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I'm a physicist trying to understand fibre bundles, and I think I'm pretty happy with the wikipedia definition "a space that is locally a product space, but globally may have a different topological structure". I think I also have a grasp, in the general and tangent cases, on the different objects in the fibre bundle: $(E, \pi, M, F, G)$ taken singularly and abstractly. However, I'm having trouble seeking what tells apart two fibre bundles $E$, the Möbius strip and the cylinder. Both have base $M=S^1$ and fibre $F=$ (a segment), so what's different in the two of them? Is it the projection $\pi$, the structure group $G$ or maybe trivially just the total space $E$?

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    $\begingroup$ The total spaces are different: a moebius strip is not homeomorphic to a cylinder $\endgroup$ – user126154 Apr 20 at 16:08
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    $\begingroup$ But they are different as a consequence of the different fibration $\pi$ $\endgroup$ – Son Gohan Apr 20 at 17:20
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The cylinder, as a fiber bundle, has a section which is never zero.

On the moebius band, such a section does not exist: if you parametrize the segment $F$, which is the fiber, with $(-1,1)$, then if a never-vanishing section starts positive, after one turn it becomes negative.

In terms of fiber bundles a section is a continuous map from the base to the fiber, that is to say a map $$s:M\to E$$ such that $\pi(s(x))=x$ for evey $x\in M$ (so $s(x)$ is in the fiber of $x$).


EDIT following a comment:

To be precise, one should make a difference between fiber bundles and vector bundles: In the case of fiber bundles, the fiber has no vectorial sturcture, so it is not defined who is the zero of the fiber, so the zero section is not defined. So, to be coherent with terminology of fiber bundles, one should say that the Moebius strip, as a fiber bundle, has not two sections that are everywhere different from each other, while the cylinder has plenty of that.

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    $\begingroup$ This answer is not quite accurate. The open Möbius band, as a vector bundle, indeed has no nonzero section. However, a fiber bundle is not the same thing as a vector bundle, and in particular "zero" is not defined (even though in this example, where the fibers are open segments, a vector bundle structure can be imposed). $\endgroup$ – Lee Mosher Apr 20 at 16:46
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    $\begingroup$ One way to phrase user126154's answer which, I suppose, would be more palatable to @Lee, is that for a cylinder, you can find two sections with disjoint images while for the Mobius band, any two sections intersect. $\endgroup$ – Jason DeVito Apr 20 at 20:24
  • $\begingroup$ I suppose it's not so much an issue of what I find palatable, as it is an issue of matching the formulation of the original post: "... a fiber $F =$ (a segment)..." $\endgroup$ – Lee Mosher Apr 20 at 20:27
  • $\begingroup$ @Lee: I apologize - I meant nothing negative by my comment, but I should have kept it focused on the actual problem statement and not your feelings about it. Sorry about that! $\endgroup$ – Jason DeVito Apr 21 at 14:04
  • $\begingroup$ not a problem :-) $\endgroup$ – Lee Mosher Apr 21 at 14:05
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The direct answer to your question about the difference in the tuple you considered is the following: usually the fibre bundle is not defined as the tuple of objects you wrote. The fiber bundle usually is defined to be just $\pi$, and the other objects are named “total space”, “basis” and “fibre”. But what is fundamental here is the map $\pi$. Indeed in the definition of fibration one has conditions on that particular map. So, yes, the main difference stands in how the map $\pi$ is defined, giving rise to different total spaces even with the same base and the same fiber.

Even topologically, they are different objects: we cannot find an homeomorphism between them. For example notice that one is orientable while the other is not (recalling that orientability of a manifold is a property that is preserved under homeomorphisms; like compactness: if $X$ is compact and $X$ is homeomorphic to $Y$, then $Y$ must be compact as well).

Locally they can be considered the same, but notice that their difference is topological, which means a difference in their global structure. Indeed, this is the same difference there is between a cartesian product of the base and the fiber and the whole structure of fiber bundle.

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    $\begingroup$ I know that they're topologically different, but I was looking for a difference in the "ingredients" of the fibre bundles. Can this topological difference be identified with a particolar property of the fibre bundle, or are global differences not present in the description of fibre bundles? It seems to me that they should be, because as you say this is precisely why we talk about fibre bundles and not just product spaces. $\endgroup$ – Mauro Giliberti Apr 20 at 16:00
  • $\begingroup$ I edited, hope it’s clearer. If not, please do not hesitate to ask more specific questions :) $\endgroup$ – Son Gohan Apr 20 at 18:38

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