Geometric features of a closed finite universe

Question

I am a student, so the question may sound silly. If the 2-sphere is the surface of a ball, that is, it is embedded in a three-dimensional space, then the 3-sphere must also be the surface of a four-dimensional ball. In this case, if our universe is a 3-sphere, does it mean that it is necessarily embedded in a 4-dimensional space?

Answer 1

No it does not.

Let me show a better reasoning for the existence of 4 dimensions of space. Usually we use 2-d space(paper) and show that how it curves and say that 4-d space-time does this in a similar fashion. But for a 2-d paper to curve it needs 3 dimensions. That means that for a 4-d space-time to curve it needs 5-d spacetime or 4 dimensions of space. Therefore 4 dimensions of space exists.

But there's a catch. When scientists use the analogy of paper or trampoline. They use extrinsic geometry to show this because it is easy for analogy. (Kind of like looking at the curvature of the earth from space)

But in reality the way we define spacetime curvature is by using intrinsic geometry, in this case there is no need for outside or a higher dimension.(kind of like trying to define the earth's curvature while living on the surface of it)

You have to remember that outside can exist but it is not necessary to our current math which describes spacetime curvature.

This all goes the same for your question where spacetime has positive curvature.

(By the way I am 14)

Answer 2

It may be convenient to consider the 3D hyper-sphere as existing in a 4D space by analogy that the 2D surface of a sphere exists in 3D space. However, I suggest the 4th dimension is not needed to be spacial. The math can get along without it. However, it may be helpful to visualize the 4th dimension of the 3D hyper-sphere space as time (t). The center of the hyper-sphere is t=0. The radius is the current time since t=0 that the universe has been expanding.