Tensor Network Theory: A formalization blueprint

15 Block-Injective Parent Hamiltonians and Degenerate Ground Spaces

For a block-injective tensor \(A^i=\bigoplus _j \mu _j A_j^i\), the local ground spaces decompose according to the normal blocks. The first sections prove the open-segment intersection relations and the trace identities that separate distinct blocks.

The remaining sections analyze block-diagonal boundary conditions across the periodic cut. Under the stated length and closure hypotheses, they identify the periodic ground space with \(\operatorname{span}\{ V^{(N)}(A_j)\} _j\), as in [ CPGSV21 , Theorem IV.16 ] and [ PGVWC07 , Theorem 12 ] .

15.1 Block injectivity and degenerate ground spaces

For an MPS with more than one block in canonical form, after blocking to block-injective canonical form, the physical–virtual correspondence is restricted to block-diagonal boundary conditions. In the notation of [ CPGSV21 , lines 2113–2117 ] , set \(\widetilde A^\omega :=(A_1^\omega ,\ldots ,A_g^\omega ) \in \prod _{j=1}^g M_{D_j}(\mathbb {C})\). Then block injectivity is the statement

\begin{align} \forall X=(X_j)_{j=1}^g\in \bigoplus _{j=1}^g M_{D_j}(\mathbb {C}), \quad \exists (c_\omega (X))_\omega , \quad X & = \sum _\omega c_\omega (X)\widetilde A^\omega . \notag \end{align}

Equivalently, with the same coefficient family, \(X_j=\sum _\omega c_\omega (X)A_j^\omega \) for every \(j\). The strengthened parent-Hamiltonian theorem in [ CPGSV21 , Theorem IV.16, lines 2126–2128 ] is

\begin{align} N\ge L_0+1 \quad \Longrightarrow \quad \ker H_N & = \operatorname{span}\{ V^{(N)}(A_j):j=1,\ldots ,g\} . \notag \end{align}

The statements below record the corresponding periodic-boundary ground-space formulation; the passage to \(\ker H_N\) is kept separate.

Definition 15.1.1 Periodic-boundary ground space for the block-diagonal tensor

For a basis of normal tensors \(A=(A_j)_{j=1}^g\) and nonzero weights \(\mu _j\), let \(B:=\bigoplus _{j=1}^g \mu _j A_j\). The ground space on the \(N\)-site ring considered in this section is \(\mathcal G_{N,L}(B)\).

Lemma 15.1.2 Block-diagonal periodic-boundary ground space

By definition, for \(B=\bigoplus _{j=1}^g \mu _j A_j\) one has \(\mathcal G_{N,L}(B) =\mathcal G_{N,L}(\bigoplus _{j=1}^g \mu _j A_j)\).

Definition 15.1.3 Vector space generated by a basis of normal tensors
#

For a basis of normal tensors \(A_1,\ldots ,A_g\), the vector space generated by their length-\(N\) matrix product vectors is \(\operatorname{span}\{ V^{(N)}(A_j):j=1,\ldots ,g\} \).

Lemma 15.1.4 BNT span as the sum of one-vector spaces

For a family \(A_0,\ldots ,A_{r-1}\) and a chain length \(N\), let \(M_j=\operatorname{span}\{ V^{(N)}(A_j)\} \). Then

\begin{align} \bigvee _{j=0}^{r-1} M_j & = \operatorname{span}\{ V^{(N)}(A_j):j=0,\ldots ,r-1\} . \notag \end{align}
Proof

Both sides are the span of the same family of vectors.

Lemma 15.1.5 Each normal-tensor vector lies in the periodic-boundary ground space

If \(1{\lt}L\) and \(N\ge L+1\), then for each normal tensor \(A_j\) in the basis, the vector \(V^{(N)}(A_j)\) lies in \(\mathcal G_{N,L}(\bigoplus _k \mu _k A_k)\).

Proof

Write \(\Gamma _N^B\) for the length-\(N\) boundary-to-state map of the tensor \(B\), and let \(\iota _j(Z)\) be the block-diagonal matrix with \(Z\) in the \(j\)-th block and \(0\) elsewhere. The direct-sum evaluation gives

\begin{align} \Gamma _N^{\oplus _k\mu _k A_k} \! \left(\iota _j(\mu _j^{-N}X)\right) & = \Gamma _N^{A_j}(X). \notag \end{align}

Taking \(X=I\) gives the vector \(V^{(N)}(A_j)\), and Lemma 13.6.8 applied to the individual block gives its membership in \(\mathcal G_{N,L}(\bigoplus _k\mu _k A_k)\).

Lemma 15.1.6 Inclusion of a block ground space

For each block index \(j\) with \(\mu _j\neq 0\), and whenever \(0{\lt}N\) and \(L\le N\), the periodic-boundary ground space of the block \(A_j\) embeds into the corresponding ground space of the direct-sum tensor:

\begin{align} \mathcal G_{N,L}(A_j) & \subseteq \mathcal G_{N,L}\! \left(\bigoplus _k \mu _k A_k\right). \notag \end{align}
Proof

The definition by length-\(L\) windows on the ring reduces the inclusion to the local block-embedding statement that each ground vector of \(A_j\) on the window lifts to the direct sum by the identity

\begin{align} \Gamma _L^{\oplus _k\mu _k A_k} \! \left(\iota _j(\mu _j^{-L}X)\right) & = \Gamma _L^{A_j}(X). \notag \end{align}

Here \(\iota _j\) inserts a matrix in the \(j\)-th diagonal block and has zero off that block. Thus \(\iota _j(\mu _j^{-L}X)\) has exactly the local vector of the \(j\)-th block on that window.

Lemma 15.1.7 Forward inclusion from the block ground spaces

Assume \(\mu _j\neq 0\) for every block index \(j\). Whenever \(0{\lt}N\) and \(L\le N\), the linear sum of the blockwise ground spaces with periodic boundary conditions is contained in the direct-sum tensor’s ground space with periodic boundary conditions:

\begin{align} \sum _j \mathcal G_{N,L}(A_j) & \subseteq \mathcal G_{N,L}\! \left(\bigoplus _k \mu _k A_k\right), \notag \end{align}

where \(j\) ranges over all block indices.

Proof

Take the linear sum of the inclusions \(\mathcal G_{N,L}(A_j) \subseteq \mathcal G_{N,L}(\bigoplus _k \mu _k A_k)\) over all block indices \(j\).

Lemma 15.1.8 Forward inclusion for the direct-sum tensor

Let \(A^i=\bigoplus _j\mu _j A_j^i\) be in block-injective canonical form, where \(A_1,\ldots ,A_g\) is a basis of normal tensors. The linear sum of the blockwise ground spaces with periodic boundary conditions is contained in the ground space of the block-diagonal tensor on the \(N\)-site ring:

\begin{align} \sum _j \mathcal G_{N,L}(A_j) & \subseteq \mathcal G_{N,L}\! \left(\bigoplus _k \mu _k A_k\right). \notag \end{align}
Proof

The block-injective canonical-form hypothesis supplies \(\mu _j\neq 0\) for every \(j\), and the block-diagonal periodic-boundary ground space is \(\mathcal G_{N,L}(\bigoplus _j \mu _j A_j)\) by definition.

15.2 Block separation and intersection

This section establishes that block-separating words make the simultaneous block-word tuples \(\omega \mapsto (A_1^\omega ,\ldots ,A_g^\omega )\) span the full product algebra \(\prod _jM_{D_j}(\mathbb {C})\) at a single length, and that this single-length span, together with the normalization \(\sum _a A^j_aA^{j\, \dagger }_a=I\), yields the one-step intersection identity \(\bigvee _jG_{n+2}(A^j)\) used for the degenerate parent-Hamiltonian ground space.

Lemma 15.2.1 Block-diagonal commutant reduction from span extension

Let \(S\) be a family of virtual matrices. If each sector projection satisfies \(P_j\in \operatorname{span}S\) and a boundary matrix \(X\) commutes with every \(M\) in \(S\), so that \(XM=MX\), then \(XP_j=P_jX\) for every sector \(j\), and hence \(P_jXP_k=0\) for distinct sectors \(j\) and \(k\). Thus \(X\) is block diagonal with respect to the sector decomposition.

Proof

The equations \(XM=MX\) are linear in \(M\), so they hold for every \(M\in \operatorname{span}S\). Substituting \(M=P_j\) gives \(XP_j=P_jX\). Multiplying this identity on the left by \(P_j\) and on the right by \(P_k\) with \(j\ne k\) gives \(P_jXP_k=0\).

Lemma 15.2.2 Block projections from product-algebra span

Let \(T_a=(T_a(i))_i\) be a family of tuples spanning the full product algebra \(\prod _i M_{n_i}(\mathbb {C})\), and let \(c_i\neq 0\) be nonzero scalars. Then, for every sector \(k\), the sector projection \(P_k\) lies in the span of the scaled dependent block-diagonal matrices \(\bigoplus _i c_i T_a(i)\).

Proof

Apply the linear map \((M_i)_i\mapsto \bigoplus _i c_i M_i\) to the full-span product-algebra element with \((c_k)^{-1}I\) in the \(k\)-th component and zero in every other component. Its image is exactly \(P_k\).

Lemma 15.2.3 Sector projections from blockwise word span

If the simultaneous length-\(m\) block-word tuples \(\omega \mapsto (A_j^\omega )_j\) span the full product algebra \(\prod _j M_{D_j}(\mathbb {C})\) and every \(\mu _j\) is nonzero, then the pulled-back length-\(m\) word products of \(\bigoplus _j\mu _j A_j\) span every virtual sector projection \(P_j\). If a virtual matrix \(X\) commutes with the pulled-back word product for every word \(\omega \) of length \(m\),

\begin{align} X\! \left(\bigoplus _j \mu _j^m A_j^\omega \right) & = \left(\bigoplus _j \mu _j^m A_j^\omega \right)X, \notag \end{align}

then \(P_jXP_k=0\) for distinct sectors \(j\) and \(k\).

Proof

First embed the product algebra linearly as dependent block-diagonal matrices, scaling the \(j\)-th block by \(\mu _j^m\). Since \(\mu _j^m\neq 0\), the product-algebra span contains the tuple with \((\mu _j^m)^{-1}I\) in one component and zero elsewhere; its scaled diagonal image is exactly \(P_j\). The word-evaluation formula for the direct-sum tensor identifies these scaled diagonal matrices with the pulled-back direct-sum word products, and the commutant conclusion follows from Lemma 15.2.1.

Definition 15.2.4 Pairwise block-separating equations

Fix a finite target set \(T\) of block indices. A length-\(S\) block-separating equation for \(k\) against \(T\) is a choice of coefficients \(c_\omega \) such that

\begin{align} \sum _{|\omega |=S} c_\omega A_\ell ^\omega & = \begin{cases} I, & \ell =k, \\ 0, & \ell \in T. \end{cases} \notag \end{align}

The pairwise version requires this equation for each ordered pair of distinct blocks, with \(T=\{ j\} \).

For two blocks \(A\) and \(B\), the length-\(S\) pair word tuple is \(w\mapsto (A^w,B^w)\). We also use the subalgebra of \(M_{D_A}(\mathbb {C})\times M_{D_B}(\mathbb {C})\) generated by the one-letter pairs \((A_i,B_i)\). The pair product-span condition says that the length-\(S\) tuples span \(M_{D_A}(\mathbb {C})\times M_{D_B}(\mathbb {C})\). Equivalently, in the trace-dual formulation of the same product-span condition, the only pair of test matrices \((\Delta _A,\Delta _B)\) whose trace pairing with every length-\(S\) tuple vanishes is \((0,0)\). The cumulative variant replaces one fixed length by all word lengths at most \(S\), and the all-length variant allows every finite word.

Let \(A\) and \(B\) be tensors with virtual bond dimensions \(D_A\) and \(D_B\). Assume that \(A\) is \(L\)-block injective. Let \(\Delta _A\in M_{D_A}(\mathbb {C})\) with \(\Delta _A\neq 0\), and let \(\Delta _B\in M_{D_B}(\mathbb {C})\). Suppose that, for all words \(u,v,t\) of length \(L\),

\begin{align} \operatorname{tr}\! \left(\Delta _A A_v A_t A_u\right) +\operatorname{tr}\! \left(\Delta _B B_v B_t B_u\right) & =0. \notag \end{align}

Then for every \(Z\in M_{D_A}(\mathbb {C})\) there is a matrix \(W\in M_{D_B}(\mathbb {C})\) such that, for all words \(t\) of length \(L\), \(\operatorname{tr}(ZA_t)+\operatorname{tr}(WB_t) =0\). This is the trace-dual algebraic step in the two-block case of the direct-sum lemma of [ PGVWC07 ] .

Proof

Since \(\Delta _A\neq 0\), Lemma D.1.5 writes \(Z\) as a linear combination of matrices \(A_u\Delta _A A_v\). For one such term, cyclicity of the trace rewrites the three-block relation as

\begin{align} \operatorname{tr}(A_u\Delta _A A_vA_t)+\operatorname{tr}(B_u\Delta _B B_vB_t) & = 0. \notag \end{align}

Taking the same linear combination of the matrices \(B_u\Delta _BB_v\) gives the required \(W\), and linearity gives the general case.

Under the hypotheses of Lemma 15.2.6, the length-\(L\) spaces \(\mathcal G_L^A\) and \(\mathcal G_L^B\) satisfy \(\mathcal G_L^A\subseteq \mathcal G_L^B\). If the corresponding nonzero middle matrix exists on both sides, then \(\mathcal G_L^A=\mathcal G_L^B\).

Proof

Write a vector in \(\mathcal G_L^A\) as \(t\mapsto \operatorname{tr}(A_tZ)\). The preceding lemma supplies \(W\) with \(\operatorname{tr}(ZA_t)+\operatorname{tr}(WB_t)=0\) for every word \(t\). Cyclicity of the trace gives \(\operatorname{tr}(A_tZ) = \operatorname{tr}(ZA_t) = -\operatorname{tr}(WB_t) = \operatorname{tr}(B_t(-W))\), so the vector lies in \(\mathcal G_L^B\). The equality statement follows by applying the same argument with the two blocks exchanged.

Lemma 15.2.8 Dimension step for two-block local spaces
#

If \(A\) and \(B\) are length-\(L\) block-injective, the map \(Z\mapsto (t\mapsto \operatorname{tr}(A_tZ))\) has image dimension \(D_A^2\), and similarly for \(B\). Therefore an inclusion \(\mathcal G_L^A\subseteq \mathcal G_L^B\), together with \(D_B\le D_A\), forces \(D_A=D_B\) and \(\mathcal G_L^A=\mathcal G_L^B\). In particular, the three-block trace relation from Lemma 15.2.7 gives this equality whenever the ordering \(D_B\le D_A\) is available. If the inclusion comes from the strictly larger block, \(D_B{\lt}D_A\), then the same dimension step already contradicts the existence of the three-block trace relation. More generally, let \(H\) be a source hypothesis which excludes the simultaneous conclusion

\begin{align} D_A & = D_B, \notag \\ \mathcal G_L^A & = \mathcal G_L^B. \notag \end{align}

Then the three-block intersection vanishes:

\begin{align} H & \Longrightarrow \mathcal G_{3L}^A\cap \mathcal G_{3L}^B=\{ 0\} . \notag \end{align}

With injectivity at the three-block length, the strict-size branch gives

\begin{align} D_B{\lt}D_A,\quad \mathcal G_{3L}^A\cap \mathcal G_{3L}^B=\{ 0\} \quad \Longrightarrow \quad \operatorname{span}\{ (A^\omega ,B^\omega ):|\omega |=3L\} & = M_{D_A}(\mathbb {C})\times M_{D_B}(\mathbb {C}). \notag \end{align}

One such source hypothesis is non-proportionality of the periodic MPS vectors at a sufficiently long chain length: \(\mathbb {C}\, V^{(N)}(A)\ne \mathbb {C}\, V^{(N)}(B)\). Indeed,

\begin{align} \mathcal G_L^A=\mathcal G_L^B & \Longrightarrow \mathcal G_{N,L}(A)=\mathcal G_{N,L}(B) \Longrightarrow \mathbb {C}\, V^{(N)}(A)=\mathbb {C}\, V^{(N)}(B), \notag \end{align}

by equality of the periodic-boundary constraint spaces, followed by injective uniqueness. For same-dimensional normalized BNT representatives that are not gauge-phase equivalent, the non-equivalence condition supplies this non-proportionality at arbitrarily large chain lengths; the direct-sum injectivity hypotheses remain explicit. Combining the equal-dimension and unequal-dimension branches over all ordered block pairs gives one common positive length \(S_0\) with

\begin{align} \operatorname{span}\{ (A_1^\omega ,\ldots ,A_g^\omega ):|\omega |=S_0\} & = \prod _{j=1}^g M_{D_j}(\mathbb {C}). \notag \end{align}

For an ordered pair of blocks, the trace-separation conclusion has the form

\begin{align} \left[\forall |\omega |=S, \operatorname{tr}(\Delta _A A^\omega )+\operatorname{tr}(\Delta _B B^\omega )=0\right] & \Longrightarrow \Delta _A=\Delta _B=0. \notag \end{align}

Equivalently, the homogeneous pair span is full:

\begin{align} \operatorname{span}\{ (A^\omega ,B^\omega ):|\omega |=S\} & = M_{D_A}(\mathbb {C})\times M_{D_B}(\mathbb {C}). \notag \end{align}

Concatenating words propagates this equality to positive multiples of \(S\), and hence to a finite period window for every ordered pair. Combining that window with the conditional period-window homogenization theorem gives a single length \(S_0\) such that

\begin{align} \operatorname{span}\{ (A_1^\omega ,\ldots ,A_g^\omega ):|\omega |=S_0\} & = \prod _{j=1}^g M_{D_j}(\mathbb {C}). \notag \end{align}

The same conclusion applies to the corresponding normal canonical-form hypotheses with a basis of normal tensors once one-site injectivity is supplied explicitly, since those hypotheses carry the irreducibility and normalization assumptions used in the direct-sum comparison.

Proof

Block injectivity makes the boundary-to-chain map injective, so \(\dim \mathcal G_L^A=D_A^2\) and \(\dim \mathcal G_L^B=D_B^2\). Inclusion gives \(D_A^2\le D_B^2\), while \(D_B\le D_A\) gives the reverse inequality. Hence \(D_A=D_B\), and equality of dimensions inside the inclusion gives \(\mathcal G_L^A=\mathcal G_L^B\). In the strict-size branch this contradicts \(D_B{\lt}D_A\). For the conditional directness statement, suppose \(\psi \in \mathcal G_{3L}^A\cap \mathcal G_{3L}^B\) has boundary representations \(\psi = \Gamma _{3L}^A(X) = \Gamma _{3L}^B(Y)\). If \(X=0\), injectivity of the \(A\) boundary-to-chain map gives \(\psi =0\). Otherwise equality of the coefficient formulas gives \(\forall |t|=3L,\qquad \operatorname{tr}(XA^t)-\operatorname{tr}(YB^t) =0\), equivalently, \(\forall |t|=3L,\qquad \operatorname{tr}(XA^t)+\operatorname{tr}((-Y)B^t) =0\). The dimension step applied to this homogeneous three-block trace relation gives \(D_A^2\le D_B^2\), contradicting \(D_B{\lt}D_A\) in the strict-size branch. In the equal-size branch, equal local spaces impose the same length-\(L\) constraints on the \(N\)-site ring:

\begin{align} \mathcal G_L^A=\mathcal G_L^B \quad \Longrightarrow \quad \mathcal G_{N,L}(A) & = \mathcal G_{N,L}(B). \notag \end{align}

The injective uniqueness theorem identifies these periodic ground spaces with the one-dimensional MPS spans,

\begin{align} \mathcal G_{N,L}(A) & = \mathbb {C}\, V^{(N)}(A), \notag \\ \mathcal G_{N,L}(B) & = \mathbb {C}\, V^{(N)}(B), \notag \end{align}

and hence \(\mathbb {C}\, V^{(N)}(A)=\mathbb {C}\, V^{(N)}(B)\), contradicting non-proportionality of the two MPS vectors.

If no nonzero trace functional vanishes on all finite pair words, then there is a finite \(S\) such that vanishing on every word of length at most \(S\) already forces the test pair to be zero. For a finite family of ordered block pairs, these bounds can be replaced by one common bound.

Proof

Write

\begin{align} W_S(A,B) & = \operatorname{span}\{ (A^w,B^w):|w|\le S\} , \notag \\ W_\infty (A,B) & = \operatorname{span}\{ (A^w,B^w):w\text{ finite}\} . \notag \end{align}

The equality \(W_\infty (A,B)=M_{D_A}(\mathbb {C})\times M_{D_B}(\mathbb {C})\) holds if and only if

\begin{align} \left[\forall w, \operatorname{tr}(\Delta _AA^w)+\operatorname{tr}(\Delta _BB^w)=0\right] & \Longrightarrow (\Delta _A,\Delta _B)=(0,0). \notag \end{align}

Since the product algebra is finite-dimensional and

\begin{align} W_0(A,B) & \subseteq W_1(A,B) \subseteq \cdots , \notag \\ \bigvee _{S\ge 0}W_S(A,B) & = W_\infty (A,B), \notag \end{align}

there is an index \(S_0\) with \(W_{S_0}(A,B) = W_\infty (A,B) = M_{D_A}(\mathbb {C})\times M_{D_B}(\mathbb {C})\). Applying the finite trace-duality criterion at \(S_0\) gives

\begin{align} \left[\forall w, |w|\le S_0 \Longrightarrow \operatorname{tr}(\Delta _AA^w)+\operatorname{tr}(\Delta _BB^w)=0\right] & \Longrightarrow (\Delta _A,\Delta _B)=(0,0). \notag \end{align}

For a finite family of ordered block pairs, choose such an \(S_{kj}\) for each pair and replace them by \(S=\max _{k\ne j}S_{kj}\). Monotonicity of \(W_S\) gives the same finite cumulative trace separation for every pair. This is the cumulative span at all lengths at most \(S\), not a single homogeneous length, since the length-zero word already contributes the identity to the cumulative span.

