Proof

The trick consists in viewing ${\displaystyle \{\psi _{t}\}_{t\in [0,1]}}$ as the flows of a time-dependent vector field, i.e. of a smooth family ${\displaystyle \{X_{t}\}_{t\in [0,1]}}$ of vector fields on ${\displaystyle M}$. Using the definition of flow, i.e. ${\displaystyle {\frac {d}{dt}}\psi _{t}=X_{t}\circ \psi _{t}}$ for every ${\displaystyle t\in [0,1]}$, one obtains from the chain rule that ${\displaystyle {\frac {d}{dt}}(\psi _{t}^{*}\omega _{t})=\psi _{t}^{*}{\Big (}{\frac {d}{dt}}\omega _{t}+{\mathcal {L}}_{X_{t}}\omega _{t}{\Big )}.}$ By hypothesis, one can always find ${\displaystyle X_{t}}$ such that ${\displaystyle {\frac {d}{dt}}\omega _{t}+{\mathcal {L}}_{X_{t}}\omega _{t}=0}$, hence their flows ${\displaystyle \psi _{t}}$ satisfies ${\displaystyle \psi _{t}^{*}\omega _{t}=\mathrm {const} =\psi _{0}^{*}\omega _{0}=\omega _{0}}$.

Proof

One implication holds by the invariance of the integral by diffeomorphisms: ${\displaystyle \int _{M}\alpha _{0}=\int _{M}\psi ^{*}\alpha _{1}=\int _{\psi (M)}\alpha _{1}=\int _{M}\alpha _{1}}$.

For the converse, we apply Moser's trick to the family of volume forms ${\displaystyle \alpha _{t}:=(1-t)\alpha _{0}+t\alpha _{1}}$. Since ${\displaystyle \int _{M}(\alpha _{1}-\alpha _{0})=0}$, the de Rham cohomology class ${\displaystyle [\alpha _{0}-\alpha _{1}]\in H_{dR}^{n}(M)}$ vanishes, as a consequence of Poincaré duality and the de Rham theorem. Then ${\displaystyle \alpha _{1}-\alpha _{0}=d\beta }$ for some ${\displaystyle \beta \in \Omega ^{n-1}(M)}$, hence ${\displaystyle \alpha _{t}=\alpha _{0}+td\beta }$. By Moser's trick, it is enough to solve the following ODE, where we used the Cartan's magic formula, and the fact that ${\displaystyle \alpha _{t}}$ is a top-degree form:

$${\displaystyle 0={\frac {d}{dt}}\alpha _{t}+{\mathcal {L}}_{X_{t}}\alpha _{t}=d\beta +d(\iota _{X_{t}}\alpha _{t})+\iota _{X_{t}}({\cancel {d\alpha _{t}}})=d(\beta +i_{X_{t}}\alpha _{t}).}$$

However, since ${\displaystyle \alpha _{t}}$ is a volume form, i.e. ${\textstyle TM\xrightarrow {\cong } \wedge ^{n-1}T^{*}M,\quad X_{t}\mapsto \iota _{X_{t}}\alpha _{t}}$, given ${\displaystyle \beta }$ one can always find ${\displaystyle X_{t}}$ such that ${\displaystyle \beta +\iota _{X_{t}}\alpha _{t}=0}$.

Proof

In order to apply Moser's trick, we need to solve the following ODE

$${\displaystyle 0={\frac {d}{dt}}\omega _{t}+{\mathcal {L}}_{X_{t}}\omega _{t}=d\sigma _{t}+\iota _{X_{t}}({\cancel {d\omega _{t}}})+d(\iota _{X_{t}}\omega _{t})=d(\sigma _{t}+\iota _{X_{t}}\omega _{t}),}$$

where we used the hypothesis, the Cartan's magic formula, and the fact that ${\displaystyle \omega _{t}}$ is closed. However, since ${\displaystyle \omega _{t}}$ is non-degenerate, i.e. ${\textstyle TM\xrightarrow {\cong } T^{*}M,\quad X_{t}\mapsto \iota _{X_{t}}\omega _{t}}$, given ${\displaystyle \sigma _{t}}$ one can always find ${\displaystyle X_{t}}$ such that ${\displaystyle \sigma _{t}+\iota _{X_{t}}\omega _{t}=0}$.

