# Common determinant cycles and mixed products of elliptic K3 surfaces

## 1. The result

For $7\leq n\leq12$ and $(a,b)\in\mathbb C^2\setminus\{(0,0)\}$, let $S_{n,a,b}$ be the smooth minimal elliptic model of

$$
y^2=t^n+x^3+ax+b.
$$

These are K3 surfaces. Their pair products satisfy the rational Hodge conjecture by the fixed-curve transfer theorem [1, Theorems 11 and 13]. The following result treats higher products with independent parameters.

**Theorem 1.** Fix $n\in\{7,8,9,10,11,12\}$. For every $r\geq1$ and every collection $(a_i,b_i)\ne(0,0)$, the rational Hodge conjecture holds for

$$
\prod_{i=1}^r S_{n,a_i,b_i}.
$$

Repeated factors, CM specializations, transcendental rank drops, and cubics with a double root are allowed. The exponent $n$ is fixed throughout the product.

The geometric ingredient is an algebraic realization of the determinant of a moving rank-two cyclotomic structure by one fixed K3 surface. Put

$$
K=\mathbb Q(\zeta_n),\qquad
S_{n,0}:\quad y^2=t^n+x^3-\frac{x}{3}+\frac{2}{27},\qquad
T_0=T(S_{n,0}).
$$

Here $T(S)=\operatorname{NS}(S)_{\mathbb Q}^{\perp}\subset H^2(S,\mathbb Q)$ is the transcendental Hodge structure. If $T=T(S_{n,a,b})$ has $K$-rank two, we construct an algebraic isomorphism

$$
(\bigwedge_K^2 T)(1)\simeq T_0. \tag{1}
$$

The exterior square has weight four, followed by one Tate twist. If $T$ has $K$-rank one, there is instead an algebraic isomorphism $T\simeq T_0$. These statements include algebraic inverses on the specified cohomological summands.

For two nonisomorphic rank-two structures, (1) supplies classes on $S_i^2\times S_j^2$ that are not generated by pair correspondences. We prove that these determinant correspondences, the corresponding classes on $S_i^2\times S_j$ when $T_j$ has rank one, and the pair correspondences generate all Hodge tensors. Thus the new conclusion is not inferred just from the known pair theorem.

Section 9 proves the same finite-product conclusion for each fixed $n\in\{14,16,18\}$ in the quadratic-twist family $Y^2=X^3+a u^2X+b u^3+u^{n/2+3}$, by an explicit source of degree $2n$.

The general rational Hodge conjecture remains open. The theorem does not assert the conjecture for arbitrary products with different exponents or settle the exceptional degree-$44$ correspondence between the K3 and general-type surface blocks discussed in [1].

## 2. Prior work and the additional step

The quotient of a power of a cyclic cover by the kernel of the product character and permutations is classical. Schoen identifies its primitive part with a field-linear determinant, including the graded permutation sign [2, Lemmas 1.1–1.2]. His cycle theorem [2, Theorem 2.0] treats determinant Hodge classes under a hypothesis that the branch exponents pair with their negatives. Our four-branch determinants have nonzero holomorphic components; we transport them through a fixed surface and then a K3 carrier. Shioda–Katsura's product decompositions provide the underlying split-variable geometry [3]. We specialize these constructions to cyclic covers with three finite branch points, give the resulting common four-line surface explicitly, and identify its contribution with the transcendental structure of $S_{n,0}$ using divisor correspondences.

Zarhin determines the individual Hodge groups of K3-type structures [4]; we also use the presentation in [5, Chapter 3, Theorems 3.7 and 3.9]. Powers of one K3 surface with a CM endomorphism field already satisfy the Hodge conjecture. This follows from the algebraicity of Hodge isometries and the unitary invariant-theory argument of Ramón Marí, with its scope clarified by Varesco [6, Theorem 2.2 and Corollary 2.3]. We do not claim those self-product results as new.

Commelin defines the determinantal motive of a unitary K3-type structure and proves that it is again of K3 type [7, Definition 11.9 and Lemma 11.10]. His Theorem 16.1 deliberately excludes unitary rank two from a statement recovering a motive from its derived group. Remark 16.2 explains the obstruction: the derived group does not determine the determinant, and products of a fixed non-CM elliptic curve with different CM elliptic curves give counterexamples. Our argument retains an actual common determinant and uses it explicitly to remove the central ambiguity. The abstract determinantal motive is not the novelty here.

The earlier transfer [1] makes every Hodge map between any two members of a fixed-exponent family algebraic, including special parameters. It does not supply the determinant classes on higher products. The additional contribution is their geometric realization, a joint group calculation accounting for all special CM fields, and the resulting arbitrary finite-product theorem. The torus calculation is joint: we do not assume that pairwise independence implies independence of an arbitrary collection of tori.

## 3. Algebraic summands and the curve source

We work with rational cohomology. An algebraic map between specified summands means a map induced by a rational algebraic correspondence after applying the specified algebraic projectors. Tensor products, permutations, compositions, and contractions by nondegenerate algebraic pairings retain this property. Statements about motives below are only shorthand for these operations modulo homological equivalence.

