This paper characterizes the automorphism groups of generalized Thompson's groups $T_{n,r}$, extending previous results on Higman-Thompson groups, and explores their structure, automorphisms, and properties like $R_{ty}$.
Contribution
It extends the automorphism characterization from Higman-Thompson groups to $T_{n,r}$, analyzes their outer automorphism groups, and investigates their algebraic properties and subgroup structures.
Findings
01
Outer automorphism groups are infinite for $n>2$ and contain Thompson's group $F$.
02
The groups $ ext{Out}(T_{n,r})$ fit into a lattice structure with specific normal subgroup relations.
03
$T_{n,r}$ groups have the $R_{ty}$ property, extending known results for Thompson's group $T$.
Abstract
The recent paper "The further chameleon groups of Richard Thompson and Graham Higman: automorphisms via dynamics for the Higman groups Gn,r" of Bleak, Cameron, Maissel, Navas and Olukoya (BCMNO) characterises the automorphisms of the Higman-Thompson groups Gn,r as the specific subgroup of the rational group Rn,r of Grigorchuk, Nekrashevych and Suchanski{\u i}'s consisting of those elements which have the additional property of being bi-synchronizing. In this article, we extend the arguments of BCMNO and characterise the automorphism group of Tn,r as a subgroup of AutGn,r. We then show that the groups OutTn,r can be identified with subgroups of the group OutTn,n−1. Extending results of Brin and Guzman, we show that the groups OutTn,r, for n>2, are all infinite and contain an isomorphic copy of…
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TopicsGeometric and Algebraic Topology · Finite Group Theory Research · Advanced Combinatorial Mathematics
Full text
Automorphisms of the generalised Thompson groups Tn,r and the R∞ property
The recent paper The further chameleon groups of Richard Thompson
and Graham Higman: automorphisms via dynamics for
the Higman groups Gn,r of Bleak, Cameron, Maissel, Navas and Olukoya (BCMNO) characterises the automorphisms of the Higman-Thompson groups Gn,r as the specific subgroup of the rational group Rn,r of Grigorchuk, Nekrashevych and Suchanskiĭ’s consisting of those elements which have the additional property of being bi-synchronizing. In this article, we extend the arguments of BCMNO and characterise the automorphism group of Tn,r as a subgroup of \mboxAut(Gn,r). We then show that the groups \mboxOut(Tn,r) can be identified with subgroups of the group \mboxOut(Tn,n−1). Extending results of Brin and Guzmán, we show that the groups \mboxOut(Tn,r), for n>2, are all infinite and contain an isomorphic copy of Thompson’s group F. For X∈{T,G}, we study the groups \mboxOut(Xn,r) and show that these fit in a lattice structure where \mboxOut(Xn,1)⊴\mboxOut(Xn,r) for all 1≤r≤n−1 and \mboxOut(Xn,r)⊴\mboxOut(Xn,n−1). This gives a partial answer to a question in BCMNO concerning the normal subgroup structure of \mboxOut(Gn,n−1). Furthermore, we deduce that for 1≤j,d≤n−1 such that d=gcd(j,n−1), \mboxOut(Xn,j)=\mboxOut(Xn,d) extending a result of BCMNO for the groups Gn,r to the groups Tn,r. We give a negative answer to the question in BCMNO which asks whether or not \mboxOut(Gn,r)≅\mboxOut(Gn,s) if and only if gcd(n−1,r)=gcd(n−1,s). We conclude by showing that the groups Tn,r have the R∞ property extending the result of Burillo, Matucci and Ventura and, independently, Gonçalves and Sankaran, for Thompson’s group T.
1 Introduction
The recent paper [3] characterises the automorphisms of the Higman-Thompson groups Gn,r as a subgroup of the rational group Rn,r consisting of those elements which have the additional property of being bi-synchronizing. In this article, we extend the arguments of [3] and characterise the automorphism group of Tn,r, generalisations of Thompson’s group T, as a subgroup of \mboxAut(G)n,r.
The automorphism group of T2,1, or T, as well as the automorphism group of Thompson’s group F, were studied in the paper [4], which also demonstrates that \mboxOut(T2) is isomorphic to the cyclic group of order 2. The paper [6] studies the metric properties of \mboxAut(F) and gives a presentation for this group, whilst the follow up paper [5] studies the automorphisms of the groups Fn and Tn,n−1, generalisations of Thompson’s group F and T (the approach there gives no information about \mboxAut(Tn,r) when r=n−1). The paper [5], amongst other things, demonstrates that \mboxOut(Tn,n−1) for n≥3 is infinite and contains an isomorphic copy of Thompson’s group F.
In this paper we extend the results of [5] to the groups Tn,r for 1≤r<n−1. Firstly, we prove the following,
Theorem 1.1**.**
The group \mboxAut(Tn,r) consists of those elements of Rn,r which can be represented by bi-synchronizing transducers such that the induced homeomorphisms on Cantor space respects the cyclic ordering.
One immediately deduces the following corollary:
Corollary 1.2**.**
\mboxAut(Tn,r)<\mboxAut(Gn,r).
We now turn our attention to the quotient group \mboxOut(Tn,r)=\mboxAut(Tn,r)/\mboxInn(Tn,r). The results below are equally true in the Gn,r context and so to avoid repetition we shall sometimes use the symbol X to represent either T or G, analogously the symbol X will represent either TO or O.
It is a result in the paper [3] that for 1≤r<n the group \mboxOut(Gn,r) is isomorphic to a subgroup On,r of a group On consisting of non-initial, bi-synchronizing transducers. Furthermore, that paper also shows that On=∪1≤r≤n−1On,r. More specifically, it is demonstrated in [3] that for 1≤i≤j≤n−1 such that i divides j in the additive group Zn−1, On,i≤On,j. As a consequence of this, one may deduce that for all 1≤i≤n−1On,1≤On,i, On,i≤On,n−1 (and so On=On,n−1) and On,i=On,d for d=gcd(n−1,i). We extend these result to the group \mboxOut(Tn,r). That is, we have:
Theorem 1.3**.**
For all 1≤r≤n−1, the group \mboxOut(Tn,r) is isomorphic to a subgroup TOn,r of On,r. Moreover for all 1≤i≤j≤n−1 such that i divides j in the additive group Zn−1, we have TOn,i≤TOn,j.
Corollary 1.4**.**
For 1≤1≤n−1, we have TOn,1≤TOn,i, TOn,i≤TOn,n−1 and TOn,i=TOn,d for d=gcd(n−1,i).
We shall subsequently use the symbol TOn for the group TOn,n−1.
The last phrase of Corollary 1.4 is perhaps to be expected, certainly when X=G, as results of Higman ([15]), Pardo ([19]), and Dicks and Martínez-Pérez ([8]) demonstrates that Gn,r≅Gm,s if and only if n=m and gcd(n−1,r)=gcd(n−1,s). In fact it is a question in [3] whether or not On,r≅On,s if and only if gcd(n−1,r)=gcd(n−1,s). We show that this question has a negative answer in both the Gn,r and Tn,r context. That is we prove the following:
Theorem 1.5**.**
There is a number n∈N, n>2, and 1≤r,s≤n−1 such that, for X=TO,O, Xn,r=Xn,s but gcd(n−1,r)=gcd(n−1,s).
Our next result demonstrates that the groups Xn,r for all 1≤r≤n−1 are normal subgroups of Xn.
Theorem 1.6**.**
For all 1≤r≤n−1 we have Xn,r⊴Xn.
We next extend a result of Brin and Guzmán [5] for TOn,n−1, and show that for n≥3, Xn,1 contains an isomorphic copy of R. Thompson’s group F:
Theorem 1.7**.**
Let n≥3, then \mboxOut(Xn,1) contains a subgroup isomorphic to R. Thompson’s group F.
From this and Corollary 1.4, we have the following:
Corollary 1.8**.**
For all 1≤r≤n−1, Xn,r contains a subgroup isomorphic to R. Thompson’s group F.
We further demonstrate (Section 8), in the case r=4, that the set TO4,3\TO4,1 is non-empty. Notice that since 3 is prime, for 1≤r<3, TO4,r=TO4,1, thus this result indicates that, in general, the group TOn,r might depend on r.
We also investigate in Section 8 the nesting properties of the groups Xn,r of Xn for 1≤r≤n−1. We show that these groups from a lattice with the ‘meet’ of Xn,r and Xn,s being the intersection of the two groups and the ‘join’ of Xn,r and Xn,s being the smallest t, 1≤t≤n−1 such that Xn,r and Xn,s are subgroups of Xn,t. We do not know if it is in fact the case that Xn,t=⟨Xn,s,Xn,r⟩ (see Question 8.12). To each element of Xn we associate a numerical invariant which yields a group homomorphism from Xn to the group of units of Zn−1 (notice that for n=2, Zn−1 is equal to its group of units), with kernel Xn,1. That is, we prove the following:
Theorem 1.9**.**
There is homomorphism from Xn to the group of units of Zn−1 with kernel Xn,1.
We should point out that the existence of this homomorphism is already a consequence of the fact that the dimension group (see [16] for the definition of the dimension group) of Xn,r is equal to the additive group Zn−1 and automorphisms of Xn,r yield automorphisms of the dimension group of Xn,r. The author is grateful to Prof. Nekrashevych for drawing these facts to his attention. Our proof of Theorem 1.9 however does not rely on these observations, instead we explicitly construct the homomorphism by making use of the synchronizing condition to associate a numerical invariant to every element of Xn,r. Our approach gives a means of resolving this question as we reduce it to one of constructing transducers which have certain properties. In particular, Theorem 1.9 demonstrates that the homomorphism from Xn,1 into the group of units of Zn−1 is the trivial homomorphism. Though we are unable show in general that the homomorphism of Theorem 1.9 is surjective, we show that under the assumption that it is surjective, then for 1≤s,r,t≤n−1 and t the smallest element of Zn such that Xn,t contains Xn,s and Xn,r, Xn,t=⟨Xn,s,Xn,r⟩. The Theorem 1.10 below, proven in Section 9, indicates that the homomorphism of Theorem 1.9 from On to the group of units of Zn−1 is surjective in many cases, indeed, elementary results in number theory indicate that there are infinitely many numbers n which satisfying the hypothesis of the theorem. We do not know if the restriction to TOn is also surjective.
Theorem 1.10**.**
If the divisiors of n generate the group of units of Zn−1 then the homomorphism of Theorem 7.15 defined on On is unto the group of units of Zn−1.
We conclude in Section 11 with the following result:
Theorem 1.11**.**
The groups Tn,r have the \mboxR∞ property.
We recall that a group G is said to have the R∞ property if for every automorphism φ of G, the equivalence relation defined on G by, for x,y∈G, x is equivalent to y if there is an element h in G such that h−1x(h)φ=y, has infinitely many equivalence classes. The question of which groups have the \mboxR∞ property has received a lot of attention. This class of groups has been shown to include, for instance, all non-elementary Gromov hyperbolic groups ([9], [17]), Baumslag-Solitar groups BS(m,n) for m,n∈Z\{0},(m,n)=(1,1) ([10]), lamplighter groups Zn≀Z where 2∣n or 3∣n ([13]), the first Grigorchuk and the Gupta-Sidki groups ([11]), and R. Thompson’s group F ([2], [7]). It was shown independently by Burillo, Matucci, and Ventura ([7]), and Gonçalves and Sankaran ([12]) that R. Thompson’s group T is in this class of groups. We thus extend this result to the family of groups Tn,r.
Interlaced through out the text are several open questions. In work in preparation we apply the techniques in the current paper and in the paper [3] to the generalizations of Thompson’s group F.
Acknowledgements
The author is grateful to Collin Bleak and Matthew Brin for helpful discussions and their comments on early drafts of this article. This work was partially supported by Leverhulme Trust Research Project Grant RPG-2017-159.
2 Some preliminaries and the groups Gn,r\mboxandTn,r
We begin by setting up some notation.
Let X be a topological space, we shall denote by H(X) the group of homeomorphisms of X, and for G≤H(X), NH(X)(G) shall denote the normaliser of G in H(X). Let Sr:=R/rZ for r∈R, the circle of length r and for n∈N\{0}, let Z[1/n]:={a/n:a∈Z}, the n-adic rationals. Though we will mainly think of Tn,r as a subgroup of Gn,r, we shall at times consider it as a subgroup of H(Sr) and we shall make it apparent at such times that we are doing so. We establish some further notation required to define the groups Gn,r and Tn,r.
Let r˙:={0˙,1˙,…,r−1˙}, and let Xn:={0,1,…,n−1}. We shall take the ordering 0˙<1˙<…<r−1˙ and 0<1<…<n−1 on r˙ and Xn respectively. For i∈{0,n−1}, if i=0 then set iˉ=n−1 otherwise set iˉ=0. Set Xn,r∗:={a˙w:a˙∈r˙\mboxandw∈Xn∗}⊔{ϵ} where ϵ denotes the empty word, and Xn,r+:={a˙w:a˙∈r˙\mboxandw∈Xn∗}. For j∈N, let Xn,rj, respectively Xnj, denote the set of all elements of Xn,r∗, respectively Xn∗, of length j. Given two words u and v in Xn∗ of Xn,r∗, we say that u is a prefix of v and denote this by u≤v. We write u<v if u is a proper prefix of v. For two words u,v in Xn∗ or Xn,r∗ if u≰v and v≰u then we say that u is incomparable to v an write u⊥v. If instead v=uν for some ν∈Xn∗⊔Xn,r∗ then we set v−u:=ν. The relation ≤ is a partial order on the set Xn,r∗ and Xn∗.
For two incomparable words ν,η in Xn+ and Xn,r+ we say ν is less than η in the lexicographic ordering, denoted ν<\mboxlexη, if there are wi,wj∈Xn,r+ or wi,wj∈Xn+ prefixes of ν and η respectively with i<j in the ordering on Xn or r˙. If i≤j then we write ν≤\mboxlexη.
We may now defined a total order of Xn,r∗ and Xn∗ as follows: for ν,η∈Xn,r∗ or ν,η∈Xn∗ we say that ν is less than η in the short-lex ordering if either ∣ν∣≤∣η∣ or if ∣ν∣=∣η∣ then ν≤\mboxlexη, with strict inequalities if ν is strictly less than η. We write ν≤\mboxslexη if ν is less than η in the short-lex ordering, and ν<\mboxslexη is the ν is strictly less than η in the short-lex ordering.
Now let Cn:=Xnω a Cantor space with the usual topology, and let Cn,r:={a˙x∣a˙∈r˙\mboxandx∈Cn}, the disjoint union of r copies of Cn. For a word ν∈Xn,r+, let Uν:={νδ∣δ∈Cn}, and Uϵ:=Cn,r. For ν∈Xn∗, set Uν:={νδ:δ∈Cn}. Let Bn,r:={Uν∣ν∈Xn,r∗} and let Bn:={Uν∣ν∈Xn}, then Bn,r and Bn are a basis for the topology on Cn,r and Cn,r respectively. For a subset Z⊆Xn∗⊔Xn,r∗ we shall denote by U(Z) the set {Uz∣z∈Z}.
The lexicographic extends to a total order on Cn and Cn,r in the natural way, thus we extend the meanings of the symbols ≤\mboxlex and <\mboxlex to this sets also. We further extend this notation to subsets of Cn and Cn,r. Given two subsets V,W⊂Cn or V,W⊂Cn,r we write V≤\mboxlexW if every element of U is less than every element of W in the lexicographic ordering. The strict inequality V<\mboxlexW is analogously defined.
Let ν,η∈Xn,r+, then we shall denote by gν,η the map from Uν to Uη that replaces the prefix ν with η.
A finite set u:={u0,…,ul} for ui∈Xn,r+ and some 1≤l∈N is called a complete antichain, if for any pair ui,uj∈u, ui⊥uj and for any word v∈Xn,r∗, there is some ui∈u such that v≤ui or ui≤v. We shall assume through out that any given antichain u is ordered according to the lexicographic ordering. If we have an antichain uˉ:={u0,…,ul} then we may form another antichain uˉ′={u0,…,ui0,ui1,…uin−1,ui+1,…,ul} where we have replaced ui with ui0,…,uin−1. We call uˉ′ a subdivision of uˉ. Notice that each time we make a subdivision the length of the antichain increases by n−1.
We now define the Higman-Thompson group Gn,r and the subgroup Tn,r.
Given two complete antichains u={u0,…,ul−1} and v={v0,…,vl−1} of equal length, we can define a homeomorphism from of Cn,r as follows. Let π be a bijection from u to v, define a map g:Cn,r→Cn,r by uix↦(ui)πx for ui∈u and x∈Xnω. Since u is a complete antichain this map is well-defined on all of Cn,r. The group Gn,r consists of all homeomorphisms of Cn,r which can be defined in this way. The group Tn,r is the subgroup of Gn,r, consisting of those element g that are derived from complete antichains u and vˉ of the same length l, and a bijection π which maps ui to vi+jmodl for some fixed j∈{0,1…,l−1}.
Define an equivalence relation on Cn,r by x∼ty if and only if there are words u and v in Xn,r+ such that x=uz and y=vz for some z∈Cn. We write [x] to denote the equivalence class of x. Let Hn,r,∼t be the subgroup of H(Cn,r) consisting of those elements of H(Cn,r) that preserve ∼t.
We shall need the following transitivity result for Tn,r
Lemma 2.1**.**
The group Tn,r fixes the equivalence classes of ∼t and acts transitively on each equivalence class.
Proof.
The first statement follows from the definition of Tn,r as a subgroup of Gn,r. The second part of the lemma follows by the following observation: if x=μz and y=νz for some μ,ν∈Xn,r+ and z∈Cn, then there is an element of Tn,r mapping x to y. To see this observe that the lengths of the complete antichains uˉ:=Xn,r∣μ∣ and vˉ:=Xn,r∣ν∣ are congruent modulo n−1. We may assume that the length of uˉ is smaller than the length of vˉ. Therefore there is a complete antichain uˉ′ such that μ is an element of uˉ′, uˉ′ is a repeated subdivision of uˉ and the length of uˉ′ is equal to the length of vˉ. It is now straightforward to see that, since ∣vˉ1′∣=∣vˉ2, there is an element of Tn,r that sends any element μz′ to the element νz′ for z′∈Cn. In particular such a map sends x to y.
∎
In the next Section, we see that elements of Tn,r induce homeomorphisms of the circle R/rZ by thinking of the circle as a quotient of Cantor space.
3 From Cantor Space to the Circle
Definition 3.1**.**
We say that a homeomorphisms h of Cn,r is orientation preserving if whenever x,y∈Cn,r and x<\mboxlexy then (x)h<\mboxlex(y)h. We say that h is orientation reversing if whenever x<\mboxlexy then (y)h<\mboxlex(x)h. We say that h is locally orientation preserving/ reversing if there is a neighbourhood of Cn,r on which h is orientation preserving/reversing.
We identify Sr with the interval [0,r] with the end points identified. Now, every point in [0,r] can be written as a˙x for x∈Cn and a˙∈r˙ in n-ary expansion. However this representation is not unique. In particular two elements x,y∈Cn,r represents the same element of [0,r] if and only if there is some integer i, 0<i≤n−1 or 0˙<i≤r−1˙, and some w∈Xn,r∗ such that x=wi00… and y=w(i−1)n−1n−1… (i.e only elements of (0,r)∩Z[1/n] have non-unique n-ary representations ). Let ≃ be the equivalence relation on Cn,r defined by x≃y if and only if there is some 0<i≤n−1 or 0˙<i≤r−1˙ and some w∈Xn,r+ such that x=wi00… and y=w(i−1)n−1n−1… or x=0˙00… and y=r˙n−1n−1… then Cn,r/≃ is homeomorphic to Sr. Let ≃I be the relation on Cn given by x≃Iy if and only if there is some 0<i≤n−1 or 0˙<i≤r−1˙, and some w∈Xn∗ such that x=wi00… and y=w(i−1)n−1n−1… then Cn/≃I is homeomorphic to the interval [0,r].
