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A diagonal Yang–Baxter operator on graded vector spaces

Example

Let $G$ be an abelian group, let $χ ⁣ : G \times G \to k^{\times}$ be any function into the units of a field $k$ , and let $X = ⨁_{g \in G} k e_{g}$ be the free $G$ -graded vector space, a direct sum with one basis vector per group element. Define a linear map

$R ⁣ : X \otimes X ⟶ X \otimes X, R (e_{g} \otimes e_{h}) = χ (g, h) e_{h} \otimes e_{g} .$

Then $R$ is invertible, with $R^{- 1} (e_{h} \otimes e_{g}) = χ (g, h)^{- 1} e_{g} \otimes e_{h}$ , and $R$ is a Yang–Baxter operator on $X$ : both sides of the cubic relation send $e_{g} \otimes e_{h} \otimes e_{l}$ to $χ (g, h) χ (g, l) χ (h, l) e_{l} \otimes e_{h} \otimes e_{g}$ , and the two scalar products agree because $k^{\times}$ is commutative. Hence A Yang–Baxter operator gives braid-group representations gives representations $ρ_{n} ⁣ : B_{n} \to Aut (X^{\otimes n})$ with $ρ_{n} (σ_{i})$ acting by scalar-weighted permutations of the graded basis. If $χ (g, h) χ (h, g) = 1$ for all $g, h$ , then $R^{2} = 1$ and the actions factor through the symmetric groups; if $χ (h, h)^{2} \neq 1$ for some $h$ with $k e_{h} \neq 0$ , then $R^{2} (e_{h} \otimes e_{h}) = χ (h, h)^{2} e_{h} \otimes e_{h}$ shows that $R$ is not involutive.

Facts & Assumptions

Given: an abelian group $G$ , a field $k$ , a function $χ ⁣ : G \times G \to k^{\times}$ , the $G$ -graded vector space $X = ⨁_{g \in G} k e_{g}$ , and the linear map $R$ of the statement.

[L1]

A Yang–Baxter operator on $X$ is an invertible $R ⁣ : X \otimes X \to X \otimes X$ satisfying the cubic equation of Yang–Baxter operators on an object using the canonical associativity identifications in ${V e c t}_{k}$ .

[L2]

A Yang–Baxter operator on $X$ yields homomorphisms $ρ_{n} ⁣ : B_{n} \to Aut (X^{\otimes n})$ with $ρ_{n} (σ_{i})$ the local operator of $R$ (A Yang–Baxter operator gives braid-group representations).

[F1]

The symmetric group has the Coxeter presentation with generators $s_{i}$ and relators $s_{i}^{2} = 1$ , the braid relations and distant commutativity (The symmetric group has the Coxeter presentation), and von Dyck's theorem extends a relator-respecting generator assignment uniquely (Von Dyck's theorem: maps of generators that satisfy the relators extend uniquely from a presented group).

Verification

1.1L1givenalgebra

Invertibility. Define $R^{- 1}$ on the graded basis by $R^{- 1} (e_{h} \otimes e_{g}) : = χ (g, h)^{- 1} e_{g} \otimes e_{h}$ and extend linearly. Then $R (R^{- 1} (e_{h} \otimes e_{g})) = χ (g, h)^{- 1} R (e_{g} \otimes e_{h}) = e_{h} \otimes e_{g}$ and $R^{- 1} (R (e_{g} \otimes e_{h})) = χ (g, h) R^{- 1} (e_{h} \otimes e_{g}) = e_{g} \otimes e_{h}$ , so $R$ and $R^{- 1}$ are mutually inverse linear bijections.

1.2L1givenalgebra

The cubic relation. Applying the left-hand composite $(R \otimes 1_{X}) (1_{X} \otimes R) (R \otimes 1_{X})$ to $e_{g} \otimes e_{h} \otimes e_{l}$ , from right to left, produces first $χ (g, h) e_{h} \otimes e_{g} \otimes e_{l}$ , then $χ (g, l) e_{h} \otimes e_{l} \otimes e_{g}$ , then $χ (h, l) e_{l} \otimes e_{h} \otimes e_{g}$ . Applying the right-hand composite $(1_{X} \otimes R) (R \otimes 1_{X}) (1_{X} \otimes R)$ produces first $χ (h, l) e_{g} \otimes e_{l} \otimes e_{h}$ , then $χ (g, l) e_{l} \otimes e_{g} \otimes e_{h}$ , then $χ (g, h) e_{l} \otimes e_{h} \otimes e_{g}$ . Both sides therefore act on $e_{g} \otimes e_{h} \otimes e_{l}$ by multiplication by $χ (g, h) χ (g, l) χ (h, l)$ , and these scalars are equal in the commutative group $k^{\times}$ ; since the pure tensors of graded basis vectors span $X^{\otimes 3}$ , the cubic equation holds.

2.1L1L2step 1.1step 1.2

The braid-group actions. By steps 1.1 and 1.2 the map $R$ is an invertible solution of the cubic equation, hence a Yang–Baxter operator on $X$ in the sense of [L1], and [L2] gives the homomorphisms $ρ_{n}$ with $ρ_{n} (σ_{i})$ acting on the graded basis by exchanging the $i$ -th and $(i + 1)$ -st entries with the scalar $χ$ attached to the two exchanged degrees.

2.2L1L2F1step 1.2algebra

Involutivity criterion. On a pure tensor, $R^{2} (e_{g} \otimes e_{h}) = R (χ (g, h) e_{h} \otimes e_{g}) = χ (g, h) χ (h, g) e_{g} \otimes e_{h}$ , so $R^{2} = 1_{X \otimes X}$ if and only if $χ (g, h) χ (h, g) = 1$ for every pair $(g, h)$ with $k e_{g}, k e_{h} \neq 0$ . If this holds, then each local operator satisfies $R_{i}^{2} = 1$ (it is a tensor product of identities with $R^{2}$ ) and the $R_{i}$ satisfy the braid relations, so the assignment $s_{i} \mapsto R_{i}$ respects the Coxeter relators of $S_{n}$ and [F1] gives homomorphisms $ψ_{n} ⁣ : S_{n} \to Aut (X^{\otimes n})$ with $ρ_{n} = ψ_{n} \circ π_{n}$ ; the actions factor through the symmetric groups. If instead $χ (h, h)^{2} \neq 1$ for some $h$ with $k e_{h} \neq 0$ , then $R^{2} (e_{h} \otimes e_{h}) = χ (h, h)^{2} e_{h} \otimes e_{h} \neq e_{h} \otimes e_{h}$ , so $R^{2} \neq 1_{X \otimes X}$ and $R$ is not involutive.

3.1step 1.1step 1.2step 2.1step 2.2∎

Conclusion. The diagonal map $R (e_{g} \otimes e_{h}) = χ (g, h) e_{h} \otimes e_{g}$ is always an invertible Yang–Baxter operator on the free graded vector space, its square is the diagonal map with coefficients $χ (g, h) χ (h, g)$ , and it is involutive exactly when those coefficients are $1$ . All computations are on a spanning set of pure tensors 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)