Lemma 1

Tag 010V, permanent. Cite it as [Bourbaki, Tag 010V].

Lie Groups and Lie Algebras, chapter VIII, § 4. Split semi-simple Lie algebra defined by a reduced root system

For all $\alpha , \beta \in B$ , we have

[X_{\alpha}^0, X_{-\alpha}^0] =-H_{\alpha}^0 \tag{10}

[H_{\alpha}^0, H_{\beta}^0] = 0 \tag{11}

[H_{\alpha}^0, X_{\beta}^0] =n(\beta , \alpha )X_{\beta}^0 \tag{12}

[H_{\alpha}^0, X_{-\beta}^0] =-n(\beta , \alpha )X_{-\beta}^0 \tag{13}

$[X_{\alpha}^0, X_{-\beta}^0] = 0$ if $\alpha \not=\beta$ . (14)

Indeed, relation (9) can be written

(X_{\alpha}^0X_{-\alpha_1}^0)(\alpha_2, . . . , \alpha_n) = (X_{-\alpha_1}^0X_{\alpha}^0)(\alpha_2, . . . , \alpha_n)-\delta_{\alpha ,\alpha_1}H_{\alpha}^0(\alpha_2, . . . , \alpha_n)

which proves (10) and (14). Relation (11) is clear. Next

[H_{\alpha}^0, X_{-\beta}^0](\alpha_1, . . . , \alpha_n) =H_{\alpha}^0(\beta , \alpha_1, . . . , \alpha_n)+(\sum_{i=1}^nn(\alpha_i, \alpha ))(\beta , \alpha_1, . . . , \alpha_n)

=-n(\beta , \alpha )(\beta , \alpha_1, . . . , \alpha_n)

=-n(\beta , \alpha )X_{-\beta}^0(\alpha_1, . . . , \alpha_n)

hence (13). Finally,

$0 = [H_{\alpha}^0,[X_{\beta}^0, X_{-\gamma}^0]]$ by $(10),(11),(14)$ (15)

= [[H_{\alpha}^0, X_{\beta}^0], X_{-\gamma}^0] + [X_{\beta}^0,[H_{\alpha}^0, X_{-\gamma}^0]]

$= [[H_{\alpha}^0, X_{\beta}^0]-n(\gamma , \alpha )X_{\beta}^0, X_{-\gamma}^0]$ by (13)

$= [[H_{\alpha}^0, X_{\beta}^0]-n(\beta , \alpha )X_{\beta}^0, X_{-\gamma}^0]$ by (14);

now, considering the empty word immediately gives

([H_{\alpha}^0, X_{\beta}^0]-n(\beta , \alpha )X_{\beta}^0)(\emptyset ) = 0

so (15) implies that

([H_{\alpha}^0, X_{\beta}^0]-n(\beta , \alpha )X_{\beta}^0)X_{-\gamma_1}^0X_{-\gamma_2}^0. . . X_{-\gamma_n}^0(\emptyset ) = 0

for all $\gamma_1, . . . , \gamma_n\in B$ ; this 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.

Cited by

Lemma 2, § 4

Cites

Nothing.