PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The PBW filtration is multiplicative and has commutative associated graded

Statement

For the PBW filtration,

$F_{m} U (g) F_{n} U (g) \subseteq F_{m + n} U (g) .$

Moreover $[F_{m}, F_{n}] \subseteq F_{m + n - 1}$ when $m + n \geq 1$ , and therefore $gr U (g)$ is commutative.

Facts & Assumptions

Given: A Lie algebra $g$ and the PBW filtration on its enveloping algebra.

[L1]

$F_{n}$ is spanned by words of length at most $n$ (PBW filtration on the enveloping algebra).

[L2]

In $U (g)$ , $ι_{g} (x) ι_{g} (y) - ι_{g} (y) ι_{g} (x) = ι_{g} ([x, y])$ (The canonical map to U(g) is a Lie homomorphism).

[L3]

Associated-graded multiplication is that of Associated graded algebra of a filtered algebra.

Proof

technique · direct

1.1L1algebra

Concatenating a word of length at most $m$ with one of length at most $n$ gives length at most $m + n$ . Taking spans and quotient images proves $F_{m} F_{n} \subseteq F_{m + n}$ .

1.2L2algebra

For a generator $x$ and a word $b_{1} \dots b_{s}$ , repeated use of $[x, a b] = [x, a] b + a [x, b]$ gives $[x, b_{1} \dots b_{s}] = \sum_{j} b_{1} \dots [x, b_{j}] \dots b_{s}$ . By [L2], every $[x, b_{j}]$ is again the image of one element of $g$ , so this commutator lies in $F_{s}$ .

2.1step 1.1step 1.2algebra

For words $a = a^{'} x$ of length $r > 0$ and $b$ of length $s > 0$ , the identity $[a^{'} x, b] = a^{'} [x, b] + [a^{'}, b] x$ , together with step 1.2 and induction on $r$ , puts both terms in $F_{r + s - 1}$ . The scalar boundary cases commute, and bilinearity therefore gives $[F_{m}, F_{n}] \subseteq F_{m + n - 1}$ .

3.1step 2.1L3∎

If $a \in F_{m}$ and $b \in F_{n}$ , step 2.1 says that $a b$ and $b a$ have the same class in $F_{m + n} / F_{m + n - 1}$ . By [L3], all homogeneous elements of $gr U (g)$ commute, hence the whole associated graded algebra is commutative.

The PBW filtration is multiplicative and has commutative associated graded

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources