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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