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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?

octopus
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  • 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)? – Jesse Madnick Apr 14 '15 at 21:09
  • As a principal $\mathbb{R}$-bundle over $S^1$. – octopus Apr 14 '15 at 21:10

3 Answers3

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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.

Andrew D. Hwang
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  • "and has no continuous nonzero section"? – Pedro May 20 '16 at 19:24
  • @PedroTamaroff: My wording was maybe suboptimal; the principal bundle has no continuous section. :) – Andrew D. Hwang May 20 '16 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 '22 at 14:21
  • @Alex Addition on the fibres amounts locally to translating a strip $(a, b) \times \mathbf{R}$ along the second factor. There's no way to do this consistently on the whole strip because the Möbius strip is non-trivial as a vector bundle. – Andrew D. Hwang Feb 24 '22 at 15:03
  • @AndrewD.Hwang I think I am misled by Nakahara (pg 363) where there is no mention of this "global action". Just to be clear, for a fiber bundle to be a principal bundle, one need to be able to pick any element $g\in G$, and define the action of this one $g$ over the entire base manifold? – Alex Feb 24 '22 at 15:33
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    @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). – Andrew D. Hwang Feb 24 '22 at 17:18
  • @AndrewD.Hwang Now I am confused again. When you say "There's no way to do this consistently" in the comment above, what do you mean exactly? I thought this would mean, for any $g\in G$, we need to be able to define its action at each local in $M$ in a continuous and non-zero manner, if the whole thing were to be called a principal bundle. But then this would give a section that would make the whole thing a trivial bundle. – Alex Feb 25 '22 at 14:05
  • @Alex Informally, make a Möbius strip out of paper and draw the central circle $C$. Pick a point $p$ on $C$ and draw an arrow based at $p$ and perpendicular to $C$. That arrow represents a local action of the additive group of reals on the fibres. Now try to extend this action continuously over the entire strip, i.e., to extend the arrow to a vector field along $C$. Somewhere the field will be discontinuous; that's the "no way to do this consistently" claim. (Or am I misunderstanding what you're asking?) – Andrew D. Hwang Feb 25 '22 at 18:32
  • @AndrewD.Hwang I understand the "drawing arrow" fully. What confuses me is the following. (a) The fact that you can't draw the arrow consistently makes the Mobius strip not a Principal bundle. (b) however, if we were able to do this, then we got a global section, then the bundle is trivial. => all principal bundles are trivial? – Alex Feb 26 '22 at 01:27
  • @Alex Ah, I see...! The issue is, a non-trivial principal bundle admits no (continuous) section. (That's why vector bundles are essentially never principal bundles for the additive group of a fibre.) – Andrew D. Hwang Feb 26 '22 at 01:51
  • @AndrewD.Hwang I must be missing something obvious here. Is a cylinder a principal bundle with Fiber $(R,+)$ and base manifold $S^1$? – Alex Feb 26 '22 at 06:39
  • @Alex Yes it is. – Andrew D. Hwang Feb 26 '22 at 14:01
  • @AndrewD.Hwang But one comment above it was stated that "vector bundles are essentially never principal bundles for the additive group of a fibre"? – Alex Feb 26 '22 at 15:06
  • @Alex I was unnecessarily qualitative. A vector bundle acquires the structure of a principal bundle for the additive group of a fibre if and only if the bundle is trivial. – Andrew D. Hwang Feb 26 '22 at 16:10
  • @AndrewD.Hwang Sorry to bother you this much but I just want to get to the bottom of this. Back to the "drawing arrow" intuition above - doesn't this mean that the Mobius strip is a non-trivial bundle, rather than that the Mobius strip is not a principal bundle? The definition of principal bundles only require local trivialization. What's wrong with that you travel one round back and the same $g$ points to a different direction? The definition of principal bundles never require that you can define the global action of $g$ on the total space, right? – Alex Feb 27 '22 at 11:44
  • @Alex A principal bundle has only local trivializations in general, but it does come equipped with a global action of the structure group that reduces in each trivializing chart to group multiplication. – Andrew D. Hwang Feb 27 '22 at 13:52
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    @AndrewD.Hwang nevermind, I think I figured out the reason why Mobius strip is not a principal bundle. Thank you! – Alex Feb 27 '22 at 16:39
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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!

mavzolej
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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? – octopus Apr 14 '15 at 21:05
  • sorry, I thought you are talking about vector bundles. –  Apr 14 '15 at 21:18