Open Research

: We study the automorphism group of an inﬁnite minimal shift ( X , σ ) such that the complexity difference function, p ( n + 1 ) − p ( n ) , is bounded. We give some new bounds on Aut ( X , σ ) / (cid:104) σ (cid:105) and also study the one-sided case. For a class of Toeplitz shifts, including the class of shifts deﬁned by constant-length primitive substitutions with a coincidence, and with height one, we show that the two-sided automorphism group is a cyclic group. We next focus on shifts generated by primitive constant-length substitutions. For these shifts, we give an algorithm that computes their two-sided automorphism group. Finally we show that with the same techniques, we are able to compute the set of conjugacies between two such shifts.


Introduction
In this article we study the automorphism groups Aut(X, σ ) of some "small" shifts (X, σ ).All terms are defined as they are needed in Sections 2 to 3.
In the brief Section 2, we show that for shifts with sublinear complexity, there are bounds on the orders of Aut(X, σ ) for one-sided shifts and Aut(X, σ )/ σ for two-sided shifts.
In [3], Coven explicitly describes the endomorphism monoid of any non-trivial two-sided constantlength substitution shift (X, σ ) on two letters.Either the shift has a metric discrete spectrum, in which case Aut(X, σ ) consists of the powers of the shift, or the shift has a partly continuous spectrum, in which case the "letter-exchanging" automorphism is also present.In both cases all endomorphisms are automorphisms.For constant-length substitutions on larger alphabets, those whose shifts have a discrete spectrum are precisely those that have coincidences, [9, Thm 7] and all other constant-length shifts, ETHAN M. COVEN, ANTHONY QUAS, AND REEM YASSAWI 2 Automorphisms of minimal shifts with sublinear complexity

Notation
Let A be a finite alphabet, with the discrete topology, and let N 0 = {0, 1, 2, . ..}.We endow A N 0 and A Z with the product topology, and let σ : A N 0 → A N 0 (or σ : A Z → A Z ) denote the shift map.We consider only infinite minimal shifts (X, σ ), which can be either one-or two-sided.An endomorphism of (X, σ ) is a map Φ : X → X which is continuous, onto, and commutes with σ ; if in addition Φ is one-to-one, then Φ is called an automorphism.We say Φ has finite order m if m is the least positive integer such that Φ m is the identity, and that Φ is a k-th root of a power of the shift if there exists an n such that Φ k = σ n .Let Aut(X, σ ) denote the automorphism group of (X, σ ).If X is two-sided, then Aut(X, σ ) contains the (normal) subgroup generated by the shift, denoted by σ .
The language of a minimal shift (X, σ ), denoted L X , is the set of all finite words that occur in points of X.We denote by p(n), n ≥ 1, the complexity function of (X, σ ): p(n) is the number of words (also called blocks) in L X of length n.A symbolic system (X, σ ) has sublinear complexity if its complexity function is bounded by a linear function.

