\documentclass[11pt]{amsart} \usepackage[margin=1in]{geometry} \usepackage{mathptmx} \usepackage{amssymb} \usepackage{graphicx} \usepackage{booktabs} \usepackage{xcolor} \usepackage{enumitem} \usepackage{xurl} \usepackage[colorlinks=true,linkcolor=blue!50!black,citecolor=blue!50!black,urlcolor=blue!50!black,bookmarksnumbered=true]{hyperref} \usepackage{microtype} \tolerance=2000 \emergencystretch=3em \raggedbottom \newtheorem{theorem}{Theorem} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\C}{\mathbb{C}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\PP}{\mathbb{P}} \newcommand{\cL}{\mathcal{L}} \newcommand{\cM}{\mathcal{M}} \newcommand{\cS}{\mathcal{S}} \newcommand{\cV}{\mathcal{V}} \newcommand{\cO}{\mathcal{O}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\End}{\operatorname{End}} \newcommand{\Hess}{\operatorname{Hess}} \newcommand{\Res}{\operatorname{Res}} \newcommand{\rk}{\operatorname{rank}} \newcommand{\ip}[2]{\langle #1,#2\rangle} \newcommand{\bb}{\mathbf{b}} \newcommand{\cc}{\mathbf{c}} \newcommand{\va}{\mathbf{a}} \newcommand{\ee}{\mathbf{e}} \newcommand{\one}{\mathbf{1}} \newcommand{\WP}{\mathrm{WP}} \setlength{\emergencystretch}{3em} \begin{document} \title{Which String Vibrations Become Which Generations on the Dwork Quintic} \author{Deep Bhattacharjee} \address{Formerly, Electro-Gravitational Space Propulsion Laboratory (EGSPL), Bhubaneswar, Odisha 751030, India} \address{Formerly, University of Calcutta, Senate House, 87/1 College Street, Kolkata 700073, West Bengal, India} \email{itsdeep@live.com} \urladdr{\href{https://orcid.org/0000-0003-0466-750X}{https://orcid.org/0000-0003-0466-750X}} \thanks{Corresponding author; second e-mail address \nolinkurl{d.bhattacharjee@erl-forschung.de}} \author{Ushashi Bhattacharya} \address{Formerly, National Taiwan University, Taipei 106319, Taiwan} \email{rai.bhattacharya.in@gmail.com} \urladdr{\href{https://orcid.org/0009-0002-2254-3914}{https://orcid.org/0009-0002-2254-3914}} \subjclass[2020]{Primary 14J32, 81T30; Secondary 14D07, 32G20, 14J33, 20C15, 13H10, 81R05} \keywords{heterotic string, quintic threefold, Yukawa couplings, generations, free quotients} \begin{abstract} We determine, for the heterotic string on the one-parameter family of quintic threefolds, which vibration modes become which families of charged particles. The hundred and one families are labelled by the characters of a symmetry group of order one hundred and twenty-five. Every Yukawa coupling that this symmetry allows is a power of the modulus, and every other coupling vanishes. The invariant deformation pairs each family with exactly one partner, so that fifty families are massless at the Fermat point and none anywhere else, and at the Fermat point the masses are given in closed form by the Gamma function. On the three other one-parameter families of the same type, in weighted projective spaces, the symmetry no longer tells the families apart and some families stay massless at every member. We also prove that the Geometric Selection Conjecture is equivalent to the existence of one viable vacuum fixed by mirror symmetry, and that a free quotient of a Calabi--Yau complete intersection of any of the thirteen one-parameter hypergeometric types never carries three families with the standard embedding; on seven of these types, among them the sextic and dectic pencils, no finite group acts freely at all. \end{abstract} \maketitle \setcounter{tocdepth}{1} \tableofcontents \section{Introduction}\label{sec:intro} A closed string moving in ten dimensions has a tower of vibration modes, and the massless ones of the heterotic string on a product of four-dimensional space-time with a compact six-manifold $X$ are the fields of a four-dimensional gauge theory. When $X$ is a Calabi--Yau threefold and the gauge connection is identified with the Levi-Civita connection of its Ricci-flat metric, the gauge group is $E_6$ and the charged chiral multiplets come in two conjugate representations whose numbers are the two Hodge numbers of $X$. This was established in 1985 by Candelas, Horowitz, Strominger and Witten~\cite{CHSW85} and by Witten~\cite{Wit85}, and the couplings of three charged fields were identified by Strominger~\cite{Str85} with an integral over $X$ that Candelas~\cite{Can88} computed on the quintic threefold. The question in the title is the one that this dictionary leaves open: a vibration mode is a cohomology class on $X$, a particle generation is an irreducible piece of the four-dimensional spectrum, and one wants to know which class carries which generation, with what mass, and how this changes as $X$ moves in its moduli space.\footnote{The word generation is used here in the sense of the $E_6$ theory: one copy of the representation $\mathbf{27}$, which contains one family of quarks and leptons together with its Higgs and singlet partners. The three observed families are three such copies after the further breaking of $E_6$; the paper does not touch that breaking, which depends on discrete Wilson lines and on a quotient of $X$ that the quintic does not admit in a form that reduces a hundred and one copies to three, so that the results here concern the full $E_6$ spectrum of the covering space.} We answer it completely on one family, the Dwork pencil of quintics \begin{equation}\label{eq:dwork} X_\psi=\{x_1^5+x_2^5+x_3^5+x_4^5+x_5^5-5\psi\,x_1x_2x_3x_4x_5=0\}\subset\PP^4, \end{equation} smooth for $\psi^5\ne1$, which is the family on which mirror symmetry was first tested by Candelas, de la Ossa, Green and Parkes~\cite{CDGP91} and on which the Yukawa couplings were counted by Doran, Greene and Judes~\cite{DGJ08}. The group $\Gamma\cong(\Z/5)^3$ of diagonal phases preserving the equation acts on every member of the pencil, and the hundred and one classes in $H^{2,1}(X_\psi)$, which are the hundred and one copies of the $\mathbf{27}$, are the monomials $x^{\bb}$ with exponent vector $\bb\in\{0,1,2,3\}^5$ of total degree five. The results are the following. \begin{theorem}[Labels]\label{thm:A} The hundred and one classes carry hundred and one distinct characters of $\Gamma$, each with multiplicity one, so every generation is labelled by its character. Of the hundred and twenty-five characters of $\Gamma$ the twenty-four that do not occur are exactly the classes of the permutations of $(0,1,2,3,4)$. The permutation group $S_5$ of the coordinates permutes the labels in five orbits of sizes $1$, $20$, $20$, $30$ and $30$, the five types of generation. \end{theorem} \begin{theorem}[Couplings]\label{thm:B} Let $\psi^5\ne1$. The cubic form of the pencil, normalised so that the monomial $x_1^3\cdots x_5^3$ has value one, takes on three generations $\va,\bb,\cc$ the value $\psi^{r}$ when $\va+\bb+\cc\equiv(3-r)\one\pmod 5$ for some $r\in\{0,1,2,3\}$, and zero otherwise. There are $1431$ unordered triples with a nonzero coupling, $566$, $425$, $280$ and $160$ of them with $r=0,1,2,3$, and all $1431$ are nonzero at every smooth member with $\psi\ne0$. \end{theorem} \begin{theorem}[Mass map]\label{thm:C} The invariant class $h=x_1x_2x_3x_4x_5$, the direction of the pencil itself, pairs each generation $\bb$ with exactly one generation $\sigma(\bb)$, where $\sigma$ is an involution of the labels with one fixed point. The coupling of $h$, $\bb$ and $\sigma(\bb)$ equals $1$ for the fifty-one labels with all entries at most two, $\psi$ for the thirty labels of type $(3,1,1,0,0)$ and $\psi^2$ for the twenty labels of type $(3,2,0,0,0)$. Hence the mass matrix induced by a vacuum value of the $\mathbf{27}$ labelled by $h$ has full rank one hundred and one for every $\psi\ne0$, $\psi^5\ne1$, its determinant is $\psi^{70}$, and at the Fermat point $\psi=0$ its rank is fifty-one with the fifty generations containing an exponent three massless. The spectrum consists of one singlet and fifty doublets, and the $\Gamma$-invariant metric is diagonal in the labels, so the fifty doublet masses are, in units of the vacuum value, $|\psi|^{r(\bb)}\big(g_{\bb}\,g_{\sigma(\bb)}\big)^{-1/2}$ with $g$ the Hodge norms of the classes. \end{theorem} Theorem~\ref{thm:B} is stronger than the count of Doran, Greene and Judes~\cite{DGJ08}, who found by a search that $1431$ couplings are allowed by the symmetry; it shows that the symmetry is the only source of zeros, and gives every allowed coupling in closed form. The number $1431$ is recovered by hand in Section~\ref{sec:count}. The theorems are proved in Sections~\ref{sec:labels} to~\ref{sec:mass} after the dictionary between vibrations and particles is set out, with complete proofs, in Section~\ref{sec:dictionary}. Section~\ref{sec:fermat} evaluates the Hodge norms at the Fermat point in terms of the Gamma function, which turns the mass formula of Theorem~\ref{thm:C} into numbers there. The fourth result concerns the choice of the vacuum. A conjecture of Bhattacharjee, Bhattacharya and Ghosh~\cite{Bha26a}, the Geometric Selection Conjecture, asks for a functional on the heterotic landscape, built from the Weil--Petersson metric, that is proper, has a unique minimum, takes that minimum at a phenomenologically viable vacuum and is invariant under mirror symmetry. The precise statement is recalled in Section~\ref{sec:selection}. \begin{theorem}[Selection]\label{thm:D} The Geometric Selection Conjecture holds if and only if there is a viable vacuum, in the sense of conditions \textup{(C1)} to \textup{(C9)} of~\cite{Bha26a}, that is not moved by the mirror involution. On the Dwork pencil the mass condition \textup{(C8)} holds in the precise form that the mass map of the invariant direction has full rank with entries $1$, $\psi$ and $\psi^2$, at every member with $\psi\ne0$; the three-family condition \textup{(C2)} fails on every smooth quintic and on every quotient of a smooth quintic by a group acting freely. \end{theorem} The first part says that the functional carries no information beyond the vacuum it selects: once a viable vacuum is given, a functional with all the required properties is written down in Section~\ref{sec:selection} from the metric alone, and conversely the minimum of any such functional is a viable vacuum. What remains of the conjecture is therefore the existence of one viable vacuum, which is where its content lies. The second part locates the Dwork pencil with respect to that question: its mass spectrum has the full rank and the hierarchy of scales that condition (C8) asks for, and the obstruction is the number of families, a hundred and one on the quintic and $100/|H|$ on a free quotient by a group $H$, never three. The same questions can be asked of the three other one-parameter families of Fermat type, the pencils \begin{equation}\label{eq:wpencil} X_\psi=\Big\{\sum_{i=1}^5x_i^{n_i}-d\psi\,x_1x_2x_3x_4x_5=0\Big\} \subset\PP(w),\qquad n_i=\frac{d}{w_i}, \end{equation} of sextics in $\PP(1,1,1,1,2)$, octics in $\PP(1,1,1,1,4)$ and dectics in $\PP(1,1,1,2,5)$, whose periods and mirror maps were computed by Morrison~\cite{Mor92}, Klemm and Theisen~\cite{KT93} and Font~\cite{Fon93}. Here $d=\sum_iw_i$ is $6$, $8$ and $10$, each member with $\psi^d\delta\ne1$, where $\delta=\prod_iw_i^{w_i}$ is $4$, $256$ and $12\,500$, is smooth, and the group $\Gamma$ of diagonal phases preserving the pencil has order $108$, $128$ and $100$. The answer is different there. \begin{theorem}[Weighted pencils]\label{thm:E} Let $X_\psi$ be the sextic, octic or dectic pencil \eqref{eq:wpencil} with $\psi^d\delta\ne1$. Then $X_\psi$ is a smooth Calabi--Yau threefold with $h^{1,1}=1$ and $h^{2,1}=103$, $149$ and $145$, and the following hold. \begin{enumerate}[label=\textup{(\roman*)},leftmargin=*,itemsep=1pt] \item The generations are the monomials $x^{\bb}$ of weighted degree $d$ with $0\le b_i\le n_i-2$, but they carry only $96$, $122$ and $100$ distinct characters of $\Gamma$: $7$, $27$ and $42$ classes of two or three generations share a character. \item The cubic form, normalised so that $\prod_ix_i^{n_i-2}$ has value one, takes on three generations $\va,\bb,\cc$ the value $\psi^r\prod_iw_i^{k_i}$ when $a_i+b_i+c_i=n_i-2-r+n_ik_i$ for every $i$, with $r\in\{0,\dots,d-1\}$ and integers $k_i\ge0$, and zero otherwise; for $\psi\ne0$ every coupling allowed by $\Gamma$ is nonzero. \item For $\psi\ne0$ the mass map of the invariant direction has rank $99$, $125$ and $109$, so that $4$, $24$ and $36$ generations stay massless at every member; at $\psi=0$ its rank is $44$, $85$ and $72$. \item No member, and no quotient of a member by a group acting freely, carries three families with the standard embedding. \end{enumerate} \end{theorem} So the two properties that make the Dwork pencil transparent are special to it. On the weighted pencils a shift of the exponents at the coordinates of weight one can be balanced by a shift at the coordinate of larger weight, the characters of $\Gamma$ no longer label the generations, and the massless generations all lie in the span of the classes that share a character. The precise form of the mass condition (C8) in Theorem~\ref{thm:D} therefore fails on all three pencils, and so does the three-family condition (C2). Theorem~\ref{thm:E} is proved in Section~\ref{sec:weighted}, and Figure~\ref{fig:weighted} compares the four pencils. The count of families reaches beyond the four pencils. Of the fourteen one-parameter families of Calabi--Yau threefolds with hypergeometric Picard--Fuchs equations, thirteen are complete intersections in weighted projective space, and Proposition~\ref{prop:freequot} shows that a finite group acting freely on a smooth member has order divisible only by the primes that divide the greatest common divisor $g_X$ of the holomorphic Euler characteristics $\chi(X,\cO_X(k))$, $k\ge1$. This number is $1$ for seven of the thirteen types, the sextic and the dectic among them, which therefore have no free quotient at all, and no free quotient of any of the thirteen carries three families. The constructions behind the four theorems were organised on the Anaxia engine of the BLLM programme~\cite{Ana26,BB26}.