Proof. A qutrit stabilizer state \(\lvert\mspace{0.5mu}\sigma \mspace{0.5mu}\rangle\) on \(m\) qutrits has the canonical form \[\lvert\mspace{0.5mu}\sigma \mspace{0.5mu}\rangle= 3^{-k/2} \sum_{y \in \mathbb{F}_{3}^k} \omega^{Q(y)} \lvert\mspace{0.5mu}x_0 + W y \mspace{0.5mu}\rangle,\] for some \(k \in \{0, \ldots, m\}\), affine offset \(x_0\), isometric embedding \(W \in \mathbb{F}_{3}^{m \times k}\), and quadratic form \(Q : \mathbb{F}_{3}^k \to \mathbb{F}_{3}\), with \(\omega = e^{2 \pi i / 3}\). Each coordinate of \(\lvert\mspace{0.5mu}\sigma \mspace{0.5mu}\rangle\) therefore lies in \(\{0, 3^{-k/2} \omega^j : j \in \{0, 1, 2\}\}\). Given a decomposition \(\lvert\mspace{0.5mu}\psi \mspace{0.5mu}\rangle^{\otimes m} = \sum_{i=1}^r c_i \lvert\mspace{0.5mu}\sigma_i \mspace{0.5mu}\rangle\) of length \(r = \chi(\lvert\mspace{0.5mu}\psi \mspace{0.5mu}\rangle^{\otimes m})\), write \(\tilde{c}_i = 3^{-k_i/2} c_i\); then for every coordinate \(x \in \mathbb{F}_{3}^m\), \[\label{eq:coord-subset-sum} \langle x | \psi^{\otimes m} \rangle = \sum_{i = 1}^{r} \varepsilon_{i, x} \tilde{c}_i.\tag{10}\] The coefficients \(\varepsilon_{i, x}\) take values in \(\{0, 1, \omega, \omega^2\}\), so each coordinate of \(\lvert\mspace{0.5mu}\psi \mspace{0.5mu}\rangle^{\otimes m}\) is a subset-sum of the \(3r\)-tuple \((\tilde{c}_i, \omega \tilde{c}_i, \omega^2 \tilde{c}_i)_{i = 1}^{r}\), yielding a subset-sum representation of length \(3r\) of the coordinate vector. The factor of \(3\) counts the three cube-roots of unity \(\{1, \omega, \omega^2\}\) appearing as \(\varepsilon_{i, x}\) values; the qubit version of this argument lifts each \(\tilde{c}_i\) to four entries \(\{\pm \tilde{c}_i, \pm i \tilde{c}_i\}\) and yields a \(4r\)-tuple instead. By the subset-sum lower bound of [5], any subset-sum representation of a \(q\)-tuple containing an exponentially increasing subsequence of length \(\ell\) (i.e., \(\ell\) entries whose absolute values each at least double the previous) has length at least \(\ell / \log_2 \ell\). Hence \(3 r \ge \ell / \log_2 \ell\), giving \(\chi(\lvert\mspace{0.5mu}\psi \mspace{0.5mu}\rangle^{\otimes m}) \ge \ell / (3 \log_2 \ell)\).

It remains to produce an exponentially increasing subsequence of length \(\ell = m + 1\) in the coordinates of \(\lvert\mspace{0.5mu}\psi \mspace{0.5mu}\rangle^{\otimes m}\). By hypothesis, there exist indices \(i_a, i_b \in \{0, 1, 2\}\) with \(a = |a_{i_a}|\), \(b = |a_{i_b}|\), \(a \ge 2 b > 0\). For each \(k \in \{0, \ldots, m\}\), set \[\label{eq:xk-coord} x^{(k)} = (\underbrace{i_a, \ldots, i_a}_{k}, \underbrace{i_b, \ldots, i_b}_{m - k}) \in \mathbb{F}_{3}^m, \qquad\text{so that}\qquad \bigl| \langle x^{(k)} | \psi^{\otimes m} \rangle \bigr| = a^k b^{m - k}.\tag{11}\] The \(m + 1\) values \(\{ a^k b^{m - k} \}_{k = 0}^{m}\) are distinct and form an exponentially increasing sequence of consecutive ratio \(a / b \ge 2\), so we may take \(\ell = m + 1\) in ?? . ◻

Proof. By exhaustive search over the symplectic-quotient representation (Lemma 1), one identifies valid protocols for \(\lvert\mspace{0.5mu}H_3 \mspace{0.5mu}\rangle\) and \(\lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle\). The raw search outputs the following two-qutrit Cliffords: \[\label{eq:two-copy-protocols} \begin{align} C_{\lvert\mspace{0.5mu}H_3 \mspace{0.5mu}\rangle} &= H_1 \mathrm{SUM} H_1 \mathrm{SUM} H_1 S_2 \mathrm{SUM} H_1 H_2^2, \\ C_{\lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle} &= H_1 H_2 \mathrm{SUM} H_1^\dagger H_2 \mathrm{SUM} H_1. \end{align}\tag{12}\]

3.1.0.1 Hadamard-eigenstate protocol.

Let \(N = \sqrt{3 - \sqrt{3}}\) and \(c = (\sqrt{3}-1)/2\) (the same constants used throughout Appendix 8.2). The two-copy input state is \(\lvert\mspace{0.5mu}H_3 \mspace{0.5mu}\rangle\otimes \lvert\mspace{0.5mu}H_3 \mspace{0.5mu}\rangle\), where the single-copy Hadamard-eigenstate is \(\lvert\mspace{0.5mu}H_3 \mspace{0.5mu}\rangle= \frac{1}{N} \big(\lvert\mspace{0.5mu}0 \mspace{0.5mu}\rangle + c\lvert\mspace{0.5mu}1 \mspace{0.5mu}\rangle + c\lvert\mspace{0.5mu}2 \mspace{0.5mu}\rangle\big)\). Applying the two-qutrit Clifford \(C_{\lvert\mspace{0.5mu}H_3 \mspace{0.5mu}\rangle}\) to this input state and projecting the ancilla (leg 2) onto the outcome \(k = 1\) yields the data-qutrit output vector \[\label{eq:H3-output} \lvert\mspace{0.5mu}\psi_{k=1} \mspace{0.5mu}\rangle = \langle\mspace{0.5mu}1 \mspace{0.5mu}\rvert_{\mathrm{anc}} C_{\lvert\mspace{0.5mu}H_3 \mspace{0.5mu}\rangle} (\lvert\mspace{0.5mu}H_3 \mspace{0.5mu}\rangle\otimes \lvert\mspace{0.5mu}H_3 \mspace{0.5mu}\rangle) = \frac{\lambda}{\sqrt{3}} \big(\lvert\mspace{0.5mu}0 \mspace{0.5mu}\rangle + e^{i\pi/2}\lvert\mspace{0.5mu}1 \mspace{0.5mu}\rangle + e^{i\pi/3}\lvert\mspace{0.5mu}2 \mspace{0.5mu}\rangle\big),\tag{13}\] where \(\lambda \in \mathbb{C}\) is a normalization scalar satisfying \(|\lambda|^2 = 3/8\). Since all three amplitudes of the post-selected state have equal modulus, it is a phase state. The relative phases are \(\arg(\psi_1 / \psi_0) = \pi/2\) and \(\arg(\psi_2 / \psi_0) = \pi/3\), corresponding to the single-qutrit diagonal gate \(\mathrm{diag}(1, i, e^{i\pi/3})\). Since a single-qutrit diagonal Clifford gate must have phases that are multiples of \(2\pi/3\), the phase \(\pi/2\) certifies that the injected gate is non-Clifford.

3.1.0.2 Norrell protocol.

The two-copy input state is \(\lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle\otimes \lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle\), where \(\lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle= \frac{1}{\sqrt{6}}(\lvert\mspace{0.5mu}0 \mspace{0.5mu}\rangle + \lvert\mspace{0.5mu}1 \mspace{0.5mu}\rangle - 2\lvert\mspace{0.5mu}2 \mspace{0.5mu}\rangle)\) in the computational basis. First, the gate \(H_1\) acts on the data qutrit as the qutrit Fourier transform, leaving the ancilla unchanged. On the amplitudes \(v_{a,b} = c_N(a) c_N(b)/6\) (where \(c_N = (1, 1, -2)^T\)), the data Fourier transform produces \[\lvert\mspace{0.5mu}\Psi_1 \mspace{0.5mu}\rangle = (H_1 \otimes \mathbb{1}) (\lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle\otimes \lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle) = \frac{1}{6\sqrt{3}} \sum_{j, b} \hat{c}(j) c_N(b) \lvert\mspace{0.5mu}j, b \mspace{0.5mu}\rangle,\] where the inner sum evaluates using \(1 + \omega + \omega^2 = 0\) as \[\hat{c}(j) = \sum_{a=0}^2 \omega^{aj} c_N(a) = 1 + \omega^j - 2\omega^{2j}.\] This evaluates to \(\hat{c}(0) = 0\), \(\hat{c}(1) = -3\omega^2\), and \(\hat{c}(2) = -3\omega\). Thus, after \(H_1\), only the data components with \(j \in \{1, 2\}\) survive.

