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Automatic Structures and Boundaries for Graphs of Groups

arXiv:math/9306205v1 [math.GR] 16 Jun 1993Automatic Structures and Boundaries for Graphs of GroupsWalter D. Neumann and Michael Shapiro*Abstract. We study the synchronous and asynchronous automatic structures onthe fundamental group of a graph of groups in which each edge group is finite.Up to a natural equivalence relation, the set of biautomatic structures on such agraph product bijects to the product of the sets of biautomatic structures on thevertex groups.

The set of automatic structures is much richer. Indeed, it is densein the infinite product of the sets of automatic structures of all conjugates of thevertex groups.

We classify these structures by a class of labelled graphs which“mimic” the underlying graph of the graph of groups. Analogous statements holdfor asynchronous automatic structures.

We also discuss the boundaries of thesestructures.1. IntroductionGiven a group G, there is a natural equivalence relation on the set of synchronous orasynchronous automatic structures on G.Namely, two such structures, L and L′ areequivalent (written L ∼L′) if there is a constant K so that whenever a word of L and a wordof L′ represent the same element of G, these two words asynchronously K-fellow travel eachother.

(For definitions, see below.) This leads [NS1] to introduce SA(G), BSA(G), A(G)and BA(G), the sets of (respectively) automatic, biautomatic, asynchronously automatic,and asynchronously biautomatic structures on G up to equivalence.Currently, information is fairly scarce about these sets.A(G) has been computedfor G virtually abelian, virtually free ([NS1]), or virtually a surface group ([B]).

In thelatter two cases it is a single point. SA(G) = BSA(G) and is a single point if G is wordhyperbolic.

BSA(G) has been computed, and SA(G) is fairly well understood when Gis a geometrically finite hyperbolic group ([NS2]). By contrast, A(G) is very large andpoorly understood when G is the fundamental group of a closed hyperbolic 3-manifoldgroup which fibers over the circle.

Notice that here G is an HNN-extension of a hyperbolicsurface group. However, its unique automatic structure does not arise from the automaticstructure on the surface group.

Indeed, the surface group is not rational in this automaticstructure.In this paper, we will study these sets when G = π1(Y) is the fundamental group of afinite graph of groups Y in which each edge group is finite. The assumption of finite edgegroups turns out to ensure that each conjugate of a vertex group is rational with respectto any (synchronous or asynchronous) automatic structure on G. Consequently, there is a* Both authors acknowledge support from the NSF for this research.1

natural mapA(G) →YH∈HA(H),where H denotes the set of conjugates of vertex groups. One of our main results is that thismap is injective with dense image (Theorem 3.9).

Moreover, SA(G) is just the inverse imageof QH∈H SA(H) under this map. In [NS2] there is an analogous injection for synchronousautomatic structures on a geometrically finite hyperbolic group with maximal parabolicsplaying the role that vertex groups play here.Now if H ∈H, we have H = gGV g−1 where GV is a vertex group.

Given such a g,it is natural to look for “minimal” h so that H = hGV h−1. We shall see that if H ̸= GVthere is an FE orbit of such values, where FE is the edge group associated to an edgeE incident at V .

Thus, to specify an asynchronous automatic structure on G it is onlynecessary to specify a choice of structure [Lh] ∈A(GV ) for each such h, for this, in turn,specifies a structure on H = hGV h−1. In a sense which we shall make clear, the choice ofLh must be equivariant with respect to the action of FE on the set of such h. This allowsus to classify A(G) in terms of maps which we call regular deployments (Theorems 3.3 and3.8).There is a more concrete way to classify A(G), and that is in terms of objects whichwe call minimal special Y-graphs.

Roughly, a Y-graph X is a finite labelled graph whichmaps onto the underlying graph of Y. Each vertex of X is labelled by an equivalenceclass of structures on the corresponding vertex group of Y.

Each edge from this vertexis labelled by a rational subset of this vertex group. The labelling must be equivariantin terms of actions of the edge groups.

In the case of a biautomatic or asynchronouslybiautomatic structure, a most efficient Y-graph is essentially the underlying graph of Ywith biautomatic or asynchronously biautomatic structures at each vertex. In particular,this gives bijectionsBA(G) →YV ∈vert(Y)BA(GV )andBSA(G) →YV ∈vert(Y)BSA(GV ),where GV denotes the vertex group at V .As classifying objects these Y-graphs have several advantages over regular deploy-ments.

They are finite objects and they are easy to construct. Unlike regular deploy-ments, Y-graphs admit “local” modifications.

Finally, a Y-graph is easily turned into ageneralized finite state automaton for the structure which it determines.In the final section of this paper we describe the boundary of an asynchronous orsynchronous automatic structure on G = π1(Y). It is a “tree completion” of the disjointunion of the boundaries for the automatic structures on the conjugates of the vertex groups.The tree in question is the tree on which G acts with quotient Y (see for example [Se]).The assumption of finite edge groups in this paper may seem restrictive.

However,as the hyperbolic 3-manifold example mentioned above shows, it is necessary to ensurethat the vertex groups are rational. In fact, even if one restricts to abelian edge groups,2

the Heisenberg group is an example of the fundamental group of such a graph of groupswhere the vertex group has plentiful automatic structures but the group itself is not evenasynchronously automatic. Another example is F2 × Z, which can be seen either as F2∗F2or Z2 ∗Z Z2.

In work in preparation we show that SA(F2 × Z) is quite large and unlikelyto yield to classification by techniques like the current ones.2. Background and definitions.We start with a finitely generated group G and a map from a finite set A = {ai} into Gdenoted by ai 7→ai.

The set of all finite strings w = ai1 . .

. ain on elements of A (includingthe empty string) forms a monoid under the operation of concatenation.

We denote thismonoid by A∗. We define the length of w = ai1 .

. .

ain ∈A∗to be n and denote thisby len(w). (The length of the empty word is 0.) The map ai 7→ai extends to a uniquemonoid homomorphism from A∗to G and we denote this extension by w 7→w.

We willassume that this map is onto. We will also assume that A is supplied with an involutiondenoted by ai 7→a−1iand that the evaluation map respects this, that is, a−1i= (ai)−1.This allows us to form the Cayley graph of G with respect to A, Γ = ΓA.

The vertices ofΓA are the elements of G. There is a directed edge from g to g′ labelled by a ∈A exactlywhen g′ = ga. Thus there is exactly an A’s worth of edges emanating from each vertexof G. Since A is finite, Γ is locally finite.

Since A generates G, Γ is path connected. Bymaking each edge of Γ isometric with the unit interval, we endow G with a metric dA = dcalled the word metric.

That is, the distance between two points of Γ is defined to bethe length of the shortest path connecting them. G acts on Γ by left translation, and thisaction preserves distance.

We take the length of an element of G to be its distance fromthe identity, that is ℓ(g) = d(1, g). A word w ∈A∗determines a path in Γ, which we alsodenote by w, as follows.

The path w maps the interval [0, len(w)] into Γ by following atunit speed along the edge path in Γ based at 1 and labelled by w. We extend this to amap of [0, ∞) by setting w(t) = w for t ≥len(w).We call a subset of A∗a language. A language L is a normal form if L = G. Notethat we do not require L →G to be an injection.

We will say that a normal L has theasynchronous fellow traveller property if there is a constant K so that given w, w′ ∈Lwith d(w, w′) ≤1, there are monotone maps φ, ψ of [0, ∞) onto itself so that for all t,d(w(φ(t)), w′(ψ(t))) ≤K. We say L has the synchronous fellow traveller property if φ andψ can be chosen to be the identity.Given two normal forms L, L′, each with the asynchronous fellow traveller property,we will say that they are equivalent and write L ∼L′ if L∪L′ has the asynchronous fellowtraveller property.

We denote the equivalence class of L by [L].Recall that a finite state automaton A with alphabet A is a finite directed graph on avertex set S (called the set of states) with each edge labelled by an element of A and suchthat different edges leaving a vertex always have different labels. Moreover, a start states0 ∈S and a subset of accepted states T ⊂S are given.

A word w ∈A∗is in the languageL accepted by A if and only if it defines a path starting from s0 and ending in an acceptstate in this graph. We may assume there is no “dead state” in S (a state not accessiblefrom s0 or from which no accepted state is accessible).

Eliminating such states does notchange the language L accepted by A.3

A language is regular if it is accepted by some finite state automaton.We will also need the concept of a non-deterministic finite state automaton.Thedifference is that a non-deterministic finite state automaton is allowed to have severalstart states instead of just one, different edges from a vertex may have the same label,and edges are allowed to have empty label (such edges are called ǫ-transitions). A wordis accepted by such an automaton if it labels a path from a start state to an accept state.This path is allowed to traverse ǫ-transitions.

It is a standard result that the language ofwords accepted by a non-deterministic finite state automaton is a regular language.We shall also have occasion to use generalized finite state automata. A generalizedfinite state automaton is defined just like a non-deterministic finite state automaton exceptthat the edges are labelled by regular sublanguages of A∗rather than by elements of A.This machine accepts a word w if w can be written as w1 .

. .

wk such that there is acorresponding directed edge path e1 . .

. ek from a start state to an accept state such thatwi is in the language labelling ei for each i.

It is a standard fact that this language ofaccepted words is regular.A (synchronous) automatic structure for G is a regular normal form with the syn-chronous fellow traveller property.It is a result of [ECHLPT] that every automaticstructure has a sublanguage which bijects to G. Notice that if L ⊂L′ and L is an auto-matic structure, then L ∼L′. We will take the following as our definition of asynchronousautomatic structure.An asynchronous automatic structure for G is a rational normalform with the asynchronous fellow traveller property.

This is not exactly equivalent tothe use of the term in [ECHLPT]. Rather, these are the non-deterministic asynchronousautomatic structures of [S1].

Since every non-deterministic asynchronous automatic struc-ture contains an equivalent asynchronous automatic structure which bijects to G, we willmake no further distinction between the two. We will call an asynchronous automaticstructure L asynchronously biautomatic if there is a constant K so that if w, w′ ∈L withw = aw′ where a ∈A ∪{1}, then there are reparameterizations φ and ψ so that for allt, d(w(φ(t)), aw′(ψ(t))) ≤K.

(Here aw(·) is the translate of w(·) by a.) We will call anautomatic structure biautomatic if there exists K so that φ, ψ can be taken to be the iden-tity.

We take SA(G), BSA(G) BA(G) and BA(G) to be respectively the sets of automatic,biautomatic, asynchronously automatic, and asynchronously biautomatic structures on Gup to equivalence.Given an asynchronous or synchronous automatic structure L, we say that S ⊂G isL-rational if {w ∈L : w ∈S} is regular. It is a result of [NS1] that L-rationality dependsonly on the equivalence class of L. Using the techniques of [GS], one sees that if H is anL-rational subgroup of G, then L induces an equivalence class of asynchronous respectivelysynchronous automatic structures on H.Convention.

We have pointed out that any asynchronous automatic structure contains anequivalent one that bijects to G, so there is certainly no loss of generality in assumingthat all our structures are finite-to-one. Since this simplifies some proofs, we will assumeit from now on.3.

Graphs of groups with finite edge groups4

Let Y be a graph of groups. We start by fixing notation.

The underlying graph Y of Y is aconnected graph made of a finite collection of vertices and a finite collection of unorientededges. We consider each unoriented edge as a pair of oriented edges and denote the initialand terminal vertices of an oriented edge E by ∂0E and ∂1E.

The reverse of an edge E isdenoted E−1. To each vertex V is associated a group GV and to each edge E is associateda group FE with FE = FE−1.

Further, to each edge E is associated a pair of injections∂0 : FE →G∂0E and ∂1 : FE →G∂1E, which are exchanged when E is replaced by E−1.Such a graph may be seen as instructions for building a group by repeated free productswith amalgamation and HNN-extensions. To do this one takes a maximal tree T ⊂Y .Inductively one forms free product with amalgamation for each edge of T.One thenperforms an HNN-extension for each edge not on T. The resulting group is determinedup to isomorphism by the graph of groups.

We refer to it as the fundamental group of Y,denoted π1(Y). For details see [Se], for example.We describe a normal form for the elements of G = π1(Y).

We take the maximal treeT to be fixed throughout. We also choose a fixed base vertex V0 ∈Y.Definition.

For each edge E of Y we have an element tE ∈G = π1(Y) as follows: tE isthe stable letter associated to the edge E if E is not in T and tE = 1 if E is in T. Inparticular, tE−1 = t−1E . Then each element of G can be written in the normal formh = g0tE1g1 .

. .

tEmgm(∗)where:(1) E1 . .

. Em is a path in Y starting at the base vertex V0;(2) g0 ∈GV0 and gi ∈G∂1Ei for i = 1, .

. ., m;(3) if Ei+1 = E−1ithen gi /∈∂1(FEi).

This expression is unique up to the following twooperations:•one can add or delete terminal words consisting of trivial tEi’s (subject to condition(3); such words are bounded in length by the diameter of the maximal subtree T);•one can replace gitEigi+1 by (gi∂0(f))tEi(∂1(f −1)gi+1) for f ∈FEi.We shall assume from now on that all edge groups FE are finite.Note that if Yincludes an edge E with ∂1FE = G∂1E then this edge can be collapsed without changingπ1(Y) unless the edge is a loop, say ∂0E = ∂1E = V . In this case we may also eliminateE by replacing GV by GV ⋊Z.

This GV ⋊Z is virtually cyclic so A(GV ⋊Z) consistsof a single point (cf. [NS1]).

Thus, from the point of view of computing asynchronousautomatic structures on G in terms of asynchronous structures on the vertex groups, thissimplification of Y is harmless. If Y has no edge with ∂1FE = G∂1E we will say Y isreduced.To simplify later notation we define an extended graph bY by adding a new base edgeE0 to Y, going from a new vertex (which we will never need to refer to) to the base vertexV0.

We put FE0 = {1}. Using the above normal form we define for each edge E of bY:GE = {⟨= }′⊔E∞.

. .

⊔E⇕}⇕as in (∗) : ⇕≥′, E⇕= E, }⇕∈∂∞(FE)}.In particular, GE′ = {∞}. We stress that the only role of the base edge E0 is to supportthe notation GE′.

We do not include the reverse edge E−10 .5

The significance of these sets is that, as we will discuss in detail in section 5, thedisjoint union ` GE/FE over E ∈edge bY is in one-one correspondence with the vertices ofthe tree on which G acts with quotient Y and vertex and edge stabilizers given by the dataof Y. If Y is reduced then each conjugate of a vertex group stabilizes precisely one vertexof this tree, so the disjoint union ` GE/FE is a set of representatives for the conjugates ofvertex groups.

For our present purposes we formulate this as the following lemma.Lemma 3.1. Suppose Y is reduced.

Let H = gGV g−1 ∈H. Then H determines an edgeE of bY with ∂1E = V and h ∈GE so that H = hGV h−1.

If E′ with ∂1E′ = V and h′ ∈GE′also satisfy h′GV h′−1 = H then E = E′ and h′ ∈h∂1FE.Proof. The edge E and h ∈GE can be found from g in the following manner.

If V = V0and g ∈GV0 we take h = 1 ∈GE′. Otherwise, write g in normal form (∗), and delete anyfinal portion of g lying in GV .

Call the resulting expression h. If the last letter of h lies insome GV ′, we take E to be the last edge in the path in T from V ′ to V . If the last letterof h is the stable letter of an edge E1 not in T, we take E = E1 if ∂1E1 = V and otherwisewe take E to be the last edge in the path in T from ∂1E1 to V .Clearly H = hGV h−1.

The uniqueness statement about E and h follows by notingthat h (up to the action of ∂1FE) is visible as the first half of the normal form of anyhxh−1 ∈hGV h−1 with x /∈∂1FE.Definition. Whether Y is reduced or not, we define a deployment to be a mapψ :aE∈edge bYGE →aV∈vert YA(GV)with finite image taking GE to A(G∂1E) for each E and with the following equivarianceproperty: the restriction of ψ to GE is FE-equivariant in the sense that ψ(h) = fψ(hf) forf ∈∂1FE.There is some ambiguity in the notation ψ(h), for a group element h can be in morethan one GE.

Thus we are implicitly thinking of h as an element of the disjoint union` GE. The particular GE intended should be clear from context.We wish to see that an arbitrary deployment determines an equivalence class of (pos-sibly non-regular) languages with the fellow traveller property which map onto G. Westart by choosing a convenient alphabet.Definition.

The above normal form for elements G gives embeddings of the vertex groupsGV ⊂G. We can take a generating set A for G which is a union of generating sets forthe vertex groups together with a generator tE for each edge E /∈T.

We denote by AVthe subset of all elements of A evaluating into GV . We may choose our generators suchthat for each edge E of Y we have a subset of A which bijects to ∂1FE.

By identifyingany duplicates, we can assume that each element of a group ∂1FE ⊂G is representedby exactly one letter of A. We refer to the subset of A evaluating into ∂1FE as AE.

Inparticular, AE = AE−1 for E ∈T. Also, there is a unique element e ∈A which evaluatesto 1 ∈G and is in every AE.

We call A a convenient alphabet for G.6

Let ψ be a deployment. For each h ∈GE, ψ(h) is a class in A(G∂1E).

Choose a languageLψ(h) ∈ψ(h) for each ψ(h). We thus have Lψ(h) = Lψ(h′) whenever ψ(h) = ψ(h′).

Weassume our alphabet A is convenient and each GV -language is over the alphabet AV . Wetake tE to be the empty word for E ∈T.

Recall that E0 denotes the “base edge” that weadded to Y with ∂1E0 = V0. We denote the unique element of GE′ by 1 and defineLψ = {u0tE1 .

. .

tEmum : m ≥0, E1 . .

. Em forms a path based at V0,u0 ∈Lψ(1), ui ∈Lψ(u0tE1...tEi) for i ≥1,if Ei+1 = E−1ithen ui /∈∂1FE}.In particular, u0 ∈GV0, ui ∈G∂1Ei for i ≥1.Lemma 3.2.

