QCA Equivalence Classes

Nov 04, 2025 Ryder McDowell

This post sets up the framework for classifying locality-preserving automorphisms (QCAs) of the $\mathbb{Z}_2$ symmetric subalgebra supported on a finite region. The punchline is that the equivalence classes form a group isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$, with the two factors measuring what happens at the left and right boundaries. We then use this to show that the fusion operators of a $G$-symmetric QCA define a 2-cocycle.

QCAs on a Subalgebra

A QCA on a subalgebra $\mathcal{S}$ is a locality-preserving (LP) $*$-automorphism of $\mathcal{S}$. Concretely, $\alpha : \mathcal{S} \to \mathcal{S}$ is a QCA if it is a $*$-algebra automorphism, has an inverse $\alpha^{-1}$ on $\mathcal{S}$, and both $\alpha$ and $\alpha^{-1}$ have finite spread $\ell$:

\[\mathrm{supp}(\alpha(O)) \subseteq \mathrm{supp}(O)^{+\ell}\]

meaning $\alpha$ can only spread the support of an operator by at most $\ell$ sites.

Locally Supported Automorphisms

Fix a region $R_{j,\rho} = [j - \rho, j + \rho] \cap \mathbb{Z}$ on the chain. A locally supported automorphism (LSA) at scale $\ell$ is an automorphism supported in $R_{j,\rho}$ whose output can spread at most $\ell$ outside:

\[\mathrm{LSA}_{j,\rho}(\ell) := \left\{ \alpha : \mathcal{A}_{\mathbb{Z}_2} \to \mathcal{A}_{\mathbb{Z}_2} \;\middle\vert \; \mathrm{supp}(\alpha) \subseteq R_{j,\rho},\;\; \alpha, \alpha^{-1} \text{ have spread} \leq \ell \right\}\]

We need $\rho \geq 1$ since the generator $Z_i Z_{i+1}$ already has range 1.

Equivalence relation. We declare $\alpha \sim \alpha'$ if $\alpha' = \mathrm{Ad}_V \circ \alpha$ for some unitary $V \in \mathcal{A}_{\mathbb{Z}_2}$ supported in $R_{j,\rho}$. This captures the idea that two QCAs are equivalent if they differ only by a local symmetric unitary (i.e. a finite-depth circuit supported on the region).

The quotient group is

\[\mathcal{G}_{j,\rho} := \mathrm{LSA}_{j,\rho}(\ell) \big/ \!\sim\]

with product $[\alpha][\beta] = [\alpha \circ \beta]$. We expect this group to have exactly 4 elements, corresponding to: trivial, left boundary defect, right boundary defect, and both.

String and Endpoint Operators

To measure what happens at each boundary, we introduce truncated $X$-strings just outside $R_{j,\rho}$:

\[\widehat{S}_L(n) := \prod_{t=0}^{n} X_{j-\rho-t} \qquad \widehat{S}_R(n) := \prod_{t=0}^{n} X_{j+\rho+t}\]

and define the endpoint operators:

\[\widehat{D}_L^{(n)}(\alpha) := \alpha(\widehat{S}_L(n))\,\widehat{S}_L(n) \qquad \widehat{D}_R^{(n)}(\alpha) := \alpha(\widehat{S}_R(n))\,\widehat{S}_R(n)\]

These were introduced as the bounded unitary operators $O_L, O_R$ in Ma–Li–Cheng [Section IV.A].

Lemma. For all $n \geq 0$, $\widehat{D}_L^{(n+1)}(\alpha) = \widehat{D}_L^{(n)}(\alpha)$ (and similarly for $R$).

