Formula for the Reeb field of the standard overtwisted contact form

Question

I'm looking for a explicit formula for the Reeb field of the standard overtwisted form on $\mathbb{R}^3$, which is written in cylindrical coordinates $(\rho, \theta, z)$ as $$ \alpha = \cos(\rho)dz + \rho\sin(\rho)d\theta. $$ I tried to calculate it myself but I got a mess of $\sin$ and $\cos$ in denominators, which doesn't make sense because the Reeb field should be uniquely defined everywhere.

I found this question where the OP gives a formula like what I'm looking for, but there's no suggestion of how one can obtain it, and the OP themselves say they are not sure if their calculations are correct.

Is there any references where one can find such calculations done?

Answer 1

The Reeb vector field $R$ is characterized by $$ \begin{cases} \alpha(R) & \equiv 1,\\ d\alpha(R,\cdot) & \equiv 0. \end{cases} $$ Write $R = f\partial_{\rho} + g \partial_{\theta} + h \partial_z$. The first equality yields

$$ \cos \rho \, h + \rho \sin\rho \, g = 1. $$

An easy calculation shows that $d\alpha = -\sin \rho \, d\rho\wedge dz + (\sin\rho + \rho\cos \rho) d\rho \wedge d\theta$. To exploit $d\alpha(R,\cdot) = 0$, let us evaluate this equality on the frame $\{\partial_{\rho},\partial_{\theta},\partial_z\}$.

1. $d\alpha(R,\partial_{\rho})=0$ gives $$ \sin \rho \, h - (\sin\rho + \rho \cos\rho) g = 0, $$
2. $d\alpha(R,\partial_{\theta}) = 0$ gives $$ (\sin\rho + \rho \cos \rho) f = 0, $$
3. $d\alpha(R,\partial_z) = 0$ gives $$ -\sin\rho\, f =0. $$

Hence, $f = 0$, and $h$ and $g$ satisfy $$ \begin{cases} \cos\rho\, h + \rho \sin \rho\, g &= 1 \\ \sin\rho \, h - (\sin\rho + \rho\cos \rho) g &= 0. \end{cases} $$

Solving for $(g,h)$ gives the desired result. First we assume values of $\rho$ such that $cos\rho$ and $\sin\rho$ are not zero. Then the system yields $$ \frac{1-\rho \sin \rho\, g}{\cos\rho} = h = \frac{\sin\rho+\rho\cos\rho}{\sin\rho}g, $$ from which we obtain $$ g =\frac{\sin\rho}{\rho + \cos\rho\,\sin\rho} = \frac{2\sin\rho}{2\rho+\sin(2\rho)}, $$ where we used the identity $\sin(2\rho) = 2\sin\rho\,\cos\rho$. We now solve for $h$ and get $$ h = \frac{2\sin\rho + 2\rho\cos\rho}{2\rho+\sin(2\rho)}. $$ Thus $$ R = \frac{2}{2\rho + \sin(2\rho)}\left(\sin(\rho)\partial_\theta + (\sin\rho + \rho\cos\rho)\partial_z\right), $$ for $\rho \neq 0$. Since $\alpha$ is a $C^\infty$ form, the Reeb field is smooth. Due to unicity, all it takes to determine $R$ at the origin is to take the limit when $\rho \to 0$. Hence $$ R = \begin{cases} \left(\frac{2}{2\rho + \sin(2\rho)}\right)[\sin(\rho)\partial_\theta + (\sin\rho + \rho\cos\rho)\partial_z], &\text{ for } \rho \neq 0 \\ \frac{1}{2}\partial_\theta + \partial_z, &\text{ for } \rho = 0 \\ \end{cases} $$