\section{Constructions at \Fiii{}: real multiplication on K3 surfaces}\label{sec:f3prime} \subsection{Where the first class is} Let $S$ be a projective K3 surface, $T=T(S)_{\QQ}$ and $E=\End_{\Hdg}T$. The Hodge classes of $S\times S$ are $H^{0}\otimes H^{4}$, $H^{4}\otimes H^{0}$ and $\End_{\Hdg}H^{2}(S)=\End(\NS_{\QQ})\oplus E$, since $T$ is irreducible with no class of type $(1,1)$ and $\Hom_{\Hdg}(T,\NS_{\QQ})=0$. The first two pieces and $\End(\NS_{\QQ})$ are spanned by products of divisor classes. So the Hodge conjecture for $S\times S$ is the algebraicity of $E$, and by Proposition~\ref{prop:abpart} the same classes decide \Fiii{} for $S\times S$ once $T$ is not reached from abelian varieties. When $E$ is a CM field it is spanned by its elements of norm one, which are Hodge isometries of $T$ and algebraic by Buskin's theorem \cite{Bus19}; see \cite[Rem.~2.16]{Var23}. When $E$ is totally real, the Rosati involution is the identity on it \cite{Zar83}, every $e\in E$ is self-adjoint for the intersection form $q$, and $q(ex,ey)=q(e^{2}x,y)$. The first class is a real multiplication. \subsection{Chains of correspondences} \begin{lemma}\label{lem:dominant} Let $f\colon S\dashrightarrow S'$ be a dominant rational map of K3 surfaces of degree $d$ and $\Gamma_{f}$ the closure of its graph. Then $f_{*}=\Gamma_{f*}$ and $f^{*}=\Gamma_{f*}^{t}$ are morphisms of Hodge structures, $f^{*}$ maps $T(S')_{\QQ}$ isomorphically onto $T(S)_{\QQ}$ with $q_{S}(f^{*}a,f^{*}b)=d\,q_{S'}(a,b)$, and $f_{*}=d\,(f^{*})^{-1}$ on $T(S)_{\QQ}$. \end{lemma} \begin{proof} The graph is an algebraic cycle, so both maps are induced by algebraic classes and are morphisms of Hodge structures. Pull-back of a nowhere vanishing $2$-form is a nonzero $2$-form, so $f^{*}(T(S')_{\QQ})$ is a sub-Hodge structure of $H^{2}(S)$ that meets $H^{2,0}(S)$; the smallest sub-Hodge structure of $H^{2}(S)$ containing $H^{2,0}$ is $T(S)_{\QQ}$, so $f^{*}T(S')_{\QQ}\supseteq T(S)_{\QQ}$, and since $T(S')_{\QQ}$ is irreducible and $f^{*}$ is injective on it ($f_{*}f^{*}=d$), equality of dimensions gives $f^{*}T(S')_{\QQ}=T(S)_{\QQ}$. The projection formula gives $q_{S}(f^{*}a,f^{*}b)=q_{S'}(f_{*}f^{*}a,b)=d\,q_{S'}(a,b)$ and $f_{*}f^{*}=d$. \end{proof} \begin{lemma}\label{lem:isometry} A Hodge isometry $\varphi\colon T(S)_{\QQ}\to T(S')_{\QQ}$ is induced by an algebraic class on $S\times S'$. \end{lemma} \begin{proof} The spaces $\NS(S)_{\QQ}$ and $\NS(S')_{\QQ}$ are the orthogonal complements of $T(S)_{\QQ}$ and $T(S')_{\QQ}$ in the two copies of $\Lambda_{\QQ}$, isometric to each other through $\varphi$; by Witt's cancellation theorem the complements are isometric, so $\varphi$ extends to a Hodge isometry $H^{2}(S)\to H^{2}(S')$, which is algebraic by Buskin's theorem \cite{Bus19} (see \cite[Thm.~0.1]{Huy19}). The projector of $H^{2}(S)$ onto $T(S)_{\QQ}$ along $\NS(S)_{\QQ}$ is algebraic, being $1$ minus the projector onto $\NS$, which is a sum $\sum_{i,j}c_{ij}\,D_{i}\otimes D_{j}$ over a basis of divisor classes with $(c_{ij})$ the inverse Gram matrix; composing gives $\varphi$ on $T$ and zero on $\NS$. \end{proof} \begin{theorem}[Chains]\label{thm:chain} Let $S$ be a projective K3 surface with $E$ totally real and $e\in E$ with $e^{2}=n\in\QQ_{>0}$. Suppose there are projective K3 surfaces $S=S_{0},S_{1},\dots,S_{r}$ and, for each $i$, either a dominant rational map $S_{i-1}\dashrightarrow S_{i}$ of degree $n_{i}$, a dominant rational map $S_{i}\dashrightarrow S_{i-1}$ of degree $n_{i}$, or a Hodge isometry $T(S_{i-1})_{\QQ}\cong T(S_{i})_{\QQ}$ (and then $n_{i}=1$), such that \[ \prod_{i}n_{i}^{\pm1}\equiv n\pmod{\QQ^{\times2}}, \] the sign being $+$ for maps leaving $S_{i-1}$ and $-$ for maps arriving at it. Then $e$ is algebraic on $S\times S$. \end{theorem} \begin{proof} Composing along the chain the maps $f_{*}$ for maps leaving $S_{i-1}$, $f^{*}$ for maps arriving, and the classes of Lemma~\ref{lem:isometry} for the isometries, gives an algebraic isomorphism $\phi\colon T(S_{r})_{\QQ}\to T(S)_{\QQ}$ of Hodge structures with $q_{S}(\phi a,\phi b)=\mu\,q_{S_{r}}(a,b)$, where $\mu=\prod n_{i}^{\pm1}$ by Lemma~\ref{lem:dominant} (a map leaving contributes $f_{*}=n_{i}(f^{*})^{-1}$, of multiplier $n_{i}$; a map arriving contributes $f^{*}$, of multiplier $n_{i}$ in the other direction, that is $n_{i}^{-1}$ for the composite in the direction of $\phi$). Write $\mu=nc^{2}$ with $c\in\QQ^{\times}$. Since $e$ is self-adjoint with $e^{2}=n$, it is a Hodge similarity of $T(S)_{\QQ}$ of multiplier $n$. So $\psi=(c^{-1}\phi)^{-1}\circ e\colon T(S)_{\QQ}\to T(S_{r})_{\QQ}$ is a Hodge isometry, algebraic by Lemma~\ref{lem:isometry}, and $e=(c^{-1}\phi)\circ\psi$ is a composite of algebraic correspondences, hence algebraic on $S\times S$. \end{proof} The case $r=1$ with $S\dashrightarrow S_{1}$ the quotient by a symplectic automorphism of prime order $p$ is Varesco's theorem \cite[Thm.~2.1]{Var23}, which gives $e=\sqrt p$ and is used for $p=2,3$. What a chain adds is a composite multiplier. \subsection{Real multiplication by \texorpdfstring{$\QQ(\sqrt6)$}{Q(sqrt 6)}} A symplectic automorphism of finite order on a K3 surface acts trivially on $T(S)$, its quotient has a K3 resolution $Y$ \cite{Nik80b} (for involutions and their quotients see also \cite{BN26}), and $T(Y)_{\QQ}\cong T(S)_{\QQ}(1/\mathrm{ord})$ by Lemma~\ref{lem:dominant}. The invariant lattice of a symplectic involution is $U^{3}\oplus E_{8}(-2)$, and a K3 surface has a symplectic involution exactly when $E_{8}(-2)$ embeds primitively in its N\'eron--Severi lattice \cite{Mor84,vGS07}. The invariant lattice of a symplectic automorphism of order three is $U\oplus U(3)^{2}\oplus A_{2}^{2}$, and such an automorphism exists exactly when the Coxeter--Todd lattice $K_{12}(-2)$ embeds primitively in the N\'eron--Severi lattice \cite[Thm.~4.1]{GS07}. Put \[ V_{2}=\bigl(U^{3}\oplus E_{8}(-2)\bigr)_{\QQ},\qquad V_{3}=\bigl(U\oplus U(3)^{2}\oplus A_{2}^{2}\bigr)_{\QQ}, \] of ranks $14$ and $10$ and signatures $(3,11)$ and $(3,7)$; over $\QQ$, $U(3)\cong U$ and $E_{8}(-2)\cong\langle-1\rangle^{8}\cong\langle-2\rangle^{8}$. Varesco \cite[Props.~2.5 and 2.11]{Var23} shows that a projective K3 surface whose transcendental space embeds isometrically in $V_{2}$ (resp.\ $V_{3}$) is Hodge isometric to one with a symplectic involution (resp.\ a symplectic automorphism of order three). \begin{lemma}[Embedding criterion]\label{lem:embed} Let $W$ and $V$ be rational quadratic spaces with $\mathrm{rk}\,V-\mathrm{rk}\,W\ge3$. Then $W$ embeds isometrically in $V$ if and only if the signature of $V$ dominates that of $W$ in both signs. In that case a complement $W'$ exists with $V\cong W\oplus W'$. \end{lemma} \begin{proof} Necessity is clear. For sufficiency put $k=\mathrm{rk}\,V-\mathrm{rk}\,W\ge3$ and let $(s_{+},s_{-})$ be the difference of signatures, $s_{+}+s_{-}=k$. A quadratic space $W'$ of rank $k$ and signature $(s_{+},s_{-})$ with prescribed discriminant $d'=d(V)d(W)$ and prescribed Hasse invariants $\varepsilon_{p}'$ exists provided the $\varepsilon'_{p}$ equal $1$ for almost all $p$, $\prod_{p\le\infty}\varepsilon'_{p}=1$, and $\varepsilon'_{\infty}$, $d'$ match the signature \cite[Chap.~IV, Prop.~7]{Ser73}, with no further condition when $k\ge3$. Take $\varepsilon'_{p}=\varepsilon_{p}(V)\varepsilon_{p}(W)(d(W),d')_{p}$, the value forced by the formula $\varepsilon_{p}(W\oplus W')=\varepsilon_{p}(W)\varepsilon_{p}(W')(d(W),d(W'))_{p}$. The product formula for these holds because it holds for $V$, for $W$ and for the Hilbert symbol. Then $W\oplus W'$ and $V$ have the same rank, signature, discriminant and Hasse invariants at every place, so they are isometric by the Hasse--Minkowski theorem \cite[Chap.~IV, Thm.~9]{Ser73}, and $W$ embeds in $V$. \end{proof} \begin{theorem}[Two-dimensional families with algebraic real multiplication by $\QQ(\sqrt6)$]\label{thm:sqrt6} Let $S$ be a projective K3 surface with $\QQ(\sqrt6)\subseteq E$ whose transcendental space $T=T(S)_{\QQ}$ embeds isometrically in $V_{3}$. \begin{enumerate}[label=\textup{(\roman*)},leftmargin=2.4em,itemsep=2pt] \item $\sqrt6\in E$ is algebraic on $S\times S$. \item The Hodge conjecture, and \Fiii{}, hold for $S\times S$ and for the Hilbert square $S^{[2]}$. \item Such surfaces exist in two-dimensional families, whose very general member has $\mathrm{rk}\,T=8$ and $E=\QQ(\sqrt6)$ exactly. One such family is the one whose transcendental space is $E^{4}$ with the trace form of $q_{E}=\langle-2,-2,-2+\sqrt6,-1+\sqrt6\rangle$ over $E=\QQ(\sqrt6)$. \end{enumerate} \end{theorem} \begin{proof} (i) By \cite[Prop.~2.11]{Var23} $S$ is Hodge isometric to a K3 surface $S'$ with a symplectic automorphism $\sigma$ of order three, and by Lemma~\ref{lem:isometry} the isometry is algebraic, so we may replace $S$ by $S'$. Let $Y$ be the minimal resolution of $S'/\sigma$ and $f_{1}\colon S'\dashrightarrow Y$ the quotient map, of degree three, so that $T(Y)_{\QQ}\cong T(1/3)$, of rank $\mathrm{rk}\,T\le8$ and signature $(2,\mathrm{rk}\,T-2)$: the rank is at most $8$ because $T$ embeds in $V_{3}$ of rank $10$ and signature $(3,7)$ with a complement that must contain the remaining positive direction, so the complement has rank at least two. Now $T(Y)_{\QQ}$ has rank at most $8$ and signature dominated by $(3,11)$, and $14-8\ge3$, so by Lemma~\ref{lem:embed} it embeds in $V_{2}$. By \cite[Prop.