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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!

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    $\begingroup$ More generally, one has the consequence of the Chern-Weil homomorphism (but already known to É. Cartan) that, if $p = c_{i_1\cdots i_k} e^{i_1}\cdots e^{i_k}$ is any $\mathrm{Ad}(G)$-invariant (symmetric) polynomial on $\frak{g}$, then $$c_{i_1\cdots i_k}\ \mathrm{d}e^{i_1}\wedge\cdots\wedge \mathrm{d}e^{i_k} = 0.$$ The OP's case is just the case when $p$ is the Killing form. For a proof of the general result, consult any treatment of Chern-Weil theory. $\endgroup$ Jul 24, 2014 at 13:47
  • $\begingroup$ @RobertBryant: Bravo! Thank you for sharing this insight! $\endgroup$
    – Piojo
    Jul 25, 2014 at 8:30

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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 $\frak{g}$).

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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.

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