Jessica- Great question. Before giving the expert answer below. I've got one simple point to make. The Lorentz contraction only affects lengths parallel to the relative motion. Even in simple 4-D Special relativity, it's easy to construct an argument that distances in the orthogonal directions can't be affected. So the curled-up dimensions, which are definitely orthogonal to any possible motion in the extended 3 dimensions, would not Lorentz contract.
from Rob Leigh:
I guess what we're talking about here is an example where spacetime is S^1xR. There is a frame in which space is S^1 and time is R (the reals). The best way to think of this is as a 'compactification' of flat spacetime, RxR. We regard the circle S^1 as R/Z (reals mod integers), which means if we take a coordinate x along R, we regard a point x and a point x+2pi m R as equivalent (for m an integer and R the radius of the circle).
This compactification breaks the Lorentz invariance of the 'covering space' RxR. To see this, start in the frame that I was discussing above, but impose the identification (x,t) ~ (x+2pi R, t). If we boost to another frame, this identification becomes
(x',t') ~ (x',t') + 2pi R (gamma,vgamma)
where v is the velocity of the frame (in units c=1). Note that (gamma, vgamma) is a unit spacelike Lorentz vector (which in the origin frame was just (1,0)). Thus, to specify the compactification, we have to select a fixed unit spacelike Lorentz vector, and this is where the notion of a 'preferred frame' comes from.
In string theory, what often happens is that we soup up this example to include extra large flat dimensions (that are not compact). So spacetime might be of the form S^1xR^3xR (compact direction, space, time). The 5-dimensional covering space has 4+1 Lorentz symmetry, but the compactified space has only 3+1 Lorentz symmetry (one can still boost along the large flat directions) -- one spatial direction is distinguishable from the other 3).
(published on 03/08/10)