# A very special Möbius Strip

## Main Question or Discussion Point

I would welcome the parametric equations for an embedding in R3 of a locally Euclidean Möbius Strip without self intersections nor singularities and of Gaussian curvature equal to zero. That it exists in R3 is trivial to prove: just get a strip of paper of appropriate length and width, twist and paste and you are done. Paper cannot be stretched so the intrinsic curvature of the animal is zero. You may perhaps appreciate looking at the ondulation of the Möbius Band while embedding in ordinary space. That one is the one I want to capture.

Regards,

Ogai

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## Answers and Replies

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The standard embedding is to take a line, your first coordinate, and revolve it around a circle while rotating it by $\pi$ radians. The circle is your second coordinate, and the rotation is a function of your position on the circle. You can model this in POV-Ray using while loops, or derive the explicit parametric equation. The parametric equations for the above using a circle of radius 2 and a line (-1,1) are (x,y,z) = (2cos(t) + s*cos(t/2)*cos(t), 2sin(t) + s*cos(t/2)*sin(t), s*sin(t/2)).

Yes, your Möbius Strip is a sound one. Unfortunately it doesn't qualify, because its curvature is not zero. The sole component of the Riemann tensor for your example is
R1212 = 16 / [s^2+4*(2 + s*cos(t/2))^2]
which is not even constant and much less zero. The nature of the problem
was to find a "Locally Euclidean" MS.
Best,
Ogai
hypermorphism said:
The standard embedding is to take a line, your first coordinate, and revolve it around a circle while rotating it by $\pi$ radians. The circle is your second coordinate, and the rotation is a function of your position on the circle. You can model this in POV-Ray using while loops, or derive the explicit parametric equation. The parametric equations for the above using a circle of radius 2 and a line (-1,1) are (x,y,z) = (2cos(t) + s*cos(t/2)*cos(t), 2sin(t) + s*cos(t/2)*sin(t), s*sin(t/2)).

benorin
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I seem to recall that glueing the edges of a Möbius Strip gives a Klein Bottle, but that such is not embeddable in R3. Correct me if need be: I am a tourist here. So, does the geometry of R3 make this so? or might a Klien Bottle fit nicely in some non-Euclidian rendition of R3?

Please notice that a Möbius strip has only ONE single edge and not two. It is true that if you get two Mobius strips one twisted clockwise and the other anti-clockwise and you glue them along their borders you get a Klein bottle but that cannot be done in R3 without self-intersection. Yes, the geometry of R3 doesn't allow it. The same thing with a flat torus. You cannot embedd it in R3, while it is trivially embedded in R4 or, what is more, in S3, the 3-dimensional sphere.

What is enervating with the problem I proposed (to embedd a flat MS in R3) is the fact that it can be done trivially but it is so hard to write down its eqs)

benorin said:
I seem to recall that glueing the edges of a Möbius Strip gives a Klein Bottle, but that such is not embeddable in R3. Correct me if need be: I am a tourist here. So, does the geometry of R3 make this so? or might a Klien Bottle fit nicely in some non-Euclidian rendition of R3?

mathwonk