diff b34550f..b1f9bc4 kaplansky_direct_finite_odd_v2.txt
diff --git a/kaplansky_direct_finite_odd_v2.txt b/kaplansky_direct_finite_odd_v2.txt
index 47a4aea..8fca700 100644
--- a/kaplansky_direct_finite_odd_v2.txt
+++ b/kaplansky_direct_finite_odd_v2.txt
@@ -48,7 +48,7 @@
# A Counterexample to Kaplansky's Direct-Finiteness Conjecture in Odd Characteristic
-\maketitle
+maketitle
Abstract:
Kaplansky conjectured in 1972 that the group algebra of any group
@@ -58,7 +58,7 @@ 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
+sums $a,bin K[G]$ with $ab=1$ and $bane 1$. The same two sums then
produce an injective cellular automaton on~$K^G$ that is not surjective,
so the group~$G$ is not surjunctive.
@@ -70,36 +70,36 @@ so the group~$G$ is not surjunctive.
Let $K$ be a field and $G$ a group. The group algebra $K[G]$
consists of all formal finite sums
-\[
+[
c_1 g_1 + c_2 g_2 + ... + c_r g_r,
- \qquad c_i\in K,\ g_i\in G,
-\]
+ qquad c_iin K, g_iin 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
+Concrete example. When $G=mathbb Z$bbZ$ and $K$ is any field, the
+group algebra $K[mathbb Z]$bbZ]$ 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 directly finite when the equation $ab=1$, for
-$a,b\in R$, always implies~$ba=1$. Fields are directly finite (the
+$a,bin R$, 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
+The $ntimes n$ 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 stably finite.
+$Mat_n(R)$, $nge1$, 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 $\Mat_n(K[G])$
+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.
@@ -113,7 +113,7 @@ 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
+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
@@ -126,16 +126,16 @@ single specific odd characteristic.
## The main theorem
-Put $m=\binom{1200}{600}$ and let $p$ be the least integer $>1$ that
+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~lem:field.
We construct:
-\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
+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 $ba\ne 1$.
+item two finite sums $a,bin K[G]$ with $ab=1$ and $bane 1$.
The construction is entirely finite: every field element, every group
@@ -145,11 +145,11 @@ is needed to define $a,b$; the point of the construction is to build them
from finite data rather than by deciding equalities in~$G$.
-Let $m=\binom{1200}{600}$ and let $p$ be the least integer $h>1$
+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
+containing torsion, and specified sums $a,bin K[G]$ satisfying
+$ab=1$ but $bane 1$. In particular Kaplansky's direct-finiteness
conjecture fails in odd characteristic.
@@ -162,7 +162,7 @@ of a finite recipe, not of an arbitrary choice among infinitely many.
## 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
+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.
@@ -171,13 +171,13 @@ 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 $\supp c$).
+$c=sum c_u uin K[G]$, 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
+$T_b$ has left inverse $T_a$, hence is injective. If $bane 1$, we can
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~$G$.
@@ -189,30 +189,30 @@ self-contained; the reader may take them in whatever order feels best,
as long as the final step receives all the pieces.
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
+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.
+group $H=A_0rtimes P$ of order $w$ that will be the local block.
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
+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
-$\ell=(p-1)/2$,
-\[
- (u(s)-1)^\ell e = 0 \quadand\quad e\,(v(s)-1)^\ell = 0
- \quadfor every s.
-\]
+$ell=(p-1)/2$,
+[
+ (u(s)-1)^ell e = 0 quadandquad 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 $v$
+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.
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
+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.
@@ -226,7 +226,7 @@ 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.
-Piece 5 (Sections~\ref{sec:compression and~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
@@ -243,26 +243,26 @@ each entry on the left by the corresponding group algebra element.
This matters because group algebras are not always commutative.
-\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
+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 $K[G]$
+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 regular rank of $e$, i.e., the
+item $ein K[H]$: an idempotent, meaning $e^2=e$.
+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 $[n]$: $\{1,...,n\}$ 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~$S$.
+item $w=|H|$: the order of the finite group~$H$.
+item $[n]$: ${1,...,n}$ for a positive integer $n$.
+item Hermitian transpose: never used here. Ordinary transpose is
+ written $X^{mathsf T}sfT$, with no scalar conjugation.
+item $Mat_{atimes b}(S)$: $atimes b$ matrices with entries in~$S$.
# 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~$\Gamma$
+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
@@ -271,65 +271,65 @@ all here, in the order needed later.
## The prime and the field
-Recall $m=\binom{1200}{600}$, a specific (very large) positive integer,
+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.
-With $m=\binom{1200}{600}$ and $p$ the least $h>1$ dividing
+With $m=binom{1200}{600}$ and $p$ the least $h>1$ dividing
$(m!)^2+1$, we have:
-\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 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.
+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 $etainF_p^*$ 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 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 $|K|=p^4$.
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 $p\ne2$.
+a smaller divisor of~$n$. So $p$ is prime. Now, if $ple m$, then
+$p$ would divide $m!$ (since $mge p$), so also $p$ would divide
+$(m!)^2$; combined with $pmid (m!)^2+1$ this gives $pmid 1$,
+impossible. Hence $p>m>600$. Since $mge600ge1$ and $p>mge600$,
+$m!$ is even (as $mge2$), so $(m!)^2+1$ is odd, and $pne2$.
Therefore $p$ is odd.
-(2) The congruence $(m!)^2\equiv -1\pmod p$ says that $-1$ is the
+(2) The congruence $(m!)^2equiv -1pmod p$ says that $-1$ is the
square of $m!$ modulo~$p$, that is, $-1$ is a quadratic residue in
-$\F_p$.
+$F_p$.
-(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
+(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 $xmapsto x^2$ 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.
-
-
-From now on, $K$ is the specific field $K=\F_{p^4}$ supplied by
-Lemma~lem:field. We descend to $\F_p$ only when we need to check
+squares, and $p-1ge2$ 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 $xicdot(-xi)=-xi^2=-eta$ 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 $yin K_0$.
+Apply the field norm $N_{K_0/F_p}(u)=ucdot u'$ (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.
+
+
+From now on, $K$ is the specific field $K=F_{p^4}$ supplied by
+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.
@@ -338,33 +338,33 @@ construction to be ``specified'' in the sense of the introduction,
we spell out an explicit finite recipe for each choice.
-\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 Representatives and order on $K$.
+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 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
+ $c_0+c_1xi+c_2nu+c_3xinu$ with $c_jinF_p$. Order $K$ by
the lexicographic order on $(c_0,c_1,c_2,c_3)$. This gives a
total order on $K$.
-\item A generator $\zeta$ of $K^*$. Put $M=p^4-1$.
- Choose the least element $\zeta$ of $K^*$ whose multiplicative
+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.
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
+multiple of all element orders in $K^*$. Conventionally $ble M$ 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,
+$Mle b$. 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 $\zeta$
+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 $K$
@@ -375,40 +375,40 @@ integer is a factor of~$M$ and hence nonzero in~$K$.
## The graph and the low-rank fit
-The second finite ingredient is a simple triangle-free graph~$\Gamma$
+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~op:compression-section
to build a ``rank paradox'' with which the odd-characteristic
counterexample becomes possible.
