Context-free pairs of groups I: Context-free pairs and graphs

Abstract

Let $G$ be a finitely generated group, $A$ a finite set of generators and $K$ a subgroup of $G$. We define what it means for $(G,K)$ to be a context-free pair; when $K$ is trivial, this specializes to the standard definition of $G$ to be a context-free group.

We derive some basic properties of such group pairs. Context-freeness is independent of the choice of the generating set. It is preserved under finite index modifications of $G$ and finite index enlargements of $K$. If $G$ is virtually free and $K$ is finitely generated then $(G,K)$ is context-free. A basic tool is the following: $(G,K)$ is context-free if and only if the Schreier graph of $(G,K)$ with respect to $A$ is a context-free graph.

1. Introduction

Let $G$ be a finitely generated group and $K$ a (not necessarily finitely generated) subgroup of $G$. We can choose a finite set $A\subset G$ of generators such that every element of $G$ is of the form $g=g_1\cdots g_n$, where $n\ge 0$ and $g_1,\dots ,g_n\in A$. Thus, $A$ generates $G$ as a semigroup. We shall say that $(G,K)$ is context-free, if–loosely spoken–the language of all words over $A$ that represent an element of $K$ is context-free.

The precise definition needs some preparation. Let $\Sigma$ be a finite alphabet and $\psi :\Sigma \to G$ be a (not necessarily injective) mapping such that $A=\psi \left(\Sigma \right)$ satisfies the above finite generation property for $G$. Then $\psi$ has a unique extension, also denoted $\psi$, as a monoid homomorphism $\psi :{\Sigma }^{\ast }\to G$. Recall that ${\Sigma }^{\ast }$ consists of all words $w=a_1\cdots a_n$, where $n\ge 0$ and $a_1,\dots ,a_n\in \Sigma$ (repetitions allowed). The number $n$ is the length $\left|w\right|$ of $w$. If $n=0$ this means that $w=ϵ$, the empty word. This is the neutral element of ${\Sigma }^{\ast }$, and ${\Sigma }^{\ast }$ is a free monoid with the binary operation of concatenation of words. The extension of $\psi$ is of course given by

$$\psi (a_1\cdots a_n)=\psi \left(a_1\right)\cdots \psi \left(a_n\right)\text{,}$$

where the product on the right hand side is taken in $G$. Given these ingredients, we shall say that $\psi :\Sigma \to G$ is a semigroup presentation of $G$, referring to the fact that $A$ generates $G$ as a semigroup. A language over $\Sigma$ is a non-empty subset of ${\Sigma }^{\ast }$.

Definition 1.1

The word problem of $(G,K)$ with respect to $\psi$ is the language

$$L(G,K,\psi )=\{w\in {\Sigma }^{\ast }:\psi \left(w\right)\in K\}\text{.}$$

We say that the triple $(G,K,\psi )$ is context-free, if $L(G,K,\psi )$ is a context-free language.

A context-free grammar is a quadruple $\mathcal{C}=(V,\Sigma ,P,S)$, where $V$ is a finite set of variables, disjoint from the finite alphabet $\Sigma$ (the terminal symbols), the variable $S$ is the start symbol, and $P\subset V\times {(V\cup \Sigma )}^{\ast }$ is a finite set of production rules. We write $T⊢u$ or $(T⊢u)\in P$ if $(T,u)\in P$. For $v,w\in {(V\cup \Sigma )}^{\ast }$, we write $v⟹w$ if $v=v_{1}T{v}_2$ and $w=v_{1}u{v}_2$, where $u,v_1,v_2\in {(V\cup \Sigma )}^{\ast }$ and $T⊢u$. This is a single derivation step, and it is called rightmost, if $v_2\in {\Sigma }^{\ast }$. A derivation is a sequence $v=w_0,w_1,\dots ,w_k=w\in {(V\cup \Sigma )}^{\ast }$ such that $w_{i-1}⟹w_i$; we then write $v\stackrel{\ast }{⟹}w$. A rightmost derivation is one where each step is rightmost. The succession of steps of any derivation $T\stackrel{\ast }{⟹}w\in {\Sigma }^{\ast }$ can be reordered so that it becomes a rightmost derivation. For $T\in V$, we consider the language $L_T=\{w\in {\Sigma }^{\ast }:T\stackrel{\ast }{⟹}w\}$. The language generated by $\mathcal{C}$ is $L\left(\mathcal{C}\right)=L_S$.

A context-free language is a language generated by a context-free grammar. As a basic reference for Language and Automata Theory, we refer to the magnificent monograph of Harrison[6].

The above definition of a context-free pair, or rather triple, $(G,K,\psi )$ makes sense when $G$ is a finitely generated monoid and $K$ is a sub-monoid, but here we are interested in groups. When in addition $K=\left\{1_G\right\}$, this leads to the notion of $G$ being a context-free group. In two celebrated papers, Muller and Schupp[11], [12] have carried out a detailed study of context-free groups and more generally, context-free graphs. In particular, context-freeness of a group is independent of the particular choice of the generating set $A$ of $G$. The main result of [11], in combination with a fundamental theorem of Dunwoody [4], is that a finitely generated group is context-free if and only if it is virtually free, that is, it contains a free subgroup of finite index. (In [11], it is assumed that $A=A^{-1}$ and that $\psi :\Sigma \to A=\psi \left(\Sigma \right)$ is one-to-one, but the results carry over immediately to the more general setting where those two properties are not required.)

Previously, Anisimov [1] had shown that the groups whose word problem $L(G,\left\{1_G\right\},\psi )$ is regular (see Section 2 for the definition) are precisely the finite groups.

The above mentioned context-free graphs are labelled, rooted graphs with finitely many isomorphism classes of cones. The latter are the connected components of the graph that remain after removing a ball around the root with arbitrary radius. See Section 4 for more precise details. As shown in [12], there is a natural correspondence between such graphs and pushdown automata, which are another tool for generating context-free languages; see Section 3.

Among subsequent work, we mention Pélecq [13] and Sénizergues [16], who studied actions on, resp. quotients of context-free graphs. Group-related examples occur also in Ceccherini-Silberstein and Woess [3].

More recently, Holt et al. [7] have introduced and studied co-context-free groups, which are such that the complement of $L(G,\left\{1_G\right\},\psi )$ is context-free, see also Lehnert and Schweitzer [9]. This concept has an obvious extension to co-context-free pairs of groups, resp. graphs, on whose examination we do not (yet) embark.

In the present notes, we collect properties and examples of context-free pairs of groups $(G,K)$.

• •The language $L(G,K,\psi )$ is regular if and only if the index $[G:K]$ of $K$ in $G$ is finite (Proposition 2.4).
• •The property that $L(G,K,\psi )$ is context-free does not depend on the specific choice of the semigroup presentation $\psi$, so that context-freeness is just a property of the pair $(G,K)$, a consequence of Lemma 3.1.
• •If $(G,K)$ is context-free then $L(G,K,\psi )$ is a deterministic context-free language (see Section 3 for the definition) for any semigroup presentation $\psi :\Sigma \to G$ (Corollary 4.8.a).
• •If $(G,K)$ is context-free and $H$ is a finitely generated subgroup of $G$, then the pair $(H,K\cap H)$ is context-free (Lemma 3.1).
• •If $[G:H]<\infty$ then $(G,K)$ is context-free if and only if $(H,K\cap H)$ is context-free (Proposition 3.3 & Lemma 4.9).
• •If $(G,K)$ is context-free and $H$ is a subgroup of $G$ with $K\le H$ and $[H:K]<\infty$ then $(G,H)$ is context-free (Lemma 4.9).
• •If $K$ is finite then $G$ is context-free if and only if $(G,K)$ is context-free (Lemma 4.11).
• •If $(G,K)$ is context-free then $(G,g^{-1}Kg)$ is context-free for every $g\in G$ (Corollary 4.8.b).
• •If $G$ is virtually free and $K$ is a finitely generated subgroup of $G$ then $(G,K)$ is context-free (Corollary 5.3).

Several of these properties rely on the following.

• •A fully deterministic, symmetric labelled graph (see Section 2 for definitions) is context-free in the sense of Muller and Schupp if and only if the language of all words which are labels of a path that starts and ends at a given root vertex is context-free (Theorem 4.2, Theorem 4.6).

  The (harder) “if” part is not contained in previous work. It implies the following.

• •The pair $(G,K)$ is context-free if and only if for some ($⟺$ any) symmetric semigroup presentation $\psi :\Sigma \to G$, the Schreier graph of $(G,K)$ with respect to $\psi$ is a context-free graph. (See again Section 2 for precise definitions).

In a second paper [20], a slightly more general approach to context-freeness of graphs via cuts and tree-sets is given. It allows to show that certain structural properties (“irreducibility”) are preserved under finite-index-modifications of the underlying pair of groups. This is then applied to random walks, leading in particular to results on the asymptotic behaviour of transition probabilities.

In concluding the Introduction, we remark that with the exception of some “elementary” cases, context-free pairs of groups are always pairs with more than one end. Ends of pairs of groups were studied, e.g., by Scott [15], Swarup [18] and Sageev [14]. This leads directly to asking about the interplay between context-freeness of pairs and decomposition as amalgamated products or HNN-extensions. An example at the end of Section 5 shows that there is no immediate answer.

2. Schreier graphs and the regular case

Let $\Sigma$ be a finite alphabet. A directed graph labelled by $\Sigma$ is a triple $(X,E,\ell )$, where $X$ is the (finite or countable) set of vertices, $E\subset X\times \Sigma \times X$ is the set of oriented, labelled edges and $\ell :E\ni (x,a,y)\mapsto a\in \Sigma$ is the labelling map.

For an edge $e=(x,a,y)\in E$, its initial vertex is $e^{-}=x$ and its terminal vertex is $e^{+}=y$, and we say that $e$ is outgoing from $x$ and ingoing into $y$. If $y=x$ then $e$ is a loop, which is considered both as an outgoing and as an ingoing edge. We allow multiple edges, i.e., edges of the form $e_1=(x,a_1,y)$ and $e_2=(x,a_2,y)$ with $a_1\ne a_2$, but here we exclude multiple edges where also the labels coincide. The graph is always assumed to be locally finite, that is, every vertex is an initial or terminal vertex of only finitely many edges. We also choose a fixed vertex $o\in X$, the root or origin. We shall often just speak of the graph $X$, keeping in mind the presence of $E$ and $\ell$.

We call $X$ fully labelled if at every vertex, each $a\in \Sigma$ occurs as the label of at least one outgoing edge. We say that $X$ is deterministic if at every vertex all outgoing edges have distinct labels, and fully deterministic if it is fully labelled and deterministic. Finally, we say that $X$ is symmetric or undirected if there is a fixed point free involution $a\mapsto a^{-1}$ of $\Sigma$ (i.e., ${\left(a^{-1}\right)}^{-1}=a$, excluding the possibility that $a^{-1}=a$) such that for each edge $e=(x,a,y)\in E$, also the reversed edge $e^{-1}=(y,a^{-1},x)$ belongs to $E$.

