17 Pure States: Renormalization of Matrix Product States

Renormalization fixed points (RFPs) for matrix product states are characterized by physical two-site blocking, transfer map idempotence, zero correlation length (ZCL), and the square-root isometry canonical form. The presentation follows the pure-state analysis of [ 13 , Section 3 ] .

17.1 Physical blocking and transfer idempotence

Definition 17.1.1 Physical blocking isometry

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An MPS tensor $A$ is a renormalization fixed point if there exists an isometry $V\colon \mathbb {C}^d\to \mathbb {C}^{d^2}$ with coefficients $V_{(i_1,i_2),j}$ satisfying $V^\dagger V=\mathbb {1}$ such that for all physical indices $i_1,i_2\in \{ 0,\ldots ,d-1\} $,

\begin{align} A^{i_1}A^{i_2} & =\sum _{j=0}^{d-1}V_{(i_1,i_2),j}A^j. \label{eq:corr_physical_blocking_isometry} \end{align}

This expresses the physical two-site blocking relation $AA=A$ of [ 13 , Definition 3.2 ] .

Definition 17.1.2 Idempotent transfer map

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The transfer map $\mathcal{E}_A$ of an MPS tensor $A$ satisfies transfer map idempotence when $\mathcal{E}_A\circ \mathcal{E}_A=\mathcal{E}_A$.

Theorem 17.1.3 Eigenvalues of an idempotent transfer map

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If $\mathcal{E}_A\circ \mathcal{E}_A=\mathcal{E}_A$, then every nonzero eigenvalue of $\mathcal{E}_A$ equals $1$, the form in which [ 17 , Section III.A ] states the renormalization fixed-point property.

Proof ▶

If $\mathcal{E}_A(X)=\mu X$ with $X\neq 0$, then $\mu ^2X=\mathcal{E}_A^2(X)=\mathcal{E}_A(X)=\mu X$, so $\mu ^2=\mu $.

Theorem 17.1.4 A physical blocking isometry gives transfer idempotence

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Suppose there exists an isometry $V\colon \mathbb {C}^d\to \mathbb {C}^{d^2}$, written as coefficients $V_{(i_1,i_2),j}$, such that

\begin{align} A^{i_1}A^{i_2} & =\sum _{j=0}^{d-1}V_{(i_1,i_2),j}A^j \label{eq:corr_kraus_isometry_decomp} \end{align}

for all $i_1,i_2\in \{ 0,\ldots ,d-1\} $. Then $\mathcal{E}_A^2=\mathcal{E}_A$.

Proof ▶

Substituting the decomposition (2) into the double Kraus sum gives, for every bond matrix $X$,

\begin{align} \mathcal{E}_A^2(X) & =\sum _{i_1,i_2}(A^{i_1}A^{i_2})X (A^{i_1}A^{i_2})^\dagger \notag \\ & =\sum _{j,k}\left(\sum _{i_1,i_2} V_{(i_1,i_2),j}\overline{V_{(i_1,i_2),k}}\right)A^jX(A^k)^\dagger \notag \\ & =\sum _{j,k}(V^\dagger V)_{j,k}A^jX(A^k)^\dagger =\sum _{j,k}\delta _{j,k}A^jX(A^k)^\dagger =\sum _jA^jX(A^j)^\dagger =\mathcal{E}_A(X). \notag \end{align}

Theorem 17.1.5 Kraus-isometry criterion for transfer idempotence

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The transfer map of $A$ is idempotent if and only if there exists an isometry $V\colon \mathbb {C}^d\to \mathbb {C}^{d^2}$, written as coefficients $V_{(i_1,i_2),j}$ with $V^\dagger V=\mathbb {1}$, such that

\begin{align} A^{i_1}A^{i_2} & =\sum _{j=0}^{d-1}V_{(i_1,i_2),j}A^j \label{eq:corr_AA_isometry} \end{align}

for all $i_1,i_2\in \{ 0,\ldots ,d-1\} $.

Proof ▶

The backward implication is Theorem 17.1.4. For the forward implication, expanding $\mathcal{E}_A^2=\mathcal{E}_A$ as a Kraus sum gives, for every $X$,

\begin{align} \sum _{i_1,i_2}(A^{i_1}A^{i_2})X(A^{i_1}A^{i_2})^\dagger & =\sum _jA^jX(A^j)^\dagger . \notag \end{align}

Rectangular Kraus freedom [ 216 , “Rectangular Kraus freedom with general finite index types” ] then supplies the isometry $V$ relating the two Kraus families.

Theorem 17.1.6 Physical and transfer-map fixed-point criteria

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The transfer map $\mathcal{E}_A$ of an MPS tensor $A$ is idempotent, $\mathcal{E}_A^2 = \mathcal{E}_A$, if and only if there exists an isometry $V \colon \mathbb {C}^d \to \mathbb {C}^d \otimes \mathbb {C}^d$ such that

\begin{align} A^{i_1} A^{i_2} = \sum _j V_{(i_1, i_2), j} A^j. \notag \end{align}

Proof ▶

Follows from Theorem 17.1.5 and Definition 17.1.1.

Definition 17.1.7 Limit of the renormalization flow

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Let $A=(A^i)_{i=0}^{d-1}$ be an MPS tensor with bond dimension $D$. The physical-isometry class of $A$ appears as a limit of the renormalization flow if there exist a physical dimension $e$ and an MPS tensor $B=(B^a)_{a=0}^{e-1}$ of bond dimension $D$ such that

\begin{align} \widehat{\mathcal{E}}_B^{\, 2^n} & \longrightarrow \widehat{\mathcal{E}}_A. \notag \end{align}

Here $\widehat{\mathcal{E}}_A$ and $\widehat{\mathcal{E}}_B$ are the transfer matrices, and convergence is taken in the standard topology on the finite-dimensional matrix space $M_{D^2}(\mathbb {C})$. The physical dimension may change under blocking, while the bond dimension $D$ is preserved. Transfer matrices represent the quotient by physical isometries [ 13 , lines 382–394 and 1205–1209 ] .

Theorem 17.1.8 A dyadic power limit is idempotent

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Let $R$ be a Hausdorff topological monoid with continuous multiplication. If $E^{2^n}\to P$, then $P^2=P$.

Proof ▶

Continuity of multiplication gives $E^{2^n}E^{2^n}\to P^2$. On the other hand, $E^{2^n}E^{2^n}=E^{2^{n+1}}\to P$. Uniqueness of limits gives $P^2=P$.

Theorem 17.1.9 Limits of the renormalization flow

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A tensor class appears as a limit of the two-site renormalization flow if and only if it has a physical blocking isometry. Equivalently, there is an isometry $V\colon \mathbb {C}^d\to \mathbb {C}^{d^2}$ such that

\begin{align} A^{i_1}A^{i_2} =\sum _{j=0}^{d-1}V_{(i_1,i_2),j}A^j \notag \end{align}

for every $i_1,i_2$. This is [ 13 , Theorem 3.1 and lines 1205–1209 ] .

Proof ▶

If $\widehat{\mathcal{E}}_B^{\, 2^n}\to \widehat{\mathcal{E}}_A$, then Theorem 17.1.8 gives $\widehat{\mathcal{E}}_A^2=\widehat{\mathcal{E}}_A$. Since the transfer-matrix representation is injective, $\mathcal{E}_A^2=\mathcal{E}_A$. Theorem 17.1.6 supplies the physical blocking isometry. Conversely, if $A$ has such an isometry, its transfer map is idempotent by Theorem 17.1.6. Setting $B=A$ makes the dyadic orbit constant: $\widehat{\mathcal{E}}_A^{\, 2^n}=\widehat{\mathcal{E}}_A\to \widehat{\mathcal{E}}_A$.

\begin{tenkz}[cols=2, west=open, east=open]
\tn[ports={90:physical:$i_1$}]{A} & \tn[ports={90:physical:$i_2$}]{A}
\end{tenkz} — Figure 17.1 The fusion identity of the renormalization fixed-point characterization: two contracted copies of $A$ equal a single copy of $A$ with the isometry $V$ merging the summed index $j$ into the physical pair $(i_1,i_2)$, as stated in (3); conversely, applying $V^\dagger $ to the two physical legs of the contracted pair recovers the single tensor, as stated in (4). The source draws the isometry as $U$ [ 13 , Theorem 3.1, lines 396–412 ] .

\begin{tenkz}[rows={op,wire}, cols=3, bonds=none]
\tn[skin=pill, wide=3, name=V,
ports={90@1:physical:$i_1$, 90@3:physical:$i_2$,
270@2:physical}]{V} \\
& \tn[name=A,
ports={180:virtual, 0:virtual, 90:physical:$j$}]{A^j} &
\tnwire{A.90}{V.270@2}
\end{tenkz} — Figure 17.1 The fusion identity of the renormalization fixed-point characterization: two contracted copies of $A$ equal a single copy of $A$ with the isometry $V$ merging the summed index $j$ into the physical pair $(i_1,i_2)$, as stated in (3); conversely, applying $V^\dagger $ to the two physical legs of the contracted pair recovers the single tensor, as stated in (4). The source draws the isometry as $U$ [ 13 , Theorem 3.1, lines 396–412 ] .

17.2 Zero-correlation-length conditions

Definition 17.2.1 Virtual one-block idempotence convention

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An MPS tensor $A$ satisfies one-block transfer idempotence if $\mathcal{E}_A^2=\mathcal{E}_A$. Diagrammatically, two doubled columns equal one:

$\begin{tenkz}[rows={ket,bra}, cols=2, west=open, east=open] \tn[label pos=n]{A} & \tn[label pos=n]{A} \\ \tn[label pos=s, conjugate]{\overline{A}} & \tn[label pos=s, conjugate]{\overline{A}} \end{tenkz}$ $ = $ $\begin{tenkz}[rows={ket,bra}, cols=1, west=open, east=open] \tn[label pos=n]{A} \\ \tn[label pos=s, conjugate]{\overline{A}} \end{tenkz}$

In [ 13 , Definition 3.5 ] , local orthogonality denotes instead the vanishing of mixed transfer maps between distinct BNT components (Definition 17.2.2).

Definition 17.2.2 BNT local orthogonality

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Let $(A_j)_j$ be a basis of normal tensors. The family is locally orthogonal if for every pair of distinct components $j\ne j'$,

\begin{align} \mathcal{E}_{j,j'} & =\sum _i A_j^i\otimes \overline{A_{j'}^i}=0. \notag \end{align}

Diagrammatically,

$\begin{tenkz}[rows={ket,bra}, cols=1, west=open, east=open] \tn[label pos=n]{A_j} \\ \tn[label pos=s, conjugate]{\overline{A_{j'}}} \end{tenkz}$ $ = 0.$

This is the mixed-sector condition of [ 13 , Definition 3.5 ] .

Definition 17.2.3 Virtual-insertion auxiliary BNT condition

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Let $(A_j)_j$ be a basis of normal tensors for $A$. The family $(A_j)_j$ satisfies virtual-insertion BNT zero correlation length for $A$ if $A$ satisfies virtual-insertion distance independence (Definition 17.5.11) and the components $(A_j)_j$ are locally orthogonal: $\mathcal{E}_{j,j'}=0$ for all $j\ne j'$ (Definition 17.2.2). In contrast, [ 13 , Definition 3.6 ] quantifies over physical observables on all disjoint regions (Definition 17.2.6).

Definition 17.2.4 Positive-gap restriction of BNT zero correlation length

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A family $(A_j)_j$ of MPS tensors has positive-gap BNT zero correlation length for $A$ if $(A_j)_j$ is a basis of normal tensors for $A$ (Definition 8.1.2), its components are locally orthogonal (Definition 17.2.2), and the physical correlations of $A$ are independent of separation whenever both complementary gaps are positive (Definition 17.5.15). Adjacent regions are excluded.

Lemma 17.2.5 Positive-gap restriction, unfolded

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A family $(A_j)_j$ has positive-gap BNT zero correlation length for $A$ if and only if $(A_j)_j$ is a basis of normal tensors for $A$ (Definition 8.1.2), the physical correlations of $A$ are independent of separation whenever both complementary gaps are positive (Definition 17.5.15), and the components $(A_j)_j$ are locally orthogonal (Definition 17.2.2).

Proof ▶

This is the defining conjunction of Definition 17.2.4.

Definition 17.2.6 Physical BNT zero correlation length

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A basis of normal tensors $(A_j)_j$ for $A$ has physical BNT zero correlation length if the generated matrix product states have physical correlations independent of distance for all disjoint regions (Definition 17.5.14) and the components $(A_j)_j$ are locally orthogonal (Definition 17.2.2). This is [ 13 , Definition 3.6 ] .

Theorem 17.2.7 Two source gaps in the zero-correlation-length characterization

Cirac et al. [ 13 , Theorems 3.8 and 3.10 ] assert that a tensor $A$ in canonical form has physical BNT zero correlation length if and only if

\begin{align} \mathcal{E}_A^2 & =\mathcal{E}_A, \notag \end{align}

and, more generally, that for such $A$ the following are equivalent: $A$ is a renormalization fixed point; $A$ has physical zero correlation length; and, for every $N{\gt}2$, the vector $|V^{(N)}(A)\rangle $ belongs to the ground space of a nearest-neighbor commuting parent Hamiltonian. Both directions of the biconditional fail as printed, and with them the three-way equivalence.

