I am reading the book "Elementary Differential Geometry" by Andrew Pressley.
I am trying to show the circular cylinder $S=\{(x,y,z) \in \mathbb{R^3} \mid x^2 + y^2 =1\}$ can be covered by a single surface patch and so is a surface.
The hint is to let $U$ be an annulus.
We have an open set $U=\{(u,v) \in \mathbb{R^2} \mid 0<u^2 + v^2 < \pi\}$.
We also let $r=(u^2+v^2)^\frac{1}{2}$ Why?
We then let $\delta :U\to \mathbb{R^3}$ defined by $\delta(u,v)=(\frac{u}{r} \cdot \frac{v}{r} \cdot \tan(r-\frac{\pi}{2}))$
How can we see that the circular cylinder $S$ can be covered by a single surface patch?
What does it mean to say "covered by a surface patch"?
What is the difference between a surface and a surface patch?
Best Answer
I think you can cover it by the two surface patchs : $$\sigma_1 : (0,2\pi)\times\mathbb{R}\to\mathbb{R^3}$$ $$\sigma_1(\theta,z)=(cos\theta,sin\theta,z)$$ and $$\sigma_2 : (-\pi,\pi)\times\mathbb{R}\to\mathbb{R^3}$$ $$\sigma_2(\theta,z)=(cos\theta,sin\theta,z)$$ "covered by surface patch" means (for example) that the support of $\sigma$ ; $\sigma((0,2\pi)\times\mathbb{R})$ is exactly the surface S, if it's not the case you must cover it by at least two surface paths.