Proof. Since for each $k \in \mathbb {N}$ we have $f(C_k(K))= f(C_k)(K)$ (cf. for example [Reference Ceccherini-Silberstein, Coornaert and Phung14, Lemma A.22(v)]), the second equality is verified. For the first equality, we have trivially $f ( \bigcap _{k \in \mathbb {N}} C_k(K) ) \subset \bigcap _{k \in \mathbb {N}} f(C_k(K))$ . Conversely, assume that $y \in \bigcap _{k \in \mathbb {N}} f(C_k(K))$ . For each $k \in \mathbb {N}$ , set

Note that $F_k$ is the set of closed points of a constructible subset of X. Remark also that, for every $k \in \mathbb {N}$ , we have $F_{k+1} \subset F_k$ and $F_k \not = \varnothing $ . Hence, by [Reference Ceccherini-Silberstein, Coornaert and Phung14, Lemma B.3], there exists $x \in \bigcap _{k \in \mathbb {N}} F_k$ . Clearly, $f(x)=y$ and $x \in \bigcap _{k \in \mathbb {N}} C_k(K)$ . Therefore, $ \bigcap _{k \in \mathbb {N}} f(C_k(K)) \subset f ( \bigcap _{k \in \mathbb {N}} C_k(K) )$ and the proof is completed.

Proof. Since I is a countable directed set, we can find a totally ordered cofinal subset ${\{i_n : n \in \mathbb {N}\} \subset I}$ . As $\varprojlim _{n \in \mathbb {N}} Z_{i_n} = \varprojlim _{i \in I} Z_i$ , we can suppose, without any loss of generality, that $I= \mathbb {N}$ .

For each $i \in \mathbb {N}$ , we can find a decreasing sequence of constructible subsets $(C_{ik})_{k \in \mathbb {N}}$ of $X_i$ such that $C_i= \bigcap _{k \in \mathbb {N}} C_{ik}$ . For $k \in \mathbb {N}$ , let $Z_{ik}= C_{ik}(K)$ . By Lemma 3.1, we have for every $i \leq j$ :

(3.1) $$ \begin{align} Z_i= \bigcap_{k = 0}^{\infty} Z_{ik} \neq \varnothing, \quad f_{ij}(Z_j) = \bigcap_{k=0}^{\infty} f_{ij} (Z_{jk}). \end{align} $$

Consider the universal inverse system $(Z^{\prime }_i, \varphi ^{\prime }_{ij})_{i,j \in \mathbb {N}}$ of the system $(Z_i, \varphi _{ij})_{i,j \in \mathbb {N}}$ , that is, for every $i \in \mathbb {N}$ , let

and let the maps $\varphi ^{\prime }_{ij} \colon Z^{\prime }_{j} \to Z^{\prime }_i$ be the restrictions of $\varphi _{ij} \colon Z_j \to Z_i$ .

Remark that $\varprojlim _{i \in \mathbb {N}} Z^{\prime }_i= \varprojlim _{i \in \mathbb {N}} Z_i$ . Hence, it suffices to check that the sets $Z^{\prime }_i$ are non-empty and the transition maps $\varphi ^{\prime }_{ij}$ are surjective for all $i \leq j$ . By equation (3.1), Chevalley’s theorem (see for example [Reference Vakil57, Theorem 7.4.2], [Reference Grothendieck22, Théorème I.8.4]) implies that each $Z^{\prime }_i$ is a countable intersection of constructible sets:

$$ \begin{align*} Z^{\prime}_i = \bigcap_{j=i}^\infty f_{ij}(Z_{j}) = \bigcap_{j = i}^\infty \bigcap_{k=0}^\infty f_{ij} (Z_{jk}). \end{align*} $$

For each $n \geq i$ , consider the diagonal set

By Chevalley’s theorem, $Y_n$ is a constructible subset of $X_i(K)$ . For every $n \geq i$ , we have $Y_{n+1} \subset Y_n$ and since $Z_n \ne \varnothing $ ,

(3.2) $$ \begin{align} Y_n \supset \bigcap_{j=i}^n \bigcap_{k = 0}^\infty f_{ij}(Z_{jk}) = \bigcap_{j=i}^n f_{ij} (Z_j) \supset f_{in}(Z_n) = \varphi_{in}(Z_n) \neq \varnothing. \end{align} $$

