On the Orbits of Automaton Semigroups and Groups
Daniele D'Angeli, Dominik Francoeur, Emanuele Rodaro, Jan Philipp Wächter
aa r X i v : . [ c s . F L ] J u l On the Orbits of Automaton Semigroupsand Groups
Daniele D’Angeli ∗1 , Dominik Francoeur †2 , Emanuele Rodaro andJan Philipp Wächter Università degli Studi Niccolò Cusano, Via Don Carlo Gnocchi, 3, 00166 Roma, Italy Unité de mathématiques pures et appliquées, ENS de Lyon, 46, allée d’Italie, 69364 Lyon Cedex 07, France Department of Mathematics, Politecnico di Milano, Piazza Leonardo da Vinci, 32, 20133 Milano, Italy Institut für Formale Methoden der Informatik (FMI), Universität Stuttgart, Universitätsstraße 38, 70569 Stuttgart, Germany
July 21, 2020
We investigate the orbits of automaton semigroups and groups to obtainalgorithmic and structural results, both for general automata but also forsome special subclasses.First, we show that a more general version of the finiteness problem forautomaton groups is undecidable. This problem is equivalent to the finite-ness problem for left principal ideals in automaton semigroups generated bycomplete and reversible automata.Then, we look at ω -word (i. e. right infinite words) with a finite orbit. Weshow that every automaton yielding an ω -word with a finite orbit alreadyyields an ultimately periodic one, which is not periodic in general, however.On the algorithmic side, we observe that it is not possible to decide whether agiven periodic ω -word has an infinite orbit and that we cannot check whether agiven reversible and complete automaton admits an ω -word with a finite orbit,a reciprocal problem to the finiteness problem for automaton semigroups inthe reversible case.Finally, we look at automaton groups generated by reversible but not bi-reversible automata and show that many words have infinite orbits under theaction of such automata. Keywords.
Automaton Groups, Automaton Semigroups, Orbits, SchreierGraphs, Orbital Graphs, Reversible ∗ The first author was supported by the Austrian Science Fund project FWF P29355-N35. † The second author was supported by a Doc.Mobility grant from the Swiss National Science Foun-dation as well as the "@raction" grant ANR-14-ACHN-0018-01 during a visit at the École NormaleSupérieure in Paris. Introduction
Starting with the realization that the famous Grigorchuk group (the first example ofa group with subexponential but superpolynomial growth) and many other groups withinteresting and peculiar properties can be generated by finite automata, the class of so-called automaton groups grew into a widely and intensively studied object. The finiteautomaton here is a finite-state, letter-to-letter transducer. It induces an action of thefinite words over its state set on the words over its alphabet and this action is used todefine the generated group.Roughly speaking, the research in this area is divided into three branches: the studyof individual, special automaton groups (such as the mentioned Grigorchuk group), thestudy of structural properties of automaton groups and the study of algorithmic prob-lems over automaton groups. The aim of this work is to contribute to the latter twoof these branches. In fact, we will not only consider automaton groups but also theirnatural generalization to automaton semigroups, in which the interest seems to haverisen lately, both for structural results (see, for example, [5, 3, 4, 11]) but also for algo-rithmic problems. They are particularly interesting for the latter point because manyimportant and classical algorithmic problem in group theory are proven or suspectedto be undecidable for automaton groups and it is usually easier to encode computationsin semigroups than in groups. Sometimes algorithmic results for automaton semigroupscould later be lifted to groups. An example for this is the order problem: first, it couldbe shown to be undecidable for automaton semigroups [12] and later this result could beextended to automaton groups [2, 13]. Similarly, a result on the complexity of the wordproblem could be lifted from the (inverse) semigroup case [10] to the groups case [21].For another important problem, the finiteness problem, the current state is that it hasbeen proven to be undecidable in the semigroup case [12] (and also in a more restrictivesetting [8]) but the decidability of the problem in the group case remains unknown.In this paper, we give a partial solution to this problem. The classical question ofthe finiteness problem is whether a given invertible automaton generates a finite or aninfinite group. This is equivalent to the question whether there are infinitely many statesequences whose actions on the words over the alphabet are pairwise distinct. We showthat the problem is undecidable if we instead ask whether there are infinitely many statesequences whose actions pairwisely differ on all words with a given prefix. If we pass tothe dual automaton (i. e. if we change the roles of states and letters), this problem is thesame as asking whether a given element s of a semigroup S generated by a complete andreversible (i. e. co-deterministic with respect to the input) automaton has an infinite leftprincipal ideal Ss ∪ { s } .Quite recently, the current authors could show that the algebraic property of an au-tomaton semigroup (and, thus, of an automaton group) to be infinite is equivalent to thefact that the action given by the generating automaton yields an ω -word with an infiniteorbit [6]. Thus, the finiteness problem is equivalent to asking whether such an infinite See [16] for an accessible introduction to Grigorchuk’s group. Two of Dehn’s fundamental problems in algorithmic group theory, the conjugacy problem and theisomorphism problems are among the problems which have been proven to be undecidable for automatongroups [20]. ω -word with afinite orbit and we show that the corresponding decision problem is undecidable evenfor complete and reversible automata. Furthermore, we show that it is algorithmicallyimpossible to test whether a given (periodic) word has a finite or infinite orbit.Structurally, we explore further consequences of the mentioned connection betweenthe semigroup being infinite and the existence of an infinite orbit as well as the dualargument underlying its proof. Here, we first look at ω -words with a finite orbit andshow that, whenever such a word exists, there is also an ultimately periodic ω -wordwith a finite orbit. We will see that this word can be assumed to be periodic if theautomaton is reversible but that this does not hold in the general case. Finally, we lookat the class of groups generated by invertible, reversible but not bi-reversible (i. e. notco-deterministic with respect to the output) automata. Here, we obtain that they alwaysadmit periodic ω -words (of a certain form) with infinite orbits and that, if the dualautomaton is additionally connected, all ω -words have infinite orbits. For semigroupsgenerated by reversible but not bi-reversible automata, we will see, however, that this isnot true: they can be infinite while the orbits of all (ultimately) periodic words are finite.This also shows that the result about the existence of a word with an infinite orbit (if thegenerated semigroup is infinite) cannot be extended to periodic or ultimately periodicwords. Fundamentals, Words and Languages.
