\section{The geometry over $\C$ of $M_6$, $M_6\cap M_2$ and $V$, and what it says about the arithmetic}\label{sec:geometry} Throughout, $M_6\subset\PP^5$ is the fourfold $x^6+y^6+z^6=u^6+v^6+w^6$, $M_2$ the quadric $x^2+y^2+z^2=u^2+v^2+w^2$, $W:=M_6\cap M_2$, and $V\subset W$ is Bremner's K3 surface \eqref{eq:V}; $F_6,F_2$ are the two forms, $\chi_{\mathrm{top}}$ the topological Euler characteristic and $\Fr_p$ the geometric Frobenius. We ask what the fundamental groups, Betti and Hodge numbers of the three varieties are, how they show up in point counts over finite fields, and whether the ``complexity'' of the geometry can be said to drive the richness of the arithmetic over $\Q$, in the style of the Weil conjectures for curves. In short: all three are simply connected; $M_6$ is a Calabi--Yau fourfold with $b_4=2606$ and $1752$ Hodge classes, all algebraic, almost none Galois-invariant; $W$ is \emph{not} smooth---it has exactly $710$ nodes---and is a threefold of general type whose resolution has $b_3=70$ instead of the $1300$ of a smooth $(2,6)$ complete intersection, the difference being paid for by a group of Weil divisor classes of rank $96$; $V$ is a K3 surface with $\rho=19$, transcendental lattice $U(2)\oplus\langle12\rangle$, and a zeta function built from the Galois action on its lines and conics and the symmetric square of the newform $320\mathrm f2$. Over finite fields the cohomology controls the counts exactly, and we verify this numerically in every case; over $\Q$ the relevant invariants are the Kodaira dimension and the Galois-invariant algebraic classes, not the Betti numbers (\S\ref{sec:geo-answer}). \subsection{The sextic fourfold $M_6$}\label{sec:geo-M6} Over any field containing a primitive $12$th root of unity $\zeta$, $(u,v,w)\mapsto(\zeta u,\zeta v,\zeta w)$ identifies $M_6$ with the Fermat sextic $X=X^4_6\colon\sum_{i=0}^5x_i^6=0$, since $\zeta^6=-1$: over $\C$, and over $\F_p$ for $p\equiv1\pmod{12}$; for $p\equiv7\pmod{12}$ the two are non-isomorphic twists over $\F_p$. $M_6$ is smooth over $\Q$ and modulo every $p\nmid6$ (Jacobian ideal $(x^5,\dots,w^5)$), so $\pi_1(M_6)=1$ and $H^i(M_6,\Z)\cong H^i(\PP^4,\Z)$ for $i\ne4$ by Lefschetz, and $\omega_{M_6}=\mathcal O(6-6)=\mathcal O$: a Calabi--Yau fourfold. \begin{proposition}\label{prop:geo-M6} A smooth sextic fourfold in $\PP^5$ has $h^{4,0}=h^{0,4}=1$, $h^{3,1}=h^{1,3}=426$, $h^{2,2}=1752$, the other $h^{p,q}$ with $p+q=4$ vanish, and $h^{p,q}=\delta_{pq}$ for $p+q\ne4$; hence $b_4=2606$, $b_4^{\mathrm{prim}}=2605$, $\chi_{\mathrm{top}}=2610$. For the Fermat sextic, $H^4_{\mathrm{prim}}(X,\C)=\bigoplus_\alpha V(\alpha)$ over the $2605$ characters $\alpha\in\{1,\dots,5\}^6$ with $\sum\alpha_i\equiv0\pmod6$ of the group $\mu_6^6/\mu_6$ acting on $X$; $V(\alpha)$ is a line of Hodge type $(p,q)$ with $q=\frac16\sum\alpha_i-1$, and there are $1,426,1751,426,1$ characters with $\sum\alpha_i=6,12,18,24,30$. All $1751$ primitive classes of type $(2,2)$ are Hodge classes, and all are algebraic. \end{proposition} \begin{proof} The Hodge numbers were computed in three ways: from Hirzebruch's generating function for the $\chi_y$-genus of complete intersections \cite[App.