Exact differentiation

Question

How it follows below after $9.4$ that "...and so $d\sigma^i=0$ for all $i$" ?

Answer 1

Since, by hypothesis, $[e_j,e_k]$ vanishes for all $j,k$, it follows that $$d\sigma^i(e_j,e_k)=-\sigma^i([e_j,e_k])=0$$ for all $j,k$, and then $d\sigma$ is identically zero as the $e_i$'s form a basis pointwisely.

The equation $$d\sigma^i(e_j,e_k)=-\sigma^i([e_j,e_k])$$ follows from the direct application of Theorem 4.25 since $\sigma^i(e_l)$ is either constant equal to $1$ (if $l=i$) or $0$ (if $l\neq i$), as $\{\sigma^i\}$ is the dual basis of $\{e_i\}$. Therefore, the terms $e_j\sigma^i(e_k)$ and $-e_k\sigma^i(e_j)$ will vanish since they are derivations of constants.