A path in $X$ is a sequence $\pi =e_1{e}_2\dots e_n$ of edges such that $e_i^{+}=e_{i+1}^{-}$ for $i=1,\dots ,n-1$. The vertices ${\pi }^{-}=e_1^{-}$ and ${\pi }^{+}=e_n^{+}$ are the initial and the terminal vertex of $\pi$. The number $\left|\pi \right|=n$ is the length of the path. The label of $\pi$ is $\ell \left(\pi \right)=\ell \left(e_1\right)\ell \left(e_2\right)\cdots \ell \left(e_n\right)\in {\Sigma }^{\ast }$. We also admit the empty path starting and ending at a vertex $x$, whose label is $ϵ$. Denote by ${\Pi }_{x,y}={\Pi }_{x,y}\left(X\right)$ the set of all paths $\pi$ in $X$ with initial vertex ${\pi }^{-}=x$ and terminal vertex ${\pi }^{+}=y$. The following needs no proof.

Lemma/Definition 2.1

Let $(X,E,\ell )$ be a labelled graph, $x\in X$ and $w\in {\Sigma }^{\ast }$. We define ${\Pi }_x\left(w\right)=\{\pi :{\pi }^{-}=x,\ell \left(\pi \right)=w\}$, the set of all paths that start at $x$ and have label $w$. The set of all terminal vertices of those paths is denoted $x^w=\{{\pi }^{+}:\pi \in {\Pi }_x\left(w\right)\}$.

Analogously, we define ${\overline{\Pi }}_x\left(w\right)=\{\pi :{\pi }^{+}=x,\ell \left(\pi \right)=w\}$, the set of all paths that terminate at $x$ and have label $w$, and write $x^{-w}=\{{\pi }^{-}:\pi \in {\overline{\Pi }}_x\left(w\right)\}$.

If $X$ is fully labelled, then ${\Pi }_x\left(w\right)$ is always non-empty.

If $X$ is deterministic, then ${\Pi }_x\left(w\right)$ has at most one element, and if that element exists, it is denoted ${\pi }_x\left(w\right)$, while $x^w$ just denotes its endpoint.

If $X$ is fully deterministic, then $x^w$ is a unique vertex of $X$ for every $x\in X$, $w\in {\Sigma }^{\ast }$.

Finally, if $X$ is symmetric (not necessarily deterministic), then ${\overline{\Pi }}_x\left(w\right)={\Pi }_x\left(w^{-1}\right)$, where for $w=a_1\cdots a_n$, one defines $w^{-1}=a_n^{-1}\cdots a_1^{-1}$.

With a labelled, directed graph as above, we can associate various languages. We can, e.g., consider the language

$$L_{x,y}=L_{x,y}\left(X\right)=\{\ell \left(\pi \right):\pi \in {\Pi }_{x,y}\left(X\right)\}\text{,}\text{where }x,y\in X\text{.}$$ (1)

Definition 2.2

Let $G$ be a finitely generated group, $K$ a subgroup and $\psi :\Sigma \to G$ a semigroup presentation of $G$. The Schreier graph $X=X(G,K,\psi )$ has vertex set

$$X=K\setminus G=\{Kg:g\in G\}$$

(the set of all right $K$-cosets in $G$), and the set of labelled, directed edges

$$E=\{e=(x,a,y):x=Kg,y=Kg\psi \left(a\right),\text{ where }g\in G,a\in \Sigma \}\text{.}$$

$X$ is a rooted graph with origin $o=K$, the right coset corresponding to the neutral element $1_G$ of the group $G$. The Schreier graph is fully deterministic. It is also strongly connected: for every pair $x,y\in X$, there is a path from $x$ to $y$. (This follows from the fact that $\psi \left(\Sigma \right)$ generates $G$ as a semigroup.) When $K=\left\{1_G\right\}$ then we write $X(G,\psi )$. This is the Cayley graph of $G$ with respect to $\psi$, or more loosely speaking, with respect to the set $\psi \left(\Sigma \right)$ of generators.

Note that $X$ can have the loop $e=(x,a,x)\in E$ with $x=Kg$. This holds if and only if $\psi \left(a\right)\in g^{-1}Kg$. It can also have the multiple edges $e_1=(x,a_1,y)$ and $e_2=(x,a_2,y)$ with $x=Kg$ and $a_1\ne a_2$. This occurs if and only if $\psi \left(a_2\right)\psi {\left(a_1\right)}^{-1}\in g^{-1}Kg$. In particular, there might be multiple loops. The following is obvious.

Lemma 2.3

Let $K$ be a subgroup of $G$ and $\psi :\Sigma \to G$ be a semigroup presentation of $G$ . Then

$$L(G,K,\psi )=L_{o,o}\left(X\right)$$

is the language of all labels of closed paths starting and ending at $o=K$ in the Schreier graph $X(G,K,\psi )$ .

A context-free grammar $\mathcal{C}=(V,\Sigma ,P,S)$ and the language $L\left(\mathcal{C}\right)$ are called linear, if every production rule in $P$ is of the form $T⊢v_{1}U{v}_2$ or $T⊢v$, where $v,v_1,v_2\in {\Sigma }^{\ast }$ and $T,U\in V$. If furthermore in this situation one always has $v_2=ϵ$ (the empty word), then grammar and language are called right linear or regular.

A finite automaton $\mathcal{A}$ consists of a finite directed graph $X=(X,E,\ell )$ with label set $\Sigma$ and labelling map $\ell$, together with a root vertex $o$ and a non-empty set $F\subset X$. The vertices of $X$ are called the states of $\mathcal{A}$, the root $o$ is the initial state, and the elements of $F$ are the final states. The automaton is called (fully) deterministic provided the labelled graph $X$ is (fully) deterministic. The language accepted by $\mathcal{A}$ is

$$L\left(\mathcal{A}\right)=\bigcup _{x\in F}{L}_{o,x}\left(X\right)\text{.}$$

If $\mathcal{A}$ is deterministic, then for each $w\in L\left(\mathcal{A}\right)$ there is a unique path $\pi \in {\bigcup }_{x\in F}{\Pi }_{o,x}\left(X\right)$ such that $\ell \left(\pi \right)=w$. A state $y\in X$ is called useful if there is some word $w\in L$ such that the vertex $y$ lies on a path in ${\bigcup }_{x\in F}{\Pi }_{o,x}\left(X\right)$ with label $w$. It is clear that we can remove all useless states and their ingoing and outgoing edges to obtain an automaton which accepts the same language and is reduced: it has only useful states.

It is well known [6, Chapter 2] that a language $L\subseteq {\Sigma }^{\ast }$ is regular if and only if $L$ is accepted by some deterministic finite automaton.

The following, which corresponds to Theorem 1 in [5], generalizes Anisimov’s [1] characterization of groups with regular word problem, and also simplifies its proof, as well as the simpler one of [11, Lemma 1].

Proposition 2.4

Let $G$ be a finitely generated group, $K$ a subgroup and $\psi :\Sigma \to G$ a semigroup presentation of $G$ . Then $(G,K)$ has regular word problem with respect to $\psi$ if and only if $K$ has finite index in $G$ .

Proof

Suppose first that the index of $K$ in $G$ is finite. Consider the finite automaton $\mathcal{A}=(X,o,\left\{o\right\})$ where $X$ is the Schreier graph $X(G,K,\psi )$, and the initial and unique final state is $o=K$ (as a vertex of $X$). Then $L(G,K,\psi )=L\left(\mathcal{A}\right)$: indeed, $w\in {\Sigma }^{\ast }$ belongs to $L(G,K,\psi )$, i.e. $\psi \left(w\right)\in K$, if and only if $K=K\psi \left(w\right)$. This shows that $L(G,K,\psi )$ is regular.

Conversely, suppose that $L=L(G,K,\psi )$ is regular and accepted by the reduced, deterministic finite automaton $\mathcal{A}=(X,o,F)$. For $y\in X$ there is some word $w\in L$ such that the vertex $y$ lies on the unique path from $o$ to $F$ with label $w$. We choose one such $w$ and let $w_y$ be the label of the final piece of the path, starting at $y$ and ending at $F$. We set $g_y=\psi {\left(w_y\right)}^{-1}\in G$.

Let $g\in G$. There are $w,\overline{w}\in {\Sigma }^{\ast }$ with $\psi \left(w\right)=g$ and $\psi \left(\overline{w}\right)=g^{-1}$. Thus, $w\overline{w}\in L=L(G,K,\psi )$, and there is a (unique) path $\pi$ with label $w\overline{w}$ from $o$ to some final state. Now consider the initial piece ${\pi }_w$ of $\pi$, that is, the path starting at $o$ whose label is our $w$ that we started with. [Thus, we have proved that such a path ${\pi }_w$ must exist in $X$ !] Let $y$ be the final state (vertex) of ${\pi }_w$. Then clearly $w{w}_y\in L\left(\mathcal{A}\right)$, which means that $g{g}_y^{-1}=\psi \left(w{w}_y\right)\in K$. Since $\psi \left({\Sigma }^{\ast }\right)=G$, it follows that

$$G=\bigcup _{y\in X}K{g}_y\text{,}$$

and $K$ has finitely many cosets in $G$. □

Corollary 2.5

Let $G$ be finitely generated and $K$ a subgroup. Then the property of the pair $(G,K)$ to have a regular word problem is independent of the semigroup presentation of $G$ .

We shall see that the same also holds in the context-free case. Another corollary that we see from the proof of Proposition 2.4 is the following.

Corollary 2.6

Let $G$ be finitely generated and $K$ a subgroup of finite index. Then for any semigroup presentation $\psi :\Sigma \to G$ , any reduced deterministic automaton $\mathcal{A}=(X,o,F)$ that accepts $L(G,K,\psi )$ has a surjective homomorphism (as a labelled oriented graph with root $o$ ) onto the Schreier graph $X(G,K,\psi )$ . Also, the labelled graph $X$ is fully deterministic.

Proof

Let $\mathcal{A}=(X,o,F)$ be deterministic and reduced, as in part 2 of the proof of Proposition 2.4.

Let $y\in X$, and recall the construction of the label $w_y$ of a path from $y$ to $F$, and $g_y=\psi {\left(w_y\right)}^{-1}\in G$. If $v$ is another path from $y$ to $F$, and $h=\psi {\left(v\right)}^{-1}$, then we can take $w\in L_{o,y}$ (which we know to be non-empty) and find that $w{w}_y,wv\in L(G,K,\psi )$, so that $\psi \left(w\right)\in K{g}_y\cap Kh$. Thus $K{g}_y=K\psi \left(w\right)=Kh$, and the map $\kappa :X\to K\setminus G,y\mapsto K{g}_y$ is well defined. It has the property that when $w\in L_{o,y}$, then $K\psi \left(w\right)=K{g}_y$. The map $\kappa$ is clearly surjective, and $\kappa \left(o\right)=K$ by construction.

Now let $y\in X$ and $a\in \Sigma$. Take $w\in L_{o,y}$ and consider the word $wa$. Again by part 2 of the proof of Proposition 2.4, there is a unique path ${\pi }_{wa}$ in $X$ starting at $o$ with label $wa$. If $y$ is its final vertex, then there is the edge $e=(y,a,z)$ in $X$. In this situation, $\kappa \left(z\right)=K\psi \left(wa\right)=K{g}_y\psi \left(a\right)=\kappa \left(y\right)\psi \left(a\right)$. This means that in the Schreier graph, there is the edge with label $a$ from $\kappa \left(y\right)$ to $\kappa \left(z\right)$. Therefore $\kappa$ is a homomorphism of labelled graphs.  □

The following simple example shows that, in general, the map $\kappa$ constructed in the proof of the previous corollary is not injective.

Example 2.7

Let $G={\mathbb{Z}}_2=\{1,t\}$ be the group of order two and $K=\left\{1\right\}$ the trivial subgroup. Let $\Sigma =\left\{a\right\}$ and consider the presentation $\psi :\Sigma \to G$ such that $\psi \left(a\right)=t$. Then $L(G,K,\psi )=\{a^{2n}:n\ge 0\}$.

In Fig. 1 we have represented, from left to right, the Schreier graph $X(G,K,\psi )$ (which is nothing but the Cayley graph of $G$ w.r. to $\psi$), and two automata ${\mathcal{A}}_1$ and ${\mathcal{A}}_2$. As usual, $o$ denotes the origin, while the sets of final states are $F_1=\left\{o\right\}$ and $F_2=\{o,f\}$, respectively. We have $L\left({\mathcal{A}}_1\right)=L\left({\mathcal{A}}_2\right)=L(G,K,\psi )$.

Fig. 1.

(From left to right) the Schreier graph $X(G,K,\psi )$ described in Example 2.7 and two automata ${\mathcal{A}}_1$ and ${\mathcal{A}}_2$ such that $L\left({\mathcal{A}}_1\right)=L\left({\mathcal{A}}_1\right)=L(G,K,\psi )$.

3. Pushdown automata

Besides grammars, we shall need another instrument for generating context-free languages. A pushdown automaton is a 7-tuple $\mathcal{A}=(Q,\Sigma ,Z,\delta ,q_0,Q_f,z_0)$, where $Q$ is a finite set of states, $\Sigma$ the input alphabet as above, $Z$ a finite set of stack symbols, $q_0\in Q$ the initial state, $Q_f\subset Q$ the set of final states, and $z_0\in Z\cup \left\{ϵ\right\}$ is the start symbol. Finally, the function $\delta :Q\times (\Sigma \cup \left\{ϵ\right\})\times (Z\cup \left\{ϵ\right\})\to {\mathcal{P}}_{\mathrm{fin}}(Q\times Z^{\ast })$ is the transition function. Here, ${\mathcal{P}}_{\mathrm{fin}}(Q\times Z^{\ast })$ stands for the collection of all finite subsets of $Q\times Z^{\ast }$.

The automaton works in the following way. At any time, it is in some state $p\in Q$, and the stack contains a word $\zeta \in Z^{\ast }$. The automaton reads a word $w\in {\Sigma }^{\ast }$ from the “input tape” letter by letter from left to right. If the current letter of $w$ is $a$, the state is $p$ and the top (=rightmost) symbol of the stack word $\zeta$ is $z$, then it performs one of the following transitions.

• (i)$\mathcal{A}$ selects some $(q,{\zeta }^{\prime })\in \delta (p,a,z)$, changes into state $q$, moves to the next position on the input tape (it may be empty if $a$ was the last letter of $w$), and replaces the rightmost symbol $z$ of $\zeta$ by ${\zeta }^{\prime }$, or
• (ii)$\mathcal{A}$ selects some $(q,{\zeta }^{\prime })\in \delta (p,ϵ,z)$, changes into state $q$, remains at the current position on the input tape (so that $a$ has to be treated later), and replaces the rightmost symbol $z$ of $\zeta$ by ${\zeta }^{\prime }$.

If both $\delta (p,a,z)$ and $\delta (p,ϵ,z)$ are empty then $\mathcal{A}$ halts.

The automaton is also allowed to continue to work when the stack is empty, i.e., when $\zeta =ϵ$. Then the automaton acts in the same way, by putting ${\zeta }^{\prime }$ on the stack when it has selected $(q,{\zeta }^{\prime })\in \delta (p,a,ϵ)$ in case (i), resp. $(q,{\zeta }^{\prime })\in \delta (p,ϵ,ϵ)$ in case (ii).

We say that $\mathcal{A}$ accepts a word $w\in {\Sigma }^{\ast }$ if starting in the state $q_0$ with only $z_0$ on the stack and with $w$ on the input tape, after finitely many transitions the automaton can reach a final state with empty stack and empty input tape. The language accepted by $\mathcal{A}$ is denoted $L\left(\mathcal{A}\right)$.

The pushdown automaton is called deterministic if for any $p\in Q$, $a\in \Sigma$ and $z\in Z\cup \left\{ϵ\right\}$, it has at most one option what to do next, that is,

$$\left|\delta (p,a,z)\right|+\left|\delta (p,ϵ,z)\right|\le 1\text{.}$$

(Here, $|\cdot |$ denotes cardinality.)

It is well known [6] that a language is context-free if and only if it is accepted by some pushdown automaton. A context-free language is called deterministic if it is accepted by a deterministic pushdown automaton. We also remark here that a deterministic context-free language $L$ is un-ambiguous, which means that it is generated by some context-free grammar in which every word of $L$ has precisely one rightmost derivation.

The following lemma is modelled after the indications of [11, Lemma 2]. For the sake of completeness, we include the full proof.

Lemma 3.1

Suppose that $G,K,\Sigma$ and $\psi :\Sigma \to G$ are as above. Let $H$ be a finitely generated subgroup of $G$ , and let ${\Sigma }^{\prime }$ be another alphabet and ${\psi }^{\prime }:{\Sigma }^{\prime }\to H$ be such that $F^{\prime }={\psi }^{\prime }\left({\Sigma }^{\prime }\right)$ generates $H$ as a semigroup.

Then, if $L(G,K,\psi )$ is context-free, also $L(H,K\cap H,{\psi }^{\prime })$ is context-free, and if in addition $L(G,K,\psi )$ is deterministic, then so is $L(H,K\cap H,{\psi }^{\prime })$ .

Proof

We start with a pushdown automaton $\mathcal{A}=(Q,\Sigma ,Z,\delta ,q_0,Q_f,z_0)$ that accepts $L(G,K,\psi )$.

For each $b\in {\Sigma }^{\prime }$, there is $u\left(b\right)\in {\Sigma }^{\ast }$ such that ${\psi }^{\prime }\left(b\right)=\psi \left(u\left(b\right)\right)$, and we may choose $u\left(b\right)$ to have length $\ge 1$. Thus,

$$w^{\prime }=b_1\cdots b_n\in L(H,K\cap H,{\psi }^{\prime })⟺u\left(b_1\right)\cdots u\left(b_n\right)\in L(G,K,\psi )\text{.}$$

With this in mind, we modify $\mathcal{A}$ in order to obtain a pushdown automaton ${\mathcal{A}}^{\prime }$ that accepts $L(H,K\cap H,{\psi }^{\prime })$. Our ${\mathcal{A}}^{\prime }$ has to translate any $w^{\prime }=b_1\cdots b_n\in {\left({\Sigma }^{\prime }\right)}^{\ast }$ into $w=u\left(b_1\right)\cdots u\left(b_n\right)\in {\Sigma }^{\ast }$ and to use $\mathcal{A}$ in order to check whether $w\in L(G,K,\psi )$.

Let $m+1=max\{\left|u\left(b\right)\right|:b\in {\Sigma }^{\prime }\}$. If $m=0$ then the only modification of $\mathcal{A}$ needed is to replace $\Sigma$ by its subset ${\Sigma }^{\prime }$ and to use the resulting restriction of the transition function.

Otherwise, we set ${\Sigma }_m=\Sigma \cup {\Sigma }^2\cup \cdots \cup {\Sigma }^m$. For $v\in {\Sigma }^{+}={\Sigma }^{\ast }\setminus \left\{ϵ\right\}$, we denote by $v_{+}$ its subword obtained by deleting the first letter. We define $Q^{\prime }=Q\cup (Q\times {\Sigma }_m)$ and ${\mathcal{A}}^{\prime }=(Q^{\prime },{\Sigma }^{\prime },Z,{\delta }^{\prime },q_0,Q_f,z_0)$ with the transition function ${\delta }^{\prime }$ as follows. For each $p\in Q$ and $z\in Z$,

$${\delta }^{\prime }(p,ϵ,z)=\delta (p,ϵ,z)\text{,}$$

$${\delta }^{\prime }(p,b,z)=\delta (p,a,z),\text{if}u\left(b\right)=a\in \Sigma \text{,}$$

$${\delta }^{\prime }(p,b,z)=\{((q,u{\left(b\right)}_{+}),\zeta ):(q,\zeta )\in \delta (p,a,z)\},\text{if }u\left(b\right)\in a{\Sigma }^{+}\text{,}$$

$${\delta }^{\prime }((p,v),ϵ,z)=\{((q,v),\zeta ):(q,\zeta )\in \delta (p,ϵ,z)\}\cup \{((q,v_{+}),\zeta ):(q,\zeta )\in \delta (p,a,z)\},\text{if }v\in a{\Sigma }^{+},{\delta }^{\prime }((p,a),ϵ,z)=\{((q,a),\zeta ):(q,\zeta )\in \delta (p,ϵ,z)\}\cup \delta (p,a,z),\text{if}a\in \Sigma \text{.}$$

Thus, the new states of the form $(p,v)$ with $1\le \left|v\right|<m$ serve to remember the terminal parts $v$ of the words $u\left(b\right)$, $b\in {\Sigma }^{\prime }$. This automaton accepts $L(G,K,{\psi }^{\prime })$, and it is deterministic, if $\mathcal{A}$ has this property. □

Corollary 3.2

Being context-free is a property of the pair $(G,K)$ that does not depend on the specific choice of the alphabet $\Sigma$ and the map $\psi :\Sigma \to G$ for which $\psi \left(\Sigma \right)$ generates $G$ as a semigroup.

Therefore, it is justified to refer to the context-free pair $(G,K)$ rather than to the triple $(G,K,\psi )$. Furthermore, whenever this is useful, we may restrict attention to the case when the graph $X(G,K,\psi )$ is symmetric: we say that $\psi$ is symmetric, if there is a proper involution $a\mapsto a^{-1}$ of $\Sigma$ such that $\psi \left(a^{-1}\right)=\psi {\left(a\right)}^{-1}$ in $G$. (Again, it is not necessary to assume that $\psi$ is one-to-one, so that we have that $a^{-1}\ne a$ even when $\psi {\left(a\right)}^2=1_G$.)

Proposition 3.3

Let $G$ be finitely generated, $H$ be a subgroup with $[G:H]<\infty$ . If $K$ is a subgroup of $H$ then $(G,K)$ is context-free if and only if $(H,K)$ is context-free.

Proof

The “only if” is contained in Lemma 3.1. (Observe that $H$ inherits finite generation from $G$, since $[G:H]<\infty$.)

For the converse, we assume that $(H,K)$ is context-free and let $\psi :\Sigma \to H$ and ${\psi }^{\prime }:{\Sigma }^{\prime }\to G$ be semigroup presentations of $H$ and $G$, respectively. There is a pushdown automaton $\mathcal{A}=(Q,\Sigma ,Z,\delta ,q_0,Q_f,z_0)$ that accepts $L(H,K,\psi )$.