As $Z^{\prime }_i = \bigcap _{n=i}^\infty Y_n$ , [Reference Ceccherini-Silberstein, Coornaert and Phung14, Lemma B.2] implies that $Z^{\prime }_i \neq \varnothing $ for $i \in \mathbb {N}$ . Now let $k, i \in \mathbb {N}$ with $k \leq i$ and let $z \in Z^{\prime }_k$ . For each $n \geq i$ , by definition of $Z^{\prime }_k$ , there exists $y \in Z_n$ such that $\varphi _{k n}(y)=z$ and thus

(3.3) $$ \begin{align} \varphi_{i n}(y)\in \varphi_{k i}^{-1} (z) \cap \varphi_{i n} (Z_n) \neq \varnothing. \end{align} $$

By equations (3.2), (3.3), and for $n \geq i$ , the constructible subset

(3.4)

is non-empty and $T_{n+1} \subset T_n$ as $Y_{n+1} \subset Y_n$ . Finally, we find that

$$ \begin{align*} (\varphi_{k i}')^{-1} (z) & = \varphi_{k i}^{-1} (z) \cap Z^{\prime}_i = \bigcap_{n=i}^\infty \varphi_{k i}^{-1} (z) \cap Y_n = \bigcap_{n=i}^\infty T_n \end{align*} $$

is non-empty by [Reference Ceccherini-Silberstein, Coornaert and Phung14, Lemma B.2]. The proof is thus completed.

Proof. The statement is proved in [Reference Phung39, Proposition 4.2].

Proof. Let $x \in \Sigma $ and $g \in G$ , then clearly $(g^{-1}x) \vert _E \in \Sigma _E$ . Thus, $\Sigma \subset \Sigma (E, \Sigma _E)$ . Conversely, let $x \in \Sigma (E, \Sigma _E)$ and $g \in G$ , then $(g^{-1}x)\vert _D = ((g^{-1}x)\vert _E)\vert _D\in (\Sigma _E)_D \subset P$ since $D \subset E$ . Therefore, $x \in \Sigma (D,P)=\Sigma $ and the conclusion follows.

Proof. Since $\Sigma _{i j} \subset A_{ij}$ for all $i, j\in \mathbb {N}$ , each $x \in \Sigma $ defines naturally an element $\Psi _j(x) = (x\vert _{M^{i+j}})_{i \in \mathbb {N}}\in \varprojlim _{i \in \mathbb {N}} (A_{ij}, p_{ijk})$ . Conversely, let $(x_i)_{i \in \mathbb {N}} \in \varprojlim _{i \in \mathbb {N}} (A_{ij}, p_{ijk})$ . Define $x \in ~A^G$ by setting, for each $g \in G$ , for any $i \in \mathbb {N}$ large enough such that $g \in M^{i + j}$ . The fact that the configuration $x \in A^G$ is well defined follows from properties (Mem1) and (Mem2). Let $g \in G$ . Take i large enough so that $g D \subset M^{i + j}$ . Then $g \in D_{i j}$ and $(g^{-1} x)\vert _D = (g^{-1} x_i)\vert _D \in P$ since $x_i \in A_{i j}$ . This shows that $x \in \Sigma $ .

Proof. Let $y \in \Sigma $ and let $x= y\vert _{M^{i+j}} \in \Sigma _{ij}$ . Let $k \geq i$ . Since $y\vert _{M^{k+j}} \in \Sigma _{M^{k+j}} \subset A_{k j}$ , it follows that

$$ \begin{align*} x = y\vert_{M^{i+j}} = p_{ijk}(y\vert_{M^{k+j}}) \subset p_{ijk}(A_{k j}). \end{align*} $$

As y is arbitrary, the proof is finished.

Proof. The inclusion $\Sigma _{ij} \subset \bigcap _{k \geq i} p_{ijk}(A_{k j})$ follows from Lemma 5.4.