Automorphisms of minimal shifts with sublinear complexity
If ( X, σ ) is a one-sided minimal shift, let (X, σ ) denote the two-sided version of ( X, σ ) i.e.X consists of all bi-infinite sequences whose finite subwords belong to L X .Given a two-sided shift (X, σ ) we define similarly its one-sided version.If ( X, σ ) is minimal, then the one-sided version of (X, σ ) is ( X, σ ) itself.Henceforth ( X, σ ) refers to a one-sided shift and (X, σ ) refers to a two-sided shift.
For k > 1, we say that x ∈ X is a branch point of order k if |σ −1 ( x)| = k.If (X, T ) is a minimal invertible dynamical system, we define an equivalence relation ∼ on orbits in X as follows: Two orbits O x = {T n (x) : n ∈ Z} and O y = {T n (y) : n ∈ Z} in X are right asymptotic, denoted O x ∼ O y , if there exists m ∈ Z such that d(T m+n x, T n y) → 0 as n → ∞.The right asymptotic equivalence class of O x will be denoted by [x].A right asymptotic equivalence class will be said to be non-trivial if it does not consist of a single orbit.Clearly in the case that X is a shift, O x and O y are right asymptotic if there exists m such that x m+n = y n for all n sufficiently large.Lemma 2.1.Let ( X, σ ) be an infinite minimal one-sided shift and let (X, σ ) be the corresponding two-sided shift.If ( X, σ ) has sublinear complexity, then ( X, σ ) has finitely many branch points, and {O x : x ∈ X} has finitely many non-trivial right asymptotic equivalence classes.
Proof.We prove that if ( X, σ ) has sublinear complexity, then ( X, σ ) has finitely many branch points, as the latter implies that {O x : x ∈ X} has finitely many non-trivial right asymptotic equivalence classes.
Define for n ≥ 1 the complexity difference function s(n) := p(n+1)− p(n).Since ( X, σ ) has sublinear complexity, s(n) is bounded [2], say s(n) ≤ L for all n ≥ 1.Then there are at most L words of length n with at least two left extensions, and so there are at most L branch points in X.
Next we have a basic lemma about automorphisms of minimal (not necessarily symbolic) dynamical systems.
1. Suppose T is not invertible.If there is a point x ∈ X and a finite subset F of X such that Φ(x) ∈ F for every automorphism Φ of (X, T ), then |Aut(X, T )| ≤ |F|.In particular, every automorphism has order at most |F|.
2. Suppose T is invertible.If there is a point x ∈ X and a finite collection F of right asymptotic equivalence classes such that [Φ(x)] ∈ F for every automorphism Φ of (X, T ), then |Aut(X, T )/ T | ≤ |F|.
In particular, the order of each element of Aut(X, T )/ T is at most |F|.
Proof.For the first part, notice that an automorphism of a minimal dynamical system is determined by its action on a single point.
For the second part, suppose that Φ and Ψ are automorphisms of (X, T ) such that O Φ(x) ∼ O Ψ(x) for some x.Let m be such that d(T n (Φ(x)), T n+m (Ψ(x))) → 0. Then we claim that Φ = Ψ • T m .For any x ∈ X, let (n i ) be an increasing sequence of integers so that T n i x → x .Now we have where for the third equality, we used that fact that O Φ(x) and O Ψ(x) are right asymptotic.Hence we deduce that an automorphism of X is determined up to composition with a power of T by the right asymptotic equivalence class of the image of a single point.
The following result tells us that for infinite minimal shifts with sublinear complexity, |Aut(X, σ )/ σ | is bounded by the number of maximal sets of mutually right asymptotic orbits.Theorem 2.3.Let ( X, σ ) be an infinite minimal one-sided shift and let (X, σ ) be the corresponding two-sided shift.For k > 1, let Mk be the number of k-branch points in X, and M k be the number of ∼-equivalence classes of size k in X. Suppose that ( X, σ ) has sublinear complexity, so that both Mk and M k are finite.Then 1. Aut( X, σ ) has at most M := min{ Mk : Mk > 0} elements, and 2. Aut(X, σ )/ σ has at most M := min{M k : M k > 0} elements.
Proof.Using Lemma 2.1, X has finitely many branch points and {O x : x ∈ X} has finitely many non-trivial right asymptotic equivalence classes.The fact that the systems are infinite implies that X (respectively X) has at least one branch point (respectively one non-trivial right asymptotic equivalence class) so that M and M are positive and finite.
Let M = Mk .Let x be a branch point of order k.An automorphism Φ of ( X, σ ) must map branch points of order k to branch points of order k, so we apply Lemma 2.2(1), to the set of branch points of order k, to obtain (1).
Similarly, let M = M and let x be such that O x belongs to an equivalence class consisting of orbits.If Φ is an automorphism (X, σ ), then O Φ(x) must also belong to an equivalence class consisting of orbits.Hence by Lemma 2.2 (2), there are at most M automorphisms of (X, σ ) up to composition with a power of σ .
As Aut( X, σ ) and Aut(X, σ )/ σ are both finite groups, the order of any element of Aut( X, σ ) divides |Aut( X, σ )|, and any element of Aut(X, σ ) is a k-th root of the shift, where k divides |Aut(X, σ )/ σ |.A substitution is a map from A to the set of nonempty finite words on A. We use concatenation to extend θ to a map on finite and infinite words from A. We say that θ is primitive if there is some k ∈ N such that for any a, a ∈ A, the word θ k (a) contains at least one occurrence of a .By iterating θ on any fixed letter in A, we obtain one-sided (right) infinite points u = u 0 . . .such that θ j (u) = u for some natural j.The pigeonhole principle implies that θ -periodic points always exist, and, for primitive substitutions, we define Xθ to be the shift orbit closure of any one of these θ -periodic points and call ( Xθ , σ ) a one-sided substitution shift.Barge, Diamond and Holton [1] show that a primitive substitution on k letters has at most k 2 right asymptotic orbits, so that we can apply Theorem 2.3 to deduce that any automorphism of (X θ , σ ) is a j-th root of a power the shift for some j ≤ k 2 .
Example 2.4.Let A = {a n , b n , c n : 1 ≤ n ≤ N}.Let W be a word which contains all of the letters in A. Define a substitution, θ , by θ (a n ) = Wa n a n , θ (b n ) = W b n a n and θ (c n ) = W c n a n for 1 ≤ n ≤ N. Now the one-sided shift ( Xθ , σ ) contains N branch points of order three, {x n : 1 ≤ n ≤ N}, where x n satisfies the equation a n θ (x n ) = x n .Also, Xθ contains only one other branch point of order N: the unique right-infinite θ -fixed point.Similarly, in the two-sided shift (X θ , σ ), there are exactly N distinct right asymptotic equivalence classes of size 3, and one right asymptotic equivalence class of size N. Now our bounds from Theorem 2.3 tell us that each of Aut( Xθ , σ ) and Aut(X θ , σ )/ σ consist of one element.
We compare our bounds to those that would be obtained using the results in [8] or [10].In the latter, Theorem 3.1 tells us that |Aut(X)/ σ | divides 3N + 1, the number of non-trivial right asymptotic equivalence classes.In the former, Theorem 1.4 tells us that if lim inf p(n)/n < k, then |Aut(X)/ σ | < k.If one takes the word W = a 1 b 1 c 1 a 2 b 2 c 2 . . .a N b N c N , a crude check shows that lim inf p(n)/n ≥ 3N/2, so that the upper bound for |Aut(X)/ σ | coming from Theorem 1.4 of [8] would be at least 3N/2.

