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This is claimed in various places. The problem seems to be with finding a free and transitive group action that has the fibers of $Mo$ as its orbits. I construct $Mo$ as $$ Mo = \mathbb{S}^1 \times \mathbb{R} / \sim$$ where $\sim$ is the equivalence relation $$ (t,x) \sim (t+2\pi,-x).$$

I am new to this topic and don't see why it would be impossible to construct an action. $(R,+)$ is certainly a Lie group.

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  • $\begingroup$ @tomasz Yes, $\mathbb{R} \times \mathbb{R}$ is also possible, but this construction is correct too. The absence of smooth sections is an obstruction to triviality of vector bundle, if I am not mistaken. The criterion for triviality of a principal bundle is existence of any global section, not neceserally nonzero. This is actually why I am asking, If $Mo$ was a principal bundle it would be trivial trough the zero section. $\endgroup$ – Timon van der Berg Jun 3 '16 at 13:23
  • $\begingroup$ @tomasz I am talking about a principal bundle. Not a vector bundle. $\endgroup$ – Timon van der Berg Jun 3 '16 at 13:28
  • $\begingroup$ Alright, I somewhat misunderstood the question. I thought principal is just another name for trivial. :) $\endgroup$ – tomasz Jun 3 '16 at 13:34
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Maybe what you are asking is: why is a principal bundle with a global section trivializable? Well, a principal bundle $G\to E\to B$ admits a free, transitive group action whose orbits are the fibers. If there is a principal section $s:B\to E$, then for every $b\in B$ we have $s(b)\in E$ and can map $g\to g\cdot s(b)$. This map provides a trivialization of $E$ by identifying $$e = g\cdot s(b) \leftrightarrow (g,b) $$ So you see, because the Mobius band has a global section --- the zero section --- if it admitted a free and transitive $\mathbb{R}$ action, that global section could be "smeared" into a trivialization. In this case, the $\mathbb{R}$ action must be addition, so to visualize this fact, ask yourself: if a rubber band is stretched around the core of a Mobius strip, can I roll it one centimeter off the core?

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  • $\begingroup$ Should $s(b) \in E$ not be $s(b) \subset E$? $\endgroup$ – Timon van der Berg Jun 3 '16 at 14:01
  • $\begingroup$ @TimonvanderBerg No: as $s:B\to E$, if $b\in B$, then the image of $b$ under $s$ will be an element of $E$, not a subset of $E$. $\endgroup$ – Neal Jun 3 '16 at 15:27
  • $\begingroup$ Ah, yes, I see now. $\endgroup$ – Timon van der Berg Jun 5 '16 at 15:36
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Well, you said it yourself, if $Mo$ would have the structure of a principal $\mathbf R$-bundle over $S^1$ then any global section of it would prove that it is trivial. Take the zero section $0$. It is a global section and would trivialize the principal bundle. Contradiction since the total space $Mo$ is not homeomorphic to $S^1\times\mathbf R$.

However, if you take out the image of the zero section, you get a total space $Mo^\star$, say, that does admit the structure of a principal $\mathbf R^\star$-bundle! And as such, it is a nontrivial principal $\mathbf R^\star$-bundle.

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  • $\begingroup$ This is a correct proof, but it does not provide any insight as to why there is no group action. The facts you state are the reason for my question. I should have stated this in the question. :) Upvoted but not accepted. $\endgroup$ – Timon van der Berg Jun 3 '16 at 13:44

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