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+% ==================================================================
+% A Counterexample to Kaplansky's Direct-Finiteness Conjecture
+% in Odd Characteristic
+%
+% Self-contained LaTeX; compile with pdflatex.
+%
+% This version is meant to be read carefully. The motivating words are
+% in ordinary English; the technical statements are the same as in the
+% original paper. The reader is assumed to know polynomials, matrices,
+% and the very basics of groups ("a set with an associative product,
+% an identity, and inverses"). Every other object is defined here.
+%
+% Reading guide:
+%   Section 1: the puzzle, and what we are going to build.
+%   Section 2: the concrete finite data (prime p, field K, graph Gamma,
+%              local group H).
+%   Section 3: the special idempotent e in K[H] that has a large rank
+%              and two opposite annihilations.
+%   Section 4: gluing copies of H into one group G_0 without collapsing.
+%   Section 5: enlarging to G_1 so ranks become rectangular matrix
+%              identities in a single central sector.
+%   Section 6: compressing to a square matrix with a one-sided inverse.
+%   Section 7: encoding matrices into scalars via a finite Heisenberg
+%              group.
+%   Section 8: the resulting injective non-surjective cellular automaton.
+% ==================================================================
+\documentclass[11pt]{article}
+\usepackage[T1]{fontenc}
+\usepackage[utf8]{inputenc}
+\usepackage{lmodern}
+\usepackage[margin=1in]{geometry}
+\usepackage{amsmath,amssymb,amsthm,mathtools}
+\usepackage{hyperref}
+\hypersetup{colorlinks=true,linkcolor=blue!45!black,citecolor=blue!45!black,urlcolor=blue!45!black}
+\newtheorem{theorem}{Theorem}[section]
+\newtheorem{proposition}[theorem]{Proposition}
+\newtheorem{lemma}[theorem]{Lemma}
+\newtheorem{corollary}[theorem]{Corollary}
+\theoremstyle{remark}
+\newtheorem{remark}[theorem]{Remark}
+\numberwithin{equation}{section}
+\newcommand{\F}{\mathbb F}
+\newcommand{\Mat}{\operatorname{Mat}}
+\newcommand{\diag}{\operatorname{diag}}
+\newcommand{\im}{\operatorname{im}}
+\newcommand{\id}{\mathrm{id}}
+\newcommand{\supp}{\operatorname{supp}}
+\title{A Counterexample to Kaplansky's Direct-Finiteness Conjecture in Odd Characteristic}
+\author{OpenAI}
+\date{September 26, 2026}
+\begin{document}
+\maketitle
+
+\begin{abstract}
+Kaplansky conjectured in 1972 that the group algebra of any group
+over any field would be ``directly finite'': if $ab=1$ then $ba=1$.
+The conjecture is true when the field has characteristic zero, but it
+fails in positive characteristic once the group is allowed to be
+infinite. This paper constructs a counterexample in odd characteristic.
+We exhibit one explicit odd prime~$p$, a field~$K$ of order~$p^4$,
+a finitely generated group~$G$ containing torsion, and two finite
+sums $a,b\in K[G]$ with $ab=1$ and $ba\ne 1$. The same two sums then
+produce an injective cellular automaton on~$K^G$ that is not surjective,
+so the group~$G$ is not surjunctive.
+\end{abstract}
+
+\section{Introduction}
+\label{sec:introduction}
+
+\subsection{The puzzle: one-sided versus two-sided inversion}
+
+Let $K$ be a field and $G$ a group. The \emph{group algebra} $K[G]$
+consists of all formal finite sums
+\[
+ c_1 g_1 + c_2 g_2 + \cdots + c_r g_r,
+ \qquad c_i\in K,\ g_i\in G,
+\]
+added coordinate by coordinate and multiplied by writing out both sides,
+using the group law $g_ig_j$ and extending linearly. Two sums are equal
+only when their coefficients agree on every~$g$. So $K[G]$ is like an
+infinite-dimensional vector space over $K$, with $G$ itself as a basis,
+plus a multiplication rule coming from the group.
+
+Concrete example. When $G=\mathbb Z$ and $K$ is any field, the
+group algebra $K[\mathbb Z]$ is (isomorphic to) the ring of Laurent
+polynomials $K[x,x^{-1}]$ in one variable: send the integer~$n$ to~$x^n$.
+That ring is commutative, so left and right behave the same there.
+Not every group algebra is commutative; for example, if $G$ is
+nonabelian, then $K[G]$ is noncommutative.
+
+A ring $R$ is \emph{directly finite} when the equation $ab=1$, for
+$a,b\in R$, always implies~$ba=1$. Fields are directly finite (the
+only nonzero element with a multiplicative inverse is the obvious one).
+The $n\times n$ matrix ring $\Mat_n(K)$ over a field is directly finite
+because linear maps on an $n$-dimensional vector space have finite
+dimension as input and output. In general, any finite-dimensional
+algebra over a field is directly finite: any one-sided inverse between
+linear maps on an $n$-dimensional space has rank~$n$, so it is
+automatically a two-sided inverse. A ring in which every matrix ring
+$\Mat_n(R)$, $n\ge1$, is directly finite is called \emph{stably finite}.
+
+Since a finite group $G$ produces a finite-dimensional algebra $K[G]$,
+it is directly finite (and stably finite; the matrix rings $\Mat_n(K[G])$
+are again finite-dimensional). So the interesting question is about
+infinite groups.
+
+\subsection{Kaplansky's question}
+
+In his 1972 book on fields and rings, Kaplansky proved that when the
+field has characteristic~$0$, the group algebra $K[G]$ is directly finite
+for every group~$G$; his proof used finite von Neumann algebras. He then
+asked whether the same conclusion holds in characteristic~$p>0$, where
+his von Neumann-algebra method does not apply directly.
+
+A large body of later work found positive partial answers. Ara, O'Meara,
+and Perera (2002) proved stable finiteness of $K[G]$ for $G$ free-by-amenable
+over any division ring. Elek and Szab\'o (2004) proved stable finiteness
+for every sofic group: informally, a group that can be approximated by
+finite permutation models. Since many familiar infinite groups are
+sofic -- all residually finite groups, all amenable groups, all free
+groups, and many more -- this covered a wide range of cases.
+These theorems constrain any counterexample: whatever group we build
+must be far from sofic.
+
+The purpose of this paper is to give an actual counterexample, in a
+single specific odd characteristic.
+
+\subsection{The main theorem}
+
+Put $m=\binom{1200}{600}$ and let $p$ be the least integer $>1$ that
+divides $(m!)^2+1$. This $p$ is an odd prime; see Lemma~\ref{lem:field}.
+We construct:
+
+\begin{itemize}
+\item a field $K$ of order $p^4$ (an explicit quadratic tower over the
+      prime field $\F_p$);
+\item a finitely generated group $G$ containing nontrivial elements of
+      finite order;
+\item two finite sums $a,b\in K[G]$ with $ab=1$ and $ba\ne 1$.
+\end{itemize}
+
+The construction is entirely finite: every field element, every group
+generator, every coefficient is produced by a finite list of operations
+from fixed finite sets. No infinite search and no word-problem oracle
+is needed to define $a,b$; the point of the construction is to build them
+from finite data rather than by deciding equalities in~$G$.
+
+\begin{theorem}\label{thm:main}
+Let $m=\binom{1200}{600}$ and let $p$ be the least integer $h>1$
+dividing $(m!)^2+1$. Then $p$ is an odd prime. There is a specified
+field $K$ of order $p^4$, a specified finitely generated group $G$
+containing torsion, and specified sums $a,b\in K[G]$ satisfying
+$ab=1$ but $ba\ne 1$. In particular Kaplansky's direct-finiteness
+conjecture fails in odd characteristic.
+\end{theorem}
+
+The word ``specified'' means: written as an explicit composition of
+finite operations on finite sets, in an order that is fixed in advance.
+Different coefficients may happen to coincide at the end; that is
+harmless. What matters is that $a$ and $b$ are objective consequences
+of a finite recipe, not of an arbitrary choice among infinitely many.
+
+\subsection{Why the failure of direct finiteness matters}
+
+Two remarks connect this algebra to other topics. First, by
+Elek--Szab\'o, the constructed group $G$ cannot be sofic; otherwise
+$K[G]$ would be stably finite, contradicting the theorem. So the
+construction is also a source of non-sofic groups of a very specific
+kind.
+
+Second, direct finiteness is equivalent to a statement about ``linear
+cellular automata.'' A cellular automaton on $K^G$ is any transformation
+that reads a finite window of the configuration around each point and
+writes a new value there. If we take, for a fixed finite sum
+$c=\sum c_u u\in K[G]$, the rule
+\[
+ (T_c x)(g) = \sum_u c_u\, x(gu),
+\]
+then $T_c$ is always a cellular automaton (the memory set is $\supp c$).
+Composition behaves well: $T_c T_d = T_{cd}$. So if $ab=1$, then
+$T_b$ has left inverse $T_a$, hence is injective. If $ba\ne 1$, we can
+show $T_b$ is not surjective on $K^G$. Section~\ref{sec:cellular} gives
+the details. The upshot is a group that fails to be \emph{surjunctive},
+refuting Gottschalk's conjecture for this particular~$G$.
+
+\subsection{The plan of the proof}
+
+The construction has five independent pieces. Each piece is fairly
+self-contained; the reader may take them in whatever order feels best,
+as long as the final step receives all the pieces.
+
+\emph{Piece 1 (Section~\ref{sec:finite}): finite data.}
+We fix the prime $p$, the field $K=\F_{p^4}$ (built as a quadratic
+tower so we can descend to $\F_p$ for checks), and a simple triangle-free
+graph~$\Gamma$ on $m$ vertices with a factorization of its adjacency
+matrix into a very short sum of rank-one pieces. We also define a finite
+group $H=A_0\rtimes P$ of order $w$ that will be the local block.
+
+\emph{Piece 2 (Section~\ref{sec:modular}): a large idempotent.}
+Inside the group algebra $K[H]$, we build an explicit idempotent~$e$
+whose regular rank (the dimension of $eK[H]$ as a $K$-vector space)
+satisfies $\dim_K eK[H]/|H| > 1/96$. It also satisfies two opposite
+annihilations: writing $u(s),v(s)$ for certain elements of~$H$ and
+$\ell=(p-1)/2$,
+\[
+ (u(s)-1)^\ell e = 0 \quad\text{and}\quad e\,(v(s)-1)^\ell = 0
+ \quad\text{for every } s.
+\]
+The asymmetry is the point: the annihilation on the left and the
+annihilation on the right use the same exponent~$\ell$, but $u$ and $v$
+sit in different subgroups of~$H$, so no symmetry forces them to be
+compatible.
+
+\emph{Piece 3 (Section~\ref{sec:gluing}): gluing without collapsing.}
+We take $m$ copies $H_1,\ldots,H_m$ of $H$ and identify one specific
+cyclic subgroup across each edge of~$\Gamma$. The resulting quotient
+$G_0$ must remain injective on each $H_i$; if it did not, a nonzero
+element of $K[H_i]$ could vanish in $K[G_0]$. A short-and-sweet
+spherical-map argument proves this injectivity.
