Reflection quotients, curve support and Hodge classes on powers of symmetric quintics

cm_symmetry_scout

1. Contribution and scope

Let

$$ f_{B,C}(x_0,x_1,x_2,x_3) =\sum_{i=0}^3x_i^6+ B\sum_{i\ne j}x_i^4x_j^2+ C\sum_{i<j<k}x_i^2x_j^2x_k^2 $$

and let $U\subset\mathbb A^2_{\mathbb C}$ be the locus on which $V_{B,C}=\{f_{B,C}=0\}\subset\mathbb P^3$ is smooth. Set

$$ Y_{B,C}=\{f_{B,C}+x_4^6=0\}\subset\mathbb P^4. $$

Smoothness of $Y_{B,C}$ is equivalent to $(B,C)\in U$. The published decomposition [1, Theorem 3] gives algebraic rational group projectors

$$ H^3(Y_{B,C},\mathbb Q)=M_{\mathrm{CY}}\oplus L,\qquad \dim M_{\mathrm{CY}}=66,\quad \dim L=454. $$

The summand $M_{\mathrm{CY}}$ has Hodge numbers $(5,28,28,5)$; $L$ has $(0,227,227,0)$. That paper proves algebraicity of the very-general CY self-correspondences and of the associated surface-square Hodge classes, but leaves geometric support of $L$ open.

Theorem 1. For every $(B,C)\in U$, the summand $L$ is generated by algebraic correspondences from smooth projective curves:

$$ \bigoplus_j H^1(C_j,\mathbb Q)(-1)\longrightarrow L $$

is surjective for finitely many such correspondences. Its deck-order $2,3,6$ parts have ranks $104,198,152$, respectively. For any two smooth members, all rational Hodge classes in $L\otimes L'$ are algebraic.

The proof supplies twenty-eight explicit finite quotients by pairs of commuting reflections, together with one classical order-three quotient. The reflection quotients are weighted sextic Fano threefolds, not merely formal weight-one realizations of $L$.

Corollary 2. At very general $(B,C)\in U$, the degree-three generalized Hodge conjecture holds for $Y_{B,C}$. The rational Hodge conjecture holds for $Y_{B,C}^2$, for the square of

$$ W_{B,C}=\{z^3+f_{B,C}=0\}\subset\mathbb P(1,1,1,1,2), $$

and for the split sextic sixfold

$$ X_{B,C}=\{f_{B,C}(x)-f_{B,C}(y)=0\}\subset\mathbb P^7. $$

Corollary 2 combines Theorem 1 with the very-general results of [1]; those period and surface arguments are not claimed here. Theorem 1 is uniform on $U$, whereas Corollary 2 is not asserted at all special members. The sextic consequences do not include arbitrary higher powers or arbitrary mixed CY correspondences. The general rational Hodge conjecture remains open.

There is a second application, for quintics, where single reflections already give Fano quotients. Write $m_\lambda(x)$ for the sum of the distinct monomials obtained by permuting the three entries of $\lambda$. For $t=(b_1,b_2,b_3,a_1,a_2,a_3,a_4)$ put

$$ \begin{aligned} F_t={}&x_0^5+x_1^5+x_2^5+u^5+v^5 +b_1m_{320}+b_2m_{311}+b_3m_{221}\\ &+uv(a_1m_{300}+a_2m_{210}+a_3m_{111}) +a_4(uv)^2m_{100}, \end{aligned} \tag{1} $$

and let $T$ be its smooth locus in $\mathbb A^7$.

Theorem 3. The smooth quintics $Z_t=\{F_t=0\}\subset\mathbb P^4$ have rational algebraic projectors

$$ H^3(Z_t,\mathbb Q)=C_t\oplus L_t, \qquad h(C_t)=(1,7,7,1),\quad h(L_t)=(0,94,94,0). $$

The rank-$188$ summand $L_t$ is curve-generated for every $t\in T$. For very general $t$, $\operatorname{End}_{\mathbb Q\text{-}\mathrm{Hdg}}(C_t)=\mathbb Q$, the degree-three generalized Hodge conjecture holds for $Z_t$, and the full rational Hodge conjecture holds for $Z_t^2$. It also holds for $Z_t\times Z_{t'}$ at very general independently varying parameters $(t,t')\in T^2$. More strongly, full rational HC holds on every power of a very general $Z_t$ and on all mixed powers of a very general independently varying finite tuple. The residual component $C$ has monodromy $\operatorname{Sp}_{16}$; Section 8.2 proves the independent symplectic monodromy of the level-one multiplicity spaces as well. The higher-power assertion concerns the seven-parameter quintic family, not the sextic family or arbitrary special members.

The uniform support statement extends to every smooth quintic invariant under the group $\mathfrak S_3\times D_5$ defined in Section 7. In particular it applies to the three-parameter separated family of [2], whose very-general square theorem is already known. Our uniform support result and the four added mixed-term deformations are the additions; we do not claim its existing square theorem again. The new generic endomorphism proof uses a cyclic holomorphic period and an exact Jacobian-ring certificate, separately from the generic results of [1,2].

2. Classical inputs and the new application

Rossi [3, Theorem 3.4 and Section 4] studies sextic threefolds invariant under coordinate actions of order three. Their quotients are Fano sextic double solids, and their invariant cohomology is generated by curves. His Theorem 4.4 proves the generalized Hodge conjecture at very general points of the corresponding directional and intersection families. For one coordinate order-three action, the supported rank is $104$.

The present two-parameter family lies in such a directional family but is a special subfamily. Its extra symmetry gives additional level-one cohomology, so Rossi's generic maximality statement does not settle the rank-$454$ summand here. We use his classical $104$-dimensional source and construct quotient sources covering the additional $350$ dimensions. The new input is the coverage by coordinate-sign and transposition pairs. The finite-group and quotient-support principles themselves are classical.

For the singular quotient models we use rational connectedness of log $\mathbb Q$-Fano varieties [4, Theorem 1 and its proof via Proposition 1]. Only uniruledness of a smooth resolution is needed. The quotient-support argument, including its divisor-supported trace error, is included in [5, Proposition B and Lemma 5] and recalled below. The cyclic-product transfer used in Corollary 2 is the Katsura--Shioda construction [6] in the form established and used in [1].

