1.5. Inverse conjecture over finite fields 85
theorem for vector spaces; see [TaVu2006, §11.3] for further discussion.
The proof is elementary but a little lengthy and would take us too far afield,
so we simply assume this proposition for now and keep going. We conclude
(1.45) |Ex∈V Δhf(x)e(Mh · x)e(ξ0 · x)| 1
for many h V .
The most difficult term to deal with here is the quadratic term Mh · x.
To deal with this term, suppose temporarily that M is symmetric, thus
Mh · x = Mx · h. Then (since we are in odd characteristic) we can integrate
Mh · x as
Mh · x = ∂h
Mx · x
Mh · h
and thus
|Ex∈V f(x + h)e(
M(x + h) · (x + h))f(x)e(−
Mx · x)e(ξ0 · x)| 1
for many h H. Taking
norms in h, we conclude that the U
inner prod-
uct between two copies of f(x)e(
Mx · x) and two copies of f(x)e(
x)e(−ξ0 · x) is 1. Applying the U
Cauchy-Schwarz-Gowers inequality,
followed by the U
inverse theorem, we conclude that f(x)e(
Mx · x) corre-
lates with e(φ) for some linear phase, and thus f itself correlates with e(ψ)
for some quadratic phase.
This argument also works (with minor modification) when M is virtually
symmetric, in the sense that there exist a bounded index subspace of V such
that the restriction of the form Mh · x to V is symmetric, by foliating into
cosets of that subspace; we omit the details. On the other hand, if M is
not virtually symmetric, there is no obvious way to “integrate” the phase
e(Mh·x) to eliminate it as above. (Indeed, in order for Mh·x to be “exact”
in the sense that it is the “derivative” of something (modulo lower order
terms), e.g., Mh · x ∂hΦ for some Φ, it must first be “closed” in the sense
that ∂k(Mh·x) ∂h(Mk·x) in some sense, since we have ∂h∂k = ∂k∂h; thus
we again see the emergence of cohomological concepts in the background.)
To establish the required symmetry on M, we return to Gowers’ argu-
ment from Lemma 1.5.7, and tweak it slightly. We start with (1.45) and
rewrite it as
|Ex∈V f(x + h)f (x)e(Mh · x)| 1
where f (x) := f(x)e(ξ0 · x). We square-average this in h to obtain
|Ex,y,h∈V f(x + h)f (x)f(y + h)f (y)e(Mh · (x y))| 1.
Now we make the somewhat unusual substitution z = x + y + h to obtain
|Ex,y,z∈V f(z y)f (x)f(z x)f (y)e(M(z x y) · (x y))| 1.
Previous Page Next Page