1R for all SA.
If the semigroups UA and TA satisfy UA E=UA and
TA E=TA, then
[TA,UA]glc E=[TA,UA]glc.
□
For example, the right Rhodes expansion (.)∧R has the 1:1R
property used above.
2.4 The R- and L-expansions of an
automaton
The R-expansion: We generalize the right
Rhodes ∧R-expansion to the automaton category
AA. In the special case of the right Cayley graph automaton
(SI,I,⋅) of a semigroup SA over the alphabet A we recover the
known Rhodes expansion SA∧R.
The definition proceeds in two steps.
Let A=(Q,i,⋅) be an automaton in AA 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>Rq1>R … >Rqk):
k≥0 and q1,…,qk∈Q} be the set of strict
reachability chains, starting at the start state. From this we build an
automaton with state set Q∧R,
start state (i), and next-state function ∙; the latter is defined
as follows. For any a∈A and q=
(i>Rq1>R … >Rqk−1
>Rqk) ∈Q∧R,
{\bf q}\bullet a\ =\
(i>Rq1>R … >Rqk−1
>Rqk>Rqk⋅a),
if qk>Rqk⋅a; and
{\bf q}\bullet a\ =\
(i>Rq1>R … >Rqk−1
>Rqk⋅a), if
qk≡Rqk⋅a.
(2) Let QA∧R⊆Q∧R be
the set of states of the above automaton that are reachable from the start
state (i); i.e.,
QA∧R={(i)∙w : w∈A∗}.
We define AA∧R to be the sub-automaton
(QA∧R, (i), ∙) of the above automaton.
The canonical natural transformation η of the expansion is defined for
each A by ηA:AA∧R→A,
where (i>_{\mathcal{R}}q_{1}>_{\mathcal{R}}\ \ldots\
>Rqk)ηA = qk.
One easily checks that this is an automaton morphism in the category
AA.
Any automaton morphism φ:A→B in AA
can be expanded to
φA∧R:
AA∧R→BA∧R
defined by
(i_{\bf A}>_{\mathcal{R}}q_{1}>_{\mathcal{R}}\ \ldots\
>Rqk)φA∧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 ≡R-chain by its rightmost
element. More formally, red(q)=q if q is a
strict chain; and
red( … ≥Rqi−1≥R
qi≡Rqi+1≥R … ) =
red( … ≥Rqi−1≥R
qi+1≥R … ).
It is easy to check that φA∧R
is an automaton morphism.
In summary, (.)A∧R is an expansion in the category
of automata AA, according to the categorial definition of
expansions.
The analogy between the Rhodes ∧R-expansion and the
automata ∧R-expansion goes beyond the similarity of
the definitions.
Let SA be the syntactic monoid of an A-automaton
A=(Q,i,⋅); hence (Q,SA) is a faithful right action.
Proposition 2.11
The Rhodes expansion SA∧R acts on the set
QA∧R by the following right action.
For q=
(i>Rq1>R … >Rqk)
∈QA∧R and s=
(1>Rs1>R>Rs2>R
… >Rsh)∈SA∧R this
action is defined by:
{\bf q}\bullet{\bf s}\ =\
red(i>Rq1>R … >R
qk≥Rqk⋅s1≥Rqk⋅s2
≥R … ≥Rqk⋅sh).
This action yields a homomorphism from SA∧R
onto the syntactic monoid of AA∧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 SA,
we have in the automaton category AA:
Proposition 2.12
For every automaton A in AA: (AA∧R)A∧R
=AA∧R.
Proof. The proof is similar to the case of
SA∧R.
□
Proposition 2.13
The reachability order among the states of
AA∧R is unambiguous.
Proof. The proof is similar to the case of
SA∧R.
□
Proposition 2.14
For any automaton A, the canonical map
ηA:AA∧R→A
is fully 1:1R and satisfies the rank condition.
Proof. [1:1R] If p≡q∈
A∧R then p and q have the form
p=(r1>…>rk−1>p), respectively
q=(r1>…>rk−1>q). Hence, when
p≡q we have: p=q iff
η(p)=η(q).
[fully 1:1R] Let p=η(p) and suppose
p≡q in A. We need to show that there is q such that
η(q)=q and p≡q.
Since p=η(p), the chain p has the form
(r1>…>rk−1>p) for some r1,…,rk−1∈Q.
Since p≡q, the chain (r1>…>rk−1>q) (let’s call it
q) exists in Q∧R.
To show that q∈QA∧R and that
p≡q in A∧R, suppose
q=p⋅a1⋅…⋅an for some a1,…,an∈A.
Then p⋅a1⋅…⋅an =
red(r1>…>rk−1>p≥p⋅a1≥ …≥
p⋅a1…an−1≥p⋅a1…an−1an=q) =
(r1>…>rk−1>q), since p≡q. Thus,
q∈QA∧R and p≡q.
[Rank condition] Suppose there is a morphism
ψ:A→T. Suppose q⋅α=q
in AA∧R, and
η(q)⋅β=η(q) in A, and
rankT(α)≥rankT(β). We need to
show that q⋅β=q in
AA∧R.
One easily verifies that η(q)⋅β=η(q)
by itself already implies q⋅β=q (without any
further hypotheses).
□
Proposition 2.15
For any automaton A in AA we have (up to isomorphism):
If AAcov ∧R=AAcov
then AAcovglc ∧R =
AAcovglc.
Proof. Since
AAcov ∧R=AAcov,
the interval [AA∧R,
AAcov ∧R] is contained in
[A, AAcov], with upper boundaries equal.
