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The Jones–Wenzl recursion gives annihilating idempotents and their trace

Statement

Let $δ \in C^{\times}$ and suppose $[2]_{δ}, \dots, [n]_{δ}$ are nonzero. The recursively defined $p_{n}$ of Jones–Wenzl recursion in the nonvanishing-denominator range is nonzero, satisfies $p_{n}^{2} = p_{n}$ , and obeys $d_{i} p_{n} = p_{n} d_{i} = 0$ for $1 \leq i < n$ . Its corner is one-dimensional, $p_{n} {T L}_{n} p_{n} = C p_{n}$ , and its normalized closure trace is

$τ_{n} (p_{n}) = \frac{[n + 1]_{δ}}{δ^{n}} .$

Facts & Assumptions

Given: The loop convention, the local generator relations, quantum integer recurrence, and normalized closure trace.

[F1]

The generators satisfy $d_{i}^{2} = δ d_{i}$ , $d_{i} d_{i \pm 1} d_{i} = d_{i}$ , and distant commutation (The elementary diagram generators satisfy the local Temperley–Lieb relations).

[F2]

The recursion is $p_{r + 1} = P_{r} - \frac{[r]}{[r + 1]} P_{r} d_{r} P_{r}$ , with $P_{r} = p_{r} \otimes 1$ (Jones–Wenzl recursion in the nonvanishing-denominator range).

[F3]

The quantum integers satisfy $[r + 1] = δ [r] - [r - 1]$ (Temperley–Lieb quantum integers and quantum characteristic).

[F4]

The closure trace is cyclic and has $τ_{r + 1} (x \otimes 1) = τ_{r} (x)$ and $τ_{r + 1} ((x \otimes 1) d_{r}) = δ^{- 1} τ_{r} (x)$ (The normalized closure trace is cyclic and satisfies the Markov rules).

[F5]

Finite words in the elementary diagram generators, including the empty word for the unit, span the diagram algebra (Elementary cup-cap diagrams generate every finite planar pairing endomorphism diagram).

[F6]

A through strand is a path joining a top point to a bottom point. Each diagram is a matching of boundary points; an elementary $d_{i}$ in ${T L}_{n}$ has fewer than $n$ through strands, and every through strand of a composite must pass through a distinct through strand at the intermediate boundary (Temperley–Lieb planar diagram data and endomorphism modules).

[F7]

Write $ε_{s} (x)$ for the coefficient of the identity diagram in the diagram basis of ${T L}_{s}$ ; in particular, a linear combination of diagrams with fewer than $s$ through strands has $ε_{s} = 0$ .

Proof

technique · Induct on the strand number, proving the recursion identities and normalized trace together

1.1F1F3F4F5F6F7givenbasealgebra

For ranks zero and one, $p_{0} = p_{1} = 1$ has all asserted properties, $ε_{s} (p_{s}) = 1$ , and trace $1 = [1] / δ^{0} = [2] / δ$ where applicable. At rank two, $p_{2} = 1 - δ^{- 1} d_{1}$ ; $d_{1}^{2} = δ d_{1}$ gives $p_{2}^{2} = p_{2}$ and $d_{1} p_{2} = p_{2} d_{1} = 0$ , $ε_{2} (p_{2}) = 1$ , its corner is $C p_{2}$ since $1, d_{1}$ span ${T L}_{2}$ , and $τ_{2} (p_{2}) = 1 - δ^{- 2} = [3] / δ^{2}$ .

2.1step 1.1ih

Fix $r \geq 2$ and assume inductively that through rank $r$ the recursively defined elements are nonzero idempotents, satisfy $ε_{s} (p_{s}) = 1$ , are annihilated on both sides by every generator, have corner $C p_{s}$ , and satisfy $τ_{s} (p_{s}) = [s + 1] / δ^{s}$ .

3.1F1F2F3step 2.1algebra

Put $P = p_{r} \otimes 1$ , $Q = p_{r - 1} \otimes 1 \otimes 1$ , $a = d_{r - 1}$ , $b = d_{r}$ , and $μ = [r - 1] / [r]$ . The recursion at rank $r$ gives $P = Q - μ Q a Q$ ; $Q$ is idempotent, commutes with $b$ , and $b a b = b$ by [F1]. Therefore $b P b = δ Q b - μ Q b a b Q = (δ - μ) Q b = ([r + 1] / [r]) Q b$ by [F3].

4.1F1F2F6F7step 2.1step 3.1algebra

Set $X = P b P$ and $c = [r] / [r + 1]$ . Since $P Q = Q P = P$ , step 3.1 gives $X^{2} = P (b P b) P = ([r + 1] / [r]) X$ , so $(P - c X)^{2} = P - c X = p_{r + 1}$ . Also $b p_{r + 1} = b P - c b P b P = 0$ by step 3.1 and $Q b P = b P$ ; and $p_{r + 1} b = P b - c P b P b = 0$ because $P b P b = P (b P b) = ([r + 1] / [r]) P Q b = ([r + 1] / [r]) P b$ . For $i < r$ , both annihilation identities follow from the induction hypothesis. Juxtaposing a vertical strand preserves the identity coefficient, so $ε_{r + 1} (P) = ε_{r} (p_{r}) = 1$ . The correction term $P b P$ has fewer than $r + 1$ through strands, since every diagram term passes through $b = d_{r}$ , which itself has fewer than $r + 1$ , and stacking cannot increase that count by [F6]. Hence $ε_{r + 1} (P b P) = 0$ , so $ε_{r + 1} (p_{r + 1}) = 1$ and $p_{r + 1} \neq 0$ .

5.1F5step 4.1algebra

Every nonempty generator word has a rightmost letter which kills $p_{r + 1}$ by step 4.1. Since generator words span ${T L}_{r + 1}$ by [F5], $p_{r + 1} {T L}_{r + 1} p_{r + 1} = C p_{r + 1}$ .

6.1F3F4step 2.1step 4.1inductiondischarge-induction∎

By [F4], $τ_{r + 1} (P) = τ_{r} (p_{r})$ and, using cyclicity, $τ_{r + 1} (X) = τ_{r + 1} (P b P) = τ_{r + 1} (b P^{2}) = τ_{r + 1} (P b) = δ^{- 1} τ_{r} (p_{r})$ . Hence $τ_{r + 1} (p_{r + 1}) = τ_{r} (p_{r}) (1 - [r] / (δ [r + 1])) = [r + 2] / δ^{r + 1}$ by [F3] and the induction hypothesis. This proves the trace recurrence and closes the induction.

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11 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

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