taylrl3 said:
In my cosmology lectures they say that a negative curvature gives an infinite space but I was thinking what about the inside of a torus. Isn't that a closed space too?
Cant any value of k apart from 0 result in a closed space??
There are theorems called local to global theorems that constrain the global topology for a given curvature:
http://en.wikipedia.org/wiki/Riemannian_geometry#Local_to_global_theorems These theorems are not so strong that they give a 100% correspondence between local and global properties.
As you've noted, a flat space can have various topologies, such as trivial, cylindrical, toroidal, Mobius... In the non-flat topologies, you get a preferred frame of reference, which is essentially the frame in which the circumference is maximized (as opposed to other frames which see the circumference as Lorentz-contracted).
There are various exotic possibilities for the global topology of the universe:
http://physicsworld.com/cws/article/news/18368
http://arxiv.org/abs/astro-ph/0403597
One thing I didn't understand properly until recently was that in models with nonzero cosmological constants, the spatial topology is not necessarily correlated with the existence of a Big Crunch.
[EDIT] I don't pretend to understand the local to global theorems, and like the OP, I would be interested in hearing more about how this applies to spaces with nonvanishing curvature.
In the case of 2-dimensional space with a positive-definite metric, I do think I understand the possibilities to some extent. In the flat case, you have Euclidean geometry, which can have a variety of topologies (trivial, cylindrical, toroidal, Mobius). In the positive-curvature case, you get elliptic geometry, and all models of elliptic geometry are closed; that is, you can start from the axioms of elliptic geometry and prove results like an upper bound on the area of any triangle. I would be interested to know whether the negative-curvature case (hyperbolic geometry) admits any topology other than the usual one; I suspect that it doesn't.
Making the analogy with 3+1 dimensions, I would conjecture that you only get wiggle room on the spatial topology if the spatial curvature is zero.