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

Abstract

For any mixing SFT X containing 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 ${{\mathrm{Aut}}}(X)$ contains a two-element subset S whose centralizer consists only of shift maps.

The work was partially supported by the Academy of Finland grant 296018 and by the Vilho, Yrjö and Kalle Väisälä Foundation.

1 Introduction

Let $X\subseteq A^\mathbb {Z}$ be a one-dimensional subshift over a symbol set A. If X contains some constant sequence $0^\mathbb {Z}$ ($0\in A$), we may say that an element $x\in X$ is finite if it differs from $0^\mathbb {Z}$ only at finitely many coordinates. In this paper we consider the problem of constructing reversible shift-commuting continuous maps (automorphisms) on X which decompose all finite configurations into collections of gliders traveling into opposing directions. As a concrete example, consider the binary full shift $X=\{0,1\}^\mathbb {Z}$ and the map $g=g_3\circ g_2\circ g_1:X\rightarrow X$ defined as follows. In any $x\in X$, $g_1$ replaces every occurrence of 0010 by 0110 and vice versa, $g_2$ replaces every occurrence of 0100 by 0110 and vice versa, and $g_3$ replace every occurrence of 00101 by 00111 and vice versa. In Fig. 1 we have plotted the sequences $x,g(x),g^2(x),\dots$ on consecutive rows for some $x\in X$. It can be seen that the sequence x eventually diffuses into two different “fleets”, the one consisting of 1 s going to the left and the one consisting of 11 s going to the right. It can be proved that this diffusion happens eventually no matter which finite initial point $x\in X$ is chosen.¹ In Sect. 3 we construct on all mixing SFTs (that contain the point $0^\mathbb {Z}$) a glider automorphism with the same diffusion property as the binary automorphism g above.

Fig. 1.

The diffusion of $x\in X$ under the map $g:X\rightarrow X$. White and black squares correspond to digits 0 and 1 respectively.

The existence of a glider automorphism g on a subshift X is interesting, because g can be used to convert an arbitrary finite $x\in X$ into another sequence $g^t(x)$ (for some $t\in {\mathbb {N}_+}$) with a simpler structure, which nevertheless contains all the information concerning the original point x because g 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 ${{\mathrm{Aut}}}(X)$, which we may consider as an abstract group. According to Ryan’s theorem [2] the center of the group ${{\mathrm{Aut}}}(X)$ is generated by the shift map $\sigma$. There may also be subsets $S\subseteq {{\mathrm{Aut}}}(X)$ whose centralizers are generated by $\sigma$. Denote the minimal cardinality of such a finite set S by k(X). In [3] it was proved that $k(X)\le 10$ when X is the full shift over the four-letter alphabet. In the same paper it is noted that k(X) is an isomorphism invariant of ${{\mathrm{Aut}}}(X)$ and therefore computing it could theoretically separate ${{\mathrm{Aut}}}(X)$ and ${{\mathrm{Aut}}}(Y)$ 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 $0^\mathbb {Z}$.

2 Preliminaries

A finite set A containing at least two elements (letters) is called an alphabet and the set $A^\mathbb {Z}$ of bi-infinite sequences (configurations) over A is called a full shift. Formally any $x\in A^\mathbb {Z}$ is a function $\mathbb {Z}\rightarrow A$ and the value of x at $i\in \mathbb {Z}$ is denoted by x[i]. It contains finite and one-directionally infinite subsequences denoted by $x[i,j]=x[i]x[i+1]\dots x[j]$, $x[i,\infty ]=x[i]x[i+1]\dots$ and $x[-\infty ,i]=\dots x[i-1]x[i]$. A factor of $x\in A^\mathbb {Z}$ is any finite sequence x[i, j] where $i,j\in \mathbb {Z}$, and we interpret the sequence to be empty if $j<i$. Any finite sequence $w=w[1] w[2]\dots w[n]$ (also the empty sequence, which is denoted by $\lambda$) where $w[i]\in A$ is a word over A. The set of all words over A is denoted by $A^*$, and the set of non-empty words is $A^+=A^*\setminus \{\lambda \}$. More generally, for any $L\subseteq A^*$, let

$$ L^*=\{w_1\dots w_n\mid n\ge 0,w_i\in L\}\subseteq A^*, $$

i.e. $L^*$ is the set of all finite concatenations of elements of L. The set of words of length n is denoted by $A^n$. For a word $w\in A^*$, $\vert w \vert$ denotes its length, i.e. $\vert w \vert =n\iff w\in A^n$. We say that the word $w\in A^n$ occurs in $x\in A^\mathbb {Z}$ at position i if $w=x[i]\dots x[i+n-1]$. We define the shift map $\sigma _A:A^\mathbb {Z}\rightarrow A^\mathbb {Z}$ by $\sigma _A(x)[i]=x[i+1]$ for $x\in A^\mathbb {Z}$, $i\in \mathbb {Z}$. The subscript A in $\sigma _A$ is typically omitted. The set $A^\mathbb {Z}$ is endowed with the product topology (with respect to the discrete topology on A), under which $\sigma$ is a homeomorphism on $A^\mathbb {Z}$. Any closed set $X\subseteq A^\mathbb {Z}$ such that $\sigma (X)=X$ is called a subshift, and the collection of words appearing as factors of elements of X is the language of X, denoted by L(X). The restriction of $\sigma$ to X may be denoted by $\sigma _X$, but typically the subscript X is omitted.