Let A and B be sets. We write A ⊔ B for theirdisjoint union and, for a partial function from A to B , we write A → p B . If the functionis total, we omit the index p . Furthermore, we use N to denote the set of natural numbersincluding .A non-empty, finite set Σ is called an alphabet , its elements are called letters and finiteor right-infinite sequence over Σ are called finite words and ω -words , respectively. A word can be a finite word or an ω -word. The set of all finite word over Σ – includingthe empty word ε – is Σ ∗ and Σ + is Σ ∗ \ { ε } . The length of a finite word w = a . . . a ℓ with a , . . . , a ℓ ∈ Σ is | w | = ℓ . Finally, the reverse of a finite word w = a . . . a ℓ with a , . . . , a ℓ ∈ Σ is ∂w = a ℓ . . . a .The set of all ω -words over an alphabet Σ is Σ ω . An ω -word is called ultimately periodic if it is of the form uv ω for some u ∈ Σ ∗ and v ∈ Σ + where v ω = vv . . . ; it is called periodic if it is of the form v ω for v ∈ Σ + . We can also take the reverse of an ω -word α = a a . . . with a , a , · · · ∈ Σ to obtain the left-infinite sequence ∂α = . . . a a over Σ .A word u is called a suffix of another word w if there is some finite word x with w = xu .Symmetrically, u is a prefix of w if there is a word x with w = ux . A language L is a setof words over some alphabet Σ . It is suffix-closed if w ∈ L implies that every suffix of w is in L as well and it is prefix-closed if w ∈ L implies that every prefix of w is also in L . By Pre w , we denote the set of finite prefixes of a word w and Pre L for a language L is Pre L = ∪ w ∈ L Pre w . Symmetrically, we define Suf w and Suf L for the finite suffixesof a finite word or left-infinite sequence w and a set L of finite words and left-infinitesequences. 3or two languages K and L of finite words, we let KL = { uv | u ∈ K, v ∈ L } .Furthermore, we define L ∗ = { w . . . w i | i ∈ N , w , . . . , w i ∈ L } and sometimes simplywrite w for the singleton language { w } . Additionally, we lift operators on words tolanguages; for example, we let ∂L = { ∂w | w ∈ L } . Semigroups, Groups and Torsion.
We assume the reader to be familiar with basicnotions from semigroup and group theory such as inverses (in the group sense) andgenerating sets. If a semigroup S or a monoid M is generated by a (finite) set Q , thenthere is a natural epimorphism from Q + to S or from Q ∗ to M . In this case, we write q in S or in M for the image of q ∈ Q + or q ∈ Q ∗ . Similarly, for K ⊆ Q ∗ , we write K in S or in M for the image of K under this homomorphism. Additionally, we use somenatural variations for this notation. For example, we write p = q in S if p and q havethe same image under the natural homomorphism.An element s of a semigroup S has torsion if there are i, j ≥ with i = j but s i = s j .If S = G is even a group, this is connected to the order of a group element g ∈ G : it isthe smallest number i ≥ such that g i is the neutral element of the group; if there is nosuch i , then the element has infinite order. Obviously, an element of a group is of finiteorder if and only if it has torsion. Automata.
The most important objects in this paper are automata, which are moreprecisely described as finite-state, letter-to-letter transducers. Formally, an automaton isa triple T = ( Q, Σ , δ ) where Q is a set of states , Σ is an alphabet and δ ⊆ Q × Σ × Σ × Q isa set of transitions . For a transition ( p, a, b, q ) ∈ Q × Σ × Σ × Q , we use a more graphicalnotation and denote it by p q a/b or, when depicting an entire automaton, by p qa/b .An automaton T = ( Q, Σ , δ ) is complete if d p,a = (cid:12)(cid:12)(cid:12) { p q a/b ∈ δ | b ∈ Σ , q ∈ Q } (cid:12)(cid:12)(cid:12) is at least one for every p ∈ Q and a ∈ Σ . If, on the other hand, all d p,a are at most one,then T is deterministic . Additionally, T is reversible if it is co-deterministic with respectto the input, i. e. if { p q a/b ∈ δ | p ∈ Q, b ∈ Σ } contains at most one element for every a ∈ Σ and q ∈ Q and it is inverse-reversible if itis co-deterministic with respect to the output, i. e. if { p q a/b ∈ δ | p ∈ Q, a ∈ Σ } contains at most one element for every b ∈ Σ and q ∈ Q . An automaton that is both,reversible and inverse-reversible is called bi-reversible .Another way of depicting transitions in automata are cross diagrams . We write4 p qb .to indicate that an automaton T = ( Q, Σ , δ ) contains the transition p q a/b ∈ δ . Wecan combine multiple transitions into a single cross diagrams. For example, the crossdiagram a , . . . a ,m q , q , . . . q ,m − q ,m a , a ,m ... ... ... ... a n − , a n − ,m q n, q n, . . . q n,m − q n,m a n, . . . a n,m states that the automaton contains all transitions q i,j − q i,ja i − ,j /a i,j for ≤ i ≤ n and ≤ j ≤ m . Sometimes, we will omit intermediate states or letters if we do notneed to assign them a name. Instead of always drawing complete cross diagrams, we alsointroduce a shot-hand notation where we do not only allow states and letters but alsostate sequences and words. For example, the above cross diagram can be abbreviate by u = a , . . . a ,m q n, . . . q , = q p = q n,m . . . q ,m v = a n, . . . a n,m .It is important to note the ordering of the state sequences here: q n, belongs to the lasttransition but is written leftmost while q , belongs to the first transition and is writtenrightmost. Automaton Semigroups.
For a deterministic automaton T = ( Q, Σ , δ ) , we can definea partial left action of Q ∗ on Σ ∗ and a partial right action of Σ ∗ on Q ∗ using crossdiagrams. Since the automaton is deterministic, there is at most one v ∈ Σ + and atmost one q ∈ Q + for every u ∈ Σ + and every p ∈ Q + such that the cross diagram u p q v holds. In this case, we define the left partial action of p on u as p ◦ T u = v and theright partial action of u on p as p · T u = q . If there are no such q and v , we let p ◦ T u This seemingly wrong ordering is justified here because we will define automaton semigroups andgroups using left actions later on. p · T u be undefined. Additionally, we let p ◦ T ε = ε , ε ◦ T u = u , p · T ε = p and ε · T p = ε . With this definition, it is easy to see that we have q ◦ T p ◦ T u = qp ◦ T u and u · T p · q = u · T pq . Whenever the automaton T is clear form the context, we simplywrite p ◦ u and p · u instead of p ◦ T u and p · T u .Now, every p ∈ Q ∗ induces a partial, length-preserving function p ◦ : Σ ∗ → p Σ ∗ whichmaps every u to p ◦ u . These partial functions are prefix-compatible in the sense that wehave p ◦ u u = ( p ◦ u ) v for some v ∈ Σ ∗ (whenever the partial action is defined on aword u u ). Naturally, we can extend p ◦ into a partial function Σ ∗ ∪ Σ ω → p Σ ∗ ∪ Σ ω : α = a a . . . with a , a , · · · ∈ Σ gets mapped to b b . . . where the b , b , · · · ∈ Σ aredefined by b . . . b m = p ◦ a . . . a m (if p ◦ is undefined on some prefix of α , then p ◦ obviously should also be undefined on α ).In the same way, we can also define the partial, length-preserving functions · u : Q ∗ → p Q ∗ with u ∈ Σ ∗ which map p to p · u and observe that they have similar properties asthe maps p ◦ .The semigroup S ( T ) generated by the deterministic automaton T is the set Q + ◦ = { q ◦ | q ∈ Q + } with the composition of partial functions as its operation. This semigroupis generated by Q ◦ = { q ◦ | q ∈ Q } . To emphasize the fact, that they generate semigroups,we will use the name S -automata for deterministic automata from now on. An automatonsemigroup is a semigroup generated by some S -automaton. Remark . We want to point out that we do not require an S -automaton to be com-plete. If an automaton semigroup is generated by a complete automaton, we call it a complete automaton semigroup to emphasize this. It is not unknown whether the class ofcomplete automaton semigroups and the class of (partial) automaton semigroups coincide(see [11] for a discussion).Clearly, if T = ( Q, Σ , δ ) is a complete S -automaton, p ◦ u and p · u are defined for all p ∈ Q ∗ and all u ∈ Σ ∗ and all functions q ◦ are total.For an S -automaton T = ( Q, Σ , δ ) , the partial action of Σ ∗ on Q ∗ is compatible withthe structure of the generated semigroup as we have p ◦ = q ◦ = ⇒ p · u ◦ = q · u ◦ (orboth undefined) for all p , q ∈ Q ∗ and u ∈ Σ ∗ (which can be seen easily). Accordingly, wecan define a partial action of Σ ∗ on S ( T ) : for an element s = q ◦ ∈ S ( T ) with q ∈ Q + ,we let s · u = p · u ◦ for u ∈ Σ ∗ . Automaton Groups and Inverse Automata.
