1 Introduction
Let đ˝p be the field of order p, for a fixed prime p. Let đŠ be the Nottingham group over đ˝p which is the group of all formal power series fâ˘(x)âđ˝pâ˘[[x]] with leading term x, under formal substitution: given g,fâđŠ set (fâ˘g)â˘(x)=fâ˘(gâ˘(x)). For brevity, f is used instead of fâ˘(x) if it is not confusing. Consider the following chain of open subgroups of đŠ:
đŠk={gâđŠ:g=xâ˘(1+âi=kâÎąiâ˘xi),Îąiâđ˝p}âfor each â˘kââ.
The depth of fâđŠâ{1}, Dâ˘(f), is the integer kâĽ1 such that fâđŠkâđŠk+1, while Dâ˘(1)=â. The distance of f and g is usually defined by
đŠâ˘dâ˘(g,f)=p-Dâ˘(fâ˘g-1)
which makes đŠ an ultrametric space. It is clear that dâ˘(f,g)=dâ˘(fj,gj) if pâ¤j. So in understanding the distance of powers the crucial case is when the exponent j is a power of p.
The following definition is by Keating [3].
Definition 1.
Let nâĽkâĽ1 and let k0 be the least nonnegative residue of k modulo p.
Set
eâ˘(k,n)={0if pâŁk and n=k,1if pâŁk,pâŁn, and n>k,0if pâŁk and pâ¤n,iif â˘pâ¤kâ˘Â and â˘nâĄ2â˘k-iâ˘(modâ˘p)â˘Â for some â˘0â¤iâ¤k0,k0if â˘pâ¤kâ˘Â and â˘nâ˘2â˘k-iâ˘(modâ˘p)â˘Â for all â˘0â¤iâ¤k0.
He defined this constant to prove the following theorem [3, Corollary 2] and to show that for m=1 the bound is sharp [3, Theorem 1â(b)].
Theorem 1.1 (Keating).
Let p be a prime and nâĽkâĽ1. Suppose f,gâN are such that Dâ˘(f)âĽk and Dâ˘(gâ˘f-1)âĽn. Then for all mâĽ1 we have
Dâ˘(gpmâ˘f-pm)âĽn+(pm-1)â˘k+pm-pp-1â˘k0+eâ˘(k,n).
He conjectured that the bound was sharp for every m>1 as well. In this paper we confirm his conjecture except for the case p=2, k=1, when it is not sharp for m>2.
Theorem 1.2.
Let p be a prime and nâĽkâĽ1. If p=2, then assume k>1. Then there exist f,gâN such that Dâ˘(f)=k and Dâ˘(gâ˘f-1)=n and for all mâĽ1 we have
Dâ˘(gpmâ˘f-pm)=n+(pm-1)â˘k+pm-pp-1â˘k0+eâ˘(k,n).
For p=2, k=1 we have Dâ˘(g2mâ˘f-2m)âĽn+2m+2-11+eâ˘(k,n) which is strictly stronger than the bound of Theorem 1.1 if m>2. This and the exact bounds for the various values of n are treated fully in [1].
Theorems 1.1 and 1.2 have the following metric version.
Theorem 1.3.
Let d1=p-nâ¤d2=p-k. Suppose that f,gâN are such that dâ˘(f,g)â¤d1 and dâ˘(f,1)â¤d2. Let j=pmâ˘jⲠbe an integer such that pâ¤jâ˛. Then
dâ˘(fj,gj)â¤d1â˘d2pm-1â˘p-(pm-pp-1â˘k0+eâ˘(k,n))
and for every choice of d1,d2,j (except for p=2, d2=p-1) there exist f,g as above such that equality holds.
Crucial in the proof is the understanding of the Engel words. The general formulas (see Lemma 2.3 and Lemma 2.4), albeit aesthetic, are not sufficient for direct constructions. But reinterpreting them in linear algebraic terms allows direct computation of the leading terms of [a,sâ˘(p-1)b] for arbitrary sâĽ0, see the formulation in Corollary 3.4.
The choice for f,g in Theorem 1.2 provides controllably increasing sequences Dâ˘(f)<Dâ˘(fp)<Dâ˘(fp2)<⯠and Dâ˘(g)<Dâ˘(gp)<Dâ˘(gp2)<âŻ. In fact, these are the slowest possible, see Lemma 2.2.
Corollary 1.4.
Let p>2 be prime, and nâĽkâĽ1 integers. Suppose that f,gâN are such that Theorem 1.2 is satisfied with them. Then for all mâĽ1 we have
(1.1)Dâ˘(fpm)=pmâ˘k+pm-1p-1â˘k0.
If n>k, then we also have
Dâ˘(gpm)=pmâ˘k+pm-1p-1â˘k0.
2 Preliminaries
First we take a closer look at the commutators in the Nottingham group. Let a,bâđŠ. The commutator of a and b is [a,b]=[a,1b]=a-1b-1ab. The Engel word [a,mb] is defined inductively as [a,mb]=[[a,m-1b],b] for mâĽ2.
Lemma 2.1.
Let g,hâN with Dâ˘(g)=n, Dâ˘(h)=k. Write g=x+Îłâ˘(x) and h=δâ˘(x), where degâĄ(Îłâ˘(x))=n+1 and degâĄ(δâ˘(x)-x)=k+1. Then
[g,h]âĄx-Îłâ˘(x)+Îłâ˘(δâ˘(x))δâ˛â˘(x)â(modâ˘x2â˘n+k+1),
where δâ˛â˘(x) is the formal derivative of δâ˘(x).
Proof.
We know that hâ˘g=δâ˘(x+Îłâ˘(x)) and gâ˘h=δâ˘(x)+Îłâ˘(δâ˘(x)). Let us denote
Îľâ˘(x)=-Îłâ˘(x)+Îłâ˘(δâ˘(x))δâ˛â˘(x).
To prove the claim, we have to verify
(2.1)δâ˘(x+Îľâ˘(x)+Îłâ˘(x+Îľâ˘(x)))âĄÎ´â˘(x)+Îłâ˘(δâ˘(x))â˘(modâ˘x2â˘n+k+1).
Note first that
Îłâ˘(δâ˘(x))âĄÎłâ˘(x)+Îłâ˛â˘(x)â˘(δâ˘(x)-x)â˘(modâ˘xn+2â˘k+1),
so
Îľâ˘(x)=Îłâ˘(δâ˘(x))-Îłâ˘(x)â˘Î´â˛â˘(x)δâ˛â˘(x)
âĄÎłâ˛â˘(x)â˘(δâ˘(x)-x)-Îłâ˘(x)â˘(δâ˛â˘(x)-1)δâ˛â˘(x)â˘(modâ˘xn+2â˘k+1).
Here the numerator has degree at least n+k+1 (with equality if and only if nâ˘kâ˘(modâ˘p)), so Îľâ˘(x) has degree at least n+k+1. Consequently,
Îłâ˘(x+Îľâ˘(x))âĄÎłâ˘(x)â˘(modâ˘x2â˘n+k+1).
The required approximation (2.1) follows:
δâ˘(x+Îľâ˘(x)+Îłâ˘(x+Îľâ˘(x)))âĄÎ´â˘(x+Îľâ˘(x)+Îłâ˘(x))
=δâ˘(x+Îłâ˘(δâ˘(x))δâ˛â˘(x))
âĄÎ´â˘(x)+Îłâ˘(δâ˘(x))â˘(modâ˘x2â˘n+k+1).â
For later use we state the following two simple corollaries of Lemma 2.1, which are well known and fundamental for the Nottingham group. The first was noted for Îľâ˘(x) in the proof, the second follows from the first:
(2.2)Dâ˘([g,h])âĽDâ˘(g)+Dâ˘(h),with â=â if and only if â˘Dâ˘(g)â˘Dâ˘(h)â˘(modâ˘p),Dâ˘(gp)âĄDâ˘(g)â˘(modâ˘p),if â˘gpâ 1.
By [2, Lemma 1], Dâ˘(gp)âĽpâ˘Dâ˘(g) if gpâ 1. Combined with (2.2) we get the form we will use.
Lemma 2.2.
Let k0 be the remainder of Dâ˘(g) divided by p. Then
Dâ˘(gp)âĽpâ˘Dâ˘(g)+k0.
By induction, for every mâĽ1,
Dâ˘(gpm)âĽpmâ˘Dâ˘(g)+pm-1p-1â˘k0.
Theorem 1.1 is, in fact, a generalisation of this lemma.
The next two results will be used in the proof of the main theorem but they are also of independent interest. The first is a generalisation of Lemma 2.1. Note that it also holds when b=1, k=â.
Lemma 2.3.
Let a,bâN with Dâ˘(a)=n and Dâ˘(b)=k.
Suppose that aâ˘(x)=x+Îąâ˘(x) and
bâ˘(x)=βâ˘(x). Set β0â˘(x)=x, β1â˘(x)=βâ˘(x), βiâ˘(x)=βi-1â˘(βâ˘(x))
for iâĽ1. Then
(2.3)[a,mb]âĄx+âi=0m(mi)(-1)m+iÎąâ˘(βiâ˘(x))βiâ˛â˘(x)(modx2â˘n+mâ˘k+1).
Moreover, if pâ¤k and either mâĽp+1 or m=p,pâ¤2â˘n-k, then (2.3) is valid modulo x2â˘n+mâ˘k+2.
Proof.
The proof is by induction on m. The m=1 case is exactly Lemma 2.1.
Suppose that we have the result for mâĽ1, that is we can write [a,mb]=cd, where
c=x+âi=0m(mi)â˘(-1)m+iâ˘Îąâ˘(βiâ˘(x))βiâ˛â˘(x)
and Dâ˘(d)âĽ2â˘n+mâ˘k. Now
[a,m+1b]=[cd,b]=[c,b][c,b,d][d,b],
and here Dâ˘([c,b,d])âĽDâ˘(c)+Dâ˘(d)+Dâ˘(b)>2â˘n+(m+1)â˘k. On the other hand we have
Dâ˘([d,b])âĽDâ˘(b)+Dâ˘(d)
with equality if and only if Dâ˘(b)â˘Dâ˘(d)â˘(modâ˘p), see (2.2). So
[a,m+1b]âĄ[c,b]â(modx2â˘n+(m+1)â˘k+1)
holds unconditionally. But even
[a,m+1b]âĄ[c,b]â(modx2â˘n+(m+1)â˘k+2)
holds if
(2.4)Dâ˘(d)>2â˘n+mâ˘k,orâk=Dâ˘(b)âĄ2â˘n+mâ˘kâ˘(modâ˘p).
If pâ¤k, then there is a unique 1â¤m0â¤p which is a solution of
kâĄ2â˘n+m0â˘kâ˘(modâ˘p).
This m0=p if only if pâŁ2n-k. For m=m0 the second condition of (2.4) holds, while for m>m0 the first does.
For [c,b] we apply Lemma 2.1. As Dâ˘(c)âĽDâ˘(a)+mâ˘Dâ˘(b)=n+mâ˘k by (2.2), the congruence is valid modulo x2â˘n+(2â˘m+1)â˘k+1
[c,b]âĄx-âi=0m(mi)â˘(-1)m+iâ˘Îąâ˘(βiâ˘(x))βiâ˛â˘(x)+âi=0m(mi)â˘(-1)m+iâ˘Îąâ˘(βiâ˘(βâ˘(x)))βiâ˛â˘(βâ˘(x))â˘Î˛â˛â˘(x)
âĄx+âi=0m(mi)â˘(-1)m+i+1â˘Îąâ˘(βiâ˘(x))βiâ˛â˘(x)+âi=0m(mi)â˘(-1)m+iâ˘Îąâ˘(βi+1â˘(x))βi+1â˛â˘(x)
âĄx+(-1)m+1â˘Îąâ˘(x)+Îąâ˘(βm+1â˘(x))βm+1â˛â˘(x)
+âi=1m((mi)+(mi-1))â˘(-1)m+i-1â˘Îąâ˘(βiâ˘(x))βiâ˛â˘(x)
âĄx+âi=0m+1(m+1i)â˘(-1)m+1+iâ˘Îąâ˘(βiâ˘(x))βiâ˛â˘(x)â˘(modâ˘x2â˘n+(2â˘m+1)â˘k+1)
Thus we are done by induction.
â
Lemma 2.4.
Let a,bâN with Dâ˘(a)=n and Dâ˘(b)=k. Then
(2.5)[a,mbpl]âĄ[a,mâ˘plb]â(modx2â˘n+mâ˘plâ˘k+1)âfor m,lâĽ1.
Moreover, if pâ¤k, then (2.5) is valid modulo x2â˘n+mâ˘plâ˘k+2, unless p>2 and m=l=1.
Proof.
Note that bpl is of depth at least plâ˘Dâ˘(b) by Lemma 2.2. Note also that
(mâ˘plj)=0âif plâ¤j
and
(mâ˘pliâ˘pl)=(mi)
in đ˝p, a field of characteristic p. We also clearly have (-1)m+j=(-1)plâ˘(m+j). So over đ˝p
x+âi=0m(mi)â˘(-1)m+iâ˘Îąâ˘(βiâ˘plâ˘(x))βiâ˘plâ˛â˘(x)=x+âj=0mâ˘pl(mâ˘plj)â˘(-1)mâ˘pl+jâ˘Îąâ˘(βjâ˘(x))βjâ˛â˘(x).
By Lemma 2.3 the first expression is [a,mbpl] modulo x2â˘n+mâ˘Dâ˘(bpl)+1, while the second is [a,mâ˘plb] modulo x2â˘n+mâ˘plâ˘k+1, as required.
If pâ¤k then Dâ˘(bpl)>plâ˘k, by Lemma 2.2, so we get the refinement from Lemma 2.3. (Note that for p=2, kâĄ2â˘nâ˘(modâ˘p) implies that k is even.)
â
Now we will describe Keatingâs constructions and what we will need from his results. Let b,uâđŠ such that
b=x+xk+1â˘Î˛â˘(x),
where βâ˘(x)=rk+rk+1â˘x+rk+2â˘x2+âŻ, and
Dâ˘(u)=nâĽk+k0.
Assume pâ¤k. For sâ¤k, hâĽ1 we construct matrices Ah:=Ah,n,sâMsĂsâ˘(đ˝p). This matrix Ah is a linear operator that takes s-many coefficients of [u,h-1b] starting from x(h-1)â˘k+n+j+1 and gives s-many coefficients of [u,hb] starting from xhâ˘k+n+j+1. This matrix is constructed formally using formula (2.9), below.
The product Î p-1:=A1â˘A2â˘âŻâ˘Ap-1 will be our main concern. It is an operator that takes s-many âleadingâ coefficients of u and gives s-many âleadingâ coefficients of [u,p-1b].
Define w1=u and wh+1=[wh,b] for hâĽ0. Following [3], we determine the coefficients of wh+1 from those of
wh=x+x(h-1)â˘k+n+1â˘Î¸â˘(x),
where θâ˘(x)=t0+t1â˘x+t2â˘x2+âŻ:
(2.6)[wh,b]âĄx+((h-2)â˘k+n)â˘xhâ˘k+n+1â˘Î˛â˘(x)â˘Î¸â˘(x)
+xhâ˘k+n+2â˘(βâ˘(x)â˘Î¸â˛â˘(x)-βâ˛â˘(x)â˘Î¸â˘(x))â˘(modâ˘x(h+1)â˘k+n+1).
It follows from the expansion above that for 0â¤jâ¤k-1 the coefficient of xhâ˘k+n+j+1 in wh+1 is
(2.7)âi=0j((h-2)â˘k+n+2â˘i-j)â˘rk+j-iâ˘ti=âi=0j(nh+2â˘i-j)â˘rk+j-iâ˘ti,
using the abbreviation
(2.8)nh=(h-2)â˘k+n.
For sâ¤k and hâĽ1 we define the matrix Ah=Ah,n,sâMsĂsâ˘(đ˝p) as
(2.9)Ah=(nhâ˘rk(nh-1)â˘rk+1(nh-2)â˘rk+2âŚ(nh-s+1)â˘rk+s-10(nh+1)â˘rknhâ˘rk+1âŚ(nh-s+3)â˘rk+s-2âŽâŽâąâŽâŽ000âŚ(nh+s-1)â˘rk).
Clearly Ah depends only on s, b and the remainder of n modulo p.
Let now νhâđ˝ps be the row vector whose entries are the coefficients of x(h-1)â˘k+n+j+1 in wh for 0â¤jâ¤s-1. It follows from (2.7) that
νh+1=νhâ˘Ah,n,s.
For hâĽ1 define a matrix Î h=Î h,n,sâMsĂsâ˘(đ˝p) by setting Î h=A1â˘A2â˘âŻâ˘Ah. Then we have νp=ν1â˘Î p-1.
Recall now the definition of k0 and eâ˘(k,n) from Definition 1. Keating introduced the matrices above for s=eâ˘(k,n)+1, and we follow his notation. He went on to show that in his eâ˘(k,n)+1Ăeâ˘(k,n)+1 matrix Î p-1 the first eâ˘(k,n) columns are 0 (see [3, Cases 3â4]). (NB: He claimed this only for n>(p-1)â˘k+p, because he used it only in that case, however it works also for arbitrary nâĽk+k0.) Using our extended notation, we may restate his result as follows:
Lemma 2.5.
For pâ¤k and e=eâ˘(k,n) we have
Î p-1,n,e=0.
From this we instantly get:
Corollary 2.6.
Let pâ¤k>k0 so 2â˘k0<k and write e=eâ˘(k,n). For a fixed s<2â˘k0+1
denote by T the top right-hand corner s-eĂs-e block of Î p-1,n,s and let νⲠbe the first s-e coefficients of ν1. The first e entries of νp are 0. The remaining s-e entries are νâ˛â˘T.
Proof.
The top left-hand corner eĂe block of Î p-1,n,s is zero by Lemma 2.5. Similarly, the bottom right-hand corner eĂe block of Î p-1,n,s, which is exactly Î p-1,n+e,e, is zero by Lemma 2.5. As the matrix Î p-1,n,s is a triangular matrix, nonzero entries of Î p-1,n,s only occur in the top right-hand corner s-eĂs-e block of Î p-1,n,s, which is called T. So, the first e entries of νp are 0 and the remaining s-e entries are νâ˛â˘T.
â
In the rest of the paper our concern will be this top right-hand corner block, T.
3 Constructions
Lemma 4 of [3] can be generalized as the following lemma, the proof being almost identical.
Lemma 3.1.
Suppose that nâ˛>nâĽk are such that Theorem 1.1 holds for (k,n), and Theorem 1.2 holds for (k,nâ˛). If
n+eâ˘(k,n)=nâ˛+eâ˘(k,nâ˛),
then Theorem 1.2 also holds for (k,n).
