Consider the following maps:

$\sigma: \mathbb{R}^2 \to \mathbb{R}^3, (u,v)\to(u,v,u^2+v^2)$, $\Sigma: (0,+∞)\times (0,2\pi)\to \mathbb{R}^3, (r,\phi)\to(r\cos \phi, r \sin \phi, r^2).$

(i) Determine the range of the two surface patches and give an atlas for $S$.

(ii) Determine the transition map $\Phi$ between the two surface patches. State precisely its domain and range.

I have shown that both $\sigma$ and $\Sigma$ are regular, and both are for the same quadric surface $S: z=x^2+y^2$, which is a paraboloid.

Below is what I am not sure:

  1. The range of $\sigma$ is $\mathbb{R}\times\mathbb{R}\times [0,+∞]$, and the range of $\Sigma$ is $[-r,r]\times[-r,r]\times[0,+∞)$?

  2. $\Sigma$ is the expression of $\sigma$ in polar coordinates, where $u=r\cos\phi$, $v=r\sin\phi$, $u^2+v^2=r^2$. But how can it help with the following steps?

Thanks in advance! :)


1 Answer 1


You're right that the surface is the quadric $S = \{ (x, y, z) \in \mathbb R^3 : z = x^2 + y^2 \}$.

The ranges of these coordinate patches should be two-dimensional. They should be subsets of $S$. For example:

  • Every point in $S$ is in the image of the map $\sigma$, so the range of $\sigma$ is the whole of $S$.

  • However, $\Sigma$ is only defined for $r > 0$ and $0 < \phi < 2\pi$, so the points $ \{ (u, 0, u^2 ) : u \geq 0 \} \subset S$ are excluded from the range of $\Sigma$.

Finally, your statement that "$\Sigma$ is $\sigma$ expressed in polar coordinates" is a statement about how you would "transition" from the $\Sigma$ coordinates to the $\sigma$ coordinates. In other words, you are describing the transition map, $$ \Phi : (0, \infty) \times (0, 2\pi) \to \mathbb R^2 \setminus \{ (u, 0) : u \geq 0\}, \\ \Phi(r, \phi) = ( r \cos \phi, r \sin \phi)$$ which takes you from the $\Sigma$ coordinate system and the $\sigma$ coordinate system.

To go the other way, from the $\sigma$ coordinate system to the $\Sigma$ coordinate system, you would use the inverse transition map, $$ \Phi^{-1} : \mathbb R^2 \setminus \{ (u, 0) : u \geq 0\} \to (0, \infty) \times (0, 2\pi), \\ \Phi^{-1}(u, v) = (\sqrt{u^2 + v^2}, \tan^{-1}(v / u) ) $$ where $\tan^{-1}$ is taken to lie between $0 $ and $2\pi$.

Notice how the set $\{ (u, 0) : u \geq 0\} \subset \mathbb R^2$ is excluded from the range of $\Phi$ and from the domain of $\Phi^{-1}$. This is because the image of $\{ (u, 0) : u \geq 0\} $ under $\sigma$ does not overlap with the range of $\Sigma$. Transition maps are only defined in the "overlap region" between the two coordinate systems.

  • $\begingroup$ Sorry I didn't quite follow your explanation for the range. Isn't $S$ three-dimensional? The co-domain of $\sigma$ and $\Sigma$ are $\mathbb{R}^3$, which is three-dimensional as well? Then for the quadric surface $z=x^2+y^2$, how could $z$ go to negative to make the entire $S$ as its range? $\endgroup$ Oct 26, 2018 at 22:46
  • 1
    $\begingroup$ @EvelynVenne The universe is three-dimensional, but the surface of the Earth is two-dimensional. In the same way, $\mathbb R^3$ is three-dimensional, but the image of $S$ is a two-dimensional surface inside $\mathbb R^3$. As for your second question, $z$ is non-negative on the whole of $S$. There's nothing wrong with that. $\endgroup$
    – Kenny Wong
    Oct 26, 2018 at 23:12

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