H Symmetries and Virtual-Boundary Nondecay: Supporting Results

This appendix supports Chapter 12. It collects permutation and gauge bookkeeping, factor-system identities, stationary-boundary twist calculations, common trace-preserving gauges, and the finite-dimensional limit arguments used in the virtual-boundary nondecay results. The notation is introduced locally in each part, and every reference to the main development is given by its stable label.

The physical index set is $I=\{ 0,\ldots ,d-1\} $, the bond algebra is $M_{D}(\mathbb {C})$, and $G$ denotes a group. All sums below are finite. The first section treats permutation actions on $I$; the second isolates scalar uniqueness for two gauges of the same injective tensor; the third records the elementary factor-system bookkeeping used in the cohomology discussion of Chapter 12.

H.1 Permutation reindexing and identity bookkeeping

The linear-combination twist $\widetilde{A}_g^i = \sum _j U(g)_{ij}\, A^j$ is the standard form used in the virtual representation theorem. The following alternative formulation replaces the matrix action by a permutation of the physical index; it is useful for symmetries that act by reordering basis states rather than by a general linear map.

Definition H.1.1 Permutation symmetry datum

✓

An on-site symmetry datum for a group $G$ on a physical space of dimension $d$ consists of a matrix-valued map $u : G \to M_{d}(\mathbb {C})$ and a physical-index action $\sigma : G \to \mathrm{End}(\{ 0,\ldots ,d{-}1\} )$.

Definition H.1.2 Permutation-twisted tensor

✓

Given an on-site symmetry datum $(u, \sigma )$, the permutation-twisted tensor at group element $g$ is

\begin{align} (A^{\sigma _g})^i & := A^{\sigma (g)(i)}. \label{eq:symmetry_perm_twist} \end{align}

In tensor-network notation the local tensor is unchanged and only the physical leg is relabelled:

\begin{align} \begin{tenkz} [physical=up] \tn[up=$i$]{A} \end{tenkz}& \qquad \xrightarrow {\sigma _g} \qquad \begin{tenkz} [physical=up] \tn[up=$\sigma(g)(i)$]{A} \end{tenkz}. \notag \end{align}

Lemma H.1.3 Identity permutation twist

✓

If $\sigma (1) = \operatorname{id}$, then $(A^{\sigma _1})^i = A^i$ for all $i$.

Proof ▶

The claim follows from $\sigma (1)(i) = i$ for all $i$.

Composition order. The linear-combination twist (Lemma 12.1.3) composes as $\widetilde{(\widetilde{A}_h)}_g = \widetilde{A}_{gh}$, applying $h$ first then $g$. The permutation twist below uses the same left-action convention $\sigma (gh) = \sigma (g) \circ \sigma (h)$, so $(A^{\sigma _g})^{\sigma _h} = A^{\sigma _{gh}}$; the inner permutation $\sigma (h)$ acts first on the index.

Lemma H.1.4 Composition of permutation twists

✓

If $\sigma (gh) = \sigma (g) \circ \sigma (h)$ for all $g, h \in G$, then

\begin{align} A^{\sigma _{gh}} & = (A^{\sigma _g})^{\sigma _h}. \label{eq:symmetry_perm_composition} \end{align}

Proof ▶

Expanding both sides gives $(A^{\sigma _{gh}})^i = A^{\sigma (g)(\sigma (h)(i))} = ((A^{\sigma _g})^{\sigma _h})^i$. Diagrammatically this is just two consecutive relabellings of the same local tensor:

\begin{align} \tnpic[physical=up]{\tn[up=$i$]{A}} & \qquad \xrightarrow {\sigma _h} \qquad \tnpic[physical=up]{\tn[up=$\sigma(h)(i)$]{A}} \qquad \xrightarrow {\sigma _g} \qquad \tnpic[physical=up]{\tn[up=$\sigma(g)(\sigma(h)(i))$]{A}}. \notag \end{align}

H.1.1 Linear physical-index rotations

The same bookkeeping applies to a general matrix acting on the physical index.

