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.
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