Using Bohr sets to locate arithmetic progressions

Question

I've just started to read about additive combinatorics and I'd like to know how I can use Bohr sets to make a statement about arithmetic progressions in a given subset $A$ of an Abelian group $Z$ (the ambient group).

For example: Green-Tao states that if $A$ is the set of all primes then we can always find an arithmetic progression of length $k$ in $A$ for all $k \in \mathbb N$.

Can I do the following: If $B(S, \rho)$ is a Bohr set with certain properties then it contains an arithmetic progression of length say, $k=10$?

If yes: how exactly does it work?

If no: why not?

A question directly related to this is: why were Bohr sets introduced? I've read that they replace the orthogonal complement $S^\bot$ of a subset $S$ of $Z$ when $S^\bot$ is not a subgroup of $Z$ because for example $Z = Z_p$ for $p$ prime.

And one more question is: if $F$ is a linear space then every linear subspace is a Bohr set (Tao/Vu, p. 166). So for some reason we want to study linear subspaces of $Z$ or subgroups if possible. How do we use this structure or in particular the closedness with respect to addition of subsets? I'm slightly confused about whether we're interested in subsets or subspaces of $Z$.

Edit

Let $S \subset \widehat{Z}$ be a set of characters of $Z$ and let $\rho > 0$ be a real number. Then a Bohr set is defined as $$ \operatorname{Bohr}(S, \rho) := \{ z \in Z \mid \sup_{\chi \in S} \left | \chi(z) - 1 \right | < \rho \}$$

Thanks for your help.

Answer 1

Why Bohr sets?

For a given ambient group $Z$ we want to know about arithmetic progressions in a subset $A$ of $Z$. If the set $A$ is a group it means that $A$ has a high degree of additive structure. Consider for example the integers with the subgroup $A = 2 \mathbb Z$. Then clearly, $A$ contains arithmetic progressions of arbitrary length.

Unfortunately, in general, the sets we study aren't (sub-)groups.

This is why we want Bohr sets: They are the next closest thing with a comparable amount of structure. To see how this works let's give the definition of a Bohr set and then consider an example.

For an ambient group $Z$ and a set of characters $S \subset \widehat{Z}$ and a radius $\rho > 0$ we define the Bohr set as follows $$ Bohr ( S, \rho) := \{ z \in Z \mid \sup_{\chi \in S} |\chi (z) - 1| \leq \rho \}$$ Note: Subgroups of $\mathbb Z_n$ are Bohr sets.

Example:

For $\mathbb Z_{12}$, $S = \{6\}$ and $\rho = 0.4$ we get $Bohr(S, \rho) = \{0, 2, 4, 6, 8, 10 \}$.

For $S = \{2\}$ we get $Bohr(S, \rho) = \{0, 6\}$.

As you can see, the Bohr set of $S$ is the orthogonal complement $S^\bot$ of $S$ and is also a subgroup of $Z$.

Now we see how we can use Bohr sets to say something about arithmetic progressions in a subset $A$ of $Z$.

Answer 2

I just saw this old question and saw that the main question hadn't actually been answered, so will do so here for future explorers.

Does a Bohr set $B(S,\rho)$ contain an arithmetic progression of length $k$?

Yes, if $\lvert S\rvert \log(k/\rho)\leq c\log N$ for some small constant $c>0$.

The trick here is note that $0\in B(S,\rho)$ always and by the triangle inequality if $t\in B(S,\rho/r)$ then $rt\in B(S,\rho)$.

In particular for $B(S,\rho)$ to contain an arithmetic progression of length $k$ it's enough to show $B(S,\rho/k)$ contains at least one non-zero element.

For this we use the lower bound $\lvert B(S,\rho)\rvert \geq (\rho/8)^{\lvert S\rvert}N$ (proved e.g. in Tao and Vu and many other places, via a simple covering argument). In particular if $\lvert S\rvert\log(k/\rho)\leq c\log N$ for some small constant $c>0$ then $\lvert B(S,\rho/k)\rvert\geq 2$ and hence $B(S,\rho/k)$ contains at least one non-zero element as required.