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Let $P:S^{3}\to S^{2}$ be the Hopf fibration. For a vector field $X$ on $S^{2}$ there is a non-vanishing vector field $\tilde{X}$ on $S^{3}$ such that $DP(\tilde{X})=X$. It is constructed in two steps:

1) (I thank D. Panazzolo for this first step) We choose a 2-dimensional subbundle $E$ of $TS^{3}$ which is transverse to one-dimensional $\ker DP$. Then $DP:E_{x}\to T_{P(x)}S^{2} $ is a linear isomorphism. Then $DP^{-1}(X)$ is a tangent vector field on $S^{3}$ which vanishes on $P^{-1}(\text{singularities of X})$

2) We choose a non-vanishing vector field $Z$ on $S^{3}$ tangent to the Hopf fibration. Then $\tilde{X}=DP^{-1}(X)+Z$ is the desired vector field. In fact $\tilde{X}$ gives us a one-dimensional foliation of $S^{3}$ whose leaves are mapped by $P$ to the solutions of $X$ on $S^{2}$.

[Edit: I just realize that a combination of the above 2 steps is mentioned in Ghys' 1994 Bourbaki seminar, see the first paragraph of page 285.

On the other hand, we know that the singularities are the main obstructions in the study of limit cycles. (For instance, see the "Non-accumulation theorem" by Ilyashenko or the subject of "finite cyclicity of polycycles" by Roussarie, Dumortier, Rousseau et al.). So the above construction is a motivation to ask the following questions:

1.Let $Y$ be a non-vanishing analytic vector field on $S^{3}$. Is it true to say that $Y$ has only a finite number of invariant attractor torus? (Motivated by "Finiteness theorem for limit cycles" by Ilyashenko.)

  1. From the view of limit cycle theory, is it useful to study the (various) lifting of quadratic systems on $S^{2}$ to non-vanishing vector fields on $S^{3}$? By quadratic system I mean the Poincaré compactification of a degree $2$ planar polynomial vector field $$ \begin{cases} \dot x=ax+by+cx^{2}+dxy+ey^{2}\\\dot y=a'x+b'y+c'x^{2}+d'xy+e'y^{2}. \end{cases}$$

3.Can we lift an algebraic vector field to a non-vanishing algebraic vector field? An algebraic vector field on $S^{3}$ is the Poincaré compactification of a polynomial vector field on $\mathbb{R}^{3}$.

Finally, it is natural to ask: Is the Hopf fibration the only $S^{1}$ fibre bundle from $S^{3}$ to $S^{2}$ (up to equivalency of fibre bundle)?

Note 1: It would be interesting to look at the behavior of $\mathrm{Div}(\tilde{X})$, since this divergence must vanish somewhere between two invariant torus which covers two limit cycles $\gamma_{1}$ inside $\gamma_{2}$. This is a motivation to ask: Can we obtain a lifting $\tilde{X}$ whose divergence is constant on each fibre? Even if the answer is negative, we still have a vanishing result for some quantity associated to a vector field on $S^{2}$. The latter statement would be more clear after looking at some thing similar to Dulac-Bendixson criterion.

Note 2: One can repeat the same initial question of non-vanishing lifting vector field for $\pi:TS^{2}\to S^{2}$ where $\pi$ is the natural projecting map (or lifting the vector fields on $S^{2}$ via circle bundle $P:T^{1} S^{2}\to S^{2}$, where $T^{1}S^{2}$ is the unit tangent bundle). For this type of lifting, we possibly need the answer of this question.

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    $\begingroup$ Answer to the last question : the isomorphism classes of $\mathbb{S}^1$-bundles on a space $X$ form a group, identified with $[X,B\mathbb{S}^1]$. When $X=\mathbb{S}^2$ this group is $\pi _2(B\mathbb{S}^1)\cong \pi _1(\mathbb{S}^1)=\mathbb{Z}$. The Hopf fibration is a generator, but you have also all its multiples. $\endgroup$
    – abx
    Commented Sep 30, 2014 at 10:26
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    $\begingroup$ @Ali: If you just divide $S^3$ by the action of the $n$-element cyclic subgroup of $S^1$ (when $n\ge 1$), you'll get the bundle $\Sigma_n\to S^2$, and these are all of the examples. $\endgroup$
    – Robert Bryant
    Commented Sep 30, 2014 at 13:52
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    $\begingroup$ @AliTaghavi: No. $S^3\to S^2$ is the only simply-connected $S^1$-fiber bundle over $S^2$: Each of the $\Sigma_n$ is also an $S^1$-fiber bundle over $S^2$. Also, you have to distinguish the two $S^1$-fiber bundles you get by reversing the $S^1$-action on any given $S^1$-fiber bundle (this makes sense because $S^1$ is abelian). Thus, there is actually a $\mathbb{Z}$-family of $S^1$-fiber bundles, with $\Sigma_n$ and $\Sigma_{-n}$ being topologically the same space but not equivalent bundles (when $n\not=0$). Check Milnor and Stasheff for example, for proofs. $\endgroup$
    – Robert Bryant
    Commented Sep 30, 2014 at 15:05
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    $\begingroup$ @AliTaghavi: In $S^3$, two different fibers are linked, but, of course, they are not knotted. In $\Sigma_n$ with $n>1$, I'm not sure what you might mean by `knotted'. $\endgroup$
    – Robert Bryant
    Commented Sep 30, 2014 at 15:08
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    $\begingroup$ @AliTaghavi: Finally, the $S^1$-bundles over $M$ correspond one-to-one, to elements in $H^2(M,\mathbb{Z})$. I think that there is a proof in Milnor-Stasheff, but another place might be Atiyah's Lectures on $K$-theory. Check in those places for proofs. $\endgroup$
    – Robert Bryant
    Commented Sep 30, 2014 at 15:11

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