For the reverse implication, let $C^0=(1)$ and form the direct sum of two scalar copies with weights $1$ and $1/2$. The resulting tensor is

\begin{align} A^0 & =\begin{pmatrix} 1 & 0 \\ 0 & \tfrac 12 \end{pmatrix}. \notag \end{align}

The scalar tensor $C$ is normal, and the weights obey the source convention $|\mu _k|\leq 1$, with at least one weight of modulus one. The singleton BNT family $(C)$ is locally orthogonal and gives $A$ physical BNT zero correlation length, whereas

\begin{align} \mathcal{E}_A^2 & \neq \mathcal{E}_A. \notag \end{align}

Thus zero correlation length does not imply transfer map idempotence under the source weight normalization.

For the forward implication, the Bell-pair chain of Theorem 17.2.9 is a normal tensor with idempotent transfer map, but its expectation on adjacent regions differs from its expectation when the regions are shifted by one site. Thus transfer map idempotence does not imply correlation independence when adjacent regions are included.

Local fix (false source claim for adjacent regions): The printed forward implication in [ 13 , Theorem 3.8 ] asserts correlation independence for all disjoint regions, including adjacent ones. However, the Bell-pair counterexample (Theorem 17.2.9) shows that transfer idempotence does not imply correlation independence for adjacent regions; the source derivation uses $\mathcal{E}_A^n=\mathcal{E}_A$, which holds only for $n\ge 1$, whereas adjacent regions insert $\mathcal{E}_A^0=\mathbb {1}$. Documented in [ 31 ] . Theorem 17.5.17 establishes the forward implication when both complementary gaps are positive.

Proof ▶

Theorem 17.2.8 refutes the reverse implication, and Theorem 17.2.9 refutes the forward implication.

Theorem 17.2.8 Auxiliary weighted-sector example

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Let $P$ be the auxiliary BNT sector decomposition with one scalar basis tensor $C^0=(1)$, two one-dimensional copies, and weights $1$ and $1/2$. Let $A=\mathcal A(P)$, so

\begin{align} A^0 & = \begin{pmatrix} 1 & 0 \\ 0 & \tfrac 12 \end{pmatrix}. \notag \end{align}

Then $P$ satisfies the auxiliary BNT sector hypotheses, $A=\mathcal A(P)$, and $A$ and $\mathcal A(P)$ generate the same matrix-product vectors. Moreover, the singleton family $(C)$ is a physical BNT zero-correlation-length family for $A$, while

\begin{align} \mathcal{E}_A^2 & \neq \mathcal{E}_A. \notag \end{align}

Its length-$N$ matrix-product-vector coefficient is $V^{(N)}(A)=1+2^{-N}$.

Proof ▶

The unique basis block has bond dimension one. It is irreducible and left-canonical, its self-overlap is identically $1$, and its singleton MPV family is linearly independent. Pairwise distinctness is vacuous. The two weights have moduli at most $1$, and the first has modulus $1$; hence $P$ satisfies the auxiliary BNT sector hypotheses.

The two one-dimensional weighted copies yield $A^0=\operatorname{diag}(1,1/2)$. This equality gives equality of the generated matrix-product vectors at every length, and therefore at every positive length. The sector power sum is

\begin{align} 1^N+\left(\tfrac 12\right)^N & =1+2^{-N}. \notag \end{align}

The scalar block is a normal tensor, the displayed coefficient spans the MPV of $A$, and the singleton family is eventually linearly independent. Since $d=1$, every physical observable is scalar and each two-point expectation depends only on the total gap length. Thus physical correlations are independent of distance, while local orthogonality is vacuous for the singleton BNT family.

Finally, if $\mathcal{E}_A$ were idempotent, the one-letter blocking equation would give $(A^0)^2=vA^0$ for some scalar $v$. Its two diagonal entries yield $1=v$ and $1/4=v/2$, respectively, which is impossible.

Theorem 17.2.9 Adjacent regions obstruct the RFP–CID forward implication

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Let $A$ be the tensor with physical index pairs $(\alpha ,\beta )\in \{ 0,1\} ^{2}$ and letters

\begin{align} A^{(\alpha ,\beta )}_{\gamma ,\delta } & =\delta _{\gamma ,\alpha }\, \delta _{\delta ,\beta }\, \sqrt{\tfrac 12}, \notag \end{align}

the canonical normal-tensor renormalization fixed point with maximally mixed spectrum $\Lambda =(1/2,1/2)$: each node carries two spins, and the pair $(|00\rangle +|11\rangle )/\sqrt2$ is shared between the second spin of one node and the first spin of the next. The transfer map is

\begin{align} \mathcal{E}_A(X) & =\operatorname{tr}(X)\operatorname {diag}(\Lambda ) =\tfrac 12\operatorname{tr}(X)\, \mathbb {1}, \notag \end{align}

Diagrammatically, the transfer map is the rank-one map whose output legs carry $\operatorname {diag}(\Lambda )$ and whose input legs are traced:

$\begin{tenkz}[rows={ket,bra}, cols=1, west=open, east=open] \tn[label pos=n]{A} \\ \tn[label pos=s, conjugate]{\overline{A}} \end{tenkz}$ $ = $ $\begin{tenkz}[rows={wire,wire}, cols=1, west=open, east={cup=$\Lambda$}, bonds=none] \tn[skin=none]{} \\ \tn[skin=none]{} \end{tenkz}$ $\; \otimes \; $ $\begin{tenkz}[rows={wire,wire}, cols=1, west=cup, east=open, bonds=none] \tn[skin=none]{} \\ \tn[skin=none]{} \end{tenkz}$

which is idempotent since $\operatorname{tr}\operatorname {diag}(\Lambda )=1$, and $A$ is a single normal tensor, so no copy weights are involved. With the Pauli matrix $Z$ on the second spin of one node and on the first spin of the next node, the two regions are adjacent, and the two-region expectation is

\begin{align} \operatorname{Tr}_{M_{2}(\mathbb {C})}\! \left( \mathcal{E}_{I\otimes Z}\circ \mathcal{E}_A^{0}\circ \mathcal{E}_{Z\otimes I}\circ \mathcal{E}_A^{2}\right) & =1, \notag \end{align}

while after a shift by one site, which leaves one free site on each complementary arc, it is

\begin{align} \operatorname{Tr}_{M_{2}(\mathbb {C})}\! \left( \mathcal{E}_{I\otimes Z}\circ \mathcal{E}_A^{1}\circ \mathcal{E}_{Z\otimes I}\circ \mathcal{E}_A^{1}\right) & =0. \notag \end{align}

Thus $A$ is a renormalization fixed point whose correlations are not independent of distance in the unrestricted sense of Definition 17.5.14. The block $\mathcal{E}_A^{0}=1$ between adjacent observables is not governed by idempotence.

Proof ▶

The letters are the rescaled matrix units $A^{(\alpha ,\beta )}=\sqrt{1/2}\, |\alpha \rangle \! \langle \beta |$, so a conjugated letter product collapses to a diagonal matrix unit and the transfer map sums to $\mathcal{E}_A(X)=\frac12\operatorname{tr}(X)\, \mathbb {1}$. Idempotence follows from $\frac12\operatorname{tr}(\mathbb {1})=1$. The one-letter observable transfer maps are $\mathcal{E}_{Z\otimes I}(X)=\frac12\operatorname{tr}(X)\, Z$ and $\mathcal{E}_{I\otimes Z}(X)=\frac12\operatorname{tr}(ZX)\, \mathbb {1}$. With zero middle gap the composition reduces by $\operatorname{tr}(Z^{2})=2$ to the transfer map itself, whose operator trace is $1$; with one idempotent block on each side the composition is the zero map, since $\operatorname{tr}(Z)=0$.

Lemma 17.2.10 Source BNT zero correlation length implies its positive-gap form

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If a BNT family $(A_j)_j$ for $A$ has zero correlation length for all disjoint physical regions ($\mathrm{ZCL}_{\mathrm{phys}}$), then it satisfies the positive-gap zero correlation length condition ($\mathrm{ZCL}_+$):

\begin{align} \mathrm{ZCL}_{\mathrm{phys}}((A_j)_j, A) \implies \mathrm{ZCL}_+((A_j)_j, A). \notag \end{align}

Proof ▶

Follows from Lemma 17.5.16 by restricting the physical distance-independence condition to positive complementary gaps.

Theorem 17.2.11 Direct-sum RFP implies positive-gap BNT zero correlation length

✓

Let $(A_j)_j$ be nonzero-dimensional, irreducible, left-canonical, pairwise gauge-phase-distinct BNT components. If the direct-sum tensor $A=\bigoplus _j A_j$ satisfies $\mathcal{E}_A^2=\mathcal{E}_A$, then $A$ has positive-gap physical correlations independent of distance and $\mathcal{E}_{j,j'}=0$ for all $j\ne j'$. Thus $A$ has positive-gap BNT zero correlation length (Definition 17.2.4). As shown in Theorem 17.2.9, this does not extend to adjacent regions.

Proof ▶

Idempotence implies positive-gap physical CID by Theorem 17.5.17. On each off-diagonal bond block, idempotence of $\mathcal{E}_A$ gives $\mathcal{E}_{j,j'}\circ \mathcal{E}_{j,j'}=\mathcal{E}_{j,j'}$. Distinct irreducible left-canonical blocks satisfy $\rho (\mathcal{E}_{j,j'}){\lt}1$. Since the spectrum of an idempotent is contained in $\{ 0,1\} $, the conditions $\mathcal{E}_{j,j'}^2=\mathcal{E}_{j,j'}$ and $\rho (\mathcal{E}_{j,j'}){\lt}1$ imply $\mathcal{E}_{j,j'}=0$.

Theorem 17.2.12 Basis direct sum is in literal CPSV canonical form

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Let $P$ be an auxiliary BNT sector decomposition with distinct basis tensors $(A_j)_j$. The multiplicity-one unit-weight direct sum $A=\bigoplus _j A_j$ is in CPSV canonical form (Definition 7.6.7), with unit retained weights, retained blocks $(A_j)_j$, and the identity ambient coisometry.

Proof ▶

Every retained dimension is positive. Each $A_j$ is irreducible and left-canonical, hence normal by Corollary 8.6.1.19. The unit-weight retained direct sum equals $A$, so the identity coisometry gives the required exact reconstruction.

Theorem 17.2.13 Auxiliary BNT basis gives a basis for its direct sum

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Let $P$ be an auxiliary BNT sector decomposition with distinct basis tensors $(A_j)_j$, and let $A=\bigoplus _j A_j$ be their direct sum with one unit-weight copy of each sector. Then $(A_j)_j$ is a basis of normal tensors for $A$: every $A_j$ is a normal tensor, at every positive system length $N$

\begin{align} V^{(N)}(A) & =\sum _j V^{(N)}(A_j), \notag \end{align}

and the states $\{ |V^{(N)}(A_j)\rangle \} _j$ are linearly independent for all sufficiently large $N$.

Proof ▶

Every basis block is irreducible and left-canonical with normalized self-overlap converging to $1$, hence a normal tensor by Corollary 8.6.1.19. The direct sum is the unit-weight block-diagonal tensor, so its length-$N$ matrix-product vector splits as the sum of the block vectors by Theorem 2.6.9. Eventual linear independence is part of the auxiliary BNT sector hypotheses.

Theorem 17.2.14 Multiplicity-one fixed point implies positive-gap ZCL

✓

Let $P$ be an auxiliary BNT sector decomposition with distinct basis tensors $(A_j)_j$, and let $A=\bigoplus _j A_j$ be their multiplicity-one direct sum. If $A$ is a renormalization fixed point ($\mathcal{E}_A^2 = \mathcal{E}_A$), then $A$ has positive-gap BNT zero correlation length: its physical correlations are independent of separation whenever both complementary gaps are positive, and for all distinct BNT components $j\ne j'$,

\begin{align} \mathcal{E}_{j,j'} & =0. \notag \end{align}

This establishes the positive-gap, multiplicity-one formulation of [ 13 , Theorem 3.8 ] .

Proof ▶

By Theorem 17.2.13, the distinct basis tensors of $P$ are a basis of normal tensors for $A$. The basis blocks are irreducible, left-canonical, and pairwise gauge-phase distinct, so Theorem 17.2.11 applies to the direct sum.

Theorem 17.2.15 Simultaneous inverse for BNT word evaluations

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Suppose the simultaneous length-$L$ word evaluations of $(A_j)_j$ span $\bigoplus _jM_{D_j}(\mathbb {C})$. There are coefficients $C_{(j,a,b),w}$ such that, for every sector $k$,

\begin{align} \sum _{|w|=L} C_{(j,a,b),w}(A_k^w)_{x,y} & =\delta _{j,k}\delta _{a,x}\delta _{b,y}. \notag \end{align}

Proof ▶

The tuple whose $j$-th component is the matrix unit $|a\rangle \! \langle b|$ and whose other components vanish belongs to the full simultaneous word span. Reading its $(k,x,y)$ entry gives the displayed identity.

Theorem 17.2.16 Two-sided virtual multiplication by physical observables

✓

Let $M,N$ belong to the span of the length-$L$ words of a tensor $A$. There is an operator $O$ on $L$ physical sites whose inserted transfer map satisfies $\mathcal{E}_O(X)=MXN^\dagger $ for every virtual matrix $X$. This is the two-sided multiplication used in the observable choices of [ 14 , Appendix C ] and [ 219 , Lemma 2 ] .

