2 Preliminaries
First we take a closer look at the commutators in the Nottingham group. Let a,b∈𝒩. 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 m≥2.
Lemma 2.1.
Let g,h∈N with D(g)=n, D(h)=k. Write g=x+γ(x) and h=δ(x), where deg(γ(x))=n+1 and deg(δ(x)-x)=k+1. Then
[g,h]≡x-γ(x)+γ(δ(x))δ′(x) (modx2n+k+1),
where δ′(x) is the formal derivative of δ(x).
Proof.
We know that hg=δ(x+γ(x)) and gh=δ(x)+γ(δ(x)). Let us denote
ε(x)=-γ(x)+γ(δ(x))δ′(x).
To prove the claim, we have to verify
(2.1)δ(x+ε(x)+γ(x+ε(x)))≡δ(x)+γ(δ(x))(modx2n+k+1).
Note first that
γ(δ(x))≡γ(x)+γ′(x)(δ(x)-x)(modxn+2k+1),
so
ε(x)=γ(δ(x))-γ(x)δ′(x)δ′(x)
≡γ′(x)(δ(x)-x)-γ(x)(δ′(x)-1)δ′(x)(modxn+2k+1).
Here the numerator has degree at least n+k+1 (with equality if and only if n≢k(modp)), so ε(x) has degree at least n+k+1. Consequently,
γ(x+ε(x))≡γ(x)(modx2n+k+1).
The required approximation (2.1) follows:
δ(x+ε(x)+γ(x+ε(x)))≡δ(x+ε(x)+γ(x))
=δ(x+γ(δ(x))δ′(x))
≡δ(x)+γ(δ(x))(modx2n+k+1).∎
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 ε(x) in the proof, the second follows from the first:
(2.2)D([g,h])≥D(g)+D(h),with “=” if and only if D(g)≢D(h)(modp),D(gp)≡D(g)(modp),if gp≠1.
By [2, Lemma 1], D(gp)≥pD(g) if gp≠1. Combined with (2.2) we get the form we will use.
Lemma 2.2.
Let k0 be the remainder of D(g) divided by p. Then
D(gp)≥pD(g)+k0.
By induction, for every m≥1,
D(gpm)≥pmD(g)+pm-1p-1k0.
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=∞.
Lemma 2.3.
Let a,b∈N with D(a)=n and D(b)=k.
Suppose that a(x)=x+α(x) and
b(x)=β(x). Set β0(x)=x, β1(x)=β(x), βi(x)=βi-1(β(x))
for i≥1. Then
(2.3)[a,mb]≡x+∑i=0m(mi)(-1)m+iα(βi(x))βi′(x)(modx2n+mk+1).
Moreover, if p∤k and either m≥p+1 or m=p,p∤2n-k, then (2.3) is valid modulo x2n+mk+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 m≥1, that is we can write [a,mb]=cd, where
c=x+∑i=0m(mi)(-1)m+iα(βi(x))βi′(x)
and D(d)≥2n+mk. Now
[a,m+1b]=[cd,b]=[c,b][c,b,d][d,b],
and here D([c,b,d])≥D(c)+D(d)+D(b)>2n+(m+1)k. On the other hand we have
D([d,b])≥D(b)+D(d)
with equality if and only if D(b)≢D(d)(modp), see (2.2). So
[a,m+1b]≡[c,b] (modx2n+(m+1)k+1)
holds unconditionally. But even
[a,m+1b]≡[c,b] (modx2n+(m+1)k+2)
holds if
(2.4)D(d)>2n+mk,or k=D(b)≡2n+mk(modp).
If p∤k, then there is a unique 1≤m0≤p which is a solution of
k≡2n+m0k(modp).
