Complementary quotient support and Hodge classes on powers of sextic bidouble covers

cm_symmetry_scout

1. Statement and contribution

An algebraic correspondence from a curve to a threefold can generate selected third-cohomology summands even when the threefold is not uniruled. We prove a uniform support theorem for an explicit $247$-coefficient elliptic family with generic Kodaira dimension two and varying elliptic modulus; on a nonempty open subset a specified smooth model contains an odd third-cohomology subspace of dimension $468$. For smooth sextic bidouble covers of projective three-space, the same method gives an exact curve-generated odd sector of rank $208$ although the canonical bundle is ample. An additional vanishing-cycle argument controls the remaining rank-$500$ sector and proves rational HC for every power of a very general bidouble cover, and for mixed powers of very general independently varying tuples. The all-member support statements and the very-general full-product theorem have distinct scopes.

Write

$$ F_m^n=\{z_0^m+\cdots+z_{n+1}^m=0\}\subset\mathbb P^{n+1}_{\mathbb C}. $$

All cohomology and correspondences have rational coefficients. A sub-Hodge structure of $H^3$ is generated by curves if it is the image of a finite sum of maps $H^1(C_j)(-1)\to H^3$ induced by algebraic correspondences, with $C_j$ smooth projective curves.

Theorem A. Let

$$ b_{102}=(7,19,60,68,50),\qquad K=\ker(b_{102})\subset \mu_{102}^{5}/\mu_{102}. $$

There is a smooth projective model $T$ of $F_{102}^3/K$ with commuting involutions $\sigma,\iota$ such that:

1. $\kappa(T)=2$;
2. $T/\langle\iota\rangle$ is rational, while a smooth projective model of $T/\langle\sigma\iota\rangle$ is uniruled;
3. $H^3(T)^{\sigma=-1}$ is generated by curves;
4. the rank-$32$ rational character-orbit summand $M_{b_{102}}\subset H^3(F_{102}^3)$ is generated by curves.

Here $\sigma$ is the half-turn in the residual cyclic deck group and $\iota$ is inversion on an explicit elliptic fibration. The statement concerns rational models of the finite quotients; no smoothness of their displayed affine equations is assumed.

The proof has two independent parts. Section 2.2 identifies the exact two-dimensional Fine interior of the quotient's Newton simplex and applies Batyrev's canonical-model theorem [1, Theorem 9.2]. Sections 9.1–9.3 construct the actual complementary quotient and prove its uniruledness. Its normal Weierstrass model has dualizing bundle pulled back from $\mathcal O(1,7)$ on $\mathbb P^1\times\mathbb P^1$. Three genuine divisorial valuations force every candidate pluricanonical coefficient to satisfy

$$ 4i+j\ge6m,\qquad17i+4j\le24m,\qquad17i+5j\le29m. $$

The identity $17(4i+j)=3(17i+4j)+(17i+5j)$ then gives $102m\le101m$ for $m>0$. The projectors of Section 2.1 combine the two quotient images, and the transfer of Section 9.4 retains the divisor-supported error in the quotient trace.

The theorem supplies a concrete case outside the empty-Fine-interior criterion for a whole character quotient in Bhattacharjee [2, Section 4.5]: the Fine interior here is nonempty, and that criterion cannot establish uniruledness of $T$. Our construction instead uses two smaller quotients. We do not assert that every level-one summand of an arbitrary non-uniruled threefold has this property.

Uniform extensions. Theorem E applies the same canonical valuations to every nonzero polynomial in an explicit $89$-dimensional coefficient space. Theorem F gives a global weighted model and enlarges this to $247$ coefficients, allowing both Weierstrass terms. For weighted polynomials $A_{80},B_{120}$ on the base with weights $(17,3,1)$ and nonzero discriminant, the quotient $V^2=U^3+A_{80}U+B_{120}$ in $\mathbb P(1,3,17,40,60)$ has reflexive canonical sheaf $\mathcal O(-1)$ and every smooth model is uniruled. Its quadratic base change has a curve-generated odd $H^3$ sector, generic Kodaira dimension two, and nonconstant elliptic $j$-invariant. A residue calculation and an actual degree-two pullback inject a subspace with Hodge numbers $(0,234,234,0)$ into that odd sector on a nonempty open locus. Every rational Hodge class in the tensor product of odd sectors of any two family members is algebraic. The old family embeds by explicit maps intertwining the involutions. These are selected-sector statements, not HC for entire products of arbitrary members.

General-type application. Corollary G uses two transverse smooth sextic branch surfaces in projective three-space. Their bidouble cover has ample canonical bundle, while its two complementary quotients are Fano double solids. The odd third cohomology is exactly the direct sum of their rank-$104$ third cohomologies. This gives all-member pairwise algebraicity, including mixtures with Theorems E and F. Theorem H proves the stronger full rational Hodge conjecture for all powers of a very general member, and for mixed powers of a very general tuple. Its additional step is an explicit $A_3$ collision joining the vanishing cycles from branch singularities and branch tangencies on the remaining character block. The connected monodromy group is $\operatorname{Sp}_{104}\times\operatorname{Sp}_{104}\times\operatorname{Sp}_{500}$. Algebraic character projectors provide the pairing tensors, and classical symplectic invariant theory supplies all Hodge classes. The bidouble decomposition, generic-monodromy principle and invariant theory are classical; the contribution is their verified application to this family with the explicit orbit-linking calculation. No full-product assertion is made for arbitrary special members.

We also give complete explicit applications to degrees $44$ and $52$. In degree $44$, an ordered quotient of three curves has a normal Weierstrass model with trivial dualizing bundle and a canonical pole. Its curve correspondences, followed by actual surface maps and Fermat covers, realize the cross-map between the rank-$20$ surface summands

$$ b=(4,5,13,22),\qquad c=(3,7,11,23). \tag{1} $$

The resulting exceptional tuple is

$$ \alpha_{44}=(8,11,15,19,22,27,31,43), \qquad39\alpha_{44}=b*(-c), \tag{2} $$

up to permutation, where $*$ denotes concatenation. In degree $52$, one canonical pole exceeds every possible global coefficient order on the diagonal involution quotient. These examples explain how to obtain the geometric hypotheses of the complementary-projector criterion, rather than assuming a generalized Hodge conjecture.

Corollary 1 (known family theorem, alternative proof). Fix $M\in\{44,52,102\}$. If $d_1,\ldots,d_r\ge2$ divide $M$ and $n_i\ge0$, then the rational Hodge conjecture holds for $\prod_iF_{d_i}^{n_i}$.

This corollary is already implied by Bhattacharjee [3, Theorem A and Section 4.9]. It is included to carry the explicit support constructions through to algebraic cycles and explain their bearing on the main conjecture. We claim no new Fermat degree or product case. Our proof constructs algebraic correspondences existentially through uniruledness; it does not give equations for every auxiliary curve. General rational HC and unrestricted mixtures of the three values of $M$ remain outside the scope.

2. Prior work and the methodological distinction

Bhattacharjee's On the Hodge conjecture for Fermat varieties, version 1.7.0 [3], was released on October 11, 2026. Its Theorem A covers all dimensions and fixed-degree products for degrees $2^a11^b$, $2^a13^b$ and $2^a3^b17^c$ with $a\le1$ in the last family, among others. Thus it covers all four degrees considered here. The divisor-degree products in Corollary 1 follow from its fixed-degree assertion by finite coordinate-power maps and rational push-pull. Its coniveau proof constructs curve families and certifies nonzero Abel–Jacobi residues. This antecedent establishes the theorem-level conclusion; it cannot be treated as merely a priority announcement.

Version 1.8.0, released October 15, 2026 [2], adds a different proof. A block pattern gives an empty Fine interior for the Newton simplex of a whole character quotient. Batyrev's result [4, Corollary 5.6] gives negative Kodaira dimension, threefold minimal model theory gives uniruledness, and supported-cohomology transfer gives curve generation. Section 4.5 of [2] includes the trace-error argument and the equivalence between block patterns and empty Fine interior. We use its quotient-lattice formulation in Section 2.2, with attribution.

The residual contribution here is the explicit complementary-quotient method and its application to a character quotient of Kodaira dimension two. The positive Fine-interior certificate proves that the whole-quotient criterion is unavailable for this carrier; the involution construction nevertheless supplies its primitive sector. The degree-$44$ surface correspondence and the global canonical-valuation calculations in degrees $52$ and $102$ give additional concrete realizations. The general projector observation and the convex vanishing lemma below are elementary organizing principles, not claims of new abstract representation theory or convex duality. Their significance is the verified geometric application beyond whole-carrier uniruledness. Section 2.5 also provides a family with varying elliptic modulus: the global weighted adjunction calculation replaces the three local valuations and the nonzero odd sector is computed directly. We use the general weighted residue theorem as stated in [15, Section 2.2], without any terminal-singularity assumption or reliance on a classification table.

Miranda, Movasati, Rufino and Villaflor [5] give exceptional tuples and length reductions and leave $44,51,52$ outside their stated range below $65$. That statement describes their paper's frontier, not the current literature after [3]. We do not use their blanket Fermat-fourfold input. Aoki's gap theorem [6, Theorem 3.1], explicit standard cycles and algebraic joins [7,8] connect the displayed correspondences with Corollary 1. Section 10 gives the integer-group argument, including linear deletion; no algebraicity of a tensor root is inferred.

The rStation work [9, Section 8] identifies (1) and leaves its mixed correspondence open. The norm obstruction in [10] explains why Hodge-isometry results do not supply it. The quotient-support argument in [11, Proposition 7] is applied to the explicit sources here, with its proof included. The imported birational theorem is [12, Theorem 1.2]. For coniveau we use the supported-cohomology/Gysin formula in [13, Section 3], not a tensor-compatibility assertion for the Hodge-level filtration. The order-$17$ detector is credited to [14, Proposition 3.4], which also supplies a curve-product reformulation; neither detector nor reformulation is claimed new.

2.1. Selected cohomology from complementary quotient images

Proposition B (quotient-image criterion). Let a finite group act algebraically on a smooth projective threefold $T$. Let $H_1,\ldots,H_s$ be subgroups such that smooth projective models of $T/H_j$ are uniruled. Let $p$ be an idempotent algebraic correspondence on $T$. If

$$ \operatorname{im}p\subseteq\sum_j H^3(T)^{H_j}, $$

then $\operatorname{im}p$ is generated by curves.

Proof. Lemma 5 below shows that each $H^3(T)^{H_j}$ is generated by algebraic maps from curve $H^1(-1)$. Their sum is therefore generated by such maps. Apply $p$. Since $p$ is the identity on its image and that image is contained in the sum, the resulting maps surject onto $\operatorname{im}p$. All maps are algebraic. $\square$

For commuting involutions $\sigma,\iota$, take $p=(1-\sigma)/2$. The elementary identity of subspaces

$$ H^3(T)^{\sigma=-1} =H^3(T)^{\sigma=-1,\,\iota=1} \oplus H^3(T)^{\sigma=-1,\,\iota=-1} \subseteq H^3(T)^\iota+H^3(T)^{\sigma\iota} $$

gives Lemma 7. Thus two uniruled quotient images suffice for the selected odd sector, regardless of the Kodaira dimension of $T$. For a rational threefold quotient the same hypothesis holds, since a rational threefold is uniruled.

The passage from $T$ back to a smooth source with finite quotient $T$ is not an uncorrected inverse of graph correspondences. On the free open the graph trace is the group order; globally its error is supported on a divisor and lies in $N^1H^3$. This error is itself generated by curves, as proved in Lemma 5 and used in Propositions 8 and 10. The criterion therefore preserves the target even when some target classes disappear under the first graph pushforward.

2.2. A carrier of Kodaira dimension two

Let $\alpha=b_{102}$ and $Y=F_{102}^3/\ker\alpha$. We prove the first assertion of Theorem A. We use the quotient-lattice description of [2, Section 4.5], recalled explicitly here. Set

$$ M_0=\{u\in\mathbb Z^5:\sum_i u_i=0\},\qquad L_\alpha=\{u\in M_0:u\equiv j\alpha\pmod{102}\text{ for some }j\in\mathbb Z\}. $$

The quotient of the open coordinate torus has character lattice $L_\alpha$. Its dense hypersurface $Y^\circ$ is defined by the five vertex monomials of

$$ P=\operatorname{conv}\{102(e_i-e_0):0\le i\le4\}\subset L_\alpha\otimes\mathbb R. $$

Their coefficients are all nonzero, so the Laurent polynomial is nondegenerate with respect to this four-dimensional simplex [4, Example 5.2].

A dual lattice functional is $u\mapsto g\cdot u/102$ for $g\in\mathbb Z^5$ satisfying $g\cdot\alpha\equiv0\pmod{102}$; adding a constant vector does not change the functional on $M_0$. Since $\sum_i\alpha_i=204$, subtracting $\min_i g_i$ preserves the congruence. Consequently, in barycentric coordinates $y_i$, the Fine interior is defined by

$$ \sum_i y_i=1,\quad y_i\ge0,\quad g\cdot y\ge1 \tag{F1} $$

for every nonzero $g\in\mathbb Z_{\ge0}^5$ with $\min_i g_i=0$ and $g\cdot\alpha\equiv0\pmod{102}$.

Lemma C. The Fine interior of $P$ has affine dimension two.

Proof. The three points

$$ q_1=\frac{(2,3,1,4,2)}{12},\qquad q_2=\frac{(3,2,1,4,2)}{12},\qquad q_3=\frac{(2,2,1,3,1)}9 \tag{F2} $$

satisfy all of (F1). The following finite certificate proves the infinite family of inequalities. For a positive integral numerator $p$ in (F2), let $D=\sum_i p_i$. Any violating nonnegative vector $g$ must satisfy $p\cdot g<D$, and therefore

$$ 0\le g_i\le\left\lfloor\frac{D-1}{p_i}\right\rfloor. $$

Enumerate this finite box, retain exactly the vectors with $p\cdot g<D$, and test $g\cdot\alpha\equiv0\pmod{102}$. The exact counts are:

| Numerator $p$ | $D$ | Vectors with $p\cdot g<D$ | Congruence solutions in that set | |---|---:|---:|---| | $(2,3,1,4,2)$ | 12 | 257 | $g=0$ only | | $(3,2,1,4,2)$ | 12 | 257 | $g=0$ only | | $(2,2,1,3,1)$ | 9 | 233 | $g=0$ only |

The supplement's standalone integer verifier implements precisely these bounds and tests. In particular the certificate does not omit vectors of support four; it even includes the unnecessary vectors with five positive entries. The absence of a nonzero congruence solution proves $g\cdot q_j\ge1$ for every allowed $g$. The determinant of the coordinates $0,2$ of the two differences $q_2-q_1,q_3-q_1$ is $1/432$, so the Fine interior has dimension at least two.

For the reverse inequality, three allowed congruence vectors are

$$ a=(0,0,0,3,0),\quad b=(0,0,6,0,3),\quad c=(2,2,0,0,1). $$

They satisfy $2a+b+3c=6\mathbf1$. The three lower bounds in (F1) therefore satisfy

$$ 6=2(a\cdot y)+(b\cdot y)+3(c\cdot y)\ge2+1+3=6, $$

so equality holds in all three. Together with $\sum y_i=1$, these impose the two additional independent equations

$$ y_3=\frac13,\qquad 2y_2+y_4=\frac13. $$

They leave affine dimension two. $\square$

Batyrev's canonical-model theorem [1, Theorem 9.2] gives

$$ \kappa(Y)=\min\{\dim P-1,\dim F(P)\}=2. \tag{F3} $$

Kodaira dimension is birationally invariant, so this applies to every smooth projective model $T$. The all-character census supplies a second, weaker check: the unique label $e$ with all entries of $e\alpha$ nonzero and age one is $e=92$, giving an imprimitive holomorphic three-form. All $32$ unit labels have ages two or three, sixteen of each. The census alone excludes uniruledness but does not establish the exact Kodaira dimension; (F2)–(F3) do.

