Show that the Ricci form of a Kähler metric is closed (using the second Bianchi identity) Let $g = (g_{i \overline{j}})$ be a Kähler metric. The Ricci form $\rho$ is a closed real $(1,1)$-form, which can locally be written as $$\rho = i \partial \overline{\partial} \log (\det(g)),$$ and from this local expression, we see that $d \rho =0$.

I want a proof of $d \rho =0$ directly from the second Bianchi identity.

To this end, let $R_{i\overline{j} k \overline{\ell}}$ denote the Riemannian curvature tensor of $g$. The Ricci curvature is given by $$\text{Ric}_{i \overline{j}} = g^{k \overline{\ell}} R_{i \overline{j}k\overline{\ell}}.$$ Hence, we can write $\rho = i \sum_{i,j} \text{Ric}_{i\overline{j}} dz_i \wedge d\overline{z}_j$, and we want to show that $$d \rho = i \sum_{\alpha, i,j} \nabla_{\alpha} \text{Ric}_{i \overline{j}} dz_{\alpha} \wedge dz_i \wedge d\overline{z}_j + \nabla_{\overline{\alpha}} \text{Ric}_{i\overline{j}} d\overline{z}_{\alpha} \wedge dz_i \wedge d\overline{z}_j =0.$$
The second Bianchi identity can be expressed as $\nabla_m R_{i\overline{j}k\overline{\ell}} = \nabla_i R_{m \overline{j} k \overline{\ell}}$. Therefore, $$\nabla_{\alpha} \text{Ric}_{i\overline{j}} = g^{k\overline{\ell}} \nabla_{\alpha} R_{i\overline{j}k\overline{\ell}}=g^{k\overline{\ell}}\nabla_iR_{\alpha\overline{j}k\overline{\ell}}= \nabla_i \text{Ric}_{\alpha\overline{j}},$$ and similarly, $$\nabla_{\overline{\alpha}} \text{Ric}_{i\overline{j}}=\nabla_{\overline{j}}\text{Ric}_{i\overline{\alpha}}.$$
At this point, I suspect that I want to use the skew-symmetry of the wedge product. But I can't see it.
 A: Now that you've corrected the indices, your observations regarding the second Bianchi identity complete the task. Note that
\begin{align*}
\sum_{\alpha,i,j}\nabla_{\alpha}\operatorname{Ric}_{i\bar{j}}dz^{\alpha}\wedge dz^i\wedge d\bar{z}^j &= \sum_{\alpha,i,j}\nabla_i\operatorname{Ric}_{\alpha\bar{j}}dz^{\alpha}\wedge dz^i\wedge d\bar{z}^j\\
&= -\sum_{\alpha,i,j}\nabla_i\operatorname{Ric}_{\alpha\bar{j}}dz^i\wedge dz^{\alpha}\wedge d\bar{z}^j\\
&=-\sum_{\alpha,i,j}\nabla_{\alpha}\operatorname{Ric}_{i\bar{j}}dz^{\alpha}\wedge dz^i\wedge d\bar{z}^j
\end{align*}
where the last equality follows from swapping the names of the indices $\alpha$ and $i$. The analagous manipulation shows that
$$\sum_{\alpha,i,j}\nabla_{\bar{\alpha}}\operatorname{Ric}_{i\bar{j}}d\bar{z}^{\alpha}\wedge dz^i\wedge d\bar{z}^j = -\sum_{\alpha,i,j}\nabla_{\bar{\alpha}}\operatorname{Ric}_{i\bar{j}}d\bar{z}^{\alpha}\wedge dz^i\wedge d\bar{z}^j.$$
Therefore both summations in the expression for $d\rho$ are zero, so $d\rho = 0$.