Consider a fixed k. By the definition of eâ˘(k,n), the sum n+eâ˘(k,n) is constant in the intervals k+tâ˘pâ¤nâ¤k+tâ˘p+k0 (with tâĽ0). Lemma 3.1 allows us to assume that k+k0+tâ˘pâ¤n<k+(t+1)â˘p. If n=k+k0+tâ˘p, then eâ˘(k,n)=0, otherwise eâ˘(k,n)=k0. We are going to assume this in the proof of Theorem 1.2, below.
Let pâ¤k and k+k0+tâ˘pâ¤nâ¤k+(t+1)â˘p for some nonnegative integer t. We define
eâ˛â˘(k,n)={0if â˘k+k0+tâ˘p<n<k+(t+1)â˘p,k0if â˘n=k+k0+tâ˘pâ˘Â or â˘n=k+(t+1)â˘p.
This describes the number of significant digits for the following definition. For uâđŠn let Îźnâ˘(u)âđ˝peâ˛â˘(k,n)+1 denote the vector consisting of the first eâ˛â˘(k,n)+1 coefficients of u.
For given n,k as above and Îťâđ˝p we define an eâ˛â˘(k,n)+1Ăeâ˛â˘(k,n)+1 matrix Tnâ˘[Îť] as
(-Îť0âŚ02â˘Îť20Îť0âą0âŽâŽâą0âŽ00âŚÎť0-10âŚ02â˘Îť)âif â˘nâĄkâ˘(modâ˘p),
and
(-10âŻ02â˘Îť00âŻ00âŽâŽâŽâŽâŽ00âŻ00)âif â˘nâĄ2â˘kâ˘(modâ˘p),
and Tnâ˘[Îť]=(Îť) if k+k0+pâ˘t<n<k+pâ˘(t+1).
Lemma 3.2.
Let k>p, pâ¤k and k+k0+pâ˘tâ¤nâ¤k+(t+1)â˘p for some nonnegative integer t. Define b=x+xk+1+xk+k0+1.
For every uâNn we have [u,p-1b]âNn* and
Îźn*([u,p-1b])=Îźn(u)Tn[-1],
where n*=n+(p-1)â˘k+eâ˘(k,n).
Proof.
We have to show that the top right-hand eâ˛â˘(k,n)+1Ăeâ˛â˘(k,n)+1 block of Î p-1,n,eâ˘(k,n)+eâ˛â˘(k,n)+1 is Tnâ˘[-1], by Corollary 2.6.
From the definition in (2.9) and the choice of our b we see that in the matrices Ah the entries (i,j) are 0 unless j-iâ{0,k0}. Consequently, in the products Î h=A1â˘âŻâ˘Ah the entries (i,j) are 0 unless j-i is a nonnegative integer multiple of k0.
First we deal with the case nâĄ2â˘kâ˘(modâ˘p). Pick some mâĄkâ˘(modâ˘p). Now eâ˘(k,n)=0 and eâ˘(k,m)=k0, while eâ˛â˘(k,n)=eâ˛â˘(k,m)=k0. From the definition in (2.9) we also see the following block decomposition of the matrices Ah,m,2â˘k0+1,Î h,m,2â˘k0+1âM(2â˘k0+1)Ă(2â˘k0+1)â˘(đ˝p) for hâĽ1:
(3.1)Ah,m,2â˘k0+1=(Ah,m,k0*0Ah,n,k0+1),Î h,m,2â˘k0+1=(Î h,m,k0*0Î h,n,k0+1).
It is enough to compute the larger matrices Ah,m,2â˘k0+1âM(2â˘k0+1)Ă(2â˘k0+1)â˘(đ˝p), and their product Î p-1,m,2â˘k0+1âM(2â˘k0+1)Ă(2â˘k0+1)â˘(đ˝p).
For convenience we index the rows and columns of Ah from 0 to 2â˘k0. By (2.9), the (i,j) entry of Ah for 0â¤iâ¤jâ¤2â˘k0 is
ahâ˘iâ˘j={mh+iif â˘i=j,mh-k0+iif â˘j=k0+i,0otherwise,
where we used again the convention (2.8).
Note that
mh+k0=(h-2)â˘k+m+k0âĄmh+1â˘(modâ˘p).
Thus Ah has the form
0k02â˘k00mh0âŚ0mh-10âŚ000mh+10âŚ0mh-1+10âŚ0âŽâąâą0âŚ0âą0âŽ00âąâąâąâŚ0âą0k00000mh+10âŚ0mhâŽâŽâŽâŽâąâąâąâŚ0âŽâŽâŽâŽâŽâŽâąâŽâŽâŽâŽâŽâŽâŽâŽâŽâą02â˘k000000000mh+2.
As remarked above, the (i,j) entry of Î h is zero if j-iâ{0,k0,2â˘k0}. Now we determine the rest.
By induction on hâĽ1, we get that the (i,i+k0) entry of Î h is
(ât=1h-1(m1+i)â˘(m2+i)â˘âŻâ˘(mt+i)2â˘(mt+3+i)â˘(mt+4+i)â˘âŻâ˘(mh+1+i))
(3.2)+(m0+i)â˘(m3+i)â˘âŻâ˘(mh+1+i).
Finally, evaluating for h=p-1, we get that the (i,i+k0) entry of Î p-1 is
(3.3)Ďi,i+k0=ât=0p-2(mt+i)â˘âj=3p(mt+j+i).
Note that each summand in (3.3) is a product of p-2 consecutive numbers from {mt+i}t=1p, the last with multiplicity 2. Pick the unique lâ{1,âŚ,p} such that ml+i=0. So âjâ l(mj+i)=-1 and hence
(ml-1+i)â˘âj=3p(ml-1+j+i)=--k0k0=1,
(ml-2+i)â˘âj=3p(ml-2+j+i)=--2â˘k0-k0=-2,
and
(3.4)Ďi,i+k0={-2+1=-1if â˘lâ 1,p,-2if â˘l=p,1if â˘l=1.
By Lemma 2.5 we know that the first k0 columns and the last k0 rows of Î p-1 are zero. The remaining diagonal entry is
Ďk0,k0=m2â˘m3â˘m4â˘âŻâ˘mp-1=(2â˘k0)â˘(2â˘k0)â˘(3â˘k0)â˘âŻâ˘((p-2)â˘k0)=-1.
