\documentclass[11pt,reqno]{amsart} \usepackage[margin=1.1in]{geometry} \usepackage[T1]{fontenc} \usepackage{lmodern} \usepackage{amsmath,amssymb,amsthm,mathtools,mathrsfs,tikz} \usepackage{xcolor} \usepackage{microtype} \usepackage[hidelinks]{hyperref} \usepackage{orcidlink} \definecolor{vegasgold}{rgb}{0.77, 0.7, 0.35} \definecolor{darkgoldenrod}{rgb}{0.72, 0.53, 0.04} \definecolor{gold(metallic)}{rgb}{0.83, 0.69, 0.22} \hypersetup{ colorlinks=true, linkcolor=darkgoldenrod, filecolor=brown, urlcolor=gold(metallic), citecolor=darkgoldenrod, } \usepackage{enumitem} \usepackage[nameinlink,capitalize]{cleveref} \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{question}[theorem]{Question} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \newcommand{\Q}{\mathbb Q} \newcommand{\Z}{\mathbb Z} \newcommand{\cO}{\mathcal O} \newcommand{\op}[1]{\operatorname{#1}} \newcommand{\Tr}{\operatorname{Tr}} \newcommand{\Fp}{\mathbb{F}_p} \author[A.~Ray]{Anwesh Ray\, \orcidlink{0000-0001-6946-1559}} \address{Chennai Mathematical Institute, H1, SIPCOT IT Park, Siruseri, Kelambakkam 603103, Tamil Nadu, India.} \email{anwesh@cmi.ac.in} \DeclareMathOperator{\CRSF}{CRSF} \DeclareMathOperator{\Jac}{Jac} \DeclareMathOperator{\ord}{ord} \DeclareMathOperator{\rank}{rank} \DeclareMathOperator{\corank}{corank} \DeclareMathOperator{\supp}{supp} \DeclareMathOperator{\Span}{Span} \DeclareMathOperator{\Gal}{Gal} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\coker}{coker} \DeclareMathOperator{\disc}{disc} \DeclareMathOperator{\val}{val} \newcommand{\F}{\mathbb F} \newcommand{\Zl}{\Z_\ell} \newcommand{\Fl}{\mathbb F_\ell} \newcommand{\Zp}{\Z_p} \newcommand{\Gm}{\mathbb G_m} \newcommand{\kappaX}{\kappa(X)} \newcommand{\cX}{c_X} \newcommand{\chiell}{\chi_\ell} \newcommand{\chip}{\chi_p} \newcommand{\ee}{\mathbf e} \newcommand{\bb}{\mathbf b} \newcommand{\ba}{\mathbf a} % Additional notation used in this paper. \usepackage{booktabs} \setlength{\emergencystretch}{2em} \raggedbottom \newcommand{\A}{\mathbb A} \newcommand{\AR}[1]{\textcolor{blue}{#1}} \newcommand{\Pp}{\mathbb P} \newcommand{\LL}{\mathbb L} \newcommand{\cD}{\mathcal D} \newcommand{\cE}{\mathcal E} \newcommand{\cF}{\mathcal F} \newcommand{\cM}{\mathcal M} \newcommand{\Gr}{\operatorname{Gr}} \newcommand{\Pic}{\operatorname{Pic}} \newcommand{\Col}{\operatorname{Col}} \newcommand{\Sym}{\operatorname{Sym}} \newcommand{\Frob}{\operatorname{Frob}} \newcommand{\ac}{\operatorname{ac}} \newcommand{\Conf}{\operatorname{Conf}} \newcommand{\lev}{\operatorname{lev}} \newcommand{\Spec}{\operatorname{Spec}} \newcommand{\rk}{\operatorname{rank}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\circled}{\mathbin{\circ}} \newcommand{\bfone}{\mathbf 1} \newcommand{\Stwelve}{S_{12}} \title[Subrings of prime-power index and modular forms] {Subrings of $\mathbb{Z}^n$ of prime-power index and modular forms} \date{8 Nov, 2026} \subjclass[2020]{Primary 11R45; Secondary 14G10, 14H10, 11F11} \keywords{Subring growth, Hermite normal form, local-field comparison, elliptic curves, point counts} \hypersetup{ pdftitle={subrings of prime power index}, pdfauthor={Anwesh Ray}, pdfsubject={Finite-index subrings and local-field comparison}, pdfkeywords={subrings, Hermite normal form, elliptic normal quartics} } \newenvironment{ReviewAddition}{% \par\begingroup\color{red}% }{% \par\endgroup } \begin{document} \begin{abstract} Let $f_n(m)$ denote the number of unital subrings of index $m$ in $\mathbb{Z}^n$, with coordinatewise multiplication, and let $g_n(p^e)$ count those whose reduction modulo $p$ is the diagonal copy of $\mathbb{F}_p$. Voll's uniformity question asks whether the local subring zeta functions can be described by a single rational function in $p$ and $p^{-s}$. Polynomial formulas for $f_n(p^e)$ are known for $e\le9$. We prove that neither $f_{10}(p^{17})$ nor $g_{10}(p^{17})$ agrees with a polynomial in $p$ for all sufficiently large primes. We introduce new geometric and cohomological methods for this problem, moving beyond the earlier combinatorial calculations. We compare the counts over $\mathbb{Z}_p$ and $\mathbb{F}_p[[t]]$ and, for all sufficiently large primes, express them as point counts on reductions of varieties over $\mathbb{Q}$. The Grothendieck--Lefschetz trace formula relates these counts to compactly supported $\ell$-adic \'etale cohomology. We identify a contribution from Tate twists of the two-dimensional Galois representation attached to the weight-twelve cusp form $\Delta$. We show that this modular contribution does not cancel when all parts of the count are combined. Consequently, there is no polynomial $P\in\mathbb{Q}[X]$ such that $f_{10}(p^{17})=P(p)$ for all but finitely many primes $p$. This gives a negative answer to Voll's uniformity question even when finitely many exceptional primes are allowed. \end{abstract} \maketitle \section{Introduction} \subsection{Motivation} Given a group, module or ring, one may ask how many subgroups, submodules or subrings respectively it has of a fixed finite index. Let $A$ be a ring whose additive group is free of finite rank over $\Z$, and let $a_n$ denote the number of unital subrings $S$ of index $[A:S]=n$. The subring zeta function of $A$ is the Dirichlet series \[ \zeta_A^R(s)=\sum_{n\ge1}a_n n^{-s}. \] Its analytic properties can be used to determine the growth of $f(x):=\sum_{n\le X}a_n$, the number of such subrings of index at most $X$. Tauberian theorems relate the poles the zeta function to the growth of $f(x)$. Suppose that $\zeta_A^R(s)$ has abscissa of convergence $\alpha>0$ and extends meromorphically to a neighbourhood of the closed half-plane $\Re(s)\ge\alpha$, with its only pole there at $s=\alpha$, of order $b$. Let $c=\lim_{s\to\alpha}(s-\alpha)^b\zeta_A^R(s)$ be its leading Laurent coefficient. Then the Ikehara--Delange Tauberian theorem implies that \[ f(x) \sim \frac{c}{\alpha(b-1)!}\, X^\alpha(\log X)^{b-1} \] as $X\rightarrow \infty$. Analogous zeta functions have been studied to study subgroup growth and submodule growth, cf. \cite{LubotzkySegal2003,Voll2011}. The word growth series of a finitely generated group counts its elements by word length with respect to a fixed finite generating set \cite{GrigorchukNagnibeda1997}. Representation zeta functions count irreducible complex representations, up to isomorphism, by their dimension, provided there are only finitely many in each dimension \cite{LarsenLubotzky2008}. Such generating functions arise naturally in number theory. The prototype is the Dedekind zeta function of a number field $K$, whose coefficients count ideals $I$ by their norm. The study of zeta functions associated to algebraic structures has led to many questions. Grunewald, Segal and Smith \cite[Section~3]{GSS1988} showed that local factors of subgroup zeta functions are $p$-adic integrals, and hence rational functions of $p^{-s}$ by the work of Denef \cite{Denef1991}. On the other hand, du Sautoy and Grunewald \cite{duSautoyGrunewald2000} studied the analytic properties of these functions, and Voll \cite{Voll2010,Voll2015} studied the functional equations they satisfy. A natural question arises from the study of local factors. For a fixed ring $A$, does there exist a rational function $W(X,Y)\in\Q(X,Y)$ such that $\zeta^R_{A,p}(s)=W(p,p^{-s})$ for every prime $p$? When such a function exists, the subring zeta function of $A$ is called \emph{uniform}. For $A=\Z^n$, this question was posed explicitly by Voll \cite[Question~3.7]{Voll2015}, and again by Atanasov, Kaplan, Krakoff and Menzel \cite[Question~1.12]{AKKM}. Let $A$ be a ring whose additive group is free of finite rank, and let $f_A(m)$ denote the number of unital subrings of index $m$. Then $\zeta^R_A(s)=\sum_{m\ge1}f_A(m)m^{-s}$ is the subring zeta function of $A$; since $f_A$ is multiplicative, it is an Euler product $\prod_p\zeta^R_{A,p}(s)$. For fixed $n$, uniformity is equivalent to the assertion that, for every $e\ge0$, there is a polynomial $P_{n,e}$ with rational coefficients for which $f_n(p^e)=P_{n,e}(p)$ for every prime $p$. Voll's uniformity question can therefore be stated as follows. \begin{question} For each fixed pair $(n,e)$, does there exist a polynomial $P_{n,e}\in\Q[X]$ such that $f_n(p^e)=P_{n,e}(p)$ for every prime $p$? \end{question} Work on this question has proceeded by establishing polynomial formulas for successively larger exponents $e$. Liu \cite{Liu} introduced a decomposition into irreducible subrings and obtained formulas for $e\le5$. Atanasov, Kaplan, Krakoff and Menzel \cite[Proposition~1.2 and Theorem~1.3]{AKKM} extended these calculations to $e\le8$, correcting Liu's formula for $e=5$. Mishra and Ray \cite[Theorem~1.4]{MishraRay} subsequently treated $e=9$. These computations have supported the expectation that the question has an affirmative answer. These arguments are combinatorial in nature. Liu \cite{Liu} represents each finite-index additive subgroup of $\Z^n$ by its unique Hermite normal form matrix, whose columns form a basis of the subgroup. The product of the diagonal entries is the index. Thus, for index $p^e$, the diagonal entries are powers of $p$ whose exponents sum to $e$. There are finitely many possible diagonals, and for each diagonal the remaining matrix entries lie in prescribed finite ranges. To determine which matrices give subrings, one checks that the subgroup contains the identity and that the coordinatewise product of every pair of columns is an integer linear combination of the columns. Isham \cite{Isham} proves that the number of matrices satisfying any one multiplication condition is polynomial in $p$. Here a multiplication condition requires the coordinatewise product of two columns to belong to the lattice spanned by the columns. Isham's calculations also illustrate a difficulty with imposing several multiplication conditions simultaneously. In \cite{IshamLowerBounds}, Isham obtains explicit lower bounds for $f_n(p^e)$ valid for $e\ge n-1$. Using constructions of Brakenhoff, she improves the earlier general lower bounds for the number of subrings of index at most $X$ when $n\ge7$, and derives consequences for the convergence of the associated zeta functions. A subring of $\Z^n$ of $p$-power index is called \emph{irreducible} if every element has all its coordinates congruent modulo $p$. Let $g_n(p^e)$ count the irreducible subrings of index $p^e$. Liu \cite[Theorem~3.4]{Liu} proves that every subring of $p$-power index decomposes uniquely as a product of irreducible subrings on disjoint blocks of coordinates. Liu also proves the formula \[f_n(p^e)=\sum_{j=1}^{n}\binom{n-1}{j-1} \sum_{i=0}^{e}g_j(p^i)f_{n-j}(p^{e-i}),\] where $f_0(1)=1$ and $f_0(p^e)=0$ for $e>0$. Since the term with $j=n$ is $g_n(p^e)$, induction on $n$ shows that polynomiality of all the counts $f_n(p^e)$ is equivalent to polynomiality of all the irreducible counts $g_n(p^e)$. \subsection{Main results} Our main result gives a negative answer to Voll's question even when finitely many exceptional primes are allowed. \begin{theorem}\label{thm:total-main} The following assertions hold. \begin{enumerate}[label=\textup{(\roman*)}] \item No polynomial $P\in\Q[T]$ satisfies $g_{10}(p^{17})=P(p)$ for every sufficiently large prime $p$. \item No polynomial $P\in\Q[T]$ satisfies $f_{10}(p^{17})=P(p)$ for every sufficiently large prime $p$. \end{enumerate} \end{theorem} The obstruction can stated more precisely. For a composition $\alpha=(\alpha_1,\ldots,\alpha_{n-1})$ of $e$, let $g_\alpha(p)$ count the irreducible subrings whose column Hermite normal form has diagonal $(p^{\alpha_1},\ldots,p^{\alpha_{n-1}},1)$. By uniqueness of the normal form, $g_n(p^e)$ is the sum of $g_\alpha(p)$ over all such compositions. \begin{theorem}\label{thm:main} Let $n=10$ and $\alpha_0=(3,3,2,2,2,2,1,1,1)$. There is no polynomial $P\in\Q[T]$ for which $g_{\alpha_0}(p)=P(p)$ for every sufficiently large prime $p$. \end{theorem} We obtain the following result for subring zeta functions. \begin{corollary}\label{cor:nonuniform-local-zeta} There is no rational function $W(X,Y)\in\Q(X,Y)$ such that $\zeta^R_{\Z^{10},p}(s)=W(p,p^{-s})$ for every sufficiently large prime $p$. \end{corollary} \subsection{Methodology} The key idea is to study the counting problem over $\F_p[[t]]$. For fixed $n$ and $e$, we prove that, for all sufficiently large primes $p$, the counts $f_n(p^e)$ and $g_n(p^e)$ agree with the corresponding counts of unital $\F_p[[t]]$-subalgebras of $\F_p[[t]]^n$ (see Section~\ref{sec:comparison}). Every such subalgebra contains $t^e\F_p[[t]]^n$, so it is determined by its image in the finite-dimensional algebra $(\F_p[t]/(t^e))^n$. Containment of the identity, stability under multiplication by $t$, and closure under coordinatewise multiplication give polynomial conditions on this image. Thus there are schemes over $\Z$ whose $\F_p$-points give the counts $f_n(p^e)$ and $g_n(p^e)$. Their reduced generic fibers are varieties over $\Q$, which we study using cohomology. For a variety $X/\Q$ and a prime $\ell$, the compactly supported \'etale cohomology groups $H_c^i(X_{\overline\Q},\Q_\ell)$ are finite-dimensional vector spaces with a continuous action of $G_\Q=\operatorname{Gal}(\overline\Q/\Q)$. These Galois representations determine the point counts. Indeed, if $\mathscr X$ is a model over $\Z[1/N]$, then, after enlarging $N$, the Grothendieck--Lefschetz trace formula gives \[ \#\mathscr X(\F_p) =\sum_i(-1)^i \operatorname{Tr}\!