Proof. Write $\widehat{S}_L(n+1) = X_{j-\rho-(n+1)} \widehat{S}_L(n)$ and expand:

\[\widehat{D}_L^{(n+1)}(\alpha) = \alpha(X_{j-\rho-(n+1)})\,\alpha(\widehat{S}_L(n))\,X_{j-\rho-(n+1)}\,\widehat{S}_L(n)\]

Then $\widehat{D}_L^{(n+1)}(\alpha)\,(\widehat{D}_L^{(n)}(\alpha))^\dagger$ simplifies (using unitarity of $\widehat{S}_L(n)$ and disjoint support) to

\[\alpha(X_{j-\rho-(n+1)})\,X_{j-\rho-(n+1)}\]

Since site $j - \rho - (n+1)$ lies outside $R_{j,\rho}$, $\alpha$ acts trivially there, so this equals $X^2 = 1$, giving $\widehat{D}_L^{(n+1)} = \widehat{D}_L^{(n)}$. $\square$

Since the endpoint operators are independent of $n$, we drop the superscript and write $D_L(\alpha)$, $D_R(\alpha)$.

Boundary Charges are $\mathbb{Z}_2$-Valued

Lemma (Boundary to $\mathbb{Z}_2$). Assume $\mathrm{supp}(\alpha) \subseteq R_{j,\rho}$ and spread $\leq \ell$. Then:

$D_L(\alpha)$ is supported in $I_L = [j - \rho - \ell,\, j - \rho + \ell]$
$D_R(\alpha)$ is supported in $I_R = [j + \rho - \ell,\, j + \rho + \ell]$

Moreover, modulo the equivalence $\alpha \sim \mathrm{Ad}_V \circ \alpha$ with $\mathrm{supp}(V) \subseteq R_{j,\rho}$, each endpoint defines a $\mathbb{Z}_2$-valued charge:

\[D_L(\alpha) \sim \mathbf{1} \text{ or } Z_{j-\rho} \qquad D_R(\alpha) \sim \mathbf{1} \text{ or } Z_{j+\rho}\]

Proof sketch. Setting $n = 0$: $D_L(\alpha) = \alpha(X_{j-\rho})\,X_{j-\rho}$. The spread bound gives the support claim. Under conjugation by $\mathrm{Ad}_V$, the endpoint operator transforms as $D_L(\alpha) \mapsto V D_L(\alpha) V_L^\dagger$ for a local unitary $V_L$. Following the analysis of Ma–Li–Cheng (Props. 2–4), $D_L(\alpha)$ is invariant up to local symmetric unitaries in $I_L$, leaving exactly two equivalence classes represented by $\mathbf{1}$ and $Z_{j-\rho}$. $\square$

Note: The proof that exactly two classes exist (and the canonical representatives are $\mathbf{1}$ and $Z$) relies on the detailed conjugation analysis of Ma–Li–Cheng, which we haven’t fully reproduced here. This step needs to be filled in.

The Map $\pi : \mathcal{G}_{j,\rho} \to \mathbb{Z}_2 \times \mathbb{Z}_2$

Define labels $\ell(\alpha), r(\alpha) \in \{0, 1\}$ by whether $D_L(\alpha)$ and $D_R(\alpha)$ are equivalent to $\mathbf{1}$ (label 0) or $Z$ (label 1). Then define

\[\pi : \mathcal{G}_{j,\rho} \to \mathbb{Z}_2 \times \mathbb{Z}_2, \qquad \pi([\alpha]) = \big((-1)^{\ell(\alpha)},\, (-1)^{r(\alpha)}\big)\]

Theorem. $\pi$ is a well-defined group isomorphism.

Proof.

Well-defined: If $\alpha' \sim \alpha$, then $D_{L/R}(\alpha')$ and $D_{L/R}(\alpha)$ are related by conjugation by a local symmetric unitary, which doesn’t change which class ($\mathbf{1}$ or $Z$) they’re in. So the labels are constant on equivalence classes.

Homomorphism: We compute how endpoint operators compose. With $S_L = X_{j-\rho}$:

\[D_L(\alpha \circ \beta) = (\alpha \circ \beta)(S_L)\,S_L = \alpha(\beta(S_L))\,S_L = \alpha(D_L(\beta)\,S_L)\,S_L = \alpha(D_L(\beta))\,D_L(\alpha)\]

This gives the addition law $\ell(\alpha \circ \beta) = \ell(\alpha) + \ell(\beta) \pmod{2}$, and the same for $r$. So $\pi([\alpha \circ \beta]) = \pi([\alpha])\,\pi([\beta])$.

