Let $M$ be a smooth compact manifold, and let $X$ be a smooth vector field of $M$ that is nowhere vanishing, thus one can think of the pair $(M,X)$ as a smooth flow with no fixed points. Let us say that a smooth $1$-form $\theta$ on $M$ is *adapted* to this flow if


1. $\theta(X)$ is everywhere positive; and
2. The Lie derivative ${\mathcal L}\_X \theta$ is an exact $1$-form.


(By the way, I'd be happy to take suggestions for a better name than "adapted". 
Most adjectives such as "calibrated", "polarised", etc. are unfortunately already taken.)



> 
> **Question.** Is it true that every smooth flow with no fixed points has at least one $1$-form adapted to it?
> 
> 
> 


At first I was sure that there must be counterexamples (perhaps many such), but every construction of a smooth flow I tried to make ended up having at least one adapted $1$-form. Some examples:


1. If the flow is isometric (that is, it preserves some Riemannian metric $g$), one can take $\theta$ to be the $1$-form dual to $X$ with respect to the metric $g$.
2. If the flow is an Anosov flow, one can take $\theta$ to be the canonical $1$-form.
3. If $M$ is the cosphere bundle of some compact Riemannian manifold $N$ and $(M,X)$ is the geodesic flow, then one can again take $\theta$ to be the canonical $1$-form. (This example can be extended to a number of other Hamiltonian flows, such as flows that describe a particle in a potential well, which was in fact the original context in which this question arose for me.)
4. If the flow is a suspension, one can take $\theta$ to be $dt$, where $t$ is the time variable (in local coordinates).
5. If there is a morphism $\phi: M \to M'$ from the flow $(M,X)$ to another flow $(M',X')$ (thus $\phi$ maps trajectories to trajectories), and the latter flow has an adapted $1$-form $\theta'$, then the pullback $\phi^\* \theta'$ of that form will be adapted to $(M,X)$.


Some simple remarks:


1. If $\theta$ is adapted to a flow $(M,X)$, then so is $(e^{tX})^\* \theta$ for any time $t$, where $e^{tX}: M \to M$ denotes the time evolution map along $X$ by $t$. In many cases this allows one to average along the flow and restrict attention to cases where $\theta$ is $X$-invariant. In the case when the flow is ergodic, this would imply in particular that we could restrict attention to the case when $\theta(X)$ is constant. Conversely, in the ergodic case one can almost (but not quite) use the ergodic theorem to relax the requirement that $\theta(X)$ be positive to the requirement that $\theta(X)$ have positive mean with respect to the invariant measure.
2. The condition that ${\mathcal L}\_X \theta$ be exact implies that $d\theta$ is $X$-invariant, and is in turn implied by $\theta$ being closed. For many vector fields $X$ it is already easy to find a closed $1$-form $\theta$ with $\theta(X) > 0$, but this is not always possible in general, in particular if $X$ is the divergence of a $2$-vector field with respect to some volume form, in which case the integral of $\theta(X)$ along this form must vanish when $\theta$ is closed. However, in all the cases in which this occurs, I was able to locate a non-closed example of $\theta$ that was adapted to the flow. (But perhaps if the flow is sufficiently "chaotic" then one can rule out non-closed examples also?)



Finding a 1-form adapted to a smooth flow
If I understand correctly, there is already a counterexample on the torus:


On the $xy$-plane $\mathbb{R}^2$, let $X$ be the vector field
$$
X = \sin x\,\frac{\partial\ }{\partial x} + \cos x\,\frac{\partial\ }{\partial y}.
$$
Now let $T^2=\mathbb{R}^2/\Lambda$ where $\Lambda$ is the lattice generated by $(2\pi,0)$ and $(0,2\pi)$. Since $X$ is invariant under this lattice, it gives rise to a well-defined, nowhere vanishing vector field on $T^2$, which I will also call $X$. It has two closed orbits $C\_0$ defined by $x\equiv0\mod 2\pi$ and $C\_1$ defined by $x\equiv \pi\mod 2\pi$, while every other orbit is nonclosed and has $C\_0$ as its $\alpha$-limit and $C\_1$ as its $\omega$-limit. In particular, the only functions on $T^2$ that are constant on the $X$-orbits are the constant functions.