Let N be the subgroup of H(Cn,r) consisting of those elements h which preserve ≃ and which satisfy the following: for all points t∈Cn,r there is a neighbourhood of t in Cn,r such that h is orientation preserving (h is orientation reversing), and there is some w∈Xn,r∗, 0<i≤n−1 or 0˙<i≤r−1˙, and points x=wi00… and y=wi−1n−1n−1… so that (y)h=r˙n−1n−1… and (x)h=0˙00… ((y)h=0˙00… and (x)h=r˙n−1n−1…). Observe that all elements of N induce homeomorphisms of Sr, and N contains Tn,r. Our aim shall be to show that NH(Cn,r)(Tn,r)≤N.
We shall need the following results from [5]. The first follows by Rubin’s Theorem and the transitivity of Tn,r on the circle Sr and the second follows by studying the germs of elements of Tn,r at a fixed point.
Theorem 3.2**.**
Let r<n∈N and let n≥2, then \mboxAut(Tn,r)≅NH(Sr)(Tn,r).
Lemma 3.3**.**
If h∈NH(Sr)(Tn,r), then ([0,r]∩Z[1/n])h=[0,r]∩Z[1/n].
Now let g∈NH(Sr)(Tn,r). Define h∈H(Cn,r) as follows. For all t∈Sr such that t∈/[0,r]∩Z[1/n] (notice that this also means (t)g∈/[0,r]∩Z[1/n] by Lemma 3.3), let x be the unique n-ary expansion of t, and y be the unique n-ary expansion of (t)g, set (x)h=y. For t∈[0,r]∩Z[1/n] let x and x′ be the n-ary expansions of t such that x<\mboxlexx′ in the lexicographic ordering of Cn,r (if t=0, then chose x and x′, satisfying x<\mboxlexx′, from the set {0˙00…,r−1˙n−1n−1…}). Also let y and y′ be the n-ary expansions of (t)g in Cn,r such that y<\mboxlexy′ (if (t)g=0 then take y=0˙00… and y′=r˙n−1n−1…). If g is an orientation preserving homeomorphism of Sr, set (x)h=y and (x′)h=y′, otherwise g is orientation reversing and we set (x)h=y′ and (x′)h=y. Thus h is now defines a bijection from Cn,r to itself. Moreover it is easy to see that since g is continuous, h is also continuous on Cn,r. Furthermore if h′ is the homeomorphism obtained from g−1 in the same way, it is easy to see that hh′=h′h=\mboxid∈H(Cn,r). Therefore h∈N. Let ϕ:NSr(Tn,r)→N such that an element g∈NSr(Tn,r) maps to the element h∈H(Cn,r), constructed as above, which induces the map g on Sr. It follows that ϕ is an injective homomorphism.
Let N(Tn,r) denote the image of ϕ. We now show that NH(Cn,r)(Tn,r)=N(Tn,r). Observe that as N(Tn,r)⊆NH(Cn,r)(Tn,r) it suffices to show only that N(Tn,r)⊇NH(Cn,r)(Tn,r).
Let τ,η∈Xn,r+ such that τ⊥η. We further assume that the points x=τ00…, y=ηn−1n−1…, z=τn−1n−1… and t=η00… of Cn,r satisfy x≃y or z≃t.
Let G denote the set of incomparable pairs (τ,η) satisfying the conditions above where τ<\mboxlexη. For a pair (τ,η) in G let uˉ:={u1,…,ul} be any complete finite antichain of Xn,r∗ containing τ and η (since τ⊥η such an antichain exists). Let ul1=τ and ul2=η. Let j=l1−l2−1. Then we say that j is the node distance between τ and η in uˉ i.e j is the number of elements of uˉ which are strictly between ul1 and ul2. Notice that j is necessarily non-zero by assumption. Moreover we have, for any v∈{0}+ and w∈{n−1}+, that the pair (0˙v,r−1˙w) is not in G.
It is straight-forward to see that if (τ,η)∈G have node distance i in some complete antichain uˉ1, then for any other complete antichain uˉ2 containing τ and η the node distance between τ and η in uˉ2 is congruent to i modulo n−1. Moreover, for any χ∈Xn∗, (τχ,ηχ)∈G and there is a complete antichain vˉ containing τχ and ηχ such that the node distance between τχ and ηχ in vˉ is congruent to i modulo n−1. Since modulo n−1 the node distance between a pair (τ,η) in G in a given complete antichain containing τ and η is independent of the complete antichain, we define the reduced node distance between τ and η to be i∈{0,1,…,n−2} such that for any complete antichain uˉ containing τ and η the node distance between τ and η in uˉ is congruent to i modulo n−1.
We require the following transitivity result of Tn,r.
Lemma 3.4**.**
Let (ν1,ν2),(η1,η2)∈G be such that the reduced node distance between ν1 and ν2 is equal to the reduced node distance between η1 and η2. Then there is a g∈Tn,r such that g↾Uν1=gν1,η1 and g↾Uν2=gν2,η2.
Proof.
Let uˉ be a complete antichain containing ν1 and ν2, and vˉ be a complete antichain containing η1 and η2. Let uˉ={u1,…,ul} and vˉ={v1,…,vm}. Let ul1=ν1 and ul2=ν2 and vm1=η1 and vm2=η2 where l1<l2 and m1<m2. By assumption i:=l2−l1−1 and j:=m2−m1−1 are both non-zero and congruent modulo n−1. Without loss of generality we may assume i<j, moreover there is an element ul1+1∈uˉ between ν1 or ν2. By replacing ul1+1∈uˉ with {ul1+10,…,ul1+1n−1} we obtain a new complete antichain uˉ′ containing ν1 and ν2 such that the node distance between ν1 and ν2 is equal to i+n−1. Replace the antichain uˉ with uˉ′
By repeatedly performing this operation we may assume that the complete antichain uˉ containing ν1 and ν2 is such that the node distance between ν1 and ν2 in uˉ is equal to j.
Observe that as uˉ and vˉ are complete antichains the difference ∣m−l∣ is congruent to [math] modulo n−1. Without loss of generality (as we may relabel to achieve this) we assume that l<m.
Now observe that u1 must be equal to 0˙w1 for some w1∈{0}∗ and ul=r−1˙w2 for some w2∈{n−1}∗. Moreover, since the pair (0˙v,r−1˙w) for v∈{0}+ and w∈{n−1}+ is not in G, it follows that either νl1=ul=0˙w1 or νl2=r−1˙w2. Suppose that νl1=ul=1˙w1 (the other case is handled similarly). Then by repeatedly expanding along the node 1˙w1, we obtain a complete antichain uˉ′′ of length m containing ν1 and ν2 such that the node distance between ν1 and ν2 is j.
Since the complete antichain uˉ′′ and vˉ have the same lengths and the node distance between ν1 and ν2 in uˉ′′ is equal to the node distance between η1 and η2 in vˉ, it is easy to construct an element g of Tn,r such that g↾Uν1=gν1,η1 and g↾Uν2=gν2,η2.
∎
Using this transitivity result we show below that N(Tn,r) is equal to NH(Cn,r)(Tn,r).
Lemma 3.5**.**
Let h∈H(Cn,r) be such that h∈NH(Cn,r)(Tn,r). Then h (and so h−1) preserves the equivalence relation ≃.
Proof.
Suppose there are x,y∈Cn,r such that x≃y but (x)h≃(y)h. By relabelling if necessary we may assume that (x)h<(y)h. Since (x)h≃(y)h there are (τ,η)∈G such that (x)h∈Uτ and (y)h∈Uη. Let j∈{0,1,…,n−2} be the reduced node distance between τ and η
Let (μ,ν)∈G and consider the clopen sets (Uμ)h and (Uν)h. It is straight-forward to see that there are μ′ and ν′ such that (μ′,ν′)∈G, the reduced node distance between μ′ and ν′ is equal to j, and Uμ′⊂(Uμ)h and Uν′⊂(Uν)h (the case where Uμ′⊂(Uν)h and Uν′⊂(Uμ)h is analogous). By Lemma 3.4 there is an element g∈Tn,r such that g↾Uτ=gτ,μ′ and g↾Uη=gη,ν′.
Now consider the product hgh−1. Notice that (x)hgh−1 is contained in the clopen set Uμ since (x)h∈Uτ and (Uτ)g∈Uμ′⊂(Uμ)h, likewise (y)hgh−1∈Uν. Therefore (x)hgh−1≃(y)hgh−1, which is a contradiction since Tn,r preserves ≃.
∎
Lemma 3.6**.**
Let h∈H(Cn,r) be such that h∈NH(Cn,r)(Tn,r), then h∈N(Tn,r).
Proof.
Since h∈NH(Cn,r)(Tn,r), then by the previous lemma h preserves ≃. Therefore h induces a continuous function g on Sr. However since h−1 is also in N(H(Cn,r)), g must be a homeomorphism and (g)ϕ=h, therefore h∈N(Tn,r).
∎
Thus we have now proved that NH(Cn,r)(Tn,r)=N(Tn,r)≅\mboxAut(Tn,r).
4 Automorphisms of Tn,r
In what follows we adapt the results of
[3] to show that elements of
NH(Cn,r)(Tn,r) can be represented by bi-synchronizing
transducers. However we shall need to define the group Rn,r introduced in [3] which is a slight modification of the Rational group Rn in [14]. The exposition in this section will largely mirror that found in [3].
First we introduce transducers generally then we introduce transducers over Cn,r.
Let XI and XO be finite sets of symbol. A transducer over the alphabet XI is a tuple T=⟨XI,XO,QTπT,λT⟩ such that:
(i)
XI is the input alphabet and XO is the output alphabet.
2. (ii)
QT is the set of states of T.
3. (iii)
πT:XI×QT→QT is the transition function and,
4. (iv)
λT:XI×QT→XI∗ is the output function.
If ∣QT∣<∞ then we say that T is a finite transducer. If XI=XO=X then we shall write T=⟨X,QT,πT,λT⟩. If we fix a state q∈QT from which we begin processing inputs then we say that T is initialised at q and we denote this by Tq and we call Tq an initial transducer. Given an initial transducer Tq0 we shall write T for the underlying transducer with no initialised states.
We inductively extend the domain of the transition and rewrite function to XI∗×QT by the following rules:
for a word w∈XI∗, i∈XI and any state q∈QT we have πT(wi,q)=πT(i,πT(w,q)) and λT(wi,q)=λT(w,q)λT(i,πT(w,q)). We then extend the domain of πT and λT to XIω×QT.
In this paper we shall insist that for δ∈XIω and any state q∈QT we have λT(δ,q)∈XOω. This means that for a state q∈QT the initial transducer Tq induces a continuous function hTq (hq if it is clear from the context that q is a state of T) from XIω to XOω. For q∈QT we denote by \mboxim(q) the image of the map hq; if hq is a homeomorphism form XIω→XOω, then we call q a homeomorphism state.
Two states q1 and q2 of T are called ω-equivalent if hq1=hq2. A state q of T is called a state of incomplete response if for some i∈XIλT(i,q) is not equal to the greatest common prefix of the set {(iδ)hq∣δ∈XIω}. If T is an initial transducer with initial state q0, then q is called accessible if there is a word w∈XI∗ such that πT(w,q0)=q. If all the states of Tq0 are accessible then Tq0 is called accessible.
An initial transducer Tq0 is called minimal if Tq0 is accessible, has no states of incomplete response and no pair of ω-equivalent states. The initial transducer Tq0 is also called invertible if the state q0 is a homeomorphism state. Given an initial transducer Tq0 there is a unique minimal transducer Sp0ω-equivalent to Tq0 ([14]).
We give below the method given in [14] for constructing the inverse of an invertible, minimal transducer Tq0. We first define the following function.
Definition 4.1**.**
Let Tq0=⟨XI,XO,QT,πT,λT⟩ be an invertible minimal transducer and q a state of Tq0. Define a function Lq:XO+→XI∗ by (ν)Lq=φ where φ is the greatest common prefix of the set (Uν)hq−1.
Observe that since Tq0 is minimal and invertible, then each state q of Tq0 induces an injective function from XIω to XOω with clopen image. From this one can deduce that for each state q of Tq0 the set of words w∈XO∗ such that (w)Lq=ϵ and Uw⊂\mboxim(q) is finite (see [14]).
We now form a transducer T(ϵ,q0)=⟨XO,XI,QT′,πT′,λT′⟩ where QT′={(w,q)∣q∈QT,(w)Lq=ϵ,Uw⊂\mboxim(q)} and πT′ and λT′ are defined, for all i∈XO and (w,q)∈QT′, by the rules:
(i)
πT′(i,(w,q))=(wi−λT((wi)Lq,q),πT((wi)Lq,q) and,
2. (ii)
[14]**
For Tq0 a minimal invertible transducer, the transducer T(ϵ,q0) is well-defined, has no states of incomplete response, and satisfies h(ϵ,q0)=hq0−1.
We observe that given a minimal invertible transducer Tq0, the transducer T(ϵ,q0) is accessible if Tq0. This is because if for any word Γ∈Xn+, (Γ)Lq0=ϵ, then πT′(Γ,(ϵ,q0))=(Γ,q0). Furthermore, for any word w∈Xn+ and state q∈QT, such that Uw⊂\mboxim(q) and (w)Lq=ϵ, picking a word Γ∈Xn+ such that πT(Γ,q0)=q, we observe that (λT(Γ,q0)w)Lq0=Γ and so πA′(λT(Γ,q0)w,(ϵ,q0))=(w,q). Note that given a minimal invertible transducer Tq0, the transducer T(ϵ,q0), even though it has no states of incomplete response and is accessible, might not be minimal.
Given two transducers A=⟨X,QA,πA,λA⟩ and B=⟨X,QB,πB,λB⟩ then the product A∗B=⟨X,QA∗B,πA∗B,λA∗B⟩ is the transducer defined as follows. The set of states QA∗B of A∗B is equal to the cartesian product QA×QB. The transition and output function of A∗B are given by the following rules: for states q∈QA , p∈QB and i∈X we have πA∗B(i,(q,p))=(πA(i,q),πB(λA(i,q),p)) and λA∗B(i,(q,p))=λB(λA(i,q),p). For two initial transducers Aq0 and Bp0, where A and B are the resulting transducers with no initialised states, the product of the initial transducer Aq0∗Bq0, is the initial transducer (A∗B)(q0,p0). It is straightforward to see that h(A∗B)(q0,p0)=hAq0∘hBp0.
A transducer (initial or non-initial) T=⟨XI,XO,QT,πT,λT⟩ is said to be synchronizing at level k if there is a natural number k∈N and a map s:XIk→QT such that for a word Γ∈XIk and for any state q∈QT we have πT(Γ,q)=(Γ)s. We will denote by Core(T) the sub-transducer of T induced by the states in the image of s. We call this sub-transducer the core of T. If T is equal to its core then we say that Tcore. Viewed as a graph Core(T) is a strongly connected transducer. If T is an initial transducer Tq0 which is invertible, then we say that Tq0 is bi-synchronizing if both Tq0 and its inverse are synchronizing. Note that when T is synchronous, then we shall say T is bi-synchronizing if T and its inverse are synchronizing.
Given a transducer T and states q1,q2 of T we shall sometimes use the phrase q1 and q2transition identically on a subset W⊂Xn∗ to mean that the functions πT(⋅,q1):W→QT and πT(⋅,q2):W→QT are identical. We might also say that q1 and q2read all elements of W to the same location.
Now we introduce transducers over Cn,r. An initial transducer for Cn,r is a tuple Aq0=(r˙,Xn,R,S,π,λ,q0) such that:
(i)
R is a finite set, and the set Q of states of A is the disjoint union R⊔S. The state q0∈R is the initial state
2. (ii)
π:r˙×{q0}⊔Xn×Q→Q\{q0} and λ:r˙×q0⊔Xn×Q→Xn,r∗⊔Xn∗
Notice that a letter from r˙ can only be read from the initial state q0 and we can never return to q0 after leaving q0. We Inductively extend the domain of π and λ to r˙×{q0}⊔Xn∗×Q by the following rules:
π(wx,q)=π(x,π(w,q)) and λ(wx,q)=λ(w,q)λ(x,π(w,q)). Where w∈Xn,r+⊔Xn+ and x∈Xn (if w∈Xn,r+ then q=q0). We also take the convention that π(ϵ,q)=q and λ(ϵ,q)=ϵ for any state q of A. By transfinite induction we may further extend the domains of π and λ to r˙×{q0}⊔Xnω×Q.
We impose the following rules on π and λ:
(1)
For a state r∈R and for i in r˙⊔Xn such that π(i,r) is defined, if π(i,r)∈R then λ(i,r)=ϵ, otherwise λ(i,r)∈Xn,r∗.
2. (2)
For x∈Xn and q∈S, λ(x,q)∈Xn∗ and π(x,q)∈S.
3. (3)
For a state s∈S and δ∈Cn we have that λ(δ,s)∈Cn.
4. (4)
If there is a word w∈Xn+ and a state q∈Q such that π(w,q)=q then q∈S.
These rules serve the purpose of ensuring that whenever an element of Cn,r is is processed through Aq0 the output is also in Cn,r.
Let Aq0 be an initial transducer on Cn,r as above and let q be a state of Aq0. Let Aq denote the initial transducer Aq0 where we process inputs from the state q. Observe that Aq0 induces a continuous function hAq0 (or hq0 if it clear that q0 is the initial state of A) from Cn,r to itself. Furthermore every non-initial state q of Aq0 which is also an element of R induces a continuous function hq from Cn to Cn,r, otherwise it induces a continuous function from Cn to itself. Once again we denote by \mboxim(q) the image of q and call q a homeomorphism state if hq is a homeomorphism from its domain to its range.
We extend, in the natural way, the definition of accessibility, accessible transducers,and states of incomplete response given in the general setting to the specific setting of transducers over Cn,r. We also extend the function Lq for a minimal invertible transducer Tq0 over Cn,r and q∈QT. Having done this, we may thus define, given Tq0 a minimal, invertible transducer, the transducer T(ϵ,q0) such that h(ϵ,q0)=hq0−1.
We say that a transducer Aq0 over Cn,r is synchronizing if there is a k∈N such that given any word Γ of length k in Xn,r∗⊔Xn∗ the active state of Aq0 when Γ is processed from any appropriate state of Aq0 is completely determined by Γ. Thus we may also extend the notions of ‘core’ for synchronizing transducers over Cn. We now introduce the notion of ω-equivalence and minimality for transducers over Cn,r. Two initial automata with the same domain and range are said to be ω-equivalent if they induce the same continuous function from their domain to their range.
An initial transducer Aq0 is called minimal if Aq0 is accessible, no states of A are states of incomplete response and for any distinct pair q1,q2 of states of Aq0, Aq1 and Aq2 are not ω-equivalent. In [3] the authors show, by slight modifications of arguments in [14], that for an initial transducer Aq0 on Cn,r there is a unique minimal transducer under ω-equivalence.
The product (A∗B)(q0,p0)=⟨r˙,Xn,RA×RB⊔SA×RB,SA×SB,πA∗B,λA∗B⟩ of the initial transducers Aq0 and Bp0 over Cn,r is defined as follows. The set of states QA∗B=RA×RB⊔SA×RB⊔SA×SB, and the state (q0,p0)∈RA×RB is the initial state. The transition and output functions are defined as follows. First for a∈r˙ we have πA∗B(a,(q0,p0))=(πA(a,q0),πB(λA(a,q0),p0), and λA∗B(a,(p0,q0))=λB(λA(a,q0),p0). Now for any pair (q,p)∈QA,B and for any i∈Xn we have πA∗B(i,(q,p))=(πA(i,q),πB(λA(1,q),p) and λA∗B(i,(q,p))=λB(λA(i,q),p). Now observe that as A and B satisfy condition (1) to (4) above then so does the product (A∗B)(q0,p0). Furthermore, as before, it is a straightforward observation that h(A∗B)(q0,p0)=hAq0∘hBp0.
Below we outline a procedure given in [3] for constructing from a homeomorphism h of Cn,r, an initial transducer Aq0 of Cn,r such that hq0=h.
We first need to define local actions.
For an arbitrary homeomorphism g:Cn,r→Cn,r, fix the notation Pg⊂Xn,r∗ for the unique maximal subset of Xn,r∗ set satisfying: (Uν)g⊆Ua˙ for ν∈Ph(a), and for any proper prefix μ of ν there are elements δ1,δ2∈Cn and a˙1,a˙2∈r˙ such that (μδ1)g∈Ua˙1 and (μδ2)g∈Ua˙2. Observe that since g is a homeomorphism and since ⊔a˙∈r˙(Ua˙)g−1 is clopen, Pg exists. Moreover maximality of Pg implies that Pg is a complete antichain for Xn,r∗.
Let h:Cn,r→Cn,r be a homeomorphism. Define θh:Xn,r∗→Xn,r∗ as follows: for μ∈Xn,r∗ if μ is a prefix of some ν∈Ph set (μ)θh:=ϵ, otherwise μ=νχ for some ν∈Ph and χ∈Xn,r∗, in this case set (μ)θh to be the greatest common prefix of the set (Uμ)h, since h is a homeomorphism and by choice of Ph, (Uμ)h∈Xn,r+.
Now for μ∈Xn,r∗ we define a map hμ on Cn by (δ)hμ=(μδ)hμ−(μ)θh. We call hμ the local action of h at μ. Observe that if μ is a prefix of an element of Ph then the range of hμ is Cn,r otherwise the range of μ is Cn, in either case hμ is continuous. The following fact is straightforward, let μ,ν∈Xn,r∗ then (μν)θh=(μ)θh(ν)θhμ.
Using the function θh we now construct an initial transducer Aq0 of Cn,r such that hq0=h.
Let h:Cn,r→Cn,r be a homeomorphism. Form a transducer
Aϵ=(r˙,Xn,RA,SA,πA,λA,ϵ). The set QA of states of A is precisely
Xn,r∗ and RA⊂QA is the set of proper
prefixes of elements of Pg and SA:=Xn∗\R. The
transition of function and output functions πA and
λA are defined as follows. For a˙∈r˙ we
have that πA(a˙,ϵ)=a˙ and λA(a˙,ϵ)=(a˙)θh; for ν∈Xn,r+
and i∈Xn we have πA(i,ν)=νi and
λA(i,ν)=(νi)θh−(ν)θh.
Observe that the transducer Aϵ satisfies the conditions (1) to (4). The following claim is straightforward to prove:
Claim 4.3**.**
Let h:Cn,r→Cn,r be a homeomorphism, and Aϵ=(r˙,Xn,RA,SA,πA,λA,ϵ) be the transducer constructed form h as above. Let ν∈Xn,r∗ then for all w∈Xn∗ if ν=ϵ or w∈Xn,r+ if ν=ϵ we have that (w)θhν=λA(w,ν).
Proof.
First suppose ν=ϵ and let x∈Xn. Then by
definition we have λA(i,ν)=(νi)θh−(ν)θh, however by an observation above we have (νi)θh−(ν)θh)=(i)θhν. Now
assume by that for all w∈Xn+ we have that λ(w,ν)=(w)θhν. Let i∈Xn and consider
λA(wi,ν). We may write λA(wi,ν)=λA(w,ν)λA(i,νw) (since πA(w,ν)=νw). Therefore λA(wi,ν)=(w)θhν(νwi)θh−(νw)θh. Observe that (νwi)θh−(νw)θh=(i)θhνw,
therefore λA(wi,ν)=(w)θhν(i)θhνw however, (w)θhν(i)θhνw=(wi)θhν.
Now suppose that ν=ϵ. Let w∈Xn,r+ and suppose w=a˙v for some v∈Xn∗ and a˙∈r˙. Consider λA(w,ϵ), this can be broken up into λA(a˙w,ϵ)=λA(a˙,ϵ)λA(v,a˙). Now λA(a˙,ϵ)=(a˙)θh by definition, and λA(v,a˙)=(v)θha˙ by the previous paragraph. Observe that (a˙v)θh=(a˙)θh(v)θha˙ therefore λA(a˙w,ϵ)=(a˙v)θh as required.
∎
Remark 4.4**.**
From the claim we deduce that for a homeomorphism h:Cn,r→Cn,r and for Aϵ=(r˙,Xn,RA,SA,πA,λA,ϵ) the transducer constructed from h, we have that hAϵ=h, moreover for any ν∈Xn,r∗ and any local action hν of h we have that hν=hAν. Therefore if h has finitely many local actions, then it follows that the minimal transducer on Cn,r, under ω-equivalence, representing h has finitely many states. Moreover, since by Claim 4.3, Aϵ has no states of incomplete response, it follows that, if Bq0 is the minimal transducer representing Aϵ, then for all ν∈Xn,rhν=hq for some state q of B.
In what follows we show that all homeomorphisms of Cn,r in the group N(Tn,r) can be represented by a minimal finite transducer. The previous paragraph demonstrates that it suffices to show that such homeomorphisms have finitely many local actions.
Our approach shall essentially mirror that taken in [3], the arguments differ only in so far as we need to make modifications to allow for the fact that the action of Tn,r on Cn,r is not as transitive as the action Gn,r on Cn,r.
First we need the following result:
Lemma 4.5**.**
Let X be a topological space and let ∼ be any equivalence relation on X. Let H∼ be the subgroup of H(X) consisting of those elements of H(X) which respect ∼. Let G≤H(X) be a subgroup which fixes each equivalence class of ∼ and acts transitively on each class. Then NH(X)(G)≤H∼.
Observe that Tn,r≤H(Cn,r), by Lemma 2.1, preserves ∼t and acts transitively on each equivalence class of ∼t. Therefore, as corollary of the lemma above, we have N(Tn,r)≤H∼t.
The following definitions and lemmas appear in [3] and introduce crucial notions and ideas in understanding local actions of homeomorphisms of Cn,r.
Definition 4.6**.**
Let V⊆Cn,r be a clopen set. Let B⊂Xn,r∗ be the minimal antichain such that UB:={Uν∣ν∈B} and for all μ a prefix of some element of B we have Uμ⊂V. We call UB the decomposition of V and denote it by Dec(V).
Definition 4.7**.**
Let h∈H(Cn,r), and Uν and Uη be elements of Bn,r for ν,η∈Xn,r∗. Then we say that h acts on Uν and Uη in the same fashion if hν=hη. If V,W⊂Cn,r are a clopen subsets then we say that h acts on V and Win the same fashion if for any Uν∈Dec(V) and Uη∈Dec(W), h acts on Uν and Uη in the same fashion.
Definition 4.8**.**
Let h∈H(Cn,r) and let V,W⊂Cn,r be clopen subsets. Then we say that h acts on V and Walmost in the same fashion if there is some k∈N such that for Uν∈Dec(V) and Uη∈Dec(W), and for any χ∈Xnkh acts on Uνχ and Uηχ in the same fashion. We call the minimal k satisfying this condition the critical level of V and W and denote it by \mboxcrith(V,W).
Definition 4.9**.**
Let h∈H(Cn,r) and let V,W⊂Cn,r be clopen subsets. We say that h acts on V and Win the same fashion uniformly if for any χ∈Xn∗h and Uν∈Dec(V) and Uη∈Dec(W), h acts on Uν and Uηχ in the same fashion. We say that h acts almost in the same fashion uniformly on U and W if there is some k∈N such that for Uν∈Dec(V) and Uη∈Dec(W) and for any χ,ξ∈Xnk, h acts on Uνχ and Uηξ in the same fashion.
Remark 4.10**.**
Let V and W be clopen subsets of cantor space and g∈Tn,r then g acts on V and W almost in the same fashion uniformly.
The following lemmas are crucial in our understanding of local actions of elements of N(Tn,r) and are taken from [3]:
Lemma 4.11**.**
Let g,h∈H(Cn,r) and let Uν,Uη∈Bn,r. Suppose there are ν′,η′∈Xn,r+ such that (Uν)g=Uν′ and (Uη)g=Uη′, g acts in the same fashion on Uν and Uη and h acts in the same fashion on Uν′ and Uη′, then gh acts in the same fashion on Uν and Uη.
Lemma 4.12**.**
Let g,h∈H(Cn,r) and let Uν,Uη∈Bn,r. Suppose g acts on Uν and Uη in the same fashion and h acts on (Uν)g and (Uη)g in almost the same fashion uniformly, then gh acts in almost the same fashion on Uν and Uη.
We shall also need the following proposition from the same source:
Propostion 4.13**.**
Let h∈H(Cn,r) and let Uτ and Uη be elements of Bn,r. Suppose that hτχ=hηχ holds for every χ∈Xn∗. Then h∈/Hn,r,∼t.
We have the following corollary:
Corollary 4.14**.**
Let h∈Hn,r∼t and let i∈{0,1,…,n−2}. Then there exists τ,η∈Xn,r such that Uτ∪Uη=Cn,r, (τ,η)∈G the reduced node distance between τ and η is i modulo n−1, and hτ=hη.
Proof.
This is a direct consequence of Proposition 4.13, since we may chose a pair (τ,η)∈G such that there is a complete antichain uˉ containing τ and η and the node distance in uˉ between τ and η is congruent to i modulo n−1. Moreover, by observations above, for any χ∈Xn∗, (τχ,ηχ)∈G and any complete antichain vˉ containing τχ and ηχ, the node distance between τχ and ηχ in vˉ is congruent to i modulo n−1.
∎
We have the following lemma:
Lemma 4.15**.**
Let h∈H(Cn,r) such that h−1Tn,rh⊆Tn,r. Then for every Uν,Uη∈Bn,r such that Uν∪Uη=Cn,r, the map h acts on Uν and Uη almost in the same fashion.
Proof.
First observe that, by Lemma 4.5,
h∈H∼t. Moreover by Corollary 4.14
for any i∈{0,1,…,n−2} there are elements ν and
η of Xn,r+ such that (ν,η)∈G,
the reduced node distance between ν and η is i,
Uν∪Uη=Cn,r, and hν=hη. Fix
i∈{0,1,…,n−2} and let (ν,η)∈G
be such that the reduced node distance between ν and η
is i and hν=hη.
Let ν′ and η′ be elements of Xn,r+ be such that (ν′,η′)∈G and the reduced node distance between ν′ and η′ is also equal to i. By Lemma 3.4 there is an element g∈Tn,r such that g↾Uν′=gν′,ν and g↾Uη′=gη′,η.
By Lemma 4.11, since h acts on Uν and Uη in the same fashion, and g acts on Uν′ and Uη′ in the same fashion, then gh acts on Uν and Uη in the same fashion. Let f=h−1gh. By assumption f∈Tn,r and so f and f−1 act on any pair of clopen sets of Cn,r in almost the same fashion. Therefore, since h=ghf−1, by Lemma 4.12 it follows that h acts on Uν and Uη in almost the same fashion.
Since i was arbitrarily chosen, we conclude that for any pair (τ,μ)∈G with reduced node distance equal to i, h acts on Uτ and Uη in almost the same fashion.
Now let ν′ and η′ be arbitrary elements of Xn,r∗ such that Uν′∪Uη′=Cn,r . Observe that ν′,η′∈Xn,r+ otherwise Uν=Cn,r or Uη=Cn,r. Since Uν′∪Uη′=Cn,r there exists an element β∈Xn,r+ such that β⊥ν′ and β⊥η′ and such that (ν′,β)∈G and (η′,β)∈G. By arguments in the previous paragraph we therefore have that h acts on Uν′ and Uβ in almost the same fashion, and h acts on Uη′ and Uβ in almost the same fashion. Let k1=\mboxcrith(Uη′,Uβ) and k2=\mboxcrith(Uν′,Uβ). Let k=max{k1,k2}. Then for any χ∈Xnk we have that hν′χ=hβχ=hη′χ. Therefore we conclude that h acts on Uη′ and Uν′ in almost the same fashion.
∎
The Corollary below demonstrates that if h∈H(Cn,r) is such that h−1Tn,rh⊆Tn,r, then h uses only finitely many types of local action. This Corollary should be compared with Corollary 6.16 of [3] and is proved almost identically; we reproduce the proof here for completeness. This is a key result in demonstrating that automorphisms of Tn,r can be represented by bi-synchronizing transducers.
Corollary 4.16**.**
Let h∈H(Cn,r) be such that h−1Tn,rh⊆Tn,r, then h uses only finitely many types of local action.
Proof.
Let A be a complete antichain for Xn,r∗ having at least 3 elements. For instance we may take A=Xn,r3. Notice that for any pair of element ν and η in A we have by Lemma 4.15 that h acts on Uν and Uη almost in the same fashion, and for i∈Xnh acts on Uν and Uνi almost in the same fashion. Set
[TABLE]
We now demonstrate that for any m≥k+3, m∈N and for any word η∈Xn,rm we have that hη=hνξ for some ν∈A and ξ∈Xnk. We proceed by induction on m.
The base case is trivially satisfied. Let m∈N be strictly greater than k+3, and assume that for all k+3<l<m we have that for any word η∈Xn,rlHη=hνξ for some ν∈A and ξ∈Xnk.
Now let τ∈Xn,rm. We may write τ=νiξ for ν∈A, some i∈Xn and ξ∈Xn∗ such that ∣ξ∣≥k (since m is greater than k+3).
Observe that as hν=hνi we have that hτ=hνξ. Now as ∣νξ∣<m, by the inductive assumption we have that there is some χ∈Xnk and μ∈A such that hνξ=hμχ and the conclusion follows.
∎
Remark 4.17**.**
The above corollary together with Remark 4.4 means that N(Tn,r) is a subgroup of the group Rn,r of those homeomorphisms of Cn,r that can be represented by minimal, finite, initial transducers. In what follows we shall demonstrate the element of N(Tn,r) can be represented by minimal, bi-synchronizing transducers in Rn,r.
5 The group N(Tn,r) is a subgroup of Bn,r
Following [3] we shall use the notation Bn,r for the group consisting of those elements of Rn,r which can be represented by bi-synchronizing transducers. In this section we demonstrate that N(Tn,r) is a subgroup TBn,r of Bn,r.
The lemma below is straight-forward and connects the property of a homeomorphism acting in the almost the same fashion (Definition 4.8), to the synchronizing property of a transducer representing the homeomorphism.
Lemma 5.1**.**
Let h∈H(Cn,r) satisfy the following two conditions,
(a)
h* has finitely many local actions,*
2. (b)
for every pair Uν,Uη∈Bn,r such that Uν∪Uη=Cn,rh acts on Uν and Uη almost in the same fashion.
Then the minimal transducer representing h is synchronizing.
Proof.
Let Aq0 be the minimal transducer representing h. As h has finitely many local actions, by Remark 4.4, Aq0 is a finite transducer. Applying Remark 4.4, there is an m∈N such that if q∈QA is accessible from q0 by a word of length at least m, then hq=hη for some η∈Xn,r∗ of length at least m. Of course by the definition of Aq0, it is the case that hν=hp, for p=πA(ν,q0) and ν∈Xn,r∗. We may also choose m bigger than 1 and long enough, so that for any pair ν,η∈Xn,r∗ of length m, Uν∪Uη=Cn,r. Fix such an m∈N.
By assumption (b) above and Definition 4.8, it follows that for every pair ν,η∈Xn,r∗ of length m, there is a minimal number kν,η∈N such that for all words ξ∈Xnkν,η, hνξ=hηξ. Set k:=maxν,η∈Xnm{kν,η}. It therefore follows that for any pair τ,μ∈Xn,r∗ of length m, and any word χ∈Xnk, hτχ=hμχ.
Now since any state q of Aq0 accessible from q0 by a word of length at least m satisfies hq=hη for some η∈Xn,r∗ of length m. It is therefore the case, by minimality of Aq0, that for any χ∈Xnk and any pair ν,η∈Xn,r∗ of length m, we have πA(νχ,q0)=πAηχ,q0. Thus we have that if QA,≥m={πA(ν,q0)∣ν∈Xn,r∗,∣ν∣≥m}, then for any ξ∈Xnk, ∣{πA(χ,p)∣p∈QA,≥m}∣=1. From this we deduce that Aq0 is synchronizing at level at least k+m.
∎
As a corollary we have the following:
Corollary 5.2**.**
Let h∈H(Cn,r) be such that h−1Tn,rh⊆Tn,r and Aq0=⟨r˙,Xn,QA,πA,λA,q0⟩ be an initial transducer such that hq0=h. Then Aq0 is synchronizing.
Proof.
By Corollary 4.16 and Lemma 4.15 it follows that h and h−1 satisfy the conditions of Lemma 5.1. By definition of bi-synchronicity, the minimal transducer Aq0 representing h is bi-synchronizing.
∎
Now we may prove the main result.
Theorem 5.3**.**
Let h∈H(Cn,r) then h∈N(Tn,r) if and only if there is a bi-synchronizing transducer Aq0 such that hq0=h and preserves ≃.
Proof.
It follows by Corollary 5.2 that if Aq0 is a transducer such that hq0∈N(Tn,r) then Aq0 is synchronizing. Since N(Tn,r) is a group it follows that Aq0 is bi-synchronizing. This proves the forward implication.
For the reverse implication, suppose that Aq0=⟨r˙,Xn,QA,πA,λA,q0⟩ is a bi-synchronising transducer such that hq0 preserves ≃. Let T∈Tn,r observe that since hq0 preserves ≃ it follows that hq0−1Thq0 also preserves ≃. Let Cp0=⟨r˙,Xn,QC,πC,λC,p0⟩ be an initial transducer such that hp0=T. Observe that Cp0 is synchronizing and Core(Cp0) is the single state identity transducer, let us denote this state by ι. Let Bq0−1=⟨r˙,Xn,QB,πB,λB,q0−1⟩ be a minimal initial transducer representing the inverse of Aq0.
Observe that since Aq0 is bi-synchronizing, that Bq0−1 is also bi-synchronizing. Let k1∈N be minimal so that Aq0 and Bq0−1 are synchronizing at level k1. let k2∈N be minimal so that for any word Γ∈Xn,rk2 we have πC(Γ,p0)=ι. Let k=max{k1,k2}. Let j∈N be the minimal number greater than k such that for any word Δ∈Xn,rj we have that ∣λB(Δ,q0−1)∣>2k. Therefore after processing any word of length j, the active state of the product transducer Bq0−1∗Cp0Aq0 is of the for (r,ι,q) for some state r of Core(Bq0−1) and some state q of Aq0. Notice that as ι is the single state identity transducer we may identify (r,ι,q) with the state (r,q) of Bq0−1∗Aq0.
Now let Λ∈Xn,r∗ be a word such that πB(Λ,q0−1)=r and let Δ∈Xn be such that λB(Δ,r) is greater than k (recall that states of Core(Bq0−1 only process elements from Xn). Let q′ be the state of Aq0 forced by λB(Δ,r) and let r′ be the state πB(Δ,r). Then observe that after processing the word ΛΔ the active state of Bq0−1∗Aq0 is (r′,q′). However after processing Δ from the state (r,q) the active state is also (r′,q′). However since Bq0−1∗Aq0 is ω-equivalent to the single state identity transducer, we see that (r′,q′) which is a state of Core(Bq0−1∗Aq0) is ω-equivalent to a map which produces some finite initial prefix and then acts as the identity.
From this we may conclude that there some N∈N such that Bq0−1∗Cp0∗Aq0 acts as the identity after processing a word of length N. This means that since hq0−1Thq0 preserves ≃ and is a homeomorphism that hq0−1Thq0∈Tn,r.