If one of the two blocks is injective, then the corresponding coordinate projection of the pair algebra generated by the simultaneous one-letter pairs is the full matrix algebra. Every finite pair word lies in this pair algebra, and the all-word linear span is the submodule generated by the pair algebra. If the pair algebra is the graph of an algebra automorphism of the matrix algebra, Skolem–Noether turns this graph case into gauge-phase equivalence, so under non-gauge-equivalence the graph branch is impossible. For two blocks of unequal bond dimensions, the graph branch is also impossible because it would identify matrix algebras with different finite dimensions over \(\mathbb {C}\).

Proof

Projecting the pair generators to one coordinate gives exactly the one-block generator family. Injectivity says that the linear span of this family is the full matrix algebra, and the linear span is contained in the unital algebra it generates. The pair-word membership follows by induction on the word, using closure of the generated algebra under multiplication. Conversely, the all-word linear span is closed under componentwise multiplication, so it contains the algebra generated by the pair generators. In the graph case, each generator satisfies \(B_i=\varphi (A_i)\); applying Skolem–Noether to \(\varphi \) gives an ordinary gauge, hence a gauge-phase equivalence with scalar \(1\).

Lemma 15.2.11 Conditional padding from exact homogeneous identity witnesses
#

Homogeneous identity-padding witnesses multiply by concatenation of words. Therefore, if one has a positive period and a full residue window of homogeneous pair spans, then the simultaneous identity pair lies in all sufficiently large homogeneous pair spans. Combined with all-length pair trace separation, this period-window input gives one homogeneous pair trace-separation length. This does not prove the block-separation span condition of [ PGVWC07 ] by itself: fullness of spans up to a bounded length is weaker than the exact homogeneous input used here.

Proof

The length-zero word gives the identity only at length zero, so it is not a positive-length padding theorem. The useful padding facts are homogeneous: if the identity pair is in the span at lengths \(m\) and \(n\), componentwise multiplication puts it in the span at length \(m+n\). A period witness and a complete residue window therefore give homogeneous padding at every sufficiently large length, which is the input needed by the all-length-to-fixed-length implication.

At a fixed length \(S\), the pair trace-separation criterion is dual to full pair product-span. Therefore a pair trace-separation witness supplies a linear combination of word evaluations that is identity on the first block and zero on the second. A common witness for all ordered distinct pairs gives pairwise block-separating words at the same length.

Proof

If the pair span were proper, a nonzero linear functional would vanish on it. Nondegeneracy of the matrix trace pairing represents that functional as \((\Delta _A,\Delta _B)\), contradicting pair trace separation. Once the pair span is the whole product algebra, the tuple \((I,0)\) is in the span. Using the same coefficients on the ambient block family gives the required pairwise block-separating equation.

If every ordered pair of distinct blocks admits a common-length pairwise block-separating equation, then for each \(k\) there are coefficients \(c_\omega ^{(k)}\) with

\begin{align} \sum _{|\omega |=(g-1)S} c_\omega ^{(k)} A_j^\omega & = \begin{cases} I, & j=k, \\ 0, & j\ne k. \end{cases} \notag \end{align}

Hence common block injectivity plus pairwise separation implies the finite-length blockwise physical–virtual correspondence of the block-injectivity proposition in [ CPGSV16 , Section 2.3 ] and the block-injective canonical-form trace-separation condition.

Proof

For the base of induction over the target blocks, the empty word gives the identity tuple on the empty target set. This does not represent a physical chain of length zero. The span of word tuples is closed under pointwise multiplication because the product of a length-\(L\) word and a length-\(S\) word is the concatenated length-\((L+S)\) word. Inducting over the finite set of blocks different from \(k\) and multiplying the corresponding pairwise equations keeps the value \(I\) on \(k\) and introduces one additional zero block at each step. Extracting coefficients from the final tuple-span membership gives the coefficient form of the displayed block-separating equation.

Lemma 15.2.14 Homogeneous padding for blockwise product spans

Suppose the length-\(L\) simultaneous block-word tuples span \(\prod _j M_{D_j}(\mathbb {C})\), and suppose the simultaneous identity tuple \((I,\ldots ,I)\) lies in the span of the length-\(S\) simultaneous block-word tuples. Then the length-\(L+S\) simultaneous block-word tuples span \(\prod _j M_{D_j}(\mathbb {C})\). In particular, two full homogeneous spans at lengths \(L\) and \(S\) give a full homogeneous span at length \(L+S\).

Proof

Let \(M=(M_j)_j\) be a target tuple. Write

\begin{align} M_j & = \sum _{|u|=L} a_u A^j_u, \notag \\ I & = \sum _{|v|=S} b_v A^j_v \notag \end{align}

simultaneously for every block \(j\). Multiplying the two expansions gives

\begin{align} M_j & = M_j I = \sum _{|u|=L}\sum _{|v|=S} a_u b_v A^j_uA^j_v = \sum _{|u|=L}\sum _{|v|=S} a_u b_v A^j_{uv}. \notag \end{align}

The words \(uv\) have length \(L+S\), so the target tuple lies in the length-\(L+S\) span.

Lemma 15.2.15 Period-window propagation for blockwise product spans

Let \(p{\gt}0\). Suppose the length-\(p\) simultaneous block-word tuples span \(\prod _j M_{D_j}(\mathbb {C})\), and suppose that for every \(0\le r{\lt}p\), the length-\(s+r\) simultaneous block-word tuples span \(\prod _j M_{D_j}(\mathbb {C})\). Then there is a length \(N\) such that, for every \(n\ge N\), the length-\(n\) simultaneous block-word tuples span \(\prod _j M_{D_j}(\mathbb {C})\). More concretely, one may take \(N=s\): writing \(n=s+r+qp\), with \(0\le r{\lt}p\), the span at length \(s+r\) propagates to the span at length \(n\) by multiplying \(q\) times by the period-\(p\) span.

Proof

The Euclidean division of \(n-s\) by \(p\) gives \(n=s+r+qp\) with \(0\le r{\lt}p\). The assumed residue-window span gives the full product algebra at length \(s+r\). Applying homogeneous padding with the length-\(p\) span once gives the full product algebra at length \(s+r+p\); iterating gives the full product algebra at length \(s+r+qp\).

Let \(J\) be a finite set of \(r\) blocks, and let \((A_j)_{j\in J}\) be a finite family of tensors with common block injectivity at length \(L\). Suppose there are length-\(S\) block-separating equations: for each block \(k\),

\begin{align} \sum _{|\omega |=S} c_\omega ^{(k)} A_j^\omega & = \begin{cases} I, & j=k, \\ 0, & j\ne k. \end{cases} \notag \end{align}

Then the simultaneous length-\((L+S)\) block-word evaluations span \(\prod _j M_{D_j}(\mathbb {C})\). If the block-separating equations are first obtained from pairwise block-separating equations at length \(S\), the same conclusion holds at length \(L+(r-1)S\).

Proof

Block injectivity writes each target block matrix as a length-\(L\) linear combination of word evaluations. For a target tuple \((M_j)_j\), write \(M_k\) in this form in block \(k\), multiply by the block-separating equation for \(k\), and sum over \(k\). Thus, if \(M_k=\sum _{|u|=L}a_u^{(k)}A_k^u\) and \(\sum _{|v|=S}b_v^{(k)}A_j^v=\delta _{j,k}I\), then for every block \(j\),

\begin{align} M_j & = \sum _{k\in J}\sum _{|u|=L}\sum _{|v|=S} a_u^{(k)}b_v^{(k)} A_j^u A_j^v. \notag \end{align}

The right side is a linear combination of length-\((L+S)\) word tuples, so these tuples span the full product algebra. If one starts with pairwise separation instead, first use Lemma 15.2.13 to multiply the pairwise equations into the full equations.

15.2.1 Trace identities for block intersections

Once the simultaneous block-word tuples span the full product algebra at a single length, nondegeneracy of the product trace pairing turns blockwise trace equalities into blockwise matrix identities \(A^j_bC^j_a=A^j_bE^jA^j_a\). The lemmas here use those identities to place left-boundary trace decompositions in the block sum \(\bigvee _jG_{n+2}(A^j)\) and to obtain the one-step block-intersection identity.

Lemma 15.2.1.1 Arbitrary interpolation on one block of a pair

Let \((A^\ell )_{\ell =1}^r\) be a finite family of tensors. Suppose that the length-\(S\) pair evaluations of blocks \(k\) and \(j\) span their product matrix algebra. For every \(X\in M_{D_k}(\mathbb {C})\), there is a tuple \(M=(M_\ell )_\ell \) in the span of the simultaneous length-\(S\) word evaluations such that \(M_k=X\) and \(M_j=0\).

Proof

Full pair span gives coefficients satisfying \((X,0) = \sum _{|w|=S}c_w(A^k_w,A^j_w)\). Define \(M_\ell =\sum _{|w|=S}c_wA^\ell _w\) for every block \(\ell \). Then \(M\) lies in the simultaneous word-tuple span, and its \(k\)- and \(j\)-components are \(X\) and \(0\), respectively.

Lemma 15.2.1.2 Blockwise matrix equality from trace pairings

Let \(A^1,\ldots ,A^g\) be block tensors whose length-\(m\) simultaneous word tuples span the product algebra. If, for every word \(w\) of length \(m\), two blockwise matrix families \(X_j\) and \(Y_j\) satisfy \(\sum _j\operatorname{tr}(X_jA^j_w) = \sum _j\operatorname{tr}(Y_jA^j_w)\), then \(X_j=Y_j\) for every block \(j\).

Proof

For every word \(w\) of length \(m\),

\begin{align} \sum _j\operatorname{tr}\! \left((X_j-Y_j)A^j_w\right) & = \sum _j\operatorname{tr}(X_jA^j_w) - \sum _j\operatorname{tr}(Y_jA^j_w) = 0. \notag \end{align}

Nondegeneracy of the product trace pairing, together with the spanning hypothesis, gives \(X_j-Y_j=0\) for every block \(j\).

Let \(A^1,\ldots ,A^g\) be block tensors whose length-\(m\) simultaneous word tuples span the product algebra. Let \(C^j_a\in M_{D_j}(\mathbb {C})\) be matrices indexed by blocks \(j\) and physical letters \(a\). Suppose that, for every pair of boundary letters \(a,b\) and every word \(w\) of length \(m\), a vector \(\psi \) has the left-boundary trace form \(\psi (a,w,b) = \sum _j\operatorname{tr}\! \left(A^j_b C^j_a A^j_w\right)\). If \(\psi \in \bigvee _j G_{m+2}(A^j)\), then there are matrices \(E_j\) such that, for every block \(j\) and all \(a,b\), \(A^j_bC^j_a = A^j_bE_jA^j_a\).

Proof

Write \(\psi =\sum _j\phi _j\) with \(\phi _j\in G_{m+2}(A^j)\). For each \(j\), choose \(E_j\) whose image under the length-\(m+2\) ground-space map for \(A^j\) is \(\phi _j\). Comparing the coefficient of the word \(awb\) in the two expressions for \(\psi \) gives

\begin{align} \sum _j\operatorname{tr}\! \left(A^j_b C^j_a A^j_w\right) & = \sum _j\operatorname{tr}\! \left(A^j_bE_jA^j_aA^j_w\right). \notag \end{align}

Lemma 15.2.1.2 gives the displayed blockwise identities.

Lemma 15.2.1.4 Blockwise boundary compatibility from traces

Let \(A^1,\ldots ,A^g\) be block tensors with the following common word-span property: for a fixed \(m\), the simultaneous tuples \(w \longmapsto \left(A^1_w,\ldots ,A^g_w\right), \qquad |w|=m\), span the full product algebra \(\prod _j M_{D_j}(\mathbb {C})\). Suppose that matrices \(C^j_a,D^j_b\in M_{D_j}(\mathbb {C})\) satisfy, for all physical indices \(a,b\) and all words \(w\) of length \(m\),

\begin{align} \sum _j \operatorname{tr}\! \left(A^j_b C^j_a A^j_w\right) & = \sum _j \operatorname{tr}\! \left(D^j_b A^j_a A^j_w\right). \notag \end{align}

Then \(A^j_bC^j_a = D^j_bA^j_a\) for every block \(j\) and all \(a,b\).

Proof

For fixed \(a,b\), set \(\Delta _j=A^j_bC^j_a-D^j_bA^j_a\). The assumed trace equality says that \(\sum _j \operatorname{tr}\! \left(\Delta _jA^j_w\right) = 0\) for every word \(w\) of length \(m\). Since the tuples \((A^1_w,\ldots ,A^g_w)\) span the product algebra, the product trace pairing is nondegenerate and hence every \(\Delta _j\) is zero.

Let \(A^1,\ldots ,A^g\) be block tensors with the common word-span property at length \(m\). Suppose that \(C^j_\rho \in M_{D_j}(\mathbb {C})\) is indexed by a block \(j\) and a word \(\rho \) of some fixed length \(M\), and that \(D^j_\beta \in M_{D_j}(\mathbb {C})\) is indexed by a block \(j\) and a word \(\beta \) of some fixed length \(K\). If, for every word \(\rho \) of length \(M\), every word \(\beta \) of length \(K\), and every middle word \(w\) of length \(m\),

\begin{align} \sum _j\operatorname{tr}\! \left(A^j_\beta C^j_\rho A^j_w\right) & = \sum _j\operatorname{tr}\! \left(D^j_\beta A^j_\rho A^j_w\right), \notag \end{align}

then \(A^j_\beta C^j_\rho = D^j_\beta A^j_\rho \) for every block \(j\) and all words \(\beta ,\rho \). In particular, taking \(D^j_\beta =X_jA^j_\beta \) gives \(A^j_\beta C^j_\rho = (X_jA^j_\beta )A^j_\rho \).

Proof

Fix \(\beta \) and \(\rho \), and set \(\Delta _j=A^j_\beta C^j_\rho -D^j_\beta A^j_\rho \). The trace equality says that \(\sum _j\operatorname{tr}(\Delta _jA^j_w) = 0\) for every word \(w\) of length \(m\). The same nondegenerate product trace-pairing argument as in Lemma 15.2.1.4 gives \(\Delta _j=0\) for every \(j\). The specialization \(D^j_\beta =X_jA^j_\beta \) is immediate.

Lemma 15.2.1.6 Fixed complementary-word compatibility from traces

Let \(A^1,\ldots ,A^g\) be block tensors with the common word-span property at length \(m\). Fix a complementary word \(\rho \) of length \(M\) and matrices \(C^j_\rho ,X_j\in M_{D_j}(\mathbb {C})\). If, for every wrapped word \(\beta \) of length \(K\) and every middle word \(w\) of length \(m\),

\begin{align} \sum _j\operatorname{tr}\! \left(A^j_\beta C^j_\rho A^j_w\right) & = \sum _j\operatorname{tr}\! \left((X_jA^j_\beta )A^j_\rho A^j_w\right), \notag \end{align}

then \(A^j_\beta C^j_\rho = (X_jA^j_\beta )A^j_\rho \) for every block \(j\) and every wrapped word \(\beta \).

Proof

Fix \(\beta \) and set \(\Delta _j=A^j_\beta C^j_\rho -(X_jA^j_\beta )A^j_\rho \). The trace equality gives \(\sum _j\operatorname{tr}(\Delta _jA^j_w) = 0\) for every word \(w\) of length \(m\). The common word-span property and the product trace pairing imply \(\Delta _j=0\) for every \(j\).

Let \((A_a)_{a\in I}\), \((C_a)_{a\in I}\), and \((D_b)_{b\in I}\) be matrices in \(M_D(\mathbb {C})\), indexed by a finite alphabet. If

\begin{align} \sum _a A_a A_a^\dagger & = \mathbb {1}, \notag \\ A_b C_a & = D_b A_a \notag \end{align}

for all \(a,b\in I\), and if \(E=\sum _a C_a A_a^\dagger \), then

\begin{align} D_b & = A_bE, \notag \\ A_bC_a & = A_bEA_a \notag \end{align}

for all \(a,b\in I\).

In particular, the same calculation applies when the alphabet is the finite set of words of a fixed length and \(A_\beta \) denotes \(A_{\beta _1}\cdots A_{\beta _K}\).

The same argument also applies when the normalized family is indexed by a finite set \(J\), while \((A_b)_{b\in I}\) and \((D_b)_{b\in I}\) are indexed by a set \(I\). If \((B_a)_{a\in J}\) and \((C_a)_{a\in J}\) are indexed by \(J\), and if, for all \(a\in J\) and \(b\in I\),

\begin{align} \sum _a B_aB_a^\dagger & = \mathbb {1}, \notag \\ A_bC_a & = D_bB_a, \notag \end{align}

then, with \(E=\sum _a C_aB_a^\dagger \), one has

\begin{align} D_b & = A_bE, \notag \\ A_bC_a & = A_bEB_a. \notag \end{align}

This includes the case where \(I\) and \(J\) are word sets of different lengths.

Proof

Multiply \(D_b\) by the normalization:

\begin{align} D_b & = D_b\sum _a A_aA_a^\dagger = \sum _a D_bA_aA_a^\dagger = \sum _a A_bC_aA_a^\dagger = A_bE. \notag \end{align}

Substituting this identity into \(A_bC_a=D_bA_a\) gives \(A_bC_a=A_bEA_a\). For two alphabets, the corresponding step is

\begin{align} D_b & = D_b\sum _a B_aB_a^\dagger = \sum _a D_bB_aB_a^\dagger = \sum _a A_bC_aB_a^\dagger = A_bE, \notag \end{align}

where \(E=\sum _a C_aB_a^\dagger \). Substituting this identity into \(A_bC_a=D_bB_a\) gives \(A_bC_a=A_bEB_a\).

Lemma 15.2.1.8 Explicit \(C^j,D^j,E^j\) identities from trace decompositions

Let \(A^1,\ldots ,A^g\) be block tensors with the common word-span property at length \(m\). Fix matrices \(X_j\), and put \(D^j_\beta =X_jA^j_\beta \). Suppose that \(C^j_\rho \) is indexed by a block \(j\) and a complementary word \(\rho \) of length \(M\), and assume that, for every wrapped word \(\beta \) of length \(K\), every complementary word \(\rho \), and every middle word \(w\) of length \(m\),

\begin{align} \sum _j\operatorname{tr}\! \left(A^j_\beta C^j_\rho A^j_w\right) & = \sum _j\operatorname{tr}\! \left(D^j_\beta A^j_\rho A^j_w\right). \notag \end{align}

If each block satisfies \(\sum _a A^j_aA^{j\, \dagger }_a = \mathbb {1}\), then, for every block \(j\), there is a matrix \(E^j = \sum _\rho C^j_\rho \left(A^j_\rho \right)^\dagger \) such that, for every wrapped word \(\beta \) and complementary word \(\rho \),

\begin{align} D^j_\beta & = A^j_\beta E^j, \notag \\ A^j_\beta C^j_\rho & = A^j_\beta E^jA^j_\rho . \notag \end{align}

This is the calculation in [ PGVWC07 , Theorem 12, proof lines 1446–1451 ] , with the adjoint on \(A^j_\rho \) required by the displayed normalization.

Proof

Lemma 15.2.1.5 gives \(A^j_\beta C^j_\rho = D^j_\beta A^j_\rho \). The single-letter normalization gives the word normalization \(\sum _\rho A^j_\rho \left(A^j_\rho \right)^\dagger = \mathbb {1}\). Lemma 15.2.1.7, applied to the two word alphabets, gives

\begin{align} E^j & = \sum _\rho C^j_\rho \left(A^j_\rho \right)^\dagger , \notag \\ D^j_\beta & = A^j_\beta E^j, \notag \\ A^j_\beta C^j_\rho & = A^j_\beta E^jA^j_\rho . \notag \end{align}
Lemma 15.2.1.9 Block-diagonal boundary \(C^j,D^j,E^j\) identities from trace decompositions

Let \(A^1,\ldots ,A^g\) be block tensors with the common word-span property at length \(m\). Fix \(N\in \mathbb {N}\), complex numbers \(\mu _j\), and block-diagonal boundary matrices \(X_j\), and put \(D^j_\beta =(\mu _j^NX_j)A^j_\beta \). Suppose that, for every wrapped word \(\beta \) of length \(K\), every complementary word \(\rho \) of length \(M\), and every middle word \(w\) of length \(m\),

\begin{align} \sum _j\operatorname{tr}\! \left(A^j_\beta C^j_\rho A^j_w\right) & = \sum _j\operatorname{tr}\! \left(((\mu _j^NX_j)A^j_\beta )A^j_\rho A^j_w\right). \notag \end{align}

If each block satisfies \(\sum _a A^j_aA^{j\, \dagger }_a = \mathbb {1}\), then, for every block \(j\), there is a matrix \(E^j = \sum _\rho C^j_\rho \left(A^j_\rho \right)^\dagger \) such that, for every wrapped word \(\beta \) and complementary word \(\rho \),

\begin{align} (\mu _j^NX_j)A^j_\beta & = A^j_\beta E^j, \notag \\ A^j_\beta C^j_\rho & = A^j_\beta E^jA^j_\rho . \notag \end{align}

This is the boundary-matrix calculation in [ PGVWC07 , Theorem 12, proof lines 1446–1451 ] , after substituting the block-diagonal boundary matrix \(D^j_\beta =(\mu _j^NX_j)A^j_\beta \).

Proof

Apply Lemma 15.2.1.8 with \(D^j_\beta =(\mu _j^NX_j)A^j_\beta \).

