Alphabeta Math
CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06
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The Shapovalov form need not be positive on its real PBW span

Statement refuted

For every real λ, the normalized Shapovalov form is positive definite on the real span of the PBW basis of M(λ).

Take sl2 and λ=−1. On MR:=span⁡R{fnv−1:n≥0}, the restricted bilinear form has fv−1≠0 but S−1(fv−1,fv−1)=−1.

Facts & Assumptions

Given: The norm product The sl2 Shapovalov norm product.

Counterexample

technique · direct
1.1

At n=1 the product formula gives S−1(fv−1,fv−1)=1!(−1)=−1.

givenalgebra
2.1

A positive-definite real bilinear form cannot take a negative value on a nonzero vector; hence this refutes positivity on the specified real PBW span, without asserting degeneracy or turning the complex-bilinear form into a Hermitian form.

givencontradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources