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

. 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 Aut( X ) contains a two-element subset S whose centralizer consists only of shift maps.


Introduction
Let X ⊆ A Z be a one-dimensional subshift over a symbol set A. If X 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) 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} Z and the map g = g 3 • g 2 • g 1 : X → X defined as follows.In any x ∈ 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 Figure 1 we have plotted the sequences x, g(x), g 2 (x), . . . on consecutive rows for some x ∈ X.It can be seen that the sequence x eventually 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 point x ∈ X is chosen. 1 In 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 automorphism g above.
The existence of a glider automorphism g on a subshift X is interesting, because g can be used to convert an arbitrary finite x ∈ X into another sequence g t (x) (for some t ∈ 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 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 map σ.There may also be subsets S ⊆ Aut(X) whose centralizers are generated by σ.Denote the minimal cardinality of such a finite set S by k(X).In [3] it was proved that k(X) ≤ 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 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 .

Preliminaries
A finite set A containing at least two elements (letters) is called an alphabet and the set A Z of bi-infinite sequences (configurations) over A is called a full shift.Formally any x ∈ A Z is a function Z → A and the value of x at i ∈ Z is denoted by x[i].It contains finite and one-directionally infinite subsequences denoted by A factor of x ∈ A Z is any finite sequence x[i, j] where i, 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 by λ) where w[i] ∈ 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 * \ {λ}.More generally, for any L ⊆ A * , let 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 ∈ A * , |w| denotes its length, i.e. |w| = n ⇐⇒ w ∈ A n .We say that the word The set A Z is endowed with the product topology (with respect to the discrete topology on A), under which σ is a homeomorphism on A Z .Any closed set X ⊆ A Z such that σ(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 σ to X may be denoted by σ X , but typically the subscript X is omitted.
If X ⊆ A Z is a subshift and z ∈ X is such that σ(z) = z (i.e.z is a fixed point), then there exists a ∈ A such that z[i] = a for all i ∈ 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] = 0} and say that x is finite if supp(x) is finite.Finite points x, y ∈ X with disjoint supports can be glued together; where V is a finite set of vertices (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 of ι(e) and an incoming edge of τ (e).
A sequence of edges e [1] [1]).We say that the path starts at e [1] and ends at e [n].A graph G is primitive if there is n ∈ N + such that for every v 1 , v 2 ∈ V there is a path of length n starting at v 1 and ending at v 2 .For any graph G = (V, E) we call the set (i.e. the set of bi-infinite paths on G) the edge subshift of G. 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 graph G = (V, E) containing at least two edges (in particular E ⊆ A).
Example 3. Let A = {0, a, b}.The graph in Figure 2 defines a mixing SFT X also known as the golden mean shift.A typical point of X looks like . . .000abab0ab00ab000 . . .i.e. the letter b cannot occur immediately after 0 or b and every occurrence of a is followed by b.
r] (such r always exists by continuity of f ).The set of all automorphisms of X is a group denoted by Aut(X).(In the case X = A Z automorphisms are also known as reversible cellular automata.)The centralizer of a set S ⊆ Aut(X) is and the subgroup generated by f ∈ Aut(X) is denoted by f .The following definition is from [3]: Definition 5.For a subshift X, let k(X) ∈ N ∪ {∞, ⊥} be the minimal cardinality of a set S ⊆ Aut(X) such that C(S) = σ if such a set S exists, and k(X) =⊥ otherwise.
The main result of [2] is that k(X) =⊥ whenever X is a mixing SFT.We say that subshifts

