16 1. Higher order Fourier analysis
Remark 1.1.15. The numerical constants can of course be improved, but
this is not our focus here.
Exercise 1.1.21. Let P : Z
Td
be a polynomial sequence P (n) :=
αsns+
· · ·+α0, let N 1, and let 0 δ 1. Suppose that the polynomial sequence
P is not totally δ-equidistributed on [N]. Show that there exists a non-zero
k
Zd
with |k|
d,s
δ−Od,s(1)
such that k · αs
T d,s
δ−Od,s(1)/N s.
(Hint:
Induct on s starting with Exercise 1.1.19 for the base case, and then using
Exercise 1.1.20 and Lemma 1.1.14 to continue the induction.)
Note the N
s
denominator; the higher-degree coefficients of a polynomial
need to be very rational in order not to cause equidistribution.
The above exercise only controls the top degree coefficient, but we can
in fact control all coefficients this way:
Lemma 1.1.16. With the hypotheses of Exercise 1.1.21, we can in fact
find a non-zero k
Zd
with |k|
d,s
δ−Od,s(1)
such that k · αi
T d,s
δ−Od,s(1)/N i
for all i = 0,...,s.
Proof. We shall just establish the one-dimensional case d = 1, as the general
dimensional case then follows from Exercise 1.1.18.
The case s 1 follows from Exercise 1.1.19, so assume inductively that
s 1 and that the claim has already been proven for smaller values of
s. We allow all implied constants to depend on s. From Exercise 1.1.21,
we already can find a positive k with k =
O(δ−O(1))
such that kαs
T
δ−O(1)/N s.
We now partition [N] into arithmetic progressions of spacing k
and length N
δCN
for some sufficiently large C; then by the pigeonhole
principle, we see that P fails to be totally
δO(1)-equidistributed
on one
of these progressions. But on one such progression (which can be identified
with [N ]) the degree s component of P is essentially constant (up to errors
much smaller than δ) if C is large enough; if one then applies the induction
hypothesis to the remaining portion of P on this progression, we can obtain
the claim.
This gives us the following analogue of Exercise 1.1.7. We say that a
subtorus T of some dimension d of a standard torus Td has complexity
at most M if there exists an invertible linear transformation L SLd(Z)
with integer coefficients (which can thus be viewed as a homeomorphism
of
Td
that maps T to the standard torus
Td
×
{0}d−d
), and such that all
coefficients have magnitude at most M.
Exercise 1.1.22. Show that every subtorus (i.e., compact connected Lie
subgroup) T of
Td
has finite complexity. (Hint: Let V be the Lie algebra
of T , then identify V with a subspace of
Rd
and T with V/(V
Zd).
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