Is product of rotation map topologically transitive?

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.

• $f$ 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? – Abdul Gaffar Khan Feb 12 '17 at 13:43
• That is not true for $\alpha$ rational. – Test123 Feb 12 '17 at 13:46
• Some definitions for topologically transitive sets do not assume that $U$ and $V$ are disjoint. Just distinct. – Michael Burr Feb 12 '17 at 13:48
• 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? – Abdul Gaffar Khan Feb 12 '17 at 14:15
• Any good reference of proof saying torus have have rational slope? – Abdul Gaffar Khan Jan 22 '18 at 9:08

$f\times f$ 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.
Irrational rotation on $S^1$ is an isometry and hence not weak mixing. That means $f \times f$ is not transitive.