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1. Then the universal R-matrix obeys the following equation in H ⊗3 :
R 12 R 13 R 23 = R 23 R 13 R 12
and we have
(ɛ ⊗ id H )(R) = 1 = (id H ⊗ ɛ)(R) .
2. If, moreover, H has an invertible antipode, then
(S ⊗ id H )(R) = R −1 = (id H ⊗ S −1 )(R)
and
(S ⊗ S)(R) = R .
Proof.
1. We calculate, using the defining properties of the R-matrix:
R 12 R 13 R 23 = R 12 (Δ ⊗ id)(R) [equation (10)]
= (Δ opp ⊗ id)(R) ∙ R 12 [equation (9)]
= (τ H,H ⊗ id)(Δ ⊗ id)(R) ∙ R 12 [Defn. of Δ opp ]
= (τ H,H ⊗ id)(R 13 R 23 ) ∙ R 12 [equation (11)]
= R 23 R 13 R 12 .
We use (ɛ⊗id)◦Δ = id, the tensoriality of the R-matrix and the fact that ɛ is a morphism
of algebras to get in H ⊗2
R = (ɛ ⊗ id ⊗ id)(Δ ⊗ id)(R) (10)
= (ɛ ⊗ id ⊗ id)(R 12 R 23 ) = (ɛ ⊗ id)(R) ⊗ 1 ∙ ɛ(1)R .
Since R is invertible and since ɛ(1) = 1, we get (ɛ ⊗ id)(R) = 1. The other equality is
derived in complete analogy from equation (11).
2. Using the definition of the antipode, we have for all x ∈ H
μ ◦ (S ⊗ id)Δ(x) = ɛ(x)1 .
Tensoring with the identity on H and using the relation just derived, we find
Thus, using equation (10)
(μ ⊗ id) ◦ (S ⊗ id ⊗ id)(Δ ⊗ id)(R) = (1ɛ ⊗ id)R = 1 ⊗ 1 .
1 ⊗ 1 = (μ ⊗ id)(S ⊗ id ⊗ id)(R 13 R 23 ) = (S ⊗ id)(R) ∙ S(1)R .
Using S(1) = 1, we get
(S ⊗ id)(R) = R −1 .
In the quasi-triangular Hopf algebra (H, μ, Δ opp , S −1 , τ H,H (R)), this relation reads
which amounts to
(S −1 ⊗ id)(τ H,H (R)) = τ H,H (R) −1
(id H ⊗ S −1 )(R) = R −1 .
Finally, we use the relations just derived to find
(S ⊗ S)(R) = (id ⊗ S)(S ⊗ id)(R)
= (id ⊗ S)(R −1 )
= (id ⊗ S)(id H ⊗ S −1 )(R) = R
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