Lψ has the asynchronous fellow traveller property and is determined up toequivalence by ψ.Proof. We first show that Lψ has the asynchronous fellow traveller property.

The fellowtraveller constant will be 1 + max{δh} where δh is a fellow traveller constant for ˆLψ(h) :=Sf∈AE fLψ(hf). This union is an automatic structure by the equivariance of ψ.So suppose w, w′ ∈Lψ with w = w′a, a ∈A.

These words determine based edgepaths p and p′ in Y up to terminal segments lying in T.First suppose we can take p= p′ so that w=u0tE1 . .

. tEmum and w′=u′0tE1 .

. .tEmu′m.

If m = 0 then w = u0 and w′ = u′0 both lie in Lψ(1) so they fellow-travel. Otherwise, u0 and u′0 differ at most by an element of ∂0FE1.

Thus u0 = u′0f0 withf0 ∈AE−11 . Hence, again, u0 and u′0 asynchronously fellow travel with the fellow travellerconstant of Lψ(1).More generally, u0tE1 .

. .

ui−1tEi = u′0tE1 . .

. u′i−1tEigi with gi ∈AEi.

We assumeinductively that the two word have asynchronously fellow-travelled to this point withfellow-traveller constant δ as above. If we put h = u′0tE1 .

. .u′i−1tEi then u′i ∈Lψ(h) ⊂ˆLψ(h) and giui ∈giLψ(hgi) ⊂ˆLψ(h).

Also, giui = u′ifi where fi ∈AE−1i+1 is such thatfitEi+1 = tEi+1gi+1. Thus giui and u′i asynchronously fellow-travel with the fellow-travellerconstant of ˆLψ(h).

Hence u0tE1 . .

. tEiui and u′0tE1 .

. .tEiu′i asynchronously δ-fellow-travel.Thus, by induction, w and w′ asynchronously δ-fellow-travel.We must now examine the case where we cannot take p = p′.

In this case, we canchoose the paths p and p′ so that (say) p′ is an initial segment of p. Then the previouscase will apply to w and w′a.Notice that the above argument also shows that [Lψ] did not depend on the choices{Lψ(h)}. For if we are given choices {L′ψ(h)} giving L′ψ, we repeat the argument using{L′′ψ(h)} where L′′ψ(h) = Lψ(h) ∪L′ψ(h) and observe that Lψ ∪L′ψ ⊂L′′ψ.Definition.

We say that ψ is a regular deployment if Lψ is regular, and hence an asyn-chronous automatic structure, for some choice of languages in the classes in Im(ψ). Wewill see in the proof of the following theorem that the word “some” in this definition canbe replaced by “any”.Theorem 3.3.

The map ψ 7→[Lψ] gives a bijection {regular deployments} →A(G).7

Proof. We start by constructing the inverse map.

That is, we construct a deployment ψLfrom [L] ∈A(G).Lemma 3.4 [BGSS]. Let L ⊂A∗be a finite to one rational structure for a group G.Then for each g ∈G there are only finitely many y ∈A∗so that for some x, z ∈A∗,xyz ∈L and y = g.Proof.

We suppose not, and let A be a finite state automaton for the language L. We thenhave x1y1z1, x2y2z2, . .

. ∈L with y1 = y2 = .

. ..

Among these we can find infinitely manyx′1y′1z′1, x′2y′2z′2, . .

. so that each of x′1, x′2, .

. .

labels a path from the start state of A to acommon state of A. Among these we can find infinitely many x′′1y′′1z′′1 , x′′2y′′2z′′2 , .

. .

so thateach of x′′1y′′1 , x′′2y′′2, . .

. labels a path from the start state of A to a common state of A. Butthen x′′1y′′1z′′1 , x′′1y′′2z′′1 , .

. .

∈L with x′′1y′′1z′′1 = x′′1y′′2z′′1 = . .

., contradicting the assumptionthat L is finite to one.Lemma 3.5. Let A be a convenient alphabet for G = π1(Y).

Given an asynchronousautomatic structure L′ on G, we can choose an asynchronous automatic structure L sothat L ∼L′ and L ⊂A∗. Moreover L can be chosen so that if y ∈A∗, xyz ∈L andy ∈∂1FE for some edge E of Y , then y has the form emfen with e, f ∈A and e = 1.L is synchronous if L′ is.

At the possible cost of turning a synchronous structure into anasynchronous one, we can assume y is the single letter f.Proof. It is an observation of [NS1], based on a result of [ECHLPT], that given arbitrarymonoid generating sets A and B for a group G and an synchronous or asynchronousautomatic structure L′ ⊂B∗for G, there is a synchronous respectively asynchronousautomatic structure L′′ ⊂A∗with L′′ ∼L′.

Choose A as in the lemma and L′′ as justdescribed. By the previous lemma, there are only finitely many y ∈A∗so that xyz ∈L′′and y is in some ∂1FE.

We consider such words that are not already in A{e}∗. For eachsuch word y there is a unique word wy ∈A{e}∗with the same length and value.Let A′′ be a finite state automaton for the language L′′.

Considering it as a finitegraph and replacing it by a finite cover if necessary, we may assume that every path inA′′ labelled by one of these words y is embedded. We now construct a nondeterministicmachine as follows: wherever we see a path labelled by one of the words y we add anew path from the beginning point of this path to its end point labelled by wy.

Call thelanguage of this machine N. The language we seek is obtained from the language of Nby removing the regular language of all words containing one of the words y as a subwordand is hence regular. Since two such subwords may be adjacent, we can only ensure thata word with value f has the form emfen.

If we wish a language in which the e’s do notoccur, we simply replace each edge of a machine labelled e by an ǫ-transition.We choose L as in Lemma 3.5. Then each element of L has a decompositionw = u0tE1 .

. .

tEmum(∗∗)where• tEi is the empty word if Ei ∈T;• E1 . .

. Em is a path in Y starting at V0;8

• ui ∈(A∂0Ei)∗for i < m, um ∈(A∂1Em)∗.• if Ei+1 = E−1ithen ui /∈FEi.This decomposition of w is unique up to the following two operations. The placementof each tEi, Ei ∈T, is determined up to one of the finitely many subwords emfen ofLemma 3.5.

Second, we can adjoin or delete a terminal reduced path of tEi’s lying in T.We call this decomposition an edge path decomposition of w.For each h ∈GE, we defineNh = {v ∈A∗: ∃w ∈L edge path decomposed as in (∗∗) above such thatfor some i, Ei = E, v = ui, u0tE1 . .

. ui−1tEi = h},and setLh =[f∈AEfNhf.

(As usual we think of h as lying in ` GE. This saves wear and tear on subscripts.

In thenext section it will be helpful to write LE,h instead. )Lemma 3.6.

For each h in GE, Lh is an asynchronous automatic structure for G∂1E.There are finitely many distinct languages Lh. The equivalence class of Lh depends onlyon h and the equivalence class of L. If L is a synchronous automatic structure, then sois Lh.

The assignment h 7→[Lh] is FE-equivariant in the following sense: for f ∈FE wehave [Lh] = f[Lhf]. In particular this assignment is a deployment ψL which depends onlyon [L].Proof.

Assume that h ∈GE. We first check that Nh is regular.

We will express Nh asthe union of two sublanguages, determined by whether the Ei+1 in the definition of Nhequals E−1 or not, and build a nondeterministic finite state automaton for each of theselanguages (the nondeterminism will consist only in possibly having several start states).The edge path decomposition of L-words induces similar decompositions for subwordsof L-words, which we will use in the following.Let A be a finite state automaton for L.There are only finitely many words inthe prefix-closure of L which evaluate to h.We take those which have an edge pathdecomposition ending with tE.We let Sh be the collection of states of A reached bythese words. Let R′E be the collection of states of A which are accept states or have apath to an accept state of A labelled by a word that is not in (A∂1E)∗and has edge pathdecomposition (as just described for L-subwords) tE′ .

. .

with ∂0E′ = ∂1E and E′ ̸= E−1.Let A′⟨be the nondeterministic machine obtained from A by making Sh the set of startstates, R′E the set of accept states, and deleting all arrows labelled by letters not in A∂1E.Let N ′h be the language accepted by the machine A⟨. Let R′′E be the collection of states ofA which have a path to an accept state of A labelled by a word that is not in (A∂1E)∗andhas edge path decomposition tE−1 .

. .. Let A′′⟨be defined like A′⟨but using R′′E instead ofR′E and let N ′′h be the corresponding language.

Then Nh = N ′h ∪(N ′′h −{e}∗AE{e}∗). Itis thus a regular language.

Moreover, it is determined by the E and the subset Sh of thestates of A, so there are only finitely many different languages Nh.9

The language Lh is now a finite union of regular languages, hence regular. Moreover,it is determined by E and the family of subsets Shf, f ∈AE, of the states of A, so thereare a finite number of these languages.We show that Lh surjects onto G∂1E.

For g ∈G∂1E let w ∈L be a word with valuehg. If v is the largest terminal segment of w which lies in (A∂1E)∗, then w decomposes asutEv with utE = hf and v = f −1g with f ∈AE.

