The automorphism group of a minimal shift of stretched exponential growth

The group of automorphisms of a symbolic dynamical system is countable, but often very large. For example, for a mixing subshift of finite type, the automorphism group contains isomorphic copies of the free group on two generators and the direct sum of countably many copies of $\mathbb{Z}$. In contrast, the group of automorphisms of a symbolic system of zero entropy seems to be highly constrained. Our main result is that the automorphism group of any minimal subshift of stretched exponential growth with exponent $<1/2$, is amenable (as a countable discrete group). For shifts of polynomial growth, we further show that any finitely generated, torsion free subgroup of Aut(X) is virtually nilpotent.


Complexity and the automorphism group
Let (X, σ) be a subshift over the finite alphabet A, meaning that X ⊂ A Z is closed and invariant under the left shift σ : A Z → A Z . The group of automorphisms Aut(X) of (X, σ) is the group of homeomorphisms of φ : X → X such that φ • σ = σ • φ. A classic result of Curtis, Hedlund, and Lyndon is that Aut(X) is always countable, but a number of results have shown that Aut(X) can be quite large. For example, for any mixing subshift of finite type, Aut(X) always contains (among others) a copy of: every finite group, the direct sum of countably many copies of Z, and the free group on two generators [11,2]; every countable, locally finite, residually finite group [13]; the fundamental group of any 2-manifold [13].
This extremely rich subgroup structure makes the problem of deciding when two shifts have isomorphic automorphism groups challenging. Moreover, Kim and Roush [13] showed that the automorphism group of any full shift is contained in the automorphism group of any other full shift (and more generally is contained in any mixing subshift of finite type), thereby dooming any strategy for distinguishing two such groups that relies on finding a subgroup of one that does not embed into the other. Even the question of whether the automorphism groups of the full 2-shift and the full 3-shift are isomorphic remains a difficult open problem [2] (although, as they remark, the automorphism groups of the full 2-shift and the full 4-shift are not isomorphic). In all of these examples, the complicated nature of Aut(X) is a manifestation of the relatively light constraints required on x ∈ A Z to be a member of the shift space X. Another consequence of this fact is that these shifts always have positive (although arbitrarily small) entropy. As a corollary, if G is a group that embeds into the automorphism group of the full 2-shift, then for any h > 0 there is a subshift of topological entropy less than h into whose automorphism group G also embeds. It is therefore natural to ask whether the automorphism group of a zero entropy subshift is more highly constrained than its positive entropy relatives. Over the past several years, the authors [4,5] and others (e.g., [3,7,14,15]) have shown that the zero entropy case is indeed significantly more constrained, and we continue this theme in the present work. Specifically, we study how the the growth rate of the factor complexity P X (n), the number of nonempty cylinder sets of length n, constrains the algebraic properties of group of automorphisms. For a shift whose factor complexity grows at most linearly, we showed [5] that every finitely generated subgroup of Aut(X) is virtually Z d for some d that depends on the growth rate. We further showed [4] that for a transitive shift of subquadratic growth, the quotient of Aut(X) by the subgroup generated by σ, is periodic. Further examples of minimal shifts with polynomial complexity and highly constrained automorphism groups were constructed by Donoso, Durand, Maass and Petite [7], and an example of a minimal shift with subquadratic growth whose automorphism group is not finitely generated was given by Salo [14].
Our main theorem provides a strong constraint on Aut(X) for any minimal subshift of stretched exponential growth with exponent < 1/2. We show: then Aut(X) is amenable. Moreover, every finitely generated, torsion free subgroup of Aut(X) has subexponential growth.
For minimal shifts of polynomial growth, we show more: then Aut(X) is amenable. Furthermore, every finitely generated, torsion free subgroup of Aut(X) is virtually nilpotent with polynomial growth rate at most d − 1.
In particular, the step of the nilpotent subgroup is at most −1+ √ 8d−7 2 .
In particular, this shows that Z d does not embed in the automorphism group of a minimal shift whose growth rate is o(n d ).
