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Two compact connected symplectic surfaces are symplectomorphic exactly when their genus of a surface and symplectic area agree. The sufficiency follows from Moser theorem applied to a positive interpolation of their symplectic forms after an orientation-preserving diffeomorphism.

A symplectomorphism of a closed symplectic surface carries the complementary components of a separating simple closed curve to those of its image, preserving their unordered symplectic areas. On the unit sphere a latitude at height $h$ has complementary areas $2\pi (1-h)$ and $2\pi (1+h)$. Thus the equator cannot be sent to any nonequatorial latitude.

A symplectomorphism is a diffeomorphism, so it preserves the genus of a surface. It also preserves the orientation determined by the symplectic form and, by integration of a pullback of a differential form, preserves the symplectic area.

Conversely, equal genus of a surface gives a diffeomorphism $f:{\Sigma }_0\to {\Sigma }_1$. We can choose $f$ to preserve the orientations determined by ${\omega }_0,{\omega }_1$: if necessary, compose with an orientation-reversing diffeomorphism of the standard closed orientable surface of that genus of a surface, obtained by reflection of a symmetric handle model. On ${\Sigma }_0$ put $\eta =f^{\ast }{\omega }_1$. Both $\eta$ and ${\omega }_0$ are positive volume forms for the same orientation, and their integrals agree. By top-degree de Rham cohomology, $[\eta ]=[{\omega }_0]$.

The family ${\omega }_t=(1-t){\omega }_0+t\eta$ consists of positive volume forms, hence symplectic forms in dimension two, with constant de Rham cohomology class. Moser theorem supplies $\psi$ with ${\psi }^{\ast }\eta ={\omega }_0$. ConsequentlyTherefore the genus of a surface and symplectic area classify compact connected symplectic surfaces.

Let $M=S^2$ with an area form and let $L$ be an equator dividing the sphere into two open hemispheres of equal area. Every curve in a symplectic surface is Lagrangian. A small normal push moves $L$ to a nearby latitude, so it is displaceable by a smooth isotopy.

Suppose a symplectic isotopy had final image $L^{\prime }$ disjoint from $L$. The curve $L^{\prime }$ must lie in one hemisphere. Of the two discs bounded by $L^{\prime }$, the one contained in that hemisphere has area strictly below half the total area, and the other has area strictly above half. On the other hand, a symplectomorphism maps the original two hemispheres to the two discs bounded by $L^{\prime }$ and preserves their areas, so both would have half the total area. This contradiction is the symplectic non-displaceability of an area bisector.

The symplectic area of a compact symplectic surface $(\Sigma ,\omega )$ is ${\int }_{\Sigma }\omega$, using the orientation determined by the symplectic form.