- Thread starter
- #1

- Jan 17, 2013

- 1,667

How can the open Mobius band be topologically the same as the space of all lines ?

- Thread starter ZaidAlyafey
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- Thread starter
- #1

- Jan 17, 2013

- 1,667

How can the open Mobius band be topologically the same as the space of all lines ?

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- #2

- Feb 15, 2012

- 1,967

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

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.