# Equivalent definitions of Kähler metric

Two different ways to define a Kähler metric on a complex manifold are:
1) The fundamental form $\omega = g(J\cdot,\cdot)$ is closed, ie, $d\omega=0$;
2)The complex structure $J$ is parallel with respect to the Levi-Civita connection of $g$, ie $\nabla J=0$. Using the formula
$d \omega(X,Y,Z) = X \omega(Y,Z) - Y \omega(X,Z) + Z \omega(X,Y) - \omega([X,Y],Z) + \omega([X,Z],Y) - \omega([Y,Z],X)$
for the exterior derivative I could easily proove that $\nabla J = 0$, ie $\nabla_XJY=J\nabla_YX.$

What I'm trying to understand is why $d \omega=0 \Rightarrow \nabla J =0$. There's a proof on Kobayashi-Nomizu vol. 2, a consequence of the formula
$4g((\nabla_X J)Y,Z) = 6 d \omega (X,JY,JZ) - 6 d \omega(X,Y,Z) + g(N(Y,Z),JX)$
but it mentions the Nijenhuis tensor $N$.
I'm looking for a proof that doesn't use that. I think a proof is possible by using the Koszul formula for the Levi-Civita connection and the formula for $d \omega$ above wisely used.
Does anyone knows a proof that doesn't mention $N$ explicitly?

-
I don't have an answer for you, only possible references, but you can try looking in Andrei Moroianu's "Lectures on Kahler geometry", or Fangyang Zheng's "Complex differential geometry". IIRC, both prove this equivalence without making use of the Nijenhuis tensor. – Gunnar Þór Magnússon Aug 6 '11 at 23:29
Alternatively, you could make your own proof based on the one in Kobayashi-Nomizu; just use that the almost complex structure on a Kahler manifold is integrable (by definition), so the Nijenhuis tensor is zero. – Gunnar Þór Magnússon Aug 6 '11 at 23:34

The equivalence $\nabla J = 0 \iff d\omega =0$ is an immediate consequence of the formulas \begin{align*} d\omega (X,Y,Z) &= \langle (\nabla_XJ)Y,Z\rangle + \langle (\nabla_YJ)Z,X\rangle + \langle (\nabla_ZJ)X,Y\rangle \\ 2\langle (\nabla_XJ)Y,Z\rangle &= d\omega (X,Y,Z) - d\omega (X,JY,JZ), \end{align*} which are the contents of Proposition 4.16 in Ballmann's Lectures on Kähler Manifolds.