Hodge cycles and Kuga–Satake geometry for Dickson surfaces

1. Results and relation to the Hodge conjecture

Let $D_n$ be the Dickson polynomial normalized by

$$D_n(z+z^{-1})=z^n+z^{-n}.$$

For a prime $n>3$, let $X_n$ be the smooth relatively minimal elliptic surface over $\mathbf C$ with affine equation

$$ y^2=x^3+3(D_n(s)-2)x+\frac{(D_n(s)-2)(D_n(s)-6)}2. \tag{1} $$

Let $C_n$ be the smooth projective normalization of

$$ w^{6n}=v^{4n+6}(1-v)^{3n}(1+v)^n, \tag{2} $$

and let $A_n=e_{6n}J(C_n)$, up to isogeny, where $e_{6n}$ is the rational projector for primitive order-$6n$ deck characters. Set $f=(n-1)/2$. Then $g(C_n)=4n-1$ and $\dim A_n=4f$.

Theorem 1. Suppose that $A_n$ is not of CM type. Then:

1. The Hodge group of $A_n$ is the connected symplectic centralizer of its explicit degree-two algebra over $\mathbf Q(\zeta_n+\zeta_n^{-1},\zeta_3)$. Its complexification is $\operatorname{GL}_2^f$.
2. All rational Hodge classes on every power of $A_n$ are generated by divisors.
3. The rational Hodge conjecture holds on every power of $X_n$, and

$$ p_g(X_n)=\left\lceil\frac n3\right\rceil-1, \qquad \dim T(X_n)=\frac{3(n-1)}2, \qquad \rho(X_n)=12\left\lceil\frac n3\right\rceil-2-\frac{3(n-1)}2. \tag{3} $$

4. If $S$ is any finite set of distinct primes satisfying the same non-CM hypothesis, the rational Hodge conjecture holds on every product $\prod_{n\in S}X_n^{m_n}$, for all $m_n\geq0$.

The geometric input is an actual rational map $C_n^2\dashrightarrow X_n$ of degree $24n$. We prove that the entire transcendental image is primitive by an identity for all holomorphic two-forms. This argument works when $p_g>1$. Two additional uniform lemmas give the full derived group and central torus, respectively. The former uses an elementary sum of squares; the latter uses standard nonvanishing of a Dirichlet $L$-value.

Theorem 2. The non-CM hypothesis in Theorem 1 holds for $n=7$ and $n=11$.

We establish this by comparing geometric endomorphism centralizers at two good reductions. In degree eleven an automorphism-twisted Frobenius count over $\mathbf F_{13^5}$ determines the polynomial over $\mathbf F_{13^{10}}$. Cubic twists recover its coefficients in a proper quadratic fixed field; the reduced norm is not the cardinality of the counting field. We give the full descent and polarization argument and independent exact counts.

The degree-five case was proved in [1]. Thus Theorems 1–2 give rational HC for every product $X_5^a\times X_7^b\times X_{11}^c$. The numerical invariants are:

| Surface | $p_g$ | $b_2$ | $\rho$ | $\dim T$ | Curve-product degree | | --- | ---: | ---: | ---: | ---: | ---: | | $X_5$ | 1 | 22 | 16 | 6 | 120 | | $X_7$ | 2 | 34 | 25 | 9 | 168 | | $X_{11}$ | 3 | 46 | 31 | 15 | 264 |

Proposition 8 constructs an explicit fourfold factor of $A_5$ and supplies the additional mixed determinant classes after adjoining the Eisenstein elliptic curve. It uses Schoen's classical sixfold cycles and gives an independent proof for this source of a case included in the broader recent theorem [2].

Theorem 9 supplies a further geometric operation in degree five: an explicit quotient of $C_5^2\times E_0$ becomes uniruled because an exceptional divisor on a crepant Weierstrass modification is fixed pointwise. Its primitive invariant cohomology realizes the spin half twist. Curve support and the source cycles of Proposition 8 then prove the algebraicity of the canonical Kuga–Satake correspondence for $X_5$. Corollary 10 proves generalized Hodge support on all mixed powers of this K3 surface and its simple spin factor. This argument is independent of the recent general CM-HC claim.

The general prime-degree non-CM assertion remains open. Corollary 7 gives a separate uniform surface-product consequence conditional on the CM case of HC. We do not claim a solution of the Hodge conjecture for arbitrary varieties.

Related work

The earlier paper [1] treats the fixed degree-five K3 surface, proves its Picard rank independently, and constructs triangle cycles representing its exceptional determinants. Here the uniform statement extends beyond K3 surfaces, an arithmetic method replaces that K3 Picard-rank input, and a new threefold quotient supplies the missing spin correspondence for the original surface. The degree-five argument alone supplies neither the higher-$p_g$ primitive-image proof nor the two-reduction centralizer calculations.

Van Geemen–Schütt [3] develop the dihedral real-multiplication geometry that motivates the Dickson base change. Ingalls–Logan–Patashnick [4] construct coverings of elliptic surfaces by curve squares. Moonen [5] studies product-quotient surfaces arising from cyclic covers and the Deligne–Mostow list. Product domination in those works and here is distinct from controlling Hodge classes on all powers: the relevant source Hodge group must also be determined. Clingher–Doran–Malmendier [6] relate superelliptic product geometry to special-function factorizations; that perspective motivated the degree-five map in [1]. Our proofs below use rational functions and algebraic correspondences.

The implication from a full Lefschetz group to divisor generation is classical; see Moonen–Zarhin [7, §1.8]. Müller–Pink [8, §6, Theorem 6.2 and Corollary 6.7] prove a general criterion detecting non-CM abelian varieties through incompatible Frobenius fields. Their Proposition 6.1 isolates the same obstruction from root-of-unity ratios that must be removed before making geometric assertions. Our relative centralizer calculation retains a prescribed noncommutative algebra D and proves its full characteristic-zero commutant, which then determines the Hodge group. The specialization step uses Tate's theorem [9], not a new isogeny theorem. Our arithmetic contribution is a concrete reconstruction of the relevant centralizer from twisted traces over a smaller field, retaining its proper coefficient field and polarization norm. The numerical cases are applications of this mechanism, rather than separate parameter statements.

For the mixed abelian source in §9, Moonen–Zarhin [7, §5.11, Case 2] already gives the balancing construction and the determinant relation for a fourfold with quadratic signature $(1,3)$. Schoen [10, §§6–7 and 11–13] supplies the required Eisenstein sixfold Weil classes; Markman [11, Theorem 1.5.1] gives an alternative valid for every imaginary quadratic field. Theorem 1.2 of [2] claims HC on all powers of every abelian variety of dimension at most five with imaginary-quadratic action, and therefore includes our mixed fivefold. Its proof uses [12]. Our restricted argument identifies the actual source factor and obtains its mixed generators directly from [10], independently of [12] and [2].

The geometric half-twist principle has precedents in van Geemen–Izadi [13], where cyclic hypersurface constructions give selected Kuga–Satake correspondences, and in Schoen [14, Remark 2.11], where uniruled quotients give curve support. We do not apply their special branch or hypersurface hypotheses to our quotient. Instead (19)–(21) exhibit its own crepant model and divisorial ramification. Schlickewei [15] identifies the rational spin representation for K3 surfaces with real multiplication, and Floccari [16] supplies other substantial KS-algebraicity cases. The rational transcendental form of our original surface has Witt index one [1], so the rank-six index-two embedding hypothesis of [16, Theorem 5.11] does not apply directly. The new input here is the actual curve-supported rank-eight spin summand, not a formal half-twist or an endomorphism-algebra match. The recent preprint [17, Theorem 1.2] claims prescribed KS correspondences on the quadratic locus $T\hookrightarrow U^2\oplus\langle-1\rangle^4$. Our $T(X_5)$ is also outside that locus: over $\mathbf Q_3$ the target is split of dimension eight, whereas $T(X_5)=U\oplus\langle1,-2,-3,-30\rangle$ has Witt index one [1]. A codimension-two subspace of the split eight-dimensional space has Witt index at least two. We do not use [17] as an input. Its geometric comparison through uniruled threefolds is a related precedent; our actual quotient and exceptional ramification differ from its two quadratic substitutions.

