diff b1f9bc4..56bc728 kaplansky_direct_finite_odd_v2.txt
diff --git a/kaplansky_direct_finite_odd_v2.txt b/kaplansky_direct_finite_odd_v2.txt
index 8fca700..0d4a039 100644
--- a/kaplansky_direct_finite_odd_v2.txt
+++ b/kaplansky_direct_finite_odd_v2.txt
@@ -52,15 +52,15 @@ maketitle
Abstract:
Kaplansky conjectured in 1972 that the group algebra of any group
-over any field would be ``directly finite'': if $ab=1$ then $ba=1$.
+over any field would be ``directly finite'': if ab=1 then ba=1.
The conjecture is true when the field has characteristic zero, but it
fails in positive characteristic once the group is allowed to be
infinite. This paper constructs a counterexample in odd characteristic.
-We exhibit one explicit odd prime~$p$, a field~$K$ of order~$p^4$,
-a finitely generated group~$G$ containing torsion, and two finite
-sums $a,bin K[G]$ 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.
+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,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.
# Introduction
@@ -68,51 +68,51 @@ so the group~$G$ is not surjunctive.
## The puzzle: one-sided versus two-sided inversion
-Let $K$ be a field and $G$ a group. The group algebra $K[G]$
+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_iin K, g_iin G,
+ 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,
+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 $G=mathbb Z$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~$x^n$.
+Concrete example. When G=mathbb ZbbZ 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.
+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
-$a,bin R$, always implies~$ba=1$. Fields are directly finite (the
+A ring R is directly finite when the equation ab=1, for
+a,bin R, always implies~ba=1. Fields are directly finite (the
only nonzero element with a multiplicative inverse is the obvious one).
-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
+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
+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
-$Mat_n(R)$, $nge1$, 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 $Mat_n(K[G])$
+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.
## Kaplansky's question
In his 1972 book on fields and rings, Kaplansky proved that when the
-field has characteristic~$0$, the group algebra $K[G]$ is directly finite
-for every group~$G$; his proof used finite von Neumann algebras. He then
-asked whether the same conclusion holds in characteristic~$p>0$, where
+field has characteristic~0, the group algebra K[G] is directly finite
+for every group~G; his proof used finite von Neumann algebras. He then
+asked whether the same conclusion holds in characteristic~p>0, where
his von Neumann-algebra method does not apply directly.
A large body of later work found positive partial answers. Ara, O'Meara,
-and Perera (2002) proved stable finiteness of $K[G]$ for $G$ free-by-amenable
+and Perera (2002) proved stable finiteness of K[G] for G free-by-amenable
over any division ring. Elek and Szab'o (2004) proved stable finiteness
for every sofic group: informally, a group that can be approximated by
finite permutation models. Since many familiar infinite groups are
@@ -126,61 +126,59 @@ single specific odd characteristic.
## 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~lem:field.
+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
+ a field K of order p^4 (an explicit quadratic tower over the
+ prime field F_p);
+ a finitely generated group G containing nontrivial elements of
finite order;
-item two finite sums $a,bin K[G]$ with $ab=1$ and $bane 1$.
+ two finite sums a,bin K[G] with ab=1 and bane 1.
The construction is entirely finite: every field element, every group
generator, every coefficient is produced by a finite list of operations
from fixed finite sets. No infinite search and no word-problem oracle
-is needed to define $a,b$; the point of the construction is to build them
-from finite data rather than by deciding equalities in~$G$.
+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 $h>1$
-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 $a,bin K[G]$ satisfying
-$ab=1$ but $bane 1$. In particular Kaplansky's direct-finiteness
+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,bin K[G] satisfying
+ab=1 but bane 1. In particular Kaplansky's direct-finiteness
conjecture fails in odd characteristic.
The word ``specified'' means: written as an explicit composition of
finite operations on finite sets, in an order that is fixed in advance.
