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Playing around with the definition of a fiber bundle, I found that while a Möbius strip (with its usual "half-twist") is a nontrivial fiber bundle, it seems that a Möbius strip with a "full-twist" is again trivial. In turn, a full-twist strip is homeomorphic to a simple cylinder.

I was wondering if it was well-known "how much more structure" is required before you can distinguish these two spaces. Intuitively, I would guess at most a reimann manifold structure allows you to do so, since there is more "curving" going on in the double-twist. Is it known if it can be done with less?

Tac-Tics
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  • Whoops! I incorrectly titled the post. What I want to distinguish is the full-twist Mobius strip and the simple cylinder. – Tac-Tics Oct 17 '13 at 23:06
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    They're homeomorphic. What do you mean they're different. –  Oct 17 '13 at 23:08
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    They are isomorphic as real vector bundles and thus as spaces, however (at least intuitively) they surely can't be isotopic as embedded copies of $S^1\times \Bbb R$ inside $\Bbb R^3$. – Olivier Bégassat Oct 17 '13 at 23:28

2 Answers2

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A cylinder and a full-twist Möbius strip are distinguishable when they are embedded in $\mathbb R^3$ and we consider equivalence only up to ambient isotopies of $\mathbb R^3$. This is very easy to see because the boundary of the cylinder is a pair of unlinked circles, while for the full-twist Möbius strip the boundary circles are linked.

Intuitively this is what I'm thinking about if I try to differentiate the two; that is to say that the term "full-twist Möbius strip" already implies that it is embedded in $\mathbb R^3$. The fact that these are diffeomorphic means that they abstractly admit exactly the same structures, so something like a Riemannian metric can't even hope to distinguish between the two unless e.g. we require that metric to come from some embedding (I'm not actually sure if this is sufficient or not).

Logan M
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  • It seems believable, but do you know a proof that pulling back the (standard) metric on $\mathbb{R}^3$ to various embeddings of cylinders and double-twisted Mobius bands always give non-isometric Riemannian metrics on $S^1\times [0,1]$? (By "cylinders" I mean any subset of $\mathbb{R}^3$ ambiently isotopic to the standard cylinder, and like wise for "double-twisted Mobius bands"). – Jason DeVito - on hiatus Oct 17 '13 at 23:35
  • @JasonDeVito No, I don't know of a proof of this, though as you say it is plausible. I didn't intend to imply that this was necessarily true, only that one would have to do something like that to have any hope of being able to distinguish the two. – Logan M Oct 17 '13 at 23:38
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    Thank you. This is the kind of answer I was groping for. – Tac-Tics Oct 19 '13 at 01:06
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The twisted cylinder and the cylinder are not isomorphic as bundles with structure group Z2. They are isomorphic as bundles with structure group SO(2). This is a classic example in Steenrod's Topology of Fiber Bundles.

lavinia
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