88 1. Higher order Fourier analysis

rank for all h in a bounded index subspace of V ; restricting to that subspace,

we will now assume that ∂hφ is low rank for all h ∈ V . Thus we have

∂hφ = Fh(Qh)

where Qh is some bounded collection of quadratic polynomials for each h,

and Fh is some function. To simplify the discussion, let us pretend that Qh

in fact consists of just a single quadratic Qh, plus some linear polynomials

Lh, thus

(1.47) ∂hφ = Fh(Qh, Lh).

There are two extreme cases to consider, depending on how Qh depends

on h. Consider first a “core” case when Qh = Q is independent of h. Thus

(1.48) ∂hφ = Fh(Q, Lh).

If Q is low rank, then we can absorb it into the Lh factors, so suppose instead

that Q is high rank, and thus equidistributed even after fixing the values of

Lh.

The function ∂hφ is cubic, and Q is a high rank quadratic. Because of

this, the function F (Q, Lh) must be at most linear in the Q variable; this

can be established by another application of equidistribution theory; see

[GrTa2009, §8]. Thus one can factorise

∂hφ = QFh(Lh) + Fh (Lh)

for some functions Fh,Fh . In fact, as ∂hφ is cubic, Fh must be linear, while

Fh is cubic.

By comparing the Q coeﬃcients Fh (Lh) in the cocycle equation (1.46),

we see that the function ρh := Fh (Lh) is itself a cocycle:

ρh+k = ρh + T

hρk.

As a consequence, we have ρh = ∂hR for some function R: V → R/Z. Since

ρh is linear, R is quadratic; thus we have

(1.49) ∂hφ = Q∂hR + Fh (Lh).

With a high characteristic assumption p 2, one can ensure R is classical.

We will assume that R is high rank, as this is the most diﬃcult case.

Suppose first that Q = R. In high characteristic, one can then integrate

Q∂hQ by expressing this as ∂h(

1

2

Q2)

plus lower order terms, thus ∂h(φ−

1

2

Q2)

is an order 1 function in the sense that it is a function of a bounded number of

linear functions. In particular, e(∂h(φ−

1

2

Q2))

has a large U

2

norm for all h,

which implies that e(φ−

1

2

Q2)

has a large U

3

norm, and thus correlates with

a quadratic phase. Since e(

1

2

Q2)

can be decomposed by Fourier analysis into

a linear combination of quadratic phases, we conclude that e(φ) correlates

with a quadratic phase and one is thus done in this case.