Lemma 4

Tag 09VG, permanent. Cite it as [Bourbaki, Tag 09VG].

Lie Groups and Lie Algebras, chapter I, § 2. Enveloping algebra of a Lie algebra

Let V be a K-module and T the tensor algebra of V. Let u be an endomorphism of V. There exists one and only one derivation of T which extends u. This derivation commutes with the symmetry operators on T.

Let F = V × V × ... × V (n factors). The mapping

(x_1, \ldots, x_n) \mapsto ux_1 \otimes x_2 \otimes \cdots \otimes x_n \\ + x_1 \otimes ux_2 \otimes \cdots \otimes x_n + \cdots + x_1 \otimes x_2 \otimes \cdots \otimes ux_n

of F into

\bigotimes^n V

is multilinear. Hence there exists an endomorphism

u_n

\bigotimes^n V

such that:

u_n(x_1 \otimes \cdots \otimes x_n) = ux_1 \otimes \cdots \otimes x_n + \cdots + x_1 \otimes \cdots \otimes ux_n

for all

x_1, \ldots, x_n

in V. Then

u_1 = u

. Let v be the endomorphism of the K-module T which coincides with

u_n

on each

T^n = \bigotimes^n V

and which is zero on

T^0 = K.1

. We show that v is a derivation of T. If

x_1, \ldots, x_n, y_1, \ldots, y_p

are elements of V, then

v((x_1 \otimes \cdots \otimes x_n) \otimes (y_1 \otimes \cdots \otimes y_p)) \\ = \sum_{i=1}^n x_1 \otimes \cdots \otimes x_{i-1} \otimes ux_i \otimes x_{i+1} \otimes \cdots \otimes x_n \otimes y_1 \otimes \cdots \otimes y_p \\ + \sum_{j=1}^p x_1 \otimes \cdots \otimes x_n \otimes y_1 \otimes \cdots \otimes y_{j-1} \otimes uy_j \otimes y_{j+1} \otimes \cdots \otimes y_p \\ = v(x_1 \otimes \cdots \otimes x_n) \otimes (y_1 \otimes \cdots \otimes y_p) + (x_1 \otimes \cdots \otimes x_n) \otimes v(y_1 \otimes \cdots \otimes y_p).

By linearity it follows that v is a derivation. The uniqueness of v is obvious. Finally, clearly

u_n

commutes with all the symmetry operators on

\bigotimes^n V

, whence the last assertion.

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

Cited by

Proposition 7, § 2

Cites

Nothing.