## Alpha-beta shifts

Given ${\beta>1}$, the interval map ${T\colon [0,1]\rightarrow [0,1]}$ given by ${T(x) = \beta x \pmod 1}$ is naturally semi-conjugate to the ${\beta}$-shift ${\Sigma_\beta \subset \{0,1,\dots,\lceil\beta\rceil-1\}^{\mathbb N}}$. The shift space ${\Sigma_\beta}$ admits a natural description in terms of the lexicographic order, and another in terms of a countable-state directed graph. These have been used to obtain a fairly complete description of the dynamics of the ${\beta}$-transformation ${T}$, including existence, uniqueness, and strong statistical properties of equilibrium states for Hölder potentials.

In this post I want to explain how one can give similar descriptions (both in terms of lexicographic order and in terms of a countable-state Markov structure) for the coding spaces associated to the interval map ${T(x) = \alpha + \beta x \pmod 1}$. Just as with the ${\beta}$-shifts, these “${\alpha}$${\beta}$ shifts” arise from piecewise expanding interval maps, and thus can be studied using the general machinery developed by Hofbauer, but I believe that they are worth studying a little more carefully as an intermediate case between the ${\beta}$-shifts, where the fact that ${0}$ is a safe symbol’ can be used to prove various results that are still open for ${\alpha}$${\beta}$ shifts, and the more general setting where the description of the shift space requires more bookkeeping and it can be harder to see what is going on.

1. Coding spaces for interval maps

We start by recalling the general notion of a coding space for a map of the interval. Say that a map ${T\colon [0,1]\rightarrow [0,1]}$ is a piecewise expanding interval map if there is ${\lambda>1}$ and a partition of ${[0,1]}$ into finitely many intervals ${I_0,\dots, I_{d-1}}$ such that ${T}$ is continuous on each ${I_j}$ and ${C^1}$ on the interior of each ${I_j}$, with ${|T'| \geq \lambda}$. Note that we do not care whether the intervals are closed, open, or half of each.

Let ${A = \{0,1,\dots, d-1\}}$, and define a map ${h\colon [0,1]\rightarrow A^{\mathbb N}}$ by ${h(x)_n = a \in A}$ whenever ${T^n(x) \in I_a}$. Let ${\Sigma = \Sigma_T = \overline{h([0,1])}}$; we say that ${\Sigma}$ is the coding space for the interval map ${T}$ relative to the partition ${\{I_j\}_j}$. We can define a map ${\pi\colon \Sigma\rightarrow [0,1]}$ by the condition that ${\pi(h(x)) = x}$ for all ${x\in [0,1]}$ and ${\pi}$ is continuous. Then we have ${\pi\circ \sigma = T\circ \pi}$, so ${\pi}$ is a semiconjugacy between ${(\Sigma,\sigma)}$ and ${([0,1],T)}$.

Another way of interpreting this is the following: for each ${a\in A}$ there is a unique map ${S_a \colon \overline{T(I_a)} \rightarrow \overline{I_a}}$ such that ${S_a(T x) = x}$ for all ${x}$ in the interior of ${I_a}$. Note that ${S_a}$ is ${C^1}$ with ${|S_a'| \leq \lambda^{-1} < 1}$. Given any set ${E\subset [0,1]}$, write ${S_a(E) = S_a(E \cap \overline{T(I_a)})}$ for convenience; note that ${S_a(E)=\emptyset}$ if ${E}$ does not intersect ${\overline{T(I_a)}}$. Given a finite word ${w\in A^* = \bigcup_{n\geq 0} A^n}$, let ${J_w = S_w([0,1])}$, where ${S_w = S_{w_1} \circ \cdots S_{w_{|w|}}}$ and ${|w|}$ is the length of ${w}$. Roughly speaking, ${J_w}$ represents the set of points ${x\in [0,1]}$ for which ${x\in I_{w_1}}$, ${T(x) \in I_{w_2}}$, and so on. (This is not quite completely true because of problems at the endpoints of the intervals.)

The language of the shift ${\Sigma}$ is the set of all ${w\in A^*}$ such that ${[w] := \{x\in \Sigma : x_1 \cdots x_{|w|} = w\} \neq\emptyset}$. Write ${\mathcal{L}}$ for this collection of words; then ${\mathcal{L} = \{ w \in A^* : J_w \neq \emptyset\}}$, and we have ${\pi([w]) = J_w}$ for all ${w\in \mathcal{L}}$. Let ${\mathop{\mathcal L}_n = \{w\in \mathop{\mathcal L} : |w| = n\}}$; given ${w\in \mathop{\mathcal L}_n}$, the interval ${J_w}$ has length ${\leq \lambda^{-n}}$, and since for every ${x\in \Sigma}$ we have ${\{x\} = \bigcap_{n\geq 0} [x_1\cdots x_n]}$, we also have ${\{\pi(x)\} = \bigcap_{n\geq 0} J_{x_1 \cdots x_n}}$.

So far these considerations have been completely general, valid for any piecewise expanding interval map. Now we consider the specific transformations ${T_\beta(x) = \beta x \pmod 1}$ and ${T_{\alpha,\beta} = \alpha + \beta x \pmod 1}$, where ${z\pmod 1 := z - \lfloor z \rfloor}$ and we assume without loss of generality that ${\alpha \in [0,1)}$. These are piecewise expanding interval maps, where the natural partition to consider is the partition into maximal intervals of continuity. Thus for ${T_\beta}$, we have ${I_0 = [0,\frac 1\beta)}$, ${I_1 = [\frac 1\beta, \frac 2\beta)}$, and so on, with ${I_{d-1} = [\frac {d-1}\beta,1]}$, where ${d = \lceil\beta\rceil-1}$. For ${T_{\alpha,\beta}}$, we have ${I_0 = [0,\frac{1-\alpha}\beta)}$, ${I_1 = [\frac{1-\alpha}{\beta},\frac{2-\alpha}\beta)}$, and so on, with ${d = \lceil\alpha+\beta\rceil - 1}$ and ${I_{d-1} = [\frac{d-1+\alpha}{\beta},1]}$. We write ${\Sigma_\beta}$ and ${\Sigma_{\alpha,\beta}}$ for the coding spaces of these transformations relative to their natural partitions.

2. Description via lexicographic order

The full shift ${A^{\mathbb N}}$ carries a natural lexicographic order inherited from the usual order on ${A = \{0,1,\dots, d-1\}}$: given ${y\neq z\in A^{\mathbb N}}$, let ${n=n(y,z)}$ be minimal such that ${y\neq z}$, and write ${y\prec z}$ if ${y_n < z_n}$. Write ${y\preceq z}$ if ${y\prec z}$ or ${y=z}$. This is a total order on ${A^{\mathbb N}}$. The key observation for our purposes is that because ${T_\beta}$ and ${T_{\alpha,\beta}}$ are order-preserving on each of their intervals of continuity, their coding maps are also order-preserving in the sense that ${\pi(y) \leq \pi(z)}$ if and only if ${y\preceq z}$. (Note that we must use non-strict inequalities because the coding maps are not 1-1.)

2.1. The ${\beta}$-shifts

Fix ${\beta>1}$ and let ${T=T_\beta}$. Let ${{\mathbf b} = \sup \Sigma_\beta}$, where supremum is w.r.t. lexicographic order; the supremum exists because ${A^{\mathbb N}}$ has the least upper bound property, and is in ${\Sigma_\beta}$ because the ${\beta}$-shift is closed. Thus ${{\mathbf b} = b_1 b_2 b_3 \cdots}$ is the lexicographically maximal element of ${\Sigma_\beta}$; it would code the trajectory of the right endpoint of the unit interval if we replaced ${T_\beta}$ with the map ${T(x) = 1 + \beta x - \lceil \beta x \rceil}$, which agrees with ${T_\beta}$ everywhere except at the endpoints of the intervals of monotonicity. The points ${T^n(1) \in [0,1]}$ will play an important role in the following result.

Proposition 1 A sequence ${y\in A^{\mathbb N}}$ is in ${\Sigma_\beta}$ if and only if

$\displaystyle \sigma^n(y) \preceq {\mathbf b} \text{ for all } n=0,1,2,\dots. \ \ \ \ \ (1)$

Similarly, a word ${w\in A^*}$ is in ${\mathop{\mathcal L}(\Sigma_\beta)}$ if and only if

$\displaystyle w_{|w|-k+1}\cdots w_{|w|} \preceq b_1 \cdots b_k \text{ for all } 1\leq k\leq |w|. \ \ \ \ \ (2)$

Proof: Because ${\Sigma_\beta}$ is closed and ${\sigma}$-invariant, every ${y\in \Sigma_\beta}$ satisfies (1), and it follows immediately that every ${w\in \mathop{\mathcal L}(\Sigma_\beta)}$ satisfies (2).

