**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 $$|\{u_ix_1\}+\dots+\{u_ix_r\}-\{u_iy_1\}+\dots+\{u_iy_r\}|\leq 2r \frac{\delta N}{4}<N$$ 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 $|z_i-\{xu_i\}|\leq |z_i|+|\{xu_i\}|\leq \frac{\delta N}{4}+\frac{N}{2}<N$ 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!

notsurjective: in fact, $\mathrm{Bohr}(\Gamma,\delta)$ is Freiman-isomorphic to order $r$ to asubsetof $\Lambda\cap[-\delta N/4,\delta N/4]^k$. $\endgroup$2more comments