I Physics Examples of the Multi-Block Compression Theorem: Supporting Results

The compression theorems of Chapter 31 rely on several explicit coordinate constructions: explicit gauge transformations and integer arithmetic certificates, adapted bases and scaled matrix units for normality certificates, and a sign-weighted pair testing the sharpness of the remainder bound.

I.1 Explicit gauges and exact arithmetic

The compression datum of Theorem 29.5.1 carries its change of bond coordinates as an invertible map onto a graded coordinate space, one summand per slot. For explicit bond coordinates, this map is given by an invertible matrix and a labelling of coordinates by slots.

Definition I.1.1 Explicit gauge and entrywise integer coercion

✓

Let $\tau $ be a bijection between the graded coordinate space of a slot family and the bond index set $\{ 0,\ldots ,D_B-1\} $, and let $G$ and $G^{-1}$ be mutually inverse matrices in $M_{D_B}(\mathbb {C})$. The explicit gauge attached to $(\tau ,G)$ is the invertible map sending a bond vector $v$ to the graded vector $b\mapsto (Gv)(\tau b)$. For an integer matrix $X$, its entrywise coercion is the complex matrix with the same entries; for integer matrix product operator tensors $M$ and $N$, their integer bond-space product is $(M\mathbin {\cdot }N)^{ik}=\sum _jM^{ij}\otimes N^{jk}$.

Theorem I.1.2 Conjugation, coercion and the compression pairs of a slot

✓

Conjugation by the explicit gauge attached to $(\tau ,G)$ sends a bond matrix $A$ to the matrix $GAG^{-1}$ relabelled along $\tau $. Entrywise coercion is multiplicative, and for integer tensors $M, N$ and $c \in \mathbb {C}$,

\begin{align} (cM) \mathbin {\cdot } (cN) = c^2 (M \mathbin {\cdot } N). \notag \end{align}

If a compression datum uses the explicit gauge $(\tau ,G)$, then the compression maps for slot $s$ are given in coordinates by $W_s = \pi _s G$ and $V_s = G^{-1} \iota _s$, where $\pi _s$ selects the rows and $\iota _s$ embeds the columns belonging to slot $s$.

Proof ▶

The matrix of the explicit gauge is $G$ relabelled along $\tau $ in its rows, and the matrix of its inverse is $G^{-1}$ relabelled along $\tau $ in its columns, so conjugation is the stated matrix conjugation followed by the relabelling. Multiplicativity of the coercion is the fact that a sum of products of integers is the same computed in $\mathbb {Z}$ or in $\mathbb {C}$, and the statement about the bond-space product follows by expanding both sides entrywise. The two compression maps are the composites of the coordinate embedding and projection of the slot with the gauge and its inverse; evaluating the matrix products at a single graded coordinate leaves one row of $G$ and one column of $G^{-1}$.

Definition I.1.3 One-slot compression data

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For a stacked product of two integer matrix product operator tensors, read over the pair alphabet, the letter at $a=do+i$ is $(M\mathbin {\cdot }N)^{oi}$. A one-slot compression datum for a tensor $B$ onto a single target $C$ is the data of Theorem 29.5.1 for the one-element slot set: an explicit gauge $(\tau ,G)$ conjugating $B$ into a matrix that is block upper triangular for a flag with $C$ at one diagonal position. Two shapes recur in the examples below: a conjugation $GB^iG^{-1}=C^i$ for an invertible $G$ gives such a datum with no zero slots ($z=0$); and, for a four-dimensional $B$ and a one-dimensional $C$, an integer gauge sending every letter to a matrix with $C$ at the third diagonal position of a four-step flag and zero at the other three gives such a datum with three zero slots ($z=3$).

Theorem I.1.4 The remainder of a conjugate datum vanishes; word traces of a one-slot datum

✓

The remainder of a conjugate one-slot datum vanishes identically: the compression is an exact change of coordinates. For every one-slot datum with target tensor $C$ and nonempty word $w \ne \emptyset $,

\begin{align} \operatorname{tr}(B^w) = \operatorname{tr}(C^w), \notag \end{align}

or $\operatorname{tr}(B^w) = c^{|w|}\operatorname{tr}(C^w)$ when the target is carried with scalar weight $c$.

Proof ▶

Conjugation preserves the trace, which decomposes as the sum of traces of the diagonal blocks. For the upper block-triangular gauge, the diagonal blocks of a word product are the word products of the individual diagonal blocks; the zero-slot diagonal block is strictly upper triangular (or zero) and therefore has vanishing trace on any nonempty word $w \ne \emptyset $, leaving only the contribution of the target block $C$. The scalar factor carries across words by the length $|w|$.

Theorem I.1.5 Stacked products of block-diagonal tensors

✓

Let $M$ and $N$ be matrix product operator tensors whose letters are, after relabellings of their bond coordinates, block diagonal with blocks $M_k$ and $N_l$. Then every letter of the stacked product $M\mathbin {\cdot }N$ is, after a relabelling of its bond coordinates, block diagonal over the pairs $(k,l)$, with the block $(M_k\mathbin {\cdot }N_l)^{ij}$ at the pair $(k,l)$. The bond coordinates of the block $(k,l)$ are the pairs of a bond coordinate of $M_k$ and a bond coordinate of $N_l$.

Proof ▶

The entry of $M^{ij}\otimes N^{jk}$ at a pair of pairs of coordinates is the product of an entry of $M^{ij}$ and an entry of $N^{jk}$. It vanishes unless both row coordinates lie in the same block as the corresponding column coordinates, and it then equals the corresponding entry of $M_k^{ij}\otimes N_l^{jk}$. Summing over $j$ gives the claim.

Lemma I.1.6 Direct sums over a subring of the complex numbers

✓

Let $f\colon R\to \mathbb {C}$ be a ring homomorphism and let $M$ and $N$ be matrix product operator tensors with entries in $R$. The direct sum of the entrywise images of $M$ and $N$ under $f$ is the entrywise image of the direct sum of $M$ and $N$ formed over $R$.

Proof ▶

Both sides are block diagonal with the images of $M^{ij}$ and $N^{ij}$ on the diagonal, and $f(0)=0$ off the diagonal blocks.

Theorem I.1.7 Direct sums of one-slot compression data

✓

Let the letters of a tensor $B$ be, after a relabelling of the bond coordinates, block diagonal with blocks $B_k$, $k$ in a finite set $K$, and let every $B_k$ carry a one-slot compression datum onto a target $C_k$ with $z_k$ zero slots and remainder $R_k$. Then the direct sum of the gauges of the blocks is a compression datum of $B$ onto the family $(C_k)_{k\in K}$ in the sense of Theorem 29.5.1, with $z=\sum _kz_k$ zero slots, and its remainder is

\begin{align} R^a & =\bigoplus _{k\in K}R_k^a, \end{align}

relabelled in the same way. Consequently a word $w$ satisfies $R^w=0$ if and only if $R_k^w=0$ for every $k$: the nilpotency order of the remainder of $B$ is the largest nilpotency order of the remainders of the blocks.

Proof ▶

In the direct sum of the gauges, the letters of $B$ are block diagonal with the letters of the blocks in their own block coordinates. Ordering the blocks of $B$ block by block, each in its own order, makes every letter block upper triangular, with the targets and the zero diagonal blocks of the summands on the diagonal. Removing the diagonal blocks of a block-diagonal matrix removes the diagonal blocks of each summand, so the remainder is the direct sum of the remainders, and so are its words.

Theorem I.1.8 Integer certificates for normality and for vanishing sitewise intertwiners

✓

Let $A \in (M_{D}(\mathbb {C}))^d$ have integer entries $A_i \in \mathrm{M}_D(\mathbb {Z})$. If there exist index pairs $(a_k, b_k)$, integer coefficients $T_{xyk} \in \mathbb {Z}$, and a nonzero integer $m \ne 0$ such that for all matrix units $E_{xy}$,

\begin{align} \sum _k T_{xyk} A_{a_k} A_{b_k} = m E_{xy}, \notag \end{align}

then $\operatorname{span}_{\mathbb {C}}\{ A_i A_j : i,j \in \{ 0,\dots ,d-1\} \} = M_{D}(\mathbb {C})$. Similarly, let $B \in (\mathrm{M}_D(\mathbb {Z}))^d$ and $c \in (\mathrm{M}_1(\mathbb {Z}))^d$. If a selection of rows from the matrices $B_i - c_i \mathbb {1}_D$ forms an integer matrix $K \in \mathrm{M}_D(\mathbb {Z})$ admitting an integer left inverse $K' K = m \mathbb {1}_D$ with $m \ne 0$, then every $v \in \mathbb {C}^D$ satisfying $B_i v = c_i v$ for all $i$ satisfies $v = 0$. Likewise, if a selection of columns from $B_i - c_i \mathbb {1}_D$ admits an integer left inverse up to $m \ne 0$, then every $u \in \mathbb {C}^D$ with $u B_i = c_i u$ satisfies $u = 0$.

