diff 56bc728..f6bf760 kaplansky_direct_finite_odd_v2.txt
diff --git a/kaplansky_direct_finite_odd_v2.txt b/kaplansky_direct_finite_odd_v2.txt
index 0d4a039..4da6182 100644
--- a/kaplansky_direct_finite_odd_v2.txt
+++ b/kaplansky_direct_finite_odd_v2.txt
@@ -80,8 +80,8 @@ 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 ZbbZ and K is any field, the
-group algebra K[mathbb Z]bbZ] is (isomorphic to) the ring of Laurent
+Concrete example. When G=Z and K is any field, the
+group algebra K[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
@@ -157,81 +157,6 @@ Different coefficients may happen to coincide at the end; that is
harmless. What matters is that a and b are objective consequences
of a finite recipe, not of an arbitrary choice among infinitely many.
-## Why the failure of direct finiteness matters
-
-Two remarks connect this algebra to other topics. First, by
-Elek--Szab'o, the constructed group G cannot be sofic; otherwise
-K[G] would be stably finite, contradicting the theorem. So the
-construction is also a source of non-sofic groups of a very specific
-kind.
-
-Second, direct finiteness is equivalent to a statement about ``linear
-cellular automata.'' A cellular automaton on K^G is any transformation
-that reads a finite window of the configuration around each point and
-writes a new value there. If we take, for a fixed finite sum
-c=sum c_u 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
-the details. The upshot is a group that fails to be surjunctive,
-refuting Gottschalk's conjecture for this particular~G.
-
-## The plan of the proof
-
-The construction has five independent pieces. Each piece is fairly
-self-contained; the reader may take them in whatever order feels best,
-as long as the final step receives all the pieces.
-
-Piece 1 (Section~sec:finite): finite data.
-We fix the prime p, the field K=F_{p^4} (built as a quadratic
-tower so we can descend to F_p for checks), and a simple triangle-free
-graph~Gamma on m vertices with a factorization of its adjacency
-matrix into a very short sum of rank-one pieces. We also define a finite
-group H=A_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,
-[
- (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
-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
-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
-matrix blocks with exact integer-scaled right and left inverses.
-This is the technical heart of the paper: a nilpotent-ideal
-change-of-basis argument turns dimension equalities over~K into
-equalities over the full ring.
-
-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
-embeds a matrix algebra into a group algebra, letting us turn that
-matrix identity into an identity between two scalar elements. Section~sec:cellular
-translates it into a statement about cellular automata.
-
## Notation summary
All matrices below act on column vectors in free right modules
@@ -249,7 +174,7 @@ This matters because group algebras are not always commutative.
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.
+ written X^{T}, with no scalar conjugation.
Mat_{atimes b}(S): atimes b matrices with entries in~S.
@@ -301,20 +226,20 @@ 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
+for every odd prime~p: the map x|-> 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
+xi and -xi. Therefore xi·(-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
+Apply the field norm N_{K_0/F_p}(u)=u· 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
@@ -392,14 +317,14 @@ simple.
The graph Gamma is simple and triangle-free. Let
[
- mathcal T=calT={T⊆[1200]:|T|le 400},
- t=|mathcal T|=calT|=sum_{j=0}^{400}binom{1200}{j}.
+ T={T⊆[1200]:|T|le 400},
+ t=|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
+indexed by TinT, such that the mtimes m matrix
F given by
[
- F_{ij}=sum_{Tinmathcal T}calTalpha_i(T),beta_j(T)
+ F_{ij}=sum_{TinT},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.
@@ -410,7 +335,7 @@ 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,
+ ge 3· 600 - 3· 199 = 1203,
]
by inclusion-exclusion. But the union is contained in [1200], which
has only 1200 elements. Contradiction. So Gamma has no triangles.
@@ -426,7 +351,7 @@ variables X_1,...,X_{1200} over F_p:
]
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]}
+with c_alphainF_p and multi-indices alphainN^{[1200]}
(so X^alpha=prod_x X_x^{alpha_x}). Define the square-free
reduction of Q_j to be
[
@@ -436,7 +361,7 @@ 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
+TinT 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
@@ -453,7 +378,7 @@ On the other hand, expanding Q_j^{sf} and evaluating at
T⊆ I_i and 0 otherwise. So
[
Q_j(1_{I_i})
- =sum_{Tinmathcal T}calTalpha_i(T),beta_j(T)=F_{ij}.
