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Counit normalization and the Yang–Baxter identity for a quasitriangular structure

Statement

Let $H$ be a Hopf algebra over a field $k$ and let $R \in H \otimes H$ be invertible and satisfy the quasitriangular axioms of Leg notation and quasitriangular Hopf algebras. Then $(ε \otimes id) (R) = 1_{H} = (id \otimes ε) (R),$ and $R_{12} R_{13} R_{23} = R_{23} R_{13} R_{12} in H^{\otimes 3} .$

Facts & Assumptions

Given: A field $k$ , a Hopf algebra $H$ over $k$ , its counit $ε$ , and an invertible quasitriangular element $R \in H \otimes H$ .

[F1]

Leg notation and quasitriangular Hopf algebras: $R$ is invertible, $R Δ (h) = Δ^{o p} (h) R$ , $(Δ \otimes id) R = R_{13} R_{23}$ , and $α^{- 1} (id \otimes Δ) R = R_{13} R_{12}$ .

[F2]

Hopf algebras and antipodes: $ε$ is a counit, so $(ε \otimes id) Δ = id = (id \otimes ε) Δ$ and $ε (1_{H}) = 1_{k}$ .

[F3]

Scalars, tensor powers, the empty tensor, opposite algebras and finite sums: each algebraic tensor is a finite sum of elementary tensors and those tensors span the tensor product.

Proof

technique · direct

1.1givenF1F2F3algebra

Write $R = \sum_{i} a_{i} \otimes b_{i}$ and set $x = (ε \otimes id) (R)$ . Apply $ε \otimes id \otimes id$ to $(Δ \otimes id) (R) = R_{13} R_{23}$ . The left side is $R$ by the left counit identity in [F2]. Expanding the right side as $\sum_{i, j} a_{i} \otimes a_{j} \otimes b_{i} b_{j}$ gives $(1_{H} \otimes x) R$ . Thus $R = (1_{H} \otimes x) R$ ; multiplying on the right by $R^{- 1}$ gives $1_{H} \otimes x = 1_{H} \otimes 1_{H}$ , and applying $ε \otimes id$ gives $x = 1_{H}$ .

1.2givenF1F2F3algebra

Set $y = (id \otimes ε) (R)$ . Apply $id \otimes id \otimes ε$ to $α^{- 1} (id \otimes Δ) (R) = R_{13} R_{12}$ . The left side is $R$ by the right counit identity in [F2]. Expanding $R_{13} R_{12}$ gives $\sum_{i, j} a_{i} a_{j} \otimes b_{j} \otimes b_{i}$ , whose image under the counit map is $(y \otimes 1_{H}) R$ . Hence $R = (y \otimes 1_{H}) R$ , so invertibility gives $y \otimes 1_{H} = 1_{H} \otimes 1_{H}$ and applying $id \otimes ε$ gives $y = 1_{H}$ .

2.1givenF1F3algebra∎

Using the first hexagon, $R_{12} R_{13} R_{23} = R_{12} (Δ \otimes id) (R)$ . Apply $R Δ (a_{i}) = Δ^{o p} (a_{i}) R$ term by term to obtain $R_{12} (Δ \otimes id) (R) = (Δ^{o p} \otimes id) (R) R_{12}$ . The map that flips the first two vector-space factors sends $R_{13} R_{23} = \sum_{i, j} a_{i} \otimes a_{j} \otimes b_{i} b_{j}$ to $\sum_{i, j} a_{j} \otimes a_{i} \otimes b_{i} b_{j} = R_{23} R_{13}$ after reindexing; this is a direct tensor calculation, not an assertion that the flip preserves multiplication. Therefore $(Δ^{o p} \otimes id) (R) = R_{23} R_{13}$ , and the desired QYBE follows.

Depends on

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Sources

Kalle Kytölä, Introduction to Hopf Algebras and Representations (Helsinki lecture notes, Spring 2011) (standard reference, not scraped)
Pavel Etingof and Mykola Semenyakin, A Brief Introduction to Quantum Groups, arXiv:2106.05252v3 (2024) (standard reference, not scraped)