In both the polynomial and stretched exponential cases, the amenability of Aut(X) stands in stark contrast to the possible behavior of the automorphism group of a shift of positive entropy. Furthermore, for shifts of positive entropy, we are far from being able to characterize the groups that can arise as automorphism groups, but we are approaching an answer for shifts with zero entropy.
This leaves open several natural questions. We are not able to provide examples of minimal shifts showing that Theorems 1.1 and 1.2 can not be improved, in the sense that the automorphism groups may be smaller. In particular, we can not rule out the possibility that the automorphism groups are virtually Z d , as raised in [7]. In fact, it suffices to show that the Heisenberg group does not embed in the automorphism group, as any torsion free, nonabelian nilpotent group contains an isomorphic copy of the Heisenberg group as a subgroup. However, for subcubic growth, the control on the step of the nilpotent group in Theorem 1.2 implies: then every finitely generated, torsion free subgroup of Aut(X) is virtually abelian.
After writing this paper, Donoso, Durand, Maass, and Petite [8] shared with us a proof that improves this corollary, showing that the same result holds with a growth rate that is o(n 5 ).
Furthermore, we believe that the amenability of Aut(X) from Theorem 1.1 should hold for 1/2 ≤ β ≤ 1 thus for all zero entropy shifts, but our methods do not cover this case. Finally, we do not know if these results generalize to transitive or general shifts.

Subshifts. Suppose
A is a finite set with the discrete topology and let A Z denote the set of bi-infinite sequences generates this topology and makes A Z into a compact metric space. When convenient, we denote an element of A Z by ( and it is easy to verify that σ is a homeomorphism on A Z . A subshift of A Z is a closed, σ-invariant subset X ⊂ A Z (endowed with the subspace topology). If X is a subshift of A Z and f ∈ A {−n+1,...,−1,0,1,...,n−1} is given, then the central cylinder set [f ] 0 of size n in X determined by f is defined to be the set A standard fact is that the collection of central cylinder sets, taken over all f and n, forms a basis for the topology of X. If g ∈ A {0,1,...,n−1} , then the one-sided cylinder set [g] + 0 of size n in X determined by g is defined to be the set ..,n−1} and is denoted by L n (A Z ). An element w ∈ L n (A Z ) is written w = (w 0 , w 1 , . . . , w n−1 ) or, when convenient, simply as a concatenation of letters: w = w 0 w 1 . . . w n−1 . A word w naturally defines a one-sided central cylinder set [w] 0 , and in a slight abuse of notation we write If X is a fixed subshift, then the set of words of length n in X is the set The set of all words L(X) in X is given by and is called the language of X. The complexity function P X : N → N is defined to be the map P X (n) := |L n (X)|, meaning it assigns to each n ∈ N the number of words of length n in the language of X. It follows from the Morse-Hedlund Theorem [12] that if P X is not increasing, then the associated subshift is periodic. If w ∈ L n (X), then we say that w extends uniquely to its right if there is a unique u ∈ L n+1 (X) such that u i = w i for all 0 ≤ i < n. Similarly, we say that w extends uniquely to its left if there is a unique u ∈ L n+1 (X) such that u i = w i for all 0 < i ≤ n. More generally, for N ∈ N, a word w ∈ L n (X) extends uniquely N times to its right if there is a unique u ∈ L n+N (X) such that u i = w i for all 0 ≤ i < n, and similarly for extensions to the left.

2.2.
Automorphisms. If (X, σ) is a subshift, then the group of automorphisms of (X, σ) is the set of all homeomorphisms of X that commute with σ and this group is denoted Aut(X). With respect to the compact open topology, Aut(X) is discrete. A map ϕ : X → X is a block code of range R if for all x ∈ X the symbol that ϕ(x) assigns to 0 is determined by the word (x −R , . . . , x 0 , . . . , x R ). The basic structure of an automorphism of (X, σ) is given by: Theorem 2.1 (Curtis-Hedlund-Lyndon Theorem [11]). If (X, σ) is a subshift and ϕ ∈ Aut(X), then there exists R ∈ N ∪ {0} such that ϕ is a block code of range R.
