Extending Reeb field from contact submanifold to ambient contact manifold

Question

Let $(Y,\lambda)$ be a contact manifold, with a codimension-2 contact submanifold $(S,\lambda|_S)$ (this requires $TS\pitchfork\text{Ker}\lambda$). On $Y$ there is a natural vector field, the Reeb field $R$, determined by $d\lambda(R,\cdot)=0$ and $\lambda(R)=1$.

Does there exist a contact form $\lambda'$ on $Y$ such that its Reeb field $R'$ is parallel to $S$, i.e. $R'|_S\subset TS$?

This is true for $\dim Y=3$, as (in the compact scenario) $S$ will be a transverse knot. Using the Moser trick, there is a tubular neighborhood $N(S)\approx S^1\times\mathbb{D}^2$ with $\text{Ker}\lambda|_{N(S)}=\text{Ker}(dz+r^2d\theta)$. Thus $\lambda|_{N(S)}=e^H(dz+r^2d\theta)$ for some function $H:S^1\times\mathbb{D}^2\to\mathbb{R}$. The Reeb vector field on $(S^1\times\mathbb{D}^2,dz+r^2d\theta)$ is $\partial_z$ with Reeb orbit $S=S^1\times\lbrace 0\rbrace$. So $\lambda'=e^{-H}\lambda$ will do the trick.

In general, I am not sure what obstructions would arise by taking the Reeb field on $S$ and trying to extend it to the ambient $Y$.

Answer 1

This is a step of the proof of the neighbourhood theorem for contact submanifolds in Geiges's book.

The idea is find a function $f \colon Y \to \mathbb{R}$ so that $f$ is identically $1$ on $S$ and so that \[ -df + d\lambda(R', \cdot) = 0 \] on $TY|_S$. If this holds, then it follows that $d( f\lambda)( R', \cdot) = 0$ on $TY|_S$, so $R = R'$ along $S$.

Note that the only condition on $f$ we have here is that $df(v) =0$ for any $v$ tangent to $S$. Note also that $d\lambda(R', v) = 0$ for any $v$ tangent to $S$ as well, since $d\lambda(R', v) = d\lambda_S(R', v)$ in that case. Thus, such a function exists.