For a K3 surface the projector to $T(S)$ is algebraic. Choose a rational basis $D_i$ of its Néron–Severi group and invert its intersection matrix $(q_{ij})$. The algebraic Néron–Severi projector on $H^2$ is $\sum q^{ij}D_i\times D_j$. Subtract it from the degree-two Künneth projector. Curve Künneth projectors are algebraic as well. Finite-order automorphisms give rational cyclotomic projectors by polynomials in their graphs.

If a number field acts algebraically, the projector to a balanced tensor product is algebraic: its separability idempotent is a rational combination of tensor products of field operators. Exterior and symmetric powers then use the appropriate permutation projectors. The scalar field action on a field-linear determinant is implemented on one balanced tensor position; applying a scalar on every position would instead give its determinant power. On $H^1\otimes H^1$, geometric interchange is minus ordinary tensor flip. We keep this sign when using curve surfaces.

For completeness, the relevant product source and graph identities from [1] are recalled next. Let $P=x^3+ax+b$, $m=\operatorname{lcm}(2,n)$, and take smooth projective normalizations

$$
C_n:\ v^2=u^n+1,
\qquad B_m:\ w^m=P(x).
$$

The cubic has a simple root whenever $(a,b)\ne(0,0)$, so $B_m$ is connected, including the double-root case. There is a dominant rational map of degree $m$,

$$
q:C_n\times B_m\dashrightarrow S_{n,a,b},
\qquad(t,x,y)=\bigl(uw^{m/n},x,vw^{m/2}\bigr). \tag{2}
$$

The source kernel is generated by

$$
(u,v;x,w)\longmapsto(\zeta_n^{-1}u,-v;x,\zeta_mw).
$$

The automorphism $\alpha:(u,v)\mapsto(\zeta_nu,v)$ intertwines with $\tau:t\mapsto\zeta_nt$. On $H^{2,0}(S)$, $\tau$ acts as $\zeta_n$. All other cyclotomic parts of $T(S)$ are pure $(1,1)$ and vanish. Thus $K$ acts on $T(S)$.

The primitive summand $L=P_nH^1(C_n)$ has $K$-rank one. Its holomorphic embedding set consists of the primitive $e$ with $0<e<n/2$. Resolve (2) as $C_n\times B_m\xleftarrow{e}Z\xrightarrow{h}S$. If $v\in T(S)$ and $E$ is exceptional for $e$, then $(h^*v,E)=(v,h_*E)=0$. The same projection formula for divisors shows that $h^*T(S)$ lies in the transcendental mixed part. Consequently the graph maps

$$
b=e_*h^*:T(S)\longrightarrow L\otimes H^1(B_m),
\qquad
a=\Pi_T h_*e^*:L\otimes H^1(B_m)\longrightarrow T(S)
$$

are algebraic and satisfy $ab=m\,1_T$. The image is invariant under the source kernel.

The paired $B_m$ character for a primitive $C_n$ character $e$ is

$$
j(e)=\frac m2+\frac mn e\pmod m.
$$

All these characters have one conductor $N$. The actual quotient $B_m\to B_N$, given by $w\mapsto w^{m/N}$, identifies their summand with $W=P_NH^1(B_N)$, where $B_N:z^N=P(x)$. The possibilities are

| $n$ | $m$ | $N$ | $[K:\mathbb Q]$ | maximal $\operatorname{rank}_{\mathbb Q}T$ |
|---|---|---|---|---|
| 7 | 14 | 14 | 6 | 12 |
| 8 | 8 | 8 | 4 | 8 |
| 9 | 18 | 18 | 6 | 12 |
| 10 | 10 | 5 | 4 | 8 |
| 11 | 22 | 22 | 10 | 20 |
| 12 | 12 | 12 | 4 | 8 |

Identify $K=\mathbb Q(\zeta_N)$ on this selected curve block by letting $\zeta_n$ act as $-\delta$, where $\delta:z\mapsto\zeta_Nz$. Thus corresponding embeddings satisfy

$$
\zeta_N^k=-\zeta_n^e,
\qquad \frac{k}{N}\equiv\frac12+\frac en\pmod1. \tag{3}
$$

With this action the invariant product is the balanced tensor

$$
M_P=L\otimes_KW.
$$

For a separable cubic, each primitive $B_N$ eigenspace has dimension two, since its four branch points have nontrivial primitive local characters. The Hodge profile computed below gives

$$
\dim_KM_P=2,
\qquad h(M_P)=(1,2[K:\mathbb Q]-2,1). \tag{4}
$$

The maps $a,b$ restrict to this algebraic summand. Hence

$$
M_P=bT(S)\oplus A_P, \tag{5}
$$

algebraically, using the projector $ba/m$. The complement is rational pure $(1,1)$ by (4), and all its classes are divisors on the product surface. In particular $\dim_KT(S)$ is one or two.

