A Lie algebra identity 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: 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, 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 $\frak{g}$).
A: I'm not sure if this is closer to a "glimpse" than brute-force but the 4-form you wrote is proportional to $dB$ where $B$ is the Cartan 3-form $B(X,Y,Z) = K(X,[Y,Z])$, which is well-known to be closed (should follow from Jacobi and properties of the Killing form).  To see this, note that (viewing $e_i$ as a left-invariant vector field) we have
$$
B(e_i,e_j,e_k) = K(e_i,[e_j,e_k]) = \sum_l K(e_i, e^l([e_j,e_k]) e_l) = \sum_l K(e_i, e_l) e^l([e_j,e_k]) \\
= -\sum_l K(e_i,e_l) de^l(e_j,e_k).
$$
So 
$$
B \propto \sum_{i,j,k} B(e_i,e_j,e_k) e^i \wedge e^j \wedge e^k = \sum_{i,j,k} \sum_l K(e_i,e_l) de^l(e_j,e_k) e^i\wedge e^j \wedge e^k \\
=\sum_{i, l} K(e_i,e_l) e^i\wedge \sum_{j,k} de^l(e_j,e_k) e^j\wedge e^k \propto\sum_{i,l}  K(e_i,e_l) e^i \wedge de^l.
$$
Now taking $d$ you get what you wrote since $K(e_i,e_l)$ is constant.
