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∧dA+23A∧A∧A,
where A is a Lie algebra valued 1-form. My question is "how is the last term A∧A∧A defined?"
As far as I know, a lot of sources (e.g. 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.
[ω∧η](X1,…,Xp+q):=(coefficient)×∑σ∈Sp+qsgn(σ)[ω(Xσ(1),…,Xσ(p)),η(Xσ(p+1),…,Xσ(p+q))],
where ω and η are Lie algebra valued p-form and q-form, respectively. The coefficient differs by authors. Other people utilises local description (e.g. http://math.stackexchange.com/questions/315235/reference-for-lie-algebra-valued-differential-forms, and also in the Wikipedia)
[ω∧η]=[ωa⊗Ta,ηb⊗Tb]:=ωa∧ηb⊗[Ta,Tb],
where Tc(c=1,…,dimg) are generators of the Lie algebra 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 [ω∧η] is Lie algebra valued (p+q)-form.
Then what about wedged ones without brackets, such as A∧A,A∧A∧A?
I can show that A∧A is equivalent to [A∧A] up to coefficient, using either matrix representation, considering g=gl(n), or universal enveloping algebra. The basic idea is
A∧A=(Aa⊗Ta)∧(Ab⊗Tb)=(Aa∧Ab)TaTb.
This time, by graded commutation relation, the multiplication of generators can be converted to commutators. This seems ok. Then what about A∧A∧A? I could not convert it to an expression only using commutators of generators...
So, what I did was calculating [A∧[A∧A]], which gave zero. I am totally confused at this stage. Could you point out some pieces that I possibly keep missing??
This post imported from StackExchange MathOverflow at 2015-06-06 21:05 (UTC), posted by SE-user N. Shimode