Glider Automorphisms on Some Shifts of Finite Type and a Finitary Ryan’s Theorem

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Glider Automorphisms on Some Shifts of Finite Type and a Finitary Ryan’s Theorem

Johan Kopra

To cite this version:

Johan Kopra. Glider Automorphisms on Some Shifts of Finite Type and a Finitary Ryan’s Theorem.

24th International Workshop on Cellular Automata and Discrete Complex Systems (AUTOMATA),

Jun 2018, Ghent, Belgium. pp.88-99, �10.1007/978-3-319-92675-9_7�. �hal-01824877�

Type and a Finitary Ryan’s Theorem

Johan Kopra

Department of Mathematics and Statistics, FI-20014 University of Turku, Finland jtjkop@utu.fi

Abstract. For any mixing SFTXcontaining a fixed point we construct a reversible shift-commuting continuous map (automorphism) which breaks any given finite point of the subshift into a finite collection of gliders traveling into opposing directions. As an application we show that the automorphism group Aut(X) contains a two-element subset S whose centralizer consists only of shift maps.

Keywords: Mixing SFTs, Automorphisms, Cellular automata

1 Introduction

Let X ⊆A^Z be a one-dimensional subshift over a symbol set A. IfX contains some constant sequence 0^Z (0 ∈ A), we may say that an element x ∈ X is finite if it differs from 0^Z only at finitely many coordinates. In this paper we consider the problem of constructing reversible shift-commuting continuous maps (automorphisms) onXwhich decompose all finite configurations into collections of gliders traveling into opposing directions. As a concrete example, consider the binary full shiftX ={0,1}^Z and the mapg =g₃◦g₂◦g₁:X →X defined as follows. In any x∈ X, g₁ replaces every occurrence of 0010 by 0110 and vice versa,g₂replaces every occurrence of 0100 by 0110 and vice versa, andg₃replace every occurrence of 00101 by 00111 and vice versa. In Figure 1 we have plotted the sequences x, g(x), g²(x), . . . on consecutive rows for some x∈X. It can be seen that the sequencexeventually diffuses into two different “fleets”, the one consisting of 1s going to the left and the one consisting of 11s going to the right.

It can be proved that this diffusion happens eventually no matter which finite initial pointx∈Xis chosen.¹In Section 3 we construct on all mixing SFTs (that contain the point 0^Z) a glider automorphism with the same diffusion property as the binary automorphismg above.

The existence of a glider automorphism g on a subshift X is interesting, becausegcan be used to convert an arbitrary finitex∈X into another sequence g^t(x) (for some t ∈ N⁺) with a simpler structure, which nevertheless contains

?The work was partially supported by the Academy of Finland grant 296018 and by the Vilho, Yrj¨o and Kalle V¨ais¨al¨a Foundation

1 This can be proved along similar lines as in the proof of Lemma 10 and Lemma 11.

Fig. 1.The diffusion ofx∈X under the mapg:X →X. White and black squares correspond to digits 0 and 1 respectively.

all the information concerning the original pointxbecauseg is invertible. Such maps have been successfully applied to other problems e.g. in [3, 4].

We also consider a finitary version of Ryan’s theorem. Let X be a mixing SFT and denote the set of its automorphisms by Aut(X), which we may consider as an abstract group. According to Ryan’s theorem [2] the center of the group Aut(X) is generated by the shift mapσ. There may also be subsetsS ⊆Aut(X) whose centralizers are generated byσ. Denote the minimal cardinality of such a finite set S byk(X). In [3] it was proved thatk(X)≤10 when X is the full shift over the four-letter alphabet. In the same paper it is noted thatk(X) is an isomorphism invariant of Aut(X) and therefore computing it could theoretically separate Aut(X) and Aut(Y) for some mixing SFTs X and Y. We use our glider automorphism construction to prove that k(X) = 2 for all mixing SFTs that contain the point 0^Z.

2 Preliminaries

A finite setAcontaining at least two elements (letters) is called analphabet and the setA^Z of bi-infinite sequences (configurations) over A is called a full shift.

Formally anyx∈A^Zis a functionZ→Aand the value ofxati∈Zis denoted byx[i]. It contains finite and one-directionally infinite subsequences denoted by x[i, j] =x[i]x[i+1]. . . x[j],x[i,∞] =x[i]x[i+1]. . . andx[−∞, i] =. . . x[i−1]x[i].