If ϕ is a block code of range R, then it is also a block code of range S for any S ≥ R. Therefore we can speak of a range for ϕ or of the minimal range for ϕ. We define Aut R (X) := ϕ ∈ Aut(X) : R is a range for ϕ and ϕ −1 .
The Curtis-Hedlund-Lyndon Theorem implies that We observe that it follows immediately from the definitions that if ϕ ∈ Aut R (X) for some R ∈ N, x ∈ X, and m < n are integers, then the restriction of ϕ(x) to {m, m + 1, . . . , n} is uniquely determined by the restriction of x to {m − R, m − R + 1, . . . , n + R − 1, n + R}. This motivates the definition of the action of an automorphism on a word: Definition 2.2. Suppose ϕ ∈ Aut R (X) and w ∈ L(X) has length at least 2R + 1. We define the word ϕ(w) to be the unique u ∈ L |w|−2R (X) such that for all Note that ϕ(w) depends on the choice of a range for ϕ, and so this definition only makes sense for the pair (ϕ, R) and not simply for ϕ.
We record another useful consequence of these observations: . . , m}, and let e 1 , . . . , e m ∈ {−1, 1}. Then Growth in groups and amenability. We recall some standard definitions and results; see, for example [6] for background. Suppose G is a finitely generated group and S = {g 1 , . . . , g m } ⊂ G is a finite symmetric generating set for G, meaning that every element of G can be written as a finite product of elements of S and that S is closed under inverses. Then the growth function of the pair (G, S) is the function . . , m}}|, which counts the number of distinct group elements that can be written as a product of at most n elements of S. If x(n), y(n) : N → N are nondecreasing functions, we write that x ≺ y if there constants λ ≥ 1 and C ≥ 0 such that x(n) ≤ λy(λn+C)+C and we write that x ∼ y if x ≺ y and y ≺ x. A standard fact is that γ S G is submultiplicative, and that if S 1 , S 2 are any two finite symmetric generating sets for G, then γ S1 G ∼ γ S2 G . Thus we define G to be a group of exponential growth if for some (equivalently, for every) finite symmetric generating set S, If the limit is zero, we say that G is a group of subexponential growth.
Similarly, G is a group of polynomial growth rate d if for some (equivalently, for every) finite symmetric generating set S, lim sup n→∞ log(γ S G (n)) log(n) ≤ d and G is a group of weak polynomial growth rate d if for some (equivalently, for every) finite symmetric generating set S, Finally, we say that G is a group of polynomial growth if it is a group of polynomial growth rate d for some d ∈ N, and it is a group of weak polynomial growth if it is a group of weak polynomial growth rate d for some d ∈ N.
A countable, discrete group G is amenable if there exists a sequence (F k ) k∈N of finite subsets of G such that: (i) for all g ∈ G, we have g ∈ F k for all but finitely many k; (ii) for all g ∈ G, In this case, the sequence (F k ) k∈N is called a Følner sequence for G.

Constraints on the automorphism group
We start with a lemma which shows that if P X (n) grows slowly then there are words that extend uniquely a large number of times to both sides: Suppose (X, σ) is a subshift and define k n := min{k ∈ N : no word w ∈ L n (X) extends uniquely k times to the right and left}.
then there exists C > 0 such that for infinitely many n ∈ N we have k n > Cn.
then for any C > 0 there are infinitely many n ∈ N such that k n > Cn 1−β .
