Given a Kähler manifold $(M,g)$ and the differential $\omega=\sqrt{-1} g_{\alpha\bar{\beta}}dz^{\alpha}\wedge dz^{\bar{\beta}}$, how to arrive the following results:
$\omega$ is closed, i.e. $d\omega=0$ iff in terms of components of $g$, $$\frac{\partial g_{\alpha\bar{\beta}}}{\partial z^{\gamma}}=\frac{\partial g_{\gamma\bar{\beta}}}{\partial z^{\alpha}}.$$
I also wonder if the following property is true in complex manifold: $dz^{i}\wedge dz^{j}=-dz^{j}\wedge dz^{i}$, $dz^{i}\wedge d\bar{z}^{j}=-d\bar{z}^{j}\wedge dz^{i}$ and $d\bar{z}^{i}\wedge d\bar{z}^{j}=-d\bar{z}^{j}\wedge d\bar{z}^{i}$.
The above questions seem to be easy but it take me a lot of time without correct derivations and answers. Could someone give detailed derivations and answer to the above problems?
Write $\omega=\sqrt{-1} g_{j\bar k}dz^j\wedge d\overline{z}^k$. By definition \begin{equation*} d\omega=\sqrt{-1} \left(\frac{\partial g_{j\bar k}}{\partial z^l}dz^l\wedge dz^j\wedge d\overline{z}^k+ \frac{\partial g_{j\bar k}}{\partial\overline{z}^l}d\overline{z}^l\wedge dz^j\wedge d\overline{z}^k\right), \end{equation*} then $d\omega=0$ if and only if \begin{equation*} \frac{\partial g_{j\bar k}}{\partial z^l}=\frac{\partial g_{l\bar k}}{\partial z^j}. \end{equation*}
Last equality follows by the definition of wedge product!