Corollary

Given two symplectic structures ${\displaystyle \omega _{0}}$ and ${\displaystyle \omega _{1}}$ on ${\displaystyle M}$ such that ${\displaystyle (\omega _{0})_{x}=(\omega _{1})_{x}}$ for some point ${\displaystyle x\in M}$, there are two neighbourhoods ${\displaystyle U_{0}}$ and ${\displaystyle U_{1}}$ of ${\displaystyle x}$ and a diffeomorphism ${\displaystyle \phi :U_{0}\to U_{1}}$ such that ${\displaystyle \phi (x)=x}$ and ${\displaystyle \phi ^{*}\omega _{1}=\omega _{0}}$.^[3][4]

This follows by noticing that, by Poincaré lemma, the difference ${\displaystyle \omega _{1}-\omega _{0}}$ is locally ${\displaystyle d\sigma }$ for some ${\displaystyle \sigma \in \Omega ^{1}(M)}$; then, shrinking further the neighbourhoods, the result above applied to the family ${\displaystyle \omega _{t}:=(1-t)\omega _{0}+t\omega _{1}}$ of symplectic structures yields the diffeomorphism ${\displaystyle \phi$ :=\psi _{1}} .

Darboux theorem for symplectic structures

The Darboux's theorem for symplectic structures states that any point ${\displaystyle x}$ in a given symplectic manifold ${\displaystyle (M,\omega )}$ admits a local coordinate chart ${\displaystyle (U,x^{1},\ldots ,x^{n},y^{1},\ldots ,y^{n})}$ such that

$${\displaystyle \omega |_{U}=\sum _{i=1}^{n}dx^{i}\wedge dy^{i}.}$$

While the original proof by Darboux required a more general statement for 1-forms,^[5] Moser's trick provides a straightforward proof. Indeed, choosing any symplectic basis of the symplectic vector space ${\displaystyle (T_{x}M,\omega _{x})}$, one can always find local coordinates ${\displaystyle ({\tilde {U}},{\tilde {x}}^{1},\ldots ,{\tilde {x}}^{n},{\tilde {y}}^{1},\ldots ,{\tilde {y}}^{n})}$ such that ${\displaystyle \omega _{x}=\sum _{i=i}^{n}(d{\tilde {x}}^{i}\wedge d{\tilde {y}}^{i})|_{x}}$. Then it is enough to apply the corollary of Moser's trick discussed above to ${\displaystyle \omega _{0}=\omega |_{\tilde {U}}}$ and ${\displaystyle \omega _{1}=\sum _{i=i}^{n}d{\tilde {x}}^{i}\wedge d{\tilde {y}}^{i}}$, and consider the new coordinates ${\displaystyle x^{i}={\tilde {x}}^{i}\circ \phi ,y^{i}={\tilde {y}}^{i}\circ \phi }$.^[3][4]

Proof

It is enough to check that ${\textstyle {\frac {d}{dt}}\omega _{t}=d\sigma _{t}}$; then the proof follows from the previous application of Moser's trick to symplectic structures. By the cohomologous hypothesis, ${\displaystyle \omega _{t}-\omega _{0}}$ is an exact form, so that also its derivative ${\textstyle {\frac {d}{dt}}(\omega _{t}-\omega _{0})={\frac {d}{dt}}\omega _{t}}$ is exact for every ${\displaystyle t}$. The actual proof that this can be done in a smooth way, i.e. that ${\textstyle {\frac {d}{dt}}\omega _{t}=d\sigma _{t}}$ for a smooth family of functions ${\displaystyle \sigma _{t}}$, requires some algebraic topology. One option is to prove it by induction, using Mayer-Vietoris sequences;^[3] another is to choose a Riemannian metric and employ Hodge theory.^[1]