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Temperley–Lieb Semisimplicity and Generic Structure: Examples

1 · Prerequisites

2 · Summary

These examples keep algebra semisimplicity, trace nondegeneracy, and projector recursion as separate calculations. At order three, rank two is semisimple even though the normalized Jones trace is degenerate. At zero loop value, rank three and rank four have different semisimplicity behavior. At loop value two, the rank-two Jones–Wenzl recursion and its normalized trace can be checked directly.

Each example uses only the definitions and proved complex results on temperley-lieb-semisimplicity-and-generic-structure. The examples are not theorem suppliers. The arbitrary-characteristic projector criterion is proved separately on the A page; none of these three complex examples uses it.

Examples

At q squared of order three, TL2 is semisimple with a degenerate Jones trace computes the rank-two semisimple algebra and its degenerate trace form at $n = ℓ - 1$ .
At zero loop value, rank three and rank four have opposite semisimplicity evaluates the odd/even split at zero loop value in ranks three and four.
The rank-two Jones–Wenzl idempotent at loop value two verifies the projector and trace at $q = 1$ .

Companion: temperley-lieb-semisimplicity-and-generic-structure.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-6-luna)Open item page →

At q squared of order three, TL2 is semisimple with a degenerate Jones trace

Example

Take $q = e^{i π / 3}$ , so $q^{2}$ has order $ℓ = 3$ , $δ = q + q^{- 1} = 1$ , and $n = 2 = ℓ - 1$ . The algebra ${T L}_{2} (C, 1)$ is semisimple, but its normalized closure trace form is degenerate.

Diagram

In the left panel the solid top cap and bottom cup are the basis diagram $u = d_{1}$ . The right panel adds the exterior closure arcs in dashed blue; each joins top point $i$ to bottom point $i$ . Thus the four arcs form one closed component, so $c (u) = 1$ in the closure-trace definition. The drawing records the matching graph only; the algebra and trace calculations below prove the claims independently.

\begin{tikzpicture}[x=1cm,y=1cm,line width=0.8pt,baseline=(current bounding box.center)]
  \fill (0,1) circle (1.5pt); \fill (1,1) circle (1.5pt);
  \fill (0,-1) circle (1.5pt); \fill (1,-1) circle (1.5pt);
  \draw (0,1) to[out=90,in=90] (1,1);
  \draw (0,-1) to[out=-90,in=-90] (1,-1);
  \node at (0.5,0) {$u$};
  \node at (0.5,-1.65) {basis diagram};

  \fill (3,1) circle (1.5pt); \fill (4,1) circle (1.5pt);
  \fill (3,-1) circle (1.5pt); \fill (4,-1) circle (1.5pt);
  \draw (3,1) to[out=90,in=90] (4,1);
  \draw (3,-1) to[out=-90,in=-90] (4,-1);
  \draw[dashed,blue] (3,1) to[out=180,in=180] (3,-1);
  \draw[dashed,blue] (4,1) to[out=0,in=0] (4,-1);
  \node at (3.5,-1.65) {closed graph: one component};
\end{tikzpicture}

Facts & Assumptions

Given: The rank-two diagram algebra with generator $u = d_{1}$ and loop parameter $δ = 1$ .

[F1]

The diagram basis of ${T L}_{2}$ is ${1, u}$ ; stacking gives $u^{2} = δ u$ , and the closure trace is normalized by $τ_{2} (1) = 1$ (Temperley–Lieb planar diagram data and endomorphism modules, Stacking is associative and horizontal juxtaposition is strict monoidal, The elementary diagram generators satisfy the local Temperley–Lieb relations, Normalized Temperley–Lieb trace from diagram closure).

[F2]

The closure trace satisfies $τ_{2} (u) = δ^{- 1} τ_{1} (1)$ (The normalized closure trace is cyclic and satisfies the Markov rules).

Verification

Given: $q = e^{i π / 3}$ , $δ = 1$ , and the basis ${1, u}$ .

1.1F1givenalgebra

The elements $u$ and $1 - u$ are nonzero idempotents: $u^{2} = u$ , $(1 - u)^{2} = 1 - u$ , and $u (1 - u) = (1 - u) u = 0$ by [F1]. They sum to $1$ and are linearly independent, so the two-dimensional algebra is $C u \oplus C (1 - u) ≅ C \times C$ , hence semisimple.

