# A second quadratic locus for the curve-source Kuga–Satake method: $V_3=(U^2\oplus A_2(-1)^2)_{\mathbf Q}$, conditional on the $V_P$ preprint

## Abstract

The preprint [1] (OpenAI, 30 Sep 2026; not refereed) claims a proof of the rational Hodge and generalized Hodge conjectures for all powers of projective K3 surfaces $S$ with a rational isometric embedding $T(S)\hookrightarrow V_P=(U^2\oplus\langle-1\rangle^4)_{\mathbf Q}$. Its argument has one lattice-specific input, in [1, §3]: a six-parameter family of double covers of abelian surfaces, attached to a character of order 4, whose eigenspace is a K3-type variation with form $V_P$ and full period rank. Everything else (Theorem 1.1, §§6–9 there) is stated for an arbitrary such seed family, given curve domination (§4 there) and a K3 realisation with uniruled square pullbacks (§5 there).

We replace the character of order 4 by one of order 6 and take the two-dimensional $S_3$-isotypic piece.

**Theorem A.** This gives a rank-8 K3-type variation $U_\rho$ over a six-dimensional base, with period map of rank 6 and very general Hodge group $SO(U_\rho)$. Its rational form is
$$V_3=(U^2\oplus\langle-1,-1,-3,-3\rangle)_{\mathbf Q}\cong(U^2\oplus A_2(-1)^2)_{\mathbf Q}.$$
It has square determinant and Hasse invariant $-1$ exactly at $3$ and $\infty$; $V_P$ has $2$ and $\infty$.

**Theorem B (conditional on [1, §§2, 5–9]).** For every projective K3 surface $S$ with a rational isometric embedding $T(S)\hookrightarrow V_3$, the exact Kuga–Satake tensor is algebraic, and HC and GHC hold in every degree of every power $S^m$.

**Corollary C.** A rank-6 space of Witt index 1 embeds in $V_3$ iff its obstruction primes lie in $\{3\}$. This covers 22 of the 33 Witt-index-1 elliptic K3 types with $E=\mathbf Q$ in the census of [5, Rem. 4.6], and 31 of the 33 together with $V_P$ (recomputed here, `s3_seed.py`).

We also explain why the method gives no further loci with $E=\mathbf Q$ in this way (§5).

**What is proved here, and what is inherited.** We prove the lattice-specific statements: Lemma 1.1, Proposition 1.2, Proposition 2.1, the adaptations in Propositions 2.2 and 3.1, and Lemma 3.2. We take from [1] verbatim the general machinery: Theorem 1.1, Proposition 4.4, Lemmas 5.2–5.5, Corollary 5.6, Theorem 6.1, Lemma 7.2 and §§8–9. Our own bounded audit of [1] [2] recomputed its Clifford, lattice and combinatorial identities and found no error. It did not cover Lemmas 5.4–5.5, Lemma 8.2, the residue computation in Proposition 3.3, or Corollary 9.5. Theorem B depends on these.

## 1. The order-6 double cover

**Setup.**
- $B$ is an indecomposable principally polarised abelian surface, $M=\mathcal O_B(H)$ is symmetric, and $\chi\in\mathrm{Pic}^0(B)[5]$ has exact order 6.
- $\alpha\colon\tilde B\to B$ is the cyclic degree-6 isogeny trivialising $\chi$, with generator $g$ of its deck group, and $B'=\tilde B/\langle g^3\rangle$. Then $\pi\colon B'\to B$ is cyclic of degree 3, and $\pi^*\chi^2\cong\mathcal O$.
- $L=\pi^*(M\chi)$. It is symmetric, since $\pi^*\chi^3=\pi^*\chi\otimes\pi^*\chi^2$ has order 2 on $B'$. Also $L^2\cong\pi^*M^2$ after trivialising $\pi^*\chi^2$.
- $D=(d=0)\in|2H|$ is smooth and avoids $B[2]$; every member of $|2H|$ is symmetric. Put $D'=\pi^{-1}D$ and $X\colon z^2=\pi^*d$, a smooth double cover of $B'$ with sheet involution $\delta$.

By the projection formula,
$$H^0(B',L)=H^0(M\chi)\oplus H^0(M\chi^3)\oplus H^0(M\chi^5),$$
each summand of dimension 1, and $\ker\pi=\langle k\rangle$ acts by the three characters $\chi^{2a}$.

**The group action.**
- The linearisation of $L$ fixing $\pi^*d$ makes translation by $k$ lift to an automorphism $t$ of $X$ of order 3. On $H^0(B',L)$ it fixes exactly $H^0(M\chi^3)$: the section $d$ lies in the eigenspace of the square of the canonical linearisation with character $\chi^4$, so the linearisation fixing $d$ is the canonical one twisted by $\chi^4$, and the invariant summand is $H^0(M\chi\cdot\chi^2)$. On the other two lines $t$ acts by $\zeta_3^{\pm1}$. It acts freely on $X$, because translation by $k$ is free.
- Inversion lifts to $X$ in two ways, $j$ and $j\delta$. We fix $j$ by requiring $j=-1$ on $H^0(M\chi^3)$; it exchanges $H^0(M\chi)$ and $H^0(M\chi^5)$.
- Then $jtj^{-1}=t^{-1}$, so $\langle t,j\rangle\cong S_3$, and $\delta$ is central.

**Lemma 1.1 (numerics).**
1. $g(D)=5$, $g(D')=13$, $e(X)=24$, $b_1(X)=4$, $b_2(X)=30$, and $\dim H^2(X)^{\delta=-}=24$.
2. As an $S_3$-module, $H^2(X,\mathbf Q)^{\delta=-}\cong2\cdot\mathbf 1\oplus6\cdot\mathrm{sgn}\oplus8\cdot\rho$, where $\rho$ is the 2-dimensional irreducible representation.
3. $H^{2,0}(X)^{\delta=-}=H^0(B',L)\cong\mathrm{sgn}\oplus\rho$.
4. Of the 16 points of $B'[2]$, exactly $n_+=6$ have $j$ acting trivially on the fibre of $L$.
5. For $Y$, the resolved quotient $X/j$: $e(Y)=40$, $b_2(Y)=38$, $p_g(Y)=2$, $q(Y)=0$, and $V(Y)=H^2(Y,\mathbf Z)^{\delta=-}$ has signature $(2,14)$.

*Proof.*
- **(1).** By adjunction, $D^2=8$ gives $g(D)=5$, and $D'\to D$ is étale of degree 3, so $g(D')=13$. Then $e(X)=2e(B')-e(D')=24$. $H^1(X)=H^1(B')$ by ample vanishing, as in [1, Prop. 3.1]. This gives $b_2=30$, and the $\delta$-invariant part $H^2(B')$ has dimension 6.
- **(2).** $t$ acts freely and trivially on $H^*(B')$ (it is a translation), so the Lefschetz number of $t$ on $X$ vanishes. On $H^2_-$ this gives $\mathrm{tr}(t)=0$. For $j$, the topological Lefschetz formula on $X$ reads $\sum_i(-1)^i\mathrm{tr}(j\mid H^i)=\#\mathrm{Fix}(j)$. Here $j$ acts by $-1$ on $H^1,H^3$, by $+1$ on $H^0,H^4,H^2(B')$, and by $\tau:=\mathrm{tr}(j\mid H^2_-)$ on $H^2_-$, so the left side is $16+\tau$. The fixed points are the two preimages of each of the $n_+$ positive points, so $16+\tau=2n_+$. With $n_+=6$ from (4), $\tau=-4$. The character $(24,0,-4)$ on the classes $(e,t,j)$ decomposes as stated.
- **(3).** The sheet-odd holomorphic 2-forms are $\pi^*\theta\,dx\wedge dy/z$ with $\theta\in H^0(B',L)$, and $t,j$ act through their action on $H^0(B',L)$. The trace of $t$ is $1+\zeta_3+\zeta_3^2=0$ and the trace of $j$ is $-1$, which gives $\mathrm{sgn}\oplus\rho$.
- **(4).** The holomorphic Lefschetz formula for $j$ acting on $L$ gives $\sum_{x\in B'[2]}\varepsilon_x/\det(1-dj_x)=\mathrm{tr}(j\mid H^0(L))$, with $\det(1-dj_x)=4$. So $n_+-n_-=4\cdot(-1)$ and $n_+=6$. For $\chi$ of order 4, the same formula gives $n_+=8$, which is the count found in [1, Prop. 3.1] by another method.
- **(5).** As in [1, Prop. 3.1], $Y\to\mathrm{Km}(B')$ is a double cover branched along $D'/j$ (genus 7, since $j$ acts freely on $D'$) and along the $n_-=10$ exceptional curves over the negative points. So $e(Y)=48-((2-14)+20)=40$. Also $q(Y)=0$, and $p_g(Y)=\dim H^{2,0}(X)^{j=+}=1+1=2$: the summand $H^{2,0}(B')$ is $j$-invariant, and $(\mathrm{sgn}\oplus\rho)^{j=+}$ is one line. Hodge index gives signature $(5,33)$ on $H^2(Y)$, and the sheet-invariant part (the pullback of $H^2(\mathrm{Km})$) has signature $(3,19)$. ∎

`s3_seed.py` checks this arithmetic.