To prove the converse direction, it suffices to show that every ${w\in A^*}$ which satisfies (2) is contained in ${\mathop{\mathcal L}}$; in other words, that ${J_w \neq \emptyset}$ for every such ${w}$. Equivalently, we can work with the following object: consider for each ${w \in A^*}$ the follower interval ${F_w = \overline{T^{|w|} (J_w)}}$; this can also be defined as the minimal interval with the property that ${S_w([0,1]\setminus F_w)=\emptyset}$. Note that ${F_w = \emptyset}$ whenever ${w\notin \mathop{\mathcal L}}$. Given ${w\in \mathop{\mathcal L}}$ and ${a\in A}$, observe that

$\displaystyle F_{wa} = \overline{T^{1+|w|}(J_{wa})} = \overline{T(T^{|w|}S_w S_a[0,1])} = \overline{T(T^{|w|}S_w I_a)} = \overline{T(F_w \cap I_a)}. \ \ \ \ \ (3)$

Now we can give a non-recursive description of ${F_w}$ that completes the proof of Proposition 1. If ${w\in A^*}$ satisfies (2), let

$\displaystyle k(w) = \max \{ k : w_{|w|-k+1} \cdots w_{|w|} = b_1 \cdots b_k \} \ \ \ \ \ (4)$

be the length of the longest prefix of ${{\mathbf b}}$ that appears as a suffix of ${w}$.

Lemma 2 If ${w\in A^*}$ satisfies (2), then ${F_w = [0,T^{k(w)}(1)]}$.

Proof: We go by induction in the length of ${w}$, using ${|w|=0}$ (so ${w}$ is the empty word and ${F_w = [0,1]}$) as the base case. Suppose we have some ${w\in A^n}$ satisfying (2), and that the lemma has been proved for all words with length ${. Write ${w = va}$ where ${|v| = n-1}$ and ${a\in A}$. Let ${j = k(v)}$, so that ${v=u b_1 \cdots b_j}$ for some word ${u}$. By the inductive hypothesis, we have ${F_v = [0,T^{j}(1)]}$ and hence ${v\in \mathop{\mathcal L}}$. Then by (3), we have

$\displaystyle F_w = F_{va} = \overline{T(F_v \cap I_a)} = \overline{T([0,T^{k(v)}(1)] \cap I_a)}.$

There are two cases to consider.

Case one: ${T^{k(v)}(1) \in I_a}$. In this case we have ${a=b_{j+1}}$ and ${k(w) = j+1}$, and ${F_w = \overline{T([\frac a\beta,T^{k(v)}(1)])} = [0,T^{k(v)+1}] = [0,T^{k(w)}(1)]}$.

Case two: ${T^{k(v)}(1) \notin I_a}$. In this case we must have ${a < b_{j+1}}$, otherwise we would have ${w=ub_1\cdots b_j a}$ and hence ${w_{|w|-j} \cdots w_{|w|} \succ b_1 \cdots b_{j+1}}$, violating (2). Thus ${I_a \subset F_v}$ and hence by (3), we have ${F_w = \overline{T(I_a)} = [0,1]}$. On the other hand, since ${a < b_{j+1}}$, we see that for every ${1\leq i\leq j}$ we have

$\displaystyle w_{|w|-j+i} \cdots w_{|w|}= b_i b_{i+1} \cdots b_j a \prec b_i \cdots b_j b_{j+1} \preceq b_1 \cdots b_{j-i+2},$

where the last inequality uses the fact that ${\sigma^{i-1}{\mathbf b} \preceq {\mathbf b}}$. We conclude that ${k(w) = 0}$, which completes the proof of Lemma 2. $\Box$

With Lemma 2 complete, Proposition 1 follows immediately since ${F_w\neq \emptyset}$ implies that ${w\in \mathop{\mathcal L}}$. $\Box$

A historical note: the transformation ${T_\beta}$ was introduced by Rényi in 1957 to study representations of real numbers in non-integer bases. The lexicographic characterization of ${\beta}$-shifts in Proposition 1 was given by Parry in 1960 (see Theorem 3 of that paper) using different methods than the proof here. This proof is essentially the same as the one given by Hofbauer in 1979 (see Theorem 2 there); Hofbauer’s approach has the advantage of dealing with all piecewise monotonic interval maps, and the argument given here for the ${\beta}$-shift is actually just a special case of his general result.

