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According to Elementary Differential Geometry by A N Pressley, a parameterization for Mobius strip is :

$\textit{Example 4.9}$

The Möbius band is the surface obtained by rotating a straight line segment $\cal L$ around its midpoint $P$ at the same time as $P$ moves around a circle $\cal C$, in such a way that as $P$ moves once around $\cal C$, $\cal L$ makes a half-turn about $P$. If we take $\cal C$ to be the circle $x^2+y^2=1$ in the $xy$-plane, and $\cal L$ to be a segment of length $1$ that is initially parallel to the $z$-axis with its midpoint $P$ at $(1,0,0)$, then after $P$ has rotated by an angle $\theta$ around the $z$-axis, $\cal L$ should have rotated by $\theta/2$ around $P$ in the plane containing $P$ and the $z$-axis. The point of $\cal L$ initially at $(1,0,t)$ is then at the point $$\boldsymbol\sigma(t,\theta)=\left(\left(1-t\sin\dfrac\theta2\right)\cos\theta,\left(1-t\sin\dfrac\theta2\right)\sin\theta,t\cos\dfrac\theta2\right).$$ We take the domain of definition of $\boldsymbol\sigma$ to be $$U=\{(t,\theta)\in\mathbf R^2\mid-1/2<t<1/2,\ 0<\theta<2\pi\}.$$

And according to Wiki another parameterization is

$x(u,v)= \left(1+\frac{v}{2} \cos \frac{u}{2}\right)\cos u\\ y(u,v)= \left(1+\frac{v}{2} \cos\frac{u}{2}\right)\sin u\\ z(u,v)= \frac{v}{2}\sin \frac{u}{2}.$

My questions are:

1- how these two 'different' parameterizations can be transformed to each other? Supposing the domains remain same ($0 ≤ u < 2π$ and $−1 ≤ v ≤ 1$) is it not possible by changing variables to do the reparameterizations (esp. $x$ and $y$ in $(x,y,z)$).

2- How (at least one of) the mentioned parameterizations be derived? Wiki and the book mentioning with no proof.

Thank you.

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Replace $u$, in the Wiki param, with $u + \pi/2$, and swap $x$ and $y$, and you'll more or less have the book's version.

As for "proving" the parameterization, that's hard to say: a parameterization isn't a statement that requires proof. But to see how such a parameterization is developed in the first place, the answers here should prove useful.

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John Hughes
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  • Thanks a lot. I read the other post that you mentioned. We know what is 'implicit form' of the cylinder and we 'replace' functions of t and s (u(t,s),v(t,s),w(t,s)) such that it gives the same equation. So it can be other functions (u'(t,s),v'(t,s),w'(t,s)) which what we say reparameterization. It's fine that v(t)=cos(t/2)(0,0,1)+sin(t/2)(cost,sint,0) is a guess only if we know the 'implicit form' of the Mobius strip. Would you please help me what is the level fiction, i.e. f(x,y,z)=c for the Mobius strip? It can be drived by the mentioned parameterization but it's not a solution since ... –  Aug 26 '15 at 11:57
  • The second comment on the bottom post at that location has a pointer to this: http://math.stackexchange.com/questions/1366639/derive-cartesian-cubic-m%C3%B6bius-strip-from-parametric which answers that question very nicely. – John Hughes Aug 26 '15 at 12:00
  • ... since I still don't know if it correct. –  Aug 26 '15 at 12:00
  • I looked at the comments and answers but all things are based on the mentioned parameterization which I don't know if it is correct (esp. since the parameterization has come by guess based on boundary conditions). Is there a proof for level equation of f(x,y,z)=c of Mobius strip not based on parameterization or if it is to be a parameterization with rigorous proof? Thank you. –  Aug 26 '15 at 12:10
  • I don't know what a "rigorous proof" would possibly look like. One can compute directly from the parameterization that the image of the parameterization has the homotopy type of a circle, and that it's nonorientable and a 2-manifold-with boundary, and the boundary is a single circle (All these are a pain, but...) By the classification theorem for compact 2-manifolds, that makes it a Mobius strip. Once you believe that part, the conversion the implicit form done in the referenced answer seems rock-solid to me. If it doesn't satisfy you, I probably cannot help any further. – John Hughes Aug 26 '15 at 18:04