Surjective: The four classes are realized by

$(\ell, r)$	Representative
$(0,0)$	$[\mathrm{id}]$
$(1,0)$	$[\mathrm{Ad}_{Z_{j-\rho}}]$
$(0,1)$	$[\mathrm{Ad}_{Z_{j+\rho}}]$
$(1,1)$	$[\mathrm{Ad}_{Z_{j-\rho} Z_{j+\rho}}]$

Injective: If $\pi([\alpha]) = (1,1)$, then $D_L = D_R = \mathbf{1}$. By the analysis of Ma–Li–Cheng, this means $\alpha$ is generated by a symmetric finite-depth circuit inside $R_{j,\rho}$, i.e. $\alpha = \mathrm{Ad}_V$ for some $V \in FDQC(\mathcal{A}_{\mathbb{Z}_2})$. By definition of the equivalence relation, $[\alpha] = [\mathrm{id}]$, so the kernel is trivial. $\square$

Fusion Operators and the 2-Cocycle

Now consider a $G$-symmetric QCA, where $G$ is a finite group. Recall from Disentangling Anomaly-free Symmetries of Quantum Spin Chains that the fusion operators are built from truncated unitaries:

\[\lambda_j(g,h) \;\propto\; U^{gh}_{\leq j}\,\Big(U^g_{\leq j-1}\,U^h_{\leq j}\Big)^\dagger\]

These are locally supported automorphisms, hence elements of $\mathrm{LSA}_{j,\rho}(\ell)$. If the symmetry is anomaly-free, the fusion operators satisfy the F-move identity in the quotient:

\[[\lambda_j(g,h)]\,[\lambda_j(gh,k)] = [\lambda_j(h,k)]\,[\lambda_j(g,hk)] \quad \text{in } \mathcal{G}_{j,\rho}\]

Theorem. Define $\eta : G \times G \to \mathbb{Z}_2 \times \mathbb{Z}_2$ by $\eta(g,h) := \pi([\lambda_j(g,h)])$. Then $\eta$ is a 2-cocycle:

\[\eta(g,h)\,\eta(gh,k) = \eta(h,k)\,\eta(g,hk) \quad \text{in } \mathbb{Z}_2 \times \mathbb{Z}_2\]

Proof. Apply the group isomorphism $\pi$ to both sides of the F-move identity and use multiplicativity of $\pi$. $\square$

Note (to return to): The F-move identity in the quotient needs to be derived more carefully. The relation $\lambda_j(g,hk)\,\lambda_{j-1}(g,1)\,\lambda_j(h,k) = F_j(g,h,k)\,\lambda_j(gh,k)\,\lambda_{j-1}(g,h)$ holds at the level of operators (with $F_j$ a $U(1)$ phase). Passing to the quotient and dropping the $F$-phase requires justification — this needs to be pinned down using the feedback from the 11/13 meeting.

Summary

The locally supported automorphisms of $\mathcal{A}_{\mathbb{Z}_2}$ on a region $R_{j,\rho}$, modulo local symmetric unitaries, form a group $\mathcal{G}_{j,\rho} \cong \mathbb{Z}_2 \times \mathbb{Z}_2$. The two $\mathbb{Z}_2$ factors measure the boundary defects at the left and right endpoints. Applying this to fusion operators of a $G$-symmetric QCA, the endpoint-parity data organizes into a 2-cocycle $\eta : G \times G \to \mathbb{Z}_2 \times \mathbb{Z}_2$.

Set-up

$\beta$ is an LP *-automorphism, $\beta:\mathcal{A}_{sym}\vert_I$ $\rightarrow\mathcal{A}_{sym}\vert_I$, where $\mathcal{A}_{sym}\vert_I$ is the symmetric subalgebra restricted to a finite interval $I$ of lattice sites on a 1D infinite spin chain.