Linearly recurrent shifts
The notions below are the same for one-sided and two-sided shifts, so we will state them only for two-sided shifts, thus avoiding the bars.If u, w ∈ L X , we that w is a return word to u if (a) u is a prefix of w, (b) wu ∈ L X , and (c) there are exactly two occurrences of u in wu.Letting (u) denote the length of u, a minimal shift (X, σ ) is linearly recurrent if there exists a constant K such that for any word u ∈ L X and any return word (to u) w, (w) ≤ K (u).Such a K is called a recurrence constant.Durand, Host and Skau show that if (X, σ ) has linear recurrence constant K, then its complexity is bounded above by Kn for large n [13, Theorem 23].Cassaigne [2] shows that if p(n) ≤ Kn + 1, then there are at most 2K(2K + 1) 2 branch points and asymptotic orbits, and we can now apply Theorem 2.3 to bound |Aut( X, σ )| and |Aut(X, σ )/ σ | above by 2(K + 1)(2K + 3) 2 .
Durand [12,Corollary 18] shows that for linearly recurrent two-sided shifts, any endomorphism is an automorphism.The corresponding one-sided version of this is: Theorem 2.5.Let ( X, σ ) be an infinite, minimal, linearly recurrent one-sided shift with recurrence constant K. Then every endomorphism of ( X, σ ) is a k-th root of a power of the shift for some positive k ≤ 2(K + 1)(2K + 3) 2 .
In Example 3.23 we describe a non-trivial endomorphism of ( X, σ ) which is not a power of the shift.

Constant length substitutions
Let θ be a substitution on the alphabet A. The substitution θ has (constant) length r if for each a ∈ A, θ (a) is a word of length r.In this section we generalise the results of Coven in [3] to primitive constant-length substitutions on a finite alphabet.We then provide algorithms to compute the automorphism group of a constant-length substitution shift, and also the set of conjugacies between two constant-length substitution shifts.

