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

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Symmetrization does not preserve products in sl_2

Statement refuted

PBW symmetrization preserves products in ${s l}_{2}$ .

Facts & Assumptions

Given: ${s l}_{2}$ over a characteristic-zero field, with $[h, e] = 2 e$ .

[L1]

Symmetrization satisfies $sym (e h) = \frac{1}{2} (e h + h e)$ and is a vector-space isomorphism (PBW symmetrization in characteristic zero).

Counterexample

technique · direct computation

1.1

In $U ({s l}_{2})$ , $h e - e h = 2 e$ , so $sym (e h) = \frac{1}{2} (e h + h e) = e h + e$ .

givenL1algebra

2.1

On degree-one factors, $sym (e) sym (h) = e h$ . PBW injectivity from [L1] gives $e \neq 0$ in the enveloping algebra, so $e h + e \neq e h$ .

step 1.1L1algebra

3.1

Therefore symmetrization does not preserve this product and is not an algebra homomorphism.

step 2.1∎

Depends on

PBW symmetrization in characteristic zero

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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

Etingof, MIT 18.745 notes, Corollary 13.7 and Example 13.9, printed p. 75 (standard reference, not scraped)