Definition H.1.1.1 Physical-index rotation

✓

Given a matrix $M \in M_{d}(\mathbb {C})$ and an MPS tensor $A$, the physical-index rotation is the tensor $(A_M)^i := \sum _j M_{ij}\, A^j$. When the matrix is a unitary $u$, the rotation is written $A_u$.

Lemma H.1.1.2 Matrix product vector under physical rotation

✓

For a length-$N$ physical configuration $s=(s_0,\ldots ,s_{N-1})$, physical rotation satisfies

\begin{align} \operatorname{tr}((A_M)^{s_0}\cdots (A_M)^{s_{N-1}}) & = \sum _{t\in \{ 0,\ldots ,d-1\} ^N} \left(\prod _{n=0}^{N-1}M_{s_n,t_n}\right) \operatorname{tr}(A^{t_0}\cdots A^{t_{N-1}}). \label{eq:symmetry_mpv_rotation} \end{align}

Proof ▶

Substitute $(A_M)^{s_n}=\sum _{t_n}M_{s_n,t_n}A^{t_n}$ at each site, distribute the finite sums through the ordered matrix product, and use linearity of the trace to obtain (??).

Lemma H.1.1.3 Word expansion for a twisted tensor

✓

For a word $b:\{ 0,\ldots ,L-1\} \to \{ 0,\ldots ,d-1\} $,

\begin{align} A_g^{b_0}\cdots A_g^{b_{L-1}} & =\sum _{v:\{ 0,\ldots ,L-1\} \to \{ 0,\ldots ,d-1\} } \left(\prod _{k=0}^{L-1}U(g)_{b_kv_k}\right) A^{v_0}\cdots A^{v_{L-1}}. \label{eq:app_symmetry_twisted_word} \end{align}

Proof ▶

Substitute $A_g^{b_k}=\sum _{v_k}U(g)_{b_kv_k}A^{v_k}$ into the ordered product. Repeated distributivity gives one summand for each function $v$, scalar coefficients commute past the matrices and multiply, and the remaining ordered product is $A^{v_0}\cdots A^{v_{L-1}}$.

H.2 Gauge ratios and scalar uniqueness

The virtual representation theorem (Theorem 12.1.16) relies on the fact that composing two gauge transformations yields another gauge up to a scalar factor.

Lemma H.2.1 Gauge ratio commutes with tensor matrices

✓

Diagrammatically, the hypothesis is that the two local gauges

$\begin{tenkz}[physical=up] \tn[up=$i$]{B} \end{tenkz}$

$ = $

$\begin{tenkz}[physical=up] \tnX{X} & \tn[up=$i$]{A} & \tnX{X^{-1}} \end{tenkz}$

and

$\begin{tenkz}[physical=up] \tn[up=$i$]{B} \end{tenkz}$

$ = $

$\begin{tenkz}[physical=up] \tnX{Y} & \tn[up=$i$]{A} & \tnX{Y^{-1}} \end{tenkz}$

produce the same target tensor $B^i$ from the same source tensor $A^i$. If $X, Y \in \mathrm{GL}_D(\mathbb {C})$ both conjugate an MPS tensor $A$ to the same tensor $B$ (i.e. $B^i = X A^i X^{-1} = Y A^i Y^{-1}$ for all $i$), then $Y^{-1} X$ commutes with every $A^i$.

Proof ▶

Direct matrix calculation: conjugating both gauge relations and composing gives $(Y^{-1} X) A^i = A^i (Y^{-1} X)$.

Lemma H.2.2 Gauge ratio is scalar for injective tensors

✓

Under the same hypotheses, if $A$ is additionally injective, then $Y^{-1} X = \lambda \mathbb {1}_D$ for some scalar $\lambda \in \mathbb {C}$.

Proof ▶

The same two-gauge picture shows that the ratio $Y^{-1}X$ acts invisibly on the local tensor. By Lemma H.2.1, $Y^{-1}X$ commutes with all $A^i$. Since $\{ A^i\} $ spans $M_{D}(\mathbb {C})$, the matrix $Y^{-1}X$ commutes with all of $M_{D}(\mathbb {C})$. By Lemma 2.1.5, it is a scalar matrix.