Different coefficients may happen to coincide at the end; that is
-harmless. What matters is that $a$ and $b$ are objective consequences
+harmless. What matters is that a and b are objective consequences
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
-$K[G]$ would be stably finite, contradicting the theorem. So the
+Elek--Szab'o, the constructed group G cannot be sofic; otherwise
+K[G] would be stably finite, contradicting the theorem. So the
construction is also a source of non-sofic groups of a very specific
kind.
Second, direct finiteness is equivalent to a statement about ``linear
-cellular automata.'' A cellular automaton on $K^G$ is any transformation
+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
-$c=sum c_u uin K[G]$, the rule
+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 $bane 1$, we can
-show $T_b$ is not surjective on $K^G$. Section~sec:cellular gives
+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 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~$G$.
+refuting Gottschalk's conjecture for this particular~G.
## The plan of the proof
@@ -189,45 +187,44 @@ 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 $H=A_0rtimes P$ 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
-annihilations: writing $u(s),v(s)$ for certain elements of~$H$ and
-$ell=(p-1)/2$,
+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 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 $v$
-sit in different subgroups of~$H$, so no symmetry forces them to be
+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
-$G_0$ 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
+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.
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
+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 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
+change-of-basis argument turns dimension equalities over~K into
equalities over the full ring.
-Piece 5 (Sections~ref{sec:compression and~sec:scalar):
-matrices, then scalars, then a cellular automaton.}
+Piece 5 (Sections~
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~sec:scalar
@@ -238,159 +235,154 @@ translates it into a statement about cellular automata.
## Notation summary
All matrices below act on column vectors in free right modules
-$S^n$. So multiplying a matrix by a column vector means multiplying
+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.
-
-
-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 $ein K[H]$: 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 $X^{mathsf T}sfT$, with no scalar conjugation.
-item $Mat_{atimes b}(S)$: $atimes b$ matrices with entries in~$S$.
+ K: a field. F_p: the prime field of order p.
+ G: a group; K[G]: the set of finite K-linear sums of
+ elements of G, with multiplication from the group law.
+ 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.
+ ein K[H]: an idempotent, meaning e^2=e.
+ 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].
+ w=|H|: the order of the finite group~H.
+ [n]: {1,...,n} for a positive integer n.
+ Hermitian transpose: never used here. Ordinary transpose is
+ written X^{mathsf T}sfT, with no scalar conjugation.
+ 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~$Gamma$
-on $m$ vertices together with an extremely compact factorization of its
-adjacency matrix; and a finite group $H$ with a natural action that will
+The finite-data stage has three independent ingredients: a prime~p and
+a specific field K of order p^4; a simple triangle-free graph~Gamma
+on m vertices together with an extremely compact factorization of its
+adjacency matrix; and a finite group H with a natural action that will
later host the crucial idempotent. Each ingredient is built once and for
all here, in the order needed later.
## 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.
-
+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
-$(m!)^2+1$, 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 $etainF_p^*$ 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 $|K|=p^4$.
+With m=binom{1200}{600} and p the least h>1 dividing
+(m!)^2+1, we have:
+ p is an odd prime, and p>m.
+ p divides (m!)^2+1, so -1 has a square root in F_p.
+ 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.
+ 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 $ple m$, 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 $(m!)^2equiv -1pmod p$ says that $-1$ is the
-square of $m!$ modulo~$p$, that is, $-1$ is a quadratic residue in
-$F_p$.
-
-(3) A polynomial $X^2-eta$ over a field is irreducible exactly when
-$eta$ is not a square in that field. Nonsquares exist in $F_p^*$
-for every odd prime~$p$: the map $xmapsto x^2$ on $F_p^*$ is
-two-to-one on its image, so only half of the $p-1$ elements are
-squares, and $p-1ge2$ 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
+(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 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!)^2equiv -1pmod p says that -1 is the
+square of m! modulo~p, that is, -1 is a quadratic residue in
+F_p.
+
+(3) A polynomial X^2-eta over a field is irreducible exactly when
+eta is not a square in that field. Nonsquares exist in F_p^*
+for every odd prime~p: the map 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-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.
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.
-
-
-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
- $c_0+c_1xi+c_2nu+c_3xinu$ 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
- order equals $M$. Such an element exists; the reasoning is
+ 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.
+ Representatives and order on K.
+ Write every element of K uniquely as
+ 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.
+ 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 $ble M$ 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
+Why does an element of order M exist? Let b be the least common
+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
-$Mle b$. 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
+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 $zeta$
-of order $M$ exists.
+pairwise coprime orders, has order exactly M. So an element zeta
+of order M exists.
-In particular $M=p^4-1$ and all of its powers are invertible in $K$
-(equivalently: the multiplicative group of $K$ is cyclic of order
-$M=p^4-1$). We will divide by various integers below; every such
-integer is a factor of~$M$ and hence nonzero in~$K$.
+In particular M=p^4-1 and all of its powers are invertible in K
+(equivalently: the multiplicative group of K is cyclic of order
+M=p^4-1). We will divide by various integers below; every such
+integer is a factor of~M and hence nonzero in~K.
## 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
-subsets $I⊆[1200]$ 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
-$I_1,...,I_m$ with $m=binom{1200}{600}=binom{1200}{600}$. Remember
-this same~$m$ is the one appearing in the definition of~$p$.
+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 $ine j$ to be adjacent exactly when
-the intersection $I_icap I_j$ has fewer than $200$ elements:
+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.
]
@@ -398,67 +390,67 @@ By convention a vertex is never adjacent to itself, so the graph is
simple.
-The graph $Gamma$ is simple and triangle-free. Let
+The graph Gamma is simple and triangle-free. Let
[
- mathcal T=calT={T⊆[1200]:|T|le 400},qquad
+ mathcal T=calT={T⊆[1200]:|T|le 400},
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
+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_{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 $t/m<1/100$.
+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.
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
+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_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.
+by inclusion-exclusion. But the union is contained in [1200], which
+has only 1200 elements. Contradiction. So Gamma has no triangles.
-The fit. Pick, for each $i$, the indicator function
+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 $F_p$:
+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
+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_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 $prod_{xin T}X_x$ in
-$Q_j^{sf}$ for each $T⊆[1200]$. Because $Q_j$ has
-total degree $400$ (a product of $400$ linear forms), only the
-$Tinmathcal T$calT$ can appear. The total degree of $Q_j$ bounds the
+in which every positive exponent has been replaced by 1. Set
+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
+Tinmathcal TcalT 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 $nge1$). 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,
+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 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_icap I_j|-k)/(600-k).
]
-On the other hand, expanding $Q_j^{sf}$ and evaluating at
-$1_{I_i}$, each monomial $prod_{xin T}X_x$ contributes $1$ if
-$T⊆ I_i$ and $0$ otherwise. So
+On the other hand, expanding Q_j^{sf} and evaluating at
+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_{Tinmathcal T}calTalpha_i(T),beta_j(T)=F_{ij}.
@@ -468,40 +460,40 @@ Equating the two evaluations gives
F_{ij}=prod_{k=200}^{599}(|I_icap I_j|-k)/(600-k).
-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 $ine j$ 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
+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 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}.
The ratio bound. Combinatorial estimates. The binomial
-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
+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 401binom{1200}{400}.
]
-For $m=binom{1200}{600}$ we can compare factorials directly:
+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
-$ule 500$, 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
+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 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 $401<40100<2^{33}$
+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,
@@ -511,358 +503,353 @@ which is weaker than needed; more precisely,
100 cdot frac tm < 40100(tfrac 57)^{100}
<frac{40100}{2^{33}}<1,
]
-so $t/m<1/100$.
+so t/m<1/100.
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~op:compression-section
+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~op:compression-section
will combine the two ratios into a matrix with strict inequality
-$wt<md$ between its shapes.
+wt<md between its shapes.
-## The finite 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
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/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 $M^{p^3}$.
+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 $sin P$ and
-$ain A_0$, define $scdot ain A_0$ by
+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.
+ (scdot a)(x)=a(x-s), xin P.
