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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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