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We take the trace Tr V0 ⊗V 0
over the relation in lemma 4.3.2 and use the cyclicity of the trace to
get the following equation in End (V ⊗N ).
C(μ) ∙ C(ν) = Tr V0 ⊗V 0
T 0 (μ)T 0 (ν)
= Tr V0 ⊗V 0
R 00 (λ) −1 T 0 (ν)T 0 (μ)R 00 (λ)
= C(ν)C(μ)
✷
Example 4.3.4.
We consider the case of two possible states for each bond. One represents the state of a bond
by assigning to it a direction, denoted by an arrow. A famous model is then the XXX model
or six vertex model. In this case, one assigns Boltzmann weight zero to all vertices, except for
those with two ingoing and two outgoing vertices. These are the following six configurations,
hence the name of the model:
Since now V is two-dimensional, the R-matrix is a 4 × 4-matrix
⎛
⎞
1 0 0 0
R(q, λ) = ⎜ 0 λ 1 − qλ 0
⎟
⎝ 0 1 − q −1 λ λ 0 ⎠
0 0 0 1
This model is integrable with the relation
λ − μ + ν + λμν = (q + q −1 )λν .
4.4 The square of the antipode of a quasi-triangular Hopf algebra
We have seen in corollary 2.5.9 that for a cocommutative Hopf algebra, the square of the
antipode obeys S 2 = id H . For quasi-triangular Hopf algebras, a weaker statement still holds. We
assume throughout this chapter that the antipode S is invertible. This condition is automatically
fulfilled by theorem 3.1.13.3 for finite-dimensional Hopf algebras. The reader might wish to check
for which propositions the existence of a skew-antipode as in remark 2.5.8 is sufficient.
Proposition 4.4.1.
Let (H, R) be a quasi-triangular Hopf algebra. Then the element
is invertible with inverse
u := S(R (2) ) ∙ R (1) ∈ H
u −1 := S −1 (R (2) ) ∙ R (1) ,
where R −1 = R (1) ⊗ R (2) . For all h ∈ H, we have
S 2 (h) = uhu −1 = (Su) −1 hS(u)
so that the square of the antipode is an inner automorphism.
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