Then fv ∈fNhf ⊂Lh and fv = g.We next show Lh has the asynchronous fellow traveller property. If not, we could findfivi ∈fiNhfi ⊂Lh, f ′iv′i ∈f ′iNhf ′i ⊂Lh, for i = 1, 2, .

. ., so that d(fivi, f ′iv′i) is bounded,but for any K, there is some i so that fivi and f ′iv′i do not asynchronously K-fellow travel.We would then have u1v1w1, u′1v′1w′1, u2v2w2, u′2v′2w′2, .

. .

∈L with hf i = ui, hf ′i = u′i foreach i. We can replace each wi and w′i with xi and x′i of bounded length.

Then for each i,uivixi, u′iv′ix′i ∈L, d(uivixi, u′i, v′ix′i) = d(hfivixi, hf ′iv′ix′i) = d(fivixi, f ′iv′ix′i) is bounded,and yet there is no K so that each of the pairs uivixi and u′iv′ix′i asynchronously K-fellowtravel. This contradicts the assumption that L is an asynchronous automatic structure.The same argument shows Lh is synchronous if L is.

If L ∼L′ we can apply theargument to L ∪L′ to see that, for a fixed h, [Lh] depends only on [L].Finally, theequivariance property is immediate from the definition of Lh.ProofofTheorem3.3continued. Itisclearthatψ7→[Lψ]maps{regulardeployments} →A(G).

We shall show below that [L] 7→ψ[L] maps A(G) →{regulardeployments}. Given this, it is easy to see that these maps are mutual inverses.For suppose we start with [L] ∈A(G).

For h ∈GE we take ˆLh = Nh ∪SLh∼Lh′ Lh′.Since this is a finite union of equivalent asynchronous automatic structures on G∂1E, it isitself an asynchronous automatic structure. We use the languages ˆLh ∈ψ[L](h) to defineLψ[L].

Then Lψ[L] contains the language L so certainly [L] = [Lψ[L]].Similarly, if ψ is a regular deployment then ψLψ(h) = [(Lψ)h] and (Lψ)h contains thelanguage eLψ(h). Thus ψ[Lψ](h) = [Lψ(h)] for each h. That is, ψ[Lψ](h) = ψ(h) for all h,so ψ[Lψ] = ψ.So, to complete the proof of 3.3, we need only prove that if L is an asynchronousautomatic structure on G then the deployment ψ[L] is regular.

Let ψ = ψ[L]. We shalldescribe a nondeterministic finite state automaton T forLψ, thus showing Lψ is a regularlanguage, so ψ is a regular deployment.We assume L is as in Lemma 3.5.

The following is our key lemma.Lemma 3.7. Given K > 0, there exists a finite state automaton S = SK with the followingproperties:1.

It accepts any word w ∈A∗;2. Suppose w is a word with value h which asynchronously K-fellow travels a word ofL with the same value.

Then the final state reached by w in the machine S tells one foreach edge E of bY whether h ∈GE, and if so, what the corresponding language Lh is. Inparticular, if L′ is the prefix closure of L, then the language {w ∈L′ : Lw = Lh} is regularfor any h ∈` GE.Proof.

Recall that the language Lh is determined by E and the map f 7→Shf of AE tothe power set of the set of states of A. Moreover, at any point along a path w, Sh is theset of A states reached by paths in A having the same value h as our path’s current value10

and decomposing as utE. Thus it behooves us to modify our machine A to make “visible”,the invisible tE’s when E ∈T.

So suppose we take the alphabet A ∪{rE : E ∈T}, andlet sE = tE if E /∈T, sE = rE if E ∈T. Then the reader can check that the language{u0sE1 .

. .sEmum : w = u0tE1 .

. .

tEmum is an edge path decomposition of w ∈L}is regular. (Here is a sketch proof.

Add a loop with label rE to every vertex of A for everyE ∈T. This machine accepts the language obtained from L by adding arbitrary subwordsin these rE’s.

The desired language is obtained from this by deleting any word that hasone of a certain finite collection of prohibited subwords; it is hence regular. )We take a deterministic machine for this language and replace each rE edge with atE edge which we take as an ǫ transition.

This is a machine for L and we assume A is ofthis form.Now suppose w K-fellow travels some path of L with the same value. Then any wordof L with the same value (K + KL)-fellow travels w, where KL is the fellow travellerconstant for L. It thus suffices to keep track at each step along w of what A states havebeen reached by paths which (K + KL)-fellow travelled ours and have final value in aK-neighborhood of our current value, and, when one of these has the same value as w,whether its final edge in A is a tE edge.

That is, the information we must keep track of isan element of Maps(B, P(S × E(A))), where B is a ball of radius K + KL in the Cayleygraph, S and E(A) are the sets of states and edges of A respectively, and P(·) denotes thepower set.More precisely, to keep track of the desired information we use a finite state automatonS with Maps(B, P(S × E(A))) as set of states. For α, β ∈Maps(B, S × P(E(A))) and ain our alphabet A, S has an edge labelled a from α to β if and only if each β(g) consistsof the set of pairs (s, e) such that there exists a path in B ∪aB from a point g1 ∈B to aglabelling a path in A from a state s′ with (s′, e′) in α(g1) to s with final edge e. As startstate we take the element σ with σ(g) equal to the set of pairs (s, e) so that s reachablein A from the start state by paths with value g and final edge e. Any word w ∈A∗thendefines a path in S from the start state.

If w asynchronously K-fellow travels an elementof L with the same value then this path ends in a state α with p(α(f)) = Swf for eachf ∈AE where p denotes projection onto the first factor. This state α thus gives the desiredinformation.We now return to the proof that Lψ is regular for ψ = ψL.

We choose regular languagesLψ(h) for each ψ(h) to define the language Lψ. By proving that this Lψ is regular, we willalso have proved the remark preceding Theorem 3.3.We have already shown that L asynchronously fellow travels one choice of Lψ (namely,the one with Lψ(h) = ˆLh).

Hence, by Lemma 3.2 it asynchronously fellow travels anychoice of Lψ. Let K be the fellow traveller constant for our particular choice.

Let S bethe machine of the above lemma.Let A′ be the disjoint union of machines for the languages Lψ(h). We shall constructa nondeterministic machine for the language Lψ by adding some arrows to the productmachine of A′ and S. Namely, for each state s of this product machine and each edge Eof Y we will add an arrow labelled tE from s to the following state t, if it exists.

The11

S component of t is the one determined by the tE-transition from the S-component of s.The A′ component is the start state of the machine for LhtE, where ψ(htE) = [LhtE] isdetermined by E and the S component of s as in the above lemma. It is easy to see thatthis machine performs as advertised.There is an entirely analogous version of Theorem 3.3 for synchronous automaticstructures.Theorem 3.8.

The bijection of Theorem 3.3 restricts to a bijection between SA(G) andthe set of regular deployments whose images lie in `V ∈vert Y SA(GV ).Proof. Given a synchronously automatic structure L on G, Lemma 3.6 ensures that thedeployment ψL takes its image in `V ∈vert Y SA(GV ).

It remains to check that if ψ is aregular deployment whose image lies in `V ∈vert Y SA(GV ) then we can find a synchronouslyautomatic structure L ∼Lψ.So suppose ψ is such a deployment. We choose synchronous automatic structureswith uniqueness Lψ(h) for each class in Im ψ.

For each such language and each edge E′, ifthe language occurs as Lψ(h) with h ∈GE and ∂1E = ∂0E′, we choose a set of FE′-cosetrepresentatives in the language Lψ(h). To do this we order our alphabet A.

This induces atotal order ≺on A∗by ordering first on length and then by lexicographic order for wordsof a given length. LetL′ψ(h),E′ = {w ∈Lψ(h) : w is ≺-minimal among w with w ∈wFE′}.By [BGSS], each of these languages is regular.

We takeL = {u0tE1 . .

. tEmum :u0 ∈GV0, ui ∈G∂1Ei for i > 0,ui ∈L′ψ(u0tE1 ...tEi),Ei+1 for i < m,um ∈Lψ(u1...tEm)}.Notice that by the normal form for graphs of groups, L bijects to G. Further, L ⊂Lψ,hence L ∼Lψ.

To see that L is regular, one builds a product machine based on S and A′,where here A′ is the disjoint union of machines for the languages L′ψ(h),E′, and modifiesthis to accept Lψ(h) in the final factor. Finally we wish to see that L has the synchronousfellow traveller property.

We repeat the argument that Lψ has the asynchronous fellowtraveller property, but with the following observation. Suppose w = u0tE1 .

. .

tEmum andw′ = u′0tE′1 . .

.tE′m′ u′m′ with w, w′ ∈L and w = w′a. Let p = E1 .

. .

Em, p′ = E′1 . .

. E′m′.If p = p′, then by our choice of coset representatives, u0tE0 .

. .

tEm = u′0tE′0 . .

.tE′m′ , andsince um and um′ synchronously fellow travel, so, too, do w and w′. On the other hand, ifwe cannot choose the decompositions so that p = p′, then (say) p′ is an initial segment ofp and we in fact have w′ = wa.Theorem 3.9.

