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Let $C$ be a figure "8" in th $xy$ plane and let $S$ be the cylindrical surface over $C$ that is,

$$S=\{(x,y,z)\in \mathbb{R}^3; (x,y)\in C\}.$$

Is the set $S$ a regular surface ?

So my answer is no because based on the proposition 2 in Do Carmo page 62 the image of local charts should be in this forms: $z=f(x,y)$ or $y=g(x,z)$ or $x=h(y,z)$, but the projection of $S$ in any of the planes doesn't have a one-to-one functions so $S$ in not a regular surface what do you think ?

Javi
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Bernstein
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    I don't get what you mean by "the projection odf $S$ in any of the planes doesn't have a one-to-one functions" – Javi Feb 19 '18 at 10:21
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    see example 5 in the page 64 of Do Carmo – Bernstein Feb 19 '18 at 10:25
  • ok, I've read it. The key in that example is that the projection is not differentiable, but he starts with a parametrization. In this case you can give a regular (differentiable) parametrization of the 8 and extend it to a parametrization of $S$. – Javi Feb 19 '18 at 10:30
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    this is the problem , if we have a parametrization it's easy , but when we have juste the surface , if we find a parametrization which is not differentiable it does not mean that our surface is not regular,and that's why i use the proposition 2 in Do Carmo page 62 – Bernstein Feb 19 '18 at 10:37
  • Read my comment on my answer, I gave you a differentiable parametrization of the 8. – Javi Feb 19 '18 at 10:42
  • How do you conclude by saying "the projection of S in any of the planes doesn't have a one-one functions so S is not a regular surface"? Because projections are almost never injective and if you are following Prop 3 of do Carmo then shouldn't you derive a contradiction by showing that it's not differentiable. –  Feb 13 '19 at 03:18
  • Doesn't the cylindrical surface over the figure 8 contain some 'boundary points'? If the answer is affirmative then I think we are done. –  Feb 13 '19 at 03:20
  • So is your argument correct? – Focus Mar 20 '19 at 10:49
  • @izimath ,thank you for your interest , i think the answer is yes – Bernstein Apr 08 '19 at 14:29

1 Answers1

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It is indeed a regular surface. Recall that the figure 8 is a smooth manifold of dimension 1 and so is the real line. The product of two 1-(smooth)manifolds is a 2-(smooth)manifold, so in this case $S=C\times\mathbb{R}$, which gives you a smooth manifold of dimension 2, which is the same as a regular surface.

EDIT In the answer, I am implicity using a particular topology for the figure 8. If your definition of regular surface requires $S$ to be an euclidean subspace of $\mathbb{R}^3$, this reasoning is not valid, since the figure 8 doesn't admit a differentiable structure with euclidean topology. If this is the case, if you assume $S$ to be regular, then you could project it onto the 8 (the projection is a differentiable map), and by composition of differentiable maps, you would have a regular parametrization of the 8, which is a contradiction.

Javi
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  • thank you @Javi but I am not sur that "8" is an injective curve , what do you think ? – Bernstein Feb 19 '18 at 10:27
  • Yes, it can be injective if you touch the origin only once. Consider the parametrization $(\sin(2s),\sin(s))$ for $s\in\mathbb{R}$. The other two segments where the 8 could touch the origin remain open. – Javi Feb 19 '18 at 10:33
  • Sorry, $s\in (0,2\pi)$ (or any other interval with the same lenght) – Javi Feb 19 '18 at 10:40
  • In this case the topology is no the Euclidean. If your definition of regular surface involves euclidean topology, then it is not valid and $S$ is not a regular surface. – Javi Feb 19 '18 at 10:43
  • thnak you i'am just a bigginer in the differential geometry , now using the book of the carmo , my answer is correct or not ? if S is a reular surface give me a regular parametrization – Bernstein Feb 19 '18 at 10:46
  • Ok, $(\sin(2s),\sin(s),t)$ for $s\in (0,2\pi)$ and $t\in\mathbb{R}$. I'll let you check that it is regular. – Javi Feb 19 '18 at 10:51
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    but you still have two line in the cylinder to cover by a local chart , and this two line are the same , so my answer is correct i think , but thank you very much @javi in maths, it's exchanging ideas that matter – Bernstein Feb 19 '18 at 11:22
  • Let us continue this discussion in chat. – Bernstein Feb 19 '18 at 11:33