Simplifying the Kahler form

Question

In the link here, p.4, it says that, given a fundamental 2-from $\mathcal{K}$ $$\mathcal{K}=\frac{\sqrt{-1}}{2\pi}g_{i\bar{j}}dz^i\wedge d\bar{z}^{\bar{j}},$$ a manifold is said to be Kahler if this form is closed, i.e., $$d\mathcal{K} = \frac{\sqrt{-1}}{2\pi} \left( \partial_ig_{j\bar{k}}dz^i\wedge dz^j \wedge d\bar{z}^{\bar{k}}+\partial_{\bar{i}}g_{j\bar{k}}dz^{\bar{i}}\wedge dz^j\wedge d\bar{z}^{\bar{k}}\right)=0.$$ Then this means that both terms are set equal to zero separately, i.e., $$\partial_ig_{j\bar{k}}=0 \hspace{1cm} \partial_{\bar{i}}g_{j\bar{k}}=0$$ The author then wrote the following which I didn't understand, he said that what preceded implies $$\partial_ig_{j\bar{k}}=\partial_jg_{i\bar{k}} \hspace{1cm} \text{and} \hspace{1cm} \partial_{\bar{i}}g_{j\bar{k}}=\partial_{\bar{k}}g_{j\bar{i}}.$$ My question is how did he conclude these last equations from what preceded?

Answer 1

Note that the two form is really $$\mathcal{K}=\frac{\sqrt{-1}}{2\pi} \sum_{j,k=1}^n g_{j\bar{k}}dz^j\wedge d\bar{z}^{\bar{k}},$$ so $$d\mathcal{K}=\frac{\sqrt{-1}}{2\pi} \left(\sum_{i,j,k=1}^n\partial_i g_{j\bar{k}} dz^i \wedge dz^j\wedge d\bar{z}^{\bar{k}} + \sum_{i,j,k=1}^n\partial_\bar{i} g_{j\bar{k}} d\bar z^\bar{i} \wedge dz^j\wedge d\bar{z}^{\bar{k}}\right).$$ Note that the first term is a $(2, 1)$-form and the second a $(1, 2)$ so both terms has to be zero if $d\mathcal K = 0$. Looking at the $(i, j, \bar k)$ component of the first term, which is $$\partial_i g_{j\bar{k}} dz^i \wedge dz^j\wedge d\bar{z}^{\bar{k}} + \partial_j g_{i\bar{k}} dz^j \wedge dz^i\wedge d\bar{z}^{\bar{k}} = \left(\partial_i g_{j\bar{k}} - \partial_j g_{i\bar{k}} \right) dz^i \wedge dz^j\wedge d\bar{z}^{\bar{k}},$$ so we have $$\partial_i g_{j\bar{k}} - \partial_j g_{i\bar{k}} = 0,$$ which is your first equation, similarly you get the second equation.