-Set $[1200]=\{1,2,...,1200\}$. Enumerate all six-hundred-element
+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
-$I_1,...,I_m$ 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 $\{1,...,m\}$ 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:
-\[
- i\sim j \quadiff\quad |I_i\cap I_j| < 200.
-\]
+Define a simple graph $Gamma$ on vertex set ${1,...,m}$ by
+declaring the distinct vertices $ine j$ to be adjacent exactly when
+the intersection $I_icap I_j$ has fewer than $200$ elements:
+[
+ isim j quadiffquad |I_icap I_j| < 200.
+]
By convention a vertex is never adjacent to itself, so the graph is
simple.
-The graph $\Gamma$ is simple and triangle-free. Let
-\[
- \mathcal T=\{T⊆[1200]:|T|\le 400\},\qquad
- t=|\mathcal T|=\sum_{j=0}^{400}\binom{1200}{j}.
-\]
-There exist coefficients $\alpha_i(T)$ and $\beta_j(T)$ in $\F_p$,
-indexed by $T\in\mathcal T$, such that the $m\times m$ matrix
+The graph $Gamma$ is simple and triangle-free. Let
+[
+ mathcal T=calT={T⊆[1200]:|T|le 400},qquad
+ t=|mathcal T|=calT|=sum_{j=0}^{400}binom{1200}{j}.
+]
+There exist coefficients $alpha_i(T)$ and $beta_j(T)$ in $F_p$,
+indexed by $Tinmathcal T$,calT$, such that the $mtimes m$ matrix
$F$ given by
-\[
- F_{ij}=\sum_{T\in\mathcal T}\alpha_i(T)\,\beta_j(T)
-\]
+[
+ F_{ij}=sum_{Tinmathcal T}calTalpha_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 $t/m<1/100$.
@@ -416,101 +416,101 @@ 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
-\[
- |I_i\cup I_j\cup I_k| \ge |I_i|+|I_j|+|I_k| - |I_i\cap I_j| - |I_j\cap I_k| - |I_k\cap I_i|
- \ge 3\cdot 600 - 3\cdot 199 = 1203,
-\]
+[
+ |I_icup I_jcup I_k| ge |I_i|+|I_j|+|I_k| - |I_icap I_j| - |I_jcap I_k| - |I_kcap I_i|
+ ge 3cdot 600 - 3cdot 199 = 1203,
+]
by inclusion-exclusion. But the union is contained in $[1200]$, which
-has only $1200$ elements. Contradiction. So $\Gamma$ has no triangles.
+has only $1200$ elements. Contradiction. So $Gamma$ has no triangles.
The fit. Pick, for each $i$, the indicator function
-\[
- \alpha_i(T)=1,&T⊆ I_i,\\0,&otherwise.
-\]
-To build $\beta_j$, define the following polynomial in the formal
-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 $\{1,2,...,400\}$, 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 square-free
+[
+ alpha_i(T)=1,&T⊆ I_i,0,&otherwise.
+]
+To build $beta_j$, define the following polynomial in the formal
+variables $X_1,...,X_{1200}$ over $F_p$:
+[
+ Q_j(X)=prod_{k=200}^{599}frac{sum_{xin I_j}X_x-k}{600-k}.
+]
+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 $c_alphainF_p$ and multi-indices $alphainmathbb N^bbN^{[1200]}$
+(so $X^alpha=prod_x X_x^{alpha_x}$). Define the square-free
reduction of $Q_j$ to be
-\[
- Q_j^{sf}(X)=\sum_\alpha c_\alpha\prod_{x:\alpha_x>0} X_x,
-\]
+[
+ Q_j^{sf}(X)=sum_alpha c_alphaprod_{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
+$beta_j(T)$ to be the coefficient of $prod_{xin T}X_x$ in
$Q_j^{sf}$ 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
+$Tinmathcal T$calT$ can appear. The total degree of $Q_j$ bounds the
support of beta_j exactly to subsets of size at most 400.
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,
-\[
+$0^n=0$ and $1^n=1$ for every $nge1$). In particular, evaluate at the
+indicator vector $1_{I_i}in{0,1}^{[1200]}$ (entry $1$ at positions
+$xin I_i$, entry $0$ elsewhere). On the one hand,
+[
Q_j(1_{I_i})
- =\prod_{k=200}^{599}(|I_i\cap I_j|-k)/(600-k).
-\]
+ =prod_{k=200}^{599}(|I_icap I_j|-k)/(600-k).
+]
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
+$1_{I_i}$, each monomial $prod_{xin T}X_x$ contributes $1$ if
$T⊆ I_i$ and $0$ otherwise. So
-\[
+[
Q_j(1_{I_i})
- =\sum_{T\in\mathcal T}\alpha_i(T)\,\beta_j(T)=F_{ij}.
-\]
+ =sum_{Tinmathcal T}calTalpha_i(T),beta_j(T)=F_{ij}.
+]
Equating the two evaluations gives
- F_{ij}=\prod_{k=200}^{599}(|I_i\cap I_j|-k)/(600-k).
+ F_{ij}=prod_{k=200}^{599}(|I_icap I_j|-k)/(600-k).
-Fit conditions. If $i=j$ then $|I_i\cap I_j|=600$, so every
+Fit conditions. If $i=j$ then $|I_icap I_j|=600$, 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,
+$F_{ii}=1$. If $ine j$ are nonadjacent then $|I_icap I_j|ge200$ and
+$le 599$, so the factor with $k=|I_icap I_j|$ 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}.
+polynomial specialization of cite[Proposition~13]{Haviv2018Theta}
+and cite[Theorem~3.2]{GolovnevHaviv2022}.
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 $j<(n-1)/2$
+coefficients $binom{1200}{j}$ increase with~$j$ as long as $jle599$
+(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:
-\[
- \frac{\binom{1200}{400}}{\binom{1200}{600}}
- =\prod_{u=401}^{600}(u)/(u+200)
- \le (\tfrac 57)^{100}.
-\]
+[
+ t le 401binom{1200}{400}.
+]
+For $m=binom{1200}{600}$ we can compare factorials directly:
+[
+ frac{binom{1200}{400}}{binom{1200}{600}}
+ =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
-$u\ge501$, the factor is at most $600/800=3/4<1$, so it contributes
+$ule 500$, the factor $u/(u+200)$ is at most $500/700=5/7$; for
+$uge501$, 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
-\[
- \frac tm\le 401(\tfrac 57)^{100}.
-\]
+[
+ frac tmle 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 $401<40100<2^{33}$
gives
-\[
- \frac tm<2^{33}\cdot \frac1{2^{33}}=1,
-\]
+[
+ frac tm<2^{33}cdot frac1{2^{33}}=1,
+]
which is weaker than needed; more precisely,
-\[
- 100 \cdot \frac tm < 40100(\tfrac 57)^{100}
- <\frac{40100}{2^{33}}<1,
-\]
+[
+ 100 cdot frac tm < 40100(tfrac 57)^{100}
+ <frac{40100}{2^{33}}<1,
+]
so $t/m<1/100$.
@@ -529,34 +529,34 @@ 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
+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 functions $a:P\to\mathbb Z/M\mathbb Z$, with
-addition defined pointwise: $(a+a')(x)=a(x)+a'(x)$ for each $x\in P$.
