The differential of Kähler form

Question

The Kähler form is defined as $$k=-\frac{i}{2}h_{ij}dz^i\wedge d\overline {z^j}$$

We differential the Kähler form to get the condition of Kähler manifold \begin{align} dk&=-\frac{i}{2}(\frac{\partial h_{ij}}{\partial z^k}dz^k+\frac{\partial h_{ij}}{\partial \overline {z^k}}d\overline {z^k})dz^i\wedge d\overline {z^j} \tag1\\ &=-\frac{i}{4}(\sum_{i,j,k}(\frac{\partial h_{ij}}{\partial z^k}-\frac{\partial h_{kj}}{\partial z^i})dz^k\wedge dz^i\wedge d\overline {z^j}+\sum_{i,j,k}(\frac{\partial h_{ij}}{\partial \overline {z^k}}-\frac{\partial h_{ik}}{\partial \overline {z^i}})d\overline {z^k}\wedge dz^i\wedge d\overline {z^j}) \tag2\\ \end{align}

But can anyone tell me how $(1)\Rightarrow (2)$?

Appreciate your feedback.

Answer 1

$$ \begin{aligned} & \frac{\partial h_{ij}}{\partial z^k}dz^k \wedge dz^i\wedge d\overline {z^j} \\ = & -\frac{\partial h_{ij}}{\partial z^k}dz^i \wedge dz^k\wedge d\overline {z^j} \\ = &-\frac{\partial h_{kj}}{\partial z^i}dz^k \wedge dz^i\wedge d\overline {z^j} \\ \end{aligned} $$