Applying the remaining gates of \(C_{\lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle}\) and projecting onto the ancilla outcome \(k = 0\) yields \[\label{eq:N-output} \lvert\mspace{0.5mu}\psi_{k=0} \mspace{0.5mu}\rangle = \langle\mspace{0.5mu}0 \mspace{0.5mu}\rvert_{\mathrm{anc}} C_{\lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle} (\lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle\otimes \lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle) = \frac{\mu}{\sqrt{3}} \big(\lvert\mspace{0.5mu}0 \mspace{0.5mu}\rangle - \lvert\mspace{0.5mu}1 \mspace{0.5mu}\rangle - \lvert\mspace{0.5mu}2 \mspace{0.5mu}\rangle\big),\tag{14}\] with \(|\mu|^2 = 1/4\), confirming a success probability of \(1/4\) to obtain a phase state. The relative phases are both \(\pi\), defining the injected diagonal gate \(\mathrm{diag}(1, -1, -1)\). Because \(\pi\) is not a multiple of \(2\pi/3\), this gate is non-Clifford.

Finally, for \(\lvert\mspace{0.5mu}M \mspace{0.5mu}\rangle = \lvert\mspace{0.5mu}\mathbb{S} \mspace{0.5mu}\rangle\), an exhaustive search confirms that no two-copy conversion protocol exists: every \((C, k)\) pair that maps \(\lvert\mspace{0.5mu}\mathbb{S} \mspace{0.5mu}\rangle\otimes \lvert\mspace{0.5mu}\mathbb{S} \mspace{0.5mu}\rangle\) to a phase state yields relative phases restricted to the Clifford-trivial ones. ◻

Proof. Using the identity \(\langle\mspace{0.5mu}k \mspace{0.5mu}\rvert X^{a} Z^{b} = \omega_3^{b(k - a)} \langle\mspace{0.5mu}k - a \mspace{0.5mu}\rvert\), \[\begin{align} E_k(C_{\mathrm{ent}}, \lvert\mspace{0.5mu}M \mspace{0.5mu}\rangle) &= \langle\mspace{0.5mu}k \mspace{0.5mu}\rvert_{\mathrm{anc}} (D_{\mathrm{data}} \otimes D_{\mathrm{anc}}) C_{\mathrm{sp}} (\mathbb{1}\otimes \lvert\mspace{0.5mu}M \mspace{0.5mu}\rangle) \\ &= D_{\mathrm{data}} (\langle\mspace{0.5mu}k \mspace{0.5mu}\rvert D_{\mathrm{anc}})_{\mathrm{anc}} C_{\mathrm{sp}} (\mathbb{1}\otimes \lvert\mspace{0.5mu}M \mspace{0.5mu}\rangle) \\ &= \omega_3^{b(k - a)} D_{\mathrm{data}} E_{k - a}(C_{\mathrm{sp}}, \lvert\mspace{0.5mu}M \mspace{0.5mu}\rangle). \end{align}\] The prefactor \(D_{\mathrm{data}}\) is a single-qutrit Pauli, and \(k \mapsto k - a\) permutes \(\mathbb{F}_{3}\), so the multisets \(\{E_k(C_{\mathrm{ent}}, M)\}_{k}\) and \(\{E_k(C_{\mathrm{sp}}, M)\}_{k}\) coincide up to a global Pauli prefactor and branch-dependent phases. Proportionality to a unitary, the Clifford-byproduct condition, and the Clifford-equivalent-to-phase-state condition are all invariant under both transformations. ◻

Proof of Proposition 5. For (i), \(\mathrm{Cl}(1, 3)\) is generated by the qutrit Pauli operators \(X\) (shift: \(X \lvert\mspace{0.5mu}j \mspace{0.5mu}\rangle = \lvert\mspace{0.5mu}j + 1 \mspace{0.5mu}\rangle\)) and \(Z\) (clock: \(Z \lvert\mspace{0.5mu}j \mspace{0.5mu}\rangle = \omega^j \lvert\mspace{0.5mu}j \mspace{0.5mu}\rangle\)), the phase gate \(S = \mathrm{diag}(1, 1, \omega)\), and the qutrit Hadamard \(H_{jk} = \omega^{jk}/\sqrt 3\). Any \(\lvert\mspace{0.5mu}\phi \mspace{0.5mu}\rangle\in \mathcal{O}_S\) has the form \(\lvert\mspace{0.5mu}\phi_{a,b,c} \mspace{0.5mu}\rangle = (\lvert\mspace{0.5mu}a \mspace{0.5mu}\rangle - \omega^c \lvert\mspace{0.5mu}b \mspace{0.5mu}\rangle)/\sqrt 2\) for some two-element support \(\{a, b\} \subset \mathbb{F}_{3}\) with \(a \ne b\) and some phase exponent \(c \in \mathbb{F}_{3}\); this pattern is preserved by each generator on every such state:

• \(X\) sends \(\lvert\mspace{0.5mu}\phi_{a, b, c} \mspace{0.5mu}\rangle\) to \(\lvert\mspace{0.5mu}\phi_{a+1, b+1, c} \mspace{0.5mu}\rangle\).

• \(Z\) sends \(\lvert\mspace{0.5mu}\phi_{a, b, c} \mspace{0.5mu}\rangle\) to \(\omega^a \lvert\mspace{0.5mu}\phi_{a, b, c + (b - a)} \mspace{0.5mu}\rangle\).

• \(S\) acts trivially on \(\lvert\mspace{0.5mu}j \mspace{0.5mu}\rangle\) for \(j \in \{0, 1\}\) and as \(\omega\) on \(\lvert\mspace{0.5mu}2 \mspace{0.5mu}\rangle\); it sends \(\lvert\mspace{0.5mu}\phi_{a, b, c} \mspace{0.5mu}\rangle\) to \(\lvert\mspace{0.5mu}\phi_{a, b, c + [b = 2] - [a = 2]} \mspace{0.5mu}\rangle\).

• \(H\) sends \(\lvert\mspace{0.5mu}\phi_{a, b, c} \mspace{0.5mu}\rangle\) to \[H \lvert\mspace{0.5mu}\phi_{a,b,c} \mspace{0.5mu}\rangle = \frac{1}{\sqrt 6} \sum_{j \in \mathbb{F}_{3}} \big(\omega^{a j} - \omega^{c + b j}\big) \lvert\mspace{0.5mu}j \mspace{0.5mu}\rangle.\] The coefficient at \(\lvert\mspace{0.5mu}j \mspace{0.5mu}\rangle\) vanishes iff \((a - b) j \equiv c \pmod 3\), which has a unique solution \(j^\ast \in \mathbb{F}_{3}\) since \(a \ne b\). For the other two indices \(j_1, j_2\), factor each coefficient as \(\omega^{a j}(1 - \omega^{k_j})\) with \(k_j = c + (b - a) j\); then \(k_{j_1}\) and \(k_{j_2}\) take the values \(1\) and \(2\) in some order. Since \(|1 - \omega| = |1 - \omega^2| = \sqrt 3\), the two nonzero amplitudes each have modulus \(1/\sqrt 2\), and their ratio takes the form \(-\omega^{c'}\) for some \(c' \in \mathbb{F}_{3}\) (using \((1 - \omega^2)/(1 - \omega) = -\omega^2\)). Therefore \(H \lvert\mspace{0.5mu}\phi_{a, b, c} \mspace{0.5mu}\rangle\) is, up to a global phase, \(\lvert\mspace{0.5mu}\phi_{j_1, j_2, c'} \mspace{0.5mu}\rangle\).

The pattern is thus closed under the generators and so under all of \(\mathrm{Cl}(1, 3)\).