Proof ▶

Dividing the linear combination by $m \ne 0$ exhibits each matrix unit $E_{xy}$ in the span of length-two words of $A$, showing they span $M_{D}(\mathbb {C})$. For the sitewise intertwiner, $B_i v = c_i v$ implies $K v = 0$, whence $m v = K' K v = 0$, so $v = 0$ since $m \ne 0$ in $\mathbb {C}$. The column argument for $u$ is identical.

The reduction of the square of the CZX tensor printed in the review (Section 31.3.2) replaces the letters by a diagonal block of an invariant subspace, following the construction of the canonical form in [ 17 , Section IV ] . When the complementary diagonal block vanishes, the replacement is a compression by rectangular matrices, and the following lemma shows that it preserves every nonempty word.

Lemma I.1.9 Compression onto an absorbing corner

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Let $B^0,\ldots ,B^{d-1}\in M_{D}(\mathbb {C})$, $W\in \mathrm{M}_{k\times D}(\mathbb {C})$ and $V\in \mathrm{M}_{D\times k}(\mathbb {C})$, and write $\widetilde B^i=WB^iV$. If $VWB^i=B^i$ for every $i$, then every nonempty word $w$ satisfies $VWB^w=B^w$,

\begin{align} \widetilde B^w & =WB^wV, \notag \end{align}

and $\operatorname{tr}(\widetilde B^w)=\operatorname{tr}(B^w)$. The same conclusions hold, with $B^wVW=B^w$ in place of $VWB^w=B^w$, if $B^iVW=B^i$ for every $i$.

Proof ▶

For the left condition, $VWB^w=B^w$ follows from the first letter of $w$. By induction on the length, for $w=aw'$ with $w'$ nonempty,

\begin{align} \widetilde B^a\widetilde B^{w'} & =WB^a(VWB^{w'})V=WB^aB^{w'}V. \notag \end{align}

Cyclicity of the trace then gives $\operatorname{tr}(WB^wV)=\operatorname{tr}(VWB^w)=\operatorname{tr}(B^w)$. The right condition is symmetric: $B^wVW=B^w$ follows from the last letter, the induction step uses $B^aVW=B^a$, and $\operatorname{tr}(WB^wV)=\operatorname{tr}(B^wVW)=\operatorname{tr}(B^w)$.

I.2 Adapted bases and scaled matrix units

The stacked bond spaces of vertical components are tensor squares. Reordering the four bond bits separates the two middle bits, which form the inner bond, from the two outer bits; the intermediate physical-index sum acts on the inner bond. The Bell vectors together with the two remaining coordinate vectors form the adapted basis used throughout. Each weighted copy of a target contributes its weight raised to the number of sites $N$, and weights are summed when they multiply the same target as in Theorem 29.7.2. The results below establish tensor decompositions and periodic trace identities, not a classification of length-dependent renormalization fixed points.

Definition I.2.1 The idempotents of an adapted basis

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For mutually inverse matrices $U$ and $U^{-1}$, define $\Pi _j=U^{-1}E_{jj}U$, the product of the $j$-th column of $U^{-1}$ with the $j$-th row of $U$. Reorder a stacked bond quadruple into its two middle bits and its two outer bits. On the inner bond choose the adapted basis with first two vectors $(1,0,0,\pm 1)$ and last two vectors the remaining coordinate vectors. The Bell idempotents carry the weighted targets. Their tensor products with target matrices are expressed in the original stacked coordinates by undoing the reordering.

Theorem I.2.2 The idempotents of an adapted basis are mutually orthogonal

✓

The idempotents satisfy $\Pi _j^2=\Pi _j$, $\Pi _j\Pi _k=0$ for $j\ne k$, and $U\Pi _jU^{-1}=E_{jj}$.

