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 ¹, $\alpha$ is also non-degenerate (for all $X$ . $\omega$ is thus non degenerate.

• $\omega$ is closed

  First note that \begin{eqnarray} X(\omega(Y,Z)) &=& \nabla_X (\omega(Y,Z)) \nonumber \\ &=& \nabla_X (g(JY,Z)) \nonumber \\ &=& g(\nabla_X(JY),Z) + g(JY, \nabla_X Z) \nonumber \\ &=& g(J\nabla_X Y, Z) + g(JY, \nabla_X Z) \nonumber \\ &=& \omega(\nabla_X Y, Z) + \omega(Y, \nabla_X Z) \nonumber \end{eqnarray} 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.

  Consider $X,Y,Z$ with the previous commutation relations at $p$ using the formulas for differential forms of small valence:

  \begin{eqnarray} (d \omega)(X,Y,Z) &=& X(\omega(Y,Z)) + Y(\omega(Z,X) + Z(\omega(X,Y)) ) \nonumber \\ &=& \omega(\nabla_X Y,Z) + \omega(Y,\nabla_X Z) + \nonumber \\ && \omega(\nabla_Y Z,X) + \omega(Z,\nabla_Y X) + \nonumber \\ && \omega(\nabla_Z X,Y) + \omega(X,\nabla_Z Y) \nonumber \\ &=& \omega(\nabla_X Y - \nabla_Y X, Z) + \omega(\nabla_Y Z - \nabla_Z Y, X) + \omega(\nabla_Z X - \nabla_X Z, Y) \nonumber \end{eqnarray}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.

Footnotes

...http://planetmath.org/encyclopedia/NonDegenerate.html ¹

no vector but the null vector is orthogonal to every other vector

... fields²

in particular this works for the canonical base of $T_p M$ associated with a local coordinate system