Is a preferred reference frame of the universe the old aether?

Recently a new answer was posted and an intense discussion ensued: Here.

One of the points discussed concerns a preferred reference frame in this universe:

The asymmetry comes from the fact that the universe itself has a reference frame, and its size will lorentz contract. This is measurable by the people themselves--all that needs to happen is to send out a light ray and wait for the light ray to go around the world. The 'diameter of the universe' will be (light orbit time)/c. This time will be observed to be smaller the faster the observer is travelling. So all observers will agree that there is a global, absolute notion of motion, and this will pick out who ages when.

My questions

• Which (mathematical) characteristics determine whether there is a preferred reference frame in a universe?
• Does our universe have a preferred reference frame?
• If a universe has a preferred reference frame is this comparable with the old aether?
• If a universe has a preferred reference frame don't we get all the problems back that seemed to be solved by RT (e.g. the 'speed limit' for light because if there were a preferred frame you should be allowed to classically add velocities and therefore also get speeds bigger than c?

(I will assume in my answer that people have read the discussion on the old question, linked to by the OP.)

No, it is not like the aether. It is still true that locally, there is no preferred reference frame. You don't even really need to think about spacetime to see what is going on. Consider a two-dimensional plane, parametrised by $(x,y)$, and roll it into a cylinder by identifying $(x,y) \sim (x + nL, y) ~\forall~ n$, where $L$ is some constant. Locally, this space is still perfectly isotropic, but globally, the $x$ direction has been picked out by the identification.

To see what this means, let's imagine drawing two straight line segments, each beginning at $(x, y) = (0,0)$ and ending at $(0,L)$. The first will just be $(0,t)~,~ 0\leq t\leq L$, and the other will be $(t,t)~,~0\leq t\leq L$ (which ends at a point equivalent to $(0,L)$ under the identification, and therefore the same point on the cylinder). Obviously the length of the first line is just $L$, but the length of the second line is $\sqrt{2}L$, by Pythagoras. Although any small patch of the cylinder is perfectly isotropic, we see here that the rotational symmetry is broken globally by the identification.

In spacetime, a similar thing happens, replacing rotational symmetry by boost symmetry.