Constructing a contact form for which a given contact vector field is Reeb

Question

Given a contact manifold $(M, H)$ and a smooth vector field $X$ on $M$, I'm trying to show that $X$ is the Reeb field of some contact form for $H$ if and only if it's a contact vector field that's nowhere tangent to $H$.

First, the definitions as I understand them:

• A Reeb field for some contact form $\theta$ is a smooth vector field $T$ such that $\theta(T) = 1$ and $i_T d\theta = 0$
• A contact vector field is a smooth vector field $X$ such that its flow preserves the contact structure $H$. Explicitly, if $\psi$ is the flow of $X$, then this means $\psi_{t*}Y \in \Gamma(H)$ for any $Y \in \Gamma(H)$ and any $t \in \mathbb{R}$.

Proving that Reeb $\Rightarrow$ contact is a straightforward application of the definitions, but proving contact $\Rightarrow$ Reeb has been more difficult.

My attempt: If $X$ is nowhere tangent to $H$, then for any contact form $\theta$, $\theta(X)$ is nowhere zero. Thus we can define a contact form $\phi = \frac{1}{\theta(X)}\theta$ which clearly satisfies $\phi(X) = 1$, and it remains only to show that $i_Xd\phi = 0$. Applying Cartan's magic formula, we can write $$ \require{cancel} i_X d\phi = \mathcal{L}_X \phi - \cancelto{0}{d(\theta(X))} = \mathcal{L}_X \phi. $$ This Lie derivative vanishes if and only if $\phi$ is invariant under the flow of $X$.

Because $X$ is contact, the pullback of any contact form by $\psi_t$ will still annihilate $H$ for any $t$, and so $\psi_t^* \phi = f\phi$ for some $f \in C^\infty(M)$. Thus, we will be done if we can show that $f = 1$, which will in turn follow if we can show that $\psi_t^*\phi(X) = \phi(X) = 1$. \begin{align} (\psi_t^* \phi)(X) &= \left(\psi_t^* \left( \frac{1}{\theta(X)} \theta \right)\right)(X) \\\ &= \left( \frac{1}{\theta(X)} \circ \psi_t\right)(\psi_t^* \theta)(X) \\\ &= \frac{\theta(\psi_{t*} X)}{\theta(X) \circ \psi_t} \\\ &= \frac{\theta(X)}{\psi_t^* (\theta(X))}. \end{align} (On the last line, we should be careful to note that the pullback $\psi_t^*$ is acting on the function $\theta(X)$, not on $\theta$ itself.)

Here is where I find myself in a corner. The result will follow if I can show that $\theta(X)$ is constant on every integral curve of $X$, which seems like it ought to be true, but I don't know how to show it. Should this be obvious? Or should I perhaps modify my definition of $\phi$?

Answer 1

I just saw this old question. Here's how to prove that $X$ contact $\implies$ $X$ Reeb.

Suppose $\theta$ is a contact form and $X$ is a contact vector field that lies nowhere in the kernel of $\theta$. As you noted, $\phi = \theta/\theta(X)$ is a contact form for the same contact structure that satisfies $\phi(X)\equiv 1$. The fact that $X$ is contact implies $\mathscr L_X\phi = f\phi$ for some scalar function $f$. Thus \begin{align*} \require{cancel} f &= f \phi(X) = (\mathscr L_X\phi)(X)\\ &= \big(i_X d\phi + \cancel{d(\phi(X))}\big)(X)\\ &= d\phi(X,X) = 0. \end{align*} This shows $i_Xd\phi = \mathscr L_X \phi = 0$, so $X$ is the Reeb field of $\phi$.