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 × Γ) ∩ G , we conclude that G /Γ

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: