If people think this question is a little basic I will move it to stackexchange, but I decided to try here first. So it is clear that if two symplectic forms are not cohomologous then they cannot be symplectomorphic. Does the opposite hold true? Because of the lack of local invariants I would assume that forms in the same cohomology class are symplectomorphic, but I haven't seen that written anywhere, so perhaps it is NOT true? If it is not, could someone please give an example?
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$\begingroup$ The diffeomorphism of two dimensional torus given by $(x,y)\mapsto (y,x)$ is a symplectomorphism of the forms $dx\wedge dy$ and $dy\wedge dx$ which are not cohomologous. $\endgroup$– Jarek KędraCommented Jul 10, 2017 at 19:51
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Any two symplectic forms on $\mathbb{R}^{2n}$ are in the same cohomology class. But the usual symplectic form on a ball of radius 1 in Darboux coordinates does not have the same volume as the usual symplectic form on a ball of radius 2 in Darboux coordinates, even though the rescaling is a diffeomorphism. But maybe you want compact examples.
A much fancier class of examples: Gromov, Pseudoholomorphic curves in symplectic manifolds, p. 313, corollary $0.4.A_2'$ proves that there are symplectic structures on $\mathbb{R}^{2n}$ which do not embed into the usual Darboux symplectic structure.
For compact manifolds, Moser's homotopy method proves that cohomologous symplectic structures are symplectomorphic; see the standard textbooks, like McDuff and Salamon, or Arnol'd.
answered Apr 4, 2017 at 10:24
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5$\begingroup$ "For compact manifolds, Moser's homotopy method proves that cohomologous symplectic structures are symplectomorphic". Not exactly; it proves that deformations through symplectic forms in the same cohomology class comes from symplectomorphisms. The symplectic class is not generally a complete invariant (e.g. Taubes has 4-dimensional counterexamples). $\endgroup$ Commented Apr 4, 2017 at 13:57