Proof ▶

Write $M=\sum _\sigma c_\sigma A^\sigma $ and $N=\sum _\tau e_\tau A^\tau $. Set $O_{\tau ,\sigma }=c_\sigma \overline{e_\tau }$. Expanding the inserted transfer map gives $\mathcal{E}_O(X)=\sum _{\sigma ,\tau }c_\sigma \overline{e_\tau } A^\sigma X(A^\tau )^\dagger =MXN^\dagger $.

Theorem 17.2.17 Sector-supported physical observable realization

✓

Let $A=\bigoplus _k A_k$, and suppose the simultaneous length-$L$ word evaluations span $\bigoplus _kM_{D_k}(\mathbb {C})$. For every sector $j$ and matrices $R,l\in M_{D_j}(\mathbb {C})$, there is an observable $O$ on $L$ sites whose inserted transfer map vanishes on every sector pair except $(j,j)$ and satisfies

\begin{align} \mathcal{E}_O(X)_{j,j} & =R\, \operatorname{tr}(lX_{j,j}). \label{eq:zcl_sector_rank_one} \end{align}

In particular, arbitrary virtual matrix-unit maps supported on one sector are realized by physical observables.

Proof ▶

Apply Theorem 17.2.15 independently to the left and right word evaluations. The observable with coefficients

\begin{align} O_{\tau ,\sigma } & =C_{(j,a,b),\sigma }\, \overline{C_{(j,c,e),\tau }} \notag \end{align}

sends the $(b,e)$ entry of $X_{j,j}$ to the $(a,c)$ entry and annihilates every other entry and sector pair. Summing these matrix-unit observables with coefficients $R_{a,c}l_{e,b}$ gives (5).

Theorem 17.2.18 Multiplicity-one BNT rank-one observables at the sharp blocking length

✓

Let $P$ be an auxiliary BNT sector decomposition of total bond dimension $D$, and let $A=\bigoplus _j A_j$ be the direct sum of its basis tensors. There exists a length $L\leq 3D^5$ such that, for every basis sector $j$ and every $R,l\in M_{D_j}(\mathbb {C})$, a physical observable on $L$ sites realizes the sector-supported insertion (5). This establishes the multiplicity-one case of the block-injective observable construction in [ 13 , proof of Theorem 3.8, lines 1252–1256 ] .

Proof ▶

Theorem 8.6.1.24 gives a positive $L\leq 3D^5$ for which the simultaneous word evaluations span the product matrix algebra. Apply Theorem 17.2.17 at this length.

Lemma 17.2.19 Sector compression calculus for the direct sum

✓

Let $A=\bigoplus _k A_k$. Denote compression to the $(j,j)$ bond block by $C_j$ and its trace-pairing adjoint by $I_j$. Then

\begin{align} C_jI_j & =\operatorname{id}. \notag \end{align}

If $\mathcal{E}_A$ and $\mathcal{E}_j$ are the transfer maps of $A$ and $A_j$, respectively, then, for every $n\geq 0$,

\begin{align} C_j\mathcal{E}_A^n & =\mathcal{E}_j^nC_j. \notag \end{align}

The sector-supported rank-one map $\mathcal R_{j;R,l}(X)=I_j(R)\operatorname{tr}(lC_j(X))$ has operator trace

\begin{align} \operatorname{Tr}_{\operatorname{End}}(\mathcal R_{j;R,l}) & =\operatorname{tr}(lR). \notag \end{align}

Finally, if $\mathcal{E}_j^*(l)=\lambda l$, then

\begin{align} \operatorname{tr}\! \bigl(l\mathcal{E}_j^n(X)\bigr) & =\lambda ^n\operatorname{tr}(lX). \notag \end{align}

Proof ▶

The inclusion is the trace-pairing adjoint of compression; expanding it in matrix units gives $C_jI_j=\operatorname{id}$ and the rank-one trace formula. The block-diagonal form of $A$ gives $C_j\mathcal{E}_A=\mathcal{E}_jC_j$, and induction gives the identity for every power. Moving one factor of $\mathcal{E}_j$ across the trace pairing and iterating proves the eigenvector formula.

Theorem 17.2.20 Subleading spectral pair excludes positive-gap CID

✓

Let $P$ be an auxiliary BNT sector decomposition, let $A=\bigoplus _k A_k$, and fix a sector $j$. Suppose there are matrices $r,l\in M_{D_j}(\mathbb {C})$ and a scalar $\lambda \ne 0$ with $|\lambda |{\lt}1$ such that

\begin{align} \bigl(\mathcal{E}_j(r),\mathcal{E}_j^*(l),\operatorname{tr}(lr)\bigr) & =\bigl(\lambda r,\lambda l,1\bigr). \notag \end{align}

Then $A$ does not have positive-gap physical correlations independent of distance (Definition 17.5.15).

Proof ▶

Let $\rho _j$ be the trace-one positive fixed point of $\mathcal{E}_j$. At the common blocking length, choose physical observables with inserted transfer maps

\begin{align} \bigl(\mathcal{E}_{O_1}(X),\mathcal{E}_{O_2}(X)\bigr) & =\bigl(\rho _j\operatorname{tr}(lX_{j,j}),r\operatorname{tr}(X_{j,j})\bigr). \notag \end{align}

The left eigenvector equation, together with $\mathcal{E}_j(\rho _j)=\rho _j$, $\operatorname{tr}(\rho _j)=1$, and $\operatorname{tr}(lr)=1$, gives

\begin{align} \langle O_1 O_2\rangle _{n_1,n_2} & =\lambda ^{n_1}. \notag \end{align}

Positive-gap CID compares $(n_1,n_2)=(1,2)$ with $(2,1)$, so $\lambda =\lambda ^2$. Since $\lambda \ne 0$, this forces $\lambda =1$, contrary to $|\lambda |{\lt}1$.

Lemma 17.2.21 Sector-supported probes without spectral assumptions

✓

Let $P$ be an auxiliary BNT sector decomposition in BNT canonical form, let $A=\bigoplus _j A_j$, and fix a sector $j$. For every $u,v\in M_{D_j}(\mathbb {C})$, there are a positive integer $L$ and observables $O_1,O_2$ on $L$ sites such that, for all $n_1,n_2\in \mathbb {N}$,

\begin{align} \operatorname {Tr}_{M_{D}(\mathbb {C})}\! \left( \mathcal{E}_{O_2}\mathcal{E}_A^{n_2}\mathcal{E}_{O_1}\mathcal{E}_A^{n_1}\right) =\operatorname{tr}\! \left(u\mathcal{E}_{A_j}^{n_1}(v)\right). \notag \end{align}

Proof ▶

Choose the trace-one stationary state $\rho _j$ of $\mathcal{E}_{A_j}$. Simultaneous block injectivity supplies observables with inserted maps

\begin{align} \mathcal{E}_{O_1} & =|\rho _j\rangle \! \langle u|, & \mathcal{E}_{O_2} & =|v\rangle \! \langle \mathbb {1}|. \notag \end{align}

Compression to sector $j$ intertwines powers of $\mathcal{E}_A$ with powers of $\mathcal{E}_{A_j}$. Hence $\mathcal{E}_A^{n_2}$ fixes the middle state $\rho _j$, and the composed inserted map is

\begin{align} |v\rangle \! \langle (\mathcal{E}_{A_j}^{n_1})^*(u)|. \notag \end{align}

The operator trace of this rank-one map is $\operatorname{tr}((\mathcal{E}_{A_j}^{n_1})^*(u)v) =\operatorname{tr}(u\mathcal{E}_{A_j}^{n_1}(v))$.

Theorem 17.2.22 Unconditional reverse implication at the multiplicity-one representative

✓

Let $P$ be an auxiliary BNT sector decomposition in BNT canonical form and $A=\bigoplus _j A_j$. If $A$ has positive-gap BNT zero correlation length, then $\mathcal{E}_A^2=\mathcal{E}_A$.

Proof ▶

Fix $j$ and arbitrary $u,v\in M_{D_j}(\mathbb {C})$. Apply positive-gap CID to the observables of Lemma 17.2.21, comparing complementary gaps $(1,2)$ and $(2,1)$. This gives

\begin{align} \operatorname{tr}\! \left(u\mathcal{E}_{A_j}(v)\right) =\operatorname{tr}\! \left(u\mathcal{E}_{A_j}^2(v)\right). \notag \end{align}

Since $u$ and $v$ are arbitrary and the trace pairing is nondegenerate, $\mathcal{E}_{A_j}^2=\mathcal{E}_{A_j}$. Local orthogonality makes every mixed map $\mathcal{E}_{A_j,A_k}$ with $j\ne k$ equal to zero. Therefore every diagonal and off-diagonal mixed map is idempotent, and the direct-sum criterion gives $\mathcal{E}_A^2=\mathcal{E}_A$.

Theorem 17.2.23 Positive-gap BNT zero correlation length at the multiplicity-one representative

✓

Let $P$ be an auxiliary BNT sector decomposition in BNT canonical form and $A=\bigoplus _j A_j$. Then

\begin{align} A\text{ has positive-gap BNT zero correlation length} \quad \Longleftrightarrow \quad \mathcal{E}_A^2=\mathcal{E}_A. \notag \end{align}

Proof ▶

The forward implication is Theorem 17.2.22. Conversely, transfer idempotence implies positive-gap physical correlation independence, while the BNT hypotheses give local orthogonality for the distinct sectors.

Theorem 17.2.24 Physical BNT zero correlation length at the multiplicity-one representative

✓

Let $P$ be an auxiliary BNT sector decomposition in BNT canonical form and $A=\bigoplus _j A_j$. If $A$ has physical BNT zero correlation length, then $\mathcal{E}_A^2=\mathcal{E}_A$.

Proof ▶

Physical correlation independence includes all positive-gap configurations. Restrict to those configurations and apply Theorem 17.2.22.

Theorem 17.2.25 Positive-gap BNT zero correlation length gives commuting parent ground spaces

✓

Let $P$ be an auxiliary BNT sector decomposition in BNT canonical form and $A=\bigoplus _j A_j$. If $A$ has positive-gap BNT zero correlation length, then for every $N{\gt}2$ its translated two-site parent interactions satisfy

\begin{align} h_i h_j & =h_j h_i, \notag \\ h_iV^{(N)}(A) & =0, \notag \\ \ker H_2^{(N)}(A) & =\operatorname{span}\bigl\{ V^{(N)}(A_j):j=1,\ldots ,g\bigr\} . \notag \end{align}

Proof ▶

Theorem 17.2.22 gives $\mathcal{E}_A^2=\mathcal{E}_A$. The multiplicity-one fixed-point theorem 12.1.32 gives the stated interactions and ground spaces.

Theorem 17.2.26 Positive-gap zero correlation length and commuting parent ground spaces for the basis direct sum

✓

Let $A_1,\ldots ,A_g$ be the distinct basis tensors of a BNT canonical form, and set $A=\bigoplus _{j=1}^g A_j$. Then

\begin{align} A\text{ has positive-gap BNT zero correlation length} \quad \Longleftrightarrow \quad A\text{ has all-chain commuting parent ground spaces spanned by }(A_j)_j. \notag \end{align}

This equivalence holds for the multiplicity-one direct sum of distinct BNT basis tensors with positive complementary gaps.

Proof ▶

By Theorem 17.2.23, positive-gap BNT zero correlation length is equivalent to transfer idempotence. The latter is equivalent to the all-chain commuting parent ground-space condition by Theorem 12.1.96.

Remark 17.2.27 Scope and obstructions in the converse

For the multiplicity-one unit-weight direct sum, sector-supported observables require no additional hypotheses. In the general canonical form of [ 13 , equations (1252) and (1256) ] , observable insertions must also cover repeated-copy sectors with their respective weights. Furthermore, the source argument infers from $\mathcal{E}_j^2\ne \mathcal{E}_j$ the existence of an eigenvalue $\lambda $ with $0{\lt}|\lambda |{\lt}1$, which omits non-idempotent channels whose non-trivial Jordan blocks occur at eigenvalue zero. At multiplicity one, Theorem 17.2.22 resolves this unconditionally using trace probes (Lemma 17.2.21), while Theorem 17.2.20 covers the spectral case when a nonzero subleading eigenpair exists.

17.3 Normal tensors and fixed-point isometries

17.3.1 Auxiliary block-family hypotheses

Definition 17.3.1.1 formulates an auxiliary condition for a family of injective left-canonical blocks with non-increasing nonzero weights and converging self-overlap, distinguishing it from the ambient canonical forms of Definitions 7.6.7 and 7.8.3.

Definition 17.3.1.1 Auxiliary injective block-family hypotheses

✓

A collection of scaling factors $(\mu _k)_{k=1}^r$ and block tensors $(A_k)_{k=1}^r$ satisfies the auxiliary block-family hypotheses if:

each $A_k$ is injective,
each $A_k$ satisfies the TP normalization $\sum _i(A_k^i)^\dagger A_k^i=\mathbb {1}$,
the moduli are non-increasing: $|\mu _1|\ge |\mu _2|\ge \cdots \ge |\mu _r|$,
all $\mu _k\neq 0$,
every block bond dimension is positive: $D_k\ge 1$ for every $k$,
the self-overlap converges: $O_{A_kA_k}(N)\to 1$ for each $k$.