The canonical map
AAcovglc ∧R
↠ AAcovglc is 1:1R and satisfies the rank condition (by the previous Prop.).
Moreover, by definition of (.)glc, AAcovglc
is maximal in the order ↠ for maps that are 1:1R and that satisfy the rank condition. The Proposition
follows.
□
The L-expansion: For any right action (Q⋅,S) (or any automaton A with
syntactic monoid S), we define a left action of S on
P(Q) by
s⋆X=Xs−1
for all s∈S and X⊆Q. Recall the standard notation
Xs−1={q∈Q:q⋅s∈X}, for any s∈S and
X⊆Q. For a singleton {q} we also write s⋆q
and (q)s−1 for s⋆{q}, respectively {q}s−1.
This is indeed a left action: (s1s2)⋆X = X(s1s2)−1
= (Xs2−1)s1−1 = s1⋆(s2⋆X).
One can prove fairly easily that this is a faithful left action
of S: If s1=s2 then, since the given right action is faithful,
there is q∈Q with q⋅s1=p1 = p2=q⋅s2;
then q∈(p1)s1−1 but q∈(p1)s2−1.
This action can be restricted to a faithful action on the set
{(q)s−1: q∈Q, s∈S1} ⊆ P(Q).
The above construction can be applied to automata.
Suppose an automaton A=(Q,i,⋅) has a final state f,
which is reachable from every state in Q, i.e.,
(∀q∈Q)(∃w∈A∗)[q⋅w=f]. Then Aleft=({(f)s−1:s∈S1}, {f}, ⋆)
is a left automaton with the same syntactic semigroup as A.
For an automaton A=(Q,i,⋅) and its left version
Aleft=
(Qleft={(f)s−1:s∈S1}, {f}, ⋆),
we can construct a left expansion AA∧L;
this generalizes the left Rhodes expansion SA∧L
of a semigroup, and yields a left action (not necessarily faithful) of
SA∧L.
The state set of AA∧L is
{(Pk< … <P1<{f}):k≥0, and
Pk,…,P1∈Qleft},
where, for any P2,P1∈Qleft, we define
P2≤P1 iff (∃s∈S1)
P2=s⋆P1=P1s−1.
The action is defined by
a\star(P_{k}<\ \ldots\ <P_{1}<\{f\})\ =\
red(a⋆Pk≤Pk< … <P1<{f}).
The left action of SA∧L is
(sm<L…<Ls1) ⋆
(P_{k}<\ \ldots\ <P_{1}<\{f\})\ =\
red(sm⋆Pk≤…≤
s1⋆Pk≤Pk< … <P1<{f}).
Here, the effect of the reduction operation red is to take
the left-most element of an ≡-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∧L and
(.)A∧R.
Proposition 2.16
When the automaton expansion (.)A∧L is applied to
the right-regular representation (S1,1,⋅) over the alphabet A,
the classical Rhodes expansion SA∧L is obtained.
Proof. We saw that SA∧L acts on the state
set of (S1,1,⋅)A∧L; we want to show that this
action is faithful: If s,t∈SA∧L act
in same way on all states of (S1,1,⋅)A∧L we
want to show that s=t.
Let s=
(sm<L … <Ls1) and t=
(tn<L … <Lt1).
We will prove by induction that s=t.
We will use the following (for any x,y∈S1):
(l) If y<Lx then
(x)y−1=∅.
Indeed, if there exists t∈(x)y−1 then ty=x, which
contradicts y<Lx. Moreover,
(i) x=y iff
1∈(x)y−1.
For the base-case of the induction we show that sn=tm
follows from s⋆({sm}))=t⋆({sm}).
Indeed, s⋆({sm})=((sm)sm−1< … ) = t⋆({sm}) = ((sm)tn−1< … ).
Since 1∈(sm)sm−1=(sm)tn−1, it follows by (i) that
sm=tn.
For the inductive step, let us assume that sm=tn, \ \ldots\,
sm−i+1=tn−i+1, for some i>1; we want to show that
sm−i=tn−i. We have
s⋆({sm−i}) =
red((sm−i)sm−1≤ … ≤(sm−i)sm−i+1−1
≤(sm−i)sm−i−1≤ … ) =
{\bf t}\star(\{s_{m-i}\})\ =\
{\sf red}((s_{m-i})s_{m}^{-1}\leq\ \ldots\
≤(sm−i)sm−i+1−1≤(sm−i)tm−i−1≤ … ).
By property (l) this becomes
s⋆({sm−i}) = (∅<(sm−i)sm−i−1<
… ) = {\bf t}\star(\{s_{m-i}\})\ =\
(∅<(sm−i)tm−i−1≤ … ).
Also, 1∈(sm−i)sm−i−1=(sm−i)tm−i−1, hence
sm−i=tm−i.
By induction we conclude that t=
(sm<L … <Ls1<L
tn−m<L … <Lt1), and m≤n.
Moreover, by letting s and t act on the states
({tn}), … ,({t1}), we similarly find that n≤m.
Hence n=m, and s=t.
□
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 Aleft will have
Qleft={(f)s−1:s∈S1,f∈F,(∃t∈S1)[(f)s−1t−1∩I=∅]}, and it will also have
a set of initial states Ileft={{f}:f∈F} and a set of
final states Fleft=
{P∈Qleft:P∩I=∅}. And both
AA∧L and AA∧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 A=(Q,⋅) be a finite automaton, let
ψ:A→B be an A-automaton morphism.