Afactor ofx∈A^Zis any finite sequencex[i, j] wherei, j∈Z, and we interpret the sequence to be empty if j < i. Any finite sequence w = w[1]w[2]. . . w[n]

(also the empty sequence, which is denoted byλ) wherew[i]∈Ais aword over A. The set of all words overAis denoted byA^∗, and the set of non-empty words isA⁺=A^∗\ {λ}. More generally, for anyL⊆A^∗, let

L^∗={w1. . . wn |n≥0, wi∈L} ⊆A^∗,

i.e. L^∗ is the set of all finite concatenations of elements of L. The set of words of length n is denoted by Aⁿ. For a word w ∈ A^∗, |w| denotes its length, i.e.

|w| = n ⇐⇒ w ∈ Aⁿ. We say that the word w ∈ Aⁿ occurs in x ∈ A^Z at position i ifw =x[i]. . . x[i+n−1]. We define the shift map σA : A^Z → A^Z by σA(x)[i] = x[i+ 1] for x ∈ A^Z, i ∈ Z. The subscript A in σA is typically

omitted. The setA^Zis endowed with the product topology (with respect to the discrete topology onA), under whichσis a homeomorphism onA^Z. Any closed setX ⊆A^Zsuch thatσ(X) =X is called asubshift, and the collection of words appearing as factors of elements of X is the language of X, denoted byL(X).

The restriction ofσtoX may be denoted by σX, but typically the subscriptX is omitted.

IfX ⊆A^Z is a subshift and z ∈ X is such that σ(z) =z (i.e. z is a fixed point), then there existsa∈Asuch thatz[i] =afor alli∈Z. For such subshifts we always fix one such point and denote a = 0, z = 0^Z. Then for x ∈ X we define its support supp(x) ={i∈Z|x[i]6= 0}and say thatxis finite if supp(x) is finite. Finite points x, y∈X with disjoint supports can be glued together; if supp(x)∩supp(y) =∅we definex⊗y∈A^Zby (x⊗y)[i] =x[i] wheni∈supp(x) and (x⊗y)[i] =y[i] otherwise.

Definition 1. A graph is a pair G = (V, E) whereV is a finite set ofvertices (or nodes or states) and E is a finite set of edges. Each edge e ∈ E starts at an initial state denoted by ι(e) ∈ V and ends at a terminal state denoted by τ(e)∈V. We say that e∈A is an outgoing edge ofι(e) and an incoming edge ofτ(e).

A sequence of edgese[1]. . . e[n] in a graphG= (V, E) is apath(of lengthn) if τ(e[i]) =ι(e[i+ 1]) for 1≤i < nand it is acycleif in additionτ(e[n]) =ι(e[1]).

We say that the path starts at e[1] and ends ate[n]. A graph G is primitive if there isn∈N+such that for everyv1, v2∈V there is a path of lengthnstarting atv₁and ending at v₂. For any graphG= (V, E) we call the set

{x∈E^Z|τ(x[i]) =ι(x[i+ 1])}

(i.e. the set of bi-infinite paths onG) theedge subshift ofG.

Definition 2. A subshift X ⊆ A^Z is a mixing subshift of finite type (mixing SFT) if it is the edge subshift of a primitive graphG= (V, E) containing at least two edges (in particular E⊆A).

Example 3. LetA={0, a, b}. The graph in Figure 2 defines a mixing SFTX also known as the golden mean shift. A typical point ofX looks like

. . .000abab0ab00ab000. . .

i.e. the letterb cannot occur immediately after 0 orb and every occurrence ofa is followed by b.

Definition 4. Anautomorphismof a subshiftX⊆A^Zis a continuous bijection f :X →X such that σ◦f =f◦σ. We say thatf is a radius-rautomorphism iff(x)[0] =f(y)[0] for allx, y∈X such thatx[−r, r] =y[−r, r] (such ralways exists by continuity off). The set of all automorphisms ofX is a group denoted by Aut(X). (In the case X =A^Z automorphisms are also known as reversible cellular automata.)

a

b 0

Fig. 2.The golden mean shift.

Thecentralizer of a setS⊆Aut(X) is

C(S) ={f ∈Aut(X)|f◦g=g◦f for every g∈S}

and the subgroup generated by f ∈ Aut(X) is denoted by hfi. The following definition is from [3]:

Definition 5. For a subshift X, let k(X) ∈N∪ {∞,⊥} be the minimal car- dinality of a set S ⊆Aut(X) such that C(S) =hσiif such a set S exists, and k(X) =⊥otherwise.