For (ii), the three-parameter family \(\{\lvert\mspace{0.5mu}\phi_{a, b, c} \mspace{0.5mu}\rangle\}\) has at most \(\binom{3}{2} \times 3 = 9\) projective elements (three unordered \(2\)-element supports, three phase exponents \(c \in \mathbb{F}_{3}\) each). The \(X\)-action shows all three supports are reached from \(\lvert\mspace{0.5mu}\mathbb{S} \mspace{0.5mu}\rangle= \lvert\mspace{0.5mu}\phi_{1, 2, 0} \mspace{0.5mu}\rangle\), and on a fixed support the three values of \(c\) give projectively distinct states: \(\lvert\mspace{0.5mu}\phi_{a, b, c} \mspace{0.5mu}\rangle\) and \(\lvert\mspace{0.5mu}\phi_{a, b, c'} \mspace{0.5mu}\rangle\) are projectively equal iff \(\omega^c = \omega^{c'}\) (the ratio of the \(\lvert\mspace{0.5mu}b \mspace{0.5mu}\rangle\) amplitudes), iff \(c = c'\) in \(\mathbb{F}_{3}\). Hence \(|\mathcal{O}_S| = 9\) and \(|\mathrm{Stab}(\lvert\mspace{0.5mu}\mathbb{S} \mspace{0.5mu}\rangle)| = 216/9 = 24\) by orbit-stabilizer.

The final claim is immediate from (i): a state with support cardinality \(2\) and equal-modulus nonzero amplitudes cannot have two amplitudes of differing moduli, so the \(|a_i|/|a_j| \ge 2\) hypothesis of Proposition 4 fails for every \(\lvert\mspace{0.5mu}\phi \mspace{0.5mu}\rangle\in \mathcal{O}_S\). Since the stabilizer rank is invariant under the single-qutrit Clifford action, testing the hypothesis on \(\mathcal{O}_S\) exhausts the freedom available, and Proposition 4 yields no \(\chi(\lvert\mspace{0.5mu}\mathbb{S} \mspace{0.5mu}\rangle^{\otimes m})\) lower bound. ◻

Proof of Theorem 2. Fix two quadratic forms on \(\mathbb{F}_{3}^2\), \[\label{eq:strange-m2-Qs} Q_1(y) = y_0^2 + y_0 y_1 + y_1^2, \qquad Q_2(y) = y_0^2 + 2 y_0 y_1 + y_1^2 \pmod 3,\tag{18}\] and let \(\lvert\mspace{0.5mu}\sigma_j \mspace{0.5mu}\rangle\) be the two-qutrit stabilizer state given by the canonical form 1 with \(k = 2\), \(x_0 = 0\), \(W = I_2\), and quadratic phase polynomial \(Q_j\), \[\label{eq:strange-m2-sigma} \lvert\mspace{0.5mu}\sigma_j \mspace{0.5mu}\rangle = \frac{1}{3} \sum_{y \in \mathbb{F}_{3}^2} \omega^{Q_j(y)} \lvert\mspace{0.5mu}y \mspace{0.5mu}\rangle.\tag{19}\] Subtracting the two states, \[\label{eq:strange-m2-diff-raw} \lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle - \lvert\mspace{0.5mu}\sigma_2 \mspace{0.5mu}\rangle = \frac{1}{3} \sum_{y \in \mathbb{F}_{3}^2} \omega^{y_0^2 + y_1^2} \big( \omega^{y_0 y_1} - \omega^{2 y_0 y_1} \big) \lvert\mspace{0.5mu}y \mspace{0.5mu}\rangle.\tag{20}\] Substituting \(\omega = -\frac{1}{2} + \frac{i \sqrt{3}}{2}\) gives \(\omega^a - \omega^{2a} = 0, +i\sqrt{3}, -i\sqrt{3}\) for \(a \equiv 0, 1, 2 \pmod 3\), so the inner bracket in 20 vanishes whenever \(y_0 = 0\) or \(y_1 = 0\). On the four remaining \(y \in \{1, 2\}^2\) one has \(y_0^2 + y_1^2 \equiv 2 \pmod 3\) uniformly, \(y_0 y_1 \equiv 1\) at \((1,1), (2,2)\), and \(y_0 y_1 \equiv 2\) at \((1,2), (2,1)\), so \[\label{eq:strange-m2-diff-collected} \lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle - \lvert\mspace{0.5mu}\sigma_2 \mspace{0.5mu}\rangle = \frac{i \sqrt{3}}{3} \omega^2 \big( \lvert\mspace{0.5mu}11 \mspace{0.5mu}\rangle - \lvert\mspace{0.5mu}12 \mspace{0.5mu}\rangle - \lvert\mspace{0.5mu}21 \mspace{0.5mu}\rangle + \lvert\mspace{0.5mu}22 \mspace{0.5mu}\rangle \big) = \frac{2 i \sqrt{3}}{3} \omega^2 \lvert\mspace{0.5mu}\mathbb{S} \mspace{0.5mu}\rangle^{\otimes 2},\tag{21}\] where the second equality uses \(\lvert\mspace{0.5mu}\mathbb{S} \mspace{0.5mu}\rangle^{\otimes 2} = \frac{1}{2}(\lvert\mspace{0.5mu}1 \mspace{0.5mu}\rangle - \lvert\mspace{0.5mu}2 \mspace{0.5mu}\rangle)^{\otimes 2} = \frac{1}{2}(\lvert\mspace{0.5mu}11 \mspace{0.5mu}\rangle - \lvert\mspace{0.5mu}12 \mspace{0.5mu}\rangle - \lvert\mspace{0.5mu}21 \mspace{0.5mu}\rangle + \lvert\mspace{0.5mu}22 \mspace{0.5mu}\rangle)\). Inverting 21 (and using \(\omega^{-2} = \omega\), \(i^{-1} = -i\)), \[\label{eq:strange-m2-decomp} \lvert\mspace{0.5mu}\mathbb{S} \mspace{0.5mu}\rangle^{\otimes 2} = -\frac{i \sqrt{3}}{2} \omega \big( \lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle - \lvert\mspace{0.5mu}\sigma_2 \mspace{0.5mu}\rangle \big),\tag{22}\] matching Theorem 2. This exhibits \(\lvert\mspace{0.5mu}\mathbb{S} \mspace{0.5mu}\rangle^{\otimes 2}\) as a \(\mathbb{C}\)-linear combination of two stabilizer states, so \(\chi(\lvert\mspace{0.5mu}\mathbb{S} \mspace{0.5mu}\rangle^{\otimes 2}) \le 2\).

For the lower bound, \(\lvert\mspace{0.5mu}\mathbb{S} \mspace{0.5mu}\rangle^{\otimes 2}\) has computational-basis support \(\{(1,1), (1,2), (2,1), (2,2)\}\), of cardinality four. Any single 2-qutrit stabilizer state has, by 1 , support contained in an affine subspace \(x_0 + W \mathbb{F}_{3}^k \subseteq \mathbb{F}_{3}^2\), hence of cardinality \(3^r\) for some \(r \in \{0, 1, 2\}\). Since \(4 \notin \{1, 3, 9\}\), \(\lvert\mspace{0.5mu}\mathbb{S} \mspace{0.5mu}\rangle^{\otimes 2}\) is not proportional to any stabilizer state, and \(\chi(\lvert\mspace{0.5mu}\mathbb{S} \mspace{0.5mu}\rangle^{\otimes 2}) \ne 1\). Combining, \(\chi(\lvert\mspace{0.5mu}\mathbb{S} \mspace{0.5mu}\rangle^{\otimes 2}) = 2\).

Machine-checked Lean formalization: StrangeM2Pointwise.lean. ◻

Proof that \(\chi(\lvert\mspace{0.5mu}\mathbb{S} \mspace{0.5mu}\rangle^{\otimes 3}) \le 4\). Set \(W = (e_0, e_2) \in \mathbb{F}_{3}^{3 \times 2}\). Define four 3-qutrit stabilizer states by 1 , all with \(k = 2\) and the matrix \(W\) above: \[\begin{align} \lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle &: x_0 = (0, 2, 0), Q_1(w) = w_0^2 + w_0 w_1 \pmod 3, \\ \lvert\mspace{0.5mu}\sigma_3 \mspace{0.5mu}\rangle &: x_0 = (0, 2, 0), Q_3(w) = w_0^2 + 2 w_0 w_1 \pmod 3, \\ \lvert\mspace{0.5mu}\sigma_2 \mspace{0.5mu}\rangle &: x_0 = (0, 1, 0), Q_2(w) = 2 w_0^2 + w_0 w_1 + w_1^2 \pmod 3, \\ \lvert\mspace{0.5mu}\sigma_4 \mspace{0.5mu}\rangle &: x_0 = (0, 1, 0), Q_4(w) = 2 w_0^2 + 2 w_0 w_1 + w_1^2 \pmod 3. \end{align}\] Set \(\alpha = \frac{\sqrt 6}{4} e^{-i \pi / 6} = \frac{3 \sqrt 2 - i \sqrt 6}{8}\) and \(\beta = -\frac{i \sqrt 6}{4}\). We claim \[\label{eq:strange-m3-id} \lvert\mspace{0.5mu}\mathbb{S} \mspace{0.5mu}\rangle^{\otimes 3} = \alpha \big( \lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle - \lvert\mspace{0.5mu}\sigma_3 \mspace{0.5mu}\rangle \big) + \beta \big( \lvert\mspace{0.5mu}\sigma_2 \mspace{0.5mu}\rangle - \lvert\mspace{0.5mu}\sigma_4 \mspace{0.5mu}\rangle \big).\tag{23}\]

