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Define rotation map on $f:S^{1}\rightarrow S^{1}$ such that $\theta \rightarrow \theta +2\pi\alpha, $ where $\alpha$ is some fixed irrational. Is $f\times f$ topologically transitive?

A function $f:X\rightarrow X$ where $(X, d) $ is a metric space, is said to be topologically transitive if for every pair of non-empty disjoint open sets $U$ and $V$ of $X$, there exist some natural number $n$ such that $f^{n} (U) \cap V$ is non empty.

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    $\begingroup$ $f$ is topologically transitive. I am trying to show what will happen to direct product. If direct product is not topologically transitive then we need to find such open sets. What will be that? $\endgroup$ – Abdul Gaffar Khan Feb 12 '17 at 13:43
  • $\begingroup$ That is not true for $\alpha$ rational. $\endgroup$ – Test123 Feb 12 '17 at 13:46
  • $\begingroup$ Some definitions for topologically transitive sets do not assume that $U$ and $V$ are disjoint. Just distinct. $\endgroup$ – Michael Burr Feb 12 '17 at 13:48
  • $\begingroup$ I edited. Please check, $\alpha $ is irrational. It is transitive as, orbit is dense and $S^{1} $ is thick space. What will happen to its direct product to itself? $\endgroup$ – Abdul Gaffar Khan Feb 12 '17 at 14:15
  • $\begingroup$ Any good reference of proof saying torus have have rational slope? $\endgroup$ – Abdul Gaffar Khan Jan 22 '18 at 9:08
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$f\times f$ is not topologically transitive: the slope on the torus is rational and so all orbits of $f\times f$ are contained in a closed curve (a geodesic in the flat metric), which thus is not dense.

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  • $\begingroup$ Weak mixing imply transitivity but converse need not true.. So even if it is not weak mixing we can not say it is transitive $\endgroup$ – Abdul Gaffar Khan Jan 22 '18 at 9:07
  • $\begingroup$ How does your comment relate to my complete answer? $\endgroup$ – John B Jan 22 '18 at 10:25
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Irrational rotation on $S^1$ is an isometry and hence not weak mixing. That means $f \times f$ is not transitive.

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