~2.5]{Var23} $Y$ is Hodge isometric to a K3 surface $Y'$ with a symplectic involution $\iota$; let $Z$ be the resolution of $Y'/\iota$ and $f_{2}\colon Y'\dashrightarrow Z$ the quotient, of degree two. The chain $S'\dashrightarrow Y\cong Y'\dashrightarrow Z$ has multiplier $3\cdot1\cdot2=6$, and Theorem~\ref{thm:chain} applies to $e=\sqrt6$ (Fig.~\ref{fig:chain}). (ii) By the first paragraph of this section it remains to make $E$ algebraic. $E$ contains $\QQ(\sqrt6)$, and either $E=\QQ(\sqrt6)$ or $E$ is a CM field: a totally real $E$ has $\dim_{E}T\ge3$, because the complex line $H^{2,0}$ lies in one eigenspace $T_{\sigma}\otimes\CC$ and $\overline{H^{2,0}}$ in the same one, so that $T_{\sigma}\otimes\RR$ has signature $(2,\ast)$ and $\dim_{E}T\ge3$ follows from $q$ being negative definite on the other eigenspaces together with $T\otimes\RR$ having exactly two positive directions; and $[E:\QQ]\cdot\dim_{E}T=\mathrm{rk}\,T\le8$ then forces $[E:\QQ]\le2$. If $E=\QQ(\sqrt6)$, (i) suffices. If $E$ is CM, it is spanned by its norm-one elements, Hodge isometries, algebraic by Buskin's theorem. So every Hodge class on $S\times S$ is algebraic, and \Fiii{} follows. For $S^{[2]}$ the Chow motive is a direct summand of twists of $h(S\times S)$, $h(S)$ and $\mathbf{1}$ \cite{dCM02}, so every Hodge class on $S^{[2]}$ is the image under an algebraic correspondence of a Hodge class on $S\times S$ or $S$, and both statements follow. (iii) Let $E=\QQ(\sqrt6)$ with real embeddings $\sigma_{1}$, $\sigma_{2}$ ($\sqrt6\mapsto\pm\sqrt6$), and let $q_{E}=\langle a_{1},a_{2},a_{3},a_{4}\rangle$ with $a_{1},a_{2}$ negative at both places and $a_{3},a_{4}$ positive at $\sigma_{1}$, negative at $\sigma_{2}$. The $\QQ$-space $T=E^{4}$ with the form $\Tr_{E/\QQ}\circ q_{E}$ has rank $8$ and signature $(2,2)+(0,4)=(2,6)$, and multiplication by $\sqrt6$ is self-adjoint of square $6$. For $q_{E}=\langle-2,-2,-2+\sqrt6,-1+\sqrt6\rangle$ one computes $d(T)=10$, $\varepsilon_{2}=1$, $\varepsilon_{3}=-1$, $\varepsilon_{5}=1$ and $\varepsilon_{p}=1$ otherwise, while $V_{3}$ has $d=-1$, $\varepsilon_{2}=1$, $\varepsilon_{3}=-1$, $\varepsilon_{5}=1$; and $T\oplus\langle1,-10\rangle$ has the invariants of $V_{3}$ at every place, so $V_{3}\cong T\oplus\langle1,-10\rangle$ by Hasse--Minkowski, with a complement containing a positive vector (Table~\ref{tab:invariants}; the Hilbert symbols are recomputed in the programs of Appendix~\ref{app:checks}). Embed $V_{3}$ in $\Lambda_{\QQ}$ as the invariant lattice of an order-three symplectic action, with orthogonal complement $K_{12}(-2)$, and let $N\subseteq\Lambda$ be the primitive sublattice with $N_{\QQ}=T^{\perp}$; it contains $K_{12}(-2)$ primitively and a positive class. The Hodge structures of K3 type on $T$ for which $\sqrt6$ is a Hodge endomorphism are the periods $\omega\in\PP(T_{\sigma_{1}}\otimes\CC)$ with $q(\omega)=0$, $q(\omega,\bar\omega)>0$, where $T_{\sigma_{1}}$ is the $\sigma_{1}$-eigenspace, of signature $(2,2)$; this is a domain of dimension two \cite[Prop.~2.3]{Var23}. For each such $\omega$ the surjectivity of the period map gives a K3 surface $S_{\omega}$ with $T(S_{\omega})_{\QQ}\subseteq T$, projective because $N$ contains a positive class, carrying an order-three symplectic automorphism because $K_{12}(-2)\subseteq N$ primitively. For $\omega$ outside a countable union of proper closed analytic subsets, $T(S_{\omega})_{\QQ}=T$ and $E=\QQ(\sqrt6)$, by the argument of \cite[Thm.~22.9]{BMB26}, which needs only $\dim_{E}T\ge3$. \end{proof} \begin{table}[t] \centering \caption{Rational invariants used in Theorem~\ref{thm:sqrt6}(iii). Hasse invariants are $+1$ at every prime not shown.} \label{tab:invariants} \begin{ruledtabular} \begin{tabular}{lcccccc} space & rank & signature & $d$ & $\varepsilon_{2}$ & $\varepsilon_{3}$ & $\varepsilon_{5}$ \\ \colrule $T=\Tr_{E/\QQ}q_{E}$ & 8 & $(2,6)$ & $10$ & $+1$ & $-1$ & $+1$ \\ $\langle1,-10\rangle$ & 2 & $(1,1)$ & $-10$ & $+1$ & $+1$ & $+1$ \\ $T\oplus\langle1,-10\rangle$ & 10 & $(3,7)$ & $-1$ & $+1$ & $-1$ & $+1$ \\ $V_{3}$ & 10 & $(3,7)$ & $-1$ & $+1$ & $-1$ & $+1$ \\ $V_{2}$ & 14 & $(3,11)$ & $-1$ & $-1$ & $+1$ & $+1$ \\ \end{tabular} \end{ruledtabular} \end{table} \begin{figure*}[t] \centering \includegraphics[width=0.84\textwidth]{figures/fig_chain} \caption{The chain of Theorem~\ref{thm:sqrt6}: the quotient of $S'$ by a symplectic automorphism of order three, a Hodge isometry (algebraic by Buskin), and the quotient of $Y'$ by a Nikulin involution. The multipliers compose to $6$.} \label{fig:chain} \end{figure*} In Varesco's families $E$ is $\QQ(\sqrt2)$ or $\QQ(\sqrt3)$ \cite[Thms.~2.9, 2.15]{Var23}; van Geemen and Sch\"utt obtain $\sqrt2$, $\sqrt3$, $\sqrt7$ from rational self-maps of elliptic K3 surfaces \cite[\S\S6--7]{vGS25}; Bayer-Fluckiger, van Geemen and Sch\"utt prove that families with real multiplication by any totally real $E$ with $[E:\QQ]\dim_{E}T\le21$ exist, without algebraicity \cite{BvGS24}. The lattice condition of Theorem~\ref{thm:sqrt6} is a genuine restriction: among the $165$ sign-admissible diagonal forms $q_{E}$ with small coefficients, $98$ embed in $V_{3}$ and $67$ do not (Appendix~\ref{app:checks}). \subsection{Isogenies of elliptic surfaces}\label{ssec:fricke} Varesco's quotients and the chains of Theorem~\ref{thm:chain} produce the square roots of the orders of symplectic automorphisms, $2$ and $3$, and their products. A different source of similarities is a rational self-map of a K3 surface, and the self-maps that exist in families come from isogenies of elliptic fibrations: van Geemen and Sch\"utt obtain $\sqrt2$ and $\sqrt3$ from isogenies of degree $2$ and $3$ in this way, and for degrees $5$ and $7$ they give one surface each \cite[\S\S6--7]{vGS25}. We show that the degrees $5$ and $7$ come in one-dimensional families, that in the families the isogeny is a complex multiplication by $\QQ(\sqrt{-5})$ or $\QQ(\sqrt{-7})$ and not a real one, the sign of its square being decided by how the twisting function behaves under the Fricke involution, and that the families satisfy the Hodge conjecture on all their powers. Let $N\in\{5,7\}$ and write $N^{k}$ for $125$ when $N=5$ and $49$ when $N=7$. The modular curve $X_{0}(N)$ is rational, with Fricke's coordinate $t$ in which the involution $(E,C)\mapsto(E/C,E[N]/C)$ is $w(t)=N^{k}/t$ \cite[\S7.6]{vGS25}.\footnote{The curve $X_{0}(N)$ parametrises elliptic curves with a cyclic subgroup of order $N$; it has genus zero exactly for $N\le10$ and $N\in\{12,13,16,18,25\}$, and for $N=5$ and $N=7$ a coordinate is the Hauptmodul $t=\bigl(\eta(\tau)/\eta(N\tau)\bigr)^{k}$ with $k=12/(N-1)$, so $k=3$ for $N=5$ and $k=2$ for $N=7$; this is why $N^{k}$ is $125$ in one case and $49$ in the other. The Fricke involution sends $\tau$ to $-1/N\tau$, which exchanges the two cusps $0$ and $\infty$ of $X_{0}(N)$ and, on the moduli interpretation, replaces $(E,C)$ by the quotient $E/C$ with the image of $E[N]$ as its distinguished subgroup. Applying it twice gives $E/E[N]\cong E$, so it is an involution, and in the coordinate $t$ it reads $t\mapsto N^{k}/t$ because $\eta(-1/\tau)=\sqrt{-i\tau}\,\eta(\tau)$ turns the quotient of eta functions into its reciprocal up to the constant $N^{k/2}$, squared to $N^{k}$ by the exponent.} The universal curve with a cyclic subgroup $C_{t}$ of order $N$ over $\PP^{1}_{t}$ is \begin{equation}\label{eq:fricke} \mathcal{E}_{t}\colon\ y^{2}=x^{3}-27c_{4}(t)\,x-54c_{6}(t), \end{equation} with, for $N=7$, \begin{align*} c_{4}&=(t^{2}+13t+49)(t^{2}+5t+1),\\ c_{6}&=(t^{2}+13t+49)(t^{4}+14t^{3}+63t^{2}+70t-7), \end{align*} and for $N=5$, \begin{align*} c_{4}&=(t^{2}+10t+5)(t^{2}+22t+125),\\ c_{6}&=(t^{2}+4t-1)(t^{2}+22t+125)^{2}. \end{align*} The discriminant is $1728\,t\,(t^{2}+13t+49)^{2}$ for $N=7$ and $1728\,t\,(t^{2}+22t+125)^{3}$ for $N=5$, of degree $5$ and $7$. So the relatively minimal model $R_{N}$ of \eqref{eq:fricke} over $\PP^{1}_{t}$ is a rational elliptic surface with fibres of type $\mathrm{I}_{1}$ at $t=0$, $\mathrm{I}_{N}$ at $t=\infty$, and at the two roots $e_{1},e_{2}$ of the quadratic factor, where $c_{4}$ vanishes, of type $\mathrm{II}$ for $N=7$ and $\mathrm{III}$ for $N=5$; the Euler numbers add up to $1+N+2\cdot2=12$ and $1+N+2\cdot3=12$. The roots satisfy $e_{1}e_{2}=N^{k}$, so $w$ exchanges them and exchanges $0$ with $\infty$. \begin{lemma}[The Fricke isogeny over the function field]\label{lem:fricke} $\mathcal{E}_{t}/C_{t}\cong\mathcal{E}_{w(t)}$ over $\CC(t)$. \end{lemma} \begin{proof} The two curves have the same $j$-invariant, because $w$ is the Fricke involution of $X_{0}(N)$. Two elliptic curves over $\CC(t)$ with the same non-constant $j$-invariant differ by a quadratic twist $\chi\in\CC(t)^{\times}/\CC(t)^{\times2}$, and a twist ramified at a point of $\PP^{1}$ changes the Kodaira type of the fibre there, exchanging $\mathrm{I}_{n}$ with $\mathrm{I}_{n}^{*}$, $\mathrm{II}$ with $\mathrm{IV}^{*}$ and $\mathrm{III}$ with $\mathrm{III}^{*}$. So it suffices to show that $\mathcal{E}_{t}/C_{t}$ has the same fibre types as $\mathcal{E}_{w(t)}$, which are $\mathrm{I}_{N}$ at $0$, $\mathrm{I}_{1}$ at $\infty$ and $\mathrm{II}$ or $\mathrm{III}$ at $e_{1},e_{2}$: then $\chi$ is unramified on $\PP^{1}$, hence constant, hence trivial over $\CC$. At $t=0$ the fibre of $\mathcal{E}$ is $\mathrm{I}_{1}$, the Tate curve $\mathbb{G}_{m}/q^{\ZZ}$ over $\CC((t))$, whose only cyclic subgroup of order $N$ rational over $\CC((t))$ is $\mu_{N}$; the quotient $\mathbb{G}_{m}/q^{N\ZZ}$ is $\mathrm{I}_{N}$. At $e_{i}$ the local monodromy $M\in\SL_{2}(\ZZ)$ of $\mathcal{E}$ has order $6$ or $4$ and irreducible characteristic polynomial; the isogeny is given on the period lattices by a matrix $A\in M_{2}(\ZZ)$ with $\det A=N$, and the monodromy of the quotient is $M'=AMA^{-1}$. The $\SL_{2}(\ZZ)$-classes with this characteristic polynomial are those of $M$ and $M^{-1}$, which are the types $\mathrm{II}$ and $\mathrm{II}^{*}$, or $\mathrm{III}$ and $\mathrm{III}^{*}$. If $M'=gM^{-1}g^{-1}$ with $g\in\SL_{2}(\ZZ)$, then $B=g^{-1}A$ conjugates $M$ to $M^{-1}$. All rational matrices doing so form the coset $B_{0}\,\QQ(M)^{\times}$, where $B_{0}=\bigl(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\bigr)$ for $M=\bigl(\begin{smallmatrix}0&-1\\1&\mathrm{tr}M\end{smallmatrix}\bigr)$ has determinant $-1$ and the elements of the imaginary quadratic field $\QQ(M)$ have positive norm; so $\det B<0$, against $\det B=N$. Hence the quotient has the type of $M$ at $e_{1},e_{2}$. At $\infty$ the $j$-invariant of the quotient has a simple pole, so the fibre is $\mathrm{I}_{1}$ or $\mathrm{I}_{1}^{*}$, and the Euler numbers of the other fibres add up to $11$; since the total is $12$ the fibre is $\mathrm{I}_{1}$. \end{proof} For $p\ne q$ in $\PP^{1}_{t}$ let $u_{p,q}=(t-p)(t-q)$, and let $X_{p,q}$ be the elliptic K3 surface whose generic fibre is the quadratic twist $\mathcal{E}_{t}\otimes u_{p,q}$: the twist multiplies $c_{4}$ by $u^{2}$ and $c_{6}$ by $u^{3}$, giving Weierstrass coefficients of degrees $8$ and $12$, and adds two fibres of type $\mathrm{I}_{0}^{*}$ at $p$ and $q$ when these are not among $0,\infty,e_{1},e_{2}$, by Tate's algorithm; the same bookkeeping of Weierstrass coefficients, fibre types and their resolutions in higher dimension is carried out in \cite{Bha26b}. For a double cover $\psi\colon\PP^{1}_{s}\to\PP^{1}_{t}$ let $X_{\psi}$ be the relatively minimal model of $\psi^{*}\mathcal{E}$. \begin{lemma}\label{lem:twists} \leavevmode \begin{enumerate}[label=\textup{(\roman*)},leftmargin=2.4em,itemsep=2pt] \item $X_{p,q}$ has Picard number at least $16$: its Néron--Severi lattice contains $U\oplus A_{6}\oplus D_{4}^{2}$ for $N=7$ and $U\oplus A_{4}\oplus A_{1}^{2}\oplus D_{4}^{2}$ for $N=5$. The same bound holds for $X_{\psi}$, whose fibres $2\,\mathrm{I}_{1}+2\,\mathrm{I}_{N}$, together with four fibres of type $\mathrm{III}$ when $N=5$, and the pulled-back Mordell--Weil group of $R_{N}$, of rank $2$, give rank at least $16$ by the Shioda--Tate formula. \item The fibrewise isogeny by $C_{t}$ is a dominant rational map of degree $N$ \[ f\colon X_{p,q}\dashrightarrow X_{w(p),w(q)},\qquad f\colon X_{\psi}\dashrightarrow X_{w\circ\psi}. \] \item The deck involution of $\psi$ times $[-1]$ on the fibres is a symplectic involution $\iota$ of $X_{\psi}$, and the minimal resolution of $X_{\psi}/\iota$ is $X_{p,q}$ with $\{p,q\}$ the branch locus of $\psi$. Hence $T(X_{\psi})_{\QQ}\cong T(X_{p,q})_{\QQ}(1/2)$ as in Lemma~\ref{lem:dominant}, and the two surfaces have the same Picard number. \end{enumerate} \end{lemma} \begin{proof} (i) Shioda--Tate with the fibres listed, and for $X_{\psi}$ the Mordell--Weil rank $8-6=2$ of $R_{N}$. (ii) By Lemma~\ref{lem:fricke} the quotient of the generic fibre is $\mathcal{E}_{w(t)}\otimes u_{p,q}(t)$, and \begin{widetext} \begin{equation}\label{eq:frickeu} u_{p,q}(w(t))=\Bigl(\frac{N^{k}}{t}-p\Bigr)\Bigl(\frac{N^{k}}{t}-q\Bigr)=\frac{pq}{t^{2}}\,u_{w(p),w(q)}(t), \end{equation} \end{widetext} a constant times a square times $u_{w(p),w(q)}$. So $\mathcal{E}_{w(t)}\otimes u_{p,q}(t)\cong\mathcal{E}_{w(t)}\otimes u_{w(p),w(q)}(w(t))$, which is the pull-back along the automorphism $w$ of the base of the generic fibre of $X_{w(p),w(q)}$; pulling back along an automorphism of the base gives an isomorphic surface. For $X_{\psi}$, $(\psi^{*}\mathcal{E})/C=\psi^{*}(\mathcal{E}/C)\cong\psi^{*}w^{*}\mathcal{E}$. (iii) The quadratic twist by the function defining the double cover is the quotient of the base change by the deck involution composed with $[-1]$ on the fibres; the involution is symplectic with eight fixed points, the $2$-torsion of the two fibres over the ramification points, and the quotient's resolution is the twist. \end{proof} \begin{lemma}[The sign of the Fricke isogeny]\label{lem:frickesign} Let $\varphi_{t}\colon\mathcal{E}_{t}\to\mathcal{E}_{w(t)}$ be an isogeny over $\CC(t)$ with kernel $C_{t}$, which exists by Lemma~\ref{lem:fricke} and is unique up to sign, and let $\varphi_{t}^{*}(dx/y)=\lambda(t)\,dx/y$ in the models \eqref{eq:fricke}. \begin{enumerate}[label=\textup{(\roman*)},leftmargin=2.4em,itemsep=2pt] \item $\varphi_{w(t)}\circ\varphi_{t}=[-N]$ on $\mathcal{E}_{t}$, and $\lambda(t)\lambda(w(t))=-N$. \item Let $u\in\CC(t)$ satisfy $u(w(t))=c(t)^{2}u(t)$ with $c\in\CC(t)$, let $X$ be the elliptic K3 surface with generic fibre $\mathcal{E}_{t}\otimes u$, and let $f\colon X\dashrightarrow X$ be the fibrewise isogeny by $C_{t}$ followed by the isomorphism $(x,y)\mapsto(c^{2}x,c^{3}y)$ from $\mathcal{E}_{w(t)}\otimes u(t)$ to $\mathcal{E}_{w(t)}\otimes u(w(t))$, the fibre of $X$ over $w(t)$. Then $f^{*}$ acts on $H^{2,0}(X)$ by a constant $\Lambda$ with \[ \Lambda^{2}=-\frac{N}{c(t)\,c(w(t))}. \] For $u=(t-p)(t-N^{k}/p)$ one has $c(t)c(w(t))=1$ and $\Lambda^{2}=-N$; for $u=t^{2}-N^{k}$ one has $c(t)c(w(t))=-1$ and $\Lambda^{2}=N$. \end{enumerate} \end{lemma} \begin{proof} (i) The composite is an endomorphism of $\mathcal{E}_{t}$ over $\CC(t)$ of degree $N^{2}$. Since $j$ is not constant, $\End_{\CC(t)}(\mathcal{E}_{t})=\ZZ$, so the composite is $[N]$ or $[-N]$, and changing $\varphi$ to $-\varphi$ does not change it. We read off the sign on the upper half plane $\mathfrak{H}$. Let $\tau\mapsto t(\tau)$ be the quotient map $\mathfrak{H}\to X_{0}(N)$. By the moduli interpretation of $X_{0}(N)$ the pair $(\mathcal{E}_{t(\tau)},C_{t(\tau)})$ is $(\CC/(\ZZ\tau+\ZZ),\langle1/N\rangle)$, analytically in $\tau$, and the Fricke involution lifts to $\tau\mapsto w(\tau)=-1/N\tau$. The map $f_{\tau}(z)=z/\tau$ carries $\ZZ\tau+\ZZ$ into $\ZZ w(\tau)+\ZZ=\ZZ/N\tau+\ZZ$, so it is an isogeny $\CC/(\ZZ\tau+\ZZ)\to\CC/(\ZZ w(\tau)+\ZZ)$, and its kernel $\tau(\ZZ w(\tau)+\ZZ)/(\ZZ\tau+\ZZ)=(\ZZ/N+\ZZ\tau)/(\ZZ\tau+\ZZ)$ is $\langle1/N\rangle$. Its composite with the map of the same form at $w(\tau)$ is $f_{w(\tau)}(f_{\tau}(z))=z/\tau w(\tau)=-Nz$. On the set $\mathfrak{H}'$ of $\tau$ with $j(\tau)\notin\{0,1728\}$ the automorphisms of $\mathcal{E}_{\tau}$ are $\pm1$, so $\varphi_{t(\tau)}=\varepsilon(\tau)f_{\tau}$ with $\varepsilon(\tau)\in\{\pm1\}$; $\varepsilon$ is continuous because $\varphi$ is, hence constant on $\mathfrak{H}'$, which is $\mathfrak{H}$ minus a discrete set and connected. So $\varphi_{w(t)}\circ\varphi_{t}=\varepsilon^{2}f_{w(\tau)}\circ f_{\tau}=[-N]$ on $\mathfrak{H}'$, hence identically. Pulling back $dx/y$ along the composite gives $\lambda(t)\lambda(w(t))\,dx/y$ on one side and $-N\,dx/y$ on the other. (ii) Over $\CC(t)(\sqrt u)$ the twist by $u$ is the same change of coordinates $(x,y)\mapsto(ux,u^{3/2}y)$ on $\mathcal{E}_{t}$ and on $\mathcal{E}_{w(t)}$, which multiplies $dx/y$ by $u^{-1/2}$ on both, so the twisted isogeny pulls $dx/y$ back to $\lambda(t)\,dx/y$. The isomorphism $(x,y)\mapsto(c^{2}x,c^{3}y)$ pulls $dx/y$ back to $c(t)^{-1}dx/y$, and the holomorphic two-form of $X$, which is $dx/y\wedge dt$ on the Weierstrass model, restricts to the fibre over $w(t)$ as $dx/y\wedge d(w(t))=w'(t)\,dx/y\wedge dt$. Hence $f^{*}(dx/y\wedge dt)=\Lambda(t)\,dx/y\wedge dt$ with $\Lambda(t)=w'(t)\lambda(t)/c(t)$. The pull-back of a holomorphic two-form along a dominant rational map of K3 surfaces is holomorphic off the finitely many points of indeterminacy, hence everywhere, so $f^{*}$ preserves the line $H^{2,0}(X)$ and $\Lambda$ is a constant. Then $\Lambda^{2}=\Lambda(t)\Lambda(w(t))=w'(t)w'(w(t))\,\lambda(t)\lambda(w(t))/c(t)c(w(t))$, where $w'(t)w'(w(t))=(w\circ w)'(t)=1$ and $\lambda(t)\lambda(w(t))=-N$ by (i). For $u=(t-p)(t-N^{k}/p)$ equation \eqref{eq:frickeu} with $q=N^{k}/p$ gives $u(w(t))=(N^{k}/t^{2})\,u(t)$, so $c(t)=N^{k/2}/t$ for a fixed square root $N^{k/2}$, and $c(t)c(w(t))=N^{k}/t\,w(t)=1$. For $u=t^{2}-N^{k}$, $u(w(t))=N^{2k}/t^{2}-N^{k}=-(N^{k}/t^{2})\,u(t)$, so $c(t)=iN^{k/2}/t$ and $c(t)c(w(t))=-1$. \end{proof} The archive \cite{BB26code} computes $\lambda$ itself: the kernel polynomial of $\varphi_{t}$ is the factor of degree $(N-1)/2$ of the $N$-th division polynomial over $\ZZ[t,x]$, V\'elu's formulas give the quotient curve, and the