Let L be an asynchronous automatic structure on G = π1(Y). Then eachsubgroup H ⊂G conjugate to a vertex group is L-rational, and hence has an inducedautomatic structure LH, determined up to equivalence by [L].Let H be the set of allconjugates of vertex groups.

Then the map [L] 7→([LH])H∈H defines mapsA(G) →YH∈HA(H),SA(G) →YH∈HSA(H),12

which are injective and have dense image in the product topology.Proof. We replace Y by a reduced graph of groups (this is defined just before Lemma 3.1).This does not affect any vertex with non-trivial A(GV ), and hence does not change thevalidity of the theorem.

Let L ⊂A∗be an asynchronous automatic structure on G chosenaccording to Lemma 3.5. Let H = hGV h−1 with h ∈GE be as in Lemma 3.1 and denoteV = ∂1E.

By considering the edge path decomposition one sees that any w ∈L withw ∈H −h(∂1FE)h−1 has the form w = xyz with x, (z)−1 ∈h∂1FE and y ∈(AV )∗. Sincethere are just finitely many possibilities for x and z, the set of L-words of this form is aregular sub-language of L. Thus H −h(∂1FE)h−1 is rational, whence H is, since h(FE)h−1is finite.We also see that {w ∈L : w ∈H} ∼hLhh−1.

Thus [LH] = hψL(h)h−1. In particular,the map H 7→[LH] determines ψL and hence [L], so the map A(G) →QH∈H A(H)is injective.

The statement that this map has dense image is the statement that, if wespecify a structure LH for finitely many H, there is a structure L on G which realizesthese LH’s. We defer the proof of this to the next section.The statement of the theorem in the synchronous case follows from the above togetherwith Theorem 3.8.Remark.

We can turn Lh itself into an automatic structure on H if we use the evaluationmap AV →H given by a 7→hah−1. With this interpretation, [Lh] = [LH].4.

Y-graphsRegular deployments are unsatisfactory as classifying objects. This is for several reasons.First is the fact that it is hard to specify an arbitrary deployment and there is no convenientway to tell a priori whether a deployment is in fact regular.

This fact is reflected stronglyin the second, namely, that regularity is non-local in following sense. If one changes thevalue of a regular deployment on one element h ∈` GE one is likely to obtain a non-regulardeployment.In this section we introduce a classifying object which avoids these deficiencies.Definition.

Given a finite graph of groups Y with finite edge groups, a Y-graph X is afinite directed labelled graph X with the following additional structure:• A map π: X →Y of underlying graphs is given. A vertex v of X with π(v) = V iscalled a V -vertex and an edge e of X with π(e) = E is called an E-edge.

This is calledthe Y-type of v or e.• A vertex v0 of X is chosen as start vertex and every vertex of X can be reached by adirected path from this start vertex. (We may assume that v0 is a V0-vertex, whereV0 is the base vertex for Y chosen in Section 3.

)• Each V -vertex v is labelled by an element [Lv] ∈A(GV ).• Each edge e out of v is labelled by an [Lv]-rational subset Se of GV . For each edgeE of Y out of V , the labels on the E-edges out of v are disjoint.

Their union is GVif v = v0 or if v has an incoming edge of Y-type other than E−1. If v ̸= v0 and allincoming edges at v are E−1-edges, their union is GV −∂0FE.• For each V -vertex v and each edge E out of V , there is a FE-action on X whichfixes all vertices except those reached by one E-edge from v. This action respects13

labels in the following sense. For a vertex v′ reached by an E-edge from v we have[Lfv′] = f[Lv′].

For an edge e departing v with label Se ⊂GV the edge fe has labelSfe = Sef −1. For an edge e departing a vertex reached by an edge from v we haveSfe = fSe.Given a Y-graph, X , the following choices determine a language LX for G = π1(Y).Choose a convenient generating set A for G. For each vertex v of X , choose an asyn-chronous automatic structure Lv ⊂(Aπ(v))∗in the class associated to v. For an edge edeparting v let Le be the sublanguage of words of Lv that represent elements of the setSe.

Let T be a maximal spanning tree in Y. For each edge E of Y let tE be as in Section3, that is, it is the corresponding stable letter if E /∈T and the empty word if E ∈T.Then LX isLX = {u0tπ(e1) .

. .

tπ(em)um : e1 . .

. em is a path in X from the start vertex,uk ∈Lek+1 for k = 0, .

. ., k −1,um ∈L∂1em,uk+1 /∈∂0(Fπ(ek+1)) if π(ek+1) = π(ek)−1}Theorem 4.1.

The above language LX is an asynchronous automatic structure on π1(Y)and depends, up to equivalence, only on the Y-graph X .Every asynchronous automatic structure on π1(Y) is equivalent to one constructed asabove.Proof. The proof that LX has the asynchronous fellow traveller property and is determinedup to equivalence by X is just like the proof of the analogous statement for a languagedetermined by a deployment (Lemma 3.2), and is left to the reader.

(In fact, it is nothard to see that LX is contained in a language determined by the following deploymentψ. Given h ∈GE, we find a path p in X from the start vertex of X whose final edge ehas π(e) = E, and the language determined by p contains a word with value h. We takeψ(h) = [L∂1e].

)We must check that LX is regular. To do this it is helpful to modify LX by redefiningtE for each E ∈T temporarily to be a new letter which evaluates to 1 ∈G, rather thanthe empty word.

We first turn X into a generalized finite state automaton AX. We dothis by subdividing each edge e of X into two edges.

We label the first of these by Leand the second by tπ(e). The start state of AX is the start vertex of X .

We take allvertices of AX to be accept states. The language of this machine contains LX .

In fact,LX is exactly the sublanguage of words containing no substring of the form tEut−1Ewithu ∈Lv, π(v) = ∂1E, u ∈∂1FE. Since there are finitely many such strings, LX is regularas required.

If we now replace each letter tE with E ∈T by the empty word we get ouroriginal LX back, and it is still regular.We must check that every asynchronous automatic structure arises as above. SupposeL is an asynchronous automatic structure on G. We assume that our language L andalphabet A are as in Lemma 3.5.

We shall construct a Y-graph X for L of a rather specialtype. The start vertex will have no incoming edges and each vertex other than the startvertex will have incoming edges all of one Y-type.14

Let bY be Y extended by a base edge as in Section 3, and for E an edge of bY, letGE be as defined in Section 3. We refer to the edge path decomposition of elements of Ldescribed before Lemma 3.6.

For h ∈GE we defineN E,h = {v ∈A∗: ∃w ∈L with edge path decomposition w = u0tE1 . .

. tEmumsuch that for some i, Ei = E, u0tE1 .

. .

ui−1tEi = hv = uitEi+1 . .

. tEmum},andLE,h =[f∈∂1fEfN E,hf.We claim that for each h ∈GE the language N E,h is regular.

Let A be a machine forL. As in the proof of Lemma 3.6, we let Sh be the set of states of A reached by words inthe prefix closure of L which evaluate to h and have an edge path decomposition endingin tE.

Then N E,h is the language accepted by the machine obtained from A by makingSh the set of start states. Thus N E,h is regular.

It is also determined by the finite set ofstates Sh, so there are finitely many different languages N E,h. Thus there are also finitelymany languages LE,h, and they are regular.

The language LE,h is determined by the mapf 7→Shf of AE to the power set of the set of states of A.The subset of G onto which the language LE,h evaluates depends only on E: it isthe set of elements of G whose normal form decomposition with base vertex ∂1E cannotstart with tE−1. Note that, except for a finite number of elements of G, if an element isdistance 1 from this set then it is also in this set.

It therefore makes sense to talk about theasynchronous fellow traveller property for LE,h, even though this language does not surjectto G. We claim that LE,h has this property. For suppose fv ∈fN E,hf and f ′v′ ∈f ′N E,hf ′with f, f ′ ∈∂1FE and d(fv, f ′v′) ≤1.

Then there exist u, u′ with u = hf and u′ = hf ′so that uv ∈L and u′v′ ∈L. Since uv, u′v′ ∈L and d(uv, u′v′) = d(fv, f ′v′) ≤1, uv andu′v′ asynchronously fellow-travel.

It follows that after reparameterization, fv and f ′v′ alsoasynchronously fellow-travel.For fixed E and h, h′ ∈GE it therefore also makes sense to ask if LE,h ∼LE,h′. WedefinebLE,h =[LE,h′∼LE,hLE,h′.Since this is a finite union of regular equivalent languages, it is also regular with theasynchronous fellow-traveller property.EachbLE,hinducesanasynchronousautomaticstructurebLE,honG∂1E.Namely, we definebLE,h = {u ∈A∗: ∃v ∈A∗, uv ∈bLE,h, u ∈G∂1E}.It is easy to see that this language is an asynchronous automatic structure on G∂1E.

Infact, it is just SLE,h′∼LE,h LE,h′, where LE,h′ is the Lh′ of Lemma 3.6 (we are now makingthe edge E explicit in our notation).15

We are now prepared to describe the Y-graph, X of the Theorem. For each E ∈edge bYit has a vertex for each bLE,h.

