I am trying to understand relations between various measures of topological complexity. I have read that expansive homeomorphisms on continua, for example, have positive entropy. But I do not know whether another property called mixing also implies positive entropy.

Let $X$ be a connected compact metric space.

Question. If a homeomorphism $f:X\to X$ is mixing, then does $f$ necessarily have positive entropy?

A homeomorphism $f:X\to X$ is mixing if for every pair of non-empty open sets $U,V$ of $X$, there exists a positive integer $M$ such that $f^m(U)\cap V\neq\varnothing$ for all $m\geq M$. That is, if $U$ is open and non-empty, then $d_H(f^n(U),X)\to 0$ as $n\to\infty$, where $d_H$ is the Hausdorff metric.

See https://en.wikipedia.org/wiki/Topological_entropy for the two equivalent definitions of the topological entropy of a map $f:X\to X$.

EDIT: It occurs to me that my question was essentially asked in:

Kato, Hisao, Continuum-wise expansive homeomorphisms, Can. J. Math. 45, No. 3, 576-598 (1993). ZBL0797.54047.

Question 6 in that paper is my question with a weaker hypothesis (it is not difficult to show that mixing implies sensitive dependence on initial conditions).

I'm not sure if anyone ever published a counterexample.

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    $\begingroup$ There are plenty of families of zero entropy systems which can be either mixing or non-mixing. What comes ot my mind first is primitive substitution systems. Pisot substitutions are never mixing, but some families of constant length substitutions can be mixing or not. There is a paper by Kenyon, Sadun, Solomyak that goes into more detail in this direction. I am sure there are many others that I'm not as familiar with. $\endgroup$
    – Dan Rust
    Oct 16, 2019 at 20:12
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    $\begingroup$ In particular, Dekking and Keane showed in Mixing Properties of Substitutions that the subshift associated with the substitution $0 \mapsto 001, 1 \mapsto 11100$ is topologically mixing. Of course, all subshifts associated with primitive substitutions are zero entropy. $\endgroup$
    – Dan Rust
    Oct 16, 2019 at 20:21
  • $\begingroup$ @DanRust What are the underlying topological spaces for these systems? $\endgroup$ Oct 16, 2019 at 20:31
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    $\begingroup$ Well a counter example can be made by considering a unipotent flow on a compact quotient of a semisimple Lie group, for concreteness one can consider the quotient of PSL2 by a uniform lattice and the action of the (horospherical) unipotent subgroup. It is ergodic and actually mixing (Howe-Moore), but one may show the mixing is polynomial, hence the topological entropy is 0. $\endgroup$
    – Asaf
    Oct 17, 2019 at 2:20
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    $\begingroup$ Solenoids usually come up in dynamics as two sided extensions of systems, which are themselves usually expanding homomorphisms. It can be shown easily that in such cases, as the fiber is compact, the metric entropy is the same for the two sided and one sided system (think about their Pinsker factors say, you don't get any new information gained from the far past) and therefore by variational principal they have the same topological entropies. So I would look otherwise rather than solenoids. $\endgroup$
    – Asaf
    Oct 17, 2019 at 2:23

2 Answers 2



I realized there is a simpler construction that achieves the same (examples of top. mixing zero-entropy homeos on $S^3$): Instead of the bi-directional flow through cuboids, have a flow from bottom to top on the cylinder $D^2 \times [0,1]$, and have it slow down to rate $0$ on the boundary. Also, instead of all the $C_i$ playing the same role, have a single special cylinder $C$ and to get mixing, thread small cylinders from its top to its bottom, making sure at all times not to introduce periods (by refining open sets similarly as in the construction below). Open sets in the existing gadget will go around these loops and a little part of them eventually ends up on a free area on the top of $C$, and you do the threading from there. (And threading through open sets not in the gadget is easy.) The details of making the flow continuous in the limit, and the calculations showing that slow enough rate in the new cylinders implies zero entropy, are the same (and I still didn't do them).


I think there are counterexamples on all path-connected closed manifolds of dimension at least $3$, or at least I don't know what special properties I could possibly be using. For concreteness, we can think about $M = S^3$. I'll describe an $\mathbb{R}$-flow whose time-$1$ map will have the desired property. Some figures I drew on the blackboard are also attached ($U_1$ and $V_1$ in the figure should be $U_2$ and $V_2$ resp.).

enter image description here

First, let's introduce for every $\epsilon > 0$ an flow $f_\epsilon : \mathbb{R} \times C \to C$ on the solid block $C = [0,1]^2 \times [0, R]$ (think of a large $R$). Think of $[0, R]$ as being the vertical axis, and we're staring at $C$ from the front. On the boundary of $C$, there is no movement. Inside $C$ pick two vertical lines from top to bottom, say $A$ and $B$. On the line $A$, the dynamics is trivial, i.e. all points are fixed. On the line $B$, points are moving upward at some positive rate, which should be considered very slow and parametrized by $\epsilon$ (the time-$1$ map on $B$ should behave roughly like $x \mapsto x^{1 + \epsilon}$ does on $[0,1]$). On the strip $S$ between $A$ and $B$, the dynamics is also an upward flow, whose rate is interpolated between those of $B$ and $A$ in a continuous way. Of course near the bottom and top boundaries, the movement has to slow down and stop.

