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The standard frame for $S^3$ consists of $X_i,X_j,X_k$ with $X_i(a)=ia, X_j(a)=ja, X_k(a)=ka$ where $i,j,k$ are standard quaternion numbers, $a\in S^3$, and the multiplication is the quaternion multiplication. This global frame on $S^3$ gives us a trivialization of $TS^3\simeq S^3\times \mathbb{R}^3$. So the Hopf map $P:S^3\to S^2\subset \mathbb{R}^3$ is counted as a unit vector field on $S^3$. This vector field is called "Hopf vector field".

Does the Hopf vector field have a periodic orbit? Is this vector field discussed in various attempts to find an analytic counter example to Seifert conjecture?

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    $\begingroup$ Isn't every orbit periodic because the Hopf fibration has fiber a circle? $\endgroup$
    – Dan Rust
    Commented Jun 21, 2019 at 23:36
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    $\begingroup$ @DanRust Why do you think the fibers are invariant under flow of this vector foeld?Please see the construction of this vector foeld, again. $\endgroup$ Commented Jun 22, 2019 at 2:46

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I think the question is perhaps confusing, since the term Hopf vector field is usually reserved for your $X_i$ vector field (which is tangent to the fibers of the standard Hopf fibration). As I understand, you are instead referring to the vector field which, at a point $p$ with $P(p) = (a,b,c) \in S^2 \subset \mathbb{R}^3$, is given by $Y := aX_i + bX_j + cX_k$. Assuming this is the correct interpretation, then the fiber over $(1,0,0)$ has $Y = X_i$ and so the fiber of $P$ over $(1,0,0)$ is a periodic orbit of $Y$.

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  • $\begingroup$ Yes thank you very much for your interesting answer. $\endgroup$ Commented Jun 22, 2019 at 20:13
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Seifert actually proved that every vector field close enough to X_i has a periodic orbit. (In other words, if you perturbate slightly X_i whose all orbits are compact, then most orbits will of course in general become noncompact, but there will remain at least one compact orbit). K. Kuperberg built on S^3 a smooth (C^infty) nonsingular vector field without periodic orbit; W. Thurston noticed that one can even make Kuperberg's example real analytic. So, it is not "attempts" any more.

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  • $\begingroup$ Thank you very much for your answer. $\endgroup$ Commented Jul 21, 2019 at 9:40
  • $\begingroup$ Do you mean that the analytic case is not confirmed yet? $\endgroup$ Commented Jul 21, 2019 at 10:52
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    $\begingroup$ I repeat that it is confirmed. See Krystyna Kuperberg's paper : A smooth counterexample to the Seifert conjecture, Ann. of Math Pages 723-732 from Volume 140 (1994), Issue 3; or if you read French, an excellent talk by Ghys at the Bourbaki seminar: Construction de champs de vecteurs sans orbite périodique Ghys, Étienne Séminaire Bourbaki : volume 1993/94, exposés 775-789, Astérisque, no. 227 (1995), Exposé no. 785, p. 283-307, online at numdam.org/book-part/SB_1993-1994__36__283_0 $\endgroup$ Commented Jul 21, 2019 at 15:34

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