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This questions arises from working through the 2015 Heraeus lectures on gravity, specifically tutorial 2 exercise 2, so this is why I post it as a physics question although it is purely mathematical.

When we represent the Moebius strip as a rectangle, we draw opposite sense arrows on two opposite sides that describe how to glue the sides, i.e. Moebius strip as a rectangle. Now if I want to show it is a 2d topological manifold, I need to find a set of charts that cover it, so basically some open set(s) in the "Moebius rectangle" together with the corresponding homeomorphism(s) from each to $\mathbb{R}^2$. Here are my questions:

  1. Why no single chart can do this;
  2. why (and what) two charts would suffice;

I can see there is a problem with the twist on the top and bottom sides, like when I try to draw a ball there it gets ripped in two, but that doesn't help me answer the above.

yann
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  • The only reason is that it's part of a (mathematical) physics lecture series, that tackles the question of spacetime, and that I plan on asking other questions as work my way through it. I also saw other topologically minded spacetime questions on physics stack exchange, so I decided to target the physics stack. Also, the pointers you gave are very general, and my question is more specific but I admit that doesn't make it less mathematical. – yann Jan 17 '17 at 22:27
  • Consider the usual physical method of tracing a pencil along a Möbius strip to demonstrate nonorientability. Now, imagine a normal vector whizzing along the strip... – J. M. ain't a mathematician Jan 18 '17 at 21:42
  • Ok, so that would prove a single chart does not suffice, but can't you do that without orientability (was not introduced in the lecture at this point) ? Isn't there a way using open sets like when we show $S^1$ cannot be covered by a single chart because it violates the open set conservation ? – yann Jan 19 '17 at 10:54
  • There is a simple proof if you use the Jordan curve theorem. – Hypatia du Bois-Marie May 04 '19 at 21:47

1 Answers1

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The fact that two suffice is easy, just find two. (Look at a picture of it.)

To show that one chart does not suffice, note that if one chart could cover it, then it could be oriented by the orientation inherited from $\Bbb R^2$. However, it is known that it is not orientable.

Ryan Unger
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  • So for the sufficient part, what mapping do we use there, identity? And how can we see that it solves any problem encountered by a single chart ? As for the single chart impossibility, how can you do that without orientability (was not introduced in the lecture at this point), isn't there a way using open sets like when we show $S^1$ cannot be covered by a single chart? – yann Jan 18 '17 at 06:37
  • I have the same question from the same place. I'm wondering if it has something to do with mapping the boundary. If you draw the strip as a single rectangle, one edge includes the boundary and so isn't an open subset of $\mathbb{R}^2$. But confirmation from someone who knows what they're talking about would be nice. – Dion Silverman Sep 15 '21 at 03:47
  • @DionSilverman I'm not sure I understand the question. You can show that the open Mobius band (delete the boundary) cannot be covered by one chart either. – Ryan Unger Sep 16 '21 at 13:36
  • By boundary, I meant the boundary of the rectangular chart (4 sides), not just the edge of the Möbius strip. If the chart were an open rectangle, then the strip pre-image would be cut somewhere. Unless there is some other reason you can't just draw the strip as a rectangle? – Dion Silverman Sep 18 '21 at 12:55