Let $x \in \bigcap _{k \geq i} p_{ijk}(A_{k j}) \subset A^{M^{i+j}}$ . We must show that x can be extended to an element of $\Sigma $ . Consider the following inverse system lying above x. Let $B_i=\{x\}$ and for each $k \geq i$ , we set

(6.3)

which is a closed algebraic subset of $A^{M^{k+j}}$ . Since $x \in p_{ijk}(A_{k j})$ , each set $B_{k}$ is non-empty. From equation (6.3), it is clear that for every $k \geq i$ , we have

$$ \begin{align*} \widetilde{p}_{kj}(B_{k+1} ) \subset B_k. \end{align*} $$

By restricting the map $\widetilde {p}_{kj}$ to $B_{k+1}$ , we have for each $k \geq i$ a well-defined algebraic map $\pi _k \colon B_{k+1} \to B_k$ . Thus, we obtain an inverse subsystem $(B_k)_{k \geq i}$ with transition maps $\pi _{nm} \colon B_m \to B_n$ , where $m \geq n \geq i$ , as compositions of the maps $\pi _k$ .

We claim that $\varprojlim _{k \geq i} B_k \neq \varnothing $ . Indeed, this follows from Lemma 3.2 if case (H1) is satisfied and from Lemma 3.3 and Remark 2.1 in case (C2). Suppose now that case (C3) is satisfied. Since $x \in \bigcap _{k \geq i} p_{ijk}(A_{k j})$ , there exists for each $k~\geq ~i$ a point $z_k \in B_k$ such that $p_{ijk}(z_k)=x$ . Let $V_k = \ker p_{ijk}$ be an algebraic subgroup of $A^{M^{k+j}}$ , then clearly ${B_k = z_k V_k}$ where the group law is written multiplicatively. For all integers $m \geq n \geq i$ , the map $\pi _{mn}$ is the restriction of a homomorphism of algebraic groups (cf. Remark 4.3). Therefore, $\pi _{nm}(B_m)$ is a translate of an algebraic subgroup of $A^{M^{n+j}}$ and thus is Zariski closed in $B_n$ (cf. Remark 2.1). Hence, the claim follows, also in case (C3), from Lemma 3.3.

Therefore, we can find $(y_k)_{k \geq i} \in \varprojlim _{k \geq i} B_k$ . Let $y \in A^{G}$ be defined as follows. Given $g \in G$ , set $y(g)= y_k(g)$ for any $k \geq i$ such that $g \in M^{k+j}$ . Then y is well defined by property (Mem2). For each $g \in G$ , choose $k \geq i$ so that $gD \subset M^{k+j}$ . Then $(g^{-1}y)\vert _D = y_k\vert _{gD} \in W(K)$ which follows from the definition of $A_{kj}$ and since $y_k \in B_{kj} \subset A_{kj}$ . Hence, $y \in \Sigma $ . By construction, $x= y \vert _{M^{i+j}}$ and we deduce that $\bigcap _{k \geq i} q_{ijk}(A_{k j}) \subset \Sigma _{M^{i+j}}$ . The proof of equation (6.2) is completed. Thus, by Chevalley’s theorem, $\Sigma _{i j}$ is a countably proconstructible subset of $A^{M^{i+j}}$ .

Finally, the last statement follows from equation (6.2) and Remark 2.1 and Noetherianity of the Zariski topology of $A^{M^{i+j}}$ . Note that the sequence $(q_{ijk}(A_{k j}))_{k \geq i}$ is trivially a descending sequence.

Proof. Let $i, j \in \mathbb {N}$ be large enough so that $E \subset M^{i+j}$ . Let $\pi \colon A^{M^{i+j}} \to A^E$ be the induced projection. It follows that $\Sigma _E = \pi (\Sigma _{ij})$ . In cases (C2) and (C3), Proposition 6.2 and Remark 2.1 imply that $\Sigma _E$ is respectively a complete subvariety and an algebraic subgroup of $A^E$ . In case (H1), we find by Lemma 3.1 that

(6.4) $$ \begin{align} \Sigma_E = \pi (\Sigma_{ij}) = \pi \bigg(\bigcap_{k \geq i} p_{ijk} (A_{kj}) \bigg) = \bigcap_{n \in \mathbb{N}} \pi (p_{ijk} (A_{kj}) ). \end{align} $$

Hence, $\Sigma _E$ is countably proconstructible by Chevalley’s theorem. The proof is completed.

