Proof. Let $T \in \mathrm {TM}(\mathbb {Z}^d,n,k)$ and let $f \colon \Sigma ^{F} \times Q \to \Sigma ^{F} \times Q \times F$ be such that $T = T_f$ . Clearly T commutes with the shift on $X_{n,0}$ since it acts as the identity map. Let then $x^{\vec v}_q \in X_{n,k}\setminus X_{n,0}$ .

Let $p = \sigma ^{-\vec v}(x)|_{F}$ and $f(p, q) = (\tilde {p}, \tilde {q}, \tilde {\vec {u}})$ . Then by the definition of $T_f$ we have $T_f(x_q^{\vec {v}}) = \tilde {x}_{\tilde {q}}^{\vec {v}+\tilde {\vec {u}}}$ where

$$ \begin{align*}\tilde{x}_{\vec u} = \begin{cases} x_{\vec{u}} & \text{ if } {\vec u} - {\vec v} \notin F \\ \tilde{p}_{{\vec u} - {\vec v}}&\text{ if } {\vec u}-{\vec v} \in F \end{cases}\end{align*} $$

Now, let us consider the application of $T_f$ to $\sigma ^{\vec w}(x_q^{\vec {v}})$ . By the definition of $\sigma $ we have $\sigma ^{\vec w}(x_q^{\vec {v}}) = \sigma ^{\vec w}(x)^{\vec v + \vec w}_q$ . Let $y = \sigma ^{\vec w}(x)$ and $\vec v' = \vec v + \vec w$ so that $T_f(\sigma ^{\vec w}(x_q^{\vec {v}})) = T_f(y^{\vec v'}_q)$ . Let $p' = \sigma ^{-\vec v'}(y)|_{F}$ and note that

$$\begin{align*}p^{\prime}_{\vec u} = \sigma^{-\vec v'}(y)_{\vec u} = y_{\vec u + \vec v'} = \sigma^{\vec w}(x)_{\vec u + \vec v'} = x_{\vec u + \vec v' - \vec w} = x_{\vec u + \vec v} = \sigma^{-\vec v}(x)_{\vec u} \end{align*}$$

for all $\vec u \in F$ so $p' = p$ . We have thus $f(p',q) = f(p,q) = (\tilde {p}, \tilde {q}, \tilde {\vec {u}})$ . Then by definition we have $T_f(y^{\vec v'}_q) = \tilde {y}^{\vec v' + \tilde {\vec u}}_{\tilde q}$ where

$$ \begin{align*}\tilde{y}_{\vec u} = \begin{cases} y_{\vec{u}} & \text{ if } {\vec u} - {\vec v'} \notin F\\ \tilde{p}_{{\vec u} - {\vec v}'}&\text{ if } {\vec u}-{\vec v}' \in F \end{cases}\end{align*} $$

We have to show that

$$\begin{align*}\tilde{y}^{\vec v' + \tilde{\vec u}}_{\tilde q} = \sigma^{\vec w}(\tilde{x}_{\tilde{q}}^{\vec{v}+\tilde{\vec{u}}}) = \sigma^{\vec w}(\tilde{x})_{\tilde{q}}^{\vec{v}+\tilde{\vec{u}} + \vec w}. \end{align*}$$

By definition, $\vec v' + \tilde {\vec u} = \vec v + \vec w + \tilde {\vec u}$ . We now show $\tilde {y} = \sigma ^{\vec w}(\tilde {x})$ , that is, $\tilde {y}_{\vec u} = \tilde {x}_{\vec u - \vec w}$ for all $\vec u \in \mathbb {Z}^d$ .

Let $\vec u \in \mathbb {Z}^d$ . First suppose ${\vec u} - {\vec v'} \notin F$ . Then $\tilde {y}_{\vec u} = y_{\vec u} = x_{\vec u - \vec w}$ . On the other hand $\sigma ^{\vec w}(\tilde {x})_{\vec u} = \tilde {x}_{\vec u - \vec w} = x_{\vec u - \vec w}$ where the last equality follows from the definition of $\tilde x$ because $\vec u - \vec w - \vec v = {\vec u} - {\vec v'} \notin F$ .

Suppose then that ${\vec u} - {\vec v'} \in F$ . Then $\tilde {y}_{\vec u} = \tilde {p}_{{\vec u} - {\vec v}'}$ . As above, by the definition of $\tilde x$ we have $\vec u - \vec w - \vec v = {\vec u} - {\vec v'} \in F$ so $\tilde {x}_{\vec u - \vec w} = \tilde {p}_{{\vec u} - \vec w - {\vec v}} = \tilde {p}_{{\vec u} - {\vec v}'}$ . This concludes the proof that $T_f$ commutes with the shift.

Take $m \geq 1$ and choose some r such that $F\subset [-r,r]^d$ . Given any $(x,c)\in X_{n,k}$ , we have by definition of $T_f$ that the values $T_f(x,c)|_{[-m,m]^d}$ depend exclusively on the values of $(x,c)$ in $[-m,m]^d+F \subset [-m-r,m+r]^d$ , from where it follows that $d(T_f(x,c),T_f(x',c'))\leq 2^{-m}$ whenever $d((x,c),(x',c'))\leq 2^{-(m+r)}$ , thus $T_f$ is continuous.

