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A quasi-Hopf structure from a normalized 3-cocycle on C2

Statement

Let $G = C_{2} = {0, 1}$ with addition modulo $2$ , let $H = k^{G}$ over $k = C$ , and set $Δ (p_{g}) = \sum_{a + b = g} p_{a} \otimes p_{b}, ε (p_{g}) = {\begin{cases} 1 & g = 0, \\ 0 & g = 1, \end{cases}$ for the delta-function basis $(p_{0}, p_{1})$ . Define $ω (a, b, c) = (- 1)^{a b c}$ for $a, b, c \in {0, 1}$ and $Φ = \sum_{a, b, c \in C_{2}} ω (a, b, c) p_{a} \otimes p_{b} \otimes p_{c} .$ $α = p_{0} - p_{1}, β = p_{0} + p_{1}, S (p_{g}) = p_{g} .$ These data define a quasi-Hopf algebra. Its associator acts by $- 1$ on the triple tensor product of three copies of the degree- $1$ simple module. This example asserts $Φ \neq 1$ as an element of $H^{\otimes 3}$ ; it does not assert that the quasi-Hopf algebra is not twist-equivalent to a Hopf algebra.

Example

The algebra $k^{G}$ has pointwise multiplication $p_{g} p_{h} = δ_{g, h} p_{g}$ and unit $p_{0} + p_{1}$ . All tensor products are algebraic over $C$ .

Facts & Assumptions

Given: $G = C_{2}$ with addition modulo $2$ , the pointwise function algebra $H = k^{G}$ , the maps $Δ, ε$ , and the elements $Φ, α, β, S$ in the Statement.

[F1]

Quasi-bialgebras, quasi-antipodes, and quasi-Hopf algebras: $Δ$ and $ε$ are unital algebra maps; quasi-coassociativity, the pentagon, normalization, and the four quasi-antipode identities are the displayed equations in this convention.

[F2]

Modules over a quasi-bialgebra form a monoidal category: the reassociator on left modules is the action of $Φ$ from $(U \otimes V) \otimes W$ to $U \otimes (V \otimes W)$ .

Verification

1.1givenF1algebraconstruct

The pointwise idempotents are orthogonal and sum to $1$ . The formula for $Δ$ is multiplicative because $Δ (p_{g}) Δ (p_{h}) = 0$ for $g \neq h$ and equals $Δ (p_{g})$ for $g = h$ ; it is unital since $Δ (p_{0} + p_{1}) = (p_{0} + p_{1}) \otimes (p_{0} + p_{1})$ . Both iterated coproducts send $p_{g}$ to $\sum_{a + b + c = g} p_{a} \otimes p_{b} \otimes p_{c}$ , and the two counit equations hold on each $p_{g}$ .

1.2givenF1step 1.1algebra

Each coefficient of $Φ$ is $1$ or $- 1$ on an orthogonal idempotent, so $Φ^{- 1} = Φ$ . The algebra $H^{\otimes 3}$ is commutative and $Δ$ is coassociative, hence $Φ$ commutes with $(Δ \otimes i d) Δ (h)$ and quasi-coassociativity follows. The normalization holds because $ω (a, 0, c) = 1$ . The pentagon, evaluated at $(a, b, c, d)$ , is the scalar identity $ω (b, c, d) ω (a, b + c, d) ω (a, b, c) = ω (a + b, c, d) ω (a, b, c + d) .$ Writing all sums modulo $2$ , both exponents are $b c d + a b d + a c d + a b c$ , so the identity holds for all sixteen quadruples.

2.1givenF1step 1.1algebra

For $g \in C_{2}$ , $Δ (p_{g}) = \sum_{x + y = g} p_{x} \otimes p_{y}$ . In the first quasi-antipode equation, a product $p_{x} α p_{y}$ vanishes unless $x = y$ ; this happens for both pairs when $g = 0$ and for neither pair when $g = 1$ . The sum is therefore $α$ for $g = 0$ and $0$ for $g = 1$ , exactly $ε (p_{g}) α$ . The second equation has the same two cases with $β = 1$ , and yields $ε (p_{g}) β$ .

3.1givenF1step 2.1algebra

In $Φ$ and $Φ^{- 1}$ , a product of three basis idempotents is nonzero only when $a = b = c$ . For $a = 0$ its coefficient times $α_{0}$ is $1$ ; for $a = 1$ it is $(- 1) (- 1) = 1$ . Hence both normalization sums are $p_{0} + p_{1} = 1$ . The map $S$ is the identity and is bijective, so all quasi-Hopf axioms hold.

4.1givenF2step 1.2step 3.1∎

The one-dimensional module $V_{1}$ on which $p_{1}$ acts as $1$ and $p_{0}$ as $0$ has reassociator scalar $ω (1, 1, 1) = - 1$ by [F2]. Since that coefficient of $Φ$ is not the coefficient of $1^{\otimes 3}$ , $Φ \neq 1$ . The calculation establishes only the stated associator and makes no claim about twist equivalence.

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Sources

P. Etingof, S. Gelaki, D. Nikshych, and V. Ostrik, Tensor Categories (standard reference, not scraped)
F. Haußer and F. Nill, Doubles of Quasi-Quantum Groups (standard reference, not scraped)