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A non-involutive one-dimensional Yang–Baxter operator

Example

Over $k = Q$ , let $X = Q e$ be one-dimensional and put $R (e \otimes e) = 2 e \otimes e$ . Then $R$ is invertible, with inverse multiplication by $\frac{1}{2}$ , and satisfies the Yang–Baxter equation: both sides act on the single basis vector $e \otimes e \otimes e$ of $X^{\otimes 3}$ by multiplication by $8$ . The representation of A Yang–Baxter operator gives braid-group representations on $X^{\otimes n}$ is therefore one-dimensional, with $ρ_{n} (σ_{i}) = 2$ for every $i$ , so $ρ_{n} (β) = 2^{w (β)}$ with $w$ the exponent sum (Exponent sum and writhe of a braid). Since $R^{2} = 4$ , the operator is not involutive: by An involutive Yang–Baxter operator factors through the symmetric group the two-strand action does not factor through $S_{2}$ , and in fact no $ρ_{n}$ with $n \geq 2$ factors through $S_{n}$ , because a factorization would force $ρ_{n} (σ_{i})^{2} = 1$ while $ρ_{n} (σ_{i})^{2} = 4$ . The character $B_{n} \to Q^{\times}$ , $β \mapsto 2^{w (β)}$ , is a genuine non-involutive braid character, and for $n \geq 2$ it fails symmetric-group factorization; its one-strand action is trivial. It separates the two Markov stabilizations since $w$ shifts by $\pm 1$ there.

Facts & Assumptions

Given: the field $k = Q$ , the one-dimensional vector space $X = Q e$ , and the endomorphism $R$ of $X \otimes X$ with $R (e \otimes e) = 2 e \otimes e$ .

[L1]

A Yang–Baxter operator on $X$ is an invertible $R ⁣ : X \otimes X \to X \otimes X$ satisfying the cubic equation (Yang–Baxter operators on an object); it yields homomorphisms $ρ_{n} ⁣ : B_{n} \to Aut (X^{\otimes n})$ sending $σ_{i}$ to the local operator at position $i$ (A Yang–Baxter operator gives braid-group representations).

[L2]

If $ρ_{2}$ factors through $π_{2} ⁣ : B_{2} \to S_{2}$ , then $R^{2} = 1_{X \otimes X}$ ; more generally, if $ρ_{n}$ factors through $π_{n}$ for some $n \geq 2$ , then $R_{i}^{2} = 1_{X^{\otimes n}}$ for every $i$ (An involutive Yang–Baxter operator factors through the symmetric group).

[F1]

The exponent sum $w ⁣ : B_{n} \to Z$ is additive on Artin words and $w (σ_{i}^{\pm 1}) = \pm 1$ (Exponent sum and writhe of a braid); in particular $ρ_{n} (β) = 2^{w (β)}$ for a one-dimensional representation with $ρ_{n} (σ_{i}) = 2$ .

Verification

1.1L1givenalgebra

Invertibility and the cubic equation. Multiplication by $2$ on the one-dimensional space $X \otimes X ≅ Q$ is invertible with inverse multiplication by $\frac{1}{2}$ . Both the left- and the right-hand side of the cubic equation are, on the single basis vector $e \otimes e \otimes e$ of the one-dimensional space $X^{\otimes 3}$ , multiplication by $2 \cdot 2 \cdot 2 = 8$ ; hence they agree, and $R$ is a Yang–Baxter operator on $X$ in the sense of [L1].

2.1L1F1step 1.1

The one-dimensional braid character. Since $X^{\otimes n}$ is one-dimensional, each local operator $R_{i}$ is multiplication by $2$ , so [L1] gives $ρ_{n} (σ_{i}) = 2$ for every $i$ , and additivity of the exponent sum [F1] on Artin words gives $ρ_{n} (β) = 2^{w (β)}$ for every $β \in B_{n}$ .

3.1L2F1step 2.1algebra

Non-involutivity and failure of factorization. From $R^{2} (e \otimes e) = 4 e \otimes e$ we get $R^{2} = 4 \cdot 1_{X \otimes X} \neq 1_{X \otimes X}$ . By [L2] the two-strand action $ρ_{2}$ does not factor through $π_{2}$ . For any $n \geq 2$ , if $ρ_{n}$ factored through $π_{n}$ then [L2] would give $R_{i}^{2} = 1_{X^{\otimes n}}$ ; but $R_{i}^{2}$ is multiplication by $4$ on the one-dimensional space $X^{\otimes n}$ , so $R_{i}^{2} \neq 1$ . Hence no $ρ_{n}$ with $n \geq 2$ factors through $S_{n}$ ; for $n = 1$ the braid and symmetric groups are trivial and the action factors through them, and the character of step 2.1 is a genuine non-involutive braid character. Finally, since $w (ι_{n} (β) σ_{n}^{\pm 1}) = w (β) \pm 1$ , the values $2^{w}$ on the two stabilizations of a braid differ by a factor $4$ , so the character separates them.

4.1step 1.1step 2.1step 3.1∎

Conclusion. The one-dimensional operator $R = 2 \cdot id$ is an invertible, non-involutive solution of the Yang–Baxter equation, and its braid actions are the one-dimensional characters $2^{w}$ that, for $n \geq 2$ , do not factor through the symmetric groups. All computations are finite and use no choice principle.

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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)