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Can someone please remind me how this goes?

Here's the idea of proof I'm trying to recall: let $S$ be a closed surface (connected, compact, without boundary) embedded in $\mathbb{R}^3$. Then one can define the "outward-pointing normal unit vector" to $S$ at any point, and subsequently an orientation of the surface.

One would like to define this vector by saying that it points towards exterior points to $S$. So we need some kind of generalization of the Jordan curve theorem saying that the surface cuts $\mathbb{R}^3$ into two pieces (interior and exterior). What is this theorem exactly?

Also, I apologize if this is silly, but is there an obvious argument that a piece of the surface cuts a small tubular neighborhood of it into interior and exterior points (this seems necessary to define the outward normal vector properly)?

Is there a "cleaner" approach to prove this fact? Thanks in advance.

Seub
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  • I think the answer below the accepted answer in this MO post does it: http://mathoverflow.net/questions/18987/why-cant-the-klein-bottle-embed-in-mathbbr3 – Tim kinsella Nov 28 '13 at 22:34
  • @Timkinsella: Thanks for that. The answer you mention (by Joe Hass) doest it indeed (it leaves quite a bit of detail to be written, not too suprisingly), not following the idea of proof I had in mind though. The accepted answer is interesting as well. I still wonder if there's a generalized version of Jordan theorem that applies here to define the interior and exterior of the surface? – Seub Nov 28 '13 at 22:58
  • I think the answer is "yes, there is." I saw the proof a while ago, and all I can remember is it used direct limits somehow... – Tim kinsella Nov 28 '13 at 23:12

2 Answers2

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What you are looking for is Alexander's duality theorem. One of its immediate corollaries is that compact non-orientble surfaces do not embed in three-space.

Mariano Suárez-Álvarez
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Another approach might be to show directly that the normal bundle must be trivial

For instance if the Klein bottle were embedded in $\mathbb{R}^3$, then its normal bundle could not be trivial since it is an unorientable manifold. If its normal bundle were non-trivial then the boundary of a tubular neighborhood would be an oriented two fold cover and therefore a torus. Further the entire tubular neighborhood would have the homotopy type of the Klein bottle.

It is not hard to show using a Meyer-Vietoris sequence that this is impossible.

$$H_2(\mathbb{S}^3)\longrightarrow H_1(\mathbb{T^2}) \longrightarrow H_1(\hbox{Tubular neighborhood}) \oplus H_1(\hbox{Tube neighborhood complement}) \longrightarrow H_1(\mathbb{S}^3)$$

Which is

$$0 \longrightarrow \mathbb{Z} \oplus \mathbb{Z} \longrightarrow (\mathbb{Z}_2 \oplus \mathbb{Z} ) \oplus H_1(\hbox{Tube neighborhood complement})\longrightarrow 0$$

A similar arguement works for the projective plane $\mathbb{P}^2$.

jimbo
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lavinia
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