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I want to parameterize a unit-cylinder $x^2+y^2=1$ with only one chart in a complete atlas (the sets must be open). The cylinder is in $\mathbb{R}^3$. One way to do the parametrization with two charts is: $$ \textbf{x}(u,v)=(\cos u, \sin u, v)$$ with $v\in \mathbb{R}$ and $0<u< 2\pi$ (this is the open set $U_1=(0,2\pi)\times \mathbb{R}$) and $$ \textbf{y}(\overline{u}, \overline{v}) = (\cos \overline{u}, \sin \overline{u}, \overline{v})$$ with $\overline{v}\equiv v\in \mathbb{R}$ and $-\pi < \overline{u} < \pi$ (this is the open set $U_2 = (-\pi,\pi)\times\mathbb{R}$). So the atlas is $\{ \textbf{x}, \textbf{y}\}$ into $U_1 \times U_2.$ So, it has two charts $\textbf{x}$ and $\textbf{y}$ defined in open sets (this is important: the set $(0,2 \pi]\times \mathbb{R}$ is not open, and a parametrization has to be defined into an open set).

Do someone know an atlas with only one chart? For example, with a reparametrization. The only thing I know is that the atlas with one chart exists in this case$^*$.

*Introduction to Topological Quantum Matter & Quantum Computation, Tudor D. Stanescu

PD: Something similar in General Relativistic to changing Schwarzchild coordinates by Kruskal–Szekeres ones to extend the well-behaved domain.

Clare Francis
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1 Answers1

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You can use the chart $$ \mathbf x(u,v) = \left(\frac{u}{\sqrt{u^2+v^2}}, \frac{v}{\sqrt{u^2+v^2}}, \log \sqrt{u^2+v^2} \right)\,,$$ defined on $\mathbf R^2 \setminus \{(0,0)\}$. The formula is simpler in polar coordinates, $$ \mathbf x(r,\theta) = \left( \cos \theta, \sin \theta, \log r\right)\,. $$

Martins Bruveris
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  • Isn't the polar coordinate map just $\mathbf{x}(r, \theta) = (\cos \theta, \sin \theta, \log r)$? – Brian Tung Mar 09 '17 at 00:59
  • Yes, of course it is. Thanks for spotting it. – Martins Bruveris Mar 09 '17 at 01:01
  • Where is $\theta$ defined? – Clare Francis Mar 09 '17 at 01:28
  • On the same domain as the variable $u$ in the OP. It just makes the formula prettier, but to get the global chart you need to use the first formula. – Martins Bruveris Mar 09 '17 at 08:10
  • If we denote $z =u + \text{i}v=r \text{e}^{\text{i}\theta}$, then why the domain of the chart changes from an open set to a non-open set in order to describe the same surface $S^1 \times \mathbb{R}$? I.e. from $z\in \mathbb{C}-{0}$ or $(u,v) \in \mathbb{R}^2-{(0,0)}$ (open set) to $(r,\theta) \in \mathbb{R}_+ \times [0,2\pi)$ (non-open set). – Clare Francis Mar 09 '17 at 19:06