∎
Thus we have the following:
Theorem 5.4**.**
The group \mboxAut(Tn,r) is isomorphic to the subgroup TBn,r≤Bn,r consisting of those elements of Rn,r which may be represented by finite, initial, bi-synchronizing transducers which preserve ≃.
Remark 5.5**.**
Notice that since all elements of Tn,r eventually act as the identity, then a minimal transducer Tq0 representing an element t∈Tn,r satisfies Core(Tq0) is the single states identity transducer over Cn which we denote id. Furthermore any element of TBn,r which may be represented by a bi-synchronizing transducer with trivial core, is an element of Tn,r. This is because Tn,r≤TBn,r is uniquely defined by the fact that all elements act as the identity after modifying an finite initial prefix.
6 \mboxOut(Tn,r)
The paper [3] introduces a group On which contains the group On,r≅\mboxOut(Gn,r) for all valid r. In this section we shall demonstrate that there is a corresponding subgroup TOn≤On containing the groups TOn,r:=\mboxOut(Tn,r) for all valid r.
Let Bn,r be the set of minimal transducer Aq0 such that hq0∈Bn,r. The set Bn,r of transducers becomes a group isomorphic to Bn,r with the product defined by for elements Aq0,Bp0∈Bn,r, AB(q0,p0) is the minimal initial transducer representing the product Aq0∗Bp0. Notice that since h(q0,p0)=hq0∘hp0 then we have that AB(q0,p0)∈Bn,r. To simplify the discussion below we shall identify Bn,r with the group Bn,r of transducers. In particular we shall no longer distinguish between an element h∈Bn,r and the minimal transducer Aq0∈Bn,r representing h, thus we shall use the symbol Bn,r for both Bn,r and Bn,r. Likewise we shall longer distinguish between elements of TBn,r and the minimal initial transducers representing them.
Lemma 6.1**.**
Let Aq0,Bp0∈TBn,r such that Core(Aq0)=Core(Bp0). Then Aq0Bp0−1∈Tn,r.
Proof.
By Remark 5.5 it suffices to show that Core(Aq0Bp0−1) is equal to the single state identity transducer id.
However it is a consequence of a result in [3], that Core(Aq0Bp0−1) is trivial.
∎
As a corollary we have that an element of \mboxOut(Tn,r) corresponds to a subset of elements of TBn,r which all have the same core. Therefore we may identify an element [Aq0]∈\mboxOut(Tn,r), for Aq0∈TBn,r, with the element Core(Aq0).
Let On,r:={Core(Aq0)∣Aq0∈Bn,r}. This set inherits a product from Bn,r as follows. For g1,g2∈On,r, let Aq0,Bp0∈Bn,r be elements such that Core(Aq0)=g1 and Core(Bq0)=g2, and set g1g2:=Core(AB(q0,p0)). Now observe that Core(AB(q0,p0)) depends only on the states of Core(g1∗g2) by a result in [3] which states that Core(Aq0∗Bq0)=Core(Core(Aq0)∗Core(Bq0)) and since removing states of incomplete response and identifying ω-equivalent depends only on Core(g1∗g2) (see [14]). Therefore it follows that for any other element Aq0′′ and Bp0′′ with cores g1 and g2 respectively we have Core(A′B(q0′p0′)′)=Core(AB(q0,p0)). Thus setting the product g1g2:=Core(AB(q0,p0)) for elements Aq0,Bq0∈Bn,r with cores g1 and g2 respectively, results in a well-defined product.
The authors of [3] show that this product is equivalent to the following. Let g,h∈On,r, observe that g and h are core non-initial transducers. Fix a state q1 and q2 of g and h respectively. Let gh:=Core((gh)(q1,q2)) where (gh)(q1,q2) is the minimal transducer representing the product gq1∗hq2. The set On,r is isomorphic to \mboxOut(Gn,r) under this product.
The inverse of an element of On,r can likewise be defined in two equivalent ways as demonstrated in [3]. The first approach is as follows. Given an element g∈On,r let Aq0∈Bn,r be such that Core(Aq0)=g. Let Bp0 be the minimal transducer representing the inverse of Aq0 then set g−1=Core(Bq0). It turns out (see [3]) that this inverse is independent of the the choice of transducer Aq0∈Bn,r with core equal to g and so is well-defined. The second approach makes use only of the states of g to construct the inverse of g, thus removing the need of finding an element of Bn,r with core equal to g.
Thus having removed the need to find elements of Bn,r with cores equal to the relevant transducers in order to compute products and inverses, the paper [3] introduces the group On:=∪1≤r<nOn,r with product of two elements constructed as in the second approach, likewise for inverses. The groups On,r are moreover subgroups of On. Another way of defining the group On (see [3]) is as the group of all bi-synchronizing transducers (note that as we have a method for computing inverses, our use of the word bi-synchronizing makes sense) all of whose states induce injective maps of Cn with clopen images.
Let TOn,r:={Core(Aq0)∣Aq0∈TBn,r}⊂On,r. Then TOn,r is a subgroup of On,r, under the product inherited from On, moreover TOn,r≅\mboxOut(Tn,r) under this product.
The following lemma essentially solves one half of the membership problem of TOn,r in On,r.
Lemma 6.2**.**
Let Aq0∈TBn,r, then for all states q of Core(Aq0) the maps hq:Cn→Cn either all preserve or all reverse the lexicographic ordering of Cn and preserve the relation ≃I on Cn.
Proof.
We consider only elements Aq0∈TBn,r such that hq0 induces an orientation preserving homeomorphism of Sr. The proof is similar in the other case.
Observe that since Aq0∈TBn,r this means that hq0:Cn,r→Cn,r is a homeomorphism which preserve ≃. Hence there is a ν∈Xn,r∗ such that hq0 preserve the lexicographic ordering on Uν. By restricting to a small enough open set contained in ν we may assume that assume that ν is longer than the synchronizing level of Aq0. Let γ∈Xn∗ be longer than the synchronizing level of Aq0 such that the state of Aq0 forced by γ is a state p∈Core(Aq0). Let q=πA(ν,q0)∈Core(Aq0). Observe that as hq0 preserves the lexicographic ordering of Uν, it must be the case that hq:Cn→Cn preserves the lexicographic ordering on Uν. Thus hp must also preserve the lexicographic ordering on Cn otherwise hq does not preserve the lexicographic ordering. Moreover since hq0 preserves ≃ we must also have that hp preserves ≃I otherwise hq0 would not preserve ≃. Now since Aq0 is synchronizing for all states p′ of Core(Aq0) there is a word γ′ in Xn∗ longer than the synchronizing length of Aq0 such that the state of Aq0 forced by γ′ is p′. This concludes the proof.
∎
This prompts the following definition:
Definition 6.3**.**
Let A be a non-initial transducer of over Cn, and let P be some relation on Cn. Then we say that A preserves P if for all states q of A the map hq preserves the relation P.
Lemma 6.4**.**
Let Aq0∈TBn,r then for all states q∈Core(Aq0) there is a finite antichain uˉ={u1,u2,…,um} such that \mboxim(q)=∪1≤i≤mUui. Moreover for 1≤i≤m−1 we have that uin−1n−1…≃Iui+100….
Proof.
Observe that by Lemma 6.2 it follows that for every state q∈Core(Aq0hq induces a continuous function from the interval to itself. However since the interval is connected, and the continuous image of a connected topological space is connected, it follows that \mboxim(q)/≃I is a connected. The lemma is now an easy consequence of this.
∎
The result below was demonstrated first by Brin in his seminal paper [4] describing the automorphisms of R. Thompson’s groups F and T and we have simple translated it into the language of transducers.
Theorem 6.5**.**
Let n=2, and let Aq0∈TBn,r, then Core(Aq0) is either the single state identity transducer or the single state transducer mapping induce the transposition (01) on X2. More specifically we have TOn≅C2.
In order to show the subgroup TOn,r≤On,r is completely characterised by the fact that all of its states preserve the relation ≃I and either all preserve or all reverse the lexicographic ordering of Cn we need the following notions from [3]. Recall that On is the group of core, bi-synchronizing transducers all of whose states induce injective maps of Cn with clopen images.
Let g∈On. A viable combinationvj for g is a pair of tuples
[TABLE]
satisfying
⋃1≤i≤jρi\mboxim(pi)=Cn
2. 2.
for 1≤i<l≤j we have ρi\mboxim(pi)∩ρl\mboxim(pl)=∅ .
where the ρi∈Xn∗, 1≤i≤j are not necessarily incomparable or distinct, and pi, 1≤i≤j are not necessarily distinct states of g.
Definition 6.7** (Single expansions of viable combinations [3]).**
Let T=(Xn,Q,π,λ)∈On and let
vj=((ρ1,…,ρj),(p1,…,pj)) be a viable combination for T. Fix an i such that 1≤i≤j, and let ρi,l:=ρiλ(l,pi) and pi,l:=π(l,pi) for l∈Xn, then
[TABLE]
is called a single expansion of
vj.
Remark 6.8**.**
Clearly a single expansion of a viable combination for an element T∈On results in a new viable combination for T. Therefore a sequence of single expansions applied to a viable combination for T also results in a new viable combination for T. See [3] for more detail.
Let g∈On and let vg denote the set of all viable combinations of g. Let 1≤r≤n−1. There is a transducer Aq0∈Bn,r whose core is equal to g if and only if there is a sequence vj1,vj2…,vjm of elements of vg such that r≡∑1≤i≤mji≡mmodn−1.
Now we have the following lemma which solves the other half of the membership problem of TOn,r∈On,r. However we make the following definition first.
Definition 6.10**.**
Let g∈On. A lexicographic viable combinationvj for g is a viable combination
[TABLE]
satisfying: for 1≤a<b≤jρa\mboxim(pa)<\mboxlexρb\mboxim(pb).
Lemma 6.11**.**
Let g∈On,r be such that g preserves ≃I and preserves (reverses) the lexicographic ordering on Cn then any viable combination of g may be re-ordered to get a lexicographic viable combination.
Proof.
Since g∈On,r there is an element Aq0∈TBn,r such that Core(Aq0)=g. Now by Lemma 6.9 it follows that there are viable combinations vj1,vj2,…,vjm of g. Thus the set vg of viable combinations of g is non-empty. Let vl∈vg. We shall now make use of Lemma 6.4 to re-order vl to obtain a lexicographic viable combination of g.
Suppose vl=((ρ1,…,ρl),(p1,…,pl)). Let Rvl:={ρk∣1≤k≤l\mboxsuchthatρt≤ρk∀1≤t≤l}. Observe that Rvl is an antichain by construction, thus we may assume that the set Rvl is totally ordered in the lexicographic order. Fix ρk∈Rvl. Let D(ρk)={ρt∣ρk≤ρt}. We assume that D(ρk) is ordered according to the short-lex ordering. Observe that ρk∈D(ρk) and in particular is the smallest element of ρk. Let dρk=∣D(ρk)∣. Now for each ρt∈D(ρk) let μρt be the number of times it occurs in the tuple (ρ1,…,ρk). Let lˉ:=∣Rvl∣.
Given a tuple (y1,y2,…,yt) we shall use the notation (y1,…,ysp,…,yt) to mean that ys,…,ys+p−1 are all equal to ys. Now let ξ1,…ξlˉ be the elements of Rvl in lexicographic ordering. For 1≤s≤lˉ let ξ(s,1),…ξ(s,dξs) be the elements of D(ξs) in short-lex ordering. Then we have
[TABLE]
is a reordering of (ρ1,…,ρl). Let (q1,…,ql) be the induced reordering on the tuple of states (p1,…,pl).
Now fix 1≤s≤lˉ and consider the sequence
ξsμξs,ξ(s,1)μξ(s,1),…,ξ(s,dξ1)μξ(s,dξs) let
[TABLE]
be the corresponding states. Then observe that
[TABLE]
by construction and the definition of viable
combinations. Now let 0≤k<dξs and let k≤k′≤dξs, for 1≤t1≤μξ(s,l) and 1≤t2≤μξ(s,k′) consider ξ(s,k)\mboxim(q(sk,t1)) and ξ(s,k′)\mboxim(q(sk′,t2)) (note that we set ξ(s,0)\mboxim(q(s0,t1)):=ξs\mboxim(qs0,t1)). Observe that by the ordering of the set D(ξs), by Lemma 6.4 and by Definition 6.6 part 2 of viable combinations, we must have that exactly one of the following holds:
(i)
ξ(s,k)\mboxim(q(sk,t1))<\mboxlexξ(s,k′)\mboxim(q(sk′,t2)) or,
2. (ii)
Thus we may reorder the tuple \tagform@1 to obtain a new tuple
[TABLE]
satisfying the following conditions. Let 1≤s≤lˉ and consider the subsequence
[TABLE]
then:
(i)
Each χ(s,r)μχ(s,r) for 0≤r≤dξs (where χ(s,0)=χs) corresponds to precisely one element ξ(s,r′)μξ(s,r′) for 0≤r′≤dξs (where ξ(s,0)=ξs).
2. (ii)
The subsequence χsμχs,χ(s,1)μχ(s,1),…,χ(s,dξs)μχ(s,dξs) is ordered so that the following is true. If q(s0,1),…,q(s0,μχs),…,q(sdξs,1),…,q(sdξs,μχ(s,dξs)) are the set of states corresponding to the sub-sequence χsμχs,χ(s,1)μχ(s,1),…,χ(s,dξ1)μχ(s,dξs) then for 0≤k<dξs, for k≤k′≤dξs, for 1≤t1≤μχ(s,k) and 1≤t2≤μχ(s,k′) such that if k=k′ and μχ(s,k)>1 then t1=t2, we have χ(s,k)\mboxim(q(sk,t1))<\mboxlexχ(s,k′)\mboxim(q(sk′,t2)).
Notice condition (i) above means all elements of the subsequence χsμχs,χ(s,1)μχ(s,1),…,χ(s,dξs)μχ(s,dξs) have prefix ξs.
Now observe that for 1≤s<s′≤lˉ, for 0≤k≤dξs, for 0≤k′≤dξs′, for 1≤t≤μχ(s,k) and for 1≤t′≤μχ(s′,k′) so that if the tth copy of χ(s,k) (χs if k=0) and the t′th copy of χ(s′,k′) (χs′ if k′=0) in the sequence \tagform@2 correspond to the states q(sk,t1) and q(sk′,t2) respectively, then χ(s,k)\mboxim(q(sk,t1))<\mboxlexχ(s′,k′)\mboxim(q(sk′′,t2)) in the lexicographic ordering on Cn. This follows from the observation in the previous paragraph and because ξs<\mboxlexξs′.
Let
[TABLE]
and let
{IEEEeqnarray*}rCl
(q_1’, …q_l’) := (& q_(1_0,1) , …, q_(1_0, μχ1), …, q(1_dξ1,1),…, q(1_d_ξ1, μχ(1, dξ_1)), …,
Observe that ((ζ1,…,ζl),(q1′,…,ql′)) is a viable combination for g. Moreover, by discussion above, for 1≤a<b≤l it satisfies ζa\mboxim(qa)<\mboxlexζb\mboxim(qb) in the lexicographic ordering of Cn i.e ((ζ1,…,ζl),(q1′,…,ql′)) is a lexicographic viable combination of g.
∎
Let πR,n∈On be the single state transducer which induces the permutation i↦n−1−i on Xn. Then πR,n has order 2, preserves ≃I, and reverses the lexicographic ordering on Cn. It is also not hard to see that πR,n∈TOn,r for all 1≤r<n−1. For instance the map fπR,n:Cn,r→Cn,r given by a˙δ↦r−a+1˙(δ)hπR,n is induces by an element of TBn,r with core equal to πR,n since it replaces some finite prefix before acting by πR,n.
The following lemma follows immediately from the fact that πR,n has order 2 and reverses the lexicographic ordering on Cn.
Lemma 6.12**.**
Multiplication by πR,n induces a bijection from the subset of On,r preserving (reversing) the lexicographic ordering, to the subset of On,r reversing (preserving) the lexicographic ordering.
Lemma 6.13**.**
Let g∈On,r be such that g preserves ≃I and preserves (reverses) the lexicographic ordering on Cn then g∈TOn,r.
Proof.
By Lemma 6.12 it suffices to prove this for elements g∈On,r which preserves ≃I and the lexicographic ordering on Cn. Since if gπ∈TOn,r, then as TOn,r is a group containing πR,n by an observation above, then gπR,nπR,n=g∈TOn,r also. Thus fix g∈On,r an element which preserves ≃I and the lexicographic ordering on Cn.
Since g∈On,r by Lemma 6.9 there are viable combinations vj1,vj2,…,vjm of g such that r≡∑1≤i≤mji≡mmodn−1. Let j=∑1≤i≤mji. By Remark 6.8 we may assume that ji>1 for 1≤i≤m. Now by Lemma 6.11 we may assume that all the vji for 1≤i≤m are lexicographic viable combinations for g.
Let 1≤i≤m, and consider the viable combination vji of g. Suppose vji=((ρ1,…,ρji),(p1,…,pji)). Let uˉ={u1,u2,…,uji} be an antichain of Xn,r∗ of length ji, and let w∈Xn,r+. Recall that all antichains are assumed to be ordered in the lexicographic ordering. Now if g preserves the lexicographic ordering of Cn then define a map fvji,w:Uu1⊔…⊔Uuji→Uw by uaδ↦wρa(δ)hpa for 1≤a≤ji and δ∈Cn. O Then fvji,w is homeomorphism from its domain to its range (by the definition of a viable combination), moreover fvji,w preserves the lexicographic ordering on Uu1⊔…⊔Uul since vji is a lexicographic viable combination and for all states q of ghq preserves the lexicographic ordering on Cn and the relation ≃I (Lemma 6.2).
Now let vˉ be a complete antichain of Xn,r∗ of length j and let wˉ be a complete antichain of Xn,r∗ of length m. These antichains exist since j≡m≡rmodn−1. Recall that as 1≤r<n−1 we must have that j and m are both non-zero. Let vˉ1,…,vˉm be disjoint subsets of vˉ such that each vˉi, 1≤i≤m is an antichain (ordered lexicographically) of length ji and for 1≤a<b≤m we have vˉa<\mboxlexvˉb. Note that the stipulation that ∣vˉi∣=ji for 1≤i≤m means that ⊔1≤i≤mvˉi=vˉ. Suppose that wˉ={w1,w2,…,wm}.
Recall that for a subset Z⊆Xn∗⊔Xn,r∗ we denote by U(Z) the set {Uz∣z∈Z}. Let f:Cn,r→Cn,r be defined such that f↾U(vˉi)=fvˉi,wi for 1≤i≤m. Then clearly f is a homeomorphism, and since each fvˉi,wi, 1≤i≤m, preserves the lexicographic ordering of U(viˉ) and ≃I then, f preserves the lexicographic ordering and the relation ≃I on Cn,r. Moreover since f replaces some initial prefix before acting as a state of g, it follows that there is a transducer Aq0∈TBn,r with hq0=f.
∎
Remark 6.14**.**
One may also prove Lemma 6.13 above for g∈On,r which preserves ≃I and reverses the lexicographic ordering by directly constructing a transducer Aq0∈TBn,r which induce an orientation reversing homeomorphism of the line. As in the proof above, we make use of the lexicographic viable combinations to induce maps on Cn,r which preserve ≃I and reverse the lexicographic ordering.
Lemma 6.15**.**
Let g∈On,r be such that g preserves or reverses the lexicographic ordering, then g also preserves ≃I.
Proof.
