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I have the following set of $R^3$:

$$ S=\{(x,y,z) \in \mathbb{R}^3: \, x^2+y^2-z^3=1\} $$

It is immediate to see that $S$ is a regular surface. How may I find an atlas? When $z \neq 0$, as $z=\sqrt[3]{x^2+y^2-1}$, we have the parameterization $x(u,v)=(u,v,\sqrt[3]{u^2+v^2-1})$ but what happens when $z=0$?

TheWanderer
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1 Answers1

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Hint: At least one of $x$, $y$ and $z$ is not zero. So the open sets $\{x< 0\}$, $\{x>0\}$, $\{y < 0 \}$, $\{y>0\}$ and $\{z\neq 0\}$ cover $S$.

Since you already found a parametrization for $z\neq 0$, I'm sure you will find parametrization for the coordinate neighborhoods $x < 0$, $y < 0$ and so on. Then you are left to show that the change of coordinates is smooth.

Babelfish
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  • So, such an atlas has 5 charts, two for both the cases $x \neq 0$ and $y \neq 0$ and that I found before? – TheWanderer Aug 31 '18 at 15:10
  • @TheWanderer Yeah, I guess the open sets are more like ${x>0}$, ${x<0}$ etc... I'll edit my answer. (Edit: Which means, yes, 5 charts) – Babelfish Aug 31 '18 at 15:23