CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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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.
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A noninvariant quadratic PBW element need not be central

Statement refuted

A quadratic PBW monomial need not be central. In $U ({s l}_{2})$ , the element $e^{2}$ is quadratic but not central.

Facts & Assumptions

Given: The universal enveloping algebra of ${s l}_{2}$ with standard generators $e, f, h$ .

Counterexample

technique · direct

1.1

Using the relations $[f, e] = - h$ and $h e - e h = 2 e$ , one computes $[f, e^{2}] = [f, e] e + e [f, e] = - h e - e h$ .

givenalgebra

2.1

The element $- h e - e h$ is nonzero in the PBW basis, so step 1.1 shows that $e^{2}$ does not commute with $f$ . Hence $e^{2}$ is not central.

step 1.1∎

Depends on

The quadratic Casimir element is central

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 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

Alexander Kleshchev, Lectures on Infinite Dimensional Lie Algebras (standard reference, not scraped)