In our $\text{sym} = \mathbb{Z}_2$ case, our total algebra $\mathcal{A} = \langle X_i, Z_i \rangle$ with the $\mathbb{Z}_2$ symmetric subalgebra $\mathcal{A}_{\mathbb{Z}_2} = \langle X_i, Z_{i,i+1} \rangle$ with $i \in \mathbb{Z}$.

We define the string-like operators:

\[\bar{S}_x = \prod_{i\in[-n,n]} X_{i} \qquad \bar{S}_{zz} = \prod_{i\in[-n,n]} Z_{i, i+1}\]

where the endpoints of the strings lie far outside the interval $I$. We claim:

\[\beta(\bar{S}_x)=\pm\bar{S}_x \qquad \beta(\bar{S}_{zz})=\pm\bar{S}_{zz}\]

Approach. We use the fact that *-automorphisms preserve centers and the identity.

Lemma 1

Lemma 1. Let $\bar{S}_x\vert_I = \prod_{i\in I} X_i$. The center of our $\mathbb{Z}_2$-symmetric subalgebra $\mathcal{A}_{\mathbb{Z}_2}\vert_I = \langle X_i, Z_{j,j+1} \mid i \in [1,n] = I,\; j \in [1,n-1] \rangle$ is generated by this restricted X string operator, i.e. $Z(\mathcal{A}_{\mathbb{Z}_2}\vert_I) = \text{span}\{\mathbf{e},\, \bar{S}_X\vert_I\}$

Proof

Step 1. With $\mathcal{A}_{\mathbb{Z}_2} = \langle X_i, Z_{j,j+1} \rangle$, we know for $a \in \mathcal{A}_{\mathbb{Z}_2}$:

\[a = \sum_{\alpha} c_{\alpha}W_{\alpha}\]

it is a linear combination of words from the presentation ($W_{\alpha} = s_1 s_2 \cdots s_m$ with $s_k \in \{X_i, Z_{i,i+1}\}$) and weights ($c_{\alpha} \in \mathbb{C}$). Now, since each $Z_{i,i+1} = Z_i Z_{i+1}$, we can further say that

\[W_{\alpha} = s_1 s_2 \cdots s_m \quad \text{with } s_k \in \{X_i, Z_i\}\]

(Since we are dealing with algebras over a field, this is just saying that the words are all possible operators formed by generators of the algebra, and any element in the algebra is some superposition of these words with some complex coefficient.)

Step 2. By commuting generators and using $X_i^2 = Z_i^2 = \mathbf{1}$, we can rewrite each word $W_\alpha$ in reduced form:

\[\widetilde{W}_\alpha = \prod_{i\in S_\alpha} X_i \prod_{i\in T_\alpha} Z_i\]

where $S, T$ denote the subsets of lattice sites with $X_i$ and $Z_i$ generators on them. In doing so we pick up at most a sign, i.e. $W_\alpha = \pm \widetilde{W}_\alpha$, after applying the relations: $X_i Z_i = -Z_i X_i, \quad X_i^2 = Z_i^2 = 1, \quad [X_i, X_j] = [Z_i, Z_j] = 0$

In sum,

\[a = \sum_{\alpha} c_{\alpha}W_{\alpha} = \sum_{\alpha} c_{\alpha}'\,\widetilde{W}_{\alpha}, \qquad c_{\alpha}' = \pm c_{\alpha}\]

Example. Let $a = c_1(X_2 X_4 Z_2 Z_3 X_5 Z_3 Z_4) + c_2(Z_4 Z_5 X_4)$.

First word: $\begin{aligned} X_2 X_4 Z_2 Z_3 X_5 Z_3 Z_4 &= X_2 Z_2 X_4 Z_3 X_5 Z_3 Z_4 && \text{(commute $X_4$ past $Z_2$)}\\ &= -Z_2 X_2 X_4 Z_3 X_5 Z_3 Z_4 && \text{($X_i Z_i = -Z_i X_i$)}\\ &= -Z_2 X_2 X_4 X_5 Z_3 Z_3 Z_4 && \text{(commute $Z_3$ past $X_5$)}\\ &= -Z_2 X_2 X_4 X_5 Z_4 && \text{($Z_3^2 = \mathbf{1}$)}\\ &= X_2 Z_2 X_4 X_5 Z_4 && \text{($X_i Z_i = -Z_i X_i$)}\\ &= X_2 X_4 X_5 Z_2 Z_4 && \text{(commute $Z_2$ past $X_4, X_5$)} \end{aligned}$

