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Oct 3, 2020 at 18:41 comment added Qiaochu Yuan $S^1$ acts freely and smoothly on every odd sphere $S^{2k-1}$ (regarded as the unit sphere in $\mathbb{C}^k$, with the action given by scalar multiplication) so generates a nowhere-vanishing vector field whose flow is very much non-ergodic.
Oct 3, 2020 at 18:06 comment added Christian Remling But there's no reason why specifically Lebesgue measure would be invariant. For example, you can move fast through some regions of $S^1$ and slowly through others.
Oct 3, 2020 at 16:26 comment added G. Panel 1) Ergodic with respect to the Lebesgue measure. 2) I take a look a that, thank you!
Oct 3, 2020 at 16:17 comment added D. Thomine 1) Ergodic with respect to which measure? 2) Beware of the Hopf fibration.
Oct 3, 2020 at 15:26 history asked G. Panel CC BY-SA 4.0