diff 3ff9228..b34550f kaplansky_direct_finite_odd_v2.txt
diff --git a/kaplansky_direct_finite_odd_v2.txt b/kaplansky_direct_finite_odd_v2.txt
index 1c316fe..47a4aea 100644
--- a/kaplansky_direct_finite_odd_v2.txt
+++ b/kaplansky_direct_finite_odd_v2.txt
@@ -24,33 +24,33 @@
% 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}
-
-\begin{document}
+
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+# A Counterexample to Kaplansky's Direct-Finiteness Conjecture in Odd Characteristic
+
+
\maketitle
-\begin{abstract}
+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
@@ -61,17 +61,17 @@ 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}
+# Introduction
-Let $K$ be a field and $G$ a group. The \emph{group algebra} $K[G]$
+
+## The puzzle: one-sided versus two-sided inversion
+
+Let $K$ be a field and $G$ a group. The group algebra $K[G]$
consists of all formal finite sums
\[
- c_1 g_1 + c_2 g_2 + \cdots + c_r g_r,
+ c_1 g_1 + c_2 g_2 + ... + c_r g_r,
\qquad c_i\in K,\ g_i\in G,
\]
added coordinate by coordinate and multiplied by writing out both sides,
@@ -87,7 +87,7 @@ 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 ring $R$ is 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
@@ -96,14 +96,14 @@ 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}.
+$\Mat_n(R)$, $n\ge1$, is directly finite is called 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}
+## 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
@@ -124,19 +124,19 @@ 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}
+## 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}.
+divides $(m!)^2+1$. This $p$ is an odd prime; see Lemma~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
@@ -144,14 +144,14 @@ 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.
@@ -159,7 +159,7 @@ 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}
+## 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
@@ -178,88 +178,88 @@ $c=\sum c_u u\in K[G]$, the rule
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},
+show $T_b$ is not surjective on $K^G$. Section~sec:cellular gives
+the details. The upshot is a group that fails to be surjunctive,
refuting Gottschalk's conjecture for this particular~$G$.
-\subsection{The plan of the proof}
+## 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.}
+Piece 1 (Section~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.}
+Piece 2 (Section~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
$\ell=(p-1)/2$,
\[
- (u(s)-1)^\ell e = 0 \quad\text{and}\quad e\,(v(s)-1)^\ell = 0
- \quad\text{for every } s.
+ (u(s)-1)^\ell e = 0 \quadand\quad e\,(v(s)-1)^\ell = 0
+ \quadfor every s.
\]
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
+Piece 3 (Section~sec:gluing): gluing without collapsing.
+We take $m$ copies $H_1,...,H_m$ 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.}
+Piece 4 (Section~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.
+matrix blocks with 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}):
+Piece 5 (Sections~\ref{sec:compression and~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}
+one-sided inverse. Then the finite Heisenberg group from Section~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}
+matrix identity into an identity between two scalar elements. Section~sec:cellular
translates it into a statement about cellular automata.
-\subsection{Notation summary}
+## Notation summary
-All matrices below act on column vectors in free \emph{right} modules
+All matrices below act on column vectors in free 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
+\item $d=\dim_K e K[H]$: the 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 $[n]$: $\{1,...,n\}$ 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 field, the graph, and the local group
+
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~$\Gamma$
@@ -268,30 +268,30 @@ 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}
+## The prime and the 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 $K_0$,
+\item The root $\xi$ of $X^2-\eta$ is 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}
+
+
+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
@@ -326,10 +326,10 @@ $N(\xi)=N(y^2)=N(y)^2$, 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
+Lemma~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.
@@ -337,21 +337,21 @@ 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.
+
+\item Representatives and order on $\F_p$.
+ Identify $\F_p$ with $\{0,1,...,p-1\}$ using the remainder
+ in $0,...,p-1$, 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 $K$.}
+\item 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 $M=p^4-1$.
+\item 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
@@ -372,35 +372,35 @@ In particular $M=p^4-1$ and all of its powers are invertible in $K$
$M=p^4-1$). 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 graph and the low-rank fit
+
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}
+of vertices. That is what allows Section~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
+Set $[1200]=\{1,2,...,1200\}$. Enumerate all six-hundred-element
+subsets $I⊆[1200]$ 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
+$I_1,...,I_m$ 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
+Define a simple graph $\Gamma$ on vertex set $\{1,...,m\}$ 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:
\[
- i\sim j \quad\text{iff}\quad |I_i\cap I_j| < 200.
+ i\sim j \quadiff\quad |I_i\cap I_j| < 200.
\]
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
\[
- \mathcal T=\{T\subseteq[1200]:|T|\le 400\},\qquad
+ \mathcal T=\{T⊆[1200]:|T|\le 400\},\qquad
t=|\mathcal T|=\sum_{j=0}^{400}\binom{1200}{j}.
\]
There exist coefficients $\alpha_i(T)$ and $\beta_j(T)$ in $\F_p$,
@@ -411,9 +411,9 @@ $F$ given by
\]
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
+
+Proof.
+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
\[
@@ -423,52 +423,52 @@ of the three pairwise intersections has at most $199$ elements (since
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
+The fit. Pick, for each $i$, the indicator function
\[
- \alpha_i(T)=\begin{cases}1,&T\subseteq I_i,\\0,&\text{otherwise.}\end{cases}
+ \alpha_i(T)=1,&T⊆ I_i,\\0,&otherwise.
\]
To build $\beta_j$, define the following polynomial in the formal
-variables $X_1,\ldots,X_{1200}$ over $\F_p$:
+variables $X_1,...,X_{1200}$ 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
+Every denominator lies in $\{1,2,...,400\}$, 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
+(so $X^\alpha=\prod_x X_x^{\alpha_x}$). Define the square-free
+reduction of $Q_j$ to be
\[
- Q_j^{\mathrm{sf}}(X)=\sum_\alpha c_\alpha\prod_{x:\alpha_x>0} X_x,
+ Q_j^{sf}(X)=\sum_\alpha c_\alpha\prod_{x:\alpha_x>0} X_x,
\]
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
+$Q_j^{sf}$ for each $T⊆[1200]$. 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
+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,
\[
Q_j(1_{I_i})
- =\prod_{k=200}^{599}\frac{|I_i\cap I_j|-k}{600-k}.
+ =\prod_{k=200}^{599}(|I_i\cap I_j|-k)/(600-k).