Let $F$ be a set of representatives of the right cosets of $H$ in $G$, with $1_G\in F$. Thus, $\left|F\right|<\infty$, and

$$G=\underset{g\in F}{⨄}Hg\text{,}$$

For every $g\in F$ and $b\in {\Sigma }^{\prime }$ there is a unique $\overline{g}=\overline{g}(g,b)\in F$ such that $g{\psi }^{\prime }\left(b\right)\in H\overline{g}$. Therefore there is a word $u=u(g,b)\in {\Sigma }^{\ast }$ such that

$$g{\psi }^{\prime }\left(b\right)=\psi \left(u(g,b)\right)\overline{g}(g,b)\text{.}$$

An input word $w=b_1\cdots b_n$ is transformed recursively into $u_1\cdots u_n$, along with the sequence $g_0,g_1,\dots ,g_n$ of elements of $F$ that indicate the current $H$-coset at each step:

$$g_0=1_G\text{;}{u}_k=u(g_{k-1},b_k)\text{and}{g}_k=\overline{g}(g_{k-1},b_k)\text{.}$$

Then ${\psi }^{\prime }\left(w\right)\in K$ if and only if $g_n=1_G$ and $\psi (u_1\cdots u_n)\in K$.

Thus, our new automaton ${\mathcal{A}}^{\prime }$ recalls at each step the current coset $H{g}_{k-1}$, which is multiplied on the right by $\psi \left(b_k\right)$, where $b_k$ is the next input letter. Then the new coset is $H\overline{g}(g_{k-1},b_k)$, and ${\mathcal{A}}^{\prime }$ simulates what $\mathcal{A}$ does next upon reading $u(g_{k-1},b_k)$. Then $w$ is accepted when at the end the coset is $H=H{1}_G$ and $\mathcal{A}$ is in a final state.

The simple task to write down this automaton in detail is left to the reader. □

4. Context-free graphs

In this section, we assume that $(X,E,\ell )$ is symmetric. We may think of each pair of oppositely oriented edges $(x,a,y)$ and $(y,a^{-1},x)$ as one non-oriented edge, so that $X$ becomes an ordinary graph with symmetric neighbourhood relation, but possibly multiple edges and loops. If it is in addition fully deterministic, then $X$ is a regular graph, that is, the number of outgoing edges (which coincides with the number of ingoing edges) at each vertex is $\left|\Sigma \right|$. Attention: if we consider non-oriented edges, then each loop at $x$ has to be counted twice, since it corresponds to two oriented edges of the form $(x,a,x)$ and $(x,a^{-1},x)$. For all our purposes it is natural to require that $X$ is connected: for any pair of vertices $x,y$ there is a path from $x$ to $y$. The distance $d(x,y)$ is the minimum length (number of edges) of a path from $x$ to $y$, which defines the integer-valued graph metric. A geodesic path is one whose length is the distance between its endpoints.

We select a finite, non-empty subset $F$ of $X$ and consider the balls $B(F,n)=\{x:d(x,F)\le n\}$ (where $d(x,F)=min\{d(x,y):y\in F\}$). If we delete $B(F,n)$ then the induced graph $X\setminus B(F,n)$ will fall apart into a finite number of connected components, called cones with respect to $F$. Each cone is a labelled, symmetric graph $C$ with the boundary $\partial C$ consisting of all vertices $x$ in $C$ having a neighbour outside $C$ (i.e., in $B(F,n)$).

The following notion was introduced in [12] for symmetric, labelled graphs and $F=\left\{o\right\}$.

Definition 4.1

The graph $X$ is called context-free with respect to $F$ if there is only a finite number of isomorphism types of the cones with respect to $F$ as labelled graphs with boundary.

This means that there are finitely many cones $C_1,\dots ,C_r$ (generally with respect to different radii $n$) such that for each cone $C$, we can fix a bijection ${\varphi }_C$ from (the vertex set of) $C$ to precisely one of the $C_i$, this bijection sends $\partial C$ to $\partial C_i$, and $(x,a,y)$ is an edge with both endpoints in $C$ if and only if its image $({\varphi }_C\left(x\right),a,{\varphi }_C\left(y\right))$ is an edge of $C_i$. In this case, we say that $C$ is a cone of type $i$.

Generally, as in [12], we are interested in the case when $F=\left\{o\right\}$ (or any other singleton), but there is at least one point where it will be useful to admit arbitrary finite, non-empty $F$.

Another natural notion of context-freeness of $X$ with respect to $o$ is to require that the language $L_{o,o}\left(X\right)$ is context-free. We shall see that for deterministic, symmetric graphs this is equivalent with context-freeness with respect to $o$ in the sense of Definition 4.1. One direction of this equivalence is practically contained in [12], but not stated explicitly except for the case of Cayley graphs of groups. The other direction (that context-freeness of $L_{o,o}$ implies that of the graph) is shown in [12] only for Cayley graphs of groups, which is substantially simpler than the general case treated below in Theorem 4.6.

Theorem 4.2

If the symmetric, labelled graph $(X,E,\ell )$ with label alphabet $\Sigma$ is context-free with respect to the finite, non-empty set $F\subset X$ , then $L_{x,y}$ is a context-free language for all $x,y\in X$ . Furthermore, if the graph $X$ is deterministic, then so is the context-free language $L_{x,y}$ .

Proof

Just for the purpose of this proof, we write $x_0,y_0$ instead of $x,y$ for the vertices for which $L_{x_0,y_0}$ will be shown to be context-free. We may assume without loss of generality that $x_0,y_0$ in $F$. Indeed, if this is not the case, then we can replace $F$ by $F^{\prime }=B(F,n)$, which contains $x_0$ and $y_0$ when $n$ is sufficiently large. The cones with respect to $F^{\prime }$ are also cones with respect to $F$, so that $X$ is also context-free with respect to $F^{\prime }$.

Similarly to [12, Lemma 2.3], we construct a deterministic pushdown automaton that accepts $L_{x_0,y_0}$.

We consider also the whole graph $X$ as a cone $C_0$ with boundary $F$, which we keep apart from the other representatives $C_1,\dots ,C_r$ of cones.

If $C$ is a cone, then as a component of $X\setminus B(F,n)$ for some $n\ge 0$ it must be a successor of another cone $C^{-}$. The latter is the unique component of $X\setminus B(F,n-1)$ that contains $C$, when $n\ge 1$, while it is $C_0=X$ when $n=0$. We also call $C^{-}$ the predecessor of $C$.

Different cones of type $j\in \{1,\dots ,r\}$ may have predecessors of different types. Conversely, a cone $C$ of type $i\in \{0,\dots ,r\}$ may have none, one or more than one successors of type $j$, and the number $d_{i,j}$ of those successors depends only on $i$ and $j$. In the representative cone $C_i$, we choose and fix a numbering of the distinct successors of type $j$ as $C_{i,j}^k$, $k=1,\dots ,d_{i,j}$. If $C$ is any cone with type $i$ then we use the isomorphism ${\varphi }_C:C\to C_i$ to transport this numbering to the successors of $C$ that have type $j$, which allows us to identify the $k$-th successor of $C$ with type $j$.

One can visualize the cone structure by a finite, oriented graph $\Gamma$ with multiple edges and root 0: the vertex set is the set of cone types $i\in \{0,\dots ,r\}$, and there are $d_{i,j}$ oriented edges, which we denote by $t_{i,j}^k$ ($k=1,\dots ,d_{i,j}$) from vertex $i$ to vertex $j$ ($i\ge 0$, $j\ge 1$).

Every vertex $x$ of $X$ belongs to the boundary of precisely one cone $C=C\left(x\right)$ with respect to $F$. We define the type $i$ of $x$ as the type of $C\left(x\right)$. Under the mapping ${\varphi }_C$, our $x$ corresponds to precisely one element of $\partial C_i$. We write $\varphi \left(x\right)$ for that element, without subscript $C$, so that $\varphi$ maps $X$ onto ${\bigcup }_i\partial C_i$. In particular, $\varphi \left(x\right)=x$ for every $x\in F$.

Let $y\in X\setminus F$ with type $j$. Then there is $i$ (depending on $y$) such that every neighbour $x$ of $y$ with $d(x,F)=d(y,F)-1$ has type $i$, and there is precisely one successor cone $C_{i,j}^k$ of $C_i$ that contains ${\varphi }_{C\left(x\right)}\left(y\right)$. In this case, we write $\tau \left(y\right)=t_{i,j}^k$, the second order type of $y$. Compare with [12]. If $y^{\prime }$ is such that $C\left(y^{\prime }\right)=C\left(y\right)$ then $\tau \left(y^{\prime }\right)=\tau \left(y\right)$.

We now finally construct the required pushdown automaton $\mathcal{A}$. (Comparing with [12], we use more states and stack symbols, which facilitates the description.) The set of states and stack symbols are

$$Q=\underset{i=0}{\overset{r}{⨄}}\partial C_i\text{and}Z=F\cup \{t_{i,j}^k:i=1,\dots ,r,j=0,\dots ,r,k=1,\dots ,d_{i,j}\}\text{.}$$

(When $d_{i,j}=0$ then there is no $t_{i,j}^k$.) Note that both sets contain $F$. In order to generate the language $L_{x_0,y_0}$, where $x_0,y_0\in F$, then we use $x_0$ as the initial state and $y_0$ as the (only) final state. We describe the transition function, which–like $Q$ and $Z$–does not depend on $x_0,y_0$.

We want to read an input word, which has to correspond to the label starting in $x_0$. Inside the subgraph of $X$ induced by $F$, our $\mathcal{A}$ behaves just like that subgraph, seen as a finite automaton.

Outside of $F$, it works as follows. At the $m$-th step, the automaton will be in a state that describes the $m$-th vertex, say $x$, of that path, by identifying $x$ as above with the element $\varphi \left(x\right)$ of $C_j$, where $j$ is the type of $x$. The current stack symbol is of the form $t_{i,j}^k$ and serves to recall that $x$ lies in the $k$-th successor cone of type $j$ of a cone with type $i$. If the next vertex along the path, say $y$, satisfies $d(y,F)=d(x,F)+1$, and $y$ has type $j^{\prime }$ then the state is changed to $\varphi \left(y\right)\in C_{j^{\prime }}$, and the symbol $t_{j,j^{\prime }}^{k^{\prime }}=\tau \left(y\right)$ is added to the stack. If $d(y,F)=d(x,F)$, then only the state is changed from $\varphi \left(x\right)$ to $\varphi \left(y\right)$. Finally, if $d(y,F)=d(x,F)-1$ then the new state is again $\varphi \left(y\right)$, while the top symbol on the stack is deleted. Formally, we get the following list of transition rules. If $x\in F=Q\cap Z$:

$$\delta (x,a,x)=\{(y,y):(x,a,y)\in E,y\in F\}\cup \{(\varphi \left(y\right),x\tau \left(y\right)):(x,a,y)\in E,d(y,F)=1\}\text{.}$$

If $x\in X\setminus F$:

$$\delta (\varphi \left(x\right),a,\tau \left(x\right))=\{(\varphi \left(y\right),a,\tau \left(x\right)\tau \left(y\right)):(x,a,y)\in E,d(y,F)=d(x,F)+1\}\cup \{(\varphi \left(y\right),\tau \left(y\right)=\tau \left(x\right)):(x,a,y)\in E,d(y,F)=d(x,F)\}\cup \{(\varphi \left(y\right),ϵ):(x,a,y)\in E,d(y,F)=d(x,F)-1\}\text{.}$$

This is a finite collection of transitions, since $\varphi (\cdot )$ and $\tau (\cdot )$ can take only finitely many different values.

