COSTUMBRES

In this section we show that the groupBn,ris equal to the normaliserNRn,r(Gn,r)ofGn,rinRn,r.

By appealing to results in [10] and a result of Rubin [47] we conclude that Aut(G_n,r)=∼Bn,r. We first begin by illustrating the idea behind the proof with synchronous transducers overCn. Recall (Example 1.7.39) that for an invertible synchronous transducerAq0overCn,hp∈Rn

for all statesp∈QA. We also recall Notation 2.3.1 for the mapg_γ,δ :Uγ→Uδ,γ,δ ∈X∗n, and Notation 1.7.42 whereby for a staterof an invertible, synchronous transducer we denote the state

(,r)ofA−1byr−1. We have the following lemma which first appears in [24]:

Lemma 2.4.1. LetAq₀ be a minimal, synchronous, invertible transducer overCnandp∈QAbe a fixed state ofA. ThenAq₀ is synchronizing if and only if for every stateq∈QA,h−1p hq ∈Gn,1.

Proof. We begin with the reverse implication.

Letqbe any state ofAq0. Recall (Example 1.7.39) thatA−1q₀ is also synchronous and minimal. By Lemma 1.7.4, the transducer product(A−1∗A)_(p−1_,q)is an initial transducer representing the

homeomorphismh−1_p hq. Sinceh−1_p hq ∈Gn,1, it must be the case that there is anm∈Nsuch that for anyγ∈Xm_n, there is aδ∈X∗_nsuch thath−1_p hqUγ=gγ,δ. We may further assume thatm∈N

is minimal with this property. Since(A−1∗A)_(p−1_,q)is a synchronous and invertible transducer, it

has no states of incomplete response and so it must be the case that|δ|=m. Now, using arguments similar to those in the proof of Proposition 2.3.22, after processing a word of lengthmfrom the state(p−1,q)of(A−1∗A)_(p−1_,q), the resulting state must beω-equivalent to the identity. Let

Γ ∈Xm_n, and supposeπ_(A−1_∗A)(Γ,(p−1,q)) = (r−1,t), it follows thath_(r−1_,t)must be the identity

homeomorphism. Since(A−1∗A)_(p−_1,q)is synchronous, we therefore have thattisω-equivalent

to the staterofAand sot=rsinceAis minimal.

Let∆=λ_A−1(Γ,p−1). The arguments above demonstrate thatπA(∆,q) =r. However, sinceq was chosen arbitrarily, it follows thatπA(∆,q0) =rfor anyq0∈QA. ThusAis synchronizing at levelm, since for any word∆0∈Xm_n, there is a wordΓ0∈X_nmsuch thatλ_A−1(Γ0,p−1) =∆0.

For the forward implication, assume that A is synchronizing. Let k be the minimal synchronizing level of A and q be any state of A. Let γ ∈ Xkn, ∆ = λ_A−1(γ,p−1), δ =

λ_A−1_∗A(γ,(p−1,q)), and r−1 = π_A−1(γ,p−1). Since πA(∆,p) = r, it must be the case, since

A is synchronizing at level k and ∆ ∈ Xk_n, that πA(∆,q) = r. Therefore, it follows that

π_A−1_∗A(γ,(p−1,q)) = (r−1,r). Henceh−1p hqUγ =gγ,δ.

Lemma 2.4.2. LetAq₀ be a minimal, invertible, synchronous, synchronizing transducer. Fix a statepof

Aq₀and letk∈Nbe minimal such that for every stateqofAand for anyγ∈Xkn, that there is aδ∈X∗n satisfyingh−1_p hqUγ =gγ,δ. The minimal synchronizing level ofAq0 is equal tok.

Proof. Assume that Ais an invertible, synchronous, synchronizing transducer with minimal synchronizing levell. Observe that by minimality ofl, there is a wordν ∈ Xl−1_n and states

q₁,q₂ ∈ Asuch thatπA(ν,q1)6= πA(ν,q2). Letµ ∈ Xl−1n be such thatλA−1(µ,p−1) = ν, and

π_A−1(µ,p−1) =t−1(and soπA(ν,p) =t). Observe that one ofπA(ν,q1),πA(ν,q2)is not equal tot. Without loss of generality we assume thatπA(ν,q1) =sforsa state ofAdistinct fromt. Sinces6=t, it follows that the mapλ_A−1_∗A(,(t−1,s)) :Xn →Xnis not trivial and so moves at least two points. Thus, there arei,j,i0,j0inXnsuch thati6=j,i0=6 j0,λA−1_∗A(i,(t−1,s)) =i0and

λ_A−1_∗A(j,(t−1,s)) =j0. Therefore,λ_A−1_∗A(µi,(p−1,q1)) = νi0andλA−1_∗A(µj,(p−1,q1)) = νj0, hence we conclude thath−1_p h−1_q does not act onUµasgµ,µ0for someµ0∈X+_n.

