Pointed ribbon categories from abelian 3-cocycles

Let $G$ be a finite abelian group, written additively. A normalized abelian 3-cocycle is a pair of functions $\omega :G^3\to {\mathbb{C}}^{\times }$ and $r:G^2\to {\mathbb{C}}^{\times }$, equal to $1$ whenever one of their arguments is $0$, satisfying the pentagon equation

$\omega (a,b,c)\omega (a,b+c,d)\omega (b,c,d)=\omega (a+b,c,d)\omega (a,b,c+d)$

and the two braided hexagon equations

$\omega (b,c,a)\omega (a,b,c)r(a,b+c)=r(a,b)r(a,c)\omega (b,a,c),$

$r(a+b,c)=\omega (a,b,c)\omega (c,a,b)\omega (a,c,b{)}^{-1}r(a,c)r(b,c)\mathrm{.}$

The associated pointed category has one simple object $g$ in each grade $g\in G$, tensor product $g\otimes h=g+h$, associator ${\alpha }_{g,h,k}:(g\otimes h)\otimes k\to g\otimes (h\otimes k)$ given by scalar $\omega (g,h,k)$, and braiding $c_{g,h}:g\otimes h\to h\otimes g$ given by scalar $r(g,h)$. For the left dual $g^{\vee }=-g$, take evaluation $g^{\vee }\otimes g\to 0$ to have scalar $1$ and coevaluation $0\to g\otimes g^{\vee }$ to have scalar $\omega (g,-g,g{)}^{-1}$. The first snake is the identity by this choice; the second is the identity because the normalized pentagon at $(g,-g,g,-g)$ gives $\omega (g,-g,g)\omega (-g,g,-g)=1$. A ribbon twist is a function $q:G\to {\mathbb{C}}^{\times }$ with $q(0)=1$, $q(-g)=q(g)$, and

$q(g+h)=r(g,h)r(h,g)q(g)q(h)\mathrm{.}$

Write $b(g,h)=r(g,h)r(h,g)$ for the double-braiding scalar. The pointed ribbon category data include a compatible pivotal structure whose twist is $q$ and whose simple dimensions are $1$; this convention is an explicit input, not a consequence of the quadratic function alone. A pointed metric-group category is modular exactly when its radical is trivial proves from the two hexagons that $b$ is a symmetric bicharacter and that the Hopf-link matrix has entries $s_{gh}=b(g,h)$. A quadratic function alone is not the category data: the associator, both hexagons, duality maps, and the stated pivotal convention are part of the definition.