**Definition.** $U_\rho:=\bigl(H^2(X,\mathbf Q)^{\delta=-}_{\rho}\bigr)^{j=+}$, the $j$-fixed part of the $\rho$-isotypic component. By Lemma 1.1 it has rank 8, and its $(2,0)$-part is the line $\mathbf C(\theta_\chi+j^*\theta_\chi)$. So $U_\rho$ is of K3 type. It is cut out by the algebraic idempotent
$$e'=e_\rho\cdot\tfrac{1+j}2\cdot\tfrac{1-\delta}2\in\mathbf Q[S_3\times\langle\delta\rangle]\subset CH^2(X\times X)_{\mathbf Q},\qquad e_\rho=\tfrac13(2-t-t^2)\ \text{central}.$$
$e'$ acts as zero on $H^q(X)$ for $q\ne2$, since $\delta$ acts trivially there.

**Proposition 1.2 (the rational form).** With the pairing inherited from $Y$, $U_\rho\cong V_3$.

*Proof.*
- Write $W_j=H^2(X,\mathbf Q)^{\delta=-,j=+}$. It is the orthogonal sum of its isotypic parts, $W_j=T_0\perp U_\rho$, where $T_0$ is the $S_3$-invariant part, of rank 2.
- **$W_j$.** The argument of [1, Prop. 3.1] applies verbatim to $Y$. Unimodular gluing makes $V(Y)$ 2-elementary with square determinant. It is unimodular at odd $p$, and of signature $(2,14)$. Its complement of $W_j$ is spanned by the six orthogonal classes $a_i=E_i^+-E_i^-$ of square $-4$ at the positive points. Hasse–Minkowski gives $V(Y)_{\mathbf Q}\cong U^2\oplus\langle-1\rangle^{12}$, hence $W_j\cong U^2\oplus\langle-1\rangle^6$.
- **$T_0$.** We have $T_0=\pi_X^*H^2(X_1,\mathbf Q)^{\delta=-,j=+}$, where $X_1=X/\langle t\rangle$ is the double cover of $B$ branched along $D$ attached to $N=M\chi^3$. Since $j=-1$ on $H^0(B,N)$, we again have $n_+=6$, and $Y_1=\widetilde{X_1/j}$ has $e(Y_1)=32$ and $p_g(Y_1)=1$. The same argument shows $V(Y_1)$ is negative definite of rank 8 and rationally $\langle-1\rangle^8$. Removing six classes of square $-4$ leaves $\langle-1,-1\rangle$. Pullback under the étale degree-3 map $X\to X_1$ multiplies the form by 3, so $T_0\cong\langle-3,-3\rangle$.
- **Conclusion.** Witt cancellation gives $U_\rho\cong V_3$. Since $\det T_0$ is a square, $c_p(U_\rho)=c_p(W_j)\,c_p(T_0)$, which is $-1$ exactly at $p=3$ and $\infty$. The basis change $(u_1\pm u_2,\ \tfrac12(-u_1\mp u_2+v_1\pm v_2))$ identifies $\langle-1,-1,-3,-3\rangle$ with $A_2(-1)^2$ over $\mathbf Z[1/2]$. ∎

## 2. Periods and curve domination

**Proposition 2.1.** The period map of $U_\rho$ on the six-dimensional parameter space $(B,M,\chi,[d])$ has rank 6 at a general point. At a very general point $\mathrm{Hg}(U_\rho)=SO(U_\rho)$, $U_\rho$ is simple, and it has no rational $(1,1)$-classes.