\footnote{Anaxia runs a failure-driven loop: each strategy for a target is a list of checks, each check is a computation or a proof with a named source, a failed check is recorded together with the change of strategy it forces, and the label of a target (open, candidate, proved) is computed by the engine from the record, never assigned by hand. For this paper the engine ran five targets. For the couplings, the first strategy was the count of couplings allowed by the symmetry, as in~\cite{DGJ08}; its check that the allowed couplings are nonzero failed, because a symmetry count only bounds the number of nonzero couplings from above, and the change it forced was to evaluate the socle functional of Section~\ref{sec:socle} on every monomial of degree fifteen, which passed. For the mass map the single strategy, the pairing by the invariant class, passed at once. For the Fermat point the engine checked the Gamma-function norms against the constant of Distler and Greene~\cite{DG88}. For the selection conjecture the first strategy was to look for the viable vacuum on the quintic itself; its check of the number of families failed for the quintic and for each of its free quotients, and the change it forced was to separate the functional from the vacuum, which gives Theorem~\ref{thm:D}; the check of the existence of a viable vacuum then failed on the literature, and that target stays open. For the weighted pencils the first strategy was to carry the proof of Theorem~\ref{thm:A} over; its check that the characters are distinct failed at once on the sextic, where $103$ generations carry $96$ characters, and the change it forced was to classify the generations that share a character by the wrapping of Lemma~\ref{lem:wwrap} and to compute the mass map block by block, which gives Theorem~\ref{thm:E}. The record of all five targets was made on release 3.1.0 of the engine. The engine proposes no mathematics of its own here: every statement is proved in the text, and the record of the loop is part of the archive of the paper~\cite{Repo}.} \subsection*{History} The dictionary between the vibrations of the string and the fields in four dimensions on a Calabi--Yau background was written down in~\cite{CHSW85}; the decomposition of the adjoint of $E_8$ under $E_6\times SU(3)$ and the identification of the massless charged fields with the Dolbeault cohomology of the tangent bundle are there, and the identification of the number of generations with half the Euler characteristic. Witten~\cite{Wit85} discussed the breaking of $E_6$ by Wilson lines and the pattern of masses from vacuum values of the $\mathbf{27}$, which is the mechanism behind Theorem~\ref{thm:C}. Strominger~\cite{Str85} identified the cubic couplings of the $\mathbf{27}$ with the cubic form on $H^1(X,T_X)$ given by the holomorphic three-form, and the normalisation by the K\"ahler metric of the matter fields was clarified by Dixon, Kaplunovsky and Louis~\cite{DKL90} and by Candelas and de la Ossa~\cite{CdlO91}. Candelas~\cite{Can88} computed the couplings on the Fermat quintic through the Jacobian ring, with the Gamma-function normalisation of the Hodge norms at that point; Greene and Plesser~\cite{GP90} introduced the quotient of the Dwork pencil by $\Gamma$ as the mirror of the quintic, and~\cite{CDGP91} computed the coupling of $h$ with itself along the pencil, $5/(1-\psi^5)$ in their normalisation, which is the case $\va=\bb=\cc=h$ of Theorem~\ref{thm:B} together with Remark~\ref{rem:hessian}. Doran, Greene and Judes~\cite{DGJ08} counted by computer the couplings not forced to vanish by $\Gamma$, found $1431$, and observed that the count is close to the total divided by the order of the group. The Jacobian-ring description of $H^{2,1}$ of a hypersurface and of its cubic form is Griffiths'~\cite{Gri69}, and the perfectness of the residue pairing with the Hessian in the socle is Macaulay's theorem in the form given by Carlson and Griffiths~\cite{CG80}. On the physical side, the holomorphic Yukawa couplings of heterotic line-bundle models were computed by Blesneag, Buchbinder, Candelas and Lukas~\cite{BBCL16}, and the physical, metric-normalised couplings of the quintic at the Fermat point by Butbaia, Mayorga Pe\~na, Tan, Berglund, H\"ubsch, Jejjala and Mishra~\cite{BMT24} with numerical Ricci-flat metrics, whose Gamma-function value for the invariant direction agrees with Section~\ref{sec:fermat}. Heterotic standard models with three families on other threefolds were constructed by Braun, He, Ovrut and Pantev~\cite{BHOP05}, Bouchard and Donagi~\cite{BD06} and Anderson, Gray, Lukas and Palti~\cite{AGLP11,AGLP12}; the question of which classes carry which families is posed there in a form that depends on the choice of bundle, and the present paper answers it in the one case where every piece of the data is explicit. Distler and Greene~\cite{DG88} computed the normalised couplings of the Gepner model at the Fermat point exactly, and Section~\ref{sec:fermat} recovers their constant from the Hodge norms. The monodromy and intersection numbers used there go back to Pham~\cite{Pha65}, Sebastiani and Thom~\cite{ST71} and Sakamoto~\cite{Sak74}. Two further aspects of the same compactifications, the suppression of the electroweak scale by singularities of the threefold and fibrations of Calabi--Yau manifolds by circles, are treated in~\cite{Bha26b,Bha25}. The weighted pencils of Section~\ref{sec:weighted} are the one-parameter models whose Picard--Fuchs equations, periods and mirror maps were found by Morrison~\cite{Mor92}, Klemm and Theisen~\cite{KT93} and Font~\cite{Fon93}, and the Hodge theory of quasi-smooth hypersurfaces in weighted projective space used there is due to Steenbrink~\cite{Ste77} and Dolgachev~\cite{Dol82}. \section{The dictionary}\label{sec:dictionary} Throughout, $X$ is a compact K\"ahler threefold with holonomy exactly $SU(3)$, so that $X$ has a Ricci-flat K\"ahler metric $g$ by Yau's theorem, a nowhere vanishing holomorphic three-form $\Omega$, and $h^{1,0}=h^{2,0}=0$. The tangent bundle $T_X$ with the Levi-Civita connection is a holomorphic $SU(3)$-bundle with a Hermitian Yang--Mills connection; this is the standard embedding. \subsection{Spinors and the Dolbeault complex} On a K\"ahler manifold of complex dimension $n$ with trivial canonical bundle the spinor bundle is $S=\bigoplus_{q}\Lambda^{0,q}T^*_X$, and the Dirac operator of the Levi-Civita connection is $D=\sqrt2(\bar\partial+ \bar\partial^*)$, with the chirality grading given by the parity of $q$ (Hitchin~\cite{Hit74}). If $V$ is a holomorphic Hermitian bundle with a connection whose curvature is of type $(1,1)$, the Dirac operator twisted by $V$ is $\sqrt2(\bar\partial_V+\bar\partial_V^*)$ on $\Lambda^{0,\bullet} \otimes V$, and its kernel is the harmonic representative of $\bigoplus_q H^{0,q}(X,V)$. This is the only input from spin geometry. \subsection{The charged spectrum} \begin{theorem}[Spectrum of the standard embedding]\label{thm:dictionary} Let $X$ be as above with the standard embedding. The massless charged chiral multiplets of the $E_8\times E_8$ heterotic string on $\R^{3,1} \times X$ in the observable sector transform under $E_6$ as \[ h^{2,1}(X)\ \text{copies of }\mathbf{27}\ \oplus\ h^{1,1}(X)\ \text{copies of }\overline{\mathbf{27}}, \] with the $\mathbf{27}$'s in natural bijection with $H^1(X,T_X)\cong H^{2,1}(X)$ and the $\overline{\mathbf{27}}$'s with $H^1(X,T_X^*)\cong H^{1,1}(X)$. The net number of generations is $|h^{2,1}-h^{1,1}|=\tfrac12|\chi(X)|$. \end{theorem} \begin{proof} The ten-dimensional gaugino is a Majorana--Weyl spinor with values in the adjoint of $E_8$. Under $E_8\supset E_6\times SU(3)$, \[ \mathbf{248}=(\mathbf{78},\mathbf1)\oplus(\mathbf1,\mathbf8)\oplus (\mathbf{27},\mathbf3)\oplus(\overline{\mathbf{27}},\bar{\mathbf3}), \] and the standard embedding identifies the $SU(3)$ factor with the holonomy group acting on $T_X$ and sends $\mathbf3$ to $T_X$, $\bar{\mathbf3}$ to $T_X^*$. A ten-dimensional Weyl spinor on $\R^{3,1}\times X$ decomposes as a four-dimensional Weyl spinor times a six-dimensional one of correlated chirality, so the four-dimensional massless chiral fermions in the $\mathbf{27}$ are the zero modes of the Dirac operator on $X$ twisted by $T_X$, counted with their six-dimensional chirality; the $SU(3)$ holonomy makes the twisted operator $\sqrt2(\bar\partial+\bar\partial^*)$ on $\Lambda^{0,\bullet}\otimes T_X$, and its chiral zero modes are $H^{0,1}(X,T_X)\oplus H^{0,3}(X,T_X)$ and $H^{0,0}(X,T_X)\oplus H^{0,2}(X,T_X)$ for the two chiralities. Since $\Omega$ trivialises $\Lambda^3T_X^*$, contraction with $\Omega$ gives $H^{0,q}(X,T_X)\cong H^{2,q}(X)$ and $H^{0,q}(X,T_X^*)=H^{1,q}(X)$. With $h^{2,0}=h^{1,0}=0$ and Serre duality, $H^{0,0}(T_X)=H^{2,0}=0$, $H^{0,3}(T_X)= H^{2,3}=\overline{H^{1,0}}^*=0$, $H^{0,2}(T_X)=H^{2,2}\cong \overline{H^{1,1}}$, and $H^{0,1}(T_X)=H^{2,1}$. A four-dimensional Weyl fermion in the $\mathbf{27}$ of one chirality and one in the $\overline{\mathbf{27}}$ of the opposite chirality are related by conjugation, so the chiral multiplets of definite four-dimensional chirality are $h^{2,1}$ copies of $\mathbf{27}$ from $H^{0,1}(T_X)$ and $h^{1,1}$ copies of $\overline{\mathbf{27}}$ from $H^{0,2}(T_X)$. The index of the twisted Dirac operator is $h^{2,1}-h^{1,1}$, which equals $-\chi(X)/2$ because $\chi=2(h^{1,1}-h^{2,1})$ on a threefold with $h^{1,0}=h^{2,0}=0$. \end{proof} \subsection{The cubic form} Let $V=H^1(X,T_X)$. The composition of cup product and contraction, \[ V\otimes V\otimes V\to H^3(X,\Lambda^3T_X)=H^3(X,K_X^{-1})\cong H^3(X,\cO_X)\cong\C, \] where the last two isomorphisms use $\Omega$, is a symmetric trilinear form $Y$ on $V$, the cubic form of $X$. In terms of the variation of Hodge structure, with $\nabla$ the Gauss--Manin connection and $\Omega$ extended holomorphically over the moduli space, \begin{equation}\label{eq:yukawa} Y(a,b,c)=\int_X\Omega\wedge\nabla_a\nabla_b\nabla_c\Omega, \end{equation} and this is the coupling of three $\mathbf{27}$'s~\cite{Str85}. Changing $\Omega$ to $f\Omega$ multiplies $Y$ by $f^2$, so only the cubic form up to a nowhere vanishing holomorphic function is intrinsic, and the physical couplings are $Y$ divided by the square roots of the Hodge norms of the three classes and by $\ip{\Omega}{\overline\Omega}$~\cite{DKL90,CdlO91}. \subsection{Mass terms} A vacuum value $v$ of the scalar in the $\mathbf{27}$ labelled by a class $h\in V$ in a direction that breaks $E_6$ gives, through the superpotential $Y(h,\cdot,\cdot)$, a Dirac mass term $v\,Y(h,b,c)$ between the fermions labelled by $b$ and $c$~\cite{Wit85}. The physical mass matrix is \begin{equation}\label{eq:massmatrix} M_{bc}=v\,\frac{Y(h,b,c)}{\sqrt{g_{b}\,g_{c}}} \end{equation} in an orthogonal basis of $V$, where $g_b=\ip{\omega_b}{\overline{\omega_b}}$ is the Hodge norm of the harmonic representative, and the masses are its singular values. We call $Y(h,\cdot,\cdot)$ the mass map of $h$. \subsection{Hypersurfaces} For a smooth hypersurface $X=\{F=0\}\subset\PP^4$ of degree five let $J_F=(\partial_1F,\dots,\partial_5F)$ be the Jacobian ideal and $R=\C[x_1,\dots,x_5]/J_F$ the Jacobian ring, graded by degree. Griffiths' theorem~\cite{Gri69} identifies $H^{3-q,q}_{\mathrm{prim}}(X)$ with $R_{5(q+1)-5}$ through residues: for a homogeneous $P$ of degree $5q$, \[ \omega_P=\Res_X\frac{P\,\Omega_0}{F^{q+1}},\qquad \Omega_0=\sum_i(-1)^{i}x_i\,dx_1\wedge\cdots\widehat{dx_i}\cdots\wedge dx_5, \] so $H^{3,0}=R_0$, $H^{2,1}=R_5$, $H^{1,2}=R_{10}$, $H^{0,3}=R_{15}$, and every class of a quintic threefold is primitive in degree three. The ring $R$ is a graded complete intersection with Hilbert series $(1+t+t^2+t^3)^5$, so $\dim R_5=101$, $\dim R_{15}=1$, and $R_{15}$ is the socle. Macaulay's theorem, as stated in~\cite{CG80}, says that the Hessian $\Hess F=\det(\partial_i\partial_jF)$ has nonzero image in $R_{15}$ and that the cubic form is multiplication in $R$ followed by the residue functional $\lambda_F:R_{15}\to\C$, $\lambda_F(\Hess F)=1$, up to a constant depending only on the normalisation of $\Omega$: \begin{equation}\label{eq:jacobianyukawa} Y(\omega_P,\omega_Q,\omega_S)=c\,\lambda_F(PQS). \end{equation} Figure~\ref{fig:hodge} shows a curve on the Fermat quintic and the Hodge numbers of a quintic threefold: by Theorem~\ref{thm:dictionary} the hundred and one classes of type $(2,1)$ are the generations and the single class of type $(1,1)$ is the antigeneration. \begin{figure}[t] \centering \includegraphics[width=0.96\textwidth]{figures/fig_hodge.png} \caption{Left: the intersection of the Fermat quintic with the plane $x_3=x_4=0$, which in an affine chart is the curve $z_1^5+z_2^5=1$ in $\C^2$, drawn with the parametrisation of Hanson~\cite{Han94} and projected to three dimensions along $(\mathrm{Re}\,z_1,\mathrm{Re}\,z_2, \cos\alpha\,\mathrm{Im}\,z_1+\sin\alpha\,\mathrm{Im}\,z_2)$. It is assembled from twenty-five patches, one for each pair of fifth roots of unity; $\Gamma$ preserves the curve and acts on it through $\mu_5\times\mu_5$, permuting the patches. Right: the Hodge numbers $h^{p,q}$ of a smooth quintic threefold, with height proportional to $1+\log_2h^{p,q}$; the six blue bars have height one, the two red bars are $h^{2,1}=h^{1,2}=101$, and the grey cells are zero.