Prime-Degree Dickson Abelian Varieties and Elliptic Surfaces: Hasse–Brauer Quaternion Dichotomy, Two-Reduction Twisted Frobenius Non-CM Certificates, and Moving-Branch Cyclotomic Weil Classes

1. Introduction and Summary of Main Results

A central testing ground for the rational Hodge conjecture in higher codimension is the interplay between curve-dominated elliptic surfaces $X$ and the primitive factors $A \subset J(C)$ of the Jacobians of their covering curves. In the recent foundational papers of cm_symmetry_scout [1, 2], building on the dihedral real-multiplication geometry of van Geemen–Schütt [8] and the pairing-to-Weil-cycle mechanism of Varesco [9], two complementary prime-degree Dickson constructions were introduced:

1. The Fixed-Branch Prime-Degree Family ($\lambda = -1$, [1]): In Hodge cycles and Kuga–Satake geometry for Dickson surfaces (hodge_conjecture_27 v1) [1], for every prime $n > 3$ the cyclic degree-$6n$ curve $$C_n : w^{6n} = v^{4n+6}(1 - v)^{3n}(1 + v)^n$$ admits an order-$4$ automorphism $J$ normalizing the $\mu_{6n}$-action via $J \sigma_{6n} J^{-1} = \sigma_{6n}^{\tau}$, where $\tau \in (\mathbb{Z}/6n\mathbb{Z})^\times$ satisfies $\tau \equiv -1 \pmod n$ and $\tau \equiv +1 \pmod 6$. Consequently, setting $f = (n-1)/2$, the $4f$-dimensional primitive abelian variety $A_n = e_{6n}J(C_n)$ carries an action of the quaternion algebra $$D_n = (K_n / L_n,\, \tau,\, -1), \qquad K_n = \mathbb{Q}(\zeta_{6n}), \qquad L_n = K_n^{\langle \tau \rangle} = \mathbb{Q}(\zeta_n + \zeta_n^{-1},\, \zeta_3),$$ and dominates the elliptic surface $X_n : y^2 = x^3 + 3(D_n(s)-2)x + \frac{1}{2}(D_n(s)-2)(D_n(s)-6)$ of geometric genus $p_g(X_n) = \lceil n/3 \rceil - 1 = \lfloor n/3 \rfloor$ via a degree-$24n$ rational curve-square map $\phi_n : C_n^2 \dashrightarrow X_n$. Crucially, [1, Theorem 1] proved uniformly for all primes $n > 3$ that the equality $\operatorname{Hg}(A_n) = G_n$ (the full connected $D_n$-symplectic centralizer, with complexification $\mathrm{GL}_2^f$) and the rational Hodge conjecture for all powers of $A_n$ and $X_n$ reduce to the non-CM condition $C_{\operatorname{End}^0(A_n)}(D_n) = L_n$. In [1, Theorem 2], this non-CM condition was established for $n = 7$ and $n = 11$ (alongside $n = 5$), while all primes $n \ge 13$ were explicitly left open.

2. The $1$-Parameter Moving-Branch Family for $n = 5$ ($\lambda \neq 0, 1$, [2]): In Curve-square domination and Hodge cycles in a moving Dickson K3 family (hodge_conjecture_28 v1) [2], for $n = 5$ the branch point $v = -1$ was replaced by a moving parameter $v = 1/\lambda$ ($\lambda \neq 0, 1$) on $C_{5, \lambda} : w^{30} = v^{26}(1-v)^{15}(1-\lambda v)^5$, constructing a degree-$120$ curve-square rational map $\phi_{5, \lambda} : C_{5, \lambda}^2 \dashrightarrow X_{5, a}$ onto a moving K3 surface $X_{5, a}$ ($a^3 = -2(\lambda-1)^2/\lambda$) to extract the $4$-dimensional $\mathbb{Q}(\zeta_5)$-Weil space $W_{\mathbb{Q}(\zeta_5)}(A_{5,\lambda}) \subset H^4(A_{5,\lambda}, \mathbb{Q})$, together with a second degree-$60$ quotient onto a fixed surface $S_5 : y^2 = x^3 + Q^4(1-Q^5)^2$ realizing the determinant variation $D_{5,\lambda} = \bigwedge^2_{\mathbb{Q}(\zeta_{30})} H^1(A_{5,\lambda}, \mathbb{Q})$.

In this paper, we establish six main theorems (Theorems 2.1, 3.1, 3.2, 4.1, 4.2, and 5.1) that resolve the global Hasse–Brauer–Noether quaternion dichotomy on $A_n$, discharge the open $n \ge 13$ non-CM problem of [1] across all 20 primes $13 \le n < 100$ (with $n = 5, 7, 11$ included as cross-checks), and generalize the moving-branch curve-square map, Shioda–Tate rank bound, Chebyshev $2$-form diagonalization, $\mathbb{Q}(\zeta_n)$-Weil class construction, and fixed determinant surface $S_n$ to all primes $n > 3$:

• Theorem 2.1 (Mod-$12$ Hasse–Brauer–Noether Dichotomy for $D_n$): On the fixed branch $A_n$, the quaternion algebra $D_n = (K_n/L_n, \tau, -1)$ is a non-split division quaternion algebra over $L_n$ if and only if $n \equiv 7 \pmod{12}$ ($n = 7, 19, 31, 43, 67, 79, \dots$), where it ramifies at the two primes $\mathfrak{p}_n, \bar{\mathfrak{p}}_n \mid n$ of $L_n$, and is globally split $M_2(L_n)$ if and only if $n \equiv 1, 5, 11 \pmod{12}$.
• Theorem 3.1 (Twisted Frobenius Symmetries, Exact Cyclic Convolution Norm, and Elementary Newton Polygon Slopes): Applying the norm-one circle parametrization $v(t) = \frac{t+c}{t-c}$ ($t \in \mathbb{P}^1(\mathbb{F}_p)$, $c^2 = d \notin (\mathbb{F}_p^\times)^2$) of [1, §6.4] at primes $p \equiv 1 \pmod{12}$ with $p \equiv -1 \pmod n$ (where prime-field Frobenius $F_p$ fixes $L_n$ elementwise), inversion $t \mapsto -t$ gives $c_{-k,r} = c_{k,r}$ and the free Cayley involution $t \mapsto d/t$ on each $k(t) = k$ fiber gives $c_{k,0} + c_{k,1} + c_{k,2} = 0$. The reduced trace of $\Pi_p = J^{-1} F_p$ in $C_{\operatorname{End}_{\mathbb{Q}_\ell}(V_\ell(A_n))}(D_n) \cong D_n^{\mathrm{op}} \otimes_{\mathbb{Q}} \mathbb{Q}_\ell$ is $\alpha_p = -S_p \in \mathcal{O}_{L_n}$, with norm $N_p = \alpha_p \bar{\alpha}_p = S_p \bar{S}_p = \sum_{m=1}^{f} (\gamma_m - \gamma_0) C_m \in 3\mathcal{O}_{L_n^+}$ ($C_m = \zeta_n^m + \zeta_n^{-m}$) and reduced determinant $b_p = p \alpha_p / \bar{\alpha}_p \in \mathcal{O}_{L_n}$ ([1, eq. (14)]). Completing the square in $z^2 - \alpha_p z + b_p$ shows that its discriminant is $\Delta(\Pi_p) = \alpha_p^2 - 4b_p = (\alpha_p / N_p)^2 \delta_p$ with $\delta_p := N_p(N_p - 4p) \in \mathcal{O}_{L_n^+}$. Whenever $N_p \not\equiv 0 \pmod{\mathfrak{p}_1}$ at the degree-$1$ prime $\mathfrak{p}_1 \mid p$ of $L_n^+$, integrality $\alpha_p, \bar{\alpha}_p \in \mathcal{O}_{L_n}$ immediately forces $v_{\mathfrak{P}_1}(\alpha_p) = v_{\mathfrak{P}_1}(\bar{\alpha}_p) = 0$ and $v_{\mathfrak{P}_1}(b_p) = 1$, so $z^2 - \alpha_p z + b_p$ has Newton polygon slopes $0$ and $1$ at $\mathfrak{P}_1 \mid p$ and its eigenvalues have non-torsion ratio.
• Theorem 3.2 (Two-Reduction Non-CM Discharge for All 20 Open Primes $13 \le n < 100$): For every prime $5 \le n < 100$, evaluating $\delta_{p_i} = N_{p_i}(N_{p_i} - 4p_i)$ at two distinct good reductions $p_1 < p_2$ ($p_i \equiv 1 \pmod{12}, p_i \equiv -1 \pmod n$) with $N_{p_i} \not\equiv 0 \pmod{\mathfrak{p}_1}$ and testing quadratic character residues at an auxiliary prime $\ell \equiv 1 \pmod{6n}$ (which splits completely in $K_n$ and $L_n$) certifies that $M_{p_1} = L_n(\sqrt{\delta_{p_1}})$ and $M_{p_2} = L_n(\sqrt{\delta_{p_2}})$ are two non-isomorphic quadratic field extensions of $L_n$. By [1, Lemma 6 and Theorem 1], for every prime $5 \le n < 100$ this proves $C_{\operatorname{End}^0(A_n)}(D_n) = L_n$, $\operatorname{Hg}(A_n) = G_n$, divisor generation of all rational Hodge classes on $A_n^k$, and the rational Hodge conjecture on $A_n^k \times X_n^m$ ($k, m \ge 0$), as well as on every mixed surface product $\prod_{5 \le n < 100} X_n^{m_n}$ across distinct degrees.
• Theorem 4.1 & Theorem 4.2 (Uniform Prime-$n$ Moving Curve-Square Map, Shioda–Tate Rank $4f$, Chebyshev Diagonalization, and $\mathbb{Q}(\zeta_n)$-Weil Classes): For every prime $n > 3$ and $\lambda \in \mathbb{C} \setminus \{0, 1\}$, the degree-$24n$ rational map $\phi_{n,\lambda} : C_{n,\lambda}^2 \dashrightarrow X_{n,a}$ onto $X_{n,a} : Y^2 = X^3 + 6aU_n X + 2U_n(2U_n - a^3)$ ($U_n = D_n(s) - 2$, $a^3 = -2(\lambda-1)^2/\lambda$) pulls back $\eta_{n,j} = s^j \frac{ds \wedge dX}{Y}$ ($0 \le j \le \lfloor n/3 \rfloor - 1$) to $$\phi_{n,\lambda}^*(\eta_{n,j}) = \frac{\lambda - 1}{3na}\, s^j \left(cQ - \frac{1}{cQ}\right) \frac{T - Q/T}{Q^{(n+1)/2}} \frac{dv_1 \wedge dv_2}{b_1 b_2} = \frac{\lambda - 1}{3na}\, s^j \left(c W_{1-n} - c^{-1} W_{-1-n}\right).$$ The inverse $M_n^{-1}$ of the unipotent Catalan expansion matrix transforms $\eta_{n,j}$ into the Chebyshev $2$-form basis $\widetilde{\eta}_{n,m} = \frac{U_{m-1}(s/2)\,ds \wedge dX}{Y}$ ($1 \le m \le \lfloor n/3 \rfloor$), simultaneously diagonalizing $\phi_{n,\lambda}^*$ and the Hodge–Riemann pairing. Moreover, the explicit algebraic divisor space $M_{n,a} \subset \operatorname{NS}(X_{n,a})_{\mathbb{Q}}$ has $\dim_{\mathbb{Q}}(M_{n,a}^\perp) = 2(n-1) = 4f$ in both congruence classes $n \equiv 1, 2 \pmod 3$, establishing the rank-$4$ constituent isomorphism $F_\tau : T(X_{n,a})_{E_n,\sigma} \xrightarrow{\sim} V_j \otimes V_{\tau_3 j} \subset \bigwedge^2 H_\tau$ at very general $a$ (where $\tau_3 \equiv 1 \pmod n, \tau_3 \equiv -1 \pmod 6$ generates $\operatorname{Gal}(K_n/F_0)$, $F_0 = \mathbb{Q}(\zeta_n)$). Transporting the inverse surface pairing $q_T$ through the four-slot projectors $\pi_{W,n}^{(k)} = \frac{1}{24}\delta^* R_1^k P_{\mathrm{same},n} \pi_1^{\otimes 4} \mu^*$ ($0 \le k \le n-2$) and specializing via finite monodromy proves that all classes in the $(n-1)$-dimensional $F_0$-Weil space $W_{F_0}(A_{n,\lambda}) = \bigwedge^4_{F_0} H^1(A_{n,\lambda}, \mathbb{Q}) \subset H^4(A_{n,\lambda}, \mathbb{Q})$ are algebraic for every $\lambda \in \mathbb{C} \setminus \{0, 1\}$.
• Theorem 5.1 (Beukers–Heckman $\mathrm{SL}_2^{2f}$ Monodromy, Very-General Mixed HC, Middle-Third Stabilizer, and Fixed Determinant Surface $S_n$): For every prime $n > 3$: (1) Beukers–Heckman irreducibility, exclusion of Borel/finite/torus-normalizer closures in $\mathrm{PGL}_2(\mathbb{C})$ via local orders $(\ge 2n, 3, \ge 2n)$, and pairwise distinct local projective angles at $\lambda = 0$ prove that the connected monodromy on $H^1(A_{n,\lambda}, \mathbb{C})$ is $\mathrm{SL}_2^{2f}$; combined with Dirichlet $L(1, \bar{\psi}) \neq 0$, this determines the generic Hodge group $\operatorname{Hg}(A_{n,\lambda}) = \{g \in U_{K_n / K_n^+}(H) : \operatorname{Nm}_{K_n/F_0}(\det_{K_n} g) = 1\}^\circ$ of $\mathbb{Q}$-dimension $3(n-1) + \frac{n-1}{2} = \frac{7(n-1)}{2}$ inside the $4(n-1)$-dimensional $K_n / K_n^+$-unitary group $U_{K_n / K_n^+}(H)$ ($K_n^+ = \mathbb{Q}(\zeta_{6n} + \zeta_{6n}^{-1})$), proving the rational Hodge conjecture unconditionally on all mixed products $X_{n,a}^r \times A_{n,\lambda}^k$ at very general $\lambda \in \mathbb{C} \setminus \{0, 1\}$. (2) For every $\lambda \in \mathbb{C} \setminus \{0, 1\}$, the order-$12n$ quotient $C_{n,\lambda}^2 / (\langle (\sigma_{6n}, \sigma_{6n}^{-1}) \rangle \rtimes \langle \iota \rangle)$ is birational to the fixed surface $S_n : y^2 = x^3 + Q^{2w_0}(1-Q^n)^2$ ($w_0 = 2((-n)\bmod 3)$) dominated by $(E_3 \times B_n)/\mu_3$, and $1$-dimensional $K_n$-linearity establishes an algebraic $K_n$-equivariant isomorphism $D_{n,\lambda} = \bigwedge^2_{K_n} H^1(A_{n,\lambda}, \mathbb{Q}) \cong M_{B_n} \subset H^2(S_n, \mathbb{Q})$. (3) At every non-CM parameter $\lambda \in \mathbb{C} \setminus \{0, 1\}$, the Middle-Third Stabilizer $\operatorname{Stab}_{(\mathbb{Z}/n\mathbb{Z})^\times}((n/3, 2n/3)) = \{1, -1\}$ restricts $\operatorname{Hg}(A_{n,\lambda})^{\mathrm{der}}_{\mathbb{C}}$ to $\mathrm{SL}_2^{2f}$ or $\mathrm{SL}_2^f$ (paired by an algebraic $\tau_2$-semilinear quasi-endomorphism $J_\lambda$), yielding an explicit algebraic surjection from tensor powers of $D_{n,\lambda} \cong M_{B_n}$ and Tate twists onto all derived-group invariants $(H^{\otimes 2m})^{\operatorname{Hg}^{\mathrm{der}}}$.

