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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.

juliho
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    This has been discussed on MathsOverflow. It seems that things such as conservation of energy and momentum would not be true if the two form were not closed. http://mathoverflow.net/questions/19932/what-is-a-symplectic-form-intuitively – Fly by Night Jan 29 '13 at 22:01
  • See also https://sbseminar.wordpress.com/2012/01/09/what-is-a-symplectic-manifold-really/ – N.Li Jul 13 '24 at 07:55

1 Answers1

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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.

Henry T. Horton
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  • Thanks! It's a very clear explanation :D – juliho Jan 31 '13 at 16:43
  • I'm not sure about your first item. A closed 2-form does not represent a cohomology class in $H^2(M;R)$, but a closed and non-exact form does. It turns out that $\omega=d\alpha$, where $\alpha=-pdq$, is exact. Right? – N.Li Jul 13 '24 at 08:23