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Let me quickly build up some background.

One way to build a hypercube is to take cubes, and start gluing them together, face to face, such that each edge is shared by $3$ cubes. You complete the hypercube with $8$ cubes. This involves rotating cubes in $4$ dimensions, but if you forget about the geometry, you can still do this abstractly.

Separately, lets take a torus that is made from 5 squares, like this:

5 squares can make a torus

where you identify opposite sides in the usual way to make a torus.

Now, lets combine these concepts. Let's take these tori and (abstractly?) glue them together, square face to square face, such that each edge is shared by 3 tori. It turns out this only takes $6$ of these tori.

My question is this: where the hypercube lives nicely in $4$-space, is there a nice, familiar, space where this clump of tori lives? Is there a nice way to think about this kind of object?

EDIT: As requested in the comments, here is a gluing diagram for what I am proposing: Gluing Diagram

The numbers and colors mean the same thing, they identify square faces to be glued together. The letters correspond to the vertices, so one knows the orientation of the squares to be glued.

Josh B.
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  • Also, is there any general way of telling whether gluing copies of an object together, face to face, with $n$ objects around an edge, will take only finitely many of that object? My guess is no. – Josh B. Sep 23 '17 at 00:27
  • If you take 6 solid squares and put them together in the cube way, you get a hollow cube (a 2d surface that you can embed in 3d). And if you take 6 square frames, you get just the frame (1-skeleton) of a cube. If you take 8 filled-in solid cubes, you can get a "hollow" tesseract (a 3d thing that you can embed in 4d). But if you take 8 hollow cubes, you get the 2-skeleton of a tesseract, which is sort of an awkward object since an edge is shared by 3 (just like the frame of a cube is awkward for a 1d thing). I'm not sure how you want the 6 tori glued, but you'll have a similar sort of ugliness. – Mark S. Sep 23 '17 at 01:48
  • @MarkS. Does it make any sense to talk about filled-in solid tori? – Josh B. Sep 23 '17 at 01:52
  • Yes and no. Certainly, a regular doughnut is like a filled-in solid torus, but your geometry-ignoring picture of the boundary of a doughnut might or might not be conveniently the boundary of some solid torus, depending on where it lives. For example, a circle (frame) can be the boundary of a disk (or a hemisphere surface, which is the same thing topologically), but it could also be the boundary of a hollow torus with a little hole in it (like a tire with a hole for air). Since what you get depends on how you embed things, you have to specify the solid tori if you want to glue them. – Mark S. Sep 23 '17 at 02:29
  • You have to provide more information about the construction: How many tori did you take? What is the exact gluing pattern between these tori? – Moishe Kohan Sep 23 '17 at 05:27
  • @MoisheCohen With cubes, is gluing, face to face with 3 around an edge is enough to specify the hypercube? I guess the idea is to not glue anything you don't have to. I guess I mean the largest thing which satisfies those requirements. And you do end up with something more symmetric than the cube itself. Similar with the tori, but the construction only requires 6. – Josh B. Sep 23 '17 at 05:48
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    @JoshB.: No, its not enough. If you know what you are proposing, it should be easy to describe in detail. – Moishe Kohan Sep 23 '17 at 10:58
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    @MoisheCohen You are right, the burden of proof is certainly on me. I have edited in a proposed gluing diagram for the 6 tori. – Josh B. Sep 23 '17 at 23:59
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    Your space embeds (say, topologically, but if you are familiar with the PL category, then also piecewise-linearly) into $S^3$, the 3-dimensional sphere. Is this what you wanted to know? – Moishe Kohan Sep 25 '17 at 00:28
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    @MoisheCohen Sure, but how did you arrive at that answer? – Josh B. Sep 25 '17 at 02:34
  • What troubles me is the fact that in your construction some pairs of the points $A,B,C,D$ and $E$ appear as two adjacent points in one of the squares, but also as two anitpodal points in one other square. If there were any nice geometric realization of this structure, I would have expected it to "preserve the square patches" somehow. But there seems no way, as we cannot distinguish between neighboring and antipodal points in any square, and so we cannot assign appropriate distances to the points. – M. Winter Jan 07 '20 at 22:07

3 Answers3

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Any geometric manifold can live in any kind of space, given a suitable injection. But there are two common ideas. Consider a polyhedral surface, say a cube. On the one hand it lives on the sphere $S^2$ in the sense that it is a topological decomposition of the sphere into polygons, each of which is topologically an $S^1$. But it also lives naturally in locally-Euclidean 3-spaces such as $E^3$, in the sense that it can be smoothly embedded.