2.2. The ${\alpha}$${\beta}$ shifts

Instead of describing the full generality of Hofbauer’s result here, we describe how Proposition 1 adapts to the ${\alpha}$${\beta}$ shifts. This family of examples already reveals some of the challenges that arise when we go from ${\beta}$-shifts to more general piecewise expanding transformations.

Fix ${\alpha\in [0,1)}$ and ${\beta>1}$, then let ${T = T_{\alpha,\beta} \colon x \mapsto \alpha + \beta x \pmod 1}$, and let ${\Sigma = \Sigma_{\alpha,\beta}}$ be the coding space for ${T}$ with respect to its natural partition. Let ${{\mathbf a} = \inf \Sigma}$ and ${{\mathbf b} = \sup \Sigma}$ be the lexicographically minimal and maximal elements of ${\Sigma}$. Then every ${y\in \Sigma}$ has the property that

$\displaystyle {\mathbf a} \preceq \sigma^n y \preceq {\mathbf b} \text{ for all } n=0,1,2,\dots, \ \ \ \ \ (5)$

and every ${w\in \mathop{\mathcal L} = \mathop{\mathcal L}(\Sigma)}$ has the property that

$\displaystyle a_1 \cdots a_k \preceq w_{|w|-k+1} \cdots w_{|w|} \preceq b_1 \cdots b_k \text{ for all } 1\leq k \leq |w|. \ \ \ \ \ (6)$

As with the ${\beta}$-shifts, these conditions are both necessary and sufficient for membership in ${\Sigma}$ and ${\mathop{\mathcal L}}$. The key to proving this is the following analogue of Lemma 2, whose proof we omit since it is similar to the proof there (with just a little more bookkeeping).

Lemma 3 Given ${w\in A^*}$ satisfying (6), let

\displaystyle \begin{aligned} k_1(w) &= \max \{k : a_1 \cdots a_k = w_{|w|-k+1} \cdots w_{|w|} \}, \\ k_2(w) &= \max \{k : b_1 \cdots b_k = w_{|w|-k+1} \cdots w_{|w|} \}. \end{aligned} \ \ \ \ \ (7)

Then ${F_w = [\underline{T}^{k_1(w)}(0),\overline{T}^{k_2(w)}(1)]}$, where ${\underline{T}}$ and ${\overline{T}}$ agree with ${T_{\alpha,\beta}}$ on the interiors of the intervals ${I_a}$, and are defined at the endpoints by the condition that ${\underline{T}}$ is continuous from the right and ${\overline{T}}$ is continuous from the left.

3. A directed graph on ${{\mathbb N}}$

The lexicographic description of ${\Sigma_\beta}$ in Proposition 1 is appealing, but for many purposes it is more useful to have a description of the ${\beta}$-shift in terms of a countable-state directed graph, which goes back to work of Takahashi in 1973 and Hofbauer in 1978.

Start by considering the notion of follower set in a shift space, which is analogous to the follower intervals considered in the previous part. Given a shift space ${\Sigma}$ with language ${\mathop{\mathcal L}}$, consider for each ${w\in \mathop{\mathcal L}}$ the follower set

$\displaystyle F^w = \{y\in \Sigma : wy\in \Sigma \} = \sigma^{|w|}([w] \cap \Sigma). \ \ \ \ \ (8)$

One could also consider the set of words ${v\in \mathop{\mathcal L}}$ such that ${wv\in \mathop{\mathcal L}}$, but the definition in (8) will be more convenient for our purposes. It was shown by Weiss in 1973 that a shift can be represented via a labeled directed graph on finitely many vertices if and only if it has finitely many follower sets; that is, if ${\{F^w : w\in \mathop{\mathcal L}\}}$ is finite. Such shifts are called sofic. Similarly, ${\Sigma}$ can be represented via a labeled irreducible directed graph on countably many vertices if and only if it has countably many follower sets, and such shifts are called coded; see Fiebig and Fiebig 1992 and 2002.