isomorphism to \eqref{eq:fricke} at $w(t)$ has $\lambda(t)^{2}=-t^{2}/25$ for $N=5$ and $-t^{2}/7$ for $N=7$, so that $\lambda(t)^{2}\lambda(w(t))^{2}=N^{2}$ and $\Lambda^{2}=-N$ on the pair twists directly; the program \path{code/fricke_sign.py} carries this out, and the Lean file checks the polynomial identity it reduces to. \begin{theorem}[Fricke families]\label{thm:fricke} Let $N\in\{5,7\}$ and $K=\QQ(\sqrt{-N})$. Let $\mathfrak{A}_{N}$ be the family of surfaces $X_{p}=X_{p,N^{k}/p}$ for $p\notin\{0,\infty,e_{1},e_{2},\pm\sqrt{N^{k}}\}$, and $\mathfrak{B}_{N}$ the family of surfaces $X_{\psi}$ with $\psi$ branched over a pair $\{b,N^{k}/b\}$. \begin{enumerate}[label=\textup{(\roman*)},leftmargin=2.4em,itemsep=2pt] \item Every member $X$ has a dominant rational self-map $f$ of degree $N$ with $f^{*}\in E=\End_{\Hdg}T(X)_{\QQ}$, $f^{*}f_{*}=N$, $(f^{*})^{2}=-N$ and $f_{*}=-f^{*}$. So $K\subseteq E$, with $\sqrt{-N}=\pm f^{*}$ algebraic on $X\times X$, and $E$ is a CM field. \item The families are not isotrivial, and every member has $\mathrm{rk}\,T(X)\in\{2,4,6\}$. At all but countably many members $E=K$ and the Hodge group of $T(X)$ is the unitary group of the $K$-hermitian form $h$ on $T(X)$ with $\Tr_{K/\QQ}h=q_{X}$; at the others $T(X)$ is of CM type. \item For every member $X$ and every $n\ge1$ the Hodge conjecture holds for $X^{n}$ and for the Hilbert scheme $X^{[n]}$, and \Fiii{} holds there. \end{enumerate} \end{theorem} Van Geemen and Sch\"utt state that the Fricke parametrisations give one-dimensional families with complex multiplication by $\QQ(\sqrt{-5})$ and $\QQ(\sqrt{-7})$, and give the details for the two surfaces at the fixed points of $w$ only \cite[\S7.6]{vGS25}; (i) and (ii) make their statement explicit, and (iii) is what the complex multiplication gives. \begin{proof} (i) $w$ exchanges $p$ and $N^{k}/p$, so Lemma~\ref{lem:twists}(ii) gives $f\colon X_{p}\dashrightarrow X_{p}$, the map of Lemma~\ref{lem:frickesign}(ii) for $u=(t-p)(t-N^{k}/p)$; for $\mathfrak{B}_{N}$, $w\circ\psi$ has the same branch locus as $\psi$, so $w\circ\psi=\psi\circ\sigma$ for a M\"obius transformation $\sigma$ and $X_{w\circ\psi}\cong X_{\psi}$. The correspondence of a dominant rational map preserves $\NS$ and its transpose does too, so it preserves $T=\NS^{\perp}$ and acts there as a morphism of Hodge structures, nonzero since $f^{*}\omega=\Lambda\omega$ with $\Lambda\ne0$; on a resolution $S$ of the graph with $p_{1}\colon S\to X$ birational and $p_{2}$ of degree $N$, both $p_{1}^{*}T$ and $p_{2}^{*}T$ are irreducible sub-Hodge structures of $H^{2}(S)$ containing its unique $(2,0)$-line, so they coincide, and $f_{*}f^{*}=p_{2*}p_{1}^{*}p_{1*}p_{2}^{*}=N$ on $T$. The field $E$ acts faithfully on the line $H^{2,0}(X)$, and $f^{*}$ acts there by $\Lambda$ with $\Lambda^{2}=-N$ by Lemma~\ref{lem:frickesign}(ii); so $(f^{*})^{2}=-N$ in $E$, and $f_{*}=N(f^{*})^{-1}=-f^{*}$. For $X_{\psi}$ the generic fibre is $\mathcal{E}_{\psi(s)}$ over $\CC(s)$, $f$ is the fibrewise isogeny over $\sigma$, and $\sigma$ is an involution: in the coordinate $s$ with $s^{2}=(t-b)/(t-N^{k}/b)$ one computes $(w(t)-b)/(w(t)-N^{k}/b)=b^{2}/N^{k}s^{2}$, so $\sigma(s)=\kappa/s$ with $\kappa^{2}=b^{2}/N^{k}$; the argument of Lemma~\ref{lem:frickesign}(ii) with $c=1$ and $\sigma$ in place of $w$ gives $\Lambda^{2}=\lambda(t)\lambda(w(t))\,\sigma'(s)\sigma'(\sigma(s))=-N$. The adjoint of $f^{*}$ for $q_{X}$ is $f_{*}=-f^{*}$, so the involution of $E$ induced by $q_{X}$ is not the identity; by Zarhin's theorem $E$ is totally real or a CM field \cite[Thm.~1.5.1]{Zar83}, so it is a CM field, and $K=\QQ+\QQ f^{*}\subseteq E$. The class of the graph of $f$ is algebraic, its action on $H^{2}(X)$ is $f^{*}$ on $T$ and an endomorphism of $\NS(X)_{\QQ}$, which is a sum of algebraic classes of $\NS\otimes\NS$; so $f^{*}$, extended by zero on $\NS$, is an algebraic class on $X\times X$. (ii) The elliptic fibration of $X_{p}$ has labelled singular fibres $\mathrm{I}_{1}$ at $0$, $\mathrm{I}_{N}$ at $\infty$, $\mathrm{II}$ or $\mathrm{III}$ at $e_{1},e_{2}$ and $\mathrm{I}_{0}^{*}$ at $p$ and $N^{k}/p$. An isomorphism of elliptic fibrations induces a M\"obius transformation preserving this configuration; the transformations preserving $\{0,\infty\}$ and $\{e_{1},e_{2}\}$ are $t\mapsto\lambda t$ and $t\mapsto\lambda/t$ with $\lambda e_{1}\in\{e_{1},e_{2}\}$, and since $e_{1}+e_{2}\ne0$ and $e_{1}^{2}\ne e_{2}^{2}$ they are the identity and $w$, which preserve $\{p,N^{k}/p\}$. So $X_{p}$ and $X_{p'}$ have isomorphic fibrations only when $p'\in\{p,N^{k}/p\}$. A K3 surface has finitely many elliptic fibrations up to automorphisms \cite{Ste85}, so the map $p\mapsto[X_{p}]$ to the moduli space has finite fibres and the family is not isotrivial; by local Torelli its period map has uncountable image. By (i) $T(X)_{\QQ}$ is a $K$-vector space, so its rank is even, and it is at most $6$ by Lemma~\ref{lem:twists}(i). It is also an $E$-vector space, so $[E:\QQ]$ divides $\mathrm{rk}\,T(X)$; if $E\ne K$ then $[E:\QQ]\in\{4,6\}$, $\dim_{E}T(X)=1$ and $T(X)$ is of CM type. A Hodge structure of CM type on a sublattice of $M^{\perp}$ is determined by its endomorphism field, a subfield of the endomorphism algebra of the sublattice, and by an eigenline of that field, so the periods of such members form a countable subset of the period domain of $M^{\perp}$; the period map of the family is a non-constant holomorphic map from a connected curve, its fibres over a countable set are countable, so all but countably many members have $E=K$. There the Hodge group of $T(X)$ is the unitary group of the $K$-hermitian form $h$ with $\Tr_{K/\QQ}h=q_{X}$ by Zarhin's theorem for CM fields \cite[Thm.~2.3.1]{Zar83}. The same holds for $\mathfrak{B}_{N}$ by Lemma~\ref{lem:twists}(iii). (iii) By (i) $E$ is a CM field containing $K$. If $E=K$ then $E=\QQ+\QQ f^{*}$ is algebraic. Otherwise every element of $E$ of norm one to its maximal real subfield is a Hodge isometry of $T$, extended by the identity on $\NS$ to a Hodge isometry of $H^{2}(X,\QQ)$, algebraic by Buskin's theorem \cite{Bus19}. The algebraic classes form a $\QQ$-subalgebra $E'\subseteq E$ containing these elements; for $x\in E$ with $\bar x\ne x$ and $t_{1}\ne t_{2}$ in $\QQ$ it contains $(t_{i}+x)/(t_{i}+\bar x)$, hence their quotient $1+(t_{2}-t_{1})/(t_{1}+\bar x)$, hence $\bar x$; so $E'=E$. In both cases $\End_{\Hdg}H^{2}(X)=\End(\NS)\oplus E$ is algebraic, and $\Hdg^{2}(X\times X)=\End_{\Hdg}H^{2}(X)\oplus H^{0}\otimes H^{4}\oplus H^{4}\otimes H^{0}$ since $H^{1}=H^{3}=0$, so the Hodge conjecture holds for $X\times X$. For a K3 surface whose endomorphism field is a CM field the Hodge classes in the tensor algebra of $T$ are generated by those of $T\otimes T$, and the Hodge conjecture holds for every power \cite[Thm.~2.2, Cor.~2.3]{Var25}, \cite[Prop.~5.2, Thm.~5.4]{RM08}. It passes to the Hilbert schemes $X^{[n]}$, whose motives are direct summands of motives of powers of $X$ \cite{dCM02}, and \Fiii{} on a variety follows from the Hodge conjecture on it. \end{proof} \begin{figure*}[t] \centering \includegraphics[width=0.92\textwidth]{figures/fig_fricke} \caption{The family $\mathfrak{A}_{N}$ of Theorem~\ref{thm:fricke}. Over the Fricke coordinate $t$ of $X_{0}(N)$ the involution $w(t)=N^{k}/t$ exchanges the cusps $0$, $\infty$ and the elliptic points $e_{1}$, $e_{2}$; the twist $X_{p}$ at a pair $\{p,N^{k}/p\}$ exchanged by $w$ is carried to itself by the fibrewise isogeny $f$ of degree $N$, and $f^{*}$ is the complex multiplication by $\sqrt{-N}$ on the transcendental lattice.} \label{fig:fricke} \end{figure*} The surface of \cite[\S7.6]{vGS25} with real multiplication by $\sqrt7$ is the twist at $\{7,-7\}$, the two fixed points of $w$, and the surface of \cite[Prop.