Working on a problem on matroids, I've already ask a question about some subtori. Here's the link to a previous problem: Topological subspace in $(S^{1})^{n}$
Anyway, here's another problem related to the previous one: Let $A\in M_{m,n}(\mathbb{Z})$ be a matrix of $m$ rows and $n$ columns with integer entries.
Let $\mathcal{P}=\{(h_{1},\ldots,h_{m})\in\mathbb{Z}^{m}|\mathrm{rank}(A)= \mathrm{rank}(A|(h_{1},\ldots,h_{m}))\}$ where $(A|(h_{1},\ldots,h_{m}))$ is the augmented* matrix associated to the inhomogeneous system with vector of coefficients $(h_{1},\ldots,h_{m}).$
*http://en.wikipedia.org/wiki/Augmented_matrix
For $(h_{1},\ldots,h_{m})\in\mathcal{P}$ let $S_{(h_{1},\ldots,h_{m})}\subseteq\mathbb{R}^{n}$ be the space of solutions of the inhomogeneous system $(A|(h_{1},\ldots,h_{m})).$
Thanks to Topological subspace in $(S^{1})^{n}$, if we denote $\pi:\mathbb{R}^{n}\longrightarrow(S^{1})^{n}$ the projection $(x_{1},\ldots,x_{n})\mapsto(e^{2\pi ix_{1}},\ldots,e^{2\pi ix_{n}}),$ we can deduce that $\pi(S_{(h_{1},\ldots,h_{m})})\subseteq(S^{1})^{n}$ is a subtorus of dimension $r=\mathrm{rank}(A).$ The question is the following:
Let $\bigcup_{(h_{1},\ldots,h_{m})\in\mathcal{P}}\pi(S_{(h_{1},\ldots,h_{m})})\subseteq(S^{1})^{n},$ is this set a disjoint union of a finite number of tori?
The answer seems to be affermative: using the Gauss-Jordan algorithm I can see that there are only a finite number of effective translations. However, I haven't found a complete and clean proof yet.
You are correct, that union is indeed a disjoint union of finitely many tori. Let me give a proof, after translating the question into the notation of linear transformations.
Your matrix $A$ represents a linear transformation $T : \mathbb{R}^n \to \mathbb{R}^m$ which is defined over the integers, meaning that $T(\mathbb{Z}^n) \subset \mathbb{Z}^n$. Your question is whether the set $\pi(T^{-1}(\mathbb{Z}^m)) \subset (S^1)^n$ is a disjoint union of finitely many tori.
To see that this is true, consider the set $T^{-1}(\mathbb{Z}^m)$, which is the disjunion union of affine subspaces $T^{-1}(h)$ over $h \in \mathbb{Z}^m$. The group $\mathbb{Z}^n$ acts by translation on the subspaces $T^{-1}(h)$ for $h \in \mathbb{Z}^m$, and it suffices to prove that there are only finitely many orbits for this action, because those orbits correspond one-to-one to the components of $\pi(T^{-1}(\mathbb{Z}^m))$. And since $\mathbb{Z}^n$ acts cocompactly on $\mathbb{R}^n$, it suffices to exhibit a positive lower bound to the distance between $T^{-1}(h)$ and $T^{-1}(h')$ for $h \ne h' \in \mathbb{Z}^m$.
To find this lower bound we work in the image $\mathbb{R}^m$, but we do not use the ordinary metric on $\mathbb{R}^m$ given by the Pythagorean formula. Instead, consider the linear subspace $V = T^{-1}(0) \subset \mathbb{R}^n$, let $V^{\perp}$ be its orthogonal complement, and note that $T$ restricts to a vector space isomorphism $T : V^\perp \to \mathbb{R}^m$. Take the $\mathbb{R}^n$ metric restricted to $V^{\perp}$, and push that forward to a metric on $\mathbb{R}^m$ denoted $d(x,x')$ (equivalently, pick an orthonormal basis of $V^\perp$, push that forward to get a basis for $\mathbb{R}^m$, and use the Pythagorean formula with respect to that basis to define the metric on $\mathbb{R}^m$).
So now with two further observations the proof will be done. First, $d(x,x')$ is equal to the perpendicular distance between the affine subspaces $T^{-1}(x)$ and $T^{-1}(x')$ (you can see this by noticing that for any common perpendicular line $L$ to that pair of subspaces, the line $L$ is parallel to $V^\perp$). It therefore suffices to find a positive lower bound to $d(h,h')$ for $h \ne h' \in \mathbb{Z}^m$. And, second, this positive lower bound exists because $\mathbb{Z}^m$ is a discrete cocompact subgroup of $\mathbb{R}^m$ which preserves the distance $d(x,x')$.