Proof ▶

Conjugation by $U$ returns $E_{jj}$, and products reduce to $E_{jj}E_{kk}=\delta _{jk}E_{jj}$.

Theorem I.2.3 From an integer matrix identity to a weighted direct sum

✓

Suppose $B=c_B\widehat B$ and $C_s=c_C\widehat C_s$, where the tensors with hats have integer entries and $s\in \{ 0,1\} $. Suppose a nonzero integer $n$ and integers $k_s$ satisfy, for each $s$ and each physical index $a$,

\begin{align} n\, \mu _s\, \frac{c_C}{2} & =k_s\, c_B, \label{eq:asymex_stacked_pair_scalar} \\ n\, \widehat B^a & =\sum _sk_s\, (\widehat\Pi _s\otimes \widehat C_s^a), \label{eq:asymex_stacked_pair_letters} \end{align}

where $\widehat\Pi _s=2\Pi _s$ and the right-hand side of (3) is in stacked bond coordinates. Then $B^a=\sum _s\mu _s(\Pi _s\otimes C_s^a)$ in those coordinates, the hypothesis of Proposition 29.5.5. Here $\mu _s$ denotes a scalar fusion weight, not a vertical multiplicity matrix.

Proof ▶

Using the idempotents of Theorem I.2.2, the $s$-th summand is $\mu _sc_C(\widehat\Pi _s\otimes \widehat C_s^a)/2$. Multiplication by $n$ and substitution of (2) and (3) give $nB^a$. Cancel the nonzero scalar $n$.

Theorem I.2.4 Normality from scaled matrix units

✓

Let a tensor be one nonzero scalar times an integer tensor. Suppose that for every bond position $(x,y)$ a letter of the integer tensor is a nonzero integer multiple of $E_{xy}$. Then its one-site matrices span the full bond matrix algebra, so it is normal at blocking length one in the sense of Definition 2.5.5. This establishes injectivity, not transfer spectral-radius normalization; a normal tensor (NT) in the RFP canonical form additionally requires that normalization.

Proof ▶

Dividing each designated letter by its nonzero factor exhibits every matrix unit in the one-site span.

I.3 A sign-weighted pair with a non-split extension

The sign-weighted pair $(A,-A)$ of bond dimension two and a bond-dimension-five source $B$ carrying both targets together with a one-dimensional zero slot illustrate the sharpness of Theorem 29.5.1: the remainder bound is attained (Theorem I.3.2.1), the periodic trace identity $\operatorname{tr}(B^w)=(1+(-1)^{|w|})\operatorname{tr}(A^w)$ (22) acts as an even-length filter (Theorem I.3.2.2), and a word-level compression need not arise from sitewise intertwiners ($B^iV=V(-A^i)\implies V=0$, Theorem I.3.2.3).

Consider a two-state physical site with basis $\{ |0\rangle ,|1\rangle \} $. Define the bond-two MPS tensor by

\begin{align} A^0=P=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \qquad A^1=X=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. \label{eq:asymex_parity_letters} \end{align}

For $N\geq 1$, write its periodic amplitudes as

\begin{align} a_N(x)=\operatorname{tr}(A^{x_0}\cdots A^{x_{N-1}}) \qquad (x\in \{ 0,1\} ^N). \label{eq:asymex_parity_amplitudes} \end{align}

The associated unnormalized state is

\begin{align} |V^{(N)}(A)\rangle =\sum _{x\in \{ 0,1\} ^N}a_N(x)|x\rangle , \label{eq:asymex_parity_state} \end{align}

where $|x\rangle =|x_0\rangle \otimes \cdots \otimes |x_{N-1}\rangle $. As in the elementary MPS examples of Refs. [ 229 , Section II ] and [ 17 , Appendix A ] , the state can be understood directly from its tensor letters, with an on-site $\mathbb {Z}_2$ symmetry reflected on the virtual bond space: $P=|0\rangle \! \langle 0|$ projects onto virtual coordinate $0$, while $X=|0\rangle \! \langle 1|+|1\rangle \! \langle 0|$ exchanges the two virtual coordinates.

Multiplying both letters of $A$ by $-1$ gives one minus sign per site, whatever the physical labels. For a word $w=(i_1,\ldots ,i_{|w|})$, write $A^w=A^{i_1}\cdots A^{i_{|w|}}$, with the identity matrix as the empty product. Then

\begin{align} (-A)^w=(-1)^{|w|}A^w. \label{eq:asymex_parity_weight} \end{align}

The sum of periodic amplitudes $\operatorname{tr}(A^w)+\operatorname{tr}((-A)^w)=(1+(-1)^{|w|})\operatorname{tr}(A^w)$ therefore vanishes on all odd lengths $|w|$. The following five-dimensional tensor realizes that sum without being a direct sum in its displayed coordinates.

Definition I.3.1 The graded pair and the five-dimensional source

✓

Let $A$ be the tensor in (4). The two target tensors are $C_+=A$ and $C_-=-A$. The source is the bond-dimension-five tensor with letters

\begin{align} B^0 & =\begin{pmatrix} -1 & 1 & -1 & 1 & 3 \\ 2 & 2 & -2 & -1 & 0 \\ 1 & 2 & -2 & 0 & 2 \\ 1 & 3 & -3 & -1 & 2 \\ -1 & 0 & 0 & 1 & 2 \end{pmatrix} \label{eq:asymex_parity_source_zero} \end{align}

and

\begin{align} B^1 & =\begin{pmatrix} 3 & 1 & 0 & -2 & -3 \\ -3 & 1 & -1 & 2 & 5 \\ -1 & 2 & -1 & 1 & 3 \\ 0 & 1 & -1 & -1 & 1 \\ 2 & 1 & 0 & -1 & -2 \end{pmatrix}. \label{eq:asymex_parity_source_one} \end{align}

Neither source letter has a zero row or a zero column. Nevertheless, a simultaneous change of bond coordinates reveals diagonal blocks $A$, $-A$, and a one-dimensional zero block. The latter contributes nothing to positive-length traces.

$\begin{tenkz}[cols=4, boundary=periodic, physical=up] \tn[label pos=135, ports={90:physical:$i_1$}]{B} & \tn[label pos=135]{B} & \tn[skin=dots]{} & \tn[label pos=135, ports={90:physical:$i_N$}]{B} \end{tenkz}$ $\operatorname{tr}(B^w)$, computed in Theorem I.3.2.2

Theorem I.3.2 Normality of the graded block

✓

The block $A$ is normal: its length-two words span $M_{2}(\mathbb {C})$.

Proof ▶

Let $E_{ab}$ denote the matrix unit with its sole nonzero entry equal to one in row $a$, column $b$. Direct multiplication gives $E_{00}=A^0A^0$, $E_{01}=A^0A^1$, and $E_{10}=A^1A^0$. The remaining unit is $E_{11}=A^1A^1-A^0A^0$, so all four matrix units lie in the length-two word span.

I.3.1 Occupation parity and periodic amplitudes

Call $n_1(x)=\sum _{v=0}^{N-1}x_v$ the occupation of the configuration $x$. Let $z=\operatorname{diag}(1,-1)$ act on the physical site, and let $Z=\operatorname{diag}(1,-1)$ act on the bond space. Although their matrices are equal, these operators act on different spaces. The physical parity operator has basis action

\begin{align} z^{\otimes N}|x\rangle =(-1)^{n_1(x)}|x\rangle . \label{eq:asymex_parity_basis} \end{align}

Theorem I.3.1.1 Occupation parity

✓

For each physical label $i\in \{ 0,1\} $, the tensor satisfies

\begin{align} ZA^iZ=(-1)^iA^i. \label{eq:asymex_parity_covariance} \end{align}

Consequently, $a_N(x)=0$ whenever $n_1(x)$ is odd, and

\begin{align} z^{\otimes N}|V^{(N)}(A)\rangle =|V^{(N)}(A)\rangle . \label{eq:asymex_parity_invariance} \end{align}

Proof ▶

Direct multiplication gives $ZPZ=P$ and $ZXZ=-X$. Since $Z^2=\mathbb {1}_2$, multiplying (11) along the word yields

\begin{align} Z(A^{x_0}\cdots A^{x_{N-1}})Z =(-1)^{n_1(x)}A^{x_0}\cdots A^{x_{N-1}}. \notag \end{align}

Taking the trace and using cyclicity gives $a_N(x)=(-1)^{n_1(x)}a_N(x)$. Thus odd occupation forces $a_N(x)=0$. Applying (10) to the expansion (6) proves (12).

