Fibonacci path projectors and categorical noninvertibility

The fusion rule gives two intermediate channels for $H_{\tau ,2;\tau ,\tau }$: the first pair may fuse to $\mathbf{1}$ or $\tau$, and each channel then fuses with the last $\tau$ to total $\tau$. Thus the ordered basis is exactly $(\mathbf{1},\tau )$ by [F2].

The matrix $F$ is real symmetric and satisfies $u^2+v^2=1$; its off-diagonal entries in $F^2$ cancel because $uv-vu=0$, so $F^{†}F=F^2=I$. Multiplication gives $P_2=F^{-1}\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)F=\left(\begin{array}{cc}{u}^2& uv\\ uv& v^2\end{array}\right)=w{w}^{\mathsf{T}}$ for $w=(u,v{)}^{\mathsf{T}}$. Since $w^{\mathsf{T}}w=1$, $P_2^2=w(w^{\mathsf{T}}w)w^{\mathsf{T}}=P_2$; also $\det (P_2)=u^2{v}^2-(uv{)}^2=0$ and $P_2\mathrm{\ne }0$, so its rank is one and it is not invertible.

The corner matrix product is $P_1{P}_2{P}_1=u^2{P}_1={\varphi }^{-2}{P}_1$. The finite pivotal and trace construction [F8] gives the canonical spherical dimension $d_{\tau }=\varphi$, and $F_{\mathbf{1},\mathbf{1}}=u=1\mathrm{/}\varphi$, so [F4] gives ${\kappa }_{\tau }=d_{\tau }{F}_{\mathbf{1},\mathbf{1}}=+1$ and ${\delta }_{\tau }=\varphi$. Thus $U_i=\varphi P_i$ satisfies $U_1{U}_2{U}_1={\varphi }^3{u}^2{P}_1=\varphi P_1=U_1$; this explicit block computation agrees with the general relation [F5]. The positive sign is the sign computed in this gauge.

By [F6], $[\tau ]([\tau ]-[\mathbf{1}])=[\mathbf{1}]$ but no object supplies the signed inverse. The identity map $I_H$ is nevertheless invertible on $H$, while $P_2$ is not by step 1.2. These are statements about distinct domains of invertibility under [F7]; no line-operator or physical Hamiltonian representation is asserted.

For three sites, use [F2]'s basis recursion once more. From $x_0=\tau$ to $x_3=\tau$, the allowed intermediate pairs are exactly $(\mathbf{1},\tau )$, $(\tau ,\mathbf{1})$ and $(\tau ,\tau )$: the branch $x_1=\mathbf{1}$ forces $x_2=\tau$, whereas $x_1=\tau$ allows both $x_2=\mathbf{1},\tau$. Thus the literal three-site space has dimension three. To group site1 with site2, reassociate the first three inputs, holding $x_2$ fixed. The total-$\tau$ block mixes basis coordinates one and three by $F$, and the total-unit block fixes coordinate two. Hence its coordinate matrix is $A=\left(\begin{array}{ccc}u& 0& v\\ 0& 1& 0\\ v& 0& -u\end{array}\right)$. The vacuum channel in that pair-first tree is coordinate one. To group site2 with site3, reassociate $((x_1\otimes \tau )\otimes \tau )$, holding $x_1$ fixed. The unit $x_1=\mathbf{1}$ case fixes coordinate one, while $x_1=\tau$ mixes coordinates two and three by $F$. Its matrix is $B=\left(\begin{array}{ccc}1& 0& 0\\ 0& u& v\\ 0& v& -u\end{array}\right)$, and the vacuum channel there is coordinate two. Unit-gauge associators are identities by [F1].

The pullback R phases in [F1] are complex unit scalars: $\mathrm{\mid }{e}^{-4\pi i\mathrm{/}5}\mathrm{\mid }=\mathrm{\mid }{e}^{3\pi i\mathrm{/}5}\mathrm{\mid }=1$, and their complex conjugates are their inverses. They are exchange phases and do not change the real vacuum F coefficient used in step 2.1.

Both $A$ and $B$ are real orthogonal because $F$ is. Applying [F3] gives $Q_1=A^{-1}\mathrm{diag}(1,0,0)A$ and $Q_2=B^{-1}\mathrm{diag}(0,1,0)B$, exactly the matrices in the Example. They are outer products of the unit vectors $r=(u,0,v{)}^{\mathsf{T}}$ and $s=(0,u,v{)}^{\mathsf{T}}$. Their overlap is $r^{\mathsf{T}}s=v^2=u$, so $Q_1{Q}_2{Q}_1=u^2{Q}_1={\varphi }^{-2}{Q}_1$ and likewise with indices exchanged. The positive loop sign is the one already computed in step 2.1. This proves all stated projectors and Temperley–Lieb relations on the actual three-site space without conflating it with the two-dimensional block.