The main result of [2] is thatk(X)6=⊥wheneverX is a mixing SFT.

We say that subshifts X ⊆ A^Z and Y ⊆ B^Z are conjugate if there is a continuous bijection ψ : X → Y such that ψ◦σX = σY ◦ψ. For conjugate subshiftsX andY it necessarily holds thatk(X) =k(Y).

3 Glider Automorphisms

In this section we define as a technical tool a subclass of mixing SFTs, and for any subshiftXfrom this class we construct an automorphism which breaks every finite point ofX into a collection of gliders traveling in opposite directions.

Note that ifX is a mixing SFT with a fixed point 0^Z, then necessarily in its graph G = (V, E) it holds that τ(0) = ι(0). For such a graph we denote G⁰= (V, E⁰) andE⁰=E\ {0}, i.e. we get G⁰ from G by removing the 0-edge.

Definition 6. A mixing SFTX with a fixed point 0^Zand defined by the graph G= (V, E) is called a 0-mixing SFT if the graphG⁰is also primitive and contains at least two edges.

The golden mean shift given by the graph in Figure 2 is an example of a mixing SFT which strictly speaking isn’t 0-mixing. Nevertheless, in the following lemma we show that the definition of a 0-mixing SFT is only technical and that it is not an actual restriction.

Lemma 7. Any mixing SFT with a fixed point is conjugate to a 0-mixing SFT.

Proof. Let X be a mixing SFT with a fixed point 0^Z defined by the graph G= (V, E) and lets=ι(0) =τ(0). Let 0, a1, . . . , atbe all the outgoing edges of s, let 0, b1, . . . , bu be all the incoming edges ofsand construct a new graph

H= (V ∪ {s⁰}, E∪ {0⁰, b⁰₁, . . . , b⁰_u})

with the starting and ending nodes ofe∈Ethe same as inGwith the exception that ι(a_i) =s⁰ for 1≤i≤t, and additionallyι(0⁰) =s,τ(0⁰) =s⁰,ι(b⁰_j) =ι(b_j) and τ(b⁰_j) =s⁰ for 1≤j ≤u.² LetY be the edge subshift ofH; it is conjugate withX via the continuous shift-commuting mapψ:X→Y defined forx∈X, i∈Zas

ψ(x)[i] =







0⁰ whenx[i] = 0 andx[i+ 1]∈ {a1, . . . , at},

b⁰_j whenx[i] =bj for some 1≤j ≤uandx[i+ 1]∈ {a1, . . . , at}, x[i] otherwise.

s

. . .

. . .

s

s⁰

. . .

. . . 0

bj

ai

0 0

0⁰ bj

b⁰_j ai

Fig. 3.Splitting the states.

Construct the graphs G⁰, H⁰ and let c[1]. . . c[n]bj be a cycle in G⁰ visiting s only at the beginning and ending. Then c[1]. . . c[n]bj0⁰ and c[1]. . . c[n]b⁰_j are distinct cycles in H⁰ of coprime length, so H⁰ has to be primitive and Y is a 0-mixing SFT.

In the rest of the section we assume thatX is a 0-mixing SFT defined by the graphG= (V, E). This means that the edge subshift ofG⁰ is also a mixing SFT.

Denote s=ι(0) =τ(0). Letv1 be a cycle inG⁰ visitingsonly at the beginning and ending, denotep=|v1|and letv0= 0^p. The words

g_`=v₀v₁, g_r=v₁v₁

will be left- and rightbound gliders of the automorphism g defined later. The languages of left- and rightbound gliders are

L`= (g`00^∗)^∗, Lr= (0^∗0gr)^∗.

We denote by^∞0 and 0^∞ left- and right-infinite sequences of zeroes and define the glider fleet sets

GF`=<sup>∞</sup>0(g`00^∗)^∗0^∞ GFr=^∞0(0^∗0gr)^∗0^∞ GF = GF`∪GFr 2 In other words we have performed anelementary state splittingofGat states. State

splitting is a well-known method to produce conjugate subshifts, see e.g. Chapter 2.4 of [1].

(note that these consist of finite configurations).

