I was watching Gowers' lectures on additive combinatorics, and in them, he proved some structural theorems about Bohr sets. You can find the lectures here. Before I ask my question, let me provide some definitions.
Let $G$ be a finite abelian group, and let $\widehat{G}$ be a a group of characters on $G$.
Definition 1. (Bohr set) Let $\Gamma\subset \widehat{G}$ and $\delta\in [0,2]$. The Bohr set with frequency set $\Gamma$ and width $\delta$ is the set $\text{Bohr}(\Gamma; \delta)= \big\{x\in G: \chi(x)\in e([-\delta,\delta])\ \text{for all} \ \chi\in \Gamma\big\}$, where $e(\alpha)=e^{2\pi i \alpha}$.
Definition 2. (Lattice) 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.
Definition 3. (Freiman homomorphism and isomorphism) Let $k\geq 1$, and let $A,B$ be additive sets with ambient groups $G$ and $H$, respectively. A Freiman homomorphism of order $k$ from $(A,G)$ to $(B,H)$ (or more succinctly from $A$ to $B$) is a map $\phi:A\to B$ with the property that $a_1+\dots+a_k=a'_1+\dots+a'_k \Longrightarrow \phi (a_1)+\dots+\phi(a_k)=\phi(a'_1)+\dots+\phi(a'_k)$ for all $a_1,\dots,a_k,a'_1,\dots,a'_k\in A$. If in addition there is an inverse map $\phi^{-1}:B\to A$ which is a Freiman homomorphism of order $k$ from $(B,H)$ to $(A,G)$, then we say that $\phi$ is a Freiman isomorphism of order $k$, and that $(A,G)$ and $(B,H)$ are Freiman isomorphic of order $k$.
Lemma. (Structure of Bohr sets) 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 $\Lambda\cap [-\delta N,\delta N]^k$ for some lattice $\Lambda$ in $\mathbb{R}^k$.
I understood the proof of the above result and it is a pretty deep result and my question is not about that. In the proof, the lattice $\Lambda$ is generated by some vector $\vec{u}=(u_1,\dots,u_k)$ and $N\vec{e}_1=(N,0,\dots,0),\dots,N\vec{e}_k=(0,0,\dots,N)$. In other words, $$\Lambda=\{n_0\vec{u}+n_1N\vec{e}_1+\dots+n_kN\vec{e}_k:n_0,n_1,\dots,n_k \in \mathbb{Z}\},$$ where coordinates $u_i$ are nonzero elements $\mathbb{Z}_N$ which can be identified with integers. Then Gowers says that after a linear transformation, the set $\Lambda \cap [-\delta N,\delta N]^k$ can be converted into the intersection of $\mathbb{Z}^k$ with a convex body.
Here is my question: I am a bit confused with the last sentence. I suppose that in order to demonstrate this, one needs to construct an isomorphism $f$ from $\mathbb{R}^k$ to $\mathbb{R}^k$ such that $f(\Lambda)=\mathbb{Z}^k$. Am I correct? If so, could anyone please show me how to do that? I believe it should be relatively straightforward, but I have no idea how to proceed.
$\def\Z{\mathbb{Z}}\def\R{\mathbb{R}}$By definition $\Lambda$ is a subgroup of $\Z^k$ containing a basis of $\R^k$. It is a basic fact of free abelian groups (following from the existence of Smith normal form) that any subgroup of $\Z^k$ is isomorphic to $\Z^\ell$ for some $\ell \le k$. Since $\Lambda$ contains a basis of $\R^k$ it follows that $\Lambda \cong \Z^k$. In other words there is an isomorphism $f : \Lambda \to \Z^k$. Since $\Lambda \le \R^k$ and a basis for $\Lambda$ is a basis for $\R^k$ we can extend $f$ to a linear isomorphism $f : \R^k \to \R^k$ such that $f(\Lambda) = \Z^k$. Now let $B_0 = [-\delta N, \delta N]^k$. Then $f(\Lambda \cap B_0) = \Z^k \cap B$ where $B = f(B_0)$ is the image of $B_0$ under $B$. Since $B_0$ is a convex body, so is $B$.