An edge is of the form (q,a,q⋅a), where
q∈Q,a∈A; we will also simply write (q,a). We say that an edge
is in a reachability class iff q≡Rq⋅a.
More generally, a walk of length n is of the form
(q,a1,q1,a2,q2, … ,ai,qi,ai+1,
… ,an,qn), where q,q1,…,qn∈Q and
a1,a2,…,an∈A, with qi=qa1…ai for
i=1,2,…n. We will also simply denote a walk by
(q,a1…an).
A walk (q,a1…an) is said to be within a reachability
class iff q⋅a1…an≡Rq.
Definition 3.1
When q≡Rq⋅a we define the edge-rank of
(q,a) by
rankB(q,a)* \ =\
min{rankB(ax): x∈A∗, q⋅ax=q}.*
For a path (q,w) within a reachability class we define the
path-rank by
rankB(q,w)* \ =\
min{rankB(wx): x∈A∗, q⋅wx=q}.*
Notation: For an A-automaton A=(Q,⋅) and
a word w∈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 ≡A on A∗ defined by
x≡Ay iff
(∀q∈Q)[q⋅x=q⋅y].
Definition 3.2
An A-automaton A has idempotent stabilizers iff for
every state q and every word w∈A∗: q⋅w=q implies
[w]2=[w].
Definition 3.3
An A-automaton A has R-trivial stabilizers iff
for every state q and every word w∈A∗: q⋅w=q implies
that the ≡R-class of [w] is a singleton (in the
syntactic monoid of A).
Proposition 3.4
(Invariance of rank under conjugation)* Let A=(Q,⋅)
be an A-automaton with idempotent stabilizers, and suppose (q,w)
is a closed walk, i.e., q⋅w=q, w∈A∗, q∈Q.
If w is factored as w=xy, then
rankB(xy)=rankB(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]≡J[yx]; indeed,
[xy]=[xyxy]≤J[yx], and
[yx]=[yxyx]≤J[xy].
Now, since [xy] and [yx] are ≡J-related
idempotents, they have the same algebraic rank. □
For Prop. 3.6 we will need an expansion with special properties;
we first make sure that such an expansion is available. Let
(.)A∧L be the Rhodes expansion, let
(.)RB be the rectangular-bands expansion RB {\scriptsize\sfm}⃝ (.),
and let (.)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 (.)RB∧L=(.)RB, i.e.,
(.)RB is stable under (.)∧L.
Proposition 3.5
For any A-generated semigroup SA, the expansions
SAIS∧L and SAISRB in the
category of A-generated semigroups have the following properties:
All right-stabilizers are R-trivial bands, and the L-order
in each right-stabilizer is unambiguous.
Proof. In [7] it is proved that in SAIS the
right-stabilizers are R-trivial bands. For any A-generated
semigroup TA the natural maps
ηL:TA∧L→TA and
ηRB:TARB→TA have the property that the inverse
image η−1(B) of any band B⊆TA is a band (where
η stands for either ηL or ηRB).
And any stabilizer Σ′⊆TA∧L is contained
in η−1η(Σ′), and η(Σ′) is contained in a stabilizer
Σ in TA; so any stabilizer Σ′⊆TA∧L
is contained in η−1(Σ) where Σ is a stabilizer of TA .
Hence any stabilizer in SAIS∧L or in
SAISRB is a band.
Moreover, for any TA, every right-stabilizer in TA∧L
is R-trivial and the L-order of TA∧L is
unambiguous.
Since TARB=TARB∧L,
every right-stabilizer is R-trivial there too, and the L-order
is unambiguous.
□
Proposition 3.6
(Path-rank vs. edge-rank).*
Let A be an A-automaton whose stabilizers are R-trivial
bands with unambiguous L-order.
Let (q0,a1…an) be a walk within an R-class,
i.e., q0a1…an≡Rq0. Let qi denote
q0a1…ai for i=1,…,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})\ =\
max{rankB(qi−1,ai):i=1,…,n}.*
If the path (q0,a1…an) is closed (i.e.,
q0a1…an=q0) 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})\ =\
rankB(a1…an).*
(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) ≥] By the definition of path-rank there exists
β∈A∗ such that rank(q0,a1…an) =
rank(a1…anβ)
and q0a1…anβ=q0.
The word β is the “x” in the definition of path-rank for which the
min is achieved. Then we have for any j=1,…,n:
{\sf rank}(q_{0},\,a_{1}\ldots a_{n})\ =\
{\sf rank}(a_{1}\ldots a_{n}\,\beta)\
= rank(ajaj+1…anβa1…aj−1)
(by Prop. 3.4)
≥ min{rank(ajx):
x∈A∗, qj−1ajx=qj−1} (since
qj−1⋅aj…anβa1…aj−1
=qj−1)
= rank(qj−1,aj)
(by the definition of edge-rank).
Therefore,
rank(q0,a1…an)≥rank(qj−1,aj)
for every j=1,…,n; hence,
rank(q0,a1…an)≥max{rank(qj−1,aj):
j=1,…,n}.
[(1) ≤] 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\
max{rank(qj−1,aj):j=1,…,n−1};
equivalently, {\sf rank}(q_{0},a_{1}\ldots a_{n-1})\ \leq\
rank(qj−1,aj) for some j=1,…,n−1.
We want to prove that rank(q0,a1…an−1an)≤
rank(qi−1,ai) for some i=1,…,n−1,n.
Let β∈A∗ be such that
q0a1…an−1β=q0 and
rank(q0,a1…an−1)=rank(a1…an−1β),
i.e., the min in the definition of path-rank is reached when x is
β. Similarly, let β′∈A∗ be such that
qn−1anβ′=qn and
rank(qn−1,an)=rank(anβ′).
Then each of the following is a closed path:
(qn−1, anβ′), (qn−1, βa1…an−1), (q0, a1…an−1β) , and (q0, a1…an−1anβ′β).
It follows that
[anβ′],
[βa1…an−1],
[a1…an−1β], and
[a1…an−1anβ′β]
belong to stabilizers, hence they are idempotents.
By the definition of rank (using min),
rank(q0,a1…an−1an)≤
rank(a1…an−1anβ′β), and by Prop. 3.4,
{\sf rank}(a_{1}\ldots a_{n-1}a_{n}\beta^{\prime}\beta)\ =\
rank(βa1…an−1anβ′).
Case 1: [βa1…an−1] ≡R
[βa1…an−1anβ′]
Then by R-triviality of stabilizers,
[βa1…an−1anβ′]=[βa1…an−1], and
these are idempotents. Hence,
rank(q0,a1…an−1an)≤
rank([βa1…an−1anβ′])=
rank([βa1…an−1])=
rank(q0,a1…an−1)≤rank(qj−1,aj)
for some j (the latter by the induction hypothesis).
Case 2: [βa1…an−1] >R
[βa1…an−1anβ′]
≤L [anβ′].
Case 2.1: [βa1…an−1anβ′]≡L[anβ′]
Then, since these are idempotents,
rank(q0,a1…an−1an)≤
rank([βa1…an−1anβ′])=
rank([anβ′])=rank(qn−1,an).
Case 2.2: [βa1…an−1] >R
[βa1…an−1anβ′]<L[anβ′].
Let e1=[βa1…an−1], e2=[anβ′];
then e1e2=[βa1…an−1anβ′], and
e1e2=(e1e2)2, and e1>Re1e2<Le2.
Since stabilizers are R-trivial bands, e1e2=e1e2e1.
Similarly, e2e1=e2e1e2.
Thus, e1>Le2e1=e2e1e2<Le2, which by
unambiguity of the L-order implies that either e1≤Le2,
i.e., e1=e1e2, or e2≤Le1, i.e., e2=e2e1.
But e1>Re1e2 contradicts e1=e1e2.
And e1e2<Le2 contradicts e2=e2e1 (since
e1e2=e1e2e1≡Le2e1).
Thus, when the syntactic monoid of the automaton A has unambiguous
L-order case 2.2 is impossible.
Proof of (2):
By definition, rank(q0,a1…an)=
min{rank(a1…anx):q0a1…anx=q0}.
We assumed that a1…an stabilizes q0.
By the properties of stabilizers, [a1…an] and [a1…anx]
are idempotents. Since a1…anx≤Ja1…an
and since these are idempotents, we have
rank(a1…anx)≥rank(a1…an)
for any [x], with equality when x is empty. Hence,
min{rank(a1…anx):q0a1…anx=q0}
=rank(a1…an).
□
For the rest of this subsection let TA be an A-generated semigroup.
Every R-class R of TA 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 Γ(TA).
We call this the R-class automaton of R.
We assume that the semigroup TA satisfies the assumptions of Prop. 3.6, i.e., that its stabilizers are R-trivial bands
with unambiguous L-order.
We consider the algebraic rank of elements of TA with respect to the
identity map on TA, denoted by rank(.).
We also consider the edge-rank function, as in Definition 3.1,
with respect to the identity map on TA. We also call this rank function
rank(.). So for an edge (q,a) with q,qa∈R we have:
rank(q,a)=min{rank(ax):x∈A∗, qax=a};
here, rank(ax) is the algebraic rank.
Definition 3.7
For the R-class R of an A-automaton we define
{\sf rank}(R)\ =\*
max{rank(q,a):q,qa∈R,a∈A}.*
For r∈R and j=0,1,… we define
{\sf paths}(r,j)\ =\*
{w∈A∗:r⋅w∈R and every edge e of the
path (r,w) in the automaton R*
has rank rank(e)≤j},
P(r,j)* (⊆R) is the set of vertices of the paths in
paths(r,j),*
E(r,j)* is the set of edges of the paths in paths(r,j).*
Proposition 3.8
For every R-class R of an A-automaton and every fixed
jo=0,1,… we have: The set of sets
{P(r,jo):r∈R} is a partition of R.
Proof. Since r∈P(r,jo) the sets P(r,jo) are obviously
non-empty and R=⋃r∈RP(r,jo).
Let us prove that non-equal sets that overlap are equal. If
r∈P(ri,jo), then there is a path in
R from ri to r, labelled by some α∈A+, with maximum
edge rank ≤jo. Then by Prop. 3.6 (Sausage Lemma), this
path can be extended to a cycle containing ri with no increase in
maximum edge rank. Hence, if r∈P(r1,jo)∪P(r2,jo) then
there are paths from r1 to r2 and from r2 to r1 of maximum
edge rank ≤jo; so P(r1,jo)=P(r2,jo).
□
Examples of P(.,.) and 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
{{q1},{q2},{q3},{q4}}, for j=1 it is
{{q1,q2},{q3,q4}}, and for j=2 it is
{{q1,q2,q3,q4}}.