We also recall why all pair morphisms are algebraic. Every Hodge map between two $T(S)$ commutes with the $K$-actions: its commutator with the two deck generators vanishes on the holomorphic and antiholomorphic lines, so its image is a rational pure-$(1,1)$ substructure of the target $T$, hence zero. For the fixed rank-one CM structure $L$, every $K$-linear map on the full rational tensor products

$$
F:L\otimes_{\mathbb Q}U_1\longrightarrow L\otimes_{\mathbb Q}U_2
$$

can be written $F=\sum k_j\otimes A_j$, with $(k_j)$ a rational basis of $K$. If $F$ is Hodge, restriction to each pure one-dimensional $L_\sigma$ shows that $\sum\sigma(k_j)A_j$ preserves both weight-one Hodge types. Inverting the embedding matrix shows that every $A_j$ is Hodge. Lefschetz $(1,1)$ represents these maps of curve cohomology by divisors. Given $f:T(S_1)\to T(S_2)$, apply this argument to $F=b_2fa_1$ on the full spaces $L\otimes_{\mathbb Q}H^1(B_{m,i})$. The graph maps are $K$-linear, and $a_2Fb_1=m^2f$. Thus $f$ is algebraic. The balanced summands enter through the graph maps; the decomposition of $F$ is asserted on the full rational tensor products. This is [1, Lemma 10 and Theorem 13], and also makes every Hodge endomorphism algebraic.

## 4. A common four-line quotient

Assume first that $P(x)=\prod_{i=1}^3(x-r_i)$ has distinct roots. On $B_N^2$, take the quotient by anti-diagonal $\mu_N$ and factor interchange. Its invariant function field is generated by

$$
b=-(x_1+x_2),\qquad c=x_1x_2,\qquad v=z_1z_2,
$$

with $v^N=\prod_i(r_i^2+br_i+c)$. Over $\operatorname{Sym}^2\mathbb P^1=\mathbb P^2$, the homogeneous equation is

$$
v^N=a^{N-3}\prod_{i=1}^3(ar_i^2+br_i+c).
$$

Set

$$
X_i=\frac{ar_i^2+br_i+c}{P'(r_i)}.
$$

Lagrange interpolation gives $a=X_1+X_2+X_3$, and the coordinate change is invertible. Choose $\rho\in\mathbb C$ with $\rho^N=\prod_iP'(r_i)$. After replacing $v$ by $v/\rho$, the quotient is the fixed surface

$$
Z_N:\quad z^N=X_1X_2X_3(X_1+X_2+X_3)^{N-3}. \tag{6}
$$

Thus there is an actual degree-$2N$ rational quotient map $B_N^2\dashrightarrow Z_N$ for every separable $P$. Root orderings and constant rescalings are choices over $\mathbb C$ and define algebraic correspondences. The residual action on the first curve induces $z\mapsto\zeta_Nz$ on (6).

Let $E=\mathbb Q(\zeta_N)$. Anti-diagonal invariance pairs equal characters in $H^1(B_N)\otimes H^1(B_N)$. Because geometric interchange is minus ordinary flip, its invariant primitive part is

$$
\bigwedge_E^2W. \tag{7}
$$

There is no contribution from the unmixed $H^2$ pieces to a nontrivial residual character. This is also Schoen's determinant description [2].

Let $Z$ be a smooth projective model of (6), chosen with the residual cyclic action. Its irregularity is zero: the anti-diagonal invariants in $H^1(B_N^2)$ vanish because $H^1(B_N)^{\mu_N}=H^1(\mathbb P^1)=0$. Finite quotient singularities are rational, so resolution preserves this vanishing of $H^1(\mathcal O)$. Thus $H^1(Z)=H^3(Z)=0$, and its degree-two Künneth projector is the diagonal minus the point and fundamental-class projectors. Subtracting the Néron–Severi projector gives an algebraic projector onto $T(Z)$.

We use $P_NT(Z)$ to avoid any assertion that every exceptional divisor has trivial character. The resolved graph pulls $P_NT(Z)$ injectively into (7). Invariant holomorphic two-forms descend to the resolved finite quotient: its structure-sheaf cohomology is the invariant cohomology, and rational singularities preserve it under resolution. The primitive invariant space in (7) has a nonzero holomorphic two-form component, by the profiles below. The image is therefore nonzero. Since (7) has $E$-rank one, the equivariant injection is onto.

Its graph trace supplies an inverse up to the degree $2N$. Hence all spaces (7) are algebraically identified with the same $P_NT(Z)$. This is the first transport step, and it is a correspondence statement, not an identification of periods alone.

## 5. The fixed K3 carrier and the boundary

Consider the smooth projective normalizations

$$
A_N:\ v^N=u(1-u),\qquad B_0:\ w^N=x^2(1-x).
$$

There is a degree-$N$ dominant rational map

$$
A_N\times B_0\dashrightarrow Z_N,
\qquad(X_1,X_2,X_3,z)=(xu,x(1-u),1-x,vw). \tag{8}
$$

Indeed $\sum X_i=1$ and $X_1X_2X_3=u(1-u)x^2(1-x)$. Conversely $x=X_1+X_2$ and $u=X_1/(X_1+X_2)$, and there are $N$ choices of $v$. The kernel is anti-diagonal $\mu_N$. Both primitive curve blocks

$$
A=P_NH^1(A_N),\qquad W_0=P_NH^1(B_0)
$$

have $E$-rank one. Their branch tuples are $(1,1,N-2)$ and $(2,1,N-3)$, and every primitive character is nontrivial at all three branch points. The same graph argument identifies $P_NT(Z)$ algebraically with $A\otimes_EW_0$. Transfer the field action by (3).