~7.7]{vGS25} is the twist at the fixed points $\pm5\sqrt5$ for $N=5$, which has Picard number $18$ and complex multiplication as well: there $u=t^{2}-N^{k}$, $c(t)c(w(t))=-1$ and $\Lambda^{2}=N$ by Lemma~\ref{lem:frickesign}(ii), so $f^{*}$ is the real multiplication by $\sqrt N$. The families $\mathfrak{A}_{N}$ are the twists at the pairs exchanged by $w$, where $c(t)c(w(t))=1$ and $\Lambda^{2}=-N$; the two signs are the two values of the constant by which the twisting function changes under $w$ once the square factor $N^{k}/t^{2}$ is removed. The chains of Theorem~\ref{thm:chain} apply to every member, and a member whose field $E$ contained $\sqrt2$, $\sqrt3$ or $\sqrt6$ would have that element algebraic as well; by Theorem~\ref{thm:fricke}(ii) such a member is of CM type, and we know none. \begin{remark}[The Picard number]\label{rem:frickepicard} Theorem~\ref{thm:fricke} does not say which of the three ranks the very general member has. The archive decides it by counting points. The surface $X_{1}$, the twist at $\{1,N^{k}\}$, has a Weierstrass model over $\ZZ$ with good reduction at $\ell=11$ for $N=7$ and at $\ell=7$ for $N=5$, primes split in $K$; the program \path{code/fricke_picard.py} counts its points over $\mathbb{F}_{\ell^{i}}$, $i\le3$, fibre by fibre, subtracts the trace of Frobenius on the trivial lattice, and completes the characteristic polynomial of Frobenius on the rank-six complement of the trivial lattice by the functional equation, whose sign is fixed by the nonzero middle coefficient. The polynomials are $x^{6}-726x^{3}+11^{6}$ for $N=7$ and $x^{6}+6x^{5}+56x^{4}+392x^{3}+2744x^{2}+14406x+7^{6}$ for $N=5$; in the variable $x/\ell$ neither has a root of unity among its roots, each is irreducible over $\QQ$ and splits over $K$ into a cubic and its conjugate, as the action of $f^{*}$ requires. By the Tate conjecture for K3 surfaces over finite fields, used in the same way in \cite[\S7]{vGS25}, $\rho(X_{1})=16$, so the very general member of $\mathfrak{A}_{N}$ and of $\mathfrak{B}_{N}$ has $\rho=16$, $\mathrm{rk}\,T=6$, $E=K$ with $\dim_{K}T=3$, and its Hodge group is the unitary group of a hermitian form of signature $(1,2)$ on $K^{3}$. At the primes inert in $K$, $\ell=5$ for $N=7$ and $\ell=11$ for $N=5$, the odd traces vanish, as they must when Frobenius exchanges the two eigenspaces of $f^{*}$. No statement of this section rests on this computation. \end{remark} \subsection{The reach of similarities} Every correspondence used so far acts on transcendental spaces as a Hodge similarity. The next proposition says that this is not an accident of the examples but a property of every correspondence whose components are small, and it measures what a correspondence must look like to act otherwise. \begin{proposition}[Structure of the transcendental action]\label{prop:structure} Let $X$, $X'$ be projective K3 surfaces, $\Gamma\in\mathrm{CH}^{2}(X\times X')_{\QQ}$, and $\gamma_{T}\colon T(X)_{\QQ}\to T(X')_{\QQ}$ the map induced on transcendental spaces by composing the action of $\Gamma$ with the orthogonal projections. \begin{enumerate}[label=\textup{(\roman*)},leftmargin=2.4em,itemsep=2pt] \item $\gamma_{T}$ is zero or an isomorphism of Hodge structures. \item If $\gamma_{T}\ne0$ then $a=\gamma_{T}^{\dagger}\gamma_{T}$ lies in the maximal totally real subfield of $E=\End_{\Hdg}T(X)_{\QQ}$ and is totally positive, and $\gamma_{T}$ is a similarity exactly when $a\in\QQ$. \item Let $\Gamma$ be irreducible, $S$ a resolution of its normalisation with projections $p_{1},p_{2}$ of degrees $d_{1},d_{2}$ to $X$, $X'$. If $p_{1}^{*}T(X)_{\QQ}=p_{2}^{*}T(X')_{\QQ}$ inside $H^{2}(S,\QQ)$, then $\gamma_{T}$ is a similarity of multiplier $d_{1}d_{2}$. This holds whenever $p_{g}(S)\le1$. \item If $\Gamma$ is irreducible and $\gamma_{T}$ is not a similarity, then $p_{1}^{*}T(X)_{\QQ}\cap p_{2}^{*}T(X')_{\QQ}=0$, so $p_{g}(S)\ge2$ and the transcendental rank of $S$ is at least twice that of $X$. \item The $\QQ$-algebra generated by the actions on $T(X)_{\QQ}$ of all correspondences from $X$ to itself whose irreducible components either fail to dominate a factor or have $p_{g}\le1$ is the multiquadratic subfield $E_{2}\subseteq E$ generated by the square roots of the products $d_{1}d_{2}$ that occur. When $[E:\QQ]$ is odd it is $\QQ$. \end{enumerate} \end{proposition} \begin{proof} (i) $T(X)_{\QQ}$ is an irreducible Hodge structure \cite{Zar83}, and the image of $\gamma_{T}$ is a sub-Hodge structure of $H^{2}(X')$; if nonzero it contains the $(2,0)$-part, hence $T(X')_{\QQ}$, and $\gamma_{T}$ is injective by irreducibility. (ii) $a$ is a Hodge endomorphism of $T(X)$, so $a\in E$, and it is self-adjoint, so it lies in the fixed field of the adjoint involution, the maximal totally real subfield. The form $q'(\gamma x,\gamma y)=q(x,ay)$ is a polarisation of the Hodge structure $T(X)$ up to sign on each real eigenspace of $E$, and comparison with $q$ gives $\sigma(a)>0$ at every real place. A similarity of multiplier $\mu$ has $a=\mu$. (iii) Both $p_{i}^{*}T$ are irreducible sub-Hodge structures of $H^{2}(S)$ with a $(2,0)$-part; if $p_{g}(S)\le1$ they share the unique $(2,0)$-line, so they meet and coincide. When they coincide, $\gamma_{T}=p_{2*}p_{1}^{*}=d_{2}(p_{2}^{*})^{-1}p_{1}^{*}$ on $T$, so $p_{2}^{*}\gamma x=d_{2}\,p_{1}^{*}x$; since $q_{S}(p_{1}^{*}x,p_{1}^{*}y)=d_{1}q(x,y)$ and $q_{S}(p_{2}^{*}x',p_{2}^{*}y')=d_{2}q'(x',y')$, this gives $d_{2}\,q'(\gamma x,\gamma y)=q_{S}(p_{2}^{*}\gamma x,p_{2}^{*}\gamma y)=d_{2}^{2}\,q_{S}(p_{1}^{*}x,p_{1}^{*}y)=d_{1}d_{2}^{2}\,q(x,y)$, that is $q'(\gamma x,\gamma y)=d_{1}d_{2}\,q(x,y)$. (iv) If the two subspaces met they would coincide by irreducibility and (iii) would apply; so they are independent, their $(2,0)$-lines are distinct, and their sum has rank $2\,\mathrm{rk}\,T(X)$ inside the transcendental part of $H^{2}(S)$. (v) A component that does not dominate a factor acts through $\NS$ or through $H^{0}$ and $H^{4}$, hence by zero on $T$. The others act by similarities by (iii), a composite of similarities returning to $X$ is an element $\Phi\in E$ with $\Phi^{\dagger}\Phi\in\QQ$, and for totally real $E$ this is $\Phi^{2}\in\QQ$; such elements generate $E_{2}$. For a CM field $E$ the elements with $\Phi^{\dagger}\Phi\in\QQ$ are the $\QQ^{\times}$-multiples of the norm-one elements, which generate $E$, so the statement is about the totally real case, which is the one that matters for \Fiii{}. An odd-degree field has $E_{2}=\QQ$. \end{proof} The correspondences of Lemmas~\ref{lem:dominant} and \ref{lem:isometry}, the Fourier--Mukai kernels between a K3 surface and its moduli spaces of sheaves, the rational Hodge isometries of hyper-K\"ahler varieties of K3$^{[n]}$ type \cite{Mar24}, the Fricke isogenies of Theorem~\ref{thm:fricke} and all their composites fall under (iii) or (v). A correspondence that realises an element of a cubic field must therefore have a component whose resolution has $p_{g}\ge2$ and transcendental rank at least twice that of the surface. Such correspondences exist, and the only ones we know come from finite groups. \begin{proposition}[Hecke correspondences]\label{prop:hecke} Let $Z$ be a smooth projective surface with an action of a finite group $G$, $H\le G$ a subgroup, and $X$ a smooth projective surface with $p_{g}(X)=1$ birational to $Z/H$; write $\pi\colon Z\dashrightarrow X$. \begin{enumerate}[label=\textup{(\roman*)},leftmargin=2.4em,itemsep=2pt] \item For $g\in G$ the correspondence $\Gamma_{g}=(\pi\times\pi)_{*}[\text{graph of }g]$ acts on $H^{2}(X,\QQ)$ as the Hecke operator of the double coset $HgH$, and these operators span a $\QQ$-algebra isomorphic to a quotient of $\End_{G}\QQ[G/H]$ acting by Hodge endomorphisms. \item Exactly one simple factor of this algebra acts nontrivially on $H^{2,0}(X)$; it is a field, it preserves $T(X)_{\QQ}$, and it embeds into $E$ by algebraic correspondences. If $H^{2,0}(Z)^{G}=0$ this factor is not the one on which $G$ acts trivially. \item Every field obtained in this way is abelian over $\QQ$. In particular a totally real cubic field $F$ whose Galois closure has group $S_{3}$ is realised by no correspondence built from a finite group acting on a cover of $X$. \end{enumerate} \end{proposition} \begin{proof} (i) Let $\tilde Z\to Z$ be an equivariant resolution and $\pi\colon\tilde Z\to X$ the quotient by $H$ followed by the resolution; then $\pi_{*}\pi^{*}=|H|$ and $\pi_{*}h^{*}=\pi_{*}$ for $h\in H$, so $\Gamma_{g}$ acts as $\pi_{*}g^{*}\pi^{*}=(\pi^{*})^{-1}\circ\sum_{h\in HgH/H}h^{*}\circ\pi^{*}$ on $H^{2}(X,\QQ)=H^{2}(\tilde Z,\QQ)^{H}$, the Hecke operator of $HgH$. The Hecke operators of $(G,H)$ span $\End_{G}\QQ[G/H]$, a semisimple algebra, and the action on $H^{2}(\tilde Z,\QQ)^{H}$ factors through it. (ii) The idempotents of the simple factors are rational combinations of Hecke operators, so they are Hodge endomorphisms of $H^{2}(X)$; they sum to $1$ and act on the line $H^{2,0}(X)$ by $0$ or $1$, so exactly one of them, $e_{V}$, acts as $1$. A simple factor with nonzero image in the commutative ring $\End_{\Hdg}H^{2,0}(X)=\CC$ through a ring homomorphism to a field is commutative, hence a field $\mathcal{H}_{V}$. For $f\in\mathcal{H}_{V}$, $f(T(X)_{\QQ})$ is a sub-Hodge structure of the transcendental part of $H^{2}(X)$; if $f\omega\ne0$ it contains $\omega$, hence $T(X)_{\QQ}$, and if $f\omega=0$ it has no $(2,0)$-part and lies in $\NS\cap T=0$. So $\mathcal{H}_{V}\to\End_{\Hdg}T(X)_{\QQ}=E$ is a homomorphism from a field, injective. The factor on which $G$ acts trivially fixes $H^{2,0}(\tilde Z)^{G}=H^{2,0}(Z/G)$, which is zero by hypothesis, so it is not $\mathcal{H}_{V}$. (iii) The centres of the simple factors of $\End_{G}\QQ[G/H]$ are the character fields $\QQ(\chi_{i})$ of the irreducible constituents of $\QQ[G/H]$, generated by sums of roots of unity, so each lies in $\QQ(\zeta_{|G|})$ and is abelian; a commutative simple factor is its centre. A non-abelian cubic field is not a subfield of an abelian field. \end{proof} For the dihedral group $D_{n}$ with $n$ odd, $H=\langle\tau\rangle$ of order two and $Z$ the elliptic surfaces $\tilde{\mathcal{E}}_{a}$ of \cite[\S4.7]{vGS25}, the algebra $\End_{D_{n}}\QQ[D_{n}/H]$ is $\QQ\times\prod_{d\mid n,\,d>1}\QQ(\zeta_{d})^{+}$, and the factor acting on $\omega$ is $\QQ(\zeta_{n})^{+}$; this is the content of \cite[\S4.8]{vGS25}, where the cycle $\Gamma_{1}+\Gamma_{-1}$ induces $\zeta_{n}+\zeta_{n}^{-1}$. The components of these correspondences are birational to $\tilde{\mathcal{E}}_{a}$, which has $p_{g}\ge2$, as Proposition~\ref{prop:structure}(iv) requires. \begin{corollary}[Squares of the dihedral families]\label{cor:dihedral} Let $X$ be an elliptic K3 surface of one of the families of \cite[Thm.