Proof. Consider a Turing machine $T \in \mathrm {TM}(\mathbb {Z}^d,n,k)$ seen as a function on $X_{n,k}$ . By Lemma 2.3 we have that $T \in \mathrm {End}(X_{n,k})$ . Also, T acts trivially on $X_{n,0}$ so $T|_Y = \mathrm {id}$ and if a configuration has a head, it can only be shifted but not disappear, thus $T^{-1}(Y) = Y$ . Moreover, if $T \in \mathrm {RTM}(\mathbb {Z}^d,n,k)$ , then T has a Turing machine inverse, thus a cellular automaton inverse, and it follows that $T \in \mathrm {Aut}(X_{n,k})$ .

Conversely, let $\phi \in \mathrm {End}(X_{n,k})$ , so that $\phi (x,y)_{\vec v} = \Phi (\sigma ^{-\vec v}(x,y)|_F)$ for some local rule $\Phi \colon (\Sigma \times \{0,\dots ,k\})^{F} \to \Sigma \times \{0,\dots ,k\}$ where $F$ is a finite subset of $\mathbb {Z}^d$ , and where we may suppose $\vec 0 \in F$ .

As $\phi |_Y = \mathrm {id}$ , we can deduce that $\Phi (u,v) = (u,v)_{\vec {0}}$ if $v = 0^F$ . Therefore if $(x,y) \in X_{n,k}$ , $y_{\vec v} \neq 0$ and we define $W_{\vec v} = \{\vec u : \vec v \in \vec u + F\} = \vec v - F$ we get that $\phi (x,y)|_{\mathbb {Z}^d \setminus W_{\vec v}} = (x,y)|_{\mathbb {Z}^d \setminus W_{\vec v}}$ . We extend $\Phi $ to $\widetilde {\Phi } \colon (\Sigma \times \{0,\dots ,k\})^{W_{\vec {0}} + F} \to (\Sigma \times \{0,\dots ,k\})^{W_{\vec {0}}}$ by pointwise application of $\Phi $ . Note that $W_{\vec {0}} = -F$ .

We can then define $f_{\phi } \colon \Sigma ^{F - F} \times Q \to \Sigma ^{-F} \times Q \times \mathbb {Z}^d$ by using $\widetilde {\Phi }$ as follows: We set $f_{\phi }(p,q) = (p',q',\vec u)$ if, after defining $r \in \{0,\dots ,k\}^{F - F}$ such that $r_{\vec {0}} = q$ and $0$ elsewhere, we have $\widetilde {\Phi }(p,r) = (p',r')$ and $r' \in \{0,\dots ,k\}^{-F}$ contains the symbol $q' \neq 0$ in position $\vec u$ (there is always a unique such position $\vec u$ as $\phi ^{-1}(Y) = Y$ ). It can be verified that the Turing machine $T_{f_{\phi }}$ is precisely $\phi $ ; therefore $\phi \in \mathrm {TM}(G,n,k)$ .

If $\phi \in \mathrm {Aut}(X_{n,k})$ then $\phi ^{-1}(Y) \supset Y$ is implied by $\phi |_Y = \mathrm {id}$ , and since the inverse automorphism $\phi ^{-1}$ satisfies $\phi ^{-1}|_Y = \mathrm {id}$ as well, we also have $\phi ^{-1}(Y) \subset Y$ . Thus, $\phi $ is a Turing machine. Similarly, the inverse map $\phi ^{-1}$ is a Turing machine. Thus, in this case $\phi \in \mathrm {RTM}(\mathbb {Z}^d,n,k)$ .

Proof. As the monoid of Turing machines can be seen as a submonoid of endomorphisms of a subshift, it follows that they commute with the shift. More precisely, for each machine T and $\vec {m} \in \mathbb {Z}$ if we interpret T acting on $X \times X_k$ by endomorphisms, then we have that $\sigma ^{\vec {m}} \circ T \circ \sigma ^{-\vec {m}} = T$ . In particular, as non- $\#$ symbols appear at a bounded distance in each configuration of $\mathrm {TM}(\sqrt [k]{X},1)$ , it suffices to define a Turing machine in $(\sqrt [k]{X} \setminus [\#]_0) \times \{1\} \times \mathbb {Z}$ to completely determine its action over $\sqrt [k]{X}\times \{1\}\times \mathbb {Z}$ .

Let $\psi \colon X \to \sqrt [k]{X}$ be defined by

$$\begin{align*}\psi(x)_i = \left\{ \begin{array}{ll} \#, & \text{if } i \not\equiv 0 \bmod k, \text{ and} \\ x_{i/k}, & \text{otherwise.} \end{array}\right. \end{align*}$$

Note that $\psi $ defines a $1$ -to- $1$ map onto $\sqrt [k]{X} \setminus [\#]_0$ .