Proof. By hypothesis, there exist in cases (H1) and (H2) an algebraic variety (respectively in case $(\mathrm {\widehat {H3}})$ an algebraic group) U over K, an algebraic (respectively algebraic group) cellular automaton $\tau ' \colon B^G \to A^G$ where $B= U(K)$ , and an algebraic (respectively algebraic group) subshift of finite type $\Sigma ' \subset B^G$ such that $\Sigma = \tau ' (\Sigma ')$ . Note that $U,V$ are complete varieties in case (H2). Let M be a memory set of $\tau '$ . By Corollary 6.3, the set $\Sigma ^{\prime }_{ME}$ is countably proconstructible. Hence, $\Sigma ^{\prime }_{ME} = \bigcap _{n \in \mathbb {N}} C_n$ where $(C_n)_{n \in \mathbb {N}}$ is some decreasing sequence of constructible subsets of $A^{ME}$ . Let $\varphi \colon B^{ME} \to A^E$ be given by $\varphi (x)(g)= \tau '(y)(g)$ for every $x \in B^{ME}$ , $g \in E$ and every $y \in B^G$ extending x. Then $\varphi $ is algebraic (cf. [Reference Ceccherini-Silberstein, Coornaert and Phung14, Lemma 3.2]) and in case $(\mathrm {\widehat {H3}})$ , it is a homomorphism of algebraic groups (cf. [Reference Phung39, Lemma 3.4]). In case (H1), we can conclude by Chevalley’s theorem since

(7.1) $$ \begin{align} \Sigma_E = (\tau'(\Sigma'))_E = \varphi(\Sigma^{\prime}_{ME}) = \varphi \bigg(\bigcap_{n \in \mathbb{N}} C_n \bigg) = \bigcap_{n \in \mathbb{N}} \varphi (C_n), \end{align} $$

where the last equality follows from Lemma 3.1. Finally, in cases (H2) and $(\mathrm {\widehat {H3}})$ , Corollary 6.3 implies that $\Sigma _E = \varphi (\Sigma ^{\prime }_{ME})$ is respectively a complete subvariety and an algebraic subgroup of $A^E$ .

Proof. It is clear that, up to replacing $\tau $ by the composition $\tau \circ \tau '$ and $\Sigma $ by $\Sigma '$ , we can suppose without loss of generality that $\Sigma $ is an algebraic subshift of finite type. The hypotheses $(\mathrm {\widetilde {H2}})$ , $(\mathrm {\widetilde {H3}})$ now become respectively:

1. (P2) $V_0$ is a complete K-algebraic variety;

2. (P3) $V_0$ and $V_1$ are K-algebraic groups, $\Sigma \subset A_0^G$ is an algebraic group subshift of finite type, and $\tau \colon A_0^G \to A_1^G$ is an algebraic group cellular automaton.

Let $D \subset G$ be a defining memory set of $\Sigma $ . Let $d \in A_1^G$ be in the closure of $\tau (\Sigma )$ . We must show that $d \in \tau (\Sigma )$ .

Suppose first that G is finitely generated. Let $M \subset G$ be a finite memory subset of $\tau $ containing $\{1_G\} \cup D$ which generates G and satisfies $M=M^{-1}$ . Consider the inverse system $(A_0^{M^i})_{i\in \mathbb {N}}$ whose transition maps $p _{ij}\colon A_0^{M^j} \to A_0^{M^i}$ , where $0 \leq i \leq j$ , are defined as the canonical projections induced by the inclusions $M^i \subset M^j$ . For every $i \geq 1$ , the induced map $q_i \colon A_0^{M^i} \to A_1^{M^{i-1}}$ is given as follows. For every $\sigma \in A_0^{M^i}$ , we set , where $x \in A_0^G$ is any configuration that extends $\sigma $ . For every $i \geq 1$ , we define

Since d belongs to the closure of $\tau (\Sigma )$ in $A_1^G$ , it follows that $Z_i \neq \varnothing $ for every $i \geq 1$ . By restricting the projections $p_{ij} \colon A^{M^j} \to A^{M^i}$ to $Z_j$ , we obtain well-defined transition maps $\pi _{ij} \colon Z_j \to Z_i$ , where $j \geq i \geq 1$ , of the inverse system $(Z_i)_{i \geq 1}$ .