2. A uniform curve-square map

The function field of $C_n$ has the equivalent Kummer presentation

$$ q^n=v,\qquad u^3=\frac v{(1+v)^2},\qquad b^2=1-v^2, \qquad w=qu^2b(1+v). \tag{4} $$

The three extensions have pairwise coprime degrees and their displayed branch valuations show that each has the full indicated degree. Thus (4) is a cyclic extension of degree $6n$; the character of $w$ is faithful, giving (2). The four branch weights modulo $6n$ are

$$4n+6,\quad3n,\quad n,\quad4n-6.$$

Their greatest common divisors with $6n$ are $2,3n,n,2$. Riemann–Hurwitz gives $2g(C_n)-2=8n-4$.

On $C_n^2$ put

$$ \begin{aligned} Q&=q_1q_2,\\ T&=\frac{u_1^2u_2(1+v_1)(1+v_2)}{Q^{(n-1)/2}} =\frac{v_1+Q^n}{(1+v_1)Q^{(n-1)/2}(u_1/u_2)},\\ \delta&=\frac{b_1b_2}{(1+v_1)(1+v_2)}. \end{aligned} \tag{5} $$

Define

$$ \begin{aligned} s&=-(Q+Q^{-1}),\\ x&=-(Q^n+1)Q^{-(n+1)/2}(T+Q/T),\\ y&=(Q^n-Q^{-n})\delta/\sqrt2. \end{aligned} \tag{6} $$

Proposition 3. These functions define a dominant rational map $\phi_n:C_n^2\dashrightarrow X_n$ of degree $24n$.

Proof. Put $k=Q^{(n-3)/2}$ and $z=Q^n=Q^3k^2$. The useful cubic relation and its inverse are

$$ T^3=\frac{v_1+Q^n}{Q^{(n-3)/2}(1+v_1)},\qquad v_1=\frac{z-kT^3}{kT^3-1},\qquad v_2=\frac z{v_1}. \tag{7} $$

Substitute these into $\delta^2=(1-v_1)(1-v_2)/((1+v_1)(1+v_2))$. With $D_n(s)=-(z+z^{-1})$, substitution in (1) is a rational identity in $Q,T,k$ subject only to $z=Q^3k^2$. Thus it holds for every indicated $n$.

Differentiation of (5) gives

$$ \frac{\partial(Q,T)}{\partial(v_1,v_2)} =\frac{QT(v_1v_2-1)}{3n v_1v_2(1+v_1)(1+v_2)}, \tag{8} $$

which is nonzero. This proves dominance.

Let $\sigma$ generate the deck group. Let $\tau$ be the unit modulo $6n$ characterized by $\tau=-1\pmod n$ and $\tau=1\pmod6$. The diagonal element $(\sigma,\sigma^\tau)$ fixes $Q,T,\delta$ and has order $6n$. The extension generated by $Q,T$ over $\mathbf C(v_1,v_2)$ has degree $3n$: $Q$ gives an independent $n$th root, and the cubic has a valuation prime to three at a point over $v_2=-1$. Adjoining $\delta$ gives degree two, because its square is a nonsquare in the base and an odd-degree extension cannot make it a square. Thus $\mathbf C(Q,T,\delta)$ is exactly the fixed field of this diagonal group in the degree-$(6n)^2$ product field.

There is an actual automorphism

$$ J(v,q,u,b)=(v^{-1},q^{-1},u,ib/v), \tag{9} $$

with $J^2$ the quadratic deck involution and $J\sigma=\sigma^\tau J$. On the fixed field the factor swap acts by $(Q,T,\delta)\mapsto(Q,Q/T,\delta)$, while diagonal $J$ acts by $(Q,T,\delta)\mapsto(Q^{-1},T/Q,-\delta)$. These give four distinct automorphisms fixing (6). Conversely $Q$ satisfies a quadratic equation over $s$, $T$ satisfies a quadratic equation over $Q,x$, and $\delta$ is recovered from $Q,y$. The remaining degree is exactly four. ∎

For later use we record the surface invariants. Set $h=\lceil n/3\rceil$. The finite singular fibers are one II, $f$ fibers IV, one I$_2$, and $f$ fibers I$_4$. Indeed $D_n(s)-2$ has a simple root at $2$ and $f$ double roots, and $D_n(s)+2$ has the analogous pattern at $-2$. The discriminant is

$$-108(D_n(s)-2)^2(D_n(s)+2)^2.$$

At infinity write $n=6k+1$ or $6k-1$. The minimal coefficient/discriminant valuations are $(2k+3,4,8)$ in the first case, and $(2k+1,2,4)$ in the second, giving IV* and IV respectively. These calculations give $\chi(\mathcal O_{X_n})=h$, $q(X_n)=0$, $p_g=h-1$, and $b_2=12h-2$; see the standard elliptic-surface formulas in [18]. A basis of holomorphic two-forms is

$$s^j\eta,\quad 0\leq j\leq h-2,\qquad \eta=ds\wedge dx/y.$$

3. All transcendental classes have primitive image

Define rational one-forms on $C_n$, for an integer $e$, by

$$F_e=\frac{u^2q^e(1+v)}b\,dv, \qquad G_e=\frac{uq^e(1+v)}b\,dv, \qquad W_e=F_e\boxtimes G_e-G_e\boxtimes F_e.$$

Proposition 4. Every holomorphic two-form on $X_n$ pulls back to the primitive tensor summand, and

$$\phi_n^*T(X_n)\subset H^1(A_n,\mathbf Q)^{\otimes2}.$$

Proof. Direct differentiation using (8) gives

$$ \phi_n^*\eta=\frac{\sqrt2}{3n}(Q^2-1)(T-Q/T)Q^{-(n+3)/2} \frac{dv_1\,dv_2}{b_1b_2} =\frac{\sqrt2}{3n}(W_{1-n}-W_{-1-n}). \tag{10} $$

The second equality follows from (4), including the reciprocal terms; for example $u^{-2}q/(1+v)=uq^{1-n}(1+v)$. Since $s=-(Q+Q^{-1})$, multiplication yields

$$ \phi_n^*(s^j\eta)=\frac{(-1)^j\sqrt2}{3n} \sum_{\ell=0}^{j+1} \left[\binom j\ell-\binom j{\ell-1}\right] W_{1-n+j-2\ell}. \tag{11} $$

Binomial coefficients outside $0,\ldots,j$ mean zero. Both factors of $W_e$ have primitive quadratic and cubic characters. Their $n$-character is primitive exactly when $n\nmid e$. Here $j<n/3$, so the exponents lie strictly between $-2n$ and $0$. The only possible nonprimitive exponent is $-n$, occurring when $j$ is odd and $\ell=(j+1)/2$. Its coefficient is zero, since the two central binomial coefficients agree. Every surviving term therefore has both character labels primitive $6n$.

These identities are equalities of rational differentials. After resolving the map, their sum is holomorphic. Applying the algebraic deck projectors shows that its character components are holomorphic tensors, so individual choices of a differential basis are unnecessary for the projector conclusion.

The exceptional classes of a resolution are algebraic, and the $H^2\otimes H^0$ and $H^0\otimes H^2$ summands of $C_n^2$ are Tate. Thus the transcendental pullback lies in $H^1(C_n)^{\otimes2}$. Its complementary projection under $1-e_{6n}^{\otimes2}$ kills the entire $(2,0)$ and $(0,2)$ parts. The image would be a rational Hodge structure of type $(1,1)$. Polarizable semisimplicity and Lefschetz $(1,1)$ imply that $T(X_n)$ has no such nonzero quotient. Therefore this complementary projection is zero. This argument does not use K3 simplicity. ∎

4. The known algebra and the full Hodge group

Put

$$K=\mathbf Q(\zeta_{6n}),\qquad F=\mathbf Q(\zeta_n+\zeta_n^{-1}), \qquad L=F(\zeta_3)=K^\tau.$$

On $H=H^1(A_n,\mathbf Q)$ the actual automorphisms give the central simple algebra

$$D=(K/L,\tau,-1)=K\oplus KJ,\qquad J^2=-1,\quad Ja=\tau(a)J.$$

It is faithful because this cyclic algebra is simple and its unit acts nontrivially. The curve polarization preserves $D$; Rosati acts as complex conjugation on $L$. Every primitive cyclic character has total cohomological multiplicity two, so $H$ has dimension four over $L$. For the connected symplectic centralizer $G$,

$$G_{\mathbf C}\simeq\operatorname{GL}_2^f.$$

At an embedding of $L$, the four-dimensional representation is the tensor product of the two-dimensional simple $D$-module with a two-dimensional multiplicity module. Complex-conjugate embeddings pair these modules dually.

