Let $(T,X)$ be a discrete dynamical system. By this I mean that $X$ is a compact Hausdorff space and $T: X \to X$ a homeomorphism.

For example, take $X$ to be the sequence space $2^{\mathbb{Z}}$ and $T$ the Bernoulli shift. Then there is a dense set of periodic points, and there is another (disjoint) dense set of points whose orbits are dense (in view of topological transitivity). However, there are also points of $X$ that belong to neither of these sets---for instance, the sequence $(\dots, 1, 1, 1, 0, 0, 0, \dots )$.

This led to me the following:

Question: Is there a discrete dynamical system $(T,X)$ such that every point has either a finite or a dense orbit? (Cf. the below caveats.)

There are a few caveats to add. First, we want both periodicity and topological transitivity to occur; this rules out examples such as rotations of the circle (where every point is of the same type, either periodic or with a dense orbit). Second, let's assume there are no isolated points.

I've been thinking on and off about this question for a couple of days, and the basic examples of dynamical systems that I learned (shift spaces, toral endomorphisms, etc.) don't seem to satisfy this condition, and intuitively it feels like the compactness condition should imply that there are points which are "almost periodic," but not, kind of like the $(\dots, 1, 1, 1, 0, 0,0, \dots)$ example mentioned earlier. Nevertheless, I don't see how to prove this.

  • $\begingroup$ Is it clear that no dynamical system satisfying your conditions can have all orbits finite or all orbits dense? (I can't seem to find a definition of "periodicity" online.) $\endgroup$ – Qiaochu Yuan Jul 20 '10 at 1:08
  • $\begingroup$ Yes, a rotation of a circle has either every point periodic (i.e. with finite orbit) or every orbit dense. The first case occurs if the angle is rational, the second if it is irrational $\endgroup$ – Akhil Mathew Jul 20 '10 at 1:16
  • 2
    $\begingroup$ I suspect that "X is connected" is not quite the criterion you want in order to rule out trivial examples. The disjoint union of a rotation with a one-point system already fails without this criterion because the orbits on the circle are no longer dense once the isolated fixed point is added. Furthermore, many interesting examples can be realised as shift spaces, which are totally disconnected, so I'd be hesitant to rule those out. The criterion I would add instead is that there be at least two distinct dense orbits, otherwise a fixed point with a homoclinic orbit works. $\endgroup$ – Vaughn Climenhaga Jul 20 '10 at 1:23
  • 2
    $\begingroup$ @Akhil: Homoclinic and heteroclinic orbits are particular examples, but they're not the most interesting (IMHO) since their alpha- and omega-limit sets are finite. There are also lots of orbits whose alpha- and omega-limit sets are invariant Cantor sets that are not all of X. For example, if X is the full shift on two symbols, and x is a sequence that contains every finite word except for those containing two consecutive 1s, then the closure of the orbit of x is a Markov shift. Thus it's intermediate between being periodic and being dense, in a rather different way than a homoclinic orbit. $\endgroup$ – Vaughn Climenhaga Jul 20 '10 at 2:38
  • 5
    $\begingroup$ Take a minimal flow on 2-torus generated by a constant vector field. Multiply the vector field by a function which is positive everywhere but at one point. You'll get a flow with one fixed point, other points are transitive (not all of them forward transitive though). I bet time one map inherits these properties. $\endgroup$ – Andrey Gogolev Jul 20 '10 at 5:44

I believe you will find such examples for $X=\mathbb{C}$ and $T$ a rational map in

Mary Rees, Ergodic rational maps with dense critical point forward orbit, Ergodic Theory and Dynamical Systems 4 (1984), 311-322. official version.

In my Ph.D. thesis, I showed that some of these even support a metric with respect to which these dynamical systems are ``hyperbolic''. This metric gives a notion of length of curves comparable to the usual metric on the Riemann sphere, but is defined by a function which is singular on a dense set of points on the sphere (the forward orbit of the critical point).

  • $\begingroup$ Thanks! By the way, do you have a reference to your PhD thesis? $\endgroup$ – Akhil Mathew Jul 20 '10 at 1:58
  • $\begingroup$ You can find the official reference at cat.inist.fr/?aModele=afficheN&cpsidt=185680, and a PDF (with the wrong date on it?!?) at cas.mcmaster.ca/~carette/publications/CaretteThesis.pdf. Never published it as I first went to industry, then switched fields before coming back to academia. [Oh, and it's in French]. $\endgroup$ – Jacques Carette Jul 20 '10 at 3:07
  • 2
    $\begingroup$ Akhil was asking for a homeomorphism. $\endgroup$ – Andrey Gogolev Jul 20 '10 at 5:40
  • $\begingroup$ @Andrey - I missed that part, thanks for pointing it out. These are indeed not homeomorphisms, since they have critical points. $\endgroup$ – Jacques Carette Jul 20 '10 at 13:03
  • $\begingroup$ I don't believe a rational map can have every orbit finite or dense, unless I am missing something. For example, there are always plenty of hyperbolic Cantor sets of nonzero Hausdorff dimension. Of course, you can have almost every point dense, but that is not the question here. $\endgroup$ – Lasse Rempe-Gillen Aug 24 '15 at 11:17

In the following paper the authors give an almost 1-1 extension for a minimal system $(X,\mathbb{Z})$ which is transitive and the only non-transitive point is a fixed point.

For $\mathbb{N}$ action they can have a similar one with positive topological entropy.

T.Downarowicz, X. Ye: When every point is either transitive or periodic, Colloq. Math. 93 (2002) pp. 137-150.

I do not know whether hese examples can exists on manifolds.

  • $\begingroup$ That's a great result! Surprisingly short proof, too. $\endgroup$ – Ian Morris Nov 27 '10 at 15:12

one can construct a self-mapping of a disk, mixing in the interior, identical on the boundary and glue together several copies using a periodic rotation of the boundary.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.