Now letqbe any state of A. Letγ ∈ Xln,∆ = λ_A−1(γ,p−1),δ = λ_A−1_∗A(γ,(p−1,q)), and

r−1 =π_A−1(γ,p−1). SinceπA(∆,p) = r, it must be the case, sinceAis synchronizing at levell and∆∈Xln, thatπA(∆,q) =r. Therefore, it follows thatπA−1_∗A(γ,(p−1,q)) = (r−1,r). Hence

h−1_p hqUγ = gγ,δ. Thuslis the minimal number such that for every stateqofAand for any

γ∈Xl_n, that there is aδ∈X∗_nsatisfyingh−1_p hqUγ=gγ,δ. Thereforel=kas required.

We modify the previous lemma by introducing the phrase ‘strictly accessible’ (Definition 1.3.12). Lemma 2.4.3. LetAq0 be a minimal, synchronous, invertible transducer overCn. Fix a statep∈QA.

ThenAq₀ is synchronizing if and only if for every strictly accessible stateq∈QA,h−1p hq ∈Gn,1. Proof. The forward implication follows exactly as in Lemma 2.4.1. For the reverse implication, letkbe minimal such that for any strictly accessible stateq∈QAand for anyγ∈Xkn, there is a

δ∈X∗_nsuch thath_p−1hqUγ=gγ,δ. As in the proof of Lemma 2.4.1 we once more conclude that for

any pairq1,q₂of strictly accessible states,πA(,q1) :Xkn→QAis equal toπA(,q2) :Xkn→QA. However, since any non-initial stateqofAis strictly accessible, we conclude thatAis synchronizing at level at mostk+1.

Lemma 2.4.4. LetAq₀be a minimal, synchronous, synchronizing transducer which is not strictly accessible. Fix a statepofAand letk∈Nbe minimal such that, for any strictly accessible stateqofAand for any

γ∈Xk_n, there is aδ∈X∗_nsatisfyingh−1_p hqUγ=gγ,δ. Then the minimal synchronizing level ofAq0is

eitherk+1ork.

Proof. It follows from Lemma 2.4.2 thatAq0 has minimal synchronizing level at leastk. The

proof of Lemma 2.4.3 demonstrates thatAq0has synchronizing level at mostk+1, thus the result

follows.

Corollary 2.4.5. LetAq₀ be an invertible, minimal, synchronous transducer overCn. Ifh−1q0Gn,1hq0 ⊆

Gn,1, thenAq₀ is synchronizing. Ifh−1q0Gn,1hq0 =Gn,1thenhq0 ∈Bn,randAq0 is bi-synchronizing.

Proof. LetAq0be an invertible, minimal, synchronous transducer overCnsuch thath−1q₀Gn,1hq0 ⊆

G_n,1. In order to show thatAq₀is synchronizing it suffices to show thatAq₀satisfies the equivalent condition stated in Lemma 2.4.3.

Fix a non-trivially accessible statepofAq₀. Then, by the definition of non-trivial accessibility (Definition 1.3.13), there is a word γ ∈ X+_n such that πA(γ,q0) = pand λA(γ,q0) 6= . Let

δ = λ_A(γ,q₀)∈ X+

n and observe thatπA−1(δ,q−1₀ ) =p−1andλ_A−1(δ,q−1₀ ) = γ. Letqbe any

strictly accessible state ofAq₀, andν∈Xn+be such thatqis strictly accessible fromq0byν. Observe that there are complete antichainsuandvof the same length such thatγ ∈ uandν ∈v. For instance, if|ν|₆|γ|, one may take the complete antichainX|nγ|and, since|Xn|γ||≡|X|nν|| modn−1, by repeatedly expanding the complete antichainX|nν|using elements not equal toν, one obtains a complete antichain of the same cardinality asXγncontainingν. Reordering if necessary, we may assume thatgu,vUγ=gγ,ν.

LetCr₀ be the minimal initial transducer representinggu,v. Sincehr₀ =gu,v, it follows that

λC(γ,r0) = νandπC(γ,r0) = id. Consider the product(A−1∗C∗A)(q₀−1,r0,q0). Observe that

π_(A−1_∗C∗A)(δ,(q−1₀ ,r0,q0)) = (p−1, id,q). This follows asλA−1(δ,q−1₀ ) =γ,λC(γ,p0) =νand

πA(ν,q0) = q. Observe that the initial transducer (A−1∗C∗A)_(p−1_,id,q) isω-equivalent to

the initial transducer(A−1∗A)_(p−1_,q). Moreover, since h_(q−1

0 ,r0,q0) = h

−1

q₀gu,vhq0 ∈ Gn,1, it

must be the case thath_(p−1_,id,q) = h(p−1_,r) must be an element of Gn,1. This is because as

h

(q−₀1,r0,q0)is a prefix replacement map, and sinceπ(A−1∗C∗A)(δ,(q

−1

0 ,r0,q0)) = (p−1, id,q), we have(Uδ)h_(q−1

0 ,r0,q0)=λ(ν,q0)(Cn)h(p−1,id,q). Sinceqwas an arbitrary strictly accessible state

ofAq₀ it follows thatAq₀ satisfies the hypothesis of Lemma 2.4.3.

Now suppose thatAq₀ is a minimal, invertible, synchronous transducer overCnsuch that

h−1_q0G_n,1hq₀ =Gn,1. Multiplying on the left byhq₀ and on the right byh−1q0 we therefore have that

G_n,1=hq₀Gn,1h−1q0. Thus by the arguments in the preceding paragraphs we conclude thatAq0

andA

We now extend Corollary 2.4.5 to cover all elements ofRn,r. The strategy remains the same, however, we need to adapt the lemmas above, in particular Lemma 2.4.3 and Lemma 2.4.1, to allow for all elements ofRn,r. The proof is slightly more involved but core idea is the same.

We recall (Construction 1.7.29) that for a finite invertible transducerAq0, the states of the

inverse transducerA_(,q0)=hr˙,Xn,R_A0 ,S_A0 ,π_A0 ,λ_A0 iare given by a pair(w,q)whereqis a state of

A, andw∈X∗_nis such thatUw⊂ im(q)and(w)Θq=(Claim 1.7.30). The mapΘqis defined in Definition 1.7.18. Further recall Notation 1.7.33 thatA_q−1

0 =h˙r,Xn,RA−1,SA−1,πA−1,λA−1iis the

minimal transducer representingA_(,q

0). Observe that any non-trivially accessible state(w,p)of

A_(,q

0), induces a continuous injectionh(w,p) : Cn→Cn. The proposition below first appears in

the author’s article [10].

Proposition 2.4.6. A minimal invertible finite transducer overCn,r,Aq₀, is synchronizing if and only if there is ak∈Nso that for any non-trivially accessible statepofA_q−1

0 , any state

q∈SA, for anyγ∈Xk_n

and anyΓ ∈Cn, there is aδ∈X∗_nsatisfying,(γΓ)hphq =δΓ.

Proof. Observe that sinceA_(,q0) has no states of incomplete response (Proposition 1.7.34), it suffices to prove the proposition withA_(,q0)in place ofA_q−1

0 . This is because, asAq−01is obtained

fromA_(,q0)by identifyingω-equivalent states, if, for any non-trivially accessible statepofA_q−1 0

and any stateq∈SA, there is akso that, for anyγ∈X_nkand anyΓ ∈Cn, there is aδ∈X∗_nsuch that,(γΓ)hphq=δΓ, then for any non-trivially accessible state(w,p0)ofA(,q0), any stateq∈SA,

anyγ∈Xk

n, and anyΓ ∈Cn, there is aδ∈X∗nsuch that,(γΓ)h(w,p0₎h_q=δΓ.