Theorem H.2.3 Projective multiplication law for gauge matrices

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Let $A$ be an injective MPS tensor with on-site symmetry under a group representation $U : G \to \mathrm{GL}_d(\mathbb {C})$. For each $g \in G$, let $X(g) \in \mathrm{GL}_D(\mathbb {C})$ satisfy $\widetilde{A}_g^i = X(g)\, A^i\, X(g)^{-1}$ for all $i$. Then for all $g, h \in G$, there exists $c(g,h) \in \mathbb {C}$ such that $X(h)\, X(g) = c(g,h)\, X(g \cdot h)$. In fact $c(g,h) \ne 0$ since both sides are invertible, so $c(g,h) \in \mathbb {C}^{\times }$. Diagrammatically, the two candidate gauges are competing local replacements for the same twisted tensor:

$\begin{tenkz}[physical=up] \tnX{X(h)X(g)} & \tn[up=$i$]{A} & \tnX{(X(h)X(g))^{-1}} \end{tenkz}$ $ = $ $\begin{tenkz}[physical=up] \tnX{X(gh)} & \tn[up=$i$]{A} & \tnX{X(gh)^{-1}} \end{tenkz}$

and both sides conjugate $A^i$ to the same twisted tensor $\widetilde{A}_{gh}^i$.

Proof ▶

By the composition law $\widetilde{A}_{gh} = \widetilde{(\widetilde{A}_h)}_g$ (Lemma 12.1.3), both $X(h)\, X(g)$ and $X(gh)$ conjugate $A$ to $\widetilde{A}_{gh}$. Lemma H.2.2 then gives the scalar factor.

H.3 Cocycle equivalence and coboundaries

Let $\omega _1,\omega _2:G\times G\to \mathbb {C}^\times $ be arbitrary scalar $2$-cochains. The definitions of coboundary and cohomology are Definitions 12.2.1 and 12.2.2.

Theorem H.3.1 Cohomology relation is an equivalence relation

✓

The relation “cohomologous to” is reflexive, symmetric, and transitive on the set of $\mathbb {C}^\times $-valued scalar $2$-cochains, and therefore also on the subset of cocycles.

Proof ▶

Reflexivity: take $\varphi =1$. Symmetry: replace $\varphi $ by $\varphi ^{-1}$. Transitivity: compose the witnesses $\varphi $ and $\psi $ pointwise in (??).

Lemma H.3.2 Coboundary iff cohomologous to the trivial cocycle

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A scalar $2$-cochain $\omega $ is a coboundary if and only if $\omega \sim \mathbb {1}$, where $\mathbb {1}(g,h)=1$ for all $g,h$.

Proof ▶

Both directions follow by absorbing or introducing the factor $\mathbb {1}(g,h)=1$ via $x\cdot 1=x$ in (??).

H.4 Stationary-boundary twist support

Fix an MPS tensor $A=(A^i)_{i=0}^{d-1}$, a physical matrix $u$, and a stationary boundary matrix $\Lambda \in M_{D}(\mathbb {C})$. The twisted transfer map $\mathcal{E}_u$ is Definition 12.3.1. The next constructions are virtual-boundary functionals; they should not be confused with the physical endpoint correlator of Definition 12.3.4.

Definition H.4.1 Twisted companion tensor

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For an MPS tensor $A$ and a physical-index matrix $u$, define the companion tensor $B_u$ by

\begin{align} B_u^n & := \sum _{n'=0}^{d-1} \overline{u_{n'n}}\, A^{n'}. \label{eq:symmetry_companion_tensor} \end{align}

Lemma H.4.2 Twisted transfer as a mixed transfer operator

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The twisted transfer map of $A$ is the mixed transfer operator of the pair $(A,B_u)$:

\begin{align} \mathcal{E}_u & = F_{A,B_u}. \label{eq:symmetry_twisted_mixed_transfer} \end{align}

Proof ▶

Expand both definitions and rearrange the double sum.

