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

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A critical five-defect standard module at order three is simple

Example

Let $q = e^{π i / 3}$ , so $δ = 1$ and $κ = 3$ . At rank $n = 7$ , take $p = 1$ arc and $t = 5$ defects. The critical standard module is simple, and its Gram determinant is $1$ .

Facts & Assumptions

Given: $q = e^{π i / 3}$ , $n = 7$ , $p = 1$ , $t = 5$ , and $κ = 3$ .

[F1]

The critical-label test is $κ ∣ (t + 1)$ (Complex Temperley–Lieb critical labels and reflection).

[F2]

At every critical label the complex link-state capping form is nondegenerate and its standard module is simple (Critical complex Temperley–Lieb standard modules are simple).

[F3]

For $p = 1$ , the Gram determinant is $[t + 2]_{q} / [1]_{q}$ (The Temperley–Lieb link-state Gram determinant).

Verification

Given: $δ = 1$ , $κ = 3$ , and the rank-seven label $(n, p) = (7, 1)$ .

1.1F1givenalgebra

The defect number is $t = 7 - 2 = 5$ , so $t + 1 = 6$ is divisible by $κ = 3$ . Hence $(7, 1)$ is critical by [F1].

2.1F2F3givenstep 1.1algebra∎

Since $p = 1$ , [F3] gives $\det G_{7, 1} = [7]_{q} / [1]_{q}$ . Here $[1]_{q} = 1$ and $[7]_{q} = \frac{\sin (7 π / 3)}{\sin (π / 3)} = 1,$ so $\det G_{7, 1} = 1 \neq 0$ . By [F2], this nondegenerate critical standard module is simple.

Depends on

Used by

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Sources

David Ridout and Yvan Saint-Aubin, Standard Modules, Induction and the Structure of the Temperley-Lieb Algebra (standard reference, not scraped)
Andrew Mathas, Hecke algebras and Schur algebras of the symmetric group, complete prepublication manuscript (standard reference, not scraped)