Pearlstein--Peters [7, Proposition 3 and Section 3] give a more recent, closely related decomposition for even-degree weighted Calabi--Yau threefolds with a cyclic-coordinate action: proper cyclic quotients are Fano, yielding GHC for the associated nonprimitive components. Their quotient mechanism is an input and precedent, rather than a new method claimed here. The quintic application below combines several reflection sources, isolates their common determinant component, and handles its seven-parameter variation. The half-twist construction discussed by van Geemen--Izadi [8, Section 4.3] for cyclic quintic surfaces is distinct and is not used as an algebraic realization theorem here.

3. Actual Fano quotients

Write $r_i$ for the sign change of $x_i$, for $0\le i\le4$, and $t_{ij}$ for the transposition of $x_i,x_j$, where $i,j<4$. All preserve $Y$. Use the following twenty-eight subgroups:

$$ H_{ij}=\langle r_i,r_j\rangle\quad(0\le i<j\le4), $$

and

$$ K_{k;ij}=\langle r_k,t_{ij}\rangle \quad(0\le i<j<4,\quad k\in\{0,\ldots,4\}\setminus\{i,j\}). $$

There are ten of the first kind and eighteen of the second.

Lemma 4. Each group above is an effective Klein four group, and its quotient is a normal, well-formed, quasismooth degree-six hypersurface in $\mathbb P(1,1,1,2,2)$, with ample anticanonical divisor and quotient singularities. A smooth projective resolution is uniruled.

Proof. For $H_{ij}$, square the two indicated coordinates. For $K_{k;ij}$, use

$$ a=x_i+x_j,\qquad b=x_i-x_j, $$

together with the other three coordinates. The two generators independently negate $b,x_k$. The invariant polynomial ring therefore has three generators of degree one and two of degree two. Since the sextic equation is invariant, exactness of invariants in characteristic zero identifies its quotient ring with this polynomial invariant ring modulo that equation. This is the actual finite quotient, not a proposed birational analogy.

A singular point of the weighted affine cone away from the origin would, by the chain rule at any preimage under the coordinate-squaring map, give a singular point of the original affine cone away from the origin. Thus the weighted hypersurface is quasismooth. The ambient weights are well-formed, and its singular stratum has dimension one, hence has intersection of codimension at least two in our threefold. The hypersurface is well-formed.

For completeness, the canonical class can be computed directly on the finite quotient $q:Y\to Q$. The two generators fix distinct hyperplane sections $D_1,D_2$ pointwise, with ramification index two. Their product has positive and negative linear eigenspaces of dimensions three and two. The latter give projective fixed spaces of dimensions two and one. A smooth sextic threefold cannot contain a plane: weak Lefschetz gives $\operatorname{Pic}(Y)=\mathbb Z h$, whereas a plane divisor would have intersection $1$ with $h^2$, not a multiple of $\int_Yh^3=6$. Consequently the product has no divisorial fixed locus on $Y$. The isolated fixed point of each generator contributes no divisor either. The ramification divisor is exactly $D_1+D_2$, and

$$ q^*K_Q=K_Y-D_1-D_2=h-2h=-h. \tag{2} $$

Finite pullback detects ampleness, so $-K_Q$ is ample. Equivalently, well-formed quasismooth weighted adjunction gives $\omega_Q=\mathcal O_Q(6-7)$ as a reflexive sheaf. The quotient of smooth $Y$ has quotient singularities and is klt. The theorem and proof in [4] give rational connectedness of a smooth resolution, hence uniruledness. $\square$

Let $\delta$ act by $x_4\mapsto\zeta_6^{-1}x_4$. There is one more source:

$$ Y/\langle\delta^2\rangle =\{u^2+f_{B,C}=0\}\subset\mathbb P(1,1,1,1,3),\qquad u=x_4^3. \tag{3} $$

It is a smooth Fano sextic double solid, since $f_{B,C}=0$ is smooth. Its invariant third cohomology is precisely the deck-order-two summand, of rank $104$. The deck-trivial part is zero, since the full cyclic quotient is $\mathbb P^3$. Formula (3) is the classical source of [3].

4. From quotient images to curves

We recall the support statement carefully because resolution errors matter.

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

Proof. Uniruledness gives a generically finite rational map $S\times\mathbb P^1\dashrightarrow T$ from a smooth surface $S$. After resolving the map by blowups, its third cohomology is generated by $H^1$ of curves: this holds for $S\times\mathbb P^1$ by hard Lefschetz on $S$, restriction to an ample curve, and Lefschetz $(1,1)$; blowing up a curve adds its $H^1(-1)$, and blowing up a point adds nothing in degree three. Push-pull for the generically finite morphism gives the assertion.

For the quotient, resolve its graph as $Y\xleftarrow{p}\widetilde Y\xrightarrow{f}T$. For $\xi\in H^3(Y)^G$, put $\beta=f_*p^*\xi$. On a common free quotient open the trace is $|G|$, so

$$ |G|\xi-p_*f^*\beta\in N^1H^3(Y). \tag{4} $$

This coniveau space is generated by Gysin images of $H^1$ of resolutions of divisors, and those weight-one spaces are generated algebraically by curves. Thus both terms on the right of the resulting expression for $|G|\xi$ are curve-generated. Choose finitely many curve maps spanning the invariant space and apply the rational group projector. This retains all supported errors and does not assume a global inverse of quotient graph correspondences. For the supported-cohomology formulation used in (4), see [5, Lemma 5 and its references]. $\square$

5. The quotient images cover the exact rank-454 summand

The full projective coordinate-sign group is

$$ A=(\mathbb Z/2)^5/\langle(1,1,1,1,1)\rangle. $$

Its characters have an even number of minus signs: weights zero, two, or four. Let $P_\chi$ be the rational sign-character projector.

A weight-zero or weight-two character has at least two plus coordinates $i,j$. Its whole subspace is invariant under $H_{ij}$ and hence supported by Lemmas 4--5.

A weight-four character $\chi_k$ has a unique plus coordinate $k$. Then $r_k$ acts trivially on that subspace. Its stabilizer among the base permutations is

$$ S_k= \begin{cases} \mathfrak S_4,&k=4,\\ \mathfrak S_{\{0,1,2,3\}\setminus\{k\}},&k<4. \end{cases} $$

For each transposition $t_{ij}\in S_k$,

$$ \left(P_{\chi_k}H^3(Y)\right)^{K_{k;ij}} =\left(P_{\chi_k}H^3(Y)\right)^{t_{ij}=1}. \tag{5} $$

The character projection commutes with this stabilizer and with $r_k$, so the space in (5) is a projected image of a supported quotient invariant.

Lemma 6. For a rational representation $V$ of a symmetric group, the sum of the transposition-plus eigenspaces is the complement of its sign-isotypic part.

Proof. Average a positive definite rational inner product over the group. The orthogonal complement of the sum consists precisely of vectors on which every transposition acts by $-1$. Transpositions generate the group with the sign character, so that complement is its sign-isotypic part. Both spaces are defined over $\mathbb Q$. $\square$

Thus only the sign-isotypic parts in the five weight-four characters can remain unsupported by the twenty-eight quotients. For $k=4$ the $x_4$ sign is positive, so only deck characters of order three occur; the sign-isotypic summand is the rank-ten block $M_{3,\mathrm{CY}}$ of [1]. For $k<4$ the $x_4$ sign is negative, giving deck orders two or six. The order-two part is supported by (3). The remaining sign parts are the four rank-fourteen blocks $M_{6,\mathrm{CY}}^{(k)}$.

To make the projectors explicit, set

$$ \operatorname{Alt}_{S_k} =\frac1{|S_k|}\sum_{\sigma\in S_k}\operatorname{sgn}(\sigma)\Gamma_\sigma $$

and let $e_d$ be the rational cyclotomic deck projector for order $d$. Then on all cohomology the CY projector is

$$ \pi_{\mathrm{CY}} =e_3P_{\chi_4}\operatorname{Alt}_{S_4} +e_6\sum_{k=0}^3P_{\chi_k}\operatorname{Alt}_{S_k}. \tag{6} $$

All summands are rational algebraic correspondences. With

$$ \pi_3=\Delta_Y-\frac16\sum_{j=0}^3h^{3-j}\times h^j,\qquad \pi_L=\pi_3-\pi_{\mathrm{CY}}, \tag{7} $$

we recover the precise $L$ of [1]. Applying $\pi_L$ to the curve sources supplies the surjection in Theorem 1.

The representation dimensions can be verified at the Fermat member. The Jacobian basis is $x^e$ with $0\le e_i\le4$ and total degrees $1,7,13,19$, corresponding to $(3,0),(2,1),(1,2),(0,3)$. The residue action includes the determinant, so its sign character is $((-1)^{e_i+1})_i$. The five weight-four blocks each have Hodge profile $(1,26,26,1)$; the ten weight-two blocks each have $(0,12,12,0)$; the weight-zero block has $(0,5,5,0)$. After the stabilizer-sign projections and order separation, the result is:

| Deck order | Total rank | CY rank | Supported rank | |---|---:|---:|---:| | 2 | 104 | 0 | 104 | | 3 | 208 | 10 | 198 | | 6 | 208 | 56 | 152 | | Total | 520 | 66 | 454 |

The finite-group representation is locally constant on the connected smooth parameter locus $U$. The image ranks and projector identities therefore hold throughout $U$, including special members. This also follows directly from the character argument and the global group projectors above.

6. Hodge classes and the published tower

For two smooth members, choose algebraic surjections $A=\bigoplus H^1(C_i)(-1)\twoheadrightarrow L$ and $A'=\bigoplus H^1(C'_j)(-1)\twoheadrightarrow L'$. Polarizable Hodge semisimplicity provides a Hodge lift of any rational Hodge class in $L\otimes L'$ to $A\otimes A'$. After removing the twists, each lifted component is a rational $(1,1)$ class on $C_i\times C'_j$. Lefschetz $(1,1)$ represents it by a divisor. Composing with the two algebraic curve correspondences gives the required codimension-three cycle. No algebraicity of an arbitrary abstract Hodge section is used.

At very general parameters, [1, Theorem 3] proves that the rank-ten and rank-fourteen CY blocks are simple rational Hodge structures of level three, with the appropriate endomorphisms represented by automorphism graphs. Hence they have no nonzero level-one rational sub-Hodge structures. It follows that $L$ is the entire maximal level-one subspace, and Theorem 1 establishes the degree-three generalized Hodge conjecture.

All nonmiddle cohomology of $Y$ is Tate. On $Y^2$ only $H^3(Y)^{\otimes2}$ requires attention beyond hyperplane products. The mixed $M_{\mathrm{CY}}$ and $L$ terms have no rational Hodge classes at very general parameters by the simplicity and levels just stated. The CY self-tensor classes are algebraic by [1], and the $L$ self-tensor classes by Theorem 1. This proves HC for $Y^2$.

The finite morphism $Y\to W$ given by $z=x_4^2$ has degree two. Both varieties are smooth projective, so pullback followed by pushforward is multiplication by the degree. HC for $Y^2$ therefore implies HC for $W^2$.

Finally the Katsura--Shioda blowup construction [1,6] relates $X$ to the diagonal $\mu_6$ quotient of a blowup of $Y\times Y$ along $V\times V$. The new middle classes in this blowup are shifts of classes from $V\times V$. Its HC is already proved at very general parameters in [1]. The finite quotient and birational morphism to $X$, with the stated blowup centers, now transfer algebraic representatives for every Hodge class of $X$. Equivalently, use the explicit decomposition in [1]: the order-one sector is furnished by the surface square; the remaining sector is the diagonal-deck-invariant part of $H^3(Y)^{\otimes2}$. Both are algebraic by the results just cited and Theorem 1. This proves Corollary 2.

The distinction between all-member support and generic maximality is essential. For example, the Fermat point can acquire extra level-one substructures in the fixed CY summand. Theorem 1 makes no uniform assertion about such additional pieces.

7. Single reflections on quintics

Let a finite linear group $G$ preserve a homogeneous quintic equation $F$ in five variables. Suppose that $G$ is generated by reflections whose linear eigenvalues are $(1,1,1,1,-1)$. Write $\epsilon=\det:G\to\{1,-1\}$ and

$$ e_\epsilon=\frac1{|G|}\sum_{g\in G}\epsilon(g)\Gamma_g. \tag{8} $$

Proposition 7. If $Z=\{F=0\}\subset\mathbb P^4$ is smooth, then $(1-e_\epsilon)H^3(Z,\mathbb Q)$ is curve-generated.

Proof. For a generating reflection choose linear coordinates $(z_0,z_1,z_2,z_3,b)$ in which it negates $b$. The invariant polynomial ring is $\mathbb C[z_0,z_1,z_2,z_3,w]$, where $w=b^2$. Exactness of invariants identifies the quotient of $Z$ with a weighted quintic $Q\subset\mathbb P(1,1,1,1,2)$.