} \label{fig:hodge} \end{figure} \section{Labels}\label{sec:labels} Let $\zeta=e^{2\pi i/5}$ and let $K=\{a\in(\Z/5)^5:\sum a_i=0\}$ act on $\PP^4$ by $x_i\mapsto\zeta^{a_i}x_i$. The diagonal subgroup $D$ acts trivially on $\PP^4$, and $\Gamma=K/D\cong(\Z/5)^3$ preserves $F_\psi= \sum x_i^5-5\psi\prod x_i$ for every $\psi$. The characters of $\Gamma$ are the characters of $K$ trivial on $D$, that is the classes \[ \widehat\Gamma=\{\bb\in(\Z/5)^5:\textstyle\sum b_i=0\}/\langle\one\rangle, \qquad |\widehat\Gamma|=125, \] with $\bb$ acting by $a\mapsto\zeta^{a\cdot\bb}$. A monomial $x^{\bb}$ transforms under $a\in K$ by $\zeta^{a\cdot\bb}$, so it lies in the isotypic component of the class of $\bb$. Let $B$ be the set of $\bb\in\{0,1,2,3\}^5$ with $|\bb|=\sum b_i=5$. The monomials $x^{\bb}$, $\bb\in B$, are a basis of $R_5$ for $\psi^5\ne1$, because the Jacobian ideal of $F_\psi$ is generated by $x_i^4-\psi\prod_{j \ne i}x_j$, and modulo these every monomial reduces to one with all exponents at most three; the reduced monomials of degree five are the $101$ elements of $B$, which matches $\dim R_5$. The count is the number of compositions of five into five parts, $\binom94=126$, less the twenty with a part equal to four and the five with a part equal to five. \begin{proof}[Proof of Theorem~\ref{thm:A}] Two elements $\bb,\bb'\in B$ define the same character when $\bb'\equiv\bb+k\one\pmod5$ for some $k$. Since the entries of $\bb'$ lie in $\{0,\dots,4\}$, $\bb'$ is the vector of residues of $\bb+k\one$ in that range: each $b'_i$ is $b_i+k$ or $b_i+k-5$, and $\sum b'_i=5$. For $k=1$ the residues are $b_i+1\le4$ unless $b_i=4$, impossible, so all residues are $b_i+1$, with sum $10\ne5$. For $k=2$ the residues are $b_i+2$ if $b_i\le2$ and $b_i-3$ if $b_i=3$; the sum is $15-5m$ where $m$ is the number of threes, so $m=2$, but two threes already sum to six. For $k=3$ the residues are $b_i+3$ if $b_i\le1$ and $b_i-2$ if $b_i\ge2$, sum $20-5m'$ with $m'$ the number of entries at least two, so $m'=3$, three entries at least two summing to at most five, impossible. For $k=4$ the residues are $b_i+4$ if $b_i=0$ and $b_i-1$ otherwise, sum $25-5m''$ with $m''$ the number of nonzero entries, so $m''=4$: four nonzero entries summing to five means $(2,1,1,1,0)$ up to order, and then $\bb'=(1,0,0,0,4)$ has a four. So only $k=0$ works and the $101$ characters are distinct; being characters of an abelian group acting on $R_5$, they are then the full decomposition, each isotypic component being one-dimensional. The group $\widehat\Gamma$ has $125$ elements and $B$ realises $101$ of them, so exactly $24$ are missed. A permutation of $(0,1,2,3,4)$ has sum ten, so its class lies in $\widehat\Gamma$, and adding $\one$ to it gives another permutation of $(0,1,2,3,4)$ modulo five; the hundred and twenty permutations therefore fall into $24$ classes of five. None of these classes meets $B$, because every representative with entries in $\{0,\dots,4\}$ is a permutation of $(0,1,2,3,4)$ and contains a four. Hence the $24$ missed classes are exactly the classes of the permutations. Finally $S_5$ permutes the exponent vectors, and the orbits in $B$ are the multisets of entries: $\{1,1,1,1,1\}$ (one element), $\{2,1,1,1,0\}$ (twenty), $\{3,2,0,0,0\}$ (twenty), $\{2,2,1,0,0\}$ (thirty) and $\{3,1,1,0,0\}$ (thirty). \end{proof} Figure~\ref{fig:labels} shows the hundred and one labels. \begin{figure}[t] \centering \includegraphics[width=0.96\textwidth]{figures/fig_labels.png} \caption{The hundred and one generations of the Dwork quintic as points of the character group of $\Gamma$. Left: $\widehat\Gamma\cong(\Z/5)^3$ drawn as a cube of $125$ points, the class of $\bb$ placed at $(b_0-b_4,b_1-b_4,b_2-b_4)\bmod5$; the $101$ labels are filled and coloured by type (the five $S_5$-orbits), and the $24$ classes of permutations of $(0,1,2,3,4)$ are hollow. Right: the involution $\sigma$ of Theorem~\ref{thm:C}, drawn as chords between paired labels, with the chord colour giving the power of $\psi$ of the pairing: black for $1$, blue for $\psi$, red for $\psi^2$; the fixed label is the invariant class $h$ at the centre.} \label{fig:labels} \end{figure} \section{The socle functional}\label{sec:socle} Fix $\psi$ with $\psi^5\ne1$, write $R=R_{F_\psi}$ and $m_0=x_1^3x_2^3 x_3^3x_4^3x_5^3$, the only monomial of degree fifteen with all exponents at most three. Define the sector of a monomial $x^{\ee}$ of degree fifteen as follows: $\ee$ is in sector $r\in\{0,1,2,3,4\}$ if $e_i\equiv 3-r\pmod5$ for every $i$; otherwise $\ee$ is in no sector. Sector $4$ is empty, because its exponents are at least four each. \begin{theorem}[Socle functional]\label{thm:socle} For $\psi^5\ne1$, $m_0\ne0$ in $R_{15}$, and the linear functional $\lambda$ on $R_{15}$ with $\lambda(m_0)=1$ satisfies, for every monomial $x^{\ee}$ of degree fifteen, \[ \lambda(x^{\ee})=\begin{cases}\psi^{r}&\text{if $\ee$ is in sector $r$,}\\ 0&\text{otherwise.}\end{cases} \] Moreover $\lambda(\Hess F_\psi)=3\,200\,000\,(1-\psi^5)$. \end{theorem} \begin{proof} The relations $x_i^4\equiv\psi\prod_{j\ne i}x_j$ in $R$ give, for every exponent vector $\ee$ with $e_i\ge4$, \begin{equation}\label{eq:reduce} x^{\ee}\equiv\psi\,x^{\ee'},\qquad \ee'=\ee+\one-5\,\ee_i, \end{equation} where $\ee_i$ is the $i$-th unit vector. If $\ee$ is in sector $r$ then $\ee'$ is in sector $r-1$: its $i$-th entry is $e_i-4\equiv 3-(r-1)$ and the other entries are $e_j+1\equiv3-(r-1)$. The sectors are listed by hand. In sector $0$ the entries are congruent to $3$ and sum to fifteen, so they are all equal to three and $\ee=(3,3,3,3,3)$. In sector $1$ the entries are congruent to two, so each is $2$, $7$ or $12$; a twelve would leave three for four entries of at least two, so the entries are $2$ or $7$, and a total of fifteen forces exactly one seven: $\ee$ is a permutation of $(7,2,2,2,2)$, and \eqref{eq:reduce} applied to the seven gives $\psi\,m_0$. In sector $2$ the entries are $1$, $6$ or $11$, and the total forces two sixes and three ones (an eleven would leave four for the other four entries, which are congruent to one and so are all one, a total of fifteen; but then $\ee=(11,1,1,1,1)$, and \eqref{eq:reduce} applied to the eleven gives $(7,2,2,2,2)$, in sector $1$ as well). In sector $3$ the entries are multiples of five, so $\ee$ is a permutation of $(15,0,0,0,0)$, $(10,5,0,0,0)$ or $(5,5,5,0,0)$, and one reduction at an entry of at least five lands in sector $2$. By induction on $r$, every monomial in sector $r$ satisfies $x^{\ee}\equiv\psi^rm_0$ in $R$.\footnote{The reduction \eqref{eq:reduce} uses only the relation $\partial_iF_\psi/5=x_i^4-\psi \prod_{j\ne i}x_j$. It lowers the $i$-th exponent by four and raises each of the other four by one, so it preserves the degree and moves the class of $\ee$ modulo $\one$ by $-5\ee_i+\one\equiv\one$; this is why the sectors, which are the classes $(3-r)\one$, are permuted cyclically and why each reduction costs exactly one factor of $\psi$.} If $\ee$ is in no sector, the monomial $x^{\ee}$ has a nontrivial character of $\Gamma$: its character is the class of $\ee$ modulo $\one$, which is trivial exactly when $\ee\equiv c\one$ for some $c$, that is, when $\ee$ is in a sector. The ideal $J_{F_\psi}$ is $\Gamma$-stable, so $R$ is graded by the characters of $\Gamma$ and $x^{\ee}$ lies in a nontrivial isotypic component of $R_{15}$. Since $R_{15}$ is one-dimensional it carries a single character, and $m_0$ carries the trivial one; so if $m_0\ne0$, every nontrivial component vanishes and $x^{\ee}\equiv0$. It remains to show that $m_0\ne0$. The Hessian $\Hess F_\psi$ has degree fifteen and is $\Gamma$-invariant, so its monomials in no sector vanish in $R$ and the others are multiples of $m_0$: $\Hess F_\psi\equiv H(\psi)\,m_0$, with $H(\psi)$ the sum over the sector monomials of the Hessian of their coefficients times $\psi^r$. By Macaulay's theorem $\Hess F_\psi\ne0$ in $R_{15}$, so $m_0\ne0$ and $H(\psi)\ne0$. The Hessian matrix has diagonal entries $20x_i^3$ and off-diagonal entries $-5\psi\prod_{k\ne i,j}x_k$, each of degree three; an off-diagonal entry omits exactly the two coordinates of its row and column. The determinant is the sum over permutations $\pi$ of the sign of $\pi$ times $\prod_i(\text{entry }(i,\pi(i)))$, and the monomial of $\pi$ depends only on its cycle type: a fixed point $a$ contributes $x_a^3$ from the diagonal and $x_a$ from every off-diagonal factor, and a moved coordinate $m$ contributes $x_m$ from every off-diagonal factor whose row and column both differ from $m$. Table~\ref{tab:hessian} lists the seven cycle types; the derivation of each row is in Appendix~\ref{app:hessian}. Summing the last column, \begin{align*} H(\psi)&=3\,200\,000-\big(2\,000\,000+1\,000\,000+375\,000\\ &\qquad+75\,000-187\,500-62\,500\big)\psi^5\\ &=3\,200\,000\,(1-\psi^5). \qedhere \end{align*} \end{proof} \begin{table}[t] \caption{The Hessian of $F_\psi$ in the socle, by cycle type of the permutation. Each row gives the number of permutations, their sign, the monomial they contribute, its sector $r$, and the total contribution to $H(\psi)$ after the reduction $x^{\ee}\equiv\psi^rm_0$.} \label{tab:hessian} \centering \begin{tabular}{lrrlrr} \toprule type & count & sign & monomial & $r$ & contribution\\ \midrule $1^5$ & 1 & $+$ & $(3,3,3,3,3)$ & 0 & $3\,200\,000$\\ $2\,1^3$ & 10 & $-$ & $(5,5,5,0,0)$ & 3 & $-2\,000\,000\,\psi^5$\\ $3\,1^2$ & 20 & $+$ & $(6,6,1,1,1)$ & 2 & $-1\,000\,000\,\psi^5$\\ $4\,1$ & 30 & $-$ & $(7,2,2,2,2)$ & 1 & $-375\,000\,\psi^5$\\ $5$ & 24 & $+$ & $(3,3,3,3,3)$ & 0 & $-75\,000\,\psi^5$\\ $2^2\,1$ & 15 & $+$ & $(7,2,2,2,2)$ & 1 & $187\,500\,\psi^5$\\ $3\,2$ & 20 & $-$ & $(3,3,3,3,3)$ & 0 & $62\,500\,\psi^5$\\ \bottomrule \end{tabular} \end{table} \begin{remark}[Normalisation]\label{rem:hessian} With the residue normalisation $\lambda_{F}(\Hess F)=1$ of \eqref{eq:jacobianyukawa}, the cubic form of the pencil is $\lambda_{F_\psi}=\lambda/(3\,200\,000(1-\psi^5))$. For $\va=\bb=\cc=h$ the monomial is $x^{3\one}=m_0$, so $Y(h,h,h)$ is a constant times $1/(1-\psi^5)$, which is the coupling $5/(1-\psi^5)$ of~\cite{CDGP91} in the gauge in which $\Omega$ is the residue of $\Omega_0/F_\psi$; the factor five is the constant $c$ of \eqref{eq:jacobianyukawa} in that gauge. All ratios of couplings at fixed $\psi$ are independent of these constants, and Theorem~\ref{thm:B} is stated for $\lambda$. \end{remark} \section{Couplings}\label{sec:couplings} \begin{proof}[Proof of Theorem~\ref{thm:B}] By \eqref{eq:jacobianyukawa} and Theorem~\ref{thm:socle}, $Y(x^{\va},x^{\bb},x^{\cc})=\lambda(x^{\va+\bb+\cc})$ up to the normalisation of Remark~\ref{rem:hessian}, and $\va+\bb+\cc$ has degree fifteen; the value is $\psi^r$ if $\va+\bb+\cc\equiv(3-r)\one$ and zero otherwise. For $\psi\ne0$ every sector value is nonzero. The count of the unordered triples is in Section~\ref{sec:count}. \end{proof} \subsection{The count}\label{sec:count} A triple $\{\va,\bb,\cc\}$ of elements of $B$, with repetitions allowed, has a nonzero coupling exactly when $\va+\bb+\cc\equiv c\one\pmod5$ for some $c$, and then $c\equiv3-r$ with $r$ the sector. We count ordered triples first. The congruence says that the character of $\cc$ is the inverse of the character of $\va+\bb$ in $\widehat\Gamma$. Every character occurs in $B$ at most once (Theorem~\ref{thm:A}), so for each ordered pair $(\va,\bb)$ there is at most one $\cc$, and there is one exactly when the class of $-(\va+\bb)$ is one of the hundred and one realised classes, that is, when it is not the class of a permutation of $(0,1,2,3,4)$. The negative of a permutation of $(0,1,2,3,4)$ is, modulo five, a permutation of $(0,1,2,3,4)$ again, so this fails exactly when $\va+\bb$ has five distinct residues modulo five. \emph{Excluded pairs.} The entries of $\va+\bb$ lie in $\{0,\dots,6\}$ and sum to ten. Five distinct residues are the residues $0,1,2,3,4$, whose sum is already ten, so no entry can exceed four and $\va+\bb$ is a permutation of $(0,1,2,3,4)$. Fix the permutation, so that the coordinate $k$ of $\va+\bb$ equals $k$ for $k=0,\dots,4$ after renaming. Write $p_k$ for the part given to $\va$ in coordinate $k$; then $0\le p_k\le\min(k,3)$ and $k-p_k\le3$, so $p_0=0$, $p_1\in\{0,1\}$, $p_2\in\{0,1,2\}$, $p_3\in\{0,1,2,3\}$ and $p_4\in\{1,2,3\}$, and $|\va|=5$ means $p_1+p_2+p_3+p_4=5$. The number of solutions is the coefficient of $t^5$ in \[ (1+t)(1+t+t^2)(1+t+t^2+t^3)(t+t^2+t^3), \] which is $16$.\footnote{Expand $(1+t)(1+t+t^2)=1+2t+2t^2+t^3$ and $(1+t+t^2+t^3)(t+t^2+t^3)=t+2t^2+3t^3+3t^4+2t^5+t^6$. The coefficient of $t^5$ in the product is $1\cdot2+2\cdot3+2\cdot3+1\cdot2=16$, pairing the terms of degree $0,1,2,3$ of the first factor with those of degree $5,4,3,2$ of the second.} With $120$ permutations there are $1920$ excluded ordered pairs, so the number of ordered triples with a nonzero coupling is $101^2-1920=8281$. \emph{Repetitions.} An ordered triple $(\va,\va,\va)$ has a nonzero coupling when $3\va\equiv c\one$, that is, when all entries of $\va$ are congruent modulo five; entries in $\{0,1,2,3\}$ summing to five then force $\va=\one$. For a triple $(\va,\va,\cc)$ the partner $\cc$ exists unless $2\va$ has five distinct residues; the residues of $2a_i$ for $a_i=0,1,2,3$ are $0,2,4,1$, and five distinct residues would need five distinct entries of $\va$ in a set of four values, which is impossible. So every $\va$ has a partner $\cc$, and $\cc=\va$ only for $\va=\one$. There are therefore $100$ unordered triples of the form $\{\va,\va,\cc\}$ with $\cc\ne\va$, giving $300$ ordered triples, one triple $\{\one,\one,\one\}$, and $8281-300-1=7980$ ordered triples with three distinct entries. The number of unordered triples is \[ \frac{7980}{6}+100+1=1330+101=1431. \] \emph{Sectors.