A Z2 fusion datum with a nontrivial associator scalar

For $a,b,c,d\in \{0,1\}$, the exponent of $-1$ on the left side of the $3$-cocycle equation is $(a+b)cd+ab(c+d)\equiv abc+abd+acd+bcd(\mathrm{m}\mathrm{o}\mathrm{d}2),$ while on the right it is $abc+a(b+c)d+bcd\equiv abc+abd+acd+bcd(\mathrm{m}\mathrm{o}\mathrm{d}2)\mathrm{.}$ Hence the displayed cocycle equation holds. If any argument is $0$, then $\omega =1$, so the displayed normalization holds.

Every graded object is a finite direct sum of copies of ${\delta }_0$ and ${\delta }_1$, so the category is semisimple with exactly these two simple isomorphism classes; the category is abelian because kernels and cokernels of grade-preserving maps inherit the grading; its Hom spaces are finite-dimensional because they are finite direct sums of linear-map spaces. The tensor bifunctor is bilinear because each homogeneous component is a direct sum of ordinary vector-space tensor products. The unit ${\delta }_0$ is simple and $\mathrm{End}({\delta }_0)=\mathbb{C}$. Every graded epimorphism splits by choosing bases in its two homogeneous components, so every object is projective and each simple has its identity as a projective cover. Only finite-dimensional basis choices are used. The tensor product preserves grading and gives ${\delta }_a\otimes {\delta }_b\cong {\delta }_{a+b}$ via $m_{a,b}$. On each homogeneous component the full associator is the rebracketing map multiplied by $\omega (a,b,c)\in \{1,-1\}$, so it is invertible and natural with respect to every grade-preserving linear map. On every homogeneous pure tensor $((w_a\otimes u_b)\otimes v_c)\otimes x_d$, the two pentagon composites have scalar factors given by the two sides of the cocycle equation in step 1.1 and the same final rebracketing, so they agree; these tensors span the fourfold tensor product. The displayed unitors are natural and invertible. Since $\omega (a,0,b)=1$, both sides of the triangle $({\mathrm{id}}_W\otimes {\lambda }_U)\circ {\alpha }_{W,{\delta }_0,U}={\rho }_W\otimes {\mathrm{id}}_U$ send $(w_a\otimes z)\otimes u_b$ to $z{w}_a\otimes u_b$. For ${\delta }_0$, both duals are ${\delta }_0$ with evaluation $m_{0,0}$ and coevaluation $m_{0,0}^{-1}$; with the stated unitors, every snake composite sends $z$ to $z$. For ${\delta }_1$, the left maps are ${\mathrm{ev}}_L:{\delta }_1\otimes {\delta }_1\to {\delta }_0$ with scalar $-1$ and ${\mathrm{coev}}_L:{\delta }_0\to {\delta }_1\otimes {\delta }_1$ with scalar $1$; the right maps ${\mathrm{ev}}_R:{\delta }_1\otimes {\delta }_1\to {\delta }_0$ and ${\mathrm{coev}}_R:{\delta }_0\to {\delta }_1\otimes {\delta }_1$ have the same respective scalars. The left snake composites have scalars $1\cdot (-1)\cdot (-1)=1$ and $1\cdot (-1)\cdot (-1)=1$, the second using ${\alpha }_{1,1,1}^{-1}$; the right snake composites have scalars $1\cdot (-1)\cdot (-1)=1$ and $1\cdot (-1)\cdot (-1)=1$, the first using ${\alpha }_{1,1,1}^{-1}$. Thus both snake identities hold in each orientation. For a finite sum of simple lines, take the sum of their specified duals, define evaluation to be the specified evaluation on matching summands and zero on distinct summands, and define coevaluation as the sum of the specified coevaluations followed by the matching inclusions. The snake composites restrict to the identity on each summand by the preceding calculation, so this also supplies duals for every object (and the empty sum gives the zero-object dual). Thus the category has all properties required by [F1].

Since $1+1=0$ in $G$, ${\delta }_1\otimes {\delta }_1\cong {\delta }_0$. But $\omega (1,1,1)=(-1{)}^1=-1$, so the associator on the triple $({\delta }_1,{\delta }_1,{\delta }_1)$ is multiplication by $-1$; for the untwisted function it is multiplication by $+1$. Both fusion tables have $N_{11}^0=1$ and all other $N_{11}^a=0$, so these integers do not by themselves specify that associator component. This comparison concerns the written structure data only and makes no equivalence claim.