Timeline for Unique equilibrium states for systems without specification
Current License: CC BY-SA 2.5
5 events
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| Dec 3, 2010 at 18:41 | history | edited | Ian Morris | CC BY-SA 2.5 |
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| Dec 3, 2010 at 15:19 | comment | added | Ian Morris | Ah, good point. It seems very unlikely to me that an irrational rotation will be disjoint from all of the invariant measures simultaneously. Introducing skewing might solve the problem, but more likely is just going to make things complicated. | |
| Dec 3, 2010 at 14:53 | comment | added | Vaughn Climenhaga | Another potential issue is that even if two maps $T_1$ and $T_2$ are uniquely ergodic with measures $\mu_1, \mu_2$, the direct product $T_1 \times T_2$ can have other invariant measures besides $\mu_1\times \mu_2$. For example, if $R_\alpha$ is an irrational circle rotation, then $R_\alpha\times R_\alpha$ has lots of invariant measures, despite the fact that $R_\alpha$ is uniquely ergodic. So you need the measure-preserving systems $(T_1,\mu_1)$ and $(T_2,\mu_2)$ to be "disjoint", which may well be true in the case you mention, but I don't immediately know how to prove. | |
| Dec 3, 2010 at 14:50 | comment | added | Vaughn Climenhaga | That might work, but I'm not entirely convinced yet. One potential issue is that some uniquely ergodic maps, such as irrational rotations, have a weak version of the specification property (for every $\epsilon>0$ there exists $\tau$ such that any collection of orbit segments can be $\epsilon$-shadowed by a single orbit that takes at most $\tau$ iterates to pass from one segment to the next, and is not required to be periodic). This version does not imply the existence of any periodic orbits. I'm not sure whether it still holds for the symbolic coding of such a rotation, which is expansive. | |
| Dec 2, 2010 at 22:34 | history | answered | Ian Morris | CC BY-SA 2.5 |