+[
+ A_0=(mathbb Z/MbbZ/Mmathbb Z)^PbbZ)^P
+]
+be the group of functions $a:Ptomathbb Z/MbbZ/Mmathbb Z$,bbZ$, with
+addition defined pointwise: $(a+a')(x)=a(x)+a'(x)$ for each $xin P$.
This is a finite abelian group of order $M^{p^3}$.
-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
-\[
- (s\cdot a)(x)=a(x-s), \qquad x\in P.
-\]
-This is a group action: $0\cdot a=a$ and $(s+t)\cdot a=s\cdot(t\cdot a)$.
+The group $P$ acts on $A_0$ by translation: for $sin P$ and
+$ain A_0$, define $scdot ain A_0$ by
+[
+ (scdot a)(x)=a(x-s), qquad xin P.
+]
+This is a group action: $0cdot a=a$ and $(s+t)cdot a=scdot(tcdot a)$.
Form the semidirect product
-\[
- H=A_0\rtimes P, \qquad
- (a,s)(a',s')=(a+s\cdot a',\ s+s').
-\]
+[
+ H=A_0rtimes P, qquad
+ (a,s)(a',s')=(a+scdot 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: $(a,s)(-(-s)\cdot a, -s)=(a-s\cdot((-s)\cdot a), 0)=(a-s\cdot(-s\cdot a), 0)=(a-0,0)=(0,0)$.
+$(-s^{-1}cdot a, -s)$, i.e., $(-(-s)cdot a,-s)$ since in characteristic
+$p$ subtraction is the same as~$+(-s)$. Concretely: $(a,s)(-(-s)cdot a, -s)=(a-scdot((-s)cdot a), 0)=(a-scdot(-scdot a), 0)=(a-0,0)=(0,0)$.
The order of $H$ is
-\[
- w:=|H|=|A_0|\cdot|P|=M^{p^3}\cdot p^3.
-\]
+[
+ 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 $K^w$).
@@ -566,62 +566,62 @@ below. Both are isomorphic to $P$, but they sit inside $H$ in different
ways.
Translation subgroup. Define
-\[
- u(s)=(0,s),\qquad U=\{u(s):s\in P\}⊆ H.
-\]
-Because $(0,s)(0,s')=(0,s+s')$, the map $s\mapsto u(s)$ is an
-isomorphism $P\to U$.
-
-Characteristic element. Let $c\in A_0$ be the function
-\[
- c(0)=1,\quad c(x)=0 for x\ne0.
-\]
+[
+ u(s)=(0,s),qquad U={u(s):sin P}⊆ H.
+]
+Because $(0,s)(0,s')=(0,s+s')$, the map $smapsto u(s)$ is an
+isomorphism $Pto U$.
+
+Characteristic element. Let $cin A_0$ be the function
+[
+ c(0)=1,quad c(x)=0 for xne0.
+]
The group law gives $(c,0)$ the exact order $M$ in $H$ (as an element
of the abelian normal subgroup $A_0$).
Conjugate subgroup. Define
-\[
- v(s)=c\,u(s)\,c^{-1},\qquad V=\{v(s):s\in P\}⊆ H,
-\]
-where $c$ in the formula denotes $(c,0)\in H$ and $c^{-1}=(-c,0)$
+[
+ v(s)=c,u(s),c^{-1},qquad V={v(s):sin 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
- v(s)=(c-s\cdot c,\ s),\qquad s\in P.
+ v(s)=(c-scdot c, s),qquad sin P.
-Indeed $(c,0)(0,s)=(c,s)$ and $(c,s)(-c,0)=(c+s\cdot(-c),\,s)=(c-s\cdot c,s)$.
+Indeed $(c,0)(0,s)=(c,s)$ and $(c,s)(-c,0)=(c+scdot(-c),,s)=(c-scdot c,s)$.
-The subgroups $U$ and $V$ satisfy $U\cap V=\{1\}$, and $U\cong V\cong P$.
+The subgroups $U$ and $V$ satisfy $Ucap V={1}$, and $Ucong Vcong P$.
Proof.
-Suppose $(0,0)=u(s)=v(s')$ for some $s,s'\in P$. Comparing second
+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
+$0=c-scdot c$, i.e., $c=scdot c$. The left side evaluates at $0$ to
+$c(0)=1$; the right side evaluates at $0$ to $(scdot c)(0)=c(-s)$.
+But $-sne 0$ when $sne0$, 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~$H$,
+$smapsto u(s)$ is a bijection $Pto U$ preserving the group law, and
+$smapsto v(s)$ is a conjugate of~$smapsto u(s)$ inside the group~$H$,
hence also isomorphic to $P$.
Why does this matter? In Section~sec:modular we shall build an
idempotent~$e$ with
-\[
- (u(s)-1)^\ell e = 0,\qquad e\,(v(s)-1)^\ell=0
- \quadfor every s\in P,
-\]
-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 $U\cap V=\{1\}$,
+[
+ (u(s)-1)^ell e = 0,qquad e,(v(s)-1)^ell=0
+ quadfor every sin 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 $Ucap V={1}$,
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
+$m=binom{1200}{600}$, $t=|mathcal T|$,calT|$, $w=|H|$, and
+$d=dim_KeK[H]$ for the regular rank of the idempotent $e$ we
are about to build in Section~sec:modular.
# The large idempotent with ordered annihilations
@@ -629,35 +629,35 @@ are about to build in Section~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:
+$K$ of order~$p^4$, and a finite group $H=A_0rtimes P$ with two
+different subgroups $U,Vcong P$ meeting trivially. Now we use all that
+data to build one particular idempotent $ein K[H]$ with two properties:
-\item Large regular rank. The dimension $d=\dim_K(eK[H])$
+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 Opposite annihilations. Writing $\ell=(p-1)/2$, we have
- $(u(s)-1)^\ell e=0$ and $e\,(v(s)-1)^\ell=0$ for every $s\in P$.
+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 $sin P$.
The construction proceeds in three stages.
-\item \Ssubsec:B-def: identify a subspace $B⊆ K^P$ 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 \Ssubsec:admissibility: use the polynomial identity test of
+item Ssubsec:admissibility: use the polynomial identity test of
Schwartz and Zippel to produce many ``admissible'' characters
- $\lambda\inA_0hat$, i.e., characters $\lambda$ for which
- a certain Gram matrix~$N_\lambda$ is nonsingular.
-\item \Ssubsec:idempotent: assemble these ingredients into an
+ $lambdainA_0hat$, i.e., characters $lambda$ for which
+ a certain Gram matrix~$N_lambda$ is nonsingular.
+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.
-A last subsection (\Ssubsec: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
@@ -667,52 +667,52 @@ $A_0⊆ H$.
## The annihilated subspace $B⊆ K^P$
-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
+Let $ell=(p-1)/2$ and $D=3(p-1)$. Work in the $K$-vector space $K^P$
+of functions $Pto K$, with standard basis $mathbf e_x$bfe_x$ (a $1$ at
$x$ and $0$ elsewhere). We want a linear subspace on which left shifts
-and right shifts both annihilate $\ell$-fold.
