Thanks for reading my question. I'm wonder why a symplectic form should be closed. I found many different answers in the internet, but it sounds like a technical requirement (if we omit this requisit, we obtain almost symplectic structures, insteresting as well). Why do yo think? I just want to have a fresh perspective. Thank you in advance.
[Math] a symplectic form be closed
symplectic-geometry
Best Answer
There are many reasons why we might want a symplectic form to be closed by definition. Here are a few:
A closed $2$-form represents a cohomology class in $H^2(M; \Bbb R)$.
When $\omega$ is closed, we get a one-to-one correspondence between $1$-parameter groups of symplectomorphisms and symplectic vector fields. We also get Hamiltonian diffeomorphisms from this, which turn out to be interesting for symplectic geometers.
The local rigidity theorems for symplectic manifolds, such as Darboux's theorem, Moser's stability theorem, Weinstein's tubular neighborhood theorem, and so on rely on the closedness of the symplectic form. I think some such results hold more generally, for example Gotay's coisotropic neighborhood theorem holds for presymplectic manifolds if I recall correctly. In any case, these results are what give symplectic geometry its global nature.
There's other reasons based on physical considerations as well. In short, closedness of a symplectic form is a rather mild condition that gives rise to a lot of nice structure we wouldn't otherwise have.