\]
-On the other hand, expanding $Q_j^{\mathrm{sf}}$ and evaluating at
+On the other hand, expanding $Q_j^{sf}$ 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
+$T⊆ I_i$ 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
+ F_{ij}=\prod_{k=200}^{599}(|I_i\cap I_j|-k)/(600-k).
+
+
+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,
@@ -478,7 +478,7 @@ 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
+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
@@ -489,8 +489,8 @@ $j=400$, and
For $m=\binom{1200}{600}$ we can compare factorials directly:
\[
\frac{\binom{1200}{400}}{\binom{1200}{600}}
- =\prod_{u=401}^{600}\frac{u}{u+200}
- \le \left(\tfrac 57\right)^{100}.
+ =\prod_{u=401}^{600}(u)/(u+200)
+ \le (\tfrac 57)^{100}.
\]
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
@@ -498,7 +498,7 @@ $u\ge501$, 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
\[
- \frac tm\le 401\left(\tfrac 57\right)^{100}.
+ \frac tm\le 401(\tfrac 57)^{100}.
\]
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 $401<40100<2^{33}$
@@ -508,22 +508,22 @@ gives
\]
which is weaker than needed; more precisely,
\[
- 100 \cdot \frac tm < 40100\left(\tfrac 57\right)^{100}
+ 100 \cdot \frac tm < 40100(\tfrac 57)^{100}
<\frac{40100}{2^{33}}<1,
\]
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}
+that is a fixed nonzero fraction of~$|H|$. Section~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 finite 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
@@ -535,11 +535,11 @@ 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
+be the group of 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
+The group $P$ acts on $A_0$ by 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.
@@ -565,36 +565,36 @@ 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
+Translation subgroup. Define
\[
- u(s)=(0,s),\qquad U=\{u(s):s\in P\}\subseteq H.
+ u(s)=(0,s),\qquad U=\{u(s):s\in P\}⊆ H.
\]
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
+Characteristic element. Let $c\in A_0$ be the function
\[
- c(0)=1,\quad c(x)=0\text{ for }x\ne0.
+ c(0)=1,\quad c(x)=0 for x\ne0.
\]
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
+Conjugate subgroup. Define
\[
- v(s)=c\,u(s)\,c^{-1},\qquad V=\{v(s):s\in P\}\subseteq H,
+ v(s)=c\,u(s)\,c^{-1},\qquad V=\{v(s):s\in P\}⊆ H,
\]
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}
+
+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
@@ -605,13 +605,13 @@ 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
+
+Why does this matter? In Section~sec:modular we shall build an
idempotent~$e$ with
\[
(u(s)-1)^\ell e = 0,\qquad e\,(v(s)-1)^\ell=0
- \quad\text{for every }s\in P,
+ \quadfor every s\in P,
\]
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 $U\cap V=\{1\}$,
@@ -622,10 +622,10 @@ 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}.
+are about to build in Section~sec:modular.
+
+# The large idempotent with ordered annihilations
-\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
@@ -633,116 +633,116 @@ $K$ 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 $d=\dim_K\bigl(eK[H]\bigr)$
+
+\item Large regular rank. The dimension $d=\dim_K(eK[H])$
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
+\item 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
+
+\item \Ssubsec:B-def: identify a subspace $B⊆ K^P$ 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
+\item \Ssubsec: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
+ $\lambda\inA_0hat$, i.e., characters $\lambda$ for which
a certain Gram matrix~$N_\lambda$ is nonsingular.
-\item \S\ref{subsec:idempotent}: assemble these ingredients into an
+\item \Ssubsec: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
+
+A last subsection (\Ssubsec: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
-$A_0\subseteq H$.
+$A_0⊆ H$.
+
+## The annihilated subspace $B⊆ K^P$
-\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
+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
\[
- X_j=L_{\varepsilon_j}-1_{K^P}, \qquad j=1,2,3,
+ X_j=L_{ε_j}-1_{K^P}, \qquad j=1,2,3,
\]
-where $\varepsilon_j$ are the standard generators of $P=\F_p^3$.
+where $ε_j$ 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
\[
- X_j^p=(L_{\varepsilon_j}-1)^p = L_{p\varepsilon_j}-1 = L_0-1 = 0.
+ X_j^p=(L_{ε_j}-1)^p = L_{pε_j}-1 = L_0-1 = 0.
\]
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}
+[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}
+
+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
+Fix $s=(s_1,s_2,s_3)\in P$. Since $1+X_j=L_{ε_j}$, we have
+$(1+X_j)^{s_j}=L_{s_jε_j}$ and
\[
\mathbf e_s=L_s\mathbf e_0
- =\prod_{j=1}^3 L_{s_j\varepsilon_j}\mathbf e_0
+ =\prod_{j=1}^3 L_{s_jε_j}\mathbf e_0
=\prod_{j=1}^3 (1+X_j)^{s_j}\mathbf e_0
- =\prod_{j=1}^3 \left(\sum_{k=0}^{s_j}\binom{s_j}{k}X_j^k\right)\mathbf e_0.
+ =\prod_{j=1}^3 (\sum_{k=0}^{s_j}\binom{s_j}{k}X_j^k)\mathbf e_0.
\]
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]_{
+
+ B := [X_1^{n_1}X_2^{n_2}X_3^{n_3}\mathbf e_0]_{
\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}
+[Two-sided annihilation of $B$]
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 $s=\sum_j s_j\varepsilon_j$.
+
+Proof.
+First identity. Fix $s\in P$ and write $s=\sum_j s_jε_j$.
Expand $L_s-\id_{K^P}$:
\[
- L_s-1 = \prod_{j=1}^3 L_{s_j\varepsilon_j} - 1
+ L_s-1 = \prod_{j=1}^3 L_{s_jε_j} - 1
= \prod_{j=1}^3 (1+X_j)^{s_j} - 1.
\]
Expanding the right side, every nonzero monomial has total $X$-degree
@@ -758,24 +758,24 @@ 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
+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
\[
- B^{\mathsf T}(L_s-1)^\ell = \bigl((L_{-s}-1)^\ell B\bigr)^{\mathsf T} = 0^{\mathsf T}=0,
+ B^{\mathsf T}(L_s-1)^\ell = ((L_{-s}-1)^\ell B)^{\mathsf T} = 0^{\mathsf T}=0,
\]
using the first identity with $-s$ in place of $s$.
-\end{proof}
-\begin{lemma}[Number of columns]\label{lem:h}
+
+[Number of columns]
The integer $h$ equals $\binom{\ell+2}{3}$, and
\[
- \frac{h}{p^3}=\frac{(p-1)(p+1)(p+3)}{48\,p^3}>\frac1{48}.
