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Jul 6, 2019 at 15:01 comment added Ali Taghavi @abx yes exactly. He consider the "revetment" de unit tangent bundle of $S^2$.
Jul 6, 2019 at 14:21 comment added abx OK, you are right. The vector field is actually on $SO(3)$ (= the unitary tangent bundle to $S^2$), hence it lifts to $S^3$.
Jul 6, 2019 at 14:07 comment added Ali Taghavi @abx in that paragraph of the paper of Ghys, however he starts the paragraph with 2 dimensional sphere but during the remaining part of the paragraph he is considering 3- sphere.Right?
Jul 6, 2019 at 14:03 comment added Ali Taghavi @abx By Poincare -Bendixon theorem(or some similar methods) one can easilly shows that $S^2$ does not admit any vector field with a dense orbit.
Jul 6, 2019 at 13:42 comment added abx The vector field mentioned in Ghys' paper is on $S^2$, not on $S^3$.
Jul 6, 2019 at 12:41 history edited Ali Taghavi CC BY-SA 4.0
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Jul 6, 2019 at 11:31 comment added Ali Taghavi @Misha Yes but that counter example does not satisfy the condition of this question: the condition is: all orbits would be dense.
Jul 6, 2019 at 10:28 comment added abx @Ali Taghavi: the difference is that $S^3=SU(2)$ is semi-simple. Every element is contained in a circle.
Jul 6, 2019 at 9:42 comment added Misha Real-analytic counterexamples to Seifert conjecture were published by Kuperberg and Kuperberg in Annals of Math., 1996.
Jul 6, 2019 at 8:54 comment added Ali Taghavi @abx Kronecker foliation is invariant in 2 torus. So in S^3 case os not possible that an orbit would be non closed?
Jul 6, 2019 at 7:54 comment added LeechLattice $SO(4)=SO(3) ⊕ SO(3)$, and pick two incommensurable circle actions.
Jul 6, 2019 at 7:39 history edited YCor CC BY-SA 4.0
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Jul 6, 2019 at 7:29 comment added abx Every orbit of a left invariant vector field is closed (it is contained in the translate of a torus).
Jul 5, 2019 at 20:22 history edited Ali Taghavi CC BY-SA 4.0
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Jul 5, 2019 at 20:13 history asked Ali Taghavi CC BY-SA 4.0