PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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The associated graded algebra of the PBW filtration is commutative

Statement

Let $g$ be a complex Lie algebra. In the filtration from The PBW filtration by tensor degree on the enveloping algebra, one has

$[F_{m} U (g), F_{n} U (g)] \subseteq F_{m + n - 1} U (g) (m, n \geq 0),$

so the graded algebra $gr U (g)$ is commutative.

Facts & Assumptions

Given: A complex Lie algebra $g$ and the PBW filtration $F_{∙} U (g)$ .

Proof

technique · direct

1.1

For $x \in g$ and a monomial $y_{1} \dots y_{n}$ of filtration degree $n$ , repeatedly using $x y_{i} - y_{i} x = [x, y_{i}]$ from The PBW filtration by tensor degree on the enveloping algebra rewrites $x (y_{1} \dots y_{n}) - (y_{1} \dots y_{n}) x$ as a sum of terms with exactly one bracket and therefore filtration degree at most $n$ .

givenalgebra

2.1

Applying step 1.1 repeatedly to products of filtration degrees $m$ and $n$ shows that every commutator of elements of $F_{m} U (g)$ and $F_{n} U (g)$ drops by at least one filtration step, so it lies in $F_{m + n - 1} U (g)$ .

step 1.1algebra

3.1

Hence the degree- $m$ and degree- $n$ symbols commute in $gr U (g)$ , and the associated graded algebra is commutative.

step 2.1∎

Depends on

The PBW filtration by tensor degree on the enveloping algebra

Used by

Dependency tree · two levels

2 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Pavel Etingof, Lie Groups and Lie Algebras I (standard reference, not scraped)