An automaton T = ( Q, Σ , δ ) is called invertible if the sets { p q a/b | a ∈ Σ , q ∈ Q } contain at most one element for all p ∈ Q and b ∈ Σ . If a complete S -automaton T = ( Q, Σ , δ ) is invertible, all functions p ◦ with p ∈ Q ∗ are bijections (and, in particular,total). In this case, we define Q − = { q − | q ∈ Q } as a disjoint copy of Q and let Q ±∗ = ( Q ⊔ Q − ) ∗ . We can extend the action of Q ∗ on Σ ∗ (and Σ ω ) into an action of Q ±∗ on Σ ∗ (and Σ ω ) by letting q − ◦ u be given by the pre-image of u under q − ◦ .The group G ( T ) generated by an complete and invertible S -automaton T = ( Q, Σ , δ ) is Q ±∗ ◦ = { q ◦ | q ∈ Q ±∗ } with the composition of functions as its operation and suchan automaton is called a G -automaton. A group generated by some G -automaton is an automaton group . 6he group G ( T ) generated by a G -automaton T = ( Q, Σ , δ ) is also an automatonsemigroup. It is the semigroup generated by the automaton T ′ = ( Q ⊔ Q − , Σ , δ ∪ δ − ) where we let δ − = { p − q − b/a | p q a/b ∈ δ } .The automaton T − = ( Q − , Σ , δ − ) is the inverse automaton of T . Example 2.2.
The typical example of an automaton is generated by the adding machine q id1 / / / / ,which we denote by T = ( { q, id } , { , } , δ ) in this example. It is deterministic, completeand invertible and the action of id is obviously the identity mapping on Σ ∗ . To understandthe action of q , we observe that we have q ◦
000 = 100 , q ◦
100 = 010 and q ◦
010 = 110 .Thus, if we interpret a word u ∈ { , } ∗ as the reverse/least-significant bit first binaryrepresentation of a natural number n , then q ◦ maps u to the reverse/least-significant bitfirst binary representation of n + 1 (with appropriately many leading zeros). Therefore,the element q ◦ of the semigroup S ( T ) can be identified with plus one in the monoidof natural numbers with addition as operation; accordingly, q i ◦ is plus i . Since we alsohave the identity as a state, the semigroup S ( T ) generated by T is isomorphic to N (with addition and including zero) or – in different words – the free monoid of rank one.Since the automaton is complete and invertible, we can also consider the group G ( T ) generated by it. The inverse of q ◦ can, obviously, be identified with minus one and weobtain that G ( T ) is isomorphic to the free group of rank one or the set of integers withaddition as operation. Dual Automaton and the Dual Action.
The dual of an automaton T = ( Q, Σ , δ ) isthe automaton ∂ T = (Σ , Q, ∂δ ) with ∂δ = { a b p/q | p q a/b ∈ δ } ,i. e. we swap the roles of Q and Σ . To obtain a cross diagram for ∂ T from one for T ,we have to mirror it at the north-west to south-east diagonal, i. e. we have the followingequivalence of cross diagrams: a . . . a m p . . . q ... ... ... ... p n . . . q n b . . . b m in T ⇐⇒ p . . . p n a . . . b ... ... ... ... a m . . . b m q . . . q n in ∂ T .If we let p = p n . . . p and q = q n . . . q as well as u = a . . . a m and v = b . . . b m , we canwrite the above equivalence in short-hand notation: u p q v in T ⇐⇒ ∂ p ∂u ∂v∂ q in ∂ T .7learly, taking the dual of an automaton is an involution and the dual of a deterministic(complete) automaton is also deterministic (complete). Additionally, the dual of aninvertible automaton is reversible. Thus, the dual of an S -automaton is an S -automa-ton and the dual of a G -automaton is a complete and reversible S -automaton (and viceversa).Therefore, if T = ( Q, Σ , δ ) is an S -automaton, T itself induces the actions ◦ T and · T and its dual induces the actions ◦ ∂ T and · ∂ T , which we simply write as ◦ ∂ and · ∂ if theautomaton is clear from the context. Because of the above equivalence of cross diagrams,there is a strong connection between ◦ and · ∂ (and, equivalently, between ◦ ∂ and · ). Wehave ∂u ◦ ∂ ∂ p = ∂ ( p · u ) (or both undefined)for all u ∈ Σ ∗ and p ∈ Q ∗ . Here, we have used the convention that ∂ has higherprecedence than the two automaton actions to avoid parentheses; for example, ∂ p ◦ u isto be understood as ( ∂ p ) ◦ u instead of ∂ ( p ◦ u ) .Since the dual of a complete and reversible S -automaton T = ( Q, Σ , δ ) is a G -automa-ton and we, thus, have that all u ◦ ∂ are bijections, we immediately obtain the followingfact from the above connection. Fact 2.3.
Let T = ( Q, Σ , δ ) be a complete and reversible S -automaton. Then, all maps · u : Q ∗ ∪ Q ω → Q ∗ ∪ Q ω , p p · u for u ∈ Σ ∗ are length-preserving bijections. Orbits.
Let T = ( Q, Σ , δ ) be an S -automaton and u ∈ Σ ∗ ∪ Σ ω . For K ⊆ Q ∗ , the K -orbit of u is K ◦ u = { q ◦ u | q ∈ K, q ◦ u defined } .The orbit of u is its Q ∗ -orbit. The orbit Q ∗ ◦ u of a word u has a natural graph structure:we have a q -labeled edge for q ∈ Q from p ◦ u to q p ◦ u whenever q p ◦ u is defined (where p ∈ Q ∗ ).If T is even a G -automaton, we can define the K -orbit of u for K ⊆ Q ±∗ analogously.It is well-known that, for a G -automaton T = ( Q, Σ , δ ) , the orbit of u is infinite if andonly if its Q ±∗ -orbit is infinite (see, e. g. [8, Lemma 2.5]).Recently, it could be shown that the existence of an ω -word with an infinite orbit isequivalent to the algebraic property that an automaton semigroup (or group) is infinite. Theorem 2.4 ([6, Corollary 3.3]) . The semigroup S ( T ) generated by some S -automa-ton T = ( Q, Σ , δ ) is infinite if and only if there is some ω -word α ∈ Σ ω with an infiniteorbit Q ∗ ◦ α . In fact, this result is only a special case of a more general result.