If $X\subseteq A^\mathbb {Z}$ is a subshift and $z\in X$ is such that $\sigma (z)=z$ (i.e. z is a fixed point), then there exists $a\in A$ such that $z[i]=a$ for all $i\in \mathbb {Z}$. For such subshifts we always fix one such point and denote $a=0$, $z=0^\mathbb {Z}$. Then for $x\in X$ we define its support ${{\mathrm{supp}}}(x)=\{i\in \mathbb {Z}\mid x[i]\ne 0\}$ and say that x is finite if ${{\mathrm{supp}}}(x)$ is finite. Finite points $x,y\in X$ with disjoint supports can be glued together; if ${{\mathrm{supp}}}(x)\cap {{\mathrm{supp}}}(y)=\emptyset$ we define $x\otimes y\in A^\mathbb {Z}$ by $(x\otimes y)[i]=x[i]$ when $i\in {{\mathrm{supp}}}(x)$ and $(x\otimes y)[i]=y[i]$ otherwise.

Definition 1

A graph is a pair ${\mathcal {G}}=(V,E)$ where V is a finite set of vertices (or nodes or states) and E is a finite set of edges. Each edge $e\in E$ starts at an initial state denoted by $\iota (e)\in V$ and ends at a terminal state denoted by $\tau (e)\in V$. We say that $e\in A$ is an outgoing edge of $\iota (e)$ and an incoming edge of $\tau (e)$.

A sequence of edges $e[1]\dots e[n]$ in a graph ${\mathcal {G}}=(V,E)$ is a path (of length n) if $\tau (e[i])=\iota (e[i+1])$ for $1\le i<n$ and it is a cycle if in addition $\tau (e[n])=\iota (e[1])$. We say that the path starts at e[1] and ends at e[n]. A graph ${\mathcal {G}}$ is primitive if there is $n\in {\mathbb {N}_+}$ such that for every $v_1,v_2\in V$ there is a path of length n starting at $v_1$ and ending at $v_2$. For any graph ${\mathcal {G}}=(V,E)$ we call the set

$$ \{x\in E^\mathbb {Z}\mid \tau (x[i])=\iota (x[i+1])\} $$

(i.e. the set of bi-infinite paths on ${\mathcal {G}}$) the edge subshift of ${\mathcal {G}}$.

Definition 2

A subshift $X\subseteq A^\mathbb {Z}$ is a mixing subshift of finite type (mixing SFT) if it is the edge subshift of a primitive graph ${\mathcal {G}}=(V,E)$ containing at least two edges (in particular $E\subseteq A$).

Example 3

Let $A=\{0,a,b\}$. The graph in Fig. 2 defines a mixing SFT X also known as the golden mean shift. A typical point of X looks like

$$ \dots 000abab0ab00ab000\dots $$

i.e. the letter b cannot occur immediately after 0 or b and every occurrence of a is followed by b.

Fig. 2.

The golden mean shift.

Definition 4

An automorphism of a subshift $X\subseteq A^\mathbb {Z}$ is a continuous bijection $f:X\rightarrow X$ such that $\sigma \circ f=f\circ \sigma$. We say that f is a radius-r automorphism if $f(x)[0]=f(y)[0]$ for all $x,y\in X$ such that $x[-r,r]=y[-r,r]$ (such r always exists by continuity of f). The set of all automorphisms of X is a group denoted by ${{\mathrm{Aut}}}(X)$. (In the case $X=A^\mathbb {Z}$ automorphisms are also known as reversible cellular automata.)

The centralizer of a set $S\subseteq {{\mathrm{Aut}}}(X)$ is

$$ C(S)=\{f\in {{\mathrm{Aut}}}(X)\mid f\circ g=g\circ f \text{ for } \text{ every } g\in S\} $$

and the subgroup generated by $f\in {{\mathrm{Aut}}}(X)$ is denoted by ${\langle {f}\rangle }$. The following definition is from [3]:

Definition 5

For a subshift X, let $k(X)\in \mathbb {N}\cup \{\infty ,\perp \}$ be the minimal cardinality of a set $S\subseteq {{\mathrm{Aut}}}(X)$ such that $C(S)= {\langle {\sigma }\rangle }$ if such a set S exists, and $k(X)=\perp$ otherwise.

The main result of [2] is that $k(X)\ne \perp$ whenever X is a mixing SFT.

We say that subshifts $X\subseteq A^\mathbb {Z}$ and $Y\subseteq B^\mathbb {Z}$ are conjugate if there is a continuous bijection $\psi :X\rightarrow Y$ such that $\psi \circ \sigma _X=\sigma _Y\circ \psi$. For conjugate subshifts X and Y it necessarily holds that $k(X)=k(Y)$.

