ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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One BMW generator takes three scalar values across representations

Example

In $B M W_{2}$ take $q = 2$ , $ρ = 3$ , $z = q - q^{- 1} = 3 / 2$ , and $δ = 25 / 9$ , which satisfies $z (δ - 1) = ρ - ρ^{- 1}$ . There are three one-dimensional representations with $(g_{1}, e_{1}) = (2, 0)$ , $(- 1 / 2, 0)$ , and $(1 / 3, 25 / 9)$ . Thus the generator takes all three distinct roots of its cubic across these representations. The third representation shows $e_{1} \neq 0$ in the full algebra, so the quotient by $e_{1}$ is proper; the Hecke quotient imposes the quadratic that fails in the third representation.

Facts & Assumptions

Given: The BMW $_{2}$ presentation from BMW algebra in a fixed Kauffman skein convention and the cubic/quotient relation from The BMW braid generators satisfy a cubic, with a Hecke quotient.

[F1]

BMW algebra in a fixed Kauffman skein convention: for $n = 2$ , the defining relations are the loop, skein, untwisting, and inverse relations for $g_{1}, e_{1}$ .

[F2]

The BMW braid generators satisfy a cubic, with a Hecke quotient: the full generator satisfies the cubic, while setting $e_{1} = 0$ gives the Hecke quadratic.

Verification

1.1F1algebra

For $(g, e) = (2, 0)$ and $(- 1 / 2, 0)$ , $g - g^{- 1} = 3 / 2 = z (1 - e)$ and the loop and untwisting relations hold because $e = 0$ . For $(g, e) = (1 / 3, 25 / 9)$ , $g - g^{- 1} = - 8 / 3 = z (1 - e)$ , $g e = e g = 25 / 27 = ρ^{- 1} e$ , and $e^{2} = 625 / 81 = δ e$ . Every $g$ is invertible, and BMW $_{2}$ has no adjacent-index relations, so these are valid one-dimensional representations.

2.1F2step 1.1algebra∎

The three representations evaluate $g_{1}$ at the distinct numbers $2, - 1 / 2, 1 / 3$ . Since the third sends $e_{1}$ to $25 / 9 \neq 0$ , the ideal generated by $e_{1}$ is nonzero, so the canonical map to the quotient by that ideal is not injective. The quadratic $(g_{1} - 2) (g_{1} + 1 / 2)$ evaluates to $- 25 / 18 \neq 0$ in the third representation, whereas setting $e_{1} = 0$ forces that quadratic by [F2].

Depends on

Used by

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Sources

Goodman and Hauschild, Affine Birman-Wenzl-Murakami Algebras and Tangles in the Solid Torus, Remark 3.2 and Proposition 3.4 (standard reference, not scraped)