+ (h)/(p^3)=((p-1)(p+1)(p+3))/(48\,p^3)>\frac1{48}.
\]
-\end{lemma}
-\begin{proof}
+
+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
@@ -788,16 +788,16 @@ balls into $3$ labelled bins) gives
\]
Now $\ell=(p-1)/2$, so
\[
- \binom{\ell+2}{3} = \frac{(\ell+2)(\ell+1)\ell}{6}
- = \frac{(p+3)(p+1)(p-1)}{48}.
+ \binom{\ell+2}{3} = ((\ell+2)(\ell+1)\ell)/(6)
+ = ((p+3)(p+1)(p-1))/(48).
\]
Divide by $p^3$: $h/p^3=(p+3)(p^2-1)/(48\,p^3)$. To compare with
$1/48$, compute the difference
\[
- (p+3)(p^2-1)-p^3=3p^2-p-3>0 \quad\text{for }p\ge3,
+ (p+3)(p^2-1)-p^3=3p^2-p-3>0 \quadfor p\ge3,
\]
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
@@ -810,70 +810,70 @@ $\ell$ 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.
-\subsection{Choosing a compatible projection}
-\label{subsec:admissibility}
+## Choosing a compatible projection
+
Our ultimate idempotent $e$ will be a block-diagonal sum, one block per
character of~$A_0$. Each block is a projection matrix onto the image
of~$B$. The two opposite annihilations we want follow automatically
-from Lemma~\ref{lem:B-annihilation} once $e$ is a projection onto
+from Lemma~lem:B-annihilation once $e$ is a projection onto
$\im B$. But $e$ must live inside $K[H]$, and there the two shifts
$u(s)$ and $v(s)$ act through matrix representations of different
shapes. This subsection shows that for most characters the
corresponding block projection exists and is nonsingular.
-\paragraph{Characters of $A_0$.}
+Characters of $A_0$.:
$A_0=(\mathbb Z/M\mathbb Z)^P$ is finite abelian of order
-$|A_0|=M^{p^3}$. A \emph{character} is a homomorphism
+$|A_0|=M^{p^3}$. A character is a homomorphism
$\lambda:A_0\to K^*$. It is uniquely determined on the coordinate
generators $c_x$ (the function with value~$1$ at $x\in P$ and $0$
elsewhere): $\lambda(c_x)=\zeta^{a_x}$ for some
-$a_x\in\{0,\ldots,M-1\}$. Because $\zeta$ has order $M$ and is an
+$a_x\in\{0,...,M-1\}$. Because $\zeta$ has order $M$ and is an
element of~$K$, every character has values in~$K$.
-The shift group $P\subseteq H$ acts on characters by
+The shift group $P⊆ H$ acts on characters by
$(s\lambda)(a)=\lambda((-s)\cdot a)$. Conjugating the Fourier
idempotent $p_\chi^{A_0}$ of a character by $u(s)$ gives
\[
u(s)\,p_\chi^{A_0}\,u(s)^{-1}=p_{s\chi}^{A_0}.
\]
-\paragraph{The diagonal matrix and the Gram form.}
-For $\lambda\in\widehat{A_0}$ abbreviate $\chi_x=x\lambda$ (the
+The diagonal matrix and the Gram form.:
+For $\lambda\inA_0hat$ abbreviate $\chi_x=x\lambda$ (the
translate of $\lambda$ by $x$) and define the diagonal
$p^3\times p^3$ matrix
-\begin{equation}\label{eq:D-lambda}
- D_\lambda=\diag\!\bigl(\chi_x(c):x\in P\bigr).
-\end{equation}
+
+ D_\lambda=\diag\!(\chi_x(c):x\in P).
+
Every entry lies in $K^*$ because $\chi_x$ is a character and $c$ is
a group element. So $D_\lambda^{-1}$ exists and is diagonal.
The matrix
-\begin{equation}\label{eq:N-lambda}
+
N_\lambda=B^{\mathsf T}\,D_\lambda^{-1}\,B
-\end{equation}
+
is a symmetric $h\times h$ matrix over~$K$.
-Call $\lambda$ \emph{admissible} if
-\begin{enumerate}
+Call $\lambda$ admissible if
+
\item its $P$-stabilizer is trivial, i.e., the orbit $P\lambda$ has
size exactly $p^3$; and
\item $N_\lambda$ is invertible.
-\end{enumerate}
+
Both conditions are checkable in finite time, and the check is
deterministic: enumerate the $p^3$ translates and compute one
$h\times h$ determinant.
-\paragraph{Most characters are admissible.}
+Most characters are admissible.:
The two modes of failure each account for less than half of the
characters. Their union is then smaller than half, so more than half
succeed. This count is the key estimate of the whole construction.
-\begin{lemma}\label{lem:good-count}
+
More than half of the $|A_0|$ characters are admissible.
-\end{lemma}
-\begin{proof}
-\emph{Stabilizers.} Fix a nonzero $s\in P$. Translation by $s$
+
+Proof.
+Stabilizers. Fix a nonzero $s\in P$. Translation by $s$
partitions $P$ (of size $p^3$) into $p^2$ orbits of size~$p$. A
character $\lambda$ is fixed by $s$ if and only if $\lambda(c_x)$ is
constant on every such orbit, so exactly $M^{p^2}$ characters are
@@ -886,43 +886,43 @@ most $(p^3-1)M^{p^2}$, i.e. a fraction at most
The last inequality uses $M=p^4-1>4p^3$, which holds for $p\ge5$;
we have $p>600$.
-\emph{Singular $N_\lambda$.} The entries of $D_\lambda^{-1}$ range
+Singular $N_\lambda$. The entries of $D_\lambda^{-1}$ range
independently over $K^*$ as $\lambda$ ranges over all characters of
$A_0$, because $\chi_x(c)$ for distinct $x$ are independent characters.
Introduce independent variables $\tau_x$ and Cauchy--Binet gives
-\begin{equation}\label{eq:weighted-gram}
- Q(\tau):=\det\!\bigl(B^{\mathsf T}\diag(\tau_x)_{x\in P}B\bigr)
- =\sum_{\substack{J\subseteq P\\|J|=h}}\det(B_J)^2\prod_{x\in J}\tau_x ,
-\end{equation}
+
+ Q(\tau):=\det\!(B^{\mathsf T}\diag(\tau_x)_{x\in P}B)
+ =\sum_{\substack{J⊆ P\\|J|=h}}\det(B_J)^2\prod_{x\in J}\tau_x ,
+
where $B_J$ is the $h\times h$ submatrix of $B$ consisting of the
rows indexed by~$J$, taken in the fixed increasing order.