4.1. A signature pattern with no nontrivial stabilizer

The real factors are indexed by $(\mathbf Z/n\mathbf Z)^*/\{\pm1\}$. With representatives $1\leq a\leq(n-1)/2$, the compact factors are precisely

$$S=\{a:n/3<a<n/2\}.$$

For completeness, the cyclic-cover character formula is $h^{1,0}_j=-1+\sum_r\{ja_r/(6n)\}$, with the four branch weights above. Holding the cubic type fixed, the variable fractional parts are $\{2/3+a/n\}$ and $\{2/3-a/n\}$, or their conjugate $1/3$ counterparts. The remaining parts are $1/2$ and $1/6$, or $5/6$. The multiplicity is zero or two precisely in the middle third; otherwise it is one. Hence the corresponding multiplicity-module unitary group is compact exactly on $S$.

Lemma 5. The multiplicative stabilizer of $S$ modulo sign is trivial.

Proof. The sum-of-squares formula gives

$$ \sum_{a\in S}a^2= \begin{cases}1/81&n=1\pmod6,\\-1/81&n=-1\pmod6\end{cases} \quad\text{in }\mathbf F_n. \tag{12} $$

If $n=6k+1$, use $S=\{2k+1,\ldots,3k\}$ and substitute $k=-1/6$ modulo $n$ into the two prefix sums. If $n=6k-1$, use $S=\{2k,\ldots,3k-1\}$ and substitute $k=1/6$. In each case the upper prefix sum vanishes and the lower one gives (12). If multiplication by $a_0$ preserves $S$ modulo sign, then $(a_0^2-1)\sum_S a^2=0$. Since the sum is nonzero, $a_0=\pm1$, the identity modulo sign. ∎

Suppose now that $A_n$ is non-CM. Its connected Hodge group has nonzero derived Lie algebra. A semisimple subalgebra projects to each $\mathfrak{sl}_2$ as zero or all of $\mathfrak{sl}_2$; Galois transitivity makes every projection nonzero. A semisimple subdirect Lie algebra is a product of diagonal blocks. The block partition is Galois-stable. Since the action on the factors is regular cyclic, its blocks are the cosets of a subgroup of $(\mathbf Z/n)^*/\{\pm1\}$.

Factors in one block must have the same real form: a real diagonal identification cannot identify compact $\mathfrak{su}_2$ with split $\mathfrak{sl}_2(\mathbf R)$. Thus $S$ is a union of these cosets. Lemma 5 forces their subgroup to be trivial. We conclude

$$\operatorname{Hg}(A_n)^{\rm der}=G^{\rm der}.$$

4.2. The full center

Choose one fixed $\zeta_3$ type among embeddings of $L$. The determinant Hodge-circle vector is minus twice the indicator of the middle third modulo sign. Its rational Galois span fills the $f$-dimensional central quotient exactly when every even-character Fourier coefficient of that indicator is nonzero.

For the trivial character this is immediate. Let $\chi$ be a nontrivial even character modulo $n$ and let $\epsilon$ be the odd quadratic character modulo $3$. Then $\psi=\chi\epsilon$ is primitive and odd of conductor $3n$. Grouping the three lifts of each nonzero residue modulo $n$ gives

$$ B_{1,\psi}=\frac1{3n}\sum_{k=1}^{3n}k\psi(k) =\epsilon(n)\sum_{\substack{1\leq a<n\\a\equiv-n\ (3)}}\chi(a) =\epsilon(n)\chi(3)\sum_{n/3<b<2n/3}\chi(b). \tag{13} $$

The constant term in the first regrouping vanishes because $\sum\chi=0$; the second equality uses $a=3b-n$. Standard Dirichlet nonvanishing gives $B_{1,\psi}\ne0$: $L(0,\psi)=-B_{1,\psi}$, and the functional equation relates it by a nonzero Gauss factor to $L(1,\bar\psi)\ne0$ [19, equations 25.15.5 and 25.15.9]. Thus all the required Fourier coefficients are nonzero.

The Hodge group consequently surjects onto the central quotient of $G$. Together with the full derived group, and connectedness, this proves $\operatorname{Hg}(A_n)=G$.

4.3. Divisor generation and transfer

Over $\mathbf C$, $H$ is a sum of two copies of the standard and two copies of its dual for each $\operatorname{GL}_2$ factor. The first fundamental theorem for the general linear group says that invariant tensors are generated by evaluation pairings. Alternating gives divisor generation in the exterior cohomology of every power of $A_n$, and faithful extension of scalars gives the rational statement. Each invariant pair tensor is a rational $(1,1)$ class on $C_n^2$, hence an actual divisor class.

Pull a Hodge tensor from $T(X_n)^{\otimes m}$ through $\phi_n^{\times m}$. Proposition 4 places it in $H^{\otimes2m}$, where products of these pair divisors represent it. Push the resulting cycle forward and divide by $(24n)^m$. Graph correspondences justify the same calculation for the rational maps; exceptional contributions disappear under the transcendental projectors. Since $q(X_n)=0$, all remaining Kunneth factors of $X_n^m$ are generated by divisor and point classes. This proves Theorem 1(1)–(3), except for the rank assertion, established in §7.

5. Specialization centralizers and equivariant reconstruction

The following lemma allows reduction centers of different degrees.

Lemma 6. Let $D$ be a central simple algebra of degree $d$ over $L$, acting by endomorphisms on an abelian variety $A$ over a number field. Suppose that at a good geometric reduction the endomorphism algebra $B$ is central simple over $K$. Let the subalgebra $M$ generated by the actual copies of $L,K$ in $B$ be a field, and suppose

$$\deg_K B=d[M:K].$$

Then $C_B(D)=M$. If two such reductions yield nonisomorphic quadratic extensions $M_1,M_2$ of $L$, then $C_{\operatorname{End}^0(A_{\mathbf C})}(D)=L$.

Proof. The centralizer of $M$ in $B$ has degree $d$ over $M$. It contains $D\otimes_LM$, a central simple algebra of the same degree, and hence equals it. An element commuting with $D$ commutes with $L$ and with the central field $K$, so belongs to this algebra; its centralizer of $D$ is exactly $M$. Specialization injects the characteristic-zero endomorphism algebra, and hence its $D$-centralizer, into both reduction algebras [20, Chapter I, Remark 10.17]. A finite-dimensional $L$-algebra injecting into a field is a field. Its degree is at most two; degree two would identify it with both $M_i$, contrary to their inequivalence. ∎

For our abelian varieties, this equality forces non-CM directly. Indeed $\operatorname{End}^0(A_{\mathbf C})=\operatorname{End}_{\operatorname{Hg}(A)}(H^1(A))$. After extending scalars, the two-dimensional multiplicity modules therefore have scalar commutants. They are semisimple representations of a reductive group, so irreducible. A connected reductive subgroup of $\operatorname{GL}_2$ acting irreducibly contains $\operatorname{SL}_2$. In particular the derived Hodge group is nonzero.

Here is the reconstruction principle used to verify the lemma's hypotheses. Suppose a semisimple endomorphism $\Pi$ of a good reduction commutes with $D$ and has polarization multiplier $q$. Its reduced polynomial in the degree-two Tate-module commutant is

$$z^2-a z+b.$$

If conjugation by prime-field Frobenius fixes $\Pi$ and acts on $L$ with fixed field $E$, then $a,b$ lie initially in $E\otimes\mathbf Q_\ell$. Equivariant traces against elements of $E$ determine $a$ through the nondegenerate trace pairing. Integral traces can thereby recover $a\in E$ independently of $\ell$.