~1.1]{vGS25}, with real multiplication by $F=\QQ(\zeta_{n})^{+}$ for $n\in\{5,7,9,11\}$, and suppose $E=F$ or $E$ is a CM field; the very general member has $E=F$. Then the Hodge conjecture holds for $X\times X$, and \Fiii{} holds for $X\times X$ and $X^{[2]}$. This covers the cubic fields $\QQ(\zeta_{7})^{+}$ and $\QQ(\zeta_{9})^{+}$ on families of dimensions $3$ and $2$. \end{corollary} \begin{proof} If $E=F$, $E$ is algebraic by Proposition~\ref{prop:hecke}(ii) and \cite[\S4.8]{vGS25}; if $E$ is CM, by Buskin's theorem as in Theorem~\ref{thm:fricke}(iii). The rest is as there. \end{proof} \begin{figure*}[t] \centering \includegraphics[width=0.92\textwidth]{figures/fig_zeta9} \caption{The proof of Theorem~\ref{thm:zeta9}. The surface is the two-dimensional period domain of the twelve-dimensional space $T_{12}$ with real multiplication by $F=\QQ(\zeta_{7})^{+}$ or $\QQ(\zeta_{9})^{+}$; the shaded patch is the image of the two-dimensional dihedral family, on which the multiplication is algebraic. The curves are the loci on which an $F$-line $N$ becomes algebraic, and the surface of Mumford type is a point on one of them. The rational isometries $x\mapsto axg$ of the quaternion algebra are dense in the real group, so one of them moves the point into the patch; Buskin's theorem turns that isometry into a cycle.} \label{fig:zeta9} \end{figure*} \subsection{The cyclic cubic fields \texorpdfstring{$\QQ(\zeta_{7})^{+}$}{Q(zeta7)+} and \texorpdfstring{$\QQ(\zeta_{9})^{+}$}{Q(zeta9)+}}\label{ssec:zeta9} The dihedral families reach the cyclic cubic fields only on their own members. For each of the two cubic fields they contain, $F=\QQ(\zeta_{n})^{+}$ with $n\in\{7,9\}$, there is a two-dimensional family on a transcendental lattice of rank twelve: for $n=9$ the family of \cite[\S5.6]{vGS25}, and for $n=7$ the Dickson deformation of the one-parameter family of Artebani, Sarti and Taki with Picard lattice $U\oplus E_{8}$ \cite[Ex.~6.1]{AST11}, which \cite{vGS25} do not use, their family for $n=7$ being the three-dimensional one of Picard number four. Two is the largest dimension a family with this real multiplication on a transcendental space of rank twelve can have, and this maximality lets the cycle travel: every K3 surface of Mumford type for $F$ and for the quaternion algebra ramified only at infinity is a Hodge isometric image of a member of the family, and Buskin's theorem carries the cycle along the isometry. The result settles the frontier case of Section~\ref{ssec:f3stop} for these two fields and this algebra. Write $\theta=\zeta_{n}+\zeta_{n}^{-1}$, so that $F=\QQ(\theta)$, $K=\QQ(\zeta_{n})=F(\sqrt{-\delta})$ with $-\delta=(\zeta_{n}-\zeta_{n}^{-1})^{2}=\theta^{2}-4$, and $\delta=4-\theta^{2}$ is totally positive. For $n=7$ one has $\theta^{3}+\theta^{2}-2\theta-1=0$ and the three real embeddings send $\theta$ to $-1.802$, $-0.445$ and $1.247$; for $n=9$ one has $\theta^{3}-3\theta+1=0$ and the embeddings send $\theta$ to $-1.879$, $0.347$ and $1.532$. Let $p_{n}$ be the prime dividing $n$; it is totally ramified in $K$, so it lies under a single place of $F$, and $K/\QQ$ is unramified outside $p_{n}$. The prime $2$ has order three in $(\ZZ/n)^{\times}/\{\pm1\}=\Gal(F/\QQ)$, so it is inert in $F$. Let $D_{\infty}$ be the quaternion algebra over $F$ ramified exactly at two real places and at no finite place. Put $\varepsilon=\theta$ for $n=7$ and $\varepsilon=-\theta$ for $n=9$; then $\varepsilon$ is a unit positive at exactly one real place, and $D_{\infty}=(-1,\varepsilon)_{F}$: the symbol is ramified at the two real places where $\varepsilon<0$, split at every odd finite place because both entries are units there, and split at the one place above $2$ by the parity of the number of ramified places. Its corestriction to $\QQ$ is unramified everywhere and split at infinity, hence is $M_{8}(\QQ)$, so $D_{\infty}$ is one of Mumford's algebras and the fourfolds of \cite{Mum69} exist for it. For $\mu\in F$ totally positive let $q_{\mu}$ be the quadratic form $\mathrm{Tr}_{F/\QQ}(\mu\,\mathrm{trd}(y^{2}))$ on the nine-dimensional $\QQ$-space $D_{\infty}^{0}$ of elements of reduced trace zero, of signature $(2,7)$; the surfaces $S_{\lambda}$ of \cite[Thm.~21.4]{BMB26} for the fourfolds of $(F,D_{\infty})$ have $T(S_{\lambda})_{\QQ}\cong(D_{\infty}^{0},q_{\lambda})$. \begin{theorem}[Real multiplication by $\QQ(\zeta_{7})^{+}$ and $\QQ(\zeta_{9})^{+}$ on the surfaces of Mumford type]\label{thm:zeta9} Let $n\in\{7,9\}$, $F=\QQ(\zeta_{n})^{+}$ and $D_{\infty}$ as above. Let $S$ be a projective K3 surface such that $T(S)_{\QQ}$ with its intersection form is isometric to $(D_{\infty}^{0},q_{\mu})$ for some totally positive $\mu\in F$, by an isometry under which the multiplication by $F$ on $D_{\infty}^{0}$ acts on $T(S)_{\QQ}$ by Hodge endomorphisms. Then $E=F$, the real multiplication by $F$ is induced by algebraic cycles on $S\times S$, and the Hodge conjecture and \Fiii{} hold for $S\times S$ and $S^{[2]}$. This applies to every surface $S_{\lambda}$ of a Mumford fourfold of $(F,D_{\infty})$, at every point of its Shimura curve. \end{theorem} \begin{proof} Step 1, the family for $n=9$. Let $\mathfrak{X}\to B$ be the two-dimensional family of \cite[\S5.6]{vGS25}: elliptic K3 surfaces with very general Picard lattice $U\oplus A_{2}^{4}$,\footnote{Families of K3 surfaces polarised by a fixed lattice also occur as the fibres of Calabi--Yau threefolds, the quartic and quintic ones being studied in \cite{Bha22}; the chains of this section act on the fibres only, and the Hodge classes of such a total space are not considered here.} deformations of the surfaces $y^{2}=x^{3}+bx+c_{1}t^{9}+c_{0}$ with a non-symplectic automorphism $\sigma$ of order nine, obtained by replacing $t^{9}$ by the Dickson polynomial $p_{9,a}(t)$. The transcendental lattice $T_{0}$ of a very general member has rank twelve and discriminant group $(\ZZ/3)^{4}$, is a $\ZZ[\zeta_{9}]$-module of rank two through $\sigma^{*}$, and carries the flat action of $F$ through $\sigma^{*}+\sigma^{*-1}$, which on every member of the family is induced by the cycle $\Gamma_{1}+\Gamma_{-1}$ of \cite[\S4.8]{vGS25} on a dihedral cover, as in Proposition~\ref{prop:hecke}. The period map of $B$ has two-dimensional image \cite[\S5.6]{vGS25}. Step 1, the family for $n=7$. The surfaces $X_{b}\colon y^{2}=x^{3}+bx+c_{1}t^{7}+c_{0}$ are the K3 surfaces of \cite[Ex.~6.1]{AST11} with $a=0$ there: $\sigma(x,y,t)=(x,y,\zeta_{7}t)$ is a non-symplectic automorphism of order seven, the fibration has a fibre of type $\mathrm{II}^{*}$ at $t=\infty$ and fourteen fibres of type $\mathrm{I}_{1}$, and for very general $b$ the Picard lattice is the invariant lattice $U\oplus E_{8}$ of $\sigma$, with $T_{0}=U^{2}\oplus E_{8}(-1)$ \cite[Table~3]{AST11}. Replacing $t^{7}$ by the Dickson polynomial $p_{7,a}(t)=t^{7}-7at^{5}+14a^{2}t^{3}-7a^{3}t$ gives the surfaces\footnote{The Dickson polynomial $p_{n,a}$ is the unique polynomial of degree $n$ with $p_{n,a}(v+a/v)=v^{n}+(a/v)^{n}$; for $a\ne0$ the map $t\mapsto p_{n,a}(t)$ from $\mathbb{P}^{1}$ to itself is a cover of degree $n$ with monodromy group the dihedral group $D_{n}$, totally ramified over $\infty$ \cite[Lem.~4.3]{vGS25}, since it is the quotient of $v\mapsto v^{n}+(a/v)^{n}$, on which $D_{n}$ acts by $v\mapsto\zeta_{n}v$ and $v\mapsto a/v$, by the involution. For $a=0$ it is the cyclic cover $t\mapsto t^{n}$. The identity $p_{n,a}(\mu t)=\mu^{n}p_{n,a/\mu^{2}}(t)$ follows from the defining equation with $v$ replaced by $\mu v$.} \begin{equation}\label{eq:dickson7} \mathfrak{X}_{a}\colon\ y^{2}=x^{3}+bx+c_{1}p_{7,a}(t)+c_{0}, \end{equation} where $p_{7,a}$ is the Dickson polynomial with $p_{7,a}(v+av^{-1})=v^{7}+a^{7}v^{-7}$;\footnote{Explicitly $p_{7,a}(t)=t^{7}-7at^{5}+14a^{2}t^{3}-7a^{3}t$, the case $n=7$ of $p_{n,a}(t)=\sum_{k\le n/2}\frac{n}{n-k}\binom{n-k}{k}(-a)^{k}t^{n-2k}$; for $a=0$ it is $t^{7}$, so the family $\mathfrak{X}_{a}$ contains the surfaces $X_{b}$.} these are the deformations (4.4) of \cite[Prop.