(2) Suppose the state diagram of R consists of
the states q1 and q2, with edges (q1,a(1),q2),
(q1,b(2),q2), and (q2,c(1),q1), with ranks in
parentheses.
Then paths(q1,1) consists of all paths from q1 with label in
(ac)∗∪(ac)∗a.
And P(q1,1)=P(q2,1)=P(q1,2)=P(q2,2)={{q1,q2}}.
[End, Examples.]
We are now going back to the covers
U_{A}=T_{A}^{\sf cov\ IS\ RB}\
↠φ1 [TA,UA]glc
↠φ2 TA, where φ2 is
1:1R and satisfies the rank condition.
In this setting we will also write UP for UA, MIDDLE for
[TA,UA]glc, and DOWN for TA.
Let R~ be an R-class of MIDDLE, let R be the
R-class of DOWN such that
φ2:R~ ↪ R.
To assign a rank to an edge
r~⟶ar~a of R~, we
apply φ2, yielding (r~)φ2
⟶a(r~a)φ2, and take the rank of
this arrow in R. Recall that rank(R~) is the max rank over
all idempotents in E(R~) (=E(R)). Notice that by Prop. 3.6, rank(R) is the same as algebraic rank of all
idempotents that fix any element of R under right-multiplication.
We also have
n~=rank(R~)≤rank(R)=n.
For an 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 paths(r,j) in the
R-class automaton R.
Proposition 3.9
For R~ and R as above and for any r~∈R~ and
r=(r~)φ2∈R we have:
E(R~) = E(r,n~) ⊆ E(R).
Proof. Let α∈A∗ be the label of a cycle in R~,
starting at r~, that traverses every edge of R~.
Then each path in paths(r,n~) can be extended to a cycle
that returns to r, and that uses only edges of edge-rank ≤n~
(by Prop. 3.6, the “Sausage Lemma”).
Now the rank condition implies the result as follows:
We have r~α=r~, and α has at least one edge
of rank n~, and passes through all vertices of R~.
If rβ=r and rank(β)≤rank(α)=n~
then r~β=r~, by the rank condition. So,
E(R~)=E(r,n~)⊆E(R).
□
Corollary 3.10
If rank(R~)=rank(R) then the restriction of φ2
to R~ is an isomorphism from R~ to R.
□
We will give a formula for rank(R) in terms of idempotents.
The following Lemma will be used.
For idempotents e,f we write e≥f iff f=ef=fe;
this is called the idempotent order, or equivalently,
the H-order between idempotents; it is a partial order. We write
e>f iff e≥f and e=f.
Lemma 3.11
(Rhodes 1966).* Let J1 and
J2 be regular J-classes of a finite semigroup, such that
J1>JJ2. Then for every idempotent e∈J1 there
exists an idempotent f∈J2 such that e>f.*
Proof. See the proof of Prop. 3.1 in [9].
□
Proposition 3.12
Suppose R is a regular R-class of TA (=DOWN) such that
the right-stabilizer of any element of R is an R-trivial band
with unambiguous L-order.
Then rank(R)=n implies that there exist r∈R, and
idempotents e0,e1, … ,en, such that
e0>e1> … >en≡Lr* ,*
where “>” is the strict idempotent order. Hence the stabilizer
(r)St contains the idempotent-order chain
e0>e1> … >en.
Proof. By definition, rank(R) is the maximum of the algebraic
ranks of all idempotents that fix any element of R under right
multiplication. So if rank(R)=n, there exists e=e2∈R with
rank(e)=n, such that
e≡Jen′<J … <Je1′<J
e0′ (a J-chain of idempotents).
Applying Lemma 3.12 n times yields e≡J
en< … <e1<e0, a chain in the idempotent order.
Hence there exists r∈R with e≡Rr and
r≡Len. So, (r)St=(en)St, which contains
en< … <e1<e0.
□
Examples.
The following examples show that the stabilizers of the elements
r∈R are not all isomorphic. And when rank(R)=n,
some of these stabilizers contains an idempotent chain
en< … <e1<e0, and some stabilizers do not.
(a) Consider the semigroup T={f,a,b} with the
multiplication table
[TABLE]
Then the rank of the R-class {a,b} is 1,
(a)St={a}, and (b)St={b,f} (with b<f).
So (b)St contains a chain of idempotents of length 1 (counting
the inequalities in the chain), while (a)St does not.
(b) Suppose the following diagram is part of the Cayley
graph of a semigroup SA. The vertices are {q1,q2,q3,q4} and the
edges are
(q1,a,q2), (q2,b,q1), both with rank 1;
(q2,c,q3), (q3,d,q2), both with rank 2;
(q3,e,q4), (q4,f,q3), both with rank 3.
Then under the hypotheses of Prop. 3.6, the ranks
of the right-stabilizers are as follows.
(q1)St has elements of ranks 1, 2, 3;
(q4)St has rank 3 only;
(q2)St has ranks 1, 2, 3.;
(q3)St has ranks 2 and 3.
This is proved as follows.
First, if a word ℓj∈{a,b,c,d,e,f}+ (for j=1,2)
fixes q∈{q1,q2,q3,q4}, i.e., qℓj=q, then ℓj
is an idempotent (by the hypotheses of Prop. 3.6); moreover,
ℓ2ℓ1=ℓ2 iff ℓ2≤Lℓ1.
