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Temperley–Lieb Diagram Algebra and Catalan Basis: Examples and Counterexamples

1 · Prerequisites

2 · Summary

These examples calculate the first four ranks, list and multiply the five normal forms in rank three, and show that at loop value zero a generator can be nonzero and square-zero. They use the A-page conventions and do not add general theorem suppliers.

Examples

The first four finite Temperley–Lieb algebras: bases in ranks zero through three and the rank-three multiplication table.
The five normal forms in the three-strand algebra: the five Jones normal forms and two adjacent triple reductions in rank three.
At loop value zero, a local generator is nilpotent but nonzero: the rank-two generator at $δ = 0$ is a nonzero nilpotent.

The TikZ diagram on the A page fixes the elementary cup-cap convention. The calculations use matching data and the proved local relations rather than relying on pictures as proof.

Companion: temperley-lieb-diagram-algebra-and-catalan-basis.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

The first four finite Temperley–Lieb algebras

Example

Let $R$ be a commutative ring and fix any loop value $δ \in R$ . For the ranks $n = 0, 1, 2, 3$ , the diagram and presentation algebras have the following $R$ -bases:

$\begin{array}{ccc} n & basis & rank \\ 0 & {1} & 1 \\ 1 & {1} & 1 \\ 2 & {1, u_{1}} & 2 \\ 3 & {1, u_{1}, u_{2}, u_{1} u_{2}, u_{2} u_{1}} & 5 \end{array}$

The rank column is $C_{n}$ for these four indices. In rank $3$ , writing $a = u_{1}$ and $b = u_{2}$ , the product table is

$\begin{array}{cccccc} \cdot & 1 & a & b & a b & b a \\ 1 & 1 & a & b & a b & b a \\ a & a & δ a & a b & δ a b & a \\ b & b & b a & δ b & b & δ b a \\ a b & a b & a & δ a b & a b & δ a \\ b a & b a & δ b a & b & δ b & b a \end{array}$

Facts & Assumptions

Given: A commutative ring $R$ , a loop value $δ \in R$ , and $n \in {0, 1, 2, 3}$ .

[F1]

$P_{n} (R, δ)$ is generated by $u_{1}, \dots, u_{n - 1}$ with $u_{i}^{2} = δ u_{i}$ , $u_{i} u_{i \pm 1} u_{i} = u_{i}$ , and distant generators commuting. (The abstract Temperley–Lieb generator presentation)

[F2]

The Jones normal forms use blocks $(u_{j_{1}} \dots u_{k_{1}}) \dots (u_{j_{r}} \dots u_{k_{r}})$ with $1 \leq k_{i} \leq j_{i} < n$ and both endpoint sequences strictly increasing. (Jones normal forms span and are counted by Catalan paths)

[F3]

The Jones normal forms are an $R$ -basis of $P_{n} (R, δ)$ . (The generator presentation is the planar diagram algebra over every commutative ring)

[F4]

The empty generator set gives $P_{0} = P_{1} = R$ . (The abstract Temperley–Lieb generator presentation)

Proof

technique · direct

1.1F2F3F4construct

For $n = 0, 1$ there are no generators, so the presentation is $R$ and the unique normal form is $1$ . For $n = 2$ , the admissible lists in [F2] give $1, u_{1}$ ; for $n = 3$ , they give $1, u_{1}, u_{2}, u_{2} u_{1}, u_{1} u_{2}$ . These words are bases by [F3], so the ranks are $1, 1, 2, 5$ ; the A-page path bijections identify those counts with $C_{0}, C_{1}, C_{2}, C_{3}$ .

2.1F1algebra∎

Set $a = u_{1}$ , $b = u_{2}$ in rank $3$ . From [F1], $a^{2} = δ a$ , $b^{2} = δ b$ , $a b a = a$ , and $b a b = b$ . The row of $a$ is $a, δ a, a b, a (a b) = a^{2} b = δ a b, a (b a) = a b a = a$ . The row of $b$ is $b, b a, δ b, b (a b) = b a b = b, b (b a) = b^{2} a = δ b a$ . The row of $a b$ is $a b, a b a = a, a b b = δ a b, a b a b = (a b a) b = a b, a b b a = a b^{2} a = δ a$ . The row of $b a$ is $b a, b a a = δ b a, b a b = b, b a a b = b a^{2} b = δ b, b a b a = (b a b) a = b a$ . These are the displayed entries and use no division by $δ$ .