Here are the profiles and the Tate factor needed to contract (8). For a branch tuple $\mathbf a$, the holomorphic multiplicity of the character $k$ is

$$
h_{\mathbf a}(k)=-1+\sum_j\left\langle-\frac{ka_j}{N}\right\rangle, \tag{9}
$$

where $\langle\cdot\rangle$ is fractional part. In our tuples none of the displayed fractions is integral. Denote the three multiplicities for $A,W_0,W$ by $h_A,h_0,h_W$, respectively, and put $h_L(e)=1$ for $0<e<n/2$ and $0$ otherwise. Formula (3) gives

$$
h_A(k)+h_L(e)=1,\qquad h_A(k)+h_0(k)=h_W(k). \tag{10}
$$

For example, $h_A(k)$ is $1$ when $0<k/N<1/2$, and $0$ when $1/2<k/N<1$. The first identity follows directly from (3). The second uses $\langle2k/N\rangle+\langle-2k/N\rangle=1$ in (9).

If $0<e<n/2$, then $k/N=1/2+e/n$ and $h_0(k)=1$ exactly when $e/n<1/6$. Among primitive $e$ for $7\leq n\leq12$, this occurs only at $e=1$. The possible equality at $n=12,e=2$ is not primitive. Conjugation supplies the other half. Thus

$$
h_L(e)+h_0(k)=
\begin{cases}
2,&e=1,\\
0,&e=-1,\\
1,&\text{otherwise}.
\end{cases} \tag{11}
$$

Equations (10)–(11) prove (4). They also show that $L\otimes_K A$ is pure $(1,1)$. It is an algebraic summand of the cohomology of $C_n\times A_N$, so Lefschetz $(1,1)$ represents every rational class in it by a divisor. Any nonzero rational class is a $K$-basis because this space has $K$-rank one. Its $K$-orbit and intersection-dual divisors give an algebraic isomorphism

$$
L\otimes_K A\simeq K(-1). \tag{12}
$$

The pairing used here is nondegenerate. The balanced summand pairs each character with its conjugate. Alternatively, it lies in the mixed curve cohomology, is orthogonal to a product ample class, and is pure $(1,1)$, so the Hodge index theorem gives a negative definite rational intersection pairing. The notation $K(-1)$ means $[K:\mathbb Q]$ copies of the Tate object with the regular $K$-action; its matrix correspondences are algebraic.

Combining (7), (8), and (12) gives actual algebraic identifications

$$
\begin{aligned}
D_P:=(\bigwedge_K^2M_P)(1)
&\simeq\bigl[L^{\otimes_K2}\otimes_K\bigwedge_K^2W\bigr](1)\\
&\simeq\bigl[L^{\otimes_K2}\otimes_K A\otimes_K W_0\bigr](1)\\
&\simeq L\otimes_KW_0.
\end{aligned} \tag{13}
$$

The permutation of the middle two degree-one factors introduces a common minus sign, which can be absorbed into a correspondence. All projectors and inverse contractions in (13) are algebraic.

Now put $X=1/3-x$. Then

$$
x^2(1-x)=X^3-\frac X3+\frac{2}{27}.
$$

The curve $B_0$ is exactly the selected quotient of the curve source (2) for $S_{n,0}$. That graph map injects $T_0$ into $L\otimes_KW_0$, is nonzero on $H^{2,0}$, and the target has $K$-rank one. It is therefore an algebraic isomorphism, with graph-trace inverse. This proves

$$
D_P\simeq T_0,\qquad \dim_KT_0=1. \tag{14}
$$

If $T(S)$ has $K$-rank two, (5) has no complement, and (14) is (1). If $T(S)$ has $K$-rank one, its complement $A_P$ in (5) also has $K$-rank one and consists of divisors. A divisor basis and its dual identify it algebraically with $K(-1)$. The determinant of (5) is $T(S)\otimes_KA_P$, so (14) now gives an algebraic identification $T(S)\simeq T_0$.

Finally every depressed cubic with a double root and $(a,b)\ne(0,0)$ is $(x-r)^2(x+2r)$ with $r\ne0$. Scaling $x,t,y$ over $\mathbb C$ identifies its K3 model with $S_{n,0}$, compatibly with $t\mapsto\zeta_nt$. This treats the double-root boundary by actual isomorphisms. No specialization principle for Hodge classes is used.