Denoteu⁰=v₁v₁v₁and letn∈N+ be a mixing constant ofG⁰(i.e. a number such that for every n⁰ ≥ n and s₁, s₂ ∈ V there is a path of length n⁰ in G⁰ from s1 tos2) chosen such thatn≥ |u⁰|= 3p. For every a∈E we may choose some pathwa ∈E⁰²ⁿ in G⁰ such thatwa begins withu⁰ and 0waa∈L(X). For every a ∈ E letW_a⁰ = {wa,1, . . . , wa,k_a} ⊆ E⁰²ⁿ be the paths of length 2n in G⁰ such thatwa,i does not have a prefix u⁰ and 0wa,ia∈L(X) for 1≤i≤ka, and let Wa =W_a⁰ ∪ {wa}. LetU⁰ ={u⁰₁, . . . , u⁰_k} ⊆(E⁰)⁺ be the cycles from s to s(which may visit sseveral times) of length at most 2n−1≥5pwhich are different fromv1andv1v1and do not haveu⁰as a prefix. Finally, these words are padded to constant length; u= 0^{2n−1−|u0|}u⁰ and ui = 0^{2n−1−|u0i|}u⁰_i. The words in Wa andU⁰ are chosen so as to allow the following structural definition.

Definition 8. Assume that x /∈ GF_` is a non-zero finite element of X. Then there is a maximali∈Zsuch that

x[−∞, i−1]∈^∞0L`,

and there is a unique word w∈ {v10} ∪ {v1v10} ∪ {u⁰} ∪(U⁰0)∪(S

a∈EW_a⁰a) such thatw is a prefix of x[i,∞]. Ifw=v1v10 or w∈U⁰0, letj =i+|w| −1 and otherwise letj=i+|v1|. We say thatxis ofleft bound type (w, j) and that it has left boundj (note thatj > i).

Similarly, ifx /∈GFris a non-zero finite element ofX, then there is a minimal j∈Zsuch that

x[j+ 1,∞]∈L_r0^∞ and we say thatxhas right boundj.

The point of this definition is that if x is of left bound type (w, j), then the glider automorphism g defined later will create a new leftbound glider at positionj and break it off from the rest of the configuration.

We define four mapsg₁, g₂, g₃, g₄:X →X as follows. In anyx∈X, – g₁replaces every occurrence of 0(v₀v₁)0 by 0(v₁v₁)0 and vice versa – g₂replaces every occurrence of 0(v₁v₀)0 by 0(v₁v₁)0 and vice versa – g₃replaces every occurrence of 0v₀(v₁v₀v₁) by 0v₀(v₁v₁v₁) and vice versa – g4replaces every occurrence of 0waa, 0wa,iaand 0wa,k_aaby 0wa,1a, 0wa,i+1a

and 0waarespectively (fora∈E and 1 ≤i < ka) and every occurrence of 0u0, 0u_i0 and 0u_k0 by 0u₁0, 0u_i+10 and 0u0 respectively (for 1≤i < k).

It is easy to see that these maps are well defined automorphisms ofX. Theglider automorphism g : X → X is defined as the composition g4◦g3◦g2◦g1. The name is partially justified by the following lemma.

Lemma 9. If x ∈ GF` (resp. x ∈ GFr), then g(x) = σ^p(x) (resp. g(x) = σ^−p(x)).

Proof. Assume thatx∈GF_{`</sub>(the proof forx∈GF<sub>r</sub>is similar) and assume that i∈Zis some position inxwhereg<sub>`} occurs. Then

x[i−1, i+ 2p] = 0g`0 = 0(v0v1)0

g₁(x)[i−1, i+ 2p] = 0(v₁v₁)0

g2(g1(x))[i−p−1, i+p] = 0v0(v10) = 0g`0 g(x) =g₄(g₃(g₂(g₁(x)))) =g₂(g₁(x))),

so every glider has shifted by distancepto the left andg(x) =σ^p(x).

In fact, the previous lemma would hold even ifg were replaced by g2◦g1. The role of the partg4◦g3is, for a given finite pointx∈X, to “erode” non-zero non-glider parts ofxfrom the left and to turn the eroded parts into new gliders.

This is the content of the following lemmas.

Lemma 10. Assume that x∈ X has left bound j. Then there exists t ∈ N+

such that the left bound ofg^t(x) is strictly greater thanj.

Proof. Letx∈X be of left bound type (w, j) withw∈ {v10} ∪ {v1v10} ∪ {u⁰} ∪ (U⁰0)∪(S

a∈EW_a⁰a). The gliders to the left of the occurrence ofwnearj move to the left at constant speed punder action ofg without being affected by the remaining part of the configuration.