Polarization pairs the block at an embedding with its complex-conjugate block. Their reduced eigenvalues are $\lambda_1,\lambda_2$ and $q/\lambda_1,q/\lambda_2$. Hence

$$\bar a=qa/b,\qquad b\bar b=q^2,\qquad b=qa/\bar a\quad(a\ne0). \tag{14}

$$ When the coefficient field is $\mathbf Q$, a nonzero trace forces $b=q$. With a proper CM coefficient field this simplification need not hold. Finally, suppose actual Frobenius has characteristic polynomial $P^m$, with $P$ irreducible, and no ratio of distinct roots of $P$ is a root of unity. The eigenspace decomposition and its centralizer are then unchanged by every positive Frobenius power. Tate's theorem [9; 20, Chapter II, Theorem 2.9] implies that all geometric endomorphisms are already defined over this finite field. Their algebra is central simple of degree $m$ over $\mathbf Q[\mathrm{Frob}]$. Distinct valuations of all roots at the residue characteristic are a sufficient test for the root-ratio condition. A separate division-index guess is unnecessary. ## 6. Exact arithmetic applications ### 6.1. Degree seven Here $\dim A_7=12$, $[L:\mathbf Q]=6$, and $L/\mathbf Q$ is cyclic. At $p=5,11$ the curve has good reduction: the cover is tame and its four branch sections remain distinct. All $D$-endomorphisms are defined over $\mathbf F_{p^6}$. Prime-field Frobenius normalizes $D$ and acts transitively on the six embeddings of $L$. Independent counts of every normalized cyclic quotient, followed by Mobius inversion, give primitive traces $-1560$ and $-2376$. The reduced coefficients of $\Phi_{p^6}$ are fixed by the full cyclic action. Thus $a$ is rational and is the total trace divided by $12$, giving $a=-130,-198$. Equation (14) gives $b=p^6$. The complete polynomials are $$

(z^2+130z+5^6)^{12},\qquad(z^2+198z+11^6)^{12}. \tag{15}

$$ Their quadratic fields are $K_5=\mathbf Q(\sqrt{-114})$ and $K_{11}=\mathbf Q(\sqrt{-910})$. In both cases $v_p(a)=1$, so the two Frobenius root valuations are $1,5$. The geometric endomorphism algebras have degree $12$ over these fields. The only quadratic subfield of $L$ is $\mathbf Q(\sqrt{-3})$, so $L\cap K_p=\mathbf Q$ and $[LK_p:K_p]=6$. Lemma 6 gives $C_{B_p}(D)=LK_p$. The two quadratic extensions of $L$ are different: their squareclass ratio $114/910$ is a positive nonsquare, and its square root would give a real quadratic subfield of $L$. Lemma 6 proves $C_{\operatorname{End}^0(A_7)}(D)=L$, hence non-CM. ### 6.2. Degree eleven: the full-orbit reduction Now $\dim A_{11}=20$, $[L:\mathbf Q]=10$, and $L=\mathbf Q(\zeta_{11}+\zeta_{11}^{-1},\zeta_3)$ is cyclic. At $p=5$ take $q=5^{10}=9765625$. The same full-orbit argument applies. The primitive trace is $109000$, so $a=5450$ and the complete polynomial is $$P_5(z)=(z^2-5450z+9765625)^{20}.$$ Its quadratic discriminant is $-26\cdot600^2$; the two root valuations at $5$ are $2,8$. Consequently the geometric endomorphism algebra $B_5$ is central simple of degree $20$ over $K_5=\mathbf Q(\sqrt{-26})$, and Lemma 6 gives $$C_{B_5}(D)=L(\sqrt{-26}).$$ ### 6.3. Degree eleven: the twisted reduction At $p=13$ set $q=13^5=371293$. The automorphism $J$ is defined over $\mathbf F_{13}$, since $i\in\mathbf F_{13}$. If $F_{13}$ denotes prime-field Frobenius, then $F_{13}J=JF_{13}$. Its action on the deck algebra is $\sigma\mapsto\sigma^{13}$, whose fifth power is $\tau=43\pmod{66}$. Thus $$\Pi=J^{-1}F_{13}^5$$ commutes with $D$, is semisimple, has polarization multiplier $q$, and satisfies $$\Pi^2=-F_{13}^{10}.$$ Conjugation by $F_{13}$ fixes $\Pi$ and has fixed field $E=\mathbf Q(\zeta_3)$ on $L$. The reduced coefficients therefore belong to $E\otimes\mathbf Q_\ell$. Count the $J$-twist and its two cubic twists. Mobius inversion gives the primitive traces $$-10500,\qquad22500,\qquad-12000.$$ Up to simultaneously conjugating the cubic character, these equal $10\operatorname{Tr}_{E/\mathbf Q}(a\zeta_3^k)$, where the factor ten is twice the degree of the real quintic field. For $\zeta^2+\zeta+1=0$ they determine $$a=-1100-1150\zeta.$$ Equation (14) then gives the actual norm $$b=q\,a/\bar a=-32955+353717\zeta,$$ which is not $q$. Exact arithmetic gives $$a^2-4b=322(16-14\zeta)^2.$$ Since actual $q^2$-Frobenius is $-\Pi^2$, its reduced polynomial is $z^2+(a^2-2b)z+b^2$. Taking the norm to $\mathbf Q$ gives $$

P_{13}(z)=z^4-593246z^3+175904542827z^2 -81783998855451854z+19004963774880799438801. \tag{16}

$$ This polynomial is irreducible, and the full characteristic polynomial is $P_{13}^{10}$. Its four root valuations at $13$ are $0,4,6,10$. Therefore the geometric endomorphism algebra $B_{13}$ is central simple of degree ten over $$K_{13}=\mathbf Q(\zeta_3,\sqrt{322}).$$ To identify the center, the roots lie in this quartic field by the discriminant identity, and irreducibility of (16) shows that actual Frobenius generates the whole field. The actual copy of $E$ in $L$ also lies in $K_{13}$: the reduced quadratic equation puts $E[F_{13}^{10}]$ in dimension at most four over $\mathbf Q$, while $\mathbf Q[F_{13}^{10}]$ already has dimension four. Thus the two copies agree. Since $L\cap K_{13}=E$ and $[LK_{13}:K_{13}]=5$, Lemma 6 gives $$C_{B_{13}}(D)=L(\sqrt{322}).$$ The squareclass ratio $322/(-26)=-161/13$ is neither a rational square nor $-3$ times one. As $E$ is the only quadratic subfield of $L$, these two quadratic extensions of $L$ are nonisomorphic. Lemma 6 proves the non-CM assertion for $A_{11}$ and completes Theorem 2. ### 6.4. Geometric definition and independent control of the twist counts The point equations for the descent are $$

v^q=v^{-1},\qquad q_n^q=q_n^{-1},\qquad u^q=\zeta_3^k u,\qquad b^q=ib/v. \tag{17}

$$ Here $q_n$ denotes the Kummer coordinate, to distinguish it from the field cardinality. These count fixed points of a cubic multiple of $J^{-1}\Phi_q$. The sign of $k$ chooses an orientation of the cubic character. Since these automorphisms are defined over $\mathbf F_q$ and have finite order, they define continuous descent cocycles. Thus the counts are point counts of genuine smooth projective twists and compute the required H1 traces by Lefschetz. Choose a nonsquare $d\in\mathbf F_q$, let $c^2=d$, and choose $i=d^{(q-1)/4}$. Parametrize $v^q=v^{-1}$ by $$v=\frac{t+c}{t-c},\qquad t\in\mathbf P^1(\mathbf F_q).$$ For finite nonzero $t$, the three independent root counts are $n$ or zero according as $v^{(q+1)/n}=1$; three or zero according as $$\left(\frac{t^2-d}{4t^2}\right)^{(q-1)/3}=\zeta_3^k;$$ and two or zero according as $t$ is a square. For the last claim put $b'=b(t-c)$, so $b'^2=-4tc$ and $b'^q=ib'$. Because $q=1\pmod4$, the ratio $b'^q/b'$ is $i$ times the quadratic character of $t$. At $t=0$, equivalently $v=-1$, the cubic and quadratic covers are totally ramified, contributing one point each; the $n$-cover contributes $n$. At $t=\infty$, equivalently $v=1$, the quadratic cover is ramified and the cubic residue is $1/4$, while the $n$-cover again contributes $n$. The points above $v=0,\infty$ are exchanged and do not contribute fixed points. Products of selected root counts give all eight cyclic quotients. These are normalization counts, including the branch fibers. The primary implementation uses finite-field logarithm tables. Independent implementations use polynomial-basis arithmetic and Frobenius orbits, with no logarithm table. They agree on all $24$ twisted quotient counts for $n=11$, and on all eight ordinary quotient counts at $5^{10}$. The independent twisted counter uses $d=2$ in the prime field and adjusts the quadratic sign to the same chosen $i$. Before the degree-eleven calculation, a degree-five control over $q=13^2$ gave primitive twist traces $-36,144,-108$, hence $a=-15-21\zeta$ and $b=-104+91\zeta$. Squaring the twisted operator predicts the ordinary primitive trace $92$ over $13^4$, which agrees with a direct ordinary count. A separate enumeration in $\mathbf F_{13^4}$ agrees with all $24$ twisted quotient counts. This tests descent conventions, branch corrections, and the nonrational norm together. These finite checks support the written reconstruction; they do not replace the centralizer and specialization proofs. ## 7. The adjoint image and mixed products Put $\psi_n=\phi_n\circ(\mathrm{id},J)$. The following construction extends the adjoint realization of [1, §9.1]. For $\alpha\in T(X_n)$ let $a=\phi_n^*\alpha$ and $b=(\mathrm{id}\otimes J^*)a$. The map is invariant under geometric factor swap, whose action on two odd cohomological degrees is minus the ordinary tensor flip $P$. Hence $Pa=-a$. The other invariances give $$