Second word: $Z_4 Z_5 X_4 = Z_4 X_4 Z_5 = -X_4 Z_4 Z_5$

So in reduced form: $a = c_1\, X_2 X_4 X_5 Z_2 Z_4 - c_2\, X_4 Z_4 Z_5$

Step 3. Since our total algebra on the interval is $\mathcal{A}\vert_I \subset \bigotimes_{i \in I} M_2(\mathbb{C})$, and the Pauli operators $\{\mathbf{e}, X_i, Z_i, Y_i = iX_iZ_i\}$ form a basis of $M_2(\mathbb{C})$ at each site, their tensor products form a basis of $\bigotimes_{i\in I} M_2(\mathbb{C})$.

Each reduced word $\widetilde{W}_\alpha$ is a tensor product:

\[W = \bigotimes_{i \in I} O_i, \qquad O_i = \begin{cases} \mathbf{1} & i \notin S \text{ and } i \notin T \\ X_i & i \in S \text{ and } i \notin T \\ Z_i & i \notin S \text{ and } i \in T \\ X_i Z_i & i \in S \text{ and } i \in T \end{cases}\]

Since these reduced words exhaust all combinations of single-site basis elements and are all distinct, they are linearly independent.

Step 4. Since $\widetilde{W}_\alpha$ form a basis, $a$ is central if and only if every $\widetilde{W}_\alpha$ with $c'_\alpha \neq 0$ is central. So it suffices to determine for which $S, T \subset I$ a single word

\[W = \prod_{i\in S} X_i \prod_{i\in T} Z_i\]

satisfies $[W, X_j] = 0$ and $[W, Z_{j,j+1}] = 0$ for all $j$ with $\{j, j+1\} \subset I$.

Commutator with $X_j$: Using that $X_j$ commutes with $X_i, Z_i$ for $i \neq j$ and anticommutes with $Z_j$,

\[X_j W = (-1)^{\mathbf{1}_{j\in T}}\, W X_j\]

Thus $[W, X_j] = 0$ for all $j \in I$ iff $T \cap I = \varnothing$, i.e. $T = \varnothing$. So any central word must be of the form:

\[W = \prod_{i\in S} X_i, \qquad S \subset I\]

Step 5. Commutator with $Z_{j,j+1} = Z_j Z_{j+1}$: Each of $Z_j$ and $Z_{j+1}$ anticommutes with $X_j, X_{j+1}$ and commutes with all other $X_i$, so

\[Z_{j,j+1}\, W = (-1)^{\mathbf{1}_{j\in S} + \mathbf{1}_{j+1\in S}}\, W Z_{j,j+1}\]

Therefore $[W, Z_{j,j+1}] = 0$ iff $\mathbf{1}_{j\in S} + \mathbf{1}_{j+1\in S} \equiv 0 \pmod{2}$, i.e. either both $j, j+1$ lie in $S$ or both lie in $S^c$. This forces either $S = \varnothing$ or $S = I$.

Thus the only words commuting with all generators are

\[W = \mathbf{e} \quad\text{and}\quad W = \prod_{i\in I} X_i = \bar{S}_X\vert_I\]

and we obtain:

\[\boxed{Z\!\left(\mathcal{A}_{\mathbb{Z}_2}\vert_I\right) = \text{span}\{\mathbf{e},\, \bar{S}_X\vert_I\}} \qquad \square\]

Claim: $\beta(\bar{S}_x) = \pm\bar{S}_x$

By Lemma 1, $Z(\mathcal{A}_{\mathbb{Z}_2}\vert_I) = \langle \mathbf{1}, \bar{S}_x\vert_I \rangle$.