The r-adic integers
Let Z r denote the set of r-adic integers, identified with the one-sided shift space consisting of sequences (x n ) n≥0 with the x n 's taking values in {0, 1, . . ., r − 1}.We think of these expansions as being written from right to left, so that Z r consists of left-infinite sequences of digits, x 0 is the rightmost digit of x, and so that addition in Z r has the carries propagating to the left in the usual way.Formally Z r is the inverse limit of rings Z/r n Z, and is itself a ring.
We make no assumption about the primality of r.If r is not a prime power, then Z r has zero divisors.Nevertheless, if k ∈ Z and gcd(k, r) = 1, then multiplication by k is an isomorphism of Z r as an additive group, so that one can make sense of expressions such as m/k for m ∈ Z r if gcd(k, r) = 1.If k ∈ Z and gcd(k, r) > 1, then multiplication by k is still an injection of Z r .A version of the standard argument shows that if m and k are integers with gcd(k, r) = 1, then m/k has an eventually periodic digit sequence; and conversely any point of Z r with an eventually periodic digit sequence can be written in the form m/k for m and k in Z with gcd(k, r) = 1.We may naturally think of Z as a subset of Z r : a point s in Z r whose digits are eventually all 0's or eventually all (r − 1)'s is the r-adic expansion of an integer: we write s ∈ Z.Note also that as an additive group, Z r is torsion free.
We need the following fact.
Lemma 3.1.Let kt ∈ Z for some t ∈ Z r and k ∈ Z \ {0}.Then there exists m|k such that mt ∈ Z and gcd(m, r) = 1.
Proof.Let k = lm, where gcd(m, r) = 1 and l|r n .Let p := r n /l and s := mt.Now r n s = pkt ∈ Z.However, r n s ∈ r n Z r , which is the set of elements of Z r with at least n trailing 0's.Dividing by r n removes these 0's, showing that s ∈ Z as required.
Lemma 3.2.Let H be a subgroup of Z r containing Z generated by a collection of elements (p i /q i ) n i=1 with gcd(p i , q i ) = gcd(q i , r) = 1.Then H is generated by a single element of the form 1/q.
Note that if m is a prime and m α |q, with α maximal, then m α |q i for some i, so that m p i and m pi .Hence ETHAN M. COVEN, ANTHONY QUAS, AND REEM YASSAWI gcd(q, p1 , . . ., pn ) = 1 and there exist integers and all of the generators are multiples of 1/q, so that H is generated by 1/q.

Maximal equicontinuous factors
If (X, T ) is a continuous dynamical system, we say that the dynamical system (Y, S) is the maximal equicontinuous factor of (X, T ) if (Y, S) is an equicontinuous factor of (X, T ) with the property that any other equicontinuous factor (Z, R) of (X, T ) is a factor of (Y, S).The maximal equicontinuous factor of a minimal transformation is a rotation on a compact monothetic topological group, that is a group G for which there exists an element a such that the subgroup generated by a is dense.Such a group is always abelian [11] and we will write the group operation additively.The systems we consider will have maximal equicontinuous factor (Z r , +1) or (Z r × {0, . . ., h − 1}, +1), where the group operation on Z r × {0, . . ., h − 1} is and where r is not necessarily prime.
Let End(X, σ ) be the set of endomorphisms of (X, σ ).Given two shifts (X, σ ) and (Y, σ ), let Conj(X,Y ) denote the (possibly empty) set of topological conjugacies between (X, σ ) and (Y, σ ), and let Fac(X,Y ) denote the set of factor maps from (X, σ ) to (Y, σ ).For completeness we prove the following, which is a straightforward generalization of work done by Coven in [3, §3].Theorem 3.3.Let (X, σ ) and (Y, σ ) be infinite minimal shifts.Suppose that the group rotation (G, R) is the maximal equicontinuous factor of both (X, σ ) and (Y, σ ) and let π X and π Y be the respective factor maps.Then there is a map κ :

and
(b) κ is at most c-to-one.In particular, if π X and π Y are somewhere one-to-one, then κ is an injection.
Proof.First we show that given any factor map Φ : X → Y , there exists κ(Φ) Thus for each factor mapping Φ : With the assumptions of (1), note that π , so that there exist distinct y 1 , . . ., y c+1 in Y with π Y (y i ) = z for each i.Since Φ is surjective, there exist x 1 , . . ., x c+1 with Φ(x i ) = y i , and we have Since a factor map of a minimal system is determined by its action on a single point, we deduce that there are at most c factor maps in Fac(X,Y ), proving (2b).