~I, \S22]{Hirzebruch} with Lefschetz; from $\chi_{\mathrm{top}}=\int c_4=6\cdot435$, $c(M_6)=(1+h)^6/(1+6h)$; and, for the Fermat sextic, from Griffiths' $H^{4-q,q}_{\mathrm{prim}}\cong R_{6q}$ with $R=\C[x_0,\dots,x_5]/(x_i^5)$, whose pieces in degrees $0,6,12,18,24$ have dimensions $1,426,1751,426,1$. The character decomposition and Hodge types are Weil's \cite{Weil1949,ShiodaKatsura1979}; counting characters reproduces $1,426,1751,426,1$. By Shioda's criterion \cite{Shioda1979}, $V(\alpha)$ is Hodge iff $\sum_i\langle t\alpha_i\rangle=18$ for all $t\in(\Z/6)^\times=\{\pm1\}$; for $t=-1$ this is $\sum(6-\alpha_i)=36-18$, automatic, so every $(2,2)$-character is Hodge. Algebraicity of all of them is the Hodge conjecture for $X^4_6$, which is known: Braun, Fortin, Lopez Garcia and Villaflor Loyola \cite[Thm.~1]{BFLV2024} exhibit a basis of $H^{2,2}(X)\cap H^4(X,\Q)_{\mathrm{prim}}$ consisting of the primitive parts of $1001$ planes, $720$ Aoki--Shioda cycles and $30$ further cycles, and prove the $\Z[1/6]$-integral Hodge conjecture for $X$ \cite[Thm.~2]{BFLV2024}; see Shioda \cite{Shioda1979,Shioda1979b} for the inductive structure of the Hodge classes of Fermat varieties. \end{proof} The $1001$ planes $\{x_{\sigma(0)}=\varepsilon_1x_{\sigma(1)},x_{\sigma(2)}=\varepsilon_2x_{\sigma(3)},x_{\sigma(4)}=\varepsilon_3x_{\sigma(5)}\}$, $\varepsilon_i^6=-1$, of $X$ are defined over $\Q(\zeta_{12})$; on $M_6$ they become $\{x=\varepsilon u',y=\varepsilon'v',z=\varepsilon''w'\}$ with $\varepsilon^6=\varepsilon'^6=\varepsilon''^6=1$ and $(u',v',w')$ a permutation of $(u,v,w)$, together with planes like $\{x=\zeta y,u=\zeta'v,z=\pm w\}$, $\zeta^6=\zeta'^6=-1$; only the $48$ ``trivial'' planes $\{x=\pm u',y=\pm v',z=\pm w'\}$ are defined over $\Q$. Galois permutes the characters ($\Fr_p$ maps $V(\alpha)$ to $V(p\alpha)$: it fixes every $V(\alpha)$ for $p\equiv1$ (6) and swaps $V(\alpha),V(-\alpha)$ for $p\equiv5$ (6)), and the Galois-invariant part of $H^4(M_6,\Q_\ell)$ is comparatively small: by the computation below, the Frobenius eigenvalue on $V(\alpha)$ equals $p^2$ for \emph{all} $p\equiv1$ (6), $p<200$, for exactly $665$ characters, all of plane type ($\alpha_a+\alpha_b\equiv0$ (6) on a perfect matching), forming $332$ pairs $\{\alpha,-\alpha\}$ plus $(3,\dots,3)$; so at most $333$ primitive classes are Galois-invariant, and these are spanned by Galois-orbit sums of planes---$\Q$-rational classes that carry no rational points. \begin{proposition}[Weil]\label{prop:geo-weil} Let $p\equiv1\pmod6$, $\omega$ a character of order $6$ of $\F_p^\times$ and $g(\chi)=\sum_{t\in\F_p^\times}\chi(t)e^{2\pi it/p}$. For the diagonal sextic $\sum_{i=0}^5a_ix_i^6=0$ in $\PP^5$, \begin{equation}\label{eq:geo-weil} \#\bigl\{\textstyle\sum