Since \(G_L(A)\) is the range of a linear map from \(M_{D}(\mathbb {C})\), one has \(\dim G_L(A)\le \dim M_{D}(\mathbb {C})=D^2\).

If \(G_L(A)\) were the whole space, its dimension would be \(d^L\) by Lemma 13.2.5; Lemma 13.2.4 would then give the contradiction \(d^L\le D^2\).

This is the defining equation \(G_L^{\mathrm{ES}}(A)=\iota _L(G_L(A))\).

This is the idempotency of the orthogonal projector onto \(G_L(A)^\perp \), transported from the Hilbert-space realization of the local coefficient space.

Since \(h_L(A)\) is the orthogonal projector onto \(G_L(A)^\perp \), it annihilates every element of \(G_L(A)\).

Case-split on whether the offset \((k-i+N)\bmod N\) lies in \([0,L)\): inside the cyclic interval the prescribed value is the original value of \(\sigma \); outside the interval both sides already agree with \(\sigma \).

On the cyclic interval both sides take the value prescribed by \(\upsilon \). Away from that interval neither second prescription changes the value.

This is the first case in (??).

If \(L{\gt}N\), the translated term is zero. Suppose \(L\le N\) and put \(f_{\sigma ,i}(\tau ) :=\psi (\sigma ^{[i,i+L-1]\leftarrow \tau })\). The pointwise formula for translated terms, together with Lemmas 13.3.8 and 13.3.10, gives

Here the second equality follows from Lemma 13.3.4.

The boundary matrix is the product of the \(A\)-matrices on the complement sites. Trace cyclicity rotates the full \(N\)-site product so that the window indices come first, matching the definition of the ground-space map.

This combines periodic-MPS-vector window membership with the fact that the parent interaction annihilates ground-space elements.

This is a sum of zeros, since each local term annihilates the periodic MPS vector.

The claim follows from Lemma 13.3.17.

The restriction simply absorbs the last tensor into the boundary matrix:

Direct computation gives

On the right restriction, one first rotates the trace and then reads the resulting boundary matrix on the shortened window:

By trace cyclicity,

If \(\Gamma _L(X) = 0\), then \(\operatorname{tr}(A^\sigma X) = 0\) for every length-\(L\) word \(\sigma \). Since \(A\) is injective, the products \(\{ A^\sigma \} _\sigma \) span \(M_{D}(\mathbb {C})\), so the trace pairing gives \(X = 0\).

If \(\Gamma _L(X) = 0\), then \(\operatorname{tr}(A^\sigma X) = 0\) for every word \(\sigma \) of length \(L\). The spanning hypothesis turns this into \(\operatorname{tr}(MX) = 0\) for every \(M \in M_{D}(\mathbb {C})\), and the trace pairing gives \(X = 0\).

The hypothesis is precisely the fixed-length spanning condition needed by Theorem 13.4.3.2.

The upper bound \(\dim G_L(A) \le D^2\) is Lemma 13.2.4. Injectivity of \(\Gamma _L\) gives \(\dim G_L(A) = \dim M_{D}(\mathbb {C}) = D^2\).

The “inverting and growing back” argument compares the left and right restrictions on their common \((L-1)\)-site overlap and then reconstructs the full \((L+1)\)-site word from a single boundary matrix:

1. From the right ground condition, for each physical index \(i\) there exists a unique \(Y_i \in M_{D}(\mathbb {C})\), by injectivity of \(\Gamma _L\), such that \(\psi (i,\sigma ) = \operatorname{tr}(A^\sigma Y_i)\).

2. Similarly, from the left ground condition, for each \(j\) there exists a unique \(Z_j\) such that \(\psi (\sigma ,j) = \operatorname{tr}(A^\sigma Z_j)\).

3. Matching on the \((L-1)\)-site overlap and using nondegeneracy of the trace pairing gives \(A^jY_i = Z_jA^i\) for all \(i,j\).

4. Since \(A\) is injective, the matrices \(\{ A^j\} \) span \(M_{D}(\mathbb {C})\) (Definition 2.4.1), so \(\mathbb {1}= \sum _j c_jA^j\) for some coefficients \(c_j \in \mathbb {C}\). Thus, with \(X' := \sum _j c_jZ_j\), \(Y_i = \mathbb {1}Y_i = \sum _j c_jA^jY_i = \sum _j c_jZ_jA^i = X'A^i\).

5. By trace cyclicity, \(\psi (\sigma ) = \operatorname{tr}(A^\sigma X')\), so \(\psi \in G_{L+1}(A)\).

The claim follows from the two forward-direction lemmas and Theorem 13.4.4.1.

The first two identities are the cyclic first- and last-site deletion identities, followed by the corresponding ground-space restriction formulas. For adjacent windows, equality of the completed outside configurations gives \(\Gamma _L(Y_0A^a)=\Gamma _L(A^bY_n)\). Injectivity of \(\Gamma _L\) gives \(Y_0A^a=A^bY_n\). Iterating the adjacent identity gives the word-product identity.

The induction repeatedly merges two neighbouring admissible windows \([i,i+L-1]\) and \([i+1,i+L]\) into the larger window \([i,i+L]\) by applying Theorem 13.4.4.1 to their common overlap \([i+1,i+L-1]\).

Write \(\psi (\rho ) = \operatorname{tr}(A^\rho X)\) for a boundary matrix \(X\). Fixing the final site gives the boundary matrix \(A^jX\), while fixing the first site gives \(XA^i\) by cyclicity of the trace. Fixing a prefix \(u\) gives \(\psi (u,\sigma )=\operatorname{tr}(A^uA^\sigma X)=\operatorname{tr}(A^\sigma XA^u)\), so the suffix restriction again has the form of a ground-space vector.

Choose \(Z_j\) from the long left-window condition and \(Y_u\) from the suffix ground condition. The two restrictions give the same vector in \(G_{L_0}(A)\), namely \(\Gamma _{L_0}(Z_jA^u)=\Gamma _{L_0}(A^jY_u)\). Injectivity of \(\Gamma _{L_0}\), obtained from \(L_0\)-block injectivity, gives \(Z_jA^u=A^jY_u\).

For \(K = 1\), the hypothesis gives \(Z_jA^i=A^jY_i\) for every letter \(i\). Every blocked word of length \(L_0\) is nonempty, so this identity extends to the coefficients of the blocked tensor \(A^{[L_0]}\). Since \(A^{[L_0]}\) is injective, the identity matrix is a linear combination of those blocked coefficients, and substituting this decomposition yields \(Z_j=A^jX\) for a common matrix \(X\). For larger \(K\), strip the first letter: for each \(i\), the matrices \(Z_jA^i\) satisfy the same compatibility hypothesis with words of length \(K-1\), so the induction hypothesis reduces the problem to the case \(K=1\).

Lemma 13.4.5.8 produces matrices \((Z_j)_j\) and \((Y_u)_u\) with \(Z_jA^u=A^jY_u\) for every prefix word \(u\) of length \(K\). Theorem 13.4.5.9 gives a common right factor \(X\) such that \(Z_j=A^jX\) for all \(j\). Therefore, each last-site restriction of \(\psi \) agrees with the corresponding last-site restriction of the ground-space vector determined by \(X\). Since every configuration is obtained by adjoining its final letter to its initial \((K+L_0)\)-tuple, the two states coincide, so \(\psi \) lies in \(G_{K+L_0+1}(A)\).

Induct on the extra length beyond \(L_0+1\). The induction hypothesis gives the long left-window ground condition after deleting the last site, while Definition 13.4.5.5 and the contiguous-interval restriction identities above identify every fixed-prefix suffix with one of the assumed length-\(L_0+1\) intervals. Theorem 13.4.5.10 then grows back the rightmost site.

Induct on the length \(m\) of a non-wrapping interval. The case \(m=L\) is the assumed local condition. If all length-\(m\) intervals lie in \(S_m\), then the two one-site restrictions of a length-\((m+1)\) interval are adjacent length-\(m\) intervals. Hence the length-\((m+1)\) interval lies in \((\mathbb {C}^d\otimes S_m)\cap (S_m\otimes \mathbb {C}^d)\), and the displayed identity places it in \(S_{m+1}\). Taking \(m=N\) and the interval starting at the first site gives \(\psi \in S_N\).

