How is it so that two Moebius Bands (MB) are considered not orientable when we can see them as two spiral areas joined with their lateral edges on the (external visible) surface of an automobile tire tube? It appears intriguing.
The doughnut surface is a connected sum of two Moebius Bands glued along their lateral spiral edges. We come back to the same point on each of the two tracks after two full polar rotations. The toroidal tube itself is orientable,but not the MBs it is composed of !.
Posted this in sci.math newsgroup with subject " Moebius Band is not homeomorphic with a Torus"
formatting link
We are allowed to stretch or compress or twist the rubber membrane in its own plane. If we cut along two meridians of a hollow doughnut, and re-attach the shortened tube without twisting, orientability is unaffected, OK. However, if we impart a half-twist (180 degrees) or whatever angle to shear the tube and re-join the ends we have lost all spatial orientation... or so we are led to believe. Two pictures of the idealized twisted surface ( mechanical model fabricated earlier actually) are included in my posts dated Oct 28 and Nov 2, 2006 to illustrate the anomaly.
formatting link
formatting link
In these I have brought and stuck together opposite areas of the torus section back to a central point back to form an MB or a flat annulus to look at the surface topological property,but that is supposed to be a disorienting maneuver.A flat annulus can be bent to be conical frustum and and a cylinder.
[If we symmetrically half-twist a machine V- Belt with four separate tracks, we get a figure of 8 deformed ring and a full-twist results in THREE (1/3 size) rings, but they are all supposed to be orientable.]
Do you think surface orientations on a two adjacent tracks on a doughnut are not determinable?
Regards Narasimham