# 'Contactization' and Symplectization

Given any manifold $M$, we can get a symplectic manifold by taking the cotangent bundle $T^\ast M$ with symplectic form $\omega=\sum dp_i\wedge dq_i$. Given any manifold $M$, we can get a contact manifold by taking the projectivization of the cotangent bundle $\mathbb{P}^\ast M=(T^\ast M-\lbrace0\text{-section}\rbrace)/{\sim}$ where the contact form arises from the tautological 1-form on $T^\ast M$.

Given any contact manifold $(N,\lambda)$, we can get a symplectic manifold by symplectization $\mathbb{R}\times N$ with symplectic form $d(e^s\lambda)$. Continuing in the same spirit:

Is there a "contactization" to pass from any given symplectic manifold to a contact one, making use of the symplectic data?

Aside: I came across a paper of Eliashberg-Hofer-Salamon (Lagrangian Intersections in Contact Geometry), and in certain scenarios we do indeed have one. If our symplectic manifold $M$ is exact, i.e. $\omega=d\alpha$, then $(M\times S^1,dz-\alpha)$ is a contact manifold. Now if we don't have exactness, there is at least a way to contactize $M$ when some positive multiple of $\omega$ represents an integral cohomology class in $H^2(M)$, and this is some principal $S^1$-bundle called ''pre-quantization''. Is ''pre-quantization'' the only way to contactize here?

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A remark: What you say in the cases where $\omega$ is exact or integral is precisely the content of Appendix 4L "Contactification" in Arnol'd's Mathematical methods of classical mechanics (1989 translation, p.368). – Francois Ziegler Sep 16 '12 at 6:22
IMO (and I am perhaps a crank about this), there are two things we call "symplectization". The first should be thought of canonically as consisting of all elements of $T^*M$ whose kernel is the contact structure. If the contact structure is co-orientable, there are two connected components, each of which is naturally an $R^+$ bundle. Choosing one and taking the log gives you the symplectization you wrote with the symplectic form you wrote. – Sam Lisi Sep 16 '12 at 10:50
The second symplectization, also $\mathbb{R}\times M$ is where a translation invariant $J$ should live -- this makes sense for a stable Hamiltonian structure, and should really be thought of as a kind of blown up or stretched out version of $(0,1) \times M$. If you consider e.g. the Hofer energy we use for pseudoholomorphic curves (or even better, the definition in the case of a stable Hamiltonian structure), you see that we take a sup over a family of forms that tame $J$, each of which have (uniformly bounded) finite volume. These comments may or may not be relevant to you. – Sam Lisi Sep 16 '12 at 10:53

You may think of the Boothby-Wang construction as constructing a contact fiber bundle over a symplectic manifold with fiber $S^1$. If we look at the construction this way, it can be generalized. See my paper Contact fiber bundles. J. Geom. Phys. 49 (2004), no. 1, 52–66.
But this contact fiber bundle need not have a symplectic manifold as its base, so I don't see an explicit connection between symplectic $M$ and contact fiber bundle $N\to M$. For instance, I can build a symplectic space out of any manifold by taking the cotangent bundle, but it defeats the spirit of what I'm after. – Chris Gerig Sep 17 '12 at 21:26
You can build a contact manifold out of any manifold $M$ by taking the projectivization $P(T^*M)$ of its cotangent bundle . Is this what you are after? – Eugene Lerman Sep 17 '12 at 21:41
Sorry but what I'm looking for is the exact opposite, i.e. something that depends on $\omega$ (in the same way that the symplectization $M\times \mathbb{R}$ required $\lambda$). – Chris Gerig Sep 18 '12 at 3:33
I agree with Chris that a contact fibre bundle doesn't have to have a symplectic base, even when the total space is contact. For example, take the Hopf fibration $S^{3} \to S^{7} \to S^{4}$ with the standard contact forms on $S^{3}$ and $S^{7}$. – Oldřich Spáčil Sep 19 '12 at 14:37