It suffices to show that $\varprojlim _{i \geq 1} Z_i \neq \varnothing $ since, by construction of $Z_i$ and $\Sigma _{ij}$ (see also [Reference Ceccherini-Silberstein, Coornaert and Phung14, Lemma 2.1]), we have $\tau (c) = d$ for every $c \in \varprojlim _{i \geq 1} Z_i \subset \varprojlim _{i \geq 1} \Sigma _{M^{i+1}} = \Sigma $ (by equation (4.6) since $\Sigma $ is closed as it is a subshift of finite type).

Thanks to Theorem 7.1, the conclusion follows by a direct application of Lemma 3.2, respectively Lemma 3.3, to the inverse system $(Z_i)_{i \in \mathbb {N}}$ if case $(\mathrm {\widetilde {H1}})$ , respectively case (P2), is satisfied. Assume now that case (P3) is satisfied. For each $i \geq 1$ , choose $z_i \in Z_i$ and let be an algebraic subgroup of $\Sigma _{M^i}$ (by Theorem 7.1). We have $Z_i= z_i V_i$ . Hence (by Remark 4.3), for $j \geq i \geq 1$ , $\pi _{ij}(Z_j) $ is a translate of an algebraic subgroup of $\Sigma _{M^i}$ and thus is Zariski closed in $Z_i$ . Therefore, case (P3) follows from Lemma 3.3.

For a general group G, consider a finite memory set M of $\tau $ containing $\{1_G\} \cup D$ and such that $M=M^{-1}$ . Let $H \subset G$ be the subgroup generated by M. As $\Sigma _H$ is clearly an algebraic (respectively in case $(\mathrm {\widetilde {H3}})$ an algebraic group) subshift of finite type, the above discussion shows that $\tau _H(\Sigma _H)$ is closed in $A_1^G$ and so is $\tau (\Sigma )$ by Lemma 2.10.

Proof. It suffices to apply Theorem 8.1 in the case $V_0=V_1$ to the identity map $\tau = \operatorname {\mathrm {Id}}_{A^G}$ , where $A = V_0(K) = V_1(K)$ .

Proof. By Proposition 2.3, we only need to check that $\tau ^n(\Sigma )$ is closed in $A^G$ for every $n \geq 1$ and that $\tau $ is uniformly continuous. The first property follows from Theorem 8.1. The second is a general property of cellular automata already mentioned in §1.

Proof. By Corollary 8.2, the subshift $\Sigma $ is closed in $A^G$ . Hence, $\varprojlim _{i,j \in \mathbb {N}} \Sigma _{ij}$ is the set of backward orbits of $\tau $ and we have a canonical map $\Phi \colon \varprojlim _{i,j \in \mathbb {N}} \Sigma _{ij} \to \Omega (\tau )$ given in Lemma 4.4. Now let $y_0 \in \Omega (\tau ) \subset \Sigma $ . We must show that there exists $x \in \varprojlim _{i,j \in \mathbb {N}} \Sigma _{ij} $ such that $\Phi (x)=y_0$ . For every $i, j \in \mathbb {N}$ , define a closed subset

By definition of $\Omega (\tau )$ , there exists for every $j \in \mathbb {N}$ an element $y_j \in \Sigma $ such that ${\tau ^j (y_j)=y_0}$ . Hence, it follows from the definition of the transition maps $q_{ik}$ and of $\Sigma _{ij}$ that $y_j\vert _{M^{i+j}} \in B_{ij}$ . In particular, $B_{ij} \neq \varnothing $ for every $i, j \in \mathbb {N}$ . By restricting the transition maps of the space-time inverse system $(\Sigma _{ij})_{i, j \in \mathbb {N}}$ to the sets $B_{ij}$ , we obtain a well-defined inverse subsystem $(B_{ij})_{i, j \in \mathbb {N}}$ .