Background results on constant-length substitutions
In this section, we collect some results that we use later to reduce the case of general constant-length substitutions to cases that can be more straightforwardly handled.
If h > 1, this means that A decomposes into h disjoint subsets: A 1 ∪ . . .∪ A h , where a symbol from A i is always followed by a symbol from A i+1 .Such a system is a constant height suspension.In fact, as shown in [9, Remark 9, Lemmas 17 and 19], such a system is conjugate to a constant height suspension of another constant-length substitution.Proposition 3.4.Let θ be a primitive, length r substitution defined on A, such that (X θ , σ ) is infinite.Then Furthermore h, θ , and the conjugacy between (X θ , σ ) and (X θ × {0, . . ., h − 1}, T ) can be determined algorithmically.
Proof.By Proposition 3.4, (X θ , σ ) is conjugate to (X θ × {0, . . ., h − 1}, T ), and so Aut(X θ , σ ) is group isomorphic to Aut(X θ × {0, . . ., h − 1}, T ).By the definition of Ψ i , we see that Conversely, given Ψ ∈ Aut(X θ × {0, . . ., h − 1}, T ), suppose that Ψ(x, 0) ∈ X θ × {i} for some x ∈ X θ .Define f (x) = π 2 (Ψ(x, 0)), where π 2 is the projection onto the second coordinate and notice that f is a continuous σ -invariant function on X θ and hence is constant.This, and the fact Ψ commutes with T implies that Ψ = Ψ i for some Ψ ∈ Aut(X θ , σ ).Proposition 3.5 tells us that in order to compute the automorphism group of a primitive constant-length substitution, or, as we shall see later, the set of conjugacies between two such substitutions, it is sufficient to work with with a pure base θ of θ .Thus, apart from the statements of our main results Theorems 3.22 and 3.27, and Corollary 3.7, we henceforth assume that our substitutions are of height one.
Let θ be a length r substitution.We write θ (a) = θ 0 (a) . . .θ r−1 (a); with this notation we see that for each 0 ≤ i ≤ r − 1, we have a map θ i : A → A where θ i (a) is the (i + 1)-st letter of θ (a).
Let θ have pure base θ .We say that θ has column number c if for some k ∈ N, and some (i 1 , . . ., i k ), and c is the least such number.In particular, if θ has column number one, then we will say that θ has a coincidence.If for each 0 ≤ i ≤ r − 1 and a = b, θ i (a) = θ i (b), we say θ is bijective.Henceforth we shall be working with primitive substitutions θ which are length r and such that (X θ , σ ) is infinite.
Theorem 3.6.Let θ be a primitive, length r substitution, of height one, and such that (X θ , σ ) is infinite.
Recall the map κ as defined in Theorem 3.3.
1. Let θ have height one.If κ is injective (in particular if θ has a coincidence), then Aut(X θ , σ ) is cyclic.

COMPUTING AUTOMORPHISM GROUPS
Proof.Let θ be as in the statement, and first suppose that θ has height one, so that its maximal equicontinuous factor is (Z r , +1).By the discussion in Section 2.3.1 and Theorem 2.3, Aut(X θ , σ ) is finitely generated.Let a set of generators be Φ 0 = σ , Φ 1 , . . ., Φ n , say.Then Theorem 2.3 implies κ(Φ 0 ) = 1 and κ(Φ i ) = p i /q i for each i for some p i ∈ Z and q i ≥ 1.By Lemma 3.2, κ(Aut(X θ , σ )) is cyclic.The result follows from the injectivity of κ.In particular, since substitutions with coincidences are precisely those whose shifts are a somewhere one-to-one extension of their maximal equicontinuous factor [9], we deduce by Theorem 2.3 that if θ has a coincidence, then Aut(X θ , σ ) is cyclic.
If θ has height h > 1, let θ be the pure base of θ .We assume that θ has a coincidence.By the above, Aut(X θ , σ ) is cyclic, and generated by some Φ with κ(Φ) = 1/k for some k ∈ N. Suppose that gcd(h, k) = 1 and let ah in the previous notation).Define an equivalence relation on X θ × Z to be the transitive closure of (σ (x), i) ∼ (x, i + h) so that X θ × {0, . . ., h − 1} is a system of representatives of the equivalence classes.In this notation, Since infinite cyclic groups are torsion-free, we see that Aut(X θ , σ ) is not cyclic.
Substitutions for which gcd(h, k) > 1 can be constructed.For, starting with a (primitive, nonperiodic) substitution with a coincidence on two letters, its automorphism group equals Z [3].We can use the techniques in Example 3.23 to obtain a substitution shift (X θ , σ ), the κ values of whose automorphism group is 1  k .Now building a tower of height h over (X θ , σ ) gives us the desired substitution shift.We remark also that Part (1) of Corollary 3.7 holds, with essentially the same proof, for linearly recurrent Toeplitz systems [11] whose maximal equicontinuous factor is Z r .
A constant-length substitution is called injective if θ (i) = θ ( j) for any distinct i and j in A. It will be convenient to deal with injective substitutions in what follows.The following theorem of Blanchard, Durand and Maass allows us to restrict our attention to that situation.
Proof.We only sketch the proof that the inverse of the conjugacy is computable, as this is not described in [15].There, we see that (X θ , σ ) is conjugate to (X θ , σ ) via τ, which is a composition of less than |A| algorithmically-determined letter-to-letter maps, τ = τ 1 • . . .τ N .By iteration, it suffices to show that a single one of these maps has a computable inverse, so we assume that N = 1 and τ 1 = τ.
By definition, θ • τ = θ , so that to explicitly describe τ −1 , we only need explicitly describe θ −1 .If θ is a constant-length substitution on A, then there exists k ≤ |A| 2 such that if θ k (a) = θ k (b), then θ k−1 (a) = θ k−1 (b).Fix any letter a.The bilateral recognizability of θ k [22, Definition 1.1 and Theorem 3.1] implies that there exists an ≥ k such that θ (a) does not appear starting at the interior of any θ k word.Such an can be obtained by a simple algorithm.Uniform recurrence of (X θ , σ ) implies that if L is large enough, then any word of length L in L X θ contains some θ (a) as a subword.Also, this L is computable [13,Proposition 25].Thus θ −1 can be explicitly described and the proof of our claim is complete.