Also, the L-order of (q)St is the L-order of the whole
semigroup since ℓ2,ℓ1 are idempotents. Since (q)St has
unambiguous L-order, either ℓ2≤Lℓ1 or
ℓ1≤Lℓ2; i.e., either ℓ2ℓ1=ℓ2 or
ℓ1ℓ2=ℓ1. The rank condition determines the direction of
the L-order: rank(ℓ2)≥rank(ℓ2) implies
ℓ2ℓ1=ℓ2, 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 ∼.
From these two observations we now prove the rank properties of the
stabilizers.
To see that St(q1) has ranks 1, 2, 3, we consider the loops
(q1,ab), (q1,acdb), (q1,acefdb) with start point q1;
these loops have rank respectively 1, 2, 3.
For example, to see that acefdb has rank 3, we observe that
(q1,acefdb)∼(q3,efdbac), and
(q3,dbac)∼(q2,cdba); moreover, cd has rank 2, and ba has
rank 1.
So, rank(cdba)=rank(cd), and
dbac∼cdba, cd∼dc; so rank(dbac)=rank(dc)=2,
Since rank(ef)=3 and rank(dbac)=2, we have
rank(efdbac)=rank((ef)(dbac))=rank(ef)=3.
So, rank(acefdb)=3.
3.2 Bottom-up construction
We construct the cover Acovglc 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 Acovglc 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 AcovGLC.
We start with a cover morphism
φ:Acov→A and the interval
[A,Acov], which is a finite lattice.
Since AcovGLC is intended to belong to the
interval [A,Acov], we want to construct morphisms
φ1:Acov→AcovGLC and
φ2:AcovGLC→A such that
(.)φ=(.)φ1∘φ2, and such that φ2 is
1:1R and satisfies the rank condition.
Moreover, φ2 should be maximal (i.e., modφ1 should
be maximally fine) with respect to these properties.
Hence the congruence modφ1 is a refinement of the congruence
modφ on the state set Qcov of Acov,
and every modφ1 class is mapped to one element of Q by
φ (where Q is the state-set of A). We will write states of
Acov with an overline to make them recognizable.
For a state q of Acov, we denote the
modφ1 congruence class of q by
[q]modφ1 or more briefly by
[q]φ1.
We will construct AcovGLC by defining the congruence
modφ1 on Qcov, and by defining φ2 on each
constructed congruence class.
The construction of [q]φ1 proceeds by induction
on the directed path-length from \i to q in
Acov, where \i is the start state of
Acov.
In this induction we assume that the R-order of each
Acov is a tree; this will hold if (.)cov is closed
under the right Rhodes expansion (.)∧R.
First, the congruence class [\i]φ1 is
{\i}, since the modφ congruence class of
\i is just {\i} (assuming that \i
is not reachable from any state).
For the inductive step we assume that for every state q1 with
a certain directed path-length from \i, a congruence class
[q1]φ1
={q1, … ,qℓ}
has been constructed. Our goal is to construct the congruence class
[q1⋅a]φ1 for each a∈A such that
the directed path-length from \i to q1⋅a
is larger than the directed path-length from \i to
q1.
Note that [qj⋅a]φ1 =
[q1⋅a]φ1, for j=1,…,ℓ; hence
{q1⋅a, … ,qℓ⋅a}
⊆ [q1⋅a]φ1.
Also, since qj is modφ1-equivalent to
q1
and since modφ1 refines modφ, we have for
all j:
(qj)φ=(q1)φ and (qj⋅a)φ=(q1⋅a)φ.
In A we consider the ≡R-class
R=[(q1a)φ]≡R, and the
corresponding ≡R-class R =
[q1a]≡R in Acov.
Since (qj)φ=(q1)φ, we have
R=[(q1a)φ]≡R=
[(qja)φ]≡R for
j=1,…,ℓ.
Recall that in Acov we define rank(R) to be
the maximum of the ranks of all the edges in R, and that the rank of
an edge (q1,a,q2) in R is defined to be the rank of the image
edge ((q1)φ,a,(q2)φ) in R.
Let E(S) be the set of edges between states in a set S. Below,
paths(.) is taken in the R-class R.
We now proceed in a number of stages.
Stage 1: Let
n1 = max{rank(e) : e∈
⋃1≤j≤ℓ
E([{\overline{q}}_{j}a]_{\equiv_{\mathcal{R}}})\ \subseteq\
E(Acov)} (where max(∅)=−1),
S_{1}\ =\ {\sf paths}\big{(}(\overline{q}_{1}a)\varphi,\ n_{1}\big{)}
∪ {ε},
P1 = {qjat∈Qcov : t∈S1,
1≤j≤ℓ}.
Stage i+1: Assuming Sh,Ph have been
defined for 1≤h≤i, let
ni+1 = max{rank(e) : e
is an edge of an R-class of Pi},
S_{i+1}\ =\ {\sf paths}\big{(}(\overline{q}_{1}a)\varphi,\ n_{i+1}\big{)}
∪ {ε},
Pi+1 = ⋃{[qjat]≡R :
\ \overline{q}_{j}at\in Q^{\sf cov},\
t∈Si+1, 1≤j≤ℓ},
where [qjat]≡R is the R-class
of qjat in Acov.
Stage End: Continuing in this way we construct chains
n_{1}\leq n_{2}\leq\ \ldots\ \leq n_{i}\leq\ \ldots\
≤n∞=max{ni:i=1,2, … },
S1⊆S2⊆ … ⊆Si⊆
… ⊆ S∞=⋃iSi ⊆A∗,
and
P1⊆P2⊆ … ⊆Pi
\subseteq\ \ldots\ \subseteq\
P∞=⋃iPi ⊆Qcov.