+
+\emph{Piece 4 (Section~\ref{sec:sector}): turning ranks into matrices.}
+Enlarge $G_0$ to a bigger group $G_1$ whose group algebra contains a
+nonzero central idempotent $f$ with special properties. In the ring
+$fK[G_1]$, the images of the idempotents $e_i$ become rectangular
+matrix blocks with \emph{exact} integer-scaled right and left inverses.
+This is the technical heart of the paper: a nilpotent-ideal
+change-of-basis argument turns dimension equalities over~$K$ into
+equalities over the full ring.
+
+\emph{Piece 5 (Sections~\ref{sec:compression} and~\ref{sec:scalar}):
+matrices, then scalars, then a cellular automaton.}
+Using the triangle-free condition and the deliberate asymmetry from
+Piece~2, we assemble all the rectangular blocks into two matrices with
+one-sided inverse. Then the finite Heisenberg group from Section~\ref{sec:scalar}
+embeds a matrix algebra into a group algebra, letting us turn that
+matrix identity into an identity between two scalar elements. Section~\ref{sec:cellular}
+translates it into a statement about cellular automata.
+
+\subsection{Notation summary}
+
+All matrices below act on column vectors in free \emph{right} modules
+$S^n$. So multiplying a matrix by a column vector means multiplying
+each entry on the left by the corresponding group algebra element.
+This matters because group algebras are not always commutative.
+
+\begin{itemize}
+\item $K$: a field. $\F_p$: the prime field of order $p$.
+\item $G$: a group; $K[G]$: the set of finite $K$-linear sums of
+      elements of $G$, with multiplication from the group law.
+\item $1_K$, or just $1$: the multiplicative identity of $K$. In $K[G]$
+      the identity is the group identity element, written $1_G$ or~$1$.
+\item $e\in K[H]$: an idempotent, meaning $e^2=e$.
+\item $d=\dim_K e K[H]$: the \emph{regular rank} of $e$, i.e., the
+      dimension of the image of left multiplication by~$e$ on $K[H]$.
+\item $w=|H|$: the order of the finite group~$H$.
+\item $[n]$: $\{1,\ldots,n\}$ for a positive integer $n$.
+\item Hermitian transpose: never used here. Ordinary transpose is
+      written $X^{\mathsf T}$, with no scalar conjugation.
+\item $\Mat_{a\times b}(S)$: $a\times b$ matrices with entries in~$S$.
+\end{itemize}
+
+\section{The field, the graph, and the local group}
+\label{sec:finite}
+
+The finite-data stage has three independent ingredients: a prime~$p$ and
+a specific field $K$ of order $p^4$; a simple triangle-free graph~$\Gamma$
+on $m$ vertices together with an extremely compact factorization of its
+adjacency matrix; and a finite group $H$ with a natural action that will
+later host the crucial idempotent. Each ingredient is built once and for
+all here, in the order needed later.
+
+\subsection{The prime and the field}
+\label{subsec:field}
+
+Recall $m=\binom{1200}{600}$, a specific (very large) positive integer,
+and let $p$ be the least integer $h>1$ that divides $(m!)^2+1$.
+The definition of~$p$ is purely arithmetic. We need four properties
+of~$p$ and one field-extension construction.
+
+\begin{lemma}\label{lem:field}
+With $m=\binom{1200}{600}$ and $p$ the least $h>1$ dividing
+$(m!)^2+1$, we have:
+\begin{enumerate}
+\item $p$ is an odd prime, and $p>m$.
+\item $p$ divides $(m!)^2+1$, so $-1$ has a square root in $\F_p$.
+\item If $\eta\in\F_p^*$ is any nonsquare, then the polynomial
+      $X^2-\eta$ is irreducible over $\F_p$ and its root field
+      $K_0=\F_p[\xi]/(\xi^2-\eta)$ is a quadratic extension of $\F_p$.
+\item The root $\xi$ of $X^2-\eta$ is \emph{not} a square in $K_0$,
+      so $X^2-\xi$ is irreducible over $K_0$ and
+      $K=K_0[\nu]/(\nu^2-\xi)$ is a further quadratic extension.
+      Consequently $|K|=p^4$.
+\end{enumerate}
+\end{lemma}
+\begin{proof}
+(1) The least divisor $>1$ of any integer $n>1$ is prime: if it were
+composite, one of its proper divisors would also be $>1$ and would be
+a smaller divisor of~$n$. So $p$ is prime. Now, if $p\le m$, then
+$p$ would divide $m!$ (since $m\ge p$), so also $p$ would divide
+$(m!)^2$; combined with $p\mid (m!)^2+1$ this gives $p\mid 1$,
+impossible. Hence $p>m>600$. Since $m\ge600\ge1$ and $p>m\ge600$,
+$m!$ is even (as $m\ge2$), so $(m!)^2+1$ is odd, and $p\ne2$.
+Therefore $p$ is odd.
+
+(2) The congruence $(m!)^2\equiv -1\pmod p$ says that $-1$ is the
+square of $m!$ modulo~$p$, that is, $-1$ is a quadratic residue in
+$\F_p$.
+
+(3) A polynomial $X^2-\eta$ over a field is irreducible exactly when
+$\eta$ is not a square in that field. Nonsquares exist in $\F_p^*$
+for every odd prime~$p$: the map $x\mapsto x^2$ on $\F_p^*$ is
+two-to-one on its image, so only half of the $p-1$ elements are
+squares, and $p-1\ge2$ gives at least one nonsquare. For such~$\eta$
+the polynomial $X^2-\eta$ has no roots in $\F_p$, hence is irreducible
+and $\F_p[X]/(X^2-\eta)$ is a field of $p^2$ elements.
