TITLE: what prevents a manifold to be symplectic?
QUESTION [11 upvotes]: Are there any obstructions known which prevent an even dimensional orientable manifold from
being symplectic? I am a novice in this area so I unfortunately I cannot make the question more precise. What I have in mind is raw and is as follows:from the point of handle body theory you can put a symplectic structure on each handle as you are building the manifold, so there must something go wrong on the boundary of the handle which is a three manifold, for example in the case of spheres of dimension >2.  I just want to know if there is something related to this question in the literature.

REPLY [16 votes]: For a closed manifold of arbitrary dimension, there are two algebraic conditions which must be satisfied. Firstly as Russell states there must be a reduction of the structure group from $GL(2n)$ to $Sp(2n)$, or equivalently from $SO(2n)$ to $U(n)$. This is also necessary for open symplectic manifolds, and is equivalent to the existence of an almost complex structure.
Secondly, for a closed manifold, there must be a homology class $[\omega] \in H^2_{dR}$ so that the $n$-fold wedge product of $[\omega]$ generates $H^{2n}_{dR}$. This only holds for closed manifolds since there are exact symplectic structures on open manifolds.
For 4-manifolds there are further known obstruction due to Taubes, which tell us that the Gromov invariant must be equal to the Seiberg-Witten invariant, in particular the Seiberg-Witten invariant must be non-zero. This shows that the connected sum of three copies of $\mathbb{CP}^2$ admits no symplectic form, even though it satisfies the previous two conditions. As far as I know this is the only known geometric obstruction. For higher dimensional manifolds, it's my understanding that it's an open question whether the two algebraic conditions suffice.