# '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?

• 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