Every ${\beta}$-shift is characterized by a sequence ${{\mathbf b}\in A^{\mathbb N}}$ with the property that ${\sigma^n{\mathbf b} \preceq{\mathbf b}}$ for all ${n\geq 0}$: as we saw in Proposition 1, we have

\displaystyle \begin{aligned} \Sigma_\beta &= \{y\in A^{\mathbb N} : \sigma^n y \preceq {\mathbf b} \text{ for all } n\geq 0 \}, \\ \mathop{\mathcal L}(\Sigma_\beta) &= \{w\in A^* : w_{|w|-k+1}\cdots w_{|w|} \preceq b_1 \cdots b_k \text{ for all } 1\leq k \leq |w| \}. \end{aligned}

Moreover, it follows from Lemma 2 that for every ${w\in \mathop{\mathcal L}(\Sigma_\beta)}$, the corresponding follower set in ${\Sigma_\beta}$ is

$\displaystyle F^w = \{y\in \Sigma : y\preceq \sigma^{k(w)}{\mathbf b}\} =: [\mathbf{0},\sigma^{k(w)}{\mathbf b}],$

where ${k(w)}$ is defined in (4) to be the length of the longest prefix of ${{\mathbf b}}$ that appears as a suffix or ${w}$. Thus ${F^w}$ is completely determined by ${k(w)}$, which can take countably many values, and since ${\beta}$-shifts are transitive this implies that they are coded. Thus they can be presented by a countable-state graph which supports a countable-state topological Markov chain (the Fischer cover).

Given any coded shift, the vertex set of the Fischer cover is the set of follower sets: ${V = \{F \subset \Sigma : F=F^w}$ for some ${w\in \mathop{\mathcal L}\}}$. A labeled edge is a triple ${(F,F',a)}$, where ${F,F'\in V}$ and ${a\in A}$; the labeled edge set of the Fischer cover is ${E = \{(F^w,F^{wa},a) : wa\in \mathop{\mathcal L}, a\in A\}}$; that is, there is an edge from ${F^w}$ to ${F^{wa}}$ whenever ${wa\in \mathop{\mathcal L}}$ and ${a\in A}$, and this edge is labeled with the symbol ${a}$. Note that there may be multiple edges between two vertices, each with a different label.

Say that a sequence ${y\in A^{\mathbb N}}$ labels a walk on the graph if there is a sequence of edges ${e_1,e_2,\dots}$ such that for every ${n}$,

1. ${e_n}$ is labeled with the symbol ${y_n}$, and
2. the target vertex of ${e_n}$ is the source vertex of ${e_{n+1}}$.

Then if ${y\in A^{\mathbb N}}$ labels a walk on this graph, there is a word ${w\in \mathop{\mathcal L}}$ such that ${w y_1 \cdots y_n \in \mathop{\mathcal L}}$ for all ${n}$, and hence ${y\in \Sigma}$. Conversely, every ${y\in \Sigma}$ labels a walk on the graph, so we can describe both ${\Sigma}$ and ${\mathop{\mathcal L}}$ in terms of walks on the graph.

For the ${\beta}$-shift, since the follower set ${F^w}$ is completely determined by ${k(w)}$, we can identify the vertex set of the Fischer cover with ${\{0,1,2,3,\dots\}}$, and observe that the two cases in Lemma 2 give the following two types of edges.

1. There is an edge from ${k}$ to ${k+1}$ whenever there are ${v\in \mathop{\mathcal L}}$ and ${a\in A}$ such that ${va\in \mathop{\mathcal L}}$, ${k(v)=k}$, and ${k(va) = k+1}$. This happens for every ${k}$ since we can take ${v=b_1 \cdots b_k}$ and ${a=b_{k+1}}$, so the edge from ${k}$ to ${k+1}$ is labeled with the symbol ${b_{k+1}}$.
2. There is an edge from ${k}$ to ${0}$ whenever there are ${v\in \mathop{\mathcal L}}$ and ${a\in A}$ such that ${va\in \mathop{\mathcal L}}$, ${k(v)=k}$, and ${k(va)=0}$. This happens whenever ${a < b_{k+1}}$, and so whenever ${b_{k+1} > 0}$ there are ${b_{k+1}}$ edges from ${k}$ to ${0}$ labeled with the symbols ${0,1,\dots, b_{k+1}-1}$.

