78 1. Higher order Fourier analysis

On the other hand, φhT

hφk

is a polynomial of order d. Also, since H is so

dense, every element l of V has at least one representation of the form l = h+

k for some h, k ∈ H (indeed, out of all |V | possible representations l = h+k,

h or k can fall outside of H for at most

O(εc|V

|) of these representations).

We conclude that for every l ∈ V there exists a polynomial φl ∈ Poly≤d(V →

R/Z) such that

(1.42) e(∂lψ) − e(φl)

L1(V

)

=

O(εc).

The new polynomial φl supercedes the old one φl; to reflect this, we abuse

notation and write φl for φl. Applying the cocycle equation again, we see

that

(1.43) e(φh+k) − e(φhT

hφk)

L1(V )

=

O(εc)

for all h, k ∈ V . Applying the rigidity of polynomials (Exercise 1.4.6), we

conclude that

φh+k = φhT

hφk

+ ch,k

for some constant ch,k ∈ R/Z. From (1.43) we in fact have ch,k =

O(εc)

for

all h, k ∈ V .

The expression ch,k is known as a 2-coboundary (see [Ta2009, §1.13]

for more discussion). To eliminate it, we use the finite characteristic to

discretise the problem as follows. First, we use the cocycle identity

p−1

j=0

e(T

jh∂hψ)

= 1

where p is the characteristic of the field. Using (1.42), we conclude that

p−1

j=0

e(T

jhφh)

− 1

L1(V )

=

O(εc).

On the other hand, T

jhφh

takes values in some coset of a finite subgroup C

of R/Z (depending only on p, d), by Lemma 1.4.1. We conclude that this

coset must be a shift of C by

O(εc).

Since φh itself takes values in some

coset of a finite subgroup, we conclude that there is a finite subgroup C

(depending only on p, d) such that each φh takes values in a shift of C by

O(εc).

Next, we note that we have the freedom to shift each φh by

O(εc)

(ad-

justing ch,k accordingly) without significantly affecting any of the properties

already established. Doing so, we can thus ensure that all the φh take val-

ues in C itself, which forces ch,k to do so also. But since ch,k =

O(εc),

we

conclude that ch,k = 0 for all h, k, thus φh is a perfect cocycle:

φh+k = φhT

hφk.