This m0=p if only if p∣2n-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 D(c)≥D(a)+mD(b)=n+mk by (2.2), the congruence is valid modulo x2n+(2m+1)k+1
[c,b]≡x-∑i=0m(mi)(-1)m+iα(βi(x))βi′(x)+∑i=0m(mi)(-1)m+iα(βi(β(x)))βi′(β(x))β′(x)
≡x+∑i=0m(mi)(-1)m+i+1α(βi(x))βi′(x)+∑i=0m(mi)(-1)m+iα(βi+1(x))βi+1′(x)
≡x+(-1)m+1α(x)+α(βm+1(x))βm+1′(x)
+∑i=1m((mi)+(mi-1))(-1)m+i-1α(βi(x))βi′(x)
≡x+∑i=0m+1(m+1i)(-1)m+1+iα(βi(x))βi′(x)(modx2n+(2m+1)k+1)
Thus we are done by induction.
∎
Lemma 2.4.
Let a,b∈N with D(a)=n and D(b)=k. Then
(2.5)[a,mbpl]≡[a,mplb] (modx2n+mplk+1) for m,l≥1.
Moreover, if p∤k, then (2.5) is valid modulo x2n+mplk+2, unless p>2 and m=l=1.
Proof.
Note that bpl is of depth at least plD(b) by Lemma 2.2. Note also that
(mplj)=0 if pl∤j
and
(mplipl)=(mi)
in 𝔽p, a field of characteristic p. We also clearly have (-1)m+j=(-1)pl(m+j). So over 𝔽p
x+∑i=0m(mi)(-1)m+iα(βipl(x))βipl′(x)=x+∑j=0mpl(mplj)(-1)mpl+jα(βj(x))βj′(x).
By Lemma 2.3 the first expression is [a,mbpl] modulo x2n+mD(bpl)+1, while the second is [a,mplb] modulo x2n+mplk+1, as required.
If p∤k then D(bpl)>plk, by Lemma 2.2, so we get the refinement from Lemma 2.3. (Note that for p=2, k≡2n(modp) implies that k is even.)
∎
Now we will describe Keating’s constructions and what we will need from his results. Let b,u∈𝒩 such that
b=x+xk+1β(x),
where β(x)=rk+rk+1x+rk+2x2+⋯, and
D(u)=n≥k+k0.
Assume p∤k. For s≤k, h≥1 we construct matrices Ah:=Ah,n,s∈Ms×s(𝔽p). This matrix Ah is a linear operator that takes s-many coefficients of [u,h-1b] starting from x(h-1)k+n+j+1 and gives s-many coefficients of [u,hb] starting from xhk+n+j+1. This matrix is constructed formally using formula (2.9), below.
The product Πp-1:=A1A2⋯Ap-1 will be our main concern. It is an operator that takes s-many “leading” coefficients of u and gives s-many “leading” coefficients of [u,p-1b].
Define w1=u and wh+1=[wh,b] for h≥0. Following [3], we determine the coefficients of wh+1 from those of
wh=x+x(h-1)k+n+1θ(x),
where θ(x)=t0+t1x+t2x2+⋯:
(2.6)[wh,b]≡x+((h-2)k+n)xhk+n+1β(x)θ(x)
+xhk+n+2(β(x)θ′(x)-β′(x)θ(x))(modx(h+1)k+n+1).
It follows from the expansion above that for 0≤j≤k-1 the coefficient of xhk+n+j+1 in wh+1 is
(2.7)∑i=0j((h-2)k+n+2i-j)rk+j-iti=∑i=0j(nh+2i-j)rk+j-iti,
using the abbreviation
(2.8)nh=(h-2)k+n.
For s≤k and h≥1 we define the matrix Ah=Ah,n,s∈Ms×s(𝔽p) as
(2.9)Ah=(nhrk(nh-1)rk+1(nh-2)rk+2…(nh-s+1)rk+s-10(nh+1)rknhrk+1…(nh-s+3)rk+s-2⋮⋮⋱⋮⋮000…(nh+s-1)rk).
Clearly Ah depends only on s, b and the remainder of n modulo p.
Let now νh∈𝔽ps be the row vector whose entries are the coefficients of x(h-1)k+n+j+1 in wh for 0≤j≤s-1. It follows from (2.7) that
νh+1=νhAh,n,s.
