# Possible illogicalness of a 3-sphere shape of the Universe's space?

• I

## Main Question or Discussion Point Metaphorical depictions of the universe in the shape of a 2-sphere are very common.
Now, let's consider mapping the three dimensions of space onto the "surface" of a 3-sphere. Like Non-Euclidean de Sitter geometry.

The surface of a sphere is positionally symmetrical. However, the positional symmetry of the surface being mapped to is distinct from the positional symmetry of the space being mapped.

We could represent the space in terms of a co-ordinate system, and likewise for its positional symmetry.

A Cartesian co-ordinate system has a high degree of positional symmetry, so we consider a three-dimensional Cartesian grid, equivalent to a honeycomb of cubes. Each cube being surrounded by and making contact with 26 cubes. I think a problem arises when we consider all the Platonic polyhedroids/polychorons. There's only one made with cubes, the tesseract.

Because there doesn't appear to be a magical number of cubes which space can be divided into. And so, if the three dimensions of space can be mapped to a 3-sphere, I daresay we might expect there to be an infinite number of cubic Platonic polyhedroids.

And a tesseract has a different configuration as well, each of its cubes only makes contact with 6 cubes, not 26.

For the cubic honeycomb, each edge is shared by 4 cubes, and each vertex by 8 cubes, yielding for 8 cubes, 24 edges and 8 vertices.
A tesseract, on the other hand, has 32 edges and 16 vertices.

Now you might be confused by the mentions of "non-Euclidean" and "Cartesian", however, I've already mentioned a distinction between the two positional symmetries.

Also, there's still only one cubic Platonic polyhedroid.

All of this appears to imply to me that it is illogical to map the three dimensions of space onto a 3-sphere.

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• PeroK

PeterDonis
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let's consider mapping the three dimensions of space onto the "surface" of a 3-sphere.
That is not what is being done in models of the universe in which it is spatially a 3-sphere. What you are calling "space" is actually just one particular 3-dimensional spatial geometry, called "Euclidean 3-space". It is not the only possible 3-dimensional spatial geometry.

All of this appears to imply to me that it is illogical to map the three dimensions of space onto a 3-sphere.
No, what it shows (to the extent it shows anything--some of what you say is just confused) is that the geometry of a 3-sphere is different from the geometry of Euclidean 3-space. Mathematicians have known this for getting on for two centuries now.

PeterDonis
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Non-Euclidean de Sitter geometry.
This is not the geometry of a 3-sphere. It's the geometry of a particular 4-dimensional spacetime, not a spatial geometry at all.

That is not what is being done in models of the universe in which it is spatially a 3-sphere. What you are calling "space" is actually just one particular 3-dimensional spatial geometry, called "Euclidean 3-space". It is not the only possible 3-dimensional spatial geometry.
So I'd like to know, is it possible to map a cubic honeycomb, each and every cube being surrounded by 26 cubes, to a 3-sphere?

Assuming that that is relevant to 3-sphere models of the universe.

PeterDonis
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is it possible to map a cubic honeycomb, each cube being surrounded by 26 cubes, to a 3-sphere?
To a small portion of a 3-sphere, yes. A small portion of a 3-sphere works the same as a small portion of Euclidean 3-space. The differences are global, not local. For one thing, the total volume of a 3-sphere is finite, but the total volume of Euclidean 3-space is infinite.

To a small portion of a 3-sphere, yes. A small portion of a 3-sphere works the same as a small portion of Euclidean 3-space. The differences are global, not local. For one thing, the total volume of a 3-sphere is finite, but the total volume of Euclidean 3-space is infinite.
I edited my reply to say "each and every cube", though I guess you replied earlier in that brief window.

So since its not "each and every", I'm puzzled and still have question(s), sorry; how do 3-sphere models apply a global, positionally symmetrical co-ordinate system to three-dimensional space?

PeterDonis
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edited my reply to say "each and every cube"

how do 3-sphere models apply a global, positionally symmetrical co-ordinate system to three-dimensional space?
They don't. As I've already said, Euclidean 3-space (which is what you are misleadingly referring to as "three-dimensional space", as if it were the only such space, which it isn't) is a different geometry than a 3-sphere, just as the Euclidean 2-plane is a different geometry than a 2-sphere. So obviously you can't use the same global coordinates for both.