Now suppose that $\theta$ is a $1$-form on $T^2$ such that $\mathcal{L}\_X\theta$ is exact. Then, by Cartan's formula,
$$
\mathrm{d}\bigl(\theta(X)\bigr) + i(X)(\mathrm{d}\theta) = \mathrm{d}h
$$
for some function $h$ on $T^2$. I.e.,
$$
i(X)(\mathrm{d}\theta) = \mathrm{d}\bigl(h-\theta(X)\bigr),
$$
implying that the function $h-\theta(X)$ is constant on the flow lines of $X$ and hence must be constant. Thus, $i(X)(\mathrm{d}\theta)=0$. Since $X$ is nowhere vanishing, $\mathrm{d}\theta$ must vanish identically, so $\theta$ must be closed.


Now, the integrals of $\theta$ over the two closed orbits (oriented so that $\mathrm{d}y>0$, say) must be equal, since, oriented this way, they are homologous in $T^2$. However, $X$ orients $C\_0$ and $C\_1$ so that they are *opposite* in homology, so the integrals using $X$ to orient them positively must have opposite signs (or vanish). Hence, it cannot be that $\theta(X)>0$ everywhere.


**Added on July 29:** A request has been made for an example of such a vector field without any closed leaves. This is easy to provide, as follows:


Let $\mathbb{T}^3 = \mathbb{R}^3/(2\pi\mathbb{Z}^3)$ be the 'square' $3$-dimensional torus, i.e., $xyz$-space where $x$, $y$, and $z$ are $2\pi$-periodic. Let
$$
X = \sin x\,\frac{\partial\ }{\partial x} + \cos x\,\frac{\partial\ }{\partial y}
 + \sqrt 2\,\cos x\,\frac{\partial\ }{\partial z}\,,
$$
which is a well-defined vector field on $\mathbb{T}^3$. The $2$-tori $C\_0$ (defined by $x\equiv0\,\mathrm{mod}\,2\pi$) and $C\_\pi$ (defined by $x\equiv\pi\,\mathrm{mod}\,2\pi$) are invariant under $X$ and all the flow-lines of $X$ in $C\_0$ and $C\_\pi$ are dense in their respective $2$-tori. Meanwhile, every other flow-line of $X$ $\alpha$-limits to $C\_0$ and $\omega$-limits to $C\_1$. In particular, any function on $\mathbb{T}^3$ that is constant on the flow-lines of $X$ is necessarily constant. Just as above, it follows that if $\theta$ is a $1$-form on $\mathbb{T}^3$ that is adapted to $X$, then $i(X)(\mathrm{d}\theta)=0$. It follows that there is a smooth function $\lambda$ on $T$ such that 
$$
\mathrm{d}\theta = \lambda\,i(X)(\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z)
= \lambda\,\bigl(\sin x\,\mathrm{d}y\wedge\mathrm{d}z 
+ \cos x\, \mathrm{d}z\wedge\mathrm{d}x 
+ \sqrt2\,\cos x\,\mathrm{d}x\wedge\mathrm{d}y \bigr)
$$
Taking the exterior derivative of both sides of this equation yields the identity
$$
0 = \mathrm{d}\lambda(X) + \lambda\,\cos x\,.
$$
Consequently, 
$$
\mathrm{d}(\lambda\,\sin x)(X) = 0
$$
Thus, $\lambda\,\sin x$ is constant along the flow lines of $X$ and hence is constant. Since it vanishes on $C\_0$ and $C\_\pi$, the function $\lambda\,\sin x$ must vanish identically. Hence $\lambda$ vanishes identically, i.e., $\mathrm{d}\theta = 0$. 


Since $\theta$ is closed on $\mathbb{T}^3$, its integral over any two homologous closed oriented curves must be equal. However, just as in the $2$-dimensional case, using the hypothesis that $\theta(X)>0$, one can easily construct a closed oriented curve $\gamma\_0$ in $C\_0$ on which $\theta$ pulls back to be positive while its translate by $\pi$ in the $x$-direction, say $\gamma\_\pi$ lies in $C\_\pi$ and has the property that $\theta$ pulls back to $\gamma\_\pi$ to be negative, making it impossible for the integrals of $\theta$ over the two curves to be equal.