If \(\psi \in \mathcal G_{N,L}(A)\), then every length-\(L\) cyclic interval of \(\psi \) lies in \(G_L(A)\), and hence in \(S_L\). A non-wrapping interval is the cyclic interval starting at the same site. Therefore, all non-wrapping length-\(L\) intervals lie in \(S_L\), and Lemma 13.4.5.12 gives \(\psi \in S_N\).

For every cyclic window vector \(v\in G_{L+1}(A)\) and every last letter \(b\), one has \(v|_{\mathrm{last}=b}\in G_L(A)\). Thus \(\mathcal G_{N,L+1}(A)\subseteq \mathcal G_{N,L}(A)\), and iteration gives the general antitonicity.

Lemma 13.4.5.14 reduces the cyclic hypotheses to range \(L_0+1\). On every non-wrapping interval, the cyclic and contiguous restriction maps agree, so Theorem 13.4.5.11 applies.

The trace identity for this cyclic interval is tested against all words of length \(L_0\). Injectivity of \(\Gamma _{L_0}\), which follows from \(L_0\)-block injectivity, turns these test-word identities into (??).

Factor the cyclic word at the second cyclic position, rotate the trace, and use injectivity of \(\Gamma _{L_0}\) to remove the length-\(L_0\) test word.

First use Lemma 13.4.5.14 to reduce the cyclic constraints from range \(L\) to range \(L_0+1\). Then apply the two cyclic-interval compatibility lemmas at the two boundary positions to obtain the one-sided identities in (??).

Substitute \(\psi =\Gamma _{M+1}(X)\) in the two cyclic restrictions at the last site and at the second boundary-crossing support. Lemmas 13.5.1 and 13.5.2 then give (??).

Substitute \(\tau ^+_\eta (\mu )\) and \(\tau ^-_\eta (\mu )\) into Lemma 13.5.4. The exposed-complement identities of Lemma 13.5.6 identify both complementary products with \(A^\mu \).

Put the same letter \(\eta \) on the remaining sites. Fill the physical sites \(L_0,\ldots ,M-1\) with \(\mu \) for the support crossing the last site, and fill the sites \(1,\ldots ,M-L_0\) with \(\mu \) for the second boundary-crossing support. The two complement-extraction identities follow from the index arithmetic. The same outside-site calculation gives \(\operatorname{Res}^\rho _{M,L_0+1}\psi =\operatorname{Res}^{\tau ^+_\eta (\mu )}_{M,L_0+1}\psi \) for any \(\rho \) with \(\rho _{k+L_0}=\mu _k\), and \(\operatorname{Res}^\rho _{M+1-L_0,L_0+1}\psi =\operatorname{Res}^{\tau ^-_\eta (\mu )}_{M+1-L_0,L_0+1}\psi \) for any \(\rho \) with \(\rho _{k+1}=\mu _k\).

Apply the first-letter restriction to both equal length-\((L_0+1)\) restrictions. The two resulting length-\(L_0\) vectors are represented by \(\Gamma _{L_0}(Y_\rho A^j)\) and \(\Gamma _{L_0}(Y_\tau A^j)\), and block injectivity makes \(\Gamma _{L_0}\) injective. The two displayed special cases follow from Lemma 13.5.6. The length-\(L_0\) word forms are obtained by multiplying these equations on the right by \(A^\sigma \).

By Lemma 13.5.7,

Multiplying on the right by \(A^\sigma \) and applying the hypothesis gives the displayed product equation.

Fix \(j\) and \(\sigma \). Lemma 13.5.7 gives

Multiply (??) and (??) on the right by \(A^\sigma \) and use the assumed product equation for the chosen pair \((j,\sigma )\).

Use Lemma 13.4.5.3 for the \(L_0-1\) adjacent restrictions starting at \(i_0=M+1-L_0\). For \(0\le r{\lt}L_0-1\), the two endpoint words are determined by

Substituting these two words in the iterated transport identity gives the displayed product equation.

Apply Lemma 13.4.5.3 to the \(L_0-1\) restrictions beginning at \(M+1-L_0+r\), with \(Y_r(\rho )=Y_{M+1-L_0+r}(\rho )\). With \(i_0=M+1-L_0\), the site labels satisfy

Hence, for \(r=0,\ldots ,L_0-2\),

Multiplying these \(L_0-1\) equations from left to right gives the displayed product equation. The right-multiplied form follows by multiplying this equality by \(R\).