For h≥1 define a matrix Πh=Πh,n,s∈Ms×s(𝔽p) by setting Πh=A1A2⋯Ah. Then we have νp=ν1Πp-1.
Recall now the definition of k0 and e(k,n) from Definition 1. Keating introduced the matrices above for s=e(k,n)+1, and we follow his notation. He went on to show that in his e(k,n)+1×e(k,n)+1 matrix Πp-1 the first e(k,n) columns are 0 (see [3, Cases 3–4]). (NB: He claimed this only for n>(p-1)k+p, because he used it only in that case, however it works also for arbitrary n≥k+k0.) Using our extended notation, we may restate his result as follows:
Lemma 2.5.
For p∤k and e=e(k,n) we have
Πp-1,n,e=0.
From this we instantly get:
Corollary 2.6.
Let p∤k>k0 so 2k0<k and write e=e(k,n). For a fixed s<2k0+1
denote by T the top right-hand corner s-e×s-e block of Πp-1,n,s and let ν′ be the first s-e coefficients of ν1. The first e entries of νp are 0. The remaining s-e entries are ν′T.
Proof.
The top left-hand corner e×e block of Πp-1,n,s is zero by Lemma 2.5. Similarly, the bottom right-hand corner e×e block of Πp-1,n,s, which is exactly Πp-1,n+e,e, is zero by Lemma 2.5. As the matrix Πp-1,n,s is a triangular matrix, nonzero entries of Πp-1,n,s only occur in the top right-hand corner s-e×s-e block of Πp-1,n,s, which is called T. So, the first e entries of νp are 0 and the remaining s-e entries are ν′T.
∎
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 n′>n≥k are such that Theorem 1.1 holds for (k,n), and Theorem 1.2 holds for (k,n′). If
n+e(k,n)=n′+e(k,n′),
then Theorem 1.2 also holds for (k,n).
Consider a fixed k. By the definition of e(k,n), the sum n+e(k,n) is constant in the intervals k+tp≤n≤k+tp+k0 (with t≥0). Lemma 3.1 allows us to assume that k+k0+tp≤n<k+(t+1)p. If n=k+k0+tp, then e(k,n)=0, otherwise e(k,n)=k0. We are going to assume this in the proof of Theorem 1.2, below.
Let p∤k and k+k0+tp≤n≤k+(t+1)p for some nonnegative integer t. We define
e′(k,n)={0if k+k0+tp<n<k+(t+1)p,k0if n=k+k0+tp or n=k+(t+1)p.
This describes the number of significant digits for the following definition. For u∈𝒩n let μn(u)∈𝔽pe′(k,n)+1 denote the vector consisting of the first e′(k,n)+1 coefficients of u.
For given n,k as above and λ∈𝔽p we define an e′(k,n)+1×e′(k,n)+1 matrix Tn[λ] as
(-λ0…02λ20λ0⋱0⋮⋮⋱0⋮00…λ0-10…02λ) if n≡k(modp),
and
(-10⋯02λ00⋯00⋮⋮⋮⋮⋮00⋯00) if n≡2k(modp),
and Tn[λ]=(λ) if k+k0+pt<n<k+p(t+1).
Lemma 3.2.
Let k>p, p∤k and k+k0+pt≤n≤k+(t+1)p for some nonnegative integer t. Define b=x+xk+1+xk+k0+1.
For every u∈Nn we have [u,p-1b]∈Nn* and
μn*([u,p-1b])=μn(u)Tn[-1],
where n*=n+(p-1)k+e(k,n).
Proof.
We have to show that the top right-hand e′(k,n)+1×e′(k,n)+1 block of Πp-1,n,e(k,n)+e′(k,n)+1 is Tn[-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-i∈{0,k0}. Consequently, in the products Πh=A1⋯Ah the entries (i,j) are 0 unless j-i is a nonnegative integer multiple of k0.