\left( \operatorname{Frob}_p\mid H_c^i(X_{\overline\Q},\Q_\ell)\right) \] for $p\nmid N\ell$, where $\operatorname{Frob}_p$ denotes geometric Frobenius; see \cite[\S29]{MilneLEC}. Let $\chi_\ell:G_\Q\to\Z_\ell^\times$ be the $\ell$-adic cyclotomic character, defined by $\sigma(\zeta)=\zeta^{\chi_\ell(\sigma)}$ for every root of unity $\zeta$ of $\ell$-power order. The one-dimensional representation $\Q_\ell(r)$ has Galois action given by $\chi_\ell^r$. Since $\chi_\ell(\operatorname{Frob}_p)=p^{-1}$ for geometric Frobenius at $p\ne\ell$, its action on $\Q_\ell(-r)$ is multiplication by $p^r$. We call $X$ Tate if each of its compactly supported cohomology groups has a filtration whose successive quotients are Tate representations. Its point count is then polynomial in $p$. For the counting problem, however, it is the alternating sum of the cohomology groups that matters and contributions in even and odd degrees may cancel. We show that there is a non-Tate contribution that survives this alternating sum. The relevant representation comes from the weight-twelve cusp form $\Delta(q)=\sum_{m\ge1}\tau(m)q^m$. Its associated Galois representation $S_{12,\ell}$ has dimension two and geometric-Frobenius trace $\tau(p)$ (see \cite{Deligne}). It is irreducible, so it cannot be built from one-dimensional Tate representations. The same holds for each Tate twist $S_{12,\ell}(-r)=S_{12,\ell}\otimes\Q_\ell(-r)$, whose Frobenius trace is $p^r\tau(p)$. These representations enter through a locus describing smooth genus-one curves with ten marked points, a degree-four line bundle, and a section nonzero at the markings. Its cohomology contains the modular contribution associated to $\Delta$, as described by Eichler--Shimura theory and Getzler's calculation \cite{Deligne,Getzler99}. This is the crucial ingredient that shows that the point counts are not eventually polynomial. \subsection{Organization of the paper} Section~\ref{sec:prelim} is preliminary in nature and recalls basic combinatorial results and the relationship with the subring zeta function. The lattice reductions and cohomological tools introduced here will be used throughout the article. We discuss Galois representations on the \'etale cohomology of a variety and its compactly supported Euler classes in the K group of representations. This trace of the geometric Frobenius at $p$ on this class gives the number of $\mathbb{F}_p$-points on the variety. Section~\ref{sec:hnf} expresses the subring conditions as congruences on the entries of Hermite normal form matrices. Separating the matrices according to their diagonals gives the counts $g_\alpha(p)$. We first study the composition $\alpha_0=(3,3,2,2,2,2,1,1,1)$, before returning to the full irreducible count. Section~\ref{sec:comparison} proves that, for each fixed diagonal $\alpha$, the counts over $\Z_p$ and $\F_p[[t]]$ agree for all sufficiently large primes. We express both counts as local integrals defined by polynomial valuation conditions and use resolution of singularities to compare them. The passage to $\F_p[[t]]$ is crucial since it allows us to describe $f_n(p^e)$ and $g_n(p^e)$ as $\mathbb{F}_p$-point counts on varieties which depend only on $n$ and $e$. Section~\ref{sec:eqchar} constructs the parameter space for the Hilbert normal form diagonal $\alpha_0$. There is a natural $\mathbb G_m$-action on this space. We consider the fixed locus of this action. Its points correspond to graded subalgebras, which have the form $k\bfone\oplus tU\oplus t^2V\oplus t^3k[[t]]^{10}$. Here $\bfone\in U\subseteq V\subseteq k^{10}$, with $\dim U=4$ and $\dim V=8$. The product of two degree-one terms must belong to the degree-two part, so every coordinatewise product of vectors in $U$ must lie in $V$. Section~\ref{sec:elliptic} explains how elliptic curves arise from these flags. The ten coordinate evaluations on $U$ define ten points in $\Pp^3$. If the quadratic forms vanishing at the ten points form a two-dimensional space, their common zero locus is the intersection of two quadric surfaces. When smooth, $C$ is a genus-one curve equipped with a degree $4$ line bundle. Varying the degree-four line bundle amounts to choosing an additional point on the curve. When this point is distinct from the ten markings, the resulting data are parametrized by $\cM_{1,11}$, the space of smooth genus-one curves with eleven ordered distinct marked points. Its cohomology contains a two-dimensional Galois representation associated to the modular form $\Delta$. The trace of geometric Frobenius at $p$ on this representation is the Fourier coefficient $\tau(p)$. This explains how the coefficients of $\Delta$ enter our point counts. Section~\ref{sec:boundary} studies the part of the fixed locus outside the smooth elliptic-curve locus described above. This includes configurations where some of the ten points coincide, where the quadrics fail to cut out a smooth curve, or where the points satisfy more than two independent quadratic relations. We show that the compactly supported Euler class of this boundary has no contribution from any Tate twist of the representation attached to $\Delta$. It therefore cannot cancel the modular contribution from the smooth elliptic-curve locus. In Section~\ref{sec:final}, we introduce the linear functional $\varepsilon_{12}$. On the compactly supported Euler class, it counts each copy of a Tate twist of the representation attached to $\Delta$ with sign $+1$ in even degree and $-1$ in odd degree, ignoring all other irreducible Galois constituents. Since every twist is counted equally, its value is unchanged by Tate twisting. Write $X$ for the parameter variety and $F$ for its $\mathbb G_m$-fixed locus. The complement $X\setminus F$ admits a finite partition into locally closed subsets $\mathbb G_m\times Y_i$. Since $[\mathbb G_m]=\LL-1$, where $\LL=[\A^1]$, this gives $[X]=[F]+(\LL-1)[Y]$ for a suitable variety $Y$. Multiplication by $\LL$ becomes Tate twisting in cohomology, so the second term has zero $\varepsilon_{12}$ value. Sections~\ref{sec:elliptic} and~\ref{sec:boundary} show that the smooth elliptic-curve locus in $F$ contributes $56$, while its complement contributes zero. Thus $\varepsilon_{12}(\chi_c^\ell(X)) =\varepsilon_{12}(\chi_c^\ell(F))=56$. Since an eventual polynomial point count would force this value to vanish, this proves Theorem~\ref{thm:main}. To study the full irreducible count, we construct a projective parameter space containing every HNF diagonal of the given rank and index. Its $\mathbb G_m$-fixed points are graded subalgebras, described by nested subspaces specifying the coefficients allowed at successive powers of $t$. In rank ten and colength seventeen, the scalar degree-zero part contributes nine to the colength. The remaining codimensions sum to eight, so their nonincreasing sequences are indexed by the twenty-two partitions of eight. The partition $(6,2)$ gives the flag incidence variety already studied. Section~\ref{total:sec-profiles} treats the other twenty-one possibilities and shows that each has zero $\varepsilon_{12}$ contribution. The value $56$ therefore survives in $g_{10}(p^{17})$, proving Theorem~\ref{thm:total-main}\textup{(i)}. Section~\ref{total:sec-low-rank} establishes the corresponding vanishing results in lower ranks, which are needed for Liu's decomposition into irreducible subrings. Finally, Section~\ref{total:sec:ordinary} passes from irreducible subrings to all subrings counted by $f_{10}(p^{17})$. If a subring splits into two or more irreducible factors, each factor involves at most nine coordinates. The exponents of their indices add up to seventeen, so at most one factor can have index $p^a$ with $a\ge9$. All the other factors fall within the range covered by the known polynomial formulas. We combine these formulas with the lower-rank results to show that the subrings which split contribute zero to $\varepsilon_{12}$. Adding them to the irreducible count proves Theorem~\ref{thm:total-main}\textup{(ii)}. \subsection{Further questions} The failure of polynomiality raises the question of how the functions $f_n(p^e)$ can be described more precisely. For the case considered here, can one find polynomials $P,Q\in\Q[X]$ such that $f_{10}(p^{17})=P(p)+Q(p)\tau(p)$ for all sufficiently large primes $p$? Our results do not establish such a formula. The functional $\varepsilon_{12}$ records the total signed multiplicity of the Tate twists of $S_{12,\ell}$, but does not determine their individual multiplicities or exclude other non-Tate constituents. More generally, one can ask which modular forms occur as $n$ and $e$ vary. Can the functions $f_n(p^e)$ be expressed using polynomials in $p$ and Fourier coefficients of a finite number of modular forms? The marked elliptic curves appearing here explain the contribution of $\Delta$. It would be interesting to identify other geometric configurations that produce further modular forms. One should also ask whether modular forms suffice to describe these counts, or whether larger values of $n$ and $e$ involve Galois representations arising from other motives. Similar questions can be posed for a wider range of algebraic structures. Examples include finite-index subrings of other rings whose additive groups are free of finite rank, orders in number fields, and finite-index subalgebras or ideals of Lie rings. There are already examples relating local normal-subgroup zeta functions of nilpotent groups to point counts on elliptic curves; see \cite[Theorem~1.1]{duSautoy2002}. \subsection*{Declarations} \begin{description} \item[Conflict of interest] There is no conflict of interest to report. \item[Availability of data and materials] No data was generated or analyzed in obtaining the results in this article. \item[Funding] No funding was received during the preparation of this article. \item[Use of artificial intelligence] The author used ChatGPT 6.1 Sol as an exploratory tool to study related papers and to assist in correcting mistakes in the article. It was not used to assist in writing the paper or in proving the results. \end{description} \section{Preliminaries}\label{sec:prelim} In this section, we discuss preliminary notions and set up notation which will be used throughout the article. \subsection{Combinatorial preliminaries on subrings of $\Z^n$} Given a commutative ring $A$, multiplication on $A^n$ is coordinatewise: \[ (x_1,\ldots,x_n)\circled(y_1,\ldots,y_n) =(x_1y_1,\ldots,x_ny_n). \] All subrings contain the identity $\bfone=(1,\ldots,1)$. For submodules $U,V$ of $A^n$, let $U\circled V$ be the submodule spanned by $u\circ v$ where $u\in U$ and $v\in V$, and write $U^{\circled2}=U\circled U$. Let $f_n(m)$ count the subrings $R\subseteq\Z^n$ with $\#(\Z^n/R)=m$. Associated with this arithmetical function is the zeta subring zeta function \[ \zeta_{\Z^n}^{R}(s)=\sum_{m\ge1}f_n(m)m^{-s}.\]Given a prime number $p$, let \[Z_{n,p}(T)=\sum_{e\ge0}f_n(p^e)T^e,\]and set \[\zeta_{\Z^n,p}^{R}(s)=Z_{n,p}(p^{-s}). \] It is easy to see that the function $f_n$ is multiplicative. Hence \[ f_n(m)=\prod_{p\mid m}f_n(p^{v_p(m)})\] and consequently,\[\zeta_{\Z^n}^{R}(s)=\prod_p\zeta_{\Z^n,p}^{R}(s). \] For $n\ge2$, the series and product converge absolutely when $\Re(s)>n-1$. To see this, each $R$ contains $\Z\bfone$, and $R/\Z\bfone$ has index $m$ in $\Z^n/\Z\bfone\simeq\Z^{n-1}$. If $a_{n-1}(m)$ counts all subgroups of this index, then $f_n(m)\le a_{n-1}(m)$ and \[ \sum_{m\ge1}f_n(m)m^{-\sigma} \le\sum_{m\ge1}a_{n-1}(m)m^{-\sigma} =\prod_{j=0}^{n-2}\zeta(\sigma-j)<\infty \] for $\sigma>n-1$. We refer to \cite[Section~1.1, equation~(1.2)]{AKKM} for further details. For $n=1$ the subring zeta function is $1$. Let $\cO$ be a discrete valuation ring with uniformizer $\varpi$ and residue field $k=\cO/(\varpi)$. We set $q=\#k$. The main examples that we work with in this paper are $\Z_p$ and $k[[t]]$. A full $\cO$-lattice $L\subseteq\cO^n$ is a free rank-$n$ submodule with finite-length quotient. Its colength is then defined to be $\operatorname{length}_{\cO}(\cO^n/L)$. A quotient of length $e$ has $e$ composition factors isomorphic to $k$, its cardinality is $q^e$. A full unital $\cO$-subalgebra of $\cO^n$ is a full lattice containing $\bfone$, which is closed under multiplication. Such an algebra is said to be \emph{irreducible} if its image \[ \overline R=(R+\varpi\cO^n)/\varpi\cO^n\subseteq k^n \] is $k\bfone$. Let $g_n(p^e)$ be the number of index-$p^e$ irreducible unital full $\Z_p$-subalgebras of $\Z_p^n$. \begin{lemma}\label{lem:direct-lattice-reduction} Let $\cO$ be a discrete valuation ring with uniformizer $\varpi$ and residue field $k$. Set \[ N=\{(x_1,\ldots,x_{n-1},0):x_i\in\cO\}\subseteq\cO^n. \] The assignment \[ M\longmapsto R_M=\cO\bfone\oplus\varpi M \] is a bijection between full lattices $M\subseteq N$ satisfying \[ \varpi(M\circled M)\subseteq M \] and irreducible full unital $\cO$-subalgebras of $\cO^n$. Its inverse is \[ R\longmapsto M_R=\{x\in N:\varpi x\in R\}. \] For corresponding $M$ and $R$, we have \[ \operatorname{length}_{\cO}(\cO^n/R) =n-1+\operatorname{length}_{\cO}(N/M). \] \end{lemma} \begin{proof} The proof of this result is straightforward and is thus ommitted. \end{proof} \begin{lemma}\label{lem:two-layer-lattices} Let $N=\cO^m$ and let $M$ be a lattice with $\varpi^2N\subseteq M\subseteq N$. Let $U$ and $V$ be the $k$-vector spaces defined by \[ U=(M+\varpi N)/\varpi N\] and \[V=\{\bar x\in N/\varpi N:\varpi x\in M\}. \] Then $U\subseteq V$, and, for $r=\dim U$ and $s=\dim V$, \[ \operatorname{length}_{\cO}(N/M)=2m-r-s. \] Fix a flag $U\subseteq V\subseteq k^m$, with $\dim U=r$ and $\dim V=s$. Choose a basis of $k^m$ whose first $r$ vectors span $U$ and whose first $s$ vectors span $V$, and choose lifts of these basis vectors to $N$. These choices give a bijection between the lattices $\varpi^2N\subseteq M\subseteq N$ inducing this flag and \[ \operatorname{Hom}_k(U,k^m/V). \] This vector space has dimension $r(m-s)$. Hence, exactly $q^{r(m-s)}$ lattices induce the prescribed flag. The multiplication condition depends only on the flag. Indeed, we have that $\varpi(M\circled M)\subseteq M$ if and only if $ U^{\circled2}\subseteq V$. Thus all $q^{r(m-s)}$ lattices satisfy this condition when $U^{\circled2}\subseteq V$, and none do otherwise. \end{lemma} \begin{proof} The containment $\varpi^2N\subseteq M$ makes $V$ well defined, and $\varpi M\subseteq M$ gives $U\subseteq V$. In the short exact sequence \[ 0\longrightarrow\varpi N/(M\cap\varpi N) \longrightarrow N/M\longrightarrow N/(M+\varpi N) \longrightarrow0, \] we have that \[\dim_k \varpi N/(M\cap\varpi N)=m-s\] and \[\dim_k N/(M+\varpi N)=m-r.\] This proves the length formula. Choose a basis $u_1,\ldots,u_r$ of $U$, extend it to bases of $V$ and $k^m$, and lift this basis to $N$. Every desired lattice contains $K=\{\varpi x:\bar x\in V\}$ and is generated modulo $K$ by $\widetilde u_i+\varpi a_i$ for $1\le i\le r$. Because $U\subseteq V$, multiplying these generators by $\varpi$ gives elements of $K$. Thus the residues of $a_i$ in $k^m/V$ are arbitrary and determine the lattice uniquely: differences of two lifts of the same $u_i$ belong to $M\cap\varpi N$, whose image is $\varpi V$ modulo $\varpi^2N$. These $r$ residues give the stated parameter space. Finally, $\varpi z\in M$ is equivalent to $\bar z\in V$ for $z\in N$. Apply this to products of elements of $M$, whose reductions fill $U$, to obtain the closure criterion. \end{proof} \subsection{Grothendieck groups and Euler characteristics} A $k$-variety is a reduced separated scheme of finite type over $k$. When a scheme $X$ over $k$ is not reduced, we replace it by its reduced scheme before taking its variety $X^{\text{red}}$. Let $K_{\mathrm{var},k}=K_0(\operatorname{Var}_k)\otimes_{\Z}\Q$ be the Grothendieck $\Q$-algebra of $k$-varieties. It is generated by isomorphism classes $[X]$, subject to the relation \[ [X]=[Z]+[X\setminus Z] \] where $Z\subseteq X$ is a closed subvariety. Multiplication is defined by $[X][Y]=[X\times_kY]$, and the identity $1=[\Spec k]$. A finite stratification of $X$ is a decomposition $X=\bigsqcup_iX_i$ into locally closed subvarieties. Applying the relation successively gives $[X]=\sum_i[X_i]$. Write $\A^a_k$ for affine $a$-space, $\Pp^a_k$ for projective $a$-space, and $\mathbb G_m=\A^1_k\setminus\{0\}$. We set $\LL=[\A^1_k]$. The decomposition of projective space into affine spaces gives \[ [\A^a_k]=\LL^a,\qquad [\mathbb G_m]=\LL-1\quad\text{and}\quad [\Pp^a_k]=1+\LL+\cdots+\LL^a. \] If $k=\F_q$, the rational point count defines a ring homomorphism $K_{\mathrm{var},\F_q}\rightarrow\Q$ defined by $[X]\mapsto\#X(\F_q)$. Fix a prime $\ell$ and set $G_\Q=\Gal(\overline\Q/\Q)$. A continuous $\ell$-adic representation of $G_\Q$ is a finite-dimensional $\Q_\ell$-vector space $V$ with a continuous homomorphism $G_\Q\to\operatorname{GL}(V)$. We use the rational Grothendieck ring \[ K_\ell= K_0(\operatorname{Rep}^{\mathrm{cont}}_{\Q_\ell}(G_\Q)) \otimes_{\Z}\Q. \] Its defining relations are \[ [V]=[V']+[V'']\] for every short exact sequence \[0\longrightarrow V'\longrightarrow V \longrightarrow V''\longrightarrow0, \] and its multiplication is induced by tensor product. A simple constituent of a representation means a simple quotient in a composition series. Its multiplicity is independent of the series, by the Jordan--H\"older theorem. The class of a representation records these multiplicities, and a virtual class records their signed multiplicities. For a variety $X/\Q$, each compactly supported cohomology group $H_c^i(X_{\overline\Q},\Q_\ell)$ is such a representation. We define \[ \chi_c^\ell(X)= \sum_i(-1)^i[H_c^i(X_{\overline\Q},\Q_\ell)]\in K_\ell. \] This induces a ring homomorphism \[ \chi_c^\ell:K_{\mathrm{var},\Q}\longrightarrow K_\ell. \] Indeed, for $Z\subseteq X$ closed, the compact-support localization long exact sequence \cite[Chapter~III, (1.30)]{MilneEtale} gives \[ \chi_c^\ell(X)=\chi_c^\ell(Z)+\chi_c^\ell(X\setminus Z). \] The compact-support K\"unneth formula \cite[Chapter~VI, Section~8]{MilneEtale} gives \[ \chi_c^\ell(X\times Y)=\chi_c^\ell(X)\chi_c^\ell(Y). \] The localization sequence for $\A^1_\Q=\Pp^1_\Q\setminus\{\infty\}$ gives \[ H_c^i(\A^1_{\overline\Q},\Q_\ell)\simeq \begin{cases} \Q_\ell(-1),&i=2,\\ 0,&i\ne2, \end{cases} \] as $G_\Q$-representations; see \cite[Chapter~III, (1.30)]{MilneEtale} and \cite[Theorem~23.2]{MilneLEC}. Since degree two contributes with positive sign, \[ \chi_c^\ell(\LL)=[\Q_\ell(-1)]. \]For ease of notation, we also set $\LL:=[\Q_\ell(-1)]$. The one-dimensional objects $\Q_\ell(-r)$ and $\Q(-r)$, for $r\in\Z$, are called Tate objects. A variety is said to be Tate if every simple constituent of every $H_c^i$ is a Tate object. A \emph{virtual Tate class} $P(\LL)$ is a finite rational linear combination of Tate classes: \[P(\LL)=\sum_{i=0}^N a_i \LL^i.