Let $G$ be a Lie group and $\mathfrak{g}$ its Lie algebra. Fix a basis $e\_1,\dots,e\_n$ of $\mathfrak{g}$ and let $e^1,\dots,e^n$ be its dual basis. We also use $e^i$ to denote the left-invariant 1-form on $G$ obtained from $e^i$. It seems that I can prove the following identity by brute force:
\begin{equation}\sum\_{i,j}de^i\wedge de^j\cdot\kappa(e\_i,e\_j)=0,\end{equation}
where $d$ is the exterior derivative and $\kappa$ is the Killing form.


I would like to know how can one tell this identity holds from a glimpse. I believe there must be some easy way to explain such a simple expression but I just cannot figure it out. Thanks!



A Lie algebra identity
**Some premilinaries.**
Let $\Omega\cong\mathfrak{g}\otimes\bigwedge\mathfrak{g}^\*$ be the space of left-invariant $\frak{g}$-valued differential forms on $G$. $\Omega$ is a graded vector space, the Maurer-Cartan form $\theta=\sum e\_i\otimes e^i$ belonging to its degree $1$ part.


Extend the Killing form $(\cdot,\cdot)$ to a bilinear form on $\Omega$ (by taking the Killing product of $\frak{g}$-parts and taking the wedge product of the $\bigwedge\mathfrak{g}^\*$ parts). Similarly, extend the Lie bracket $[\cdot,\cdot]$ to a bilinear map $\Omega\times \Omega\rightarrow\Omega$. Then $(\Omega, [\cdot,\cdot])$ is a [Lie superalgebra](https://en.wikipedia.org/wiki/Lie_superalgebra#Examples), and the extended $(\cdot,\cdot)$ still satisfies
$$
(\star)\quad\quad([a,b],c)=(a,[b,c]).
$$
Moreover, we have the Maurer-Cartan identity
$$
d\theta+\frac{1}{2}[\theta,\theta]=0.
$$




---


**A coordinate-free proof of your identity.** The identity you want is just
$(d\theta,d\theta)=0$. By the Maurer-Cartan identity, this is equivalent to 
$$
([\theta,\theta],[\theta,\theta])=0.
$$
But by $(\star)$ we have
$([\theta,\theta],[\theta,\theta])=(\theta,[\theta,[\theta,\theta]])$, so the identity you want follows from the fact that 
$$
[\theta,[\theta,\theta]]=0
$$ (essentially, this is just the Jacobi identity in $\frak{g}$).