First we deal with the case n≡2k(modp). Pick some m≡k(modp). Now e(k,n)=0 and e(k,m)=k0, while e′(k,n)=e′(k,m)=k0. From the definition in (2.9) we also see the following block decomposition of the matrices Ah,m,2k0+1,Πh,m,2k0+1∈M(2k0+1)×(2k0+1)(𝔽p) for h≥1:
(3.1)Ah,m,2k0+1=(Ah,m,k0*0Ah,n,k0+1),Πh,m,2k0+1=(Πh,m,k0*0Πh,n,k0+1).
It is enough to compute the larger matrices Ah,m,2k0+1∈M(2k0+1)×(2k0+1)(𝔽p), and their product Πp-1,m,2k0+1∈M(2k0+1)×(2k0+1)(𝔽p).
For convenience we index the rows and columns of Ah from 0 to 2k0. By (2.9), the (i,j) entry of Ah for 0≤i≤j≤2k0 is
ahij={mh+iif i=j,mh-k0+iif j=k0+i,0otherwise,
where we used again the convention (2.8).
Note that
mh+k0=(h-2)k+m+k0≡mh+1(modp).
Thus Ah has the form
0k02k00mh0…0mh-10…000mh+10…0mh-1+10…0⋮⋱⋱0…0⋱0⋮00⋱⋱⋱…0⋱0k00000mh+10…0mh⋮⋮⋮⋮⋱⋱⋱…0⋮⋮⋮⋮⋮⋮⋱⋮⋮⋮⋮⋮⋮⋮⋮⋮⋱02k000000000mh+2.
As remarked above, the (i,j) entry of Πh is zero if j-i∉{0,k0,2k0}. Now we determine the rest.
By induction on h≥1, we get that the (i,i+k0) entry of Πh is
(∑t=1h-1(m1+i)(m2+i)⋯(mt+i)2(mt+3+i)(mt+4+i)⋯(mh+1+i))
(3.2)+(m0+i)(m3+i)⋯(mh+1+i).
Finally, evaluating for h=p-1, we get that the (i,i+k0) entry of Πp-1 is
(3.3)πi,i+k0=∑t=0p-2(mt+i)∏j=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 l∈{1,…,p} such that ml+i=0. So ∏j≠l(mj+i)=-1 and hence
(ml-1+i)∏j=3p(ml-1+j+i)=--k0k0=1,
(ml-2+i)∏j=3p(ml-2+j+i)=--2k0-k0=-2,
and
(3.4)πi,i+k0={-2+1=-1if l≠1,p,-2if l=p,1if l=1.
By Lemma 2.5 we know that the first k0 columns and the last k0 rows of Πp-1 are zero. The remaining diagonal entry is
πk0,k0=m2m3m4⋯mp-1=(2k0)(2k0)(3k0)⋯((p-2)k0)=-1.
To determine π0,2k0 note that ap-1,2k0,2k0=mp+1=0 and ap-1,k0,2k0=mp-1 in Ap-1. So π0,2k0 is mp-1 times the (0,k0) entry of Πp-2. Using (3.2), we get
π0,2k0=m0m3m4⋯mp-1mp-1=(-k0)(2k0)(3k0)⋯((p-2)k0)2=2.
Thus, we have determined the top right-hand corner k0+1×k0+1 matrix of Πp-1,m,2k0+1
corresponding to m≡k(modp), it is
(3.5)Tm[-1]=(10…020-10…0⋮⋮⋱0⋮00…-10-10…0-2).
By formulas (3.1) and (3.5), the top right-hand corner k0+1×k0+1 matrix of Πp-1,n,k0+1∈M(k0+1)×(k0+1)(𝔽p) corresponding to n≡2k(modp), is
Tn[-1]=(-10⋯0-200⋯00⋮⋮⋱⋮⋮00⋯00)
We now turn to the case of k+k0+pt<n<k+p(t+1), that is, we have n≢2k-i(modp) for any 0≤i≤k0. So e(k,n)=k0 and e′(k,n)=0. Then the corresponding matrix Ah,n,k0+1 is
0k00nh0…0nh-10nh+10…0⋮⋱⋱0000⋱⋱⋮k00000nh+1.
