January 01, 1970
The stabilizer fragment of the ZX calculus is amongst the most important fragments of the theory. The closely related Clifford+T fragment is approximately universal [@jeandel2018complete]. Additionally, the stabilizer calculus can be described by a small collection of rewrites, most of which have been shown to be necessary [@backens2020towards]. However, two rules, describing the red/green compact-structure coincidence and the important bialgebra law, had not been shown to be necessary. We present a countermodel-style argument showing that both of these rules are individually necessary relative to the connectivity meta-rule of Backens–Perdrix–Wang [@backens2020towards], and hence establish that the rule set presented in [@backens2020towards] has no redundant rewrite rule.
The ZX calculus is a graphical language for reasoning about quantum computation [@coecke2011interacting], arising out of the study of Categorical Quantum Mechanics (CQM) [@abramsky2004categorical]. The stabilizer formalism [@gottesman1997stabilizercodesquantumerror] is central in quantum error correction [@nielsen2010quantum] and measurement-based quantum computation [@raussendorf2001one], and the ZX calculus has also recently been applied to photonic quantum computing [@de2022quantum]. The ZX calculus and CQM have also found their place as a means of communicating quantum mechanical ideas to a broader audience, without the need for matrix formalism [@dundar2023quantum]. We refer the interested reader to the following survey by van de Wetering for an overview of the ZX calculus [@van2020zx], and the books by Coecke–Kissinger [@coecke2018picturing], or Heunen–Vicary [@heunen2019categories] for a categorical introduction to CQM.
The stabilizer fragment of the ZX calculus is an important cornerstone of the broader theory, both historically and practically. The stabilizer fragment restricts spider phases to integer multiples of \(\pi/2\), and was the first version of the calculus to be found complete [@backens2014zx]. Wider completeness results have since been established [@ng2017universal], and the closely related Clifford+\(T\) fragment, with phases in integer multiples of \(\pi/4\), is approximately universal. However, the stabilizer fragment remains an important object of study. The rule set for the stabilizer ZX calculus also captures, in some sense, some of the most fundamental quantum properties among its rules.
Backens et al. [@backens2020towards] established a significantly smaller complete rule set for the stabilizer ZX calculus with just nine rules (Table 1) and one meta-rule “only connectivity matters.” Previous rule sets usually contained more explicit rules and conventions allowing for flipping of diagrams or switching colors, whereas this rule set allowed for their derivation instead. Backens et al. proved that seven of the nine rules were necessary (that is, they could not be derived from the other rules) for the overall rule set to be complete, and amongst the remaining two rules, (S3’R) and (B2’), at least one was necessary. The rule (S3’R) essentially says that the ‘red’ compact structure coincides with the ‘green’ one (which coincides with the ambient compact structure of the category by (S3’L)). It’s interesting that proving independence of the (B2’) bialgebra rule was so elusive, since this rule describes an important feature of the ZX calculus: complementarity. Notably absent from the ZX_simp rule set is the Hopf rule, which lets one ‘chop off’ double wires connecting red and green spiders. This is usually derived from the bialgebra law (or vice versa) so it would be surprising to be able to derive either of these rules from the remaining axioms.
In this paper, we prove that both of the remaining rules, (S3’R) and (B2’) are, in fact, individually necessary relative to the connectivity meta-rule of Backens–Perdrix–Wang [@backens2020towards]. This means that ZX_simp + the meta-rule has no redundant rewrite rule. This also reinforces that the complementarity property encoded by the bialgebra law, and the compact structure induced by the ‘red’ spider are distinct properties from each other, as well as the rest of the rule set.
We now describe the stabilizer ZX calculus in more depth, and outline some of the conventions we use. In this paper, we mirror the setup of Backens et al. [@backens2020towards] for continuity.
In the stabilizer ZX calculus, a diagram \(D\) with \(k\) inputs and \(l\) outputs \(D:k\to l\) is composed of the following generators:
| Name | Diagram | Typed Version |
|---|---|---|
| Green Spider | \(\vcenter{}\) | \(Z^\alpha_{n,m}:n\to m\) |
| Red Spider | \(\vcenter{}\) | \(X^\alpha_{n,m}:n\to m\) |
| Hadamard | \(\vcenter{}\) | \(H:1\to 1\) |
| Identity | \(\vcenter{}\) | \(1_1:1\to 1\) |
| Braiding | \(\vcenter{}\) | \(\sigma:2\to 2\) |
| Cup | \(\vcenter{\begin{figure}\includegraphics[width=0.8\textwidth]{_pdflatex/oajfvchd.png}\label{fouqdzrn}\end{figure} }\) | \(\eta:0\to 2\) |
| Cap | \(\vcenter{}\) | \(\epsilon:2\to 0\) |
| Unit (empty diagram) | \(\vcenter{}\) | \(e:0\to 0\) |
We can compose diagrams in two ways:
Sequentially: If \(D_1:a \to b\) and \(D_2:b \to c\) are two diagrams, we can compose \(D_2 \circ D_1 : a \to c\) by placing \(D_1\) above \(D_2\). We can think of this new process as performing the process \(D_1\) first, and then applying the process \(D_2\) next.
Spatially: If \(D_1:a\to b\) and \(D_2:c \to d\) are two diagrams, we can compose \(D_1 \otimes D_2: a+c \to b+d\) by placing \(D_1\) to the left of \(D_2\). We can think of this new process as performing the processes \(D_1\) and \(D_2\) side-by-side.
These operations satisfy the coherence conditions for a strict monoidal category. See [@heunen2019categories] for details.
Importantly, we can define the standard interpretation functor [@coecke2011interacting]:
For any ZX diagram \(D:n\to m\), \(\left\llbracket D \right\rrbracket:(\mathbb{C}^2)^{\otimes n}\to(\mathbb{C}^2)^{\otimes m}\) is defined inductively in the following way:
\[\left\llbracket D_1\otimes D_2 \right\rrbracket =\left\llbracket D_1 \right\rrbracket\otimes\left\llbracket D_2 \right\rrbracket, \quad \left\llbracket D_2\circ D_1 \right\rrbracket =\left\llbracket D_2 \right\rrbracket\circ\left\llbracket D_1 \right\rrbracket.\]
\[\begin{array}{rcl@{\qquad\qquad}rcl} \left\llbracket \vcenter{} \right\rrbracket &=& \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} & \left\llbracket \vcenter{} \right\rrbracket &=& \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} \\[2em] \left\llbracket \vphantom{\vcenter{}}\vcenter{\begin{figure}\includegraphics[width=0.8\textwidth]{_pdflatex/vagurtky.png}\label{urelsqmp}\end{figure} } \right\rrbracket &=& \ket{00}+\ket{11} & \left\llbracket \vcenter{} \right\rrbracket &=& \bra{00}+\bra{11} \end{array}\tag{1}\] \[\left\llbracket \vcenter{} \right\rrbracket = \left\{ \begin{array}{rcl} \ket{0\cdots 0} &\mapsto& \ket{0\cdots 0}\\ \ket{1\cdots 1} &\mapsto& e^{i\alpha}\ket{1\cdots 1}\\ \text{others} &\mapsto& 0 \end{array} \right.\] \[\left\llbracket \vcenter{} \right\rrbracket = \left\{ \begin{array}{rcl} \ket{+\cdots +} &\mapsto& \ket{+\cdots +}\\ \ket{-\cdots -} &\mapsto& e^{i\alpha}\ket{-\cdots -}\\ \text{others} &\mapsto& 0 \end{array} \right.\] \[\left\llbracket \vcenter{} \right\rrbracket =\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}\]
Now, we introduce the rule set ZX_simp, along with some facts previously established [@backens2020towards].