+ =sum_{TinT},beta_j(T)=F_{ij}.
]
Equating the two evaluations gives
@@ -496,11 +421,11 @@ 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}· frac1{2^{33}}=1,
]
which is weaker than needed; more precisely,
[
- 100 cdot frac tm < 40100(tfrac 57)^{100}
+ 100 · frac tm < 40100(tfrac 57)^{100}
<frac{40100}{2^{33}}<1,
]
so t/m<1/100.
@@ -525,29 +450,29 @@ 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
+ A_0=(Z/MbbZ/MZ)^P
]
-be the group of functions a:Ptomathbb Z/MbbZ/Mmathbb Z,bbZ, with
+be the group of functions a:PtoZ/MbbZ/MZ, 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
+ain A_0, define s· ain A_0 by
[
- (scdot a)(x)=a(x-s), xin P.
+ (s· 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: 0· a=a and (s+t)· a=s·(t· a).
Form the semidirect product
[
H=A_0rtimes P,
- (a,s)(a',s')=(a+scdot a', s+s').
+ (a,s)(a',s')=(a+s· 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).
+(-s^{-1}· a, -s), i.e., (-(-s)· a,-s) since in characteristic
+p subtraction is the same as~+(-s). Concretely: (a,s)(-(-s)· a, -s)=(a-s·((-s)· a), 0)=(a-s·(-s· 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|·|P|=M^{p^3}· 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
@@ -561,7 +486,7 @@ Translation subgroup. Define
[
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
+Because (0,s)(0,s')=(0,s+s'), the map s|-> u(s) is an
isomorphism Pto U.
Characteristic element. Let cin A_0 be the function
@@ -579,9 +504,9 @@ 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), sin P.
+ v(s)=(c-s· 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+s·(-c),,s)=(c-s· c,s).
The subgroups U and V satisfy Ucap V={1}, and Ucong Vcong P.
@@ -589,13 +514,13 @@ 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).
+0=c-s· c, i.e., c=s· c. The left side evaluates at 0 to
+c(0)=1; the right side evaluates at 0 to (s· 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,
+s|-> u(s) is a bijection Pto U preserving the group law, and
+s|-> v(s) is a conjugate of~s|-> u(s) inside the group~H,
hence also isomorphic to P.
@@ -612,7 +537,7 @@ 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
+m=binom{1200}{600}, t=|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.
@@ -723,7 +648,7 @@ So B is a p^3times h matrix with K-linearly independent columns
[Two-sided annihilation of B]
For every sin P,
[
- (L_s-1)^ell,B=0, B^{mathsf T}sfT(L_s-1)^ell=0.
+ (L_s-1)^ell,B=0, B^{T}-1)^ell=0.
]
Proof.
@@ -748,11 +673,11 @@ 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
+x+s=y and 0 otherwise. Hence (L_s^{T}{mathbf e_y,bfe_y,mathbf e_x}bfe_x
+is 1 when x-s=y, so L_s^{T}{-s}. Consequently
+(L_s-1)^{T}{-s}-1, and
[
- B^{mathsf T}sfT(L_s-1)^ell = ((L_{-s}-1)^ell B)^{mathsf T}sfT = 0^{mathsf T}sfT=0,
+ B^{T}-1)^ell = ((L_{-s}-1)^ell B)^{T} = 0^{T},
]
using the first identity with -s in place of s.
@@ -768,7 +693,7 @@ 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
+Thus h is the number of triples (a_1,a_2,a_3)inN^3 with
sum a_jleell-1. Stars and bars (distribute leell-1 identical
balls into 3 labelled bins) gives
[
@@ -811,7 +736,7 @@ 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)^PbbZ)^P is finite abelian of order
+A_0=(Z/MbbZ/MZ)^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
@@ -820,7 +745,7 @@ 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
+(slambda)(a)=lambda((-s)· 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}.
