1.1. Equidistribution in tori 7

Proposition 1.1.5 (Equidistribution for abelian linear sequences). Let T

be a torus, and let x(n) := nα + β for some α, β ∈ T . Then there exists

a decomposition x = x + x , where x (n) := nα is totally asymptotically

equidistributed on Z in a subtorus T of T (with α ∈ T , of course), and

x (n) = nα + β is periodic (or equivalently, that α ∈ T is rational).

Proof. We induct on the dimension d of the torus T . The claim is vacuous

for d = 0, so suppose that d ≥ 1 and that the claim has already been proven

for tori of smaller dimension. Without loss of generality we may identify T

with

Td.

If α is irrational, then we are done by Exercise 1.1.5, so we may assume

that α is not irrational; thus k · α = 0 for some non-zero k ∈

Zd.

We then

write k = mk , where m is a positive integer and k ∈

Zd

is irreducible (i.e.,

k is not a proper multiple of any other element of

Zd);

thus k ·α is rational.

We may thus write α = α1 + α2, where α2 is rational, and k · α1 = 0. Thus,

we can split x = x1 + x2, where x1(n) := nα1 and x2(n) := nα2 + β. Clearly

x2 is periodic, while x1 takes values in the subtorus T1 := {y ∈ T : k ·y = 0}

of T . The claim now follows by applying the induction hypothesis to T1

(and noting that the sum of two periodic sequences is again periodic).

As a corollary of the above proposition, we see that any linear sequence

n → nα + β in a torus T is equidistributed in some union of finite cosets

of a subtorus T . It is easy to see that this torus T is uniquely determined

by α, although there is a slight ambiguity in the decomposition x = x + x

because one can add or subtract a periodic linear sequence taking values in

T from x and add it to x (or vice versa).

Having discussed the linear case, we now consider the more general sit-

uation of polynomial sequences in tori. To get from the linear case to the

polynomial case, the fundamental tool is

Lemma 1.1.6 (van der Corput inequality). Let a1,a2,... be a sequence of

complex numbers of magnitude at most 1. Then for every 1 ≤ H ≤ N, we

have

|En∈[N]an|

(

Eh∈[H]|En∈[N]an+han|

)1/2

+

1

H1/2

+

H1/2

N

1/2

.

Proof. For each h ∈ [H], we have

En∈[N]an = En∈[N]an+h + O

H

N

and hence, on averaging,

En∈[N]an = En∈[N]Eh∈[H]an+h + O

H

N

.