Since \(\lvert\mspace{0.5mu}\mathbb{S} \mspace{0.5mu}\rangle= (\lvert\mspace{0.5mu}1 \mspace{0.5mu}\rangle - \lvert\mspace{0.5mu}2 \mspace{0.5mu}\rangle) / \sqrt 2\), the amplitude \([\lvert\mspace{0.5mu}\mathbb{S} \mspace{0.5mu}\rangle^{\otimes 3}]_y\) vanishes whenever some \(y_i = 0\), and equals \((-1)^{n_2} / (2 \sqrt 2)\) at \(y \in \{1, 2\}^3\), where \(n_2 = |\{i : y_i = 2\}|\). The shared \(W\) sends \(w \mapsto (w_0, 0, w_1)\), so \(\lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle, \lvert\mspace{0.5mu}\sigma_3 \mspace{0.5mu}\rangle\) have support on the plane \(\{y : y_1 = 2\}\) and \(\lvert\mspace{0.5mu}\sigma_2 \mspace{0.5mu}\rangle, \lvert\mspace{0.5mu}\sigma_4 \mspace{0.5mu}\rangle\) on \(\{y : y_1 = 1\}\); both sides of 23 vanish on the remaining plane \(y_1 = 0\), so the identity decouples into two independent verifications.

Consider the plane \(y_1 = 2\). On this plane the canonical-form parameter \(w = (w_0, w_1)\) satisfies \(w_0 = y_0\) and \(w_1 = y_2\), so \([\sigma_j]_y = \omega^{Q_j(y_0, y_2)} / 3\) for \(j \in \{1, 3\}\). The polynomials \(Q_1, Q_3\) differ only in the cross-term \(\pm w_0 w_1 = \pm y_0 y_2\), which vanishes whenever \(y_0 = 0\) or \(y_2 = 0\); on those lines \(\lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle_y = \lvert\mspace{0.5mu}\sigma_3 \mspace{0.5mu}\rangle_y\), so \(\alpha (\lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle - \lvert\mspace{0.5mu}\sigma_3 \mspace{0.5mu}\rangle)_y = 0\), matching \(\lvert\mspace{0.5mu}\mathbb{S} \mspace{0.5mu}\rangle^{\otimes 3}_y = 0\). At \((y_0, y_2) \in \{1, 2\}^2\) one has \(y_0^2 \equiv 1 \pmod 3\), and direct evaluation yields \[(Q_1, Q_3)(y_0, y_2) = \begin{cases} (2, 0), & y_0 = y_2, \\ (0, 2), & y_0 \neq y_2, \end{cases} \qquad \text{whence} \qquad \omega^{Q_1} - \omega^{Q_3} = \epsilon(y_0, y_2) (\omega^2 - 1),\] with \(\epsilon(y_0, y_2) = (-1)^{[y_0 = 2] + [y_2 = 2]}\) equal to \(+1\) when \(y_0 = y_2\) and \(-1\) otherwise. Meanwhile \([\lvert\mspace{0.5mu}\mathbb{S} \mspace{0.5mu}\rangle^{\otimes 3}]_{(y_0, 2, y_2)} = -\epsilon(y_0, y_2) / (2 \sqrt 2)\) (the leading minus sign coming from the fixed \(y_1 = 2\) factor). The four-point identity therefore reduces to the single scalar equation \(\alpha (\omega^2 - 1) / 3 = -1 / (2 \sqrt 2)\). Using \(\omega^2 - 1 = -\frac{3}{2} - \frac{i \sqrt 3}{2} = -\sqrt 3 e^{i \pi / 6}\), \[\alpha = \frac{-3}{2 \sqrt 2 (-\sqrt 3 e^{i \pi / 6})} = \frac{3}{2 \sqrt 6} e^{-i \pi / 6} = \frac{\sqrt 6}{4} e^{-i \pi / 6},\] which matches the stated value of \(\alpha\).

Consider the plane \(y_1 = 1\). The same argument applies: \(Q_2\) and \(Q_4\) agree whenever \(y_0 = 0\) or \(y_2 = 0\), so \(\beta(\lvert\mspace{0.5mu}\sigma_2 \mspace{0.5mu}\rangle - \lvert\mspace{0.5mu}\sigma_4 \mspace{0.5mu}\rangle)\) vanishes there. At \((y_0, y_2) \in \{1, 2\}^2\), \(y_0^2 \equiv y_2^2 \equiv 1\) reduces \(Q_2 \equiv y_0 y_2\) and \(Q_4 \equiv 2 y_0 y_2 \pmod 3\), giving \[(Q_2, Q_4)(y_0, y_2) = \begin{cases} (1, 2), & y_0 = y_2, \\ (2, 1), & y_0 \neq y_2, \end{cases} \quad \text{whence} \quad \omega^{Q_2} - \omega^{Q_4} = i \sqrt 3 \epsilon(y_0, y_2).\] Now \([\lvert\mspace{0.5mu}\mathbb{S} \mspace{0.5mu}\rangle^{\otimes 3}]_{(y_0, 1, y_2)} = \epsilon(y_0, y_2) / (2 \sqrt 2)\), so the identity reduces to \(\beta i \sqrt 3 / 3 = 1 / (2 \sqrt 2)\), giving \[\beta = \frac{3}{2 \sqrt 2 i \sqrt 3} = -\frac{i \sqrt 6}{4},\] which matches the stated value of \(\beta\).

Machine-checked Lean formalization: StrangeM3.lean, StrangeM3Pointwise.lean. ◻

Proof that \(\chi(\lvert\mspace{0.5mu}H_3 \mspace{0.5mu}\rangle^{\otimes 2}) \le 3\). Define three 2-qutrit stabilizer states by 1 : \[\label{eq:h3-m2-states} \begin{align} \lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle &: k = 1, x_0 = 0, W = e_1, Q_1 = 0, \quad \text{so } \lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle = \lvert\mspace{0.5mu}0, + \mspace{0.5mu}\rangle, \\ \lvert\mspace{0.5mu}\sigma_2 \mspace{0.5mu}\rangle &: k = 1, x_0 = 0, W = e_0, Q_2 = 0, \quad \text{so } \lvert\mspace{0.5mu}\sigma_2 \mspace{0.5mu}\rangle = \lvert\mspace{0.5mu}+, 0 \mspace{0.5mu}\rangle, \\ \lvert\mspace{0.5mu}\sigma_3 \mspace{0.5mu}\rangle &: k = 2, x_0 = 0, W = I_2, Q_3(y) = 2 y_0^2 + y_1^2 \pmod 3, \end{align}\tag{26}\] and set \[\label{eq:h3-m2-coeffs} \alpha_1 = \frac{c \sqrt 3}{N^2}\big( 1 - c \omega \big), \qquad \alpha_2 = \frac{c \sqrt 3}{N^2}\big( 1 - c \omega^2 \big), \qquad \alpha_3 = \frac{3 c^2}{N^2}.\tag{27}\] We claim \[\label{eq:h3-m2-id} \lvert\mspace{0.5mu}H_3 \mspace{0.5mu}\rangle^{\otimes 2} = \alpha_1 \lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle + \alpha_2 \lvert\mspace{0.5mu}\sigma_2 \mspace{0.5mu}\rangle + \alpha_3 \lvert\mspace{0.5mu}\sigma_3 \mspace{0.5mu}\rangle.\tag{28}\]

By 24 , \([\lvert\mspace{0.5mu}H_3 \mspace{0.5mu}\rangle^{\otimes 2}]_y = c^{|y|_*}/N^2\) with \(|y|_* = [y_0 \neq 0] + [y_1 \neq 0]\). The components \([\sigma_1]_y = (1/\sqrt 3)[y_0 = 0]\), \([\sigma_2]_y = (1/\sqrt 3)[y_1 = 0]\), and \([\sigma_3]_y = \omega^{Q_3(y)}/3\) all depend on \(y\) only through the pair \((a, b) = ([y_0 \ne 0], [y_1 \ne 0]) \in \{0, 1\}^2\), since \(y_i^2 \equiv [y_i \ne 0] \pmod 3\). In particular \(Q_3(y) \equiv 2a + b \pmod 3\), and 28 reduces to four scalar identities, one per class.

