Temperley-Lieb Modules

From exponential state sums to polynomial-time matrix traces
Demo 3 · Building on Demos 1 & 2
Recap from Demos 1 & 2

In Demo 1 we computed the Kauffman bracket by brute force: resolve every crossing as A or B, count loops, sum 2n weighted terms. In Demo 2 we showed that braids give a different encoding of the same knots, and the same state sum on the braid closure produces the same bracket. But both approaches cost 2n.

This demo introduces the Temperley-Lieb algebra — the algebraic structure that lurks behind the smoothing operation. By representing TL generators as matrices and braid generators as matrix products, we can compute the bracket via traces of matrix products: a polynomial-time operation. The key verification: the matrix trace agrees with the state-sum oracle for every test braid.

Contents
1. The Temperley-Lieb algebra 2. TL diagrams and the cup-cap picture 3. Standard modules: W₁ and W₃ 4. Explicit matrices for TL₃ 5. The braid representation 6. Inverses and the key cancellation 7. Full bracket via module decomposition 8. What the C code does 9. Interactive explorer

1. The Temperley-Lieb algebra

Behind every smoothing operation in the Kauffman bracket lies an algebraic structure. When we resolve a crossing as an A-smoothing, we are applying a specific operation to the strands: cupping two adjacent strands together (joining them with a cap on top and a cup on the bottom). The Temperley-Lieb algebra TLn is the algebra generated by these cup-cap operations.

For n strands, TLn has generators U₁, U₂, …, Un−1, where Ui cups together strands i and i+1. These generators satisfy three fundamental relations:

Ui² = δ · Ui     (idempotent up to δ)
Ui Uj = Uj Ui   when |ij| ≥ 2     (far-apart generators commute)
Ui Ui±1 Ui = Ui     (the Jones relation)

Here δ = −A² − A⁻² is the same loop value from the bracket. The first relation says: if you cup strands i and i+1, then cup them again, you get the same configuration but with an extra loop — and each loop is worth δ. The third relation is deeper: it says that cupping, then doing a neighbouring cup, then cupping again, just gives the original cup back. This "absorption" property is what makes the TL algebra finite-dimensional despite being generated by repeated compositions.

2. TL diagrams and the cup-cap picture

Each element of TLn can be drawn as a diagram: n points on top, n points on bottom, connected by non-crossing curves. The identity has n vertical through-lines. The generator Ui has a cap connecting points i and i+1 on top, a cup connecting them on the bottom, and through-lines elsewhere.

TL₃ diagrams. Left: identity (3 through-lines). Center: U₁ (cap on 1-2, through-line on 3). Right: U₂ (through-line on 1, cap on 2-3).

Multiplication is composition: to multiply diagram A by diagram B, stack A on top of B and connect the bottom points of A to the top points of B. Any closed loops that form in the middle are removed and contribute a factor of δ each. The remaining connected curves form the product diagram.

The number of through-lines (strands that go from top to bottom without being cupped) is an important invariant of each diagram. The identity has n through-lines; Ui has n−2 through-lines. The through-line count can only decrease (or stay the same) under multiplication — it never increases.

3. Standard modules: W₁ and W₃

To turn the abstract algebra into concrete matrices, we need a representation: a way to assign a matrix to each generator such that the TL relations are satisfied. The natural representations of TLn are its standard modules Wk, indexed by the number of "defects" (through-lines in the basis diagrams).

For TL₃, the defect count must have the same parity as n = 3 (odd). So the modules are:

Basis: the single diagram with 3 through-lines (the identity). Since Ui reduces through-lines by 2, and you cannot go from 3 to 1 defect within W₃, both U₁ and U₂ act as zero. This is the trivial module. Dimension = 1.

Basis: diagrams with exactly 1 through-line and 1 cap. There are two such diagrams:
ea: cap on positions 1-2, through-line at position 3.
eb: cap on positions 2-3, through-line at position 1.
This is the interesting module where the TL algebra acts non-trivially. Dimension = 2.

The dimensions follow from the ballot-number formula: dim(Wk) = C(n, (nk)/2) − C(n, (nk)/2 − 1). For W₁: C(3,1) − C(3,0) = 3 − 1 = 2. For W₃: C(3,0) − C(3,−1) = 1 − 0 = 1.

4. Explicit matrices for TL₃

Working out the action of U₁ and U₂ on the basis {ea, eb} of W₁ by composing diagrams gives:

U₁ = [[δ, 1], [0, 0]]      U₂ = [[0, 0], [1, δ]]

These matrices encode how each generator transforms the basis diagrams. For instance, U₁ acting on ea (cap on 1-2, through at 3) gives δ · ea — the cap absorbs itself with a loop factor. U₁ acting on eb (cap on 2-3, through at 1) gives ea — the cup-cap at positions 1-2 reconnects the through-line.