\] Here, $P(X)$ denotes the polynomial $\sum_{i=0}^N a_i X^i\in \Q[X]$. The Grothendieck--Lefschetz trace formula relates compactly supported cohomology to point counts. Let $X$ be a variety over $\F_q$, where $q$ is a power of a prime different from $\ell$, and write $\overline X=X\times_{\F_q}\overline{\F}_q$. Then \[ \#X(\F_q)=\sum_i(-1)^i \Tr\bigl(\Frob_q\mid H_c^i(\overline X,\Q_\ell)\bigr); \] see \cite[Theorem~29.8]{MilneLEC}. Here $\Frob_q$ denotes geometric Frobenius, the inverse of the automorphism $x\mapsto x^q$ of $\overline{\F}_q$. Arithmetic Frobenius acts on $\Q_\ell(1)$ by $q$, so geometric Frobenius acts on $\Q_\ell(-r)$ by $q^r$. Consequently, the trace of geometric Frobenius on a virtual Tate class $P(\LL)$ is $P(q)$. A virtual Euler class is an alternating sum of cohomology groups. Consequently, an equality $\chi_c^\ell(X)=P(\LL)$ does not by itself say that each $H_c^i(X_{\overline\Q},\Q_\ell)$ has only Tate constituents, since constituents in adjacent degrees may cancel. We distinguish this assertion from the stronger statement that $X$ is Tate. For a variety $X/\Q$, choose a model $\mathscr X$ over $\Z[1/N]$. After increasing $N$, base change for compactly supported cohomology gives \[ \#\mathscr X(\F_p) =\Tr\bigl(\Frob_p\mid\chi_c^\ell(X)\bigr) \] where $p\nmid N\ell$; see \cite{SGA4half}. The trace on a virtual representation is defined by extending the usual trace linearly. \par The Hodge Euler class on the other hand is defined as \[\chi_c(X)=\sum_i(-1)^i[H_c^i(X(\mathbb C),\Q)]\] in the Grothendieck ring of rational mixed Hodge structures. Additivity and the K\"unneth formula make this a ring homomorphism, just as above; see \cite{DeligneHodgeII,DeligneHodgeIII}. We refer to this homomorphism and to $\chi_c^\ell$ as the two \emph{realizations}. The terminology ``in both realizations'' means one assertion about mixed Hodge structures and a second assertion about $G_\Q$-representations. In these rings, $\LL$ denotes $[\Q(-1)]$ and $[\Q_\ell(-1)]$, respectively. \section{Hermite normal forms}\label{sec:hnf} We abbreviate the Hermite Normal form of a lattice by HNF. We always use the column convention, i.e., the columns form a basis of the lattice, and the entries in row $i$ above the diagonal are reduced modulo the $i$th diagonal entry. \begin{lemma} \label{lem:DVR-HNF} Let $\cO$ be a discrete valuation ring with uniformizer $\varpi$. For each $a\ge0$, fix a set $\mathcal R_a\subseteq\cO$ of representatives for $\cO/\varpi^a\cO$, with $\mathcal R_0=\{0\}$. Every full lattice $L\subseteq\cO^m$ has a unique basis matrix of the form \[ A= \begin{pmatrix} \varpi^{e_1}&a_{12}&\cdots&a_{1m}\\ 0&\varpi^{e_2}&\cdots&a_{2m}\\ \vdots&\ddots&\ddots&\vdots\\ 0&\cdots&0&\varpi^{e_m} \end{pmatrix}, \] where $e_i\ge0$ and $a_{ij}\in\mathcal R_{e_i}$ for $i0$, $0\le b_{ij}r$, the row-$r$ equation gives \[ R_{jk,r}(t)=t^{d_r}v_r(t). \] The coefficients of degrees less than $d_r$ on the right are zero, so all the remainder coefficients at row $r$ vanish. Dividing by $t^{d_r}$ then gives $v_r=C_{jk,r}$, since the quotient of the polynomial $R_{jk,r}$ is the polynomial already defined above. This completes the descending induction. In summary, we find that $w_{jk}\in\mathcal Bk[[t]]^9$ if and only if $E_{jk;r,q}=0$ for every $r,q$. Let $I\subseteq A$ be the ideal generated by the polynomials $E_{jk;r,q}$ for $1\le j\le k\le9$, $1\le r\le9$, and $0\le qr$: \[ C^\lambda_{jk,s}(t)= \lambda^{d_s-d_j-d_k-1}C_{jk,s}(\lambda t). \] Each product $\mathcal B^\lambda_{rs}C^\lambda_{jk,s}$ then has the common factor $\lambda^{-d_j-d_k-1}$. Equation~\eqref{eq:R-rec} gives \[ R^\lambda_{jk,r}(t)= \lambda^{-d_j-d_k-1}R_{jk,r}(\lambda t). \] Taking the coefficient of $t^q$ proves \eqref{eq:E-semi-invariant}. Removing the low-degree terms and dividing by $t^{d_r}$ gives \[ C^\lambda_{jk,r}(t)= \lambda^{d_r-d_j-d_k-1}C_{jk,r}(\lambda t), \] which is the required formula for this row and completes the descending induction. The defining ideal is homogeneous for the grading by these weights, so the action restricts to $X_{\alpha_0}$. Its radical is homogeneous as well, and the action restricts to the reduced scheme $\mathscr X$. \end{proof} For $i=1,2$, write $b_{ij}=x_{ij}+ty_{ij}$. For $3\le i\le6$, write $b_{ij}=x_{ij}$. The list of weights is \begin{center} \begin{tabular}{c|c|c} coordinates & weight & number\\ \hline $x_{12}$ & $-2$ & $1$\\ $x_{ij}$ with $3\le j\le6$, and $y_{12}$ & $-1$ & $15$\\ $x_{ij}$ with $7\le j\le9$, and $y_{ij}$ with $3\le j\le6$ & $0$ & $26$\\ $y_{ij}$ with $7\le j\le9$ & $1$ & $6$ \end{tabular} \end{center} Only existing digit variables are included in this table: $i0. \] Choose $c=(c_j)_{j\in J}\in\Z^J$ such that $\sum_{j\in J}a_jc_j=d$, and let \[ L=\left\{u\in\Z^J:\sum_{j\in J}a_ju_j=0\right\}. \] Then $\Z^J=\Z c\oplus L$: for every $n\in\Z^J$, \[ n=\frac{\sum_{j\in J}a_jn_j}{d}\,c +\left(n-\frac{\sum_{j\in J}a_jn_j}{d}\,c\right), \] where the second term belongs to $L$, and the decomposition is unique. Choose a basis $u^{(1)},\ldots,u^{(m-1)}$ of $L$. Define Laurent monomials \[\begin{split} & s=\prod_{j\in J}z_j^{c_j},\\ & v_\ell=\prod_{j\in J}z_j^{u_j^{(\ell)}} \end{split} \]for $1\le\ell\le m-1$. Since $c,u^{(1)},\ldots,u^{(m-1)}$ form an integral basis, every $z_j$ is a Laurent monomial in these new coordinates. They therefore identify \[ T_J\simeq\mathbb G_m\times(\mathbb G_m)^{m-1}, \] with action \[ \lambda\cdot(s,v_1,\ldots,v_{m-1}) =(\lambda^d s,v_1,\ldots,v_{m-1}). \] Let $Y_J\subseteq(\mathbb G_m)^{m-1}$ be the reduced closed subvariety obtained by setting $s=1$ in $X_J$. For any $s\in\overline\Q^\times$, choose $\lambda\in\overline\Q^\times$ with $\lambda^d=s^{-1}$. We find that $(s,v)\in X_J(\overline\Q)$ if and only if $(1,v)\in X_J(\overline\Q)$ if and only if $v\in Y_J(\overline\Q)$. Thus $X_J$ and $\mathbb G_m\times Y_J$ have the same $\overline\Q$-points. Since both are reduced closed subvarieties of $T_J$, they are equal over $\Q$. It follows that \[ [X_J]=(\LL-1)[Y_J]. \] Taking $Y$ to be the finite disjoint union of these $Y_J$ and adding the classes of all the strata proves the affine case. For the projective case, write the coordinates as $z_0,\ldots,z_M$, with weights $a_0,\ldots,a_M$. Decompose $\Pp^M_\Q$ into the invariant affine strata \[ U_i=\{z_0=\cdots=z_{i-1}=0,\ z_i\ne0\}, \] for $0\le i\le M$. The coordinates $z_j/z_i$, $j>i$, identify $U_i$ with $\A^{M-i}_\Q$, and have weights $a_j-a_i$. Apply the affine case to each reduced closed subvariety $(X\cap U_i)_{\mathrm{red}}\subseteq U_i$. Adding the resulting identities gives \eqref{total:eq:torus-localization}, since these strata also decompose the reduced fixed locus. The required $Y$ is the disjoint union of the varieties obtained from the affine case. \end{proof} In what follows, we let $X=\mathscr X_{\Q}$ and set $F=(X^{\mathbb G_m})_{\mathrm{red}}$. \begin{proposition}\label{prop:fixed-factor} There is a variety $Y/\Q$ such that \begin{equation}\label{eq:K0-factor} [X]=[F]+(\LL-1)[Y] \end{equation} in $K_0(\operatorname{Var}_{\Q})$. \end{proposition} \begin{proof} The affine variety $X\subseteq\A^{48}_{\Q}$ is reduced and is preserved by the diagonal action in \eqref{eq:weights-general}. Lemma~\ref{total:lem:torus-localization} therefore gives \eqref{eq:K0-factor}. \end{proof} In the $\alpha_0$ HNF chart, the loop-rotation weights in \eqref{eq:weights-general} give the fixed-point equations \begin{equation}\label{eq:fixed-coords} x_{ij}=0\quad\text{for}\quad 1\le i0$, and \[h^0(E,M)= \begin{cases} 1,&M\simeq\cO_E,\\ 0,&M\not\simeq\cO_E \end{cases} \] when $\deg M=0$ (see \cite[Chapter~IV, Theorem~1.3 and Example~1.3.6]{Hartshorne}). \begin{lemma}\label{lem:elliptic-quartic}\label{lem:prelim-elliptic-quartic} A smooth complete intersection of two quadrics in $\Pp^3$ is a geometrically connected genus-one curve of degree four. Conversely, a degree-four line bundle $L$ on a smooth proper geometrically connected genus-one curve $E/k$ embeds $E$ in $\Pp^3$. The multiplication map \[ \Sym^2H^0(E,L)\longrightarrow H^0(E,L^2) \] is surjective, with two-dimensional kernel, and the two quadrics in this kernel cut out $E$ scheme-theoretically. \end{lemma} \begin{proof} These assertions may be checked after extending $k$ to an algebraic closure. The first assertion follows from \cite[Lemma~5.16(ii) and Proposition~5.10(i)]{FisherInvariants}, with $n=4$. Conversely, Riemann--Roch and \cite[Chapter~IV, Corollary~3.2(b)]{Hartshorne} show that $L$ embeds $E$ in $\Pp^3$. The multiplication map is surjective by \cite[Lemma~5.5]{FisherInvariants}. Since $h^0(E,L)=4$ and $h^0(E,L^2)=8$, its kernel has dimension two. By \cite[Proposition~5.3]{FisherInvariants}, these two quadrics generate the homogeneous ideal of the image. The assertions over $k$ follow by faithfully flat descent and cohomology base change. More explicitly, for an extension $K/k$, the natural maps $H^0(E,L)\otimes_kK\to H^0(E_K,L_K)$ and $H^0(E,L^2)\otimes_kK\to H^0(E_K,L_K^2)$ are isomorphisms. The multiplication map and its kernel therefore commute with extension of scalars. If its cokernel becomes zero over $\bar k$, it was already zero over $k$, since $\bar k/k$ is faithfully flat. The morphism defined by $L$ is a closed immersion for the same reason. Indeed, being a closed immersion can be checked after faithfully flat base change. Finally, let $Z$ be the common zero scheme of the kernel quadrics. The inclusion $E\subseteq Z$ becomes an equality over $\bar k$, so its ideal quotient becomes zero after extension of scalars and is already zero over $k$. This gives the asserted scheme-theoretic equality over $k$. \end{proof} For $n\ge1$, let $\cM_{1,n}$ denote the moduli stack of smooth proper geometrically connected genus-one curves with $n$ ordered distinct marked points. Choosing the first marking as the origin identifies its objects with elliptic curves carrying $n-1$ further ordered distinct nonzero points. For $n=10,11$, every automorphism is trivial, since a nonidentity elliptic-curve automorphism fixes at most four points. To obtain these spaces as varieties over $k$, choose $N\ge3$ invertible in $k$. The full-level moduli scheme $Y(N)$ and its universal elliptic curve $\mathcal E_N\to Y(N)$ exist by \cite[Corollary~4.7.2]{KatzMazur}. Let $T_n(N)$ be the open subset of the $(n-1)$-fold fiber product of $\mathcal E_N$ over $Y(N)$ where the chosen points are pairwise distinct and nonzero. Since $\mathcal E_N$ is projective over the affine scheme $Y(N)$, this open subset is quasi-projective. For $n=10,11$, changing the level structure gives a free action of $\operatorname{GL}_2(\Z/N\Z)$, and \[ \cM_{1,n}\simeq T_n(N)/\operatorname{GL}_2(\Z/N\Z). \] The quotient is a scheme by \cite[Tag~07S7]{StacksProject}. Thus $\cM_{1,10}$ and $\cM_{1,11}$ are fine moduli varieties. Let $B$ be a variety and let $E$ be a locally free $\cO_B$-module of rank $m$. We use the convention \[ \Pp(E)=\operatorname{Proj}_B\operatorname{Sym}(E^\vee). \] For a morphism $g:T\to B$, its $T$-points correspond to line subbundles $\mathcal L\subseteq g^*E$. Here a subbundle means a locally free subsheaf whose quotient is locally free. Thus the fiber over $b\in B$ is the projective space of lines in $E_b$. For integers $00$. For coefficients $\mathcal A$ in either realization, pullback gives a split injection \[ H_c^i(B,\mathcal A)\longrightarrow H_c^i(B',\pi^*\mathcal A). \] \end{lemma} \begin{proof} Pullback followed by trace is multiplication by $h$, so $h^{-1}\operatorname{Tr}_\pi$ is a retraction. These maps respect the Hodge structures and, in the $\ell$-adic realization, the $G_\Q$-action. \end{proof} For $f:X\to B$, the stalk of $R^vf_!$ at a geometric point $b$ is $H_c^v(X_b)$. A constant Tate object on $B$ is pulled back from $\Q(-r)$ or $\Q_\ell(-r)$ on the ground field. In the $\ell$-adic case, this includes the arithmetic Galois action. \begin{lemma}\label{revprelim:relative-hodgelevel} Fix either realization, and let $f:X\to B$ be defined over $\Q$, with $\dim B\le d$. Suppose $B$ admits a finite stratification such that, on each stratum, the objects $R^vf_!$ have finite filtrations whose quotients become sums of constant Tate objects after finite \'etale covers defined over $\Q$. Then every simple constituent of $H_c^i(X)$ is a subquotient of a Tate twist of $H_c^j(B')$ for some variety $B'/\Q$ of dimension at most $d$. Consequently, in the Hodge realization, $\lev H_c^i(X)\le d$. In the $\ell$-adic realization, each simple constituent has Hodge--Tate spread at most $d$. The same conclusions hold if $X$ has a finite stratification by varieties satisfying these hypotheses. \end{lemma} \begin{proof} Work first over one stratum. Let $\mathcal A$ be a filtration quotient, and choose a finite \'etale cover $\pi:B'\to B$ on which it becomes a sum of constant Tate objects. Lemma~\ref{revprelim:finite-etale-trace} makes $H_c^u(B,\mathcal A)$ a direct summand of $H_c^u(B',\pi^*\mathcal A)$. The latter is a sum of Tate twists of $H_c^u(B')$, and $\dim B'=\dim B\le d$. The filtrations and the compact-support Leray spectral sequence \[ H_c^u(B,R^vf_!)