# Structural description of Bohr sets in $\mathbb{Z}_N$

Definition 1. Let $$\Gamma\subset \widehat{G}$$ and $$\delta\in [0,2]$$. The Bohr set with frequency set $$\Gamma$$ and width $$\delta$$ is the set $$\text{Bohr}(\Gamma; \delta)= \big\{x\in G: |\chi(x)-1|\leq \delta \ \text{for all} \ \chi\in \Gamma\big\}$$. The size of $$\Gamma$$ is called the rank of the Bohr set.

Definition 2. A subgroup $$\Lambda$$ of $$\mathbb{R}^k$$ is called a lattice if it is a discrete subgroup that generates all of $$\mathbb{R}^k$$ as a vector space.

Lemma. Let $$N$$ be a prime, and let $$\Gamma\subset \widehat{\mathbb{Z}_N}$$ be a set of nontrivial characters of size $$k$$. Also, let $$r\in \mathbb{N}$$ and $$\delta\in (0,\frac{1}{2r})$$. Then the Bohr set $$\text{Bohr}(\Gamma;\delta)$$ is Freiman isomorphic of order $$r$$ to $$\Lambda\cap [-\frac{\delta N}{4},\frac{\delta N}{4}]^k$$ for some lattice $$\Lambda$$ in $$\mathbb{R}^k$$.

Proof. Let $$\Gamma=\{\chi_{u_1},\dots, \chi_{u_k}\}$$ and since $$\Gamma$$ is a set of nontrivial characters, then $$u_1,\dots,u_k$$ are nonzero elements in $$\mathbb{Z}_N$$. Let $$\vec{u}=(u_1,\dots,u_k)$$ and $$N\vec{e_1}=(N,0\dots,0),\dots,$$ $$N\vec{e_k}=(0,0\dots,N)$$ and consider the lattice $$\Lambda=\{n_0\vec{u}+n_1N\vec{e_1}+\dots+n_kN\vec{e_k}: n_0,n_1,\dots,n_k\in \mathbb{Z}\}.$$ Define the map $$\phi: \text{Bohr}(\Gamma;\delta)\to \Lambda \cap \left[-\frac{\delta N}{4},\frac{\delta N}{4}\right]^k$$ as follows: $$x\mapsto(\{u_1x\},\dots,\{u_kx\})$$, where $$\{u_ix\}$$ is the least absolute value residue of $$u_ix \pmod{N}$$, i.e. $$\{u_ix\}\in (-\frac{N}{2},\frac{N}{2}]$$. Indeed, the map $$\phi$$ is well-defined and obviously $$\phi(x)=(\{u_1x\},\dots,\{u_kx\})\in \Lambda$$. Also, if $$x\in \text{Bohr}(\Gamma; \delta)$$. then $$|\chi_{u_i}(x)-1|\leq \delta$$ for each $$i\in [k]$$. In other words, $$|e(\frac{u_ix}{N})-1|=|e(\frac{\{u_ix\}}{N})-1|=2|\sin(\frac{\pi \{u_ix\}}{N})|\leq \delta$$. But by Jordan's inequality, we have $$|\{u_ix\}|\leq \frac{\delta N}{4}$$. Hence $$\phi(x)\in \Lambda \cap \left[-\frac{\delta N}{4},\frac{\delta N}{4}\right]^k$$.

It is not difficult to see that it is a Freiman homomorphism of order $$r$$. Indeed, let $$x_1,\dots x_r,y_1,\dots y_r\in \text{Bohr}(\Gamma;\delta)$$ such that $$x_1+\dots+x_r=y_1+\dots+y_r$$. Then $$u_ix_1+\dots+u_ix_r=u_iy_1+\dots+u_iy_r$$ and hence $$\{u_ix_1\}+\dots+\{u_ix_r\}\equiv \{u_iy_1\}+\dots+\{u_iy_r\} \pmod N$$. But note that $$|\{u_ix_1\}+\dots+\{u_ix_r\}-\{u_iy_1\}+\dots+\{u_iy_r\}|\leq 2r \frac{\delta N}{4} and so $$\{u_ix_1\}+\dots+\{u_ix_r\}=\{u_iy_1\}+\dots+\{u_iy_r\}$$ for each $$i\in [k]$$ and this implies that $$\phi(x_1)+\dots+\phi(x_r)=\phi(y_1)+\dots+\phi(y_r)$$.

