Construct two examples of covering spaces path connected with 6 sheets over the Torus T that are not isomorphic. And prove that they are not isomorphic.
Lets consider $G=\mathbb{Z}^2$ acting with translation in $\mathbb{R}^2$. In this case, $\pi_1(\mathbb{R}^2/\mathbb{Z}^2 )= \mathbb{Z}^2$, thats abelian.
So I need subgoups of index 6 in $\mathbb{Z}^2$.
For example I have $K=\langle(3,0),(0,2)\rangle \subset \mathbb{Z}^2 $, this is the covering space $p_k: \mathbb{R}^2/K \to \mathbb{R}^2/\mathbb{Z}^2 $.
In this case $\mathbb{R}^2/K $ its homeomorphic to the Torus and its a covering space of $6$-sheets.
And for example $B=\langle(3,1),(0,2)\rangle$ this also works ?, such as
$p_B: \mathbb{R}^2/B \to \mathbb{R}^2/\mathbb{Z}^2 $.
Thanks for the help.
And how can I show that they are not isomorphic?
Best Answer
Since isomorphism classes of covering spaces correspond to conjugacy classes of subgroups, it suffices to show that the two subgroups are not conjugate. This is easy in this case, because $\mathbb{Z}^2$ is abelian.
Edit: To be more explicit, $\mathbb{Z}^2$ is abelian, so for any $g \in \mathbb{Z}^2$, $gBg^{-1} = Bgg^{-1} = B$, so you just have to show that $B \neq K$. Clearly, $(3,1) \in B$. If we assume $(3,1) \in K$, we know
$$a(3,0) + b(0,2) = (3,1)$$
for integers $a,b$. But this is impossible. Then, $(3,1) \not \in K$, so $K \neq B$, meaning $gBg^{-1} \neq K$ for all $g$.