Proof. Without loss, it suffices to assume that P X (n) is nondecreasing, as otherwise (X, σ) is a periodic system. First assume that (X, σ) satisfies (3). For contradiction, suppose that for all C > 0 there exists N = N (C) ∈ N such that k n ≤ Cn for all n ≥ N . Pick a constant C such that 0 < C < (2 1/d − 1)/2. By the definition of k n , any word of length n extends in at least two ways to a word of length n + 2k n (adding k n letters on each side), and any two words of length n extend to different words of length n+2k n by this procedure. So we have P X (n+2k n ) ≥ 2P X (n) for all n ∈ N. Since P X is nondecreasing, it follows that for all n ≥ N we have P X (⌈(1 + 2C)n⌉) ≥ 2P X (n). By induction, it follows that for any i ∈ N, But by (3), for all sufficiently large i. By choice of C, we have (1 + 2C) d < 2 and so (5) and (6) are incompatible for all sufficiently large i. The first statement follows. If (X, σ) satisfies (4), define d n := min{m ∈ N : P X (n + m) ≥ 2P X (n)} to be the sequence of doubling times for P X . As in the case of polynomial growth, we have the trivial inequality that d n ≤ 2k n for all n ∈ N. Again we proceed by contradiction and assume that there exists C > 0 and N = N (C) ∈ N such that for all n ≥ N we have k n ≤ Cn 1−β . Thus d n ≤ (2C)n 1−β for all n ≥ N . If λ > 1, then the doubling time for λ n β is D n (λ) := min{m ∈ N : λ (n+m) β ≥ 2λ n β } = n · 1 + log(2)/(n β log(λ)) 1/β − n .
Thus we can find 1 < λ < 2 1/2βC and M ≥ N such that d n < D n (λ) for all n ≥ M . We recursively define a sequence {a i } by setting a 0 := M and a i+1 = a i + d ai . If P X (a i ) ≥ λ a β i −a β 0 · P X (a 0 ), then where the last inequality holds because a i+1 < a i +D ai (λ). By induction, P X (a i ) ≥ λ a β i −a β 0 · P X (a 0 ) for all i. But then log(P X (a i )) ≥ log(λ) · a β i − log(λ) · a β 0 − log(P X (a 0 )) for all i, a contradiction of (4). . LetÑ be the largest multiple of 3k less than or equal to N . Then for n ≤ 2Ñ /3k, we have Taking n = 2Ñ /3k, But 1 2−2β > β 1−β , and so for all sufficiently large N , and thus sufficiently largeÑ , this contradicts Lemma 3.3. Suppose (X, σ) is minimal and w ∈ L(X). Then the subgroup G w of Aut(X) generated by S w , where Proof. Let ϕ ∈ S w be fixed. By the definition of S w , if u ∈ L(X) and x ∈ [wuw] + 0 , then there exists v x ∈ L |u| (X) such that ϕ(x) ∈ [wv x w] + 0 . However, since the range of ϕ is at most ⌊|w|/2⌋, the word v x does not depend on choice of x ∈ [wuw] + 0 , meaning there exists v ∈ L |u| (X) such that for all x ∈ [wuw] + 0 we have v x = v. In other words, ϕ[wuw] + 0 ⊆ [wvw] + 0 . Since (X, σ) is minimal, we have that the quantity K w := max{K ≥ 1 : there exists x ∈ [w] + 0 such that σ k x / ∈ [w] + 0 for all 0 < k < K} is finite. Thus if x ∈ X, then there is a bi-infinite sequence of words of bounded length . . . , u −2 , u −1 , u 0 , u 1 , u 2 , . . . ∈ Kw k=0 L k (X) such that x has the form · · · wu −2 wu −1 wu 0 wu 1 wu 2 w · · · .
Since P X is nondecreasing, it follows from these observations that ϕ PX (Kw)! (x) = x. Since ϕ PX (Kw)! is an automorphism of (X, σ) and the orbit of x is dense in X, ϕ PX (Kw)! is the identity map. Thus ϕ −1 [w] + 0 ⊆ [w] + 0 . Moreover, since ϕ ∈ Aut ⌊|w|/2⌋ (X), we have ϕ −1 ∈ Aut ⌊|w|/2⌋ (X), and so S w is closed under the operation of taking inverses. Let G w denote the subgroup of Aut(X) generated by S w . It is immediate that if ψ ∈ G w , then ψ[w] + 0 ⊆ [w] + 0 (but the range of ψ may not be bounded by ⌊|w|/2⌋ and so it is not necessarily the case that ψ ∈ S w ). As before, if u ∈ L(X) and x ∈ [wuw] + 0 , then there exists v x ∈ L |u| (X) such that ϕ(x) ∈ [wv x w] + 0 . We claim that the word v x is independent of x for x ∈ [wuw] + 0 . Writing ψ as a product of elements of S w , where ϕ 1 , . . . , ϕ k ∈ S w , there exist words v 1 , . . . , v k ∈ L |u| (X) such that Therefore, an element of G w is determined once we know the image of every cylinder set of the form [wuw] + 0 with |u| ≤ K w . There are only finitely many such sets and only finitely many possible images for each, and so G w is finite.
Then Aut(X) is amenable.
Proof. We show the amenability of Aut(X) by constructing a Følner sequence.
Step 1 (Extendable Words). By Lemma 3.1, there are infinitely many integers n for which there exists a word of length n that extends uniquely at least n 1−β many times to both the right and the left. Let R ∈ N be fixed and find n > (4R) 1/(1−β) such that there exists a word w ∈ L n (X) which extends uniquely at least 2R times to both the right and to the left. Letw ∈ L n+4R (X) be the word obtained by extending w exactly 2R times to both the right and to the left. Then if ϕ 1 , ϕ 2 ∈ Aut R (X) are such that ϕ 1 (w) = ϕ 2 (w) (recall that ϕ i (w) is a word of length n + 2R), we have ϕ −1 1 (ϕ 2 (w)) = w (a word of length n). But by construction of w, we have Since R ≤ ⌊|w|/4⌋, it follows from Lemma 3.3 that ϕ −1 1 • ϕ 2 is an element of the set Sw. If Gw is the subgroup of Aut(X) generated by Sw, then we have shown that for ϕ 1 , ϕ 2 ∈ Aut R (X), ϕ 1 and ϕ 2 are in the same coset of Gw if and only if ϕ 1 (w) = ϕ 2 (w).
Consequently, Aut R (X) is contained in the union of at most P X (|w| − 2R) many cosets of Gw.
Step 2 (Estimating Coset Growth). We claim that for any fixed k ∈ N, there exists a finite set F k ⊂ Aut(X) such that Aut k (X) ⊂ F k and furthermore, if ϕ ∈ Aut k (X) then The sequence (F k ) k∈N constructed in this way is then the desired Følner sequence.
To construct the sets, fix k ∈ N and find N > k sufficiently large that P X (n) ≤ exp n β 4 β/(1−β) for all n ≥ N . By Lemma 3.1, there are infinitely many n ≥ N for which there exists a word w ∈ L n (X) that extends uniquely at least n 1−β times to both the right and the left. By Lemma 3.2, for all sufficiently large R and for any increasing f : Fix n ≥ N such that there exists w ∈ L n (X) that extends at least n 1−β times to both the right and the left, and sufficiently large that if R := ⌊n 1−β /9⌋ then (9) holds and 6R < n 1−β . Then w extends uniquely at least 6R times to both the right and the left, and moreover we have |w| ≤ (9R) 1/(1−β) . Letw be the word obtained by extending w exactly 6R times to both the right and the left. Recall that if ϕ ∈ Aut R (X), then ϕ may also be thought of as a range R + 1 automorphism, meaning there is a natural embedding Aut R (X) ֒→ Aut R+1 (X). (The distinction between Aut R (X) and its embedded image is simply that an element of Aut R (X) takes a word of length n to a word of length n − 2R, whereas an element of Aut R+1 (X) takes a word of length n to a word of length n − 2R − 2.) We use the notation Aut R (X) ֒→ Aut S (X) to refer to the embedded image of Aut R (X) in Aut S (X) for S ≥ R.
for all 1 ≤ i ≤ k (note that we only make use of this for i = k). Since for all 1 ≤ i ≤ k, the set Aut M+i (X) is covered using (8) by at most f (M + i) cosets of Gw, we can define F k to be the set where ϕ i ∈ Aut M (X) for all i. It is immediate that Aut M (X) ⊂ F k and hence Aut k (X) ⊂ F k . If ϕ ∈ Aut k (X), then ϕ • ϕ i ∈ Aut M+k (X) for all i. Since Aut M+k (X) can be covered by f (M + k) ≤ f (M ) · exp M (2β−1)/(2−2β) cosets of Gw, it follows that at most f (M ) · −1 + exp M (2β−1)/(2−2β) additional cosets (beyond those that already appear in the definition of F k ) are needed to cover Aut M+k (X), and hence to also cover ϕ · F k . Consequently, Since β < 1/2, this quantity tends to 0 as k → ∞. Therefore, (F k ) k∈N is a Følner sequence and Aut(X) is amenable.
