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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

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

I could easily prove that $\nabla J = 0$, i.e. $\nabla_XJY=J\nabla_YX$.

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?

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    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.2011-08-06

2 Answers 2

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A complete proof is given in Voisin's book: Hodge Theory and Complex Algebraic Geometry: Theorem 3.13.

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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.