The description of
We continue to assume that θ is a length r substitution of height one, so that Z r is the maximal equicontinuous factor of (X θ , σ ).We let c be the column number of θ .We show that all points of Z r have at least c preimages and most points of Z r have exactly c preimages.We study the structure of the subset of Z r having excess preimages.
If P 0 := θ n (X θ ), then P 0 generates a σ r n -cyclic partition of size r n [9, Lemma II.7].We use the notation of [18], [9], using Λ r n to denote the equivalence relation whose classes are the members of this cyclic σ r n -partition, i.e. we write Λ r n (x) = i if x ∈ σ i (P 0 ).
Using the maps Λ n , we can define a maximal equicontinuous factor map π : X θ → r .Note that if Λ n (x) = i, then Λ n+1 (x) ≡ i mod r n .Using this we can define π(x) := . . .x 2 x 1 x 0 where for each n ∈ N, Λ n (x) = ∑ n−1 i=0 r i x i .Definition 1.Let θ be a length r substitution with column number c and of height one.Let Σ θ be the one-sided shift on {0, 1, . . ., r − 1}, whose points are left-infinite, and whose set of forbidden words is Note that F θ = / 0. It can happen that Σ θ = / 0: this is the case for the Thue-Morse substitution: θ (0) = 01, θ (1) = 10.Lemma 3.9.Let θ be a primitive, length r substitution on A with column number c, of height one, and such that (X θ , σ ) is infinite.Then Σ θ is either empty or a sofic shift.
This is the vertex set of a directed labelled graph.If A ∈ V and θ i (A) ∈ V , then we add an edge from θ i (A) to A labelled with the symbol i.We claim that the set Σ θ is the set of left-infinite one-sided sequences in (Z/rZ) N such that for every m < n and finite segment, w n−1 w n−2 . . .w m , the sequence (w m , w m+1 , . . .w n−1 ) is the sequence of labels of a path in the graph.To see this, notice that (w m , w m+1 , . . ., w n−1 ) is the sequence of labels of a path in the graph if and only if there are elements A m , A m+1 , . . ., The reason for the reversal of the order of the indices is that we want to maintain compatibility with standard definitions of sofic: Σ θ is a collection of left-infinite sequences, which is precisely the collection of edge-labellings of infinite paths in the graph listed in reverse order.Depending on θ , we may have to slightly alter the sofic shift that we work with.We write a to denote . . .aaa.Definition 2. Let θ be a length r substitution with column number c. Let Σ θ be as above.Let P = {x ∈ X θ : θ n (x) = x for some n}, that is the set of (two-sided) periodic points under the substitution.
if Σ θ is sofic, then Σθ is sofic.We shall show below (in Corollary 3.13) that Σθ is not empty and hence by the definition above, Σθ is always sofic.Define Σθ = {x : σ n (x) ∈ Σθ for some n ≥ 0}.
Lemma 3.12.Let θ be a primitive, length r substitution, with column number c, of height one and such that Proof.We separate the proof into two parts: z ∈ Z and z ∈ Z.
First suppose z ∈ Z and |π −1 (z)| > c.Then there are more than c θ -periodic points, so that z ∈ Z ⊂ Σθ .
For the converse, notice that if either 0 j ∈ F θ for all j or (r − 1) j ∈ F θ for all j, then there are more than c one-sided right-or leftθ -periodic points respectively.By primitivity, each one-sided θ -periodic point extends to at least one two-sided θ -periodic point, so that we deduce that if Z ∩ Σθ is non-empty then there are more than c θ -periodic points.Hence if z ∈ Z ∩ Σθ , we see |π −1 (z)| > c.
For i = 1, . . ., c + 1, take x (n),i ∈ X such that for each i, x (n),i 0 = āi , π(x (n),i ) ends with z n−1 . . .z 0 and the segment of Taking limits gives at least c + 1 distinct points of π −1 (z ).Since θ is injective and If z ∈ Σθ ∪ Z, then for infinitely many n, there exist k n such that z n+k n . . .z n ∈ F θ .Let x = (x j ) satisfy π(x) = z and fix n.Since |θ z n • . . .• θ z n+kn (A)| = c, there are c possible choices for the block of x j 's for j ∈ [0, r n − 1] − ∑ n−1 i=0 r i z i .Since z ∈ Z, the union of these intervals exhausts all of Z.
Corollary 3.13.Let θ be a primitive, length r substitution defined on A with column number c, of height one, and such that (X θ , σ ) is infinite.Then the function |π −1 (z)| : Z r → N has minimum value c, is non-constant, and Σθ is a nonempty and proper subset of Z r .
Proof.Let z ∈ Z r and for each a ∈ A, let x (a) ∈ X θ satisfy x Since F θ = / 0, we take a word w ∈ F θ , and then a point z ∈ Z r \Z which contains w infinitely often.Then z ∈ Σθ , and by Lemma 3.12, |π −1 (z)| = c.
To see that there exist points z with |π −1 (z)| > c, we note that our shifts will always have at least one non-trivial right asymptotic orbit equivalence class.This follows from the fact that any continuous, positively expansive map on an infinite compact metric space cannot be invertible [7].We pick x and x that are right asymptotic, and we suppose that x n = x n for n ≥ 0. Let z = π(x) = π(x ).The sets θ z 0 • θ z 1 . . .θ z n (A) are decreasing in n and so equal A 0 , a set of cardinality at least c, for all large n.For each a ∈ A 0 , there is a y in π −1 z with y 0 = a.As x and x are two elements in π −1 z with the same 0 coordinate, we conclude that |π −1 z| ≥ c + 1.