@@ -838,7 +763,7 @@ a group element. So D_lambda^{-1} exists and is diagonal.
The matrix
- N_lambda=B^{mathsf T}sfT,D_lambda^{-1},B
+ N_lambda=B^{T},D_lambda^{-1},B
is a symmetric htimes h matrix over~K.
@@ -878,7 +803,7 @@ 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)
+ Q(tau):=det!(B^{T}{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
@@ -936,21 +861,21 @@ 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.
+Mat_{p^3}(K), via u(s)|-> L_s and c|-> D_lambda.
Explicit projection.:
Define
- T_lambda=B,N_lambda^{-1}B^{mathsf T}sfTD_lambda^{-1}
+ T_lambda=B,N_lambda^{-1}B^{T}{-1}
inMat_{p^3}(K).
-Because N_lambda=B^{mathsf T}sfTD_lambda^{-1}B, we have
+Because N_lambda=B^{T}{-1}B, we have
[
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,
+im B⊆im T_lambda; and it factors through B^{T},
so im T_lambda⊆im B. Hence im T_lambda=im B and
rankT_lambda=h.
@@ -966,7 +891,7 @@ 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}sfTD_lambda^{-1}
+ =(L_s-1)^ell B;N_lambda^{-1}B^{T}{-1}
=0,
]
using (L_s-1)^ell B=0 from Lemma~lem:B-annihilation. The
@@ -974,10 +899,10 @@ 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^{T}-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.
+ B^{T}{-1}(D_lambda L_s D_lambda^{-1}-1)^ell
+ =B^{T}-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].
@@ -985,10 +910,10 @@ 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 RscrR be the set
+in each P-orbit that contains one, and let R be the set
of chosen characters. Define
- e=sum_{lambdainmathscr R}scrRe_lambda in K[H].
+ e=sum_{lambdainR} in K[H].
By orthogonality the sum is an idempotent: e^2=e. The two
annihilations persist on~e.
@@ -998,17 +923,17 @@ 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|.
+ d:=dim_K(eK[H])=p^3,h,|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).scrR|>(|A_0|)/(2p^3).
+ |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|)
+ frac dw=(p^3h|R|)/(|A_0|p^3)scrR|)/(|A_0|p^3)
+ =(h|R|)/(|A_0|)scrR|)/(|A_0|)
>(h)/(2p^3)>frac1{96},
]
using h/p^3>1/48 from Lemma~lem:h.
@@ -1179,7 +1104,7 @@ 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
+ε,phiin 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.
@@ -1190,7 +1115,7 @@ 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
+εphi 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.
@@ -1198,13 +1123,13 @@ 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
- (εvarphi)^{-1}DE
- =varphi^{-1}(ε^{-1}D),varphi,(varphi^{-1}E).
+ (εphi)^{-1}DE
+ =phi^{-1}(ε^{-1}D),phi,(phi^{-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:
- p=q=1: merged value varphi^{-1}cdot 1cdotvarphicdot 1=1;
- pne1, q=1: merged value varphi^{-1}pvarphine1 (conjugate
+Let p=ε^{-1}D (old X's value) and q=phi^{-1}E (old
+Y's value). The merged value is phi^{-1}pphi· q:
+ p=q=1: merged value phi^{-1}· 1·phi· 1=1;
+ pne1, q=1: merged value phi^{-1}pphine1 (conjugate
of nonidentity);
p=1, qne1: merged value qne1;
pne1, qne1: excluded (at most one exceptional face).
@@ -1232,10 +1157,10 @@ 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
+Let A=(εphi)^{-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:
+ε^{-1}Dphi^{-1}E, whose cyclic shift starting at barbar ff
+is (phi^{-1}Bphi)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
@@ -1344,7 +1269,7 @@ completes the proof of Lemma~lem:triangle-free-gluing.
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),
+P with A_0=(Z/MbbZ/MZ)^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
@@ -1364,11 +1289,11 @@ vectors. We must verify the short-word condition.
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
+ is u(lambda·(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
+ is v(lambda·(1,barbar j,j,barbar j^2))j^2)) for some
lambdainF_p^*.