These sequences are actually finite. Indeed,
Pi⊆Qcov, which is a fixed finite set. Hence,
ni is bounded since it is defined in terms of Pi−1. Hence the
sequence Si is of bounded, being defined in terms of ni.
To define φ1, φ2, and AcovGLC,
we want to construct the modφ1 congruence class
[qja]φ1 so that
qja⟼φ1[qja]φ1
⟼φ2(q1a)φ, where
φ1φ2=φ, and where qja and
(qja)φ=(q1a)φ are known. And we want
φ2 to be injective on each R-class. So
([(q1a)φ]≡R)φ−1 is a union of
R-classes of Acov, each of which maps into
[(q1a)φ]≡R.
By construction, P∞ is a union of R-classes of
Acov.
Therefore, for qja∈P∞ we define
[qja]φ1 to be (q1a)φφ−1 ∩ P∞.
More generally, for any q∈P∞ we define
[q]φ1 to be
(q)φφ−1 ∩ P∞.
Then φ2 is defined by
[q]φ1 \ \longmapsto\
([q]φ1)φ ∈Q
where Q is the state set of A.
This completes the inductive step of the construction of
AcovGLC.
3.3 Proof of correctness
Proposition 3.13
Suppose that A is the Cayley graph of an
A-monoid MA such that the stabilizers of Acov are
R-trivial bands with unambiguous L-order, and such that
the R-order of MA is a tree.
Then glc=GLC.
Proof. By construction φ, φ1, φ2 are A-automaton
morphisms satisfying φ=φ1∘φ2.
We want to show that φ2:AcovGLC→A is
1:1R and obeys the rank condition; and we want to show
that φ2 is maximal among all the right-factors of φ that
are 1:1R and that satisfy the rank condition.
Proof that φ2 is 1:1R and
satisfies the rank condition
Let q∈Qcov. If (qa)φ is in the
R-class R, and (qa)φ1 is in the R-class
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 paths(r,j).
Then, clearly, φ2 is 1:1R 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 (.)covGLC with
respect to the 1:1R property and the rank condition
We want to show that φ2 is the maximal right-factor of φ
that is 1:1R and satisfies the rank condition.
Let B be an A-automaton, and let
θ1:Acov↠B and
θ2:B↠A be automaton morphisms, where
θ2 is 1:1R and satisfies the rank condition, and such
that (.)φ=(.)θ1θ2. Hence the congruence
modθ1 refines modφ. We want to show
that the congruence modφ1 refines modθ1.
Claim. For any q∈Acov,
[q]φ1⊆[q]θ1.
Proof of the Claim: We follow the inductive construction of
modφ1. First, q∈[q]θ1,
obviously.
We will use E(V) to denote the set of edges between vertices in V
(including loops).
We have: [q]≡R⊆
⋃p∈[q]θ1[p]≡R
= [[q]θ1]≡R;
hence, since θ2 is 1:1R, we obtain E1o⊆
E\big{(}\big{(}[[{\overline{q}}]_{\theta_{1}}]_{\equiv_{\mathcal{R}}}\big{)}\varphi\big{)}.
Since θ2 satisfies the rank condition we also have E1⊆
E\big{(}\big{(}[[{\overline{q}}]_{\theta_{1}}]_{\equiv_{\mathcal{R}}}\big{)}\varphi\big{)}.
Thus, P1⊆[q]θ1.
Continuing along the inductive construction of modφ1 we
obtain P∞⊆[q]θ1. Hence,
[q]φ1⊆[q]θ1.
This proves the Claim.
Since the Claim holds for every state p in [q]θ1,
we conclude that [q]θ1 is a union of congruence classes
[p]φ1. □
4 The Key Lemma
We will use =+ to denote equality in A+ or A∗, i.e., literal
equality of words.
Definition 4.1
A cover (.)cov of A-automata has the backwards-k
property (for k≥1) iff for every A-automaton
A=(Q,\i,⋅) and every α,β∈A+, the following
holds. If αβ=α in the syntactic semigroup
SAcov of AAcov, then α can be
factored in A+ as
α=+α~β1…βk (with
α~,β1, …,βk∈A+) such that in
the syntactic semigroup SA of A we have α~=α
and β1=β2= … =βk=β=β2.
Proposition 4.2
For A-generated semigroups, let (.)AIS be an expansion with
R-trivial idempotent stabilizers, let (.)A∧(k+1)
be the Henckell (k+1)-factor expansion, and let (.)ARB
be the rectangular-bands expansion.
Then for any A-generated semigroup S, the triply expanded semigroups
SAIS∧(k+1)RB and
SA∧(k+1)IS RB
have the backwards-k property.
Proof. If αβ=α holds in
SAIS∧(k+1)RB or in
SA∧(k+1)IS RB then
αβ=α also holds in
SAIS∧(k+1), respectively
SA∧(k+1).
The existence of a factorization
α=α~β1…βk in SA then follows from
the basic properties of the expansion (.)A∧(k+1).
□
For an A-automaton A and an expansion (.)exp, let
AexpGLC be the glc-cover in the interval
[A,Aexp], with start state \i~ and state
set Qglc. From AexpGLC and a string
s=a1a2 … an, where ai∈A, we define
a string automaton str(s) as follows.
