Timeline for Transformations whose product with the odometer are ergodic
Current License: CC BY-SA 3.0
4 events
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| Jan 29, 2016 at 4:38 | comment | added | Anthony Quas | So the general statement is that if $S$ and $T$ are ergodic, then $S\times T$ is ergodic unless $S$ and $T$ have a common eigenvalue (other than 1). | |
| Jan 29, 2016 at 2:17 | comment | added | Stéphane Laurent | Ok, thank you @AnthonyQuas. I see: saying that $e^{2\pi im/2^n}$ is an eigenvalue is the same as saying that $T^{2^n}$ is not ergodic. I still have to study your answer to the other post. I link it here by the way: mathoverflow.net/questions/229435/… | |
| Jan 29, 2016 at 1:38 | comment | added | Anthony Quas | Same proof as I showed you the other day. $O$ has eigenvalues $e^{2\pi im/2^n}$ for each $m$ and $n$. If $T$ has any of these eigenvalues then you have non-trivial invariant functions. If not, then $T\times O$ is ergodic for the same reason, by taking the same decomposition of $L^2(X)$ into continuous and discrete spectrum. | |
| Jan 29, 2016 at 1:17 | history | asked | Stéphane Laurent | CC BY-SA 3.0 |