Lemma 13.5.11 gives

The one-sided equation for the window beginning at \(M\) gives \(Y_M(\rho )A^j =A^{\rho _{L_0}}\cdots A^{\rho _{M-1}}A^jX\). Substitution gives the displayed formula.

Apply Lemma 13.4.5.3 to the \(M+2-L_0\) restrictions beginning at \(M+r\) modulo \(M+1\). With \(i_0=M\), the site labels satisfy

The final window is \(i_0+M+2-L_0\equiv M+1-L_0\pmod{M+1}\). Substituting these site labels in the iterated product identity gives the displayed equation.

Define \(Y_\mu \) to be the first boundary matrix evaluated on the boundary condition \(\tau ^+_\eta (\mu )\). Lemma 13.5.6 rewrites the first cyclic-window complement as \(\mu \), while the assumed comparison at \(\tau ^-_\eta (\mu )\) rewrites the second boundary matrix to the same \(Y_\mu \).

Multiply the second identity in (??) on the right by \(A^b\) and the first identity on the left by \(A^a\): \(X A^aA^\mu A^b =A^aY_\mu A^b =A^aA^\mu A^bX\).

The periodic boundary condition produces matrix identities of the form \(XMW=MY\) for arbitrary test matrices \(M\) and complementary words \(W\). Spanning first over the length-\((L-1)\) tails and then over the complementary words, injectivity forces \((XM-MX)W=0\) for every \(W\), hence \(XM=MX\) for every matrix \(M\). In particular, \(X\) commutes with each letter \(A^j\).

For every \(\sigma \), \(V^{(N)}(A)(\sigma )=\operatorname{tr}(A^\sigma )=\operatorname{tr}(A^\sigma \mathbb {1}) =\Gamma _N(\mathbb {1})(\sigma )\).

Apply the standard finite-dimensional criterion that a nonzero space has dimension one exactly when all of its vectors are proportional to a single nonzero vector.

This is the defining cyclic-window condition for membership in \(\mathcal G_{N,L}(A)\), together with the definition of the local ground space as the range of \(\Gamma _L\).

Suppose for contradiction that \(V^{(N)}(A)=0\), i.e., \(\operatorname{tr}(A^\sigma )=0\) for every word \(\sigma \) of length \(N\). For any words \(w\) and \(u\) with \(|w|=N-L_0\) and \(|u|=L_0\), we have \(\operatorname{tr}(A^wA^u)=0\). Since \(A\) is \(L_0\)-block-injective, \(\{ A^u:|u|=L_0\} \) spans \(M_{D}(\mathbb {C})\). By nondegeneracy of the trace pairing, \(A^w=0\) for every \(|w|=N-L_0\). Repeatedly strip a length-\(L_0\) suffix: if every product of length \(m+L_0\) vanishes, then the spanning property of the length-\(L_0\) products implies that every product of length \(m\) vanishes. Eventually all products of some length \(r{\lt}L_0\) vanish. If \(r=0\), this gives \(\mathbb {1}=0\). If \(r{\gt}0\), every length-\(L_0\) product has a zero length-\(r\) prefix, so all such products vanish, contradicting that they span \(M_{D}(\mathbb {C})\).

Induct on \(K\). For \(K = 0\), the empty suffix gives \(Z_u = A^u Y_{\emptyset }\). For the successor step, fix the first physical index \(i\) and apply the induction hypothesis to the shortened suffix. This gives a matrix \(X_i\) satisfying \(Z_u A^i = A^u X_i\) for every word \(u\) of length \(L_0\). Extending this letter equation to all length-\(L_0\) words and decomposing \(\mathbb {1}\) in the spanning family \(\{ A^u : |u| = L_0\} \) yields a common factor \(X\).

Suppose \(Z_bM_1 = F_bY_1\) and \(Z_bM_2 = F_bY_2\) for every \(b\in B\). Then \(Z_b(M_1+M_2) = Z_bM_1+Z_bM_2 = F_b(Y_1+Y_2)\). For a scalar \(c\), one similarly has \(Z_b(cM_1) = cZ_bM_1 = F_b(cY_1)\). Thus the identity holds for every matrix in the span of the length-\(K\) word products. Since this span is all of \(M_{D}(\mathbb {C})\), apply it to the identity matrix.

Write each length-\(m\) word as a concatenation \(uw\) with \(|u| = L_0\) and \(|w| = m - L_0\). The hypothesis gives \(X A^u A^w = A^u A^w X = A^u(A^w X)\). Thus the family \(Z_u := X A^u\) satisfies the hypothesis of Theorem 13.7.1. One obtains \(X A^u = A^u R\) for all \(|u| = L_0\). Decomposing \(\mathbb {1}\) in the span of the length-\(L_0\) words shows \(R = X\), hence \(X A^u = A^u X\).

By Theorem 13.7.3, \(X\) commutes with every length-\(L_0\) word product. These products span \(M_{D}(\mathbb {C})\) by block injectivity, so linearity gives \(XM = MX\) for every \(M \in M_{D}(\mathbb {C})\).

Proceed by induction on \(q\). For \(q=0\), the claim is trivial. For the inductive step, let a word of length \((q+1)m\) be given and write it as a length-\(m\) prefix followed by a length-\(q\, m\) suffix. The prefix commutes with \(X\) by hypothesis, and the suffix commutes by the induction hypothesis.

The hypothesis \(\operatorname{span}\{ A^u : |u|=L_0\} =M_{D}(\mathbb {C})\) implies the same spanning statement at every positive multiple \(qL_0\). Each length-\(qL_0\) word factors as a length-\(k\) prefix followed by padding, so \(Z\) annihilates all generators of the full span and hence annihilates the identity. For the left-handed form, factor each length-\(qL_0\) word as a prefix followed by a length-\(k\) suffix and use the same spanning argument.

If \(Y_1 A^j = Y_2 A^j\) for every physical index \(j\), then induction on the word gives \(Y_1 A^w = Y_2 A^w\) for every nonempty word \(w\). Since \(A\) is \(L_0\)-block-injective, \(\operatorname{span}\{ A^w : |w|=L_0\} =M_{D}(\mathbb {C})\). Thus the left multiplication maps \(M\mapsto Y_1 M\) and \(M\mapsto Y_2 M\) are equal on \(M_{D}(\mathbb {C})\), and evaluating them on the identity gives \(Y_1=Y_2\). The other one-sided statement follows by the same argument, with right multiplication in place of left multiplication.

Apply Theorem 13.7.4 with \(M = A^i\).

Theorem 13.7.4 makes \(X\) commute with the whole matrix algebra. The center of \(M_{D}(\mathbb {C})\) consists of scalar matrices, so \(X=c\mathbb {1}\), and therefore \(\Gamma _N(X)=cV^{(N)}(A)\).

Lemma 13.7.5 amplifies the length-\(m\) commutation hypothesis to length \(L_0\, m \ge L_0\). Then Theorem 13.7.9 applies.

The identities in (??) give commutation with all words of length \(|\mu |+2\), a positive length. The positive-length containment theorem then applies.

Lemma 13.5.14 gives a family \(Y_\mu \) satisfying

Theorem 13.7.11 applies.

The cyclic-to-open-chain reduction writes any chain ground state as \(\Gamma _N(X)\). The two boundary-crossing cyclic windows give

Theorem 13.7.12 finishes.

Each length-\(L_0\) word \(s\) is the word \(w(b)\) of its block letter \(b\), and each length-\(L_0K\) word \(c\) regroups into \(K\) blocks, recovering it as the concatenated complement word \(\widetilde{w(c_b)}\) of an iterated block index \(c_b\); both identifications are bijective. Substituting these representations into the hypothesis and setting \(Y'_c=Y_{c_b}\) gives the asserted equation for every \(s\) and \(c\).

By Lemma 8.1.1.13, \(\operatorname{span}\{ A^u : |u|=L_0\} = M_{D}(\mathbb {C})\). Hence, for every \(M\in M_{D}(\mathbb {C})\) and every complementary word \(v\) of length \(K\), the displayed hypothesis extends by linearity in the length-\(L_0\) block to \(XMA^v=MY_v\). Taking \(M=\mathbb {1}\) gives

Taking \(M=A^j\) and substituting (??) gives \((X A^j-A^jX)A^v=0\) for every word \(v\) of length \(K\). Since \(K\le (K+1)L_0\) and \(K+1\ge 1\), Lemma 13.7.6 gives \(X A^j-A^jX=0\).

