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The flip operator gives the permutation representation

Example

Take $C = {V e c t}_{k}$ , $X = k^{n}$ for $n \geq 0$ , with standard basis $e_{1}, \dots, e_{n}$ , and $R (e_{i} \otimes e_{j}) = e_{j} \otimes e_{i}$ . Then $R^{2} = 1$ and $R$ is a Yang–Baxter operator on $X$ : both sides of the cubic relation act on $e_{i} \otimes e_{j} \otimes e_{l}$ by the permutation of the three basis vectors reversing the order. By An involutive Yang–Baxter operator factors through the symmetric group the action $ρ_{m} ⁣ : B_{m} \to Aut (X^{\otimes m})$ of A Yang–Baxter operator gives braid-group representations factors through $S_{m}$ , and on the basis $e_{i_{1}} \otimes \dots \otimes e_{i_{m}}$ the generator $σ_{j}$ acts by exchanging the entries in positions $j$ and $j + 1$ . Thus $ρ_{m}$ is the place-permutation representation of $B_{m}$ through $S_{m}$ : for $m = 2$ and $n \geq 2$ , $R$ swaps $e_{1} \otimes e_{2}$ with $e_{2} \otimes e_{1}$ and fixes $e_{1} \otimes e_{1}$ and $e_{2} \otimes e_{2}$ .

For $m = 0, 1$ , use the trivial action on $X^{\otimes m}$ , with $X^{\otimes 0} = k$ ; it is also the place-permutation action of the trivial group $S_{m}$ . If $n = 0$ and $m \geq 1$ , the tensor power is the zero space and its unique automorphism is its identity, so the same conclusion holds.

Facts & Assumptions

Given: the field $k$ , the vector space $X = k^{n}$ with basis $e_{1}, \dots, e_{n}$ , and the linear flip $R (e_{i} \otimes e_{j}) = e_{j} \otimes e_{i}$ on $X \otimes X$ .

[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), and it gives homomorphisms $ρ_{m} ⁣ : B_{m} \to Aut (X^{\otimes m})$ with $ρ_{m} (σ_{j})$ the local operator at position $j$ (A Yang–Baxter operator gives braid-group representations).

[L2]

If $R^{2} = 1_{X \otimes X}$ , then for every $m \geq 2$ the homomorphism $ρ_{m}$ factors through $π_{m} ⁣ : B_{m} \to S_{m}$ , and $ψ_{m} (s_{j}) = ρ_{m} (σ_{j})$ (An involutive Yang–Baxter operator factors through the symmetric group).

Verification

1.1L1givenalgebra

The flip is an involutive Yang–Baxter operator. On the basis, $R^{2} (e_{i} \otimes e_{j}) = R (e_{j} \otimes e_{i}) = e_{i} \otimes e_{j}$ , so $R^{2} = 1_{X \otimes X}$ . For the cubic relation, the left-hand composite applied to $e_{i} \otimes e_{j} \otimes e_{l}$ reverses the order of the three factors: $R \otimes 1$ exchanges the first two, then $1 \otimes R$ exchanges the last two, then $R \otimes 1$ the first two, giving $e_{l} \otimes e_{j} \otimes e_{i}$ ; the right-hand composite produces the same by the mirror computation. Since the pure tensors span $X^{\otimes 3}$ , the cubic equation holds and $R$ is a Yang–Baxter operator on $X$ .

2.1L1step 1.1

The action on pure tensors. For $m \geq 2$ , by [L1] the local operator at position $j$ is $1^{\otimes (j - 1)} \otimes R \otimes 1^{\otimes (m - j - 1)}$ , which on the basis vector $e_{i_{1}} \otimes \dots \otimes e_{i_{m}}$ exchanges the entries in positions $j$ and $j + 1$ ; thus each $ρ_{m} (σ_{j})$ is the corresponding place permutation.

3.1L2step 1.1step 2.1

Factorization and identification of the representation. By [L2] and $R^{2} = 1$ the action $ρ_{m}$ factors as $ψ_{m} \circ π_{m}$ with $ψ_{m} (s_{j}) = ρ_{m} (σ_{j})$ ; by step 2.1 the value $ψ_{m} (s_{j})$ is the place permutation exchanging positions $j$ and $j + 1$ . Since the $s_{j}$ generate $S_{m}$ , $ψ_{m}$ is the place-permutation representation of $S_{m}$ on $X^{\otimes m}$ , and $ρ_{m}$ is that representation composed with $π_{m}$ .

4.1step 1.1step 3.1given

The two-strand case. For $m = 2$ and $n \geq 2$ the operator $R$ swaps $e_{1} \otimes e_{2}$ with $e_{2} \otimes e_{1}$ and fixes $e_{i} \otimes e_{i}$ for $i = 1, 2$ ; this is the place-permutation representation of $S_{2}$ , in agreement with step 3.1.

5.1step 1.1step 3.1step 4.1∎

Conclusion and small strand counts. For $m = 0, 1$ , the braid and symmetric groups are trivial and their actions send the sole element to the identity, the place permutation on $X^{\otimes m}$ . If $n = 0$ and $m \geq 1$ , the tensor power is zero and its unique endomorphism is its identity. The flip operator is an involutive Yang–Baxter operator, and its braid actions are exactly the place-permutation representations of the symmetric groups, pulled back along the canonical surjections $B_{m} \to S_{m}$ . All computations are finite and linear 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)