In view of the above explanations, $\mathcal{A}$ accepts $L_{x_0,y_0}$. Also, when the graph $X$ is deterministic, then so is $\mathcal{A}$. □

Before proving a converse of Theorem 4.2, we first need some preliminaries, and start by recalling a fact proved in [11], [12], see also Woess [21] and Berstel and Boasson [2].

Lemma 4.3

If$L_{o,o}$is context-free then there is a constant$M$such that for each cone$C$with respect to$o$, one has$\mathrm{diam}\left(\partial C\right)\le M$.

(The diameter is of course taken with respect to the graph metric.) We shall see below how to deduce this, but it is good to know it in advance.

A context-free grammar $\mathcal{C}=(V,\Sigma ,P,S)$ is said to have Chomsky normal form (CNF), if (i) every production rule is of the form $T⊢U\hat{U}$ or $T⊢a$, where $U,\hat{U}\in V$ (not necessarily distinct), resp. $a\in \Sigma$, and (ii) if $ϵ\in L\left(\mathcal{C}\right)$, then there is the rule $S⊢ϵ$, and $S$ is not contained in the right hand side of any production rule.

With a slight deviation from [11], we associate with each $w=a_1\cdots a_n\in L\left(\mathcal{C}\right)$, $n\ge 2$ a labelled (closed) polygon $P\left(w\right)$ with length $n+1$. As a directed graph, it has distinct vertices $t_0,t_1,\dots ,t_n$ and labelled edges $(t_{i-1},a_i,t_i)$, $i=1,\dots ,n$, plus the edge $(t_0,S,t_n)$. A (diagonal) triangulation of $P\left(w\right)$ is a plane triangulation of $P\left(w\right)$ obtained by inserting only diagonals. Here, we specify those diagonals as oriented, labelled edges $(t_i,T,t_j)$, where $t_i,t_j$ are not neighbours in $P\left(w\right)$ and $T\in V$. Furthermore, we will never have two diagonals between the same pair of vertices of $P\left(w\right)$. (If $\left|w\right|\le 2$ we consider $P\left(w\right)$ itself triangulated.) The proof of the following Lemma may help to make the construction of [11] (used for Cayley graphs of groups) more transparent.

Lemma 4.4

If $\mathcal{C}=(V,\Sigma ,P,S)$ is in CNF and $w=a_1\cdots a_n\in L\left(\mathcal{C}\right)$ with $n\ge 2$ then there is a diagonal triangulation of $P\left(w\right)$ with the property that whenever $(t_i,T,t_j)$ is a diagonal edge, then $T$ occurs in a derivation $S\stackrel{\ast }{⟹}w$ , $j-i\ge 2$ and $T\stackrel{\ast }{⟹}{a}_{i+1}\cdots a_j$ .

Proof

We start with a fixed derivation $S\stackrel{\ast }{⟹}w$, and explain how to build up the triangles step by step. Suppose that $T\in V$ occurs in our derivation, and that we have a “sub-derivation” $T⊢U\hat{U}\stackrel{\ast }{⟹}{a}_{i+1}\cdots a_k$, where $U,\hat{U}\in V$. Then there is $j\in \{i+1,\dots ,k-1\}$ such that $U\stackrel{\ast }{⟹}{a}_{i+1}\cdots a_j$ and $\hat{U}\stackrel{\ast }{⟹}{a}_{j+1}\cdots a_k$. In this case, we draw a triangle with three oriented, labelled edges, namely the ‘old’ edge $(t_i,T,t_k)$ and the two ‘new’ edges $(t_i,U,t_j)$ and $(t_j,\hat{U},t_k)$.

If we have the derivation $S\stackrel{\ast }{⟹}{a}_1\cdots a_n$, then it uses successive steps of the form $T⊢U\hat{U}$ with $U\hat{U}\stackrel{\ast }{⟹}{a}_{i+1}\cdots a_k$ as above. We work through these steps one after the other, starting with $S⊢T_1{\hat{T}}_1$, where $T_1\stackrel{\ast }{⟹}{a}_1\dots a_k$ and ${\hat{T}}_1\stackrel{\ast }{⟹}{a}_{k+1}\cdots a_n$. The first triangle has the ‘old’ edge $(t_0,S,t_n)$ and the ‘new’ edges $(t_0,T_1,t_k)$ and $(t_k,{\hat{T}}_1,t_n)$.

At any successive step, we take one of the ‘new’ edges $(t_i,T,t_k)$, where $k-i\ge 2$ and proceed as explained at the beginning, so that we add two ‘new’ edges that make up a triangle together with $(t_i,T,t_k)$, which is then declared ‘old’. We continue until all derivation steps of the form $T⊢U\hat{U}$ in our derivation $S\stackrel{\ast }{⟹}w$ are exhausted. At this point, we have obtained a tiling of triangles that constitute a diagonal triangulation of its outer polygon, whose edges have the form $(t_0,S,t_n)$ and $(t_{i-1},U_i,t_i)$ with $U_i\in V$, $i=1,\dots ,n$. The only steps of our derivation that we have not yet considered are the terminal ones $U_i⊢a_i$. Thus, we conclude by replacing the label $U_i$ of $(t_{i-1},U_i,t_i)$ by $a_i$. □

The construction is best understood by considering an example: suppose our rightmost derivation is

$$S⊢T_1{\hat{T}}_1⟹T_1\left(T_2{\hat{T}}_2\right)⟹T_1\left(T_2\left(T_3{\hat{T}}_3\right)\right)⟹T_1\left(T_2\left(T_3{a}_6\right)\right)⟹T_1\left(T_2\left(\left(T_4{\hat{T}}_4\right)a_6\right)\right)⟹T_1\left(T_2\left(\left(T_4{a}_5\right)a_6\right)\right)⟹T_1\left(T_2\left(\left(a_4{a}_5\right)a_6\right)\right)⟹T_1\left(a_3\left(\left(a_4{a}_5\right)a_6\right)\right)⟹T_1\left(a_3\left(\left(a_4{a}_5\right)a_6\right)\right)⟹\left(T_5{\hat{T}}_5\right)\left(a_3\left(\left(a_4{a}_5\right)a_6\right)\right)⟹\left(T_5{a}_2\right)\left(a_3\left(\left(a_4{a}_5\right)a_6\right)\right)⟹\left(a_1{a}_2\right)\left(a_3\left(\left(a_4{a}_5\right)a_6\right)\right)\text{.}$$

(We have inserted the parentheses to make the rules that we used in each step more visible.) The associated triangulation is as in Fig. 2.

Fig. 2.

The polygon $P\left(w\right)$, with $w=a_1{a}_2{a}_3{a}_4{a}_5{a}_6$, and its (diagonal) triangulation associated with the rightmost derivation $S⊢\left(a_1{a}_2\right)\left(a_3\left(\left(a_4{a}_5\right)a_6\right)\right)$.

The variables of the terminal rules $T_5⊢a_1$, ${\hat{T}}_5⊢a_2$, $T_2⊢a_3$, $T_4⊢a_4$, ${\hat{T}}_4⊢a_5$ and ${\hat{T}}_3⊢a_6$ are not visible in this figure (but we might add them to the boundary edges). Apart from this, one can read the derivation $S\stackrel{\ast }{⟹}w$ from the diagonalization in a similar way as it can be read from the so-called derivation tree (see e.g. [6, Section 1.6] for the latter).

The following goes back to [11] in the case of (Cayley graphs of) finitely generated groups (recall from Lemma/Definition 2.1 that in case $X$ is deterministic and symmetric, if $x\in X$ and $w=a_1{a}_2\dots a_n\in {\Sigma }^{\ast }$, then $x^{-w}=x^{a_n^{-1}\cdots a_2^{-1}{a}_1^{-1}}\in X$ denotes the initial vertex of the path $\pi$ in $X$ terminating at $x$ with label $\ell \left(\pi \right)=w$).

Lemma 4.5

Let$\mathcal{C}=(V,\Sigma ,P,S)$be in CNF and$L\left(\mathcal{C}\right)=L_{x,y}\left(X\right)$, where$X$is a deterministic, symmetric graph. If$w=a_1\cdots a_n\in L_{x,y}\left(X\right)$and$(t_i,T,t_j)$is a diagonal edge in a triangulation of$P\left(w\right)$as in Lemma 4.4, then the vertices$\overline{x}=x^{a_1\cdots a_i}$and$\overline{y}=x^{a_1\cdots a_j}$of$X$satisfy$d(\overline{x},\overline{y})\le m\left(T\right)$, where

$$m\left(T\right)=min\{\left|w\right|:w\in L_T\}\text{.}$$ (2)

Proof

Since $X$ is deterministic, Lemma 2.3 implies that ${\pi }_x\left(w\right)$ exists as the unique path with initial vertex $x$ and label $w$. In particular, $\overline{x}$ and $\overline{y}$ lie on that path. Furthermore, we have $\overline{y}=y^{-a_{j+1}\cdots a_n}$.

Now let $v\in L_T$ with $\left|v\right|=m\left(T\right)$. Then by Lemma 4.4, $T$ arises in a derivation $S\stackrel{\ast }{⟹}{a}_1\cdots a_{i}T{a}_{j+1}\cdots a_n\stackrel{\ast }{⟹}w$. But then we also have $S\stackrel{\ast }{⟹}{a}_1\cdots a_{i}v{a}_{j+1}\cdots a_n$, a word in $L_{x,y}$. By Lemma 2.3, again using that $X$ is symmetric and deterministic, ${\overline{x}}^v=y^{-a_{j+1}\cdots a_n}=\overline{y}$. Therefore, $\overline{x}$ and $\overline{y}$ are connected by a path with label $v$. Its length is $m\left(T\right)$. □

Theorem 4.6

Let $(X,E,\ell )$ be a fully deterministic, symmetric graph with label alphabet $\Sigma$ and root $o$ . If $L_{o,o}$ is a context-free language, then $X$ is a context-free graph with respect to $o$ , and in particular, $L_{o,o}$ is deterministic.

Proof

There is a reduced grammar $\mathcal{C}=(V,\Sigma ,P,S)$ in CNF that generates $L_{o,o}$. Each of the languages $L_T$, $T\in V$, is non-empty, only $L_S$ contains $ϵ$, and we define

$$m=max\{m\left(T\right):T\in V\}\text{,}$$ (3)

where $m\left(T\right)$ is as in (2).

Let $C$ be a cone with respect to $o$ such that $k=d(o,\partial C)>m$.

Construction of$\tilde{D}\left(C\right)$. We define $D\left(C\right)$ as the subgraph of $X$ induced by all vertices $y\in X$ with

$$d(o,x)=d(o,y)+d(x,y)\text{and}d(x,y)\le m\text{for some }x\in \partial C\text{.}$$

In particular, $y$ lies on some geodesic path from $o$ to $\partial C$.

