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Corollary 4.4.3.
We have uS(u) = S(u)u. This element is central in H.
Proof.
One directly sees that the element Su ∙ u is central: h ∙ S(u) ∙ u = S(u) ∙ u ∙ h. For h = u, we get
u ∙ Su ∙ u = Su ∙ u ∙ u and, since u is invertible, the claim.
✷
We recall from proposition 2.5.13 that a finite-dimensional module V over a Hopf algebra
with invertible antipode has a right dual V ∨ defined on V ∗ = Hom K (V, K) with action S(h) t
and a left dual ∨ V defined on the same K-vector space V ∗ with action by S −1 (h) t .
Proposition 4.4.4.
Let (H, R) be a quasi-triangular Hopf algebra and u the Drinfeld element. Then we have, for
any H-module V , an isomorphism of H-modules
j V : V ∨ → ∨ V
α ↦→ α(u.−)
where by α(u.−) we mean the linear form v ↦→ α(u.v) on V .
Proof.
The map j V is bijective, because the Drinfeld element u is invertible. We have to show that it
is a morphism of H-modules: we have for all a ∈ H and α ∈ H ∗ , v ∈ V
〈j V (a.α), v〉 = 〈a.α, u.v〉 = 〈α, S(a)u.v〉 = 〈α, S 2 (S −1 (a))u.v〉
(∗)
= 〈α, uS −1 (a)v〉 = 〈j V (α), S −1 (a).v〉 = 〈a.j V (α), v〉 .
✷
We comment on the relation to Radford’s formula:
Remarks 4.4.5.
1. We have for the coproduct of the Drinfeld element u = S(R (2) )R (1)
Δ(u) = R R 21 (u ⊗ u) .
2. One can show that g := u ∙ (Su) −1 is a group-like element of H and that
S 4 (h) = ghg −1 .
For a proof, we refer to [Montgomery, p. 181].
Now denote by a ∈ H and α ∈ H ∗ the distinguished group-like elements. Set
Then g = a −1 ˜α = ˜αa −1 .
˜α := (α ⊗ id H )(R) ∈ H .
We now turn to a subclass of quasi-triangular Hopf algebras for which the braiding obeys
additional constraints.
Definition 4.4.6
Let (H, R) be a quasi-triangular Hopf algebra.
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