3 Glider Automorphisms

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

Note that if X is a mixing SFT with a fixed point $0^\mathbb {Z}$, then necessarily in its graph ${\mathcal {G}}=(V,E)$ it holds that $\tau (0)=\iota (0)$. For such a graph we denote ${\mathcal {G'}}=(V,E')$ and $E'=E\setminus \{0\}$, i.e. we get ${\mathcal {G'}}$ from ${\mathcal {G}}$ by removing the 0-edge.

Definition 6

A mixing SFT X with a fixed point $0^\mathbb {Z}$ and defined by the graph ${\mathcal {G}}=(V,E)$ is called a 0-mixing SFT if the graph ${\mathcal {G'}}$ is also primitive and contains at least two edges.

The golden mean shift given by the graph in Fig. 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 $0^\mathbb {Z}$ defined by the graph ${\mathcal {G}}=(V,E)$ and let $s=\iota (0)=\tau (0)$. Let $0,a_1,\dots ,a_t$ be all the outgoing edges of s, let $0,b_1,\dots ,b_u$ be all the incoming edges of s and construct a new graph

$$ {\mathcal {H}}=(V\cup \{s'\},E\cup \{0',b_1',\dots ,b_u'\}) $$

with the starting and ending nodes of $e\in E$ the same as in ${\mathcal {G}}$ with the exception that $\iota (a_i)=s'$ for $1\le i\le t$, and additionally $\iota (0')=s$, $\tau (0')=s'$, $\iota (b_j')=\iota (b_j)$ and $\tau (b_j')=s'$ for $1\le j\le u$.² Let Y be the edge subshift of ${\mathcal {H}}$; it is conjugate with X via the continuous shift-commuting map $\psi :X\rightarrow Y$ defined for $x\in X$, $i\in \mathbb {Z}$ as

$$\begin{aligned} \psi (x)[i]= \left\{ \begin{array}{l} 0' \text{ when } x[i]=0 \text{ and } x[i+1]\in \{a_1,\dots ,a_t\}, \\ b_j' \text{ when } x[i]=b_j \text{ for } \text{ some } 1\le j\le u \text{ and } x[i+1]\in \{a_1,\dots ,a_t\}, \\ x[i] \text{ otherwise. } \end{array} \right. \end{aligned}$$

Fig. 3.

Splitting the state s.

Construct the graphs ${\mathcal {G}}'$, ${\mathcal {H}}'$ and let $c[1]\dots c[n]b_j$ be a cycle in ${\mathcal {G}}'$ visiting s only at the beginning and ending. Then $c[1]\dots c[n]b_j 0'$ and $c[1]\dots c[n]b_j'$ are distinct cycles in ${\mathcal {H'}}$ of coprime length, so ${\mathcal {H'}}$ has to be primitive and Y is a 0-mixing SFT (Fig. 3).    $\square$

In the rest of the section we assume that X is a 0-mixing SFT defined by the graph ${\mathcal {G}}=(V,E)$. This means that the edge subshift of ${\mathcal {G'}}$ is also a mixing SFT. Denote $s=\iota (0)=\tau (0)$. Let $v_1$ be a cycle in ${\mathcal {G'}}$ visiting s only at the beginning and ending, denote $p=\vert v_1 \vert$ and let $v_0=0^p$. The words

$$ g_\ell =v_0v_1, \qquad g_r=v_1v_1 $$

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

$$ L_\ell =(g_\ell 0 0^*)^*, \qquad L_r=(0^*0g_r)^*. $$

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

$$ \mathrm {GF}_{\ell }= {}^{\infty }0(g_\ell 0 0^*)^*0^\infty \qquad \mathrm {GF}_r={}^\infty 0(0^* 0 g_r)^*0^\infty \qquad \mathrm {GF}=\mathrm {GF}_\ell \cup \mathrm {GF}_r $$

(note that these consist of finite configurations).

Denote $u'=v_1v_1v_1$ and let $n\in {\mathbb {N}_+}$ be a mixing constant of ${\mathcal {G'}}$ (i.e. a number such that for every $n'\ge n$ and $s_1,s_2\in V$ there is a path of length $n'$ in ${\mathcal {G'}}$ from $s_1$ to $s_2$) chosen such that $n\ge \vert u' \vert =3p$. For every $a\in E$ we may choose some path $w_a\in E'^{2n}$ in ${\mathcal {G'}}$ such that $w_a$ begins with $u'$ and $0w_a a\in L(X)$. For every $a\in E$ let $W_a'=\{w_{a,1},\dots ,w_{a,k_a}\}\subseteq E'^{2n}$ be the paths of length 2n in ${\mathcal {G'}}$ such that $w_{a,i}$ does not have a prefix $u'$ and $0w_{a,i}a\in L(X)$ for $1\le i\le k_a$, and let $W_a=W_a'\cup \{w_a\}$. Let $U'=\{u_1',\dots ,u_k'\}\subseteq (E')^+$ be the cycles from s to s (which may visit s several times) of length at most $2n-1\ge 5p$ which are different from $v_1$ and $v_1v_1$ and do not have $u'$ as a prefix. Finally, these words are padded to constant length; $u=0^{2n-1-\vert u' \vert }u'$ and $u_i=0^{2n-1-\vert u'_i \vert }u_i'$. The words in $W_a$ and $U'$ are chosen so as to allow the following structural definition.