~4.6]{vGS25} with $n=7$, $\alpha=b$ and $\beta(u)=c_{1}u+c_{0}$; by that proposition each $\mathfrak{X}_{a}$ is a K3 surface on whose transcendental space $F$ acts by Hodge endomorphisms, induced by the cycle $\Gamma_{1}+\Gamma_{-1}$ on the $D_{7}$-cover as before. The polynomial $c_{1}p_{7,a}+c_{0}$ has degree seven, so at $t=\infty$ the coefficients of the Weierstrass equation vanish to orders $8$ and $5$ and the fibre of type $\mathrm{II}^{*}$ survives; hence $\Pic(\mathfrak{X}_{a})$ contains the unimodular lattice $U\oplus E_{8}$ of rank ten, and since the Picard number of a very general member of a family is at most that of a very general member of any subfamily, here the locus $a=0$, it equals $U\oplus E_{8}(-1)$, with $T_{0}=U^{2}\oplus E_{8}(-1)$, unimodular of rank twelve. The parameters $(b,c_{1},c_{0},a)$ modulo the scalings $(x,y,t)\mapsto(\lambda^{2}x,\lambda^{3}y,\mu t)$, which send $a$ to $a/\mu^{2}$, leave two, and the image of the period map is two-dimensional: otherwise it is the closure of the one-dimensional image of the locus $a=0$, and a very general $\mathfrak{X}_{a}$ is isomorphic by the Torelli theorem to some $X_{b'}$, so carries a non-symplectic automorphism $\sigma'$ of order seven. Such a $\sigma'$ has no invariant vector in $T_{0}\otimes\QQ$, so its invariant lattice is $\Pic(\mathfrak{X}_{a})$; it fixes the class of a fibre and the class of the zero section $O$, hence preserves the fibration and $O$, the only effective curve in its class, and it acts on the base with order seven, since an automorphism over the identity of the base that fixes $O$ is a group automorphism of the fibres, of order dividing six. It fixes $t=\infty$, the only reducible fibre, so it acts by $t\mapsto\zeta t+\beta$ with $\zeta$ a primitive seventh root of unity, and since the Weierstrass equation of a fibration with a chosen section is unique up to $(x,y)\mapsto(\lambda^{2}x,\lambda^{3}y)$, the polynomial $c=c_{1}p_{7,a}+c_{0}$ satisfies $c(\zeta t+\beta)=\lambda^{6}c(t)$. The coefficients of $t^{7}$, $t^{6}$ and $t^{5}$ give $\lambda^{6}=1$, $\beta=0$ and $7ac_{1}(\zeta^{5}-1)=0$, so $a=0$. Thus the direction $a$ is not isotrivial and the image is two-dimensional. If $E\supsetneq F$ then $E$ has degree six or twelve and $\dim_{E}T_{0}\otimes\QQ\le2$, which allows at most one modulus by the count of \cite[\S5.4]{vGS25}; so $E=F$ on the very general member. In both cases the family is maximal: $\dim_{F}T_{12}=4$ for $T_{12}=T_{0}\otimes\QQ$, the period domain of Hodge structures of K3 type on $T_{12}$ with $F$ acting by Hodge endomorphisms has dimension $4-2=2$, and the period map has two-dimensional image in its quotient by the arithmetic group, so by the open mapping theorem the image contains a nonempty open set $U$ of one of the two components of that domain. Step 2, the quadratic structure. Since the intersection form $q$ of $T_{12}$ is $\sigma$-invariant there is a unique hermitian form $h$ on the two-dimensional $K$-space $T_{12}$ with $q=\mathrm{Tr}_{K/\QQ}h$, and a unique $F$-bilinear form $\Phi=\mathrm{Tr}_{K/F}h$ with $q=\mathrm{Tr}_{F/\QQ}\Phi$. Diagonalise $h=\langle h_{1},h_{2}\rangle$ with $h_{i}\in F^{\times}$. For $x=u+v\sqrt{-\delta}$ one has $\mathrm{Tr}_{K/F}(h_{1}x\bar x)=2h_{1}(u^{2}+\delta v^{2})$, so \begin{equation}\label{eq:phiform} \Phi\cong2h_{1}\langle1,\delta,h,h\delta\rangle,\qquad h=h_{2}/h_{1}, \end{equation} and $\langle1,\delta,h,h\delta\rangle$ is the reduced norm form of the quaternion algebra $Q=(-\delta,-h_{1}h_{2})_{F}$; thus $(T_{12},\Phi)\cong(Q,2h_{1}\mathrm{nrd})$ as $F$-quadratic spaces. The signature of $q$ is $(2,10)$ and splits over the three real places of $F$ as $(2,2)$ at the place $\sigma_{1}$ whose eigenspace carries $H^{2,0}$ and $(0,4)$ at the two others. From \eqref{eq:phiform}, signature $(2,2)$ at $\sigma_{1}$ forces $h_{1}h_{2}<0$ there, and $(0,4)$ at $\sigma_{2},\sigma_{3}$ forces $h_{1},h_{2}<0$ there; so $Q$ is split at $\sigma_{1}$ and ramified at $\sigma_{2}$ and $\sigma_{3}$, and after exchanging $h_{1}$ and $h_{2}$ we may take $h_{1}>0$ at $\sigma_{1}$ and $h_{1}<0$ at the other two places. Step 3, $Q=D_{\infty}$. The lattice $T_{0}$ is unimodular at every prime $p\ne p_{n}$, and $K/\QQ$ is ramified only at $p_{n}$, so for a place $v$ of $F$ not above $p_{n}$ the ring $\mathcal{O}_{K}\otimes\ZZ_{p}$ is a product of unramified extensions whose trace forms are perfect, and the hermitian $\mathcal{O}_{K,v}$-lattice $T_{0}\otimes\ZZ_{p}$ is unimodular for $h$; its determinant $h_{1}h_{2}$ is therefore a unit of $F_{v}$ times a norm from $K_{v}$. Units are norms from an unramified extension, so $(-\delta,-h_{1}h_{2})_{v}=1$ at every finite $v\nmid p_{n}$, including the place above $2$. The ramification set of $Q$ has even cardinality, contains $\sigma_{2},\sigma_{3}$ and is otherwise contained in the single place above $p_{n}$; so that place is unramified as well, and $Q\cong D_{\infty}$. As a check independent of this argument, the rational invariants of $T_{0}\otimes\QQ$, computed from the Picard lattice by Witt cancellation in the K3 lattice, and those of $\mathrm{Tr}_{F/\QQ}(\varepsilon\,\mathrm{nrd}_{D_{\infty}})$ agree for both $n$: signature $(2,10)$, discriminant $1$, Hasse invariant $-1$ at $2$ and $+1$ at every odd prime, while the trace forms of the algebras over $F$ with the same real ramification and ramification at finite places differ from them at $7$ or $43$ for $n=7$ and at $3$ or $19$ for $n=9$. Step 4, the surfaces of Mumford type inside $T_{12}$. For $n_{0}\in Q^{\times}$ the $F$-line $N=Fn_{0}$ has $\Phi(n_{0},n_{0})=2h_{1}\mathrm{nrd}(n_{0})$, and its orthogonal complement is $n_{0}Q^{0}$, with $\Phi(n_{0}y,n_{0}y')=-h_{1}\mathrm{nrd}(n_{0})\,\mathrm{trd}(yy')$ for $y,y'\in Q^{0}$; so $(N^{\perp},\Phi)\cong(D_{\infty}^{0},\mu\,\mathrm{trd})$ with $\mu=-h_{1}\mathrm{nrd}(n_{0})$, and $N$ is negative definite exactly when $\mu$ is totally positive. By the Hasse--Schilling norm theorem \cite[Thm.~33.15]{Rei03} the reduced norms of $Q^{\times}$ are the elements of $F^{\times}$ positive at $\sigma_{2}$ and $\sigma_{3}$, and $-h_{1}$ is negative at $\sigma_{1}$ and positive at the other two places, so every totally positive $\mu$ occurs. Given $S$ as in the statement, choose $n_{0}$ with $\mu=-h_{1}\mathrm{nrd}(n_{0})$ and an $F$-linear isometry $T(S)_{\QQ}\cong n_{0}^{\perp}$; the period $\omega_{0}$ of $S$ becomes a point of the period domain of $T_{12}$ at which $Fn_{0}$ is algebraic and $F$ acts by Hodge endomorphisms, because $\omega_{0}$ lies in the $\sigma_{1}$-eigenspace. Replacing $S$ by its complex conjugate, whose period is $\bar\omega_{0}$ and for which the conclusion is equivalent, we may take $\omega_{0}$ on the component containing $U$. Step 5, moving the point. The group $\mathrm{SO}(\Phi)(\RR)$ at the place $\sigma_{1}$ is $(\SL_{2}(\RR)\times\SL_{2}(\RR))/\{\pm1\}$ acting on $Q\otimes_{F,\sigma_{1}}\RR\cong M_{2}(\RR)$ by $x\mapsto axg$, and it acts transitively on each component of the two-dimensional period domain. The group $Q^{1}\times Q^{1}$ of pairs of elements of reduced norm one is simply connected, so by weak approximation \cite[Thm.~7.8]{PR94} its rational points are dense in its real points; hence there are $a,g\in Q^{1}(\QQ)$ with $x\mapsto axg$ carrying $\omega_{0}$ into $U$. This map is an $F$-linear isometry of $(T_{12},\Phi)$, it carries $n_{0}^{\perp}$ onto $(an_{0}g)^{\perp}$ with $\mathrm{nrd}(an_{0}g)=\mathrm{nrd}(n_{0})$, and it carries the Hodge structure of $T(S)$ to the Hodge structure on $(an_{0}g)^{\perp}$ determined by the image point, which is the period of a member $X_{b}$ of the family. The rational classes of type $(1,1)$ in $T_{12}$ at that period are exactly $F\cdot an_{0}g$, since a rational class orthogonal to the period decomposes into a component in the line and a component in $(an_{0}g)^{\perp}$ that is a rational class of type $(1,1)$ in a copy of $T(S)$, which has none. So $T(X_{b})_{\QQ}=(an_{0}g)^{\perp}$ with the restricted action of $F$, and $x\mapsto axg$ is a Hodge isometry $\varphi\colon T(S)_{\QQ}\to T(X_{b})_{\QQ}$ that commutes with $F$. Step 6, the cycle. $H^{2}(S,\QQ)$ and $H^{2}(X_{b},\QQ)$ are both isometric to $\Lambda_{\QQ}$, so by Witt cancellation $\NS(S)_{\QQ}\cong\NS(X_{b})_{\QQ}$, and $\varphi$ extends to a rational Hodge isometry $H^{2}(S,\QQ)\to H^{2}(X_{b},\QQ)$, which is algebraic by Buskin's theorem \cite{Bus19}. The action of $e\in F$ on $T(X_{b})$ is induced by the cycle of Step~1, so $m_{e}=\varphi^{-1}\circ m_{e}\circ\varphi$ on $T(S)$ is induced by a cycle on $S\times S$. Since $T(S)$ has rank nine and $F\subseteq E$, $E$ is $F$ or a totally real field of degree nine with $\dim_{E}T=1$, which \cite{vG08} excludes; so $E=F$ is algebraic, and the Hodge conjecture for $S\times S$ and the rest follow as in Theorem~\ref{thm:fricke}(iii). \end{proof} The next proposition says exactly which pairs $(F,D)$ the mechanism reaches and where it stops. \begin{proposition}[What the dihedral mechanism reaches]\label{prop:cubicstop} Let $n\in\{7,9\}$ and $F=\QQ(\zeta_{n})^{+}$. \begin{enumerate}[label=\textup{(\roman*)},leftmargin=2.4em,itemsep=2pt] \item Let $X$ be a K3 surface with a purely non-symplectic automorphism of order $n\in\{7,9\}$ whose transcendental lattice has rank twelve and is unimodular at every prime other than $p_{n}$. Then $(T(X)_{\QQ},\Phi)$ is isometric to $(D_{\infty},c\,\nrd)$ for some $c\in F^{\times}$, and every surface of Mumford type obtained from $X$ or from a rational isometric image of $X$ as in Step~4 has the algebra $D_{\infty}$. The Mumford algebras over $F$ ramified at two finite places are reached by no family of rank twelve with such an automorphism. \item Let $F=\QQ(\zeta_{7})^{+}$ and let $T_{18}=U^{2}\oplus A_{6}\oplus E_{8}(-1)$ be the transcendental lattice of the three-dimensional family of \cite[\S5.4]{vGS25}, with its $F$-bilinear form $\Phi_{6}$ of dimension six. For every Mumford algebra $D$ over $F$ and every totally positive $\lambda$ the space $(D^{0},\lambda\,\mathrm{trd})$ embeds $F$-isometrically