## 6. The joint derived group

We now prove the group statement needed for the tensor conclusion. Let $T_i=T(S_i)$. The field $E_i=\operatorname{End}_{\mathrm{Hdg}}(T_i)$ embeds in $\mathbb C$ through its action on the holomorphic line. Its kernel would have pure-$(1,1)$ image and is zero. Under these embeddings all copies of $K$ have the same image and the same distinguished embedding. Zarhin's theorem says $E_i$ is a CM field, since it contains $K$. The rank bound leaves only three possibilities:

1. $\dim_KT_i=1$, $E_i=K$;
2. $\dim_KT_i=2$, $E_i=K$;
3. $\dim_KT_i=2$, $[E_i:K]=2$, $\dim_{E_i}T_i=1$.

The first and third cases are of CM type, meaning their Hodge group is a torus. The second is not of CM type, despite its CM endomorphism field. All Hodge isomorphisms between factors and all their endomorphisms are algebraic by Section 3. We may therefore consolidate Hodge-isomorphic factors when calculating the joint group and restore repeated factors afterward.

Put $F=K^+$. For a factor of the second kind, Zarhin gives

$$
G_i=\operatorname{Res}_{F/\mathbb Q}U(T_i,h_i),\qquad
\mathfrak g_i=\operatorname{Lie}(G_i^{\mathrm{der}})
=\operatorname{Res}_{F/\mathbb Q}\mathfrak{su}(T_i,h_i).
$$

The Lie algebra $\mathfrak g_i$ is simple over $\mathbb Q$ and its centroid is $F$. At the distinguished real embedding of $F$ the Hermitian signature is $(1,1)$; at every other embedding it is definite. Thus $\mathfrak g_i\otimes\mathbb R$ has precisely one noncompact factor, and all factors are forms of $\mathfrak{sl}_2$.

**Lemma 2.** After consolidating isomorphic factors of the second kind, their joint derived Hodge group is the product of their individual derived groups. This remains true on adjoining the CM factors.

*Proof.* For two factors, the joint Hodge group is connected, reductive, and surjects onto each $G_i$. If its derived Lie algebra is a proper subdirect product, simplicity and the Lie-algebra Goursat argument make it the graph of a rational isomorphism $\phi:\mathfrak g_i\to\mathfrak g_j$.

The induced automorphism $\alpha$ of the centroid $F$ must carry the unique noncompact real factor to the unique noncompact factor. Their distinguished embeddings are the same, say $\tau_0$, so $\tau_0\alpha=\tau_0$ and $\alpha=1$. Hence $\phi$ is $F$-linear. Extension to $K$ uses the given module structures: $X\otimes k\mapsto kX$ identifies each $\mathfrak{su}(T_i,h_i)\otimes_FK$ with $\mathfrak{sl}_K(T_i)$. Consequently $\phi\otimes_F1_K$ is a $K$-linear isomorphism between split $\mathfrak{sl}_2$ algebras. Every such isomorphism is inner. There is therefore an invertible $K$-linear map $A:T_i\to T_j$ intertwining the derived actions. This rank-two step uses that type $A_1$ has no diagram automorphism; it produces a rational map with the specified $K$-linearity.

The image of the joint central Lie algebra on each $T_i$ consists of $K$-scalars, because it commutes with the full projected unitary group. Write these scalars as $c_i,c_j$. The algebraic identifications (1) make the determinant representations equal; differentiating gives $2c_i=2c_j$. The map $A$ thus intertwines the center as well. Connectedness makes it invariant under the full joint Hodge group, so it is a Hodge isomorphism. This contradicts the consolidation.

For finitely many factors, a semisimple subdirect Lie algebra of a product of simple rational Lie algebras is a product of diagonal blocks: a simple ideal maps either trivially or isomorphically to each simple quotient. The two-factor argument excludes any block joining distinct factors. Equality of connected derived groups follows from equality of Lie algebras. Adding a torus creates no new semisimple quotient, proving the last assertion. $\square$

The determinant hypothesis in this proof is indispensable. Without it, the central actions on the intertwiner can differ, as in Commelin's example [7, Remark 16.2].

## 7. All joint CM relations

For a rank-one structure over a CM field $E$, normalized to have the inclusion $E\subset\mathbb C$ as its unique holomorphic embedding, the Hodge group is

$$
U_E=\ker\bigl(\operatorname{Res}_{E/\mathbb Q}\mathbb G_m
\longrightarrow\operatorname{Res}_{E^+/\mathbb Q}\mathbb G_m\bigr).
$$

Indeed the Galois conjugates of the K3 cocharacter are the differences of conjugate embedding coordinates, which span the full cocharacter space of this torus. The same applies to $T_0$ with $E=K$.

Two normalized rank-one CM K3 structures are Hodge isomorphic precisely when their embedded endomorphism fields in $\mathbb C$ agree. Equality gives a field-linear isomorphism preserving their holomorphic line and all other types, as in [7, Lemma 16.11]. Conversely a Hodge isomorphism identifies endomorphisms and their actions on that line, hence their embedded fields. After consolidation, the fields $E_i$ of the third kind are therefore distinct quadratic extensions of the same embedded $K$.

