$\DeclareMathOperator\Aut{Aut}$Let $X$ and $Y$ be two subshifts of finite type and $X\subset Y$, and $\phi:X\rightarrow X$ be a homeomorphism commuting with the shift map. Is there any homeomorphism $\tilde{\phi}:Y\rightarrow Y$ commuting with the shift map such that $\tilde{\phi}|_X=\phi$?
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- $\begingroup$ The two questions are not equivalent. The question asks, denoting $\mathrm{Aut}(Y,X)$ the automorphisms of $Y$ leaving $X$ invariant, whether the restriction map $\mathrm{Aut}(Y,X)\to\mathrm{Aut}(X)$ is surjective. When this is true, this map is probably not bijective, and hence it doesn't yield an embedding of $\mathrm{Aut}(X)$ into $\mathrm{Aut}(Y)$. (By the way, there are several interpretations of "is $\mathrm{Aut}(X)\subset\mathrm{Aut}(Y)$".) $\endgroup$YCor– YCor2024-04-18 11:34:29 +00:00Commented Apr 18, 2024 at 11:34
- $\begingroup$ Thank you for your answer. You are right, I edited the question. $\endgroup$Ali Ahmadi– Ali Ahmadi2024-04-19 08:35:35 +00:00Commented Apr 19, 2024 at 8:35
- $\begingroup$ Not in general, see mathoverflow.net/questions/452585/… (a finite shift-invariant set is a subshift of finite type) $\endgroup$Ville Salo– Ville Salo2024-04-19 13:58:58 +00:00Commented Apr 19, 2024 at 13:58
- $\begingroup$ @VilleSalo thanks for your answer. Therefore an automorphism of $\{0,1\}^{\mathbb{Z}} $ can not be extended to an automorphism of $\{0,1,2\}^{\mathbb{Z}} $. Is it true? $\endgroup$Ali Ahmadi– Ali Ahmadi2024-04-20 05:33:11 +00:00Commented Apr 20, 2024 at 5:33
- $\begingroup$ I think every automorphism of 2-shift can be extended to 3-shift: kill the dimension rep by composing with shift (which extends), then inerts extend. $\endgroup$Ville Salo– Ville Salo2024-04-20 08:17:43 +00:00Commented Apr 20, 2024 at 8:17
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