Timeline for Decomposition of a dynamical system into ergodic componenents
Current License: CC BY-SA 2.5
12 events
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| Nov 8, 2010 at 10:14 | comment | added | Łukasz Grabowski | R W: No, I indeed meant just the standard product action. However, I accepted coudy's answer because of the information about measure decomposition - this is enough to solve the question I had. | |
| Nov 6, 2010 at 23:36 | comment | added | R W | @coudy: Thank you for making the correction as I am not sure whether the author had indeed meant skew products (appearance of subsets $U_i$ in his question makes me doubtful). I guess the issue is closed now. | |
| Nov 6, 2010 at 23:29 | comment | added | R W | @Lukasz Grabowski: (1) Take the full shift on 2 symbols and consider the family of $(p,1−p)$ Bernoulli measures with $0<p<1/2$. (2) You are right - it is ergodicity which is used here. (3) By now you probably know how to complete the proof. The sets $A_i$ must be such that the intersection with a.e. ergodic component has non-zero conditional measure (which can be done by Rohlin's theorem). By the way, it is enough to take not all elements of $H$, but only its generators - so that if $H$ is finitely generated, then the graphings arising in the proof have uniformly bounded vertex degrees. | |
| Nov 6, 2010 at 18:52 | vote | accept | Łukasz Grabowski | ||
| Nov 6, 2010 at 18:52 | |||||
| Nov 6, 2010 at 15:32 | comment | added | coudy | @R W. The quick question asks for a product action, without giving a precise meaning to the term "product". The technical term is "skew-product" here, the action on the ith level C_i depends on i. So the answer is yes if the term "product" is understood that way. Does it answer your question ? | |
| Nov 6, 2010 at 14:35 | comment | added | Łukasz Grabowski | @R W : Thanks for noting there are problems with atoms. I tend to forget about atoms, although for this particular application they're not necessary, since cost isintersting mainly for infinite groups, which can't act freely on a psace with atoms. | |
| Nov 6, 2010 at 14:31 | comment | added | Łukasz Grabowski | @R W : 1) "which can be easily put together as ergodic components" - how do you do that? It's not immediately clear to me how to do it in a measurable way. 2) "However, you don't really need ergodicity for the quoted argument, because what is actually used there is recurrence of the action (which follows in this case from the Poincare theorem)" I beg to differ :-). In the argument I need to choose a measurable set which intersects almost all orbits. This has nothing to do with Poincare theorem as I know it, which says that for a given sets almost all orbits intersecting it are recurrent. | |
| Nov 6, 2010 at 14:10 | comment | added | R W | Of course, it is not a problem - one just had to mention it - as otherwise the way you formulate Rohlin's theorem in this case is wrong. However, I repeat my question: do you still claim that the answer to the original query is "yes"? | |
| Nov 6, 2010 at 13:56 | comment | added | coudy | Yes, I forgot to remove the periodic points, but I don't think this is a problem here. The action on the set of periodic points of period n is isomorphic to {1,..n}xB, where the action on B is just identity. So the result also holds for the set of periodic points. | |
| Nov 6, 2010 at 13:37 | comment | added | R W | Sure (and this is explicitly mentioned in my answer) - but even in this case one has to take care of the periodic points (as you did it with the atomic part of the action) - right? By the way, it would be good if we came to a common opinion about author's original "quick question". | |
| Nov 6, 2010 at 13:20 | comment | added | coudy | May I point out that the question is about probability measure preserving actions (pmp) ? | |
| Nov 5, 2010 at 17:21 | history | answered | R W | CC BY-SA 2.5 |