Condition (6) corresponds to channel primitivity: for a primitive channel $T$, $T^n(\rho )\to \rho _\infty $ for every state $\rho $ [ 244 , Theorem 6.7(3) ] , which implies convergence of the self-overlap to $1$ (see also [ 244 , Theorem 6.8 ] and Chapter 5).

Theorem 17.3.1.2 Auxiliary block-family hypotheses from peripheral primitivity

✓

Let $(\mu _k,A_k)_{k=1}^r$ be nonzero weights and injective left-canonical blocks with positive bond dimensions, non-increasing moduli $|\mu _1|\ge \cdots \ge |\mu _r|{\gt}0$, and $\sigma _\partial (\mathcal{E}_{A_k})=\{ 1\} $. Then $(\mu _k,A_k)_{k=1}^r$ satisfies the auxiliary hypotheses of Definition 17.3.1.1, and $O_{A_kA_k}(N)\to 1$ for every $k$.

Proof ▶

The injectivity, left-canonical identity, weight ordering, and nonzero-weight clauses are exactly the hypotheses. For each block, peripheral primitivity gives the complementary spectral gap $\rho _{\operatorname{spec}}(\mathcal{E}_{A_k}-P){\lt}1$ by the complementary transfer-map gap for primitive maps [ 216 , “Complementary transfer-map gap for primitive maps” ] . Theorem 5.10.2 then yields $O_{A_kA_k}(N)\to 1$.

Theorem 17.3.1.3 RFP normal tensor is injective

✓

If $A$ is a normal RFP tensor, then $A$ is injective.

Proof ▶

The RFP condition $\mathcal{E}_A^2=\mathcal{E}_A$ gives coefficients $V_{i_1i_2,j}$ such that

\begin{align} A^{i_1}A^{i_2} & =\sum _j V_{i_1i_2,j}A^j. \notag \end{align}

Induction on the word length gives $A^{i_1}\cdots A^{i_n}\in \operatorname{span}\{ A^j:0\le j{\lt}d\} $ for $n\ge 1$. Normality implies that products of some length $N$ span $M_{D}(\mathbb {C})$, so the one-site matrices $\{ A^j\} _{j=0}^{d-1}$ already span $M_{D}(\mathbb {C})$, establishing injectivity.

Theorem 17.3.1.4 Left-canonical normal RFP tensor is injective

✓

Assume $D {\gt} 0$. Let $A$ be a normal renormalization fixed point such that $\sum _i (A^i)^\dagger A^i = \mathbb {1}$. Then $A$ is injective. This is the left-canonical injectivity step in [ 13 , Appendix B ] .

Proof ▶

Follows from Theorem 17.3.1.3.

Theorem 17.3.1.5 Unitary diagonal fixed point for a left-canonical normal RFP

✓

Assume $D {\gt} 0$. Let $A$ be a normal renormalization fixed point such that $\sum _i (A^i)^\dagger A^i = \mathbb {1}$. Then there exist a unitary $U$ and a diagonal positive definite matrix $\Lambda $ such that, for $B^i := U^\dagger A^i U$, $\sum _i (B^i)^\dagger B^i = \mathbb {1}$ and $\mathcal{E}_B(\Lambda ) = \Lambda $. This is the diagonal fixed-point reduction in [ 13 , Appendix B ] .

Proof ▶

By Theorem 17.3.1.4, $A$ is injective. By Theorem 2.5.2, its transfer map $\mathcal{E}_A$ is irreducible. Applying canonical form II (Theorem 7.8.6) yields a unitary matrix $U$ preserving the left-canonical condition and a diagonal positive-definite matrix $\Lambda $ satisfying $\mathcal{E}_B(\Lambda ) = \Lambda $ for $B^i = U^\dagger A^i U$.

Theorem 17.3.1.6 Rank-one classification for RFP transfer maps

✓

Let $A$ be an injective left-canonical RFP tensor with positive-definite fixed point $\rho $ of the transfer map $\mathcal{E}_A$. Then $\mathcal{E}_A = P_\rho $, where $P_\rho (X) = \frac{\operatorname{tr}(X)}{\operatorname{tr}(\rho )} \rho $ is the rank-one fixed-point projection.

Proof ▶

Exponential convergence of primitive channels (Theorem 14.3.1) gives $\lVert \mathcal{E}_A^n(X) - P_\rho (X)\rVert \le C(1-\delta )^n\lVert X\rVert $. Transfer map idempotence implies $\mathcal{E}_A^n = \mathcal{E}_A$ for all $n \ge 1$, so $\lVert \mathcal{E}_A(X) - P_\rho (X)\rVert \le C(1-\delta )^n\lVert X\rVert $ for all $n\ge 1$. Taking $n \to \infty $ yields $\mathcal{E}_A = P_\rho $.

Theorem 17.3.1.7 Diagonal Kraus form for a normal fixed point

✓

A normal left-canonical renormalization fixed-point tensor $A$ admits a decomposition $A^i=X\Lambda U^iX^{-1}$, where $\Lambda $ is diagonal positive, $\sum _i (U^i)^\dagger U^i=I$, and

\begin{align} \sum _i \overline{(U^i)_{\alpha ,\beta }} (U^i)_{\alpha ',\beta '} & = D^{-1}\delta _{\alpha ,\alpha '}\delta _{\beta ,\beta '}. \notag \end{align}

Proof ▶

After the diagonal fixed-point reduction, write the conjugated tensor as $B$ and its positive diagonal fixed point as $\rho $. The rank-one classification gives $\mathcal{E}_B(Y)=P_\rho (Y)=\frac{\operatorname{tr}(Y)}{\operatorname{tr}(\rho )}\rho $. Set

\begin{align} \Lambda _\alpha & =\sqrt{\frac{D\rho _{\alpha ,\alpha }}{\operatorname{tr}(\rho )}}, \notag \\ E_{\alpha ,\beta } & =D^{-1/2}e_{\alpha ,\beta }, \notag \\ K_{\alpha ,\beta } & =\Lambda E_{\alpha ,\beta }. \notag \end{align}

Direct multiplication of matrix units yields

\begin{align} \sum _{\alpha ,\beta }K_{\alpha ,\beta }Y K_{\alpha ,\beta }^{\dagger } & =P_\rho (Y). \notag \end{align}

Kraus freedom therefore gives an isometry $V$ such that $B^i=\sum _{\alpha ,\beta }V_{i,(\alpha ,\beta )}K_{\alpha ,\beta }$. With $U^i=\sum _{\alpha ,\beta }V_{i,(\alpha ,\beta )}E_{\alpha ,\beta }$, one has $B^i=\Lambda U^i$ and

\begin{align} U^i_{\alpha ,\beta } & =D^{-1/2}V_{i,(\alpha ,\beta )}, \notag \\ \sum _i\overline{U^i_{\alpha ,\beta }}U^i_{\alpha ',\beta '} & =D^{-1}\delta _{\alpha ,\alpha '}\delta _{\beta ,\beta '}. \notag \end{align}

Conjugating back gives $A^i=X\Lambda U^iX^{-1}$.

Theorem 17.3.1.8 Diagonal Kraus form with square-sum normalization

✓

In the decomposition of Theorem 17.3.1.7, the diagonal weights satisfy $\sum _\alpha \Lambda _\alpha ^2 = D$.

Proof ▶

From $\Lambda _\alpha = \sqrt{D\, \rho _{\alpha ,\alpha }/\operatorname{tr}\rho }$, squaring and summing over $\alpha $ gives $\sum _\alpha \Lambda _\alpha ^2 = \frac{D}{\operatorname{tr}\rho }\sum _\alpha \rho _{\alpha ,\alpha } = D$.

Theorem 17.3.1.9 Unit pair-index form for a normal fixed point

✓

A normal left-canonical renormalization fixed-point tensor $A$ admits a decomposition $A^i=X\Lambda U^iX^{-1}$, where $\Lambda $ is diagonal positive and

\begin{align} \sum _i \overline{(U^i)_{\alpha ,\beta }} (U^i)_{\alpha ',\beta '} & = \delta _{\alpha ,\alpha '}\delta _{\beta ,\beta '}. \notag \end{align}

This is the unit pair-index convention of [ 13 , Section 3 ] .

Proof ▶

Apply Theorem 17.3.1.12, which gives $A^i = X\sqrt{\Lambda }\, U^i X^{-1}$ with $U$ already a unit pair-index isometry. Set $\widetilde\Lambda _\alpha = \sqrt{\Lambda _\alpha }$, so that $A^i = X\widetilde\Lambda \, U^i X^{-1}$ with $\widetilde\Lambda $ diagonal positive; the decomposition and the pair-index condition are immediate.

Definition 17.3.1.10 Square-root fixed-point form

✓

Let $\lambda _\alpha {\gt}0$ satisfy $\sum _\alpha \lambda _\alpha =1$, and set $\rho =\operatorname{diag}(\lambda _\alpha )$. A normal tensor $A$ has the square-root fixed-point form if there exist an invertible matrix $X$ and a tensor $U$ satisfying the unit pair-index isometry

\begin{align} \sum _i \overline{(U^i)_{\alpha ,\beta }}\, (U^i)_{\alpha ',\beta '} & = \delta _{\alpha ,\alpha '}\delta _{\beta ,\beta '}, \notag \end{align}

such that

\begin{align} A^i & =X\sqrt{\rho }\, U^iX^{-1}. \notag \end{align}

Local fix (square-root factor in canonical form): The definition uses the square-root diagonal $\sqrt{\rho }$ rather than the bare diagonal printed in [ 13 , line 1278 ] , as required by the unit pair-index isometry and reference-tensor construction; documented in [ 85 ] .

Explicitly, the reference tensor is

\begin{align} \widehat A^{(\alpha ',\beta ')}_{\alpha ,\beta } & = \delta _{\alpha ,\alpha '}\delta _{\beta ,\beta '} \sqrt{\lambda _\alpha }. \notag \end{align}

Thus $\rho =\operatorname{diag}(\lambda _\alpha )$ is the trace-one diagonal fixed point. This records the $j=j'$ part of the joint isometry condition [ 13 , Appendix B, lines 1278, 1281–1283, 1300 ] .

Lemma 17.3.1.11 Gauge invariance of the square-root fixed-point form

✓

If $A \sim _{\mathrm{gauge}} B$ (so $B^i = Y A^i Y^{-1}$ for invertible $Y$), then $A$ has the square-root fixed-point form if and only if $B$ does.

Proof ▶

Write $A^i=X\sqrt\rho \, U^iX^{-1}$ and $B^i=Y A^iY^{-1}$. Then

\begin{align} B^i & =(YX)\sqrt\rho \, U^i(YX)^{-1}. \notag \end{align}

Thus the diagonal weight and pair-index isometry are unchanged. Applying the same argument to the inverse gauge proves the converse.

Theorem 17.3.1.12 Trace-normalized square-root form for a normal fixed point

✓

Every normal left-canonical renormalization fixed-point tensor has the square-root fixed-point form (Definition 17.3.1.10): setting $\lambda _\alpha =\rho _{\alpha ,\alpha }/\operatorname{tr}\rho $ and $\rho _0=\operatorname{diag}(\lambda _\alpha )$, there exist an invertible matrix $X$ and a unit pair-index isometry $U$ such that

\begin{align} A^i & =X\sqrt{\rho _0}\, U^iX^{-1}. \notag \end{align}

This is the normal-tensor characterization of [ 13 , Appendix B, lines 1278–1300 ] , incorporating the square-root weight correction.

Proof ▶

By Theorem 17.3.1.8, $A$ admits a decomposition $A^i = X\widetilde\Lambda U_{\mathrm{raw}}^i X^{-1}$ with $\sum _\alpha \widetilde\Lambda _\alpha ^2 = D$. Rescaling to the unit pair-index convention via $U^i = \sqrt{D}\, U_{\mathrm{raw}}^i$ and setting

\begin{align} \rho _0 & =\operatorname{diag}\left(\left(\frac{\widetilde\Lambda _\alpha }{\sqrt D}\right)^2\right), \notag \end{align}

one has $\operatorname{tr}(\rho _0) = D^{-1}\sum _\alpha \widetilde\Lambda _\alpha ^2 = D^{-1}\cdot D = 1$, while $\sqrt{(\rho _0)_{\alpha ,\alpha }} =\widetilde\Lambda _\alpha /\sqrt D$ recovers the decomposition $A^i = X\sqrt{\rho _0}\, U^i X^{-1}$.

\begin{tenkz}[rows={wire}, cols=1, west=open, east=open]
\tn[label pos=s, ports={90:physical:$i$}]{A}
\end{tenkz} — Figure 17.2 The square-root form of a normal renormalization fixed point: $\rho =\operatorname{diag}(\lambda _\alpha )$ and $A^i=X\sqrt\rho \, U^iX^{-1}$. The pair-index isometry is the one in lines 1281–1283, and the square root follows from the reference tensor at line 1300 [ 13 , Appendix B ] .

\begin{tenkz}[rows={wire}, cols=4, west=open, east=open]
\tn[skin=ring]{X} & \tn[skin=ring]{\sqrt\rho} &
\tn[label pos=s, ports={90:physical:$i$}]{U} & \tn[skin=ring]{X^{-1}}
\end{tenkz} — Figure 17.2 The square-root form of a normal renormalization fixed point: $\rho =\operatorname{diag}(\lambda _\alpha )$ and $A^i=X\sqrt\rho \, U^iX^{-1}$. The pair-index isometry is the one in lines 1281–1283, and the square root follows from the reference tensor at line 1300 [ 13 , Appendix B ] .