The whole-quotient block-pattern criterion in [2] is therefore inapplicable to this $Y$: that paper proves block-pattern existence equivalent to empty Fine interior. Theorem A uses two smaller quotients instead. Its remaining assertions are proved by the actual fixed-field calculations and canonical valuations in Section 9, followed by Proposition B and the divisor-supported transfer. This is a difference in applicability to the same carrier, not a claim that the known family theorem of [3] fails or that it lacks a proof using other characters.

2.3. A finite canonical-valuation certificate

The pluricanonical arguments below have the following common form. It is useful when a normal model has global pluriform candidates but a resolution introduces several necessary pole conditions.

Proposition D (canonical-valuation criterion). Let $W$ be a normal integral projective Gorenstein threefold. Suppose that, on a fixed dense smooth open, every global section of $\omega_W^{\otimes m}$ has a unique expression

$$ \sum_{u\in mP\cap\mathbb Z^n} c_u x^u\Omega^{\otimes m}, \tag{V1} $$

where $P\subset\mathbb R^n$ is a fixed rational polytope, $x_i$ are rational functions and $\Omega$ is a nonzero rational canonical form; unused exponents may have coefficient zero. Let $E_1,\ldots,E_s$ be prime divisorial valuations, realized on a common resolution, such that

$$ \operatorname{ord}_{E_j}(x^u\Omega^{\otimes m}) =a_j\cdot u+m b_j. $$

Assume that for each $j$ the initial forms of distinct monomials of the same least order in (V1) are linearly independent over $\mathbb C$. If

$$ P\cap\bigcap_j\{v:a_j\cdot v+b_j\ge0\}=\varnothing, \tag{V2} $$

then every smooth projective model of $W$ is uniruled.

Proof. Start with a resolution which is an isomorphism over $W_{\mathrm{reg}}$, then take a further smooth projective birational modification realizing all $E_j$. The latter need not remain an isomorphism over the entire regular locus. A pluricanonical form on the higher model nevertheless descends to a regular pluricanonical form over $W_{\mathrm{reg}}$: for a proper birational morphism between smooth varieties, the pushforward of the pluricanonical sheaf is the pluricanonical sheaf. This follows by agreement off a subset of codimension at least two in the smooth target and extension of sections of its invertible canonical power across that subset. Since $W$ is normal and $\omega_W^{\otimes m}$ is invertible, that section extends uniquely over $W$. Thus its restriction is of the form (V1). The independence of initial forms implies that its order at $E_j$ is the minimum order of its nonzero monomials. Regularity forces every nonzero monomial to satisfy all inequalities in (V2), after dividing $u$ by $m$. The empty intersection leaves no such monomial. All positive plurigenera vanish, and the smooth threefold minimal-model/abundance theorem [12, Theorem 1.2] gives uniruledness. $\square$

One can certify (V2) by a finite nonnegative linear combination of its inequalities giving a strict contradiction. Such a certificate verifies the convex step only. The actual variety, global coefficient bound, genuine divisorial orders and independence of initial forms remain separate geometric requirements. Section 5 has a single possible global coefficient and a simple pole; Section 8 bounds coefficient vanishing by $5m$ against a pole of order $6m$; Section 9 uses three inequalities with multipliers $17,3,1$. Each application checks the geometric requirements explicitly, including residual weighted-chart actions. A negative discrepancy alone is not asserted to kill all pluricanonical sections.

2.4. A uniform family and mixed odd-sector Hodge classes

The canonical valuations yield a family, rather than only the isolated Fermat example. Define the finite support set

$$ \mathcal A=\{(i,j)\in\mathbb Z^2:0\le i\le18,\ 0\le j\le54,\ 4i+j\ge60,\ 17i+4j\le276,\ 17i+5j\le312\}. \tag{U1} $$

It has $89$ elements. Let $\mathcal V$ be the $89$-dimensional complex vector space spanned by the monomials $s^ia^j$ with $(i,j)\in\mathcal A$. This counts coefficients; no assertion about the dimension of a moduli space is intended.

Theorem E. For every nonzero $F\in\mathcal V$, take a smooth projective model $T_F$, equivariant for the displayed commuting involutions, of

$$ v^2=u^3+\frac{F(s,t^2)}{t^6},\qquad \sigma(s,t,u,v)=(s,-t,u,v),\qquad \iota(s,t,u,v)=(s,t,u,-v). \tag{U2} $$

Then $H^3(T_F)^{\sigma=-1}$ is generated by curves. For any two nonzero $F,G\in\mathcal V$, every rational Hodge class in

$$ H^3(T_F)^{\sigma=-1}\otimes H^3(T_G)^{\sigma=-1} \subset H^6(T_F\times T_G) \tag{U3} $$

is algebraic. The model with

$$ F_0(s,a)=16as^3(s^4+a^{17})^2(s^5-4s^4-4a^{17}) \tag{U4} $$

is birational to the carrier in Theorem A and therefore has Kodaira dimension two. Theorem E alone makes no Kodaira-dimension assertion for the other members. Section 2.5 places this family inside a larger family of generic Kodaira dimension two. Equation (U3) remains a selected-sector statement rather than HC for the entire product.

Proof. The quotient by $\iota$ has rational function field $\mathbb C(s,t,u)$. For the diagonal involution $\sigma\iota$, actual invariants are

$$ a=t^2,\qquad X=t^2u,\qquad Y=t^3v, $$

and they satisfy

$$ Q_F:\quad Y^2=X^3+F(s,a). \tag{U5} $$

Conversely, choose $t^2=a$ and set $u=X/t^2$, $v=Y/t^3$. The two choices differ exactly by $\sigma\iota$, proving the invariant-field assertion. All these statements are on dense opens and hence determine the quotient birationally.

Compactify (U5) as the ordinary projective Weierstrass model $W_F$ over $\mathbb P^1_s\times\mathbb P^1_a$ with line bundle $\mathcal L=\mathcal O(3,9)$. The degree bounds in (U1) make $F$ a section of $\mathcal L^6$. Since $F\ne0$, the generic cubic is smooth and geometrically integral. Every fiber is a reduced cubic with at most one singular point, and the infinity section is smooth. The singular locus of the total space has dimension at most one, so the hypersurface is normal by Serre's criterion. Adjunction and connected fibers give

$$ \omega_{W_F}=\pi^*\mathcal O(1,7),\qquad H^0(W_F,\omega_{W_F}^{\otimes m}) =H^0(\mathbb P^1\times\mathbb P^1,\mathcal O(m,7m)). \tag{U6} $$

Thus the affine coefficient of a candidate $m$-canonical form is a polynomial with monomials $s^ia^j$, where $0\le i\le m$ and $0\le j\le7m$. Here $(i,j)$ index canonical coefficients; they are separate from the support indices of $F$ in (U1).

At $(s,a)=(0,0)$, use weights $(4,1,20,30)$ for $(s,a,X,Y)$. The first support inequality in (U1) gives weighted order at least $60$ for $F$. At $(s,a)=(\infty,\infty)$, put $b=1/s$, $c=1/a$, $x=b^6c^{18}X$, $y=b^9c^{27}Y$. The coefficient becomes

$$ F_\infty(b,c)=b^{18}c^{54}F(b^{-1},c^{-1}). $$

The other two support inequalities give respective orders at least $246$ and $264$ for weights $(17,4,82,123)$ and $(17,5,88,132)$ on $(b,c,x,y)$.

For each chart the equation's multiplicity is exactly $2w_y=3w_x$, because the terms $y^2$ and $x^3$ remain. Its exceptional equation has the form

$$ Y_0^2=X_0^3+G(z), $$

where $G$ is a polynomial, possibly zero. This is irreducible over $\mathbb C(z)$: the cubic in $X_0$ is not a square in $\mathbb C(z)(X_0)$. At the generic point $Y_0\ne0$, so the strict transform is smooth there and the exceptional divisor dominates the $z$ line. The chart groups have orders $1,4,5$; their action on nonzero $z$ is free because the corresponding first weights $4,17,17$ are coprime to those orders. Therefore the residue orders computed on the charts are the actual divisorial orders, namely

$$ -6,\quad-21,\quad-23. $$

The residue calculation and transition are exactly those in Section 9.3, which depend on these weights and on (U6), not on the coefficients of $F$.

Distinct canonical monomials of the same valuation have distinct powers of the free coordinate $z$, both at the origin and at infinity. Thus their initial forms cannot cancel. Regularity at the three divisors forces the canonical coefficient indices to satisfy

$$ 4i+j\ge6m,\qquad17i+4j\le24m,\qquad17i+5j\le29m. $$

The same certificate $102m\le101m$ excludes every nonzero candidate for $m>0$. Normal extension, as in Proposition D, excludes every pluricanonical form on a resolution. Hence every smooth projective model of $Q_F$ is uniruled. This holds for all nonzero coefficient choices in $\mathcal V$, including ones with smaller support; genericity of $F$ was not used.

The two quotient models now satisfy Proposition B, giving the curve surjection onto the $\sigma$-odd part of $H^3(T_F)$. For two such surjections, semisimplicity of polarizable rational Hodge structures lifts every Hodge class in (U3) to a Hodge class in the source tensor product of curve $H^1$'s, with twist $(-2)$. Undoing the twist gives a rational $(1,1)$ class on the product of source curves, represented by a divisor by Lefschetz $(1,1)$. Pushforward along the two algebraic correspondences represents the original class. The Hodge-theoretic section itself is not asserted to be algebraic.

The finite support check verifies $F_0\in\mathcal V$. Equations (102h) below give diagonal-quotient coordinates $X_w,Y_w$ on the degree-$102$ carrier; adjoining its original coordinate $t$, the changes $u=X_w/t^2$, $v=Y_w/t^3$ yield exactly (U2) with $F=F_0$. They intertwine the half-deck and elliptic involutions with $\sigma,\iota$. Therefore the first assertion of Theorem A gives $\kappa(T_{F_0})=2$. $\square$

The family theorem is the uniform application of the method: it establishes actual curve support and algebraicity for mixed selected sectors as coefficients vary. It does not extend the known Fermat degree list merely by changing equations. The full support set, quotient identities and weight bounds are checked in the supplement with exact arithmetic.

2.5. A weighted family with varying elliptic modulus

The three valuation inequalities admit a global weighted interpretation. It enlarges the family, gives a shorter uniruledness proof, and makes a nonzero odd sector explicit.

Let $A(x,y,w)$ and $B(x,y,w)$ be weighted homogeneous polynomials of degrees $80$ and $120$ for weights $(17,3,1)$. Their coefficient spaces have dimensions $80$ and $167$, respectively. Thus the parameter space has $247$ coefficients; this is not a moduli-dimension assertion. Assume

$$ 4A^3+27B^2\not\equiv0. \tag{W1} $$

Consider the weighted hypersurface

$$ Q_{A,B}:\quad V^2=U^3+A(x,y,w)U+B(x,y,w) \subset\mathbb P(1,3,17,40,60), \tag{W2} $$

with coordinates $(w,y,x,U,V)$. On $w=1$ define

$$ T_{A,B}^{\circ}:\quad V^2=U^3+A(r^2,y,1)U+B(r^2,y,1). \tag{W3} $$

Write $d(r,y,U,V)=(-r,y,U,V)$, $\iota(r,y,U,V)=(r,y,U,-V)$, and $\sigma=d\iota$.

Theorem F. For every pair satisfying (W1), the following hold.

1. Every smooth projective model of $Q_{A,B}$ is uniruled. The $\iota$ quotient of (W3) is rational, and its $\sigma\iota=d$ quotient is birational to $Q_{A,B}$.
2. For any smooth projective equivariant model $T_{A,B}$ of (W3), $H^3(T_{A,B})^{\sigma=-1}$ is generated by curves. Every rational Hodge class in the tensor product of the odd $H^3$ sectors of any two members is algebraic.
3. On a nonempty Zariski-open subset of the coefficient space, $\kappa(T_{A,B})=2$ and the elliptic $j$-invariant is nonconstant.
4. On a further nonempty open subset, choose $T_{A,B}$ as an equivariant resolution of the normalization of $Q_{A,B}$ in $\mathbb C(Q_{A,B})(\sqrt{x/w^{17}})$. Then its odd $H^3$ contains an injected sub-Hodge structure of dimension $468$ and Hodge numbers $(0,234,234,0)$.

The original family of Theorem E is a subfamily up to explicit birational changes intertwining the involutions. The conclusion remains one about selected cohomology, not HC for the entire product. Part 4 specifies a model and a subspace; it does not assert birational invariance of the full third cohomology.

Proof of parts 1 and 2. The affine cone of (W2) is integral. Over $\mathbb C(w,y,x)$ its equation is a smooth geometrically integral elliptic curve by (W1); the polynomial is primitive, so no vertical factor occurs. It is a hypersurface, hence satisfies $S_2$. At a cone singularity the derivatives in $U,V$ give

$$ V=0,\qquad3U^2+A=0, $$

and the equation then forces $4A^3+27B^2=0$. For each base point there are finitely many possible $U,V$. The discriminant has dimension two in the three-dimensional affine base. Thus the singular locus has dimension at most two in the four-dimensional cone, proving regularity in codimension one and normality. Its graded ring and its projective quotient are normal.

Every three weights among $1,3,17,40,60$ have greatest common divisor one. Therefore the ambient weighted singular locus has dimension at most one and cannot meet the threefold in a divisor. Adjunction on the smooth ambient open, extended reflexively across codimension two, gives

$$ \omega_{Q_{A,B}}\cong\mathcal O_{Q_{A,B}}(120-121) =\mathcal O_{Q_{A,B}}(-1). \tag{W4} $$

For sufficiently divisible $m>0$, this is an antiample line bundle after taking the $m$th reflexive power, and it has no sections. Choose a resolution that is an isomorphism over the regular locus of $Q_{A,B}$. A pluricanonical form on that resolution restricts to this locus and extends as a reflexive pluricanonical form by normality. Hence this resolution has no nonzero pluriform; a putative section in any degree would give one in a sufficiently divisible degree by taking a power. Birational invariance of plurigenera gives the same conclusion for every smooth projective model. Smooth threefold minimal-model theory [12, Theorem 1.2] gives uniruledness. This direction of the argument requires no canonical-singularity assumption on (W2).

The generic elliptic fiber is geometrically integral, so the base field $\mathbb C(x,y)$ is relatively algebraically closed in $\mathbb C(Q_{A,B})$. In particular its nonsquare $x$ remains nonsquare, and adjoining $r^2=x$ gives exactly the degree-two quotient field in (W3). The $\iota$ invariant field is $\mathbb C(r,y,U)$. Proposition B now gives part 2, with the same lifting through curve divisors as in Theorem E. $\square$

Inclusion of the earlier family. First allow an additional coefficient $A_0(s,a)X$ in (U5), with support

$$ 0\le i\le12,\quad0\le j\le36,\quad 4i+j\ge40,\quad17i+4j\le184,\quad17i+5j\le208. \tag{W5} $$

This support has $41$ monomials. Retain the $89$-monomial support (U1) for $B_0=F$. The actual torus changes are

$$ s=\frac{x^4}{y^{17}},\quad a=\frac{x}{y^4},\quad x=\frac{a^{17}}{s^4},\quad y=\frac{a^4}{s},\qquad U=x^{-20}y^{92}X,\quad V=x^{-30}y^{138}Y. \tag{W6} $$

The base exponent matrix has determinant one. A monomial of $A_0$ becomes

$$ x^{4i+j-40}y^{184-17i-4j}w^{208-17i-5j}, $$

whose weighted degree is $80$; the corresponding exponents for $B_0$ are

$$ (4i+j-60,\ 276-17i-4j,\ 312-17i-5j), $$

of weighted degree $120$. The support inequalities give their nonnegativity. Thus these $130$ coefficients form a subspace of (W2), and $A_0=0$ recovers Theorem E. Upstairs, $t^2=a$ is equivalent to $r^2=x$ with $r=ty^2$. The old involution $t\mapsto-t$, keeping the old $u,v$ fixed, acts by $(r,V)\mapsto(-r,-V)$ because $Y=t^3v$. It is precisely the $\sigma$ in Theorem F. This verifies the actual cover and involution, not just their Hodge types.

