# 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.

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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.

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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. –  user3138766 Apr 16 at 2:10
And by the way does that mean the surface of the 3 sphere is 3 dimensional? –  user3138766 Apr 16 at 2:12
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. –  Gerald Apr 16 at 2:14
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? –  user3138766 Apr 16 at 2:16
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. –  Gerald Apr 16 at 2:18