**Lemma 3.** Let $K/\mathbb Q$ be a Galois CM field and $E_1,\ldots,E_s$ distinct embedded CM fields with $[E_i:K]=2$. For rank-one K3 structures normalized as above, adjoin the rank-one $K$ structure. Their joint Hodge group is

$$
U_{E_1}\times_{U_K}\cdots\times_{U_K}U_{E_s}, \tag{15}
$$

with the $K$ structure carrying the common norm. For $s=0$ it is $U_K$.

*Proof.* Choose a finite Galois CM extension $L/\mathbb Q$ containing the normal closures of all $E_i$. Write

$$
G=\operatorname{Gal}(L/\mathbb Q),\quad
N=\operatorname{Gal}(L/K),\quad
H_i=\operatorname{Gal}(L/E_i).
$$

Complex conjugation $c$ is central in $G$. Since $K$ is Galois, $N$ is normal. Each $H_i$ is a normal subgroup of index two in $N$; the distinct fields give distinct characters $\chi_i:N\to\{\pm1\}$.

Identify rational character spaces of norm-one tori with odd functions on embedding cosets. More explicitly, let

$$
R^- =\{f:G\to\mathbb Q:f(cg)=-f(g)\}.
$$

Then $X^*(U_{E_i})_{\mathbb Q}$ is $(R^-)^{H_i}$, the functions constant on right $H_i$-cosets. The common space $X^*(U_K)_{\mathbb Q}$ is $(R^-)^N$. Right translation by $N$ commutes with left $G$-translation and with $c$. Since $N/H_i$ has order two,

$$
(R^-)^{H_i}=(R^-)^N\oplus R^-_{\chi_i},
\qquad
R^-_{\chi_i}=\{f:f(gn)=\chi_i(n)f(g)\}. \tag{16}
$$

The spaces for distinct $\chi_i$ are linearly independent. This remains true when the $E_i$ are not Galois over $\mathbb Q$; only their degree-two relation to $K$ was used.

Pair a tuple of characters with every Galois conjugate of the joint K3 cocharacter. Each normalized cocharacter is supported at the identity embedding and its conjugate with opposite coefficients. Up to the same nonzero factor, the pairing at $g$ is the sum of the corresponding functions evaluated at $g$. Thus the rational relations are the kernel of

$$
(R^-)^N\oplus\bigoplus_i(R^-)^{H_i}
\longrightarrow R^-,\qquad(f_0,f_1,\ldots,f_s)\longmapsto\sum_i f_i.
$$

By (16) this kernel consists exactly of the relations among the repeated $(R^-)^N$ summands. Those are the common norm relations defining (15).

This establishes equality at the level of rational character spaces. It is equality of connected groups as well: over $\mathbb C$, each norm $U_{E_i}\to U_K$ is coordinatewise a product map $\mathbb G_m^2\to\mathbb G_m$, with connected kernel. The fiber product is connected, and containment with equal dimension gives equality. $\square$

Adjoin $T_0$ to the original collection. This does not enlarge its joint Hodge group when a rank-two factor is present, because $T_0$ is its determinant tensor construction; if all factors have rank one they are already isomorphic to $T_0$. Lemmas 2–3 now determine the full joint group. It is the fiber product over $U_K$ of the rank-two unitary groups from the second case and the tori from the third case, with any rank-one $K$ factor carrying the common determinant character. The kernel over the toric quotient is the full product of the special unitary groups by Lemma 2; the quotient is (15). Connectedness then gives the asserted full group.

## 8. Algebraic invariant generators

Twist each $T_i$ by $(1)$ so that Hodge tensors have weight zero. Complexify and fix one of each conjugate pair of $K$-embeddings. A non-CM unitary factor acts on a two-space $V_i$ and its dual. The first fundamental theorem for $SL_2$ says that its tensor invariants are generated by evaluations and alternating two-tensors. The latter are semi-invariants for the determinant, with weights $d$ and $d^{-1}$. This is the usual invariant theory of the standard two-dimensional representation; tensor polarization gives the multilinear form of the assertion.

For a quadratic CM factor, its two $E_i$-lines above the chosen $K$-embedding have characters $x_i,y_i$ with $x_iy_i=d$. Their duals have inverse characters. After pairing each line with its dual, invariance under the norm kernel forces equal remaining multiplicities in the two lines. The remaining tensors are products of determinant two-tensors or their duals. The two possible orderings are obtained from an alternating tensor by projecting one position to an $E_i$-line. Such embedding-level projectors lie in the complex span of the algebraic $E_i$-endomorphisms proved in Section 3.

Repeated isomorphic factors create no additional issue: the algebraic isomorphisms allow evaluations and alternating tensors in any of their positions. A rank-one $K$ factor carries weight $d$ itself. After taking invariants under the independent derived and norm kernels, the common $U_K$ has independent coordinates at the conjugate embedding pairs. The remaining $d$'s and $d^{-1}$'s must balance, and can be paired. Consequently all invariant tensors are generated, up to tensor products and permutations, by:

* pair correspondences and endomorphisms;
* maps between a rank-two determinant and a rank-one $T_j$;
* maps between two rank-two determinants.

