Periodic Ising paths, two sectors, and the full line representation

By [F2], the carrier ${\mathcal{H}}_L$ is the direct sum of the Hom spaces $H_{D,L;b,b}$, and by [F3] the path composites in each summand form an orthonormal basis. A basis vector in the $b$ summand is exactly a cyclic path with $x_1=b$, $x_{L+1}=x_1$, and all vertex indices retained; conversely every cyclic path belongs to exactly one such summand. Since the Ising fusion rules are $D\otimes D=\mathbf{1}\oplus \eta$ and $D\otimes g=D$ for $g\in G=\{\mathbf{1},\eta \}$, the labels alternate between $D$ and $G$. Thus the cyclic path space is zero for odd $L$. For $L=2N$, $N\ge 1$, it is the orthogonal sum $H\oplus \bar{H}$, with $H$ given by $x_{2j-1}=D$, $x_{2j}=b_j\in G$ and $\bar{H}$ by $x_{2j-1}=b_j\in G$, $x_{2j}=D$. The $N$ bits are independent even across the wrap edge, giving dimension $2^N$ in each sector. The maps ${\mathrm{\Phi }}_H,{\mathrm{\Phi }}_{\bar{H}}$ send the paths to the standard orthonormal basis $\mathrm{\mid }{b}_1\cdots b_N⟩$, with inverse obtained by inserting the fixed $D$ labels. This includes both sectors without a total-flux or global-spin-parity projection; at $N=1$ there are four states.

Define $P_i$ to project the two reference edges from $x_i$ to $x_{i+2}$ to vacuum, changing only the intermediate label $x_{i+1}$. Its local F matrix is $F_{x_{i+2}}^{x_{i}DD}$, with source channel $x_{i+1}$ and target $DD$ output. Thus its index differs from a convention naming the intermediate link by one. If both neighbors are $D$, the F matrix is $\kappa$ times Hadamard and $F^{-1}{\mathrm{\Pi }}_{0}F=(I+X)\mathrm{/}2$ on the intermediate bit. If they are group labels $g,h$, the intermediate label is forced to be $D$ and the $DD$ output is $g+h$, so the vacuum projector is ${\delta }_{g,h}$. The overall $\kappa$ cancels, so both coherent sign conventions give the same projectors.

With bit indices modulo $N$, the complete sector images are ${\mathrm{\Phi }}_H{P}_{2j-1}{\mathrm{\Phi }}_H^{-1}=(I+X_j)\mathrm{/}2$, ${\mathrm{\Phi }}_H{P}_{2j}{\mathrm{\Phi }}_H^{-1}=(I+Z_j{Z}_{j+1})\mathrm{/}2$, ${\mathrm{\Phi }}_{\bar{H}}{P}_{2j-1}{\mathrm{\Phi }}_{\bar{H}}^{-1}=(I+Z_j{Z}_{j+1})\mathrm{/}2$, and ${\mathrm{\Phi }}_{\bar{H}}{P}_{2j}{\mathrm{\Phi }}_{\bar{H}}^{-1}=(I+X_{j+1})\mathrm{/}2$. This includes the wraparound edge. At $N=1$ the bond is $(I+Z_1^2)\mathrm{/}2=I$ and the field is $(I+X_1)\mathrm{/}2$. No three-distinct-site cyclic TL relation is asserted for that length-two identification; the ordinary interior theorem keeps its own applicable-index hypothesis.

Define the path shift by $(Tx{)}_i=x_{i+1}$ cyclically and let $C\mathrm{\mid }{b}_1\cdots b_N⟩=\mathrm{\mid }{b}_2\cdots b_N{b}_1⟩$. On $H\to \bar{H}$ the bit coordinates are unchanged, and on $\bar{H}\to H$ they shift by $C$. Therefore, in the order $H\oplus \bar{H}$, $\mathrm{\Phi }T{\mathrm{\Phi }}^{-1}=\left(\begin{array}{cc}0& C\\ I& 0\end{array}\right)$, $T^2=\mathrm{diag}(C,C)$ and $T^{2N}=I$. It is unitary. Since $C{X}_j{C}^{-1}=X_{j-1}$ and the same relabelling holds for $Z$, substituting all four formulas of step 3.1 proves $T{P}_i{T}^{-1}=P_{i-1}$ at every cyclic index, including $i=1$ and $i=2N$. Thus the basis identifications intertwine the specified operators, not merely their dimensions.

The represented operator ${\sum }_i(1-2P_i)$ is two copies of $-{\sum }_j(X_j+Z_j{Z}_{j+1})$. Unequal odd/even couplings give $-g{\sum }_j{X}_j-h{\sum }_j{Z}_j{Z}_{j+1}$ on $H$, and $-g{\sum }_j{Z}_j{Z}_{j+1}-h{\sum }_j{X}_j$ on $\bar{H}$. Only $g=h$ is invariant under one-step $T$, while both are invariant under $T^2$. These Hamiltonian statements follow from the specified projector representation, rather than from fusion rules alone.

