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Temperley–Lieb Link-State Modules and Cellularity: Examples and Counterexamples

1 · Prerequisites

2 · Summary

This companion verifies concrete calculations from the A-page definitions. The rank-three example gives the two one-defect states, their generator actions and Gram matrix. The rank-two counterexample shows that an identically zero cell form can have a nonzero simple underlying module. Neither item supplies a general theorem.

Examples

Reading route

The A-page temperley-lieb-cellularity-and-representation-theory is the sole prerequisite for these finite examples.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

The rank-three one-defect link-state module

Example

Let $k$ be any field, $δ \in k$ , and consider $W_{3, 1} (k, δ)$ . Its basis consists of the state $a$ with an arc between sites $1$ and $2$ and a defect at site $3$ , and the state $b$ with a defect at site $1$ and an arc between sites $2$ and $3$ . For the elementary diagrams $d_{1}, d_{2}$ and the Gram form $ϕ_{3, 1}$ ,

$a \cdot d_{1} = δ a, a \cdot d_{2} = b, b \cdot d_{1} = a, b \cdot d_{2} = δ b,$

and

$[ϕ_{3, 1}]_{(a, b)} = (\begin{array}{cc} δ & 1 \\ 1 & δ \end{array}) .$

At $δ = 0$ the form is nondegenerate, so $W_{3, 1} (k, 0)$ is simple by Nonzero Temperley–Lieb cell forms give all simple heads.

Facts & Assumptions

Given: A field $k$ , a parameter $δ \in k$ , and $W_{3, 1} (k, δ)$ .

[F1]

The basis of $W_{n, t}$ consists of link states with $n$ sites and $t$ defects; its action is stacking on the right (Temperley–Lieb link-state standard modules by defect number, The link-state quotient is a right Temperley–Lieb module).

[F2]

$d_{i}$ has a cap at top sites $i, i + 1$ , a cup at bottom sites $i, i + 1$ , and vertical strands elsewhere; stacking removes each closed loop with coefficient $δ$ (Temperley–Lieb planar diagram data and endomorphism modules).

[F3]

The Gram form is computed by reflecting and gluing link states, and is the cellular form (The link-state Gram form is symmetric and invariant, The Temperley–Lieb link-state diagram basis is cellular).

[F4]

A nonzero link-state cell form gives a simple head, and its quotient is simple; if the form is nondegenerate then that quotient is the whole module (Nonzero Temperley–Lieb cell forms give all simple heads).

Verification

technique · direct

1.1F1givenconstruct

For $n = 3$ and $t = 1$ there is one arc. The only noncrossing placements are $a$ , with arc $(1, 2)$ and defect $3$ , and $b$ , with defect $1$ and arc $(2, 3)$ . The dimension formula of Link states are counted by a ballot-number difference gives $(\binom{3}{1}) - (\binom{3}{0}) = 2$ , so these two distinct states are the full basis.

2.1F1F2step 1.1

For $a \cdot d_{1}$ , the cap of $d_{1}$ joins the two endpoints of $a$ 's arc, making one closed loop; the bottom cup of $d_{1}$ leaves the same arc, so $a \cdot d_{1} = δ a$ . For $a \cdot d_{2}$ , tracing from the defect at site $3$ through the top cap of $d_{2}$ , then through the arc of $a$ and the vertical strand at site $1$ , leaves a defect at site $1$ and a cup at sites $2, 3$ , so $a \cdot d_{2} = b$ . For $b \cdot d_{1}$ , tracing from the defect at site $1$ through $d_{1}$ 's cap, then through $b$ 's arc and the vertical strand at site $3$ , leaves the arc at sites $1, 2$ and a defect at site $3$ , so $b \cdot d_{1} = a$ . For $b \cdot d_{2}$ , the cap closes $b$ 's arc into one loop and the bottom cup leaves $b$ , so $b \cdot d_{2} = δ b$ .

2.2F2F3step 1.1

Gluing $a$ with its reflection creates one loop, as does gluing $b$ with its reflection, so $ϕ (a, a) = δ = ϕ (b, b)$ . Gluing $a$ to the reflection of $b$ creates one through path and no loop, so $ϕ (a, b) = 1$ ; by symmetry $ϕ (b, a) = 1$ . These four values give the displayed Gram matrix.

3.1F3F4step 2.2algebra∎

At $δ = 0$ , if $x a + y b$ lies in the radical then pairing with $a$ gives $y = 0$ and pairing with $b$ gives $x = 0$ . Thus the radical is zero and the cell-form quotient is $W_{3, 1}$ itself. The form is nonzero, so [F4] makes this module simple.

Remarks

The two basis half-diagrams use the same boundary convention as the A-page definition; defects are the strands meeting the top boundary.

CounterexampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

A zero Temperley–Lieb cell form can have a nonzero simple module

Statement refuted

For every Temperley–Lieb link-state module over a field, an identically zero cell form implies that the underlying module has no nonzero simple quotient.

Facts & Assumptions

Given: A field $k$ , the rank-two algebra ${T L}_{2} (k, 0)$ , and the link-state module $W_{2, 0} (k, 0)$ .

[F1]

W_{n,t} has link states as a basis and the right action is defined by stacking a rank-n diagram below the state (Temperley–Lieb link-state standard modules by defect number, The link-state quotient is a right Temperley–Lieb module).

[F2]

The rank- $n$ algebra has as a basis all noncrossing pairings of its $2 n$ boundary points; each removed loop contributes $δ$ (Temperley–Lieb planar diagram data and endomorphism modules, Stacking is associative and horizontal juxtaposition is strict monoidal).

[F3]

The cell form is the loop-counting reflection/capping form (The link-state Gram form is symmetric and invariant).

Counterexample

technique · direct

1.1F1F2givenconstruct

There is exactly one rank-two link state with zero defects, namely the arc joining sites $1$ and $2$ . Call it $x$ . Thus $W_{2, 0} (k, 0) = k x$ . The four cyclic boundary points have three pairings; the pairing with alternating endpoints is crossing, while the other two are the identity matching and $d_{1}$ with its top cap and bottom cup. Hence those two diagrams are the full basis of ${T L}_{2} (k, 0)$ .

2.1F1F2step 1.1

The identity acts as the identity on $x$ . Stacking $x$ above $d_{1}$ closes the top cap of $d_{1}$ with the arc of $x$ , producing one loop and leaving the same link state, so $x \cdot d_{1} = δ x = 0$ . Since $1$ and $d_{1}$ span ${T L}_{2} (k, 0)$ , every algebra element acts on the one-dimensional space $k x$ by a scalar. The module is nonzero and simple.

3.1F3step 1.1step 2.1∎

Reflecting $x$ and gluing it to itself creates one loop, so $ϕ_{2, 0} (x, x) = δ = 0$ . Since $x$ is the sole basis vector, bilinearity makes the cell form identically zero and its cell-form quotient $W_{2, 0} / rad (ϕ_{2, 0})$ is zero. But $W_{2, 0} = k x$ is itself a nonzero simple module by step 2.1. This refutes the statement.