Let ρ be a finite dimensional linear representation of a, and m an integer ≥0. The function x→ Tr(ρ(x)m) on a is an invariant polynomial function.
Put g(x1,...,xm)= Tr(ρ(x1)...ρ(xm)) for x1,...,xm∈a. If x∈a, we have
−(θ∗(x)g)(x1,...,xm)
=∑1≤i≤m Tr(ρ(x1)...ρ(xi−1)[ρ(x),ρ(xi)]ρ(xi+1)...ρ(xm))
= Tr(ρ(x)ρ(x1)...ρ(xm))− Tr(ρ(x1)...ρ(xm)ρ(x)) =0,
so θ∗(x)g=0. Let h be the symmetric multilinear form defined by
h(x1,...,xm)=m1! ∑g(xσ(1),...,xσ(m)).
σ∈Sm
For all x∈a, we have θ∗(x)h=0 and Tr(ρ(x)m) =h(x,...,x), hence the lemma.
Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.
Nothing.