0

In this question covering spaces of a torus are discussed. There, it is said that the only (I guess connected) covering spaces of the torus are the torus itself, $\mathbb R^2$ and $\mathbb R\times S^1$, being $\mathbb R^2$ the universal cover.

I think I can understand how $\mathbb R\times S^1$ covers the torus: it is an analogue for the covering $\mathbb R\rightarrow S^1$, but now instead of a helix we have got an infinite cylinder. Thus, it would be a $\mathbb Z$-sheeted covering space.

But I do not know how to visualise $\mathbb R^2\rightarrow S^1\times S^1$. Could you give some suggestions, please? And am I right with the visualisation of $S^1\times \mathbb R^2$?

Dog_69
  • 1,877
  • Maybe (I am not sure wether this is what really happens) you can think of a sheet of paper, rolled up infinitely many times to a cylinder and then stuffed into itself infinitely many times. But this may be harder to visualize than the „flattened“ variant with a tesselation of the plane by squares... – Jonas Linssen Mar 31 '20 at 12:59
  • You can think in terms of the explicit covering map $(x, y) \mapsto (e^{i x}, e^{i y})$, where a fundamental domain is the square $[0, 2\pi]\times [0, 2\pi]$. To help you visualize, maybe try thinking about what the image of the line $x=y$ looks like in $S^1 \times S^1$. – William Mar 31 '20 at 13:23
  • 1
    It's going to be hazrd to visualize, because you have "$4$ dimensions" at play (two dimensions of $\mathbb R^2$, and "two dimensions" of $\mathbb Z^2$ acting on it). One way to picture it is to just look at individual squares in $\mathbb Z^2$, and glue their sides so as to actually make a torus. You can also visualize it in two steps: first vertically to get an infinite cylinder, and then fold horizontally, to get a torus – Maxime Ramzi Mar 31 '20 at 15:17
  • @William what do you mean by "the fundamental domain"? I guess that you was traying to explain me that the covering map is defined by your formula on the squares of the form $[2\pi (m-1),2\pi m]\times[2\pi(n-1),2\pi n]$, $m,n\in\mathbb Z$. – Dog_69 Mar 31 '20 at 22:17
  • @PrudiiArca, I have some problems to image the second part, once you have the cylinder, but it isbnot a bad idea. Thank you! – Dog_69 Mar 31 '20 at 22:22
  • @MaximeRamzi just to be sure. Are you saying the same as PrudiiArca? – Dog_69 Mar 31 '20 at 22:23
  • The covering map I described (call it $\varphi$) is defined on all of $\mathbb{R}^2$, but it is not injective in general. $\varphi$ is injective on the interior of the square $[0,2\pi] \times [0,2\pi]$, and in general $\varphi(x,y) = \varphi(x + 2\pi n, y + 2\pi m)$ for any $n, m \in \mathbb{Z}$. The term "fundamental domain" I was using is in reference to the action of $\mathbb{Z}^2$ on $\mathbb{R}^2$, look at this link for a more elaborate explanation: https://en.wikipedia.org/wiki/Fundamental_domain (more accurately the fundamental domain would be $[0,2\pi)\times[0,2\pi)$). – William Mar 31 '20 at 23:34
  • @William Thank you, that was what I understood. – Dog_69 Apr 01 '20 at 09:41

0 Answers0