ZX_simp↩︎We follow the phase convention of Backens–Perdrix–Wang throughout. Stabilizer phases are integer-labelled: the label \(k\in\mathbb{Z}\) denotes the angle \(k\pi/2\). Thus \(Z^k_{n,m}\) and \(X^k_{n,m}\) denote spiders with phase \(k\pi/2\). The addition in (S1) is ordinary integer addition, not addition modulo \(4\). In particular, \(k\) and \(k+4\) are not identified at the level of syntax. The \(2\pi\)-periodicity is a theorem of the full rule set of [@backens2020towards], not a syntactic quotient that we impose here.
| Rule | Diagram | Typed version |
|---|---|---|
| Rule | Diagram | Typed version |
| (S1) | \(\begin{aligned} &(1_q\otimes Z^\ell_{r+1,s})\circ (Z^k_{p,q+1}\otimes 1_r) = Z^{k+\ell}_{p+r,q+s}: p+r\to q+s,\\ &p,q,r,s\in\mathbb{N},\quad k,\ell\in\mathbb{Z}. \end{aligned}\) | |
| (S3\(^{\prime}\)L) | \(\vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=none] (l) at (0, -0.25) {}; \node [style=none] (r) at (1, -0.25) {}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw [in=90, out=90, looseness=1.75] (l.center) to (r.center); \end{pgfonlayer} \end{tikzpicture}}=\vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=none] (l) at (0, -0.25) {}; \node [style=none] (r) at (1, -0.25) {}; \node [style=gn] (g) at (0.5, 0.25) {}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw [in=90, out=90, looseness=1.75] (l.center) to (r.center); \end{pgfonlayer} \end{tikzpicture}}\) | \(\eta=Z^0_{0,2}:0\to 2.\) |
| (S3\(^{\prime}\)R) | \(\vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=none] (l) at (0, -0.25) {}; \node [style=none] (r) at (1, -0.25) {}; \node [style=gn] (g) at (0.5, 0.25) {}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw [in=90, out=90, looseness=1.75] (l.center) to (r.center); \end{pgfonlayer} \end{tikzpicture}}=\vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=none] (l) at (0, -0.25) {}; \node [style=none] (r) at (1, -0.25) {}; \node [style=rn] (x) at (0.5, 0.25) {}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw [in=90, out=90, looseness=1.75] (l.center) to (r.center); \end{pgfonlayer} \end{tikzpicture}}\) | \(Z^0_{0,2}=X^0_{0,2}:0\to 2.\) |
| (B1) | \((Z^0_{1,0}\circ X^0_{0,1})\otimes (Z^0_{1,2}\circ X^0_{0,1})=X^0_{0,1}\otimes X^0_{0,1}:0\to 2.\) | |
| (B2\(^{\prime}\)) | \(\begin{aligned} &Z^0_{0,0}\otimes\big((X^0_{2,1}\otimes X^0_{2,1}) \circ(1_1\otimes\sigma\otimes 1_1) \circ(Z^0_{1,2}\otimes Z^0_{1,2})\big)\\ &\qquad =(Z^0_{1,0}\circ X^0_{0,1})\otimes (Z^0_{1,2}\circ X^0_{2,1}):2\to 2. \end{aligned}\) | |
| (EU\(^{\prime}\)) | \(H=Z^1_{1,1}\circ X^0_{2,1}\circ (Z^1_{1,1}\otimes Z^{-1}_{0,1}):1\to 1.\) | |
| (H) | \(\begin{aligned} H^{\otimes m}\circ X^k_{n,m}\circ H^{\otimes n} &=Z^k_{n,m}:n\to m,\quad n,m\in\mathbb{N},\quad k\in\mathbb{Z}. \end{aligned}\) | |
| (IV\(^{\prime}\)) | \(Z^0_{0,0}\otimes (Z^0_{3,0}\circ X^0_{0,3})\otimes (Z^0_{3,0}\circ X^0_{0,3})=e:0\to 0.\) | |
| (ZO\(^{\prime}\)) | \(\begin{aligned} Z^2_{0,0}\otimes Z^0_{0,1} =Z^2_{0,0}\otimes X^0_{0,1}:0\to 1. \end{aligned}\) |
Definition 1 (Meta-rule: only connectivity matters). Two diagrams represent the same matrix whenever one can be transformed into the other by moving components around without changing their connections.
Theorem 1 ([@backens2020towards], Theorem 2.10). \(\mathrm{ZX}_{\mathrm{simp}}\) is sound and complete, i.e. for any two stabilizer ZX-calculus diagrams \(D_1\) and \(D_2\), we have: \[\mathrm{ZX}_{\mathrm{simp}}\vdash D_1=D_2\quad\Longleftrightarrow\quad \left\llbracket D_1 \right\rrbracket=\left\llbracket D_2 \right\rrbracket.\]
Theorem 2 ([@backens2020towards], Lemmas 3.1–3.8). Every rule in \(\mathrm{ZX}_{\mathrm{simp}}\setminus\{(\mathrm{S3}'\mathrm{R}),(\mathrm{B2}')\}\) is necessary, and at least one of (S3’R), (B2’) is necessary.
As we will see, it turns out that both of these rules are individually necessary, meaning every rule in \(\mathrm{ZX}_{\mathrm{simp}}\) is necessary relative to the connectivity meta-rule of Backens–Perdrix–Wang [@backens2020towards].
S3’R is necessary, as we argue below.
Theorem 3. The (S3’R) rule is necessary: \(\mathrm{ZX}_{\mathrm{simp}}\setminus\{(\mathrm{S3}'\mathrm{R})\}\nvdash(\mathrm{S3}'\mathrm{R})\).
A countermodel here means an interpretation satisfying every non-target rule while falsifying the target equation.
Proof. Work in ordinary complex matrices on qubits, with object \(1\) interpreted as \(\mathbb{C}^2\).
Let \(\left\llbracket - \right\rrbracket\) be the standard interpretation functor.
