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In order to understand the question of the title I need to understand another thing first. If we consider the Mobius band, locally, for a $U_i \subset S^1$, where $S^1$ is the base space, the bundle looks like $U_i \times [0,1] \to U_i$. What is the obstruction that makes the Mobius band a non trivial fibration? I mean, I understand that there is some sort of twisting but how can I see this twisting practically? I assume that it exhibits itself in the transition function between intersections of local patches. Can you give me an example on how such a transition function would differ from the one of the normal band (i.e. globally $S^1 \times \mathbb{R}$)?

Let us fix a base space $S^1$ and then fiber over each point on $S^1$ an $\mathbb{R}$. Then, this would look like an infinite cylinder.But, in principle I can give some sort of twisting as I move across $S^1$ patches. Now, although in the case of Mobius strip we would identify 0 with 1 and 1 with 0 after a $2\pi$ walk around the band I am not sure if we can do the same with $\mathbb{R}$. So the question is: Is there a construction $$ S^1\times \mathbb{R} $$ with a $U(1)$ action on $\mathbb{R}$ such that after a rotation around $S^1$ the "direction" of the fiber is reversed?

If so what do transition functions look like? Of course this surface would have a lot of self-intersections, so another question is does there exist a smooth embedding to some $\mathbb{R}^n$ (for proper $n$)?

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  • $\begingroup$ a klein bottle? $\endgroup$ – Doug M Apr 29 '16 at 7:44
  • $\begingroup$ Indeed. The bundle will self-wrap to a Klein bottle. $\endgroup$ – Yiannis Galidakis Apr 29 '16 at 8:01
  • $\begingroup$ @DougM I am not sure because the problem, at least for me, is that $\mathbb{R}$ is not bounded if you understand what I mean. $\endgroup$ – Marion Apr 29 '16 at 9:50
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The transition functions between neighboring patches do not reveal that globally there is something wrong. Now locally there are two classes of coordinate patches. Two patches belong to the same class if the Jacobian between them is positive, and belong to different classes if the Jacobian between them is negative. Locally you can arbitrarily declare one class as the positive class, but globally you can't.

As an abstract manifold the Moebius band is the same whether you consider the fibers as copies of $\ ]{-1},1[\ $ or as copies of ${\mathbb R}$. If the fibers are copies of $[{-1},1]$ then you obtain a Moebius band with boundary, which is compact.

As to your last question: Here is the double cover of a Moebius band of "infinite width" and without selfintersections in ${\mathbb R}^4$:

$$S^1\times{\mathbb R}\to{\mathbb R}^4,\qquad(\phi,t)\mapsto\bigl(\cos(2\phi),\sin(2\phi), t\cos\phi,t\sin\phi\bigr)\ .$$

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  • $\begingroup$ so the normal band differs to the Mobius band only on the fact of the existence of these negative Jacobians of the transformations of coordinates from one patch to the other patch? I would assume that some kind of boundary condition on the interval $[0,1]$ apply. Also, is there some way to describe the twisting of this unit interval as it goes around the circle? $\endgroup$ – Marion Apr 29 '16 at 9:53

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