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Suppose the dynamical system $(X,T)$ has only proper factors (i.e. not $(X,T)$ itself) of zero topological entropy. Does the system $(X,T)$ also have zero entropy?

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  • $\begingroup$ If $(X,T)$ is minimal, then your hypothesis is vacuously true, at least interpreting it as "there are no positive-entropy subsystems". But the topological entropy of a minimal system can be whatever. So I guess you should clarify the question. $\endgroup$ Jul 20 at 14:03
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    $\begingroup$ @AlessandroDellaCorte A "factor" in this context means the image of a factor map between dynamical systems (a continuous surjection $f : (X,T) \to (Y,S)$ such that $f \circ T = S \circ f$). A minimal system can have many nontrivial factors. $\endgroup$ Jul 20 at 14:28
  • $\begingroup$ Sure. I took it as a synonym of subsystem. I was wrong. $\endgroup$ Jul 20 at 14:35
  • $\begingroup$ One way to produce such systems would indeed be to find a "cominimal" system with positive entropy, i.e. no nontrivial factors. I don't know If these exist off the top of my head. Such systems are sometimes called "prime" I think. Prime is implied by doubly minimal, and doubly minimal implies zero entropy, maybe under some additional conditions (Weiss). Am not at computer, so take with grain of salt. $\endgroup$
    – Ville Salo
    Jul 20 at 15:40

2 Answers 2

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This question is very related to the question of lowering topological entropy, introduced in ``Can one always lower topological entropy?'' by Shub and Weiss and then very nearly solved by Lindenstrauss in "Lowering topological entropy" and "Mean Dimension, Small Entropy Factors, and an Embedding Theorem." There's too much to state everything here, but here's a quick overview:

The question is: does every $(X,T)$ of positive entropy possess a nontrivial (i.e. not one point) factor $(Y,S)$ with $h(Y) < h(X)$?

  1. Shub and Weiss showed that the answer is "yes" for subshifts and more generally, systems with the so-called small boundary property.

  2. Lindenstrauss showed that the answer is "yes" for systems $(X,T)$ where $X$ is finite-dimensional. He then showed that the same is true when $(X,T)$ has zero mean dimension and a nontrivial minimal factor. Mean dimension is too long to define here, but in particular, $(X,T)$ has zero mean dimension if $(X,T)$ has finite entropy!

  3. Germane to your question: Lindenstrauss also gives examples of $(X,T)$ (with infinite entropy and $X$ infinite-zero dimension) for which every nontrivial factor $(Y,S)$ factors onto the original system $(X,T)$! The easiest is $X = [0,1]^{\mathbb{Z}}$ and $T$ the shift $\sigma$, but he also constructs a minimal example.

To summarize: any finite-entropy system (with a nontrivial minimal factor) has a nontrivial factor of smaller entropy, thus not conjugate to the original. So for systems with a nontrivial minimal factor, the only possibilities for your condition are zero or infinite entropy.

For infinite entropy, we can't yet rule out your condition, since technically the example above doesn't satisfy your criteria (the nontrivial factors are not isomorphic to the original system). However, the factors are in a way "bigger" than the original system, which is perhaps even more surprising!

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  • $\begingroup$ Very interesting stuff. Besides the zero-dimensional case, I had a feeling it's true in the expansive case, but there was some glitch in my proof. I think Lindenstrauss' result gives this, because Mañé's theorem states you can only have expansive actions on finite-dimensional spaces. (Ok, it's also covered by finite entropy...) $\endgroup$
    – Ville Salo
    Jul 21 at 12:36
  • $\begingroup$ You don't want to resort to finite entropy, since that requires this technical assumption of having a non-trivial minimal factor. So your finite-dimensionality reason is definitely the better one to use. $\endgroup$ Jul 21 at 12:54
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    $\begingroup$ The way I'm reading your answer, you just need that technical assumption for the zero mean dimension case. $\endgroup$
    – Ville Salo
    Jul 21 at 12:59
  • $\begingroup$ Right, but I don't know any proof of the finite entropy case without just applying zero mean dimension case. I made a summary forgetting this assumption in my post, which I've fixed now; sorry! $\endgroup$ Jul 21 at 13:19
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YES under strong additional assumptions.

