All-label unitary Ising normal data and ribbon coherence

Let $G=\mathbb{Z}\mathrm{/}2=\{0,1\}$, with $0=\mathbf{1}$, $1=\eta$, and a third simple $D$. Set $ab=a+b$ for $a,b\in G$, $aD=Da=D$, and $DD=0\oplus 1$. Objects are multiplicity triples $m=(m_0,m_1,m_D)\in {\mathbb{N}}_0^3$, representing the canonical sums $0^{\oplus m_0}\oplus 1^{\oplus m_1}\oplus D^{\oplus m_D}$; arbitrary reorderings are not additional objects. Set $\mathrm{Hom}(m,n)={⨁}_{c\in \{0,1,D\}}{\mathrm{Mat}}_{n_c\times m_c}(\mathbb{C})$, with block composition, matrix-adjoint dagger, and maximum block operator norm. An isomorphism forces $m_c=n_c$ for every $c$, so this category is skeletal. Direct sum adds triples and $(m\otimes n{)}_c={\sum }_{a,b}{m}_a{n}_b{N}_{ab}^c$. In each output block order the contributing coordinates $(a,r,b,t)$ lexicographically using $0<1<D$; tensor on morphisms has entries $f_a[r^{\mathrm{\prime }},r]g_b[t^{\mathrm{\prime }},t]$ on the matching $(a,b,c)$ blocks and zero elsewhere. This is a dagger-preserving bifunctor by the matrix product identities. Order $DD$ as $(0,1)$ and use the canonical coordinate inclusions $i_{ab}^c$ and retractions $p_{ab}^c=(i_{ab}^c{)}^{†}$. Every nonzero elementary multiplicity has coordinate $\mu =1$; impossible channels have multiplicity zero. Extend all subsequent simple-input maps over these coordinates by direct sums. This fixes the skeletal carrier and bases before any coefficient is calculated.

Put $d=\sqrt{2}$, $\chi (a,b)=(-1{)}^{ab}$, $\kappa \in \{-1,1\}$ and $\tau =\kappa \mathrm{/}d$. Take identity unit constraints. For $a,b,c\in G$, define the entire associator by ${\alpha }_{a,b,c}={\alpha }_{a,b,D}={\alpha }_{D,a,b}=1$, ${\alpha }_{a,D,b}=\chi (a,b)$, ${\alpha }_{D,a,D}{\mathrm{\mid }}_z=\chi (a,z)$ for $z\in G$, and ${\alpha }_{D,D,D}[f,e]=\tau \chi (e,f)$ for $e,f\in G$. The maps ${\alpha }_{a,D,D}$ and ${\alpha }_{D,D,a}$ have coefficient one on the equal simple output, after relabelling the $DD$ summand by addition of $a$. This specifies the direct-sum permutations as well as their scalars. For fixed total $z$, the target row is $(f,{\mu }_{bc}^f=1,{\mu }_{af}^z=1)$ and the source column is $(e,{\mu }_{ab}^e=1,{\mu }_{ec}^z=1)$; a coefficient exists only when all four fusion multiplicities are one.

In the ordered channels $(0,1)$ the unique two-dimensional block is $F_D^{DDD}=\frac{\kappa }{d}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)$. Every other admissible block is $1\times 1$ and equals $+1$, except $F_D^{\eta D\eta }=F_{\eta }^{D\eta D}=-1$. Unit-input blocks are identities, and an impossible fixed-total space has the unique $0\times 0$ matrix. These real matrices and the summand permutations are orthogonal, so their inverse-transposes give exactly the same coefficients. Hence the object associators of step 2.1 are also the F coefficients in the library's inverse-precomposition convention, with no suppressed transpose, inverse, channel, or multiplicity coordinate.

