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Similar question here, but it doesn't quite address what I'm hoping to have illuminated.

I've recently started watching a series of lectures on topology. The material I've gone through thus far is mostly presented geometrically. Frankly, I'm not sure that's the technically correct term, so to clarify, I mean "not very rigorous" and "lots of drawings are used". Therefore, I'd like any answers or interpretations to be presented in a similar light, in - dare I say - layman's terms.

In this video (between 0:34 and 11:50), prof. Tokieda outlines the process of

  • considering a sphere
  • drilling a hole into it by cutting a disk from its boundary
  • filling in that void with a Möbius strip and connecting it to the rest of the sphere

which is permitted since the boundary of a disk and the boundary of a Möbius strip are homeomorphic (to circles, as I understand it), thus putting a "lid" on the sphere, and calling it the (real) projective plane, $\mathbb RP^2$. This manifold is then referred to by a symbol that strongly resembles the Death Star.

In the question linked above, one of the comments directed me to this interesting animation that seems to project a point onto $\mathbb RP^2$, and tracks its position along $\mathbb RP^2$, but the surface itself is split into a disk that is homeomorphic to the drilled sphere and the Möbius lid.

What I would like to do is somehow be able to visualize this completed "Death Star" in the same way I can visualize the Klein bottle as a self-intersecting surface in 3 dimensions, like in this image. Is this possible?

Inspired by the first 3 or 4 videos in the playlist, I embarked on a brief investigation into making this "Death Star" out of paper. Here's what I attempted, step-by-step:

  • apply a twist to strip 1 and tape its ends together; leave strip 2 alone
  • starting at an arbitrary point along strip 1, begin lining up one of the ends of strip 2 with the boundary of strip 1
    • after a few moments of this, the boundary of strip 1 exceeds the length of the boundary of strip 2, so I "deform" strip 2 by cutting another strip of paper with the same dimensions, then taping it to the end of strip 2 and effectively doubling its length
  • continue until the free end of strip 2 can be taped to the end taped to strip 1 at the start

If this process is unclear, I'll be happy to upload a sequence of pictures I took at various points showing its progress.

At the end of this admittedly frustrating process, I end up with strip 3, what appears to just be a wider (by a factor of $3$) Möbius strip than strip 1 was. If it were any wider, I'm certain the paper would tear immediately, which makes me think completing the process is impossible or requires a material much stronger than cheap notebook paper. (Unfortunately, I'm all out of infinitely pliable, self-permeable rubber.)

Is there a way to visualize this complete surface?

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    $\begingroup$ You might want to look up “cross-cap” and “Roman surface” $\endgroup$ – G Tony Jacobs Apr 19 '18 at 20:45
  • $\begingroup$ Keep in mind, that linked picture of a Klein bottle does not depict an embedding of the Klein bottle, in other words the Klein bottle is not homeomorphic to the subset being depicted. That depiction is what topologists call an immersion. The projective plane also has an immersion, the "Boy's surface" linked in the answer of @AloizioMacedo. The only problem is, the Boy's surface immersion of the projective plane is definitely a less helpful visualization than the immersion of the Klein bottle in your linked picture. Oh well, that's just how it goes. $\endgroup$ – Lee Mosher Apr 22 '18 at 21:19
  • $\begingroup$ @LeeMosher Thanks for pointing out the distinction between "embedding" and "immersion", both are terms I have not yet covered in Tokieda's lectures. $\endgroup$ – user170231 Apr 27 '18 at 19:25
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It depends on what you mean by "visualize".

The standard mathematical interpretation of "visualizable in three dimensions" is that there exists an embedding in three dimensions. Such embedding cannot exist: a compact non-orientable $m-$manifold cannot embed in $\mathbb{R}^{m+1}$. There are several proofs for this: via Alexander duality, intersection number etc, and it is a non-trivial result.

Another interpretation can be that there exists an immersion in three dimensions, which allows self-intersections, for instance. Under this interpretation, it indeed is visualizable: the Boy's surface is an example.

But restricting to those concepts of visualizable is doing the enhancement of abstract geometrical thinking a disservice, imho. I am very content of saying that I visualize the projective plane by, say, visualizing $D^2$ with certain identifications on boundary, and this is arguably a two-dimensional visualization.

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  • $\begingroup$ I suppose I was hoping to see something like the computer-generated image of the Klein bottle, as in the image I linked in the question. In terms of embeddings (I haven't been formally introduced to the term yet, so forgive me if I'm misusing it), is that particular image of the Klein bottle not an embedding of it in $\mathbb R^3$? Also, what is $D^2$? A disk of some kind? $\endgroup$ – user170231 Apr 20 '18 at 22:37
  • $\begingroup$ Indeed that is not an embedding of the Klein bottle in $\mathbb{R}^3$. There is no embedding of the Klein bottle in $\mathbb{R}^3$ for the same reason that there is no embedding of the projective plane in $\mathbb{R}^3$: namely, because of the theorem in this answer. $\endgroup$ – Lee Mosher Apr 22 '18 at 21:16
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$\Bbb{R}\Bbb{P}^2$ is best visualised as the set of lines through the origin in $\Bbb{R}^3$ (under the topology that says two lines are close if the angle between them is small).

Each line corresponds to two points on the unit sphere. If you choose some small angle $\delta > 0$, the lines within $\delta$ of the equator form the Möbius strip $M$ that you get by looking at all the points in the northern hemisphere within $\delta$ of the equator and thinking about how things join up when you rotate a line through $360^o$ about the $z$-axis. The lines that are not within $\delta$ of the equator have a unique representative in the northern hemisphere and those representatives form a a disc $D$ such that $\Bbb{R}P^2 = M \cup D$.

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  • $\begingroup$ I'm afraid I don't see the connection between this description and the "sphere-glued-to-a-Möbius-strip" one. Would you kindly be able to elaborate, or refer me to sources that do? $\endgroup$ – user170231 Apr 20 '18 at 23:06
  • $\begingroup$ I hope my updated answer helps. $\endgroup$ – Rob Arthan Apr 21 '18 at 20:17
  • $\begingroup$ Yes, much appreciated. I see now it's essentially capturing what's going on in that animation I linked in the OP. Thanks! $\endgroup$ – user170231 Apr 27 '18 at 19:24

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