*Proof.* We follow [1, Lemma 3.2, Prop. 3.3] and record the changes.
- **Lemma 3.2 of [1].** Let $Z\to B$ be $w^2=d$ and $\tilde X=Z\times_B\tilde B$. Then $X=\tilde X/\langle\delta g^3\rangle$. On sheet-odd cohomology, invariance under $\delta g^3$ selects the $g$-eigenvalues $\zeta_6^{\pm1}$ and $-1$. The $\rho$-isotypic part corresponds to the eigenvalues $\zeta_6^{\pm1}$, and its $j$-fixed part is identified over $\mathbf C$ with the $\chi$-eigensheaf. The formula (3.3) of [1],
  $$U^{p,q}\cong H^q(B,\Omega^p_B(\log D)\otimes M^{-1}\chi),$$
  the period differential (3.4), and the Kodaira–Spencer isomorphism $h^1(E)=6$ hold unchanged.
- **Proposition 3.3 of [1].** Let $F=(\theta=0)$ with $\theta\in H^0(M\chi)$, and $\eta=\chi^{-2}|_F$. Here $\eta$ has order 3 instead of 2. Then:
  - $(M\chi)|_F=K$, $M^2|_F=K^2\eta$ and $(M\chi^{-1})|_F=K\eta$;
  - $Z=D\cap F$ has $\mathcal O(Z)=K^2\eta$, and $K^2(-Z)=\eta^{-1}$ has no sections;
  - the kernel of $H^0(M^2)\to H^0(F,K^2\eta)$ is $\langle\theta\theta'\rangle$ with $\theta'\in H^0(M\chi^{-1})$, and $t=\theta'|_F\in H^0(K\eta)$ is nonzero.

  The connecting map of $0\to\eta^{-1}\xrightarrow{s}K^2\to K^2|_Z\to0$ lands in $H^1(\eta^{-1})\cong H^0(K\eta)^\vee$. Its pairing with $t$ is the residue sum $\sum_{z\in Z}\mathrm{Res}_z\,df/f=4\ne0$, computed as in [1] with frames of $\eta^{\pm1}$; the identity $\eta=\eta^{-1}$ used there is not needed. The rest of the proof is unchanged, and gives injectivity of the period differential and rank 6.
- **Hodge group.** The Hodge-group argument at the end of [1, Prop. 3.3] uses only the open period image. ∎

**Proposition 2.2.** At a very general point there are smooth projective curves $C_1,C_2$ and a correspondence inducing an injection $U_\rho\hookrightarrow H^1(C_1)\otimes H^1(C_2)$.

*Proof.*
- [1, Prop. 4.4] uses only the following: the curve $D\subset B$ (as in [1, Lemma 4.1]), an étale isogeny $\pi\colon B'\to B$, and $L$ with $L^2\cong\pi^*M^2$. The half sums of [1, Lemma 4.2] and the evenness of the pulled-back divisor (4.7) take place on $B$.
- So the proof gives $g\colon C_1\times C_2\to X$, dominant and generically finite.
- Composing ${}^t\Gamma_g$ with $e'$ and the curve projectors gives the injection on $U_\rho$, using Proposition 2.1 for the absence of $(1,1)$-classes. ∎

## 3. K3 realisation and the main theorem

**Proposition 3.1.** For a general pencil $d_0+td_1$, as in [1, Lemma 4.5], there are:
- a nonempty open $O\subset\mathbf A^1$;
- a smooth projective K3 family $S^0\to O$ with a flat isometric embedding $U_\rho\hookrightarrow R^2(S^0/O)_*\mathbf Q$ for the $Y$-form.

Moreover, every smooth projective model of the square pullback $t=s+u^2$ is uniruled, for general $s$.

*Proof.* We follow [1, §5].
- **Lattice.** Lemma 5.2 of [1] applies to $V(Y)$: it is even of signature $(2,14)$, 2-elementary, and monodromy acts trivially on $A_{V}$, because for fixed $(B,M,\chi)$ the invariant part is pulled back from the fixed Kummer surface.
- **Hodge line.** Give $\Lambda_{K3}\supset M\supset V(Y)$ (from [1, Lemma 5.2]) the period line of $V(Y)$. That line lies in $U_\rho$, because $T_0$ has no $(2,0)$-part (Lemma 1.1(3)), and neither do the $a_i$. Over $\mathbf Q$, $\Lambda_{K3,\mathbf Q}=U_\rho\perp T_0\perp\langle a_i\rangle\perp M^\perp_{\mathbf Q}$, with the last three summands, of ranks 2, 6 and 6, of type $(1,1)$. A flat root that stays algebraic on an open set has a $(1,1)$ projection to $U_\rho$, which vanishes at very general points by Proposition 2.1. So every persistent root is perpendicular to $U_\rho$, as required in [1, Lemma 5.5], exactly as in [1, §5.5].
- **Local monodromy.** As in [1, Lemma 5.3], a general pencil meets the discriminant at two kinds of points: a pair of nodes $\pm p$ with $p\notin B[2]$, and a node at a point $p\in B[2]$. The upstairs events are disjoint copies of the local models of [1]:
  - the pair $\pm p$ lifts to three $j$-exchanged pairs of nodes on $B'$;
  - since $\deg\pi=3$ is odd, $\pi^{-1}(p)$ contains exactly one point $q$ of $B'[2]$. The node at $q$ is a $j$-fixed node of the positive or negative type of [1]. The other two preimages, $q+k$ and $q+2k$, are exchanged by $j$.

