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Temperley–Lieb Jones–Wenzl Projectors, Traces, and Quotients: Examples and Counterexamples

1 · Prerequisites

2 · Summary

These examples check the A-page conventions at the zero-loop, denominator, first-root, and low-level fusion boundaries. Each computation states the exact parameter and uses the A-page result it illustrates; none supplies a general theorem.

Examples

At loop value zero the normalized closure trace is undefined distinguishes the unnormalized diagram closure from the normalized trace when $δ = 0$ .
The two-strand Jones-Wenzl projector at loop value two computes the first nontrivial projector at $δ = 2$ .
The normalized Hecke generator requires inverting v plus one records where the normalized Hecke generators cease to be defined.
At the order-three root the trace degenerates on a semisimple block works out the semisimple-but-trace-degenerate $ℓ = 3$ boundary and its quotient.
Truncated fusion tables for ell four and ell five evaluates the truncated rule at $ℓ = 4$ and $ℓ = 5$ .

Companion: temperley-lieb-jones-wenzl-traces-and-quotients.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedOpen item page →

At loop value zero the normalized closure trace is undefined

Example

Set $R = C$ and $δ = 0$ . The unnormalized planar closure ${Tr}_{n}$ remains defined, but no normalized closure trace $τ_{n} = δ^{- n} {Tr}_{n}$ exists. Already at $n = 1$ , the identity has closure value ${Tr}_{1} (1_{1}) = 0$ , so a scalar normalization of this closure cannot satisfy $τ_{1} (1_{1}) = 1$ .

Facts & Assumptions

Given: The loop value $δ = 0$ and the one-strand identity diagram.

[F1]

The unnormalized closure assigns $δ^{c (D)}$ to each closed diagram and requires no unit hypothesis; normalization is defined only when $δ$ is a unit (Unnormalized diagram closure and partial trace).

[F2]

The closure of $1_{1}$ has one circle, so ${Tr}_{1} (1_{1}) = δ$ (Diagram closure trace is cyclic and satisfies the tower rules, Temperley–Lieb planar diagram data and endomorphism modules).

Verification

technique · direct calculation

1.1F1F2given

At $n = 1$ , closing the single vertical strand joins its top and bottom endpoints to make exactly one circle. By [F1] and [F2], ${Tr}_{1} (1_{1}) = δ^{1} = 0$ .

2.1F1step 1.1algebra∎

Any normalized trace must send the algebra identity to $1$ , but the proposed scalar multiple $δ^{- 1} {Tr}_{1}$ is undefined because $0$ has no multiplicative inverse in $C$ . Even replacing it by any scalar multiple $c {Tr}_{1}$ gives $c {Tr}_{1} (1_{1}) = c \cdot 0 = 0$ , never $1$ . Hence the unnormalized closure exists while the normalized closure does not.

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The two-strand Jones-Wenzl projector at loop value two

Example

Take $R = C$ , $δ = 2$ , so $[1] = 1$ , $[2] = 2$ , and $[3] = 3$ . The recursion gives

$p_{2} = 1_{2} - \frac{1}{2} U_{1} .$

It is a nonzero idempotent annihilated by $U_{1}$ on both sides, has partial trace $\frac{3}{2} p_{1}$ , and has unnormalized closure trace $3$ .

Facts & Assumptions

Given: The two-strand Temperley–Lieb algebra with $U_{1}^{2} = 2 U_{1}$ .

[F1]

The recursion is $p_{2} = 1_{2} - ([1] / [2]) U_{1}$ when $[2]$ is a unit (Jones–Wenzl recursion over unit quantum integers).

[F2]

For unit quantum integers the projector is idempotent, annihilates the generator, and has partial trace $([3] / [2]) p_{1}$ and closure trace $[3]$ (Jones–Wenzl projectors: annihilation, uniqueness, partial trace, and closure).

[F3]

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

Verification

technique · direct calculation

1.1F1F3givenalgebra

Substituting $δ = 2$ into [F3] yields $[2] = 2$ and $[3] = 2 \cdot 2 - 1 = 3$ . Thus [F1] gives $p_{2} = 1_{2} - \frac{1}{2} U_{1}$ ; the coefficient denominator is a unit in $C$ .

2.1F1step 1.1algebra

Directly, $p_{2}^{2} = 1 - U_{1} + \frac{1}{4} U_{1}^{2} = 1 - U_{1} + \frac{1}{2} U_{1} = p_{2}$ and $U_{1} p_{2} = p_{2} U_{1} = U_{1} - \frac{1}{2} (2 U_{1}) = 0$ . The identity and cup-cap diagrams are distinct basis vectors, so the coefficient $1$ on the identity makes $p_{2} \neq 0$ .

3.1F2F3step 1.1algebra∎

From [F2], ${ptr}_{2} (p_{2}) = ([3] / [2]) p_{1} = \frac{3}{2} p_{1}$ and ${Tr}_{2} (p_{2}) = [3] = 3$ . These agree with direct closure: ${Tr}_{2} (1_{2}) = 4$ and ${Tr}_{2} (U_{1}) = 2$ , so $4 - \frac{1}{2} (2) = 3$ .