The right side is a sum of distinct monomials (different $J$ give
different exponent vectors), so it is a nonzero polynomial. Its total
degree is exactly~$h$. Because $B$ has full column rank by
-Lemma~\ref{lem:monomial-basis}, some maximal minor $B_J$ is nonzero.
+Lemma~lem:monomial-basis, some maximal minor $B_J$ is nonzero.
The polynomial $Q$ therefore has a nonzero leading monomial and does
not cancel.
-Now apply Schwartz's grid bound~\cite{Schwartz1980} (see also
-Zippel~\cite{Zippel1979}): a nonzero polynomial of total degree at
+Now apply Schwartz's grid bound~[cite] (see also
+Zippel~[cite]): a nonzero polynomial of total degree at
most~$a$ vanishes on at most an $a/M$ fraction of a product grid whose
coordinates each range over $M$ values. Here $Q$ has degree $h$ and
the grid is $(K^*)^{p^3}$ of side $M$. Thus $Q$ vanishes on at most
the fraction $h/M$ of characters. Since $h\le p^3$ (it counts a
subset of $K^P$) and $M=p^4-1$, this is at most
\[
- \frac{h}{M}\le\frac{p^3}{p^4-1}<\frac14 ,
+ (h)/(M)\le(p^3)/(p^4-1)<\frac14 ,
\]
because $p^4-1>4p^3$ for $p\ge5$.
Combining: fewer than $1/4+1/4=1/2$ of characters fail either
condition. So more than half admit a trivial stabilizer and an
invertible Gram form simultaneously.
-\end{proof}
-\subsection{Realizing the projection in $K[H]$}
-\label{subsec:idempotent}
+
+## Realizing the projection in $K[H]$
+
Each admissible character $\lambda$ gives a projection $T_\lambda$ on
the $p^3$-dimensional block attached to~$\lambda$. The task is to
@@ -931,17 +931,17 @@ can compute. The Fourier idempotents do this job: for each character
$\lambda$, the products $p_{x\lambda}u(s)$ (for $x,s\in P$) form a
$K$-basis of one matrix block of $K[H]$. Then we sum over $\lambda$.
-\paragraph{One character, one block.}
-Fix $\lambda\in\widehat{A_0}$ with trivial $P$-stabilizer. For
+One character, one block.:
+Fix $\lambda\inA_0hat$ with trivial $P$-stabilizer. For
$x,y\in P$ set
-\begin{equation}\label{eq:E-xy}
+
E_{xy}^{\lambda}=p_{x\lambda}^{A_0}\,u(x-y).
-\end{equation}
+
Because $u(s)\,p_{x\lambda}u(s)^{-1}=p_{(x+s)\lambda}$, the row
indices line up with the shift. Fourier orthogonality gives
-\begin{equation}\label{eq:E-mult}
+
E_{xy}^{\lambda}E_{zt}^{\lambda}=\delta_{yz}\,E_{xt}^{\lambda}.
-\end{equation}
+
Indeed, the Fourier product survives when $(x-y+z)\lambda=x\lambda$,
i.e. $(y-z)\lambda=0$, which by the trivial stabilizer forces $y=z$.
@@ -951,38 +951,38 @@ central idempotent. The basis $\{p_{x\lambda}u(s)\}_{x,s\in P}$
identifies the ideal $q_\lambda K[H]$ with the matrix algebra
$\Mat_{p^3}(K)$, via $u(s)\mapsto L_s$ and $c\mapsto D_\lambda$.
-\paragraph{Explicit projection.}
+Explicit projection.:
Define
-\begin{equation}\label{eq:T-lambda}
+
T_\lambda=B\,N_\lambda^{-1}B^{\mathsf T}D_\lambda^{-1}
\in\Mat_{p^3}(K).
-\end{equation}
+
Because $N_\lambda=B^{\mathsf T}D_\lambda^{-1}B$, we have
\[
T_\lambda B=B,\qquad
T_\lambda^2=T_\lambda.
\]
So $T_\lambda$ is a projection. It fixes each column of $B$, so
-$\im B\subseteq\im T_\lambda$; and it factors through $B^{\mathsf T}$,
-so $\im T_\lambda\subseteq\im B$. Hence $\im T_\lambda=\im B$ and
-$\operatorname{rank}T_\lambda=h$.
+$\im B⊆\im T_\lambda$; and it factors through $B^{\mathsf T}$,
+so $\im T_\lambda⊆\im B$. Hence $\im T_\lambda=\im B$ and
+$rankT_\lambda=h$.
Define an element of the $\lambda$-block of $K[H]$:
-\begin{equation}\label{eq:e-lambda}
+
e_\lambda=\sum_{x,y\in P}(T_\lambda)_{xy}\,E_{xy}^{\lambda}.
-\end{equation}
-By the matrix-unit relation~\eqref{eq:E-mult},
+
+By the matrix-unit relation~(eq:E-mult),
$e_\lambda^2=e_\lambda$. The action of $H$ on this block is the
matrix action of $T_\lambda$.
-\paragraph{The two annihilations.}
+The two annihilations.:
Left multiplication by $u(s)$ in this block is $L_s$. Therefore
\[
(L_s-1)^\ell\,T_\lambda
=(L_s-1)^\ell B\;N_\lambda^{-1}B^{\mathsf T}D_\lambda^{-1}
=0,
\]
-using $(L_s-1)^\ell B=0$ from Lemma~\ref{lem:B-annihilation}. The
+using $(L_s-1)^\ell B=0$ from Lemma~lem:B-annihilation. The
conjugate $v(s)$ acts by $D_\lambda L_sD_\lambda^{-1}$, so
\[
D_\lambda L_s D_\lambda^{-1}-1=D_\lambda(L_s-1)D_\lambda^{-1}.
@@ -995,38 +995,38 @@ Combining with the transpose identity $B^{\mathsf T}(L_s-1)^\ell=0$:
Hence $(u(s)-1)^\ell e_\lambda=0$ and
$e_\lambda(v(s)-1)^\ell=0$ in $K[H]$.
