LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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The elementary diagram generators satisfy the local Temperley–Lieb relations

Statement

Fix $n \geq 0$ , a commutative ring $R$ , and $δ \in R$ . For $1 \leq i < n$ , let $d_{i} \in {T L}_{n} (R, δ)$ be the elementary cup-cap diagram of Temperley–Lieb planar diagram data and endomorphism modules. Then

$d_{i}^{2} = δ d_{i} (1 \leq i < n), d_{i} d_{i + 1} d_{i} = d_{i}, d_{i + 1} d_{i} d_{i + 1} = d_{i + 1} (1 \leq i \leq n - 2), d_{i} d_{j} = d_{j} d_{i} (1 \leq i, j < n, ∣ i - j ∣ > 1) .$

Consequently the assignment $u_{i} \mapsto d_{i}$ extends to a unital $R$ -algebra homomorphism

$π_{n} : P_{n} (R, δ) ⟶ {T L}_{n} (R, δ) .$

Facts & Assumptions

Given: $n \geq 0$ , a commutative ring $R$ , $δ \in R$ , the endomorphism algebra ${T L}_{n} (R, δ)$ , its product convention, and the presented algebra $P_{n} (R, δ)$ .

[F1]

$d_{i}$ has a cap joining top points $i, i + 1$ , a cup joining bottom points $i, i + 1$ , and vertical through strands at all other positions. (Temperley–Lieb planar diagram data and endomorphism modules)

[F2]

Stacking removes each closed cycle with one coefficient $δ$ and leaves the exterior matching. (Temperley–Lieb planar diagram data and endomorphism modules)

[F3]

Products $x y = x \circ y$ stack the left factor above the right factor; composition is associative and bilinear. (Temperley–Lieb planar diagram data and endomorphism modules, Stacking is associative and horizontal juxtaposition is strict monoidal)

[F4]

$P_{n} (R, δ)$ is the quotient of the free associative $R$ -algebra by the two-sided ideal generated by the listed square, adjacent triple, and distant commutation relations; maps satisfying them factor uniquely through the quotient. (The abstract Temperley–Lieb generator presentation)

Proof

technique · direct

1.1F1F2F3

In the stack $d_{i}^{2}$ , all strands outside positions $i, i + 1$ pass straight through. In the two affected positions, the cup of the upper copy joins to the cap of the lower copy to form exactly one closed cycle; the remaining exterior path connections are the same cap and cup as in $d_{i}$ . The stack rule [F2] therefore gives $d_{i}^{2} = δ d_{i}$ .

1.2F1F2F3

Restrict to positions $i, i + 1, i + 2$ in the stack with upper factor $d_{i}$ , middle factor $d_{i + 1}$ and lower factor $d_{i}$ . The upper cap and lower cup are the exterior cap and cup of the result. The remaining path starts at the exterior top point $i + 2$ : it follows the upper through strand to the first interface, the middle cap to position $i + 1$ , the upper cup to position $i$ , the middle through strand to the second interface, the lower cap to position $i + 1$ , the middle cup to position $i + 2$ , and the lower through strand to exterior bottom point $i + 2$ . Thus the two remaining exterior endpoints are joined exactly as in $d_{i}$ , and all six interface points lie on this path; no closed component is formed. The product is therefore $d_{i} d_{i + 1} d_{i} = d_{i}$ . Reflecting this three-position stack horizontally about its middle strand preserves stacking and loop counts while interchanging $d_{i}$ and $d_{i + 1}$ , so it gives $d_{i + 1} d_{i} d_{i + 1} = d_{i + 1}$ .

1.3F1F2F3

If $∣ i - j ∣ > 1$ , the two cup-cap diagrams modify disjoint pairs of positions. In either stack order, each affected pair has the same cap and cup and every other strand is vertical; no component joins the two disjoint position intervals and no closed cycle forms. Hence the resulting boundary matching is the same and $d_{i} d_{j} = d_{j} d_{i}$ .

2.1F3F4step 1.1step 1.2step 1.3∎

The images $d_{i}$ satisfy every defining relation of $P_{n} (R, δ)$ by steps 1.1–1.3. The quotient universal property in [F4] gives the unique unital $R$ -algebra homomorphism $π_{n}$ with $π_{n} (u_{i}) = d_{i}$ . For $n = 0$ and $n = 1$ the generator and relation sets are empty, and the unique map sends the scalar unit to the diagram identity.

Remarks

The three-strand calculation only follows connected paths and cycles in the finite stacked graph. It makes no appeal to a drawing as a proof and uses no invertibility assumption on $δ$ .

Depends on

Used by

Dependency tree · two levels

13 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, §2 (standard reference, not scraped)
Dana C. Ernst, Michael G. Hastings, and Sarah K. Salmon, Factorization of Temperley–Lieb diagrams (standard reference, not scraped)
Vaughan F. R. Jones, The Jones Polynomial, author-hosted notes (standard reference, not scraped)