We claim that $\varprojlim _{i, j \in \mathbb {N}} B_{ij} \neq \varnothing $ . Indeed, by Theorem 7.1, case (H1) is implied by Lemma 3.2. In case (H2), Theorem 7.1 implies that $B_{ij}$ is a complete algebraic subvariety of $\Sigma _{ij}$ and thus of $A^{M^{i+j}}$ . Hence, case (H2) follows from Lemma 3.3 and Remark 2.1. In case (H3), a similar argument as in the proof of Proposition 6.2 shows that the transition maps of the system $(B_{ij})_{i, j \in \mathbb {N}}$ have Zariski closed images. Therefore, case (H3) follows immediately from Lemma 3.3. Thus, we can find

$$ \begin{align*} x \in \varprojlim_{i, j \in \mathbb{N}} B_{ij} \subset \varprojlim_{i, j \in \mathbb{N}} \Sigma_{ij}. \end{align*} $$

It is clear from the constructions of the inverse system $(B_{ij})_{i, j \in \mathbb {N}}$ and of the map $\Phi $ (see the proof of Lemma 4.4) that $\Phi (x)= y_0$ . The proof of the lemma is completed.

Proof. Let $M\subset G$ be a finite subset containing $1_G$ , a memory set of $\tau $ , and a memory set of $\Sigma $ and such that $M=M^{-1}$ . Let $H \subset G$ be the subgroup generated by M. Since $\tau = \prod _{c \in G/H} \tau _c$ and $\Omega (\tau )= \prod _{c \in G /H} \Omega (\tau _c)$ (cf. Lemma 2.9), we can suppose without loss of generality that $G=H$ . Let $x {\kern-1pt}\in{\kern-1pt} \Omega (\tau )$ , then $x \in \tau ^n(X)$ for every $n {\kern-1pt}\geq{\kern-1pt} 0$ . Thus, $\tau (x) \in \tau ^{n+1}(X)$ for every $n \geq 0$ and it follows that $\tau (x) \in \Omega (\tau )$ . Therefore, $\tau (\Omega (\tau )) \subset \Omega (\tau )$ . For the converse inclusion, let $y \in \Omega (\tau )$ . By Theorem 9.1, there exists ${x = (x_{ij}) \in \varprojlim _{i,j \in \mathbb {N}} \Sigma _{ij}}$ such that $ \Phi (x)=y$ . However, equation (4.7) tells us that $\Phi ^{-1}(y) \subset \varprojlim _{i,j \in \mathbb {N}} \Sigma _{ij}$ is the set of backward orbits of y under $\tau $ . Hence, we can find $z \in \Omega (\tau )$ such that $\tau (z)= y$ . Thus, $\Omega (\tau ) \subset \tau (\Omega (\tau ))$ and the conclusion follows.

Proof. It is trivial that (b) $\implies $ (a). Assume that $\Sigma $ is a subshift of finite type. Hence, $\Sigma = \Sigma (D, W)$ , where $D \subset G$ is finite and $W \subset A^D$ is some subset. Let $\Sigma _0 \supset \Sigma _1 \supset \cdots $ be a descending sequence of algebraic (respectively algebraic group) sofic subshifts of $A^G$ whose intersection is $\Sigma $ . Let $M \subset G$ be a finite generating subset containing ${\{1_G\} \cup D}$ and such that $M=M^{-1}$ . Consider the inverse system $(X_{ij})_{i,j \in \mathbb {N}}$ defined by . Remark that $X_{i,j+1} \subset X_{ij}$ since $\Sigma _{j+1} \subset \Sigma _j$ for all $i, j \in \mathbb {N}$ . We define the unit transition maps $p_{ij} \colon X_{i+1, j} \to X_{ij}$ by $p_{ij}(x)= x\vert _{M^{i}}$ for every $x \in X_{i+1, j}$ and $q_{ij} \colon X_{i, j+1} \to X_{ij}$ simply as the inclusion maps.

For all $i, j \in \mathbb {N}$ , Theorem 7.1 implies that every $X_{ij}$ is a complete variety (respectively an algebraic group) over K. By Noetherianity of the Zariski topology, the decreasing sequence $(X_{0j})_{j \in ~\mathbb {N}}$ of algebraic closed subsets of $A^{M}$ eventually stabilizes, say, ${X_{0j}= X_{0m}}$ for all $j \in \mathbb {N}$ for some $m \in \mathbb {N}$ . Let , then is an algebraic (respectively algebraic group) subshift of finite type. It is clear that $\Sigma _m \subset \Sigma '$ and hence $\Sigma \subset \Sigma '$ . We shall prove the converse inclusion.

