Timeline for Minimal, uniquely ergodic but not Lebesgue-ergodic?
Current License: CC BY-SA 4.0
29 events
| when toggle format | what | by | license | comment | |
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| S Oct 28, 2018 at 22:36 | history | bounty ended | Selim G | ||
| S Oct 28, 2018 at 22:36 | history | notice removed | Selim G | ||
| Oct 28, 2018 at 22:36 | vote | accept | Selim G | ||
| Oct 28, 2018 at 20:06 | answer | added | Andres Koropecki | timeline score: 3 | |
| S Oct 22, 2018 at 21:38 | history | bounty started | Selim G | ||
| S Oct 22, 2018 at 21:38 | history | notice added | Selim G | Improve details | |
| Oct 22, 2018 at 21:38 | vote | accept | Selim G | ||
| Oct 22, 2018 at 21:38 | |||||
| S Oct 22, 2018 at 20:00 | history | bounty ended | CommunityBot | ||
| S Oct 22, 2018 at 20:00 | history | notice removed | CommunityBot | ||
| Oct 15, 2018 at 6:19 | history | edited | Selim G | CC BY-SA 4.0 |
added 311 characters in body
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| Oct 15, 2018 at 6:09 | history | edited | Selim G | CC BY-SA 4.0 |
added 34 characters in body
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| Oct 15, 2018 at 1:55 | answer | added | R W | timeline score: 5 | |
| Oct 14, 2018 at 22:30 | comment | added | R W | @Christian Remling As Anthony says, the definitions used by the OP are absolutely standard and the question is completely clear. The notion of ergodicity is routinely defined and used for quasi-invariant measures (and their classes) as well, and OP tacitly referred to the fact that the Lebesgue measure class (one shouldn't really talk about the Lebesgue measure on a smooth manifold) is quasi-invariant with respect to any diffeomorphism. | |
| S Oct 14, 2018 at 18:10 | history | bounty started | Selim G | ||
| S Oct 14, 2018 at 18:10 | history | notice added | Selim G | Draw attention | |
| Oct 14, 2018 at 16:12 | comment | added | Anthony Quas | @ChristianRemling: this is actually a standard definition in non-singular ergodic theory, although I agree the juxtaposition of this terminology with the term uniquely ergodic is quite unfortunate. | |
| Oct 14, 2018 at 16:03 | comment | added | Christian Remling | @AnthonyQuas: That (this non-standard definition of "ergodic") may have been the OP's intention, but it seems a bad idea: for example, then any measure $\delta_x$ is "ergodic," so how could a system ever be uniquely "ergodic" ? It really would have been better to use the standard definitions and say clearly and explicitly what is meant. | |
| Oct 14, 2018 at 1:29 | comment | added | Anthony Quas | So the definition of ergodic as I understand the OP is $T^{-1}A=A$ implies $\lambda(A)=A$. This makes sense for non-invariant measures. If one starts with a circle rotation and conjugated by something smooth, the new map preserves the push-forward of the original measure (which will generally not be Lebesgue). Nonetheless, Lebesgue measure is ergodic for the new map according to the definition that the OP is using. | |
| Oct 13, 2018 at 15:10 | comment | added | Christian Remling | @AnthonyQuas: Actually, I have no idea what the second part of your comment is about. The transformed measure is invariant because it's the same as before, just expressed in the new coordinates, and, in general, ergodic measures are invariant as part of the definition (at least for me). | |
| Oct 13, 2018 at 14:32 | comment | added | Selim G | @ChristianRemling No I didn't meant that. Although item 2 and 3 together imply that the invariant measure is not equivalent to Lebesgue, 3. demands that Lebesgue is not ergodic which is much stronger. | |
| Oct 13, 2018 at 12:35 | comment | added | Christian Remling | @AnthonyQuas: Yes, of course I know all that. The OP asked (item 3) for an invariant measure that is not Lebesgue measure, and my trivial example does that. (It now turns out the OP really meant: the ergodic measure is not equivalent to Lebesgue measure.) | |
| Oct 13, 2018 at 6:59 | comment | added | Selim G | @ChristianRemling I am making no assumption on the ergodic measure. It seems to me that the question makes sense formulated as it is... | |
| Oct 13, 2018 at 5:36 | comment | added | Anthony Quas | @ChristianRemling: If you do a smooth change of coordinates, then the Lebesgue measure is (of course) mapped to something mutually absolutely continuous with respect to Lebesgue. This measure remains ergodic (even though it is not invariant): for any invariant set $A$, if you pull it back by the change of coordinates has measure 0 or 1, and so $A$ itself has 0 or 1 with respect to Lebesgue. | |
| Oct 12, 2018 at 19:53 | comment | added | Christian Remling | So you mean you want the ergodic measure to have a singular component? You didn't say this though. | |
| Oct 12, 2018 at 19:44 | comment | added | Selim G | In that case you won’t be able to produce any example. | |
| Oct 12, 2018 at 18:08 | comment | added | Christian Remling | By "non-standard coordinates," I meant a smooth change of variable, so the map stays $C^{\infty}$. | |
| Oct 12, 2018 at 16:45 | comment | added | Selim G | Hi Christian. If you use "non-standard coordinates", then you loose the fact that your map is a diffeo. For any smooth manifold, the class of the Lebesgue measure is well-defined and it makes sense to say that a diffeo is ergodic (when the measure of any invariant set or its complement is zero for any representative of the class). It is standard that any sufficiently regular ($\mathcal{C}^2$) minimal circle diffeo is ergodic with respect to Lebesgue. | |
| Oct 12, 2018 at 16:37 | comment | added | Christian Remling | Just take something like a shift on the circle, but use non-standard coordinates so that Lebesgue measure becomes something else in the new coordinates. (More to the point perhaps, what is "Lebesgue measure" anyway on a manifold?) | |
| Oct 12, 2018 at 14:15 | history | asked | Selim G | CC BY-SA 4.0 |