\documentclass[11pt]{amsart} \usepackage[T1]{fontenc} \usepackage{lmodern} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage[letterpaper,margin=1.02in]{geometry} \usepackage{microtype,enumitem,booktabs,longtable} \usepackage[dvipsnames]{xcolor} \usepackage[colorlinks=true,linkcolor=MidnightBlue,citecolor=MidnightBlue,urlcolor=MidnightBlue,bookmarksnumbered=true]{hyperref} \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \DeclareMathOperator{\rank}{rank} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\Sym}{Sym} \DeclareMathOperator{\Span}{span} \DeclareMathOperator{\coker}{coker} \DeclareMathOperator{\Gr}{Gr} \DeclareMathOperator{\codim}{codim} \DeclareMathOperator{\SL}{SL} \setlist[enumerate]{label=\textup{(\roman*)},itemsep=3pt,topsep=5pt} \emergencystretch=2em \hypersetup{pdftitle={Froberg's conjecture through degree 2d in sufficiently many variables},pdfauthor={Vlado Vojdanovski},pdfsubject={Generic forms, Koszul homology, and Hilbert functions}} \title[Fr\"oberg's conjecture through degree $2d$]{Fr\"oberg's conjecture through degree $2d$\\in sufficiently many variables} \author{Vlado Vojdanovski} \date{} \begin{document} \begin{abstract} Let $k$ be any infinite field. We prove that, for every fixed $d\geq1$ and all sufficiently large $n$, the quotient of $k[x_1,\ldots,x_n]$ by any number of general forms of degree $d$ has the Hilbert function predicted by Fr\"oberg's conjecture through degree $2d$. The threshold for $n$ is independent of the field and the number of generators. For generating degrees at least three, a direct construction with two groups of variables gives a nonincreasing recurrence for the two defects at the critical generator counts. Divisibility of these bounded defects and polynomial equidistribution then force their eventual vanishing. \end{abstract} \maketitle \section{Introduction} Let $k$ be an infinite field, let $S=k[x_1,\ldots,x_n]$ with its standard grading, and let $I=(f_1,\ldots,f_r)$, where the $f_i$ are general forms of the same degree $d$. Fr\"oberg's conjecture predicts \begin{equation}\label{intro:series} \operatorname{HS}_{S/I}(t)= \left[\frac{(1-t^d)^r}{(1-t)^n}\right]_+. \end{equation} Here $[\sum_{j\geq0}a_jt^j]_+$ retains the coefficients preceding the first nonpositive coefficient and replaces that coefficient and all subsequent coefficients by zero. The conjecture originates in~\cite{froberg1985}. We write $[t^j]F(t)$ for the coefficient of $t^j$ in a power series $F(t)$. A statement about general forms means that it holds on a nonempty Zariski open subset of their coefficient space. Hochster and Laksov proved the expected behavior in degree $d+1$ \cite{hochsterlaksov1987}. More recently, Boij, Dannetun and Lundqvist proved that, for each fixed $d>1$ and sufficiently large $n$, the prediction holds through degree $2d-1$, uniformly in the number of generators~\cite[Theorem~2]{boijdannetunlundqvist2026}. At the next degree, the products $f_if_j=f_jf_i$ give the first Koszul relations. The expected coefficient is therefore \[ \binom{n+2d-1}{2d}-r\binom{n+d-1}{d}+\binom r2, \] provided that positive truncation has not already occurred. The purpose of this paper is to treat this endpoint by a construction that retains these relations throughout the proof. \begin{theorem}\label{intro:main} For every integer $d\geq1$ there is an integer $n_0(d)$ such that, for every infinite field $k$, every $n\geq n_0(d)$ and every $r\geq0$, the quotient by $r$ general degree-$d$ forms in $k[x_1,\ldots,x_n]$ has Hilbert function \[ \dim_k(S/I)_j=[t^j]\left[\frac{(1-t^d)^r}{(1-t)^n}\right]_+ \qquad(0\leq j\leq2d). \] \end{theorem} \subsection*{Outline of the proof} The proof has three parts: a reduction to two generator counts, a comparison between different numbers of variables, and an arithmetic argument that turns this comparison into vanishing. The quadratic case is treated separately in Appendix~\ref{low:section}; the same comparison construction then works for every $d\geq3$. The main step compares the two critical defects in $m$ variables with those in $m+h$ variables, for a fixed $h=h(d)$. Write $H_n(r)$ for the dimension of the degree-$2d$ first Koszul homology and $C_n(r)$ for the degree-$2d$ quotient dimension. If $r_n^-,r_n^+$ are the integers on either side of the smaller zero of the expected coefficient, put \[ E_d(n)=\max\{H_n(r_n^-),C_n(r_n^+)\}. \] Monotonicity reduces the endpoint assertion for every number of generators to $E_d(n)=0$. For $d\geq3$ we prove the recurrence \begin{equation}\label{intro:recurrence} E_d(m+h)\leq E_d(m) \end{equation} by splitting the variables into groups of sizes $h$ and $m$. Monomial matchings supply spaces with independent symmetric products. A hyperplane replacement reduces the homology from the smaller ring exactly, and a final scalar perturbation gives the comparison of defects. No vanishing of the smaller-ring defects is assumed in this step. The recurrence bounds the defects. If $g$ is the largest divisor of $\gcd(n,dr)$ prime to the characteristic, with $g=\gcd(n,dr)$ in characteristic zero, then \[ g\mid 2dH_n(r),\qquad g\mid 2dC_n(r). \] These divisibilities and polynomial equidistribution force eventual vanishing, uniformly in the characteristic. The main results that carry the proof are Proposition~\ref{uf:two-counts} (the two-count reduction), Theorem~\ref{normal:transfer} (the comparison inequalities), and Theorem~\ref{uc:eventual-zero} (eventual vanishing). Section~\ref{complete:section} combines them with the quadratic endpoint, Corollary~\ref{low:quadratic-theorem}, to prove the theorem. The comparison is presented first using four explicit properties of the constructed family. Appendices~\ref{appendix:multiplication} and~\ref{normal:section} establish these properties; the product spaces they use are constructed in Appendix~\ref{unified:products}. Appendix~\ref{unified:section} supplies the numerical estimates for degrees three through eight. \subsection*{Notation and conventions} Throughout, $k$ is an infinite field and all vector spaces are over $k$. Rank conditions may be checked over an algebraic closure: their nonvanishing minors define nonempty opens over $k$, which have $k$-rational points because $k$ is infinite. The theorem of Boij, Dannetun and Lundqvist supplies the required multiplication statements below twice the relevant generating degree. The quadratic endpoint is proved in Appendix~\ref{low:section}. We write $S(V)=\operatorname{Sym}(V)$ and $S_aV=\operatorname{Sym}^aV$, with $S_aV=0$ for $a<0$. For subspaces of forms, $AL$ denotes the span of the products $a\ell$, $a\in A$, $\ell\in L$; this multiplication image is also called the shadow of $L$ under $A$. \section{The endpoint and the two critical counts}\label{uf:section} Fix an infinite field $k$ and an integer $d\geq2$. Write $S=\operatorname{Sym}(k^n)$ and \[ N_t(n)=\dim_k S_t=\binom{n+t-1}{t}. \] For $0\leq r\leq N_d(n)$ let $W\subset S_d$ be a general $r$-dimensional subspace. The two dimensions relevant to the endpoint are \begin{equation}\label{uf:defects} H_n(r)=\dim_k H_1(W;S)_{2d},\qquad C_n(r)=\dim_k(S/(W))_{2d}. \end{equation} Here ``general'' means that the multiplication matrix has its largest rank; this is a nonempty Zariski open condition on the Grassmannian. The notation records these generic dimensions, rather than the dimensions at an arbitrary special choice of $W$. In degree $2d$, the Koszul complex on $W$ has the form \begin{equation}\label{uf:complex} 0\longrightarrow\bigwedge^2W \xrightarrow{\delta_2} W\otimes_k S_d \xrightarrow{\mu_W} S_{2d}\longrightarrow0, \qquad \delta_2(f\wedge g)=f\otimes g-g\otimes f. \end{equation} The first map is injective, and there is no preceding term because its coefficient degree would be $-d$. Consequently, with \begin{equation}\label{uf:chi} \chi_n(r)=N_{2d}(n)-rN_d(n)+\binom r2, \end{equation} the Euler characteristic gives \begin{equation}\label{uf:euler} C_n(r)-H_n(r)=\chi_n(r). \end{equation} Thus maximal rank after removing the constant Koszul relations is equivalent to \[ C_n(r)=\max\{\chi_n(r),0\},\qquad H_n(r)=\max\{-\chi_n(r),0\}. \] The monotonicity needed below concerns formal symmetric products. We distinguish them from their polynomial images: if $V=S_d$, set \[ W\mathbin{\odot}V =\operatorname{im}\bigl(W\otimes V\longrightarrow \operatorname{Sym}^2V\bigr). \] The map $\operatorname{Sym}^2V\to S_{2d}$ in the following lemma is ordinary polynomial multiplication. \begin{lemma}\label{uf:monotonicity} For an inclusion $W\subseteq W'\subseteq S_d$, there is a natural injection \[ H_1(W;S)_{2d}\longrightarrow H_1(W';S)_{2d}. \] In particular, $H_n(r)$ is nondecreasing in $r$, and $C_n(r)$ is nonincreasing in $r$. \end{lemma} \begin{proof} Choose a vector-space complement $V=W\oplus U$. Under this decomposition, the kernel of $W\otimes V\to\operatorname{Sym}^2V$ is exactly $\bigwedge^2W$: its restriction to $W\otimes W$ has that kernel, and its restriction to $W\otimes U$ is injective into the separate summand $W\otimes U$ of $\operatorname{Sym}^2V$. Hence \begin{equation}\label{uf:formal-homology} H_1(W;S)_{2d} \simeq (W\mathbin{\odot}V)\cap \ker\bigl(\operatorname{Sym}^2V\longrightarrow S_{2d}\bigr). \end{equation} The asserted injection follows from $W\mathbin{\odot}V\subseteq W'\mathbin{\odot}V$. The quotient map $S/(W)\twoheadrightarrow S/(W')$ gives the opposite monotonicity for the cokernel. To pass from particular subspaces to generic dimensions, use the irreducible flag variety of pairs of the prescribed dimensions. Both of its projections to the corresponding Grassmannians are dominant. The inverse images of the two nonempty rank opens therefore meet, so the same inequalities hold for their generic values. \end{proof} The smaller real zero of $\chi_n$ is \begin{equation}\label{uf:kappa} \kappa_d(n)=N_d(n)+\frac12- \sqrt{\left(N_d(n)+\frac12\right)^2-2N_{2d}(n)}. \end{equation} It lies in $(0,N_d(n)]$. Indeed, polynomial multiplication $\operatorname{Sym}^2S_d\to S_{2d}$ is surjective, since every monomial of degree $2d$ is a product of two degree-$d$ monomials. Thus $\chi_n(N_d(n))\leq0$, whereas $\chi_n(0)>0$; moreover, $\chi_n$ is strictly decreasing on $[0,N_d(n)]$. Define \begin{equation}\label{uf:endpoint-counts} r_n^- =\lfloor\kappa_d(n)\rfloor,\qquad r_n^+ =\lceil\kappa_d(n)\rceil,\qquad E_d(n)=\max\{H_n(r_n^-),C_n(r_n^+)\}. \end{equation} These definitions also cover an integral root, in which case the two counts coincide. \begin{proposition}\label{uf:two-counts} If $E_d(n)=0$, then \[ C_n(r)=\max\{\chi_n(r),0\} \qquad(0\leq r\leq N_d(n)). \] For fixed $d$ and sufficiently large $n$, this is exactly the degree $2d$ coefficient prescribed by the positive truncation of $(1-t^d)^r/(1-t)^n$. For $r\geq N_d(n)$, the quotient by general degree-$d$ forms is already zero in every degree at least $d$. \end{proposition} \begin{proof} If $r\leq r_n^-$, Lemma~\ref{uf:monotonicity} gives $H_n(r)=0$, so \eqref{uf:euler} gives $C_n(r)=\chi_n(r)\geq0$. If $r\geq r_n^+$, it gives $C_n(r)=0$. There are no integer counts strictly between $r_n^-$ and $r_n^+$. At an integral root, \eqref{uf:euler} shows that the two endpoint defects are equal, so the same argument applies. For clarity, we also check the effect of positive truncation. Put \[ H=\binom{2d-1}{d-1},\qquad \rho_d=1-\sqrt{1-\frac1H}. \] The leading terms in \eqref{uf:kappa} give \begin{equation}\label{uf:critical-asymptotics} \kappa_d(n)=\frac{\rho_d}{d!}n^d+O(n^{d-1}),\qquad \frac{N_{2d-1}(n)}{N_{d-1}(n)} =\frac{1}{H d!}n^d+O(n^{d-1}). \end{equation} Since $0<\rho_d<1/H$, the first quantity is strictly smaller than the second for all sufficiently large $n$. For $n>1$, the ratio \[ \frac{N_t(n)}{N_{t-d}(n)} =\prod_{i=0}^{d-1}\frac{n+t-d+i}{t-d+1+i} \] decreases as $t$ runs from $d$ to $2d-1$. Thus if $r\leq r_n^-$, every coefficient $N_t(n)-rN_{t-d}(n)$ with $d\leq t<2d$ is positive. No earlier truncation occurs, and the coefficient in degree $2d$ is $\chi_n(r)$. If $r\geq r_n^+$, either truncation has occurred earlier, or the nonpositive coefficient $\chi_n(r)$ causes it in degree $2d$. In both cases the prescribed coefficient is zero. Finally, $N_d(n)$ general forms span $S_d$, which proves the last assertion, including all larger counts. \end{proof} It therefore remains to make the single nonnegative integer $E_d(n)$ vanish eventually. The geometric construction will give a fixed step $h$ for which $E_d(n+h)\leq E_d(n)$ for all sufficiently large $n$. The arithmetic argument in Section~\ref{uc:section} shows why that recurrence suffices, without any separately prescribed initial value of zero. \section{An exact hyperplane replacement}\label{core:hyperplane} The formal-product description of homology permits an exact replacement of the homology contributed by a subspace of generators. The situation to keep in mind is that $A$ consists of forms in a smaller polynomial ring and $Q\subset A$ is the family already chosen there. We replace one generator $f_*$ by $f_*+\epsilon M$, with $M$ outside $W+A$. The lemma shows that this removes precisely the relations involving $f_*$: the remaining homology is the homology contributed by $Q_-$. \begin{lemma}[Hyperplane replacement]\label{normal:child-repair} \label{normal:formal-intersection} Let $T$ be a commutative graded algebra with $T_0=k$, put $V_T=T_d$ and $\mathcal K_T=\ker(S_2(V_T)\to T_{2d})$, and suppose $W\cap A=Q=Q_-\oplus kf_*$ for subspaces $W,A\subset V_T$. Products denoted by $\odot$ are formal products in $S_2(V_T)$. Suppose $M\notin W+A$ and, under the natural formal-product identification, \begin{equation}\label{normal:enlarged-homology} H_1(W+kM;T)_{2d}=\mathcal K_T\cap(Q\odot A). \end{equation} Choose $W=W_0\oplus kf_*$ with $Q_-\subset W_0$. For every $\epsilon\ne0$, the space $W_\epsilon=W_0+k(f_*+\epsilon M)$ has $\dim W_\epsilon=\dim W$ and \begin{equation}\label{normal:hyperplane-homology} H_1(W_\epsilon;T)_{2d}=\mathcal K_T\cap(Q_-\odot A). \end{equation} \end{lemma} \begin{proof} The complement argument in Lemma~\ref{uf:monotonicity} applies unchanged to $T$ and gives the natural identification \begin{equation}\label{normal:formal-T} H_1(L;T)_{2d}=\mathcal K_T\cap(L\odot V_T). \end{equation} For any $L\subset V_T$, one also has \begin{equation}\label{normal:intersection-T} (L\odot V_T)\cap S_2(A)=(L\cap A)\odot A. \end{equation} Indeed, $L\odot V_T$ is the kernel of $S_2(V_T)\to S_2(V_T/L)$, and $A/(A\cap L)\hookrightarrow V_T/L$ stays injective on symmetric squares, as is seen by extending a basis. The condition on $M$ gives $W_\epsilon\cap A=Q_-$ and preserves the dimension of $W$: a nonzero coefficient of $f_*+\epsilon M$ in an element of $A$ would put $M$ in $W+A$. Inclusion in $W+kM$ places $H_1(W_\epsilon;T)_{2d}$ in $\mathcal K_T\cap S_2(A)$ by \eqref{normal:enlarged-homology}. Intersecting with $W_\epsilon\odot V_T$ and using \eqref{normal:intersection-T} proves one containment in \eqref{normal:hyperplane-homology}; $Q_-\subset W_\epsilon$ proves the other. \end{proof} \section{Spaces with independent symmetric products}\label{core:products} We first construct a large space of forms whose symmetric products are independent and have a single bidegree. The argument uses only matchings between products and monomials. \begin{lemma}\label{products:matching} Let $s\geq1$, $w\geq s$, and $a=\lfloor s/2\rfloor$. In the paired variables $x_1,\ldots,x_w,y_1,\ldots,y_w$, there is a space \[ C\subset S_a\langle x_1,\ldots,x_w\rangle \otimes S_{s-a}\langle y_1,\ldots,y_w\rangle, \qquad \dim C=\binom ws, \] whose symmetric multiplication $S_2C\to S_{2s}(k^{2w})$ is injective. In particular, $\dim C=N_s(w)-O_s(w^{s-1})$. \end{lemma} \begin{proof} For each $s$-subset $S\subset[w]$, choose a form \begin{equation}\label{products:forms} f_S=\sum_{\substack{A\subset S\\|A|=a}} c_{S,A}x_Ay_{S\setminus A}, \qquad x_A=\prod_{i\in A}x_i, \end{equation} with independent coefficient parameters. Give both $x_i$ and $y_i$ index degree $e_i\in\mathbb Z^w$. Products with different index degrees are independent. Fix one product degree, specified by a set $I$ of doubled indices and a disjoint set $R$ of single indices. Write $i=|I|$ and $t=s-i$, so $|R|=2t$. For $t>0$, its source columns are the unordered partitions $R=P\sqcup Q$ with $|P|=|Q|=t$, representing $f_{I\cup P}f_{I\cup Q}$. Each form in this fiber has a unique partner. Thus distinct columns use disjoint sets of coefficient parameters. Suppose first that $s$ is even. Use target monomials with $x_jy_j$ on $I$, with $x_j$ on a $t$-subset $D\subset R$, and with $y_j$ on $R\setminus D$. Such a target factors into an allowed term of each source form exactly when \[ a-i\leq |D\cap P|\leq a. \] Indeed, the first factor must receive $a-|D\cap P|$ of the $x$ variables on $I$, and the second receives the rest. If $s$ is odd and $i\geq1$, choose one fixed index of $I$ to carry $y_j^2$, let the others carry $x_jy_j$, and use