To determine Ď0,2â˘k0 note that ap-1,2â˘k0,2â˘k0=mp+1=0 and ap-1,k0,2â˘k0=mp-1 in Ap-1. So Ď0,2â˘k0 is mp-1 times the (0,k0) entry of Î p-2. Using (3.2), we get
Ď0,2â˘k0=m0â˘m3â˘m4â˘âŻâ˘mp-1â˘mp-1=(-k0)â˘(2â˘k0)â˘(3â˘k0)â˘âŻâ˘((p-2)â˘k0)2=2.
Thus, we have determined the top right-hand corner k0+1Ăk0+1 matrix of Î p-1,m,2â˘k0+1
corresponding to mâĄkâ˘(modâ˘p), it is
(3.5)Tmâ˘[-1]=(10âŚ020-10âŚ0âŽâŽâą0âŽ00âŚ-10-10âŚ0-2).
By formulas (3.1) and (3.5), the top right-hand corner k0+1Ăk0+1 matrix of Î p-1,n,k0+1âM(k0+1)Ă(k0+1)â˘(đ˝p) corresponding to nâĄ2â˘kâ˘(modâ˘p), is
Tnâ˘[-1]=(-10âŻ0-200âŻ00âŽâŽâąâŽâŽ00âŻ00)
We now turn to the case of k+k0+pâ˘t<n<k+pâ˘(t+1), that is, we have nâ˘2â˘k-iâ˘(modâ˘p) for any 0â¤iâ¤k0. So eâ˘(k,n)=k0 and eâ˛â˘(k,n)=0. Then the corresponding matrix Ah,n,k0+1 is
0k00nh0âŚ0nh-10nh+10âŚ0âŽâąâą0000âąâąâŽk00000nh+1.
As above, we determine the (i,j) entries of Î p-1=Î p-1,n,k0+1:
Ď0,k0=(ât=1p-2n1â˘n2â˘âŻâ˘nt2â˘nt+3â˘nt+4â˘âŻâ˘np)+n0â˘n3â˘âŻâ˘np=ât=0p-2ntâ˘âj=3pnt+j.
Note that we get n1=n-kâ 0, np=(p-2)â˘k+nâ 0 by our assumptions, so Ď0,k0=-1 as in (3.4). For the diagonal entries we again get
Ďi,i=ât=1p-1(nt+i)â{0,-1},
with Ďi,i=-1 if and only if 0âĄn0+i=n-2â˘k+iâ˘(modâ˘p). In our case we have nâ˘2â˘k-iâ˘(modâ˘p) for any 0â¤iâ¤k0 so all Ďi,i=0. We cut the first k0 columns and last k0 rows and we get the matrix
Tnâ˘[-1]=(-1).
â
Lemma 3.3.
Let p>2. Suppose k=k0<p and k+k0+pâ˘tâ¤nâ¤k+(t+1)â˘p for some nonnegative integer t. If k<p-1, then let b=x+xk+1, otherwise let b=x+xk+1+x2â˘k+1. For every uâNn we have [u,p-1b]âNn* and
Îźn*([u,p-1b])=Îźn(u)Tn[Îť],
where n*=n+(p-1)â˘k+eâ˘(k,n) and the nonzero ÎťâFp is
Îť={k-12=-1if â˘k=p-1,k+12if â˘k<p-1.
Proof.
The proof follows the argument of Lemma 3.2. However, as k is small, we have to be concerned with more terms than in (2.6) and in (2.7). We extend the dimension of the matrices to eâ˘(k,n)+eâ˛â˘(k,n)+1â¤2â˘k+1. Then we can use Lemma 2.5 and it remains to show that the top right-hand eâ˛â˘(k,n)+1Ăeâ˛â˘(k,n)+1 block of Î p-1,n,eâ˘(k,n)+eâ˛â˘(k,n)+1 is Tnâ˘[Îť].
The refinement of (2.6) (using nâĽ2â˘k) for uh+1 reads as follows:
(3.6)[uh,b]âĄx+((h-2)â˘k+n)â˘Î˛â˘(x)â˘Î¸â˘(x)â˘xhâ˘k+n+1
+(βâ˘(x)â˘Î¸â˛â˘(x)-βâ˛â˘(x)â˘Î¸â˘(x))â˘xhâ˘k+n+2
+(((h-1)â˘k+n+12)-(k+1)â˘((h-2)â˘k+n))
Ăβâ˘(x)2â˘Î¸â˘(x)â˘x(h+1)â˘k+n+1
+(((h-2)k+n)β(x)2θâ˛(x)
+((3-h)k-n+1)β(x)βâ˛(x)θ(x))x(h+1)â˘k+n+2
+(12β(x)2θâ˛â˛(x)+βâ˛(x)2θ(x)
-β(x)βâ˛(x)θâ˛(x))x(h+1)â˘k+n+3
+Eh,nâ˘(modâ˘x(h+2)â˘k+n+2).
Here, for some qh,n,qh,nâ˛âđ˝p,
Eh,n={qh,nâ˘x(h+2)â˘k+n+1â˘Î˛â˘(x)3â˘Î¸â˘(x)if â˘(h-1)â˘k+n>2â˘k,qh,nâ˘x(h+2)â˘k+n+1â˘Î˛â˘(x)3â˘Î¸â˘(x)+qh,nâ˛â˘Î˛â˘(x)â˘Î¸â˘(x)2if â˘(h-1)â˘k+n=2â˘k.
The coefficient of xhâ˘k+n+j+1 in uh+1 for 0â¤j<k is the same as in (2.7), and for kâ¤j<2â˘k it is
(3.7)âi=1k+a((h-2)â˘k+n+2â˘i-j)â˘rk+j-iâ˘ti
+((nh+a2)+(k+12))rk2ta+âi=01c1â˘irk+irk+1-i)ta-1
+(âi=02c2â˘iâ˘rk+iâ˘rk+2-i)â˘ta-2+âŻ+(âi=0acaâ˘iâ˘rk+iâ˘rk+a-i)â˘t0,
where a=j-k and csâ˘wâđ˝p for 1â¤sâ¤a, 0â¤wâ¤a, and wâ¤s. For j=2â˘k there will be an extra term in the coefficient of t0 and the rest will satisfy the above formula. But its value is irrelevant for our proof.