Fix q any state of g. We observe that, by definition q induces a continuous injection function from Cn to itself with clopen image. Let x,y∈Cn be such that x=y, x<\mboxlexy, x≃Iy and there is a word ν∈Xn,r+ such that (x)h,(y)h∈Uν. As there is no point y′∈Cn not equal to x or y satisfying x<\mboxlexy′<\mboxlexy, then it must be the case that (x)h≃(y′)h. Now as (Uν)hq−1 is open, there is a μ∈Xn,r+ such that (Uμ)hq=Uν and so hq preserves the relation ≃I on Uμ. Finally, observe that as g is synchronizing for any other state p of g, there is a word Γ∈Xn,r+ such that πg(Γ,q)=p. Therefore, we deduce that all states of g preserve ≃I as required.
∎
Putting together Lemmas 6.9, 6.11, Lemma 6.15, and 6.13, we obtain the following result.
Theorem 6.16**.**
Let g∈On. The following are equivalent:
(a)
g∈TOn,r**
2. (b)
g* preserves or reverses the lexicographic ordering, and there are lexicographic viable combinations vj1,…,vjm such that r≡∑1≤i≤mji≡mmodn−1.*
3. (c)
g* preserves or reverses the lexicographic ordering, and there are viable combinations vj1,…,vjm such that r≡∑1≤i≤mji≡mmodn−1.*
4. (d)
g∈On,r* and preserves or reverses the lexicographic ordering on Cn.*
7 The enveloping group TOn
We make the following definition.
Definition 7.1**.**
Let TOn⊂On consists of those element which reverse or preserve the lexicographic ordering on Cn.
Remark 7.2**.**
By an observation in [3], for an element g∈On the set vg of viable combinations of g is non-empty. From this it follows that TOn=∪1≤r≤n−1TOn,r.
Propostion 7.3**.**
The set TOn is a subgroup of On.
Proof.
It suffices to show that TOn is closed under inverses and products. It is clear that the single state identity transducer is an element of TOn.
Let g∈TOn. Then there is an r∈{1,2,…,n−1} such that there is an Aq0∈TBn,r with Core(Aq0)=g. Let Bp0 be the minimal transducer representing the inverse of Aq0, then Bp0∈TBn,r and g−1=Core(Bp0)∈TOn,r⊂TOn.
Now let h∈TOn. Let q be a state of g and p be a state of h. Let gh(p,q) be the minimal transducer representing the product gp∗hq. Then observe that gh(p,q) induces the function g∘h:Cn→Cn. Thus since g and h either preserve or reverse the lexicographic ordering on Cn it follows that the states of gh(p,q) either all reverse or all preserve the lexicographic ordering on Cn. Therefore gh=Core(gh(p,q))∈TOn.
∎
Definition 7.4**.**
Let TOn be the subset of TOn consisting of all those elements which preserve the lexicographic ordering on Cn. Set TOn,r:=TOn∩TOn,r. Then we call elements of TOnorientation preserving, and elements of TOn\TOnorientation reversing. Likewise set TBn,r to be those elements of TBn,r with core in TOn, then TBn,r are precisely those elements of TBn,r which induce orientation preserving maps of Sr. We will also call elements of TBn,rorientation preserving and the elements of TBn,r\TBn,rorientation reversing.
The following proposition is straightforward and so we omit its proof.
Propostion 7.5**.**
The set TOn is an index 2 subgroup of TOn and so a normal subgroup of TOn. The set TOn,r is an index 2 subgroup of TOn,r and so a normal subgroup of TOn,r.
We now investigate how the groups TOn intersect each other. For this we require the following definition, which in fact applies to the elements of the group On.
Definition 7.6** (Signature).**
Let T∈On, for each state q∈QT let mq be the size of the smallest subset V of Xn∗ such that U(V)={Uv∣v∈V} is a clopen cover of \mboxim(q) and Uv⊂\mboxim(q) for all v∈V. Let k∈N be the minimal synchronizing level of T an order the elements of the set Xnk lexicographically as follows: x1<x2<…<xnk. Let (qx1,qx2,…,qxnk)∈QTnk be such that, for all 1≤i≤nk, qxi is the unique state of T forced by xi. Set (T)sig=∑1≤i≤nkmqxi, we call (T)sig the signature of T; set (T)sig=(T)sigmodn−1, we call (T)sig the reduced signature of T.
We have the following proposition. We prove the proposition below for the group TOn noting that a similar result holds, with almost identical proof, in the group On.
Propostion 7.7**.**
Let T∈TOn[T∈On], and 1≤r<n, then T∈TOn,r[T∈On,r] if and only if r(T)sig≡rmodn−1 (equivalently r((T)sig−1)≡0modn−1.
Proof.
We begin with the forward implication. First suppose that T∈TOn,r and T has minimal synchronizing level k. Since T∈TOn,r, there is an element Aq0∈TBn,r with Core(Aq0)=T. Let j∈N be minimal such that after reading a word of length j from the state q0 of A, the resulting state is a state of T. Let {μi∣1≤i≤rnk+j−1} be the set of all words of length j+k in Xn,r∗ ordered lexicographically. For 1≤i≤rnk+j−1, set νi=λA(μi,q0) and qμi to be the state of T forced by μi. Observe that the state qμi depends only on the last k letters of μi, hence if the elements of Xnk are ordered lexicographically as x1<x2<…<xnk, the sequence (qμi)1≤i≤rnj+k−1, where qxi, 1≤i≤nk, is the state of T forced by xi, is precisely the sequence qx1,…,qxnk repeated rnj−1 times. Since ⋃1≤i≤rnj+k−1νi\mboxim(qμi)=Cn,r, it must therefore be the case that rnj−1(∑1≤i≤nkmqxi)≡rmodn−1. This is because if, for each qμi, 1≤i≤rnk+j−1, Vqμi is the smallest subset of Xn∗ such that U(Vqμi) is a cover of \mboxim(q) and Uv⊂\mboxim(q) for all v∈Vqμi, then ⋃1≤i≤rnj+k−1{νiv∣v∈V(qμi)} must be a complete antichain of Cn,r (otherwise Aq0 is not a homeomorphism). Therefore, setting mqxi=∣Vqxi∣ we have,
[TABLE]
Since n≡1modn−1 we therefore have that r(T)sig≡rmodn−1 as required.
For the reverse implication let T∈TOn and 1≤r<n−1 be such that r(T)sig≡rmodn−1. Let k∈N be such that T is synchronizing at level k, once more assume that the set Xnk is ordered lexicographically as x1<x2<…<xnk. For 1≤i≤nk, let qxi be the state of T forced by xi. For each 1≤i≤nk, let Vqxi be the smallest subset of Xn∗, with size mqxi, such that U(Vqxi) is a clopen cover of \mboxim(q) consisting of clopen subsets of \mboxim(q); let Mi=max{∣v∣∣v∈Vqxi} and set M=max1≤i≤nkMi. Let j∈N be minimal such that for any word Γ∈Xnj and any state q of T, ∣λT(Γ,q)∣>M. Order the set Xnj lexicographically as y1<y2<…<ynj. For each state qxi, 1≤i≤nk, order the set Vqxi lexicographically as νi,1<νi,2<…<νi,mqxi, and for all 1≤l≤nj let μi,lφi,l=λT(yl,qxi), for some μi,l∈Vqxi, φi,l∈Xn+, and pi,l=πT(yl,qxi). Now since, for 1≤i≤nk, ∣Vqxi∣=mqxi, each μi,l=νi,a for some 1≤a≤mqxi and we may write, for all 1≤l≤nj, μi,lφi,l=νi,aρi,la where νi,a=μi,l and ρi,la=φi,l for some 1≤a≤mqxi, we also adopt the same notation for the set of pi,l and write pi,la, for 1≤a≤mqxi, where νi,aρi,la=μi,lφi,l.
Now let uˉ be a maximal antichain of Cn,r of length rnj+k and let vˉ be a maximal antichain of Cn,r of length r(T)sig. Write uˉ=∪1≤t≤ruˉt where each uˉt is ordered lexicographically, ∣uˉt∣=nj+k and for 1≤t1<t2≤r all elements of uˉt1 are strictly less than all elements of uˉt2 in the lexicographic ordering on Xn∗. Let vˉ=∪1≤t≤rvˉt where each vˉt is ordered lexicographically, ∣vˉt∣=(T)sig and for 1≤t1<t2≤r all elements of vˉt1 are strictly less than all elements of vˉt2 in the lexicographic ordering on Xn∗.
Fix 1≤t≤r and consider uˉt. Write uˉt={ut,i,l∣1≤i≤nk,1≤l≤nj}. We further assume that, for a fixed 1≤i≤nk, the set {ut,i,l∣1≤l≤nj} is ordered lexicographically and, for 1≤i1<i2≤nk and for all 1≤l1,l2≤nj, ut,i1,l1<\mboxlexut,i2,l2. Likewise write vˉt={vt,i,a∣1≤i≤nk,1≤a≤mqxi}. We further assume that, for a fixed 1≤i≤nk, the set {vt,i,a∣1≤a≤mqxi} is ordered lexicographically and, for 1≤i1<i2≤nk, vt,i1,a<\mboxlexvt,i2,b for all 1≤a≤mqxi1 and 1≤b≤mqxi2. Furthermore, whenever, for 1≤a≤mqxi and 1≤l≤nj, we have νi,aρi,la=μi,lφi,l we set ηt,i,l:=vt,i,a.
Define a map f from Cn,r to itself as follows. For 1≤t≤r, f acts on elements with prefix in the set uˉt as follows. For 1≤i≤nk and 1≤l≤nj, and Γ∈Cn, ut,i,lΓ↦ηt,i,lφi,l(Γ)pi,l. Observe that for a fixed 1≤i≤nk, and a fixed 1≤a≤mqxi, since we have ∪{μi,lφi,l\mboxim(pi,l)∣μi,l=νi,a}=Uνi,a then it is the case that, for a fixed 1≤t≤r, ∪{ηt,i,lφi,l\mboxim(pi,l)∣ηi,l=vt,i,a}=Uvt,i,a. Now since, for a given 1≤i≤nk, ⋃1≤l≤njμi,lφi,l\mboxim(pi,l)=\mboxim(q) and as q is injective and preserves the lexicographic ordering of Cn, we see that f↾U(uˉt) is a bijection unto the set U(vˉt) which preserves the lexicographic ordering of Cn. More specifically, since f acts by a state of T after a finite depth, we see that f is a homeomorphism of Cn,r which is in fact an element of TBn,r.
The proof of the other reading proceeds in an analogous fashion only here we do not have to worry about preserving the lexicographic ordering on Cn,r.
∎
Remark 7.8**.**
It follows from results in [14] that given an element T∈TOn[T∈On] and 1≤r<n, then it is possible to decide in finite time if T∈TOn,r[T∈On,r]. Furthermore, by the above proposition, T∈TOn,1[T∈On,1] if and only if (T)sig≡1modn−1.
The following result is a consequence of Proposition 7.7:
Lemma 7.9**.**
Let n∈N and suppose that n≥2. Let i,j,m∈Zn such that, i and j are non-zero and mi≡jmodn−1 then TOn,i⊆TOn,j [On,i⊆On,j].
Proof.
Let i,j,m be as in the statement of the lemma. Let T∈TOn,i[T∈On,i], then by Proposition 7.7i(T)sig≡imodn−1, therefore mi(T)sig≡mimodn−1 and so j(T)sig≡jmodn−1 and T∈TOn,j[T∈On,j] again by Proposition 7.7.
∎
Remark 7.10**.**
Observe that the lemma above implies that whenever j∈Zn is co-prime to n−1 then TOn,j=TOn,1 [On,j=On,1]. Thus it follows that for n a natural number bigger than 2 such that n−1 is prime, if T∈TOn [T∈On], satisfies (T)sig≡1modn−1 then T∈TOn\{TOn,1}[T∈On\{On,1}]. The following result generalises this observation.
An immediate corollary of the Lemma 7.9 is the following result, an analogous result appears in [3] for the groups On,r:
Theorem 7.11**.**
Let n∈N, n≥2, then for all non-zero i∈Zn we have TOn,i⊆TOn,n−1. Hence TOn,n−1=TOn and TOn,1⊆TOn,i for all 1≤i≤n−1, hence ∩1≤r≤n−1TOn,r=TOn,1.
Proof.
Let i be non-zero in Zn, then observe that (n−1)∗i≡n−1modn−1 and i∗1≡imodn−1. Therefore by Lemma 7.9 above, we have TOn,i⊆TOn,n−1 and TOn,1⊆TOn,i. Thus TOn,n−1=TOn and ∩1≤r≤n−1TOn,r=TOn,1.
∎
The following result is again a corollary of Lemma 7.9, we observe that the result for On,r was proved in [3]:
Corollary 7.12**.**
Let n be a natural number bigger than 2, j∈Zn\{0} and d be the greatest common divisor of n−1 and j, then TOn,j=TOn,d [On,j=On,d].
Proof.
Let n,j,d be as in the statement of the corollary. Since d is the greatest common divisor of n−1 and j, there are co-prime numbers a,b∈Zn\{0} such that j=da and n−1=db. Since a and b are co-prime, there are numbers u,v∈Z such that ua=1+vb. Multiplying both sides of the equation by d it follows that uad=d+vbd and so uj=d+v(n−1). Therefore there is some m1∈Zn−1 such that m1j≡dmodn−1. Moreover, since d divides j, there is some m2∈Zn−1 such that m2d=jmodn−1. It therefore follows by Lemma 7.9 that TOn,j=TOn,d[On,j=On,d].
∎
Corollary 7.12 should be compared with the result of Pardo [19] showing that Gn,r≅Gm,s if and only if n=m and gcd(n−1,r)=gcd(n−1,s). It is a question in [3] whether or not On,r≅On,s if and only if gcd(n−1,r)=gcd(n−1,s). Below (Remark 8.2) we show that this question has a negative answer.
The following Lemma can be thought of as a partial converse to Lemma 7.9.
Lemma 7.13**.**
Let T∈TOn [T∈On] and let r be minimal such that T∈TOn,r [T∈On,r], then T∈TOn,j [T∈On,j] for some 1≤j≤n−1 if and only if there is some m∈Zn such that mr=j.
Proof.
The forward implication is a consequence of Lemma 7.9. Therefore let T∈TOn and r be minimal such that T∈TOn,r Let 1≤j≤n−1 be such that T∈TOn,j. By assumption we must have that r<j. Since T∈TOn,r then by Proposition 7.7, r(T)sig≡rmodn−1 and since T∈TOn,j, then j(T)sig≡jmodn−1. Thus we deduce that (j−r)(T)sig≡(j−r)modn−1. If (j−r)>r, then we may repeat the process otherwise j−r=r, by the minimality assumption on r and Proposition 7.7, in which case j=2r. Inductively there is some k∈N such that j−kr=r and so j=(k+1)r which concludes the proof.
The other reading of the lemma is proved analogously, simply replace TOn with On in the paragraph above.
∎
We now show that the map sig:On→Zn−1 is a homomorphism from On to the group of units of Zn−1 with kernel On,1.
We begin with the following result:
Propostion 7.14**.**
Let T∈On, q be any state of T and mq be the size of the smallest subset V of Xn∗ such that U(V)={Uv∣v∈V} is a clopen cover of \mboxim(q) and Uv⊂\mboxim(q) for all v∈V, then mqmodn−1=(T)sig.
Proof.
For any state p of T let V(p)={v1,p,v2,p,…,vmp,p}⊂Xn∗ be such that \mboxim(p)=∪1≤i≤mpUvi,p. Now fix a state q of T, let k be the minimal synchronizing level of T and j∈Nk be such that for any word Γ∈Xnj and any state p of T, ∣λT(Γ,p)∣≥max{∣v1,q∣∣v1,q∈V(q)}. Order the set Xnj in the lexicographic ordering as follows x1<x2<x3…<xnj, for 1≤a≤nj let qxa be the unique state of T forced by xa and ρxa=λA(xa,q). Observe that for all 1≤a≤nj, ∣ρxa∣≥max{∣v1,q∣∣v1,q∈V(q)} and so ρxa has a prefix in V(q). Now since the state q is induces a homeomorphism from Cn unto its image, it follows that the set ∪1≤a≤nj{ρxavi,qxa∣1≤i≤mqxa} is a complete antichain for the subset of Xn∗ consisting of all elements with prefix in V(q). From this we deduce that ∣∪1≤a≤nj{ρxavi,qxa∣1≤i≤mqxa}∣≡mqmodn−1.
On the other hand
[TABLE]
However, since T is synchronizing at level k, the sequence (qxa)1≤a≤nj is in fact equal to the sequence (qxa)(1≤a≤nk) repeated nj−k times. Therefore we have,
[TABLE]
and so we conclude that mqmodn−1=nj−k∑1≤a≤nkmqxamodn−1=(T)sig.
∎
Theorem 7.15**.**
The map sig:On→Zn−1 is a homomorphism from On into the group of units of Zn−1 with kernel On,1.
Proof.
Let T,U∈On, j,k∈N be the minimal synchronizing levels of T and U. Consider the transducer product T∗U. Let (p,q) be any state in the core of T∗U, and for ♯∈{p,q} let V(♯)={v1,♯,v2,♯,…,vmp,♯}⊂Xn∗ be such that \mboxim(♯)=∪1≤i≤m♯Uvi,♯. We may assume that Vq is the smallest subset of Xn∗ with this property. Let m=max{∣vi,q∣∣vi,q∈V(q)} and l∈N be such that for any state p′ of T and any word Γ∈Xnl, ∣λT(Γ,p′)∣≥m. We may further assume that V(p) is the smallest subset of Xnl satisfying \mboxim(p)=∪1≤i≤mpUvi,p. For each 1≤i≤mp let qi=πU(vi,p,q) and ρi=λU(vi,p,q). Since \mboxim(p)=∪1≤i≤mpUvi,p, it follows that \mboxim((p,q))=⊔1≤i≤mp{ρi\mboxim(qi)}. For each 1≤i≤mp, let mqi be the size of the smallest subset V(qi)⊂Xn∗ such that ∪v∈V(qi)Uv=\mboxim(qi), it follows that V:=∪1≤i≤mp{ρiv∣v∈V(qi)} satisfies ∪v∈V(Uv)=\mboxim((p,q)). Notice that since both p and q are injective, then ∣V∣=∑1≤i≤mpmqi.
Observe for any other set V′⊂Xn∗ such that ∪v∈V′Uv=\mboxim((p,q)) it must be the case that ∣V′∣≡∣V∣modn−1. Thus if V′ is the smallest subset of Xn∗ with this property, we have ∣V′∣≡∣V∣modn−1. Furthermore we observe that removing incomplete response from the states of T∗U simply removes the greatest common prefix of \mboxim(p,q). Therefore if the state s of TU is equal to the state (p,q) after removing the incomplete response, then for V′′⊂Xn∗ minimal such that ∪v∈V′′Uv=\mboxim(s), we have ∣V′′∣≡∣V∣modn−1. By Proposition 7.14 it follows that
[TABLE]
Since (\mboxid)sig=1, it follows that sig is a homomorphism from On to the group of units of Zn−1.
To see that ker(sig)=On,1, observe that by Proposition 7.7(T)sig=1 if and only if T∈On,1.
∎
Given T∈On the following result enables us to compute the reduced signature of T−1 directly from T i.e. without computing the inverse.
Propostion 7.16**.**
Let T∈On and q be any state of T. Let ν∈Cn be such that Uν⊂\mboxim(q) and j∈N be such that for any word Γ∈Xnj, ∣λT(Γ,q)∣≥∣ν∣. Let W⊂Xnj be maximal such that for any word Δ∈W, ν is a prefix λT(Δ,Γ). Let 1≤w≤n−1 be such that w≡∣W∣modn−1, then w depends only on T, in particular w=(T−1)sig.
Proof.
By Proposition 7.14 it suffices to show that there is a state p′ of T−1 and a subset V⊂Xn∗ such that ∪μ∈VUμ=\mboxim(p) and ∣V∣=∣W∣.