I have some issues to show that this map is surjective. Let $$(z_1,\dots,z_k)\in \Lambda \cap \left[-\frac{\delta N}{4},\frac{\delta N}{4}\right]^k$$, then $$z_i\equiv xu_i \pmod N$$ and $$|z_i|\leq \frac{\delta N}{4}$$. Since $$|z_i-\{xu_i\}|\leq |z_i|+|\{xu_i\}|\leq \frac{\delta N}{4}+\frac{N}{2} and hence $$z_i=\{u_ix\}$$ for each $$i\in [k]$$. This implies that $$(z_1,\dots,z_k)=(\{u_1x\},\dots, \{u_kx\})$$. But in order to claim that $$\phi(x)=z$$ we need to be sure that $$x\in \text{Bohr}(\Gamma;\delta)$$. In order words, we need to show that $$|\chi_{u_i}(x)-1|\leq \delta$$. But $$|\chi_{u_i}(x)-1|=\left|e\left(\frac{u_ix}{N}\right)-1\right|=\left|e\left(\frac{\{u_ix\}}{N}\right)-1\right|=2\left|\sin\left(\frac{\pi\{u_ix\}}{N}\right)\right|\leq \frac{2\pi}{N}|\{u_ix\}|=\frac{2\pi}{N}|z_i|\leq \frac{\pi \delta}{2}.$$

For some reason I cannot show that $$|\chi_{u_i}(x)-1|\leq \delta$$ and actually this is not true.

Can someone tell me how to fix that issue please? Thank you!

• What is $\mathbb Z_{N}$? Is it $\mathbb Z/N \mathbb Z$ ? Commented Oct 7, 2023 at 18:15
• @NickS, yes exactly.
– RFZ
Commented Oct 7, 2023 at 18:42
• Not that I've checked it carefully, but I think the map is not surjective: in fact, $\mathrm{Bohr}(\Gamma,\delta)$ is Freiman-isomorphic to order $r$ to a subset of $\Lambda\cap[-\delta N/4,\delta N/4]^k$.
– Seva
Commented Oct 7, 2023 at 19:27
• Your definition of a Bohr set is not quite the one that Gowers uses (see youtu.be/… ); he uses the condition $\chi(x) \in e([-\delta,\delta])$ rather than $|\chi(x)-1| \leq \delta$. Commented Oct 7, 2023 at 23:44
• $\chi(x) \in e([-\delta,\delta])$ is equivalent to $|\chi(x)-1| \leq 2 \sin(\pi \delta)$, so one can prove a similar result after adjusting the value of $\delta$ appropriately. Commented Oct 7, 2023 at 23:59

The map does not have to be surjective.

Let $$N$$ be even. Choose $$\delta\in(0,1)$$ arbitrarily, and suppose that $$\Gamma\subseteq(N/2)^\perp$$; that is, $$\Gamma$$ contains only those characters with $$N/2$$ in the kernel. In this case $$\mathrm{Bohr}(\Gamma;\delta)$$ contains $$0$$ and $$N/2$$, while all other elements come in pairs: $$g\in\mathrm{Bohr}(\Gamma;\delta)$$ if and only if $$-g\in\mathrm{Bohr}(\Gamma;\delta)$$. Therefore, $$|\mathrm{Bohr}(\Gamma;\delta)|$$ is even.

On the other hand, the intersection $$\Lambda\cap[-\delta N/4,\delta N/4]^k$$ has an odd size (it contains $$0$$ and is symmetric). It follows that $$\mathrm{Bohr}(\Gamma;\delta)$$ and $$\Lambda\cap[-\delta N/4,\delta N/4]^k$$ are not isomprphic.

• I apologize for the delayed response to your answer. Unfortunately, I did not have the opportunity to read it until now. So, if I understand correctly, $\Gamma$ is the set of all characters $\chi$ such that $\chi(N/2) = 1$, is that right? I am a bit confused by the notation $\Gamma \subset (N/2)^{\perp}$.
– RFZ
Commented Nov 23, 2023 at 3:41
• @RFZ: Correct - $(N/2)^\perp$ is the set of all characters $\chi$ such that $\chi(N/2)=1$.
– Seva
Commented Nov 23, 2023 at 6:15