Lemma 15.2.1.10 Complementary-word boundary identities from a compatibility hypothesis

Let \(A_\beta \) be the products over wrapped words of length \(K\), let \(A_\rho \) be the products over complementary words of length \(M\), and let \(X\in M_D(\mathbb {C})\) be a matrix. Suppose that the complementary-word products satisfy \(\sum _\rho A_\rho A_\rho ^\dagger = \mathbb {1}\) and that there are matrices \(C_\rho \), indexed by complementary words, such that, for every wrapped word \(\beta \) and complementary word \(\rho \), \(A_\beta C_\rho = (X A_\beta )A_\rho \). Then, for every complementary word \(\rho \), with

\begin{align} E_\rho & = \left(\sum _\gamma C_\gamma A_\gamma ^\dagger \right)A_\rho , \notag \end{align}

one has, for every wrapped word \(\beta \), \((X A_\beta )A_\rho = A_\beta E_\rho \).

Proof

Apply Lemma 15.2.1.7 with \(D_\beta =X A_\beta \). It gives \(A_\beta C_\rho = A_\beta E A_\rho \) for \(E=\sum _\gamma C_\gamma A_\gamma ^\dagger \). With \(E_\rho =EA_\rho \), the compatibility hypothesis gives

\begin{align} (X A_\beta )A_\rho & = A_\beta C_\rho = A_\beta E A_\rho = A_\beta E_\rho . \notag \end{align}
Lemma 15.2.1.11 Complementary-word boundary identities from trace decompositions

Let \(A^1,\ldots ,A^g\) be block tensors with the common word-span property at length \(m\). Fix matrices \(X_j\). Suppose that \(C^j_\rho \) is indexed by a block \(j\) and a complementary word \(\rho \) of length \(M\). Assume that, for every wrapped word \(\beta \) of length \(K\), every complementary word \(\rho \), and every middle word \(w\) of length \(m\),

\begin{align} \sum _j\operatorname{tr}\! \left(A^j_\beta C^j_\rho A^j_w\right) & = \sum _j\operatorname{tr}\! \left((X_jA^j_\beta )A^j_\rho A^j_w\right). \notag \end{align}

If each block satisfies \(\sum _a A^j_aA^{j\, \dagger }_a = \mathbb {1}\), then, for every block \(j\) and every complementary word \(\rho \), there is a matrix \(E_{j,\rho }\) such that, for every wrapped word \(\beta \), \((X_jA^j_\beta )A^j_\rho = A^j_\beta E_{j,\rho }\).

Proof

Lemma 15.2.1.5, applied with \(D^j_\beta =X_jA^j_\beta \), gives \(A^j_\beta C^j_\rho = (X_jA^j_\beta )A^j_\rho \). The hypothesis \(\sum _a A^j_a A^{j\, \dagger }_a=\mathbb {1}\) extends to words of length \(M\): \(\sum _\rho A^j_\rho \left(A^j_\rho \right)^\dagger = \mathbb {1}\). Lemma 15.2.1.10 gives the matrices \(E_{j,\rho }\).

Lemma 15.2.1.12 A left-boundary summand is a ground-space vector

Let \(A\) be a block tensor, and let \(C_a,E\in M_D(\mathbb {C})\) satisfy \(A_b C_a = A_bEA_a\) for all physical indices \(a,b\). For a word \(\sigma =(\sigma _1,\ldots ,\sigma _{n+2})\), define

\begin{align} \alpha (\sigma ) & = \operatorname{tr}\! \left(A_{\sigma _{n+2}} C_{\sigma _1} A_{\sigma _2}\cdots A_{\sigma _{n+1}}\right). \notag \end{align}

Then \(\alpha \in G_{n+2}(A)\).

Proof

The boundary identity gives

\begin{align} A_{\sigma _{n+2}} C_{\sigma _1} & = A_{\sigma _{n+2}}EA_{\sigma _1}. \notag \end{align}

Hence, by cyclicity of the trace,

\begin{align} \alpha (\sigma ) & = \operatorname{tr}\! \left(A_{\sigma _1}A_{\sigma _2}\cdots A_{\sigma _{n+2}}E\right), \notag \end{align}

which is the ground-space parametrization of \(E\).

Lemma 15.2.1.13 Left-boundary summands are ground-space vectors

Let \(A^1,\ldots ,A^g\) be block tensors, and let \(C^j_a,E^j\in M_{D_j}(\mathbb {C})\) satisfy \(A^j_b C^j_a = A^j_bE^jA^j_a\) for every block \(j\) and all physical indices \(a,b\). For a word \(\sigma =(\sigma _1,\ldots ,\sigma _{n+2})\), define

\begin{align} \alpha _j(\sigma ) & = \operatorname{tr}\! \left(A^j_{\sigma _{n+2}} C^j_{\sigma _1} A^j_{\sigma _2}\cdots A^j_{\sigma _{n+1}}\right). \notag \end{align}

Then \(\sum _j \alpha _j \in \bigvee _j G_{n+2}(A^j)\).

Proof

Fix \(j\). By the assumed boundary identity and cyclicity of the trace,

\begin{align} \operatorname{tr}\! \left(A^j_{\sigma _{n+2}} C^j_{\sigma _1} A^j_{\sigma _2}\cdots A^j_{\sigma _{n+1}}\right) & = \operatorname{tr}\! \left(A^j_{\sigma _1}A^j_{\sigma _2}\cdots A^j_{\sigma _{n+2}}E^j\right). \notag \end{align}

Thus \(\alpha _j\) is the image of \(E^j\) under the parametrization of \(G_{n+2}(A^j)\). Hence each \(\alpha _j\) lies in \(G_{n+2}(A^j)\), and the finite sum lies in the supremum of these subspaces.

Lemma 15.2.1.14 Left-boundary trace decompositions lie in the block ground spaces

Let \(A^1,\ldots ,A^g\) be block tensors with common word-span separation at length \(n\), and suppose that \(\sum _a A^j_a A^{j\, \dagger }_a = \mathbb {1}\) for every block \(j\). Suppose a vector \(\psi \in (\mathbb {C}^d)^{\otimes (n+2)}\) has the left-boundary trace decomposition

\begin{align} \psi & = \sum _j\alpha _j, \notag \\ \alpha _j(i_1,\ldots ,i_{n+2}) & = \operatorname{tr}\! \left(A^j_{i_{n+2}} C^j_{i_1} A^j_{i_2}\cdots A^j_{i_{n+1}}\right), \notag \end{align}

and suppose that the two trace decompositions agree for every middle word:

\begin{align} \sum _j \operatorname{tr}\! \left(A^j_b C^j_a A^j_w\right) & = \sum _j \operatorname{tr}\! \left(D^j_b A^j_a A^j_w\right). \notag \end{align}

Then \(\psi \in \bigvee _j G_{n+2}(A^j)\).

Proof

The equality of the two boundary trace expressions and the common word-span hypothesis give \(A^j_bC^j_a = D^j_bA^j_a\) for every \(j,a,b\). By the normalization,

\begin{align} D^j_b & = D^j_b\sum _a A^j_aA^{j\, \dagger }_a = \sum _a A^j_bC^j_aA^{j\, \dagger }_a = A^j_bE^j, \notag \\ E^j & := \sum _a C^j_aA^{j\, \dagger }_a. \notag \end{align}

Hence \(A^j_bC^j_a=A^j_bE^jA^j_a\). Lemma 15.2.1.13 then gives \(\psi = \sum _j\alpha _j \in \bigvee _j G_{n+2}(A^j)\).

Lemma 15.2.1.15 Trace decompositions from fixed-boundary restrictions

Let \(A^1,\ldots ,A^g\) be block tensors. If \(\psi \in (\mathbb {C}^d)^{\otimes (n+2)}\) satisfies

\begin{align} \psi (a,-)& \in \bigvee _jG_{n+1}(A^j), \notag \\ \psi (-,b)& \in \bigvee _jG_{n+1}(A^j) \notag \end{align}

for every physical index \(a,b\), then there are matrices \(C^j_a\) and \(D^j_b\) such that

\begin{align} \psi & =\sum _j\alpha _j, \notag \\ \alpha _j(i_1,\ldots ,i_{n+2}) & = \operatorname{tr}\! \left(A^j_{i_{n+2}}C^j_{i_1} A^j_{i_2}\cdots A^j_{i_{n+1}}\right), \notag \end{align}

and, for every middle word \(w\) of length \(n\),

\begin{align} \sum _j\operatorname{tr}\! \left(A^j_bC^j_aA^j_w\right) & = \sum _j\operatorname{tr}\! \left(D^j_bA^j_aA^j_w\right). \notag \end{align}
Proof

Choose finite decompositions of the fixed-boundary vectors:

\begin{align} \psi (a,i_2,\ldots ,i_{n+2}) & = \sum _j \operatorname{tr}\! \left(A^j_{i_2}\cdots A^j_{i_{n+2}}C^j_a\right), \notag \\ \psi (i_1,\ldots ,i_{n+1},b) & = \sum _j \operatorname{tr}\! \left(A^j_{i_1}\cdots A^j_{i_{n+1}}D^j_b\right). \notag \end{align}

For every middle word \(w\),

\begin{align} \psi (a,w,b) & = \sum _j\operatorname{tr}\! \left(A^j_wA^j_bC^j_a\right) = \sum _j\operatorname{tr}\! \left(A^j_bC^j_aA^j_w\right), \notag \end{align}

and the second decomposition gives

\begin{align} \psi (a,w,b) & = \sum _j\operatorname{tr}\! \left(A^j_aA^j_wD^j_b\right) = \sum _j\operatorname{tr}\! \left(D^j_bA^j_aA^j_w\right). \notag \end{align}

Hence

\begin{align} \sum _j\operatorname{tr}\! \left(A^j_bC^j_aA^j_w\right) & = \sum _j\operatorname{tr}\! \left(D^j_bA^j_aA^j_w\right). \notag \end{align}

For an arbitrary word \((i_1,\ldots ,i_{n+2})\), the first decomposition gives

\begin{align} \psi (i_1,\ldots ,i_{n+2}) & = \sum _j \operatorname{tr}\! \left(A^j_{i_2}\cdots A^j_{i_{n+2}}C^j_{i_1}\right) = \sum _j \operatorname{tr}\! \left(A^j_{i_{n+2}}C^j_{i_1} A^j_{i_2}\cdots A^j_{i_{n+1}}\right), \notag \end{align}

by cyclicity of the trace.

Lemma 15.2.1.16 One-step block intersection from block restrictions

Let \(A^1,\ldots ,A^g\) be block tensors with common word-span separation at length \(n\), and suppose that \(\sum _a A^j_aA^{j\, \dagger }_a=\mathbb {1}\) for every block \(j\). If \(\psi \in (\mathbb {C}^d)^{\otimes (n+2)}\) satisfies

\begin{align} \psi (a,-)& \in \bigvee _jG_{n+1}(A^j), \notag \\ \psi (-,b)& \in \bigvee _jG_{n+1}(A^j) \notag \end{align}

for every physical index \(a,b\), then \(\psi \in \bigvee _jG_{n+2}(A^j)\).

Proof

Choose finite decompositions of the fixed-boundary vectors:

\begin{align} \psi (a,i_2,\ldots ,i_{n+2}) & = \sum _j \operatorname{tr}\! \left(A^j_{i_2}\cdots A^j_{i_{n+2}}C^j_a\right), \notag \\ \psi (i_1,\ldots ,i_{n+1},b) & = \sum _j \operatorname{tr}\! \left(A^j_{i_1}\cdots A^j_{i_{n+1}}D^j_b\right). \notag \end{align}

Evaluating these two decompositions on \((a,w,b)\), where \(w\) has length \(n\), gives

\begin{align} \sum _j\operatorname{tr}\! \left(A^j_bC^j_aA^j_w\right) & = \sum _j\operatorname{tr}\! \left(D^j_bA^j_aA^j_w\right). \notag \end{align}

The first decomposition also gives

\begin{align} \psi & =\sum _j\alpha _j, \notag \\ \alpha _j(i_1,\ldots ,i_{n+2}) & = \operatorname{tr}\! \left(A^j_{i_{n+2}}C^j_{i_1} A^j_{i_2}\cdots A^j_{i_{n+1}}\right). \notag \end{align}

Lemma 15.2.1.14 therefore gives \(\psi \in \bigvee _jG_{n+2}(A^j)\).

Lemma 15.2.1.17 One-step block intersection characterization

Under the hypotheses of Lemma 15.2.1.16, an \((n+2)\)-site vector satisfies \(\psi \in \bigvee _jG_{n+2}(A^j)\) if and only if

\begin{align} \psi (a,-)& \in \bigvee _jG_{n+1}(A^j), \notag \\ \psi (-,b)& \in \bigvee _jG_{n+1}(A^j) \notag \end{align}

for every physical index \(a,b\).

Proof

The reverse implication is Lemma 15.2.1.16. For the forward implication, write

\begin{align} \psi & =\sum _j\gamma _j, \notag \\ \gamma _j& \in G_{n+2}(A^j). \notag \end{align}

Each \(\gamma _j\) has fixed-boundary restrictions in \(G_{n+1}(A^j)\):

\begin{align} \gamma _j(a,-)& \in G_{n+1}(A^j), \notag \\ \gamma _j(-,b)& \in G_{n+1}(A^j). \notag \end{align}

Taking the finite sum over \(j\) gives the two restrictions of \(\psi \) in \(\bigvee _jG_{n+1}(A^j)\).

Lemma 15.2.1.18 Subspace form of the one-step block intersection

Under the hypotheses of Lemma 15.2.1.16, set \(S_n:=\bigvee _jG_{n+1}(A^j)\). Then

\begin{align} \left\{ \psi :\forall b,\ \psi (-,b)\in S_n\right\} \cap \left\{ \psi :\forall a,\ \psi (a,-)\in S_n\right\} & = \bigvee _jG_{n+2}(A^j). \notag \end{align}
Proof

Lemma 15.2.1.17 gives the displayed equivalence for each vector \(\psi \). Interpreting the two sides of that equivalence as subspaces gives the stated equality.

15.2.2 Eventual block-intersection consequences

Propagating the single-length product span through a period window gives the block-intersection identity for all sufficiently large \(n\), both from a general period window and from the normalized BNT representative hypotheses. The full product span at length \(n\) also makes the sum of the local spaces \(G_n(A^j)\) direct.

Lemma 15.2.2.1 Period-window form of the block intersection identity

Let \(p{\gt}0\). Suppose the length-\(p\) simultaneous block-word tuples span \(\prod _j M_{D_j}(\mathbb {C})\), and suppose that, for every \(0\le r{\lt}p\), the length-\((s+r)\) simultaneous block-word tuples span \(\prod _j M_{D_j}(\mathbb {C})\). Assume also the normalization \(\sum _a A^j_aA^{j\, \dagger }_a=\mathbb {1}\) for every block \(j\). Then there is a length \(N\) such that, for every \(n\ge N\), if \(S_n:=\bigvee _jG_{n+1}(A^j)\), then

\begin{align} \left\{ \psi :\forall b,\ \psi (-,b)\in S_n\right\} \cap \left\{ \psi :\forall a,\ \psi (a,-)\in S_n\right\} & = \bigvee _jG_{n+2}(A^j). \notag \end{align}
Proof

Lemma 15.2.15 gives a length \(N\) such that the simultaneous block-word tuples span \(\prod _j M_{D_j}(\mathbb {C})\) at every length \(n\ge N\). For each such \(n\), Lemma 15.2.1.18 applies and gives the displayed subspace equality.

Lemma 15.2.2.2 BNT block separation gives one block-intersection step

Under the hypotheses of Lemma 10.3.2, assume in addition the normalization \(\sum _a A^j_aA^{j\, \dagger }_a=\mathbb {1}\) for every block \(j\). Let

\begin{align} n& =(L_0+1)+(r-1)3(L_0+1), \notag \\ S_n& =\bigvee _jG_{n+1}(A^j). \notag \end{align}

Then

\begin{align} \left\{ \psi :\forall b,\ \psi (-,b)\in S_n\right\} \cap \left\{ \psi :\forall a,\ \psi (a,-)\in S_n\right\} & = \bigvee _jG_{n+2}(A^j). \notag \end{align}
Proof

Lemma 10.3.2 gives \(\operatorname{span}\{ (A^1_w,\ldots ,A^r_w): |w|=n\} = \prod _jM_{D_j}(\mathbb {C})\). Substituting this span into Lemma 15.2.1.18, together with the normalization, gives the stated subspace equality.

Lemma 15.2.2.3 BNT block separation with a block-injectivity window

Let \(A^1,\ldots ,A^r\) be normalized representatives of a basis of normal tensors: each representative is irreducible, the family is left-canonical, the self-overlaps are normalized, and no two distinct same-dimensional representatives are gauge-phase equivalent. Assume \(L{\gt}1\), and assume that, for every block \(j\),

\begin{align} \operatorname{span}\{ A^j_a:a=0,\ldots ,d-1\} & =M_{D_j}(\mathbb {C}), \notag \\ \operatorname{span}\{ A^j_u:|u|=L\} & =M_{D_j}(\mathbb {C}), \notag \\ \operatorname{span}\{ A^j_v:|v|=3L\} & =M_{D_j}(\mathbb {C}). \notag \end{align}

Set \(S=3L\) and \(q=(r-1)S\). Suppose that \(p{\gt}0\) and let \(s_0\) be a starting length. Assume that, for every block \(j\), \(\operatorname{span}\{ A^j_u:|u|=p\} =M_{D_j}(\mathbb {C})\), and that, for every \(0\le s{\lt}p+q\) and every block \(j\), \(\operatorname{span}\{ A^j_v:|v|=s_0+s\} =M_{D_j}(\mathbb {C})\). Assume also the normalization \(\sum _a A^j_aA^{j\, \dagger }_a=\mathbb {1}\) for every block \(j\). Then there is \(N\) such that, for every \(n\ge N\), if \(S_n:=\bigvee _jG_{n+1}(A^j)\), then

\begin{align} \left\{ \psi :\forall b,\ \psi (-,b)\in S_n\right\} \cap \left\{ \psi :\forall a,\ \psi (a,-)\in S_n\right\} & = \bigvee _jG_{n+2}(A^j). \notag \end{align}
Proof

The direct-sum dimension step gives, for each ordered pair of distinct blocks \(k\ne j\), coefficients \(c^{k,j}_\omega \) of length \(S\) such that

\begin{align} \sum _{|\omega |=S} c^{k,j}_\omega A^k_\omega & =\mathbb {1}, \notag \\ \sum _{|\omega |=S} c^{k,j}_\omega A^j_\omega & =0. \notag \end{align}

Multiplying these equations over the \(r-1\) blocks \(j\ne k\) gives coefficients \(c^{(k)}_\omega \) of total length \(q\) satisfying

\begin{align} \sum _{|\omega |=q}c^{(k)}_\omega A^j_\omega & = \begin{cases} \mathbb {1}, & j=k, \\ 0, & j\ne k. \end{cases} \notag \end{align}

Combining this block-separating equation with block injectivity at length \(p\) gives \(\operatorname{span}\{ (A^1_w,\ldots ,A^r_w):|w|=p+q\} = \prod _j M_{D_j}(\mathbb {C})\). Combining the same separating equations with the assumed block-injectivity window gives the full product span at lengths \(s_0+q+s\), for \(0\le s{\lt}p+q\). The period-window form of Lemma 15.2.2.1 then gives the displayed equality for all sufficiently large \(n\).

Lemma 15.2.2.4 Unitality propagates simultaneous product spans

Suppose that the length-\(q\) simultaneous products span the product algebra, \(\operatorname{span}\{ (A^1_w,\ldots ,A^r_w):|w|=q\} = \prod _jM_{D_j}(\mathbb {C})\), and that every block satisfies \(\sum _a A^j_aA^{j\, \dagger }_a=\mathbb {1}\). Then the simultaneous products span the product algebra at every length \(n\ge q\). This is the simultaneous-tuple form of the unital propagation in [ PGVWC07 , lines 893–898 ] .

Proof

Given matrices \(M_j\), first represent the tuple \(((A^j_a)^\dagger M_j)_j\) by words of length \(q\). Prefixing the letter \(a\) and summing over \(a\) gives \(\left(\sum _aA^j_aA^{j\, \dagger }_aM_j\right)_j =(M_j)_j\). This proves the step from \(q\) to \(q+1\); induction gives every \(n\ge q\).

Let \(A^1,\ldots ,A^r\) be blocks satisfying the normalization \(\sum _a A^j_aA^{j\, \dagger }_a=\mathbb {1}\) and assume that every block is injective at length \(L\): \(\operatorname{span}\{ A^j_u:|u|=L\} =M_{D_j}(\mathbb {C})\). If there are block-separating equations of length \(S\), then, for every \(n\ge L+(r-1)S\), \(\operatorname{span}\{ (A^1_w,\ldots ,A^r_w):|w|=n\} = \prod _j M_{D_j}(\mathbb {C})\). In particular, for normalized BNT representatives satisfying the displayed unital normalization, if \(L_0{\gt}0\) and \(\Gamma _{L_0}^{A^j}\) is injective for every block \(j\), the same conclusion holds with \(S=3(L_0+1)\) and \(n\ge (L_0+1)+(r-1)3(L_0+1)\).

Proof

By Theorem 8.8.1, the normalization propagates the length-\(L\) span equality to every larger homogeneous length:

\begin{align} m\ge L \quad \Longrightarrow \quad \operatorname{span}\{ A^j_u:|u|=m\} & =M_{D_j}(\mathbb {C}). \notag \end{align}

For \(n\ge L+(r-1)S\), take \(m=n-(r-1)S\). Combining the length-\(m\) block span with the \(r-1\) block-separating equations gives the displayed product span at length \(n\). In the BNT case, the injectivity of \(\Gamma _{L_0}^{A^j}\) gives the full span at length \(L_0\), and the normalization propagates it to the lengths \(L_0+1\) and \(3(L_0+1)\). Lemma 10.3.2 then supplies the block-separating equations with \(S=3(L_0+1)\).

Let \(r\ge 2\). Suppose that the normalized BNT blocks are injective at the common length \(L_0{\gt}0\) and satisfy \(\sum _aA^j_aA^{j\, \dagger }_a=\mathbb {1}\). For every \(n\ge 3(r-1)(L_0+1)\), one has \(\operatorname{span}\{ (A^1_w,\ldots ,A^r_w):|w|=n\} =\prod _jM_{D_j}(\mathbb {C})\). Consequently the spaces \(G_n(A^j)\) form an internal sum. This is the simultaneous spanning estimate used in [ PGVWC07 , Theorem 12, proof lines 1424–1456 ] .

Proof

The block-separation argument gives the full simultaneous span at \(q=3(r-1)(L_0+1)\). Lemma 15.2.2.4 propagates this equality to every \(n\ge q\). To prove independence, write \(\phi _j=\Gamma _n^{A^j}(X_j)\) and suppose that \(\sum _j\phi _j=0\). Comparison of the coefficient of each word, followed by the simultaneous spanning equality, gives

\begin{align} \sum _j \Gamma _n^{A^j}(X_j)=0 & \quad \Longrightarrow \quad \left(\forall (M_j)_j, \sum _j\operatorname{tr}(X_jM_j)=0\right) \quad \Longrightarrow \quad \forall j,\ X_j=0. \notag \end{align}

Indeed, in the last implication one takes a tuple supported in one block and uses nondegeneracy of the trace pairing. Hence every \(\phi _j\) vanishes, so the spaces \(G_n(A^j)\) form an internal sum.

Theorem 15.2.2.7 Direct-sum restriction intersection at the source length

Under the hypotheses of Lemma 15.2.2.6, let \(m\ge 3(r-1)(L_0+1)+1\). Then

\begin{align} \mathbb {C}^d\otimes \left(\sum _jG_m(A^j)\right) \cap \left(\sum _jG_m(A^j)\right)\otimes \mathbb {C}^d & = \sum _jG_{m+1}(A^j). \notag \end{align}

Lemma 15.2.2.6 also shows that the sums at lengths \(m\) and \(m+1\) are internal. This is the direct-sum lemma in the proof of [ PGVWC07 , Theorem 12, proof lines 1346–1456 ] .

Proof

Let a vector belong to both restricted spaces. The two endpoint decompositions give boundary matrices \(C_a^j\) and \(D_b^j\). For every pair of physical indices \(a,b\) and every middle word \(w\) of length \(m-1\), their coefficients obey

\begin{align} \sum _j\operatorname{tr}\! \left((A_b^jC_a^j)A_w^j\right) & = \sum _j\operatorname{tr}\! \left((D_b^jA_a^j)A_w^j\right). \notag \end{align}

The simultaneous product span at length \(m-1\) and nondegeneracy of the trace pairing therefore give, block by block, \(A_b^jC_a^j=D_b^jA_a^j\). Set \(E^j=\sum _a C_a^jA_a^{j\, \dagger }\). The preceding identities and the right-unital normalization imply \(A_b^jC_a^j=A_b^jE^jA_a^j\). Thus the component in block \(j\) lies in \(G_{m+1}(A^j)\). The reverse inclusion follows by splitting a word of length \(m+1\) at either endpoint.

Lemma 15.2.2.8 Product span makes the sum of local spaces direct

Suppose \(\operatorname{span}\{ (A^1_w,\ldots ,A^r_w):|w|=n\} = \prod _jM_{D_j}(\mathbb {C})\). Then the sum of the local spaces \(G_n(A^j)\) is direct: if

\begin{align} \phi _j& \in G_n(A^j), \notag \\ \sum _j\phi _j& =0, \notag \end{align}

then \(\phi _j=0\) for every \(j\). In particular, for the normalized BNT hypotheses above, this directness holds for every \(n\ge (L_0+1)+(r-1)3(L_0+1)\).

Proof

Write \(\phi _j(\sigma )=\operatorname{tr}(A^j_\sigma X_j)\). Evaluating \(\sum _j\phi _j=0\) on each word \(w\) of length \(n\) gives \(\sum _j\operatorname{tr}(A^j_wX_j)=0\). Since \(\operatorname{tr}(A^j_wX_j)=\operatorname{tr}(X_jA^j_w)\), the product-span equation gives, for every \((M_1,\ldots ,M_r)\in \prod _jM_{D_j}(\mathbb {C})\), \(\sum _j\operatorname{tr}(X_jM_j)=0\). Taking only the \(j\)-th component nonzero and using nondegeneracy of the trace pairing yields \(X_j=0\), hence \(\phi _j=0\). The BNT statement follows by substituting Lemma 15.2.2.5.

15.3 Block-diagonal boundary conditions for periodic ground spaces

For a block-diagonal tensor \(B=\bigoplus _{j=1}^g\mu _jA_j\) with \(\mu _j\ne 0\), this section proves implications from explicit block-diagonal boundary identities to the component membership \(\Gamma _N^{A_j}(\mu _j^NX_j)\in \mathcal G_{N,L}(A_j)\). These implications supply the boundary-condition step in the periodic ring used by [ CPGSV21 , Theorem IV.16 ] .

Let \(A^1,\ldots ,A^r\) be normalized representatives of a basis of normal tensors: each representative is irreducible, the family is left-canonical, the self-overlaps are normalized, and no two distinct same-dimensional representatives are gauge-phase equivalent. Suppose \(L_0{\gt}0\), and, for every block \(j\), the map

\begin{align} \Gamma _{L_0}^{A^j}:M_{D_j}(\mathbb {C}) & \longrightarrow (\mathbb {C}^d)^{\otimes L_0} \notag \end{align}

is injective. Assume also the normalization \(\sum _a A^j_aA^{j\, \dagger }_a=\mathbb {1}\). Set \(N=(L_0+1)+(r-1)3(L_0+1)\). For every \(n\ge N\), set \(S_n:=\bigvee _jG_{n+1}(A^j)\). The two consecutive sums are internal direct sums:

\begin{align} \bigvee _jG_{n+1}(A^j) & =\bigoplus _jG_{n+1}(A^j), \notag \\ \bigvee _jG_{n+2}(A^j) & =\bigoplus _jG_{n+2}(A^j). \notag \end{align}

With this notation,

\begin{align} & \left\{ \psi :\forall b,\ \psi (-,b)\in S_n\right\} \cap \left\{ \psi :\forall a,\ \psi (a,-)\in S_n\right\} =\bigvee _jG_{n+2}(A^j). \notag \end{align}
Proof

Lemma 15.2.2.5 gives \(\operatorname{span}\{ (A^1_w,\ldots ,A^r_w):|w|=n\} =\prod _jM_{D_j}(\mathbb {C})\) for every \(n\ge N\). Substituting this product-span equation into Lemma 15.2.1.18, together with the normalization equation, gives the displayed intersection identity. Lemma 15.2.2.8 at lengths \(n+1\) and \(n+2\) gives the two directness statements.

Lemma 15.3.2 Normalized BNT representatives give an eventual block intersection

Let \(A^1,\ldots ,A^r\) be normalized representatives of a basis of normal tensors: each representative is irreducible, the family is left-canonical, the self-overlaps are normalized, and no two distinct same-dimensional representatives are gauge-phase equivalent. Suppose \(L_0{\gt}0\), and, for every block \(j\), the map

\begin{align} \Gamma _{L_0}^{A^j}:M_{D_j}(\mathbb {C}) & \longrightarrow (\mathbb {C}^d)^{\otimes L_0} \notag \end{align}

is injective. Assume also the normalization \(\sum _a A^j_aA^{j\, \dagger }_a=\mathbb {1}\). Then there is a length \(N\) such that, for every \(n\ge N\), with \(S_n:=\bigvee _jG_{n+1}(A^j)\), one has

\begin{align} & \left\{ \psi :\forall b,\ \psi (-,b)\in S_n\right\} \cap \left\{ \psi :\forall a,\ \psi (a,-)\in S_n\right\} =\bigvee _jG_{n+2}(A^j). \notag \end{align}
Proof

Apply Lemma 15.3.1 and take \(N=(L_0+1)+(r-1)3(L_0+1)\).

Lemma 15.3.3 Block-diagonal boundary-condition formula

Let \(B=\bigoplus _{j=1}^g\mu _jA_j\). For block-diagonal boundary conditions \(X_j\),

\begin{align} \Gamma _L^B\! \left(\bigoplus _{j=1}^g X_j\right) & =\sum _{j=1}^g\Gamma _L^{A_j}(\mu _j^LX_j). \notag \end{align}
Proof

This is the diagonal-block trace formula specialized to a block-diagonal boundary matrix. The \(j\)-th diagonal block of \(\bigoplus _kX_k\) is \(X_j\), giving the displayed sum.

Lemma 15.3.4 One block lies in the local parent space of the direct-sum tensor

Let \(B=\bigoplus _{k=1}^g\mu _kA_k\). If \(\mu _j\ne 0\), then, for every length \(L\), one has \(G_L(A_j)\subseteq G_L(B)\).

Proof

A vector in \(G_L(A_j)\) has the form \(\Gamma _L^{A_j}(X)\). Put \(\mu _j^{-L}X\) in the \(j\)-th diagonal block of a boundary matrix for \(B\) and put zero in every other block. If \(\iota _j(Z)\) denotes this block insertion, Lemma 15.3.3 gives

\begin{align} \Gamma _L^B\! \left(\iota _j(\mu _j^{-L}X)\right) & = \sum _{k=1}^g \Gamma _L^{A_k} \! \left(\mu _k^L(\iota _j(\mu _j^{-L}X))_{kk}\right) =\Gamma _L^{A_j}(X). \notag \end{align}

Hence \(\Gamma _L^{A_j}(X)\in G_L(B)\).

Lemma 15.3.5 Local ground space of a block-diagonal tensor

Let \(B=\bigoplus _{j=1}^g\mu _jA_j\), where \(\mu _j\ne 0\) for \(1\le j\le g\). Then, for every length \(L\), \(G_L(B)=\bigvee _{j=1}^gG_L(A_j)\).

Proof

A boundary matrix \(X\) for \(B\) has diagonal blocks \(X_j\), and block-diagonality gives \(\Gamma _L^B(X) =\sum _{j=1}^g\Gamma _L^{A_j}(\mu _j^LX_j)\). This proves \(G_L(B)\subseteq \bigvee _jG_L(A_j)\). The reverse inclusion is Lemma 15.3.4 for each block \(j\).

Lemma 15.3.6 Periodic constraints for a block-diagonal tensor

Let \(B=\bigoplus _{j=1}^g\mu _jA_j\), where \(\mu _j\ne 0\) for \(1\le j\le g\), and set \(S_M:=\bigvee _{j=1}^gG_M(A_j)\). If, for every \(M\ge L\), \(\mathbb {C}^d\otimes S_M\cap S_M\otimes \mathbb {C}^d =S_{M+1}\), then \(\mathcal G_{N,L}(B)\subseteq S_N\).

Proof

Lemma 15.3.5 gives \(G_L(B)=S_L\). Apply Lemma 13.4.5.13 to the sequence \(S_M\).

Lemma 15.3.7 Periodic constraints lie in the block local-space sum

Let \(A_1,\ldots ,A_g\) be normalized representatives of a basis of normal tensors: each representative is irreducible, the family is left-canonical, the self-overlaps are normalized, and no two distinct same-dimensional representatives are gauge-phase equivalent. Suppose \(L_0{\gt}0\), and, for every block \(j\), the map

\begin{align} \Gamma _{L_0}^{A_j}:M_{D_j}(\mathbb {C}) & \longrightarrow (\mathbb {C}^d)^{\otimes L_0} \notag \end{align}

is injective. Assume also the normalization \(\sum _a A^j_aA^{j\, \dagger }_a=\mathbb {1}\). For nonzero scalars \(\mu _j\), set

\begin{align} B& =\bigoplus _{j=1}^g\mu _jA_j, \notag \\ S_M& :=\bigvee _{j=1}^gG_M(A_j). \notag \end{align}

If \(L\ge (L_0+1)+(g-1)3(L_0+1)+1\), then, for every \(N\ge L\), \(\mathcal G_{N,L}(B)\subseteq S_N\).

Proof

Put \(T=(L_0+1)+(g-1)3(L_0+1)\). For \(M\ge L\), one has \(M-1\ge T\). Applying Lemma 15.3.1 at \(n=M-1\) gives \(\mathbb {C}^d\otimes S_M\cap S_M\otimes \mathbb {C}^d =S_{M+1}\) for every \(M\ge L\). Lemma 15.3.6 then propagates the local constraint from \(L\) to \(N\).

Lemma 15.3.8 Periodic constraints and internality of the block local-space sum

With the hypotheses and notation of Lemma 15.3.7, set \(S_N:=\bigvee _{j=1}^gG_N(A_j)\). Then, for every \(N\ge L\), \(\mathcal G_{N,L}(B)\subseteq S_N\), and the sum defining \(S_N\) is internal: if

\begin{align} \phi _j& \in G_N(A_j), \notag \\ \sum _{j=1}^g\phi _j& =0, \notag \end{align}

then \(\phi _j=0\) for every \(j\).

Proof

The inclusion is Lemma 15.3.7. Since \(N\ge L\) and \(L\ge (L_0+1)+(g-1)3(L_0+1)+1\), the product-span directness range of Lemma 15.2.2.8 applies at length \(N\):

\begin{align} \operatorname{span}\{ (A^1_w,\ldots ,A^g_w):|w|=N\} & =\prod _j M_{D_j}(\mathbb {C}) \notag \\ & \quad \Longrightarrow \ker \! \left(\bigoplus _{j=1}^g G_N(A_j) \longrightarrow (\mathbb {C}^d)^{\otimes N}, \quad (\phi _j)_j\longmapsto \sum _j\phi _j\right)=0. \notag \end{align}
Lemma 15.3.9 Periodic parent-space vectors decompose in the local block spaces

With the hypotheses and notation of Lemma 15.3.7, every vector \(\psi \in \mathcal G_{N,L}(\bigoplus _{j=1}^g\mu _jA_j)\) has a unique family \((\psi _j)_{j=1}^g\) such that

\begin{align} \psi & =\sum _{j=1}^g\psi _j, \notag \\ \psi _j& \in G_N(A_j). \notag \end{align}
Proof

Lemma 15.3.8 gives

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _{j=1}^g\mu _jA_j\right) & \subseteq \bigvee _{j=1}^gG_N(A_j), \notag \end{align}

and the spaces \(G_N(A_j)\) are independent: for every \(j\) and every \(x_j\in G_N(A_j)\), \(\sum _jx_j=0 \quad \Longrightarrow \quad x_j=0\). Thus every \(\psi \in \mathcal G_{N,L}(\bigoplus _j\mu _jA_j)\) has a unique decomposition \(\psi =\sum _j\psi _j\) with \(\psi _j\in G_N(A_j)\).

Lemma 15.3.10 Block-diagonal boundary conditions for periodic parent-space vectors

With the hypotheses and notation of Lemma 15.3.7, every vector \(\psi \in \mathcal G_{N,L}(\bigoplus _{j=1}^g\mu _jA_j)\) admits block-diagonal boundary conditions \(X_j\) such that

\begin{align} \psi & =\Gamma _N^B\! \left(\bigoplus _{j=1}^gX_j\right), \notag \\ \Gamma _N^{A_j}(\mu _j^NX_j) & \in G_N(A_j) \notag \end{align}

for every \(j\).

Proof

Lemma 15.3.9 gives

\begin{align} \psi & =\sum _{j=1}^g\phi _j, \notag \\ \phi _j& \in G_N(A_j). \notag \end{align}

Choose \(Y_j\) with \(\phi _j=\Gamma _N^{A_j}(Y_j)\), and set \(X_j=\mu _j^{-N}Y_j\). Since \(\mu _j\ne 0\), \(\Gamma _N^{A_j}(\mu _j^NX_j)=\phi _j\). The block-diagonal boundary formula then gives

\begin{align} \Gamma _N^B\! \left(\bigoplus _{j=1}^gX_j\right) & =\sum _{j=1}^g\Gamma _N^{A_j}(\mu _j^NX_j) =\psi . \notag \end{align}

Let \(g\ge 2\). Let \(A_j\) be irreducible left-canonical blocks with normalized self-overlap, pairwise inequivalent up to gauge and phase, injective at the common length \(L_0{\gt}0\), and satisfying \(\sum _aA^j_aA^{j\, \dagger }_a=\mathbb {1}\). Let every weight \(\mu _j\) be nonzero. If \(0{\lt}L\le N\) and \(L\ge 3(g-1)(L_0+1)+1\), then

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _j\mu _jA_j\right) & \subseteq \bigoplus _jG_N(A_j). \notag \end{align}

In particular, every vector \(\psi \) in the left-hand side admits matrices \(X_j\) such that

\begin{align} \psi & =\Gamma _N^{\oplus _j\mu _jA_j} \left(\bigoplus _jX_j\right), \notag \\ \Gamma _N^{A_j}(\mu _j^NX_j) & \in G_N(A_j). \notag \end{align}

This is the open-boundary decomposition used in [ PGVWC07 , Theorem 12, proof lines 1424–1456 ] .

Proof

Put \(S_M=\sum _jG_M(A_j)\). For every \(M\) with \(L\le M{\lt}N\), Theorem 15.2.2.7 gives

\begin{align} \left(\bigcap _b\operatorname{Res}_{-,b}^{-1}S_M\right) \cap \left(\bigcap _a\operatorname{Res}_{a,-}^{-1}S_M\right) & =S_{M+1}. \notag \end{align}

Induction through the cyclic local constraints therefore gives

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _j\mu _jA_j\right) & \subseteq S_N. \notag \end{align}

Decompose the vector into its block components, choose one boundary matrix for each component, and absorb the nonzero factor \(\mu _j^N\) into that matrix.

The lemma gives \(\Gamma _N^{A_j}(\mu _j^NX_j)\in G_N(A_j)\). This is not enough for periodic-boundary membership when a cyclic window crosses the chosen cut. The boundary-cut comparison of Lemma 15.3.1.24 concerns the separate assertion \(\Gamma _N^{A_j}(\mu _j^NX_j)\in \mathcal G_{N,L}(A_j)\).

15.3.1 Cyclic windows at the boundary cut

A cyclic length-\(L\) window either stays inside the chosen linear interval (\(i+L\le N\)) or crosses the boundary cut (\(N{\lt}i+L\)). The non-wrapping case follows from trace cyclicity, while the crossing case requires a boundary-matrix identity \(\mu _j^NX_jA^j_\beta =A^j_\beta Y\). The lemmas below establish the single-block periodic membership \(\Gamma _N^{A_j}(\mu _j^NX_j)\in \mathcal G_{N,L}(A_j)\) in both cases.

Lemma 15.3.1.1 Cyclic restrictions of block-diagonal boundary vectors

Let \(B=\bigoplus _{j=1}^g\mu _jA_j\), and fix block-diagonal boundary conditions \(X_j\). For every cyclic restriction \(R_{i,\tau }\) from \(N\) sites to \(L\) sites,

\begin{align} R_{i,\tau }\! \left(\Gamma _N^B\! \left(\bigoplus _{j=1}^gX_j\right)\right) & = \sum _{j=1}^g R_{i,\tau }\! \left(\Gamma _N^{A_j}(\mu _j^NX_j)\right). \notag \end{align}
Proof

Apply Lemma 15.3.3 at length \(N\), then use linearity of the cyclic restriction \(R_{i,\tau }\).

Lemma 15.3.1.2 Local direct-sum constraint from block-diagonal boundary conditions

Let \(B=\bigoplus _{j=1}^g\mu _jA_j\), with \(\mu _j\ne 0\), and suppose

\begin{align} \psi & = \Gamma _N^B\! \left(\bigoplus _{j=1}^gX_j\right), \notag \\ \psi & \in \mathcal G_{N,L}(B). \notag \end{align}

Then every local constraint of \(\psi \) gives

\begin{align} \sum _{j=1}^g R_{i,\tau }\! \left(\Gamma _N^{A_j}(\mu _j^NX_j)\right) & \in \bigvee _{j=1}^gG_L(A_j). \notag \end{align}
Proof

The hypothesis \(\psi \in \mathcal G_{N,L}(B)\) says that the left-hand side before decomposition lies in \(G_L(B)\). Lemma 15.3.1.1 rewrites this local vector as the displayed sum, and Lemma 15.3.5 identifies \(G_L(B)\) with \(\bigvee _jG_L(A_j)\).

Fix block-diagonal boundary conditions \(X_j\). If the cyclic window beginning at \(i\) stays inside the chosen linear interval, \(i+L\le N\), then for every block \(j\),

\begin{align} R_{i,\tau }\! \left(\Gamma _N^{A_j}(\mu _j^NX_j)\right) & \in G_L(A_j). \notag \end{align}
Proof

When the window stays inside the chosen linear interval, the boundary matrix remains outside the window. Write \(A_{\mathrm{left}}=A^j_{\tau _0}\cdots A^j_{\tau _{i-1}}\) and \(A_{\mathrm{right}}=A^j_{\tau _{i+L}}\cdots A^j_{\tau _{N-1}}\). Trace cyclicity gives

\begin{align} R_{i,\tau }\! \left(\Gamma _N^{A_j}(\mu _j^NX_j)\right) & = \Gamma _L^{A_j}\! \left(A_{\mathrm{right}} \mu _j^NX_j\, A_{\mathrm{left}}\right). \notag \end{align}

Hence the restricted vector lies in \(G_L(A_j)\).