Every polytope is a topological decomposition of some manifold, but not every polytope can be locally embedded in a higher space. For example every polyhedral decomposition of the projective plane $P^2$ must self-intersect as an immersion when injected into $E^3$, it cannot be smoothly embedded. On the other hand a toroidal polyhedron can be embedded in $E^3$ and will have holes through it like a donut or a pretzel.

Similarly, the hypercube is a polyhedral decomposition of $S^3$ and can be embedded in $E^4$. Your 6-torus polychoron is a decomposition of a far more complicated toroidal 3-manifold. Whether or not it can be embedded in $E^4$ would take a lot of heavy maths to discover, although I expect that somebody already has.

There is no easy road to visualising objects in higher dimensions, but treating them as manifolds and studying their topology is perhaps the most accessible.

Guy Inchbald
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2

I have a more straightforward answer which doesn't require any calculations.

These tori you have there are actually Clifford tori which live in 4D. Look at the vertices, each of them has degree 4 and they are connected each to each. You have to take into account the wrapped edges also. And all of these edges are the same length which means that the vertices and edges form a regular pentachoron (4-simplex). And 6 regular pentachora form a regular hexateron (5-simplex) which lives in 5D.

Hume2
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tl;dr: It lives in 5 dimensions.

First, note that the square faces are not planar. The figures you have drawn there are projections of Clifford tori. Clifford torus lives in 4 dimensions. If you project the central square back onto the Clifford torus, the vertices will have these coordinates: $$ A=\left[\cos-\frac35\pi;\sin-\frac35\pi;\cos-\frac15\pi;\sin-\frac15\pi\right] \\ B=\left[\cos-\frac15\pi;\sin-\frac15\pi;\cos\frac35\pi;\sin\frac35\pi\right] \\ C=\left[\cos\frac35\pi;\sin\frac35\pi;\cos\frac15\pi;\sin\frac15\pi\right] \\ D=\left[\cos\frac15\pi;\sin\frac15\pi;\cos-\frac35\pi;\sin-\frac35\pi\right] \\ $$ It is obvious that the sides of the square are equal and the diagonals are equal. Let's show that the diagonal is the same length as the side, namely $||A-B||=||A-C||$. We can evaluate the sines and cosines first: $$ A=\left[\frac14\left(1-\sqrt5\right);-\sqrt{\frac58+\frac{\sqrt5}8};\frac14\left(1+\sqrt5\right);-\sqrt{\frac58-\frac{\sqrt5}8}\right]\\ B=\left[\frac14\left(1+\sqrt5\right);-\sqrt{\frac58-\frac{\sqrt5}8};\frac14\left(1+\sqrt5\right);\sqrt{\frac58+\frac{\sqrt5}8}\right]\\ C=\left[\frac14\left(1-\sqrt5\right);\sqrt{\frac58+\frac{\sqrt5}8};\frac14\left(1+\sqrt5\right);\sqrt{\frac58-\frac{\sqrt5}8}\right] $$ First, we calculate $||A-C||^2$: $$ ||A-C||^2=\left(2\sqrt{\frac58+\frac{\sqrt5}8}\right)^2+\left(2\sqrt{\frac58-\frac{\sqrt5}8}\right)^2=\frac52+\frac{\sqrt5}2+\frac52-\frac{\sqrt5}2=5 $$ And now $||A-B||^2$ $$ ||A-B||^2=\left(\frac14(1-\sqrt5-1-\sqrt5)\right)^2+\left(\sqrt{\frac58+\frac{\sqrt5}8}-\sqrt{\frac58+\frac{\sqrt5}8}\right)^2+\left(\frac14(1+\sqrt5-1+\sqrt5)\right)^2+\left(\sqrt{\frac58+\frac{\sqrt5}8}+\sqrt{\frac58+\frac{\sqrt5}8}\right)^2=\\ =\frac54+\frac58+\frac{\sqrt5}8+\frac58-\frac{\sqrt5}8-2\sqrt{\frac58+\frac{\sqrt5}8}\sqrt{\frac58-\frac{\sqrt5}8}+\frac54+\frac58+\frac{\sqrt5}8+\frac58-\frac{\sqrt5}8+2\sqrt{\frac58+\frac{\sqrt5}8}\sqrt{\frac58-\frac{\sqrt5}8}=5 $$ So $||A-B||=||A-C||=\sqrt5$.

This means that the sides and diagonals of these squares are the same so they are not planar. Their vertices actually form regular tetrahedra. 5 tetrahedra form a pentachoron (4-simplex) and 6 pentachora form a hexateron (5-simplex) which lives in 5 dimensions.

Hume2
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