Siefert-Weber space is a compact connected hyperbolic 3 manifold without boundary (closed hyperbolic 3 manifold). It can be constructed by gluing faces of a dodecahedron.
The other two consistent ways of gluing the faces of the dodecahedron yield $ \mathbb{RP}^3 $ and the Poincare homology sphere, both of which are spherical 3 manifolds admitting a transitive action by the group of rotations of 3 space $ SO_3(\mathbb{R}) $.
Is something similar true for Siefert-Weber space? Does there exist a Lie group $ G $ and a closed subgroup $ H $ such that $ G/H $ is Siefert-Weber space?
Please note: I am already aware that hyperbolic 3 space $ \mathbb{H}^3 $ is a homogeneous space. There are lots of essentially equivalent ways of doing this that all involve taking a 6 dimensional Lie group $ G $ that is essentially the isometries of $ \mathbb{H}^3 $ and moding out by a maximal compact subgroup $ K $ to get a contractible 3 manifold with constant sectional curvature -1. For example one can take $ G/K $ to be $ PSL_2(\mathbb{C})/SO_3(\mathbb{R}) $ or $ SL_2(\mathbb{C}) / SU_2 $ or $ O_{3,1}/O_3 $ or $ SO_{3,1}/SO_3 $ etc... Once you have some model of hyperbolic space $ G/K \cong \mathbb{H}^3 $ then you can take an appropriate discrete closed subgroups $ \Gamma $ of $ G $ ($ \Gamma $ is a lattice or is Zariski dense or is a Kleinian group or large in some other suitable sense) and then the double coset space $$ \Gamma \backslash G/K $$ will be a nice hyperbolic manifold, going by names like locally symmetric space, Clifford Klein space form etc... Just wanted to clarify that I already know that Seifert-Weber space is $ \Gamma \backslash G/K $ where $ \Gamma $ is the fundamental group of Seifert-Weber space and $ G/K \cong \mathbb{H}^3 $. Returning to my original question, what I am looking for is a way to realize Seifert-Weber space as $ G/H $ where $ G $ is a Lie group and $ H $ is a closed subgroup.
Mostow, G. D., A structure theorem for homogeneous spaces, Geom. Dedicata 114, 87-102 (2005). ZBL1086.57024.
i. either $M$ itself fibers, or
ii. (The exceptional case.) Either $G$ is solvable or is semisimple and in this case $H$ has finitely many connected components. Unfortunately, this exceptional case is poorly explained in the statement of the theorem in the introduction: In this case Mostow should have allowed one more subgroup in his sequence. In Mostow's notation, $k=1$, causing appearance of a subgroup $F_{-1}$ which has to be equal to $\{1\}$, while $F_0=H$. Then the fibration in the statement of Theorem C reads: $$ H/F_{-1}=H \to G/F_{-1}=G\to G/H. $$
Then $G_k=G_1=G$, $\Gamma_k=\Gamma_1=H$ and we are in Mostow's case 3, where $G=G_1$ is semisimple or solvable and, in the semisimple case, $\Gamma_k$ has finitely many components. Mostow explains this better when he repeats the formulation of Theorem C in the end of section 4 (where this theorem is actually proven).
Case i. If a compact connected 3-manifold fibers, then either the fiber is a surface and the base is a circle or vice-versa.
(a) If the fiber is a surface, it has to be of nonnegative Euler characteristic (see part 1). Hyperbolic 3-manifolds do not admit such fibrations:
Namely, asphericity of $M$ implies that the fiber $F$ cannot be $S^2, P^2$. The long exact sequences of homotopy groups of a fibration implies that we have a short exact sequence $$ 1\to \pi_1(F)\to \pi_1(M)\to \pi_1(S^1)= {\mathbb Z}\to 1. $$ If $F$ is the torus or the Klein bottle then $\pi_1(M)$ would contain a nontrivial normal virtually abelian subgroup which is known to be impossible.
(b) On the other hand, a compact hyperbolic manifold also cannot fiber over a surface $B$ with circular fibers $F$. This again follows from asphericity of $M$ which results in a short exact sequence $$ 1\to \pi_1(F)= {\mathbb Z}\to \pi_1(M)\to \pi_1(B)\to 1. $$ Then, again we would obtain a normal infinite cyclic subgroup in $\pi_1(M)$, which is impossible.
Thus, compact hyperbolic 3-manifolds cannot admit transitive actions of Lie groups.
With more work, one can relax the compactness assumption in (2) to finiteness of volume and prove a similar theorem in higher dimensions as well.
ii. Consider now the exceptional case. If $G$ is semisimple and $H$ has finitely many components, $\pi_1(M)=\pi_1(G/H)$ is finite. This is impossible for a hyperbolic manifold. Lastly, if $G$ is solvable, so is $H$, hence, $H/H^c$ is solvable too (I am using Mostow's notation where $H^c$ denotes the identity component with respect to the Lie group topology). But then $\pi_1(M)$ is solvable, which is impossible for a hyperbolic manifold of finite volume.