Van der Corput sets with respect to compact
groupsMichael Kelly and Thái Hoàng Lê
Abstract. We study the notion of van der Corput sets with respect togeneral compact groups.
Mathematics Subject Classification (2010). Primary 11K06; Secondary
11B05, 37A45.
Keywords. van der Corput’s theorem, van der Corput sets, intersective
sets, uniform distribution, compact groups.

1. Introduction
Given a sequence (xn )∞n=1 ⊂ T = R/Z. The classical van der Corput’s difference theorem in uniform distribution theory states that if (xn+h − xn )∞n=1 isuniformly distributed in T for all h ∈ Z+ , then (xn )∞n=1 itself is also uniformly
distributed in T.
In [5], Kamae and Mendès France made the important observation thatin order for (xn )∞n=1 to be uniformly distributed in T, it suffices to have theuniform distribution in T of (xn+h − xn )∞n=1 for h in a certain subset H of
Z+ . Such a set H is called a van der Corput set. A prototype result of thiskind had already been proven by Delange, where one can take H to be theset of all multiples of a positive integer a. Other examples of van der Corput
sets areH1 = {n2 : n ∈ Z+ }H2 = {p − 1 : p prime}
H3 = {i − j : i, j ∈ I, i > j}where I is any infinite set of integers.

On the other hand, it is known that the set of all odd numbers is not van der
Corput. Also, no lacunary set is van der Corput.
Thanks to works of Kamae-Mendès France [5], Ruzsa [8, 9], Montgomery[7], Bergelson-Lesigne [1], Ninčević-Rabar-Slijepčević [10], many criteria forvan der Corput sets are known. Extensive accounts of van der Corput sets

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Michael Kelly and Thái Hoàng Lê

can be found in [1] and [7]. For a modern treatment of van der Corput’s
difference theorem, see [2].
Let G be a compact (not necessarily abelian) group and µG be thenormalized Haar measure on G. We recall that a sequence (xn )∞n=1 is said tobe uniformly distributed in G if
lim µG ({1 ≤ n ≤ N : xn ∈ C}) = µG (C)N →∞for any open set C ⊂ G with boundary measure 0. Van der Corput’s difference
theorem has been generalized to any compact group by Hlawka [4] (see also∞
[6, Chapter 4, Section 2]), namely that if the sequences (xn+h x−1n )n=1 are+
uniformly distributed in a compact group G for all h ∈ Z , then the sequence(xn )∞
n=1 is also uniformly distributed in G. Naturally, the notion of van der
Corput sets also makes sense in any compact group. We make the following:
Definition 1. Let G be a compact topological group. We say a set H ⊂ Z+ is
G-van der Corput (G-vdC for short) if the following is true. For any sequence−1 ∞(xn )∞
n=1 ⊂ G, if the sequence (xn+h xn )n=1 is uniformly distributed for each∞h ∈ H, then the sequence (xn )n=1 is also uniformly distributed in G.
Given this definition, from now on usual van der Corput sets are referred
to as T-vdC sets. Presumably, the property of G-van der Corput depends on
G. We will, however, prove the following:
Theorem 1. If a set H ⊂ Z+ is T-vdC, then it is also G-vdC for any compact
group G.
Kamae and Mendès France [5] found a connection between T-vdC sets
and intersective sets which are much studied in combinatorial number theory
and ergodic Ramsey theory. A set H ⊂ Z+ is called intersective if for any}|
dense subset A of the integers (that is, limN →∞ |A∩{1,...,N> 0), there existN
two elements of A whose difference is in H. Alternatively, H is intersective
if and only if it is a set of recurrence, that is, for any probability measurepreserving dynamical system (X, B, µ, T ), for any A ∈ B with µ(A) > 0, thereis h ∈ H such that µ(A ∩ T −h A) > 0. Kamae and Mendès France showedthat any T-vdC set is intersective. The converse is not true: Bourgain [3]constructed a set that is intersective but not T-vdC. Furthermore, he showedthat the generic density conditions for T-vdC and intersective sets are thesame. We will extend Kamae and Mendès France’s argument to prove thefollowing.
Theorem 2. If G is a compact, second countable group (that is, its topology
has a coutable base), then every G-vdC set is intersective.
As a consequence of Theorems 1 and 2, if G is a compact, second countable group then the class of all G-vdC sets lies in between the class of all
T-vdC sets and the class of all intersective sets. It is an interesting problem todetermine if these two inclusions are strict, even for a specific choice of G, e.g.
G = Z/2Z. This could be a difficult problem since Bourgain’s constructionof a set that is intersective but not T-vdC is difficult.