Write $A_z^{abc}[f,e]$ for those associator coefficients, and set them to zero on inadmissible channels. Start at $(((ab{)}_{e}c{)}_{f}t{)}_z$ and end at $(a(b(ct{)}_g{)}_h{)}_z$. The root/root route and the left/root/right route around the pentagon give, respectively, the two sides of $A_z^{ect}[g,f]A_z^{abg}[h,e]={\sum }_u{A}_f^{abc}[u,e]A_z^{aut}[h,f]A_h^{bct}[g,u]$. This follows by applying each rotation to the indicated subtrees, retaining every spectator output and multiplicity coordinate. We verify all sixteen placements of group objects and $D$ in this indexed equation, rather than using a classification theorem.

With zero $D$ inputs both pentagon routes have coefficient one. With one $D$ input, the four placements reduce in order to the following equalities: $\begin{array}{cl}(D,a,b,c)& 1=1\\ (a,D,b,c)& \chi (a,b+c)=\chi (a,b)\chi (a,c)\\ (a,b,D,c)& \chi (a+b,c)=\chi (a,c)\chi (b,c)\\ (a,b,c,D)& 1=1\end{array}$. They hold for all free group labels by bicharacter multiplicativity.

For two $D$ inputs the six placements reduce to $\begin{array}{cl}(D,D,a,b)& 1=1\\ (D,a,D,b)& \chi (a,x+b)=\chi (a,x)\chi (a,b)\\ (D,a,b,D)& \chi (a,z)\chi (b,z)=\chi (a+b,z)\\ (a,D,D,b)& 1=1\\ (a,D,b,D)& \chi (b,a+y)=\chi (a,b)\chi (b,y)\\ (a,b,D,D)& 1=1\end{array}$. Here $x$ is the preceding $DD$ output, and $z$ or $a+y$ is the total group output; admissibility forces these labels in the indexed equation. There is exactly one admissible term in its sum over $u$. All six identities follow from symmetry and multiplicativity of $\chi$.

With three $D$ inputs, let $a$ be the group input and $x,y$ the initial and final corresponding two-$D$ output labels. The four placements give $\begin{array}{cl}(D,D,D,a)& \tau \chi (x,a)\chi (x,y)=\tau \chi (x,a+y)\\ (D,D,a,D)& \tau \chi (x,y)=\tau \chi (x+a,y)\chi (a,y)\\ (D,a,D,D)& \tau \chi (x,y)=\chi (a,x)\tau \chi (x,a+y)\\ (a,D,D,D)& \tau \chi (x,y)\chi (a,y)=\tau \chi (x+a,y)\end{array}$. The middle associator forces each shifted label $x+a$ or $y+a$. Multiplicativity, symmetry and ${\chi }^2=1$ prove every entry for all free channel labels.

With four $D$ inputs, let $x$ be the initial $DD$ channel, $y$ the final $DD$ channel and $z\in G$ the total. The short route has coefficient ${\delta }_{y,x+z}$. The long route has coefficient ${\tau }^2{\sum }_{u\in G}\chi (x,u)\chi (u,z)\chi (u,y)={\tau }^2{\sum }_u(-1{)}^{u(x+y+z)}=2{\tau }^2{\delta }_{y,x+z}={\delta }_{y,x+z}$, since $2{\tau }^2=1$. The four placement calculations exhaust all sixteen placements, every free group/channel label, and every pentagon entry; a nonexistent fusion path gives an empty equality. Every unit-input coefficient agrees with the identity unitor, so the triangle holds. The orthogonal F matrices and ordinary block dagger now give a unitary monoidal category.

Set $u(0)=1$, $u(1)=i$ and choose a unit-modulus $\beta$ with ${\beta }^2=\tau (1+i)$. Define all object braiding coefficients by $c_{a,b}=\chi (a,b)$ for $a,b\in G$, $c_{a,D}=c_{D,a}=u(a)$, and $c_{D,D}^g=v(g)=\beta \mathrm{/}u(g)$ for $g\in G$. This covers every channel, including units, and all entries have modulus one. Expanding the two hexagons in the normal summand bases reduces their independent equations to $c_{a,b}=\chi (a,b)$, $u(a+b)=u(a)u(b)\chi (a,b)$, $u(a)v(a+b)=\chi (a,b)v(b)$, $u(a{)}^2=\chi (a,a)$, and $\tau {\sum }_{c\in G}\chi (a+b,c)u(c)=\chi (a,b)v(a)v(b)$.