For any diagram \(D:n\to m\), define a new interpretation \(\left\llbracket D \right\rrbracket^{(\mathrm{S3}'\mathrm{R})}\in\mathbb{C}^{2^n\times 2^m}\) inductively: \[\left\llbracket D_1\otimes D_2 \right\rrbracket^{(\mathrm{S3}'\mathrm{R})} =\left\llbracket D_1 \right\rrbracket^{(\mathrm{S3}'\mathrm{R})}\otimes\left\llbracket D_2 \right\rrbracket^{(\mathrm{S3}'\mathrm{R})}, \quad \left\llbracket D_2\circ D_1 \right\rrbracket^{(\mathrm{S3}'\mathrm{R})} =\left\llbracket D_2 \right\rrbracket^{(\mathrm{S3}'\mathrm{R})}\circ\left\llbracket D_1 \right\rrbracket^{(\mathrm{S3}'\mathrm{R})},\] \[\left\llbracket \vcenter{} \right\rrbracket^{(\mathrm{S3}'\mathrm{R})} =\left\llbracket Z^\alpha_{n,m}:n\to m \right\rrbracket^{(\mathrm{S3}'\mathrm{R})} := i^{n+m-2}\cdot \left\llbracket Z^\alpha_{n,m} \right\rrbracket,\] \[\left\llbracket {\let\alpha\beta\vcenter{}} \right\rrbracket^{(\mathrm{S3}'\mathrm{R})} =\left\llbracket X^\beta_{n,m}:n\to m \right\rrbracket^{(\mathrm{S3}'\mathrm{R})} := -1\cdot \left\llbracket X^\beta_{n,m} \right\rrbracket,\] \[\left\llbracket \vcenter{} \right\rrbracket^{(\mathrm{S3}'\mathrm{R})} =\left\llbracket H:1\to 1 \right\rrbracket^{(\mathrm{S3}'\mathrm{R})} := i\cdot \left\llbracket H \right\rrbracket.\] We leave \(I,\sigma,\eta,\epsilon,e\) unchanged.
Lemma 1. For any ZX diagram \(D\), \[\left\llbracket D \right\rrbracket^{(\mathrm{S3}'\mathrm{R})}=c(D)\cdot\left\llbracket D \right\rrbracket,\] where \[\label{eq:scalar-cD} c(D)=(i)^{\sum_{v\in Z(D)}(\deg(v)-2)}\cdot(-1)^{|X(D)|}\cdot(i)^{|H(D)|}.\tag{2}\] Here \(\deg(v)\) is the number of incident half-edges, with self-loops counted twice, and \(Z(D)\), \(X(D)\), and \(H(D)\) are the collections of \(Z\)-spider, \(X\)-spider, and Hadamard vertices in the diagram.
Proof. See 5.1. ◻
The scalar \(c(D)\) is invariant under graph isomorphism, so the connectivity meta-rule is sound.
Now, we can just compare the scalar \(c(D)\) on both sides of each rewrite, since the standard interpretation agrees with each of them.
\[\begin{array}{@{}c|c@{}} \toprule \text{Rule} & \text{Scalar (verified on both sides)}\\ \midrule (\mathrm{S1}) & i^{p+q+r+s-2}\\ (\mathrm{S3}'\mathrm{L}) & 1\\ (\mathrm{B1}) & 1\\ (\mathrm{B2}') & 1\\ (\mathrm{EU}') & i\\ (\mathrm{H}) & -1\cdot i^{n+m}\\ (\mathrm{IV}') & 1\\ (\mathrm{ZO}') & \text{both sides zero}\\ \bottomrule \end{array}\]
However, \(\left\llbracket \vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=none] (l) at (0, -0.25) {}; \node [style=none] (r) at (1, -0.25) {}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw [in=90, out=90, looseness=1.75] (l.center) to (r.center); \end{pgfonlayer} \end{tikzpicture}} \right\rrbracket^{(\mathrm{S3}'\mathrm{R})}=\eta\), and \(\left\llbracket \vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=none] (l) at (0, -0.25) {}; \node [style=none] (r) at (1, -0.25) {}; \node [style=rn] (x) at (0.5, 0.25) {}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw [in=90, out=90, looseness=1.75] (l.center) to (r.center); \end{pgfonlayer} \end{tikzpicture}} \right\rrbracket^{(\mathrm{S3}'\mathrm{R})}=-\eta\). Hence, (S3’R) is necessary. ◻
Physically, one can interpret this result as saying that we cannot assume the compact structure on the X-spider holds a priori.
Consider the ring of dual numbers over \(\mathbb{F}_5\), \(R_5:=\mathbb{F}_5[\delta]/\delta^2\).
Here, \(\delta\ne 0\), \(\delta^2=0\). (\(4\equiv -1\), \(1/2\equiv 3\).)
\(R_5\) is commutative.
Now consider the free \(R_5\)-module \(V:=(R_5)^4\), with the canonical basis \(e_0,e_1,e_2,e_3\).
Finally, define our target category \(\mathcal{C}_5\) as follows. It is the strict symmetric monoidal category whose objects are natural numbers, with \[\mathcal{C}_5(n,m)=\operatorname{Hom}_{R_5}(V^{\otimes n},V^{\otimes m}),\] where \(V^{\otimes 0}=R_5\). Composition is ordinary composition of \(R_5\)-linear maps, and tensor product is the usual Kronecker tensor product. We use the standard ordered-basis identification \(V^{\otimes n}\otimes V^{\otimes m}=V^{\otimes(n+m)}\), so \(\mathcal{C}_5\) is being used as a strict PROP-style target category.
Our structural generators are defined as: \[\begin{array}{rcl} \text{identity:} & I &= \operatorname{id}_V,\\ \text{braiding:} & \sigma(v\otimes w) &= w\otimes v,\\ \text{cup:} & \eta(1) &= \displaystyle\sum_{r=0}^{3}e_r\otimes e_r,\\ \text{cap:} & \epsilon(e_r\otimes e_s) &= \delta_{rs},\\ \text{unit:} & e &= 1_{R_5}. \end{array}\]
These maps give the usual compact structure: for each basis vector \(e_a\), \[(\epsilon\otimes I)(I\otimes\eta)(e_a)=e_a, \qquad (I\otimes\epsilon)(\eta\otimes I)(e_a)=e_a.\] Because \(V\) is finite free, the usual basis cup and cap give the expected compact structure.
Consider phase units \(t=(t_0,t_1,t_2,t_3)=(1,3+2\delta,3+3\delta,4)\).
They have inverses \(t^{-1}=y=(y_0,y_1,y_2,y_3)=(1,2+2\delta,2+3\delta,4)\), since each \(t_ry_r=1\) in \(R_5\) for each \(r=0,1,2,3\).
