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 dim 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 vanish on $P^{-1}(\text{singularities of X})$

2)We chose 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 $S^{3}$ which leaves are mapped by $P$ to the solutions of $X$ on $S^{2}$.

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 constructions is a motivation to ask the following questions:

1.Let $Y$ be a non vanishing analytic vec. 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)

2.From the view of limit cycle theory, is it usefull to study the (various) lifting of quadratic systems on $S^{2}$ to non vanishing vec. fields on $S^{3}$? By quadratic system I mean the Poincare compactification of a $2$ degree 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 vec. field? An algebraic vec. field on $S^{3}$ is the Poincare 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 $Div(\tilde{X})$, 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 $\tilde{X}$ which 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-Bendixon criterion

Note2: One can repeat the same initial question of nonvanishing 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 to 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 Sep 30 '14 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 Sep 30 '14 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 Sep 30 '14 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 Sep 30 '14 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 Sep 30 '14 at 15:11

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