For completeness, the all-group hexagon is bicharacter multiplicativity; the placements $abD$ and $aDb$ give $c_{a,b}=\chi (a,b)$; $Dab$ gives the quadratic equation for $u$; $DDa$ and $DaD$ give the transport equation for $v$; $aDD$ gives $u(a{)}^2=\chi (a,a)$; and $DDD$ gives the final equation of step 9.1 entry by entry. In that last case the full matrix equality is $A\mathrm{diag}(u)A=\mathrm{diag}(v)A\mathrm{diag}(v)$ with $A=\tau (\chi (a,b){)}_{a,b}$; cancelling one $\tau$ and combining the character factors gives the displayed sum. The other hexagon reverses these placements and replaces $\chi$ by ${\chi }^{-1}$, so it gives the same five equations because $\chi ={\chi }^{-1}$ and both mixed braidings are $u$. This accounts for all eight typed summand patterns of each hexagon.

The only nontrivial quadratic case is $u(1+1)=1=i^2(-1)$, and the nontrivial transport cases are $iv(1)=v(0)$ and $iv(0)=-v(1)$. For the last equation of step 9.1 the four pairs $(a,b)$ give $\begin{array}{cl}(0,0)& \tau (1+i)={\beta }^2\\ (1,0),(0,1)& \tau (1-i)=-i{\beta }^2\\ (1,1)& \tau (1+i)=-(-i\beta {)}^2={\beta }^2\end{array}$. Each follows from ${\beta }^2=\tau (1+i)$. Thus both hexagons hold in every channel; none of the braiding data has been inferred solely from a two-channel braid relation.

Let $s=e^{\pi i\mathrm{/}8}$. At $\kappa =1$, ${\beta }_{+}=s$, inverse pullback gives the complete table $R_{0a}^a=R_{a0}^a=1$, $R_{\eta \eta }^0=-1$, $R_{\eta D}^D=R_{D\eta }^D=-i$, $R_{DD}^0=s^{-1}=e^{-\pi i\mathrm{/}8}$ and $R_{DD}^{\eta }=i{s}^{-1}=e^{3\pi i\mathrm{/}8}$. At $\kappa =-1$, ${\beta }_{-}=-s^{-3}$, the unit/group/mixed entries are identical, while $R_{DD}^0=-s^3$ and $R_{DD}^{\eta }=s^{-1}$. Every displayed elementary multiplicity index is one, and a nonexistent channel is empty. These are pullback matrices of $f\mapsto f\circ c_{a,b}^{-1}$, rather than object coefficients with only selected phases inverted.

For a group simple $a$, take ${\mathrm{coev}}_a=i_{aa}^0$ and ${\mathrm{ev}}_a=p_{aa}^0$. For $D$, take ${\mathrm{coev}}_D=\sqrt{d}{i}_{DD}^0$ and ${\mathrm{ev}}_D=\kappa \sqrt{d}{p}_{DD}^0$. The $D$ snake scalar is $\kappa d(F_D^{DDD}{)}_{0,0}=\kappa d\tau =1$, and reflection proves the other snake; group snakes have scalar one. Thus every object is rigid and the normalized FS sign of $D$ is $\kappa$. Choose right coevaluation $\kappa \sqrt{d}{i}_{DD}^0$ and right evaluation $\sqrt{d}{p}_{DD}^0$ for $D$; both positive left/right dimensions are $d$, although ${\mathrm{ev}}_D{\mathrm{coev}}_D=\kappa d$. The dimension and FS of $\eta$ are one. On a finite sum, the candidate left/right trace is ${\sum }_a{d}_a\mathrm{Tr}(M_a)$, where $d_0=d_{\eta }=1$ and $d_D=d$; we verify its pivotal compatibility explicitly below.

