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given is the following hyperboloid: $$H = \{(x,y,z) \mid \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1\},$$ where a,b,c are free parameters. I have to find an $C^\infty$-atlas for H. In order to do this, I have to find one (or more) $C^\infty$-chart, which overlays H completely. In addition to this, I should use cylindrical coordinates.

So, first of all I tried to transform H into cylindrical coordinates: $$H = \{(r,\phi,z) \mid r^2 \left(\frac{\cos(\phi)^2}{a^2} + \frac{\sin(\phi)^2}{b^2}\right) - \frac{z^2}{c^2} = 1\}$$ But now I am having heavy problems to find such a chart for this set.

I tried to do this for the special case a=b=c=1, just to get more clear about the whole thing. In this case, our Hyperboloid is $H^* = \{(r,\phi,z) \mid r^2 - z^2 = 1\}$. I think a chart for this will look like this sketch:

sketch

But I don't know how I can transform this idea to a Hyperboloid with arbitrary parameters.

I am very thankfull for any help or new idea I can get.

Best regards!

Paul85
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  • How many parameters do you expect you will need in one of your charts? – Lee Mosher Jun 19 '14 at 17:24
  • Sorry, I think I don't get your question. – Paul85 Jun 19 '14 at 17:41
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    A chart is a certain function which takes values in a Euclidean space. The dimension of that Euclidean space $=$ the number of parameters should $=$ the dimension of the hyperboloid. Your picture, with arrows pointing to the $z$-axis, seems to indicate that your chart has just one parameter, namely $z$. So that's why I asked: how many parameters do you expect you will need, in a chart for the hyperboloid ${(r,\theta,z) | r^2-z^2=1}$? From your picture, it seems you expect to need just one parameter, and I wanted to check whether or not that was your intent. – Lee Mosher Jun 19 '14 at 19:06
  • I am sorry, I should have mentioned that. My sketch is actually in 3D space, with $\phi$-axis pointing out of picture. So, the arrows don't only pointing to the $z$-axis, but pointing to the $\phi$-$z$-plane. Since $\phi$ doesn't matter in this special case, I decided to draw it 2-dimensionally. – Paul85 Jun 19 '14 at 20:58
  • If you have a good sketch in $r,\phi,z$-space, then by projecting on to two of those coordinates you should have your chart. But, it might not work if you pick the wrong two. For example, it definitely does not work if you pick $r$ and $z$: the projection onto the $r,z$ plane is not a chart, as your sketch clearly indicates. – Lee Mosher Jun 19 '14 at 21:14
  • I could do an orthogonal projection of points from Hyperboloid to the $\phi-z$-plane. But then I need two charts, one for points with $r>0$ and one for points with $r<0$, otherwise the chart wouldn't be bijective. One of these charts looks like: $\psi(r,\phi,z)+ = (0,\phi,z)$ with inverse $\psi(u,v)+^{-1} = (\sqrt({(1 + \frac{v^2}{c^2})(\frac{a^2b^2}{b^2\cos(u)^2 + a^2\sin(u)^2}}),u,v)$. Then I think $D\phi(u,v)_+^{-1}$ has rank of 2, so the chart is $C^\infty$. But I don't know how I can build such a map like descriped in my sketch, so that I only need one chart. – Paul85 Jun 20 '14 at 10:12
  • There is no requirement that an atlas have one chart. As you write, you have to find one or more charts that cover the hyperboloid. It's sounds like you've done that successfully, and the problem is solved. – Lee Mosher Jun 20 '14 at 13:24
  • Yeah, but I thought only one chart is more elegant. But nevermind, two are fine too ;-) The last problem I am facing is to show that $\psi(u,v)^{-1}$ is infinity times continuously differentiable, so of class $C^\infty$. Only if $\psi(u,v)^{-1}$ is satisfying this condition, $\psi$ is a $C^\infty$-chart (by my definition). But I don't know how I can show that. – Paul85 Jun 20 '14 at 14:13

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Here is a summary of the discussion in comments. The hyperboloid, being 2-dimensional, will have charts taking values in $\mathbb{R}^2$. To put it another way, each chart will have two parameters.

For the special case of the hyperboloid $r^2-z^2=1$ expressed using cyclindrical coordinates $(r,\phi,z)$, the idea is to pick two of the three coordinates $r,\phi,z$. But not any two will give a chart: picking $r$ and $z$ does not work because of the equation $r^2 = z^2+1$. Instead, the pair of coordinates $(\phi,z)$ works fine and does define charts, as long as $\phi$ has been properly restricted. Two such charts will cover the hyperboloid: one chart defined by restricting $0 < \phi < 2\pi$; and the other chart obtained by restricting $-\pi < \phi < \pi$. The actual functions that define these charts are obtained from the standard formulas expressing cylindrical coordinates in terms of Euclidean coordinates.

Lee Mosher
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  • Hey, first of all thank you! But from where do you get $0<\phi<2\pi$ for the first, and $-\pi<\phi<\pi$ for the second chart - and why they are overlapping? As I see it in the sketch above, I would make a chart for the "left side" with $r<0$ and one for the "right side" with $r>0$. – Paul85 Jun 21 '14 at 10:36
  • @Paul85---The $r$ coordinate in cylindrical coordinates should always be positive, never negative. To better visualize the $\phi$-coordinate, imagine looking down the $z$-axis from high above. What you see is the $x,y$-plane, on top of which is superimposed the $r,\phi$ polar coordinate system. The inequality $0<\phi<2\pi$ covers everthing except the positive $x$-axis in the $x,y$ plane, and therefore it covers everything except the positive $x$ half of the $x,z$ plane in cylindrical coordinates. Similarly, $-\pi < \phi < \pi$ covers everything except the negative $x$ half of the $x,z$ plane. – Lee Mosher Jun 22 '14 at 13:43
  • "The $r$ coordinate in cylindrical coordinates should always be positive, never negative." - But if you transform the set H into cylindrical coordinates, you get $H = {(r,\phi,z) \mid r^2 \left(\frac{\cos(\phi)^2}{a^2} + \frac{\sin(\phi)^2}{b^2}\right) - \frac{z^2}{c^2} = 1}$. So if you choose a point with positive $r$ from $H$, you always have the same point with negative $r$ in $H$, because in the equality of $H$ you square $r$. Or do I get something wrong? – Paul85 Jun 22 '14 at 14:05
  • In that situation, rather than changing putting a negative sign on $r$ you should add $\pi$ to $\phi$, just like in polar coordinates. – Lee Mosher Jun 22 '14 at 19:30