Symplectic space and symplectic manifold I'm struggling with some basics and I hope you can explain this to me. 
I know what if we have blinear, antisymmetric, nondegenerate form $\omega$ on vector space $\mathcal{V}$, we can tell about this space as symplectic space. 
I read somewhere, that pair ($\mathcal{V}$, $\omega$) is symplectic manifold. But they also said that here $\omega$ must be some 2-form. 
What's the difference? Why this $\omega$ have to be 2 form if we want to talk about symplectic manifold?
I hope you can help me. Thanks in advanced.
 A: Manifolds in general are lines, surfaces... they don’t have linear structures, so you can't define a bilinear antisymmetric form on them. 
The definition of a symplectic manifold is a couple $(M,\omega)$ where $M$ is a smooth manifold and $\omega$ is a closed 2-form which is degenerate at no point. 
A 2-form on a manifold is a collection of bilinear antisymmetric forms on the tangent spaces at various points of the manifold, which is “smooth” in some way.
Now, a (finite-dimensional) vector space $V$ clearly is a manifold. Now, let $\omega$ be a nondegenerate bilinear antisymmetric form on $V$. 
How can $\omega$ be seen as a 2-form on $V$? At each point $x \in V$, the abstract tangent space at $x$, $T_xV$, can be identified with $V$ in a natural way. So we can say that $\omega_x$ acts on $T_xV$ as $\omega$ on $V$. 
Thus the collection $\omega_{mfd}$ of the $\omega_x$ is a 2-form. It is “constant” in some way, which entails that it is closed. By definition, it is nowhere degenerate, so $(V,\omega_{mfd})$ is a symplectic manifold. 
