PlanetMath: a Kähler manifold is symplectic

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a Kähler manifold is symplectic

(Result)

Let $\omega(X,Y) = g(JX,Y)$ on a Kähler manifold. We will prove that $\omega$ is a symplectic form.

$\omega$ is anti-symmetric
$\omega(X,Y) = g(JX,Y) = g(Y,JX) = g(JY, J^2 X) = g(JY,-X) = -g(JY,X) = -\omega(Y,X)$ . Here we used the fact that is an Hermitian tensor on a Kähler manifold ( )
$\omega$ is linear
Due to anti-symmetry, we just need to check linearity on the second slot. Since $g(JX,\cdot)$ is by definition linear, $\omega$ will also be linear.
$\omega$ is non degenerate
On a given point on the manifold, pick a non null vector , $\alpha_X(\cdot) = \omega(X, \cdot ) = g(JX, \cdot)$ . Since is non-degenerate ¹, $\alpha$ is also non-degenerate (for all ). $\omega$ is thus non degenerate.
$\omega$ is closed
First note that

$\displaystyle X(\omega(Y,Z))$ $\displaystyle =$ $\displaystyle \nabla_X (\omega(Y,Z))$

$\displaystyle =$ $\displaystyle \nabla_X (g(JY,Z))$

$\displaystyle =$ $\displaystyle g(\nabla_X(JY),Z) + g(JY, \nabla_X Z)$

$\displaystyle =$ $\displaystyle g(J\nabla_X Y, Z) + g(JY, \nabla_X Z)$

$\displaystyle =$ $\displaystyle \omega(\nabla_X Y, Z) + \omega(Y, \nabla_X Z)$

Here we used the fact that both and are covariantly constant ( $\nabla g = 0$ and $\nabla J = 0$ )

We aim to prove that $d \omega = 0$ which is equivalent to proving $(d \omega)(X,Y,Z) = 0$ for all vector fields .

Since this is a tensorial identity, WLOG we can assume that at a specific point in the Kähler manifold and prove the indentity for these vector fields ².

Consider with the previous commutation relations at , using the formulas for differential forms of small valence:

$\displaystyle (d \omega)(X,Y,Z)$ $\displaystyle =$ $\displaystyle X(\omega(Y,Z)) + Y(\omega(Z,X) + Z(\omega(X,Y)) )$

$\displaystyle =$ $\displaystyle \omega(\nabla_X Y,Z) + \omega(Y,\nabla_X Z) +$

$\displaystyle \omega(\nabla_Y Z,X) + \omega(Z,\nabla_Y X) +$

$\displaystyle \omega(\nabla_Z X,Y) + \omega(X,\nabla_Z Y)$

$\displaystyle =$ $\displaystyle \omega(\nabla_X Y - \nabla_Y X, Z) + \omega(\nabla_Y Z - \nabla_Z Y, X) + \omega(\nabla_Z X - \nabla_X Z, Y)$

The Levi-Civita connection is torsion-free, $\nabla_X Y - \nabla_Y X = [X,Y]$ thus:

$\displaystyle (d \omega)(X,Y,Z) = \omega([X,Y],Z) + \omega([Y,Z],X) + \omega([Z,X],Y)$

And since all the commutators are null at (by assumption) we get that:

$\displaystyle (d \omega)(X,Y,Z) = 0$

$\omega$ is therefore closed.

Footnotes

...¹: no vector but the null vector is orthogonal to every other vector
... fields ²: in particular this works for the canonical base of associated with a local coordinate system

"a Kähler manifold is symplectic" is owned by cvalente.

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See Also: Kähler manifold

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Cross-references: commutators, torsion-free, Levi-Civita connection, formulas for differential forms of small valence, relations, local coordinate, base, canonical, WLOG, identity, tensorial, vector fields, equivalent, closed, orthogonal, non-degenerate, vector, null, manifold, point, tensor, Hermitian, anti-symmetric, symplectic form, Kähler manifold

This is version 12 of a Kähler manifold is symplectic, born on 2006-07-31, modified 2007-04-27.
Object id is 8203, canonical name is AKahlerManifoldIsSymplectic.
Accessed 689 times total.

Classification:

53D99 (Differential geometry :: Symplectic geometry, contact geometry :: Miscellaneous)

Pending Errata and Addenda

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