Now let $x_1,x_2\in \partial C$, and consider some path $\pi \in {\Pi }_{x_1,x_2}\left(C\right)$ (i.e., it lies in $C$). Choose a geodesic path ${\pi }_1$ from $o$ to $x_1$ and a geodesic path ${\pi }_2$ from $x_2$ to $o$. Then we can concatenate the three paths to a single path ${\pi }_1\pi {\pi }_2\in {\Pi }_{o,o}$. Its label is the word $w=\ell \left({\pi }_1\right)\ell \left(\pi \right)\ell \left({\pi }_2\right)\in L_{o,o}$. Set $n=\left|w\right|$ and write

$$w=(a_1\cdots a_k)(a_{k+1}\cdots a_{n-k})(a_{n-k+1}\cdots a_n)$$

where the 3 pieces in the parentheses are (in order) $\ell \left({\pi }_1\right)$, $\ell \left(\pi \right)$ and $\ell \left({\pi }_2\right)$. The words $\ell \left({\pi }_1\right)$, $\ell \left(\pi \right)$ and $\ell \left({\pi }_2\right)S$ are the labels of three consecutive arcs that fill the boundary of the polygon $P\left(w\right)$. (To be precise, along the last edge of the $3^{\text{rd}}$ arc, we are reading the label $S$ in the reversed direction.) By [11, Lemma 5], its triangulation has a triangle which meets each of those arcs. (It may also occur that one corner of the triangle meets two arcs.) Thus, there are $i\in \{0,\dots ,k\}$ and $i^{\prime }\in \{k,\dots ,n-k\}$ such that the vertices $t_i$ and $t_{i^{\prime }}$ of $P\left(w\right)$ lie on that triangle. They correspond to the vertices $y_1=o^{a_1\cdots a_i}$ and $y^{\prime }=o^{a_1\cdots a_{i^{\prime }}}$ of $X$. We either have $i^{\prime }-i\le 1$, or else a diagonal $(t_i,U,t_{i^{\prime }})$ is a side of our triangle. By Lemma 4.5, we get $d(y_1,y^{\prime })\le m\left(U\right)\le m$. Thus $k\le i^{\prime }\le d(o,y^{\prime })\le i+m$, that is, $i\ge k-m>0$. In particular, $t_i$ does not lie on the third arc. In the same way, there is $j\in \{n-k,\dots ,n-k+m\}$ (and not larger) such that $t_j$ is a corner of our triangle. This yields that there must be a “true” diagonal $(t_i,T,t_j)$ of $P\left(w\right)$. We set $v_1=a_{i+1}\cdots a_k$ and $v_2=a_{n-k+1}\cdots a_j$, so that $x_1=y_1^{v_1}$, and let $y_2=x_2^{a_{n-k+1}\cdots a_j}$. The points $y_1$ and $y_2$ are in $D\left(C\right)$, and by Lemma 4.4, $T\stackrel{\ast }{⟹}{v}_1\ell \left(\pi \right)v_2$.

[It is here that we can see Lemma 4.3, since we deduced that $d(x_1,x_2)\le 3m$ for all $x_1,x_2\in \partial C$.]

By Lemma 4.4, we also have

$$S\stackrel{\ast }{⟹}{a}_1\cdots a_{i}T{a}_{j+1}\cdots a_n\text{,}$$

so that $v\in L_T$ implies $a_1\cdots a_{i}v{a}_{j+1}\cdots a_n\in L_{o,o}$ and consequently $v\in L_{y_1,y_2}$, that is, $y_1^v=y_2$.

We now insert into $D\left(C\right)$ the additional labelled edge $(y_1,v_{1}T{v}_2,y_2)$, whose label is the word $v_{1}T{v}_2\in {\Sigma }^{\ast }V{\Sigma }^{\ast }$. We insert all diagonals of the same type that can be obtained in the same way, and write $\tilde{D}\left(C\right)$ for the resulting “edge-enrichment” of $D\left(C\right)$.

Subsuming, we have an edge $(y_1,v_{1}T{v}_2,y_2)$ in $\tilde{D}\left(C\right)$ if and only if the following properties hold.

• •$\left|v_i\right|\le m$ ($i=1,2$) and $T\in V$,
• •the path with label $v_1$ starting at $y_1$ and ending at $x_1=y_1^{v_1}\in \partial C$ is part of a geodesic from $o$ to $x_1$,
• •the path with label $v_2$ starting at $x_2=y_2^{-v_2}\in \partial C$ and ending at $y_2$ is part of a geodesic from $x_2$ to $o$, and
• •there is a path $\pi$ in $C$ from $x_1$ to $x_2$ such that $T\stackrel{\ast }{⟹}{v}_1\ell \left(\pi \right)v_2$,
• •if $T\stackrel{\ast }{⟹}v\in {\Sigma }^{\ast }$ then $v$ is the label of a path in ${\Pi }_{y_1,y_2}$.

Now, there are only finitely many cones $C$ with respect to $o$ with $d(\partial C,o)\le m$. On the other hand, for all cones $C$ with $d(\partial C,o)\ge m$, there is a bound on the number of vertices of $\tilde{D}\left(C\right)$, as well as on the number of possible labels on its edges. In particular, there are only finitely many possible isomorphism types of the labelled graphs $(\tilde{D}\left(C\right),\partial C)$ with “marked” boundary $\partial C\subset \tilde{D}\left(C\right)$.

We now suppose that $C$ and $C^{\prime }$ are two cones at distance $\ge m$ from $o$, such that $(\tilde{D}\left(C\right),\partial C)$ and $(\tilde{D}\left(C^{\prime }\right),\partial C^{\prime })$ are isomorphic. We claim that $C$ and $C^{\prime }$ are isomorphic, and this will conclude the proof that there are only finitely many isomorphism types of cones with respect to $o$.

Let $\varphi :\tilde{D}\left(C\right)\to \tilde{D}\left(C^{\prime }\right)$ be an isomorphism with $\varphi \left(\partial C\right)=\partial C^{\prime }$, and ${\varphi }^{\prime }$ its inverse mapping. We extend $\varphi$ to a mapping from $C$ to $C^{\prime }$, also denoted $\varphi$.

Claim 1

Let$x\in \partial C$and$v\in {\Sigma }^{+}$such that the path${\pi }_x\left(v\right)$lies in$C$and meets$\partial C$only in its initial point$x$. Then the path${\pi }_{x^{\prime }}\left(v\right)$lies in$C^{\prime }$and meets$\partial C^{\prime }$only in its initial point$x^{\prime }=\varphi \left(x\right)\in \partial C^{\prime }$.

Proof

If $a$ is the initial letter of $v$ then (always using the notation of Definition 1.1) the first edge of ${\pi }_x\left(v\right)$ is $(x,a,x^a)$. We now consider the path ${\pi }_{x^{\prime }}\left(v\right)$ with label $v$ starting at $x^{\prime }\in \partial C^{\prime }$. We first claim that the latter lies in $C^{\prime }$ and only its initial point $x^{\prime }$ is in $\partial C^{\prime }$. Let $(x^{\prime },a,{\left(x^{\prime }\right)}^a)$ be the first edge of the path. Then ${\left(x^{\prime }\right)}^a$ cannot lie in $\tilde{D}\left(C^{\prime }\right)$, since otherwise $(x,a,x^a)=({\varphi }^{\prime }\left(x^{\prime }\right),a,{\varphi }^{\prime }{\left(x^{\prime }\right)}^a)$ would be an edge in $\tilde{D}\left(C\right)$, a contradiction. Thus, the path ${\pi }_{x^{\prime }}\left(v\right)$ goes at least initially into $C^{\prime }\setminus \partial C$.

So now suppose that ${\pi }_{x^{\prime }}\left(v\right)$ ever returns to $\partial C^{\prime }$, and let ${\pi }^{\prime }$ be its initial part up to the first return. Then $v^{\prime }=\ell \left({\pi }_{x^{\prime }}\left(v\right)\right)$ is an initial part of $v$ with $\left|v^{\prime }\right|\ge 2$, and ${\pi }^{\prime }$ is a path within $C^{\prime }$ from $x_1^{\prime }=x^{\prime }$ to $x_2^{\prime }={\left(x^{\prime }\right)}^{v^{\prime }}\in \partial C^{\prime }$. But then, by construction, $\tilde{D}\left(C^{\prime }\right)$ must contain an edge $(y_1^{\prime },v_{1}T{v}_2,y_2^{\prime })$ such that $x_1^{\prime }={\left(y_1^{\prime }\right)}^{v_1}$, $y_2^{\prime }={\left(x_2^{\prime }\right)}^{v_2}$, and $T\stackrel{\ast }{⟹}{v}_1{v}^{\prime }{v}_2$. Using the isomorphism ${\varphi }^{\prime }:\tilde{D}\left(C^{\prime }\right)\to \tilde{D}\left(C\right)$, we set $y_i={\varphi }^{\prime }\left(y_i^{\prime }\right)$, $i=1,2$, and $x_2={\varphi }^{\prime }\left(x_2^{\prime }\right)\in \partial C$. We have of course $x_1={\varphi }^{\prime }\left(x_1^{\prime }\right)$. Now we must have the edge $(y_1,v_{1}T{v}_2,y_2)$ in $\tilde{D}\left(C\right)$. But then $v_1{v}^{\prime }{v}_2\in L_{y_1,y_2}$, and consequently $v^{\prime }\in L_{x_1,x_2}$, that is, $x_1^{v^{\prime }}\in \partial C$. But this contradicts the fact that ${\pi }_x\left(v\right)$ meets $\partial C$ only in its initial point. We conclude that also the path ${\pi }_{x^{\prime }}\left(v\right)$ lies in $C^{\prime }$ and meets $\partial C^{\prime }$ only in its initial point, and Claim 1 is verified. □

Now let $z\in C\setminus \partial C$. Then there are $x\in \partial C$ and $v\in {\Sigma }^{+}$ such that $z=x^v$ and the path ${\pi }_x\left(v\right)$ from $x$ to $z$ meets $\partial C$ only in its initial point $x$. By Claim 1, the analogous statement holds for the path ${\pi }_{x^{\prime }}\left(v\right)$ in $C^{\prime }$, where $x^{\prime }=\varphi \left(x\right)$. The only choice is to define $\varphi \left(z\right)=z^{\prime }={\left(x^{\prime }\right)}^v$, which lies in $C^{\prime }\setminus \partial C^{\prime }$ as required. We have to show that $\varphi$ is well-defined. This will follow from the next claim.

Claim 2

Let$x_1,x_2\in \partial C$,$v,w\in {\Sigma }^{+}$such that the paths${\pi }_{x_1}\left(v\right)$and${\pi }_{x_2}\left(w\right)$lie in$C$, meet$\partial C$only in their initial points and end at the same point of$C\setminus \partial C$. Then, setting$x_i^{\prime }=\varphi \left(x_i\right)$, also${\pi }_{x_1^{\prime }}\left(v\right)$and${\pi }_{x_2^{\prime }}\left(w\right)$end at the same point of$C^{\prime }\setminus \partial C^{\prime }$.

