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An involutive Yang–Baxter operator factors through the symmetric group

Statement

Let $C$ be a monoidal category, let $X \in C$ , and let $R$ be a Yang–Baxter operator on $X$ with $R^{2} = 1_{X \otimes X}$ . Then for every $n \geq 2$ the homomorphism $ρ_{n} ⁣ : B_{n} \to Aut (X^{\otimes n})$ of A Yang–Baxter operator gives braid-group representations factors through the canonical surjection $π_{n} ⁣ : B_{n} \to S_{n}$ of The braid group surjects onto the symmetric group: there is a unique homomorphism $ψ_{n} ⁣ : S_{n} \to Aut (X^{\otimes n})$ with $ρ_{n} = ψ_{n} \circ π_{n}$ .

Conversely, if $ρ_{n}$ factors through $π_{n}$ for some $n \geq 2$ , then $R_{i}^{2} = 1_{X^{\otimes n}}$ for every local operator $R_{i}$ , because $σ_{i}^{2}$ lies in the kernel of $π_{n}$ ; and since $ρ_{2} (σ_{1}) = R$ , if $ρ_{2}$ factors through $π_{2}$ then $R^{2} = 1_{X \otimes X}$ . Consequently an involutive Yang–Baxter operator is exactly one whose two-strand braid action factors through $S_{2}$ , and an involutive Yang–Baxter operator has its braid actions factoring through $S_{n}$ for every $n$ .

Facts & Assumptions

Given: a monoidal category $C$ , an object $X$ , a Yang–Baxter operator $R$ on $X$ , the homomorphisms $ρ_{n}$ of A Yang–Baxter operator gives braid-group representations with $ρ_{n} (σ_{i}) = R_{i}$ the local operator of $R$ at position $i$ , and the surjection $π_{n} ⁣ : B_{n} \to S_{n}$ .

[L1]

The local operators satisfy the Artin relations of [L3] (Local Yang–Baxter operators satisfy the Artin relations).

[L2]

The symmetric group $S_{n}$ has the Coxeter presentation with generators $s_{1}, \dots, s_{n - 1}$ and relations $s_{i}^{2} = 1$ , the braid relations and the distant-commutativity relations (The symmetric group has the Coxeter presentation).

[L3]

The braid group has the Artin presentation (The braid group by Artin presentation as used in A Yang–Baxter operator gives braid-group representations), and the canonical surjection $π_{n}$ sends $σ_{i}$ to $s_{i}$ (The braid group surjects onto the symmetric group).

[L4]

A generator assignment that respects the relators of a presented group extends uniquely to a homomorphism; precomposition with a surjection is injective on homomorphisms (Von Dyck's theorem: maps of generators that satisfy the relators extend uniquely from a presented group).

[L5]

The local operator at position $i$ is $1_{X}^{\otimes (i - 1)} \otimes R \otimes 1_{X}^{\otimes (n - i - 1)}$ in a strict model and its bracket-corrected conjugate in general (A Yang–Baxter operator gives braid-group representations, Local Yang–Baxter operators satisfy the Artin relations).

Proof

technique · direct

1.1L1L2L4L5givenconstruct

The direct implication. Assume $R^{2} = 1_{X \otimes X}$ . By [L5] the local operator satisfies $R_{i}^{2} = 1_{X^{\otimes n}}$ : in the strict model $R_{i}^{2}$ is the tensor product of identities with $R^{2}$ , and the bracket correction of [L5] is by a common canonical isomorphism, so it preserves the identity. By [L1] the operators $R_{i}$ also satisfy the braid relations and the distant-commutativity relations. Hence the assignment $s_{i} \mapsto R_{i}$ from the Coxeter generators of $S_{n}$ satisfies all relators of [L2], and [L4] gives a homomorphism $ψ_{n} ⁣ : S_{n} \to Aut (X^{\otimes n})$ with $ψ_{n} (s_{i}) = R_{i}$ .

1.2L2L3givenalgebra

The converse. Suppose $ρ_{n} = ψ \circ π_{n}$ for some homomorphism $ψ$ and some $n \geq 2$ . Then for every $i$ , $R_{i}^{2} = ρ_{n} (σ_{i})^{2} = ψ (π_{n} (σ_{i}))^{2} = ψ (π_{n} (σ_{i})^{2}) = ψ (π_{n} (σ_{i}^{2})) = ψ (1) = 1_{X^{\otimes n}}$ , using that $s_{i}^{2} = 1$ in $S_{n}$ by [L2] and that $π_{n}$ is a homomorphism [L3].

2.1L3L4step 1.1algebra

Factorization. Both $ρ_{n}$ and $ψ_{n} \circ π_{n}$ are homomorphisms $B_{n} \to Aut (X^{\otimes n})$ , and on every Artin generator $σ_{i}$ they agree: $ρ_{n} (σ_{i}) = R_{i} = ψ_{n} (s_{i}) = ψ_{n} (π_{n} (σ_{i}))$ by [L3] and step 1.1. By the uniqueness clause of [L4] applied to the Artin presentation, $ρ_{n} = ψ_{n} \circ π_{n}$ . Since $π_{n}$ is surjective, $ψ_{n}$ is unique with this property: two such homomorphisms agree on the image of $π_{n}$ , which is all of $S_{n}$ .

3.1L5step 1.2step 2.1algebra

The two-strand converse. For $n = 2$ the only local operator is $R_{1} = R$ , so step 1.2 gives $R^{2} = 1_{X \otimes X}$ as soon as $ρ_{2}$ factors through $π_{2}$ . Hence an involutive Yang–Baxter operator is exactly one whose two-strand braid action factors through $S_{2}$ : one direction is step 1.1 with $n = 2$ together with step 2.1, the other is the present step. For $n \geq 3$ , step 1.2 gives the weaker identity $(R^{2}) \otimes 1_{X^{\otimes (n - 2)}} = 1_{X^{\otimes n}}$ ; in a general monoidal category this whiskering does not by itself imply $R^{2} = 1$ , which is why the criterion is stated at $n = 2$ .

4.1step 1.1step 1.2step 2.1step 3.1∎

Conclusion. Step 1.1 with step 2.1 shows that an involutive Yang–Baxter operator has all its braid actions factoring through the symmetric groups, and steps 1.2 and 3.1 give the converse at the level of the two-strand action: $ρ_{2}$ factors through $π_{2}$ exactly when $R^{2} = 1$ . This proves the proposition. The argument uses only the Coxeter and Artin presentations and von Dyck's theorem, so no choice principle is used.

Depends on

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Sources

P. Etingof, S. Gelaki, D. Nikshych, and V. Ostrik, Tensor Categories (AMS Mathematical Surveys and Monographs 205), author's final version (standard reference, not scraped)