Corollary

Tag 0A28, permanent. Cite it as [Bourbaki, Tag 0A28].

Lie Groups and Lie Algebras, chapter II, § 1. Enveloping bigebra of a Lie algebra

The canonical mapping $\sigma$ defines an isomorphism of $\mathfrak{g}$ onto the Lie algebra $P(U)$ of primitive elements of $U$ .

As $c^+$ is a graded homomorphism of degree 0,

P(U) = \sum_{n \geq 1} (P(U) \cap U^n).

It suffices to prove that if $n > 1$ and $a \in U^n$ is primitive, then $a = 0$ . Now $a$ can be written as $\sum_i \lambda_i a_i^n$ , where $\lambda_i \in K$ , $a_i \in \sigma(g)$ . By (12), the term of bi-degree $(1, n-1)$ in $c^+(a)$ is $n \sum_i \lambda_i a_i \otimes a_i^{n-1}$ . Hence $\sum_i \lambda_i a_i \otimes a_i^{n-1} = 0$ . If $\mu : U \otimes U \to U$ is the linear mapping defined by multiplication on $U$ , then

a = \sum_i \lambda_i a_i^n = \mu \left( \sum_i \lambda_i a_i \otimes a_i^{n-1} \right) = 0.

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

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