Hence, \(t_r^k\) is defined for all \(k\in\mathbb{Z}\).
The interpretation below is an interpretation of the integer-labelled syntax of Backens–Perdrix–Wang [@backens2020towards] described in Section 2. It does not factor through the quotient identifying \(k\sim k+4\). Indeed, \[(3+2\delta)^4 = 1+\delta \neq 1, \qquad (3+3\delta)^4 = 1+4\delta \neq 1.\] This is compatible with the present countermodel, since \(\mathrm{ZX}_{\mathrm{simp}}\setminus\{(B2')\}\) is a reduced rule system in which \(2\pi\)-periodicity is not assumed syntactically.
Definition 2 (\(\mathcal{C}_5\)-spider). For each \(k\in\mathbb{Z}\), define \[(Z')^k_{n,m}:V^{\otimes n}\to V^{\otimes m}\] by \[(Z')^k_{n,m} = \sum_{r=0}^{3} t_r^k \bigl|e_r^{\otimes m}\bigr\rangle \bigl\langle e_r^{\otimes n}\bigr|,\] with the convention \(e_r^{\otimes 0}=1\). Equivalently, on basis elements, \[(Z')^k_{n,m}(e_{a_1}\otimes\cdots\otimes e_{a_n}) = \begin{cases} t_r^k e_r^{\otimes m}, & a_1=\cdots=a_n=r,\\ 0, & \text{otherwise.} \end{cases}\] Thus \((Z')^k_{n,m}\) is the usual basis-copying spider for the basis \((e_r)\), decorated by the pointwise coefficient tuple \(t^k\). Spider fusion is pointwise multiplication of decorations: \[t_r^k t_r^\ell=t_r^{k+\ell}.\]
In particular, with the convention \(e_r^{\otimes 0}=1\), we have \[(Z')^k_{0,0}=\sum_{r=0}^{3}t_r^k,\quad (Z')^0_{0,0}=4,\quad (Z')^2_{0,0}=1+(3+2\delta)^2+(3+3\delta)^2+4^2=0.\]
Definition 3 (\(\mathcal{C}_5\)-Hadamard). Let \[K= \begin{pmatrix} 1 & 1 & 1 & 1\\ 1 & 4+4\delta & 1 & 4+\delta\\ 1 & 1 & 4+\delta & 4+4\delta\\ 1 & 4+\delta & 4+4\delta & 1 \end{pmatrix},\] and define the “\(\mathcal{C}_5\) Hadamard” \[H':=\frac{1}{2}K=3K.\] By direct computation, we find that: \[K^T=K\quad\&\quad K^2=4\cdot I_4\] The key structural properties are symmetry and \(K^2=4I\): the remaining rule checks are coordinate computations. \[\Longrightarrow\quad (H')^T=(H')\quad\&\quad (H')^2=I_4.\]
Remark 4. The matrix \(K\) has a useful first-order decomposition. Since \(R_5=\mathbb{F}_5[\delta]/(\delta^2)\), we can write \[K=K_0+\delta K_1,\] where \[K_0= \begin{pmatrix} 1&1&1&1\\ 1&4&1&4\\ 1&1&4&4\\ 1&4&4&1 \end{pmatrix}\] and \[K_1= \begin{pmatrix} 0&0&0&0\\ 0&4&0&1\\ 0&0&1&4\\ 0&1&4&0 \end{pmatrix}.\] Here \(4=-1\) in \(\mathbb{F}_5\). Thus \(K_0\) is the usual \(4\times4\) Sylvester, or Walsh–Hadamard, matrix [@cameron_hadamard] \[K_0 = \begin{pmatrix} 1&1\\ 1&4 \end{pmatrix} \otimes \begin{pmatrix} 1&1\\ 1&4 \end{pmatrix}.\] Equivalently, after identifying \(4\) with \(-1\), this is the character table of the elementary abelian group \((\mathbb{Z}/2)^2\).
The term \(\delta K_1\) should be viewed as a first-order deformation of this ordinary Walsh–Hadamard matrix. Since \(\delta^2=0\), the identity \(K^2=4I_4\) separates into the ordinary identity \[K_0^2=4I_4\] and the first-order condition \[K_0K_1+K_1K_0=0.\] Intuitively, this construction is such that the matrix \(K\) produces an involutive \(H'=\frac{1}{2}K\). The later failure of \((B2')\) is therefore detected only in the first-order \(\delta\)-part of the model.
Definition 4 (\(\mathcal{C}_5\)-Red spiders). Now, we define \(\mathcal{C}_5\) Red spiders by conjugation with \(H'\): for all \(k\in\mathbb{Z}\), \[(X')^k_{n,m}:=(H')^{\otimes m}\circ (Z')^k_{n,m}\circ (H')^{\otimes n}:n\to m.\]
We define the strict monoidal interpretation functor \(\left\llbracket - \right\rrbracket^{(\mathrm{B2}')}\) inductively on diagrams as below, and by sending each ZX generator to their \(\mathcal{C}_5\) counterpart. \[\left\llbracket D_1\otimes D_2 \right\rrbracket^{(\mathrm{B2}')} =\left\llbracket D_1 \right\rrbracket^{(\mathrm{B2}')}\otimes\left\llbracket D_2 \right\rrbracket^{(\mathrm{B2}')}, \quad \left\llbracket D_2\circ D_1 \right\rrbracket^{(\mathrm{B2}')} =\left\llbracket D_2 \right\rrbracket^{(\mathrm{B2}')}\circ\left\llbracket D_1 \right\rrbracket^{(\mathrm{B2}')}.\]
Lemma 2 (Connectivity). The interpretation \(\left\llbracket - \right\rrbracket^{(\mathrm{B2}')}\) respects the connectivity meta-rule of Backens–Perdrix–Wang [@backens2020towards] “only connectivity matters.”
Proof. The structural maps are the usual compact structure for the basis \((e_0,e_1,e_2,e_3)\): \[\eta(1)=\sum_{r=0}^3 e_r\otimes e_r, \qquad \epsilon(e_r\otimes e_s)=\delta_{rs},\] so the yanking equations hold: \[(\epsilon\otimes I)(I\otimes \eta)(e_a)=e_a, \qquad (I\otimes \epsilon)(\eta\otimes I)(e_a)=e_a.\]
For green spiders, \[(Z')^k_{n,m} = \sum_{r=0}^3 t_r^k \ket{e_r^{\otimes m}}\bra{e_r^{\otimes n}}.\] This formula is unchanged by permuting input legs or output legs. It is also unchanged by bending wires, since the cup and cap identify \(e_r\) with the dual basis vector \(e_r^\ast\).
The Hadamard satisfies \[(H')^T=H', \qquad (H')^2=I.\] Hence Hadamard boxes can also be bent through the compact structure. Finally, red spiders are defined by \[(X')^k_{n,m} = (H')^{\otimes m}(Z')^k_{n,m}(H')^{\otimes n},\] so the same permutation and bending properties pass from green spiders to red spiders.
