Is it possible for topological dynamical systems to be semiconjugate if one of the dynamical systems is minimal but the other one is not? (minimal means every orbit of the dynamical system is dense) If so, is there any simple example of this?
1 Answers
Yes, it is certainly possible. One of the best known examples is the Denjoy construction. Full details of this construction are given in Milnor's notes, section 15B, under the title "Denjoy counterexamples". Let me give a brief outline.
Start with an irrational rotation of the circle $f : S^1 \to S^1$, $f(z) = e^{2\pi i \theta}z$ where $\theta$ is irrational. This map $f$ is minimal.
Pick an orbit, say $A = \{z_n \mid n \in \mathbb Z\}$ where $z_n = e^{2 \pi i \theta n}$. Pick a bi-infinite convergent sequence of reals $(\ell_n)_{n=-\infty}^\infty$, $\sum_{n=-\infty}^\infty \ell_n = 1$.
You then cut open $S^1$ at each point $e^{2\pi i \theta n} \in A$, and insert an open interval $I_n$ of length $\ell_n$ between the resulting two points (Milnor explains what happens with those two points carefully). The resulting space $X$, with the obvious circular ordering, is homeomorphic to the circle. The union of the open intervals $I_n$ is an open subset $W \subset X$, and its complement $X-W$ is a Cantor set.
You then define $\tilde f : X \to X$ to be the homeomorphism that takes each $I_n$ to $I_{n+1}$ by the obvious similarity map of expansion factor $\ell_{n+1} / \ell_n$. This map $\tilde f$ is not minimal, because the union of the $I_n$'s is a wandering set. Also, the map $X \mapsto S^1$ which collapses (the closure of) each $I_n$ back to the point $z_n$ is a semiconjugacy from $\tilde f$ to $f$.
As a side note, the map $\tilde f$ takes the Cantor set $X-W$ to itself, and the restricted map $\tilde f : X-W \to X-W$ is minimal.
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Thank you. I think there's a typo, it should be the map $f$ is minimal? – pops Mar 09 '20 at 23:55
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1Actually it wasn't a typo, but it was confusing, and I rewrote that part. – Lee Mosher Mar 10 '20 at 00:41
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Come to think of it, couldn't we get a simpler example by considering an irrational rotation and the one sided shift map on the set, $S$ say,of infinite binary sequences? The semi-conjugacy would then be the map from $S$ to the circle which we identify with the interval $I=[0, 1]$. The semi conjugacy can then map each binary sequence $(x_n) \in S$ to the number $x\in I$ with binary expansion given by $(x_n)$. – pops Mar 10 '20 at 00:53
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I just noticed that this doesn't work... the defined semi conjugacy fails to be one. – pops Mar 10 '20 at 01:05