# Computing the cohomology class of a symplectic form

If a symplectic manifold $(M,\omega)$ is given,

is there any efficient method to compute $[\omega]$, the cohomology class of the symplectic form?

I do know that efficient is not a good word to use, but for instance: if one knows for sure that the symplectic form is integral -- the class of $\omega$ belongs to the image of $\check{H}^2(M;\mathbb{Z})\to H_{DR}^2(M;\mathbb{R})$ -- one could choose a set of orientable smooth surfaces $N$ representing the generators of $H_2(M)$ and compute $\int_N\omega$. The value of this integral determines $[\omega]$. However, not knowing ahead this information (but conjecturing that it is the case) implies that one must prove that all the periods have the same value (is that right?) and this seems a hard task.

When the symplectic manifold is a cotangent bundle with the canonical form, it is trivial to compute its cohomology class. This is also the case for a exact symplectic manifold, when one knows a potential 1-form for the symplectic form.

Another "easy" case is $(\mathbb{C}P^{n}, \omega_{FS})$. The only problem is that I do not know how to prove that the Fubini-Study form come from $\mathcal{O}(1)$. I have seen this stated everywhere but no proof was provided.

Bonus question: can someone give me references for that result about the Fubini-Study form?

There are at least two, related, motivations for this question:

1) Given a symplectic manifold, does it admit a hermitian line bundle with connection such that its curvature is proportional to the symplectic form?

2) Given a closed $2n$-dimensional symplectic manifold, is it possible to find a symplectic embedding of it in $(\mathbb{C}P^{2n+1}, \omega_{FS})$ ?

They are both related because if the symplectic form is integral, then there exists an affirmative answer for them (Kostant, Weil and Tischler).

It would be also interesting to know if

is there any kind of classification (or attempts to classify) integral symplectic manifolds?

This is maybe too much, since to my knowledge not even exact symplectic manifolds are classified.

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I'm not sure what you mean by contractible disk'' but you'll certainly need to know more than just $\int_D\omega$ to determine $[\omega]$. Perhaps you are assuming $\pi_1M=0$ and you want $\int_D\omega$ over all disks? –  Paul Sep 27 '12 at 16:26
I dont think this is a really well defined question until you tell us how the manifold and symplectic form are described. As regards 1 and 2: these are obviously impossible if the form isn't integral. –  Ben Webster Sep 27 '12 at 16:41
I was thinking about Stokes... Yes, if the manifold is simply connected and I compute that integral over all disks it will give me the answer. Now let me rethink about what happens in general for a integral one and two different disks. –  Romero Solha Sep 27 '12 at 16:48
@Ben 1 and 2 are motivations, not questions, as I know their answers from Kostant, Weil and Tischler. About how are the symplectic manifold described: If I give you a sphere with its area form as the symplectic form. You just need to compute its total area to know if it is integral, for instance. But it is true that you will need extra data to deal with this problem, and I am expecting this from the method. I mean: what is the extra data needed? –  Romero Solha Sep 27 '12 at 16:56
@Ben In the case of coadjoint orbits of compact Lie groups with the KKS form, I think that there is a way to tell which are the integral orbits by looking at the Weyl chamber, but I do not know exactly how it goes... This would be another efficient method when one has this extra data about how the manifold and symplectic form are described. –  Romero Solha Sep 27 '12 at 18:12

As a partial answer I can give you a reference how to describe the cohomology class of the symplectic form for symplectic toric manifolds, i.e. symplectic manifolds $(M^{2n}, \omega)$ with an effective Hamiltonian action of the torus $T^{n}$.
As you probably know, the image of the corresponding moment map is a (Delzant) polytope $\triangle \subset \mathbb{R}^{n}$ - suppose it is given by the inequalities $$\langle x, u_{i} \rangle \geq \lambda_{i}, \quad i=1, \ldots, d,$$ where $u_{i} \in \mathbb{Z}^{n}$ are primitive and $d$ is the number of facets, i.e. $(n-1)$-dimensional faces of $\triangle$. The preimage of the $i$-th facet under the moment map is a codimension 2 submanifold of $M$, let $c_{i} \in H^{2}(M; \mathbb{Z})$ be the cohomology class Poincare dual to this submanifold. Then $$[\omega] = -\sum_{i=1}^{d} \lambda_{i}c_{i}.$$ The proof can be found in V. Guillemin: Moment maps and combinatorial invariants of Hamiltonian $T^{n}$ spaces, Appendix 2.
In fact, the above classes $c_{i}$ generate the cohomology ring $H^{*}(M, \mathbb{Z})$, the proof of which and a complete description of the cohomology ring can be found in V. Guillemin, S. Sternberg: Supersymmetry and equivariant de Rham theory, section 9.8.2. There's also a bit on this in D. McDuff, D. Salamon: J-holomorphic curves and symplectic topology, section 11.3.1.