Bohr sets in sumsets I: Compact abelian groups

Bohr sets in sumsets I: Compact abelian groups, Discrete Analysis 2025:11, 37 pp.

Since Bogolyubov’s 1939 result concerning Bohr structure inside sets of the form $A+A−A−A$, much has been written about how “smooth” sumsets are, where smoothness is generally measured in
terms of additive structure. This paper features both partition and density results in some discrete
abelian groups ($\mathbb Z$, $\mathbb Z[i]$, and $\mathbb F_q[t]$) and in all compact abelian groups. The main results are accomplished by counting certain linear patterns; for the density results the authors establish a Fourier-analytic
regularity lemma (following an argument of Bourgain), while for the partition results their main tool is the Hales-Jewett theorem.

Let $G$ be an Abelian group, let $\Lambda$ be a finite set of characters on $G$ and let $\eta>0$. The Bohr set $B(\Lambda,\eta)$ is defined to be the set
$\{x\in G:\forall \chi\in\Lambda\ |\chi(x)-1|<\eta\}$. The rank of $B(\Lambda,\eta)$ is defined to be $\Lambda$ and the radius is $\eta$. (Strictly speaking these are not well-defined, since two different choices of $(\Lambda,\eta)$ can give rise to the same set, but this is an unimportant detail that can easily be remedied.) An important argument of Bogolyubov shows that if $A$ is a dense subset of $G$ (in various contexts where that notion makes sense), then the set $A+A-A-A$ contains a Bohr set of bounded rank and radius bounded away from 0. This result is extremely useful in additive combinatorics, because iterated sumsets are ubiquitous in the area and Bohr sets turn out to be highly structured. Thus, one can use Bogolyubov’s argument as a tool to prove many structural results from surprisingly weak hypotheses. A notable example of this is Ruzsa’s proof of Freiman’s theorem.

It is natural to ask whether four sets are necessary to obtain this structure. Bogolyubov’s argument depends on the decay of Fourier coefficients of convolutions, and the decay is faster the more (characteristic functions of) sets are convolved together. Two sets are not enough for the argument to work, though an “almost all” version can be formulated that is sometimes useful. However, for three sets there is rapid decay.

That said, the proof $A+A-A-A$ contains a Bohr set also depends on the fact that this iterated sum/difference set is centred at 0, and in particular contains 0, which is an element of every Bohr set. If one considers a set such as $A+A-A$ this is no longer the case: it is easy to come up with dense sets $A$ such that $A+A-A$ does not contain 0. One way of dealing with this is to weaken the conclusion and show instead that $A+A-A$ may contains a Bohr neighbourhood (that is, a translate of a Bohr set). Another way, which is the way followed in this paper, is to look instead at sums of dilates of $A$. Writing $r.A$ for $\{ra:a\in A\}$, one can consider sets of the form $r.A+s.A+t.A$ with $r+s+t=0$. As long as $r,s$ and $t$ are not too big, one can show that under suitable hypotheses such sets contain large Bohr sets.

The set-up of the paper is, however, more general. In any Abelian group, the map $a\mapsto ra$ is a homomorphism and has index at most $r$. Here the authors let $G$ be a compact Abelian group and take $\phi_1,\phi_2$ and $\phi_3$ to be commuting homomorphisms from $G$ to $G$ such that each $\phi_i(G)$ is of finite index. They prove results of two kinds. One kind is that if $\phi_1+\phi_2+\phi_3=0$, then for every dense subset $A$ (with respect to the Haar measure on $G$), $\phi_1(A)+\phi_2(A)+\phi_3(A)$ contains a Bohr set with rank and radius that depend only on the density of $A$ indexes of the images $\phi_i(G)$. For the other kind they specialize to the case $\phi_1=-\phi_2$, drop the condition that $\phi_1+\phi_2+\phi_3=0$ and prove a colouring result instead. While it is certainly not true that a sumset of the form $\phi(A)-\phi(A)+\psi(A)$ must contain a Bohr set whenever $A$ is dense, if $G$ is partitioned into sets $A_1,\dots,A_k$, then at least one of the sumsets $\phi(A_i)-\phi(A_i)+\psi(A_i)$ must contain a Bohr set (with the rank and radius now depending on $k$ instead of the density of $A$).

This last result lies in the vicinity of a well-known and long-standing open problem of Katznelson and Ruzsa, who asked whether if $\mathbb Z$ is finitely coloured with sets $A_1,\dots,A_k$, then one of the difference sets $A_i-A_i$ must contain a Bohr set. The density version of this question is known to have a negative answer, thanks to a result of Kriz, but it does not seem to be easy to convert this construction into one that works for colourings as well. A variant of the argument in this paper shows that for any $s$, we can at least say that one of the sets $A_i-A_i+s.A_i$ contains a Bohr set.