Writing \(\alpha_1/\sqrt 3 = (c/N^2)(1 - c\omega)\) and \(\alpha_2/\sqrt 3 = (c/N^2)(1 - c\omega^2)\), the right side of 28 at any \(y\) in class \((a, b)\) equals \[\label{eq:h3-m2-class-eval} [\alpha_1 \sigma_1]_y + [\alpha_2 \sigma_2]_y + [\alpha_3 \sigma_3]_y = \frac{c}{N^2}[a = 0] (1 - c \omega) + \frac{c}{N^2}[b = 0] (1 - c \omega^2) + \frac{c^2}{N^2} \omega^{2a + b}.\tag{29}\] Evaluating 29 in each class (using \(\omega + \omega^2 = -1\) and \(c(1 + c) = 1/2\)):

This establishes 28 and hence \(\chi(\lvert\mspace{0.5mu}H_3 \mspace{0.5mu}\rangle^{\otimes 2}) \le 3\).

Machine-checked Lean formalization: H3M2Pointwise.lean. ◻

Proof that \(\chi(\lvert\mspace{0.5mu}H_3 \mspace{0.5mu}\rangle^{\otimes 3}) \le 4\). Define four 3-qutrit stabilizer states by 1 : \(\lvert\mspace{0.5mu}\sigma_j \mspace{0.5mu}\rangle\) for \(j \in \{1, 2\}\) has \(k = 3\), \(x_0 = 0\), \(W = I_3\), and \(Q_j(y) \pmod 3\) given by \[\label{eq:h3-Qs} Q_1(y) = 2 y_0^2 + 2 y_1^2 + y_2^2, \qquad Q_2(y) = y_0^2 + y_1^2 + 2 y_2^2,\tag{30}\] and \[\label{eq:h3-S3S4} \lvert\mspace{0.5mu}\sigma_3 \mspace{0.5mu}\rangle = \lvert\mspace{0.5mu}0, 0, + \mspace{0.5mu}\rangle, \qquad \lvert\mspace{0.5mu}\sigma_4 \mspace{0.5mu}\rangle = \lvert\mspace{0.5mu}+, +, 0 \mspace{0.5mu}\rangle,\tag{31}\] recognizing the canonical form with \(k = 1\), \(W = e_2\), \(Q_3 = 0\) and \(k = 2\), \(W = (e_0, e_1)\), \(Q_4 = 0\) respectively. Set \[\label{eq:h3-coeffs} \alpha_1 = \frac{3 c}{4 N}(1 + i), \quad \alpha_2 = \frac{3 c}{4 N}(1 - i), \quad \alpha_3 = \alpha_4 = \frac{3}{4 N}.\tag{32}\] We claim \[\label{eq:h3-id} \lvert\mspace{0.5mu}H_3 \mspace{0.5mu}\rangle^{\otimes 3} = \alpha_1 \lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle + \alpha_2 \lvert\mspace{0.5mu}\sigma_2 \mspace{0.5mu}\rangle + \alpha_3 \lvert\mspace{0.5mu}\sigma_3 \mspace{0.5mu}\rangle + \alpha_4 \lvert\mspace{0.5mu}\sigma_4 \mspace{0.5mu}\rangle.\tag{33}\]

Both \(Q_1\) and \(Q_2\) depend on \(y\) only through \(y_i^2 \pmod 3 = [y_i \ne 0]\), and 31 likewise depends only on which coordinates of \(y\) vanish, so each component of the right side of 33 depends only on the pair \[(n, z) = \big( |\{ i \in \{0, 1\} : y_i \ne 0 \}|, [y_2 \ne 0] \big) \in \{0,1,2\} \times \{0,1\}.\] Each component of \(\lvert\mspace{0.5mu}H_3 \mspace{0.5mu}\rangle^{\otimes 3}\) depends only on \(|y|_* = n + z\) via 24 , namely \([\lvert\mspace{0.5mu}H_3 \mspace{0.5mu}\rangle^{\otimes 3}]_y = c^{n + z} / N^3\). Hence 33 reduces to six scalar identities, one per class.

For the \(\lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle, \lvert\mspace{0.5mu}\sigma_2 \mspace{0.5mu}\rangle\) contribution at \(y\) in class \((n, z)\), the canonical form gives \(\alpha_1 [\sigma_1]_y + \alpha_2 [\sigma_2]_y = (3 \sqrt 3)^{-1} (\alpha_1 \omega^{2n + z} + \alpha_2 \omega^{n + 2z})\). Substituting \(\omega = -\frac{1}{2} + \frac{i \sqrt 3}{2}\), the three values of \((\alpha_1 \omega^A + \alpha_2 \omega^B)\) that arise are \[\label{eq:h3-three-values} \begin{align} \alpha_1 + \alpha_2 &= \frac{3 c}{2 N}, \\ \alpha_1 \omega + \alpha_2 \omega^2 &= -\frac{3 c (1 + \sqrt 3)}{4 N} = -\frac{3}{4 N} && \text{(using c(1 + \sqrt 3) = 1)}, \\ \alpha_1 \omega^2 + \alpha_2 \omega &= \frac{3 c (\sqrt 3 - 1)}{4 N} = \frac{3 c^2}{2 N} && \text{(using \sqrt 3 - 1 = 2 c)}. \end{align}\tag{34}\] Combining with \([\sigma_3]_y = (1/\sqrt 3) [n = 0]\) and \([\sigma_4]_y = (1/3)[z = 0]\), the right side of 33 at any \(y\) in class \((n, z)\) is the entry below; the left side is \(c^{n+z}/N^3\):

Equality of the right-hand entries with \(c^{n+z}/N^3\) in each row is the identity \[\label{eq:h3-collapse} \frac{c^k}{N^3} = \frac{c^k (3 + \sqrt 3)}{6 N}, \qquad k \in \{0, 1, 2, 3\},\tag{35}\] which follows from \((3 + \sqrt 3) N^2 = (3 + \sqrt 3)(3 - \sqrt 3) = 6\). The class-by-class identities then reduce to elementary arithmetic in \(c, \sqrt 3\). As an example, the \((n, z) = (0, 0)\) row asks for \[\label{eq:h3-row00} \frac{c}{2 N \sqrt 3} + \frac{3 + \sqrt 3}{4 N \sqrt 3} = \frac{2 c + 3 + \sqrt 3}{4 N \sqrt 3} = \frac{(\sqrt 3 - 1) + 3 + \sqrt 3}{4 N \sqrt 3} = \frac{2 + 2\sqrt 3}{4 N \sqrt 3} = \frac{3 + \sqrt 3}{6 N} = \frac{1}{N^3},\tag{36}\] using \(2c = \sqrt 3 - 1\) in the third equality and rationalizing \(1/\sqrt 3\) in the fourth. The \((1, 0)\) row similarly uses \(2c^2 = 2 - \sqrt 3\) to collapse \(2 c^2 + \sqrt 3 = 2\), giving \(1/(2 N \sqrt 3) = c/N^3\); the remaining four rows are the analogous one-line manipulations. This proves 33 and hence \(\chi(\lvert\mspace{0.5mu}H_3 \mspace{0.5mu}\rangle^{\otimes 3}) \le 4\).

Machine-checked Lean formalization: H3M3.lean, H3M3Pointwise.lean. ◻

Proof that \(\chi(\lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle^{\otimes 2}) \le 3\). Define three 2-qutrit stabilizer states by 1 : \[\label{eq:norrell-m2-states} \begin{align} \lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle &: k = 2, x_0 = 0, W = I_2, Q_1(y) = y_0 + y_1 + y_0 y_1 \pmod 3, \\ \lvert\mspace{0.5mu}\sigma_2 \mspace{0.5mu}\rangle &: k = 0, x_0 = (2, 2), \quad \text{so } \lvert\mspace{0.5mu}\sigma_2 \mspace{0.5mu}\rangle = \lvert\mspace{0.5mu}2, 2 \mspace{0.5mu}\rangle, \\ \lvert\mspace{0.5mu}\sigma_3 \mspace{0.5mu}\rangle &: k = 2, x_0 = 0, W = I_2, Q_3(y) = 2 Q_1(y) \pmod 3, \end{align}\tag{40}\] so \(\lvert\mspace{0.5mu}\sigma_3 \mspace{0.5mu}\rangle = \overline{\lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle}\) (entrywise complex conjugate in the computational basis). Set \(\alpha = -\omega/2\). We claim \[\label{eq:norrell-m2-id} \lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle^{\otimes 2} = \alpha \lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle + \lvert\mspace{0.5mu}\sigma_2 \mspace{0.5mu}\rangle + \overline{\alpha} \lvert\mspace{0.5mu}\sigma_3 \mspace{0.5mu}\rangle,\tag{41}\] equivalently \(\lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle^{\otimes 2} = -\frac{\omega}{2} \lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle + \lvert\mspace{0.5mu}2, 2 \mspace{0.5mu}\rangle - \frac{\omega^2}{2} \lvert\mspace{0.5mu}\sigma_3 \mspace{0.5mu}\rangle\).

