Folklore

In this section we study the quotient groupBn,r/Gn,r =∼ Out(Gn,r). We show thatBn,r/Gn,r is isomorphic to a groupOn,rof non-initial transducers with an appropriately defined product. We highlight a groupHn6∩₁₆r<nOn,rof particular interest. The results and exposition in this section are based on the paper [10]. The set of minimal transducers inducing homeomorphisms

ofRn,r, by Lemma 1.7.11, is a group isomorphic toRn,runder the binary operation which takes two elementsAq0,Bp0and returns the minimal transducerAB(p₀,q0)representing(A∗B)(p0,q0).

Therefore, for convenience and to ease the exposition, we will henceforth identify elements ofRn,r with the set{Aq0 |Aq0is minimal andhq0 ∈Rn,r}. We might still sometimes distinguish between

the two objects for emphasis.

The following lemma demonstrates that two elements ofg₁,g₂∈Bn,rare in the same coset of

Bn,r/Gn,rif and only if the initial transducersAq₀ andBp₀ representingg1andg2respectively have the same cores. We recall (Notation 1.7.12) for initial transducersAq₀,Bp₀overCnorCn,r, thatAB(p₀,q0)is the minimal transducer representing the product(A∗B)(p0,q0).

Lemma 2.5.1([10]). LetAq₀ andBp₀ be transducers representing elements g,h ∈ Bn,r respectively. ThenCore(Aq0)=∼ωCore(Bp0)if and only ifg−1h∈Gn,r.

Proof. Letc=Core(Aq₀)=∼ωCore(Bp₀),k∈Nbe such thatAq₀,Bp₀ andA_q−1

0 are synchronizing

at level k. Since Core(Aq₀) = Core(Bp₀), we may choose k such that, ifγ ∈ Xkn,r, then the state of Aq₀ forced by γ(a state of c) is equal to the state of Bp₀ forced by γ. Consider the product (A

q−₀1 ∗B)(q−₀1,p0). Letj ∈ Nmaxk,1be minimal such that for any γ ∈ X

j

n,r we have,

|λ_A−1(γ,q0))|>k. Letγ∈Xjn,rbe arbitrary,qbe the state ofAq−₀1 forced byγ, andpbe the state ofBp₀forced byλ_A−1(γ,q0). We observe that(q,p)is a state of(A_q−1

0 ∗Aq0)(q−01,q0)since the state

ofAq0 forced byλA−1(γ,q−1₀ )is equal to the state ofBp0 forced byλA−1(γ,q−1₀ ). Therefore, as

γ∈Xj_n,rwas arbitrary, as in the proof of Theorem 2.4.9, we conclude thath−1_q

0hp0 ∈Gn,r.

For the reverse implication let Aq0 and Bp0 be transducers respectively representing

homeomorphismsg,h ∈ B_n,r such that g−1h ∈ Gn,r. Let A

q−₀1 be the minimal transducer representingg−1,C= Core(Aq0),D = Core(Bp0)andE = Core(Aq₀−1). Sinceg−1h ∈ Gn,r, it must be the case that the minimal transducer representing the product(A_q−1

0 Bp0))(q−01,p0)has

trivial core. Clearly(A

q−₀1Aq0)(q−₀1,q0)also has trivial core.

Letqbe a state ofEand notice that im(q)is a clopen subset ofCn. FixΓ ∈ X+_n such that

UΓ ⊆ im(q)and letj ∈ Nbe minimal such that for all wordsν ∈ Xn,j |λA−1(ν,q)| > |Γ|. Let

w₁, . . . ,wmbe the set of all words inXjnsuch that, for all 16i6m,λA−1(wi,q) =Γ ρifor some

ρi ∈X∗n. SinceUΓ ⊂ im(q), we have,∪16i6mΓ ρiim(πA−1(wi,q)) =UΓ. Letwbe the greatest common prefix of the set{wi | 1 6 i 6 m}and,p1 andp2be states of CandD respectively such that(q,p1)and(q,p2)are states in the core of(A_q−1

0 Bp0))(q−01,p0) and(Aq−01Aq0)(q−01,p0) respectively. Since(A_q−1 0 Bp₀))_(q−1 0 ,p0) and(Aq−01 Aq₀)_(q−1

0 ,p0) have trivial core, it must be the

case that, for any wordν ∈ X∗_n,(ν)θhqhp₁ − ()θhqhp₁ = νand(ν)θhqhp₂ − ()θhqhp₂ = ν.

