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Can we provide the set $\{(x,y,z)\in\mathbb{R^3}|x^2+y^2=1\}$ with a 2-dimensional manifold structure involving only 1 chart? I can see it with 2 charts with cylindrical coordinates, but not with only one...

Lisa
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    iAccording to your definition, Is a chart a map from $M$ to any open subset of $\mathbb{R}^2$? Or is is a map from $M$ to any open ball of $\mathbb{R}^2$? – Jason DeVito - on hiatus Nov 23 '12 at 16:24
  • @Jason: I'll go for any open subset...what kind of difference does it make in the global theory? – Lisa Nov 23 '12 at 16:30
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    Well, Hans's answer now handles it. If you restrict to open balls, then you need at least 2 charts. If you restrict to general open subsets of $\mathbb{R}^2$ an annulus (or a unit disc with origin removed) works. – Jason DeVito - on hiatus Nov 23 '12 at 16:38

1 Answers1

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You can't do it if you want a chart that is homeomorphic to (e.g.) the unit disk in $\mathbb{R}^2$. The latter is homotopy equivalent to a point, but your manifold is not. But you can do it with the unit disk with the origin removed.

Hans Engler
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  • I see the problem. How would you do it in that case, then? – Lisa Nov 23 '12 at 16:31
  • Map points in $\mathbb{R}^2$ with polar coordinates $(r, \theta), , 0 < r < 1$, to points with cylindrical coordinates $(1, \theta, \tan \left( \pi r - \frac{\pi}{2} \right))$, or something like that – Hans Engler Nov 23 '12 at 16:58
  • But then you're taking teta in (0,2.pi)? In which case it doesn't cover the space. Or in [0,2.pi)? In which case it is not open...I am confused! – Lisa Nov 23 '12 at 17:43
  • No, I'm taking points from the punctured disk. In Cartesian coordinates, the map is $(u,v) \mapsto (\frac{u}{\sqrt{u^2 + v^2}}, \frac{v}{\sqrt{u^2 + v^2}}, \tan \left( \pi \sqrt{u^2+v^2} - \pi/2 \right))$. – Hans Engler Nov 23 '12 at 22:11