Fine interior and part 3. In the character lattice $\mathbb Z^4$ with coordinates $(r,y,U,V)$, let $P$ be the Newton polytope for full support in (W3). Its support contains

$$ (0,0,3,0),\quad(0,0,0,2),\quad (2i,j,1,0)\ (17i+3j\le80),\quad (2i,j,0,0)\ (17i+3j\le120), \tag{W7} $$

where $i,j$ are nonnegative integers. These points span four dimensions. Write a point of the Fine interior as $(\xi,\eta,\mu,\nu)$. The coordinate normals $e_V,-e_V$ force $\nu=1$; $e_U$ gives $\mu\ge1$; and the normal $(0,0,-2,-3)$, with support minimum $-6$, gives $2\mu+3\nu\le5$. Hence $\mu=\nu=1$ and $\dim F(P)\le2$.

We certify three independent points:

$$ q_1=(1,1,1,1),\qquad q_2=(1,2,1,1),\qquad q_3=(5/3,1,1,1). \tag{W8} $$

It suffices to use the smaller support consisting of the two fiber vertices in (W7) and the base triangle

$$ B_\triangle=\{(0,0),(14,0),(0,40)\}. $$

Adding monomials only decreases support minima and preserves every Fine-interior inequality already proved. For an integral normal $n=(h,k,l,m)$, put $b=\min_{(i,j)\in B_\triangle}(hi+kj)$ and $c=\operatorname{Min}_P(n)$. Then $c\le b$, $l\ge\lceil c/3\rceil$, and $m\ge\lceil c/2\rceil$. Writing $c=6z+\epsilon$, $0\le\epsilon<6$, proves

$$ \left\lceil\frac c3\right\rceil+ \left\lceil\frac c2\right\rceil-c \ge-\left\lfloor\frac c6\right\rfloor. $$

Consequently, for $q=(\xi,\eta,1,1)$,

$$ \langle q,n\rangle-\operatorname{Min}_P(n) \ge h\xi+k\eta-\left\lfloor b/6\right\rfloor. \tag{W9} $$

For $(h,k)\ne(0,0)$, any violation of the desired lower bound one would lie in the bounded polar polygon

$$ (6(\xi,\eta)-b_0)\cdot(h,k)\le6 \quad\text{for every }b_0\in B_\triangle. \tag{W10} $$

Boundedness follows because $6(\xi,\eta)$ is strictly inside the base triangle. The positive barycentric weights and exact containing boxes are:

| Point | Barycentric weights in the displayed triangle | Box for $(h,k)$ | Nonzero normals | |---|---|---|---| | $q_1$ | $(59/140,3/7,3/20)$ | $[-2,1]\times[-1,1]$ | $11$ | | $q_2$ | $(19/70,3/7,3/10)$ | $[-2,1]\times[-1,1]$ | $11$ | | $q_3$ | $(19/140,5/7,3/20)$ | $[-4,1]\times[-2,1]$ | $23$ |

Pairwise intersections of the three lines in (W10) give these boxes by exact arithmetic. Every nonzero integral pair in each box satisfies $h\xi+k\eta-\lfloor b/6\rfloor\ge1$, as the standalone verifier enumerates. All potential violations have been included. For $(h,k)=(0,0)$ and $(l,m)\ne(0,0)$, the required inequality is $l+m-\min(0,3l,2m)\ge1$, since $(1,1)$ is strictly inside the integral triangle with vertices $(0,0),(3,0),(0,2)$. This proves every lattice-normal inequality. The base-coordinate determinant of $q_2-q_1,q_3-q_1$ is $-2/3$, so $\dim F(P)=2$.

Nondegeneracy for fixed full Newton support is a dense open condition [1, Introduction]. The two nonzero fiber coefficients can be normalized by torus scaling, so this condition meets our family. Batyrev's Theorem 9.2 gives $\kappa(T)=\min(3,2)=2$ on that open subset. Nonconstant $j$ is another nonempty open condition: the sparse choice $A=w^{80}$, $B=w^{120}+xw^{103}$ gives

$$ j=\frac{6912}{4+27(1+x)^2}. $$

This example certifies nonconstancy only; its own Kodaira dimension is not being asserted. Irreducibility of the coefficient space makes the two open subsets intersect, proving part 3.

The nonzero subspace in part 4. The quasismooth locus of (W2) is nonempty. For

$$ f=V^2-U^3-x^7w-y^{40}-w^{120}, \tag{W11} $$

the cone derivatives force $V=U=y=0$, $x^6w=0$, and $x^7+120w^{119}=0$, hence $x=w=0$. Thus the cone is smooth off its vertex. For any quasismooth member, weighted Lefschetz and the weighted residue theorem, in the form stated in [15, Section 2.2], give

$$ H^{2,1}(Q)\cong R_{119},\qquad H^{3,0}(Q)\cong R_{-1}=0, $$

where $R$ is the graded Milnor algebra. Its Hilbert series is

$$ \frac{(1-z^{119})(1-z^{117})(1-z^{103})(1-z^{80})(1-z^{60})} {(1-z)(1-z^3)(1-z^{17})(1-z^{40})(1-z^{60})}. \tag{W12} $$

The coefficient of $z^{119}$ is $234$. Independently, a lexicographic Gröbner basis for (W11)'s Jacobian ideal is

$$ V,\ U^2,\ y^{39},\ x^7+120w^{119},\ wx^6,\ w^{120}. $$

Counting its standard monomials of weighted degree $119$ gives the same $234$. Weighted Lefschetz gives $H^1(Q)=0$, and duality gives $H^5(Q)=0$, so all of $H^3(Q)$ is primitive. In every member, $\partial f/\partial V=2V$, so residue representatives have $V$-exponent zero. The involution $\iota$ changes the sign of the weighted volume form and fixes $f$, and therefore acts as $-1$ on $H^3(Q)$, which has dimension $468$.

A quasismooth weighted hypersurface is a compact complex orbifold. Locally it is a finite quotient of a smooth ball, so its local rational homology is that of a manifold: average the relative cohomology of the ball over the finite, orientation-preserving group. It therefore satisfies rational Poincaré duality. Let $\pi:T\to Q$ be the resolution of the finite normalization specified in part 4. For $\alpha,\beta\in H^3(Q,\mathbb Q)$,

$$ \int_T\pi^*\alpha\smile\pi^*\beta =2\int_Q\alpha\smile\beta. \tag{W13} $$

The pairing on the right is nondegenerate; hence $\pi^*$ is injective. Pullback is a morphism of pure Hodge structures. On its image $d$ acts trivially, $\iota$ acts by $-1$, and thus $\sigma=d\iota$ acts by $-1$. This gives the claimed $468$-dimensional subspace. Quasismoothness is open, and its nonempty locus intersects the nondegenerate, nonconstant-$j$ loci used above. $\square$

The change (W6) shows why the three local valuation bounds fit together: their base normals satisfy $17(4,1)+3(-17,-4)+(-17,-5)=0$, the relation for the weighted plane $\mathbb P(1,3,17)$. The global negative canonical class of (W2) extends the support mechanism to both Weierstrass coefficients. Neither the constant-$j$ splitting nor a fixed CM elliptic factor is needed for Theorem F.

2.6. General-type carriers and very general higher tensors

The quotient-image criterion also applies to threefolds with ample canonical bundle. The following elementary application uses classical bidouble-cover geometry and classical monodromy. It illustrates a different range of the same method; the decomposition of a bidouble cover and the monodromy criterion are not new assertions of general theory. Related decompositions and Hodge applications for surfaces of general type are studied in [16]. For a general sextic double solid, Ceresa--Verra [17, Theorem 3.3] give the stronger description of its intermediate Jacobian as the Albanese of its surface of conics. We use uniruledness for curve support uniformly over all the smooth branch pairs below.

Let $W$ be the open subset of $H^0(\mathbb P^3,\mathcal O(6))\times H^0(\mathbb P^3,\mathcal O(6))$ consisting of pairs $(f,g)$ with smooth transverse zero surfaces. Define

$$ T_{f,g}:\quad u^2=f(x_0,x_1,x_2,x_3),\qquad v^2=g(x_0,x_1,x_2,x_3) \subset\mathbb P(1,1,1,1,3,3). \tag{G1} $$

Let $d$ change the sign of $u$, let $\iota$ change the sign of $v$, and put $\sigma=d\iota$ and $V_T=H^3(T,\mathbb Q)^{\sigma=-1}$.

Corollary G. The following hold.

1. Every member of (G1) is a smooth threefold with ample canonical bundle, $\kappa(T)=3$ and $h^{3,0}(T)=10$. Its selected sector $V_T$ has dimension $208$ and Hodge numbers $(0,104,104,0)$ and is generated by curves.
2. For every two members $T,T'$ of (G1), every rational Hodge class in $V_T\otimes V_{T'}$ is algebraic. The same pairwise conclusion holds when either factor is replaced by an odd sector of Theorem E or F.
3. Fix a positive integer $s$. For a very general ordered tuple $(T_1,\ldots,T_s)\in W^s$ and any nonnegative integers $a_1,\ldots,a_s$, every rational Hodge class in

$$ \bigotimes_{i=1}^s V_{T_i}^{\otimes a_i} \subset H^{3\sum_i a_i}\!\left(\prod_{i=1}^s T_i^{a_i},\mathbb Q\right) \tag{G2} $$

is algebraic. In particular this applies to all tensor powers of the selected sector of a very general member.

Part 3 concerns very general tuples and their repetitions. The all-member curve-support argument does not cover the complementary even third cohomology, which contains the holomorphic three-forms. Section 2.7 separately treats that sector by monodromy at very general points and proves full HC there. No higher-tensor statement for every special member follows from Corollary G.

Geometry and the all-member statement. The map $\pi:T\to\mathbb P^3$ is finite of degree four, with

$$ \pi_*\mathcal O_T=\mathcal O\oplus\mathcal O(-3)\oplus \mathcal O(-3)\oplus\mathcal O(-6),\qquad K_T=\pi^*\mathcal O(2). \tag{G3} $$

Off the branch locus the cover is etale. Near a point on exactly one branch surface it is a smooth double cover; near their transverse intersection its local map is $(u,v,z)\mapsto(u^2,v^2,z)$. Thus $T$ is smooth. It avoids the ambient weighted singular locus, since setting all four $x_i$ to zero in (G1) would force $u=v=0$. Adjunction, or the finite-cover canonical formula, gives (G3); the canonical bundle is ample. Projection formula gives $h^{3,0}=h^0(\mathbb P^3,\mathcal O(2))=10$. A local canonical form has denominator $uv$, so both $d$ and $\iota$ act by $-1$ on the holomorphic forms, and $\sigma$ acts by $+1$. The open $W$ is nonempty: take $f=\sum x_i^6$ and $g=\sum c_i x_i^6$ with distinct nonzero $c_i$. Linear dependence of the two gradients at a common zero would force just one $x_i$ to be nonzero, contradicting $f=0$.

The quotient maps $p_f:T\to Q_f=\{u^2=f\}$ and $p_g:T\to Q_g=\{v^2=g\}$ give smooth sextic double solids. Their canonical bundles are $\mathcal O(-1)$; hence they are Fano and uniruled. Finite-quotient rational cohomology, and $H^3(\mathbb P^3)=0$, give the exact splitting

$$ V_T=p_f^*H^3(Q_f)\oplus p_g^*H^3(Q_g). \tag{G4} $$

Indeed these images have characters $(-,+)$ and $(+,-)$ for $(d,\iota)$, respectively; the $(+,+)$ part is zero. The projectors belong to the rational group algebra, so they are algebraic. Proposition B gives curve generation. Applying the curve-divisor lifting proof after (U3) to any two such generated sectors gives part 2, including the asserted mixtures with Theorems E and F.

For a sextic double solid in $\mathbb P(1,1,1,1,3)$, the weighted residue theorem [15, Section 2.2] gives

$$ h^{2,1}(Q_f)=\dim R_5=\binom{8}{3}-4=52, \qquad h^{3,0}(Q_f)=0. \tag{G5} $$

Here the derivative in the weight-three variable removes that variable, and the four independent quintic derivatives impose four relations in degree five. As an independent check, a smooth sextic surface has Euler characteristic $108$, so $e(Q_f)=2e(\mathbb P^3)-108=-100$. The smooth sextic hypersurface $a^6=f(x)$ in $\mathbb P^4$ maps to $Q_f$ with degree three by $u=a^3$. Ordinary weak Lefschetz and finite pullback therefore give $b_1=0,b_2=1$, whence $b_3=104$. Equation (G4) proves the exact rank and Hodge numbers in part 1. The number $168=2\binom93$ counts affine coefficients, not moduli.

Monodromy and higher tensors. We record the classical monodromy argument to make the generic scope precise. Let $U\subset H^0(\mathbb P^3,\mathcal O(6))$ parametrize smooth sextics. Its discriminant is irreducible: the incidence of a point of $\mathbb P^3$ and a sextic singular there is a projective bundle after projectivizing coefficients, and its image is the discriminant. A general discriminant member has one ordinary node. The associated double solid also has one ordinary threefold node, so its local cohomological monodromy is a Picard--Lefschetz transvection. Meridians normally generate $\pi_1(U)$, since the ambient affine coefficient space is simply connected; all nodal vanishing cycles belong to one transported orbit up to sign.

These cycles span $H^3(Q_f)$. To see this without an assumption about invariant cycles at a boundary compactification, use the scalar loop $f_t=e^{2\pi it}f$, $0\le t\le1$. The trivialization $u_t=e^{\pi it}u$ has endpoint the deck involution. Its cohomological action is $-\mathrm{Id}$ on $H^3(Q_f)$, because the invariant part is $H^3(\mathbb P^3)=0$. Thus the monodromy invariant subspace is zero. By Picard--Lefschetz and normal generation, the orthogonal complement of all transported nodal cycles is exactly this invariant subspace. Nondegeneracy of the pairing proves spanning. This is the scalar-loop argument used in the corrigendum to [18, Proposition 0.1 and Remark 0.2].

The single-orbit transvection criterion, in the form of [18, Proposition 4.3], now gives connected algebraic monodromy $\operatorname{Sp}_{104}$. For completeness, the Zariski closure of a group generated by transvections is connected: the powers of each transvection are Zariski dense in its one-parameter unipotent subgroup. In the criterion's alternative, a nontrivial invariant subspace and its orthogonal complement would contain the one algebraic orbit of vanishing cycles. That orbit is irreducible because the group is connected, so it would be contained in one of the two subspaces, contradicting spanning. We use this linear-algebra criterion from [18], not its theorem about hyperplane arrangements.

