Remark

Tag 0A4K, permanent. Cite it as [Bourbaki, Tag 0A4K].

Lie Groups and Lie Algebras, chapter II, § 4. Central filtrations

When the filtration $(G_\alpha)$ is integral, $\operatorname{gr}_\alpha(G) = \{0\}$ for non-integral $\alpha$ and $\operatorname{gr}_n(G) = G_n / G_{n-1}$ for every integer $n$ . The definition of the associated graded group therefore coincides essentially with that of Commutative Algebra, Chapter III, § 2, no. 3.

Let $A$ be an algebra (resp. unital algebra) and $(A_\alpha)_{\alpha \in \mathbf{R}}$ a real filtration compatible with the algebra (resp. unital algebra) structure (no. 1). Then

A_\alpha \cdot A_\beta \subset A_{\alpha + \beta}, \qquad A_\alpha^+ \cdot A_\beta + A_\alpha \cdot A_\beta^+ \subset A_{\alpha + \beta}^+

and the bilinear mapping of $A_\alpha \times A_\beta$ into $A_{\alpha + \beta}$ the restriction of multiplication on $A$ defines on taking quotients a bilinear mapping

\operatorname{gr}_\alpha(A) \times \operatorname{gr}_\beta(A) \to \operatorname{gr}_{\alpha + \beta}(A).

We derive a bilinear mapping of gr(A) × gr(A) into gr(A) which makes it a graded algebra (resp. unital graded algebra) of type $\mathbf{R}$ . If A is an associative (resp. commutative, resp. Lie) algebra, so is gr(A).

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

Cited by

Nothing.

Cites

*Commutative Algebra*, Chapter III, § 2, no. 3 not yet on this site: AC III