]
-This is a group action: $0cdot a=a$ and $(s+t)cdot a=scdot(tcdot a)$.
+This is a group action: 0cdot a=a and (s+t)cdot a=scdot(tcdot a).
Form the semidirect product
[
- H=A_0rtimes P, qquad
+ H=A_0rtimes P,
(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: $(a,s)(-(-s)cdot a, -s)=(a-scdot((-s)cdot a), 0)=(a-scdot(-scdot a), 0)=(a-0,0)=(0,0)$.
-The order of $H$ is
+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-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.
]
-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 $K^w$).
+So H is finite. The group algebra K[H] is thus a
+finite-dimensional algebra over K, of dimension w as a K-vector
+space (and w^2 as a K-algebra of linear maps on K^w).
-The two subgroups of $H$ that matter now are $U$ and $V$, defined
-below. Both are isomorphic to $P$, but they sit inside $H$ in different
+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.
Translation subgroup. Define
[
- u(s)=(0,s),qquad U={u(s):sin P}⊆ H.
+ u(s)=(0,s), U={u(s):sin P}⊆ H.
]
-Because $(0,s)(0,s')=(0,s+s')$, the map $smapsto u(s)$ is an
-isomorphism $Pto U$.
+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
+Characteristic element. Let cin A_0 be the function
[
- c(0)=1,quad c(x)=0 for xne0.
+ c(0)=1, 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 $A_0$).
+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):sin P}⊆ H,
+ v(s)=c,u(s),c^{-1}, 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
+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-scdot c, s),qquad sin P.
+ v(s)=(c-scdot c, s), sin P.
-Indeed $(c,0)(0,s)=(c,s)$ and $(c,s)(-c,0)=(c+scdot(-c),,s)=(c-scdot c,s)$.
+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 $Ucap V={1}$, and $Ucong Vcong P$.
+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
-coordinates gives $s=s'$. Comparing first coordinates gives
-$0=c-scdot c$, 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.
+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-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:
-$smapsto u(s)$ 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 $P$.
+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
+idempotent~e with
[
- (u(s)-1)^ell e = 0,qquad e,(v(s)-1)^ell=0
+ (u(s)-1)^ell e = 0, 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 $Ucap V={1}$,
+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
-$m=binom{1200}{600}$, $t=|mathcal T|$,calT|$, $w=|H|$, 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
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 $H=A_0rtimes P$ 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 $d=dim_K(eK[H])$
- 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 $sin P$.
+section we built three pieces of finite data: a prime p, a field
+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:
+ 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|.
+ 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
- which both $u(s)$-shifts and $v(s)$-shifts have the same
+ 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
+ calculus (the operators X_j) and care with degrees.
+ Ssubsec:admissibility: use the polynomial identity test of
Schwartz and Zippel to produce many ``admissible'' characters
- $lambdainA_0hat$, 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 $K[H]$,
+ lambdainA_0hat, i.e., characters lambda for which
+ a certain Gram matrix~N_lambda is nonsingular.
+ 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
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
-$A_0⊆ H$.
+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⊆ H.
-## The annihilated subspace $B⊆ K^P$
+## 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 $Pto K$, 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.
+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_xbfe_x (a 1 at
+x and 0 elsewhere). We want a linear subspace on which left shifts
+and right shifts both annihilate ell-fold.
-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
+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,
+ X_j=L_{ε_j}-1_{K^P}, j=1,2,3,
]
-where $ε_j$ 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 $1le kle p-1$;
-so $(1+X_j)^p=1+X_j^p$ and
+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 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
+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 $P$.
+generators of P.
[Monomial basis]
-The $p^3$ vectors
+The p^3 vectors
[
- 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,
+ X_1^{n_1}X_2^{n_2}X_3^{n_3}mathbf e_0,bfe_0, 0le n_1,n_2,n_3< p,
]
-form a $K$-basis of $K^P$.
+form a K-basis of K^P.
Proof.
-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.
+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_sbfe_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
-$(1+X_j)^{s_j}=L_{s_jε_j}$ 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_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
-$X_1^{n_1}X_2^{n_2}X_3^{n_3}mathbf e_0$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.