The vertex corresponding to bLE,h projects to ∂1E underπ, and is labelled by [bLE,h]. We take the vertex corresponding to bLE0,1 to be the startvertex.

Suppose that E and E′ are edges of Y with ∂1E = ∂0E′. There is an edge E′ fromthe vertex for bLE,h to the vertex for bLE′,h′ if the setSE′ = {g ∈G∂1E : bLE′,h′ = bLE′,hg}is not empty.

In this case E′ is labelled by SE′ and projects to E′ under π.We must check that this defines a Y-graph. That is, we must see that the set SE′ iswell defined, that it is bLE,h-rational, that the labels on the E′ edges out of the vertex forbLE,h partition G∂1E if E′ ̸= E−1 and partition G∂1E −∂1FE if E′ = E−1, and that X hasthe appropriate equivariance properties.So suppose g ∈G∂1E.We check that bLE′,hg depends only on bLE,h and g, anddoes not depend on h.We first show that LE′,hg depends only on LE,h and g.Noww ∈LE′,hg = Sf∈AE′ fN E′,hgf if and only if w = fv with f ∈AE′ and there is u withu = hgf and uv ∈L, and u ending with tE′ in some edge path decomposition of uv.

Ifthere is such a u, it has the form u = xy with x = hf ′, y = f ′−1gf, where f ′ in ∂1FE.Thus w = fv ∈LE′,hg if and only if we find f ′yv ∈LE,h with f ′y = gf. (The readermight want to draw a picture.) Thus LE′,hg depends only on LE,h and g, and not onh.

A similar argument shows that LE′,hg ∼LE′,h′g if LE,h ∼LE,h′. So by taking theappropriate unions we see that bLE′,hg depends only on bLE,h and g, and thus SE′ is welldefined.The fact that SE′ is bLE,h-rational will follow from the existence of the machine Sof Lemma 3.7.

Recall that that machine does the following: if w is a word with valueh ∈GE which fellow travels some L-word with the same value then the state of S reachedby w determines the language LE,h (called Lh in Lemma 3.7). The way it does this is bydetermining the map f 7→Shf of AE to the power set of the set of states of A.

But, as wesaw above, this map also determines LE,h, and hence bLE,h. Thus LE,h can be replaced bybLE,h in Lemma 3.7.Now let u be an L-word for h and v be a bLE,h-word for g. Then, by constructionof bLE,h, the word w = uv fellow travels an L-word for hg.

We can thus test w with themachine S to see if bLE′,hg = bLE′,h′. That is, g ∈SE′ if and only if the word v for g labelsa path in S from the state reached by u to a state that determines the language bLE′,h′.This is a regular condition, so SE′ is bLE,h-rational, as required.The labels on the E′-edges out of the vertex for bLE,h are disjoint, since bLE,h and gdetermine bLE′,hg, and their union is clearly G∂1E −∂1FE if E′ = E−1 and G∂1E otherwise.We must construct the FE′ action and show that it respects labels.

To this end, let E′′be an edge of Y with ∂1E′ = ∂0E′′. We suppose that there is an edge E′′ℓfrom the vertexfor bLE′,h′ to the vertex for bLE′′,h′′.

If f ∈FE′, we let f carry the vertex for bLE′,h′ = bLE′,hgto the vertex bLE′,hgf −1. As we have seen, this is well defined.

The label at this vertex is[LE′,h′f −1], which is [fLE′,h′] as required. It now follows that the action of f on verticesinduces an action on edges out of the vertex for bLE,h, and this action respects edge labels,16

for if g ∈SE′ then gf −1 labels an edge from the vertex for bLE,h to the f image of thevertex for bLE′,h′. In particular, we have SfE′ = SE′f −1 as required.

In the same way, thef action on vertices induces an action on edges whose initial vertex is moved by f carrying(say) E′′ℓto (say) fE′′ℓso that SfE′′ℓ= fSE′′ℓ. This completes the proof that X is indeeda Y-graph.Finally, we must check that L is equivalent to LX .

It is an easy induction on freeproduct length that L and LX determine the same deployment. Alternatively, one maynote that LX is equivalent to L since it contains L as a sublanguage.Remark.

Call a Y-graph X special if it satisfies:• the start vertex v0 has no incoming edges;• each vertex v ̸= v0 has incoming edges of just one Y-type.Then the Y-graph X for L constructed in the above proof is special, and it is nothard to verify that it is minimal with this property, in the sense that any other specialY-graph X ′ defining a language equivalent to L can be mapped to X by a graph mappingthat respects the Y-type of vertices and edges, respects vertex labels, and also respectsedge labels in the sense that the rational set associated to an edge of X ′ is contained inthe rational set associated to corresponding edge of X . Thus minimal special Y-graphsactually classify asynchronous automatic structures on G = π1(Y).

They are, however, notalways efficient classifying objects, in that one can often find a much smaller non-specialY-graph to describe the same structure, as we will now describe.Let A be a convenient alphabet for G = π1(Y). Suppose X is a Y-graph and letLX ⊂A∗be, as in Theorem 4.1, the language of words labelling paths in X from the startvertex.

For any vertex v of X we can define similarly the language Lv of words in A∗thatlabel paths starting at v. (If X is as constructed in the above proof and v is the vertexcorresponding to bLE,h then Lv ∼bLE,h.) Now suppose that for some vertex V of Y we haveV -vertices v and v′ of X such that Lv ∪Lv′ has the asynchronous fellow-traveller property.We can then attempt to create a smaller Y-graph by identifying the vertices v and v′ ofX .

If E is an incoming edge at V then we have a FE-action on X which permutes the V -vertices, so we must do this identification equivariantly. There is no guarantee that we cando this, for if Lv ∪Lv′ and Lv′ ∪Lv′′ have the asynchronous fellow-traveller property, wecannot deduce that Lv ∪Lv′′ does.

However, if we can do this identification equivariantly,we obtain a smaller Y-graph for [L].Even if one can collapse X as above, there may be several inequivalent ways of doingso. Indeed, it is not hard to find an example of a graph of groups Y with finite edgegroups for which the Y-graph X constructed in the proof of Theorem 4.1 can be collapsedto several inequivalent “minimal” Y-graphs.There are some situations in which the above collapse is clearly possible.

For example,if one has a Y-graph X with no “special” vertices — that is, every vertex v has more thanone Y-type of incoming edge — then every Lv surjects to G so the condition that Lv ∪Lv′have the asynchronousfellow traveller property defines an equivalence relation on the vertices of X . It is theneasy to see that there is a Y-graph that can be obtained by collapsing X as above.Another case is when the language L is asynchronously biautomatic.Each Lv is17

equivalent to a sublanguage of a translate of L. But, by definition of biautomaticity, anytranslate of L is equivalent to L. We can thus collapse all V -vertices to a single vertex foreach V . This collapse is clearly equivariant and extends trivially to edges, so we see thatthere is a Y-graph X whose underlying graph X is isomorphic to the underlying graph Yof Y.

This proves part of:Theorem 4.2. L is an asynchronously biautomatic structure on π1(Y) if and only if Lhas a Y-graph X for which π is an isomorphism of the underlying graphs of X and Y andthe structure at each vertex v of X is an asynchronously biautomatic structure for Gπ(v).The corresponding statement holds also with “asynchronously biautomatic” replacedby “biautomatic”.Proof.

We first point out that a asynchronous or synchronous biautomatic structure on Ginduces asynchronous or synchronous biautomatic structures on the vertex groups, sincethey are rational subgroups. Thus the labels on the vertices of the above Y-graph X areas claimed.Now suppose we have a Y-graph X as in the theorem.

We check that the ensuingstructure L is asynchronously biautomatic. So suppose A is a convenient alphabet we havechosen languages at each vertex, and suppose also that w = u0tE1 .

. .

tEmum is an edgepath decomposition of a word in the resulting language. We must show that if a ∈A, theword for aw asynchronously fellow travels w. There are several cases.

We use the samesymbol for a vertex of Y and the corresponding vertex of X .We suppose first that u0 ∈(AV0)∗is not empty and a ∈GV0. We take u′0 to be theword in LV0 for au0.

Then w′ = u′0tE1 . .

.tEmum is an accepted word for aw. Since LV0 isasynchronously biautomatic, w′ and aw asynchronously fellow travel as required.

If LV0 isbiautomatic, they synchronously fellow travel.We now suppose that u0 is non-empty and that a /∈GV0).Then we have w′ =au0tE1 . .

. tEmum an accepted word and again aw and w′ appropriately fellow travel, unlessit happens that a = tE, u0 ∈∂1FE, and the first letter after u0 in w is tEi = tE−1.

In thiscase w′ = f ′ui+1 . .

. um, for suitable f ′ ∈AE−1, is the word we seek, and again aw and w′synchronously or asynchronously fellow travel as required.The remaining cases are similar and are left to the reader.Proof of Theorem 3.9 (completed).

We need to show the map of Theorem 3.9 has denseimage. Suppose that every vertex group GV has an asynchronous automatic structure.We will first show that G = π1(Y) has at least one asynchronous automatic structure (thisfollows from the methods of [S2], but we give a proof here for completeness).

