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We consider (topological) dynamical systems $(\Omega, S)$, where $S$ is the shift $(Sx)_n=x_{n+1}$, and $\Omega\subset[0,1]^{\mathbb Z}$ is a compact, shift invariant subspace. I call such a system $(\Omega, S)$ deterministic if $x_n=y_n$ for all $n<0$ implies that $x=y$ for any two points $x,y\in\Omega$. (Of course, this is a non-standard use of the term.)

We now consider conjugacies between shifts, that is, homeomorphisms $\varphi: \Omega_1\to\Omega_2$ such that $\varphi S= S\varphi$. Determinism in the above sense is a property of the space $\Omega$, not of the dynamical system, and it is not invariant under conjugacy: for an easy example, take the full shift on two symbols $\Omega=\{0,1\}^{\mathbb Z}$. This is clearly non-deterministic, but we can map $f:\Omega\to C\subset[0,1]$ homeomorphically and then define $\varphi(x)$ as the sequence whose $n$th entry is $f(S^nx)$. Then $\varphi$ is an isomorphism (of dynamical systems) onto a deterministic system; in fact, in this new realization, $x$ is already determined by $x_0$.

Two questions: (1) Are there systems of this type that are not isomorphic to a deterministic shift? Put differently, is the full shift on $[0,1]^{\mathbb Z}$ isomorphic to a deterministic shift?

(2) Can one say non-trivial things about systems that have only deterministic realizations in this sense? (For example, distal systems have this property, but that's quite obvious.)

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  • $\begingroup$ There are a few strange points in your question, possibly misprints (e.g. in the conjugacy, I guess there is a $S_1$ and a $S_2$). With regard to question (1), why can't you use the same construction as in the previous §, taking $\Omega_2=\Omega_1^\mathbb{Z}$? $\endgroup$ Commented Apr 12, 2014 at 18:52
  • $\begingroup$ @Benoit: All my $\Omega$'s are by assumption contained in $[0,1]^{\mathbb Z}$ (the question disappears otherwise, as you pointed out). That probably answers the first question, too: all shifts are restrictions of the one on that space. $\endgroup$ Commented Apr 12, 2014 at 18:59
  • $\begingroup$ The property that you call "deterministic shift" seems to be a stronger variant of having zero entropy. In particular, any system with positive entropy cannot be isomorphic to a deterministic shift. $\endgroup$ Commented Apr 13, 2014 at 21:17
  • $\begingroup$ Moreover, to be isomorphic to a subshift of the shift $[0,1]^\mathbb{Z}$ is a property of coding of your system, which seems to me to be totally independent of your property of "determinism". This "determinism" property can be expressed in terms of stable/unstable manifolds, without referring to any coding: if $W^u(v)=W^u(w)$ then $W^s(v)=W^s(w)$ which is an extremely rigid property. $\endgroup$ Commented Apr 13, 2014 at 21:20
  • $\begingroup$ @Barbara: The example I gave (Bernoulli shift) has positive entropy. I'm pretty sure in fact that there are subshifts with infinite entropy that can be realized as deterministic shifts. $\endgroup$ Commented Apr 13, 2014 at 23:39

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