54 1. Higher order Fourier analysis
Exercise 1.3.11. Obtain a complete solution to the functional equation
(1.29) in the case when |G| is allowed to be divisible by 2 or 3. (This is an
open-ended and surprisingly tricky exercise; it of course depends on what
one is willing to call a “solution” to the problem. Use your own judgement.)
Exercise 1.3.12. Call a map φ: G R/Z a polynomial of degree d if
one has ∂h1 . . . ∂hd+1 φ(x) = 0 for all x, h1,...,hd+1 G. Show that if k 1
and φ1,...,φk obey the functional equation
φ1(x) + φ2(x + y) + · · · + φk(x + (k 1)y) = c
and |G| is not divisible by any integer between 2 and k 1, then φ1,...,φk
are polynomials of degree k 2.
We are now ready to turn to the general case of solving equations of
the form (1.27). We relied on two main tricks to solve these equations:
differentiation, and change of variables. When solving an equation such as
(1.29), we alternated these two tricks in turn. To handle the general case,
it is more convenient to rearrange the argument by doing all the change of
variables in advance. For instance, another way to solve (1.29) is to first
make the (non-injective) change of variables
(x, y) := (b + 2c + 3d, −a b c d)
for arbitrary a, b, c, d G, so that
(x, x+y, x+2y, x+3y) = (b+2c+3d, −a+c+2d, −2a−b+d, −3a−2b−c)
and (1.29) becomes
φ1(b+2c+3d)+φ2(−a+c+2d)+φ3(−2a−b+d)+φ4(−3a−2b−c) = const
for all a, b, c, d G. The point of performing this change of variables is
that while the φ4 term (for instance) involves all three variables, a, b, c, the
remaining terms only depend on two of the a, b, c variables at a time. If we
now pick h, k, l G arbitrarily, and then differentiate in the a, b, c variables
by the shifts h, k, l, respectively, then we eliminate the φ1,φ2,φ3 terms and
arrive at
(∂−l∂−2k∂−3hφ4)(−3a 2b c) = 0
which soon places us back at (1.30) (assuming as before that |G| is not
divisible by 2 or 3).
Now we can do the general case, once we put in place a definition (from
Definition 1.3.2 (Cauchy-Schwarz complexity). A system ψ1,...,ψt :

G of affine-linear forms (with linear coefficients in Z) have Cauchy-Schwarz
complexity at most s if, for every 1 i t, one can partition [t]\{i} into
s + 1 classes (some of which may be empty), such that ψi does not lie in
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