Let $w \in W'$ . We construct an inverse subsystem $(Z_{ij})_{i \geq m, j \geq 0}$ of $(X_{ij})_{i \geq m, j \geq 0}$ as follows. For $i \geq m$ , let which is clearly an algebraic closed subvariety (respectively a translate of an algebraic subgroup) of $X_{i0}$ . For $i \geq m, j \geq 0$ , we define an algebraic closed subvariety (respectively a translate of an algebraic subgroup (by Theorem 7.1)) of $X_{ij}$ as follows:

The transition maps of $(Z_{ij})_{i \geq m, j \geq 0}$ are well defined as the restrictions of the transition maps of the system $(X_{ij})_{i \geq m, j \geq 0}$ . These transition maps have Zariski closed images (by Remark 2.1).

By our construction, each $Z_{ij}$ is clearly non-empty. Hence, Lemma 3.3 implies that there exists $x = (x_{ij})_{i \geq m,j \geq 0} \in \varprojlim Z_{ij}$ . Let $y \in A^G$ be defined by $y(g)= x_{i0}(g)$ for every $g \in G$ and any large enough $i\geq m$ such that $g \in M^{i}$ . Observe that $x_{ij} = x_{ik}$ for every $i \geq m$ and $0 \leq j \leq k$ since the vertical transition maps $X_{ik} \to X_{ij}$ are simply inclusions. Consequently, for every $n \in \mathbb {N}$ , we have $y \in \Sigma _n$ by equation (4.6) since $\Sigma _n$ is closed in $A^G$ (cf. Corollary 8.2). Hence, $y \in \Sigma $ . By construction, $y\vert _{M^m} = w$ . Since w was arbitrary, this shows that $W' \subset \Sigma _{M^m}$ . Hence, $\Sigma ' = \Sigma (M^m, W') \subset \Sigma (M^m, \Sigma _{M^m}) = \Sigma $ . The last equality follows from Lemma 5.1 as $D \subset M^m$ . Therefore, $\Sigma ' = \Sigma $ and $\Sigma _n= \Sigma $ for all $n \geq m$ . This proves that (a) $\implies $ (c).

Suppose now that $\Sigma \subset A^G$ is an algebraic (respectively algebraic group) sofic subshift which is not a subshift of finite type. Let $M \subset G$ be a finite generating subset containing $\{1_G\}$ such that $M=M^{-1}$ . For every $n \in \mathbb {N}$ , consider (as in §4). Theorem 7.1 tells us that $W_n$ is a complete algebraic subvariety (respectively an algebraic subgroup) of $A^{M^n}$ . Set for every $n \in \mathbb {N}$ , then $\Sigma _n$ is an algebraic (respectively algebraic group) subshift of finite type. As $(\Sigma _{M^{n+1}})_{M^n} = \Sigma _{M^n}$ , it is clear that ${\Sigma \subset \Sigma _{n+1} \subset \Sigma _n}$ for every $n \in \mathbb {N}$ . We claim that $\Sigma = \bigcap _{n \in \mathbb {N}} \Sigma _n$ . Indeed, we only need to prove that $\bigcap _{n \in \mathbb {N}} \Sigma _n \subset \Sigma $ . Let $x \in \bigcap _{n \in \mathbb {N}} \Sigma _n$ . Then by definition of $\Sigma _n$ , we find that $x\vert _{M^n} \in W_n= \Sigma _{M^n}$ for every $n \in \mathbb {N}$ . Thus, since $\Sigma $ is closed (cf. Corollary 8.2), ${x \in \varprojlim _{n \in \mathbb {N}} \Sigma _{M^n} =\Sigma }$ (cf. equation (4.6)) and hence $\bigcap _{n \in \mathbb {N}} \Sigma _n \subset \Sigma $ . However, the descending sequence $(\Sigma _n)_{n \in \mathbb {N}}$ cannot stabilize since otherwise the subshift $\Sigma $ would be of finite type. This shows that (c) $\implies $ (a) if $\Sigma $ is an algebraic (respectively algebraic group) sofic subshift. The proof is complete.

Examples 10.2. (Markov properties)

The examples below provide the original sources and motivations for our main result in this section (Theorem 10.1).