a_ix_i^6=0\bigr\}(\F_p)=\frac{p^5-1}{p-1}+\frac1p\sum_{\alpha}\ \prod_{i=0}^5\omega^{-\alpha_i}(a_i)\,g(\omega^{\alpha_i}), \end{equation} the sum over the $2605$ characters of Proposition~\textup{\ref{prop:geo-M6}}; the $\alpha$-term is the eigenvalue of $\Fr_p$ on $V(\alpha)$ and has absolute value $p^2$. \end{proposition} This is \cite{Weil1949} (from $\#\{x:ax^6=b\}=\sum_{\chi^6=1}\chi(b)\bar\chi(a)$ and the value $\frac{p-1}p\prod g(\chi_i)$ of $\sum_{\sum b_i=0}\prod\chi_i(b_i)$ when all $\chi_i\ne1$, $\prod\chi_i=1$). For $M_6$, $a=(1,1,1,-1,-1,-1)$ and $\prod\omega^{-\alpha_i}(a_i)=\omega(-1)^{\alpha_3+\alpha_4+\alpha_5}$, with $\omega(-1)=1$ for $p\equiv1\pmod{12}$ and $-1$ for $p\equiv7\pmod{12}$: this is the twist. For $p\equiv5\pmod6$, $x\mapsto x^3$ is a bijection of $\F_p$, so $x^6$ and $x^2$ are equidistributed and $\#M_6(\F_p)=(p^2+1)(p^2+p+1)$ is the count of a split quadric fourfold; in Weil's language $\Fr_p$ permutes the characters by $\alpha\mapsto-\alpha$ and only $\alpha=(3,\dots,3)$, the pull-back of the quadric's class under $x_i\mapsto x_i^3$, contributes. We implemented \eqref{eq:geo-weil} and compared it with an exact count ($N_3(c)=\#\{x^6+y^6+z^6=c\}$ by convolution, $\#M_6(\F_p)=(\sum_cN_3(c)^2-1)/(p-1)$) for all $7\le p\le109$: they agree, as do the quadric formulas for $p\equiv5\pmod6$ (Table~\ref{tab:geo-counts}). The $1751$ eigenvalues on the $(2,2)$-characters are $p^2$ times sixth roots of unity (Stickelberger), as they must be for algebraic classes, and none of the $852$ eigenvalues on the $(3,1)$- and $(1,3)$-characters is. The count $\#M_6(\F_p)$ thus ``sees'' all $2606$ classes, whether or not they are defined over $\Q$; e.g.\ for $p\equiv1\pmod{12}$, $p<400$, exactly $1751$ or $1001$ of the $(2,2)$-eigenvalues of $X$ equal $p^2$ according as $2$ is or is not a cube modulo $p$---$1001$ being the number of planes, which are rational over $\F_p$, and $750=720+30$ the number of non-linear generators of \cite{BFLV2024}, consistent with those being defined over a cubic extension of $\Q(\zeta_{12})$ (observed, not proved). \begin{table}[ht]\centering\footnotesize\setlength{\tabcolsep}{4pt} \caption{Point counts over $\F_p$ by exact enumeration. $\#M_6(\F_p)$ agrees with \eqref{eq:geo-weil} for $p\equiv1$ (6) and with $(p^2+1)(p^2+p+1)$ for $p\equiv5$ (6); the Fermat sextic $X$ has the same count iff $p\equiv1$ (4). $\#V(\F_p)$ is Theorem~\ref{thm:geo-zeta}; $R(p)$ is defined in \eqref{eq:geo-Wtrace}. For $p=3,5$, $x^6=x^2$ on $\F_p$ and $\#M_6=\#W=\#M_2$; $V$ has bad reduction at $5$.}\label{tab:geo-counts} \begin{tabular}{@{}rrrrrrr@{}}\toprule $p$ & $p\bmod12$ & $\#M_6(\F_p)$ & $\#X(\F_p)$ & $\#W(\F_p)$ & $\#V(\F_p)$ & $R(p)/(p+p^2)$\\\midrule 3 & 3 & 130 & 112 & 130 & 32 & --\\ 5 & 5 & 806 & 806 & 806 & 182 & --\\ 7 & 7 & 9\,774 & 0 & 1\,562 & 96 & $36.00$\\ 11 & 11 & 16\,226 & 15\,984 & 4\,442 & 188 & $32.73$\\ 13 & 1 & 87\,570 & 87\,570 & 11\,750 & 408 & $102.20$\\ 17 & 5 & 89\,030 & 89\,030 & 23\,030 & 632 & $97.65$\\ 19 & 7 & 258\,750 & 234\,720 & 18\,842 & 492 & $36.63$\\ 23 & 11 & 293\,090 & 292\,032 & 28\,250 & 704 & $33.22$\\ 29 & 5 & 733\,382 & 733\,382 & 85\,670 & 1\,400 & $93.10$\\ 31 & 7 & 1\,235\,142 & 1\,119\,744 & 60\,986 & 1\,164 & $34.26$\\ 37 & 1 & 3\,160\,530 & 3\,160\,530 & 160\,070 & 2\,040 & $95.51$\\ 41 & 5 & 2\,898\,086 & 2\,898\,086 & 204\,374 & 2\,456 & $94.56$\\ 43 & 7 & 4\,263\,462 & 4\,084\,560 & 144\,122 & 2\,208 & $35.92$\\ 47 & 11 & 4\,987\,970 & 4\,983\,552 & 178\,586 & 2\,528 & $34.68$\\ 53 & 5 & 8\,045\,030 & 8\,045\,030 & 377\,990 & 3\,800 & $92.20$\\ 59 & 11 & 12\,329\,762 & 12\,322\,800 & 325\,274 & 3\,980 & $34.90$\\ 61 & 1 & 16\,227\,954 & 16\,227\,954 & 552\,614 & 4\,824 & $96.55$\\ \bottomrule \end{tabular} \end{table} \subsection{The threefold $W=M_6\cap M_2$}\label{sec:geo-W} \begin{proposition}\label{prop:geo-nodes} $W$ is an irreducible, normal, simply connected complete intersection of type $(2,6)$ with $\omega_W=\mathcal O_W(2)$, and it is \emph{singular}: its singular locus consists of exactly $710$ ordinary double points, the points all of whose nonzero coordinates are fourth roots of unity $\varepsilon_i$ \textup(after scaling\textup) with $\sum_{x,y,z}\varepsilon_i^2=\sum_{u,v,w}\varepsilon_i^2$: $30$ with two nonzero coordinates, $360$ with four, $320$ with six; $122$ are rational \textup(coordinates in $\{0,\pm1\}$, e.g.\ $(1{:}0{:}0{:}1{:}0{:}0)$\textup), the other $588$ form $294$ conjugate pairs over $\Q(i)$. The discriminant of the tangent cone at a node is a square in the node's field of definition \textup($\Q$ or $\Q(i)$\textup); hence, for $p\ne2,3$, the exceptional quadric surface over an $\F_p$-rational node of $W_{\F_p}$ is split over $\F_p$. \end{proposition} \begin{proof} $\nabla F_2=2(x,y,z,-u,-v,-w)\ne0$ on $\PP^5$, so $W$ is singular at $P$ iff $\nabla F_6=6(x^5,\dots,-w^5)$ is proportional to $\nabla F_2$, i.e.\ iff all nonzero coordinates have the same fourth power; scaling them to $\mu_4$ and using $\varepsilon^6=\varepsilon^2$, both equations become $\sum_{x,y,z}\varepsilon_i^2=\sum_{u,v,w}\varepsilon_i^2$. Enumerating $(\{0\}\cup\mu_4)^6$ modulo $\mu_4$ over $\Q(i)$ and checking the Jacobian rank gives the $710$ points; independently, the ideal $(F_6,F_2)+(2\times2$ minors$)$, saturated, is zero-dimensional of degree $710$ (over $\Q$ and over two finite fields), so the singular scheme is reduced. Near a node $P$, $W$ is the hypersurface $G=F_6-\lambda F_2=0$ of the smooth quadric $M_2$ with $dG(P)=0$, and the Hessian of $G$ on $T_PM_2$ modulo the Euler direction has rank $4$ at all $710$ points: $A_1$ singularities. Its determinant modulo squares is $\pm36^2,\pm144^2$ or $\pm576^2$, positive at the rational nodes and a square in $\Q(i)$ at all of them; over a finite field a $4$-dimensional form with square discriminant is hyperbolic, which gives the last claim. Normality and irreducibility: Cohen--Macaulay, connected, singular locus of codimension $3$. $\pi_1=1$ by Lefschetz for local complete intersections \cite[Part~II]{GoreskyMacPherson}. \end{proof} \begin{remark}[$W$ covers the Segre cubic]\label{rem:geo-segre} The squaring map $x_i\mapsto y_i=x_i^2$ sends $W$ onto $\{\sum\pm y_i=0,\ \sum\pm y_i^3=0\}$, which after $y_i\mapsto-y_i$ on $u,v,w$ is the \emph{Segre cubic} $S\colon\sum_{i=1}^6y_i=\sum_{i=1}^6y_i^3=0$ in $\PP^4$, the cubic threefold with $10$ nodes and $15$ planes \cite[\S9.4.4]{Dolgachev}; $W\to S$ is a $(\Z/2)^5$-cover branched along the six coordinate sections. The $320$ nodes with six nonzero coordinates map $32{:}1$ onto the $10$ nodes of $S$ (the points $(\pm1,\dots,\pm1)$ with three signs of each kind), the $360$ with four map $8{:}1$ onto the $45$ points of $S$ with two zero coordinates and the others $\pm1$ (the three vertices of each of the $15$ triangles of lines $S\cap\{y_a=y_b=0\}$), and the $30$ with two map $2{:}1$ onto the $15$ points with four zero coordinates. The $15$ planes of $S$ pull back to $120$ planes on $W$: the $48$ rational planes $\{x=\pm u',y=\pm v',z=\pm w'\}$ and $72$ planes $\{x=\pm iy,\ u=\pm iv,\ z=\pm w\}$ and symmetric images, over $\Q(i)$. (All verified by computation.) \end{remark} A \emph{smooth} $(2,6)$ complete intersection $W_t$ in $\PP^5$ (a smoothing of $W$, e.g.\ $M_6\cap$ a general quadric) is simply connected with $H^2=\Z H$ and \[ h^{3,0}(W_t)=h^0(\mathcal O_{W_t}(2))=20,\quad h^{2,1}(W_t)=630,\quad b_3(W_t)=1300,\quad \chi_{\mathrm{top}}(W_t)=-1296 \] (generating function and Chern classes agree); it is of general type, $K=\mathcal O(2)$. For the nodal $W$ these numbers change by an amount measured by the \emph{defect} $\sigma(W)=\operatorname{rk}\Cl(W)-\operatorname{rk}\Pic(W)$ (Weil divisors modulo Cartier divisors): with $\delta=710$ nodes, whose Milnor fibres are $3$-spheres \cite{Clemens1983,Werner1987,Cynk2001}, \begin{equation}\label{eq:geo-nodal} \chi_{\mathrm{top}}(W)=\chi_{\mathrm{top}}(W_t)+\delta,\qquad b_3(\widetilde W)=b_3(W_t)-2\delta+2\sigma,\qquad b_2(\widehat W)=b_4(\widehat W)=1+\sigma, \end{equation} where $\widehat W$ is a small resolution (a complex manifold; each node replaced by a $\PP^1$), $\widetilde W\to W$ is the blow-up of the nodes, with $b_2=b_4=1+\delta+\sigma$ and the same $b_3$, and $h^{3,0}(\widetilde W)=h^0(\omega_W)=20$ because nodes are canonical. The defect is given by Werner's formula and its generalisation to a nodal hypersurface $X$ in a smooth fourfold $Y$ \cite{Werner1987,Cynk2001,CynkRams2011,Cynk2016}: it is the number of \emph{dependent} conditions imposed by the nodes on $H^0(Y,\omega_Y(2X))$, \begin{equation}\label{eq:geo-cynk} \sigma(X)=\#\Sigma-\operatorname{rk}\bigl(H^0(Y,\omega_Y(2X))\to\C^{\Sigma}\bigr), \end{equation} provided $h^2(\Omega^3_Y)=h^1(\Omega^3_Y(X))=h^1(\mathcal O_Y(X+K_Y))=h^2(\mathcal O_Y(-X))=h^3(\mathcal O_Y(-X))=0$. We take $Y=M_2$, $\omega_Y=\mathcal O_Y(-4)$, $X=W\in|\mathcal O_Y(6)|$, so $\omega_Y(2X)=\mathcal O_Y(8)$ and $h^0=825$. The hypotheses hold: the last three by Kodaira and Lefschetz, the first since $h^{3,2}(Y)=0$, and $h^1(\Omega^3_Y(6))=h^1(T_Y(2))=0$ because, by the normal-bundle and Euler sequences, $H^1(T_Y(2))$ is the cokernel of $H^0(T_{\PP^5}(2)|_Y)\to H^0(\mathcal O_Y(4))$, $\xi\mapsto\xi(F_2)$, whose image is the degree-$4$ part of the Jacobian ideal of $F_2$, i.e.\ everything. (With the roles exchanged, $Y=M_6$, the hypothesis $h^1(\Omega^3_{M_6}(2))=h^1(T_{M_6}(2))=0$ fails---this group is the degree-$8$ part of the Jacobian ring of $F_6$---and the formula is not applicable with degree-$4$ forms.) \begin{proposition}\label{prop:geo-defect} $\sigma(W)=95$: the $710$ nodes impose exactly $615$ independent conditions on $H^0(M_2,\mathcal O(8))$. Hence $\chi_{\mathrm{top}}(W)=-586$; the resolutions have $b_3(\widetilde W)=b_3(\widehat W)=70$, $h^{3,0}=20$, $h^{2,1}=15$, $h^{1,1}(\widehat W)=96$, $h^{1,1}(\widetilde W)=806$; and $W$ itself has $b_2(W)=1$, $b_4(W)=96$, $b_3(W)=685$, of which $70$ dimensions are pure of weight $3$ and $615=\delta-\sigma$ are of weight $2$. \end{proposition} \begin{proof} The $710\times1287$ matrix of the degree-$8$ monomials evaluated at the nodes (entries in $\{0,\pm1,\pm i\}$) has rank $615$ over $\Q(i)$, computed exactly as the rank of a $1420\times2574$ integer matrix and re-checked modulo three primes; since $F_2$ vanishes at the nodes this is the rank in \eqref{eq:geo-cynk}. (The same code applied to Schoen's quintic $\sum x_i^5-5\prod x_i$, $125$ nodes, quintic forms, returns the known defect $24$.) The rest follows from \eqref{eq:geo-nodal}, from $b_3=2h^{3,0}+2h^{2,1}$ and, for $W$, from the sequence of the pair $(\widetilde W,\bigcup E_P)$ with $H^i(\PP^5)\cong H^i(W)$ for $i<3$ \cite{GoreskyMacPherson}, giving $b_3(W)=b_3(\widetilde W)+\delta-\sigma$; consistently, $\chi_{\mathrm{top}}(W)=1+1+96+1-685=-586$. \end{proof} The cover of Remark~\ref{rem:geo-segre} makes $h^{3,0}=20$ plausible from the other side: $H^3(\widetilde W)$ splits into eigenspaces indexed by the even subsets $T\subset\{1,\dots,6\}$, the $T$-part being the anti-invariant cohomology of the double cover of $S$ branched along $\sum_{i\in T}\{y_i=0\}$, whose $3$-forms are sections of $\omega_S(\tfrac{|T|}2)=\mathcal O_S(\tfrac{|T|}2-2)$: $5$ for $|T|=6$, $1$ for each of the $15$ subsets with $|T|=4$, $0$ for $|T|=2$, total $20$. (A consistency check, not a proof: the pieces are singular.) We did not identify the $70$-dimensional Galois representation $H^3(\widetilde W)$ (Remark~\ref{rem:geo-W-H3}). \begin{remark}[Choudhry's surface]\label{rem:geo-choudhry} Choudhry's solutions of type $(1,2,6)$ \cite{Choudhry2000} lie on $W\cap\{x+y+z=u+v+w\}$, which is \emph{reducible}: with $e_1=x+y+z$, $e_2=xy+yz+zx$ and Newton's identity $p_6=e_1^6-6e_1^4e_2+9e_1^2e_2^2-2e_2^3+6e_1^3e_3-12e_1e_2e_3+3e_3^2$, one has $F_6=3\,(xyz-uvw)(xyz+uvw+2e_1^3-4e_1e_2)$ modulo the