Then every finitely generated and torsion free subgroup of Aut(X) has subexponential growth. Moreover, there are infinitely many n for which the growth function of any such subgroup is at most exp(n β/(1−β) ).
Proof. Let G be a finitely generated, torsion free subgroup of Aut(X) and let R ∈ N be fixed. We estimate the growth rate of |Aut R·m (X)∩G| as m → ∞. By the second part of Lemma 3.1, there are infinitely many integers n for which there exists a word w n ∈ L n (X) that extends uniquely at least n 1−β times to both the right and the left. For each such n, let m n ∈ N be the largest integer for which R · m n ≤ n 1−β /2. Then w extends uniquely at least 2R · m n times to both the right and the left, and |w n | ≤ (2R·m n +2R) 1/(1−β) . Letw n be the word obtained by extending w n exactly 2R · m n times to both the right and the left. Then, as in first step (Extendable Words) of the proof of Theorem 3.4, we have that if ϕ 1 , ϕ 2 ∈ Aut R·mn (X) ∩ G and ϕ 1 (w) = ϕ 2 (w), then ϕ 1 and ϕ 2 lie in the same coset of Gw. But by Lemma 3.3, Gw is finite and since G is torsion free, it follows that ϕ 1 = ϕ 2 . Hence the size of Aut R·mn (X) ∩ G is bounded by the number of words of the form ϕ(w n ), where ϕ ∈ Aut R·mn (X) ∩ G. So, log |Aut R·mn (X) ∩ G| ≤ log (P X (|w n | + R)) But m n → ∞ when n → ∞, and so this condition holds infinitely often. If ϕ 1 , . . . , ϕ m ∈ Aut(X) is a (symmetric) set of generators for G and if R is chosen such that ϕ 1 , . . . , ϕ m ∈ Aut R (X), then for any k ∈ N and any i 1 , . . . , i k ∈ {1, . . . , m}, we have ϕ i1 • ϕ i2 • · · · • ϕ i k ∈ Aut R·k (X).
There are infinitely many k for which there exists n such that k = m n , and for all such k we have that log of the number of reduced words of length at most Rk that may be written in ϕ 1 , . . . , ϕ m is also o k β/(1−β) . If γ : N → N is the growth function of G, then lim inf k→∞ log γ(Rk) Rk = 0.
But since γ is submultiplicative, the limit converges, and so lim k→∞ γ(Rk) Rk = 0 and G has subexponential growth. The second claim in the theorem follows immediately from the estimate of γ(Rm n ).
We complete the proof of Theorem 1.2 with: Proof. The proof is almost identical to that of Theorem 3.5. This time, by the first part of Lemma 3.1, there exists C > 0 such that for infinitely many n there is a word w n ∈ L n (X) that extends uniquely at least Cn times to the right and the left. As before, fix R ∈ N and for each such n define m n = ⌊Cn/2R⌋. Then w n extends uniquely at least 2R · m n times to both the right and the left and |w n | ≤ 2R · m n /C + 2R. Then proceeding in the same way, if G is a finite subgroup of Aut(X), then |Aut R·mn (X) ∩ G| ≤ P X (|w n | + R) ≤ P X (⌈2R · m n /C + 2R⌉) = o ⌈2R · m n /C + 2R⌉ d = o((m n ) d ).
Proceeding as in the proof of Theorem 3.5, we have that lim inf k→∞ log γ(Rk) (Rk) d = 0.
By van den Dries and Wilkie's [16] strengthening of Gromov's Theorem [9] that assumes that the complexity bound need only hold for infinitely many values, G is virtually nilpotent and the nilpotent subgroup has polynomial growth rate at most d − 1. By the Bass-Guivarc'h Formula [1,10], the step is at most −1+ √ 8d−7 2 .