the same $t$-subsets $D$. The condition is now \[ a-i+1\leq |D\cap P|\leq a. \] In either case the interval contains an integer nearest $t/2$. When its endpoints coincide, $t$ is even. The graph joining source partitions to the allowed target subsets is therefore nonempty. Permutations of $R$ act transitively on each vertex class and preserve adjacency. The graph is biregular, and its target class has twice as many vertices as its source class. Counting edges from any set of source vertices proves Hall's inequality. There is consequently a matching covering every source column. If $s$ is odd and $i=0$, use instead the $(s-1)$-subsets $D\subset R$, with $x$ on $D$ and $y$ elsewhere. Adjacency means $|D\cap P|=|D\cap Q|=a$. The same transitivity and edge-counting argument applies, with target-to-source ratio \[ \frac{\binom{2s}{s-1}}{\tfrac12\binom{2s}{s}} =\frac{2s}{s+1}\geq1. \] If $t=0$, the fiber has the single column $f_I^2$, and any one allowed monomial gives a nonzero entry. For a matching in a fiber with $t>0$, specialize the two forms in each column to the factors of its matched monomial. Unique partners make these choices consistent. The selected square submatrix is a permutation matrix, so its determinant is a nonzero polynomial in the parameters. For $t=0$, specialize the one form to one monomial. There are finitely many fibers, and the product of their nonzero determinants is nonzero. Choose parameters in this common open set. The forms $f_S$ themselves have distinct index degrees, so they are independent and span the required space. \end{proof} The degrees three through eight use two larger spaces in the fixed group of variables. \begin{proposition}\label{core:enhanced-products} Let $k$ be any infinite field. For $s\in\{2,4\}$ and $w\geq s$, there is a space $C_s\subset S_s(k^{2w})$ of dimension $2\binom ws$ whose symmetric multiplication map is injective. In particular, there are spaces \[ H_2\subset S_2(k^h),\qquad \dim H_2=\lfloor5h^2/32\rfloor \quad(h\geq65), \] and \[ H_4\subset S_4(k^h),\qquad \dim H_4=\lfloor h^4/256\rfloor \quad(h\geq144,\ 4\mid h), \] with injective symmetric multiplication. \end{proposition} \begin{proof} Use two general multilinear forms on every $s$-element set of variable pairs. Lemmas~\ref{products:quadratic} and~\ref{products:quartic} prove product independence by three quadratic and five quartic intersection patterns, and verify the stated dimension bounds. The full spaces have dimensions $h^2/4+O(h)$ and $h^4/192+O(h^3)$, respectively, for $w=\lfloor h/2\rfloor$. \end{proof} The space in Lemma~\ref{products:matching} has one product bidegree and dimension $N_s(w)-O_s(w^{s-1})$. These are the two properties used in the multiplication estimates. All the capacity inequalities there are strict, so their lower-order terms are absorbed by first choosing the fixed block large and then increasing $m$. \section{The construction and its generator counts} \label{construct:parameters} We parametrize each family by an ordered list of independent forms; all assertions depend only on its span. The choices below are made on nonempty open subsets of the corresponding parameter spaces. The construction takes place in $S(X\oplus Y)$, where $\dim X=h$ is fixed before $\dim Y=m$ tends to infinity. We use bidegrees with respect to $(X,Y)$. A row of $X$-degree $b$ is the component $S_bX\otimes S_{2d-b}Y$ of the degree-$2d$ target. The following families will be chosen so that successive bidegrees of the multiplication map can be treated separately: \[ \begin{array}{c|c|l} \text{family}&\text{bidegree}&\text{role}\\ \hline Q&(0,d)&\text{forms in the smaller polynomial ring}\\ F&(1,d-1)&\text{comparison of the two defects}\\ E_j&(j,d-j)&\text{intermediate multiplication maps}\\ U&(d,0)&\text{the highest target bidegrees}. \end{array} \] We call $F$ the outer family and the $E_j$ the intermediate families; only even $j$ occur. First one chooses $Q,E_j,U$, with suitable lower-bidegree terms, so that their first homology is exactly the homology contributed by $Q$. Replacing one member of $Q$ then reduces this to the homology at the smaller generator count. The forms $F$ are adjoined and varied by adding forms in $S_dY$. The induced map from homology to cokernel is the normal map studied in Appendix~\ref{normal:section}; its rank gives the comparison of defects. All auxiliary forms are chosen within the exact count $r=q+f+\sum_j e_j+u$; the count is proved below. Fix $d\geq3$. The following single count construction is used throughout the comparison. We first specify fixed constants depending on $d$, then choose the size $h$ of the first block, and finally choose the integer counts as the second block size $m$ grows. This order of choices ensures that rounding does not affect any of the strict dimension inequalities. Write \[ H=\binom{2d-1}{d-1},\qquad \rho=\rho_d=1-\sqrt{1-H^{-1}},\qquad \rho_2=1-\sqrt{2/3}. \] \[ \mathcal J=\begin{cases} \{2,4\}\cap\{1,\ldots,d-1\},&3\leq d\leq8,\\ \{2,4,\ldots,2\lceil d/4\rceil\},&d\geq9. \end{cases} \] Put \[ \tau_2=\frac9{32(d-2)!\binom{2d-2}{d-2}},\quad \tau_3=\frac1{6(d-2)!\binom{2d-3}{d-2}},\quad \tau_4=\frac{\rho_2}{2(d-2)!\binom{2d-4}{d-2}}, \] and define \begin{equation}\label{unified:A} \gamma_2=\begin{cases} \max\{\tau_2,\tau_3,\tau_4\},&3\leq d\leq8,\\ \tau_4,&d\geq9. \end{cases} \end{equation} For the other active indices set \begin{align*} \gamma_4&=\frac{(d-1)!}{120(2d-5)!},\\ \gamma_j&=\frac{(j-4)!(d-j+4)!}{(2j-4)!(2d-2j+4)!} &&(j\in\mathcal J,\ j\geq6). \end{align*} Set \begin{equation}\label{unified:densities} \alpha=(101/100)\gamma_2,\qquad \beta=(1011/1000)\gamma_2. \end{equation} Choose a sufficiently large integer $h$, divisible by $180180$ if $d\leq8$ and by $4H$ otherwise, put $X=k^h$, and let $m=\dim Y$ tend to infinity with $h$ fixed. In particular, constants implicit in $O_h(\cdot)$ may depend on this already fixed $h$. Put \[ K=h/H,\qquad L=\left\lceil\frac{11}{10}2^d\rho h\right\rceil, \qquad c=\binom{L+1}{2}. \] The choice of $h$ can and will satisfy \begin{equation}\label{construct:detector-size} L\leq h,\qquad c\geq2h-1,\qquad cd(d-1)\rho/(2d!)$. This follows from $\rho\binom{2d-4}{d-2}<\rho_2$; indeed $H\rho<13/25$ and $\binom{2d-4}{d-2}/H<1/4$, whereas $\rho_2>2/11$. Consequently \eqref{construct:reserve} is uniformly positive for our whole interval once $h$ is large. Additional scalar columns cost $O_h(m^{2d-3})$. Adding a fixed number $v$ of outer columns changes $T-qA$ by $-v(N_d(m)-q)=O_h(m^d)$, also lower order. \end{proof} \section{The local comparison of defects}\label{normal:repair-section} Fix $d\geq3$, using the quadratic endpoint of Corollary~\ref{low:quadratic-theorem}, and take the parameters of Section~\ref{construct:parameters}. The purpose of this section is to deduce the comparison inequalities from four properties of a family of forms. The appendices establish those properties for the counts already chosen, so the argument here can be read independently of their dimension estimates. The construction uses forms \[ Q\subset S_dY,\qquad F\subset X\otimes S_{d-1}Y,\qquad E_j\subset S_jX\otimes S_{d-j}Y, \] with the prescribed dimensions $q,f,e_j$. The quadratic outputs of $E_2$ lie in $D$. Write $a_j$ for scalar forms in $S_dY$ with the same labels as $E_j$. The background, before adjoining $F$, is \begin{equation}\label{core:background} W=\begin{cases} \langle Q,E_j+a_j,U+P\rangle,&d\text{ odd},\\ \langle Q,E_j+a_j,U+P_2+b\rangle,&d\text{ even}. \end{cases} \end{equation} Here $U\subset S_dX$ has dimension $u$; $P$ is a list in $X\otimes S_{d-1}Y$, while $P_2$ and $b$ are lists in $D\otimes S_{d-2}Y$ and $S_dY$. Each tail has the labels of $U$. The terms ``even'' and ``odd'' refer to total $X$-degree modulo two. Every background generator is homogeneous for this parity grading. The families $Q$ and $E_j+a_j$ are even; $U+P$ is odd when $d$ is odd, whereas $U+P_2+b$ is even when $d$ is even. The family $F$ is odd. The final deformation adds forms in $S_dY$ to the generators of $F$. For a varying multiplication map $\mu_t$, its first-order effect on a relation in $\ker\mu_0$, taken modulo $\operatorname{im}\mu_0$, defines a map from the kernel to the cokernel. We call this the \emph{normal map}. In the present construction it descends through the Koszul relations. Its role is to remove any excess homology and cokernel left after the background has been chosen. We use the following four properties of this family. \begin{enumerate} \item The background, also with one additional generator $M=M_2+a_M$ of the quadratic mixed type, has exactly the naturally included homology of $Q$. This remains true after quotienting the degree-$2d$ target by a complement to $QS_dY$ in $S_{2d}Y$. \item The symmetric products of $F$ are independent modulo the background ideal. The construction allows one fixed scalar family containing every scalar tail, including $a_M$. \item After adjoining $F$ and projecting as in (i), the even target is filled and the odd first homology is zero. The remaining homology modulo that of $Q$ injects into the space of $F$ coefficients. \item The scalar-contraction maps on the odd cokernel satisfy the covector-stratum bound of Proposition~\ref{normal:strata}. Consequently, if the homology in (iii), modulo the included homology of $Q$, has dimension $v$ and the odd cokernel has dimension $j$, general scalar motions of the $F$ generators induce a normal map of rank $\min(v,j)$. \end{enumerate} Here (i) is Proposition~\ref{construct:background}; (ii) is Proposition~\ref{normal:detector}; (iii) follows from Propositions~\ref{construct:targets} and~\ref{normal:rank}; and (iv) is the covector-incidence calculation in Propositions~\ref{normal:strata} and~\ref{normal:rank}. Appendices~\ref{appendix:multiplication} and~\ref{normal:section} prove these assertions with the parameter bounds for $d\geq9$. Appendix~\ref{unified:section} verifies the same hypotheses for $3\leq d\leq8$. The comparison below uses just these four properties. \subsection{Replacing one generator} There are two operations in this step. We first remove a complement to the smaller ring's multiplication image from the target. This makes its cokernel zero without changing its relations. We then apply the hyperplane replacement to pass from the homology of $Q$ to that of its codimension-one subspace $Q_-$. Put $q=\lceil\kappa_d(m)\rceil$, $c_0=C_m(q)$ and $k_0=H_m(q-1)$. Choose a general flag $Q_-\subset Q=Q_-\oplus kf_*$ realizing these generic dimensions, and a complement $Z_0$ to $QS_d(Y)$ in $S_{2d}(Y)$. Set \begin{equation}\label{normal:projected-ring} T=S(X\oplus Y)/(Z_0),\qquad A=S_d(Y)\subset T_d. \end{equation} This leaves degree $d$ unchanged and removes $c_0$ target dimensions. Since $Z_0\cap QS_d(Y)=0$, it changes neither the homology of $Q$ nor that of $Q_-$ with coefficients in $S(Y)$. Let $W$ be the prepared background and take the additional generator allowed by Proposition~\ref{construct:background}, written $M=M_2+a_M$ with $M_2\in D\otimes S_{d-2}(Y)$ and $a_M\in S_d(Y)$. Its high component may be chosen independent of those of $E_2$. The positive highest-$X$-degree components of the other families are independent degree by degree. Projecting successively to those degrees gives \begin{equation}\label{normal:independent-high-components} W\cap A=Q,\qquad M\notin W+A. \end{equation} In particular, the independent degree-$d$ components first eliminate the pure generators, including any quadratic tails they carry. The background theorem, with its extra-generator and projected versions, gives \eqref{normal:enlarged-homology}, since $H_1(Q;S(Y))_{2d}=\mathcal K_T\cap(Q\odot A)$. Take $W_0$ to consist of $Q_-$ and the non-$Q$ background columns. Lemma~\ref{normal:child-repair} then gives, for every $\epsilon\ne0$, \begin{equation}\label{normal:repaired-background} H_1(W_\epsilon;T)_{2d}=H_1(Q_-;S(Y))_{2d}, \qquad \dim W_\epsilon=\dim W. \end{equation} \begin{lemma}[Compatibility with the final normal map] \label{normal:mark-compatibility} For a nonempty open set of nonzero $\epsilon$, adjoining $F$ to $W_\epsilon$ gives the product injection of Proposition~\ref{normal:detector}, zero odd first homology, the covector bound \eqref{normal:final-strata}, and surjectivity onto the even degree-$2d$ target of $T$. The final scalar motion of $F$ then has the maximal normal rank of Proposition~\ref{normal:rank}. \end{lemma} \begin{proof} We preserve the four properties in two ways. Product separation is proved in one quotient that works for every replacement parameter; the remaining rank and covector conditions persist on an open neighborhood of the original family. Before choosing $\epsilon$, include $a_M$ in the fixed scalar family $\mathcal S^+=\mathcal S\cup\{a_M\}$ used for the product-separation map. This single extra column is allowed by the strict estimates. Its product-injection open meets the earlier background, covector and target opens in their common parameter space, so choose the data there. The product injection now holds for every $\epsilon$: in $X$-degree two, $\epsilon M_2$ is killed by the quotient by $D$, and $f_*+\epsilon a_M$ belongs to the fixed span of $\mathcal S^+$. In odd degree the same fixed $P$ family is included in the denominator. The projection by $Z_0$ affects only $X$-degree zero. Thus this argument requires no constancy of the even background rank. The replacement is even in $X$, so the odd degree-$d$ coefficient space $V$ in \eqref{normal:even-source} or \eqref{normal:odd-source} is fixed. At $\epsilon=0$ odd homology after adjoining $F$ is zero. While the generators remain independent, the mandatory odd Koszul image has constant dimension. The odd multiplication map therefore has constant maximal rank on an open neighborhood of zero, and its cokernels form a vector bundle $\mathcal J$. Scalar contraction with source $V$ is regular on $\mathbb P(\mathcal J^*)$. For each $1\leq k\leq\dim V$, the condition $\dim\ker C_\lambda\geq k$ is closed and its projection to the parameter line is proper. Upper semicontinuity of fiber dimensions preserves \eqref{normal:final-strata} on an open neighborhood of zero. Intersecting these finitely many neighborhoods preserves every stratum. At zero, Proposition~\ref{construct:targets} fills the positive even rows by decreasing-$X$-degree elimination, and $Q$ fills the residual pure-$Y$ target after projection by $Z_0$. Surjectivity onto the full even target is open, so it persists as well. Choose a nonzero $\epsilon$ in these neighborhoods, which is possible over the infinite field $k$. Equation~\eqref{normal:repaired-background} supplies the exact remaining background homology. Product detection then gives the faithful coefficient map \eqref{normal:coefficient-inclusion}, and the preserved covector bounds give maximal normal rank by Proposition~\ref{normal:rank}. \end{proof} \subsection{The two inequalities} The remaining calculation compares the homology and cokernel after replacement. Maximal rank of the normal map cancels their excess dimensions against one another; the Euler identity determines which one can remain. \begin{theorem}[Transfer inequalities]\label{normal:transfer} For either exact count $r=\lfloor\kappa_d(m+h)\rfloor$ or $r=\lceil\kappa_d(m+h)\rceil$, put $q=\lceil\kappa_d(m)\rceil$, $c_0=C_m(q)$ and $k_0=H_m(q-1)$. The constructed family gives \begin{align} H_{m+h}(r)&\leq\max\{k_0,c_0-\chi_{m+h}(r)\}, \label{normal:transfer-H}\\ C_{m+h}(r)&\leq\max\{c_0,k_0+\chi_{m+h}(r)\}. \label{normal:transfer-C} \end{align} \end{theorem} \begin{proof} Choose the nonzero replacement parameter of Lemma~\ref{normal:mark-compatibility} and adjoin $F$. In $T$ the even target is filled and odd homology is zero. Write $k_0+a_0$ for the even homology dimension, distinguishing the naturally included homology of $Q_-$, and $j_0$ for the odd cokernel dimension. Removing the $c_0$ dimensions of $Z_0$ from the target gives \begin{equation}\label{normal:transfer-euler} \chi_{m+h}(r)=c_0-k_0+j_0-a_0. \end{equation} The final scalar motion of $F$ has first normal rank $\min(a_0,j_0)$ and keeps $Q_-$ fixed. Lemma~\ref{normal:rank-lifting} bounds the resulting homology in $T$ by \[ k_0+\max\{a_0-j_0,0\}=\max\{k_0,c_0-\chi_{m+h}(r)\}. \] Restoring the unprojected target cannot increase the kernel. These rank minors therefore bound the homology of general independent $r$-tuples by the same number, proving \eqref{normal:transfer-H}. The Euler identity \eqref{uf:euler} gives \eqref{normal:transfer-C}. The replacement preserves the number of generators, so the total remains exactly $r$. The endpoint inputs are those of Proposition~\ref{construct:targets}; the transfer argument applies whenever the background and its one-generator extension, target maps, product injection and covector estimate hold. \end{proof} The transfer inequalities \eqref{normal:transfer-H}--\eqref{normal:transfer-C} have