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
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) :=
∂h∂kφ(x), but there
is a difficulty because the operation of dividing by
is not well defined on
R/Z. It is, however, well defined on
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.
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