Let \(Y_\nu \) be the matrix representing the cyclic restriction whose window begins at the last site. The cyclic word is \(\alpha \, \nu \, j\) after deleting the first letter from the window and reading the complement, so the ground-space representation and cyclicity of the trace give

For every physical letter \(j\), Lemma 13.8.3 gives

Put \(Z=X A^\alpha A^\nu -A^\alpha Y_\nu \). For every physical letter \(j\), Then (??) gives \(\operatorname{tr}(A^jZ)=0\). Since the matrices \(A^j\) span the full matrix algebra, the trace pairing gives \((\forall j,\ \operatorname{tr}(A^jZ)=0) \quad \Longrightarrow \quad Z=0\). Hence \(X A^\alpha A^\nu =A^\alpha Y_\nu \).

Choose representation matrices for every cyclic restriction of length \(L_0+1\). Fix \(\alpha \) and \(\nu \). The boundary-crossing equations move the matrix \(X\) through the first boundary letter. In local boundary notation, this has the form

The adjacent boundary-window product then transports this comparison through the remaining \(L_0-1\) boundary letters:

Finally, the outside-label uniqueness lemma identifies the final boundary matrix with the matrix \(Y_\nu \) attached to the complementary word \(\nu \). Thus \(X A^\alpha A^\nu =A^\alpha Y_\nu \). Multiplying on the left by \(A^\beta \) and taking the trace gives the asserted identity for every length-\(L_0\) word \(\beta \).

The trace identities say exactly that, for each fixed pair \((\alpha ,\nu )\),

Since \(A\) is \(L_0\)-block-injective, \(\Gamma _{L_0}\) is injective by Lemma 13.4.3.3. Hence \(X A^\alpha A^\nu =A^\alpha Y_\nu \).

Theorem 13.8.5 gives, for every word \(\beta \) of length \(L_0\),

Theorem 13.8.6 then gives the displayed matrix equation.

The commutation relation \(X A^j=A^jX\) extends by induction to \(X A^w=A^wX\) for every word \(w\). Hence the second one-sided equation gives, for every \(k\), \(A^kV_\eta = X A^kA^\mu = A^kX A^\mu = A^kA^\mu X\). Lemma 13.7.7 gives \(V_\eta =A^\mu X\). The first one-sided equation then gives \(W_\eta A^j = A^\mu A^jX = A^\mu X A^j = V_\eta A^j\).

Apply Lemma 13.7.7 to the two matrices \(Y^+_{\tau ^+_\eta (\mu )}\) and \(Y^-_{\tau ^-_\eta (\mu )}\).

Equality of the cyclic restrictions gives, for every physical letter \(j\), \(Y_\rho A^j = Y_{\tau ^\pm _\eta (\mu )}A^j\). Lemma 13.7.7 identifies the two matrices from these one-site products.

At \((\sigma _0,\sigma _1,\ldots ,\sigma _L)\), the required equality is

Apply Lemma 13.8.11 to the two length-\((L_0+1)\) restrictions, using the displayed family of first-letter equalities.

Fix \(j\) and take the first-letter restriction of the two displayed length-\((L_0+1)\) restrictions. These are \(\Gamma _{L_0}(Y^+_{\eta ,\mu }A^j)\) and \(\Gamma _{L_0}(Y^-_{\eta ,\mu }A^j)\). The one-site product equality gives, for every \(j\),

Lemma 13.8.11 identifies the two original restrictions.

Apply Lemma 13.8.13 for each boundary letter \(\eta \), using the corresponding first-letter product equations.

The cyclic-window representation lemma gives matrices \(Y_M(\tau ^+_\eta (\mu ))\) and \(Y_{M+1-L_0}(\tau ^-_\eta (\mu ))\) such that

Theorem 13.8.1.8 gives, for every \(j\), \(Y_M(\tau ^+_\eta (\mu ))A^j = Y_{M+1-L_0}(\tau ^-_\eta (\mu ))A^j\). Lemma 13.8.13 identifies the two length-\((L_0+1)\) restrictions.