-\paragraph{Summing over characters.}
+Summing over characters.:
Blocks for distinct orbits are orthogonal: if $P\lambda\ne P\lambda'$
then $e_\lambda e_{\lambda'}=0$. Choose the least admissible character
in each $P$-orbit that contains one, and let $\mathscr R$ be the set
of chosen characters. Define
-\begin{equation}\label{eq:e-full}
+
e=\sum_{\lambda\in\mathscr R}e_\lambda \in K[H].
-\end{equation}
+
By orthogonality the sum is an idempotent: $e^2=e$. The two
annihilations persist on~$e$.
-\paragraph{The regular rank.}
+The regular rank.:
Left multiplication by $T_\lambda$ on $\Mat_{p^3}(K)$ has image the
matrices whose $p^3$ columns all lie in~$\im B$, of dimension $p^3h$.
Summing over the blocks:
\[
d:=\dim_K(eK[H])=p^3\,h\,|\mathscr R|.
\]
-The admissible characters form more than half of $\widehat{A_0}$
-(Lemma~\ref{lem:good-count}), and each $P$-orbit has size $p^3$, so
+The admissible characters form more than half of $A_0hat$
+(Lemma~lem:good-count), and each $P$-orbit has size $p^3$, so
\[
- |\mathscr R|>\frac{|A_0|}{2p^3}.
+ |\mathscr R|>(|A_0|)/(2p^3).
\]
Recall $w=|H|=|A_0|\,p^3$. Therefore
\[
- \frac dw=\frac{p^3h|\mathscr R|}{|A_0|p^3}
- =\frac{h|\mathscr R|}{|A_0|}
- >\frac{h}{2p^3}>\frac1{96},
+ \frac dw=(p^3h|\mathscr R|)/(|A_0|p^3)
+ =(h|\mathscr R|)/(|A_0|)
+ >(h)/(2p^3)>\frac1{96},
\]
-using $h/p^3>1/48$ from Lemma~\ref{lem:h}.
+using $h/p^3>1/48$ from Lemma~lem:h.
+
-\begin{proposition}\label{prop:modular}
There is a specified idempotent $e\in K[H]$ and positive integers
$d,w$ with $w=|H|$, $d=\dim_KeK[H]$, $d/w>1/96$, satisfying for
every $s\in P$
@@ -1035,44 +1035,44 @@ every $s\in P$
e\,(v(s)-1)^\ell=0 .
\]
Its coefficients are obtained by finite operations in the specified
-field $K$ of Lemma~\ref{lem:field}.
-\end{proposition}
+field $K$ of Lemma~lem:field.
+
Remark on $\ell$. Choosing $\ell$ smaller would shrink $B$'s fraction
below what we need; choosing $\ell$ larger would not annihilate the
top-degree columns of $B$, so the identities would fail. The exponent
$\ell=(p-1)/2$ is exactly the right size.
-\subsection{Ordered annihilations at edge level}
-\label{subsec:edge-zero}
+## Ordered annihilations at edge level
+
-Now take $m$ copies $H_1,\ldots,H_m$ of $H$, one per vertex of
+Now take $m$ copies $H_1,...,H_m$ of $H$, one per vertex of
$\Gamma$. For each edge $i<j$ identify the cyclic subgroup generated
by $v(1,\bar j,\bar j^2)$ in $H_i$ with the one generated by
$u(1,\bar i,\bar i^2)$ in $H_j$, where the bar denotes residues
in~$\F_p$. Let $G_0$ be the resulting free-product quotient. We prove
-in Section~\ref{sec:gluing} that $H_i\to G_0$ is injective; here we
+in Section~sec:gluing that $H_i\to G_0$ is injective; here we
only need the algebraic fact that the edge products vanish.
Denote by $e_i\in K[G_0]$ the image of the idempotent $e$ under the
map $H_i\to G_0$.
-\begin{corollary}\label{op:edge-zero}
+
For every edge $i<j$ of $\Gamma$, $e_ie_j=0$ in $K[G_0]$.
-\end{corollary}
-\begin{proof}
-Let $y\in K[H_i]\subseteq K[G_0]$ denote the common generator of
+
+Proof.
+Let $y\in K[H_i]⊆ K[G_0]$ denote the common generator of
order $p$ across the edge, and put $X=y-1$. On the $H_i$ side we
have $e_iX^\ell=0$ (as $y=v(s)$ is the conjugating shift); on the
$H_j$ side we have $X^\ell e_j=0$ (as $y=u(s)$ is the incoming
shift).
-Work inside $K[H_i]$, which is a free right $K[\langle y\rangle]$-module
-of rank $w/p$: pick representatives $a_1,\ldots,a_{w/p}$ of the left
-cosets of $\langle y\rangle$ in $H_i$. Then
-$e_i=\sum_a a\,q_a$ with $q_a\in K[\langle y\rangle]$.
+Work inside $K[H_i]$, which is a free right $K[< y>]$-module
+of rank $w/p$: pick representatives $a_1,...,a_{w/p}$ of the left
+cosets of $< y>$ in $H_i$. Then
+$e_i=\sum_a a\,q_a$ with $q_a\in K[< y>]$.
-The cyclic group algebra is $K[\langle y\rangle]=K[X]/(X^p)$ because
+The cyclic group algebra is $K[< y>]=K[X]/(X^p)$ because
$(1+X)^p=1+X^p$ in characteristic $p$. Inside this quotient the right
annihilator of $X^\ell$ is precisely the principal ideal
$(X^{p-\ell})$, since $X^{\ell+j}=0$ in $K[X]/(X^p)$ iff
@@ -1080,7 +1080,7 @@ $\ell+j\ge p$.
Thus $q_aX^\ell=0$ forces $q_a=q_a'X^{p-\ell}$, and
\[
- e_i=q\,X^{p-\ell}\quad\text{for}\quad q=\sum_a a\,q_a'\in K[H_i].
+ e_i=q\,X^{p-\ell}\quadfor\quad q=\sum_a a\,q_a'\in K[H_i].
\]
Because $p-\ell=\ell+1$,
@@ -1089,9 +1089,9 @@ Because $p-\ell=\ell+1$,
\]
No factor is commuted past $X$, no symmetry is used, and no reverse
product $e_je_i$ is asserted.
-\end{proof}
-\paragraph{Summary so far.}
+
+Summary so far.:
We have a minimal exceptional spherical map: connected, on the sphere,
with exactly one face of nonidentity value, and minimal among all such.
We now perform local reductions. Each reduction either produces a
@@ -1100,7 +1100,7 @@ forces a local restriction on the configuration. After the reductions,
the restrictions combine with Euler's formula to give a final
contradiction.
