Lemma 3

Tag 0181, permanent. Cite it as [Bourbaki, Tag 0181].

Lie Groups and Lie Algebras, chapter VIII, § 12. Chevalley orders

Let $x, h\in A$ and $\lambda \in k$ be such that (ad $h$ ) $x=\lambda x$ . For all $n\in \mathbf{N}$ , and all $P\in k[X]$ , we have

P(h)x^{(n)}=x^{(n)}P(h+n\lambda ) \tag{12}

Since ad $h$ is a derivation of A and since (ad $h$ ) $x$ commutes with $x$ , we have

(ad $h$ ) $x^n=nx^{n-1}$ ((ad $h$ ) $x$ ) $=n\lambda x^n$ , (13) so

(ad $h$ ) $x^{(n)}=n\lambda x^{(n)}$ .

Thus, formula (12) follows from the special case

P(h)x=xP(h+\lambda ) \tag{14}

by replacing $x$ by $x^{(n)}$ and $\lambda$ by $n\lambda$ . It suffices to prove (14) when $P = X^m$ , by induction on $m$ . It is clear when $m= 0,1$ . If (14) is true for $P = X^m$ , then

h^{m+1}x=h.h^mx=hx(h+\lambda )^m=x(h+\lambda )^{m+1}

which proves (12).

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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