### Our universe the surface of a 4-dimensional sphere?

The cosmic microwave background that we observe uniformly around us is usually explained by assuming that our universe is the surface of a four dimensional sphere. That way the uniformity makes sense since there is no center. My question is if this is true then what is the explanation that describes the fact that the farther we look into space, the further we look back in time. I can't perfectly picture this and see how it would coexist. Help me out.

s on. The four dimension including time as the fourth is a surface of a 5-sphere? And it goes on and on and on? Think about it. It should make sense. And I fancy a way to beat speed of light by finding the fifth dimension. lol!-Nitin (Mathematician and not Astronomer) P.S.: I will love to hear what you have to say to that.

@dotancohen Oh, shoot. I didn't check the numbers after "user". My bad; deleting comment.

The surface of the 4-dimensional ball (called 3-sphere) is a slice through the universe as a whole for a fixed cosmic time. This slice describes just three spatial dimensions. The observable universe is a tiny part of this 3-sphere; hence it looks flat (3-dimensional Euclidean space) up to measurement precision (about 0.4% at the moment).

Adding time makes the universe 4-dimensional. The observable part is similar to a 3+1-dimensional Minkowski space-time. The universe as a whole may be a de Sitter space-time. A de Sitter space-time is the analogon of a sphere ( = surface of a ball), embedded in a Minkowski space, instead of an Euclidean space, but it's not literally a sphere.

If time would be taken as an additional spatial dimension, the de Sitter space would ressemble a hyperboloid of revolution, if embedded in a 5-dimensional hyperspace.

The difference to an Euclidean space is due to the different definition of the distance: In a 4-dimensional Euclidean space the distance between two points is defined by

$l = \sqrt{\Delta x^2+\Delta y^2+\Delta z^2+\Delta t^2};$ for a 3+1-dimensional Minkowski space it's $l = \sqrt{\Delta x^2+\Delta y^2+\Delta z^2-\Delta t^2}.$ For simplicity the speed of light has been set to $1$.This model of the universe as a whole can hold, if it actually originated (almost) as a (0-dimensional) singularity (a point); but our horizon is restricted to the observable part, so everything beyond is a theoretical model; other theoretical models could be defined in a way to be similar in the observable part of the universe, but different far beyond, see e.g. this Planck paper.

But isn't it necessary for us to view our universe as the surface of a 4-D sphere in order to explain the CMBR appearing uniformly in all directions? That is the only way we would be seeing this background radiation in all directions, uniformly regardless of our position in space.

And by the way does that mean the surface of the 3 sphere is 3 dimensional?

A 3-sphere can be seen as the surface of a 4-ball. The 4-ball is 4-dimensional; its surface, the 3-sphere, is 3-dimensional, and has no surface.

That's pretty intense! So our universe itself would be this 3-D surface of a 4-D sphere or some other shape possibly? Can you see how hard this is to visualize, I'm trying to see it in my head and the question that keeps popping up is, if the surface is 3-D then how can it be a surface?

The assumption of a 3-sphere isn't necessary to see the CMB appearing uniformly. It could also be an infinite, and expanding, 3d Euclidean space. The observable universe would look like a 3-ball, too, as it does.

Take the surface of a usual everyday 3d-ball, and consider its surface. It's a 2-sphere; it hasn't a surface. This can be generalized to higher-dimensional spaces.

But if the universe is this infinite 3-D space, then why would we expect to see uniform background radiation from the Big Bang? If the Big Bang happened, then it happened everywhere but how does that guarantee that we would see this CMBR uniformly around us? It seems as if you would have to assume a 3-sphere because then our position in space can be neglected.

If the universe started with an infinite Euclidean space, expanding uniformly at least to the horizon of the observable universe, we couldn't distinguish this from a universe starting as a point. Similar considerations would hold for a universe the shape of the surface of an expanding 4-d torus, as long as it is large enough to be locally indistiguishable from the surface of a 4-d ball.

I have no experience in topology at all so I really am having a hard imminent understanding this but you have helped me a lot! I guess I am just having a hard time imagining why the CMB should be uniform if our universe is not the surface of a 3-sphere. If it's not, I am having a hard time understand why we would expect uniform radiation as our position in space would determine whether or not we would receive uniform radiation. Please correct me if I am wrong.

Almost perfect isotropy of the CMB is explained in current models by inflation, eliminating amost any primoridial heterogeneities. -- Thinking about the universe as the surface of a 4-ball is probably the easiest way. But no significant difference to the infinite Euclidean model could be detected thus far; hence any non-trivial topology (including the 3-sphere) - if present - must most likely be beyond the horizon of the recombination (CMB).

That makes perfect sense, thanks so much. You're very good at explaining things, I appreciate the conversation!

I believe your confusion is from combining two popular simplifications of our Universe. As we look further away we see further back in time because of the finite speed of light. So these distant objects are also evolving with us, but that light hasn't hit Earth yet.

It might help to know that the observable universe may only be 14 Gyr old, but its radius is 46 Gly, not 14 Gly. If the speed of light was infinite, then we wouldn't be observing back in time as we look at more distant objects.

Could you please expand a bit on the latter paragraph of your answer? I realize where you're going with it, but I'm not sure all the readers will, and judging by the lack of upvotes, that seems to be the case. Cheers!

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user116034 8 years ago

Here is a crazy thing. I was always a believer that it is wrong to think of the space as a non-compact thing. That is just unimaginable. But then I started comparing space with the surface of the earth. Now, you can imagine that a non-compact thing. So, why should the space be one? I can't. So, I started asking myself, what if the space is not a non-compact R^3 but a compact set the surface of a 4-sphere like you guys are talking about. Now, that is compact. And I am happy to see that you have other reasons to claim that it is. But now I have food for your thoughts. What if the whole thing goe