As in Lemma 4, the chain rule proves quasismoothness of this actual quotient from smoothness of $Z$. The ambient singular locus is its single weight-two point, so $Q$ is well-formed. Weighted adjunction gives $K_Q=\mathcal O_Q(5-6)$. Equivalently, the only divisorial fixed locus is the hyperplane section $D=\{b=0\}\cap Z$; the other fixed locus is one point. Thus $q^*K_Q=K_Z-D=-H$. The finite quotient is klt and its anticanonical divisor is ample. A smooth resolution is uniruled by [4], and Lemma 5 makes $H^3(Z)^{r=1}$ curve-generated.

Average a positive definite rational inner product over $G$. The perpendicular to the sum of the generating reflection-plus eigenspaces is the subspace on which every generator acts by $-1$. Since the generators have determinant $-1$, this is exactly $e_\epsilon H^3(Z)$. Hence their sum is $(1-e_\epsilon)H^3(Z)$. This is a rational group-projector statement independent of the auxiliary inner product. Applying the complementary algebraic projector to the curve sources proves the proposition. $\square$

For the application, let $\mathfrak S_3$ permute $x_0,x_1,x_2$, and let $D_5$ denote the group of order ten generated by

$$ r_a:(u,v)\longmapsto(\zeta_5^a v,\zeta_5^{-a}u), \qquad 0\le a<5. \tag{9} $$

Together with the three $x$-transpositions these give eight reflections in $G=\mathfrak S_3\times D_5$. Its polynomial invariant ring has generators $e_1(x),e_2(x),e_3(x),uv,u^5+v^5$, of degrees $1,2,3,2,5$. Consequently every invariant quintic has the form

$$ c(u^5+v^5)+f_5(x)+uvf_3(x)+(uv)^2f_1(x), \tag{10} $$

where each $f_i$ is symmetric of degree $i$. This coefficient space has dimension ten, and its connected smooth locus contains the Fermat member.

At Fermat, the residue ring is

$$ R=\mathbb Q[x_0,x_1,x_2,u,v]/(x_0^4,x_1^4,x_2^4,u^4,v^4), $$

with Hodge grades $R_0,R_5,R_{10},R_{15}$. The action on residues includes the determinant of the ambient linear action. Being negative for every $x$-transposition means that the polynomial is symmetric in $x$. Being negative for all the reflections (9) means invariance under their rotations, hence equal exponents of $u,v$; the ordinary swap then acts negatively on the residue. Counting the symmetric monomial orbits gives

$$ \dim(e_\epsilon R_{5q})=1,7,7,1\quad(q=0,1,2,3). \tag{11} $$

Finite-group representations on the Hodge bundles are locally constant over the connected smooth family. Thus the same profiles hold throughout (10). Define $C=e_\epsilon H^3(Z)$ and $L=(1-e_\epsilon)H^3(Z)$. Proposition 7 proves uniform support for $L$, with rank $204-16=188$. On the separated family of [2], these are its $M_{1,\mathrm{III}}$ and complement. That paper already realizes sixty dimensions by curves; the reflection coverage adds the other $128$.

For completeness the algebraic middle projector of a smooth quintic is

$$ \pi_3^Z=\Delta_Z-\frac15\sum_{j=0}^3h^{3-j}\times h^j. $$

Its determinant projector is $\pi_C=\pi_3^Z\circ e_\epsilon$, and $\pi_L=\pi_3^Z-\pi_C$. These expressions will provide the polarization class needed for the square theorem.

8. The cyclic holomorphic period in the enlarged family

We prove the remaining assertions of Theorem 3 on the smooth locus $T\subset\mathbb A^7$ of (1), which contains $t=0$. The seven deformation polynomials, in the following order, are

$$ \begin{aligned} Q_0&=(uv)^2m_{100},& Q_1&=uvm_{300},& Q_2&=uvm_{210},\\ Q_3&=m_{320},&Q_4&=uvm_{111},&Q_5&=m_{311},&Q_6&=m_{221}. \end{aligned} \tag{12} $$

Their classes form a basis of $R_5$ in the determinant-residue part. Their orbit representatives, in the coordinate order $(x_0,x_1,x_2,u,v)$, are

$$ \begin{gathered} (0,0,1,2,2),\ (0,0,3,1,1),\ (0,1,2,1,1),\ (0,2,3,0,0),\\ (1,1,1,1,1),\ (1,1,3,0,0),\ (1,2,2,0,0). \end{gathered} $$

All orbit sums use distinct monomials with coefficient one.

In the degree-ten symmetric orbit basis ordered lexicographically by its exponent representatives, the seven products

$$ Q_0Q_1,\ Q_0Q_2,\ Q_0Q_3,\ Q_0Q_4,\ Q_0Q_5,\ Q_0Q_6,\ Q_1Q_3 $$

have coefficient matrix

$$ A=\begin{pmatrix} 1&1&0&0&0&0&0\\ 0&2&0&0&0&0&0\\ 0&0&2&0&0&0&0\\ 0&2&0&1&0&0&0\\ 0&0&1&0&1&1&0\\ 0&0&0&0&0&3&0\\ 0&0&0&0&0&0&2 \end{pmatrix},\qquad \det A=24. \tag{13} $$

The final product is

$$ Q_0Q_1Q_3=6x_0^3x_1^3x_2^3u^3v^3\quad\text{in }R_{15}. \tag{14} $$

Equations (12)--(14) are direct calculations modulo the five fourth powers; the supplement gives a short exact reconstruction. They show that products of the actual parameter polynomials span every grade in (11). In particular, these are allowed family deformations, not unrealized infinitesimal directions in a larger Jacobian ring.

Set $\omega_t=\operatorname{Res}(\Omega/F_t)$, a local generator of the holomorphic line in $C_t$. Griffiths' derivative formula [9] gives the order-$r$ leading symbol

$$ \operatorname{gr}^{3-r}_F \nabla_{i_1}\cdots\nabla_{i_r}\omega\big|_0 =(-1)^r r! [Q_{i_1}\cdots Q_{i_r}],\qquad 0\le r\le3. \tag{15} $$

The nonzero determinants $1,24,6$ prove that $\omega$ and its first three derivatives span all sixteen dimensions of $C_0$. Lower-filtration terms do not affect this conclusion: successive leading symbols span the associated graded spaces of dimensions $1,7,7,1$.

Lemma 8. Suppose a polarized rational variation $\mathbb V$, admitting an integral lattice, over a connected smooth algebraic base has a one-dimensional top Hodge piece, and a local generator of that line is cyclic for the flat connection at one point. Then its very-general rational Hodge endomorphisms are scalars.