The support of the state is further constrained by the relative positions of the zeros.

Theorem I.3.1.2 Amplitudes from cyclic runs

✓

Suppose $x$ contains $m\geq 1$ zeros. Read the sites cyclically, and let $r_1,\ldots ,r_m$ be the numbers of ones between successive zeros. Define $b(r)=1$ for even $r$ and $b(r)=0$ for odd $r$. Then

\begin{align} a_N(x)=\prod _{j=1}^{m}b(r_j). \label{eq:asymex_parity_runs} \end{align}

For the word with no zeros,

\begin{align} a_N(1,\ldots ,1)=1+(-1)^N. \label{eq:asymex_parity_all_one} \end{align}

Proof ▶

The identities $X^2=\mathbb {1}_2$ and $PXP=0$ imply $PX^rP=b(r)P$ and $\operatorname{tr}(PX^r)=b(r)$. Rotate the word to start at a zero, using cyclicity of the trace. Successive contractions give

\begin{align} a_N(x) & =\operatorname{tr}(PX^{r_1}PX^{r_2}\cdots PX^{r_m}) \notag \\ & =\left(\prod _{j=1}^{m-1}b(r_j)\right)\operatorname{tr}(PX^{r_m}) =\prod _{j=1}^{m}b(r_j). \notag \end{align}

For the all-one word, $X^N=\mathbb {1}_2$ when $N$ is even and $X^N=X$ when $N$ is odd. Their traces are $2$ and $0$, respectively, which proves (14).

Thus a word containing a zero has amplitude one exactly when every cyclic run of ones has even length. For example, $0101$ has even occupation but zero amplitude, since both runs have length one. The all-zero word has amplitude one at every positive length, so $|V^{(N)}(A)\rangle $ is always nonzero in the specified computational basis.

I.3.2 Asymmetric compression and length cancellation

Theorem I.3.2.1 Compression of the graded pair

✓

The source of Definition I.3.1 carries the block triangular form of Theorem 29.5.1 for $S=\{ +,-\} $, with targets $C_+=A$, $C_-=-A$, one zero slot $z=1$, and dimension count $5=2+2+1$. There are compression maps $V_s\in \mathbb {C}^{5\times 2}$ and $W_s\in \mathbb {C}^{2\times 5}$ such that, for every $s,t\in S$,

\begin{align} W_sV_t=\delta _{st}\mathbb {1}_2. \label{eq:asymex_parity_biorthogonal} \end{align}

For every word $w$ and each $s\in S$,

\begin{align} W_sB^wV_s=C_s^w. \label{eq:asymex_parity_reduction} \end{align}

Define the remainder letters by

\begin{align} R^i=B^i-V_+A^iW_+-V_-(-A^i)W_-. \label{eq:asymex_parity_remainder} \end{align}

Every product $R^w$ with $|w|\geq 3$ vanishes, whereas $R^0R^0\neq 0$. Thus the bound $|S|+z=3$ in clause (vi) of Theorem 29.5.1 is attained.

Proof ▶

Use the integer change of coordinates

\begin{align} G=\begin{pmatrix} 0 & -1 & 1 & 0 & 0 \\ 1 & 1 & 0 & -1 & -1 \\ -1 & 0 & 0 & 1 & 1 \\ 1 & 0 & 0 & 0 & -1 \\ 0 & 1 & -1 & 0 & 1 \end{pmatrix}. \label{eq:asymex_parity_gauge} \end{align}

Its inverse is

\begin{align} G^{-1}=\begin{pmatrix} 1 & 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 0 & 1 \end{pmatrix}. \label{eq:asymex_parity_gauge_inv} \end{align}

Direct multiplication verifies both inverse identities and gives

\begin{align} GB^0G^{-1} =\begin{pmatrix} 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & -1 & 0 & 2 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}. \label{eq:asymex_parity_conj_zero} \end{align}

For the second letter,

\begin{align} GB^1G^{-1} =\begin{pmatrix} 0 & 1 & 0 & 1 & 0 \\ 1 & 0 & 2 & 0 & 1 \\ 0 & 0 & 0 & -1 & 1 \\ 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}. \label{eq:asymex_parity_conj_one} \end{align}

With block sizes $(2,2,1)$, the matrices in (20) and (21) are block upper triangular, with diagonal blocks $A^i$, $-A^i$, and $0$. As in Theorem I.1.2, take $W_+$ and $W_-$ to be rows $(0,1)$ and $(2,3)$ of $G$, respectively, and $V_+$ and $V_-$ to be the corresponding columns of $G^{-1}$. The inverse identities give (15). Diagonal blocks multiply under word evaluation, giving (16).