We again get that the (i,j) entry of Î h is 0 unless j-iâ{0,k,2â˘k} as in Lemma 3.2.
As before, we assume first that nâĄ2â˘kâ˘(modâ˘p) and pick some mâĄkâ˘(modâ˘p) and hence eâ˘(k,n)=0 and eâ˘(k,m)=k. As a subcase assume that k<p-1 so b=x+xk+1. Now from (3.7) we compute the entries of the matrix Ah,m,2â˘k+1:
ahâ˘iâ˘j={mh+iif â˘i=j,(mh+i2)+(k+12)if â˘j=i+k,Ch,mif â˘(i,j)=(0,2â˘k),0otherwise.
The value of Ch,mâđ˝p will turn out to be irrelevant. By induction on hâĽ1, we get that the (i,i+k) entry of Î h is
(3.8)(ât=1h(m1+i)(m2+i)âŻ(mt+i)(mt+2+i)(mt+3+i)
âŻ(mh+1+i)mt+i-12)
+(k+12)â˘ât=1h(m1+i)â˘(m2+i)
âŻâ˘(mt-1+i)â˘(mt+2+i)â˘âŻâ˘(mh+1+i).
Consider h=p-1. For i=0 the first sum is 0, because m1=0 is a factor in each product. For i>0 pick the unique lâ{2,âŚ,p} such that ml+i=0 and observe that the first sum has a single nonzero term, that for t=l-1. Its value is k+12. For i=0,k
the second sum has a single nonzero summand, otherwise two. For i=0 the value is -k-12, for i=k it is k+12 and for iâ 0,k the two terms add up to 0. So finally
(3.9)Ďi,i+k={-k-12if â˘i=0,k+1if â˘i=k,k+12otherwise.
For the diagonal entries we again have
Ďi,i=âh=1p-1(mh+i)â{0,-1},
with Ďi,i=-1 if and only if 0âĄm0+i=m-2â˘k+iâĄ-k+iâ˘(modâ˘p). Note that Î p-2 has (0,0) entry 0 since m1=0 and (0,k) entry k+12â˘k by (3.8). Also Ap-1 has (k,2â˘k) entry kâ˘(k+1) and (2â˘k,2â˘k) entry 0. So
Ď0,2â˘k=(k+1)22.
Put Îť=k+12â 0 as k<p-1. Now we have the claimed matrix
Tmâ˘[Îť]=(-Îť0âŚ02â˘Îť20Îť0âŚ0âŽâŽâą0âŽ00âŚÎť0-10âŚ02â˘Îť).
As (3.1) is still valid, we also have
Tnâ˘[Îť]=(-10âŻ02â˘Îť00âŻ00âŽâŽâąâŽâŽ00âŻ00).
Now suppose that k=p-1 so take b=x+xp+x2â˘p-1. We continue to assume that nâĄ2â˘kâ˘(modâ˘p) and mâĄkâ˘(modâ˘p) and hence eâ˘(k,n)=0,eâ˘(k,m)=k. By using (3.6), (3.7) we have the corresponding matrix Ah,m,2â˘k+1 whose (i,j) entries for 0â¤i,jâ¤2â˘k are
ahâ˘iâ˘j={mh+iif â˘i=j,mh+i-k+(mh+i2)+(k+12)if â˘j=k+i,Dh,mif â˘(i,j)=(0,2â˘k),0otherwise.
The value of Dh,mâđ˝p is uninteresting, as before. Again the (i,j) entry of Î h,m,2â˘k+1 is zero if j-iâ{0,k,2â˘k}. By the value of ahâ˘iâ˘i+k we get that the (i,i+k) entry of Î h is the sum of (3.8) and (3.2), so
Ďi,i+k={-k-12+1=-k-12if i=0,k+1-2=k-1if i=k,k+12-1=k-12otherwise.
For the diagonal entries we again have
Ďi,i=âh=1p-1(mh+i)â{0,-1},
with Ďi,i=-1 if and only if 0âĄm0+i=m-2â˘k+iâĄ-k+iâ˘(modâ˘p). Note that Î p-2 has (0,0) entry 0 and (0,k) entry k+12â˘k+-1k=k-12â˘k. On the other hand, Ap-1 has (k,2â˘k) entry -2â˘k+kâ˘(k+1)=kâ˘(k-1) and (2â˘k,2â˘k) entry 0. So
Ď0,2â˘k=(k-1)22.
Put Îť=k-12.
(We use the assumption p>2 to claim Îťâ 0.) We obtained the claimed matrix
Tmâ˘[Îť]=(-Îť0âŚ02â˘Îť20Îť0âŚ0âŽâŽâą0âŽ00âŚÎť0-10âŚ02â˘Îť).
We get Tnâ˘[Îť] from this exactly as above.
Assume finally that k+k0+pâ˘t<n<k+pâ˘(t+1) that is nâ˘2â˘k-iâ˘(modâ˘p) for any 0â¤iâ¤k0, so eâ˘(k,n)=k0. Of course k<p-1. So we take the element b=x+xk+1 and obtain the corresponding matrix Tnâ˘[Îť]=(k+12) (from (3.9)) exactly as in Lemma 3.2. â
We collect in the following corollary what we need for the proof of Theorem 1.2.
Corollary 3.4.
Let pâ¤k and k>1 if p=2. In addition, let k+k0+pâ˘tâ¤nâ¤k+pâ˘(t+1) for some nonnegative integer t. Define
(3.10)b=x+xk+1+xk+k0+1,Îť=-1if â˘kâĽp-1,
b=x+xk+1,Îť=k+12if â˘k<p-1.
Then for every uâNn and for sâĽ1 we have [u,sâ˘(p-1)b]âNn* and
Îźn*([u,sâ˘(p-1)b])=Îťs-1Îźn(u)Tn[Îť],
where n*=n+sâ˘(p-1)â˘k+(s-1)â˘k0+eâ˘(k,n).
Proof.
Suppose that k+k0+pâ˘t<n<k+(t+1), for some integer t. Now eâ˛â˘(k,n)=0. Recall the matrix Tnâ˘[Îť]=(Îť), where Îť is as above, exactly as in Lemmas 3.2 and 3.3. If uâđŠn is such that Îźnâ˘(u)=(Îą), then by these lemmas,
Îźn*([u,sâ˘(p-1)b])=(ÎťsÎą)=Îťs-1Îźn(u)Tn[Îť],
as claimed.