For the product indexed by \(j\) and \(\sigma \), specialize the preceding equality at the outside letter \(\eta =j\). Take \(\rho ^+_{j,\sigma }=\tau ^+_j(\mu )\) and \(\rho ^-_{j,\sigma }=\tau ^-_j(\mu )\). The complement equations are the two identities of Lemma 13.5.6. Fix \(j\). Applying the first-letter restriction to the displayed equality gives

where the two sides are identified by Lemma 13.4.5.3. By Lemma 13.4.3.3, \(Y_M(\tau ^+_j(\mu ))A^j = Y_{M+1-L_0}(\tau ^-_j(\mu ))A^j\). Right multiplication by \(A^\sigma \) gives the asserted product equation.

For each boundary letter \(\eta \), Lemma 13.8.13 gives

Applying these equalities, Lemma 13.8.16 gives the displayed boundary conditions and product equation.

Fix \(\eta \) and \(j\), and set \(Z_{\eta ,j} = Y_{M+1-L_0}(\tau ^-_\eta (\mu ))A^j-A^\mu A^jX\). The hypothesis gives \(Z_{\eta ,j}A^\sigma =0\) for every \(\sigma \) with \(|\sigma |=L_0\). Since \(A\) is \(L_0\)-block-injective, the products \(A^\sigma \) with \(|\sigma |=L_0\) span the full matrix algebra. Applying Lemma 13.7.6 gives \(Z_{\eta ,j}=0\).

Fix \(\eta \), \(j\), and \(\sigma \), and set

The hypothesis says that \(A^\alpha Z_{\eta ,j,\sigma }=0\) for every word \(\alpha \) of length \(L_0\). The left-handed form of Lemma 13.7.6 gives \(Z_{\eta ,j,\sigma }=0\).

Lemma 13.8.18 gives, for every \(\eta \) and \(j\), \(Y_{M+1-L_0}(\tau ^-_\eta (\mu ))A^j = A^\mu A^jX\). Combining this equation with \(Y_M(\tau ^+_\eta (\mu ))A^j=A^\mu A^jX\) gives \(Y_M(\tau ^+_\eta (\mu ))A^j = Y_{M+1-L_0}(\tau ^-_\eta (\mu ))A^j\). Lemma 13.8.17 then gives the displayed boundary conditions and product equation.

Lemma 13.8.19 gives

Lemma 13.8.20 then gives the displayed boundary conditions and product equation.

Lemma 13.5.5 gives the equation for the boundary-crossing window beginning at \(M\), \(Y_M(\tau ^+_\eta (\mu ))A^j = A^\mu A^jX\). The assumed left-multiplied coordinate equation and Lemma 13.8.21 then give the displayed boundary conditions and product equation.

The assumed equations and Lemma 13.8.19 give, for every \(\eta \), \(j\), and \(\sigma \),

Lemma 13.8.18 removes the right word: \(Y_{M+1-L_0}(\tau ^-_\eta (\mu ))A^j = A^\mu A^jX\). The equation for the boundary-crossing window beginning at \(M\), from Lemma 13.5.5, gives \(Y_M(\tau ^+_\eta (\mu ))A^j = A^\mu A^jX\). Hence the first-letter products agree, and Lemma 13.8.14 identifies the two restrictions.

Choose matrices \(Y_i(\tau )\) representing the length-\((L_0+1)\) restrictions. The one-sided boundary-product lemma gives, for every outside letter \(\eta \) and physical letter \(j\),

Lemma 13.8.7 gives matrices \(Y_\nu \) satisfying, for every word \(\alpha \) of length \(L_0\) and every complementary word \(\nu \), \(XA^\alpha A^\nu = A^\alpha Y_\nu \). Lemma 13.8.2 then gives, for every physical letter \(k\), the commutation identity \(XA^k=A^kX\). Lemma 13.8.8 gives \(Y_M(\tau ^+_\eta (\mu ))A^j = Y_{M+1-L_0}(\tau ^-_\eta (\mu ))A^j\). Lemma 13.8.14 then identifies the two boundary-crossing restrictions, giving the periodic-boundary comparison of [ CPGSV21 , lines 2078–2079 ] .

Restricting both sides to the first physical letter \(j\) gives

Since \(A\) is \(L_0\)-block-injective, \(\Gamma _{L_0}\) is injective, and the displayed product equality follows.

The preceding lemma gives \(Y_M(\tau ^+_\eta (\mu ))A^j = Y_{M+1-L_0}(\tau ^-_\eta (\mu ))A^j\). Multiplying on the left by \(A^\alpha \) and on the right by \(A^\sigma \) gives the desired comparison.

The matrices \(Y_i(\tau )\) give the local ground-space representations needed in the preceding lemma. Once the preceding boundary-restriction equality is proved, it gives

Applying the left-multiplied comparison for equal boundary restrictions gives

Once the comparison of the two cyclic restrictions is available, it gives

The one-sided equation for the boundary-crossing window beginning at \(M\) gives \(Y_M(\tau ^+_\eta (\mu ))A^j = A^\mu A^jX\). Substitution gives the displayed identity.

Let \(Y_i(\tau )\) be the representation matrices supplied by Lemma 13.6.7. By Theorem 13.8.1.8, \(Y_M(\tau ^+_\eta (\mu ))A^j = Y_{M+1-L_0}(\tau ^-_\eta (\mu ))A^j\). Fixing the first physical letter in the two length-\((L_0+1)\) windows gives

The two restrictions are therefore equal.

Theorem 13.8.1.4 gives, for every boundary letter \(\eta \), physical letter \(j\), and words \(\alpha ,\sigma \) of length \(L_0\),

Theorem 13.8.22 applies this left-multiplied comparison together with the one-sided boundary products and gives the displayed boundary conditions \(\rho ^+_{j,\sigma }\) and \(\rho ^-_{j,\sigma }\), including

Theorem 13.8.1.6 gives boundary conditions \(\rho ^+_{j,\sigma }\) and \(\rho ^-_{j,\sigma }\) satisfying

and

Theorem 13.5.9 replaces these auxiliary conditions by the displayed boundary conditions:

which is the product equality needed for the periodic-boundary comparison.

Fix \(j\) and set

Theorem 13.8.1.7 gives, after subtracting the two sides, \(Z_jA^\sigma =0\) for every word \(\sigma \) of length \(L_0\). Since \(A\) is \(L_0\)-block-injective, \(\operatorname{span}\{ A^\sigma :|\sigma |=L_0\} = M_D(\mathbb {C})\). Applying Lemma 13.7.6 gives \(Z_j=0\), hence

The reduced cyclic-window compatibilities at the boundary give, for every physical letter \(j\),

By Lemma 13.8.9, it is enough to prove, for every \(j\),

Write \(N=M+1\). The two boundary-crossing supports start at \(M\) and \(M+1-L_0\). Let \(Y_i(\tau )\) denote the cyclic-window representation matrix given by Lemma 13.6.7 at the window beginning at \(i\). The first step identifies the two abstract boundary-condition families with the two boundary-restriction matrices:

Together with the preceding identifications, Theorem 13.8.1.8 gives, for every physical letter \(j\),

Lemma 13.8.9 then gives the desired equality of boundary matrices.

Let \(\psi \in \mathcal G_{N,L}(A)\). The non-wrapping window conditions and Lemma 13.4.5.4 imply \(\psi \in G_{N}(A)\), so \(\psi = \Gamma _{N}(X)\) for some boundary matrix \(X\). The periodic boundary condition now applies Theorem 13.5.16 and forces \(X\) to commute with every letter \(A^{j}\). Since \(A\) is injective, the letters generate the full matrix algebra, hence \(X\) is scalar. Therefore \(\psi \) is proportional to \(V^{(N)}(A)\), while the reverse inclusion is Lemma 13.6.8.

For a chain ground state, the open-chain containment gives \(\psi =\Gamma _N(X)\). The two boundary-crossing cyclic-window constraints give

The boundary-crossing comparison is \(Y^+_{\tau ^+_\eta (\mu )}=Y^-_{\tau ^-_\eta (\mu )}\). These equations imply \(\psi \in \operatorname{span}\{ V^{(N)}(A)\} \).

By Theorem 13.8.2.1, the ground space is the line \(\operatorname{span}\bigl\{ V^{(N)}(A)\bigr\} \). It remains to show that the MPV is nonzero. If \(V^{(N)}(A)=0\), then Lemma 13.6.2 gives \(\Gamma _{N}(\mathbb {1})=\Gamma _{N}(0)\). Since \(A\) is injective and \(N {\gt} 0\), Theorem 13.4.3.1 forces \(\mathbb {1}= 0\), a contradiction. Therefore the spanning line is one-dimensional.