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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. 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). So assume:

  1. there is at least a point with dense orbit;
  2. there is at least a periodic point;
  3. also, assume there is no isolated point.

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.

YCor
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Akhil Mathew
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  • 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.) – Qiaochu Yuan Jul 20 '10 at 01:08
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    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 – Akhil Mathew Jul 20 '10 at 01:16
  • Do you mean to assume that, instead of being connected, X has no isolated points? Otherwise you rule out the sequence spaces like the one mentioned in your question. – Robin Tucker-Drob Jul 20 '10 at 01:22
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    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.

    – Vaughn Climenhaga Jul 20 '10 at 01:23
  • Or, as Robin says, you could rule out isolated points, with similar effect. I think this is actually equivalent to my suggestion. – Vaughn Climenhaga Jul 20 '10 at 01:26
  • (Also: Hi, Robin! Good to see you on here!) – Vaughn Climenhaga Jul 20 '10 at 01:27
  • @Vaughn: Oh yeah, that's right. So leaving out isolated points is probably better. – Akhil Mathew Jul 20 '10 at 01:45
  • [I edited the question to correspond to that.] – Akhil Mathew Jul 20 '10 at 01:48
  • Am I missing something or does the cat map work? – Steve Huntsman Jul 20 '10 at 02:00
  • The cat map is an Anosov diffeo, hence semi-conjugate to a Markov shift. Thus it contains many orbits which are neither dense nor periodic -- just consider any non-periodic sequence that does not contain a particular word of your choosing. – Vaughn Climenhaga Jul 20 '10 at 02:18
  • Also, there are homoclinic points (which I'm pretty sure corresponds to what Vaughn said). – Akhil Mathew Jul 20 '10 at 02:20
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    @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. – Vaughn Climenhaga Jul 20 '10 at 02:38
  • Fair enough, that is a good point -- I haven't yet looked to carefully at the Markov partition semiconjugacy between symbolic dynamics and toral endomorphisms. – Akhil Mathew Jul 20 '10 at 03:04
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    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. – Andrey Gogolev Jul 20 '10 at 05:44
  • @Andrey: Thanks! If you posted that as an answer, I'd definitely upvote it. – Akhil Mathew Jul 20 '10 at 07:15

5 Answers5

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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.

Pengfei
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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).

  • Thanks! By the way, do you have a reference to your PhD thesis? – Akhil Mathew Jul 20 '10 at 01:58
  • You can find the official reference at http://cat.inist.fr/?aModele=afficheN&cpsidt=185680, and a PDF (with the wrong date on it?!?) at http://www.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]. – Jacques Carette Jul 20 '10 at 03:07
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    Akhil was asking for a homeomorphism. – Andrey Gogolev Jul 20 '10 at 05:40
  • @Andrey - I missed that part, thanks for pointing it out. These are indeed not homeomorphisms, since they have critical points. – Jacques Carette Jul 20 '10 at 13:03
  • 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. – Lasse Rempe Aug 24 '15 at 11:17
  • With my "complex dynamics" hat on, I read the question to be about critical point orbits -- which was not what the question was about. That is surprising enough as it is that I thought I would point it out! – Jacques Carette Aug 24 '15 at 21:12
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This reminds me of the theorem of Le Calvez and Yoccoz: There is no minimal homeomorphism on the multipunctured sphere, i.e. there is no homeomorphism on the 2-sphere such that every orbit is dense except a finite set. Clearly, the finite set consists of periodic points.

Now, to question. Google leads to https://arxiv.org/abs/1605.08873 This article is exactly the idea of Andrey Gogolev. This technique shows that there are diffeomorphisms on all orientable surfaces (of course, except the sphere) such that the non dense orbits form a finite set. I am sure that there are older proofs.

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Here is a really simple example, based on the proposer's comments. Let $X$ consist of two fixed points and a single orbit, whose forward limit point is one of the fixed points and whose backward limit point is the other. This is a compact zero-dimensional system, easily embeddable in the full 2-shift: the fixed points are $0^{\infty}, 1^{\infty}$, and the single orbit is that of $0^{-N}1^N$. Or did I miss something?

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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.

whyambee
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