Theorem 2.5 ([6, Theorem 3.2]) . Let T = ( Q, Σ , δ ) be some S -automaton and let K ⊆ Q ∗ be suffix-closed. The image of K in S ( T ) is infinite if and only if there is some ω -word α whose K -orbit K ◦ α is infinite. S -automaton T = ( Q, Σ , δ ) and a language L ⊆ Σ ∗ , we define the relation ≡ T ,L ⊆ Q ∗ × Q ∗ by p ≡ T ,L q ⇐⇒ ∀ u ∈ L : p ◦ u = q ◦ u (or both undefined).As is the case with the two automaton actions, we do not write the index T wheneverthe automaton is clear from the context. It is easy to verify that ≡ L is an equivalencerelation for all languages L ⊆ Σ ∗ . We write [ p ] L for the equivalence class of p ∈ Q ∗ under ≡ L and K/L = { [ p ] L | p ∈ K } for the set of classes with a representative in K ⊆ Q ∗ . The set K/L is a generalizationof a couple of other notions (including the automaton semigroup itself and co-sets withrespect to stabilizers in the group case; see [6] for these examples) but, most importantly,it is closely related to the K -orbit of a word. Proposition 2.6 ([6, Proposition 2.4]) . Let T = ( Q, Σ , δ ) be an S -automaton and let K ⊆ Q ∗ be suffix-closed. For all α ∈ Σ ω , we have | K/ Pre α | = ∞ ⇐⇒ | K ◦ α | = ∞ . In particular, this result also holds for non-complete S -automata.For an S -automaton T = ( Q, Σ , δ ) and a set K ⊆ Q ∗ , we also have the equivalence ≡ K,∂ T belonging to the dual of T . We also simply write ≡ K for this relation and L/K for the classes of ≡ K with a representative in L ⊆ Σ ∗ . That these are to be understoodwith respect to ∂ T (and not with respect to T itself) can be seen from the fact that K is a set of state sequences of T .The above-mentioned result shows a close relation between the sets K/L and
L/K : Proposition 2.7 ([6, Proposition 3.1]) . Let T = ( Q, Σ , δ ) be an S -automaton, let K ⊆ Q ∗ be suffix-closed and let L ⊆ Σ ∗ be prefix-closed. Then, we have: | K/L | = ∞ ⇐⇒ | ∂L/∂K | = ∞ An important special case of Proposition 2.7 is when K and L are both given by asingle ω -word. Combined with Proposition 2.6, this case yields the following dualityresult for orbits (which we will also use below). Corollary 2.8 ([6, Corollary 3.11]) . Let T = ( Q, Σ , δ ) be an S -automaton and let π ∈ Q ω and α ∈ Σ ω . Then, we have | ∂ Pre π ◦ α | = ∞ ⇐⇒ | ∂ Pre α ◦ ∂ π | = ∞ Theorem 2.9 ([6, Theorem 3.12]) . Let T = ( Q, Σ , δ ) be an S -automaton and let q ∈ Q + .Then, the statements1. ∂ q has torsion in S ( T ) .2. The orbit Σ ∗ ◦ ∂ q ω of q ω under the action of the dual of T is finite.3. The orbit Σ ∗ ◦ ∂ pq ω of pq ω under the action of the dual of T is finite for all p ∈ Q ∗ .are equivalent. The question whether the finiteness problem for automaton groups
Input: a G -automaton T Question: is G ( T ) finite?is undecidable is an important open question in the algorithmic study of automatongroups [15, 7.2 b)]. In this section, we will show that a more general version of theproblem is undecidable. We will show this by using Gillibert’s result that there is anautomaton group whose order problem Constant: a G -automaton T = ( Q, Σ , δ ) Input: a finite state sequence q ∈ Q ∗ Question: has q finite order in G ( T ) ?is undecidable [13]. In fact, this result was also obtain by Bartholdi and Mitrofanov [2]but we specifically use the construction given by Gillibert. Theorem 3.1.
The decision problem
Constant: a G -automaton T = ( Q, Σ , δ ) Input: a finite word w ∈ Σ ∗ Question: is G ( T ) · w = { g · w | g ∈ G ( T ) } finite?is undecidable for some G -automaton T .Proof. Although it is not explicitly stated in his proof, Gillibert actually shows theundecidability of the decision problem
Constant: a G -automaton R = ( P, Γ , τ ) anda state $ ∈ P Input: a finite sequence p ∈ P ∗ of states Question: has $Λ( p ) finite order in G ( R ) ?where Λ : P ∗ → P ∗ is given by Λ( ε ) = ε and Λ(ˆ p p ) = Λ( p )ˆ p Λ( p ) [13]. In Gillibert’s paper, the function is called Q , actually. However, this notation clashes with theconvention of using Q to denote the state sets of automata followed in this work. Therefore, we use Λ instead. Additionally, Gillibert uses right actions to define automaton groups. Therefore, we mirror theordering here. t p p $id ∗ / ∗ ( a p , / ( a p , a p , / ( a p , / a q , / ( a q , a q , / ( a q ,
1) / a/a Figure 1: New transitions of T We take the G -automaton R and extend it into a new G -automaton T = ( Q, Σ , δ ) .Then, we reduce the above version of the order problem of R to the generalized finitenessproblem for T from the theorem statement.As the alphabet of T , we use Σ = Γ ⊎ { a p | p ∈ P } × { , } ⊎ {∗ , } , i. e. we add twonew special letters ∗ and as well as two new letters ( a p , and ( a p , for every state p ∈ P . Similarly, we use Q = P ⊎ { s, t, id } ⊎ { p | p ∈ P } for the state set, i. e. we addthree new states s , t and id as well as a new state p for every old state p ∈ P . Ofcourse, we also add new transitions δ ′ = τ ∪ { s t ∗ / ∗ , t $ / } ∪ { t t ( a p , / ( a p , , t p ( a p , / ( a p , | p ∈ P }∪ { p p ( a q , i ) / ( a q , i ) , p p / | p, q ∈ P, i ∈ { , }}∪ { id id a/a | a ∈ Σ } ,which are depicted schematically in Figure 1, and make the automaton complete byadding a transition to the identity state whenever some transition is missing: δ = δ ′ ∪ { q id a/a | q ∈ Q, a ∈ Σ , ∄ a ′ ∈ Σ , q ′ ∈ Q : q q ′ a/a ′ ∈ δ ′ } Note that the resulting automaton is indeed a G -automaton!For the reduction of the strengthened version of the order problem to the generalizedversion of the finiteness problem, we map the input sequence p = p ℓ . . . p to the finiteword w = ∗ w ′ = ∗ ( a p , . . . ( a p ℓ , , which is obviously computable. In the remainderof this proof, we show that $Λ( p ) has finite order in G ( R ) if and only if G ( T ) · w is finite.First, we show that $Λ( p ) has finite order in G ( R ) if and only if it has in G ( T ) . We