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Let $(M,\xi)$ be a transversally orientable contact manifold, that is, there exists a form $\alpha \in \Omega^1(M)$ such that $\xi = \ker \alpha$. Then we can associate to $(M,\xi)$ its symplectisation $(\mathbb{R} \times M,d(e^t\alpha))$, a symplectic manifold. I wondered, if there is a categorical setting for this process. I mean, naively, we could consider symplectisation as a map on objects $$S \colon \mathsf{TOCont} \to \mathsf{Symp}$$ where $\mathsf{TOCont}$ denotes the category of transversally orientable contact manifolds as objects and maps $F \in C^\infty(M,\widetilde{M})$ such that there exists a nowhere vanishing function $f \in C^\infty(M)$ with $F^* \widetilde{\alpha} = f\alpha$ as morphisms $F \colon (M,\xi = \ker \alpha) \to (\widetilde{M},\widetilde{\xi} = \ker \widetilde{\alpha})$. Likewise, $\mathsf{Symp}$ denotes the category with objects symplectic manifolds and morphisms $F \colon (M,\omega) \to (\widetilde{M},\widetilde{\omega})$ such that $F \in C^\infty(M,\widetilde{M})$ with $F^*\widetilde{\omega} = \omega$.

Now the problem I am facing is the following: I would define $S$ on morphisms $$S(F) \colon (\mathbb{R} \times M,d(e^t\alpha)) \to (\mathbb{R} \times \widetilde{M},d(e^t\widetilde{\alpha}))$$ by $$S(F) := \operatorname{id}_{\mathbb{R}} \times F.$$ But then, if $F^* \widetilde{\alpha} = f\alpha$, we compute $$S(F)^* d(e^t\widetilde{\alpha}) = d(e^tf\alpha),$$ that is, $S(F)$ is not a morphism in $\mathsf{Symp}$. If $f > 0$, we could use the definition $$S(F)(t,x) := (t - \log(f(x)),F(x))$$ and things would work out fine. However, this would impose a restriction on orientation.

I think everything boils down to the fact that if $(M,\xi = \ker \alpha)$ is a contact manifold, then also $\xi = \ker f\alpha$ for every nowhere vanishing smooth function $f$. But I guess the symplectisations are not symplectomorphic in general in this case, that is, a single t.o. contact manifolds admits different non-symplectomorphic symplectisations. Is that right? Do you have any idea how to turn symplectisation into a functor between appropriate categories?

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first of all I think your $S(F)$ can be modified into \begin{align*} S(F)(t,x)=(t-\log(|f(x)|), F(x)) \end{align*} since $f$ is non-vanishing, this is always smooth. Nevertheless, there is a more conceptual way to see the symplectization: the symplectization $S$ is a functor from contact manifolds into homogeneous symplectic manifolds. The latter is the category of pairs $(P,\omega)$ consisting of a $\mathbb{R}^\times$-principal bundle $P$ and a symplectic structure $\omega\in \Omega^2(P)$, such that \begin{align*} h_r^*\omega=r\omega \end{align*} for the principal action $h\colon \mathbb{R}^\times\times P\to P$. The morphisms are equivariant symplectomorphisms. This functor is even an equivalence of categories and does not work just for co-orientable contact structures. Everything what I said is (more or less) done in Remarks on Contact and Jacobi Geometry (Bruce, Grabowska, Grabowski 2015).

HD

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    $\begingroup$ I think it's friendlier to link to the abstract instead of (or as well as) the PDF. People are sometimes on slow connections and they may wish to see what the paper is about before downloading. $\endgroup$ – José Figueroa-O'Farrill Jul 28 '19 at 19:16
  • $\begingroup$ I also thought about modifying the $S(F)$ in the way you did, but unfortunately, at least as far as I can tell, it doesn't work since $$S(F)^*d(e^t\widetilde{\alpha}) = d(e^t \operatorname{sgn}(f) \alpha) \neq d(e^t\alpha)$$ in general. Thank you for the suggested paper! I will check it out. $\endgroup$ – TheGeekGreek Jul 28 '19 at 19:19
  • $\begingroup$ @JoséFigueroa-O'Farrill thanks for the suggestion, I already edited my post. $\endgroup$ – Heinz Doofenschmirtz Jul 28 '19 at 19:38
  • $\begingroup$ @TheGeekGreek I think you made a mistake. Note that you have $d\log(|f|)=\frac{df}{f}$ for all non-vanishing functions $f$. $\endgroup$ – Heinz Doofenschmirtz Jul 28 '19 at 19:41
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    $\begingroup$ @TheGeekGreek You are right. I see now, where the sign issue arises. Take a contact manifold $(M,\alpha)$, then $d(t\alpha)$ is a symplectic structure on $M\times\mathbb{R}^\times $. For a contactopmorphism $F\colon M\to\tilde M$ ($F^*\tilde\alpha=f \alpha$ ), then the map $S(F)(x,t)=(F(x),\frac{t}{f})$ is a symplectomorphism. "Your" symplectization basically chooses the open subset with positive reals, but a morphism with with negative $f$ doesn't preserve this choice. $\endgroup$ – Heinz Doofenschmirtz Jul 28 '19 at 21:24
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The right notion of the category $\mathsf{TOCont}$ is not transversally orientable, but rather transversally oriented, that is, we choose an orientation of the hyperplane distribution and the $1$-form compatible with this orientation. Then everything works well.

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