Definition 8

Assume that $x\notin \mathrm {GF}_\ell$ is a non-zero finite element of X. Then there is a maximal $i\in \mathbb {Z}$ such that

$$ x[-\infty ,i-1]\in {}^\infty 0 L_\ell , $$

and there is a unique word $w\in \{v_1 0\}\cup \{v_1v_10\}\cup \{u'\}\cup (U'0)\cup (\bigcup _{a\in E} W_a'a)$ such that w is a prefix of $x[i,\infty ]$. If $w=v_1v_10$ or $w\in U'0$, let $j=i+\vert w \vert -1$ and otherwise let $j=i+\vert v_1 \vert$. We say that x is of left bound type (w, j) and that it has left bound j (note that $j>i$).

Similarly, if $x\notin \mathrm {GF}_r$ is a non-zero finite element of X, then there is a minimal $j\in \mathbb {Z}$ such that

$$ x[j+1,\infty ]\in L_r 0^\infty $$

and we say that x has right bound j.

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 position j and break it off from the rest of the configuration.

We define four maps $g_1,g_2,g_3,g_4:X\rightarrow X$ as follows. In any $x\in X$,

• $g_1$ replaces every occurrence of $0(v_0v_1)0$ by $0(v_1v_1)0$ and vice versa

• $g_2$ replaces every occurrence of $0(v_1v_0)0$ by $0(v_1v_1)0$ and vice versa

• $g_3$ replaces every occurrence of $0v_0(v_1v_0v_1)$ by $0v_0(v_1v_1v_1)$ and vice versa

• $g_4$ replaces every occurrence of $0w_a a$, $0w_{a,i}a$ and $0w_{a,k_a}a$ by $0w_{a,1}a$, $0w_{a,i+1}a$ and $0w_a a$ respectively (for $a\in E$ and $1\le i<k_a$) and every occurrence of 0u0, $0u_i0$ and $0u_k0$ by $0u_10$, $0u_{i+1}0$ and 0u0 respectively (for $1\le i<k$).

It is easy to see that these maps are well defined automorphisms of X. The glider automorphism $g:X\rightarrow X$ is defined as the composition $g_4\circ g_3\circ g_2\circ g_1$. The name is partially justified by the following lemma.

Lemma 9

If $x\in \mathrm {GF}_\ell$ (resp. $x\in \mathrm {GF}_r$), then $g(x)=\sigma ^p(x)$ (resp. $g(x)=\sigma ^{-p}(x)$).

Proof

Assume that $x\in \mathrm {GF}_\ell$ (the proof for $x\in \mathrm {GF}_r$ is similar) and assume that $i\in \mathbb {Z}$ is some position in x where $g_\ell$ occurs. Then

$$\begin{aligned}&x[i-1,i+2p]=0g_\ell 0=0(v_0v_1)0 \\&g_1(x)[i-1,i+2p]=0(v_1v_1)0 \\&g_2(g_1(x))[i-p-1,i+p]=0v_0(v_1 0)=0g_\ell 0 \\&g(x)=g_4(g_3(g_2(g_1(x))))=g_2(g_1(x))), \end{aligned}$$

so every glider has shifted by distance p to the left and $g(x)=\sigma ^{p}(x)$.    $\square$

In fact, the previous lemma would hold even if g were replaced by $g_2\circ g_1$. The role of the part $g_4\circ g_3$ is, for a given finite point $x\in X$, to “erode” non-zero non-glider parts of x from the left and to turn the eroded parts into new gliders. This is the content of the following lemmas.

Lemma 10

Assume that $x\in X$ has left bound j. Then there exists $t\in {\mathbb {N}_+}$ such that the left bound of $g^t(x)$ is strictly greater than j.

Proof

Let $x\in X$ be of left bound type (w, j) with $w\in \{v_1 0\}\cup \{v_1 v_1 0\}\cup \{u'\}\cup (U'0)\cup (\bigcup _{a\in E} W_a'a)$. The gliders to the left of the occurrence of w near j move to the left at constant speed p under action of g without being affected by the remaining part of the configuration.

