closed dual of vector fields

Question

Given a closed manifold and a nowhere zero vector field $X$, can one always find a closed 1-form $\alpha$ such that $\alpha(X)=1$ as the constant function? Or under what kind of conditions on $X$ can this be true? Thanks!

Answer 1

It is rarely possible.

First of all, if the manifold is simply connected, then there are no nowhere vanishing closed 1-forms at all. Indeed, every closed 1-form on such a manifold is a derivative of a function and this function must attain a maximum where the derivative is zero. So, for example on $S^3$ there exist nonvanishing vector fields but no novaninshing closed 1-form. The same argument applies to any closed manifold $M$ with $H^1(M,\mathbb R)=0$ (so every closed 1-form has an antiderivative) and $\chi(M)=0$ (so there exists a nonzero vector field).

On some manifolds, such as tori, there are nowhere zero closed 1-forms. But a vector field $X$ has to be very special to admit such a form $\alpha$. For example, if $X$ has a closed orbit, this orbit cannot be contractible (more generally, homological to 0 in $H_1(M;\mathbb R)$.) Because otherwise a closed form $\alpha$ would integrate to zero along this loop. And if two closed orbits are from the same homology class, then they must have the same period (because the integrals of $\alpha$ must be equal). A small perturbation of the coordinate vector field on the torus may fail this test. And if you have closed orbits in many homology classes, the period must depend linearly on the homology class. With these observation, you can prescribe $X$ on a tiny subset of the manifold and guarantee that no such $\alpha$ exists.

Even one periodic orbit may present an obstruction. Indeed, a local antiderivative of $\alpha$ grows with unit rate along $X$. This implies that $L_X\alpha=0$ where $L_X$ is the Lie derivative. Now imagine that $X$ has a periodic orbit through $x_0$ with period $T$ and let $\Phi:M\to M$ be the $T$-shift along $X$. Since $L_X\alpha=0$, $\alpha$ is invariant under $\Phi$. On the other hand, $\Phi(x_0)=x_0$, so the kernel of $\alpha_{x_0}$ must be preserved by $d_{x_0}\Phi$. But it may easily happen that $d_{x_0}\Phi$ does not have any invariant subspaces transverse to $X$.

Even without periodic orbits, there are problems. Having fixed $\alpha_{x_0}$ for some point $x_0$, the condition $L_X\alpha=0$ uniquely determines $\alpha$ along the orbit. It may happen that the orbit is dense, so $\alpha$ is determined everywhere by its value at $x_0$. But there is no reason for this uniquely determined extension to be continuous. There are examples (coming from hyperbolic dynamics) where the form exists and unique, is continuous, but not $C^1$.

Answer 2

The answer is NO.

Consider torus $\mathbb R^2/\mathbb Z^2$ with the vector field $v_{(x,y)}=(0,2+\cos x)$.

Assume there is a closed form $\omega$ such that $\omega(v)\equiv 1$. Then you arrive to a contradiction by integrating around rectangles which are long in direction of $y$-axis.

Answer 3

On the other hand, there are situations where the answer is positive. Actually Schwartzman and Fried gave a very satisfactory criterion for the existence of forms which are positive along the flow (see for example Fried's The geometry of cross sections to flows).

For $X$ a vector field on a compact manifold $M$, denote by $\phi^t$ the associated flow. For $p$ a point in $M$, let $k(p,t)$ be a loop obtained by following $X$ from $p$ to $\phi^t(p)$, and closed by an arbitrary (short) segment. Consider the classes $[k(p,t)]/t$ in $H_1(M, \mathbb R)$, denote by $C_\phi$ the set of such classes, and by $S_\phi$ the set of accumulation points of $C_\phi$. As long as the length of the closing segments is bounded, $S_\phi$ depends on $X$ only. Schwartzman-Fried's theorem states that $X$ admits a closed $1$-form $\alpha$ with $\alpha(X)>0$ everywhere if and only if $S_\phi$ lies in an open half-space in $H_1(M, \mathbb R)$.

Answer 4

Fix a Riemann metric metric $g$ on the manifold. Denote by $\omega$ the $1$-form dual to $X$ defined by

$$\omega(Y)= g(X,Y), $$

for any vector field $Y$. The $1$-form

$$\alpha :=\frac{1}{|X|^2_g}\omega =\frac{1}{g(X,X)}\omega $$

will do the trick.

Mea Culpa I missed the word closed in the question. Obviously the $\alpha$ above need not be closed.

If $\alpha$ is closed and $\alpha(X)=i_X\alpha=1$ ($i_X$ denoting the contraction with $X$), then the identity

$$L_X=di_X+i_Xd $$

implies that $L_X\alpha=0$. Conversely, if $\alpha$ is a closed $1$-form such that $L_X\alpha=0$ then the above identity implies that $i_X\alpha$ is constant. Thus you need to find closed forms which are invariant under the flow generated by $X$ and such that $\alpha(X)$ is nonzero at one point of the manifold.

Over each contractible open subset $U\subset M$ a closed form $\alpha$ is exact so that

$$\alpha|_U= d f_U $$.

The condition $\alpha(X)=1$ implies that

$$ X\cdot f_U = df_U(X) =1. $$.

Thus $f_U$ increases linearly along the orbits of $X$ in $U$. For example, if $X$ has a periodic orbit contained in a simply connected set your question does not have solution.

For example, if $X$ is the generator of the usual action of $S^1$ on $S^3$, then you cannot find a closed $1$-form on $S^3$ such that $\alpha(X)=1$.