They don't. As I've already said, Euclidean 3-space (which is what you are misleadingly referring to as "three-dimensional space", as if it were the only such space, which it isn't) is a different geometry than a 3-sphere, just as the Euclidean 2-plane is a different geometry than a 2-sphere. So obviously you can't use the same global coordinates for both.
I was not specifying Euclidean 3-space in #6, but nevermind, to ensure clarity, although non-Euclidean, cosmologists still have to work with some kind of 3-space, three dimensions of space, three different numbers.

what sort of global, positionally symmetrical co-ordinate system do 3-sphere theorists apply to the kind of space they work with?

PeterDonis
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what sort of global, positionally symmetrical co-ordinate system do 3-sphere theorists apply to the kind of space they work with?
3-dimensional spherical coordinates are the most common ones.

3-dimensional spherical coordinates are the most common ones.
hmm, though it isn't really positionally symmetric, requiring a central radial position and it can be readily converted to Cartesian co-ordinates, and therefore might lead back to the original problem

PeterDonis
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though it isn't really positionally symmetric, requiring a central radial position
The coordinate origin is an artifact of the choice of coordinates. You could equally well choose any point you like to be the origin; they're all the same as far as the actual geometry is concerned, just as all points on a 2-sphere are the same as far as the geometry is concerned, even though if we put 2-spherical coordinates on them we have to pick one such point to be the origin.

it can be readily converted to Cartesian co-ordinates
No, it can't. We're not talking about spherical coordinates on Euclidean 3-space. We're talking about spherical coordinates on a 3-sphere. The transformation you show is only valid between spherical and Cartesian coordinates in Euclidean 3-space. It is not valid on a 3-sphere.

It is possible to have coordinates on a 3-sphere that are like Cartesian coordinates in a small patch of the 3-sphere, but they can't be extended globally, just as it's possible to have coordinates on a 2-sphere that are like Cartesian coordinates in a small patch of the 2-sphere, but they can't be extended globally.

I strongly suggest that you take some time to think carefully about how things work on a 2-sphere. A 3-sphere is similar, just with one more dimension.

PeterDonis
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3-dimensional spherical coordinates
Note that by this I did not actually mean the coordinates ##\rho##, ##\theta##, ##\phi##, although you can use those on a 3-sphere (but they won't work the same as they do in Euclidean 3-space). I meant the 3-sphere analogue of latitude and longitude, i.e., three angular coordinates (instead of the two we use on a 2-sphere).

Ibix
Am I right in understanding that the rather artistic diagram at the top of the OP is of a sequence of 2-spheres embedded in a 3d space - the universe in this picture is the surface of each sphere, not the body of the enclosed ball? But this is just a diagram, and the real model is the surface of a 3-sphere that would need to be embedded in a 4d space if we wanted to draw it (but I guess the model doesn't include an embedding space). So the problem with the coordinate transform in #10 is that it relates coordinates on the 2-sphere to Cartesian coordinates in the embedding space - which doesn't exist in the model, only the diagram?

If so then the transform needed is between coordinates on a 3-sphere, which presumably has three angular coordinates, and Cartesian coordinates in a 3-space. This will have all the same issues that maps of the world have - they can be reasonably good representations over small regions, but must be distorted over large regions and actually torn if you try to cover the globe. But you can draw an honest flat map of any small region - and that's the relevant symmetry.

The coordinate origin is an artifact of the choice of coordinates. You could equally well choose any point you like to be the origin; they're all the same as far as the actual geometry is concerned, just as all points on a 2-sphere are the same as far as the geometry is concerned, even though if we put 2-spherical coordinates on them we have to pick one such point to be the origin.
This will have all the same issues that maps of the world have - they can be reasonably good representations over small regions, but must be distorted over large regions and actually torn if you try to cover the globe.
we know that each point on the 3-sphere is the same as any other point, a symmetry of each point in 3-D space

Imagine that we can materialize identical solid metal cubes out of thin air. We start materializing cubes and packing them together into a cubic honeycomb, a solid honeycomb ironically.

eventually we will have so many cubes that we will fill up almost all of the volume of the 3-sphere universe, and it will have to terminate. What will the termination of this process look like?

And we can also consider a 3-sphere universe which has a really tiny volume.

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Ibix
What will the termination of this process look like?
Depends on what process you use to force things as the mismatch between your cube and the underlying geometry grows, I would imagine. Your fundamental problem is that a Cartesian approach doesn't reflect the symmetry of a spherical space - so your statement in the OP that a "Cartesian co-ordinate system has a high degree of positional symmetry" is where you went wrong. It has translational symmetry in a flat space, but you are not operating in a flat space.