Van der Corput sets with respect to compact groups

3

In Section 2 we will recall some preliminaries on uniform distribution.
In Section 3 and Section 4 we will prove Theorems 1 and 2.

2. Prelimiaries
We first recall Weyl’s criterion for general compact groups. Let G be a compact group with normalized Haar measure µG . A representation of G of degree
k is a continuous homomorphism D from G to the multiplicative group GL(k)of all nonsingular complex matrices of order k. A representation D is called
unitary if D(x) is unitary for all x ∈ G. Two representations D1 , D2 of thesame degree k are said to be equivalent if there exists a nonsingular k × k
matrix S such that
D2 (x) = SD1 (x)S −1
for all x ∈ G.
A representation D of degree k is called reducible if there exists a subspace V of Ck of dimension 0 < dim(V ) < k such that D(x)V ⊂ V for allx ∈ G. D is called irreducible if it is not reducible.
Let {D(λ) : λ ∈ Λ} be a system of representations of G that is obtained by choosing exactly one representation from each equivalence class ofirreducible unitary representation. Let D(0) be the trivial representation. We
then have:
Proposition 1 (Weyl’s criterion, [6, Theorem 4.1.3]). The sequence (xn )∞n=1 ⊂G is uniformly distributed in G if and only if for any λ ∈ Λ, λ 6= 0, we haveN

1 X (λ)lim
D (xn ) = 0.
N →∞ Nn=1
We will also need the following simple fact, whose proof we will omit:
Lemma 1. If D is a non-trivial irreducible representation of G, thenZ
D(x)dµG = 0.
G

Next, we recall some criteria for T-vdC sets. Though we will only need(A), (C) and (F), we will list all of them for completeness. For a set H ⊂ Z+ ,the following conditions are all equivalent to H being T-vdC.
(A) (Ruzsa) For any sequence (un )∞n=1 of bounded complex numbers, if
N

1 X
un+h un = 0
N →∞ N
n=1
lim

for any h ∈ H, thenN

1 X
un = 0.
N →∞ Nn=1
lim

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Michael Kelly and Thái Hoàng Lê

(B) (Ruzsa) For any sequence (un )∞n=1 of complex numbers satisfying
PN
limN →∞ N1 n=1 |un |2 < ∞, ifN

1 X
un+h un = 0lim
N →∞ N
n=1
for any h ∈ H, thenN

1 X
un = 0.
N →∞ Nn=1
lim(C) (Kamae-MendèsR France, Ruzsa) For any nonnegative, finite measure µ
on T, if µb(h) = T e(−hx)dµ(x) = 0 for any h in H, then µ is continuous
at 0 (that is, µ({0}) = 0).
(D) P(Bergelson-Lesigne) For any nonnegative, finite measure µ on T, ifµ(h)| < ∞, then µ is continuous at 0.
h∈H |b(E) (Kamae-Mendès France, Ruzsa) For any  > 0, there exists a nonnegative real trigonometric polynomialX
T (x) =an e(nx)n∈Z

supported on H ∪ (−H) ∪ {0} (that is, an =0 for n 6∈ H ∪ (−H) ∪ {0})satisfying T (0) = 1 and a0 ≤ .
(F) (Bergelson-Lesigne) ForPany  > 0, there exists a finite, positive-definite
sequence (an ) (that is, n,n0 ∈Z an−n0 zn zn0 ≥ 0 for any sequence (zn ) ⊂C) supported on H ∪ (−H) ∪ {0} (that is, an =0 for n 6∈ H ∪ (−H) ∪ {0})satisfyingX
an = 1anda0 ≤ .
n∈Z(G) (Ninčević-Rabar-Slijepčević) H is operator recurrent, that is, for any
Hilbert space H, for any unitary operator U on H, for any x ∈ H whose
orthogonal projection on Ker(U − I) is non-zero, there exists h ∈ Hsuch thathU h x, xi =6 0.

