Is product of rotation map topologically transitive?

Question

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.

Answer 1

$f\times f$ 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.

Answer 2

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