We begin with the reverse implication. For this, our strategy is to show that the subtransducer

Aq₀SA =hXn,SA,πASA,λASAiconsisting of all states inSA, is synchronizing. To do this we

show that there is a baseS forAq₀SAoverCn(Definition 2.1.15). From this it will follow that

Aq₀is synchronizing at a level equal to the maximum length of a word in this base plus the length of a minimal path from a state ofAq₀ to a state inSA(recall that by Restriction (R.1) such a path always exists). This is because after reading the initial prefix of such a word guaranteeing that the remaining suffix is processed from a state inSA, the resulting final state is determined by this suffix.

Fixk∈Nsuch that for any stateq∈SA, any non-trivially accessible state(w0,p0)∈QA_(,q

0),

anyγ∈Xk_nandΓ ∈Cn, we have,(γΓ)h_(w0_,p0₎h_q =δΓfor someδ∈X∗_n. Letl∈Nbe minimal

such that whenever there is aγ∈Xkn,q∈SA, and a non-trivially accessible state(w0,p0)∈Q_A,q

0

such that(γΓ)h_(w0_,p0₎h_q = δΓ, then|δ| 6l. Letk >kbe such that for any wordγ ∈Xkn, any q∈SA, and non-trivially accessible(w0,p0)∈Q_A,q

0, we have,|(γ)A(w0,p0)Aq|>l.

Now fix a non-trivially accessible state (w,p) of QA(,q₀). Note that w satisfies that

Uw ⊆ im(p) and (w)Θp = . Let x1x2. . .xkxk+1. . .xk ∈ X k

n andy1, . . .yt ∈ X∗n be such that, (x₁x₂. . .x_k)A_(w,p) = y₁. . .yt. Let q₁ and q₂ be any arbitrary states of SA, and let

(y₁. . .yt)Aq₁ = z1. . .zl₁xk1+1. . .xk1+i and(y1. . .yt)Aq2 = u1. . .ul2xk2+1. . .xk2+j. We may

assume thatul₂ 6=xk₂ andzk₁ 6= xk₁. Hereka6kandla6lfora∈ {1, 2}, moreover we may assume that thelaare minimal such that(x1x2. . .xkχ)h(w,p)hqa=∗1. . .∗laxka+1. . .xrχwhere

(∗,a)∈{(z, 1),(u, 2)}.

It cannot be the case that(y₁. . .yt)Aq₁ =z1. . .zl₁xk₁+1. . .xkρfor someρ∈X+n, since picking a wordρ0which is incomparable toρ, we then have,(x₁. . .x_kρ0)A_(w,p)Aq1 =z1. . .zl1xk1+1. . .xkρδ for someδ, however, by minimality ofl1,ρ =ρ0which is a contradiction. A similar argument demonstrates that(y₁. . .yt)Aq₂ 6=u1. . .ul2xk2+1. . .xkρfor someρ∈X

+ n. Therefore we may assume thatk₁+i₆kandk₂+j₆k.

Letπ_A0 (x₁. . .x_k,(w,p)) = (v,s). Recall that, by Construction 1.7.29 and Lemma 1.7.32, we have,

wx₁. . .x_k−λA(y1. . .yt,p) = v andπA(y1. . .yt,p) = s. SinceA(,q0) has only finitely many

states (Lemma 1.7.28) we may choosekso thatvis a suffix ofx₁. . .x_k. Moreover, we also have

(wx₁. . .x_k)Θp=y1. . .yt. Letwx1. . .xκ=λA(y1. . .yt,p)for someκ6k. Letmbe minimal such that for any stateqofAand any wordµ∈Xm_n, we have,|λA(µ,q)|>k−κ. Now notice that there is a maximal setα={α₁, . . .αm₁}⊆Xmn such that the greatest common of prefix of the elements of

because(wx₁. . .x_k)Θp=y1. . .ytandπA(y1. . .yt,p) =s. Further notice that the setαdepends only onx₁. . .x_k,(w,p)and(v,s), that is,αis independent of the choice ofq₁andq2.