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The normalized Hecke generator requires inverting v plus one

Example

Over $D = C [v, v^{- 1}, (v + 1)^{- 1}]$ , the Jones idempotent generator $e_{i} = (T_{i} + 1) / (v + 1)$ is defined. The specialization $v = - 1$ is not a ring map from $D$ to $C$ , because it sends the inverted element $v + 1$ to zero. Moreover, in the specialized two-strand Hecke algebra the element $T_{1} + 1$ is a nonzero square-zero element, so no scalar multiple of it is an idempotent generator.

Facts & Assumptions

Given: The rank-two Hecke quadratic relation and the normalized generator definition.

[F1]

$T_{i}^{2} = (v - 1) T_{i} + v$ and $1, T_{i}$ are distinct standard-basis elements (The generic type-A Hecke algebra).

[F2]

The element $e_{i} = (T_{i} + 1) / (v + 1)$ is defined in the localization where $v + 1$ is a unit (The Temperley-Lieb quotient and the Jones specialization).

Verification

technique · direct calculation

1.1F2givenalgebra

A unital map from $D$ to $C$ specializing $v$ to $- 1$ would have to send the unit $(v + 1)^{- 1}$ to an inverse of $(- 1) + 1 = 0$ . No element of $C$ is an inverse of zero, so the localization has no such specialization. Thus the formula in [F2] does not define $e_{i}$ at $v = - 1$ .

2.1F1F2step 1.1algebra∎

In the specialized Hecke algebra, put $x = T_{1} + 1$ . The quadratic relation at $v = - 1$ gives $T_{1}^{2} = - 2 T_{1} - 1$ , hence $x^{2} = T_{1}^{2} + 2 T_{1} + 1 = 0$ . The standard basis in [F1] shows $x \neq 0$ because $1$ and $T_{1}$ are linearly independent. If $c \in C$ and $c x$ were idempotent, then $(c x)^{2} = c x$ would force $0 = c x$ ; for $c \neq 0$ this contradicts $x \neq 0$ . Therefore a scalar multiple cannot supply a normalized idempotent at this specialization.

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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 $C$ .

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Truncated fusion tables for ell four and ell five

Example

Apply the truncated rule of The rank-one tilting quotient is the truncated fusion category to list the full fusion tables for $ℓ = 4$ and $ℓ = 5$ . For $ℓ = 4$ the simple labels are $0, 1, 2$ and

$1 \otimes 1 = 0 \oplus 2, 1 \otimes 2 = 1, 2 \otimes 2 = 0.$

For $ℓ = 5$ the simple labels are $0, 1, 2, 3$ and

$\begin{aligned} 1 \otimes 1 & = 0 \oplus 2, & 1 \otimes 2 & = 1 \oplus 3, & 1 \otimes 3 & = 2, \\ 2 \otimes 2 & = 0 \oplus 2, & 2 \otimes 3 & = 1, & 3 \otimes 3 & = 0. \end{aligned}$

In both tables $0$ is the tensor unit and the products with $0$ are $0 \otimes a = a$ .

Facts & Assumptions

Given: The truncated fusion rule and the retained simple-label ranges for each root order.

[F1]

At order $ℓ$ , the retained labels are $0, 1, \dots, ℓ - 2$ and $a \otimes b$ contains $c = ∣ a - b ∣, ∣ a - b ∣ + 2, \dots, \min (a + b, 2 ℓ - 4 - a - b)$ (The rank-one tilting quotient is the truncated fusion category, The root negligible quotient is semisimple with truncated fusion).

[F2]

The conversion theorem identifies the pivotal/ribbon rank-one labels with the same retained Temperley–Lieb fusion labels (The rank-one tilting quotient matches the Temperley-Lieb ribbon convention).

Verification

technique · direct calculation

1.1F1givenalgebra

For $ℓ = 4$ , the cutoff is $2 ℓ - 4 = 4$ . Substituting $(a, b) = (1, 1), (1, 2), (2, 2)$ into [F1] gives upper bounds $\min (2, 2) = 2$ , $\min (3, 1) = 1$ , and $\min (4, 0) = 0$ , respectively; the lower bounds and parity give exactly ${0, 2}$ , ${1}$ , and ${0}$ . The unit rule supplies products with $0$ .

2.1F1F2step 1.1algebra∎

For $ℓ = 5$ , the cutoff is $6$ . The substitutions $(1, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 3)$ give intervals and parity sets ${0, 2}$ , ${1, 3}$ , ${2}$ , ${0, 2}$ , ${1}$ , and ${0}$ , respectively. These exhaust unordered pairs among $0, 1, 2, 3$ ; commutativity of the fusion rule supplies the reversed pairs. The conversion in [F2] preserves the labels and cutoff.

Temperley–Lieb Jones–Wenzl Projectors, Traces, and Quotients: Examples and Counterexamples

1 · Prerequisites

2 · Summary

Examples

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

At loop value zero the normalized closure trace is undefined

Example

Facts & Assumptions

Verification

The two-strand Jones-Wenzl projector at loop value two

Example

Facts & Assumptions

Verification

The normalized Hecke generator requires inverting v plus one

Example

Facts & Assumptions

Verification

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

Example

Facts & Assumptions

Verification

Truncated fusion tables for ell four and ell five

Example

Facts & Assumptions

Verification

Sources