Bourbaki

Lemma 3

Tag 0ASX, permanent. Cite it as [Bourbaki, Tag 0ASX].

Lie Groups and Lie Algebras, chapter V, § 5. Invariants in the symmetric algebra

Let K, V and S be as in Lemma 2, G a finite group of automorphisms of V, q the order of G, and R the graded subalgebra of S consisting of the elements invariant under G. Assume that K is of characteristic 0. Then the Poincaré series of R is

q−1∑g∈G(det⁡(1−gT))−1. q^{-1} \sum_{g \in G} (\det(1 - gT))^{-1}.

Indeed, the endomorphism f=q−1∑g∈Gg(n)f = q^{-1} \sum_{g \in G} g^{(n)} is a projection of SnS_n onto RnR_n, so Tr(f)=dim⁡KSnG\mathrm{Tr}(f) = \dim_K S_n^G. Thus, the Poincaré series of R is

q−1∑g∈G(∑n=0∞(Trg(n))Tn), q^{-1} \sum_{g \in G} \left( \sum_{n=0}^{\infty} (\mathrm{Tr} g^{(n)}) T^n \right),

and it suffices to apply Lemma 2.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Transcribed from the printed volume. Not translated.