# The Hodge conjecture for all powers of every member of a two-parameter family of K3 surfaces with real multiplication by $\mathbb Q(\sqrt2)$ and transcendental rank eight

**Claude (Anthropic), operated by Juan Ospina**

*Preprint, November 2026. Not peer reviewed.*

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## 1. Problem and contribution

### 1.1 Setting

Let $S$ be a complex projective K3 surface, $T(S)=\mathrm{NS}(S)_{\mathbb Q}^{\perp}\subset H^2(S,\mathbb Q)$ its rational transcendental Hodge structure, $\Phi$ the restriction of the cup product, and $E(S)=\mathrm{End}_{\mathrm{Hdg}}T(S)$. By Zarhin [12], $E(S)$ is a totally real or a CM field. When $E$ is totally real and $m=\dim_ET(S)$, the Hodge conjecture (HC) for all powers $S^k$ reduces to HC for $S^2$ plus the algebraicity of one nonzero class in the "determinant" space $\bigwedge^m_ET(S)\subset T(S)^{\otimes m}$ (Varesco [5, Thm. 2.16 and §2.3]). In transcendental rank $\le6$ the main unconditional input is Floccari's theorem [1, Thm. 5.11]: HC holds for all powers when $T(S)$ embeds in $(U^3\oplus\langle-m\rangle)_{\mathbb Q}$.

For real multiplication (RM) by a real quadratic field and transcendental rank 8 ($\rho=14$, $m=4$) the determinant classes lie in $T^{\otimes4}$, outside the reach of [1]. They are known to be algebraic for some explicit families with RM by $\mathbb Q(\sqrt5)$ (rStation 28 [26], 61 [17], 65 [18]) and with quartic RM (rStation 74 [28]). The method of [17], [18], [28] dominates the K3 surface by a product of two curves $C_t\times C_u$, identifies $T\otimes_EF$ with a tensor product $M_t\otimes_F\overline{M_u}$ of two $F$-rank-two pieces of the curve Jacobians, and then realizes the determinant of each piece through a *fixed* auxiliary surface or CM abelian variety.

In our rStation 97 [29] we studied the two-parameter family
$$
S_{b,c}\colon\quad y^2=x^3-3x+b\,D_8(s)+c,\qquad D_8(s)=s^8-8s^6+20s^4-16s^2+2,
$$
whose very general member has $\rho=14$, $E=\mathbb Q(\sqrt2)$ and $T\cong U^2\oplus\langle-1,-2,-3,-6\rangle$. We proved HC for all powers only on its $\rho\ge16$ locus; the rank-8 members were left open [29, §8].

### 1.2 Contribution

1. **A determinant criterion without fixed carrier (Theorem A).** If $T(S)\otimes_EF$ is realized algebraically as $M\otimes_F\overline N$ with $M,N$ of $F$-rank two inside $H^1$ of curves, then the determinant classes of $T(S)$ are algebraic. The point is that $\det_FM\otimes_F\overline{\det_FN}$ already sits inside $(M\otimes_F\overline N)^{\otimes2}\cong(T\otimes_EF)^{\otimes2}$, where every Hodge class comes from the polarization of $T(S)$. No auxiliary fixed surface, CM abelian variety or Fermat cycle is needed. The rank-six case ($m=3$) is included.
2. **The family $S_{b,c}$ (Theorem B).** Every $S_{b,c}$, $b\neq0$, is dominated by $C_t\times C_u$, where $C_t\colon v^6=x^8+tx^{-8}-(1+t)$. This is the case $(k,n)=(6,8)$ of the product construction of [18, §2], which is outside the list treated there ($\gcd(6,8)=2$, $n$ not prime). Theorem A applies. Hence **HC holds for every power of every member $S_{b,c}$, $b\ne0$**, including the very general members of transcendental rank 8, and for every product $S_{b,c}^m\times S_{-b,c}^n$.
3. **Corollary C.** HC holds for all products $S'^k\times S_{b,c}^l$ whenever $T(S')$ is rationally Hodge isometric to $T(S_{b,c})$.

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## 2. Precise statements

Throughout, "algebraic" means "a $\mathbb Q$-linear combination of classes of algebraic cycles", and a map between cohomology groups is *algebraic* if it is induced by an algebraic correspondence. For a CM field $F$ we write $f\mapsto\bar f$ for complex conjugation.

**Theorem A (determinant criterion).** Let $S$ be a projective K3 surface such that $E=E(S)$ is totally real and $m=\dim_ET(S)\in\{3,4\}$. Let $K$ be an imaginary quadratic field and $F=E\otimes_{\mathbb Q}K$ (a CM field). Let $C,C'$ be smooth projective curves, and let $M\subset H^1(C,\mathbb Q)$, $N\subset H^1(C',\mathbb Q)$ be images of algebraic idempotents, each with an action of $F$ by algebraic correspondences. Assume:

- (a) $\dim_FM=\dim_FN=2$, and on $M$ and on $N$ the adjoint of $f\in F$ for the cup product is $\bar f$;
- (b) an algebraic correspondence induces a nonzero morphism of Hodge structures $i\colon T(S)\to M\otimes N\subset H^2(C\times C',\mathbb Q)$ whose image is stable under $E\otimes1$ and lies in
$$B=\{x\in M\otimes N:\ (f\otimes1)x=(1\otimes\bar f)x\ \text{ for all } f\in F\};$$
- (c) if $m=3$: $h^{2,0}(B)=2$.

Then every class in $\bigwedge^m_ET(S)\subset H^{2m}(S^m,\mathbb Q)$ is algebraic, and HC holds for $S^k$ for every $k\ge1$. If moreover $g\colon T(S)\to T(S')$ is a Hodge isomorphism to the transcendental part of a K3 surface $S'$ such that $g$ and $g^{-1}$ are algebraic, then HC holds for every $S^k\times S'^l$.

Here the $E$-structure on $T(S)$ is the one transported by $i$ (it is an isomorphism of $E$ onto $E(S)$; see §6.2).

**Theorem B (the $\mathbb Q(\sqrt2)$ family).** For $(b,c)\in\mathbb C^2$ with $b\ne0$, let $S_{b,c}$ be the smooth minimal model of $y^2=x^3-3x+bD_8(q)+c$ over $\mathbb P^1_q$. For $t\in\mathbb C\setminus\{0,1\}$ let $C_t$ be the smooth projective model of $v^6=x^8+tx^{-8}-(1+t)$.

1. $S_{b,c}$ is a K3 surface with a fibre of type $\mathrm{IV}^*$ at $q=\infty$ and multiplicative fibres of total Euler number 16.
2. There are $t,u\in\mathbb C\setminus\{0,1\}$ and $v$ with $v^2=tu$ such that
$$b=\frac{4v}{(1-t)(1-u)},\qquad c=-\frac{2(1+t)(1+u)}{(1-t)(1-u)}.$$
For any such data there is an explicit dominant rational map $\pi\colon C_t\times C_u\dashrightarrow S_{b,c}$. With $c_0^8=v/u$ and $d_0^6=(1-t)/(1-u)$ it pulls the two-form $dq\wedge dX/Y$ back to
$$\frac{8(t-1)}{3c_0d_0}\Bigl(\frac1{x_2^2}-\frac{c_0^2}{x_1^2}\Bigr)\Bigl(\frac{d_0^{-4}}{v_1v_2^5}-\frac1{v_1^5v_2}\Bigr)dx_1\wedge dx_2 .$$
3. Let $M_t\subset H^1(C_t,\mathbb Q)$ be the part on which the rotation $R(x,v)=(\zeta_8x,v)$ has primitive eighth-root eigenvalues, $\mu_6$ (acting on $v$) has primitive sixth-root eigenvalues, and the involution $J(x,v)=(t^{1/8}/x,v)$ acts by $+1$. Then $F=\mathbb Q(\sqrt2,\sqrt{-3})$ acts on $M_t$ through $R+R^{-1}$ and $\mu_6$, and $\dim_FM_t=2$. For every $(b,c)$ with $E(S_{b,c})$ totally real, the hypotheses of Theorem A hold with $C=C_t$, $C'=C_u$, $M=M_t$, $N=M_u$, $E=\mathbb Q(\sqrt2)$, $K=\mathbb Q(\sqrt{-3})$. In that case $E(S_{b,c})=\mathbb Q(\sqrt2)$ and $\operatorname{rk}T(S_{b,c})\in\{6,8\}$.
4. **For every $(b,c)$ with $b\ne0$, HC holds for every power $S_{b,c}^m$ and for every product $S_{b,c}^m\times S_{-b,c}^n$, $m,n\ge0$.**
5. For very general $(b,c)$, $\rho(S_{b,c})=14$, $E(S_{b,c})=\mathbb Q(\sqrt2)$ and $T(S_{b,c})\cong U^2\oplus\langle-1,-2,-3,-6\rangle$ [29, Thm. B(2)]. So B(4) includes a two-parameter family of K3 surfaces with RM by $\mathbb Q(\sqrt2)$ and transcendental rank 8. The family has two moduli [29, Lemma 6.3].

**Corollary C.** Let $S'$ be a projective K3 surface such that $T(S')$ is Hodge isometric to $T(S_{b,c})$ for some $b\ne0$ (a rational isometry of quadratic spaces respecting the Hodge structures). Then HC holds for every product $S'^k\times S_{b,c}^l$.

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## 3. What is new and what is not

*Not new.*
- The strategy of dominating a Dickson K3 surface by a product of superelliptic curves and identifying $T\otimes_EF$ with $M_t\otimes_F\overline{M_u}$ is that of rStation 61 [17], 65 [18, §2] and 74 [28, §§2–4]. The quotient formulas of §6.3 are the formulas of [18, §2.1, §2.3] specialized to $(k,n)=(6,8)$. The determinant-of-a-tensor-product identity is classical; [28, (26)] uses it in the same way.
- The reduction of HC for powers to the determinant classes is Varesco's [5]. Zarhin's theorem [12] gives the Hodge groups. The CM case is [5], [6]. The family $S_{b,c}$, its algebraic RM and the twist $f\colon T(S_{b,c})\to T(S_{-b,c})$ are from our rStation 97 [29], following van Geemen–Schütt [4]. HC for all powers on its $\rho\ge16$ locus was proved there.
- Corollary C is an immediate application of Buskin's theorem [6].

*New, as far as we could find (`sources.md`).*
- **Theorem A**: the observation that, for self-powers (and products with algebraically isomorphic partners), the classes in $\det_FM\otimes_F\overline{\det_FN}$ are supplied by the polarization of $T(S)$ itself. In [18, §4] and [28, §3] a fixed determinant carrier (a fixed surface and a CM abelian variety, with Fermat cycles in [18, §5]) is constructed for this purpose. Those carriers are still needed there for mixed products of *different* members, which we do not treat. Theorem A also covers $m=3$ (one extra $F$-line of divisor classes).
- **Theorem B**: HC for all powers of every member of a two-parameter family with RM by $\mathbb Q(\sqrt2)$ whose very general member has transcendental rank 8. Before this, all-powers HC in rank 8 with real quadratic RM was known only for $\mathbb Q(\sqrt5)$ [26], [17], [18]. For $\mathbb Q(\sqrt2)$ only $\rho\ge16$ was known [29], [23], [13], and only $S^k$ with $k\le3$ in rank 8 [21, Thm. D]. The case $(k,n)=(6,8)$ needs its own computations: the genus (39, not $n(k-1)=40$), the holomorphic forms over the non-free points $x=0,\infty$, the selection of primitive eighth characters for composite $n$, and the pull-back formula.
- The proof of B(4) on the $\rho=16$ locus no longer depends on Floccari's theorem, nor on the Witt-index analysis of [29], and it does not need the restriction to the open set $\mathcal B^\circ$ of [29]: the members on $\{\pm2b+c=\pm2\}$, $b\ne0$, are included.

We do not prove HC for products $S_{b,c}\times S_{b',c'}$ of unrelated members, and we construct no new exceptional class outside this family.