The inclusion $W\subset U\times U$ induces a surjection on fundamental groups: a loop in the smooth ambient space can be perturbed away from the removed complex algebraic locus of positive codimension. The two quotient variations in (G4) therefore have product algebraic monodromy $\operatorname{Sp}_{104}\times\operatorname{Sp}_{104}$. For $s$ independently varying pairs the group is $\operatorname{Sp}_{104}^{2s}$.

At very general points the Hodge group contains the connected algebraic monodromy group [13, Lemma 6.1; 18, Corollary 1.2 and its preceding discussion]. It preserves the rational block projectors and their polarizations, and so is contained in the same product of symplectic groups. Equality follows. The exceptional sets can be taken to be the countable union of proper Hodge loci for tensor classes; the assertion is simultaneous for all tensor degrees.

The first fundamental theorem for symplectic invariants [19, Theorem 4.1 and Corollary 4.2] says that tensor invariants are generated by pairwise contractions, or dually by inverse-polarization tensors. Apply it separately to each of the $2s$ blocks. Every such generating Hodge tensor belongs to a pair of selected sectors and is algebraic by part 2. Taking external products, permuting factors and applying the algebraic block projectors represents every Hodge class in (G2). This proves part 3. $\square$

The generic conclusion uses monodromy to exclude exceptional higher tensors. At special coefficient pairs the proof supplies curve support and algebraicity for two factors, but does not assert that Hodge classes in arbitrary powers of the supporting curves are algebraic. This distinction is essential.

2.7. Full Hodge classes on very general bidouble-cover products

The third character of the bidouble family can also be controlled at very general points. This uses local vanishing cycles in addition to the curve-support argument. It gives a full-product conclusion for that generic family, while Corollary G retains its all-member selected-sector scope.

Theorem H. Fix $s\ge1$. For a very general tuple $(T_1,\ldots,T_s)\in W^s$ of the sextic bidouble covers (G1), the rational Hodge conjecture holds for

$$ \prod_{i=1}^s T_i^{a_i} \tag{H1} $$

for every choice of nonnegative integers $a_i$. In particular, it holds for every power of a very general member, which has ample canonical bundle. The connected algebraic monodromy group on the third cohomology of one member is

$$ \operatorname{Sp}_{104}\times\operatorname{Sp}_{104} \times\operatorname{Sp}_{500}. \tag{H2} $$

The three blocks and the nonmiddle cohomology. Write $A=H^3(T)^{(-,+)}$, $B=H^3(T)^{(+,-)}$ and $C=H^3(T)^{(-,-)}$ for the characters of $(d,\iota)$. There is no invariant third cohomology. The ordinary complete intersection

$$ X:\quad a^6=f(x),\qquad b^6=g(x)\subset\mathbb P^5 \tag{H3} $$

is smooth by the same branch and transversality check as (G1). It maps finitely to $T$ with degree nine by $u=a^3,v=b^3$. Ordinary weak Lefschetz gives $H^1(X)=0$ and $H^2(X)=\mathbb Q$, so injectivity of finite pullback gives $H^1(T)=0$ and $H^2(T)=\mathbb Q H$, where $H=\pi^*c_1(\mathcal O_{\mathbb P^3}(1))$. Duality shows that all nonmiddle cohomology is Tate, with one generator in degrees $0,2,4,6$ and none in degrees $1,5$.

The complete-intersection curve $D_f\cap D_g$ has genus $145$ and Euler characteristic $-288$. Stratifying the degree-four cover by its branch loci gives

$$ e(T)=4e(\mathbb P^3)-2e(D_f)-2e(D_g)+e(D_f\cap D_g) =-704. \tag{H4} $$

Hence $b_3(T)=708$. The two quotient blocks have ranks $104$ each, and therefore $\dim C=500$, with Hodge numbers $(10,240,240,10)$. Each block is nondegenerate for the alternating intersection pairing, since distinct group characters are orthogonal.

Three kinds of nodal degeneration. In the affine coefficient space $H^0(\mathcal O(6))^2$, the complement of $W$ has three irreducible divisorial components: $\Delta_f$, where $f$ is singular; $\Delta_g$, where $g$ is singular; and $\Delta_t$, where their smooth zero surfaces are tangent. For the last assertion, choose $p\in\mathbb P^3$ and $\lambda\in\mathbb C^*$ and impose

$$ f(p)=g(p)=0,\qquad dg_p=\lambda\,df_p,\qquad df_p\ne0. $$

The five linear first-jet conditions on coefficients give an irreducible incidence over the four-dimensional space of $(p,\lambda)$. Its image closure is a divisor, whose general point has one ordinary tangency and no other singularities. Sextic coefficients prescribe the needed second jets, so ordinary tangencies form a nonempty open subset. The analogous singular-point incidence gives $\Delta_f,\Delta_g$. Worse loci have higher codimension or lie in these closures. Meridians about these components normally generate $\pi_1(W)$.

Near a general point of $\Delta_t$, use $f=z$, $g=z+x^2+y^2-a$. Eliminating $z$ gives the ordinary threefold node

$$ v^2-u^2-x^2-y^2+a=0. \tag{H5} $$

Both $d$ and $\iota$ change the sign of one quadratic coordinate, and hence act by $-1$ on its vanishing three-sphere. The cycle lies in $C$, and the meridian is its Picard--Lefschetz transvection.

At a general point of $\Delta_f$, the node of $f$ lies away from $g=0$. There are two disjoint nodes on $T$, exchanged by $\iota$. Orient their vanishing cycles $\delta_1,\delta_2$ so that $\iota\delta_1=\delta_2$. Both are $d$-odd. The meridian is the product of their two commuting transvections. On $C$ its direction is $\delta_1-\delta_2$, with nonzero coefficient $1/2$ relative to the restricted intersection pairing. On $A$ its direction is $\delta_1+\delta_2$, also with coefficient $1/2$; it is the identity on $B$. The description for $\Delta_g$ is symmetric. The coefficients can be absorbed by rescaling the directions over $\mathbb C$ for the monodromy criterion.

An $A_3$ collision joins the three orbits on $C$. Let a node of $f$ meet the smooth surface $g=0$. The local two-parameter model

$$ f=x^2+y^2+z^2-a,\qquad g=z-b $$

gives, after $z=v^2+b$,

$$ u^2-x^2-y^2-(v^2+b)^2+a=0. \tag{H6} $$

At $a=b=0$ this is an $A_3$ singularity. Its even deformation has discriminant branches $a=0$, with two nodes, and $a=b^2$, with one node. They are $\Delta_f$ and $\Delta_t$, respectively. The identity

$$ \operatorname{disc}_v(v^4+2bv^2+b^2-a)=256a^2(b^2-a) \tag{H7} $$

also checks the multiplicities.

Choose four real ordered roots in a nearby smoothing. The vanishing two-sphere in the quadratic fiber over each interval between consecutive roots collapses at both endpoints, producing matching three-spheres $e_1,e_2,e_3$. Nonadjacent spheres are disjoint, while adjacent spheres have one transverse intersection after a local perturbation. Orienting the chain gives the alternating matrix

$$ \begin{pmatrix}0&1&0\\-1&0&1\\0&-1&0\end{pmatrix}. \tag{H8} $$

The middle sphere is the tangency cycle of (H5), so $\iota e_2=-e_2$. The map $v\mapsto-v$ exchanges the outer spheres; preservation of (H8) fixes their relative signs as $\iota e_1=e_3$. The involution $d$ reflects one coordinate of every quadratic fiber and acts by $-1$ on each matching sphere. Consequently $e_1-e_3$ and $e_2$ lie in $C$ and

$$ \langle e_1-e_3,e_2\rangle=2\ne0. \tag{H9} $$

Local intersection numbers are preserved under inclusion of the Milnor fiber into a global smooth fiber. Thus these cycles survive globally, and the $\Delta_f$ and $\Delta_t$ orbits on $C$ have nonzero pairing. The symmetric collision links the $\Delta_g$ orbit to the $\Delta_t$ orbit.

These collisions occur in the sextic family, with no extra singularities needed. For an explicit nodal branch, take

$$ f_0=x_0^4(x_1^2+x_2^2+x_3^2)+x_1^6+4x_2^6+16x_3^6. \tag{H10} $$

Its only singular point is $p=[1,0,0,0]$. No singular point lies on $x_0=0$. On $x_0=1$, the gradient equations give $x_1^2+x_2^2+x_3^2=0$ and, for each nonzero coordinate, $x_i^4=-1/(3c_i)$ with $(c_1,c_2,c_3)=(1,4,16)$. They would force a nonzero signed sum of $1,1/2,1/4$ to vanish, which is impossible. The quadratic part at $p$ is nondegenerate. Choose $g$ generally with first jet $x_3$ at $p$. Bertini with this prescribed jet gives a smooth surface transverse to $f_0$ away from $p$; the restricted quadratic form on $dg=0$ is nondegenerate. To see the equivariant normal form, first use $g$ as the coordinate $z$ and apply the Morse lemma in the two transverse variables. The remaining term has the form $h(z)=c z^2+O(z^3)$ with $c\ne0$. Substituting $z=v^2$ and rescaling $v$ by an invertible even analytic function gives (H6) at the origin, compatibly with both involutions. The even part of the $A_3$ deformation space has basis $1,v^2$. Varying the constant jets of $f,g$ supplies these two directions: in (H6) the coefficient map is $(a,b)\mapsto(b^2-a,2b)$, with invertible derivative at zero. The same derivative remains invertible in the above normal form. Thus this slice is versal for the even deformations, so the local discriminant and vanishing-cycle calculation persist with the general higher jets.

Full symplectic monodromy on $C$. Let $R_{\mathrm{mon}}$ consist of the transported $C$ directions from all three discriminant components. Picard--Lefschetz and normal generation give $R_{\mathrm{mon}}^\perp=C^{\mathrm{mon}}$. The scalar loop scaling $f$ alone stays in $W$ and acts by $d=-\mathrm{Id}$ on $C$. Thus $R_{\mathrm{mon}}$ spans $C$. Let $G$ be the Zariski closure of the monodromy on $C_{\mathbb C}$. Choose one direction for each discriminant component, rescaled so its transvection has coefficient one, and replace the transported set by the union $R$ of their $G$-orbits and their negatives. Then $R$ spans $C_{\mathbb C}$ and is a finite union of algebraic $G$-orbits. Every associated transvection belongs to $G$, by conjugation, and they generate $G$ as an algebraic group because they include the original meridian transvections.

Apply [18, Proposition 4.3] to $G$, which is connected as in Section 2.6. If its alternative held, there would be a nontrivial proper $G$-invariant subspace $L$ with every direction in $L\cup L^\perp$. Each algebraic $G$-orbit is irreducible, so it lies in one of these subspaces. An orbit in $L\cap L^\perp$ would pair trivially with every other orbit, contrary to the links (H9). Those links force the $f,t,g$ orbits to lie on the same side; taking their negatives does not change the side. Their spanning then contradicts properness of $L$ or $L^\perp$. Therefore $G=\operatorname{Sp}_{500}$.

The product group and algebraic invariants. Section 2.6 gives full symplectic projections on $A$ and $B$, and the preceding paragraph gives it on $C$. The connected monodromy Lie algebra is a subdirect product of three simple symplectic Lie algebras. Its solvable radical maps to zero in every factor, so it is semisimple. A proper subdirect product could only identify isomorphic simple factors diagonally. The rank-$500$ factor is not isomorphic to either rank-$104$ factor. The latter two cannot be diagonally identified: a $\Delta_f$ meridian has nontrivial unipotent projection on $A$ and trivial projection on $B$. This proves (H2). Independently varying $s$ members gives the product of $s$ copies of (H2).

At very general points the Hodge group contains this connected monodromy group, and it is contained in it because it preserves the three rational character projectors and polarizations. Thus equality holds. Each nontrivial character projector is an algebraic linear combination of automorphism graphs. Since it kills all nonmiddle cohomology, its cohomology class lies entirely in the corresponding $H^3\otimes H^3$ block and represents its inverse-pairing tensor, with the conventional duality sign. These tensors, hyperplane classes, their external products and permutations are algebraic. By symplectic invariant theory [19, Theorem 4.1], they span every rational Hodge class in every product (H1). The Hodge-locus theorem [20, Theorem 1.1 and Corollaries 1.2--1.3] makes the exceptional tensor loci algebraic. There are countably many rational tensor classes and powers, so a countable union of proper closed algebraic subsets can be excluded simultaneously. This proves Theorem H. $\square$

Theorem H is a very-general result obtained by classical monodromy and invariant theory with the explicit collision (H6). At special members the Hodge group can drop and acquire additional tensors; the theorem does not assert their algebraicity. The all-member statements of Theorems E, F and Corollary G concern the specified curve-generated sectors and pairs of them.

3. Three curves and their invariant cohomology

Let $A,D,C$ be the smooth projective normalizations of

$$ A:\ a^2=l^{11}+1,\qquad D:\ w^{22}=x^3-x,\qquad C:\ v^4=u^3(1+u^{11}). \tag{3} $$

Their genera are $5,21,16$, respectively. Fix compatible primitive roots of unity and automorphisms

$$ \begin{aligned} \gamma(l,a)&=(\zeta_{11}l,-a),\quad &&\operatorname{ord}(\gamma)=22,\\ k(x,w)&=(-x,\zeta_{44}w),&&\operatorname{ord}(k)=44,\\ h(u,v)&=(\zeta_{44}^{16}u,\zeta_{44}v),&&\operatorname{ord}(h)=44. \end{aligned} \tag{4} $$

On $X=A\times D\times D$, set

$$ G=\{(i,j,k)\in\mathbb Z/22\times(\mathbb Z/44)^2: 18i+31j+7k=0\pmod {44}\}. \tag{5} $$

The order of $G$ is $968$.

Lemma 2. The invariant space $M=H^3(X)^G$ is a rank-one module over $E=\mathbb Q(\zeta_{44})$, with Hodge numbers $(0,10,10,0)$. Its complex character lines are

$$ (9e\bmod22,31e\bmod44,7e\bmod44), \qquad e\in(\mathbb Z/44)^*. \tag{6} $$

The residual automorphism $k_2^{19}$ acts on the line indexed by $e$ as $\zeta_{44}^e$. The $(1,0)$ embeddings of $M(1)$ are

$$ \{7,13,19,23,27,29,35,39,41,43\}. \tag{7} $$

Proof. A basis of $H^{1,0}(A)$ is $l^{r-1}dl/a$, $1\leq r\leq5$, with $\gamma$ character $2r+11$ modulo $22$. Together with their negatives these are precisely the odd residues other than $11$, each with multiplicity one.

On $D$, the holomorphic forms are $x^s dx/w^j$ for $1\leq j\leq21$ and $s\geq0$ satisfying

$$ 3j-22(s+1)-1\geq0. $$

Their $k$ characters are $22(s+1)-j$ modulo $44$. This gives the sets $\{1,\ldots,14\}$ and $\{23,\ldots,29\}$, each with multiplicity one. The full $H^1(D)$ has every character other than $0,22$, once.

The annihilator of $G$ consists of the multiples of $(9,31,7)$ in these character conventions. The $A$ factor permits exactly the units $e$ modulo $44$, yielding (6). All other degree-three Künneth terms have only one $H^1$ factor; the two zero characters force the multiple to be zero, and these terms have no invariants. Since $7\cdot19\equiv1\pmod {44}$, the asserted residual field action follows.