+
+(4) In $K_0=\F_p[X]/(X^2-\eta)$, the two roots of $X^2-\eta$ are
+$\xi$ and $-\xi$. Therefore $\xi\cdot(-\xi)=-\xi^2=-\eta$ is the
+field norm of $\xi$ down to $\F_p$. By (2), $-1$ is a square in $\F_p$.
+Since $\eta$ is chosen to be a nonsquare, the product $-\eta$ is a
+product of a square and a nonsquare, hence a nonsquare in $\F_p$.
+
+Now suppose, toward a contradiction, that $\xi=y^2$ for some $y\in K_0$.
+Apply the field norm $N_{K_0/\F_p}(u)=u\cdot u'$ (where $u'$ is the
+conjugate, negating the $\xi$-coordinate). Because $N$ is multiplicative,
+$N(\xi)=N(y^2)=N(y)^2$, which would exhibit the nonsquare~$-\eta$
+as a square in $\F_p$. That is impossible. So $\xi$ is not a square
+in~$K_0$, and hence $X^2-\xi$ is irreducible over~$K_0$. The quotient
+$K=K_0[\nu]/(\nu^2-\xi)$ is therefore a field of $|K_0|^2=p^4$ elements.
+\end{proof}
+
+From now on, $K$ is the specific field $K=\F_{p^4}$ supplied by
+Lemma~\ref{lem:field}. We descend to $\F_p$ only when we need to check
+things like ``$-1$ is a square.'' All our formulas over $K$ are
+already fixed.
+
+Several additional choices are needed below. Because we want the
+construction to be ``specified'' in the sense of the introduction,
+we spell out an explicit finite recipe for each choice.
+
+\begin{itemize}
+\item \emph{Representatives and order on $\F_p$.}
+      Identify $\F_p$ with $\{0,1,\ldots,p-1\}$ using the remainder
+      in $0,\ldots,p-1$, so that $\F_p$ has a natural total order.
+      Choose $\eta$ as the least nonsquare in $\F_p^*$ under this order.
+\item \emph{Representatives and order on $K$.}
+      Write every element of $K$ uniquely as
+      $c_0+c_1\xi+c_2\nu+c_3\xi\nu$ with $c_j\in\F_p$. Order $K$ by
+      the lexicographic order on $(c_0,c_1,c_2,c_3)$. This gives a
+      total order on $K$.
+\item \emph{A generator $\zeta$ of $K^*$.} Put $M=p^4-1$.
+      Choose the least element $\zeta$ of $K^*$ whose multiplicative
+      order equals $M$. Such an element exists; the reasoning is
+      standard but we include it for completeness below.
+\end{itemize}
+
+Why does an element of order $M$ exist? Let $b$ be the least common
+multiple of all element orders in $K^*$. Conventionally $b\le M$ because
+every element of $K^*$ has order at most $M$. On the other hand, every
+element of $K^*$ is a root of $X^b-1$ (because each element's order
+divides~$b$), so $X^b-1$ has at least $M$ distinct roots. A nonzero
+polynomial over a field has at most as many roots as its degree, so
+$M\le b$. Hence $b=M$. Now, by the definition of least common multiple,
+we can choose elements of $K^*$ whose orders are the prime powers
+occurring in the factorization of $M$; raising appropriates elements to
+the right powers gives an element of order exactly each prime power
+factor. Their product, being the product of commuting elements of
+pairwise coprime orders, has order exactly $M$. So an element $\zeta$
+of order $M$ exists.
+
+In particular $M=p^4-1$ and all of its powers are invertible in $K$
+(equivalently: the multiplicative group of $K$ is cyclic of order
+$M=p^4-1$). We will divide by various integers below; every such
+integer is a factor of~$M$ and hence nonzero in~$K$.
+
+\subsection{The graph and the low-rank fit}
+\label{subsec:graph}
+
+The second finite ingredient is a simple triangle-free graph~$\Gamma$
+on many vertices whose adjacency matrix admits a factorization into
+a number of rank-one pieces that is very small compared to the number
+of vertices. That is what allows Section~\ref{op:compression-section}
+to build a ``rank paradox'' with which the odd-characteristic
+counterexample becomes possible.
+
+Set $[1200]=\{1,2,\ldots,1200\}$. Enumerate all six-hundred-element
+subsets $I\subseteq[1200]$ in increasing lexicographic order (comparing
+subsets as increasing lists of integers) and denote them
+$I_1,\ldots,I_m$ with $m=\binom{1200}{600}=\binom{1200}{600}$. Remember
+this same~$m$ is the one appearing in the definition of~$p$.
+
+Define a simple graph $\Gamma$ on vertex set $\{1,\ldots,m\}$ by
+declaring the distinct vertices $i\ne j$ to be adjacent exactly when
+the intersection $I_i\cap I_j$ has fewer than $200$ elements:
+\[
+ i\sim j \quad\text{iff}\quad |I_i\cap I_j| < 200.
+\]
+By convention a vertex is never adjacent to itself, so the graph is
+simple.
+
+\begin{proposition}\label{prop:graph}
+The graph $\Gamma$ is simple and triangle-free. Let
+\[
+ \mathcal T=\{T\subseteq[1200]:|T|\le 400\},\qquad
+ t=|\mathcal T|=\sum_{j=0}^{400}\binom{1200}{j}.
+\]
+There exist coefficients $\alpha_i(T)$ and $\beta_j(T)$ in $\F_p$,
+indexed by $T\in\mathcal T$, such that the $m\times m$ matrix
+$F$ given by
+\[
+ F_{ij}=\sum_{T\in\mathcal T}\alpha_i(T)\,\beta_j(T)
+\]
+satisfies $F_{ii}=1$ for every $i$ and $F_{ij}=0$ for every pair of
+distinct nonadjacent vertices~$i,j$. Moreover $t/m<1/100$.