Let Aq0 a bi-synchronizing transducer with Core(Aq0)=T and k∈N1 be the minimal bi-synchronizing level of Aq0. Let φ=(ν)Lq i.e. φ is the greatest common prefix of the set hq−1(Uν). Let ν1=λT(φ,q), p=πT(φ,q) and ν2=ν−ν1. We claim that (ν2,p) is a state of A(ϵ,q0) and is ω-equivalent to a state of T−1.
For, let l∈N be such that for any word Γ∈Xnl, ∣λA(Γ,q0)∣≥k.
Let Γ∈Xnl be such that πA(Γ,q0)=q∈Core(Aq0). Such a word Γ exists because of the bi-synchronizing condition. Let Δ=λA(Γ,q0). Then observe that (Δν)Lq0=Γ(ν)Lq=Γφ since Uν⊂\mboxim(q). Thus, in the transducer A(ϵ,q0), πA(Δν,q0)=(Δν−λA(Γφ,q0),πA(Γφ,q0))=(ν2,p). Now since A(ϵ,q0) has no states of incomplete response, is ω-equivalent to the minimal transducer Bp0 representing hq0−1, and by the choice of l, we therefore have that (ν2,p) is ω-equivalent to a state in Core(Bp0)=T−1.
Now, suppose W={ρ1′,ρ2′,…,ρm′}. We observe that by definition of W, φ is the greatest common prefix of W, thus for 1≤i≤m, let ρi∈Xn∗ be such that φρi=ρi′. Observe that for any δ∈Cn, there is a unique 1≤i≤m, and ξ∈Cn such that λA(ρi′ξ,q)=νδ, therefore λA(ρiξ,p)=ν2δ. Moreover, for any 1≤i≤m and ξ∈Cn, λA(ρiξ,p)=ν2δ for some δ∈Cn. Since \mboxim((ν2,p))={(ν2δ)Lp∣δ∈Cn}=(Uν2)hp−1, it follows that \mboxim((ν2,p))=∪1≤i≤mUρi′. Therefore it follows, from Proposition 7.14, that m=∣W∣≡(T−1)sigmodn−1.
∎
The corollary below follows straight-forwardly from Theorem 7.15
Corollary 7.17**.**
Let 1<r<n, then the following hold:
(a)
On,r* is a normal subgroup of On,n−1=On, in particular for non-zero i,j∈Zn such that i divides j in the additive group Zn−1, On,i⊴On,j,*
2. (b)
[On,On]≤On,r,
3. (c)
On,r/On,1* is isomorphic to a subgroup of the group of units of Zn−1 and On/On,r is isomorphic to a quotient of the group of units of Zn−1.*
Proof.
Since the group of units of Zn−1 is abelian, it follows that [On,On]≤On,1 and by Theorem 7.11[On,On]≤On,r.
To see that On,r is normal in On, we observe that given T∈On,r, and U∈On, then (U−1TU)sig≡(U−1)sig(T)sig(U)sigmodn−1 and so (U−1TU)sig=(T)sig and by Proposition 7.7, we have U−1TU∈On,r.
That On,r/On,1 is isomorphic to a subgroup of the group of units of Zn−1, follows by the correspondence theorem since On,1≤On,r.
That On/On,r is isomorphic to a quotient of the group of units of Zn−1, follows from the third isomorphism theorem: On/On,r≅(On/On,1)/(On,r/On,1).
∎
Question 7.18**.**
Is it the case that [On,On]=On,1? Is the map sig from On to the group of units of Zn−1 surjective?
Remark 7.19**.**
Since TOn≤On the results above and questions can be restated with TOn in place of On.
8 Nesting properties of the groups TOn,r
In this section we focus on the group TOn however, all the results below are equally valid when all occurrences of TOn are replaced with On.
The following result is a direct consequence corollary of Proposition 7.7 and generalises Corollary 7.12:
Corollary 8.1**.**
Let 1≤r,s<n be natural numbers and let (TOn)sig[(On)sig] denote the image of TOn [On] under the map sig. Then TOn,r=TOn,s [On,r=On,s] if and only if for all j∈(TOn)sig [j∈(On)sig], r(j−1)≡s(j−1)≡0modn−1.
Remark 8.2**.**
Observe that when n=7, then (On)sig⊆{1,5}. Further observe, by Corollary 7.12, that for 1≤r≤6, O7,r is equal to O7,a for a∈{1,2,3,6}. Now notice that as 12 is divisible by 6, then O7,3=O7,6 and, since 10 is not a multiple of 6, O7,1=O7,2. However gcd(1,6)=gcd(2,6) and gcd(3,6)=gcd(6,6). Therefore it is not the case that for 1≤r,s≤nOn,r≅On,s if and only if gcd(r,n−1)=gcd(s,n−1). This yields a negative solution to a question in the paper [3]. We notice that the same result holds if On is replaced with TOn. More specifically, TO7,a=TO7,b for (a,b)∈{(1,2),(3,6)}.
There are numbers n∈N, n>2, and 1≤r,s≤n−1 such that, for X=TO,O, Xn,r=Xn,s but gcd(n−1,r)=gcd(n−1,s).
Partition the set Zn\{0} as follows. For i∈Zn\{0}, set [i]:={j∈Zn\{0}∣TOn,j=TOn,i}⊂Zn, [Zn0]:={[i]∣i∈Zn\{0}} and, for [i]∈[Zn0], TOn,[i]:=TOn,i. Observe that the set [Zn0] inherits an ordering from Zn where [i]<[j] if the smallest element of [i] is less than the smallest element of [j]. Further observe that if n−1 is prime, then [Zn0] has size at most 2 by Theorem 7.11.
Definition 8.4** (Atoms).**
Let [r]∈[Zn0], then we say [r] is an atom (of [Zn0]) if there is an element T∈TOn,[r] which is not an element of TOn,[s] for any [s]<[r]. Let [i]∈[Zn0], an atom of [i] is an atom [r] of [Zn0] such that TOn,[r]≤TOn,[i].
Remark 8.5**.**
Let [r]∈[Zn0] be an atom with r the minimal element of r, then as a consequence of Lemma 7.13 and Theorem 7.11, we have that r∣n−1. Observe that by Remark 8.2 it is not always the case that for every element r dividing n−1 that [r] is an atom of [Zn0].
Observe that if the map sig is surjective, then for each i∈Zn\{0}, we may completely determine the elements of [i].
Question 8.6**.**
Is it the case that all elements [i]∈[Zn0] are atoms?
In the interim we make the following definitions.
Definition 8.7**.**
Let [i],[j]∈[Zn0] where i,j∈Zn are the minimal elements of [i] and [j] respectively. We say that [i]divides[j] if r∣j for any atom [r] of [i] with r the minimal element of [r].
Definition 8.8**.**
Given two elements [i],[j]∈[Zn0], the lowest common multiple of [i] and [j] is the smallest element [l] of [Zn0] such that [i] and [j] divide [l]. Notice that by Remark 8.5 the lowest common multiple of any pair of numbers [i],[j]∈[Zn0] always exists. We extend the definition of lowest common multiple to tuples of elements of [Zn0] in the usual way.
Definition 8.9**.**
Let i,j∈Zn\{0} be the minimal elements of [i] and [j] respectively, then we define the greatest common divisor of [i] and [j] to be [r]∈[Zn0] such that r, the minimal element of [r], is the greatest common divisor of i and j.
Remark 8.10**.**
Notice that for [i],[j],[r]∈[Zn0] where i,j,r are the minimal elements of [i], [j] and [r] respectively, if [r] is an atom of [i] then, by Lemma 7.13, r∣i and so if i∣j then r∣j and [i]∣[j]. Further observe that for [i]∈[Zn0] either [i] is an atom or TOn,[i] is a union of groups TOn,[r] for atoms [r] of [i]. This is because for any element T∈TOn,[i], either T is not an element of TOn,[r] for any [r]<[i] and so [i] is an atom, or there is a minimal [j]∈[Zn0] such that T∈TOn,[j] and so [j] is an atom of [i] by Lemma 7.13 and Lemma 7.9.
As a consequence of Lemma 7.13 we have the following result.
Propostion 8.11**.**
Let 1≤i,≤j≤n−1 be integers such i is the smallest element of [i] and j is the smallest element of [j], then the following things hold:
(a)
TOn,[i]≤TOn,[j]* if and only if [i] divides [j] ,*
2. (b)
if [r] is the lowest common multiple of [i],[j], then [r] is minimal such that ⟨TOn,[i],TOn,[j]⟩≤TOn,[r] ,
3. (c)
if [r], where r∈Zn is the smallest element of [r], is the greatest common divisor of [i] and [j] then TOn,[i]∩TOn,[j]=TOn,[r].
Proof.
For part (a), let i,j be elements of Zn\{0} which are the minimal elements of sets [i] and [j] respectively. Suppose [i] divides [j]. If [i] is an atom then i∣j, and so by Lemma 7.9 we have that TOn,i≤TOn,j, in particular TOn,[i]≤TOn,[j]. If [i] is not an atom, then TOn,[i] is the union of the groups TOn,[r] over all atoms [r] of [i]. Therefore let [r] be any atom of [i], where r is the smallest element of [r], since [i] divides [j] then r∣j and so TOn,r≤TOn,j by Lemma 7.9 once more. Since [r] was an arbitrary atom of [i], we conclude that TOn,[i]≤TOn,[j]. On the other hand suppose that TOn,i≤TOn,j.If [i] is an atom then i∣j and so [i] divides [j]. If [i] is not an atom, then let [r] be any atom of [i] where r is the minimal element of [r]. Since TOn,[i]≤TOn,[j], we also have TOn,[r]≤TOn,[j] by definition of an atom of [i]. Now making use of the definition of an atom together with Lemma 7.13, we have that r∣j. Therefore, since [r] was an arbitrarily chose atom of [i], we conclude that [i] divides [j].
Part (b) follows from Part (a) since for any s∈Zn\{0} such that ⟨TOn,[i],TOn,[j]⟩≤TOn,[s], then [i] and [j] divide [s].
For Part (c) let [r] be the greatest common divisor of [i] and [j]. First observe that if [l] is an atom of [r], where l is the minimal element of [l], then l∣i and l∣j. Therefore [r] divides [i] and [j] and by Part (a), TOn,[r]≤TOn,[i]∩TOn,[j]. Now let [s], with s the minimal element of [s] be an atom of [i] and [j]. This means that s∣i and s∣j and so s∣r, since r is the greatest common divisor of i and j. By Part (a) once more, we conclude that s is an atom of [r] and so TOn,[s]≤TOn,[r]. Now since every element T∈TOn,[i]∩TOn,[j] lies in some atom of both [i] and [j], we conclude that TOn,[i]∩TOn,[j]=TOn,[r].
∎
The proposition above prompts the following question:
Question 8.12**.**
Let n be a natural number and let [i],[j]∈[Zn0] is it true that ⟨TOn,[i],TOn,[j]⟩=TOn,[r] where [r] is the lowest common multiple of [i] and [j]?
The proposition below addresses this question by showing that in the case where the map sig is unto the group of units of Zn−1, then Question 8.12 has an affirmative answer.
Propostion 8.13**.**
Let X=TO,O and suppose the map sig:Xn→Zn−1 is unto the group of units of Zn−1. Then for [i],[j]∈[Zn0], ⟨Xn,[i],Xn,[j]⟩=Xn,[r] for [r] the lowest common multiple of [i] and [j].
Proof.
The result is essentially a consequence of the following claim, as we have not been able to find a reference for it, we also provide a proof.
Claim 8.14**.**
Let n∈N2 and consider the integer ring Zn. Let Zn∗ denote the group of units, and for each i∈Zn let (Zn∗)i:={a∈Zn∗∣ai≡imodn}≤Zn∗. Let i,j∈Zn let i1=gcd(i,n), j1=gcd(j,n) and r=lcm(i1,j1) then ⟨(Zn∗)i1,(Zn∗)j1⟩=⟨(Zn∗)i,(Zn∗)j⟩=(Zn∗)r.
Proof.
We first observe that for i∈Zn the group (Zn∗)i=(Zn∗)d where d=gcd(i,n). This is because for any a∈Zn∗ such that ad≡dmodn−1 then ai≡imodn−1 as well. If, on the other hand ai≡imodn−1, then observe that, as, by Bezout’s lemma, there is a u∈Zn such that iu≡dmodn−1, then ad≡dmodn−1 also.
Thus let i,j∈Zn be divisors of n with r=lcm(i,j) (a divisor of n). First assume that n=pα for a prime p. Then i=pβ and j=pγ for β,γ≤α. Without loss of generality we may assume that i≤j, and so, in particular, that i divides j. However, it then follows that (Zn∗)i≤(Zn∗)j, from which we conclude that ⟨(Zn∗)i,(Zn∗)j⟩=(Zn)j∗ noting that lcm(i,j)=j.
Now suppose that n=p1α1p2α2…pmαm where the pl’s, 1≤l≤m are distinct primes. We may further assume that gcd(i,j)=1. This is because if gcd(i,j)=d, then setting i1=i/d and j1=j/d we observe that (Zn∗)i1≤(Zn∗)i, (Zn∗)j1≤(Zn∗)j and lcm(i,j)=lcm(i1,j1). Thus, noting that ⟨(Zn∗)i,(Zn∗)j⟩≤⟨(Zn∗)lcm(i,j)⟩, the result holds for i1 and j1 precisely if it holds for i and j.
Making use of the Chinese remainder theorem, there is a ring isomorphism from Zn→Zp1α1×Zp2α2×…×Zpmαm defined by k↦(kmodp1α1,kmodp2α1,…,kmodpmαm). For 1≤l≤m let il=imodplαm likewise define the sequence jl and rl, where r=ij. It follows that (Zn∗)i≅(Zp1α1∗)i1×(Zp1α2∗)i2×…×(Zpmαm∗)im, likewise (Zn∗)j≅(Zp1α1∗)j1×(Zp1α1∗)j2×…×(Zpmαm∗)jm. Observe that
[TABLE]
is precisely the group
[TABLE]
Since i and j are divisors of n then i=p1β1p2β2…pmβm, j=p1γ1p2γ2…pmγm where for 1≤l≤m, 0≤βl,γl≤αl and γl+βl=max{γl,βl} (since i,j are assumed co-prime). It therefore follows that imodplαl is either a unit of Zplαl otherwise it is equal to a unit of Zplαl times a non-trivial (appropriate) power of pl and likewise for j. We note that if one of il or jl is equal to a unit of Zplαl times a non-trivial (appropriate) power of pl, then the other must be equal to a unit of Zplαl. Therefore for 1≤l≤m, ⟨(Zplαl∗)il,(Zplαl∗)jl⟩ is either equal to the trivial group or it is equal to (Zplαl∗)plδ where plδ=1 is the maximum power of pl dividing ij. In the case that ⟨(Zplαl∗)il,(Zplαl∗)jl⟩ is the trivial group, then il and jl are both units of Zplαl and so rl=iljlmodplαl is also a unit of Zplαl. In the case that ⟨(Zplαl∗)il,(Zplαl∗)jl⟩ is equal to (Zplαl∗)plδ where plδ=1 is the maximum power of pl dividing ij, then it follows that one of il or jl is a unit and the other is equal to a unit times plδ. In this case we have that rl is equal to a unit time plδ and so (Zplαl∗)plδ=(Zplαl∗)rl. In either case we see that (Zplαl∗)rl=⟨(Zplαl∗)il,(Zplαl∗)jl⟩. Thus we conclude that that ⟨(Zn∗)i,(Zn∗)j⟩=(Zn∗)r as required.
∎
We can now prove the proposition. We observe that for 1≤r<n since sig is surjective, then (Xn,r)sig=(Zn−1∗)r. Let 1≤i,j≤n−1 such that i the the minimal element of [i] and j is the minimal element of [j] so that both i and j are divisors of n. Let r=lcm(i,j). By Claim 8.14 we have that ⟨(Zn−1∗)i,(Zn−1∗)j⟩=(Zn−1∗)r. Thus, let T∈Xn,r be any element. By Proposition 7.7(T)sig∈(Zn−1∗)r and so there are elements u∈(Zn−1∗)i and v∈(Zn−1∗)j such that (T)sig=uvmodn−1. Let U\iXn,i and V∈Xn,j be such that u=(U)sig and v=(V)sig, then (TU−1V−1)sig=1 and so TU−1V−1∈Xn,1≤⟨Xn,i,Xn,j⟩. It therefore follows that Xn,r≤⟨Xn,i,Xn,j⟩. Proposition 7.7 guarantees that ⟨Xn,i,Xn,j⟩≤Xn,r Thus by Proposition 8.11[r] is the lowest common multiple of [i] and [j].
∎
We conclude our investigation of the nesting properties of the groups TOn,r for 1≤r<n by making use of Proposition 7.7 to construct an element of TO4 which is not an element of TO4,r for any 1≤r≤3. This indicates that in general the group TOn,r depends on r.
Let g be the transducer below:
We make the following observations about g which the reader may verify:
(i)
For all states α of g the map gα:Cn→Cn preserves the lexicographic ordering on Cn. Thus, since g2=\mboxid under the product defined for TOn, by Theorem 6.16 there is some non-zero r∈Z4\ such that g∈TOn,r.
2. (ii)
We have \mboxim(a)∩\mboxim(b)=∅ and \mboxim(a)⊔\mboxim(b)=C4. In particular we have \mboxim(a)=U0∪U1 and \mboximb=U2∪U3.
3. (iii)
The following fact is not essential to our discussion here but is nevertheless worth mentioning. The paper forthcoming article [1] shows that there is subgroup Ln≤On which is isomorphic to the quotient \mboxAut(XnZ,σn)/⟨σn⟩ of the automorphisms of the two-sided full shift on n letters by the group generated by the shift (σn denotes the shift map on n letters). The transducer g is in fact an element of L4.
Lemma 8.15**.**
Let 1≤r≤3 and let Aq0∈TB4,r be such that Core(Aq0) is equal to the transducer g of Figure 1, then r=3.
Proof.
We compute (g)sig. First observe that g is synchronizing at level 1 and moreover the states of g forced by [math] and 2, q0 and q2 respectively are both equal to a and the states of g forced by 1 and 3, q1 and q3 respectively are both equal to b. Further observe that ma=2 and mb=2 (where ma and mb are as in Definition 7.6). Therefore (g)sig=2.4=8≡2mod3. The only number r∈Z4\{0} such that 2r≡rmodn−1 is 3. Therefore by Proposition 7.7 we conclude that if g∈TB4,r then r must be equal to 3.
∎
In the next section we address the question of the surjectivity of the homomorphism sig.
9 On the surjectivity of sig
In this section we show that there are infinitely many numbers n∈N such that the map sig from On to the group of units of Zn−1 is surjective. More specifically, for n∈N we show that for every divisor d of n, 1≤d<n, there is an element T∈On with (T)sig=d. We then argue that there are infinitely many numbers n∈N such that the divisors of n generate the group of units of Zn−1.
We first define some subgroups of On highlighted in the paper [3] and outline an algorithm, ‘the collapsing procedure’, given in that paper for determining whether or not an automata (or transducer) is synchronizing.
Definition 9.1**.**
Let Ln be the subgroup of On consisting of all elements of T∈On, such that if q∈QT and Γ∈Xn∗ satisfies, πT(Γ,q)=q, then ∣λT(Γ,q)∣=∣Γ∣. Let Hn be the subgroup of Ln consisting of synchronous, invertible transducers.