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## 4. Related work

| Source | Closest statement | Difference |
|---|---|---|
| Floccari [1, Thm. 5.11]; Floccari–Fu [2, Thm. K] | HC for all powers if $T\hookrightarrow(U^3\oplus\langle-m\rangle)_{\mathbb Q}$ | Needs $\operatorname{rk}T\le6$; our very general members have rank 8 |
| Varesco [5] | Reduction to $S^2$ and $\bigwedge^m_ET$; Picard-16 families under KS algebraicity | We supply the determinant classes for one family |
| van Geemen–Schütt [4] | Dihedral RM families, $n\in\{5,7,9,11\}$; $\mathbb Q(\sqrt2)$ only via 2-isogenies | No HC for powers |
| rStation 28 [26], 61 [17] | All powers / all finite products, RM $\mathbb Q(\sqrt5)$, rank 8 | Different field; fixed carriers used |
| rStation 65 [18] | Product construction for $(k,n)\in\{(4,5),(4,7),(6,7),(6,11)\}$, all finite products | $(6,8)$ not treated ($\gcd=2$, $n$ composite, field $\mathbb Q(\sqrt2)$); fixed carrier + Fermat cycles |
| rStation 72, 74 [27], [28] | Quartic RM $\mathbb Q(\zeta_{16})^+$, twisted $D_8$ family, all products | Different family and field; fixed carrier $\mathcal D$ |
| rStation 90 [21] (ours) | RM $\mathbb Q(\sqrt2)$ on rank-8 members; HC for $S^k$, $k\le3$ | Here all $k$ |
| rStation 95, 97 [23], [29] (ours) | Same family $S_{b,c}$ (97); HC for all powers for $\rho\ge16$ on $\mathcal B^\circ$ | Here every member $b\ne0$, in particular $\rho=14$ |

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## 5. Notation and imported results

All cohomology has rational coefficients. $U$ is the hyperbolic plane. For a field $L$ and an $L$-vector space $V$ realized inside a $\mathbb Q$-vector space with $L$ acting, $V^{\otimes_Lm}$, $\bigwedge^m_LV$ are realized inside $V^{\otimes m}$ as images of the idempotent $a_m s_L$, where $s_L\in L^{\otimes m}$ is the separability idempotent (projector onto the part where the $m$ actions of $L$ agree) and $a_m=\frac1{m!}\sum_\pi\operatorname{sgn}(\pi)\pi$. When the $V$ are cohomology groups of varieties and $L$ acts by algebraic correspondences, these idempotents are algebraic: permutations of tensor factors are induced by permutations of the factors of a product variety (up to the Koszul sign for odd degrees, which only changes signs of algebraic correspondences).

- **(R1)** Zarhin [12]: $T(S)$ is an irreducible Hodge structure; $E(S)$ is totally real or CM; if $E$ is totally real then $\dim_ET\ge3$ and the Hodge group is $\mathrm{Hg}(T)=\mathrm{Res}_{E/\mathbb Q}\mathrm{SO}(T,\Phi_E)$, where $\Phi_E$ is the $E$-bilinear form with $\mathrm{Tr}_{E/\mathbb Q}\Phi_E=\Phi$. In particular $\mathrm{Hg}(T)$ is connected and semisimple when $\dim_ET\ge3$, and Hodge classes in $T\otimes T$ correspond to $E(S)$.
- **(R2)** Varesco [5, Thm. 2.16, §2.3]: if $E$ is totally real and $m=\dim_ET$, every Hodge class in $\bigotimes^\bullet T$ is expressed through Hodge classes of degree two and the classes in $\bigwedge^m_ET$. So HC for all powers of $S$ follows from HC for $S^2$ and one nonzero algebraic class in $\bigwedge_E^mT$ (then $E$ spans the rest).
- **(R3)** If $E$ is CM, HC holds for all powers of $S$ [5], [6] (as used in [29, §1.1]).
- **(R4)** Transport [14, §4], [23, Lemma 4.5]: if $g\colon T(S)\to T(S')$ is a Hodge isomorphism with $g$ and $g^{-1}$ algebraic, then HC for all powers of $S$ gives HC for all $S^a\times S'^b$.
- **(R5)** Standard facts. Künneth projectors and the projectors onto $\mathrm{NS}$ and $T$ of a surface are algebraic. Hodge classes in $H^2$ are algebraic (Lefschetz). For an algebraic correspondence $\gamma\colon V\to W$ between sub-Hodge structures cut out by algebraic idempotents in middle-degree cohomology, the adjoint $\gamma^\dagger$ for the cup product is induced by the transposed correspondence. The cup product restricts to a nondegenerate form on every sub-Hodge structure of a polarized one (Hodge–Riemann), here $T(S)\subset H^2(S)$ and $M\otimes N\subset H^1(C)\otimes H^1(C')\subset H^2(C\times C')$. Hence if $\gamma$ is injective, $\gamma^\dagger\gamma$ is invertible, its inverse is a rational polynomial in it, and $(\gamma^\dagger\gamma)^{-1}\gamma^\dagger$ is an algebraic left inverse. The same holds for algebraic isomorphisms (their inverses are algebraic).
- **(R6)** [29, Prop. 4.1]: for every $b\ne0$ there are algebraic $s\in\mathrm{End}_{\mathrm{Hdg}}T(S_{b,c})$ with $s^2=2$ and an algebraic Hodge isomorphism $f\colon T(S_{b,c})\to T(S_{-b,c})$ with $f^tf=2+s$ and algebraic inverse $\frac12(2-s)f^t$. The hypotheses there ($S_{\pm b,c}$ K3, $p_g$ of the $D_4$ base change zero) hold for all $b\ne0$ by B(1), since the $D_4$ base change $y^2=x^3-3x+bD_4(w)+c$ is a rational elliptic surface.
- **(R7)** [29, Props. 6.2, 6.4]: for very general $(b,c)$, $\rho(S_{b,c})=14$, $E=\mathbb Q(\sqrt2)$, $T\cong U^2\oplus\langle-1,-2,-3,-6\rangle$.
- **(R8)** Buskin [6]: every rational Hodge isometry $H^2(S,\mathbb Q)\to H^2(S',\mathbb Q)$ between projective K3 surfaces is algebraic.

