# Symplectisation as a functor between appropriate categories

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?

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).
• 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. – TheGeekGreek Jul 28 '19 at 19:19
• @TheGeekGreek I think you made a mistake. Note that you have $d\log(|f|)=\frac{df}{f}$ for all non-vanishing functions $f$. – Heinz Doofenschmirtz Jul 28 '19 at 19:41
• @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. – Heinz Doofenschmirtz Jul 28 '19 at 21:24
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.