Also, define $\eta \colon \mathbb {Z} \times Q \to \mathbb {Z}$ by $\eta (\vec {v},q) = k \vec {v}+q-1$ which is clearly bijective. We can thus finally define a bijection $\delta \colon X \times Q \times \mathbb {Z} \to (\sqrt [k]{X} \setminus [\#]_0) \times \{1\} \times \mathbb {Z}$ by $\delta (x_q^{\vec {v}}) = \psi (x)^{\eta (\vec {v},q)}_1$ .

Now, given a machine $T \in \mathrm {TM}(\sqrt [k]{X},1)$ we define $\varphi (T) \in \mathrm {TM}(X,k)$ by $\varphi (T) = \delta ^{-1} \circ T \circ \delta $ . Note that this is well-defined since $T \in \mathrm {TM}(\sqrt [k]{X},1)$ implies that $T(\sqrt [k]{X} \setminus [\#]_0 \times X_1) = \sqrt [k]{X} \setminus [\#]_0 \times X_1$ again because non- $\#$ -symbols appear with bounded gaps, and since T is the identity map on points where the head does not appear. By definition it is then clear that $\varphi (T \circ T') = \varphi (T) \circ \varphi (T')$ and that $\varphi $ is $1$ -to- $1$ . We only need to show that $\varphi (T) \in \mathrm {TM}(X,k)$ and that $\varphi $ is onto.

Firstly, it is clear that $\delta ^{-1} \circ T \circ \delta $ maps $X \times Q \times \mathbb {Z}$ to itself; therefore, the head cannot disappear. We have that $\delta \circ \sigma ^{\vec {m}} = \sigma ^{\vec {km}} \circ \delta $ and thus $\varphi (T)$ is shift commuting.

$$ \begin{align*} \varphi(T) \circ \sigma^{\vec{m}} &= \delta^{-1} \circ T \circ \delta \circ \sigma^{\vec{m}}\\ &= \delta^{-1} \circ T \circ \sigma^{\vec{km}} \circ\delta\\ &= \delta^{-1} \circ \sigma^{\vec{km}} \circ T \circ \delta\\ &= \sigma^{\vec{m}} \circ\delta^{-1} \circ T \circ \delta\\ &= \sigma^{\vec{m}} \circ \varphi(T). \end{align*} $$

Since $\delta $ and T are continuous, and $\delta ^{-1}$ is continuous on the image of $\delta $ , we have that $\varphi (T)$ is continuous and shift invariant and therefore defines an endomorphism of $X \times X_k$ which is an element of $\mathrm {TM}(X,k)$ . Conversely, an analogous argument shows that for each $T \in \mathrm {TM}(X,k)$ then the map $T'$ defined as $T' = \delta \circ T \circ \delta ^{-1}$ on $\sqrt [k]{X} \setminus [\#]_0 \times \{1\} \times \mathbb {Z}$ (and by conjugating with a suitable power of $\sigma $ on other points) is in $\mathrm {TM}(\sqrt [k]{X},1)$ and thus $\varphi (T') = T$ , showing that $\varphi $ is onto.

Finally, if $T \in \mathrm {RTM}(\sqrt [k]{X},1)$ then $\varphi (T)\circ \varphi (T^{-1}) = \mathrm {id}$ and thus $\varphi (T) \in \mathrm {RTM}(X,k)$ .

Proof. It is easy to see that $T_f$ for any local rule $f \colon \Sigma ^{F_{\text {in}}} \times Q \to \Sigma ^{F_{\text {out}}} \times Q \times \mathbb {Z}^d$ is continuous. The projection to the third component of f gives the function $-s$ , and one can take F as the minimal neighborhood of $T_f$ .

For the converse, since s is a continuous function from a compact space to a discrete one we conclude that the image of s is bounded. Furthermore it only depends on a finite set $F_0$ of coordinates of x. Since T is continuous, $T(x,q)|_{F + s(\Sigma ^{\mathbb {Z}^d} \times Q)}$ depends only on a finite set of coordinates $F_1$ of x. It is then easy to extract a local rule

$$\begin{align*}f \colon \Sigma^{F_0 \cup F_1} \times Q \to \Sigma^{F - s(\Sigma^{\mathbb{Z}^d} \times Q)} \times Q \times {\mathbb{Z}^d}, \end{align*}$$

for T.

Proof. If T is not injective, then $T(c_1)=T(c_2)$ for some $c_1\neq c_2$ . Because T is injective in each $\equiv $ -class, $c_1\not \equiv c_2$ . Hence there are two distinct $\equiv $ -classes with the same image so that $\Psi (T)$ is not injective. Conversely, if $\Psi (T)$ maps two distinct $\equiv $ -classes into the same class, there are $c_1\not \equiv c_2$ such that $T(c_1)\equiv T(c_2)$ . Because T is surjective in $\equiv $ -classes, there is $c^{\prime }_1\equiv c_1$ such that $T(c^{\prime }_1)=T(c_2)$ , so T is not injective.

If T is surjective, then clearly every $\equiv $ -class has a preimage so that $\Psi (T)$ is surjective. And if $\Psi (T)$ is surjective then every $\equiv $ -class has a preimage, and because T is surjective in $\equiv $ -classes, every element of every $\equiv $ -class has a preimage, that is, T is surjective.

