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In Manfredo do Carmo's Riemannian Geometry, the Möbius band is defined as the quotient of the cyllinder $S = \{ (x, y, z) \in \Bbb{R}^3 \ : \ x^2 + y^2 = 1, \ - 1 < z < 1\}$ by the group $\{A, Id\}$ where $A$ is the antipodal map.

Why is this definition equivalent, and in what sense equivalent, to the definition given in basic Differential Geometry books, such as in page 71 of Montiel and Ros or in page 108 of do Carmo? Are they diffeomorphic? If so, how to find a diffeomorphism?

The context is Exercise 3 of chapter 0 of do Carmo's Riemannian Geometry:

Show that: (a) A regular surface of $\Bbb{R}^3$ is orientable if and only if there exists a differentiable map $N: S \to \Bbb{R}^3$ such that $N(p) \perp T_pS$ and $|N(p)| = 1$ for all $p \in S$. (b) The Möbius strip is not orientable.

I believe that given item (a) the author wants us to use the standard argument given in Differential Geometry books. The difficulty then resides on showing that the two definitions are equivalent.

Thanks in advance and kind regards.

EDIT Here is the parametrization of the Möbius strip as given in do Carmo's Differential Geometry of Curves and Surfaces:

enter image description here

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  • $\begingroup$ You'll need to let us know what those other definitions are, otherwise it's pretty much impossible to show they are equivalent to the one you've given. $\endgroup$ – Lee Mosher May 18 '20 at 19:57
  • $\begingroup$ @LeeMosherThank you for you attention. I have edited the post to contain the description of the Möbius band via parametrizations. Does it suffice? $\endgroup$ – Danilo Gregorin Afonso May 18 '20 at 20:17
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Let me first focus on the topological aspects of your question, and then I'll address briefly the diffeomorphical aspects.

The topological principles to do this proof are the concept of quotient maps, and of universality properties of quotient maps. These are things you can find in most any topology book, for example Munkres' "Topology". Here's a basic outline.

In your setting you have a topological space $S$, a partition $D$ of $S$ into disjoint subsets, namely the set of orbits of the group action, namely $$D = \{\{A(x),x\} \mid x \in S\} $$ The quotient map $q : S \to D$ is the map defined by $q(x) = \{A(x),x\}$. The quotient space has underlying set $D$, with quotient topology defined so that a subset $U \subset D$ is open if and only if $q^{-1}(U)$ is open.

The universality property of $q$ tells you, among other things, that for any topological space $M$ and for any surjective function $f : S \to M$, if $f$ has the property that a subset $V \subset M$ is open if and only if $f^{-1}(V)$ is open, and if $D = \{f^{-1}(m) \mid m \in M\}$, then the obvious induced bijection $D \mapsto M$ is a homeomorphism.

So, what you need to do is to construct a surjective function $f : S \to M$ with those properties, and then you'll know that $D$ is homeomorphic to $M$.

I'll give you a hint that the "system of coordinates" that is given to you for $M$ should guide you on how to construct a formula for $f$: your given surface $S$ is equal to $S^1 \times (-1,1) \subset \mathbb R^2 \times \mathbb R = \mathbb R^3$, and you should define the map $f : S \to M$ so that the the $S^1$ coordinate corresponds to one of $u$ or $v$, and the $(-1,1)$ coordinate of $S$ corresponds to the other one.


As for the diffeomorphic aspects of your question, it's not too hard to prove that the quotient map $f : S \to M$ is a local diffeomorphism. The original quotient map $q : S \to D$ is also a covering map. From this it follows that there exists a unique differential manifold atlas on $D$ satisfying either of the following two properties: the homeomorphism $D \to M$ is a diffeomorphism; $q$ is local diffeomorphism.

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  • $\begingroup$ Thank you very much. Only today I had time to study your answer. I am confused with one aspect of it. You are using $D$ to denote different things, aren't you? Because you start using $D$ to denote the quotient $S/G$, but in the fourth paragraph $D = \{f^{-1}(m) | m \in M\}$, which is precisely $S$ since $f$ is a bijection. Or am I mistaken? $\endgroup$ – Danilo Gregorin Afonso May 21 '20 at 19:04
  • $\begingroup$ I tried setting $D = \{[f^{-1}(m)] | m \in M\}$, so that now the definitions of $D$ coincide. But in this case we get that the map $D \to M$ is not well defined, since for every $[f^{-1}(m)] \in D$ we should get two points in $M$. I do not believe, however, that your answer is incorrect. What am I missing? $\endgroup$ – Danilo Gregorin Afonso May 21 '20 at 19:09
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    $\begingroup$ The map $f : S \to M$ is not a bijection, it is only a surjection. But because it is a surjection, there is a bijective induced map $\{f^{-1}(m) \mid m \in M\} \to S$, defined by the formula $f^{-1}(m) \mapsto m$; here I'm using the standard set theoretic definition $f^{-1}(m) = \{s \in S \mid f(s)=m\}$. What's confusing, I think, is an ambiguity in the symbol $f^{-1}$: it is used as the inverse function of a bijection; but for any function $f : S \to M$ the symbol $f^{-1}$ is also used for the function that inputs a subset of the range of $f$ and outputs a subset of the domain of $f$. $\endgroup$ – Lee Mosher May 21 '20 at 19:13
  • $\begingroup$ Here is how I built $f$: let $g: S^1 \to [0, 2 \pi)$ be the trivial bijection. Then $f(x, y, z) = \mathbf{x}(g(x, y),z)$, that is, $u = g(x, y)$ and $v = z$. That means we are just extending $\mathbf{x}$ so that it is not defined on an open set, but rather on $[0, 2\pi) \times (-1, 1)$. Isn't this correct? $\endgroup$ – Danilo Gregorin Afonso May 21 '20 at 19:29

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