Proof

Let $w^{-1}$ be the “inverse” of $w$, as defined in Definition 1.1. Then $x_2^{-w^{-1}}=x_2^w$, and $v{w}^{-1}$ is the label of the path from $x_1$ to $x_2$ that we obtain by first following ${\pi }_{x_1}\left(v\right)$ and then the “inverse” of ${\pi }_{x_2}\left(w\right)$. It lies entirely in $C$, and only its endpoints are in $\partial C$. By construction, $\tilde{D}\left(C\right)$ has an edge $(y_1,v_{1}T{v}_2,y_2)$ such that $y_1^{v_1}=x_1$, $x_2^{v_2}=y_2$ and $T\stackrel{\ast }{⟹}{v}_{1}v{w}^{-1}{v}_2$. We set $y_i^{\prime }=\varphi \left(y_i\right)$, $i=1,2$. Then $(y_1^{\prime },v_{1}T{v}_2,y_2^{\prime })$ is an edge of $\tilde{D}\left(C^{\prime }\right)$. Therefore $v_{1}v{w}^{-1}{v}_2\in L_{y_1^{\prime },y_2^{\prime }}$. But this implies that $v{w}^{-1}$ is the label of a path from $x_1^{\prime }$ to $x_2^{\prime }$, and we know from Claim 1 that it lies in $C$ and has only its endpoints in $\partial C$. Thus ${\left(x_1^{\prime }\right)}^v={\left(x_2^{\prime }\right)}^{-w^{-1}}={\left(x_2^{\prime }\right)}^w$, and Claim 2 is true. □

Thus, $\varphi$ is well defined, and the same works of course also for ${\varphi }^{\prime }$ by exchanging the roles of $C$and $C^{\prime }$.

Claim 3

The map$\varphi :C\to C^{\prime }$is bijective.

Proof

We know that $\varphi :\partial C\to \partial C^{\prime }$ is bijective and that $\varphi (C\setminus \partial C)\subset C^{\prime }\setminus \partial C$. Let $z\in C\setminus \partial C$, and let $x\in \partial C$, $v\in {\Sigma }^{+}$ such that ${\pi }_x\left(v\right)$ is a path from $x$ to $z$ that intersects $\partial C$ only at the initial point. Setting $x^{\prime }=\varphi \left(x\right)$, $z^{\prime }=\varphi \left(z\right)$, we know from the construction of $\varphi$ and Claim 1 that ${\pi }_{x^{\prime }}\left(v\right)$ is a path in $C^{\prime }$ from $x^{\prime }$ to $z^{\prime }$ that meets $\partial C^{\prime }$ only in its initial point. Now the way how ${\varphi }^{\prime }$ is constructed yields that ${\varphi }^{\prime }\left(z^{\prime }\right)=z$. Therefore ${\varphi }^{\prime }\varphi$ is the identity on $C$. Exchanging roles, we also get the $\varphi {\varphi }^{\prime }$ is the identity on $C^{\prime }$. This proves Claim 3. □

It is now immediate from the construction that $\varphi$ also preserves the edges and their labels, so that it is indeed an isomorphism between the labelled graphs $C$ and $C^{\prime }$ that sends $\partial C$ to $\partial C^{\prime }$. This concludes the proof of Theorem 4.6. □

[12, Cor. 2.7] says that if a symmetric labelled graph is context-free with respect to one root $o$, then it is context-free with respect to any other vertex chosen as the root $x$. In view of Theorem 4.2, Theorem 4.6, this is also obtained from the following, when the graph is fully deterministic.

Corollary 4.7

Let $(X,E,\ell )$ be a fully deterministic, strongly connected graph with label alphabet $\Sigma$ . If $L_{o,o}$ is context-free then $L_{x,y}$ is deterministic context-free for all $x,y\in X$ .

Theorem 4.2, Theorem 4.6, together with Lemma 3.1 also imply the following.

Corollary 4.8

Let $G$ be a finitely generated group and $K$ a subgroup.

• (a) The pair $(G,K)$ is context-free if and only if for any symmetric $\psi :\Sigma \to G$ , the Schreier graph $X(G,K,\psi )$ is a context-free graph. In this case, the language $L(G,K,\psi )$ is deterministic for every (not necessarily symmetric) semigroup presentation $\psi :\Sigma \to G$ .
• (b) If $(G,K)$ is context-free, then also $(G,g^{-1}Kg)$ is context-free for every $g\in G$ .

Proof

(a) is clear. Regarding (b), for the Schreier graph $X(G,K,\psi )$, we have $L(G,K,\psi )=L_{o,o}$ and $L(G,g^{-1}Kg,\psi )=L_{x,x}$ with $x=Kg$, $g\in G$. Thus, the statement follows from Corollary 4.7. □

Lemma 4.9

Let $G$ be a finitely generated group and $K,H$ be subgroups with $K\le H$ and $[H:K]<\infty$ .

If $(G,K)$ is context-free then also $(G,H)$ is context-free.

Proof

In the context-free graph $X(G,K,\psi )$, consider the finite set of vertices $F=\{Kh:h\in H\}$, containing the root vertex $o=o_K=K$. Then $L(G,H,\psi )={\bigcup }_{x\in F}{L}_{o,x}$ is a finite (disjoint) union of context-free languages. Therefore it is context-free by standard facts. □

Remark 4.10

In terms of Schreier graphs, we have the mapping $Kg\mapsto Hg$ which is a homomorphism of labelled graphs from $X=X(G,K,\psi )$ onto $Y=X(G,H,\psi )$ which is finite-to-one. The lemma says that in this situation, if $X$ is a context-free graph then so is $Y$. We do not see an easy direct proof of this fact in terms of graphs, the main problem being how the homomorphism $X\to Y$ interacts with the isomorphisms between the cones of $X$ with respect to the set $F$. On the other hand, reformulating this in terms of the associated “path languages” with the help of Theorem 4.2, Theorem 4.6, it has become straightforward.

The converse of Lemma 4.9 is not true, that is, when $(G,H)$ is context-free and $[H:K]<\infty$ then $(G,K)$ is not necessarily context-free. See Example 5.6 in the last section. However, we have the following.

Lemma 4.11

If $K$ is a finite subgroup of $G$ then $(G,K)$ is context-free if and only if $G$ is a context-free (i.e. virtually free) group.

Proof

Fix $\Sigma$ and $\psi$. Let $X=X(G,\psi )$ be the associated Cayley graph of $G$, and $Y=X(G,K,\psi )$. We let $o$ be the root of $Y$, that is, $o=K{1}_G$ as an element of $Y$ (a coset). The group $K$ acts on $X$ by automorphisms of that labelled graph. It leaves the set $F=K$ (now as a set of vertices of $X$) invariant. The factor graph of $X$ by this action is $Y$. Write $\pi$ for the factor mapping. It is $\left|K\right|$-to-one. Each cone of $X$ with respect to $F$ is mapped onto a cone of $Y$ with respect to $o$, and this mapping sends boundaries of cones of $X$ to boundaries of cones of $Y$. By assumption, $Y$ is a context-free graph. By Lemma 4.3, there is an upper bound on the number of elements in the latter boundaries. Therefore there also is an upper bound on the number of elements of any of the boundaries of the cones of $X$ with respect to $F$.

Without going here into the details of the definition of the space of ends of $X$, we refer to the terminology of Thomassen and Woess [19] and note that the above implies that all ends of $X$ are thin. But then, as proved in [19], $G$ must be a virtually free group. □

One should not tend to believe that in the situation of the last lemma, the Cayley graphs of $G$ are quasi-isometric with the Schreier graphs of $(G,K)$. As a simple counter-example, take for $G$ the infinite dihedral group $\langle a,b\mid a^2=b^2\rangle$ and for $K$ the 2-element subgroup generated by $a$.

5. Covers and Schreier graphs

We assume again that $(X,E,\ell )$ is symmetric and fully deterministic. Recall the involution $a\mapsto a^{-1}\ne a$ of $\Sigma$. A word in ${\Sigma }^{\ast }$ is called reduced if it contains no subword of the form $a{a}^{-1}$, where $a\in \Sigma$. We write ${\mathbb{T}}_{\Sigma }$ for the set of all reduced words in ${\Sigma }^{\ast }$. We can equip ${\mathbb{T}}_{\Sigma }$ with the structure of a labelled graph, whose edges are of the form

$$(v,a,w)\text{and}(w,a^{-1},v)\text{,}\text{where }v,w\in {\mathbb{T}}_{\Sigma },a\in \Sigma ,va=w\text{.}$$ (4)

Thus, the terminal letter of $v$ must be different from $a^{-1}$. Then ${\mathbb{T}}_{\Sigma }$ is fully deterministic, and it is a tree, that is, it has no closed path whose label is a (non-empty) reduced word. As the root of ${\mathbb{T}}_{\Sigma }$, we choose the empty word $ϵ$. Then ${\mathbb{T}}_{\Sigma }$ is the universal cover of $X$. Namely, if we choose (and fix) any vertex $o\in X$ as the root, then the mapping

$$\Phi :{\mathbb{T}}_{\Sigma }\to X\text{,}\Phi \left(w\right)=o^w\text{,}$$ (5)

is a covering map: it is a surjective homomorphism between labelled graphs which is a local isomorphism, that is, it is one-to-one between the sets of outgoing (resp. ingoing) edges of any element $w\in {\mathbb{T}}_{\Sigma }$ and its image $\Phi \left(w\right)$. (Note that this allows the image of an edge to be a loop.) “Universal” means that it covers every other cover of $X$, but this is not very important for us. The property of $w\in {\mathbb{T}}_{\Sigma }$ to be reduced is equivalent with the fact that the path ${\pi }_o\left(w\right)$ in $X$ is non-backtracking, that is, it does not contain two consecutive edges which are the reversal of each other.

We now realize that ${\mathbb{T}}_{\Sigma }$ is the standard Cayley graph of the free group ${\mathbb{F}}_{\Sigma }$, where $\Sigma$ is the set of free generators together with their inverses. The group product is the following: if $v,w\in {\mathbb{T}}_{\Sigma }\equiv {\mathbb{F}}_{\Sigma }$, then $v\cdot w$ is obtained from the concatenated word $vw$ by step after step deleting possible subwords of the form $a{a}^{-1}$ that can arise from that concatenation. The group identity is $ϵ$, and the inverse of $w$ is $w^{-1}$ as at the end of Definition 1.1. With $\Phi$ as in (5), let

$$\mathbb{K}=\mathbb{K}\left(X\right)={\Phi }^{-1}\left(o\right)=\{w\in {\mathbb{T}}_{\Sigma }:{\pi }_o\left(w\right)\text{ is a closed path from }o\text{ to }o\text{ in }X\}\text{.}$$ (6)

Then, under the indentification ${\mathbb{T}}_{\Sigma }\equiv {\mathbb{F}}_{\Sigma }$, we clearly have that $\mathbb{K}$ is a subgroup of ${\mathbb{F}}_{\Sigma }$. The following is known, see e.g. Lyndon and Schupp [10, Ch. III] or (our personal source) Imrich [8].

Proposition 5.1

The graph$X$is the Schreier graph of the pair of groups$({\mathbb{F}}_{\Sigma },\mathbb{K}\left(X\right))$with respect to the semigroup presentation$\psi$given by$\psi \left(a\right)=a$,$a\in \Sigma$.

In $\psi \left(a\right)=a$, we interpret $a$ simultaneously as a letter from the alphabet and as a generator of the free group.

Thus, in reality the study of context-free pairs of groups is the same as the study of fully deterministic, symmetric context-free graphs under a different viewpoint.

