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?
Best Answer
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.