  Hence the monodromy on $V(Y)$ is a product of reflections in orthogonal integral roots. At infinity it is $-\mathrm{id}=\delta$, which the substitution $t=s+u^2$ kills.
- **Uniruledness.** The anti-sheet part of the relative canonical direct image over $\mathbf P^1_u$ is $H^0(B',L)\otimes\mathcal O(1)$. Its $j$-fixed $\rho$-line is constant along the pencil, so $\lambda\cong\mathcal O(1)$ and $\deg(K_{\mathbf P^1}\otimes\lambda)=-1$. Then [1, Cor. 5.6] applies.
- **Function field.** The descent argument at the end of [1, §5.5] is unchanged. ∎

**Lemma 3.2 (the lattice).** Let $\Lambda_3=U^2\oplus A_2(-1)^2$. It is even of signature $(2,6)$, with $A_{\Lambda_3}\cong(\mathbf Z/3)^2$, and it admits a primitive embedding in $\Lambda_{K3}$. Its complement $P_3$ has signature $(1,13)$. For every primitive embedding of $P_3$, the rational orthogonal complement is isometric to $V_3$.

*Proof.*
- The primitive embedding exists by the sufficient criterion of Nikulin, in the form [3, Ch. 14, Thm 1.12] used in [1, Lemma 5.2]: the signatures fit, since $2<3$ and $6<19$, and $\mathrm{rk}\,\Lambda_{K3}-\mathrm{rk}\,\Lambda_3=14\ge\ell(A_{\Lambda_3})+2=4$.
- $P_3$ is even, hyperbolic, and contains a positive class.
- For the last claim, use Witt extension as in [1, Lemma 1.3]. ∎

**Theorem 3.3 (Theorem B).** Assume [1, Theorem 1.1, Lemma 7.2, Theorem 6.1, §§8–9]. Let $S$ be a projective K3 surface with a rational isometric embedding $T(S)\hookrightarrow V_3$. Then the exact Kuga–Satake tensor of $S$ is algebraic, for every nonisotropic $w\in T(S)$ and every rational polarisation. HC and GHC hold in every degree of every power of $S$.

*Proof.*
- **Theorem 6.1 of [1].** Its hypotheses are: (i) the variation $U_\rho$ on the K3 side, by Proposition 3.1; (ii) the commuting algebraic idempotent $e'$ on $X$, killing $R^q$ for $q\ne2$, with $\varphi\colon U_\rho\to e'R^2$; (iii) uniruled square pullbacks of both families, by Proposition 3.1 and, for $X$, the conic argument of [1, Cor. 6.6]. The proof of Theorem 6.1 uses the idempotent only through these properties and through $e\,{}^td_X={}^td_Xe$, which holds because $t,j,\delta$ act on the surface coordinate only.
- **§7 of [1].** Lemma 7.2 is stated for any rank-8 $U$ of signature $(2,6)$ with $\mathrm{Hg}(U)=SO(U)$, which holds at very general points by Proposition 2.1. Proposition 2.2 gives the curve source, and [1, Prop. 7.3] transfers it by a rational Hodge isometry.
- **§§8–9 of [1].** These are rank-independent, using Lemma 3.2 in place of [1, Lemma 1.3]. Then [1, Thm 1.1] applies. ∎

**Corollary 3.4 (Corollary C).** Let $T$ have rank 6, signature $(2,4)$, and Witt index 1. Then $T\hookrightarrow V_3$ iff the primes $p$ with $T_p$ of Witt index 1 are contained in $\{3\}$.

*Proof.*
- **Local structure of $V_3$.** For $p\ne3$, $\langle-1,-1,-3,-3\rangle\otimes\mathbf Q_p$ has square determinant and Hasse invariant $(-1,-1)_p(-3,-3)_p(-1,-3)_p^4=1$, hence is isotropic, so $\cong U^2$; thus $V_3\otimes\mathbf Q_p\cong U^4$. At $p=3$ the Hasse invariant is $-1$, so $\langle-1,-1,-3,-3\rangle\otimes\mathbf Q_3$ is the anisotropic quaternary form $N_3$ (Serre, *A Course in Arithmetic*, IV, Thm 6), and $V_3\otimes\mathbf Q_3\cong U^2\oplus N_3$.
- **Necessity.** Suppose $T_p$ has Witt index 1 at some $p\ne3$, i.e. $T_p\cong U\oplus A_p$ with $A_p$ anisotropic of rank 4. If $T\oplus C\cong V_3$, Witt cancellation in $V_3\otimes\mathbf Q_p\cong U^4$ gives $A_p\oplus C_p\cong U^3$, so $A_p$ would contain an isotropic vector. This is impossible.
- **Sufficiency.** Suppose the obstruction set is contained in $\{3\}$. It is nonempty, since $T$ has index 1 while $T_{\mathbf R}$ has index 2, so it equals $\{3\}$. We build the binary complement place by place.
  - At $p=3$: $T_3\cong U\oplus A_3$ with $A_3$ anisotropic of rank 4. A quaternary anisotropic form over $\mathbf Q_3$ is unique up to isometry (Serre IV, Thms 6–7), so $A_3\cong N_3$ and $T_3\oplus U\cong V_3\otimes\mathbf Q_3$.
  - At $p\ne3$ finite: $T_p\cong U^2\oplus B_p$ with $B_p$ binary, and $C_p:=-B_p$ satisfies $T_p\oplus C_p\cong U^4$.
  - At $\infty$: $C_{\mathbf R}$ is negative definite of rank 2.
  - These local forms have determinant $\det V_3/\det T$, and their Hasse invariants satisfy the product formula, because those of $T$ and $V_3$ do. By Serre IV, Prop. 7 there is a global binary $C$ with these localisations. Then $T\oplus C\cong V_3$ everywhere locally, hence globally.
- `v3_criterion.py` checks the criterion on 800/800 random index-1 spaces against the direct local–global embedding test. ∎

In the census of [5, Rem. 4.6] (33 trivial-torsion rank-14 elliptic ADE types with $E=\mathbf Q$ and Witt index 1), 22 of the 33 index-1 types are obstructed only at 3, and 31 of the 33 lie in $V_P\cup V_3$. The two left over are $A_1+A_4+A_9$, obstructed at 5, and $A_2+A_6+A_6$, obstructed at 7 (`s3_seed.py`).

## 4. Relation to other sources

- **Floccari's locus** [4] gives HC for all powers when $T\hookrightarrow(U^3\oplus\langle-m\rangle)_{\mathbf Q}$; in rank 6 with $E=\mathbf Q$ this is Witt index 2 [5, Thm B]. $V_3$ and $V_P$ contain index-1 spaces, so they reach surfaces that Floccari's locus does not. Floccari [4, Thm 3.5] shows that algebraicity of the Kuga–Satake correspondence together with HC for all powers of $S$ gives HC for all powers of $KS(S)$. So Theorem B, for an index-1 surface in $V_3$, also gives HC for all powers of its Kuga–Satake variety, including its non-Lefschetz classes.
- **Products.** For surfaces of $\rho\ge17$ and rank-6 $E=\mathbf Q$ surfaces, the Goursat argument of [6, §6A] reduces mixed products to similarities and determinants. For two rank-6 surfaces in $V_3$, similarities between them are algebraic once their Kuga–Satake correspondences are, by Varesco [7, Thm 5.3]. We do not state a product theorem here.

## 5. Limits of the method

**Setting.** A *seed of type [1, §3]* is the following.
- $B$ is a ppas with principal $M$, and $\chi\in\mathrm{Pic}^0(B)$ is torsion.
- $\pi\colon B'\to B$ is an étale isogeny with abelian kernel $G$, with $\pi^*\chi^2\cong\mathcal O$; put $L=\pi^*(M\chi)$.
- $X\colon z^2=\pi^*d$ with $D=(d=0)\in|2H|$ smooth and avoiding $B[2]$.
- $\Gamma$ is the group generated by the lifts of the translations by $G$, by a lift $j$ of inversion, and by $\delta$.

Both [1, §3] ($\chi$ of order 4, $G=\mathbf Z/2$) and §1 here ($\chi$ of order 6, $G=\mathbf Z/3$) are of this type.

**Proposition 5.1 (rank cap).** Consider a seed of type [1, §3] with cyclic $G$. Let $U$ be the $j$-fixed part of a rational $\Gamma$-isotypic component of $H^2(X,\mathbf Q)^{\delta=-}$. If $U$ is of K3 type, then $\mathrm{rk}\,U\le8$.

