Now I’ll actually give the proof of the Darboux theorem that a symplectic manifold is locally symplectomorphic to with the usual form.

**Proof of the Darboux theorem **

We will prove the equivalent:

**Theorem 1** *Let be a manifold with closed symplectic forms , and with . Then there are neighborhoods of and a diffeomorphism with . *

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

where

We will consider a small neighborhood of and a smooth map such that is a diffeomorphism, , and

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

Now we want to choose such that

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

How do we know this? Well, first, if is fixed, I claim:

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

The more general result

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 ) vanishes by closedness. The first term becomes

so if we choose such that , then we have the appropriate time-dependent vector field. is uniquely determined because is symplectic on .

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

So shrinking if necessary, is the diffeomorphism in question.

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