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

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$\omega $ is antisymmetric
$\omega (X,Y)=g(JX,Y)=g(Y,JX)=g(JY,{J}^{2}X)=g(JY,X)=g(JY,X)=\omega (Y,X)$. Here we used the fact that $g$ is an Hermitian tensor on a Kähler manifold ($g(X,Y)=g(JX,JY)$)

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$\omega $ is linear
Due to antisymmetry, we just need to check linearity on the second slot. Since $g(JX,\cdot )$ is by definition linear, $\omega $ will also be linear.

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$\omega $ is non degenerate
On a given point on the manifold, pick a non null vector $X$, ${\alpha}_{X}(\cdot )=\omega (X,\cdot )=g(JX,\cdot )$. Since $g$ is nondegenerate^{1}^{1}no vector but the null vector is orthogonal^{} to every other vector, $\alpha $ is also nondegenerate (for all $X$). $\omega $ is thus non degenerate.

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$\omega $ is closed
First note that
$X(\omega (Y,Z))$ $=$ ${\nabla}_{X}(\omega (Y,Z))$ $=$ ${\nabla}_{X}(g(JY,Z))$ $=$ $g({\nabla}_{X}(JY),Z)+g(JY,{\nabla}_{X}Z)$ $=$ $g(J{\nabla}_{X}Y,Z)+g(JY,{\nabla}_{X}Z)$ $=$ $\omega ({\nabla}_{X}Y,Z)+\omega (Y,{\nabla}_{X}Z)$ Here we used the fact that both $g$ and $J$ 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 $X,Y,Z$.
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:
$(d\omega )(X,Y,Z)$ $=$ $X(\omega (Y,Z))+Y(\omega (Z,X)+Z(\omega (X,Y)))$ $=$ $\omega ({\nabla}_{X}Y,Z)+\omega (Y,{\nabla}_{X}Z)+$ $\omega ({\nabla}_{Y}Z,X)+\omega (Z,{\nabla}_{Y}X)+$ $\omega ({\nabla}_{Z}X,Y)+\omega (X,{\nabla}_{Z}Y)$ $=$ $\omega ({\nabla}_{X}Y{\nabla}_{Y}X,Z)+\omega ({\nabla}_{Y}Z{\nabla}_{Z}Y,X)+\omega ({\nabla}_{Z}X{\nabla}_{X}Z,Y)$ The LeviCivita connection^{} is torsionfree, ${\nabla}_{X}Y{\nabla}_{Y}X=[X,Y]$ thus:
$$(d\omega )(X,Y,Z)=\omega ([X,Y],Z)+\omega ([Y,Z],X)+\omega ([Z,X],Y)$$ And since all the commutators are null at $p$ (by assumption^{}) we get that:
$$(d\omega )(X,Y,Z)=0$$ $\omega $ is therefore closed.
Title  a Kähler manifold is symplectic 

Canonical name  AKahlerManifoldIsSymplectic 
Date of creation  20130322 16:07:54 
Last modified on  20130322 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 