1.3. Linear patterns 53

In preparation for then eliminating φ2, we shift x backwards by y, obtaining

(∂hφ2)(x) + (∂2hφ3)(x + y) + (∂3hφ4)(x + 2y) = 0.

Differentiating in the y direction by another arbitrary shift k ∈ G, we obtain

(∂k∂2hφ3)(x + y) + (∂2k∂3hφ4)(x + 2y) = 0.

We shift x backwards by y again:

(∂k∂2hφ3)(x) + (∂2k∂3hφ4)(x + y) = 0.

One final differentiation in y by an arbitrary shift l ∈ G gives

(∂l∂2k∂3hφ4)(x + y) = 0.

For simplicity, we now make the assumption that the order |G| of G is not

divisible by either 2 or 3, so that the homomorphisms k → 2k and h → 3h

are automorphisms of G. We conclude that

(1.30) ∂l∂k∂hφ4 ≡ 0

for all l, k, h. Such functions will be called quadratic functions from G to

R/Z, thus φ4 is quadratic. A similar argument shows that φ1,φ2,φ3 are

quadratic.

Just as (aﬃne-)linear functions can be completely described in terms of

homomorphisms, quadratic functions can be described in terms of bilinear

forms, as long as one avoids the characteristic 2 case:

Exercise 1.3.9. Let G be a finite abelian group with |G| not divisible

by 2. Show that a map φ: G → R/Z is quadratic if and only one has a

representation of the form

φ(x) = B(x, x) + L(x) + c

where c ∈ R/Z, L: G → R/Z is a homomorphism, and B : G × G →

R/Z is a symmetric bihomomorphism (i.e., B(x, y) = B(y, x), and B is

a homomorphism in each of x, y individually (holding the other variable

fixed)). (Hint: Heuristically, one should set B(h, k) :=

1

2

∂h∂kφ(x), but there

is a diﬃculty because the operation of dividing by

1

2

is not well defined on

R/Z. It is, however, well defined on

|G|th

roots of unity, thanks to |G| not

being divisible by two. Once B has been constructed, subtract it off and use

Exercise 1.3.8.) What goes wrong when |G| is divisible by 2?

Exercise 1.3.10. Show that when |G| is not divisible by 2, 3, that the

complete solution to (1.29) is given by

φi(x) = Bi(x, x) + Li(x) + ci

for i = 1, 2, 3, 4, ci ∈ R/Z, homomorphisms Li : G → R/Z, and symmetric

bihomomorphisms Bi : G × G → R/Z with B2 = −3B1,B3 = 3B1,B4 =

−B1 and L1 + L2 + L3 + L4 = L2 + 2L3 + 3L4 = 0.