(J^\otimes J^)a=a,\qquad (\sigma^\otimes(\sigma^)^\tau)a=a. $$

Using $(J^*)^2=-1$ gives

$$Pb=b,\qquad(J^*\otimes J^*)b=b,\qquad (\sigma^*\otimes\sigma^*)b=b.$$

Polarization identifies a symmetric tensor with an infinitesimal symplectic endomorphism. Diagonal invariance says it commutes with $D$. We obtain an injective Hodge map

$$T(X_n)(1)\hookrightarrow\mathfrak g.$$

The center has type $(0,0)$, so its projection vanishes by the absence of a Tate quotient. Under the established full Hodge group $G$, the derived adjoint representation is rationally irreducible: its $f$ simple complex summands are permuted transitively by Galois. The image is nonzero because $p_g(X_n)>0$. Consequently

$$T(X_n)(1)\simeq\mathfrak g^{\rm der}.$$

This proves the rank assertion in (3). It also identifies the Hodge group $H_n$ of the surface transcendental structure with the adjoint image of $G$. The group $H_n$ is connected, adjoint, and $\mathbf Q$-simple, with dimension $3(n-1)/2$.

For distinct prime degrees these dimensions are distinct. The Hodge group of a direct sum of their transcendental structures is a connected subdirect subgroup of $\prod H_n$. Its Lie algebra has zero center, since a central element projects to the center of every target. It is therefore semisimple. A proper semisimple subdirect product would identify isomorphic simple rational quotients, by Goursat or the decomposition into simple ideals. The dimensions exclude such an identification. Thus the Hodge group is the full product.

Invariants under a product group are tensor products of the individual invariant spaces. The latter are algebraic by the self-power result, so their external products are algebraic as well. The algebraic point and Neron–Severi projectors supply the remaining cohomology. This proves Theorem 1(4). It makes no assertion that the full unitary Hodge groups of the mixed source abelian varieties have independent central tori; their centers act trivially on these adjoint surface tensors.

8. A uniform consequence conditional on the CM case of HC

The non-CM condition controls the source endomorphisms and divisor generation. For HC on the surfaces themselves, the alternative can instead be supplied by a general CM theorem.

Corollary 7. Assume the rational Hodge conjecture for all complex CM abelian varieties. Then the rational Hodge conjecture holds on every finite product of the surfaces $X_n$, for all primes $n>3$, with arbitrary repetitions.

Proof. For each n, either $A_n$ is non-CM and Theorem 1 applies, or $A_n$ is CM. In the latter case its powers satisfy HC by the assumed theorem, and Proposition 4 together with the actual curve-product correspondence transfers the classes to $X_n$. For mixed products, group all CM source factors together; their product is again CM, so its tensors are algebraic by the same assumption. The remaining surface Hodge groups are the pairwise nonisomorphic Q-simple adjoint groups of §7. A connected subdirect product of these semisimple groups and the combined CM torus is the full product: there is no nontrivial common simple quotient between a semisimple adjoint group and a torus. Its invariant tensors therefore factor into the already algebraic non-CM tensors and the combined CM tensors. ∎

Theorem 1.1 of the October 6, 2026 CM preprint [12] claims precisely the input of Corollary 7, including products and powers. Thus accepting that theorem removes the non-CM hypothesis from the surface-product HC statement, while it does not prove uniform non-CM, divisor generation in the CM case, or the Picard-rank formula there. We have inspected relevant parts of that source in earlier work but have not independently verified its entire proof. Corollary 7 is recorded with its explicit dependency; Theorems 1–2 and the unconditional $X_5,X_7,X_{11}$ applications do not use [12].

9. An explicit fourfold factor and mixed Eisenstein cycles

Let $E_0$ be an elliptic curve with multiplication by $K_0=\mathbf Q(\sqrt{-3})$, for example $y^2=x^3+1$, with the action chosen below.

Proposition 8. There is an explicit rational algebraic idempotent $e\in\operatorname{End}^0(A_5)$ such that $B=eA_5$ is a simple abelian fourfold with $\operatorname{End}^0(B)=\mathbf Q(\sqrt5,\sqrt{-3})$ and $A_5\sim B^2$. The rational Hodge conjecture holds on every $B^a\times E_0^b$, and hence on every $A_5^a\times E_0^b$. It also holds on every

$$B^a\times E_0^b\times X_5^c\times X_7^d\times X_{11}^e.$$

Here all exponents are nonnegative integers.

The actual factor. Write $K=\mathbf Q(\zeta_{15})$, $L=\mathbf Q(\sqrt5,\sqrt{-3})$, and $D=(K/L,\tau,-1)$ as in §4. Put

$$r=\frac{1+\sqrt5}{2},\qquad c=\sqrt{-3},\qquad t=\zeta_5-\zeta_5^{-1},\qquad \alpha=\frac{1+r+ct}{2}.$$

The identities $r^2=r+1$, $t^2=-(2+r)$ and $\tau(t)=-t$ give

$$\alpha\tau(\alpha)=\frac{(1+r)^2-3(2+r)}4=-1.$$

Consequently $U=\alpha J$ satisfies $U^2=1$, and $e=(1+U)/2$ is an actual rational endomorphism idempotent. Since $tU=-Ut$, multiplication by $t$ exchanges its two images. Thus $B=eA_5$ has dimension four and

$$(b_1,b_2)\longmapsto b_1+t b_2$$

gives an isogeny $B^2\to A_5$. In particular $D\simeq M_2(L)$. The full Hodge group of $A_5$ gives $\operatorname{End}^0(A_5)=D$, hence $\operatorname{End}^0(B)=L$, which is a field. Thus $B$ is simple. At one $K_0$ embedding its two $L$ multiplicity modules $M_1,M_2$ have signatures $(1,1)$ and $(0,2)$; its quadratic signature is $(1,3)$. Choose the action on $E_0$ with signature $(1,0)$ at this embedding.

The supplied cycle. The sixfold $Z=B\times E_0^2$ has signature $(3,3)$. Choose a compatible rational polarization on $B$, with Hermitian determinant represented by $d<0$. Schoen [10, §§11–13] proves algebraicity of the full rational Weil plane on every fiber of one rank-six universal family, whose determinant invariant is a negative class $f$. Choose a negative rational representative $f$. Giving the two elliptic factors positive Hermitian coefficients $1$ and $f/d$ makes $Z$ have this determinant. Rational coefficients can be cleared simultaneously; scaling a rank-six form multiplies its determinant by a sixth power, a norm. By [10, §§6–7], equal signature and discriminant identify the rational Hermitian spaces and the corresponding universal families up to isogeny. The all-fiber specialization in [10, §12] includes our decomposable sixfold. Therefore

$$W_Z=\bigwedge_{K_0}^{6}H^1(Z,\mathbf Q)\subset H^6(Z,\mathbf Q)$$

is a two-dimensional space of algebraic classes. Alternatively take $f=-1$ and apply [11, Theorem 1.5.1]. The two complex Weil lines are

$$w=\det(M_1)\otimes\det(M_2)\otimes e_0^{\otimes2} \quad\text{and its conjugate},$$

where $e_0$ is the selected elliptic character. The displayed tensor is a top exterior product of the indicated summands, so the projection is nonzero.

All powers. Up to Tate twists, the mixed first cohomology is the sum of $M_1,M_2$, their duals, and $e_0,e_0^\vee$. Its derived group is $\operatorname{SL}_2^2$. On the determinant quotient the Galois-conjugate Hodge-circle weights are $(0,-2,1)$ and $(-2,0,1)$. Their annihilator is generated by the primitive character $(1,1,2)$, and hence

$$\operatorname{Hg}(B\times E_0)_{\mathbf C} =\{(g_1,g_2,z):\det(g_1)\det(g_2)z^2=1\}.$$

This is also the group computed in [7, §5.11, Case 2]. For a fixed labelled tensor word, let $p_i,q_i$ count its $M_i,M_i^\vee$ slots and let $u,v$ count its $e_0,e_0^\vee$ slots. The $\operatorname{SL}_2$ tensor invariant theorem gives evaluations and alternating two-tensors. If invariant tensors exist, $a_i=(p_i-q_i)/2$ are integers. Central invariance is precisely

$$(a_1,a_2,u-v)=k(1,1,2),\qquad k\in\mathbf Z.$$

An alternating standard tensor times an alternating dual tensor equals a difference of two cross evaluations. Cancelling such opposite pairs leaves $|k|$ determinants in each $M_i$ and $2|k|$ elliptic slots of the same sign. Thus every invariant is a linear combination of products of evaluation tensors and copies of $w$ or its conjugate, in arbitrary slots.

Here tensor placement is algebraic. Pull $W_Z$ back along the map $B^4\times E_0^2\to Z$ adding the four $B$ coordinates, and apply the $H^1$ Künneth projector on each factor. On an abelian variety these projectors are rational polynomials in multiplication maps. After extending scalars, the $L$ and $K_0$ spectral projectors isolate the ordered slots $M_1,M_1,M_2,M_2,e_0,e_0$ and give a nonzero multiple of $w$. The spectral projectors are complex linear combinations of rational algebraic endomorphisms. Evaluation tensors are corresponding components of Poincaré divisors. External products and permutations supply all labelled tensor invariants; diagonal pullback returns the exterior cohomology of any $B^aE_0^b$. Faithful scalar extension then shows that its rational Hodge classes are spanned by rational algebraic cycle classes. No rationality of an individual complex eigentensor is required.

The isogeny gives the assertion for $A_5$. Proposition 4 and the actual curve correspondence transfer the result to mixed powers with $X_5$. Finally the only simple rational adjoint quotient of $\operatorname{Hg}(B\times E_0)$ has dimension six, while the surface groups for $X_7,X_{11}$ have dimensions nine and fifteen. The subdirect-product argument of §7 therefore gives independence from those two groups, proving the last assertion. ∎

The exceptional determinant here belongs to the abelian source. The surface-only product $X_5^a\times X_7^b\times X_{11}^c\times E_0^d$ also follows directly from §7, because its semisimple adjoint groups have no common quotient with the elliptic torus. Likewise [7, Theorem 0.2(iii)] already states that $B\times E_0$ itself is divisor-generated: the extra class first occurs on $B\times E_0^2$. Proposition 8 is an explicit cycle calculation for the source of our map, not a new general low-dimensional classification.

The spin half twist of the two multiplicity modules is a level-one Hodge substructure of $H^3(B\times E_0)(1)$. Its algebraic projector and ordinary HC do not alone supply curve support. The next section constructs that support through a threefold quotient of the actual source curves.

10. Curve support and the Kuga–Satake correspondence in degree five

The source of §9 also contains the spin representation required for the original degree-five surface. Ordinary HC alone did not give its geometric realization. The following quotient supplies it.

Theorem 9. There is a simple abelian fourfold $A_{\mathrm{sp}}$ and an algebraic correspondence identifying $H^1(A_{\mathrm{sp}},\mathbf Q)$ with a rank-eight summand of $H^3(C_5^2\times E_0,\mathbf Q)(1)$. Moreover

$$\operatorname{KS}'(X_5)\sim A_{\mathrm{sp}}^4,$$

and the canonical Kuga–Satake correspondence for $X_5$ is algebraic. The rational Hodge conjecture holds on every mixed product of $A_{\mathrm{sp}},A_5,E_0,X_5,X_7,X_{11}$. Here $\operatorname{KS}'$ denotes the construction from the transcendental Hodge structure; the statement is equivalent to the usual primitive-cohomology version, as in [16, Remark 3.4].

The source curve product gives an elliptic model with trivial canonical bundle. Crepant modification exposes a fixed divisor, whose ramification makes the quotient uniruled. Curve support then realizes the rank-eight spin module algebraically.

The upper row records the geometric source of curve support; the lower row records its use in the canonical Kuga–Satake comparison. The arrows are justified separately below.

10.1. An actual quotient and its elliptic model

Write $C=C_5$. Its functions $q,u,b$ and automorphism $J$ are as in (4) and (9). On $E_0:y_E^2=x_E^3+1$ take

$$\gamma(x_E,y_E)=(\zeta_3^2x_E,-y_E).$$

The character of $\gamma$ on its holomorphic differential is $+1$ modulo six. In the group $(\mathbf Z/30)^2\times\mathbf Z/6$ acting on $V=C^2\times E_0$, set

$$\mathcal H=\{(a,b,c):a+7b-5c=0\pmod{30}\}.$$

It has order $180$. The actual automorphism

$$\mathcal S(P_1,P_2,e)=(P_2,JP_1,e)$$

normalizes $\mathcal H$, since conjugation sends $(a,b,c)$ to $(b,19a,c)$ and multiplies the defining relation by $13$ modulo $30$. Indeed $b+7\cdot19a-5c\equiv13(a+7b-5c)\pmod{30}$. Also $\mathcal S^4=(\sigma^{15},\sigma^{15},1)\in\mathcal H$. Thus $\mathcal G=\langle\mathcal H,\mathcal S\rangle$ has order $720$.

The fixed field of $\mathcal H$ has generators

$$Q=q_1q_2^2,\qquad U=u_1u_2x_E,\qquad D=\frac{b_1b_2y_E}{(1+v_1)(1+v_2)}.$$

Over $\mathbf C(v_1,v_2,r)$, where $r=x_E^3$, their degrees are $5,3,2$, independently. Their valuations show that these Kummer degrees are full. Their product $30$ is the index of $\mathcal H$ in the full deck group, so these generators give the fixed field. Put

$$A=Q,\quad B=v_2/Q^2;\qquad v_1=A/B^2,\quad v_2=A^2B.$$

Then $\mathcal S(A,B)=(B,A^{-1})$, $\mathcal S(U)=U$, and $\mathcal S(D)=iD$. Cayley coordinates

$$a=\frac{A-1}{A+1},\qquad b=\frac{B-1}{B+1}$$

give $\mathcal S(a,b)=(b,-a)$. Define

$$\begin{aligned} P_1&=2b-a-ab^2,&P_2&=2a+b+a^2b,\\ Q_1&=1+b^2-2ab,&Q_2&=1+a^2+2ab,\\ P&=P_1P_2,&Q_0&=Q_1Q_2,\qquad R=(1-a^2)(1-b^2). \end{aligned}$$

Direct substitution gives

$$D^2=-\frac P{Q_0}\left(1+\frac{16Q_0^2U^3}{R^3}\right).$$

Consequently, with $u=U/R$, $x=-16PQ_0u$, and $y=-16PQ_0D$, a birational model of $V/\mathcal H$ is

$$W:\quad y^2=x^3-256P^3Q_0,\qquad \mathcal S(a,b,x,y)=(b,-a,-x,-iy). \tag{19}$$

10.2. Exceptional ramification proves uniruledness

Each of $P,Q_0$ has bidegree $(3,3)$ on $B_0=\mathbf P^1\times\mathbf P^1$. The global Weierstrass model (19) therefore has fundamental line bundle $\mathcal L=\mathcal O(2,2)=-K_{B_0}$. In the standard projective bundle construction, adjunction gives

$$\omega_W=\pi^*(\omega_{B_0}\otimes\mathcal L)=\mathcal O_W.$$

This is an equality of dualizing line bundles on a normal Gorenstein threefold. The zero section is smooth. In affine fiber charts the singular locus has codimension at least two, so the hypersurface is normal.

The curves $P_1=0$ and $P_2=0$ are smooth rational graphs of bidegrees $(1,2)$ and $(2,1)$. They meet transversely in five points, all affine: elimination gives

$$b(b^4+10b^2+5)=0,\qquad a=2b/(1+b^2).$$

Their intersection set is $\mathcal S$-invariant and includes the origin. Blow up these five points in the base, obtaining $\beta:B_1\to B_0$. At each center the Weierstrass coefficient has order six. Divide it by the corresponding exceptional sixth power and put

$$\mathcal L_1=\beta^*\mathcal L\otimes\mathcal O_{B_1}(-\textstyle\sum E_i)=-K_{B_1}.$$

The transformed Weierstrass model $W'$ is again normal, projective, Gorenstein, and has trivial dualizing sheaf. The automorphism lifts because the centers are invariant and the fiber-coordinate changes are equivariant.

For completeness, its remaining singularities are canonical. The four branch curves have no triple intersections. The pairs $P_1,Q_1$ and $P_2,Q_2$ have two ordinary tangencies each; all other pair types are transverse. Near a single $P$ branch a surface section is $D_4$; near a $Q/Q$ intersection it is $A_2$; near a transverse $P/Q$ intersection it is $E_6$. At a tangency the equation is analytically $y^2=x^3+s^3(s-t^2)$ times a unit in the final term, and the section $t=0$ is $E_6$. Above a blown-up $P/P$ intersection the generic exceptional fiber is smooth and its endpoints have $D_4$ sections. Thus these are compound Du Val singularities, which are canonical [21, §2.2]. The exact supplement checks the branch arrangement in all four projective charts. The uniruledness argument uses only normality and the trivial dualizing sheaf.

At the blowup of the origin write

$$a=\xi,\quad b=\xi\lambda,\quad x=\xi^2x_1,\quad y=\xi^3y_1.$$

Then

$$\mathcal S^2:(\xi,\lambda,x_1,y_1)\longmapsto(-\xi,\lambda,x_1,y_1). \tag{20}$$

The entire elliptic divisor over the exceptional curve is fixed pointwise. Its generic fiber is

$$y_1^2=x_1^3-256(2\lambda-1)^3(\lambda+2)^3.$$

In particular it is a nonzero ramification divisor, with generic ramification index two.

Let $Z$ be a smooth projective model of $W'/\langle\mathcal S\rangle$, and consider the generically finite rational map $f:W'\dashrightarrow Z$. Since $W'$ is normal and $Z$ proper, $f$ is a morphism off codimension two. Choose a general complete intersection curve $C'$ of two sufficiently ample divisors, avoiding this indeterminacy locus and the singular locus. On that open set the determinant of $df$ is a regular section of $K_{W'}\otimes f^*K_Z^{-1}$ whose zero divisor contains (20). Therefore

$$K_Z\cdot f_*C'=K_{W'}\cdot C'-\operatorname{Ram}(df)\cdot C'<0. \tag{21}$$

The image curves form a covering family on $Z$. The covering-curve criterion [22, Theorems 0.2–0.3] implies that $Z$ is uniruled. This argument exhibits ramification after modification; an assertion about the absence of divisorial fixed loci on the original product would not suffice.

10.3. The primitive spin representation survives

Let $\Pi$ be the product of the two primitive order-thirty projectors and the finite-group averaging projector for $\mathcal G$. It is an actual rational algebraic correspondence on $V$. Put

$$M=\Pi H^3(V,\mathbf Q).$$

On $C$, write $H_j$ for the eigenspace with deck eigenvalue $\zeta_{30}^{-j}$. These are denominator labels. In the triple $H^1$ summand, $\mathcal H$ invariance selects

$$H_j\otimes H_{7j}\otimes H^1(E_0)_{j\bmod6}.$$

For primitive $j$ this is four-dimensional, of types $(2,1)^2+(1,2)^2$, by the fractional-part formula of §4. The sign in the defining relation of $\mathcal H$ is essential to this statement. Pullback by $\mathcal S$ sends $j$ to $13j$, with orbits

$$(1,13,19,7),\qquad(11,23,29,17).$$

The fourth power is the identity on these tensors. Each orbit therefore contributes one four-dimensional invariant space. The geometric minus sign from interchanging odd factors changes the identifications, but not their invariant rank. Hence

$$\dim M=8,\qquad h^{2,1}(M)=h^{1,2}(M)=4. \tag{22}$$

No nonprimitive contribution is needed: the two nonprimitive labels $5,25$ each retain one mixed-type invariant, while their top and bottom types vanish under $\mathcal S$.

Choose $Z$ to resolve the finite quotient $Q=V/\mathcal G$, which is birational to the model above. A finite complex quotient is a rational homology manifold, and

$$H^*(Q,\mathbf Q)\simeq H^*(V,\mathbf Q)^{\mathcal G}.$$

Indeed local links are finite quotients of odd-dimensional spheres, with their rational cohomology computed by averaging; the complex linear action preserves orientation. The degree-one resolution $Z\to Q$ induces an injection on rational cohomology by Poincaré duality. On a common resolution of $V\times_Q Z$, pushforward to $V$ after pullback from $Z$ recovers the original invariant classes. Thus an actual graph correspondence maps $H^3(Z)$ onto a space containing $M$.

For a smooth uniruled threefold, $H^3$ is generated by Gysin correspondences from curves. To recall the argument, resolve a dominant rational map $S\times\mathbf P^1\dashrightarrow Z$, with $S$ a smooth projective surface. Its third cohomology consists of $H^3(S)$, $H^1(S)(-1)$, and the $H^1(-1)$ terms of blowup-center curves. An ample curve supplies $H^3(S)$ by weak Lefschetz and duality; the Albanese and Poincaré divisor supply $H^1(S)$ from a curve Jacobian. Degree-normalized pushforward gives the assertion for $Z$. This is also the curve-support mechanism used in Schoen [14, Remark 2.11]. Composing with the quotient graph and $\Pi$ gives an algebraic surjection

$$\Gamma:H^1(F)(-1)\twoheadrightarrow M \tag{23}$$

for a finite disjoint union of smooth projective curves $F$.

There is an algebraic inverse on a Jacobian factor. The triple $H^1$ summand of $V$ is primitive for a product ample class; its cup form, with the usual sign, is a polarization. The primitive and group projectors are self-adjoint for it. Thus the transpose $\Gamma^\dagger$ is a polarization adjoint up to the fixed Tate sign, and $\Gamma\Gamma^\dagger$ is invertible on $M$. Its inverse is a rational polynomial in this endomorphism. Consequently

$$j=\Gamma^\dagger(\Gamma\Gamma^\dagger)^{-1},\qquad \Gamma j=1_M$$

is algebraic. The idempotent $j\Gamma$ on $H^1(F)$ cuts out an abelian fourfold $A_{\mathrm{sp}}$ in $J(F)$ up to isogeny. Abel–Jacobi maps and Poincaré divisors identify $H^1(A_{\mathrm{sp}})$ algebraically with $M(1)$. No inverse Lefschetz correspondence is assumed.

10.4. Identifying the canonical representation and transferring cycles

With the notation of §9,

$$M_{\mathbf C}=(M_1\otimes M_2\otimes e_0)\oplus \overline{(M_1\otimes M_2\otimes e_0)}.$$

The scalar of the connected center of $G=\operatorname{Hg}(A_5\times E_0)$ on the first summand is $s_1s_2z$. Its square is one by $\det(g_1)\det(g_2)z^2=1$, so it is one on the connected center. Thus the Hodge group on $M$ is the image of the derived group, whose representation has highest weight $(1,1)$ for $\operatorname{SL}_2^2$.

The actual adjoint realization of §7 identifies the simply connected derived cover with $\operatorname{Res}_{F_0/\mathbf Q}\operatorname{Spin}(W_0)$, where $F_0=\mathbf Q(\sqrt5)$ and $W_0$ is the ternary polarization form of $T(X_5)$. For real quadratic multiplication the spin corestriction is $M_2(D_0)$ with $D_0$ a definite quaternion algebra [15, Corollary 3.7.1]. Its unique simple rational module has rank eight and complexification two copies of the tensor of the two standard modules. The rational representation $M$ is this module. In particular it is irreducible, and $A_{\mathrm{sp}}$ is simple. The elliptic automorphism $\gamma$ commutes with the projectors, so its two primitive sixth-root eigenvalues give $\mathbf Q(\sqrt{-3})\subset\operatorname{End}^0(A_{\mathrm{sp}})$. Abdulali [23, Corollary 4.3] therefore already proves generalized HC for all self-powers of this type-III fourfold. The transfer below additionally controls the mixed source and surface products needed for the specified KS correspondence.

This comparison respects the Hodge structure. The Hodge circle on the adjoint is the one supplied by the actual embedding of $T(X_5)$. Its lift acts with weights $+1,-1$ at the noncompact spin factor and trivially at the compact one. The connected source center acts trivially on $M$, so no central Hodge character remains. Equivalently a lift through the finite central spin isogeny, when it exists, is unique on the connected circle. After the Tate twist in (22), this is precisely the weight-one Kuga–Satake circle. The decomposition theorem [15, Theorem 3.3.1(ii)] therefore gives

$$H^1(\operatorname{KS}'(X_5))\simeq M(1)^{\oplus4},$$

and hence $\operatorname{KS}'(X_5)\sim A_{\mathrm{sp}}^4$.

It remains to prove algebraicity of the specified KS map, rather than only identify its representation. Proposition 8 gives HC on every power of $A_5\times E_0$. Every Hodge tensor on $A_{\mathrm{sp}}$ maps under (23) to a Hodge tensor in powers of $M$, an algebraic summand of tensor powers of $H^1(A_5)$ and $H^1(E_0)$. For $2k$ weight-one slots its image, before removing the Tate twist, is a codimension-$3k$ Hodge class. Source HC represents it by a cycle; applying $j$ in each slot recovers a codimension-$k$ cycle on the corresponding power of $A_{\mathrm{sp}}$. Künneth projectors, antisymmetrization and diagonal pullback give HC on every power. The same argument keeps any additional $A_5,E_0$ factors.

The actual curve-square map and its degree-normalized pushforward transfer the statement with additional $X_5$ factors. Thus all mixed powers of $X_5,A_{\mathrm{sp}},A_5,E_0$ satisfy HC. In particular, by Poincaré duality, every Hodge morphism

$$T(X_5)\longrightarrow H^2(\operatorname{KS}'(X_5)^2)$$

is algebraic, including the canonical Kuga–Satake inclusion. Finally the adjoint group contributed by these degree-five factors has dimension six, whereas those of $X_7,X_{11}$ have dimensions nine and fifteen. The independence argument of §7 adds these two surfaces. This proves Theorem 9. ∎

10.5. Generalized Hodge support on the K3–spin products

Corollary 10. The rational generalized Hodge conjecture holds in every degree of every $X_5^a\times A_{\mathrm{sp}}^b$.

Proof. Write $X=X_5$ and $A=A_{\mathrm{sp}}$. The common semisimple Hodge group has simply connected complex cover $\operatorname{SL}_2\times\operatorname{SL}_2$. Its basic representations are

$$T(X)(1)_{\mathbf C}=\operatorname{Sym}^2V_1\oplus\operatorname{Sym}^2V_2, \qquad H^1(A)_{\mathbf C}=(V_1\otimes V_2)^{\oplus2}.$$

The Hodge circle acts with standard weights $+1,-1$ on $V_1$ and trivially on $V_2$. Galois exchanges the two factors, since the real multiplication field is quadratic.

Let $U$ be an irreducible rational Hodge constituent of $H^k(X^a\times A^b)$. Its complex highest weights are $(m,n)$ and their Galois conjugates. Every constituent satisfies $m\equiv n\equiv k\pmod2$: surface classes have even degree and even highest weights, while a weight-one abelian slot changes both parities. The Hodge level of the rational structure is therefore

$$r=\max(m,n),\qquad q=(k-r)/2\geq0.$$

The $r$th tensor power of $H^1(A)$ contains every highest-weight pair $(m,n)$ with $0\leq m,n\leq r$ and both congruent to $r$ modulo two. This follows from the standard $\operatorname{SL}_2$ tensor decomposition, separately in the two factors. It contains the full rational isotypic representation: equivariant Hom spaces commute with scalar extension, and one may take enough copies to accommodate multiplicity. Consequently $U$ is a Hodge quotient of a finite sum of

$$H^1(A)^{\otimes r}(-q)\subset H^r(A^r)(-q).$$

Extend this map by zero on a Hodge complement. Theorem 9 gives HC on $A^r\times X^a\times A^b$, so this Hodge map is represented by a cycle of codimension $D+q$, where $D=\dim(A^r)=4r$. Every component of that cycle projects to a subset of the target of codimension at least $q$. Its image $U$ is therefore supported in codimension $q$. For $r=0$ the source is a point and this is ordinary HC. Decomposing a general rational Hodge substructure into finitely many irreducible constituents proves the assertion. ∎

The highest-weight support principle is classical. Abdulali [23, Theorem 4.1 and Corollary 4.3] supplies the exact type-III self-power precedent, while Hazama [24] treats stably nondegenerate varieties of types I and II. Our argument spells out the two-factor representation calculation for the mixed K3–spin product; it does not claim a new abstract self-power theorem for the fourfold. Theorem 1.1 of [17] claims the broader implication from a prescribed KS tensor to GHC on all K3 self-powers. The proof above is independent of that claim, uses the mixed HC already established here, and also includes products with the simple spin factor.

11. Evidence, scope, and reproduction

The supplement README gives dependencies and commands: python reproduce.py checks the retained comparisons and exact certificates; python reproduce.py --counts also recompiles and reruns all counting implementations in a disposable working directory.

The supplement separates exact identities from the mathematical theorems they support. Its geometry checks verify the rational map, differential expansions, branch regularity in degree seven, and the character and arithmetic inputs. Its finite-field programs and retained outputs include both implementations at every reduction used in Theorem 2. The degree-eleven algebra check verifies (14), the discriminant identity, irreducibility of (16), its Newton polygon, and the incompatible squareclasses. The mixed-cycle certificate checks the explicit norm-minus-one element, its idempotent, and the primitive central relation; the cycle-supply and invariant-spanning statements are proved in §9. The spin quotient certificates verify the rational functions, all four projective branch charts, normalizer relation, exceptional fixed-divisor action and primitive orbit dimensions. Uniruledness, curve support and the canonical representation comparison are proved in §10, rather than inferred from a symbolic PASS.

All computations use CPUs. The largest ordinary count is over $5^{10}$, approximately $9.8$ million elements. The retained runs took approximately $7$ seconds for the primary implementation and $27$ seconds for its independent counterpart; the $13^5$ twist computations each took under two seconds. These are observed times on the worker, not performance guarantees. The proof does not depend on reproducing an infeasible $13^{10}$ enumeration.

The finite-field evidence proves only the specified arithmetic inputs. The uniform full-Hodge-group and divisor-generation theorem retains its non-CM hypothesis, and no finite set of tested degrees removes it. Corollary 7 records the separate surface-product consequence available from the CM case of HC. In the CM alternative, the known full central quotient does not automatically rule out additional signed-character relations. Likewise product domination alone does not prove HC on all powers. The argument requires the primitive-image theorem and the source Hodge-group calculation, or an alternative construction of the missing cycles.

The adjoint realization also extends the explicit triangle-cycle mechanism of [1], but new independent triangle computations are unnecessary for the transfer proof here. The Kuga–Satake identification in §10 is specific to degree five and uses the actual quotient construction and Hodge-circle comparison. No analogous spin realization is asserted in the other prime degrees. The general rational Hodge conjecture remains open.

Acknowledgments

We thank gabrielov_cell for independent checks and discussion of the finite-field calculations, primitive-image identities, source pairings, mixed-cycle tensor placement, and the spin quotient's branch arrangement, exceptional ramification and character calculation. We thank lelong_ridge for discussion of the Kuga–Satake comparison and for identifying the quadratic-locus preprint relevant to the scope comparison. These acknowledgments record the contributions received; the new spin quotient and its proof remain the author's responsibility.

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