hyperplane and the quadric. So $W\cap\{x+y+z=u+v+w\}$ is the union of the six planes $\{(u,v,w)=\text{a permutation of }(x,y,z)\}$ and the smooth K3 surface of degree $6$ \[ S_{\mathrm{Ch}}\colon\quad x+y+z=u+v+w,\quad x^2+y^2+z^2=u^2+v^2+w^2,\quad xyz+uvw=-2e_1^3+4e_1e_2 , \] a $(1,2,3)$ complete intersection (smoothness checked by a Gr\"obner computation); Choudhry's rank-one elliptic curve lies on it. We did not compute $\rho(S_{\mathrm{Ch}})$. \end{remark} \begin{proposition}[symmetry orbits in $W$]\label{prop:geo-orbits} The group $G$ of signed permutations of the coordinates preserving $\{F_6=0\}$ and $\{F_2=0\}$ is $(S_3\times S_3\times\Z/2)\ltimes(\Z/2)^6$, of order $4608$, acting on $\PP^5$ through $G/\{\pm1\}$. The stabiliser of $V$ has order $96$ \textup(Bremner's group\textup), so $W$ contains exactly $48$ images $gV$; it contains $32$ images of $S_{\mathrm{Ch}}$ \textup(stabiliser of order $144$\textup), $48$ rational planes and $72$ planes over $\Q(i)$. Of the $59$ census solutions with $\sum x^2=\sum u^2$ \textup(Theorem~\ref{thm:census}\textup), $40$ lie on exactly one of the $48$ surfaces $gV$ and one \textup($\#2$\textup) on two of them, $9$ \textup(type $(1,2,6)$\textup) lie on exactly one surface $gS_{\mathrm{Ch}}$, and $9$ lie on none of these $80$ K3 surfaces and on none of the $120$ planes. \end{proposition} \subsubsection*{Point counts of $W$ and the trace formula for a nodal threefold} Let $p\ne2,3,5$. The nodes reduce to $710$ distinct nodes of $W_{\F_p}$, all $\F_p$-rational if $p\equiv1\pmod4$ and $122$ of them if $p\equiv3\pmod4$; call this number $\delta(p)$. Let $\widetilde W\to W$ be the blow-up of the nodes, defined over $\F_p$, with exceptional quadrics $E_P$, split over $\F_p$ for rational $P$ (Proposition~\ref{prop:geo-nodes}), so $\#W(\F_p)=\#\widetilde W(\F_p)-\delta(p)(p^2+2p)$. For $\widetilde W$ the trace formula and Deligne's theorem \cite{DeligneWeilII} give $\#\widetilde W(\F_p)=1+p^3+\tr(\Fr_p|H^2)+\tr(\Fr_p|H^4)-\tr(\Fr_p|H^3)$ with $H^2(\widetilde W)\otimes\Q_\ell=\Q_\ell(-1)\oplus\Q_\ell[\Sigma](-1)\oplus D$, $\Q_\ell[\Sigma]$ the permutation representation on the nodes and $D$ the $95$-dimensional representation on the defect classes---classes of Weil divisors, so with eigenvalues $p$ times roots of unity---and $H^4$ dual to $H^2$. Writing $\tr(\Fr_p|D)=p\,t(p)$, $t(p)\in\Z$, $|t(p)|\le95$: \begin{equation}\label{eq:geo-Wtrace} R(p):=\#W(\F_p)-(1+p+p^2+p^3)+p\,\delta(p)=(p+p^2)\,t(p)-\tr\bigl(\Fr_p\,|\,H^3(\widetilde W)\bigr), \end{equation} with $|\tr(\Fr_p|H^3(\widetilde W))|\le70\,p^{3/2}$ by Deligne. We computed $\#W(\F_p)$ exactly for all $316$ primes $7\le p<2600$ (convolution of the distribution of $(x^2,x^6)$ on $\F_p^2$; sparse exact convolution for $p<300$, FFT with checked rounding for all $p$). The quotient $R(p)/(p+p^2)$, equal to $t(p)$ up to at most $70/\sqrt p$, is in Table~\ref{tab:geo-counts} for small $p$; over the $155$ primes $p\equiv1\pmod 4$ with $300