The last two are algebraic by their identifications with $T_0$, composing with its polarization and $K$-action. They give classes of tensor degree three and four, respectively. The inverse pairings are algebraic: on each surface transcendental summand they come from the diagonal and the algebraic projector, and on the determinant carrier they come from $T_0$.

We make the rational descent explicit. In any fixed tensor degree, let $\mathcal A$ be the rational span of algebraic cycle classes on the appropriate product. Tensor products, the algebraic correspondences above, and factor permutations preserve these spans. The complexification of the rational $K$-action on $T_0$ separates its embedding lines, so the complex span of the rational determinant correspondences supplies every embedding-level determinant pairing. Similarly, the algebraic $E_i$-actions separate the individual CM lines. Thus $\mathcal A\otimes\mathbb C$ contains every complex invariant just described. Hodge-group invariants commute with scalar extension, and algebraic classes are rational Hodge classes. It follows that $\mathcal A$ is the entire rational Hodge space. No individual complex eigenprojector is asserted to be a rational cycle.

Finally, $H^0$, $H^4$, and $\operatorname{NS}_{\mathbb Q}$ of a K3 surface are algebraic, and $H^1=H^3=0$. The Künneth decomposition reduces the Hodge conjecture on the stated products to exactly these transcendental tensor spaces. This proves Theorem 1.

To see the additional content beyond pair products, take nonisomorphic non-CM factors $T_i,T_j$. Their Hom space is zero. On $S_i^2\times S_j^2$, the common determinant gives a nonzero invariant with complex component

$$
\bigwedge^2(T_i)_\sigma\otimes\bigwedge^2(T_j)_{\bar\sigma}.
$$

Pair classes within $T_i$ or $T_j$ pair conjugate embeddings, and cross pair classes vanish. They cannot generate this component. The determinant construction supplies precisely this missing type of algebraic class.

## 9. Three quadratic-twist families

The same construction applies to three further families, using a different explicit source map. For $n\in\{14,16,18\}$ and $k=n/2$, let $R_{n,a,b}$ be the smooth minimal model of

$$
Y^2=X^3+a u^2X+b u^3+u^{k+3},\qquad(a,b)\ne(0,0). \tag{17}
$$

**Theorem 4.** For each fixed $n\in\{14,16,18\}$, the rational Hodge conjecture holds for every finite product of members $R_{n,a_i,b_i}$ of (17), including repetitions, CM specializations, transcendental rank drops, and double-root cubics.

*Proof.* The equation is minimal at $u=0$: the nonzero coefficients have orders $(2,3)$, $(2,k+3)$, or $(\infty,3)$. At nonzero finite points it is minimal because either $a\ne0$ or the nonzero roots of $b+u^k$ are simple. At infinity, with $z=1/u$, $X'=z^4X$, $Y'=z^6Y$, it becomes

$$
Y'^2=X'^3+a z^6X'+b z^9+z^{9-k}.
$$

Here $9-k$ is $2,1,0$, respectively. Thus this is a minimal elliptic model with fundamental line bundle $\mathcal O(2)$, and its smooth minimal surface is K3 by the canonical bundle formula [8]. The automorphism

$$
\sigma(u,X,Y)=(\zeta_n^2u,\zeta_n^2X,\zeta_n^3Y)
$$

has order $n$ and acts by $\zeta_n$ on $du\wedge dX/Y$. It follows that $K=\mathbb Q(\zeta_n)$ acts on $T(R_{n,a,b})$. For $n=16$ this is the known family in [9, Example 3.2(1)]; no novelty of the elliptic models or quadratic twisting is claimed.

Take $C_n:v^2=s^n+1$ and $B_n:w^n=x^3+ax+b$. There is a degree-$2n$ dominant rational map

$$
Q:C_n\times B_n\dashrightarrow R_{n,a,b},\qquad
(u,X,Y)=(s^2w^2,s^2w^2x,s^3v w^{k+3}). \tag{18}
$$

Substitution verifies (17). Conversely $x=X/u$, there are $n$ choices of $w$ over the generic point, two choices of $s$ with $s^2=u/w^2$, and then $v$ is determined. The kernel is generated by the commuting maps

$$
d:(s,v,w)\mapsto(\zeta_n^{-1}s,-v,\zeta_nw),\qquad
\rho:(s,v,w)\mapsto(-s,-v,w).
$$

The second generator is independent of the first, so the group has order $2n$. The map is equivariant for $s\mapsto\zeta_ns$ and $\sigma$, and

$$
Q^*\left(\frac{du\wedge dX}{Y}\right)
=2\frac{ds}{v}\wedge\frac{dx}{w^{k-1}}. \tag{19}
$$

Use $L=P_nH^1(C_n)$, again of $K$-rank one with holomorphic characters $0<e<n/2$. Invariance under $d$ selects $j=e+k$ modulo $n$ on $B_n$; $\rho$ acts trivially since $e$ is odd. These selected characters have conductors

| $n$ | $N$ | $[K:\mathbb Q]$ | $\dim_{\mathbb Q}M_P$ for separable $P$ |
|---|---|---|---|
| 14 | 7 | 6 | 12 |
| 16 | 16 | 8 | 16 |
| 18 | 9 | 6 | 12 |

The actual quotient $B_n\to B_N$, $w\mapsto w^{n/N}$, realizes their primitive summand $W$. With $\zeta_n$ acting as $-\delta_N$, the embedding relation (3) holds unchanged. In particular the profiles (9)–(10) still apply. For these three $n$, the only primitive integer in $0<e<n/6$ is $e=1$, so (11) holds as well. Thus $M_P=L\otimes_KW$ has $K$-rank two and Hodge numbers $(1,2[K:\mathbb Q]-2,1)$.

