Noninvertible Symmetries and Anyonic Chains: Examples and Counterexamples

The fusion rule gives $N_{xa}^y=\dim \mathrm{Hom}(x\otimes a,y)=2$. Apply the $n=1$ case of [F2] with boundary $x$, site charge $a$, and endpoint $y$: it gives $\dim \mathrm{Hom}(y,x\otimes a)=N_{xa}^y=2$. Therefore the chosen basis of this split space consists of exactly $i_1,i_2$.

Inserting $i_1$ and $i_2$ into the same surrounding path produces two different vertex-index tuples. By [F2] both composites are members of a path basis and are linearly independent. The local two-dimensional multiplicity contributes exactly two basis vectors; identifying the indices would lose one dimension.

Take $\kappa =1$ and $\beta =s$. Since $s^2=(1+i)\mathrm{/}\sqrt{2}$, the normal-data condition ${\beta }^2=\kappa (1+i)\mathrm{/}\sqrt{2}$ holds. Thus [F1] gives an actual coherent unitary category, not just a single sector matrix. Its full pullback R table and positive-FS Hadamard block are exactly the displayed entries, and its curl is ${\theta }_D=\kappa \mathrm{/}\beta =s^{-1}$.

The ell-four root supplier has FS $-1$ and twist $s^3$. Apply [F2]'s parity associator $(-1{)}^{\mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }b\mathrm{\mid }\mathrm{\mid }c\mathrm{\mid }}$ and odd-odd crossing factor $-i$, reversing the odd evaluation as specified there. This produces the category of step 1.1. It proves the exact relationship to the root construction without identifying distinct pivotal or chirality conventions.

Fusion forces alternation between $D$ and a group label. Odd length is therefore impossible. For $2N$, choose the $N$ group labels independently in each of the two parity patterns, yielding both orthonormal bit bases and dimension $2^{N+1}$. Apply [F3] to the local projections specified by [F4]: the Hadamard vacuum conjugate is $(I+X)\mathrm{/}2$, while a group-group endpoint pair has vacuum channel exactly when its bits agree, giving $(I+ZZ)\mathrm{/}2$. These prove all four displayed projector formulas at every cyclic index.

With the displayed order $H\oplus \bar{H}$, the path shift sends $H$ to $\bar{H}$ without changing the bits and sends $\bar{H}$ to $H$ by $C$. Therefore it is the displayed block matrix and $T^2=\mathrm{diag}(C,C)$. [F3]'s complete cyclic F contraction at $\kappa =1$ gives exactly the displayed $Y_{\eta },Y_D$, not an inferred fusion-only representation. Its Fourier identities $K^{†}K=K{K}^{†}=I+Q$, $KQ=QK=K$ and $K^{†}=CK=KC$ prove the line fusion relations, commutation with all projectors and $T$, and the stated distinct single-chain square. Consequently any Hamiltonian formed from these specified projectors commutes with these line maps.

For $N=1$, the cyclic bond is $(I+Z_1^2)\mathrm{/}2=I$, the field remains $(I+X_1)\mathrm{/}2$, and both two-dimensional sectors remain present. Thus the wraparound formulas include the smallest length without adding an inapplicable three-site relation or projecting away either sector.

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.