Therefore, for(i,T)∈{(1,A),(2,B)}, we have,λT(Γ,pi)has a prefix equal to()θhqh_piw. This is

because, as∪_16i6mΓ ρiim(πA−1(wi,q)) =UΓ and(wi)θhqh_pi − ()θhqh_pi =wi, any element

of(UΓ)hpi has prefix()θhqh_piw, and asAq0 andBq0 have no states of incomplete response,

we conclude thatλT(Γ,pi)has a prefix equal to()θhqh_piw. In fact it is actually the case that

λT(Γ,pi) = ()θhqh_piw, since if there was someϕ ∈ X

+

n such that, λT(Γ,pi) = ()θhqh_piwϕ,

then choosingδ∈X∗_nand 16i₆msuch thatwiδ⊥wϕ, we have,λT(Γ ρi,pi)⊥()θhqh_piwiδ

contradicting the fact that(wiδ)θhqhpi− ()θhqh_pi =wiδ.

Now let ∆ ∈ X+_n be arbitrary, and l ∈ N be minimal such that, for all words µ ∈ Xl_n,

|λ_A−1(µ,q)| >|Γ ∆|. By repeating the argument above withΓ ∆in place ofΓ, we again conclude

that, for(i,T)∈{(1,A),(2,B)},λ_T(Γ ∆,p_i) = ()θ_hqh

piξwhereξis the greatest common prefix of

the set of all wordsµ∈Xl_nsuch that,λ_A−1(µ,q) =Γ ∆φfor someφ∈X∗n. Notice thatµ=wµ0 for someµ0 ∈X∗_n, and soξ = wξ0for someξ0 ∈X_n∗. Therefore, sinceλT(Γ,pi) = ()θhqh_piw,

we must have, λ_T(∆,π_T(Γ,p_i)) = ξ0. Hence, we conclude, since ∆ ∈ X∗_n was arbitrary, that

λA(∆,πA(Γ,p1)) =λB(∆,πB(Γ,p2))for all∆∈X∗n. This implies that, settingp10:=πA(Γ ∆,p1)and

p₂0:=πB(Γ ∆,p2),Ap0

1andBp20 areω-equivalent. SinceEandDare strongly connected and have

no pair ofω-equivalent states,A_p0

1 =∼ωBp20, and soEandDmust be isomorphic, that is they are

Implicit in the proof of the reverse implication of Lemma 2.5.1 is the following result:

Lemma 2.5.2. Let E,C,D be core synchronizing transducers over Cn without states of incomplete response and no pair of ω-equivalent states. Let e,c,d be states of E, C and D respectively. If Core(EC_(e,c)) =Core(ED_(e,d)) =id, thenC=∼ωD.

Lemma 2.5.1 means that we may identify elements ofBn,r/Gn,r with the set{Core(Aq0) |

Aq₀is minimal andhq₀ ∈Bn,r}.

Notation 2.5.3. SetOn,r={Core(Aq₀)|Aq₀is minimal andhq₀ ∈Bn,r}. Forn=2, there is only one choice ofr, in this case we writeO2for the groupO2,1.

Remark 2.5.4. Observe that by definition, forT ∈On,r, and any statetofT, the initial transducer

Ttis minimal (it is accessible sinceT is synchronizing).

The following definition gives a multiplication of core synchronizing, automata overCnand is taken from [10].

Definition 2.5.5. LetCandDbe core synchronizing transducers overCn, andpandqbe states ofCandDrespectively. Set CD = Core(CD_(p,q)), where CD_(p,q) is the minimal transducer representing the product(C∗D)p,qof the initial transducersCpandDq. We callCDthecore productofCandD.

Remark 2.5.6. Observe that for core, synchronizing transducersCandDoverCn with statesp andq, the productCD_(p,q)is synchronizing also by Proposition 2.1.30 and Proposition 2.1.32.

The lemma below demonstrates that for core synchronizing transducersCandDoverCn, and transducersAq₀,Bp₀ ∈Bn,rsuch that Core(Aq₀) =Cand Core(Bp₀) =D, the core productCD, for any choice of states ofCandD, coincides with Core(AB_(p

0,q0)).

Lemma 2.5.7. LetAp₀ andBq₀ be minimal transducers such thathp₀,hq₀ ∈Bn,r ,C=Core(Ap₀), andD=Core(Bq₀). For any choice of statep∈QCandq∈QD,Core(AB(p₀,q0)) =Core(CD(p,q)).

Proof. Letk∈Nbe such thatAp₀,Bq₀,CandDare all synchronizing at levelk. Letj∈Nkbe minimal satisfying the following conditions for allγ∈Xjn,r:

i.) |λA(γ,p0)|>kand,

ii.) for any statep0∈QC,|λA(γ,p0)|>k.

For any wordν ∈ Xkn,r, we haveπB(ν,q0) ∈ QD, since Bq0 is synchronizing at level kwith

Core(Bq0) =D.

Let∆ ∈ Xjn be arbitrary,p0 be the state ofCforced by∆, andq0 be the state ofD forced byλC(∆,p). Observe thatπC∗D(∆,(p,q)) = (p0,q0). We now show that (p0,q0)is a state of

(A∗B)_(p

0,q0)). LetΓ ∈X

j

n,rbe arbitrary, andΛ∈Xjnbe such that the state ofCforced byΛisp (Λexists sinceCis core and synchronizing). The states ofAp0 forced byΓ Λ∆andΓ Λare equal to

p0andprespectively, sinceπA(Γ,p0)∈QCandCis synchronizing at levelk. Thus,λA(Γ Λ∆,p0) has length greater than or equal to 2k, by assumptions placed onj. Moreover, the lengthksuffix ofλA(Γ Λ∆,p0)is equal toλC(∆,p). Therefore, it must be the case thatπB(λA(Γ Λ∆,p0),q0) =q0, since reading the initial lengthkprefix fromq₀guarantees that the lengthksuffix,λC(∆,p), is processed from a state ofDto the stateq0. In total we have,π(A∗B)(Γ Λ∆,(p0,q0)) = (p0,q0).

The arguments above show that for any wordµ∈X∗_nof long enough length (length at least

j),π(C∗D)(µ,(p,q))is a state of(A∗B)(p₀,q0). From this it follows that Core(A∗B)(p0,q0) =

Core((C∗D)_(p,q))since the core of any synchronizing transducer is strongly connected.

The conclusion of the proof follows from the following observation. LetEbe a synchronizing transducer, then ifE0is the transducer obtained from applying the procedureM2to the transducer

E, thenπ_E0_(Core(E0₎₎=π_E(Core(E)). Moreover, the procedureM2modifies the transition function

of a stateeofEusing only information about the functionhe. Therefore, after applying procedures M2andM3to(A∗B)_(p

0,q0)and(C∗D)(p,q)to obtain minimal transducersAB(p0,q0)andCD(p,q),

The following Proposition follows from Lemma 2.5.7.

Proposition 2.5.8. The setOn,requipped with the binary operation core product is a group. Moreover

On,r=∼ Out(Gn,r).

Proof. That the setOn,r is closed under the binary operation core product is a consequence of Lemma 2.5.7. That this binary operation is associative is a consequence of the associativity of the multiplication of minimal transducers, and Lemma 2.5.7 once more. We now show that inverses exist and that they are unique.