In the state graph of AexpGLC we consider any walk
\i~⟶a1u1
⟶∗v1
\stackrel{{\scriptstyle a_{2}}}{{\longrightarrow}}\ \ \ldots\ \
⟶aiui
⟶∗vi
\stackrel{{\scriptstyle a_{i+1}}}{{\longrightarrow}}\ \ \ldots\ \
⟶anun∈Rn,
where
u1,v1, …, un−1,vn−1,un∈Qglc are such
that ui≡Rvi for i=1, …, n−1; there is no
vn.
Let Ri be the reachability class of ui (and of vi) in
AexpGLC.
We assume Ri=Rj (hence they are vertex-disjoint) when i=j;
hence Ri>RRi+1 for all i.
The state set of str(s) is
{\i~} ∪ ⋃1≤i≤nRi,
and the arrows of str(s) are just the
AexpGLC-arrows between states of str(s).
We will use \i~ as the start state, and Rn as the set of accept
states of str(s).
Conversely, str(s) uniquely determines the above walk: ui is
the entry point into Ri, and vi is the exit point.
Any A-generated monoid MA will be viewed as an A-automaton via
its right Cayley graph.
Theorem 4.3
(“Key Lemma”).*
Let MA be an A-generated monoid, viewed as an A-automaton A.
We consider the A-generated expansion
MA∧(k+1)IS RB (for some
k>1), viewed as an A-automaton Aexp.
Let AexpGLC be the glc-cover of A-automata in
the interval [A,Aexp].
Let str(a1a2…an) be a string automaton in
AexpGLC, with R-classes Ri for
i=1, …, n. Finally, let
rn=max{rankA(e):e∈E(Rn)}, where
E(Rn) is the edge-set of the reachability class Rn.*
Then there exist d1, …, dn−1,dn∈A∗ such that di
labels a path ui⟶divi in Ri (in
AexpGLC), 1≤i≤n−1, and such that the word
w=+a1d1a2 … an−1dn−1andn
satisfies:
(1)* There is ℓn∈A∗ such that in
Aexp: wℓn=w.*
(2)* rankA(ℓn) = rn > rankA(dn).*
(3)* (Backwards-k property): There exist
w′,t1,…, tk∈A+ (where k is as in ∧(k+1))
such that w=+w′tk …t2t1 in A+,
and in MA: w=w′, tk= … =t1=ℓn.*
*The path labeled by t1, when read backwards from the state
q~n that ℓn loops on, eventually visits the source
state vn−1 of the edge labeled by an in
str(a1a2…an); i.e., andn is a suffix of t1,
as seen in the next graph \ \ \ \ldots\
⟶∗vn−1
⟶anun
⟶dnq~n
↺ℓn (in AexpGLC).
*
Proof. Let us pick d1, …, dn−1 arbitrarily so that
ui⟶divi in Ri, for 1≤i≤n−1.
Reading w1=+a1d1a2 … an−1dn−1an from the
start state \i in Aexp, we reach the state
\i⋅w1, which maps into Rn by the map
Aexp↠AexpGLC.
But \i⋅w2 does not map into Rn, where w2=w1an−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
q⋅a, where for q we take \i⋅w2 in
Aexp, and for a we take an.
Then we run the inductive Stage i+1 for increasing i until we
reach n∞.
If n1=n∞ then as we enter the R-class
[qa]≡R we have maximum edge-rank equal to
n∞; we let dn be the empty word, and let ℓn be a loop
at the entrance of [qa]≡R, passing through all the
edges in E([qa]≡R).
If n1<n∞ we go to the 2nd stage of the induction, i.e.,
we find an R-class R of Aexp with rank
n2>n1. So we have a path
… ⟶dnR2,
and this path up to (but excluding) R2 has edges of rank n1.
Then we take dn as indicated, and we take ℓn to be a loop at the
entrance of R2, passing through all the edges in E(R2).
We continue with n3, etc., until n∞ is first encountered.
Now, item (3) follows from Prop. 4.2, applied to the
ℓn-loop in …
⟶dn∙↺ℓn.
□
The Key Lemma can be generalized by replacing k by two parameters,
k1 and k2. For this we use the expansion (.)A
∧k1RB ∧(k2+1) IS RB.
Theorem 4.4
(Key Lemma with k1 and k2).* Under the assumptions of Theorem 4.3, except that the expansion is
now based on (.)A
∧k1RB ∧(k2+1) IS RB,
we have the same conclusions, except for changes in item (3):*
(3)* (Backwards-k1-k2 property): There exist
w′,t1,…, tk2∈A+ such that
w=+w′tk2 …t2t1 in A+,
and in MA∧k1RB: w=w′, tk2= … =t1=ℓn.*
The path labeled by t1, when read backwards from the state
q~n that ℓn loops on, eventually visits the source
state vn−1 of the edge labeled by an in
str(a1a2…an); i.e., andn is a suffix of t1,
as seen in the next graph \ \ \ \ldots\
⟶∗vn−1
⟶anun
⟶dnq~n
↺ℓn (in AexpGLC).
Proof. The proof is the same as for Theorem 4.3, but we use
the morphisms MA∧k1RB ∧(k2+1) IS RB
↠ MA∧k1RB
↠ MA∧k1 for item (3).
□
Notation: Let (GM,A) be the starting group
mapping semigroup whose complexity we are trying to compute, let
MA=(GM,A)ISRB, and let (.)exp be
(.)∧k1RB ∧(k2+1) IS RB.
Then the semigroup automaton AAexp GLC will be called
PreFF(k1,k2) (for “pre-funny fractal”).
For any finite semigroup S, let ω(S) be the smallest positive
integer m such that for all s∈S, sm is an idempotent.
To be continued.