Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 140 / 1

Moser theorem says that if $M$ is a compact smooth manifold without boundary and ${\omega }_t$, $0\le t\le 1$, is a smooth family of symplectic forms with constant de Rham cohomology class, then there is a smooth isotopy ${\psi }_t$, with ${\psi }_0=\mathrm{id}$, such thatThus the endpoints are symplectomorphic. The hypothesis is a whole path of symplectic forms, not merely equality of the endpoint de Rham cohomology classes.

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.

There is an orientation convention in this statement: the two volume forms must be positive for one fixed orientation, as is usual in Moser theorem for volume forms. Without that convention, the stated necessity is false. For example, on an oriented two-sphere, take a positive volume form ${\beta }_0$ and an orientation-reversing diffeomorphism $r$, and set ${\beta }_1=r^{\ast }{\beta }_0$. Then $(r^{-1}{)}^{\ast }{\beta }_1={\beta }_0$ but $\int {\beta }_1=-\int {\beta }_0$. We prove the intended result with a common orientation.

If ${\varphi }^{\ast }{\beta }_1={\beta }_0$, positivity forces $\varphi$ to preserve the orientation, and integration of the pullback of a differential form gives equality of the integrals.

For sufficiency, top-degree de Rham cohomology gives a differential form $\alpha$ of degree $n-1$ with ${\beta }_1-{\beta }_0=d\alpha$. Put ${\beta }_t=(1-t){\beta }_0+t{\beta }_1$. Positivity makes every ${\beta }_t$ a volume form. The interior product of a differential form gives an linear isomorphismThus there is a unique smooth time-dependent vector field $X_t$ satisfying ${\iota }_{X_t}{\beta }_t=-\alpha$. Since $M$ is compact, its local flow exists throughout $0\le t\le 1$ and defines a smooth isotopy ${\varphi }_t$ starting at the identity. By Cartan's magic formula, and since a top-degree differential form has zero exterior derivative,HenceThe zero-dimensional case, a single point, is immediate from equality of the integrals.