Let $E_c=i_{ab}^c{p}_{ab}^c$. Inserting the normalized $a$ cup, rotating by ${\alpha }_{a,a,b}$, applying $E_c$, rotating back and using the positive $a$ cap gives ${\mathrm{ptr}}_a(E_c)=d_a\mathrm{\mid }{A}_b^{aab}[c,0]{\mathrm{\mid }}^2{1}_b=(d_c\mathrm{/}{d}_b)1_b$. Closing $b$ similarly gives ${\mathrm{ptr}}_b(E_c)=d_b\mathrm{\mid }{A}_a^{abb}[0,c]{\mathrm{\mid }}^2{1}_a=(d_c\mathrm{/}{d}_a)1_a$. The entire nonunit calculation is $\begin{array}{ccc}(a,b;c)& {\mathrm{ptr}}_a& {\mathrm{ptr}}_b\\ (\eta ,\eta ;0)& 1& 1\\ (\eta ,D;D)& 1& d\\ (D,\eta ;D)& d& 1\\ (D,D;0)& 1\mathrm{/}d& 1\mathrm{/}d\\ (D,D;\eta )& 1\mathrm{/}d& 1\mathrm{/}d\end{array}$. The first three rows use scalar F entries of modulus one; the last two use $\mathrm{\mid }\tau {\mathrm{\mid }}^2=1\mathrm{/}2$ and $d\mathrm{/}2=1\mathrm{/}d$. Unit rows are identity traces. In particular all FS signs occur twice and cancel.

Relative to these left duals the standard-conjugate comparison is $j_a=({\mathrm{coev}}_a^{†}\otimes 1_{a^{\ast \ast }}){\alpha }^{-1}(1_a\otimes {\mathrm{coev}}_{a^{\ast }})$, with typed dual comparisons and unitors inserted. It equals one on a group simple and $d\tau =\kappa$ on $D$. Direct sums use their orthogonal summand inclusions. Iterating the last-vertex calculation of step 14.1 on fusion-path matrix units gives zero off the diagonal and the endpoint dimension on the diagonal; the pentagon makes it tree-independent. These units span every Hom block, and the weighted matrix pairing is nondegenerate. The fusion-dimension equations $d_a{d}_b={\sum }_c{N}_{ab}^c{d}_c$ and those same units prove tensor multiplicativity of the trace. Thus the left/right closures both equal ${\sum }_a{d}_a\mathrm{Tr}(M_a)$; this is the trace produced by the dualities, rather than an independently assigned dimension function. The following typed tensorator calculation verifies monoidality of $j$ without assuming a canonical positive pivot.

Choose the left dual of a direct sum by its orthogonal simple summands. Its contravariant tensorator $q_{a,b}:b^{\ast }\otimes a^{\ast }⟶(a\otimes b{)}^{\ast }$ is uniquely characterized by ${\mathrm{ev}}_{a\otimes b}(q_{a,b}\otimes 1_{a\otimes b})=E_{a,b}$, where $E_{a,b}$ is folded evaluation: reassociate $(b^{\ast }\otimes a^{\ast })\otimes (a\otimes b)$ to $b^{\ast }\otimes ((a^{\ast }\otimes a)\otimes b)$, evaluate the middle $a$ pair and then the outer $b$ pair. Uniqueness follows from the two snakes. This fixes all types before taking its channel scalars.

