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The BMW braid generators satisfy a cubic, with a Hecke quotient

Statement

For the convention in BMW algebra in a fixed Kauffman skein convention, every generator satisfies $(g_{i} - q) (g_{i} + q^{- 1}) (g_{i} - ρ^{- 1}) = 0$ . In the cup-cap-zero quotient, it instead satisfies the Hecke quadratic $(g_{i} - q) (g_{i} + q^{- 1}) = 0$ .

Facts & Assumptions

Given: The BMW relations in BMW algebra in a fixed Kauffman skein convention, with $z = q - q^{- 1}$ .

[F1]

BMW algebra in a fixed Kauffman skein convention: $g_{i} e_{i} = e_{i} g_{i} = ρ^{- 1} e_{i}$ and $g_{i} - g_{i}^{- 1} = z (1 - e_{i})$ .

Proof

1.1givenF1algebra

Fix $i$ . Right-multiplying the skein relation by $g_{i}$ and using $e_{i} g_{i} = ρ^{- 1} e_{i}$ gives $g_{i}^{2} - z g_{i} - 1 + z ρ^{- 1} e_{i} = 0$ . Multiplying this identity on the right by $g_{i} - ρ^{- 1}$ removes its last term because $e_{i} (g_{i} - ρ^{- 1}) = 0$ . Hence $(g_{i}^{2} - z g_{i} - 1) (g_{i} - ρ^{- 1}) = 0$ .

2.1step 1.1algebra∎

Since $z = q - q^{- 1}$ , $g_{i}^{2} - z g_{i} - 1 = (g_{i} - q) (g_{i} + q^{- 1})$ , which proves the cubic. In the quotient $e_{i} = 0$ , the skein relation gives $g_{i} - g_{i}^{- 1} = z$ ; right-multiplying by the invertible $g_{i}$ yields $g_{i}^{2} - z g_{i} - 1 = 0$ , the Hecke quadratic. The cubic statement did not impose this quadratic on the full BMW algebra.

Depends on

BMW algebra in a fixed Kauffman skein convention

Used by

One BMW generator takes three scalar values across representations Example

Dependency tree · two levels

2 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

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