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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?

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  • $\begingroup$ I think you need two surface patches, don't you? $\endgroup$
    – TonyK
    Commented Feb 20, 2016 at 16:40
  • $\begingroup$ "S is covered by a single surface patch" means, that $S$ is homeomorphic to an open subset $U$ in $\mathbb{R}^2$. A homeomorphism of $U$ and $S$ is called a surface patch. $\endgroup$
    – user128245
    Commented Apr 29, 2017 at 13:24
  • $\begingroup$ The unit cylinder $S^1 \times \mathbb{R}$ is homeomorphic to the punctured plane $\mathbb{R}^2 \setminus \{ 0 \}$ or any annulus. In your case, where $U=\{(u,v) \in \mathbb{R^2} \mid 0<u^2 + v^2 < \pi\}$, you can take $\sigma : U \to S, \, (u,v) \to (\frac{u}{\sqrt{u^2 + v^2}}, \frac{v}{\sqrt{u^2 + v^2}}, \ln(\frac{\pi}{\sqrt{u^2 + v^2}} - 1))$ or $\sigma : U \to S, \, (u,v) \to (\frac{u}{\sqrt{u^2 + v^2}}, \frac{v}{\sqrt{u^2 + v^2}}, \tan(\sqrt{u^2 + v^2} - \frac{pi}{2}))$. To see that these patches cover the whole $S$, take any point $p \in S$ and find $(u,v) \in U$ s.t. $\sigma(u,v) = p$ $\endgroup$
    – user128245
    Commented Apr 29, 2017 at 13:26

1 Answer 1

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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.

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  • $\begingroup$ I am new to this and struggling to wrap my head around this concept. Why is the function you have mentioned special?And how can you rigorously check if a surface patch covers the whole surface? $\endgroup$
    – Al jabra
    Commented Feb 20, 2016 at 16:37
  • $\begingroup$ You must show that every point of S is in the image of at least one of the surface patches $\sigma_1$ or $\sigma_2$.˜ $\endgroup$ Commented Feb 20, 2016 at 16:40
  • $\begingroup$ Thank you, lastly what is the difference between surface and a surface patch? $\endgroup$
    – Al jabra
    Commented Feb 20, 2016 at 16:44
  • $\begingroup$ A surface patch is a parametrization of a part of a surface, this part can be the entire surface. $\endgroup$ Commented Feb 20, 2016 at 16:47
  • $\begingroup$ @Mathstudent this is not the answer. Moreover, it can be easily covered by a single surface patch. $\endgroup$
    – user128245
    Commented Apr 29, 2017 at 12:45

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