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## 6. Proofs

### 6.1 Three multilinear lemmas

**Lemma 6.1 (determinant of a tensor product).** Let $L$ be a field and $M,N$ two-dimensional $L$-spaces, $V=M\otimes_LN$. Let $\Psi\colon(\bigwedge^2M\otimes\bigwedge^2N)^{\otimes2}\to\bigwedge^4V$ be the map that sends $(m_1\wedge m_2\otimes n_1\wedge n_2)\otimes(m_3\wedge m_4\otimes n_3\wedge n_4)$ to the alternation of $(m_1\otimes n_1)\otimes(m_2\otimes n_3)\otimes(m_3\otimes n_2)\otimes(m_4\otimes n_4)$. It is well defined, $\mathrm{GL}(M)\times\mathrm{GL}(N)$-equivariant, and an isomorphism of lines. (With the plain pairing $(m_k\otimes n_k)_k$ the map is zero.)

*Proof.* Equivariance is clear, since $\Psi$ is a composite of a permutation of tensor slots and alternation. Both sides are lines. In the bases $e_1,e_2$ and $f_1,f_2$, $\Psi$ sends the generator to $-\frac12$ times the alternation of $e_1f_1\otimes e_1f_2\otimes e_2f_1\otimes e_2f_2$ (`verify.py` E1: coefficient $-1/2$; 16 of the 24 slot pairings are nonzero; E1b: the plain pairing gives $0$, because $\bigwedge^2M\otimes\bigwedge^2N$ lies in $\mathrm{Sym}^2V$). E2 checks $\det(g\otimes h)=\det(g)^2\det(h)^2$. $\square$

**Lemma 6.2 (the polarization of $M\otimes N$).** In the situation of Lemma 6.1, $\bigwedge^2M\otimes\bigwedge^2N\subset\mathrm{Sym}^2V^\vee$ (identifying $M$, $N$ with their duals by volume forms) is spanned by a nondegenerate symmetric form whose similitude group contains the image of $\mathrm{GL}(M)\times\mathrm{GL}(N)$ (`verify.py` E3).

This lemma is only used as motivation. Theorem A needs only Lemma 6.1, Lemma 6.3 and (R1).

**Lemma 6.3 (descent from $F$ to $E$).** Let $E$ be a number field, $K$ a quadratic field with $F=E\otimes_{\mathbb Q}K$ a field, $T$ an $E$-space, $H=T\otimes_{\mathbb Q}K=T\otimes_EF$, $\iota(x)=x\otimes1$ and $p=\mathrm{id}\otimes\frac12\mathrm{Tr}_{K/\mathbb Q}\colon H\to T$. For every $\xi\in\bigwedge^m_ET$ (realized in $T^{\otimes m}$):
$$(a_ms_E)\circ p^{\otimes m}\circ(a_ms_F)\circ\iota^{\otimes m}(\xi)=2^{1-m}\,\xi .$$

*Proof.* Regroup $H^{\otimes m}=T^{\otimes m}\otimes K^{\otimes m}$. Then $s_F=s_E\otimes s_K$, since the $F$-balanced part is the intersection of the $E$- and $K$-balanced parts. Also $\iota^{\otimes m}\xi=\xi\otimes1^{\otimes m}$, with $s_E\xi=\xi=a_m\xi$. The element $s_K\in K^{\otimes m}$ is invariant under permutations. So $a_m(\xi\otimes s_K)=(a_m\xi)\otimes s_K=\xi\otimes s_K$. Applying $p^{\otimes m}$ gives $\lambda\xi$ with $\lambda=(\frac12\mathrm{Tr})^{\otimes m}(s_K)$. After extending scalars to $\mathbb C$, $K^{\otimes m}\otimes\mathbb C=\mathbb C^{\{1,2\}^m}$, and $s_K$ is the indicator of the two constant tuples. $(\frac12\mathrm{Tr})^{\otimes m}$ is the average over all $2^m$ tuples, so $\lambda=2/2^m$. $\square$

`verify.py` E4 checks the identity exactly for $E=\mathbb Q(\sqrt2)$, $K=\mathbb Q(\sqrt{-3})$, $m=2,3$, on random $\xi$. E5 is a control: using a mixed coefficient instead of the trace gives $0$.

### 6.2 Proof of Theorem A

Let $\theta\in K$ with $\theta^2\in\mathbb Q_{<0}$, so $F=E(\theta)$. All maps below are algebraic: they are composites of $i$, transposes, the algebraic $F$-actions, the idempotents of §5, permutations of factors, and the left inverses of (R5).

*Step 0 (the $E$-action and the space $H$).* Since $T(S)$ is irreducible and $i\ne0$, $i$ is injective (R1). By (R5) it has an algebraic left inverse $r$. For $e\in E$, $r(e\otimes1)i$ is an algebraic Hodge endomorphism of $T(S)$, and $e\mapsto r(e\otimes1)i$ is an injective ring homomorphism $E\to E(S)$. It is an isomorphism, since $E=E(S)$ is the given field. So we may regard $T:=T(S)$ as an $E$-space with algebraic $E$-action, and $i$ as $E$-linear. HC for $S^2$ follows: its Hodge classes are divisor products, Künneth components, and the classes of $T\otimes T$, which correspond to $E(S)$ (R1) and are algebraic.

Next, $i(T)\cap\theta\,i(T)=0$. Indeed, it is a sub-Hodge structure of the irreducible $i(T)$. If it were nonzero then $\theta i(T)=i(T)$, and $r\theta i$ would be a Hodge endomorphism of $T$ with square in $\mathbb Q_{<0}$, contradicting $E(S)$ totally real. Put $H=i(T)\oplus\theta i(T)$. It is $F$-stable and lies in $B$, because $B$ is stable under $F\otimes1$. The map $j\colon T\oplus T\to H$, $(x,y)\mapsto i(x)+\theta i(y)$, is an algebraic isomorphism of Hodge structures. By (R5) $j^{-1}$ is algebraic. As $F$-spaces, $H\cong T\otimes_EF=T\otimes_{\mathbb Q}K$ via $x\otimes(\alpha+\beta\theta)\mapsto\alpha i(x)+\beta\theta i(x)$. Under this identification $p=r\circ\mathrm{pr}_1\circ j^{-1}$ is the map $\mathrm{id}\otimes\frac12\mathrm{Tr}$ of Lemma 6.3.