For example, when ${{\mathbf b} = 21201\dots}$, the first part of the graph looks like this:

The vertex labeled 0 can be thought of as the base vertex; it corresponds to the follower set ${\Sigma}$, so every sequence ${y\in \Sigma}$ labels a walk starting at this vertex. The vertex labeled 1 corresponds to the follower set ${F^2 = \{y\in \Sigma : y\preceq \sigma{\mathbf b}\}}$; a sequence ${y\in \Sigma}$ labels a walk starting at this vertex if and only if ${2y\in \Sigma}$. For a more typical sort of example, let ${w=21121200212}$. Then ${w}$ corresponds to the walk which starts at vertex 0, goes out to vertex 2 before returning to vertex 0 (thanks to the fact that ${w_3=1 < 2 = b_3}$), then goes out to vertex 4 before returning to vertex 0 (because ${w_8=0 < 1 = b_5}$), then goes out to vertex 3 and stops. We see that ${k(w) = 3}$ and that

$\displaystyle F^w = F^{212} = \{y\in \Sigma : 212y\in \Sigma\} = \{y\in \Sigma : y\preceq \sigma^3{\mathbf b}\}.$

4. A directed graph on ${{\mathbb N}^2}$

In order to give an analogous description of ${\Sigma_{\alpha,\beta}}$ via a countable-state Markov shift, we first observe that ${\Sigma_{\alpha,\beta}}$ is determined lexicographically by two sequences ${{\mathbf a},{\mathbf b}\in A^{\mathbb N}}$ with the property that

$\displaystyle {\mathbf a} \preceq \sigma^n{\mathbf a} \preceq {\mathbf b} \text{ and } {\mathbf a}\preceq \sigma^n{\mathbf b} \preceq{\mathbf b} \text{ for all } n\geq 0.$

Given such an ${{\mathbf a}}$ and ${{\mathbf b}}$, the corresponding shift space ${\Sigma}$ and language ${\mathop{\mathcal L}}$ are given by

\displaystyle \begin{aligned} \Sigma &= \{y\in A^{\mathbb N} : {\mathbf a} \preceq \sigma^n y \preceq {\mathbf b} \text{ for all } n\geq 0 \}, \\ \mathop{\mathcal L} &= \{w\in A^* : a_1 \cdots a_k \preceq w_{|w|-k+1}\cdots w_{|w|} \preceq b_1 \cdots b_k \text{ for all } 1\leq k \leq |w| \}. \end{aligned}

From Lemma 3, the follower sets are given by

$\displaystyle F^w = [\sigma^{k_1(w)}{\mathbf a},\sigma^{k_2(w)}{\mathbf b}],$

where ${k_1,k_2}$ are given by (7) and represent the lengths of the longest prefixes of ${{\mathbf a}}$ and ${{\mathbf b}}$, respectively, that appear as suffixes of ${w}$. Thus the follower set of ${w}$ is completely determined by ${k_1(w)}$ and ${k_2(w)}$, and so we can represent ${\Sigma}$ by a graph whose vertices lie in ${\{0,1,2,\dots\}^2}$. As we will see momentarily, not every pair ${(k_1,k_2) \in \{0,1,2,\dots\}^2}$ corresponds to a follower set, so we will not use all such vertices in building the graph, but thinking of the integer lattice still provides a useful way to visualize that situation.

It is useful to look at a concrete example. Let ${{\mathbf b} = 212012112\dots}$ and ${{\mathbf a} = 012101\dots}$. Then some of the follower sets, together with their corresponding values of ${(k_1,k_2)}$, are as follows:

\displaystyle \begin{aligned} \Sigma &= [{\mathbf a},{\mathbf b}] \quad &&(0,0), \qquad\qquad & F^{01} &= [\sigma^2{\mathbf a},{\mathbf b}] \quad &&(2,0), \\ F^0 &= [\sigma{\mathbf a},{\mathbf b}] &&(1,0), & F^{012} &= [\sigma^2{\mathbf a},\sigma {\mathbf b}] &&(3,1), \\ F^1 &= [{\mathbf a},{\mathbf b}] &&(0,0), & F^{0121} &= [\sigma^4{\mathbf a},\sigma^2{\mathbf b}] && (4,2), \\ F^2 &= [{\mathbf a},\sigma{\mathbf b}] &&(0,1), & F^{01210} &= [\sigma^5{\mathbf a},{\mathbf b}] && (5,0). \end{aligned}