+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_0bfe_0 with 0le n_jle s_j<p.
+So mathbf e_sbfe_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.
-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 $X_j^p=0$.
+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.
-Define $B$ as the matrix with columns the basis vectors whose total
+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]_bfe_0]_{
substack{0le n_1,n_2,n_3<p n_1+n_2+n_3ge D-ell+1}}.
-So $B$ is a $p^3times h$ matrix with $K$-linearly independent columns
-(they are a subset of a basis). The integer $h$ counts the 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 $sin P$,
+[Two-sided annihilation of B]
+For every sin P,
[
- (L_s-1)^ell,B=0,qquad B^{mathsf T}sfT(L_s-1)^ell=0.
+ (L_s-1)^ell,B=0, B^{mathsf T}sfT(L_s-1)^ell=0.
]
Proof.
-First identity. Fix $sin P$ and write $s=sum_j s_jε_j$.
-Expand $L_s-id_{K^P}$:
+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
-of total $X$-degree $d$, the result lies in the $K$-span of monomials
-of total $X$-degree at least $d+ell$.
-
-Now take a column $X_1^{n_1}X_2^{n_2}X_3^{n_3}mathbf e_0$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,
+Expanding the right side, every nonzero monomial has total X-degree
+at least one (because the constant term 1 cancels). Therefore the
+ellth power (L_s-1)^ell has the property: applied to any monomial
+of total X-degree d, the result lies in the K-span of monomials
+of total X-degree at least d+ell.
+
+Now take a column X_1^{n_1}X_2^{n_2}X_3^{n_3}mathbf e_0bfe_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
-$(L_s)_{mathbf e_x,bfe_x,mathbf e_y}bfe_y$ (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
+Second identity. Compute the transpose of L_s: its matrix entry
+(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 $s$.
+using the first identity with -s in place of s.
[Number of columns]
-The integer $h$ equals $binom{ell+2}{3}$, and
+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 $a_jge0$ 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
+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^3bbN^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
+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
-$1/48$, compute the difference
+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 pge3,
]
-which gives $h/p^3>1/48$.
+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
-$h/p^3$ is at least $1/48$ of the ambient dimension.
+To summarize the situation in one line: B is a tall matrix whose
+columns are the ``high-degree part'' of K^P, and both u(s)-shifts
+and their transposes kill B after ell applications. The fraction
+h/p^3 is at least 1/48 of the ambient dimension.
-We should remark what is special about the value $ell=(p-1)/2$.
-Choosing $ell$ smaller would shrink $B$'s fraction too much; choosing
-$ell$ larger would not annihilate $B$ at all (the top-degree columns
-would survive). The exponent $ell$ is exactly the right size to
-simultaneously annihilate on both sides while keeping $B$ fat.
+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
-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
-$u(s)$ and $v(s)$ act through matrix representations of different
+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
+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 $A_0$.:
-$A_0=(mathbb Z/MbbZ/Mmathbb Z)^P$bbZ)^P$ is finite abelian of order
-$|A_0|=M^{p^3}$. A character is a homomorphism
-$lambda:A_0to K^*$. 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
-$(slambda)(a)=lambda((-s)cdot a)$. Conjugating the Fourier
-idempotent $p_chi^{A_0}$ of a character by $u(s)$ gives
+Characters of A_0.:
+A_0=(mathbb Z/MbbZ/Mmathbb Z)^PbbZ)^P is finite abelian of order
+|A_0|=M^{p^3}. A character is a homomorphism
+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
+(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 $lambdainA_0hat$ abbreviate $chi_x=xlambda$ (the
-translate of $lambda$ by $x$) and define the diagonal
-$p^3times p^3$ 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):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}sfT,D_lambda^{-1},B
-is a symmetric $htimes h$ matrix over~$K$.
-
-Call $lambda$ admissible if
+is a symmetric htimes h matrix over~K.