\Longrightarrow H_c^{u+v}(X) \] give the asserted constituent bound. Tate twists preserve Hodge level and Hodge--Tate spread, so the dimension bounds above apply. Localization combines the finitely many strata. \end{proof} \begin{lemma}\label{lem:low-dimensional-class} Suppose a variety $Z/\Q$ satisfies \[ [Z]=\sum_i P_i(\LL)[B_i] \]in $K_0(\operatorname{Var}_{\Q})$, where $P_i\in\Z[T]$, $\dim B_i\le10$. Then $[\chi_c(Z)]_{S_{12}}=0$ in either realization. Thus for each integer $r$, the total multiplicity of $S_{12}(r)$ in even-degree compactly supported cohomology equals its total multiplicity in odd-degree cohomology. In the $\ell$-adic realization, we use $S_{12,\ell}(r)$ instead. \end{lemma} \begin{proof} The dimension bounds exclude every Tate twist of $S_{12}$ from the cohomology of each $B_i$: these twists have Hodge level and Hodge--Tate spread eleven. Apply the realization map. Multiplication by $P_i(\LL)$ introduces only Tate twists, so every summand has zero $S_{12}$-part. \end{proof} Under the identifications $U\simeq H^0(E,L)$ and $V\simeq H^0(E,L^2)$, the kernels of the two coordinate projections in \eqref{eq:bigcells-linear} are $H^0(E,L(-q_7-\cdots-q_{10}))$ and $H^0(E,L^2(-q_3-\cdots-q_{10}))$, respectively. Dividing evaluations by the nonzero values of $s$ or $s^2$ does not change these kernels. The projection conditions in \eqref{eq:bigcells-linear} require the last four coordinates to determine a vector of $U$ uniquely, and the last eight coordinates to determine a vector of $V$ uniquely. To express the first condition in coordinates, choose a basis $u_1,\ldots,u_4$ of $U$. Form the $4\times4$ matrix whose $j$th column consists of coordinates $7,8,9,10$ of $u_j$. This is the matrix of the projection $U\to k^{\{7,8,9,10\}}$, so the projection is an isomorphism precisely when its determinant is nonzero. Similarly, choosing a basis of $V$ and retaining coordinates $3,\ldots,10$ gives an $8\times8$ matrix whose determinant must be nonzero. For the spaces constructed from $(E,L,[s])$, these matrix entries are the normalized evaluations of sections described above. Changing either basis multiplies the corresponding determinant by a nonzero scalar, so these conditions depend only on $U$ and $V$. They define open subsets of the two Grassmannians, usually called their standard big cells. Both line bundles have degree zero, and each projection has source and target of the same dimension. Since a degree-zero line bundle has a nonzero section precisely when it is trivial, the first projection fails to be an isomorphism exactly when \begin{equation}\label{eq:U-fail} L\simeq\cO_E(q_7+q_8+q_9+q_{10}), \end{equation} and the second fails exactly when \begin{equation}\label{eq:V-fail} L^2\simeq\cO_E(q_3+\cdots+q_{10}). \end{equation} \begin{proposition}\label{prop:bigcell-does-not-change} Let $\cE_4^{\mathrm{big}}\subset F$ be the smooth elliptic-quartic locus satisfying both HNF big-cell conditions. In either realization, \begin{equation}\label{eq:E4big-S12} [\chi_c(\cE_4^{\mathrm{big}})]_{S_{12}} = [\chi_c(\cE_4)]_{S_{12}} = -B_4(\LL)S_{12}. \end{equation} \end{proposition} \begin{proof} Let $B_U,B_V\subseteq\mathcal P^4$ be the loci defined by \eqref{eq:U-fail} and \eqref{eq:V-fail}, and set $B_{UV}=B_U\cap B_V$. On $B_U$, the markings determine $L$, so $B_U\simeq\cM_{1,10}$. The map $\Pic^4(E)\to\Pic^8(E)$ given by $L\mapsto L^2$ is, after identifying both Picard schemes with $E$ using the origin, multiplication by two. In characteristic zero it is finite \'etale of degree four. Hence $B_V$ is finite \'etale of degree four over $\cM_{1,10}$. Thus $B_U,B_V$, and $B_{UV}$ all have dimension at most ten. The elliptic curve varies in a one-dimensional moduli space, and the nine further marked points each add one parameter. Removing the diagonals and the origin does not change this dimension and we find that $\dim\cM_{1,10}=10$. A finite \'etale cover has the same dimension as its base, and a closed subvariety cannot have larger dimension. This explains why both projection conditions fall within Lemma~\ref{lem:low-dimensional-class}, even though $\mathcal P^4$ itself has dimension $11$. For any of these bases $B$, let $Z_B$ parametrize the projective sections $[s]$ nonzero at all ten markings. For each four-element subset $J$, set \[ B_J= \left\{b\in B: L_b\simeq \cO_{E_b}\Bigl(\sum_{j\in J}q_j(b)\Bigr)\right\}. \] The inclusion--exclusion calculation of Lemma~\ref{lem:fiber-class}, applied to the family over $B$, gives \[ [Z_B]=B_4(\LL)[B]+\sum_{|J|=4}[B_J]. \] Here the four-point intersections occur precisely over $B_J$. Each $B_J$ is closed in $B$ and has dimension at most ten. Therefore Lemma~\ref{lem:low-dimensional-class} gives $[\chi_c(Z_B)]_{S_{12}}=0$. The excluded locus is $Z_{B_U}\cup Z_{B_V}$, with intersection $Z_{B_{UV}}$. Additivity shows that its $S_{12}$-part is zero. The result follows from \eqref{eq:unrestricted-S12}. \end{proof} \section{The complete boundary of the fixed locus} \label{sec:boundary} The preceding section computes the $S_{12}$-part of the smooth elliptic-quartic locus $\cE_4^{\mathrm{big}}$. To obtain the contribution of the complete fixed locus $F$, we must account for its complement. We show that this complement has zero $S_{12}$-part in its compactly supported Euler class. In what follows, the \emph{boundary} means the reduced closed complement $F\setminus\cE_4^{\mathrm{big}}$. The description by flags turns this into a question about ten points in projective three-space and the quadratic equations they satisfy. The smooth locus considered above has exactly two independent quadratic relations, cutting out a smooth elliptic quartic, and the ten points are distinct. The remaining configurations have more quadratic relations, a singular intersection, a common plane, or coincident points. We treat these possibilities separately. In each case, the class formulas reduce the calculation to parameter spaces of dimension at most ten. This dimension bound excludes the contribution of $S_{12}$, whose Hodge level and Hodge--Tate spread are eleven. Dimension estimates and classifications of geometric components may be checked after extending the ground field to an algebraic closure. \begin{lemma}\label{lem:dimension-no-S12} Let $X/\Q$ admit a finite stratification whose strata are affine-space or projective-space bundles over varieties of dimension at most $d$. Then $\lev H_c^i(X)\le d$, and every simple $\ell$-adic constituent has Hodge--Tate spread at most $d$. If $d\le10$, these cohomology groups contain no Tate twist of $S_{12}$ in either realization. \end{lemma} \begin{proof} Apply Lemmas~\ref{revprelim:tate-bundles} and~\ref{revprelim:relative-hodgelevel}. Tate twists preserve Hodge level and Hodge--Tate spread, whereas every Tate twist of $S_{12}$ has both invariants equal to eleven. \end{proof} Let $e_i^*:k^{10}\to k$ be the $i$th coordinate functional. For a four-plane $U$ containing $\bfone$, set $q_i=[e_i^*|_U]\in\Pp(U^\vee)$. These points are well defined because $e_i^*(\bfone)=1$. \begin{lemma}\label{lem:canonical-frame} Fix an ordered quadruple $I=(i_0,i_1,i_2,i_3)$ of labels. The four-planes $U\subseteq k^{10}$ containing $\bfone$ for which these four coordinate evaluations are independent are parametrized by six points in the affine hyperplane \[ \{(X_0,X_1,X_2,X_3)\in k^4: X_0+X_1+X_2+X_3=1\}. \] The points with labels in $I$ are the coordinate vertices; the remaining six points may coincide. Equivalently, these four-planes parametrize labelled projective configurations with the four specified points independent, together with a hyperplane avoiding all ten points, up to projective transformations. In these coordinates, set \[ H=\Span_k\{X_\alpha X_\beta: 0\le\alpha<\beta\le3\}. \] Set $\mathcal K_U=\ker(\Sym^2U\to k^{10})\subseteq H$. Then, \begin{equation}\label{eq:K-dimension} \dim\mathcal K_U=10-\dim U^{\circled2}. \end{equation} These identifications are regular and hold in families. \end{lemma} The word \emph{exact} means that $\mathcal K_U$ contains all the quadratic relations among the ten evaluation points. We are not choosing a subspace of their relations. In particular, specifying $A=\mathcal K_U$ requires both that every member of $A$ vanish at the points and that there be no further independent quadratic relation. \begin{proof} Choose the basis $f_0,\ldots,f_3$ of $U$ dual to the four specified evaluations. The evaluation matrix has identity rows at the labels in $I$. Since $\bfone=f_0+\cdots+f_3$, every row has coordinate sum one. Conversely, a matrix whose rows indexed by $I$ form the identity matrix has rank four. If every row sums to one, the sum of its columns is $\bfone$. Its column space therefore contains $\bfone$, and the projection $\pi_I:U\to k^I$ is an isomorphism. These constructions identify the matrices above with the open subset of $\Gr(3,k^{10}/k\bfone)$ on which $\pi_I$ is invertible. Both maps are regular. Taking the column space gives the forward map, while the inverse is obtained by choosing the unique basis of $U$ whose projections are the standard basis of $k^I$. After fixing the four identity rows, each of the six remaining rows has three free entries, with the fourth determined by its row sum. Thus this open subset is isomorphic to $\A^{18}_k$. The hyperplane associated to $U$ is $\{\ell:\ell(\bfone)=0\}$, which avoids every $q_i$. Conversely, send the four independent points to the coordinate vertices. The avoided hyperplane then has four nonzero coefficients. A unique diagonal change of coordinates, modulo scalar, makes its equation $X_0+\cdots+X_3=0$. Each remaining point has a unique representative with coordinate sum one. This recovers the preceding matrix. The normalizations divide only by quantities assumed nonzero, so they are regular in families. The quadratic evaluation map has image $U^{\circled2}$. A quadric vanishing at the four vertices has no square terms, so its kernel lies in $H$. Rank--nullity gives \eqref{eq:K-dimension}. \end{proof} For the HNF chart, take $I=(7,8,9,10)$. The remaining six points give its eighteen affine coordinates. Define \[ \begin{aligned} \mathcal K_{3\ldots10} &=\{Q\in H:Q(q_3)=\cdots=Q(q_6)=0\},\\ \mathcal K_{1\ldots10} &=\{Q\in\mathcal K_{3\ldots10}: Q(q_1)=Q(q_2)=0\} =\mathcal K_U. \end{aligned} \] The conditions at $q_7,\ldots,q_{10}$ already hold for every $Q\in H$. \begin{lemma}\label{lem:second-bigcell-kernel} Put $a=\dim\mathcal K_{1\ldots10} =10-\dim U^{\circled2}$. There exists an eight-dimensional space $V\supseteq U^{\circled2}$ whose projection onto coordinates $3,\ldots,10$ is an isomorphism if and only if \begin{equation}\label{eq:kernel-equality-bigcell} \mathcal K_{3\ldots10} =\mathcal K_{1\ldots10}. \end{equation} When this holds, the choices of $V$ form an affine space of dimension \begin{equation}\label{eq:V-affine-dim} 2(a-2). \end{equation} On a locus where the kernel equality holds and $a$ is constant, these spaces form a Zariski locally trivial affine-space bundle. \end{lemma} \begin{proof} Put $R=U^{\circled2}$, and let $p:k^{10}=k^2\oplus k^8\to k^8$ be the projection onto coordinates $3,\ldots,10$. The full evaluation map $\mu:\Sym^2U\to k^{10}$ has image $R$ and kernel $\mathcal K_{1\ldots10}$. The kernel of $p\mu$ is $\mathcal K_{3\ldots10}$: vanishing at the four vertices first forces a quadric to lie in $H$, and the remaining conditions are vanishing at $q_3,\ldots,q_6$. Hence \[ \ker(p|_R)\simeq \mathcal K_{3\ldots10}/ \mathcal K_{1\ldots10}. \] Thus \eqref{eq:kernel-equality-bigcell} is equivalent to injectivity of $p|_R$. Every eight-plane with the required projection is the graph of a map $\phi:k^8\to k^2$. If $p|_R$ is injective, $R$ is the graph of a map $\phi_0:p(R)\to k^2$, and the required choices are exactly its extensions to $k^8$. They form an affine space under \[ \Hom\!\left(k^8/p(R),k^2\right), \] of dimension $2(a-2)$, since $\dim p(R)=10-a$. On a constant-rank locus, $p(R)$ is a vector subbundle. Locally choose a complement and extend $\phi_0$ by zero on it. This identifies the family of extensions locally with the displayed Hom bundle. \end{proof} For $a\ge3$, let $B_a$ be the locus of six affine points for which \[ A:=\mathcal K_{1\ldots10} =\mathcal K_{3\ldots10}, \]and $\dim A=a$. These are locally closed matrix conditions. Indeed, use the six monomials spanning $H$ as a basis. Evaluation at $q_3,\ldots,q_6$ gives a matrix with four rows and six columns, and evaluation at $q_1,\ldots,q_6$ gives a matrix with six rows and six columns. The two kernels already have the required inclusion. Thus the conditions defining $B_a$ say that both matrices have rank $6-a$. A specified rank is expressed by the vanishing of all larger minors and the nonvanishing of at least one minor of that rank. By Lemma~\ref{lem:second-bigcell-kernel}, the choices of $V$ form an affine-space bundle of rank $2(a-2)$ over $B_a$. For a space $A$ of quadrics, its \emph{curve part} is the union of the one-dimensional components of the reduced common zero locus $Z(A)$. Thus $C$ is a reduced curve and does not include any isolated base points or any multiplicity carried by the equations. Its degree is the sum of the degrees of its irreducible components. The isolated base points will be counted separately. \begin{lemma}\label{lem:quadratic-base-curves} Suppose $\dim A\ge3$ and the quadrics in $A$ have no common plane. Let $C$ be their curve part. Then $\deg C\le3$. If $\deg C=3$, then, over an algebraic closure, $C$ is a twisted cubic, a conic together with an incident line outside its plane, a nonplanar chain of three lines, or three noncoplanar concurrent lines. In each case, $\dim H^0(I_C(2))=3$, $A=H^0(I_C(2))$ and $Z(A)=C$. \end{lemma} \begin{proof} We work over an algebraic closure. There is no common quadratic factor, since that would make $A$ one-dimensional, and no common linear factor by hypothesis. We may therefore choose $Q_1,Q_2\in A$ with no common factor: after fixing $Q_1$, avoid the finitely many proper subspaces of elements divisible by one of its irreducible factors. Their intersection $D$ is a complete intersection of degree four, so $\deg C\le4$. If $\deg C=4$, every component of $D$ occurs with multiplicity one and there are no other components. Thus $D$ is generically reduced. A complete intersection is Cohen--Macaulay, and a generically reduced Cohen--Macaulay scheme is reduced; see \cite[Tags~02JN and~031R]{StacksProject}. Hence $D=C$. The Koszul resolution gives $h^0(I_D(2))=2$, contradicting $\dim A\ge3$. Therefore $\deg C\le3$. Suppose $\deg C=3$. If $C$ were planar, every containing quadric would contain its plane, contrary to the hypothesis. An irreducible nonplanar cubic is a twisted cubic, whose ideal is generated by three quadrics; see \cite[Theorem~1 and \S1, p.