} In sector $r$ the sum $\ee=\va+\bb+\cc$ is one of the vectors listed in the proof of Theorem~\ref{thm:socle} with entries at most nine: $(3,3,3,3,3)$ for $r=0$, the five permutations of $(7,2,2,2,2)$ for $r=1$, the ten permutations of $(6,6,1,1,1)$ for $r=2$ and the ten permutations of $(5,5,5,0,0)$ for $r=3$. The number $O(\ee)$ of ordered triples with sum $\ee$ is the coefficient of $x^5y^5$ in $\prod_{i=1}^5P_{e_i}(x,y)$, where $P_e(x,y)=\sum x^ay^b$ over the pairs $0\le a,b\le3$ with $0\le e-a-b\le3$. These coefficients are \begin{gather*} O(3,3,3,3,3)=3391,\quad O(7,2,2,2,2)=480,\\ O(6,6,1,1,1)=162,\quad O(5,5,5,0,0)=87, \end{gather*} so the ordered counts by sector are $3391$, $2400$, $1620$ and $870$, with total $8281$. The hundred unordered triples $\{\va,\va,\cc\}$ with $\cc\ne\va$ split as $0$, $50$, $20$ and $30$ by sector (the sector of $2\va+\cc$), and $\{\one,\one,\one\}$ lies in sector $0$, so the unordered counts are \[ N_r=\frac{O_r-3D_r-T_r}{6}+D_r+T_r=566,\ 425,\ 280,\ 160, \] with $O_r$, $D_r$, $T_r$ the ordered, repeated and constant counts of sector $r$. The coefficient extractions are written out in Appendix~\ref{app:count}. \section{The mass map}\label{sec:mass} \begin{proof}[Proof of Theorem~\ref{thm:C}] For $\bb\in B$ the coupling $\lambda(x^{\bb+\cc+\one})$ is nonzero exactly when $\bb+\cc+\one\equiv(3-r)\one$, that is $\cc\equiv(2-r)\one-\bb\pmod5$, with $\cc\in B$. For each $r$ the residues of $(2-r)\one-\bb$ are determined, and $\cc$ exists in $B$ exactly when those residues, taken in $\{0,1,2,3\}$ (a residue $4$ excludes), sum to five. Go through the types. If all $b_i\le2$ then $r=0$ gives $\cc=2\one-\bb\in\{0,1,2\}^5$ of sum five: this is the partner. For $r=1$ the residues of $\one-\bb$ are $1,0,4$ for $b_i=0,1,2$, so a two in $\bb$ excludes, and $\bb\in\{0,1\}^5$ of sum five means $\bb=\one$, whose $r=1$ residues are all zero, sum zero: excluded. For $r=2$ the residues of $-\bb$ are $0,4,3$: a one excludes, and $\bb\in\{0,2\}^5$ has even sum: excluded. For $r=3$ the residues of $-\one-\bb$ are $4,3,2$: a zero excludes, so $\bb\in\{1,2\}^5$ of sum five means $\bb=\one$, residues all $3$, sum fifteen: excluded. So a label with entries at most two has exactly one partner, $2\one-\bb$, with coupling $1$; this covers the fifty-one labels of the types $\{1,1,1,1,1\}$, $\{2,1,1,1,0\}$ and $\{2,2,1,0,0\}$, and $\sigma(\bb)=2\one-\bb$ is of the same type, with the single fixed point $\bb=\one$. If $\bb$ has type $\{3,1,1,0,0\}$: $r=0$ gives residue $4$ at the three, excluded; $r=1$ gives residues $3,0,0,1,1$, sum five, so $\cc$ is the label with $3$ where $\bb$ has $3$, zeros where $\bb$ has ones and ones where $\bb$ has zeros, of the same type, with coupling $\psi$; $r=2$ gives residues $2,4,\dots$, excluded by the ones; $r=3$ gives residues $1,3,3,4,4$, excluded. If $\bb$ has type $\{3,2,0,0,0\}$: $r=0$ excluded by the three; $r=1$ gives residues $3,4,\dots$, excluded by the two; $r=2$ gives residues $2,3,0,0,0$, sum five, so $\cc$ swaps the three and the two of $\bb$, of the same type, coupling $\psi^2$; $r=3$ gives residues $1,2,4,4,4$, excluded. This proves that every label has exactly one partner, that $\sigma$ is an involution preserving type with the one fixed point $\one$, and the three values of the coupling. The matrix $M=(\lambda(x^{\bb+\cc+ \one}))_{\bb,\cc}$ is therefore a monomial matrix with entries $1$ (fifty-one times), $\psi$ (thirty times) and $\psi^2$ (twenty times), so its determinant is $\psi^{30+40}=\psi^{70}$, the sign of $\sigma$, a product of fifty transpositions, being $+1$, and its rank is $101$ for $\psi\ne0$. At $\psi=0$ the rows of the fifty labels containing a three vanish and the remaining fifty-one rows are a permutation matrix on the fifty-one labels with entries at most two: rank fifty-one. The $\Gamma$-invariant Hermitian metric on $H^{2,1}$, the Hodge metric, pairs the isotypic components of distinct characters to zero because a unitary representation of a finite abelian group is an orthogonal sum of its isotypic components; by Theorem~\ref{thm:A} the components are the lines spanned by the $x^{\bb}$, so the metric is diagonal with entries $g_{\bb}$. The mass matrix \eqref{eq:massmatrix} is then $v$ times a monomial matrix with entries $\psi^{r(\bb)}(g_{\bb}g_{\sigma\bb})^{-1/2}$ at $(\bb,\sigma \bb)$. The singular values of a symmetric monomial matrix with an involution are the absolute values of its entries, each two-cycle giving a doubly degenerate pair and the fixed point a singlet; this is the stated spectrum. \end{proof} \begin{figure}[t] \centering \includegraphics[width=0.96\textwidth]{figures/fig_masses.png} \caption{The mass spectrum of Theorem~\ref{thm:C} along the pencil. Left: the three mass scales $1$, $|\psi|$ and $|\psi|^2$ against $|\psi|$ for $0\le|\psi|<1$, with the multiplicities of the labels (fifty-one, thirty, twenty) that carry them; the singlet $h$ sits in the first group. Right: the rank of the mass map over the $\psi$-plane, fifty-one at the Fermat point and a hundred and one elsewhere (the drop at $\psi=0$ is drawn with a small width); the conifold points $\psi^5=1$, where the Jacobian ring degenerates, are marked on the unit circle.} \label{fig:masses} \end{figure} \begin{remark}[Degenerations] The determinant $\pm\psi^{70}$ shows that the mass map of the invariant direction degenerates on the pencil only at the Fermat point, where the group of automorphisms of $X_\psi$ jumps from $\Gamma\rtimes S_5$ to the full group of order $5^4\cdot120$ of the Fermat quintic. The fifty massless generations at $\psi=0$ are the labels with an exponent three. The larger group acts on a monomial through its full exponent vector, not only through its class modulo $\one$, so at $\psi=0$ a coupling of $h$, $\bb$ and $\cc$ survives only if $\one+\bb+\cc=(3,3,3,3,3)$; for a label with an exponent three the required partner $2\one-\bb$ has a negative entry. At the conifold points $\psi^5=1$ the matrix $M$ has nonzero determinant, but the normalisation of Remark~\ref{rem:hessian} has a pole: the physical couplings diverge as $1/(1-\psi^5)$ in the gauge of $\Omega$ chosen there, and the Hodge norm of $h$ diverges logarithmically, as in~\cite{CDGP91}. \end{remark} \section{The Fermat point}\label{sec:fermat} At $\psi=0$ the Hodge norms are explicit. \begin{proposition}[Hodge norms at the Fermat point]\label{prop:fermat} Let $X_0$ be the Fermat quintic and $\omega_{\bb}$ the class of $x^{\bb}$, $\bb\in B$. Write $k_i=b_i+1$ and $\gamma(t)=\Gamma(t)/\Gamma(1-t)$. There is a constant $C>0$ independent of $\bb$ such that \begin{equation}\label{eq:gamma} g_{\bb}=\ip{\omega_{\bb}}{\overline{\omega_{\bb}}}=C\prod_{i=1}^5 \gamma(k_i/5). \end{equation} \end{proposition} \begin{proof} The full group $\mu_5^5/\mu_5$ of diagonal phases acts on $X_0$, and $\omega_{\bb}$ spans the eigenline of the character $\mathbf k=(k_i)$, with $1\le k_i\le4$ and $\sum k_i=10$. Eigenlines of distinct characters are orthogonal for the Hodge metric, so $g_{\bb}$ is the only entry of the metric on that line, and it is computed from the periods of $\omega_{\bb}$ by the second Riemann bilinear relation: if $P$ is an integral cycle and $P_{\mathbf k}$ its component in the eigenspace of $\mathbf k$ in $H_3(X_0,\C)$, then $g_{\bb}=|\int_P\omega_{\bb}|^2/ |P_{\mathbf k}\cdot P_{-\mathbf k}|$ up to a factor depending only on the Hodge type. For the cycle of Pham, the join of the five sets of fifth roots of unity, the period is a product of beta integrals, and the intersection number is a product of factors $1-\zeta^{k_i}$; combining them with $\Gamma(t)\Gamma(1-t)=\pi/\sin\pi t$ gives \eqref{eq:gamma}. This is the computation of Candelas~\cite{Can88} on the Fermat quintic and, in the language of the Gepner model, of Distler and Greene~\cite{DG88}; the steps are written out in Appendix~\ref{app:fermat}. \end{proof} The proposition fixes the ratios of the Hodge norms inside $H^{2,1}$, which is all that the masses of Theorem~\ref{thm:C} need. Since $\gamma(1-t)= 1/\gamma(t)$, every norm is a monomial in the two numbers $\gamma(1/5)=3.94325$ and $\gamma(2/5)=1.48951$. For the singlet, $\mathbf k=(2,2,2,2,2)$ and $g_{\one}=C\gamma(2/5)^5$; for the type $\{2,1,1,1,0\}$, $\mathbf k$ is a permutation of $(3,2,2,2,1)$ and $g_{\bb}=C\gamma(2/5)^2\gamma(1/5)$; for the type $\{2,2,1,0,0\}$, $\mathbf k$ is a permutation of $(3,3,2,1,1)$ and $g_{\bb}=C\gamma(1/5)^2/ \gamma(2/5)$. The ratio of the second to the first is $\Gamma(3/5)^3\Gamma(1/5)/\Gamma(2/5)^3\Gamma(4/5)=1.19324$, the square of the constant $\kappa=1.09236$ in which Distler and Greene~\cite{DG88} expressed the normalised couplings of the Gepner model (see~\cite[eq.~(4.3)]{BMT24}). The norm of $\Omega$, the class of $1$ with $\mathbf k=(1,1,1,1,1)$, has the same shape with a different constant, and the Weil--Petersson metric of the pencil at the Fermat point, computed from the periods by Candelas, de la Ossa, Green and Parkes~\cite{CDGP91}, is \[ G_{\psi\bar\psi}(0)=25\,\frac{\Gamma(2/5)^5\,\Gamma(4/5)^5} {\Gamma(1/5)^5\,\Gamma(3/5)^5}=0.19226, \] the value that Butbaia and collaborators~\cite{BMT24} reproduce with numerical Ricci-flat metrics.\footnote{In~\cite{BMT24} the value is given for the mirror quintic, whose complex structure moduli space is the $\psi$-line of the Dwork pencil divided by the symmetry $\psi\mapsto\zeta\psi$. The Weil--Petersson metric is $-\partial\bar\partial\log\int\Omega\wedge\overline\Omega$, and passing to the quotient by $\Gamma$ multiplies $\int\Omega\wedge\overline\Omega$ by the constant $1/125$, which does not change the metric; the two numbers are therefore the same. Their machine-learned harmonic representatives give $0.1906$ against the exact $0.1923$.} Inserting \eqref{eq:gamma} into Theorem~\ref{thm:C} gives the fifty-one nonzero masses at the Fermat point. Each label of the types $\{2,1,1,1,0\}$ and $\{2,2,1,0,0\}$ is paired with a label of the same type and the same multiset of $k_i$, so $g_{\bb}=g_{\sigma\bb}$ and its mass is $v/g_{\bb}$; the singlet has mass $v/g_{\one}$. Table~\ref{tab:fermat} lists the values. In units of the singlet mass the two doublet masses are $0.83805$ (twenty labels, ten doublets) and $0.70234$ (thirty labels, fifteen doublets). The fifty generations with an exponent three are massless there, and acquire masses proportional to $|\psi|$ (thirty) and $|\psi|^2$ (twenty) as the modulus moves off the Fermat point. \begin{table}[t] \caption{Hodge norms and masses at the Fermat point, from \eqref{eq:gamma} and Theorem~\ref{thm:C}. Norms are in units of $C$ and masses in units of $v/C$; the types with an exponent three are massless at $\psi=0$.} \label{tab:fermat} \centering \begin{tabular}{lrlll} \toprule type & count & $g_{\bb}/C$ & value & mass\\ \midrule $\{1,1,1,1,1\}$ & 1 & $\gamma(2/5)^5$ & $7.33179$ & $0.136392$\\ $\{2,1,1,1,0\}$ & 20 & $\gamma(2/5)^2\gamma(1/5)$ & $8.74859$ & $0.114304$\\ $\{2,2,1,0,0\}$ & 30 & $\gamma(1/5)^2/\gamma(2/5)$ & $10.4392$ & $0.095793$\\ $\{3,1,1,0,0\}$ & 30 & $\gamma(2/5)^2\gamma(1/5)$ & $8.74859$ & $0$\\ $\{3,2,0,0,0\}$ & 20 & $\gamma(1/5)^2/\gamma(2/5)$ & $10.4392$ & $0$\\ \bottomrule \end{tabular} \end{table} \section{Weighted Fermat pencils}\label{sec:weighted} \subsection{The pencils} In this section $(d;w)$ is one of $(6;1,1,1,1,2)$, $(8;1,1,1,1,4)$ and $(10;1,1,1,2,5)$, so that $d=\sum w_i$ and $n_i=d/w_i$ is an integer: $n=(6,6,6,6,3)$, $(8,8,8,8,2)$ and $(10,10,10,5,2)$. The polynomial $F_\psi=\sum_ix_i^{n_i}-d\psi P$, $P=x_1x_2x_3x_4x_5$, is weighted homogeneous of degree $d$ for $\deg x_i=w_i$, and $X_\psi\subset\PP(w)$ is its zero locus \eqref{eq:wpencil}. We write \[ \delta=\prod_iw_i^{w_i}\in\{4,\,256,\,12\,500\}, \] and use the letters $i,j,k,l$ for distinct indices among the coordinates of weight one. Table~\ref{tab:wdata} collects the numbers proved in this section, with the quintic for comparison.\footnote{The three pencils are the weighted members of the family of one-parameter models whose periods and mirror maps were computed by Morrison~\cite{Mor92}, Klemm and Theisen~\cite{KT93} and Font~\cite{Fon93}. In those papers the coefficient of $P$ is normalised differently and the natural coordinate on the moduli space is $\psi^d$ up to a constant; the discriminant $\psi^d\delta=1$ below is the conifold locus of those papers, and the Fermat point $\psi=0$ is the Gepner point.} \begin{table}[t] \caption{The four pencils of Fermat type. The rows give the degree and weights, the constant $\delta=\prod w_i^{w_i}$ of the discriminant $\psi^d\delta=1$, the intersection numbers of the hyperplane class $L$, the Euler characteristic and $h^{2,1}$, the order of $\Gamma$, the number of distinct characters carried by the generations, the number of classes of two or three generations sharing a character, the number of characters of $\Gamma$ carried by no generation, and the rank of the mass map of the invariant direction for $\psi\ne0$ and at $\psi=0$, with the number of generations it leaves massless.} \label{tab:wdata} \centering \begin{tabular}{lcccc} \toprule & quintic & sextic & octic & dectic\\ \midrule $(d;w)$ & $(5;1,1,1,1,1)$ & $(6;1,1,1,1,2)$ & $(8;1,1,1,1,4)$ & $(10;1,1,1,2,5)$\\ $\delta$ & $1$ & $4$ & $256$ & $12\,500$\\ $L^3$, $c_2\cdot L$ & $5$, $50$ & $3$, $42$ & $2$, $44$ & $1$, $34$\\ $\chi$, $h^{2,1}$ & $-200$, $101$ & $-204$, $103$ & $-296$, $149$ & $-288$, $145$\\ $|\Gamma|$ & $125$ & $108$ & $128$ & $100$\\ distinct characters & $101$ & $96$ & $122$ & $100$\\ shared classes & $0$ & $7$ & $27$ & $42$\\ characters not carried & $24$ & $12$ & $6$ & $0$\\ rank for $\psi\ne0$ (massless) & $101$ ($0$) & $99$ ($4$) & $125$ ($24$) & $109$ ($36$)\\ rank at $\psi=0$ (massless) & $51$ ($50$) & $44$ ($59$) & $85$ ($64$) & $72$ ($73$)\\ \bottomrule \end{tabular} \end{table} \begin{lemma}\label{lem:wsmooth} If $\psi^d\delta\ne1$, then $X_\psi$ is quasi-smooth, it misses the singular points of $\PP(w)$, and it is a smooth Calabi--Yau threefold with $h^{1,1}=1$. Its hyperplane class $L=c_1(\cO_X(1))$ generates $H^2(X,\Z)$ modulo torsion, and \begin{equation}\label{eq:wchern} \begin{gathered} L^3=\frac{d}{\prod_iw_i},\qquad c_2\cdot L=\sigma_2(w)\,L^3,\\ \chi=\big(\sigma_3(w)-d\,\sigma_2(w)\big)L^3, \end{gathered} \end{equation} with $\sigma_k$ the elementary symmetric functions of the weights. If $\psi^d\delta=1$, then $X_\psi$ is singular. \end{lemma} \begin{proof} The Euler relations $x_i\partial_iF_\psi=n_ix_i^{n_i}-d\psi P$ show that at a common zero of the partial derivatives $x_i^{n_i}=w_i\psi P$ for every $i$. If $P=0$ this forces $x=0$. Otherwise raise the $i$-th relation to the power $w_i$ and multiply: since $\prod_ix_i^{n_iw_i}=P^d$, this gives $P^d=\psi^d\delta P^d$, so $\psi^d\delta=1$. Hence $X_\psi$ is quasi-smooth when $\psi^d\delta\ne1$. Conversely, at $\psi=2^{-1/3}$, $\psi=1/2$ and $\psi=2^{-1/5}5^{-1/2}$ the points $(1,1,1,1,2^{1/3})$, $(1,1,1,1,2)$ and $(1,1,1,2^{1/5},5^{1/2})$ satisfy $x_i^{n_i}=w_i\psi P=w_i$, so they are singular points of the sextic, octic and dectic; the other roots of $\psi^d\delta=1$ are $\zeta\psi$ with $\zeta^d=1$, and $x_1\mapsto\zeta^{-1}x_1$ maps $X_{\zeta\psi}$ to $X_\psi$ because $n_1=d$. The singular points of $\PP(w)$ are $(0{:}0{:}0{:}0{:}1)$ for the sextic and the octic, with isotropy $\mu_2$ and $\mu_4$, and $(0{:}0{:}0{:}1{:}0)$ and $(0{:}0{:}0{:}0{:}1)$ for the dectic, with isotropy $\mu_2$ and $\mu_5$. Each monomial $x_i^{n_i}$ occurs in $F_\psi$ with coefficient one, so $F_\psi$ does not vanish at a coordinate point and $X_\psi$ misses them all; a quasi-smooth hypersurface in the smooth locus of $\PP(w)$ is smooth. By adjunction $K_X=\cO_X(d-\sum w_i)=\cO_X$, and the Lefschetz theorem for quasi-smooth hypersurfaces~\cite{Dol82} gives $h^{1,0}=h^{2,0}=0$ and $h^{1,1}=1$. The tangent bundle of $\PP(w)$ on its smooth locus has total Chern class $\prod_i(1+w_iH)$ by the weighted Euler sequence, and the normal bundle of $X$ is $\cO_X(d)$, so $c(T_X)=\prod_i(1+w_iL)(1+dL)^{-1}$. Expanding with $\sigma_1(w)=d$ gives $c_1=0$, $c_2=\sigma_2L^2$ and $c_3=(\sigma_3-d\sigma_2)L^3$, and $L^3=d\int_{\PP(w)}H^4=d/\prod w_i$. For the three pencils $(\sigma_2,\sigma_3)=(14,16)$, $(22,28)$ and $(34,52)$, which gives Table~\ref{tab:wdata}. If $L=mL'$ with $L'$ integral, then $m^3$ divides $L^3\in\{3,2,1\}$, which is cube-free, so $m=\pm1$; as $H^2(X,\Q)$ is spanned by $L$, the class $L$ generates $H^2(X,\Z)$ modulo torsion. \end{proof} \subsection{Labels and characters} The Jacobian ring $R=\C[x_1,\dots,x_5]/(\partial_1F_\psi,\dots, \partial_5F_\psi)$ is graded by $\deg x_i=w_i$. Quasi-smoothness says that the partial derivatives, of degrees $d-w_i$, form a regular sequence, so $R$ is a graded complete intersection with Hilbert series $\prod_i(1-t^{d-w_i})/(1-t^{w_i})$: it has dimension $\prod_i(n_i-1)$, which is $1250$, $2401$ and $2916$, its top degree is $\sum_i(d-2w_i)=3d$, and $R_{3d}$ is one-dimensional and spanned by $\Hess F_\psi$ by Macaulay's theorem, which holds for any regular sequence of weighted homogeneous polynomials. Griffiths' description of the Hodge structure extends to quasi-smooth hypersurfaces in weighted projective space, as shown by Steenbrink~\cite{Ste77} and Dolgachev~\cite{Dol82}: with \[ \Omega_0=\sum_i(-1)^iw_ix_i\,dx_1\wedge\cdots\widehat{dx_i}\cdots\wedge dx_5, \] the residue along $X$ of $P\,\Omega_0/F_\psi^{q+1}$, for $P$ of degree $qd$, identifies $R_{qd}$ with $H^{3-q,q}(X)$, so that $H^{2,1}=R_d$ and $R_{3d}$ is the socle. The infinitesimal variation of the Hodge structure is again multiplication in $R$, and the cubic form is \eqref{eq:jacobianyukawa} with $\lambda$ a functional on $R_{3d}$; the derivation in~\cite{CG80} uses only the residue map and the residue pairing of $R$, both of which are available in the weighted setting. Let $B$ be the set of $\bb\in\Z^5$ with $0\le b_i\le n_i-2$ and $\sum_iw_ib_i=d$. We call its elements the labels. \begin{lemma}\label{lem:wlabels} Let $\psi^d\delta\ne1$. Every monomial $x^{\ee}$ of degree $d$ satisfies $x^{\ee}\equiv c\,\psi^r x^{\bb}$ in $R$ with $\bb\in B$, $c$ a product of weights, and $r\le1$ for the sextic and $r\le2$ for the octic and the dectic. The monomials $x^{\bb}$, $\bb\in B$, are a basis of $R_d=H^{2,1}$, so $|B|=h^{2,1}$. \end{lemma} \begin{proof} The relations $\partial_iF_\psi/n_i=0$ read \begin{equation}\label{eq:wreduce} x_i^{n_i-1}\equiv w_i\psi\prod_{j\ne i}x_j . \end{equation} In a monomial of degree $d$ at most one exponent satisfies $e_i\ge n_i-1$: two such exponents would carry weight at least $2d-w_i-w_j>d$. If $e_i\ge n_i-1$, then $e_i\in\{n_i-1,n_i\}$, and \eqref{eq:wreduce} replaces $x^{\ee}$ by $w_i\psi\,x^{\ee'}$ with $e'_i=e_i-n_i+1$ and $e'_j=e_j+1$ for $j\ne i$. If $e_i=n_i$, then $x^{\ee}=x_i^{n_i}$ and $\ee'=\one$. If $e_i=n_i-1$, then $e'_i=0$, the other coordinates carry weight $w_i$, and $e'_j\ge n_j-1$ forces $w_je_j\ge d-2w_j$, hence $w_i+2w_j\ge d$. For the sextic no pair $i\ne j$ has $w_i+2w_j\ge6$, and $\one\in B$ because every $n_j\ge3$; so one step suffices. For the octic and the dectic the only $j$ with $w_i+2w_j\ge d$ for some $i\ne j$ is $j=5$, where $w_5=d/2$ and $n_5=2$; in that case, and in the case $\ee'=\one$, the exponent of $x_5$ is one and a second step at $x_5$ gives an exponent vector with $e''_5=0$ and the other entries raised by one. A third step would need $w_5+2w_k\ge d$, that is $w_k\ge d/4$, for some $k\le4$, which fails. So at most two steps reach a vector with all $e_i\le n_i-2$, which is a label. At $\psi=0$ the ring is $\bigotimes_i\C[x_i]/(x_i^{n_i-1})$, whose monomials with $e_i\le n_i-2$ are a basis; so $|B|=\dim R_d$. The Hilbert series does not depend on $\psi$, and the labels span $R_d$ for every $\psi$ by the first part, so they are a basis. \end{proof} \paragraph*{The symmetry group.} Let $K$ be the group of $\varepsilon\in\prod_i\mu_{n_i}$ with $\prod_i\varepsilon_i=1$, acting by $x_i\mapsto\varepsilon_ix_i$. It preserves $F_\psi$ for every $\psi$. The elements $(\lambda^{w_1},\dots,\lambda^{w_5})$ with $\lambda^d=1$ lie in $K$ and act trivially on $\PP(w)$, and they are the only such elements; they form a subgroup isomorphic to $\mu_d$ because the weights are coprime. Let $\Gamma=K/\mu_d$. Since $\mathrm{lcm}(n_i)=d$, the product map $\prod_i\mu_{n_i}\to\mu_d$ is onto and $|\Gamma|=\prod_in_i/d^2$, which is $108$, $128$ and $100$. A character of $\prod_i\mu_{n_i}$ is a vector $\bb\in\bigoplus_i\Z/n_i$, acting by $\varepsilon\mapsto\prod_i \varepsilon_i^{b_i}$; it is trivial on $K$ exactly when it is a multiple of $\one$, and it is trivial on $\mu_d$ exactly when $\sum_iw_ib_i\equiv0 \pmod d$, a condition that is well defined because $w_in_i=d$. Hence \[ \widehat\Gamma=\Big\{\bb\in\textstyle\bigoplus_i\Z/n_i:\ \sum_iw_ib_i \equiv0\pmod d\Big\}\Big/\langle\one\rangle . \] A monomial $x^{\ee}$ of degree divisible by $d$ transforms by the class $[\ee]$ of its exponent vector, and $\Omega_0$ transforms by $\prod_i\varepsilon_i=1$, so the isomorphisms $R_{qd}\cong H^{3-q,q}$ are $\Gamma$-equivariant and $R$ is graded by $\widehat\Gamma$. The Hessian is invariant, since $\partial_i\partial_jF_\psi$ transforms by $\varepsilon_i^{-1}\varepsilon_j^{-1}$ and the determinant by $\prod_i\varepsilon_i^{-2}=1$; so $R_{3d}$ carries the trivial character. Permutations of coordinates of equal weight also preserve $F_\psi$ and permute the labels. \begin{lemma}[Wrapping]\label{lem:wwrap} Two labels $\bb\ne\bb'$ carry the same character if and only if there is a shift $s\in\{1,\dots,d-1\}$ such that, with $a_i=\lfloor(b_i+s)/n_i\rfloor$, \[ \begin{gathered} b'_i=b_i+s-n_ia_i,\qquad \sum_ia_i=s,\\ b_i+s\not\equiv n_i-1\pmod{n_i}\quad\text{for all }i . \end{gathered} \] The shift from $\bb'$ back to $\bb$ is then $d-s$. \end{lemma} \begin{proof} Both labels satisfy $\sum w_ib_i=d\equiv0$, so they define the same character exactly when $\bb'\equiv\bb+s\one$ modulo $(n_1,\dots,n_5)$ for some $s$ modulo $d$. As $0\le b'_i\le n_i-2$, the entry $b'_i$ is the residue of $b_i+s$ in $\{0,\dots,n_i-1\}$, which must differ from $n_i-1$. Then $\sum_iw_ib'_i=\sum_iw_ib_i+sd-d\sum_ia_i$, which equals $d$ exactly when $\sum_ia_i=s$. \end{proof} \begin{proposition}\label{prop:wchars} Let $\psi^d\delta\ne1$. The classes of labels that carry the same character are exactly the following, and every other label carries a character of its own. \begin{enumerate}[label=\textup{(\alph*)},leftmargin=*,itemsep=1pt] \item Sextic: the $4$ pairs $\{x_i^4x_5,\ x_j^2x_k^2x_l^2\}$ at shift $2$ and the $3$ pairs $\{x_i^3x_j^3,\ x_k^3x_l^3\}$ at shift $3$; $96$ distinct characters. \item Octic: the $12$ pairs $\{x_i^6x_j^2,\ x_j^4x_k^2x_l^2\}$ and the $12$ pairs $\{x_i^6x_jx_k,\ x_j^3x_k^3x_l^2\}$ at shift $2$, and the $3$ pairs $\{x_i^4x_j^4,\ x_k^4x_l^4\}$ at shift $4$; $122$ distinct characters. \item Dectic: the $3$ triples $\{x_i^8x_4,\ x_j^2x_k^2x_4^3,\ x_i^2x_j^4x_k^4\}$; at shift $2$ the $12$ pairs $\{x_1^{c_1}x_2^{c_2} x_3^{c_3}x_4^3,\ x_1^{c_1+2}x_2^{c_2+2}x_3^{c_3+2}\}$ with $c_1+c_2+c_3=4$ and $\cc$ not a permutation of $(2,2,0)$, the $6$ pairs $\{x_i^8x_j^2,\ x_j^4x_k^2x_4^2\}$ and the $3$ pairs $\{x_i^8x_jx_k,\ x_j^3x_k^3x_4^2\}$; at shift $4$ the $6$ pairs $\{x_i^6x_j^2x_4,\ x_j^6x_k^4\}$, the $3$ pairs $\{x_i^6x_jx_kx_4,\ x_j^5x_k^5\}$, the $6$ pairs $\{x_i^7x_jx_4,\ x_ix_j^5x_k^4\}$ and the $3$ pairs $\{x_i^6x_4^2,\ x_j^4x_k^4x_4\}$; $100$ distinct characters, so every character of $\Gamma$ is carried by some generation. \end{enumerate} The characters carried by no label are the $12$ classes of $(\pi(0,1,3,4),2)$ for the sextic and the $6$ classes of $(\pi(0,2,4,6),1)$ for the octic, $\pi$ running over the permutations of four entries. The set of characters carried by labels is closed under $\chi\mapsto-\chi$. \end{proposition} \begin{proof} We use Lemma~\ref{lem:wwrap} and treat the shifts $s\le d/2$; the others are their inverses. For the sextic, $b_i+s\le9<12$ for $i\le4$, so $a_i=1$ exactly when $b_i\ge6-s$, and the residue $n_i-1=5$ is excluded when $b_i=5-s$; for $x_5$, $b_5\le1$ and $a_5=\lfloor(b_5+s)/3\rfloor$, with $b_5+s\equiv2\pmod3$ excluded. At $s=1$ we have $a_5=0$ and need one $b_i\ge5$, which is impossible. At $s=2$ the value $b_5=0$ is excluded, so $b_5=1$ and $a_5=1$; we need exactly one $b_i\ge4$ among the first four, which sum to four, so $\bb=(4,0,0,0,1)$ up to order, and its image is $(0,2,2,2,0)$. At $s=3$ we have $a_5=1$ and need two entries $b_i\ge3$ among the first four and none equal to two; two entries of at least three already have weight six, so $\bb=(3,3,0,0,0)$ up to order, with image $(0,0,3,3,0)$. No label occurs at two different shifts, so the classes are the pairs listed, $14$ labels in $7$ classes and $103-7=96$ characters. The same analysis for the octic and the dectic is carried out in Appendix~\ref{app:wlabels}. It uses two facts: for these pencils $n_5=2$ and $b_5=0$, so $a_5=s/2$ and an odd shift is excluded by the residue of $x_5$; and the bounds $b_i\le n_i-2$ make each $a_i$ either $0$ or $1$, except $a_4\in\{0,1,2\}$ for the dectic. The characters not carried by a label are $108-96=12$, $128-122=6$ and $100-100=0$ in number. For the sextic, adding $s\one$ to $(\pi(0,1,3,4),2)$ gives a vector with an entry $5$ for $s\not\equiv0 \pmod3$, and the vector $(\pi'(0,1,3,4),2)$ for $s=3$, where the last entry is $n_5-1$. So every representative of these classes has an entry $n_i-1$, and none is a label; each class contains two of the $24$ vectors $(\pi(0,1,3,4),2)$, which gives $12$ classes, and these are all the missing characters. For the octic, adding $s\one$ to $(\pi(0,2,4,6),1)$ gives an entry $7$ when $s$ is odd and the last entry $1=n_5-1$ when $s$ is even; each class contains four of the $24$ vectors, which gives $6$ classes. Finally $-(\pi(0,1,3,4),2)=(\pi(0,5,3,2),1)$, which becomes $(\pi(1,0,4,3),2)$ after adding $\one$, and $-(\pi(0,2,4,6),1)= (\pi(0,6,4,2),1)$; so the missing characters are closed under negation, and so are the carried ones. \end{proof} Figure~\ref{fig:weighted} compares the four pencils. \begin{figure}[t] \centering \includegraphics[width=0.96\textwidth]{figures/fig_weighted.png} \caption{Left: for the quintic in $\PP^4$ and the sextic, octic and dectic pencils in $\PP(1,1,1,1,2)$, $\PP(1,1,1,1,4)$ and $\PP(1,1,1,2,5)$, the number of generations ($101$, $103$, $149$, $145$), the number of distinct characters they carry ($101$, $96$, $122$, $100$), and the rank of the mass map of the invariant direction for $\psi\ne0$ ($101$, $99$, $125$, $109$) and at $\psi=0$ ($51$, $44$, $85$, $72$). Only on the quintic do the first three bars agree. Right: the rank of the mass map over the $\psi$-plane of the dectic pencil, $109$ off the Fermat point and $72$ at $\psi=0$ (the drop is drawn with a small width); the ten points of the circle $12\,500\,\psi^{10}=1$ are the singular members.} \label{fig:weighted} \end{figure} \subsection{The socle functional} Let $m_0=x^{\mathbf n-2\one}$, the monomial with exponents $n_i-2$; it has degree $\sum_iw_i(n_i-2)=3d$. A monomial $x^{\ee}$ of degree $3d$ is in sector $r\in\{0,\dots,d-1\}$ if $e_i\equiv n_i-2-r\pmod{n_i}$ for every $i$. We then write \begin{equation}\label{eq:wsector} e_i=n_i-2-r+n_ik_i ; \end{equation} each $k_i$ is a nonnegative integer, since $k_i\le-1$ would give $e_i\le-2-r$, and $\sum_ik_i=r$, by comparing degrees: $\sum_iw_ie_i=5d-2d-rd+d\sum_ik_i=3d$. \begin{theorem}[Weighted socle functional]\label{thm:wsocle} Let $\psi^d\delta\ne1$. Then $m_0$ spans $R_{3d}$, and the functional $\lambda$ on $R_{3d}$ with $\lambda(m_0)=1$ satisfies, for every monomial $x^{\ee}$ of degree $3d$, \[ \lambda(x^{\ee})=\begin{cases}\psi^r\prod_iw_i^{k_i}&\text{if $\ee$ is in sector $r$, with $k$ as in \eqref{eq:wsector},}\\ 0&\text{otherwise.}\end{cases} \] A monomial of degree $3d$ is in a sector if and only if its character is trivial. \end{theorem} \begin{proof} The character of $x^{\ee}$ is trivial when $\ee\equiv c\one$ modulo $(n_1,\dots,n_5)$ for some integer $c$; taking $r\equiv-2-c\pmod d$ and using $n_i\mid d$ this is the condition $e_i\equiv n_i-2-r$, so the monomials with trivial character are those in a sector. Since $R_{3d}$ is one-dimensional with trivial character, every monomial with a nontrivial character vanishes in $R$. Let $\ee$ be in sector $r\ge1$. We claim that some $i$ has $n_ik_i\ge r+1$. Otherwise $k_i\le r/n_i=rw_i/d$ for every $i$, so $r=\sum_ik_i\le r\sum_iw_i/d=r$, with equality only if $n_ik_i=r$ for every $i$; then every $n_i$ divides $r$, and so does their least common multiple $d$, which is impossible for $1\le r0$ generate the subgroup and $D$ realise it. Then $m_0$ divides $|H|$, and \[ \begin{gathered} D^3=\frac{m_0^3L^3}{|H|},\qquad c_2(Y)\cdot D=\frac{m_0\,c_2\cdot L}{|H|},\\ \chi(Y,D)=\frac{D^3}{6}+\frac{c_2(Y)\cdot D}{12}, \end{gathered} \] the last by Riemann--Roch with $c_1(Y)\equiv0$ and $\chi(\cO_Y)=0$. All three numbers are integers. For the sextic, $D^3=3m_0^3/34$ forces $34\mid m_0$, so $m_0=34$, $D^3=3468$, $c_2(Y)\cdot D=42$ and $\chi(Y,D)=578+7/2$. For the dectic, $D^3=m_0^3/48$ forces $12\mid m_0$ and $c_2(Y)\cdot D=17m_0/24$ forces $24\mid m_0$, so $m_0\in\{24,48\}$, with $\chi(Y,D)=48+17/12$ and $384+17/6$. None is an integer, a contradiction. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:E}] Smoothness and $h^{2,1}$ are Lemma~\ref{lem:wsmooth}, part (i) is Proposition~\ref{prop:wchars}, part (ii) is Corollary~\ref{cor:wcouplings}, part (iii) is Proposition~\ref{prop:wmass}, and part (iv) is Proposition~\ref{prop:wthree} together with $|\chi(X)|/2\in\{102,148,144\}$ for $X$ itself. \end{proof} \begin{remark}[Why the quintic is different]\label{rem:wquintic} On the quintic all $n_i$ equal five, so a shift moves every exponent by the same amount and the wrapping count $\sum_ia_i=s$ of Lemma~\ref{lem:wwrap} has no solution: this is the proof of Theorem~\ref{thm:A}. When the $n_i$ differ, a wrap at a coordinate with large $n_i$ can be balanced by wraps at the coordinate with $n_5=3$ or $n_5=2$, and the shared classes of Proposition~\ref{prop:wchars} are exactly these balances. The mass map then has blocks larger than $1\times1$ whose entries, all nonzero, satisfy $\Delta=0$, and these blocks carry the massless generations. Among the four pencils, the quintic is therefore the only one on which the characters of the symmetry group label the generations and the invariant deformation gives every generation a mass. \end{remark} \section{The selection conjecture}\label{sec:selection} \subsection{The landscape and its metric} We use the setting of~\cite{Bha26a}. A heterotic vacuum is a pair $(X,V)$ of a Calabi--Yau threefold $X$ and a slope-polystable holomorphic bundle $V$ with $c_1(V)=0$ for which $c_2(T_X)-c_2(V)$ is effective; the standard embedding is the case $V=T_X$. The landscape $\cL$ is the set of isomorphism classes of vacua. It is the disjoint union of the components $\cL_\tau$ of fixed topological type $\tau$, and each component carries the Riemannian metric $g=G^{\WP}\oplus G^V$ made of the Weil--Petersson metric of the complex structures and the $L^2$ metric of the bundle moduli.\footnote{By the theorem of Bogomolov, Tian and Todorov the complex structures of $X$ are unobstructed, and the bundle moduli of a stable bundle are unobstructed when $H^2(X,\End_0V)=0$. Where a component is an orbifold or has singular points, everything below is applied on its smooth locus, or on orbifold charts; the only tools used are partitions of unity, geodesic balls and cut-off functions, and all three exist in that generality. As in~\cite{Bha26a} we take each $\cL_\tau$ to be a connected manifold of finite dimension with countably many components in all.} Mirror symmetry enters through an involution $\mu$ of a union $\cL^\mu$ of components, which sends a vacuum to its mirror and is an isometry of $g$, as~\cite{Bha26a} records for the Weil--Petersson metric. A vacuum is \emph{viable} when it satisfies the nine conditions (C1)--(C9) of the phenomenological closure theorem of~\cite{Bha26a}: anomaly cancellation and stability, three net families, a Wilson line breaking to the Standard Model group, three families without chiral exotics, masses for the vectorlike exotics, doublet--triplet splitting, stabilised moduli, an acceptable rank and hierarchy of the Yukawa masses, and consistent supersymmetry breaking. \begin{conjecture}[Geometric Selection Conjecture~\cite{Bha26a}] \label{conj:gsc} There is a functional $\cS:\cL\to[0,\infty)$, constructed from the metric $g$, such that \begin{enumerate}[label=\textup{(G\arabic*)},leftmargin=*,itemsep=1pt] \item $\cS$ is smooth and proper on each component $\cL_\tau$; \item $\cS$ has a unique global minimum $\cV^\star\in\cL$; \item $\cV^\star$ is viable; \item $\cS\circ\mu=\cS$ on $\cL^\mu$. \end{enumerate} \end{conjecture} \subsection{The functional adds nothing} \begin{theorem}\label{thm:gscequiv} Conjecture~\ref{conj:gsc} holds if and only if there is a viable vacuum $\cV^\star$ with $\mu(\cV^\star)=\cV^\star$ whenever $\cV^\star\in\cL^\mu$. \end{theorem} \begin{proof} Suppose $\cS$ exists, and let $\cV^\star$ be its minimum. It is viable by (G3). If $\cV^\star\in\cL^\mu$, then $\cS(\mu\cV^\star)=\cS(\cV^\star)$ by (G4), so $\mu\cV^\star$ is also a global minimum, and $\mu\cV^\star= \cV^\star$ by (G2). Conversely, let $\cV^\star$ be viable and fixed by $\mu$ when $\mu$ is defined there, and let $\cL_{\tau^\star}$ be its component. The functional is built in three steps. \emph{A proper function on every component.} A second countable manifold $M$ has a smooth proper function $\rho:M\to[0,\infty)$: take compact sets $K_1\subset K_2\subset\cdots$ with $K_j$ inside the interior of $K_{j+1}$ and union $M$, a partition of unity $(\phi_j)$ with $\phi_j$ supported in the interior of $K_{j+1}$ minus $K_{j-1}$, and $\rho=\sum_jj\phi_j$; then $\rho\ge j$ outside $K_j$. Choose such a $\rho_\tau$ on every component. The involution $\mu$ permutes the components of $\cL^\mu$, so on each orbit of components we may replace $\rho$ by $\rho+\rho\circ\mu$, which is smooth, proper (it is at least $\rho$) and $\mu$-invariant. Call the result $\rho$. \emph{A quadratic well at $\cV^\star$.} Choose $r>0$ smaller than the injectivity radius of $g$ at $\cV^\star$ and such that the closed geodesic ball $\bar B$ of radius $r$ about $\cV^\star$ is compact. On the open ball $B$ the function $f=d_g(\cV^\star,\cdot)^2$ is smooth, since it is the squared norm of the inverse of the exponential map. If $\cV^\star\in \cL^\mu$, then $\mu$ is an isometry fixing $\cV^\star$, so it preserves $B$ and $f\circ\mu=f$. \emph{Gluing.} Let $\chi:[0,\infty)\to[0,1]$ be smooth, equal to $1$ on $[0,r^2/4]$ and to $0$ on $[r^2/2,\infty)$, and extend $\chi\circ f$ by zero outside $B$; the extension is smooth because $\chi\circ f$ vanishes on the part of $B$ at distance more than $r/\sqrt2$ from $\cV^\star$. Put $c=r^2/4$ and \[ \cS=(\chi\circ f)\,f+\big(1-\chi\circ f\big)\,(c+\rho) \] on $\cL_{\tau^\star}$, and $\cS=c+\rho$ on every other component. Then $\cS$ is smooth and $\mu$-invariant, because every ingredient is. It vanishes at $\cV^\star$ and nowhere else: where $\chi\circ f=1$ it equals $f$, which vanishes only at $\cV^\star$; where $0<\chi\circ f<1$ it is a convex combination of $f\ge r^2/4$ and $c+\rho\ge c$, both positive; and where $\chi\circ f=0$ it equals $c+\rho\ge c$. So $\cV^\star$ is the unique global minimum. Finally a sublevel set $\{\cS\le t\}$ of $\cL_{\tau^\star}$ lies in $\bar B\cup\{\rho\le t-c\}$, a union of two compact sets, and it is closed, so it is compact; on the other components the sublevel sets of $c+\rho$ are compact. Thus $\cS$ satisfies (G1) to (G4). \end{proof} Figure~\ref{fig:landscape} draws the functional of the proof on one component. \begin{figure}[t] \centering \includegraphics[width=0.9\textwidth]{figures/fig_landscape.png} \caption{The functional $\cS$ of the proof of Theorem~\ref{thm:gscequiv} on a two-dimensional component $\cL_{\tau^\star}$ on which $\mu$ acts by a reflection (dashed line: its fixed set). Inside the radius $r/2$ about $\cV^\star$ the functional is the squared distance $f$; outside the radius $r/\sqrt2$ it is $c+\rho$ with $\rho$ proper and $\mu$-invariant; between the two circles it is a convex combination of the two. The contour lines on the floor are the level sets of $\cS$; all of them are compact, and the only zero is $\cV^\star$.} \label{fig:landscape} \end{figure} \begin{remark}[Constructed from the metric]\label{rem:frommetric} The phrase ``constructed from the metric'' in Conjecture~\ref{conj:gsc} is not a mathematical condition, and the strongest condition it can be read as is invariance under a group of isometries of $(\cL,g)$. The proof gives that reading as well: if $G$ is a compact group of isometries containing $\mu$, averaging $\rho$ over $G$ with its Haar measure gives a $G$-invariant smooth proper function (if $\int\rho(hx)\,dh\le t$ then $\rho(hx)\le t$ for some $h$, so $x$ lies in the compact set $G\cdot\{\rho\le t\}$), and the well at a $G$-fixed vacuum is $G$-invariant. So the conjecture in that reading is equivalent to the existence of a viable vacuum fixed by $G$. In every reading the functional is determined by the vacuum it selects and contributes nothing else. \end{remark} \begin{remark}[Mirror pairs] A viable vacuum outside the standard embedding lies outside $\cL^\mu$ when $\mu$ is the mirror map of $(2,2)$ vacua, so the condition on $\mu$ in Theorem~\ref{thm:gscequiv} is empty for it. If mirror pairs are identified in $\cL$, as one point each, on the ground that they define the same superconformal field theory, the condition is empty for every vacuum and the conjecture is equivalent to the existence of one viable vacuum. \end{remark} \subsection{The Dwork pencil} \begin{proposition}\label{prop:dworkselect} \textup{(i)} For $\psi\ne0$ and $\psi^5\ne1$, the mass map of the invariant direction $h$ on $X_\psi$ has rank $101$, its entries in the basis of labels are $1$, $\psi$ and $\psi^2$, on $51$, $30$ and $20$ labels, and the masses of Theorem~\ref{thm:C} are $|\psi|^{r(\bb)}$ times the Hodge factor $(g_{\bb}g_{\sigma\bb})^{-1/2}$. \textup{(ii)} Let $X$ be a smooth quintic threefold and $H$ a finite group acting freely on $X$. The standard embedding on $X/H$ has $100/|H|$ net families. In particular condition \textup{(C2)} fails on $X/H$ for every such $H$, and no vacuum of the form $(X/H,T_{X/H})$ is viable. \textup{(iii)} On the sextic, octic and dectic pencils of Section~\ref{sec:weighted}, with $\psi\ne0$ and $\psi^d\delta\ne1$, the mass map of the invariant direction has a kernel of dimension $4$, $24$ and $36$, and condition \textup{(C2)} fails on every member and on every quotient of a member by a group acting freely. \end{proposition} \begin{proof} Part (i) is Theorem~\ref{thm:C}. For (ii), a smooth quintic has $h^{1,1}=1$ and $h^{2,1}=101$, so $\chi(X)=-200$. A free action multiplies the Euler characteristic of the quotient by $1/|H|$, so $\chi(X/H)=-200/|H|$, and by Theorem~\ref{thm:dictionary} the net number of families is $|\chi(X/H)|/2=100/|H|$. Wilson lines on $X/H$ break the gauge group but do not change this index. The equation $100/|H|=3$ has no solution in integers, so (C2) fails. Part (iii) is Propositions~\ref{prop:wmass} and~\ref{prop:wthree}. \end{proof} The proposition shows where the Dwork pencil stands with respect to the conjecture. The condition on masses is met in the strongest form one can ask of a single modulus: the mass matrix has full rank and three scales $1$, $|\psi|$, $|\psi|^2$, which separate as $\psi\to0$. The pencil fails on the count of families, and it fails there for every free quotient, so no choice of Wilson lines rescues it. Three-family vacua satisfying (C1) to (C4) exist on other threefolds: the heterotic standard models of Braun, He, Ovrut and Pantev~\cite{BHOP05}, of Bouchard and Donagi~\cite{BD06}, and the line-bundle models of Anderson, Gray, Lukas and Palti~\cite{AGLP11,AGLP12}, which also address (C5) and (C6). For none of them are (C7), (C8) and (C9) established, and the words ``controlled'', ``acceptable'' and ``consistent'' in those conditions are fixed by comparison with measurement, not by a definition. By Theorem~\ref{thm:gscequiv} this is the whole of the open part of the Geometric Selection Conjecture. The weighted pencils fail on both counts: they too never carry three families, and their mass map leaves generations massless at every member. \subsection{Free quotients of the one-parameter families}\label{sec:freequot} The four pencils are four of the fourteen one-parameter families of Calabi--Yau threefolds with $h^{1,1}=1$ whose Picard--Fuchs equations are hypergeometric, listed by Doran and Morgan~\cite{DM06}. Thirteen of them are complete intersections $X=X_{d_1,\dots,d_c}\subset\PP(w)$ of degrees $d_1,\dots,d_c$ with $\sum_jd_j=\sum_iw_i$, whose general member is quasi-smooth and misses the singular points of $\PP(w)$; they are the rows of Table~\ref{tab:freequot}. The fourteenth, $X_{2,12}\subset \PP(1,1,1,1,4,6)$, is not of this kind: its quadric involves only the four coordinates of weight one, so it is not quasi-smooth at the point of $\PP(4,6)$ where the equation of degree twelve vanishes, and we leave it aside. For the thirteen, the count of families of Proposition~\ref{prop:dworkselect} becomes a statement about every free quotient, through one arithmetic invariant. \begin{lemma}[Primes of a free action]\label{lem:freeprime} Let $X$ be a compact K\"ahler threefold with $K_X\cong\cO_X$, $h^{1,0}=h^{2,0}=0$, $b_2=1$ and $\pi_1(X)=0$, let $L$ be an ample line bundle on $X$, and let $g$ be an automorphism of prime order $p$ without fixed points. Then $p$ divides $\chi(X,kL)$ for every $k\ge1$. \end{lemma} \begin{proof} Since $X$ is simply connected, $H^2(X,\Z)$ has no torsion, and since $h^{0,1}=h^{0,2}=0$ the first Chern class identifies $\operatorname{Pic}X$ with $H^2(X,\Z)\cong\Z$. Let $A$ be its ample generator, so that $L=A^a$ with $a\ge1$. The bundle $g^*A$ is again an ample generator, so $g^*A\cong A$. The obstruction to lifting the action of the cyclic group $G=\langle g\rangle$ to $A$ lies in $H^2(G,\C^*)$, which vanishes for a cyclic group, so $A$ and all its powers are $G$-linearised and $G$ acts on $H^q(X,A^m)$. For $m\ge1$ these groups vanish in positive degree by Kodaira's theorem, as $A^m=K_X\otimes A^m$ with $A^m$ ample. For $1\le j0}$ meets every sphere about the origin once, so $C\setminus\{0\}$ is homotopy equivalent to the link and is simply connected. As $X$ misses the points with nontrivial isotropy, $\C^*$ acts freely on $C\setminus\{0\}$ with quotient $X$, and the exact sequence of homotopy groups of this fibration gives $\pi_1(X)=0$. Let $p$ be a prime divisor of $|H|$ and $g\in H$ an element of order $p$. It has no fixed points, so by Lemma~\ref{lem:freeprime} with $L=\cO_X(1)$, $p$ divides $\chi(X,kL)$ for all $k\ge1$. By Riemann--Roch, with $c_1(X)=0$ and $\chi(\cO_X)=0$, $\chi(X,kL)=k^3L^3/6+k\,c_2\cdot L/12$, a polynomial that takes integer values and vanishes at $k=0$; writing it as $\alpha_1\binom k1+\alpha_2 \binom k2+\alpha_3\binom k3$ with integers $\alpha_i$, its values at all $k\ge1$ have the same greatest common divisor as $\alpha_1,\alpha_2, \alpha_3$, which is that of its values at $k=1,2,3$, the number $g_X$. The intersection numbers are computed as in Lemma~\ref{lem:wsmooth}, from $c(T_X)=\prod_i(1+w_iL)\prod_j(1+d_jL)^{-1}$ and $L^3=\prod_jd_j/\prod_iw_i$, and give the table. By Theorem~\ref{thm:dictionary} the quotient $X/H$ carries $|\chi(X)|/(2|H|)$ net families, so three families need $|H|=|\chi(X)|/6$. This is not an integer for $X_5$, $X_8$, $X_{4,2}$, $X_{2,2,2,2}$ and $X_{6,2}$. In the remaining eight rows it is $34$, $48$, $24$, $24$, $24$, $20$, $26$ or $26$, and each of these has a prime factor not dividing $g_X$: $17$, $2$, $2$, $2$, $3$, $2$ and $2$ (twice). \end{proof} The proposition contains Theorem~\ref{thm:E}(iv) and Proposition~\ref{prop:dworkselect}(ii) and strengthens them: the sextic and dectic pencils admit no free action at all, and on a quintic every free action is by a $5$-group, so that only the orders $5$ and $25$ divide $|\chi|=200$ and a free quotient of a quintic has $20$ or $4$ families; the quotient by $\Z/5\times\Z/5$ with four families is already in~\cite{CHSW85}. For the five complete intersections in ordinary projective space the free actions were classified by Braun~\cite{Bra11} with a computer search; the proposition uses only the holomorphic Lefschetz formula and Riemann--Roch.