Let $T$ be a uniquely ergodic transformation in a compact space with respect to a measure $\mu$, is it true that $T^k$ is ergodic with respect to the same $\mu$?
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What do you mean by "$T^k$ is ergodic"? A transformation is or isn't ergodic for a particular measure. Do you mean ergodic for the unique Borel probability measure for which $T$ is ergodic? – Robert Israel Aug 20 '24 at 13:41
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Sorry I will correct. If it is Uniquely ergodic, respec $\mu$ then I am asking if its ergodic also with respect the same $\mu$ – GatitoPochoton Aug 20 '24 at 13:45
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In dynamics sometimes the adverb "totally" is used to signify that all iterates of the map satisfies the property, as in https://math.stackexchange.com/q/4431241/169085 (More generally an action of a group has property $P$ if all its (reasonable) subgroups has the property $P$). – Alp Uzman Aug 20 '24 at 20:11
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... has property $P$ totally ... – Alp Uzman Aug 20 '24 at 20:19
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Consider the discrete space $\{1,2,3,4\}$ and let $T$ be the cyclic permutation $(1,2,3,4)$, i.e. $T(1) = 2$, $T(2) = 3$, $T(3) = 4$, $T(4) = 1$. It is easy to see that this is uniquely ergodic for the Borel probability measure that gives measure $1/4$ to each of $1,2,3,4$. But $\{1,3\}$ and $\{2,4\}$ are invariant under $T^2$, so $T^2$ is not ergodic for that measure. There are two Borel probability measures for which $T^2$ is ergodic, namely the one that gives each of $\{1\}$ and $\{3\}$ measure $1/2$ and $\{2\}$ and $\{4\}$ measure $0$, and the one that gives $\{2\}$ and $\{4\}$ measure $1/2$ and $\{1\}$ and $\{3\}$ measure $0$.
Robert Israel
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