Covering space

**kierkegaard** · December 1st 2011, 02:19 PM

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 $(n,0), n>0$ .

**Drexel28** · December 1st 2011, 07:39 PM

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.

**kierkegaard** · December 1st 2011, 08:03 PM

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?

**Drexel28** · December 1st 2011, 08:06 PM

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.

**kierkegaard** · December 1st 2011, 08:36 PM

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.

**Drexel28** · December 1st 2011, 09:42 PM

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.

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