the same form for both critical counts. Since $q-1\leq r_m^-$, even when $\kappa_d(m)$ is integral, monotonicity gives $k_0\leq E_d(m)$ and $c_0\leq E_d(m)$. At $r^-_{m+h}$ the Euler characteristic is nonnegative, and at $r^+_{m+h}$ it is nonpositive. The two inequalities therefore give \begin{equation}\label{uc:recurrence} E_d(m+h)\leq E_d(m) \qquad\text{for every sufficiently large }m. \end{equation} In characteristic different from two, the quadratic construction in Appendix~\ref{low:section} gives the same recurrence with $h=3$. The characteristic-two quadratic input is established there by a separate induction. The arithmetic argument below applies to every field for which the indicated recurrence has been proved. \section{The recurrence and eventual vanishing}\label{uc:section} The recurrence makes the two critical defects bounded. We combine equivariant divisibility with polynomial equidistribution to show that they vanish, using a common threshold over all fields to which the recurrence applies uniformly. The argument is as follows. A symmetry of the multiplication complex forces each defect to be divisible by certain factors of $\gcd(n,dr)$. Equidistribution supplies a dimension in every residue class modulo $h$ where these factors are larger than the bound on the defects. Both defects must vanish there, and the recurrence propagates vanishing to all later dimensions in that residue class. \begin{lemma}[Equivariant divisibility]\label{uc:divisibility} For $d\geq1$, $n\geq1$, and $0\leq r\leq N_d(n)$, let $g$ be the largest divisor of $\gcd(n,dr)$ prime to $\operatorname{char}k$; in characteristic zero take $g=\gcd(n,dr)$. The generic dimensions in \eqref{uf:defects} satisfy \begin{equation}\label{uc:divisibility-equations} g\mid 2d\,H_n(r),\qquad g\mid 2d\,C_n(r). \end{equation} \end{lemma} \begin{proof} Extension to an algebraic closure changes none of these dimensions, so assume $k$ algebraically closed. For $r=0$ the assertions are $H_n(0)=0$ and \[ 2d\,N_{2d}(n)=n\binom{n+2d-1}{2d-1}, \] and are immediate. We may therefore assume $r\geq1$. Let $V=k^n$, $F=k^r$, and $\mathcal A=\operatorname{Hom}(F,\operatorname{Sym}^dV)$. The group $G=\operatorname{SL}(V)\times\operatorname{SL}(F)$ acts by \[ (g,a)\cdot B=(\operatorname{Sym}^d g)\,B\,a^{-1}. \] There is a nonempty invariant open set $\mathcal U\subset\mathcal A$ on which $B$ is injective and the multiplication map has constant maximal rank. On this open set the universal degree-$2d$ complex is \[ \bigwedge^2F\otimes\mathcal O_{\mathcal U} \longrightarrow (F\otimes\operatorname{Sym}^dV)\otimes\mathcal O_{\mathcal U} \longrightarrow \operatorname{Sym}^{2d}V\otimes\mathcal O_{\mathcal U}. \] Its first homology and its cokernel are $G$-equivariant vector bundles of ranks $H_n(r)$ and $C_n(r)$, respectively. Their determinant lines are ordinary trivial line bundles on $\mathcal U$. Here is a direct justification of the last assertion and of the equivariance information it preserves. Every line bundle on a nonempty open subset of affine space has a rational section, hence is represented by a Cartier divisor. The closure in affine space of each prime divisor in that open set is a hypersurface, because the polynomial ring is factorial. The product of their defining polynomials, with the divisor's integer multiplicities, is a rational function having the prescribed divisor on the open set. The line bundle is therefore trivial. If $L$ is one of our determinant lines and $s$ is a nowhere vanishing section, write its linearization as \[ g\cdot s(x)=a(g,x)s(gx),\qquad (g,x)\in G\times\mathcal U. \] The function $a$ is regular and invertible. For fixed $x$, $a(-,x)$ is a regular unit on $G$, and every such unit is constant. Indeed, its restriction to any translate of an elementary unipotent subgroup is a unit of $k[t]$, and hence is constant. Elementary unipotent subgroups generate each special linear factor, proving the assertion; a factor $\operatorname{SL}_1$ is trivial. Since $a(1,x)=1$, we obtain $a(g,x)=1$. Thus every stabilizer acts trivially on each determinant line. Let $\ell=g$. Since $\ell$ is prime to the characteristic, there exists $\zeta\in k$ of order $\ell$. The central pair \[ z=(\zeta I_V,\zeta^d I_F) \] belongs to $G$ and fixes every point of $\mathcal A$. It acts on each term of the displayed complex by the same scalar $\zeta^{2d}$: the three weights are respectively $(\zeta^d)^2$, $\zeta^d\zeta^d$, and $\zeta^{2d}$. It follows that it acts on the two determinant lines by \[ \zeta^{2dH_n(r)}\quad\text{and}\quad \zeta^{2dC_n(r)}. \] Both scalars are $1$ by the preceding paragraph. Since $\zeta$ has order $\ell$, these are precisely the two divisibilities. \end{proof} \begin{lemma}\label{uc:polynomial-root} For each fixed $d\geq2$ there is a real polynomial $P_d(x)$ of degree $d$ such that \begin{equation}\label{uc:root-expansion} \kappa_d(n)=P_d(n)+O(n^{-1}). \end{equation} Its leading coefficient is the irrational number $\rho_d/d!$, where $\rho_d=1-\sqrt{1-1/H}$ and $H=\binom{2d-1}{d-1}$. \end{lemma} \begin{proof} The expression under the square root in \eqref{uf:kappa} is a polynomial $\Delta(n)$ of degree $2d$, with leading coefficient \[ \frac1{(d!)^2}-\frac2{(2d)!} =\frac{1-1/H}{(d!)^2}>0. \] Write $\Delta(n)=n^{2d}D(n^{-1})$, where $D$ is a real polynomial and $D(0)>0$. The positive square root of $D(z)$ is real analytic near $z=0$. Its Taylor polynomial through degree $d$, multiplied by $n^d$, differs from $\sqrt{\Delta(n)}$ by $O(n^{-1})$. Subtracting this polynomial from $N_d(n)+1/2$ gives $P_d$ and the stated leading coefficient. Finally, $H\geq3$ and $H,H-1$ are coprime. If $\sqrt{(H-1)/H}$ were rational, the numerator and denominator of this reduced fraction would both be squares. Two positive consecutive integers cannot both be squares: if $b>a\geq1$, then $b^2-a^2\geq2a+1>1$. This contradiction proves irrationality. \end{proof} We recall the precise elementary form of polynomial equidistribution needed here; see~\cite{weyl1916}. \begin{lemma}[Weyl's polynomial equidistribution theorem] \label{uc:weyl} If a real polynomial $P$ has positive degree and irrational leading coefficient, the fractional parts of $P(1),P(2),\ldots$ are uniformly distributed in $[0,1)$. The same conclusion holds for $P(b+Ls)/A$, as $s$ tends through the positive integers, whenever $b$ is an integer and $L,A$ are positive integers. \end{lemma} \begin{proof} The first assertion is Weyl's polynomial equidistribution theorem \cite{weyl1916}. For the second, substitution $x=b+Ls$ and division by $A$ multiply the leading coefficient by the nonzero rational number $L^{\deg P}/A$. This preserves irrationality, so the first assertion applies to the resulting polynomial. \end{proof} \begin{theorem}[Eventual vanishing from the recurrence] \label{uc:eventual-zero} Fix $d\geq2$ and a class $\mathcal F$ of infinite fields. Suppose there are integers $h>0$ and $m_0$ such that \eqref{uc:recurrence} holds for every $m\geq m_0$ over every field in $\mathcal F$. Then there is an integer $N$, independent of the field in $\mathcal F$, such that $E_d(n)=0$ for every $n\geq N$. Consequently the degree-$2d$ prediction holds for every number of general degree-$d$ generators in these dimensions. \end{theorem} \begin{proof} The two initial defects are bounded by the target dimension. Indeed, at $r=r_n^-$ one has $\chi_n(r)\geq0$, and the dimension of the source after the constant Koszul boundaries are removed is \[ rN_d(n)-\binom r2=N_{2d}(n)-\chi_n(r)\leq N_{2d}(n). \] The upper-endpoint cokernel has the same bound. Thus, with $M=N_{2d}(m_0+h-1)$, iteration of the recurrence gives, over every field in $\mathcal F$, \begin{equation}\label{uc:bounded} 0\leq E_d(n)\leq M\qquad(n\geq m_0). \end{equation} Choose three distinct primes $p_1,p_2,p_3>\max\{2dM,2d\}$, none dividing $h$. For each pair $p=p_i$, $q=p_j$ with $i0. \] Thus $2\binom w2\geq\lfloor5h^2/32\rfloor$. Embed the $2w$ variables in the $h$ variables and restrict to a subspace of the required dimension. \end{proof} \begin{lemma}[A quartic product space]\label{products:quartic} For every $w\geq4$, there is a space $C_4\subset S_4(k^{2w})$ of dimension $2\binom w4$ whose symmetric multiplication is injective. Consequently, if $h\geq144$ and $4\mid h$, there is a space $H_4\subset S_4(k^h)$ of dimension $\lfloor h^4/256\rfloor$ with injective symmetric multiplication. \end{lemma} \begin{proof} Use \eqref{products:paired-forms} with $s=4$. Set $r=|I|$. After relabelling the indices, every nonempty product component has \[ I=\{0,\ldots,r-1\},\qquad U=\{0,\ldots,7-r\}, \qquad 0\leq r\leq4. \] The numbers of unordered products in these five components are, respectively, $140,40,12,4,3$. For $r<4$, each decomposition $U=S\cup T$, $S\cap T=I$ gives the four products $f_{S,\alpha}f_{T,\beta}$, with $\alpha,\beta\in\{0,1\}$. Three cases follow directly from the quadratic construction. If $r=2$, apply its twelve-product specialization to the four indices of $U\setminus I$ and multiply every quadratic form by $x_I$. If $r=3$, write $S=I\cup\{a\}$ and $T=I\cup\{b\}$; take $x_Ix_a,x_Iy_a$ on $S$ and $x_Ix_b,x_Iy_b$ on $T$. If $r=4$, the two forms $x_I,y_I$ have three distinct unordered products. For the remaining two cases, explicit monomial choices are given in Tables~\ref{products:quartic-disjoint} and~\ref{products:quartic-one-common}. In each row set $T=I\cup(U\setminus S)$, and specialize \[ f_{S,\alpha}=x_{A_\alpha}y_{S\setminus A_\alpha},\qquad f_{T,\beta}=x_{B_\beta}y_{T\setminus B_\beta}. \] Each support occurs in exactly one row, so the tables assign one monomial to each form. To check the products, it suffices to add the characteristic vectors of the displayed subsets: all vectors \begin{equation}\label{products:quartic-vector-check} \mathbf1_{A_\alpha}+\mathbf1_{B_\beta}, \qquad \alpha,\beta\in\{0,1\}, \end{equation} are distinct within each table. There are $35\cdot4=140$ such vectors in the first table and $10\cdot4=40$ in the second. These are the $x$-exponent vectors; the fixed product degree determines the $y$-exponents. Thus the tables give distinct monomials for every unordered product in the two remaining components. The common specialization criterion now yields the space $C_4$. \begin{table}[htbp] \centering \setlength{\tabcolsep}{3pt} \begin{tabular}{c|cccc@{\qquad}c|cccc} $S$&$A_0$&$A_1$&$B_0$&$B_1$& $S$&$A_0$&$A_1$&$B_0$&$B_1$\\ \hline $0123$ & $\varnothing$ & $013$ & $46$ & $457$ & $0237$ & $023$ & $7$ & $46$ & $1456$ \\ $0124$ & $02$ & $024$ & $67$ & $567$ & $0245$ & $2$ & $4$ & $1$ & $17$ \\ $0125$ & $0$ & $15$ & $7$ & $47$ & $0246$ & $6$ & $026$ & $3$ & $137$ \\ $0126$ & $02$ & $16$ & $347$ & $57$ & $0247$ & $4$ & $024$ & $13$ & $135$ \\ $0127$ & $0$ & $7$ & $35$ & $356$ & $0256$ & $25$ & $26$ & $34$ & $1347$ \\ $0134$ & $\varnothing$ & $1$ & $25$ & $67$ & $0257$ & $027$ & $0257$ & $36$ & $146$ \\ $0135$ & $015$ & $135$ & $2$ & $47$ & $0267$ & $7$ & $067$ & $1$ & $13$ \\ $0136$ & $6$ & $06$ & $2$ & $57$ & $0345$ & $05$ & $35$ & $2$ & $127$ \\ $0137$ & $3$ & $0137$ & $2$ & $456$ & $0346$ & $36$ & $0346$ & $12$ & $27$ \\ $0145$ & $\varnothing$ & $45$ & $6$ & $236$ & $0347$ & $0$ & $34$ & $6$ & $126$ \\ $0146$ & $01$ & $4$ & $5$ & $2357$ & $0356$ & $05$ & $036$ & $7$ & $47$ \\ $0147$ & $014$ & $147$ & $26$ & $356$ & $0357$ & $05$ & $0357$ & $124$ & $1246$ \\ $0156$ & $01$ & $06$ & $3$ & $24$ & $0367$ & $037$ & $067$ & $4$ & $14$ \\ $0157$ & $05$ & $07$ & $23$ & $24$ & $0456$ & $045$ & $6$ & $13$ & $127$ \\ $0167$ & $7$ & $0167$ & $34$ & $25$ & $0457$ & $7$ & $0457$ & $23$ & $36$ \\ $0234$ & $0$ & $024$ & $\varnothing$ & $17$ & $0467$ & $4$ & $47$ & $\varnothing$ & $23$ \\ $0235$ & $0$ & $05$ & $16$ & $146$ & $0567$ & $56$ & $0567$ & $1$ & $123$ \\ $0236$ & $2$ & $3$ & $\varnothing$ & $7$ & & & & & \\ \end{tabular} \caption{Quartic monomial choices when $I=\varnothing$ and $U=\{0,\ldots,7\}$.} \label{products:quartic-disjoint} \end{table} \begin{table}[htbp] \centering \begin{tabular}{c|cccc} $S$&$A_0$&$A_1$&$B_0$&$B_1$\\ \hline $0123$ & $012$ & $123$ & $45$ & $46$ \\ $0124$ & $24$ & $024$ & $05$ & $36$ \\ $0125$ & $01$ & $15$ & $3$ & $4$ \\ $0126$ & $06$ & $16$ & $034$ & $045$ \\ $0134$ & $3$ & $14$ & $02$ & $6$ \\ $0135$ & $03$ & $5$ & $24$ & $06$ \\ $0136$ & $0$ & $03$ & $0$ & $02$ \\ $0145$ & $04$ & $5$ & $0$ & $3$ \\ $0146$ & $1$ & $46$ & $035$ & $235$ \\ $0156$ & $056$ & $156$ & $023$ & $034$ \\ \end{tabular} \caption{Quartic monomial choices when $I=\{0\}$ and $U=\{0,\ldots,6\}$.} \label{products:quartic-one-common} \end{table} \begin{samepage} For the required count, put $w=h/2$. Since $w\geq72$, the elementary inequality \[ 32\binom w4-w^4 =\frac w3\bigl(w^2(w-24)+44w-24\bigr)\geq0 \] gives \[ 2\binom w4\geq\frac{w^4}{16}=\frac{h^4}{256}. \] Restricting $C_4$ to the required dimension proves the assertion.\qedhere \end{samepage} \end{proof} \begin{lemma}[Separated coefficient subspaces]\label{low:separated-products} Suppose $O\subset S_j(X)$ and $C\subset S_t(Y)$ have injective symmetric-product maps. There is a subspace $E\subset O\otimes C$ of every dimension at most $(\dim O)\lfloor\dim C/2\rfloor$ whose symmetric products are independent. \end{lemma} \begin{proof} Choose a basis $o_1,\ldots,o_b$ of $O$ and integers $r_i$ differing by at most one, with prescribed sum and $0\leq r_i\leq\lfloor\dim C/2\rfloor$. General $r_i$-planes $C_i\subset C$ have pairwise zero intersection. Put $E=\bigoplus_i o_iC_i$. Independent products in $O$ separate the unordered pairs of indices. The diagonal products are injective by the hypothesis on $C$. For $i\ne j$, the map $C_i\otimes C_j\to S_2(C)$ is injective, as is seen from a decomposition $C=C_i\oplus C_j\oplus C'$. Composing with $S_2(C)\to S_{2t}(Y)$ proves the assertion for every pair. \end{proof} \begin{remark}\label{unified:product-restrictions} Product injectivity is an open condition on the Grassmannian of subspaces of a fixed dimension. The flag variety of pairs $H\subset D\subset S_a(X)$ projects onto both Grassmannians. Thus, whenever a product-injective $r$-plane exists, a general $D$ of dimension at least $r$ contains one: the inverse image of the nonempty rank open has image containing a nonempty open in the Grassmannian of $D$. For a prescribed output space $O\subset S_a(X)$ and a subspace $D$ of codimension $c$, one instead has $\dim(O\cap D)\geq\dim O-c$. Replacing $O$ by $O\cap D$ preserves every multiplication injection already proved on $O$ or on its tensor products, as well as any prescribed bidegree. \end{remark} \section{Multiplication estimates for the prepared family} \label{appendix:multiplication} This appendix supplies the multiplication statements used in Section~\ref{normal:repair-section}. We first prove a normalized multiplication estimate and a stronger estimate for the quotient module of the outer family. We then use the prescribed counts to construct the intermediate families, identify the background homology, and fill the positive target degrees. The normalized estimate and the outer-module construction apply for every $d\geq3$. The capacity inequalities and polynomial row maps in Section~\ref{construct:rows} are proved here for $d\geq9$; Appendix~\ref{unified:section} supplies the corresponding bounds for $3\leq d\leq8$. \subsection{A normalized multiplication estimate} We begin with the multiplication estimate used to impose the required rank conditions. It applies equally to one set of variables and to a tensor product of polynomial rings. \begin{lemma}[Normalized multiplication and incidence] \label{construct:incidence} Let $A=S_aV$, $B=S_bV$, and $T=S_{a+b}V$, or let these three spaces be tensor products of such spaces for disjoint sets of variables, with multiplication in each factor. Every subspace $L\subset B$ satisfies \[ \dim(AL)\geq \frac{\dim T}{\dim B}\dim L. \] Suppose $A_iB_i\subset T$ are finitely many multiplication maps having bounds $\dim(A_iL)\geq\sigma_i\dim L$. For general lists of $r_i$ elements of $A_i$, the sum of their coefficient maps from $\bigoplus_i k^{r_i}\otimes B_i$ is injective if \begin{equation}\label{construct:incidence-criterion} \sum_i\frac{r_i+\dim B_i}{\sigma_i}<1. \end{equation} One may replace this sufficient inequality by the sharper rank-by-rank condition obtained by replacing $r_i+\dim B_i$ with $r_i+\dim B_i-k_i$ for coefficient rank $k_i$. \end{lemma} \begin{proof} Degenerate $L$ to its initial monomial subspace. For uniformly weighted monomials of degrees $b$ and $a+b$ there is a coupling supported on divisibility: conditionally on a target monomial $x^\beta$, choose $x^\alpha$ with probability \[ \frac{\prod_\nu\binom{\beta_\nu}{\alpha_\nu}} {\binom{a+b}{b}}. \] Its source marginal is uniform. This follows, for example, by summing the displayed weights using the coefficient identity for $\prod_\nu(1-z_\nu)^{-1}$, or by first choosing a uniformly random point of the standard simplex