A *-automorphism preserves the center and the identity, so

\[\beta(\bar{S}_x\vert_I) = a\,\mathbf{1} + b\,\bar{S}_x\vert_I\]

for some $a, b \in \mathbb{C}$. Set $S := \bar{S}_x\vert_I$. Since $S^* = S$ and $S^2 = \mathbf{1}$,

\[\beta(S)^2 = \beta(S^2) = \beta(\mathbf{1}) = \mathbf{1}\]

Thus

\[\beta(S)^2 = (a\mathbf{1} + bS)^2 = (a^2 + b^2)\mathbf{1} + 2ab\,S = \mathbf{1}\]

So $2ab = 0$ and $a^2 + b^2 = 1$. Since $S$ is not a scalar multiple of the identity and $\beta$ is an automorphism, we must have $b \neq 0$, forcing $a = 0$ and $b = \pm 1$. Therefore:

\[\boxed{\beta(\bar{S}_x\vert_I) = \pm\,\bar{S}_x\vert_I} \qquad \square\]

Lemma 2

To approach the effect of $\beta$ on $\bar{S}_{zz}$, we consider a slightly different $\mathbb{Z}_2$ subalgebra:

\[\mathcal{A}_{\mathbb{Z}_2}'\vert_I = \langle X_i, Z_{i,i+1} \mid i \in [2,n-1],\; j \in [1,n-1] \rangle\]

Lemma 2. The center of $\mathcal{A}_{\mathbb{Z}_2}'\vert_I$ is generated by the restricted Z string operator: $Z(\mathcal{A}_{\mathbb{Z}_2}'\vert_I) = \text{span}\{\mathbf{e},\, \bar{S}_{zz}\vert_I\}$

Proof

For any $a \in \mathcal{A}_{\mathbb{Z}_2}'$ we have the reduced form

\[\widetilde{W}_\alpha = \prod_{i\in S_\alpha} X_i \prod_{i\in T_\alpha} Z_i\]

with $S_\alpha \subset \{2, \ldots, n-1\}$ and $T_\alpha \subset \{1, \ldots, n\}$. By the same argument as before, these reduced words are linearly independent.

X commutation condition. For $j \in \{2, \ldots, n-1\}$:

\[X_j W = (-1)^{\mathbf{1}_{j\in T}}\, W X_j\]

So $[W, X_j] = 0$ for all $j = 2, \ldots, n-1$ iff $T \subset \{1, n\}$.

Z bond commutation condition. By the same analysis:

\[Z_{j,j+1} W = (-1)^{\mathbf{1}_{j\in S} + \mathbf{1}_{j+1\in S}}\, W Z_{j,j+1}\]

So $[W, Z_{j,j+1}] = 0$ for all $j = 1, \ldots, n-1$ iff $\mathbf{1}_{j\in S} + \mathbf{1}_{j+1\in S} \equiv 0 \pmod{2}$ for all $j$.

But $S \subset \{2, \ldots, n-1\}$, so:

For $j = 1$: $\mathbf{1}_{1\in S} + \mathbf{1}_{2\in S} = 0$. Since $1 \notin S$, this forces $2 \notin S$.
For $j = 2$: now $2 \notin S$ forces $3 \notin S$.
Continuing inductively: $S = \varnothing$.

Hence no $X_i$ can appear in a central word, and any central $W$ has the form $W = \prod_{i\in T} Z_i$ with $T \subset \{1, n\}$ and $\vert T \vert$ even (since the generators are Z-bonds, not on-site $Z_i$). The only possibilities are:

\[W = \mathbf{e} \quad\text{or}\quad W = Z_1 Z_n\]

Now $Z_1 Z_n \in \mathcal{A}_{\mathbb{Z}_2}'\vert_I$ since:

\[\prod_{j=1}^{n-1} Z_{j,j+1} = \prod_{j=1}^{n-1}(Z_j Z_{j+1}) = Z_1 Z_n\]

as each interior $Z_2, \ldots, Z_{n-1}$ appears twice and squares to $1$. So

\[Z_1 Z_n = \bar{S}_{zz}\vert_I = \prod_{j=1}^{n-1} Z_{j,j+1} \in \mathcal{A}_{\mathbb{Z}_2}'\vert_I\]