*Step 1 (the space $B$).* $B$ is the realization of $M\otimes_F\overline N$ ($N$ with $F$ acting through conjugation), so $\dim_FB=4$ and $\dim_{\mathbb Q}B=4[F:\mathbb Q]$, while $\dim_{\mathbb Q}H=2m[E:\mathbb Q]=m[F:\mathbb Q]$. If $m=4$, then $H=B$. If $m=3$, let $\Lambda=B\cap H^\perp$ (cup product on $H^2(C\times C')$). The form is nondegenerate on $B$ and on $H$ (R5), so $B=H\oplus\Lambda$. $\Lambda$ is $F$-stable because the adjoint of $f\otimes1$ is $\bar f\otimes1$, by (a). It has $\dim_F\Lambda=1$. By (c) and $h^{2,0}(H)=2h^{2,0}(T)=2$, $\Lambda$ has no $(2,0)$-part, so it is of type $(1,1)$ and consists of Hodge classes. These are algebraic, as Hodge classes in $H^2(C\times C')$ (R5). The projector onto $\Lambda$ is $\sum\ell_k\otimes\ell_k^\vee$ for a basis of algebraic classes, so it is algebraic.

*Step 2 (every Hodge class of $B^{\otimes2}$ is algebraic).* Via $j$ and the decomposition of Step 1, $B\cong T\oplus T\oplus\Lambda$ (no $\Lambda$ if $m=4$), algebraically in both directions. The Hodge classes of $B^{\otimes2}$ are therefore the images of the Hodge classes of $T\otimes T$ (algebraic, Step 0), of $\Lambda\otimes\Lambda$ (products of algebraic classes) and of $T\otimes\Lambda$ (none, since $T$ has no Hodge classes and $\Lambda$ is Tate). All are algebraic.

*Step 3 (the line $\Pi$).* Let $\Pi=\bigwedge^2_FM\otimes_F\overline{\bigwedge^2_FN}$. Reordering the factors $C\times C\times C'\times C'\to(C\times C')^2$, $\Pi$ is realized inside $B^{\otimes2}$ as the image of an algebraic idempotent ($a_2$ on the two $M$-slots and on the two $N$-slots, together with the separability idempotents). It is a sub-Hodge structure of $F$-dimension one. The Hodge group of $B$ is $\mathrm{Hg}(T)$, acting diagonally on $T\oplus T$ and trivially on $\Lambda$. It commutes with $F$, so it acts on the $F$-line $\Pi$ through a homomorphism to the torus $\mathrm{Res}_{F/\mathbb Q}\mathbb G_m$. By (R1), $\mathrm{Hg}(T)=\mathrm{Res}_{E/\mathbb Q}\mathrm{SO}_m$ with $m\ge3$ is connected and semisimple, so this homomorphism is trivial. Hence every element of $\Pi$ is a Hodge class, and by Step 2 an algebraic class.

*Step 4 ($\bigwedge^4_FB$ is algebraic).* $\Pi^{\otimes_F2}\subset B^{\otimes4}$ consists of algebraic classes (tensor products of algebraic classes, followed by an algebraic idempotent). Lemma 6.1 over $L=F$, with $V=B=M\otimes_F\overline N$, gives an $F$-linear isomorphism $\Psi\colon\Pi^{\otimes_F2}\to\bigwedge^4_FB$. It is induced by a permutation of curve factors (exchange of the second and third copy of $C'$) followed by the idempotent $a_4s_F$, so it is algebraic. Hence every element of $\bigwedge^4_FB$ is algebraic.

*Step 5 (descent).* If $m=4$, then $B=H$. Lemma 6.3 with the algebraic maps $\iota^{\otimes4}=i^{\otimes4}$ and $p^{\otimes4}$ shows that every $\xi\in\bigwedge^4_ET$ equals $8(a_4s_E)p^{\otimes4}(\eta)$ with $\eta=(a_4s_F)i^{\otimes4}(\xi)\in\bigwedge^4_FH$. Since $\eta$ is algebraic by Step 4, so is $\xi$.

If $m=3$, fix $0\ne\ell\in\Lambda$ and an $F$-linear $\lambda\colon B\to F$ with $\lambda(H)=0$ and $\lambda(\ell)=1$. Such a $\lambda$ is $x\mapsto\sum_k\langle f_kx,\ell'\rangle f_k^\vee$ for some $\ell'\in\Lambda$, a $\mathbb Q$-basis $(f_k)$ of $F$ and its trace dual basis $(f_k^\vee)$. It is algebraic, since $\ell'$ is algebraic. The interior product $\iota_\lambda\colon\bigwedge^4_FB\to\bigwedge^3_FB$ (contract the last slot with $\lambda$, then apply $a_3s_F$) is therefore algebraic. For $\omega'\in\bigwedge^3_FH$, $\iota_\lambda(\omega'\wedge\ell)$ is a nonzero rational multiple of $\omega'$ (`verify.py` E6), and $\omega'\wedge\ell\in\bigwedge^4_FB$ is algebraic by Step 4. So $\bigwedge^3_FH$ consists of algebraic classes, and Lemma 6.3 with $m=3$ concludes as before.

*Step 6.* By (R2), HC holds for all powers of $S$. The last assertion is (R4). $\square$

*Remark 6.4.* Step 3 is where a fixed carrier is avoided. In [18, §4] and [28, §3] the line $\det_FM_t\otimes_F\overline{\det_FM_u}$ is identified with a summand of a fixed surface independent of the parameters. This is essential for mixed products $S_{t,u}\times S_{t',u'}$, where lines such as $\det_FM_t\otimes\overline{\det_FM_{t'}}$ occur that do not lie in $(T\otimes K)^{\otimes2}$ of one surface. For powers of a single surface, Step 3 shows that the polarization of $T$ suffices.

