Möbius strip as a non-trivial principal bundle There is a well-known theorem that a principal bundle is trivial if and only if it admits a global section.  I'm trying to get a good picture of what this theorem means.
The Möbius Strip can be regarded as a principal bundle of $\mathbb{R}$ over $S^1$.  My intuition is that it is not the trivial bundle, but I can imagine drawing a line along the centre of the strip.  It seems to me that this line satisfies the definition of a global section.
What's wrong with this intuition?
 A: Actually, the Möbius strip $M$ is not trivial as an $(\mathbf{R}, +)$ principal bundle over the circle $S^{1}$, because $M$ is not the total space of a principal bundle with structure group $G = (\mathbf{R}, +)$ at all: There's no continuous action of $G$ on $M$ reducing to addition in the fibres.
It's likely the intuition you're seeking is to view $S^{1}$ as the total space of a principal bundle over $S^{1}$ with structure group $O(1) = \bigl(\{\pm1\}, \cdot\bigr)$. The projection map is the double covering $\pi:S^{1} \to S^{1}$, defined by $\pi(e^{it}) = e^{2it}$.
The Möbius strip $M$ may be viewed as the vector bundle associated to the multiplicative representation of $O(1)$ on $\mathbf{R}$ viewed as a one-dimensional vector space. As expected, this "double-covering" principal bundle is non-trivial, and has no continuous section.
Generally, if the total space $E$ of an $n$-plane bundle admits an action of the additive group $(\mathbf{R}^{n}, +)$ as translation in the fibres, i.e., if $E$ admits the structure of an $(\mathbf{R}^{n}, +)$ principal bundle, then $E$ is trivial; the group action defines a global frame in an obvious way.
A: If you want to think of a Möbius strip $M$ as of something like the principle $G$-bundle, you can do the following. Let $G$ be the $\mathbb{Z}_2$ group consisting of two elements: $\mathbb{Z}_2=\{e,\,a\}$. Now, the typical fiber $F$ of $M$ is just a torsor of $G$ (a set without a group operation). In other words, a set of two points, let's call them $1$ and $-1$.
What we've just done is replacing of a traditional $[-1,\,1]$ fiber by a discrete $F=\{-1,\,1\}$.
Think of that new Möbius strip as of the "edge" of a traditional one (which is the $S^1$, but this is not important for us; you can make a whole revolution around this "edge" only once you make two revolutions around the base).
For our "discrete Möbius strip", it is clear why the cross-section ($=$the global section) cannot be defined for such a bundle. Indeed, after one revolution you will come to the "opposite point" which is not allowed to be taken, since, by definition, the section has just a single point from each fiber.
Next time you want to show a Möbius strip to your fiends, you won't have to glue anything. Just tale a soft ring made of something, twist it once and that's it!
A: It is a global section, but the zero section! The global section should be always non zero to obtain the triviality.
