a Kähler manifold is symplectic

a Kähler manifold is symplectic

Let

ω

⁢

(

X

,

Y

)

=

g

⁢

(

J

⁢

X

,

Y

)

on a Kähler

manifold

. We will prove that

ω

is a

symplectic form

.

•

ω

is

anti-symmetric

ω

⁢

(

X

,

Y

)

=

g

⁢

(

J

⁢

X

,

Y

)

=

g

⁢

(

Y

,

J

⁢

X

)

=

g

⁢

(

J

⁢

Y

,

J

2

⁢

X

)

=

g

⁢

(

J

⁢

Y

,

-

X

)

=

-

g

⁢

(

J

⁢

Y

,

X

)

=

-

ω

⁢

(

Y

,

X

)

. Here we used the fact that

g

is an

Hermitian

tensor on a Kähler manifold (

g

⁢

(

X

,

Y

)

=

g

⁢

(

J

⁢

X

,

J

⁢

Y

)

)

•

ω

is linear

Due to anti-symmetry, we just need to check linearity on the second slot. Since

g

⁢

(

J

⁢

X

,

⋅

)

is by definition linear,

ω

will also be linear.

•

ω

is non degenerate

On a given point on the manifold, pick a non

null vector

X

,

α

X

⁢

(

⋅

)

=

ω

⁢

(

X

,

⋅

)

=

g

⁢

(

J

⁢

X

,

⋅

)

. Since

g

is

non-degenerate

1

1

no vector but the null vector is

orthogonal

to every other vector

,

α

is also non-degenerate (for all

X

).

ω

is thus non degenerate.

•

ω

is closed

First note that

X

⁢

(

ω

⁢

(

Y

,

Z

)

)

=

∇

X

⁡

(

ω

⁢

(

Y

,

Z

)

)

=

∇

X

⁡

(

g

⁢

(

J

⁢

Y

,

Z

)

)

=

g

⁢

(

∇

X

⁡

(

J

⁢

Y

)

,

Z

)

+

g

⁢

(

J

⁢

Y

,

∇

X

⁡

Z

)

=

g

⁢

(

J

⁢

∇

X

⁡

Y

,

Z

)

+

g

⁢

(

J

⁢

Y

,

∇

X

⁡

Z

)

=

ω

⁢

(

∇

X

⁡

Y

,

Z

)

+

ω

⁢

(

Y

,

∇

X

⁡

Z

)

Here we used the fact that both

g

and

J

are covariantly

constant

(

∇

⁡

g

=

0

and

∇

⁡

J

=

0

)

We aim to prove that

d

⁢

ω

=

0

which is

equivalent

to proving

(

d

⁢

ω

)

⁢

(

X

,

Y

,

Z

)

=

0

for all

vector fields

X

,

Y

,

Z

.

Since this is a

tensorial

identity

, WLOG we can assume that at a specific point

p

in the Kähler manifold

[

X

,

Y

]

p

=

[

Y

,

Z

]

p

=

[

Z

,

X

]

p

=

0

and prove the indentity for these vector fields

2

2

in particular this works for the canonical base of

T

p

⁢

M

associated with a local coordinate system

.

Consider

X

,

Y

,

Z

with the previous commutation

relations

at

p

, using the

formulas for differential forms of small valence

:

(

d

⁢

ω

)

⁢

(

X

,

Y

,

Z

)

=

X

⁢

(

ω

⁢

(

Y

,

Z

)

)

+

Y

⁢

(

ω

⁢

(

Z

,

X

)

+

Z

⁢

(

ω

⁢

(

X

,

Y

)

)

)

=

ω

⁢

(

∇

X

⁡

Y

,

Z

)

+

ω

⁢

(

Y

,

∇

X

⁡

Z

)

+

ω

⁢

(

∇

Y

⁡

Z

,

X

)

+

ω

⁢

(

Z

,

∇

Y

⁡

X

)

+

ω

⁢

(

∇

Z

⁡

X

,

Y

)

+

ω

⁢

(

X

,

∇

Z

⁡

Y

)

=

ω

⁢

(

∇

X

⁡

Y

-

∇

Y

⁡

X

,

Z

)

+

ω

⁢

(

∇

Y

⁡

Z

-

∇

Z

⁡

Y

,

X

)

+

ω

⁢

(

∇

Z

⁡

X

-

∇

X

⁡

Z

,

Y

)

The

Levi-Civita connection

is torsion-free,

∇

X

⁡

Y

-

∇

Y

⁡

X

=

[

X

,

Y

]

thus:

(

d

⁢

ω

)

⁢

(

X

,

Y

,

Z

)

=

ω

⁢

(

[

X

,

Y

]

,

Z

)

+

ω

⁢

(

[

Y

,

Z

]

,

X

)

+

ω

⁢

(

[

Z

,

X

]

,

Y

)

And since all the

commutators

are null at

p

(by

assumption

) we get that:

(

d

⁢

ω

)

⁢

(

X

,

Y

,

Z

)

=

0

ω

is therefore closed.

Title

a Kähler manifold is symplectic

Canonical name

AKahlerManifoldIsSymplectic

Date of creation

2013-03-22 16:07:54

Last modified on

2013-03-22 16:07:54

Owner

cvalente (11260)

Last modified by

cvalente (11260)

Numerical id

15

Author

cvalente (11260)

Entry type

Result

Classification

msc 53D99

Related topic

KahlerManifold

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