12
$\begingroup$

I find the sphere example underwhelming. Sure I can see that one open patch will not cover it, but it still manages to cover it mostly. So much so that you can go ahead and, say, calculate the area of a sphere using only one patch

$$\sigma(u,v) = (r \cos(u) \sin(v), r \sin(u) \sin(v), r \cos(v))$$ with $$ u \in \Omega_u = (0,\pi), \qquad v \in \Omega_v = (0, 2\pi) $$ Then $$\sigma_{Area} = \int_{\Omega_u}\int_{\Omega_v}\sqrt{\sigma_u^2 \sigma_v^2-\sigma_u \cdot\sigma_v} du dv = \int_{\Omega_u}\int_{\Omega_v}\sqrt{r^4 \sin(v)^2} du dv \\ \\= r^2 \int_0^\pi \int_0^{2\pi}|\sin(v)| du dv \\ \\ = 2 \pi r^2 \int_0^{\pi} \sin(v) dv \\ \\ = 4 \pi r^2$$

The problem for me when trying to understand differential geometry is that the books all too often mention the sphere as an example of something needing an atlas (which seems, to me, to be pragmatically false) then move on to generalized theorems in $n$ dimensions and very quickly loose me.

I'm sure that for doing something you may sometimes need more than one patch on the sphere but I would appreciate an example that I can easily understand, and compute things with, but that requires some extra care because I cannot use only one patch and get away with it. (The torus also allows me to cheat).

In particular I'd like an example of a bounded smooth surface that is easy to understand but requires at least two coordinate patches to, say, compute the total surface area. With it's patches given explicitly. Pretty Please?!

Thank you so much in advance.

Edit: I appreciate all the answers below. I guess the question was not well formed in the first place, however so I selected the answer that addresses the cheating that I brought up in the best way.

$\endgroup$
  • $\begingroup$ Many standard examples are somewhat like the sphere in that they consist only of large patches and adding points at the boundary of the patch. For example, this describes most examples in algebraic geometry. For more patches, think about a projective space. $\endgroup$ – Michael Burr Jul 21 '15 at 20:54
  • $\begingroup$ Yeah, that's my problem. I tried to construct a smooth surface like the one I'm asking for, but it was harder than I thought to come up with the surface patches. I was just hoping someone had done this already and could provide it to me. $\endgroup$ – amcalde Jul 21 '15 at 20:57
  • $\begingroup$ Try a hyperbolic surface. $\endgroup$ – Neal Jul 21 '15 at 20:58
  • 4
    $\begingroup$ Your question is not clear to me. Are you after a manifold which does not admit a dense coordinate chart? Is that it? There are many features of the sphere which require more than one patch. For example, in order to describe a vector field on the sphere, you have to take into account two patches. (Note: every vector field on the sphere is uniquely determined by its restriction to one patch, but nevertheless, a generic vector field on one patch does not extend to the whole sphere). $\endgroup$ – Amitai Yuval Jul 21 '15 at 21:13
  • $\begingroup$ @AmitaiYuval that's a good point. Maybe my question was ill posed. I'm not sure I could say it better without some one-on-one face-to-face time with someone who really understands this stuff. Like a teacher (the DG teacher I never had!) $\endgroup$ – amcalde Jul 21 '15 at 21:23
12
$\begingroup$

You can compute volume of any connected manifold using just one patch - a connected manifold admits a Morse function with 1 maximum (Any manifold admits a morse function with one minimum and one maximum) and thus is a union of a large ball (the unstable manifold of the maximum) and a subset of strictly smaller dimension (the union of all other unstable submanifolds).

However a) the patch would in general be quite complicated (not easy to write explicitly) b) computing volumes is very far from the only thing one wants to do.

$\endgroup$
  • 2
    $\begingroup$ Thanks. This is very interesting. So you can always cheat in this way. This leads me to believe that I need to rewrite my question. $\endgroup$ – amcalde Jul 21 '15 at 21:26
5
$\begingroup$

If we start from a sphere and keep attaching handles, we get a surface with high genus.

The number of coordinate patches so needed are at least the genus $g$ (plus two).

enter image description here

If we identify a detached handle with an open cylinder, it is clear that we need at most two extra patches every handle. On the other hand, we need at least one extra patch to compensate the change of the Euler characteristic. Hence the number $\eta$ of the needed coordinate patches is:

$$ g+2 \leq \eta \leq 2g+2.$$

$\endgroup$
  • $\begingroup$ How do I attach a handle? I've never seen this done except (only) through pictures. $\endgroup$ – amcalde Jul 21 '15 at 20:59
  • $\begingroup$ @amcalde: you have to take two disjoint disks in your surface and take them as "endpoints" of the handle. As an alternative, you may just keep drilling holes through the surface. Every hole change the Euler characteristic by one, and the needed coordinate patches accordingly. $\endgroup$ – Jack D'Aurizio Jul 21 '15 at 21:03
  • $\begingroup$ Nice picture, can you tell me how to construct patches for this surface (or give a reference)? Also I think I can cheat here at least: any surface where I can arrange the "holes" to lie in the plane like that, I can mostly cover with 2 patches only. No matter the genus. This is more than one though. I just need to get the patches! $\endgroup$ – amcalde Jul 21 '15 at 21:04
  • $\begingroup$ Hmm. I'll think on your comment. $\endgroup$ – amcalde Jul 21 '15 at 21:05
  • 1
    $\begingroup$ A patch with holes is bad. OK. $\endgroup$ – amcalde Jul 21 '15 at 21:33
3
$\begingroup$

"Glue" the lower hemisphere of the sphere $x^2+y^2+z^2 = 1$ with the part of $x^2+y^2-z^2 = 1$ with $z \geq 0$ (one-sheeted hyperboloid). It will overlap smoothly in the circle $\{ (x,y,z) \in \Bbb R^3 \mid x^2+y^2 = 1,\,z = 0\}$. For the lower part use the usual parametrization of the sphere: $${\bf x}(u,v) = (\cos u \cos v, \cos u \sin v, \sin u), \quad 0 \leq v \leq 2\pi, \,\pi \leq u \leq 2\pi,$$ and for the upper part use: $${\bf x}(u,v) = (\cosh u \cos v, \cosh u \sin v, \sinh u), \quad u \geq 0, \,0 \leq v \leq 2\pi$$

This surface is not bounded, but if you really want to, you can try to come up with a third surface to "glue" smoothly to the intersection of the hyperboloid with a horizontal plane, say $z = 1$, to get a bounded surface.

I think you get my point: take surfaces you know well and "glue" them.


Aha! A nice shuttlecock it is.

enter image description here

From another angle:

enter image description here


Edit: if we rotate this surface around the vertical axis, then we get a surface that can't be expressed globally as $z = f(x,y)$ for some function $f$. I know the image is crappy but I'm stupid with computers and can't do better I just wanted to give the idea here, the overlaps are smooth.

enter image description here

$\endgroup$
  • $\begingroup$ Lemme see if I can make an image with Mathematica. $\endgroup$ – amcalde Jul 21 '15 at 21:08
  • $\begingroup$ I'm doing it. I'm just too slow.. $\endgroup$ – Ivo Terek Jul 21 '15 at 21:09
  • $\begingroup$ I did it :D ${}{}$ $\endgroup$ – Ivo Terek Jul 21 '15 at 21:18
  • 1
    $\begingroup$ I think I have an image in Mathematica, but I can't put it in a comment. I have a quibble: This is good, two patches, but they don't overlap at all, so hrmm, not sure if there is a normal patch that would cover the transition. $\endgroup$ – amcalde Jul 21 '15 at 21:20
  • $\begingroup$ Yeah, you're right. I wonder how that patch would be like. $\endgroup$ – Ivo Terek Jul 21 '15 at 21:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.