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I read that glueing together two Möbius strips along their edges creates a surface that is equivalent to the so-called Klein bottle.

The Möbius strip comes in two versions that are mirrored versions of each other (wrt the chirality of the half-turn in the strip).

So glueing together two of them, does it matter whether they are mirrored versions of each other or not? Is there a different result?

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I remember when I had this issue the answer was to look at the classification of nonorientable surfaces. –  Frost Boss Aug 23 at 21:20

3 Answers 3

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If you are careful with your deformations when you draw topological examples, you can prove it by making a very crude drawing, like the one below. The cut occurs at the vertical plane of symmetry of the bottle. It is fairly clear that the final objects are Mobius bands.

enter image description here

foudn here: www.ifp.illinois.edu/~sdickson/Klein/Klein.html

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I added a picture with the same idea where you can clearly see that the two versions are mirrored versions of each other. Source: ifp.illinois.edu/~sdickson/Klein/Klein.html –  Gerard Aug 24 at 14:12
    
@Gerard: Ah, excellent! Thanks for the improvement. My mind didn't go to searching the net for this. –  ioannis galidakis Aug 24 at 14:42

They are only in two version if you consider the Mobius strip as being an object embedded in three-dimensional ambient space. This becomes pretty moot when you consider the Klein bottle can't be embedded into $3$-space, so instead we usually only consider the Mobius band and the Klein bottle as topological spaces, in which case they don't come with 'chiralities' or indeed orientations in this case because they are non-orientable.

To put it succinctly: All Mobius bands are homeomorphic.

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Your answer is helpful. The Klein bottle kan be embedded into 3-space with an intersection, does that make a difference? –  Gerard Aug 23 at 20:30
    
Sure, that's called an immersion (allowed to self intersect but only 'at an angle'). I think in this case, the two halves of the Klein bottle would be a chiral pair of Mobius bands, although I'm not totally certain/would have to check. Although thinking about it, the bands themselves would not be embedded but again only immersed, so how do you define chirality in this case? –  Daniel Rust Aug 23 at 20:35
    
It beats me, sorry ... –  Gerard Aug 23 at 20:55
    
The point that @DanielRust is making, loosely speaking, is that if you can allow self-intersection in an immersion of a Klein bottle in $\mathbb R^3$, then the constituent Möbius strips are also allowed to self-intersect. In such a case, then there is a deformation of a strip that changes its apparent handedness as it is immersed in 3-space. –  heropup Aug 24 at 2:25
    
@heropup I'm even now doubting the well-definedness of 'handedness' or 'chirality' of an embedding of a Mobius band. There are a least as many isotopy classes of embeddings of the Mobius band into $\mathbb{R}^3$ as there are knots, and ambient isotopy seems like the only reasonable tool for determining 'chirality'. So what does it even mean for an embedding of a Mobius band to be left or right handed? –  Daniel Rust Aug 24 at 2:31

I quickly threw this animation together to demonstrate that the "figure-8" immersion of the Klein bottle admits a decomposition into two Möbius bands with the same "apparent handedness," whatever that means.

enter image description here

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Is it possible to change the animation into two Möbius bands with opposite "apparent handedness whatever that means"? –  Gerard Aug 24 at 13:46

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