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Two closed, connected symplectic surfaces are symplectomorphic exactly when their genus of a surface and total symplectic area agree. Necessity follows from invariance of topology and the integral of the pulled-back symplectic form.

For sufficiency, the classification of compact surfaces supplies an orientation-preserving diffeomorphism between surfaces of the same genus. Pull both forms back to one surface. Their equal integrals make their classes in $H^2(\Sigma ;\mathbb{R})$ equal. The interpolation ${\omega }_t=(1-t){\omega }_0+t{\omega }_1$ remains positive, since in two dimensions the two positive forms are pointwise positive multiples of each other. Moser theorem now gives an isotopy identifying them.^[ref]

The boundaryless hypothesis matters. A disk and an annulus can have the same genus and the same total area, but different numbers of boundary components prevent even a diffeomorphism. Boundary versions of the classification require the corresponding topology and boundary conditions.

Contact topology studies global properties of contact structures on odd-dimensional smooth manifolds. Locally such a structure is a hyperplane distribution $\xi =\ker \alpha$ with $\alpha \wedge (d\alpha {)}^n$ nowhere zero in dimension $2n+1$. Multiplying $\alpha$ by a nowhere-zero function leaves the hyperplane distribution unchanged: the structure is the field of hyperplanes, not a preferred one-form.^[ref]

In ${\mathbb{R}}^3$, take $\alpha =dz-ydx$. Then $d\alpha =dx\wedge dy$ and $\alpha \wedge d\alpha =dz\wedge dx\wedge dy\ne 0$. The spanning vector fields ${\partial }_x+y{\partial }_z$ and ${\partial }_y$ have Lie bracket $-{\partial }_z$, outside the distribution. Thus the planes cannot be tangent to a foliation by surfaces. The local normal form is universal, while global topology distinguishes contact structures and constrains their relation to symplectic geometry.

Let $M$ be a closed smooth manifold, and let ${\omega }_t$ be a smooth family of symplectic forms with constant de Rham cohomology class. Choose a smooth family of differential one-forms ${\sigma }_t$ with ${\dot{\omega }}_t=d{\sigma }_t$. Nondegeneracy uniquely solvesCompactness without boundary gives its flow ${\varphi }_t$ throughout $0\le t\le 1$. By Cartan's magic formula,so ${\varphi }_t^{\ast }{\omega }_t={\omega }_0$. This converts a constant-cohomology deformation into an actual isotopy of forms by diffeomorphisms.^[ref]

On a manifold with boundary one additionally needs a boundary-preserving flow, for instance by choosing $X_t$ tangent to the boundary. On a noncompact manifold the flow may fail to exist for the full interval. For example, on ${\mathbb{R}}^2$ the positive forms $dx\wedge dy$ and $e^{-(x^2+y^2)}dx\wedge dy$ are cohomologous, and their affine interpolation stays symplectic. They cannot be symplectomorphic, because their total symplectic volumes are respectively infinite and $\pi$.

A Poisson algebra has two compatible products on one vector space: a commutative, associative multiplication and a Lie bracket $\{f,g\}$. Compatibility is the Leibniz ruleThus fixing $f$ makes $g\mapsto \{f,g\}$ an algebraic derivation. The Lie bracket also satisfies the Jacobi identity; an arbitrary bilinear bracket satisfying antisymmetry of a Lie bracket and the product rule need not do so.^[ref]

For example, the polynomial ring $\mathbb{R}[q,p]$ becomes a Poisson algebra with the canonical Poisson bracketThe partial derivatives give $\{q,p\}=1$ and $\{q,p^2\}=2p$, illustrating how the Leibniz rule extends the bracket from generators to their products. More generally, smooth functions on a symplectic manifold form a Poisson algebra. A finite-dimensional Lie algebra of moment map components need not be closed under ordinary multiplication: adding all their products normally gives a much larger Poisson algebra.