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?)
  • $\begingroup$ Take a closed $1$-form $\theta$. Form Cartna's formula we deduce $\endgroup$ Commented Jul 4, 2017 at 13:07
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    $\begingroup$ @LiviuNicolaescu For some manifolds, say $S^3$, the first part of the question, the positivity of $\theta (X)$ is not satisfied if we apply your comment.Since the first coholomology vanish. $\endgroup$ Commented Jul 4, 2017 at 13:25
  • $\begingroup$ @TerryTao In the last paragraph what do you mean by "....$X$ is the divergence of a 2-vectorfield"? $\endgroup$ Commented Aug 27, 2017 at 1:08
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    $\begingroup$ Using the Hodge star operation mapping $k$-vector fields to $d-k$-forms and vice versa (which is well defined once one selects a volume form), I mean that $X = * d * \alpha$ for some $2$-vector field $\alpha$. $\endgroup$
    – Terry Tao
    Commented Aug 27, 2017 at 2:17

5 Answers 5


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.

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    $\begingroup$ The key here seems to be that there are no non-constant $X$-invariant functions. My intuition previously was that this basically only happened when the flow was ergodic with respect to some volume form, but now I see that there are many other vector fields with this property also. $\endgroup$
    – Terry Tao
    Commented Jul 5, 2017 at 2:13
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    $\begingroup$ @TerryTao Yes, similarly, one can construct a vector field $X$ on $S^3$ that has two closed orbits, $C_0$ and $C_1$, while all the other orbits $\alpha$-limit to $C_0$ and $\omega$-limit to $C_1$ and intersect an separating torus transversely. Again, any function on $S^3$ that is constant on the $X$-orbits must be constant, and, again, one can show that $\theta$ must be closed (though the argument is a little more subtle), which, since $S^3$ is simply connected, implies that $\theta(X)$ cannot be positive everywhere. $\endgroup$ Commented Jul 5, 2017 at 6:10
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    $\begingroup$ @AliTaghavi You should just draw a picture of the flow lines of X in the plane; then it should be clear. I don't have any better advice to help you see what is happening. $\endgroup$ Commented Jul 5, 2017 at 10:48
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    $\begingroup$ Picture's worth a thousand words. goo.gl/chZMZS (It was surprisingly difficult to use WolframAlpha for this.) $\endgroup$ Commented Jul 5, 2017 at 16:33
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    $\begingroup$ This flow, thought of as a foliation, is a very popular one in low dimensional topology. I entered "Reeb foliation" into wikipedia and found an appropriate picture of the flow restricted to the annulus defined by $0 \le x \le \pi$: en.wikipedia.org/wiki/Reeb_foliation#/media/… $\endgroup$
    – Lee Mosher
    Commented Jul 19, 2017 at 12:52

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:


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)

  • $\begingroup$ @MatinSleziak Thank you very much for your edit. $\endgroup$ Commented Jul 4, 2017 at 15:57
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    $\begingroup$ Thanks for this! Looks like being geodesible is equivalent to the existence of a 1-form obeying the stronger conditions $\theta(X)=1$ and ${\mathcal L}_X(\theta) = 0$. The paper of Sullivan ams.org/mathscinet-getitem?mr=508734 referenced in Tondeur's book suggests that horocycle flows may be a source of counterexamples to the original question (will have to think about this...). $\endgroup$
    – Terry Tao
    Commented Jul 4, 2017 at 16:28
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    $\begingroup$ @TerryTao You are well come. I think if we do not require $X$ to be unit vector, we can replace $\theta (X) =1$ by $\theta (X) >0$. BTW, I am interested in the concept of geodesible flow since many years ago. I mentioned it at this post: mathoverflow.net/questions/160945/… $\endgroup$ Commented Jul 4, 2017 at 17:15

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 $L_X(w) = df$.


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.

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    $\begingroup$ Note, however, that merely having no nonconstant first integrals is not enough to rule out the existence of an adapted $1$-form in the OP's sense. For example, the constant coefficient vector field on the torus with irrational slope has every flow line dense in the torus; hence, only the constants are first integrals, but it has an adapted $1$-form. $\endgroup$ Commented Jul 15, 2017 at 19:45

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.


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