Therefore the interpreted map depends only on the labelled multigraph of the diagram, with the external input and output labels fixed. This is exactly the connectivity meta-rule in [@backens2020towards]. ◻
Theorem 5. The (B2’) rule is necessary: \(\mathrm{ZX}_{\mathrm{simp}}\setminus\{(\mathrm{B2}')\}\nvdash(\mathrm{B2}')\).
Proof. We evaluate \(\left\llbracket - \right\rrbracket^{(\mathrm{B2}')}\) on each rule to find agreement on all \(\mathrm{ZX}_{\mathrm{simp}}\setminus\{(\mathrm{B2}')\}\) rules (see 5.2 for details).
It is enough to find one basis vector and one coefficient on which the two sides disagree.
However, in the case of (B2’), evaluating both sides on \(e_1\otimes e_1\) and comparing the \(e_0\otimes e_2\) coefficient gives \[\left\llbracket \vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=rn] (1) at (0.5, -0.25) {}; \node [style=none] (3) at (1.25, 1) {}; \node [style=none] (5) at (0.5, -0.75) {}; \node [style=none] (6) at (0.5, 1) {}; \node [style=gn] (7) at (1.25, 0.5) {}; \node [style=rn] (9) at (1.25, -0.25) {}; \node [style=gn] (11) at (0.5, 0.5) {}; \node [style=none] (12) at (1.25, -0.75) {}; \node [style=gn] (17) at (0, 0) {}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw [style=none] (12.center) to (9); \draw [style=none] (5.center) to (1); \draw [style=none] (7) to (3.center); \draw [style=none, bend right=23, looseness=1.00] (9) to (7); \draw [style=none] (11) to (6.center); \draw [style=none, bend left=23, looseness=1.00] (1) to (11); \draw (11) to (9); \draw (7) to (1); \end{pgfonlayer} \end{tikzpicture}} \right\rrbracket^{(\mathrm{B2}')}\text{ has }e_0\otimes e_2\text{ coefficient }\delta,\quad\text{but}\quad\left\llbracket \vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=none] (0) at (0.75, 0.75) {}; \node [style=none] (2) at (0.25, -0.75) {}; \node [style=none] (4) at (0.25, 0.75) {}; \node [style=gn] (10) at (0.5, -0.25) {}; \node [style=none] (13) at (0.75, -0.75) {}; \node [style=rn] (14) at (0.5, 0.25) {}; \node [style=rn] (15) at (0, 0.25) {}; \node [style=gn] (16) at (0, -0.25) {}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw [style=none, bend right=23, looseness=1.00] (13.center) to (10); \draw [style=none] (10) to (14); \draw [style=none, bend left=23, looseness=1.00] (14) to (4.center); \draw [style=none, bend right=23, looseness=1.00] (14) to (0.center); \draw [bend right=23, looseness=1.00] (10) to (2.center); \draw (15) to (16); \end{pgfonlayer} \end{tikzpicture}} \right\rrbracket^{(\mathrm{B2}')}\text{ has }e_0\otimes e_2\text{ coefficient }0.\] But, \(\delta \neq 0\) in \(R_5\), so the two sides do not agree. Therefore, \(\mathrm{LHS}\neq\mathrm{RHS}\) and (B2’) fails.
Thus, \(\mathrm{ZX}_{\mathrm{simp}}\setminus\{(\mathrm{B2}')\}\nvdash(\mathrm{B2}')\). ◻
Corollary 1. Every rule in \(\mathrm{ZX}_{\mathrm{simp}}\) is necessary relative to the connectivity meta-rule of Backens–Perdrix–Wang [@backens2020towards]. Hence, relative to the meta-rule, \(\mathrm{ZX}_{\mathrm{simp}}\) has no redundant rewrite rule.
Proof. Immediate by [@backens2020towards] and the two theorems above. ◻
I thank Miriam Backens and Simon Perdrix for helpful discussions about this work and for checking the computations in an earlier version of this paper.
Proof. Let \(D\) be a ZX diagram.
We write \(\left\llbracket g \right\rrbracket^{(\mathrm{S3}'\mathrm{R})}=c(g)\cdot\left\llbracket g \right\rrbracket\) for each generator \(g\). By definition of \(\left\llbracket - \right\rrbracket^{(\mathrm{S3}'\mathrm{R})}\), \[c(Z^\alpha_{n,m})=i^{n+m-2},\quad c(X^\beta_{n,m})=-1,\quad c(H)=i,\] and \[c(I)=c(\sigma)=c(\eta)=c(\epsilon)=c(e)=1.\] If \(D=D_2\circ D_1\), then \[\begin{align} \left\llbracket D \right\rrbracket^{(\mathrm{S3}'\mathrm{R})} &=\left\llbracket D_2\circ D_1 \right\rrbracket^{(\mathrm{S3}'\mathrm{R})} =\left\llbracket D_2 \right\rrbracket^{(\mathrm{S3}'\mathrm{R})}\left\llbracket D_1 \right\rrbracket^{(\mathrm{S3}'\mathrm{R})}\\ &=c(D_2)c(D_1)\left\llbracket D_2 \right\rrbracket\circ\left\llbracket D_1 \right\rrbracket\\ &=c(D_2)c(D_1)\left\llbracket D \right\rrbracket. \end{align}\] If \(D=D_1\otimes D_2\), then \[\begin{align} \left\llbracket D \right\rrbracket^{(\mathrm{S3}'\mathrm{R})} &=\left\llbracket D_1\otimes D_2 \right\rrbracket^{(\mathrm{S3}'\mathrm{R})} =\left\llbracket D_1 \right\rrbracket^{(\mathrm{S3}'\mathrm{R})}\otimes\left\llbracket D_2 \right\rrbracket^{(\mathrm{S3}'\mathrm{R})}\\ &=c(D_1)\cdot c(D_2)\cdot\left\llbracket D_1 \right\rrbracket\otimes\left\llbracket D_2 \right\rrbracket\\ &=c(D_1)c(D_2)\left\llbracket D \right\rrbracket. \end{align}\] So, the total scalar for \(D\) is the product of the local scalars of the non-structural vertices of \(D\).
That is, we can simply count & multiply together the factors from our nontrivial generators.
Each green spider \(v=Z^\alpha_{n,m}\) contributes \(i^{n+m-2}=i^{\deg(v)-2}\).
As a collection they contribute: \[\prod_{v\in Z(D)} i^{\deg(v)-2} =i^{\sum_{v\in Z(D)}(\deg(v)-2)}.\] Each red spider contributes a factor of \(-1\), so as a collection they contribute: \[(-1)^{|X(D)|}.\] Each Hadamard contributes \(i\), so as a collection they contribute: \[(i)^{|H(D)|}.\] Combining these results, we get: \[c(D)=i^{\sum_{v\in Z(D)}(\deg(v)-2)}\cdot(-1)^{|X(D)|}\cdot(i)^{|H(D)|}.\] ◻
Note: Maps \(0\to m\) are determined by their value on \(1\). In the other cases, we decide equality by evaluating on the relevant basis vectors or by direct matrix computation.
Proof.
(S1). By definition 4.1, \((Z')_{n,m}^k\) is the basis-copying spider with pointwise coefficient \(t_a^k\).
So, we can apply the generalized spider theorem for Frobenius algebras [@coecke2011interacting].
Therefore, \[\begin{align} \left\llbracket \vcenter{\begin{tikzpicture}[font={}] \begin{pgfonlayer}{nodelayer} \node [style=gn] (3) at (-0.75, -0.25) {~\ell~}; \node [style=none] (4) at (-1.75, -0.5) {\raisebox{2mm}{...}}; \node [style=none] (6) at (-1, -0.75) {}; \node [style=none] (8) at (-0.5, -0.75) {}; \node [style=none] (10) at (-2, -0.5) {}; \node [style=none] (11) at (-1.5, -0.5) {}; \node [style=none] (12) at (-0.75, -0.75) {\raisebox{2mm}{...}}; \node [style=none] (15) at (-2, 0.75) {}; \node [style=none] (16) at (-0.5, 0.5) {}; \node [style=none] (17) at (-1.5, 0.75) {}; \node [style=gn] (19) at (-1.75, 0.25) {~k~}; \node [style=none] (20) at (-0.75, 0.5) {\raisebox{-2mm}{...}}; \node [style=none] (21) at (-1, 0.5) {}; \node [style=none] (22) at (-1.75, 0.75) {\raisebox{-2mm}{...}}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw[bend right=23] (3) to (16.center); \draw[bend right=23] (3) to (6.center); \draw[bend left=23] (3) to (8.center); \draw[bend right=23] (19) to (10.center); \draw[bend left=23] (19) to (11.center); \draw (19) to (3); \draw[bend left=23] (19) to (15.center); \draw[bend right=23] (19) to (17.center); \draw[bend left=23] (3) to (21.center); \end{pgfonlayer} \end{tikzpicture}} \right\rrbracket^{(\mathrm{B2}')} &=(\mathrm{id}_q\otimes (Z')_{r+1,s}^\ell)\circ \big((Z')_{p,q+1}^k\otimes \mathrm{id}_r\big)\\ &=(Z')_{p+r,q+s}^{k+\ell} =\left\llbracket \vcenter{\begin{tikzpicture}[font={}] \begin{pgfonlayer}{nodelayer} \node [style=gn] (z) at (0, 0) {k+\ell}; \node [style=none] (t1) at (-0.5, 0.75) {}; \node [style=none] (t2) at (0.5, 0.75) {}; \node [style=none] (td) at (0, 0.75) {\raisebox{-2mm}{...}}; \node [style=none] (b1) at (-0.5, -0.75) {}; \node [style=none] (b2) at (0.5, -0.75) {}; \node [style=none] (bd) at (0, -0.75) {\raisebox{2mm}{...}}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw[bend left=23] (z) to (t1.center); \draw[bend right=23] (z) to (t2.center); \draw[bend right=23] (z) to (b1.center); \draw[bend left=23] (z) to (b2.center); \end{pgfonlayer} \end{tikzpicture}} \right\rrbracket^{(\mathrm{B2}')}. \end{align}\] Concretely, the connecting wire forces a common basis label \(a\), & the scalar on that label is \[t_a^k t_a^\ell=t_a^{k+\ell},\] while all non-matching labels vanish on both sides.
(S3’L). Evaluating on \((1)\) on both sides, \[\left\llbracket \vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=none] (l) at (0, -0.25) {}; \node [style=none] (r) at (1, -0.25) {}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw [in=90, out=90, looseness=1.75] (l.center) to (r.center); \end{pgfonlayer} \end{tikzpicture}} \right\rrbracket^{(\mathrm{B2}')}(1) =\eta(1)=\sum_{r=0}^3 e_r\otimes e_r,\] and \[\left\llbracket \vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=none] (l) at (0, -0.25) {}; \node [style=none] (r) at (1, -0.25) {}; \node [style=gn] (g) at (0.5, 0.25) {}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw [in=90, out=90, looseness=1.75] (l.center) to (r.center); \end{pgfonlayer} \end{tikzpicture}} \right\rrbracket^{(\mathrm{B2}')}(1) =(Z')_{0,2}^0(1) =\sum_{r=0}^3 t_r^0 e_r^{\otimes 2} =\sum_{r=0}^3 e_r\otimes e_r.\]
(S3’R). Evaluating on \((1)\) on both sides,
Note that \((H')^T=H'\) & \((H')^2=I\). \[\begin{align} \left\llbracket \vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=none] (l) at (0, -0.25) {}; \node [style=none] (r) at (1, -0.25) {}; \node [style=rn] (x) at (0.5, 0.25) {}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw [in=90, out=90, looseness=1.75] (l.center) to (r.center); \end{pgfonlayer} \end{tikzpicture}} \right\rrbracket^{(\mathrm{B2}')}(1) &=(X')_{0,2}^0(1)\\ &=(H'\otimes H')(Z')_{0,2}^0(1)\\ &=(H'\otimes H')\sum_{r=0}^3 t_r^0 e_r^{\otimes 2}\\ &=(H'\otimes H')\eta(1)\\ &=(I\otimes H'(H')^T)\eta(1)\\ &=(I\otimes I)\eta(1)=\eta(1)\\ &=\left\llbracket \vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=none] (l) at (0, -0.25) {}; \node [style=none] (r) at (1, -0.25) {}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw [in=90, out=90, looseness=1.75] (l.center) to (r.center); \end{pgfonlayer} \end{tikzpicture}} \right\rrbracket^{(\mathrm{B2}')}(1). \end{align}\]
(B1). We use as shorthand: \[\rho:=(X')_{0,1}^0,\qquad u:=(Z')_{0,1}^0=\sum_{r=0}^3 e_r,\qquad \bar u:=(Z')_{1,0}^0,\qquad \Delta:=(Z')_{1,2}^0.\] Note that the row sums of \(K\) are \((4,0,0,0)\), so \(K\circ u=(4,0,0,0)\).
Now, \[\rho=H'\circ u=\frac{1}{2}K\circ u =\frac{1}{2}(4,0,0,0)=2\cdot e_0.\] Then, \[(\bar u\circ\rho)=2(\bar u\cdot e_0)=2.\] And, \[(\Delta\rho)=2(\Delta e_0)=2(e_0\otimes e_0).\] Therefore, \[\left\llbracket \vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=rn] (r1) at (0, 0.35) {}; \node [style=gn] (g1) at (0, -0.25) {}; \node [style=rn] (r2) at (0.85, 0.35) {}; \node [style=gn] (g2) at (0.85, -0.25) {}; \node [style=none] (a) at (0.6, -0.8) {}; \node [style=none] (b) at (1.1, -0.8) {}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw (r1) to (g1); \draw (r2) to (g2); \draw[bend right=23] (g2) to (a.center); \draw[bend left=23] (g2) to (b.center); \end{pgfonlayer} \end{tikzpicture}} \right\rrbracket^{(\mathrm{B2}')}=4(e_0\otimes e_0),\] \[\left\llbracket \vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=rn] (r1) at (0, 0.25) {}; \node [style=rn] (r2) at (0.55, 0.25) {}; \node [style=none] (b1) at (0, -0.45) {}; \node [style=none] (b2) at (0.55, -0.45) {}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw (r1) to (b1.center); \draw (r2) to (b2.center); \end{pgfonlayer} \end{tikzpicture}} \right\rrbracket^{(\mathrm{B2}')}=\rho\otimes\rho=4(e_0\otimes e_0).\]
(ZO’). \[(Z')_{0,0}^2=0,\] so \[\left\llbracket \vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=gn] (pi) at (0, 0) {\pi}; \node [style=gn] (g) at (0.7, 0.25) {}; \node [style=none] (out) at (0.7, -0.25) {}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw (g) to (out.center); \end{pgfonlayer} \end{tikzpicture}} \right\rrbracket^{(\mathrm{B2}')}=0=\left\llbracket \vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=gn] (pi) at (0, 0) {\pi}; \node [style=rn] (r) at (0.7, 0.25) {}; \node [style=none] (out) at (0.7, -0.25) {}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw (r) to (out.center); \end{pgfonlayer} \end{tikzpicture}} \right\rrbracket^{(\mathrm{B2}')}.\]
(H). Note that \((H')^2=I\). \[\begin{align} \left\llbracket \vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=rn] (x) at (0, 0) {k}; \node [style={H box}] (h1) at (-0.45, 0.6) {}; \node [style={H box}] (h2) at (0.45, 0.6) {}; \node [style={H box}] (h3) at (-0.45, -0.6) {}; \node [style={H box}] (h4) at (0.45, -0.6) {}; \node [style=none] (t1) at (-0.45, 1.0) {}; \node [style=none] (t2) at (0.45, 1.0) {}; \node [style=none] (b1) at (-0.45, -1.0) {}; \node [style=none] (b2) at (0.45, -1.0) {}; \node [style=none] (td) at (0, 1.0) {\raisebox{-2mm}{...}}; \node [style=none] (bd) at (0, -1.0) {\raisebox{2mm}{...}}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw[bend left=23] (x) to (h1); \draw[bend right=23] (x) to (h2); \draw[bend right=23] (x) to (h3); \draw[bend left=23] (x) to (h4); \draw (h1) to (t1.center); \draw (h2) to (t2.center); \draw (h3) to (b1.center); \draw (h4) to (b2.center); \end{pgfonlayer} \end{tikzpicture}} \right\rrbracket^{(\mathrm{B2}')} &=(H')^{\otimes m}\circ (X')_{n,m}^k\circ (H')^{\otimes n}\\ &=(H')^{\otimes m}\circ (H')^{\otimes m}\circ (Z')_{n,m}^k\circ (H')^{\otimes n}\circ (H')^{\otimes n}\\ &=(Z')_{n,m}^k =\left\llbracket \vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=gn] (z) at (0, 0) {k}; \node [style=none] (t1) at (-0.45, 0.75) {}; \node [style=none] (t2) at (0.45, 0.75) {}; \node [style=none] (b1) at (-0.45, -0.75) {}; \node [style=none] (b2) at (0.45, -0.75) {}; \node [style=none] (td) at (0, 0.75) {\raisebox{-2mm}{...}}; \node [style=none] (bd) at (0, -0.75) {\raisebox{2mm}{...}}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw[bend left=23] (z) to (t1.center); \draw[bend right=23] (z) to (t2.center); \draw[bend right=23] (z) to (b1.center); \draw[bend left=23] (z) to (b2.center); \end{pgfonlayer} \end{tikzpicture}} \right\rrbracket^{(\mathrm{B2}')}. \end{align}\]
(IV’). Consider each “dumbbell” \[\chi:=(Z')_{3,0}^0\circ (X')_{0,3}^0.\] Let \(H'=(h_{ab})_{a,b}\). Then \[\chi = \sum_{b=0}^3\sum_{a=0}^3 h_{ab}^3 =8+(10+5\delta)+(10+5\delta)+(10+5\delta) =3\] in \(R_5\). Also, \[(Z')_{0,0}^0=\sum_{r=0}^3 t_r^0=1+1+1+1=4.\] So, \[\left\llbracket \vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=gn] (s) at (-0.75, 0) {}; \node [style=rn] (r1) at (-0.25, 0.45) {}; \node [style=gn] (g1) at (-0.25, -0.35) {}; \node [style=rn] (r2) at (0.35, 0.45) {}; \node [style=gn] (g2) at (0.35, -0.35) {}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw (r1) to (g1); \draw[bend left=38] (r1) to (g1); \draw[bend right=38] (r1) to (g1); \draw (r2) to (g2); \draw[bend left=38] (r2) to (g2); \draw[bend right=38] (r2) to (g2); \end{pgfonlayer} \end{tikzpicture}} \right\rrbracket^{(\mathrm{B2}')}=4\cdot 3\cdot 3=36=1,\] \[\left\llbracket \vcenter{} \right\rrbracket^{(\mathrm{B2}')}=1.\]
