symplectic manifolds I know sometimes they use advanced methods to prove a given 4-manifold is not symplectic. for instance by Seiberg-Witten theory. But for a manifold to be symplectic we just need to check that there is an element in the second cohomology which is closed and when we cup product with itself gives us a nonzero element. Isn't this easy to check by hand? for instance I guess connected sum of two copies of $CP^2$ is not symplectic, can't we check this by hand without using any S-W theory? 

On Mike Miller's suggestion, I've turned my comments into an answer:
From a more analytic perspective, you're asking for a global solution $\omega \in \Omega^2(M)$ to the differential equation $d\omega = 0$ on $M^{2n}$ which also satisfies $\omega^n \neq 0$. In general, I don't think it's easy to know when a differential equation has global solutions or not.
Whether there exist non-degenerate 2-forms (not necessarily closed) is a purely topological question; a complete characterization of such manifolds can be given in terms of characteristic classes. But it's not clear which of those also support a non-degenerate 2-form which solves $d\omega = 0$. (Just as it is not clear, for example, which smooth manifolds admit integrable complex structures.)