Suppose now that n=k+k0+pâ˘t or n=k+pâ˘(t+1) for some nonnegative integer t. Hence eâ˛â˘(k,n)=k0. Note that n+(p-1)â˘k+eâ˘(k,n)âĄkâ˘(modâ˘p) in both cases, so from the second step there will be no difference between the two cases. Recall the matrices X=Tkâ˘[Îť],Y=T2â˘kâ˘[Îť]âMk0+1Ăk0+1â˘(đ˝p), where Îť is as above, exactly as in Lemmas 3.2 and 3.3. Note that Yâ˘X=Îťâ˘Y and X2=Îťâ˘X.
If n=k+pâ˘(t+1), then these lemmas imply by induction on s that
Îźn*([u,sâ˘(p-1)b])=Îźn(u)Xs=Îťs-1Îźn(u)Tn[Îť],
as claimed.
If n=k+k0+pâ˘t, then these lemmas imply first that
Îźn+(p-1)+eâ˘(k,n)([u,p-1b])=Îźn(u)Y.
Then (using eâ˘(k,k)=k0) we obtain by induction on s that
Îźn*([u,sâ˘(p-1)b])=Îźn(u)YXs-1=Îťs-1Îźn(u)Tn[Îť],
as claimed.
â
Proof of Theorem 1.2.
Suppose that pâ¤k. By the remark before Lemma 3.1 we may assume that k+k0+pâ˘tâ¤n<k+(t+1)â˘p, for some integer tâĽ0. If n=k+k0+pâ˘t, then eâ˘(k,n)=0, otherwise eâ˘(k,n)=k0.
Let g-1=b=x+xk+1+xk+k0+1, or x+xk+1 as in Lemmas 3.2 and 3.3. We pick some u=u0âđŠn of depth n, its leading coefficient, Îąâ 0 will be chosen appropriately later. Set f-1=g-1â˘u and then um=gpmâ˘f-pm.
Let nm=n+(pm-1)â˘k+pm-pp-1â˘k0+eâ˘(k,n). We prove that Dâ˘(um)=nm by induction on m. For that, assume mâĽ1 and that for every smaller index the theorem holds. Note that
nm=nm-1+(p-1)â˘pm-1â˘k+pm-1â˘k0.
If nâĄ2â˘kâ˘(modâ˘p), then for the inductive proof to work we will need information on the subsequent k0=eâ˛â˘(k,nm)=eâ˛â˘(k,n) coefficients of um.
To this end, we recall some facts about formal basic commutators for two generators, a,b. These are defined inductively and are totally ordered. First, a and b are basic commutators of weight 1 and a>b. Second, if x>y are basic commutators of weight p and q, then [x,y] is called a basic commutator of weight p+q provided that if x=[z,w], then yâĽw. The total ordering is arbitrary among basic commutators of equal weights but if x has weight larger than y, then x>y.
Every basic commutator has naturally defined weights, Wa and Wb in the variables: the number of times that variable occurs in the commutator.
More about the formal basic commutators and relevant descriptions of weight can be found in [4, pp.â9â12]. The following finer approximation is from there [4, Proposition 1.1.32â(i)]:
umâĄum-1p[um-1,g-pm-1](p2)[um-1,2g-pm-1](p3)
âŻ[um-1,p-1g-pm-1](modK(g-pm-1,um-1)),
where Kâ˘(g-pm-1,um-1) is the normal closure in đŠ of the set of all formal basic commutators in {g-pm-1,um-1} of weight at least p and of weight at least 2 in um-1 and also the p-th powers of all basic commutators of weight at least 3 and at most p-1 and of weight at least 2 in um-1. The only weight p commutator with weight 2 in um-1 is w=[[um-1,p-2g-pm-1],um-1]. We denote its multiplicity by δ0. (The exact value of δ0 is p2-p-1, but it is irrelevant for our proof.) Now our goal is to show we can approximate um by the commutator [um-1,p-1g-pm-1].
If m=1 and nm-1=n=k+k0, then eâ˘(k,n)=0 and eâ˛â˘(k,n)=k0 so
D([u,p-1g-1])=n+(p-1)k=n+(p-1)k+e(k,n),
by (2.2). The element w in Kâ˘(g-1,u) has depth exactly
2â˘n+(p-2)â˘k=n+(p-1)â˘k+eâ˘(k,n)+eâ˛â˘(k,n),
every other element in Kâ˘(g-1,u) has larger depth. The leading coefficient of w is
(n-k)â˘Îąâ˘nâ˘(n+k)â˘âŻâ˘(n+(p-4)â˘k)â˘Îąâ˘(p-2)â˘k=-2â˘Îą2,
where Îą is the leading coefficient of u.
If mâĽ1 is arbitrary with nm-1>k+k0, then every element in Kâ˘(g-pm-1,um-1) has depth at least
2â˘nm-1+(p-2)â˘k>nm-1+(p-1)â˘k+eâ˘(k,nm-1)+eâ˛â˘(k,nm-1).
From now let mâĽ1 arbitrary. Also note that
D([um-1,ig-pm-1](pi+1))âĽpâ˘(nm-1+k)>nm-1+(p-1)â˘k+eâ˘(k,nm-1)+eâ˛â˘(k,nm-1)
for 1â¤iâ¤p-2 for every nm-1âĽk+k0. Let nm-1ÂŻ denote the remainder of nm-1 modulo p. Then
Dâ˘(um-1p)âĽpâ˘nm-1+nm-1ÂŻ=nm-1+(p-1)â˘k+k0+(p-2)â˘k0+(p-1)â˘(nm-1-k-k0)+nm-1ÂŻ>nm-1+(p-1)â˘k+eâ˘(k,nm-1)+eâ˛â˘(k,nm-1).
We obtain a first version for the coefficient vector
(3.11)Îźnmâ˘(um)
={Îźnm([um-1,p-1g-pm-1])if â˘nm-1>k+k0,Îźn1([u,p-1g-1])+(0,âŚ,0,-2δ0Îą2)if â˘m=1,n=k+k0.
Here δ0 is the exponent of w in u1âĄ[u,p-1g-1]wδ0(modxn1+eâ˛â˘(k,n)+1) in the second case.
