Derivative of Left invariant differential form

In this, we see that for a left-invarinat differential form $\omega$ on $G$,

$d\omega(X,Y)=-\frac{1}{2}\omega([X,Y])$

for vector fields $X, Y$ in $G$. Let $\omega:G\rightarrow \Lambda^1 T^*G$ be a Left-invariant differential form on $G$ i.e., $(L_g)^*\omega=\omega$ for all $g\in G$ i.e.,

$\omega(g)(v)=\omega(e)((L_{g^{-1}})_{*,g}(v))$

for all $g\in G$ and $v\in T_gG$. Given $A\in \mathfrak{g}$ we have vector field $A^*:G\rightarrow TG$ defined as $A^*(g)=(L_g)_{*,e}(A)$. Then, $\omega(A^*):G\rightarrow \mathbb{R}$ is constant function, $\omega(A^*)(g)=\omega(e)(A)$ for all $g\in G$. As $\omega(A^*):G\rightarrow \mathbb{R}$ is constant map, $X( \omega(A^*))=0$ for any vector field $X:G\rightarrow TG$ on $G$. In particular, $B^*(X(\omega(A^*)))=0$ for $B\in \mathfrak{g}$. Interchanging $A$ and $B$ we have $A^*(\omega(B^*))=0$ As

$(d\omega)(A^*,B^*)=\frac{1}{2}\left[ A^*(\omega(B^*))-B^*(\omega(A^*))-\omega([A^*,B^*])\right]$

and $A^*(\omega(B^*))=0, B^*( \omega(A^*))=0$ we have $(d\omega)(A^*,B^*)=-\frac{1}{2}\omega([A^*,B^*])$. Let $X:G\rightarrow TG$ be vector fields on $G$. Fix a basis $\{A_1,\cdots,A_n\}$ for the Lie algebra $\mathfrak{g}$ of $G$. Given $g\in G$ we have $X(g)\in T_gG$. Given $v\in T_gG$, we can find $A\in \mathfrak{g}$ such that $v=A^*(g)$. Here, $X(g)=A^*(g)$. As $\{A_1,\cdots,A_n\}$ is basis for $\mathfrak{g}$ and $A\in \mathfrak{g}$, there exists $a_i\in \mathbb{R}$ such that $A=\sum_{i=1}^n a_i A_i$. Then, $X(g)=A^*(g)$ says that

$X(g)=(\sum_{i=1}^n a_i A_i)^*(g)=\sum_{i=1}^n a_i A_i^*(g)$.

Given $g\in G$ we have $a_i\in \mathbb{R}$ such that $X(g)=\sum_{i=1}^n a_iA_i^*(g)$. Varying $g$ over $G$ gives smooth functions $a_i:G\rightarrow \mathbb{R}$ such that $X=\sum a_i A_i^*$. For a vector field $Y:G\rightarrow TG$ of $G$, we have $b_i:G\rightarrow \mathbb{R}$ such that $Y=\sum b_i A_i^*$. Then,

$X(\omega(Y))=\sum_{i,j} a_ib_jA_i^*(\omega(A_j^*))$

As $A_i^*(\omega(A_j^*))=0$ for $i, j$, we see that $X(\omega(Y))=0$. Interchanging $X,Y$ we see that $Y(\omega(X))=0$. As

$(d\omega)(X,Y)=\frac{1}{2}\left[ X(\omega(Y))-Y(\omega(X))-\omega([X,Y])\right]$

and $X(\omega(Y))=0=Y(\omega(X))$ we see that

$d\omega(X,Y)=-\frac{1}{2}\omega([X,Y])$.

In particular, for Maurer-Cartan form on a Lie group $\theta:G\rightarrow TG$ which is a left invarinat $1$-form, we see that $d\theta(X,Y)=-\frac{1}{2}\theta([X,Y])$ for vector fields $X,Y$ on $G$. In literature, this property is denoted by $d\theta=-\frac{1}{2}[\theta,\theta]$.