and then independent multinomial samples of sizes $a$ and $b$. Thus the normalized mass of a set of source monomials is at most the mass of its upper shadow. Products of these couplings prove the tensor-product assertion. Initial monomials cannot increase the dimension of a product space. For the incidence assertion, stratify each coefficient space by rank. A coefficient tuple of rank $k_i$ in $k^{r_i}\otimes B_i$ has dimension at most $k_i(r_i+\dim B_i-k_i)$. For fixed tuples, their equation on the independently variable lists in $A_i$ has image $\sum_i A_iL_i$, hence codimension at least $\max_i\sigma_i k_i$. Under \eqref{construct:incidence-criterion} this is greater than the sum of the coefficient-tuple dimensions for every nonzero tuple. Projectivizing the coefficient tuples and taking the incidence projection proves the assertion. This also proves the sharper version. \end{proof} \subsection{An outer multiplication module with a strict shadow estimate} \label{construct:outer} The following estimate strengthens normalized multiplication for the quotient module associated to the outer family. Its extra term is positive for every nonzero proper source subspace. This strict gain will give both maximal rank for scalar multiplication and the covector bounds needed in the normal-map argument. \begin{theorem}[Outer module] \label{construct:outer-strata} Let $d\geq3$ and $K\geq1$ be integers. Put $s=d-1$ and $n=2d-1$ for the source and target degrees, respectively, and write $H=\binom ns$ and $h=KH$. Take $\dim X=h$, $\dim Y=m$, and split $Y=C\oplus Z$ with $\dim C=a=\sigma m+O(1)$, $0<\sigma<1$. For each degree-$s$ monomial $u$ in $C$, choose $K$ general vectors $v_{u,1},\ldots,v_{u,K}$ in $X$, and let $F\subset X\otimes S_sY$ be the span of the forms $v_{u,\ell}u$. Its dimension is $f=KN_s(a)$, and multiplication \[ F\otimes S_dY\longrightarrow X\otimes S_nY \] is injective. For the module $\mathcal N=(X\otimes S(Y))/(F\cdot S(Y))$, put $A=\dim\mathcal N_s$ and $T=\dim\mathcal N_n$. There is $\eta>0$ such that, for all sufficiently large $m$ and all $M\subset\mathcal N_s$, \begin{equation}\label{construct:strong-shadow} \dim(S_dY\cdot M)\geq \frac TA\dim M +\eta m^d\min\{\dim M,A-\dim M\}. \end{equation} General scalar forms of every count $q$ act with maximal rank $k^q\otimes\mathcal N_s\to\mathcal N_n$. If $qA\leq T$ and $j=T-qA=o(m^n)$, put $J=\operatorname{coker}(k^q\otimes\mathcal N_s\to\mathcal N_n)$. For $\lambda\in J^*$ define $C_\lambda:\mathcal N_s\to(S_dY)^*$ by $C_\lambda(v)(p)=\lambda(pv)$. Then, for a constant $\eta'>0$, \begin{equation}\label{construct:outer-covectors} \dim\{[\lambda]\in\mathbb P(J^*):\dim\ker C_\lambda\geq k\} \leq j-\eta'm^dk-1\quad(k\geq1). \end{equation} For the choice $\sigma^s=H\rho$, the same assertions hold for a general $(f+v)$-plane when $v$ is fixed as $m$ grows, and the limiting slope is $(T/A)/m^d\longrightarrow\rho/d!$. \end{theorem} \begin{proof} We first prove the strict estimate for $K=1$. The argument begins with the monomial fibers of the quotient and a bound on exceptional projections. A transport between the degree profiles then gives a positive gain, which we transfer to the exact quotient dimensions. Finally we deduce the rank and covector statements and treat general $K$ and the extra columns. \smallskip \noindent\emph{Step 1: the monomial fibers.} For now write $h=\binom ns$. The core degree of a monomial is its degree in the variables of $C$. Each target monomial of core degree $j$ has at most $\binom js$ distinct core divisors of degree $s$. Its attached vectors are therefore independent. This proves polynomial injectivity and \[ A=hN_s(m)-N_s(a),\qquad T=hN_n(m)-N_s(a)N_d(m). \] For each monomial, its fiber is the quotient of $X$ by the span of the vectors attached to its degree-$s$ core divisors. The exact source fiber dimensions are $d_i=h$ for $i0$. Summing squared differences over these transitions gives \begin{equation}\label{construct:outer-variance} \mathbb E\operatorname{Var}(u_\alpha\mid\text{target}) \geq (\kappa/2)\operatorname{Var}(u_\alpha). \end{equation} There is also a positive-probability decrease of capacity from every source profile: use target core degree $s$ for $i0$. Choose $\delta>0$ with $h\delta<\lambda/4$. If $k\leq\delta m^s$, then \[ hmk^2\leq h\delta m^d k \leq (T/A-\eta m^d)k \] for all sufficiently large $m$, with any fixed $0<\eta\leq\lambda/4$. If $k\geq\delta m^s$, replacing $T_0$ by $T$ in \eqref{construct:coarse-shadow} costs only $O_\delta(m^{d-1})\min(\ell,k)$, which is absorbed by its strict gain. Decreasing $\eta$ if necessary proves \eqref{construct:strong-shadow} in both ranges. \smallskip \noindent\emph{Step 6: scalar multiplication and covectors.} We now apply the strict growth estimate to the two incidence problems in the statement. For scalar injectivity, stratify coefficient tuples by rank $r$. Their projective dimension is at most $r(q+A-r)-1$, while their condition on the scalar forms has codimension at least the shadow in \eqref{construct:strong-shadow}. The excess term dominates $r(A-r)$ because $A=O(m^{d-1})$; rank $A$ uses $T\geq qA$. For scalar surjectivity, stratify target covectors by the dimension $k$ of the kernel of their contraction map. Choose this $k$-plane and impose the equations requiring the covector to annihilate its shadow. The remaining $q$ scalar equations have codimension $q(A-k)$. The same estimate proves surjectivity when $qA\geq T$. For $qA\leq T$, this calculation gives the sharper bound \[ j+k(A-k)-(j/A)k-\eta m^d\min(k,A-k)-1. \] For $k\leq A/2$ it implies \eqref{construct:outer-covectors} after decreasing the constant; for $k>A/2$, write $t=A-k$ to obtain $t[j/A+k-\eta m^d]-1<0$, so these strata are empty. The full kernel $k=A$ is impossible because $\mathcal N_s$ generates $\mathcal N_n$. \smallskip \noindent\emph{Step 7: general $K$ and the limiting slope.} The proof for general $K$ is unchanged: all source and coarse target capacities multiply by $K$, the profile transport is the same, and only fixed constants in the exceptional-set argument change. For reference, direct leading-term calculation gives \[ \frac{T/A}{m^d}\longrightarrow \frac{1-\sigma^s}{d!(H-\sigma^s)}; \] this is $\rho/d!$ when $\sigma^s=H\rho$. \smallskip \noindent\emph{Step 8: finitely many additional columns.} Finally append $v$ fixed columns $w_i z_i^s$, where the $z_i$ are distinct free variables and the $w_i$ are their attached vectors in $X$. Every target involving a private power has at most $Kd+2\leq h$ attached vectors, proving polynomial injectivity. Put $A'=A-v$, $T'=T-vN_d(m)$, $\nu=T/A$, and $\nu'=T'/A'$. We compare the extended quotient with the one already constructed, separately for subspaces of dimension at most $A'/2$ and at least $A'/2$. In the first range, discard targets divisible by any private power from the regular-source estimate; this loses only $O_v(m^{d-1})$ per source dimension. On a private source fiber retain targets of core degree at most $s-1$ and with no other private power. They give the identity on $X/kw_i$, on a proportion \[ p_{\rm good}=1-\sigma^d-d\sigma^{d-1}(1-\sigma)>\rho \] of all multipliers. To check the strict inequality, set $p=\sigma^s=H\rho$. Since $H\geq10$, $1/229/250>\rho$. These private targets are pairwise disjoint and disjoint from the retained regular targets. For a subspace of dimension $\ell\geq A'/2$, lift it to an $(\ell+v)$-plane in the old source. Its shadow contains all private column products; subtracting their exact dimension $vN_d(m)$ from \eqref{construct:strong-shadow} gives \[ \dim(S_dYM)\geq\nu'\ell+ \left[\eta m^d-\frac{v(N_d(m)-\nu)}{A'}\right](A'-\ell). \] The term subtracted from $\eta m^d$ is $O_v(m)$. Thus the strict estimate persists. Failure of any required subspace estimate is the projection of a closed rank condition on a relative Grassmannian. Properness therefore promotes the witness to an open set of general $(f+v)$-planes. Scalar rank and covector statements follow exactly as before. \end{proof} \subsection{Products of the intermediate families} \label{construct:rows} We now apply the multiplication estimates to the intermediate families. First we check that the prescribed counts fit within the available dimensions; then we construct the row maps with precisely their required Koszul kernels. We work here with $d\geq9$. Appendix~\ref{unified:section} verifies the same rank conditions for $3\leq d\leq8$, so the subsequent coefficient and deformation arguments apply in either range. For $j\in\mathcal J$ let $E_j$ be a labelled $e_j$-dimensional space of forms in $S_jX\otimes S_{d-j}Y$; for $j=2$ require its outputs to lie in $D$. Write $a_j$ for a list of scalar degree-$d$ forms with the same labels. All scalar lists used below, including $Q\subset S_dY$, are contained in one common list $\mathcal S$. Its total length is \begin{equation}\label{construct:scalar-density} \#\mathcal S=(\rho/d!)m^d+O_h(m^{d-1}). \end{equation} Its additional columns beyond $Q$ number $O_h(m^{d-2})$. For an even integer $R$, define the formal product space \[ \Pi_R=\bigoplus_{\substack{j1/4$ for $d\geq9$. There are finitely many source profiles, so the same strict lower bounds hold for large finite $m$ after a small decrease. The monomial coupling proof of Lemma~\ref{construct:incidence}, with its target restricted to retained profiles, gives this fraction of the usual normalized shadow for arbitrary subspaces as well. When products occur, $R\geq4$, and the scalar dimension ratio is $\binom{2d-R}{d-R}\rho<1/8$. Coefficient-rank incidence therefore makes the entire common scalar list injective in this quotient. Its image is disjoint from all products, simultaneously, in the original target. \smallskip \noindent\emph{The new layer.} When a new layer $E_R$ occurs with $R\geq4$, put its outputs in \[ K_R=\bigoplus_{a\ \mathrm{odd}}S_aX_1\otimes S_{R-a}X_2, \qquad \dim K_R=(1/(2R!)+o(1))h^R. \] This separates its image from all product outputs. On each monomial of degree $t_R$ in $Y$ place $b$ general output vectors, where $b=\lceil\eta\gamma_Rt_R!h^R\rceil$; for sufficiently large $h$ this provides more than the required count while satisfying \[ B_Rb/\dim K_R+B_R\rho<1. \] A target monomial has at most $B_Rb$ labels, proving polynomial injectivity of $E_R\otimes S_dY$. Form the graded quotient module $\mathcal M_R=(K_R\otimes S(Y))/(E_R\cdot S(Y))$, graded by $Y$-degree. Uniformly for all subspaces $M\subset(\mathcal M_R)_{t_R}$, its multiplication satisfies the shadow estimate \begin{equation}\label{construct:sparse-shadow} \dim(S_dYM)\geq \left[\frac{\dim K_R-B_Rb}{\dim K_R} \frac{N_{d+t_R}(m)}{N_{t_R}(m)}-O_h(m^{d-1})\right]\dim M. \end{equation} Indeed, degenerate to monomial fibers and project a source fiber to each target fiber. Apart from exceptional projections, each retains at least the displayed fraction of its dimension. The disjoint-set Grassmannian argument in Theorem~\ref{construct:outer-strata} bounds exceptional additional-label sets by a constant. An extra degree-$t_R$ monomial has least common multiple of degree at least $t_R+1$ with the source, so each exceptional label affects $O_h(m^{d-1})$ targets. Summing the normalized divisibility coupling proves \eqref{construct:sparse-shadow}. The strict scalar inequality and coefficient-rank incidence now give injectivity of $\mathcal S$ on this quotient. Before taking the quotient this says exactly that its kernel is the constant $\mathcal S$--$E_R$ Koszul source. Deleting surplus labelled columns preserves this kernel assertion. For $R=2$ there are no products; use output space $D$ and the first inequality of \eqref{construct:scalar-capacity}. Unused output complements are free modules and satisfy a stronger estimate. \smallskip \noindent\emph{The fixed odd family.} It remains to include $P$. Reserve distinct variables $z_i$ outside the variables used above and put $P_i=w_i z_i^{d-1}$, with general nonzero $w_i\in X$. The scalar/new-layer map has no kernel one degree below the degree just considered. To see this when $E_R$ is present, multiply a hypothetical lower relation by two independent core variables. Both resulting relations are constant Koszul relations. Comparing their scalar-side coefficients gives a linear-coefficient relation among the $E_R$ columns; it vanishes because $E_R\otimes S_dY$ is injective, hence so is $E_R\otimes Y$. The two constant Koszul matrices therefore vanish. Lower degrees follow by multiplication. When $E_R$ is absent this assertion is immediate. This proves exactness after adjoining the private variables in every component not involving $P$. Impose also, for each $i$, the injection \[ E_R\otimes Y_{\rm core}\longrightarrow [S_RX/(w_iS_{R-1}X)]\otimes S_{t_R+1}Y_{\rm core}. \] The same sparse construction proves it: a target has at most $(t_R+1)b$ labels, while the output quotient loses only $N_{R-1}(h)=O(h^{R-1})$ dimensions; its strict capacity is weaker than the inequality already imposed. Finitely many such opens intersect. The scalar coefficient degree is $d-R2d-R$. For $R=2$ two distinct powers meet only in the target $z_i^{d-1}z_j^{d-1}$; the intersection $w_iX\cap w_jX=kw_iw_j$ gives exactly the alternating $P$--$P$ relation. Multiplication by a single $w_i$ is injective. This proves precisely the asserted additional kernel. \smallskip \noindent\emph{Simultaneous choice.} Every assertion proved is a rank condition on a polynomial matrix with a known, constant-dimensional mandatory kernel. Its witness therefore gives a nonempty open in the same irreducible parameter space. Intersecting the finitely many row opens proves simultaneity. A bounded number of added columns does not affect any strict asymptotic inequality, proving the final assertion. \end{proof} \subsection{Prepared backgrounds and their exact homology} \label{construct:background-section} The row maps now allow us to eliminate all background relations except those already present among the forms in $Q$. The proposition also records the projected and enlarged versions needed for the replacement argument. \begin{proposition}[Prepared background] \label{construct:background} Let $Q\subset S_dY$ be general of dimension $q$, without any assumption that its endpoint homology or cokernel vanishes. There are simultaneous nonempty opens of the preceding parameter spaces on which the following statements hold. For odd $d$, take $U\subset S_dX$ of dimension $u=\lceil2h^d/(d+1)!\rceil$ and an equally labelled family $P\subset X\otimes S_{d-1}Y$, and set \[ W_{\rm bg}=\langle Q,\ E_j+a_j\ (j\in\mathcal J),\ U+P\rangle. \] For even $d$, let $U$ be a basis of $S_dX$, take labelled columns $P_2\in D\otimes S_{d-2}Y$ and $b\in S_dY$, and set \[ W_{\rm bg}=\langle Q,\ E_j+a_j\ (j\in\mathcal J),\ U+P_2+b\rangle. \] In both cases \begin{equation}\label{construct:bg-homology} H_1(W_{\rm bg};S)_{2d}=H_1(Q;S(Y))_{2d} \end{equation} by natural inclusion. Equality continues to hold after projecting the degree-$2d$ target by any complement to $Q S_dY$ in $S_{2d}Y$, and after adding one further generator of the form $E_2+a_2$. Odd homology is zero both before and after adjoining $F$. Consequently, in the background quotient $R=S/(W_{\rm bg})$, \begin{equation}\label{construct:relative-odd} F\otimes R_{d,\mathrm{even}}\longrightarrow R_{2d,\mathrm{odd}} \quad\hbox{is injective}. \end{equation} The pure family may simultaneously be chosen with $U X=S_{d+1}X$ in odd degree, and with $U=S_dX$ in even degree. \end{proposition} \begin{proof} We first identify the even homology by eliminating coefficients in increasing $X$-degree. We then treat projection of the scalar target, prove the odd-parity assertion by a separate witness, and choose the pure family to fill the highest degrees. \smallskip \noindent\emph{Even homology when $d$ is odd.