# Following curves on S^n

Suppose $V$ is a no-where zero vector field on $S^n$ ($n$ odd). Let $p \in S^n$. Let $\gamma_p$ be the unique curve on $S^n$ through $p$ and tangential to $V$ everywhere along it. Is it true that $\gamma_p$ is a closed curve $\forall p \in S^n$? If so, is it true that the length of $\gamma_p$ must be finite?

Alternatively, is it possible to find a curve $\gamma$ on $S^n$ of infinate length such that the tangent vectors of $\gamma$ can be extended to form a continuous, no-where zero tangential vector field of $S^n$?

This is the Seifert conjecture. There are nowhere-zero vector fields on $S^3$ with no closed orbits, and there are both smooth and real-analytic constructions.

• He only asks for one non-closed orbit. A counter-example to Seifert conjecture answers the question but this is an overkill. – Sergei Ivanov Mar 11 '10 at 19:39
• I guess I was aiming for maximal failure. I suppose I could have offered an irrational winding in the Hopf foliation... – S. Carnahan Mar 11 '10 at 19:48
• But I liked your answer even so, hence +1. – Harald Hanche-Olsen Mar 11 '10 at 21:17

Here is an explicit version of Ryan's example. Consider $S^3$ as the unit sphere in $\mathbb C^2$ and define a vector field $V$ on it by $V(z_1,z_2)=(iz_1,\sqrt 2 i z_2)$. Here $z_1,z_2\in \mathbb C$, $|z_1|^2+|z_2|^2=1$.

Only two trajectories of this field are closed (namely those where $z_1=0$ or $z_2=0$). Apart from these two, every trajectory is dense in a torus of the form $\{|z_1|=C_1, |z_2|=C_2\}$.

No, your solutions do not need to be closed curves. For example, take the vector field "multiplication by $i$" on the $3$-sphere, thought of as the sphere of unit quaternions. That has closed solutions (Hopf fibration fibres). With a little bump function you can perturb this vector field to ensure there's only two closed solutions -- make the perturbation "away" from one closed solution and "towards" the other. This can be made concrete in many ways but I hope you get the idea.

edit: If you're interested in minimizing the number of closed orbits you can readily get down to one, on $S^3$. The meta-idea above is that the normal bundle to the Hopf fibration is the pull-back of the tangent bundle to $S^2$. $S^2$ has a flow with only one fixed point, so pull that vector field back to the normal bundle of the Hopf fibration, add it to "multiplication by $i$" and now you have a vector field with only one closed orbit (the Hopf fibre over the fixed point to your vector field on $S^2$).