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2. The Mod-$12$ Hasse–Brauer–Noether Splitting Dichotomy on the Fixed Branch

On the fixed branch $C_n = C_{n, -1}$, let $K_n = \mathbb{Q}(\zeta_{6n})$, let $\tau \in \mathrm{Gal}(K_n/\mathbb{Q})$ be the unique involution with $\tau(\zeta_n) = \zeta_n^{-1}$ and $\tau(\zeta_3) = \zeta_3$, and let $L_n = K_n^{\langle \tau \rangle} = \mathbb{Q}(\zeta_n + \zeta_n^{-1}, \zeta_3)$ with totally real subfield $L_n^+ = \mathbb{Q}(\zeta_n + \zeta_n^{-1})$ of degree $f = \frac{n-1}{2}$. Because the order-$4$ curve automorphism $J(v, q, u, b) = (v^{-1}, q^{-1}, u, ib/v)$ ([1, eq. (9)]) satisfies $J^2 = \sigma_{6n}^{3n} = -1$ on $H^1(A_n, \mathbb{Q})$ and $J \sigma_{6n} J^{-1} = \sigma_{6n}^\tau$, $A_n$ carries the cyclic quaternion algebra $$D_n = (K_n / L_n,\, \tau,\, -1) = K_n \oplus K_n J, \qquad J^2 = -1, \qquad J x = \tau(x) J \quad (x \in K_n).$$

Theorem 2.1 (Mod-$12$ Hasse–Brauer–Noether Splitting Dichotomy). For every prime $n > 3$, the quaternion algebra $D_n = (K_n / L_n, \tau, -1)$ is a non-split division quaternion algebra over $L_n$ if and only if $n \equiv 7 \pmod{12}$ ($n = 7, 19, 31, 43, 67, 79, \dots$), where it ramifies at the two primes $\mathfrak{p}_n, \bar{\mathfrak{p}}_n$ of $L_n$ above the unique ramified prime of $L_n^+$ dividing $n$ and splits at all other places of $L_n$; and $D_n \cong M_2(L_n)$ is globally split if and only if $n \equiv 1, 5, 11 \pmod{12}$.

Proof. By the Hasse–Brauer–Noether theorem, $D_n = (K_n/L_n, \tau, -1)$ is split at a place $w$ of $L_n$ if and only if $-1 \in \mathrm{Nm}(K_{n, w}^\times / L_{n, w}^\times)$:

• Archimedean places: Every place $w \mid \infty$ of $L_n = L_n^+(\sqrt{-3})$ is complex, so $K_{n, w} \cong \mathbb{C} \times \mathbb{C}$ and $\mathrm{inv}_w(D_n) = 0$.
• Unramified finite places ($w \nmid n$): Because $K_n = L_n(\zeta_n)$ is ramified over $L_n$ only at primes dividing $n$, for every finite place $w \nmid n$ of $L_n$ the extension $K_{n, w} / L_{n, w}$ is unramified, so the unit $-1 \in \mathcal{O}_{L_{n, w}}^\times$ is a local norm and $\mathrm{inv}_w(D_n) = 0$.
• $n$-adic places ($w \mid n$): In $L_n^+ = \mathbb{Q}(\zeta_n + \zeta_n^{-1})$, the rational prime $n$ is totally ramified: $(n) = \mathfrak{p}_{n, +}^{(n-1)/2}$ with residue field $\mathbb{F}_n$, and $\mathfrak{p}_{n, +}$ is totally ramified in $\mathbb{Q}(\zeta_n)$ with residue field $\mathbb{F}_n$. Since $\mathrm{Nm}_{K_{n, \mathfrak{P}}^\times / L_{n, w}^\times}$ on units reduces modulo the maximal ideal to the squaring map $x \mapsto x^2$ on the residue field $\mathbb{F}_{n^r}^\times$ of $L_n$ at $w \mid n$ (where $r = 1$ if $\left(\frac{-3}{n}\right) = +1$, i.e. $n \equiv 1 \pmod 3$, and $r = 2$ if $\left(\frac{-3}{n}\right) = -1$, i.e. $n \equiv 2 \pmod 3$), $-1$ is a local norm at $w \mid n$ if and only if $-1 \in (\mathbb{F}_{n^r}^\times)^2$: $$-1 \in (\mathbb{F}_{n^r}^\times)^2 \iff \frac{n^r - 1}{2} \text{ is even} \iff n \equiv 1 \pmod 4 \text{ or } r = 2 \iff n \not\equiv 7 \pmod{12}.$$ When $n \equiv 7 \pmod{12}$, $n \equiv 1 \pmod 3$ ($r = 1$, so $\mathfrak{p}_{n,+}$ splits into two primes $\mathfrak{p}_n, \bar{\mathfrak{p}}_n$ in $L_n$) and $n \equiv 3 \pmod 4$ ($-1 \notin (\mathbb{F}_n^\times)^2$), so $\mathrm{inv}_{\mathfrak{p}_n}(D_n) = \mathrm{inv}_{\bar{\mathfrak{p}}_n}(D_n) = \frac{1}{2}$ and $D_n$ is a division quaternion algebra over $L_n$ (test_01). $\square$

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3. Two-Reduction Twisted Frobenius Non-CM Certificates for $5 \le n < 100$

To discharge the open non-CM condition $C_{\operatorname{End}^0(A_n)}(D_n) = L_n$ of [1] for primes $n \ge 13$, we apply the twisted Frobenius descent and norm-one circle parametrization of [1, §§5 and 6.3–6.4] at prime fields $\mathbb{F}_p$ with $$p \equiv 1 \pmod{12} \quad \text{and} \quad p \equiv -1 \pmod n.$$ Because $p \equiv 1 \pmod 4$, $i = \sqrt{-1} \in \mathbb{F}_p^\times$, so the automorphism $J(v, q_n, u, b) = (v^{-1}, q_n^{-1}, u, ib/v)$ of $C_n$ ([1, eqs. (4), (9)]) is defined over $\mathbb{F}_p$ and commutes with prime-field Frobenius $F_p$. Because $p \equiv 1 \pmod 6$ and $p \equiv -1 \pmod n$, we have $p \equiv \tau \pmod{6n}$, so $F_p \sigma_{6n} = \sigma_{6n}^\tau F_p$ and $F_p$ fixes $L_n = \mathbb{Q}(\zeta_n + \zeta_n^{-1}, \zeta_3)$ elementwise (in particular, $p$ splits completely in $L_n$ into $n - 1$ degree-$1$ prime ideals $(\mathfrak{P}_a, \bar{\mathfrak{P}}_a)$, $1 \le a \le f = \frac{n-1}{2}$). Consequently, the twisted Frobenius endomorphism $$\Pi_p := J^{-1} F_p$$ commutes with all of $D_n = K_n \oplus K_n J$, has polarization multiplier $p$, and satisfies $\Pi_p^2 = -F_p^2 = -F_{p^2}$ ([1, §6.3]).

3.1. Exact Norm-One Circle Character Sum, Polarization Determinant, and Newton Polygon Slopes

Following [1, §6.4, eq. (17)], choose a quadratic non-residue $d \in \mathbb{F}_p^\times \setminus (\mathbb{F}_p^\times)^2$, let $c^2 = d$ in $\mathbb{F}_{p^2}$, set $i = d^{(p-1)/4} \in \mathbb{F}_p^\times$, and parametrize the norm-one circle $v^p = v^{-1}$ in $\mathbb{P}^1(\mathbb{F}_{p^2})$ by $$v(t) = \frac{t + c}{t - c}, \qquad t \in \mathbb{P}^1(\mathbb{F}_p).$$ For $t \in \mathbb{F}_p^\times$ ($v(t) \neq \pm 1$), on the Kummer coordinates $q_n^n = v$, $u^3 = \frac{v}{(1+v)^2}$, $b^2 = 1 - v^2$ of $C_n$:

1. $(q_n)^p / q_n^{-1} = q_n^{p+1} = v(t)^{(p+1)/n} = \zeta_n^{k(t)} \in \mu_n(\mathbb{F}_{p^2})$;
2. $u^3 = \frac{v}{(1+v)^2} = \frac{t^2 - d}{4t^2} \in \mathbb{F}_p^\times$, so $u^p / u = \left(\frac{t^2 - d}{4t^2}\right)^{(p-1)/3} = \omega^{r(t)} \in \mu_3(\mathbb{F}_p)$;
3. Writing $b' = b(t-c)$, we have $(b')^2 = -4tc$ and $(b')^p / b' = (-4tc)^{(p-1)/2} = i \left(\frac{t}{p}\right)$, while $(t-c)^p / (t-c) = v(t)$, so $b^p = \left(\frac{t}{p}\right) \frac{ib}{v(t)} = \left(\frac{t}{p}\right) J(b)$.

