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What is the motivation behind studying smooth manifolds with a non-degenerate closed two-form?

The subject certainly originated from physics, but is there a deeper reason for why it is still an active subject of research in mathematics? What lead mathematicians to put so much effort to understand these objects?

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    $\begingroup$ Physics is probably why people first looked at symplectic manifolds (namely cotangent bundles), but why we continue to study them is deeper. Somewhat, it is just to see the consequences of a simple condition (e.g., there are actual topological consequences to having a symplectic form on a manifold), somewhat it is because that condition appears in surprising places (e.g., co-adjoint orbits carry a symplectic structure), somewhat it is because there are continuing applications to physics (e.g., understanding quantization of classical systems), but there is more out there. $\endgroup$
    – Aaron
    Commented Apr 29, 2016 at 8:04
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    $\begingroup$ Thanks to all who commented. A question for those who closed the question: What is the difference between my question and this well-received question? Isn't that the same question but for a different topic (GIT)? $\endgroup$
    – Ron
    Commented Apr 29, 2016 at 12:40
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    $\begingroup$ Symplectic geometry can be seen as a generalization of complex geometry. To each symplectic structure one finds a (unique up to homotopy) compatible almost-complex structure, which in general is not integrable. So sympl. manifolds are much more flexible than complex manifolds. For example, a 4-manifold is symplectic if and only if it is a Lefschetz pencil. In particular, every fin. pres. group is the fundamental group of a symplectic 4-manifold. By contrast, there are strong restrictions for a group to be the fundamental group of a Kähler manifold. Also, sympl. mfd. occur in mirror symmetry. $\endgroup$
    – ThiKu
    Commented Apr 29, 2016 at 13:06
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    $\begingroup$ One motivation is just differential geometry: to each smooth manifold we can canonically associate a symplectic manifold, its cotangent bundle. So symplectic invariants (of the cotangent bundle) yield differential invariants (of the original manifolds). $\endgroup$
    – YCor
    Commented Apr 29, 2016 at 21:25
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    $\begingroup$ Questions about applications are very important, and, by the way, they inspire research. It is not good to close them. $\endgroup$
    – Sergei Akbarov
    Commented Apr 29, 2016 at 22:13

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