-\paragraph{Reduction 1: every edge is nonneutral.}
+Reduction 1: every edge is nonneutral.:
Suppose first that a nonloop edge is neutral. Contract it: edges with
value $1$ and their darts disappear from the boundary-word description,
and the graph stays connected. Face evaluations are unchanged, since
@@ -1112,7 +1112,7 @@ evaluations, and the discarded loop disappears. Minimality fails.
Hence every edge carries a nontrivial label.
-\paragraph{Reduction 2: every edge is a nonbridge.}
+Reduction 2: every edge is a nonbridge.:
Suppose an edge is a bridge with value $s$. Deleting it splits the map
into two connected components, and the unique boundary walk of each
adjacent face splits into two walks, one per component. After suitable
@@ -1133,26 +1133,26 @@ results.
So every edge is a nonbridge.
-\paragraph{Reduction 3: the two faces across an edge have distinct types.}
+Reduction 3: the two faces across an edge have distinct types.:
Suppose an edge separates two faces $\Phi,\Phi'$ of the same type $i\in
V(\Delta)$. Delete the edge and merge the two faces into one of type
$i$. If the boundary walk of $\Phi$ near the edge reads $sD$ and that
of $\Phi'$ reads $s^{-1}E$ (with the shared edge contributing $s$ on
one side and $s^{-1}$ on the other, and $D,E$ the remaining letters),
the merged walk reads $DE$. Direct computation gives
-\begin{equation}\label{eq:gluing-merge}
+
DE=s^{-1}(sD)s\,(s^{-1}E).
-\end{equation}
+
If $sD$ evaluates to $1$, then $s^{-1}(sD)s$ also evaluates to $1$; if
$sD$ evaluates to $g\ne1$, then its conjugate $s^{-1}gs$ also has value
$\ne1$. Conjugation preserves identity-versus-nonidentity. Hence:
-\begin{itemize}
+
\item ordinary + ordinary $\to$ ordinary;
\item ordinary + exceptional (in either order) $\to$ exceptional.
-\end{itemize}
-We are assuming the map has \emph{exactly one} exceptional face, so
+
+We are assuming the map has exactly one exceptional face, so
never both-exceptional. In every case, the merge produces a smaller
map with at most one exceptional face, and thus a smaller exceptional
map. Minimality fails.
@@ -1160,10 +1160,10 @@ map. Minimality fails.
Therefore the two face types across an edge labelled $\gamma$ are
exactly the two endpoints of~$\gamma$.
-\paragraph{Reduction 4: at most one loop, bounding the exceptional monogon.}
+Reduction 4: at most one loop, bounding the exceptional monogon.:
Suppose there is a loop. Its two sides are faces (possibly with other
stuff in between, but the loop has just its two immediate face-sides).
-Consider the side \emph{not} containing the exceptional face. Retain
+Consider the side not containing the exceptional face. Retain
the loop and everything on that side; erase everything on the side
that contains the exceptional face. Retained faces are all ordinary
(they were ordinary before). The erased side becomes a single monogon
@@ -1174,29 +1174,29 @@ face of the new, smaller map.
Minimality fails, unless the erased side consisted only of that
isolated monogon originally. Hence every loop bounds the exceptional
monogon in the original map. Since there is only one exceptional face
-(it bounds only one walk), \emph{there is at most one loop}, and if it
+(it bounds only one walk), there is at most one loop, and if it
exists it bounds exactly the exceptional monogon.
-\paragraph{Reduced configuration.}
+Reduced configuration.:
Combining the four reductions, a minimal exceptional map satisfies:
-\begin{itemize}
+
\item every edge is nonneutral;
\item every edge is a nonbridge;
\item the two faces across an edge labelled $\gamma$ have the two
distinct endpoint types of $\gamma$;
\item at most one loop exists, and it bounds the exceptional monogon.
-\end{itemize}
+
We continue in the next reduction. The one remaining configuration to
rule out is that a face has two consecutive nonloop darts with the
same label. Doing so will give the final contradiction with Euler's
formula.
-\paragraph{Reduction 5: no two consecutive nonloop darts with the same label.}
+Reduction 5: no two consecutive nonloop darts with the same label.:
Suppose a face $\Phi$ has consecutive nonloop darts $e:a\to b$ and
$f:b\to d$ with a common label $\gamma\in E(\Delta)$ and values
-$\epsilon,\varphi\in C_\gamma\setminus\{1\}$. The edges $e$ and $f$
+$ε,\varphi\in C_\gamma∖\{1\}$. The edges $e$ and $f$
are distinct: otherwise the same edge would occur on both sides of
$\Phi$, making it a bridge, contradicted by Reduction 2.
@@ -1207,56 +1207,56 @@ The faces $X$ and $Y$ might coincide.
Remove $e,f$. Insert a new edge $g:a\to d$ along their arcs, smoothed
through the empty corner of $\Phi$ at~$b$. Give $g$ value
-$\epsilon\varphi$ and label $\gamma$ (or make it neutral if this
+$ε\varphi$ and label $\gamma$ (or make it neutral if this
equals $1$; Reduction 1 then applies if needed). The walk of $\Phi$
replaces $ef$ by $g$: same face type $i$, same evaluation. Only $X$
and $Y$ change; every other face keeps its type and value.
-\emph{Case $X\ne Y$.} Write $X=\bar e D$, $Y=\bar f E$, with $D:a\to b$
+Case $X\ne Y$. Write $X=\bar e D$, $Y=\bar f E$, with $D:a\to b$
and $E:b\to d$. The merged face has walk $\bar gDE$. Its evaluation is
-\begin{equation}\label{eq:gluing-smooth-merge}
- (\epsilon\varphi)^{-1}DE
- =\varphi^{-1}(\epsilon^{-1}D)\,\varphi\,(\varphi^{-1}E).
-\end{equation}
-Let $p=\epsilon^{-1}D$ (old $X$'s value) and $q=\varphi^{-1}E$ (old
+
+ (ε\varphi)^{-1}DE
+ =\varphi^{-1}(ε^{-1}D)\,\varphi\,(\varphi^{-1}E).
+
+Let $p=ε^{-1}D$ (old $X$'s value) and $q=\varphi^{-1}E$ (old
$Y$'s value). The merged value is $\varphi^{-1}p\varphi\cdot q$:
-\begin{itemize}
+
\item $p=q=1$: merged value $\varphi^{-1}\cdot 1\cdot\varphi\cdot 1=1$;
\item $p\ne1$, $q=1$: merged value $\varphi^{-1}p\varphi\ne1$ (conjugate
of nonidentity);
\item $p=1$, $q\ne1$: merged value $q\ne1$;
\item $p\ne1$, $q\ne1$: excluded (at most one exceptional face).
-\end{itemize}
+
So the merged face is ordinary when both old faces were ordinary, and
exceptional when exactly one was. A smaller exceptional map results,
contradicting minimality.
-\emph{Case $X=Y$.} Now $X$ is a single face whose boundary walk
+Case $X=Y$. Now $X$ is a single face whose boundary walk
encounters $e$ and $f$ on their opposite sides. Reading from the dart
$\bar e$: the walk is $\bar eD\bar fE$, with $D:a\to d$ and
$E:b\to b$.
-After smoothing, the single walk becomes \emph{two} walks
+After smoothing, the single walk becomes two walks
$\bar gD$ and $E$. The rules are easiest to check by slots: at $a$
and $d$ the new darts replace the old slots of $e,\bar f$; at $b$ the
consecutive slots $\bar e,f$ are deleted. Counting $v$, $e$, $F$:
$v$ unchanged, $e$ decreased by one, $F$ increased by one. Euler's
-formula gives $2c_{\text{new}}=v-(e-1)+(F+1)=2+2=4$, so the new
-drawing has \emph{exactly two} components.
+formula gives $2c_{new}=v-(e-1)+(F+1)=2+2=4$, so the new
+drawing has exactly two components.
Every old face except $X$ survives in one component; $X$ was the only
face shared between the two. One component contains $a,d$ (with $g$),
the other contains $b$ (isolated if $E$ is empty, otherwise with the
walk $E$). Both new faces have type $j$.
-Let $A=(\epsilon\varphi)^{-1}D$ be the value of walk $\bar gD$, and
+Let $A=(ε\varphi)^{-1}D$ be the value of walk $\bar gD$, and
$B=E$ the value of walk $E$. The old walk of $X$ evaluated to
-$\epsilon^{-1}D\varphi^{-1}E$, whose cyclic shift starting at $\bar f$
+$ε^{-1}D\varphi^{-1}E$, whose cyclic shift starting at $\bar f$
is $(\varphi^{-1}B\varphi)A$:
-\begin{itemize}
+
\item $X$ exceptional: the old value was $\ne1$, so at least one of
$A,B$ is $\ne1$. Keep the component whose walk is that one.
\item $X$ ordinary with $A=B=1$: both new walks are ordinary, so
@@ -1268,57 +1268,57 @@ is $(\varphi^{-1}B\varphi)A$:
exception), or in the $E$-component. Keep the component
without the face $\Phi$, whose unique exception is its
sole nonidentity boundary.
-\end{itemize}
+
In every case, a smaller exceptional map results, contradicting
minimality. So the reduced configuration forbids consecutive nonloop
darts with a common label, on any face.
-\subsection{Euler deficit and the final contradiction}
-\label{subsec:gluing-euler-deficit}
+## Euler deficit and the final contradiction
+
Recall the reduced configuration: no neutral edges, no bridges,
distinct endpoint types across every edge, at most one loop (bounding
the exceptional monogon), and no face has two consecutive nonloop
darts with the same label.
-\paragraph{Vertices away from the loop.}
+Vertices away from the loop.:
Let $u$ be a vertex that is not the base of any loop. Read the face
types around $u$ cyclically; call the resulting sequence
-$i_1,i_2,\ldots,i_r$, where $i_a$ is the type of the $a$-th face,
+$i_1,i_2,...,i_r$, where $i_a$ is the type of the $a$-th face,
and $r=\deg u$. Successive $i_a$ differ: the edge between them has
endpoints $i_{a}$ and $i_{a+1}$, by Reduction 3. No immediate
reversal by Reduction 5 (which says the labels, hence the ordered
endpoints, are distinct).
-So we get a closed walk $i_1,i_2,\ldots,i_r,i_1$ in $\Delta$ with
+So we get a closed walk $i_1,i_2,...,i_r,i_1$ in $\Delta$ with
successive edges distinct and length $r\ge1$. Since $\Delta$ is
simple and triangle-free, no such closed walk has $r<4$: length $1$
or $2$ would be a loop or a doubling of an edge in a simple graph;
length $3$ would be a triangle. Thus $\deg u\ge4$.
-\paragraph{Faces away from the loop.}
+Faces away from the loop.:
Let $F$ be an ordinary face not incident to the loop. Its boundary is
nonempty: any connected graph with at least one edge has nonempty
dart cycles. Successive labels differ by Reduction 5, so the closed
-walk in $\Delta$ has length at least $4$. Thus $|\partial F|\ge4$.
+walk in $\Delta$ has length at least $4$. Thus $|∂ F|\ge4$.
-\paragraph{Euler deficit.}
+Euler deficit.:
For a connected spherical drawing,
-\begin{equation}\label{eq:gluing-deficit}
- \sum_{u}(4-\deg u)+\sum_{F}(4-|\partial F|)=8,
-\end{equation}
-which follows from Euler's formula and $\sum_u\deg u=\sum_F|\partial
+
+ \sum_{u}(4-\deg u)+\sum_{F}(4-|∂ F|)=8,
+
+which follows from Euler's formula and $\sum_u\deg u=\sum_F|∂
F|=2e$ (every edge contributes two to each side).
-\emph{No loop present.} Every vertex satisfies $\deg u\ge4$, so
-contributes $\le0$. Every ordinary face has $|\partial F|\ge4$, so
+No loop present. Every vertex satisfies $\deg u\ge4$, so
+contributes $\le0$. Every ordinary face has $|∂ F|\ge4$, so
contributes $\le0$. Only the exceptional face can contribute
-positively: its walk is nonempty, so $|\partial F|\ge1$ and the
+positively: its walk is nonempty, so $|∂ F|\ge1$ and the
contribution is at most $3$. Total $\le3<8$, contradicting
-\eqref{eq:gluing-deficit}.
+(eq:gluing-deficit).
-\emph{Loop present.} The base vertex $u_0$ of the loop has $\deg u_0
+Loop present. The base vertex $u_0$ of the loop has $\deg u_0
\ge2$ (the loop contributes two), so contributes at most $2$. The
exceptional monogon has walk of length $1$, so contributes $3$. The
ordinary face on the other side of the loop sees the loop's dart with
@@ -1327,65 +1327,65 @@ a monogon (a monogon would have just the loop's dart, giving value
$\ne1$, contradicting ordinari-ness); hence its length is at least
$2$, contributing at most $2$. All other vertices and faces
contribute at most zero. Total $\le 2+3+2=7<8$, contradicting
-\eqref{eq:gluing-deficit}.