We shallneed the following lemma.Lemma 4.3. Suppose that E is an edge of Y, with ∂0E = V , and suppose we are given[LV ] ∈A(GV ).

Then there is a partition of GV into distinct [LV ]-rational sets Sf, f ∈FEso that FE acts on the right to permute these sets {Sf}.Proof. We assume LV ∈[LV ] is a structure with uniqueness.

We take L′ to be those wordsw ∈LV that are least in dictionary order among the words that evaluate into wFE. SinceFE is finite, and it is easy to check dictionary order by means of a finite state automaton,18

the languageL′′ = {(u, v) ∈LV × LV : u ∈vFE and u precedes v}is the language of an asynchronous two tape automaton. It follows that L′ = LV −p2(L′′)is regular.

(Here p2 denotes projection onto the second factor.) We take S1 = L′, and foreach f ∈FE, we take Sf = S1f.

It is easy to check that each of these is LV -rational.We now construct a special Y-graph X ′. This graph will have a vertex for each edgeE of bY and each element of ∂1FE.

For fixed E these vertices will constitute a ∂1FE-orbit,and they will be labelled by the orbit of structures [∂1fLV ] ∈A(GV ), f ∈FE. The abovelemma allows us to put in edges in an equivariant fashion to complete the Y-graph X ′.Let [L′] be the structure determined by X ′.Recall that H is the set of conjugates of vertex groups in G = π1(Y) and we are tryingto show that the mapL 7→(LH)H∈H: SA(G) →YH∈HSA(H)has dense image.

We must show that if H1, . .

., Hn are distinct groups in H and we aregiven [LH1], . .

., [LHn] in A(H1), . .

., A(Hn), we can find [L] ∈A(G) so that [L] induces[LH1], . .

., [LHn] on H1, . .

., Hn. We shall modify the structure L′ described above to dowhat is required.As discussed at the beginning of the proof of Theorem 3.9, we may assume that Yis reduced.

For each i = 1, . .

.n choose Ei and hi ∈GE⟩as in Lemma 3.1 with Hi =hiG∂1Eih−1i. For each i the normal form for hi determines a path in Y starting at the basevertex V0.

Inclusion of these paths in each other induces a partial order on the hi andhence on the Hi. We may assume the ordering H1, .

. ., Hn respects this partial order.

Wewill describe a modification of X ′ to make the structure on Hi equal to the desired onewithout changing the structure on any Hj which is earlier in the partial order. Repeatingthis iteratively for i = 1, .

. ., n then proves the theorem.Thus suppose i is chosen and write h = hi, H = Hi.By taking a cover of Y ifnecessary, we may assume that the path σ in Y determined by h is embedded.

There isan induced covering of X ′ and we replace X ′ by this covering. We choose a lift σ′ of thepath σ to X ′.Suppose σ′ has length at least 2 (we leave the case that it is shorter to the reader).Let the final two edges of σ′ be e′ and e, so ∂1e′ = ∂0e = w say.

Let π(e′) = E′, π(e) = E,π(w) = W. There is a word u0 . .

. tE′utE labelling the path σ′ and evaluating to h. Thenu ∈Se.

We delete u from Se and establish a new edge out of v to a new vertex. We labelthe new edge with {u} and the new vertex with [h−1LHh].

We let the edges out of thisnew vertex duplicate the edges out of ∂1e. Likewise, for each f ∈∂1FE we delete uf fromSf(e) and establish a new edge with label {uf} to a new vertex labelled [(hf)−1LHhf].For each of the vertices f(w) with f ∈∂1FE′ we perform the same operation, constructingnew edges to the vertices we have just added.

This produces a new Y-graph for a structurewhich induces the desired structure on H = hGV h−1 and has not changed the inducedstructure on any earlier Hj.19

Remark. In the proof of Theorem 4.1, we were required to show that a Y-graph X deter-mines a regular language LX , and to do this, we turned X into a generalized finite stateautomaton AX which almost accepted the language in question.

To obtain the desiredlanguage, we only needed to delete those words containing subwords of the form tEft−1Ewhere f ∈∂1FE. In fact, there is a straight forward procedure for turning an Y-graphX into a generalized finite state automaton which accepts LX itself.

The method hereis to build an generalized finite state automaton BX whose underlying graph projects tothat of AX. For each vertex of v of X , and each edge E into v, there are two vertices inBX .

One of these is reached only by elements of ∂1FE, and there are no π(E−1) edges outof this vertex. The other is reached by all elements not in ∂1FE.

This latter has a fullarmamentarium of edges out of it. The interested reader may wish to fill in the detailsalong the lines of the proofs of Lemmas 1.1 and 3.1 in [S2].5.

The boundaryWe recall the boundary of an asynchronous automatic structure, as defined in [NS1]. LetL ⊂A∗be an asynchronous automatic structure on a group G. As usual, we assume L isfinite to one.

An L-ray is an infinite word w ∈AN, all of whose initial segments are initialsegments of L-words. Two rays are equivalent if they asynchronously fellow travel (at adistance that may depend on the rays).

The boundary of L is the set ∂L of equivalenceclasses of rays with the following topology. For an L-rational subset R of G, define ∂R tobe the set of rays which fellow travel R (that is, travel in a bounded neighborhood of R;the bound may depend on the ray).

These sets form a basis of closed sets for a topologyon ∂L. This boundary can be attached to G: the sets cl R := R ∪∂R are a basis of closedsets for a topology on cl G := G ∪∂L which has ∂L as a closed subspace and G as an opendiscrete subspace.

(This topology is the “rational topology” of [NS1]. Other topologieson ∂L are also discussed there.

)In [NS1] the “rehabilitated boundary” c∂L is also discussed, which appears to bean appropriate notion for groups with large abelian subgroups.A subset σ of ∂L iscalled an abstract simplex if, for any choice of a neighborhood in cl G for each point of σ,the intersection of these neighborhoods is non-empty. This makes ∂L into a topologicalabstract simplicial complex, the geometric realization of which is the rehabilitated boundaryc∂L.Let B be a tree and Bˆ= B ∪C be its end compactification.

Given a continuous mapp of a space X to B, we define the tree completion of X with respect to p as the disjointunionXˆ= X ∪C,with the smallest topology for which X is a subspace and the induced map pˆ: Xˆ→Bˆ iscontinuous. It is an easy exercise to see that Xˆis compact if and only if p is a proper mapand is Hausdorf if and only if X is Hausdorf.Now let Y be a graph of groups with finite edge stabilizers and G = π1(Y).

By [Se],there is a G-tree B with B/G equal to the underlying graph Y of Y and with edge andvertex stabilizers given by the data of Y. Let π: B →Y be the projection.

The stabilizerof a vertex v of B is therefore a conjugate Hv of the vertex group Gπ(v).20

As in section 3, for [L] ∈A(G) and H a conjugate of a vertex group, [LH] denotes theinduced structure on H.Theorem 5.1. Let X be the disjoint union of boundaries ∂LHv, indexed by the vertices ofB, and p: X →vert B ⊂B the obvious map.

Then the boundary ∂L is the tree completionXˆ. The analogous statement holds also for rehabilitated boundaries.Proof.

We first describe B, following Serre [Se]. It has vertices `V ∈vert Y G/GV .Wedenote the vertex determined by V and [g] ∈G/GV by g eV .

For each E ∈edge Y andeach [g] ∈G/∂0FE there is an edge, denoted g eE, from g g∂0E to gtE g∂1E. (Thus the reverseof the edge g eE is the edge determined by gtE gE−1; this is slightly different notation from[Se].) Serre shows that B is a tree and the quotient by the obvious action of G is Y .We sketch the proof that B is a tree.We can choose a base vertex for B as thevertex v0 = 1eV0, where V0 is the base vertex for Y.A normal form representationh = g0tE1g1 .

. .

tEmgm with Em = E for an element of G, as defined early in section3, determines a path in B from the base vertex v0 of B to the vertex v = h g∂1E. Thispath consists of the sequence of edges g0fE1, .

. .

, gm−1gEm. Thus B is a connected graph.Moreover, since we can right-multiply h by an element of G∂1E without changing v, we canassume gm ∈∂1FE.

Then h ∈GE and h is determined up to the right-action of FE. Thuswe see that v is actually determined by E and an element of GE/FE.

Now it not hard tosee that any path without back-tracking from v0 to v gives a normal form representationfor this h, and uniqueness of normal forms up to the operations mentioned in section 3leads to uniqueness of such paths, showing that B is a tree.This also shows that the vertices of B are in one-one correspondence with`E∈bY GE/FE. From this point of view the stabilizer of the vertex v = h g∂1E correspondingto h ∈GE/FE is Hv = hG∂1Eh−1, and the induced language LHv is equivalent to thelanguage Lh on G∂1E (see Remark at end of section 3).Now suppose L is an asynchronous automatic structure on G. Since the boundary ∂Lonly depends on the equivalence class of L, we may assume that L is chosen as in Lemma3.5.

