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Temperley–Lieb Jones–Wenzl Projectors, Traces, and Quotients

1 · Prerequisites

2 · Summary

This page uses the supplier-first Jones–Wenzl, root-path and full rank-one tilting constructions on temperley-lieb-root-of-unity-and-modular-representation-theory. Their canonical carriers have a single home there; the full root-of-unity, fusion, trace-radical, centralizer and ribbon conclusions are retained here by reuse. The local result proves the exact Jones specialization of the Hecke Markov trace.

Results

The trace-radical interface is Negligible morphisms for a pivotal trace and Partial-trace closure of negligible morphisms. Companion: temperley-lieb-jones-wenzl-traces-and-quotients-examples.

3 · Logical flowchart

4 · Definitions, theorems and proofs

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

The specialized Hecke Markov trace descends to the diagram quotient

Statement

Let ℓ≥3, q=eπi/ℓ, v=q2, δ=q+q−1 and z0=−1/(v+1). Specialize the type-A Hecke tower at v,z=z0. For each n, let Jn be the two-sided ideal generated by the six-term parabolic sums

Ei=1+Ti+Ti+1+TiTi+1+Ti+1Ti+TiTi+1Ti(1≤i≤n−2).

The assignment Ti↦qUi−1 defines a surjective tower map ϕn:H(n)→TLn(C,δ) that kills Jn. The specialized Ocneanu Markov trace therefore vanishes on Jn and descends to H(n)/Jn. Under the induced isomorphism with the diagram algebra presentation, the descended trace is the normalized closure trace τn=δ−nTr⁡n.

Facts & Assumptions

Given: ℓ≥3, q=eπi/ℓ, the Hecke parameter v=q2, the tower trace parameter z0=−1/(q2+1), and the planar convention Ui2=δUi, UiUi+1Ui=Ui.

[F1]

The generic type-A Hecke generators obey Ti2=(v−1)Ti+v, the braid relation, and distant commutation (The generic type-A Hecke algebra).

[F2]

tr⁡ is cyclic, restricts along the tower, and satisfies tr⁡n+1(xTn)=ztr⁡n(x); its recursive proof on the free tower basis works after any scalar specialization (The Markov trace on the type-A Hecke tower, The Ocneanu Markov trace exists and is unique).

[F3]

For n≥0, the diagram closure has τn(1n)=1 and τn+1(x⊗1)=τn(x). For n≥1, it also has τn+1((x⊗1)Un)=δ−1τn(x) because δ≠0 (Unnormalized diagram closure and partial trace, Diagram closure trace is cyclic and satisfies the tower rules).

[F4]

The diagram algebra has the presentation on generators Ui with relations Ui2=δUi, UiUi±1Ui=Ui, and distant commutation; its generator presentation is isomorphic to the planar diagram algebra over every commutative ring (The abstract Temperley–Lieb generator presentation, Elementary cup-cap diagrams generate every finite planar pairing endomorphism diagram, The generator presentation is the planar diagram algebra over every commutative ring).

[F5]

The Hecke tower is free over its preceding rank, with H(n+1)=H(n)⊕H(n)TnH(n) as an H(n)-bimodule, and the standard Hecke basis persists after scalar specialization (The Hecke tower is free over the previous level, The standard basis of the generic type-A Hecke algebra).

Proof

1.1F1F4givenalgebra

Since v+1=q2+1=qδ≠0, the assignment Ti↦qUi−1 is defined. Its quadratic relation follows from (qUi−1−v)(qUi)=q2(Ui−δ)Ui=0, using v=q2 and Ui2=δUi. For adjacent generators, expanding both braid words gives difference q3(UiUi+1Ui−Ui+1UiUi+1)−q2δ(Ui−Ui+1)+q(Ui−Ui+1)=0, since q2δ=q3+q. Distant commutation is inherited, so the assignment defines a tower homomorphism; it is surjective by [F4].

2.1F1step 1.1algebra

Put Ei=∑w∈⟨si,si+1⟩Tw, the six-term expression in the Statement. Expanding with the Hecke quadratic and braid relations gives (Ti+1)(Ti+1+1)(Ti+1)=Ei+v(Ti+1). Under Tj+1=qUj, the left side maps to q3UiUi+1Ui=q3Ui and the second term maps to q3Ui, so ϕn(Ei)=0. Since Jn is the two-sided ideal generated by the Ei, ϕn(Jn)=0 and ϕn factors through H(n)/Jn.

2.2F2F3step 1.1algebra

The pullback τ~n:=τn∘ϕn is cyclic and has τ~n(1)=1 by [F3]. For n≥1 and x∈H(n), its image in TLn+1 is ϕn(x)⊗1, and Tn maps to qUn−1, so τ~n+1(xTn)=(q/δ−1)τn(ϕn(x))=−τ~n(x)/(qδ)=−τ~n(x)/(q2+1)=z0τ~n(x). The tower restriction rule follows from [F3]; hence τ~ satisfies the specialized Markov rules.

3.1F1F4step 1.1step 2.1algebra

The presentation [F4] gives ψn:TLn→H(n)/Jn by Ui↦q−1(Ti+1). The Hecke quadratic relation gives Ui2=δUi for these images, while step 2.1 in the quotient gives (Ti+1)(Ti+1+1)(Ti+1)=q2(Ti+1) and hence UiUi+1Ui=Ui; the other adjacent relation follows by interchanging indices and distant commutation is inherited. Thus ψn is well defined. Its composites with the map induced by ϕn fix each generator: Ui↦q−1(qUi−1+1)=Ui and Ti↦q(q−1(Ti+1))−1=Ti. Therefore H(n)/Jn≅TLn, proving the quotient isomorphism.

4.1F2F5step 2.2step 3.1inductionalgebra∎

The decomposition in [F5] writes every element of H(n+1) as a sum of an element of H(n) and terms xTny with x,y∈H(n). For any Markov trace, restriction determines the first term; cyclicity and (M4) force tr⁡n+1(xTny)=ztr⁡n(xy) on each remaining term. Induction from H(1)=C therefore shows at most one specialized trace can satisfy (M1)--(M4), even when v,z have been specialized: this uses only the free decomposition [F5], not division by a parameter. By step 2.2 the pullback τ~ is such a trace, so it is the specialized Ocneanu trace. Since ϕn(Jn)=0, the trace vanishes on all of the two-sided ideal Jn. The quotient isomorphism of step 3.1 identifies the descended functional with the normalized closure trace τn.

5 · Examples, counterexamples and false statements

None yet.

Sources