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 Z , then necessarily in its graph G = (V, E) it holds that τ (0) = ι(0).For such a graph we denote G = (V, E ) and E = E \ {0}, i.e. we get G from G by removing the 0-edge.Definition 6.A mixing SFT X with a fixed point 0 Z and defined by the graph G = (V, E) is called a 0-mixing SFT if the graph G is 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 let s = ι(0) = τ (0).Let 0, a 1 , . . ., a t be all the outgoing edges of s, let 0, b 1 , . . ., b u be all the incoming edges of s and construct a new graph with the starting and ending nodes of e ∈ E the same as in G with the exception that ι(a i ) = s for 1 ≤ i ≤ t, and additionally ι(0 ) = s, τ (0 ) = s , ι(b j ) = ι(b j ) and τ (b j ) = s for 1 ≤ j ≤ u. 2 Let Y be the edge subshift of H; it is conjugate with X via the continuous shift-commuting map ψ : X → Y defined for x ∈ X, i ∈ Z as In the rest of the section we assume that X is a 0-mixing SFT defined by the graph G = (V, E).This means that the edge subshift of G is also a mixing SFT.Denote s = ι(0) = τ (0).Let v 1 be a cycle in G visiting s only at the beginning and ending, denote p = |v 1 | and let v 0 = 0 p .The words will be left-and rightbound gliders of the automorphism g defined later.The languages of left-and rightbound gliders are We denote by ∞ 0 and 0 ∞ left-and right-infinite sequences of zeroes and define the glider fleet sets (note that these consist of finite configurations).Denote u = v 1 v 1 v 1 and let n ∈ N + be a mixing constant of G (i.e. a number such that for every n ≥ n and s 1 , s 2 ∈ V there is a path of length n in G from s 1 to s 2 ) chosen such that n ≥ |u | = 3p.For every a ∈ E we may choose some path w a ∈ E 2n in G such that w a begins with u and 0w a a ∈ L(X).For every a ∈ E let W a = {w a,1 , . . ., w a,ka } ⊆ E 2n be the paths of length 2n in G such that w a,i does not have a prefix u and 0w a,i a ∈ L(X) for 1 ≤ i ≤ k a , and let W a = W a ∪ {w a }.Let U = {u 1 , . . ., u k } ⊆ (E ) + be the cycles from s to s (which may visit s several times) of length at most 2n − 1 ≥ 5p which are different from v 1 and v 1 v 1 and do not have u as a prefix.Finally, these words are padded to constant length; u = 0 2n−1−|u | u and u i = 0 2n−1−|u i | u i .The words in W a and U 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 maximal i ∈ Z such that and there is a unique word w ∈ {v 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 / ∈ GF r is a non-zero finite element of X, then there is a minimal 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 → X as follows.In any x ∈ X, g 1 replaces every occurrence of 0(v 0 v 1 )0 by 0(v 1 v 1 )0 and vice versa g 2 replaces every occurrence of 0(v 1 v 0 )0 by 0(v 1 v 1 )0 and vice versa g 3 replaces every occurrence of 0v 0 (v 1 v 0 v 1 ) by 0v 0 (v 1 v 1 v 1 ) and vice versa g 4 replaces every occurrence of 0w a a, 0w a,i a and 0w a,ka a by 0w a,1 a, 0w a,i+1 a and 0w a a respectively (for a ∈ E and 1 ≤ i < k a ) and every occurrence of 0u0, 0u i 0 and 0u k 0 by 0u 1 0, 0u i+1 0 and 0u0 respectively (for 1 ≤ i < k).
It is easy to see that these maps are well defined automorphisms of X.The glider automorphism g : X → X is defined as the composition The name is partially justified by the following lemma.
Proof.Assume that x ∈ GF (the proof for x ∈ GF r is similar) and assume that i ∈ Z is some position in x where g occurs.Then so every glider has shifted by distance p to the left and g(x) = σ p (x).
In fact, the previous lemma would hold even if g were replaced by g 2 • g 1 .The role of the part g 4 • g 3 is, for a given finite point x ∈ 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 ∈ X has left bound j.Then there exists t ∈ N + such that the left bound of g t (x) is strictly greater than j.
Proof.Let x ∈ X be of left bound type (w, j) with w ∈ {v 1 0} ∪ {v 1 v 1 0} ∪ {u } ∪ (U 0) ∪ ( a∈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.
Lemma 11.Assume that x ∈ X has right bound j.Then there exists t ∈ 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 ∈ N + .Assume first that the right bound of g t (x) is equal to j for every t ∈ N + .By the previous lemma there is t ∈ N + such that the left bound of g t (x) is at least j + 3n, which means that g t (x) contains only g -gliders to the left of j + n and only g r -gliders to the right of j.This can happen only if 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 ∈ N + and fix the minimal such t.This can happen only if g 1 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 ∈ N such that g

