Does a Möbius strip have only one shape? Or may it have different shapes?

Question

I'm reading a book about geometry, and after thinking and viewing the Möbius strip, I want to know whether the book is right or not.

The book says with a little description (that I can't write here despite of my little english...) that the Möbius strip is given by

$f(u) + v g(u), \{v, -1, 1\}, \{u, 0, 2 \pi \}$

where

$f(u) = \{2 sin (u), 2 cos(u), 0\}$

$g(u) = \{0, sin(u/2), cos(u/2)\}$

(You can see that $g(u)$ is always parallel to plane ${X=0}$).

You can see the shape in this file I programmed in Mathematica

but

when I try to view in my head the Möbius strip with the description of a piece of paper or any other material... I view any similar to this

$f(u) + v g(u), \{v, -1, 1\}, \{u, 0, 2 \pi \}$

where

$f(u) = \{2 sin(u), 2 cos(u), 0\}$

$g(u) = \{sin(u/2)cos(u/2), sin(u/2)sin(u/2), cos(u/2)\}$

You can see the shape in this file I programmed in Mathematica

To view the cdf files, you need the CDF Player http://www.wolfram.com/cdf-player/

or in a easy way vieweing

I´m interested on this question considering the strip as a figure.

Thank you very much.

Answer 1

When discussing topological spaces we are usually only interested in them up to homeomorphism (bijections that are continuous in both directions). "Mild" deformations can be ignored. The two subsets of $\mathbb R^3$ that you describe are homeomorphic, which is best seen by the fact that they both are homeomorphic (via the $(u,v)\mapsto f(u)+vg(u)$) to the topological space $[-1,1]\times[0,2\pi]$ where the points $(x,0)$ and $(-x,2\pi)$ are identified.

Answer 2

I try to answer your question as best as I can " considering the strip as a figure" in your query, looking at the Band as internally inflexible unlike any flexible rubbery surface.

RotatingLineChristianBlatter

In the explanation he gives how $v$ vector second part g(u) of rotating stick (relative spherical vector rotation, my addition) motion is added around a fixed director circular ring described by first part f(u) when generating the Moebius strip.( Read $ a=2, \phi=u, t=v $)

$$M:\quad(\phi,t)\mapsto(a\cos\phi,a\sin\phi,0)+ t\left(\sin{\phi\over2}\cos\phi,\ \sin{\phi\over2}\sin\phi,\ \cos{\phi\over2}\right)$$ with $0\leq\phi\leq 2\pi$, $\>-b\leq t\leq b$.

$$M: \left(( a+t \sin{\phi\over2})\cos\phi,(a+t\ \sin{\phi\over2})\sin\phi,t\ \cos{\phi\over2}\right) $$ with $0\leq\phi\leq 2\pi$, $\>-b\leq t\leq b$.

$f(u) = \{2 sin (u), 2 cos(u), 0\}; g(u) = \{0, sin(u/2), cos(u/2)\}$

$f(u) = \{2 sin(u), 2 cos(u), 0\}; g(u) = \{sin(u/2)cos(u/2), sin(u/2)sin(u/2), cos(u/2)\}$

The first g(u) is a unit magnitude vector in cylindrical coordinates with half argument u/2.

The second g(u) is a unit magnitude vector in spherical coordinates with half argument u/2.

They should essentially be modified with half arguments to induce non-orientabilty.

The first cylindrical modification perhaps the simplest ( may be most quoted) parametrization, the rest move the vector forward and backwards resulting in a fold that you can see in Mathematica 3D plots.

In topological consideration either of the above two is viewed same because the rubber membrane can stretch and, upto homeo/diffeomorphism they are topologically the same.

It appears that there are several parametrizations, because topologically it perhaps "needs no standard parametrization".

Conserved invariants like Euler characteristic $\chi$ and genus $g$ are invariants associated with second fundamental form. ( Not fully sure, I have yet to verify this .)

However, if you are looking to local quantities that are conserved by isometry like Gauss curvature $K$, geodesic curvature $k_g$ etc.(scalar quantities dependent on first fundamental form of surface theory) they are different after homeomorphism.

The Gauss-Bonnet theorem brings these together in an elegant way, as stated by KF Gauss.