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Imagine a grid in 3D made of pipes smoothed so that it forms one continuous infinite surface. The surface is 2D but it fills 3D space.

Like this (at one instant): enter image description here

Could any surface like this be a solution of Einstein's equations for (2+1)D but actually on the large scale act more like a (3+1)D space?

This doesn't look Ricci flat but perhaps with curvature in space and time it could be made Ricci flat?

One could imagine particles on such a space would act more like being in 3D than in 2D until you got down to short distances.

Why I'm asking is because the dimension of space seems to depend on what scale you're looking at things.

Qmechanic
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1 Answers1

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The structure you have described is one of the infinite generalizations of the regular polyhedra. Rather than having three squares meet at a vertex, giving each vertex a positive solid angle, there is a meeting six squares at each vertex, with a negative solid angle. These negative solid angles are conical points; they are positions of concentrated Ricci curvature, leaving the remainder of the surface (the interiors of the connected squares) flat. It is possible to distort the surface so that the negative curvature is not confined to conical points of zero size; however, it is not possible to eliminate the negative curvature entirely.

To summarize, because the vertices are conical points with negative curvature, this surface is not spatially Ricci flat, nor can it be made flat by any kind of continuous distortion.

Buzz
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  • Yes it's not spacially flat. But the Schwarschield solution is not spacially flat. But it is space-time flat. Also, what about if you put some black holes at the negative curvature points. In other words you had singularities of mass at these points. –  Dec 15 '18 at 03:19
  • @zooby That makes no sense. To be spacetime flat means that all the components of the (3+1)-dimensional Riemann tensor vanish. This immediately implies that all spatial sections are also flat. There are no $g_{0j}$ elements of the metric, but that does not mean that there exist any flat sections. – Buzz Dec 15 '18 at 03:23
  • I think you are mistaken because the 3x3 section of the 4D Ricci tensor depends also on $g_{00}$ and $g_{0j}$. So it is not the same as saying 3D dimensional Ricci tensor is $0$. One can split the 3x3 section of the 4D Ricci tensor into the 3D Ricci tensor and another term. Both can be non-zero and the total can be zero. –  Dec 15 '18 at 03:29
  • It is well known "space is curved around a star" and yet the (3+1)D Ricci curvature is 0. –  Dec 15 '18 at 03:34
  • @zooby The Ricci tensor is zero, but the Riemann tensor is not, because of its Weyl part. See here: https://physics.stackexchange.com/questions/295814/non-zero-components-of-the-riemann-tensor-of-the-schwarzschild-metric – Buzz Dec 15 '18 at 03:39
  • OK, so you're saying that the model above is not even Ricci flat? –  Dec 15 '18 at 03:52
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    @zooby In any case, the conical points in that structure carry Ricci curvature. – Buzz Dec 15 '18 at 03:54
  • OK. Would that discount all such space-filling 2D spaces do you think? I guess it would because you would need negative curvature somewhere. –  Dec 15 '18 at 03:55
  • BTW. I thought of a simpler example. Take a 2D sphere cut from a 3D solid cube. The 3D solid cube is flat yet the 2D sphere is curved. Or indeed cut a 2D cone from a 3D solid cube. –  Dec 15 '18 at 04:00
  • @zooby I don't know enough about space-filling surfaces in general (as opposed to that specific one made up of squares) to say for sure; however, I would guess that other surfaces would not work either. – Buzz Dec 15 '18 at 04:13