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Local Yang–Baxter operators satisfy the Artin relations

Statement

Let C be a monoidal category, let X∈C, let R be a Yang–Baxter operator on X, let n≥2, and let R1,…,Rn−1 be the local operators of Local Yang–Baxter operators on tensor powers, so that Ri∈Aut⁡C(X⊗n) is the local operator acting on the i-th and (i+1)-st tensor factors. Then

RiRj=RjRiwhenever ∣i−j∣>1,

RiRi+1Ri=Ri+1RiRi+1for 1≤i≤n−2.

In the non-strict model the identities are those of the bracket-corrected local operators defined in Local Yang–Baxter operators on tensor powers.

Facts & Assumptions

Given: a monoidal category C, an object X, a Yang–Baxter operator R on X, an integer n≥2, and the local operators R1,…,Rn−1.

[L1]

In a strict model the local operator is Ri=1X⊗(i−1)⊗R⊗1X⊗(n−i−1), and in general it is the bracket-corrected conjugate of that word; every Ri is invertible (Local Yang–Baxter operators on tensor powers).

[L2]

The Yang–Baxter operator satisfies the cubic equation (R⊗1X)(1X⊗R)(R⊗1X)=(1X⊗R)(R⊗1X)(1X⊗R) (Yang–Baxter operators on an object).

Proof

technique · direct
1.1L1givenalgebra

Far commutativity in the strict model. Suppose first that C is strict and ∣i−j∣>1. The endomorphisms Ri and Rj are tensor products of the identity with the single factor R inserted at positions {i,i+1} respectively {j,j+1}; these supports are disjoint because ∣i−j∣>1. Tensoring the two words and using functoriality of the tensor product, both RiRj and RjRi are the same tensor product of identities with two copies of R at positions {i,i+1} and {j,j+1} (in the two possible orders of composition); hence RiRj=RjRi.

1.2L1L2givenalgebra

The adjacent relation in the strict model. Suppose C is strict and 1≤i≤n−2. Every tensor factor outside positions i,i+1,i+2 carries only identities in each of the words RiRi+1Ri and Ri+1RiRi+1, so both sides are 1X⊗(i−1) tensored with an endomorphism of the three middle factors tensored with 1X⊗(n−i−2). On those three middle factors the two sides are (R⊗1X)(1X⊗R)(R⊗1X) and (1X⊗R)(R⊗1X)(1X⊗R), which are equal by the cubic equation [L2]. Since tensoring equal morphisms with identities gives equal morphisms, RiRi+1Ri=Ri+1RiRi+1.

2.1L1step 1.1step 1.2algebra

The non-strict model. Let E:C→C′ be the strong monoidal equivalence used in [L1], put X′=E(X), and let Jn:X′⊗n→E(X⊗n) be its iterated tensor constraint. Transport R as R′=J2−1E(R)J2. Naturality and associativity coherence of the constraints give E(Ri)=JnRistrJn−1: apply E to the bracket-corrected composite of [L1] and use the strong monoidal constraint at each tensor product. Thus E sends each proposed Artin identity to the corresponding identity of steps 1.1 and 1.2, conjugated by the single isomorphism Jn. Since an equivalence is faithful, the two identities hold in C. Every comparison here is a morphism in C′.

3.1step 1.1step 1.2step 2.1∎

Conclusion. Steps 1.1 and 1.2 prove the two families of identities in the strict model, and step 2.1 transports them to the bracket-corrected operators of the non-strict model. This proves the lemma. The argument is a finite computation in the tensor product and uses no choice principle.

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