+and right shifts both annihilate $ell$-fold.
-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
-\[
- X_j=L_{ε_j}-1_{K^P}, \qquad j=1,2,3,
-\]
-where $ε_j$ are the standard generators of $P=\F_p^3$.
+Translation calculus. For $sin P$ let $L_s:K^Pto K^P$ be the
+shift operator $L_smathbf e_x=bfe_x=mathbf e_bfe_{x+s}$. Define three operators
+[
+ X_j=L_{ε_j}-1_{K^P}, qquad j=1,2,3,
+]
+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 $1\le k\le p-1$;
+binomial coefficients $binom pk$ are divisible by $p$ for $1le kle p-1$;
so $(1+X_j)^p=1+X_j^p$ and
-\[
+[
X_j^p=(L_{ε_j}-1)^p = L_{pε_j}-1 = L_0-1 = 0.
-\]
+]
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 $P$.
[Monomial basis]
The $p^3$ vectors
-\[
- X_1^{n_1}X_2^{n_2}X_3^{n_3}\mathbf e_0, \qquad 0\le n_1,n_2,n_3< p,
-\]
+[
+ X_1^{n_1}X_2^{n_2}X_3^{n_3}mathbf e_0,bfe_0, qquad 0le n_1,n_2,n_3< p,
+]
form a $K$-basis of $K^P$.
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
+The standard basis ${mathbf e_x:xbfe_x:xin P}$ already has cardinality
+$p^3=dim_KK^P$. So it suffices to show every $mathbf e_s$bfe_s$ is in the
$K$-span of the displayed vectors.
-Fix $s=(s_1,s_2,s_3)\in P$. Since $1+X_j=L_{ε_j}$, we have
+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
-\[
- \mathbf e_s=L_s\mathbf e_0
- =\prod_{j=1}^3 L_{s_jε_j}\mathbf e_0
- =\prod_{j=1}^3 (1+X_j)^{s_j}\mathbf e_0
- =\prod_{j=1}^3 (\sum_{k=0}^{s_j}\binom{s_j}{k}X_j^k)\mathbf e_0.
-\]
+[
+ mathbf e_s=L_sbfe_s=L_smathbf e_0bfe_0
+ =prod_{j=1}^3 L_{s_jε_j}mathbf e_0bfe_0
+ =prod_{j=1}^3 (1+X_j)^{s_j}mathbf e_0bfe_0
+ =prod_{j=1}^3 (sum_{k=0}^{s_j}binom{s_j}{k}X_j^k)mathbf e_0.bfe_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 $\{\mathbf e_s\}$
+$X_1^{n_1}X_2^{n_2}X_3^{n_3}mathbf e_0$bfe_0$ with $0le n_jle s_j<p$.
+So $mathbf e_s$bfe_s$ lies in the span of the display. Since ${mathbf e_sbfe_s}$
spans, the display spans, and it has $p^3$ elements, so it is a basis.
@@ -726,88 +726,88 @@ truncated by $X_j^p=0$.
Define $B$ as the matrix with columns the basis vectors whose total
degree is large enough:
- 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}}.
+ B := [X_1^{n_1}X_2^{n_2}X_3^{n_3}mathbf e_0]_bfe_0]_{
+ substack{0le n_1,n_2,n_3<p n_1+n_2+n_3ge D-ell+1}}.
-So $B$ is a $p^3\times h$ matrix with $K$-linearly independent columns
+So $B$ is a $p^3times h$ matrix with $K$-linearly independent columns
(they are a subset of a basis). The integer $h$ counts the columns.
[Two-sided annihilation of $B$]
-For every $s\in P$,
-\[
- (L_s-1)^\ell\,B=0,\qquad B^{\mathsf T}(L_s-1)^\ell=0.
-\]
+For every $sin P$,
+[
+ (L_s-1)^ell,B=0,qquad B^{mathsf T}sfT(L_s-1)^ell=0.
+]
Proof.
-First identity. Fix $s\in P$ and write $s=\sum_j s_jε_j$.
-Expand $L_s-\id_{K^P}$:
-\[
- L_s-1 = \prod_{j=1}^3 L_{s_jε_j} - 1
- = \prod_{j=1}^3 (1+X_j)^{s_j} - 1.
-\]
+First identity. Fix $sin P$ and write $s=sum_j s_jε_j$.
+Expand $L_s-id_{K^P}$:
+[
+ 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
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
+$ell$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 $d+\ell$.
+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
+Now take a column $X_1^{n_1}X_2^{n_2}X_3^{n_3}mathbf e_0$bfe_0$ of $B$,
+with $d=n_1+n_2+n_3ge D-ell+1$. 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.
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
-\[
- B^{\mathsf T}(L_s-1)^\ell = ((L_{-s}-1)^\ell B)^{\mathsf T} = 0^{\mathsf T}=0,
-\]
+$(L_s)_{mathbf e_x,bfe_x,mathbf e_y}bfe_y$ (row $x$, column $y$) is $1$ when
+$x+s=y$ and $0$ otherwise. Hence $(L_s^{mathsf T}sfT)_{mathbf e_y,bfe_y,mathbf e_x}bfe_x$
+is $1$ when $x-s=y$, so $L_s^{mathsf T}sfT=L_{-s}$. Consequently
+$(L_s-1)^{mathsf T}sfT=L_{-s}-1$, and
+[
+ B^{mathsf T}sfT(L_s-1)^ell = ((L_{-s}-1)^ell B)^{mathsf T}sfT = 0^{mathsf T}sfT=0,
+]
using the first identity with $-s$ in place of $s$.
[Number of columns]
-The integer $h$ equals $\binom{\ell+2}{3}$, and
-\[
- (h)/(p^3)=((p-1)(p+1)(p+3))/(48\,p^3)>\frac1{48}.
-\]
+The integer $h$ equals $binom{ell+2}{3}$, and
+[
+ (h)/(p^3)=((p-1)(p+1)(p+3))/(48,p^3)>frac1{48}.
+]
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
+Make the substitution $a_j=p-1-n_j$, $j=1,2,3$. Then $a_jge0$ and
+$sum a_j=D-sum n_j$. The condition $sum n_jge D-ell+1$ becomes
+$sum a_jle ell-1$. Each $a_jle p-1$ 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 $(a_1,a_2,a_3)inmathbb N^3$bbN^3$ with
+$sum a_jleell-1$. Stars and bars (distribute $leell-1$ identical
balls into $3$ labelled bins) gives
-\[
- h=\binom{(\ell-1)+3}{3}=\binom{\ell+2}{3}.
-\]
-Now $\ell=(p-1)/2$, so
-\[
- \binom{\ell+2}{3} = ((\ell+2)(\ell+1)\ell)/(6)
+[
+ h=binom{(ell-1)+3}{3}=binom{ell+2}{3}.
+]
+Now $ell=(p-1)/2$, so
+[
+ binom{ell+2}{3} = ((ell+2)(ell+1)ell)/(6)
= ((p+3)(p+1)(p-1))/(48).
