In a 1947 *Comptes Rendus* note (T224, p. 448), Koszul makes the following claim (paraphrased, hopefully correctly), which seems like it should have a simple proof I am missing.


Given a compact, connected Lie group, choose a basis $\alpha, \theta^1,\ldots,\theta^n$ of the dual Lie algebra $\mathfrak g^\vee$ which is orthonormal with respect to an invariant inner product, the closed, invariant Cartan form $$\omega = \frac 1 3 \big( \alpha \wedge d\alpha + \sum \theta^i \wedge d\theta^i\big) \in \bigwedge{}^3 \mathfrak g^\vee.$$ 
decomposes as $\omega = \omega\_2 + \omega\_3$,
where
$$
\omega\_2 \in \mathbb R\alpha \otimes \Lambda^2[\vec \theta],
\qquad
\omega\_3 \in \Lambda^3[\vec \theta].
$$
**How do we know**, as Koszul claims, **that
$$d(\alpha \wedge d\alpha) = 3 d\omega\_2?$$**


This would be clear if we knew $d\omega\_3$ were $\sum\theta^i \wedge d\theta^i$, but this does not seem immediately obvious as the $d\theta^i$ may nontrivially involve $\alpha$.


**What am I missing?**



Simple identity on Lie algebras in a note of Koszul
Are you sure about the factor of $3$ in Koszul's formula? The computation below does not give such a factor, and it gives a stronger result.


Let $X\_i$ be any basis of the left-invariant vector fields (where the indices run from $0$ to $n$), with $[X\_i,X\_j] = c^k\_{ij}X\_k$, where, of course, $c^k\_{ij}=-c^k\_{ji}$ (and where here, as below, we use the summation convention that we sum over repeated indices in any given term if no summation is explicitly given). Let $\langle,\rangle$ be an $\mathrm{ad}$-invariant inner product, with $\langle X\_k,X\_l\rangle = g\_{kl}$. Then $\mathrm{ad}$-invariance is just the equation
$$
g\_{lk}c\_{ij}^k + g\_{jk}c\_{il}^k = 0.
$$
If we assume that the $X\_i$ are $\langle,\rangle$-orthonormal, then $g\_{kl} = \delta\_{kl}$, so $c^k\_{ij} = - c^j\_{ik}$, i.e., $c^i\_{jk}$ is skew-symmetric in *all* its indices. 


For simplicity, let's agree to write $c^i\_{jk} = c\_{ijk}$, when the $X\_i$ are orthonormal with respect to an $\mathrm{ad}$-invariant $\langle,\rangle$, which I'll assume from now on.


Now consider the dual $1$-forms $\theta\_i$ to the vector fields $X\_i$. By the formula relating exterior derivative and Lie bracket, they satisfy
$$
\mathrm{d}\theta\_i = -\tfrac12\,c\_{ijk}\,\theta\_j\wedge\theta\_k
= -\sum\_{j<k} c\_{ijk}\,\theta\_j\wedge\theta\_k\,.
$$
By definition, the Cartan form is just
$$
\omega = -\tfrac16\,c\_{ijk}\,\theta\_i\wedge\theta\_j\wedge\theta\_k
= -\sum\_{i<j<k} c\_{ijk}\,\theta\_i\wedge\theta\_j\wedge\theta\_k\,.
$$  

Let $\alpha = \theta\_0$. Then, according to your definition, 
$$
\omega\_2 = -\sum\_{0<j<k}\,c\_{0jk}\,\theta\_0\wedge\theta\_j\wedge\theta\_k
= \theta\_0\wedge\mathrm{d}\theta\_0 = \alpha\wedge\mathrm{d}\alpha\,,
$$
which is a stronger result than Koszul claimed (although the factor of $3$ in his formula is mysterious), while
$$
\omega\_3 = -\sum\_{0<i<j<k}\,c\_{ijk}\,\theta\_i\wedge\theta\_j\wedge\theta\_k\,.
$$


*Remark:* By the way, the definition that you give for 'the' Cartan form is a bit odd, because this form definitely depends on choice of the inner product, as you can see, since, if you double the metric, you'll double the Cartan form. This isn't so serious in the simple case because it's just a matter of a scale, but in the general semi-simple case, you run the risk of having many different 'Cartan forms'. However, maybe this isn't such a bad defect, since it's not possible, even in the simple case, to define 'the' Cartan form for all simple groups so that pullback to every simple subgroup $H$ of $G$ yields 'the' Cartan form of $H$. (Just look at the various $\mathrm{SU}(2)$ subgroups of $G$ when there is more than one such subgroup up to conjugacy.)