Proof. Apply the Hodge-locus theorem [10] to the weight-zero tensor variation $\operatorname{End}(\mathbb V)$. There are countably many algebraic component images. Exclude the proper ones. A rational Hodge endomorphism at a remaining point extends by flat transport to a Hodge morphism on the universal cover: locally its Hodge condition is an analytic equation; a component dominating the base makes that equation hold on an open set, and the identity theorem continues it. This also explains why the exceptional set is formulated using component images, rather than a single algebraic set for each marked rational map.

On a simply connected neighborhood of the cyclic point, the resulting flat endomorphism $f$ satisfies $f\omega(t)=\lambda(t)\omega(t)$. The scalar is holomorphic and belongs to the finite spectrum of the fixed flat matrix $f$, so it is constant. Therefore $f$ acts by this same scalar on every iterated derivative of $\omega$. Cyclicity implies $f=\lambda\mathrm{Id}$. Its rationality implies $\lambda\in\mathbb Q$. $\square$

Applied to (15), Lemma 8 gives $\operatorname{End}_{\mathbb Q\text{-}\mathrm{Hdg}}(C_t)=\mathbb Q$ very generally. A polarized sub-Hodge structure has a rational Hodge orthogonal projector, so this equality makes $C_t$ simple. Since it contains $H^{3,0}$, no nonzero Hodge morphism can join it to a level-one structure. Thus $L_t$ is the maximal level-one part, and its uniform curve support proves degree-three GHC for very general $t$.

The argument of Section 6 proves algebraicity of all Hodge classes in $L_t\otimes L_{t'}$ for arbitrary smooth parameters, even in the full coefficient family (10). At very general $t$, the mixed $C_t$--$L_t$ Hodge tensors vanish and the Hodge tensors of $C_t\otimes C_t$ form a one-dimensional space generated by its polarization projector. This is the algebraic correspondence $\pi_C$ written above. Hence all $H^6(Z_t^2)$ Hodge classes are algebraic. Weak Lefschetz gives only hyperplane classes outside middle cohomology on each factor, so all other even cohomology of the square is already algebraic. This proves the full square assertion.

For a very general independently varying pair, apply the same countability and Hodge-locus argument to $\operatorname{Hom}(\operatorname{pr}_1^*C,\operatorname{pr}_2^*C)$ on $T^2$. A nonzero generic morphism would be an isomorphism by simplicity. Its flat extension to a product neighborhood, with the second parameter fixed, would carry the first holomorphic line to a fixed line. This contradicts its rank-seven first derivative at $0$. Therefore the $C_t\otimes C_{t'}$ Hodge tensors vanish at very general independent pairs. The mixed terms again vanish and the $L$ terms are algebraic. This proves the remaining product assertion of Theorem 3.

8.1. A nodal orbit and full residual monodromy

The cyclic-period calculation also combines with a concrete degeneration to control every tensor power of the selected component $C$.

Theorem 9. The connected algebraic monodromy of $C$ in (1) is $\operatorname{Sp}(C)\simeq\operatorname{Sp}_{16}$. For very general $t$, every rational Hodge class in $C_t^{\otimes 2r}(3r)$, for every $r\ge1$, is algebraic on $Z_t^{2r}$. The corresponding assertion holds for all selected $C$-tensors of a very general independently varying finite tuple, with contractions within each parameter's tensor factors.

Proof. A fixed scaling of $u,v$ identifies the family (1) with the one having coefficient $6$ on $u^5+v^5$. In that normalization consider

$$ F_*=x_0^5+x_1^5+x_2^5+6(u^5+v^5)-5(uv)^2(x_0+x_1+x_2). \tag{16} $$

Its complete projective singular locus consists of the five points

$$ p_k=[\zeta^{3k},\zeta^{3k},\zeta^{3k},\zeta^k,1], \qquad k=0,\ldots,4,\quad\zeta^5=1. \tag{17} $$

To prove completeness, a singular point with $u=0$ or $v=0$ would have all $x_i=0$ by the three $x$ derivatives, and the remaining pure fifth-power derivative excludes a projective point. Set $v=1$. The identity $uF_u-vF_v=30(u^5-v^5)$ gives $u^5=1$. Then $x_i^4=u^2$, so $x_i=u^3\epsilon_i$ with $\epsilon_i^4=1$, and the $u$ derivative gives $\sum\epsilon_i=3$. Equality in the triangle inequality forces all $\epsilon_i=1$, proving (17).

At $p_0$, the Hessian in coordinates $(x_0,x_1,x_2,u)$ is

$$ \begin{pmatrix} 20&0&0&-10\\ 0&20&0&-10\\ 0&0&20&-10\\ -10&-10&-10&90 \end{pmatrix},\qquad \det=600000. \tag{18} $$

The determinant is the same at all five points after reduction by $u^5=1$. All are ordinary double points. The allowed deformation $Q=(uv)^2\sum x_i$ takes value $3$ at each. Thus $F_*+sQ$ is a transverse simultaneous smoothing. For sufficiently small $s\ne0$ there are no other singularities, by compactness and (17).

The rotations $(u,v)\mapsto(\zeta^j u,\zeta^{-j}v)$ send $k$ to $k+2j$; the nodes form one orbit. The stabilizer of $p_0$ is $\mathfrak S_3$ times the untwisted swap of $u,v$. Its action on the node's local vanishing three-sphere is the determinant on the four-dimensional affine tangent space, hence the character $\varepsilon=\det$ of $G$. Indeed, the $x$ permutations have their usual determinants; the swap has derivative $(\dot x_i,\dot u)\mapsto(\dot x_i-\dot u,-\dot u)$ and determinant $-1$. Equivariant holomorphic Morse coordinates identify this with the usual determinant action on the vanishing sphere of a nondegenerate quadratic form.

Orient vanishing cycles $\delta_k$ by transporting $\delta_0$ with the rotations. The local module is $\varepsilon$ times the permutation module of the five nodes, and its $\varepsilon$ part is a line. Its image in global cohomology is generated by

$$ q=e_\varepsilon\delta_0=\frac15\sum_{k=0}^4\delta_k. \tag{19} $$

This global image is nonzero. The holomorphic residue $\omega_s$ has character $\varepsilon$. In local holomorphic Morse coordinates the smoothing is $\sum_{j=1}^4 y_j^2=a(s)$ with $a'(0)\ne0$, and the residue numerator has nonzero constant term. Scaling $y=\sqrt{a(s)}z$ in the standard quadratic residue integral gives

$$ \int_{\delta_0}\omega_s=c\,a(s)+O(a(s)^2),\qquad c\ne0. $$

The period is nonzero for small $s\ne0$. Since $\omega_s$ is $\varepsilon$-isotypic, its period on $q$ is the same. Thus $q\ne0$; this argument does not presume that every local vanishing cycle always survives in global cohomology.