} Suppose $d=2s+1$. Reindex the even layers by half their $X$-degrees, so $I=\{j/2:j\in\mathcal J\}$ and $k=\max I$. In an even cycle write $q_t,e_{i,t}$ for its $Q,E_{2i}$ coefficients of $X$-degree $2t$, and $u_t$ for its $U+P$ coefficient of $X$-degree $2t+1$. The row of degree $2r$ is exactly \begin{equation}\label{construct:triangular-equation} Qq_r+\sum_i a_{2i}e_{i,r}+\sum_iE_{2i}e_{i,r-i} +Pu_{r-1}+Uu_{r-s-1}=0, \end{equation} with out-of-range terms zero. Process $r=1,\ldots,\max(2k,s+1)$ in increasing order. The normalization maintained at each processed column $t>0$ is as follows. Put the $Q$ coefficient equal to zero. If $t\notin I$, all even-family coefficients in that column are zero. If $t\in I$, put $e_{i,t}=0$ for $i>t$, retain $e_{i,t}$ for $ir$ subtract $(E_{2i}+a_{2i})$--$(E_{2r}+a_{2r})$ boundaries to kill $e_{i,r}$; and subtract boundaries within $E_{2r}+a_{2r}$ to put the diagonal coefficient in the chosen upper-triangular form. Only scalar coefficients and future positive columns change. The coefficients with $iN_{d+1}(h)/h+h$. The full kernel is impossible. This proves the pure cutoff, and it can be intersected with all the preceding nonempty opens. \end{proof} \subsection{Surjectivity onto the positive target degrees} \label{construct:targets-section} It remains to fill the positive target rows. The convolution lemma below converts multiplication in the $X$-variables into vector-valued multiplication in the $Y$-variables, with an exact number of generators. We then apply it to the prescribed families. \begin{lemma}[Symmetric-power convolution] \label{construct:convolution} Let $a,e,m\geq1$. On bases $v_0,\ldots,v_{a-1}$ of $V$, $y_0,\ldots,y_{m-1}$ of $Y$, and $l_0,\ldots,l_{m+a-2}$ of $L$, define $\phi(l_j)=\sum_{i+k=j}v_i\otimes y_k$. Multiplication gives an isomorphism \[ S_eL\otimes S_{a-1}Y\longrightarrow S_eV\otimes S_{a+e-1}Y. \] Thus $\binom{m+a+e-2}{e}$ vector forms of degree $e$, in $\binom{a+e-1}{e}$ output directions, fill the entire indicated target. \end{lemma} \begin{proof} If $N=a+e-1$ vectors in $V$ are in linear general position, their products over all $e$-element subsets form a basis of $S_eV$. For a subset $I$, a functional vanishing on its $a-1$ complementary vectors and nonvanishing on its own vectors, raised to its $e$th power, separates precisely that product. Write $v(t)=\sum v_it^i$, $y(t)=\sum y_kt^k$, and $l(t)=\sum l_jt^j$; then $\phi(l(t))=v(t)y(t)$. A functional annihilating the claimed image annihilates \[ v(t_1)\cdots v(t_e)\otimes y(t_1)\cdots y(t_N) \] and, by permuting the $t_i$, the analogous expression for every $e$-element subset. Distinct $t_i$ give the preceding general-position basis by the Vandermonde determinant. Polynomial continuation then shows that the functional annihilates all $S_eV\otimes S_NY$. The dimensions agree by the factorial identity \[ \binom{m+a+e-2}{e}\binom{m+a-2}{a-1} =\binom{a+e-1}{e}\binom{m+a+e-2}{a+e-1}.\qedhere \] \end{proof} \begin{proposition}[The target maps] \label{construct:targets} Use the quadratic endpoint and the theorem of Boij, Dannetun and Lundqvist, stated in the introduction. For the exact counts of Proposition~\ref{construct:inventory}, the homogeneous positive components of $F,E_j,U$ fill every target of $X$-degree $b\geq2$ in total degree $2d$. Their deformed families in Proposition~\ref{construct:background} consequently fill all positive even target degrees, modulo lower even degrees. After projecting a complement to the image in $S_{2d}Y$, all even targets are filled. These assertions hold simultaneously with every background and product assertion above. \end{proposition} \begin{proof} We treat the middle rows first, then rows three and four, and finally row two, where two convolution constructions are needed. The pure family fills the remaining highest rows. \smallskip \noindent\emph{The middle rows.} Suppose $j(b-j)!/b!$. This follows from that theorem: there is no degree-$b$ Koszul contribution, and $N_b(h)-\tau h^jN_{b-j}(h)<0$ for large $h$. Put $e=d-j$ and $a=d-b+j+1$. The convolution lemma supplies blocks with $\binom{2d-b}{d-j}$ output directions. Round the number of chosen output forms up to a multiple of this integer and apply convolution independently to each block. Multiplication by $S_{b-j}X$ then fills $S_bX\otimes S_{2d-b}Y$ using \[ \left[\frac{\tau h^j} {(d-j)!\binom{2d-b}{d-j}}+O_d(1)+O_h(m^{-1})\right]m^{d-j} \] generators. The $O_d(1)$ term accounts for rounding the output count and is absorbed by first choosing $h$ sufficiently large. Strictly greater counts preserve surjectivity by extension and rank openness. Use $j=4$ for $b=5,6,7$. Use each even $j\geq6$ for $2j-4\leq b\leq\min(2j-1,d)$ in odd degree, or through $d-1$ in even degree. These intervals cover every required row from five onwards. The largest threshold in an interval is its first one: the ratio of consecutive thresholds is \[ \frac{b-j+1}{b+1}\frac{2d-b}{d+j-b}<1 \quad(b\leq2j-2). \] At the first row it is precisely $\gamma_j$. The prescribed factor $101/100$ therefore gives the desired strict margin. Only target degrees below $2j$ occur in this application. \smallskip \noindent\emph{Rows three and four.} For $j=2$, the quadratic theorem supplies output spaces filling $S_3X$ at density $1/6$ and $S_4X$ at density $\rho_2/2$. Convolution gives thresholds $\tau_3$ and $\tau_4$ for rows three and four, respectively. For $3\leq d\leq8$, both are at most $\gamma_2$ by definition. For $d\geq9$, we have $\gamma_2=\tau_4$ and $\tau_4/\tau_3=3\rho_2(2d-3)/(d-1)>1$. The ratio increases with $d$, and at $d=7$ the inequality is exactly $\rho_2>2/11$, which follows from $2/3<(9/11)^2$. Both successful quadratic output spaces have dimension less than $\dim D$. Extend a successful one to a subspace of dimension $\dim D$. The flags with a successful smaller space form a nonempty open incidence locus, dominating a nonempty open in the Grassmannian of such $D$. Thus a general $D$ admits the required output space, and the convolution construction stays inside $D$. This proves rows three and four with the imposed output restriction. \smallskip \noindent\emph{Row two.} Two convolution constructions fill complementary parts of this row. Write $X=A\oplus B$ with $p=\dim A=h/4$, and put \[ q_1=\binom{2d-2}{d-1},\qquad b_1=\left\lceil\frac{p}{q_1}\right\rceil. \] Apply Lemma~\ref{construct:convolution} with $a=d$ and $e=d-1$ to $b_1$ blocks, and project their $b_1q_1$ output directions surjectively onto $A$. This gives \begin{equation}\label{construct:row-two-linear-count} b_1\binom{m+2d-3}{d-1} \end{equation} forms in $A\otimes S_{d-1}Y$ whose multiples by $S_{d-1}Y$ fill $A\otimes S_{2d-2}Y$. Their further multiples by $X$ fill $(AX)\otimes S_{2d-2}Y$. These forms fit inside the prescribed $F$ family. Indeed, \begin{equation}\label{construct:row-two-margin} \frac{1}{4\binom{2d-2}{d-1}}<\rho_d. \end{equation} To see this, use $H=\binom{2d-1}{d-1}=\frac{2d-1}{d}q_1<2q_1$ and $H\rho_d>1/2$. Since $b_1=h/(4q_1)+O_d(1)$, first taking $h$ large makes $b_12/11$. Thus the same order of choices, first $h$ and then $m$, absorbs all rounding and gives at most the prescribed $e_2$ generators. The subspace $D$ can be required to project surjectively onto $S_2B$. Its codimension $c$ is smaller than $\dim(AX)$: use \eqref{construct:detector-size} for $d\geq9$ and \eqref{unified:D} for $3\leq d\leq8$. A general subspace of this dimension therefore has a surjective restriction of the quotient map. Lift the convolution forms through $D\twoheadrightarrow S_2B$ and extend to the prescribed $e_2$-dimensional family. The lifted forms fill the quotient target, while the chosen $F$ forms fill its kernel $(AX)\otimes S_{2d-2}Y$. Together they fill row two. \smallskip \noindent\emph{Simultaneity and the highest rows.} These are witnesses for rank-open conditions on the same irreducible parameter spaces as the background and outer conditions. Their nonempty opens therefore intersect. For odd $d$, the pure cutoff $UX=S_{d+1}X$ fills every row $b\geq d+1$. For even $d$, $U=S_dX$ fills every row $b\geq d$. This completes the homogeneous target assertions. Their nonvanishing rank minors give nonempty opens in the same parameter spaces used above. For the deformed families, their lower components have strictly smaller $X$-degree. Descending elimination therefore gives surjectivity onto positive even rows modulo the lower rows. The projection of a complement to $QS_dY$ makes the residual degree-zero target equal to $QS_dY$, which $Q$ fills. All claims follow by finite intersection of the established opens. \end{proof} \section{Product separation and the normal map}\label{normal:section} The purpose of this appendix is to prove that a general scalar motion of the outer generators has maximal normal rank. There are two ingredients. First, we separate the symmetric products of those generators from the background relations. This gives an injective coefficient map on the remaining homology. Second, we bound the loci of target covectors for which scalar contraction has a large kernel. An incidence argument then gives the required rank. We retain the exact counts and the notation of Proposition~\ref{construct:inventory}. In particular, $X$ and $Y$ have dimensions $h$ and $m$, respectively, $s=d-1$, and \[ H=\binom{2d-1}{d-1},\qquad \rho=1-\sqrt{1-H^{-1}},\qquad f=\frac{h\rho}{(d-1)!}m^{d-1}+O_h(m^{d-2}). \] Every assertion below is made for fixed $d$ and $h$, and sufficiently large $m$. The numerical estimates in this appendix are for $d\geq9$; Appendix~\ref{unified:section} supplies their replacements for $3\leq d\leq8$, with the same algebraic constructions and comparison arguments. Products of subspaces always mean their images under polynomial multiplication. A tensor product in the source of a map, on the other hand, denotes a formal source and is not identified with its image without an injectivity assertion. \subsection{Separation of products} We will place the symmetric products of $F$ in one bidegree in the $Y$-variables. The next lemma ensures that the scalar relations remain independent after that bidegree is removed. Thus the two kinds of products can be separated in a quotient. \begin{lemma}[A deleted bidegree]\label{normal:deleted-profile} Split $Y=A\oplus B$ into two subspaces whose dimensions differ by at most one, and let $\mathcal P$ be any one full bidegree summand of $S_{2d-2}(Y)$. For all sufficiently large $m$ and every $L_0\subset S_{d-2}(Y)$, \begin{equation}\label{normal:deleted-shadow} \dim\bigl((S_d(Y)L_0+\mathcal P)/\mathcal P\bigr) \geq \frac25\frac{N_{2d-2}(m)}{N_{d-2}(m)}\dim L_0. \end{equation} Consequently, a general scalar family $Q^+\subset S_d(Y)$ of dimension $(\rho/d!+o(1))m^d$ satisfies \[ Q^+\otimes S_{d-2}(Y) \lhook\joinrel\longrightarrow S_{2d-2}(Y)/\mathcal P. \] \end{lemma} \begin{proof} We first prove the shadow bound, then use it to exclude scalar relations. Put $s=d-1$. Give a pair of monomials $y^\alpha\mid y^\beta$, of degrees $s-1$ and $2s$, the weight $\prod_i\binom{\beta_i}{\alpha_i}$. Every source row has total weight $\binom{m+2s-1}{s+1}$ and every target column total weight $\binom{2s}{s-1}$; their ratio is $N_{2s}(m)/N_{s-1}(m)$. If $y^\alpha$ has $A$-degree $i$, the fraction of its row weight in target $A$-degree $j$ is \[ \binom{s+1}{j-i} \frac{(\dim A+i)_{j-i} (\dim B+s-1-i)_{s+1-j+i}}{(m+s-1)_{s+1}}, \] where $(a)_r=a(a+1)\cdots(a+r-1)$. Uniformly in the finitely many possible $i,j$, this tends to $2^{-s-1}\binom{s+1}{j-i}\leq 1/2$. Deleting one target bidegree therefore retains at least $2/5$ of every source row for large $m$. Summing the weights proves \eqref{normal:deleted-shadow} for a monomial subspace. For general $L_0$, pass to leading monomials: the products of a leading monomial which survive the deleted bidegree are leading monomials of the projected product image. This proves the same bound for $L_0$. To deduce scalar injectivity, compare this shadow with the number of scalar parameters. The coefficient of $m^d$ on the right, per source dimension, is $\sigma_d=\frac25(d-2)!/(2d-2)!$. It is strictly greater than $\rho/d!$, since \[ d!\sigma_d=\frac25\frac{2d-1}{(d-1)H}, \qquad H\rho=\frac1{1+\sqrt{1-H^{-1}}}<\frac35. \] If a putative relation has coefficient span of dimension $\ell$, its projective coefficient tuples have dimension at most $\ell(\dim Q^++N_{d-2}(m)-\ell)-1$. The equations on the scalar family have rank at least $\frac25 N_{2d-2}(m)\ell/N_{d-2}(m)$. The strict leading inequality makes this rank larger than the parameter count. Projective incidence excludes every such relation for a general family. \end{proof} \begin{samepage} \begin{proposition}[Separation in the quadratic quotient] \label{normal:detector} Choose $L\leq h$ with $L>2^d\rho h$, put $c=L(L+1)/2$, and let \[ \Lambda:S_2(X)\longrightarrow k^c,\qquad D=\ker\Lambda \] be general. Let $Q^+$ contain $Q$ and all the scalar families in the prepared background, including one extra auxiliary generator when needed. Then $\operatorname{Sym}^2 F$ injects into \begin{equation}\label{normal:detector-quotient} \frac{S_2(X)\otimes S_{2d-2}(Y)} {D\otimes S_{2d-2}(Y)+Q^+S_2(X)S_{d-2}(Y)}. \end{equation} If $c\geq2h-1$, the same assertion holds after adding $\sum_{i=1}^v P_i(XS_{d-1}(Y))$ to the denominator, where $P_i\in XS_{d-1}(Y)$ are general and $v$ is fixed independently of $m$. In particular, \begin{equation}\label{normal:actual-product} \operatorname{Sym}^2F\lhook\joinrel\longrightarrow (S(X\oplus Y)/(W_{\rm bg}))_{2d} \end{equation} for either prepared background $W_{\rm bg}$ in Proposition~\ref{construct:background}. \end{proposition} \end{samepage} \begin{proof} We construct a witness for the quotient injection, then include the private columns and pass to the prepared background. Use Lemma~\ref{products:matching} on equal-sized subspaces of the two halves of $Y$, discarding one variable if necessary. It gives a space $C$ of dimension $((2^{d-1}(d-1)!)^{-1}+o(1))m^{d-1}$ with independent symmetric products, all in one bidegree summand $\mathcal P$. Choose subspaces $A_1,\ldots,A_L\subset C$, of dimensions differing by at most one and adding to $f$. The inequality $L>2^d\rho h$ ensures $\dim A_i<\frac12\dim C$. General choices have $A_i\cap A_j=0$ for $i\ne j$. For independent $x_1,\ldots,x_L\in X$, set $F=\bigoplus_i x_iA_i$. Its symmetric multiplication is injective: diagonal products use the injection of $\operatorname{Sym}^2C$, and a cross term uses the injection $A_i\otimes A_j\longrightarrow\operatorname{Sym}^2C$. Choose $\Lambda$ to be an isomorphism on $\operatorname{Sym}^2\langle x_1,\ldots,x_L\rangle$. The images of the products are still independent and lie in $k^c\otimes\mathcal P$. Lemma~\ref{normal:deleted-profile} makes the scalar multiplication map injective modulo $\mathcal P$. Its image therefore meets $k^c\otimes\mathcal P$ trivially, which separates the products from the scalar denominator. The scalar count is unchanged to leading order by all the extra families, so the lemma applies to the entire $Q^+$. For the assertion with $P_i$, reserve variables $z_1,\ldots,z_v$, construct $F,Q^+$ on the other variables, and specialize $P_i=w_i z_i^{d-1}$. A general $\Lambda$ is simultaneously injective on every $w_iX$ and every $w_iX+w_jX$: the latter has dimension $2h-1$ when $w_i,w_j$ are independent. A monomial of degree $2d-2$ is divisible by at most two of the private powers $z_i^{d-1}$. When two occur it is exactly $z_i^{d-1}z_j^{d-1}$, and the two detected output spaces meet only in $k\Lambda(w_iw_j)$. Hence the $P$ multiplication map has exactly its constant alternating relations. It remains to separate these private-column products from the other terms. Every scalar product has total private-variable degree at most $d-2$, whereas every $P$ product has a private exponent at least $d-1$. These images are disjoint. Scalar injectivity in each private-variable degree follows from Lemma~\ref{normal:deleted-profile}, or its lower-degree restriction. The $F$ products have private-variable degree zero and are separated from the scalar image by the same lemma. Thus a single polynomial matrix has precisely the obligatory alternating kernel and an independent $\operatorname{Sym}^2F$ summand. This is its largest possible rank, so the conclusion holds on a nonempty open set of all the parameters. To obtain the assertion for the actual background quotient, extract $X$-degree two from an alleged relation in \eqref{normal:actual-product}. Every quadratic high component of the background lies in $D$; components of higher $X$-degree do not contribute. The scalar contributions all belong to $Q^+$. For odd $d$, the only additional term is $\sum P_i(XS_{d-1}(Y))$, already included above. For even $d$, the quadratic tails of the pure generators lie in $D$ as well. The asserted quotient injection therefore excludes the relation. \end{proof} \subsection{The odd coefficient and target spaces} All parities in what follows refer to $X$-degree. By Proposition~\ref{construct:background}, the prepared background has only even first homology, namely the included homology from $S(Y)$, and adjoining $F$ creates no odd first homology. The normal map will take the remaining even homology to the odd cokernel. We now describe the source of scalar contraction and the dual of its target. In even generating degree these spaces split by $X$-degree. In odd generating degree the relations $U+P$ couple the lowest and top components; we keep that coupling explicit. \smallskip\noindent\emph{Even generating degree.} Put $I_0=(Q,F)$ and $R_0=S(X\oplus Y)/I_0$. The odd part of its degree-$d$ component is \begin{equation}\label{normal:even-source} V=V_1\oplus\bigoplus_{\substack{3\leq t\leq d-1\\t\text{ odd}}}V_t, \qquad V_1=\frac{XS_{d-1}(Y)}{F},\qquad V_t=S_t(X)S_{d-t}(Y). \end{equation} Define, for every odd $1\leq t\leq2d-1$, \begin{equation}\label{normal:even-targets} T_t=\frac{S_t(X)S_{2d-t}(Y)} {F S_{t-1}(X)S_{d+1-t}(Y)+Q S_t(X)S_{d-t}(Y)}, \end{equation} omitting any summand having a negative index. Homogeneity gives the exact decomposition \begin{equation}\label{normal:even-dual} \Lambda_0=((R_0)_{2d}^{\rm odd})^* =\bigoplus_{\substack{1\leq t\leq2d-1\\t\text{ odd}}}T_t^*. \end{equation} In particular, the summands with $t>d-1$ are retained, even though they do not occur in the coefficient space $V$. Every remaining generator has the form $B_i+p_i$, with $B_i$ of positive even $X$-degrees and $p_i\in S_d(Y)$. The $B_i$ include the mixed pure generators $U_i+(P_2)_i$. \smallskip\noindent\emph{Odd generating degree.} Put instead $I_0=(Q,F,U+P)$. Choose a vector-space complement $C_d$ of $U$ in $S_d(X)$, and write $B_d=S_d(X)/U$. Eliminating the degree-$d$ relations $U_i=-P_i$ gives \begin{equation}\label{normal:odd-source} V=\frac{XS_{d-1}(Y)}{F} \oplus\bigoplus_{\substack{3\leq t\leq d-2\\t\text{ odd}}} S_t(X)S_{d-t}(Y)\oplus C_d. \end{equation} This is a vector-space decomposition of the quotient. The target equations couple its first and last terms. To control those equations, we first establish a shadow estimate after projection modulo $U$. \begin{lemma}[The projected top degree]\label{normal:top} For odd $d\geq9$, put $a=N_{d-1}(h)$, $u=\dim U$, and $b=N_d(h)-u$. For general $U$ and sufficiently large $h$, \begin{equation}\label{normal:projected-shadow} \dim(XL_0\bmod U)\geq\frac ba\dim L_0 \quad\text{for every }L_0\subset S_{d-1}(X), \qquad \frac ba\geq\frac h{2d}. \end{equation} With $\delta=1/(100(2d)!)$, the map \begin{equation}\label{normal:top-injection} \bigl(F\otimes S_{d-1}(X)Y\bigr) \oplus\bigl(Q^+\otimes B_d\bigr) \longrightarrow B_d\otimes S_d(Y) \end{equation} is generically injective when $Q^+$ consists of $Q$ and $\lfloor\delta m^d\rfloor$ additional general scalar forms. Consequently, if \begin{align*} T_d^{\rm raw} &=\frac{S_d(X)S_d(Y)}{FS_{d-1}(X)Y+QS_d(X)},\\ J_d&=\frac{B_d\otimes S_d(Y)} {\overline{FS_{d-1}(X)Y}+Q B_d}, \end{align*} there is an exact sequence \begin{equation}\label{normal:top-sequence} 0\longrightarrow U\otimes(S_d(Y)/Q) \longrightarrow T_d^{\rm raw}\longrightarrow J_d \longrightarrow0. \end{equation} Moreover, every $L_0\subset B_d$ satisfies \begin{equation}\label{normal:top-scalar-shadow} \dim(S_d(Y)L_0\text{ in }J_d) \geq\lfloor\delta m^d\rfloor\dim L_0. \end{equation} \end{lemma} \begin{proof} We first prove the projected shadow estimate. The ordinary normalized multiplication estimate gives $\dim XL_0\geq N_d(h)\dim L_0/a$. We use the following elementary projection observation. Suppose a family of subspaces indexed by $\operatorname{Gr}(\ell,a)$ has dimension at least $(u+b)\ell/a$ in an ambient space of dimension $u+b$. For one such subspace of dimension $w$, the condition that projection modulo a $u$-plane have rank less than $b\ell/a$ is a Schubert condition of codimension greater than \[ \frac{u\ell}{a}\frac{b(a-\ell)}a =\frac{ub}{a^2}\ell(a-\ell). \] Indeed, writing the intersection dimension as $r$, its codimension is $r(b-w+r)$, and rank failure gives $r>w-b\ell/a\geq u\ell/a$. If $ub>a^2$, this exceeds the dimension $\ell(a-\ell)$ of the indexing Grassmannian, so a general $u$-plane has no such failure. The case $\ell=a$ is immediate. In the present situation \[ \frac{ub}{a^2} =\left(\frac{2(d-1)}{d^2(d+1)^2}+o(1)\right)h^2, \qquad \frac{b}{ah}\longrightarrow\frac{d-1}{d(d+1)}> \frac1{2d}. \] This proves \eqref{normal:projected-shadow}. Its nonempty open set meets the open where $UX=S_{d+1}(X)$. We next use this estimate to prove injectivity of \eqref{normal:top-injection}. Apply the two-family coefficient incidence estimate of Lemma~\ref{construct:incidence}. By torus degeneration in $Y$ and the divisibility coupling, the projected $X$ estimate tensorizes with multiplication in $Y$. The $F$ parameter action therefore has normalized shadow at least $(b/a)N_d(m)/m$, and the scalar parameter action has shadow $N_d(m)$. With $\sigma=b/a$, the limiting sum of the dimension ratios is \[ \frac{dh\rho}{\sigma}+\rho+d!\delta \leq(2d^2+1)\rho+d!\delta<1. \] For example, $H>4d^2+3$ for $d\geq9$ and $\rho<1/(2H-1)$ give $(2d^2+1)\rho<1/4$, while $d!\delta<1/100$. The coefficient-space overhead is of smaller order in $m$. Thus the displayed polynomial map is injective. At $Q^+=Q$ this says exactly that its image has no unintended intersection with $U\otimes S_d(Y)$, giving \eqref{normal:top-sequence}. Finally, fix one successful additional scalar family. Its tensor product with every $L_0\subset B_d$ injects into $J_d$, proving \eqref{normal:top-scalar-shadow} simultaneously for all $L_0$. \end{proof} We can now describe the coupling between the lowest and top target components. Let $T_1$ be the outer quotient \[ T_1=\frac{XS_{2d-1}(Y)}{FS_d(Y)+QXS_{d-1}(Y)}. \] For odd $d$, a covector on the quotient by $I_0$ must satisfy \begin{equation}\label{normal:affine-top} \lambda_d(U_i p)=-\lambda_1(P_i p), \qquad p\in S_d(Y)/Q. \end{equation} Sequence~\eqref{normal:top-sequence} shows that every $\lambda_1\in T_1^*$ extends across these equations. The space of extensions is an affine translate of $J_d^*$, and its direction has the scalar shadow \eqref{normal:top-scalar-shadow}. Thus the top covector is not chosen independently of $\lambda_1$. Multiplying the same relations by positive even coefficients gives the remaining equations. For every $v\in S_{2r}(X)S_{d-2r}(Y)$, they are \begin{equation}\label{normal:higher-top-equations} \lambda_{d+2r}(U_i v) =-\lambda_{1+2r}(P_i v), \qquad1\leq r\leq(d-1)/2. \end{equation} Their number is at most \begin{equation}\label{normal:higher-cost} H_0=u\sum_{r=1}^{(d-1)/2}N_{2r}(h)N_{d-2r}(m) =O_h(m^{d-2}). \end{equation} For the dimension estimate below, we first omit these higher equations. Allow the higher odd target covectors to be free, impose all the low $Q,F$ quotient conditions and \eqref{normal:affine-top}, and call the resulting vector space $\Lambda_{\rm amb}$. If $\Lambda_0=((S(X\oplus Y)/I_0)_{2d}^{\rm odd})^*$, then \begin{equation}\label{normal:ambient-cost} \dim\Lambda_{\rm amb}\leq\dim\Lambda_0+H_0. \end{equation} Thus omitting the higher equations increases the dimension by at most $H_0$. No extension assertion for \eqref{normal:higher-top-equations} is needed: their possible inconsistency can only reduce the dimension. \subsection{Scalar contraction on the full odd cokernel} In either parity let $V$ be the space just defined and put $D_V=\dim V=O_h(m^{d-1})$. For $\lambda\in\Lambda_0$, define the scalar contraction \begin{equation}\label{normal:contraction} C_\lambda:V\longrightarrow S_d(Y)^*,\qquad C_\lambda(v)(p)=\lambda(pv). \end{equation} All products here take place in the quotient by $I_0$. The remaining even generators are denoted $B_i+p_i$, $1\leq i\leq e$. For odd $d$ these are the $E_j+a_j$; for even $d$ they include the mixed pure generators. In both cases $e=O_h(m^{d-2})$. Let $J$ be the odd degree-$2d$ quotient after their addition, and write $j_0=\dim J$ at general parameters. The next proposition bounds the covectors for which this contraction has a large kernel. Its proof has two incidence counts: first over the source components, then over the scalar terms $p_i$. Keeping the second count in the common parameter space gives the exact cancellation needed in the final bound. \begin{proposition}[All covector strata]\label{normal:strata} There is $\xi >0$ such that, for every integer $k\geq1$ and all sufficiently large $m$, \begin{equation}\label{normal:final-strata} \dim\{[\lambda]\in\mathbb P(J^*): \dim\ker C_\lambda\geq k\} \leq j_0-\xi m^d k-1. \end{equation} The assertion holds on a nonempty open set of the common parameter space of the construction. \end{proposition} \begin{proof} \emph{Scalar estimates on the separate components.} The strong outer estimate (Theorem~\ref{construct:outer-strata}) supplies $c_1>0$ with \begin{equation}\label{normal:outer-stratum} \dim\{[\lambda_1]\in\mathbb P(T_1^*): \dim\ker(C_{\lambda_1}|_{V_1})=k_1\} \leq\dim T_1-c_1m^d k_1-1. \end{equation} For every other non-top source layer $V_t$, the quotient $T_t$ defined by \eqref{normal:even-targets} has a uniform scalar shadow of order $m^d$. Indeed, append $\lfloor\delta m^d\rfloor$ general scalar forms to $Q$, with $\delta=1/(100(2d)!)$. The two-family incidence estimate has unaugmented limiting dimension ratio \[ \rho\left[ \frac{th}{h+t-1}\binom{2d-t}{d-1}+\binom{2d-t}{d} \right]0$ such that \begin{equation}\label{normal:initial-strata} \dim\{[\lambda]\in\mathbb P(\Lambda_0):\lambda_1\ne0, \dim\ker C_\lambda=k\} \leq\dim\Lambda_0-c_2m^d k-1 \qquad(k\geq1). \end{equation} For $k=0$ we use $\dim\Lambda_0-1$. \smallskip\noindent\emph{Imposing the remaining scalar terms.} We now pass from $\Lambda_0$ to $J^*$ by varying all the scalar terms $p_i$ in their common parameter space. First record the codimension of $J^*$ at general parameters. Odd homology vanishes both before and after adding the $e$ remaining even generators. Their relative odd coefficient map is consequently injective: \[ k^e\otimes V\lhook\joinrel\longrightarrow (S(X\oplus Y)/I_0)_{2d}^{\rm odd}. \] Here all old--new constant Koszul relations have already been removed by passing to $V$, and products of two new even generators have even parity. We therefore have the exact identity \begin{equation}\label{normal:baseline} \dim\Lambda_0=j_0+eD_V. \end{equation} For fixed $\lambda$ with contraction kernel dimension $k$, its annihilation of the $i$th new generator means \[ \lambda(B_i v)+C_\lambda(v)(p_i)=0\qquad(v\in V). \] If consistent, this imposes exactly $D_V-k$ independent affine conditions on the scalar form $p_i$. Thus the entire common scalar family imposes $e(D_V-k)$ conditions. Subtracting these from \eqref{normal:initial-strata} and using \eqref{normal:baseline} proves, for general scalar parameters, \[ \dim\{[\lambda]\in\mathbb P(J^*):\lambda_1\ne0, \dim\ker C_\lambda=k\} \leq j_0-(c_2m^d-e)k-1. \] \smallskip\noindent\emph{The locus with zero lowest component.} To complete the count, we show that there is no nonzero covector in $J^*$ with $\lambda_1=0$. Use the target surjections of Proposition~\ref{construct:targets} in increasing odd $X$-degree. If $a_i$ is the largest $X$-degree of $B_i$, its equation in target degree $t$ has the form \[ \lambda_t(B_i^{[a_i]}v) +\sum_{00$. Taking the finite union of the exact strata of dimensions at least $k$ gives \eqref{normal:final-strata}. All polynomial rank conditions used above hold on nonempty opens. The scalar-incidence bounds also hold on a nonempty open by projective fiber-dimension semicontinuity. Their finite intersection supplies the final assertion. \end{proof} \subsection{From coefficient maps to maximal normal rank} We now turn the covector estimate into a rank statement for a first-order deformation. We begin with the elementary fact that rank gained by the normal map lowers the dimension of homology. Given independent degree-$d$ generators $g_i$ and a motion $g_i(\epsilon)=g_i+\epsilon p_i$, the first derivative induces \begin{equation}\label{normal:derivative} \partial_p:H_1(g;S)_{2d}\longrightarrow(S/(g))_{2d}, \qquad \left[\sum_i e_i\otimes a_i\right]\longmapsto \left[\sum_i p_i a_i\right]. \end{equation} It is well-defined: a constant Koszul boundary maps to an element of the ideal $(g)$. \begin{lemma}[First derivative and rank]\label{normal:rank-lifting} If \eqref{normal:derivative} has rank $r_0$, the degree-$2d$ first homology of a general member of the motion has dimension at most $\dim H_1(g;S)_{2d}-r_0$. \end{lemma} \begin{proof} The constant Koszul image is a subbundle while the generators stay independent. Divide the coefficient bundle by it and choose local complements to the kernel and image of the resulting matrix at $\epsilon=0$. An invertible block may be eliminated. The remaining block vanishes at zero and its coefficient of $\epsilon$ is the map from the old kernel to the old cokernel in \eqref{normal:derivative}. A nonzero $r_0$-minor of this coefficient gives an $r_0$-minor of the remaining block with nonzero leading term. The generic rank increases by at least $r_0$, as asserted. \end{proof} \begin{proposition}[Maximal rank of the first normal map]\label{normal:rank} Let $K$ be the naturally retained homology from $S(Y)$ in a prepared background, let $\mathcal H$ be the even homology after adjoining $F$, modulo $K$, and put $a_0=\dim\mathcal H$. Scalar motions $F_i\mapsto F_i+\epsilon z_i$, $z_i\in S_d(Y)$, induce a map \[ \partial_z:\mathcal H\longrightarrow J \] of rank $\min(a_0,j_0)$ for general $z$. The same conclusion holds after the even perturbation in Section~\ref{normal:repair-section}. \end{proposition} \begin{proof} We first identify the remaining homology with a subspace of the coefficients of $F$, and then apply the covector estimate to scalar motions of those generators. The coefficient map is \begin{equation}\label{normal:coefficient-inclusion} \mathcal H\lhook\joinrel\longrightarrow k^f\otimes V. \end{equation} To define it, take the coefficients of the $F$ generators in a cycle and reduce them modulo the degree-$d$ relations of the other generators and $F$. Background cycles and constant Koszul boundaries give zero, so the map is well-defined. To prove injectivity, suppose that the resulting coefficients vanish. Then subtract old--new boundaries to make all $F$ coefficients constant combinations of the $F_i$. Modulo alternating combinations the remaining relation is a symmetric $F$ product. Proposition~\ref{normal:detector} makes that product zero only when the symmetric coefficient is zero. Subtracting the remaining alternating boundary leaves a background cycle, hence an element of $K$. This proves injectivity of \eqref{normal:coefficient-inclusion}. It remains to prove maximal rank. For $[\lambda]\in\mathbb P(J^*)$ with $\dim\ker C_\lambda=k$, the kernel of $\operatorname{id}_{k^f}\otimes C_\lambda$ has dimension $fk$. Its restriction to \eqref{normal:coefficient-inclusion} therefore has rank at least $a_0-fk$. Equivalently, the equations $\lambda\partial_z=0$ impose at least $a_0-fk$ linear conditions on the scalar motion $z=(z_1,\ldots,z_f)$. Consider the incidence of scalar motions and nonzero target covectors that annihilate their image. Together with \eqref{normal:final-strata}, the preceding count bounds its dimension by \[ \dim(S_d(Y)^f)+j_0-a_0-1-(\xi m^d-f)k. \] The same formula without the last term holds for $k=0$. Since $f=O_h(m^{d-1})$, its last coefficient is positive for large $m$. Thus a general motion has no nonzero annihilator when $a_0\geq j_0$; when $a_00$. Choose $[v]\in\mathcal X$ with $\dim T(v)=b$ and put $T_0=\{H\in T:H(v)=0\}$. Then $\operatorname{codim}_T T_0=b$, and every member of $T_0$ descends to a symmetric form on $Z/kv$. By symmetry, $T_0(w)\subset v^\perp$ for every $w$. For general $[w]\in\mathcal X$, the space $T(w)$ is not contained in $v^\perp$. Indeed, choose $H\in T$ with $H(v)\ne0$. Nondegeneracy of $\mathcal X$ gives $H(v)(w)=H(w)(v)\ne0$ for general $w$. It follows that \[ \dim T_0(w)\le\dim\bigl(T(w)\cap v^\perp\bigr)\le b-1. \] If $\dim Z=1$, then $T_0=0$. Otherwise the closure of the projection of $\mathcal X$ in $\mathbb P(Z/kv)$ is irreducible and nondegenerate. Apply induction to the descended space $T_0$ on this projection. It gives $\dim T_0\le\binom b2$, and hence $\dim T\le b+\binom b2$. \end{proof} \begin{proposition}\label{low2:cv:linear-image} For every $i$-dimensional subspace $L\subset M_1$, \begin{equation}\label{low2:cv:sharp-linear} \dim(UL)\ge b(i):= B-\binom{p-\lceil i/2\rceil+1}{2} \ge\frac BA\,i. \end{equation} For each $0\le i\le A$, the first bound is attained by some $L$. \end{proposition} \begin{proof} Put $Z=V_{t-2}$, so $\dim Z=p$. Lemma~\ref{low2:cv:hilbert} identifies $M_1^*=Z\otimes V_1$ and $M_2^*=\operatorname{Inv}^2(Z)$. Let $K=L^\perp\subset Z\otimes V_1$, with $k_0=\dim K=2p-i$, and let $T\subset\operatorname{Inv}^2(Z)$ annihilate $UL$. Write a member of $M_2^*$ as a symmetric polynomial $H(z,a)$ of degree $t-2$ in both binary slots. Its inverse expression is $\det(s,z)\det(s,a)H(z,a)$. In the dual of multiplication $U\otimes M_1\to M_2$, removing the degree-one inverse factor $\det(s,z)$ leaves \[ \det(s,a)H(z,a). \] Thus all coefficient contractions in the $a$ slot belong to $K$ when $H\in T$. In particular, for each binary point $a$, \begin{equation}\label{low2:cv:evaluated-contraction} H(z,a)\otimes\det(s,a)\in K. \end{equation} Let $\lambda_a=k\det(s,a)\subset V_1$ and define \[ Z_a=\{f\in Z:f\otimes\det(s,a)\in K\}. \] Write $b_0$ for the dimension of $Z_a$ at a general binary point. For two distinct general points $a,a'$, the quotient maps \[ K\longrightarrow Z\otimes(V_1/\lambda_a),\qquad K\longrightarrow Z\otimes(V_1/\lambda_{a'}) \] have rank $k_0-b_0$ and are jointly injective. Their kernels lie in the disjoint subspaces $Z\otimes\lambda_a$ and $Z\otimes\lambda_{a'}$. Consequently $k_0\le2(k_0-b_0)$, or $b_0\le\lfloor k_0/2\rfloor$. The evaluations at binary points form a nondegenerate rational normal curve in $\mathbb P(Z^*)$; for $t=2$ this is the single point of $\mathbb P^0$. Regard $T\subset\operatorname{Inv}^2(Z)$ as symmetric maps $Z^*\to Z$. Equation~\eqref{low2:cv:evaluated-contraction} shows that its general evaluation image lies in $Z_a$. Lemma~\ref{low2:cv:symmetric-evaluation} therefore gives \[ \dim T\le\binom{b_0+1}{2} \le\binom{\lfloor(2p-i)/2\rfloor+1}{2} =\binom{p-\lceil i/2\rceil+1}{2}. \] Since $\dim(UL)=B-\dim T$, this proves the first inequality. For $a_0=\lceil i/2\rceil\le p$, \[ b(i)=\frac{a_0(2p+1-a_0)}2 \ge\frac{a_0(p+1)}2\ge\frac{p+1}{4}i=\frac BA i. \] For sharpness, choose $Z_0\subset Z$ of dimension $p-\lceil i/2\rceil$. Take $K$ to contain $Z_0\otimes V_1$ and to have dimension $2p-i$, adding one dimension if $i$ is odd. Every $H\in\operatorname{Inv}^2(Z_0)$ has all the displayed contractions in $Z_0\otimes V_1$, so it annihilates $UL$ for $L=K^\perp$. Its dimension equals the preceding upper bound for $T$, which proves equality. \end{proof} \subsubsection{Images of the output space} The preceding estimate starts with a subspace of $M_1$. When free variables are adjoined, we will also need the images of subspaces of $M_0=D$. The next proposition supplies both their linear and quadratic multiplication bounds. \begin{proposition}\label{low2:cv:output} For every subspace $D'\subset D$ of dimension $r$, \begin{equation}\label{low2:cv:output-linear} \dim(UD')\ge A\min(r/2,1). \end{equation} If $D'\ne0$, then \begin{equation}\label{low2:cv:output-quadratic} \operatorname{Sym}^2(U)D'=M_2. \end{equation} \end{proposition} \begin{proof} Let $F=(D')^\perp\subset D^*=V_2$. Project the $(t+2)$-dimensional column space of \eqref{low2:cv:presentation} to $(D/D')\otimes U$. The transpose of this projection is binary multiplication $F\otimes V_{t-1}\to V_{t+1}$. For $r=1,2$, its image has dimension at least $t+2-r$. To see this, choose a basis of $F$ with distinct leading binary exponents and multiply by all $t$ monomial exponents of $V_{t-1}$. The resulting sumset contains at least $t+\dim F-1=t+2-r$ exponents, which give that many independent products. The intersection of the column space with $D'\otimes U$ therefore has dimension at most $r$. Hence $\dim(UD')\ge rt-r=r(t-1)$. This is \eqref{low2:cv:output-linear} for $r=1,2$; the cases $r=0,3$ are immediate. It remains to prove \eqref{low2:cv:output-quadratic} for a nonzero vector $e\in D'$. The inverse expression of $H\in M_2^*$ is $\det(s,z)\det(s,a)H(z,a)$. Contraction in the $s$ slot by $e$ produces \[ B_e(z,a)H(z,a), \] where $B_e$ is a nonzero symmetric bilinear polynomial. Explicitly, if $e$ takes values $\alpha,\beta,\gamma$ on $s_0^2,s_0s_1,s_1^2$, then \[ B_e(z,a)=\alpha z_1a_1-\beta(z_1a_0+z_0a_1)+\gamma z_0a_0. \] Thus $B_e\ne0$ whenever $e\ne0$. Multiplication by $B_e$ is injective in the polynomial ring, so the dual of $\operatorname{Sym}^2(U)e\to M_2$ is injective. The original map is therefore onto. \end{proof} \subsubsection{Adjoining free variables} We now pass from the $t$ variables in the presentation to the full space of $m$ variables. Let $W$ have dimension $w$, set $Y=U\oplus W$ and $m=t+w$, and put $N=M\otimes\operatorname{Sym}(W)$. The column count is $c=t+2$, so $t=c-2$. The relevant pieces are \begin{align} N_1&=M_1\oplus(D\otimes W),\label{low2:cv:free-pieces}\\ N_3&=(M_2\otimes W)\oplus(M_1\otimes\operatorname{Sym}^2W) \oplus(D\otimes\operatorname{Sym}^3W).\notag \end{align} In particular, \begin{equation}\label{low2:cv:free-dimensions} \begin{split} a:=\dim N_1&=A+3w=3m-c,\\ T:=\dim N_3&=Bw+A\binom{w+1}{2}+3\binom{w+2}{3}. \end{split} \end{equation} Extension by the polynomial ring in $W$ preserves the absence of coefficient syzygies through degree two. Equivalently, decompose each coefficient polynomial by its $W$-monomial and apply Lemma~\ref{low2:cv:hilbert} to its $U$-coefficient. Hence \begin{equation}\label{low2:cv:free-euler} T=3\binom{m+2}{3}-c\binom{m+1}{2}. \end{equation} \subsection{Quadratic coefficients and conditioned covectors}\label{low2:inc:section} We now convert multiplication-image bounds into generic rank statements. There are two related incidences. The first excludes relations among products with the quadrics in $S_2Y$. The second bounds the covectors that annihilate those products and have a prescribed coefficient-kernel dimension. This second bound will control the cokernel of the deformation in Section~\ref{low2:transfer-section}. In both arguments, the quadrics in $S_2Y$ vary as one common $q$-dimensional space. Let $X$ and $Y$ have dimensions $3$ and $m$, respectively, and let $E\subset X\otimes Y$ be a $c$-plane, with $4\le c\le m$. Its polynomial presentation has cokernel \[ N=\operatorname{coker}\left(E\otimes\operatorname{Sym}(Y)(-1) \longrightarrow X\otimes\operatorname{Sym}(Y)\right). \] Work on the nonempty open where the presentation has no coefficient syzygies in degrees zero, one, or two. The convolution presentation provides such a point. Set \begin{equation}\label{low2:inc:dimensions} a=\dim N_1=3m-c,\qquad T=\dim N_3=3\binom{m+2}{3}-c\binom{m+1}{2}. \end{equation} Suppose that, for every $v$-plane $L\subset N_1$, a specified function $\mathcal B_m(v)$ satisfies \begin{equation}\label{low2:inc:expansion} \dim\bigl(\operatorname{Sym}^2Y\,L\bigr)\ge \mathcal B_m(v). \end{equation} \begin{proposition}\label{low2:inc:outer} If \begin{equation}\label{low2:inc:outerinequality} \mathcal B_m(v)\ge v(q+a-v)\qquad(1\le v\le\min(q,a)), \end{equation} then a general $q$-plane $Q\subset\operatorname{Sym}^2Y$ makes $Q\otimes N_1\to N_3$ injective. If also $Q\otimes Y\to\operatorname{Sym}^3Y$ is injective, then, writing $A=\operatorname{Sym}(Y)/(Q)$, the induced map \begin{equation}\label{low2:inc:outermap} F_{13}:E\otimes A_2\longrightarrow X\otimes A_3 \end{equation} is injective and has cokernel $J=N_3/(QN_1)$ of dimension \begin{equation}\label{low2:inc:j} j=T-qa=3\left[\binom{m+2}{3}-mq\right] -c\left[\binom{m+1}{2}-q\right]. \end{equation} \end{proposition} \begin{proof} Parametrize $Q$ first by ordered quadratic coefficients $f_1,\ldots,f_q$. A nonzero tuple $(v_1,\ldots,v_q)\in N_1^{\oplus q}$ whose entries span a $v$-plane has projective parameter dimension $v(q+a-v)-1$. At a fixed tuple, the equation $\sum_i f_i v_i=0$ imposes exactly $\dim(\operatorname{Sym}^2Y\,\langle v_i\rangle)$ linear conditions on the ordered quadrics. Indeed, after a constant change among the entries, the nonzero $v_i$ include a basis of their span. By \eqref{low2:inc:outerinequality}, every such relation incidence has dimension smaller than the quadratic coefficient space. The union over the finitely many possible $v$ is proper. Restricting to linearly independent ordered quadrics proves injectivity for general $Q$. Both $Q\otimes N_1\to N_3$ and \eqref{low2:inc:outermap} have cokernel \[ \frac{X\otimes\operatorname{Sym}^3Y} {E\operatorname{Sym}^2Y+X\otimes QY}. \] Cubic independence gives their common Euler difference \eqref{low2:inc:j}. The first map is injective, so its cokernel has dimension $j$. The second therefore has zero kernel as well. \end{proof} We next refine the incidence count by recording the kernel of a covector's coefficient map. These are the ``conditioned covectors'' of the subsection title. \begin{proposition}\label{low2:inc:conditioned} Assume \eqref{low2:inc:expansion} and \eqref{low2:inc:outerinequality}. For a general pair $(E,Q)$, put $J=N_3/(QN_1)$ and, for $0\ne\lambda\in J^*$, define \[ R_\lambda:N_1\longrightarrow (\operatorname{Sym}^2Y/Q)^*,\qquad R_\lambda(v)(f)=\lambda(fv). \] Let $\Sigma_v\subset\mathbb P(J^*)$ be the locus where $\dim\ker R_\lambda=v$. Then \begin{equation}\label{low2:inc:conditionedbound} \dim\Sigma_v\le U_m(v):= j-\mathcal B_m(v)+qv+v(a-v)-1\qquad(0\le v2449/1000$ already proves these inequalities. On the first interval use the auxiliary quantity $D=\ell$, on the last use $D=a_0-\ell$, and on the two middle intervals use $D=a_0/2$. In each case $D\geq\min(\ell,a_0-\ell)$. On each interval $D$ is affine in $x$ and in the profile coordinates. Subtracting $a_0D/1000$ therefore preserves the concavity and affine dependence used in the vertex reduction above. Also $a_0D<9$ at all vertices. The strict table therefore implies \[ a_0I-T_0\ell>\frac{a_0D}{1000} \] at every nontrivial vertex. The empty and full vertices have both sides identically zero, including before passage to the limit. The normalized finite-dimensional quantities depend continuously on $t/m$ and $1/m$. Thus the same weak inequalities hold for all sufficiently large $m$. Concavity in the ordered triple, followed by affine interpolation in $x$ on each interval, proves them for every $L_0$. Restoring the powers of $m$ gives \eqref{low2:strong-shadow}. Finally work on the coefficient open where $N_1$ and $N_3$ have dimensions $a,T$. Failure of any specified image-rank lower bound is closed on the relative Grassmannian of $v$-planes. Its projection to the coefficient space is closed by properness. The convolution point avoids it. Intersecting the finitely many resulting opens, for $0\leq v\leq a$, gives the last assertion. \end{proof} \subsection{The quadratic deformation and the two defects} \label{low2:transfer-section} In this subsection assume $\operatorname{char}k\ne2$. We now combine the block calculations and incidence bounds to prove the recurrence for the two defects. We first check the numerical hypotheses and record the homology of the split system. We then construct its leading deformation map to $J$ and use the conditioned covector bound to prove that this map has maximal rank. The resulting rank increase gives the recurrence. \subsubsection{Numerical hypotheses and the mixed maps} Retain the counts $q,c$ from the preceding subsection. Put $\alpha=N_2(m)-q$, $\beta=N_3(m)-mq$, and \[ j=3\beta-c\alpha=T-qa,\qquad H=3mc-2\alpha-\binom c2,\qquad k_{31}=2m+2c. \] For all sufficiently large $m$ the preceding dimensions satisfy \begin{equation}\label{low2:counts} \frac m20,\quad j>0,\quad j=O(m^2),\quad \alpha>\binom{c+1}{2}+3. \end{equation} We check these inequalities directly from the critical-root formula, which gives \[ \tau m-41/4$ and $3\tau-2/\sqrt6-\tau^2/2>2/3$. These prove \eqref{low2:counts}. The cubic multiplication map $Q\otimes Y\to S_3Y$ is injective for general $Q$. We verify this by counting possible linear relations among the quadrics. A nonzero $q$-tuple of linear coefficients spanning a $b$-plane has projective dimension $b(m+q-b)-1$, and its relation imposes \[ N_3(m)-N_3(m-b) \] linear conditions on the ordered quadrics. If $mq\leq N_3(m)$, the difference between this condition count and the tuple dimension is positive: divided by $b$, it is \[ F(b)=\frac{m^2}{2}-\frac{mb}{2}+\frac{b^2}{6} +\frac b2+\frac13-q+\frac1b, \] which decreases on $1\leq b\leq m$ and satisfies $F(m)=N_3(m)/m-q+1/m>0$. The required inequality $mq0$, and $D_m>a$ eventually, \[ \mathcal B_m(v)-v(q+a-v) =\frac ja v-v(a-v)+D_m\min(v,a-v)\geq0. \] Thus the outer map $F_{13}$ is injective by Proposition~\ref{low2:inc:outer}; the middle map $F_{22}$ is onto by Proposition~\ref{low2:tb:middle}. \subsubsection{The distinguished coefficient and the split complex} To isolate the pure $Y$ homology affected by the distinguished generator, choose a general flag $Q_-=\langle q_1,\ldots,q_{q-1}\rangle \subset Q=Q_-\oplus kf_*$, and write \[ c_0=C_m(q),\qquad k_0=H_m(q-1),\qquad \eta=-\chi_m(q). \] We may require $f_*^2\notin Q_-S_2Y$: the latter product space is proper because $q-1<\kappa_2(m)$, and squares of quadrics span $S_4Y$: for quadratic forms $a,b$, the identity $2ab=(a+b)^2-a^2-b^2$ expresses every product as a linear combination of squares. Reading the coefficient of $f_*$ modulo $Q$ gives \[ \theta:H_1(Q)_{4}\longrightarrow A_2=S_2Y/Q, \qquad \ker\theta=H_1(Q_-)_{4}. \] Indeed, if that coefficient vanishes modulo $Q$, constant Koszul boundaries reduce it to a scalar multiple of $f_*$. The cycle equation and the excluded-square condition force that scalar to be zero. The formal symmetric-square description in \eqref{uf:formal-homology} makes the remaining inclusion of homology from $S(Y)$ injective. Consequently the image $W$ of $\theta$ has dimension \[ w_*=\eta+c_0-k_0. \] The next projection removes the pure $Y$ cokernel while we study the deformation; we will restore the full target at the end. Take a complement $Z_0$ of $QS_2Y$ in $S_4Y$ and first project the degree-four target modulo $Z_0$. For the split generator space $Q_X\oplus E\oplus Q$, the projected Koszul complex has the following homology and cokernel: \[ \begin{array}{c|cc} \text{bidegree}&\text{homology}&\text{cokernel}\\\hline (4,0)&3&0\\ (3,1)&k_{31}&0\\ (2,2)&H&0\\ (1,3)&0&j\\ (0,4)&\eta+c_0&0. \end{array} \] For $(3,1)$ the source, target, and incoming boundary dimensions are $12m+6c$, $10m$, and $4c$, respectively. For $(2,2)$, eliminating the pure images leaves $F_{22}$ and the independent alternating source $\bigwedge^2E$. The outer injection handles $(1,3)$; the remaining rows are the complexes in the pure $X$ and pure $Y$ degrees. Thus all entries follow from the preceding rank statements. \subsubsection{The leading deformation map} Let $r_1,\ldots,r_4\in A_2$ be the directions in which the pure quadrics $h_1,\ldots,h_4$ move, and use the direction $s_X=\omega u-v\in P$ for the distinguished generator. Proposition~\ref{low2:tb:augmented} applies to this particular $W$, without a genericity assumption on $W$. If $W_{\mathbf r}$ denotes the three-dimensional image of $\xi\mapsto\sum_bR_b(\xi)\otimes r_b$, it gives an injection \[ \operatorname{Sym}^2E\oplus W_{\mathbf r}\oplus(s_X\otimes W) \longrightarrow P\otimes A_2. \] Consequently \[ T_{\mathbf r,s_X} =F_{22}^{-1}\bigl(W_{\mathbf r}\oplus(s_X\otimes W)\bigr) /\bigwedge^2E \] has dimension $t_0=H+3+w_*$ and injects into $E\otimes P_1$, where $P_1=(X\otimes Y)/E$. To check injectivity, a vector in its kernel is represented in $E\otimes E$. Its product lies both in the symmetric-product image and in the trace image; the displayed directness makes the product zero, leaving exactly the alternating space already divided out. Choose lifts of the $r_b$ to $S_2Y$, and lift $s_X$ to $\omega x^2-y^2\in S_2X$; we use the same notation for these lifts. Deform the quadratic generators by \[ g_i\mapsto g_i+\varepsilon p_i,\qquad h_b\mapsto h_b+\varepsilon r_b,\qquad f_*\mapsto f_*+\varepsilon s_X, \qquad p_i\in S_2Y, \] and keep $Q_-$ fixed. After eliminating the split image, the first-order terms on the middle and $(3,1)$ homology, together with the second-order terms on the pure $X$ homology and the distinguished homology from $S(Y)$, give the map \begin{equation}\label{low2:leading} \begin{split} T_{\mathbf r,s_X}\oplus K_{31}&\longrightarrow J,\\ (z,[(a,b)])&\longmapsto \left[\sum_i p_i z_i+\sum_{b=1}^4r_b b_b\right]. \end{split} \end{equation} Here $K_{31}$ is represented by full relations $\sum_i g_i a_i+\sum_b h_b b_b=0$, and its map to $P_1^4$ is injective by Lemma~\ref{low2:tb:k31injective}. To justify that this leading map detects a rank increase, write multiplication as $M_0+\varepsilon M_1$. A cycle $a$ in the pure $X$ or pure $Y$ degree first maps to the surjective $(2,2)$ row. Choose a complete source correction $a_1$ with $M_0a_1=-M_1a$. The coefficient of $\varepsilon^2$ is then $M_1a_1$, whose only component in $J$ is its mixed coefficient contribution $\sum_i p_i z_i$. The other derivatives have pure $X$ or pure $Y$ bidegree. They are zero or lie in the image of $M_0$ after the projection; eliminating them contributes only at a later order. The same calculation for the middle and $(3,1)$ cycles gives their first-order terms in \eqref{low2:leading}. Quotient by the moving constant Koszul image on a fixed nonzero minor chart, eliminate the rows and columns on which $M_0$ is an isomorphism, and divide the first-order and second-order columns by $\varepsilon$ and $\varepsilon^2$, respectively. Their constant matrix is \eqref{low2:leading}, up to signs and additions of first-order columns. Hence each nonzero minor of this map yields a rank increase of the polynomial multiplication matrix. This argument also shows why projecting $Z_0$ does not change the selected leading terms: a correction entering a pure row can return to the $(1,3)$ row only after another deformation. \subsubsection{Maximal rank and the two defect bounds} It remains to prove that \eqref{low2:leading} has maximal rank. We bound its annihilating covectors by stratifying them according to the dimension of their coefficient kernels. In the numerical bounds below, $x_+=\max\{x,0\}$ denotes the positive part of a real number, distinct from the truncation notation for power series. For a covector in the stratum $\Sigma_v$ of Proposition~\ref{low2:inc:conditioned}, the $K_{31}$ equations impose at