4Dspace wrote:Thank you quickfur for your reply. Sorry I was not clear -- it's cause my understanding is evolving. I am new to topology.
I understand your example with threads in a piece of fabric and agree that 3D space can be distorted within 3D, without leaving its constrains. But. Think about it: if that were true with gravity in our world, then space would not appear to us perfectly flat, especially here on Earth. Spacetime is heavily curved around the planet, all around us. Yet we do not perceive this curvature in the refraction of light. (I'm aware that there is a lensing effect around stars, that's not what I'm talking about here.) What I'm saying is that space all around us on the planet is condensed and curved, yet we do not see it in all sorts of halos, mirages and lensings. This is possible ONLY if the distortion is in fact in the 4th spatial dimension, while the 3D is practically perfect.
I don't follow your reasoning. A "fabric distortion", if we may call it that, does not necessarily mean halos and mirages, etc., if the distortion is only slight. Even a 4D-style distortion would cause all sorts of halos and distortions (as you point out with the gravitational lensing example, except on a more extreme scale) if the gravitational constant were much larger. A fabric that is only slightly distorted would show almost no visible effects, until you extend it to a very much larger piece of fabric, in which case the accumulated effect of the slight distortions will start showing up in a visible way.
The analogy here is with a thin 2D sheet of Flatland that is gently curved in 3D, appearing perfectly orderly and regular for the Flatlanders.
So, when I analyzed the situation in regard to curvature in (n+1) dimension, I saw that what curves is the surface plane from which the (n+1)th dimension arises. And to determine if it's curved, for 2D seen from 3D, I need to make orthogonal lines from 2 arbitrary points on the plane and see if they are parallel or not.
Correct. What you just described is the idea behind a "manifold" - locally Euclidean, but globally not necessarily so. So on the small scale, it looks like Euclidean space, but slightly distorted so that over the large scale, it actually forms a curved surface.
But anyway, with regard to your orthogonal lines, note that those lines would lie
outside the 2D space, which is a bit troublesome if you're confined to working within the 2-manifold itself (since you would have no access to the 3rd dimension where the lines extend). A more convenient means of measuring curvature, incidentally, is to observe how the parallel postulate fails to hold within the 2D space itself. So hyperbolic spaces will see parallel lines eventually diverge, whereas elliptic spaces will see parallel lines eventually intersect. These are the two idealized cases, of course, but one can easily have position-dependent violations of the parallel postulate, in which case you'll have other kinds of global shapes for the space.
One advantage of this approach is that you don't need to add the assumption that (n+1)-dimensional space exists; one can imagine a space "fabric" of uneven "spatial density", like the fabric distortion I mentioned, that can also give rise to the same kind of effects (violations of the parallel postulate) without needing n+1 dimensions. So by using this approach, it can apply to both situations, whereas conclusions drawn by assuming n+1 dimensions may no longer apply in the "fabric distortion" cases.
When I went from 3D to 4D, I noticed that the 3 planes that bind a cube from one side, all belong to the same plane from which the 4th dimension emerges (<-- as seen from the 4th D point of view).
It would help to use unambiguous terminology, so that we're all on the same page. Keep in mind that in 4D, you have both 2-planes (planes spanned by two linearly independent vectors), which do not divide 4D space, and 3-planes, which do divide 4D space. One needs to distinguish between them in order to not get into trouble.
A cube (3D cube) in 4D unambiguously defines a hyperplane, and there is exactly one vector orthogonal to this hyperplane. However, each face of the cube only defines a 2-plane, and in 4D, there is another 2-plane that is orthogonal to this 2-plane. So a 2-plane is not sufficient to define an orthogonal vector in 4D.
I hope this is right -?
I want to learn to visualize this 4D thing well. Actually, I need only to see a 3D surface from the 4th D perspective, which is much easier. That's how I realized that the 3 faces of a cube lie on the same plane in 4D. - is that right? [...]
This is correct if by "plane" you mean "hyperplane", that is, the 3-plane defined by the cube. It is not to be confused with the 2-planes that each face of the cube lies in.
As for 4D visualization, here's my shameless plug:
http://eusebeia.dyndns.org/4d/vis/vis I use 4d->3D projections as my primary method of 4D visualization, because I found that it is most intuitive to me. YMMV.