Curve support from complementary uniruled quotients: an elliptic family and a carrier of Kodaira dimension two

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 selected-sector support theorem for an explicit elliptic family and exhibit a member whose carrier has Kodaira dimension two. Two complementary involution quotients supply the required cohomology. This separates support of a summand from the birational type of its whole carrier.

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 extension. Theorem E below applies the same canonical valuations to every nonzero polynomial in the explicit $89$-dimensional coefficient space (U1). For the threefolds $T_F$ in (U2), the odd sector is generated by curves, and every Hodge class in the tensor product of odd sectors of any two members is algebraic. The degree-$102$ carrier is one member. The family theorem is not restricted to Fermat equations and does not assert HC for the entire product of arbitrary 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.

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. Choose a resolution which is an isomorphism over the regular locus and realizes all $E_j$. A pluricanonical form on it restricts to a regular section on $W_{\mathrm{reg}}$. 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. No Kodaira-dimension assertion is made for the other members, and (U3) is 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.

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, 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 eleven 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.

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 statement of Theorem E. 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). 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.

References

[1] V. V. Batyrev. Canonical models of toric hypersurfaces. Algebraic Geometry 10 (2023), 394–431. Theorem 9.2. https://arxiv.org/abs/2008.05814v4

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