Consider m>1. By the last line of Lemma 2.4 (which holds in our case, as p-1=1 would imply p=2),
[um-1,p-1g-pm-1]âĄ[um-1,(p-1)â˘pm-1g-1](modx2â˘nm-1+(p-1)â˘pm-1â˘k+2).
Now
2â˘nm-1+(p-1)â˘pm-1â˘k+2=nm+nm-1-pm-1â˘k0+2=nm+n+(pm-1-1)â˘k+(-pm-1+pm-1-pp-1)â˘k0+eâ˘(k,n)+2>nm+eâ˛â˘(k,nm)+1.
So we obtain Îźnm(um)=Îźnm([um-1,(p-1)â˘pm-1g-1]) for m>1. By induction and by (3.11), we get a second version for the coefficient vector
(3.12)Îźnmâ˘(um)={Îźnm([u1,pm-pg-1])for â˘m>1,Îźnm([u,pm-1g-1])for â˘mâĽ1â˘Â if â˘n>k+k0.
After these preliminary observations we turn to the proof.
First let k+k0+pâ˘t<n<k+(t+1)â˘p. Then eâ˛â˘(k,n)=0 and eâ˘(k,n)=k0, and so nmâĄnâ˘k,2â˘kâ˘(modâ˘p) for every mâĽ0.
By Corollary 3.4 and
(3.12),
D(um)=D([um-1,p-1g-pm-1])=D([u,pm-1g-1])=nm,
as required.
Now suppose that n=k+k0+tâ˘p. Then eâ˛â˘(k,n)=k0 and eâ˘(k,n)=0, and so nmâĄkâ˘(modâ˘p) for every mâĽ1. Recall that Îąâ 0 denotes the leading coefficient of uâđŠn.
By the definition of g-1=b and Îť in (3.10), Corollary 3.4 shows that for the first k0 coefficients of v=[u,p-1g-1] we have
Îźn1â˘(v)=Îźnâ˘(u)â˘Tnâ˘[Îť]=(-Îą,0,âŚ,0,2â˘Îťâ˘Îą).
By (3.11),
Îźn1â˘(gpâ˘f-p)=(-Îą,0,âŚ,0,2â˘Îťâ˘Îą-2â˘Î´â˘Îą2),
where δ=δ0 if n=k+k0 and δ=0 otherwise. The leading coefficient is -Îąâ 0, so
Dâ˘(u1)=Dâ˘(gpâ˘f-p)=n+(p-1)â˘k+eâ˘(k,n)=n1
holds. Now n1âĄkâ˘(modâ˘p) so eâ˛â˘(k,n1)=eâ˘(k,n1)=k0.
We apply Corollary 3.4
again to get
D(u2)=D([u1,p-1g-p])=D([u1,(p-1)â˘pg-1])=n2.
We also obtain
(3.13)Îźn2â˘(gp2â˘f-p2)=(-Îťâ˘Îą+2â˘Î´â˘Îą2,0,âŚ,0,2â˘Îť2â˘Îą-4â˘Îťâ˘Î´â˘Îą2).
The leading coefficient is nonzero if Îąâ 0 and 2â˘Îąâ˘Î´â Îť. Such an Îą exists because Îťâ 0. Fix this Îą. The vector at (3.13) is an eigenvector of Tnmâ˘[Îť] with eigenvalue Îťâ 0 for all mâĽ1.
By Corollary 3.4 and (3.12), we obtain
D(um)=D([um-1,p-1g-pm-1])=D([u1,pm-pg-1]=nm,
as required.
Now it remains to consider the case of pâŁk. Our arguments are similar to those of [3]. Suppose that pâ¤n. Let g,u be arbitrary and set f-1:=g-1â˘u. Then
D(gpf-p)=D([u,p-1g-1])=n+(p-1)kâĄn(modp)
since 2â˘n+(p-2)â˘k>n+(p-1)â˘k. We can proceed with induction to get
D(gpmf-pm)=D([u,pm-1â˘(p-1)g-1])=n+(pm-1)k
for mâĽ1. Suppose finally that pâŁn. If n=k, then let g be arbitrary of depth k and f=gp. Then
gpmâ˘fp-m=gpmâ˘(1-p)
and so
Dâ˘(gpmâ˘fp-m)=pmâ˘k
by [2, Lemma 1]. If n>k, then pick g and f to work for the pair k,n+1. Then eâ˘(k,n)=1=eâ˘(k,n+1)+1 so
we have the result by Lemma 3.1.
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Proof of Corollary 1.4.
Assume the contrary: for a particular mâĽ1 we have
(p-1)â˘Dâ˘(fm)>(pm+1-pm)â˘k+(pm-1)â˘k0.
By Theorem 1.2, we have
Dâ˘(gpmâ˘f-pm)=n1=n+(pm-1)â˘k+pm-pp-1â˘k0+eâ˘(k,n),
where n1âĄnâ˘(modâ˘p) if nâ˘2â˘k-iâ˘(modâ˘p) for any 0â¤iâ¤k0, and otherwise n1âĄk0â˘(modâ˘p). Therefore, in any case eâ˘(k,n1)=k0.
Now apply Theorem 1.1 for gpm,fpm to get
Dâ˘((gpm)pâ˘(fpm)-p)âĽn1+(p-1)â˘Dâ˘(fpm)+k0
>n+(pm-1)â˘k+pm-pp-1â˘k0+eâ˘(k,n)
+(pm+1-pm)â˘k+(pm-1)â˘k0+k0
=n+(pm+1-1)â˘k+pm+1-pp-1â˘k0+eâ˘(k,n)
=Dâ˘(gpm+1â˘f-pm+1).
Here the last equation is the statement of Theorem 1.2 for m+1. The strict inequality is a contradiction, so (1.1) holds.
Now suppose that n>k and hence n-k+eâ˘(k,n)-k0>0.
Since Theorem 1.2 holds for f,g, we have
Dâ˘(g-pmâ˘fpm)=n+(pm-1)â˘k+pm-pp-1â˘k0+eâ˘(k,n)
=n-k+eâ˘(k,n)-k0+pmâ˘k+pm-1p-1â˘k0
>pmâ˘k+pm-1p-1â˘k0
=Dâ˘(fpm),
by the first part. Therefore,
Dâ˘(gpm)=Dâ˘(fpm)=pmâ˘k+pm-1p-1â˘k0,
as claimed.
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