As above, we determine the (i,j) entries of Πp-1=Πp-1,n,k0+1:
π0,k0=(∑t=1p-2n1n2⋯nt2nt+3nt+4⋯np)+n0n3⋯np=∑t=0p-2nt∏j=3pnt+j.
Note that we get n1=n-k≠0, np=(p-2)k+n≠0 by our assumptions, so π0,k0=-1 as in (3.4). For the diagonal entries we again get
πi,i=∏t=1p-1(nt+i)∈{0,-1},
with πi,i=-1 if and only if 0≡n0+i=n-2k+i(modp). In our case we have n≢2k-i(modp) for any 0≤i≤k0 so all πi,i=0. We cut the first k0 columns and last k0 rows and we get the matrix
Tn[-1]=(-1).
∎
Lemma 3.3.
Let p>2. Suppose k=k0<p and k+k0+pt≤n≤k+(t+1)p for some nonnegative integer t. If k<p-1, then let b=x+xk+1, otherwise let b=x+xk+1+x2k+1. For every u∈Nn we have [u,p-1b]∈Nn* and
μn*([u,p-1b])=μn(u)Tn[λ],
where n*=n+(p-1)k+e(k,n) and the nonzero λ∈Fp is
λ={k-12=-1if k=p-1,k+12if 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 e(k,n)+e′(k,n)+1≤2k+1. Then we can use Lemma 2.5 and it remains to show that the top right-hand e′(k,n)+1×e′(k,n)+1 block of Πp-1,n,e(k,n)+e′(k,n)+1 is Tn[λ].
The refinement of (2.6) (using n≥2k) for uh+1 reads as follows:
(3.6)[uh,b]≡x+((h-2)k+n)β(x)θ(x)xhk+n+1
+(β(x)θ′(x)-β′(x)θ(x))xhk+n+2
+(((h-1)k+n+12)-(k+1)((h-2)k+n))
×β(x)2θ(x)x(h+1)k+n+1
+(((h-2)k+n)β(x)2θ′(x)
+((3-h)k-n+1)β(x)β′(x)θ(x))x(h+1)k+n+2
+(12β(x)2θ′′(x)+β′(x)2θ(x)
-β(x)β′(x)θ′(x))x(h+1)k+n+3
+Eh,n(modx(h+2)k+n+2).
Here, for some qh,n,qh,n′∈𝔽p,
Eh,n={qh,nx(h+2)k+n+1β(x)3θ(x)if (h-1)k+n>2k,qh,nx(h+2)k+n+1β(x)3θ(x)+qh,n′β(x)θ(x)2if (h-1)k+n=2k.
The coefficient of xhk+n+j+1 in uh+1 for 0≤j<k is the same as in (2.7), and for k≤j<2k it is
(3.7)∑i=1k+a((h-2)k+n+2i-j)rk+j-iti
+((nh+a2)+(k+12))rk2ta+∑i=01c1irk+irk+1-i)ta-1
+(∑i=02c2irk+irk+2-i)ta-2+⋯+(∑i=0acairk+irk+a-i)t0,
where a=j-k and csw∈𝔽p for 1≤s≤a, 0≤w≤a, and w≤s. For j=2k 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 Πh is 0 unless j-i∈{0,k,2k} as in Lemma 3.2.
As before, we assume first that n≡2k(modp) and pick some m≡k(modp) and hence e(k,n)=0 and e(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,2k+1:
ahij={mh+iif i=j,(mh+i2)+(k+12)if j=i+k,Ch,mif (i,j)=(0,2k),0otherwise.
The value of Ch,m∈𝔽p will turn out to be irrelevant. By induction on h≥1, we get that the (i,i+k) entry of Πh is
(3.8)(∑t=1h(m1+i)(m2+i)⋯(mt+i)(mt+2+i)(mt+3+i)
⋯(mh+1+i)mt+i-12)
+(k+12)∑t=1h(m1+i)(m2+i)
⋯(mt-1+i)(mt+2+i)⋯(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 l∈{2,…,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 i≠0,k the two terms add up to 0. So finally
(3.9)πi,i+k={-k-12if i=0,k+1if i=k,k+12otherwise.
