why symplectic form should be closed when we work on a manifold

Question

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?

Answer 1

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.

Now

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

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 $d\omega=0$

Answer 2

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.