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I am fairly new to symplectic geometry but I started wondering about this :

From the definition of a symplectic manifold we can extract that $M$ has to be even dimensional , orientable , and in the case that it's compact $H^2_{dR}(M)\neq 0$. Now is there extra conditions that we can impose on even-dimensional compact manifold with $H^2_{dR}(M)\neq 0$ such that $M$ will be symplectic , other than having a symplectic form of a $2-$form $\omega$ such that $\omega^n$ is a volume form ? And does anybody know if there exists an example of a compact manifold that satisfies all those 3 conditions such that it is not symplectic ? Or if it will automatically be a symplectic manifold?

Thanks in advance.

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In general there are no purely topological conditions one can impose to ensure the existence of a symplectic structure; it necessarily involves the smooth structure. There are examples of smooth 4-dimensional manifolds that are homeomorphic to symplectic (even K\”ahler) manifolds but which admit no symplectic structure themselves. Many such examples were constructed by Fintushel and Stern in their paper “Knots, links and 4-manifolds,” but I’m not sure if these were the first.

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  • $\begingroup$ Alright , thanks for the answer. $\endgroup$
    – Someone
    Commented Mar 8, 2021 at 8:01
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Try $S^2\times S^4$. What happens then? You need $H^2(M)$ to generate $H^{2n}(M)$

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