Outside $S$, the vertical movement quickly dies off, and turns into horizontal movement parallel to the strip $S$, so that if $A$ is on the left and $B$ on the right, the dynamics moves points say from left to right, so that the closer they are to $S$, the slower they go (and very close they also shift up). We want to introduce some horizontal movement immediately after leaving $S$, so that all points close to $S$ but not on its affine hull will eventually reach the right boundary of $C$. (On the affine hull of $S$ you have to stop movement altogether when you hit $S$.)

In the blackboard photo, see the leftmost figure for a front "perspective" view of $C$ and some indications of the vector field on $S$, and see the bottommost "top view" figure for indications of the horizontal flow.

The point is now that if you go into (properly inside) $C$ from the bottom, somewhere between the bottom points of $A$ and $B$, then you walk up $C$, and you can control how long it takes to reach the top from the bottom of $C$. (Of course the bottom and top have no flow because they are on the boundary, so this is indeed only true if you step properly inside $C$, but that'll change later when we start embedding copies of this in $M$.)

We need some niceness properties from the flow, which we call the splotch properties, because they describe how the dynamics splotches open sets to the boundary of $C$. If you take an open set inside $C$, then we want that almost all points (all but the ones in the two-dimensional affine hull of $S$) will eventually stop moving vertically and start tending towards the right side of $C$. We want that the limit of these points on the right boundary contains a relative open set, i.e. as the open set is squeezed to the right side, the splotch you get in the $\omega$-limit always contains a square (homeomorphic copy of $[0,1]^2$). For the inverse dynamics, we want the same to happen on the left. Assuming far enough from $S$ there is no vertical movement whatsoever, this should be more or less automatic, I didn't write any formulas though. Another thing we need that if $U$ tends to the splotch $U^+$ on the right hand side and we pick a small square $D$ inside $U^+$, then some small open ball inside $U$ has its splotch ($\omega$-limit) contained in $D$.

Now, having $f_{\epsilon} : \mathbb{R} \times C \to C$ with these properties, let's think of $C$ as very flexible and as carrying the dynamics of $f_{\epsilon}$. So when I stretch $C$ around the manifold $M$, the conjugate dynamics follows along.

Now, enumerate a sequence of pairs of open sets $(U_i, V_i)$ in $M^2$ so that for any pair of open sets $U, V$ in $M^2$ there exists $i$ such that $U_i \subset U$ and $V_i \subset V$. We may assume $U_i$ and $V_i$ are open balls whose radius tends to $0$ very quickly. We build a sequence of flows inductively. To get the first one, called $g_1$, take a path from the center of $U_1$ to the center of $V_1$ and position an elongated $C$ called $C_1$ along this path so that its bottom is in $U_1$ and its top is in $V_1$. Now, the $f_{\epsilon_1}$-flow in $C$ (pick some tiny $\epsilon_1$) turns into a flow $g_1$ on $M$ which in $C_1$ uses the flow conjugate to $f_{\epsilon_1}$, and fixes all other points of $M$. Observe that for any large enough $m$, in the time-$1$ flow of $g_1$ we can get from $U_1$ to $V_1$ in exactly $m$ steps, by picking a suitable position between the lines $A$ and $B$.

Now, we continue the construction process. We have some $U_2$ and $V_2$, and want to do the same for them. If they are disjoint from $C_1$, then this can be done in exactly the same way, and if they are not entirely inside $C_1$ they can be refined to be completely disjoint from $C_1$. So consider the case where one or both are inside $C_1$; suppose for concreteness that both are inside $C_1$. Follow the dynamics forward from $U_2$. It travels according to the flow of $C_1$ until most of it gets very close to the side of $C_1$ that corresponds to the right side of $C$, and it tends to some splotch $U_2^+$ in the $\omega$-limit. Follow it also backward to obtain an $\alpha$-limit $U_2^-$ on the left side. Then do the same for $V_2$ to get $V_2^+$ and $V_2^-$. Now, it's possible that $U_2^+$ and $V_2^+$ intersect, then refine these sets to be smaller so that they don't, using the splotch properties. Do the same in the backward direction.

