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The canonical R-matrix for the finite double of the group of order two

Example

Let $G = {e, s}$ with $s^{2} = e$ , and let $p_{x}$ ( $x \in G$ ) be the characteristic functions on $G$ . Let $u_{g}$ denote the group basis symbols. Use the finite-group double convention with basis $p_{x} # u_{g}$ and product $(p_{x} # u_{g}) (p_{y} # u_{h}) = δ_{x, g y g^{- 1}} p_{x} # u_{g h}$ . For this abelian group the product reduces to $δ_{x, y} p_{x} # u_{g h}$ . The evaluation pairing $⟨ p_{y}, u_{x} ⟩ = δ_{y, x}$ makes $(p_{x})_{x \in G}$ and $(u_{x})_{x \in G}$ dual bases. Its coproduct, counit, and antipode are $Δ (p_{x} # u_{g}) = \sum_{a b = x} (p_{a} # u_{g}) \otimes (p_{b} # u_{g}), ε (p_{x} # u_{g}) = δ_{x, e}, S (p_{x} # u_{g}) = p_{x} # u_{g} .$ Set $q_{g} : = \sum_{z \in G} p_{z} # u_{g}$ and $R = \sum_{x \in G} (p_{x} # u_{e}) \otimes q_{x} .$ Then $D (k G)$ is a four-dimensional Hopf algebra and $R$ is its canonical quasitriangular structure.

Facts & Assumptions

Given: The two-element group $G = {e, s}$ with multiplication $s^{2} = e$ , the function basis $p_{e}, p_{s}$ , the group basis $u_{e}, u_{s}$ , and the displayed group-double structure maps.

[F1]

Hopf algebras and antipodes: a Hopf algebra has a unital associative multiplication, algebra-map coproduct and counit satisfying coassociativity/counitality, and an antipode satisfying both contractions.

[F2]

Leg notation and quasitriangular Hopf algebras: $R$ is quasitriangular when it is invertible and satisfies the intertwining identity and both leg hexagons.

Verification

Given: The finite set $G = {e, s}$ with its group law and the four basis vectors $p_{x} # u_{g}$ .

1.1givenF1constructalgebra

The group law is addition of exponents modulo two, so it is associative. For basis vectors, either association of a triple product is $δ_{x, y} δ_{y, z} p_{x} # u_{g h ℓ}$ ; bilinearity therefore makes the displayed multiplication associative. The element $1_{D} = \sum_{x} p_{x} # u_{e}$ is a unit because $1_{D} (p_{y} # u_{h}) = p_{y} # u_{h} = (p_{y} # u_{h}) 1_{D}$ . Thus the vector space with the stated product is a unital algebra with basis of size four.

2.1step 1.1F1algebra

The formula for $Δ$ is coassociative because both $(Δ \otimes id) Δ (p_{x} # u_{g})$ and $(id \otimes Δ) Δ (p_{x} # u_{g})$ sum once over all triples $a, b, c \in G$ with $a b c = x$ . The counit contractions retain only $a = e$ or $b = e$ , respectively, and return $p_{x} # u_{g}$ . The coproduct preserves multiplication: expanding $Δ (p_{x} # u_{g}) Δ (p_{y} # u_{h})$ leaves only terms with equal first and second function indices, so it is zero if $x \neq y$ and is $\sum_{a b = x} (p_{a} # u_{g h}) \otimes (p_{b} # u_{g h})$ if $x = y$ , which equals $Δ ((p_{x} # u_{g}) (p_{y} # u_{h}))$ . The counit is multiplicative by the same orthogonal-idempotent multiplication, and $Δ (1_{D}) = 1_{D} \otimes 1_{D}$ , $ε (1_{D}) = 1$ .

2.2givenstep 1.1F2algebra

The elements $q_{g}$ satisfy $q_{e} q_{g} = q_{g} = q_{g} q_{e}$ and $q_{g} q_{h} = q_{g h}$ . Also $q_{e} = 1_{D}$ . Therefore $R^{2} = \sum_{x, y} (p_{x} # u_{e}) (p_{y} # u_{e}) \otimes q_{x} q_{y} = \sum_{x} p_{x} # u_{e} \otimes q_{x^{2}} = 1_{D} \otimes 1_{D},$ so $R$ is invertible.

3.1step 2.1F1algebra

Since every element of $G$ is its own inverse, $S (p_{x} # u_{g}) = p_{x} # u_{g}$ . In $m (S \otimes id) Δ (p_{x} # u_{g})$ , a summand survives exactly when $a = b$ in $a b = x$ ; this is impossible when $x = s$ , and for $x = e$ the two surviving terms sum to $p_{e} # u_{e} + p_{s} # u_{e} = 1_{D}$ . The same calculation gives $m (id \otimes S) Δ (p_{x} # u_{g})$ . These values equal $ε (p_{x} # u_{g}) 1_{D}$ , so both antipode identities hold and $D (k G)$ is a Hopf algebra.

3.2givenstep 2.1F2algebra

For $d = p_{y} # u_{g}$ , expanding the two sides with the displayed product gives $R Δ (d) = \sum_{a b = y} (p_{a} # u_{g}) \otimes (p_{b} # u_{a g})$ and $Δ^{o p} (d) R = \sum_{a b = y} (p_{b} # u_{g}) \otimes (p_{a} # u_{g b})$ . Since $G$ is abelian, $u_{g b} = u_{b g}$ ; interchanging $a$ and $b$ preserves the condition $a b = y$ and makes the latter sum equal to the former. Bilinearity extends equality from the basis to all $d \in D (k G)$ .

4.1givenstep 2.1step 2.2step 3.2F2algebra∎

For each $g \in G$ , the formula for $Δ$ gives $Δ (q_{g}) = \sum_{x \in G} \sum_{a b = x} (p_{a} # u_{g}) \otimes (p_{b} # u_{g}) = q_{g} \otimes q_{g}$ . Hence $(Δ \otimes id) (R) = \sum_{a, b \in G} (p_{a} # u_{e}) \otimes (p_{b} # u_{e}) \otimes q_{a b} = R_{13} R_{23}$ , using $q_{a} q_{b} = q_{a b}$ . Also $(id \otimes Δ) (R) = \sum_{x} (p_{x} # u_{e}) \otimes q_{x} \otimes q_{x}$ . Expanding $R_{13} R_{12}$ and using $p_{x} p_{y} = δ_{x, y} p_{x}$ gives the same sum. Thus both quasitriangular hexagons hold. Together with invertibility from step 2.2 and the intertwining identity from step 3.2, this proves that the canonical finite tensor $R$ is quasitriangular.

Remarks

This is the explicit $C_{2}$ instance only. The general finite-dimensional double construction is the later HH-10 result Finite double and its canonical quasitriangular structure ↗; no part of its proof or acceptance is assumed here.

Depends on

Used by

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Sources

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