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Dominik Kwietniak
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Does conjugacy preservespreserve the set of synchronizing blocks?

A synchronized system is a transitive shift space $X$ which has a synchronizing block $v$, that is $v$ is an admissible block for $X$ and whenever $vw$ and $uv$ are admissible blocks in $X$, then $uvw$ is also admissible. Let $\Psi\colon X\to Y$ be a conjugacy of synchronized systems.

Since $\Psi$ is continuous and shift-commuting there is an $n\in\mathbb{N}$, a map from admissible $n$-bloks in $X$ to $1$-blocks in $Y$, denoted $\psi\colon \mathscr{B}_n(X ) \to \mathscr{B}_1(Y)$, and an $m \in\mathbb{Z}$ such that, in standard notation, $(\Psi \circ \sigma^m(x))_i = \psi(x_{[i,i+n)})$, for all $i \in \mathbb{Z}$ and all $x \in X$, cf. [LM]Lind and Marcus book. The map $\psi$ extends to maps $\psi^{(k)} \colon \mathscr{B}_k(X ) \to \mathscr{B}_{k-n+1}(Y)$, $k \ge n$, in the natural way.

Is it true that a $k$-block $v$ ($k\ge n$) is a synchronizing block for $X$ if and only if $\psi^{(k)}(v)$ is a synchronizing block for $Y$?

In general, a factor of a synchronized system may not be synchronized.

Does conjugacy preserves the set of synchronizing blocks?

A synchronized system is a transitive shift space $X$ which has a synchronizing block $v$, that is $v$ is an admissible block for $X$ and whenever $vw$ and $uv$ are admissible blocks in $X$, then $uvw$ is also admissible. Let $\Psi\colon X\to Y$ be a conjugacy of synchronized systems.

Since $\Psi$ is continuous and shift-commuting there is an $n\in\mathbb{N}$, a map from admissible $n$-bloks in $X$ to $1$-blocks in $Y$, denoted $\psi\colon \mathscr{B}_n(X ) \to \mathscr{B}_1(Y)$, and an $m \in\mathbb{Z}$ such that, in standard notation, $(\Psi \circ \sigma^m(x))_i = \psi(x_{[i,i+n)})$, for all $i \in \mathbb{Z}$ and all $x \in X$, cf. [LM]. The map $\psi$ extends to maps $\psi^{(k)} \colon \mathscr{B}_k(X ) \to \mathscr{B}_{k-n+1}(Y)$, $k \ge n$, in the natural way.

Is it true that a $k$-block $v$ ($k\ge n$) is a synchronizing block for $X$ if and only if $\psi^{(k)}(v)$ is a synchronizing block for $Y$?

In general, a factor of a synchronized system may not be synchronized.

Does conjugacy preserve the set of synchronizing blocks?

A synchronized system is a transitive shift space $X$ which has a synchronizing block $v$, that is $v$ is an admissible block for $X$ and whenever $vw$ and $uv$ are admissible blocks in $X$, then $uvw$ is also admissible. Let $\Psi\colon X\to Y$ be a conjugacy of synchronized systems.

Since $\Psi$ is continuous and shift-commuting there is an $n\in\mathbb{N}$, a map from admissible $n$-bloks in $X$ to $1$-blocks in $Y$, denoted $\psi\colon \mathscr{B}_n(X ) \to \mathscr{B}_1(Y)$, and an $m \in\mathbb{Z}$ such that, in standard notation, $(\Psi \circ \sigma^m(x))_i = \psi(x_{[i,i+n)})$, for all $i \in \mathbb{Z}$ and all $x \in X$, cf. Lind and Marcus book. The map $\psi$ extends to maps $\psi^{(k)} \colon \mathscr{B}_k(X ) \to \mathscr{B}_{k-n+1}(Y)$, $k \ge n$, in the natural way.

Is it true that a $k$-block $v$ ($k\ge n$) is a synchronizing block for $X$ if and only if $\psi^{(k)}(v)$ is a synchronizing block for $Y$?

In general, a factor of a synchronized system may not be synchronized.

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Dominik Kwietniak
  • 1.7k
  • 1
  • 13
  • 22

Does conjugacy preserves the set of synchronizing blocks?

A synchronized system is a transitive shift space $X$ which has a synchronizing block $v$, that is $v$ is an admissible block for $X$ and whenever $vw$ and $uv$ are admissible blocks in $X$, then $uvw$ is also admissible. Let $\Psi\colon X\to Y$ be a conjugacy of synchronized systems.

Since $\Psi$ is continuous and shift-commuting there is an $n\in\mathbb{N}$, a map from admissible $n$-bloks in $X$ to $1$-blocks in $Y$, denoted $\psi\colon \mathscr{B}_n(X ) \to \mathscr{B}_1(Y)$, and an $m \in\mathbb{Z}$ such that, in standard notation, $(\Psi \circ \sigma^m(x))_i = \psi(x_{[i,i+n)})$, for all $i \in \mathbb{Z}$ and all $x \in X$, cf. [LM]. The map $\psi$ extends to maps $\psi^{(k)} \colon \mathscr{B}_k(X ) \to \mathscr{B}_{k-n+1}(Y)$, $k \ge n$, in the natural way.

Is it true that a $k$-block $v$ ($k\ge n$) is a synchronizing block for $X$ if and only if $\psi^{(k)}(v)$ is a synchronizing block for $Y$?

In general, a factor of a synchronized system may not be synchronized.