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It is well known (using for instance sheaf cohomology) that there exist only two possible one-$\mathbb R$-dimensional vector bundles over $S^1$: the trivial bundle and the Möbius bundle. But what about the Möbius bands with not one, but $n\in \mathbb N$ half-twists? It is not obvious to me why these are not distinct line bundles. I am also not entirely comfortable with sheaf cohomology arguments.

This is also physically motivated; if you make a Möbius band, it's easy to see that two half-twists don't "cancel" to make a trivial bundle; so why do they "cancel" topologically?

Dwagg
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    If I cut my Möbius band along a line, along a fiber, I can unwind two half-twists and then reglue my Möbius band back together. Because nothing changes along the cutting line, this is a homeomorphism of these two spaces. – Santana Afton Apr 25 '19 at 03:38
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    The point is that your picture of the trivial bundle and the two half twist depends on an embedding in $\mathbb{R}^3$. Concretely, let $X_0=\mathbb{S^1}\times\mathbb{R}$ be the trivial bundle and $i_0:X_0\to\mathbb{R}^3$ an embedding as a cylinder. Let $i_1:X_1\to\mathbb{R}^3$ an embedding of the two half twist bundle. Then there is no topological space $X$ with maps $p:X\to[0,1]$ and $i:X\to\mathbb{R}^3$ such that for all $t$, $X_t:=p^{-1}(t)$ is homeomorphic to the total space of a line bundle, and $i_t=i|_{X_t}:X_t\to\mathbb{R}^3$ is an embedding. But such a thing exists in $\mathbb{R}^4$ ! – Roland Apr 25 '19 at 08:03
  • The key word here is isotopy : https://en.wikipedia.org/wiki/Homotopy#Isotopy – Roland Apr 25 '19 at 08:05

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As Roland explained in his comments, you have to distinguish between a line bundle $L$ over $S^1$ and embeddings of $L$ into $\mathbb R ^3$. In fact, any subspace $L' \subset \mathbb R ^3$ which is homeomorphic to $L$ is a geometric instantiation of $L$, and it may be impossible to deform (inside $\mathbb R ^3$) one instantiation $L'_1$ into another instantiation $L'_2$ without rupturing. The mathematical concept of an "admissible" deformation process inside an ambient space is isotopy.

Take for example the trivial line bundle $L = S^1 \times \mathbb R$. Since $S^1 \subset \mathbb R ^2$, it is a genuine (unbounded) subset of $\mathbb R ^3$. However, there are many other instantiations of $L$, for example $L' = S^1 \times (-\frac{1}{4},\frac{1}{4})$ or twisted versions $L'_n$ of $L'$ (with $n$ full twists). All these are homeomorphic copies of $L$, but no two of them can be deformed into each other.

As an analogue you may also consider the circle $S^1$. A homeomorphic copy of $S^1$ in $\mathbb R ^3$ is a knot, and knot theory is a highly non-trivial field devoted to the classification of knots.

Paul Frost
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