﻿ oipapio

### mobius band - Derive Cartesian cubic Möbius strip from parametric

The following link:http://mathworld.wolfram.com/MoebiusStrip.htmlshows the Möbius strip parametrized as\begin{eqnarray}x = [ R + s \cos \left ( \frac{1}{2} t \right ) ] \cos t \\y = [ R + s \cos \left ( \frac{1}{2} t \right ) ] \sin t \\z = s \sin \left ( \frac12 t \right )\end{eqnarray}The symbols for $R$ and $s$ and angle $t$ are explained there.Then they say that from this parametrization we can derive the cubic.$$-R^2 y + x^2 y + y^3 - 2 R x z - 2 x^2 z + y z^2 = 0.$$Any ideas about how to do this? I have tried wit...Read more

### mobius band - Möbius bundle no global trivialization

I there! I am trying to write down why the Möbius bundle has no global trivialization. I just read that there is none but I want to see a written prove for this. I am not even sure which definition of the Möbius strip would be the best. First let me define this relation: $(a,b) \tilde{} (c,d)$ if $b = 1-d$ and $a = 0, c = 1$ or $a=1, c = 0$. Now let me define the bundle: Let ($E$, $\pi$, $S^1$) be a vector bundle where $E = [0,1] \times \mathbb{R} / \tilde{}$, $S^1$ is the unit sphere and $\pi([x,y]) = e^{2\pi i x}$.So far I have the following ...Read more