Proposition 6

Tag 0A21, permanent. Cite it as [Bourbaki, Tag 0A21].

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

Let $E$ be a filtered bigebra and $(E_n)_{n \geq 0}$ its filtration. For every integer $n \geq 0$ , let $E_n^+ = E_n \cap E^+$ . Then $E_0^+ = \{0\}$ and

c^+(E_n^+) \subset \sum_{i=1}^{n-1} \operatorname{Im}(E_i^+ \otimes E_{n-1}^+) \quad \text{for } n \geq 0.\dagger

As $E_0 = K.1, E_0^+ = 0$ . If $x \in E_n, \pi(x) = x - \varepsilon(x).1$ (formula (1)), whence $\pi(x) \in E_n^+$ and $\pi(E_n) \subset E_n^+$ . It follows that $\pi \otimes \pi$ maps $\operatorname{Im}(E_i \otimes E_j)$ into

\dagger If A and B are two submodules of E, we denote by $\operatorname{Im}(A \otimes B)$ the image of the canonical mapping $A \otimes B \to E \otimes E$ .

$Im(E_i^+ \otimes E_j^+)$ for $i \geq 0, j \geq 0$ . As $c^+ = (\pi \otimes \pi) \circ c$ in $E^+$ (no. 1, Proposition 2), by (6)

c^+(E_n^+) \subset \sum_{i=0}^n \operatorname{Im}(E_i^+ \otimes E_{n-i}^+) = \sum_{i=1}^{n-1} \operatorname{Im}(E_i^+ \otimes E_{n-i}^+).

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

Cited by

Cites

no. 1, Proposition 2