least $(k_{31}-4v)_+$ conditions on the four $r_b$. After fixing these motions and $s_X$ subject to the injectivity condition above, the remaining equations impose at least $(t_0-cv)_+$ independent conditions on the $p_i$, because $T_{\mathbf r,s_X}\hookrightarrow E\otimes P_1$. These are successive fiber bounds for the same motions, so their sum is valid even though $T_{\mathbf r,s_X}$ depends on $\mathbf r$. Put $h_0=k_{31}+t_0$; the sum is at least $h_0-(c+4)v$. For $0\leq v0$, we use the strengthened assertion \begin{equation}\label{low2:square-property} H_n(r_n^-)=0, \qquad \zeta^2\notin Q_-S_2 \quad\text{for some quadratic form }\zeta. \end{equation} At an integral critical root only ordinary endpoint exactness is required. We shall prove the two endpoints and \eqref{low2:square-property} in every dimension $n\geq28$. The strengthened assertion is open. Indeed, if $R=r_n^-N_2(n)-\binom{r_n^-}{2}$, it is expressed by a rank-$R+1$ minor of the multiplication matrix with the column $\zeta^2$ adjoined. The rank of the ordinary matrix is at most $R$, so such a minor proves both assertions at once. Nonvanishing of a coefficient of this minor, as a polynomial in $\zeta$, gives an open condition on $Q_-$. Consequently, on the common flag space, a general extension $Q=Q_-+kf$ satisfies $f^2\notin Q_-S_2$ whenever \eqref{low2:square-property} holds. The coefficient reader used in the preceding subsection then has kernel $H_1(Q_-)_4=0$. Use the scalar choice $s_X=\omega u-v$ and the functional $\ell$ from \eqref{low2:tb:scalar}. Their three nonzero trace coefficients are $-(1+\omega),\omega,1$. \begin{lemma}\label{low2:extra-square-product} Suppose $c0$ and append the fixed column $\zeta^2$ to the moving multiplication matrix. Its split bidegree is $(2,2)$, where the quotient is zero, so this does not increase the split rank. Replace the middle preimage space by \[ T^+=F_{22}^{-1} \bigl(W_{\mathbf r}\oplus(s_X\otimes W)\oplus k[\zeta^2]\bigr) /\bigwedge^2E. \] Lemma~\ref{low2:extra-square-product} gives $\dim T^+=t_0+1$ and an injection $T^+\hookrightarrow E\otimes P_1$. Indeed, a vector killed by this injection has mixed coefficients in $E\otimes E$. Its product lies both in the symmetric-product image and in the trace-plus-square image; their intersection is zero, leaving only the alternating boundaries. Write the moving multiplication matrix as $M_0+\varepsilon M_1$. If $M_0b=\zeta^2$, the additional split kernel vector is $(b,-1)$, whose response is $\varepsilon M_1b$. Its $(1,3)$ component is $\sum_i p_i b_i$. Its other first-response components are in pure rows and are eliminated by the split image. Thus, after the same row elimination as in \eqref{low2:leading}, the augmented leading map is \[ T^+\oplus K_{31}\longrightarrow J,\qquad (z,[(a,b)])\longmapsto \left[\sum_i p_i z_i+\sum_{b=1}^4r_b b_b\right]. \] The added square column is fixed throughout this calculation. For a covector with $v$-dimensional coefficient kernel, impose its equations first on the four pure motions and then, with these fixed, on the mixed motions. Their total codimension is at least \begin{equation}\label{low2:augmented-codimension} C^+(v)=(k_{31}-4v)_++(t_0+1-cv)_+. \end{equation} The dependence of $T^+$ on the pure motions is therefore respected. The numerical verification below shows \begin{equation}\label{low2:augmented-incidence} \dim\Sigma_v-C^+(v)\leq e-2 \end{equation} for every nonempty covector stratum. The projective annihilator of the general leading image has dimension at most $e-2$, so this image has dimension $h+1$. The augmented polynomial matrix consequently has rank at least $R+1$, where $R=rN_2(m+3)-\binom r2$ is the upper bound for the ordinary matrix. Deleting the fixed column proves that the ordinary rank is $R$ and that $\zeta^2$ is outside its image. This is the strengthened lower assertion. When $e=0$, and at the upper endpoint, the unaugmented calculation proves ordinary exactness. \end{proof} \subsubsection{The transfer inequalities} We give the finite bounds and the uniform estimate used in the proposition. In addition to the preceding counts, put \[ t=c-2,\quad p=t-1,\quad w=m-t,\quad A=2p,\quad B=\binom{p+1}{2},\quad T=Bw+A\binom{w+1}{2}+3\binom{w+2}{3}. \] Thus $a=A+3w$ and $j=T-qa$. For $k=2,3$ write \[ H_k(n)=\binom{w+k-1}{k}-\binom{w-n+k-1}{k}. \] For each endpoint, the required structural inequalities are \begin{equation}\label{low2:finite-structure} \begin{gathered} 4\leq c\leq m,\quad mq\leq N_3(m),\quad \alpha\geq\binom{c+1}{2}+3,\\ q\geq c w_0+\binom{w_0+1}{2},\qquad w_0=(m-2c)_+, \end{gathered} \end{equation} together with $j>0$, $H\geq0$, and $0\leq\delta\leq\alpha$. At a strengthened lower endpoint, $c0$, $j/a\leq2m/3+1$, and $a<5m/2+4$. With $D_m=m^2/100$, for $m\geq320$ one has \[ D_m>3m+4\geq c+4+a-j/a,\qquad D_m>19m/6+5>a+j/a. \] For $0m$, the cubic margin is at least $(1/6-\rho)m^3-5m/3$, and $H>2m^2/3-12m-10$. These verify the structural conditions, including the extra square. \subsubsection{Three initial dimensions} We give a deterministic construction of the required forms over $\mathbf F_2$. The following integer recurrence specifies all their coefficients. Let $a\mathbin{\oplus}b$ denote exclusive-or of the binary digits of nonnegative integers. Define \[ \begin{aligned} C_0&=11400714819323198485,\\ C_1&=13787848793156543929,\\ C_2&=10723151780598845931. \end{aligned} \] For each $n\in\{28,29,30\}$, restart the following recurrence at $s_0=n$. All residues are chosen in $\{0,\ldots,2^{64}-1\}$: \[ \begin{aligned} s_j&=(s_{j-1}+C_0)\bmod 2^{64},\\ a_j&=C_1\bigl(s_j\mathbin{\oplus}\lfloor s_j/2^{30}\rfloor\bigr) \bmod 2^{64},\\ b_j&=C_2\bigl(a_j\mathbin{\oplus}\lfloor a_j/2^{27}\rfloor\bigr) \bmod 2^{64},\\ w_j&=b_j\mathbin{\oplus}\lfloor b_j/2^{31}\rfloor \qquad(j\geq1). \end{aligned} \] List the quadratic monomials $m_0,\ldots,m_{b-1}$ in the lexicographic order of their nondecreasing index pairs, where $b=N_2(n)$, and put $L_n=\lceil b/64\rceil$. Write $\operatorname{bit}_t(w)=\lfloor w/2^t\rfloor\bmod2$. The $i$th quadratic form is \[ f_i=\sum_{\ell=0}^{b-1} \operatorname{bit}_{\ell\bmod64} \bigl(w_{(i-1)L_n+\lfloor\ell/64\rfloor+1}\bigr)m_\ell \qquad(1\leq i\leq r_n^+). \] The lower list is the prefix $f_1,\ldots,f_{r_n^-}$ of the upper list. Let $M_r$ be the matrix whose columns are the coefficient vectors of $f_i m_\ell$, $1\leq i\leq r$, $0\leq\ell\frac{d(d-1)\rho}{2d!}. \end{equation} All the additional scalar lists have size $O_h(m^{d-2})$, and the pure family has a fixed number of labels as $m$ grows. \subsection{The intermediate multiplication maps} Choose $E\subset D\otimes S_{d-2}Y$ of dimension $e$. If $d\geq5$, also choose $G\subset S_4X\otimes S_{d-4}Y$ of dimension $g$. The formal product spaces to be considered are just \begin{equation}\label{unified:product-rows} \Pi_4=S_2E,\qquad \Pi_6=E\otimes G,\qquad \Pi_8=S_2G, \end{equation} with the last two omitted when $G$ is absent. In particular, each product row has a single source. Put $P_j=d!/[j!(2d-j)!]$ and $B_j=\binom{2d-j}{d-j}$. The parameter inequalities used below are \begin{align} \frac{\beta_d}{P_2(1-\ell_d^2)}+B_2\rho&<1, \label{unified:sparse-E}\\ \frac{2\gamma_d}{P_4}+B_4\rho&<1\quad(d\geq5), \label{unified:sparse-G}\\ \binom{2d-R}{d-R}\rho&<\frac25 \quad(R\in\{4,6,8\},\ R\leq d). \label{unified:scalar-profile} \end{align} We also need the product capacities in the following table. A capacity is the leading coefficient after division by $h^2m^{d-2}$ for $E$, or by $h^4m^{d-4}$ for $G$. \begin{equation}\label{unified:capacity-table} \begin{array}{c|c|c} \text{product}&\text{range}&\text{capacity}\\ \hline S_2E&d=3&5/64\\ S_2E&d=4&5/512\\ S_2E&5\leq d\leq8& \displaystyle\frac{1/8-\ell_d^2/2}{2^{d-1}(d-2)!}\\[5pt] E\otimes G\text{, }E\text{ factor}&5\leq d\leq8& \displaystyle\frac{2/25-\ell_d^2/2}{2^{d-2}(d-2)!}\\[5pt] E\otimes G\text{, }G\text{ factor}&5\leq d\leq8& \displaystyle\frac{(3/5)^4}{24\cdot2^{d-4}(d-4)!}\\[5pt] S_2G&d=5&1/512\\ S_2G&6\leq d\leq8& \displaystyle\frac1{512\cdot2^{d-4}(d-4)!} \end{array} \end{equation} Every $E$ capacity is strictly greater than $\beta_d$, and every $G$ capacity is strictly greater than $\gamma_d$. These, \eqref{unified:D}, \eqref{unified:reserve}, and \eqref{unified:sparse-E}--\eqref{unified:scalar-profile} follow by rational substitution. Explicitly, put $H_d=\binom{2d-1}{d-1}$, $M=10^8$, and \[ (q_2,q_3,q_4,q_5,q_6,q_7,q_8) =(18350341,5131670,1438923,397615,108283,29141,7770). \] Then $q_d/M<\rho_d<(q_d+1)/M$. These bounds are verified by $H_d(M-q_d-1)^2<(H_d-1)M^2(5/32+o(1))h^2$. Use $Y$ itself as the scalar space in degree three and the degree-two space of Lemma~\ref{products:matching} in degree four. Lemma~\ref{low:separated-products} gives the first two capacities in \eqref{unified:capacity-table}. When $G$ is present, use the degree-two space $O\subset S_2X$ of Lemma~\ref{products:matching}, associated with equal halves of $X$, and replace it by $O\cap D$. It still has independent symmetric products, and \[ \dim(O\cap D)\geq\dim O-c_D =\left(\frac18-\frac{\ell_d^2}{2}+o(1)\right)h^2. \] Use the degree-$(d-2)$ space of Lemma~\ref{products:matching} in $Y$. Lemma~\ref{low:separated-products} gives the third capacity. Every $E$ product has bidegree $(2,2)$ with respect to this splitting of $X$. Thus the components of $S_4X$ with odd degree in the first summand are available for the new $G$ family. For $E\otimes G$, split $X=A\oplus B$ with dimensions $2h/5,3h/5$ and split $Y=C\oplus C'$ into nearly equal halves. Place the two families in \[ (D\cap S_2A)\otimes S_{d-2}C, \qquad S_4B\otimes S_{d-4}C', \] respectively. Their product map is injective because the variable sets are disjoint. The bound $\dim(D\cap S_2A)\geq N_2(2h/5)-c_D$ gives exactly the two cross-product capacities in the table. For $S_2G$, use the quartic output space of Lemma~\ref{products:quartic}, together with the full linear scalar space if $d=5$ and the corresponding space of Lemma~\ref{products:matching} otherwise. The separated-product lemma gives the last two capacities. Whenever the product row also has a scalar source, its product image lies in a single bidegree for the chosen splitting of $Y$. Delete that bidegree component from the scalar target. The divisibility coupling in the proof of Lemma~\ref{normal:deleted-profile}, now with source degree $d-R$ and multiplier degree $d$, retains at least $2/5$ of every source row for large $m$. Indeed the removed fraction tends to an atom of $\operatorname{Bin}(d,1/2)$, which is at most $1/2$. Passing to initial subspaces proves the same estimate for every coefficient subspace. Inequality~\eqref{unified:scalar-profile} and coefficient incidence separate the entire scalar image from the product image. The full linear scalar spaces used for $d=3,5$ occur only in rows $R>d$, where there is no scalar source. At row four, put the new $G$ outputs in the components of $S_4X$ with odd degree in the first summand, whose combined dimension is $(1/48+o(1))h^4$. They are disjoint from the $E$ products. The sparse quotient estimate with \eqref{unified:sparse-G} gives exactly the scalar--$G$ Koszul kernel on these outputs; on their complement use the preceding scalar-avoidance estimate. There are no other new-family/product collisions. We have thus established all the row maps of Lemma~\ref{construct:row-maps} for the present parameters. \subsection{The background and the target maps} Use the two backgrounds \begin{equation}\label{unified:backgrounds} W_{\rm bg}=\begin{cases} \langle Q,E+a_2,G+a_4,U+P\rangle,&d\text{ odd},\\ \langle Q,E+a_2,G+a_4,U+P_2+b\rangle,&d\text{ even}, \end{cases} \end{equation} omitting the $G$ family when absent. Here $a_2,a_4,b$ are scalar degree-$d$ lists; $P\subset XS_{d-1}Y$ and $P_2\subset DS_{d-2}Y$ have $u$ labels. All scalar forms belong to the same list $\mathcal S=Q,a_2,a_4,b$, with absent terms omitted. Its count is $(\rho/d!)m^d+O_h(m^{d-1})$. For odd $d$, the private-variable argument in Lemma~\ref{construct:row-maps} includes $P$ without changing any leading inequality. More explicitly, specialize $P_i=w_i z_i^{d-1}$ and use only the other $Y$ variables in the preceding witnesses. Projecting an output space by $w_iS_{R-1}X$ loses $O(h^{R-1})$ directions. The sparse inequality has a strict margin of order $h^R$, so the necessary injection of the linear $Y$ multiples of the new family persists in this quotient. At lower coefficient degrees there is no kernel: multiply a putative relation by two coprime core monomials to reach the established row map. The resulting constant Koszul matrices vanish by comparison and the coefficient injection of the new family. The only private-power overlap is consequently $\bigwedge^2P$ at row two, as in that lemma. The coefficient elimination of Proposition~\ref{construct:background} now applies with $\mathcal J=\{2,4\}\cap\{1,\ldots,d-1\}$. The retained symmetric and cross coefficients are removed by the three product rows \eqref{unified:product-rows}. The alternating $P$ coefficients are removed by the $(U+P)$ Koszul boundaries; their higher components are removed at the later $U$ column. For even $d$, first set $U=0$ and absorb the fixed $P_2$ list into $E$ with its scalar list $b$. After proving the required rank, restore $U$ on the open where it is a basis of $S_dX$. This proves \begin{equation}\label{unified:homology} H_1(W_{\rm bg};S)_{2d}=H_1(Q;S(Y))_{2d}. \end{equation} The equality holds also with one extra $E$ generator and after projection of a complement to $QS_dY$ in $S_{2d}Y$. For completeness, the odd-parity specialization in that proposition uses, in every row $t\geq3$, the strict bound \[ \rho\left[\frac{th}{h+t-1}\binom{2d-t}{d-1} +\binom{2d-t}{d}\right]N_{d+1}(h)+h^2$ and Lemma~\ref{construct:incidence} give $UX=S_{d+1}X$; for even $d$ we have $U=S_dX$. Every remaining target is filled. \subsection{The normal map and the recurrence} The product injection of Proposition~\ref{normal:detector} applies because $L>2^d\rho h$ and $c_D\geq2h-1$. Its denominator contains the entire common scalar list and, in odd degree, the fixed $P$ list. It therefore separates $S_2F$ in the prepared quotient \eqref{unified:backgrounds}. Together with \eqref{unified:homology}, it gives the injective coefficient map \eqref{normal:coefficient-inclusion}. We check the numerical hypotheses for the odd cokernel argument. Its coefficient space is exactly \eqref{normal:even-source} or \eqref{normal:odd-source}, according to parity. In odd degree put $a_X=N_{d-1}(h)$, $b_X=N_d(h)-u$. The projection argument of Lemma~\ref{normal:top} applies since $ub_X/a_X^2\to\infty$. For $d=5,7$, one has $b_X/a_X\geq h/(2d)$ for large $h$ by its displayed limit. For $d=3$, write $u=h^3/12+\theta$, $0\leq\theta<1$; then \[ b_X-\frac h6a_X=\frac{5h^2}{12}+\frac h3-\theta>0. \] With $\delta=1/[100(2d)!]$, the remaining top-degree inequality is \begin{equation}\label{unified:top} (2d^2+1)\rho+d!\delta<1\qquad(d=3,5,7). \end{equation} In particular the degree-three case is $19\rho_3+1/12000<1$. Thus the exact sequence \eqref{normal:top-sequence} and the affine equation \eqref{normal:affine-top} remain valid; the relation $U=-P$ is retained throughout. For each intermediate coefficient degree, the inequality needed after adding $\lfloor\delta m^d\rfloor$ scalar columns is \begin{equation}\label{unified:middle} H\rho+\delta\frac{(2d-t)!}{(d-t)!}<1 \qquad(3\leq t\leq d-1,\ t\text{ odd}). \end{equation} Indeed $H\rho<13/25$ for every $d\geq3$, and the second term is at most $1/100$. This supplies the scalar shadow \eqref{normal:middle-scalar-shadow} on every such layer. The outer strict shadow supplies the first layer; the preceding top-degree calculation supplies the last layer in odd degree. Consequently the proof of Proposition~\ref{normal:strata}, including the exact common-scalar dimension identity \eqref{normal:baseline}, gives \[ \dim\{[\lambda]\in\mathbb P(J^*):\dim\ker C_\lambda\geq k\} \leq\dim J-c_1m^dk-1 \] for some $c_1>0$. The target surjections proved above exclude a nonzero covector with zero outer component. The graph parameters cost only $O_h(m^{d-1})$ per kernel direction, and the higher affine equations cost $O_h(m^{d-2})$, so both are absorbed by the strict $m^d$ shadow. Proposition~\ref{normal:rank} now gives the maximal rank of the first normal map. Lemmas~\ref{normal:child-repair} and~\ref{normal:mark-compatibility} apply because the background equality, its one-generator extension and its projected version have all been established. Perturbing the distinguished generator therefore leaves precisely the homology from the lower count in $S(Y)$. Theorem~\ref{normal:transfer} gives the two inequalities in Proposition~\ref{unified:transfer} at the exact counts \eqref{construct:exact-count}. Their signs at $r^-_{m+h}$ and $r^+_{m+h}$ give the recurrence. 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