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For defining the symplectic space $(V, \omega)$ where $V$ is a vector space, it doesn't necessary to add the condition $d\omega=0$. But, when we work on a manifold instead of vector space, then we need $\omega$ be closed. So why?

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I guess that "symplectic vector space" and "symplectic manifold" are two different notions. In particular not every symplectic vector space is a symplectic manifold. –  user10001 Jun 28 '13 at 15:55
Thanks @user10001 , can you give an example for more effectiveness of your comment. In fact every vector space is a manifold, so your example would be good? –  hassan joulani Jun 28 '13 at 16:09
Now when you are asking for an example I think my statement is not correct;) It should be other way around - every symplectic vector space is also a symplect manifold. Doesn't the condition $d\omega=0$ follow from bilinearity? Since matrix elements of $\omega$ will have no dependence on coordinates. –  user10001 Jun 28 '13 at 16:31
Yes, of course, thinking of $V$ as a manifold, the derivative of a constant $2$-form is $0$. Note that the Darboux Theorem says that on any symplectic manifold there are local coordinates in which the symplectic form becomes the standard symplectic form $\sum_{i=1}^n dx_i\wedge dy_i$ on $\mathbb R^{2n}$. –  Ted Shifrin Jun 28 '13 at 16:45
We have also a theorem which says :Every symplectic manifold $M$ is locally isomorphic to a symplectic vector space $V$ –  hassan joulani Jun 28 '13 at 16:58

2 Answers 2

up vote 2 down vote accepted

Suppose $\omega$ be a bilinear form on a vector space $V$. Then consider a basis $e_1,...,e_n$ of $V$ wrt to which coordinates of points are $x_1,...,x_n$. Wrt these coordinates we can define one forms $dx_i,i=1,...,n$ and vector fields $\partial_i,i=1,...,n$ in the usual way.


$\omega(a_1\partial_1+\dots a_n\partial_n,b_1\partial_1+\dots+b_n\partial_n)=\displaystyle\sum_{i,j}a_ib_j\omega_{ij}\tag 1$

where $\omega_{ij}=\omega(\partial_i,\partial_j)$ are constants independent of a's and b's.

From (1)

$\omega=\displaystyle\sum_{i,j}\omega_{ij}dx_i\otimes dx_j$

So $d\omega=0$

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When you have an alternating tensor on a vector space, there's nothing to differentiate. The point of closedness of the symplectic $2$-form on a manifold is that it therefore represents a cohomology class and induces a generator of the top cohomology (in the compact case). [My view, as a complex geometer, is that the symplectic form is quite analogous to the Kähler form in Kähler geometry.] Most of the interplay with the Poisson bracket and Lie derivatives depend on closedness, as well.

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Ted Shifrin@ what do you mean of top cohomology? –  hassan joulani Jun 28 '13 at 16:02
If $\dim M = 2n$, then I mean $H^{2n}(M,\mathbb R)$. –  Ted Shifrin Jun 28 '13 at 16:10

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