The Milnor neighborhoods are disjoint. The Picard--Lefschetz formula therefore restricts to

$$ T(x)=x\mathbin{\pm}5\langle x,q\rangle q\quad(x\in C). \tag{20} $$

The polarization restricted to $C$ is nondegenerate, since $e_\varepsilon$ is a self-adjoint polarization projector. Hence $T-I$ has rank one and square zero. The Zariski closure of its integer powers contains the associated additive one-parameter subgroup, which lies in connected monodromy.

Let $M$ be the generic Mumford--Tate group of $C$, and $H$ its connected algebraic monodromy. Lemma 8 makes the commutant of $M$ equal to $\mathbb Q$. The commutation equations are rational linear equations, so extension of scalars makes the complex commutant equal to $\mathbb C$. Reductivity of $M$ and complete reducibility then imply that $C_{\mathbb C}$ is irreducible for $M_{\mathbb C}$. Andr'e's normality theorem [11, Theorem 1] gives $H\triangleleft M^{\mathrm{der}}$ and thus $H\triangleleft M$. The restriction to this connected normal reductive subgroup is isotypic: its isotypic idempotents are permuted by the connected group $M_{\mathbb C}$, hence are fixed; see also [11, Lemma 1]. Write it as $U\otimes\mathbb C^m$, with $H$ acting trivially on the multiplicity factor. The rank of $h-I$ is divisible by $m$ for every $h\in H$. Formula (20) forces $m=1$. Thus $H$ acts irreducibly. This step uses both normality and the transvection; scalar Hodge endomorphisms alone would not suffice.

We give the needed symplectic group argument. The Lie algebra of a connected irreducible subgroup $H\subset\operatorname{Sp}(V)$ containing a transvection contains $N_w:x\mapsto\langle x,w\rangle w$ for every $w$ in the orbit $O$ of one nonzero transvection vector. The orbit is irreducible and spans $V$. Put

$$ S(a,b)x=\langle x,a\rangle b+\langle x,b\rangle a. $$

For $a,b\in O$ with $\langle a,b\rangle\ne0$, the bracket $[N_a,N_b]$ is a nonzero multiple of $S(a,b)$. For an orthogonal pair, choose $w\in O$ pairing nontrivially with both; the two nonempty Zariski opens intersect because $O$ is irreducible. The bracket $[S(a,w),N_b]$ is then a nonzero multiple of $S(a,b)$. These operators span $\operatorname{Sym}^2V\simeq\mathfrak{sp}(V)$, since $O$ spans $V$. Thus $H=\operatorname{Sp}(V)$. Applied to (20), this proves $H=\operatorname{Sp}_{16}$. The generic Hodge group is contained in this symplectic group and contains $H$, so it too is the full group.

The symplectic invariant theorem [12, Theorem 4.1 and Corollary 4.2] says that all even invariant tensors are spanned by pair contractions; there are no odd invariant tensors. The inverse polarization tensor of $C$ is its diagonal component, given by the algebraic projector $\pi_C$. Products and pullbacks of that class give every pairing tensor on $Z_t^{2r}$, proving the algebraicity assertion.

For independently varying finite tuples the parameter fundamental group is a product and each factor acts only on its own variation. The connected monodromy is consequently the product of the respective symplectic groups. Its invariants contract only within the tensor factors belonging to a given parameter. There are countably many orders and rational tensors; [10] supplies the countable union of proper algebraic exceptional loci, simultaneously for all orders. $\square$

Theorem 9 alone concerns the algebraically projected $C$ sectors. Curve support and the Lefschetz $(1,1)$ theorem on products of two curves do not by themselves control higher tensors of $L$. The next section supplies the additional monodromy argument.

8.2. The level-one components and full powers

We now control the remaining tensors of $L$, using the actual seven Higgs multiplication maps and two further nodal orbits. Let $K=\mathbb Q(\sqrt5)$, the character field of the two-dimensional representations of $D_5$.

Theorem 10. For the seven-parameter family (1), the rational Hodge conjecture holds on every power of a very general $Z_t$, and on all mixed powers of a very general independently varying finite tuple. These assertions hold simultaneously for all tensor orders. They are not asserted at arbitrary special parameters.

We give the finite certificates and the geometric argument, including the passage from the finite-field calculation to characteristic zero.

Multiplicity spaces. Use polynomial types $s,a,b$ for the symmetric, alternating and standard representations of $\mathfrak S_3$. These are polynomial labels: residue cohomology includes the determinant twist. The rotation weight is $k=a_u-a_v\pmod5$. We identify weights $k,-k$ by the $D_5$ reflection. A primitive standard projector is obtained by taking the $(01)$-even polynomial line and removing the symmetric part. The multiplicity spaces, suppressing their finite-group factors, are as follows:

| Label | Polynomial type, weight | $h^{2,1}$ | Total rank over $\mathbb C$ | |---|---|---:|---:| | $A$ | $a,0$ | 2 | 4 | | $B$ | $b,0$ | 8 | 16 | | $D_1,D_2$ | $s,1$ and $s,2$ | 6,6 | 12,12 | | $E_1,E_2$ | $a,1$ and $a,2$ | 1,1 | 2,2 | | $J_1,J_2$ | $b,1$ and $b,2$ | 6,6 | 12,12 |

The remaining $s,0$ component is $C$. The $k=1,2$ pairs are the two real embeddings of $K$-linear multiplicity spaces. The dihedral representations are orthogonal over $K$, as is the standard $\mathfrak S_3$ representation over $\mathbb Q$; consequently their multiplicity polarizations are alternating. All displayed components of $L$ have only Hodge types $(2,1),(1,2)$. The ranks follow from the Fermat residue action as in Section 7 and remain constant. Counting finite-group multiplicities gives $\dim L=188$.

