Timeline for Invariant mesures for expanding maps of the circle
Current License: CC BY-SA 3.0
8 events
when toggle format | what | by | license | comment | |
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Jul 3, 2015 at 13:11 | comment | added | D. Thomine | @chaut: no, this is false. Let $T$ be the $\times 2$ map, and let $(u_n)$ be an enumeration of the dyadic rationals. Take $x = (0.u_1 v_1 u_2 v_2...)$, where the $v_n$ are finite sequences of $0$. Then, if the length of the $v_n$ grow fast enough, the orbit of $x$ spends most of its time close to $0$, so $\mu_x = \delta_0$. The orbit of $x$ will still be dense, as it goes very close to each dyadic rational. | |
May 11, 2015 at 14:13 | comment | added | chaut | Excuse me,@IanMorris but I have another question: Is it truth that if x has a dense positive orbit and $x$ generates an ergodic measure $\mu_x=\lim \frac{1}{n}\sum \delta_{T^jx}$ then its support is total? | |
Apr 27, 2015 at 18:52 | comment | added | Pietro Majer | ah, ok thanks -I assumed T was a diffeo :) | |
Apr 27, 2015 at 18:38 | comment | added | Ian Morris | @PietroMajer: let $T$ have degree two or higher, for example $T \colon \mathbb{R}/\mathbb{Z} \to \mathbb{R}/\mathbb{Z}$ defined by $T(x):=2x \mod \mathbb{Z}$. | |
Apr 11, 2015 at 22:02 | comment | added | Asaf | Even for the $\times p$ map, there are very wild sets who are the support for an invariant and ergodic measure, basically you can get Cantor sets of every possible dimension. | |
Apr 11, 2015 at 22:02 | comment | added | Pietro Majer | I don't understand... how can $T:\mathbb{S}^1\to \mathbb{S}^1$ have $T'(x)>1$ for all $x$ ? | |
Apr 11, 2015 at 21:43 | review | First posts | |||
Apr 11, 2015 at 22:08 | |||||
Apr 11, 2015 at 21:41 | history | asked | chaut | CC BY-SA 3.0 |