Let A be an automaton, the collapsing procedure constructs a new automaton A1 as follows. For each q∈QA set [q]:={p∈QA∣πA(x,p)=πA(x,q)\mboxforallx∈Xn} and set QA1:={[q]∣q∈QA}. Let πA1:Xn×QA1→QA1 be defined by πA(x,[q])=[πA(x,q)] for all x∈Xn and let A1=⟨Xn,QA1,πA1⟩. Let A=A0, A1,A2… be a sequence of automaton, where Al+1, for l∈N is obtained from Al by applying the collapsing procedure. We observe that for l∈N either ∣Al∣>∣Al+1∣ or Al=Al+1. Thus let k be minimal in N such that Ak=Ak+1. The automaton A is synchronizing if and only if Ak is the single state automaton over Xn ([3]). If T=⟨Xn,QT,πT,λT⟩ is a transducer, then T is synchronizing if and only if the automaton A(T)=⟨Xn,QT,πT⟩ is synchronizing. Typically we when applying the collapsing procedure, we shall not distinguish between T and A(T).
We have the following general construction, which is in some sense an extension of a construction given in the paper [18] for embedding direct sums of the group of automorphisms of one-sided shift over alphabets sizes summing to n into the group of automorphisms of the one-sided shift on n letters.
Let n∈N2 and d be a divisor of n not equal to n. Let m∈N be such that md=n and fix an element T∈Hd. For 0≤i≤m−1, partition the set Xn into sets Xn,i:={di,di+1,…,d(i+1)−1}, likewise form sets QT,i={q(i)∣q∈QT}. Form a transducer ⊕dT=⟨Xn,∪0≤i≤m−1QT,i,π⊕dT,λ⊕dT⟩ with transition and output defined such that:
(1.)
the restriction ⊕dT↾QT,i=⟨Xn,i,QT,i,π⊕dT↾QT,i,λ⊕dT↾QT,i⟩ of ⊕dT to the set of states QT,i is a transducer equal to T up to relabelling the alphabet and the set of states,
2. (2.)
for 0≤i,j≤m−1 such that i=j, for any b∈Xd, and for any q∈QT, π⊕dT(dj+b,q(i))=(qb)(j), where πT(b,qb)=b, and λ⊕dT(dj+b,q(i))=di+b.
We have the following result:
Propostion 9.2**.**
Let n∈N2, d be a divisor of n such that n=md for some m∈N2, and T∈Hd. Then, the transducer ⊕dT is bi-synchronizing, and is fact an element of Ln. Moreover, for 0≤i≤m−1, and any state q(i)∈QT,i, ⋃0≤b≤d−1Udi+b=\mboxim(qi). In particular, (⊕dT)sig=d.
Proof.
We begin by arguing that the transducer ⊕dT is synchronizing. However, this follows straight-forwardly from the observation that for 0≤i,j≤m−1 such that i=j, any pair of states q(i),p(i)∈QT,i, and any 0≤b<d, π⊕dT(dj+b,q(i))=π⊕dT(dj+b,p(i))=(qb)(i). Thus since, each transducer ⊕d(T)↾QT,i for 0≤i≤m−1 at the same level k∈N, then after applying the collapsing procedure at most k+1 times we have reduced ⊕dT to the single state automaton.
Further observe that as T is in fact a synchronous transducer, then if T∈On it is in fact an element of Ln.
Therefore in order to show that ⊕dT∈Ln it suffices to show that each state is injective and has clopen image and that it has a synchronizing inverse.
We begin by showing that each state of ⊕dT is injective and has clopen image. Let 0≤i≤m−1 and q(i) be a state of T. Let μ,ν∈Xn+ such that ∣μ∣=∣ν∣≥2 and μ=ν. We now show that either λ⊕dT(μ,q(i))=λ⊕dT(ν,q(i)) or for any x∈Xn, λ⊕dT(μ,q(i))=λ⊕dT(ν,q(i)). If μ,ν∈Xn,i+, then this follows from the fact that T∈Hn and so each state of T induces a permutation of Xd. Thus, we may assume that there are 1≤j1,j2≤m−1 with j1 and j2 not both equal to i, a1∈Xn,j1, a2∈Xn,j2, and μ1,ν1∈Xn+ such that μ=a1μ1 and ν=a2ν1. If j2=j2, then as π⊕dT(a1,q(i))∈QT,j1 and π⊕dT(a2,q(i))∈QT,j2, it follows that the first letter of λ⊕dT(μ1,π⊕dT(a1,q(i))) is an element of Xn,j1 and the first letter of λ⊕dT(ν1,π⊕dT(a2,q(i))) is an element of Xn,j2. Thus λ⊕dT(μ,q(i))=λ⊕dT(ν,q(i)). Therefore we may assume that j1=j2=j. In this case, since μ⊥ν, there are words μ2,ν2,ϕ∈Xn∗ and a=b∈Xn such that μ=ϕaμ2 and ν=ϕbν2. If a,b∈Xn,l for some 0≤l≤m−1, then since for any state t∈∪0≤i≤m−1QT,i, λ⊕dT(a,t)=λ⊕dT(b,t), we have λ⊕dT(μ,q(i))=λ⊕dT(ν,q(i)). Thus, we may assume that a∈Xn,l1, b∈Xn,l2, by adding a letter to μ2 and ν2 if necessary, we may assume they are non-empty. In this case, the fact that λ⊕dT(μ,q(i))=λ⊕dT(ν,q(i)) follows from the observation above that for any state t∈∪0≤i≤m−1QT,i, the first letter of λ⊕dT(μ2,π⊕dT(a,t)) is an element of Xn,l1 and the first letter of λ⊕dT(ν2,π⊕dT(b,t)) is an element of Xn,l2. Therefore, since T is synchronous, each state must induce an injective map from Cn to itself.
Let q(i)∈QT,i be an arbitrary state of T, b∈Zd be arbitrary. We now show, by induction, that Udi+b⊂\mboxim(qi).
Let j∈Zm and a∈Zd be arbitrary so that jd+a∈Xn,j⊂Xn is arbitrary. There is a letter x∈Xn,j such that λ⊕dT(x,q(i))=di+b and π⊕dT(x,q(i))∈QT,j. Since π⊕dT(x,q(i))∈QT,j, by construction, for any l∈Zm, there is a letter y∈Xn,l such that λA(y,π⊕dT(x,q(i)))=jd+a and π⊕dT(y,π⊕dT(x,q(i)))∈QT,l. Thus, for any j∈Zm, a∈Zd and any l∈Zm, there is a word xy∈Xn2 such that π⊕dT(xy,q(i))∈QT,l and λ⊕dT(xy,q(i))=(di+b)(jd+a).
Assume by induction that there for any ν∈XnM and any l∈Zm, there is a word μ∈XnM+1, such that λ⊕dT(μ,q(i))=(di+b)ν and π⊕dT(μ,q(i))∈QT,l. Let ξ∈XnM+1 be arbitrary. Let j∈Zm, a∈Zd be such that ξ=χ(jd+a) for some χ∈XnM. By the inductive assumption, there is a word ζ∈XnM+1 such λ⊕dT(ζ,q(i))=(di+b)χ and π⊕dT(ζ,q(i))=p(j)∈QT,j. Now by construction, for any l∈Zm, there is a word y∈Xn,l such that λ⊕dT(y,p(j))=dj+a and π⊕dT(y,p(j))∈QT,l. Therefore, we have that for any l∈Zm, there is a y∈Xn,l such that λ⊕dT(ζy,q(i))=(di+b)χ(dj+a) and π⊕dT(ζy,q(i))∈QT,l. Therefore we see, by transfinite induction, that Udi+b⊂\mboxim(qi). Therefore every state of T is injective and has clopen image.
We now show that ⊕dT has an inverse in On. We do this by showing that for any transducer Aq0, with Core(Aq0)=⊕dT, then set of states of A(ϵ,q0) which are reached by arbitrarily long words, form a synchronizing transducer. To this end, fix a transducer Aq0 with Core(Aq0)=⊕dT.
Let 0≤i≤m−1 and q(i)∈QT,i, we observe that for a∈Zd, (di+a)Lq(i)=ϵ. Set QU={(di+a,q(i))∣i∈Zm,a∈Zd}. Define λU:Xn×QU→Xn by λU(dj+b,(di+a,q(i)))=((di+a)(dj+b))Lq(i) for all i,j∈Zm and a,b∈Zd. Further define πU:Xn×QU→QU by πU(dj+b,(di+a,q(i)))=((di+a)(dj+b))Lq(i)=((di+a)(dj+b)−λA(((di+a)(dj+b))Lq(i),q(i)),πA(((di+a)(dj+b))Lq(i),q(i))). Now, observe that ((di+a)(dj+b))Lq(i) is precisely the element x∈Xn,j such that λ⊕dT(x,q(i))=di+a, thus if p(j)=π⊕dT(x,(q(i))), then πU(dj+b,(di+a,q(i)))=(dj+b,p(j)). Let U=⟨Xn,QU,πU,λU⟩. We now show that U is synchronizing, then we argue that the set of states of A(ϵ,q0) reached by arbitrarily long words forms a transducer precisely equal to U.
We partition the sets of U as follows: for 0≤i≤m−1 and a∈Zd, set QU,i,a:={(di+a,q(i))∣q(i)∈QT,i}. Fix 0≤i≤m−1 an a∈Zd.
Let 0≤j≤n−1 such that i=j and (di+a,q(i)),(di+a,p(i))∈QU,i,a and dj+b, b∈Zd, be arbitrary. We observe that, by construction, if x,y∈Xn,j satisfy λ⊕dT(x,q(i))=λ⊕dT(y,p(i))=di+a, then x=y=dj+a. Observe, moreover, that π⊕dT(dj+a,q(i))=π⊕dT(dj+a,p(i))=(qa)(j). Therefore it follows that πU(dj+b,(di+a,q(i)))=(dj+b,(qa)(j))=πU(dj+b,(di+a,p(i))). Thus we conclude that for any 0≤i≤m−1 and any a∈Zd, the set of elements QU,i,a:={(di+a,q(i))∣q(i)∈QT,i} transition identically on all elements of Xn,j for j=i.
Now let, T−1=⟨Xn,QT−1,πT−1,λT−1⟩, a synchronous and synchronizing transducer, denote the inverse of T and denote by (a)q−1, for a∈Zd, the element of Xd such that λT((a)q−1,q)=a. Consider a letter di+c, we observe that πU(di+c,(di+a,q(i)))=(di+c,π⊕dT(di+(a)q−1,q(i))). Note moreover that πT−1(a,q−1)=(πT((a)q−1,q))−1. Thus,
[TABLE]
A simple induction argument now shows that given a word ν=v1v2…vl∈Xd+, with corresponding word ν(i)=(di+v+1)(di+v2)…(di+vl)∈Xn,i+, if πT−1(ν,q−1)=p−1, for a state q∈QT, then, setting ν(i)1=(di+v2)…(di+vl), πU(ν(i)1(di+b),(di+v1,q(i)))=(di+b,p(i)). Let k be the synchronizing level of T−1, it therefore follows that, all elements of QU,i,a transition identically on all words in Xn,ik.
Finally, since by an observation above we have that for any state (dj+b,q(j))∈QU, and any letter dl+c∈XnπU(dl+c,(dj+b,q(j)))∈QU,l,c, it therefore follows that U is synchronizing at level k+1.
To conclude the proof we have to demonstrate that (⊕dT)−1=U. We do this by showing that A(ϵ,q0) is synchronizing and has core equal to U. Let q(i) be an arbitrary state of ⊕dT. Since Aq0 is synchronizing, there is a word Γ∈Xn+ such that Δ=λA(Γ,q0)=ϵ and πA(Γ,q0)=q(i). It therefore follows that, for 0≤a≤d−1, (Γ(di+a))Lq0=Δ(di+a)Lq and so πA′(Γ(di+a),(ϵ,q0))=(di+a,q(i)). Thus we see that A(ϵ,q0) is synchronizing and has Core(A(ϵ,q0))=U. Therefore T−1=U. This concludes the proof.
∎
Corollary 9.3**.**
Let n∈N2, then the image of the map sig from On to Zn−1∗, the group of units of Zn−1, contains the subgroup of Zn−1∗ generated by the divisors of n. ∎
Corollary 9.4**.**
Let n∈N2 be such that the group of units of Zn−1 is generated by the divisors of n, then the map sig:On→Zn−1∗ is surjective. ∎
Corollary 9.5**.**
There are infinitely many number n∈N such that the map sig from On to the group of units of Zn−1 is surjective.
Proof.
It is an elementary result in number theory that for an m≥1, 2 generates the group of units of 3m, thus we may take n=3m+1.
∎
Thus, we are left with the question:
Question 9.6**.**
for n∈N2 and p in the group Zn−1∗ of units of Zn−1 such that p is not an element of the subgroup of Zn−1 generated by the divisors of n, is there an element T∈On such that (T)sig=p?
We observe that in order to answer this question, it suffices to show that for any element p∈Zn−1 co-prime both to n and n−1, there is an element T∈On such that (T)sig=p.
10 The group TOn,r and TBn,r for n>2 and 1≤r<n contain an isomorphic copy of Thompson’s group F
We now show that for n>2, n∈N, and for all valid r, the group TOn,r is infinite. This extends the result of Brin and Guzmán stating that the group \mboxOut(Tn,n−1) is infinite whenever n>3 ([5]) to the groups \mboxOut(Tn,r). We observe that the group \mboxOut(T2) is the cyclic group of order 2 and so finite. By Theorem 7.11 it suffices to demonstrate that TOn,1 is infinite. First we have the following straight-forward lemma.
Lemma 10.1**.**
Let T∈TOn be an element with a homeomorphism state, then T∈TOn,1.
Proof.
Let q∈QT be an homeomorphism state of T, then the initial transducer Tq is an element of TBn,1.
∎
Therefore to show that TOn,1 is infinite, for n>2, it suffices to find an element of infinite order of TOn which has a homeomorphism state.
Let T be the element of TOn depicted in Figure 2. In this figure we use the symbol x to represent an element of Xn strictly greater than zero and less than or equal to n−2.
The states a and b of T are homeomorphism states, and so T is in fact an element of TOn,1.
In order to show that T has infinite order we make use of the action of the group On, as introduced in the paper [3], on the space XnZ/⟨σn⟩ where σn is the shift map on XnZ. Let U∈On and suppose that k∈N is the minimal synchronizing level of U. The action of U on XnZ/⟨σn⟩ is given as follows: let y=…y−1y−1y0y1y2⋯∈XnZ represent an equivalence class of XnZ/⟨σn⟩, the image of this class under U is the equivalence class of the bi-infinite word w defined as follows: for i∈Z, let y(i+1,k)=yi+1yi+1…yi+k and qy(i+1,k) be the state of U forced by y(i+1,k), then set wi=λU(yi,qy(i+1,k) and let w=…w−2w−1w−1w0w1w2…. Consider the bi-infinite word z:=…(xn−1)(xn−1)… where x∈Xn\{0,n−1}, then (z)T=…(xn−1n−1)(xn−1n−1)…, (z)T2=…x(n−1)3x(n−1)3…, and (z)Ti=…x(n−1)i+1x(n−1)i+1…. Therefore we see that z is on an infinite orbit under the action of T, demonstrating that T is an element of infinite order. We have shown the following:
Theorem 10.2**.**
For n>3 and 1≤r≤n−1, the group Tn,r is infinite.
Let U be the following transducer where, once more, n>2 and x∈Xn is any element strictly bigger than [math] and strictly less than n−1:
The state p of U is a homeomorphism state and moreover U is an element of TOn,1 of infinite order.
In order to state our next result we require the following notion and result from [3].
Definition 10.3**.**
Given an element g=⟨Xn,Qg,πg,λg⟩∈On a state q of g is called a loop state if there is some i∈Xn such that πg(i,q)=q.
Let g=⟨Xn,Qg,πg,λg⟩∈On and let w∈Xn+ then there is a unique state qw∈Qg such that πg(w,qw)=qw. Moreover for any periodic equivalence class of the tail equivalence ∼t with minimal period w there exists a unique j∈N1 and a unique circuit, decorated by some prime word v, in A with output w.
Lemma 10.5**.**
The elements T and U of TOn,1 depicted in Figures 2 and 3 generate a subgroup of TOn,1 isomorphic to R. Thompson’s group F.
Proof.
First we make some observations about T and U. Let q be a state of T or of U,let x∈Xn\{0,n−1}, and let (D,d)∈{(T,a),(U,p)}, then all transitions πD(x,q)=d and λD(x,q) end in the symbol x. The output of all other transitions in T or U do not involve the symbol x. Therefore the transducers A and B below are in fact sub-transducers of T and U respectively.
Notice that Aa and Bp induce self-homeomorphisms of the Cantor space {0,n−1}ω equal, respectively, to the restrictions of the homeomorphisms Ta and Up to the space {0,n−1}ω. It is not hard to see that F≅⟨Aa,Bp⟩≤H({0,n−1}ω). In fact Aa−1 and Aa−2BpAa are the standard generators for F acting on the space {0,n−1}ω. Moreover Aa and Bp satisfy the defining relations for F, and so we see that ⟨Aa,Bp⟩=⟨Aa,Bp∣[Bp−1Aa,AaBpAa−1],[Bp−1Aa,Aa2BpAa−2]⟩. Moreover Ta−1↾{0,n−1}ω=Aa−1 and Up−1↾{0,n−1}ω=Bp−1. We also observe (one can verify this by direct computation) that the transducers T−1 and U−1 representing the inverses of T and U in TOn respectively, have states a−1,b−1 of T−1 and p−1,s−1 of U−1 such that Ta−1−1=Ta−1, Ts−1−1=Ts−1, Up−1−1=Up−1 and Us−1−1=Us−1.
We now define a homomorphism from ⟨T,U⟩ to ⟨Aa,Bp⟩. To do this we make the following observation. Let W(T,U)=w1w2…wl be a word in {T,U,T−1,U−1}+. For each letter wi of W(T,U) let αi be the state of wi so that αi is the unique loop state of x in wi. Observe that, for all 1≤i≤l, αi∈{a,a−1,p,p−1}. Let SW(T,U) be the element of TOn,1 representing the word W(T,U)∈⟨T,U⟩. Under the product defined on TOn,1, we have that SW(T,U) is ω-equivalent to the core of the minimal transducer representing the product of the initial transducers (w1∗w2∗…∗wl)(α1,α2,…,αl). However, since αi is the unique loop state state of x in wi, λwi(x,αi) ends in x and for any state q of T, U, T−1 or U−1, the resulting state when x is read from q is the appropriate element of the set {a,a−1,p,p−1}, we must have that the state (α1,α2,…,αl) is the unique loop state of x in the product (w1∗w2∗…∗wl)(α1,α2,…,αl). Therefore, since the core is strongly connected, (w1∗w2∗…∗wl)(α1,α2,…,αl) is equal to its core. Therefore if s is the unique loop state of x in SW(T,U), we must have that (SW(T,U))s is ω-equivalent to (w1∗w2∗…∗wl)(α1,α2,…,αl). Now let D(A,B)=d1d2…dl be the word in {Aa,Bp,Aa−1,Bp−1}+ defined as follows, for all 1≤i≤l, if wi=T±1 then di=Aa±1 and if wi=U±1 then di=Bp±1. Let CD(A,B) be the element of ⟨Aa,Bp⟩ representing the word D(A,B), then it follows that CD(A,B)=(SW(T,U))s↾{0,n−1}ω since for all i, (wi)αi↾{0,n−1}ω=di. Therefore the map ϕ:⟨T,U⟩→⟨Aa,Bp⟩ by S↦Ss↾{0,1}ω, where s is the unique loops state of x in S, is a surjective homomorphism.
In order to conclude that ⟨T,U⟩≅⟨Aa,Bp⟩ it suffices to show that the map φ which maps Aa to T and Bp to U extends to a group homomorphism φ:⟨T,U⟩→⟨Aa,Bp⟩. To see this, it suffices to show that [U−1T,TUT−1]=1 and [U−1T,T2UT−2]=1. This is easily verified by direct computation.