Hence:

\[\boxed{Z(\mathcal{A}_{\mathbb{Z}_2}'\vert_I) = \text{span}\{\mathbf{e},\, \bar{S}_{zz}\vert_I\}} \qquad \square\]

Claim: $\beta(\bar{S}_{zz}) = \pm\bar{S}_{zz}$

By Lemma 2, $Z(\mathcal{A}_{\mathbb{Z}_2}'\vert_I) = \text{span}\{\mathbf{e},\, \bar{S}_{zz}\vert_I\}$.

A *-automorphism preserves the center and the identity, so

\[\beta(\bar{S}_{zz}\vert_I) = a\,\mathbf{e} + b\,\bar{S}_{zz}\vert_I\]

for some $a, b \in \mathbb{C}$. Set $S := \bar{S}_{zz}\vert_I$. Since $S^* = S$ and $S^2 = \mathbf{1}$,

\[\beta(S)^2 = \beta(S^2) = \beta(\mathbf{e}) = \mathbf{e}\]

Thus

\[\beta(S)^2 = (a\,\mathbf{e} + bS)^2 = (a^2 + b^2)\,\mathbf{e} + 2ab\,S = \mathbf{e}\]

So $2ab = 0$ and $a^2 + b^2 = 1$. Since $S$ is not a scalar multiple of the identity and $\beta$ is an automorphism, $b \neq 0$, forcing $a = 0$ and $b = \pm 1$. Therefore:

\[\boxed{\beta(\bar{S}_{zz}\vert_I) = \pm\,\bar{S}_{zz}\vert_I} \qquad \square\]

By Lemma 2, $Z(\mathcal{A}_{\mathbb{Z}_2}'\vert_I)$ = $\text{span}\{\mathbf{e}, \bar{S}_{zz_{I}}\}$.

A *-automorphism preserves the center and the identity, so

\[\beta(\bar{S}_{zz}\vert_I) = a\,\mathbf{e} + b\,\bar{S}_{zz}\vert_I\]

for some $a, b \in \mathbb{C}$. Set $S := \bar{S}_{zz}\vert_I$. Since $S^* = S$ and $S^2 = \mathbf{1}$,

\[\beta(S)^2 = \beta(S^2) = \beta(\mathbf{e}) = \mathbf{e}\]

Thus

\[\beta(S)^2 = (a\,\mathbf{e} + bS)^2 = (a^2 + b^2)\,\mathbf{e} + 2ab\,S = \mathbf{e}\]

So $2ab = 0$ and $a^2 + b^2 = 1$. Since $S$ is not a scalar multiple of the identity and $\beta$ is an automorphism, $b \neq 0$, forcing $a = 0$ and $b = \pm 1$. Therefore:

\[\boxed{\beta(\bar{S}_{zz}\vert_I) = \pm\,\bar{S}_{zz}\vert_I} \qquad \square\]

\((\ell, r)\)	Representative
\((0,0)\)	\([\mathrm{id}]\)
\((1,0)\)	\([\mathrm{Ad}_{Z_{j-\rho}}]\)
\((0,1)\)	\([\mathrm{Ad}_{Z_{j+\rho}}]\)
\((1,1)\)	\([\mathrm{Ad}_{Z_{j-\rho} Z_{j+\rho}}]\)

QCA Equivalence Classes

QCAs on a Subalgebra

Locally Supported Automorphisms

String and Endpoint Operators

Boundary Charges are \(\mathbb{Z}_2\)-Valued

The Map \(\pi : \mathcal{G}_{j,\rho} \to \mathbb{Z}_2 \times \mathbb{Z}_2\)

Fusion Operators and the 2-Cocycle

Summary

Set-up

Lemma 1

Proof

Claim: \(\beta(\bar{S}_x) = \pm\bar{S}_x\)

Lemma 2

Proof

Claim: \(\beta(\bar{S}_{zz}) = \pm\bar{S}_{zz}\)