dothis, by showing that ($Λ( p )) i and ($Λ( p )) j are distinct in G ( R ) if and only if they aredistinct in G ( T ) for i, j ∈ N . If they are distinct in G ( R ) , there is some witness u ∈ Γ ∗ which they act differently on. Since we have τ ⊆ δ ′ ⊆ δ , this is also a witness for theirdifference in G ( T ) . For the other direction, suppose that ($Λ( p )) i is different to ($Λ( p )) j in G ( T ) . Then, there must be some witness u ∈ Σ ∗ which they act differently on. Weare done if u is already in Γ ∗ . Otherwise, we can factorize u = u au with u ∈ Γ ∗ , a ∈ Σ \ Γ and u ∈ Σ ∗ . By the construction of T , we remain in states from P if we start11n P and read letters from Γ . If we read a letter from Σ \ Γ , we go to id , which yieldsthe cross diagrams u a u ($Λ( p )) i id i | $Λ( p ) | v a u and u a u ($Λ( p )) j id j | $Λ( p ) | v ′ a u for T .Thus, ($Λ( p )) i and ($Λ( p )) j must already act differently on u , which is from Γ ∗ and,thus, also a witness for R .Next, we observe that ∗ is not changed by the action of any state and that we have q · ∗ = id for all q ∈ Q except s and s · ∗ = t . Thus, G ( T ) · ∗ is the subgroup T generatedby t in G ( T ) and we obtain G ( T ) · w = T · w ′ . To understand the elements in T · w ′ , wewill show that we have the cross diagram w ′ t k | $Λ( p ) | ($Λ( p )) k w ′ in T ( † )for all k ∈ N . This shows that G ( T ) · w = T · w ′ is given by the state sequences (Suf $Λ( p )) ($Λ( p )) ∗ and their inverses in G ( T ) . These form a finite set in G ( T ) if and only if ($Λ( p )) ∗ isfinite in G ( T ) and this is the case if and only if $Λ( p ) has finite order in G ( T ) (or,equivalently, in G ( R ) ).The easiest way to establish the cross diagrams ( † ) is by calculation. For example, for p = p p p , we have w ′ = ( a p , a p , a p , and the cross diagram: ( a p ,
0) ( a p ,
0) ( a p ,
0) t p p p p ( a p ,
1) ( a p ,
0) ( a p ,
0) t t p p p ( a p ,
0) ( a p ,
1) ( a p ,
0) t p p p p ( a p ,
1) ( a p ,
1) ( a p ,
0) t t t p p ( a p ,
0) ( a p ,
0) ( a p ,
1) t p p p p ( a p ,
1) ( a p ,
0) ( a p ,
1) t t p p p ( a p ,
0) ( a p ,
1) ( a p ,
1) t p p p p ( a p ,
1) ( a p ,
1) ( a p ,
1) t t t t $( a p ,
0) ( a p ,
0) ( a p ,
0) q q )Λ( q q ) Λ( q q q ) t implements a binary increment (in the same wayas the adding machine in Example 2.2). This is what creates the pattern of Λ( p ) .For a formal proof, we first define the shorthand notations ( a p , i ) = ( a p , i ) . . . ( a p ℓ , i ) for i ∈ { , } and p = p ℓ . . . p as well as ε = ε and Λ(ˆ p p ) = Λ( p ) ˆ p Λ( p ) for p ∈ P ∗ and ˆ p ∈ P . We start by showing the cross diagram(s) ( a p , t | Λ( p ) | Λ( p ) ( a p , t t ( a p , for every p ∈ P + by induction on the length of p . For p = p ∈ P , this is easily verifiedfrom the definition of T (note that Λ( p ) = Λ( p ) = p in this case). For p ′ = ˆ p p with ˆ p ∈ P , we have | Λ(ˆ p p ) | = 2 | Λ( p ) | + 1 and the cross diagram ( a p ,
0) ( a ˆ p , t | Λ( p ) | Λ( p ) Λ( p ) ( a p ,
1) ( a ˆ p , t t ˆ p ( a p ,
0) ( a ˆ p , t | Λ( p ) | Λ( p ) Λ( p ) ( a p ,
1) ( a ˆ p , t t t ( a p ,
0) ( a ˆ p ,
0)
Λ(ˆ p p ) where the shaded part is obtained from using the induction hypothesis twice. For thepart on the right, notice that we have p ◦ ( a q , i ) = ( a q , i ) and p · ( a q , i ) = p for all p, q ∈ P and i ∈ { , } by construction. The two transactions on the right involving t can directly be verified, which concludes the induction.Finally, we can extend this to prove the cross diagrams ( † ) required above: ( a p ,
0) t | Λ( p ) | Λ( p ) Λ( p )( a p ,
1) t t $( a p ,
0)
The only point to notice here is that we indeed have Λ( p ) · p ) ; however, this isstraight-forward to verify. 13 rbital and Dual Formulation. The finiteness problem for automaton groups and thegeneralized problem from Theorem 3.1 can also be formulated in other ways.The first one is a re-formulation based on orbits. Using Theorem 2.4, we immediatelyobtain that the finiteness problem for automaton groups is equivalent to (the complementof) the problem:
Input: a G -automaton T = ( Q, Σ , δ ) Question: ∃ α ∈ Σ ω : | Q ∗ ◦ α | = ∞ ?For the problem in Theorem 3.1, this view yields the following formulation. Corollary 3.2.
The decision problem
Constant: a G -automaton T = ( Q, Σ , δ ) Input: a finite word w ∈ Σ ∗ Question: ∃ α ∈ Σ ω : | Q ∗ ◦ wα | = ∞ ?is undecidable for some G -automaton T .Proof. We have to show that G ( T ) · w is infinite if and only if the is some ω -word α ∈ Σ ω such that the orbit Q ∗ ◦ wα is infinite.In G ( T ) , the elements of G ( T ) · w are given by Q ±∗ · w , which is suffix-closed and, thus,by Theorem 2.5, infinite in G ( T ) if and only if there is some α ∈ Σ ω with | Q ±∗ · w ◦ α | = ∞ .We claim that Q ±∗ · w ◦ α is infinite if and only if Q ±∗ ◦ wα is. Since the latter is thecase if and only if Q ∗ ◦ wα is infinite, we are done when we have shown this claim.Clearly, we can map Q ±∗ ◦ wα surjectively onto Q ±∗ · w ◦ α by removing the prefix oflength | w | . Thus, if Q ±∗ · w ◦ α is infinite, so must be Q ±∗ ◦ wα . On the other hand, wehave Q ±∗ ∗ ◦ wα ⊆ ( Q ±∗ ◦ w ) ( Q ±∗ · w ◦ α ) and the first of the two sets on the right isalways finite. Thus, if Q ±∗ ◦ wα is infinite, Q ±∗ · w ◦ α must also be infinite.Another re-formulation is based on the dual automaton. A G -automaton generatesan infinite group if and only if its dual generates an infinite semigroup (see, e. g., [1,Proof of Lemma 5] combined with [1, Proposition 10]). Thus, the finiteness problem forautomaton groups is equivalent to the problem Input: a complete and reversible S -automaton T = ( Q, Σ , δ ) Question: is S ( T ) finite?If we re-formulate the problem from Theorem 3.1 under this view, we basically obtainthe finiteness problem for left principal ideals for semigroups generated by complete andreversible S -automata. Corollary 3.3.
