# 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?

• Are you regarding the Mobius strip as a principal $G$-bundle for some Lie group $G$ (with fibers being copies of $G$), or as a vector bundle (with fibers being vector spaces)? Apr 14, 2015 at 21:09
• As a principal $\mathbb{R}$-bundle over $S^1$. Apr 14, 2015 at 21:10

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.

• "and has no continuous nonzero section"?
– Pedro
May 20, 2016 at 19:24
• @PedroTamaroff: My wording was maybe suboptimal; the principal bundle has no continuous section. :) May 20, 2016 at 19:47
• What do you mean by "There's no continuous action of G on M reducing to addition in the fibres"?
– Alex
Feb 24, 2022 at 14:21
• @Alex There are multiple definitions of principal bundles that may differ in nuances (or may be equivalent but the proof of equivalence is technical), but to me a principle $G$-bundle over a manifold $M$ is a fibre bundle $p:E \to M$ locally modeled on $U \times G$ and equipped with a right $G$-action that locally is right multiplication. (The transition functions act on the left.) A section is a mapping $\sigma:M \to E$ such that $p\circ\sigma$ is the identity on $M$. A principal bundle with a section is therefore trivial (globally a product, with the obvious right action). Feb 24, 2022 at 17:18
• @AndrewD.Hwang nevermind, I think I figured out the reason why Mobius strip is not a principal bundle. Thank you!
– Alex
Feb 27, 2022 at 16:39

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!

It is a global section, but the zero section! The global section should be always non zero to obtain the triviality.

• Thanks for your answer. I still don't quite understand. In Nakahara, it just states the theorem as: a principal bundle is trivial if and only if it admits a global section. Also, if you are just given the bundle without a choice of local trivialisations, how do you know where zero is? We only need to view $\mathbb{R}$ as an additive group to make the principal bundle. What if we replace $\mathbb{R}$ with $S^1$ and get the Klein bottle. $S^1$ doesn't have a zero element? Apr 14, 2015 at 21:05
• sorry, I thought you are talking about vector bundles.
– user151873
Apr 14, 2015 at 21:18