+\end{proposition}
+\begin{proof}
+\emph{Triangle-free.} Suppose $i,j,k$ are pairwise adjacent. Then each
+of the three pairwise intersections has at most $199$ elements (since
+``fewer than $200$'' and integer valued). So
+\[
+ |I_i\cup I_j\cup I_k| \ge |I_i|+|I_j|+|I_k| - |I_i\cap I_j| - |I_j\cap I_k| - |I_k\cap I_i|
+ \ge 3\cdot 600 - 3\cdot 199 = 1203,
+\]
+by inclusion-exclusion. But the union is contained in $[1200]$, which
+has only $1200$ elements. Contradiction. So $\Gamma$ has no triangles.
+
+\emph{The fit.} Pick, for each $i$, the indicator function
+\[
+ \alpha_i(T)=\begin{cases}1,&T\subseteq I_i,\\0,&\text{otherwise.}\end{cases}
+\]
+To build $\beta_j$, define the following polynomial in the formal
+variables $X_1,\ldots,X_{1200}$ over $\F_p$:
+\[
+ Q_j(X)=\prod_{k=200}^{599}\frac{\sum_{x\in I_j}X_x-k}{600-k}.
+\]
+Every denominator lies in $\{1,2,\ldots,400\}$, hence is invertible
+in $\F_p$ because~$p>600>400$. Expand $Q_j$ as $\sum_\alpha c_\alpha X^\alpha$
+with $c_\alpha\in\F_p$ and multi-indices $\alpha\in\mathbb N^{[1200]}$
+(so $X^\alpha=\prod_x X_x^{\alpha_x}$). Define the \emph{square-free
+reduction} of $Q_j$ to be
+\[
+ Q_j^{\mathrm{sf}}(X)=\sum_\alpha c_\alpha\prod_{x:\alpha_x>0} X_x,
+\]
+in which every positive exponent has been replaced by $1$. Set
+$\beta_j(T)$ to be the coefficient of $\prod_{x\in T}X_x$ in
+$Q_j^{\mathrm{sf}}$ for each $T\subseteq[1200]$. Because $Q_j$ has
+total degree $400$ (a product of $400$ linear forms), only the
+$T\in\mathcal T$ can appear. The total degree of $Q_j$ bounds the
+support of beta_j exactly to subsets of size at most 400.
+
+\emph{Evaluation on $0$-$1$ vectors.} The square-free reduction does
+not change the value of the polynomial at any $0$-$1$ point (because
+$0^n=0$ and $1^n=1$ for every $n\ge1$). In particular, evaluate at the
+indicator vector $1_{I_i}\in\{0,1\}^{[1200]}$ (entry $1$ at positions
+$x\in I_i$, entry $0$ elsewhere). On the one hand,
+\[
+ Q_j(1_{I_i})
+ =\prod_{k=200}^{599}\frac{|I_i\cap I_j|-k}{600-k}.
+\]
+On the other hand, expanding $Q_j^{\mathrm{sf}}$ and evaluating at
+$1_{I_i}$, each monomial $\prod_{x\in T}X_x$ contributes $1$ if
+$T\subseteq I_i$ and $0$ otherwise. So
+\[
+ Q_j(1_{I_i})
+ =\sum_{T\in\mathcal T}\alpha_i(T)\,\beta_j(T)=F_{ij}.
+\]
+Equating the two evaluations gives
+\begin{equation}\label{eq:threshold-F}
+ F_{ij}=\prod_{k=200}^{599}\frac{|I_i\cap I_j|-k}{600-k}.
+\end{equation}
+
+\emph{Fit conditions.} If $i=j$ then $|I_i\cap I_j|=600$, so every
+numerator $600-k$ cancels the corresponding denominator, giving
+$F_{ii}=1$. If $i\ne j$ are nonadjacent then $|I_i\cap I_j|\ge200$ and
+$\le 599$, so the factor with $k=|I_i\cap I_j|$ is exactly zero,
+forcing $F_{ij}=0$. For adjacent distinct pairs the product is
+generally nonzero, but we do not need any particular value; only the
+diagonal and the nonedge vanishing matter. This is the threshold-Kneser
+polynomial specialization of \cite[Proposition~13]{Haviv2018Theta}
+and \cite[Theorem~3.2]{GolovnevHaviv2022}.
+
+\emph{The ratio bound.} Combinatorial estimates. The binomial
+coefficients $\binom{1200}{j}$ increase with~$j$ as long as $j\le599$
+(because $\binom{n}{j+1}/\binom{n}{j}=(n-j)/(j+1)>1$ when $j<(n-1)/2$
+and $n=1200$). So on the sum defining $t$, the largest term is at
+$j=400$, and
+\[
+ t \le 401\binom{1200}{400}.
+\]
+For $m=\binom{1200}{600}$ we can compare factorials directly:
+\[
+ \frac{\binom{1200}{400}}{\binom{1200}{600}}
+ =\prod_{u=401}^{600}\frac{u}{u+200}
+ \le \left(\tfrac 57\right)^{100}.
+\]
+Here the inequality is a crude but adequate estimate: for
+$u\le 500$, the factor $u/(u+200)$ is at most $500/700=5/7$; for
+$u\ge501$, the factor is at most $600/800=3/4<1$, so it contributes
+less than one. Multiplying all one hundred factors gives the displayed
+bound. Then
+\[
+ \frac tm\le 401\left(\tfrac 57\right)^{100}.
+\]
+The cube of $5/7$ is $125/343<1/2$, so $(5/7)^{99}<(1/2)^{33}$. Thus
+$ (5/7)^{100}<(1/2)^{33} $ as well. Multiplying by $401<40100<2^{33}$
+gives
+\[
+ \frac tm<2^{33}\cdot \frac1{2^{33}}=1,
+\]
+which is weaker than needed; more precisely,
+\[
+ 100 \cdot \frac tm < 40100\left(\tfrac 57\right)^{100}
+ <\frac{40100}{2^{33}}<1,
+\]
+so $t/m<1/100$.