-\]
-Divide by $p^3$: $h/p^3=(p+3)(p^2-1)/(48\,p^3)$. To compare with
+]
+Divide by $p^3$: $h/p^3=(p+3)(p^2-1)/(48,p^3)$. To compare with
$1/48$, compute the difference
-\[
- (p+3)(p^2-1)-p^3=3p^2-p-3>0 \quadfor p\ge3,
-\]
+[
+ (p+3)(p^2-1)-p^3=3p^2-p-3>0 quadfor pge3,
+]
which gives $h/p^3>1/48$.
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
+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
+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.
## Choosing a compatible projection
@@ -817,52 +817,52 @@ 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~lem:B-annihilation once $e$ is a projection onto
-$\im B$. But $e$ must live inside $K[H]$, and there the two shifts
+$im B$. 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.
Characters of $A_0$.:
-$A_0=(\mathbb Z/M\mathbb Z)^P$ is finite abelian of order
+$A_0=(mathbb Z/MbbZ/Mmathbb Z)^P$bbZ)^P$ is finite abelian of order
$|A_0|=M^{p^3}$. 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
-$a_x\in\{0,...,M-1\}$. Because $\zeta$ has order $M$ and is an
+$lambda:A_0to K^*$. It is uniquely determined on the coordinate
+generators $c_x$ (the function with value~$1$ at $xin P$ and $0$
+elsewhere): $lambda(c_x)=zeta^{a_x}$ for some
+$a_xin{0,...,M-1}$. Because $zeta$ has order $M$ and is an
element of~$K$, every character has values in~$K$.
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
-\[
- u(s)\,p_\chi^{A_0}\,u(s)^{-1}=p_{s\chi}^{A_0}.
-\]
+$(slambda)(a)=lambda((-s)cdot a)$. 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_{schi}^{A_0}.
+]
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
+For $lambdainA_0hat$ abbreviate $chi_x=xlambda$ (the
+translate of $lambda$ by $x$) and define the diagonal
+$p^3times p^3$ matrix
- D_\lambda=\diag\!(\chi_x(c):x\in P).
+ D_lambda=diag!(chi_x(c):xin 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.
+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
- N_\lambda=B^{\mathsf T}\,D_\lambda^{-1}\,B
+ N_lambda=B^{mathsf T}sfT,D_lambda^{-1},B
-is a symmetric $h\times h$ matrix over~$K$.
+is a symmetric $htimes h$ matrix over~$K$.
-Call $\lambda$ admissible if
+Call $lambda$ admissible if
-\item its $P$-stabilizer is trivial, i.e., the orbit $P\lambda$ has
+item its $P$-stabilizer is trivial, i.e., the orbit $Plambda$ has
size exactly $p^3$; and
-\item $N_\lambda$ is invertible.
+item $N_lambda$ is invertible.
Both conditions are checkable in finite time, and the check is
deterministic: enumerate the $p^3$ translates and compute one
-$h\times h$ determinant.
+$htimes h$ determinant.
Most characters are admissible.:
The two modes of failure each account for less than half of the
@@ -873,28 +873,28 @@ succeed. This count is the key estimate of the whole construction.
More than half of the $|A_0|$ characters are admissible.
Proof.
-Stabilizers. Fix a nonzero $s\in P$. Translation by $s$
+Stabilizers. Fix a nonzero $sin P$. 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
+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
fixed. The union over the $p^3-1$ nonzero $s$ therefore has size at
most $(p^3-1)M^{p^2}$, i.e. a fraction at most
-\[
- (p^3-1)\,M^{-p^2(p-1)}
- \le (p^3-1)/M < 1/4 .
-\]
-The last inequality uses $M=p^4-1>4p^3$, which holds for $p\ge5$;
+[
+ (p^3-1),M^{-p^2(p-1)}
+ le (p^3-1)/M < 1/4 .
+]
+The last inequality uses $M=p^4-1>4p^3$, which holds for $pge5$;
we have $p>600$.
-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
+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
- 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 ,
+ Q(tau):=det!(B^{mathsf T}sfTdiag(tau_x)_{xin P}B)
+ =sum_{substack{J⊆ P|J|=h}}det(B_J)^2prod_{xin J}tau_x ,
-where $B_J$ is the $h\times h$ submatrix of $B$ consisting of the
+where $B_J$ is the $htimes h$ 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
@@ -909,12 +909,12 @@ 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
+the fraction $h/M$ of characters. Since $hle p^3$ (it counts a
subset of $K^P$) and $M=p^4-1$, this is at most
-\[
- (h)/(M)\le(p^3)/(p^4-1)<\frac14 ,
-\]
-because $p^4-1>4p^3$ for $p\ge5$.
+[
+ (h)/(M)le(p^3)/(p^4-1)<frac14 ,
+]
+because $p^4-1>4p^3$ for $pge5$.
Combining: fewer than $1/4+1/4=1/2$ of characters fail either
condition. So more than half admit a trivial stabilizer and an
@@ -924,169 +924,169 @@ invertible Gram form simultaneously.
## 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
+Each admissible character $lambda$ gives a projection $T_lambda$ on
+the $p^3$-dimensional block attached to~$lambda$. The task is to
write that projection as an element of $K[H]$ whose regular rank we
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 $\lambda$.
+$lambda$, the products $p_{xlambda}u(s)$ (for $x,sin P$) form a
+$K$-basis of one matrix block of $K[H]$. Then we sum over $lambda$.
One character, one block.:
-Fix $\lambda\inA_0hat$ with trivial $P$-stabilizer. For
-$x,y\in P$ set
+Fix $lambdainA_0hat$ with trivial $P$-stabilizer. For
+$x,yin P$ set
- E_{xy}^{\lambda}=p_{x\lambda}^{A_0}\,u(x-y).
+ E_{xy}^{lambda}=p_{xlambda}^{A_0},u(x-y).
-Because $u(s)\,p_{x\lambda}u(s)^{-1}=p_{(x+s)\lambda}$, the row
+Because $u(s),p_{xlambda}u(s)^{-1}=p_{(x+s)lambda}$, the row
indices line up with the shift. Fourier orthogonality gives
- E_{xy}^{\lambda}E_{zt}^{\lambda}=\delta_{yz}\,E_{xt}^{\lambda}.
+ E_{xy}^{lambda}E_{zt}^{lambda}=delta_{yz},E_{xt}^{lambda}.
-Indeed, the Fourier product survives when $(x-y+z)\lambda=x\lambda$,
-i.e. $(y-z)\lambda=0$, which by the trivial stabilizer forces $y=z$.
+Indeed, the Fourier product survives when $(x-y+z)lambda=xlambda$,
+i.e. $(y-z)lambda=0$, which by the trivial stabilizer forces $y=z$.
-Put $q_\lambda=\sum_{x\in P}p_{x\lambda}^{A_0}$. Because $q_\lambda$
-commutes with $A_0$ and $P$ permutes its summands, $q_\lambda$ is a
-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 $c\mapsto D_\lambda$.
+Put $q_lambda=sum_{xin P}p_{xlambda}^{A_0}$. Because $q_lambda$
+commutes with $A_0$ and $P$ permutes its summands, $q_lambda$ is a
+central idempotent. The basis ${p_{xlambda}u(s)}_{x,sin P}$
+identifies the ideal $q_lambda K[H]$ with the matrix algebra
+$Mat_{p^3}(K)$, via $u(s)mapsto L_s$ and $cmapsto D_lambda$.