The C code verifies all four TL relations for these matrices: U₁² = δU₁, U₂² = δU₂, U₁U₂U₁ = U₁, and U₂U₁U₂ = U₂. All four pass.

5. The braid representation

The Kauffman bracket resolves each crossing as A · (identity) + A⁻¹ · (cup-cap). In algebraic terms, the braid generator σi maps to:

ρ(σi) = A⁻¹ · I + A · Ui
ρ(σi⁻¹) = A · I + A⁻¹ · Ui

On W₁ these are 2×2 matrices with polynomial entries. On W₃ (where Ui = 0), the positive generator becomes the scalar A⁻¹ and the negative generator becomes A.

For a braid word σi₁e₁ · σi₂e₂ · … · σiₖeₖ, the representation is simply the matrix product ρ(σi₁e₁) · ρ(σi₂e₂) · … · ρ(σiₖeₖ). This is a sequence of k matrix multiplications — linear in the number of crossings, not exponential.

6. Inverses and the key cancellation

A natural worry: does ρ(σi) have an inverse? If σi and σi⁻¹ are inverses in the braid group, their images should multiply to the identity. Let us check:

ρ(σi) · ρ(σi⁻¹) = (A⁻¹I + AU)(AI + A⁻¹U)
= I + A⁻²U + A²U + U²
= I + (A² + A⁻² + δ) · U
= I + 0 · U = I

The miraculous cancellation: A² + A⁻² + δ = A² + A⁻² + (−A² − A⁻²) = 0. This is not a coincidence — it is why δ has the value it does. The loop value δ = −A² − A⁻² is precisely the value that makes the braid representation well-defined (invertible). The C code verifies g₁ · g₁⁻¹ = I and g₂ · g₂⁻¹ = I.

The code also verifies the braid relation: g₁g₂g₁ = g₂g₁g₂ (the Yang-Baxter equation). This confirms that ρ is a genuine representation of the braid group, not just a formal construction.

7. Full bracket via module decomposition

The bracket of a 3-strand braid closure is recovered by combining the traces from both modules W₁ and W₃, weighted by their quantum dimensions:

⟨closure(b)⟩ = 1 · trW₁(ρ(b)) + (A⁻⁴ + A⁴) · trW₃(ρ(b))

The coefficient 1 for W₁ and (A⁻⁴ + A⁴) for W₃ are the quantum dimensions of these modules. On W₃, since all Ui = 0, the representation reduces to scalars: each positive crossing contributes A⁻¹ and each negative crossing contributes A. So the W₃ trace is simply A−(#pos) + (#neg).

The C code verifies this decomposition for five different 3-strand braids: the figure-eight, the trefoil, the mirror trefoil, the identity (3-component unlink), and a single crossing. In every case, the module formula matches the state-sum oracle from Demo 2 exactly.

This is the central result of Demo 3: the exponential state sum and the polynomial-time matrix trace compute the same invariant. The matrix approach is not an approximation — it is an exact algebraic reformulation.

8. What the C code does

Constructs the 2×2 matrices U₁ and U₂ for TL₃ and verifies all four defining relations: U₁² = δU₁, U₂² = δU₂, U₁U₂U₁ = U₁, U₂U₁U₂ = U₂.

Constructs gi = A⁻¹I + AUi and gi⁻¹ = AI + A⁻¹Ui, then verifies g₁·g₁⁻¹ = I and g₂·g₂⁻¹ = I.

Verifies the braid relation g₁g₂g₁ = g₂g₁g₂ and the inverse version g₁⁻¹g₂⁻¹g₁⁻¹ = g₂⁻¹g₁⁻¹g₂⁻¹.

For each test braid, computes trW₁(product) and trW₃(product), combines them with quantum dimension weights, and compares against the state-sum oracle. All five tests pass.

The interactive tool below implements the same 2×2 polynomial matrix arithmetic and module decomposition. The state-sum oracle is also computed for verification.


Interactive Explorer

Watch the matrix product build step by step, verify against the state-sum oracle
Figure-eight: σ₁σ₂⁻¹σ₁σ₂⁻¹ on 3 strands. Writhe = 0.
Trace of product (W₁)
W₁ (dim 2)
coefficient: 1
trace =
contribution =
+
W₃ (dim 1)
coefficient: A⁻⁴ + A⁴
trace =
contribution =
Bracket from modules
State-sum oracle (Demo 2)
TL algebra braid word polynomial verified