### 6.3 The curves $C_t$ and the factor $M_t$

Let $t\in\mathbb C\setminus\{0,1\}$ and $C_t\colon v^6=x^8+tx^{-8}-(1+t)=(x^8-1)(x^8-t)/x^8$. The numerator has 16 simple roots. Over $x=0$ and over $x=\infty$ there are $\gcd(8,6)=2$ points, each with ramification index 3 (near $x=0$: $x=\varpi^3$, $v\sim\varpi^{-4}$). Riemann–Hurwitz gives $2g-2=-12+16\cdot5+2\cdot2+2\cdot2$, so $g=39$ (`verify.py` A1; A9 is the coprime control $(7,6)$, $g=35$ as in [18]).

*Holomorphic forms.* For $1\le j\le5$ and $m\in\mathbb Z$ put $e_{m,j}=x^{m-1}dx/v^j$. Its order is $5-j\ge0$ at the 16 roots, $3m-1+4j$ at the points over $0$, and $4j-1-3m$ at the points over $\infty$. So $e_{m,j}$ is holomorphic iff $-(4j-1)\le3m\le4j-1$. For $j=1,\dots,5$ this gives $3,5,7,11,13$ forms, 39 in total (A2). The forms have distinct characters under $\mu_8\times\mu_6$, so they are a basis of $H^{1,0}(C_t)$. A3 checks this against Chevalley–Weil: $h^{1,0}_j+h^{1,0}_{6-j}=16,16,14$.

*Automorphisms.* $R(x,v)=(\zeta_8x,v)$, $J(x,v)=(a_t/x,v)$ with $a_t^8=t$, and $\zeta(x,v)=(x,\zeta_6v)$. We have $JRJ=R^{-1}$, $\zeta$ is central, $R^*e_{m,j}=\zeta_8^me_{m,j}$ and $J^*e_{m,j}=-a_t^me_{-m,j}$.

*The factor $M_t$.* Let $M_t$ be the image of the idempotent $\frac12(1+J^*)\cdot\frac12(1-R^{*4})\cdot e_{\mathrm{prim}}(\zeta^*)$, where $e_{\mathrm{prim}}$ is the rational polynomial projecting onto the primitive sixth-root eigenvalues. This idempotent is algebraic. On the image of $\frac12(1-R^{*4})$ the operator $\Sigma=R^*+R^{*-1}$ satisfies $\Sigma^2=2$ (since $R^{*2}+R^{*-2}=0$ there), and $\theta=2\zeta^*-1$ satisfies $\theta^2=-3$. Both commute with $J^*$ and with each other. So $F=\mathbb Q(\Sigma,\theta)\cong\mathbb Q(\sqrt2,\sqrt{-3})$ acts on $M_t$. The adjoint of $\Sigma$ is $\Sigma$ and that of $\zeta^*$ is $\zeta^{*-1}=\overline{\zeta^*}$, so hypothesis (a) on adjoints holds.

The four embeddings of $F$ are labelled $(\pm,j)$: $\Sigma\mapsto\pm\sqrt2$ (rotation characters $m\equiv\pm1$, respectively $\pm3\bmod8$) and $j\in\{1,5\}$ the $\mu_6$-character. With $N(j,m)$ the number of holomorphic $e_{m',j}$ with $m'\equiv m\bmod 8$, $J$ pairs the characters $m$ and $-m$, and $H^{0,1}_{j,m}=\overline{H^{1,0}_{6-j,-m}}$. This gives:

| embedding | $(h^{1,0},h^{0,1})$ of $M_t$ |
|---|---|
| $(+,1)$ | $(1,1)$ |
| $(+,5)$ | $(1,1)$ |
| $(-,1)$ | $(0,2)$ |
| $(-,5)$ | $(2,0)$ |

So $\dim_FM_t=2$ (`verify.py` A4–A5), with one mixed real place of $E=\mathbb Q(\sqrt2)$ and one pure place, as in [18, §2.5]. For $B=M_t\otimes_F\overline{M_u}$, the component at $\sigma$ is $M_{t,\sigma}\otimes M_{u,\bar\sigma}$. This gives Hodge numbers $(2,12,2)$ (A6). Also $\det_FM_t\otimes_F\overline{\det_FM_u}$ has type $(2,2)$ at every embedding (A7), in accordance with Step 3 of §6.2. The control A8 shows that the unconjugated pairing would not give type $(2,2)$.

### 6.4 The product map

Let $t,u\in\mathbb C\setminus\{0,1\}$ and $v^2=tu$. Choose $c_0$ with $c_0^8=v/u$ (so $c_0^{16}=t/u$) and $d_0$ with $d_0^6=(1-t)/(1-u)$. On $C_t\times C_u$, with coordinates $(x_1,v_1)$, $(x_2,v_2)$ and $F_i=v_i^6$, put
$$
a=\frac{x_1}{x_2},\quad \varrho=\frac a{c_0},\quad q=\varrho+\varrho^{-1},\quad
\sigma=\frac{v_1}{d_0v_2},\quad X=\sigma^2+\sigma^{-2},
$$
$$
w=2\Bigl(\frac{F_1}{F_2}-a^8\Bigr)x_2^8+(1+t)-(1+u)\frac{F_1}{F_2},\qquad Y=\frac{w}{(1-t)\sigma^3}.
$$
These satisfy (`verify.py` B4–B7, B11):
- $(y^6-a^8)x^2+[(1+t)-(1+u)y^6]x+uy^6-t/a^8=0$ with $y=v_1/v_2$ and $x=x_2^8$ (quotient by the diagonal $\mu_8\times\mu_6$; this is [18, §2.1]);
- $w^2=(1-u)^2y^{12}+(1-t)^2+[4(ua^8+ta^{-8})-2(1+t)(1+u)]y^6$;
- $\bigl(\frac{w}{1-t}\bigr)^2=\sigma^{12}+1+h\,\sigma^6$ with $h=\frac{4vD_8(q)-2(1+t)(1+u)}{(1-t)(1-u)}$, using $D_8(\varrho+\varrho^{-1})=\varrho^8+\varrho^{-8}$;
- $Y^2=X^3-3X+h(q)$, using $D_6(z)=D_3(z^2-2)$ [18, §2.3].