Lemma 15.3.1.4 Boundary-crossing coefficient formula

Assume \(0{\lt}N\), \(L\le N\), and let the cyclic interval beginning at \(i\) cross the boundary cut, so \(N{\lt}i+L\). Write \(a=i+L-N\). For a local word \(\sigma \) of length \(L\), set

\begin{align} \beta & =\sigma _{N-i}\cdots \sigma _{L-1}, \notag \\ \rho & =\tau _a\cdots \tau _{i-1}, \notag \\ \alpha & =\sigma _0\cdots \sigma _{N-i-1}. \notag \end{align}

Then the corresponding coefficient of the \(j\)-th block component is

\begin{align} R_{i,\tau }\! \left(\Gamma _N^{A_j}(\mu _j^NX_j)\right)(\sigma ) & = \operatorname{tr}\! \left(A^j_\beta A^j_\rho A^j_\alpha \, \mu _j^NX_j\right). \notag \end{align}
Proof

Let \(\omega \) be the \(N\)-site word obtained by inserting \(\sigma \) in the cyclic interval beginning at \(i\) and by using \(\tau \) outside that interval. Since the interval crosses the cut, \(\omega =\beta \rho \alpha \). Hence

\begin{align} \Gamma _N^{A_j}(\mu _j^NX_j)(\omega ) & = \operatorname{tr}\! \left(A^j_\beta A^j_\rho A^j_\alpha \, \mu _j^NX_j\right). \notag \end{align}

This is exactly \(R_{i,\tau }\! \left(\Gamma _N^{A_j}(\mu _j^NX_j)\right)(\sigma )\).

Lemma 15.3.1.5 Boundary-crossing trace decomposition from the local constraint

Assume \(0{\lt}N\), \(L\le N\), and let the cyclic interval beginning at \(i\) cross the boundary cut, so \(N{\lt}i+L\). Put \(a=i+L-N\). For a local word \(\sigma \) of length \(L\), write

\begin{align} \beta & =\sigma _{N-i}\cdots \sigma _{L-1}, \notag \\ \rho & =\tau _a\cdots \tau _{i-1}, \notag \\ \alpha & =\sigma _0\cdots \sigma _{N-i-1}. \notag \end{align}

These are the word coordinates obtained by opening the periodic boundary at the chosen cut, in the comparison with the source expressions involving \(C^j_{i_1}\) and \(D^j_{i_{m+1}}\). If

\begin{align} \sum _j R_{i,\tau }\! \left(\Gamma _N^{A_j}(\mu _j^NX_j)\right) & \in \bigvee _jG_L(A_j), \notag \end{align}

then there are matrices \(C^j_{i,\tau }\) such that, for every \(\sigma \),

\begin{align} \sum _j\operatorname{tr}\! \left(A^j_\beta C^j_{i,\tau }A^j_\alpha \right) & = \sum _j\operatorname{tr}\! \left(((\mu _j^NX_j)A^j_\beta )A^j_\rho A^j_\alpha \right). \notag \end{align}
Proof

Decompose the local vector in \(\bigvee _jG_L(A_j)\) as a sum of vectors \(\Gamma _L^{A_j}(C^j_{i,\tau })\). Evaluating at \(\sigma \) gives \(\sum _j\operatorname{tr}\! \left(A^j_{\alpha \beta }C^j_{i,\tau }\right)\). Trace cyclicity rewrites this as \(\sum _j\operatorname{tr}\! \left(A^j_\beta C^j_{i,\tau }A^j_\alpha \right)\). On the other hand, Lemma 15.3.1.4 gives the second trace expression for the same local coefficient sum.

Lemma 15.3.1.6 Boundary representation gives fixed-tail boundary traces

Assume that the physical dimension \(d\) is positive, \(0{\lt}N\), \(L\le N\), and \(\mu _j\ne 0\) for every block \(j\). Let the cyclic interval beginning at \(i\) cross the boundary cut, so \(N{\lt}i+L\). Let

\begin{align} \psi & \in \mathcal G_{N,L}\! \left(\bigoplus _j\mu _jA_j\right), \notag \\ \psi & = \Gamma _N^{\oplus _j\mu _jA_j}\! \left(\bigoplus _jX_j\right). \notag \end{align}

For every outside word \(\rho \) of length \(N-L\), there are matrices \(C^j_{i,\rho }\) such that, for every word \(\beta \) of length \(i+L-N\) before the cut and every wrapped tail word \(\alpha \) of length \(N-i\),

\begin{align} \sum _j\operatorname{tr}\! \left(A^j_\beta C^j_{i,\rho }A^j_\alpha \right) & = \sum _j\operatorname{tr}\! \left(((\mu _j^NX_j)A^j_\beta )A^j_\rho A^j_\alpha \right). \notag \end{align}
Proof

Since \(d{\gt}0\), extend the outside word \(\rho \) to a full outside configuration \(\tau \) whose entries between the end of the interval and the cut are the word \(\rho \). By the nonzero-weight hypothesis, Lemma 15.3.1.2 applies and puts the sum of the corresponding block restrictions in \(\bigvee _jG_L(A_j)\). Applying Lemma 15.3.1.5 gives the displayed trace equality after rewriting the opened cyclic word as \(\beta \rho \alpha \).

Lemma 15.3.1.7 Boundary representation gives all fixed-tail boundary traces

Assume that the physical dimension \(d\) is positive, \(0{\lt}N\), \(L\le N\), and \(\mu _j\ne 0\) for every block \(j\). Let

\begin{align} \psi & \in \mathcal G_{N,L}\! \left(\bigoplus _j\mu _jA_j\right), \notag \\ \psi & = \Gamma _N^{\oplus _j\mu _jA_j}\! \left(\bigoplus _jX_j\right). \notag \end{align}

Then there are matrices \(C^j_{i,\rho }\), indexed by the boundary-crossing start \(i\) and by an outside word \(\rho \) of length \(N-L\), such that whenever \(N{\lt}i+L\),

\begin{align} \sum _j\operatorname{tr}\! \left(A^j_\beta C^j_{i,\rho }A^j_\alpha \right) & = \sum _j\operatorname{tr}\! \left(((\mu _j^NX_j)A^j_\beta )A^j_\rho A^j_\alpha \right) \notag \end{align}

for every word \(\beta \) of length \(i+L-N\) before the cut and every wrapped tail word \(\alpha \) of length \(N-i\).

Proof

Apply Lemma 15.3.1.6 to each boundary-crossing start \(i\) and each outside word \(\rho \), and choose the resulting matrices \(C^j_{i,\rho }\).

Theorem 15.3.1.8 Fixed boundary-crossing \(C^j,D^j\) comparison from the local block-sum constraint

Assume \(0{\lt}N\), \(L\le N\), and let the cyclic interval beginning at \(i\) cross the boundary cut, so \(N{\lt}i+L\). Put \(a=i+L-N\) and \(\rho =\tau _a\cdots \tau _{i-1}\). Suppose also that the block tensors have the common word-span property at length \(N-i\). If

\begin{align} \sum _j R_{i,\tau }\! \left(\Gamma _N^{A_j}(\mu _j^NX_j)\right) & \in \bigvee _jG_L(A_j), \notag \end{align}

then there are matrices \(C^j_{i,\tau }\) such that, for every block \(j\) and every word \(\beta \) of length \(a\),

\begin{align} A^j_\beta C^j_{i,\tau } & = ((\mu _j^NX_j)A^j_\beta )A^j_\rho . \label{eq:ph_fixed_boundary_crossing_comparison} \end{align}
Proof

Lemma 15.3.1.5 gives matrices \(C^j_{i,\tau }\) with

\begin{align} \sum _j\operatorname{tr}\! \left(A^j_\beta C^j_{i,\tau }A^j_\alpha \right) & = \sum _j\operatorname{tr}\! \left(((\mu _j^NX_j)A^j_\beta )A^j_\rho A^j_\alpha \right) \notag \end{align}

for every word \(\alpha \) of length \(N-i\) and every word \(\beta \) of length \(a\). The fixed complementary-word trace comparison, applied with \(X_j\) replaced by \(\mu _j^NX_j\), gives (??).

Theorem 15.3.1.9 \(C^j,D^j\) comparison for a block-diagonal boundary representation

Let

\begin{align} \psi & \in \mathcal G_{N,L}\! \left(\bigoplus _j\mu _jA_j\right), \notag \\ \psi & = \Gamma _N^{\oplus _j\mu _jA_j}\! \left(\bigoplus _jX_j\right). \notag \end{align}

Assume that the physical alphabet is nonempty, that every \(\mu _j\) is nonzero, that \(0{\lt}N\), and that \(L\le N\). For every boundary-crossing interval beginning at \(i\), assume that the simultaneous products

\begin{align} w & \longmapsto (A^1_w,\ldots ,A^r_w), \notag \\ |w| & =N-i, \notag \end{align}

span the product algebra \(\prod _jM_{D_j}(\mathbb {C})\). Then there are matrices \(C^j_{i,\rho }\) such that, for every boundary-crossing interval \(i\), every outside word \(\rho \) of length \(N-L\), every block \(j\), and every word \(\beta \) before the cut, \(A^j_\beta C^j_{i,\rho } = ((\mu _j^NX_j)A^j_\beta )A^j_\rho \).

Proof

For a boundary-crossing interval beginning at \(i\) and an outside word \(\rho \), choose a configuration whose outside segment is \(\rho \). The local constraint on \(\psi \) gives a vector in \(\bigvee _jG_L(A_j)\) by Lemma 15.3.1.2. Theorem 15.3.1.8 applied with the stated span at length \(N-i\) gives the displayed \(C^j,D^j\) comparison for that interval and outside word.

Theorem 15.3.1.10 Fixed boundary-crossing component in the local block space

Assume \(0{\lt}N\), \(L\le N\), and let the cyclic interval beginning at \(i\) cross the boundary cut, so \(N{\lt}i+L\). Suppose that the block tensors have the common word-span property at length \(N-i\) and that

\begin{align} \sum _j R_{i,\tau }\! \left(\Gamma _N^{A_j}(\mu _j^NX_j)\right) & \in \bigvee _jG_L(A_j). \notag \end{align}

Then each block component of the fixed boundary-crossing restriction lies in its local block space:

\begin{align} R_{i,\tau }\! \left(\Gamma _N^{A_j}(\mu _j^NX_j)\right) & \in G_L(A_j). \notag \end{align}
Proof

Theorem 15.3.1.8 gives matrices \(C^j_{i,\tau }\) satisfying (??). Thus \(E=C^j_{i,\tau }\) satisfies the boundary matrix identity required in Lemma 15.3.1.11, and the displayed membership follows for every block \(j\).

Lemma 15.3.1.11 Boundary-crossing cyclic intervals from the boundary matrix identity

Assume \(0{\lt}N\), \(L\le N\), and let the cyclic interval beginning at \(i\) cross the boundary cut, so \(N{\lt}i+L\). Put \(a=i+L-N\). Suppose that there is a matrix \(E\) such that for every word \(\beta \) of length \(a\),

\begin{align} \mu _j^N X_j A^j_\beta A^j_{\tau _a}\cdots A^j_{\tau _{i-1}} & = A^j_\beta E. \label{eq:ph_boundary_crossing_matrix_identity} \end{align}

Then

\begin{align} R_{i,\tau }\! \left(\Gamma _N^{A_j}(\mu _j^NX_j)\right) & \in G_L(A_j). \notag \end{align}
Proof

Write a length-\(L\) word in the boundary-crossing interval as \(\alpha \beta \), where \(\alpha \) is the segment inside \(\{ i,\ldots ,N-1\} \) and \(\beta \) is the segment after the interval wraps past the cut, inside \(\{ 0,\ldots ,a-1\} \). The full \(N\)-site word appearing in the restriction is then \(\beta \, \tau _a\cdots \tau _{i-1} \alpha \). Trace cyclicity gives

\begin{align} \operatorname{tr}\! \left(A^j_\beta A^j_{\tau _a}\cdots A^j_{\tau _{i-1}} A^j_\alpha \mu _j^NX_j\right) & = \operatorname{tr}\! \left(A^j_\alpha \mu _j^NX_j A^j_\beta A^j_{\tau _a}\cdots A^j_{\tau _{i-1}}\right) = \operatorname{tr}\! \left(A^j_\alpha A^j_\beta E\right), \notag \end{align}

where the second equality follows from (??). The last expression is the coefficient of \(\Gamma _L^{A_j}(E)\) at the word \(\alpha \beta \). Hence the restricted vector lies in \(G_L(A_j)\).

Write \(N=M+1\) and let the local length be \(m+2\le M+1\). Consider the cyclic interval beginning at the last site \(M\). For an outside configuration \(\tau \), put

\begin{align} \rho & =\tau _{m+1}\cdots \tau _{M-1}, \notag \\ C^j_a & =A^j_\rho A^j_a(\mu _j^{M+1}X_j). \notag \end{align}

Then the \(j\)-th component of the last crossing-window restriction has the left-boundary trace form

\begin{align} R_{M,\tau }\! \left(\Gamma _{M+1}^{A_j}(\mu _j^{M+1}X_j)\right) & = \Phi ^j, \notag \\ \Phi ^j(a,w,b) & = \operatorname{tr}\! \left(A^j_b C^j_a A^j_w\right). \notag \end{align}
Proof

The cyclic interval beginning at \(M\) reads the last site first and then wraps to the sites \(0,\ldots ,m\). Thus, for the local word \(awb\), the full word in the trace is \(w\, b\, \rho \, a\). Hence the \(j\)-th coefficient is

\begin{align} \operatorname{tr}\! \left(A^j_wA^j_bA^j_\rho A^j_a(\mu _j^{M+1}X_j)\right) & = \operatorname{tr}\! \left(A^j_bA^j_\rho A^j_a(\mu _j^{M+1}X_j)A^j_w\right), \notag \end{align}

where the equality follows from trace cyclicity. This is the displayed left-boundary trace form.

Lemma 15.3.1.13 Last crossing-window trace form

Write \(N=M+1\) and let the local length be \(m+2\le M+1\). Consider the cyclic interval beginning at the last site \(M\). For an outside configuration \(\tau \), put

\begin{align} \rho & =\tau _{m+1}\cdots \tau _{M-1}, \notag \\ C^j_a & =A^j_\rho A^j_a(\mu _j^{M+1}X_j). \notag \end{align}

Then the block sum of the last crossing-window restrictions has the left-boundary trace form

\begin{align} \sum _j R_{M,\tau }\! \left(\Gamma _{M+1}^{A_j}(\mu _j^{M+1}X_j)\right) & = \sum _j \Phi ^j, \notag \\ \Phi ^j(a,w,b) & = \operatorname{tr}\! \left(A^j_b C^j_a A^j_w\right). \notag \end{align}
Proof

Sum Lemma 15.3.1.12 over the block index.

Lemma 15.3.1.14 Last crossing-window coefficient comparison

In the setting of Lemma 15.3.1.13, assume that the length-\(m\) simultaneous word tuples span the product algebra and that

\begin{align} \sum _j R_{M,\tau }\! \left(\Gamma _{M+1}^{A_j}(\mu _j^{M+1}X_j)\right) & \in \bigvee _j G_{m+2}(A^j). \notag \end{align}

Then there are matrices \(E_j\) such that, for every block \(j\) and all boundary letters \(a,b\),

\begin{align} A^j_b\bigl( A^j_\rho A^j_a(\mu _j^{M+1}X_j) \bigr) & = A^j_bE_jA^j_a. \label{eq:ph_last_crossing_coefficient_comparison} \end{align}
Proof

Lemma 15.3.1.13 writes the local block sum in the form

\begin{align} \Phi (a,w,b) & = \sum _j\operatorname{tr}\! \left(A^j_bC^j_aA^j_w\right), \notag \\ C^j_a & = A^j_\rho A^j_a(\mu _j^{M+1}X_j). \notag \end{align}

Applying Lemma 15.2.1.3 gives (??).

In the setting of Lemma 15.3.1.13, assume that the length-\(m\) simultaneous word tuples span the product algebra and that

\begin{align} \sum _j R_{M,\tau }\! \left(\Gamma _{M+1}^{A_j}(\mu _j^{M+1}X_j)\right) & \in \bigvee _j G_{m+2}(A^j). \notag \end{align}

Then, for every block \(j\),

\begin{align} R_{M,\tau }\! \left(\Gamma _{M+1}^{A_j}(\mu _j^{M+1}X_j)\right) & \in G_{m+2}(A^j). \notag \end{align}
Proof

Lemma 15.3.1.14 gives matrices \(E_j\) satisfying (??). By Lemma 15.3.1.12, the \(j\)-th cyclic restriction is the left-boundary component determined by \(C^j_a\). The displayed identity makes that component equal to \(\Gamma _{m+2}^{A_j}(E_j)\), hence it lies in \(G_{m+2}(A^j)\).

Lemma 15.3.1.16 Last crossing-window coefficients from a block-diagonal chain vector

Let \(B=\bigoplus _j\mu _jA^j\), and assume \(\mu _j\ne 0\) for every block \(j\). Suppose

\begin{align} \psi & = \Gamma _{M+1}^{B}\! \left(\bigoplus _jX_j\right), \notag \\ \psi & \in \mathcal G_{M+1,m+2}(B). \notag \end{align}

Assume also that the length-\(m\) simultaneous word tuples span the product algebra. Then, for every outside configuration \(\tau \), with \(\rho =\tau _{m+1}\cdots \tau _{M-1}\), there are matrices \(E_j\) such that \(A^j_b(A^j_\rho A^j_a(\mu _j^{M+1}X_j)) = A^j_bE_jA^j_a\) for every block \(j\) and all boundary letters \(a,b\).

Proof

The local constraint of \(\psi \) at the cyclic interval beginning at \(M\) gives

\begin{align} \sum _j R_{M,\tau }\! \left(\Gamma _{M+1}^{A_j}(\mu _j^{M+1}X_j)\right) & \in \bigvee _j G_{m+2}(A^j) \notag \end{align}

by Lemma 15.3.1.2. Applying Lemma 15.3.1.14 gives the displayed equations.

Lemma 15.3.1.17 Last crossing-window component membership from a chain vector

Let \(B=\bigoplus _j\mu _jA^j\), and assume \(\mu _j\ne 0\) for every block \(j\). Suppose

\begin{align} \psi & = \Gamma _{M+1}^{B}\! \left(\bigoplus _jX_j\right), \notag \\ \psi & \in \mathcal G_{M+1,m+2}(B). \notag \end{align}

If the length-\(m\) simultaneous word tuples span the product algebra, then for every outside configuration \(\tau \) and every block \(j\),

\begin{align} R_{M,\tau }\! \left(\Gamma _{M+1}^{A_j}(\mu _j^{M+1}X_j)\right) & \in G_{m+2}(A^j). \notag \end{align}
Proof

Lemma 15.3.1.2, applied to the window beginning at the last site, gives

\begin{align} \sum _j R_{M,\tau }\! \left(\Gamma _{M+1}^{A_j}(\mu _j^{M+1}X_j)\right) & \in \bigvee _j G_{m+2}(A^j). \notag \end{align}

Lemma 15.3.1.15 then applies to each block component.

Lemma 15.3.1.18 Boundary-crossing intervals from a right boundary-matrix identity

Assume \(0{\lt}N\) and \(L\le N\), and let the cyclic interval beginning at \(i\) cross the boundary cut, so \(N{\lt}i+L\). Put \(a=i+L-N\). Suppose that there is a matrix \(Y\) such that, for every word \(\beta \) of length \(a\), \(\mu _j^N X_j A^j_\beta = A^j_\beta Y\). Then

\begin{align} R_{i,\tau }\! \left(\Gamma _N^{A_j}(\mu _j^NX_j)\right) & \in G_L(A_j). \notag \end{align}
Proof

Let \(E:=Y A^j_{\tau _a}\cdots A^j_{\tau _{i-1}}\). Then, for every word \(\beta \) on the segment \(0,\ldots ,a-1\),

\begin{align} \mu _j^N X_j A^j_\beta A^j_{\tau _a}\cdots A^j_{\tau _{i-1}} & = A^j_\beta E. \notag \end{align}

Lemma 15.3.1.11 gives the stated local constraint.

Lemma 15.3.1.19 Boundary-crossing cyclic intervals suffice

Assume \(0{\lt}N\) and \(L\le N\). Fix block-diagonal boundary conditions \(X_j\). Suppose that, for every block \(j\), every cyclic interval beginning at \(i\) with \(N{\lt}i+L\), and every configuration \(\tau \),

\begin{align} R_{i,\tau }\! \left(\Gamma _N^{A_j}(\mu _j^NX_j)\right) & \in G_L(A_j). \notag \end{align}

Then, for every block \(j\), \(\Gamma _N^{A_j}(\mu _j^NX_j) \in \mathcal G_{N,L}(A_j)\).

Proof

By definition, membership in \(\mathcal G_{N,L}(A_j)\) is the collection of all cyclic length-\(L\) local constraints. For an interval beginning at \(i\), either \(i+L\le N\) or \(N{\lt}i+L\). In the first case, use Lemma 15.3.1.3. In the second case, use the displayed hypothesis.

Lemma 15.3.1.20 Right boundary-matrix identities give single-block periodic-boundary constraints

Assume \(0{\lt}N\) and \(L\le N\). Fix block-diagonal boundary conditions \(X_j\). Suppose that, for every block \(j\), every boundary-crossing interval beginning at \(i\), and every configuration \(\tau \), there is a matrix \(Y_{j,i,\tau }\) such that, for every word \(\beta \) of length \(a=i+L-N\), \(\mu _j^N X_j A^j_\beta = A^j_\beta Y_{j,i,\tau }\). Then, for every block \(j\), \(\Gamma _N^{A_j}(\mu _j^NX_j) \in \mathcal G_{N,L}(A_j)\).

Proof

Lemma 15.3.1.18 turns the displayed boundary identity into the local constraint for each boundary-crossing interval. Lemma 15.3.1.19 then yields \(\Gamma _N^{A_j}(\mu _j^NX_j) \in \mathcal G_{N,L}(A_j)\), handling both boundary-crossing intervals \(N{\lt}i+L\) and non-crossing intervals \(i+L\le N\).

Lemma 15.3.1.21 Spanning matrix identities give single-block constraints

Assume \(0{\lt}N\) and \(L\le N\), and assume that, for every block \(j\), the length-\(N-L\) word products of \(A_j\) span the full matrix algebra. Suppose that, for every block \(j\), every boundary-crossing interval beginning at \(i\), and every outside word \(\rho \) of length \(N-L\), there is a matrix \(E_{j,i,\rho }\) such that, for every word \(\beta \) before the cut of length \(i+L-N\), \(\mu _j^N X_j A^j_\beta A^j_\rho = A^j_\beta E_{j,i,\rho }\). Then, for every block \(j\), \(\Gamma _N^{A_j}(\mu _j^NX_j) \in \mathcal G_{N,L}(A_j)\).