Your fundamental problem is that a Cartesian approach doesn't reflect the symmetry of a spherical space
yeah, so i'm wondering how it works out logically

Depends on what process you use to force things as the mismatch between your cube and the underlying geometry grows, I would imagine.
what sort of mismatch would we start to observe?

because i was thinking about the positional symmetry, every point in space being identical to any other point.

Ibix
yeah, so i'm wondering how it works out logically
You have global rotational symmetries instead of translational ones.
what sort of mismatch would we start to observe?
Your cubes would need to be squashed at their outside edges because the circumference of a circle of radius ##r## is less than ##2\pi r## in this space.
because i was thinking about the positional symmetry, every point in space being identical to any other point.
It is. You can start piling up cubes at any point and you'll get the same failure mode. Contrast, for example, a black hole spacetime where you can pile cubes happily for a long time if you start distant from the hole, but will rapidly run into problems if you start near the hole.

Your cubes would need to be squashed at their outside edges because the circumference of a circle of radius ##r## is less than ##2\pi r## in this space.

It is. You can start piling up cubes at any point and you'll get the same failure mode.
Squashed because the leftmost side of our honeycomb pile bumps into the rightmost side?

Ibix
Squashed because the leftmost side of our honeycomb pile bumps into the rightmost side?
Before that. Imagine tiling the surface of the Earth. Build a complete single row of tiles around the equator. Fine. Now try to add another row, touching it on the north edge. You can't do it, because the line of constant latitude at the north edge of the new row is shorter than the one at the south edge - there isn't room for square tiles. The problem gets worse as you get further away from the first ring, but it's present right there.

Now let's go back and look at just that first equatorial ring. The problem actually applies even to this - the corners of each tile must be slightly lifted off the surface if the middle touches. The underlying reason for this is that you cannot build a Euclidean square on the surface of a sphere because it isn't a Euclidean surface. Thus your Cartesian grid fails right from the word "go", because it doesn't reflect the symmetries of the space in which you are attempting to define it.

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@Ibix @PeterDonis

Per Ibix's suggestion, I'm considering the POV of a flatlander on (or in) the surface of a 2-sphere.

A flatlander has no conception of the 3rd dimension of space, so to him the 2-sphere is a flat plane in which he can move about infinitely in any direction, though he might end up looping around back to where he started.

I start with a cube, its a Platonic solid and topologically similar to a 2-sphere. The cube has to be squished onto a flat plane.

Say said cube is a dice, and our flatlander is located on the 1-dot face. After drawing 5 faces, its unclear how we ought to proceed further.
The 6-dot face is a convex quadrilateral, but it is also connected to all of the four outer quadrilaterals.

Though this resembles the Schlegel diagram of the cube.

We can repeat it with polyhedra which have more faces and are rounder in shape, like the truncated icosahedron.  Once we've reached the stage of the right image, its also unclear how to proceed further, and it also resembles the Schlegel diagram of the truncated icosahedron.

How do you think we could proceed further?

Ibix
to him the 2-sphere is a flat plane
No it isn't. He can make measurements on a relatively small scale to show that the space is non-Euclidean. Triangle angles don't quite add to 180, for example, and the larger you make the triangle the worse the difference becomes.

This is different from a cube, which is mostly flat with some very curved regions at the corners.

No it isn't. He can make measurements on a relatively small scale to show that the space is non-Euclidean. Triangle angles don't quite add to 180, for example, and the larger you make the triangle the worse the difference becomes.
yeah, but he observes all this within two dimensions of space, he can't visualize the curvature in the third dimension

so what do you think of the "Schlegel diagrams"?

PeterDonis
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Contrast, for example, a black hole spacetime where you can pile cubes happily for a long time if you start distant from the hole, but will rapidly run into problems if you start near the hole.
Note, however, that the issue with a black hole spacetime is different from the issue with a 3-sphere. In a black hole spacetime, you can fit more cubes radially than you would expect based on tangential distances, whereas in a 3-sphere you can fit fewer.

Ibix
yeah, but he observes all this within two dimensions of space, he can't visualize the curvature in the third dimension
And that's why we apply non-Euclidean geometry - to provide a mathematical description of something we cannot visualise. The same tools are available to a Flatlander. Or Curvedlander, I suppose.
so what do you think of the "Schlegel diagrams"?
I don't see what you are trying to achieve. Why not just use a Mercator projection, or stereographic projection, or whatever?

Ibix