(a) Let G be a group and let A be a finite group. Equip the configuration space $A^G$ with the product group structure (thus, given two configurations $x,y \in A^G$ , their product is defined as the configuration $xy \in A^G$ given by for all $g \in G$ ). A closed subshift $X \subset A^G$ which is also a subgroup of $A^G$ is called a group subshift. Group subshifts $X \subset A^{\mathbb {Z}}$ are called Markov subgroups in [Reference Kitchens28, §6.3], and were studied and classified up to topological conjugacy by Kitchens in [Reference Kitchens27] (see also [Reference Kitchens28, Theorem 6.3.3]).

One says that a group G is of finite Markov type if for any finite group A, every group subshift $\Sigma \subset A^G$ is of finite type. The finite Markov type property is a weakening of the Markov type property introduced by Schmidt in [Reference Schmidt54, Definition 4.1].

The following hold:

1. (i) every finite group is of finite Markov type;

2. (ii) the additive group $\mathbb {Z}$ is of finite Markov type. This result was established by Kitchens in [Reference Kitchens27, Proposition 4] (see also [Reference Kitchens28, Lemma 6.3.5], [Reference Lind and Marcus30, Exercise 2.1.11], and [Reference Ceccherini-Silberstein and Coornaert13, Exercise 1.114]);

3. (iii) every subgroup of a group of finite Markov type is finitely generated (this is also expressed by saying that groups of finite Markov type are Noetherian). As a consequence, every group of finite Markov type is countable and contains no non-abelian free subgroups;

4. (iv) every quotient of a group of finite Markov type is of finite Markov type;

5. (v) every group containing a finite index subgroup of finite Markov type is itself of finite Markov type;

6. (vi) a group G is of finite Markov type if and only if G is countable and, for any finite group A, every descending sequence of group subshifts of $A^G$

   $$ \begin{align*} X_0 \supset X_1 \supset X_2 \supset \cdots \supset X_n \supset X_{n+1} \supset \cdots \end{align*} $$
   eventually stabilizes, that is, there exists $n_0 \in \mathbb {N}$ such that $X_n = X_{n_0}$ for all ${n \geq n_0}$ (cf. [Reference Ceccherini-Silberstein and Coornaert13, Exercise 1.112]).

In other words, the class of groups of finite Markov type is closed under the operations of taking subgroups, quotients, and extensions by finite or cyclic groups. As a consequence, all finitely generated abelian groups are of finite Markov type (a result observed by Kitchens and Schmidt [Reference Kitchens and Schmidt26, Remark 3.10(2)]). In fact, more generally, the class of groups of finite Markov type contains all polycyclic-by-finite groups (cf. [Reference Ceccherini-Silberstein and Coornaert13, Exercise 4.37]), a particular case of [Reference Schmidt54, Theorem I.4.2]. The question whether or not every group of finite Markov type is polycyclic-by-finite remains, at our present knowledge, open.

(b) Let G be a group, let K be a field, and let A be a finite-dimensional vector space over K. Equip the configuration space $A^G$ with the product vector space structure (thus, given a scalar $\unicode{x3bb} \in K$ and two configurations $x,y \in A^G$ , one defines the configuration ${\unicode{x3bb} x \in A^G}$ (respectively $x + y \in A^G$ ) by setting (respectively ) for all $g \in G$ . A closed subshift $\Sigma \subset A^G$ which is also a vector subspace of $A^G$ is called a linear subshift. One says that a group G is of K-linear Markov type if for any finite-dimensional vector space A over K, every linear subshift $\Sigma \subset A^G$ is of finite type. Analogous properties to items (i)–(vi) in point (a), for groups of K-linear Markov type, are shown in [Reference Ceccherini-Silberstein, Coornaert and Phung16, §6] (see also [Reference Phung42] for some more general results). In other words, the class of K-linear Markov groups is closed under the operations of taking subgroups, quotients, and extensions by finite or cyclic groups, and contains all polycyclic-by-finite groups (cf. [Reference Ceccherini-Silberstein, Coornaert and Phung16, Corollary 1.4]). In addition, one has the following characterization: a group G is of K-linear Markov type if and only if its group ring $K[G]$ is one-sided Noetherian [Reference Ceccherini-Silberstein, Coornaert and Phung16, Theorem 1.3].