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I was watching “Who cares about topology? (the inscribed rectangle problem)” by $3$Blue$1$Brown, and it explained how the unit square, when considered as representing all unordered $2$-tuplets, maps to the Mobius Strip.

I noticed that it was folded in half because it has one axis of symmetry, with two equivalent sides, which is equal to the number of permutations of a two-element set.

When we consider the unit cube, it has three sides, or three coordinates. This gives $3!$, or $6$ permutations per set, or $6$ axes of symmetry. This means that $6$ volumes within the cube can be considered equivalent if we consider the points as unordered, instead of ordered, sets.

Now, in the case of the directed unit square as derived from sets of two points around a circle, to preserve the manifold structure, one side had to be given a twist (thus resulting in a Mobius Strip).

Take the unit cube, as derived from the set of unordered triples between zero and one, which correspond to points around a circle. If we were to somehow bend each (in our case, considered equivalent) volume in the fourth dimension so that all faces of these volumes formed a different, contiguous and continuous volume of some kind, what would we create?

Essentially, what volume (closed manifold, I think) does this particular unit cube map to?

Edit: It has been brought to my attention that, in the original context, the coordinates of the points corresponded to individual numbers on a circle, (basically, distributed across the circumference of a circle and identified with a single number) and were then mapped onto the unit square. As such, my question really regards the unit cube as derived from all unordered triples of points around that circle. The question has been modified to match this.

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    $\begingroup$ You have to be a lot more explicit than this. First of all, for a Möbius strip, we identify one pair of edges oppositely. A cube has three pairs of opposite faces. You're identifying one pair? How are you identifying them? At any rate, you're going to build a $3$-dimensional topological object; a Klein bottle is still a surface. $\endgroup$ Jan 24, 2023 at 20:23
  • $\begingroup$ @TedShifrin I only included a Klein bottle as a naïve guess, and I can tell now how insufficient it would be. I really have no idea how I would identify the faces, because I have no formal training in topology, but if anyone could suggest how I might begin to answer this question, I’d be grateful. $\endgroup$ Jan 24, 2023 at 20:43
  • $\begingroup$ In the case of a square, you can build a cylinder, a Möbius strip, or a Klein bottle (and stranger things, too). Your saying that you had to introduce a half-twist to preserve manifold structure seems wrong. In the case of the cube, you can certainly build a solid torus or solid Klein bottle, a solid cylinder, and lots more bizarre spaces. $\endgroup$ Jan 24, 2023 at 21:08
  • $\begingroup$ My vague recollection of that 3Blue1Brown video is that the set of unordered pairs of points on a circle maps to the Möbius strip. Since the set of ordered pairs of points on a circle corresponds to a torus (equivalently, a square with opposite sides identified), this essentially means the Möbius strip is the torus folded on itself. $\endgroup$ Jan 24, 2023 at 21:26
  • $\begingroup$ In that case, are you then asking for the set of unordered triplets of points on a circle? $\endgroup$ Jan 24, 2023 at 21:26

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