I am confused with the wedging operations of Lie algebra valued differential forms. Especially, for instance, I have some problems with the Chern-Simons 3-form


$$A \wedge dA + \frac{2}{3}A \wedge A \wedge A,$$


where $A$ is a **Lie algebra valued** 1-form. My question is "how is the last term $A \wedge A \wedge A$ defined?"


As far as I know, a lot of sources (e.g. [Wedge Product of Lie Algebra Valued One-Form](https://mathoverflow.net/questions/123632/wedge-product-of-lie-algebra-valued-one-form), <http://en.wikipedia.org/wiki/Lie_algebra-valued_differential_form>) define the wedge of Lie algebra valued 1-forms as follows.


$$[\omega \wedge \eta](X\_1,\dots,X\_{p+q}) := \text{(coefficient)}\times \sum\_{\sigma \in S\_{p+q}} \text{sgn}(\sigma) [\omega(X\_{\sigma(1)},\dots,X\_{\sigma(p)}),\eta(X\_{\sigma(p+1)},\dots,X\_{\sigma(p+q)})],$$


where $\omega$ and $\eta$ are Lie algebra valued $p$-form and $q$-form, respectively. The coefficient differs by authors. Other people utilises local description (e.g. <https://math.stackexchange.com/questions/315235/reference-for-lie-algebra-valued-differential-forms>, and also in the Wikipedia)


$$[\omega \wedge \eta] = [\omega^a \otimes T^a, \eta^b\otimes T^b] := \omega^a \wedge \eta^b \otimes [T^a,T^b],$$


where $T^c (c=1,\dots,\dim \mathfrak{g})$ are generators of the Lie algebra $\mathfrak{g}$, and the implicit sums understood.


These definitions, as the notations suggest, force you to take Lie bracket explicitly. Therefore it is obvious that wedged one $[\omega\wedge\eta]$ is **Lie algebra valued** $(p+q)$-form. 


Then what about wedged ones **without brackets**, such as $A\wedge A, A\wedge A \wedge A$?


I can show that $A \wedge A$ is equivalent to $[A \wedge A]$ up to coefficient, using either matrix representation, considering $\mathfrak{g}=\mathfrak{gl}(n)$, or universal enveloping algebra. The basic idea is 


$$A \wedge A = (A^a \otimes T^a) \wedge (A^b \otimes T^b) = (A^a \wedge A^b) T^a T^b.$$


This time, by graded commutation relation, the multiplication of generators can be converted to commutators. This seems ok. Then what about $A\wedge A \wedge A$? I could not convert it to an expression only using commutators of generators...


So, what I did was calculating $[A \wedge [A \wedge A]]$, which gave **zero**. I am totally confused at this stage. Could you point out some pieces that I possibly keep missing??



$A \wedge A \wedge A$ in Chern-Simons
**Option (1)** Use the definition $(\omega \otimes S) \wedge (\eta \otimes S) = (\omega \wedge \eta) \otimes (S\otimes T)$ of the wedge product for Lie algebra valued forms. Define Lie bracket and Killing form as bilinear maps $[S\otimes T] = [S,T]$ and $\langle S \otimes T \rangle = \langle S, T\rangle$. Then the formula that you want is $$\langle A \wedge [A \wedge A] \rangle,$$ where the commutator and Killing form apply only to the Lie algebra factors, ignoring the differential form factors.


**Option (2)** Use the definition $(\omega \otimes S) \wedge (\eta \otimes S) = (\omega \wedge \eta) \otimes ST$ of the wedge product of forms valued in a particular matrix representation of a Lie algebra. Then the formula that you want is $$\operatorname{tr} (A \wedge A \wedge A),$$ where again the trace applies only to the matrix factors ignoring the differential form factors.


The two formulas agree up to a constant factor, as long as your Lie algebra is simple.