We may also assume it is prefix-closed. Note that if R is an L-rational subset of Gand R′ is its “prefix-closure” (i.e., R′ = N′, where N ′ is the prefix closure of the set N ofL-words evaluating into R), then R′ lies in a bounded neighborhood of R, so ∂R′ = ∂R.As bound one may take the diameter of a finite state automaton for N. Thus, in discussingthe topology on ∂L we need only consider “prefix-closed” rational subsets of G.For any word u ∈L, the shortest edge path decomposition (see definition precedingLemma 3.6) determines a shortest normal form representative for u, and hence, as above, asimple path γu from the base vertex v0 in B.

If u1 is a subword of u then γu1 is a subpathof γu. It follows that an L-ray w determines a simple path γw in B, which is a finite orinfinite path according as longer and longer initial segments of w eventually all evaluateinto a fixed hGV or not.

We shall need the following Lemma.Lemma 5.2. 1.

For any k > 0 there exists K > 0 such that if u, u′ ∈L satisfy d(u, u′) ≤kthen, by deleting at most the last K letters from u and u′ one may obtain words u0 andu′0 with γu0 = γu′0.21

2.If the L-ray w fellow travels a subset S ⊂G then every initial segment of γwappears as an initial segment of some γg, g ∈S. The converse holds if γw is infinite andS = R with R ⊂L prefix-closed.Proof.

1. Since L is finite-to-one, there exists a function φ: N →N such that any terminalsegment u1 of an L-word with len(u1) > φ(k) satisfies d(u1, 1) > k. Let k1 = max{d(f, 1) :f ∈FE for some edge E of Y} and define K = φ(k + k1).

Now suppose that u, u′ ∈Lsatisfy d(u, u′) ≤k but do not satisfy the conclusion of the lemma. Let γ be the longestcommon segment of γu and γu′, and let u0 and u′0 be the longest initial segments of u andu′ with γu0 = γu′0 = γ.

Write u = u0u1, u′ = u′0u′1. At least one of u1 and u′1, say u1, haslength greater than K. By choice of K, the distance of u0 to u = u0u1 exceeds k + k1,so the distance from u to u0f exceeds k for any f in an edge group.

But, by consideringthe normal form of u−1u′ one sees that the shortest path in the Cayley graph from u′ tou must pass through u0f for some f ∈∂0FE, where E is the first edge of u1 in the edgepath decomposition of u = u0u1. This path hence has length exceeding k, contradictingd(u, u′) ≤k.2.

Part 1 of the lemma shows that if w fellow travels a subset S of G then there existarbitrarily long initial segments u of w with γu equal to an initial segment of a path γgwith g ∈S. But every initial segment of γw is an initial segment of some such γu, so thefirst sentence of Lemma 5.2.2 is proved.Conversely, suppose γw is infinite and S = R with R ⊂L prefix-closed and supposeevery initial segment of γw is an initial segment of some γg with g ∈S.

For a given initialsegment γ of γw, choose such a g = u with u ∈R and let u0 and w0 be the initial segmentsof u and w corresponding to γ. Then u0 and w0 differ by an element of the edge group forthe final edge of γ, so u0 and w0 fellow travel.

Since u0 travels in S and w0 is an arbitrarilylong initial segment of w, the result follows.We return to the proof of Theorem 5.1 for the boundary ∂L. If two rays w and w′fellow travel then their paths γw and γw′ in B are equal by Lemma 5.2.1.

Moreover, if γwand γw′ are equal and infinite then w and w′ do fellow travel by Lemma 5.2.2. Thus raysw with γw infinite determine a subset of ∂L that bijects to the set C of infinite rays in B.This is the same as the set of ends of B.Suppose now γw is finite, say it ends at the vertex of B determined by [h] ∈GE/FE.Then by cutting w at the point where it has determined the whole path γw, we write win the form w0u with w0 = hf for some f ∈AE, and fu a ray in Lh.

If w′ is another raywith the same path γw we decompose it likewise as w′0u′ with w′0 = hf ′ so that f ′u′ is aray in Lh. Then w and w′ fellow travel if and only if the rays fu and f ′u′ fellow travel.We thus get a copy of ∂Lh in ∂L.

We have thus shown that, as a set, ∂L is as claimed inthe theorem.It remains to verify that the topology is correct. Consider v ∈vert B.

Let the simplepath in B from v0 to v be γ. The set Rv := {u ∈L : γu = γ} is regular.

Hence, Rv ⊂Gand its complement are both rational. The rays which fellow-travel Rv define the imageof ∂LHv in ∂L = X ∪C and the rays that fellow travel the complement of Rv define thecomplement of ∂LHv.

It follows that ∂LHv is an open and closed subset of ∂L. Moreover,rational subsets of Rv correspond to rational subsets of Hv, so ∂LHv carries the appropriate22

topology as a subspace of ∂L.Now suppose w is a ray for which γw is infinite, so w represents an element of C ⊂∂L.Suppose S = R ⊂G is a rational subset with R ⊂L prefix-closed, and suppose [w] is in theset U of equivalence classes of rays that fail to fellow-travel S. Then Lemma 5.2.2 impliesthat there is some initial segment γ of γw which does not appear as an initial segment ofany γg, g ∈S. Let Sγ be the set of g ∈G such that γg has γ as an initial segment and letUγ ⊂∂L be the set of equivalence classes of rays which fail to fellow-travel G −Sγ.

Then,[w] ∈Uγ ⊂U, so these sets Uγ form a neighborhood basis for [w] ∈∂L. But Uγ is the setof equivalence classes of rays w such that γw has γ as an initial segment.

This defines thetopology on ∂L claimed in the theorem.To see the analogous statement for the rehabilitated boundary we must show thatevery non-trivial abstract simplex in ∂L is an abstract simplex of some ∂LHv ⊂∂L andvice versa. It is easy to see that an abstract simplex of ∂LHv is one for ∂L.

Thus, wemust show that if x, y are points of ∂L which do not lie in some common ∂LHv, then theydo not form an abstract simplex, that is, they have disjoint neighborhoods in cl G. Bywhat was said above, a set of the form cl Rv is an open and closed subset of cl G whoseintersection with ∂L is ∂LHv. These sets are disjoint for different v’s, so they providedisjoint neighborhoods for x and y lying in distinct sets ∂LHv.

Suppose just one of x andy lies in a ∂LHv, say x ∈∂LHv and y ∈C. Then y = [w] with γw infinite, so we canchoose an initial segment γ of γw which is not an initial segment of the path in B fromv0 to v. The set cl Sγ is an open and closed neighborhood of y which is disjoint from theneighborhood cl Rv of x.

Finally, if x = [w] and y = [w′] are distinct points of C and γand γ′ are initial segments of γw and γw′ which are longer than the longest common initialsegment of γw and γw′ then cl Sγ and cl Sγ′ are disjoint open and closed neighborhoods ofx and y.Remark. cl G can also be seen as a tree completion.

For g ∈G the path γg ends in a vertexvg of B, so we get a map p: G →vert B ⊂B. If v is the vertex determined by [h] ∈GE/FE,then p−1(v) = h(G∂1E −∂1FE) (except that p−1(v0) = GV0 rather than GV0 −{1}).

Thatis, up to a finite set p−1(v) is just a translate of the group Gπ(v) on which the languageLh (equivalent to LHv) is defined, so we can attach to p−1(v) the boundary ∂Lh ≃∂LHv.This gives us a topology on G ∪`v∈vert B ∂LHv and a map of this space to vert B ⊂B.Then cl G is its tree completion. We leave the details to the reader.References[B]N. Brady, Asynchronous automatic structures on closed hyperbolic surface groups,preprint, University of California 1993.

[BGSS] G. Baumslag, S. M. Gersten, M. Shapiro and H. Short, Automatic groups andamalgams, Journal of Pure and Applied Algebra 76 (1991), 229–316. [ECHLPT] D.B.A.

Epstein, J.W. Cannon, D.F.

Holt, S.V.F. Levy, M.S.

Paterson, andW.P. Thurston, “Word Processing in Groups,” Jones and Bartlett Publishers, Boston,1992.[GS]S.

M. Gersten and H. Short, Rational sub-groups of biautomatic groups, Annals ofMath. 134 (1991), 125–158.23

[N]W. D. Neumann, Asynchronous combings of groups, Internat. J. Alg.

Comp. 2(1992), 179–185.

[NS1] W. D. Neumann and M. Shapiro, Equivalent automatic structures and their bound-aries, Internat. J. Alg.

Comp. 2 (1992), 443–469.

[NS2] W. D. Neumann and M. Shapiro, Automatic structures and geometrically finitehyperbolic groups, preprint.[Se]J. P. Serre, Trees, Graduate Texts in Mathematics (Springer Verlag, 1980).

[S1]M.Shapiro,Nondeterministicanddeterministicasynchronousautomaticstructures, Internat. J. Alg.

and Comp. (1992).[S2]M.

Shapiro, Automatic structures and graphs of groups, in: “Topology ‘90, Pro-ceedings of the Research Semester in Low Dimensional Topology at Ohio State,”(Walter de Gruyter Verlag, Berlin - New York 1992), 355–380.The Ohio State UniversityDepartment of MathematicsColumbus, OH 43210City CollegeDepartment of MathematicsNew York, NY 1003124

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