Generic Hodge maps from Higgs matrices. At the coefficient vector $(2,3,5,7,11,13,17)$ in the order $Q_0,\ldots,Q_6$, the supplement constructs the actual Jacobian quotients in degrees $5,10$ modulo $1009$. It applies the polynomial projectors

$$ \sum_{g\in\mathfrak S_3}g,\qquad \sum_{g\in\mathfrak S_3}\operatorname{sgn}(g)g,\qquad 3(1+(01))-\sum_{g\in\mathfrak S_3}g, \tag{21} $$

respectively. Each is six times the desired projector. The weight condition and these rational matrices select the primitive polynomial spaces. Projecting every monomial multiple of the five actual partial derivatives gives all Jacobian relations in each space. Row reduction chooses quotient coordinates. The resulting multiplication maps are

$$ M_i^V:R_5(V)\longrightarrow R_{10}(V),\qquad P\longmapsto Q_iP, \quad 0\le i\le6. $$

All eight quotient dimensions in each degree are, in order $(A,B,D_1,E_1,J_1,D_2,E_2,J_2)$,

$$ (2,8,6,1,6,6,1,6). $$

For every ordered pair $V,W$, solve the linear equations

$$ Y M_i^V=M_i^W X\quad(0\le i\le6). \tag{22} $$

The computed kernel dimensions are zero off the diagonal and

$$ (2,1,1,1,1,1,1,1) \tag{23} $$

on the diagonal. The supplement records the quotient pivots, all seven matrices for every space, and the pivots of all sixty-four systems.

Here is why this finite calculation bounds the characteristic-zero problem. The smooth characteristic-zero coefficient locus is nonempty and irreducible. Its Jacobian dimensions are the stated constant ones. Every polynomial projector and Macaulay matrix is defined over $\mathbb Z[1/30]$. The nonzero pivots modulo $1009$ supply nonzero minors over $\mathbb Q$ at the same coefficient vector. Thus the quotient bases and matrix formulas are valid on a nonempty rational open locus; the generic relation ranks cannot exceed those on the smooth locus, where the dimensions are known. This open meets the smooth locus. Clear the nonzero pivot denominators in (22). Its finite-prime ranks give lower bounds for the generic rational ranks, so (23) gives upper bounds for the generic kernel dimensions over $\mathbb C$. No claim that a random finite-field sample determines an analytic period domain is used here.

A generic rational Hodge map extends flatly over the universal cover, by the argument of Lemma 8. Differentiating its Hodge-filtration compatibility gives (22) on the associated graded spaces. This passage is injective on Hodge maps: they preserve both Hodge types, so zero maps on both graded pieces mean the original map is zero.

We also spell out the scalar extension used here. Let $M$ be the generic Mumford--Tate group of the full rational variation. Its rational commutant is the algebra of generic rational Hodge endomorphisms, and its complex commutant is obtained by scalar extension, since the commutation equations are rational linear equations. Every primitive complex finite-group projector commutes with $M_{\mathbb C}$. A map between two primitive multiplicity spaces that commutes with $M_{\mathbb C}$ therefore extends, by inclusion and projection, to an element of this full complex commutant. It is a complex linear combination of the flat rational Hodge maps just considered. It too satisfies (22) and is detected by its associated graded map. This justifies the bounds separately for the two embeddings of $K$, even though the individual embedding spaces are not rational subvariations. The scalar maps supply the lower bound one on every diagonal; for $A$ the upper bound is still two. All cross commutants vanish. These statements concern generic Hodge maps; an arbitrary solution of (22) need not be flat.

Two complete nodal orbits. Both of the following members belong to (1) after a fixed scaling of $u,v$:

$$ \begin{aligned} F_{15}={}&\sum x_i^5-\frac{46}{3}(u^5+v^5) -\frac{25}{3}uv\sum x_i^3+20(uv)^2\sum x_i,\\ F_{30}={}&\sum x_i^5-648(u^5+v^5)-25uv\sum x_i^3 -50uvx_0x_1x_2+370(uv)^2\sum x_i. \end{aligned} \tag{24} $$

Their complete singular loci consist respectively of the permutations of $(1,1,2)$ and $(1,2,3)$ in the $x$ coordinates at $u=v=1$, followed by the five rotations used in (17). There are fifteen and thirty nodes, each set a single $G$ orbit. The Hessian determinants at the seeds are $-10350000$ and $1312200000$. The deformation $Q_0$ has values $4$ and $6$ respectively, so it smooths all the nodes transversely.

For completeness, both singularity classifications reduce first to $uv\ne0$ and $u^5=v^5$ by the same derivative identity as for (16). After the rotation and projective normalization take $u=v=1$. For $F_{15}$ the remaining derivatives give

$$ x_i\in\{-2,-1,1,2\},\qquad \sum_i(24x_i-5x_i^3)=46, $$

whose only solutions are the permutations of $(1,1,2)$. For $F_{30}$ a repeated-coordinate solution $(r,r,t)$ would satisfy

$$ \begin{aligned} r^4-15r^2-10rt+74&=0,\\ t^4-15t^2-10r^2+74&=0,\\ -648-5(2r^3+t^3)-10r^2t+148(2r+t)&=0. \end{aligned} $$

Their rational ideal is the unit ideal, as the small exact elimination in the supplement verifies. If the three coordinates are distinct, put $h(T)=T^3-e_1T^2+e_2T-e_3$. Their derivative equations imply

$$ h(T)\mid 5T^5-75T^3+370T-50e_3. $$

The three remainder coefficients and the $u$ derivative have ideal $(e_1-6,e_2-11,e_3-6)$, again by explicit rational elimination. Thus the coordinates are the three roots $1,2,3$. This also treats possible zero coordinates and completes the classification.

We must check that the local vanishing cycles survive independently in global cohomology. Evaluate all degree-five monomials on these nodes over $\mathbb Z[\zeta]$. Reduction modulo $101$ with $\zeta\mapsto95$ gives ranks $15,30$; the listed square witness minors have determinants $17,84$ modulo $101$. Hence their ranks are full in characteristic zero. On a nearby smooth fiber, the periods of $\operatorname{Res}(P\Omega/F^2)$ on the local vanishing cycles have leading terms $c_jP(p_j)$, with every $c_j\ne0$. This follows by differentiating the quadratic holomorphic vanishing period used above, allowing all degree-five directions in the full hypersurface space. The evaluation minors remain nonzero for small smoothing parameter. Thus all fifteen, respectively thirty, local cycles are independent in global $H^3$.

The node stabilizers are $\mathfrak S_2\times C_2$ for $F_{15}$ and $C_2$ for $F_{30}$, with $C_2$ the $D_5$ reflection. They fix the chosen projective representative itself, so their local vanishing character is the determinant. The local representation of the first orbit is consequently

$$ (\operatorname{sgn}_{\mathfrak S_3}\oplus\operatorname{std}_{\mathfrak S_3}) \otimes(\operatorname{sgn}_{D_5}\oplus\rho_1\oplus\rho_2). \tag{25} $$

Every multiplicity in (25) is one. Thus it supplies a rank-one Picard--Lefschetz logarithm on $B,D_1,D_2,J_1,J_2$, as well as $C$. For $F_{30}$, the $\mathfrak S_3$ factor is its regular representation; its polynomial alternating component has multiplicity one. It supplies rank-one logarithms on $A,E_1,E_2$.

