if the algebra of functions on a manifold is a Poisson ring then the manifold is symplectic

if the algebra of functions on a manifold is a Poisson ring then the manifold is symplectic

Let

M

be a

smooth manifold

and let

A

be the

algebra

of smooth functions from

M

to

ℝ

. Suppose that there exists a bilinear operation

[

,

]

:

A

×

A

→

A

which makes

A

a

Poisson ring

.

For this proof, we shall use the fact that

T

*

⁢

(

M

)

is the

sheafification

of the

A

-module generated by the set

{

d

⁢

f

|

f

∈

A

}

modulo the

relations

•

d

⁢

(

f

+

g

)

=

d

⁢

f

+

d

⁢

g

•

d

⁢

f

⁢

g

=

g

⁢

d

⁢

f

+

f

⁢

d

⁢

g

Let us define a map

ω

:

T

*

⁢

(

M

)

→

T

⁢

(

M

)

by the following conditions:

•

ω

⁢

(

d

⁢

f

)

⁢

(

g

)

=

[

f

,

g

]

for all

f

⁢

g

∈

A

•

ω

⁢

(

f

⁢

X

+

g

⁢

Y

)

=

f

⁢

ω

⁢

(

X

)

+

g

⁢

ω

⁢

(

Y

)

for all

f

,

g

∈

A

and all

X

,

Y

∈

T

*

⁢

(

M

)

For this map to be well-defined, it must respect the relations:

ω

⁢

(

f

+

g

)

⁢

(

h

)

=

[

f

+

g

,

h

]

=

[

f

,

h

]

+

[

g

,

h

]

=

ω

⁢

(

f

)

⁢

(

h

)

+

ω

⁢

(

g

)

⁢

(

h

)

ω

⁢

(

f

⁢

g

)

⁢

(

h

)

=

[

f

⁢

g

,

h

]

=

f

⁢

[

g

,

h

]

+

g

⁢

[

f

,

h

]

=

f

⁢

ω

⁢

(

g

)

⁢

(

h

)

+

g

⁢

ω

⁢

(

g

)

⁢

(

h

)

These two equations show that

ω

is a well-defined map from the

presheaf

hence, by general nonsense, a well defined map from the sheaf. The fact that

ω

⁢

(

f

⁢

d

⁢

g

)

is a

derivation

readily follows from the fact that

[

,

]

is a derivation in each slot.

Since

[

,

]

is

non-degenerate

,

ω

is

invertible

. Denote its

inverse

by

Ω

. Since our manifold is

finite-dimensional

, we may naturally regard

Ω

as an element of

T

*

⁢

(

M

)

⊗

T

*

⁢

(

M

)

. The fact that

Ω

is an

antisymmetric tensor

field (in other words, a 2-form) follows from the fact that

Ω

⁢

(

d

⁢

f

)

⁢

(

g

)

=

[

f

,

g

]

=

-

[

g

,

f

]

=

-

Ω

⁢

(

d

⁢

g

)

⁢

(

f

)

.

Finally, we will use the

Jacobi identity

to show that

Ω

is
closed. If

u

,

v

,

w

∈

T

⁢

(

M

)

then, by a general

identity

of

differential geometry

,

⟨

d

⁢

Ω

,

u

∧

v

∧

w

⟩

=

⟨

u

,

d

⁢

⟨

Ω

,

v

∧

w

⟩

⟩

+

⟨

v

,

d

⁢

⟨

Ω

,

w

∧

u

⟩

⟩

+

⟨

w

,

d

⁢

⟨

Ω

,

u

∧

v

⟩

⟩

Since this identity is trilinear in

u

,

v

,

w

, we can restrict attention
to a

generating set

. Because of the non-degeneracy assumption,

vector
fields

of the form

a

⁢

d

f

where

f

is a function form such a set.

By the definition of

Ω

, we have

⟨

Ω

,

a

⁢

d

f

∧

a

⁢

d

g

⟩

=

[

f

,

g

]

. Then

⟨

a

d

f

,

d

⟨

Ω

,

a

d

g

∧

a

d

h

⟩

=

[

f

,

[

g

,

h

]

]

so the Jacobi identity is satisfied.

Title

if the algebra of functions on a manifold is a Poisson ring then the manifold is symplectic

Canonical name

IfTheAlgebraOfFunctionsOnAManifoldIsAPoissonRingThenTheManifoldIsSymplectic

Date of creation

2013-03-22 14:46:34

Last modified on

2013-03-22 14:46:34

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

18

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 53D05

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