Counting holomorphic factors in (6) gives two exactly for the ten entries in (7), and one for their negatives. This proves the profile and the weight-one type. These finite computations are also reproduced by the supplement. $\square$

The primitive-$44$ projector on $H^1(C)$ is the rational cyclotomic idempotent for $h$. Denote its image by $L$. It is another $E$-line. To check its type on the actual curve, use the forms $u^sdu/v^j$, $1\leq j\leq3$. Regularity at zero and infinity is equivalent to

$$ 4s+3-3j\geq0,\qquad -2s-3+7j\geq0. \tag{8} $$

At the eleven other branch points the orders are $3-j\geq0$. The resulting sixteen forms have $h$ character $16(s+1)-j$. The primitive holomorphic characters are

$$ \{1,3,5,9,15,17,21,25,31,37\}, \tag{9} $$

which is the negative of (7). Thus

$$ L^{\mathrm{inv}}\simeq M(1) \tag{10} $$

as $E$-linear Hodge structures, where the superscript indicates that $E$ acts on $L$ through $h^{-1}$. At this stage (10) is only a Hodge identification.

4. The actual quotient and its Weierstrass equation

Write $P_i=x_i^3-x_i$. On a dense open of $X$, define

$$ A_0=\frac{x_1}{x_2},\quad B=x_2^2,\quad T=l w_1^{-20}w_2^4,\quad V=\frac{aP_2w_1^{11}w_2^{11}}{P_1^5}. \tag{11} $$

These functions are $G$-invariant. Put

$$ K=A_0B(A_0^2B-1)(B-1),\qquad F=\frac{B-1}{A_0^5B^2(A_0^2B-1)^5}. $$

Direct substitution gives

$$ V^2=K(F^2+T^{11}). \tag{12} $$

Lemma 3. Equation (12) is a birational model of $X/G$.

Proof. In addition to invariance, it is necessary to check the degree of the function-field extension. From a generic point of (12), choose $x_2^2=B$, set $x_1=A_0x_2$, choose $w_i^{22}=P_i$, and set

$$ l=T\frac{w_1^{20}}{w_2^4},\qquad a=\frac{VP_1^5}{P_2w_1^{11}w_2^{11}}. $$

Equation (12) implies $a^2=l^{11}+1$. There are exactly $2\cdot22^2=968$ choices on this open. This equals $|G|$, so the invariant functions generate the quotient field. $\square$

Set $c_0=B^2(B-1)$ and make the birational change

$$ S=TA_0(A_0^2B-1),\qquad X_w=c_0A_0,\qquad Y_w=c_0V/F. \tag{13} $$

Its inverse is $A_0=X_w/c_0$, $T=S/[A_0(A_0^2B-1)]$, $V=FY_w/c_0$. Equation (12) becomes

$$ Y_w^2=X_w^3-B^3(B-1)^2X_w+S^{11}B^9(B-1). \tag{14} $$

Both directions of this change are identities of rational functions.

5. A canonical pole and uniruledness

Proposition 4. Every smooth projective model of (14) is uniruled.

Proof. Compactify (14) as the ordinary cubic Weierstrass model $\pi:W\to Q$, where

$$ Q=\mathbb P^1_B\times\mathbb P^1_S,\qquad \mathcal L=\mathcal O_Q(2,2). $$

The coefficients $a_4=-B^3(B-1)^2$ and $a_6=S^{11}B^9(B-1)$ are global sections of $\mathcal L^4$ and $\mathcal L^6$ after homogenization. The model is the cubic $y^2z=x^3+a_4xz^2+a_6z^3$ in the associated projective bundle. Adjunction gives

$$ \omega_W=\pi^*(\omega_Q\otimes\mathcal L)=\mathcal O_W. \tag{15} $$

Here $W$ is integral and normal. Every Weierstrass fiber is an integral plane cubic and the generic one is smooth, since its discriminant is not zero. The hypersurface has no vertical irreducible component, because every fiber has dimension one. Its singular locus is contained in the relative singular locus, which has finite fibers over the discriminant divisor; hence its dimension is at most one. The infinity section is smooth. The hypersurface is $S_2$ and regular in codimension one, proving normality. These arguments also justify using its invertible dualizing sheaf in (15).

We exhibit a genuine pole of its canonical generator on a birational model. At $B=0,S=\infty$, use $b=B,t=1/S,x=t^4X_w,y=t^6Y_w$. Then

$$ y^2=x^3-b^3(b-1)^2t^8x+b^9(b-1)t. \tag{16} $$

In the $t$ chart of the weighted blowup with weights $(b,t,x,y)=(2,1,6,9)$, put

$$ b=r^2b_0,\qquad t=r,\qquad x=r^6x_0,\qquad y=r^9y_0. $$

After division by $r^{18}$ the strict transform is

$$ y_0^2-x_0^3+r^2b_0^3(r^2b_0-1)^2x_0 -rb_0^9(r^2b_0-1)=0. \tag{17} $$

Its exceptional divisor $r=0,y_0^2=x_0^3$ is irreducible and the strict transform is generically smooth there; the derivative with respect to $r$ is $b_0^9$. At its generic point $y_0\ne0$ as well.

The canonical generator is locally a nonzero constant multiple of $db\wedge dt\wedge dx/(2y)$. Its pullback is

$$ \frac1r\frac{db_0\wedge dr\wedge dx_0}{2y_0}. \tag{18} $$

Since $(b_0,r,x_0)$ are coordinates at the generic point, (18) has a simple pole along this divisor. Equivalently its discrepancy is $2+1+6+9-1-18=-1$.

Choose a smooth projective resolution $Y\to W$ dominating this weighted blowup and an isomorphism over $W_{\mathrm{reg}}$. A section of $mK_Y$, restricted to $W_{\mathrm{reg}}$, is $f\Omega^m$, where $\Omega$ is the generator in (15). The regular function $f$ extends to $W$ by normality, and is constant by projectivity and integrality. Formula (18) excludes every nonzero constant. Thus

$$ H^0(Y,mK_Y)=0\quad\text{for every }m>0. \tag{19} $$

The threefold minimal model and abundance theorem [12, Theorem 1.2] says that a smooth projective non-uniruled threefold has a good minimal model and hence nonnegative Kodaira dimension. Applied to the smooth $Y$, it contradicts (19). Therefore $Y$ is uniruled. We do not apply that theorem to the noncanonical model $W$. $\square$

Alternative global calculation. The following interpretation of the same valuation was suggested by gabrielov_cell in the discussion and is checked independently in the supplement. Blow up the point $(b,t)=(0,0)$ on the base, with exceptional curve $E_1$. In the chart $b=tb_1$, divide the two Weierstrass coefficients by $t^4,t^6$, respectively. In the equation $y^2-x^3+g_2x+g_3=0$, they become

$$ g_{2,1}=b_1^3t^7(b_1t-1)^2,\qquad g_{3,1}=-b_1^9t^4(b_1t-1). $$

The fundamental line bundle becomes $\mathcal L_1=\sigma_1^*\mathcal L-E_1$, and $K_{Q_1}+\mathcal L_1=0$. Blow up $(b_1,t)=(0,0)$, with exceptional curve $E_2$. The coefficient orders there are $(10,13)$, so dividing by the second exceptional parameter to powers $8,12$ gives regular coefficients and fundamental bundle

$$ \mathcal L_2=\sigma_2^*\mathcal L_1-2E_2,\qquad K_{Q_2}+\mathcal L_2=-E_2. $$

In the chart $b_1=tb_0$ these coefficients are

$$ b_0^3t^2(b_0t^2-1)^2,\qquad -b_0^9t(b_0t^2-1), $$

and in the other chart $t=b_1t_0$ they are

$$ b_1^2t_0^7(b_1^2t_0-1)^2,\qquad -b_1t_0^4(b_1^2t_0-1). $$

The unchanged chart of the first blowup, $t=bt_1$, has coefficients $b^7t_1^8(b-1)^2,-b^4t_1(b-1)$. Thus the resulting global cubic Weierstrass model $\pi_2:W_2\to Q_2$ is regular in codimension one and normal by the same fiber argument as above, and

$$ \omega_{W_2}=\pi_2^*\mathcal O_{Q_2}(-E_2). $$

For a resolution, restriction to the regular locus and extension of sections of this invertible sheaf across the codimension-two complement give

$$ H^0(\widetilde W_2,mK_{\widetilde W_2}) \hookrightarrow H^0(W_2,\omega_{W_2}^{\otimes m}) =H^0(Q_2,\mathcal O_{Q_2}(-mE_2))=0. $$

Here $\pi_{2*}\mathcal O_{W_2}=\mathcal O_{Q_2}$, since the fibers are connected and the base is normal. This proves the same vanishing globally; it still uses normality for the extension step. The two calculations are complementary descriptions of the same birational modification.

6. Curve generation and realization of the Hodge identification

We say that $H^3(Z)$ is generated by curves if a finite sum of algebraic correspondences induces a surjection $\bigoplus_jH^1(U_j)(-1)\to H^3(Z)$, with $U_j$ smooth projective curves.

Lemma 5. If a smooth projective threefold $Y$ is uniruled, then $H^3(Y)$ is generated by curves. If a smooth projective threefold $X$ has a finite group $G$ and a smooth projective model of $X/G$ is uniruled, then $H^3(X)^G$ is generated by curves.

Proof. Uniruledness gives a dominant generically finite rational map $R\times\mathbb P^1\dashrightarrow Y$ for a smooth projective surface $R$. Resolve its indeterminacy by blowups in smooth centers. The third cohomology of $R\times\mathbb P^1$ is curve-generated: hard Lefschetz on $R$ identifies $H^3(R)$ with an ample divisor times $H^1(R)$, and $H^1(R)$ is an algebraic image of the first cohomology of a curve. For the latter assertion, restrict to a smooth ample curve, split the resulting injection as a map of polarizable weight-one Hodge structures, and use Lefschetz $(1,1)$ on the curve times $R$ to realize the retraction. Blowups in curve centers add $H^1$ of those curves with twist $(-1)$; point centers add nothing in degree three. Push-pull for the resulting generically finite morphism is its degree, proving the first assertion.

For the second, first replace $G$ by its effective image in $\operatorname{Aut}(X)$. This leaves the quotient and the invariant cohomology unchanged; henceforth $G$ denotes that effective group, whose order is the generic quotient degree. Resolve the graph of $X\dashrightarrow Y$ as

$$ X\xleftarrow{p}X'\xrightarrow{f}Y. $$

For $\xi\in H^3(X)^G$ put $\beta=f_*p^*\xi$. On a common free quotient open, finite étale trace gives $p_*f^*\beta=|G|\xi$. Globally the difference

$$ |G|\xi-p_*f^*\beta \tag{20} $$

is supported on a divisor, so lies in $N^1H^3(X)$. The supported-cohomology description of coniveau expresses it using Gysin images of $H^1$ of divisor resolutions, with twist $(-1)$ [13, Section 3, displayed formula for $N^pH^i$]. These weight-one spaces are algebraically generated by curves by the preceding ample-curve and Lefschetz $(1,1)$ argument. The other term in (20) is curve-generated by the first assertion. Choose finitely many such curve maps spanning the finite-dimensional invariant space and apply the algebraic group projector. This proves the claim. In particular no global graph inverse or vanishing of exceptional cohomology is assumed. $\square$

Apply Lemma 5 to the quotient of Lemma 3 and Proposition 4. There is an algebraic surjection

$$ \bigoplus_jH^1(U_j)(-1)\longrightarrow M. \tag{21} $$

Choose a Hodge section of (21), and compose the Hodge isomorphism (10), with twist $(-1)$, with this section. After undoing the twist it is a map from the projected $H^1(C)$ to a sum of curve $H^1$ spaces. Each component is represented by a divisor on the corresponding curve product, by Lefschetz $(1,1)$. Composing these divisors with the cycles in (21) realizes the prescribed map

$$ L^{\mathrm{inv}}(-1)\xrightarrow{\ \sim\ }M \tag{22} $$

by a codimension-two cycle on $C\times A\times D\times D$. Its every embedding component is nonzero, since it is an $E$-linear isomorphism of $E$-lines. The auxiliary Hodge section itself is not claimed to be algebraic; the lift is algebraic because its source and target are curve $H^1$ structures.

7. The nonzero surface correspondence

The degree-two map

$$ C\longrightarrow A,\qquad (l,a)=(1/u,v^2/u^7) \tag{23} $$

intertwines $h$ with $\gamma^7$. Thus the $A$ character $9e$ pulls back to the $C$ character $2\cdot7\cdot9e=38e\pmod {44}$.

In (22), the input line has original $h$ character $-e$, whereas the corresponding cycle class has character $e$ in that curve position, by Poincaré duality. Pull (22) back through (23) in its $A$ position and reorder the four curves. Its nonzero character component on $C_b\times D_b\times C_c\times D_c$ is

$$ (38e,31e;e,7e)=(-6e,-13e;e,7e). \tag{24} $$

All these operations, including the degree-one curve projectors, are algebraic.

Consider the smooth projective models of

$$ K_b:\ Y^2=X^3-X+T^{11},\qquad S_c:\ r^4=x_s^3-tx_s+t^7. \tag{25} $$

There are dominant rational maps of degree $44$ from $C\times D$:

$$ \begin{aligned} q_b(u,v,x,w)&=(T,X,Y)=(w^2/u,x,w^{11}v^2/u^7),\\ q_c(u,v,x,w)&=(t,x_s,r)=(w^4u^2,w^2xu,w^7v). \end{aligned} \tag{26} $$

Substitution verifies the equations. For $q_b$, a generic target fixes $x$, permits $22$ choices of $w$, then fixes $u$ and permits two choices of $v$. For $q_c$, first $x$ has two choices from $x^2=x_s^2/t$, then $w$ has $22$ choices; $u=x_s/(w^2x)$ and $v=r/w^7$ are fixed.

Their holomorphic two-form pullbacks are

$$ \begin{aligned} q_b^*(dT\wedge dX/Y)&=-\frac{u^5du}{v^2}\wedge\frac{dx}{w^9},\\ q_c^*(dt\wedge dx_s/r^3)&=2\frac{u^2du}{v^3}\wedge\frac{dx}{w^{15}}. \end{aligned} \tag{27} $$

They are nonzero regular forms, with $(h,k)$ characters $(6,13)$ and $(1,7)$, respectively. Their Galois translates span rank-$20$ summands with embedding types $(1,18,1)$. These types can be checked either from the curve bases in Section 3 or from the Fermat covers below. Rational graph pullback and trace identify the corresponding transcendental summands of (25) with these subspaces of $H^2(C\times D)$; any resolution error in degree two consists of divisor classes. The character-orbit projectors on the curve products, their polarization adjoints and trace inverses are algebraic.

Here the graph identification can be seen directly from the quotient groups, so no stability of the image is implicit. On $K_b$ put

$$ g_b(T,X,Y)=(\zeta_{44}^2T,-X,\zeta_{44}^{11}Y). $$

The maps induced by $h$ and $k$ are $g_b^{14}$ and $g_b$. Hence $q_b$ is the quotient by the order-$44$ group $\langle(h,k^{-14})\rangle$, as its already computed degree confirms. Its primitive character pairs are $(6e,13e)$. On $S_c$ put

$$ g_c(t,x_s,r)=(\zeta_{44}^{32}t,\zeta_{44}^{16}x_s,\zeta_{44}r). $$

The maps induced by $h$ and $k$ are $g_c$ and $g_c^7$. Thus $q_c$ is the quotient by $\langle(h^{-7},k)\rangle$, with primitive pairs $(e,7e)$. These rational character orbits have no nonzero rational $(1,1)$ substructure: as $E$-lines with a nonzero $(2,0)$ embedding, a nonzero rational vector has a nonzero coordinate at that embedding. Semisimplicity then proves simplicity. Divisor-supported graph errors cannot meet them, and the quotient trace acts as $44$ there. This proves the claimed injectivity and algebraic inverses on exactly the summands used in (24).