Remarks

The rank is the number of basis diagrams, not the number of generators. The table remains valid at $δ = 0$ and over rings with zero divisors.

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

The five normal forms in the three-strand algebra

Example

In rank $3$ , the two generators are $u_{1}, u_{2}$ . The admissible Jones normal forms from Jones normal forms span and are counted by Catalan paths are

$1, u_{1}, u_{2}, u_{2} u_{1}, u_{1} u_{2} .$

The two adjacent triple relations reduce alternating length-three words:

$u_{1} u_{2} u_{1} = u_{1}, u_{2} u_{1} u_{2} = u_{2} .$

For example, $u_{2} u_{1} u_{2} u_{1} = (u_{2} u_{1} u_{2}) u_{1} = u_{2} u_{1}$ , while $u_{1} u_{2} u_{1} u_{2} = (u_{1} u_{2} u_{1}) u_{2} = u_{1} u_{2}$ . The five listed words are an $R$ -basis for every commutative coefficient ring and every $δ \in R$ .

Facts & Assumptions

Given: A commutative ring $R$ , $δ \in R$ , and rank $n = 3$ .

[F1]

The parameters of a Jones normal form satisfy $1 \leq k_{i} \leq j_{i} < 3$ , with both $j_{i}$ and $k_{i}$ strictly increasing from block to block. (Jones normal forms span and are counted by Catalan paths)

[F2]

$u_{1}^{2} = δ u_{1}$ , $u_{2}^{2} = δ u_{2}$ , $u_{1} u_{2} u_{1} = u_{1}$ , and $u_{2} u_{1} u_{2} = u_{2}$ . (The abstract Temperley–Lieb generator presentation)

[F3]

The Jones normal forms are linearly independent and span the presented algebra because the presentation map to the diagram algebra is an isomorphism. (The generator presentation is the planar diagram algebra over every commutative ring)

Proof

technique · direct

1.1F1construct

For an empty parameter list the word is $1$ . For one block, the pairs $(j, k) = (1, 1), (2, 1), (2, 2)$ give $u_{1}, u_{2} u_{1}, u_{2}$ . For two blocks, strict increase forces $(j_{1}, j_{2}) = (1, 2)$ and $(k_{1}, k_{2}) = (1, 2)$ , giving $u_{1} u_{2}$ ; three blocks are impossible because only $j = 1, 2$ are available. This lists exactly five forms.

2.1F2F3algebra∎

The reductions in the Example follow by one use of the corresponding triple relation in [F2]; the basis claim follows from [F3]. Since no step divides by $δ$ , these statements include $δ = 0$ and all other ring specializations.

CounterexampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

At loop value zero, a local generator is nilpotent but nonzero

Statement refuted

The assertion “if the loop value is $0$ , then every local generator is $0$ ” is false for $R = Z$ , loop value $δ = 0$ , and rank $n = 2$ .

Facts & Assumptions

Given: $R = Z$ , $δ = 0$ , and $n = 2$ .

[F1]

The rank-two diagram module is free on its loop-free noncrossing pairings; its identity and cap-cup diagram are distinct basis elements. (Temperley–Lieb planar diagram data and endomorphism modules, Stacking is associative and horizontal juxtaposition is strict monoidal)

[F2]

At loop value $0$ , the local relation is $d_{1}^{2} = 0$ , and $u_{1} \mapsto d_{1}$ under the presentation map. (The elementary diagram generators satisfy the local Temperley–Lieb relations)

[F3]

The presentation map $P_{2} (Z, 0) \to {T L}_{2} (Z, 0)$ is an isomorphism. (The generator presentation is the planar diagram algebra over every commutative ring)

Counterexample

technique · direct

1.1F1algebra

In cyclic boundary order the four points have exactly three pairings: $(1, 2) (3, 4)$ , $(1, 3) (2, 4)$ , and $(1, 4) (2, 3)$ . The middle pairing crosses, while the other two are the identity and $d_{1}$ . They are distinct members of the diagram indexing set, so $d_{1}$ is a nonzero standard basis vector in the free abelian module [F1].

2.1F2F3step 1.1∎

By [F2], $d_{1}^{2} = δ d_{1} = 0$ . Since the isomorphism [F3] sends $u_{1}$ to this nonzero $d_{1}$ , injectivity gives $u_{1} \neq 0$ and $u_{1}^{2} = 0$ . This is the claimed counterexample.

Remarks

This boundary case does not affect the diagram basis: the loop parameter changes multiplication coefficients, not which pairings index the free module.