Is there an example of a smooth vector field $v$ on $S^3$ such that $v$ preserves a volume form and $v$ is not a Reeb vector field?


Recall that $v$ is a *Reeb vector field* if there exists a contact $1$-form $\alpha$ such that $\alpha(v)=1$ and $v$ belongs to the kernel of $d\alpha$.



Reeb flows on $S^3$ versus volume preserving flows
I took the time to give a clean version of the answer and to eliminate the dependence on really hard theorems.


**Construction of geodesible, volume-preserving flows on $S^3$ that are not Reeb flows for any contact form.**


Ingredients: 


1. A contact form $\alpha$ whose Reeb vector field $X$ has no nonconstant integrals of motion and which carries a closed characteristic. Such a form can be obtained from a Finsler or Riemannian metric on the two-sphere with ergodic geodesic flow (examples by Katok and Donnay).
2. A smooth function $f$ that is nowhere zero and such that for some closed characteristic $\gamma$ in $(S^3, \alpha)$ we have that
$$
{ \int\_{S^3} f^{-1} \ \alpha \wedge d\alpha \over \int\_{S^3} \alpha \wedge d\alpha}
\neq 
{\int\_\gamma f^{-1} \ \alpha \over \int\_\gamma \alpha}.
$$


**Theorem.**
*The vector field $fX$ preserves the volume form $f^{-1} \alpha \wedge d\alpha$, but it is not the Reeb vector field of any contact form on $S^3$.* 


The proof is by contradiction and we will need the following


**Claim.**
*If $fX$ is the Reeb vector field of a contact form $\beta$, then for some nonzero constant $\lambda$ and some smooth function $g$ we have that $\beta = \lambda \alpha + dg$.*


*Proof of the claim.* Since the kernels of the $2$-forms
$d\alpha$ and $d\beta$ coincide and the space is three-dimensional, the forms are multiples:
$d\beta = \lambda d\alpha$, with $\lambda$ some smooth function. Moreover 
$\lambda$ must be an integral of motion of $X$,
$$
0 = {\cal L}\_X d\beta = {\cal L}\_X \lambda d\alpha = X(\lambda) \alpha .
$$
By hypothesis $\lambda$ must be a constant and, therefore, 
$d(\beta - \lambda \alpha) = 0$ or $\beta = \lambda \alpha + dg$.


Now we proceed with the proof of the theorem. First we show that $\lambda$ is the
average of the function $f^{-1}$ over $S^3$. Indeed, 
$$
1 = \beta(fX) = \lambda \alpha (fX) + dg(fX) = \lambda f + fX(g),
$$
and so 
$$
\int\_{S^3} f^{-1} \ \alpha \wedge d\alpha = 
\int\_{S^3} (\lambda f + fX(g))f^{-1} \ \alpha \wedge d\alpha. 
$$
Since $X$ preserves the form $\alpha \wedge d\alpha$, the integral of 
$X(g)\alpha \wedge d\alpha$ is zero and we obtain
$$
\int\_{S^3} f^{-1} \ \alpha \wedge d\alpha = \lambda \int\_{S^3} \alpha \wedge d\alpha .
$$


Now we notice that $\lambda$ is also the average of $f^{-1}$ over *every* closed leaf of the characteritic foliation of $(S^3, \alpha)$. If $\gamma$ is a closed leaf, then 
$$
\int\_\gamma f^{-1} \ \alpha = \int\_\gamma (\lambda f + fX(g)) f^{-1} \ \alpha.
$$
As before, using that $X$ preserves $\alpha$, we obtain that the integral of 
$X(g)\alpha$ along $\gamma$ is zero, and hence
$$
\int\_\gamma f^{-1} \ \alpha = \lambda \int\_\gamma \alpha.
$$


This contradicts our assumption of the existence of a closed characteristic for which 
$$
{ \int\_{S^3} f^{-1} \ \alpha \wedge d\alpha \over \int\_{S^3} \alpha \wedge d\alpha}
\neq 
{\int\_\gamma f^{-1} \ \alpha \over \int\_\gamma \alpha}.
$$


%%%%%%%%%%%%%%%%%% **Old, messy answer** %%%%%%%%%%%%%


Edit. Long answer and a bit "thinking aloud" or "typing while thinking" in its organization (I'll reorganize it when I'll have the time), but at the end you have the following result:


