Now I’ll actually give the proof of the Darboux theorem that a symplectic manifold is locally symplectomorphic to $\mathbb{R}^{2n}$ with the usual form.

Proof of the Darboux theorem

We will prove the equivalent:

Theorem 1 Let ${M}$ be a manifold with closed symplectic forms ${\omega_0, \omega_1}$, and ${p \in M}$ with ${\omega_0(p) = \omega_1(p)}$. Then there are neighborhoods ${U,V}$ of ${p}$ and a diffeomorphism ${f: U \rightarrow V}$ with ${f^*\omega_1 = \omega}$.

The idea is to consider the continuously varying family of 2-forms

$\displaystyle \omega_t = (1-t) \omega_0 + t \omega_1 = \omega_0 + t \alpha$

where

$\displaystyle \alpha = \omega_1 - \omega_0 .$

We will consider a small neighborhood ${U}$ of ${p}$ and a smooth map ${ G: U \times [0,1] \rightarrow M }$ such that ${G_t:=G(\cdot, t)}$ is a diffeomorphism, ${G_0 = id}$, and

$\displaystyle G_t^*(\omega_t) = \omega_0 \ (*).$

So we will need to find an appropriate time-dependent vector field ${X_t}$. To choose ${U}$, make it so small that it is a coordinate neighborhood (diffeomorphic to a ball) and such that ${\omega_t}$ is symplectic (i.e., nondegenerate) on ${U}$ for all ${t}$. This can be done since ${\omega_t(p)}$ is constant!

Now we want to choose ${X_t}$ such that

$\displaystyle \frac{d}{dt} G_t^*(\omega_t) = 0.$

This will prove (*). So, this equals (see below)

$\displaystyle \frac{d}{dt} G_t^*(\omega_t) =G_t^* ( L_{X_t} (\omega_t) ) + G_t^*(\alpha)= G_t^*(\alpha + d i_{X_t} \omega_t + i_{X_t} d\omega_t ).$

How do we know this? Well, first, if ${\omega}$ is fixed, I claim:

$\displaystyle \boxed{ \frac{d}{dt} G_t^*(\omega) = \lim_{h \rightarrow 0} \frac{ G_{t + h}^* \omega - G_t^* \omega}{h}. }$

One way to see this is first to check the identity on functions, for which it is ok (both sides are equal to ${X_t f}$). Next, it can be checked that both sides of the equation for ${\frac{d}{dt} G_t^*(\omega)}$ are derivations of the algebraof forms (alternating tensors) under the wedge product. Finally, both sides commute with the exterior derivative ${d}$ since the pull-back does. Therefore, by induction on the size ${k}$ of the ${k}$-form ${\omega}$, we get the boxed result.

The more general result

$\displaystyle \frac{d}{dt} G_t^*(\omega_t) =G_t^* ( L_{X_t} (\omega_t) ) + G_t^*\left( \frac{d \omega_t}{dt} \right)$

follows from this using the same trick as in the product rule.

Now back to the Darboux theorem.

The last term on the left (with ${i_{X_t} d\omega_t}$) vanishes by closedness. The first term becomes

$\displaystyle G_t^*( d( \beta + i_{X_t} \omega_t))$

so if we choose ${X_t}$ such that ${i_{X_t} \omega_t = -\beta}$, then we have the appropriate time-dependent vector field. ${X_t}$ is uniquely determined because ${\omega_t}$ is symplectic on ${U}$.

Of course, we don’t yet know whether the flow of ${X_t}$ lasts all the way until 1, i.e. if ${G}$ is defined. But the flow does last past 1 at ${p}$ (where ${X_t(p) \equiv 0}$, all ${t}$), so it must in some neighborhood of ${p}$ by the openness of the domain of the flow.

So shrinking ${U}$ if necessary, ${G_1}$ is the diffeomorphism in question.