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We can distinguish between a (closed) Mobius strip and 'regular' (untwisted) strip by examining the set of points which have no neighborhood homeomorphic to a disk (intuitively, the 'boundary' of the strip). For a Mobius strip, this set is connected, while for an untwisted strip the set has two connected components.

However, the above criteria cannot distinguish between, say, a Mobius strip with 1 half-twist and a Mobius strip with 3 half-twists; it can only tell us if the number of twists is even or odd.

Can we tell how many half-twists a strip has in a purely topological way that avoids the embedding of the strip in $\mathbb{R}^3$? The more I think about it, the less convinced I am that we can. I can show all strips with an even number of half-twists are homeomorphic to $[0,1] \times S^1$ (by coordinatizing one of the circles as $\{0\} \times S^1$, the other as $\{1\}\times S^1$, and interpolating in between), but I'm not sure how or if I can adapt this approach to the case when there is an odd number of half-twists.

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    Yes, Mobius strips with an odd number of twists are homeomorphic to the standard strip. –  Jul 30 '15 at 16:05
  • @MikeMiller Is this true if the strips are closed? If so, why does my argument in the first paragraph not work? –  Jul 30 '15 at 16:08
  • @MikeMiller Or do you mean the standard Mobius strip? If so, I'd be interested to see an explicit homeomorphism. –  Jul 30 '15 at 16:09

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You are right in not being convinced: the parity of the number of twists is a topological invariant, but the actual number is not. To see this think of the strip as $I \times I$ (where $I = [0, 1]$) with opposite edges $\{0\} \times I$ and $\{1\} \times I$ identified either by $(0, x) \mapsto (1, x)$ (for an even number of twists) or by $(0, x) \mapsto (1, 1-x)$ (for an odd number of twists).

Rob Arthan
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