∎
Remark 10.6**.**
Notice that, for T and U as in Figures 2 and 3, the proof above shows that the subgroup ⟨Ta,Up⟩≤TBn,1, for n≥3 is isomorphic to F.
As a corollary we have the following theorem generalising the result of Brin and Guzman ([5]) stating that the groups \mboxOut(Tn,n−1) and \mboxAut(Tn,n−1) contain an isomorphic copy of F.
Theorem 10.7**.**
For n≥3 and 1≤r≤n−1, the groups TOn,r≅\mboxOut(Tn,r) and TBn,r≅\mboxAut(Tn,r) contain an isomorphic copy of Thompson’s group F.
Proof.
That TOn,r, for n≥3 and 1≤r≤n−1, contains a group isomorphic to F is a consequence of Theorem 7.11. To deduce implications for TBn,r, let T and U be the transducers in Figures 2 and 3. Observe that the map f defined by, for all d˙∈r˙ and ξ∈Cn, d˙ξ↦d˙(ξ)Ta, and the map g given by, for all d˙∈r˙ and ξ∈Cn, d˙ξ↦d˙(ξ)Up, are elements of TBn,r since Core(f)=T and Core(g)=U. Moreover, they generate a subgroup isomorphic to F by Remark 10.6.
∎
11 Further properties of TOn and the R∞ property for Tn
In this section we deduce some properties of elements of TOn which will enable us to demonstrate that the groups Tn,r for 1<r<n have the R∞ property. The case n=2 has been dealt with in the paper [7].
Throughout this section we shall identify elements of [0,r], 1≤r≤n−1, with their n-ary expansions.
Remark 11.1**.**
A consequence of Lemma 10.4 is that for g∈On, and letting Xn+ be the set of equivalences classes of Xn+ under the relation that two words are related if they are rotations of each other, there is a bijection gˉ:Xn+→Xn+ by [w]↦[λg(w,qw)]. Let \mboxSym(Xn+) be the group consisting of bijections from the set Xn+ to itself. The map from On to \mboxSym(Xn+) given by g↦gˉ is an injective homomorphism from On into \mboxSym(Xn+) (this follows from results in [3] and the forthcoming article [1]). We may fix a choice of representatives for the equivalence classes of Xn+ such that, for all i∈Xn, [i]=i.
The following Lemma explores properties of the map gˉ when g∈TOn.
Lemma 11.2**.**
Let g∈TOn then [math] and n−1 are fixed by the map gˉ.
Proof.
We proceed by contradiction. We first consider the case that (0)gˉ=0. By Lemma 10.4 there is some prime word w∈Xn+ such that (w)gˉ=0. Let qw be the unique state such that πg(w,qw)=w. Then since hqw preserves the lexicographic ordering on Cn we must have that (00…)hqw<\mboxlex(ww…)hqw however as (ww…)hqw=00… and hqw is injective, we have the desired contradiction.
The case (1)gˉ=1 is dealt with analogously.
∎
By applying Lemma 6.12 we immediately deduce the following result:
Lemma 11.3**.**
Let g∈TOn then if g∈TOn, (0)gˉ=0 otherwise (0)gˉ=n−1.
As a corollary we have a different proof that for Aq0∈TBn,r, ([0,r]∩Z[1/n])hq0=[0,r]∩Z[1/n].
Corollary 11.4**.**
Let Aq0∈TBn,r then ([0,r]∩Z[1/n])hq0=[0,r]∩Z[1/n]
Proof.
Observe that since Aq0 is synchronizing then after reading a sufficiently long string of [math]’s or n−1’s the next input if processed from the state q0 or qn−1 respectively. However these states either swap [math] with n−1 and vice versa or fix [math] and n−1 (depending on whether or not Aq0 is orientation preserving or reversing) by Lemma 11.3. Hence since elements of [0,r]∩Z[1/n] are precisely those elements of Cn,r of the form w00… or wn−1n−1… for w∈Xn,r+, the result follows.
∎
From this we may deduce the following:
Lemma 11.5**.**
Let Aq0∈TBn,r then there is an element h∈Tn,r such that hq0∘h fixes the point 0˙0…≃r−1˙n−1….
The following constructions demonstrate that Tn,r has the R∞ property for any 1≤r<n. and naturally breaks up into two cases. We first require the following standard lemma.
Lemma 11.6**.**
Let Aq0,Bp0∈TBn,r and g∈Tn,r be such that hq0g=hp0 then Tn,r has infinitely many hq0-twisted conjugacy classes if and only if it has infinitely many hp0-twisted conjugacy classes.
Proof.
Let f1 and f2 be elements of Tn,r such that f1 is hp0-twisted conjugate to f2 by an element f3. The computation below demonstrates that f1g−1 and f2g−1 are hq0-twisted conjugate to each other by the element f3.
[TABLE]
In a similar way we see that if f1′ and f2′ are elements of Tn,r such that f1′ and f2′ are hq0-twisted conjugate to each other by an element f3′, then f1′g and f2′g are hp0 twisted conjugate to each other by f3′.
∎
We also observe that given Aq0∈TBn,r, by straight-forward manipulations, there are infinitely many hq0-twisted conjugacy classes if and only if there is an infinite subset J of Tn,r such that for any pair f1,f2∈J, f1hq0−1 is not conjugate to f2hq0−1 by an element of Tn,r.
Orientation preserving case: First let Aq0∈TBn,1 be an orientation preserving element. By Lemma 11.6 and Lemma 11.5 we may assume that Aq0 fixes the points 00… and n−1n−1… of Cn. Let uˉ be a complete antichain of Xn∗ of length l>0. Suppose that uˉ={u0,u1,…,ul−1} (recall that antichains are always assumed to be ordered lexicographically). We form a homeomorphism of Cn as follows. Let σ:uˉ→uˉ be given by ui↦u(i+1)modl for 1≤i≤l−1. Let hl:Cn→Cn be given by (uiδ)hl=(ui)σ(δ)hAq0. Then since Aq0 is a homeomorphism of Cn, hl is also homeomorphism of Cn. Moreover, as hq0 is orientation preserving, and preserves the relation ≃I, by choice of the permutation σ it follows that hl is an orientation preserving element of TBn,1. Moreover, if Bp0 is the transducer such that hp0=hl, then Core(Bp0)=Core(Aq0). Thus there is an element of f∈Tn,1 such that hlf=hq0. Furthermore, observe that as Aq0 fixes 00…, we have that the point 00… is on an orbit of length precisely l under hl. Moreover, any point on a finite orbit under hl must have orbit length a multiple of l. Thus for every l∈N such that l≅1modn−1, let hl∈TBn,1 be the map constructed as above, and let fl be such that hl=flhq0. Therefore we conclude that for any Aq0∈TBn,1 there are infinitely may hq0-twisted conjugacy classes.
Now fix r>1, and let rˉ be minimal such that TOn,rˉ=TOn,r, by Lemma 7.13rˉ divides r and rˉ divides n−1. Let Aq0∈TBn,rˉ be an element which fixes 0˙00… and r−1˙n−1n−1…. Let uˉ be a complete antichain of Xn,r∗ of length l. Observe that as l≅rmodn−1 hence there is an m∈N such that mrˉ=l. For 1≤i≤m let uˉi:={ui,0,…,ui,r−1}⊂uˉ be such that for 1≤i1<i2≤m all elements of uˉi1 are less than (in the lexicographic ordering) all elements of uˉi2. Let σ:{1,2…,m}→{1,2…,m} be given by (i)σ=i+1modm. We construct an element hl∈TBn,r as follows: for Γ∈Cn and 1≤i≤m and 0≤a≤r−1 let (ui,aΓ)hl=u(i)σ,bδ if and only if (a˙Γ)Aq0=b˙δ. Therefore since Aq0 induces an orientation preserving homeomorphism on Crˉ, it follows that the ma hl induces a orientation preserving homeomorphism of Cn,r. Moreover since Core(Aq0)∈TOn,r it follows that hl is in fact and element of TOn,r and moreover (identifying hl with the initial transducer inducing inducing the homeomorphism hl) Core(hl)=Core(Aq0). Lastly observe that the point 0˙00… is on an orbit of length m under the action of hl and any other finite orbit of hl has length at least l.
Now let Aq0∈TBn,r be arbitrary. Since Core(Aq0)∈TOn,rˉ=TOn,r, there is an element Bp0∈TBn,rˉ such that Core(Bp0)=Core(Aq0). As before for each l∈N such that l≡rmodn−1, there is an element hl with a minimal finite orbit of length l and such that Core(hl)=Core(Aq0). Thus, there is a fl∈Tn,r such that hq0fl=hl. Therefore as in the case r=1, we conclude that for any there are infinitely many hq0-twisted conjugacy classes for any element Aq0∈TBn,r.
Orientation reversing case: We now consider the orientation reversing case. Let As0∈TBn,1 be orientation reversing. By Lemma 11.5 we may assume that (000…)As0=n−1n−1n−1… and (n−1n−1n−1…)As0=n−1n−1n−1…. Let i be minimal such that πA(0i,s0)=q0 and πA((n−1)i,s0)=qn−1 (where q0 and qn−1 are the zero and one loop state of As0 respectively). Let (n−1)l1=(0i+1)Aq0 and 0l2=((n−1)i+1)Aq0. Observe that 0<min{l1,l2} since λA(0,q0)=n−1 and λA(n−1,q0)=n−1 by Lemma 11.3. There is an element h∈Tn such that for any δ∈Cn, ((n−1)l1δ)h=(n−1)iδ and (0l2δh)=0iδ. Let Cs0′∈TBn,r be such that hs0′=hs0h. Observe that Cs0′ satisfies, πC(0i,s0′)=q0′, πC((n−1)i,s0′)=qn−1′, where q0′ and q(n−1)′ are the [math] and n−1 loop states of Cs0′ respectively, and, for j≥i, λC(0j,s0′)=(n−1)j and λC((n−1)j,s0′)=0j. As before we may replace As0 with Cs0′ without loss of generality, since there infinitely many As0-twisted conjugacy classes if and only if there are infinitely many Cs0′-twisted conjugacy classes. Therefore we may assume that there is an i∈N, i>0 such that As0 satisfies πA(0i,s0)=q0, πA((n−1)i,s0)=qn−1, where q0 and q(n−1) are the [math] and n−1 loop states of As0 respectively, and, for j≥i, λA(0j,s0)=(n−1)j and λA((n−1)j,s0)=0j.
We now construct an element g∈Tn such that ghs0 such (ghs0)2 has the point 000… and n−1n−1… as attracting fixed points.
By a branch of length k, for k∈N, k>0, we shall mean the finite rooted n-ary tree with 1+k(n−1) leaves and k the length of the geodesic from the root to the left-most leaf.
Let fˉ∈Tn be defined as follows. Let E and F be the finite trees where the tree E is obtained from the single caret, by attaching a branch of length i+1 to the first and penultimate leaves of the branch and the tree F is obtained from the single caret by attaching a branch to the first and last leaves of the caret. Label the leaves of E left to right by 1,2,…,1+(2k+1)(n−1) and the leaves of F from right to left by 1,2,…,1+(2k+1)(n−1). The map fˉ is now defined by, for Γ∈Cn and μj and νj the addresses of the j’th leaves of E and F respectively, μjΓ↦νjΓ.
Let f∈TBn,1, be defined by, for Γ∈Cn, μjΓ↦νj(Γ)As0 where μj and νj are the addresses of the j’th leaves of E and F respectively. Essentially f is obtained from fˉ by attaching the state s0 to all the leaves of F the range tree of fˉ. Moreover since Core(f)=Core(As0) there is an element g∈Tn such that f=gAs0.
Below we shall consider the orbit of cones Uν for ν∈Xn+ under f, and to simplify notation, we identify the cone Uν with the finite word ν. Let us consider the orbit of the cone (n−1) (a leaf of E) under f:
[TABLE]
This is because for (α,αˉ)∈{(0,n−1),(n−1,0)} and j∈Nj≥i, λA(αj,s0)=αˉj and πA(αj,s0)=qα. Hence we see that the point n−1n−1… is an attracting fixed point of f2 and in a similar way the point 00… is an attracting fixed point of f. In particular we have, for j∈N1, (n−1)f2j=(n−1)(n−1)ji and (00i)f2j=00i0ji. We call the words (n−1)i and 0i the attracting paths (in f2) of 00… and n−1n−1… respectively.
Let h∈Tn be arbitrary and consider h−1fh. Now as h is a prefix exchange map, by an abuse of notation for a finite word Γ∈Xn∗ such that h acts as a prefix exchange on the cone UΓ⊂Cn, we shall write, (Γ)h for the word Δ∈Xn∗ such that (UΓ)h=UΔ. There is an l∈N such that ((n−1)l)h is a finite word in Xn+ and we may assume that l≥i+1. Thus we have,
[TABLE]
Hence we deduce that ((n−1)l)hh−1f2jh=((n−1)l)h(n−1)ji and we see that conjugation of f by an element of Tn maps the attracting fixed point n−1n−1… to an attracting fixed point and preserves the path (n−1)i. We also observe that since an orientation reversing homeomorphism of the circle has precisely two fixed points (on the circle), then h must map the pair of fixed points of f to the pair of fixed of h−1fh. Notice that thinking of the fixed points as elements of Cn, since Tn elements Z[1/n]→Z[1/n], this means that either h fixes the points 00… and n−1n−1… otherwise, the other fixed point of f is an element of Z[1/n] of the form η1≃Iη2 (where η1>η2 in the lexicographic ordering on Cn), (00…)h=η1 and (n−1n−1…)h=η2. We make use of these facts to show that there are infinitely many hs0-twisted conjugacy classes.
Let j be any integer greater than i, as before we may construct a map fj∈TBn,1 such that fj=gjhs0 for some gj∈Tn, and 00… and n−1n−1… are attracting fixed points of fj2 with attracting paths 0j and (n−1)j. We have the following claim:
Claim 11.7**.**
There is an infinite subset J⊂Ni for which the set {fj∣j∈J} are pairwise not conjugate by an element of Tn.
Proof.
Suppose there is an N∈N such that every element fj, j∈Ni is conjugate to some element of {fi+1,…fi+N} by an element of Tn. Since Ni is infinite, there is an infinite subset I⊂Ni and 1≤M≤N such that every element of {fl∣l∈I} is conjugate to fi+M by an element of Tn. Let l>i+M. Since conjugation by an element of Tn preserves attracting paths and l>M it must be the case that any conjugator h∈Tn such that h−1flh=fi+M must satisfy (n−1n−1…)h=n−1n−1…. Hence the other fixed point of fi+m must be a dyadic rational of the form η1≃Iη2∈Cn/≃I where η1 is greater than η2 in the lexicographic ordering of Cn. This means by arguments above that (n−1n−1…)h=η2 and η2 is an attracting fixed point of fi+M with attracting path (n−1)i. However for l′∈I such that l′>l, it must be the case that fl′ is not conjugate to fi+M since the attracting path in fl′2 of n−1n−1… is longer than the attracting path in fi+M2 of both attracting fixed points of fi+M.
∎
Now fix 1<r<n. Let As0∈TBn,r be an orientation reversing element, rˉ be minimal such that TOn,rˉ=TOn,r and Bp0 be an orientation reversing element of TBn,rˉ with Core(Bp0)=Core(As0). As in the case r=1 we may assume that there is an i∈N, i>0 such that Bp0 satisfies πB(0˙0i,p0)=q0, πB(r−1˙(n−1)i,p0)=qn−1, where q0 and q(n−1) are the [math] and n−1 loop states of Core(As0)=Core(Bp0) respectively, and, for j≥i, λB(0˙0j,p0)=r−1˙(n−1)j and λB(r−1˙(n−1)j,p0)=0˙0j. Now as in the orientation preserving case we construct elements fj, for j∈Ni with desirable properties by simulating the element Bp0 appropriately on cones.
Let uˉ={a˙∣1≤a≤r}, that is, uˉ corresponds to the roots of the disjoint union of r copies of the n-ary tree. Since rˉ divides r there is an M∈N such that Mrˉ=r. Let uˉk:={uk,1,…,uk,r}, 1≤k≤M be subsets of uˉ, such that for 1≤k1<k2<M, all elements of uˉk1 are less than (in the lexicographic ordering) all elements of uˉk2. Observe that uˉ1 corresponds to the roots of the disjoint union of the rˉ copies of the n-ary tree. Replace uˉ1, still retaining the symbol uˉ1 for the resulting antichain, with the antichain corresponding to attaching a branch of length i+1 to the root u1,1. Likewise replace uˉM with the antichain corresponding to attaching a branch of length i+1 to the root uM,r−1. In a similar way let vˉM be the antichain obtained from uˉM corresponding to attaching a branch of length i+1 to the root uˉ1, and let vˉk:=uˉk for 1≤k<M. Observe that uˉ1, uˉM and vˉM all have equal length d congruent to rˉ modulo n−1. Since rˉ divides n−1 let m∈N be such that mrˉ=d. Let uˉ1=∪1≤a≤muˉ1,a where uˉ1,a={u1,a,b∣1≤b≤r}, likewise let uˉM=∪1≤a≤muˉM,a where uˉM,a={uM,a,b∣1≤b≤r} and vˉM=∪1≤a≤mvˉM,a where vˉM,a={vM,a,b∣1≤b≤r}.
We construct a homeomorphism fi of Cn,r as follows. Let σ:{1,2,…,M}→{1,2,…M} by k↦M−k+1 and let ρ:{1,2,…,m}→{1,2,…m} by a↦m−a+1. For Γ∈Cn, 1<k<M and 1≤b≤r, (uk,bΓ)fi=v(k)σ,b′δ if and only if λB(b˙Γ,p0)∈Ub′˙ and δ=λB(Γ,πB(b˙,p0)). For Γ∈Cn1≤a≤m and 1≤b≤r, (u1,a,bΓ)fi=vM,((a)ρ),b′δ if and only if λB(b˙Γ,p0)∈Ub′˙δ=λB(Γ,πB(b˙,p0)). For Γ∈Cn1≤a≤m and 1≤b≤r, (uM,a,bΓ)fi=v1,((a)ρ),b′δ if and only if λB(b˙Γ,p0)∈Ub′˙δ=λB(Γ,πB(b˙,p0)). Since Bp0 induces an orientation reversing homeomorphism on Cr˙, we see that fi is in fact an element of TBn,rr. Moreover, Core(fi)=Core(Bp0)=Core(As0) and so there is an element gi∈Tn,r such that fi=hs0gi.
We now argue that the points 0˙0… and r−1˙n−1n−1… are attracting fixed points of fi with attracting paths 0i and (n−1)i. We consider the orbit of the cone r−1˙(n−1)i:
[TABLE]
this follows by making use of the definition of fi and the facts: πB(0˙0i,p0)=q0, πB(r−1˙(n−1)i,p0)=qn−1, where q0 and q(n−1) are the [math] and n−1 loop states of Core(As0)=Core(Bp0) respectively, and, for j≥i, λB(0˙0j,p0)=r−1˙(n−1)j and λB(r−1˙(n−1)j,p0)=0˙0j. In general we see that, for l∈N1, (r−1˙(n−1)i)fi2l=r−1˙(n−1)i(n−1)l and (0˙0i0i)fi2l=0˙0i0i0l.
For each j>i we may repeat the construction above to get elements hj such that fj=hs0gj for some gj∈Tn,r and the points 0˙0… and r−1˙n−1n−1… are attracting fixed points of fj with attracting paths 0j and (n−1)j. Since fj for j∈Ni induces a homeomorphism on circle of length r, [0,r] with end points identified, we may repeat the arguments as in the case r=1 to conclude that there is an infinite subset J⊂Ni for which the set {fj∣j∈J} are pairwise not conjugate by an element of Tn,r.
Thus we have demonstrated the following:
Theorem 11.8**.**
The group Tn,r for 1≤r<n−1 has the R∞ property.
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