The decision problem
Constant: a complete and reversible S -automaton R = ( P, Γ , τ ) Input: a finite state sequence p ∈ P ∗ Question: is P ∗ p finite in S ( R ) ?is undecidable.Proof. We reduce the problem from Corollary 3.2 to this problem. As the automaton R ,we choose the dual ∂ T of the G -automaton T = ( Q, Σ , δ ) and, for the reduction, we map14 ∈ Σ ∗ to ∂w as the input sequence p . We have to show that there is some α ∈ Σ ω with | Q ∗ ◦ wα | = ∞ if and only if Σ ∗ ∂w is infinite in S ( ∂ T ) .If the orbital graph Q ∗ ◦ wα is infinite (for some α ∈ Σ ω ), it must contain an infinite sim-ple path starting in wα . In other words, there is some π ∈ Q ω with π = p p . . . (where p , p , · · · ∈ Q ) such that all p i . . . p ◦ wα are pairwise distinct and that, thus, ∂ Pre π ◦ wα is infinite. From Corollary 2.8, we obtain that ∂ Pre( wα ) ◦ ∂ π = Suf (( ∂w )( ∂α )) ◦ ∂ π ⊆ Σ ∗ ∂w ◦ ∂ π is also infinite, which shows that Σ ∗ ∂w is infinite in S ( ∂ T ) .On the other hand, if Σ ∗ ∂w is infinite in S ( ∂ T ) , we have in particular that L =Σ ∗ ∂w ∪ Suf ∂w is infinite in S ( ∂ T ) . Since L is suffix-closed, we obtain by Theorem 2.5that there is some π ∈ Q ω such that L ◦ ∂ π is infinite. This is only possible if Σ ∗ ∂w ◦ ∂ π is infinite. Therefore, we have an infinite path in the orbital graph Σ ∗ ∂w ◦ ∂ π whichstarts in π , goes to ∂w ◦ ∂ π and then continues as an infinite simple path. In other words,there has to be some α ∈ Σ ω with α = a a . . . (where a , a , · · · ∈ Σ ) such that all a i . . . a i ( ∂w ) ◦ ∂ π are pairwise distinct. In particular, ∂ Pre( wα ) ◦ ∂ π is infinite, whichimplies that ∂ Pre π ◦ wα ⊆ Q ∗ ◦ wα is also infinite by Corollary 2.8. So far, we have looked at infinite orbits. In this section, we look at the opposite end andstudy ω -words with finite orbits.While the existence of an infinite orbit is coupled to the algebraic property of anautomaton semigroup S ( T ) to be infinite (by Theorem 2.4), the existence of an ω -wordwith a finite orbit depends on the generating automaton T (i. e. it is a property of theway the semigroup is presented, not an algebraic property). Indeed, if an S -automaton T does not admit an ω -word whose orbit is finite, we can add a new letter a to thealphabet of T and loops q q a/a to every state q . Obviously, this does not change thegenerated semigroup; however, now a ω has a finite orbit. Periodic and Ultimately Periodic Words.
We have just seen that we can add transitionsto any S -automaton to even obtain a periodic and, thus, ultimately periodic ω -word witha finite orbit (without changing the generated semigroup). We will see next that, in manycases, we do not even need to change the automaton: if there is an ω -word with a finiteorbit, then there is already an ultimately periodic word with finite orbit for every S -au-tomaton. If the S -automaton is complete and reversible, we even have a periodic wordwith a finite orbit. Proposition 4.1.
Let T = ( Q, Σ , δ ) be an S -automaton. If there is an ω -word α ∈ Σ ω such that its orbit Q ∗ ◦ α is finite, then there are u ∈ Σ ∗ and v ∈ Σ + such that Q ∗ ◦ uv ω isfinite and v can be chosen in such a way that it contains all letters that appear infinitelyoften in α .If, in addition, T is complete and reversible, then we already have that the orbit Q ∗ ◦ v ω is finite. Contrasting this, we will later see in Counter-Example 5.7 that there are semigroups where the onlyinfinite orbits belong to words that are not ultimately periodic. roof. Suppose we have | Q ∗ ◦ α | < ∞ for the ω -word α = a a . . . with a , a , · · · ∈ Σ and the S -automaton T . By Proposition 2.6, this is equivalent to | Q ∗ / Pre α | < ∞ ,which, by Proposition 2.7, is equivalent to | ∂ Pre α/Q ∗ | < ∞ . Thus, there is an infiniteset I ⊆ N with a i . . . a ≡ Q ∗ a j . . . a for all i, j ∈ I . Let k = min I and ℓ = min I \ { k } and define u = a . . . a k and v = a k +1 . . . a ℓ .For this choice, we have that ∂ ( uv ∗ ) /Q ∗ contains only one element and that ∂ Pre uv ω /Q ∗ = { [ ∂w ] Q ∗ | w ∈ Pre uv } is still finite. By Proposition 2.7, this implies that Q ∗ / Pre uv ω is finite, which is the case if and only if Q ∗ ◦ uv ω is finite by Proposition 2.6.If T is additionally complete and reversible, there is a surjective function Q ∗ ◦ uv ω → Q ∗ ◦ v ω , which shows | Q ∗ ◦ v ω | ≤ | Q ∗ ◦ uv ω | < ∞ . This function maps a word wβ ∈ Q ∗ ◦ uv ω with prefix w of length | w | = | u | to β . Clearly, wβ ∈ Q ∗ ◦ uv ω implies β ∈ Q ∗ ◦ v ω and,to show that the function is surjective, consider an arbitrary element β ∈ Q ∗ ◦ v ω . Then,there is some q ∈ Q ∗ with q ◦ v ω = β . Since T is complete and reversible, there is some p ∈ Q | q | with p · u = q (as the map · u is a length-preserving permutation in this caseby Fact 2.3 and we can choose p as the pre-image of q ). This yields the cross diagram u v ω p q w β for w = p ◦ u and, thus, that wβ ∈ Q ∗ ◦ uv ω is a pre-image of β for our function.In the general case, when the automaton is not reversible, however, the existence ofan ω -word with finite orbit does not imply the existence of a periodic ω -word with finiteorbit; not even, if the automaton is complete and invertible. Proposition 4.2.
Let T = ( Q, Σ , δ ) be the G -automaton p id q / ′ / ′ ′ / ′ / ′ / ′ / / ′ / ′ where id acts as the identity.Then, there exist an (ultimately periodic) ω -word α ∈ Σ ω with finite orbit Q ±∗ ◦ α butevery periodic ω -word u ω with u ∈ Σ + has an infinite orbit Q ∗ ◦ u ω .Proof. Let α = 1 ′ ω and α ′ = 0 ′ ω . It is easy to see that p ◦ α = α , p ◦ α ′ = α ′ , q ◦ α = α ′ and q ◦ α ′ = α . Thus, we have Q ±∗ ◦ α = { α, α ′ } , which is finite.To see that the orbit of every periodic word is infinite, let u ∈ Σ + be arbitrary. Wedistinguish two cases: u ∈ { , } + or u contains a ′ or a ′ . For the first case, observethat we obtain the adding machine (see Example 2.2) if we remove the state q and theletters ′ and ′ from T . Thus, for u ∈ { , } + , the orbit of u ω is infinite; in fact, wealready have | p ∗ ◦ u ω | = ∞ . 16n the other case, we can factorize u = u a u a . . . a n u n with a , a , . . . , a n ∈ { ′ , ′ } and u , u , . . . , u n ∈ { , } ∗ . Similarly to the other case, we observe that we obtain theadding machine (with letters ′ and ′ ) from T if we remove the state p and the letters and . Thus, we have | q ∗ ◦ ( u ′ ) ω | = ∞ where u ′ = a a . . . a n is obtained by removing allletters in { , } from u . By the construction of the automaton, reading any of the blocks u i ∈ { , } does not change the state as long as we are in q or in id . Therefore, it followseasily that q ∗ ◦ u ω and, thus, the orbit of u ω remain infinite. Undecidability of Orbit Finiteness.
