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


\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.