Proof

For fixed \(j,i\), apply Theorem 13.7.2 to the family

\begin{align} Z_\beta & = \mu _j^NX_jA^j_\beta , \notag \\ F_\beta & = A^j_\beta . \notag \end{align}

Since the length-\(N-L\) outside-word products span the full matrix algebra, there is a single matrix \(Y_{j,i}\) such that, for every \(\beta \), \(\mu _j^NX_jA^j_\beta = A^j_\beta Y_{j,i}\). Lemma 15.3.1.20 then gives the periodic-boundary constraints for each block.

Lemma 15.3.1.22 Matrix identities under block injectivity give single-block constraints

Assume \(0{\lt}N\), \(L\le N\), and \(L+L_0\le N\). Suppose each block \(A_j\) is \(L_0\)-block-injective and satisfies the normalization equation \(\sum _a A^j_aA^{j\dagger }_a=\mathbb {1}\). Suppose also that the boundary-crossing matrix identities of Lemma 15.3.1.21 hold for every outside word \(\rho \) of length \(N-L\). Then, for every block \(j\), \(\Gamma _N^{A_j}(\mu _j^NX_j) \in \mathcal G_{N,L}(A_j)\).

Proof

Since \(L+L_0\le N\), the complementary length satisfies \(L_0\le N-L\). Theorem 8.8.1 therefore gives, for every block \(j\), \(S_{N-L}(A_j) = M_{D_j}(\mathbb {C})\). Apply Lemma 15.3.1.21.

Lemma 15.3.1.23 Normalization for words

If \(\sum _a A_aA_a^\dagger =\mathbb {1}\), then, for every length \(M\), \(\sum _\rho A_\rho A_\rho ^\dagger = \mathbb {1}\), where \(\rho \) ranges over all words of length \(M\).

Proof

The case \(M=0\) is the identity matrix. For the induction step, write a word of length \(M+1\) as \(a\rho \). Since \(A_{a\rho }=A_aA_\rho \),

\begin{align} \sum _{a,\rho } A_aA_\rho A_\rho ^\dagger A_a^\dagger & = \sum _a A_a \left(\sum _\rho A_\rho A_\rho ^\dagger \right) A_a^\dagger = \sum _a A_aA_a^\dagger = \mathbb {1}. \notag \end{align}
Lemma 15.3.1.24 Compatibility hypotheses under block injectivity give single-block constraints

Assume \(0{\lt}N\), \(L\le N\), and \(L+L_0\le N\). Suppose each block \(A_j\) is \(L_0\)-block-injective and satisfies \(\sum _a A^j_aA^{j\dagger }_a=\mathbb {1}\). Fix block-diagonal boundary conditions \(X_j\). For every boundary-crossing interval beginning at \(i\), suppose there are matrices \(C^j_{i,\rho }\), indexed by outside words \(\rho \) of length \(N-L\), such that, for every word \(\beta \) before the cut of length \(i+L-N\), \(A^j_\beta C^j_{i,\rho } = ((\mu _j^NX_j)A^j_\beta )A^j_\rho \). After opening the periodic boundary at the chosen cut, this is the source comparison

\begin{align} A^j_{i_{m+1}}C^j_{i_1} & = D^j_{i_{m+1}}A^j_{i_1}, \notag \\ D^j_\beta & = (\mu _j^NX_j)A^j_\beta . \notag \end{align}

Then, for every block \(j\), \(\Gamma _N^{A_j}(\mu _j^NX_j) \in \mathcal G_{N,L}(A_j)\).

Proof

The normalization equation gives, by induction on word length, \(\sum _\rho A^j_\rho A^{j\dagger }_\rho = \mathbb {1}\) for words \(\rho \) of length \(N-L\). Applying Lemma 15.2.1.10 with \(X=\mu _j^NX_j\) gives, for every outside word \(\rho \), a matrix \(E_{j,i,\rho }\) such that, for every word \(\beta \) before the cut, \(((\mu _j^NX_j)A^j_\beta )A^j_\rho = A^j_\beta E_{j,i,\rho }\). Lemma 15.3.1.22 then gives the periodic-boundary constraint for each block.

Lemma 15.3.1.25 Boundary representation and crossing constraints give periodic components

Assume \(0{\lt}N\), \(L\le N\), and \(L+L_0\le N\). Suppose each block \(A_j\) is \(L_0\)-block-injective and satisfies \(\sum _a A^j_aA^{j\dagger }_a=\mathbb {1}\). Let

\begin{align} \psi & \in \mathcal G_{N,L}\! \left(\bigoplus _j\mu _jA_j\right), \notag \\ \psi & = \Gamma _N^{\oplus _j\mu _jA_j}\! \left(\bigoplus _jX_j\right). \notag \end{align}

If every boundary-crossing interval has the corresponding tail-word spanning property, then there are block-diagonal boundary conditions \(X_j\) with the displayed representation of \(\psi \) and, for every \(j\), \(\Gamma _N^{A_j}(\mu _j^NX_j) \in \mathcal G_{N,L}(A_j)\).

Proof

Theorem 15.3.1.9 supplies matrices \(C^j_{i,\rho }\) satisfying \(A^j_\beta C^j_{i,\rho } = ((\mu _j^NX_j)A^j_\beta )A^j_\rho \). Lemma 15.3.1.24 then gives the single-block periodic constraints.

Lemma 15.3.1.26 Boundary comparison implies trace comparison

Fix block-diagonal boundary matrices \(X_j\). Suppose the opened-boundary matrices \(C^j_{i,\rho }\) satisfy, for every boundary-crossing interval \(i\), every outside word \(\rho \), and every word \(\beta \) before the cut, \(A^j_\beta C^j_{i,\rho } = ((\mu _j^NX_j)A^j_\beta )A^j_\rho \). Then, for every middle word \(w\),

\begin{align} \sum _j\operatorname{tr}\! \left(A^j_\beta C^j_{i,\rho }A^j_w\right) & = \sum _j \operatorname{tr}\! \left(((\mu _j^NX_j)A^j_\beta )A^j_\rho A^j_w\right). \notag \end{align}
Proof

Substituting the displayed comparison identity in each summand gives

\begin{align} \operatorname{tr}\! \left(A^j_\beta C^j_{i,\rho }A^j_w\right) & = \operatorname{tr}\! \left(((\mu _j^NX_j)A^j_\beta )A^j_\rho A^j_w\right). \notag \end{align}

Summing over \(j\) gives the conclusion.

Lemma 15.3.1.27 Boundary trace comparisons give single-block constraints

Assume \(0{\lt}N\), \(L\le N\), and \(L+L_0\le N\). Suppose each block \(A_j\) is \(L_0\)-block-injective and satisfies \(\sum _a A^j_aA^{j\dagger }_a=\mathbb {1}\). Assume also that the simultaneous tuples \(w\mapsto (A^1_w,\ldots ,A^g_w)\) at the middle-word length \(m\) span the full product algebra \(\prod _jM_{D_j}(\mathbb {C})\). Fix complex numbers \(\mu _j\) and block-diagonal boundary conditions \(X_j\). For every boundary-crossing interval beginning at \(i\), suppose there are matrices \(C^j_{i,\rho }\), indexed by outside words \(\rho \) of length \(N-L\), such that, for every word \(\beta \) before the cut, outside word \(\rho \), and middle word \(w\),

\begin{align} \sum _j\operatorname{tr}\! \left(A^j_\beta C^j_{i,\rho }A^j_w\right) & = \sum _j \operatorname{tr}\! \left(((\mu _j^NX_j)A^j_\beta )A^j_\rho A^j_w\right). \notag \end{align}

Then, for every block \(j\), \(\Gamma _N^{A_j}(\mu _j^NX_j) \in \mathcal G_{N,L}(A_j)\).

Proof

Applying Lemma 15.2.1.9 with the same scalars \(\mu _j\) and boundary matrices \(X_j\), for every block \(j\) and every boundary-crossing interval \(i\), gives a matrix \(E^j_i\) such that, for every word \(\beta \) before the cut, \((\mu _j^NX_j)A^j_\beta = A^j_\beta E^j_i\). Hence, for every outside word \(\rho \), taking \(E_{j,i,\rho }=E^j_iA^j_\rho \) gives \(((\mu _j^NX_j)A^j_\beta )A^j_\rho = A^j_\beta E_{j,i,\rho }\). These are precisely the block-diagonal boundary-condition equations of Lemma 15.3.1.22, which gives the periodic-boundary constraint for each block.

Lemma 15.3.1.28 Boundary trace comparisons with a block-diagonal representation

Assume \(0{\lt}N\), \(L\le N\), and \(L+L_0\le N\). Suppose each block \(A_j\) is \(L_0\)-block-injective and satisfies \(\sum _a A^j_aA^{j\dagger }_a=\mathbb {1}\). Assume also that the simultaneous tuples \(w\mapsto (A^1_w,\ldots ,A^g_w)\) at the middle-word length \(m\) span the full product algebra \(\prod _jM_{D_j}(\mathbb {C})\). Let

\begin{align} \psi & = \Gamma _N^{\oplus _j\mu _jA_j}\! \left(\bigoplus _jX_j\right) \notag \end{align}

for some block-diagonal boundary conditions \(X_j\). Suppose that whenever \(\psi \) has such a representation, there are matrices \(C^j_{i,\rho }\) such that, for every boundary-crossing interval \(i\), every outside word \(\rho \) of length \(N-L\), every word \(\beta \) of length \(i+L-N\) before the cut, and every word \(w\) of length \(m\),

\begin{align} \sum _j\operatorname{tr}\! \left(A^j_\beta C^j_{i,\rho }A^j_w\right) & = \sum _j \operatorname{tr}\! \left(((\mu _j^NX_j)A^j_\beta )A^j_\rho A^j_w\right). \notag \end{align}

Then there are block-diagonal boundary conditions \(X_j\) with

\begin{align} \psi & = \Gamma _N^{\oplus _j\mu _jA_j}\! \left(\bigoplus _jX_j\right) \notag \end{align}

and, for every \(j\), \(\Gamma _N^{A_j}(\mu _j^NX_j) \in \mathcal G_{N,L}(A_j)\).

Proof

Choose a block-diagonal boundary representation of \(\psi \). For that representation, the boundary trace comparison supplies the matrices \(C^j_{i,\rho }\). Applying Lemma 15.3.1.27 gives the desired containment \(\Gamma _N^{A_j}(\mu _j^NX_j)\in \mathcal G_{N,L}(A_j)\) for every block \(j\).

Lemma 15.3.1.29 \(C^j\) boundary comparison with a block-diagonal representation

Assume \(0{\lt}N\), \(L\le N\), and \(L+L_0\le N\). Suppose each block \(A_j\) is \(L_0\)-block-injective and satisfies \(\sum _a A^j_aA^{j\dagger }_a=\mathbb {1}\). Assume also that the simultaneous tuples \(w\mapsto (A^1_w,\ldots ,A^g_w)\) at the middle-word length \(m\) span the full product algebra \(\prod _jM_{D_j}(\mathbb {C})\). Let \(\psi =\Gamma _N^{\oplus _j\mu _jA_j}(\bigoplus _jX_j)\) for some block-diagonal boundary conditions \(X_j\). Suppose that whenever \(\psi \) has such a representation, there are matrices \(C^j_{i,\rho }\) such that, for every boundary-crossing interval \(i\), every outside word \(\rho \) of length \(N-L\), and every word \(\beta \) of length \(i+L-N\) before the cut, \(A^j_\beta C^j_{i,\rho } = ((\mu _j^NX_j)A^j_\beta )A^j_\rho \). Then there are block-diagonal boundary conditions \(X_j\) such that the displayed representation holds and, for every \(j\), the displayed component lies in the corresponding single-block ground space:

\begin{align} \psi & = \Gamma _N^{\oplus _j\mu _jA_j}\! \left(\bigoplus _jX_j\right), \notag \\ \Gamma _N^{A_j}(\mu _j^NX_j) & \in \mathcal G_{N,L}(A_j). \notag \end{align}
Proof

Choose a block-diagonal boundary representation of \(\psi \). For that representation, Lemma 15.3.1.26 gives, for every boundary-crossing interval \(i\), outside word \(\rho \), word \(\beta \) before the cut, and middle word \(w\),

\begin{align} \sum _j\operatorname{tr}\! \left(A^j_\beta C^j_{i,\rho }A^j_w\right) & = \sum _j\operatorname{tr}\! \left(((\mu _j^NX_j)A^j_\beta )A^j_\rho A^j_w\right). \notag \end{align}

Applying Lemma 15.3.1.28 to these trace comparisons gives \(\Gamma _N^{A_j}(\mu _j^NX_j)\in \mathcal G_{N,L}(A_j)\) for every \(j\).

Lemma 15.3.1.30 Boundary trace comparisons give periodic-boundary single-block states

With the hypotheses and notation of Lemma 15.3.10, assume in addition that \(L+L_0\le N\). Assume also that, for some middle-word length \(m\), the simultaneous tuples \(w\mapsto (A^1_w,\ldots ,A^g_w)\), where \(w\) ranges over words of length \(m\) in \(\{ 0,\ldots ,d{-}1\} \), span the full product algebra \(\prod _jM_{D_j}(\mathbb {C})\). Let \(\psi \in \mathcal G_{N,L}(\bigoplus _j\mu _jA_j)\). Suppose that whenever \(\psi =\Gamma _N^{\oplus _j\mu _jA_j}(\bigoplus _jX_j)\), there are matrices \(C^j_{i,\rho }\) such that, for every boundary-crossing interval \(i\), every outside word \(\rho \) of length \(N-L\), every word \(\beta \) before the cut, and every word \(w\) of length \(m\),

\begin{align} \sum _j\operatorname{tr}\! \left(A^j_\beta C^j_{i,\rho }A^j_w\right) & = \sum _j\operatorname{tr}\! \left(((\mu _j^NX_j)A^j_\beta )A^j_\rho A^j_w\right). \notag \end{align}

Then there are block-diagonal boundary conditions \(X_j\) such that the displayed representation holds and, for every \(j\), the displayed component lies in the corresponding single-block ground space:

\begin{align} \psi & = \Gamma _N^{\oplus _j\mu _jA_j}\! \left(\bigoplus _jX_j\right), \notag \\ \Gamma _N^{A_j}(\mu _j^NX_j) & \in \mathcal G_{N,L}(A_j). \notag \end{align}
Proof

By Lemma 15.3.10, there are matrices \(X_j\) with \(\psi =\Gamma _N^{\oplus _j\mu _jA_j}(\bigoplus _jX_j)\). Lemma 15.3.1.28 applies to this representation and the assumed equality of boundary trace expressions. Thus, \(\Gamma _N^{A_j}(\mu _j^NX_j)\in \mathcal G_{N,L}(A_j)\) for every \(j\).

Lemma 15.3.1.31 Basis-of-normal-tensors product span gives boundary-trace periodic components

With the hypotheses and notation of Lemma 15.3.10, assume in addition that \(L+L_0\le N\) and \(L \ge (L_0+1) +(g-1)((L_0+1)+((L_0+1)+(L_0+1)))+1\). Put \(m = (L_0+1) +(g-1)((L_0+1)+((L_0+1)+(L_0+1)))\). Let \(\psi \in \mathcal G_{N,L}(\bigoplus _j\mu _jA_j)\). Suppose that whenever \(\psi =\Gamma _N^{\oplus _j\mu _jA_j}(\bigoplus _jX_j)\), there are matrices \(C^j_{i,\rho }\) such that, for every boundary-crossing interval \(i\), every outside word \(\rho \) of length \(N-L\), every word \(\beta \) before the cut, and every word \(w\) of length \(m\),

\begin{align} \sum _j\operatorname{tr}\! \left(A^j_\beta C^j_{i,\rho }A^j_w\right) & = \sum _j\operatorname{tr}\! \left(((\mu _j^NX_j)A^j_\beta )A^j_\rho A^j_w\right). \notag \end{align}

Then there are block-diagonal boundary conditions \(X_j\) such that the displayed representation holds and, for every \(j\), the displayed component lies in the corresponding single-block ground space:

\begin{align} \psi & = \Gamma _N^{\oplus _j\mu _jA_j}\! \left(\bigoplus _jX_j\right), \notag \\ \Gamma _N^{A_j}(\mu _j^NX_j) & \in \mathcal G_{N,L}(A_j). \notag \end{align}
Proof

Lemma 15.2.2.5 gives \(\operatorname{span}\{ (A^1_w,\ldots ,A^g_w):|w|=m\} = \prod _jM_{D_j}(\mathbb {C})\). Applying Lemma 15.3.1.30 with this middle-word span gives the stated block-diagonal boundary conditions and the periodic-boundary constraint in each block.

Lemma 15.3.1.32 \(C^j\) boundary comparison gives periodic single-block states

With the hypotheses and notation of Lemma 15.3.10, assume in addition that \(L+L_0\le N\). Let \(\psi \in \mathcal G_{N,L}(\bigoplus _j\mu _jA_j)\). Suppose that whenever \(\psi =\Gamma _N^{\oplus _j\mu _jA_j}(\bigoplus _jX_j)\), there are matrices \(C^j_{i,\rho }\) such that, for every boundary-crossing interval \(i\), every outside word \(\rho \) of length \(N-L\), and every word \(\beta \) before the cut, \(A^j_\beta C^j_{i,\rho } = ((\mu _j^NX_j)A^j_\beta )A^j_\rho \). Then there are block-diagonal boundary conditions \(X_j\) such that the displayed representation holds and, for every \(j\), the displayed component lies in the corresponding single-block ground space:

\begin{align} \psi & = \Gamma _N^{\oplus _j\mu _jA_j}\! \left(\bigoplus _jX_j\right), \notag \\ \Gamma _N^{A_j}(\mu _j^NX_j) & \in \mathcal G_{N,L}(A_j). \notag \end{align}
Proof

By Lemma 15.3.10, there are matrices \(X_j\) with \(\psi =\Gamma _N^{\oplus _j\mu _jA_j}(\bigoplus _jX_j)\). For these boundary conditions, the assumed \(C^j\) boundary comparison supplies the matrices \(C^j_{i,\rho }\). Lemma 15.3.1.24 gives \(\Gamma _N^{A_j}(\mu _j^NX_j)\in \mathcal G_{N,L}(A_j)\) for every \(j\).

Theorem 15.3.1.33 Crossing-tail spans give periodic single-block states

With the hypotheses and notation of Lemma 15.3.10, assume in addition that \(L+L_0\le N\) and that every boundary-crossing interval beginning at \(i\) has the simultaneous tail-word spanning property: the products

\begin{align} w & \longmapsto (A^1_w,\ldots ,A^g_w), \notag \\ |w| & =N-i, \notag \end{align}

span the product algebra \(\prod _jM_{D_j}(\mathbb {C})\). Let \(\psi \in \mathcal G_{N,L}(\bigoplus _j\mu _jA_j)\). Then there are block-diagonal boundary conditions \(X_j\) such that the displayed representation holds and, for every \(j\), the displayed component lies in the corresponding single-block ground space:

\begin{align} \psi & = \Gamma _N^{\oplus _j\mu _jA_j}\! \left(\bigoplus _jX_j\right), \notag \\ \Gamma _N^{A_j}(\mu _j^NX_j) & \in \mathcal G_{N,L}(A_j). \notag \end{align}
Proof

Lemma 15.3.10 supplies block-diagonal boundary conditions \(X_j\), and Lemma 15.3.1.25, applied with the assumed tail-word spans, places each \(\Gamma _N^{A_j}(\mu _j^NX_j)\) in \(\mathcal G_{N,L}(A_j)\).

Theorem 15.3.1.34 Short crossing-tail spans give periodic single-block states

With the hypotheses and notation of Lemma 15.3.10, assume in addition that \(L+L_0\le N\). Assume also that, for every boundary-crossing interval beginning at \(i\) whose tail length lies below the block-separation bound, \(N-i {\lt} (L_0+1) +(g-1)((L_0+1)+((L_0+1)+(L_0+1)))\), the simultaneous products

\begin{align} w & \longmapsto (A^1_w,\ldots ,A^g_w), \notag \\ |w| & =N-i, \notag \end{align}

span the product algebra \(\prod _jM_{D_j}(\mathbb {C})\). Let \(\psi \in \mathcal G_{N,L}(\bigoplus _j\mu _jA_j)\). Then there are block-diagonal boundary conditions \(X_j\) such that the displayed representation holds and, for every \(j\), the displayed component lies in the corresponding single-block ground space:

\begin{align} \psi & = \Gamma _N^{\oplus _j\mu _jA_j}\! \left(\bigoplus _jX_j\right), \notag \\ \Gamma _N^{A_j}(\mu _j^NX_j) & \in \mathcal G_{N,L}(A_j). \notag \end{align}

This derives the boundary-condition comparison of [ PGVWC07 , Theorem 12, proof lines 1446–1456 ] at every crossing tail whose length reaches the displayed block-separation bound.

Proof

For a boundary-crossing interval beginning at \(i\), either \(N-i {\lt} (L_0+1) +(g-1)((L_0+1)+((L_0+1)+(L_0+1)))\) or \(N-i \ge (L_0+1) +(g-1)((L_0+1)+((L_0+1)+(L_0+1)))\). In the first case, the assumed short-tail span applies; in the second case, Lemma 15.2.2.5 gives the simultaneous product span at length \(N-i\). Hence every boundary-crossing interval has the tail-word spanning property, and Theorem 15.3.1.33 gives the block-diagonal boundary conditions and \(\Gamma _N^{A_j}(\mu _j^NX_j)\in \mathcal G_{N,L}(A_j)\) for every \(j\).

Definition 15.3.1.35 Cyclic translation of configurations
#

For a periodic configuration \(\sigma \) on \(N\) sites and \(s\in \mathbb {Z}/N\mathbb {Z}\), define \((T_s\sigma )(k)=\sigma (k+s)\).

Definition 15.3.1.36 Cyclic translation of states
#

For a state \(\psi \) on a periodic chain, define \((T_s\psi )(\sigma )=\psi (T_s\sigma )\).

Lemma 15.3.1.37 Cyclic translation preserves the periodic chain space

If \(\psi \in \mathcal G_{N,L}(A)\), then \(T_s\psi \in \mathcal G_{N,L}(A)\) for every cyclic translation \(s\).

Proof

A length-\(L\) interval of \(T_s\psi \) beginning at \(i\) is the corresponding interval of \(\psi \) beginning at \(i-s\), with the outside configuration translated by \(s\).

Lemma 15.3.1.38 Translation exchanges the two words at a cut

If \(\beta \) has length \(K\) and \(\alpha \) has positive length \(S\), then \(T_K(\beta \alpha )=\alpha \beta \) as configurations on the periodic chain of length \(K+S\).

Proof

This is addition modulo \(K+S\), separated according to whether the site lies before or after the cut.

Lemma 15.3.1.39 Comparison across a global cut

Let \(X_j,Y_j\in M_{D_j}(\mathbb {C})\), and suppose

\begin{align} T_K\! \left(\sum _j\Gamma _{K+S}^{A_j}(X_j)\right) & = \sum _j\Gamma _{K+S}^{A_j}(Y_j). \notag \end{align}

If the simultaneous block words of length \(K\) span \(\prod _jM_{D_j}(\mathbb {C})\), then, for every block \(j\) and every word \(\alpha \) of length \(S\), \(X_jA^j_\alpha = A^j_\alpha Y_j\).

Proof

Evaluate the translated equality on the word \(\beta \alpha \), where \(|\beta |=K\). Cyclicity of the trace gives

\begin{align} \sum _j\operatorname{tr}(X_jA^j_\alpha A^j_\beta ) & = \sum _j\operatorname{tr}(A^j_\alpha Y_jA^j_\beta ). \notag \end{align}

The simultaneous spanning hypothesis separates the blocks and gives the asserted matrix identities.

With the hypotheses and notation of Lemma 15.3.11, assume in addition that \(L+L_0\le N\). Thus

\begin{align} g & \ge 2, \notag \\ L & \ge 3(g-1)(L_0+1)+1, \notag \\ N & \ge L+L_0. \notag \end{align}

Let \(\psi \in \mathcal G_{N,L}(\bigoplus _j\mu _jA_j)\). Then there are block-diagonal boundary conditions \(X_j\) such that the displayed representation holds and, for every \(j\), the displayed component lies in the corresponding single-block ground space:

\begin{align} \psi & = \Gamma _N^{\oplus _j\mu _jA_j}\! \left(\bigoplus _jX_j\right), \notag \\ \Gamma _N^{A_j}(\mu _j^NX_j) & \in \mathcal G_{N,L}(A_j). \notag \end{align}

No simultaneous spanning assumption is imposed at the short crossing tails.

Proof

Lemma 15.3.11 gives boundary matrices \(X_j\). Translate the periodic state so that the cut moves past the final \(L_0\) sites. By Lemma 15.3.1.37, the translated state remains in the periodic chain space, so applying Lemma 15.3.11 again yields boundary matrices \(Y_j\) for the translated state. For a word \(\alpha \) of length \(L_0\), comparison of the two decompositions against every complementary word \(w\) of length \(N-L_0\) gives

\begin{align} \sum _j\operatorname{tr}\! \left((\mu _j^NX_j)A^j_\alpha A^j_w\right) & = \sum _j\operatorname{tr}\! \left(A^j_\alpha (\mu _j^NY_j)A^j_w\right). \notag \end{align}

The length bound and Lemma 15.2.2.6 imply that the simultaneous products of length \(N-L_0\) span the product algebra. Therefore, Lemma 15.3.1.39 gives, block by block, \((\mu _j^NX_j)A^j_\alpha = A^j_\alpha (\mu _j^NY_j)\). Since the words of length \(L_0\) span each block matrix algebra, evaluation at the identity gives

\begin{align} \mu _j^NX_j & = \mu _j^NY_j. \label{eq:ph_boundary_crossing_scaled} \end{align}

Thus (??) says that \(\mu _j^NX_j\) commutes with every word of length \(L_0\), and consequently with every one-site matrix. The boundary condition may then be moved across every interval crossing the cut, which places each block component in its periodic chain space. This is the change-of-cut argument of [ PGVWC07 , Theorem 12, proof lines 1276–1289 and 1454–1456 ] .

Theorem 15.3.1.41 Periodic block decomposition from a global cut

Let \(g\ge 2\). Let \(A_j\) be irreducible left-canonical blocks with normalized self-overlap, pairwise inequivalent up to gauge and phase, injective at a common length \(L_0{\gt}0\), and satisfying \(\sum _aA^j_aA^{j\, \dagger }_a=\mathbb {1}\). Let \(\mu _j\ne 0\) for every \(j\), and assume

\begin{align} L & \ge 3(g-1)(L_0+1)+1, \notag \\ N & \ge L+L_0. \notag \end{align}

Then

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _j\mu _jA_j\right) & = \sum _j\mathcal G_{N,L}(A_j), \notag \end{align}

as asserted in [ PGVWC07 , Theorem 12, proof lines 1424–1456 ] .

Proof

This is the equality component of Theorem 15.3.1.42.

Under the same hypotheses,

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _j\mu _jA_j\right) & = \sum _j\mathcal G_{N,L}(A_j). \notag \end{align}

The open-boundary spaces \(G_N(A_j)\) also form an internal sum.

Proof

Theorem 15.3.1.40 gives the forward inclusion in the displayed equality. Every periodic single-block vector is a periodic vector for the block-diagonal tensor, which gives the reverse inclusion. For independence, write \(\phi _j=\Gamma _N^{A^j}(X_j)\) and suppose that \(\sum _j\phi _j=0\). Lemma 15.2.2.6 and nondegeneracy of the trace pairing give

\begin{align} \sum _j\Gamma _N^{A^j}(X_j)=0 & \quad \Longrightarrow \quad \left(\forall (M_j)_j, \sum _j\operatorname{tr}(X_jM_j)=0\right) \quad \Longrightarrow \quad \forall j,\ X_j=0. \notag \end{align}

The last implication follows by choosing a tuple supported in one block. Thus every \(\phi _j\) vanishes, and the open-boundary spaces form an internal sum.

Under the same hypotheses, suppose additionally that every periodic single-block chain space is contained in its matrix-product-vector line. Then

\begin{align} \ker H_L^{(N)}\! \left(\bigoplus _j\mu _jA_j\right) & \subseteq \operatorname{span}\{ V^{(N)}(A_j):1\le j\le g\} . \notag \end{align}
Proof

The parent-Hamiltonian kernel is the periodic chain space. Apply the preceding block decomposition and then the assumed single-block containments.

Lemma 15.3.1.44 Matrix identities give periodic single-block states

With the hypotheses and notation of Lemma 15.3.10, assume in addition that \(L+L_0\le N\). Let \(\psi \in \mathcal G_{N,L}(\bigoplus _j\mu _jA_j)\). Suppose that whenever \(\psi =\Gamma _N^{\oplus _j\mu _jA_j}(\bigoplus _jX_j)\), the corresponding boundary conditions satisfy the block-diagonal boundary-condition equations: for every boundary-crossing interval \(i\), every word \(\beta \) before the cut, and every outside word \(\rho \) of length \(N-L\), \(\mu _j^N X_jA^j_\beta A^j_\rho = A^j_\beta E_{j,i,\rho }\). Then there are block-diagonal boundary conditions \(X_j\) such that the displayed representation holds and, for every \(j\), the displayed component lies in the corresponding single-block ground space:

\begin{align} \psi & = \Gamma _N^{\oplus _j\mu _jA_j}\! \left(\bigoplus _jX_j\right), \notag \\ \Gamma _N^{A_j}(\mu _j^NX_j) & \in \mathcal G_{N,L}(A_j). \notag \end{align}
Proof

By Lemma 15.3.10, there are matrices \(X_j\) with \(\psi =\Gamma _N^{\oplus _j\mu _jA_j}(\bigoplus _jX_j)\). These \(X_j\) satisfy the assumed identities, so Lemma 15.3.1.22 gives \(\Gamma _N^{A_j}(\mu _j^NX_j)\in \mathcal G_{N,L}(A_j)\) for every \(j\).

15.3.2 Periodic-boundary ground space for a block-diagonal tensor

The periodic ground-space theorem states that, for sufficiently long periodic systems, the periodic-boundary ground space of a parent Hamiltonian is generated by a basis of normal tensors of the initial tensor \(A\)  [ CPGSV21 , Theorem IV.16 ] . The lemmas in this subsection are conditional reductions toward that statement. Their hypotheses explicitly include a boundary inclusion, a block-diagonal boundary representation, or periodic-boundary membership for the single-block states; in the source theorem these conditions are obtained from the boundary-condition comparison in the periodic ring. The source proof [ PGVWC07 ] writes this comparison with \(C^j_{i_1}\), \(D^j_{i_{m+1}}\), and the normalized matrix \(E^j\); the symbols \(\beta \) and \(\rho \) below are the words obtained after opening the periodic boundary at the chosen cut. The source theorem writes the number of blocks as \(b\) and assumes \(b\ge 2\); here the same number is denoted by \(g\). In these coordinates the source comparison input becomes \(A^j_\beta C^j_{i,\rho }=(\mu _j^N X_j A^j_\beta )A^j_\rho \), which is the form obtained after opening the boundary from \(A^j_{i_{m+1}}C^j_{i_1}=D^j_{i_{m+1}}A^j_{i_1}\) with \(D^j_\beta =(\mu _j^N X_j)A^j_\beta \). Once this compatibility is known, the normalized \(E^j\) calculation is the matrix identity of Lemma 15.2.1.9; the local boundary form used after opening the periodic boundary is Lemma 15.2.1.10.

Lemma 15.3.2.1 The block-diagonal periodic-boundary space lies between two block sums

With the hypotheses and notation of Lemma 15.3.7, one has

\begin{align} \bigvee _{j=1}^g\mathcal G_{N,L}(A_j) & \subseteq \mathcal G_{N,L}\! \left(\bigoplus _{j=1}^g\mu _jA_j\right) \subseteq \bigvee _{j=1}^gG_N(A_j), \notag \end{align}

and the sum defining the right-hand side is internal.

Proof

The first inclusion is Lemma 15.1.7. The second inclusion and internal directness are Lemma 15.3.8.

Lemma 15.3.2.2 Boundary inclusion gives the block-diagonal periodic-boundary equality

Let \(B=\bigoplus _{j=1}^g\mu _jA_j\), where \(\mu _j\ne 0\) for \(1\le j\le g\). If the periodic-boundary comparison with block-diagonal boundary conditions gives

\begin{align} \mathcal G_{N,L}(B) & \subseteq \bigvee _{j=1}^g\mathcal G_{N,L}(A_j), \notag \end{align}

then \(\mathcal G_{N,L}(B) = \bigvee _{j=1}^g\mathcal G_{N,L}(A_j)\).

Proof

The reverse inclusion is Lemma 15.1.7; combine the two inclusions.

Lemma 15.3.2.3 Componentwise periodic-boundary decomposition gives the reverse inclusion

Let \(B=\bigoplus _{j=1}^g\mu _jA_j\). Suppose that every vector \(\psi \in \mathcal G_{N,L}(B)\) can be written as

\begin{align} \psi & = \sum _{j=1}^g\psi _j, \notag \\ \psi _j & \in \mathcal G_{N,L}(A_j). \notag \end{align}

Then

\begin{align} \mathcal G_{N,L}(B) & \subseteq \bigvee _{j=1}^g\mathcal G_{N,L}(A_j). \notag \end{align}
Proof

For \(\psi \in \mathcal G_{N,L}(B)\), choose the displayed decomposition. Each summand lies in the corresponding subspace \(\mathcal G_{N,L}(A_j)\), hence their finite sum lies in \(\bigvee _j\mathcal G_{N,L}(A_j)\).

Lemma 15.3.2.4 Block-diagonal boundary vectors in the periodic-boundary block sum

Let \(B=\bigoplus _{j=1}^g\mu _jA_j\). Fix block-diagonal boundary conditions \(X_j\). Suppose that, for every \(j\), \(\Gamma _N^{A_j}(\mu _j^N X_j)\in \mathcal G_{N,L}(A_j)\). Then

\begin{align} \Gamma _N^B\! \left(\bigoplus _{j=1}^gX_j\right) & \in \bigvee _{j=1}^g\mathcal G_{N,L}(A_j). \notag \end{align}
Proof

The block-diagonal boundary formula gives

\begin{align} \Gamma _N^B\! \left(\bigoplus _{j=1}^gX_j\right) & = \sum _{j=1}^g\Gamma _N^{A_j}(\mu _j^N X_j). \notag \end{align}

Each summand belongs to the corresponding block periodic-boundary space by assumption, so the finite sum lies in the linear span of these spaces.

Lemma 15.3.2.5 Block-diagonal boundary representation gives the reverse inclusion

Let \(B=\bigoplus _{j=1}^g\mu _jA_j\). Suppose every \(\psi \in \mathcal G_{N,L}(B)\) admits block-diagonal boundary conditions \(X_j\) such that \(\psi = \Gamma _N^B\! \left(\bigoplus _{j=1}^gX_j\right)\), and, for every \(j\), \(\Gamma _N^{A_j}(\mu _j^N X_j)\in \mathcal G_{N,L}(A_j)\). Then

\begin{align} \mathcal G_{N,L}(B) & \subseteq \bigvee _{j=1}^g\mathcal G_{N,L}(A_j). \notag \end{align}
Proof

Apply Lemma 15.3.2.4 to the block-diagonal boundary representation of each \(\psi \in \mathcal G_{N,L}(B)\).

Lemma 15.3.2.6 Block-diagonal boundary representation reduces to periodic-boundary equality

With the hypotheses and notation of Lemma 15.3.7, suppose every vector \(\psi \in \mathcal G_{N,L}(B)\) has block-diagonal boundary conditions \(X_j\) such that \(\psi = \Gamma _N^B\! \left(\bigoplus _{j=1}^gX_j\right)\), and, for every \(j\), \(\Gamma _N^{A_j}(\mu _j^N X_j)\in \mathcal G_{N,L}(A_j)\). Then \(\mathcal G_{N,L}(B) = \bigvee _{j=1}^g\mathcal G_{N,L}(A_j)\), and the sum defining \(\bigvee _jG_N(A_j)\) is internal.

Proof

The block-diagonal boundary representation gives

\begin{align} \mathcal G_{N,L}(B) & \subseteq \bigvee _{j=1}^g\mathcal G_{N,L}(A_j) \notag \end{align}

by Lemma 15.3.2.5. The equality follows from Lemma 15.3.2.2; the internality statement is Lemma 15.3.8.

Lemma 15.3.2.7 Complementary-word matrix identities reduce to periodic-boundary equality

With the hypotheses and notation of Lemma 15.3.2.6, assume in addition that \(L+L_0\le N\). Assume further that, for every \(\psi \in \mathcal G_{N,L}\! \left(\bigoplus _j\mu _jA_j\right)\) and every block-diagonal boundary representation

\begin{align} \psi & = \Gamma _N^{\oplus _j\mu _jA_j}\! \left(\bigoplus _jX_j\right), \notag \end{align}

the complementary-word matrix identities hold: for every block \(j\), boundary-crossing interval \(i\), word \(\beta \) before the cut, and outside word \(\rho \) of length \(N-L\), there is a matrix \(E_{j,i,\rho }\) such that \(\mu _j^N X_jA^j_\beta A^j_\rho = A^j_\beta E_{j,i,\rho }\). Then

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _j\mu _jA_j\right) & = \bigvee _j\mathcal G_{N,L}(A_j), \notag \end{align}

and the \(N\)-site ground spaces \(G_N(A_j)=\mathcal G_{N,N}(A_j)\) have internal sum.

Proof

For each \(\psi \), Lemma 15.3.1.44 gives matrices \(X_j\) with

\begin{align} \psi & = \Gamma _N^{\oplus _j\mu _jA_j}\! \left(\bigoplus _jX_j\right) \notag \end{align}

and, for every \(j\), the membership \(\Gamma _N^{A_j}(\mu _j^N X_j)\in \mathcal G_{N,L}(A_j)\) holds. Lemma 15.3.2.6 then gives

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _j\mu _jA_j\right) & = \bigvee _j\mathcal G_{N,L}(A_j), \notag \end{align}

and the internality of the sum of the \(G_N(A_j)\).

Lemma 15.3.2.8 \(C^j,D^j\) boundary-condition comparisons reduce to periodic-boundary equality

With the hypotheses and notation of Lemma 15.3.2.6, assume in addition that \(L+L_0\le N\). Assume further that, for every \(\psi \in \mathcal G_{N,L}\! \left(\bigoplus _j\mu _jA_j\right)\) and every block-diagonal boundary representation

\begin{align} \psi & = \Gamma _N^{\oplus _j\mu _jA_j}\! \left(\bigoplus _jX_j\right), \notag \end{align}

there are matrices \(C^j_{i,\rho }\) such that, for every boundary-crossing interval \(i\), every outside word \(\rho \) of length \(N-L\), and every word \(\beta \) before the cut, \(A^j_\beta C^j_{i,\rho } = ((\mu _j^N X_j)A^j_\beta )A^j_\rho \). This is the \(C^j,D^j\) compatibility input; the subsequent normalized \(E^j\) identities follow from Lemma 15.2.1.10. Then

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _j\mu _jA_j\right) & = \bigvee _j\mathcal G_{N,L}(A_j), \notag \end{align}

and the \(N\)-site ground spaces \(G_N(A_j)=\mathcal G_{N,N}(A_j)\) have internal sum.

Proof

For each \(\psi \), Lemma 15.3.1.32 gives matrices \(X_j\) with

\begin{align} \psi & = \Gamma _N^{\oplus _j\mu _jA_j}\! \left(\bigoplus _jX_j\right) \notag \end{align}

and, for every \(j\), \(\Gamma _N^{A_j}(\mu _j^N X_j)\in \mathcal G_{N,L}(A_j)\). Lemma 15.3.2.6 then gives

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _j\mu _jA_j\right) & = \bigvee _j\mathcal G_{N,L}(A_j), \notag \end{align}

and the internality of the sum of the \(G_N(A_j)\).

Lemma 15.3.2.9 \(C^j,D^j\) comparisons give the periodic block ground-space equality

With the hypotheses and notation of Lemma 15.3.2.8, one has

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _j\mu _jA_j\right) & = \bigvee _j\mathcal G_{N,L}(A_j). \notag \end{align}

This records only the ground-space equality conclusion obtained after the \(C^j,D^j\) boundary-condition comparison input in [ PGVWC07 , Theorem 12, proof lines 1446–1456 ] ; the local normalized \(E^j\) calculation is Lemma 15.2.1.10. The independence of the \(N\)-site spaces is a separate conclusion of Lemma 15.3.2.8. The statement is still in the length-\(L_0\) injectivity range displayed in that lemma, not the shorter range in [ PGVWC07 , Theorem 12 ] .

Proof

The claim is the equality part of Lemma 15.3.2.8.

Lemma 15.3.2.10 \(C^j,D^j\) comparisons at boundary-crossing intervals reduce to periodic-boundary equality

With the hypotheses and notation of Lemma 15.3.2.6, assume in addition that \(L+L_0\le N\). Assume also that, for every boundary-crossing interval beginning at \(i\), the simultaneous products

\begin{align} w & \longmapsto (A^1_w,\ldots ,A^g_w), \notag \\ |w| & = N-i, \notag \end{align}

span the product algebra \(\prod _jM_{D_j}(\mathbb {C})\). Then

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _j\mu _jA_j\right) & = \bigvee _j\mathcal G_{N,L}(A_j), \notag \end{align}

and the \(N\)-site ground spaces \(G_N(A_j)=\mathcal G_{N,N}(A_j)\) have internal sum. The present statement is the span-dependent consequence, after opening the boundary, of the \(C^j,D^j\) compatibility input in [ PGVWC07 , Theorem 12, proof lines 1436–1451 ] in the length-\(L_0\) injectivity range \(L \ge (L_0+1)+3(g-1)(L_0+1)+1\). The source theorem has the shorter bound \(L \ge 3(g-1)(L_0+1)+1\).

Proof

Lemma 15.3.2.6 reduces the claim to producing, for each \(\psi \), block-diagonal boundary conditions \(X_j\) whose single-block components lie in \(\mathcal G_{N,L}(A_j)\). The crossing comparison theorem supplies this: it first uses the boundary-comparison-free BNT representation to obtain \(X_j\), and then uses the stated tail-word spans and Theorem 15.3.1.9 to produce matrices \(C^j_{i,\rho }\) satisfying \(A^j_\beta C^j_{i,\rho } = ((\mu _j^N X_j)A^j_\beta )A^j_\rho \). Lemma 15.3.1.24 then gives \(\Gamma _N^{A_j}(\mu _j^N X_j)\in \mathcal G_{N,L}(A_j)\) for every \(j\). Applying Lemma 15.3.2.6 to these boundary conditions gives the displayed equality and internality.

Lemma 15.3.2.11 Boundary-crossing comparisons give the periodic block ground-space equality

With the hypotheses and notation of Lemma 15.3.2.10, one has

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _j\mu _jA_j\right) & = \bigvee _j\mathcal G_{N,L}(A_j). \notag \end{align}

This records the ground-space equality conclusion of the conditional crossing-span version above. It is the corresponding part of [ PGVWC07 , Theorem 12 ] , but only under the visible tail-word span hypothesis and in the longer stated range. The independence of the \(N\)-site spaces is a separate conclusion of Lemma 15.3.2.10. Thus the statement remains conditional on the short tail-word spans; it is not the unconditional boundary-condition comparison in the source theorem.

Proof

The claim is the equality part of Lemma 15.3.2.10.

Lemma 15.3.2.12 Boundary trace comparisons reduce to periodic-boundary equality

With the hypotheses and notation of Lemma 15.3.2.6, assume in addition that \(L+L_0\le N\). Assume also that, for some middle-word length \(m\), the simultaneous tuples \(w\mapsto (A^1_w,\ldots ,A^g_w)\), where \(w\) ranges over words of length \(m\) in \(\{ 0,\ldots ,d{-}1\} \), span the full product algebra \(\prod _jM_{D_j}(\mathbb {C})\). Assume further that, for every \(\psi \in \mathcal G_{N,L}\! \left(\bigoplus _j\mu _jA_j\right)\) and every block-diagonal boundary representation

\begin{align} \psi & = \Gamma _N^{\oplus _j\mu _jA_j}\! \left(\bigoplus _jX_j\right), \notag \end{align}

there are matrices \(C^j_{i,\rho }\) such that, for every boundary-crossing interval \(i\), every outside word \(\rho \) of length \(N-L\), every word \(\beta \) before the cut, and every word \(w\) of length \(m\),

\begin{align} \sum _j\operatorname{tr}\! \left(A^j_\beta C^j_{i,\rho }A^j_w\right) & = \sum _j \operatorname{tr}\! \left(((\mu _j^N X_j)A^j_\beta )A^j_\rho A^j_w\right). \notag \end{align}

Then

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _j\mu _jA_j\right) & = \bigvee _j\mathcal G_{N,L}(A_j), \notag \end{align}

and the \(N\)-site ground spaces \(G_N(A_j)=\mathcal G_{N,N}(A_j)\) have internal sum.

Proof

For each \(\psi \), Lemma 15.3.1.30 gives matrices \(X_j\) with

\begin{align} \psi & = \Gamma _N^{\oplus _j\mu _jA_j}\! \left(\bigoplus _jX_j\right) \notag \end{align}

and, for every \(j\), \(\Gamma _N^{A_j}(\mu _j^N X_j)\in \mathcal G_{N,L}(A_j)\). Lemma 15.3.2.6 then gives

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _j\mu _jA_j\right) & = \bigvee _j\mathcal G_{N,L}(A_j), \notag \end{align}

and the internality of the sum of the \(G_N(A_j)\).

Lemma 15.3.2.13 Basis-of-normal-tensors product span gives boundary-trace ground-space equality

With the hypotheses and notation of Lemma 15.3.2.6, assume in addition that \(L+L_0\le N\) and \(L \ge (L_0+1)+(g-1)((L_0+1)+((L_0+1)+(L_0+1)))+1\). Put \(m = (L_0+1)+(g-1)((L_0+1)+((L_0+1)+(L_0+1)))\). Assume that, for every \(\psi \in \mathcal G_{N,L}\! \left(\bigoplus _j\mu _jA_j\right)\) and every block-diagonal boundary representation

\begin{align} \psi & = \Gamma _N^{\oplus _j\mu _jA_j}\! \left(\bigoplus _jX_j\right), \notag \end{align}

there are matrices \(C^j_{i,\rho }\) such that, for every boundary-crossing interval \(i\), every outside word \(\rho \) of length \(N-L\), every word \(\beta \) before the cut, and every word \(w\) of length \(m\),

\begin{align} \sum _j\operatorname{tr}\! \left(A^j_\beta C^j_{i,\rho }A^j_w\right) & = \sum _j \operatorname{tr}\! \left(((\mu _j^N X_j)A^j_\beta )A^j_\rho A^j_w\right). \notag \end{align}

Then

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _j\mu _jA_j\right) & = \bigvee _j\mathcal G_{N,L}(A_j), \notag \end{align}

and the \(N\)-site ground spaces \(G_N(A_j)=\mathcal G_{N,N}(A_j)\) have internal sum.

Proof

Lemma 15.2.2.5 gives \(\operatorname{span}\{ (A^1_w,\ldots ,A^g_w):|w|=m\} = \prod _jM_{D_j}(\mathbb {C})\). Applying Lemma 15.3.2.12 with this middle-word span gives the equality of periodic-boundary ground spaces and the internality of the sum of the \(N\)-site spaces.

Lemma 15.3.2.14 Boundary decomposition reduces to periodic-boundary equality

Let \(A_1,\ldots ,A_g\) be normalized representatives of a basis of normal tensors: each representative is irreducible, the family is left-canonical, the self-overlaps are normalized, and no two distinct same-dimensional representatives are gauge-phase equivalent, and let \(B=\bigoplus _{j=1}^g\mu _jA_j\), where \(\mu _j\ne 0\). Assume each block is injective at length \(L_0\), assume \(\sum _aA^j_aA^{j\dagger }_a=\mathbb {1}\) for every \(j\), and suppose

\begin{align} L & \ge (L_0+1)+(g-1)3(L_0+1)+1, \notag \\ N & \ge L. \notag \end{align}

If every vector \(\psi \in \mathcal G_{N,L}(B)\) can be written as

\begin{align} \psi & = \sum _{j=1}^g\psi _j, \notag \\ \psi _j & \in \mathcal G_{N,L}(A_j), \notag \end{align}

then \(\mathcal G_{N,L}(B) = \bigvee _{j=1}^g\mathcal G_{N,L}(A_j)\), and the sum \(\bigvee _jG_N(A_j)\) is internal.

Proof

Lemma 15.3.2.3 gives

\begin{align} \mathcal G_{N,L}(B) & \subseteq \bigvee _{j=1}^g\mathcal G_{N,L}(A_j). \notag \end{align}

The blockwise inclusion gives the reverse containment, hence equality. Internality follows from the directness conclusion of Lemma 15.3.8.

15.4 Periodic-boundary ground space generated by a basis of normal tensors

For a block-diagonal tensor \(\bigoplus _j\mu _j A_j\) with \(\mu _j\ne 0\), this section relates three ground-space statements. Each block \(A_j\) contributes the local space \(G_m(A_j)\) on an open length-\(m\) segment, and we write \(S_m:=\sum _{j=1}^g G_m(A_j)\) for their linear sum; internal directness is proved only under the BNT separation hypotheses and the stated length ranges. On the \(N\)-site ring the parent Hamiltonian has the periodic-boundary ground space \(\mathcal G_{N,L}(\bigoplus _j\mu _j A_j)\). The results below establish the open-segment recursion \(\mathbb {C}^d\otimes S_m\cap S_m\otimes \mathbb {C}^d=S_{m+1}\) and, conditional on the block-diagonal boundary-condition comparison in the periodic ring, reduce the periodic-boundary ground space to the span \(\operatorname{span}\{ V^{(N)}(A_j):j=1,\ldots ,g\} \) of the vectors \(V^{(N)}(A_j)\) associated with the chosen basis of normal tensors.

Theorem 15.4.1 Block-diagonal local recursion and boundary-condition comparison

Let \(A_1,\ldots ,A_g\) be the blocks in a block-injective canonical form, and assume that their MPV families are pairwise different:

\begin{align} j\ne k \quad \Longrightarrow \quad \exists n\ge 1, V^{(n)}(A_j)\ne V^{(n)}(A_k). \notag \end{align}

Let \(L_0\) be a common injectivity length. Assume the multi-block range used in [ PGVWC07 , Theorem 12 ] :

\begin{align} g& \ge 2, \notag \\ L_0& \ge 1, \notag \\ L& \ge 3(g-1)(L_0+1)+1. \notag \end{align}

Set \(S_m:=\bigoplus _{j=1}^gG_m(A_j)\). The open-segment recursion in [ PGVWC07 , Theorem 12 ] says that, for every \(m\ge L\),

\begin{align} \mathbb {C}^d\otimes S_m \cap S_m\otimes \mathbb {C}^d & = S_{m+1}. \label{eq:parent_bnt_open_recursion} \end{align}

Let \(H_{S_L,N}\) denote the periodic parent Hamiltonian whose local ground space is \(S_L\). The periodic-boundary conclusion of the same theorem is

\begin{align} \ker H_{S_L,N} & = \operatorname{span}\{ V^{(N)}(A_j):j=1,\ldots ,g\} \label{eq:parent_bnt_periodic_kernel} \end{align}

for every \(N\ge L+L_0\).

Proof

The source proof takes a vector \(\psi \) in the left-hand side and writes its two boundary descriptions as

\begin{align} \psi & = \sum _{j=1}^g \alpha _j = \sum _{j=1}^g \beta _j, \notag \\ \alpha _j & \in \mathbb {C}^d\otimes G_m(A_j), \notag \\ \beta _j & \in G_m(A_j)\otimes \mathbb {C}^d. \notag \end{align}

In the source notation, the two components have boundary matrices \(C^j_{i_1}\) and \(D^j_{i_{m+1}}\): the first family leaves the initial physical index open, and the second family leaves the final physical index open.

\begin{align} \alpha _j & = \sum _{i_1,\ldots ,i_{m+1}} \operatorname{tr}\! \left(A^j_{i_{m+1}} C^j_{i_1} A^j_{i_2}\cdots A^j_{i_m}\right) |i_1\cdots i_{m+1}\rangle , \notag \\ \beta _j & = \sum _{i_1,\ldots ,i_{m+1}} \operatorname{tr}\! \left(D^j_{i_{m+1}} A^j_{i_1} A^j_{i_2}\cdots A^j_{i_m}\right) |i_1\cdots i_{m+1}\rangle . \notag \end{align}

The direct-sum separation lemma used in [ PGVWC07 , Theorem 12 ] , together with injectivity of each block, gives the blockwise matrix identity

\begin{align} A^j_{i_{m+1}} C^j_{i_1} & = D^j_{i_{m+1}} A^j_{i_1} \notag \end{align}

for every \(j\), \(i_1\), and \(i_{m+1}\). For fixed \(j\), set \(E^j:=\sum _a C^j_a A^{j\, \dagger }_a\). The printed formula in [ PGVWC07 , Theorem 12, proof lines 1449–1451 ] omits the adjoint on \(A^j_a\); the adjointed form is forced by the displayed normalization below. In the normalization used in the source proof, \(\sum _a A^j_a A^{j\, \dagger }_a=\mathbb {1}\), so Lemma 15.2.1.7 gives

\begin{align} D^j_b & = \sum _a D^j_b A^j_a A^{j\, \dagger }_a = \sum _a A^j_b C^j_a A^{j\, \dagger }_a = A^j_b E^j. \notag \end{align}

Hence \(A^j_b C^j_a=A^j_b E^j A^j_a\) for every \(a,b\), and cyclicity of the trace gives

\begin{align} \alpha _j & = \sum _{i_1,\ldots ,i_{m+1}} \operatorname{tr}\! \left(A^j_{i_1}\cdots A^j_{i_m} A^j_{i_{m+1}}E^j\right) |i_1\cdots i_{m+1}\rangle = \Gamma _{m+1}^{A_j}(E^j) \in G_{m+1}(A_j). \notag \end{align}

Since \(\psi =\sum _j\alpha _j\), this gives \(\psi \in \bigoplus _j G_{m+1}(A_j)\). For one block the preceding implication is

\begin{align} \mathbb {C}^d\otimes G_m(A_j) \cap G_m(A_j)\otimes \mathbb {C}^d & \subseteq G_{m+1}(A_j). \notag \end{align}

The reverse inclusion is blockwise:

\begin{align} G_{m+1}(A_j) & \subseteq \mathbb {C}^d\otimes G_m(A_j) \cap G_m(A_j)\otimes \mathbb {C}^d. \notag \end{align}

Thus \(\mathbb {C}^d\otimes G_m(A_j) \cap G_m(A_j)\otimes \mathbb {C}^d = G_{m+1}(A_j)\). With \(S_m:=\bigoplus _{j=1}^g G_m(A_j)\), the preceding calculation is the open-segment recursion \(\mathbb {C}^d\otimes S_m \cap S_m\otimes \mathbb {C}^d = S_{m+1}\) for every \(m\ge L\), as stated in (??). For \(B=\bigoplus _j\mu _j A_j\) with every \(\mu _j\ne 0\), Lemma 15.3.5 gives \(G_L(B)=\bigvee _jG_L(A_j)\). In the range where the sums \(\bigvee _jG_m(A_j)\) are direct, this identifies \(G_L(B)\) with the subspace denoted in the source by \(S=\bigoplus _j\mathcal G_L^{A^j}\). The passage from the open-segment recursion to the \(N\)-site ring is the separate boundary-condition comparison in the periodic ring in [ PGVWC07 , Theorem 12 ] . The source periodic-boundary conclusion is \(\ker H_{S_L,N} = \operatorname{span}\{ V^{(N)}(A_j):j=1,\ldots ,g\} \), in agreement with (??).

Let \(B^i=\bigoplus _j\mu _j A_j^i\), and assume \(1{\lt}L\), \(N\ge L+1\), and \(0{\lt}m\). Assume that every virtual sector projection \(P_j\) lies in the pulled-back span of the length-\(m\) words \(B^\omega \), and assume that every vector \(\psi \in \mathcal G_{N,L}(\bigoplus _j\mu _j A_j)\) has a boundary matrix \(X\) with

\begin{align} \psi & = \Gamma _N^B(X), \notag \\ XB^\omega & = B^\omega X \notag \end{align}

for the boundary-crossing words \(\omega \). Then

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _j\mu _j A_j\right) & \subseteq \sum _j \mathcal G_{N,L}(A_j), \notag \end{align}

with equality after the forward inclusion. This is a conditional form of the block-diagonal boundary-condition step in [ CPGSV21 , lines 2126–2128 ] .

Proof

The commutant reduction makes the pulled-back boundary matrix block diagonal. Restricting the same word-commutation identities to a diagonal block and using the injectivity of that block shows, by the scalar-commutant lemma, that each diagonal block is scalar. Write \(\Gamma _N^B\) for the length-\(N\) boundary-to-state map of the tensor \(B\). For a block-diagonal boundary matrix,

\begin{align} \Gamma _N^{\oplus _j\mu _j A_j}\! \left(\bigoplus _j X_j\right) & = \sum _j \mu _j^N \Gamma _N^{A_j}(X_j). \notag \end{align}

If \(X_j=\lambda _j\mathbb {1}\), this is a finite sum of scalar multiples of the vectors \(V^{(N)}(A_j)\). The reverse inclusion and equality statements are reformulations of this calculation, combined with the forward inclusion and the blockwise uniqueness theorem.

Lemma 15.4.3 Block-diagonal boundary reduction from blockwise word span

Let \(A^i=\bigoplus _j\mu _j A_j^i\) be in block-injective canonical form, and assume \(1{\lt}L\), \(N\ge L+1\), and \(0{\lt}m\). If

\begin{align} \operatorname{span}\{ (A_1^\omega ,\ldots ,A_g^\omega ):|\omega |=m\} & = \prod _j M_{D_j}(\mathbb {C}) \notag \end{align}

and every vector in \(\mathcal G_{N,L}(\bigoplus _j \mu _j A_j)\) comes with a boundary matrix representation commuting with all length-\(m\) direct-sum words, then

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _j \mu _j A_j\right) & \subseteq \sum _j \mathcal G_{N,L}(A_j). \notag \end{align}

Combined with the forward inclusion, this gives the corresponding conditional equality. In the parent-ground-space formulation, the same hypotheses imply containment in \(\operatorname{span}\{ V^{(N)}(A_j):j=1,\ldots ,g\} \).

Proof

The blockwise word-span hypothesis gives

\begin{align} P_j & \in \operatorname{span}\left\{ \left(\bigoplus _r\mu _r A_r\right)^\omega :|\omega |=m\right\} . \notag \end{align}

If \(X\) commutes with all displayed word products, then \(XP_j=P_jX\) for every \(j\), hence \(X=\bigoplus _jX_j\). The direct-sum boundary formula

\begin{align} \Gamma _N^{\oplus _j\mu _j A_j}\! \left(\bigoplus _jX_j\right) & = \sum _j\mu _j^N\Gamma _N^{A_j}(X_j) \notag \end{align}

gives the reverse inclusion, and the blockwise uniqueness theorem gives the stated span containment.

Lemma 15.4.4 Block-diagonal boundary reduction from block-separating equations

Let \(A^i=\bigoplus _j\mu _j A_j^i\) be in block-injective canonical form, with \(A_1,\ldots ,A_g\) a basis of normal tensors. Assume \(1{\lt}L\) and \(N\ge L+1\), and suppose that for each block \(k\) there are coefficients \(c_\omega ^{(k)}\) such that

\begin{align} \sum _{|\omega |=S} c_\omega ^{(k)} A_j^\omega & = \begin{cases} \mathbb {1}, & j=k,\\ 0, & j\ne k. \end{cases} \notag \end{align}

If every vector in \(\mathcal G_{N,L}(\bigoplus _j \mu _j A_j)\) has a boundary matrix representation commuting with all length-\((1+S)\) direct-sum words, then

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _j \mu _j A_j\right) & \subseteq \sum _j \mathcal G_{N,L}(A_j). \notag \end{align}

Together with the forward inclusion, this gives the corresponding conditional equality. In the parent-ground-space formulation, the same hypotheses imply containment in \(\operatorname{span}\{ V^{(N)}(A_j):j=1,\ldots ,g\} \). This is the block-injective ground-space argument of [ CPGSV21 , lines 2120–2128 ] , where the re-growing step is restricted to block-diagonal boundary conditions.

Proof

Finite-length injectivity gives a block length at which each block has the required physical–virtual correspondence. In the length-one specialization, Lemma 2.4.3 supplies the corresponding one-block injectivity. The displayed separation equations imply

\begin{align} \operatorname{span}\{ (A_1^\omega ,\ldots ,A_g^\omega ):|\omega |=1+S\} & = \prod _jM_{D_j}(\mathbb {C}). \notag \end{align}

This is Lemma 15.2.16 with \(L=1\). Therefore Lemma 15.4.3 applies with \(m=1+S\). Its conclusion is the displayed containment, and the equality follows after adding the forward inclusion.

Theorem 15.4.5 Periodic-boundary containment in the BNT span

If \(A^i=\bigoplus _j\mu _j A_j^i\) is in block-injective canonical form, where \(A_1,\ldots ,A_g\) is a basis of normal tensors, assume \(g\ge 2\) and \(\mu _j\ne 0\) for every \(j\). Let \(L_0\ge 1\) be a common injectivity length for the blocks, and suppose

\begin{align} L_0& {\lt}L, \notag \\ L& \ge 3(g-1)(L_0+1)+1, \notag \\ N& \ge L+L_0, \notag \end{align}

and assume the block-diagonal boundary-condition comparison in the periodic ring,

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _j \mu _j A_j\right) & \subseteq \bigvee _{j=1}^g\mathcal G_{N,L}(A_j). \label{eq:parent_bnt_boundary_comparison} \end{align}

Then

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _j \mu _j A_j\right) & \subseteq \operatorname{span}\{ V^{(N)}(A_j):j=1,\ldots ,g\} . \notag \end{align}
Proof

The length hypotheses are the conservative range used in [ PGVWC07 , Theorem 12 ] . The shorter range stated in [ CPGSV21 , lines 2126–2128 ] requires the strengthened block-diagonal re-growing step. These hypotheses do not by themselves give the displayed boundary-condition comparison in the periodic ring. That comparison follows only after the block-diagonal boundary conditions have been compared across the windows crossing the chosen periodic-boundary cut. Since \(g\ge 2\) and \(L_0\ge 1\),

\begin{align} L & \ge 3(g-1)(L_0+1)+1 \ge 3L_0+4 {\gt} L_0, \notag \\ N & \ge L+L_0 \ge L+1. \notag \end{align}

Because \(A_1,\ldots ,A_g\) is a basis of normal tensors, the block MPV families are non-redundant:

\begin{align} j\ne k \quad \Longrightarrow \quad \exists n\ge 1, V^{(n)}(A_j)\ne V^{(n)}(A_k). \notag \end{align}

By the assumed block-diagonal boundary-condition comparison (??),

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _j \mu _j A_j\right) & \subseteq \bigvee _j \mathcal G_{N,L}(A_j). \notag \end{align}

Theorem 13.8.2.3 gives \(\mathcal G_{N,L}(A_j) = \operatorname{span}\{ V^{(N)}(A_j)\} \) for every \(j\). Substituting these one-block identities into the join gives the displayed containment.

Theorem 15.4.6 Periodic-boundary equality with the BNT span

Under the hypotheses of Theorem 15.4.5, the corresponding periodic-boundary ground space is exactly the span of the BNT component vectors:

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _j \mu _j A_j\right) & = \operatorname{span}\{ V^{(N)}(A_j):j=1,\ldots ,g\} . \label{eq:parent_bnt_ground_space} \end{align}
Proof

We prove the two inclusions separately. The inclusion

\begin{align} \mathcal G_{N,L}\! \left(\bigoplus _j \mu _j A_j\right) & \subseteq \operatorname{span}\{ V^{(N)}(A_j):j=1,\ldots ,g\} \notag \end{align}

is exactly Theorem 15.4.5. For the reverse inclusion, the nonzero-weight hypothesis and Lemma 15.1.5 show that each vector \(V^{(N)}(A_j)\) lies in \(\mathcal G_{N,L}(\bigoplus _j \mu _j A_j)\). Since this ground space is linear, it contains every linear combination of such vectors, and hence contains \(\operatorname{span}\{ V^{(N)}(A_j):j=1,\ldots ,g\} \). Thus

\begin{align} \operatorname{span}\{ V^{(N)}(A_j):j=1,\ldots ,g\} & \subseteq \mathcal G_{N,L}\! \left(\bigoplus _j \mu _j A_j\right), \notag \end{align}

and the two spaces are equal, as stated in (??).

Remark. A separate kernel-identification statement for these direct-sum block families would convert the periodic-boundary ground-space statement into the corresponding statement for \(\ker (H_N)\).