The explicit Fibonacci model has a positive spherical pivotal structure

For a finite direct sum $A={⨁}_{s,p}s$, let $I_{s,p}:s\to A$ be its fixed copy inclusions. Set $A^{\vee }=A$ and $h_A={\sum }_{s,p}(I_{s,p}\otimes I_{s,p})h_s$, $e_A=h_A^{†}$, where $h_{\mathbf{1}}=e_{\mathbf{1}}={\mathrm{id}}_{\mathbf{1}}$. Each cross-copy term in a snake vanishes, and each diagonal term is the simple snake from [F1], so these are duality maps on every finite sum. In the evaluation mate identity, the coefficient for copies $p,q$ of simple $s$ is $\sqrt{d_s}(f_s{)}_{pq}$ on one side and $\sqrt{d_s}(f_s^{\vee }{)}_{qp}$ on the other, where $d_{\mathbf{1}}=1$ and $d_{\tau }=\varphi$ denote the cup/cap normalization constants at this stage. Thus $f_s^{\vee }=f_s^{\mathsf{T}}$, without complex conjugation, by uniqueness of mates. The double dual is therefore literally the identity on objects and morphisms. This alone does not yet establish its tensor comparison. The empty sum uses the unique zero maps and satisfies the same identities.

Define ${\theta }_A$ to be the identity on every unit copy and $x{\mathrm{id}}_{\tau }$ on every $\tau$ copy. These central scalar blocks are natural for arbitrary morphism matrices. Of the four ordered simple pairs, the three containing a unit satisfy the twist law by the strict unit gauge. On $\tau \otimes \tau =\mathbf{1}\oplus \tau$ its two channel equations are $x^2(x^{-1}{)}^2=1,x^2(y^{-1}{)}^2=x,$ the latter because $x=y^2$. Tensoring copy inclusions and then inserting the simple channel inclusions exhausts every finite direct sum, so these equations prove the twist law on all objects, including zero. Also ${\theta }_{\mathbf{1}}=1$. Duality is block transpose by step 1.1, so ${\theta }_A^{\vee }={\theta }_{A^{\vee }}$; hence this twist is ribbon.

In [F2]'s typed formula for ${\mathfrak{u}}_{\tau }$, the map ${\mathrm{id}}_{\tau }\otimes h_{\tau }$ starts in the right tree $\tau \otimes (\tau \otimes \tau )$, with its last pair in the vacuum channel. Apply ${\alpha }^{-1}$ to reach $(\tau \otimes \tau )\otimes \tau$, then $c_{\tau ,\tau }\otimes {\mathrm{id}}_{\tau }$, and finally $e_{\tau }\otimes {\mathrm{id}}_{\tau }$. Let $R_{\mathbf{1}}={\mathrm{id}}_{\tau }\otimes w$ be the right-vacuum inclusion, and let $L_{\mathbf{1}}=w\otimes {\mathrm{id}}_{\tau },L_{\tau }$ be the orthonormal left-tree inclusions. The object block is $F^{-1}=F$, so ${\alpha }^{-1}{R}_{\mathbf{1}}=u{L}_{\mathbf{1}}+v{L}_{\tau }$. The middle braiding gives $u{x}^{-1}{L}_{\mathbf{1}}+v{y}^{-1}{L}_{\tau }$. The final cap is on this same first pair: $L_{\mathbf{1}}^{†}{L}_{\mathbf{1}}=1$ and $L_{\mathbf{1}}^{†}{L}_{\tau }=0$. With $e_0=(1,0{)}^{\mathsf{T}}$, the complete two-channel compression is therefore ${\mathfrak{u}}_{\tau }=\varphi e_0^{\mathsf{T}}\mathrm{diag}(x^{-1},y^{-1})F{e}_0{\mathrm{id}}_{\tau }=\varphi (u{x}^{-1}+0\cdot v{y}^{-1}){\mathrm{id}}_{\tau }=x^{-1}{\mathrm{id}}_{\tau },{\mathfrak{u}}_{\mathbf{1}}={\mathrm{id}}_{\mathbf{1}}\mathrm{.}$ Naturality at all copy inclusions gives precisely these scalar blocks on every finite sum. No dagger conjugation of $x$ or extra F move is inserted.

By steps 2.1–2.2 and [F3], $j=\mathfrak{u}\theta$ is a monoidal natural isomorphism, and its blocks are all identity. To check the tensor comparison explicitly, the three unit-input cases in [F2]'s relation give $d_{a,b}=1$. In the two $\tau \tau$ channels that relation gives $d_{\tau ,\tau }{\mathrm{\mid }}_{\mathbf{1}}=\frac{1\cdot x^{-2}}{x^{-2}}=1,d_{\tau ,\tau }{\mathrm{\mid }}_{\tau }=\frac{x^{-1}{y}^{-2}}{x^{-2}}=x{y}^{-2}=1.$ Thus the four ordered simple pairs, including both nonunit root channels, have identity comparison. Naturality at all tensor products of copy inclusions extends this to every finite sum; the unit comparison is identity in the strict unit gauge. This proves the required pivotal monoidality rather than merely objectwise double-dual isomorphisms.

Since $A^{\vee }=A$ and $j_A=1$, the left closure is $e_A(f\otimes 1)h_A$ and the right closure is $e_A(1\otimes f)h_A$. Orthogonal copy inclusions kill all off-diagonal terms; the diagonal term for a copy of $s$ is $d_s(f_s{)}_{pp}$ on either side. Summing proves both displayed trace formulas. In particular ${\dim }_j(A)=m_1+\varphi m_{\tau }={\dim }_j(A^{\vee })$, so [F4] makes $j$ spherical with simple dimensions $1,\varphi >0$. For $f=g^{†}g$, the formula is a sum of these positive weights times squared absolute matrix entries, and vanishes only for $g=0$. The zero object has trace and dimension zero.

If $j^{\mathrm{\prime }}$ is another pivotal structure, $\gamma =j^{-1}{j}^{\mathrm{\prime }}$ is a monoidal natural automorphism of the identity. Naturality at copy inclusions makes it scalar $1$ on unit copies and $t\mathrm{\ne }0$ on $\tau$ copies. Monoidality on the unit and $\tau$ summands of $\tau \otimes \tau$ gives respectively $t^2=1$ and $t^2=t$. Hence $t=1$, and all components of $\gamma$ are identity. This proves uniqueness and the asserted canonical positive spherical structure, with every trace and ribbon convention already established.