Bounding denominators of κ(Φ)
We have established that if θ has column number c, then the set {z : |π −1 (z)| > c} is the set of points whose tail lies in a sofic shift that is a proper subshift of the full one-sided shift on r letters, that is Σθ .Although we already know that κ(Φ) is rational for Φ ∈ Aut(X θ , σ ), Part (2) of Theorem 3.3 tells us that Σθ ⊂ Σθ + κ(Φ), which will allow us to give bounds on the denominator of κ(Φ).We also recover a special case of a result in [13], namely that for our shifts, all endomorphisms are automorphisms.
We write d for the r-adic metric: d(x, y) = r −k , where k = min{ j : x j = y j } (or d(x, y) = 0 if x = y).We also need a metric on the circle.We identify the circle with [0, 1) and define d • (x, y) = min n∈Z |n+(x −y)|.
If X is a subshift of Z r , we say that x has a tail belonging to X if there exists k ∈ N such that σ k (x) ∈ X.
ETHAN M. COVEN, ANTHONY QUAS, AND REEM YASSAWI Now let k be chosen so that the block B appears in the coordinate range n i − j to n i − 1 for each i in w := σ k (z).Let α = Φ(B)/r j .The fact that B appears in these blocks is equivalent to the assertion that ψ n i (w) ∈ [α, α + r − j ), while the appearance of B + 1 in locations n − j to n − 1 of x ∈ X is equivalent to the assertion that ψ n (x) ∈ [α + r − j , α + 2r − j ).Now notice that for each i, either ψ n i (w+mt) or ψ n i (w+2mt) lies in [α +r − j , α +2r − j ).In particular, one of w + mt and w + 2mt contains infinitely many B + 1 blocks, and hence does not have a tail lying in X. Hence we have shown that if t is irrational, or t is rational with denominator exceeding r j , then the hypotheses of the theorem cannot be satisfied.Theorem 3.16.Let θ be a primitive, length r substitution on A, with column number c, of height one, and such that (X θ , σ ) is infinite.Then End(X θ , σ ) = Aut(X θ , σ ).
Proof.Let θ be a substitution as in the statement and let Σ θ , Σθ and Σθ be as constructed above.Let Φ be an endomorphism of X θ and let t = κ(Φ).By Theorem 3.3, Lemma 3.12 and Corollary 3.13, we have t + Σθ ⊂ Σθ .By Lemma 3.14, we deduce t is rational, /n say, with denominator coprime to r and at most r j − 1.Now κ(Φ n σ − ) = 0, so that Φ n σ − is a self-map of π −1 (z) for any z in Z r .Choosing z so that |π −1 (z)| = c, we see that there is a 1 ≤ k ≤ c such that (Φ n σ − ) k has a fixed point.By minimality of X θ , Φ nk = σ k and we see that Φ is bijective, so Φ ∈ Aut(X θ , σ ).
Example 3.17.We continue with Example 3.11, which we already noted has a coincidence and is of height one.Since the word 1 belongs to F θ , Theorem 3.16 implies for any Φ ∈ Aut(X θ ), κ(Φ) has denominator 1 or 3.If there were an automorphism with denominator 3, then by taking powers and composing with a power of the shift, we could find an automorphism Φ with κ(Φ) = −1/3 = 1.However, from the proof of Theorem 3.16, this would imply 1 + Σθ ⊂ Σθ , which is false as 0 ∈ Σθ , but 1 ∈ Σθ .Hence κ(Φ) ∈ Z for all Φ ∈ Aut(X θ , σ ) and such Φ are powers of the shift by the theorem.

