# Would a closed universe with special relativity violate causality? Does the universe have to be simply connected?

This question may be more appropriate for physics.stackexchange.com, but it would be helpful to get feedback from experts in Minkowski geometry.

The classic twin paradox is a false thought experiment trying to disprove special relativity. Special relativity says that an observer at rest watching a passerby at a high velocity will see the other person's clocks running slower. However, relativity also says there is no preferred frame of uniform motion, so the passerby sees the clock of the original observer running slow. So if a twin leaves the earth at a high speed and returns, leaving her sister on earth the whole time, both should find the other younger, a contradiction.

This is resolved in flat, open spacetime by the fact that the passenger twin must accelerate to leave and return, and acceleration is a preferred frame in the sense that the laws of physics are different for an accelerating person.

Now, though, imagine a closed universe (I.e. one with closed geodesics). In this universe, two objects in relative uniform motion will encounter each other repeatedly. Even if a twin must accelerate to get into uniform motion, she will pass her sister on earth over and over again if she keeps on a fixed closed geodesic. Now both frames are equivalent, and so time is slower for both observers. Thus, past and future are not well-defined, and causality is violated.

Is there something wrong with this reasoning? Is there any way to handle special relativity and a closed universe so the above paradox does not happen?

Edit: For geometrists, I'm really asking if the invariant quadratic form becomes poorly-behaved if we take a quotient of $\mathbb{E}^{3,1}$ by a proper action of a subgroup of the Lorenz group.

Edit: By closed universe, I meant a universe compact in space dimensions for some observer. By a poorly behaved quadratic form, I was just thinking of how proper time is the root of the quadratic form for events with time like separation, so any paradox with proper time is an inconsistency with the quadratic form.

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Brian: I could not quite understand the geometric question: What do you mean by "poorly behaved"? Are you asking if nontrivial fundamental group of the space-time implies existence of a closed geodesic, provided that the space-time is geodesically complete and locally-flat? Then the answer is positive. –  Misha Apr 2 '13 at 16:30
If two twins follow different paths to the same point, their odometers will in general differ. So will their clocks. Why is either of these paradoxical? –  Steven Landsburg Apr 2 '13 at 16:43
This should probably be closed because it's pretty standard. First of all, you don't really mean closed geodesics (which would be examples of closed timelike curves, ie, time travel). You want a universe with some foliation of spacelike compact surfaces. The easiest example is a cylinder: S^1 x R. For the twin paradox to hold, you would need the Lorentz group to act isometrically on this space for a given flat metric. But, it's easy to see that the full Lorentz group won't act here. In fact, the choice of a metric defines for you a distinguished frame, solving the 'paradox'. –  Aaron Bergman Apr 2 '13 at 16:56
Was a little too quick there -- there are closed spacelike curves; just not the first things that come to mind. –  Aaron Bergman Apr 2 '13 at 17:10
A very old question. The answer, as Aaron suggests, is standard. See, for instance, dx.doi.org/10.1119/1.16373. –  Igor Khavkine Apr 2 '13 at 18:28