# Exercise in Do Carmo's “Riemannian Geometry”: the Möbius band is nonorientable.

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

$$\pi: C \to M$$
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)
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