LetT ∈ On,r, then there is someAq₀ ∈ Bn,r such that Core(Aq₀) = T. However, since

Aq₀ ∈Bn,r, there is someBp₀ ∈Bn,rsuch thatAB(q₀,p0) = BA(p0,q0) =id. LetS=Core(Bp0),

then by Lemma 2.5.7 we have thatT S=ST =id. Now by Lemma 2.5.2, or the associativity of the product, it follows thatSis unique.

The map from Bn,r/Gn,r → On,r mapping an element [Aq0] ∈ Bn,r/Gn,r to Core(Aq0)

is surjective, by the definition of On,r, injective by Lemma 2.5.1, and a homomorphism by Lemma 2.5.7. Therefore we conclude thatOn,r=∼ Bn,r/Gn,r=Out(Gn,r).

Definition 2.5.9. LetT ∈_On,rwe say thatT has ahomeomorphism stateif there is stateuofT such that the maphu: Cn→Cnis a homeomorphism; the stateuis called ahomeomorphism state.

The question arises if all elements ofOn,ror even ofOnpossess homeomorphism states. Before we answer this question, we demonstrate why possessing a homeomorphism state is important. Proposition 2.5.10. Letr,r0∈{1, 2, . . . ,n−1}be distinct. IfT ∈O_n,r0has a homeomorphism state, then T ∈On,ralso.

Proof. LetCq0be a transducer representing an element ofGn,r(and so Core(Cq0) =id). LetTbe an

element ofOn,r0with a homeomorphism statetandB_R

0 ∈Bn,r0be such that Core(BR0) =T. We

form a new transducerDq₀ satisfying, Core(Dq₀) =T andhDq₀ ∈Bn,r, by replacing Core(Cq0)

withTt.

We setQD :=QC\{id}tQT. Define the transition functionπDand output functionλDofDq₀ as follows:πDXn×QT =πT,λDXn×Qt =λT; fora∈Xn,rtXnandq∈QCsuch thatπC(a,q)

is defined we have:

π_D(a,q) =

πC(a,q) ifπC(a,q)6=id

t otherwise andλD(a,q) =λC(a,q).

IfCq₀ is synchronizing at levelk, then after processing any input of lengthkfrom any state of

Dq₀, the active state is a state ofT. SinceT is synchronizing, it follows thatDq₀is synchronizing also. Further observe that the setINof minimal paths inCq0, fromq0to the state id is a complete

antichain forX+n,r. Moreover the set of outputsOUTof the setINwhen processed fromq0, is also a complete antichain forX+n,r. Notice thatINcoincides with the set of minimal paths inDq0 from

q₀to the statet, likewiseOUTcoincides with the outputs of the setINwhen processed fromq₀in

Dq0. SinceTtis a homeomorphism state, we therefore conclude thathDq₀ is a homeomorphism of

Cn,r.

Let γ ∈ IN, and δ = λD(γ,q0) ∈ OUT. Observe thatδ ∈ X+n,r since Cq₀ is a minimal transducer overCn,r. Consider(δ)Θq₀, sinceπD(γ,q0) =t, a homeomorphism state, it follows that(δ)Θq₀ =γotherwisehDq₀ is not a homeomorphism. Thusπ

0

D(δ,(,q0)) = (,t). Moreover, sinceγ∈INwas arbitrary, we deduce that for anyδ0∈OUT,π_D0 (δ0,(,q₀)) = (,t). Letν∈X∗_n,r0

satisfyπB(ν,R0) = t and let µ = λB(ν,R0) ∈ X+_n,r0. By a similar argument, we once more

conclude thatπ_B0(µ,(,R₀) = (,t). Sincetis in the core ofBR₀, by choosing a large enoughν we may makeµas long as we like. Therefore, ifB_R−1

0 is the minimal transducer representing

B_(,R

0)andDq−₀1is the minimal transducer representingD(,q0), thenDq−₀1 is synchronizing and Core(D_q−1

0

) =Core(B_R−1

0 . This is becauseOUTis a complete antichain of

X∗_n,rand after processing any word ofOUT from the state(,q0), the resulting state is (,t). Hence, we conclude that

h_Dq

0 ∈Bn,ras required.