 

Case 1.:

Assume that $w=v_1v_10$. Then $g_1(x)[j-(p+1),j]=0v_1 0$. If $g_1(x)[j-(p+1),j+p]=0v_1v_0 0$, then $g(x)[j-(p+1),j+p]=g_2(g_1(x))[j-(p+1),j+p]=0v_1v_10$, g(x) is of left bound type $(v_1v_1 0, j+p)$ and we are done. Otherwise $g_2(g_1(x))[j-(p+1),j]=0v_10$. Denote $y=g_3(g_2(g_1(x)))$. If $y[j-(p+1),j]=0v_10$, then $g(x)=g_4(y)$ is of left bound type $(v_1 0,j)$ and we proceed as in Case 3. Otherwise $y[j-(p+1),j+(2p-1)]=0u'$. If $y[(j+2p)-2n,j+2p]=0u0$, then $g(x)[(j+2p)-2n,j+2p]=0u_1 0$, g(x) is of left bound type $(u_1'0,j+2p)$ and we are done. On the other hand, if $y[(j+2p)-2n,j+2p]\ne 0u0$, then g(x) is of left bound type $(w',j)$ for some $w'\in W_a'a\cup \{u'\}$ ($a\in E$) and we proceed as in Case 4 or Case 5.

Case 2.:

Assume that $w=u_i'0$ for $1\le i\le k$. There is a minimal $t\in \mathbb {N}$ such that $g_3(g_2(g_1(g^t(x))))[j-2n,j]=0u_i0$. 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 $(u',j-2p)$. Then $g(y)[j-5p,j]=g_3(g_2(g_1(y)))[j-5p,j]=v_0(v_0v_1)v_0v_10$ is of left bound type $(v_1 0,j)$ and we proceed as in Case 3.

Case 3.:

Assume that $w=v_1 0$. Then $x[j-(2p+1),j]\ne 0v_0v_10=0g_\ell 0$ because otherwise the left bound of x would already be greater than j, so $g_1(x)[j-(p+1),j]=0v_10$. If moreover $g_1(x)[j-(p+1),j+p]=0v_1v_0 0$, then $g(x)[j-(p+1),j+p]=g_2(g_1(x))[j-(p+1),j+p]=0(v_1v_1)0$ so g(x) is of left bound type $(v_1v_1 0,j+p)$ and we are done. Let us therefore assume that $g_1(x)[j-(p+1),j+p]\ne 0v_1v_0 0$, in which case $g_2(g_1(x))[j-(2p+1),j]=0v_0v_10$.

If $g_2(g_1(x))[j-(2p+1),j+2p-1]\ne 0v_0v_1v_0v_1$, then $g(x)[j-(2p+1),j]=g_3(g_2(g_1(x)))[j-(2p+1),j]=0v_0v_10$. The left bound of g(x) is now greater than j and we are done. Otherwise $g_3(g_2(g_1(x)))[j-(2p+1),j+2p-1]=0v_0u'$. If $g_3(g_2(g_1(x)))[(j+2p)-2n,j+2p]=0u0$, then $g(x)[(j+2p)-2n,j+2p]=0u_1 0$ and the left bound of g(x) equals $j+2p$. Finally, if $g_3(g_2(g_1(x)))[(j+2p)-2n,j+2p]\ne 0u0$, then g(x) is of left bound type $(w',j)$ for some $w'\in W_a'a\cup \{u'\}$ ($a\in E$) and we proceed as in Case 4 or Case 5.

Case 4.:

Assume that $w=w_{a,i}a$ for $a\in E$ and $1\le i\le k_a$. Then $g^{k_a-i+1}(x)[j-p,j+(2p-1)]=u'$ and we proceed as in Case 5.

Case 5.:

Assume that $w=u'$. Then $g_2(g_1(x))[j-(2p+1),j+(2p-1)]=0v_0u'$, $g_3(g_2(g_1))[j-(2p+1),j+(2p-1)]=0v_0v_1v_0v_1$ and the left bound of g(x) is at least $j+2p$.

     $\square$

Lemma 11

Assume that $x\in X$ has right bound j. Then there exists $t\in {\mathbb {N}_+}$ such that the right bound of $g^t(x)$ is strictly less than j.

Proof

Let us assume to the contrary that the right bound of $g^t(x)$ is at least j for every $t\in {\mathbb {N}_+}$.

Assume first that the right bound of $g^t(x)$ is equal to j for every $t\in {\mathbb {N}_+}$. By the previous lemma there is $t\in {\mathbb {N}_+}$ such that the left bound of $g^t(x)$ is at least $j+3n$, which means that $g^t(x)$ contains only $g_\ell$-gliders to the left of $j+n$ and only $g_r$-gliders to the right of j. This can happen only if $g^t(x)[j+1,n-1]=0^{n-1}$. Then the right bound of $g^{t+1}(x)$ is at least $j-p$, a contradiction.

Assume then that the right bound of $g^t(x)$ is strictly greater that j for some $t\in {\mathbb {N}_+}$ and fix the minimal such t. This can happen only if $g_1(g^{t-1}(x))[j-p,j+p+1]=0v_1 v_0 0$, in which case $g^t(x)[j-p,j+p+1]=0 v_1v_10=0g_r0$. But then the right bound of $g^t(x)$ is less than $j-p$, a contradiction.    $\square$

Together these two lemmas yield the following theorem.