Case 1. Assume thatw=v₁v₁0. Theng₁(x)[j−(p+ 1), j] = 0v₁0. Ifg₁(x)[j− (p+ 1), j+p] = 0v₁v₀0, theng(x)[j−(p+ 1), j+p] = g₂(g₁(x))[j−(p+ 1), j+p] = 0v₁v₁0, g(x) is of left bound type (v₁v₁0, j+p) and we are done. Otherwiseg₂(g₁(x))[j−(p+ 1), j] = 0v₁0. Denote y=g₃(g₂(g₁(x))).

Ify[j−(p+ 1), j] = 0v10, theng(x) = g4(y) is of left bound type (v10, j) and we proceed as in Case 3. Otherwisey[j−(p+ 1), j+ (2p−1)] = 0u⁰. If y[(j+ 2p)−2n, j+ 2p] = 0u0, theng(x)[(j+ 2p)−2n, j+ 2p] = 0u10,g(x) is of left bound type (u⁰₁0, j+ 2p) and we are done. On the other hand, if y[(j+ 2p)−2n, j+ 2p]6= 0u0, theng(x) is of left bound type (w⁰, j) for some w⁰∈W_a⁰a∪ {u⁰} (a∈E) and we proceed as in Case 4 or Case 5.

Case 2. Assume thatw=u⁰_i0 for 1≤i≤k. There is a minimalt∈Nsuch that g₃(g₂(g₁(g^t(x))))[j−2n, j] = 0u_i0. Because g^t+k−i+1(x)[j−2n, j] = 0u0, it follows that y = g^t+k−i+1(x) is of left bound type (u⁰, j−2p). Then g(y)[j−5p, j] =g₃(g₂(g₁(y)))[j−5p, j] =v₀(v₀v₁)v₀v₁0 is of left bound type (v₁0, j) and we proceed as in Case 3.

Case 3. Assume thatw=v10. Thenx[j−(2p+ 1), j]6= 0v0v10 = 0g`0 because otherwise the left bound ofxwould already be greater thanj, sog1(x)[j− (p+ 1), j] = 0v10. If moreoverg1(x)[j−(p+ 1), j+p] = 0v1v00, theng(x)[j− (p+ 1), j+p] = g2(g1(x))[j −(p+ 1), j+p] = 0(v1v1)0 so g(x) is of left bound type (v1v10, j+p) and we are done. Let us therefore assume that g1(x)[j−(p+ 1), j+p]6= 0v1v00, in which caseg2(g1(x))[j−(2p+ 1), j] = 0v0v10.

Ifg₂(g₁(x))[j−(2p+ 1), j+ 2p−1]6= 0v0v₁v₀v₁, theng(x)[j−(2p+ 1), j] =g₃(g₂(g₁(x)))[j−(2p+ 1), j] = 0v₀v₁0. The left bound ofg(x) is now greater than j and we are done. Otherwise g₃(g₂(g₁(x)))[j−(2p+ 1), j+ 2p−1] = 0v₀u⁰. Ifg₃(g₂(g₁(x)))[(j+ 2p)−2n, j+ 2p] = 0u0, theng(x)[(j+ 2p)−2n, j+ 2p] = 0u10 and the left bound of g(x) equalsj+ 2p. Finally, ifg3(g2(g1(x)))[(j+ 2p)−2n, j+ 2p]6= 0u0, theng(x) is of left bound type

(w⁰, j) for some w⁰ ∈W_a⁰a∪ {u⁰} (a∈ E) and we proceed as in Case 4 or Case 5.

Case 4. Assume thatw=wa,iafora∈Eand 1≤i≤ka. Theng^ka−i+1(x)[j− p, j+ (2p−1)] =u⁰ and we proceed as in Case 5.

Case 5. Assume thatw=u⁰. Theng2(g1(x))[j−(2p+ 1), j+ (2p−1)] = 0v0u⁰, g3(g2(g1))[j−(2p+ 1), j+ (2p−1)] = 0v0v1v0v1and the left bound ofg(x) is at leastj+ 2p.

Lemma 11. Assume that x∈X has right boundj. Then there existst∈N⁺ such that the right bound ofg^t(x) is strictly less thanj.

Proof. Let us assume to the contrary that the right bound ofg^t(x) is at leastj for everyt∈N+.

Assume first that the right bound ofg^t(x) is equal tojfor everyt∈N⁺. By the previous lemma there ist∈N+ such that the left bound ofg^t(x) is at least j+ 3n, which means thatg^t(x) contains onlyg`-gliders to the left ofj+nand onlygr-gliders to the right ofj. This can happen only ifg^t(x)[j+1, n−1] = 0ⁿ⁻¹. Then the right bound ofg^t+1(x) is at leastj−p, a contradiction.