For the diagonal entries we again have
πi,i=∏h=1p-1(mh+i)∈{0,-1},
with πi,i=-1 if and only if 0≡m0+i=m-2k+i≡-k+i(modp). Note that Πp-2 has (0,0) entry 0 since m1=0 and (0,k) entry k+12k by (3.8). Also Ap-1 has (k,2k) entry k(k+1) and (2k,2k) entry 0. So
π0,2k=(k+1)22.
Put λ=k+12≠0 as k<p-1. Now we have the claimed matrix
Tm[λ]=(-λ0…02λ20λ0…0⋮⋮⋱0⋮00…λ0-10…02λ).
As (3.1) is still valid, we also have
Tn[λ]=(-10⋯02λ00⋯00⋮⋮⋱⋮⋮00⋯00).
Now suppose that k=p-1 so take b=x+xp+x2p-1. We continue to assume that n≡2k(modp) and m≡k(modp) and hence e(k,n)=0,e(k,m)=k. By using (3.6), (3.7) we have the corresponding matrix Ah,m,2k+1 whose (i,j) entries for 0≤i,j≤2k are
ahij={mh+iif i=j,mh+i-k+(mh+i2)+(k+12)if j=k+i,Dh,mif (i,j)=(0,2k),0otherwise.
The value of Dh,m∈𝔽p is uninteresting, as before. Again the (i,j) entry of Πh,m,2k+1 is zero if j-i∉{0,k,2k}. By the value of ahii+k we get that the (i,i+k) entry of Πh is the sum of (3.8) and (3.2), so
πi,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
πi,i=∏h=1p-1(mh+i)∈{0,-1},
with πi,i=-1 if and only if 0≡m0+i=m-2k+i≡-k+i(modp). Note that Πp-2 has (0,0) entry 0 and (0,k) entry k+12k+-1k=k-12k. On the other hand, Ap-1 has (k,2k) entry -2k+k(k+1)=k(k-1) and (2k,2k) entry 0. So
π0,2k=(k-1)22.
Put λ=k-12.
(We use the assumption p>2 to claim λ≠0.) We obtained the claimed matrix
Tm[λ]=(-λ0…02λ20λ0…0⋮⋮⋱0⋮00…λ0-10…02λ).
We get Tn[λ] from this exactly as above.
Assume finally that k+k0+pt<n<k+p(t+1) that is n≢2k-i(modp) for any 0≤i≤k0, so e(k,n)=k0. Of course k<p-1. So we take the element b=x+xk+1 and obtain the corresponding matrix Tn[λ]=(k+12) (from (3.9)) exactly as in Lemma 3.2. ∎
We collect in the following corollary what we need for the proof of Theorem 1.2.
Corollary 3.4.
Let p∤k and k>1 if p=2. In addition, let k+k0+pt≤n≤k+p(t+1) for some nonnegative integer t. Define
(3.10)b=x+xk+1+xk+k0+1,λ=-1if k≥p-1,
b=x+xk+1,λ=k+12if k<p-1.
Then for every u∈Nn and for s≥1 we have [u,s(p-1)b]∈Nn* and
μn*([u,s(p-1)b])=λs-1μn(u)Tn[λ],
where n*=n+s(p-1)k+(s-1)k0+e(k,n).
Proof.
Suppose that k+k0+pt<n<k+(t+1), for some integer t. Now e′(k,n)=0. Recall the matrix Tn[λ]=(λ), where λ is as above, exactly as in Lemmas 3.2 and 3.3. If u∈𝒩n is such that μn(u)=(α), then by these lemmas,
μn*([u,s(p-1)b])=(λsα)=λs-1μn(u)Tn[λ],
as claimed.
