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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.

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    $\begingroup$ One correction: I see in your question and one or two of your comments below that you are equating "solvgeometry" with the geometry of the universal cover of $\text{SL}(2,\mathbb{R})$. This is incorrect. Solvgeometry is the geometry of the 3-dimensional simply connected solvable non-nilpotent Lie group $\mathbb{R}^2 \rtimes \mathbb{R}$, where $\mathbb{R}$ acts on $\mathbb{R}^2$ taking $t$ to a matrix with eigenvalues $e^t,e^{-t}$. And this geometry is distinct from $\widetilde{\text{SL}}(2,\mathbb{R})$. $\endgroup$
    – Lee Mosher
    Commented Dec 7, 2015 at 22:07
  • $\begingroup$ I am sorry for confusing Lee and many others. When I was typing ``Solvgeometry'' what I had in mind was actually $SL(2,\mathbb{R})$-geometry. They both start with `S' : ) $\endgroup$
    – PhysicsMath
    Commented Dec 9, 2015 at 0:17

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Yes, the answer is that you want a simply-connected manifold. All $SL_2(\mathbb R)$ manifolds are covered by the universal cover. Not all $SL_2(\mathbb R)$ manifolds are covered by $SL_2(\mathbb R)$.

Using the universal cover helps simplify the language of the "classification" of geometric manifolds.

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    $\begingroup$ As a specific example, consider the circle bundle over a compact surface of genus $g> 1$ such that the Euler number of the bundle is $1$. This is a manifold modeled on $\widetilde{SL}(2,R)$, which is not covered by $SL(2,R)$. $\endgroup$
    – Misha
    Commented Dec 5, 2015 at 23:58
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    $\begingroup$ To elaborate on the previous answers: $PSL_2(\mathbb R)$ is the unit tangent bundle of ${\mathbb H}^2$ and its quotient by a surface group $\Gamma=\pi_1\Sigma_g$ is the unit tangent bundle of a hyperbolic surface $\Sigma_g$. The latter has Euler number $2-2g$. The circle bundles over $\Sigma_g$ are (via the Euler class) in one-to-one correspondence with $H^2(\Sigma_g;{\mathbb Z})={\mathbb Z}$. $\endgroup$
    – ThiKu
    Commented Dec 6, 2015 at 7:23
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    $\begingroup$ In particular, the circle bundle of Euler number $-1$ is a fibre-wise $(2g-2)$-fold covering of the unit tangent bundle and so its total space obtained as $\Gamma\backslash \widehat{PSL_2(\mathbb R)}$, where $\widehat{PSL_2(\mathbb R)}$ means the $(2g-2)$-fold covering of $PSL_2(\mathbb R)$. So you should need at least that many coverings of $SL_2(\mathbb R)$ to produce the closed $3$-manifolds locally modelled after $SL_2(\mathbb R)$. $\endgroup$
    – ThiKu
    Commented Dec 6, 2015 at 7:25
  • $\begingroup$ I see. Thanks to Ryan for your concise answer that not all $SL_2(\mathbb{R})$ manifolds are covered by $SL_2(\mathbb{R})$! Thanks to Misha and ThiKu for your specific examples of such $SL_2(\mathbb{R})$ manifolds that would not be included if I use $PSL(2,\mathbb{R})$! $\endgroup$
    – PhysicsMath
    Commented Dec 7, 2015 at 19:01
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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.

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  • $\begingroup$ Actually, every flat manifold is finitely covered by a torus, so starting with the universal cover does not yield any new examples. $\endgroup$
    – Igor Belegradek
    Commented Dec 7, 2015 at 20:34
  • $\begingroup$ @IgorBelegradek, cylinders have a place under the sun also :-) $\endgroup$
    – Mikhail Katz
    Commented Dec 8, 2015 at 8:10
  • $\begingroup$ @PhysicsMath, look at it this way: PSL(2,R) is homotopy equivalent to a circle, whereas the Heisenberg group is already simply connected. Another relevant example may be the Weil-Heisenberg group where you place elements from R/Z in the upper-right entry. What I am trying to say is that there is no more reason to use PSL(2,R) as a basic model for solvmanifolds than there is to use the Weil-Heisenberg manifold as a basic model for nilmanifolds. $\endgroup$
    – Mikhail Katz
    Commented Dec 8, 2015 at 8:22
  • $\begingroup$ I see your point, Katz. Thanks for pointing out this to avoid triviality! $\endgroup$
    – PhysicsMath
    Commented Dec 9, 2015 at 0:20

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