Explicit projection.:
Define
- T_\lambda=B\,N_\lambda^{-1}B^{\mathsf T}D_\lambda^{-1}
- \in\Mat_{p^3}(K).
+ T_lambda=B,N_lambda^{-1}B^{mathsf T}sfTD_lambda^{-1}
+ inMat_{p^3}(K).
-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
-$\im B⊆\im T_\lambda$; and it factors through $B^{\mathsf T}$,
-so $\im T_\lambda⊆\im B$. Hence $\im T_\lambda=\im B$ and
-$rankT_\lambda=h$.
+Because $N_lambda=B^{mathsf T}sfTD_lambda^{-1}B$, 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
+$im B⊆im T_lambda$; and it factors through $B^{mathsf T}sfT$,
+so $im T_lambda⊆im B$. Hence $im T_lambda=im B$ and
+$rankT_lambda=h$.
-Define an element of the $\lambda$-block of $K[H]$:
+Define an element of the $lambda$-block of $K[H]$:
- e_\lambda=\sum_{x,y\in P}(T_\lambda)_{xy}\,E_{xy}^{\lambda}.
+ e_lambda=sum_{x,yin P}(T_lambda)_{xy},E_{xy}^{lambda}.
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 $T_\lambda$.
+$e_lambda^2=e_lambda$. The action of $H$ on this block is the
+matrix action of $T_lambda$.
The two annihilations.:
Left multiplication by $u(s)$ in this block is $L_s$. Therefore
-\[
- (L_s-1)^\ell\,T_\lambda
- =(L_s-1)^\ell B\;N_\lambda^{-1}B^{\mathsf T}D_\lambda^{-1}
+[
+ (L_s-1)^ell,T_lambda
+ =(L_s-1)^ell B;N_lambda^{-1}B^{mathsf T}sfTD_lambda^{-1}
=0,
-\]
-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
-\[
- D_\lambda L_s D_\lambda^{-1}-1=D_\lambda(L_s-1)D_\lambda^{-1}.
-\]
-Combining with the transpose identity $B^{\mathsf T}(L_s-1)^\ell=0$:
-\[
- B^{\mathsf T}D_\lambda^{-1}(D_\lambda L_s D_\lambda^{-1}-1)^\ell
- =B^{\mathsf T}(L_s-1)^\ell D_\lambda^{-1}=0.
-\]
-Hence $(u(s)-1)^\ell e_\lambda=0$ and
-$e_\lambda(v(s)-1)^\ell=0$ in $K[H]$.
+]
+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}.
+]
+Combining with the transpose identity $B^{mathsf T}sfT(L_s-1)^ell=0$:
+[
+ B^{mathsf T}sfTD_lambda^{-1}(D_lambda L_s D_lambda^{-1}-1)^ell
+ =B^{mathsf T}sfT(L_s-1)^ell D_lambda^{-1}=0.
+]
+Hence $(u(s)-1)^ell e_lambda=0$ and
+$e_lambda(v(s)-1)^ell=0$ in $K[H]$.
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
+Blocks for distinct orbits are orthogonal: if $Plambdane Plambda'$
+then $e_lambda e_{lambda'}=0$. Choose the least admissible character
+in each $P$-orbit that contains one, and let $mathscr R$scrR$ be the set
of chosen characters. Define
- e=\sum_{\lambda\in\mathscr R}e_\lambda \in K[H].
+ e=sum_{lambdainmathscr R}scrRe_lambda in K[H].
By orthogonality the sum is an idempotent: $e^2=e$. The two
annihilations persist on~$e$.
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 $p^3h$.
+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:
-\[
- d:=\dim_K(eK[H])=p^3\,h\,|\mathscr R|.
-\]
+[
+ d:=dim_K(eK[H])=p^3,h,|mathscr R|.scrR|.
+]
The admissible characters form more than half of $A_0hat$
(Lemma~lem:good-count), and each $P$-orbit has size $p^3$, so
-\[
- |\mathscr R|>(|A_0|)/(2p^3).
-\]
-Recall $w=|H|=|A_0|\,p^3$. Therefore
-\[
- \frac dw=(p^3h|\mathscr R|)/(|A_0|p^3)
- =(h|\mathscr R|)/(|A_0|)
- >(h)/(2p^3)>\frac1{96},
-\]
+[
+ |mathscr R|>(|A_0|)/(2p^3).scrR|>(|A_0|)/(2p^3).
+]
+Recall $w=|H|=|A_0|,p^3$. Therefore
+[
+ frac dw=(p^3h|mathscr R|)/(|A_0|p^3)scrR|)/(|A_0|p^3)
+ =(h|mathscr R|)/(|A_0|)scrR|)/(|A_0|)
+ >(h)/(2p^3)>frac1{96},
+]
using $h/p^3>1/48$ from Lemma~lem:h.
-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 $s\in P$
-\[
- (u(s)-1)^\ell\,e=0,\qquad
- e\,(v(s)-1)^\ell=0 .
-\]
+There is a specified idempotent $ein K[H]$ and positive integers
+$d,w$ with $w=|H|$, $d=dim_KeK[H]$, $d/w>1/96$, satisfying for
+every $sin P$
+[
+ (u(s)-1)^ell,e=0,qquad
+ e,(v(s)-1)^ell=0 .
+]
Its coefficients are obtained by finite operations in the specified
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
+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.
+$ell=(p-1)/2$ is exactly the right size.
## Ordered annihilations at edge level
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~sec:gluing that $H_i\to G_0$ is injective; here we
+$Gamma$. For each edge $i<j$ identify the cyclic subgroup generated
+by $v(1,barbar j,j,barbar j^2)$j^2)$ in $H_i$ with the one generated by
+$u(1,barbar i,i,barbar i^2)$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~sec:gluing that $H_ito G_0$ 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 $H_i\to G_0$.
+Denote by $e_iin K[G_0]$ the image of the idempotent $e$ under the
+map $H_ito G_0$.
-For every edge $i<j$ of $\Gamma$, $e_ie_j=0$ in $K[G_0]$.
+For every edge $i<j$ of $Gamma$, $e_ie_j=0$ in $K[G_0]$.
Proof.
-Let $y\in K[H_i]⊆ K[G_0]$ denote the common generator of
+Let $yin K[H_i]⊆ K[G_0]$ 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
+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 $K[< y>]$-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 $q_a\in K[< y>]$.
+$e_i=sum_a a,q_a$ with $q_ain K[< y>]$.
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
-$\ell+j\ge p$.
-
-Thus $q_aX^\ell=0$ forces $q_a=q_a'X^{p-\ell}$, and
-\[
- e_i=q\,X^{p-\ell}\quadfor\quad q=\sum_a a\,q_a'\in K[H_i].
-\]
-
-Because $p-\ell=\ell+1$,
-\[
- e_ie_j=q\,X^{\ell+1}e_j=q\,X\,(X^\ell e_j)=0.
-\]
+annihilator of $X^ell$ is precisely the principal ideal
+$(X^{p-ell})$, since $X^{ell+j}=0$ in $K[X]/(X^p)$ iff
+$ell+jge p$.