Lemma H.4.3 The companion tensor has the same transfer map

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If $uu^\dagger =\mathbb {1}_d$, then for every $X\in M_{D}(\mathbb {C})$ the companion family $B_u$ satisfies

\begin{align} \mathcal{E}_{B_u}(X) & =\mathcal{E}_A(X). \notag \end{align}

Proof ▶

Theorem 4.5.9 applies to the unitary combination defining $B_u$. Equivalently, expand $B_u^n=\sum _{n'}\overline{u_{n'n}}A^{n'}$. The coefficient of $A^jX(A^k)^\dagger $ in $\mathcal{E}_{B_u}(X)$ is

\begin{align} \sum _n\overline{u_{jn}}u_{kn} & =(uu^\dagger )_{kj}=\delta _{kj}. \notag \end{align}

All off-diagonal terms vanish and the diagonal terms sum to $\mathcal{E}_A(X)$.

Definition H.4.4 Virtual-boundary twist functional

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For a boundary state $\Lambda $ and boundary matrices $X,Y \in M_{D}(\mathbb {C})$, the virtual-boundary twist functional is

\begin{align} R_L(u;X,Y) & := \operatorname{tr}\! (\Lambda X \mathcal{E}_u^L(Y)). \label{eq:symmetry_boundary_string} \end{align}

The special case $X=Y=\mathbb {1}$ is the stationary-boundary block-twist functional of Definition 12.3.2. The matrices $X,Y$ act on the virtual level. In the physical string correlator of [ PGWS$^{+}$08 ] the string of twists is instead flanked by local operators $x,y$ acting on physical sites; expanding such operators in the transfer picture produces particular virtual matrices, and [ PGWS$^{+}$08 ] argues that for injective tensors, physical operators on sufficiently many sites produce every virtual matrix. The results below are stated directly in this virtual-boundary form.

Lemma H.4.5 Virtual-boundary nondecay prevents convergence to zero

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If virtual-boundary nondecay holds for $A,u,\Lambda $, then for some boundary matrices $X,Y$ the sequence $R_L(u;X,Y)$ does not tend to $0$ as $L\to \infty $.

Proof ▶

A uniform lower bound $c \le \| R_L(u;X,Y)\| $ is incompatible with convergence to $0$.

H.5 Kraus mixing, transfer expansions, and a common TP gauge

The spectral estimates in the main chapter compare $A$ with $B_u$. They are placed in one trace-preserving gauge obtained from their common transfer map.

Lemma H.5.1 Unitary Kraus mixing

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If $u$ is unitary, then replacing Kraus operators $\{ A_i\} $ by their unitary mixtures $\{ \sum _j u_{ij} A_j\} $ preserves the channel:

\begin{align} \sum _i \left(\sum _j u_{ij} A_j\right)\, Y \left(\sum _j u_{ij} A_j\right)^\dagger & = \sum _i A_i\, Y\, A_i^\dagger . \label{eq:symmetry_unitary_kraus_mixing} \end{align}

Proof ▶

This is exactly the usual invariance of a Kraus decomposition under a unitary change of Kraus operators; see Theorem 4.5.9.

Definition H.5.2 Common trace-preserving gauge for a twisted pair

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A common trace-preserving gauge setup for $(A,u)$ consists of a tensor $B$ equal to the companion $B_u$ and satisfying $\mathcal{E}_B(X)=\mathcal{E}_A(X)$ for every $X\in M_{D}(\mathbb {C})$. It also includes a positive definite matrix $\sigma $ fixed by the adjoint transfer map of $B$, its positive square root $S$, and gauged tensors $A',B'$ satisfying

\begin{align} B & =B_u, & \mathcal{E}_B(X) & =\mathcal{E}_A(X), & \mathcal{E}_{B^\dagger }(\sigma ) & =\sigma , & \sigma & {\gt}0, \\ S & =\sigma ^{1/2}, & S^\dagger & =S, & A’^i & =SA^iS^{-1}, & B’^i & =SB^iS^{-1}. \notag \end{align}

In addition,

\begin{align} SS^{-1} & =S^{-1}S=\mathbb {1}, & S^\dagger (S^\dagger )^{-1} & =\mathbb {1}, & (S^{-1})^\dagger & =S^{-1}, \\ \sum _i(A’^i)^\dagger A’^i & =\mathbb {1}, & \sum _i(B’^i)^\dagger B’^i & =\mathbb {1}, \notag \end{align}

together with irreducibility of both $A'$ and $B'$. These identities complete the common trace-preserving gauge data.

