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

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At zero loop value, rank three and rank four have opposite semisimplicity

Example

Take $q = i$ , so $δ = 0$ and $q^{2}$ has order two. Then ${T L}_{3} (C, 0)$ is semisimple, whereas ${T L}_{4} (C, 0)$ is not.

Facts & Assumptions

Given: The quantum integers at $q = i$ and the link-state Gram determinants.

[F1]

For $q = i$ , $[2 k]_{q} = 0$ and $[2 k + 1]_{q} = (- 1)^{k}$ ; the determinants and semisimplicity test are given by The Temperley–Lieb link-state Gram determinant and Temperley–Lieb cell forms detect semisimplicity over the complex numbers.

Verification

Given: $q = i$ , $ℓ = 2$ , and ranks $n = 3, 4$ .

1.1F1givenalgebra

For $n = 3$ , the allowed arc counts are $p = 0, 1$ . The empty product gives $\det G_{3, 0} = 1$ , while the determinant formula gives $\det G_{3, 1} = [3]_{q} / [1]_{q} = - 1$ . Both link-state forms are nondegenerate, so Temperley–Lieb cell forms detect semisimplicity over the complex numbers implies ${T L}_{3} (C, 0)$ is semisimple.

2.1F1givenstep 1.1algebra∎

For $n = 4$ , choose $p = 1$ . Then $\det G_{4, 1} = [4]_{q} / [1]_{q} = 0$ , so one cell form is degenerate. The reverse implication of Temperley–Lieb cell forms detect semisimplicity over the complex numbers gives that ${T L}_{4} (C, 0)$ is not semisimple.

Depends on

Used by

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Dependency tree · two levels

15 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

David Ridout and Yvan Saint-Aubin, Standard Modules, Induction and the Structure of the Temperley-Lieb Algebra (standard reference, not scraped)