Bianchi Identity in Chern's book

Question

I'm reading Chern's book "complex manifolds without potential theory".

In chapter 5 he claims the following: Let $E\to M$ be a vector bundle, $D$ a connection, $\omega$ the corresponding connection form and $\Omega := D\omega = d\omega -\omega \land \omega$ the curvature form. Then he claims that taking exterior derivative for the equation $\Omega = d\omega - \omega\land \omega$ yields: $$D\Omega = d\Omega + \Omega \land \omega - \omega \land \Omega = 0 $$which is Bianchi identity.

I'm not familiar with these notations, and therefore struggling to understand this equation:

1. Shouldn't $D\Omega$ be just $d\Omega + \Omega \land \omega$, by the definition of exterior covariant derivative $D\left(\alpha \otimes s \right) = d\alpha \otimes s + \left( -1\right)^\left|\alpha\right|\alpha\land Ds$?
2. Why is the RHS equals zero?

Answer 1

I'll answer in reverse order.

First, recall the definition of curvature: $\Omega = d\omega-\omega\wedge\omega$, so that $$d\omega = \Omega + \omega\wedge\omega.$$ If you differentiate $$\Omega - d\omega + \omega\wedge\omega = 0,$$ you get $$0=d\Omega + (\Omega+\omega\wedge\omega)\wedge\omega - \omega\wedge(\Omega+\omega\wedge\omega) = d\Omega +\Omega\wedge\omega-\omega\wedge\Omega,$$ since the $\omega\wedge\omega\wedge\omega$ terms cancel.

Second, $\Omega$ is a $2$-form with values in $\text{Hom}(E,E)$, so you have to use the formula (5.28) to define the covariant derivative.