Suppose now that n=k+k0+pt or n=k+p(t+1) for some nonnegative integer t. Hence e′(k,n)=k0. Note that n+(p-1)k+e(k,n)≡k(modp) in both cases, so from the second step there will be no difference between the two cases. Recall the matrices X=Tk[λ],Y=T2k[λ]∈Mk0+1×k0+1(𝔽p), where λ is as above, exactly as in Lemmas 3.2 and 3.3. Note that YX=λY and X2=λX.
If n=k+p(t+1), then these lemmas imply by induction on s that
μn*([u,s(p-1)b])=μn(u)Xs=λs-1μn(u)Tn[λ],
as claimed.
If n=k+k0+pt, then these lemmas imply first that
μn+(p-1)+e(k,n)([u,p-1b])=μn(u)Y.
Then (using e(k,k)=k0) we obtain by induction on s that
μn*([u,s(p-1)b])=μn(u)YXs-1=λs-1μn(u)Tn[λ],
as claimed.
∎
Proof of Theorem 1.2.
Suppose that p∤k. By the remark before Lemma 3.1 we may assume that k+k0+pt≤n<k+(t+1)p, for some integer t≥0. If n=k+k0+pt, then e(k,n)=0, otherwise e(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=u0∈𝒩n of depth n, its leading coefficient, α≠0 will be chosen appropriately later. Set f-1=g-1u and then um=gpmf-pm.
Let nm=n+(pm-1)k+pm-pp-1k0+e(k,n). We prove that D(um)=nm by induction on m. For that, assume m≥1 and that for every smaller index the theorem holds. Note that
nm=nm-1+(p-1)pm-1k+pm-1k0.
If n≡2k(modp), then for the inductive proof to work we will need information on the subsequent k0=e′(k,nm)=e′(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 y≥w. 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. 9–12]. The following finer approximation is from there [4, Proposition 1.1.32 (i)]:
um≡um-1p[um-1,g-pm-1](p2)[um-1,2g-pm-1](p3)
⋯[um-1,p-1g-pm-1](modK(g-pm-1,um-1)),
where K(g-pm-1,um-1) is the normal closure in 𝒩 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 e(k,n)=0 and e′(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 K(g-1,u) has depth exactly
2n+(p-2)k=n+(p-1)k+e(k,n)+e′(k,n),
every other element in K(g-1,u) has larger depth. The leading coefficient of w is
(n-k)αn(n+k)⋯(n+(p-4)k)α(p-2)k=-2α2,
where α is the leading coefficient of u.
If m≥1 is arbitrary with nm-1>k+k0, then every element in K(g-pm-1,um-1) has depth at least
2nm-1+(p-2)k>nm-1+(p-1)k+e(k,nm-1)+e′(k,nm-1).
From now let m≥1 arbitrary. Also note that
D([um-1,ig-pm-1](pi+1))≥p(nm-1+k)>nm-1+(p-1)k+e(k,nm-1)+e′(k,nm-1)
for 1≤i≤p-2 for every nm-1≥k+k0. Let nm-1¯ denote the remainder of nm-1 modulo p. Then
D(um-1p)≥pnm-1+nm-1¯=nm-1+(p-1)k+k0+(p-2)k0+(p-1)(nm-1-k-k0)+nm-1¯>nm-1+(p-1)k+e(k,nm-1)+e′(k,nm-1).
We obtain a first version for the coefficient vector
(3.11)μnm(um)
={μnm([um-1,p-1g-pm-1])if nm-1>k+k0,μn1([u,p-1g-1])+(0,…,0,-2δ0α2)if m=1,n=k+k0.
Here δ0 is the exponent of w in u1≡[u,p-1g-1]wδ0(modxn1+e′(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]≡[um-1,(p-1)pm-1g-1](modx2nm-1+(p-1)pm-1k+2).
Now
2nm-1+(p-1)pm-1k+2=nm+nm-1-pm-1k0+2=nm+n+(pm-1-1)k+(-pm-1+pm-1-pp-1)k0+e(k,n)+2>nm+e′(k,nm)+1.