Define the annular line action by its full cyclic contraction $(Y_a{)}_{x^{\mathrm{\prime }},x}={\prod }_{i=1}^{2N}{F}_{x_{i+1}^{\mathrm{\prime }}}^{a{x}_{i}D}[x_{i+1},x_i^{\mathrm{\prime }}]$ when each $x_i^{\mathrm{\prime }}$ occurs in $a{x}_i$, and set it to zero otherwise. The row and column include all elementary multiplicity indices, each one here. This is the associator moving $(a{x}_i)D$ to $a(x_{i}D)$ around the circle, so the formula specifies the wrap contraction. The unit line gives $I$. For $a=\eta$ every admissible output flips all group bits and fixes $D$, and each local F factor is one. Hence $Y_{\eta }=\mathrm{diag}(Q,Q)$ with $Q={\prod }_j{X}_j$.

For $a=D$, an input $H$ bit string $b$ maps to an output $\bar{H}$ bit string $c$. The odd local factors are $(\kappa \mathrm{/}d)(-1{)}^{b_j{c}_j}$, while the even factors are $(-1{)}^{b_j{c}_{j+1}}$. Their product is ${\kappa }^N{K}_{c,b}$, where $K_{c,b}=2^{-N\mathrm{/}2}(-1{)}^{{\sum }_j{c}_j(b_{j-1}+b_j)}$. On the reverse sector the cyclic contraction is ${\kappa }^N{K}^{†}$. All exponents are modulo two and $b_0=b_N$. Thus $Y_D={\kappa }^N\left(\begin{array}{cc}0& K^{†}\\ K& 0\end{array}\right)$. This derives the kernel and its normalization from all-label F data, including every wrap factor.

Orthogonality of finite binary characters gives $(K^{†}K{)}_{c,b}=(K{K}^{†}{)}_{c,b}=2^{-N}{\sum }_y(-1{)}^{{\sum }_j{y}_j((b_j+c_j)+(b_{j-1}+c_{j-1}))}={\delta }_{c,b}+{\delta }_{c,b+\mathbf{1}}$, where $\mathbf{1}$ is the all-ones bit string. Indeed the character is trivial exactly when $b+c$ is constant around the cycle. Consequently $K^{†}K=K{K}^{†}=I+Q$. Also $KQ=QK=K$, because complementing all bits in either string changes the exponent by an even integer. This is an exact character-sum proof for every $N$, not a finite-size inference.

Substitution in the two-sector matrices proves $Y_D^2=I+Y_{\eta }$, $Y_{\eta }^2=I$, and $Y_{\eta }{Y}_D=Y_D{Y}_{\eta }=Y_D$. Each $K$ has rank $2^{N-1}$, so $Y_D$ has rank $2^N$ on the full $2^{N+1}$-dimensional space and is noninvertible. At $N=1$, $K$ is $2^{-1\mathrm{/}2}$ times the all-ones matrix, so these identities and ranks still hold without deleting either sector.

Flipping input bit $b_j$ changes the kernel phase by $c_j+c_{j+1}$, and flipping output bit $c_{j+1}$ changes it by $b_j+b_{j+1}$. Therefore $K{X}_j=Z_j{Z}_{j+1}K$ and $K{Z}_j{Z}_{j+1}=X_{j+1}K$ for all periodic $j$. Their adjoints prove the reverse-sector equations. Comparison with all four projector formulas of step 3.1 gives $[Y_D,P_i]=[Y_{\eta },P_i]=0$ for every $i$, hence commutation with every Hamiltonian of step 5.1. In particular the same line intertwines the two opposite-coupling spin Hamiltonians. This is the actual local operator intertwining, including wraparound.

Transposing the kernel interchanges ${\sum }_j{c}_j(b_{j-1}+b_j)$ with ${\sum }_j{c}_j(b_j+b_{j+1})$, which is exactly a cyclic shift of its input or output. Thus $K^{†}=CK=KC$. Substitution into the full block formulas for $Y_D$ and $T$ proves their commutation; $Q$ also commutes with $C$, so $Y_{\eta }$ commutes with $T$. These are statements on the doubled path space with the explicitly defined translation.

On one bit-chain space the same identities instead give $K^2=(I+Q)C^{-1}$ and $(K\mathrm{/}\sqrt{2}{)}^2=((I+Q)\mathrm{/}2)C^{-1}$. The latter normalized map has spin-lattice translation $T_{\mathrm{I}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}}=C^{-1}$ in this orientation. Its translated square differs from the doubled-space fusion square and its domain has half as many states. Thus both the full two-sector fusion algebra and the specified translated single-chain algebra have been proved, together with all local projectors, line maps, wrapping, ranks, Hamiltonian consequences and intertwining assertions.