MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $M$ be a contact manifold, and let $F$ be an oriented 1-dimensional foliation that is transverse to the contact structure.

Is there a contact form $\alpha$ whose associated Reeb vector field generates the foliation $F$?

share|cite|improve this question
up vote 7 down vote accepted

Answer is negative. A possible solution is to construct a transversal vector field without closed trajectories. By Taubes theorem (see his paper "The Seiberg–Witten equations and the Weinstein conjecture") it is impossible. One could easily prove that such a field exists for a standard $3$-torus with contact structure given by $cos\theta dx + sin \theta dy$.

share|cite|improve this answer

I think one can construct easily local counter-examples (please correct me if I am wrong):

If $\xi$ is the Reeb vector field of some background contact form $\alpha_0$ and the distribution $F$ is generated by $\xi+X$ for some $X\in ker(\alpha_0)$, then the question is whether there exists a function $f$ such that $\xi+X$ is in the kernel of $d(f\alpha_0)$. This is equivalent to $$\xi(f)\alpha_0-df+X(f)\alpha_0+fd\alpha_0(X)=0.$$ Applying this to $X$ yields $X(f)=0$, so $$\xi(f)\alpha_0-df+fd\alpha_0(X)=0.\qquad (*)$$ Take now in the half-space $\{x>0\}$ of $\mathbb{R}^3$ $$\alpha_0=xdy +dz,\ \ \xi=\partial_z,\ \ X=z\partial_x.$$ We are thus looking for a function $f=f(y,z)$ (because $X(f)=0$) such that $$\partial_z f(xdy+dz)+fzdy=df,$$ i.e. $zf+x\partial_z f=\partial_y f$. This is impossible: one gets $\partial_z f=0$ and then $zf=\partial_y f$, so $f=0$.

share|cite|improve this answer

I think the answer is no.

Since a Reeb vector field must be in the Kernel of $d\alpha$, the foliation would need to have very special holonomy. The restriction of $d \alpha$ to transversals would give a symplectic form invariant under the holonomy.

If we start with a Reeb vector field having a periodic orbit we can perturb it to obtain a new foliation with the same periodic orbit but now with holonomy having linear part not in $Sp(2n)$.

share|cite|improve this answer
I think André asked about a contact form whose Reeb vector field is proportional to $X$, not equal to it. – Andrei Moroianu Dec 10 '10 at 20:52
@Andrei. You are of course right. Thanks. I have edited my answer. – Jorge Vitório Pereira Dec 10 '10 at 21:08

I believe the answer is no even if isotopy of the contact structure is allowed. Consider the following scenario:

OB is an open book decomposition of a 3-manifold M, xi is a contact structure on M which is not compatible with OB, and v is a contact vector field for xi (i.e. a vector field whose flow preserves xi) which is positively transverse to the pages of OB and positively tangent to its binding (and, hence, positively transverse to the contact structure compatible with OB). This can happen -- xi is said to be quasi-compatible with OB (see arxiv:0803.0758).

Here, v is not proportional to the Reeb vector field associated to any contact form for xi -- otherwise, xi would be compatible with OB. xi is obtained from the contact structure compatible with OB by adding Giroux torsion.

share|cite|improve this answer

If there is already a contact structure given say $\eta$ one might try to see if there exists a function $f$ so that Reeb vector field $R_f$ of $\eta_f = f\eta$ lies inside the given line bundle $\ell$. Assume that you can give $\ell$ as the kernel of two independent differential 1-forms $\eta_1,\eta_2$. Then we want $\eta_1(R_f)=\eta_2(R_f)=0$ which is if and only if $d\eta_f = \theta_1 \wedge \theta_2$. Let $E_i = ker(\theta_i)$ and $\Delta = ker(\eta)$ which is iff $d\eta_f|_{E_i}=0$ for both $i$. This condition could be written as a system of PDE.

Take $X_i \subset E_i \cap \Delta$ a vector field. Since $\ell$ is transverse to $\Delta$, $X_1$ and $X_2$ are distinct. And take $Z_i$ any other vector fields independent from $X_i$ inside $E_i$. Then the conditions above are true iff

$$(fd\eta + df \wedge \eta)(X_i,Z_i)=0$$

which is iff

$$X_i(ln(f)) = d\eta(X_i,\frac{Z_i}{\eta(Z_i)}) = \eta([X_i,\frac{Z_i}{\eta(Z_i)}])$$

or in more suggestive notation

$$\mathcal{L}_{X_i}(ln(f)) = \eta(\mathcal{L}_{X_i}W_i)$$

where $W_i=\frac{Z_i}{\eta(Z_i)}$. Hence $f$ must be a solution to this system of PDE.

share|cite|improve this answer

Your Answer


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.