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We know by Darboux theorem that any symplectic form on a manifold $W^{2n}$ is locally symplectomorophic to the standard symplectic form $dx\wedge dy$ on $R^{2n}$. Is it true that any symplectic form on a ball $B^{2n}$ of arbitrary radius is symplectomorphic to the standard one? or -I guess equivalently- can we extend this local symplectomorphsim as long as we stay on a ball (say handles of different indices in a Morse decomposition of the manifold) in the manifold $W^{2n}$?

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    No. Symplectic volume is a global invariant and it changes as the radius of the ball changes. – Qiaochu Yuan May 19 '15 at 00:49
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    Gromov showed that there are symplectic structures on $\mathbb R^{2n}$ such that they are not the pullback under an embedding $\mathbb R^{2n}\to\mathbb R^{2n}$ of the standard symplectic form (priovided $n>1$) – Mariano Suárez-Álvarez May 19 '15 at 00:54

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