Divergence invariant lifting of a vector field via a submersion

Question

What is an example of a smooth submersion $P:S^{3}\to S^{2}$ for which the following statment is Not true:

For every vector field $X$ on $S^{2}$ there is a non vanishing vector field $\tilde{X}$ on $S^{3}$ with the following properties:

1. $P$ maps the solutions of $\tilde{X}$ to solutions of $X$

2.$Div(\tilde{X})$ is constant on each level set $P^{-1}(a),\;\;\forall a \in S^{2}$

Does the answer of this question depend on a particular metric on our manifolds $S^{2}$ and $S^{3}$, as divergence is a geometric quantity? I mean that if we change the metric, would the answer change?

The Note 1 of the following post is a motivation for this question.

One can repeat the same question by replacing the above spheres with two compact reimannian manifolds $M,N$ of dimension $m>n$ respectively, such that every $m-n$ dimensional subbundle of $TM$ possess a nonvanishing smooth global section. Is such lifting $\tilde{X}$ always possible? Am I greedy if I ask for exsistence of a lifting $\tilde{X}$ with the above property with the additional condition $Div(\tilde{X})=Div(X)\circ P$. What type of obstruction would appear in this latest version?

To start, perhaps we need to have a zero- divergence non vanishing vector field tangent to $\ker DP$. Can we find such global vec. field?

Answer 1

The only thing one really needs to define the operation $\mathrm{Div}:{\frak{X}}(S^3)\to C^\infty(S^3)$, i.e., mapping vector fields on $S^3$ to functions on $S^3$, is a volume form on $S^3$. (One can get away with even less, of course; really, one only needs a flat connection on the top exterior power of the cotangent bundle of the manifold in question, but I'll leave that refinement to the interested.)

If $\Omega$ is a volume form on $M=S^3$, then one defines the operator $\mathrm{Div}$ so that $$ \mathrm{d}(\iota_X\Omega) = \mathrm{Div}(X)\ \Omega, $$ where $\iota_X\Omega$ denotes the interior product of $X$ with $\Omega$.

In the case of a submersion $\pi:S^3\to S^2$ (though this can obviously be done in much greater generality), one can choose volume forms $\Omega$ on $S^3$ and $\omega$ on $S^2$, and there will exist a unique vector field $V$ on $S^3$ such that $\iota_V\Omega = \pi^*\omega$. This $V$ will be non vanishing and will be tangent to the fibers of $\pi$.

Now, given any vector field $X$ on $S^2$, there will be some $\pi$-related vector field $\tilde X$ on $S^3$, and any other vector field $\hat X$ on $S^3$ that is $\pi$-related to $X$ will satisfy $\hat X = \tilde X + \lambda V$ for some unique function $\lambda$ on $S^3$. One then finds, by computation, that $$ \mathrm{Div}(\hat X) = \mathrm{Div}(\tilde X) + \mathrm{d}\lambda(V). $$

Finally, since the fibers of $\pi$ must be compact and connected (and, hence, be circles), one can introduce a smooth family of metrics on the $\pi$-fibers in which $V$ has constant length and use the smooth dependence of the Greens operator on the metric to show that there will always exist a choice of $\lambda$ (unique up to the addition of a function of the form $\pi^*f$ for $f\in C^\infty(S^2)$) such that the function $\mathrm{Div}(\hat X)$ is constant on each fiber.