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The first four finite Temperley–Lieb algebras

Example

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

nbasisrank0{1}11{1}12{1,u1}23{1,u1,u2,u1u2,u2u1}5

The rank column is Cn for these four indices. In rank 3, writing a=u1 and b=u2, the product table is

⋅1ababba11ababbaaaδaabδababbbaδbbδbaababaδababδababaδbabδbba

Facts & Assumptions

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

[F1]

Pn(R,δ) is generated by u1,…,un−1 with ui2=δui, uiui±1ui=ui, and distant generators commuting. (The abstract Temperley–Lieb generator presentation)

[F2]

The Jones normal forms use blocks (uj1⋯uk1)⋯(ujr⋯ukr) with 1≤ki≤ji<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 Pn(R,δ). (The generator presentation is the planar diagram algebra over every commutative ring)

[F4]

The empty generator set gives P0=P1=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,u1; for n=3, they give 1,u1,u2,u2u1,u1u2. These words are bases by [F3], so the ranks are 1,1,2,5; the A-page path bijections identify those counts with C0,C1,C2,C3.

2.1F1algebra∎

Set a=u1, b=u2 in rank 3. From [F1], a2=δa, b2=δb, aba=a, and bab=b. The row of a is a, δa, ab, a(ab)=a2b=δab, a(ba)=aba=a. The row of b is b, ba, δb, b(ab)=bab=b, b(ba)=b2a=δba. The row of ab is ab, aba=a, abb=δab, abab=(aba)b=ab, abba=ab2a=δa. The row of ba is ba, baa=δba, bab=b, baab=ba2b=δb, baba=(bab)a=ba. 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.

Depends on

Used by

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