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).

Show