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False statementConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14
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PBW symmetrization is generally not multiplicative

Statement

For a nonabelian Lie algebra g in characteristic zero, PBW symmetrization S(g)→U(g) is an algebra isomorphism.

Facts & Assumptions

Given: A characteristic-zero nonabelian Lie algebra with a supplied basis.

[L1]

Symmetrization is a vector-space isomorphism with sym⁡(x)=ιg(x) and sym⁡(xy)=12(ιg(x)ιg(y)+ιg(y)ιg(x)) (PBW symmetrization in characteristic zero).

Refutation

technique · direct computation
1.1

Choose x,y with [x,y]≠0. The enveloping relation gives ιg(y)ιg(x)=ιg(x)ιg(y)−ιg([x,y]), so [L1] yields sym⁡(xy)=ιg(x)ιg(y)−12ιg([x,y]).

givenL1choosealgebra
2.1

But sym⁡(x)sym⁡(y)=ιg(x)ιg(y), and PBW injectivity, contained in [L1], makes ιg([x,y])≠0. Hence the two expressions differ and symmetrization is not multiplicative.

step 1.1L1algebra∎

Depends on

Used by

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Dependency tree · two levels

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Sources