Theorem 12

If $x\in X$ is a finite configuration, then for every $N\in \mathbb {N}$ there exists $t\in \mathbb {N}$ such that $g^t(x)[-N,N]=0^{2N+1}$, $g^t(x)[\infty ,-(N+1)]\in {}^\infty 0 L_\ell$ and $g^t(x)[N+1,\infty ]\in L_r0^\infty$.

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 $g:X\rightarrow X$ 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 both g and f. We make a simple choice of such f.

First we define maps $f_1,f_2:X\rightarrow X$ for a 0-mixing SFT X as follows. In any $x\in X$,

• $f_1$ replaces every occurrence of $0(v_1v_1)v_0v_0v_0(v_1)0$ by $0(v_1v_1)v_0v_0(v_1)v_0 0$ and vice versa

• $f_2$ replaces every occurrence of $0(v_1v_1)v_0v_0(v_1)0$ by $0v_0(v_1v_1)v_0(v_1)0$ and vice versa,

where $v_0$ and $v_1$ are as in the previous section. It is easy to see that these maps are well defined automorphisms of X. The automorphism $f:X\rightarrow X$ is then defined as the composition $f_2\circ f_1$. The map f has two important properties. First, it replaces any occurrence of $0(v_1v_1)v_0v_0v_0(v_1)0$ by $0v_0(v_1v_1)v_0(v_1)v_0 0$. Second, if $x\in X$ is a configuration containing only gliders $g_\ell$ and $g_r$ and every occurrence of $g_\ell$ is sufficiently far from every occurrence of $g_r$, then $f(x)=x$.

To prove our main result we need the following lemma.

Lemma 13

([3], Lemma 7.5). If X is a mixing SFT containing a fixed point $0^\mathbb {Z}$ and $h:X\rightarrow X$ is an automorphism which is not a power of $\sigma$, then there exists a finite configuration $x\ne 0^\mathbb {Z}$ such that $h(x)\notin \mathcal {O}(x)\doteqdot \{\sigma ^i(x)\mid i\in \mathbb {Z}\}$.

Theorem 14

Let $X\subseteq A^\mathbb {Z}$ be a 0-mixing SFT and $g,f:X\rightarrow X$ as above. The only automorphisms of X which commute with both g and f are powers of $\sigma$.

Proof

Assume to the contrary that $h:X\rightarrow X$ is a radius-r automorphism whose inverse is also a radius-r automorphism and which commutes with g and f but is not a power of $\sigma$. Let us first show that $h(0^\mathbb {Z})=0^\mathbb {Z}$. Namely, if it were that $h(0^\mathbb {Z})=a^\mathbb {Z}$, for some $a\in A\setminus \{0\}$, consider $x=\dots 000 g_\ell 000\dots$ with the glider $g_\ell$ at the origin and note that $h(x)[i]\ne a$ for some $-r\le i\le (2p-1)+r$ (recall: $g_\ell =v_0v_1$, $\vert v_0 \vert =\vert v_1 \vert =p$) and $h(x)[-\infty ,i-jp]=\dots aaa$ for some $j\in {\mathbb {N}_+}$. Then $g^t(h(x))[-\infty ,i-jp]=\dots aaa$ for every $t\in \mathbb {Z}$ but $h(g^{j}(x))[i-jp]=h(\sigma ^{jp}(x))[i-jp]=h(x)[i]\ne a$, contradicting the commutativity of h and g. Thus h maps finite configurations to finite configurations.

We have $h(\mathrm {GF}_\ell )\subseteq \mathrm {GF}_\ell$. To see this, assume to the contrary that there exists $x\in \mathrm {GF}_\ell$ such that $h(x)\notin \mathrm {GF}_\ell$. Recall that g is reversible and $g(\mathrm {GF}_\ell )=\mathrm {GF}_\ell$, so $g^t(h(x))\notin \mathrm {GF}_\ell$ for all $t\in \mathbb {N}$. Combining this with Theorem 12 it follows that $g^t(h(x))$ contains an occurrence of $0 v_1 v_1 0$ to the right of coordinate r for all sufficiently large t and therefore $h^{-1}(g^t(h(x)))[i]$ is non-zero for some $i\ge 0$ which depends on t. This contradicts the fact that $h^{-1}(g^t(h(x)))[i]=g^t(x)[i]=0$ for all $i\ge 0$ given that t is sufficiently big. Similarly $h(\mathrm {GF}_r)\subseteq \mathrm {GF}_r$.

