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.

• $\omega$ 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 $g$ is an Hermitian tensor on a Kähler manifold ($g(X,Y)=g(JX,JY)$)

• $\omega$ 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 $X$, ${\alpha }_X(\cdot )=\omega (X,\cdot )=g(JX,\cdot )$. Since $g$ is non-degenerate, $\alpha$ is also non-degenerate (for all $X$). $\omega$ is thus non degenerate.

• $\omega$ 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.

  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 Levi-Civita connection is torsion-free, ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]$ 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.