(B2’). Now, let \[F:=(X')^0_{2,1} = H'\circ (Z')^0_{2,1}\circ(H'\otimes H').\] We compute \(F(e_1\otimes e_1)\). First, \[H'e_1 = 3e_0 + (2+2\delta)e_1 +3e_2 + (2+3\delta)e_3.\] Thus, \[(Z')^0_{2,1}(H'e_1\otimes H'e_1) = 3^2e_0+(2+2\delta)^2e_1+3^2e_2+(2+3\delta)^2e_3\] \[= 4e_0+(4+3\delta)e_1+4e_2+(4+2\delta)e_3.\] Applying \(H'\), we get \[F(e_1\otimes e_1) =3e_0+3\delta e_2+2\delta e_3.\] (note that \(\sigma\) does nothing to \(e_1\otimes e_1\)). \[\begin{align} \left\llbracket \vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=rn] (1) at (0.5, -0.25) {}; \node [style=none] (3) at (1.25, 1) {}; \node [style=none] (5) at (0.5, -0.75) {}; \node [style=none] (6) at (0.5, 1) {}; \node [style=gn] (7) at (1.25, 0.5) {}; \node [style=rn] (9) at (1.25, -0.25) {}; \node [style=gn] (11) at (0.5, 0.5) {}; \node [style=none] (12) at (1.25, -0.75) {}; \node [style=gn] (17) at (0, 0) {}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw [style=none] (12.center) to (9); \draw [style=none] (5.center) to (1); \draw [style=none] (7) to (3.center); \draw [style=none, bend right=23, looseness=1.00] (9) to (7); \draw [style=none] (11) to (6.center); \draw [style=none, bend left=23, looseness=1.00] (1) to (11); \draw (11) to (9); \draw (7) to (1); \end{pgfonlayer} \end{tikzpicture}} \right\rrbracket^{(\mathrm{B2}')}(e_1\otimes e_1) &=(Z')_{0,0}^0\otimes F(e_1\otimes e_1)\otimes F(e_1\otimes e_1)\\ &=4(F(e_1\otimes e_1)\otimes F(e_1\otimes e_1))\\ &=4\big((3e_0+3\delta e_2+2\delta e_3)\otimes(3e_0+3\delta e_2+2\delta e_3)\big) \end{align}\] Comparing the \(e_0\otimes e_2\) coefficient, we get \[4\cdot 3\cdot 3\delta = 36\delta=\delta \quad\text{in } R_5.\] Now, \[(X')_{0,1}^0(1)=H'(e_0+e_1+e_2+e_3)=2e_0,\] so, \[(Z')_{1,0}^0\circ (X')_{0,1}^0(1)=2.\] Thus, \[\begin{align} \left\llbracket \vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=none] (0) at (0.75, 0.75) {}; \node [style=none] (2) at (0.25, -0.75) {}; \node [style=none] (4) at (0.25, 0.75) {}; \node [style=gn] (10) at (0.5, -0.25) {}; \node [style=none] (13) at (0.75, -0.75) {}; \node [style=rn] (14) at (0.5, 0.25) {}; \node [style=rn] (15) at (0, 0.25) {}; \node [style=gn] (16) at (0, -0.25) {}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw [style=none, bend right=23, looseness=1.00] (13.center) to (10); \draw [style=none] (10) to (14); \draw [style=none, bend left=23, looseness=1.00] (14) to (4.center); \draw [style=none, bend right=23, looseness=1.00] (14) to (0.center); \draw [bend right=23, looseness=1.00] (10) to (2.center); \draw (15) to (16); \end{pgfonlayer} \end{tikzpicture}} \right\rrbracket^{(\mathrm{B2}')}(e_1\otimes e_1) &=2\cdot (Z')_{1,2}^0\circ F(e_1\otimes e_1)\\ &=2(Z')_{1,2}^0(3e_0+3\delta e_2+2\delta e_3)\\ &=e_0\otimes e_0+\delta e_2\otimes e_2+4\delta e_3\otimes e_3. \end{align}\] Thus the \(e_0\otimes e_2\) coefficient on the right-hand side is \(0\). But, \(\delta \neq 0\) in \(R_5\), so the two sides do not agree. \[\therefore\quad (\mathrm{B2}')\;\text{fails under}\;\left\llbracket - \right\rrbracket^{(\mathrm{B2}')}.\]
(EU’). Note that \((Z')_{2,1}^0\) just multiplies matching coordinates.
For vectors \(u,v\in V\): \[(Z')_{2,1}^0(u\otimes v)=\sum_{r=0}^3 u_rv_r e_r.\] Define \(D=\operatorname{diag}(t_0,t_1,t_2,t_3)\), the diagonal matrix of our phases: \[t= \begin{pmatrix} t_0\\ t_1\\ t_2\\ t_3 \end{pmatrix} = \begin{pmatrix} 1\\ 3+2\delta\\ 3+3\delta\\ 4 \end{pmatrix}, \qquad t^{-1}= \begin{pmatrix} 1\\ 2+2\delta\\ 2+3\delta\\ 4 \end{pmatrix}.\] \[H't^{-1}= \begin{pmatrix} 2\\ 1+4\delta\\ 1+\delta\\ 3 \end{pmatrix} =2 \begin{pmatrix} 1\\ 3+2\delta\\ 3+3\delta\\ 4 \end{pmatrix} =2t \quad\Longrightarrow\quad \operatorname{diag}(H't^{-1})=2D.\] Then, \[(Z')_{1,1}^1=D \quad\&\quad (Z')_{0,1}^{-1}=t^{-1}.\] So, \[\begin{align} \left\llbracket \vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=none] (top) at (0, 1.0) {}; \node [style=gn] (g1) at (0, 0.55) {1}; \node [style=rn] (x) at (0, 0.1) {}; \node [style=gn] (g2) at (0, -0.35) {1}; \node [style=none] (bot) at (0, -0.85) {}; \node [style=gn] (g3) at (0.55, 0.35) {-1}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw (top.center) to (g1); \draw (g1) to (x); \draw (x) to (g2); \draw (g2) to (bot.center); \draw (g3) to (x); \end{pgfonlayer} \end{tikzpicture}} \right\rrbracket^{(\mathrm{B2}')} &=(Z')_{1,1}^1\circ (X')_{2,1}^0\circ \big((Z')_{1,1}^1\otimes (Z')_{0,1}^{-1}\big)\\ &=D\circ H'\circ (Z')_{2,1}^0\circ (H'\otimes H')\circ(D\otimes t^{-1})\\ &=D\circ H'\circ (Z')_{2,1}^0\circ(H'D\otimes H't^{-1})\\ &=D\circ H'\circ \operatorname{diag}(H't^{-1})\circ H'\circ D\\ &=2(DH'DH'D)=18(DKDKD)=3K=H', \end{align}\] where \(DKDKD=K\) by direct computation and \(18=3\) in \(R_5\). \[\therefore\quad \left\llbracket \vcenter{} \right\rrbracket^{(\mathrm{B2}')} =\left\llbracket \vcenter{\begin{tikzpicture} \begin{pgfonlayer}{nodelayer} \node [style=none] (top) at (0, 1.0) {}; \node [style=gn] (g1) at (0, 0.55) {1}; \node [style=rn] (x) at (0, 0.1) {}; \node [style=gn] (g2) at (0, -0.35) {1}; \node [style=none] (bot) at (0, -0.85) {}; \node [style=gn] (g3) at (0.55, 0.35) {-1}; \end{pgfonlayer} \begin{pgfonlayer}{edgelayer} \draw (top.center) to (g1); \draw (g1) to (x); \draw (x) to (g2); \draw (g2) to (bot.center); \draw (g3) to (x); \end{pgfonlayer} \end{tikzpicture}} \right\rrbracket^{(\mathrm{B2}')}.\]
◻