Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!

share|improve this question
add comment

4 Answers

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 $C^1$.

share|improve this answer
add comment

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.

share|improve this answer
add comment

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 $H_1(M, \mathbb R)$.

share|improve this answer
add comment

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 $\alpha(X)=1$.

share|improve this answer
    
Assuming you can construct a Riemmanian metric! (though I grant that one can in most cases of interest) –  Greg Friedman Sep 20 '12 at 17:44
1  
Another answerer and I had written this and then deleted it, because I (and presumably the other answerer) see no reason for $\alpha$ to be closed. My answer actually included a second wrong approach, namely to find such $\alpha$'s locally and patch them together with a partition of unity. Again, the result need not be closed. –  Andreas Blass Sep 20 '12 at 18:09
1  
As soon as I read the question I was about to post exactly the same answer, but then I asked myself if the resulting form would be closed or not. It may depend on the metric, so probably a (nontrivial?) differential equation would be involved. –  Qfwfq Sep 20 '12 at 18:41
    
Greg: Riemannian metrics always exist. –  Kofi Sep 20 '12 at 18:57
    
@ all commenters: Apologies for the dumb mistake. Please read the Mea Culpa included in my updated answer. –  Liviu Nicolaescu Sep 20 '12 at 19:07
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.