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The Hecke generators satisfy the Artin relations and are units

Statement

Let H(n) be the Hecke tower over Λ=Z[v±1,z] of The Markov trace on the type-A Hecke tower. Then: (1) the elements T1,…,Tn−1 are units of H(n) with Ti−1=v−1Ti+(v−1−1); (2) the assignment σi↦Ti descends to a group homomorphism πn:Bn⟶H(n)×,πn(σi)=Ti, where H(n)× is the group of units of H(n); (3) for every Artin word β=σi1ε1⋯σikεk one has πn(β)=Ti1ε1⋯Tikεk, and πn+1(ιnBr(β))=ιn(πn(β)) for all β∈Bn under the standard inclusion of braid groups Bn→Bn+1.

Facts & Assumptions

Given: The Hecke tower over Λ=Z[v±1,z] and an integer n≥1. No choice principle is used.

[F1]

H(n) is the Λ-algebra with generators T1,…,Tn−1 and the relations TiTi+1Ti=Ti+1TiTi+1 for 1≤i≤n−2 and TiTj=TjTi for ∣i−j∣>1 (The generic type-A Hecke algebra, The Markov trace on the type-A Hecke tower).

[F2]

Each generator Ti is a unit of H(n) with Ti−1=v−1Ti+(v−1−1) (The Markov trace of an inverse Hecke generator).

[F3]

Bn=⟨σ1,…,σn−1∣Artin relators⟩ with the braid and far-commutation relations, B0,B1 trivial (The braid group by Artin presentation).

[F4]

Von Dyck: a function from the generators of a presented group to a group G that sends every defining relator to the identity extends to a unique group homomorphism (Von Dyck's theorem: maps of generators that satisfy the relators extend uniquely from a presented group); a group homomorphism satisfies φ(x1⋯xk)=φ(x1)⋯φ(xk) and φ(x−1)=φ(x)−1 (Monoid homomorphism and group homomorphism).

Proof

1.1F2F3F4

The assignment kills the relators. Define u(σi):=Ti∈H(n)×, using [F2] to regard each Ti as an element of the unit group. At the braid relator σiσi+1σi=σi+1σiσi+1 both sides are sent to the equal elements TiTi+1Ti=Ti+1TiTi+1 of [F1]; at a far-commutation relator both sides are sent to TiTj=TjTi by [F1]. Hence every defining relator of [F3] is sent to the identity and [F4] applies, giving a unique group homomorphism πn:Bn→H(n)× with πn(σi)=Ti. For n=1 the domain B1 is trivial and its unique homomorphism into H(1)× sends the identity to the unit of H(1).

2.1F3F4step 1.1∎

Words and compatibility. For an Artin word β=σi1ε1⋯σikεk, [F4] gives πn(β)=πn(σi1)ε1⋯πn(σik)εk=Ti1ε1⋯Tikεk, since negative exponents are the inverses from step 1.1; this also shows that the value does not depend on the chosen word, being the value of the homomorphism πn at the element β. The standard inclusion Bn→Bn+1 sends each Artin generator σi with i≤n−1 to the generator with the same name (The braid group by Artin presentation), and ιn:H(n)→H(n+1) sends Ti to Ti (The Markov trace on the type-A Hecke tower); hence πn+1(ιnBr(β)) and ιn(πn(β)) are both the product of the Tiεi computed in H(n+1), and they agree.

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