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 both g and f .We make a simple choice of such f .First we define maps f 1 , f 2 : X → X for a 0-mixing SFT X as follows.In any x ∈ X, 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 → X is then defined as the composition f 2 • f 1 .The map f has two important properties.First, it replaces any occurrence of 0(v x ∈ X is a configuration containing only gliders g and g r and every occurrence of g 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 Z and h : X → X is an automorphism which is not a power of σ, then there exists a finite configuration Theorem 14.Let X ⊆ A Z be a 0-mixing SFT and g, f : X → X as above.The only automorphisms of X which commute with both g and f are powers of σ.
Proof.Assume to the contrary that h : X → 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 σ.Let us first show that h(0 Z ) = 0 Z .Namely, if it were that h(0 Z ) = a Z , for some a ∈ A \ {0}, consider x = . . .000g 000 . . .with the glider g at the origin and note that h(x)[i] = a for some −r ≤ i ≤ (2p − 1) + r (recall: = a, contradicting the commutativity of h and g.Thus h maps finite configurations to finite configurations.We have h(GF ) ⊆ GF .To see this, assume to the contrary that there exists x ∈ GF such that h(x) / ∈ GF .Recall that g is reversible and g(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 0v 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 ≥ 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 ≥ 0 given that t is sufficiently big.Similarly h(GF r ) ⊆ GF r .
For all nonzero x ∈ GF and x r ∈ GF r we have off (x )−off r (x r ) = 0.If this did not hold, we could assume without loss of generality that off (x ) − off r (x r ) > 0 (by replacing h with h −1 if necessary) and that min{supp(x )} = (r + 2)p, max{supp(x r )} = −(r + 1)p − 1 (by shifting x and x r suitably).Then consider x = x r ⊗x and note that from min{supp(x )} = (r +2)p > r, max{supp(x r )} = −(r+1)p−1 < −r it follows that h(x) = h(x r )⊗h(x ).Then g r (x)[−3p−1, 3p] = 0(v 1 v 1 )v 0 v 0 v 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 off (x ) − off r (x r ) > 0, so f (g r (x)) = g r (x) and f (g r (h(x))) = g r (h(x)).Now a contradiction.It also follows that there is a fixed s ∈ Z such that off (x ) = off r (x r ) = s for all nonzero x ∈ GF , x r ∈ GF r .If x ∈ GF and x r ∈ GF r are configurations containing exactly one occurrence of g and g r respectively, then h(x ) = σ −s (x ) and h(x r ) = σ −s (x ).To see this, assume to the contrary (without loss of generality) that min{supp(x )} = (r+2)p (i.e. the occurrence of g in x is at (r+1)p), max{supp(x r )} = −(r+1)p−1 and h(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 which h(x r ) contains at least two occurrences of g r being similar).Then consider x = x r ⊗ x and note that it follows that h(x) = y, a contradiction.
By Lemma 13 there exists a finite configuration x = 0 Z such that h(x) / ∈ O(x) and h(x) is finite.Use Theorem 12 to get t ∈ N such that g t (x)[−r, r] = 0 2r+1 and g t (x) = y ⊗ y r where y ∈ GF has max{supp(y )} < −r and y r ∈ GF r has min{supp(y r )} > r (it is possible that either y or y r is equal to 0 Z ).Then also h(g t (x)) = h(y ) ⊗ h(y r ) / ∈ O(g t (x)), and combining this with off (y ) = off r (y r ) it follows that h(y ) / ∈ O(y ) or h(y r ) / ∈ O(y r ).Without loss of generality assume that h(y ) / ∈ O(y ) (the case h(y r ) / ∈ O(y r ) is similar), that y contains a minimal number of occurrences of g (at least two by the previous paragraph) and that the distance from the leftmost g to the second-to-leftmost g in y is maximal (at most 2r + 2p since otherwise by dropping the leftmost g we would get a new configuration y such that h(y ) / ∈ O(y ), contradicting the minimal number of occurrences of g in y ).Let x r ∈ GF r contain exactly one occurrence of g r and assume that min{supp(y )} = (r + 2)p, max{supp(x r )} = −(r + 1)p − 1.
Decompose y = x ⊗ x so that x contains only the leftmost g from y and x contains all the other occurrences of g from y .In a similar way as in the previous paragraph we see that g −(r+1) (f (g r ))(x r ⊗ y ) = x r ⊗ (x ⊗ (σ −p (x ))).
Proof.The fact that k(X) ≥ 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 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 ∈ Aut(X) only if h is a power of the shift.Because h commutes with itself, it follows that h = σ i for some i ∈ Z.But all h ∈ Aut(X) commute with σ i and so Aut(X) contains again only powers of the shift, a contradiction.

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

Fig. 1 .
Fig.1.The diffusion of x ∈ X under the map g : X → X. White and black squares correspond to digits 0 and 1 respectively.