He guys,
I have two questions regarding the following:
Consider the three-manifold $\mathbf{T}^3 = S^1 \times S^1 \times S^1$ and let $S_n$ be the permutation group acting on $n$ letters. Let $\phi:\pi_1(\mathbf{T}^3) \to S_n$ be an homomorphism hence it is also a permutation representation of the fundamental group of the three-torus. Also, let $\mathcal{O}(\phi)$ be the set of orbits of $\phi(\pi_1(\mathbf{T}^3))$ acting on $X = \{1,\dots,n\}$.
My question is the following. Does this map $\phi$ also determine an unramified covering of $\mathbf{T}^3$? If so, is this again a sum of three-tori which are in one-to-one correspondence with the orbits in $\mathcal{O}(\phi)$? How do I see this / what references should I consult?
What I do know is that the stabilizer $G_{\xi}$ on a point of an orbit $\xi \in \mathcal{O}(\phi)$ is a finite index subgroup of $\pi_1(T^3) = \mathbf{Z}^3$ and from that I can see that these are again three-tori. Denote by $|\xi|$ the size of the orbit $\xi$, then $|\xi| = [G:G_{\xi}]$ and so $G_{\xi}$ corresponds to a $|\xi|$-fold covering of the three-torus. However, I don't know whether this is enough / most general.
So my other question would be: Do all finite index subgroups of $\mathbf{Z}^3$ always correspond to a three-torus? Intuitively it is clear because these are just sublattices and hence are again three-tori.
In two dimensions, we know that the above statements are true and that due to Riemann-Hurwitz these unramified coverings are again two-tori.
Best,
Every finite index subgroup of $\mathbb{Z}^3$ corresponds to a 3-torus exactly as you suspect. Think of it this way: You obtain the torus by looking at $\mathbb{R}^3 / \mathbb{Z}^3$. A finite index subgroup is generated by basis elements $v_1, v_2, v_3 \in \mathbb{Z}^3$. However, we can write a(n invertible) linear map $f: \mathbb{R}^3 \to \mathbb{R}^3$ so that $f(v_i)$ are the standard basis vectors. It follows that $\mathbb{R}^3/\langle v_1, v_2, v_3\rangle \cong \mathbb{R}^3/\mathbb{Z}^3$, and so it is a 3-torus.