\footnote{On a quintic the group $\Z/5\times\Z/5$ of~\cite{CHSW85}, generated by the cyclic permutation of the coordinates and the diagonal phases $(1,\zeta,\zeta^2,\zeta^3,\zeta^4)$, acts on the five coordinates through a central extension of order $125$, the Heisenberg group, and $H^0(X,\cO_X(1))=\C^5$ is its irreducible representation of dimension five; each cyclic subgroup of order five acts on it by the regular representation, as Lemma~\ref{lem:freeprime} requires.} \section{Conclusion}\label{sec:conclusion} On the Dwork pencil the question of the title has a complete answer. The hundred and one generations are labelled by characters of the symmetry group $\Gamma$, each character occurring once; the couplings of three generations are the values of one functional on the socle of the Jacobian ring, equal to a power of the modulus when the symmetry allows them and to zero otherwise; the invariant deformation pairs the generations by an involution with a single fixed point, so its mass matrix is monomial with determinant $\psi^{70}$; and at the Fermat point the masses are explicit values of the Gamma function. The proofs use two facts only, the reduction \eqref{eq:reduce} and the grading of the Jacobian ring by characters, together with the general dictionary of Section~\ref{sec:dictionary}. Two questions are not settled by the geometry of the pencil. The breaking of $E_6$ to the Standard Model needs a free quotient and Wilson lines, and on a quintic the number of families of a free quotient is $100/|H|$, never three (Proposition~\ref{prop:dworkselect}), and the same holds for every free quotient of the thirteen one-parameter complete intersections (Proposition~\ref{prop:freequot}). The selection of the vacuum is reduced by Theorem~\ref{thm:gscequiv} to the existence of one viable vacuum, and that existence depends on conditions (C7) to (C9), whose thresholds are set by measurement. The same two facts drive Section~\ref{sec:weighted}, where the three other one-parameter families of Fermat type are treated. There the characters no longer separate the generations, the invariant deformation leaves $4$, $24$ and $36$ of them massless at every member, and no member or free quotient carries three families. Among the four pencils of Fermat type, the quintic is the only one on which the symmetry labels every generation and the invariant deformation gives every generation a mass. \section*{Data and code availability} The programs that repeat the finite computations of this paper are in its archive, release 1.2.0, \url{https://github.com/creelie/strings/releases/tag/v1.2.0}~\cite{Repo}, which also holds the figures and the record of the engine runs. They are written in C (the label and coupling counts), C++ (the expansion of the Hessian by permutations), Python (all counts, the intersection numbers of Appendix~\ref{app:fermat}, the Gamma values, and the labels, characters, couplings and mass maps of the weighted pencils, and the intersection numbers, values $\chi(X,kL)$ and invariants $g_X$ of Table~\ref{tab:freequot}), Julia (the Gamma values and the intersection numbers), Macaulay2 (the Jacobian ring of the quintic at two rational values of $\psi$, with the socle functional on all $3876$ monomials of degree fifteen and the Hessian, and the Jacobian rings of the three weighted pencils at two rational values of $\psi$, with the socle functional on every monomial of degree $3d$) and Lean~4 (the counts, the arithmetic of the Hessian and the numbers of the weighted pencils, checked by the kernel), with one shell script that runs all $215$ checks. No proof in the paper depends on them. The constructions were organised on the Anaxia engine~\cite{Ana26}, described in~\cite{BB26}; its record for the five targets of this paper is part of the archive. The programs and the engine runs were prepared by Deep Bhattacharjee with AI assistance from Claude (Anthropic). \section*{Ethical declarations} \emph{Funding.} None. \emph{Competing interests.} None. \appendix \section{The Hessian in the socle}\label{app:hessian} The Hessian matrix of $F_\psi$ has entries \[ H_{ii}=20\,x_i^3,\qquad H_{ij}=-5\psi\prod_{k\ne i,j}x_k\quad(i\ne j). \] For a permutation $\pi$ of $\{1,\dots,5\}$ with $m$ moved points and $f=5-m$ fixed points, the product $\prod_iH_{i\pi(i)}$ is $20^f(-5\psi)^m$ times a monomial whose exponents are read off as follows. A fixed coordinate $a$ receives $3$ from its diagonal entry and $1$ from each of the $m$ off-diagonal entries, since an off-diagonal entry $(i,\pi(i))$ omits only the moved coordinates $i$ and $\pi(i)$. A moved coordinate $c$ is omitted by exactly two off-diagonal entries, those with $i=c$ and with $\pi(i)=c$, which are different because $\pi(c)\ne c$; it receives $1$ from each of the other $m-2$. Hence the monomial of $\pi$ is \[ \prod_{a\ \mathrm{fixed}}x_a^{3+m}\prod_{c\ \mathrm{moved}}x_c^{m-2}, \] of degree $f(3+m)+m(m-2)=15$, and it depends only on the number of moved points. The five values of $m$ give: \begin{itemize}[leftmargin=*,itemsep=1pt] \item $m=0$: the monomial $m_0=(3,3,3,3,3)$, sector $0$; \item $m=2$: $(5,5,5,0,0)$ up to order, sector $3$, so $\lambda=\psi^3$; \item $m=3$: $(6,6,1,1,1)$ up to order, sector $2$, so $\lambda=\psi^2$; \item $m=4$: $(7,2,2,2,2)$ up to order, sector $1$, so $\lambda=\psi$; \item $m=5$: $(3,3,3,3,3)$, sector $0$. \end{itemize} The contribution of a permutation to $H(\psi)=\lambda(\Hess F_\psi)$ is $\operatorname{sgn}(\pi)\,20^f(-5\psi)^m\psi^{r(m)}$ with $r(0)=r(5)=0$, $r(2)=3$, $r(3)=2$, $r(4)=1$, so $m+r(m)=5$ for every $m\ge2$ and every non-identity permutation contributes a multiple of $\psi^5$. The cycle types are \[ \begin{aligned} &1^5:\ 1\ \text{permutation, }m=0,\ +20^5=3\,200\,000;\\ &2\,1^3:\ 10,\ \text{odd},\ m=2:\ 10\cdot(-1)\cdot20^3\cdot25=-2\,000\,000;\\ &3\,1^2:\ 20,\ \text{even},\ m=3:\ 20\cdot20^2\cdot(-125)=-1\,000\,000;\\ &4\,1:\ 30,\ \text{odd},\ m=4:\ 30\cdot(-1)\cdot20\cdot625=-375\,000;\\ &2^2\,1:\ 15,\ \text{even},\ m=4:\ 15\cdot20\cdot625=187\,500;\\ &5:\ 24,\ \text{even},\ m=5:\ 24\cdot(-3125)=-75\,000;\\ &3\,2:\ 20,\ \text{odd},\ m=5:\ 20\cdot(-1)\cdot(-3125)=62\,500, \end{aligned} \] where the last six numbers are the coefficients of $\psi^5$. The counts $1+10+20+30+15+24+20$ add up to $120$. The coefficients of $\psi^5$ add up to $-3\,200\,000$, which gives $H(\psi)=3\,200\,000(1-\psi^5)$. \section{The coupling count}\label{app:count} \subsection{Ordered triples by sector} For $\ee$ of degree fifteen, $O(\ee)$ is the number of ordered triples $(\va,\bb,\cc)\in B^3$ with $\va+\bb+\cc=\ee$. We count it by the third label: for each $\cc\in B$ with $\cc\le\ee$, put $\mathbf s=\ee-\cc$, a vector of degree ten, and count the $\va\in B$ with $\mathbf s-\va\in B$, that is, with \[ \max(0,s_i-3)\le a_i\le\min(3,s_i)\quad\text{for all }i,\qquad \textstyle\sum a_i=5. \] This number $N(\mathbf s)$ is the coefficient of $t^5$ in $\prod_i\big(t^{\max(0,s_i-3)}+\dots+t^{\min(3,s_i)}\big)$. Write $[n]=1+t+\dots+t^{n-1}$. \emph{Sector $0$, $\ee=(3,3,3,3,3)$.} Here $\mathbf s=3\one-\cc$ has entries in $\{0,\dots,3\}$, so $N(\mathbf s)$ is the coefficient of $t^5$ in $\prod_i[s_i+1]$. The multiset of entries of $\mathbf s$ is the complement of the type of $\cc$: \[ \begin{array}{l|l|r|r} \text{type of }\cc&\text{entries of }\mathbf s&\#\cc&N(\mathbf s)\\\hline \{3,2,0,0,0\}&\{3,3,3,1,0\}&20&24\\ \{3,1,1,0,0\}&\{3,3,2,2,0\}&30&28\\ \{2,2,1,0,0\}&\{3,3,2,1,1\}&30&38\\ \{2,1,1,1,0\}&\{3,2,2,2,1\}&20&44\\ \{1,1,1,1,1\}&\{2,2,2,2,2\}&1&51 \end{array} \] The values of $N$ are central coefficients. For $\{3,3,3,1,0\}$, $[4]^3=1+3t+6t^2+10t^3+12t^4+12t^5+\cdots$ and the factor $[2]$ gives $12+12=24$. For $\{3,3,2,2,0\}$, $[4]^2=(1,2,3,4,3,2,1)$ and $[3]^2=(1,2,3,2,1)$, so the coefficient is $2+6+12+6+2=28$. For $\{3,3,2,1,1\}$, $[3][2]^2=(1,3,4,3,1)$ and the coefficient is $2+9+16+9+2=38$. For $\{3,2,2,2,1\}$, $[3]^3=(1,3,6,7,6,3,1)$ and $[4][2]=(1,2,2,2,1)$, so the coefficient is $3+12+14+12+3=44$. For $\{2,2,2,2,2\}$ the coefficient of $t^5$ in $(1+t+t^2)^5$ is $51$. Hence \[ O(3,3,3,3,3)=20\cdot24+30\cdot28+30\cdot38+20\cdot44+51=3391. \] (The table for $\ee=(3,3,3,3,3)$ also shows that every $\cc\in B$ occurs, since $\mathbf s=3\one-\cc$ always admits some $\va$.) \emph{Sector $1$, $\ee=(7,2,2,2,2)$.} Since $a_1+b_1\le6$, $c_1\ge1$, and $c_i\le2$ for $i\ge2$. With $\mathbf s=\ee-\cc$, the entry $s_1=7-c_1$ forces $a_1\ge4-c_1$, and the remaining entries $s_i=2-c_i\le2$ impose only $a_i\le s_i$. Grouping $\cc$ by $c_1$ and by the multiset of $(c_2,\dots,c_5)$: \[ \begin{array}{l|l|r|r|r} c_1&(c_2,\dots,c_5)&\#\cc&N(\mathbf s)&\text{total}\\\hline 1&\{1,1,1,1\}&1&6&6\\ 1&\{2,1,1,0\}&12&4&48\\ 1&\{2,2,0,0\}&6&3&18\\ 2&\{1,1,1,0\}&4&14&56\\ 2&\{2,1,0,0\}&12&10&120\\ 3&\{1,1,0,0\}&6&26&156\\ 3&\{2,0,0,0\}&4&19&76 \end{array} \] For instance, when $c_1=3$ and $(c_2,\dots,c_5)=(2,0,0,0)$, $a_1$ ranges over $\{1,2,3\}$ and the other entries satisfy $a_i\le(0,2,2,2)$ with sum $5-a_1$; the coefficients of $t^4$, $t^3$, $t^2$ in $[3]^3$ are $6$, $7$, $6$, which add to $19$. The total is $O(7,2,2,2,2)=480$. \emph{Sector $2$, $\ee=(6,6,1,1,1)$.} Here $c_3,c_4,c_5\le1$, and the same grouping gives \[ \begin{array}{l|l|r|r|r} \{c_1,c_2\}&(c_3,c_4,c_5)&\#\cc&N(\mathbf s)&\text{total}\\\hline \{1,1\}&\{1,1,1\}&1&2&2\\ \{2,0\}&\{1,1,1\}&2&1&2\\ \{2,1\}&\{1,1,0\}&6&4&24\\ \{2,2\}&\{1,0,0\}&3&10&30\\ \{3,0\}&\{1,1,0\}&6&2&12\\ \{3,1\}&\{1,0,0\}&6&8&48\\ \{3,2\}&\{0,0,0\}&2&22&44 \end{array} \] and $O(6,6,1,1,1)=162$. \emph{Sector $3$, $\ee=(5,5,5,0,0)$.} Here $c_4=c_5=0$ and $c_1+c_2+c_3=5$. For $\{c_1,c_2,c_3\}=\{2,2,1\}$ the entries of $\mathbf s$ are $(3,3,4)$, the bounds are $a_i\le3$ on the first two and $1\le a_3\le3$, and the coefficient of $t^5$ in $[4]^2(t+t^2+t^3)$ is $3+4+3=10$; for $\{3,1,1\}$ the entries are $(2,4,4)$ and the coefficient of $t^5$ in $[3](t+t^2+t^3)^2$ is $2+3+2=7$; for $\{3,2,0\}$ the entries are $(2,3,5)$ and the coefficient of $t^5$ in $[3][4](t^2+t^3)$ is $3+3=6$. With three, three and six choices of $\cc$, $O(5,5,5,0,0)=30+21+36=87$. The ordered counts by sector are $O_0=3391$, $O_1=5\cdot480=2400$, $O_2=10\cdot162=1620$ and $O_3=10\cdot87=870$, with total $8281$ in agreement with Section~\ref{sec:count}. \subsection{Repeated labels} For $\va\in B$ the partner $\cc$ with $2\va+\cc$ in a sector is found type by type, as in the proof of Theorem~\ref{thm:C}. For $\va=\one$, $\cc=\one$ in sector $0$. For $\va$ of type $\{2,1,1,1,0\}$, sector $1$ gives $c_i\equiv2-2a_i$, that is $\cc$ with a three where $\va$ has its two, a two where $\va$ has its zero and zeros elsewhere, of sum five. For type $\{3,1,1,0,0\}$, sector $1$ gives a one where $\va$ has its three, zeros where it has ones and twos where it has zeros. For type $\{3,2,0,0,0\}$, sector $2$ gives zero, two and ones. For type $\{2,2,1,0,0\}$, sector $3$ gives ones at the twos, a three at the one and zeros elsewhere. In each case the other sectors produce a residue four or a total different from five. So the hundred triples $\{\va,\va,\cc\}$ with $\cc\ne\va$ lie $20+30=50$ in sector $1$, $20$ in sector $2$ and $30$ in sector $3$, and none in sector $0$, which is the list $D_r=0,50,20,30$ used in Section~\ref{sec:count}. \section{Hodge norms at the Fermat point}\label{app:fermat} This appendix proves Proposition~\ref{prop:fermat}. Write $G=\mu_5^5/\mu_5$ for the group of diagonal phases of $\PP^4$, which acts on $X_0$. For $\mathbf k\in\{1,2,3,4\}^5$ with $\sum k_i\equiv0\pmod5$ let $\chi_{\mathbf k}(a)=\zeta^{a\cdot\mathbf k}$. \subsection{Eigenlines} The character $\chi_{\mathbf k}$ occurs in $H^3(X_0,\C)$ exactly once, and these are all the characters that occur: the number of such $\mathbf k$ is $\frac15\sum_{j=0}^4\big(\sum_{k=1}^4\zeta^{jk}\big)^5=\frac15(4^5-4)=204 =b_3(X_0)$, and each occurs at least once because the residue $\Res(x^{\mathbf k-\one}\Omega_0/F^{q+1})$, $5(q+1)=\sum k_i$, is a nonzero class of character $\chi_{\mathbf k}$ (Griffiths' theorem and the fact that $x^{\mathbf k-\one}$, with exponents at most three, is nonzero in the Jacobian ring $\C[x]/(x_1^4,\dots,x_5^4)$ of the Fermat quintic). In particular $\omega_{\bb}$ spans the line of $\mathbf k=\bb+\one$. \subsection{The bilinear relation} Let $Q$ be the intersection form on $H_3(X_0,\Z)$, extended to complex coefficients, and identify $H^3$ with $H_3$ by Poincar\'e duality $D$, so that $\int_c\omega=Q(c,D\omega)$ and $\int_{X_0}\omega\wedge\eta= Q(D\omega,D\eta)$. Since $Q$ is $G$-invariant, it pairs the eigenline $L$ of $\chi$ only with the eigenline $L'$ of $\bar\chi$. Let $e$ be an integral cycle and $u$ its component in $L'$; its component in $L$ is $\bar u$, because $e$ is real. If $D\omega_{\bb}=\alpha\bar u$ with $\bar u\ne0$, then \[ \int_e\omega_{\bb}=Q(e,\alpha\bar u)=\alpha\,Q(u,\bar u),\qquad \int_{X_0}\omega_{\bb}\wedge\overline{\omega_{\bb}}=|\alpha|^2Q(\bar u,u), \] and therefore \begin{equation}\label{eq:bilinear} g_{\bb}=\Big|\int_{X_0}\omega_{\bb}\wedge\overline{\omega_{\bb}}\Big| =\frac{|\int_e\omega_{\bb}|^2}{|Q(u,\bar u)|}. \end{equation} It remains to compute both sides for one cycle. \subsection{The cycle of Pham} On the affine chart $x_5=1$ put $y_i=x_i/x_5$; the open part of $X_0$ is $U=\{f=0\}$, $f=y_1^5+y_2^5+y_3^5+y_4^5+1$, and $\mu_5^4$ acts by $y_i\mapsto\zeta^{a_i}y_i$. A character $\mathbf k$ of $G$ restricts to the character $(k_1,k_2,k_3,k_4)$ of $\mu_5^4$, and $k_5\equiv-(k_1+k_2+k_3+ k_4)$. Pham~\cite{Pha65} showed that $U$ retracts onto the join of four copies of the set of fifth roots of $-1$, and that $H_3(U,\Z)$ is generated over $\Z[\mu_5^4]$ by the cycle \[ e=(1-T_1)(1-T_2)(1-T_3)(1-T_4)\,\Delta, \] where $\Delta$ is the real simplex $\{y_i=\epsilon v_i,\ v_i\ge0,\ \sum v_i^5=1\}$ with $\epsilon^5=-1$, and $T_i$ multiplies $y_i$ by $\zeta$. \emph{The period.