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Inspired by the similar question regarding a torus, imagine that you're a flatlander walking in your world. How could you distinguish between your world being a sphere versus being a projective plane?

It seems like this would be somewhat harder because you can't necessarily use an argument about non-positive curvature, and you can't (easily?) take advantage of the fact that the projective plane is non-orientable and not embeddedable in $\mathbb{R}^3$.

I would also be interested in any methods that could be used in this case, would wouldn't tell you anything substantial if you were on a torus.

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Mike Pierce
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  • Unless you can define precisely what "walking on a surface" means (and as you can see in the comments below this is indeed an issue), then this is a soft question. – Najib Idrissi Nov 03 '16 at 10:06

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Since the projective plane is $\mathbb{Z}_2$ quotient of the sphere, it is impossible to distinguish the two if you confine yourself to a simply-connected region.

If you don't, then you can take two identical, oriented objects, leave one where you are, and walk with the other one straight until you get back to the first object. On a sphere you will find that the objects still look identical, and on the projective plane you will find that the objects are now oriented differently.

Peter Kravchuk
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  • If I'm thinking about it alright, you won't be able to find the first object when you get back to it because you will be on the "other side" of the surface. As a flatlander walking on the surface of the projective plane, you would have to walk over that same loop twice before you made it back to where you started and you could see the object. – Mike Pierce Nov 03 '16 at 06:53
  • @MikePierce, your interpretation then is not intrinsic, but rather uses an extra structure on the projective plane (the "two sides"). That is actually equivalent to living on a sphere, so you will not be able to distinguish the two cases at all. – Peter Kravchuk Nov 03 '16 at 06:59
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    @MikePierce, more precisely, if you say that "you would have to walk over that same loop twice...", given that the fundamental group of projective plane is $\mathbb{Z}_2$, you are literally saying that we live on the universal covering of the projective plane, which is the sphere. – Peter Kravchuk Nov 03 '16 at 07:05
  • Alright, I get what you're saying. If you allow for there to be "two sides" of a single point on the surface, its like you're looking at the sphere before you quotient it to the projective plane, which really defeats the purpose. But then I keep thinking of the projective plane as the result of contracting the boundary of a mobius strip. So does this mean it's not quite correct to think that a point on the mobius strip has "two sides?" – Mike Pierce Nov 03 '16 at 07:19
  • And in your solution I suppose you shouldn't have to confine yourself to a simply connected region. See some of the answers to the related question. – Mike Pierce Nov 03 '16 at 07:23
  • @MikePierce Depends on the purpose, when you think about sides, you think about the embedding, which is extrinsic to the surface. It is an extra structure. I don't get your last point. I only said that you cannot tell the difference in a simply connected region. You could tell the difference between the geometrists torus and sphere in a simply connected region, by measuring the curvature, but not in this case. – Peter Kravchuk Nov 03 '16 at 08:04
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If you go for a walk and when you return find that your wife is suddenly left-handed, you know you are in the projective plane.

Mikhail Katz
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