These rank assertions use the proved global independence. If $D$ is the injective map from the space of local cycles to global cohomology, the Picard--Lefschetz logarithm is, up to sign, $DD^{\mathsf T}J$, where $J$ is the global nondegenerate alternating pairing. Its rank is the number of nodes because $D^{\mathsf T}J$ is surjective. Its image as a $G$ representation is exactly the local-cycle representation. Therefore its rank on each primitive multiplicity space is the multiplicity in (25), or in the corresponding regular representation. This avoids inferring a nonzero projection merely from a local character match.

The remaining endomorphism bound for $A$. This rank-four rational variation has generic endomorphism dimension at most two by (23). At Fermat, simultaneous rotation $(u,v)\mapsto(\zeta u,\zeta v)$ commutes with $G$ and acts on its four embedding lines with eigenvalues $\zeta^2,\zeta^4$ in $H^{2,1}$ and $\zeta,\zeta^3$ in $H^{1,2}$. Thus it is rank one over $E=\mathbb Q(\zeta_5)$, with shifted CM type $\{2,4\}$. Its Hodge endomorphism algebra is exactly $E$: decompose $\operatorname{End}_{\mathbb Q}(E)=\bigoplus_{g\in\operatorname{Gal}(E/\mathbb Q)}Eg$; only the identity $g$ preserves this type. In particular it is simple.

Generic rational Hodge endomorphisms specialize injectively to Fermat by flat continuation. Their algebra is thus a subalgebra of $E$, and its dimension is at most two. It is either $\mathbb Q$ or the only quadratic subfield $K$. The rational rank-one transvection on $A$ just constructed commutes with this algebra: connected monodromy is contained in the generic Mumford--Tate group. A $K$-linear map has even rational rank, so the second possibility is excluded. Therefore $\operatorname{End}_{\mathrm{Hdg}}(A)=\mathbb Q$ very generally.

Full monodromy and tensors. Each primitive complex multiplicity space now has generic scalar Hodge commutant and a rank-one transvection. Normality, isotypy, and the Lie-algebra argument in Theorem 9 apply to each. The respective connected groups are the full symplectic groups. The joint group has complex form

$$ \operatorname{Sp}_{16}^{\,2}\times\operatorname{Sp}_4 \times\operatorname{Sp}_{12}^{\,4}\times\operatorname{Sp}_2^{\,2}. \tag{26} $$

Here one $\operatorname{Sp}_{16}$ acts on $C$ and the other on $B$. We justify the product assertion. A connected subdirect subgroup of a product of simple symplectic groups can fail to be the full product only through a diagonal identification of isomorphic factors. The connected normalizer of such a graph in the product of the two symplectic groups is that graph itself, up to the finite centers. Connected monodromy is normal in the generic Hodge group, and the latter lies in the product of the multiplicity symplectic groups because the finite-group projectors and polarizations are Hodge tensors. A diagonal identification would therefore force a complex Hodge intertwiner between the two standard representations. Automorphisms of the complex symplectic group are inner, so this is an ordinary linear intertwiner. For two $L$ components it is excluded by the off-diagonal bounds (22). For $B$ and $C$ it is excluded by their different Hodge profiles. This proves (26).

Finally, invariant theory for the product in (26) expresses every invariant tensor of $H^3(Z_t)$ as a sum of products of invariant pairings, allowing arbitrary vectors in its constant finite-group multiplicity factors. All rational two-factor Hodge classes are already algebraic by Theorem 3: $L\otimes L$ by curve support and Lefschetz, $C\otimes C$ by its projector, and the mixed terms vanish. Taking rational spans commutes with extension to $\mathbb C$, so products and pullbacks of these rational algebraic pair classes span every rational higher Hodge tensor. Nonmiddle cohomology is Tate by weak Lefschetz. This proves full rational HC on every power. For independent parameter tuples the monodromy is the corresponding external product; the same proof works, with contractions within each parameter's factors. The countable Hodge-locus argument supplies one very-general set for all orders. This proves Theorem 10. $\square$

9. Verification and limitations

The supplement contains seven finite inputs to the proof.

1. The commuting-quotient verifier constructs all twenty-eight groups on Fermat residues, includes the determinant character, constructs (6), and verifies $C^2=C$, $C^{\mathsf T}=C$ and $MC=0$ exactly over $\mathbb Q$. Here $M$ is a positive sum of Reynolds operators, with the order-two source added. Its rank modulo the prime $1009$ is a lower bound for its rational rank; the complementary projector gives the matching upper bound. The certified ranks are $454$ and $66$.

2. The equation verifier forms the actual $r_4,t_{01}$ quotient in reflected coordinates, substitutes back exactly, and checks weighted degree six and the dimensions of the linear fixed spaces.

3. The quintic verifier constructs the signed action of all eight reflections, sums the five cyclotomic phases exactly, checks the determinant projector and complementary rank $188$, and verifies quotient equations for both reflection classes for (10).

4. The quintic period verifier reconstructs the orbit bases and products in (12)--(14), checks full graded ranks $7,7,1$, and supplies the nonzero witness determinants $1,24,6$.

5. The nodal-orbit verifier checks the complete singularity reduction for (16), all $64$ fourth-root tuples, the five Hessians, the smoothing values, the node orbit and all twelve stabilizer tangent actions.

6. The generic Higgs verifier constructs the projected Macaulay quotients and all seven multiplication maps at the stated coefficient vector modulo $1009$, records pivot certificates for all sixty-four Hom systems, and checks the genus-two Fermat CM-type stabilizer.

7. The larger nodal-orbit verifier checks the complete fifteen/thirty-node classifications, Hessians, smoothing values and full degree-five evaluation ranks, with the explicit finite-prime witness minors.

Run the seven Python scripts from the supplement root. They require SymPy; the two new matrix verifiers also require NumPy. The arithmetic checks do not prove rational connectedness, the quotient-support lemma, generic Hodge simplicity or HC by themselves: those steps are the written arguments and the cited theorems above.

The proof constructs auxiliary curves existentially through uniruledness and resolution; it does not give equations for every curve or cycle. The higher-power theorem applies to very general members and independent tuples of the seven-parameter quintic family, using all of Section 8.2. No corresponding full sextic higher-power theorem, arbitrary special-member full-square theorem, new Fermat degree, or resolution of general rational HC is claimed.

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