1.6. Inverse conjecture over the integers 105
and ˜: g Z G × G is the sequence
˜(n) g := (g(n),∂hg(n)g(n)).
Now we give a filtration on G × G by setting
(G × G)≥j := G≥j ×G≥j+1 G≥j
for j 0, where G≥j ×G≥j+1 G≥j is the subgroup of G≥j ×G≥j+1 G≥j gen-
erated by G≥j+1 × G≥j+1 and the diagonal group G≥j
Δ
:= {(gj,gj) : gj
G≥j. One easily verifies that this is a filtration on G × G. The sequences
(g(n),g(n)) and (id,∂hg(n)) are both polynomial with respect to this fil-
tration, and hence by the Lazard-Leibman theorem (Theorem 1.6.8), ˜ g is
polynomial also.
Next, we use the hypothesis that F has a vertical frequency to conclude
that F is invariant with respect to the action of the diagonal group Gs
Δ
=
(G × G)≥s. If we then define G to be the Lie group G := (G ×
G)≥0/GsΔ
with filtration G≥j := (G × G)≥j/Gs
Δ,
then G is a degree s 1 filtered
nilpotent Lie group; setting Γ := × Γ) G , we conclude that G
is a degree s 1 nilmanifold and
Δhψ(n) = F (g (n)Γ )
where F , g are the projections of
˜
F, ˜ g from G × G to G . The claim
follows.
We now prove Theorem 1.6.12 by induction on s. The claim is trivial for
s = 0, so we assume that s 1 and that the claim has already been proven
for smaller values of s.
Let f, δ, ψ be as in Theorem 1.6.12. From Exercise 1.6.20 we see (after
modifying δ, M) that we may assume that ψ has a vertical frequency. Next,
we use the identity
|En∈Z/NZ
f(n)ψ(n)|2
= Eh∈Z/N
Z
En∈Z/N
Z
Δhf(n)Δhψ(n)
(extending f by zero outside of [N], and extending ψ arbitrarily) to conclude
that
|En∈[N]Δhf(n)Δhψ(n)|
δ
1
for N values of h [−N, N]. By induction hypothesis and Lemma 1.6.13,
we conclude that
Δhf
Us[N] δ,M
1
for N values of h [−N, N]. Using the identity
f
2s+1
Us+1(Z/N Z)
= Eh∈Z/N
Z
Δhf
2s
Us(Z/N
Z)
we close the induction and obtain the claim.
In the other direction, we have the following recent result:
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