Definition H.5.3 Construction of the common trace-preserving gauge

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Let $D{\gt}0$. If $A$ is injective, $uu^\dagger =\mathbb {1}_d$, and $\mathcal{E}_A(\mathbb {1})=\mathbb {1}$, define the common trace-preserving gauge setup for $(A,u)$ by choosing the unique positive definite stationary state of the adjoint channel and using the data in Definition H.5.2.

Proof ▶

Injectivity makes $\mathcal{E}_A$ irreducible. The adjoint channel therefore has a positive definite stationary state $\sigma $. By Lemma H.4.3, $A$ and $B_u$ have the same transfer map, hence the same adjoint fixed point. Put $S=\sigma ^{1/2}$. Positivity makes $S$ invertible, and direct substitution gives

\begin{align} \sum _i(SA^iS^{-1})^\dagger (SA^iS^{-1}) & =\mathbb {1}, \\ \sum _i(SB_u^iS^{-1})^\dagger (SB_u^iS^{-1}) & =\mathbb {1}. \notag \end{align}

Similarity preserves irreducibility, so the two gauged families have all the asserted properties.

Theorem H.5.4 Twisted eigenvalues survive the common gauge

✓

If $V\ne 0$ and $\mathcal{E}_u(V)=\lambda V$, then $\lambda $ is an eigenvalue of the mixed transfer map $F_{A',B'}$ in any common trace-preserving gauge setup.

Proof ▶

By Lemma H.4.2, the original equation is $F_{A,B_u}(V)=\lambda V$. Multiplying on the left and right by $S$ and using $S^\dagger =S$ gives

\begin{align} F_{A',B'}(SVS) & =\lambda SVS. \notag \end{align}

Since $S$ is invertible, $SVS\ne 0$.

H.6 Boundary and limit support

Lemma H.6.1 Spectral radius ${\lt}1$ forces virtual-boundary decay

✓

If $\rho (\mathcal{E}_u) {\lt} 1$, then, as $L \to \infty $, for all boundary matrices $X,Y,\Lambda \in M_{D}(\mathbb {C})$ one has $R_L(u;X,Y) \longrightarrow 0$.

Proof ▶

In finite dimension, $\rho (\mathcal{E}_u) {\lt} 1$ implies $\mathcal{E}_u^L \to 0$ in operator norm. Composing with the fixed trace pairing $Z \mapsto \operatorname{tr}(\Lambda X Z)$ gives the stated scalar convergence.

H.7 Routine universality corollary

The following consequence uses only the virtual MPV-family symmetry predicate and the canonical fixed-point hypotheses from the main chapter.

Theorem H.7.1 Virtual-boundary nondecay agrees for two injective MPV-symmetric tensors

✓

If $A$ and $B$ are injective, both on-site symmetric under a unitary representation $U$, and if $\mathcal{E}_A(\mathbb {1})=\mathbb {1}$, $\mathcal{E}_B(\mathbb {1})=\mathbb {1}$, $\Lambda _A$ and $\Lambda _B$ are positive definite, $\operatorname{tr}(\Lambda _A)=1$, $\operatorname{tr}(\Lambda _B)=1$, $\mathcal{E}_A^\dagger (\Lambda _A)=\Lambda _A$, and $\mathcal{E}_B^\dagger (\Lambda _B)=\Lambda _B$, then for every $g \in G$, virtual-boundary nondecay holds for $A$ with twist $U(g)$ if and only if it holds for $B$. In particular, since Definition 12.4.1 implies on-site symmetry for both tensors, this theorem applies to tensors in the same SPT phase; it applies equally to symmetric tensors in different SPT phases.

Proof ▶

Both directions follow from Theorem 12.4.4, which shows that virtual-boundary nondecay holds universally for injective symmetric tensors with canonical normalisations.