A non-vanishing vector field on $S^3$ whose flow does not preserve any transversal foliation Is there a non-vanishing vector field $X$ on $S^3$ which does not admit a transversal $2$-dimensional foliation? if the answer is negative, is there a non-vanishing vector field $X$ on $S^3$ which does not admit a transversal foliation whose leaves are invariant under the flow of $X$? If the answer is positive, is there an example of this situation with the extra assumption that $X$ is invariant under the obvious action of $S^1$ on $S^3$?
The motivation is described in the following post and the paper linked in that post. (The property "Flow invariant foliation" is frequently used in the paper and closed orbits of the flow involve in a trace formula consisting of  Leaf De Rham cohomology.)
 A: There is a well-known fiber bundle $\pi: S^3\to S^2$, the "Hopf fibration". Its fibers are circles, and they are the leaves of a one-dimensional foliation of $S^3$. There is of course a nowhere vanishing vector field $X$ that is tangent to this foliation. A two-dimensional foliation transverse to $X$ is the same as a two-dimensional foliation transverse to the fibers of $\pi$. I will show that such a foliation cannot exist. Suppose for contradiction that $\mathcal F$ is such a foliation.
Let $x\in S^2$ be any point. I claim that there is an open neighborhood $U$ of $x$ such that for every point $y\in \pi^{-1}(x)$ there is a section of $\pi$ defined in $U$, taking $x$ to $y$, and going into a leaf of $\mathcal F$. First, for every $z\in \pi^{-1}(x)$ there exist an open neighborhood $U_z$ of $x$ in $S^2$ and an open neighborhood $V_z$ of $z$ in $\pi^{-1}(x)$ such that for every $y\in V_z$ there is a section of $\pi$ defined in $U_z$, taking $x$ to $y$, and going into a leaf of $\mathcal F$. Now, the open sets $V_z$ form a cover of the (compact) fiber $\pi^{-1}(x)$, and therefore finitely many of them, say $V_{z_i}$, cover it. Let $U$ be the intersection of the $U_{z_i}$. That proves the claim.
But now by the claim we can say that for every leaf $L$ of $\mathcal F$ the projection of $L$ to $S^2$ by $\pi$ is a covering space. (Of course I am considering $L$ with the topology which makes $L$ a manifold and makes the inclusion $L\to S^3$ an immersion.) But because $S^2$ is simply connected this means that each leaf maps to $S^2$ by a homeomorphism. In particular one leaf does so, and therefore $\pi$ has a right inverse: the bundle has a section. Of course this is impossible for any of a number of reasons.
A: This answer is in opposite way to the answer by Prof. Goodwillie. His answer is based on a vector field whose all orbits are closed orbit. Our answer is based on a vector field which does not have any closed orbit.
According to a corollary of a  theorem by S.P. Novikov, if a vector field on $S^3$  is transversal to a $2$ dimensional foliation of $S^3$, then it must have a periodic orbit. So every vector field on $S^3$ which does not have any periodic orbit can not be transverse to a 2 dimensional foliation. So every counter example to the Seifert conjecture is an example of a vector field which can not be transveral to any 2 dimensional foliation.
The above theorem of Novikov is mentioned in page 285 of the following paper(page 285, 2 paragraph before section 3)
http://perso.ens-lyon.fr/ghys/articles/constructionchamps.pdf