Theorem. Let $X$ be compact metrizable and zero-dimensional, and let $T : X \to X$ be an aperiodic homeomorphism. If $(X, T)$ has only zero-entropy proper factors, then $(X, T)$ has zero entropy.

I assume the asker already had "compact metrizable" in mind, as it's a standard category for this type of question. I guess "homeomorphism" is non-essential and you could take $T$ just continuous, but I didn't check. The assumption that $T$ is aperiodic is just to simplify the proof a little, I believe it is non-essential. From the tags I guess the zero-dimensional case is of interest, but (as I'll discuss after the proof) I believe it is essential for this proof.

Proof. For a contradiction, suppose $(X, T)$ satisfies the assumptions, but has positive entropy. We may first take $X \subset \{0,1\}^{\mathbb{N}}$ where $\{0,1\}^{\mathbb{N}}$ is the Cantor set, and then after the topological conjugacy $x \mapsto (T^i(x))_{i \in \mathbb{Z}} \in (\{0,1\}^{\mathbb{N}})^{\mathbb{Z}}$, we may assume $T$ is a shift map and $X$ a shift-invariant closed subset of $(\{0,1\}^{\mathbb{N}})^{\mathbb{Z}}$.

Restricting to the first $n$ coordinates is a factor map, and by the definition of entropy one of these factors has positive entropy. By the assumption that proper factors have zero entropy, we deduce that $(X, T)$ is topologically conjugate to a subshift: we have a finite discrete alphabet $A$, $X \subset A^{\mathbb{Z}}$ is closed and shift-invariant, and $T$ is the shift map.

Now take an entropy pair $(x, y) \in X^2$ [1], which exists by the assumption of positive entropy. This means $x \neq y$ and if $(U, V)$ is a cover of $X$ by non-dense open sets such that $x \in (U^c)^{\circ}, y \in (V^c)^{\circ}$, then this cover has positive entropy. Since the entropy pair relation is $T \times T$-invariant, we may assume $x_0 \neq y_0$.

For any $n \in \mathbb{N}$, define now a factor map onto a subshift of $\{0,1\}^{\mathbb{Z}}$ by $f(z)_i = 1 \iff z_{[i, i+n]} = x_{[0, n]}$. It is easy to see that the image subshift has positive entropy for any $n$. Namely, consider the cover $\{[0], [1]\}$, where for a word $w$, $[w]$ denotes the cylinder set $\{z \;|\; z_{[0,|w|-1]} = w\}$. Computing the entropy of this is equivalent to computing the entropy of the clopen cover $(U, V) = ([x_{[0, n]}]^c, [x_{[0, n]}])$ of $X$. This cover satisfies $x \in (U^c)^{\circ}, y \in (V^c)^{\circ}$ because $x_0 \neq y_0$, and clearly neither $U$ nor $V$ is dense, thus this cover has positive entropy since $(x, y)$ is an entropy pair, and we conclude the image has positive entropy.

Since these are positive entropy factors, $(X, T)$ is topologically conjugate to each of them. By aperiodicity, the point $x$ is aperiodic, so density of $1$s in the image subshift tends to $0$ as $n \rightarrow \infty$. From this it is easy to deduce an upper bound on their entropies, which tends to $0$. All these upper bounds apply to $X$, so it is of zero entropy. Thus $(X, T)$ must have zero entropy. Square.

Specifically what fails for general systems is that the factors need to be taken in some type of continuum. I tried subsystems of variants of $[0,1]^{\mathbb{Z}}$, and the problem is that unlike our discrete alphabet $\{0, 1\}$, every symbol $a \in [0,1]$ can transmit an infinite amount of information, so to speak. On the other hand, I have no idea how you could possibly construct a system that actually forces this sort of thing for all factor maps, let alone for all other factors, so I don't really have a conjecture about the general case at the moment.

[1] Blanchard, François, A disjointness theorem involving topological entropy, Bull. Soc. Math. Fr. 121, No. 4, 465-478 (1993). ZBL0814.54027.

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