Proposition 1

Tag 00XX, permanent. Cite it as [Bourbaki, Tag 00XX].

Lie Groups and Lie Algebras, chapter VIII, § 1. The Lie algebra sl(2, k) and its representations

Let E be an $\mathfrak{s}\mathfrak{l}(2, k)$ -module, and $e$ a primitive element of E of weight $\lambda$ . Put $e_n=\frac{(-1)^n}{n}X_-^ne$ for $n\geq 0$ , and $e_{-1}= 0$ . Then

He_n= (\lambda -2n)e_n

X_-e_n=-(n+ 1)e_{n+1} \tag{2}

X_+e_n= (\lambda -n+ 1)e_{n-1}

The first formula follows from Lemma 3, and the second from the definition of the $e_n$ . We prove the third by induction on $n$ . It is satisfied for $n= 0$ since $e_{-1}= 0$ . If $n >0$ ,

nX_+e_n=-X_+X_-e_{n-1}=-[X_+, X_-]e_{n-1}-X_-X_+e_{n-1}

=He_{n-1}-X_-(\lambda -n+ 2)e_{n-2}

= (\lambda -2n+ 2 + (n-1)(\lambda -n+ 2))e_{n-1}

=n(\lambda -n+ 1)e_{n-1}

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

Exercise 2, § 1
Proposition 4, § 1
Theorem 1, § 1
Lemma 4, § 2

Cites

Lemma 3