In the graph for the ${\beta}$-shifts, remember that the outgoing edges from vertex ${k}$ came in two types: from ${k}$ to ${0}$ with label ${i}$ for each ${0\leq i < b_{k+1}}$, and from ${k}$ to ${k+1}$ with label ${i}$ for ${i=b_{k+1}}$, depending on whether ${k(wi) = k(w)+1}$ or ${k(wi)=0}$. (We switch to using ${i}$ for a generic element of ${A}$ now that ${{\mathbf a}}$ is in play.)

For the ${\alpha}$${\beta}$ shifts, there are more possibilities for ${k_1(wi)}$ and ${k_2(wi)}$:

1. ${k_1(wi) = k_2(wi)=0}$, which happens if ${a_{k_1 + 1} < i < b_{k_2 + 1}}$, and gives an edge from ${(k_1,k_2)}$ to ${(0,0)}$ labeled ${i}$;
2. ${k_1(wi)=k_1(w)+1}$ and ${k_2(wi)=0}$, which happens if ${a_{k_1 +1} = i < b_{k_2 + 1}}$, and gives an edge from ${(k_1,k_2)}$ to ${(k_1+1,0)}$ labeled ${i}$;
3. ${k_1(wi)=0}$ and ${k_2(wi)=k_2(w)+1}$, which happens if ${a_{k_1+1} < i = b_{k_2+1}}$, and gives an edge from ${(k_1,k_2)}$ to ${(0,k_2+1)}$ labeled ${i}$;
4. ${k_1(wi)=k_1(w)+1}$ and ${k_2(wi) = k_2(w)+1}$, which happens if ${a_{k_1 + 1} = i = b_{k_2 + 1}}$, and gives an edge from ${(k_1,k_2)}$ to ${(k_1+1,k_2+1)}$ labeled ${i}$.

For our concrete example with ${{\mathbf b} = 212012112\dots}$ and ${{\mathbf a} = 012101\dots}$, the first part of the graph is as follows. Note that our coordinate conventions follow standard matrix notation (down, then right) instead of the usual cartesian coordinates, and that we write the vertices as ${03}$ instead of ${(0,3)}$, and so on, in order to save space.

Observe that there are two types of vertices: on the one hand there are those such as ${(2,0)}$, ${(3,1)}$, ${(0,3)}$, ${(1,4)}$, with exactly one outgoing edge, which goes down and to the right; on the other hand there are vertices which have at least two outgoing edges, each of which resets’ in either the horizontal or vertical directions (or both).

Let us conclude by formulating an open problem. In a previous post, I gave a proof of the (well-known) result that if ${\Sigma}$ is a shift space with the specification property, then every Hölder continuous potential ${\varphi\colon \Sigma \rightarrow {\mathbb R}}$ has the property that every equilibrium state for ${\varphi}$ has positive entropy (“every Hölder potential is hyperbolic”). The ${\beta}$-shifts do not have specification in general (Schmeling showed in 1997 that the set of ${\beta>1}$ such that ${\Sigma_\beta}$ has specification is a null set for Lebesgue measure), but it turns out that they still have the property that every Hölder potential is hyperbolic; Dan Thompson and I showed this in a 2013 paper (arXiv:1106.3575).

Our argument there (see Proposition 3.1 and Section 6.1) relies on the fact that 0 is a `safe symbol’ for the ${\beta}$-shifts; if ${w\in \mathop{\mathcal L}(\Sigma_\beta)}$ and ${w'}$ is obtained from ${w}$ by replacing a single symbol with 0, then ${w'\in \mathop{\mathcal L}(\Sigma_\beta)}$ as well. This is no longer true for ${\Sigma_{\alpha,\beta}}$, and so our proof does not generalize. It would be interesting to know whether ${\Sigma_{\alpha,\beta}}$ still has the property that every Hölder potential is hyperbolic; so far as I know this question is wide open. Buzzi conjectured in 2004 that this property holds not just for the ${\alpha}$${\beta}$ shifts, but for every coding space of a piecewise monotonic interval map. Attacking this problem for the ${\alpha}$${\beta}$ shifts seems like a reasonable first step towards resolving this conjecture.