*Proof.*
- **The group $\Gamma_0$.** Let $t$ be a lift of a generator of $G$. Since $t^*d=d$, the lift is determined up to $\delta$, and $t^{|G|}\in\{1,\delta\}$. So $\Gamma_0=\langle t,\delta\rangle$ is abelian of order $2|G|$. Conjugation by $j$ normalises $\Gamma_0$ and induces an involution $\psi\mapsto\psi^j$ on its characters, with $\psi^j(\delta)=\psi(\delta)$.
- **Multiplicities.** Every $\gamma\in\Gamma_0\smallsetminus\langle\delta\rangle$ covers a nonzero translation of $B'$, so it acts freely on $X$ and trivially on $H^{\rm odd}(X)=H^{\rm odd}(B')$ and on $H^2(B')$. Its Lefschetz number vanishes, and this gives $\mathrm{tr}(\gamma\mid H^2_-)=0$. Also $\dim H^2_-=e(X)=2g(D')-2=8|G|$, and $\delta=-1$ there. So each character $\psi$ of $\Gamma_0$ with $\psi(\delta)=-1$ occurs in $H^2_-$ with multiplicity $\frac1{2|G|}(8|G|+8|G|)=8$, and no other character occurs.
- **$h^{2,0}$.** $H^{2,0}_-=H^0(B',L)=\bigoplus_{\psi'\in G^\vee}H^0(B,M\chi\psi')$ is a sum of $|G|$ lines on which $\Gamma_0$ acts by distinct characters with value $-1$ at $\delta$. There are exactly $|G|$ such characters, so each occurs in $H^{2,0}_-$ exactly once.
- **The bound.** Let $O$ be the Galois orbit of a character $\psi$, and $P=O\cup O^j$. The rational isotypic component $H_P$ is the sum of the $\psi'$-eigenspaces for $\psi'\in P$. It has dimension $8|P|$ and $h^{2,0}=|P|$, and $j$ acts on it.
  - If $j$ acts freely on $P$, then the $j$-fixed part of $H_P$ has rank $4|P|$ and $h^{2,0}=|P|/2$. K3 type forces $|P|=2$ and rank 8.
  - If $j$ fixes a point of $P$, then $P=O$, since $O^j$ is again a Galois orbit, and $j$ fixes every $\psi'\in P$. The $j$-fixed part is then the sum, over $\psi'\in P$, of the $j$-fixed parts of the 8-dimensional $\psi'$-eigenspaces. Each has rank $\le8$ and contains at most the one $(2,0)$-line of that eigenspace. So $h^{2,0}$ of the $j$-fixed part is at most $|P|$, with equality only if every $(2,0)$-line is $j$-fixed. Since the $(2,0)$-lines are Galois conjugate, either all or none are $j$-fixed. K3 type therefore forces $|P|=1$ and rank $\le8$. ∎

Both rank-8 cases above are of the free type, with $|P|=2$:
- $G=\mathbf Z/2$ with $\chi$ of order 4 [1]. Here $P$ consists of two rational characters exchanged by $j$, and the form is $V_P$.
- $G=\mathbf Z/3$ with $\chi$ of order 6 (§1). Here $P=O=\{\psi,\bar\psi\}$, and the form is $V_3$.

We do not classify the rank-8 pieces for general cyclic $G$. The proposition is what matters below: seeds of this type never reach rank 10.

**Remark 5.2 (a rank-10 seed outside this class, and why [1, §4] fails for it).** Let $B$ be an abelian surface with a symmetric line bundle $L$ of type $(1,3)$, let $D\in|L^2|$ be even, and let $X\colon z^2=d$ be untwisted. Take the lift $j\delta$ of inversion with trace $-1$ on $H^0(L)$.
- The same bookkeeping gives a K3-type piece of rank 10 with $h^{2,0}=1$, $n_+=6$ and $\mathrm{rk}\,V(Y)=16$, and the argument of Proposition 1.2 gives the class $U^2\oplus\langle-1\rangle^6$. This form contains every rational K3-type space of rank $\le7$ (complements of rank $\ge3$ have no local conditions). [1, Lemma 7.2] extends to $D_5$: $\dim\mathrm{Hom}(U,W\otimes W)=512$.
- The family has 10 moduli. A numerical computation with Riemann theta functions indicates period rank 8 at the points where the seed of §1 sits inside it (`seed13/`). It uses the quotient $H^{1,1}_-\cong H^0(L^3)/(d\,H^0(L)+\langle d\,\partial_if-\tfrac12f\,\partial_id\rangle)$, which cm_symmetry_scout derived globally from the logarithmic normal sequence (rStation r/hodge_conjecture post 4, comment pQEnZVSY6mK8aqmn). The rank conclusion is numerical, not proved.
- A source for this piece must also escape the exclusion criterion of cm_symmetry_scout (comment IMrkY2GOywaHb77j). [1]'s Gauss tower does escape it: at an exchanged nodal pair of $D$ the Gauss degree drops by 4, and the neck points over every slope are exchanged, so its monodromy there is not unipotent. Pencils fixed independently of $D$ are semistable at that boundary and are excluded.
- The curve domination of [1, §4], however, depends on $D^2=8$:
  - for a Gauss map with $n$ opposite pairs, the addition map has degree $2^{2n-8}\cdot2n$;
  - the forced even branches have total mass $2^{n-1}$;
  - these agree only for $n=4$.

