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A non-invertible solution of the Yang–Baxter equation does not represent the braid group

Statement refuted

The claim refuted is: every solution of the Yang–Baxter equation on an object gives a representation of the braid group. Take $C = {V e c t}_{k}$ , $X = k$ one-dimensional, and $R = 0$ , the zero endomorphism of $X \otimes X ≅ k$ . The cubic relation holds because both sides are zero, so the cubic equation alone does not force invertibility. But $R$ is not invertible, so for $n \geq 2$ the local operators $R_{i} = 0$ are not automorphisms of $X^{\otimes n}$ , and the assignment $σ_{i} \mapsto R_{i}$ cannot define a homomorphism $B_{n} \to Aut (X^{\otimes n})$ : a Yang–Baxter operator in the sense of Yang–Baxter operators on an object is required to be invertible, and a homomorphism from a group lands in a group. The zero operator therefore witnesses, for every braid group with at least two strands, that invertibility in the definition of a Yang–Baxter operator is not redundant.

Facts & Assumptions

Given: $n \geq 2$ , the field $k$ , the one-dimensional vector space $X = k$ , the object $X \otimes X ≅ k$ , the zero endomorphism $R = 0$ of $X \otimes X$ , and the defining cubic equation of Yang–Baxter operators on an object read using the canonical associativity identifications in ${V e c t}_{k}$ .

[L1]

A Yang–Baxter operator on $X$ is an invertible morphism $R ⁣ : X \otimes X \to X \otimes X$ satisfying $(R \otimes 1_{X}) (1_{X} \otimes R) (R \otimes 1_{X}) = (1_{X} \otimes R) (R \otimes 1_{X}) (1_{X} \otimes R)$ (Yang–Baxter operators on an object).

[F1]

In ${V e c t}_{k}$ the tensor product $X \otimes X$ is again one-dimensional, hence nonzero, and the zero endomorphism of a nonzero vector space is not invertible: from $0 \circ f = id$ one gets $id = 0$ , which fails on a nonzero vector.

Counterexample

The witness is the pair $(X, R) = (k, 0)$ .

1.1L1givenalgebra

The cubic equation holds for $R = 0$ . Every factor in both composites $(R \otimes 1_{X}) (1_{X} \otimes R) (R \otimes 1_{X})$ and $(1_{X} \otimes R) (R \otimes 1_{X}) (1_{X} \otimes R)$ is a tensor product containing the zero morphism $R$ ; a composite with a zero factor is zero, so both sides are the zero endomorphism of $X \otimes X \otimes X$ and the cubic relation holds.

1.2F1algebra

The morphism $R$ is not invertible. By [F1] the object $X \otimes X$ is nonzero and $0$ has no inverse as an endomorphism of it. The local operator $R_{i} = 1^{\otimes (i - 1)} \otimes R \otimes 1^{\otimes (n - i - 1)}$ is the tensor product of $R = 0$ with identities, hence is the zero endomorphism of the nonzero object $X^{\otimes n}$ , so it too has no inverse.

2.1step 1.2given

No braid-group representation arises. The assignment $σ_{i} \mapsto R_{i}$ does not even land in $Aut (X^{\otimes n})$ , because $R_{i} = 0$ is not invertible by step 1.2. Moreover there is no homomorphism $ρ ⁣ : B_{n} \to End (X^{\otimes n})$ with $ρ (σ_{i}) = R_{i}$ : from $σ_{i} σ_{i}^{- 1} = 1$ one would get $R_{i} ρ (σ_{i}^{- 1}) = id$ , contradicting $R_{i} ρ (σ_{i}^{- 1}) = 0$ . So the non-invertible solution $R = 0$ of the cubic equation gives no representation of the braid group.

3.1step 1.1step 2.1∎

Conclusion. The zero solution satisfies the Yang–Baxter equation but is not a Yang–Baxter operator in the sense of [L1], and it produces no braid-group action; the claim that the cubic equation alone suffices is therefore refuted, and invertibility is a genuine part of the definition.

Remarks

The group $B_{1}$ has no braid generator, so its trivial action exists independently of the chosen cubic-equation solution. The counterexample concerns $n \geq 2$ .

Depends on

Yang–Baxter operators on an object

Used by

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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)