Lemma 17.3.1.13 Reference-tensor transfer identity for a unit pair-index isometry

✓

Let $U$ be a family of bond matrices satisfying the unit pair-index isometry condition

\begin{align} \sum _i \overline{(U^i)_{\alpha ,\beta }}\, (U^i)_{\alpha ',\beta '} & = \delta _{\alpha ,\alpha '}\delta _{\beta ,\beta '}. \notag \end{align}

Then for every bond matrix $Z$,

\begin{align} \sum _i U^i\, Z\, (U^i)^\dagger & = \operatorname{tr}(Z)\, I. \notag \end{align}

Proof ▶

Reading off the $(x,y)$ entry, the pair-index condition gives $\sum _i (U^i)_{x,\alpha } \overline{(U^i)_{y,\beta }} = \delta _{x,y} \delta _{\alpha ,\beta }$, so the entry collapses to $\delta _{x,y}\sum _\alpha Z_{\alpha ,\alpha } = \delta _{x,y} \operatorname{tr}(Z)$, which is the $(x,y)$ entry of $\operatorname{tr}(Z)\, I$.

Theorem 17.3.1.14 The square-root form implies the fixed-point property

✓

If an MPS tensor $A$ has the square-root fixed-point form, then its transfer map $\mathcal{E}_A$ is idempotent:

\begin{align} \mathcal{E}_A^2 = \mathcal{E}_A. \notag \end{align}

In particular, $A$ is a renormalization fixed point [ 13 , Section 3 ] .

Proof ▶

Write $A^i = X\sqrt\rho \, U^i X^{-1}$. Substituting into the transfer map and pulling the index-independent factors $X\sqrt\rho $ and $\sqrt\rho \, X^\dagger $ outside the sum, the unit pair-index isometry identity (Lemma 17.3.1.13) gives the rank-one form

\begin{align} \mathcal{E}_A(Y) & = \operatorname{tr}\! (X^{-1} Y (X^{-1})^\dagger )\, R, \notag \\ R & = X\, \rho \, X^\dagger . \notag \end{align}

Idempotence then reduces to $\operatorname{tr}\! (X^{-1} R (X^{-1})^\dagger ) = \operatorname{tr}(\rho ) = 1$, the trace normalization, so $\mathcal{E}_A \circ \mathcal{E}_A = \mathcal{E}_A$.

Theorem 17.3.1.15 Normal-tensor fixed-point characterization

✓

Let $A$ be a normal tensor. Then its transfer map satisfies $\mathcal{E}_A^2 = \mathcal{E}_A$ if and only if $A$ has the square-root fixed-point form

\begin{align} A^i = X\sqrt\rho \, U^i X^{-1}. \notag \end{align}

In particular, the bond dimension of $A$ is positive. This establishes the corrected form of [ 13 , Lemma B.1, lines 1274–1301 ] .

Proof ▶

Normality first gives positive bond dimension by Theorem 8.2.1. For the forward implication, choose the pure trace-preserving Perron gauge from Theorem 8.2.3. Idempotence and algebraic normality are invariant under this gauge, so Theorem 17.3.1.12 gives the square-root form for the gauged tensor. Lemma 17.3.1.11 gives the form for the original tensor. The reverse implication is Theorem 17.3.1.14.

Across all blocks, the repaired trace-normalized decomposition has the following representation.

\begin{tenkz}[rows={wire}, cols=1, west=open, east=open]
\tn[skin=box, ports={90:physical}]{A}
\end{tenkz} — Figure 17.3 Block and multiplicity indices in the fixed-point decomposition. The independent rails $j$ and $q$ pass through the coefficient network without being fused: $X$ and $X^{-1}$ depend on both, $\sqrt\rho $ and $U$ carry the block label $j$, and $M$ carries the multiplicity label $q$. The upper leg of $U$ is the physical index, as in the source figure [ 13 , Section 3.4, lines 543–563 ] .

\begin{tenkz}[rows={wire,wire,wire}, cols=5, bonds=none]
\tn[at=(1,1), skin=box, name=X,
ports={180:virtual, 0:virtual, 270:virtual}]{X}
\tn[at=(1,2), skin=box, name=rho,
ports={180:virtual, 0:virtual, 270:virtual}]{\sqrt\rho}
\tn[at=(1,3), name=mid, ports={90:virtual, 0:virtual, 270:virtual}]{}
\tn[at=(1,4), skin=box, name=M,
ports={180:virtual, 0:virtual, 270:virtual}]{M}
\tn[at=(1,5), skin=box, name=Xinv,
ports={180:virtual, 0:virtual, 135:virtual, 270:virtual}]{X^{-1}}
\tn[at={1 n of (1,3)}, skin=box, name=U,
ports={225:virtual, 270:virtual, 315:virtual, 90:physical}]{U}
\tn[at=(2,1), name=jX, ports={180:virtual, 0:virtual, 90:virtual, 270:virtual}]{}
\tn[at=(2,2), name=jrho, ports={180:virtual, 0:virtual, 90:virtual}]{}
\tn[at=(2,3), name=jU, ports={180:virtual, 0:virtual, 90:virtual}]{}
\tn[at=(2,5), name=jXinv, ports={180:virtual, 0:virtual, 90:virtual, 270:virtual}]{}
\tn[at=(3,1), name=qX, ports={180:virtual, 0:virtual, 90:virtual}]{}
\tn[at=(3,4), name=qM, ports={180:virtual, 0:virtual, 90:virtual}]{}
\tn[at=(3,5), name=qXinv, ports={180:virtual, 0:virtual, 90:virtual}]{}
\tnwire{X.180}{open w}
\tnwire{X.0}{rho.180}
\tnwire{U.225}{rho.0}
\tnwire{U.270}{mid.90}
\tnwire{U.315}{Xinv.135}
\tnwire{mid.0}{M.180}
\tnwire{M.0}{Xinv.180}
\tnwire{Xinv.0}{open e}
\tnwire{U.90}{open n}
\tnwire{X.270}{jX.90}
\tnwire{rho.270}{jrho.90}
\tnwire{mid.270}{jU.90}
\tnwire{Xinv.270}{jXinv.90}
\tnwire{jX.270}{qX.90}
\tnwire[crossing=over]{M.270}{qM.90}
\tnwire{jXinv.270}{qXinv.90}
\tnwire[name=jw]{jX.180}{open w}
\tnwire{jX.0}{jrho.180}
\tnwire{jrho.0}{jU.180}
\tnwire{jU.0}{jXinv.180}
\tnwire{jXinv.0}{open e}
\tnwire[name=qw]{qX.180}{open w}
\tnwire{qX.0}{qM.180}
\tnwire{qM.0}{qXinv.180}
\tnwire{qXinv.0}{open e}
\tnmark[form=label, label pos=w]{on jw 0.8}{$j$}
\tnmark[form=label, label pos=w]{on qw 0.8}{$q$}
\end{tenkz} — Figure 17.3 Block and multiplicity indices in the fixed-point decomposition. The independent rails $j$ and $q$ pass through the coefficient network without being fused: $X$ and $X^{-1}$ depend on both, $\sqrt\rho $ and $U$ carry the block label $j$, and $M$ carries the multiplicity label $q$. The upper leg of $U$ is the physical index, as in the source figure [ 13 , Section 3.4, lines 543–563 ] .

Here $j$ indexes the distinct normal blocks and $q$ their repeated copies. The joint isometry condition across distinct blocks is established in Theorem 17.4.16.

17.4 Direct sums and the joint isometry condition

Lemma 17.4.1 Block decomposition of the direct-sum transfer map

✓

Let $(B_k)_k$ be a family of tensors and let $\bigoplus _k B_k$ be their direct sum on the total bond space. For every bond matrix $X$ and every pair of blocks, the $(j,j')$ bond block of the transfer sum

\begin{align} \sum _i \left(\bigoplus _k B_k^i\right)X \left(\bigoplus _k B_k^i\right)^\dagger \notag \end{align}

equals $\mathcal{E}_{j,j'}$ applied to the $(j,j')$ bond block of $X$:

$\begin{tenkz}[rows={ket,bra}, cols=1, west=open, east={cup=$X_{j,j'}$}] \tn[label pos=n]{B_j} \\ \tn[label pos=s, conjugate]{\overline{B_{j'}}} \end{tenkz}$

Proof ▶

A block-diagonal factor connects only matching blocks, so the $(j,j')$ block of the product is

\begin{align} \left(\sum _i \left(\bigoplus _k B_k^i\right)X \left(\bigoplus _k B_k^i\right)^\dagger \right)_{j,j'} & = \sum _i B_j^i X_{j,j'}(B_{j'}^i)^\dagger = \mathcal{E}_{j,j'}(X_{j,j'}). \notag \end{align}

Theorem 17.4.2 Reindexing the direct-sum transfer map

✓

Let $(B_k)_k$ be a family of tensors, let $\Phi $ be the canonical identification of the block-labelled bond space with the total bond space, and let $\mathcal S_B$ be the block transfer sum. Then, for every block matrix $Y$,

\begin{align} \mathcal{E}_{\oplus _k B_k}(\Phi (Y)) & =\Phi (\mathcal S_B(Y)). \notag \end{align}

Proof ▶

Since $\Phi $ preserves products and adjoints, expanding the transfer map gives

\begin{align} \mathcal{E}_{\oplus _k B_k}(\Phi (Y)) & =\sum _i\Phi \! \left(\bigoplus _k B_k^i\right)\Phi (Y) \Phi \! \left(\bigoplus _k B_k^i\right)^\dagger \notag \\ & =\Phi \! \left(\sum _i \left(\bigoplus _k B_k^i\right)Y \left(\bigoplus _k B_k^i\right)^\dagger \right) \notag \\ & =\Phi (\mathcal S_B(Y)). \notag \end{align}

Theorem 17.4.3 Direct-sum quasi-idempotence on the block transfer sum

✓

Let $(B_k)_k$ be a family of tensors and let $c\in \mathbb {C}$. If

\begin{align} \mathcal{E}_{\oplus _k B_k}^2 & =c\mathcal{E}_{\oplus _k B_k}, \notag \end{align}

then the corresponding block transfer sum $\mathcal S_B$ satisfies, for every bond matrix $Y$,

\begin{align} \mathcal S_B(\mathcal S_B(Y)) & =c\mathcal S_B(Y). \notag \end{align}

Proof ▶

Reindex the total bond space by its block labels. By Theorem 17.4.2, the direct-sum transfer map becomes $\mathcal S_B$. Transporting the assumed equation through this identification gives the displayed identity.

Theorem 17.4.4 Direct-sum quasi-idempotence on every mixed corner

✓

Let $(B_k)_k$ be a family of tensors and let $c\in \mathbb {C}$. If

\begin{align} \mathcal{E}_{\oplus _k B_k}^2 & =c\mathcal{E}_{\oplus _k B_k}, \notag \end{align}

then, for every ordered pair of blocks $(j,j')$ whose two bond dimensions are positive,

\begin{align} \mathcal{E}_{j,j'}^2 & =c\mathcal{E}_{j,j'}. \notag \end{align}

Proof ▶

Let $Z$ be an arbitrary rectangular matrix and realize it as the $(j,j')$ block of a matrix $Y$ on the total bond space. By Lemma 17.4.1, $[\mathcal S_B(Y)]_{j,j'}=\mathcal{E}_{j,j'}(Z)$. Theorem 17.4.3 gives $\mathcal S_B^2=c\mathcal S_B$, and hence

\begin{align} \mathcal{E}_{j,j'}(\mathcal{E}_{j,j'}(Z)) & =[\mathcal S_B(\mathcal S_B(Y))]_{j,j'} =[c\mathcal S_B(Y)]_{j,j'} =c\mathcal{E}_{j,j'}(Z). \notag \end{align}

Lemma 17.4.5 Pairwise idempotence of the block transfer sum

✓

Let $(B_k)_k$ be a family of tensors. If every mixed transfer operator $\mathcal{E}_{j,j'}$ is idempotent, then the block transfer sum $\mathcal S_B$ is idempotent: $\mathcal S_B(\mathcal S_B(X))=\mathcal S_B(X)$.

Proof ▶

By Lemma 17.4.1, the $(j,j')$ block satisfies

\begin{align} (\mathcal S_B^2(X))_{j,j'} & = \mathcal{E}_{j,j'}^2(X_{j,j'}) = \mathcal{E}_{j,j'}(X_{j,j'}) = (\mathcal S_B(X))_{j,j'}. \notag \end{align}

Equality on every block gives the asserted equality on the total bond space.