~6] {EisenbudHarris1987}. For reducible $C$, first exclude three pairwise skew lines. Using the first two as complementary two-dimensional subspaces, their quadrics are the bilinear forms $x^{\mathsf t}My$. The third line is the graph $y=\phi x$ of an invertible map. Vanishing on it requires $M\phi$ to be alternating, leaving only one independent quadric. Every other reduced nonplanar cubic has the form $C=C_0\cup l$, where $C_0$ is a plane conic, possibly a pair of intersecting lines, and $l$ lies outside its plane. To see that these are all the remaining reducible possibilities, note that a reduced reducible cubic has either components of degrees two and one or three line components. In the latter case, unless the three lines are pairwise skew, two of them meet and form a planar conic. The third lies outside that plane because $C$ is nonplanar. Write the plane as $X_3=0$ and the conic equation as $Q=0$. Its containing quadrics are $X_3L+cQ$. If $l$ meets the plane outside $C_0$, vanishing at that intersection forces $c=0$. Vanishing along $l$ then forces $L|_l=0$, leaving only two independent quadrics. This case is also excluded. If $l$ meets $C_0$, choose coordinates with $l=\{X_1=X_2=0\}$. Then $Q\in(X_1,X_2)$ and \[ I_C=(X_3,Q)\cap(X_1,X_2) =(Q,X_3X_1,X_3X_2). \] Indeed, modulo $(X_1,X_2)$, multiplication by $X_3$ is injective. The three displayed quadrics are independent and generate the union ideal. When $C_0$ is reducible, this gives either a nonplanar chain or three noncoplanar concurrent lines. In every admissible case, $H^0(I_C(2))$ has dimension three and generates $I_C$. Since it contains $A$, whose dimension is at least three, the two spaces are equal. Thus $Z(A)=C$, with no residual points. \end{proof} \begin{proposition}\label{prop:higher-kernel-boundary} Fix an ordered quadruple $I$ of labels. Let $X_{a,I}$ parametrize four-planes $U\subset\Q^{10}$ containing $\bfone$ for which these four coordinate evaluations are independent, the exact quadratic kernel $A=\mathcal K_U$ has dimension $a\ge3$, and the quadrics in $A$ have no common plane. Then $\dim X_{a,I}\le10$. Over every locally closed sublocus $B\subseteq X_{a,I}$, the choices of an eight-plane $V\supset U^{\circled2}$ form a Grassmannian bundle with fiber $\Gr(a-2,a)$. Its compactly supported Euler class has zero $S_{12}$-part in both realizations. The same holds for the HNF affine bundle over the no-common-plane part of $B_a$. \end{proposition} \begin{proof} We make the dimension estimates over $\overline{\Q}$. Choose four coordinate labels $I=\{i_1,i_2,i_3,i_4\}$ such that the coordinate projection $\pi_I:U\to\Q^I$ is an isomorphism. The spaces $U$ satisfying this condition form an open Grassmannian chart. On this chart, choose the unique basis whose coordinates at these four labels form the identity matrix. The corresponding four projective evaluation points are then the coordinate vertices of $\Pp^3$. Let $H$ be the six-dimensional space of quadrics through the four vertices. For each configuration, $A\subseteq H$ is the full space of quadrics vanishing at all ten points. Thus a quadric belongs to $A$ precisely when it vanishes at every evaluation point. We shall use this exactness to restrict the possible configurations. If the common zero set $Z(A)$ is finite, first choose $A\in\Gr(a,H)$. Each of the six remaining points then has only finitely many choices. The resulting incidence has dimension at most $a(6-a)\le9$, and the same bound holds for its image in $X_{a,I}$. Suppose now that $Z(A)$ has a curve part. Let $C$ be the union of its reduced one-dimensional components. By Lemma~\ref{lem:quadratic-base-curves}, we need only consider a line, a reduced planar conic, two skew lines, or a curve of degree three. We first treat a line or planar conic. Fix which of the four coordinate vertices lie on $C$, and let their number be $k$. Put \[ W_C=H^0(I_C(2))\cap H,\qquad m=\dim W_C. \] Thus $W_C$ consists of the quadrics vanishing on $C$ and at all four fixed vertices. Let $D_k$ be an upper bound for the dimension of the family of possible curves $C$. Let $\mathcal Z_{k,a}$ denote the incidence locus of choices of $C$, of an $a$-plane $A\subseteq W_C$, and of the six remaining evaluation points, with $A$ the exact quadratic kernel and $C$ its entire reduced curve part. Choosing $C$ and then $A$ contributes at most $D_k+a(m-a)$ parameters. To count the remaining point choices, fix which of the six labels lie outside $C$, and let their number be $b$. Once $C$ and $A$ have been chosen, these outside points lie in the finite residual base locus $Z(A)\setminus C$. The other $6-b$ points lie on $C$, so each contributes at most one parameter. There is also a lower bound on $b$. Every quadric in $W_C$ already vanishes at the four vertices and at the variable points on $C$. Thus the evaluation map $W_C\to\overline{\Q}^{\,b}$ at the outside points has kernel exactly $A$. Its rank is $m-a$, and therefore $b\ge m-a$. Consequently, \begin{equation}\label{eq:curve-dimension-bound} \dim\mathcal Z_{k,a} \le D_k+a(m-a)+6-(m-a) =D_k+6+(a-1)(m-a). \end{equation} There are finitely many choices of the outside labels. Forgetting the curve and the kernel maps these incidences onto the corresponding configuration loci, so the same dimension bound applies to those loci. We now compute $D_k$ and $m$. A line contains at most two coordinate vertices. The family of all lines has dimension four, the lines through one fixed point form $\Pp^2$, and two points determine a line. Thus $D_k=4-2k$ for $0\le k\le2$. Quadrics through a line form a seven-dimensional space. The $4-k$ vertices outside the line impose independent conditions on this space. Indeed, for each such vertex $e_i$, choose a linear form $\ell_i$ vanishing on the line but not at $e_i$. The quadric $X_i\ell_i$ vanishes on the line and at every other coordinate vertex, but not at $e_i$. It follows that $m=7-(4-k)=3+k$. For a reduced planar conic, let $\ell=0$ be its plane and let $Q$ be a quadratic equation restricting to its equation in that plane. Every quadric through the conic has a unique expression $\ell L+cQ$, where $L$ is linear and $c$ is a scalar. This space has dimension five. A fixed vertex in the plane but outside the conic would force $c=0$ for every quadric in $A$. All members of $A$ would then be divisible by $\ell$, giving the common plane excluded in the statement. Hence every fixed vertex in the conic plane lies on the conic. At each of the other $4-k$ vertices, $\ell$ is nonzero, and the vanishing condition independently determines the corresponding coordinate value of $L$. Thus $m=5-(4-k)=1+k$. Since $a\le m$ and $a\ge3$, we have $k\ge2$. The four vertices are noncoplanar, so $k\le3$. The planes through the $k$ specified vertices form a family of dimension $3-k$. Within such a plane, conics through those vertices form a family of dimension $5-k$: the conditions are independent because the specified vertices are noncollinear. Therefore $D_k\le8-2k$. Substituting these values into \eqref{eq:curve-dimension-bound} gives \[ \begin{array}{c|c|c|c|c|c} C&k&D_k&m&a&\text{dimension bound}\\ \hline \text{line}&0&4&3&3&10\\ \text{line}&1&2&4&3,4&10,8\\ \text{line}&2&0&5&3,4,5&10,9,6\\ \text{planar conic}&2&4&3&3&10\\ \text{planar conic}&3&2&4&3,4&10,8 \end{array} \] In particular, all these loci have dimension at most ten. Next suppose that the curve part consists of two skew lines $L_1,L_2$. Choose coordinates $[x:y]$, with $x$ and $y$ each having two coordinates, such that $L_1=\{y=0\}$ and $L_2=\{x=0\}$. The quadrics vanishing on both lines are exactly the bilinear forms in $x,y$, and form a four-dimensional space. There can be no base point outside these two lines. Indeed, at such a point $[x:y]$ both $x$ and $y$ are nonzero. For every quadric $Q\in A$, bilinearity gives $Q(sx,ty)=stQ(x,y)=0$. Thus the entire joining line $\{[sx:ty]:[s:t]\in\Pp^1\}$ would lie in $Z(A)$. It is distinct from $L_1$ and $L_2$, contradicting the assumption that they form the entire curve part. All ten evaluation points therefore lie on $L_1\cup L_2$. Every quadric through these lines vanishes at all ten points, so exactness forces $A$ to be their full four-dimensional quadratic ideal. Hence $a=4$. Each line contains at most two of the four fixed vertices, so each must contain exactly two and is determined by them. There are finitely many such pairs. The six remaining points contribute at most six parameters. Finally, suppose $\deg C=3$. Lemma~\ref{lem:quadratic-base-curves} gives $A=H^0(I_C(2))$ and $Z(A)=C$. In particular, all ten evaluation points lie on $C$. We first bound the families of possible curves before fixing the four vertices. Twisted cubics form a family of dimension $15-3=12$: $\operatorname{PGL}_4$ acts transitively, with stabilizer $\operatorname{PGL}_2$. A conic and an incident line outside its plane form a family of dimension at most eleven. Indeed, the conic contributes eight parameters, its intersection point with the line contributes one, and the line through that point contributes two. The remaining cubic types in the lemma are unions of three lines, whose families have dimension at most $3\dim\Gr(2,4)=12$. Let $\mathcal F$ be any one of these $\operatorname{PGL}_4$-invariant families, of dimension $d\le12$. Consider the incidence of a curve in $\mathcal F$ together with four ordered independent points on it. Its dimension is at most $d+4$. The space of independent ordered quadruples in $\Pp^3$ is an open subset of $(\Pp^3)^4$, of dimension twelve. The projection from the incidence to this space is equivariant for $\operatorname{PGL}_4$. If the incidence is nonempty, transitivity on independent quadruples makes this projection surjective, with isomorphic fibers. Its fiber over our four fixed vertices, which parametrizes the curves through them, therefore has dimension at most $d+4-12\le4$. Choosing the six remaining points on the curve adds at most six parameters. Thus the cubic case also has dimension at most ten. These estimates cover $X_{a,I}$. There are only finitely many curve types and label assignments, so $\dim X_{a,I}\le10$. It remains to include the choices of $V$. Let $B$ be a locally closed stratum of $X_{a,I}$. Quadratic evaluation has constant rank $10-a$, so its image $\mathcal R$ is a vector subbundle of the trivial rank-ten bundle on $B$. The condition $\mathcal R\subseteq V$, with $\dim V=8$, is equivalent to choosing an $(a-2)$-plane in the rank-$a$ quotient. The Grassmannian bundle formula gives \[ [\text{flags over }B]=[\Gr(a-2,a)][B]. \] For the HNF calculation, restrict $B$ to the locus satisfying the kernel-equality condition of Lemma~\ref{lem:second-bigcell-kernel}. That lemma gives an affine bundle of rank $2(a-2)$, and hence the class $\LL^{2(a-2)}[B]$. Every such base still has dimension at most ten. Both fiber factors are polynomials in $\LL$, so Lemma~\ref{lem:low-dimensional-class} gives the asserted vanishing in both realizations. \end{proof} \begin{proposition}\label{prop:common-plane-boundary} The compactly supported Euler class of the common-plane part of $B_a$, together with its choices of $V$, has zero $S_{12}$-part in both realizations. \end{proposition} \begin{proof} Write $A=\ell W$, where $\ell=0$ defines a plane $\Pi$ and $W$ is an $a$-dimensional space of linear forms. Its reduced base is $\Pi\cup Z(W)$, and $a=3$ or $4$. Multiplication by the nonzero linear form $\ell$ identifies $W$ with $A$. Since the space of linear forms on $\Pp^3$ has dimension four, the assumption $a\ge3$ leaves only these two possibilities. When $a=3$, if the point $Z(W)$ lay on $\Pi$, the whole reduced base would lie in $\Pi$. This is excluded by the four independent evaluation points. If $a=4$, the base is $\Pi$, which cannot contain the four independent vertices. If $a=3$, then $Z(W)$ is a point $r$ outside $\Pi$. Exactly one vertex equals $r$; the other three determine $\Pi$. Thus there are four possibilities, each with $A=\ell H^0(I_r(1))$. In the fixed affine chart, its base is $\A^2\sqcup\{r\}$. Let $T_A$ parametrize $q_3,\ldots,q_6$ in this base for which the quadratic kernel, together with the four vertices, is exactly $A$. This is locally closed in $(\A^2\sqcup\{r\})^4$, so $\dim T_A\le8$. The remaining points $q_1,q_2$ may be chosen arbitrarily in the same base: they impose no further condition on the already exact kernel. The choices of $V$ form an affine bundle of rank two. Hence the resulting locus $Z_A$ has class \[ [Z_A]=\LL^2(\LL^2+1)^2[T_A]. \] Apply Lemma~\ref{lem:low-dimensional-class} and sum over the four possibilities. \end{proof} Let $Q_0,Q_1$ be linearly independent quadratic forms on $\Pp^3$. Their span is called a \emph{pencil of quadrics}: its members, up to scalar, are the forms $aQ_0+bQ_1$ indexed by $[a:b]\in\Pp^1$. The \emph{base} of the pencil is the closed subscheme defined by $Q_0=Q_1=0$. Its points lie on every quadric in the pencil, and its definition is independent of the chosen generators. We call the pencil \emph{proper} if $Q_0$ and $Q_1$ have no common surface component, or equivalently, no nonconstant common factor. Their intersection is then a complete intersection curve of degree four. This curve need not be smooth or reduced. The configurations considered below consist of a pencil together with ten ordered points in its base. Two configurations are projectively equivalent if a projective transformation carries one pencil to the other and carries each marked point to the point with the same label. We want parameter spaces in which equivalent configurations are counted once. To construct these spaces, first send four specified independent points to the coordinate vertices of $\Pp^3$. The remaining coordinate changes are diagonal transformations, with three independent scaling parameters, since a common scaling acts trivially on projective space. Under the hypotheses of the following lemma, we can divide the configurations into finitely many locally closed strata and use three nonzero coordinate ratios to fix these remaining scalings uniquely. The resulting normalized locus is called an \emph{algebraic slice}. Each projective equivalence class in the stratum has exactly one representative on this slice. The normalization is regular because all coordinate ratios used in its construction are invertible on the chosen stratum. \begin{lemma}\label{lem:pencil-quotient-slices} Let $Z$ be a $\operatorname{PGL}_4$-invariant locally closed locus, defined over $\Q$, of pencils in $\Pp^3$ with ten ordered points in their bases. Require that the pencil be the exact quadratic kernel and that a fixed ordered quadruple of labels be independent. If $\dim Z\le25$, the configurations modulo projective transformations admit a finite stratification by algebraic slices of dimension at most ten. After adjoining a hyperplane avoiding all ten points, their compactly supported Euler classes have zero $S_{12}$-part in both realizations. By Lemma~\ref{lem:canonical-frame}, the same holds for the corresponding $U$-loci. These conclusions remain valid after imposing further projectively invariant locally closed conditions defined over $\Q$, including ranks of evaluation maps on subsets of labels. \end{lemma} \begin{proof} Send the four independent points to the coordinate vertices, and let $Z_0$ be the resulting locus. Independent ordered quadruples form a twelve-dimensional homogeneous space for $\operatorname{PGL}_4$, so \[ \dim Z_0=\dim Z-12\le13. \] The residual coordinate changes form the three-dimensional diagonal torus $D=(\mathbb G_m)^4/\mathbb G_m$. Stratify $Z_0$ by the nonzero coordinate positions of the marked points. On each stratum, form a graph on $\{0,1,2,3\}$, joining two indices when some marked point has both coordinates nonzero. This graph is connected. Otherwise every point would be supported in one connected component, and all cross monomials between different components would vanish at every point. A nontrivial partition of four indices gives at least three such independent quadrics, contradicting the exact kernel dimension two. Choose a spanning tree of this graph and a marked point witnessing each edge. Orient the tree from a root, and let $r_{ij}$ be the ratio of its $j$th to its $i$th coordinate at the chosen point. Under a diagonal transformation, \[ r_{ij}\longmapsto(t_j/t_i)r_{ij}. \] Set the root coordinate equal to one and recursively choose $t_j=t_i/r_{ij}$. This gives the unique element of $D$ making the three chosen ratios equal to one. It is regular on the stratum, since the ratios are invertible regular functions. If $Y$ is the locus with these ratios equal to one, this normalization gives an isomorphism from the support stratum to $D\times Y$. Thus $\dim Y\le13-3=10$, and each orbit meets $Y$ exactly once. The finitely many support strata therefore give the required slices, all defined over $\Q$. Indeed, let $t(z)\in D$ be the unique normalizing element just constructed for a point $z$ of a support stratum. The product decomposition sends $z$ to $(t(z)^{-1},t(z)z)\in D\times Y$; its inverse sends $(d,y)$ to $dy$. Both maps are regular because the chosen coordinate ratios never vanish on this stratum. This also explains why subtracting the three-dimensional torus orbit is legitimate. It remains to adjoin the avoided hyperplane. On a slice $Y$, let $W=\Q^4$ and let $\mathcal L_j\subset W\otimes\cO_Y$ be the line represented by the $j$th point. For each subset $J$, stratify by the rank $r_J$ of \[ W^\vee\otimes\cO_Y \longrightarrow\bigoplus_{j\in J}\mathcal L_j^\vee. \] There are finitely many simultaneous rank strata $B$. On each, hyperplanes containing the points indexed by $J$ form a projective bundle with fiber $\Pp^{3-r_J}$, interpreted as empty when $r_J=4$. Consequently, the locus $Z_B$ of hyperplanes avoiding all ten points satisfies \[ [Z_B]=[B]\sum_{J\subseteq\{1,\ldots,10\}} (-1)^{|J|}[\Pp^{3-r_J}], \] with rank-four terms zero. The sum is a polynomial in $\LL$ and $\dim B\le10$. Apply Lemma~\ref{lem:low-dimensional-class}. Lemma~\ref{lem:canonical-frame} identifies these data with the corresponding $U$-loci. Further invariant locally closed conditions restrict the same slices and rank strata. \end{proof} \begin{proposition}\label{prop:pencil-incidence-boundary} Fix an ordered independent quadruple among ten ordered points in $\Pp^3$, and require their exact quadratic kernel to be a pencil $A$. Each of the following incidence loci has dimension at most $25$: \begin{enumerate} \item $A$ is proper and its base is singular; \item $A$ is nonproper; \item the base is smooth and proper, and two labels coincide. \end{enumerate} After imposing any further projectively invariant locally closed conditions defined over $\Q$, the configurations modulo projective transformations admit finite algebraic slices of dimension at most ten. Adjoining an avoided hyperplane gives zero $S_{12}$-part in both realizations, as do the corresponding $U$-loci. For configurations only required to span $\Pp^3$, the same conclusions hold after stratifying by the first independent quadruple in a fixed ordering. \end{proposition} \begin{proof} Pencils form $\Gr(2,10)$, of dimension sixteen. The required bounds are \[ \begin{array}{c|c|c|c} \text{pencil locus}& \text{pencil parameters}& \text{point parameters}& \text{total}\\ \hline \text{proper singular}&15&10&25\\ \text{nonproper}&7&18&25\\ \text{proper smooth with a collision}&16&9&25 \end{array} \] By Bertini's theorem, smooth complete intersections of two quadrics form a nonempty open locus; see \cite[Tag~0FD6]{StacksProject}. One can also check nonemptiness directly. Choose distinct rational numbers $b_0,\ldots,b_3$ and take the pencil spanned by $\sum_i X_i^2$ and $\sum_i b_iX_i^2$. At a singular point of their intersection, dependence of the two gradients would force all nonzero coordinates to have the same $b_i$. There would then be at most one nonzero coordinate, which is incompatible with $\sum_iX_i^2=0$. Hence this pencil has smooth base over $\overline{\Q}$. Since $\Gr(2,10)$ is irreducible, the complement of the nonempty open locus of smooth complete intersections has dimension at most fifteen. Thus proper singular pencils have parameter dimension at most fifteen. Their reduced bases are curves, so ten labels add at most ten parameters. A nonproper pencil has the form $A=\ell M$, where $\ell=0$ defines a plane $\Pi$ and the two-dimensional space $M$ of linear forms vanishes on a line $l$. Its reduced base is $\Pi\cup l$. Choosing the plane and line uses at most $3+4=7$ parameters. Independence of four labels excludes $l\subseteq\Pi$. There must be two distinct marked points outside $\Pi$. Otherwise the four-dimensional space $\ell H^0(\Pp^3,\cO(1))$ would satisfy at most one independent vanishing condition outside $\Pi$, leaving at least three quadrics in the exact kernel. Choose two labels at distinct outside points. They contribute at most two parameters on $l$, and the other eight contribute at most sixteen on $\Pi\cup l$. This gives the bound eighteen for the point choices. For a smooth proper pencil, a collision between two specified labels leaves at most nine point parameters on the curve. Taking the finite union over pairs of labels gives the third row. The independence, exact-kernel, and additional rank conditions only restrict these incidence loci. Apply Lemma~\ref{lem:pencil-quotient-slices}. If the points are only required to span, stratify by the first independent quadruple. \end{proof} Let $Z$ denote the locus of spanning configurations $(q_1,\ldots,q_{10})$. Order the four-element subsets of $\{1,\ldots,10\}$ as $I_1,\ldots,I_{210}$. Choose nonzero homogeneous representatives $v_i\in k^4$ of the points $q_i$. For $I=\{i_10$. Fix $e\ge1$ and set $A_{n,e}=(\Z[t]/(t^e))^n$ and $N=ne$. The algebra $A_{n,e}$ is free over $\Z$ with basis $t^re_i$, where $1\le i\le n$ and $0\le r1$, and $\mathscr I_{1,0}=\Spec\Z$. \begin{lemma}\label{total:lem:projective-count} Let $k$ be a field and $e\ge1$. The points of $\mathscr Q_{n,e}(k)$ correspond to the unital $k[[t]]$-subalgebras of $k[[t]]^n$ of colength $e$. The subset $\mathscr I_{n,e}(k)$ corresponds to those whose reduction image is $k\bfone$. For fixed $n,e$ and every sufficiently large prime $p$, \begin{equation}\label{total:eq:irreducible-point-count} g_n(p^e)=\#\mathscr I_{n,e}(\F_p). \end{equation} \end{lemma} \begin{proof}The result follows from the above construction. In greater detail, set $\cO=k[[t]]$ and let $\pi:\cO^n\to(\cO/(t^e))^n=A_k$ be the quotient map. If $R\subseteq\cO^n$ has colength $e$, the elementary-divisor decomposition of $\cO^n/R$ has the form $\bigoplus_i\cO/(t^{a_i})$, where $a_i\ge0$ and $\sum_i a_i=e$. Since each $a_i\le e$, this quotient is annihilated by $t^e$. Hence $t^e\cO^n\subseteq R$, and $S=\pi(R)$ is a subspace of $A_k$ of codimension $e$. The algebra conditions on $R$ descend to the three conditions on $S$ above. Conversely, let $S$ satisfy those conditions and set $R=\pi^{-1}(S)$. Stability under $t$ makes $S$ a $k[t]/(t^e)$-submodule, so its inverse image is an $\cO$-submodule. It contains $t^e\cO^n$, and $\cO^n/R\simeq A_k/S$ has $k$-dimension $e$. Thus $R$ is a full lattice of colength $e$. The unit and multiplication conditions on $S$ imply the same conditions on $R$, because $\pi$ is an algebra homomorphism. These two constructions are inverse. Since $e\ge1$, the kernel $t^e\cO^n$ has zero reduction modulo $t$; the reduction images of $R$ and $S$ therefore agree. For fixed $n,e$, there are only finitely many positive compositions in \eqref{total:eq:composition-sum}. Proposition~\ref{prop:general-classical-comparison} identifies the counts over $\Q_p$ and $\F_p((t))$ for all of these compositions outside one finite set of primes. Summing over them proves \eqref{total:eq:irreducible-point-count}. This proof concerns lattices over fields and their rational points. We do not require the same truncation assertion for lattice families over arbitrary nonreduced bases. \end{proof} Loop rotation gives an action on $A_{n,e}$ by \[ \rho_\lambda\!\left( \sum_{r=0}^{e-1}t^rx_r\right) =\sum_{r=0}^{e-1}\lambda^rt^rx_r. \] It gives an action on both $\mathscr Q_{n,e}$ and $\mathscr I_{n,e}$. Replacing $t$ by $\lambda t$ fixes $\bfone$, respects multiplication, and preserves $t$-stable subspaces. It also leaves constant terms unchanged, so both $\mathscr Q_{n,e}$ and $\mathscr I_{n,e}$ are preserved. To apply Lemma~\ref{total:lem:torus-localization}, use the Pl\"ucker embedding of $G$. Rotation multiplies each basis vector $t^re_i$ by $\lambda^r$. It therefore multiplies a wedge of these vectors by $\lambda^{r_1+\cdots+r_{N-e}}$, where the $r_j$ are their exponents. Thus each Pl\"ucker coordinate is scaled separately, giving the required diagonal action on the ambient projective space. Set $I_{n,e}=(\mathscr I_{n,e,\Q})_{\mathrm{red}}$. Lemma~\ref{total:lem:torus-localization} and \eqref{eq:epsilon-L-1} give \begin{equation}\label{total:eq:epsilon-fixed} \varepsilon_{12}(\chi_c^\ell(I_{n,e})) = \varepsilon_{12}\!\left( \chi_c^\ell((I_{n,e}^{\mathbb G_m})_{\mathrm{red}}) \right). \end{equation} \begin{proposition}\label{total:prop:fixed-filtrations} For $e\ge1$, the reduced fixed locus of $I_{n,e}$ is the finite disjoint union of the reduced projective varieties of filtrations \begin{equation}\label{total:eq:filtration} \begin{gathered} F_0=\Q\bfone\subseteq F_1\subseteq\cdots \subseteq F_e=\Q^n,\\ F_i\circled F_j\subseteq F_{i+j}\quad \text{and}\quad \sum_{r=0}^{e-1}(n-\dim F_r)=e, \end{gathered} \end{equation} where $F_r=\Q^n$ for $r\ge e$. The positive terms of $c_r=n-\dim F_r$, $r\ge1$, form a partition of $e-(n-1)$, with each part at most $n-1$. \end{proposition} \begin{proof} The weight decomposition of $A_{n,e,\Q}$ is $\bigoplus_{r=0}^{e-1}t^r\Q^n$. Over any base with trivial $\mathbb G_m$-action, an invariant subbundle and its locally free quotient decompose into weight subbundles; see \cite[Tag~0EKL]{StacksProject}. Hence every invariant subbundle has the form \[ S=\bigoplus_{r=0}^{e-1}t^rF_r, \] where each $F_r$ is a subbundle of the trivial rank-$n$ bundle. For fixed ranks $f_r$, these are parametrized by $\prod_{r=0}^{e-1}\Gr(f_r,\Q^n)$. The unit and reduction conditions give $F_0=\Q\bfone$, while $t$-stability gives $F_r\subseteq F_{r+1}$. For $i+jl$. Thus only products with $i,j\ge1$ and $i+j\le l$ impose conditions. \par The parameter space $\mathcal T_c$ is the reduced closed subvariety of the corresponding partial flag variety defined by these product inclusions. For example, the inclusion $F_iF_j\subseteq F_{i+j}$ means that the multiplication map from the two tautological subbundles has zero image in the quotient bundle $A/F_{i+j}$. Repeated dimensions in a flag force equality of the corresponding spaces. This explains why repeated parts of $c$ give the algebra and stability conditions used below. We first treat conditions of the form $U^2\subseteq T$, where $T$ is a hyperplane. Recall that a subspace $P$ is totally isotropic for a symmetric bilinear form $b$ if $b(x,y)=0$ for every $x,y\in P$. \begin{lemma}\label{revprelim:split-og-tate} Let $W$ be a split nondegenerate quadratic space over $\Q$. The variety $\operatorname{OG}(a,W)$ of totally isotropic $a$-planes has only Tate cohomology. This includes both components when $\dim W=2a$. \par Here "split" means that $W$ is an orthogonal sum of hyperbolic planes, with one additional nondegenerate line if $\dim W$ is odd. A hyperbolic plane has a basis $e,f$ with $b(e,e)=b(f,f)=0$ and $b(e,f)=1$. The variety $\operatorname{OG}(a,W)$ is a closed subvariety of $\Gr(a,W)$: its defining condition is that the form restrict to zero on the tautological subbundle. \end{lemma} \begin{proof} Each connected component is a projective homogeneous space for the split special orthogonal group. The Bruhat decomposition for its parabolic quotient gives a paving by affine spaces defined over $\Q$; see \cite[\S21h]{MilneAlgebraicGroups}. Localization therefore gives the assertion. In the even-dimensional maximal case, apply this to each of the two components. \end{proof} Let $E\to B$ be a vector bundle and $L\to B$ a line bundle. A symmetric bilinear form on $E$ with values in $L$ is an $\cO_B$-linear map \[ b:E\otimes_{\cO_B}E\longrightarrow L, \]such that $b(u,v)=b(v,u)$. \begin{lemma}\label{lem:relative-quadratic-flags} Let $E\to B$ be a vector bundle with a symmetric bilinear form taking values in a line bundle, and suppose $\dim B\le d$. Let $Z\to B$ parametrize its totally isotropic $a$-planes. Every simple constituent of $H_c^*(Z)$ is a subquotient of a Tate twist of the cohomology of a variety of dimension at most $d$. Consequently its Hodge level and individual Hodge--Tate spread are at most $d$. If $d\le10$, no Tate twist of $S_{12}$ occurs. \end{lemma} \begin{proof} We trivialize the line bundle in which the form takes values, and stratify $B$ by the rank of the resulting map $E\to E^\vee$. \par The strata are given their reduced induced structures. On a constant-rank stratum, elementary row operations on an invertible minor identify the kernel and image with locally free summands. Thus the radical used below is a vector bundle. We may replace a finite open cover by a finite disjoint decomposition into locally closed pieces, so all subsequent local choices can be combined by localization. On each rank stratum, its kernel \[ R=\{v\in E:b(v,w)=0\text{ for every }w\in E\} \] is a vector subbundle, called the radical. Locally choose a complementary subbundle $M$, so that $E=R\oplus M$. The restriction of the form to $M$ is nondegenerate, because its entire radical has been removed. After a further finite decomposition, we may trivialize $R$ and choose an orthogonal basis $m_1,\ldots,m_r$ of $M$, with $b(m_i,m_i)=a_i$ invertible and $b(m_i,m_j)=0$ for $i\ne j$. This is the usual diagonalization of a symmetric form, carried out locally where each chosen diagonal coefficient is invertible. For an invertible function $a_i$, adjoining a square root $u_i$ gives a finite \'etale cover: the equation is $u_i^2=a_i$, and its derivative $2u_i$ is invertible. Take the fiber product of these covers and also adjoin $v$ with $v^2=-1$. Denote the resulting cover by $B'\to B_0$, where $B_0$ is the piece under consideration. Rescaling $m_i$ by $u_i^{-1}$ makes every diagonal coefficient equal to one. Using $v$, each pair of these coordinates becomes a hyperbolic plane. Thus, over $B'$, the form is isometric to \[ E_{B'}=R\oplus(M_0\otimes_\Q\cO_{B'}), \] where $R$ is trivial and $M_0$ is a fixed split nondegenerate quadratic space over $\Q$. We now describe the isotropic $a$-planes for this form. Fix a point of $B'$ and an isotropic plane $P$, and set $D=P\cap R$ and $Q=\operatorname{im}(P\longrightarrow M_0)$. Stratify by $c=\dim D$. Then $\dim Q=a-c$. The form vanishes on $R$, so $P$ is isotropic exactly when $Q$ is isotropic in $M_0$. There is no restriction on the $c$-plane $D\subseteq R$. For fixed $D$ and $Q$, the projection $P/D\to Q$ is an isomorphism. Consequently $P/D$ is the graph of a unique linear map $\phi:Q\longrightarrow R/D$. Conversely, every such map gives the isotropic plane \[ P_\phi= \{(r,q)\in R\oplus Q:r\bmod D=\phi(q)\}. \] It has intersection $D$ with $R$ and image $Q$ in $M_0$. Thus the choices of $P$ with fixed $D,Q$ form the vector space $\operatorname{Hom}(Q,R/D)$. Writing $\rho=\operatorname{rank}R$, the stratum with intersection dimension $c$ is a vector bundle of rank $(a-c)(\rho-c)$ over \[ B'\times\Gr(c,\rho) \times\operatorname{OG}(a-c,M_0). \] The last two factors have Tate cohomology. The vector-bundle formula and K\"unneth therefore show that every simple constituent of the cohomology of this stratum