into $(T_{18}\otimes\QQ,\Phi_{6})$, so the four-dimensional period domain of this family contains the Shimura curve of the surfaces $S_{\lambda}$ of every pair $(F,D)$. The image of the family is a threefold in it, and Step~5 would need a rational isometry carrying the period of $S_{\lambda}$ into that threefold. \item A finite group acting on the base $\mathbb{P}^{1}$ of an elliptic surface is cyclic, dihedral, tetrahedral, octahedral or icosahedral, and the centres of the simple factors of its rational group algebra are the fields $\QQ(\zeta_{m})$, $\QQ(\zeta_{m})^{+}$, $\QQ(\sqrt{-3})$, $\QQ$ and $\QQ(\sqrt5)$. The only cubic fields among them are $\QQ(\zeta_{7})^{+}$ and $\QQ(\zeta_{9})^{+}$, so the base change of elliptic surfaces along covers of $\mathbb{P}^{1}$ reaches no cyclic cubic field of conductor at least $13$. \end{enumerate} \end{proposition} \begin{proof} (i) Steps~2 and~3 of the theorem use only the order of the automorphism, the rank, the signature and the unimodularity away from $p_{n}$, so they apply to $X$ and give $\Phi\cong2h_{1}\mathrm{nrd}_{D_{\infty}}$. A rational isometry preserves $(T,\Phi)$ up to the $F$-similarity class, which determines the algebra, and Step~4 exhibits the orthogonal complement of every negative definite $F$-line as $(D_{\infty}^{0},\mu\,\mathrm{trd})$; an algebra $D\ne D_{\infty}$ has $(D^{0},\lambda\,\mathrm{trd})\not\cong(D_{\infty}^{0},\mu\,\mathrm{trd})$ for all $\lambda,\mu$, since the even Clifford algebra of the ternary form $\lambda\,\mathrm{trd}$ is $D$. (ii) $T_{18}\otimes\QQ$ is a three-dimensional hermitian $K$-space and $\Phi_{6}=2\langle h_{1},h_{1}\delta,h_{2},h_{2}\delta,h_{3},h_{3}\delta\rangle$ has signature $(2,4)$ at $\sigma_{1}$ and $(0,6)$ at $\sigma_{2},\sigma_{3}$, by the same count as in Step~2. The form $q_{3}=\lambda\,\mathrm{trd}$ on $D^{0}$ has signature $(2,1)$ at $\sigma_{1}$, where $D$ is split, and $(0,3)$ at the two ramified places. The Witt class of $\Phi_{6}-q_{3}$ is represented at every real place by a negative definite ternary form and at every finite place by a ternary form, because a class of odd dimension has an anisotropic representative of dimension one or three; these local ternary forms have the global discriminant $d(\Phi_{6})d(q_{3})$, Hasse invariants $+1$ at almost every place and satisfying the product formula, since they are products of the invariants of $\Phi_{6}$ and $q_{3}$ and of Hilbert symbols of their discriminants; so by \cite[Thm.~72:1]{OMe00} there is a global ternary form $q_{3}'$ with these localisations, and $\Phi_{6}\cong q_{3}\perp q_{3}'$ by the Hasse--Minkowski theorem. The image $N$ of $q_{3}'$ is a negative definite $F$-subspace of dimension three with $N^{\perp}\cong(D^{0},\lambda\,\mathrm{trd})$, and the Hodge structures of K3 type on $N^{\perp}$ with $F$ acting form the Shimura curve of the $S_{\lambda}$ of $(F,D)$. The family has three moduli \cite[\S5.4]{vGS25} in a domain of dimension $6-2=4$. (iii) The list of finite subgroups of $\mathrm{PGL}_{2}(\CC)$ is Klein's. The simple factors of $\QQ[G]$ have centres the fields $\QQ(\chi)$ generated by the values of the irreducible characters: $\QQ(\zeta_{m})$ for the cyclic group of order $m$, $\QQ(\zeta_{m})^{+}$ for the dihedral group of order $2m$ (its two-dimensional characters take the values $\zeta_{m}^{k}+\zeta_{m}^{-k}$), $\QQ$ and $\QQ(\sqrt{-3})$ for $A_{4}$, $\QQ$ for $S_{4}$, and $\QQ$ and $\QQ(\sqrt5)$ for $A_{5}$. By Proposition~\ref{prop:hecke}(iii) the correspondences cut out of the action act through these fields; $\QQ(\zeta_{m})$ is never cubic, and $\QQ(\zeta_{m})^{+}$ is cubic exactly for $m\in\{7,9,14,18\}$, where it is $\QQ(\zeta_{7})^{+}$ or $\QQ(\zeta_{9})^{+}$. \end{proof} \begin{figure*}[t] \centering \includegraphics[width=0.92\textwidth]{figures/fig_cubicstop} \caption{Where the dihedral mechanism stops, Proposition~\ref{prop:cubicstop}. (i)~Over $F=\QQ(\zeta_{7})^{+}$ or $\QQ(\zeta_{9})^{+}$ the Mumford algebras are $D_{\infty}$ and the algebras $D_{p}$ ramified at two real places and at two places above a split prime $p$; a rank-twelve family unimodular away from $p_{n}$ has square discriminant and carries only $D_{\infty}$. (ii)~The period domain of the rank-eighteen family is four-dimensional, the image of the family is a threefold, and the Shimura curves of the surfaces of Mumford type for the other algebras lie in the domain; whether the threefold meets them is the open transport condition. (iii)~The finite subgroups of $\mathrm{PGL}_{2}$ acting on the base produce only $\QQ(\zeta_{m})$, $\QQ(\zeta_{m})^{+}$, $\QQ$, $\QQ(\sqrt{-3})$ and $\QQ(\sqrt5)$; a cubic field arises only from the dihedral groups $D_{7}$, $D_{9}$, $D_{14}$ and $D_{18}$.} \label{fig:cubicstop} \end{figure*} For the fourfolds $X$ of $(F,D_{\infty})$ the theorem leaves \Fii{} where it was: by Theorem~\ref{thm:link}(v) their exceptional classes are algebraic as soon as one algebraic cycle on $S_{\lambda}\times X^{r}$ acts nontrivially on $T(S_{\lambda})$, and the real multiplication, now algebraic, supplies no such cycle. \subsection{Powers of K3 surfaces and Kummer type}\label{ssec:powers} The theorems of this section decide $S\times S$ and $S^{[2]}$. On the higher powers the Hodge classes are described by Varesco \cite[Thm.~2.16]{Var25}. Let $E$ be totally real, $m=\dim_{E}T$, and for each real embedding $\sigma$ of $E$ let $V_{\sigma}=T\otimes_{E,\sigma}\CC$, so that $T_{\CC}=\bigoplus_{\sigma}V_{\sigma}$. The line $\bigwedge^{m}V_{\sigma}\subset T_{\CC}^{\otimes m}$ is spanned by a class $\det V_{\sigma}$ of type $(m,m)$, and the span of the $\det V_{\sigma}$ over all $\sigma$ is defined over $\QQ$: it is $\bigwedge^{m}_{E}T\subset\bigwedge^{m}T\subset T^{\otimes m}$, a space of Hodge classes of dimension $[E:\QQ]$ in $H^{2m}(S^{m})$, the exceptional classes. Varesco's theorem is that every Hodge class in the tensor algebra of $T$ is obtained from the Hodge classes of $T\otimes T$, which are the elements of $E$, and from the exceptional classes by tensor products, permutations of the factors, linear combinations and contractions with $q_{S}$. Each of these operations carries algebraic classes on powers of $S$ to algebraic classes: a tensor product is an exterior product, a permutation is the pull-back along an automorphism of $S^{k}$, and the contraction of two factors is the push-forward along the projection $S^{k}\to S^{k-2}$ after cupping with the pull-back of the class of the diagonal of $S\times S$, since $q_{S}(a,b)=\int_{S\times S}(a\otimes b)\cup[\Delta_{S}]$. \begin{proposition}[Hodge classes on the powers]\label{prop:powers} Let $S$ be a projective K3 surface with $E$ totally real and $m=\dim_{E}T$, so that $m\ge3$ as in the proof of Theorem~\ref{thm:sqrt6}(ii). \begin{enumerate}[label=\textup{(\roman*)},leftmargin=2.4em,itemsep=2pt] \item The Hodge conjecture holds for every power $S^{n}$ if and only if the elements of $E$ are algebraic on $S\times S$ and the exceptional classes $\bigwedge^{m}_{E}T\subset H^{2m}(S^{m})$ are algebraic. When $E=\QQ$, $m=r=\mathrm{rk}\,T$ and the exceptional class is the determinant $\det T$, which spans $\bigwedge^{r}T\subset H^{2r}(S^{r})$. \item For a surface of Theorems~\ref{thm:sqrt6}, \ref{thm:zeta9} or Corollary~\ref{cor:dihedral} with $E$ totally real, $E$ is algebraic, and the Hodge conjecture for all powers of $S$ is equivalent to the algebraicity of $\bigwedge^{m}_{E}T$, $m=\mathrm{rk}\,T/[E:\QQ]$: $m=3$ for the surfaces $S_{\lambda}$ of Theorem~\ref{thm:zeta9}, with $\mathrm{rk}\,T=9$; $m=4$ for the very general members of the two-dimensional family of Theorem~\ref{thm:zeta9}, with $\mathrm{rk}\,T=12$. \end{enumerate} \end{proposition} \begin{proof} (i) If the Hodge conjecture holds on all powers, both kinds of classes are algebraic, being Hodge classes on $S\times S$ and on $S^{m}$. Conversely, $H^{2}(S)=\NS(S)_{\QQ}\oplus T$, and $H^{0}(S)$, $\NS(S)_{\QQ}$ and $H^{4}(S)$ are spanned by algebraic classes and are Tate twists of trivial Hodge structures. So a Hodge class of $H^{*}(S^{n})=\bigotimes^{n}\bigl(H^{0}\oplus\NS_{\QQ}\oplus T\oplus H^{4}\bigr)$ is a sum of exterior products of algebraic classes with Hodge classes of the tensor powers $T^{\otimes k}$, $k\le n$, pulled back along the projections of $S^{n}$ onto $k$ of its factors. By Varesco's theorem and the remark preceding the proposition, the Hodge classes of $T^{\otimes k}$ are algebraic as soon as $E$ and the exceptional classes are. When $E=\QQ$ there is one embedding, $V_{\sigma}=T_{\CC}$ and $\bigwedge^{m}_{E}T=\bigwedge^{r}T$. (ii) The theorems cited prove that $E$ is algebraic; $m=\mathrm{rk}\,T/[E:\QQ]$ since $T$ is an $E$-vector space. The ranks are those of the nine-dimensional space $D_{\infty}^{0}$ and of the lattice $T_{0}$ in Step~1 of the proof of Theorem~\ref{thm:zeta9}. \end{proof} \begin{proposition}[K3 surfaces of Kummer type over $\QQ$]\label{prop:kummerq} Let $S$ be a projective K3 surface, $r=\mathrm{rk}\,T$, let $\langle a_{1},\dots,a_{r}\rangle$ be a diagonalisation of $q_{S}$ on $T$, $d(T)=\prod_{i}a_{i}\in\QQ^{\times}/\QQ^{\times2}$ its discriminant and $\varepsilon_{p}(T)=\prod_{i