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At the first root boundary the Jones trace degenerates before semisimplicity fails

Statement

Let q2 have finite order ℓ≥3, put δ=q+q−1, and let n=ℓ−1. Then TLn(C,δ) is semisimple but the bilinear trace form (x,y)↦τn(xy) of Normalized Temperley–Lieb trace from diagram closure is degenerate.

Facts & Assumptions

Given: The complex order criterion, the normalized closure trace, and the Jones–Wenzl recursion.

[F1]

Since ℓ≥3, the complex semisimplicity theorem gives semisimplicity at rank n=ℓ−1 (Complex Temperley–Lieb semisimplicity by the order of q squared).

[F2]

The Jones–Wenzl idempotent pn is nonzero and primitive, and τn(pn)=[n+1]/δn when the required denominators are nonzero (The Jones–Wenzl recursion gives annihilating idempotents and their trace).

[F4]

If q2 has order ℓ≥3, then [r]q≠0 for 1≤r<ℓ and [ℓ]q=0 (Temperley–Lieb quantum integers and quantum characteristic).

Proof

1.1F2F4given

Since δ≠0 and [r]q≠0 for 1≤r≤n=ℓ−1, [F2] supplies the nonzero primitive idempotent pn and gives τn(pn)=[ℓ]q/δn=0 by [F4].

2.1F2F3step 1.1algebra

For any x∈TLn, cyclicity [F3] and pn2=pn imply τn(pnx(1−pn))=τn((1−pn)pnx)=0, so τn(pnx)=τn(pnxpn). The one-dimensional corner in [F2] gives pnxpn=λxpn for some λx∈C. Hence τn(pnx)=λxτn(pn)=0 for every x, while pn≠0. Thus pn is a nonzero vector in the radical of the trace form, so the form is degenerate.

3.1F1step 2.1∎

Semisimplicity is the independent assertion in [F1]; combining it with step 2.1 proves that trace degeneracy occurs at rank ℓ−1 while the algebra is still semisimple.

Depends on

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