At $t = 0$ ($v = -1$) and $t = \infty$ ($v = 1$), the quadratic cover $b^2 = 1 - v^2$ is ramified ($b = 0$), so the primitive quadratic projector $\frac{1 - J^2}{2}$ vanishes identically at both points. Summing over $t \in \mathbb{F}_p^\times$, define the character counts $$c_{k, r} := \sum_{\substack{t \in \mathbb{F}_p^\times \\ k(t) = k,\, r(t) = r}} \left(\frac{t}{p}\right) \in \mathbb{Z} \qquad (k \in \mathbb{Z}/n\mathbb{Z},\; r \in \{0, 1, 2\}),$$ and the character sum $S_p := \sum_{k=0}^{n-1} (c_{k,0} + c_{k,1}\omega + c_{k,2}\omega^2)\zeta_n^k$.

Theorem 3.1 (Twisted Frobenius Symmetries, Exact Cyclic Convolution Norm, and Elementary Newton Polygon Slopes). Let $n > 3$ be prime and let $p \equiv 1 \pmod{12}$ with $p \equiv -1 \pmod n$.

1. (Inversion and Free Cayley Involution Symmetries): On $\mathbb{F}_p^\times$, inversion $t \mapsto -t$ sends $v(t) \mapsto v(t)^{-1}$ ($k \mapsto -k \bmod n$) while preserving $u^3 = \frac{t^2-d}{4t^2}$ and $\left(\frac{-t}{p}\right) = \left(\frac{t}{p}\right)$ (since $p \equiv 1 \pmod 4$), forcing $c_{-k, r} = c_{k, r}$. The Cayley involution $t \mapsto d / t$ acts freely on $\mathbb{F}_p^\times$ (since $d \notin (\mathbb{F}_p^\times)^2$), sends $v(t) \mapsto -v(t)$, so that $v(d/t)^{(p+1)/n} = (-v(t))^{(p+1)/n} = (-1)^{(p+1)/n} v(t)^{(p+1)/n} = v(t)^{(p+1)/n} = \zeta_n^{k(t)}$ (since $p \equiv 1 \pmod 4 \implies p + 1$ is even while $n$ is odd, so $\frac{p+1}{n}$ is an even integer), and reverses $\left(\frac{d/t}{p}\right) = -\left(\frac{t}{p}\right)$ on each order-$n$ fiber $\{t \in \mathbb{F}_p^\times : k(t) = k\}$, forcing the fiber sum over all three cubic labels $r \in \{0, 1, 2\}$ to vanish: $$c_{k, 0} + c_{k, 1} + c_{k, 2} = 0 \qquad (\forall k \in \mathbb{Z}/n\mathbb{Z}).$$ Consequently, setting $A_k := c_{k,0} - c_{k,2} = A_{-k}$ and $B_k := c_{k,1} - c_{k,2} = B_{-k}$, because $H^1(A_n, \mathbb{Q})$ is a free rank-$1$ left $D_n$-module ($\dim_{L_n} H^1(A_n, \mathbb{Q}) = 4 = \dim_{L_n} D_n$), its rational $D_n$-commutant is the central simple quaternion algebra $C_{\operatorname{End}_{\mathbb{Q}}(H^1(A_n,\mathbb{Q}))}(D_n) \cong D_n^{\mathrm{op}}$ of degree $2$ over $L_n$, and the reduced trace of $\Pi_p = J^{-1} F_p$ in $C_{\operatorname{End}_{\mathbb{Q}_\ell}(V_\ell(A_n))}(D_n) \cong D_n^{\mathrm{op}} \otimes_{\mathbb{Q}} \mathbb{Q}_\ell$ ($\ell \nmid pn$) is $$\alpha_p = -S_p = -\sum_{k=0}^{n-1} (A_k + B_k \omega)\zeta_n^k \in (1 - \omega)\mathcal{O}_{L_n},$$ and its norm $N_p := \alpha_p \bar{\alpha}_p = S_p \bar{S}_p \in 3\mathcal{O}_{L_n^+}$ has the exact cyclic convolution expansion in the integral basis $C_m := \zeta_n^m + \zeta_n^{-m}$ ($1 \le m \le f = \frac{n-1}{2}$): $$N_p = \sum_{m=1}^{f} (\gamma_m - \gamma_0) C_m, \qquad \gamma_m := \sum_{k=0}^{n-1} \left(A_k A_{k-m} + B_k B_{k-m} - A_k B_{k-m}\right).$$
2. (Polarization Determinant, Discriminant Square-Completion, and Elementary Newton Polygon Slopes): By [1, eq. (14)], the reduced characteristic polynomial of $\Pi_p$ over $L_n$ in $C_{\operatorname{End}_{\mathbb{Q}_\ell}(V_\ell(A_n))}(D_n) \cong D_n^{\mathrm{op}} \otimes_{\mathbb{Q}} \mathbb{Q}_\ell$ (and hence in $\mathcal{C}_p := C_{\operatorname{End}^0(A_{n,\overline{\mathbb{F}}_p})}(D_n) \hookrightarrow D_n^{\mathrm{op}} \otimes_{\mathbb{Q}} \mathbb{Q}_\ell$) is $$P_{\Pi_p}(z) = z^2 - \alpha_p z + b_p, \qquad b_p = \frac{p \alpha_p}{\bar{\alpha}_p} = \frac{p \alpha_p^2}{N_p} \in \mathcal{O}_{L_n},$$ while actual $p^2$-Frobenius $F_{p^2} = -\Pi_p^2$ has reduced polynomial $P_{F_{p^2}}(z) = z^2 + (\alpha_p^2 - 2b_p)z + b_p^2$. Completing the square gives $$\Delta(\Pi_p) = \alpha_p^2 - 4b_p = \left(\frac{\alpha_p}{N_p}\right)^2 \delta_p, \qquad \Delta(F_{p^2}) = \left(\frac{\alpha_p^2}{N_p}\right)^2 \delta_p, \qquad \text{where } \delta_p := N_p(N_p - 4p) \in \mathcal{O}_{L_n^+}.$$ If $\mathfrak{p}_1 \mid p$ is any degree-$1$ prime ideal of $L_n^+$ at which $N_p \not\equiv 0 \pmod{\mathfrak{p}_1}$, then $\alpha_p, \bar{\alpha}_p \in \mathcal{O}_{L_n}$ and $N_p = \alpha_p \bar{\alpha}_p \notin \mathfrak{p}_1$ immediately force $v_{\mathfrak{P}_1}(\alpha_p) = v_{\mathfrak{P}_1}(\bar{\alpha}_p) = 0$ and $v_{\mathfrak{P}_1}(b_p) = v_{\mathfrak{P}_1}(p) = 1$. The Newton polygon of $P_{\Pi_p}(z)$ at $\mathfrak{P}_1 \mid p$ therefore has slopes $0$ and $1$, so the two eigenvalues $\lambda_1, \lambda_2$ of $\Pi_p$ at $a = 1$ have $\mathfrak{P}_1$-adic valuations $0$ and $1$, $\lambda_1/\lambda_2$ is not a root of unity, and $\Pi_p^m \notin L_n$ for every $m \ge 1$. When $\delta_p \notin (L_n^\times)^2$, $M_p := L_n[\Pi_p] \subset \mathcal{C}_p$ is the quadratic field extension $L_n(\sqrt{\delta_p})$, and for every $m \ge 1$ the centralizer of $\Pi_p^m$ in $D_n^{\mathrm{op}} \otimes_{\mathbb{Q}} \mathbb{Q}_\ell$ (and in $\mathcal{C}_p$) is $M_p \otimes_{\mathbb{Q}} \mathbb{Q}_\ell$ (respectively $M_p$).

Proof.

1. By the Lefschetz trace formula on the $2$-dimensional eigenspace $H^1_{\text{ét}}(C_{n,\overline{\mathbb{F}}_p}, \mathbb{Q}_\ell)_{\chi_j}$ (where $J^2$ acts as $-1$, annihilating $H^0$ and $H^2$, and the $6n$ sheet summation cancels the $\frac{1}{6n}$ character projector normalization), the eigenvalue sum of $\Pi_p = J^{-1} F_p$ is $\alpha_p = -S_p$. Using $1 + \omega + \omega^2 = 0$, we have $c_{k,0} + c_{k,1}\omega + c_{k,2}\omega^2 = (c_{k,0}-c_{k,2}) + (c_{k,1}-c_{k,2})\omega = A_k + B_k\omega$, and $c_{k,0}+c_{k,1}+c_{k,2}=0 \implies A_k + B_k \equiv 0 \pmod 3 \implies S_p \equiv 0 \pmod{1-\omega} \implies 3 \mid N_p$. Multiplying $S_p = \sum_k (A_k + B_k\omega)\zeta_n^k$ by $\bar{S}_p = \sum_\ell (A_\ell + B_\ell\omega^2)\zeta_n^{-\ell}$ and using $\omega + \omega^2 = -1$ and $A_{-k}=A_k, B_{-k}=B_k$ gives $S_p \bar{S}_p = \sum_{m=0}^{n-1} \gamma_m \zeta_n^m = \gamma_0 + \sum_{m=1}^{f} \gamma_m C_m = \sum_{m=1}^{f} (\gamma_m - \gamma_0) C_m$ since $1 = -\sum_{m=1}^{f} C_m$. In test_02 of the supplement, at $n = 5, p = 109, d = 2$, this formula gives $\gamma = (48, 15, -21, -21, 15)$ and $N_{109} = -33 C_1 - 69 C_2 = 51 \pm 18\sqrt{5} > 0$, $b_{109} = \frac{109 \alpha_{109}}{\bar{\alpha}_{109}} = \frac{133}{2} \pm 15\sqrt{5} - \frac{35\sqrt{-3}}{2} \pm 19\sqrt{-15} \in \mathcal{O}_{L_5}$, and predicts the untwisted $109^2$-Frobenius trace $-(\alpha_{109}^2 - 2b_{109}) = \frac{179}{2} \pm 12\sqrt{5} - \frac{85\sqrt{-3}}{2} \pm 32\sqrt{-15}$, which matches an independent direct enumeration of all $109^2 = 11{,}881$ points of $\mathbb{F}_{109^2}$ on $C_5$ to exact equality (test_02).
2. Since $N_p = \alpha_p \bar{\alpha}_p$, we have $\alpha_p^2 - 4b_p = \alpha_p^2 - \frac{4p\alpha_p}{\bar{\alpha}_p} = \frac{\alpha_p^2}{\alpha_p \bar{\alpha}_p}(\alpha_p \bar{\alpha}_p - 4p) = \left(\frac{\alpha_p}{N_p}\right)^2 N_p(N_p - 4p) = \left(\frac{\alpha_p}{N_p}\right)^2 \delta_p$. Because $\alpha_p, \bar{\alpha}_p \in \mathcal{O}_{L_n}$, both $v_{\mathfrak{P}_1}(\alpha_p) \ge 0$ and $v_{\mathfrak{P}_1}(\bar{\alpha}_p) \ge 0$; thus $N_p \not\equiv 0 \pmod{\mathfrak{p}_1} \implies v_{\mathfrak{P}_1}(\alpha_p) + v_{\mathfrak{P}_1}(\bar{\alpha}_p) = v_{\mathfrak{p}_1}(N_p) = 0 \implies v_{\mathfrak{P}_1}(\alpha_p) = v_{\mathfrak{P}_1}(\bar{\alpha}_p) = 0$. Since $p$ splits completely in $L_n$, $v_{\mathfrak{P}_1}(p) = 1$, so $v_{\mathfrak{P}_1}(b_p) = 1 + 0 - 0 = 1$. The Newton polygon of $z^2 - \alpha_p z + b_p$ at $\mathfrak{P}_1$ therefore has slopes $0$ and $1$, so $v_{\mathfrak{P}_1}(\lambda_1) = 0 \neq 1 = v_{\mathfrak{P}_1}(\lambda_2)$. For every $m \ge 1$, $v_{\mathfrak{P}_1}(\lambda_1^m) = 0 \neq m = v_{\mathfrak{P}_1}(\lambda_2^m)$, so $\lambda_1^m \neq \lambda_2^m$ and $\Pi_p^m \notin L_n$. Since the $\ell$-adic Tate representation embeds $\mathcal{C}_p = C_{\operatorname{End}^0(A_{n,\overline{\mathbb{F}}_p})}(D_n) \hookrightarrow C_{\operatorname{End}_{\mathbb{Q}_\ell}(V_\ell(A_n))}(D_n) \cong D_n^{\mathrm{op}} \otimes_{\mathbb{Q}} \mathbb{Q}_\ell = \prod_{v \mid \ell} Q_v$ (where each $Q_v = D_n^{\mathrm{op}} \otimes_{L_n} L_{n,v}$ is a $4$-dimensional central simple quaternion algebra over $L_{n,v}$), $\Pi_p \in \mathcal{C}_p$ satisfies $P_{\Pi_p}(\Pi_p) = 0$ inside $\mathcal{C}_p$. When $\delta_p \notin (L_n^\times)^2$, $P_{\Pi_p}(z)$ is irreducible over $L_n$, so $M_p := L_n[\Pi_p] \subset \mathcal{C}_p$ is a field isomorphic to $L_n(\sqrt{\delta_p})$, and $\Pi_p^m \in M_p \setminus L_n$ generates $M_p = L_n[\Pi_p^m]$ over $L_n$ for every $m \ge 1$. In each $4$-dimensional quaternion factor $Q_v$ over $L_{n,v}$, the centralizer of the non-central element $\Pi_p^m \in Q_v \setminus L_{n,v}$ is the $2$-dimensional commutative subalgebra $L_{n,v}[\Pi_p^m] = M_p \otimes_{L_n} L_{n,v}$, so $C_{D_n^{\mathrm{op}} \otimes \mathbb{Q}_\ell}(\Pi_p^m) = M_p \otimes_{\mathbb{Q}} \mathbb{Q}_\ell$ and $C_{\mathcal{C}_p}(\Pi_p^m) = \mathcal{C}_p \cap (M_p \otimes_{\mathbb{Q}} \mathbb{Q}_\ell) = M_p$. (In test_02, we also Hensel-lift $\zeta_n$ and $\omega$ modulo $p^4$ to verify $v_{\mathfrak{P}_a}(b_p) = 1 + v_{\mathfrak{P}_a}(\alpha_p) - v_{\mathfrak{P}_a}(\bar{\alpha}_p) = h^{1,0}(M_a) \in \{0, 1, 2\}$ across all places $1 \le a \le f$ and all 23 primes $5 \le n < 100$.) $\square$

3.2. Two-Reduction Centralizer Comparison Across All 23 Primes $5 \le n < 100$

Theorem 3.2 (Unconditional Non-CM Discharge for All 20 Open Primes $13 \le n < 100$ and $n = 5, 7, 11$). For every prime $5 \le n < 100$ (discharging the open primes $13 \le n < 100$ of [1] alongside $n = 5, 7, 11$):

1. The characteristic-zero centralizer satisfies $C_{\operatorname{End}^0(A_{n,\overline{\mathbb{Q}}})}(D_n) = L_n$, $A_n = e_{6n}J(C_n)$ is not of CM type, $\operatorname{Hg}(A_n) = G_n$, all rational Hodge classes on every self-power $A_n^k$ ($k \ge 1$) are generated by divisor classes, and the rational Hodge conjecture holds unconditionally on $A_n^k \times X_n^m$ ($k, m \ge 0$).
2. Across distinct primes $5 \le n < 100$, the rational Hodge conjecture holds unconditionally on every mixed product of the Dickson elliptic surfaces $$\prod_{5 \le n < 100} X_n^{m_n} \qquad (m_n \ge 0).$$

Proof. Because $\dim_{L_n} H^1(A_n, \mathbb{Q}) = 4 = \dim_{L_n} D_n$, the rational Betti commutant $C_{\operatorname{End}_{\mathbb{Q}}(H^1(A_n,\mathbb{Q}))}(D_n) \cong D_n^{\mathrm{op}}$ is a central simple quaternion algebra of degree $2$ over $L_n$ (which is a division algebra when $n \equiv 7 \pmod{12}$ and $M_2(L_n)$ otherwise by Theorem 2.1), and the characteristic-zero geometric endomorphism centralizer $\mathcal{C} := C_{\operatorname{End}^0(A_{n,\overline{\mathbb{Q}}})}(D_n) \hookrightarrow D_n^{\mathrm{op}}$ is an $L_n$-subalgebra. At any good reduction prime $p \equiv 1 \pmod{12}$ with $p \equiv -1 \pmod n$ and $N_p \not\equiv 0 \pmod{\mathfrak{p}_1}$ such that $\delta_p = N_p(N_p - 4p) \notin (L_n^\times)^2$, let $\mathrm{sp}_p : \mathcal{C} \hookrightarrow \mathcal{C}_p := C_{\operatorname{End}^0(A_{n,\overline{\mathbb{F}}_p})}(D_n)$ be the injective specialization homomorphism of $L_n$-algebras ([1, §5, Lemma 6]). Every $\phi \in \mathrm{sp}_p(\mathcal{C})$ is defined over a finite extension $\mathbb{F}_{p^m}$ ($m \ge 1$) and therefore commutes with $\Pi_p^m = (J^{-1} F_p)^m \in \mathcal{C}_p$. By Theorem 3.1(2), $C_{\mathcal{C}_p}(\Pi_p^m) = M_p := L_n[\Pi_p] \cong L_n(\sqrt{\delta_p})$, so $\mathrm{sp}_p$ embeds $\mathcal{C}$ as an $L_n$-subalgebra into the quadratic field extension $M_p \cong L_n(\sqrt{\delta_p})$. For each prime $5 \le n < 100$, we evaluate the two smallest primes $p_1 < p_2$ satisfying $p_i \equiv 1 \pmod{12}$, $p_i \equiv -1 \pmod n$, and $N_{p_i} \not\equiv 0 \pmod{\mathfrak{p}_1}$, and select a prime $\ell \equiv 1 \pmod{6n}$. Because $\ell \equiv 1 \pmod{6n}$, $\ell$ splits completely in $K_n = \mathbb{Q}(\zeta_{6n})$ and hence splits completely in $L_n = \mathbb{Q}(\zeta_n + \zeta_n^{-1}, \zeta_3)$, so every prime ideal $\mathfrak{L} \mid \ell$ of $L_n$ has residue field $\mathcal{O}_{L_n}/\mathfrak{L} \cong \mathbb{F}_\ell$. In the table below (certified in test_03 of the supplement in $0.5\text{ s}$), the witness embeddings $(a_1, a_2, a_{12}) \in \{1, \dots, f\}^3$ satisfy $$\left(\frac{\sigma_{a_1}(\delta_{p_1})}{\ell}\right) = -1, \qquad \left(\frac{\sigma_{a_2}(\delta_{p_2})}{\ell}\right) = -1, \qquad \left(\frac{\sigma_{a_{12}}(\delta_{p_1}\delta_{p_2})}{\ell}\right) = -1,$$ proving that $\delta_{p_1}, \delta_{p_2}, \delta_{p_1}\delta_{p_2} \notin (L_n^\times)^2$. Thus $M_{p_1} = L_n(\sqrt{\delta_{p_1}})$ and $M_{p_2} = L_n(\sqrt{\delta_{p_2}})$ are two non-isomorphic quadratic field extensions of $L_n$. Since $\mathcal{C}$ embeds as an $L_n$-subalgebra into both $M_{p_1}$ and $M_{p_2}$, we must have $\mathcal{C} = L_n$, completing the proof of (1) and (2) by [1, Theorem 1(1)–(4) and §7]:

| $n$ | $(p_1, d_1)$ | $[N_{p_1, 1}, N_{p_1, 2}]$ | $(p_2, d_2)$ | $[N_{p_2, 1}, N_{p_2, 2}]$ | $\ell \equiv 1 \pmod{6n}$ | Witnesses $(a_1, a_2, a_{12})$ | | :--- | :--- | :--- | :--- | :--- | :--- | :--- | | $5$ | $(109, 2)$ | $[-33, -69]$ | $(229, 2)$ | $[-555, -255]$ | $31$ | $(2, 1, 1)$ | | $7$ | $(13, 2)$ | $[-9, -9]$ | $(97, 5)$ | $[-231, -285]$ | $43$ | $(1, 2, 1)$ | | $11$ | $(109, 2)$ | $[-147, -99]$ | $(241, 7)$ | $[-216, -246]$ | $67$ | $(1, 1, 5)$ | | $13$ | $(181, 2)$ | $[-345, -306]$ | $(337, 5)$ | $[-342, -402]$ | $79$ | $(2, 1, 2)$ | | $17$ | $(373, 2)$ | $[-579, -552]$ | $(577, 5)$ | $[-1083, -885]$ | $103$ | $(1, 2, 1)$ | | $19$ | $(37, 2)$ | $[-27, -27]$ | $(1861, 2)$ | $[-1557, -864]$ | $229$ | $(1, 1, 7)$ | | $23$ | $(229, 2)$ | $[-132, -132]$ | $(1609, 7)$ | $[-1857, -2934]$ | $139$ | $(2, 1, 1)$ | | $29$ | $(2029, 2)$ | $[-2871, -1974]$ | $(2377, 5)$ | $[-1842, -1371]$ | $349$ | $(4, 1, 1)$ | | $31$ | $(61, 2)$ | $[-51, -60]$ | $(433, 5)$ | $[-396, -438]$ | $373$ | $(1, 1, 2)$ | | $37$ | $(73, 5)$ | $[-57, -81]$ | $(2293, 2)$ | $[-3033, -2499]$ | $223$ | $(1, 8, 1)$ | | $41$ | $(409, 7)$ | $[-510, -456]$ | $(2377, 5)$ | $[-1128, -1299]$ | $739$ | $(2, 5, 2)$ | | $43$ | $(601, 7)$ | $[-738, -789]$ | $(1117, 2)$ | $[-1593, -1401]$ | $1033$ | $(1, 10, 1)$ | | $47$ | $(1033, 5)$ | $[-1107, -873]$ | $(1597, 2)$ | $[-2664, -2178]$ | $283$ | $(2, 1, 1)$ | | $53$ | $(1801, 11)$ | $[-1965, -2490]$ | $(2437, 2)$ | $[-2385, -1704]$ | $3181$ | $(4, 6, 4)$ | | $59$ | $(1297, 5)$ | $[-1482, -1746]$ | $(2713, 5)$ | $[-2136, -1842]$ | $709$ | $(6, 1, 1)$ | | $61$ | $(853, 2)$ | $[-951, -939]$ | $(3049, 11)$ | $[-3588, -3090]$ | $367$ | $(2, 2, 9)$ | | $67$ | $(937, 5)$ | $[-837, -729]$ | $(1741, 2)$ | $[-1623, -1482]$ | $1609$ | $(1, 1, 2)$ | | $71$ | $(709, 2)$ | $[-699, -600]$ | $(4969, 7)$ | $[-6810, -6375]$ | $853$ | $(2, 3, 2)$ | | $73$ | $(1021, 2)$ | $[-675, -858]$ | $(6277, 2)$ | $[-5913, -5955]$ | $439$ | $(1, 2, 1)$ | | $79$ | $(157, 2)$ | $[-192, -147]$ | $(2053, 2)$ | $[-2466, -2259]$ | $1423$ | $(2, 1, 1)$ | | $83$ | $(829, 2)$ | $[-1047, -1038]$ | $(4813, 2)$ | $[-6915, -7977]$ | $499$ | $(1, 4, 1)$ | | $89$ | $(4093, 2)$ | $[-3525, -4533]$ | $(6229, 2)$ | $[-8082, -7656]$ | $1069$ | $(8, 1, 1)$ | | $97$ | $(193, 5)$ | $[-186, -186]$ | $(2521, 11)$ | $[-3081, -3483]$ | $1747$ | $(3, 1, 1)$ |

$\square$

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4. The Uniform Moving Curve-Square Map, Shioda–Tate Rank $4f$, and $\mathbb{Q}(\zeta_n)$-Weil Classes

We now turn to the $1$-parameter moving family $$C_{n, \lambda} : w^{6n} = v^{4n+6}(1 - v)^{3n}(1 - \lambda v)^n \qquad (\lambda \in \mathbb{C} \setminus \{0, 1\})$$ and its $4f$-dimensional primitive abelian variety $A_{n, \lambda} = e_{6n}J(C_{n, \lambda})$ ($f = \frac{n-1}{2}$), generalizing the $n = 5$ moving construction of [2] (hodge_conjecture_28 v1) to all primes $n > 3$.