Assume then that the right bound ofg^t(x) is strictly greater thatj for some t∈N+and fix the minimal sucht. This can happen only ifg₁(g^t−1(x))[j−p, j+ p+ 1] = 0v₁v₀0, in which caseg^t(x)[j−p, j+p+ 1] = 0v₁v₁0 = 0g_r0. But then the right bound ofg^t(x) is less than j−p, a contradiction.

Together these two lemmas yield the following theorem.

Theorem 12. If x ∈X is a finite configuration, then for every N ∈ N there exists t∈Nsuch thatg^t(x)[−N, N] = 0^2N+1,g^t(x)[∞,−(N+ 1)]∈^∞0L` and g^t(x)[N+ 1,∞]∈Lr0^∞.

4 Finitary Ryan’s Theorem

In this section we prove a finitary version of Ryan’s theorem. The idea is that only very specific automorphisms commute with the glider map g : X → X defined in the previous section, so it will be relatively easy to choose another automorphism f on X such that only powers of the shift map commute with bothg andf. We make a simple choice of suchf.

First we define mapsf₁, f₂ : X → X for a 0-mixing SFT X as follows. In anyx∈X,

– f₁replaces every occurrence of 0(v₁v₁)v₀v₀v₀(v₁)0 by 0(v₁v₁)v₀v₀(v₁)v₀0 and vice versa

– f2 replaces every occurrence of 0(v1v1)v0v0(v1)0 by 0v0(v1v1)v0(v1)0 and vice versa,

wherev₀ andv₁ are as in the previous section. It is easy to see that these maps are well defined automorphisms of X. The automorphism f : X → X is then defined as the composition f₂◦f₁. The map f has two important properties.

First, it replaces any occurrence of 0(v1v1)v0v0v0(v1)0 by 0v0(v1v1)v0(v1)v00.

Second, ifx∈X is a configuration containing only gliders g` andgr and every occurrence of g`is sufficiently far from every occurrence ofgr, thenf(x) =x.

To prove our main result we need the following lemma.

Lemma 13([3], Lemma 7.5). IfX is a mixing SFT containing a fixed point 0^Z andh:X →X is an automorphism which is not a power ofσ, then there exists a finite configurationx6= 0^Z such thath(x)∈ O(x)/ +{σⁱ(x)|i∈Z}.

Theorem 14. LetX⊆A^Zbe a 0-mixing SFT andg, f:X→Xas above. The only automorphisms ofX which commute with both gandf are powers ofσ.

Proof. Assume to the contrary that h : X → X is a radius-r automorphism whose inverse is also a radius-rautomorphism and which commutes withg and fbut is not a power ofσ. Let us first show thath(0^Z) = 0^Z. Namely, if it were that h(0^Z) =a^Z, for some a∈A\ {0}, considerx=. . .000g`000. . . with the glider g` at the origin and note thath(x)[i]6=afor some−r≤i≤(2p−1) +r(recall:

g` = v0v1, |v0| = |v1| = p) and h(x)[−∞, i−jp] = . . . aaa for some j ∈ N+. Then g^t(h(x))[−∞, i−jp] = . . . aaa for every t ∈ Z but h(g^j(x))[i−jp] = h(σ^jp(x))[i−jp] = h(x)[i] 6= a, contradicting the commutativity of h and g.

Thushmaps finite configurations to finite configurations.

We haveh(GF`)⊆GF`. To see this, assume to the contrary that there exists x∈GF` such thath(x)∈/ GF`. Recall thatg is reversible andg(GF`) = GF`, so g^t(h(x)) ∈/ GF` for all t ∈ N. Combining this with Theorem 12 it follows that g^t(h(x)) contains an occurrence of 0v1v10 to the right of coordinater for all sufficiently large t and therefore h⁻¹(g^t(h(x)))[i] is non-zero for some i≥0 which depends ont. This contradicts the fact thath⁻¹(g^t(h(x)))[i] =g^t(x)[i] = 0 for alli≥0 given thattis sufficiently big. Similarly h(GF_r)⊆GF_r.

For any finitex6= 0^Zdefine its left and right offsets off`(x) = min{supp(h(x))} −min{supp(x)}, offr(x) = max{supp(h(x))} −max{supp(x)}.

For all nonzerox_{`</sub>∈GF<sub>`}andx_r∈GF_rwe have off_{`</sub>(x<sub>`})−off_r(x_r) = 0. If this did not hold, we could assume without loss of generality that off_{`</sub>(x<sub>`})−off_r(x_r)>0 (by replacing h with h⁻¹ if necessary) and that min{supp(x<sub>`</sub>)} = (r+ 2)p, max{supp(xr)}=−(r+ 1)p−1 (by shiftingx`andxrsuitably). Then consider x=xr⊗x`and note that from min{supp(x`)}= (r+ 2)p > r, max{supp(xr)}=

−(r+1)p−1<−rit follows thath(x) =h(xr)⊗h(x`). Theng<sup>r</sup>(x)[−3p−1,3p] = 0(v1v1)v0v0v0(v1)0 andg<sup>r</sup>(h(x)) contains no occurrence of the words mentioned in the definition of f1 and f2 by the assumption off`(x`)−offr(xr) > 0, so f(g^r(x))6=g^r(x) andf(g^r(h(x))) =g^r(h(x)). Now

x6=g^−r(f(g^r(x))) =h⁻¹(g^−r(f(g^r(h(x)))))

=h⁻¹(g^−r(g^r(h(x)))) =x,

a contradiction. It also follows that there is a fixed s∈Z such that off_{`</sub>(x<sub>`}) = off_r(x_r) =sfor all nonzerox_{`</sub>∈GF<sub>`},x_r∈GF_r.

Ifx_{`</sub>∈GF<sub>`} and x_r ∈GF_r are configurations containing exactly one occur- rence ofg`andgrrespectively, thenh(x`) =σ^−s(x`) andh(xr) =σ<sup>−s</sup>(x`). To see this, assume to the contrary (without loss of generality) that min{supp(x`)}= (r+2)p(i.e. the occurrence ofg`inxis at (r+1)p), max{supp(xr)}=−(r+1)p−1 andh(x`)[(r+1)p+s,(r+3)p−1+s] =h(x`)[(r+1)p+s+k,(r+3)p−1+s+k] =g`

for some k >2p(i.e.h(x`) contains at least two occurrences of g`, the case in whichh(xr) contains at least two occurrences ofgrbeing similar). Then consider x=xr⊗x`and note that

g^r(x)[−3p−1,3p] = 0(v1v1)v0v0v0(v1)0 f(g^r(x))[−3p−1,3p] = 0v0(v1v1)v0(v1)v00 g⁻¹(f(g^r(x)))[−3p−1,3p] = 0(v₁v₁)v₀v₀v₀(v₁)0 g^−(r+1)(f(g^r(x))) =x,

therefore alsoh(x) =g^−(r+1)(f(g^r(h(x))))+y. On the other hand, f(g^r(h(x)))[p+s+k,3p−1 +s+k]

=g^r(h(x))[p+s+k,3p−1 +s+k] =g_`

g^−(r+1)(f(g^r(h(x))))[(r+ 2)p+s+k,(r+ 4)p−1 +s+k] =g`, so in particulary[(r+ 2)p+s+k,(r+ 3)p−1 +s+k] =v0. Because we assumed that

h(x)[(r+ 2)p+s+k,(r+ 3)p−1 +s+k]

=h(x`)[(r+ 2)p+s+k,(r+ 3)p−1 +s+k] =v1, it follows that h(x)6=y, a contradiction.

By Lemma 13 there exists a finite configurationx6= 0^Zsuch thath(x)∈ O(x)/ and h(x) is finite. Use Theorem 12 to gett∈N such thatg^t(x)[−r, r] = 0^2r+1 andg^t(x) =y`⊗yrwherey`∈GF`has max{supp(y`)}<−randyr∈GFrhas min{supp(yr)}> r(it is possible that eithery` or yr is equal to 0<sup>Z</sup>). Then also h(g<sup>t</sup>(x)) =h(y`)⊗h(yr)∈ O(g/ ^t(x)), and combining this with off`(y`) = offr(yr) it follows thath(y<sub>`</sub>)∈ O(y/ `) orh(y_r)∈ O(y/ r). Without loss of generality assume thath(y<sub>`</sub>)∈ O(y/ `) (the caseh(y_r)∈ O(y/ r) is similar), thaty_{`</sub>contains a minimal number of occurrences ofg<sub>`} (at least two by the previous paragraph) and that the distance from the leftmost g_{`</sub> to the second-to-leftmost g<sub>`} in y_{`</sub> is maximal (at most 2r+ 2psince otherwise by dropping the leftmostg<sub>`}we would get a new configurationy_{`</sub><sup>0</sup> such thath(y<sub>`}⁰)∈ O(y/ ⁰<sub>`</sub>), contradicting the minimal number of occurrences of g` in y`). Let xr ∈ GFr contain exactly one occurrence of gr

and assume that min{supp(y`)} = (r+ 2)p, max{supp(xr)} = −(r+ 1)p−1.