For any finite $x\ne 0^\mathbb {Z}$ define its left and right offsets

$$\begin{aligned} {{\mathrm{off}}}_\ell (x)&=\min \{{{\mathrm{supp}}}(h(x))\}-\min \{{{\mathrm{supp}}}(x)\}, \\ {{\mathrm{off}}}_r(x)&=\max \{{{\mathrm{supp}}}(h(x))\}-\max \{{{\mathrm{supp}}}(x)\}. \end{aligned}$$

For all nonzero $x_\ell \in \mathrm {GF}_\ell$ and $x_r\in \mathrm {GF}_r$ we have ${{\mathrm{off}}}_\ell (x_\ell )-{{\mathrm{off}}}_r(x_r)=0$. If this did not hold, we could assume without loss of generality that ${{\mathrm{off}}}_\ell (x_\ell )-{{\mathrm{off}}}_r(x_r)>0$ (by replacing h with $h^{-1}$ if necessary) and that $\min \{{{\mathrm{supp}}}(x_\ell )\}=(r+2)p$, $\max \{{{\mathrm{supp}}}(x_r)\}=-(r+1)p-1$ (by shifting $x_\ell$ and $x_r$ suitably). Then consider $x=x_r\otimes x_\ell$ and note that from $\min \{{{\mathrm{supp}}}(x_\ell )\}=(r+2)p>r$, $\max \{{{\mathrm{supp}}}(x_r)\}=-(r+1)p-1<-r$ it follows that $h(x)=h(x_r)\otimes h(x_\ell )$. Then $g^r(x)[-3p-1,3p]=0(v_1v_1)v_0v_0v_0(v_1)0$ and $g^r(h(x))$ contains no occurrence of the words mentioned in the definition of $f_1$ and $f_2$ by the assumption ${{\mathrm{off}}}_\ell (x_\ell )-{{\mathrm{off}}}_r(x_r)>0$, so $f(g^{r}(x))\ne g^{r}(x)$ and $f(g^{r}(h(x)))=g^{r}(h(x))$. Now

$$\begin{aligned} x\ne g^{-r}(f(g^{r}(x)))\,=\,&h^{-1}(g^{-r}(f(g^{r}(h(x))))) \\ =\,&h^{-1}(g^{-r}(g^{r}(h(x))))=x, \end{aligned}$$

a contradiction. It also follows that there is a fixed $s\in \mathbb {Z}$ such that ${{\mathrm{off}}}_\ell (x_\ell )={{\mathrm{off}}}_r(x_r)=s$ for all nonzero $x_\ell \in \mathrm {GF}_\ell$, $x_r\in \mathrm {GF}_r$.

If $x_\ell \in \mathrm {GF}_\ell$ and $x_r\in \mathrm {GF}_r$ are configurations containing exactly one occurrence of $g_\ell$ and $g_r$ respectively, then $h(x_\ell )=\sigma ^{-s}(x_\ell )$ and $h(x_r)=\sigma ^{-s}(x_\ell )$. To see this, assume to the contrary (without loss of generality) that $\min \{{{\mathrm{supp}}}(x_\ell )\}=(r+2)p$ (i.e. the occurrence of $g_\ell$ in x is at $(r+1)p$), $\max \{{{\mathrm{supp}}}(x_r)\}=-(r+1)p-1$ and $h(x_\ell )[(r+1)p+s,(r+3)p-1+s]=h(x_\ell )[(r+1)p+s+k,(r+3)p-1+s+k]=g_\ell$ for some $k>2p$ (i.e. $h(x_\ell )$ contains at least two occurrences of $g_\ell$, the case in which $h(x_r)$ contains at least two occurrences of $g_r$ being similar). Then consider $x=x_r\otimes x_\ell$ and note that

$$\begin{aligned}&g^{r}(x)[-3p-1,3p]=0(v_1v_1)v_0v_0v_0(v_1)0 \\&f(g^{r}(x))[-3p-1,3p]=0v_0(v_1v_1)v_0(v_1)v_00 \\&g^{-1}(f(g^{r}(x)))[-3p-1,3p]=0(v_1v_1)v_0v_0v_0(v_1)0 \\&g^{-(r+1)}(f(g^{r}(x)))=x, \end{aligned}$$

therefore also $h(x)=g^{-(r+1)}(f(g^{r}(h(x))))\doteqdot y$. On the other hand,

$$\begin{aligned}&f(g^r(h(x)))[p+s+k,3p-1+s+k] \\&=g^r(h(x))[p+s+k,3p-1+s+k]=g_\ell \\&\quad g^{-(r+1)}(f(g^r(h(x))))[(r+2)p+s+k,(r+4)p-1+s+k]=g_\ell , \end{aligned}$$

so in particular $y[(r+2)p+s+k,(r+3)p-1+s+k]=v_0$. Because we assumed that

$$\begin{aligned}&h(x)[(r+2)p+s+k,(r+3)p-1+s+k] \\&\quad =\,h(x_\ell )[(r+2)p+s+k,(r+3)p-1+s+k]=v_1, \end{aligned}$$

it follows that $h(x)\ne y$, a contradiction.

