# Thread: Covering space

1. ## Covering space

Give an explicit description of the covering space of the torus $T^{2}$ corresponding to the subgroup of $\mathbb{Z}\times\mathbb{Z}$ generated by $(n,0), n>0$.

2. ## Re: Covering space

Originally Posted by kierkegaard
Give an explicit description of the covering space of the torus $T^{2}$ corresponding to the subgroup of $\mathbb{Z}\times\mathbb{Z}$ generated by $(n,0), n>0$.
You know that the universal cover of $\mathbb{T}^2$ is $\mathbb{R}^2$ and $\text{Aut}_D(\mathbb{R}^2)$ consists of translations by elements of the lattice $\mathbb{Z}^2$. Your subgroup $G$ corresponds to the subgroup $\Gamma_G\subseteq\text{Aut}_D(\mathbb{R}^2)$ consisting of $(x,y)\mapsto (x+n,y)$. Your covering space $\widetilde{\mathbb{T}^2}$ then corresponds to the quotient $\mathbb{R}^2/\Gamma_G$, where (as I hope you know) this means that $(x,y)\sim (x',y')$ if they differ by a $\Gamma_G$-action.

3. ## Re: Covering space

Originally Posted by Drexel28
You know that the universal cover of $\mathbb{T}^2$ is $\mathbb{R}$ and $\text{Aut}_D(\mathbb{R}^2)$ consists of translations by elements of the lattice $\mathbb{Z}^2$. Your subgroup $G$ corresponds to the subgroup $\Gamma_G\subseteq\text{Aut}_D(\mathbb{R}^2)$ consisting of $(x,y)\mapsto (x+n,y)$. Your covering space $\widetilde{\mathbb{T}^2}$ then corresponds to the quotient $\mathbb{R}^2/\Gamma_G$, where (as I hope you know) this means that $(x,y)\sim (x',y')$ if they differ by a $\Gamma_G$-action.
So, and please,correct me if I am wrong, can we say that the covering space corresponding this subgroup is the torus itself?

4. ## Re: Covering space

Originally Posted by kierkegaard
So, and please,correct me if I am wrong, can we say that the covering space corresponding this subgroup is the torus itself?
Ok, well, why would you think that? What is your evidence? Write something out for us.

5. ## Re: Covering space

Wait, I'm a little confused. It's a cylinder, isn't it? If we consider the relation $(x,y)\mapsto(x+n,y)$ and the quotient map $p:\mathbb{R}\mapsto\mathbb{R}/\sim$, the image of $p$ is a cylinder identifying a pair of opposite sides of a square. I hope to be right.

6. ## Re: Covering space

Originally Posted by kierkegaard
Wait, I'm a little confused. It's a cylinder, isn't it? If we consider the relation $(x,y)\mapsto(x+n,y)$ and the quotient map $p:\mathbb{R}\mapsto\mathbb{R}/\sim$, the image of $p$ is a cylinder identifying a pair of opposite sides of a square. I hope to be right.
Right! Intuitively what we've done is decomposed $\pi_1(\mathbb{T}^2)=G_1\oplus G_2$ ( $G_1=\mathbb{Z}\times\{0\}$ and $G_2=\{0\}\times\mathbb{Z}$) which induces a decomposition $\text{Aut}_D(\mathbb{R}^2)=\Gamma_{G_1}\oplus \Gamma_{G_2}$ (where $\Gamma_{G_1},\Gamma_{G_2}$ are the deck transformation $(x,y)\mapsto (x+n,y)$ and $(x,y)\mapsto (x,y+m)$ respectively). Intuitively you can think then of $\mathbb{T}^2=\mathbb{R}^2/(\text{Aut}_D(\mathbb{R}^2))$ as $(\mathbb{R}^2/\Gamma_{G_1})/\Gamma_{G_2}$ where the identifcation $\mathbb{R}^2/\Gamma_{G_1}$ (our identification) corresponds to taking $\mathbb{R}^2$ and wrapping it around itself (horizontally) to get a cylinder, and then modding by $\Gamma_{G_2}$ gives you the vertical identifications giving us the torus.

So, in short, yes. It's the cylinder.