-item its $P$-stabilizer is trivial, i.e., the orbit $Plambda$ has
- size exactly $p^3$; and
-item $N_lambda$ is invertible.
+Call lambda admissible if
+ its P-stabilizer is trivial, i.e., the orbit Plambda has
+ size exactly p^3; and
+ N_lambda is invertible.
Both conditions are checkable in finite time, and the check is
-deterministic: enumerate the $p^3$ translates and compute one
-$htimes h$ determinant.
+deterministic: enumerate the p^3 translates and compute one
+htimes h determinant.
Most characters are admissible.:
The two modes of failure each account for less than half of the
@@ -870,86 +857,86 @@ characters. Their union is then smaller than half, so more than half
succeed. This count is the key estimate of the whole construction.
-More than half of the $|A_0|$ characters are admissible.
+More than half of the |A_0| characters are admissible.
Proof.
-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
-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
+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
+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 $pge5$;
-we have $p>600$.
+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}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 $htimes h$ submatrix of $B$ consisting of the
-rows indexed by~$J$, taken in the fixed increasing order.
+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
+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~lem:monomial-basis, some maximal minor $B_J$ is nonzero.
-The polynomial $Q$ therefore has a nonzero leading monomial and does
+degree is exactly~h. Because B has full column rank by
+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] (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 $hle p^3$ (it counts a
-subset of $K^P$) and $M=p^4-1$, this is at most
+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 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 $pge5$.
+because p^4-1>4p^3 for pge5.
-Combining: fewer than $1/4+1/4=1/2$ of characters fail either
+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.
-## Realizing the projection in $K[H]$
+## 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
-write that projection as an element of $K[H]$ whose regular rank we
+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 $p_{xlambda}u(s)$ (for $x,sin P$) 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 $lambdainA_0hat$ with trivial $P$-stabilizer. For
-$x,yin P$ set
+Fix lambdainA_0hat with trivial P-stabilizer. For
+x,yin P set
E_{xy}^{lambda}=p_{xlambda}^{A_0},u(x-y).
-Because $u(s),p_{xlambda}u(s)^{-1}=p_{(x+s)lambda}$, 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}.
-Indeed, the Fourier product survives when $(x-y+z)lambda=xlambda$,
-i.e. $(y-z)lambda=0$, which by the trivial stabilizer forces $y=z$.
+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 $q_lambda=sum_{xin P}p_{xlambda}^{A_0}$. 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 $cmapsto D_lambda$.
+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
@@ -957,138 +944,138 @@ Define
T_lambda=B,N_lambda^{-1}B^{mathsf T}sfTD_lambda^{-1}
inMat_{p^3}(K).
-Because $N_lambda=B^{mathsf T}sfTD_lambda^{-1}B$, we have
+Because N_lambda=B^{mathsf T}sfTD_lambda^{-1}B, we have
[
- T_lambda B=B,qquad
+ T_lambda B=B,
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
-$rankT_lambda=h$.
+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,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
+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}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
+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 $B^{mathsf T}sfT(L_s-1)^ell=0$:
+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 $K[H]$.
+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 $Plambdane Plambda'$
-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
+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 RscrR be the set
of chosen characters. Define
e=sum_{lambdainmathscr R}scrRe_lambda in K[H].
-By orthogonality the sum is an idempotent: $e^2=e$. The two
-annihilations persist on~$e$.
+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|.scrR|.
]
-The admissible characters form more than half of $A_0hat$
-(Lemma~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|>(|A_0|)/(2p^3).scrR|>(|A_0|)/(2p^3).
]
-Recall $w=|H|=|A_0|,p^3$. Therefore
+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.
+using h/p^3>1/48 from Lemma~lem:h.
-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 $sin P$
+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
+ (u(s)-1)^ell,e=0,
e,(v(s)-1)^ell=0 .
]
Its coefficients are obtained by finite operations in the specified
-field $K$ of Lemma~lem:field.
+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.
+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.
## 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 $v(1,barbar j,j,barbar j^2)$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
+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,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 $e_iin K[G_0]$ the image of the idempotent $e$ under the
-map $H_ito G_0$.
