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

that

(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

1

2

Mx · x −

1

2

Mh · h

and thus

|Ex∈V f(x + h)e(

1

2

M(x + h) · (x + h))f(x)e(−

1

2

Mx · x)e(ξ0 · x)| 1

for many h ∈ H. Taking

L2

norms in h, we conclude that the U

2

inner prod-

uct between two copies of f(x)e(

1

2

Mx · x) and two copies of f(x)e(

1Mx

2

·

x)e(−ξ0 · x) is 1. Applying the U

2

Cauchy-Schwarz-Gowers inequality,

followed by the U

2

inverse theorem, we conclude that f(x)e(

1

2

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.