Algorithmically, it is not possible to decide whethera given ω -word has a finite or infinite orbit. This can be seen from the connection statedin Theorem 2.9. Proposition 4.3.
There is some complete and reversible S -automaton whose orbit finite-ness problem for periodic ω -words Constant: a complete and reversible S -automaton T = ( Q, Σ , δ ) Input: a finite word u ∈ Σ + Question: is Q ∗ ◦ u ω finite?is undecidable.Proof. There is an automaton group with an undecidable order problem [2, 13], i. e. thereis a G -automaton T such that the problem Constant: a G -automaton T = ( Q, Σ , δ ) Input: a state sequence q ∈ Q ∗ Question: is q of finite order in G ( T ) ?is undecidable. We reduce this problem to the one in the proposition: as the automaton,we use ∂ T , the dual of T , which – as the dual of a G -automaton– is reversible andcomplete; the word is ( ∂ q ) ω , which is periodic. By Theorem 2.9, the orbit of ( ∂ q ) ω under the action of the dual ∂ T is finite if and only if q ◦ has torsion (i. e. is of finiteorder).By Theorem 2.4, the question whether a given (complete) S -automaton admits aword with an infinite orbit or not is equivalent to the finiteness problem for automatonsemigroups and, thus, undecidable [12]. Here, we show a dual result: checking theexistence of an ω -word with a finite orbit is undecidable, even for complete and reversible S -automata. Proposition 4.4.
The decision problem
Input: a complete and reversible S -automaton T = ( Q, Σ , δ ) Question: is there an ω -word α ∈ Σ ω such that | Q ∗ ◦ α | < ∞ ?is undecidable. We have already discussed this for the finiteness problem for automaton groups above. roof. By [8, Theorem 1], the problem
Input: a G -automaton T = ( Q, Σ , δ ) Question: is there a state sequence q ∈ Q + such that q ◦ is the identity?is undecidable. We reduce this problem to the one in the proposition by taking the dualautomaton. Obviously, the dual of a G -automaton is complete and reversible.Now, suppose that there is some q ∈ Q + such that q ◦ is the identity. Then, q ◦ , inparticular, has torsion. By Theorem 2.9, this implies that ( ∂q ) ω has a finite orbit underthe action of the dual.If, on the other hand, there is some word with a finite orbit under the action of thedual, then, by Proposition 4.1, this implies that there already is some periodic word q ω with a finite orbit (where q ∈ Q + ). Again, by Theorem 2.9, this implies that ∂ q ◦ hastorsion in the group generated by the original automaton. In other words, there is k ≥ such that ∂ q k ◦ is the identity. In this section, we have a closer look at the class of automaton groups generated byreversible but not bi-reversible G -automata.First, we show that every such G -automaton T admits a periodic word u ω with aninfinite orbit. The main idea is to take the dual of T , which is a reversible but notbi-reversible G -automaton as well, and to find elements without torsion in the semigroupgenerated by the dual. By Theorem 2.9, these elements correspond to periodic ω -wordswith infinite orbits. For the special case that the dual is connected (or only contains non-bi-reversible connected components), we could use [14, Theorem 23] or [7, Proposition 7]to obtain that none of the elements have torsion. However, the result is also true in thegeneral case.In a reversible and complete S -automaton T = ( Q, Σ , δ ) , all maps · u : Q → Q for u ∈ Σ ∗ are bijections (see Fact 2.3). It is not difficult to see that, therefore, in suchautomata, every connected component is already strongly connected.The central argument for our proof is that the semigroup generated by a reversiblebut not bi-reversible G -automaton cannot contain the (group) inverse of any functioninduced by a state from a non-bi-reversible connected component. Lemma 5.1.
Let T = ( Q, Σ , δ ) be a reversible G -automaton with a non-bi-reversible(strongly) connected component consisting of the states P ⊆ Q . Then, S ( T ) does notcontain the inverse p − ◦ for any p ∈ P .Proof. We first show that S ( T ) must contain q − ◦ for all q ∈ P if it contains p − ◦ fora single p ∈ P . Therefore, assume the latter to be true. Since q and p are in the same(strongly) connected component, there is some u ∈ Σ ∗ with p · u = q . Then, for v = p ◦ u ,we have q − ◦ = p − · v ◦ . Since we have p − ◦ = p ′ ◦ for some p ′ ∈ Q + by assumption,we obtain q − ◦ = p ′ · v ◦ , which is in S ( T ) .18ow, assume to the contrary that S ( T ) contains p − ◦ for one, and thus for all, p ∈ P .Since P is the state set of some non-bi-reversible component, there are q, p, r ∈ P and a, b, c ∈ Σ with p = q or a = b (or both) and the transitions p rq a/cb/c (i. e. p · a = q · b = r and p ◦ a = q ◦ b = c ). Notice that p = q (which implies a = b ) isnot possible since T is a G -automaton and that neither is a = b because T is reversible.We have p − · c ◦ = r − ◦ = q − · c ◦ and, by assumption, there are p ′ , q ′ ∈ Q + with p − ◦ = p ′ ◦ and q − ◦ = q ′ ◦ . Together, this yields p ′ · c ◦ = r − ◦ = q ′ · c ◦ . Since T iscomplete and reversible, the map · c : Q + → Q + is a bijection (see Fact 2.3) and there issome k ≥ with p ′ · c k = p ′ and q ′ · c k = q ′ . For this k , we have p − ◦ = p ′ ◦ = ( p ′ · c ) · c k − ◦ = ( q ′ · c ) · c k − ◦ = q ′ ◦ = q − ◦ and, thus, a = p − ◦ c = q − ◦ c = b , which is a contradiction.Actually, Lemma 5.1 allows for a stronger formulation: Lemma 5.2.
Let T = ( Q, Σ , δ ) be a reversible G -automaton and let P be the non-emptystate set of some non-bi-reversible connected component of T . Then, P contains at leasttwo elements and no element q p ◦ with q ∈ Q + and p ∈ P has an inverse in S ( T ) . Inparticular, no element q p ◦ has torsion in G ( T ) .Proof. Since the connected component belonging to P is non-bi-reversible (but needs tobe reversible since T is), it contains transitions q p a/b and q ′ p a ′ /b with a = a ′ and q = q ′ , as q = q ′ contradicts the invertibility of T . Therefore, { q, q ′ , p } ⊆ P containsat least two elements.Now, let p ∈ P and q ∈ Q ∗ be arbitrary and suppose that there is some r ∈ Q + such that rq p ◦ is the identity on Σ ∗ . Then, rq ◦ ∈ S ( T ) would be an inverse of p ◦ contradicting Lemma 5.1.Finally, if q p ◦ was of torsion, then ( q p ) i ◦ for some i would be its inverse, which wouldalso constitute a contradiction because of ( q p ) i ∈ Q + .We can now apply Lemma 5.2 to obtain periodic ω -words with infinite orbits. Theorem 5.3.
Let T = ( Q, Σ , δ ) be an S -automaton such that its dual ∂ T = (Σ , Q, ∂δ ) contains a reversible G -automaton D = (∆ , R, κ ) as a sub-automaton.Then, every non-bi-reversible connected component of D with state set Γ contains atleast two elements and Q ∗ ◦ u ( av ) ω is infinite for all a ∈ Γ and all u, v ∈ ∆ ∗ . roof. Let Γ be the state set of some non-bi-reversible connected component of D . ByLemma 5.2, no element va ◦ from S ( D ) with a ∈ Γ and v ∈ ∆ ∗ has torsion. ByTheorem 2.9, this means that all orbits R ∗ ◦ u ( a∂v ) ω ⊆ Q ∗ ◦ u ( a∂v ) ω for u ∈ ∆ ∗ areinfinite. Corollary 5.4.
Let T = ( Q, Σ , δ ) be a reversible but not bi-reversible G -automaton.Then, there are two distinct letters a, b ∈ Σ such that all Q ∗ ◦ u ( av ) ω and all Q ∗ ◦ u ( bv ) ω for u, v ∈ Σ ∗ are infinite.Proof. Notice that ∂ T is a reversible but not bi-reversible G -automaton as well. Thus, thecorollary follows from Theorem 5.3 since ∂ T must contain at least one non-bi-reversibleconnected component.Interestingly, we can combine Theorem 5.3 with the result about ω -words with a finiteorbit from Proposition 4.1 to obtain that many orbits in groups generated by reversiblebut not bi-reversible G -automata are infinite . Corollary 5.5.
Let T = ( Q, Σ , δ ) be a reversible but not bi-reversible G -automaton andlet Γ ⊆ Σ denote the set of states in the dual automaton ∂ T belonging to a non-bi-reversible connected component.Then, every α ∈ Σ ω which contains at least one letter from Γ infinitely often has aninfinite orbit: | Q ∗ ◦ α | = ∞ .Proof. Suppose to the contrary that there is some α ∈ Σ ω with a finite orbit such that α contains a letter a ∈ Γ infinitely often. Then, by Proposition 4.1, there is some w ∈ Σ + with w = w aw for some w , w ∈ Σ ∗ such that the orbit of w ω = w ( aw w ) ω is finite.However, from Theorem 5.3 follows that all words of the form u ( av ) ω must have infiniteorbit; a contradiction.The previous corollary directly implies that no infinite word has a finite orbit underthe action of a reversible but not bi-reversible G -automaton with a connected dual: Corollary 5.6.
Let T = ( Q, Σ , δ ) be a reversible but not bi-reversible G -automaton whosedual ∂ T is connected. Then, every ω -word α ∈ Σ ω has an infinite orbit Q ∗ ◦ α .Proof. Obviously, all letters from Σ belong to a non-bi-reversible connected componentof the dual. Therefore, any α ∈ Σ ω must, in particular, contain at least one of theminfinitely often and the result follows from Corollary 5.5.As a side remark, we note that, in Lemma 5.2, we can neither drop the completenessnor the invertibility requirement, which also has consequences for the other results above.To see the former, consider the S -automaton T q pa/ba/a b/b ,which is strongly connected, reversible and invertible but neither bi-reversible nor com-plete. One may observe that q ◦ is undefined on all words (except ε ) and that q ◦ ,20 ◦ q ◦ qp ◦ p ◦ pq ◦ q ◦ qp ◦ p ◦ q ◦ p ◦ q ◦ p ◦ q ◦ p ◦ q ◦ p ◦ q ◦ p ◦ , q ◦ p ◦ q ◦ Figure 2: Left Cayley graph of S ( T ) .therefore, has torsion as we have q ◦ = q ◦ . In fact, it turns out that the semigroupgenerated by the automaton is finite and, thus, that all its elements have torsion.To see that the automaton in the statements of Lemma 5.2 and the corollaries needsto be invertible, we use the following counter-example based on the connection fromTheorem 2.9. Counter-Example 5.7.
Let T = ( Q, Σ , δ ) be the G -automaton b ad c id / / / / / / / / / / ,which generates the Grigorchuk group. The Grigorchuk group G ( T ) is infinite (see e. g.[19, Theorem 1.6.1]) and, thus, so is S ( T ) (see, for example, [1, Proof of Lemma 5]) .The dual ∂ T of T id / id b/ac/ad/ id a/ id id / id b/cc/dd/ba/ idis complete and reversible but neither invertible nor bi-reversible. It generates an infinite This can, for example, be seen by computing its powers (up to the fourth one); the left Cayleygraph of the generated semigroup can be found in Figure 2. In fact, we have S ( T ) = G ( T ) since the group is a Burnside group (see e. g. [19, Theorem 1.6.1]). S ( ∂ T ) since an S -automaton generates a finite semigroup if and only ifits dual does [1, Proposition 10]. However, the orbits of all ultimately periodic words Σ ∗ ◦ ∂ pq ω under its action with p ∈ Q ∗ and q ∈ Q + are finite.This is the case because all elements of the Grigorchuk group G ( T ) have finite order(see e. g. [19, Theorem 1.6.1]), i. e. G ( T ) is a Burnside group, and because Theorem 2.9yields that the orbit Σ ∗ ◦ ∂ pq ω for any p ∈ Q ∗ and q ∈ Q + is finite if and only if ∂ q ◦ has torsion in S ( T ) .Another important consequence of Counter-Example 5.7 is that an infinite automatonsemigroup does not always admit a periodic or ultimately periodic word with an infiniteorbit. Thus, Theorem 2.4 cannot be extended in this way. However, since the automaton ∂ T is not a G -automaton, this still leaves the following question for automaton groupsopen. Open Problem 5.8.
Does every G -automaton generating an infinite group admit aperiodic or ultimately periodic ω -word with an infinite orbit?In fact, for arbitrary S -automata T and T ′ , an isomorphism between S ( T ) and S ( T ′ ) does not imply that S ( ∂ T ) and S ( ∂ T ′ ) are isomorphic. In other words, thedual is not an algebraic property of an automaton semigroup but only a property of thepresentation by a specific automaton. The dual of the Grigorchuk automaton depictedin Counter-Example 5.7 generates the free semigroup of rank two . This semigroup canalso be generated by a different automaton (see [5, Proposition 4.1]) which admits aperiodic ω -word with an infinite orbit (in fact, every ω -word has an infinite orbit underits action).Therefore, we have not completely settled the semigroup case either. It could still bethe case that every infinite automaton semigroup is generated by some S -automatonadmitting a periodic word with infinite orbit. References [1] Ali Akhavi, Ines Klimann, Sylvain Lombardy, Jean Mairesse, and Matthieu Picantin.On the finiteness problem for automaton (semi)groups.
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