The same is not true without assuming symmetry. Indeed, given a semigroup presentation $\psi$ of $G$, for every $a\in \Sigma$ there must be $w_a\in {\Sigma }^{\ast }$ such $\psi \left(w_a\right)=\psi {\left(a\right)}^{-1}$, the inverse in $G$. But then in the Schreier Graph $X(G,K,\psi )$, for any subgroup $K$ of $G$, we have the following: if $(x,a,y)\in E$ then $y^{w_a}=x$, that is, there is the oriented path from $y$ to $x$ with label $w_a$. In a general fully deterministic graph this property does not necessarily hold, even if it has the additional property that for each $a\in \Sigma$, there is precisely one incoming edge with label $a$ at every vertex. As an example, consider $X=\{x,y,z\}$, $\Sigma =\{a,b\}$ and labelled edges $(x,a,y),(x,b,y),(y,a,z),(y,b,x),(z,a,x),(z,b,z)$.

We return to the situation of Proposition 5.1. As a subgroup of the free group, the group $\mathbb{K}\left(X\right)$ is itself free. There is a method for finding a set of free generators. First recall the notion of a spanning tree of $X$. This is a tree $T$, which as subgraph of $X$ is obtained by deleting edges (but no vertices) of $X$. Every connected (non-oriented) graph has a spanning tree, for locally finite graphs it can be constructed inductively. Now let $T$ be a spanning tree of $X$, and consider all edges of $X$ that are not edges of $T$. They must come in pairs $(e,e^{-1})$. For each pair, we choose one of the two partner edges, and we write $E_0$ for the chosen (oriented) edges. For each $e\in E_0$, we choose non-backtracking paths in $T$ from $o$ to $e^{-}$ and from $e^{+}$ to $o$. Together with $e$ (in the middle), they give rise to a non-backtracking path in $X$ that starts and ends at $o$. Let $w\left(e\right)$ be the label on that path. Then the following holds [10], [8].

Proposition 5.2

As elements of ${\mathbb{F}}_{\Sigma }$ , the $w\left(e\right)$ , $e\in E_0$ , are free generators of $\mathbb{K}\left(X\right)$ .

Corollary 5.3

Let $G$ be a virtually free group and $K$ a finitely generated subgroup. Then $(G,K)$ is context-free.

Proof

Let $\mathbb{F}={\mathbb{F}}_{\Sigma }$ be a free subgroup of $G$ of finite index. Then $\mathbb{K}=K\cap \mathbb{F}$ is a free subgroup of $K$ with $[K:\mathbb{K}]<\infty$. Since $K$ is finitely generated, also $\mathbb{K}$ is finitely generated. In the Schreier graph $X$ of $(\mathbb{F},\mathbb{K})$ with respect to the standard labelling by $\Sigma$, choose a spanning tree and remaining set $E_0$ of edges, as described above. Since all sets of free generators of $\mathbb{K}$ must have the same cardinality, $E_0$ is finite. Thus, $X$ is obtained by adding finitely many edges to a tree. If $o$ is the root vertex of $X$ and $n$ is the largest distance between $o$ and an endpoint of some edge in $E_0$, then every cone $C$ of $X$ with $d(\partial C,o)>n$ is a rooted, labelled tree that is isomorphic to one of the cones of ${\mathbb{T}}_{\Sigma }$. Thus, the Schreier graph, resp. $(\mathbb{F},\mathbb{K})$ are context-free. It now follows from Proposition 3.3 and Lemma 4.9 that also $(G,K)$ is context-free. □

We remark here that one can always reduce the study of context-free pairs to free groups and their subgroups. Given $(G,K)$, let $\mathbb{F}$ be a finitely generated free group that maps by a homomorphism onto $G$. Let $\mathbb{K}$ be the preimage of $K$ under that homomorphism. Then clearly $(G,K)$ is context-free if and only $(\mathbb{F},\mathbb{K})$ has this property. (This reduction, however, is not very instructive.)

Of course, there are context-free pairs with $G$ free beyond the situation of Corollary 5.3.

Example 5.4

Consider the free group $\mathbb{F}=\langle a,b\mid \rangle$ and the subgroup $\mathbb{K}$ with the infinite set of free generators $\{a^k{b}^{l}a{b}^{-l}{a}^{-k}:k,l\in \mathbb{Z},l\ne 0\}$. The associated Schreier graph with respect to $\{a^{\pm 1},b^{\pm 1}\}$ is the comb lattice.

Its vertex set is the set of integer points in the plane. The edges labelled by $a$ are along the $x$-axis, from $(k,0)$ to $(k+1,0)$, and there is a loop with label $a$ at each point $(k,l)$ with $l\ne 0$. The edges labelled by $b$ are all the upward edges of the grid, that is, all edges from $(k,l)$ to $(k,l+1)$, where $(k,l)\in {\mathbb{Z}}^2$. To these, we have to add the oppositely oriented edges whose labels are the respective inverses (in Fig. 3, the oppositely oriented edges together with the corresponding labels are omitted for simplicity). The comb lattice is clearly a context-free graph (tree).

Fig. 3.

The comb lattice described in Example 5.4.

We proceed giving some simple examples. It is very easy to see that context-freeness is not “transitive” in the following sense: if $(G,H)$ and $(H,K)$ are context-free (with $G,H$ finitely generated and $K\le H\le G$) then in general $(G,K)$ will not be context-free.

Example 5.5

Let $G={\mathbb{Z}}^2$, $H=\mathbb{Z}\times \left\{0\right\}\cong \mathbb{Z}$ and $K=\left\{(0,0)\right\}$. Then $H$ (i.e., $(H,K)$) is context-free. Of course, this also holds for $(G,H)$, whose Schreier graphs are just the Cayley graphs of $\mathbb{Z}$. But ${\mathbb{Z}}^2$ (i.e., $(G,K)$) is not context-free.

This also shows that the converse of Lemma 3.1 does not hold in general (while we know that it does hold when $[G:H]<\infty$). Finally, we construct examples of three groups $K\le H\le G$, where $(G,H)$ is context-free, $[H:K]<\infty$, and $(G,K)$ is not context-free.

Example 5.6

We construct a family of fully deterministic, symmetric labelled graphs $X_W$, $W\subset \mathbb{Z}$(non-empty), and one such graph $Y$, so that $Y$ is the factor graph with respect to the action of a 2-element group of automorphisms of each of the labelled graphs $X_W$. While $Y$ will be a context-free graph, many of the graphs $X_W$ in our family are not context-free. We then translate this back into the setting of pairs of groups.

The vertex set of $X_W$ is $\mathbb{Z}\times \{0,1\}$. The set of labels is $\Sigma =\{a,b,a^{-1},b^{-1}\}$. The edges are as follows:

$$((k,0),a,(k+1,0))\text{and}((k,1),a,(k+1,1))\text{for all}k\in \mathbb{Z}\text{,}$$

$$((k,0),b,(k+1,0))\text{and}((k,1),b,(k+1,1))\text{for all}k\in \mathbb{Z}\setminus W\text{,}\text{and}$$

$$((k,0),b,(k+1,1))\text{and}((k,1),b,(k+1,0))\text{for all}k\in W\text{.}$$

The reversed edges carry the respective inverse labels (in Fig. 4, these reversed edges together with the corresponding labels are omitted for simplicity). Since $W\ne 0̸$, there is at least one of the “crosses” (pair of the third type of edges). Therefore $X_W$ is connected. In general, it does not have finitely many cone types, i.e., it is not context-free. For example, it is not context-free when $W=\{k(\left|k\right|+1):k\in \mathbb{Z}\}$

For arbitrary $W$, the two-element group that exchanges each $(k,0)$ with $(k,1)$ acts on $X_W$ by label preserving graph automorphisms. The factor graph $Y$ (see Fig. 5) has vertex set $\mathbb{Z}$ and edges

$$(k,a,k+1)\text{and}(k,b,k+1)\text{for all}k\in \mathbb{Z}\text{,}$$

plus the associated reversed edges (in Fig. 5, these edges together with the corresponding labels are omitted for simplicity). It is clearly a context-free graph.

Now let $\mathbb{F}={\mathbb{F}}_{\Sigma }$ be the free group (universal cover of $X_W$ and $Y$), and for given $W$, let ${\mathbb{K}}_W$ be the fundamental group of $X_W$ at the vertex $(0,0)$. Furthermore, let $\mathbb{K}$ be the fundamental group of $Y$ at the vertex 0. Then it is straightforward that ${\mathbb{K}}_W$ has index 2 in $\mathbb{K}$. The mapping $\psi$ is the embedding of $\Sigma$ into ${\mathbb{F}}_{\Sigma }$, as above. We then have $Y=X(\mathbb{F},\mathbb{K},\psi )$ and $X_W=X(\mathbb{F},{\mathbb{K}}_W,\psi )$, providing the required example.

Fig. 4.

The fully deterministic, symmetric labelled graph $X_W$, with $W\subset \mathbb{Z}$ described in Example 5.6 (here $W=\{0,1,-3,\dots \}$). The reverse edges, together with the corresponding labels, are omitted for simplicity.

Fig. 5.

The factor graph $Y$ of the graph $X_W$ from Fig. 4 (cf. Example 5.6). The reverse edges, together with the corresponding labels, are omitted for simplicity.

Example 5.7

At the end of the Introduction, we mentioned the possible interplay with ends. The number of ends $e\left(X\right)$ of a symmetric, connected graph is the supremum of the number of connected components of the complement of any finite subgraph. Via Stallings’ [17] celebrated structure theorem, ends of groups (i.e., ends of Cayley graphs) are closely related with amalgamated free products and HNN-extensions. Thus, it is natural to ask the following question.

Let $(G_1,K)$ and $(G_2,K)$ be two context-free pairs of groups sharing the same subgroup $K$. Let $G=G_1{\ast }_K{G}_2$ be the amalgamated free product of $G_1$ and $G_2$ over the group $K$. Is it then true that $(G,K)$ is context-free ? When $K$ is finite, the answer is of course “yes”, because then $G_1,G_2$ and $G$ are virtually free. When $K$ is infinite, we have a counter-example. Here is a brief outline.

Let $G=\langle a_1,a_2,b_1,b_2\mid [a_1,b_1][a_2,b_2]\rangle$ be the fundamental group of an orientable surface of genus 2. Let $K$ be the infinite cyclic subgroup generated by the commutator $[a_1,b_1]={[a_2,b_2]}^{-1}$, and for $i=1,2$, let $G_i$ be the free group with free generators $a_i$ and $b_i$. Then $G$ is the amalgamated free product of $G_1$ and $G_2$ over $K$.

By Corollary 5.3, the pairs $(G_1,K)$ and $(G_2,K)$ are context-free. But $(G,K)$ is not context-free. Indeed, let $X$ be the Schreier graph of $(G,K)$ with respect to the above generators and their inverses. It has two ends, see e.g. the outline in the Introduction of [14]. Thus, there is a finite subgraph $F$ of $X$ such that $X\setminus B(F,n)$ has exactly two infinite cones for any $n$. If $X$ were context-free, then the finite upper bound on the number of boundary elements of any cone would yield that $X$ has linear growth, that is $\left|B(F,n)\right|\le C\cdot n$ for all $n$. This contradicts the fact that $G$, as well as the Schreier graphs of $(G_1,K)$ and $(G_2,K)$, have exponential growth.