+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 $yin K[H_i]⊆ K[G_0]$ 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
+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
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 $q_ain K[< y>]$.
+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_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
-$ell+jge p$.
+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+jge p.
-Thus $q_aX^ell=0$ forces $q_a=q_a'X^{p-ell}$, and
+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 $p-ell=ell+1$,
+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.
+No factor is commuted past X, no symmetry is used, and no reverse
+product e_je_i is asserted.
Summary so far.:
@@ -1102,7 +1089,7 @@ contradiction.
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,
+value 1 and their darts disappear from the boundary-word description,
and the graph stays connected. Face evaluations are unchanged, since
the two neutral darts contributed nothing. Minimality fails.
@@ -1113,43 +1100,41 @@ evaluations, and the discarded loop disappears. Minimality fails.
Hence every edge carries a nontrivial label.
Reduction 2: every edge is a nonbridge.:
-Suppose an edge is a bridge with value $s$. Deleting it splits the map
+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,
+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
-exactly one exceptional face (that of $W_a$). Minimality fails.
+exactly one exceptional face (that of W_a). Minimality fails.
-If both $W_a$ are trivial, the original face was either ordinary or
+If both W_a are trivial, the original face was either ordinary or
exceptional. If it was ordinary, the exceptional face is on one of the
two components; keep that one. If the original face was the unique
-exceptional one, both $W_a$ are trivial (this was assumed), so $sWs^{-1}$
+exceptional one, both W_a are trivial (this was assumed), so sWs^{-1}
was the exceptional word; keep the component whose walk is the whole
-$sWs^{-1}$ from the other side. In every case, a smaller exceptional map
+sWs^{-1} from the other side. In every case, a smaller exceptional map
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 $iin
-V(Delta)$. 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
+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).
-If $sD$ evaluates to $1$, then $s^{-1}(sD)s$ also evaluates to $1$; if
-$sD$ 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.
+If sD evaluates to 1, then s^{-1}(sD)s also evaluates to 1; if
+sD evaluates to gne1, then its conjugate s^{-1}gs also has value
+ne1. Conjugation preserves identity-versus-nonidentity. Hence:
+ ordinary + ordinary to ordinary;
+ ordinary + exceptional (in either order) to exceptional.
We are assuming the map has exactly one exceptional face, so
@@ -1157,8 +1142,8 @@ 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
@@ -1179,13 +1164,11 @@ exists it bounds exactly the exceptional monogon.
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.
+ every edge is nonneutral;
+ every edge is a nonbridge;
+ the two faces across an edge labelled gamma have the two
+ distinct endpoint types of gamma;
+ at most one loop exists, and it bounds the exceptional monogon.
We continue in the next reduction. The one remaining configuration to
@@ -1194,79 +1177,75 @@ 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 $e:ato b$ and
-$f:bto d$ with a common label $gammain E(Delta)$ and values
-$ε,varphiin C_gamma∖{1}$. The edges $e$ and $f$
+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 $jin V(Delta)$,
-where $j$ is the other endpoint of $gamma$ from $Phi$'s type~$i$.
-The faces $X$ and $Y$ might coincide.
+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 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 $g:ato d$ 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.
+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 $Xne Y$. 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
+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
(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 $varphi^{-1}pvarphicdot q$:
-
-
-item $p=q=1$: merged value $varphi^{-1}cdot 1cdotvarphicdot 1=1$;
-item $pne1$, $q=1$: merged value $varphi^{-1}pvarphine1$ (conjugate
+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:
+ p=q=1: merged value varphi^{-1}cdot 1cdotvarphicdot 1=1;
+ pne1, q=1: merged value varphi^{-1}pvarphine1 (conjugate
of nonidentity);
-item $p=1$, $qne1$: merged value $qne1$;
-item $pne1$, $qne1$: excluded (at most one exceptional face).
+ p=1, qne1: merged value qne1;
+ pne1, qne1: excluded (at most one exceptional face).
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.
