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In Surgery on Contact Manifolds and Stein Surfaces there is the following exercise [below $Y$ is a 3-manifold, and $e(\xi)$ is the Euler class of $\xi$]:

Exercise 4.2.6. Find a contact structure $(Y,\xi)$, a Legendrian knot $L\subset (Y,\xi)$ and surfaces $\Sigma_1,\Sigma_2$ such that $rot_{\Sigma_1}(L)\neq rot_{\Sigma_2}(L)$. [Hint: Start with a contact structure $\xi$ and a closed surface $\Sigma$ such that $\langle e(\xi),[\Sigma]\rangle \neq 0$ and find $L$ on $\Sigma$ separating it.)

To define $rot_{\Sigma}(L)$ where $L\subset (Y,\xi)$ is a Legendrian knot and $\Sigma$ is a Seifert surface for $L$, one first chooses a trivialization of $\xi\mid_{\Sigma}$. [Any oriented $\mathbb{R}^2$-bundle over a compact surface with nonempty boundary is trivial.] This induces a trivialization of $\xi\mid_L$. One then chooses a nonvanishing tangent vector field on $L$ [compatible with the orientation on $L$] and measures the winding number of the vector field with respect to this trivialization. This is independent of the choice of trivialization $\xi\mid_{\Sigma}$, and if $e(\xi) = 0$ is also independent of the choice of $\Sigma$.

When I started working on this exercise, I realized all the examples of contact structures given earlier in the textbook have trivial Euler class [or equivalently are trivial as $\mathbb{R}^2$-bundles]. When I tried googling, I found this overflow post where someone mentions in the comments that there is a contact structure on $S^1\times S^2$ with nontorsion Euler class, but it seems the expression they give for the contact 1-form is slightly off [doesn't define a contact structure].

I would greatly appreciate if someone could point me towards an example, or give a hint as to how to construct an example.

Hrhm
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It is easy to construct an example where $\xi$ is overtwisted. By Eliashberg's classification of overtwisted contact structures (see Theorem 4.3.4 of the book you referenced), every oriented $2$-plane field on a $3$-manifold can be homotoped to an overtwisted contact structure. You can proceed from here. Note if $\Sigma=S^2$ then every simple closed curve on it is separating.

Qiuyu Ren
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  • Thanks for the answer! I might not have thought this all the way through but I'm still a bit stuck. If I let $Y=\mathbb{R}\times S^2$ with projection $\pi\colon Y\to S^2$, then I can find a contact structure $\xi$ on $Y$ in the same homotopy class as $\pi^*TS^2\subset TY$. If I then let $\Sigma = {\text{pt}}\times S^2$, then $\langle e(\xi),\Sigma\rangle =2$. However, am I guaranteed to find a Lagrangian curve on $S^2$? I guess this is the same as asking whether the characteristic foliation on $\Sigma$ has a closed leaf? – Hrhm Jul 08 '24 at 06:50
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    You can start with a Legendrian knot instead. For example, there is always a local unknot that is Legendrian. – Qiuyu Ren Jul 08 '24 at 09:07
  • Ahh I see, thanks! – Hrhm Jul 08 '24 at 14:14