It follows from (24) and (27) that pushing the cycle along $q_b\times q_c$ is nonzero on the dual of the first summand tensored with the second. Indeed, its component pairs nontrivially with the pulled-back first summand and lands in the pulled-back second one, where the finite traces are nonzero. We have obtained an algebraic isomorphism of their rank-$20$ Hodge structures. No matching-type assertion was used as a replacement for (21)--(22).

For completeness, the following explicit Fermat covers identify the two summands with (1). On the affine chart

$$ z_0^{44}+z_1^{44}+1+z_3^{44}=0 $$

of $F_{44}^2$, choose $\eta^{11}=-i$ and $\rho^2=i$. The map

$$ T=\eta z_3^4z_1^2,\qquad X=iz_1^{22},\qquad Y=\rho z_0^{22}z_1^{11} \tag{28} $$

lands on $K_b$. The pullback of its two-form is a nonzero constant times

$$ \frac{z_3^3z_1^{12}}{z_0^{22}}\,dz_3\wedge dz_1, $$

whose full Fermat character is $(22,13,5,4)$, a permutation of $b$.

Choose $\lambda^{11}=-1$, $\mu=-\lambda^6$ and $\nu^4=\lambda\mu$. Then

$$ t=\lambda z_1^4z_3^8,\quad x_s=\mu z_1^{24}z_3^4,\quad r=\nu z_0^{11}z_1^7z_3^3 \tag{29} $$

lands on $S_c$. The exponent determinant for $(t,x_s)$ is $4\cdot4-8\cdot24=-176$, proving dominance and nonzero differential. The two-form pulls back to a nonzero constant times

$$ \frac{z_1^6z_3^2}{z_0^{33}}\,dz_1\wedge dz_3, $$

with character $(11,7,23,3)$, a permutation of $c$.

Pulling back the constructed surface correspondence through (28)--(29), and applying the rational diagonal-character projectors, therefore gives a nonzero algebraic tensor on $F_{44}^2\times F_{44}^2$ in the orbit of $(-b,c)$. Conjugation gives the orbit of $(b,-c)$. Field operators coming from Fermat automorphisms span its full rational orbit from any nonzero component. The algebraic binary join maps this tensor nontrivially to $H^6(F_{44}^6)$ with character $b*(-c)$. Since $39$ is a unit modulo $44$, $\alpha_{44}$ in (2) is algebraic.

8. Degree 52: complementary involution quotients

The degree-$52$ construction starts from two threefold characters

$$ b_{52}=(7,15,23,26,33),\qquad c_{52}=(27,31,35,44,19). \tag{52a} $$

Write $M_b,M_c\subset H^3(F_{52}^3)$ for their rational orbit summands and set $\gamma_{52}=(14,15,23)$. For every unit $e$ modulo $52$, direct calculation gives

$$ \operatorname{age}(eb_{52})=1+\operatorname{age}(e\gamma_{52}),\qquad \operatorname{age}(ec_{52})=4-\operatorname{age}(e\gamma_{52}), \tag{52b} $$

where $\operatorname{age}(a)=\sum_i\langle a_i/52\rangle$. Each summand has rank $24$ and Hodge numbers $(0,12,12,0)$. The twists $M_b(1)$ and $M_c(1)$ have conjugate $\mathbb Q(\zeta_{52})$ types. We will prove that both are algebraically generated by curves with the indicated Tate twist. This is a geometric assertion beyond (52b).

8.1. The actual cyclic quotient

On the chart $z_0=1$ of $X=F_{52}^3$, put

$$ x=z_1^{52},\quad y=z_2^{52},\quad z=z_3^{52},\quad h=z_4^{52}=-1-x-y-z,\quad Z=z_1^{31}z_2^{35}z_3^{44}z_4^{19}. $$

The diagonal group is $(\mu_{52})^4$. The character $c_{52}$ is surjective, so its kernel $K$ has order $52^3$, and the quotient field is exactly

$$ \mathbb C(x,y,z,Z),\qquad Z^{52}=x^{31}y^{35}z^{44}h^{19}. \tag{52c} $$

Indeed the diagonal quotient has field $\mathbb C(x,y,z)$, and the monomial $Z$ has character of exact order $52$, so it generates the intermediate degree-$52$ extension fixed by $K$.

Set

$$ U=\frac{yh}{x^3},\quad V=\frac{zh^2}{x^4},\quad q=\frac{x}{h},\quad R=\frac{Z}{q^6h^4}. $$

Then $x=qh$, $y=Uq^3h^2$, $z=Vq^4h^2$, and the base relation becomes

$$ 1+(q+1)h+q^3(U+Vq)h^2=0. $$

With $\eta=2q^3(U+Vq)h+q+1$, this is the birational quadratic presentation

$$ \eta^2=(q+1)^2-4Uq^3-4Vq^4,\qquad R^{52}=U^{35}V^{44}. $$

The inverse is $h=(\eta-q-1)/(2q^3(U+Vq))$ on a dense open. The toric base cover is rational, with mutually inverse maps

$$ U=s^4,\quad V=t^{13}/s^2,\quad R=st^{11},\qquad s=U^9V^{11}/R^{13},\quad t=R^6/(U^4V^5). $$

Thus the actual quotient $X/K$ is birational to

$$ \eta^2=(q+1)^2-4s^4q^3-\frac{4t^{13}q^4}{s^2}. \tag{52d} $$

The residual deck generator $\delta:Z\mapsto\zeta_{52}Z$ acts by

$$ \delta(s,t,q,\eta)=(\zeta_{52}^{-13}s,\zeta_{52}^{6}t,q,\eta). $$

Its order-two element $\sigma=\delta^{26}$ sends $s$ to $-s$ and fixes the other coordinates. It commutes with $\iota:\eta\mapsto-\eta$, whose quotient has rational field $\mathbb C(s,t,q)$. The invariant field for the diagonal involution $\sigma\iota$ is generated by

$$ a=s^2,\qquad u=1/q,\qquad w=s\eta/q^2,\qquad t, $$

with relation

$$ w^2=au^2(u+1)^2-4a^3u-4t^{13}. \tag{52e} $$

These functions generate that field because $s$ is a quadratic choice and then $\eta=wq^2/s$ is determined. The remaining ambiguity is precisely $(s,\eta)\mapsto(-s,-\eta)$. We work with this direct double cover; no identification of its genus-one fibers with a Jacobian is needed.

8.2. A global pluriform bound and a canonical pole

Proposition 6. Every smooth projective model of (52e) is uniruled.

Proof. Let $Q=\mathbb P^1_a\times\mathbb P^1_u\times\mathbb P^1_t$ and $\mathcal L=\mathcal O_Q(2,2,7)$. Homogenize the right side of (52e) to a section $F$ of $\mathcal L^2$ and let $\pi:W\to Q$ be the associated finite double cover.

Its branch divisor is reduced. On the affine chart, a repeated prime factor of $F$ would divide its $t$ derivative, which is $-52t^{12}$, hence would be $t$; but $F$ is not divisible by $t$. At infinity the divisors $a=\infty$ and $t=\infty$ have multiplicity exactly one, and $u=\infty$ is not a component. Every other prime divisor meets the affine chart. The nonempty reduced branch makes $F$ nonsquare in $\mathbb C(Q)$, so $W$ is integral. It is a hypersurface in the total space of $\mathcal L$, hence Cohen--Macaulay and Gorenstein. Away from the branch it is étale; at a branch generic point its equation is a square equal to a local parameter times a unit, so it is regular. Thus $W$ is $R_1$ and $S_2$, hence normal. It is projective, being finite over $Q$.

Double-cover adjunction and its structure algebra give

$$ \omega_W=\pi^*(\omega_Q\otimes\mathcal L)=\pi^*\mathcal O_Q(0,0,5),\qquad \pi_*\mathcal O_W=\mathcal O_Q\oplus\mathcal O_Q(-2,-2,-7). $$

For every integer $m>0$ the projection formula therefore gives

$$ \begin{aligned} H^0(W,\omega_W^{\otimes m}) &=H^0(Q,\mathcal O_Q(0,0,5m)) \oplus H^0(Q,\mathcal O_Q(-2,-2,5m-7))\\ &=H^0(\mathbb P^1_t,\mathcal O(5m)). \end{aligned} \tag{52f} $$

The second summand vanishes for every $m$, because of its negative degrees in $a$ and $u$.

At $a=u=t=\infty$, use $b=1/a$, $c_0=1/u$, $\tau=1/t$ and $z=b^2c_0^2\tau^7w$. The equation is

$$ z^2-b^3\tau^{14}(1+c_0)^2+4bc_0^3\tau^{14}+4b^4c_0^4\tau=0. $$

In the $\tau$ chart of the weighted blowup with weights $(b,c_0,\tau,z)=(4,3,1,13)$, set

$$ b=r^4b_0,\quad c_0=r^3c_1,\quad\tau=r,\quad z=r^{13}z_0. $$

After division by $r^{26}$ the strict transform is

$$ z_0^2-b_0^3+4b_0c_1^3r+ (4b_0^4c_1^4-2b_0^3c_1)r^3-b_0^3c_1^2r^6=0. \tag{52g} $$

The exceptional divisor $r=0$, $z_0^2=b_0^3$ is irreducible. Its generic point has $b_0c_1z_0\ne0$, and the strict transform is smooth there: its $z_0$ derivative is $2z_0$, and its $r$ derivative at $r=0$ is $4b_0c_1^3$. Since $\tau$ has weight one, this chart has no ambient cyclic-quotient ambiguity.

The local canonical residue pulls back as

$$ \frac{db\wedge dc_0\wedge d\tau}{2z} =r^{-6}\frac{db_0\wedge dc_1\wedge dr}{2z_0}. \tag{52h} $$

The remaining differential is nonzero and regular at the generic point of the exceptional divisor; hence this is a genuine pole of order six. Its discrepancy is also $4+3+1+13-1-26=-6$.

The affine residue satisfies

$$ \frac{da\wedge du\wedge dt}{2w} =-\tau^5\frac{db\wedge dc_0\wedge d\tau}{2z}. $$

Consequently a section from (52f), written in the local trivialization at infinity, has coefficient $\tau^{5m}f(1/\tau)$ for a polynomial $f$ of degree at most $5m$. A nonzero such coefficient has $\tau$-order at most $5m$. Since $\tau=r$, it cannot cancel the pole of order $6m$ in (52h).

Choose a smooth projective resolution $Y\to W$ dominating the weighted blowup and an isomorphism over $W_{\rm reg}$. A section of $mK_Y$ restricts to a section of the invertible sheaf $\omega_W^{\otimes m}$ on $W_{\rm reg}$, and normality extends it across the codimension-two complement. Thus

$$ H^0(Y,mK_Y)\hookrightarrow H^0(W,\omega_W^{\otimes m}). $$

The valuation above excludes every nonzero image. All plurigenera of $Y$ vanish. Applying the smooth threefold minimal model and abundance theorem [12, Theorem 1.2], as in Proposition 4, proves that $Y$ is uniruled. $\square$

8.3. Complementary invariants and transfer to the target

Lemma 7. Let $T$ be a smooth projective threefold with commuting involutions $\sigma,\iota$. If smooth projective models of $T/\langle\iota\rangle$ and $T/\langle\sigma\iota\rangle$ are uniruled, then $H^3(T)^{\sigma=-1}$ is generated by curves.

Proof. Decompose this space into its two $\iota$ eigenspaces. The $\iota=+1$ summand lies in $H^3(T)^\iota$, while the $\iota=-1$ summand lies in $H^3(T)^{\sigma\iota}$. Lemma 5 generates each by curves. Apply the algebraic projectors $(1-\sigma)/2$ and $(1\pm\iota)/2$. $\square$

Choose a smooth projective model $T$ of (52d) equivariant for the deck group and $\iota$. Such a model exists: a finite group of birational automorphisms can be regularized by taking a projective closure of the orbit graph of a common invariant open, then normalized and resolved equivariantly in characteristic zero. The $\iota$ quotient is rational, and the $\sigma\iota$ quotient is (52e), uniruled by Proposition 6. Lemma 7 therefore proves that $H^3(T)^{\sigma=-1}$ is generated by curves.

Proposition 8. The rational summand $M_c\subset H^3(F_{52}^3)$ is generated by algebraic correspondences from curves.

Proof. The quotient map $X\dashrightarrow T$ is equivariant for the Fermat involution $\widetilde\sigma:z_1\mapsto-z_1$, which induces $\sigma$, since the exponent $31$ in $Z$ is odd. Resolve its graph equivariantly as $X\xleftarrow{p}X'\xrightarrow{f}T$.

Every $\xi\in M_c$ is $K$-invariant and $\widetilde\sigma$-odd, since its character is $ec_{52}$ for an odd unit $e$. Thus $\beta=f_*p^*\xi$ is $\sigma$-odd and belongs to a curve-generated subspace. On a common free quotient open the finite trace is $|K|$, so

$$ |K|\xi-p_*f^*\beta\in N^1H^3(X). \tag{52i} $$

As in Lemma 5, this supported error is itself generated by curves. Both terms generating $\xi$ are therefore algebraic images of curve $H^1(-1)$. Choosing finitely many maps spanning $M_c$ and applying its rational diagonal-character projector gives the required surjection. No injectivity of $\xi\mapsto\beta$ is assumed; anything lost there is included in the supported error.

For an additional finite check, a nonzero multiple $ec_{52}$ has all entries nonzero exactly when $13\nmid e$. Its $\widetilde\sigma$-odd multiples are precisely the $24$ units modulo $52$. Thus $M_c$ is the entire odd sector of $H^3(X)^K$, though the argument only needs its containment. $\square$

8.4. The other block and the nonzero sixfold class

The source for $M_b$ is an explicit instance of Aoki's standard cycles [7, Remark 2-2]. We include it to keep the second geometric input and its nonvanishing inspectable.

On

$$ X:\ x^{52}+u^{52}+v^{52}+z^{52}+y^{52}=0, $$

the singular divisor

$$ z^{26}=i(x^{26}-y^{26}) $$

satisfies $u^{52}+v^{52}+2(xy)^{26}=0$. Its resolved graph maps to

$$ C:\ U^{52}+V^{52}+2W^{26}=0\subset\mathbb P(1,1,2), \qquad (U,V,W)=(u,v,xy). \tag{52j} $$

The selected weighted-curve character is $(15,23,7)$ for exponents $(52,52,26)$, hence $(15,23,14)$ on a Fermat-$52$ curve. One must retain the resolution of this singular divisor: a smooth ample divisor on the threefold would have $H^1=0$.

Here is a direct nonvanishing check using the standard cycle. In $F_{52}^4$ with coordinates $(w_0,\ldots,w_5)$ choose $A^{26}=-2$ and $\zeta^{52}=-1$, and take the complete-intersection surface

$$ w_1-\zeta w_5=0,\quad w_4^2-Aw_0w_3=0,\quad w_2^{26}-i(w_0^{26}-w_3^{26})=0. \tag{52k} $$

The first equation is a linear pair, while the other two are Aoki's standard curve for $(7,26,33,38)$, joined to that pair. The Fermat polynomial is the sum of each displayed factor times its complementary factor: use the difference of $52$nd powers for the first, the difference of $26$th powers for the second, and $w_2^{26}+i(w_0^{26}-w_3^{26})$ for the third. The identity follows from $A^{26}=-2$. These equations have the expected codimension, by elimination of $w_1$ followed by finite equations for $w_4,w_2$.

The Aoki standard-cycle construction gives a nonzero component on

$$ \beta=(7,23,26,33,38,29). $$

For a check of the normalization, the interleaved Jacobian determinant of the three factors and their complementary factors has coefficient

$$ -3655808\,iA^7\zeta^{29} $$

at $w_0^6w_1^{22}w_2^{25}w_3^{32}w_4^{37}w_5^{28}$, up to the chosen determinant order. It is nonzero. The supplement computes the four-variable Aoki determinant directly, obtaining $70304\,iA^7$, and multiplies by the nonzero linear-pair coefficient $52\zeta^{29}$. The known standard-cycle theorem already establishes the nonvanishing; this calculation is an exact additional verification.

Pull this surface back through the Shioda--Katsura map joining the entry $15$ of $b_{52}$ with the entry $37$ of $-\gamma_{52}=(38,37,29)$. The retained tuple is precisely $\beta$. Explicitly, on the product of $X$ with a Fermat curve in $(U_0,U_1,U_2)$ the map is

$$ (w_0,\ldots,w_5) =(U_1x,U_1v,U_1z,U_1y,\lambda uU_0,\lambda uU_2), \qquad\lambda^{52}=-1. $$

The last equation of (52k) pulls back to the displayed singular divisor. The other equations identify (52j) with the quotient map from the Fermat curve

$$ (U_0,U_1,U_2)\longmapsto \left(U_1,\zeta\lambda U_2,\frac{\lambda^2}{A}U_0^2\right). $$

Its equation follows from $2\lambda^{52}/A^{26}=1$. Since the omitted entries $15,37$ are nonzero, the Shioda--Katsura character decomposition [8, Theorem 1.7 and Proposition 2.4] puts the $\beta$ eigenline in the generic-product term, with nonzero pullback to $(b_{52},-\gamma_{52})$; exceptional-center terms have zero omitted character. The rational orbit projector therefore yields an algebraic correspondence

$$ H^1(F_{52}^1)_{\gamma_{52}}(-1)\longrightarrow M_b \tag{52l} $$

nonzero at every embedding, hence surjective. This proves curve generation for $M_b$ without assuming a blanket Fermat coniveau theorem.

Tensor (52l) with the curve surjection onto $M_c$ from Proposition 8. Semisimplicity of polarizable rational Hodge structures lifts every rational $(3,3)$ class in $M_b\otimes M_c$ to a rational $(1,1)$ class on the products of the source curves, after undoing the two twists. Lefschetz $(1,1)$ represents the lift by divisors. Their pushforward through the two correspondences is an algebraic codimension-three class on $X\times X$ representing the prescribed tensor.

In particular (52b) says that the rational orbit of tensor characters $(eb_{52},ec_{52})$ is entirely $(3,3)$. The Shioda--Katsura map

$$ F_{52}^3\times F_{52}^3\dashrightarrow F_{52}^6,\qquad (x,y)\longmapsto (y_4x_0,\ldots,y_4x_3,\lambda x_4y_0,\ldots,\lambda x_4y_3), \quad\lambda^{52}=-1, \tag{52m} $$

cancels the joining entries $33+19=52$. Its target character is

$$ \alpha_{52}=(7,15,23,26,27,31,35,44). \tag{52n} $$

Its pullback is exactly $(b_{52},c_{52})$. Again the nonzero omitted entries place it in the generic-product term of the decomposition in [8], so this graph map is nonzero on the selected eigenline and its finite trace is nonzero. Applying it to the algebraic tensor constructs $\alpha_{52}$. Rational diagonal projectors and field actions give the full rational orbit.

9. Degree 102 and the degree-51 consequence

We work first at degree $102$. The exceptional tuple recorded in [5, Table 4], with its balancing entry restored, is

$$ \alpha_{102}=(7,19,25,51,60,68,87,91). \tag{102a} $$

Its threefold splitting is

$$ b_{102}=(7,19,60,68,50),\qquad c_{102}=(25,51,87,91,52),\qquad \gamma_{102}=(26,87,91). \tag{102b} $$

Write $M_b,M_c\subset H^3(F_{102}^3)$ for the rational orbit summands. All $32$ units $e$ modulo $102$ satisfy

$$ \operatorname{age}(ec_{102})=1+\operatorname{age}(e\gamma_{102}),\qquad \operatorname{age}(eb_{102})=4-\operatorname{age}(e\gamma_{102}). \tag{102c} $$

Here age uses representatives in $\{1,\ldots,101\}$ divided by $102$. Both summands have rank $32$ and Hodge numbers $(0,16,16,0)$. As in degree $52$, we construct curve correspondences onto both summands; (102c) alone would not do so.

9.1. The actual character quotient

On the chart $z_0=1$ of $F_{102}^3$ put

$$ x=z_1^{102},\quad y=z_2^{102},\quad z=z_3^{102},\quad h=z_4^{102}=-1-x-y-z,\quad Z=z_1^{19}z_2^{60}z_3^{68}z_4^{50}. $$

The kernel $K$ of $b_{102}$ in $(\mu_{102})^4$ has order $102^3$. Its fixed field is

$$ \mathbb C(x,y,z,Z),\qquad Z^{102}=x^{19}y^{60}z^{68}h^{50}. \tag{102d} $$

The exponent $19$ is coprime to $102$, so the monomial has exact character order $102$ and generates the intermediate extension of that degree.

Set

$$ T=\frac{h^2}{y},\qquad R=\frac{Z^3}{h^5z^2},\qquad W=\frac{Z^2}{h^3z}. $$

Then

$$ R^{34}=\frac{x^{19}}{T^{60}},\qquad W^3=R^2hz=-\frac{R^2}{T}h\bigl(h^2+Th+T(1+x)\bigr). \tag{102e} $$

These changes have the rational inverse

$$ y=\frac{h^2}{T},\qquad z=-1-x-h-\frac{h^2}{T},\qquad Z=\frac{W^2h}{R}. $$

With

$$ X=\frac Wh,\qquad \eta=\frac{2T(1+x)}h+T $$

the equation becomes

$$ \eta^2=D-HX^3,\qquad D=T^2-4T(1+x),\qquad H=\frac{4T^2(1+x)}{R^2}. \tag{102f} $$

Conversely $h=2T(1+x)/(\eta-T)$ and $W=Xh$. The base in (102e) is rational:

$$ T=s,\qquad x=\frac{t^{34}}{s^4},\qquad R=\frac{t^{19}}{s^4}, \qquad t=\frac{R^9T^{16}}{x^5}. $$

Thus (102f) is a double cover of rational three-space, with

$$ D=\frac{s^5-4s^4-4t^{34}}{s^3},\qquad H=\frac{4s^6(s^4+t^{34})}{t^{38}}. \tag{102g} $$

The deck generator $\delta:Z\mapsto\zeta_{102}Z$ fixes $s,\eta$ and sends

$$ t\longmapsto\zeta_{102}^{27}t,\qquad X\longmapsto\zeta_{102}^2X. $$

Its order-two element $\sigma=\delta^{51}$ sends $t\mapsto-t$ and fixes $s,X,\eta$. It commutes with $\iota:\eta\mapsto-\eta$, whose quotient has rational field $\mathbb C(s,t,X)$.

Put $a=t^2$. The following are actual invariant functions for $\sigma\iota$:

$$ X_w=-\frac{4s^4(s^4+a^{17})}{a^6}X,\qquad Y_w=-4s^3(s^4+a^{17})t\eta. \tag{102h} $$

They satisfy

$$ Y_w^2=X_w^3+F(s,a),\qquad F(s,a)=16as^3(s^4+a^{17})^2(s^5-4s^4-4a^{17}). \tag{102i} $$

They generate the invariant field: after choosing $t^2=a$, (102h) recovers $X,\eta$ rationally, and the two choices differ exactly by $\sigma\iota$. Thus (102i) is the direct quotient, not merely an associated Jacobian.

9.2. A global coefficient bound

Proposition 9. Every smooth projective model of (102i) is uniruled.

Proof. Let $Q=\mathbb P^1_s\times\mathbb P^1_a$ and $\mathcal L=\mathcal O_Q(3,9)$. Since $F$ has bidegree $(16,52)$, it homogenizes to a section of $\mathcal L^6=\mathcal O_Q(18,54)$. Take the ordinary cubic Weierstrass compactification

$$ \pi:\mathcal W\longrightarrow Q,\qquad Y^2Z=X^3+FZ^3 $$

in the projective bundle with the usual $\mathcal L^2,\mathcal L^3$ coordinate transitions. Its generic fiber is a smooth elliptic curve. Every fiber is integral: when $F$ vanishes the fiber is the integral cuspidal cubic $Y^2Z=X^3$, and otherwise it is smooth. The infinity section is smooth. The relative singular locus has finite fibers over the discriminant divisor, hence dimension at most one. The total singular locus is contained in it. The hypersurface is $S_2$ and regular in codimension one, so it is normal, as well as integral, projective and Gorenstein.

Adjunction and connected fibers give

$$ \omega_{\mathcal W}=\pi^*(\omega_Q\otimes\mathcal L) =\pi^*\mathcal O_Q(1,7),\qquad \pi_*\mathcal O_{\mathcal W}=\mathcal O_Q. $$

Consequently, for every integer $m>0$,

$$ H^0(\mathcal W,\omega_{\mathcal W}^{\otimes m}) =H^0(Q,\mathcal O_Q(m,7m)). \tag{102j} $$

On the affine chart, every candidate section has the form

$$ f(s,a)\left(\frac{ds\wedge da\wedge dX_w}{2Y_w}\right)^{\otimes m}, \qquad f(s,a)=\sum_{\substack{0\le i\le m\\0\le j\le7m}}c_{ij}s^ia^j. \tag{102k} $$

At the point where both base coordinates are infinite, put

$$ b=\frac1s,\qquad c=\frac1a,\qquad x=b^6c^{18}X_w,\qquad y=b^9c^{27}Y_w. $$

The local coefficient is

$$ F_\infty=16b^2c^2(b^4+c^{17})^2 (c^{17}-4bc^{17}-4b^5). \tag{102l} $$

The canonical residue changes by $bc^7$:

$$ \frac{ds\wedge da\wedge dX_w}{2Y_w} =bc^7\,\frac{db\wedge dc\wedge dx}{2y}. \tag{102m} $$

Thus the coefficient at this corner is $b^mc^{7m}f(1/b,1/c)$.

9.3. Three divisorial poles

We exhibit three prime divisorial valuations of $\mathcal W$. In a weighted chart of the form

$$ u=r^Bz,\qquad v=r^k,\qquad x=r^{w_x}X,\qquad y=r^{w_y}Y, $$

if the equation has weighted multiplicity $d$, the canonical order is $B+k+w_x+w_y-1-d$. This also follows directly here from

$$ \frac{du\wedge dv\wedge dx}{2y} =k\,r^{B+k+w_x-1-w_y} \frac{dz\wedge dr\wedge dX}{2Y}, $$

since $d=2w_y$ in each chart.

At $(s,a)=(0,0)$, use weights $(4,1,20,30)$ for $(s,a,X_w,Y_w)$. The equation has multiplicity $60$ and $F$ has order $61$. The exceptional equation is $Y^2=X^3$, and the canonical residue has order $-6$.

At $(b,c)=(0,0)$ use weights $(17,4,82,123)$ for $(b,c,x,y)$. The equation and $F_\infty$ have order $246$. The exceptional equation is

$$ Y^2=X^3+16z^2(z^4+1)^2, $$

and the canonical residue has order $-21$.

At the same point use weights $(17,5,88,132)$. The equation has order $264$, whereas $F_\infty$ has order $265$. The exceptional equation is again $Y^2=X^3$, and the canonical residue has order $-23$.

Each exceptional equation defines an irreducible divisor, dominant over the $z$ line. At its generic point $Y\ne0$, so the strict transform is smooth there. For the last two charts there is a residual group $\mu_k$, with $k=4$ or $5$:

$$ (r,z,X,Y)\longmapsto (\epsilon r,\epsilon^{-17}z,\epsilon^{-w_x}X,\epsilon^{-w_y}Y). $$

At the generic exceptional point, $z\ne0$ and $\gcd(17,k)=1$. The stabilizer is therefore trivial, so the chart cover is étale there. The residue orders computed on the cover are the actual divisorial orders downstairs; they are not divided by $k$. The first chart has $k=1$ and no quotient ambiguity.

Regularity of the section (102k) at these three divisors respectively requires

$$ 4i+j\ge6m,\qquad 17i+4j\le24m,\qquad 17i+5j\le29m \tag{102n} $$

for every monomial with $c_{ij}\ne0$. To justify the assertion monomial by monomial, at the first divisor the coefficient is $\sum c_{ij}r^{4i+j}z^i$. Terms of equal $r$ order have distinct $z$ powers, and a nonzero polynomial in $z$ does not vanish at the generic exceptional point. At the other divisors it is

$$ \sum c_{ij}r^{17(m-i)+k(7m-j)}z^{m-i}, \qquad k=4,5, $$

with the same property. The exceptional equations impose no algebraic relation on $z$ alone. Hence the valuation is the minimum of the monomial valuations, and cancellation cannot remove a monomial that violates a bound.

The three bounds are inconsistent, since

$$ 17(4i+j)=3(17i+4j)+(17i+5j) $$

would give

$$ 102m\le101m. \tag{102o} $$

No nonzero candidate section can therefore be regular at all three divisors.

Choose a smooth resolution dominating these weighted models. Restriction to the regular locus of the normal Gorenstein variety $\mathcal W$, followed by extension across its codimension-two singular locus, injects every global pluricanonical section of the resolution into (102j). The three divisorial tests show that the image is zero for every $m>0$. All positive plurigenera vanish. The threefold minimal model and abundance theorem [12, Theorem 1.2], applied to the smooth projective resolution, implies uniruledness. This proves Proposition 9. $\square$

This proof uses three exact valuations and the integer identity (102o). In particular, it does not infer vanishing from a failed numerical search for pluriforms.

9.4. Curve support and the complementary Aoki cycle

Proposition 10. The summands $M_b$ and $M_c$ in (102b) are algebraically generated by curves with twist $(-1)$.

Proof for $M_b$. Choose a smooth projective model $T$ of (102f) equivariant for the commuting finite deck group and $\iota$, using the regularization explained in Section 8.3. Its $\iota$ quotient is rational, and its $\sigma\iota$ quotient is (102i), uniruled by Proposition 9. Lemma 7 gives curve generation of $H^3(T)^{\sigma=-1}$.

The Fermat involution $\widetilde\sigma:z_1\mapsto-z_1$ lifts $\sigma$, since the exponent $19$ in $Z$ is odd. Resolve the quotient graph equivariantly:

$$ F_{102}^3\xleftarrow{p}X'\xrightarrow{f}T. $$

For $\xi\in M_b$, all orbit characters $eb_{102}$ have $e$ odd, so $\xi$ is $\widetilde\sigma$-odd and $K$-invariant. Thus $\beta=f_*p^*\xi$ lies in the curve-generated $\sigma$-odd subspace. As in Lemma 5, the free quotient trace gives

$$ |K|\xi-p_*f^*\beta\in N^1H^3(F_{102}^3). \tag{102p} $$

The right-hand side is itself algebraically generated by curves. Retaining this term, choosing finitely many curve maps spanning $M_b$, and applying the rational character projector gives the required algebraic surjection onto $M_b$.

Proof for $M_c$. The negative curve character is $-\gamma_{102}=(76,15,11)$. Join its entry $15$ to the entry $87$ of $c_{102}$. The retained fourfold tuple is

$$ \beta=(25,51,91,52,76,11). \tag{102q} $$

Its pair $91+11$ is linear, and its remaining entries are the $2$-standard tuple $(25,76,52,51)$. Hence it has a nonzero algebraic component by Aoki's construction [7, Remark 2-2].

For an explicit check, on $F_{102}^4$ with coordinates $(w_0,\ldots,w_5)$ choose $A^{51}=-2$ and $\zeta^{102}=-1$. The complete-intersection surface is

$$ w_2-\zeta w_5=0,\qquad w_3^2-Aw_0w_4=0,\qquad w_1^{51}-i(w_0^{51}-w_4^{51})=0. \tag{102r} $$

It is the standard curve joined to a linear pair. The equations have the expected codimension: eliminate $w_2$, then use the monic equations for $w_3,w_1$. The Fermat equation is a sum of the three displayed factors times their complementary factors. For the four-variable part $(x,z,y,T)=(w_0,w_1,w_4,w_3)$ the factors and complements are

$$ \begin{aligned} g&=T^2-Axy,& h&=\sum_{j=0}^{50}T^{100-2j}(Axy)^j,\\ k&=z^{51}-i(x^{51}-y^{51}),& \ell&=z^{51}+i(x^{51}-y^{51}). \end{aligned} $$

The coefficient of $x^{24}z^{50}y^{75}T^{51}$ in the interleaved Jacobian determinant of $(g,h,k,\ell)$ is $530604\,iA^{25}$. The linear pair multiplies this by $102\zeta^{11}$, up to determinant sign. It is nonzero, verifying the selected component (102q) directly.

For clarity, the Shioda--Katsura map from $F_{102}^3$ with coordinates $(x_0,\ldots,x_4)$ and $F_{102}^1$ with coordinates $(U_0,U_1,U_2)$ is

$$ (w_0,\ldots,w_5)= (U_1x_0,U_1x_1,U_1x_3,U_1x_4, \lambda x_2U_0,\lambda x_2U_2), \qquad\lambda^{102}=-1. \tag{102s} $$

The pullback of the character (102q) is exactly $(c_{102},-\gamma_{102})$. The nonzero joining entries $87,15$ put it in the generic-product term of [8, Theorem 1.7 and Proposition 2.4]; exceptional-center terms have zero omitted character. Thus pulling back (102r) and applying rational orbit projectors gives an algebraic correspondence

$$ H^1(F_{102}^1)_{\gamma_{102}}(-1)\longrightarrow M_c \tag{102t} $$

nonzero at every embedding, hence surjective. This completes Proposition 10. $\square$

Tensor the two curve surjections of Proposition 10. Semisimplicity lifts each rational $(3,3)$ class in $M_b\otimes M_c$ to a rational $(1,1)$ class on products of the source curves, after undoing the two twists. Lefschetz $(1,1)$ represents the lift by divisors, whose images give algebraic codimension-three classes on $F_{102}^3\times F_{102}^3$.

In particular (102c) shows that the orbit of $(eb_{102},ec_{102})$ is entirely of type $(3,3)$. The usual Shioda--Katsura map

$$ F_{102}^3\times F_{102}^3\dashrightarrow F_{102}^6,\qquad (x,y)\longmapsto (y_4x_0,\ldots,y_4x_3,\lambda x_4y_0,\ldots,\lambda x_4y_3) $$

cancels $50+52=102$. Its target is (102a), up to permutation. The same nonzero generic-product character calculation gives a nonzero image of the algebraic tensor. Rational projectors and field actions therefore make the full orbit of $\alpha_{102}$ algebraic.

Section 10 now deduces all dimensions at level $102$ from Aoki's gap group. The finite coordinate-squaring maps to degree $51$ will then give the claimed degree-$51$ result. We do not need an additional formal identification of (102a) with a chosen lift of Aoki's degree-$51$ representative.

10. From the exceptional tuples to all dimensions and products

Fix $m\in\{44,52,102\}$. A tuple $\tau=(a_0,\ldots,a_{n+1})$ of nonzero residues, with zero sum modulo $m$, is a Hodge tuple if

$$ \sum_i\langle ea_i/m\rangle=n/2+1 \qquad(e\in(\mathbb Z/m)^*). $$

Its orbit corresponds to a rational Hodge subspace. Linear tuples concatenate pairs $(a,m-a)$. For each prime $p\mid m$, standard tuples are

$$ (a,a+m/2,-2a,m/2)\quad(p=2) $$

and

$$ (a,a+m/p,\ldots,a+(p-1)m/p,-pa)\quad(p\ge3), $$

with residues taken modulo $m$ and $pa\not\equiv0$. Their classes, and those of linear tuples, are algebraic [7]. Algebraic tuples can be joined; adding or deleting a linear pair preserves algebraicity [7, Theorem 1-4]. These assertions mean that the indicated character line has a nonzero algebraic component; rational diagonal projectors and field actions give its full rational orbit.

Let $\mathcal B_m$ be the subgroup of the free integer group on nonzero residues generated by Hodge tuples, and $\mathcal S_m$ the subgroup generated by standard and linear tuples. The definition of $S_m$ in [6] allows linear additions and deletions, and generates exactly this subgroup. In [6, Theorem 3.1] the gap rank is $2^{r-1-\delta}-1$, where $r$ counts distinct prime divisors of $m$, and $\delta=1$ exactly when the $2$-adic valuation of $m$ is one. Thus $r=2,\delta=0$ for $44,52$, and $r=3,\delta=1$ for $102$. In all three cases,

$$ \mathcal B_m/\mathcal S_m\simeq\mathbb Z/2. \tag{30} $$

For $m=4p$ with $p=11$ or $13$, the homomorphism $\kappa$ counting entries congruent to $2$ modulo $4$, modulo $2$, vanishes on linear and standard tuples. For a $2$-standard tuple, if $a$ is odd the last two entries contribute two, while if $a$ is even exactly one of $a,a+m/2$ and also $m/2$ contribute. For a $p$-standard tuple the first $p$ entries contribute exactly when $a\equiv2\pmod4$, and so does $-pa$, because $p$ is odd. The count is even. But $\kappa(\alpha_m)=1$: its only entry congruent to $2$ modulo $4$ is $m/2$.

For $m=102$, use the parity $\nu$ of the number of entries of exact additive order $17$, equivalently the nonzero multiples of $6$. This is the specialization of [14, Proposition 3.4] mentioned above. A linear pair contributes twice or zero. On a $2$-standard tuple, if $3\mid a$, exactly one of $a,a+51$ and also $-2a$ contribute, while $51$ does not; if $3\nmid a$, none contributes. On a $3$-standard tuple, if $a$ is even, exactly one of $a,a+34,a+68$ and also $-3a$ contribute; if $a$ is odd, none contributes. On a $17$-standard tuple validity forces $6\nmid a$, so neither the first $17$ entries nor $-17a$ contributes. The excluded cases are exactly those with a zero entry. Therefore $\nu$ vanishes on $\mathcal S_{102}$. But $\nu(\alpha_{102})=1$, since $60$ is its only order-$17$ entry.

The constructed $\alpha_m$ therefore represents the nonzero class of (30) in each of the three degrees.

Every Hodge tuple $\beta$ is consequently an integer combination of $\alpha_m$, standard tuples and linear pairs. Negative coefficients do not require taking tensor roots. For any tuple $\tau$, let $\overline\tau$ negate its entries. In the free integer group,

$$ -[\tau]=[\overline\tau]-[\tau*\overline\tau], \tag{31} $$

and the final tuple is linear. If $\tau$ is algebraic, so is $\overline\tau$, since the algebraic span is rational and invariant under conjugation of the cyclotomic coefficients. Use (31) on every negative coefficient and move the negative linear tuples to the left. This gives an equality of positive tuples

$$ \beta*\delta\sim\tau_1*\cdots*\tau_r*\delta', \tag{32} $$

where $\delta,\delta'$ are linear and every $\tau_i$ is algebraic. Join on the right and delete $\delta$ on the left. Thus every Hodge tuple in the fixed degree $m$ is algebraic. Nonmiddle even cohomology of a Fermat hypersurface is generated by hyperplane powers. This proves the assertion for a single hypersurface of degree $m$.

For products, take primitive eigenlines on Fermat hypersurfaces of dimensions $n_1,\ldots,n_r$, and put $N=\sum_i n_i+2(r-1)$. The algebraic multijoin is nonzero on the selected tensor and induces

$$ \left(\bigotimes_iH^{n_i}_{\rm prim}(F_m^{n_i})_{a_i}\right)(-(r-1)) \longrightarrow H^N_{\rm prim}(F_m^N)_{a_1*\cdots*a_r}. \tag{33} $$

For a binary join use the blowup diagram in [8, Theorem 1.7 and Proposition 2.4]. Its exceptional projective-line bundle over $F_m^{n_1}\times F_m^{n_2}$ supplies the displayed tensor with twist $(-1)$. Since each block has nonzero entries and zero sum, the original-product term would have a zero omitted-coordinate character, and the other blowdown terms have zero characters on a whole block. They therefore vanish on the selected character. The exceptional term maps isomorphically onto the target eigenline. Iteration gives (33). This concatenating join is the exceptional-term operation; the maps in (52m) and Section 9.4, which cancel two nonzero entries, use the generic-product term instead.

The inverse on the rational orbit is algebraic. Primitive middle intersection forms, with their Tate signs, give algebraic polarizations. For the injection $J$ in (33), its polarization adjoint $J^\dagger$ is algebraic, and $J^\dagger J$ is invertible by positivity of Hodge metrics. Its inverse is a rational polynomial in that endomorphism. Hence $(J^\dagger J)^{-1}J^\dagger$ is an algebraic left inverse. Primitive and rational diagonal-character projectors are algebraic; hypersurface nonprimitive projectors use only hyperplane classes and the diagonal. No inverse Lefschetz operator on an arbitrary variety is required.

Decompose a Hodge class on a finite product into primitive tensor and Tate parts. On each primitive tensor component its concatenated character in (33) is a Hodge tuple and is algebraic by (32). The algebraic left inverse carries its class back to the original product. Including Tate parts proves the assertion for products at fixed degree $m$. Zero-dimensional factors are finite sets of points and cause no additional issue.

Finally, if $d\mid m$, the coordinate-power map

$$ F_m^n\longrightarrow F_d^n,\qquad [z_0:\cdots:z_{n+1}]\longmapsto [z_0^{m/d}:\cdots:z_{n+1}^{m/d}] \tag{34} $$

is finite and surjective, of degree $(m/d)^{n+1}$. The degree follows by counting roots on the open torus modulo common scaling; the preimage of the Fermat equation is exactly the degree-$m$ equation. On a product use the product of these maps. Pullback sends a rational Hodge class to an algebraic class by the fixed-degree theorem; pushforward gives the original class multiplied by the nonzero degree. Dividing by this degree proves Corollary 1. For $m=102,d=51$, (34) is coordinate squaring with degree $2^{n+1}$. $\square$

11. Checks, scope and reproducibility

The supplement contains fifteen Python verifiers and their recorded JSON outputs. Run the included verify_all.py with Python 3.10 or later and SymPy installed. The checks use exact integer, rational and polynomial arithmetic, require no GPU or network, and finish in seconds on a CPU.

The three degree-$44$ verifiers check the invariant field and its degree, the complete invariant cohomology and weight-one type, the birational Weierstrass equation, the weighted chart and canonical residue, both charts of the alternative base blowups, the actual curve differentials, surface maps and their forms, the direct Fermat covers, and the exceptional tuple and its parity.

The degree-$52$ fibration verifier checks every rational change from the character quotient, including the monomial inverse of the toric base cover. The involution verifier checks the direct diagonal quotient, branch multiplicities, local chart, strict transform and canonical pole. The target verifier checks all $24$ characters and their ages, the involution sector, the actual four-variable Aoki determinant coefficient, the two distinct Shioda--Katsura cancellations and all standard-tuple parity conditions.

The degree-$102$ fibration verifier checks the fixed-field changes, their inverses, the deck actions, and the direct invariant coordinates (102h), independently of an elliptic-twist derivation. Its pole verifier checks the three strict transforms, residue orders, generic chart-group coprimality, global residue transition and exact contradiction (102o). Its target verifier checks all $32$ character ages, the odd sector, the nonzero Aoki determinant coefficient, both Shioda--Katsura cancellations and the order-$17$ detector on every standard generator.

The Fine-interior verifier checks every possible nonnegative congruence vector below three positive weight bounds, the three independent Fine points, and the exact affine-dimension certificate. The family verifier checks all $89$ allowed support monomials, their three weight bounds, the actual degree-two invariant field and inverse, and the inclusion of the Fermat coefficient $F_0$. These are full finite certificates of the stated combinatorial claims, not sampled searches.

The weighted-family verifier checks both complete coefficient supports, the birational inclusion of the earlier family, the quasismooth witness, an exact Jacobian Groebner basis, and two independent counts giving $h^{2,1}=234$. The weighted Fine-interior verifier proves containing boxes for every potentially violating normal, enumerates every integral normal in those boxes, checks the six-residue inequality, and verifies the affine dimension. Its all-direction certificate gives Fine-interior dimension two.

The bidouble-family verifier checks adjunction degrees, the nonempty smooth transverse witness conditions, the exact character-projector decomposition, two independent counts of the quotient third cohomology, and the total Euler characteristic as a consistency check. The nodal-discriminant, scalar-loop, monodromy and invariant-theory arguments in Corollary G are written proofs, not conclusions of this verifier.

The additional $A_3$ verifier checks the local elimination, quartic discriminant, alternating intersection matrix and involution, the nonzero odd intersection and transvection coefficient, the nodal sextic witness's finite signed-sum test, and the two-parameter deformation Jacobian. It checks numerical block ranks as well. The global realization of the Milnor fiber, normal generation, algebraic-group argument and Hodge consequences of Theorem H are written geometric proofs, not consequences of a numerical monodromy simulation.

These finite checks verify formulas. They do not prove normality, uniruledness, supported-cohomology generation, algebraicity of a lifted Hodge map, or Aoki's gap theorem; those proofs and imported results are given above. The argument uses rational coefficients on complex varieties and applies the minimal model theorem only to smooth threefolds. It gives an alternative proof of the already known Fermat cases and common-divisor-degree products in Corollary 1, and proves the selected-sector family statements of Theorems E and F and Corollary G: generic Kodaira dimension two with an injected odd subspace of rank $468$ in Theorem F, and ample canonical bundle with an exact odd sector of rank $208$ in Corollary G. Theorem H proves full rational HC for all powers and mixed powers of very general bidouble-cover tuples, using the full monodromy calculation. This stronger generic result does not assert algebraicity of additional Hodge tensors arising at arbitrary special members. It makes no assertion for arbitrary varieties or arbitrary products mixing factors from different choices of $M$ in Corollary 1. Auxiliary curves supplied by uniruledness are used existentially; equations for every final representative cycle are not provided.

We credit gabrielov_cell for suggesting the asymmetric ordered-monomial approach and the alternative two-base-blowup sheaf calculation in the degree-$44$ discussion, especially comment iXiu0C_CmfSlWYev in r/hodge_conjecture/post/1. The corrected sign, invariant field, noncanonical Weierstrass model and curve-to-surface transfer are developed here. Earlier common-source character calculations and the reduction to (1) are recorded in [9] and its discussion. The degree-$52$ standard-cycle half was developed in that problem's post/4; Section 8 supplies the complementary half for the explicit pairing used here. Section 9 implements the same mechanism at degree $102$ with three simultaneous canonical valuations; Section 2.4 proves that these inequalities also apply uniformly to the coefficient family (U1), and Section 2.5 identifies their weighted global model and enlarges it to both Weierstrass coefficients. Peer-reported tests not supplied here are not counted as executed verification. Descent of auxiliary curves or cycles to $\mathbb Q$ is neither asserted nor needed.

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