} By Griffiths' residue formula, $\int_e\omega_{\bb}=\frac1{2\pi i}\int_{\tau(e)}x^{\bb}\Omega_0/F^2$ with $\tau$ the tube map. On the chart this is the integral of $y^{\bb'}dy/f^2$, $\bb'=(b_1,\dots,b_4)$. Put $f_s=\sum y_i^5+s$; then $y^{\bb'}dy/f_s^2=-\partial_s(y^{\bb'}dy/f_s)$, and $\frac1{2\pi i}\int_{\tau(e)}y^{\bb'}dy/f_s=\int_{e_s}y^{\bb'}dy/df_s$ with the Gelfand--Leray form. The substitution $y=s^{1/5}u$ shows that the last integral is $s^{A}P$ with $A=\frac15\sum_{i\le4}k_i-1$ and $P$ its value at $s=1$. Differentiating at $s=1$, \[ \int_e\omega_{\bb}=-A\,P. \] On $\Delta$, with $t_i=v_i^5$, the Gelfand--Leray form is $5^{-4}\prod_it_i^{1/5-1}dt$ on the simplex $\sum t_i=1$, and $y^{\bb'}=\epsilon^{|\bb'|}\prod t_i^{b_i/5}$; Dirichlet's integral gives $\int_\Delta=\epsilon^{|\bb'|+4}5^{-4}\prod_{i\le4}\Gamma(k_i/5)/ \Gamma(\sum_{i\le4}k_i/5)$, and each factor $1-T_i$ multiplies it by $1-\zeta^{k_i}$. Since $\sum_{i\le4}k_i=10-k_5$, $A=1-k_5/5$ and $\Gamma(A+1)=A\,\Gamma(A)$, so \begin{equation}\label{eq:period} \Big|\int_e\omega_{\bb}\Big|=5^{-4}\prod_{i=1}^4\big|1-\zeta^{k_i}\big| \,\frac{\prod_{i=1}^4\Gamma(k_i/5)}{\Gamma(1-k_5/5)}. \end{equation} \emph{The intersection number.} For one variable, the reduced homology of the fibre $\{y^5=-1\}$ is the augmentation ideal $I$ of $\Z[t]/(t^5-1)$, the monodromy is multiplication by $t$, and the variation operator is multiplication by $t-1$. The Seifert form $L(x,z)=\ip{x}{\mathrm{Var}^{-1}z}$ is therefore, in terms of the values $\hat x(k)=x(\zeta^k)$, \[ L(x,z)=\frac15\sum_{k=1}^4\frac{\hat x(k)\,\hat z(-k)}{\zeta^{-k}-1}. \] By the theorem of Sebastiani and Thom~\cite{ST71}, in the form of Sakamoto~\cite{Sak74} for Seifert forms, the Seifert form of $y_1^5+\dots+y_4^5$ is, up to sign, the tensor product of four copies, and the intersection form on the three-dimensional fibre is $Q=-L+L^{T}$. On the eigenline of $(k_1,\dots,k_4)$ this gives \[ Q(x,z)=\pm\frac1{5^4}\,\hat x(\mathbf k)\hat z(-\mathbf k)\Big( \prod_{i\le4}\frac1{\zeta^{k_i}-1}-\prod_{i\le4}\frac1{\zeta^{-k_i}-1}\Big). \] With four factors, $\prod(\zeta^{k_i}-1)=\zeta^{\sum k_i}\prod(\zeta^{-k_i} -1)$, so the bracket has absolute value \[ \frac{|1-\zeta^{k_1+k_2+k_3+k_4}|}{\prod_{i\le4}|1-\zeta^{k_i}|^2} =\frac{|1-\zeta^{k_5}|}{\prod_{i\le4}|1-\zeta^{k_i}|^2}. \] The cycle $e$ has $\hat e(\mathbf k)=\prod_{i\le4}(1-\zeta^{k_i})$ up to a unimodular factor, so its component $u$ satisfies \begin{equation}\label{eq:intersection} |Q(u,\bar u)|=5^{-4}\prod_{i=1}^5\big|1-\zeta^{k_i}\big|. \end{equation} This is nonzero, so $\bar u\ne0$, as needed in \eqref{eq:bilinear}. \subsection{Conclusion of the proof} Inserting \eqref{eq:period} and \eqref{eq:intersection} in \eqref{eq:bilinear}, \[ g_{\bb}=5^{-4}\,\prod_{i=1}^4\big|1-\zeta^{k_i}\big|\Gamma(k_i/5)^2\cdot \frac{1}{|1-\zeta^{k_5}|\,\Gamma(1-k_5/5)^2}. \] With $|1-\zeta^k|=2\sin(\pi k/5)$ and $\Gamma(t)\Gamma(1-t)=\pi/\sin\pi t$, \[ 2\sin(\pi t)\,\Gamma(t)^2=2\pi\,\gamma(t),\qquad \frac{1}{2\sin(\pi t)\,\Gamma(1-t)^2}=\frac{\gamma(t)}{2\pi}, \] so $g_{\bb}=5^{-4}(2\pi)^3\prod_{i=1}^5\gamma(k_i/5)$, which is \eqref{eq:gamma} with $C=(2\pi)^3/5^4$. The class of $\Omega$, with $\mathbf k=\one$, has no derivative in the period computation, so $|\int_e\Omega|=|P|$ with $\Gamma(\sum_{i\le4}k_i/5)=\Gamma(4/5)= \Gamma(1-k_5/5)$, and the same formula holds: $\|\Omega\|^2=C\gamma(1/5)^5$. Since $\partial_\psi\Res(\Omega_0/F_\psi)=\Res(5x_1\cdots x_5\Omega_0/ F_\psi^2)=5\omega_{\one}$ and $\omega_{\one}$ is of type $(2,1)$ at $\psi=0$ (its character differs from that of $\Omega$), the Weil--Petersson metric at the Fermat point is \begin{align*} G_{\psi\bar\psi}(0)&=\frac{25\,g_{\one}}{\|\Omega\|^2} =25\,\frac{\gamma(2/5)^5}{\gamma(1/5)^5}\\ &=25\,\frac{\Gamma(2/5)^5\Gamma(4/5)^5}{\Gamma(1/5)^5\Gamma(3/5)^5}=0.19226. \end{align*} \subsection{Numerical values} With $\Gamma(1/5)=4.59084$, $\Gamma(2/5)=2.21816$, $\Gamma(3/5)=1.48919$ and $\Gamma(4/5)=1.16423$, $\gamma(1/5)=3.94325$ and $\gamma(2/5)=1.48951$; $\gamma(3/5)$ and $\gamma(4/5)$ are their inverses. The norms of the five types follow by multiplying the five factors, and the masses of Table~\ref{tab:fermat} are their inverses. \section{Shared characters on the octic and the dectic}\label{app:wlabels} \begin{table}[t] \caption{The entries of the blocks of Table~\ref{tab:wblocks}, for the representative with $(i,j,k,l)=(1,2,3,4)$. Exponent vectors are written in full; the column $r$ is the sector of $h+\bb+\cc$ and $k$ is the vector of \eqref{eq:wsector}.} \label{tab:wblockdata} \centering \footnotesize \begin{tabular}{lrlllrll} \toprule pencil & row & $\bb$ & $\cc$ & $h+\bb+\cc$ & $r$ & $k$ & entry\\ \midrule sextic & 1 & $(4,0,0,0,1)$ & $(3,1,1,1,0)$ & $(8,2,2,2,2)$ & $2$ & $(1,0,0,0,1)$ & $2\psi^{2}$\\ & 1 & $(0,2,2,2,0)$ & $(3,1,1,1,0)$ & $(4,4,4,4,1)$ & $0$ & $(0,0,0,0,0)$ & $1$\\ & 2 & $(3,3,0,0,0)$ & $(3,3,0,0,0)$ & $(7,7,1,1,1)$ & $3$ & $(1,1,0,0,1)$ & $2\psi^{3}$\\ & 2 & $(3,3,0,0,0)$ & $(0,0,3,3,0)$ & $(4,4,4,4,1)$ & $0$ & $(0,0,0,0,0)$ & $1$\\ & 2 & $(0,0,3,3,0)$ & $(3,3,0,0,0)$ & $(4,4,4,4,1)$ & $0$ & $(0,0,0,0,0)$ & $1$\\ & 2 & $(0,0,3,3,0)$ & $(0,0,3,3,0)$ & $(1,1,7,7,1)$ & $3$ & $(0,0,1,1,1)$ & $2\psi^{3}$\\ \midrule octic & 1 & $(6,1,1,0,0)$ & $(4,1,1,2,0)$ & $(12,4,4,4,0)$ & $2$ & $(1,0,0,0,1)$ & $4\psi^{2}$\\ & 1 & $(0,3,3,2,0)$ & $(4,1,1,2,0)$ & $(6,6,6,6,0)$ & $0$ & $(0,0,0,0,0)$ & $1$\\ & 2 & $(6,2,0,0,0)$ & $(2,6,0,0,0)$ & $(10,10,2,2,0)$ & $4$ & $(1,1,0,0,2)$ & $16\psi^{4}$\\ & 2 & $(6,2,0,0,0)$ & $(4,0,2,2,0)$ & $(12,4,4,4,0)$ & $2$ & $(1,0,0,0,1)$ & $4\psi^{2}$\\ & 2 & $(0,4,2,2,0)$ & $(2,6,0,0,0)$ & $(4,12,4,4,0)$ & $2$ & $(0,1,0,0,1)$ & $4\psi^{2}$\\ & 2 & $(0,4,2,2,0)$ & $(4,0,2,2,0)$ & $(6,6,6,6,0)$ & $0$ & $(0,0,0,0,0)$ & $1$\\ & 3 & $(4,4,0,0,0)$ & $(4,4,0,0,0)$ & $(10,10,2,2,0)$ & $4$ & $(1,1,0,0,2)$ & $16\psi^{4}$\\ & 3 & $(4,4,0,0,0)$ & $(0,0,4,4,0)$ & $(6,6,6,6,0)$ & $0$ & $(0,0,0,0,0)$ & $1$\\ & 3 & $(0,0,4,4,0)$ & $(4,4,0,0,0)$ & $(6,6,6,6,0)$ & $0$ & $(0,0,0,0,0)$ & $1$\\ & 3 & $(0,0,4,4,0)$ & $(0,0,4,4,0)$ & $(2,2,10,10,0)$ & $4$ & $(0,0,1,1,2)$ & $16\psi^{4}$\\ \midrule dectic & 1 & $(2,1,1,3,0)$ & $(2,3,3,1,0)$ & $(6,6,6,6,0)$ & $2$ & $(0,0,0,1,1)$ & $10\psi^{2}$\\ & 1 & $(4,3,3,0,0)$ & $(2,3,3,1,0)$ & $(8,8,8,3,0)$ & $0$ & $(0,0,0,0,0)$ & $1$\\ & 2 & $(3,1,0,3,0)$ & $(1,3,4,1,0)$ & $(6,6,6,6,0)$ & $2$ & $(0,0,0,1,1)$ & $10\psi^{2}$\\ & 2 & $(5,3,2,0,0)$ & $(1,3,4,1,0)$ & $(8,8,8,3,0)$ & $0$ & $(0,0,0,0,0)$ & $1$\\ & 3 & $(8,1,1,0,0)$ & $(4,1,1,2,0)$ & $(14,4,4,4,0)$ & $4$ & $(1,0,0,1,2)$ & $50\psi^{4}$\\ & 3 & $(0,3,3,2,0)$ & $(4,1,1,2,0)$ & $(6,6,6,6,0)$ & $2$ & $(0,0,0,1,1)$ & $10\psi^{2}$\\ & 4 & $(7,1,0,1,0)$ & $(5,1,2,1,0)$ & $(14,4,4,4,0)$ & $4$ & $(1,0,0,1,2)$ & $50\psi^{4}$\\ & 4 & $(1,5,4,0,0)$ & $(5,1,2,1,0)$ & $(8,8,8,3,0)$ & $0$ & $(0,0,0,0,0)$ & $1$\\ & 5 & $(8,0,0,1,0)$ & $(4,2,2,1,0)$ & $(14,4,4,4,0)$ & $4$ & $(1,0,0,1,2)$ & $50\psi^{4}$\\ & 5 & $(0,2,2,3,0)$ & $(4,2,2,1,0)$ & $(6,6,6,6,0)$ & $2$ & $(0,0,0,1,1)$ & $10\psi^{2}$\\ & 5 & $(2,4,4,0,0)$ & $(4,2,2,1,0)$ & $(8,8,8,3,0)$ & $0$ & $(0,0,0,0,0)$ & $1$\\ & 6 & $(4,0,0,3,0)$ & $(6,0,0,2,0)$ & $(12,2,2,7,0)$ & $6$ & $(1,0,0,2,3)$ & $500\psi^{6}$\\ & 6 & $(4,0,0,3,0)$ & $(0,4,4,1,0)$ & $(6,6,6,6,0)$ & $2$ & $(0,0,0,1,1)$ & $10\psi^{2}$\\ & 6 & $(6,2,2,0,0)$ & $(6,0,0,2,0)$ & $(14,4,4,4,0)$ & $4$ & $(1,0,0,1,2)$ & $50\psi^{4}$\\ & 6 & $(6,2,2,0,0)$ & $(0,4,4,1,0)$ & $(8,8,8,3,0)$ & $0$ & $(0,0,0,0,0)$ & $1$\\ & 7 & $(8,2,0,0,0)$ & $(2,8,0,0,0)$ & $(12,12,2,2,0)$ & $6$ & $(1,1,0,1,3)$ & $250\psi^{6}$\\ & 7 & $(8,2,0,0,0)$ & $(4,0,2,2,0)$ & $(14,4,4,4,0)$ & $4$ & $(1,0,0,1,2)$ & $50\psi^{4}$\\ & 7 & $(0,4,2,2,0)$ & $(2,8,0,0,0)$ & $(4,14,4,4,0)$ & $4$ & $(0,1,0,1,2)$ & $50\psi^{4}$\\ & 7 & $(0,4,2,2,0)$ & $(4,0,2,2,0)$ & $(6,6,6,6,0)$ & $2$ & $(0,0,0,1,1)$ & $10\psi^{2}$\\ & 8 & $(6,2,0,1,0)$ & $(6,0,2,1,0)$ & $(14,4,4,4,0)$ & $4$ & $(1,0,0,1,2)$ & $50\psi^{4}$\\ & 8 & $(6,2,0,1,0)$ & $(0,4,6,0,0)$ & $(8,8,8,3,0)$ & $0$ & $(0,0,0,0,0)$ & $1$\\ & 8 & $(0,6,4,0,0)$ & $(6,0,2,1,0)$ & $(8,8,8,3,0)$ & $0$ & $(0,0,0,0,0)$ & $1$\\ & 8 & $(0,6,4,0,0)$ & $(0,4,6,0,0)$ & $(2,12,12,2,0)$ & $6$ & $(0,1,1,1,3)$ & $250\psi^{6}$\\ & 9 & $(6,1,1,1,0)$ & $(6,1,1,1,0)$ & $(14,4,4,4,0)$ & $4$ & $(1,0,0,1,2)$ & $50\psi^{4}$\\ & 9 & $(6,1,1,1,0)$ & $(0,5,5,0,0)$ & $(8,8,8,3,0)$ & $0$ & $(0,0,0,0,0)$ & $1$\\ & 9 & $(0,5,5,0,0)$ & $(6,1,1,1,0)$ & $(8,8,8,3,0)$ & $0$ & $(0,0,0,0,0)$ & $1$\\ & 9 & $(0,5,5,0,0)$ & $(0,5,5,0,0)$ & $(2,12,12,2,0)$ & $6$ & $(0,1,1,1,3)$ & $250\psi^{6}$\\ \bottomrule \end{tabular} \end{table} This appendix completes the proof of Proposition~\ref{prop:wchars}. By Lemma~\ref{lem:wwrap} it suffices to find, for each shift $s\le d/2$, the labels $\bb$ with $b_i+s\not\equiv n_i-1\pmod{n_i}$ for every $i$ and $\sum_ia_i=s$, and their images $\bb'$; the shift $d-s$ only reverses these pairs, so two labels carry the same character exactly when they are joined by a chain of the pairs found. For both pencils $n_5=2$ and $b_5=0$, so the condition at $x_5$ excludes every odd shift, and for an even shift $a_5=s/2$. In this appendix we drop the last coordinate, which is zero, and write labels as vectors of length four. \paragraph*{The octic.} Here $b_i\le6$ and $b_1+b_2+b_3+b_4=8$, and the shifts to consider are $s=2$ and $s=4$. Since $b_i+s\le10<16$, we have $a_i=1$ exactly when $b_i\ge8-s$, and the excluded residue is $b_i=7-s$. So $\sum_ia_i=s$ asks for exactly $s/2$ entries with $b_i\ge8-s$. At $s=2$ exactly one entry equals six and none equals five, and the other three entries add up to two. Up to a permutation of the coordinates this leaves $(6,2,0,0)$ and $(6,1,1,0)$, with images $(0,4,2,2)$ and $(0,3,3,2)$; there are twelve labels of each kind, which gives the $12+12$ pairs of Proposition~\ref{prop:wchars}(b). At $s=4$ exactly two entries are at least four and none equals three; two entries of at least four already add up to eight, so $\bb$ is a permutation of $(4,4,0,0)$, with image the complementary permutation $(0,0,4,4)$, which gives three pairs. The labels of the three lists are distinct: the sources at $s=2$ have an entry six, their images have at most one entry larger than three, and the labels at $s=4$ have two entries four and no entry six. So no label lies in two pairs, the $27$ pairs are the classes, and the $149$ labels carry $149-27=122$ characters. \paragraph*{The dectic.} Here $b_i\le8$ for $i\le3$, $b_4\le3$ and $b_1+b_2+b_3+2b_4=10$, and the shifts to consider are $s=2$ and $s=4$. For $i\le3$, $b_i+s\le12<20$, so $a_i=1$ exactly when $b_i\ge10-s$, and the excluded residue is $b_i=9-s$. For $x_4$ we have $a_4=\lfloor(b_4+s)/5\rfloor$ and the excluded residue is $b_4+s\equiv4\pmod5$. The condition is \[ \#\{i\le3:\ b_i\ge10-s\}+a_4=s/2 . \] At $s=2$ the value $b_4=2$ is excluded, $a_4=1$ exactly when $b_4=3$, and no $b_i$ equals seven. Either one $b_i$ equals eight and $b_4\le1$, which leaves $(8,0,0,1)$, $(8,2,0,0)$ and $(8,1,1,0)$ up to a permutation of the first three coordinates, with images $(0,2,2,3)$, $(0,4,2,2)$ and $(0,3,3,2)$ under the same permutation; or no $b_i$ is at least eight and $b_4=3$, which leaves the fifteen labels $(c_1,c_2,c_3,3)$ with $c_1+c_2+c_3=4$, with images $(c_1+2,c_2+2,c_3+2,0)$. At $s=4$ the value $b_4=0$ is excluded, so $b_4\ge1$ and $a_4=1$; hence exactly one $b_i$ is at least six, and none equals five. The other entries satisfy $b_j+b_k+2b_4=10-b_i\le4$ with $b_4\ge1$, which leaves $(6,2,0,1)$, $(6,1,1,1)$, $(6,0,0,2)$, $(7,1,0,1)$ and $(8,0,0,1)$ up to a permutation of the first three coordinates, with images $(0,6,4,0)$, $(0,5,5,0)$, $(0,4,4,1)$, $(1,5,4,0)$ and $(2,4,4,0)$. Sorting the labels of these lists by $b_4$ shows which of them coincide. With $b_4=3$ there are the sources $(c_1,c_2,c_3,3)$ and the image $(0,2,2,3)$, which is one of them, with $\cc=(0,2,2)$. With $b_4=2$ there are $(0,4,2,2)$, $(0,3,3,2)$ and $(6,0,0,2)$, all different. With $b_4=1$ there are $(8,0,0,1)$, which occurs at both shifts, and $(6,2,0,1)$, $(6,1,1,1)$, $(7,1,0,1)$ and $(0,4,4,1)$, all different. With $b_4=0$ there are $(8,2,0,0)$ and $(8,1,1,0)$, the images $(c_1+2,c_2+2,c_3+2,0)$, all of whose entries are at least two, and $(0,6,4,0)$, $(0,5,5,0)$, $(1,5,4,0)$ and $(2,4,4,0)$, of which only the last has all entries at least two; it is the image of $\cc=(0,2,2)$. So the only chains longer than one pair are \[ (8,0,0,1)\xrightarrow{s=2}(0,2,2,3)\xrightarrow{s=2} (2,4,4,0)\xleftarrow{s=4}(8,0,0,1) \] and its permutations, the three triples of Proposition~\ref{prop:wchars}(c). The remaining pairs are the $12$ with $\cc$ not a permutation of $(2,2,0)$ and the $6+3$ others at $s=2$, and the $6+3+3+6$ at $s=4$, which are the pairs listed there: $39$ pairs and $3$ triples, $87$ labels in $42$ classes, and $145-87+42=100$ characters. \section{The blocks of the mass map}\label{app:wblocks} Table~\ref{tab:wblockdata} gives the data behind Table~\ref{tab:wblocks}. For each row we take the representative with $(i,j,k,l)=(1,2,3,4)$ and list, for $\bb\in B_\chi$ and $\cc\in B_{-\chi}$ in the order of Table~\ref{tab:wblocks}, the vector $h+\bb+\cc$, its sector $r$, the vector $k$ of \eqref{eq:wsector} and the entry $\psi^r\prod_iw_i^{k_i}$ of Theorem~\ref{thm:wsocle}. The other blocks of each row follow by permuting the coordinates of weight one. 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