Proof. Consider again the epimorphism $\Psi \colon \mathrm {TM}(\mathbb {Z}^d,n,k) \to \mathrm {TM}_{\mathrm {fix}}(\mathbb {Z}^d,n,k)$ and suppose there exists a pair $T \neq T'$ in $\mathrm {TM}(X,k)$ such that $\Psi (T) = \Psi (T')$ . Let $x_q^{\vec {0}}$ be such that $T(x_q^{\vec {0}}) \neq T'(x_q^{\vec {0}})$ . Denoting by $W = F \cup F'$ the union of the neighborhoods F of T and $F'$ of $T'$ we get that T and $T'$ can be described by rules of the form $f_T(x|_{W},q) = (p,r,\vec {v})$ and $f_{T'}(x|_{W},q) = (p',r',\vec {v}')$ . Denote by $x_{[p]}$ and $x_{[p']}$ the configuration where the symbols of x in the support W have been replaced by p and $p'$ , respectively. Clearly $\Psi (T) = \Psi (T')$ implies that $r = r'$ and $\sigma ^{-\vec {v}}(x_{[p]}) = \sigma ^{-{\vec {v}'}}(x_{[p']})$ .

If $\vec {v}=\vec {v}'$ then also $x_{[p]}=x_{[p']}$ , which contradicts $T(x_q^{\vec {0}}) \neq T'(x_q^{\vec {0}})$ . So we must have $\vec {v}\neq \vec {v}'$ . As X is strongly locally aperiodic, there exists $y\in [x|_{W}]$ such that $y_{\vec {u}}\neq y_{\vec {u}+\vec {v'}-\vec {v}}$ for some position $\vec {u}$ that satisfies $\vec {u},\vec {u}+\vec {v'}-\vec {v}\not \in W$ . As $y|_{W}=x|_{W}$ , we have that $\Psi (T)(y,q)=(\sigma ^{-\vec {v}}(y_{[p]}),r)$ and $\Psi (T')(y,q)=(\sigma ^{-\vec {v}'}(y_{[p']}),r')$ . Then $\Psi (T) = \Psi (T')$ implies that $y_{[p]}=\sigma ^{\vec {v}-\vec {v}'}(y_{[p']})$ , which is not true in position $\vec {u}$ .

Proof. Let T be arbitrary, and let F be its minimal neighborhood. Consider the cylinders $C_i = [p_i] \times \{q\}$ where $p_i \in \Sigma ^F, q \in Q$ . These cylinders form a clopen partition of $\Sigma ^{\mathbb {Z}^d} \times Q$ into $kn^{|F|}$ cylinders of measure $\frac {1}{kn^{|F|}}$ .

Now, because F is the minimal neighborhood of T, T is a homeomorphism from $C_i$ onto $D_i = T(C_i)$ , and $D_i$ is a cylinder set of the form $[p'] \times \{q'\}$ for some $p' \in \Sigma ^{\vec {v} + F}$ , $q' \in Q$ , which must be of the same measure as $C_i$ as the domain $\vec {v} + F$ of $p'$ has as many coordinates as the domain F of p. Note that $D_i$ is not necessarily a cylinder centered at the origin, and the offset $\vec v$ is given by the shift-indicator. Now, observe that injectivity is equivalent to the cylinders $D_i = T(C_i)$ being disjoint. Namely, they must be disjoint if T is injective, and if they are disjoint, then T is injective because $T|_{C_i} : C_i \to D_i$ is a homeomorphism. Surjectivity, on the other hand, corresponds to the equality $\Sigma ^{\mathbb {Z}^d} \times Q = \bigcup _i D_i$ , since $\bigcup _i D_i = \bigcup _i T(C_i) = T(\Sigma ^{\mathbb {Z}^d} \times Q)$ .

Now, it is easy to show that injectivity and surjectivity are equivalent. If T is injective, then the $D_i$ are disjoint, and $\mu (\bigcup _i D_i) = \sum _i \mu (D_i) = 1$ , so we must have $\bigcup _i D_i = \Sigma ^{\mathbb {Z}^d} \times Q$ because $\Sigma ^{\mathbb {Z}^d} \times Q$ is the only clopen set of full measure. If T is not injective, then for some $i \neq j$ we have $D_i \cap D_j \neq \varnothing $ . Then $D = D_i \cap D_j$ is a nonempty clopen set, and thus has positive measure. It follows that $\mu (\bigcup _i D_i) \leq \sum _i \mu (D_i) - \mu (D) < 1$ , so $\bigcup _i D_i \subsetneq \Sigma ^{\mathbb {Z}^d} \times Q$ . Of course, since injectivity and surjectivity are equivalent, they are both equivalent to bijectivity and thus to reversibility of T by Corollary 2.5.

The argument given above in fact shows that reversibility is equivalent to preserving the uniform Bernoulli measure in the forward sense – if T is reversible, then $\mu (T(A)) = \mu (A)$ for all clopen sets A, and thus for all Borel sets, while if T is not reversible, then there is a disjoint union of cylinders $C \cup D$ such that $\mu (T(C \cup D)) < \mu (C \cup D)$ .

For measure-preservation in the usual (backward) sense, observe that the reverse of a reversible Turing machine is reversible and thus measure-preserving in the forward sense, so a reversible Turing machine must itself be measure-preserving in the traditional sense. If T is not reversible, then $\mu (T(C \cup D)) < \mu (C \cup D)$ for some disjoint cylinders C and D large enough that $T|_C$ and $T|_D$ are measure-preserving homeomorphisms. Then for $E = T(C) \cap T(D)$ we have $\mu (T^{-1}(E)) \geq \mu ((T^{-1}(E) \cap C) \cup (T^{-1}(E) \cap D)) = 2\mu (E)$ .

Proof. If $T_1,T_2 \in \mathrm {RTM}_{\mathrm {fix}}(\mathbb {Z}^d,n,k)$ , then since reversibility implies measure-preservation, we have

$$ \begin{align*} \alpha(T_1 \circ T_2) &= \mathbb{E}_{\mu}(s_{T_1 \circ T_2}) \\ &= \mathbb{E}_{\mu}(s_{T_1}\circ T_2+s_{T_2}) \\ &= \mathbb{E}_{\mu}(s_{T_1}\circ T_2) + \mathbb{E}_{\mu}(s_{T_2}) \\ &= \mathbb{E}_{\mu}(s_{T_1})+ \mathbb{E}_{\mu}(s_{T_2}) \\ &= \alpha(T_1) + \alpha(T_2), \end{align*} $$

where $\mathbb {E}_{\mu }(s_{T_1 \circ T_2}) = \mathbb {E}_{\mu }(s_{T_1}\circ T_2+s_{T_2})$ holds because $s_{T_1 \circ T_2}(x, q) = s_{T_1}(T_2(x, q)) + s_{T_2}(x, q)$ for all $(x,q)\in \Sigma ^{\mathbb {Z}^d}\times Q$ .

Proof. For even n, by the note above, $[C_4^{(2)}]_m \subset \mathrm {Alt}(Q\times \Sigma ^{m})$ , and for odd n there are odd permutations in $[C_4^{(2)}]_m$ . So in both cases it is enough to show $\mathrm {Alt}(Q\times \Sigma ^{m}) \subset G_m$ . We also note that, obviously, $[G_{m-1}]_m\subset G_m$ .

Based on the decomposition in Figure 3, we first conclude that any controlled 3-cycle f of $Q\times \Sigma ^m$ is a composition of four applications of controlled swaps of $Q\times \Sigma ^{m-2}$ . In the figure, the components of $Q\times \Sigma ^m$ have been ordered in parallel horizontal wires, with the Q-component being among the topmost three wires. Referring to the symbols in the illustration, the gate on the left is a generic 3-cycle $(pszabcdw\; ptzabcdw\; qszabcdw)$ where one of the first three wires is the Q-component, $a,b,c,d\in \Sigma $ and $w\in \Sigma ^{m-6}$ . The proposed decomposition consists of two different controlled swaps $p_1$ and $p_2$ applied twice in the order $f=p_1p_2p_1p_2$ . Because $p_1$ and $p_2$ are involutions, the decomposition amounts to identity unless the input is of the form $xyz abcd w$ where $x\in \{p,q\}$ and $y\in \{s,t\}$ . When the input is of this form, it is easy to very that the circuit on the right indeed amounts to the required 3-cycle. We conclude that $C_m^{(3)}\subset \langle [C_{m-2}^{(2)}]_m \rangle $ , for all $m\geq 6$ . By Lemma 3.15,

(3.1) $$ \begin{align} \mathrm{Alt}(Q\times \Sigma^m) = \langle C_m^{(3)} \rangle \subset \langle [C_{m-2}^{(2)}]_m \rangle. \end{align} $$

Figure 3

A decomposition of a controlled 3-cycle of $Q\times \Sigma ^m$ on the left into a sequence of four applications of controlled swaps of $Q\times \Sigma ^{m-2}$ on the right. The ordering of the wires is such that topmost three wires contain the Q-component and the two wires changed by the 3-cycle (one of which may or may not be the Q-component). Black circles are control points: the gate computes the identity unless the wire carries the symbol indicated at the left of the wire or next to the control point.

We proceed by induction on m. The base case $m=6$ is clear: By (3.1),

$$\begin{align*}\mathrm{Alt}(Q\times \Sigma^6) \subset \langle [C_{4}^{(2)}]_6 \rangle = G_6.\end{align*}$$

Consider then $m>6$ and suppose that $G_{m-1}$ is as claimed. If n is odd, then by the inductive hypothesis,

$$ \begin{align*} [C_{m-2}^{(2)}]_m \subset [\mathrm{Sym}(Q\times \Sigma^{m-1})]_m \subset [G_{m-1}]_m \subset G_m. \end{align*} $$

By (3.1) then $\mathrm {Alt}(Q\times \Sigma ^m)\subset \langle [C_{m-2}^{(2)}]_m \rangle \subset G_m$ . As pointed out above, $G_m$ contains odd permutations (all elements of $[C_4^{(2)}]_m$ are odd), so $G_m=\mathrm {Sym}(Q\times \Sigma ^{m})$ as claimed.