3. Proof of Theorem 1
In proving Theorem 1, we use the following result due to Bergelson-Lesigne,
which is a Hilbert space generalization of criterion (A).
Proposition 2 ([1, Corollary 1.31]). Let H ⊂ Z+ be a T-vdC set. Let (un )∞n=1
be any bounded sequence in a Hilbert space H. IfN

1 X
hun+h , un i = 0
N →∞ N
n=1
lim

Van der Corput sets with respect to compact groups

5

for any h ∈ H, thenN

1 X
un = 0.
N →∞ N
n=1
lim

Bergelson-Lesigne deduced this from a generalized van der Corput inequality ([1, Proposition 1.30]) and criterion (F). We remark that in the casewhere H is finite-dimensional (which is all we need), this can also be provedby the method of correlation functions ([6, Section 2, Chapter 4]) and criterion (C).
For each k, GL(k) is naturally a Hilbert space under the inner productthA, Bi = Tr(B A).
Let H be any T-vdC set. Let (xn )∞n=1 ⊂ G be a sequence such that∞(xn+h x−1)
is
uniformly
distributedin
G for any h ∈ H. Let D be any
n n=1
unitary representation of degree k of G. By Weyl’s criterion (Proposition 1),N

1 X
D(xn+h x−1n ) = 0.
N →∞ Nn=1
lim

We have
N
N
1 X
1 X
−1
hD(xn+h xn ), Ik i = lim
hD(xn+h ), D(xn )i.
0 = limN →∞ NN →∞ Nn=1
n=1

Since D is unitary, the sequence (D(xn )) is bounded. Proposition 2 implies
that
N
1 X
lim
D(xn ) = 0.
N →∞ N
n=1
By Weyl’s criterion, this implies that (xn )∞n=1 is uniformly distributed in G
and H is G-vdC.

4. Proof of Theorem 2
Since G is second-countable, it follows from the Peter-Weyl theorem that thesystem of representatives {D(λ) : λ ∈ Λ} of unitary representations of G iscountable (since, in this case, L2 (G) is separable).
In proving Theorem 2, we will generalize Kamae-Mendès France’s probabilistic argument in [5]. Suppose H ⊂ Z+ is G-vdC but not intersective. Then
there is a set A ⊂ Z+ of positive upper density such that A ∩ (A − h) = ∅ forall h ∈ H.
Let {un }n∈Z+ be a sequence of random variables taking values in G
as follows. Put un = 1G for all n ∈ A. For n ∈ Z+ \ A, we select un ∈ G6Michael Kelly and Thái Hoàng Lê

uniformly and independently (with respect to µG ). Fix h ∈ H. We have
1,
if n ∈ A and n + h ∈ A;un+h ,
if n ∈ A and n + h 6∈ A;−1un+h un =−1u,if
n 6∈ A and n + h ∈ A;
 nun+h u−1,
ifn6∈ A and n + h 6∈ A.
nSince A ∩ (A − h) = ∅, the first case does not occur. From here it canbe shown using standard probablistic arguments that the random variables
un+h u−1
n are independent and uniform in G. Indeed, uniformity is immediate.
As for independence, one only needs to verify independence of families ofIrandom variables {uni +h u−1ni }i=1 where the pairs of indices {ni , ni + h} are
not pairwise disjoint. We leave the details to the reader.
By the law of large numbers and Lemma 1, for any non-trivial λ ∈ Λ,
we haveZN
1 X (λ)D (un+h u−1)=
D(λ) (x)dµG = 0limn
N →∞ NG
n=1
almost surely.
Since Λ is countable, the above equation in conjunction with Weyl’s
criterion (Proposition 1) almost surely implies that for any h ∈ H, the se∞quence (un+h u−1n )n=1 is uniformly distributed in G. Since H is G-vdC, we
have almost surely, (un )∞n=1 is uniformly distributed in G.
Let D be any non-trivial irreducible unitary representation of G. On
the one hand, by Weyl’s criterion, we have almost surelyN

1 X
D(un ) = 0.
N →∞ N
n=1
lim

On the other hand, by the law of large numbers, and Lemma 1
Z
N
X
1
lim
D(un ) =D(x)dµG = 0
N →∞ N − |AN |
G
n=1,

(1)

(2)

n6∈A

almost surely, where AN = A ∩ {1, 2, . . . , N }. From (1) and (2) and thefact that un = 1G for all n ∈ A, we see that limN →∞ |ANN | = 0, which is acontradiction since A has positive upper density.

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Michael Kelly
Department of Mathematics, University of Michigan, 530 Church St, Ann Arbor,
MI 48109, USAe-mail: michaesk@umich.edu
Thái Hoàng Lê
Department of Mathematics, University of Mississippi, Hume Hall 305, University,
MS 38677, USAe-mail: leth@olemiss.edu