The polynomial \(Q_1\) factors as \[\label{eq:norrell-m2-Qfactor} Q_1(y) \equiv (1 + y_0)(1 + y_1) - 1 \pmod 3,\tag{42}\] so \(Q_1(y) \equiv 2 \pmod 3\) iff \((1 + y_0)(1 + y_1) \equiv 0\), iff \(y_0 = 2\) or \(y_1 = 2\). Equivalently, \[\label{eq:norrell-m2-Qind} [Q_1(y) = 2] = [y_0 = 2] + [y_1 = 2] - [y_0 = y_1 = 2].\tag{43}\] By 1 , the contribution of \(\alpha \lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle + \overline{\alpha} \lvert\mspace{0.5mu}\sigma_3 \mspace{0.5mu}\rangle\) at \(y\) is \[\label{eq:norrell-m2-S13} [\alpha \sigma_1 + \overline{\alpha} \sigma_3]_y = \frac{1}{3}\big(-\frac{\omega}{2}\big) \omega^{Q_1(y)} + \frac{1}{3}\big(-\frac{\omega^2}{2}\big) \omega^{2 Q_1(y)} = -\frac{1}{6} \big( \omega^{r} + \omega^{2 r} \big), \qquad r = 1 + Q_1(y) \pmod 3.\tag{44}\] Since \(1 + \omega + \omega^2 = 0\), \(\omega^r + \omega^{2 r}\) equals \(2\) when \(r = 0\) and \(-1\) when \(r \in \{1, 2\}\); noting \(r = 0\) iff \(Q_1(y) = 2\), \[\label{eq:norrell-m2-S13-cN} [\alpha \sigma_1 + \overline{\alpha} \sigma_3]_y = \frac{1}{6} c_N(Q_1(y)).\tag{45}\] Adding the contribution \(\delta_{y, (2,2)}\) from \(\lvert\mspace{0.5mu}\sigma_2 \mspace{0.5mu}\rangle\), \[\label{eq:norrell-m2-sum} [\alpha \sigma_1 + \sigma_2 + \overline{\alpha} \sigma_3]_y = \frac{1}{6} c_N(Q_1(y)) + \delta_{y, (2, 2)}.\tag{46}\] Comparing with \([\lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle^{\otimes 2}]_y = (1/6)c_N(y_0) c_N(y_1)\) from 38 , 41 reduces to the single identity \[\label{eq:norrell-m2-key} c_N(y_0) c_N(y_1) - c_N(Q_1(y)) = 6 [y = (2, 2)].\tag{47}\] Using \(c_N(y) = 1 - 3 [y = 2]\) and 43 , \[\begin{align} c_N(y_0) c_N(y_1) &= 1 - 3 [y_0 = 2] - 3 [y_1 = 2] + 9 [y_0 = y_1 = 2], \\ c_N(Q_1(y)) &= 1 - 3 [Q_1(y) = 2] = 1 - 3 [y_0 = 2] - 3 [y_1 = 2] + 3 [y_0 = y_1 = 2], \end{align}\] so their difference is \(6 [y_0 = y_1 = 2] = 6 [y = (2,2)]\), establishing 47 and hence 41 . Therefore \(\chi(\lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle^{\otimes 2}) \le 3\).

Machine-checked Lean formalization: NorrellM2Pointwise.lean. ◻

Proof that \(\chi(\lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle^{\otimes 3}) \le 4\). Define four 3-qutrit stabilizer states by 1 : \[\begin{align} \lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle &: k = 2, x_0 = (2,0,0), W = (e_1, e_2), Q_1(w) = w_0 + 2 w_0^2 + 2 w_1 + w_1^2 \pmod 3, \\ \lvert\mspace{0.5mu}\sigma_2 \mspace{0.5mu}\rangle &: k = 2, x_0 = (0,2,0), W = (e_0, e_2), Q_2(w) = 2 w_0 + w_0^2 + w_1 + 2 w_1^2 \pmod 3, \\ \lvert\mspace{0.5mu}\sigma_3 \mspace{0.5mu}\rangle &: k = 3, x_0 = 0, W = I_3, Q_3 = 0, \quad \text{so } \lvert\mspace{0.5mu}\sigma_3 \mspace{0.5mu}\rangle = \lvert\mspace{0.5mu}+ \mspace{0.5mu}\rangle^{\otimes 3}, \\ \lvert\mspace{0.5mu}\sigma_4 \mspace{0.5mu}\rangle &: k = 2, x_0 = (0,0,2), W = (e_0, e_1), Q_4(w) = w_0 + 2 w_0^2 + 2 w_1 + w_1^2 \pmod 3. \end{align}\] We claim \[\label{eq:norrell-id} \lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle^{\otimes 3} = -\frac{\sqrt 6}{4} \big( \lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle + \lvert\mspace{0.5mu}\sigma_2 \mspace{0.5mu}\rangle + \lvert\mspace{0.5mu}\sigma_4 \mspace{0.5mu}\rangle \big) + \frac{\sqrt 2}{4} \lvert\mspace{0.5mu}\sigma_3 \mspace{0.5mu}\rangle.\tag{48}\]

By the reduction identities 39 , on the supports \[\label{eq:norrell-m3-supp} \mathrm{supp}(\sigma_1) = \{(2, y_1, y_2)\}, \quad \mathrm{supp}(\sigma_2) = \{(y_0, 2, y_2)\}, \quad \mathrm{supp}(\sigma_4) = \{(y_0, y_1, 2)\},\tag{49}\] the phase polynomials reduce to \[Q_1(y) = [y_1{=}2] + 2 [y_2{=}2], \quad Q_2(y) = 2 [y_0{=}2] + [y_2{=}2], \quad Q_4(y) = [y_0{=}2] + 2 [y_1{=}2] \pmod 3.\] Setting \(a_i = [y_i = 2] \in \{0, 1\}\) for \(i \in \{0, 1, 2\}\) (consistent with the Norrell m=4 proof below), the right side of 48 at any \(y\) depends only on \(a = (a_0, a_1, a_2)\) and equals \[\label{eq:norrell-rhs} \text{RHS at } y = \frac{\sqrt 6}{36} - \frac{\sqrt 6}{12} B, \qquad B = a_0 \omega^{a_1 + 2 a_2} + a_1 \omega^{2 a_0 + a_2} + a_2 \omega^{a_0 + 2 a_1},\tag{50}\] the leading \(\sqrt 6 / 36\) coming from \(\alpha_3 [\sigma_3]_y = \frac{\sqrt 2}{4 \sqrt{27}}\), and \(B\) collecting the contributions from \(\sigma_1, \sigma_2, \sigma_4\) each scaled by \(\alpha_j / 3 = -\sqrt 6 / 12\).

The sum \(B\) depends only on \(n_2 = a_0 + a_1 + a_2 = |\{i : y_i = 2\}|\): it vanishes when \(a = (0,0,0)\); for each of the three \(n_2 = 1\) subsets exactly one term contributes \(\omega^0 = 1\); for each of the three \(n_2 = 2\) subsets two terms contribute \(\omega^A + \omega^B\) with \(\{A, B\} = \{1, 2\}\), summing to \(\omega + \omega^2 = -1\); and for \(a = (1,1,1)\) all three terms equal \(\omega^3 = 1\), summing to \(3\). Therefore \[B(n_2) = (0, 1, -1, 3) \quad \text{for } n_2 = 0, 1, 2, 3,\] and substituting into 50 , \[\label{eq:norrell-rhs-collapsed} \text{RHS}_{n_2} = \frac{\sqrt 6}{36} \big( 1 - 3 B(n_2) \big) = \frac{\sqrt 6}{36} (-2)^{n_2},\tag{51}\] which matches \([\lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle^{\otimes 3}]_y\) from 38 term-for-term. This establishes 48 .