-Case $X=Y$. Now $X$ is a single face whose boundary walk
-encounters $e$ and $f$ on their opposite sides. Reading from the dart
-$barbar e$:e$: the walk is $barbar eDeDbarbar fE$,fE$, with $D:ato d$ and
-$E:bto b$.
+Case X=Y. Now X is a single face whose boundary walk
+encounters e and f on their opposite sides. Reading from the dart
+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
-$barbar gD$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
+barbar gDgD 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,fe,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.
-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 $A=(εvarphi)^{-1}D$ be the value of walk $barbar gD$,gD$, and
-$B=E$ the value of walk $E$. The old walk of $X$ evaluated to
-$ε^{-1}Dvarphi^{-1}E$, 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
+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=(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}Dvarphi^{-1}E, whose cyclic shift starting at barbar ff
+is (varphi^{-1}Bvarphi)A:
+ X exceptional: the old value was ne1, so at least one of
+ A,B is ne1. Keep the component whose walk is that one.
+ 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 $Ane1$ and $Bne1$: both new walks are
+ X ordinary with Ane1 and Bne1: both new walks are
exceptional. The exceptional case sits either in the
- $barbar gD$-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
+ 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
sole nonidentity boundary.
@@ -1283,122 +1262,120 @@ the exceptional monogon), and no face has two consecutive nonloop
darts with the same label.
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
-endpoints $i_{a}$ and $i_{a+1}$, by Reduction 3. No immediate
+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
+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 $rge1$. 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 $deg uge4$.
+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 uge4.
Faces away from the loop.:
-Let $F$ be an ordinary face not incident to the loop. Its boundary is
+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,
-which follows from Euler's formula and $sum_udeg u=sum_F|∂
-F|=2e$ (every edge contributes two to each side).
+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 $deg uge4$, 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
+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
-exceptional monogon has walk of length $1$, so contributes $3$. 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
-$2$, contributing at most $2$. All other vertices and faces
-contribute at most zero. Total $le 2+3+2=7<8$, contradicting
+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
(eq:gluing-deficit).
-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~$i$.
+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 $D_i^0to Q$ to $D_ito Q$. 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 $Ω=Ltimesℕ$, 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
+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 $D_itimesℕ$.
-
-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 $phi_i:D_itimesℕtoΩ$.
-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 $x_gammain L$ 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 $D_itimesℕ$ is faithful and
-$phi_i$ is a bijection.
-
-Therefore $D_ito Q$ is injective for every vertex $i$. This
+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_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_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_itimesℕ is faithful and
+phi_i is a bijection.
+
+Therefore D_ito Q is injective for every vertex i. This
completes the proof of Lemma~lem:triangle-free-gluing.
-## Applying the criterion to $H_i$
+## Applying the criterion to H_i
-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 $p$.
+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 $H_ito G_0$ 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
-$D_i=H$. The vertex groups $H_i$ are finite, the cyclic edge groups
-have order $p$, and the edge embeddings are given by the displayed
+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:
-
-
-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 $v(lambdacdot(1,barbar j,j,barbar j^2))$j^2))$ for some
- $lambdainF_p^*$.
+Fix a vertex i. The incident edges split into two kinds:
+ 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.
+ 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 $(1,barbar j,j,barbar j^2)j^2)inF_p^3$. 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
+Within each of U or V, distinct neighbors j correspond to the
+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_3j_3 are linearly
+independent over F_p: the determinant
[
det
1 & barbar j_1j_1 & barbar j_1^2j_1^2
@@ -1408,28 +1385,28 @@ det
= (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)
+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
-distinct labels, evaluates nontrivially in $F_p^3$ and hence in $H_i$.
+So a word of at most 3 syllables, all in U or all in V, with
+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 $V$
+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 $Ucap V={1}$
-and the $U$-element is nonidentity, contradiction. For lengths $2$
-and $3$, cyclically distinct labels are just distinct labels, so
+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 $H_ito G_0$. 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 $e_iin K[G_0]$ satisfy
+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_iin K[G_0] satisfy
Corollary~op:edge-zero.