Now, since $U_2^+$ contains a square $D \cong [0,1]^2$ on the boundary of $C_1$, we can glue another copy $C_2$ of $C$ so that $D$ becomes the bottom square of $C_2$, and on $C_2$ use the flow $f_{\epsilon_2}$. Of course, since there is no movement on the boundary of $C_1$ nor the boundary of $C_2$, the dynamics are not in any way connected. But we can distort the dynamics near the common boundary of $C_1$ and $C_2$ slightly so that the flow drags points of $C_1$ into $C_2$, in particular we want that some points are dragged into the $S$-strip of $C_2$ and start moving upward along $C_2$.

Now glue the top of $C_2$ to $V_2^-$ and distort the flow on the boundary of $C_1$ and $C_2$ as we did with the bottom. Observe that the distortion can be made arbitrarily small and made to affect an arbitrarily small area, by making the flow along $C_2$ arbitrarily slow by decreasing $\epsilon_2$ and making $C_2$ very thin. Observe that in this new flow $g_2$, in the time-$1$ map we can still get from $U_1$ to $V_1$ in $m$ steps, for any large enough $m$ (as we didn't modify the dynamics on the $C_1$-copy of $S$), but now we can also get from $U_2$ to $V_2$ in any large enough number of steps by following the dynamics to the $S$-strip of $C_2$, and picking a point on the $S$-strip with a suitable rate (the speed at the beginning and the end of these orbits cannot be controlled, but we can freely control the length of time in the middle of the orbit where we go through $C_2$). Note that in the flow $g_2$, the dynamics of every point that is not on (what corresponds to) the affine hull of $S$ in $C_1$ or $C_2$ tends to the boundary of $C_1 \cup C_2$. (We made sure $U_2^+ \cap V_2^+ = \emptyset$ to ensure no periodic behavior is introduced, and every point has singleton $\alpha$ and $\omega$-limit sets.)

Now, the idea is to continue by induction, keeping roughly the following characteristic: we have in $M$ a finite set of very thin elongated blocks $C_i$. Outside the sets $C_i$, there is no movement, and on $C_i$ points move according to $f_{\epsilon_i}$, with $\epsilon_i$ tending to $0$ very fast, plus some linking behavior at their common boundaries. Every point that is not on (a set correspond to) the affine hull a copy of $S$ in some $C_i$ will eventually tend to the boundary of their union, so in particular every open set will contain an open ball which has such movement and a well-defined connected splotch as its $\omega$-limit. To get the next step of the construction, for $U_{i+1}, V_{i+1}$ again follow their forward and backward iterates until you hit splotches on some boundaries (the sets may split finitely many times as you travel between the sets $C_j$, but just refine them; againmake sure $U_{i+1}^+ \cap V_{i+1}^+ = \emptyset$ and $U_{i+1}^- \cap V_{i+1}^- = \emptyset$). Now drag a very thin copy of $C_{i+1}$ between these splotches. Observe that the existing $C_j$s don't disconnect $M$ (if you made them thin enough), so it is indeed possible to position $C_{i+1}$ this way. Finally add a bit of additional flow to connect the dynamics of $C_{i+1}$ to whereever you glued them on the boundary of $\bigcup_{j = 1}^i C_j$. Observe that no periodic points are introduced at any finite level.

Now, take the pointwise limit of these flows as $i \rightarrow \infty$, say $g$. This should tend to a flow assuming $\epsilon_i$ tends to zero fast enough, and everything is regular enough, and all the added $C_i$s are small enough and so on. Just do the math and you'll see (obviously I haven't done it or I'd show it).

On the other hand, if $\epsilon_i$ goes to zero very fast, clearly there is no entropy. This is true on finite levels of the construction just because every point has singleton limit sets. To get it for $g$, do the calculations, make sure not to introduce too many new partial orbits with respect to any fixed resolution $\epsilon > 0$ at finite steps.

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    $\begingroup$ Two important points: 1. I'm not very used to describing flows or geometric constructions in text since I usually only work in one dimension (namely, zero), so this is probably not so easy to read. Hopefully the idea is possible to extract (and hopefully it's ok). 2. I saw this question just before going to bed, and in my dream I got a 20 point bounty for this construction. A man can dream, as they say. $\endgroup$
    – Ville Salo
    Oct 17, 2019 at 7:25
  • $\begingroup$ You may get more than bonus points; see the edit to my question. Possibly no example like this has been published. $\endgroup$ Oct 19, 2019 at 23:36
  • $\begingroup$ Ok, but two important points 1. Google scholar does not give you a green +10 when you get a citation, so what's the point really? 2. My paper will probably conclude with "I still didn't do the math". $\endgroup$
    – Ville Salo
    Oct 20, 2019 at 6:21
  • $\begingroup$ After some quick googling, I feel skeptical that this is really new, but I'll give it some thought next week when I have better access to the literature. $\endgroup$
    – Ville Salo
    Oct 20, 2019 at 10:14

The simplest counterexample would be the identity map on a one-point space.

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    $\begingroup$ good point, I should specify that I only think about the non-degenerate spaces $\endgroup$ Oct 17, 2019 at 21:04

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