2.1F1F2step 1.1algebra∎

By [F2], $τ_{2} (u) = 1$ , while $τ_{2} (1) = 1$ . The trace-pairing matrix in the basis ${1, u}$ is therefore $(\begin{array}{cc} τ_{2} (1) & τ_{2} (u) \\ τ_{2} (u) & τ_{2} (u^{2}) \end{array}) = (\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}),$ whose nonzero vector $1 - u$ pairs to zero with both basis vectors. Thus the trace form is degenerate although the algebra is semisimple.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-6-luna)Open item page →

At zero loop value, rank three and rank four have opposite semisimplicity

Example

Take $q = i$ , so $δ = 0$ and $q^{2}$ has order two. Then ${T L}_{3} (C, 0)$ is semisimple, whereas ${T L}_{4} (C, 0)$ is not.

Facts & Assumptions

Given: The quantum integers at $q = i$ and the link-state Gram determinants.

[F1]

For $q = i$ , $[2 k]_{q} = 0$ and $[2 k + 1]_{q} = (- 1)^{k}$ ; the determinants and semisimplicity test are given by The Temperley–Lieb link-state Gram determinant and Temperley–Lieb cell forms detect semisimplicity over the complex numbers.

Verification

Given: $q = i$ , $ℓ = 2$ , and ranks $n = 3, 4$ .

1.1F1givenalgebra

For $n = 3$ , the allowed arc counts are $p = 0, 1$ . The empty product gives $\det G_{3, 0} = 1$ , while the determinant formula gives $\det G_{3, 1} = [3]_{q} / [1]_{q} = - 1$ . Both link-state forms are nondegenerate, so Temperley–Lieb cell forms detect semisimplicity over the complex numbers implies ${T L}_{3} (C, 0)$ is semisimple.

2.1F1givenstep 1.1algebra∎

For $n = 4$ , choose $p = 1$ . Then $\det G_{4, 1} = [4]_{q} / [1]_{q} = 0$ , so one cell form is degenerate. The reverse implication of Temperley–Lieb cell forms detect semisimplicity over the complex numbers gives that ${T L}_{4} (C, 0)$ is not semisimple.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-6-luna)Open item page →

The rank-two Jones–Wenzl idempotent at loop value two

Example

Let $q = 1$ , so $δ = 2$ and $[r]_{q} = r$ . In rank two the recursion gives $p_{2} = 1 - \frac{1}{2} d_{1}$ , an idempotent annihilated by $d_{1}$ , with normalized closure trace $τ_{2} (p_{2}) = 3 / 4$ .

Facts & Assumptions

Given: $q = 1$ , $δ = 2$ , $[r]_{q} = r$ , and the rank-two generator $d_{1}$ .

[F1]

$d_{1}^{2} = 2 d_{1}$ , and $τ_{2} (1) = 1$ , $τ_{2} (d_{1}) = 1 / 2$ (The elementary diagram generators satisfy the local Temperley–Lieb relations, Normalized Temperley–Lieb trace from diagram closure, The normalized closure trace is cyclic and satisfies the Markov rules).

[F2]

The recursion is $p_{2} = 1 - \frac{[1]}{[2]} d_{1}$ , and its trace formula is $τ_{2} (p_{2}) = [3] / 2^{2}$ (Jones–Wenzl recursion in the nonvanishing-denominator range, The Jones–Wenzl recursion gives annihilating idempotents and their trace).

Verification

Given: The stated parameter and rank-two algebra.

1.1F1F2givenalgebra

Since $[1] = 1$ and $[2] = 2$ , the recursion gives $p_{2} = 1 - \frac{1}{2} d_{1}$ . By [F1], $p_{2}^{2} = 1 - d_{1} + \frac{1}{4} d_{1}^{2} = 1 - \frac{1}{2} d_{1} = p_{2}, d_{1} p_{2} = p_{2} d_{1} = 0.$ The identity coefficient is $1$ , so $p_{2} \neq 0$ .

2.1F1F2step 1.1∎

The trace formula in [F2] gives $τ_{2} (p_{2}) = [3] / 2^{2} = 3 / 4$ . This calculation uses the closure trace normalization $τ_{2} (1) = 1$ , not the unnormalized categorical trace.

Temperley–Lieb Semisimplicity and Generic Structure: Examples

1 · Prerequisites

2 · Summary

Examples

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

At q squared of order three, TL2 is semisimple with a degenerate Jones trace

Example

Diagram

Facts & Assumptions

Verification

At zero loop value, rank three and rank four have opposite semisimplicity

Example

Facts & Assumptions

Verification

The rank-two Jones–Wenzl idempotent at loop value two

Example

Facts & Assumptions

Verification

Sources