+\end{proof}
+
+Now we have our two finite combinatorial ratios. The first is
+$t/m<1/100$: the low-rank fit uses far fewer summands than there are
+vertices. The second, coming from the group $H$ below, is
+$d/w>1/96$: the idempotent each vertex carries has a regular rank
+that is a fixed nonzero fraction of~$|H|$. Section~\ref{op:compression-section}
+will combine the two ratios into a matrix with strict inequality
+$wt<md$ between its shapes.
+
+\subsection{The finite group $H$}
+\label{subsec:group-H}
+
+The third finite ingredient is the local group. It is where the large
+idempotent lives, and it is where the crucial ``one-sided annihilation
+in each direction'' happens.
+
+Let $P=(\F_p)^3$ be the additive group of three-dimensional column
+vectors over $\F_p$. It has $p^3$ elements. Let $M=p^4-1$, as in the
+field construction, and let
+\[
+ A_0=(\mathbb Z/M\mathbb Z)^P
+\]
+be the group of \emph{functions} $a:P\to\mathbb Z/M\mathbb Z$, with
+addition defined pointwise: $(a+a')(x)=a(x)+a'(x)$ for each $x\in P$.
+This is a finite abelian group of order $M^{p^3}$.
+
+The group $P$ acts on $A_0$ by \emph{translation}: for $s\in P$ and
+$a\in A_0$, define $s\cdot a\in A_0$ by
+\[
+ (s\cdot a)(x)=a(x-s), \qquad x\in P.
+\]
+This is a group action: $0\cdot a=a$ and $(s+t)\cdot a=s\cdot(t\cdot a)$.
+Form the semidirect product
+\[
+ H=A_0\rtimes P, \qquad
+ (a,s)(a',s')=(a+s\cdot a',\ s+s').
+\]
+The identity of $H$ is $(0,0)$. The inverse of $(a,s)$ is
+$(-s^{-1}\cdot a, -s)$, i.e., $(-(-s)\cdot a,-s)$ since in characteristic
+$p$ subtraction is the same as~$+(-s)$. Concretely: $(a,s)(-(-s)\cdot a, -s)=(a-s\cdot((-s)\cdot a), 0)=(a-s\cdot(-s\cdot a), 0)=(a-0,0)=(0,0)$.
+The order of $H$ is
+\[
+ w:=|H|=|A_0|\cdot|P|=M^{p^3}\cdot p^3.
+\]
+So $H$ is finite. The group algebra $K[H]$ is thus a
+finite-dimensional algebra over $K$, of dimension $w$ as a $K$-vector
+space (and $w^2$ as a $K$-algebra of linear maps on $K^w$).
+
+The two subgroups of $H$ that matter now are $U$ and $V$, defined
+below. Both are isomorphic to $P$, but they sit inside $H$ in different
+ways.
+
+\emph{Translation subgroup.} Define
+\[
+ u(s)=(0,s),\qquad U=\{u(s):s\in P\}\subseteq H.
+\]
+Because $(0,s)(0,s')=(0,s+s')$, the map $s\mapsto u(s)$ is an
+isomorphism $P\to U$.
+
+\emph{Characteristic element.} Let $c\in A_0$ be the function
+\[
+ c(0)=1,\quad c(x)=0\text{ for }x\ne0.
+\]
+The group law gives $(c,0)$ the exact order $M$ in $H$ (as an element
+of the abelian normal subgroup $A_0$).
+
+\emph{Conjugate subgroup.} Define
+\[
+ v(s)=c\,u(s)\,c^{-1},\qquad V=\{v(s):s\in P\}\subseteq H,
+\]
+where $c$ in the formula denotes $(c,0)\in H$ and $c^{-1}=(-c,0)$
+uses the additive inverse in $A_0$. A short computation with the
+group law in $H$ gives the explicit form
+\begin{equation}\label{eq:v(s)-formula}
+ v(s)=(c-s\cdot c,\ s),\qquad s\in P.
+\end{equation}
+Indeed $(c,0)(0,s)=(c,s)$ and $(c,s)(-c,0)=(c+s\cdot(-c),\,s)=(c-s\cdot c,s)$.
+
+\begin{proposition}\label{prop:UV-triv}
+The subgroups $U$ and $V$ satisfy $U\cap V=\{1\}$, and $U\cong V\cong P$.
+\end{proposition}
+\begin{proof}
+Suppose $(0,0)=u(s)=v(s')$ for some $s,s'\in P$. Comparing second
+coordinates gives $s=s'$. Comparing first coordinates gives
+$0=c-s\cdot c$, i.e., $c=s\cdot c$. The left side evaluates at $0$ to
+$c(0)=1$; the right side evaluates at $0$ to $(s\cdot c)(0)=c(-s)$.
+But $-s\ne 0$ when $s\ne0$, and $c$ vanishes away from the origin, so
+$c(-s)=0$. Thus $s=0$, and the only common element is the identity.
+The isomorphism statements follow from the displayed definitions:
+$s\mapsto u(s)$ is a bijection $P\to U$ preserving the group law, and
+$s\mapsto v(s)$ is a conjugate of~$s\mapsto u(s)$ inside the group~$H$,
+hence also isomorphic to $P$.
