TITLE: Thurston geometries---the geometry of the universal cover of $SL(2, \mathbb{R})$
QUESTION [6 upvotes]: In one of the eight Thurston geometries there is the geometry of the universal cover of $SL(2, \mathbb{R})$.  But from the algebraic point of view $PSL(2,\mathbb{R})$ is sufficient for building 3-manifolds i.e. we let the group act on itself and then quotient out a discrete subgroup of it.  This is what we do for the nilgeometry where we use the 2-step nilpotent Heisenberg group instead of $PSL(2,\mathbb{R})$.
Thus my question is, why passing to the universal cover?  Professor told me that it is for obtaining a simply-connected 3-manifold so as to include as many manifolds as possible for the classification i.e. there is a preference for starting with simply-connected manifolds.  But I do not really understand the reason behind it although I guess there should be a quick answer to it.

Let me try to clarify my question a bit.  My intention is to make a comparison between nilgeometry and $SL(2,\mathbb{R})$-geometry.  Their algebraic construction is the same except that one starts with the 2-step nilpotent Heisenberg group and the other with $PSL(2,\mathbb{R})$.  However for $SL(2,\mathbb{R})$-geometry there is an additional step of passing to the universal cover of $SL(2,\mathbb{R})$.  What I am asking is the motivation for this extra step.

REPLY [2 votes]: Your question "why pass to the universal cover?" is really a topological question that has little to do with the special case of solvgeometry, so it may be helpful to point out that you wouldn't want to study flat manifolds by starting with the tori, but rather you start with euclidean spaces to get more examples as quotients, including cylinders.