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](https://mathoverflow.net/questions/332979/irrational-closed-orbits-of-vector-fields-on-s2limit-cycles-and-trace-formul) and [the paper](https://arxiv.org/pdf/math/0204110.pdf) 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 non-vanishing vector field on $S^3$ whose flow does not preserve any transversal foliation
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.


