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Of course there are many ways to prove this. However, I came across the following exercise (Ch. 0 #3).

Prove that: (a) a regular surface $S\subset \mathbb{R}^3$ is an orientable manifold if and only if there exists a differentiable mapping of $N:S\rightarrow \mathbb{R}^3$ with $N(p)\perp T_p(S)$ and $|N(p)|=1$, for all $p\in S$. (b) the Möbius band (Example 4.9 (b)) is non-orientable.

In Example 4.9 (b), he constructs the Möbius band as the quotient by the antipodal map of the cylinder $C=\{(x,y,z)\in \mathbb{R}^3:x^2+y^2=1,|z|<1\}$. The problem, of course, is that this isn't given as a surface in $\mathbb{R}^3$! I was thinking for a second that maybe I should try and construct a map $C\rightarrow S^2$ with the right properties and check that it doesn't descend to a map on $M$, but that's stupid because if I were to embed $M\subset \mathbb{R}^3$, I'm pretty sure it couldn't possibly have those tangent planes anyways.

Does anyone have any insight? Presumably the solution to (b) should use the fact given in (a).

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I think you need to find an embedding into $\mathbb{R}^3$ to make this work. You can't use this condition with an abstract manifold. (I presume you've been given an abstract definition of orientability?) –  Zhen Lin Feb 28 '11 at 7:38
    
Yes, that there's a coherent atlas. I agree that it seems like I'd need an embedding for this to work. Given (a), it's clear that analyzing any embedding (or even immersion, I think) will determine for once and for all whether the manifold is orientable. Mainly it's weird to me that he specifically referenced that construction. –  Aaron Mazel-Gee Feb 28 '11 at 8:23
    
Have you done part (a)? –  Glen Wheeler Feb 28 '11 at 10:40
    
Yes I have. I guess I didn't say that explicitly. –  Aaron Mazel-Gee Feb 28 '11 at 17:33
    
Sorry I didn't get your reply @Aaron and check back on this question earlier. Using part (a), there must not exist a differentiable map $N$. The intuitive idea is that following a curve around the strip gives a contradiction. I have to say, I don't think do Carmo had the particular construction that he gave only a few pages earlier in mind when he wrote the exercise. –  Glen Wheeler Apr 4 '11 at 19:49

1 Answer 1

What doCarmo might have had in mind:

I didn't check, but I think the quotient map

$$ \pi: C \to M $$

is precisely the orientation covering of $M$. And one can prove that this covering is connected iff $M$ is not orientable (see Lee "Intro to Smooth Manifolds", p. 331 for example).

If the above is not useful: Maybe you could try to pull back the orientation of $M$ to $C$ and get a contradiction or so. (although this really has not much to do with part a) of the exercise, so it might be that doCarmo wants you to imbed the Möbius strip)

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That's a nice way to prove it. It's true that $\pi$ is the orientation covering. Unfortunately, this is precisely Exercise 12! So I really doubt it's what he's looking for. Pulling back a hypothetical "orientating map to $S^2$ of $M$ to $C$ was my first guess, but yes the problem is that it can't really be so related to (a). –  Aaron Mazel-Gee Feb 28 '11 at 17:39

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