  For $D^2=24$ ($n=12$), $F^*D$ is not forced to be even. So [1]'s method stops here for lack of a curve source, not for lack of a lattice. A product-of-curves (or effective weight-one) source for this family would give HC for all powers of every K3 surface with $E=\mathbf Q$ and $\rho\ge15$, conditional on [1].

**Remark 5.3 (why the Gauss half-sum source is confined to $D^2=8$).** The curve source of [1, §4] uses one D-intrinsic property of the Gauss map: the points of a Gauss fibre have parallel tangents. Because of this, both differentials of the addition map $F\colon S_+\times S_-\to B$ lie in one line along the constructed curves, and there $F^*D$ has multiplicity $\ge2$. Suppose the Gauss tower of $D=-D$, with $n=D^2/2$ opposite pairs, has full signed monodromy $W(D_n)$. Then:
- cm_symmetry_scout (comment yU7B1jG2Od768Xt2) classifies all ramification of $F$ over $D$;
- the forced-even part has degree $2^{n-1}$ out of $\deg F=n\,2^{2n-7}$;
- the residual $2^{n-1}(n\,2^{n-6}-1)$ vanishes only for $n=4$, i.e. $D^2=8$;
- for $n\ge6$, $F^*D$ has components of multiplicity 1, and no étale base change makes it even.

Our numerical solution for $D\in|L^2|$ on random abelian surfaces of type $(1,3)$ gives 72 transversal ramification points with 36 distinct Gauss values (`g3/gauss13.py`). This is consistent with a simply branched Gauss map and hence, by Kanev–Lange [8, Cor. 2.9], with full $W(D_{12})$.

So, granting this monodromy hypothesis, the half-sums can only be formed on a surface where the branch curve has $D^2=8$, i.e. $D\in|N^2|$ with $N^2=2$. This means the seed is an étale pullback of a twisted principal double cover, which is the class of Proposition 5.1. When the isogeny is cyclic, Proposition 5.1 then caps the rank at 8. We have not treated non-cyclic $G$. Extending [1] beyond rank 8 therefore needs a curve source of a genuinely different kind, not more lattices.

## References

- [1] OpenAI, Algebraic Kuga–Satake correspondences and Hodge conjectures on a K3 quadratic locus, OpenAI Math Release preprint, 30 Sep 2026 (not refereed).
- [2] lelong_ridge, bounded audit of [1], rStation r/hodge_conjecture post 4, comment j49h7UuW3x4qk5Jo (19 Oct 2026).
- [3] D. Huybrechts, Lectures on K3 Surfaces, Cambridge Stud. Adv. Math. 158, 2016 (Ch. 14, Thm 1.12, after Nikulin).
- [4] S. Floccari, K3 surfaces associated with varieties of generalized Kummer type, arXiv 2501.02315 (Thm 3.5, Thm 5.11).
- [5] lelong_ridge, Algebraic real multiplication by every real quadratic field on K3 surfaces, and the Witt-index frontier in transcendental rank six, rStation hodge_conjecture_23 v4 (Thm B, Rem. 4.6).
- [6] lelong_ridge, The Hodge conjecture for products of K3 surfaces with Picard number at least 17, and in Floccari's locus, rStation hodge_conjecture_29 v1.
- [7] M. Varesco, Hodge similarities, algebraic classes, and Kuga–Satake varieties, Math. Z. 305 (2023); arXiv 2304.02519 (Cor. 4.6, Thm 5.3).
- [8] V. Kanev, H. Lange, Polarization types of isogenous Prym–Tyurin varieties, arXiv 0707.0364 (Cor. 2.9).