If n is even, then an application of a permutation of $Q\times \Sigma ^{m-2}$ on $Q\times \Sigma ^m$ is also an application of an even permutation of $Q\times \Sigma ^{m-1}$ on $Q\times \Sigma ^m$ . (For this reason we left two noncontrolling wires for the gates on the right side of Figure 3.) By this and the inductive hypotheses,

$$\begin{align*}[C_{m-2}^{(2)}]_m \subset [\mathrm{Alt}(Q\times \Sigma^{m-1})]_m\subset [G_{m-1}]_m \subset G_m,\end{align*}$$

so, by (3.1), we have the required $\mathrm {Alt}(Q\times \Sigma ^m)\subset G_{m}$ .

Proof. If n is even then $[\mathrm {Sym}(Q\times \Sigma ^{m})]_{m+1} \subset \mathrm {Alt}(Q\times \Sigma ^{m+1})$ and if n is odd then $[\mathrm {Sym}(Q\times \Sigma ^{m})]_{m+1} \subset \mathrm {Sym}(Q\times \Sigma ^{m+1})$ . In either case, the claim follows from Lemma 3.16 and $C_4^{(2)}\subset \mathrm {Sym}(Q\times \Sigma ^4)$ .

Proof. We construct a finite generating set $A_1\cup A_2\cup A_3$ . Let $A_1$ contain the one step moves $T_{e_i}$ for $i=1,\dots ,d$ . These clearly generate $\mathrm {SHIFT}(\mathbb {Z}^d,n,k)$ .

Each $T\in \mathrm {LP}(\mathbb {Z}^d,n,k)$ is defined by a local rule $f\colon \Sigma ^F\times Q \rightarrow \Sigma ^F\times Q\times \{\vec {0}\}$ with a finite $F\subset \mathbb {Z}^d$ . To have injectivity, we clearly need that $\pi \colon (p,q)\mapsto (f_1(p,q), f_2(p,q))$ is a permutation of $\Sigma ^F\times Q$ . We denote $T=P_\pi $ . Let us fix an arbitrary $E\subset \mathbb {Z}^d$ of size $4$ , and let $A_2$ be the set of all $P_\pi \in \mathrm {LP}(\mathbb {Z}^d,n,k)$ determined by $\pi \in \mathrm {Sym}(\Sigma ^E\times Q)$ .

For any permutation $\alpha $ of $\mathbb {Z}^d$ with finite support, we define the cell permutation machine $C_\alpha \colon (p,q) \mapsto (p',q)$ , where $p^{\prime }_{\vec v}=p_{\alpha ({\vec v})}$ for all ${\vec v}\in \mathbb {Z}^d$ . These are clearly in $\mathrm {LP}(\mathbb {Z}^d,n,k)$ . We take $A_3$ to consist of the cell permutation machines $C_i=C_{(0\; e_i)}$ that, for each $i=1,\dots ,d$ , swaps the contents of the currently scanned cell and its neighbor with offset $e_i$ .

Observe that $A_1$ and $A_3$ generate all cell permutation machines $C_\alpha $ . First, conjugating $C_i$ with ${T_{\vec v}\in \mathrm {SHIFT}(\mathbb {Z}^d,n,k)}$ gives the cell permutation machine $C_\alpha =T_{\vec v}^{-1}C_iT_{\vec v}$ for the transposition ${\alpha =({\vec v}\; {\vec v} + e_i)}$ . Such transpositions generate all permutations of $\mathbb {Z}^d$ with finite support. This follows from Lemma 3.14(a) by considering a finite connected grid graph containing the support of the permutation.

Consider then an arbitrary $P_\pi \in \mathrm {LP}(\mathbb {Z}^d,n,k)$ , where $\pi \in \mathrm {Sym}(\Sigma ^F\times Q)$ . We can safely assume $|F|\geq 5$ . Let us pick one ancilla $v\in \mathbb {Z}^d\setminus F$ and denote $F'=F\cup \{v\}$ . By Corollary 3.17, $P_\pi $ is a composition of machines of type $P_\rho $ for $\rho \in \mathrm {Sym}(\Sigma ^H\times Q)$ where $H\subset F'$ has size $|H|=4$ . It is enough to be able to generate these. Let $\alpha $ be a permutation of $\mathbb {Z}^d$ that exchanges E and H, two sets of cardinality four. Then $C_\alpha ^{-1} P_\rho C_\alpha \in A_2$ , which implies that $P_\rho $ is generated by $A_1\cup A_2\cup A_3$ .

