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.

For an exact symplectic manifold with $\omega =d\lambda$, the Liouville vector field is defined by ${\iota }_Z\omega =\lambda$. Cartan's magic formula gives ${\mathcal{L}}_Z\omega =\omega$. On a cotangent bundle using $\omega =-d\alpha$, take $\lambda =-\alpha$; then $Z={\sum }_i{p}_i{\partial }_{p_i}$ generates the cotangent fiber-dilation flow.

The vertical Liouville vector field on a cotangent bundle has complete flow $D_s(x,p)=(x,e^{s}p)$. Its pullbacks scale both the Liouville one-form and the canonical symplectic form by $e^s$. As $s\to -\infty$, each orbit tends to its base point on the zero section. This contraction proves rigidity of cotangent Liouville-form preservation.

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.