**Theorem.** *Consider a contact form $\alpha$ on the three-sphere and so that its Reeb vector field $X$ does not admit any smooth integral of motion except constants (e.g., the lift of some ergodic Finslerian or Riemannian geodesic flow on the two-sphere). If $f$ is a nowhere zero smooth function on $S^3$ such that it is not constant, and such there exist two closed Reeb orbits of $\alpha$ over which the averages of $1/f$ are distinct, then the vector field $fX$ preserves the volume form $(1/f) \alpha \wedge d\alpha$, but it is not the Reeb vector field of any contact form on $S^3$.* 


Consider a contact form $\alpha$ on the three-sphere and so that its Reeb vector field $X$ does not admit any smooth integral of motion except constants (e.g., the lift of some ergodic Finslerian or Riemannian geodesic flow on the two-sphere). Let $f$ be a nowhere zero smooth function on $S^3$ that is **not** constant. 


**Claim.** *The vector field $fX$ preserves the volume form $(1/f) \ \alpha \wedge d\alpha$, but it is not the Reeb vector field of any contact form DEFINING THE SAME CONTACT STRUCTURE AS $\alpha$*


The proof is by contradiction:


Assume $fX$ is the Reeb vector field of the contact form $\beta$. This means that the characteristic distributions of $d\alpha$ and $d\beta$ coincide. Since these are $2$-forms in a three-dimensional space, we conclude that these forms are multiples: $d\alpha = \lambda d\beta$. Moreover $\lambda$ must be an integral of motion of $X$,
$$
0= \mathcal{L}\_X d\alpha = \mathcal{L}\_X \lambda d\beta = X(\lambda) d\beta .
$$
By hypothesis, this means that $\lambda$ is a constant and, *since both forms define the same contact structure,* that $\beta = \lambda \alpha$. In this case, the Reeb vector field of $\beta$ is $fX = \lambda X$ and $f = \lambda$, contrary to the assumption that $f$ was not constant. $\square$


**Remark.** If $\alpha$ and $\beta$ do not define the same contact structure, the above proof still shows that $\alpha$ and $\beta$ differ only in a multiplicative constant and the addition of the differential of some function. I have to finish grading so I'll come back to this later if someone has not given some nice explicit example.


**Continuation.** If we drop the assumption that both forms define the same contact structure, the preceding proof give us that $\beta = \lambda \alpha + dg$, where $\lambda$ is a nonzero constant and $g$ is some smooth function on the three-sphere. The relation between $\lambda$ and $f$ is easy to establish. Indeed, if we note that
$$
1 = \beta(fX) = \lambda \alpha (fX) + dg(fX) = \lambda f + fX(g) ,
$$
multiply both sides by the volume form $(1/f) \ \alpha \wedge d\alpha$, and integrate over the three-sphere, we obtain that 
$$
\int (1/f) \ \alpha \wedge d\alpha = \lambda \int \ \alpha \wedge d\alpha
$$.
Note that the "missing" integral is zero because $X$ preserves the form $ \alpha \wedge d\alpha$. *So $\lambda$ is the average of $1/f$ with respect to the volume form $ \alpha \wedge d\alpha$.*


Going back to the equation
$$
1 = \beta(fX) = \lambda f + fX(g) = f(\lambda + X(g)) \ \ {\rm or} \ 1/f = \lambda + X(g),
$$
observe that if we multiply both sides by $\alpha$ and integrate over any closed characteristic $\gamma$, we obtain
$$
\int\_\gamma (1/f) \alpha = \lambda \int\_\gamma \alpha
$$
Therefore $\lambda$ is also the average of $1/f$ over every single closed characteristic of $(S^3,\alpha)$. That is a bit too much to ask of $\lambda$!


Thanks to D. Cristofaro-Gardiner and M. Hutchings we know that every Reeb vector field in the sphere has alt least two closed characteristics. We can be more restrictive with our original choice of $f$ and require that the average of $1/f$ be different for two closed characteristics of $(S^3,\alpha)$. The form $\beta$ then does not exist. 


We have then:


**Theorem.** *Consider a contact form $\alpha$ on the three-sphere and so that its Reeb vector field $X$ does not admit any smooth integral of motion except constants (e.g., the lift of some ergodic Finslerian or Riemannian geodesic flow on the two-sphere). If $f$ is a nowhere zero smooth function on $S^3$ such that it is not constant, and such there exist two closed Reeb orbits of $\alpha$ over which the averages of $1/f$ are distinct, then the vector field $fX$ preserves the volume form $(1/f) \alpha \wedge d\alpha$, but it is not the Reeb vector field of any contact form on $S^3$.* 