Decompose y_{`</sub> =x<sub>`}⊗x⁰_{`</sub> so that x<sub>`} contains only the leftmost g_{`</sub> from y<sub>`} and x⁰_{`</sub> contains all the other occurrences of g<sub>`} from y_`. In a similar way as in the previous paragraph we see that

g^−(r+1)(f(g^r))(x_r⊗y_{`</sub>) =x<sub>r</sub>⊗(x<sub>`}⊗(σ^−p(x⁰_`))).

Denoteh⁰= (g^−(r+1)◦f◦g^r)⁻¹. By the maximality of distance from the leftmost g` to the second-to-leftmostg` in y` we know that

h(x`⊗(σ<sup>−p</sup>(x<sup>0</sup><sub>`</sub>)))∈ O(x`⊗(σ<sup>−p</sup>(x<sup>0</sup><sub>`</sub>))), and this is possible only if

h(x_{`</sub>⊗(σ<sup>−p</sup>(x<sup>0</sup><sub>`}))) =σ^−s(x_{`</sub>⊗(σ<sup>−p</sup>(x<sup>0</sup><sub>`}))),

h(x_r⊗x_{`</sub>⊗(σ<sup>−p</sup>(x<sup>0</sup><sub>`}))) =σ^−s(x_r⊗x_{`</sub>⊗(σ<sup>−p</sup>(x<sup>0</sup><sub>`}))), h(h⁰(x_r⊗x_{`</sub>⊗(σ<sup>−p</sup>(x<sup>0</sup><sub>`})))) =σ^−s(h⁰(x_r⊗x_{`</sub>⊗(σ<sup>−p</sup>(x<sup>0</sup><sub>`})))), h(x_r⊗y_{`</sub>) =σ<sup>−s</sup>(x<sub>r</sub>⊗y<sub>`}),

so in particularh(y`)∈ O(y`), a contradiction.

Corollary 15 (Finitary Ryan’s theorem). k(X) = 2 for every mixing SFTX with a fixed point.

Proof. The fact thatk(X)≥2 follows from the previous theorem and Lemma 7.

To see thatk(X) = 2, assume to the contrary thatk(X)<2. Fromk(X) = 0 it would follow that Aut(X) contains only powers of the shift, which is evidently false. Assume then that k(X) = 1 and thath is a single automorphism which commutes with h⁰ ∈Aut(X) only if h⁰ is a power of the shift. Because hcom- mutes with itself, it follows that h=σⁱ for some i ∈Z. But all h⁰ ∈Aut(X) commute with σⁱ and so Aut(X) contains again only powers of the shift, a contradiction.

5 Conclusions

We have constructed glider automorphisms g for mixing SFTs X which have a fixed point, and we have applied these glider maps to prove for suchX that k(X) = 2. It seems that our construction of g should generalize to arbitrary mixing SFTs X which do not necessarily have any fixed points. In this case instead of a fixed point 0^Zwe need to fix some periodic configurationp∈X (i.e.

σ^k(p) =pfor somek∈N+) and we consider pointsx∈X which are finite (in some sense) with respect to pinstead of 0^Z. In light of this it is probable that k(X) = 2 for all mixing SFTsX.

Acknowledgments. The author thanks Ville Salo for helpful discussions con- cerning these topics.

References

1. D. Lind, B. Marcus:An introduction to symbolic dynamics and coding.Cambridge University Press, Cambridge (1995)

2. J. Patrick Ryan:The shift and commutativity.Mathematical systems theory 6 (1-2), 82–85 (1972)

3. V. Salo:Transitive action on finite points of a full shift and a finitary Ryan’s theo- rem.arXiv:1610.05487v2 (2017)

4. V. Salo, I. T¨orm¨a:A One-Dimensional Physically Universal Cellular Automaton.

In: Kari J., Manea F., Petre I. (eds) Unveiling Dynamics and Complexity. CiE 2017.

Lecture Notes in Computer Science, vol 10307. Springer, Cham (2017)