+\end{proof}
+
+Why does this matter? In Section~\ref{sec:modular} we shall build an
+idempotent~$e$ with
+\[
+ (u(s)-1)^\ell e = 0,\qquad e\,(v(s)-1)^\ell=0
+ \quad\text{for every }s\in P,
+\]
+where $\ell=(p-1)/2$. The first annihilation uses $u(s)\in U$ on the
+left; the second uses $v(s)\in V$ on the right. Because $U\cap V=\{1\}$,
+the two sides do not have to agree, and no symmetry forces either one
+onto the other. This asymmetry is what makes the later triangular
+compression work.
+
+That completes the finite data. In later sections, we always write
+$m=\binom{1200}{600}$, $t=|\mathcal T|$, $w=|H|$, and
+$d=\dim_KeK[H]$ for the regular rank of the idempotent $e$ we
+are about to build in Section~\ref{sec:modular}.
+
+\section{The large idempotent with ordered annihilations}
+\label{sec:modular}
+
+This section is the technical heart of the paper. In the previous
+section we built three pieces of finite data: a prime $p$, a field
+$K$ of order~$p^4$, and a finite group $H=A_0\rtimes P$ with two
+different subgroups $U,V\cong P$ meeting trivially. Now we use all that
+data to build one particular idempotent $e\in K[H]$ with two properties:
+
+\begin{itemize}
+\item \emph{Large regular rank.} The dimension $d=\dim_K\bigl(eK[H]\bigr)$
+      of the image of left multiplication by $e$ satisfies
+      $d/w>1/96$, where $w=|H|$.
+\item \emph{Opposite annihilations.} Writing $\ell=(p-1)/2$, we have
+      $(u(s)-1)^\ell e=0$ and $e\,(v(s)-1)^\ell=0$ for every $s\in P$.
+\end{itemize}
+
+The construction proceeds in three stages.
+
+\begin{enumerate}
+\item \S\ref{subsec:B-def}: identify a subspace $B\subseteq K^P$ on
+      which both $u(s)$-shifts and $v(s)$-shifts have the same
+      annihilation behavior. This requires a piece of translation
+      calculus (the operators $X_j$) and care with degrees.
+\item \S\ref{subsec:admissibility}: use the polynomial identity test of
+      Schwartz and Zippel to produce many ``admissible'' characters
+      $\lambda\in\widehat{A_0}$, i.e., characters $\lambda$ for which
+      a certain Gram matrix~$N_\lambda$ is nonsingular.
+\item \S\ref{subsec:idempotent}: assemble these ingredients into an
+      idempotent $e$ spread over many matrix-block summands of $K[H]$,
+      compute its regular rank exactly, and verify the annihilations.
+\end{enumerate}
+
+A last subsection (\S\ref{subsec:edge-zero}) turns the above
+annihilations into the ``edge products vanish'' identity used later.
+
+Throughout, we do all linear algebra over~$K$ first, then lift to
+$K[H]$ using the Fourier idempotents of the abelian normal subgroup
+$A_0\subseteq H$.
+
+\subsection{The annihilated subspace $B\subseteq K^P$}
+\label{subsec:B-def}
+
+Let $\ell=(p-1)/2$ and $D=3(p-1)$. Work in the $K$-vector space $K^P$
+of functions $P\to K$, with standard basis $\mathbf e_x$ (a $1$ at
+$x$ and $0$ elsewhere). We want a linear subspace on which left shifts
+and right shifts both annihilate $\ell$-fold.
+
+\emph{Translation calculus.} For $s\in P$ let $L_s:K^P\to K^P$ be the
+shift operator $L_s\mathbf e_x=\mathbf e_{x+s}$. Define three operators
+\[
+ X_j=L_{\varepsilon_j}-1_{K^P}, \qquad j=1,2,3,
+\]
+where $\varepsilon_j$ are the standard generators of $P=\F_p^3$.
+
+The $L_s$ commute because the group $P$ is abelian. So the $X_j$ commute
+as well. A stronger fact: because we are in characteristic~$p$, the
+binomial coefficients $\binom pk$ are divisible by $p$ for $1\le k\le p-1$;
+so $(1+X_j)^p=1+X_j^p$ and
+\[
+ X_j^p=(L_{\varepsilon_j}-1)^p = L_{p\varepsilon_j}-1 = L_0-1 = 0.
+\]
+Thus each $X_j$ is nilpotent of index at most $p$. These are the
+``differential operators of order 1'' associated to the three
+generators of $P$.
+
+\begin{lemma}[Monomial basis]\label{lem:monomial-basis}
+The $p^3$ vectors
+\[
+ X_1^{n_1}X_2^{n_2}X_3^{n_3}\mathbf e_0, \qquad 0\le n_1,n_2,n_3< p,
+\]
+form a $K$-basis of $K^P$.
+\end{lemma}
+\begin{proof}
+The standard basis $\{\mathbf e_x:x\in P\}$ already has cardinality
+$p^3=\dim_KK^P$. So it suffices to show every $\mathbf e_s$ is in the
+$K$-span of the displayed vectors.
+
+Fix $s=(s_1,s_2,s_3)\in P$. Since $1+X_j=L_{\varepsilon_j}$, we have
+$(1+X_j)^{s_j}=L_{s_j\varepsilon_j}$ and
+\[
+ \mathbf e_s=L_s\mathbf e_0
+ =\prod_{j=1}^3 L_{s_j\varepsilon_j}\mathbf e_0
+ =\prod_{j=1}^3 (1+X_j)^{s_j}\mathbf e_0
+ =\prod_{j=1}^3 \left(\sum_{k=0}^{s_j}\binom{s_j}{k}X_j^k\right)\mathbf e_0.
+\]
+The last product expands to a $K$-linear combination of the vectors
+$X_1^{n_1}X_2^{n_2}X_3^{n_3}\mathbf e_0$ with $0\le n_j\le s_j<p$.
