# Can you really construct a Möbius strip from this model?

I was reminded by question 77569 about something that has bothered me off and on for a while now. Consider the model of the projective plane given by the last diagram on page 37 of these notes (it's just a lens-shape in the plane, with opposite sides identified in a "chasing" way): Now if you remove an open disk from it, what's left, call it $M$, is homeomorphic to a Möbius strip. Unlike the projective plane, the Möbius strip can be embedded in ${\bf R}^3$ with no self-intersections. So it's natural to ask whether you can use the model $M$ to construct a physical Möbius strip. That is, does the presence of the hole make it possible to carry out in 3-space the indicated identification of edges, without tearing or self-intersection, provided the material being used is sufficiently elastic.

Experiments with paper have not been encouraging. Is there some obstruction to making a Möbius strip from $M$?

Another plane model for a Möbius strip is a triangle with 2 edges identified "chasing" and one edge unidentified. The 2nd figure on page 40 of those notes will do as an illustration. Again, I haven't been able to carry out the identification with paper in 3-space; is it possible? or is there some obstruction?

I should add that of course I know that one can make a Möbius strip from the usual model, a rectangle with one pair of opposite edges identified chasing. And I know that these other models are homeomorphic to the Möbius strip. The questions are about using the other models to make a Möbius strip.

• As for your second example of a triangle, that is very close to the usual rectangle model. If you have a trapezoid with two opposite sides of length $1$ and $\epsilon$ unidentified, and the other two opposite sides identified with the "chasing" orientation, then the the triangle model is the limit as $\epsilon\to 0$. The usual strip model happens when $\epsilon=1$. Nov 1, 2011 at 1:04
• That seems to be a strange direction to go in -- since the model in the notes is not an actual embedding into $\mathbb R^3$ (they say "we’re not too worried about where these surfaces live, we won’t pursue that point"), you couldn't construct it to start with and cut out a disk from it. I suppose you could remove one of the "eardrums" from a Boy's surface and get a partially self-intersecting Möbius strip, though. Nov 1, 2011 at 1:07
• @Jim, yes, and now I am reminded that I think I saw a discussion somewhere of how short the unidentified sides of a rectangle could be (relative to the identified pair) and still permit making a Mobius strip. I'll have to see if I can find it. But are you suggesting your comment answers my question? Nov 1, 2011 at 1:12
• @Henning, we have a failure of commutativity here. I didn't mean, construct the projective plane, then cut out a disk; I meant, cut out a disk (from the diagram), then construct a Mobius strip (if possible). Nov 1, 2011 at 1:15
• @t.b., thanks for the edit. Nov 1, 2011 at 1:15

Note that if you cut a disk out of your lens shape, the remaining annulus is embedded in $\mathbb{R}^3$ without any "twists". If this could be "glued" to itself to form a Möbius band in $\mathbb{R}^3$, then the (external) boundary of your lens-shape region would be mapped to the "soul" of the Möbius band; namely, the curve running along the middle of the Möbius band for one of the standard embeddings.