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Is it true than an aperiodic, ergodic, minimal and equicontinuous dynamical system on a compact metric space is totally ergodic ?

According to some results I found in some books, a rotation on a compact metric group is equicontinous, and it is minimal and totally ergodic whenever it is ergodic.

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  • $\begingroup$ You should probably add some connectedness hypothesis, since otherwise the claim fails for the non-identity transformation on the two point set. $\endgroup$
    – pavel
    Commented Aug 5, 2015 at 16:49
  • $\begingroup$ @pavel Thank you for this remark. Is it better with "aperiodic" ? $\endgroup$
    – Stéphane Laurent
    Commented Aug 5, 2015 at 16:55
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    $\begingroup$ I think if the powers of $T$ are equicontinuous, then $T$ has to be a group rotation (look up "maximal equicontinuous factor"). $\endgroup$
    – Anthony Quas
    Commented Aug 5, 2015 at 17:13
  • $\begingroup$ @AnthonyQuas I have just seen in Petersen's book that a minimal transformation on a compact metric space is equicontinuous if and only if it is a group rotation. $\endgroup$
    – Stéphane Laurent
    Commented Aug 5, 2015 at 17:29
  • $\begingroup$ @pavel I am under the impression your remark applies to this Proposition 18.37. The last statement is not true for the transformation you mention, isn't it ? $\endgroup$
    – Stéphane Laurent
    Commented Aug 5, 2015 at 17:29

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The answer is no in this generality! If you consider the classical odometer (i.e. addition by 1 on 2-adic integers) then its second power (addition by 2) is not minimal. This second power preserves the numbers starting with 0 (the "even numbers") and the numbers starting with 1 (the "odd numbers"). But of course this example is totally disconnected.

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  • $\begingroup$ I said in the comments that the answer is yes. If you are right, that would mean that theorem 2.11 in Petersen's book is wrong. Are you sure the odometer is minimal and equicontinuous ? BTW I have also shown this result in a different way (actually I posted this question to check my result). $\endgroup$
    – Stéphane Laurent
    Commented Aug 5, 2015 at 21:08
  • $\begingroup$ very strange... the odometer is for sure minimal (not hard to see), equicontinuous (it is a rotation on a compact group). I do not understand the argument of the last sentence of the proof of the Proposition 18.37 that you mentioned: "the last assertion follows from the fact that $T_\phi$ has nontrivial periodic points if and only if $T_{\phi^k}$ does for some/all $k \in \mathbb{N}$". This fact is right, but I do not see how it proves anything about total ergodicity. $\endgroup$
    – V. Delecroix
    Commented Aug 5, 2015 at 21:19
  • $\begingroup$ Thank you for this remark. You are right, there's a problem somewhere. $\endgroup$
    – Stéphane Laurent
    Commented Aug 5, 2015 at 21:52
  • $\begingroup$ If the fact you mention in the proof of Proposition 18.37 is right, then the conclusion follows from the first assertion of the theorem. No ? $\endgroup$
    – Stéphane Laurent
    Commented Aug 5, 2015 at 22:01
  • $\begingroup$ The same trap as here ! Proposition 18.37 is about a measure-preserving group automorphism ! $\endgroup$
    – Stéphane Laurent
    Commented Aug 5, 2015 at 22:52
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Just knowing that a transformation $T$ is minimal is no guarantee that $T^n$ is also minimal. For example, let $T_1$ be the non-identity homeomorphism of a two-point metric space and let let $T_2$ be an aperiodic, minimal, equicontinuous, uniquely ergodic transformation of a compact metric space (for example, an irrational circle rotation). Then $T_1 \times T_2$ is minimal and uniquely ergodic, but $T_1^2 \times T_2^2$ has two nonempty, disjoint, closed invariant sets, and the unique invariant measure for $T_1 \times T_2$ is not ergodic for $T_1^2 \times T_2^2$, since it gives both "halves" of the phase space positive measure, but those "halves" are invariant sets for $T_1^2 \times T_2^2$.

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