Fix an αa ∈ α. Let Pa = πA(αa,s) and let im(Pa), as usual, represent the image of the map hPa from Cn to itself. Since hq0 is injective, (vρaim(Pa))Θs = αa and

(ρaim(Pa))h(v,s) = Uαa. Let T1 := πA(y1. . .yt,q1) and T2 := πA(y1. . .yt,q2) . Observe

that(x₁. . .x_kρaim(Pa))h(w,p) = Uy1...ytαa. Since(y1. . .yt)Aq1 = z1. . .zl1xk1+1. . .xk1+i and

(y₁. . .yt)Aq₂ =u1. . .ul₂xk₂+1. . .xk₂+j, it must be the case, for anyδ∈Cn, thatλA(αaδ,T1)has a prefixxk₁+i+1. . .xkρa, likewise,λA(αaδ,T2)has a prefixxk₂+j+1. . .xkρa. As we assumed that

Aq₀has no states of incomplete response, it must be the case thatλA(αa,T∗) =xk∗+]+1. . .xkρa for(∗,])∈{(1,i),(2,j)}, otherwise the statesπA(αa,T∗)will be states of incomplete response for ∗ ∈{1, 2}. Letδ0∈ im(Pa)be arbitrary, and let(x1. . .xkρaδ0)(w,p) =y1. . .ytαaδ, then we must have,(δ)h_π

A(αa,T1) = (δ)hπA(αa,T2) =δ

0_{. Sinceδ}0_{was arbitrary and(ρ}

aim(Pa))A(v,s) =Uαa,

the previous equality holds for anyδ in Cn. This means that π_A(αa,T₁)and π_A(αa,T₂) are

ω-equivalent.

Now asq₁andq₂were arbitrary states ofAq₀SA, we must have, for any pairq1,q2of states

ofAq₀SA, that πA(y1. . .ytαa,q1)andπA(y1. . .ytαa,q2) areω-equivalent (and so equal by

minimality ofAq0) for allαa∈α. Therefore, the set of words{y1. . .ytαa|16a6m1}is a set of

synchronizing words forAq0SA.

We now consider the setβ:=Xm_n\α. Once againβis independent of the choice ofq₁andq2. Letβ:={β1, . . . ,βm2}. Letβb∈βbe arbitrary. By assumption we have that|λ

0 A(βb,(v,s))|>k−κ, thus letx_κ+10 . . .x0 kρ 0 b:=λA0 (βb,(v,s)).

The word x₁. . .xκx_κ+10 . . .x_k0ρ_b0 has length greater than or equal to k, therefore we may repeat the arguments above with x₁. . .xκ. . .x_κ+10 x_k0ρb0 in place of x1. . .xk. Notice that

λ_A0 (x₁. . .xκx_κ+10 . . .x_k0ρ_b0,(w,p)) is either a prefix of y1. . .yt or contains y1. . .yt as a prefix. Letλ_A0 (x₁. . .xκx_κ+10 . . .x_k0ρ_b0,(w,p)) :=y₁0. . .y_t00 and(v0,s0) :=π_A0 (x₁. . .xκx_κ+10 . . .x_k0ρ_b0,(w,p)). After repeating the arguments above, we end up with a new set of synchronizing words forAq₀SA.

However, this new set of synchronizing words contains as a subset{y₁. . .ytβbηc|16c 6m₁0} where theηc’s all have the same size and form a maximal antichain ofX∗n. Take the union of the sets{y₁. . .ytβbηc|16c6m₁0}and{y1. . .ytαa|16a6m1}and letS(y1. . .yt)denote the union. Continuing in this way across all theβb∈β, and lettingS(y1. . .yt)denote the union at each stage, we see thatS(y₁. . .yt)is finite and is a base forAq0SA over the clopen setUy1...yt.

Now to finish to proof it suffices to construct a finite setM⊂ X∗_n satisfying the following conditions:

1.) for every element ofν∈X∗_nof long enough length there is an element ofMwhich is a prefix of a (possibly trivial) rotation ofν,

2.) for every wordγ∈M,S(γ)exists and is a base forAq₀SA over the clopen setUγ.

To see that this suffices, observe that the conditions above imply that there is someD∈N, such that for alld∈NDand for any wordν∈Xdn, there is a rotationν0ofνwhich has a prefix inM. Therefore, for large enoughD,ν0has a prefix inS(γ)for someγ∈M, and so is a synchronizing word forAq0, we then appeal to Lemma 2.2.20 to conclude thatAq0SAis synchronizing.

To do this letPbe the set of all non-trivially accessible states ofA_(,q

0). For each(w,p)∈P,

letO((w,p)) :={λ_A0 (ϕ,(w,p))|ϕ∈Xk_n}. The arguments above demonstrate that for each word

In document La progresión temática en la obra Yawar Fiesta de José María Arguedas (página 39-44)