proof of Darboux’s theorem (symplectic geometry)

proof of Darboux’s theorem (symplectic geometry)

We first observe that it suffices to prove the theorem for

symplectic
forms

defined on an open neighbourhood of

0

∈

ℝ

2

⁢

n

.

Indeed, if we have a

symplectic manifold

(

M

,

η

)

, and a point

x

0

, we can take a (smooth)

coordinate chart

about

x

0

. We can then
use the

coordinate

function to push

η

forward to a symplectic form

ω

on a neighbourhood of

0

in

ℝ

2

⁢

n

. If the result holds on

ℝ

2

⁢

n

,
we can compose the coordinate chart with the resulting symplectomorphism to
get the theorem in general.

Let

ω

0

=

∑

i

=

1

n

d

⁢

x

i

∧

d

⁢

y

i

. Our goal is then to find a (local)

diffeomorphism

Ψ

so that

Ψ

⁢

(

0

)

=

0

and

Ψ

*

⁢

ω

0

=

ω

.

Now, we recall that

ω

is a non–degenerate two–form. Thus, on

T

0

⁢

ℝ

2

⁢

n

, it is a non–degenerate anti–

symmetric bilinear form

. By a linear

change of basis

, it can be put in the standard form. So, we
may assume that

ω

⁢

(

0

)

=

ω

0

⁢

(

0

)

.

We will now proceed by the “Moser trick”. Our goal is to find a
diffeomorphism

Ψ

so that

Ψ

⁢

(

0

)

=

0

and

Ψ

*

⁢

ω

=

ω

0

.
We will obtain this diffeomorphism as the time–

1

map of the
flow of an

ordinary differential equation

. We will see this as the
result of a

deformation

of

ω

0

.

Let

ω

t

=

t

⁢

ω

0

+

(

1

-

t

)

⁢

ω

.
Let

Ψ

t

be the time

t

map of the

differential equation

d

d

⁢

t

⁢

Ψ

t

⁢

(

x

)

=

X

t

⁢

(

Ψ

t

⁢

(

x

)

)

in which

X

t

is a

vector field

determined by a condition to be stated later.

We will make the ansatz

Ψ

t

*

⁢

ω

=

ω

t

.

Now, we differentiate this :

0

=

d

d

⁢

t

⁢

Ψ

t

*

⁢

ω

t

=

Ψ

t

*

⁢

(

L

X

t

⁢

ω

t

+

d

d

⁢

t

⁢

ω

t

)

.

(

L

X

t

⁢

ω

t

denotes the

Lie derivative

of

ω

t

with respect
to the vector field

X

t

.)

By applying Cartan’s identity and recalling that

ω

is closed, we
obtain :

0

=

Ψ

t

*

⁢

(

d

⁢

ι

X

t

⁢

ω

t

+

ω

-

ω

0

)

Now,

ω

-

ω

0

is closed, and hence, by Poincaré’s Lemma,
locally exact. So, we can write

ω

-

ω

0

=

-

d

⁢

λ

.

Thus

0

=

Ψ

t

*

⁢

(

d

⁢

(

i

X

t

⁢

ω

t

-

λ

)

)

We want to require then

i

X

t

⁢

ω

t

=

λ

.

Now, we observe that

ω

0

=

ω

at

0

, so

ω

t

=

ω

0

at

0

. Then, as

ω

0

is non–degenerate,

ω

t

will be non–degenerate on an
open neighbourhood of

0

. Thus, on this neighbourhood, we may use this to
define

X

t

(uniquely!).

We also observe that

X

t

⁢

(

0

)

=

0

. Thus, by choosing a sufficiently
small neighbourhood of

0

, the flow of

X

t

will be defined for time
greater than

1

.

All that remains now is to check that this resulting flow has the desired
properties. This follows merely by reading our of the ODE,
backwards.

Title

proof of Darboux’s theorem (

symplectic geometry

)

Canonical name

ProofOfDarbouxsTheoremsymplecticGeometry

Date of creation

2013-03-22 14:09:55

Last modified on

2013-03-22 14:09:55

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

8

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 53D05

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