By Lemma 13 there exists a finite configuration $x\ne 0^\mathbb {Z}$ such that $h(x)\notin \mathcal {O}(x)$ and h(x) is finite. Use Theorem 12 to get $t\in \mathbb {N}$ such that $g^t(x)[-r,r]=0^{2r+1}$ and $g^t(x)=y_\ell \otimes y_r$ where $y_\ell \in \mathrm {GF}_\ell$ has $\max \{{{\mathrm{supp}}}(y_\ell )\}<-r$ and $y_r\in \mathrm {GF}_r$ has $\min \{{{\mathrm{supp}}}(y_r)\}>r$ (it is possible that either $y_\ell$ or $y_r$ is equal to $0^\mathbb {Z}$). Then also $h(g^t(x))=h(y_\ell )\otimes h(y_r)\notin \mathcal {O}(g^t(x))$, and combining this with ${{\mathrm{off}}}_\ell (y_\ell )={{\mathrm{off}}}_r(y_r)$ it follows that $h(y_\ell )\notin \mathcal {O}(y_\ell )$ or $h(y_r)\notin \mathcal {O}(y_r)$. Without loss of generality assume that $h(y_\ell )\notin \mathcal {O}(y_\ell )$ (the case $h(y_r)\notin \mathcal {O}(y_r)$ is similar), that $y_\ell$ contains a minimal number of occurrences of $g_\ell$ (at least two by the previous paragraph) and that the distance from the leftmost $g_\ell$ to the second-to-leftmost $g_\ell$ in $y_\ell$ is maximal (at most $2r+2p$ since otherwise by dropping the leftmost $g_\ell$ we would get a new configuration $y_\ell '$ such that $h(y_\ell ')\notin \mathcal {O}(y_\ell ')$, contradicting the minimal number of occurrences of $g_\ell$ in $y_\ell$). Let $x_r\in \mathrm {GF}_r$ contain exactly one occurrence of $g_r$ and assume that $\min \{{{\mathrm{supp}}}(y_\ell )\}=(r+2)p$, $\max \{{{\mathrm{supp}}}(x_r)\}=-(r+1)p-1$. Decompose $y_\ell =x_\ell \otimes x_{\ell }'$ so that $x_\ell$ contains only the leftmost $g_\ell$ from $y_\ell$ and $x_{\ell }'$ contains all the other occurrences of $g_\ell$ from $y_\ell$. In a similar way as in the previous paragraph we see that

$$ g^{-(r+1)}(f(g^{r}))(x_r\otimes y_\ell )=x_r\otimes (x_\ell \otimes (\sigma ^{-p}(x_{\ell }'))). $$

Denote $h'=(g^{-(r+1)}\circ f\circ g^r)^{-1}$. By the maximality of distance from the leftmost $g_\ell$ to the second-to-leftmost $g_\ell$ in $y_\ell$ we know that

$$ h(x_\ell \otimes (\sigma ^{-p}(x_{\ell }')))\in \mathcal {O}(x_\ell \otimes (\sigma ^{-p}(x_{\ell }'))), $$

and this is possible only if

$$\begin{aligned}&h(x_\ell \otimes (\sigma ^{-p}(x_{\ell }')))=\sigma ^{-s}(x_\ell \otimes (\sigma ^{-p}(x_{\ell }'))), \\&h(x_r\otimes x_\ell \otimes (\sigma ^{-p}(x_{\ell }')))=\sigma ^{-s}(x_r\otimes x_\ell \otimes (\sigma ^{-p}(x_{\ell }'))), \\&h(h'(x_r\otimes x_\ell \otimes (\sigma ^{-p}(x_{\ell }'))))=\sigma ^{-s}(h'(x_r\otimes x_\ell \otimes (\sigma ^{-p}(x_{\ell }')))), \\&h(x_r\otimes y_\ell )=\sigma ^{-s}(x_r\otimes y_\ell ), \end{aligned}$$

so in particular $h(y_\ell )\in \mathcal {O}(y_\ell )$, a contradiction.    $\square$

Corollary 15

(Finitary Ryan’s theorem). $k(X)=2$ for every mixing SFT X with a fixed point.

Proof

The fact that $k(X)\ge 2$ follows from the previous theorem and Lemma 7. To see that $k(X)=2$, assume to the contrary that $k(X)<2$. From $k(X)=0$ it would follow that ${{\mathrm{Aut}}}(X)$ contains only powers of the shift, which is evidently false. Assume then that $k(X)=1$ and that h is a single automorphism which commutes with $h'\in {{\mathrm{Aut}}}(X)$ only if $h'$ is a power of the shift. Because h commutes with itself, it follows that $h=\sigma ^i$ for some $i\in \mathbb {Z}$. But all $h'\in {{\mathrm{Aut}}}(X)$ commute with $\sigma ^i$ and so ${{\mathrm{Aut}}}(X)$ contains again only powers of the shift, a contradiction.    $\square$

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 such X 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 $0^\mathbb {Z}$ we need to fix some periodic configuration $p\in X$ (i.e. $\sigma ^k(p)=p$ for some $k\in {\mathbb {N}_+}$) and we consider points $x\in X$ which are finite (in some sense) with respect to p instead of $0^\mathbb {Z}$. In light of this it is probable that $k(X)=2$ for all mixing SFTs X.