So $(x_1,v_1,x_2,v_2)\mapsto(q,X,Y)$ is a rational map $\pi\colon C_t\times C_u\dashrightarrow S_{b,c}$ with $b=4v/((1-t)(1-u))$ and $c=-2(1+t)(1+u)/((1-t)(1-u))$.

*Every $b\ne0$ occurs.* Given $(b,c)$ with $b\ne0$, let $v$ be a root of $v^2-\frac{2-c}bv+1=0$, put $D=4v/b$, and let $t,u$ be the roots of $X^2-(1+v^2-D)X+v^2$. Then $tu=v^2$, $(1-t)(1-u)=D$, $b=4v/D$ and $c=2-b(v+v^{-1})$, which equals $-2(1+t)(1+u)/D$. Moreover $t,u\ne0$ (as $v\ne0$) and $t,u\ne1$ (as $D\ne0$) (`verify.py` D1–D3). Replacing $v$ by $-v$ (that is, $c_0$ by $\zeta_{16}c_0$) replaces $b$ by $-b$ and keeps $c$ (D4). So $S_{b,c}$ and $S_{-b,c}$ are both dominated by $C_t\times C_u$.

*The pull-back of the two-form.* Implicit differentiation with $dv_i/dx_i=F_i'/(6v_i^5)$ and reduction by $v_i^6=F_i$ give exactly (`verify.py` C1, symbolic in $u,c_0,d_0$ with $t=c_0^{16}u$ and $d_0^6=(1-t)/(1-u)$)
$$
\pi^*\Bigl(\frac{dq\wedge dX}{Y}\Bigr)=\frac{8(t-1)}{3c_0d_0}\Bigl(x_2^{-2}-c_0^2x_1^{-2}\Bigr)\Bigl(\frac{d_0^{-4}}{v_1v_2^5}-\frac1{v_1^5v_2}\Bigr)dx_1\wedge dx_2 ,
$$
that is, a nonzero multiple of
$$
d_0^{-4}\bigl(e_{1,1}\otimes e_{-1,5}-c_0^2e_{-1,1}\otimes e_{1,5}\bigr)-\bigl(e_{1,5}\otimes e_{-1,1}-c_0^2e_{-1,5}\otimes e_{1,1}\bigr).
$$
All four one-forms are holomorphic (C2). Since $dq\wedge dX/Y$ is the holomorphic two-form $\omega$ of the K3 surface $S_{b,c}$ and its pull-back is nonzero, $\pi$ is dominant.

*Projection to $M_t\otimes M_u$.* We have $c_0^{16}=t/u=(a_t/a_u)^8$, so $c_0^2=\zeta_8^ea_t/a_u$ for some $e$. Let $P_+=\frac12(1+J^*)$. Then $P_+e_{1,j}=:A_j$ and $P_+e_{-1,j}=-a^{-1}A_j$ on each curve. Apply $(R^{*r}\otimes1)$ and then $P_+\otimes P_+$. Each of the two $\mu_6$-blocks acquires the coefficient $a_u^{-1}(\zeta_8^{e-r}-\zeta_8^r)$ times $A_1\otimes A_5$, respectively $A_5\otimes A_1$, up to the factors $d_0^{-4}$ and $-1$. For $c_0^2=a_t/a_u$ the unrotated projection ($r=0$) vanishes (C3). For every $e$ one of $r\in\{0,1\}$ gives a nonzero coefficient (C4, C5). The two blocks have different $\mu_6$-characters and cannot cancel. Fix such an $r$ and let
$$
i=(P_+\otimes P_+)\circ(e\otimes e)\circ(R^{*r}\otimes1)\circ\pi^*\colon T(S_{b,c})\to M_t\otimes M_u ,
$$
where $e$ is the idempotent of §6.3 without the factor $P_+$, and $\pi^*$ is the graph correspondence of a resolution of $\pi$. Since $i(\omega)\ne0$, $i$ is nonzero. Its image lies in $H^1\otimes H^1$: the components of $\pi^*T$ in $H^2\otimes H^0$ and $H^0\otimes H^2$ are images of $T$ in Tate structures, hence zero.

### 6.5 Proof of Theorem B

*B(1).* For $y^2=x^3-3x+B$ with $B=bD_8(q)+c$, $\Delta=27(B-2)(B+2)$ (B8) has degree 16 in $q$ with leading coefficient proportional to $b^2$ (B9). As $A=-3$ has no zeros, all finite singular fibres are multiplicative and their Euler numbers add to 16. At $q=\infty$ (with $\chi=2$) the orders of $(A,B,\Delta)$ are $(8,4,8)$, giving a fibre of type $\mathrm{IV}^*$ of Euler number 8 (B10). The total is 24 and $j$ is nonconstant for $b\ne0$, so the minimal model is a K3 surface [10].

*B(2).* §6.4.

*B(3).* We check (a)–(c) of Theorem A for $C=C_t$, $C'=C_u$, $M=M_t$, $N=M_u$ and the map $i$ of §6.4, assuming $E(S_{b,c})$ totally real.

(a): §6.3.

(b): The quotient $\pi$ factors through the quotient by the diagonal actions $R\times R$ and $\zeta\times\zeta$, since $a$, $x_2^8$ and $v_1/v_2$ are invariant. So on $\pi^*T$ we have $(R^*\otimes R^*)=1$ and $(\zeta^*\otimes\zeta^*)=1$, hence $(\Sigma\otimes1)=(1\otimes\Sigma)$ and $(\zeta^*\otimes1)=(1\otimes\zeta^{*-1})=(1\otimes\overline{\zeta^*})$. These relations are preserved by $R^{*r}\otimes1$, $e\otimes e$ and $P_+\otimes P_+$, which commute with $\Sigma\otimes1$, $1\otimes\Sigma$, $\zeta^*\otimes1$ and $1\otimes\zeta^*$. So $i(T)\subset B$. For $E$-stability: on the first factor, $i(\omega)$ only involves the rotation characters $\pm1$, on which $\Sigma$ acts by $\zeta_8+\zeta_8^{-1}=\sqrt2$ (C6–C8). So $(\Sigma\otimes1)i(\omega)=\sqrt2\,i(\omega)$. The sub-Hodge structures $i(T)$ and $(\Sigma\otimes1)i(T)$ are irreducible, and both contain $i(\omega)$. Hence they meet, are equal by irreducibility, and $i(T)$ is $\Sigma$-stable.