So we obtain μnm(um)=μnm([um-1,(p-1)pm-1g-1]) for m>1. By induction and by (3.11), we get a second version for the coefficient vector
(3.12)μnm(um)={μnm([u1,pm-pg-1])for m>1,μnm([u,pm-1g-1])for m≥1 if n>k+k0.
After these preliminary observations we turn to the proof.
First let k+k0+pt<n<k+(t+1)p. Then e′(k,n)=0 and e(k,n)=k0, and so nm≡n≢k,2k(modp) for every m≥0.
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+tp. Then e′(k,n)=k0 and e(k,n)=0, and so nm≡k(modp) for every m≥1. Recall that α≠0 denotes the leading coefficient of u∈𝒩n.
By the definition of g-1=b and λ in (3.10), Corollary 3.4 shows that for the first k0 coefficients of v=[u,p-1g-1] we have
μn1(v)=μn(u)Tn[λ]=(-α,0,…,0,2λα).
By (3.11),
μn1(gpf-p)=(-α,0,…,0,2λα-2δα2),
where δ=δ0 if n=k+k0 and δ=0 otherwise. The leading coefficient is -α≠0, so
D(u1)=D(gpf-p)=n+(p-1)k+e(k,n)=n1
holds. Now n1≡k(modp) so e′(k,n1)=e(k,n1)=k0.
We apply Corollary 3.4
again to get
D(u2)=D([u1,p-1g-p])=D([u1,(p-1)pg-1])=n2.
We also obtain
(3.13)μn2(gp2f-p2)=(-λα+2δα2,0,…,0,2λ2α-4λδα2).
The leading coefficient is nonzero if α≠0 and 2αδ≠λ. Such an α exists because λ≠0. Fix this α. The vector at (3.13) is an eigenvector of Tnm[λ] with eigenvalue λ≠0 for all m≥1.
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 p∣k. Our arguments are similar to those of [3]. Suppose that p∤n. Let g,u be arbitrary and set f-1:=g-1u. Then
D(gpf-p)=D([u,p-1g-1])=n+(p-1)k≡n(modp)
since 2n+(p-2)k>n+(p-1)k. We can proceed with induction to get
D(gpmf-pm)=D([u,pm-1(p-1)g-1])=n+(pm-1)k
for m≥1. Suppose finally that p∣n. If n=k, then let g be arbitrary of depth k and f=gp. Then
gpmfp-m=gpm(1-p)
and so
D(gpmfp-m)=pmk
by [2, Lemma 1]. If n>k, then pick g and f to work for the pair k,n+1. Then e(k,n)=1=e(k,n+1)+1 so
we have the result by Lemma 3.1.
∎
Proof of Corollary 1.4.
Assume the contrary: for a particular m≥1 we have
(p-1)D(fm)>(pm+1-pm)k+(pm-1)k0.
By Theorem 1.2, we have
D(gpmf-pm)=n1=n+(pm-1)k+pm-pp-1k0+e(k,n),
where n1≡n(modp) if n≢2k-i(modp) for any 0≤i≤k0, and otherwise n1≡k0(modp). Therefore, in any case e(k,n1)=k0.
Now apply Theorem 1.1 for gpm,fpm to get
D((gpm)p(fpm)-p)≥n1+(p-1)D(fpm)+k0
>n+(pm-1)k+pm-pp-1k0+e(k,n)
+(pm+1-pm)k+(pm-1)k0+k0
=n+(pm+1-1)k+pm+1-pp-1k0+e(k,n)
=D(gpm+1f-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+e(k,n)-k0>0.
Since Theorem 1.2 holds for f,g, we have
D(g-pmfpm)=n+(pm-1)k+pm-pp-1k0+e(k,n)
=n-k+e(k,n)-k0+pmk+pm-1p-1k0
>pmk+pm-1p-1k0
=D(fpm),
by the first part. Therefore,
D(gpm)=D(fpm)=pmk+pm-1p-1k0,
as claimed.
∎