Theorem 17.4.6 Pairwise mixed-transfer criterion for a direct sum

✓

Suppose every bond dimension is positive. The transfer map of $\bigoplus _k B_k$ is idempotent if and only if, for every pair $j,j'$, the mixed transfer operator is idempotent:

\begin{align} \mathcal{E}_{\oplus _k B_k}^2=\mathcal{E}_{\oplus _k B_k} \quad \Longleftrightarrow \quad \mathcal{E}_{j,j'}^2=\mathcal{E}_{j,j'}. \notag \end{align}

Proof ▶

By Lemma 17.4.1, the $(j,j')$ block of the direct-sum transfer map depends only on the $(j,j')$ block of its argument and is obtained by applying $\mathcal{E}_{j,j'}$. Thus pairwise idempotence gives

\begin{align} (\mathcal{E}_{\oplus _k B_k}^2(X))_{j,j'} & = \mathcal{E}_{j,j'}^2(X_{j,j'}) = \mathcal{E}_{j,j'}(X_{j,j'}) = (\mathcal{E}_{\oplus _k B_k}(X))_{j,j'} \notag \end{align}

for every pair of blocks, and hence gives idempotence on the total bond space, as in Lemma 17.4.5. Conversely, every matrix $M$ of the appropriate rectangular size occurs as the $(j,j')$ block of some $Y$ on the total bond space. Applying Lemma 17.4.1 and whole-tensor idempotence gives

\begin{align} \mathcal{E}_{j,j'}^2(M) & = (\mathcal{E}_{\oplus _k B_k}^2(Y))_{j,j'} = (\mathcal{E}_{\oplus _k B_k}(Y))_{j,j'} = \mathcal{E}_{j,j'}(M). \notag \end{align}

Lemma 17.4.7 Scaling of a rectangular mixed transfer operator

✓

For complex scalars $c,e$ and tensors $A,B$ of possibly different bond dimensions, $\mathcal{E}_{cA,eB}=c\overline e\, \mathcal{E}_{A,B}$.

Proof ▶

Conjugate transposition changes the second scalar to its complex conjugate: $\sum _i(cA^i)X(eB^i)^\dagger =c\overline e\sum _iA^iX(B^i)^\dagger $.

Theorem 17.4.8 Phase coherence for literal repeated blocks

✓

Let $A$ have positive bond dimension and a nonzero idempotent transfer map. Let $\mu _q$ have unit modulus. If the literal bond direct sum $\bigoplus _q \mu _qA$ has an idempotent transfer map, then $\mu _q=\mu _{q'}$ for every pair $q,q'$.

Proof ▶

The pairwise criterion (Theorem 17.4.6) gives that $\mathcal{E}_{\mu _qA,\mu _{q'}A}$ is idempotent. Lemma 17.4.7 and $\mathcal{E}_{A,A}=\mathcal{E}_A$ give $\mathcal{E}_{\mu _qA,\mu _{q'}A} =\mu _q\overline{\mu _{q'}}\mathcal{E}_{A,A} =\mu _q\overline{\mu _{q'}}\mathcal{E}_A$. Hence $(\mu _q\overline{\mu _{q'}})\mathcal{E}_A$ is idempotent. Since $\mathcal{E}_A$ is nonzero and $|\mu _q\overline{\mu _{q'}}|=|\mu _q||\mu _{q'}|=1$,

\begin{align} (\mu _q\overline{\mu _{q'}})^2\mathcal{E}_A =\mu _q\overline{\mu _{q'}}\mathcal{E}_A & \quad \Longrightarrow \quad \mu _q\overline{\mu _{q'}} (\mu _q\overline{\mu _{q'}}-1)=0 \quad \Longrightarrow \quad \mu _q\overline{\mu _{q'}}=1. \notag \end{align}

Since $|\mu _{q'}|=1$ gives $\overline{\mu _{q'}}\mu _{q'}=1$, one has $\mu _q =\mu _q(\overline{\mu _{q'}}\mu _{q'}) =(\mu _q\overline{\mu _{q'}})\mu _{q'} =\mu _{q'}$.

Theorem 17.4.9 Cross-block transfer vanishing for a direct-sum fixed point

✓

Let $(A_j)_j$ be a family of distinct irreducible left-canonical blocks, no two of which are gauge-phase equivalent, and suppose the direct sum $\bigoplus _j A_j$ is a renormalization fixed point. Then for every pair of distinct components $j\ne j'$, the mixed transfer operator vanishes:

\begin{align} \mathcal{E}_{j,j'}=\sum _i A_j^i\otimes \overline{A_{j'}^i}=0. \notag \end{align}

This establishes the cross-block ($\delta _{j,j'}$) orthogonality of [ 13 , Theorem 3.11 and Corollary 3.12 ] . The passage to the residual tensors $U_j$ is given by Theorem 17.4.16.

Proof ▶

By Lemma 17.4.1 the $(j,j')$ bond block of the direct-sum transfer map acts as $\mathcal{E}_{j,j'}$. Whole-tensor idempotence $\mathcal{E}^2=\mathcal{E}$ therefore restricts to each block as $\mathcal{E}_{j,j'}^2=\mathcal{E}_{j,j'}$. For distinct irreducible left-canonical blocks, Theorem 5.6.2 gives $\rho _{\operatorname{spec}}(\mathcal{E}_{j,j'}){\lt}1$ when the bond dimensions agree, because the blocks are not gauge-phase equivalent; Theorem 5.6.4 gives the same inequality when the bond dimensions differ. Consequently, Lemma 5.5.3 gives

\begin{align} \mathcal{E}_{j,j'}^2=\mathcal{E}_{j,j'},\qquad \rho _{\operatorname{spec}}(\mathcal{E}_{j,j'}){\lt}1 \quad \Longrightarrow \quad \mathcal{E}_{j,j'}=0. \notag \end{align}

Definition 17.4.10 Joint isometry condition

✓

A family $(U_j)_j$ satisfies the joint isometry condition if

\begin{align} \sum _i(U_j^i)_{\alpha ,\beta } \overline{(U_{j'}^i)_{\alpha ',\beta '}} & =\delta _{j,j'}\delta _{\alpha ,\alpha '}\delta _{\beta ,\beta '}. \notag \end{align}

Diagrammatically, contracting $U_j$ with the conjugate of $U_{j'}$ over the physical index identifies the virtual legs pairwise:

$\begin{tenkz}[rows={ket,bra}, cols=1, west=open, east=open] \tn[skin=box, label pos=n]{U_j} \\ \tn[skin=box, label pos=s, conjugate]{\overline{U_{j'}}} \end{tenkz}$ $ = \delta _{j,j'}\, $ $\begin{tenkz}[rows={wire,wire}, cols=1, west=open, east=cup, bonds=none] \tn[skin=none]{} \\ \tn[skin=none]{} \end{tenkz}$ $ \otimes $ $\begin{tenkz}[rows={wire,wire}, cols=1, west=cup, east=open, bonds=none] \tn[skin=none]{} \\ \tn[skin=none]{} \end{tenkz}$

Each block satisfies unit pair-index orthonormality ($j=j'$), and the cross-block sums between distinct blocks vanish ($j\ne j'$), as in [ 13 , Theorem 3.11 ] .

Theorem 17.4.11 One-letter Gram matrix under the joint isometry condition

✓

Let $\mathcal V=\bigsqcup _j\{ j\} \times [D_j]\times [D_j]$ be the disjoint union of the within-sector virtual pairs. For $x=(j,\alpha ,\beta )$ in $\mathcal V$, set $u_x(i)=(U_j^i)_{\alpha ,\beta }$. Then $\sum _i u_x(i)\overline{u_y(i)}=\delta _{x,y}$.

Proof ▶

If $x$ and $y$ belong to the same sector, the assertion is the within-sector part of the joint isometry condition. If they belong to distinct sectors, it is the cross-sector vanishing condition.

Lemma 17.4.12 Covariance of the mixed transfer operator

✓

For decompositions $A^i=X_AD_AU^iX_A^{-1}$ and $B^i=X_BD_BV^iX_B^{-1}$, the mixed transfer operator of $A,B$ is the mixed transfer operator of the tensors $U,V$ conjugated by the outer factors:

\begin{align} \mathcal{E}_{A,B}(Y) & =X_AD_A\, \mathcal{E}_{U,V}\! \left(X_A^{-1}Y(X_B^{-1})^\dagger \right)(X_BD_B)^\dagger . \notag \end{align}

Proof ▶

Substitute the two decompositions into $\mathcal{E}_{A,B}(Y)=\sum _iA^iY(B^i)^\dagger $ and collect the outer factors, which are independent of the summation index, outside the sum.

Lemma 17.4.13 Vanishing after removing the diagonal factors

✓

With decompositions $A^i=X_AD_AU^iX_A^{-1}$ and $B^i=X_BD_BV^iX_B^{-1}$ whose factors $X_A,D_A,X_B,D_B$ are invertible, if $\mathcal{E}_{A,B}=0$ then $\mathcal{E}_{U,V}=0$.

Proof ▶

The covariance identity Lemma 17.4.12 expresses $\mathcal{E}_{A,B}$ as $\mathcal{E}_{U,V}$ conjugated by the invertible outer factors $X_AD_A$ and $(X_BD_B)^\dagger $, so $\mathcal{E}_{A,B}=0$ forces $\mathcal{E}_{U,V}=0$.

Lemma 17.4.14 Entrywise cross-block isometry sum

✓

If $\mathcal{E}_{U,V}=0$, then, for all virtual indices,

\begin{align} \sum _i(U^i)_{\alpha ,\beta } \overline{(V^i)_{\alpha ',\beta '}} & =0. \notag \end{align}

Proof ▶

Apply $\mathcal{E}_{U,V}=0$ to the matrix unit supported at $(\beta ,\beta ')$ and read the $(\alpha ,\alpha ')$ entry.

Lemma 17.4.15 Each block of a direct-sum fixed point is a fixed point

✓

If the direct sum $\bigoplus _k B_k$ is a renormalization fixed point, with $\dim _k\ge 1$ for all $k$, then each block $B_j$ is a renormalization fixed point.

Proof ▶

The diagonal mixed transfer operator $\mathcal{E}_{j,j}$ is the transfer map of $B_j$, and whole-tensor idempotence makes it idempotent, which is exactly the renormalization fixed-point condition for $B_j$.

Theorem 17.4.16 Joint isometry condition for a direct-sum fixed point

✓

Let $(A_j)_j$ be a family of normal, irreducible, left-canonical blocks with bond dimensions $D_j\ge 1$, no two of which are gauge-phase equivalent. If the direct sum $\bigoplus _j A_j$ is a renormalization fixed point, then there exist invertible matrices $X_j$, trace-one positive diagonal matrices $\rho _j=\operatorname{diag}(\lambda _{j,\alpha })$, and tensors $U_j$ with

\begin{align} A_j^i & =X_j\sqrt{\rho _j}\, U_j^iX_j^{-1}, \notag \end{align}

that is,

$\begin{tenkz}[rows={wire}, cols=1, west=open, east=open] \tn[label pos=s, ports={90:physical:$i$}]{A_j} \end{tenkz}$ $ = $ $\begin{tenkz}[rows={wire}, cols=4, west=open, east=open] \tn[skin=ring]{X_j} & \tn[skin=ring]{\sqrt{\rho_j}} & \tn[label pos=s, ports={90:physical:$i$}]{U_j} & \tn[skin=ring]{X_j^{-1}} \end{tenkz}$

such that $(U_j)_j$ satisfies the joint isometry condition (Definition 17.4.10). This formalizes the isometry equations of [ 13 , Corollary 3.12 ] for distinct normal blocks.

Proof ▶

Whole-tensor idempotence makes the diagonal mixed transfer operator of each block its own transfer map, hence each block is a renormalization fixed point; the square-root fixed-point form (Theorem 17.3.1.12) supplies the decomposition $A_j^i=X_j\sqrt{\rho _j}U_j^iX_j^{-1}$ with the within-block orthonormality, which is the $j=j'$ case. For $j\ne j'$, Theorem 17.4.9 gives $\mathcal{E}_{A_j,A_{j'}}=0$; by the covariance Lemma 17.4.12 and invertibility of the outer factors $\mathcal{E}_{U_j,U_{j'}}=0$, and reading the entry of this operator at a matrix unit yields the cross-block sum $\sum _i(U_j^i)_{\alpha ,\beta } \overline{(U_{j'}^i)_{\alpha ',\beta '}}=0$.

Theorem 17.4.17 Direct sum of square-root fixed-point blocks

✓

Let $(B_k)_k$ be a family of tensors in square-root fixed-point form whose cross-block mixed transfer operators vanish ($\mathcal{E}_{j,j'}=0$ for all $j\ne j'$). Then the direct sum $\bigoplus _k B_k$ is a renormalization fixed point. This establishes the converse direction of the pure-state fixed-point characterization [ 13 , Section 3 ] for distinct blocks.

Proof ▶

The direct-sum transfer map decouples block by block: by Lemma 17.4.1 its $(j,j')$ bond block acts as $\mathcal{E}_{j,j'}$ on the $(j,j')$ bond block of the argument. Each diagonal block $\mathcal{E}_{j,j}$ is the transfer map of $B_j$, which is idempotent because $B_j$ is a renormalization fixed point (Theorem 17.3.1.14); each off-diagonal block $\mathcal{E}_{j,j'}$ with $j\ne j'$ vanishes by hypothesis. Therefore, for every bond block,

\begin{align} [\mathcal{E}_{\bigoplus B}^2(Y)]_{j,j'} & =\mathcal{E}_{j,j'}\! (\mathcal{E}_{j,j'}(Y_{j,j'})) =\mathcal{E}_{j,j'}(Y_{j,j'}) =[\mathcal{E}_{\bigoplus B}(Y)]_{j,j'}, \notag \end{align}

so $\mathcal{E}_{\bigoplus B}^2=\mathcal{E}_{\bigoplus B}$ and the direct sum is a renormalization fixed point.

