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Let $M=S^1\times \mathbb{R}^2$ and $\alpha_1, \alpha_2$ be a pair of contact one-forms on $M$ such that the restrictions $\alpha_1|_{S^1\times \{0\}}$, $\alpha_2|_{S^1\times \{0\}}$ coincide and satisfy $T(S^1\times \{0\})\subset \ker \alpha_i$. Let $\xi_1:= \ker \alpha_1 \subset TM$, $\xi_2:= \ker \alpha_2 \subset T M$ be the associated contact structures.

Question: does there necessarily exist a diffeomorphism $f:M\to M$ such that the tangent map $Tf:TM\to TM$ satisfies $Tf(\xi_1)=\xi_2$? If not, what is a counterexample?

An h-principle implies that $\alpha_1$, $\alpha_2$ are homotopic through contact forms, so it is tempting to deduce from Gray's stability theorem the existence of an isotopy whose time-1 diffeomorphism $f$ has the required property. But Gray's stability theorem does not hold on noncompact manifolds (as far as I know), and $M$ is noncompact.

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    $\begingroup$ No. Take any tight contact structure satisfying your conditions. "Tight" means that there is no embedded disk that is tangent to contact structure along the boundary of the disk. For example, one can take a tubular neighborhood of any Legendrian knot in the standard contact 3-space which is tight by the result of Bennequin. Then apply Lutz twist outside some neighborhood of $S^1\times\{0\}$. The resulting contact structure is not tight. Thus it is not diffeomorphic to the initial one. $\endgroup$
    – Maxim Prasolov
    Commented May 29 at 14:45
  • $\begingroup$ @MaximPrasolov many thanks for your clear explanation, which I would be happy to accept as an answer if you would like to paste it below. $\endgroup$
    – Matthew Kvalheim
    Commented May 30 at 18:17

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