Equivalence between Kähler condition and $\partial_kg_{i\bar j} = \partial_ ig_{k\bar j}$ Let $(M,\omega)$ be a Kähler manifold. Why is the Kähler condition $$d \omega = 0$$ equivalent to
$$\partial_kg_{i\bar j} = \partial_ ig_{k\bar j}$$
for all $i, j, k$?
I am looking for a reference.
 A: I'm assuming that $(U, (z^1, \dots, z^n))$ is a complex coordinate chart and in these coordinates the Kähler form is $\omega|_U = \dfrac{i}{2}\displaystyle\sum_{i=1}^n\sum_{j=1}^ng_{i\bar{j}}dz^i\wedge dz^{\bar{j}}$. 
Note that $d\omega = \partial\omega + \bar{\partial}\omega$ and if $d\omega = 0$, then $\partial\omega = 0$ and $\bar{\partial}\omega = 0$ (because $\partial\omega$ and $\bar{\partial}\omega$ have different bidegrees). As $\omega$ is a real $(1, 1)$-form, $\partial\omega = \partial\overline{\omega} = \overline{\bar{\partial}\omega}$, so the converse is also true. In fact, 
$$d\omega = 0 \Longleftrightarrow \partial\omega = 0 \Longleftrightarrow \bar{\partial}\omega = 0.$$
Note that
$$\partial\omega|_U = \frac{i}{2}\sum_{k=1}^n\sum_{i=1}^n\sum_{j=1}^n\partial_kg_{i\bar{j}}dz^k\wedge dz^i\wedge dz^{\bar{j}}.$$
As $dz^k\wedge dz^i = 0$ for $i = k$, we have
\begin{align*}
-2i\partial\omega|_U &= \sum_{k=1}^n\sum_{i\neq k}\sum_{j=1}^n\partial_kg_{i\bar{j}}dz^k\wedge dz^i\wedge dz^{\bar{j}}\\
&= \sum_{k=1}^n\sum_{i < k}\sum_{j=1}^n\partial_kg_{i\bar{j}}dz^k\wedge dz^i\wedge dz^{\bar{j}} + \sum_{k=1}^n\sum_{i > k}\sum_{j=1}^n\partial_kg_{i\bar{j}}dz^k\wedge dz^i\wedge dz^{\bar{j}}\\
&= \sum_{k=1}^n\sum_{i < k}\sum_{j=1}^n\partial_kg_{i\bar{j}}dz^k\wedge dz^i\wedge dz^{\bar{j}} - \sum_{k=1}^n\sum_{i > k}\sum_{j=1}^n\partial_kg_{i\bar{j}}dz^i\wedge dz^k\wedge dz^{\bar{j}}\\
&= \sum_{k=1}^n\sum_{i < k}\sum_{j=1}^n\partial_kg_{i\bar{j}}dz^k\wedge dz^i\wedge dz^{\bar{j}} - \sum_{i=1}^n\sum_{k > i}\sum_{j=1}^n\partial_ig_{k\bar{j}}dz^k\wedge dz^i\wedge dz^{\bar{j}}\\
&= \sum_{k=1}^n\sum_{i < k}\sum_{j=1}^n\partial_kg_{i\bar{j}}dz^k\wedge dz^i\wedge dz^{\bar{j}} - \sum_{k=1}^n\sum_{i < k}\sum_{j=1}^n\partial_ig_{k\bar{j}}dz^k\wedge dz^i\wedge dz^{\bar{j}}\\
&= \sum_{k=1}^n\sum_{i < k}\sum_{j=1}^n(\partial_kg_{i\bar{j}}-\partial_ig_{k\bar{j}})dz^k\wedge dz^i\wedge dz^{\bar{j}}.
\end{align*}
As $\{dz^k\wedge dz^i\wedge dz^{\bar{j}} \mid 1 \leq i < k \leq n, j = 1, \dots, n\}$ is a basis for the $C^{\infty}(U)$-module $\Omega^{2,1}(U)$, we see that $\partial\omega|_U = 0$ if and only if $\partial_kg_{i\bar{j}}-\partial_ig_{k\bar{j}} = 0$. Combining with the logical equivalences above, we see that $d\omega|_U = 0$ if and only if $\partial_kg_{i\bar{j}} = \partial_ig_{k\bar{j}}$ for all $i, j,$ and $k$.
Note, by using the equivalence $d\omega|_U = 0 \Longleftrightarrow \bar{\partial}\omega|_U = 0$, we also see that $d\omega|_U = 0$ if and only if $\partial_{\bar{k}}g_{i\bar{j}} = \partial_\bar{j}g_{i\bar{k}}$ for all $i, j,$ and $k$.
