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Apr 20 at 9:56 comment added Ali Ahmadi Very interesting point, thank you so much for sharing it.
Apr 20 at 8:20 comment added Ville Salo Point here being, for a full shift on prime alphabet, the dimension group representation is generated by the shift map.
Apr 20 at 8:17 comment added Ville Salo 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.
Apr 20 at 5:33 comment added Ali Ahmadi @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?
Apr 19 at 14:43 history edited YCor CC BY-SA 4.0
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Apr 19 at 13:58 comment added Ville Salo Not in general, see mathoverflow.net/questions/452585/… (a finite shift-invariant set is a subshift of finite type)
Apr 19 at 8:36 history edited Ali Ahmadi CC BY-SA 4.0
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Apr 19 at 8:36 history edited Ali Ahmadi CC BY-SA 4.0
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Apr 19 at 8:35 comment added Ali Ahmadi Thank you for your answer. You are right, I edited the question.
Apr 19 at 8:34 history edited Ali Ahmadi CC BY-SA 4.0
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Apr 18 at 11:34 comment added YCor 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 $\mathrm{Aut}(X)\subset\mathrm{Aut}(Y)$".)
Apr 18 at 11:31 history edited YCor CC BY-SA 4.0
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S Apr 18 at 10:41 review First questions
Apr 18 at 10:52
S Apr 18 at 10:41 history asked Ali Ahmadi CC BY-SA 4.0