# Find the atlas for the quadric surfaces and the transition map between its two surface patches

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?

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.

• 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? Oct 26, 2018 at 22:46
• @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. Oct 26, 2018 at 23:12