Machine-checked Lean formalization: NorrellM3.lean, NorrellM3Pointwise.lean. ◻

Proof that \(\chi(\lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle^{\otimes 4}) \le 7\). Let \(\xi = e^{i \pi / 6}\), a primitive 12th root of unity, so that \(\xi^4 = \omega\), \(\xi^8 = \omega^2\), \(\xi^3 = i\), \(\xi^6 = -1\), and \(\xi^{12} = 1\). Define seven 4-qutrit stabilizer states by 1 ; the canonical-form data \((k_j, x_{0,j}, W_j, Q_j)\) for \(j = 0, 1, \ldots, 6\) are \[\label{eq:norrell-m4-states} \begin{align} \lvert\mspace{0.5mu}\sigma_0 \mspace{0.5mu}\rangle &: k = 3, x_0 = (0, 2, 0, 0), W = (e_0, e_2, e_3), Q_0(w) = 2 w_0 + w_0^2 + w_1 + 2 w_1^2 + 2 w_2 + w_2^2, \\ \lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle &: k = 4, x_0 = 0, W = I_4, Q_1(y) = \sum_{i=0}^{1} (2 y_i + y_i^2) + \sum_{i=2}^{3} (y_i + 2 y_i^2), \\ \lvert\mspace{0.5mu}\sigma_2 \mspace{0.5mu}\rangle &: k = 2, x_0 = (2, 0, 0, 2), W = (e_1, e_2), Q_2(w) = w_0 + 2 w_0^2 + w_1 + 2 w_1^2, \\ \lvert\mspace{0.5mu}\sigma_3 \mspace{0.5mu}\rangle &: k = 2, x_0 = (0, 0, 2, 2), W = (e_0, e_1), Q_3(w) = 2 w_0 + w_0^2 + w_1 + 2 w_1^2, \\ \lvert\mspace{0.5mu}\sigma_4 \mspace{0.5mu}\rangle &: k = 3, x_0 = (2, 0, 0, 0), W = (e_1, e_2, e_3), Q_4(w) = 2 w_0 + w_0^2 + 2 w_1 + w_1^2 + w_2 + 2 w_2^2, \\ \lvert\mspace{0.5mu}\sigma_5 \mspace{0.5mu}\rangle &: k = 2, x_0 = (0, 2, 2, 0), W = (e_0, e_3), Q_5(w) = w_0 + 2 w_0^2 + 2 w_1 + w_1^2, \\ \lvert\mspace{0.5mu}\sigma_6 \mspace{0.5mu}\rangle &: k = 4, x_0 = 0, W = I_4, Q_6 = 0, \quad \text{so } \lvert\mspace{0.5mu}\sigma_6 \mspace{0.5mu}\rangle = \lvert\mspace{0.5mu}+ \mspace{0.5mu}\rangle^{\otimes 4}. \end{align}\tag{52}\] Setting \(\tau = \sqrt 3 / 4\), we claim \[\label{eq:norrell-m4-id} \begin{align} \lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle^{\otimes 4} &= \tau\omega \lvert\mspace{0.5mu}\sigma_0 \mspace{0.5mu}\rangle + \tau\xi \lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle + \tau\xi \lvert\mspace{0.5mu}\sigma_2 \mspace{0.5mu}\rangle + \tau\xi \lvert\mspace{0.5mu}\sigma_3 \mspace{0.5mu}\rangle + \tau\omega \lvert\mspace{0.5mu}\sigma_4 \mspace{0.5mu}\rangle + \tau\xi \lvert\mspace{0.5mu}\sigma_5 \mspace{0.5mu}\rangle + \frac{1}{4}\omega^2 \lvert\mspace{0.5mu}\sigma_6 \mspace{0.5mu}\rangle. \end{align}\tag{53}\] The seven coefficients take only three distinct values (\(\tau\omega\), \(\tau\xi\), \(\omega^2/4\)), each a \(12\)th-root-of-unity multiple of a real number.

By the reduction identities 39 , on each basis state’s support the polynomial \(Q_j\) collapses to an indicator-only expression. Setting \(a_i = [y_i = 2] \in \{0, 1\}\), \[\label{eq:norrell-m4-Qreductions} \begin{align} \mathrm{supp}(\sigma_0) = \{y_1 = 2\}\colon\quad & Q_0(y) \equiv 2 a_0 + a_2 + 2 a_3 \pmod 3, \\ \mathrm{supp}(\sigma_1) = \mathbb{F}_{3}^4\colon\quad & Q_1(y) \equiv 2 a_0 + 2 a_1 + a_2 + a_3 \pmod 3, \\ \mathrm{supp}(\sigma_2) = \{y_0 = y_3 = 2\}\colon\quad & Q_2(y) \equiv a_1 + a_2 \pmod 3, \\ \mathrm{supp}(\sigma_3) = \{y_2 = y_3 = 2\}\colon\quad & Q_3(y) \equiv 2 a_0 + a_1 \pmod 3, \\ \mathrm{supp}(\sigma_4) = \{y_0 = 2\}\colon\quad & Q_4(y) \equiv 2 a_1 + 2 a_2 + a_3 \pmod 3, \\ \mathrm{supp}(\sigma_5) = \{y_1 = y_2 = 2\}\colon\quad & Q_5(y) \equiv a_0 + 2 a_3 \pmod 3. \end{align}\tag{54}\] Every component of the right side of 53 therefore depends on \(y\) only through \(a = (a_0, a_1, a_2, a_3) \in \{0,1\}^4\). Writing \(\mathbf{1}_j(a) \in \{0, 1\}\) for the indicator that \(a\) lies in \(\mathrm{supp}(\sigma_j)\), the right-hand side amplitude is \[\label{eq:norrell-m4-rhs} \mathrm{RHS}(a) = \sum_{j=0}^{6} \alpha_j 3^{-k_j/2} \omega^{Q_j(a)} \mathbf{1}_j(a),\tag{55}\] with the \(\alpha_j\) as in 53 and \(k_j\) as in 52 .

8.3.3.1 Endpoints.

At \(a = (0, 0, 0, 0)\) only \(\lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle\) and \(\lvert\mspace{0.5mu}\sigma_6 \mspace{0.5mu}\rangle\) have \(a\) in their support, so \[\label{eq:norrell-m4-rhs0} \mathrm{RHS}(0,0,0,0) = \frac{\tau\xi}{9} \omega^0 + \frac{\omega^2}{36} \omega^0 = \frac{1}{36} \big( \sqrt 3\xi + \omega^2 \big).\tag{56}\] Since \(\sqrt 3\xi = \frac{3}{2} + i \frac{\sqrt 3}{2}\) and \(\omega^2 = -\frac{1}{2} - i \frac{\sqrt 3}{2}\), the sum equals \(1\), so \(\mathrm{RHS}(0,0,0,0) = 1/36\).

At \(a = (1, 1, 1, 1)\) all seven basis states contribute. After evaluating \(Q_j\) in each case (using \(\xi^4 = \omega\) and \(\tau\omega \omega^2 = \tau\)), the seven contributions collect to \[\label{eq:norrell-m4-rhs1} \begin{align} \mathrm{RHS}(1,1,1,1) &= \frac{1}{12} + \frac{\sqrt 3 \xi}{36} + \frac{(-i)\tau}{3} + \frac{\sqrt 3 \xi}{12} + \frac{1}{12} + \frac{\sqrt 3 \xi}{12} + \frac{\omega^2}{36} \\ &= \frac{1}{36} \big( 6 + 7\sqrt 3 \xi - 3 i \sqrt 3 + \omega^2 \big). \end{align}\tag{57}\] Substituting \(7\sqrt 3 \xi = \frac{21}{2} + i \frac{7\sqrt 3}{2}\) and collecting reals/imaginaries yields \(6 + \frac{21}{2} - \frac{1}{2} - 0 = 16\) and \(\frac{7\sqrt 3}{2} - 3 \sqrt 3 - \frac{\sqrt 3}{2} = 0\), so \(\mathrm{RHS}(1,1,1,1) = 16/36 = 4/9\).

8.3.3.2 Intermediate cases.

Direct enumeration over the remaining \(14\) patterns \(a \in \{0, 1\}^4\) with \(n_2(a) = \sum_i a_i \in \{1, 2, 3\}\) analogously reduces \(\mathrm{RHS}(a)\) to a polynomial identity in \(\xi\) over \(\mathbb{Z}[\sqrt 3]\), decided by \(\xi^4 = \omega\), \(\xi^6 = -1\), and \(1 + \omega + \omega^2 = 0\). Across the 14 patterns, \(\mathrm{RHS}(a)\) depends only on \(n_2(a) \in \{1, 2, 3\}\): the coefficient pattern in Eq. 53 and the reduced phases in 54 are jointly invariant under the symmetric-group action that permutes the four coordinates of \(a\) within each \(n_2\)-class, so the 4, 6, and 4 patterns in classes \(n_2 = 1, 2, 3\) each yield the same value. The two representatives below cover \(n_2 = 1\) and \(n_2 = 2\); the same mechanism handles \(n_2 = 3\) and the remaining patterns in each class.

Case \(n_2 = 1\), \(a = (1, 0, 0, 0)\). Only \(\lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle, \lvert\mspace{0.5mu}\sigma_4 \mspace{0.5mu}\rangle, \lvert\mspace{0.5mu}\sigma_6 \mspace{0.5mu}\rangle\) have \(a\) in their support; reading off the reduced phases 54 , \(Q_1(a) = 2\), \(Q_4(a) = 0\), \(Q_6 = 0\), so the contributions are \(\tau\xi \omega^2 / 9 = -i \sqrt 3 / 36\), \(\tau\omega 1 / (3 \sqrt 3) = \omega / 12\), and \(\omega^2 / 4 1 / 9 = \omega^2 / 36\), respectively. Summing, \[\label{eq:norrell-m4-rhs-n21} \mathrm{RHS}(1,0,0,0) = \frac{1}{36} \big(- i \sqrt 3 + 3\omega + \omega^2\big) = -\frac{1}{18},\tag{58}\] using \(-i \sqrt 3 + 3\omega + \omega^2 = -i \sqrt 3 + 3(-\frac{1}{2} + i \frac{\sqrt 3}{2}) + (-\frac{1}{2} - i \frac{\sqrt 3}{2}) = -2\).

Case \(n_2 = 2\), \(a = (1, 1, 0, 0)\). The supports containing this pattern are \(\lvert\mspace{0.5mu}\sigma_0 \mspace{0.5mu}\rangle, \lvert\mspace{0.5mu}\sigma_1 \mspace{0.5mu}\rangle, \lvert\mspace{0.5mu}\sigma_4 \mspace{0.5mu}\rangle, \lvert\mspace{0.5mu}\sigma_6 \mspace{0.5mu}\rangle\); \(Q_0(a) = 2\), \(Q_1(a) = 4 \equiv 1\), \(Q_4(a) = 2\), \(Q_6 = 0\), so the four contributions are \(\tau\omega \omega^2 / (3\sqrt 3) = 1/12\), \(\tau\xi \omega/9 = \xi\omega \sqrt 3 / 36\), \(\tau\omega \omega^2 / (3\sqrt 3) = 1/12\), and \(\omega^2 / 4 1/9 = \omega^2 / 36\). Summing, \[\label{eq:norrell-m4-rhs-n22} \mathrm{RHS}(1,1,0,0) = \frac{1}{36} \big(6 + \xi \omega \sqrt 3 + \omega^2\big) = \frac{1}{9},\tag{59}\] using \(\xi \omega = e^{i 5\pi/6}\), hence \(\xi \omega \sqrt 3 = -\frac{3}{2} + i \frac{\sqrt 3}{2}\), which combines with \(\omega^2\) to give \(-2\).

8.3.3.3 All 16 cases.

The remaining 12 patterns are verified by the same mechanism. The \(\mathrm{RHS}(a)\) value depends only on \(n_2(a)\), and matches the target \((-2)^{n_2(a)}/36\) in each \(n_2\)-class:

All \(16\) scalar identities are checked at exact rational precision and the pointwise identity is independently formalized in Lean 4 (Section 4). Comparing with the target amplitude \([\lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle^{\otimes 4}]_y = (-2)^{n_2(y)}/36\) from 38 establishes 53 , and hence \(\chi(\lvert\mspace{0.5mu}\mathbb{N} \mspace{0.5mu}\rangle^{\otimes 4}) \le 7\).

Machine-checked Lean formalization: NorrellM4Pointwise.lean. ◻

Proof. The amplitude of \(\lvert\mspace{0.5mu}T \mspace{0.5mu}\rangle^{\otimes 4}\) at \(\lvert\mspace{0.5mu}x \mspace{0.5mu}\rangle\) depends only on the Hamming weight \(w = x_1 + x_2 + x_3 + x_4\) and equals \(\cos^{4-w}\!\beta \sin^w\!\beta e^{i \pi w / 4}\). Reducing via \(\cos^2\beta = (1+1/\sqrt 3)/2\), \(\sin^2\beta = (1-1/\sqrt 3)/2\), \(\cos\beta \sin\beta = 1/\sqrt 6\), \[\label{eq:qubit-t-m4-targets} \bigl[\lvert\mspace{0.5mu}T \mspace{0.5mu}\rangle^{\otimes 4}\bigr]_x = \begin{cases} (2 + \sqrt 3)/6 & w = 0,\\ (\sqrt 3 + 1)(1 + i)/12 & w = 1,\\ i/6 & w = 2,\\ (\sqrt 3 - 1)(-1 + i)/12 & w = 3,\\ (\sqrt 3 - 2)/6 & w = 4. \end{cases}\tag{61}\]

Set \(a = [x_1 = x_3]\) and \(b = [x_2 = x_4]\). From the support conditions in ?? , \(\sigma_1\) vanishes off \(\{a = 1\}\) and \(\sigma_3\) off \(\{b = 1\}\), while \(\sigma_2\) has full support. The right-hand side of ?? therefore splits into four support classes by \((a, b) \in \{0, 1\}^2\). Throughout we use \[\label{eq:qubit-t-m4-trig} \cos\frac{\pi}{12} = \frac{\sqrt 6 + \sqrt 2}{4}, \quad \sin\frac{\pi}{12} = \frac{\sqrt 6 - \sqrt 2}{4}, \quad \cos\frac{\pi}{12} - \sin\frac{\pi}{12} = \frac{1}{\sqrt 2},\tag{62}\] which together give the useful identity \[\label{eq:qubit-t-m4-prefactor} \frac{e^{i \pi / 12}}{3 \sqrt 2} = \frac{\sqrt 3 + 1}{12} + i \frac{\sqrt 3 - 1}{12}.\tag{63}\]

8.4.0.1 Case \((a, b) = (0, 0)\).

Every \(x\) in this class has \(w = 2\), \(x_1 x_3 = 0\), \(x_2 x_4 = 0\), and \(x_2 + x_4 = 1\), so the only contributing term is \[[c_2 \sigma_2]_x = \frac{2}{3} \frac{1}{4} i (-1)^0 = \frac{i}{6},\] matching the \(w = 2\) target.

8.4.0.2 Case \((a, b) = (1, 1)\).

Write \(x_1 = x_3 = p\) and \(x_2 = x_4 = q\) with \(p, q \in \{0, 1\}\); \(w = 2(p + q)\). The four sub-classes are listed below; in each row the right-most column collapses via 62 .

Each row matches the \(w \in \{0, 2, 2, 4\}\) target.

8.4.0.3 Case \((a, b) = (1, 0)\).

Only \(\sigma_1\) and \(\sigma_2\) contribute. Writing \(x_1 = x_3 = p\) and \(x_2 \ne x_4\) gives \([\sigma_1]_x = i^p / (2\sqrt 2)\) and \([\sigma_2]_x = i (-1)^p / 4\), so \[\begin{align} p = 0 (w = 1):\quad & [c_1 \sigma_1 + c_2 \sigma_2]_x = \frac{e^{i \pi / 12}}{3\sqrt 2} + \frac{i}{6} = \frac{(\sqrt 3 + 1)(1 + i)}{12},\\ p = 1 (w = 3):\quad & [c_1 \sigma_1 + c_2 \sigma_2]_x = \frac{i e^{i \pi / 12}}{3\sqrt 2} - \frac{i}{6} = \frac{(\sqrt 3 - 1)(-1 + i)}{12}, \end{align}\] using 63 in each line; these match the \(w = 1\) and \(w = 3\) targets.

8.4.0.4 Case \((a, b) = (0, 1)\).

Only \(\sigma_2\) and \(\sigma_3\) contribute. Writing \(x_2 = x_4 = q\) and \(x_1 \ne x_3\) gives \(i^{x_2 + x_4} = i^{2q} = (-1)^q\), \((-1)^{x_1 x_3 + x_2 x_4} = (-1)^{0 + q} = (-1)^q\) (since \(x_1 \ne x_3 \in \{0, 1\}\) forces \(x_1 x_3 = 0\)), so \([\sigma_2]_x = (-1)^q (-1)^q / 4 = 1/4\) and \([\sigma_3]_x = i^{1 + q} / (2\sqrt 2)\), so \[\begin{align} q = 0 (w = 1):\quad & [c_2 \sigma_2 + c_3 \sigma_3]_x = \frac{1}{6} + \frac{i e^{-i \pi / 12}}{3\sqrt 2} = \frac{(\sqrt 3 + 1)(1 + i)}{12},\\ q = 1 (w = 3):\quad & [c_2 \sigma_2 + c_3 \sigma_3]_x = \frac{1}{6} - \frac{e^{-i \pi / 12}}{3\sqrt 2} = \frac{(\sqrt 3 - 1)(-1 + i)}{12}, \end{align}\] using the conjugate \(e^{-i \pi / 12} / (3\sqrt 2) = (\sqrt 3 + 1)/12 - i (\sqrt 3 - 1)/12\) of 63 ; again matching the \(w = 1\) and \(w = 3\) targets.

The four cases cover \(\mathbb{F}_{2}^4\), so ?? holds pointwise. The exponent bound is then 6 . ◻