+So $\mathbf e_s$ lies in the span of the display. Since $\{\mathbf e_s\}$
+spans, the display spans, and it has $p^3$ elements, so it is a basis.
+\end{proof}
+
+\begin{remark}
+The lemma can be re-read as ``polynomials in $X_1,X_2,X_3$ of degree
+$<p$ in each variable are a basis of $K^P$''. This is precisely the
+finite-dimensional analogue of the polynomial ring $K[X_1,X_2,X_3]$,
+truncated by $X_j^p=0$.
+\end{remark}
+
+Define $B$ as the matrix with columns the basis vectors whose total
+degree is large enough:
+\begin{equation}\label{eq:B-def}
+ B := \bigl[X_1^{n_1}X_2^{n_2}X_3^{n_3}\mathbf e_0\bigr]_{
+ \substack{0\le n_1,n_2,n_3<p\\ n_1+n_2+n_3\ge D-\ell+1}}.
+\end{equation}
+So $B$ is a $p^3\times h$ matrix with $K$-linearly independent columns
+(they are a subset of a basis). The integer $h$ counts the columns.
+
+\begin{lemma}[Two-sided annihilation of $B$]\label{lem:B-annihilation}
+For every $s\in P$,
+\[
+ (L_s-1)^\ell\,B=0,\qquad B^{\mathsf T}(L_s-1)^\ell=0.
+\]
+\end{lemma}
+\begin{proof}
+\emph{First identity.} Fix $s\in P$ and write $s=\sum_j s_j\varepsilon_j$.
+Expand $L_s-\id_{K^P}$:
+\[
+ L_s-1 = \prod_{j=1}^3 L_{s_j\varepsilon_j} - 1
+ = \prod_{j=1}^3 (1+X_j)^{s_j} - 1.
+\]
+Expanding the right side, every nonzero monomial has total $X$-degree
+at least one (because the constant term $1$ cancels). Therefore the
+$\ell$th power $(L_s-1)^\ell$ has the property: applied to any monomial
+of total $X$-degree $d$, the result lies in the $K$-span of monomials
+of total $X$-degree at least $d+\ell$.
+
+Now take a column $X_1^{n_1}X_2^{n_2}X_3^{n_3}\mathbf e_0$ of $B$,
+with $d=n_1+n_2+n_3\ge D-\ell+1$. Applying $(L_s-1)^\ell$ gives a
+combination of monomials of total degree at least $D+1$. But every
+monomial of total degree greater than $D=3(p-1)$ has some exponent
+at least $p$, and $X_j^p=0$ kills it. So each column is annihilated,
+proving the first identity.
+
+\emph{Second identity.} Compute the transpose of $L_s$: its matrix entry
+$(L_s)_{\mathbf e_x,\mathbf e_y}$ (row $x$, column $y$) is $1$ when
+$x+s=y$ and $0$ otherwise. Hence $(L_s^{\mathsf T})_{\mathbf e_y,\mathbf e_x}$
+is $1$ when $x-s=y$, so $L_s^{\mathsf T}=L_{-s}$. Consequently
+$(L_s-1)^{\mathsf T}=L_{-s}-1$, and
+\[
+ B^{\mathsf T}(L_s-1)^\ell = \bigl((L_{-s}-1)^\ell B\bigr)^{\mathsf T} = 0^{\mathsf T}=0,
+\]
+using the first identity with $-s$ in place of $s$.
+\end{proof}
+
+\begin{lemma}[Number of columns]\label{lem:h}
+The integer $h$ equals $\binom{\ell+2}{3}$, and
+\[
+ \frac{h}{p^3}=\frac{(p-1)(p+1)(p+3)}{48\,p^3}>\frac1{48}.
+\]
+\end{lemma}
+\begin{proof}
+Make the substitution $a_j=p-1-n_j$, $j=1,2,3$. Then $a_j\ge0$ and
+$\sum a_j=D-\sum n_j$. The condition $\sum n_j\ge D-\ell+1$ becomes
+$\sum a_j\le \ell-1$. Each $a_j\le p-1$ is automatic, because
+$\ell-1=(p-1)/2-1<p-1$ forces $\sum a_j<p-1$, making each $a_j<p-1$.
+Thus $h$ is the number of triples $(a_1,a_2,a_3)\in\mathbb N^3$ with
+$\sum a_j\le\ell-1$. Stars and bars (distribute $\le\ell-1$ identical
+balls into $3$ labelled bins) gives
+\[
+ h=\binom{(\ell-1)+3}{3}=\binom{\ell+2}{3}.
+\]
+Now $\ell=(p-1)/2$, so
+\[
+ \binom{\ell+2}{3} = \frac{(\ell+2)(\ell+1)\ell}{6}
+ = \frac{(p+3)(p+1)(p-1)}{48}.
+\]
+Divide by $p^3$: $h/p^3=(p+3)(p^2-1)/(48\,p^3)$. To compare with
+$1/48$, compute the difference
+\[
+ (p+3)(p^2-1)-p^3=3p^2-p-3>0 \quad\text{for }p\ge3,
+\]
+which gives $h/p^3>1/48$.
+\end{proof}
+
+To summarize the situation in one line: $B$ is a tall matrix whose
+columns are the ``high-degree part'' of $K^P$, and both $u(s)$-shifts
+and their transposes kill $B$ after $\ell$ applications. The fraction
+$h/p^3$ is at least $1/48$ of the ambient dimension.
+
+We should remark what is special about the value $\ell=(p-1)/2$.
+Choosing $\ell$ smaller would shrink $B$'s fraction too much; choosing
+$\ell$ larger would not annihilate $B$ at all (the top-degree columns
+would survive). The exponent $\ell$ is exactly the right size to
+simultaneously annihilate on both sides while keeping $B$ fat.