1.1. Equidistribution in tori 7
Proposition 1.1.5 (Equidistribution for abelian linear sequences). Let T
be a torus, and let x(n) := + β for some α, β T . Then there exists
a decomposition x = x + x , where x (n) := is totally asymptotically
equidistributed on Z in a subtorus T of T (with α T , of course), and
x (n) = + 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 + 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
.
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