+
+Thus $q_aX^ell=0$ forces $q_a=q_a'X^{p-ell}$, and
+[
+ e_i=q,X^{p-ell}quadforquad q=sum_a a,q_a'in K[H_i].
+]
+
+Because $p-ell=ell+1$,
+[
+ e_ie_j=q,X^{ell+1}e_j=q,X,(X^ell e_j)=0.
+]
No factor is commuted past $X$, no symmetry is used, and no reverse
product $e_je_i$ is asserted.
@@ -1117,7 +1117,7 @@ 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
choice of starting point and orientation, the old walk is
-$W_1\,s\,W_2\,s^{-1}$, where $W_1,W_2$ are the two split walks.
+$W_1,s,W_2,s^{-1}$, where $W_1,W_2$ are the two split walks.
If some $W_a$ has nonidentity value, keep the component containing it,
with the same face type as before. This component is connected and has
@@ -1134,22 +1134,22 @@ results.
So every edge is a nonbridge.
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
+Suppose an edge separates two faces $Phi,Phi'$ of the same type $iin
+V(Delta)$. 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
- DE=s^{-1}(sD)s\,(s^{-1}E).
+ DE=s^{-1}(sD)s,(s^{-1}E).
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:
+$sD$ evaluates to $gne1$, then its conjugate $s^{-1}gs$ also has value
+$ne1$. Conjugation preserves identity-versus-nonidentity. Hence:
-\item ordinary + ordinary $\to$ ordinary;
-\item ordinary + exceptional (in either order) $\to$ exceptional.
+item ordinary + ordinary $to$ ordinary;
+item ordinary + exceptional (in either order) $to$ exceptional.
We are assuming the map has exactly one exceptional face, so
@@ -1157,8 +1157,8 @@ 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.
-Therefore the two face types across an edge labelled $\gamma$ are
-exactly the two endpoints of~$\gamma$.
+Therefore the two face types across an edge labelled $gamma$ are
+exactly the two endpoints of~$gamma$.
Reduction 4: at most one loop, bounding the exceptional monogon.:
Suppose there is a loop. Its two sides are faces (possibly with other
@@ -1181,11 +1181,11 @@ Reduced configuration.:
Combining the four reductions, a minimal exceptional map satisfies:
-\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.
+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.
We continue in the next reduction. The one remaining configuration to
@@ -1194,39 +1194,39 @@ same label. Doing so will give the final contradiction with Euler's
formula.
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
-$ε,\varphi\in C_\gamma∖\{1\}$. The edges $e$ and $f$
+Suppose a face $Phi$ has consecutive nonloop darts $e:ato b$ and
+$f:bto d$ with a common label $gammain E(Delta)$ and values
+$ε,varphiin C_gamma∖{1}$. 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.
+$Phi$, making it a bridge, contradicted by Reduction 2.
Let $X$ be the face on the opposite side of $e$, $Y$ the face on the
-opposite side of $f$. By Reduction 3, both have type $j\in V(\Delta)$,
-where $j$ is the other endpoint of $\gamma$ from $\Phi$'s type~$i$.
+opposite side of $f$. By Reduction 3, both have type $jin V(Delta)$,
+where $j$ is the other endpoint of $gamma$ from $Phi$'s type~$i$.
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
-$ε\varphi$ and label $\gamma$ (or make it neutral if this
-equals $1$; Reduction 1 then applies if needed). The walk of $\Phi$
+Remove $e,f$. Insert a new edge $g:ato d$ along their arcs, smoothed
+through the empty corner of $Phi$ at~$b$. Give $g$ value
+$εvarphi$ 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.
-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
+Case $Xne Y$. Write $X=barbar ee D$, $Y=barbar ff E$, with $D:ato b$
+and $E:bto d$. The merged face has walk $barbar gDE$.gDE$. Its evaluation is
- (ε\varphi)^{-1}DE
- =\varphi^{-1}(ε^{-1}D)\,\varphi\,(\varphi^{-1}E).
+ (εvarphi)^{-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 $\varphi^{-1}p\varphi\cdot q$:
+Let $p=ε^{-1}D$ (old $X$'s value) and $q=varphi^{-1}E$ (old
+$Y$'s value). The merged value is $varphi^{-1}pvarphicdot q$:
-\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
+item $p=q=1$: merged value $varphi^{-1}cdot 1cdotvarphicdot 1=1$;
+item $pne1$, $q=1$: merged value $varphi^{-1}pvarphine1$ (conjugate
of nonidentity);
-\item $p=1$, $q\ne1$: merged value $q\ne1$;
-\item $p\ne1$, $q\ne1$: excluded (at most one exceptional face).
+item $p=1$, $qne1$: merged value $qne1$;
+item $pne1$, $qne1$: excluded (at most one exceptional face).
So the merged face is ordinary when both old faces were ordinary, and
@@ -1235,13 +1235,13 @@ contradicting minimality.
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
-$E:b\to b$.
+$barbar e$:e$: the walk is $barbar eDeDbarbar fE$,fE$, with $D:ato d$ and
+$E:bto b$.
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 $v$, $e$, $F$:
+$barbar gD$gD$ and $E$. The rules are easiest to check by slots: at $a$
+and $d$ the new darts replace the old slots of $e,barbar f$;f$; at $b$ the
+consecutive slots $barbar e,f$e,f$ are deleted. Counting $v$, $e$, $F$:
$v$ unchanged, $e$ decreased by one, $F$ increased by one. Euler's
formula gives $2c_{new}=v-(e-1)+(F+1)=2+2=4$, so the new
drawing has exactly two components.
@@ -1251,22 +1251,22 @@ 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 $A=(ε\varphi)^{-1}D$ be the value of walk $\bar gD$, and
+Let $A=(εvarphi)^{-1}D$ be the value of walk $barbar gD$,gD$, and
$B=E$ the value of walk $E$. The old walk of $X$ evaluated to
-$ε^{-1}D\varphi^{-1}E$, whose cyclic shift starting at $\bar f$
-is $(\varphi^{-1}B\varphi)A$:
+$ε^{-1}Dvarphi^{-1}E$, whose cyclic shift starting at $barbar f$f$
+is $(varphi^{-1}Bvarphi)A$:
-\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
+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
the exceptional face is elsewhere. Keep the component
containing it.
-\item $X$ ordinary with $A\ne1$ and $B\ne1$: both new walks are
+item $X$ ordinary with $Ane1$ and $Bne1$: both new walks are
exceptional. The exceptional case sits either in the
- $\bar gD$-component (if that also contains the original
+ $barbar gD$-componentgD$-component (if that also contains the original
exception), or in the $E$-component. Keep the component
- without the face $\Phi$, whose unique exception is its
+ without the face $Phi$, whose unique exception is its
sole nonidentity boundary.
@@ -1286,99 +1286,99 @@ 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
$i_1,i_2,...,i_r$, where $i_a$ is the type of the $a$-th face,
-and $r=\deg u$. Successive $i_a$ differ: the edge between them has
+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 $i_1,i_2,...,i_r,i_1$ in $\Delta$ with
-successive edges distinct and length $r\ge1$. Since $\Delta$ is
+So we get a closed walk $i_1,i_2,...,i_r,i_1$ in $Delta$ with
+successive edges distinct and length $rge1$. 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 $\deg u\ge4$.
