Closed Dimensions: Metric Tensor, Curvature & Time

[thehangedman]

If we assumed an empty space, but also assumed space dimensions are closed ( repeat after some distance D ), what would the metric tensor look like? Is this just equivalent to a space with a constant curvature R? If so, how does R relate to D? Would the time dimension also necessarily be closed if space dimensions are closed?

 

[bcrowell]

You don't need any curvature in order to have a topology that wraps around. This is similar to the fact that you can wrap a piece of paper into a cylinder without introducing any Gaussian curvarure. The metric tensor can be the same flat metric as in Minkowski space. No, you don't need time to wrap around (in closed timelike curves, CTCs) in order to have the spatial dimensions wrap around.

One odd thing about these topologies is that although they locally have the same symmetry as special relativity, globally they have a lower symmetry. There is a preferred frame in which the circumference of the universe is maximized. In other frames its length-contracted.

I don't know the best possible introduction to this topic, but one possibility I found is this paper: The Shape and Topology of the Universe, Jean-Pierre Luminet, http://arxiv.org/abs/0802.2236 Luminet seems to be someone who has done a lot of work on this idea. E.g., he co-authored a Scientific American article on the topic.

 

[thehangedman]

Ok cool. Are there any decent references you know of so I can read about this? Surely the closed nature of the space would manifest somewhere in the equations...

 

[tom.stoer]

A simple question regarding these compact, flat extradimensions, e.g. Calabi-Yau: is there any reason why they are stable, and not subject to Penrose's singularity theorem?

 

[Bill_K]

One odd thing about these topologies is that although they locally have the same symmetry as special relativity, globally they have a lower symmetry. There is a preferred frame in which the circumference of the universe is maximized. In other frames its length-contracted.

Also global rotational invariance is lost, right? If I identify points at intervals D along the x, y and z axes, then the interval between points along some diagonal direction will be greater.

Plus: even though the various patches may have been identical to begin with, quantum indeterminacy will cause them to evolve differently. E.g. a nucleus may have decayed in one patch while its image in some other patch has not.

Or are images in different patches entangled?