Subtracting the two diagonal target blocks leaves a strictly upper triangular three-block remainder in these coordinates. The compression remainder bound therefore gives $R^w=0$ for $|w|\geq 3$. To check sharpness in the original coordinates, substitution into (17) gives

\begin{align} R^0=\begin{pmatrix} -1 & 2 & -2 & 1 & 3 \\ 1 & 2 & -2 & 0 & 1 \\ 0 & 3 & -3 & 1 & 3 \\ 0 & 3 & -3 & 0 & 3 \\ -1 & 1 & -1 & 1 & 2 \end{pmatrix}. \notag \end{align}

Its square has $(0,1)$ entry $2$, so $R^0R^0\neq 0$.

Theorem I.3.2.2 Periodic trace of the sign-weighted pair

✓

For every nonempty word $w$,

\begin{align} \operatorname{tr}(B^w)=(1+(-1)^{|w|})\operatorname{tr}(A^w). \label{eq:asymex_parity_trace} \end{align}

Proof ▶

Proposition 29.5.3 applied to the datum of Theorem I.3.2.1 gives $\operatorname{tr}(B^w)=\operatorname{tr}(A^w)+\operatorname{tr}((-A)^w)$. By (7), the second summand is $(-1)^{|w|}\operatorname{tr}(A^w)$, proving (22).

Define $|V^{(N)}(B)\rangle $ by the same periodic contraction as in (6), with $A$ replaced by $B$. Applying (22) to each basis coefficient gives, for $N\geq 1$,

\begin{align} |V^{(N)}(B)\rangle =(1+(-1)^N)|V^{(N)}(A)\rangle . \label{eq:asymex_parity_length} \end{align}

The source acts as a length filter: at odd length the periodic vector vanishes, while at even length every amplitude of $A$ is doubled. For instance, at odd length the all-zero amplitude of $A$ is one but its amplitude in $B$ vanishes; at even length the run restrictions of Theorem I.3.1.2 apply with doubled amplitudes.

The positive-length hypothesis is necessary. The empty-word traces are $\operatorname{tr}(\mathbb {1}_5)=5$ and $\operatorname{tr}(\mathbb {1}_2)=2$, whereas substituting $|w|=0$ into (22) would give $5=4$. The missing contribution is the empty trace of the one-dimensional zero block, which is one; its traces vanish only at positive length.

Theorem I.3.2.3 The weight $-1$ block has no sitewise intertwiner

✓

If $V\in \mathbb {C}^{5\times 2}$ satisfies $B^iV=V(-A^i)$ for both letters, then $V=0$; if $W\in \mathbb {C}^{2\times 5}$ satisfies $WB^i=(-A^i)W$ for both letters, then $W=0$.

Proof ▶

Each condition is a homogeneous linear system in ten unknown entries. For the right intertwiner, select entries $(3,0)$, $(4,0)$, $(2,1)$, $(3,1)$, and $(4,1)$ of $B^0V=V(-A^0)$, and all five entries of column $0$ of $B^1V=V(-A^1)$. Rational linear combinations of these ten equations force all entries of column $0$ of $V$ to vanish. The selected equations for letter $1$ then force column $1$ to vanish.

For the left intertwiner, select entries $(0,4)$, $(1,0)$, $(1,2)$, $(1,3)$, and $(1,4)$ of $WB^0=(-A^0)W$, and all five entries of row $0$ of $WB^1=(-A^1)W$. Rational linear combinations force row $0$ of $W$ to vanish, after which the letter-$1$ equations force row $1$ to vanish.

The nonzero maps $W_-$ and $V_-$ in (16) therefore give a word-level compression onto $-A$ without either map being a sitewise intertwiner. The trace identity and the sharp remainder bound hold despite this obstruction.