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If the universe is compact then there is a twin paradox that is resolvable only by selecting a preferred inertial reference frame (arXiv). I was under the impression that the lack of a preferred inertial reference frame was an empirical result (e.g. from early experiments designed to measure Earth's movement against the aether).

But if I've understood these first points correctly then we should not still be entertaining the notion of a compact universe at all, as in the top voted answer to this question.

So: Is a compact universe theory still viable? If so, what might account for the lack of evidence supporting a preferred inertial reference frame (or is there actually now evidence of a preferred frame)?

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up vote 3 down vote accepted

If the universe is compact then its scale is at least 100 times greater than the observable universe (because according to experiment the observable universe is flat to within 1%). No experiment of a length scale much shorter than this scale would be able to detect the compact topology, so the Michelson Morley experiment is not incompatible with a compact universe. It just means the universe is locally non-compact, which of course we already know just by looking around.

Even if the universe is compact its scale is increasing as the universe expands, and indeed the expansion is probably exponential due to dark energy. If so, no experiment we can ever do will show the universe to be compact.

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I though flatness (geometry) and compactness (topology) are independent... – QuantumDot Sep 11 '12 at 18:58
@QuantumDot, that confuses me too. For example one could realize a topological torus as a cube where opposite sides are identified with each other. In such a space the geodesics would be Euclidean straight lines, but according to the paper in the question there would still be a preferred inertial reference frame. However, if the observable universe failed to intersect a side of the cube then I don't see how observations could distinguish that space from an infinite flat space. So maybe there would be a preferred inertial reference frame but still no way to observe it. – Aaron Golden Sep 11 '12 at 20:37
Compact is not the same as curved and topological compactness is not incompatible with flatness. Compactness is a global property, so what is "locally non-compact" supposed to mean? (Edit: sorry, posted without refreshing and spotting the previous comments, of which this essentially a repeat) – James Sep 11 '12 at 20:53
Yes, OK, I assumed a compact universe would be a closed FLRW (or similar) universe because that's well established Physics. If you want to make the universe a torus, icoshedron or whatever my answer wouldn't apply but I know of no physical justification for this assumption. "Locally non-compact" is an inelegant way of saying the scale of the experiment is much less than the scale on which the universe is compact. – John Rennie Sep 12 '12 at 6:32
@John: I didn't know when I wrote the question that there was this constraint on, shall we say, "reasonable" universes. Let me see if I understand. For various reasons the only metrics worth considering for empty space are of the FLRW form, and a consequence is that the global geometry is either elliptical, Euclidean, or hyperbolic. Only the elliptical global geometry might be compact, so if the universe is compact we should be able to measure the uniform (positive) curvature of empty space, unless the radius of the universe is much larger than the observable universe. Is that right? – Aaron Golden Sep 12 '12 at 20:57

The problem topological compactness poses to Lorentz invariance only shows up globally. That is to say, a preferred frame induced by having a compact universe only shows up when you perform a global experiment (as in the Twin Paradox problem linked to by Qmechanic). Local experiments like Michaelson-Morley only tell you about local properties and they can be perfectly compatible with full Lorentz invariance - they wouldn't tell you anything about global properties unless your apparatus were large enough to see the topology.

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