Let $(\Sigma^{\mathbb{Z}},S)$ be a left-shift system, where $\Sigma$ is a metrizable compact set. Consider the automorphism group of it (bijective factor maps of itself), denoted by $G$.

Now let $(A,S)$ and $(B,S)$ be two isomorphic shift invariant subsystems, let $f$ be an isomorphism between them.

The question is, does there always exist a $g\in G$ such that $g$ restricted on $A$ is equal to $f$? I mean, can we extend $f$ to the whole system? I think this must be known but not easy to prove.. Thank you!

  • $\begingroup$ This is discussed in works of Boyle and others, it's very classical and I think there is discussion here on MO too. IIRC on the binary full shift, there is an automorphism of the 2-periodic points (a 4-point system) which doesn't extend, though I can't reconstruct the argument so maybe it wasn't quite this simple. $\endgroup$
    – Ville Salo
    Aug 12, 2023 at 9:45
  • $\begingroup$ Sorry yeah that example is about inert automorphisms. Anyway the answer to your questions is definitely "no". $\endgroup$
    – Ville Salo
    Aug 12, 2023 at 9:53
  • $\begingroup$ I agree, even if we took A,B to be two singletons $\endgroup$
    – Bo Peng
    Aug 14, 2023 at 6:42
  • 1
    $\begingroup$ If you take two singletons, you can always extend a bijection to an automorphism, just permute symbols. $\endgroup$
    – Ville Salo
    Aug 14, 2023 at 9:33

1 Answer 1


Let $X = \{0,1\}^{\mathbb{Z}}$, and $\sigma$ the left shift. Let $\mathcal{Q}_n(X)$ be the set of $\sigma$-orbits of cardinality $n$, i.e. the orbits of points of least period $n$ in $X$. If $f \in \mathrm{Aut}(X)$, then $f$ naturally acts on $\mathcal{Q}_n$ i.e. orbits are mapped to orbits. The signs of $f$ are defined as the signs of these permutations, i.e. $s_n(f) = -1$ if $f$ is an odd permutation on $\mathcal{Q}_n$, and $s_n(f) = 1$ otherwise.

The gyration numbers of $f$ are defined as follows: For each $\gamma \in \mathcal{Q}_n(X)$, pick a representative $x_\gamma \in \gamma$. Let $k_\gamma$ be such that $f(x_\gamma) = \sigma^{k_\gamma}(x_{f(\gamma)})$. Then $g_n(f) = \sum_{\gamma \in \mathcal{Q}_n(X)} k_\gamma$. It is easy to check that this does not depend on the choice of representatives, since if you shift the representative you decrease one $k$-value and increase another.

Finally we say $f$ satisfies the sign-gyration compatibility condition if for all $q$ odd and all $m$, we have $g_{2^m q}(f) = \left\{\begin{array}{ll} 0 & \mathrm{if} \prod_{j = 0}^{m-1} s_{2^j q}(\phi) = 1 \\ 2^{m-1} q & \mathrm{if} \prod_{j = 0}^{m-1} s_{2^j q}(\phi) = -1 \end{array}\right.$

All we need to notice in this formula is that $g_n(f)$ is always equal to $0$ or $n/2$ (when this condition holds).

Now, according to [1], [2] proves that there is a shift-commuting permutation of the points of least period 6 of $X$, which is not the restriction of an automorphism of $X$. This solves your question in the negative, taking $A = B$ the set of points of least period $6$.

I cannot access [2], but it seems that this follows immediately from a (to me) better-known result also from [2] (though I don't know the proof of this one either). Namely, again according to [1], Kim and Roush prove in [2] that the SGCC holds for every inert automorphism of every mixing SFT. Inert automorphisms are the ones in the kernel of the dimension representation, i.e. they act trivially on the dimension group.

I won't define the dimension representation here (see e.g. [1] for the definition), but it is known that the dimension representation of the binary full shift $X$ is isomorphic to $\mathbb{Z}$, and the shift map $\sigma$ generates it.

From the sign-gyration result of [2] it then follows that the gyration numbers for period $6$ are generated by $g_6(\sigma) = 3$ and whatever SGCC allows (i.e. $0$ and $3$), since gyration is a group homomorphism. So we deduce that no automorphism $f$ can satisfy $g_6(f) = 1$. But there is a permutation of the $6$-periodic points which would give $g_6(f) = 1$ for any automorphism extending this, for example the automorphism that rotates the orbit of $000001$, but does not rotate any other orbit.


[1] Lind, Douglas; Marcus, Brian, An introduction to symbolic dynamics and coding, Cambridge: Cambridge University Press (ISBN 0-521-55124-2/hbk; 0-521-55900-6/pbk). 484 p. (1995). ZBL1106.37301.

[2] Kim, K. H.; Roush, F. W., On the structure of inert automorphisms of subshifts, PU.M.A., Pure Math. Appl., Ser. B 2, No. 1, 3-22 (1991). ZBL0766.54041.

  • $\begingroup$ Thank you so much! $\endgroup$
    – Bo Peng
    Aug 14, 2023 at 15:17

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