Exact symplectic manifold

An exact symplectic manifold has a symplectic form that is an exact differential form, written here $\omega =-d\lambda$. A positive-dimensional example without boundary cannot be compact, by the Generalized Stokes theorem applied to the top wedge product of differential forms.

On a nonempty compact $2n$-dimensional symplectic manifold without boundary, with $n\ge 1$, the class $[\omega {]}^n$ in top-degree de Rham cohomology is nonzero, since its integral in the symplectic orientation is positive. In particular $\omega$ cannot be an exact differential form: if $\omega =d\alpha$, then ${\omega }^n=d(\alpha \wedge {\omega }^{n-1})$ and the Generalized Stokes theorem would make that integral zero. Thus vanishing second de Rham cohomology obstructs a closed positive-dimensional symplectic manifold.