Proof. Let $u', v'$ be two words of the same length in $\sqrt [k]{X}$ . Then possibly by extending $u'$ and $v'$ we may write them as subwords of words $u"$ and $v"$ of the form

$$\begin{align*}u' \sqsubset u" = \#^{n-1} u_1 \#^{k-1} u_2 \#^{k-1} u_3 \cdots u_{n} \#^{k-1},\end{align*}$$

$$\begin{align*}v' \sqsubset v" = \#^{k-1} v_1 \#^{k-1} v_2 \#^{k-1} v_3 \cdots v_{n} \#^{k-1},\end{align*}$$

for some $n \in \mathbb {N}$ and words $u,v$ in $L_n(X)$ . Since X is weakly chain-transitive, there is a finite chain $u = w_0 \sim w_1 \sim \cdots \sim w_k = v$ such that $w_i$ and $w_{i+1}$ occur in the same point of X for all i. Then the corresponding interspersed versions of the $w_i$ give a chain between $u"$ and $v"$ , and thus between $u'$ and $v'$ .

Proof. Suppose that $f = g \circ \pi $ for some $g \colon \widehat X \to \mathbb {Z}$ . Then f is continuous as the composition of two continuous functions, and

$$\begin{align*}f(x) = g(\pi(x)) = g(\pi(\sigma(x))) = f(\sigma(x)), \end{align*}$$

so f is an abstract orbitwise shift.

On the other hand, if $f \colon X \to \mathbb {Z}$ is continuous and $f(x) = f(\sigma (x))$ , then the map $g \colon \widehat X \to \mathbb {Z}$ given by the formula $g(\pi (x)) = f(x)$ for all $x \in X$ is well defined: suppose not, and that $\pi (x) = \pi (y)$ and $f(x) \neq f(y)$ . Since f is continuous and X is compact, $f(z)$ depends only on finitely many coordinates of z, that is, there exists n such that $f(z) = f(z')$ whenever $z_{\{-n,\dots ,n\}} = z_{\{-n,\dots ,n\}}$ . Since $\pi (x) = \pi (y)$ , $u = x_{\{-n,\dots ,n\}}$ and $v = y_{\{-n,\dots ,n\}}$ are connected by a finite chain $u = w_0 \sim w_1 \sim \cdots \sim w_k = v$ such that $w_i$ and $w_{i+1}$ occur in the same configuration of X for all $0\leq i<k$ .

Since the image of f is determined by the restriction to the central $2n+1$ coordinates, we can write $n_i = f(w_i)$ for this unique value. Now, since $f(z) = f(\sigma (z))$ for all $z \in X$ , it is easy to see that in fact $n_i = n_{i+1}$ for all i, by moving along orbits of points connecting the words $w_i$ and $w_{i+1}$ . This shows that g is well-defined. Furthermore, as the image of f is determined by the restriction to the central $2n+1$ coordinates, it follows easily that g is continuous.

Proof. Consider an arbitrary $\phi $ in the permutation model acting on $X \times (\mathbb {Z}/2\mathbb {Z})^{\mathbb {Z}}$ with biradius r. We first claim that if $|u| \geq 4r$ and $au \sqsubset X$ , then $L_{\phi }(au) = L_{\phi }(u)$ . To see this, note that all heads except the leftmost new one map exactly as before, and thus $R^*_{\phi }(au) = R^*_{\phi }(u)$ . As $R^*_{\phi }(v) + L^*_{\phi }(v) = 2r$ for every $v \in L(X)$ we conclude that $L^*_{\phi }(au)=L^*_{\phi }(u)$ and thus $L_{\phi }(au) = L_{\phi }(u)$ . Symmetrically, one shows that $L_{\phi }(ua) = L_{\phi }(u)$ .

Figure 4

For $T = \sigma ^2$ its permutation model $\phi _{\sigma ^2}$ moves the heads to the right twice. If we choose $r = 4$ we observe that the sum of the red zones is $2r$ , the blue zone remains unchanged, $L_{\phi }(u) = -2$ and $R_{\phi }(u)=2$ .

Suppose now that $u, v \sqsubset x \in X$ . Let $w \sqsubset x$ be any word containing both u and v, and apply the observation of the previous paragraph repeatedly to get $L_{\phi }(u) = L_{\phi }(w) = L_{\phi }(v)$ . It follows that $H_T(x) = H_T(\sigma (x))$ for all x. The function $H_T$ is continuous because the limit stabilizes after $n> 4r$ . We conclude that $H_T \in \mathrm {AOS}(X)$ .

To see that $\gamma (T) = H_T$ is a homomorphism, we need the following stronger fact about $L_{\phi }$ : fix $m \in \mathbb {N}$ and let $(x, y) \in X \times (\mathbb {Z}/2\mathbb {Z})^{\mathbb {Z}}$ be a configuration where $x_{[0,|u|-1]} = u$ where $|u| \geq m+4r$ , and let $y = ...000.0^rv1^{|u|-|v|-r}000...$ where $v\in (\mathbb {Z}/2\mathbb {Z})^m$ is any word of length m. Then the number of heads in ${\phi }_2(x, y)|_{\{0,\dots ,m+2r-1\}}$ is $|v^{-1}(1)| + r + H_{T}(x)$ . To see this, simply fill in the gaps of v (turn its $0$ -symbols to $1$ -symbols). At each step, the number of heads in the image increases by one, and the head is always added among the coordinates in $\{0,\dots ,m+2r-1\}$ since ${\phi }$ has biradius r. After filling the gaps in v, we have $m+r+H_T(x)$ heads in ${\phi }_2(x, y)|_{\{0,\dots ,m+2r-1\}}$ by the definition of $L_{\phi }$ .