Theorem 17.4.18 Distinct-block fixed-point characterization

✓

Let $(B_k)_k$ be a family of normal, irreducible, left-canonical blocks with $D_k\ge 1$ for all $k$, no two of which are gauge-phase equivalent. Then the direct sum $\bigoplus _k B_k$ is a renormalization fixed point if and only if each block has the square-root fixed-point form and all cross-block mixed transfer operators vanish:

\begin{align} \left(\bigoplus _k B_k\text{ is an RFP}\right) & \iff \left(\forall k,\ B_k\text{ in square-root form}\right) \wedge \left(\forall j\ne j’,\ \mathcal{E}_{j,j'}=0\right). \notag \end{align}

This is the distinct-block characterization of pure-state renormalization fixed points [ 13 , Theorem 3.11 ] .

Proof ▶

For the forward direction, whole-tensor idempotence makes each diagonal mixed transfer operator the transfer map of its block (Lemma 17.4.15), so each block is a renormalization fixed point and hence in the square-root fixed-point form (Theorem 17.3.1.12), while the off-diagonal operators vanish, $\mathcal{E}_{j,j'}=0$ for $j\ne j'$, by Theorem 17.4.9. Conversely, a direct sum of square-root fixed-point blocks whose cross-block operators vanish is a renormalization fixed point (Theorem 17.4.17).

Corollary 17.4.19 BNT basis direct-sum characterization

✓

Let $P$ satisfy the auxiliary BNT sector hypotheses, with distinct basis tensors $(B_j)_j$. Then the multiplicity-one direct sum $\bigoplus _j B_j$ is a renormalization fixed point if and only if every $B_j$ has the square-root fixed-point form and $\mathcal{E}_{j,j'}=0$ for all $j\ne j'$.

In this case, there exist invertible matrices $X_j$, trace-one positive diagonal matrices $\rho _j=\operatorname{diag}(\lambda _{j,\alpha })$, and tensors $U_j$ such that $B_j^i=X_j\sqrt{\rho _j}\, U_j^iX_j^{-1}$, where $(U_j)_j$ satisfies the joint isometry condition (Definition 17.4.10).

Proof ▶

Positive basis dimensions supply the nonzero bond spaces. For each basis tensor, irreducibility and left-canonical normalization are part of the auxiliary BNT sector hypotheses. If $m_j{\gt}0$ is the peripheral period of $B_j$, the normalized self-overlap hypothesis and the periodic self-overlap formula used in Theorem 16.4.20 give, along the same subsequence,

\begin{align} \lim _{k\to \infty }O_{B_jB_j}(m_jk) & =1, \notag \\ \lim _{k\to \infty }O_{B_jB_j}(m_jk) & =m_j. \notag \end{align}

Hence $m_j=1$. The transfer map is primitive by the corresponding theorem in the companion quantum-channel volume [ 216 , “Primitivity equals period one” ] , and normality follows from Theorem 7.11.1.1. Gauge-phase distinctness is also part of the auxiliary BNT sector hypotheses. The equivalence now follows from Theorem 17.4.18; the same derived hypotheses in Theorem 17.4.16 gives the joint isometry condition.

Theorem 17.4.20 Transfer idempotence under coisometric reconstruction

✓

Let $A$ and $B$ be tensors with bond dimensions $D$ and $R$. Suppose that $U\in \mathbb {C}^{R\times D}$ satisfies $UU^\dagger =I_R$ and, for every physical index $i$,

\begin{align} A^i & =U^\dagger B^iU. \notag \end{align}

Then $A$ is a renormalization fixed point if and only if $B$ is a renormalization fixed point.

Proof ▶

For every bond matrix $X$,

\begin{align} \mathcal{E}_A(X) & =U^\dagger \mathcal{E}_B(UXU^\dagger )U. \notag \end{align}

The identity $UU^\dagger =I_R$ gives $U(U^\dagger YU)U^\dagger =Y$. Applying these two identities twice proves each implication.

Theorem 17.4.21 A fixed nonzero scaling of a normal tensor

✓

Let $B$ be a normal tensor and let $c\in \mathbb {C}\setminus \{ 0\} $. If $cB$ is a renormalization fixed point, then

\begin{align} |c| & =1, & \mathcal{E}_B^2 & =\mathcal{E}_B. \notag \end{align}

Proof ▶

Normality gives $r(\mathcal{E}_B)=1$, while $\mathcal{E}_{cB}=|c|^2\mathcal{E}_B$. Since $c\ne 0$, this operator is nonzero. Theorem 5.5.2 gives $r(\mathcal{E}_{cB})=1$, and Theorem 5.5.1 gives

\begin{align} 1=r(\mathcal{E}_{cB})=|c|^2r(\mathcal{E}_B)=|c|^2. \notag \end{align}

Thus $|c|=1$, so $\mathcal{E}_{cB}=\mathcal{E}_B$ and the idempotence of $\mathcal{E}_B$ follows.

Theorem 17.4.22 Blocks of a CPSV fixed point

✓

Suppose $A$ has CPSV canonical-form data

\begin{align} A^i & =U^\dagger \! \left(\bigoplus _{k=1}^r\mu _kA_k^i\right)U, & UU^\dagger & =I, \notag \end{align}

where every $A_k$ is normal. If $A$ is a renormalization fixed point, then every block $k$ satisfies

\begin{align} |\mu _k| & =1, & \mathcal{E}_{A_k}^2 & =\mathcal{E}_{A_k}. \notag \end{align}

Proof ▶

By Theorem 17.4.20, the direct sum $\bigoplus _k\mu _kA_k$ is a renormalization fixed point. Lemma 17.4.15 implies that each block $\mu _kA_k$ is also a renormalization fixed point. Since $\mu _k\ne 0$, Theorem 17.4.21 gives both conclusions.

Theorem 17.4.23 Phase-class residual isometries

✓

Let $A$ be a CPSV canonical-form renormalization fixed point. Choose one representative $C_j$ from each matrix-product-vector phase class. Then there are invertible matrices $X_j$, positive numbers $\lambda _{j,\alpha }$ with $\sum _\alpha \lambda _{j,\alpha }=1$, and tensors $U_j$ such that

\begin{align} C_j^i & =X_j\operatorname{diag}\! \left(\sqrt{\lambda _{j,\alpha }}\right) U_j^iX_j^{-1}, \notag \\ \sum _i(U_j^i)_{\alpha ,\beta } \overline{(U_{j'}^i)_{\alpha ',\beta '}} & =\delta _{j,j'}\delta _{\alpha ,\alpha '}\delta _{\beta ,\beta '}. \notag \end{align}

If no classes are present, the statement holds vacuously.

Local fix (square-root factor in canonical form): The formula uses the square-root weights $\sqrt{\lambda _{j,\alpha }}$ rather than the bare diagonal printed in [ 13 , line 547 ] , as required by the reference-tensor construction; documented in [ 85 ] .

Proof ▶

The retained weighted direct sum is a renormalization fixed point. For each representative, Theorem 17.4.22 gives $|\mu _j|=1$ and $\mathcal{E}_{C_j}^2=\mathcal{E}_{C_j}$. By Theorem 17.3.1.15, each normal fixed point $C_j$ admits the displayed square-root form. Putting the representatives independently in left-canonical gauge, distinct phase-class representatives are not gauge-phase equivalent. Since the direct sum is a fixed point, Theorem 17.4.9 gives

\begin{align} \mathcal{E}_{\mu _jC_j,\mu _{j'}C_{j'}} & =0 \qquad (j\ne j’). \notag \end{align}

Cancelling the unit-modulus scalars and gauge transformations yields $\mathcal{E}_{C_j,C_{j'}}=0$. Removing the positive diagonal factors and evaluating on matrix units gives the off-diagonal residual orthogonality. Combined with the within-block orthonormality of each $U_j$, this yields the joint isometry condition.

Theorem 17.4.24 Square-root characterization of fixed points

Cirac et al. [ 13 , Theorem 3.11 ] assert that a tensor $A$ in canonical form is a renormalization fixed point if and only if it can be written as

\begin{align} A^i & =\bigoplus _{j=1}^g\bigoplus _{q=1}^{r_j} \mu _{j,q}X_{j,q}\sqrt{\rho _j}\, U_j^iX_{j,q}^{-1}, \notag \end{align}

where $|\mu _{j,q}|=1$, each $\rho _j$ is positive diagonal with $\operatorname{tr}(\rho _j)=1$, and

\begin{align} \sum _i(U_j^i)_{\alpha ,\beta } \overline{(U_{j'}^i)_{\alpha ',\beta '}} & =\delta _{j,j'}\delta _{\alpha ,\alpha '}\delta _{\beta ,\beta '}. \notag \end{align}

The source specifies that $\bigoplus _{j,q}$ is a simultaneous direct sum in both the physical and virtual spaces [ 13 , lines 561–562 ] . However, it does not provide an explicit coordinate embedding for the physical routing of $q$, a dimension constraint, or a $q$-indexed isometry equation. Consequently, this statement remains unformalized pending clarification of the coordinate-level routing in the source.

Local fix (square-root factor in canonical form): CPSV prints a bare $\Lambda _j$ in Theorem 3.11, whereas the reference tensor at line 1300 has coefficients $\sqrt{(\rho _j)_{\alpha ,\alpha }}$, which forces the factor $\sqrt{\rho _j}$ used above. This repair is documented in [ 85 ] .

17.5 Renormalization flow and physical correlations

One renormalization step blocks two sites of a tensor $B$ and compresses the blocked physical index by an isometry $U$ with $U^\dagger U=\mathbb {1}$, producing a tensor $A$ with

\begin{align} B^{i_1}B^{i_2} & =\sum _j U_{(i_1,i_2),j}A^j \notag \end{align}

[ 13 , lines 389–394 ] . Since $U^\dagger U=\mathbb {1}$, the isometry cancels in the doubled contraction, so the step acts on transfer maps as $\mathcal{E}_A=\mathcal{E}_B^2$ [ 13 , line 1209 ] ; repeated steps produce the dyadic powers $\mathcal{E}_B^{2^n}$.

\begin{tenkz}[cols=2, west=open, east=open]
\tn[ports={90:physical:$i_1$}]{B} & \tn[ports={90:physical:$i_2$}]{B}
\end{tenkz} — Figure 17.4 One renormalization step. Top: blocking two sites of $B$ and compressing the pair index by the isometry $U$ gives the tensor $A$. Bottom: on the doubled network the isometry cancels, so $\mathcal{E}_B^2=\mathcal{E}_A$ [ 13 , lines 389–394 and 1209 ] .

\begin{tenkz}[rows={op,wire}, cols=3, bonds=none]
\tn[skin=pill, wide=3, name=U,
ports={90@1:physical:$i_1$, 90@3:physical:$i_2$,
270@2:physical}]{U} \\
& \tn[name=A,
ports={180:virtual, 0:virtual, 90:physical:$j$}]{A} &
\tnwire{A.90}{U.270@2}
\end{tenkz} — Figure 17.4 One renormalization step. Top: blocking two sites of $B$ and compressing the pair index by the isometry $U$ gives the tensor $A$. Bottom: on the doubled network the isometry cancels, so $\mathcal{E}_B^2=\mathcal{E}_A$ [ 13 , lines 389–394 and 1209 ] .

Theorem 17.5.1 Primitive-transfer convergence for each auxiliary block

✓

Fix one block $A_k$ in a family satisfying Definition 17.3.1.1. Then the dyadic powers of its transfer map converge pointwise to an idempotent linear map: there exists $\mathcal{E}_{k,\infty }$ with $\mathcal{E}_{k,\infty }^2=\mathcal{E}_{k,\infty }$ such that $\mathcal{E}_{A_k}^{2^n}(\rho )\to \mathcal{E}_{k,\infty }(\rho )$ as $n\to \infty $ for every virtual matrix $\rho $. This formalizes the per-block primitive convergence of [ 13 , Appendix B, lines 1211–1244 ] .

Proof ▶

Each block is injective and left-canonical, so its transfer map is a primitive channel with a unique positive-definite fixed point $\rho _0$. The exponential convergence bound $\lVert \mathcal{E}^n(X)-P(X)\rVert \le C(1-\delta )^n\lVert X\rVert $ (Theorem 14.3.1) yields pointwise convergence $\mathcal{E}^n(X)\to P(X)$, and restricting to the dyadic subsequence $2^n\to \infty $ yields the claim with idempotent projection $\mathcal{E}_{k,\infty }=P_k$.

Theorem 17.5.2 Phase obstruction to canonical-form flow convergence

✓

Let $\omega =e^{2\pi i/3}$ and consider the tensor with one physical component

\begin{align} A^1 & = \begin{pmatrix} 1 & 0 \\ 0 & \omega \end{pmatrix}. \notag \end{align}

This tensor is in canonical form, with two one-dimensional normal components of weights $1$ and $\omega $. For the matrix unit $e_{21}$,

\begin{align} \mathcal{E}_A^{2^n}(e_{21}) & =\omega ^{2^n}e_{21}. \notag \end{align}

The right-hand side alternates between $\omega e_{21}$ and $\omega ^2 e_{21}$, so the dyadic renormalization flow does not converge. This refutes the unrestricted canonical-form convergence assertion of [ 13 , lines 417 and 1211–1244 ] ; it does not affect the single-component convergence of Theorem 17.5.1.

Proof ▶

The two one-dimensional components are normal and both weights have modulus one. Direct calculation gives $\mathcal{E}_A(e_{21})=\omega e_{21}$. Since $2^n$ is congruent alternately to $1$ and $2$ modulo $3$, the sequence alternates between $\omega e_{21}$ and $\omega ^2 e_{21}$, precluding convergence.