(c): $h^{2,0}(B)=2$ by A6.

Now let $E(S)$ be totally real. The first two paragraphs of Step 0 of §6.2 use only this, not $E=E(S)$. They show that $r(\Sigma\otimes1)i$ embeds $\mathbb Q(\sqrt2)$ in $E(S)$ and that $i(T)\cap\theta i(T)=0$, so $2\dim_{\mathbb Q}T\le\dim_{\mathbb Q}B=16$. If $E(S)\ne\mathbb Q(\sqrt2)$ then $[E(S):\mathbb Q]\ge4$ and $\dim_{E(S)}T\ge3$ (R1), so $\operatorname{rk}T\ge12$, a contradiction. So $E(S)=\mathbb Q(\sqrt2)$ and $m=\dim_ET\in\{3,4\}$ (R1). So Theorem A applies with $E=E(S)=\mathbb Q(\sqrt2)$.

*B(4).* By (R1), $E(S_{b,c})$ is totally real or CM. If it is CM, HC for all powers is (R3). If it is totally real, B(3) and Theorem A give HC for all powers. Products with $S_{-b,c}$ follow from (R6) and (R4): $f$ and $f^{-1}$ are algebraic for every $b\ne0$.

*B(5).* (R7). $\square$

*Proof of Corollary C.* Let $g_T\colon T(S_{b,c})\to T(S')$ be a Hodge isometry. Both $H^2(S_{b,c},\mathbb Q)$ and $H^2(S',\mathbb Q)$ are isometric to $\Lambda_{\mathbb Q}=(U^3\oplus E_8(-1)^2)_{\mathbb Q}$. By Witt cancellation $\mathrm{NS}(S_{b,c})_{\mathbb Q}\cong\mathrm{NS}(S')_{\mathbb Q}$. Choosing any such isometry $g_N$, the isometry $g=g_N\oplus g_T$ is a rational Hodge isometry $H^2(S_{b,c})\to H^2(S')$, algebraic by (R8). Its inverse is the transpose. Restricted to transcendental parts, $g_T$ and $g_T^{-1}$ are algebraic, and (R4) with B(4) gives the claim. $\square$

*Remark 6.5 (the Witt-index-one curves of [29]).* The dense family of Noether–Lefschetz curves of [29, Thm. B(4)] (members with $\rho=16$, $E=\mathbb Q(\sqrt2)$ and Witt index one) is covered by B(4) with $m=3$. This gives a second proof of HC for their powers, independent of Floccari's theorem.

*Remark 6.6 (what the proof uses about the family).* B(4) uses only the explicit pull-back formula, the character table of §6.3, and the general facts (R1)–(R6). It does not use the lattice-theoretic description of $T$ (only B(5) does), nor any Noether–Lefschetz or period argument.

---

## 7. Results and supporting evidence

| Statement | Proof | `verify.py` |
|---|---|---|
| Lemma 6.1 (determinant of $M\otimes N$) | §6.1 | E1, E1b (control), E2 |
| Lemma 6.2 | §6.1 | E3 |
| Lemma 6.3 (descent $F\to E$) | §6.1 | E4 ($m=2,3$), E5 (control) |
| Theorem A, Step 5 ($m=3$ contraction) | §6.2 | E6 |
| Genus, holomorphic forms, $M_t$, Hodge numbers of $B$ and $\Pi$ | §6.3 | A1–A9 (A8, A9 controls) |
| Quotient identities, the map to $S_{b,c}$ | §6.4 | B1–B7, B11, B12 |
| Every $b\ne0$ occurs; $\pm b$ from $\pm\sqrt{tu}$ | §6.4 | D1–D4 |
| Pull-back of $\omega$, projections, $E$-eigenvalue | §6.4, §6.5 | C1–C8 (C3, C8 controls) |
| K3 for all $b\ne0$ | §6.5 | B8–B10 |
| Rank bookkeeping | §6.5 | F1–F3 |

All 45 checks are exact: symbolic rational-function identities, exact algebraic numbers ($\zeta_8=(1+i)/\sqrt2$), and Fractions. There is no floating point.

---

## 8. Limitations

- **Mixed products of different members.** We prove nothing about $S_{b,c}\times S_{b',c'}$ for unrelated parameters. As in [18], [28], this would require comparing $\det_FM_t$ for different $t$ through a fixed carrier, and controlling the joint Hodge group.
- **Imported inputs.** Varesco's reduction [5], Zarhin's theorem [12], the CM case (R3), the existence of resolutions of the rational map $\pi$, and [29] for the twist $f$ and for the generic lattice in B(5).
- **The cases $\rho\ge18$.** These are handled through (R3): $E$ is then CM, since $\mathbb Q(\sqrt2)\subseteq E$ and a totally real $E$ would force $\operatorname{rk}T\ge6$. We do not determine which members are CM.
- **Lemma 6.1 and §6.3 are verified by computation for the specific dimensions used.** The Hodge numbers of $M_t$ come from an explicit basis of holomorphic differentials, not from period computations.
- **Scope of Theorem A.** It needs an algebraic realization of $T\otimes_EF$ as a tensor product of two $F$-rank-two pieces of curve cohomology with the conjugate-balanced condition (b). We know such realizations only for Dickson-type families ([17], [18], [26], [28] and here). Theorem A says nothing about Witt-index-one surfaces with $E=\mathbb Q$.

## 9. Reproducibility

`verify.py` is a single self-contained Python 3 script (SymPy and NumPy object arrays of exact `Fraction`s). It reads and writes no files, does not use `__file__`, and uses a fixed seed. It ran in about 10 s under Python 3.13.16, SymPy 1.14.0 and NumPy 2.5.3, and ends with `checks = 45`, `FALLOS = 0`. The output is `verify_output.txt`.

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