Timeline for Connection between properties of dynamical and ergodic systems
Current License: CC BY-SA 4.0
22 events
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| S Mar 20, 2022 at 1:54 | history | suggested | The Amplitwist | CC BY-SA 4.0 |
fixed broken link to springerlink.com
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| Mar 19, 2022 at 20:46 | review | Suggested edits | |||
| S Mar 20, 2022 at 1:54 | |||||
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Mar 10, 2017 at 9:42 | history | edited | CommunityBot |
replaced http://www.math.uh.edu/ with https://www.math.uh.edu/
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| Sep 30, 2016 at 19:48 | comment | added | Vaughn Climenhaga | The definition you give for "topologically ergodic" is what I called "topologically transitive" in my answer. "Uniquely ergodic" means that there is exactly one invariant measure for the system (X,T). The fact that there is always at least one invariant measure is the Krylov-Bogolyubov theorem; in many cases, however, there is more than one. If there's only one, then we say (X,T) is uniquely ergodic. So the definition does involve measures, but it's a definition about a topological dynamical system, not about a measure-preserving system. | |
| Sep 30, 2016 at 19:44 | comment | added | Leo | Could you please clarify what you mean by "uniquely ergodic"? I had problems finding a definition that doesn't involve measures. Would it be synonymous to "topologically ergodic"? I.e. for any non-empty open $U$ and $V$ there is an $n>0$ s.t. $T^n(U)\cup V \neq \emptyset$. | |
| Dec 3, 2012 at 14:45 | comment | added | Vaughn Climenhaga | It does raise the further point that transitivity depends on whether you consider $n\in \mathbb{N}$ or $n\in \mathbb{Z}$. Your examples are transitive in the latter sense but not in the former. So I suppose really one should add another element to the diagram, distinguishing between these two possibilities (I think I implicitly used forward transitivity without clearly specifying that). Now we have (at least) 4 definitions of transitivity (dense orbit/open sets, $\mathbb{N}$/$\mathbb{Z}$). I think the examples in the answer all satisfy either all or none of those definitions. | |
| Dec 3, 2012 at 14:42 | comment | added | Vaughn Climenhaga | @Ian: Ah yes, I see that indeed I did manage to misunderstand your examples. When you said "homoclinic orbit" I pictured a continuous loop based at the fixed point (likely because I'm teaching ODEs this term so I pictured a homoclinic trajectory for an ODE). Of course that's not what you meant, and now that I understand your examples as being one or more fixed points together with a discrete orbit, I agree there is a dense orbit. | |
| Dec 3, 2012 at 11:18 | comment | added | Ian Morris | Topological transitivity is equivalent to the existence of a dense orbit. Indeed, the books by Walters and by Katok and Hasselblatt take this as the definition. My suggestions for 1 and 8 both have a dense orbit. Using the other popular version of the definition, if $U$ and $V$ are nonempty open sets then it is clear that both must contain a point from the connecting homoclinic or heteroclinic orbit (call them $x \in U$ and $y \in V$). Since by definition $T^nx=y$ for some $n \in \mathbb{Z}$ we have $y \in T^nU \cap V \neq \emptyset$. | |
| Dec 3, 2012 at 1:32 | comment | added | Vaughn Climenhaga | @Ian: I'm not sure about your homo/heteroclinic orbit examples for 8, 1 - I don't see where transitivity should come from, unless I'm misunderstanding the example. Your suggestion for 9 amounts to gluing together two copies of the present example, but one copy seems to do the trick. Certainly you're correct on 6, though I find it marginally unsatisfying to cite the trivial example. | |
| Dec 3, 2012 at 1:29 | comment | added | Vaughn Climenhaga | @Andres: As you say, the north-south map should have endpoints identified, I've edited. | |
| Dec 3, 2012 at 1:28 | comment | added | Vaughn Climenhaga | @Andres: Thanks for the examples, I've added them to the answer, even if the example for 8 is not completely satisfying for the reasons you say. | |
| Dec 3, 2012 at 1:28 | comment | added | Vaughn Climenhaga | @R W: Yes, certainly the world is broader if one considers measures that are not necessarily finite. Good point. Everything here is for finite invariant measures. | |
| Dec 3, 2012 at 1:27 | history | edited | Vaughn Climenhaga | CC BY-SA 3.0 |
Added examples suggested in the comments, clarified a couple points
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| Nov 29, 2012 at 16:22 | comment | added | Ian Morris | 1: Two fixed points connected by a heteroclinic orbit. 9: One fixed point together with two homoclinic orbits. | |
| Nov 29, 2012 at 16:19 | comment | added | Ian Morris | 6: A set with one point in it. | |
| Nov 29, 2012 at 16:11 | comment | added | Ian Morris | 8: A set with two points in it, and a transformation which maps both points to the same point. If you want a homeomorphism, use a map with one point and one homoclinic orbit. | |
| Nov 25, 2012 at 15:34 | comment | added | Andres Koropecki | In 9: I assume you are identifying the two endpoints of the interval, right? So rather than a north-south map this is a south-south map on the circle (with a saddle node). | |
| Nov 25, 2012 at 15:32 | comment | added | Andres Koropecki | For 8: You could do something similar to what you did in 1: Let $X=\mathbb{T}^2\times \{a,b\}$, where the dynamics on $X$ is given by the map from my previous example on the first coordinate and a swap of $a$ and $b$ on the second coordinate (a period two orbit). But I consider this cheating: I would like to see an example on a connected manifold. | |
| Nov 25, 2012 at 15:27 | comment | added | Andres Koropecki | For 5: take the vector field on $\mathbb{T}^2=\mathbb{R}^2/\mathbb{Z}^2$ given by $V(x) = \phi(x)v$ where $v$ is a vector of irrational slope and $\phi\colon \mathbb{T}^2\to \mathbb{R}$ is a map such that $\phi(x)>0$ for all $x\neq x_0$, $\phi(x_0)=0$. If $\phi$ is smooth (or at least if it goes to $0$ fast enough near $x_0$) you can show that the time-one map of $V$ is uniquely ergodic (the unique invariant measure being the Dirac measure at $x_0$). This type of map is topologically mixing as well, but not minimal. | |
| Nov 24, 2012 at 23:38 | comment | added | R W | Wow - that's impressive! I would just add that there is ergodic theory beyond finite invariant measures - even for a single transformation. In particular, there are ergodic properties which don't make much sense for transformations with a finite invariant measure - like conservativity vs dissipativity dichotomy (the latter corresponds to proper discontinuity in the topological category). | |
| Nov 24, 2012 at 22:33 | history | answered | Vaughn Climenhaga | CC BY-SA 3.0 |