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Given a contact manifold $(M,\lambda)$ we can pass to the symplectization $(\mathbb{R}\times M,\omega=d(e^s\lambda))$ and this is great to bring the machinery of symplectic geometry into the contact world. [Edit]: As pointed out in comments, there is another notion of symplectization (although less used). I am then naturally curious about the other direction.
In particular, for Seiberg-Witten theory and Embedded Contact Homology you can talk about symplectic cobordisms (4-manifolds) with contact boundary. But this assumes that I already have some contact geometry at the ends of my symplectic world.

The immediate question in my mind was: Is there a contactization to pass from a given symplectic manifold to a contact one? Or something in the spirit of it?

And then I come 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 $\ast$exact$\ast$, 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 $\ast$integral$\ast$ cohomology class in $H^2(M)$, and this is some principal $S^1$-bundle called ''pre-quantization''.

So my ultimate two questions:
1) Is ''pre-quantization'' the only way to contactize here?
2) Is there some useful notion of contactization in other scenarios?
I should clarify that I am not interested in getting contact manifolds in general; I'm not looking for an analog of $T^*M$ (which is a symplectic manifold for any $M$), or more explicitly $\mathbb{P}T^*M$ (which is a contact manifold for any $M$). So while symplectization requires the contact form, contactization should make use of the symplectic form.

edited Sep 19 at 19:46

asked Sep 16 at 5:50

Chris Gerig
5,526●1●12●40

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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 at 6:22

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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 at 10:50

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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 at 10:53

Eugene Lerman · Answer 1 · 2012-09-17 14:43:09Z UTC

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The "pre-quantization" construction of a contact manifold out of symplectic manifold predates prequantization by a couple of decades: see Boothby, W. M.; Wang, H. C. On contact manifolds. Ann. of Math. (2) 68 1958 721–734. The analogue of the theorem for symplectic orbifolds is due to Thomas: Thomas, C. B. Almost regular contact manifolds. J. Differential Geometry 11 (1976), no. 4, 521–533.

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.

answered Sep 17 at 14:43

Eugene Lerman
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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 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 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 at 3:33

The contact structure on the contact fiber bundle does depend on the symplectic form on the base. Contact fiber bundles have natural connections and symplectic forms on the base are part of the curvature, if I remember correctly. – Eugene Lerman Sep 18 at 11:07

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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 at 14:37

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‘Contactization’ and Symplectization

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