every symplectic manifold has even dimension

every symplectic manifold has even dimension

All we need to prove is that every finite dimensional

vector space

V

with an

anti-symmetric

non-degenerate

linear form

ω

has an even

dimension

2

⁢

k

. This is only a

linear algebra

result. In the case of a symplectic manifold

V

is just the tangent space at a point, and thus its dimension equals the manifold’s dimension.

Pick any not

null vector

v

0

∈

V

. Since

ω

is non-degenerate

ω

⁢

(

v

0

,

⋅

)

is a not null linear form. Therefore there exists a not null vector

u

0

such that

ω

⁢

(

v

0

,

u

0

)

=

1

Now

v

0

and

u

0

are

linearly independent

because if

v

0

=

λ

⁢

u

0

then

ω

⁢

(

v

0

,

u

0

)

=

ω

⁢

(

λ

⁢

u

0

,

u

0

)

=

λ

⁢

ω

⁢

(

u

0

,

u

0

)

=

0

(by anti-symmetry).

Let

V

0

=

span

⁡

{

v

0

,

u

0

}

. Consider a space

V

1

of ”

orthogonal

” elements to

V

0

under

ω

. That is:

V

1

=

{

v

1

∈

V

:

for all

v

∈

V

0

,

ω

⁢

(

v

,

v

1

)

=

0

}

We now prove

V

=

V

0

⁢

⊕

V

1

:

•

V

0

⁢

⋂

V

1

=

{

0

}

Suppose

w

∈

V

0

⁢

⋂

V

1

is not null, then it can be written

w

=

α

⁢

v

0

+

β

⁢

u

0

because it belongs to

V

0

. Since it also belongs to

V

1

is is ”orthogonal” to both

v

0

and

u

0

. That is:

ω

⁢

(

v

0

,

w

)

=

0

⟹

β

⁢

ω

⁢

(

v

0

,

u

0

)

=

0

⟹

β

=

0

similarly

ω

⁢

(

u

0

,

w

)

=

0

⟹

α

⁢

ω

⁢

(

u

0

,

v

0

)

=

0

⟹

α

=

0

So

w

must be null.

•

V

=

V

0

⁢

⊕

V

1

Suppose

w

∈

V

. Let

α

=

ω

⁢

(

v

0

,

w

)

,

β

=

ω

⁢

(

u

0

,

w

)

,

w

0

=

α

⁢

u

0

-

β

⁢

v

0

.

Then

ω

⁢

(

v

0

,

w

0

)

=

α

=

ω

⁢

(

v

0

,

w

)

and

ω

⁢

(

u

0

,

w

0

)

=

β

=

ω

⁢

(

u

0

,

w

)

.

Considering

w

1

=

w

-

w

0

we have

w

=

w

0

+

w

1

(by construction) and

ω

⁢

(

v

0

,

w

1

)

=

ω

⁢

(

v

0

,

w

-

w

0

)

=

ω

⁢

(

v

0

,

w

)

-

ω

⁢

(

v

0

,

w

0

)

=

ω

⁢

(

v

0

,

w

)

-

ω

⁢

(

v

0

,

w

)

=

0

and similarly for

ω

⁢

(

u

0

,

w

1

)

So

w

1

∈

V

1

,

w

0

∈

V

0

and

w

=

w

0

+

w

1

and thus

V

=

V

0

⁢

⊕

V

1

So the

matrix representation

of

ω

is block-diagonal in

V

0

⁢

⊕

V

1

and a restriction anti-symmetric

bilinear

for of

ω

to

V

1

exists.

If

V

1

is not null we can repeat the procedure with the restriction. Since

dim

⁡

(

V

)

=

dim

⁡

(

V

0

)

+

dim

⁡

(

V

1

)

and

V

is finite dimensional the procedure must stop at a finite step.

At the end we get a

decomposition

V

=

⊕

i

=

0

k

-

1

V

i

, where

dim

⁡

(

V

i

)

=

2

and

dim

⁡

(

V

)

=

2

⁢

k

is even.

Title

every symplectic manifold has even dimension

Canonical name

EverySymplecticManifoldHasEvenDimension

Date of creation

2013-03-22 15:44:05

Last modified on

2013-03-22 15:44:05

Owner

cvalente (11260)

Last modified by

cvalente (11260)

Numerical id

18

Author

cvalente (11260)

Entry type

Theorem

Classification

msc 53D05

Related topic

AlternatingForm

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