Finding a 1-form adapted to a smooth flow 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


*

*$\theta(X)$ is everywhere positive; and

*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:


*

*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 $g$.

*If the flow is an Anosov flow, one can take $\theta$ to be the canonical $1$-form.

*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.)

*If the flow is a suspension, one can take $\theta$ to be $dt$, where $t$ is the time variable (in local coordinates).

*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:


*

*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.

*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?)

 A: 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.
A: It is worth to mention that a conjecture by Thom asserts that generic vector fields in $C^r$ ($r\geq 1$) topology do not admit non-constant first integrals. There is a short note about that by Hurley in PAMS, 1986.
A: If the  vector  field  is  geodesible then  such   $1$ - form exists.

A  geodesible  flow on a manifold $M$  is  a one  dimensional  foliation associated with a non vanishing vector field  such that  all leaves  of  the  foliation are  geodesics  for  some    Riemannian  metric on $M$. This  is  mentioned  in the  book by   Philippe Tondeur,  Geometry  of Foliations, Page  71 Proposition 6.8.

I  quote  the Proposition here:
6.8 Proposition: Let  $V$  be  a  non singular  vector  field on  $M$. Then the  following  conditions are equivalent:
(i) The  flow  of  $V$ is  geodesible,
(ii) there  exist  a  $1$-form $\chi\in \Omega^1(M)$  such that $\chi (V)>0$  and $i(V)d\chi=0$

Another equivalent  condition:  From the  Proposition 6.7 of the  same  page of the  above book we learn that a non vanishing vector field $V$ is  geodesible if  and only if there is  a subbundle $E$ of $TM$, complementary  to  $V$,  such that for  every vector  field $X$ tangent to $E$ we  have $[X,V]$ is  tangent  to $E$. This  obviously implies the  following:

Obvious  Fact: Assume that $U$ is  an open  subset  of  $\mathbb{C}$ and  $V: U \to \mathbb{C}$ is  a  non vanishing holomorphic  function. Then the  vector  field  on $U$ determined by $V$ generates  a  geodesible foliation. The reason is that in the  above  equivalent  formulation on can put $E=$ the  direction determined by $W=iV$. Note that $[V, fW]=(V.f)W$ since $[V,W]=0$. This is a continuational research question on this topic.. This question search for a possible counter example of a  planar vector field whose linear part has orthogonal columns.
In the  following  "Edited"  part,  we  explain about a possible relation between the concept  "geodesible  flow" and the  problem of the  number of  limit cycles  of  a  polynomial vector  field:
EDITED:
This very interesting  answer  contains an  example of  a  non geodesible  foliation  of  $\mathbb{R}^2 \setminus \{0\}$ associated  with a  polynomial vector  field  of  degree $5$.

But  in degree $2$, the situation is different. The  key  point is  that every  limit  cycle  of  a  quadratic  vector  field  is  necessarily a  convex curve. We explain in details as  follows::

Put $$\begin{cases} x'=P(x,y)\\y'=Q(x,y) \end{cases}\;\;\;\; (V)$$
where $P.Q$ are polynomials  of  degree $2$  with $P(0,0)=Q(0,0)=0$. It  is  well known that every  closed  orbit of $(V)$ is  a  convex  curve.( See 4.6,page 108 of this book or see directly theorem $1$ of this paper).
This  shows  that if  a  limit  cycle $\gamma$ of  $(V)$  surrounds origin then it  does  not  intersect the algebraic  curve $$C=\{(x,y)\mid yP-xQ=0\}$$ This  is an  immediate  consequence  of the following  fact:
The  limit  cycle $\gamma$  is  a  convex closed curve and  surrounds  origin. Thus no point $(x,y)$ of  $\gamma $ satisfy $yP-xQ=0$.
Now put  $$\chi=\frac{1}{x^2+y^2}(ydx-xdy)$$ then $\chi$ is  a  closed  $1$_form on the  punctured plane which satisfies $\chi(V) \neq 0$ on $\mathbb{R}^2 \setminus C$.
In fact  $\chi= d \theta$ where $\theta$  is  the standard "argument" in the  polar  coordinate $(r, \theta)$ ( My serious thanks to  Ben MacKay for  his  valuable suggestion of  $d\theta$ as a possible  candidate  to find  a  $1$-form $\chi$  which (globally ) satisfies  the  above Theorem  of  Sullivan,  that is global  satisfaction of  $\chi (V) \neq 0$ ).
But in reality $\chi=d\theta$ does not  satisfies  $\chi (V) \neq 0$ golabally on the  phase  space , since it  vanishes on  $C$.  But, fortunately, no  limit cycle of  a quadratic  vector  field as $(V)$ can intersect this obstruction curve $C$. However $C$ need   not be  necessarily a  transversal  curve!(It  may be  tangent to the  vector  field  $(V)$ at  some  points).
So with $(V)$  and  $C$  described above,  we  summarize the above statements  as  follows:

Every  quadratic vector  field  $(V)$  is  geodesible  on $\mathbb{R}^2 \setminus C$. A  limit  cycle  which  surrounds   origin,  it never  intersect $C$.

So  if  we  can  control  the  sign   of  the  curvature of  the  corresponding  Riemannian metrics which  make the  flow  of $(V)$ geodesible, then we can  count the  number  of  limit  cycles  of  $(V)$  which  surround origin. But  if  a  limit  cycle  surround  a  singularity other than origin, we translate the  singularity to the  origin. This  procedure  would  count  all  possible  limit  cycles  of an  arbitrary  quadratic  vector  field.
In the  above, by "Control of sign of  the  curvature"  we mean that we  need   that  the  following  situation would  occur :

Either the curvature  is  identically zero, corresponding to the center  singularity, or the  curve consisting of all points with   zero  curvature would  be  a  transversal  curve, hence  no  limit  cycle can intersect it. This  is  explained  here:

Limit cycles as closed geodesics (in negatively or positively curved space)
As an updated related post please see the following
Limit cycles as closed geodesics(2)
A: For a slightly different perspective:  The Reeb foliation of an annulus (https://en.wikipedia.org/wiki/Reeb_foliation) can be doubled to give a flow on a torus, generated by a nowhere zero vector field $X$, such that all but one trajectory spirals towards a closed loop in forward time.  This flow has the property that (i) all but 2 trajectories are nonrecurrent and (ii) the foliation has no closed transversal (it is the standard example of a "non-taut" foliation, see e.g. Thurston, "A Norm on the Homology of a 3-Manifold").  
It follows that there is no $X$-invariant 1-form $w$ with $w(X)>0$.  Indeed, if $dw \neq 0$ then $|dw|$ gives a smooth X-invariant measure, contradicting (i), and if $dw = 0$ then $w$ is a closed, nowhere zero 1-form, and the leaves of the measured foliation defined by $Ker(w)$ are recurrent and transverse to the leaves of $X$, giving a closed transversal, contradicting (ii).  
This is the same as Bryant's example; as he observes, one can uses the additional fact that every $X$-invariant function is constant to even rule out $w$ with $L_X(w) = df$.  
A: Sure. In the case of a vector field $X$ admitting only constant first integrals, the existence of an adapted $1$-form for $X$ is equivalent to the existence of a "Lyapunov" $1$-form for $X$, that is a closed $1$-form $\theta$ such that $\theta(X)$ is everywhere positive. In the case of the torus for example, if you lift $X$ into a periodic vector field $Y$ then $\theta$ gives rise to a smooth time function for $Y$, that is a function $F$ such that $dF(Y)$ is everywhere positive. Conversely, if $Y$ admits a smooth time function then, in some cases, you can average it into a smooth time function with periodic differential in such a way to get an adapted $1$-form for $X$ on the torus. 
In Lorentzian geometry and general relativity, the existence of smooth time functions with respect to cone structures is now well understood. Following A. Fathi and A. Siconolfi (Cambridge, 2012), given a smooth manifold $M$, a cone structure is a multivalued mapping $x\mapsto C_x \subset T_xM$ where $C_x$ is a convex cone. A trajectory (also called $C$-causal curve) with respect to this cone structure is a Lipschitz curve $\xi$ such that 
$$
\dot{\xi}(t) \in C_{\xi(t)} \quad \mbox{for a.e. } t.
$$
The cone structure $C$ is called causal if it is continuous and does not contain closed trajectories and it is called stably causal if $C$ admits a small enlargement (a cone structure that contains $C$ in its interior) which is causal. A smooth function $F:M \rightarrow \mathbb{R}$ is a time function for $C$ if $d_xF(v)>0$ for every $x\in M$ and every $v\in C_x \setminus \{0\}$. The Theorem of Fathi and Siconolfi is the following:
If $C$ is a stably causal cone structure on $M$, there there exists a smooth time function for $C$.
In the case of a smooth periodic vector field $Y$ on $\mathbb{R}^n$, the existence of a smooth time function for $Y$ will follow, roughly speaking, from the non-chain recurrence of the flow of $Y$. In the case of your example on the $xy$-plane in $\mathbb{R}^2$,
$$
X = \sin x \, \partial_x + \cos x \, \partial_y,
$$
by the Fathi-Siconolfi, there should be a smooth time function. But this function cannot be averaged into a  adapted $1$-form for $X$ on the torus. Averaging procedure can be performed for example from a periodic vector field admitting a globally Lipschitz and uniform (smooth) time function ( a time function which is globally Lipschitz and such that $d_xF(X)\geq 1$ for all $x$.  
Maybe, a condition on a periodic vector field which could guarantee the existence of a globally Lipschitz and uniform time function could be the so-called « globally hyperbolic » property which says that for a enlargement of the cone structure given by the vector field the set of trajectories connecting two points remain in a compact set (which depends upon the two given points). It is not the case of your vector field.