Put $k_a=1$ for $a\in G$ and $k_D=\kappa$. On a canonical $c$ summand, evaluation of both sides of step 16.1 after the $c$ cup gives $q_{a,b}^c=\frac{k_a{k}_b}{k_c}\sqrt{\frac{d_a{d}_b}{d_c}}{A}_0^{bac}[b,c]A_b^{aab}[c,0]$. Indeed the first rotation takes $((ba{)}_c(ab{)}_c{)}_0$ to $b(a(ab{)}_c)$ with coefficient $A_0^{bac}[b,c]$; the inverse right rotation makes the inner $aa$ channel vacuum with coefficient $A_b^{aab}[c,0]$. The two evaluations contribute $k_a\sqrt{d_a}{k}_b\sqrt{d_b}$, the initial $c$ cup contributes $\sqrt{d_c}$, and its comparison cap/cup has scalar $k_c{d}_c$. Their quotient is the stated formula, derived entirely from the specified dualities and F data.

The complete channel computation is $\begin{array}{ccccc}(a,b;c)& q_{a,b}^c& {\nu }_{a,b}^c& j_c& j_a{j}_b\\ (0,0;0)& 1& 1& 1& 1\\ (0,\eta ;\eta )& 1& 1& 1& 1\\ (\eta ,0;\eta )& 1& 1& 1& 1\\ (0,D;D)& 1& 1& \kappa & \kappa \\ (D,0;D)& 1& 1& \kappa & \kappa \\ (\eta ,\eta ;0)& 1& 1& 1& 1\\ (\eta ,D;D)& 1& 1& \kappa & \kappa \\ (D,\eta ;D)& 1& 1& \kappa & \kappa \\ (D,D;0)& \kappa & 1& 1& 1\\ (D,D;\eta )& \kappa & 1& 1& 1\end{array}$. For either $DD$ row the two F factors of step 17.1 are $1,\tau$, and the dimension factor is $d$, so $q=d\tau =\kappa$. In mixed rows both F factors, the dimension quotient, and $k_a{k}_b\mathrm{/}{k}_c$ are one; unit/group rows are immediate. All entries are real unit scalars.

The typed double-dual tensorator is ${\nu }_{a,b}:a^{\ast \ast }\otimes b^{\ast \ast }⟶(a\otimes b{)}^{\ast \ast }$ with ${\nu }_{a,b}=(q_{a,b}^{\ast }{)}^{-1}{q}_{b^{\ast },a^{\ast }}$. On a $c$ summand its coefficient is $q_{b,a}^c\mathrm{/}{q}_{a,b}^c=1$, as recorded in step 18.1. The equality of that table's last columns therefore proves $j_{a\otimes b}={\nu }_{a,b}(j_a\otimes j_b)$ on all ten admissible channels. Scalar simple components give naturality, chosen sum duals give additivity, and pairwise monoidality extends to arbitrary products by coherence of the canonical dual tensorator. Together with steps 13.1–15.1 this gives a positive spherical pivotal structure with the asserted FS signs and trace.

The positive pivotal curl on $D$ is fixed by partial trace: ${\theta }_D=(v(0)+v(1))\mathrm{/}d=\beta (1-i)\mathrm{/}d=\kappa \mathrm{/}\beta$, since ${\beta }^2=\kappa (1+i)\mathrm{/}d$. The group twists are ${\theta }_0=1$ and ${\theta }_{\eta }=-1$. Balancing holds in every channel: group-group crossings square to one; mixed crossings square to $-1={\theta }_D\mathrm{/}({\theta }_{\eta }{\theta }_D)$; and on $DD$ one has $v(g{)}^2={\theta }_g\mathrm{/}{\theta }_D^2$ because $u(g{)}^2=\chi (g,g)={\theta }_g$. All labels are self-dual and scalar twist equals its dual, so the ribbon condition holds. In particular the negative-sign parameters ${\beta }_{-}=-s^{-3}$ give ${\theta }_D=-1\mathrm{/}{\beta }_{-}=s^3$, while the positive-sign parameters ${\beta }_{+}=s$ give ${\theta }_D=1\mathrm{/}{\beta }_{+}=s^{-1}$; in both cases ${\theta }_{\eta }=-1$. This proves all normal-data, coherence, duality, pivotal, FS and ribbon claims.