Welcome to our community

Be a part of something great, join today!

Mobius band and space of all lines

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
How can the open Mobius band be topologically the same as the space of all lines ?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Have you seen the explanation here (it comes about half way through the section on Open Möbius Band)?
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
A more "naive" explanation: since we are talking about the OPEN mobius band, we can stretch the open edges all the way to infinity without changing its topological properties (in much the same way as we can map the open interval (-1,1) homeomorphically to all of $\Bbb R$ using the mapping:

$t \mapsto \dfrac{t}{1-t^2}$).

For ease of visualization, let's say the original rectangle we created the open Mobius band from was a unit square of side length 2 (open on the lines $y = \pm 1$), centered at the origin. As above, we stretch the "height" to infinity, creating a "twisted cylinder". Any line in our original plane (besides the y-axis, itself) will now cross the "reverse-identified" edges at some point, coming back on itself in the opposite quadrant. It is important that we use UN-oriented lines to identify with, because after crossing the "twist", the orientation gets reversed (now heading TO the origin, instead of away).

Note that this space of unoriented lines is NOT the Euclidean plane, as it is not possible to define a consistent orientation on it. This is because rotating 180 degrees returns the same line, instead of the customary 360 degrees needed in the (oriented) Euclidean plane.