Bourbaki

Lemma 7

Tag 017B, permanent. Cite it as [Bourbaki, Tag 017B].

Lie Groups and Lie Algebras, chapter VIII, § 11. Classes of nilpotent elements and sl2 -triplets

Let K be a commutative field with at least 4 elements. Let G be the group of matrices (α0αβ−1)(\alpha 0\alpha \beta_{-1}) where α∈K∗,β∈K\alpha \in K^*,\beta \in K. Let G′G' be the group

of such matrices such that α=1\alpha = 1. Then G′=(G,G)G'= (G,G).

If α,α′∈K∗\alpha , \alpha '\in K^* and β,β′∈K\beta , \beta '\in K,

(αβ)((α′β′)((αβ)−1(α′β′)−1 (\alpha \beta )((\alpha '\beta ')((\alpha \beta )^{-1}(\alpha '\beta ')^{-1}

0 α−1\alpha^{-1} 0 α′−1\alpha^{'-1} 0 α−1\alpha^{-1} 0 α′−1\alpha^{'-1}

=(1−α′β′−αβα′2+α2α′β′+αβ) =(1-\alpha '\beta '-{\alpha \beta \alpha '}^2+\alpha^2\alpha '\beta '+\alpha \beta )

0 1

In particular,

(1β)((α′(1\beta )((\alpha ' 0 )((1β)−1(α′)((1\beta )^{-1}(\alpha ' 0 )−1=(1β(1−α′2)))^{-1}=(1\beta (1-{\alpha '}^2)).

0 1 0 α′−1\alpha^{'-1} 0 1 0 α′−1\alpha^{'-1} 0 1

But there exists α0′∈K∗\alpha '_0\in K^* such that α0′≠1\alpha '_0\not= 1 and α0′≠−1\alpha '_0\not=-1, and then k.(1−α0′2)k.(1-{\alpha '_0}^2) =k=k, 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.

Cited by

Cites

Nothing.

Transcribed from the printed volume. Not translated.