+length $3$ would be a triangle. Thus $deg uge4$.
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 $|∂ F|\ge4$.
+walk in $Delta$ has length at least $4$. Thus $|∂ F|ge4$.
Euler deficit.:
For a connected spherical drawing,
- \sum_{u}(4-\deg u)+\sum_{F}(4-|∂ F|)=8,
+ sum_{u}(4-deg u)+sum_{F}(4-|∂ F|)=8,
-which follows from Euler's formula and $\sum_u\deg u=\sum_F|∂
+which follows from Euler's formula and $sum_udeg u=sum_F|∂
F|=2e$ (every edge contributes two to each side).
-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 $|∂ F|\ge1$ and the
-contribution is at most $3$. Total $\le3<8$, contradicting
+No loop present. Every vertex satisfies $deg uge4$, so
+contributes $le0$. Every ordinary face has $|∂ F|ge4$, so
+contributes $le0$. Only the exceptional face can contribute
+positively: its walk is nonempty, so $|∂ F|ge1$ and the
+contribution is at most $3$. Total $le3<8$, contradicting
(eq:gluing-deficit).
-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
+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
-value $\ne1$ (by Reduction 1), so its walk is nonempty and cannot be
+value $ne1$ (by Reduction 1), so its walk is nonempty and cannot be
a monogon (a monogon would have just the loop's dart, giving value
-$\ne1$, contradicting ordinari-ness); hence its length is at least
+$ne1$, 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
+contribute at most zero. Total $le 2+3+2=7<8$, contradicting
(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~$i$.
+Hence no minimal exceptional spherical map exists. So $Lto D_i^0$
+is injective, i.e., $D_i^0to Q$ is injective, for every vertex~$i$.
## Extending to the full vertex group
-We now extend $D_i^0\to Q$ to $D_i\to Q$. The argument is a
+We now extend $D_i^0to Q$ to $D_ito Q$. The argument is a
permutation-action construction.
-Fix $i$. Let $Ω=L\timesℕ$, with $L$ acting by left
+Fix $i$. Let $Ω=Ltimesℕ$, with $L$ acting by left
multiplication on the first coordinate. The action of $D_i^0$ 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 $D_i\timesℕ$.
+$D_i^0$-action on $D_itimesℕ$.
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 $\phi_i:D_i\timesℕ\toΩ$.
-Transport the $D_i$ left-multiplication along $\phi_i$: this gives a
+$D_i^0$-equivariant bijection $phi_i:D_itimesℕtoΩ$.
+Transport the $D_i$ left-multiplication along $phi_i$: this gives a
$D_i$-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 $Ω$ by the
+Do this independently for each $i$. On a shared edge $gamma={i,j}$,
+both copies of the generator $x_gammain L$ act on $Ω$ by the
same element, so the resulting actions glue to a $Q$-action on
$Ω$. 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.
+the original left action on $D_itimesℕ$ is faithful and
+$phi_i$ is a bijection.
-Therefore $D_i\to Q$ is injective for every vertex $i$. This
+Therefore $D_ito Q$ is injective for every vertex $i$. This
completes the proof of Lemma~lem:triangle-free-gluing.
## Applying the criterion to $H_i$
-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~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 $p$.
+We now apply the lemma to our concrete setting. Recall $H=A_0rtimes
+P$ with $A_0=(mathbb Z/MbbZ/Mmathbb Z)^P$,bbZ)^P$, $P=F_p^3$, $U=u(P)$,
+$V=cUc^{-1}$, and $Ucap V={1}$ from Proposition~prop:UV-triv.
+For each edge $i<j$ of $Gamma$ we identify the cyclic subgroup
+generated by $v(1,barbar j,j,barbar j^2)$j^2)$ in $H_i$ with the one generated
+by $u(1,barbar i,i,barbar i^2)$i^2)$ in $H_j$. Here the bars denote residues in
+$F_p$, and both elements have exact order $p$.
-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.
+Every natural homomorphism $H_ito G_0$ is injective. Consequently
+the induced maps $K[H_i]to K[G_0]$ on group algebras are injective.
Proof.
-Apply Lemma~lem:triangle-free-gluing with $\Delta=\Gamma$ and
+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.
@@ -1386,30 +1386,30 @@ vectors. We must verify the short-word condition.
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~$U$ 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
+item $j<i$: the syllable from edge ${i,j}$ in $H_i$ lies in~$U$ and
+ is $u(lambdacdot(1,barbar i,i,barbar i^2))$i^2))$ for some scalar
+ $lambdainF_p^*$; the vector $(1,barbar i,i,barbar i^2)$i^2)$ is attached
via the $H_j$ side.
-\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
- $\lambda\in\F_p^*$.
+item $j>i$: the syllable from edge ${i,j}$ in $H_i$ lies in~$V$ and
+ is $v(lambdacdot(1,barbar j,j,barbar j^2))$j^2))$ for some
+ $lambdainF_p^*$.
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
-\[
-\det
- 1 & \bar j_1 & \bar j_1^2\\
- 1 & \bar j_2 & \bar j_2^2\\
- 1 & \bar j_3 & \bar j_3^2
-
-= (\bar j_2-\bar j_1)(\bar j_3-\bar j_1)(\bar j_3-\bar j_2)
-\ne 0,
-\]
-because $p>m>3$, so the three residues are distinct in $\F_p$. The
-same argument handles two vectors (a $2\times 2$ minor of the above)
+vectors $(1,barbar j,j,barbar j^2)j^2)inF_p^3$. Three of these vectors with
+three distinct residues $barbar j_1,j_1,barbar j_2,j_2,barbar j_3$j_3$ are linearly
+independent over $F_p$: the determinant
+[
+det
+ 1 & barbar j_1j_1 & barbar j_1^2j_1^2
+ 1 & barbar j_2j_2 & barbar j_2^2j_2^2
+ 1 & barbar j_3j_3 & barbar j_3^2j_3^2
+
+= (barbar j_2-j_2-barbar j_1)(j_1)(barbar j_3-j_3-barbar j_1)(j_1)(barbar j_3-j_3-barbar j_2)j_2)
+ne 0,
+]
+because $p>m>3$, so the three residues are distinct in $F_p$. The
+same argument handles two vectors (a $2times 2$ minor of the above)
and one vector (nonzero).
So a word of at most $3$ syllables, all in $U$ or all in $V$, with
@@ -1418,18 +1418,18 @@ distinct labels, evaluates nontrivially in $F_p^3$ and hence in $H_i$.
For a mixed word of length $2$ or $3$: exactly one of $U$ or $V$
supplies a single syllable, and the other supplies one or two. After
a cyclic permutation, a putative relation would equate a nonidentity
-element of $U$ to a product of elements of~$V$. But $U\cap V=\{1\}$
+element of $U$ to a product of elements of~$V$. But $Ucap V={1}$
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~lem:triangle-free-gluing hold, giving
-the injection $H_i\to G_0$. The induced map $K[H_i]\to K[G_0]$ is
+the injection $H_ito G_0$. 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.
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
+each $H_i$ as a subgroup, and the $e_iin K[G_0]$ satisfy
Corollary~op:edge-zero.