Definition 17.5.3 One-letter doubled-MPO virtual-gauge condition

✓

For an MPO tensor $M$ with one physical component, the Appendix D blocking diagram holds up to virtual gauge if $M^{[2]}$ and $M$ are related by similarity on the virtual indices. For a pure tensor $A$, this condition is imposed on the doubled tensor

\begin{align} M & =A\otimes \overline A. \notag \end{align}

Since the physical algebra is one-dimensional, every physical unitary channel in the diagram is trivial, recovering the relation in [ 13 , Appendix D, lines 2100–2107 ] .

Definition 17.5.4 One-letter MPS blocking witness

✓

A tensor $A$ with one physical component has a blocking witness if there exist an invertible matrix $X$ and a scalar $\zeta $ with $|\zeta |=1$ such that

\begin{align} A & =\zeta X A^{[2]}X^{-1}. \notag \end{align}

Definition 17.5.5 Doubled virtual gauge

✓

Given an invertible virtual matrix $X$, the induced gauge on the doubled bond space is $X\otimes \overline X$.

Definition 17.5.6 Cube-phase swap gauge

✓

The permutation matrix

\begin{align} S & =\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \notag \end{align}

defines an invertible virtual gauge that exchanges the $1$ and $\omega $ coordinates of the cube-phase tensor.

Theorem 17.5.7 An MPS blocking witness gives the doubled-MPO condition

✓

If $A = \zeta X A^{[2]} X^{-1}$ is a one-letter MPS blocking witness with $|\zeta |=1$, then the doubled tensor $M = A \otimes \overline{A}$ satisfies

\begin{align} M & = (X \otimes \overline{X}) M^{[2]} (X \otimes \overline{X})^{-1}, \notag \end{align}

satisfying the doubled-MPO virtual-gauge condition (Definition 17.5.3).

Proof ▶

Tensoring $A=\zeta X A^{[2]}X^{-1}$ with its complex conjugate yields $(X\otimes \overline X) M^{[2]} (X\otimes \overline X)^{-1}$, with scalar factor $\zeta \overline\zeta =|\zeta |^2=1$.

Theorem 17.5.8 The printed pure blocking-up-to-gauge equivalence is false

✓

Let $\omega =e^{2\pi i/3}$, and set

\begin{align} A & =\begin{pmatrix} 1 & 0 \\ 0 & \omega \end{pmatrix}, & S & =\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. \notag \end{align}

Then $\omega ^3=1$, $\omega \ne 1$, and

\begin{align} A & =\omega S A^2S^{-1}. \notag \end{align}

The doubled MPO tensor generates positive semidefinite operators on every nonempty chain, is in CPSV canonical form, and has four scalar-block weights satisfying the source normalization convention. It also satisfies the literal one-letter virtual-gauge condition in the source diagram, but its transfer map is not idempotent. Hence the canonical-restricted pure-state equivalence claimed in [ 13 , Appendix D, lines 2100–2107 ] is false.

Proof ▶

On every nonempty chain, $M=A\otimes \overline A$ generates the rank-one operator $|V^{(N)}(A)\rangle \! \langle V^{(N)}(A)|$, which is positive semidefinite. Moreover, $M$ is the direct sum of four bond-one scalar blocks with weights $\mu _p\overline{\mu _q}$, where $\mu _0=1$ and $\mu _1=\omega $. Each block is normal and every weight has modulus one, proving both the CPSV canonical-form condition and its normalization.

Swapping the two virtual coordinates transforms $\operatorname {diag}(1,\omega ^2)$ into $\operatorname {diag}(\omega ^2,1)$; multiplication by $\omega $ gives $A$. Writing $M=A\otimes \overline A$, doubling cancels the phase:

\begin{align} (\omega S A^{[2]}S^{-1})\otimes (\overline\omega \, \overline S\, \overline{A^{[2]}}\, \overline S^{-1}) & =(S\otimes \overline S)M^{[2]}(S\otimes \overline S)^{-1}. \notag \end{align}

However, idempotence $\mathcal{E}_A^2=\mathcal{E}_A$ would imply $\mathcal{E}_A^{2^n}=\mathcal{E}_A$ for all $n\geq 0$, contradicting the non-convergent alternating dyadic orbit of Theorem 17.5.2.

Remark 17.5.9

Theorem 17.5.8 refutes the pure-state equivalence assertion of [ 13 , Appendix D ] ; the general mixed-state case is treated separately in MPO theory.

Remark 17.5.10

When the transfer map is idempotent ($\mathcal{E}^2=\mathcal{E}$), connected correlations become separation-independent for all separations $n\ge 1$. In the rank-one (primitive) case where the peripheral spectrum consists only of the simple eigenvalue $1$, all subleading spectral values vanish, yielding zero correlation length ($\xi =0$).

Definition 17.5.11 Virtual-insertion auxiliary condition

✓

An MPS tensor satisfies virtual-insertion distance independence if for every positive-definite right fixed point $\rho ^R{\gt}0$, every pair of virtual bond matrices $X,Y$, and all separations $n,m\geq 1$, the connected correlator satisfies $C(X,Y;n)=C(X,Y;m)$.

Definition 17.5.12 Physical observable transfer

✓

For an observable $O$ on a block of $L$ physical spins, define the linear map on the virtual matrix space by

\begin{align} \mathcal{E}_O(X) & =\sum _{\boldsymbol i,\boldsymbol j} \langle \boldsymbol j|O|\boldsymbol i\rangle A^{\boldsymbol i}X(A^{\boldsymbol j})^\dagger . \notag \end{align}

This transfer map also depends linearly on the observable $O$. Diagrammatically, $O$ is inserted between the two layers of the doubled chain on its $L$ sites:

$\begin{tenkz}[rows={ket,op,bra}, cols=3, bonds=none] \tn[name=k1, label pos=n, ports={180:virtual, 0:virtual, 270:physical}]{A} & \tn[name=kd, skin=dots, ports={180:virtual, 0:virtual}]{} & \tn[name=k3, label pos=n, ports={180:virtual, 0:virtual, 270:physical}]{A} \\ \tn[name=O, skin=pill, wide=3, ports={90@1:physical, 90@3:physical, 270@1:physical, 270@3:physical}]{O} \\ \tn[name=b1, label pos=s, conjugate, ports={180:virtual, 0:virtual, 90:physical}]{\overline{A}} & \tn[name=bd, skin=dots, ports={180:virtual, 0:virtual}]{} & \tn[name=b3, label pos=s, conjugate, ports={180:virtual, 0:virtual, 90:physical}]{\overline{A}} \tnwire{k1.0}{kd.180} \tnwire{kd.0}{k3.180} \tnwire{b1.0}{bd.180} \tnwire{bd.0}{b3.180} \tnwire{k1.270}{O.90@1} \tnwire{k3.270}{O.90@3} \tnwire{O.270@1}{b1.90} \tnwire{O.270@3}{b3.90} \tnwire{k1.180}{open w} \tnwire{b1.180}{open w} \tnwire{k3.0}{open e} \tnwire{b3.0}{open e} \end{tenkz}$

This is the observable transfer map of [ 13 , lines 490–496 ] .

Definition 17.5.13 Two-observable periodic expectation

✓

Let $O_1,O_2$ act on blocks of $L_1,L_2$ physical spins, separated around the periodic chain by complementary gaps of lengths $n_1,n_2$. Their two-observable expectation is

\begin{align} \operatorname{Tr}_{M_{D}(\mathbb {C})}\! \left(\mathcal{E}_{O_2}\circ \mathcal{E}_A^{n_2}\circ \mathcal{E}_{O_1}\circ \mathcal{E}_A^{n_1}\right). \notag \end{align}

Diagrammatically, the four maps are composed around the ring:

$\begin{tenkz}[rows={wire,wire}, cols=4, west=trace, east=trace] \tn[skin=ring, wires=2]{\E_{O_2}} & \tn[skin=ring, wires=2]{\E_A^{n_2}} & \tn[skin=ring, wires=2]{\E_{O_1}} & \tn[skin=ring, wires=2]{\E_A^{n_1}} \end{tenkz}$

The trace is the operator trace on the virtual matrix space, as in [ 13 , lines 490–496 ] .

Definition 17.5.14 Physical correlations independent of distance

✓

Let $O_1,O_2$ act on two disjoint contiguous regions of positive lengths $L_1,L_2\geq 1$ in a periodic chain, with complementary gap lengths $n_1,n_2\geq 0$. Physical correlations are independent of distance if, for all $m_1,m_2\geq 0$ satisfying $n_1+n_2=m_1+m_2$,

This allows either region to be translated along the chain without crossing the other, including adjacent regions ($n_i=0$) [ 13 , Definition 3.3 ] .

Definition 17.5.15 Positive-gap restriction of physical distance independence

✓

Let $O_1,O_2$ be observables on disjoint blocks of a periodic chain, with observable transfer maps $\mathcal{E}_{O_1},\mathcal{E}_{O_2}$ (Definition 17.5.12). The tensor has positive-gap physical correlations independent of distance if for all positive complementary gaps $n_1,n_2,m_1,m_2{\gt}0$ satisfying $n_1+n_2=m_1+m_2$,

This restricts Definition 17.5.14 to strictly positive gaps, excluding adjacent regions [ 13 , Definition 3.3 ] .

Lemma 17.5.16 Physical CID implies positive-gap physical CID

✓

If an MPS tensor satisfies physical distance independence (Definition 17.5.14), then it satisfies the positive-gap condition (Definition 17.5.15).

Proof ▶

Restrict the gap lengths in Definition 17.5.14 to strictly positive integers.

Theorem 17.5.17 Idempotent transfer implies positive-gap physical CID

✓

If $\mathcal{E}_A^2=\mathcal{E}_A$, then $A$ satisfies positive-gap physical distance independence (Definition 17.5.15).

Proof ▶

Idempotence yields $\mathcal{E}_A^n=\mathcal{E}_A$ for all $n\ge 1$. Since $n_1,n_2,m_1,m_2\ge 1$, substituting $\mathcal{E}_A$ for each power in the periodic expectation gives

\begin{align} \operatorname{Tr}_{M_{D}(\mathbb {C})}\! \left(\mathcal{E}_{O_2}\circ \mathcal{E}_A^{n_2}\circ \mathcal{E}_{O_1}\circ \mathcal{E}_A^{n_1}\right) & = \operatorname{Tr}_{M_{D}(\mathbb {C})}\! \left(\mathcal{E}_{O_2}\circ \mathcal{E}_A\circ \mathcal{E}_{O_1}\circ \mathcal{E}_A\right) \notag \\ & = \operatorname{Tr}_{M_{D}(\mathbb {C})}\! \left(\mathcal{E}_{O_2}\circ \mathcal{E}_A^{m_2}\circ \mathcal{E}_{O_1}\circ \mathcal{E}_A^{m_1}\right). \notag \end{align}

Definition 17.5.18 Virtual-insertion zero correlation length

✓

A tensor $A$ satisfies the virtual-insertion zero-correlation-length condition if its transfer map is idempotent ($\mathcal{E}_A^2=\mathcal{E}_A$) and it satisfies virtual-insertion distance independence (Definition 17.5.11).

Theorem 17.5.19 Virtual-insertion distance independence implies RFP

✓

If the virtual-insertion condition (Definition 17.5.11) holds for every positive-definite right fixed point $\rho ^R{\gt}0$ of the transfer map, then $\mathcal{E}_A^2=\mathcal{E}_A$.

Proof ▶

Let $\rho {\gt}0$ be a right fixed point. For any virtual matrix $Z$, set $X=Z\rho ^{-1}$, so that $X\rho =Z$. Applying distance independence with virtual insertions $X$ and $Y$ at separations $n=2$ and $m=1$, the disconnected terms cancel, yielding

\begin{align} \operatorname{tr}\! \left(Y\mathcal{E}_A^2(Z)\right) = \operatorname{tr}\! \left(Y\mathcal{E}_A(Z)\right) \notag \end{align}

for all $Y$. Nondegeneracy of the trace pairing gives $\mathcal{E}_A^2(Z)=\mathcal{E}_A(Z)$ for all $Z$, concluding $\mathcal{E}_A^2=\mathcal{E}_A$.

Theorem 17.5.20 Redundancy of the one-block auxiliary condition

✓

A tensor satisfies the virtual-insertion zero-correlation-length condition (Definition 17.5.18) if and only if its transfer map is idempotent:

\begin{align} \text{IsZCL}(A) \iff \mathcal{E}_A^2 = \mathcal{E}_A. \notag \end{align}

Proof ▶

The forward direction is immediate from Definition 17.5.18. Conversely, idempotence gives $\mathcal{E}_A^n=\mathcal{E}_A$ for all $n\ge 1$. For every positive-definite right fixed point $\rho ^R$, all virtual matrices $X,Y$, and all $n\ge 1$,

\begin{align} C(X,Y;n) & =\operatorname{tr}\! \left(Y\mathcal{E}_A^n(X\rho ^R)\right) -\operatorname{tr}(X\rho ^R)\operatorname{tr}(Y\rho ^R) \notag \\ & =\operatorname{tr}\! \left(Y\mathcal{E}_A(X\rho ^R)\right) -\operatorname{tr}(X\rho ^R)\operatorname{tr}(Y\rho ^R) \notag \\ & =C(X,Y;1). \notag \end{align}

Hence connected correlators are independent of distance.