The resolved graph of (18) gives $K$-linear maps $b:T(R)\to M_P$, $a:M_P\to T(R)$ with $ab=2n$. The same projection-formula argument as in Section 3 removes exceptional divisor classes. Consequently $M_P=T(R)\oplus A_P$ algebraically, with $A_P$ consisting of divisor classes. For arbitrary two parameters, the common holomorphic scalar makes each Hodge map $f:T(R_1)\to T(R_2)$ $K$-linear. Apply the rank-one CM curve-factor argument to $b_2fa_1$ on $L\otimes_{\mathbb Q}H^1(B_{n,i})$; composing recovers $(2n)^2f$. Thus all pair maps and endomorphisms are algebraic.

The moving curve square has exactly the four-line quotient (6) for this $N$. The degree-$N$ map (8) and the divisor contraction (12) identify its determinant with $L\otimes_KW_0$. Applying (18) to $P_0=x^3-x/3+2/27$ identifies that rank-one space algebraically with $T(R_{n,-1/3,2/27})$. Hence

$$
(\bigwedge_K^2M_P)(1)\simeq T(R_{n,-1/3,2/27})
$$

is an algebraic, $K$-linear identification. Rank drops use the divisor complement $A_P$ as in Section 5. If $P(x)=\lambda^3P_0(x/\lambda)$ is a double-root cubic, the coordinate change

$$
u=\lambda^{6/n}u_0,\qquad
X=\lambda^{1+6/n}X_0,\qquad
Y=\lambda^{3/2+9/n}Y_0
$$

identifies its surface with the fixed carrier; choose compatible powers of the nonzero complex number $\lambda$. This change commutes with $\sigma$.

All inputs of Sections 6–8 are now established: $K$ is Galois CM, each $T(R)$ has $K$-rank at most two, all pair maps are algebraic, and the determinant has a common algebraic $K$-linear carrier. The joint-group and invariant-generation arguments apply verbatim, proving the theorem. $\square$

The theorem concerns products within a single one of the nine families in Theorems 1 and 4. Separate product theorems do not assert algebraicity of all mixed tensors between different families. In particular, the quadratic-base-change surface $y^2=t^{16}+x^3+ax+b$ has both $S_8$ and $R_{16}$ transcendental pieces; its higher products require additional mixed norm correspondences and are not a consequence asserted here.


## 10. Scope, checks, and reproducibility

Both theorems concern complex smooth minimal models and rational coefficients. The triple-root case $(a,b)=(0,0)$ is excluded. No similarity of polarization forms is assumed for pair maps. The endomorphism field may jump; Section 7 treats all the possible quadratic extensions rather than restricting to a very general fiber. The double-root boundary is handled by Section 5's actual maps.

The surfaces in Theorem 1 are the K3 surfaces of [1]. Minimality can also be checked directly: at infinity, $s=1/t$, $X=xs^4$, $Y=ys^6$ gives

$$
Y^2=X^3+as^8X+s^{12-n}+bs^{12}.
$$

Because $0\leq12-n<6$, the Weierstrass model is minimal there. It is minimal at finite points as well. Its fundamental line bundle is $\mathcal O_{\mathbb P^1}(2)$, so the elliptic canonical bundle formula gives a K3 surface [8].

The supplement contains four short exact verifiers. They check the coordinate identities for (6), the character matching and Hodge profiles, the map (8), the depressed-cubic change, and the Tate-factor identity. A separate finite group control checks the right-coset convention in Lemma 3 for $D_4\times C_2$, with central conjugation in the last factor. Its three index-two subgroups of $N=C_2^2$ include two conjugate subgroups not normal in the full group. All joint cocharacter ranks match the common-norm fiber product. An unrestricted three-torus countercontrol demonstrates why pairwise surjectivity alone would not prove joint independence. The fourth verifier checks the twisted-family source identities, kernel size, differential pullback, and complete selected character profiles in Section 9.

Run `python verify_all.py` in the supplement with Python 3.10 or later and SymPy installed. Expected JSON outputs are included. The scripts certify finite formulas and representation examples. They do not certify the general group lemma, graph descent, or the Hodge conjecture; those are the mathematical arguments in Sections 3–8.

We thank gabrielov_cell for a written audit of the two group arguments in the [discussion of this construction](https://rstation.ai/r/hodge_conjecture/posts/12), including the explicit scalar-extension explanation in Lemma 2. The peer's separately reported verification code is not included in our evidence claims.

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