Lemma 2.5.11. LetT ∈O_n,rpossess a homeomorphism statet, thenTtis a bi-synchronizing transducer, in particularTt∈Bn,1.

The example which follows is from the author’s article [10] and demonstrates that elements of

On,rneed not possess a homeomorphism state:

Example 2.5.12. Consider the elementAq₀ ∈B4,3below

q_{0 R} q₁ q₄ q₅ q3 q₂ ˙1| ˙2|˙3 ˙3|˙3 0|˙1,2|˙2 1|˙1, 3|˙2 0|0 1|0, 3|1 2|1 1|2 3|3 0|2 2|3 3|1 2|1 1|2 0|2 0|0 3|0,1|1 2|1 1|0, 3|3 0|0 2|3

Figure 2.12: An example of an element ofB4,3whose core has no homeomorphism state One can observe that Core(Aq₀)is the subtransducer induced by the state{q1,q2,q3,q4,q5} none of which are homeomorphism states. That{q1,q2,q3,q4,q₅}is the Core(Aq₀)follows since

Aq₀is synchronizing at level 3. That none of{q1,q2,q3,q4,q5}are homeomorphisms states follows since the functionλA(,qa) :X4→X4is not injective for any 16a65.

On the other extreme, one could ask about elements of On,r all of whose states are homeomorphism states. These are characterised by the following lemma from [10]:

Lemma 2.5.13. Let T ∈ On,r be such that all states of T are homeomorphism states, then T ∈

∩₁₆r0_<nO_n,r0andT is synchronous.

Proof. Fix a stateqofT, and observe that since all states ofTare homeomorphism states then the set{λ_T(i,q)|i∈Xn}must form a complete antichain forX∗n. However, there is only one complete antichain ofX∗_nconsisting ofnelements and that is precisely the setXn. Thus, for any stateqofT

the mapλT(,q)with domainXn, is a bijection fromXnto itself. ThatT ∈ ∩₁₆r0_6nO_n,r0follows

from Proposition 2.5.10.

Notation 2.5.14. Let Hn be the set of minimal, invertible, synchronous, bi-synchronizing transducers overCn. From Lemma 2.5.13 and the proof of Lemma 2.5.10, it follows thatHn is precisely the set of all those elementsT ∈ ∩₁₆r<nOn,rall of whose states are homeomorphism states. We also denote byHenthe set of invertible, synchronous, synchronizing (but not necessarily

bi-synchronizing) transducers overCn. Note thatHen is not a subset ofOn,rfor anyrsince it contains invertible transducers whose inverses are not synchronizing.

The setHnis closed under the core product given in Definition 2.5.5, and so it is a subgroup of

∩₁₆r<nOn,ras it is also closed under inversion. The setHenis also closed under the core product, however it is not closed under inversion. The next chapter focuses on the monoidHenand the groupHn. Some of the results in the next chapter also concern the monoidPenconsisting of all synchronous, synchronizing transducers overCnwith binary operation a modification of the core product. We close this chapter with a few more observations about the groupOn,r, some of which may be found in the author’s article [10].

Notation 2.5.15. LetKn=∩16r<n−1On,r; in the casen=2 we have,Kn=O2. Notice that asKn is the intersection of the groupsOn,r, with the binary operation core product, it is a subgroup of

On,rfor all 16r < n−1.

A natural question which arises is the following: do all elements of Kn possess a homeomorphism state? Notice that forn = 2 this reduces to answering the question: do all elements ofO2 possess a homeomorphism state? The example below demonstrates that the answer to this question is no, and shows that, in general,Knneed not consist only of elements with homeomorphism states. In fact this example also demonstrates that the subset ofKnof all elements which possess a homeomorphism state is not closed under the binary operation. Example 2.5.16. The transducersAandBhave homeomorphism statesp₀andq₀respectively, such thatAp₀ andBq₀ are bi-synchronizing. Therefore, by Proposition 2.5.10 and its proof, we have thatA,B∈K2. However, the core product ofAandBis a transducerCwhich, as may be

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