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

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- Dec 1st 2011, 02:19 PMkierkegaardCovering space
Give an explicit description of the covering space of the torus $\displaystyle T^{2}$ corresponding to the subgroup of $\displaystyle \mathbb{Z}\times\mathbb{Z}$ generated by $\displaystyle (n,0), n>0$.

- Dec 1st 2011, 07:39 PMDrexel28Re: Covering space
You know that the universal cover of $\displaystyle \mathbb{T}^2$ is $\displaystyle \mathbb{R}^2$ and $\displaystyle \text{Aut}_D(\mathbb{R}^2)$ consists of translations by elements of the lattice $\displaystyle \mathbb{Z}^2$. Your subgroup $\displaystyle G$ corresponds to the subgroup $\displaystyle \Gamma_G\subseteq\text{Aut}_D(\mathbb{R}^2)$ consisting of $\displaystyle (x,y)\mapsto (x+n,y)$. Your covering space $\displaystyle \widetilde{\mathbb{T}^2}$ then corresponds to the quotient $\displaystyle \mathbb{R}^2/\Gamma_G$, where (as I hope you know) this means that $\displaystyle (x,y)\sim (x',y')$ if they differ by a $\displaystyle \Gamma_G$-action.

- Dec 1st 2011, 08:03 PMkierkegaardRe: Covering space
- Dec 1st 2011, 08:06 PMDrexel28Re: Covering space
- Dec 1st 2011, 08:36 PMkierkegaardRe: Covering space
Wait, I'm a little confused. It's a cylinder, isn't it? If we consider the relation $\displaystyle (x,y)\mapsto(x+n,y)$ and the quotient map $\displaystyle p:\mathbb{R}\mapsto\mathbb{R}/\sim $, the image of $\displaystyle p$ is a cylinder identifying a pair of opposite sides of a square. I hope to be right.

- Dec 1st 2011, 09:42 PMDrexel28Re: Covering space
Right! Intuitively what we've done is decomposed $\displaystyle \pi_1(\mathbb{T}^2)=G_1\oplus G_2$ ($\displaystyle G_1=\mathbb{Z}\times\{0\}$ and $\displaystyle G_2=\{0\}\times\mathbb{Z}$) which induces a decomposition $\displaystyle \text{Aut}_D(\mathbb{R}^2)=\Gamma_{G_1}\oplus \Gamma_{G_2}$ (where $\displaystyle \Gamma_{G_1},\Gamma_{G_2}$ are the deck transformation $\displaystyle (x,y)\mapsto (x+n,y)$ and $\displaystyle (x,y)\mapsto (x,y+m)$ respectively). Intuitively you can think then of $\displaystyle \mathbb{T}^2=\mathbb{R}^2/(\text{Aut}_D(\mathbb{R}^2))$ as $\displaystyle (\mathbb{R}^2/\Gamma_{G_1})/\Gamma_{G_2}$ where the identifcation $\displaystyle \mathbb{R}^2/\Gamma_{G_1}$ (our identification) corresponds to taking $\displaystyle \mathbb{R}^2$ and wrapping it around itself (horizontally) to get a cylinder, and then modding by $\displaystyle \Gamma_{G_2}$ gives you the vertical identifications giving us the torus.

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