-2
$\begingroup$

I would appreciate if you introduce me a reference (paper or book) who address the concept of hyperbolic dynamics but with emphasis on absence of periodic orbits. a possible title something like 'hyperbolicity in absence of periodic orbits.

My motivation was my following answer to certain post in MO https://mathoverflow.net/a/476303/36688

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Do you have an example in mind of a hyperbolic system without periodic orbits? $\endgroup$
    – Martin M. W.
    Commented Aug 18 at 16:56
  • 2
    $\begingroup$ @MartinM.W. on the $\mathbb{R}^2 \setminus \{0\}$ the map $(2x,y/3)$ $\endgroup$
    – Ali Taghavi
    Commented Aug 18 at 18:08

1 Answer 1

Reset to default
5
+50
$\begingroup$

I'm not sure why people are downvoting this question—I think it's reasonable—but here are a couple of reasons there might not be a rich literature on this topic. One is simply that in many natural situations, especially when there's a compactness condition, hyperbolicity implies the existence of periodic points.

To be sure, one can construct hyperbolic systems with no periodic points, as in the map in the comments: $(x, y) \rightarrow (2x, y/3)$ on $\mathbb{R}^2 \backslash \{0\}$. But it's not clear what the hyperbolicity condition is good for in this simple setting, and it feels a little like "cheating": this is actually a system with a fixed point, but in disguise.

For what it's worth, it is possible to do the same thing in a slightly less trivial setting. The paper "Hyperbolic Nonwandering Sets Without Dense Periodic Points" (M. Kurata, Nagoya Math. J. Vol. 74, 1979) gives an example of a system as described in the title. Looking at their example, it seems that one could remove a finite number of points and have nontrivial hyperbolic recurrence on a non-compact manifold, without any periodic points.

Overall, it seems like hyperbolicity without periodic points ends up being a fairly restrictive condition, so there aren't a lot of natural and interesting examples. That said, who knows, there may be rich structure here that's still undiscovered!

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Thank you very much for your answer $\endgroup$
    – Ali Taghavi
    Commented Aug 20 at 20:58
  • $\begingroup$ One more point is that without recurrence, there's may not be a link between metric conditions and topology. Simplest example: the map $x \rightarrow x + 1$ could be seen as uniformly expanding or uniformly contracting, given an appropriate choice of metric on $\mathbb{R}$. $\endgroup$
    – Martin M. W.
    Commented Aug 21 at 18:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .