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this is the exercice 2-2 of Do Carmo's page 67

1- is the set $\{(x,y,z)\in \mathbb{R}^3 ; \;z=0 \text{ and } \;x^2+y^2\leq 1 \}$ a regular surface ?

2-is the set $\{(x,y,z)\in \mathbb{R}^3 ; \;z=0 \text{ and }\;x^2+y^2<1 \}$ a regular surface ?

my answer

1- the closed disk is not a regular surface,if it's the case the closed disk wich is compacte in $\mathbb{R}^3 $ is homeomorphic to an open set of $\mathbb{R}^2$ and this is contradiction

2- the open disk is a regular surface because the paramtrisation $X$from the open uniatry disk from $\mathbb{R}^2$ to the open disk in $\mathbb{R}^3 $ verify the definition of a regular sufraces $(x(u,v)=(u,v,0))$

For the second i 'am sur but for my first answer i have some doubt what do you think ?

Michael Hardy
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Bernstein
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  • I have the same question and my reasoning is similar to yours. I just do not understand the accepted answer. If you have had any further insight on this in the past 1+ year, I would appreciate hearing it. Thanks. – bob.sacamento Oct 17 '19 at 19:23

1 Answers1

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You are correct that the closed unit disk is not a regular surface (aka a 2-manifold), but the reasoning is not quite correct. For example, the 2-sphere (embedded in $\mathbb{R}^3$, if you like) is a regular surface which is not homeomorphic to an open subset of $\mathbb{R}^2$ due to the compactness obstruction that you mention.

To argue that the closed disk is not a regular surface you will have to find a point in it around which there cannot possibly be a neighbourhood homeomorphic to an open subset of $\mathbb{R}^2$. Given what you know about the open disk (namely, that it is a regular surface), this point will clearly have to be a boundary point.

For your reference, the closed disk is an example of an object called a "manifold with boundary" (which is exactly what it sounds like). The charts on such spaces are homeomorphisms from (relatively) open sets of the space to (relatively) open subsets of $\mathbb{H}^n$ (the closed upper half plane).

Good luck!

user345730091
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  • thanks @itinerantleopard , yes but i totaly agree with you , but my intuition was that the closed disk can be paramitrized by one chard ... but the reasoning is not true , i want to ask you one other questions idid you know a references when can i foun the dfinition of embedded disk ? – Bernstein Aug 04 '18 at 15:43
  • As in, the definition of an embedding? An embedding is a smooth map which is a diffeomorphism onto its image. – user345730091 Aug 04 '18 at 15:49
  • ok thanks , but if i need more properties of this kind of surfaces where can i look ? – Bernstein Aug 04 '18 at 15:52