comes from a Tate twist of $H_c^j(B')$. Localization over the finitely many values of $c$ gives the same conclusion for $Z'=Z\times_{B_0}B'$. Finally, $p:Z'\to Z|_{B_0}$ is finite \'etale. Its pullback and trace satisfy $\operatorname{Tr}\circ p^*=(\deg p)\operatorname{id}$. With rational coefficients, pullback therefore identifies $H_c^i(Z|_{B_0})$ with a direct summand of $H_c^i(Z')$. These maps respect the Hodge structures and the arithmetic Galois action. Applying localization once more to the pieces $B_0$ proves the asserted description of the constituents of $H_c^*(Z)$. Each finite cover $B'$ has dimension at most $d$. The dimension bounds established earlier therefore give Hodge level and individual Hodge--Tate spread at most $d$. Since Tate twists preserve these quantities and $S_{12}$ has level and spread eleven, its Tate twists are excluded when $d\le10$. \end{proof} \begin{lemma}\label{total:lem-one-hyperplane} Let $N\ge2$ and $1\le d\le N-1$. Define \[ X_{N,d}= \{(U,T):\bfone\in U\subseteq T\subseteq\Q^N,\ \dim U=d,\ \dim T=N-1,\ U^2\subseteq T\}. \] Every simple constituent of $H_c^*(X_{N,d})$ is a subquotient of a Tate twist of the cohomology of a variety of dimension at most $N-2$. In particular, no Tate twist of $S_{12}$ occurs if $N\le12$. \end{lemma} \begin{proof} Write $T=\ker\lambda$. The hyperplanes containing $\bfone$ form \[ B=\Pp\{\lambda\in(\Q^N)^\vee:\lambda(\bfone)=0\} \simeq\Pp^{N-2}. \] The form $b_\lambda(x,y)=\lambda(x\circled y)$ takes values in $\cO_B(1)$. On $T$, the vector $\bfone$ belongs to its radical, so the form descends to $T/\Q\bfone$. The condition $U^2\subseteq T$ says exactly that $U/\Q\bfone$ is a totally isotropic $(d-1)$-plane in this quotient. Apply Lemma~\ref{lem:relative-quadratic-flags}. \end{proof} Let $\pi=\{I_1,\ldots,I_h\}$ be a partition of $\{1,\ldots,N\}$ into nonempty disjoint blocks. Its coordinate-partition algebra is \[ H_\pi= \{x\in\Q^N:x_i=x_j \text{ whenever }i,j\text{ belong to the same block}\}. \] Thus a vector in $H_\pi$ has one common value on $I_1$, another on $I_2$, and so on. For example, the partition $\{\{1,3\},\{2,4,5\}\}$ gives $H_\pi=\{(a,b,a,b,b):a,b\in\Q\}$. Coordinatewise addition and multiplication preserve these equalities, so $H_\pi$ is a unital subalgebra of $\Q^N$. For a block $I$, let $\bfone_I$ denote the vector whose $i$th coordinate is $1$ if $i\in I$ and $0$ otherwise. Every vector in $H_\pi$ has a unique expression \[ x=a_1\bfone_{I_1}+\cdots+a_h\bfone_{I_h}, \] where $a_j$ is its common coordinate value on $I_j$. Hence the block indicators form a basis of $H_\pi$, and $\dim H_\pi=h$. We may identify $H_\pi$ with $\Q^h$ as an algebra. \par For a subspace $U\subseteq H$, write $\Q[U]$ for the smallest unital subalgebra of $H$ containing $U$; it consists of finite $\Q$-linear combinations of products of elements of $U$, including the empty product $\bfone$. A subspace $V\subseteq\Q^N$ is $H$-stable if $h\circled v\in V$ for every $h\in H$ and $v\in V$. \begin{lemma}\label{total:lem-partition-algebras} Every unital subalgebra of $\Q^N$ is a coordinate-partition algebra. In particular, there are finitely many such algebras. Given an algebra $H$, set \[ A_d(H)= \{U\subseteq H:\bfone\in U,\ \dim U=d,\ \Q[U]=H\}. \] Then $A_d(H)$ has only Tate cohomology. The same holds for the varieties of $H$-stable spaces. \end{lemma} \begin{proof} The classification follows from \cite[Lemma~2.7]{MirandolaZemor2015}. In particular, $H$ has only finitely many proper unital subalgebras, say $H_1,\ldots,H_m$. To study $A_d(H)$, begin with the Grassmannian $G=\Gr(d-1,H/\Q\bfone)$ of all unital $d$-planes in $H$. Such a plane fails to generate $H$ precisely when it is contained in one of the $H_i$. Thus, setting $Y_i=\Gr(d-1,H_i/\Q\bfone)$, we have $A_d(H)=G\setminus Y$, where $Y=\bigcup_iY_i$. For every nonempty set of indices $J$, \[ \bigcap_{j\in J}Y_j = \Gr\left(d-1, \left(\bigcap_{j\in J}H_j\right)/\Q\bfone\right), \] or the intersection is empty. Every such intersection therefore has Tate cohomology, by the Schubert decomposition. We first show that $Y$ has Tate cohomology. For two closed subvarieties, the compact-support Mayer--Vietoris sequence contains \[ \cdots\longrightarrow H_c^{i-1}(Y_1\cap Y_2) \longrightarrow H_c^i(Y_1\cup Y_2) \longrightarrow H_c^i(Y_1)\oplus H_c^i(Y_2) \longrightarrow\cdots. \] By exactness, every simple constituent of the middle group occurs in one of the adjacent groups. Hence the union has only Tate constituents whenever both pieces and their intersection do. Applying this argument inductively proves the assertion for $Y$. At each step, the intersection of the new piece with the preceding union is itself a union of intersections of the $Y_i$, so the same induction applies. Now use the localization sequence \[ \cdots\longrightarrow H_c^{i-1}(Y) \longrightarrow H_c^i(A_d(H)) \longrightarrow H_c^i(G) \longrightarrow\cdots. \] The adjacent groups have only Tate constituents, and exactness gives the same conclusion for $H_c^i(A_d(H))$. For the assertion about $H$-stable spaces, the block idempotents separate the choices on the individual coordinate blocks. Indeed, if the blocks of $H$ are $I_1,\ldots,I_h$, stability gives $V=\bigoplus_{j=1}^h V_j$, where $V_j=\bfone_{I_j}V\subseteq\Q^{I_j}$. The condition $H\subseteq V$ is equivalent to $\bfone_{I_j}\in V_j$ for every $j$. Thus, for prescribed dimensions $v_j=\dim V_j$, the choices form $\prod_j\Gr(v_j-1,|I_j|-1)$. These are products of Grassmannians and therefore have Tate cohomology. \par If an $H$-stable space $V$ is not required to contain $H$, the same decomposition gives $\prod_j\Gr(v_j,|I_j|)$ instead. Taking the finite union over the possible block dimensions proves the assertion for all $H$-stable spaces. The version with $\Gr(v_j-1,|I_j|-1)$ is the one used when $H\subseteq V$. \end{proof} \begin{lemma}\label{total:lem-generated-algebra} Let $\bfone\in B\subseteq C\subseteq\Q^N$ and set $H=\Q[B]$. Then $BC\subseteq C$ if and only if $C$ is $H$-stable. In this case $H\subseteq C$ and $B^2\subseteq C$. For fixed $H$ and prescribed dimensions, the choices of $B$ generating $H$ and of $H$-stable $C$ containing $H$ are independent and have only Tate cohomology. \end{lemma} \begin{proof} If multiplication by $B$ preserves $C$, so does multiplication by every sum of products of elements of $B$. These elements form $H$. Since $\bfone\in C$, this also gives $H\subseteq C$. The converse is immediate. For fixed $H$, the parameter space is the product of $A_{\dim B}(H)$ with the $H$-stable-space variety in Lemma~\ref{total:lem-partition-algebras}. \end{proof} For a subspace $R\subseteq\Q^N$, set $\operatorname{Stab}(R)=\{a\in\Q^N:aR\subseteq R\}$. This is a unital subalgebra. Thus $\operatorname{Stab}(R)$ consists of the vectors whose coordinatewise multiplication maps preserve $R$. The notation refers to this algebra of multiplication operators, with the vector $\bfone$ acting as the identity. \begin{lemma}\label{total:lem-hyperplane-stabilizer} Let $T=\ker\lambda$ contain $\bfone$, and set $S=\{i:\lambda_i\ne0\}$ and $s=|S|$. Then \[ H_S:=\operatorname{Stab}(T) =\{x\in\Q^N:x_i\text{ is constant for }i\in S\} \]and $\dim H_S=N+1-s$. Further, $H_S\subseteq T$ and $\operatorname{rad}(b_\lambda|_T)=H_S$. The hyperplanes with support $S$ form \[ B_S= \{[\lambda_i]_{i\in S}: \textstyle\sum_{i\in S}\lambda_i=0,\ \lambda_i\ne0\text{ for }i\in S\}. \] Here $s\ge2$, and $B_S$ is a hyperplane-arrangement complement in $\Pp^{s-2}$ with only Tate cohomology. A totally isotropic $d$-plane for $b_\lambda$ satisfies \begin{equation}\label{total:eq-isotropic-support-bound} d\le N-s+\lfloor s/2\rfloor. \end{equation} \end{lemma} \begin{proof} The condition $xT\subseteq T$ says that $y\mapsto\lambda(x\circled y)$ is a scalar multiple of $\lambda$. Thus $\lambda_i x_i=c\lambda_i$ for every $i$. This gives the description of $H_S$. It also gives $\lambda(x)=c\sum_{i\in S}\lambda_i=0$, so $H_S\subseteq T$. The same calculation describes the radical of the restricted form. The assertion about $B_S$ follows by removing coordinate hyperplanes from $\sum_{i\in S}\lambda_i=0$ in $\Pp^{s-1}$. All their intersections are projective linear spaces, so localization gives Tate cohomology. Finally, the radical of $b_\lambda$ on $\Q^N$ has dimension $N-s$. Its nondegenerate quotient has dimension $s$, and an isotropic subspace of that quotient has dimension at most $\lfloor s/2\rfloor$. This proves \eqref{total:eq-isotropic-support-bound}. \end{proof} \begin{lemma}\label{total:lem-pure-hyperplane-tail} Let $N\ge2$, $1\le d\le N-1$, and $r\ge1$. Consider multiplicative filtrations with $F_1=U$, $F_2=\cdots=F_{r+1}=T$, $\dim U=d$, and $\dim T=N-1$. Their cohomology has Hodge level and individual Hodge--Tate spread at most $N-2$. If $r\ge2$, it has only Tate cohomology. \end{lemma} \begin{proof} Recall that $F_1=U$, that $F_2,\ldots,F_{r+1}$ are all equal to $T$, and that every later term is $\Q^N$. Thus only products whose indices sum to at most $r+1$ impose a condition. For $r=1$, the proper terms are $U,T$. The only required product inclusion is $F_1F_1\subseteq F_2$, namely $U^2\subseteq T$. The parameter space is therefore $X_{N,d}$, and Lemma~\ref{total:lem-one-hyperplane} applies. For $r=2$, the proper terms are $U,T,T$. Besides $U^2\subseteq T$, we require $F_1F_2\subseteq F_3$, or $UT\subseteq T$. This says that multiplication by every element of $U$ preserves $T$. Write $T=\ker\lambda$ and fix the support $S$ of $\lambda$. By Lemma~\ref{total:lem-hyperplane-stabilizer}, the condition $UT\subseteq T$ is equivalent to $U\subseteq H_S$. Since $H_S$ is an algebra contained in $T$, this also gives $U^2\subseteq H_S\subseteq T$. Thus no separate square condition remains. For fixed $S$, the algebra $H_S$ is fixed and has dimension $N+1-s$, where $s=|S|$. We may therefore choose $T\in B_S$ and, independently, a unital $d$-plane in $H_S$. The resulting parameter space is \[ B_S\times\Gr(d-1,H_S/\Q\bfone) \simeq B_S\times\Gr(d-1,N-s). \] It is empty if $d>N+1-s$. Both factors have Tate cohomology. For $r\ge3$, the term $F_4$ is still $T$. Hence $F_2F_2\subseteq F_4$ requires $T^2\subseteq T$. Since $T$ contains $\bfone$, it is a unital subalgebra, and therefore one of finitely many coordinate-partition hyperplanes. For each such $T$, any unital $d$-plane $U\subseteq T$ is allowed: all products among $U$ and $T$ already lie in $T$. Its choices form $\Gr(d-1,N-2)$. Thus these cases also have Tate cohomology. \end{proof} \begin{lemma}\label{total:lem-codim-two-hyperplane-tail} Let $N\ge5$ and $1\le d\le N-2$. Consider multiplicative filtrations with proper terms $U,V,T,\ldots,T$, where $T$ occurs $k\ge2$ times, $\bfone\in U\subseteq V\subseteq T\subseteq\Q^N$, and $(\dim U,\dim V,\dim T)=(d,N-2,N-1)$. Every simple cohomology constituent is a subquotient of a Tate twist of the cohomology of a variety of dimension at most $N-3$. In particular, no Tate twist of $S_{12}$ occurs if $N\le13$. \end{lemma} \begin{proof} For $k=2$, the conditions reduce to $U^2\subseteq V$, $UT\subseteq T$ and $V^2\subseteq T$. Write $T=\ker\lambda$ and fix its support $S$. The last condition makes $V$ an isotropic $(N-2)$-plane for $b_\lambda$. By \eqref{total:eq-isotropic-support-bound}, $N-2\le N-s+\lfloor s/2\rfloor$, so $2\le s\le4$. We have that $U\subseteq H_S$. If $s=2$, then $B_S$ is a point and $T=H_S\simeq\Q^{N-1}$ as an algebra. The remaining condition is $U^2\subseteq V$, giving $X_{N-1,d}$. Lemma~\ref{total:lem-one-hyperplane} gives the bound $N-3$. If $s=3$, the form on $T/H_S$ is a nondegenerate form on a line. It has no nonzero isotropic vectors, so $V=H_S$. The choices of $U$ form $\Gr(d-1,N-3)$ over the one-dimensional base $B_S$. If $s=4$, the form on $T/H_S$ is nondegenerate of dimension two. An isotropic $(N-2)$-plane $V$ must contain $H_S$ and choose an isotropic line in this quotient. These lines form a finite \'etale double cover of $B_S$, of dimension two. Over this cover, $U$ is any unital $d$-plane in $H_S$, with choices $\Gr(d-1,N-4)$. Its square lies in $H_S\subseteq V$. Thus the required constituent bound follows from the bundle formula. The double cover need not itself have Tate cohomology. For $k=3$, the additional condition is $VT\subseteq T$. It forces $V\subseteq H_S$, so only $s=2,3$ remain. For $k\ge4$, the condition $T^2\subseteq T$ also appears, forcing $s=2$. The preceding descriptions remain valid. Localization over the finite support decomposition proves the assertion. \end{proof} \begin{lemma}\label{total:lem-kneser-parameter-bound} Let $1\le u\le q\le N$, and let $\bfone\in U\subseteq\Q^N$ satisfy $\dim U=u$ and $r=\dim U^2\le q$. Set $H=\operatorname{Stab}(U^2)$, $h=\dim H$, $W=HU$, and $d=\dim W$. Then \begin{equation}\label{total:eq-kneser-constraints} r\ge2d-h\quad \text{and}\quad \max(u,h)\le d\le r\le q. \end{equation} Fix the coordinate blocks of $H$, of sizes $n_1,\ldots,n_h$, and the dimensions $w_j$ of the corresponding summands of $W$. The locus of these $U$ has dimension at most \begin{equation}\label{revkneser:parameter-bound} \sum_{j=1}^h(w_j-1)(n_j-w_j)+(u-1)(d-u) \le\Phi_{N,u}(h,d), \end{equation} where \[ \Phi_{N,u}(h,d) =(d-h)(N-d)+(u-1)(d-u). \] Fixing these ranks and algebras gives a finite locally closed stratification. \end{lemma} \begin{proof} Put $R=U^2$. Since $\bfone\in U$ and $HR=R$, we have $U,H\subseteq W\subseteq R$ and $W^2=H^2U^2=R$. Apply \cite[Theorem~3.3]{MirandolaZemor2015} to $W,W$ to obtain $r\ge2d-h$. This proves \eqref{total:eq-kneser-constraints}. \par The reason for enlarging $U$ to $W=HU$ is that multiplication by $H$ then preserves $W$, while the square remains $R$. Indeed, $H$ contains $\bfone$, so $H^2=H$ and $W^2=H^2U^2=HR=R$. The stabilizer in the cited inequality is therefore precisely the algebra $H$ already chosen. For fixed $H$, the choices of $W$ form $\prod_j\Gr(w_j-1,n_j-1)$. For fixed $W$, the unital $u$-planes form $\Gr(u-1,d-1)$. The dimension of this incidence is the first expression in \eqref{revkneser:parameter-bound}. \par Here every $w_j$ is at least one, since $W$ contains the block indicator $\bfone_{I_j}$. The factors have dimensions $(w_j-1)(n_j-w_j)$ and $(u-1)(d-u)$, respectively. We count all such pairs $(U,W)$, allowing pairs that do not have the prescribed square or stabilizer. This can only increase the dimension and hence gives an upper bound for the required locus. Since $\sum_j(w_j-1)=d-h$ and $\sum_j(n_j-w_j)=N-d$, it is at most $\Phi_{N,u}(h,d)$. To apply this bound to the actual $U$-locus, first fix the rank of $\Sym^2U\to\Q^N$. Its image $R$ then varies as a vector subbundle. The stabilizer is the kernel of $\Q^N\to\operatorname{Hom}(R,\Q^N/R)$ sending $a$ to the linear map $x\mapsto a\circled x\bmod R$. On constant-rank strata it varies regularly and takes values among finitely many coordinate-partition algebras. Fixing $H$ and the ranks of its block projections on $U$ makes $W=HU$ vary regularly as well. The graph of $U\mapsto W$ lies in the incidence just counted. This proves both the dimension bound and the stratification assertion. \end{proof} \begin{lemma}\label{total:lem-small-squares} The loci \[ \begin{aligned} D_5&=\{U\subseteq A:\bfone\in U,\ \dim U=5,\ \dim U^2\le7\},\\ D_6&=\{U\subseteq A:\bfone\in U,\ \dim U=6,\ \dim U^2\le7\} \end{aligned} \] satisfy $\dim D_5\le10$ and $\dim D_6\le5$. \end{lemma} \begin{proof} In Lemma~\ref{total:lem-kneser-parameter-bound}, take $N=10$ and $q=7$. Then $u\le d\le7$ and $h\ge2d-7$. For fixed $d$, the bound $\Phi_{10,u}(h,d)$ decreases with $h$, giving \[ \begin{array}{c|c|c} u&d&\Phi_{10,u}(2d-7,d)\\\hline 5&5,6,7&10,8,8\\ 6&6,7&4,5. \end{array} \] Take the maximum over the finitely many strata. \end{proof} \begin{proposition}\label{total:prop-large-first-space} The profiles $(5,3)$ and $(4,3,1)$ contain no Tate twist of $S_{12}$ or $S_{12,\ell}$ in compactly supported cohomology. \end{proposition} \begin{proof} For $(5,3)$, the proper terms have dimensions $(5,7)$ and the only condition is $U^2\subseteq V$. On the stratum where $r=\dim U^2\le7$, the choices of $V$ form $\Gr(7-r,A/U^2)$. Its base lies in $D_5$ and has dimension at most ten. For $(4,3,1)$, the proper dimensions are $(6,7,9)$, with conditions $U^2\subseteq V$ and $UV\subseteq T$. If $\dim U^2=7$, then $V=U^2$ and the base of choices of $U$ lies in $D_6$, of dimension at most five. If $\dim U^2=6$, then $U^2=U$, so $U$ is one of finitely many partition algebras; the choices of $V$ form $\Pp(A/U)=\Pp^3$. In either case, stratify by $s=\dim UV$. The hyperplanes $T$ containing $UV$ form $\Pp((A/UV)^\vee)$, or the empty set if $s=10$. Since $\bfone\in U$, we have $V\subseteq UV$, so this also imposes $V\subseteq T$. The bundle formula and Lemma~\ref{revprelim:relative-hodgelevel} give the required exclusions. \end{proof} The next argument treats three-step flags. When the first space is two-dimensional, its products are controlled by four elementary moment equations. \begin{lemma}\label{total:lem-cubic-moment-base} For $s\ge3$, let $M_s$ parametrize pairs $([\lambda],P)$, where $[\lambda]\in\Pp((\Q^s)^\vee)$, $P$ is a two-plane containing $\bfone$, and $\lambda(P^3)=0$. Then \[ \dim M_s\le\max\{s-3,2s-7\}. \] \end{lemma} \begin{proof} Write $P=\langle\bfone,z\rangle$. Stratify by the partition of the coordinate labels on which the values of $z$ agree, and let $v\ge2$ be the number of blocks. Replacing $z$ by $az+b\bfone$ does not change $P$. Normalize its values on two specified blocks to $0$ and $1$. \par More precisely, two choices $z,z'$ spanning $P$ with $\bfone$ satisfy $z'=az+b\bfone$ with $a\ne0$. On a fixed equality partition, choose two of its blocks in a fixed order. Their values are distinct, so there is a unique such change giving values $0$ and $1$ on those two blocks. The remaining block values are then genuine coordinates for $P$ on this stratum. The remaining distinct block values then vary in an open subset of $\A^{v-2}$. The condition on $\lambda$ is \[ \sum_{i=1}^s\lambda_i z_i^j=0 \]for $0\le j\le3$. The coefficient matrix has rank $\min(4,v)$, by the Vandermonde determinant. Thus the projective space of possible $\lambda$ is empty or has dimension $s-\min(4,v)-1$. The stratum has dimension at most \[ v-2+s-\min(4,v)-1 = \begin{cases} s-3,&2\le v\le3,\\ s+v-7\le2s-7,&v\ge4. \end{cases} \] There are finitely many equality partitions. \end{proof} \begin{lemma}\label{total:lem-three-step-moments} Let $N\ge4$ and $2\le d\le N-2$. Define \[ Y_{N,d}=\left\{(U,V,T): \begin{gathered} \bfone\in U\subseteq V\subseteq T\subseteq\Q^N,\\ (\dim U,\dim V,\dim T)=(d,N-2,N-1),\\ U^2\subseteq V,\qquad UV\subseteq T \end{gathered}\right\}. \] Every simple constituent of $H_c^*(Y_{N,d})$ is a subquotient of a Tate twist of the cohomology of a variety of dimension at most \[ L(N,d)=\max\{N-3,2N-2d-3\}. \] In particular, no Tate twist of $S_{12}$ occurs if $L(N,d)\le10$. \end{lemma} \begin{proof} Write $T=\ker\lambda$ and fix its support $S$, of size $s$. Put $P=\operatorname{pr}_S(U)$ and $t=\dim P$. Under the ordinary coordinate pairing, $UV\subseteq T$ means \[ \operatorname{diag}(\lambda)U\subseteq V^\perp. \] The two sides have dimensions $t$ and two, respectively. Since $\bfone_S\in P$, we have $t=1$ or $2$. These are locally closed rank conditions. Suppose first that $t=2$. The inclusion is then an equality, so \[ V=(\operatorname{diag}(\lambda)P)^\perp \] is uniquely determined. The condition $U^2\subseteq V$ says exactly $\lambda(P^3)=0$. If $P=\langle\bfone_S,z\rangle$, these are \begin{equation}\label{total:eq-cubic-moments} \sum_{i\in S}\lambda_i z_i^j=0. \end{equation} They also ensure $U\subseteq V\subseteq T$. \par To see the equivalence, a vector $u_1\circled u_2$ lies in $V$ exactly when $\lambda(u_1\circled u_2\circled u_3)=0$ for every $u_3\in U$. The restrictions of the $u_i$ to $S$ fill $P$, so these are precisely the cubic moment conditions. Taking one of the $u_i$ equal to $\bfone$ also gives $U\subseteq V$; taking $u_3=\bfone$ in the definition of $V$ gives $V\subseteq T$. The case $s=2$ is empty, since then $P=\Q^2$ and $\lambda(P)=0$ forces $\lambda=0$. For $s\ge3$, the pair $([\lambda],P)$ lies in the open part of $M_s$ where every supported coefficient is nonzero. Also $D=U\cap\Q^{S^c}$ has dimension $d-2$, so $s\le N-d+2$. Lemma~\ref{total:lem-cubic-moment-base} gives the bound \[ \dim\{([\lambda],P)\} \le\max\{s-3,2s-7\}\le2N-2d-3. \] Over this pair base, choose $D\in\Gr(d-2,\Q^{S^c})$. Modulo $D$, a lift $U$ is the graph of a map $\phi:P\to\Q^{S^c}/D$ satisfying \[ \phi(\bfone_S)=\bfone_{S^c}\bmod D. \] These maps form an affine torsor of rank $N-s-d+2$: a prescribed map on the line $\Q\bfone_S$ extends locally to $P$. \par The difference of two extensions vanishes on $\Q\bfone_S$, so these extensions form a torsor under $\Hom(P/\Q\bfone_S,\Q^{S^c}/D)$. The first bundle has rank one and the second has rank $N-s-d+2$. This gives the stated rank and also proves Zariski local triviality of the choices of lifts. There are no further conditions, since the multiplication equations depend only on the supported coordinates. Grassmannian and affine bundles therefore add only Tate twists of the cohomology of the pair base. Now suppose $t=1$. Put $E=\Q\bfone_S\oplus\Q^{S^c}$ and $M=N+1-s$. Then $E=H_S$ is a fixed coordinate-partition algebra and $U\subseteq E$. Since $E$ stabilizes $T$, the condition $UV\subseteq T$ is automatic. If $E\subseteq V$, the condition $U^2\subseteq V$ is automatic too. The choices are $U\in\Gr(d-1,M-1)$ and a hyperplane $V/E$ in $T/E$. These give Grassmannian and projective bundles over $B_S$, and hence Tate cohomology. \par This case is empty when $s=2$, because then $E=T$ and an $(N-2)$-plane cannot contain $E$. For $s\ge3$, the quotient $T/E$ has rank $s-2$, so its hyperplanes form a projective bundle with fiber $\Pp^{s-3}$. This is the projective factor in the preceding description. Otherwise, set $W=V\cap E$. It is a hyperplane in $E$, and $(U,W)$ belongs to $X_{M,d}$ after identifying $E$ with $\Q^M$. Conversely, every such pair extends to a flag: the choices of $V/W$ are the complements to the line $E/W$ in $T/W$. They form an affine torsor of rank $\dim(T/E)=s-2$. Thus this stratum is an affine bundle over $B_S\times X_{M,d}$. The variety $B_S$ has only Tate cohomology. Lemma~\ref{total:lem-one-hyperplane} therefore gives the constituent bound $M-2\le N-3$. If $d=M$, this second stratum is empty. Combining the two projection-rank cases and the finitely many supports proves the lemma. \end{proof} \begin{lemma}\label{total:lem-two-plane-moments} Let $N\ge3$ and $2\le v\le N-1$. The flags $\bfone\in U\subseteq V\subseteq T\subseteq\Q^N$ with \[(\dim U,\dim V,\dim T)=(2,v,N-1),\] $U^2\subseteq V$, and $UV\subseteq T$ have the constituent bound of Lemma~\ref{total:lem-three-step-moments} with base dimension $\max\{N-3,2N-7\}$. In particular, no Tate twist of $S_{12}$ occurs if $N\le8$. \end{lemma} \begin{proof} Write $T=\ker\lambda$. The product conditions imply $\lambda(U^3)=0$, so $(U,[\lambda])$ lies in $M_N$. Over this base, set $K=(\operatorname{diag}(\lambda)U)^\perp$. The moment condition gives $U^2\subseteq K$, and $\bfone\in U$ gives $K\subseteq T$. Stratify by $q=\dim U^2$ and $t=\dim\operatorname{diag}(\lambda)U$. On each stratum these spaces vary as vector bundles, and the admissible $V$ form $\Gr(v-q,K/U^2)$, or the empty set. Apply the bundle formula and Lemma~\ref{total:lem-cubic-moment-base}. \end{proof} \begin{proposition}\label{total:prop-521} The profile $(5,2,1)$ contains no Tate twist of $S_{12}$ or $S_{12,\ell}$ in compactly supported cohomology. \end{proposition} \begin{proof} Its proper dimensions are $(5,8,9)$, and its conditions are $U^2\subseteq V$ and $UV\subseteq T$. Thus it is $Y_{10,5}$. Lemma~\ref{total:lem-three-step-moments} gives $L(10,5)=7$. \end{proof} It remains to treat the profiles with repeated terms. Equalities among consecutive spaces often force a space to be an algebra, or force the next space to be stable under that algebra. \begin{proposition}\label{total:prop-plateaux} For every partition $c$ of $8$ other than $(8)$, $(7,1)$, $(6,2)$, $(5,3)$, $(5,2,1)$, and $(4,3,1)$, the compactly supported cohomology of $\mathcal T_c$ contains no Tate twist of $S_{12}$ or $S_{12,\ell}$. \end{proposition} \begin{proof} First consider the seven profiles $(4,4)$, $(3,3,2)$, $(3,3,1,1)$, $(2,2,2,2)$, $(2,2,2,1,1)$, $(2,2,1,1,1,1)$, and $(1^8)$. Here $1^r$ denotes $r$ successive parts equal to one. In each case $F_1=F_2=U$, so $U^2\subseteq U$ and $U$ is one of finitely many coordinate-partition algebras. For $(4,4)$, $(2,2,2,2)$, and $(1^8)$ there are no further choices. For $(3,3,2)$, the final space $V$ is an arbitrary eight-plane containing the seven-plane $U$, with choices $\Pp^2$. For $(3,3,1,1)$ and $(2,2,2,1,1)$, the remaining condition is $UV\subseteq V$. Lemma~\ref{total:lem-partition-algebras} gives products of block Grassmannians. For $(2,2,1,1,1,1)$, the condition is $V^2\subseteq V$, so both spaces are partition algebras. All seven profiles therefore have Tate cohomology. The five profiles $(8-r,1^r)$, with $2\le r\le6$, have a first space followed by a hyperplane repeated at least twice. Lemma~\ref{total:lem-pure-hyperplane-tail} gives Tate cohomology for these as well. For $(4,2,2)$, the proper terms are $U,V,V$, of dimensions $6,8,8$. The conditions reduce to $UV\subseteq V$, which also implies $U^2\subseteq V$. Fix $H=\Q[U]$. By Lemma~\ref{total:lem-generated-algebra}, the choices of $U$ generating $H$ and of $H$-stable $V$ containing $H$ are independent and have Tate cohomology. There are finitely many possible $H$. For $(3,2,2,1)$, the proper terms are $U,V,V,W$, of dimensions $7,8,8,9$. The conditions reduce to $UV\subseteq V$ and $V^2\subseteq W$. Again set $H=\Q[U]$. Then $H\subseteq V$, so $h=\dim H$ is seven or eight. For fixed $H$, omit the remaining product equations. The choices are contained in the incidence with $U\in\Gr(6,h-1)$, $V/H\in\Gr(8-h,10-h)$ and $W/V\in\Pp^1$. Its dimension is \[ 6(h-7)+2(8-h)+1 = \begin{cases} 3,&h=7,\\ 7,&h=8. \end{cases} \] The actual locus has dimension at most seven, so Lemma~\ref{lem:dimension-no-S12} applies. Finally, $(4,2,1,1)$ and $(3,2,1,1,1)$ have proper dimensions $(6,8,9,9)$ and $(7,8,9,9,9)$. Apply Lemma~\ref{total:lem-codim-two-hyperplane-tail} with $(N,d,k)=(10,6,2)$ and $(10,7,3)$. Both have constituent bound seven. These groups account for all sixteen profiles in the statement. Localization over their finite decompositions proves the assertion. \par In each decomposition used here, the ranks of the multiplication maps are fixed before taking a Grassmannian or projective bundle. Their images and quotients are therefore vector bundles. The bundle formulas apply on each such stratum. Localization then shows that every simple constituent for the whole profile occurs among those of its strata; it cannot introduce a new simple constituent. Thus the argument excludes the twists in each cohomological degree, as asserted. \end{proof} \begin{proposition}\label{total:prop-all-noncritical} For every partition $c$ of $8$ other than $(6,2)$, the compactly supported cohomology of $\mathcal T_c$ contains no Tate twist of $S_{12}$ or $S_{12,\ell}$. \end{proposition} \begin{proof} For $(8)$, the only proper term is a unital two-plane, with parameter space $\Pp^8$. For $(7,1)$, the proper dimensions are $(3,9)$ and the only condition is $F_1^2\subseteq F_2$. This is $X_{10,3}$, so Lemma~\ref{total:lem-one-hyperplane} applies. Proposition~\ref{total:prop-large-first-space} treats $(5,3)$ and $(4,3,1)$, and Proposition~\ref{total:prop-521} treats $(5,2,1)$. The sixteen remaining noncritical profiles are covered by Proposition~\ref{total:prop-plateaux}. These are all twenty-one partitions other than $(6,2)$. \end{proof} \begin{theorem}\label{total:thm-irreducible-nonpolynomial} Let $I_{10,17}$ be the projective parameter variety over $\Q$ defined in the preceding section. Then \[ \varepsilon_{12}\bigl(\chi_c^\ell(I_{10,17})\bigr)=56. \] Consequently no polynomial in $\Q[T]$ agrees with $g_{10}(p^{17})$ for every sufficiently large prime $p$. \end{theorem} \begin{proof} The loop-rotation identity \eqref{total:eq:epsilon-fixed} and Proposition~\ref{total:prop:fixed-filtrations} give \[ \varepsilon_{12}\bigl(\chi_c^\ell(I_{10,17})\bigr) = \sum_{c\vdash8} \varepsilon_{12}\bigl(\chi_c^\ell(\mathcal T_c)\bigr). \] By Proposition~\ref{total:prop-all-noncritical}, only $c=(6,2)$ can contribute. Proposition~\ref{total:prop:critical-augmentation} gives its contribution as $-B_4(1)=56$. For sufficiently large primes, Lemma~\ref{total:lem:projective-count} gives $g_{10}(p^{17})=\#\mathscr I_{10,17}(\F_p)$. If these counts agreed with a polynomial $P(p)$, the trace criterion of Lemma~\ref{total:lem-trace-criterion} would give $\chi_c^\ell(I_{10,17})=P(\LL)$. The right side is a virtual Tate class and has $\varepsilon_{12}$-value zero, contradicting the value $56$. \end{proof} \section{Lower-rank irreducible counts} \label{total:sec-low-rank} To pass from irreducible subrings to arbitrary subrings, we must also consider irreducible factors of ranks less than ten. We show that, up to colength seventeen, these lower-rank spaces have zero $\varepsilon_{12}$-value. This will ensure that the contribution found in rank ten survives when we sum over all decompositions into irreducible factors. As before, we work with the graded fixed loci. Most of them have dimension at most ten, which excludes every Tate twist of $S_{12}$. Some have larger dimension but are Grassmannian bundles over smaller bases, or are covered by the hyperplane and moment calculations of the preceding section. The estimates below keep track of both the choices of the first space $U$ and the choices of the spaces containing its square. \begin{proposition}\label{total:low-rank-vanishing} For $1\le n\le9$ and $0\le e\le17$, we have \[ \varepsilon_{12}\bigl(\chi_c^\ell(I_{n,e})\bigr)=0. \] For $n\le6$, the same assertion holds for every $e\ge0$. \end{proposition} We first record the dimension estimates used in the proof. Write $U=F_1$, $u=\dim U$, $R=U^2$, and $q=\dim R$. Since $\bfone\in U$, we have $U\subseteq R$. Every term from $F_2$ onward contains $R$. On a constant-square-rank stratum, omitting the remaining multiplication conditions therefore gives a partial flag bundle in $\Q^n/R$. If the distinct dimensions in this quotient are $d_1<\cdots0$. For $n\ge2$, scalar reduction gives a surjection $\cO^n/L\to k^n/k\bfone$ for every irreducible lattice $L$. Hence its colength is at least $n-1$, and $I_{n,e}$ is empty if $e