In Donaldson's papers on symplectic hypersurfaces and Lefschetz pencils for symplectic manifolds, he considers symplectic manifolds $(M, \omega)$ where the cohomology class $[\omega/2\pi] \in H^2(M; \mathbb{R})$ lies in the integral lattice $H^2(M; \mathbb{Z})$.
How restrictive of a condition is this on the symplectic manifold? In particular, I'd be interested in knowing if there is or is not a smooth manifold which admits no symplectic forms representing an integral cohomology class in this way.
Best Answer
Suppose that $M$ is compact. Let $\omega_1,...,\omega_k\in \Omega^2(M)$ be closed forms which project to a basis of $H^2_{dR}(M)$. Let $\omega$ be a symplectic form on $M$. Then for $a_1,...,a_k\in {\mathbb R}$ sufficiently close to $0$, the linear combination $$ \eta= \omega+ \sum_{i=1}^k a_i \omega_i $$ is nondegenerate; it is of course, closed no matter what $a_i$'s are. Hence, the form $\eta$ is symplectic (if all $a_i$'s are close to zero). The projection of the set of such forms to $H^2_{dR}(M)$ forms a neighborhood of $[\omega]$ since $[\omega_1],...,[\omega_k]$ was a basis of $H^2_{dR}(M)$. Thus, for a dense subset of parameters $a_i$, the cohomology class of $\eta$ is rational. Hence, after rescaling, it is integral. Thus, every compact symplectic manifold admits an integral symplectic form. I am not sure about noncompact case, but you do not care about it anyhow.