Now, computation of $\gamma (T \circ T')(x)$ can be done by first applying $\phi '=\phi _{T'}$ on the left side of a long word u in x, with heads positioned suitably on coordinates of u, and then applying $\phi =\phi _T$ to the resulting scattered set of heads. By the previous paragraph, we see that $\gamma $ is a homomorphism, that is, $\gamma (T \circ {T'})(x) = \gamma (T)(x) + \gamma (T')(x)$ for all $x \in X$ . The map $\beta $ is well-defined essentially by the definition of $\mathrm {AOS}(X)$ , and it is clearly a section for $\gamma $ .

Proof. Since $\widehat {X}$ is a singleton, Lemma 3.22 implies that $\mathrm {AOS}(X) \cong \mathbb {Z}$ and therefore $\beta \circ \gamma $ must map every element of $\mathrm {RFA}(X)$ to a power of $\sigma $ by the definition of $\beta $ .

Proof. The cylinders $[u.av]$ and $[ua.v]$ have empty intersection when $uav$ is nonunary. Thus, the conditions under which we move the head do not overlap, and the finite-state machine $T_{u,a,v}$ is well-defined. It is clearly an involution, thus invertible.

Proof. For $T \in \mathrm {RFA}(X)$ , let $P_T \subset \{(0,0)\} \cup \mathbb {Z}_+^2$ be the set of all pairs $(m, k)$ such that there exists $x \in X$ and $i \in \mathbb {Z}$ such that $T(x^{i+j}_1) = x^{i+j+m}_1$ for all $0 \leq j < k$ . Order these pairs lexicographically (m is more significant). The order type is a suborder of $1 + \omega ^2 \cong \omega ^2$ , thus well-founded. It is now enough to prove that, whenever T has zero average movement,

1. 1. $P_T$ has a maximal element $M_T$ ,

2. 2. if $M_T> (0,0)$ , there exists controlled position swap $T'$ such that $M_{T \circ T'} < M_T$ , and

3. 3. T is the identity map if and only if $M_T = (0,0)$ .

Namely, by well-foundedness we must reach $(0,0)$ in finitely many steps by iterating the second item, and by the third we have reached the identity map.

Fix now $T \in \mathrm {RFA}(X)$ . To see that a maximum $M_T = (m, k)$ exists, first observe that $P_T \neq \emptyset $ since $(0,0) \in P_T$ . Next, for $(m, k) \in P_T$ the local rule of T clearly gives a finite upper bound on m. Finally, if m is maximal and $(m,k) \in P_T$ for arbitrarily large k, then $m> 0$ (since $P_T \subset \{(0,0)\} \cup \mathbb {Z}_+^2$ by choice), and by compactness there exists a configuration $x \in X$ satisfying $T(x^i_1) = x^{i+m}_1$ for all $i \in \mathbb {Z}$ , thus average movement on x is clearly nonzero.

Suppose now that $M_T = (m,k)> (0,0)$ . Now, let $C \subset X$ be the set of configurations x such that $\forall i \in [-k+1, 0]: T(x^i_1) = x^{i+m}_1$ . Observe that $x \in C \implies T(x^1_1) = x^{1+j}_1$ where $j \leq m-2$ , namely, $j \leq m-1$ by the maximality of k, and $j = m-1$ would contradict bijectivity of T. Clearly C is a clopen set, and $C \cap \sigma (C) = \varnothing $ because $k> 0$ . We let $T' = T_C$ .

Figure 5

The composition $T \circ T'$ illustrated with $m = 3$ , $k = 4$ . The movement of heads in the permutation model on a configuration x is shown, and values at the nodes are those of y (central row) and z (top row). The four nodes forming $m^k$ are highlighted. The configuration x determines these moves, but its contents are not shown.

Consider now an arbitrary point $x \in X$ and $i \in \mathbb {Z}$ .

1. 1. If $\sigma ^{-i}(x) \notin C$ and $\sigma ^{-i+1}(x) \notin C$ , then $(T \circ T') (x^i_1) = T(x^i_1)$ .

2. 2. If $\sigma ^{-i}(x) \in C$ , then $(T \circ T')(x^i_1) = T(x^{i+1}_1)$ .

3. 3. If $\sigma ^{-i+1}(x) \in C$ , then $(T \circ T')(x^i_1) = T(x^{i-1}_1)$ .

Defining $y, z \in \mathbb {Z}^{\mathbb {Z}}$ by $y_i = j$ where $T(x^i_1) = x^{i+j}_1$ and $z_i = j$ where $(T \circ T')(x^i_1) = x^{i+j}_1$ , the difference between y and z is precisely that all subsequences of the form $m^k j \in \mathbb {Z}^{k+1}$ are replaced by $m^{k-1} (j+1) (m-1)$ . Since $j \leq m-2$ , the run of ms become strictly shorter, and no symbols larger than m are introduced, thus $M_{T \circ T'} < M_T$ are required. See Figure 5 for an illustration.