+(eq:gluing-deficit).
Hence no minimal exceptional spherical map exists. So $L\to D_i^0$
is injective, i.e., $D_i^0\to Q$ is injective, for every vertex~$i$.
-\subsection{Extending to the full vertex group}
-\label{subsec:gluing-extension}
+## Extending to the full vertex group
+
We now extend $D_i^0\to Q$ to $D_i\to Q$. The argument is a
permutation-action construction.
-Fix $i$. Let $\Omega=L\times\mathbb{N}$, with $L$ acting by left
+Fix $i$. Let $Ω=L\timesℕ$, with $L$ acting by left
multiplication on the first coordinate. The action of $D_i^0$ on
-$\Omega$, transported through $D_i^0\hookrightarrow L$, is free on
+$Ω$, transported through $D_i^0↪ L$, is free on
countably infinitely many orbits. The same is true of the left
-$D_i^0$-action on $D_i\times\mathbb{N}$.
+$D_i^0$-action on $D_i\timesℕ$.
Pair the $D_i^0$-orbits bijectively between the two sets and choose
one point in each pair. Extending $D_i^0$-equivariantly gives a
-$D_i^0$-equivariant bijection $\phi_i:D_i\times\mathbb{N}\to\Omega$.
+$D_i^0$-equivariant bijection $\phi_i:D_i\timesℕ\toΩ$.
Transport the $D_i$ left-multiplication along $\phi_i$: this gives a
-$D_i$-action on $\Omega$ whose restriction to $D_i^0$ is exactly the
+$D_i$-action on $Ω$ whose restriction to $D_i^0$ is exactly the
original $L$-action.
Do this independently for each $i$. On a shared edge $\gamma=\{i,j\}$,
-both copies of the generator $x_\gamma\in L$ act on $\Omega$ by the
+both copies of the generator $x_\gamma\in L$ act on $Ω$ by the
same element, so the resulting actions glue to a $Q$-action on
-$\Omega$. The action restricted to each $D_i$ is faithful, since
-the original left action on $D_i\times\mathbb{N}$ is faithful and
+$Ω$. The action restricted to each $D_i$ is faithful, since
+the original left action on $D_i\timesℕ$ is faithful and
$\phi_i$ is a bijection.
Therefore $D_i\to Q$ is injective for every vertex $i$. This
-completes the proof of Lemma~\ref{lem:triangle-free-gluing}.
+completes the proof of Lemma~lem:triangle-free-gluing.
+
+## Applying the criterion to $H_i$
-\subsection{Applying the criterion to $H_i$}
-\label{subsec:gluing-application}
We now apply the lemma to our concrete setting. Recall $H=A_0\rtimes
P$ with $A_0=(\mathbb Z/M\mathbb Z)^P$, $P=\F_p^3$, $U=u(P)$,
-$V=cUc^{-1}$, and $U\cap V=\{1\}$ from Proposition~\ref{prop:UV-triv}.
+$V=cUc^{-1}$, and $U\cap V=\{1\}$ from Proposition~prop:UV-triv.
For each edge $i<j$ of $\Gamma$ we identify the cyclic subgroup
generated by $v(1,\bar j,\bar j^2)$ in $H_i$ with the one generated
by $u(1,\bar i,\bar i^2)$ in $H_j$. Here the bars denote residues in
$\F_p$, and both elements have exact order $p$.
-\begin{proposition}\label{prop:first-gluing-injective}
+
Every natural homomorphism $H_i\to G_0$ is injective. Consequently
the induced maps $K[H_i]\to K[G_0]$ on group algebras are injective.
-\end{proposition}
-\begin{proof}
-Apply Lemma~\ref{lem:triangle-free-gluing} with $\Delta=\Gamma$ and
+
+Proof.
+Apply Lemma~lem:triangle-free-gluing with $\Delta=\Gamma$ and
$D_i=H$. The vertex groups $H_i$ are finite, the cyclic edge groups
have order $p$, and the edge embeddings are given by the displayed
vectors. We must verify the short-word condition.
Fix a vertex $i$. The incident edges split into two kinds:
-\begin{itemize}
+
\item $j<i$: the syllable from edge $\{i,j\}$ in $H_i$ lies in~$U$ and
is $u(\lambda\cdot(1,\bar i,\bar i^2))$ for some scalar
$\lambda\in\F_p^*$; the vector $(1,\bar i,\bar i^2)$ is attached
@@ -1393,18 +1393,18 @@ Fix a vertex $i$. The incident edges split into two kinds:
\item $j>i$: the syllable from edge $\{i,j\}$ in $H_i$ lies in~$V$ and
is $v(\lambda\cdot(1,\bar j,\bar j^2))$ for some
$\lambda\in\F_p^*$.
-\end{itemize}
+
Within each of $U$ or $V$, distinct neighbors $j$ correspond to the
vectors $(1,\bar j,\bar j^2)\in\F_p^3$. Three of these vectors with
three distinct residues $\bar j_1,\bar j_2,\bar j_3$ are linearly
independent over $\F_p$: the determinant
\[
-\det\begin{pmatrix}
+\det
1 & \bar j_1 & \bar j_1^2\\
1 & \bar j_2 & \bar j_2^2\\
1 & \bar j_3 & \bar j_3^2
-\end{pmatrix}
+
= (\bar j_2-\bar j_1)(\bar j_3-\bar j_1)(\bar j_3-\bar j_2)
\ne 0,
\]
@@ -1423,13 +1423,13 @@ and the $U$-element is nonidentity, contradiction. For lengths $2$
and $3$, cyclically distinct labels are just distinct labels, so
the check is exhaustive.
-All hypotheses of Lemma~\ref{lem:triangle-free-gluing} hold, giving
+All hypotheses of Lemma~lem:triangle-free-gluing hold, giving
the injection $H_i\to G_0$. The induced map $K[H_i]\to K[G_0]$ is
injective because distinct elements of an embedded subgroup remain
distinct basis elements of the group algebra.
-\end{proof}
+
Consequently $G_0$ is a single group, finitely generated (by the
finite sets of generators of the finitely many $H_i$), containing
each $H_i$ as a subgroup, and the $e_i\in K[G_0]$ satisfy
-Corollary~\ref{op:edge-zero}.
+Corollary~op:edge-zero.