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At the order-three root the trace degenerates on a semisimple block

Example

Set $ℓ = 3$ , $q = e^{π i / 3}$ and $δ = q + q^{- 1} = 1$ . Then $[2] = 1$ and $[3] = 0$ , so $p_{2} = 1_{2} - U_{1}$ is defined and is a nonzero idempotent of trace zero. In the basis $(1_{2}, U_{1})$ , the unnormalized trace pairing has matrix

$(\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}) .$

Thus the trace radical is $C (1_{2} - U_{1})$ , although ${T L}_{2} (C, 1) ≅ C \oplus C$ is semisimple. Its negligible quotient is one-dimensional.

Facts & Assumptions

Given: The order-three parameter, the rank-two generator relation, and the closure trace.

[F1]

At order $ℓ = 3$ , the complex semisimplicity criterion gives semisimplicity of ${T L}_{ℓ - 1} = {T L}_{2}$ (Complex Temperley–Lieb semisimplicity by the order of q squared).

[F2]

The quantum integer recurrence has $[1] = 1$ , $[2] = δ$ , $[3] = δ [2] - 1$ (Temperley–Lieb quantum integers and quantum characteristic).

[F3]

The two-strand recursion and trace identities give $p_{2} = 1 - ([1] / [2]) U_{1}$ , $p_{2}^{2} = p_{2}$ , and ${Tr}_{2} (p_{2}) = [3]$ (Jones–Wenzl recursion over unit quantum integers, Jones–Wenzl projectors: annihilation, uniqueness, partial trace, and closure).

[F4]

Closure assigns $δ^{2}$ to $1_{2}$ and $δ$ to $U_{1}$ (Unnormalized diagram closure and partial trace).

[F5]

The rank-two diagrams $1_{2}, U_{1}$ are distinct basis vectors and $U_{1}$ generates the algebra (Temperley–Lieb planar diagram data and endomorphism modules, Elementary cup-cap diagrams generate every finite planar pairing endomorphism diagram).

Verification

technique · direct calculation

1.1F2F3givenalgebra

Since $q = e^{π i / 3}$ , $q + q^{- 1} = 2 \cos (π / 3) = 1$ . Substituting $δ = 1$ into [F2] gives $[2] = 1$ and $[3] = 0$ , so the recursion [F3] is defined and yields $p_{2} = 1 - U_{1}$ . The relation $U_{1}^{2} = U_{1}$ gives $p_{2}^{2} = 1 - 2 U_{1} + U_{1} = 1 - U_{1} = p_{2}$ ; the diagram basis makes $p_{2} \neq 0$ . Its trace is $[3] = 0$ by [F3].

2.1F1F2F3F4F5step 1.1algebra∎

By [F4], ${Tr}_{2} (1_{2}) = 1$ and ${Tr}_{2} (U_{1}) = 1$ . Also $U_{1}^{2} = U_{1}$ , so every entry of the trace-pairing matrix in the basis [F5] is $1$ . Its kernel is the span of $(1, - 1)$ , corresponding to $1_{2} - U_{1}$ . By [F5], ${1_{2}, U_{1}}$ spans the full two-dimensional algebra; $U_{1}$ and $1 - U_{1}$ are nonzero orthogonal idempotents summing to $1$ , so the algebra is $C \oplus C$ . Quotienting by the block $C (1 - U_{1})$ leaves $C$ .

Depends on

Used by

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Sources

K. Iohara, G. I. Lehrer, and R. B. Zhang, The Jones Quotients of the Temperley-Lieb Algebras (standard reference, not scraped)
Joshua Chen, The Temperley-Lieb Categories and Skein Modules (standard reference, not scraped)