Computing Aut(X θ , σ )
In this section, we use the bounds on the denominator appearing in κ together with bounds on the radius of the block code to obtain an algorithm to compute the automorphism group of a constant-length substitution.
The argument that Conditions (3) and (4) characterize elements of ker κ is similar.
We can now deduce Theorem 3.22.Let θ be a primitive, length r substitution on A such that (X θ , σ ) is infinite.Then there is an algorithm to compute Aut(X θ , σ ).
Proof.If θ does not have height one, we use Proposition 3.4 to compute a pure base of θ .Proposition 3.5 tells us how to retrieve Aut(X θ , σ ) from the automorphism group of its pure base.Similarly, if the pure base is not injective, we use Theorem 3.8 to compute its automorphism group from the automorphism group of a conjugate injective substitution shift.Hence it suffices to give an algorithm for injective substitutions of height one.We can algorithmically compute the column number, c and also j, the length of the shortest word in F θ from the transition graphs described in Section 3.4.Recall from Lemma 3.2 that κ(Aut(X θ , T )) is a cyclic subgroup of Z r generated by 1/d for some integer d.By Theorem 3.16, d satisfies gcd(d, r) = 1 and d ≤ r j − 1.Now Aut(X θ , T ) is the semi-direct product of ker κ with the cyclic subgroup generated by any element of κ −1 (−1/d).
To find ker κ, one applies Proposition 3.21 to test all of the finitely many 3-block maps.Note that one can algorithmically list the words of length two and three that belong to L X θ .One then needs to find an element of κ −1 (−1/d) for the largest d so that this set is non-empty.By the above, there are finitely many values of d to check.Since gcd(d, r) = 1, we see d|r p − 1 for some p and then −1/d can be expressed as k/(1 − r p ) for some 0 < k < r p − 1.By Proposition 3.21, there are finitely many 2-block maps to check for each potential d.

ETHAN M. COVEN, ANTHONY QUAS, AND REEM YASSAWI
We do not address the question of efficiency.From Corollary 3.7, we know that for substitutions with a coincidence, Aut(X θ , σ )/ σ is a finite cyclic group.In Example 3.17, this group was trivial, and it is natural to wonder if this is always the case.
The following example shows that the quotient can indeed be non-trivial.
The techniques of Section 3.6 can be generalized to compute the set of conjugacies between two substitution shifts generated by primitive constant-length substitutions, one of which generates an infinite shift.We use κ to also denote the restriction to Conj(X θ , X θ ) of the map in Theorem 3.3.We continue to assume that our maximal equicontinuous factor mappings are those that were defined in Section 3.4, so that r n |π(x) if and only if x ∈ θ n (X θ ).Proposition 3.24.Let θ and θ be primitive, length r substitutions, both with column number c and of height one, and such that (X θ , σ ) and (X θ , σ ) are infinite.Suppose that Φ ∈ Conj(X θ , X θ ).Then κ(Φ) is rational, and if F θ contains a word of length j, then nκ(Φ) ∈ Z for some 0 ≤ n ≤ (r − 1)(r j − 1).
The proofs of the following two propositions are a straightforward generalization of those of Propositions 3.19 and 3.21.