Bourbaki

Lemma 2

Tag 0157, permanent. Cite it as [Bourbaki, Tag 0157].

Lie Groups and Lie Algebras, chapter VIII, § 8. Symmetric invariants

Let ρ\rho be a finite dimensional linear representation of a\mathfrak{a}, and mm an integer ≥0\geq 0. The function x→x \rightarrow Tr(ρ(x)m\rho (x)^m) on a\mathfrak{a} is an invariant polynomial function.

Put g(x1,...,xm)=g(x_1, . . . , x_m) = Tr(ρ(x1)...ρ(xm)\rho (x_1). . . \rho (x_m)) for x1,...,xm∈ax_1, . . . , x_m\in \mathfrak{a}. If x∈ax\in \mathfrak{a}, we have

−(θ∗(x)g)(x1,...,xm) -(\theta^*(x)g)(x_1, . . . , x_m)

=∑1≤i≤m=\sum_{1\leq i\leq m} Tr(ρ(x1)...ρ(xi−1)[ρ(x),ρ(xi)]ρ(xi+1)...ρ(xm)\rho (x_1). . . \rho (x_{i-1})[\rho (x), \rho (x_i)]\rho (x_{i+1}). . . \rho (x_m))

= Tr(ρ(x)ρ(x1)...ρ(xm)\rho (x)\rho (x_1). . . \rho (x_m))−- Tr(ρ(x1)...ρ(xm)ρ(x)\rho (x_1). . . \rho (x_m)\rho (x)) =0= 0,

so θ∗(x)g=0\theta^*(x)g= 0. Let hh be the symmetric multilinear form defined by

h(x1,...,xm)=m1h(x_1, . . . , x_m) =m1! ∑g(xσ(1),...,xσ(m))\sum g(x_{\sigma(1)}, . . . , x_{\sigma(m)}).

σ∈Sm\sigma \in \mathfrak{S}_m

For all x∈ax\in \mathfrak{a}, we have θ∗(x)h=0\theta^*(x)h= 0 and Tr(ρ(x)m\rho (x)^m) =h(x,...,x)=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.

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Transcribed from the printed volume. Not translated.