On $C_{n, \lambda}$, use the Kummer coordinates $q^n = v$, $u^3 = \frac{v}{(1 - \lambda v)^2}$, $b^2 = (1 - v)(1 - \lambda v)$, and $w = q u^2 b (1 - \lambda v)$ ([2, eq. (11)]). On the curve square $C_{n, \lambda}^2$, define $$Q = q_1 q_2, \qquad k = Q^{(n-3)/2}, \qquad z = Q^n = Q^3 k^2 = v_1 v_2, \qquad T = \frac{u_1^2 u_2 (1 - \lambda v_1)(1 - \lambda v_2)}{Q^{(n-1)/2}}, \qquad \delta = \frac{b_1 b_2}{(1 - \lambda v_1)(1 - \lambda v_2)},$$ and choose $c, a \in \mathbb{C}^\times$ satisfying $c^n = \lambda$ and $a^3 = -\frac{2(\lambda - 1)^2}{\lambda}$.

For any integer $e \in \mathbb{Z}$, define the rational $1$-forms on $C_{n,\lambda}$ and alternating $2$-forms on $C_{n,\lambda}^2$: $$F_e := \frac{u^2 q^e (1 - \lambda v)}{b}\,dv, \qquad G_e := \frac{u q^e (1 - \lambda v)}{b}\,dv, \qquad W_e := F_e \boxtimes G_e - G_e \boxtimes F_e.$$ Let $F_0 := \mathbb{Q}(\zeta_n) \subset K_n = \mathbb{Q}(\zeta_{6n})$ and $E_n := \mathbb{Q}(\zeta_n + \zeta_n^{-1}) \subset F_0$, and let $\tau_3 \in (\mathbb{Z}/6n\mathbb{Z})^\times$ be the unique involution generating $\operatorname{Gal}(K_n/F_0)$: $$\tau_3 \equiv 1 \pmod n, \qquad \tau_3 \equiv -1 \pmod 6.$$

4.1. Exact Differential Pullback, Chinese Remainder Characters, and Chebyshev Diagonalization

Theorem 4.1 (Uniform Moving Curve-Square Map, Chinese Remainder Character Formula, and Chebyshev Diagonalization). For every prime $n > 3$ and every $\lambda \in \mathbb{C} \setminus \{0, 1\}$:

1. (Rational Map $\phi_{n, \lambda} : C_{n, \lambda}^2 \dashrightarrow X_{n, a}$ and Exact Differential Identity): Setting $s = cQ + (cQ)^{-1}$ and $U_n = D_n(s) - 2 = \frac{(\lambda z - 1)^2}{\lambda z}$, the rational functions $$X = -\frac{2(\lambda - 1)(1 - \lambda Q^n)}{a \lambda Q^{(n+1)/2}}\left(T + \frac{Q}{T}\right), \qquad Y = \frac{2(1 - \lambda Q^n)(1 - \lambda^2 Q^n)}{\lambda Q^n}\,\delta$$ define a dominant rational map $\phi_{n, \lambda} : C_{n, \lambda}^2 \dashrightarrow X_{n, a}$ of degree $24n$ onto the moving Dickson elliptic surface $$X_{n, a} : Y^2 = X^3 + 6aU_n X + 2U_n(2U_n - a^3)$$ of geometric genus $p_g(X_{n,a}) = \lceil n/3 \rceil - 1 = \lfloor n/3 \rfloor$. The logarithmic Jacobian $$\frac{\partial(\log Q, \log T)}{\partial(v_1, v_2)} = \frac{\lambda^2 v_1 v_2 - 1}{3n\,v_1 v_2(1 - \lambda v_1)(1 - \lambda v_2)}$$ cancels $(1 - \lambda^2 Q^n)(1 - \lambda v_1)(1 - \lambda v_2)$ in $dX / Y$, pulling back the holomorphic $2$-forms $\eta_{n, j} = \frac{s^j\,ds \wedge dX}{Y}$ ($0 \le j \le \lfloor n/3 \rfloor - 1$) to $$\phi_{n, \lambda}^*(\eta_{n, j}) = \frac{\lambda - 1}{3na}\,s^j\left(cQ - \frac{1}{cQ}\right)\frac{T - Q/T}{Q^{(n+1)/2}}\,\frac{dv_1 \wedge dv_2}{b_1 b_2} = \frac{\lambda - 1}{3na}\,s^j\left(c W_{1-n} - c^{-1} W_{-1-n}\right).$$
2. (Chinese Remainder Character Formula and $\tau_3$-Pairing): On $C_{n, \lambda}$, the differential $\frac{u^p q^e (1-\lambda v)\,dv}{b}$ ($p \in \{1, 2\}$) transforms under $\sigma_{6n} : w \mapsto \zeta_{6n} w$ with denominator character $w^{-j(e, p)}$ given by the unique Chinese Remainder solution $$\begin{cases} j(e, p) \equiv -e \pmod n, \\ j(e, p) \equiv 4p + 3 \pmod 6 \end{cases} \iff j(e, p) \equiv -6(6^{-1} \bmod n)\,e + n(n \bmod 6)(4p + 3) \pmod{6n}.$$ In particular, $j(e, 2) \equiv \tau_3 \cdot j(e, 1) \pmod{6n}$, $(j(-1-n, 1), j(-1-n, 2)) = (1, \tau_3)$, and $(j(1-n, 1), j(1-n, 2)) = (6n-\tau_3, 6n-1)$. For every $j \in (\mathbb{Z}/6n\mathbb{Z})^\times$, $h^{1,0}(V_j) + h^{1,0}(V_{\tau_3 j}) = 2$, and for $1 \le m \le f$, setting $j_1(m) := j(-n-m, 1)$ and $j_2(m) := \tau_3 j_1(m)$, we have $h^{1,0}(V_{j_1(m)}) = h^{1,0}(V_{j_2(m)}) = 1 \iff 1 \le m \le \lfloor n/3 \rfloor = p_g(X_{n,a})$.
3. (Unipotent Catalan Matrix $M_n$ and Simultaneous Chebyshev Diagonalization): Expanding $s^j = (cQ + (cQ)^{-1})^j$ gives $$\phi_{n, \lambda}^*(\eta_{n, j}) = \frac{\lambda - 1}{3na} \sum_{m=1}^{\lfloor n/3 \rfloor} M_n(j, m)\left(c^m W_{m-n} - c^{-m} W_{-m-n}\right),$$ where $M_n(j, m) = \binom{j}{(j+1-m)/2} - \binom{j}{(j-1-m)/2}$ ($1 \le m \le j+1,\, m \equiv j+1 \pmod 2$) is unipotent lower-triangular ($\det M_n = 1$). Its inverse $M_n^{-1}$ transforms $\eta_{n, j}$ into the Chebyshev $2$-form basis $$\widetilde{\eta}_{n, m} := \frac{U_{m-1}(s/2)\,ds \wedge dX}{Y} \in H^{2,0}(X_{n,a}) \qquad \left(1 \le m \le \lfloor n/3 \rfloor\right),$$ which simultaneously diagonalizes the Hodge–Riemann pairing $q_{X_{n,a}}(\widetilde{\eta}_{n, m}, \overline{\widetilde{\eta}_{n, m'}}) = C_{n, m}(\lambda)\,\delta_{m, m'}$ ($C_{n, m}(\lambda) > 0$) and the pullback $\phi_{n, \lambda}^*(\widetilde{\eta}_{n, m}) = \frac{\lambda - 1}{3na}(c^m W_{m-n} - c^{-m} W_{-m-n})$ (test_04).

Proof.

1. From $u_i^3 = \frac{v_i}{(1 - \lambda v_i)^2}$ and $v_1 v_2 = Q^n = Q^3 k^2$, we have $(u_1^2 u_2 (1-\lambda v_1)(1-\lambda v_2))^3 = \frac{Q^n(v_1 - \lambda Q^n)}{1 - \lambda v_1}$. Dividing by $(Q^{(n-1)/2})^3 = Q^n k$ gives $T^3 = \frac{v_1 - \lambda z}{k(1 - \lambda v_1)}$, hence $v_1 = \frac{\lambda z + k T^3}{1 + \lambda k T^3}$ and $v_2 = z / v_1$. Likewise, $u_1^3 u_2^3(1-\lambda v_1)^2(1-\lambda v_2)^2 = v_1 v_2 = Q^n$ implies $\frac{Q}{T} = \frac{u_1 u_2^2(1-\lambda v_1)(1-\lambda v_2)}{Q^{(n-1)/2}}$. Substituting $v_1, v_2$ into $\delta^2 = \frac{(1-v_1)(1-v_2)}{(1-\lambda v_1)(1-\lambda v_2)}$ gives $\delta^2 = A_n(T^3 + Q^3/T^3) + B_n$ with $A_n = -\frac{k(\lambda - 1)(\lambda z - 1)}{(\lambda^2 z - 1)^2}$ and $B_n = \frac{\lambda^2 z^2 + (\lambda^2 - 4\lambda + 1)z + 1}{(\lambda^2 z - 1)^2}$, which yields $Y^2 = X^3 + 6aU_n X + 2U_n(2U_n - a^3)$ identically (test_04). Differentiating $\log Q = \frac{1}{n}(\log v_1 + \log v_2)$ and $\log T = \frac{2}{3}\log v_1 + \frac{1}{3}\log v_2 - \frac{1}{3}\log(1-\lambda v_1) + \frac{1}{3}\log(1-\lambda v_2) - \frac{n-1}{2}\log Q$ gives $\frac{\partial(\log Q, \log T)}{\partial(v_1, v_2)} = \frac{\lambda^2 v_1 v_2 - 1}{3n v_1 v_2(1-\lambda v_1)(1-\lambda v_2)}$. Since $ds = (cQ - (cQ)^{-1})\,d\log Q$ and $\left.\frac{\partial X}{\partial \log T}\right|_Q = -\frac{2(\lambda-1)(1-\lambda Q^n)}{a\lambda Q^{(n+1)/2}}(T - Q/T)$, dividing $ds \wedge dX$ by $Y$ cancels $(1 - \lambda Q^n)$, $(1 - \lambda^2 Q^n)$, and $Q^n = v_1 v_2$, leaving $Q^{(n+1)/2}$ in the denominator. Finally, substituting $T - Q/T = Q^{-(n-1)/2}(u_1^2 u_2 - u_1 u_2^2)(1-\lambda v_1)(1-\lambda v_2)$ into $\frac{T - Q/T}{Q^{(n+1)/2}}$ combines $Q^{-(n-1)/2} Q^{-(n+1)/2} = Q^{-n}$, so $(cQ - (cQ)^{-1})Q^{-n} = c Q^{1-n} - c^{-1} Q^{-1-n}$, giving $c W_{1-n} - c^{-1} W_{-1-n}$ (test_04). Because the complementary primitive projector $1 - e_{6n}^{\otimes 2}$ kills all $(2,0)$ and $(0,2)$ forms and $T(X_{n,a})$ has no pure $(1,1)$ rational quotient by polarizable semisimplicity and Lefschetz $(1,1)$, $\phi_{n,\lambda}^* T(X_{n,a}) \subset H^1(A_{n,\lambda}, \mathbb{Q})^{\otimes 2}$ ([1, Proposition 4]).
2. On $C_{n,\lambda}$, the branch weights $(4n+6, 3n, n, 4n-6)$ divided by $6n$ are $(\frac{2}{3}+\frac{1}{n}, \frac{1}{2}, \frac{1}{6}, \frac{2}{3}-\frac{1}{n})$. Matching the fractional exponents of $\frac{u^p q^e(1-\lambda v)\,dv}{b}$ with $w^{-j}$ modulo $1$ gives $j \equiv -e \pmod n$ and $j \equiv 4p + 3 \pmod 6$. Since $\tau_3 \equiv 1 \pmod n$ and $\tau_3 \equiv -1 \pmod 6$, multiplying $j$ by $\tau_3$ fixes $j \bmod n$ and exchanges $p = 1 \leftrightarrow p = 2$ ($1 \leftrightarrow 5 \pmod 6$). Moreover, because $\tau_3(4n + 6) \equiv -(4n - 6) \pmod{6n}$, $\tau_3(4n - 6) \equiv -(4n + 6) \pmod{6n}$, and $\tau_3 n \equiv -n \pmod{6n}$, the $4$ fractional parts of $\tau_3 j$ at $(0, 1, 1/\lambda, \infty)$ are $(1 - r_\infty(j),\, \frac{1}{2},\, 1 - r_{1/\lambda}(j),\, 1 - r_0(j))$, which pair complementarily with $(r_0(j),\, \frac{1}{2},\, r_{1/\lambda}(j),\, r_\infty(j))$ to sum to $(r_0(j) + 1 - r_0(j)) + (\frac{1}{2} + \frac{1}{2}) + (r_{1/\lambda}(j) + 1 - r_{1/\lambda}(j)) + (r_\infty(j) + 1 - r_\infty(j)) = 4$, giving $h^{1,0}(V_j) + h^{1,0}(V_{\tau_3 j}) = 4 - 2 = 2$.
3. Writing $\tilde{z} = cQ$, we have $s^j ds = (\tilde{z} + \tilde{z}^{-1})^j (\tilde{z} - \tilde{z}^{-1})\frac{d\tilde{z}}{\tilde{z}} = \sum_{m=1}^{j+1} M_n(j, m)(\tilde{z}^m - \tilde{z}^{-m})\frac{d\tilde{z}}{\tilde{z}}$ (where the $m = 0$ central binomial difference at odd $j$ vanishes). Since $U_{m-1}\!\left(\frac{\tilde{z}+\tilde{z}^{-1}}{2}\right) = \frac{\tilde{z}^m - \tilde{z}^{-m}}{\tilde{z} - \tilde{z}^{-1}}$, we have $U_{m-1}(s/2)\,ds = (\tilde{z}^m - \tilde{z}^{-m})\frac{d\tilde{z}}{\tilde{z}}$. On the double cover $\tilde{z} \mapsto s = \tilde{z} + \tilde{z}^{-1}$, the deck rotation $\tilde{z} \mapsto \zeta_n \tilde{z}$ preserves $U_n = \tilde{z}^n + \tilde{z}^{-n} - 2$ and $(X, Y)$ while acting with distinct conjugate eigenvalue pairs $\{\zeta_n^m, \zeta_n^{-m}\}$ on $\widetilde{\eta}_{n,m}$ for $1 \le m \le \lfloor n/3 \rfloor < n/2$, forcing orthogonality $q_{X_{n,a}}(\widetilde{\eta}_{n,m}, \overline{\widetilde{\eta}_{n,m'}}) = C_{n,m}(\lambda)\,\delta_{m,m'}$. $\square$

4.2. Uniform Shioda–Tate Rank $4f$ and Algebraicity of the $\mathbb{Q}(\zeta_n)$-Weil Space

For each embedding $\tau : F_0 = \mathbb{Q}(\zeta_n) \hookrightarrow \mathbb{C}$, the $4$-dimensional $F_0$-eigenspace of $H = H^1(A_{n,\lambda}, \mathbb{Q})$ is $H_\tau = V_j \oplus V_{\tau_3 j}$. Because $h^{1,0}(V_j) + h^{1,0}(V_{\tau_3 j}) = 2$ at every parameter, the $(n-1)$-dimensional $F_0$-Weil subspace $$W_{F_0}(A_{n,\lambda}) := \bigwedge^4_{F_0} H^1(A_{n,\lambda}, \mathbb{Q}) = \bigoplus_{\tau : F_0 \hookrightarrow \mathbb{C}} \bigwedge^4 H_\tau \subset H^4(A_{n,\lambda}, \mathbb{Q})$$ consists purely of $(2, 2)$ Hodge classes for every $\lambda \in \mathbb{C} \setminus \{0, 1\}$.

Theorem 4.2 (Uniform Shioda–Tate Transcendental Rank $4f$ and Algebraicity of $W_{F_0}(A_{n,\lambda})$). For every prime $n > 3$:

1. (Uniform Shioda–Tate Rank $\dim_{\mathbb{Q}}(M_{n,a}^\perp) = 4f$ and Constituent Isomorphism): For $a \neq 0$ and $a^3 \neq 8$ ($\lambda \neq 0, 1, -1$), the elliptic surface $X_{n,a}$ has singular fiber configuration $\mathrm{II} + f \cdot \mathrm{IV} + n \cdot \mathrm{I}_2 + \mathrm{IV}_\infty$ if $n \equiv 2 \pmod 3$, and $\mathrm{II} + f \cdot \mathrm{IV} + n \cdot \mathrm{I}_2 + \mathrm{IV}^*_\infty$ if $n \equiv 1 \pmod 3$, with Euler number $e(X_{n,a}) = 12\lceil n/3 \rceil$ and $b_2(X_{n,a}) = 12\lceil n/3 \rceil - 2$. The zero section, fiber, reducible fiber components, and the section $P = (a^2, a^3 + 2U_n)$ of Shioda height $\frac{n}{6} > 0$ span a non-degenerate algebraic divisor subspace $M_{n,a} \subset \operatorname{NS}(X_{n,a})_{\mathbb{Q}}$ of rank $12\lceil n/3 \rceil - 2n$, whose orthogonal complement $M_{n,a}^\perp \subset H^2(X_{n,a}, \mathbb{Q})$ satisfies in both congruence classes $n \equiv 1, 2 \pmod 3$: $$\dim_{\mathbb{Q}}\left(M_{n,a}^\perp\right) = b_2(X_{n,a}) - \operatorname{rank}(M_{n,a}) = 2(n - 1) = 4f.$$ For very general $a$, $T(X_{n,a}) = M_{n,a}^\perp$ has $\dim_{\mathbb{Q}} T(X_{n,a}) = 4f$, $\rho(X_{n,a}) = 12\lceil n/3 \rceil - 2n$, $\operatorname{End}_{\mathrm{Hdg}} T(X_{n,a}) = E_n = \mathbb{Q}(\zeta_n + \zeta_n^{-1})$, and the algebraic correspondence $F := \operatorname{cup} \circ \iota : T(X_{n,a}) \hookrightarrow H^2(A_{n,\lambda}, \mathbb{Q})$ induces an isomorphism of $4$-dimensional spaces at every embedding $\sigma : E_n \hookrightarrow \mathbb{R}$ and extending $\tau : F_0 \hookrightarrow \mathbb{C}$: $$F_\tau : T(X_{n,a})_{E_n, \sigma} \xrightarrow{\sim} V_j \otimes V_{\tau_3 j} \subset \bigwedge^2 H_\tau.$$
2. (All-Parameter Algebraicity of the $\mathbb{Q}(\zeta_n)$-Weil Space): Let $R := \sigma_{6n}^6$ be the order-$n$ automorphism induced by $\zeta_n \in F_0$ on $A_{n,\lambda}$, and define on $H^{\otimes 4}$ the equal-embedding projector $$P_{\mathrm{same}, n} := \frac{1}{n^3} \sum_{b, c, d = 0}^{n - 1} R_1^{b+c+d} R_2^{-b} R_3^{-c} R_4^{-d}, \qquad \pi_{W, n}^{(k)} := \frac{1}{24} \delta^* R_1^k P_{\mathrm{same}, n} \pi_1^{\otimes 4} \mu^* \quad (0 \le k \le n - 2).$$ For very general $\lambda$, transporting the algebraic inverse intersection tensor $q_T \in T(X_{n,a})^{\otimes 2}$ produces $n - 1$ linearly independent algebraic cycle classes $\beta_{n, k} := \pi_{W, n}^{(k)} \operatorname{cup}(F \otimes F)(q_T)$ that span $W_{F_0}(A_{n,\lambda})$ over $\mathbb{Q}$. By finite-monodromy specialization, all classes in $W_{F_0}(A_{n,\lambda}) \subset H^4(A_{n,\lambda}, \mathbb{Q})$ are algebraic for every $\lambda \in \mathbb{C} \setminus \{0, 1\}$.

Proof.

1. Over the $U_n$-line, $Y^2 = X^3 + 6aU_n X + 2U_n(2U_n - a^3)$ is a rational elliptic surface with fibers $\mathrm{II}$ at $0$, $\mathrm{I}_2$ at $-a^3/2$, $\mathrm{IV}^*$ at $\infty$, and section $P = (a^2, a^3+2U_n)$ of height $2 - \frac{1}{2} - \frac{4}{3} = \frac{1}{6}$ ([2, §2]; [6]). Under the degree-$n$ base change $U_n = D_n(s) - 2$, $U_n = 0$ has one simple root at $s = 2$ (giving $1 \times \mathrm{II}$) and $f = (n-1)/2$ double roots (giving $f \times \mathrm{IV}$, contributing $2f$ reducible components $A_2(-1)^f$); $U_n = -a^3/2$ ($a \neq 0, a^3 \neq 8$) has $n$ simple roots (giving $n \times \mathrm{I}_2$, contributing $n$ reducible components $A_1(-1)^n$); and at $s = \infty$, minimalizing the $\mathrm{IV}^*$ fiber of valuation $(\ge 3n, 4n, 8n)$ leaves valuations $(\ge 3, 2, 4)$ (type $\mathrm{IV}$, $e = 4$, $2$ reducible components) when $n = 3h - 1 \equiv 2 \pmod 3$, and $(\ge 3, 4, 8)$ (type $\mathrm{IV}^*$, $e = 8$, $6$ reducible components) when $n = 3h - 2 \equiv 1 \pmod 3$ ($h = \lceil n/3 \rceil$). In both cases, $e(X_{n,a}) = 2 + 4f + 2n + e_\infty = 12h$, $b_2(X_{n,a}) = 12h - 2$, the pulled-back section $P$ has height $\frac{n}{6} > 0$, and $\operatorname{rank}(M_{n,a}) = 2 + 2f + n + (e_\infty - 2) + 1 = 12h - 2n$, so $\dim_{\mathbb{Q}}(M_{n,a}^\perp) = (12h - 2) - (12h - 2n) = 2(n - 1) = 4f$ (test_05). By [8, §§4.6–4.8] and [2, §2], the dihedral correspondences on $s = t + t^{-1}$ induce a self-adjoint algebraic $E_n$-action on $M_{n,a}^\perp$. As in [2, §5], the Abel–Jacobi isomorphism on $H^1(C_{n,\lambda}, \mathbb{Q})$, the primitive projector $e_{6n}$, and the Poincaré/inverse-Lefschetz correspondences identify the primitive curve summand with $H = H^1(A_{n,\lambda}, \mathbb{Q})$, giving an algebraic injection $\iota : T(X_{n,a}) \hookrightarrow H \otimes H$. Since $\phi_{n,\lambda}$ is invariant under factor interchange (which acts on $H^1 \otimes H^1$ as minus tensor flip), $\iota(T(X_{n,a}))$ lies in the alternating subspace, where cup product is injective; thus $F = \operatorname{cup} \circ \iota : T(X_{n,a}) \hookrightarrow H^2(A_{n,\lambda}, \mathbb{Q})$ is injective. By Theorem 4.1(1)–(2), $F(\eta_{n,0}) \neq 0$ projects non-trivially onto $V_1 \otimes V_{\tau_3} \subset \bigwedge^2 H_{\tau_1}$. By Theorem 5.1(1) below, the $f$ Galois conjugates $V_j \otimes V_{\tau_3 j}$ ($j \in (\mathbb{Z}/6n\mathbb{Z})^\times / \{\pm 1, \pm \tau_3\}$) are pairwise non-isomorphic irreducible rank-$4$ monodromy modules of total dimension $4f = \dim_{\mathbb{Q}}(M_{n,a}^\perp)$. Semisimplicity and the dimension bound force $T(X_{n,a}) = M_{n,a}^\perp$ at very general $a$ and make each $F_\tau$ an isomorphism.
2. Following [2, Lemma 5 and Proposition 6] (adapting Varesco [9, Proposition 4.2]), on any $F_0$-embedding tuple $(r_1, r_2, r_3, r_4) \in (\mathbb{Z}/n\mathbb{Z})^{\times 4}$, each geometric series in $P_{\mathrm{same}, n}$ averages to $1$ if $r_1 = r_i$ and $0$ otherwise, so $P_{\mathrm{same}, n}$ projects $H^{\otimes 4}$ onto $\bigoplus_\tau H_\tau^{\otimes 4}$, and $\pi_{W,n}^{(k)}$ is an algebraic correspondence mapping $H^4(A_{n,\lambda}, \mathbb{Q})$ onto $W_{F_0}(A_{n,\lambda})$ that acts on each complex Weil line $\bigwedge^4 H_\tau$ by $\tau(\zeta_n)^k$. Since $E_n$ acts self-adjointly on $T(X_{n,a})$, the inverse pairing $q_T$ (the transcendental component of $[\Delta_{X_{n,a}}]$) is non-degenerate on each $T(X_{n,a})_{E_n,\sigma}$ and transports via $F_\tau$ to a non-zero multiple of the symmetric invariant $\epsilon \otimes \epsilon$ on $V_j \otimes V_{\tau_3 j}$, whose exterior product in $\bigwedge^4(V_j \oplus V_{\tau_3 j})$ is $-4 u_1 \wedge u_2 \wedge v_1 \wedge v_2 \neq 0$ ([2, eq. (25)]; test_05). Because the cyclotomic polynomial $\Phi_n(t)$ is irreducible of degree $n - 1$ over $\mathbb{Q}$, the Vandermonde matrix $(\tau_a(\zeta_n)^k)_{a \in (\mathbb{Z}/n\mathbb{Z})^\times, 0 \le k \le n-2}$ is invertible, so $\{\beta_{n,0}, \dots, \beta_{n,n-2}\}$ spans $W_{F_0}(A_{n,\lambda})$ over $\mathbb{Q}$ at very general $\lambda$. Finally, because $W_{F_0}(A_{n,\lambda})$ has pure type $(2, 2)$ everywhere over $\mathbb{C} \setminus \{0, 1\}$, its polarization is definite and its monodromy group is finite; the relative Chow spreading and specialization argument of [2, §7] and [9, Proposition 2.22] specializes these $n - 1$ basis cycles to every $\lambda \in \mathbb{C} \setminus \{0, 1\}$. $\square$

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5. Generic $\mathrm{SL}_2^{2f}$ Monodromy, the Middle-Third Stabilizer, and the Fixed Determinant Surface $S_n$

Let $K_n^+ := \mathbb{Q}(\zeta_{6n} + \zeta_{6n}^{-1})$ be the maximal totally real subfield of the CM field $K_n = \mathbb{Q}(\zeta_{6n})$ ($[K_n : \mathbb{Q}] = 2(n-1) = 4f$, $[K_n^+ : \mathbb{Q}] = n - 1 = 2f$), and let $\tau_2 := -\tau_3 \in (\mathbb{Z}/6n\mathbb{Z})^\times$ ($\tau_2 \equiv -1 \pmod n$, $\tau_2 \equiv 1 \pmod 6$, generating $\operatorname{Gal}(K_n/L_n)$, while $-1 \in (\mathbb{Z}/6n\mathbb{Z})^\times$ is complex conjugation generating $\operatorname{Gal}(K_n / K_n^+)$). The polarization on $H = H^1(A_{n,\lambda}, \mathbb{Q})$ induces a rank-$2$ $K_n / K_n^+$-Hermitian form $h$ whose ambient unitary group $$U_{K_n / K_n^+}(H) := \operatorname{Res}_{K_n^+ / \mathbb{Q}} U(H, h)$$ has $\mathbb{Q}$-dimension $4[K_n^+ : \mathbb{Q}] = 4(n-1) = 8f$, with derived subgroup $\operatorname{Res}_{K_n^+ / \mathbb{Q}} \mathrm{SU}_2(H, h)$ of $\mathbb{Q}$-dimension $3(n-1) = 6f$ and $K_n / K_n^+$-norm-one determinant torus $T_{K_n / K_n^+} := \ker(\operatorname{Nm}_{K_n / K_n^+} : \operatorname{Res}_{K_n/\mathbb{Q}}\mathbb{G}_m \to \operatorname{Res}_{K_n^+/\mathbb{Q}}\mathbb{G}_m)$ of $\mathbb{Q}$-dimension $2(n-1) - (n-1) = n - 1 = 2f$. Since $h^{1,0}(V_{\tau_2 j}) = 2 - h^{1,0}(V_{\tau_3 j}) = h^{1,0}(V_j)$, the centered Hodge-circle weight vector $w_{\det}(j) := h^{1,0}(V_j) - 1 \in \{-1, 0, +1\}$ is $\tau_2$-invariant ($w_{\det}(\tau_2 j) = w_{\det}(j)$) and satisfies $w_{\det}(j) + w_{\det}(\tau_3 j) = 0$.

Theorem 5.1 (Beukers–Heckman $\mathrm{SL}_2^{2f}$ Monodromy, Very-General Mixed HC, Fixed Determinant Surface $S_n$, and Non-CM Source Descent). For every prime $n > 3$:

1. (Unconditional $\mathrm{SL}_2^{2f}$ Connected Monodromy and Very-General Mixed HC): On $H = H^1(A_{n,\lambda}, \mathbb{Q})$, the connected monodromy group of the primitive variation over $\lambda \in \mathbb{P}^1 \setminus \{0, 1, \infty\}$ is $\operatorname{Res}_{K_n^+ / \mathbb{Q}} \mathrm{SU}_2(H, h)$, with complexification $\mathrm{SL}_2^{2f}$ of dimension $3(n-1) = 6f$. For very general $\lambda$, the Hodge group is $$\operatorname{Hg}(A_{n,\lambda}) = \left\{ g \in U_{K_n / K_n^+}(H) : \operatorname{Nm}_{K_n / F_0}(\det_{K_n} g) = 1 \right\}^\circ \qquad \left(\dim_{\mathbb{Q}} \operatorname{Hg}(A_{n,\lambda}) = 3(n-1) + \frac{n-1}{2} = \frac{7(n-1)}{2}\right),$$ and the rational Hodge conjecture holds unconditionally on every mixed product $X_{n,a}^r \times A_{n,\lambda}^k$ ($r, k \ge 0$).
2. (Uniform Fixed Determinant Surface $S_n$): For every $\lambda \in \mathbb{C} \setminus \{0, 1\}$, the order-$12n$ quotient $C_{n, \lambda}^2 / (\langle (\sigma_{6n}, \sigma_{6n}^{-1}) \rangle \rtimes \langle \iota \rangle)$ is birational (via $e_n = (-n)\bmod 3 \in \{1, 2\}$, $m = \frac{n+e_n}{3}$, $w_0 = 2e_n \in \{2, 4\}$, $k = (1-\lambda v_1)(1-\lambda v_2) = q^{e_n} R^3$, and $Q = cq$) to the $\lambda$-independent surface $$S_n : y^2 = x^3 + Q^{2w_0}(1 - Q^n)^2,$$ dominated via $(x, y) = (t^2 x_E, t^3 y_E)$ by $(E_3 \times B_n)/\mu_3$ ($E_3 : y_E^2 = x_E^3 + 1$, $B_n : t^{3n} = v^{w_0}(1 - v)^n$). The determinant variation $D_{n,\lambda} := \bigwedge^2_{K_n} H^1(A_{n,\lambda}, \mathbb{Q}) = \bigoplus_{j \in (\mathbb{Z}/6n\mathbb{Z})^\times} \det V_j \subset H^2(A_{n,\lambda}, \mathbb{Q})$ is algebraically and $K_n$-equivariantly isomorphic to the $4f$-dimensional primitive summand $M_{B_n} := \bigoplus_{k \in (\mathbb{Z}/3n\mathbb{Z})^\times} \left(H^1(B_n)_k \otimes H^1(E_3)_{-k\bmod 3}\right) \subset H^2(S_n, \mathbb{Q})$.
3. (Middle-Third Stabilizer and Non-CM Source Descent): On $\mathfrak{sl}(V_j)$, the adjoint Hodge-circle action is trivial for the compact $(2,0)/(0,2)$ labels $j \bmod n \in I_n := \{r \in (\mathbb{Z}/n\mathbb{Z})^\times : n/3 < r < 2n/3\}$ and non-trivial (weights $-2, 0, 2$) for the non-compact $(1,1)$ labels $j \bmod n \notin I_n$. For every prime $n > 3$, $$\operatorname{Stab}_{(\mathbb{Z}/n\mathbb{Z})^\times}(I_n) = \{1, -1\} = \{1, \tau_2 \bmod n\}.$$ Consequently, for every $\lambda \in \mathbb{C} \setminus \{0, 1\}$ such that $A_{n,\lambda}$ is not of CM type, the complex derived Hodge group $\operatorname{Hg}(A_{n,\lambda})^{\mathrm{der}}_{\mathbb{C}}$ is either $\mathrm{SL}_2^{2f}$ or $\mathrm{SL}_2^f$ (embedded diagonally across the $f$ pairs $\{j, \tau_2 j\}$, which are intertwined by an algebraic $\tau_2$-semilinear quasi-endomorphism $J_\lambda \in \operatorname{End}^0(A_{n,\lambda})$), and there is an explicit $\operatorname{Hg}(A_{n,\lambda})$-equivariant algebraic surjection from a finite direct sum of tensor powers of $D_{n,\lambda} \cong M_{B_n}$ and Tate structures onto the derived-invariant subspace $(H^{\otimes 2m})^{\operatorname{Hg}(A_{n,\lambda})^{\mathrm{der}}}$ for every $m \ge 1$.

Proof.

1. For each of the $2f = n - 1$ conjugate-pair representatives $j \in (\mathbb{Z}/6n\mathbb{Z})^\times / \{\pm 1\}$ ($1 \le j < 3n$), the Euler period integrals of $V_j$ have Gauss hypergeometric parameters $(A_j, B_j, C_j) = (r_{1/\lambda}, 1 - r_0, \frac{3}{2} - r_0)$, where $r_0 = \left\{\frac{(4n+6)j}{6n}\right\}$ and $r_{1/\lambda} = \left\{\frac{j}{6}\right\} \in \{\frac{1}{6}, \frac{5}{6}\}$ ([2, eq. (18)]). We follow the three-step method of [2, Lemma 4]:
   • Irreducibility: Since $\gcd(j, 6n) = 1$ and $n > 3$, $A_j \in \{\frac{1}{6}, \frac{5}{6}\} \notin \mathbb{Z}$, $B_j = 1 - r_0 \notin \mathbb{Z}$ (as $r_0$ has denominator divisible by $n$), $C_j - B_j = \frac{1}{2} \notin \mathbb{Z}$, and $C_j - A_j = \frac{3}{2} - r_{1/\lambda} - r_0 \notin \mathbb{Z}$ (since $\frac{3}{2} - r_{1/\lambda} \in \{\frac{2}{3}, \frac{4}{3}\}$ has denominator $3$ while $r_0$ has denominator divisible by $n$). By the Gauss / Beukers–Heckman criterion [2, ref. 10], every rank-$2$ local system $V_j$ is irreducible.
   • Single-factor closure $\mathrm{SL}_2$: At $\lambda = 0, 1, \infty$, the local exponent differences are $1 - C_j = r_0 - \frac{1}{2}$ (denominator $2n$ or $6n$), $C_j - A_j - B_j = \frac{1}{2} - r_{1/\lambda} = \pm \frac{1}{3}$ (denominator $3$), and $A_j - B_j = r_{1/\lambda} + r_0 - 1$ (denominator $2n$ or $6n$), so the local projective monodromy orders are $(\ge 2n, 3, \ge 2n)$ with $2n \ge 10$ (test_06). Irreducibility excludes a Borel subgroup of $\mathrm{PGL}_2(\mathbb{C})$; no primitive finite subgroup ($A_4, S_4, A_5$) has an element of order $\ge 10$; and in a torus normalizer $N(\mathbb{G}_m) \subset \mathrm{PGL}_2(\mathbb{C})$ every element outside $\mathbb{G}_m$ has order $2$, whereas all three local generators around $0, 1, \infty$ have orders $\ge 3$ (and their product is $1$, so if any generator lay outside $\mathbb{G}_m$ it would have order $2$). Thus the projective closure is $\mathrm{PGL}_2(\mathbb{C})$, and since $\det V_j$ has finite order dividing $6n$, the connected linear closure is $\mathrm{SL}_2(\mathbb{C})$.
   • Pairwise inequivalence and Goursat's lemma: If two representatives $1 \le j_1, j_2 < 3n$ in $(\mathbb{Z}/6n\mathbb{Z})^\times$ have the same local projective eigenvalue-ratio angle $\min(|1-C_j|, 1-|1-C_j|)$ at $\lambda = 0$, then $(4n+6)(j_1 \mp j_2) \equiv 0 \pmod{6n}$. Since $\gcd(4n+6, 6n) = 2$, this implies $3n \mid (j_1 \mp j_2)$; since $j_1$ and $j_2$ are both odd, $2 \mid (j_1 \mp j_2)$, and since $\gcd(2, 3n) = 1$, $6n \mid (j_1 \mp j_2)$, so $j_1 = j_2$. Thus all $2f$ projective representations are pairwise inequivalent at $\lambda = 0$ (test_06). They cannot become equivalent after restriction to a finite-index subgroup, since the full joint closure in $\mathrm{PGL}_2^2$ would then normalize a diagonal graph, which is its own normalizer ([2, Lemma 4]). Goursat's lemma gives connected monodromy $\mathrm{SL}_2^{2f}$. On the $(n-1)$-dimensional $K_n / K_n^+$-norm-one determinant torus $T_{K_n / K_n^+}$ (parametrized by the $2f = n-1$ conjugate pairs $\{j, -j\}$), the relative norm $\operatorname{Nm}_{K_n / F_0}(z) = z \cdot \tau_3(z)$ to the CM subfield $F_0 = \mathbb{Q}(\zeta_n) = K_n^{\langle \tau_3 \rangle}$ maps $T_{K_n / K_n^+}$ onto the $f$-dimensional $F_0 / E_n$-norm-one torus $T_{F_0 / E_n}$ ($[F_0 : \mathbb{Q}] - [E_n : \mathbb{Q}] = 2f - f = f = \frac{n-1}{2}$), so $\ker(\operatorname{Nm}_{K_n / F_0}|_{T_{K_n / K_n^+}})^\circ$ has $\mathbb{Q}$-dimension $2f - f = f = \frac{n-1}{2}$. By [1, §4.2, eq. (13)], the non-vanishing of the Dirichlet $L$-values $L(0, \psi) = -B_{1,\psi} \neq 0$ for all odd characters $\psi = \chi \epsilon$ of conductor $3n$ proves that the $\mathrm{Gal}(K_n/\mathbb{Q})$-span of $w_{\det}(j) = h^{1,0}(V_j) - 1$ has full rank $f = (n-1)/2$, while $w_{\det}(j) + w_{\det}(\tau_3 j) = 0$ imposes $\operatorname{Nm}_{K_n/F_0}(\det_{K_n} g) = 1$. Together with $\operatorname{Res}_{K_n^+ / \mathbb{Q}} \mathrm{SU}_2(H, h)$ of dimension $3(n-1) = 6f$, this proves the formula for $\operatorname{Hg}(A_{n,\lambda})$ of dimension $6f + f = 7f = \frac{7(n-1)}{2}$ ([2, Proposition 7]). At very general $\lambda$, the $\mathrm{SL}_2$ First Fundamental Theorem reduces $\operatorname{Hg}(A_{n,\lambda})$-invariant tensors on $A_{n,\lambda}^k$ to evaluation pairings (divisors) and determinant monomials satisfying $d_j = d_{\tau_3 j}$ ($1 \le j < 3n$), which are products of the $n - 1$ complex Weil lines $\det V_j \otimes \det V_{\tau_3 j} \subset W_{F_0}(A_{n,\lambda})$ ([2, Proposition 7]). Since $W_{F_0}(A_{n,\lambda})$ is algebraic by Theorem 4.2(2) and tensor placement is algebraic, HC holds on all powers $A_{n,\lambda}^k$, and transfers to $X_{n,a}^r \times A_{n,\lambda}^k$ via the degree-$24n$ graph correspondence of $\phi_{n,\lambda}$ and its transpose ([2, §6]).
2. On $C_{n,\lambda}^2$, the invariants $x = v_1 + v_2$, $q = q_1 q_2$ ($q^n = v_1 v_2$), $u = u_1 u_2$, $b = b_1 b_2$, and $k = (1-\lambda v_1)(1-\lambda v_2) = 1 - \lambda x + \lambda^2 q^n$ generate the fixed field of $\langle (\sigma_{6n}, \sigma_{6n}^{-1}) \rangle \rtimes \langle \iota \rangle$ (since $z = w_1 w_2 = q u^2 b k$ has weight $1 \bmod 6n$), and eliminating $x$ from $b^2 = (1 - x + q^n)k$ gives $\lambda b^2 = k^2 - D_0 k$ with $D_0 = (1-\lambda)(1-\lambda q^n)$ ([2, §6.1]). Setting $k = q^{e_n} R^3$, $\beta = q^{e_n} D_0 / \lambda$, $x_{\mathrm{ell}} = -\beta / R$, and $y_{\mathrm{ell}} = -\beta b / R^3$ yields $y_{\mathrm{ell}}^2 = x_{\mathrm{ell}}^3 + q^{4e_n} D_0^2 / \lambda^3$, which scales via $Q = cq$ ($c^n = \lambda$, $w_0 = 2e_n$) to $S_n : y^2 = x^3 + Q^{2w_0}(1 - Q^n)^2$, dominated by $(E_3 \times B_n)/\mu_3$ (test_06). The residual transformation $(\sigma_{6n}, 1)$ induces a $K_n$-action on both $D_{n,\lambda}$ and $M_{B_n}$ with character bijection $j(k) = (k w_t \mp n) \bmod 6n$ (test_06). Because $K_n = \mathbb{Q}(\zeta_{6n})$ is a field of degree $2(n-1) = 4f$ and both $D_{n,\lambda}$ and $M_{B_n}$ have $\mathbb{Q}$-dimension $4f$ with faithful $K_n$-actions, both are $1$-dimensional $K_n$-vector spaces. Since the $K_n$-linear correspondence induced by the dominant quotient map is non-zero on the holomorphic $2$-form $\frac{dQ \wedge dx}{y}$ of $S_n$, it is an isomorphism of $K_n$-Hodge structures.
3. By [1, Lemma 5, eq. (12)], in $\mathbb{F}_n$ we have $\sum_{a \in I_n \cap [1, f]} a^2 = \pm \frac{1}{81} \neq 0$, so any $g \in \operatorname{Stab}_{(\mathbb{Z}/n\mathbb{Z})^\times}(I_n)$ satisfies $(g^2 - 1)(\pm \frac{1}{81}) \equiv 0 \pmod n \implies g \equiv \pm 1 \pmod n$. At any non-CM parameter $\lambda \in \mathbb{C} \setminus \{0, 1\}$, Goursat's lemma describes $G'_{\mathbb{C}} = \operatorname{Hg}(A_{n,\lambda})^{\mathrm{der}}_{\mathbb{C}}$ by a $\mathrm{Gal}(K_n/\mathbb{Q})$-stable block partition of $(\mathbb{Z}/6n\mathbb{Z})^\times / \{\pm 1\}$ on which the adjoint Hodge-circle types agree. Because the subgroup of $(\mathbb{Z}/6n\mathbb{Z})^\times / \{\pm 1\}$ preserving $I_n$ and the balanced $(2,2)$ condition has order at most $2$ (generated by $\tau_2$), $G'_{\mathbb{C}}$ has either $2f$ blocks of size $1$ ($\mathrm{SL}_2^{2f}$) or $f$ blocks $\{j, \tau_2 j\}$ of size $2$ ($\mathrm{SL}_2^f$). In the latter case, exactly as in [2, Proposition 8, eqs. (28g)–(28h)], polarization $\theta_{-j} = -\theta_j$ and the differentiated norm relation $\theta_{\tau_3 j} = -\theta_j$ give $\theta_{\tau_2 j} = \theta_{-\tau_3 j} = \theta_j$ on the connected center of $\operatorname{Hg}(A_{n,\lambda})$, so $V_j$ and $V_{\tau_2 j}$ are isomorphic $\operatorname{Hg}(A_{n,\lambda})_{\mathbb{C}}$-modules and the $1$-dimensional $K_n$-space of $\tau_2$-semilinear $\operatorname{Hg}(A_{n,\lambda})$-intertwiners furnishes an invertible algebraic quasi-endomorphism $J_\lambda \in \operatorname{End}^0(A_{n,\lambda})$ identifying the paired modules. Applying the $\mathrm{SL}_2$ First Fundamental Theorem in each block yields the stated algebraic surjection from tensor powers of $D_{n,\lambda} \cong M_{B_n}$ and Tate structures onto $(H^{\otimes 2m})^{G'}$. $\square$

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6. Verification Suite and Reproducibility

The supplemental file test_prime_dickson_fixed_and_moving_vhs.py provides an automated, self-contained CPU verification suite (pytest test_prime_dickson_fixed_and_moving_vhs.py -v, executing in $< 3\text{ s}$ with Python 3 and SymPy). The scripts verify the symbolic identities, finite arithmetic instances, and character bookkeeping used in Theorems 2.1–5.1, while the geometric, monodromy, and specialization proofs are given in the text above:

1. test_01_hasse_brauer_mod12_quaternion_dichotomy: verifies the mod-$12$ local norm criterion at $w \mid n$ across all 23 primes $5 \le n < 100$.
2. test_02_twisted_frobenius_symmetries_and_direct_fp2_jacobi_crosscheck: verifies the inversion and Cayley involution symmetries, exact cyclic convolution $N_{109} = -33 C_1 - 69 C_2 = 51 \pm 18\sqrt{5}$, polarization determinant $b_{109} = 109\alpha_{109}/\bar{\alpha}_{109} \in \mathcal{O}_{L_5}$, cross-checks $-(\alpha_{109}^2 - 2b_{109})$ against a direct $109^2$-element enumeration on $C_5(\mathbb{F}_{109^2})$, and verifies the Hensel-lifted valuation identity $v_{\mathfrak{P}_a}(b_p) = 1 + v_{\mathfrak{P}_a}(\alpha_p) - v_{\mathfrak{P}_a}(\bar{\alpha}_p) = h^{1,0}(M_a)$ modulo $p^4$ across all 23 primes $5 \le n < 100$.
3. test_03_two_reduction_non_cm_discharge_all_23_primes_below_100: computes $\delta_{p_i} = N_{p_i}(N_{p_i} - 4p_i)$ at two good reductions $p_1 < p_2$ for each of the 23 primes $5 \le n < 100$, verifies $N_{p_i} \not\equiv 0 \pmod{\mathfrak{p}_1}$, and certifies the split prime witnesses $\ell \equiv 1 \pmod{6n}$ and embeddings $(a_1, a_2, a_{12})$ in the table of §3.2.
4. test_04_moving_curve_square_differential_crt_characters_and_chebyshev_diagonalization: symbolically verifies the moving curve-square Weierstrass identity, logarithmic Jacobian, exact differential pullback with $Q^{(n+1)/2}$, expansion into $c W_{1-n} - c^{-1} W_{-1-n}$, Chinese Remainder character formula, $\tau_3$-pairing, and Chebyshev $U_{m-1}(s/2)$ inverse matrix $M_n^{-1}$.
5. test_05_shioda_tate_rank_4f_four_slot_projector_and_weil_cycle_spanning: verifies the singular fiber Euler sum $e(X_{n,a}) = 12\lceil n/3 \rceil$, $\dim_{\mathbb{Q}}(M_{n,a}^\perp) = 2(n-1) = 4f$ across all primes $5 \le n < 100$, the four-slot projector $P_{\mathrm{same},n}$, and the $-4 \neq 0$ exterior wedge contraction.
6. test_06_beukers_heckman_sl2_monodromy_middle_third_stabilizer_and_determinant_surface: verifies the Gauss irreducibility parameters, local projective orders $(\ge 2n, 3, \ge 2n)$, pairwise distinct local angles at $\lambda = 0$, middle-third stabilizer $\operatorname{Stab}(I_n) = \{1, -1\}$, Segre monoid degree-$2$ generation, and the second quotient $S_n$ Weierstrass identity and $K_n$-equivariant character bijection.

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